Given a set of $ n \plus{} 1$ integers between $ 1$ and $ 2n$, prove that one number must divide
another. Prove that this is not necessarily true for $ n$ integers between $ 1$ and $ 2n$

For the first one, I looked at $ 2n$ as $ n\plus{}n$

= $ 1, 2, ...,n$ and $ n\plus{}1 ,n\plus{}2, ...,2n$ and looked at this $ mod n$ and saw how the two sides were similar in this respect. Sice we're picking $ n\plus{}1$ numbers, there must at least one number picked from either one of the sides. The residues $ mod n$ are $ 1,2,...0$ so there will be two numbers have the same remainder $ mod n$ But then I noticed that just because two numbers have the same remainder $ mod n$ doesn't mean that one divides the other.  What are the conditions "modwise" in order for one number to divide another, and how can use it to solve the problem?

I also noticed that every nth number is divisible by $ n$ and since we're picking one more than half of the total numbers, one number must forcedly divide the other. How do I show this in a proof?

Unfortunately, there is no "modwise" method to determine divisibility. For your second idea, you should use a different letter than $ n$ because $ n$ is already used. Also, it merely states what needs to be proved. Avoid using "must" or, in your case, "forcedly", unless it is a direct consequence ("we're picking one more than half of the total numbers" does not directly result in "one number must forcedly divide the other"). Here is the proof, if you wish to see it. It is truly ingenious proof by Bollobás.
[hide]
Form the $ n$ sets $ A_i \equal{} \{2p(2i \minus{} 1)\}$, consisting of the $ i$th odd integer less than $ 2n$ together with its multiples by powers of 2. Then the union of these $ n$ sets contains $ \{1, 2, ..., 2n\}$. Hence some two of the selected integers belong to $ A_i$, for some $ i$; and so one of them divides the other. QED

For example, let $ n\equal{}12$. Then this proof suggests forming the following partitions:
$ \{1, 2, 4, 8, 16\}, \{3, 6, 12, 24\}, \{5, 10, 20\}, \{7, 14\}, \{9, 18\}, \{11, 22\}, \{13\}, \{15\}, \{17\}, \{19\}, \{21\}, \{23\}$.
Since there are $ 12$ odd integers, there are $ 12$ sets. We choose $ 13$ elements, for example, $ \{4, 5, 6, 7, 10, 13, 15, 17, 18, 19, 21, 22, 23\}$. Since there are $ 12$ partitions and $ 13$ elements, $ 2$ must be in the same subset, in this case, $ 5$ and $ 10$ are in the same partition. Note that in each partition, each element differs by a power of $ 2$. That also means that any element in the partition divides another, larger element in the same partition. In this case, $ 5$ does divide $ 10$, and the proposition holds.
[/hide]


geometry
perimeter
calculus
integration
geometry unsolved

This is a question of math olympiad of india.It was held today.Please solve the problem.I want to see if my solution were correct.

 Find the number of all integer sided isosceles obtuse angled triangles with perimeter 2008.

[b]Let the two equal sides be 'a' and the greatest side be 'b'. And, Perimeter be denoted by 'p'.
As per the question, b>a>0
and a+a>b [sum of two sides is greater than third side]
i.e. 2a>b>a
So, 2a+2a > b+2a > a+2a
      4a>p>3a
      4a>2008>3a
Solving we get a $ \in$ [503,669]
For every integral value of a, there is a different value of b
And so a different $ \bigtriangleup$ is formed.
So, the no of such triangle is the no of different values that 'a' can possess.
i.e. 669 - 502 = 167 triangles
 [/b]

You forgot the condition that  the triangle is obtuse angled
Please stop creating new topics about the same problems...this must be the third topic of this question

Daniel

i'm getting a wierd answer.....0 ...yes nil=0=none...anyone with the same ans?

i'm doubtful abt it....i'll try it once again and post it in a day or two...

For obtuse angle we will get b^2 > 2*a^2

swapnilshwetank
Let the two equal sides be 'a' and the greatest side be 'b'. And, Perimeter be denoted by 'p'.
As per the question, b>a>0
and a+a>b [sum of two sides is greater than third side]
i.e. 2a>b>a
So, 2a+2a > b+2a > a+2a
4a>p>3a
4a>2008>3a
Solving we get a [503,669]
For every integral value of a, there is a different value of b
And so a different is formed.
So, the no of such triangle is the no of different values that 'a' can possess.
i.e. 669 - 502 = 167 triangles

See; what you've calculated is the number of triangles possible in which the third side is gr8er than the equal sides but you should realize that even if the angle betn the equal sides is 61 degrees then the third side will be gr8er.

i guess that's why ur solution is wrong.


Find the minimum value of the expression:

\[f(a,b,c)= (a+b)^4+(b+c)^4+(c+a)^4 - \frac{4}{7} \cdot (a^4+b^4+c^4).\]

Posted: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=2177]www.mathlinks.ro/Forum/viewtopic.php?t=2177[/url]

Note that $a$, $b$, and $c$ run over all reals.  Also, this is an inequalities problem.

I posted much more beautiful solution on inequalities forum.

Could anyone give the link in which Harazi posted the beautiful solution ?Please

i solved this problem when i found it in old and new inequalities and it was similar with their solution.just denote 2x=a+b 2y=b+c 2z=a+c.i just calculated after but they used some identities

Let's solve with $pqr$ methods .
If $p=0$ the problem is easy .
In other case that $p \neq 0$ , we can let $p=1$ . Then we can write inequality like this in $pqr$ form :

$$3q^4+4q^2+10-108r \geq 0 \to 3q^4+4q^2+10-108r \geq 3q^4+4q^2=q^2(q-4)^2+2q^4+6 \geq 0  $$


In the first two-mile race, Kendi beat Jerri by $ \frac{1}{3}$ mile, and in the second two-mile race, Jerri beat Micah by $ \frac{1}{2}$ mile. If Kendi and Micah run a third two-mile race, and all three runners run at the same speed in every race, by how many mules will Kendi beat Micah?

[hide]3/4

Pick a speed for Kendi say 60 mph. You can find Jerri's speed from speed = distance/time. Jerri would run 50 mph. Now you can find Micahs speed using the speed formula. Micah runs 75/2 mph and so in 2 minutes you can find he runs 1 1/4 miles so he is 3/4 miles behind. Their is probably an easier way.[/hide]

Ok.. I'm not sure about this, but I got a different answer than Echen. I got 5/6.

The reason for this is that if they are still running the same speed, then the same results will come out. Micah-last Jerri-second Kendi-first. So if they already give you the distance between two of them, just imagine it like a x-axis
         1/2              1/3
 l____________l__________l

So doesn't this look simple now?? It's just 1/2 + 1/3!!! Which you get: 5/6.

(Sorry about the messy format. I don't know how to use that Latex thing.)

Do you have the source where you got the problem so I can look it up?

[hide="solution"]

The best way is to set up a proportion table.

Kendi runs 2 miles while Jerri runs 5/3 miles.
Jerry runs 2 miles while Micah runs 3/2 miles.

So when Kendi runs 2 miles, Micah runs $ \frac{5}{3}\cdot\frac{3}{2}$ divided by 2 which is $ \frac{5}{4}$ miles. Thus, Kendi wins by 3/4 of a mile.[/hide]

Um i'm not sure if this will change anything but

"third two-mile race"

The correct answer is $ \frac {3}{4}$. Jonathan, in the second race, Jerri finishes the race, but in the third race, if he were in it, he wouldn't finish, so it's not just $ \frac {1}{2} \plus{} \frac {1}{3}$...


function
superior algebra
superior algebra unsolved

I am trying to find an example of an endomorphism of $\mathbb{R}^{n}$ (or just $\mathbb{R}$ of course) that isn't just a linear transformation (so I'd like to find an endomorphism that does not preserve scalar multiplication), but I can't think of one. I am sure it is easy to give one. Can you?

So you just want $f(x+y)=f(x)+f(y)$¿
One example are the non-continuous solutions of the Cauchy-Equation.

Well, is it possible to constructively give such a function? I just can't think of a construction that works.

No. The existence of such maps depends on the Axiom of Choice, and objects with that dependence are intrinsically impossible to construct with ordinary methods.

but the construction is very simple: since Q is a Q-subspace of R, the AC implies that Q is a direct summand of R. The projection R -> Q yields the desired map.

So... how exactly do you plan to write down the other factor? You can't project without knowing both factors.


inequalities
integration
real analysis
real analysis unsolved

Let $ a_1,\ldots,a_n \in \mathbf C$. Prove that there exists $ x\in[0,1]$ such that
	\[ \left|1 \minus{} \sum_{k\equal{}1}^n a_ke^{2\pi ikx}\right| \ge 1.\]

Let $ p(u)\equal{}1\minus{}\sum_{k\equal{}1}^{n}a_ku^k$. $ sup_{\parallel{}u\parallel{}\leq1}\parallel{}p(u)\parallel{}\equal{}\parallel{}p(v)\parallel{}$ where $ \parallel{}v\parallel{}\equal{}1$  (cf. maximum principle). $ p(0)\equal{}1$ implies the request result.

Here's a solution not using complex analysis.

Set
	\[ f(x) \equal{} \left|1 \minus{} \sum_{k\equal{}1}^n a_ke^{2\pi ikx}\right|^2\]
	then the statement is equivalent to the inequality
	\[ \int_0^1 f(x) dx \ge 1.\]
	Let us prove it. Since
	\[ f(x) \equal{} \left(1 \minus{} \sum_{k\equal{}1}^n a_ke^{2\pi ikx}\right) \left(1 \minus{} \sum_{k\equal{}1}^n \bar a_ke^{\minus{}2\pi ikx}\right),\]
	we have that
	\[ \int_0^1 f(x) dx \equal{} 1 \plus{} \sum_{k\equal{}1}^n |a_k|^2 \ge 1.\]


inequalities
number theory
relatively prime

What's the greatest integer and positive number such that it can't be expressed as the sum of two compossite odd numbers :?:

[quote="vvr1590"]wouldn't it just be the greatest odd integer? no odd integer can be expressed as the sum of two odd integers.[/quote]

The integer has to be even. Forgot that :roll:

[hide]We claim that 38 is the greatest even integer that cannot be expressed as the sum of two composite odd numbers.

Assume for the sake of contradiction that 38 can be expressed as the sum of two composite odd numbers $a$ and $b$, WLOG $a\le 19$.  The only composite odd numbers less than or equal to 19 are 9 and 15, both of which lead to prime $b$, a contradiction.  Therefore 38 cannot be expressed as the sum of two composite odd numbers.

Now we will show that every even integer $n$ greater than 38 can be expressed as the sum of an odd composite multiple of 3 and an odd composite multiple of 5.  Let $n=3(2a+3)+5(2b+3)$, where $a$ and $b$ are nonnegative.  Then we have $n=2(3a+5b)+24$, and $\frac{n-24}{2}=3a+5b$.  By the Chicken Mcnugget Theorem, for an equation $n=k_{1}a+k_{2}b$ with fixed relatively prime positive integers $k_{1}$ and $k_{2}$, and nonnegative integers $a$ and $b$, the greatest number that is not a solution for $n$ is $k_{1}k_{2}-k_{1}-k_{2}$.  Therefore, there is a solution for any $n$ such that $\frac{n-24}{2}\ge 8$.  This inequality is satisfied for all even $n$ greater than $38$, so we're done.[/hide]

EDIT: Made a small change in wording.

[quote="M4RI0"][quote="matt276eagles"][hide] By the Chicken Mcnugget Theorem, for an equation $n=k_{1}a+k_{2}b$ with fixed relatively prime positive integers $k_{1}$ and $k_{2}$, and nonnegative integers $a$ and $b$, the greatest number that is not a solution for $n$ is $k_{1}k_{2}-k_{1}-k_{2}$. [/hide][/quote]

Do you have a link for this theorem :?:[/quote]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=105458]Here[/url] is a topic where it was proved.

[quote="vvr1590"]What is the significance / how did you get the (2a+3) and (2b+3) in the equation for n?[/quote]

In my solution, I proved that every number greater than the answer could be expressed as the sum of an odd multiple of 3 and an odd multiple of 5.  Using $2a+3$ and $2b+3$ guarantees that they are odd multiples, and also guarantees that for nonnegative $a,b$ the multiples are not $1\cdot 3$ or $1\cdot 5$ (which would be bad because those multiples are not composite).

[hide="A related method"] We know that $38$ satisfies our conditions.

Suppose we have $n > 38$.  Consider $n-3, n-9, n-21, n-27, n-33 \bmod 5$.  They are distinct, so one is divisible by $5$.  For $n > 38$ we have $n-33 > 5$ so if one is divisible by $5$, it will be odd and composite.

Equivalently, consider $n-5, n-25, n-35 \bmod 3$. [/hide]


Given f(0) =3, f(1) = 01, and f(n) = f(n-2) - f(n-1), find f(100) in terms of the Fibonacci numbers, where
$ F_0 = F_1 = 1$

[hide=" My solution"]Write out the terms of f(n). You get:
3, -1,  4,  -5,  9,  -14,  23,  -37,  60
The absolute value of this is very similar to the Fibonacci sequence.
From observing, I wrote, for all $ n \ge 3, f(n) = \abs{F(n+1) + n -3$
Also, note that all even terms are positive. Thus, $ f(100) = F(101) + 97.$
[/hide]

[hide=" But the book gives it as"] $ 3F_{100} - 2F_{99}$ [/hide]

[quote="georgiamathdude"]From observing, I wrote, for all $ n \ge 3, f(n) = \abs{F(n + 1) + n - 3$[/quote]
1. You intended to write $ |f(n)| = F(n + 1) + n - 3$ -- there is no such thing as \abs, and you put it in the wrong place.
2. "From observing" is not a very convincing explanation.
3. Apparently, you only observed the cases $ n = 3$ and $ n = 4$?  Because $ f(5) = -14$ while $ F(6) + 5 - 3 = 13 + 2 = 15$.

Heh, I just did this problem like a few hours ago.

Actually, what ends up happening is if you give each f(x) for the small values of x a value in terms of a Fibonacci number (like $ F_2 \plus{} 1$, $ F_5 \plus{} 8$ etc.)

What ends up happening is that you get two Fibonacci sequences...one in terms of the terms itself, and one that is apparent in the adding of the excess numbers (1, 1, 2, 3, 5...)

hmm, this is hard to understand. If you want, I'll post a solution.

When I write out the sequence and observe patterns I get that
for all $ n\ge 4, |f(n)| \equal{} F(n\plus{}1)\plus{}F(n\minus{}4)$

(is this possibly what veezbo was talking about?)

so the answer would be $ f(100)\equal{}F(101)\plus{}F(96)$
this is not the book's answer either


calculus
derivative
function
search
real analysis
real analysis unsolved

If derivatives of all orders exist then is the function necessarily analytic? If not, can you please give a counter-example.

no.
Examples:
http://en.wikipedia.org/wiki/Non-analytic_smooth_function
http://en.wikipedia.org/wiki/Bump_function

The example given in the wikipedia link is

$ f(x)\equal{}\begin{cases}e^{\minus{}1/x}&x>0\\0&x\le0\end{cases}$

which is $ C^{\infty}$ everywhere but fails to be analytic at one point.

Then there's this example:

$ g(x)\equal{}\sum_{k\equal{}1}^{\infty}2^{\minus{}2^{n/2}}\cos(2^nx)$

which is $ C^{\infty}$ everywhere and analytic nowhere (on the real line).

In general how can we define a smooth function on R such that its derivatives of all orders at a fixed point are arbitrarily specified?

[quote="Kent Merryfield"]

Then there's this example:

$ g(x) \equal{} \sum_{k \equal{} 1}^{\infty}2^{ \minus{} 2^{n/2}}\cos(2^nx)$

which is $ C^{\infty}$ everywhere and analytic nowhere (on the real line).[/quote]

Kent, could you please give more details why $ g$ is analytic nowhere? thanks

@§outh§tar

Aha, we came to the Borel theorem! :lol:. You start with constructing for each $ n\in \mathbb Z_ \plus{}$, $ \delta > 0$ and $ a\in\mathbb R$, a $ C^\infty$-function $ F \equal{} F_{n,\delta,a}$ such that $ |F^{(k)}|\le \delta$ on the entire real line and $ F^{(k)}(0) \equal{} 0$ for $ k \equal{} 0,1,\dots,n \minus{} 1$ and $ F^{(n)}(0) \equal{} a$ (can you do it yourself?). Then you just put $ F \equal{} \sum_{n\ge 0}F_{n,2^{ \minus{} n},a_n}$.

fedja I will work on it and tell you what I find.

 :)


trigonometry
logarithms
inequalities unsolved
inequalities

I think all of them can be solved directly by differentiating them?

yes, i have try it out but it is not simple like that

You can see them here:

http://www.mathlinks.ro/Forum/viewtopic.php?t=43122

As darkmaster has pointed out, you have already posted these problems.  Please do not double-post; pick the right forum first.


Refer to the first post of the second page in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=259275]this thread.[/url] Post here saying something along the lines of "I would like to be the moderator for [whatever the next available Tourney is.]

[u][b]MODERATOR SIGN-UPS[/b][/u]

FTW Tourney 1: [color=red]AIME15 [ abandoned ][/color]
FTW Tourney 2: [color=red]isabella2296 [ cancelled ][/color]
FTW Tourney 3: [color=green]jjx1 [ FINISHED!!! - congrats to AIME15 for 1st, isabella2296 for 2nd, and r15s11z55y89w21 and David2143 for 3rd ][/color]
FTW Tourney 4: [color=red]nt811581147 [ cancelled due to ban ][/color]
FTW Tourney 5: [color=red]lotsofmath [ abandoned ][/color]
FTW Tourney 6: [color=red]Meta_Knight [ cancelled due to inactivity ][/color]
FTW Tourney 7:  [color=green]BOGTRO [ FINISHED!!! - congrats to AIME15 for 1st, isabella 2296 for 2nd, and vallon22 for 3rd ][/color]
FTW Tourney 8: [color=green]theprodigy [ FINISHED!!! - congrats to Jongao for 1st, AIME15 for 2nd, and isabella2296 for 3rd ][/color]
FTW Tourney 9: [color=red]ReseesPieces [ abandoned ][/color]
FTW Tourney 10: [color=orange]Dojo [ playing ] [/color]
FTW Tourney 11: [color=orange]vallon22 [ playing - round 2 ][/color]
FTW Tourney 12: [color=red]jonathanchou711 [ cancelled ][/color]
FTW Tourney 13: [color=red]akhilupneja [ inactive ][/color]
FTW Tourney 14: [color=red]andrewnycpops [ inactive ][/color]
FTW Tourney 15: [color=orange]math07 [ sign-ups started ][/color]
FTW Tourney 16: mewto55555
FTW Tourney 17: the pink panther
FTW Tourney 18: cgyao15
FTW Tourney 19: jjx1
FTW Tourney 20: BOGTRO

Hm, I would like to be moderator of FTW Tourney II, FTW Tourney III, etc.

Um, maybe a limit, like a person can only mod one out of every 5 or so?

I would like to be the moderator of the next tourney possible.

Yeah, I was thinking about that. To maximize the number of people who get to host tournies, I was thinking you can't sign up to mod another tourney until the one you are or will be modding is over. What do you guys think?

By the way, I sign-up to mod FTW Tourney 2.

I sign up to mod next available tourney. possibly 3??

[quote="isabella2296"]Yeah, I was thinking about that. To maximize the number of people who get to host tournies, I was thinking you can't sign up to mod another tourney until the one you are or will be modding is over. What do you guys think? [/quote]\

kool
i like it

meta knight are you active enough. I don't see you on a lot......

Just because someone doesn't log on when you are online doesn't mean they're inactive.

Meta Knight is active.  Whether he will be when that tourney starts, I don't know.

Keep the number of signups down, please, or we'll have a list of 30? 40? 50? tourneys pretty soon.

Keep under 15, or 10 even better.

Hmmm... if possible, I'll sign up for next slot.

Might as well sign up to mod the next one.

ill mod one, can someone unmute me for that time slot? or u think my stupid 30 days will be over by then......gah anyways next open slot?

FTW Tourney 1: AIME15 [ currently going on ]
FTW Tourney 2: isabella2296
FTW Tourney 3: jjx1
FTW Tourney 4: nt811581147
FTW Tourney 5: lotsofmath
FTW Tourney 6: Meta_Knight
FTW Tourney 7: ernie
FTW Tourney 8: BOGTRO
FTW Tourney 9: theprodigy
FTW Tourney 10: Dojo

because of the steady decline in activity in the recent weeks and with school starting up, anyone who wishes to start a tourney can now do so by themselves
however, if you do start one, plz try your best to finish <-- as usual, we don't want a bunch of unfinished threads filling up the forum

since this system works well when people are active, we have decided that it will be restarted when there is more activity (in other words, during the holidays)

good luck to everyone who participates and have a good year!

Since summer has started, I decided to open the sign-ups again so we can start having some tournies. even though chat is disabled(time is TBA), we can still have some tournaments going on. I would like to make the first tourney.

Sign- up if you would like to make tournaments. Also, remember to make sure that you are active.

As a thread in the FTW Other Suff, as I'm sure you have figured out now.

[size=1][quote="eagle2"]i made a tourney...[/quote]
okay um gfy eagle2
[/size]
Anyways most tournaments are lacking players and I think 8 is a better idea now

I've made a tournament and I have around 8-10 people so far. May I sign up for the tournament moderator.

This system of organizing tournaments is pretty much nonexistant during the school year when people are busy and do not have the time to create/play in tournaments. As you can see in the forum, there are few active tournaments. Therefore, if you want to start a tournament, feel free to do so, but please do attempt to complete it.

If the number of active tournaments goes back up to where it was during summer, we will likely implement this system again.

I would like to be a Moderator on the next tourny if possible?

Ty, JayZz428xX

[quote="JayZz428xX"]I would like to be a Moderator on the next tourny if possible?[/quote]

Please look at the post right above yours.....

[quote="jeff10"]What is this?[/quote]

This was a system designed during the summer to reduce the number of unfinished tournaments. During that time, many people apparently had little to do, so there was quite a bit of activity on FTW. People started tournaments but due to the immense amount of tournaments going at any one time, few of them actually finished. This system was an attempt to reduce the number of tournaments going on at any one time, and thus promote the finishing of tournaments.

I no, OOPS! I thought my post got deleted so i doubled posted. Now I see the above post.
Thank you,
JayZz428xX

well once this ones over could i be one?

This system of hosting tournaments is suspended. Feel free to host one whenever you want.


Let $A_{1}, A_{2},...A_{63}$ be the nonempty subsets of $[{1,2,3,4,5,6}]$.  For each of these sets $A_{i}$ let $\pi(A_{i})$ deonote the product of all the elements in $A_{i}$.  Then the value of $\pi(A_{1})+\pi(A_{2})+...+\pi(A_{63})$ is?

[quote="p4fn2w"][hide="my attempt"]
I reinterpret the question as finding the sum of the first through sixth symmetric sum of the subset $[1, 2, 3, 4, 5, 6]$.

To find this, we first make a polynomial with all the elements as roots.

$P(x) = (x-1)(x-2)(x-3)(x-4)(x-5)(x-6)$

And unless I did a miscalculation (I sort of brute forced this), the polynomial is equivalent to

$P(x) = x^{6}-21x^{5}+175x^{4}-735x^{3}+1624x^{2}-1762x+720$

Let $S_{i}$ denote the $i^{th}$ symmetric sum of the roots.

With vieta's formula, we know that:
$S_{1}= 21$
$S_{2}= 175$
$S_{3}= 735$
$S_{4}= 1624$
$S_{5}= 1762$
$S_{6}= 720$

So the sum of the first through sixth symmetric sum of the group is
$21+175+735+1624+1762+720 = 5037$[/hide][/quote]
i think you're off by 2...
[hide="hint"]
if you add 1, you get a very nice expression  :D
[/hide]

[hide="Solution"]
It is equivalent to finding $\sum_{i=1}^{6}\sigma_{i}$ of the set, i.e. $\prod_{k=1}^{6}(1+k)-1 = \boxed{5039}$. [/hide]

[hide="Generalization"]The product of all subsets of $\{1,2,3,\ldots n\}$ (including the empty set) is $(1+1)(1+2)(1+3)\ldots(1+n)$, or $(n+1)!$.  Then we must subtract the product of the empty set, which would simply be the empty product: $1$.  So then it would be $\boxed{(n+1)!-1}$.[/hide]

would this work?
[hide]
(1+1)(2+1)(3+1)(4+1)(5+1)(6+1)-1
since every memeber can either be in the set or not, but the empty subest isn't included?
[/hide]

[quote="lingomaniac88"]The product of all subsets of $\{1,2,3,\ldots n\}$ (including the empty set) is $(1+1)(1+2)(1+3)\ldots(1+n)$, or $(n+1)!$.  Then we must subtract the product of the empty set, which would simply be the empty product: $1$.  So then it would be $\boxed{(n+1)!-1}$.[/quote]

I believe that works for any subset of any number its $(1+n_{1})(1+n_{2})...(1+n_{k})-1$ for any subset in the form $(n_{1},n_{2},...n_{k})$ and yea your answers are all correct.


Let x,y,z be non-negative real numbers such that $ x\plus{}y\plus{}z\equal{}1$.Find the maximum value of the following expression
$ P\equal{}(x\plus{}2y\plus{}3z)(6x\plus{}3y\plus{}2z)$

[quote="WMB-NP"]Let x,y,z be non-negative real numbers such that $ x \plus{} y \plus{} z \equal{} 1$.Find the maximum value of the following expression
$ P \equal{} (x \plus{} 2y \plus{} 3z)(6x \plus{} 3y \plus{} 2z)$[/quote]
Applying the AM-GM Inequality, we have
$ P\le \frac{1}{8}[2(x\plus{}2y\plus{}3z)\plus{}(6x\plus{}3y\plus{}2z)]^2\equal{}\frac{1}{8}(8x\plus{}7y\plus{}8z)^2 \le \frac{1}{8}(8x\plus{}8y\plus{}8z)^2\equal{}8,$
with equality holds iff $ 2(x\plus{}2y\plus{}3z)\equal{}6x\plus{}3y\plus{}2z,y\equal{}0$ and $ x\plus{}y\plus{}z\equal{}1,$ i.e. $ x\equal{}z\equal{}1/2,y\equal{}0.$


function
limit
real analysis
real analysis unsolved

If $ f(x)$ is differentiable on $ \mathbb{R}$, prove that $ f'(x)$ is Borel measurable.

Is arguing as follows correct?
Differentiability implies continuity of f, which implies that the set $ [x: f(x) >\alpha]$ is open and hence a Borel set so that f is Borel measurable. The sequence of functions $ {g_{n}} \equal{} n[f(x \plus{} \frac{1}{n}) \minus{} f(x)]$ is measurable. Since $ f'(x) \equal{} \lim \sup g_{n}(x)$ is measurable, we are done.

Yes, this will work (provided that you have covered all the theorems you quoted in class).


vector
symmetry
geometry
circumcircle
complex numbers
geometric series
complex analysis

Prove that the sum of all n roots of $ z^{\frac{1}{n}}$ is zero.  Give a vector interpretation of your result.

for $ n\geq 2$ we are considering the roots of $ x^n\minus{}z\equal{}0$. As the coefficient of $ x^{n\minus{}1}$ is $ 0$, by Viete's formula we get that the sum of the roots is $ 0$. for the vector interpretation notice that our roots lie in the circle centered at the origin with radius $ |z|^{\frac{1}{n}}$
and are equidistributed with separation angle $ \frac{2\pi}{n}$. If we consider these complex numbers as vectors the result follows by symmetry :wink:

[quote="Albanian Eagle"]for $ n\geq 2$ we are considering the roots of $ x^n \minus{} z \equal{} 0$. As the coefficient of $ x^{n \minus{} 1}$ is $ 0$, by Viete's formula we get that the sum of the roots is $ 0$. for the vector interpretation notice that our roots lie in the circle centered at the origin with radius $ |z|^{\frac {1}{n}}$
and are equidistributed with separation angle $ \frac {2\pi}{n}$. If we consider these complex numbers as vectors the result follows by symmetry :wink:[/quote]

Can this be proved without Viete's formula?
http://en.wikipedia.org/wiki/Vi%C3%A8te's_formulas

It's equivalent to the following statement:  the centroid of a regular $ n$-gon inscribed in a circle centered at the origin is the origin.  And this is obvious by considering the symmetries of a regular $ n$-gon, which leave its centroid (which is therefore its circumcenter) fixed.

Another way to prove it:

We know that the $ n$th roots of unity can be written as $ e^{2\pi i k/n} , 1\le k\le n$. Hence we want to prove that
$ e^{2\pi i /n} + e^{4\pi i/n} + \ldots + e^{2\pi i (n - 1)/n} + 1 = 0$.

Assume not. We have in general that if $ a\not = 0, b\not = 0,1$, then $ ab\not = a$. But

$ (e^{2\pi i /n} + e^{4\pi i/n} + \ldots + e^{2\pi i (n - 1)/n} + 1)e^{2\pi i /n} = (e^{2\pi i /n} + \e^{6\pi i/n} + \ldots + 1 + e^{2\pi i /n})$ and hence we must have
$ (e^{2\pi i /n} + e^{4\pi i/n} + \ldots + e^{2\pi i (n - 1)/n} + 1)e^{2\pi i /n} = 0$.

Another proof without Viete:

The n-th roots of $ 1$ are $ z_k \equal{} \zeta^k$ where $ \zeta \equal{} \exp\left(\frac {2\pi}{n}\right)$ and $ k \equal{} 0,\ldots,n \minus{} 1$. Now sum a geometric series:
$ \sum z_k \equal{} 1 \plus{} \zeta \plus{} \zeta^2 \plus{} \ldots \plus{} \zeta^{n \minus{} 1} \equal{} \frac {\zeta^n \minus{} 1}{\zeta \minus{} 1} \equal{} 0$

The result for $ z^{1/n}$ follows by scaling.


function
calculus
derivative
real analysis
real analysis unsolved

Suppose $f$ differentiable on $[a,b]$ and $f'(x)>0$. Prove that $f^{-1}(x)$ is absolutely continuous on $[f(a),f(b)]$.

I think just the condition $f$ is strictly increasing will ensure the conclusion

There is a book titled "Counterexamples in Real Analysis". I don't have it now but I have an index of the book.
There is a item "a strictly increasing function, it is absolutely continuous, but its inverse function is not absolutely continuous"

On the other hand, the stated condition makes it quite easy. By the Inverse Function theorem, $(f^{-1})^{\prime}(f(x))=\frac1{f^{\prime}(x)}$. Since the inverse is differentiable, it is continuous.

sorry...i forget to say that we want to prove $f^{-1}(x)$ is ABSOLUTELY continuous...
(it has happened so many times :( )

There's one thing I don't understand about this formulation: we know that $f:[a,b]\to[f(a),f(b)]$ is a differentiable and strictly increasing bijection. Doesn't $f^{-1}:[f(a),f(b)]\to[a,b]$ share the same properties? Why not just say that the problem asks for the absolute continuity of a differentiable function with positive derivative defined on a compact interval? Why go through the trouble of introducing $f^{-1}$? :?

Am I missing something here? :?


I think the sign must be $\geq $
can it be useful as following?
$ a_1,a_2,b_1,b_2.c_1,c_1,d_1,d_2$ are positive numbers.
if we have $\frac{a_1}{b_1} \leq \frac{c_1}{d_1},\frac{a_2}{b_2} \leq \frac{c_2}{d_2},a_1 \leq c_1,a_2 \leq c_2$
then we have $\frac{b_1+b_2}{a_1 a_2}  \geq \frac{d_1+d_2}{c_1 c_2} $ :)

[quote="harazi"]I would like very much to see a complete and correct solution for this problem. I did not manage to solve it.  :([/quote]
didn't you notice my $lemma$:
for $ a_1,a_2,b_1,b_2.c_1,c_1,d_1,d_2$  are positive numbers.
if we have $\frac{b_1}{a_1} \geq \frac{d_1}{c_1},\frac{b_2}{a_2} \geq \frac{d_2}{c_2},a_1 \leq c_1,a_2 \leq c_2 $,then we have $ \frac{b_1+b_2}{a_1 a_2} \geq \frac{d_1+d_2}{c_1 c_2}$
proof for the $lemma$:because $b{}_1c{}_1c{}_2 \geq a{}_1d{}_1a{}_2$ and $b{}_2c{}_2c{}_1 \geq a{}_2d{}_2 a_1$
so $(b_1+b_2) c_1 c_2-(d_1+{}_d{}_2)a_1 a_2 \geq 0$.over.
for our problem:it is clear that $ \frac{x_1}{x_1} \geq \frac{1-x_1}{1-x_1}$ , $ \frac{x_2}{x_2} \geq \frac{1-x_2}{1-x_2}$ ,$x_1 \geq 1-x_1$ and $x_2 \geq 1-x_3$
so we have  $\ds \frac {\sum_{i=1}^2 x_i^4}{\prod_{i=1}^2 x_i} \geq \frac{\sum_{i=1}^2(1-x_i)^4}{\prod_{i=1}^2(1-x_i)}$
and also   $\ds \frac {\sum_{i=3}^4 x_i^4}{\prod_{i=3}^4 x_i} \geq \frac{\sum_{i=3}^4(1-x_i)^4}{\prod_{i=3}^4(1-x_i)}$
and with the $lemma$ we have $\ds \frac {\sum_{i=1}^4 x_i^4}{\prod_{i=1}^4 x_i}>\frac{\sum_{i=1}^4(1-x_i)^4}{\prod_{i=1}^4(1-x_i)}$

forgive my careless.I don't notice $4$ in $x_{i}^4$ :blush:
so I was wrong.


(a) Let $D_{n}$ denote the determinant of the $n$-by-$n$ matrix whose $(i, j)$th element ($i$th row, $j$th column) is the absolution value of the difference of $i$ and $j$. Show that $D_{n}=(-1)^{n-1}(n-1)2^{n-2}.$
(b) Let $F_{n}$ denote the determinant of the $n$-by-$n$ matrix with $a$ on the main diagonal, $b$ on the superdiagonal (immediately above main diagonal w/ $n-1$ entries), and $c$ on the subdiagonal (the diagonal immediately below the main diagonal , w/ $n-1$ entries). Show that $F_{n}=aF_{n-1}-bcF{n-2}, n>2$.
(d) Evaluate the $n$-b-$n$ determinant $A_{n}$ whose $(i,j)$th entry is $a^{|i-j|}$ by finding a recursive relationship between $A_{n}$ and $A_{n-1}$.

[quote="me@home"]Phelpedo, you keep posting these problems from a book called "Larson". What book is this? Could you post a link or something??[/quote]

problem solving through problems, at the rate he's posting the problems, i don't even think i have to buy the book :D

Yes, the book is called Problem Solving through Problems by Loren Larson.

I guess you can kinda chart my progress through it  :lol:


[u]       [b]              hi!
             every one, I have a question?
            Can we use lagranje in IMO?
                        thans![/b][/u]
 :blush:

As far as I remember, as long as your solution is [b]flawless[/b], it may use Lagrange multipliers or any other result from standard university courses. However, if it has mistakes, grading will be very strict if you use advanced stuff, and believe me, it is really easy to make mistakes when applying calculus.

  Darij

By the way,

 I have in paper (printed) the IMO -  short list 2006 , but does anyone have this list in pdf ?

  Babis


Find the smallest positive integer n such that 2n −1 is divisible by 47.

$ n \equal{} 24$ Or did you mean $ 47|2^{n} \minus{} 1$

In the second case.

We note $ 2^{47}\equiv 2 (mod 47)$
Hence $ 2^{46}\equal{}47a\plus{}1$

We claim no smaller value exists.

Supposed c is the smallest number such that $ 47|2^c\minus{}1$. $ 47\equal{}mc\plus{}r$

$ 2^c\equal{}47d\plus{}1$, so $ 2^{mc} \equal{} (47d\plus{}1)^{m} \equal{} 47k\plus{}1$ However, $ 2^{46}\equal{}41a\plus{}1\equal{}2^{mc\plus{}r}\equal{}(47d\plus{}1)(47f\plus{}s)$

Hence $ s\equal{}1 \Rightarrow 47|2^{r}\minus{}1$ Contradiction

Hence $ r\equal{}0, c|46$

Actually, $ 2^{23}\equiv 1\pmod{47}$. (A simple way to see this is that $ 2\equiv 7^2$, so $ 2^{23}\equiv 7^{46}$.)

Now 23 is the smallest power, since the order of 2 modulo 47 must divide 23 and clearly can't be 1.

[quote="SimonM"]$ r \equal{} 0, c|46$[/quote]

This is true.  What you concluded was that $ c \equal{} 46$, which isn't.


A book has 500 pages numbered 1, 2, 3, and so on.  How many times does the digit 1 appear in the page numbers?

[hide]  There are 5 * 10 *10 values in all.  When the 1 is in the units place there are 5*10=50 combos.  When the 1 is in the tens place there are 5*10=50 combos.  And when the 1 is in the hundreds palce there are 10*10=100 combos.  [b] Altogether there are 50+50+100=200 cases in which the digit "1" appears [/b] [/hide]

[hide]There are 20 ones total in each of the numbers 0 - 99, 200 - 299, 300 - 400, and 400 - 500, so there are 80 ones total there.
Then in the numbers from 100 - 199, an extra one is added to the each of the numbers from 0 - 99, so there are 100 more ones than the number of ones in the number set 0 - 99, so there are 120 ones.
120 + 80 = 200.[/hide]


Three points are randomly and uniformly chosen on a circle's circumference.  What is the probability that no two of these points will form an obtuse triangle with the circle's center?

[hide]These three points form 3 angles with the center of the circle. They sum to 360 degrees. Therefore at least one has to be obtuse (otherwise the sum would be at most 270 degrees), so the answer is 0.[/hide]

[hide]Not true. The points could all be within a 90-degree sector.[/hide]

[hide="my solution"]Based on Jerod Mayo's excellent solution to a different problem:
Place point $a$ arbitrarily.  For the sake of the rest of this example, every other point will be in relation to it, between -180 degree- and 180 degree- arcs.

Some quick sketches show that for $a$ and $b$ to not form an obtuse triangle, $b$ can be no further than a 90 degree arc away from $a$.  That covers half the circle.

Now we create a Cartesian probability map, again from -180 to 180 to represent the positions of both points.

On the x-axis, let's map the location of $b$, where $-90<=x<=90$ is an acute or right triangle.

Let's now look at $x<=90$.  $c$ has to be within that same 90 degree arc, but now from both of the points that are already placed.  At one edge of the allowable region, where $x=90$, the legal region is $0<=y<=90$ (0 is the lower bound of how far $c$ can be from $b$, and 90 is the upper bound of how far $c$ can be from $a$).  Same for the other edge.  At the center of the allowable region, where $x=0$, the legal region is the same as when we were dealing with just $a$, because they're at the same point.  That's $-90<=y<=90$.

Map those points to indicate the ranges of the legal possibilities: $(90,0)$, $(90, 90)$, $(-90,0)$,$(-90,-90)$, $(0,90)$, $(0,-90)$.  Draw lines between them, forming an irregular hexagon, and scratch out everything outside the allowed region.  At this point, your "map" should clearly show $\boxed{3/16}$ of the total region--two $1/16$ squares and two $1/32$ triangles.
[/hide]

Nice solution MindWanderer! My solution is similar. I decided to fix one of the points at the top of the circle, so in relation each of the other two points would form an $-180<x<180$ angle. Then I graphed the possibilities. Note that I did casework on which Quadrant the other two points are in.

Here's it visually:

What do you mean by $(x,y) < 90$? Do they represent coordinates?

[hide=Solution]Using a geometric probability approach similar to shiningsunnyday's, we get that for general $n \le 120$, the probability that if we choose $3$ points randomly on a circle's circumference, the probability that they all lie on a $n$ degree arc is $\left(\frac{3}{4}\right)\left(\frac{n}{180}\right)^2$. Apply this to $n = 90$ and get $\boxed{\frac{3}{16}}$[/hide]

[hide=sketch of my solution]WLOG we will place the first point somewhere.  There is a $\left(\frac12\right)^2=\frac14$ chance that both of the other points do not form arcs greater than $\frac{\pi}{2}$ with the first point.  Then a basic graph shows that there is a $\frac34$ chance that the second and third points placed fall within $\frac{\pi}{2}$ radians of each other on the circle.  We multiply these together to get $\boxed{\frac{3}{16}}$.
$\boxed{}$[/hide]

[quote=coolmath_2018]This is similar to 2003 AMC12B Problem 25

Why is 3/8 wrong?
[hide=My Solution]Note that for not creating obtuse triangles we need the angle at the center to be less than 90 degrees.

Call the first point we pick point A. We can pick A anywhere we want. Now call the second point we pick, B. We can pick B on either side of A(90 degrees left or right). So on the circumference, this is $\frac{1}{4} \cdot 2$. Then let's pick point C. C has to be 90 degrees from A and 90 degrees from B. So we have a 135-degree sector. This means it is $\frac{3}{4}$ of the circumference.

Therefore we have $\frac{1}{2} \cdot \frac{3}{4} = \frac{3}{8}$.....
[/hide][/quote]

Please help

[hide=My solution]Without loss of generality, let's assume that the circle has its center at the origin of the co-ordinate axes such that the first of the points passes through the x-axis and the second is above the x-axis. The third point can in any quadrant. Of the 8 cases where the second and the third points can lie in a quadrant (1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4). All but the cases (1,1) and (1,4) create obtuse angles. In  the case (1,4), we get an acute angle half the time and an obtuse angle the other half of the time. You can realize this by rotating the third point which is on the fourth quadrant counterclockwise by 90 degrees. If it lies between the x-axis and the second point, there is an obtuse angle and if it lies between the y-axis and the second point, there is no obtuse angle. Therefore, the probability is 1/8+ 1/2 of 1/8 = 3/16[/hide]

Cheese the Level 25 Counting & Probability +125 XP :clap2:

[hide]
Choose a random point $A$ on the circle; the probability that $\angle AOB \leq 90^\circ$ is $\tfrac{1}{2}.$ Now, when $\angle AOB = 90^\circ,$ the probability of $C$ satisfying the conditions of the problem is $\tfrac{1}{4},$ and when $\angle AOB = 0^\circ,$ the probability of $C$ satisfying the conditions of the problem is $\tfrac{1}{2}.$ From there, assume that the probability of $C$ satisfying the conditions of the problem is linear as $C$ goes from $A$ to $P$ s.t. $\angle AOP = 90^\circ;$ this makes the requested probability
$$\frac{1}{2} \left(\frac{\frac{1}{2} + \frac{1}{4}}{2}\right) = \frac{1}{2} \cdot \frac{\frac{3}{4}}{2} = \frac{1}{2} \cdot \frac{3}{8},$$
or $\boxed{\tfrac{3}{16}}.$
[/hide]

[hide=yooo free level 25] because you don’t know where first point is it really doesn’t matter so you label it A. There is a 1/2 chance it will be not obtuse if you just take those triangles. Next, you realize that this next point has to be in the 1/2 of the circumference the other point was on, so the minimum it could has a chance to be on is 1/4, the maximum is 1/2, average to get 3/8. 1/2*3/8=3/16, so that is our answer.


geometry
trigonometry
analytic geometry
vector
geometric transformation
rotation
similar triangles

That's a vague question, so I respond with a vague answer:

[i]Review/Learn fundamental concepts [/i]
* similar triangles, congruence, collinearity, concurrency (for synthetic methods)
* trigonometry, coordinate systems, complex numbers, vectors (analytic methods)
* homotheties, rotations, spiral similarity (more advanced, transformational methods)

[i]Do lots of problems[/i]; if you can't solve them, make sure you've spent a lot of time (days even) before looking at any solutions, and if you do, try to figure out how you could have solved the problem from the solution -- figure out main ideas and come back to the problem in a few days to see if you can rework the solution without it in front of you.

For geometry problems in particular, you want to [i] have the right tools [/i]. At home, I'd usually sketch diagrams using Geometer's Sketchpad, or kseg, or some dynamic geometry software. But for actual competitions, those are, of course, not allowed. You want to have a good straightedge and compass, and make sure you make a few large, clean diagrams (I'd use a sheet of paper for each), so you can make and test conjectures on the fly.

[quote="cnyd"]Hello,I'm studying for math olympiad. But my geometry is very bad. I can't solve any problem. What can I do?
What do you advise me?[/quote]
Do you have a teacher?


inequalities
induction
function
calculus
derivative
rearrangement inequality
inequalities open

I'm sorry, I've done a mistake in my last post:

Determine the n-somes of [color=red]positive[/color] real numbers (x_1, x_2, ..., x_n) that the next inequality is held for.

x_1/(x_2 + x_3) + x_2/(x_3 + x_4) + ... + x_n/(x_1 + x_2)  \geq n/2

I know that it's called "Shapiro inequality" and it's valid for every n-some of real positive numbers if n is even \leq  12 and odd \leq  23.

But I think that I should look for some special cases of n-somes (x_1, x_2, ..., x_n) such that the inequality is held for each n. For example I have an idea that if

 x_1 \leq  x_2 \leq  .... \leq  x_n

the inequality is valid (but without proof, it's only an idea).
Does anybody have another idea, what do you think - what could be the aim of the task?

I have tried to find a counter example, but failed. Maybe you are right ... :?

you realise that the example you gave is wrong :) cuz 2n/3 > n/2 :D

of course I just now noticed that Dapet has wrote the ineq. in the reverse order. I have edited his post.

Of course Shapiro's inequality is true for $x_1 \leq x_2 \leq \dots \leq x_n$ (and for
$x_1 \geq x_2 \geq \dots \geq x_n$ too).  It's only induction and monotony.

The first case.
Let $f_n = \sum \frac{x_k}{x_{k+1} + x_{k+2}}$.
We prove that $f_{n+1} \geq f_n + 1/2$.
Concider
\[ f_{n+1} - f_n = \frac{x_{n-1}}{x_n + x_{n+1}} + \frac{x_n}{x_{n+1} + x_1} +
     \frac{x_{n+1}}{x_1 + x_2} - \frac{x_{n-1}}{x_n + x_1} - \frac{x_n}{x_1 + x_2} .\]
Terms with $x_{n-1}$ are
\[ \frac{x_{n-1}}{x_n + x_{n+1}} - \frac{x_{n-1}}{x_n + x_1} =
  x_{n-1} \cdot \frac{x_1 - x_{n+1}}{(x_n + x_{n+1})(x_n + x_1)} ,\]
it's $x_{n-1}$ multiplied by nonpositive number. Hence we can put $x_{n-1} = x_n$. Terms with $x_2$ are $\frac{x_{n+1} - x_n}{x_1 + x_2}$. Trerefore we can put $x_2 = x_{n-1} = x_n$.
Consequently we need to prove inequality
\[ \frac{x_{n+1} - 2 x_n}{x_1 + x_n} + \frac{x_n}{x_n + x_{n+1}} +
     \frac{x_n}{x_1 + x_{n+1}} \geq \frac{1}{2} .\]
Concider the terms with $x_1$:
\[ \frac{x_{n+1} - 2 x_n}{x_1 + x_n} + \frac{x_n}{x_1 + x_{n+1}} =
     \frac{(x_{n+1} - x_n)(x_{n+1} + x_1 - x_n)}{(x_1 + x_n)(x_1 + x_{n+1})} .\]
Numerator is positive, so fraction is decrease with increasing $x_1$.
Thus we can put $x_1 = x_n$. And so we must to prove inequality
\[ \frac{x_{n+1}}{2 x_n} + 2 \cdot \frac{x_n}{x_n + x_{n+1}} \geq \frac{3}{2} ,\]
but it's Shapiro inequality for three numbers.

The second case is similar but a little bit easier.

well U can't say that $ x_1\equal{}x_n$ , numerator is positive but it depends on $ x_1$ if we look at the
diferential we can see that fraction isn't decrease when x_1 increasing...

just put $ x_1\equal{}1$ $ x_n\equal{}10$ i $ x_{n\plus{}1}\equal{}11$ then induction is failed.

Let $ 0 < x_{1}\leqslant x_{2}\leqslant\ldots \leqslant x_{n}, \ x_{n \plus{} 1} \equal{} x_{1}, \ x_{n \plus{} 2} \equal{} x_{2}$
$ \sum_{i \equal{} 1}^{n}\frac {x_{i}}{x_{i \plus{} 1} \plus{} x_{i \plus{} 2}} \equal{} \sum_{i \equal{} 1}^{n} \frac {x_{i} \plus{} x_{i \plus{} 1}}{x_{i \plus{} 1} \plus{} x_{i \plus{} 2}} \minus{} \sum_{i \equal{} 1}^{n}\frac {x_{i \plus{} 1}}{x_{i \plus{} 1} \plus{} x_{i \plus{} 2}}$
From AM-GM:
$ \sum_{i \equal{} 1}^{n} \frac {x_{i} \plus{} x_{i \plus{} 1}}{x_{i \plus{} 1} \plus{} x_{i \plus{} 2}} \geqslant n$
Now we will show that:
$ \sum_{i \equal{} 1}^{n}\frac {x_{i \plus{} 1}}{x_{i \plus{} 1} \plus{} x_{i \plus{} 2}} \leqslant \frac {n}{2}$
This inequality is equivalent to:
$ f(t_{1},\ldots,t_{m}): \equal{} \sum_{i \equal{} 1}^{m}\frac {1}{1 \plus{} t_{i}} \plus{} \frac {t_{1}\cdot\ldots\cdot t_{m}}{1 \plus{} t_{1}\cdot\ldots\cdot t_{m}}\leqslant \frac {m \plus{} 1}{2}$
for $ m \equal{} n \minus{} 1$, where $ t_{i} \geqslant 1,\ i \equal{} 1,\ldots, m$
(just put $ t_{i} \equal{} \frac {x_{i \plus{} 1}}{x_{i}}$).
For $ t_{1} \equal{} \ldots \equal{} t_{m} \equal{} 1$ (in)equality holds. Since $ f$ is symmetric then it suffices to prove that for every $ t_{2},\ldots, t_{m}\geqslant 1$ function
$ [1, \plus{} \infty)\ni x\stackrel{\varphi}{\mapsto} f(x, t_{2},\ldots t_{m})$ is decreasing.
But for $ t_{2},\ldots, t_{n} \geqslant 1$
$ \varphi'(x) \equal{} \minus{} \frac {1}{(1 \plus{} x)^{2}} \plus{} \frac {t_{2}\cdot\ldots\cdot t_{m}}{(1 \plus{} xt_{2}\cdot\ldots\cdot t_{m})^{2}} \equal{} \\
\equal{} \frac {(x^{2}t_{2}\cdot\ldots\cdot t_{m} \minus{} 1)(1 \minus{} t_{2}\cdot\ldots\cdot t_{m})}{(1 \plus{} xt_{2}\cdot\ldots\cdot t_{m})^{2}(1 \plus{} x)^{2}}\leqslant 0$
and that's all:)

edit:
If someone doesn't like my proof (for example because I used derivative;)), here is a brilliant idea which I've found in [i]Słynne nierówności[/i] by Lev Kourliandtchik (the title may be translated as [i]Famous Inequalities[/i], but I don't think this book is avaliable in english):
From AM-GM:
$ \sum_{i \equal{} 1}^{n} \frac {x_{i} \plus{} x_{i \plus{} 1}}{x_{i \plus{} 1} \plus{} x_{i \plus{} 2}} \geqslant n \equal{} \sum_{i \equal{} 1}^{n}\frac {x_{i} \plus{} x_{i \plus{} 1}}{x_{i} \plus{} x_{i \plus{} 1}}$
hence:
$ (*)\ \sum_{i \equal{} 1}^{n}\frac {x_{i}}{x_{i \plus{} 1} \plus{} x_{i \plus{} 2}}\geqslant \sum_{i \equal{} 1}^{n}\frac {x_{i \plus{} 1}}{x_{i} \plus{} x_{i \plus{} 1}}$
Using Rearrangement Inequality we get:
$ (**)\ \sum_{i \equal{} 1}^{n}\frac {x_{i}}{x_{i \plus{} 1} \plus{} x_{i \plus{} 2}} \geqslant \sum_{i \equal{} 1}^{n \minus{} 2}\frac {x_{i}}{x_{i \plus{} 1} \plus{} x_{i \plus{} 2}} \plus{} \frac {x_{n \minus{} 1}}{x_{1} \plus{} x_{2}} \plus{} \frac {x_{n}}{x_{n} \plus{} x_{1}} \geqslant \sum_{i \equal{} 1}^{n}\frac {x_{i}}{x_{i} \plus{} x_{i \plus{} 1}}$
Adding $ (*)$ and $ (**)$ we obtain:
$ 2\sum_{i \equal{} 1}^{n}\frac {x_{i}}{x_{i \plus{} 1} \plus{} x_{i \plus{} 2}} \geqslant n$
$ \blacksquare$
(the same reasoning shows that the inequality also holds for $ x_{1}\geqslant \ldots\geqslant x_{n} > 0$)


Find all functions $ f: \mathbb R \longrightarrow \mathbb R$ such that
\[ xf(y) \minus{} yf(x) \equal{} f\left(\frac {y}{x}\right),\] for all $ y\in \mathbb R$ and $ x\in \mathbb R\
\{0\}$.

[quote="mathVNpro"]Let $ f: \mathbb R \longrightarrow \mathbb R$. Find all $ f$ such that:
$ xf(y) \minus{} yf(x) \equal{} f(\frac {y}{x})$, for all $ y\in \mathbb R$ and $ x\in \mathbb R\ \{0\}$.[/quote]

Let $ P(x,y)$ be the assertion $ xf(y)\minus{}yf(x)\equal{}f(\frac{y}{x})$

$ P(1,1)$ $ \implies$ $ f(1)\equal{}0$
$ P(2,0)$ $ \implies$ $ f(0)\equal{}0$

Let $ x\neq 0$. Then :
$ P(x,1)$ $ \implies$ $ f(\frac{1}{x})\equal{}\minus{}f(x)$

$ P(\frac{1}{x},2)$ $ \implies$ $ f(2x)\equal{}\frac{1}{x}f(2)\minus{}2f(\frac{1}{x})$ $ \equal{}\frac{1}{x}f(2)\plus{}2f(x)$

$ P(\frac{1}{2},x)$ $ \implies$ $ f(2x)\equal{}\frac{1}{2}f(x)\minus{}xf(\frac{1}{2})$ $ \equal{}\frac{1}{2}f(x)\plus{}xf(2)$

So $ \frac{1}{x}f(2)\plus{}2f(x)\equal{}\frac{1}{2}f(x)\plus{}xf(2)$

And $ f(x)\equal{}\frac{2f(2)}{3}(x \minus{} \frac{1}{x})$

So we have :
$ f(0)\equal{}0$
$ f(x)\equal{}a(x \minus{} \frac{1}{x})$ $ \forall x\in \mathbb{R}^*$, and for any real $ a$.

And it's easy to verify that this function match the original assertion.


limit
trigonometry
calculus
calculus computations

Another limit: $ L \equal{}\lim_{(x,y)\to(0,0)}\frac{x^{2}y^{2}}{x^{2}\plus{}y^{2}}$.

Here's my attempt:

Computing the limit along the trajectory $ y \equal{} mx$ we get

$ L \equal{}\lim_{x\to 0}\frac{x^{2}\cdot m^{2}x^{2}}{x^{2}\plus{}m^{2}x^{2}}\equal{}\lim_{x\to 0}\frac{m^{2}x^{2}}{1\plus{}m^{2}}\equal{} 0$.

Now along the trajectory $ y \equal{} x^{2}$:

$ L \equal{}\lim_{x\to 0}\frac{x^{2}\cdot x^{4}}{x^{2}\plus{}x^{4}}\equal{}\lim_{x\to 0}\frac{x^{4}}{1\plus{}x^{2}}\equal{} 0$.

So the limit appears to be 0. Let's use the definition: we want to show that, for every $ \epsilon > 0$ there is a $ \delta > 0$ such that $ 0 <\sqrt{x^{2}\plus{}y^{2}}<\delta$ implies that $ \left|\frac{x^{2}y^{2}}{x^{2}\plus{}y^{2}}\right| <\epsilon$. But:

$ \left|\frac{x^{2}y^{2}}{x^{2}\plus{}y^{2}}\right| \equal{}\frac{x^{2}y^{2}}{x^{2}\plus{}y^{2}}\le y^{2}$
$ \le x^{2}\plus{}y^{2}$.

As $ \delta > 0$ then $ \sqrt{x^{2}\plus{}y^{2}}<\delta\Rightarrow  x^{2}\plus{}y^{2}<\delta^{2}$, and so

$ \left|\frac{x^{2}y^{2}}{x^{2}\plus{}y^{2}}\right|\le\delta^{2}$.

Now choosing $ \delta \equal{}\sqrt{\epsilon}$, then $ \left|\frac{x^{2}y^{2}}{x^{2}\plus{}y^{2}}\right|\le\epsilon$ as we wanted. So L = 0.

Is this reasoning correct?

Yes, although too long  :wink: . $ (x\minus{}y)^{2}\geq 0,\frac{1}{x^{2}\plus{}y^{2}}\leq\frac{1}{2xy}$, and then use the "squeeze" theorem.
All other examples are more or less the same.

Edit: Computing limit along some trajectories doesn't mean anything, it's better used as a counterexample (that a limit doesn't exists), rather then a indication of what limit should be.

Let $ (x, y) \equal{} (r\cos t, r\sin t)$.

(1) $ \lim_{(x,y)\to(0,0)}\frac{3x^{2}y}{x^{2}\plus{}y^{2}}\equal{}\lim_{r\to 0}3 r\cos^{2}t\sin t \equal{} 0$ by squeezing theorem.

Similarly,

$ \frac{2xy^{4}}{(x^{2}\plus{}y^{2})^{2}}\equal{} 2r\cos t\sin^{4}t$

$ \frac{x^{2}\minus{}y^{2}}{\sqrt{x^{2}\plus{}y^{2}}}\equal{} r\cos 2t$

$ \frac{2x^{3}\minus{}y^{3}}{x^{2}\plus{}y^{2}}\equal{} r(2\cos^{3}t\minus{}\sin^{3}t)$

and all these limits converge to $ 0$.

[quote="Mr.Doe"]and then use the "squeeze" theorem.[/quote]

Yes, thanks.

[quote="Mr.Doe"]Computing limit along some trajectories doesn't mean anything[/quote]

Ok, so can you provide an example where the limits along some trajectories are the same but the true limit is different?

[quote="Sos440"]and all these limits converge to 0.[/quote]

Ok, thanks, but the purpose was to use explicitly the definition.

I'll adjust your example;

$ f(x,y) \equal{}\frac{x^{5}}{(y\minus{}x^{2})^{2}\plus{}x^{8}},(x,y)\neq (0,0), f(0,0)\equal{}0$ ,doesn't have limit in $ (0,0)$ although restriction to all $ y \equal{} ax,\lim_{x\rightarrow 0}f(x,ax) \equal{} 0$, and also restriction to $ y \equal{} 0$.

I didn't meant your approach was wrong but it can trick you sometimes :wink: .

Nice example: however, with the trajectory y = x^2 the limit would be infinity, and so the limit doesn't exist.

In the limits 3) and 4) (using the definition) above I am having difficulties in relating |f(x,y)| with something usefull. Any hint?

3) $ x^{2}\plus{}y^{2}\minus{}2y^{2}$ (just adding and taking away $ y^{2}$ makes it simple to evaluate ) , first term is obvious for the second use $ x^{2}\plus{}y^{2}\geq y^{2},\frac{1}{\sqrt{x^{2}\plus{}y^{2}}}\leq\frac{1}{y}$ , 4) is more or less the same way, you should be to work it out.

Ok, let me see if I understood: we want to show that for every $ \epsilon > 0$ there is a $ \delta > 0$ such that $ 0 <\sqrt{x^{2}\plus{}y^{2}}<\delta$ implies $ \left|\frac{x^{2}\minus{}y^{2}}{\sqrt{x^{2}\plus{}y^{2}}}\right| <\epsilon$. So:

$ \left|\frac{x^{2}\minus{}y^{2}}{\sqrt{x^{2}\plus{}y^{2}}}\right| \equal{}\frac{|x^{2}\minus{}y^{2}|}{\sqrt{x^{2}\plus{}y^{2}}}\equal{}\frac{|x^{2}\plus{}y^{2}\minus{}2y^{2}|}{\sqrt{x^{2}\plus{}y^{2}}}\leq\frac{|x^{2}\plus{}y^{2}|\plus{}|2y^{2}|}{\sqrt{x^{2}\plus{}y^{2}}}$

$ \leq\frac{x^{2}\plus{}y^{2}}{\sqrt{x^{2}\plus{}y^{2}}}\plus{}\frac{2y^{2}}{\sqrt{x^{2}\plus{}y^{2}}}$

$ \leq\sqrt{x^{2}\plus{}y^{2}}\plus{}\frac{2(x^{2}\plus{}y^{2})}{\sqrt{x^{2}\plus{}y^{2}}}$

$ \leq 3\sqrt{x^{2}\plus{}y^{2}}$

$ \leq 3\delta$

and so choosing $ \delta \equal{}\frac{\epsilon}{3}$ we have prooved that the limit is zero.


is there an easier way to do this problem besides listing out the possiblities?

Julian wants to label the five fingers on his left
hand with the integers 1 through 5, such that no
two adjacent ringers have consecutive integers,
and each integer is used exactly once. In how
many ways can Julian label his fingers? One such
way is 42531. :huh:

[hide="Hint"]Divide the problem into two cases: one where one of the middle three fingers gets labeled first and one where one of the other two fingers are labeled first.[/hide]


Prove that for any positive integer $n$,
\[\sqrt[n]{n}< 1+\sqrt{\frac{2}{n}}\]


calculus
integration
trigonometry
calculus computations

What is the value of the integral  :
$\int_0^{2\pi} \frac{xsin^{100}x}{sin^{100}x+cos^{100}x} dx$

[quote="kokosai"][quote="socrates"][quote="kunny"]
$\Longleftrightarrow I=\pi \int_0^{2\pi }{\frac{\sin ^ {100} t}{\sin ^ {100} t+\cos ^ {100} t}dt.}$[/quote]

can we compute this one??
 :)[/quote]$\int_0^{2\pi}{\frac{\sin^{2n}t}{\sin^{2n}t+\cos^{2n}t}dt=\pi}$ for all integral $n$.[/quote]

why??
 :)

[quote="kunny"]Let $J=\int_0^{2\pi} \frac{\cos ^ {100} t}{\sin ^ {100} t +\cos ^ {100} t}dt,$ we have $I-J=$ .....  and by $I+J=$ ....
Thus .....[/quote]

thanks so much!!
 :)


I would like to define all the real values that are greater or equal than all the elements of a given set. For instance:

A=]-00;-2]

{x E IR:x>=yEA} = [-2;+00[

Is the previous notation correct? Can it be read as "all the values of xEIR that are equal or greater than all the elements (y) of A"?

[quote="JC1"]I would like to define all the real values that are greater or equal than all the elements of a given set. For instance:

$A=\ ]-\infty;-2]$

$\{x \in\mathbb{R}: x\ge y\in A\}= [-2;+\infty[$

Is the previous notation correct? Can it be read as "all the values of $x\in\mathbb{R}$ that are equal or greater than all the elements (y) of A"?[/quote]

This looks better.  :wink:

Hmm...is $]$ the same as $($?

Yes, it looks much better.  :)

[quote]"all the values of xEIR that are equal or greater than all the elements (y) of A"?[/quote]y here represents all the elements of the A set.
I suppose it is correct.


limit
ratio
calculus
calculus computations

[hide="Contest Style Approach"]
For $ k \in \{ 2 \ldots n \minus{} 2 \}$, $ \binom{n}{k} \geq \binom{n}{2} \equal{} \frac {n(n \minus{} 1)}{2} > \frac {n(n \minus{} 3)}{2}$
We thus have that $ \binom{n}{k} ^{ \minus{} 1} < \frac {2}{n(n \minus{} 3)}$ for $ k \in \{ 2 \ldots n \minus{} 2 \}$

Thus, the sum of the terms from $ k \in \{ 2 \ldots n \minus{} 2 \}$ does not exceed $ (n \minus{} 3)\frac {2}{n(n \minus{} 3)} \equal{} \frac {2}{n}$

Thus, the total sum does not exceed $ 1 \plus{} \frac {1}{n} \plus{} \frac {2}{n} \plus{} \frac {1}{n} \plus{} 1 \equal{} 2 \plus{} \frac {4}{n}$

At the same time, we know the terms for $ k \in \{ 1 \ldots n \minus{} 1 \}$ are all positive, thus the sum is greater than $ 2$.

Hence, $ 2 < S_n < 2 \plus{} \frac {4}{n}$, and thus by squeeze, as $ n \to \infty$, $ S_n \to 2$
[/hide]

Let $ a_{n} \equal{} \sum_{k \equal{} 0}^n\binom{n}{k}^{ \minus{} 1}$. then $ a_{n} \equal{} \frac {s_{n}}{n!},$ where $ s_{n} \equal{} \sum_{k \equal{} 0}^n k!(n \minus{} k)!$

$ s_{n \plus{} 1} \minus{} s_{n} \equal{} (n \plus{} 1)! \plus{} \sum_{k \equal{} 0}^n k! ((n \plus{} 1 \minus{} k)! \minus{} (n \minus{} k)!) \equal{} (n \plus{} 1)! \plus{} \sum_{k \equal{} 0}^n k!(n \minus{} k)!(n \minus{} k) \equal{} (n \plus{} 1)! \plus{} n*s_{n} \minus{} \sum_{k \equal{} 0}^{n} k!(n \minus{} k)!k \equal{} (n \plus{} 1)! \plus{} n*s_{n} \minus{} \sum_{k \equal{} 0}^{n} (k \plus{} 1)!(n \minus{} k)! \plus{} s_{n} \equal{} (n \plus{} 1)! \plus{} n*s_{n} \minus{} s_{n \plus{} 1} \plus{} (n \plus{} 1)! \plus{} s_{n} \equal{} 2(n \plus{} 1)! \plus{} (n \plus{} 1)s_{n} \minus{} s_{n \plus{} 1}$
$ \implies (*) 2s_{n \plus{} 1} \equal{} 2(n \plus{} 1)! \plus{} (n \plus{} 2)s_{n}$
Denote $ L \equal{} \lim_{n\rightarrow\infty}\sum_{k \equal{} 0}^n\binom{n}{k}^{ \minus{} 1} \equal{} \lim_{n\rightarrow\infty} \frac {s_{n}}{n!}$
From $ (*)$:
$ L \equal{} \lim_{n\rightarrow\infty} \frac {s_{n \plus{} 1}}{(n \plus{} 1)!} \equal{} \lim_{n\rightarrow\infty} 1 \plus{} \frac {s_{n}(n \plus{} 2)}{2(n \plus{} 1)!} \equal{} 1 \plus{}\lim_{n\rightarrow\infty} \frac {1}{2}\frac {s_n}{n!}\frac {n \plus{} 2}{n \plus{} 1} \equal{} 1 \plus{} \frac {L}{2}$
$ \implies L \equal{} 2$

It only remains to show that the limit does exist. It's true because $ a_{n}$ is decreasing for large n:
$ \frac {a_{n \plus{} 1}}{a_{n}} \equal{} \frac {n!}{s_{n}} \plus{} \frac {n \plus{} 2}{2(n \plus{} 1)} ( < 1)$

It's sufficient to show that the  $ s_{n} > \frac {2(n \plus{} 1)!}{n} \equal{} 2(n)! \plus{} 2(n \minus{} 1)!$, which is trivial because $ s_{n} > 0!n! \plus{} 1!(n \minus{} 1)! \plus{} .... \plus{} (n \minus{} 1)!1! \plus{} n!0!$
Hence the ratio is less than 1 for large n (actually, n>3), and $ a_{n}$ is decreasing and positive, hence converges.

Let's change a little bit the statement of the problem.

Find  the limit:

$ \lim\limits_{n\rightarrow \infty}\sum\limits_{k\equal{}0}^{n}\frac{(\minus{}1)^{k}}{\binom{n}{k}}.$

[quote="didilica"]Let's change a little bit the statement of the problem.

Find  the limit:

$ \lim\limits_{n\rightarrow \infty}\sum\limits_{k \equal{} 0}^{n}\frac {( \minus{} 1)^{k}}{\binom{n}{k}}.$[/quote]
it does not exist. for $ n \equal{} 2p \plus{} 1$ you will have $ 0$ and for $ n \equal{} 2p$ you will have $ \approx 2$ for $ n \equal{} 2p$ solution is $ \approx$ as The Zuton Force  solution.
It's not so interesting question.
What's about this one:
find limit :
$ \lim_{n \to \infty} \frac {\sum\limits_{1\leq k \leq n} \sqrt {C_{n}^{k}}}{2^{\frac {n}{2}}n^{\frac {1}{4}}}$

[quote="worthawholebean"]Determine the value of $ \lim_{n\rightarrow\infty}\sum_{k \equal{} 0}^n\binom{n}{k}^{ \minus{} 1}$.[/quote]

This problem was also proposed in Vojtech Jarnik 2008 Longlist (which was decided to be secret till IMC 2008).

[quote="didilica"]Let's change a little bit the statement of the problem.

Find  the limit:

$ \lim\limits_{n\rightarrow \infty}\sum\limits_{k \equal{} 0}^{n}\frac {( \minus{} 1)^{k}}{\binom{n}{k}}.$[/quote]

well ummm... it looks to me like

$ \sum\limits_{k \equal{} 0}^{n}\frac {( \minus{} 1)^{k}}{\binom{n}{k}}\;\equal{}\;\boxed{\left[1 \plus{} ( \minus{} 1)^n\right] \cdot \dfrac{n \plus{} 1}{n \plus{} 2}}$ ...  :maybe:

[quote="TZF"]
For $ k \in \{ 2 \ldots n \minus{} 2 \}$, $ \binom{n}{k} \geq \binom{n}{2} \equal{} \frac {n(n \minus{} 1)}{2} > \frac {n(n \minus{} 3)}{2}$
We thus have that $ \binom{n}{k} ^{ \minus{} 1} < \frac {2}{n(n \minus{} 3)}$ for $ k \in \{ 2 \ldots n \minus{} 2 \}$

[b]ok, i get this[/b]

Thus, the sum of the terms from $ k \in \{ 2 \ldots n \minus{} 2 \}$ does not exceed $ (n \minus{} 3)\frac {2}{n(n \minus{} 3)} \equal{} \frac {2}{n}$

[b]a little confused here[/b]

Thus, the total sum does not exceed $ 1 \plus{} \frac {1}{n} \plus{} \frac {2}{n} \plus{} \frac {1}{n} \plus{} 1 \equal{} 2 \plus{} \frac {4}{n}$

[b]why? i don't get this.[/b]

At the same time, we know the terms for $ k \in \{ 1 \ldots n \minus{} 1 \}$ are all positive, thus the sum is greater than $ 2$.

Hence, $ 2 < S_n < 2 \plus{} \frac {4}{n}$, and thus by squeeze, as $ n \to \infty$, $ S_n \to 2$
[/quote]


number theory
greatest common divisor
algebra
polynomial
induction
trigonometry
function

Let $ n$ be odd and $ \omega$ be a $ n$-th primitive root of unity.  Let $ A(n)\equal{}\{r\equal{}1,2,\ldots,n\minus{}1 : gcd (r,n)\equal{}1 \}$. Prove that

\[ \Pi_{j\in A(n)}(1\plus{}\omega^j)\equal{}1.\]

Thanks in advance!

consider the polynomial

$ \Phi_n(x) = \prod_{k=1}^{\phi(n)}(x-z_k) \in \mathbb{Z}[x]$

where $ z_1,...,z_{\phi(n)}$ are the primitive $ n\textsuperscript{th}$ roots of unity. The LHS of your equation is the same as $ \Phi_n(-1)$ because $ \phi(n)$ is always even for odd $ n \geq 3$.

and also the identity  $ x^n-1 = \prod_{d\mid n} \Phi_d(x)$

now induction on odd $ n$. for $ n=3$, we have $ \Phi_3(-1) = \prod_{k=1}^{2}(-1-z_k) = (-1-\zeta_3)(-1-\zeta_3^2) = 1-(\zeta_3 x + \zeta_3^2) + \zeta_3^3 = 1  -1 +1 = 1$

assume true for $ d$ such that $ 3 \leq d < n$, then we get that

\begin{align*}(-1)^{n} - 1  =& \prod_{d\mid n}\Phi_d(-1) \\
-2 =& (-2) \prod_{ d \mid n, \; d > 1 } \Phi_d(-1) \\
1 =& \Phi_n(-1) \tag{1}
\end{align*}

$ (1)$ follows from the application of the inductive hypothesis that $ \Phi_d(-1) = 1$ for $ 3 \leq d < n$.

this induction might not be considered entirely proper since its over just the odd numbers, but this can be altered by taking $ n = 2k+1$ and proceeding with induction on $ k$ instead, i just find it easier to see this way.

Thanks for the reply!

Out of curiosity, let $ f(n)\equal{}\prod_{k\in A(n)}(1\plus{}\omega^k)$ for any $ n$.

We've shown $ f(n)\equal{}1$ when $ n$ is odd.  Like to check if there's any formulation for even $ n$.

For info, $ f(2)\equal{}0$, $ f(4)\equal{}2$, $ f(6)\equal{}3$, $ f(8)\equal{}2$, $ f(10)\equal{}4\sin^2(\frac{\pi}{5})$, $ f(12)\equal{}1$.

not sure about all even $ n$, but for $ n \equal{} 2^r$ we have that $ \Phi_n(x) \equal{} x^{2^{r\minus{}1}}\plus{}1$. So $ f(2^r) \equal{} 2$ for $ r > 1$. i dont know if there are any other easy ones like that.

these identities might be useful

$ \Phi_{2n}(x) \equal{} \Phi_n(x^2)$ for even $ n$.
$ \Phi_{2n}(x) \equal{} \Phi_n(\minus{}x)$ for odd $ n$.

btw, if you didn't already know, these functions are called the cyclotomic polynomials and are the irreducible factors of polynomials of the form $ x^n\minus{}1$ as you partially saw in the previous post. they are largely used to generate field extensions.


Let $a,b$ be the smaller sides of a right triangle. Let $c$ the hypothenuse and $h$ be the altitude from the right angle to hypothenuse. Find the maximal value of

$\frac{c+h}{a+b}$

[b]Answer for the Siluan's proposed problem:[/b] $\frac{c+h}{a+b}\le \frac{3\sqrt 2}{4}$ (with equality iff $a=b$).
[u]Indication:[/u] $\frac{c+h}{a+b}=f(t)$, where $f(t)=\frac{t^2+t+1}{(t+1)\sqrt{t^2+1}}\ ,\ t=\frac bc>0\Longrightarrow$
$\left(\forall\right)t>0\ ,\ 1=f(0)<f(t)\le f(1)=\frac{3\sqrt 2}{4}\ .$

[b]Another nice exercises [/b](with the same notations):

\begin{eqnarray*}\triangle ABC\ ,\ CA\perp CB & \Longrightarrow & 1.\blacktriangleright h+\max\{a,b\}\le \frac{3\sqrt 3}{4} c\ .\\ & \Longrightarrow & 2.\blacktriangleright a^3+b^3+abc\le \frac{3\sqrt 3}{4}\cdot c^3\ .\\ \end{eqnarray*}

h = a * b / c. :)
Also, c = a / sinA, b = a / tanA. Then the expression simplifies to:
(1 + sinAcosA)/(sinA + cosA).
And what do we do now? :?

Sorry,  :wallbash_red:
Squaring the expression I got and if sinAcosA = k, get:
(1 + 2k + k ^ 2)/(2k + 1)
= 1 + k ^ 2 / (2k + 1)
= 1 + (sin2A)^2/4(sin2A+1). I think this is a maximum when sin2A = 1. The answer is 9/8. We need to take the root to get 3sqroot(2)/4 which is the answer.


trigonometry
search
real analysis
real analysis unsolved

Without using Picard-Lindelof show that $ y'' \plus{} y \equal{} 0$ on $ (\minus{}\infty,\infty)$ has solutions $ y\equal{}a\sin x \plus{} b\cos x$.
Is it hard? I wasn't able to come up with anything. I can however reduce it to a IVP problem and then if we can show uniqueness this will follow.

Just a question: aren't we just supposed to solve the diff. eq.? The general solution can also be written as

$ y \equal{} A \sin(x \plus{} k)$

where A and k are constants (that diff. eq. is pratically the harmonic oscillator diff. eq.).

$ y'y''+y'y=0$ Integrating gives $ y'^2+y^2=A^2$, yielding $ y'=A\cos (\theta +\alpha),\　y=A\sin (\theta +\alpha).$

[quote="phils"]First off what you have is a solution. Here's a hint to show all solutions have this form: note that if $ u$ and $ v$ are solutions then $ u'v \minus{} uv'$ (yup, this is the Wronskian) is a constant. From there you should be able to conclude using some linear algebra.[/quote]
You cant do that.  Because the proof behind Wronskian being zero implies linear independence and all of that stuff rests on Picard-Lindelof theorem.

[quote="kunny"]$ (y')^2\plus{}y^2 \equal{} 0$[/quote]
What did you do here? Did you factorize $ (y'\plus{}iy)(y'\minus{}iy)\equal{}0$?

The special case of uniqueness in the equation $ f'(x) \equal{} g(x)$ (with $ g$ continuous, and constant in this case) is easier than the general uniqueness theorem, and that special case is what's being used here. (Oops- it's not. The approach kunny used is what uses that.) The special case is doable with nothing more advanced than the Mean Value Theorem.

[quote="sylow_theory"]
You cant do that.  Because the proof behind Wronskian being zero implies linear independence and all of that stuff rests on Picard-Lindelof theorem.[/quote]

Yes you can. Let $ u$ and $ v$ be solutions such that $ u'v \minus{} uv' \equal{} a$ is non zero, and $ f$ be a general solution. Then, owing to $ u'f \minus{} uf' \equal{} b$, $ v'f \minus{} vf' \equal{} c$, we have (multiply and subtract) $ (u'v \minus{} uv')f \equal{} bv \minus{} cu$. $ a$ being non zero, we conclude $ f$ is a linear combination of $ u$ and $ v$. No Picard-Lindelöf here.

There is a very special method which seems to work only for these type of d.e.

[hide="Proof using Taylor theorem"]Consider i.v.p. $ y'' \plus{} y\equal{}0$ with $ y(0) \equal{} 0, \ y'(0) \equal{} 0$. We claim this equation has the unique solution $ y \equal{} 0$ on $ \mathbb{R}$.

Existence of solution is trivial. Now let $ y \equal{} f(x)$ be a solution for this problem on $ I \equal{} (\minus{}2a, 2a)$, where $ a > 0$. By Taylor's theorem, for all $ |x| \leq a$, there is $ |c| \leq |x|$ such that $ f(x) \equal{} \frac{f^{(n)}(c)}{n!}$.

If $ n \equal{} 2k$, then $ f(x) \equal{} (\minus{}1)^{k}\frac{f(c)}{n!}$ and if $ n \equal{} 2k\plus{}1$, then $ f(x) \equal{} (\minus{}1)^{k}\frac{f'(c)}{n!}$.

Since both $ f$ and $ f'$ are bounded on $ [\minus{}a, a]$, there is a common bound $ M$ and $ |f(x)| \leq \frac{M}{n!}$.

Therefore, taking $ n \to \infty$, $ f(x) \equal{} 0$ on $ [\minus{}a, a]$. Since $ a$ is arbitrary, $ f(x) \equal{} 0$.[/hide]

Then the uniqueness for the i.v.p $ y'' \plus{} y \equal{} 0$ and $ y(0) \equal{} A, \ y'(0) \equal{} B$ directly follows.

There's also this [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=733993707&t=109071]"trick"[/url]. Plain ugly, but I think it works for higher orders too :D


inequalities
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inequalities unsolved

Hello, can anyone find a [b]nice solution[/b] to this inequality? Thanks.

The real numbers [i]a, b, c, x, y,[/i] and [i]z[/i] are such that $a>b>c>0$ and $x>y>z>0$.  Prove that

\[\frac{a^2x^2}{(by+cz)(bz+cy)}+\frac{b^2y^2}{(cz+ax)(cx+az)}+\frac{c^2z^2}{(ax+by)(ay+bx)}\geq\frac 34.\] :?

I think I have a nice solution, but this is a matter of taste. Of course, it is enough to have $ \sum\frac {a^2\cdot x^2}{(b \plus{} c)^2\cdot (y \plus{} z)^2}\geq\frac {3}{16}$ because of AM-GM. From Cauchy it is enough to have $ \sum\frac {ax}{(b \plus{} c)(y \plus{} z)}\geq\frac {3}{4}$. This follows from the fact that the systems $ \left(\frac {a}{b \plus{} c},\frac {b}{a \plus{} c},\frac {c}{a \plus{} b}\right)$ and $ \left(\frac {x}{y \plus{} z},\frac {y}{z \plus{} x},\frac {z}{x \plus{} y}\right)$ have the same order and thus we can apply Cebasev. Then we use that $ \sum\frac {m}{n \plus{} p}\geq\frac {3}{2}$.

first, by+cz>=bz+cy, cz+ax>=cx+az, ax+by>=ay+bx by rearrengement inequality. thus we only have to prove: (putting ax=p, by=q, cz=r)

S=p 2/(q+r) 2+q 2 /(p+r) 2 +r 2 /(p+q) 2 >=3/4

but by power mean: S>=1/3(p/(q+r)+q/(p+r)+r/(p+q)) 2   but by Nesbitt we know that p/(q+r)+q/(p+r)+r/(p+q)) >=3/2 so result follows


Ok many hours have been spent on trying to solve this problem.

You know:
length "L"
angle "a"

you are solving for:
length "x1"
length "x2"

x1 is solvable, but I can't seem to figure out x2.
Is x2 even solvable?

There is not enough information to compute x_2.

Before spending too much time on a problem, you should ask, if the result can be changed, while all problem conditions are satisfied. Clearly, moving the point $P$ on the line $k$ leads to different segment lengths $x_2$, while the 2 parallels $x_1 \parallel x_2$ are perpendicular to the line $k$, the segment lenghts $L, x_1$ and the angle $\alpha$ are fixed.


Moderator, this is definitely not for this forum -- send it up to Pre-Olympiad, I guess.  (I don't know where in advanced they put questions like this.)

Almost all of solution:
[hide]We get f(0) = 0 by plugging in $x = 0, y = -f(0)$ which gives $f(0) = f(f(0)) - 6f(0)^3 - 2f(f(0))$ and then putting in $x = 0, y = 0$ gives $f(f(0)) = -f(0)$, and together we get $6f(0)^3 = 0$ so $f(0) = 0$.  Then, putting in $x = 0$, we get $f(y) = -f(-y)$.  (You had gotten that far.)

Now, put in $y = 0$ to get (1)$f(f(x)) = f(x^3)$.
Put in $y = x^3$ to get (2)$f(f(x)+x^3) = 6x^3f(x)^2 + 2f(x^3)$
Put in $y = -f(x)$ to get (3)$0 = f(x^3 + f(x)) -6f(x)^3 -2f(f(x))$
Add (2) to (3) and substitute from (1) to get
$f(f(x)+x^3) = 6x^3f(x)^2 + 2f(x^3)+f(x^3 + f(x)) -6f(x)^3 -2f(f(x))$
$0= 6x^3f(x)^2 -6f(x)^3$
$x^3f(x)^2 =f(x)^3$
so for all x, $f(x) = 0$ or $f(x) = x^3$.  I haven't tried to show that it always must be one or the other.[/hide]

May I ask the moderators AGAIN to please move this problem up to Pre-Olympiad? Thank you.


Prove that
$ 2^{\frac{(n\minus{}1)}{2}}\left(\sin\frac{2\pi}{2n}\sin\frac{4\pi}{2n}\ldots\sin\frac{n\minus{}2}{2n}\pi\right) \equal{}\sqrt{n}$

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1804129935&t=12775]here[/url]. It follows easily from $ \sin{x} \equal{} \sin(\pi \minus{} x)$.


I believe we're starting up our practices again sometime soon, so if you're interested in trying to be part of the Ohio ARML team, you should e-mail [url]AYANG@cscc.edu[/url]

:huh: I thought determining the ARML team was the point of OHMIO? :maybe:

Yeah, I thought that the OCTM and AMC scores determine elligibility for ARML.  :huh:

How would they know a person have enough credentials to get into such a competition like ARML if it isn't determined by contests such as AMC and OCTM?

who (arml coach, octm person, etc.) owns the email address given, and what exactly is used in the ohio arml selection process? I heard that it was a combination of OCTM and AMC scores (not OHMIO) but im not too sure...

That e-mail address is owned by Tzu-yi Alan Yang, but that is about the limit of my knowledge.

Actually, I'm not exactly sure what the selection process is, but I think they invite you if you have high scores on the OCTM or the AMC.

But last year, I don't think they were able to get the addresses for the high scorers on the AMC, so I was kind of posting this for people who might not have been able to get the invitation.

And yea, you'd probably have to show them scores from previous contests and stuff, and they would have to be good enough for you to be on the team.  And this is basically what I did last year (my school doesn't do the OCTM), except it took me forever to find the e-mail address of the coach.

Yeah, I just received an email and it said for me to wait for an invitation in the mail because I got pretty good scores on the OCTM and the AMC.

Ok, update on the ARML stuff

If you've made OHMIO, then at the second half of the ice cream social, there will be an ARML meeting.

It looks like I'm one of three students in my school on the ARML team, and we are generally only a powerhouse at the regional level :roll: .

Alex Anderson had posted [url=http://www.artofproblemsolving.com/Forum/files/arml_advice_550.pdf]this[/url] on his blog to help the Chicago ARML team, but when you post something on AoPS, anyone can read it (like the Ohio team(s) for example :ninja: ).  Anyway, I encourage people to read it.

Darn it, I got invited to do ARML, but I'm taking the SAT that Saturday, so I can't...

Why the heck do they schedule it that weekend anyway?  A lot of seniors have graduation that weekend and can't do it...

I think one of the seniors in my school is taking the ARML despite the fact that he will miss his graduation.

[quote="DanK"]Darn it, I got invited to do ARML, but I'm taking the SAT that Saturday, so I can't... [/quote]

I thought you can request an alternate SAT testing date.
See [url]http://www.arml.com/SAT%20Alternate%20Test%20Day%20Policy%202006-07.pdf[/url]

I'm also going, but apparently Tong Zhan isn't :?:  :!:

I wonder what percent of the ARML team currently frequents this forum?  Oh, correction, team[b][u][i]s[/i][/u][/b], we have [b]3[/b] of them this year :o !

Yea, and he wasn't able to go last year either (Tong Zhan).

The only person I know of that is on AoPS but hasn't posted at all on this forum is nr1337, Carlos.  I don't think he does the OHMIO competitions or whatever.

And, how did the ice cream reception thing do?

The reception went quite well; apparently there were still two open slots between the three teams, and one of them was filled there.

Oh man, 3 teams?

I hope I can make A this year.  But I'm kind of sucking at math this year, so I probably won't.

how is it determined who makes what team?

[quote="roadnottaken"]It is based on your OCTM score, your AMC score, and how well you do in the practices (don't ask me when those are; I won't know until Professor Yang announces them).[/quote]

Ahhhh

I didn't take the OCTM.

Hopefully that won't count against me.

I made a Ohio ARML facebook group.

You guys should join.

I know what facebook is (sort of), but I haven't used it, and don't plan to start now (this statement includes myspace).

so ARE YOU GUYS READY TO OWN AT THE ARML???? OH YEAH

By the way, tjhance, congratulations on being the only Ohioan going to MOSP! :clap:  :clap:  :clap:


In the laboratories of the firm "Halides 'R Us" has been found a compound A in a vial labeled only "achiral alkyl halide $ C_{10}H_{17}Br$". The management feels that the compound might be useful as a pesticide, but they need to know its structure. You have been called in as a consultant at a handsome fee. Compound A, when treated with KHO in ethanol, yields two compounds, B and C, each with molecular formula $ C_{10}H_{16}$. Compound A rapidly reacts in aqueous ethanol to give an acidic solution which, in turn, gives a precipitate of AgBr when tested with AgNO3 solution. Ozonolysis of A followed by treatment with $ (CH_{3})_{2}S$ gives acetone as one of the products plus an unidentified halogen-containing product. Hydrogenation of either B or C gives a mixture of both trans- and cis-1-isopropyl-4-methylcyclohexane. Compound A reacts with one equivalent of bromine to give a mixture of two separable compounds D and E, both of which can be shown to be achiral. Finally, ozonolysis of B gives acetone and cyclohexane-1,4-dione.
Propose structures for compounds A through E.

Yes, the halogen is attached to an sp3 carbon.

well are halides in which a halogen is attached to s carbon whihc is sp3 hybrididsed called alkyl..
well i studied [url=http://en.wikipedia.org/wiki/Haloalkane]this[/url] definition...

That's not the definition I have studied, and I don't agree with that wikipedia article. How would you classify something like 1-chloro-2-phenylethane? I think that the definition given in the wikipedia article is a more technical one, more close to industrial "environments". But the "pure" definition from the organic chemistry point of view (which is based on reactivity in SN reactions) is the one I gave. We have two groups of halogen containing compounds: 1) alkyl halydes (where the halogen is attached to an sp3 carbon), and 2) vinyl and aryl halides (where the halogen is attached to an sp2 carbon). The halydes from the first group are reactive in "normal" nucleophilic substitution reactions. Those belonging to the second group don't react in "normal" SN reactions, but do have a special reactivity when certain groups are also present ion the molecule or when they react with special reagents.

In conclusion, the alkyl halide in this problem have an halogen attached to an sp3 carbon, but of course it can also be present in its molecule a double or triple bond.


The fast development of computer technology and the internet has changed many things in our daily life. We can easily find everything we need, from groceries to cars and everything in between. It can all take place in the blink of an eye, and we take many things for granted which were beyond our imagination 10 years ago.

These modern technologies have changed our life style, but have they also changed our view on art and the art format itself?

Instead of being in front of the easel, spending time painting with brush and oil, pictures can be made using computer software and a mouse in only a few minutes. This can be demonstrated by the potential to allow vast time and labor consuming art work to be simplified and easily produced. An example of this would be the animated cartoon industry, where computer animation is slowly replacing the traditional cell method.

In addition to it’s useful role in film production, one of other advantages of digital art is that it has helped people with little skill, experience, judgment or perspective to create fairly sophisticated-looking and often interesting images. The fact that less training is often required to work in this creative world is one of the reasons that more and more people are moving towards it, in their goal to become an “artist”.

However, in the eyes of traditional artists whether this kind of work could be called “art” is still in doubt. Some artists worry that in the future people will become so dependent upon this tool, that they may eventually lose their ability to use oil colors to create the magic.

But even digital art still needs the training and skill used in traditional painting. Like some famous abstract artists, many digital artists are also capable of painting realistic depictions in oil.

Digital art is more than just computer skill, it is the integration of artistic talent and computer technology. Making a great painting in a digital format is as same as making it on the canvas, requiring the artist to have the same creative design, passion and skills; it can be seen as just another powerful tool to express yourself. A great painter/draftsman can use the same skills in the digital format, of course learning how transfer these skills to a computer is also essential.

For me, as an art graduate, although I use computer in my daily life and work, I still prefer to rely on my natural abilities to draw, design, and paint pictures. I still prefer to use my brush, knife and oil to make myself dirty in front of my easel for several hours or even days and weeks. I still enjoy the moment when the knife touches the canvas, and the satisfaction at the moment of the final touch.
About the author: Cherry Xu is an Art Design graduate, and a teacher and practitioner of traditional Chinese art. She works as the Director’s Assistant at Aspect Art Ltd

Interesting view :) I love both traditional and digital art. I've only done a small amount of traditional artwork with anything but pencils. Things like paints, watercolors, and other obscure tools are things I either can't afford, wouldn't use, or end up making a huge mess out of them. So I draw alot. Cartoons are my favorite style, and I thought it was cool that you mentioned them. I think too many animators are jumping onto the 3d style, just because it's "3d Oooh technology!", when really, the concept of the idea matters so much more than the tools.

But I love digital art, for the main reason that it's so much easier to fix your mistakes. To me it seems like one misguided line or stroke of paint will ruin my "big creations" (the kind I do in art class.) I felt about ready to strangle something when I counted 10 huge blots of ink on a pen+ink assignment.  :wallbash: But with a digital painting, all it takes to fix something is a ctrl+z.

Not every digital picture can be done in few hours. In digital  art too you have to find references, models, to adjust lightning...If we are talking about quality, more important is artist's style then used technics.


algebra
polynomial
trigonometry
algebra proposed

Determine the lowest degree of a polynomial with integer coefficient which has a root of $ tan \frac{p}{q}\pi$ and $ p,q$ are positive integer with $ (p,q)\equal{}1$
I create this problem when I think about http://www.mathlinks.ro/viewtopic.php?t=254507
The original has been solved successfully.How about this?I think the new one will be much tougher.
I am curious about your idea :)

Let $ \tan \frac{p}{q} \pi \equal{} t$.  Then $ \frac{1 \minus{} t^2}{1 \plus{} t^2} \equal{} \cos \frac{2p}{q} \pi$ and now I believe we can apply the result of the first problem.

This ia also what I thought.Define $ f$ to be the polynomial whose degree is the lowest and has the root$ \cos \frac{2p}{q}\pi$
But how to proof $ f(\frac{1\minus{}t^2}{1\plus{}t^2})*(1\plus{}t^2)^m$ irreducible?Maybe $ f(x)$ is irreducible but $ f(g(x))$ not.
So can you give me details?I think it won't be easy to "directly" apply what we proved :)

[quote="t0rajir0u"]I didn't mean to suggest that the resulting polynomial was irreducible; my intuition suggests that it splits into two irreducible factors, but I'll have to get back to you on a nice way to prove this.[/quote]
Maybe you are right.But I think the new polynomial will be irreducible though I can't prove it,for $ \tan \frac{1}{3} \pi$ is a root of a polynomial generated
by $ \cos \frac{2}{3} \pi$ whose degree is 1.And the algebraic degree of $ \tan \frac{1}{3} \pi$ is 2.That suggests the degree may become double.So I weakly conjectured the irreducibility.

Remark: I am going to school today.So I don't have time to surf the Internet.I will come back next week.I wish the problem would be solved in this forum during the week's time.


Someone finally replied. Anyways, another [hide="hint."]Its a picture. Try quoting me and remove the size modifications.[/hide]

Umm... a very messy jumble of text? Or maybe a traffic jam...

[quote="pythag011"]EDIT:[quote="dragon96"]
~ (__) ~
~ (oo) ~
~ /-------\/ ~
~ / | || ~
~ * ||----||~
 ~ ^^ ^^ ~
~ ~

Hint: Somethings may have been modified.[/quote]

EDIT: it's a personish thing.[/quote]

Very close.Would you like another hint (I'm very impatient, but I'm not sure about you)?

is it ... an owl or something?
or maybe some kind of bird?

None of your are right, but it is an animal.

Yes, it is a cow. Since AoPS wont allow multiple spaces, quote me to see the actual picture.

~             (__) ~
~             (oo) ~
~   /-------\/   ~
~  / |        ||   ~
~ *  ||----||   ~
~    ^^    ^^   ~
~                    ~


[hide]\[ \frac{x+1}{x}\leq-2  \]
\[ x+1\leq-2x  \]
\[ x+2x+1\leq0  \]
\[ 3x+1\leq0  \]
\[ x\leq\frac{-1}{3}  \]
[/hide]
That seems too easy ... I must have done something wrong :S

[quote="shobber"]$\frac{x+1}{x}+2 \leq 0 \Longleftrightarrow \frac{3x+1}{x} \leq 0 \Longleftrightarrow x(3x+1) \leq 0$.

Therefore: $-\frac{1}{3} \leq x < 0$.[/quote]

this is what I got. I think there is a typo in the solution manual of aops 1. For some reason they turned it into a quadratic half way through.  :huh:

[quote="Hikaru79"]That seems too easy ... I must have done something wrong :S[/quote]
This is indeed very easy, but you forget that $x$ maybe negative so you cannot simply multiply both sides with it without discussing the sign of it.

omg torajirou made a mistake

...probably a typo though :(

lim x approaches zero from the left is less than -3, but it's weird when x actually [i]is[/i] 0... I think...

[hide="my convoluted solution"]
x is obviously less than 0, so we can multiply through by x and switch the sign...

$x + 1 \geq -2x$

$0 > 3x \geq -1$

$-1/3 \leq x < 0$
[/hide]

[quote="maniacman384"]If you cross-multiply, then simplify and solve for x, you get  -1/3<=x
X has to  be less than zero to make the inequality negative, therefore you get an answer of

[b] -1/3<=x<0[/b] Sorry, I don't know tex[/quote]

Note that cross-multiplying normally does not work with these types of problems, I just did it because I knew that x would always be negative. If really really really want to cross-multiply for some odd reason, you'd have to go into fairly messy casework in most cases.

[quote="Iron Fist"]
lim x approaches zero from the left is less than -3, but it's weird when x actually [i]is[/i] 0... I think...[/quote]

Right, I should've written strict inequality...  :blush:


trigonometry
induction
number theory
algebra unsolved
algebra

For $ n\ge 2$ prove ($ \csc\equiv\frac1\sin$) \[{ \sum_{k=1}^n\csc\frac{{2^k\pi}}\displaystyle{2^n-1}}=0.\] :) :D

Nobody?

Please try it!
This is almost as innocent as it looks at [i]first [/i]glance (not at second though)  :lol:

[hide="hint"]There is an easy proof by induction  :D  no need for number theory or algebra  :D   [/hide]

Still nobody?

[hide="ultimate hint"] Start with writing $ \cot(x) \minus{} \cot(2x)$ in an easier way...  :D

Isn't that CUTE?  :lol:  [/hide]


Two circles with radii of 24 and 54 are tangent to one another, and one of their common external tangents is drawn. What is the radius of a third circle inscribed tangent to both original circles and their common external tangent?

EDIT: Oops, now I understand. I'll solve the problem tomorrow; as it's late...

You can see my crummy image [url=http://img407.imageshack.us/img407/236/contributionkn1.png]here.[/url] Sorry about not labeling.

I'll have a full solution (including an immaculately labeled diagram) posted by sunrise. On that note, I should probably mention that I don't intend on using the green triangle, so disregard its presence. I'll just give you some info on that image:

[hide="A General Layout"]
The larger circle has center $ p\equal{}(96,54)$ and is described by $ (y\minus{}54)^2\plus{}(x\minus{}96)^2\equal{}54^2$
The smaller circle is described by $ (y\minus{}24)^2\plus{}(x\minus{}24)^2\equal{}24^2$ and is centered at $ q\equal{}(24,24)$

Thus, the circles are mutually tangent at $ j\equal{}\displaystyle\left(\frac{600}{13}, \frac{432}{13}\right)$. Note that $ p, q, j$ all lie on $ \ell$, defined to be $ 12y\equal{}5x\plus{}168$.[/hide]

Cheers, I'll post the remainder soon!  :)

[hide="Solution"]
Apply pythagoras' theorem twice, here x is the distance from intersection of tangent to circle of radius 26 to the intersection of the tangent with the circle of unknown radius.

You can also use pythagoras' theorem to calculate the length of the tangent segment. ($ \sqrt{80^2 \minus{} 28^2})$.

$ (26\minus{}r)^2 \plus{} x^2 \equal{} (26 \plus{} r)^2$
$ (54\minus{}r)^2 \plus{} (12\sqrt{39} \minus{} x)^2 \equal{} (54\plus{}r)^2$

$ \implies r\equal{} \frac{78}{7}(9\minus{} \sqrt{39})$
[/hide]


Does anyone have stats on the national scores.   I don't care what rank and do not need any names.  I would like to have an idea what it is like at the nationals.  If you could, please post

rank, score

These do not have to be yours.

Thanks

i'll post mine
33- 14 target, 19 sprint
team- 7 right (after i screwed up two easy problems)

I wasted a good target round with a really sucky sprint

I thought 2nd place was a 45 and 10th was a 41.

top 10 aren't 100% accurate due to Countdown round.  I had 45,44 listed on my computer at home, but the actual printouts they give us are at school, so top 10 could be off.  My list does say #11 is a 40, which means #10 was a 41.  I will try to post top 10 more correctly later,

frost

Thanks for the responses.  AMC puts out statistics by grade.  Does MC have anything like that too?

Here are the top 10 scores at Nationals last year before countdown:

45,45,44,43,43,43,42,41,41,41

As for grades, a couple years ago they posted the grade level as well, but not last year.

frost


Find $a$ and $b$ such that the graph $y=x+k$ intersects the graph $(x-3)^2+(y+2)^2 = 50$ in one or more points if and only if $a\leq k\leq b$

[hide="hint"]Just draw the tangents that have slopes of 1 above and below the circle. Find where they intersect with the y-axis, and this is your range because if the y intercept is between those intersections, it will intersect the circle 2 times[/hide]

[hide]We want to find the two values of $k$ that make the line tangent to the circle, and all values in between will satisfy the requirements of the problem.  From the equation of the circle, the center is $(3,-2)$ and the radius is $\sqrt{50}=5\sqrt{2}$.  The slope of the line is $1$, so the slope of the line connecting the center and the points of tangency is $-1$.  To get to the points of tangency, we must move from the center an equal number of units up and left, or down and right.  We must move 5 units in each direction so that the distance to the center of $5\sqrt{2}$, which is the radius.  The two points of tangency are then $(-2,3)$ and $(8,-7)$, and the two tangent lines are $y=x+5$ and $y=x-15$.  Therefore, $a=-15$ and $b=5$.[/hide]


geometry inradius trigonometry parameterization incenter inequalities

Let $ ABC$ be a right-angled triangle with a hypotenuse $ AC\equal{}1$. The points M and N are where the inscribed circle touches the sides AC and AB, respectively. Prove, that $ MN\leq \frac{2\sqrt{3}}{9}$.

[hide="Stuck..."]
I tried expressing in terms of the sides and trying to find a way to maximize the inradius, but that didn't really help.
Using $ rs \equal{} A \equal{} bc$, we have $ r\left(\frac {b \plus{} c \plus{} 1}{2}\right) \equal{} bc$. Expanding and simplyfying, we get $ r \equal{} \frac {2bc}{b \plus{} c \plus{} 1}$. Using Pythagorean on $ MN$ and $ r$ and simplifying a lot, we get $ MN \equal{} \frac {2bc\sqrt {2}}{b \plus{} c \plus{} 1}$.
I can't seem to finish the problem. That's the best I've come up with, and that doesn't seem to lead to much...[/hide]

[quote="Temperal"]$ rs \equal{} A \equal{} bc$[/quote]That's where you got a mistake. Area of triangle is $ A\equal{}\frac12bc$.

[hide="Used trig but leads to nowhere..."]I assigned the angle $ \alpha$ as our parameter for the angle $ \angle BAC$ and after a lot of algebra and simplification...

$ MN\equal{}\frac{2AC\sin(\frac12\alpha)\cos\alpha(1\plus{}\cos\alpha)}{(1\plus{}\sin\alpha\plus{}\cos\alpha)}$

Eh... looks like I didn't prove anything here.[/hide]

Oops... I guess I can't remember the first formula for the area of a triangle that everyone learns.

Let $ I$ be the incenter of $ \triangle{ABC}$, quadrilateral $ CMIN$ are on a circle, we have $ \angle{MIN} \equal{} \frac {\pi}{2} \plus{} \theta$,yielding $ MN \equal{} 2r\sin \left(\frac {\pi}{4} \plus{} \frac {\theta}{2}\right),$
where $ r \equal{} \frac {\cos \theta \plus{} \sin \theta \minus{} 1}{2}$ is the inradius of $ \triangle{ABC}, \theta \equal{}\angle{A}$. $ \therefore MN \equal{} \frac {1}{\sqrt {2}}(\sin \theta \plus{} \cos \theta \minus{} 1)\left(\sin \frac {\theta}{2} \plus{} \cos \frac {\theta}{2}\right).$

But the inequality is false!
It is equivalent to proving$ 2r^2\plus{}2r^2\sin C\equal{}2(s\minus{}b)^2(1\plus{}c)\le \frac {2\sqrt 3}{9}\Leftrightarrow (s\minus{}1)^2(1\plus{}c)\le \frac {\sqrt 3}{9}$
Put the the sides equal to $ \frac {\sqrt 2}{\sqrt 5}$ and $ \frac {\sqrt 3}{\sqrt 5}$

Yes it is false. By judging from the form of this I think the intended inequality was:

$ MN\le\frac {2 \minus{} \sqrt {2}}{2}$

 :blush:  sorry the above is wrong it would be true if $ M$ and $ N$ where on $ AB$ and $ BC$ (the legs).

 :)

What am I missing here?
[hide] Let $ \angle A \equal{} \theta$, then inraidus $ r \equal{} \frac {\sin\theta \plus{} \cos\theta\minus{}1}{2}$.

Connect $ AI$ , then clearly $ \angle IMN \equal{} \angle IAM \equal{} \frac {\theta}{2}$. Hence,
$ MN \equal{} 2r\cdot \cos\frac {\theta}{2} \equal{} (\sin\theta \plus{} \cos\theta \minus{} 1)\cdot \cos\frac {\theta}{2} \equal{} \left(\sqrt {2}\cdot \sin(\theta \plus{} \frac {\pi}{4}) \minus{} 1 \right)\cdot \cos\frac {\theta}{2}$.

However, from here, I can only prove that $ MN \leq \sqrt {2} \minus{} 1$, far weaker than either of the statements...[/hide]

@hell_ever:

I think you forget to square, the inequality is equivalent to $ MN^2 \equal{} 2r^2 \plus{} 2r^2\cos C\le \frac {4}{27}$

and also (I had to check with a calculator) those values you gave satisfy the inequality, supposing I'm not completele stupid...

Anyway:
$ MN^2 \equal{} 2r^2(1 \plus{} \cos A) \equal{} 2(\frac {\sin A \plus{} \cos A \minus{} 1}{2})^2(1 \plus{} \cos A) \equal{}$
$ \frac {1}{2}(2 \minus{} 2\cos A \minus{} 2\sin A \plus{} 2\sin A\cos A)(1 \plus{} \cos A) \equal{}$
$ (1 \minus{} \cos A)(1 \plus{} \cos A)(1 \minus{} \sin A) \equal{} \sin^2A(1 \minus{} \sin A)\le\frac {\sin A}{4} < \frac {1}{4}$

So $ MN<\frac{1}{2}$

This is the best I could reach.
 :)

$ MN^2\equal{}\sin^2A \left ( 1\minus{}\sin A \right ) \\ \sin A\equal{}t, \ t \in (0;1) \\ f(t)\equal{}t^2\minus{}t^3 \\ f'(t)\equal{}2t\minus{}3t^2\equal{}t(2\minus{}3t) \\ f'(t)>0 \ \ for  \ \ t\in (0; \frac23) \\ f'(t)<0 \ \  for  \ \ t \in (\frac 23 ; 1) \\ \Rightarrow \max_{t \in (0;1)}{f(t)}\equal{}f\left ( \frac23 \right ) \equal{} \frac{4}{27} \\ MN\equal{} \sqrt{f(t)} \leq \sqrt {\frac{4}{27}}\equal{}\frac{2\sqrt{3}}{9}$

 :)

yes, i forgot to square....
and nightwish, please forgive me if im doing something really stupid but isn't
$ MN^2\equal{}(1\minus{}\cos A)(\cos^2 A)$???

Look at the post of ElChapin. The result follows from the law of cosines for $ \triangle MNI$, where I is the incenter.

[quote="hell_ever"]yes, i forgot to square....
and nightwish, please forgive me if im doing something really stupid but isn't
$ MN^2 \equal{} (1 \minus{} \cos A)(\cos^2 A)$???[/quote]

acording to the problem notation I think it is $ MN^2 \equal{} (1 \minus{} \sin A)\sin^2 A$

but if not, and $ MN^2 \equal{} (1 \minus{} \cos A)(\cos^2 A)$ you can change the angle($ A \plus{} C \equal{} 90$) a proceed as nightwish.

 :)


Dear Friends :

I want to travel to your country (travel to new delhi)about next 20 days!

I Want to buy this books from your country:

1.C. R. Pranesachar, B. J. Venkatachala, and C. S. Yogananda Problem Primer for the Olympiad Interline publishing, Bangalore.

2.PROBLEM PRIMER FOR THE OLYMPIAD by C. R. Pranesachar, B. J. Venkatachala, C. S. Yogananda, Prism Books Pvt. Ltd., #1865, 32nd. Cross, BSK II Stage, Bangalore 560 070. or 49, Sardar Sankar Road, Kolkata 700029. Phone: 24633890/24633944.

3.MATHEMATICAL CHALLENGES FROM OLYMPIADS, by C. R. Pranesachar, S. A. Shirali, B. J. Venkatachala, C. S. Yogananda, Prism Books Pvt. Ltd.

4..N.Kapur, Mathematical Olympiad Problems Book I (1991), Books II, III (1992), Books IV, V, VI (1993), Mathematical Sciences Trust Soc., New Delh

5.M. R. Modak, S. A. Katre, V. V. Acharya An Excursion in Mathematics, RMO committe (Maharastra and Goa)

I want you for helping me:

i) Which of the books is available now?

ii)Please tell me the Address of some bookshop in new delhi ,which i probabily can find this books there!

iii)If you can please tell me an estimate about the total price of this book!(in US Dollar please)

I'l try to amend your great favour!

                                                                                Regards
                                                                                Navid.s

i think most of these books will be available at Central Book Store ,Nai Sarak,New Delhi

Total cost should be within 25-30 US$ :)

[quote="rituraj007"]Total cost should be within USD 25 to 30 :)[/quote]

I suppose INR is the currency in use throughout India.

but he certainly must b wanting his own currency amount :P

Thanks and when i can find exact Address of this bookshop?

[quote="rituraj007"]Central Book Store ,Nai Sarak,New Delhi
[/quote]

actually thats the exact address :)
if u r coming in person

Hi
All the books may not be available in Delhi.

You could send request to 'Bhaskracharcharya pratishthana' Pune http://www.bprim.org/publication.php for 'An Excursion in Mathematics' and they will send the book by post.

Aternatively you could contact 'Universal book stall' in Pune Tel 91-20-24455188,
address-27 Budhwar Peth,Pune 411002.They have all the books and will send the books by post.

Wow! I've lived in India for 17 years and I was dyiing to lay my hands on those books!  :rotfl:
I guess I'll get them by post. Thanks swanand

yo swan& u frm pune huh ?

m prakash academy by any chance ?

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coz ur advertisin univ book store  :rotfl:
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And where is Pune? ,Do they send book to delhi until 10 days?Do they send book out of india?
thanks!

pune is near mumbai
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and where is mumbai ???   :rotfl:
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try callin universal book store on the number.

they are generally very cooperative. i have a very good experience buying books from them. but i don't know if they send things over post

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and i can't tell whether the books will reach via post in 10 days, given the exceptional efficiency of indian postal services
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The Universal store will send you all the books by post, either to Delhi or to Iran.

Their alternative tel nos are 91-20-24451780/24450976.

For plaing your firm order you can send them an e-mail on:
universalbooks@eth.net

They will do te needful!

[quote="swanand"]All the books may not be available in Delhi.
[/quote]

How can u b sure abt that?? :rotfl:

i m sure abt atleast 4 of those 5 books

(in fact i myself have 3 of them...though they were sent by RMO guys :P )

All that the RMO 'guys'(at  60 yrs) gave us what an ancient edition of some sanskrit maths/astronomy book. Basically unsold yellow pages printed decades ago.  :rotfl:

but my RMO coordinator had given us RMO ques of Maha and Karnataka region prev year and photocopy of d same this yr.

And they provided me wid most of the book i have at this point of time(including 1,2 n 5)and the books were mew  :rotfl:  :rotfl:

navid has very little time in India. So,lets provide all info we all have.He will take the most suitable option or options!

Thanks alot dear friends !for your great favours you do for me by posting your comments!

First i'l try to the book shop:Central Book Store ,Nai Sarak,New Delhi
i hope to find most of the books i want in it!
is there any other bookstore which may have books in new delhi?

Second i'l try by the Universal store (via contacting!) and try to find rest of the books!

third If i cant find any of this books esp if they were out of stock! ,Can any of you help me
 to obaining it as Photo-copy version and so in?

fourth Thanks again!

[quote="navid"]is there any other bookstore which may have books in new delhi?
[/quote]
nai sarak itself has many book stores....i think u urself will find wen u reach there :)

@navid,
what are your dates of travel to Delhi and back? If you remain in touch, we can see how and where to send you Photocopies if you dont get the books.
At which place in Iran do you teach and to what level of students? What other books do you refer to for training Olympiad students?