function
algorithm
parameterization
algebra
binomial theorem

I was wondering if someone could help me with this recursive algorithm.  The goal is to find the number of unique ways in which the numbers of the array num can be used to sum to val.
This works the only problem is that
1+1+2+1 = 5
and
1+1+1+2 = 5
are seen as different solutions, can anyone help me fix this?  Thanks.
[code]
void find_add(int val)
{
    int i = 0;

    if(sum == val)
    {
        ways++;
        return;
    }
    else if(sum > val)
    {
         return;
    }
    else
    {
        for(i = 0; i < units; i++)
        {
              sum += num[i];
              find_add(val);
              sum -= num[i];
        }
    }

    return;
}
[/code]

Maybe this will help, but I dont know your array size:

Find the number of solutions to :

 :Sigma: (Pi * Ei)  = val,
where Ei is an element at position 'i' and Pi is a natural number with which Ei should be multiplied

You could pass 'i' as the only parameter to the recursive function to denote that you will consider only the multiples of Ei in that iteration.

Please tell me if it works or you find a better solution, as I am too lazy to  code it out myself :P

Hey, can anyone please tell me if this can be implemented in C++?

In my 10+2 class we used to use binomial theorem to solve such type of problems:

In this problem maybe , suppose val = 10, and the only elements are 1,2,5;
then the answer is coefficient of x^10 in :

(1+x+x^2+x^3+x^4.......)(1+x^2+x^4+x^6+x^8.......(1+x^5+x^10+x^15.....)

What does the sigma denote in your equation?

:Sigma: stands for summation of the multiples of numbers in the array.

HI,
As i said in one post, my exam is running so i can't check this section regularly.

Anyhow, the problem is NICE one. To say the truth i thought of this problem to propose in some online contest.

Yes htmnssn, your recursive idea was good. and well, it can't be solved by that bionomial process. If u think a bit then you will understand that!

Now i think if u r a bit experienced then u must know that all the experienced programmers think of DP before backtraking. :D  yes it can also be solved by DP.

Before saying anything further i want to give u a chance for thinking a bit on it :lol:

OK?

-Mahbub

I am still unclear.  Ei * Pi will not yield an array

Wait, say the val=5

and the given dataset is: {1,2,5} and all the numbers can be used multiple time. like this?

or

given:
{1,1,2,5} all the number can be used once?

well, i think u r not convinced with what i said in the last post.

best will be if u post the exact problem statement. :?

Couldn't you sort the numbers, then store them to an array, and count the unique members of the array at the end?

mahbub - yes numbers can be used multiple times
sola - i was going to do that, but i wanted to see if anyone had an easier way

ahaaaaaa DP! Can't believe missed it. Is this the money system problem?

If it is multiple time (as snow said) then it is quite ez by DP. But if it was one time thed it would be a bit tricky.

yes msn, it is some kinda coin-changing algo. But i need sometime to be sure about it. i need to think. but i am pretty confident that it is possible. (as my exam is running so i can't get time to think. so just wait let me think)

Well, I think it can’t be solved by DP. Because as snow said it also matters even for ordering. Such as 1+2+2=5 and 2+2+1=5 is different. But my DP can’t take care of this kinds of order. But it can say how many solutions are there (not considering the ordering. Such as the above two are same). It can also say which numbers can be made by the given numbers. So a fair backtracking code should be written to solve it.

Note: msn’s backtracking method will not work here.

The function will be like this:

Func(sum)
{
if(sum>desired_value) return;
if(sum==desired_value) then do as you want

for(i=1 to n)
func(sum+number[i])
}

it can be a bit improved if you use the if() inside for.

And undoubtedly time complexity is very bad here as you can understand. It may be worse than 2^n.

Anyhow, if you have no problem will you please say where you got this problem?

Well, i got a DP solution. But i can't understand why i id not get it earlier!

Anyhow, say the given numbers are in number[] array and the desired sum is val. we need an auxiliary array for flagging say flag[0 to val] initialize allall flag[] to 0 except flag[0]=1.

now start like this:

for(i=0 to val-1)
for(j=1 to sizeof array (i mean #of the given numbers))
{
if(i+number[j] <= val)
flag[i+number[j]] += flag[i];
}

That's all.

 :lol:

Thanks.
And yes it is the usaco money problem.  I was thinking of finding all the ways to sum to each given money unit with the other money units, so I only have to find multiples of the target value and then multiple the number i get by the number of ways there are to sum to each value.

Well, for ur practice can u write up a DP code which will also print the answers in sorted way. By sorted it means:
say for 1st soln:
n[1][1], n[1][2],...,n[1][x]
n[2][1], n[2][2],...,n[2][y]

is there is no val for say n[x][5] then it will be considered as infinity.

well OK. u don't need to think much on order just print all the possible soln. :D

I'm still having trouble understanding dynamic programming.  I know that it basically is attempting to calculate all of the possibilities at once rather than one at a time as recursion does.  But I'm having trouble seeing how it can be applied to a problem like this.

i think this was a problem with coins, so i'll call them coins, even if they're not...

create an array, called something like numways[][]
and let numways[value][numcoins] be the number of ways to use the first numcoins coins to reach the value
so if your coins are 1,3,5   numways[10][2] gives the number of ways to reach 10 using coins of 1 and 3
then if you want to calculate numways[x][3], its very easy (say x=13 as an example):
numways[13][3]=numways[8][2]+numways[3][2]
because if you can make 8 or 3 with the first two coins, all you have to do is add some 5-coins.
you dont really care about how the lower coins are used, as long as you know the number of ways.

I am not 100% sure with kmod's soln. Anyhow,

back to my given soln:

for(i=0 to val-1)
for(j=1 to sizeof array (i mean #of the given numbers))
{
if(i+number[j]  <=  val)
flag[i+number[j]] += flag[i];
}

see it goes from flag[0] to flag[1] to flag[2] to .... flag[val]

initially flag[0]=1 (it is just convention and will help us to go for the next flags)

say val=10 and numbers={NIL,1,3,5}

now see how it progresses:
flag[0+1]+=flag[0]
flag[0+3]+=flag[0]
flag[0+5]+=flag[0]

now come to flag[1]
flag[1+1]+=flag[1]
flag[1+3]+=flag[1]
flag[1+5]+=flag[1]

and so on...

now think why does it work. It will be sufficient to prove that every possible seq say S1, S2,...,Sn can be gained by this process, obvious way to prove is Induction.

Say Every number x=1 to n-1 can be made in flag[x] ways.
S'=S1+S2+....+Sn-1 can be expressed in flag[S'] ways.

So obviously we are including for flag[S] all those possible way that ended with Sn at last. Right? [note S=S'+Sn]

Understand?

our two solutions are pretty much the same, except mine uses extra space (which can easily be avoided), and goes about the computation in a slightly different order; both of which are superficial differences.  i personally think that mine is more intuitive, but thats probably because it's well...mine.

[quote]
numways[13][3]=numways[8][2]+numways[3][2]
because if you can make 8 or 3 with the first two coins, all you have to do is add some 5-coins.
you dont really care about how the lower coins are used, as long as you know the number of ways.
[/quote]
Why isn't this
numways[13][3] = numways[8][2] + numways[5][1] ?

i think that was a typo from kmod. may be u r correct snow.

well, i guess we're thinking of different problems.  if you have a set of coins that you can only use once, it would be something more like what you said.  i thought the problem was that you can take as many as you want of each kind.
so there are two ways for reaching 13 then:
make 8 somehow with the first two types of coins, then take a coin of value 5
or make 3 somehow with the first two types of coins, then take two coins of value 5

so that second number doesnt keep track of how many coins are used, but how many different types of coins could've been used (1 and 3, in my example, so thats why it was 2.  when you add the possibility of using a 5-coin, it becomes 3)

hope that helps

Oh, I see what you are saying now.  We are still talking about the same problem.
Like you could also make 4 with 1 or 2 coins and make 9 with 1 or 2 if you had those coins.  I just didn't understand your wording.

Look, i know two problems.

One is famous one called coin changing algorithm (CC)
in this problem ordering is not a prob. Say 1+1+2=4 and 2+1+1=4 and 1+2+1=4 is all counted once.

And as snow wrote in his first and aother post, for his problem ordering is a big fact. that means those three given just up, will count 3, not 1.

Now clarify me snow, which one is ur problem? first one CC? or the second one? my algo works for second one and the algo for the first one is quite simple. :D

The one where 1+2+1+1 = 5 and 1+1+2+1 = 5 only count as 1.

Oh and kmod, is this sort of what you were describing for your solution?

[code]
for(i = 1; i <= val; i++)
    {
          for(j = 1; j <= units; j++)
          {
                if(i-numbers[j] > 0 && j-1 > 0)
                    flag[i][j] += flag[i-numbers[j]][j-1] + flag[numbers[j]][1];
          }
    }
[/code]
Where numbers contains each coin type possible.  units is the number of coin types and val is what you are trying to sum to.

I haven't tested this yet, so I'm not quite sure whether it will work or not.[/code]

Well, snow i am really puzzled. I request u to post the problem statement excatly as it is in the source. :?

kmod's solutio uses extra memory it is right. and my soln for both of those problems don't require it. So if u r interested to the soln of the problem where ordering is not the matter then i can post my soln.

but plz don't forget to post the problem statement snow.

Here you go:
[quote]
Money Systems

The cows have not only created their own government but they have chosen to create their own money system. In their own rebellious way, they are curious about values of coinage. Traditionally, coins come in values like 1, 5, 10, 20 or 25, 50, and 100 units, sometimes with a 2 unit coin thrown in for good measure.

The cows want to know how many different ways it is possible to dispense a certain amount of money using various coin systems. For instance, using a system of {1, 2, 5, 10, ...} it is possible to create 18 units several different ways, including: 18x1, 9x2, 8x2+2x1, 3x5+2+1, and many others.

Write a program to compute how many ways to construct a given amount of money using supplied coinage. It is guaranteed that the total will fit into both a signed long long (C/C++) and Int64 (Free Pascal).
PROGRAM NAME: money
INPUT FORMAT

The number of coins in the system is V (1 <= V <= 25).

The amount money to construct is N (1 <= N <= 10,000).
Line 1: 	Two integers, V and N
Lines 2..: 	V integers that represent the available coins (no particular number of integers per line)
SAMPLE INPUT (file money.in)

3 10
1 2 5

OUTPUT FORMAT
A single line containing the total number of ways to construct N money units using V coins.
SAMPLE OUTPUT (file money.out)

10
[/quote]

Oh and you don't need to post your solution.  I want to solve it and understand it myself.

Aha, I think I got it
for(i=0 to units-1)
   for(j=numbers[i] to val)
          if j == numbers[i] then increment flag[j]
       flag[j] += flag[j-numbers[i]];

Thinking about it with a 1 dimensional array made it a lot simpler I think.

Ok. so if u feel any trouble with it or with ur algo just drop a note here! :lol:


LaTeX
algebra
polynomial
Rational Root Theorem

the only method that i know of is using the rational root theorem ( i think that's what its called) where one of the roots is of the form $ x\equal{}\frac{p}{a_{0}}$ where $ p$ is the last constant term and $ a_{0}$ is the coefficient of the leading term.

You can also use Descartes' Rule of signs, where you count the number of sign changes in $ f(x)$ for positive roots and $ f(\minus{}x)$ for negative roots.

For example, in this case, $ 2x^{3}\plus{}15x^{2}\minus{}34x\plus{}24$ changes sign twice (from $ \plus{}15x^{2}$ to $ \minus{}34x$ to $ \plus{}24$), so there are either $ 2$ or $ 0$ positive roots.

Substituting in $ \minus{}x$, $ \minus{}2x^{3}\plus{}15x^{2}\plus{}34x\plus{}24$ changes sign once, so there is exactly $ 1$ negative root.

After using the Rational Root Theorem, try for a negative root since you know there is $ 1$ negative root.

[quote="linzk3"]Find the zeros of y = 2x^3 + 15x^2 - 34x +24
 or how could I go about finding them?[/quote]

Why don't you learn [[LaTeX]]?

[hide="Solution"]$ y = 2x^{3}+15 x^{2}-34 x+24$

Start playing around with RRT, which in this case, limits the rational solutions to:
$ \pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 8,\pm 12,\pm 24$

Unfortunately, none of these work, which you can easily see by plugging in. So, we have to resort to the [url=http://planetmath.org/encyclopedia/CubicFormula.html]cubic formula[/url].

Mathematica gives me one real (messy) solution:
$ \frac{1}{2}\left(-5-\frac{\left(3087-2\sqrt{189237}\right)^{1/3}}{3^{2/3}}-\frac{143}{\left(3\left(3087-2\sqrt{189237}\right)\right)^{1/3}}\right)\approx-9.43632$

If you really need the complex solutions as well, they are below:

[hide]\begin{eqnarray*}-\frac{5}{2}+\frac{\left(1+i\sqrt{3}\right)\left(3087-2\sqrt{189237}\right)^{1/3}}{4 3^{2/3}}+\frac{143\left(1-i\sqrt{3}\right)}{4\left(3\left(3087-2\sqrt{189237}\right)\right)^{1/3}}&\approx& 0.968158-0.578233 i\end{eqnarray*}

\begin{eqnarray*}-\frac{5}{2}+\frac{\left(1-i\sqrt{3}\right)\left(3087-2\sqrt{189237}\right)^{1/3}}{4 3^{2/3}}+\frac{143\left(1+i\sqrt{3}\right)}{4\left(3\left(3087-2\sqrt{189237}\right)\right)^{1/3}}&\approx& 0.968158+0.578233 i\end{eqnarray*}[/hide][/hide]

[quote="Kyron"][quote="vishalarul"]Start playing around with RRT, which in this case, limits the rational solutions to:
$ \pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 8,\pm 12,\pm 24$[/quote]
You missed $ \pm\frac{1}{2},\pm\frac{3}{2}$. But in this case it really doesn't matter, they weren't solutions anyway  :lol: .[/quote]

Actually, I missed a lot more than that. RRT actually implies that the only rational solutions are:

$ \pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 8,\pm 12,\pm 24$

$ \pm\frac{1}{2},\pm\frac{3}{2}$

$ \pm\frac{1}{3},\pm\frac{2}{3},\pm\frac{4}{3},\pm\frac{8}{3}$

$ \pm\frac{1}{4},\pm\frac{3}{4}$

$ \pm\frac{1}{6}$

$ \pm\frac{1}{8},\pm\frac{3}{8}$

$ \pm\frac{1}{12}$

$ \pm\frac{1}{24}$

Yuck.

No no, you didn't miss that many. RRT says that a rational root is of the form $ \frac{p}{q}$ where $ p$ divides the constant term and $ q$ divides the leading coefficient. An interesting result is that any root of a monic integer polynomial is an integer. This is a result of the integers having unique factorization, and the property is called being "integrally closed".


I saw the answer in the answer key but didn't know how to do it.[hide="The Problem"]
A bowl contains fewer than $50$ marbles, and each is either red, green, or blue. The probability of drawing a red marble is $\frac{2}{5}$ and the probability of drawing a green marble is $\frac{3}{7}$. If two marbles are drawn without replacement, what is the probability that both are blue? Express your answer as a common fraction.[/hide]

Does the answer involve knowing that the fraction of blue marbles to total marbles is $\frac{6}{35}$?

-Jorian Hoover

Since no one has posted a solution yet... well no one has posted at all...

[hide]Yes, you are correct in getting that the ratio of blue:total is 6:35, so I'll skip the explanation for that part.

The probability of getting a blue marble on your first draw, therefore, is $6/35$. The next draw, that one blue marble is out. There are 5 blue marbles left and 34 marbles total. Thus, the probability of drawing another blue marble is $5/34$. $5/34 * 6/35$ is $3/119$[/hide]

Ok, after I first did it, I read it wrong. When I saw the answer sheet, I read the problem correctly, but still did it wrong. I thought 2 marbles were randomly picked, then u had to figure out the chance that 2 more were picked and they were both blue. Thank you Fanatic.
-Jorian Hoover


function
real analysis
real analysis unsolved

Suppose $ U$ is connected and $ u\in W^{1,p}(U)$ satisfies $ Du \equal{} 0$ a.e. in $ U$. Prove $ u$ is constant a.e. in $ U$.

The main idea is, like most of the time when you state things like that for Sobolev-Spaces, approximating $ u$ by smooth functions:

Let $ \eta$ be some standard mollifier and $ \eta_\epsilon(x) : \equal{} \frac{1}{\epsilon^n} \eta(\frac{x}{\epsilon})$.
Then $ D(\eta_\epsilon*u) \equal{} \eta_\epsilon * (Du) \equal{} 0$ on sufficently small open subsets $ V\subset U$. Now use the fact, that $ \eta_\epsilon * u$ is smooth and  $ \eta_\epsilon * u \to u$ a.e.


What is the basic concept of modular arithmetic?

What is it used for?

[url=http://www.math.rutgers.edu/~erowland/modulararithmetic.html]Here[/url] is a pretty cool site I used to learn mods.

Modular arithmetic is basically doing math not on a line but kinda like on a circle; the values "wrap around", always staying less than a fixed number called the modulus.

To answer your second question, modular arithmetic is used very heavily in number theory. It is especially applicable to cryptology.

"Wrap around?"

I'm not sure I'm getting this, but would something like modulo 12 be kind of like time on a clock?

Basically. Like, say it's 9:00 now and you add 72 hours. It isn't 81:00, it's 9:00. If it's 11:00 and you wait 2:00, it isn't really 13:00, it's 1:00. $ 11 \plus{} 2 \equiv 1$.

Based on what you said, that's why 13 is congruent to 1 (mod 12)?


search
combinatorics proposed
combinatorics

There are $ 100$ guys in a prison. The security members of the prison have decided the following. Some evening, all guys are gathered in a room and allowed to discuss whatever they want. The following day they all go to their cells and aren't allowed to talk / see each other anymore. A special room $ S$ is built with a lamp inside. Initially the lamp is off. Then the convicts are brought in random order in this room and when leaving it they either leave the lamp as it is, or switch it ON/OFF. If at some point one of them says that all the covicts have been at least once in the room $ S$ and the information is correct, all of them are released. If the information is false, they're sentenced to death. The convicts are provided the information that as long as they're silent each of them will be infinitely many times in $ S$. Can the convicts find a strategy so that they will surely be released?

Can all the convicts see the status of the lamp at every moment?
if so, the strategy is to change the status if it is the first time in that room, then if you notice exactly 100  changes you are sure that all the people have been there otherwise keep counting...

http://www.ocf.berkeley.edu/~wwu/riddles/hard.shtml and search "100 PRISONERS AND A LIGHT BULB".

mszew, the convicts can't see the lamp at all times.

One of the less optimal solutions to this problem is this: Designate a leader. The other 99 people do the following: if they are entering the room for the first time with the lamp off, they turn it on. They do nothing in no other case. The leader, if he goes in the room, turns off the light if it is on, and does nothing otherwise. Once the leader flips the light off the 99th time, he announces everyone has been inside.


for number 12, ab means a*b, not 10a+b, right?
i just want to make sure because i'll have time to do this one.

NO!  You are supposed to send a "private message" to rep123max.  When you are logged in, click on "private messages" and work from there.

[quote="Chigr"]sorry im new to this site
are you supposed to post the answers in the forum or what?[/quote]

Welcome to the Art of Problem Solving!  Since it is a contest, you don't post your answers in the forum, because then people might cheat.  You have to private message (PM) them.  You can click "Private Messages" near the top of the screen like JS1527 said, or you can just click the PM button right below Rep123Max's post.

[quote="tetrahedr0n"]15 is so tedious! Rep, can I get credit if I just describe how to solve it? :)[/quote]
There is actually a quite elegant way to solve it. I didn't see it until the second time i did it, but its pretty cool.

I think I know what you mean. I found the roots in under 15 seconds. The geometry is tedious!

I just did it, and I think I did it the hard way.  Maybe Rep123max will post my solution for kicks - an example of how to spend WAY too much time on a problem...

No, I mean that I found a slick way to find the area. I will pm it to you so you can see.

thx porkchop ... im ready to be humbled  i had a creative way of getting roots, but ahem ... the way i found the area [hide]subtracting a trapezoid with three equal sides from a regular dodecagon[/hide] was tedious.

[hide]You can do it with affine transformations. [/hide]


Assume that the sequence $\{p_n\}_{n \geq 1}$ primes satisfies a linear recurrence. Prove that $\{p_n\}_{n \geq 1}$ is a periodic sequence.

I think the following stronger version holds:

Let $(u_{n})_{n\ge 0}$ be an integer linear recursive sequence with the property that for every integer $m\ge 2$, the set of values $\{u_{n},\ n\ge 0\}$ contains only finitely many multiples of $m$. Then, $(u_{n})$ is periodic.

Oh, yes, it does -- but you meant "only finitely many proper multiples of $m$" (that is to say, $km$ with $k>1$), right? :)


geometry
3D geometry
tetrahedron
triangle inequality
IMO
IMO 1968

Prove that every tetrahedron has a vertex whose three edges have the right lengths to form a triangle.

[hide]Suppose the contrary that there is a tetrahedron $ ABCD$ that does not have any such vertex. So $ AB\ge AC\plus{}AD$ and $ AB\ge BC\plus{}BD$. Thus we have $ 2AB\ge AC\plus{}AD\plus{}BC\plus{}BD$. On the other hand, $ AB<AC\plus{}BC$ and $ AB<AD\plus{}BD$, which imply $ 2AB<AC\plus{}AD\plus{}BC\plus{}BD$. Contradiction. Q.E.D.[/hide]


calculus
integration
topology
function
advanced fields
advanced fields unsolved

Prove that every closed $ 1$-form on $ S^{2}$ is exact.

The integral of a closed $ 1$-form over a curve $ \gamma$ depends only on the homotopy class of $ \gamma$. But $ S^{2}$ is simply connected. Therefore, we can define $ F(p) \equal{}\int_{\gamma}\omega$, where $ \gamma$ is some path from the North Pole to $ p$. It follows that $ dF \equal{}\omega$.

A bit more down-to-earth: stereographic projection from north (resp. south) pole allows application of Poincaré's lemma, so we get a function $ F_{N}$ on $ S^{2}\setminus\{N\}$ satisfying $ dF_{N}\equal{}\omega$ (same for south pole). Since $ d(F_{N}\minus{}F_{S})\equal{}\omega\minus{}\omega\equal{}0$ on the intersection (which is connected!), $ F_{N}$ and $ F_{S}$ differ by a constant $ c$, i.e., $ F_{N}\equal{}F_{S}\plus{}c$. Gluing these together, we get a global function satisfying $ dF\equal{}\omega$. owk


modular arithmetic
number theory unsolved
number theory

I have a problem with the following problem.

I want to solve the following equation:$ p^a \minus{} 1 \equal{} 2^n (p \minus{} 1)$ where p is a prime number.
I showed that $ a$ must be power of $ 2$ but now I'd like to prove that $ a$ can't be divisible by $ 4$ but I can't.

Zsigmondy's Theorem states that there is prime divisor of $ a^n \minus{} b^n$ which doesn't divide $ a^i \minus{} b^i$ when $ i < n$, with 2 exceptions:
a) $ a \equal{} 2 ,b \equal{} 1 , n \equal{} 6$
b) $ a \plus{} b$ a power of 2 and $ n \equal{} 2$
[ [url]http://www.problem-solving.be/pen/viewtopic.php?t=669[/url] ]
Thus, if $ a \equal{} st$, where $ s,t > 1$, then $ \frac {p^a \minus{} 1}{p \minus{} 1} \equal{} 2^n$ must be divisible by 2 different primes, unless we have one of the exceptions. So either $ a$ is a prime (and then it follows that $ a \equal{} 2$, because, as you said, $ a$ is a power of 2), or we have one of the exceptions:
a) $ a \equal{} 2 ,b \equal{} 1 , n \equal{} 6 \implies p^2 \minus{} 1 \equal{} 64(p \minus{} 1) \implies p \equal{} 65\equal{}5 \cdot 13$, a contradiction.
b) $ a \plus{} b$ a power of 2 and $ n \equal{} 2$, and it is OK because we want $ n$ to equal $ 2$.

So indeed, $ a$ is not divisible by $ 4$.
Zsigmondy's Theorem is very strong (though there's a short proof), it is possible that another proof exists.

$ p^a \minus{} 1 \equal{} (p \minus{} 1)2^n$

If $ a$ is divisible by $ 4$,$ L.H.S.$ is divisible by $ p^4 \minus{} 1$ , $ p^2 \minus{} 1$.
So $ p \plus{} 1$ is a power of $ 2$ , $ p^2 \plus{} 1$ is a power of $ 2$ .

$ (p^2 \plus{} 1,p \plus{} 1) \equal{} (p \minus{} 1,p \plus{} 1) \equal{} 2$ for odd prime $ p$. So if $ p > 3$ , one of them have other divisors.

It is easy to see that $ p\geq 3$. Now, suppose that $ a$ is a power of 2 greater than 4, thus $ (p \minus{} 1)(p \plus{} 1)(p^2 \plus{} 1)(p^4 \plus{} 1)\cdot\dots\cdot (p^{2^k} \plus{} 1) \equal{} 2^n(p \minus{} 1)$, so  $ (p \plus{} 1)(p^2 \plus{} 1)(p^4 \plus{} 1)\cdot\dots\cdot (p^{2^k} \plus{} 1) \equal{} 2^n$. Here the only factor which can be $ \equiv 0 \pmod{4}$ is $ (p \minus{} 1)$. The other factors are  $ \equiv 2 \pmod{4}$. But, for example, factor $ (p^2 \plus{} 1) > 2$, so there must be also odd prime factors in $ LHS$. This is absurd since the $ RHS$ is a power of 2.

EDIT: Sorry Makoto Kohno! I have seen your post too late   :maybe:


Lep $ p$ and $ q$ be positive rational numbers for which:
\[ \frac{1}{p}+\frac{1}{q}=1\]
Prove that for any positive numbers $ x$ and $ y$ the following inequality holds:
\[ xy \le \frac{x^{p}}{p}+\frac{y^{q}}{q}\]

[hide]It's trivial, weighted $ AM-GM$.[/hide]


algebra polynomial function LaTeX quadratics

[hide]1. a+b=2

ab=7

a ^2+b^2 =-10

a^3+b^3=-34

2.  a+b+c=-1

abc=-1001

ab+bc+ca=-333

a^2+b^2+c^2=667

[/hide]

I don't realy have time to solve the rest right now.

[quote="Rep123max"]
3. Given three numbers [tex]\displaystyle x_1,\; x_2,\; x_3[/tex], let [tex]\displaystyle a=x_1+x_2+x_3[/tex], [tex]\displaystyle b=x_1x_2+x_2x_3+x_3x_1[/tex], [tex]\displaystyle c=x_1x_2x_3[/tex] and [tex]\displaystyle d=x_1^3+x_2^3+x_3^3[/tex]. If [tex]\displaystyle a=3[/tex], [tex]\displaystyle b=7[/tex], and [tex]\displaystyle c=10[/tex]. What is d?
[/quote]

Given the same conditions, what is the value of [tex]\displaystyle e=x_1^6+x_2^6+x_3^6[/tex] ?

[hide]i believe (a + b +c) :^3:  = a :^3:  + b  :^3:  + c  :^3:  + 3(ab + bc + ac)(a + b + c) - 3abc.    (you can get this equation by manipulating (a + b + c) :^3: )

using this,

3 :^3:  = d + 3 * 7 * 3 - 3 * 10

so d  = - 6

[/hide]

Rep123max wrote:If [tex]\displaystyle 3x^3+7x^2-5x-2=0[/tex] has roots r, s, and t, then what does [tex]\displaystyle\frac{1}{rs}+\frac{1}{st}+\frac{1}{rt}[/tex]?

For some reason I got stuck, but here's my work:

[hide]

[tex]\displaystyle\frac{1}{rs}+\frac{1}{st}+\frac{1}{rt}=\frac{r+s+t}{rst}[/tex]

Now since r, s, and t are zeros of the function, it's equation can be expressed as:

f(x)=3(x-r)(x-s)(x-t)=(3)x^3+(-3r-3s-3t)x^2+(3rs+3rt+3st)x+(-3rst)

So you want [tex]\frac{r+s+t}{rst}[/tex], and you already know that rst=2/3 by equating the constant terms.

-3(r+s+t)=7

r+s+t=-7/3

[tex]\displaystyle\frac{r+s+t}{rst}=\frac{\frac{-7}{3}}{\frac{2}{3}}=\frac{-7}{2}[/tex]

[/hide]

Edited.  Is it correct now?

The LaTeX kinda undermines the spoiler function... anyway here's my (LaTeX free) attempt:

1.

[hide]x - 2x + 7 ... roots are (2  √(4-4*7)]/2 = 1  √24√-1 = 1  i√6

So roots are:

a = 1 + i√6

b = 1 - i√6

(note these are complex conjunctuates)

a+b = (1 + i√6) + (1 - i√6) = 2

ab = (1 + i√6)(1 - i√6) = (1 + 6i) = 1 - 6 = -5

a+b = (1 + i√6) + (1 - i√6)

= 1 + 2i√6 + 6i + 1 - 2i√6 + 6i

= 2 - 12 = -10

a+b = (1 + i√6)(1 + i√6) + (1 - i√6)(1 - i√6)  (he does neglecting his newfound knowledge of the binomial expansion)

= (1 + i√6)(1 + 2i√6 + 6i) + (1 - i√6)(1 - 2i√6 + 6i)

= (1 + i√6)(2i√6 - 5) + (1 - i√6)(-2i√6 - 5)

= (-17 - 3i√6) + (-17 + 3i√6)

= -17 - 17 = -34[/hide]

I'm not sure what you mean, if (x-r)(x-s)=0? Then the roots are simply x=r, x=s.

right.

If you multiply it all out, you get x^2-(r+s)x + rs*x.  So if you have a general polynomial x^2+bx+c with roots r and s, -b=r+s and c=rs.
Then you can generalize this to other degrees like max suggested.

[quote="Rep123max"]Expand it and see what happens.[/quote]

Ok.

(x-r)(x-s) = x - (s + r)x + sr

I'm not sure what you're trying to tell me...

What he is getting at is that the sum of the roots to the general quadratic equation, [tex]ax^2+bx+c=0[/tex] is -b/a and the product of those roots is c/a.  This is a very useful shortcut to know.  And of course you can expand these "symmetric sums" to polynomials of higher degrees.

(px + q)(rx + s)

roots are -q/p and -s/r

If we multiply this out,

px*rx + px*s + q*rx + q*s

= prx + (ps + qr)x + qs

In the form ax + bx + c,

a = pr
b = (ps + qr)
c = qs

The sum of the roots is:

-q/p - s/r = -qr/pr - sp/pr = -(qr + sp)/pr
-(qr + sp)/pr = -b/a (wohoo)

The product of the roots is:

(-q/p)(-s/r) = qs/pr = c/a

Heh! Thankyou, I never even thought to do this before!


I don't know how to do subscripts and math symbols but hopefully it is clear in understanding.  Anyone having a start to this problem or ideas I would greatly appreciate it.

Let An be n x n matrix with 2's on the main diagonal, 1's directly above and below the main diagonal, and 0's elsewhere. Let a(lower n) = det(A(lower n)).

a) For n(greater than or equal)3, find a formula expressing a(lower n) in terms of a(lower n-1) and a(lower n-2).

b) Find a(lower i), i = 1, 2,…, 5

c) Give a closed formula for an in terms of n.
[/hide][/list][/i]

Try some row/column reduction.  If you can get it into an ordinary diagonal matrix, then the determinant is easy to evaluate.

Here's the statement, made visible:

$ A_n$ is an $ n\times n$ matrix, and $ A_n \equal{} \begin{bmatrix}2 & 1 & 0 & 0 & \cdots & 0 \\
1 & 2 & 1 & 0 & \cdots & 0 \\
0 & 1 & 2 & 1 & \cdots & 0 \\
0 & 0 & 1 & 2 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & 0 & \cdots & 2\end{bmatrix}.$

Let $ a_n \equal{} \det(A_n).$

a) For $ n\ge 3,$ find a formula expressing $ a_n$ in terms of $ a_{n \minus{} 1}$ and $ a_{n \minus{} 2}.$

And so on.

I would suggest approaching a) by looking at the determinant in terms of minors of the first row.

Kent, I think the last row looks easier because it's almost entirely 0's.  Why did you opt for the first?

The last row has two nozero elements. Maybe I should have included the next to last column in the display. Using the last row would be exactly the same as using the first row.

Thank you both so much for the replies. I believe I got an answer for a and b.

For a.  I got an equation 2a(lower n-1)-a(lower n-2).  I tested it out with 3 different ones and got the correct answer so I am hoping/guessing that is right.

I believe I have b figured out.

However c, I have no idea where to begin.  I know of the traditional closed formula as...under the operation of addition if a and b are in A then a+b is also in A.  I just don't know how to take that and apply it to matrices and determinants.

I can confirm that $ a_n\equal{}2a_{n\minus{}1}\minus{}a_{n\minus{}2}.$

Do you have a full proof of that? Minors of the first row followed by a single row-elimination step in the second term should give such a proof.

Now, as for turning that into a formula, what do you know about linear recursions (also known as constant-coefficient linear difference equations)? That's what you got for the answer to a), so that's what we can use.

In parallel to the theory of linear constant-coefficient differential equations, we can assume that this has a solution of the form $ a_n\equal{}\lambda^n,$ Then we have $ \lambda^n\equal{}2\lambda^{n\minus{}1}\minus{}\lambda^{n\minus{}2}.$ This simplifies to $ \lambda^2\minus{}2\lambda\plus{}1\equal{}0.$ That gives us $ \lambda\equal{}1$ as a double root. Double roots can be a bit tricky; we wind up assuming that $ a_n\equal{}(c_0\plus{}c_1n)\cdot 1^n.$

We do have initial conditions: $ a_1\equal{}2,a_2\equal{}3,a_3\equal{}2\cdot 3\minus{}2\equal{}4.$ The only linear expression to satisfy that will be $ a_n\equal{}n\plus{}1.$

We can verify that this works: $ (n\plus{}1)\equal{}2n\minus{}(n\minus{}1).$


Is there any way to get LaTex, or some math typsetting into an instant messenger?

If so, could someone tell me where to find one and help me get it up and running?

THANKS!  :lol:


Shouldn't this be in Number Theory?

Also, if I'm not mistaken, this is quite an easy problem.

Note that your equation is equivalent to $ (z-2^x)(z+2^x)=3^y$. So, $ z-2^x=3^a$ and $ z+2^x=3^b$ ($ a+b=y$, $ b>a\geq 0$). Subtracting these two equations we get $ 2^{x+1}=3^b-3^a$. Obviously, the left side isn't divisible by $ 3$, while the right is for $ a>0$. So, $ a=0$ and $ z=2^x+1$. Our original equation is, therefore, equivalent to $ 4^x+3^y=1+2^{x+1}+4^x$, which, in turn, is equivalent to ${ 3^y-1=2^{x+1}}$. However, since $ x>0$, observing $ \mod 4$, we gather $ y=2y_1$. Needless to say, all variables so far are integers. From that, we get $ (3^{y_1}-1)(3^{y_1}+1)=2^{x+1}$. However, only two powers of two which differ by two are $ 2$ and $ 4$. The only solution to that is $ y_1=1$. that means $ y=2$ and we have $ 8=2^{x+1}$, $ x=2$, obviously. That's one and the only solution: $ (2,2,5)$.

I'm writing this in a hurry, so there can be some minor problems, which I will happily explain, if I can, tommorow.

Maybe you should just notify a moderator instead of double posting. I've already posted a solution here, too.


Of the following
(1) $ a(x\minus{}y)\equal{}ax\minus{}ay$
(2) $ a^{x\minus{}y}\equal{}a^x\minus{}a^y$
(3) $ \log (x\minus{}y)\equal{}\log x\minus{}\log y$
(4) $ \frac {\log x}{\log y}\equal{} \log{x}\minus{} \log{y}$
(5) $ a(xy)\equal{}ax\times ay$

$\textbf{(A)}\ \text{Only 1 and 4 are true} \qquad\\
\textbf{(B)}\ \text{Only 1 and 5 are true} \qquad\\
\textbf{(C)}\ \text{Only 1 and 3 are true} \qquad\\
\textbf{(D)}\ \text{Only 1 and 2 are true} \qquad\\
\textbf{(E)}\ \text{Only 1 is true}$

[hide="Solution"]
1. $ a(x\minus{}y) \equal{} ax \minus{} ay$, true by the distributive property
2. $ a^{x \minus{} y} \equal{} \frac{a^{x}}{a^{y}} \ne a^{x} \minus{} a^{y}$
3. $ \log x \minus{} \log y \equal{} \log \frac{x}{y} \ne \log (x \minus{} y)$ unless $ x \equal{} \frac{y^{2}}{y \minus{} 1}$
4. $ \frac{\log x}{\log y} \equal{} \log_{y} x \ne \log x \minus{} \log y$
5. $ ax \cdot ay \equal{} a^{2}xy \ne axy$ unless $ a \equal{} 0, 1$

So E.
[/hide]

$log_c \frac{x}{y} = log_c x - log_c y$ made me think that 4 was right. However, how did SnipedYou get that $\frac{log x}{log y} = log_y x$.

(please explain well since I'm not super familiar with logs.)

The statement we want is:

$ \frac{log_c(x)}{log_c(y)}= log_y(x) $

Suppose $x=y^a$.

$ \frac{log_c(y^a)}{log_c(y)}= log_y(x) $

$ \frac{a\cdot log_c(y)}{log_c(y)}= log_y(x) $

$ a= log_y(x) $

Which is the definition, if $x=y^a$, so this is true.


AMC AIME MIT college Columbia Putnam Harvard

As of Wendsday, Canada's not seeming like a bad idea.  I've decided I should apply to two Canadian colleges.  What's the best Canadian math school?  What's a Canadian school with a good math department that's easy to get into (quick stats: 1500 SAT, 33 ACT, nearly straight a's, 4 on the aime last year, math team oralist, scholastic bowl captain)

Well, Waterloo has quite a reputation for math and computing.

Yeah, I checked out Waterloo and Mcgill, but neither seemed to have very many classes.  At all.

The University of Waterloo and University of Toronto are probably head and shoulders above all the others.  Waterloo is especially well known for their Computer Science department (they created Maple there).

I believe Toronto is considered to be the preeminent Canadian university, although this may be a misconception on my part due to the fact that most Canadians I know are from Toronto.

Toronto IS the pre-eminent Canadian university (although some would (incorrectly) say that McGill is), but Waterloo is more well-known for Math and CS.

I have friends who went to Dalhousie, Universite de Montreal, and Western Ontario, and they all ended up getting Ph.D.'s from MIT; also Simon Fraser and University of British Columbia are strong.

I go to Simon Fraser, the math program here is not that strong.  I don't think that UBC's is much better, if at all - they were pretty much equivalent in my view and the views of most others I consulted.

I think that UofT and Waterloo, hands down, offer far more rigorous mathematical programs, and the only ones in Canada which are on par with the more prestigious universities in the states.  Of course, how rigorous you want to make it at, say, SFU or UBC is up to you.

Late reply here, but I'd say Waterloo is the place to be for Math in Canada.  (This is the biased opinion of an alumna.)  They even have a Math faculty, so your degree will be a BMath.   fwiw, when I went to MIT for grad school, they looked very favorably upon Waterloo, and I felt very well prepared.   Check recent Putnam scores, but Waterloo used to always do quite well.  They have a co-op program too which is very worthwhile, IME.  (I went to grad school for CS, my undergrad was joint Pure Math and CS.)  The person who said they didn't think there were many math classes at all -- I suspect you were looking in the wrong place or something.  There are *lots* of math classes of all varieties -- pure, applied, statistics, combinatorics/optimization, etc.

Waterloo's a great school.  The fact that they have the largest math faculty in the world and one of the most cutting-edge CS programs in the world is also helpful.

However, for Pure math, I feel that U of T has surpassed Waterloo, and certainly is a better institution as a whole.  (Oddly enough, I applied to Waterloo and not U of T)

[quote="mandy_pal"]Waterloo has consistently placed in the top 10 for Putnam in the previous 12 years. It came first in 2000, beating Harvard, MIT, Princeton, and other great math schools. It's most famous for its CS programs, but any math degree from Waterloo will be very favourably looked upon in Canada (I don't know how much prestige Waterloo has in the US - though judging by the heavy recruiting of Waterloo students by IBM and Microsoft, i'd say it's got a very good reputation).

Hope this helps. Cheers.[/quote]

All the Putnam statistic means is that they get a larger chunk of the Canadian IMO pie than does any other Canadian university, save for the University of Toronto (I suspect the numbers are close, although UofT probably pulled ahead by getting Jacob and Janos this year)

[quote="Magnara"][quote="ComplexZeta"]Keep in mind that Canada will probably do nothing to shelter you in case the US wants you back for some reason.[/quote]

Yeah, I gave up on that whole thing after I was underwhelmed by the Canadian math programs, so next year, I'll be going to the University of Chicago.[/quote]

Uh, there would certainly be more than enough math courses for you at both Waterloo and Toronto, no matter how gung ho you might be.  If it's worth anything, there would also be a lot more of the IMO-er type at Waterloo and Toronto than at anywhere in the States save for Princeton, Harvard, and MIT.

See http://www.adm.uwaterloo.ca/infoucal/COURSE/course-PMATH.html#PMATH354

[quote="Magnara"]Yeah, there definately are far more classes there than I remember.  Hmmmm.  I guess I just checked McGill and Waterloo.[/quote]

Err, that was Waterloo.


a,b,c>0
prove that $ (a\plus{}b)(b\plus{}c)(c\plus{}a)$≥$ \sqrt{abc(a\plus{}b\plus{}2c)(a\plus{}2b\plus{}c)(2a\plus{}b\plus{}c)}$

[quote="SUPERMAN2"]a,b,c>0
prove that $ (a \plus{} b)(b \plus{} c)(c \plus{} a)$≥$ \sqrt {abc(a \plus{} b \plus{} 2c)(a \plus{} 2b \plus{} c)(2a \plus{} b \plus{} c)}$[/quote]
Because $ (a\plus{}b)^2(a\plus{}c)(b\plus{}c)\equal{}(a^2\plus{}ab\plus{}ac\plus{}bc)(ab\plus{}b^2\plus{}bc\plus{}ac)\geq$
$ \geq\left(\sqrt{a^3b}\plus{}\sqrt{b^3a}\plus{}2c\sqrt{ab}\right)^2\equal{}ab(a\plus{}b\plus{}2c)^2.$

[quote="SUPERMAN2"]a,b,c>0
prove that $ (a \plus{} b)(b \plus{} c)(c \plus{} a)$≥$ \sqrt {abc(a \plus{} b \plus{} 2c)(a \plus{} 2b \plus{} c)(2a \plus{} b \plus{} c)}$[/quote]
Use:$ (a\plus{}b)(b\plus{}c)(c\plus{}a) \ge\ \frac{8(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)}{9}$
      $ (ab\plus{}bc\plus{}ca)^2 \ge\ 3abc(a\plus{}b\plus{}c)$
      $ (a\plus{}b\plus{}2c)(a\plus{}2b\plus{}c)(2a\plus{}b\plus{}c) \le\ \frac{64(a\plus{}b\plus{}c)^3}{27}$.
we have this inequality :)

[quote="Agr_94_Math"]$ 9 (b \plus{} c)(c \plus{} a) (a \plus{} b) \ge 8 (\sum a )(\sum ab)$ by using the identity  $ (b \plus{} c)(c \plus{} a) (a \plus{} b) \equal{} (\sum a )(\sum ab) \minus{} abc$
and now by Cauchy Schwatrz $ (ab \plus{} bc \plus{} ca) \ge \sqrt {3 abc(a \plus{} b \plus{} c)}$.

So in our $ LHS$, we get that $ LHS \ge \frac {8(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)}{9} \ge \frac {8(a \plus{} b \plus{} c)(\sqrt {3abc(a \plus{} b \plus{} c)})}{9}$.
So our inequality is equivalent to proving that
$ (a \plus{} b \plus{} 2c)(a \plus{} 2b \plus{} c)(2a \plus{} b \plus{} c) \le\ \frac {64(a \plus{} b \plus{} c)^3}{27}$ which is true  because
if $ a \plus{} b \plus{} 2c \equal{} x, b \plus{} c \plus{} 2a \equal{} y, c \plus{} a \plus{} 2b \equal{} z$ then the inequality is equivalent to proving that  $ 27 xyz \le (x \plus{} y \plus{} z)^3$. $ QED$.[/quote]
it similar with my solution   :)


Let $ f(n)\equal{}\frac{4n\plus{}\sqrt{4n^{2}\minus{}1}}{\sqrt{2n\plus{}1}\plus{}\sqrt{2n\minus{}1}}$
$ n$ belonging to set of natural numbers.Find the value of $ f(1)\plus{}f(2)\plus{}...\plus{}f(60)$

$ f(n) \equal{} \frac {4n \plus{} \sqrt {4n^{2} \minus{} 1}}{\sqrt {2n \plus{} 1} \plus{} \sqrt {2n \minus{} 1}}$

$ f(n) \equal{} \frac {(4n \plus{} \sqrt {4n^2 \minus{} 1})(\sqrt {2n \plus{} 1} \minus{} \sqrt {2n \minus{} 1})}{2n \plus{} 1 \minus{} (2n \minus{} 1)}$

$ f(n) \equal{} \frac {4n\sqrt {2n \plus{} 1} \minus{} 4n\sqrt {2n \minus{} 1} \plus{} (2n \plus{} 1)\sqrt {2n \minus{} 1} \minus{} (2n \minus{} 1)\sqrt {2n \plus{} 1}}{2}$
$ f(n) \equal{} \frac {(2n \plus{} 1)^\frac {3}{2} \plus{} (2n \minus{} 1)^\frac {3}{2}}{2}$
Summing this to $ f(60)$ everything telescopes and you are left with
$ \frac {121^\frac {3}{2} \minus{} 1}{2} \equal{} \frac {11^{3} \minus{} 1}{2} \equal{} 665$

[b]Another Solution(Got It)[/b]
Write $ 4n\equal{}(2n\plus{}1)\plus{}(2n\minus{}1)$ and $ \sqrt{4n^{2}\minus{}1}\equal{}\sqrt{(2n\plus{}1)(2n\minus{}1)}$ rest all is easy..


can u tell me what the basic thing one should think of when solving questions like these:If f{x+y}=f{x}+f{y},f{y} is differentiable at x=0,then proove that f{x} is differentiable everywhere.I know it is a fairly easy sum but still there can be questions based on this that r fairly difficult.
Can u tell me any book in which the types of questions r available in plenty.

The idea, with a somewhat significant lack of rigor here, is that if $f$ is differentiable at zero, then it certainly is also continuous at zero and thus continuous everywhere (from here it's obvious that $f$ is now linear, so its derivative is trivial to find now)


Suppose that $ ABC$ is an acute triangle such that $ \tan A$, $ \tan B$, $ \tan C$ are the three roots of the equation $ x^3 \plus{} px^2 \plus{} qx \plus{} p \equal{} 0$, where $ q\neq 1$. Show that $ p \le \minus{} 3\sqrt 3$ and $ q > 1$.

Let tanA=x, tanB=y,tanc=z, so x,y,z are positive
   Then -p =x+y+z=xyz
  $ \frac{xyz}{3} \equal{} \frac{x\plus{}y\plus{}z}{3} \ge xyz^{\frac {1}{3}}$ by AM-GM
  $ xyz \ge 3 \sqrt{3}$
$ q\equal{}xy\plus{}yz\plus{}zx \ge 3{xyz}^{2/3} \equal{} 9$
So q>1 , but I am not sure as it should not be so easy.


projective geometry
geometry unsolved
geometry

Given circles $ \omega_1$ and $ \omega_2$ intersecting at points $ X$ and $ Y$, let $ l_1$ be a line through the center of $ \omega_1$ intersecting $ \omega_2$ at points $ P$ and $ Q$ and let $ l_2$ be a line through the center of $ \omega_2$ intersecting $ \omega_1$ at points $ R$ and $ S$.
Prove that if $ P, Q, R$ and $ S$ lie on a circle then the center of this circle lies on line $ XY$.

Dear Mathlinkers,
a rapid synthetic proof is based of an application of
1. Monge's theorem i.e. the three chords are concurrent
2. the altitude of a triangle are concurrent
Sorry for the rapidity...
Sincerely
Jean-Louis

I believe you mean the Radical Axis Theorem. Monge's theorem is for tangents.


geometry
incenter
circumcircle
ratio
angle bisector
power of a point
radical axis

let G  be the Gergonne point and I be the incenter of the acute triangle ABC .
Consider  the incircle of ABC touches BC,AC,AB at X,Y ,Z respectively.
D is the point in the ABC  such that :

XDY = A/2 + B/2 + 60
YDZ = B/2 + C/2 + 60
ZDX = C/2 + A/2 + 60

prove that D, I and G are collinear.

(this is the hardest and the most beautiful geometry problem that i have seen!!)

The requirement that the triangle $\triangle ABC$ has all angles acute is not necessary, the proposition holds for any triangle.

The contact triangle $\triangle XYZ$ is the pedal triangle with respect to the incenter $I$ (see the 1st figure). The quadrilaterals $AYIZ$, $BZIX$, $CXIY$  are all cyclic, because each has 2 opposite angles right. From the circumcircles of these cyclic quadrilaterals, the following angles are equal:

$\frac{\measuredangle A}{2} = \measuredangle IAZ = \measuredangle IYZ,\ \frac{\measuredangle A}{2} = \measuredangle IAY = \measuredangle IZY$

$\frac{\measuredangle B}{2} = \measuredangle IBX = \measuredangle IZX,\ \frac{\measuredangle B}{2} = \measuredangle IBZ = \measuredangle IXZ$

$\frac{\measuredangle C}{2} = \measuredangle ICY = \measuredangle IXY,\ \frac{\measuredangle C}{2} = \measuredangle ICX  = \measuredangle IYX$

Thus the internal angles of the contact triangle are equal to

$\measuredangle ZXY = \measuredangle IXZ + \measuredangle IXY = \frac{\measuredangle B}{2} + \frac{\measuredangle C}{2}$

$\measuredangle XYZ = \measuredangle IYX + \measuredangle IYZ = \frac{\measuredangle C}{2} + \frac{\measuredangle A}{2}$

$\measuredangle YZX = \measuredangle IZX + \measuredangle IZX = \frac{\measuredangle A}{2} + \frac{\measuredangle B}{2}$

The incenter $I$ is the circumcenter of the contact triangle $\triangle XYZ$. By definition, the point $D$ is the 1st isodynamic point of the contact triangle. The Gergonne point $G_e$ is the perspector of the original triangle $\triangle ABC$ and the contact triangle $\triangle XYZ$. The cevians $XG_e, YG_e, ZG_e$ of the contact triangle pass through the intersections $A, B, C$ of the tangents to the circumcircle $(I)$ of the triangle $\triangle XYZ$ at the opposite vertex pairs $Y, Z;\ Z, X;\ X, Y$. Hence, they are symmedians of the contact triangle and their intersection $G_e$ is its symmedian point (a well known fact). Thus we have to prove the following theorem (see the 2nd figure):

[color=blue]The circumcenter $O$, the 1st isodynamic point $D_1$ and the symmedian point $K$ of a triangle $\triangle ABC$ are collinear. [/color]

The contact triangle of an incircle always has all angles acute, but the theorem holds for any triangle $\triangle ABC$ - the contact triangle of an excircle can be obtuse. First, we will show that the pedal triangle with respect to the 1st isodynamic point $D_1$ is equilateral (also a well known fact). The following angular sums are all equal to $60^o$:

$\measuredangle D_1AB + \measuredangle D_1CB = 180^o - (\measuredangle D_1AC + \measuredangle D_1CA + \measuredangle B) =$

$= 180^o - \left[180^o - (\measuredangle B + 60^o) + \measuredangle B \right] = 60^o$

and similarly,

$\measuredangle D_1BC + \measuredangle D_1AC = 60^o$

$\measuredangle D_1CA + \measuredangle D_1BA = 60^o$

As before, the quadrilaterals $AVD_1W$, $BWD_1U$, $CUD_1V$  are all cyclic, because each has 2 opposite angles right. From the circumcircles of these cyclic quadrilaterals, the following angles are equal:

$\measuredangle D_1AW = \measuredangle D_1VW,\ \measuredangle D_1AV = \measuredangle D_1WV$

$\measuredangle D_1BU = \measuredangle D_1WU,\ \measuredangle D_1BW = \measuredangle D_1UW$

$\measuredangle D_1CV = \measuredangle D_1UV,\ \measuredangle D_1CU  = \measuredangle D_1VU$

Thus the internal angles of the pedal triangle $\triangle UVW$ are all equal to $60^o$:

$\measuredangle WUV = \measuredangle D_1UW + \measuredangle D_1UV = \measuredangle D_1BW + \measuredangle D_1CV \equiv$

$\equiv \measuredangle D_1BA + \measuredangle D_1CA = 60^o$

$\measuredangle UVW = \measuredangle D_1VU + \measuredangle D_1VW = \measuredangle D_1CU + \measuredangle D_1AW \equiv$

$\equiv \measuredangle D_1CB + \measuredangle D_1AB = 60^o$

$\measuredangle VWU = \measuredangle D_1WV + \measuredangle D_1WU = \measuredangle D_1AV + \measuredangle D_1BU \equiv$

$\equiv \measuredangle D_1AC + \measuredangle D_1BC = 60^o$

Now we can use the solution of problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31831]Locus of point inside a triangle[/url]: Let $L, M, N$ be the intersections of the internal bisectors of the angles $\measuredangle A, \measuredangle B, \measuredangle C$ with the opposite sides $a = BC, b = CA, c = AB$ and $L', M', N'$ the intersections of the external bisectors of these angles with their opposite sides (cannot use exactly the same notation, because $K$ is now the symmedian point). The 3 circles $(R), (S), (T)$ with diameters $LL', MM', NN'$ intersect at the point with an equilateral pedal triangle, i.e., at the 1st isodynamic point $D_1$.

Let $K_A, K_B, K_C$ be the feet of symmedians $AK, BK, CK$. An internal angle bisector cuts the opposite side internally in the ratio of the adjacent sides, an external angle bisector cuts the opposite side externally in this ratio, and a symmedian cuts the opposite side in the square of this ratio. WLOG, assume that $a = BC > b = CA$. Calculating the power of the point $K_C$ to the circumcircle $(O)$ and the circle $(T)$:

$p(K_C, (O)) = -K_CA \cdot K_CB = -\frac{cb^2}{a^2 + b^2} \cdot \frac{ca^2}{a^2 + b^2} = -\frac{a^2b^2c^2}{(a^2 + b^2)^2}$

$p(K_C, (T)) = -K_CN \cdot K_CN' = -\left(\frac{cb}{a + b} - \frac{cb^2}{a^2 + b^2}\right) \cdot \left(\frac{cb}{a - b} + \frac{cb^2}{a^2 + b^2}\right) =$

$= -\frac{abc(a - b)}{(a + b)(a^2 +b^2)} \cdot \frac{abc(a + b)}{(a - b)(a^2 + b^2)} = -\frac{a^2b^2c^2}{(a^2 + b^2)^2}$

The power of the point $K_C$ to the circles $(O), (T)$ is the same and this point is on their radical axis. Since the circles $(O), (T)$ meet at the vertex $C$, their radical axis is the symmedian $CK_C$. Similarly, the symmedians $AK_A, BK_B$ are the radical axes of the circle pairs $(O), (R);\ (O), (S)$. The 3 radical axes $AK_A, BK_B, CK_C$ meet at the symmedian point $K$. Thus this point is either the radical center of the circles $(R), (S), (T)$, or these 3 circles belong to the same pencil and the symmedian point $K$ lies on the radical axis of this pencil. Since the circles $(R), (S), (T)$ meet at the 1st isodynamic point $D_1$, the 1st possibility is elliminated and the line $KD_1$ is the radical axis of the pencil containing the circles $(R), (S), (T)$.

As the last step, we are going to determine the angles formed by the circles $(R), (S), (T)$ with the circumcircle $(O)$. From the triangle $\triangle ANC$, we have

$\measuredangle ANC = 180^o - \left(\measuredangle A + \frac{\measuredangle C}{2}\right) = \measuredangle B + \frac{\measuredangle C}{2}$

from the isosceles triangle $\triangle CTN$ with $CT = NT$,

$\measuredangle CTN = 180^o - 2 \measuredangle ANC = 180^o - 2 \left(\measuredangle B + \frac{\measuredangle C}{2}\right) = \measuredangle A - \measuredangle B$

and from the triangle $\triangle ACT$,

$\measuredangle ACT = 180^o - (180 - \measuredangle A + \measuredangle CTN) = 180^o - (180^o - \measuredangle A + \measuredangle A - \measuredangle B) = \measuredangle B$

But this means that the line $CT$ is a tangent to the circumcircle $(O)$ at the vertex $C$. Since the tangent to the circle $(O)$ at the intersection $C$ of the circles $(O), (T)$ passes through the center of the circle $(T)$, the circles $(O), (T)$ are perpendicular. Similarly, it can be shown that the circle pairs $(O), (R);\ (O), (S)$ are also perpendicular. Consequently, the circle $(O)$ is perpendicular to 3 circles $(R), (S), (T)$ of the same pencil, it belongs to a pencil perpendicular to the given pencil and its center $O$ must be on the radical axis $KD_1$ of the given pencil. This concludes the proof of the blue theorem. Beautiful. :cool:

To [b]noshad68_ir[/b]: If your proof is different, please, could you post it ? Thanks.

Yetti.

[quote="pestich"] It used to be called Brocard line ...

   Maj.  Pestich[/quote]
You are wrong, as usual.


geometry
parallelogram
geometry proposed

Consider $ \rho$ a semicircle of diameter $ AB$. A parallel to $ AB$ cuts the semicircle at $ C, D$ such that $ AD$ separates $ B, C$. The parallel at $ AD$ through $ C$ intersects the semicircle the second time at $ E$. Let $ F$ be the intersection point of the lines $ BE$ and $ CD$. The parallel through $ F$ at $ AD$ cuts $ AB$ in $ P$. Prove that $ PC$ is tangent to $ \rho$.

[i]Author: Cosmin Pohoata[/i]

Because $FDAP$ is a parallellogram, we have that $\left|FD\right|=\left|PA\right|$.
Now construct $U \in CD$ such that $BU \parallel OD$.

We have that \[\left|PO\right|=\left|PA\right|+\left|AO\right|=\left|FD\right|+\left|DU\right|=\left|FU\right| \hspace{1cm}(1)\]
and obviously also that \[\left|BU\right|=\left|OD\right|=\left|OC\right| \hspace{1cm}(2)\]
and finally also \[\angle DUB = \angle DOB = \angle COP \hspace{1cm}(3)\]

From $(1),(2)$ and $(3)$ follows that $\Delta PCO \cong \Delta FBU$

Because $\left|BD\right| = \left|CA\right|=\left|ED\right|$, it follows that $OD \bot EB$.

Therefore, $\angle FBU = \frac{\pi}{2}= \angle PCO$.

[img]http://img469.imageshack.us/img469/1226/trivialtangentromaniazu2.jpg[/img]

Very nice solution, Jan!

Although the problem is easy, I will post my solution.

Because $ AD \| PF$, we have that $ \angle{PFC} = \angle{ADC} = \angle{BCD}$, thus the quadrilateral $ PBFC$ is an issoscel trapezoid.

Therefore $ \angle{PCB} = \angle{PFB}\iff 90 + \angle{PCA} = 90 + \angle{DBF}$

$ \iff \angle{PCA} = \angle{DBF}$.

But $ \angle{DBF} = \angle{DAE}$, because the quadrilateral $ ABDE$ is cyclic, and $ \angle{DAE} = \angle{ADC}$.

Therefore $ \angle{PCA} = \angle{ADC}$, so $ PC$ is tangent to $ \rho$.


Consider the triangle $ ABC, AD \perp BC, D \in (BC)$ and $ DE \perp AC, E \in (AC).$ Prove that $ A\equal{}90^{\circ}$ if and only if: $ BD \cdot DC \equal{} AE \cdot AC.$

[quote="moldovan"]Consider the triangle $ ABC, AD \perp BC, D \in (BC)$ and $ DE \perp AC, E \in (AC).$ Prove that $ A \equal{} 90^{\circ}$ if and only if: $ BD \cdot DC \equal{} AE \cdot AC.$[/quote]
If $ A\equal{}90$ then $ AD^2\equal{}DB\cdot DC$
But in the $ \triangle ADC$ , $ D\equal{}90$ so $ AD^2\equal{}AE \cdot AC$.
So $ AD^2\equal{} DB\cdot DC\equal{}AE \cdot AC$.

Now if
$ DB\cdot DC\equal{}AE \cdot AC$ then by $ AD^2\equal{}AE \cdot AC$ we have that $ AD^2\equal{}DB\cdot DC$.
From this we have that the $ \triangle{DBA},\triangle{DAC}$ are similars,so $ \angle B\equal{}\angle DAC$.So $ \angle B\plus{}\angle C\equal{}90$.


Our school team took a practice test on an old contest to see where we stand.  We trounced the 1st place team by almost 30 points out of 150, and would have easily won in 1992.  In any other year (save the last 3) our team would have been the best in state.  But this year, we will be competing for a spot in the top 3.  I believe this is partly due to Art of Problem Solving.  It is making math more competitive, and making it so that individuals don't need a good coach to do well.  As a result:

[quote="JSteinhardt"]Note how Haitao won the Duke Math Meet.

Okay so we got 5th as a team, I think basically that in the past TJ has basically been the best organized math team by a factor of about $\infty$, and then AoPS came along and other teams have access to a bunch of information as well, so it is getting pretty tough for us to always stay on top.

Maybe that just means we need to practice more, I guess we'll see.[/quote]

Opinions?

Here is my opinion, though I doubt it has any value.

yes. art of problem solving is making math a lot more competitive. Last year I didn't care about math, nor did I do any math in very competitive tournaments like MC or AMC 8 or anything. This year, I really care about math and adore it. Now, with this forum, I can post all I want and ask questions, and communicate on AOL with people. The books also help, and I have yet to try any classes. I might do so this upcoming year.

And, I wonder if I would be enjoying math and getting better without this forum and the help I have been getting.

Obviously I agree with my own comment.

Another interesting thing, though, is how this has not yet happened with computer science (there is no real "AoPS" for CompSci, and additionally a few select schools such as Exeter, Jefferson, and Gunn seem to be enjoying mostly monopolies in terms of USAICO qualification, and at least in our case on high school ACM competitions and TopCoder High School).

How long do you think it will be until CompSci takes off like this as well? Although obviously it's cool for TJ to win all the time, I think that nationally it is far better for us to have friendly, viable competition and so look forward to seeing this happen.

Yeah, like looking back to the 1990s, people who won contests back then would be hard-pressed to win the same contests with this new generation of mathletes.

yea i totally agree

and what was up with TJ getting 5th? Did you guys have your best  people? I totally would've expected TJ to beat Vestavia

I'm not really sure what happened. We apparently did really poorly on the power round (20/50 according to Peng Shi), although based on how we did I would've thought we would've scored higher.

[quote="JSteinhardt"]Another interesting thing, though, is how this has not yet happened with computer science (there is no real "AoPS" for CompSci, and additionally a few select schools such as Exeter, Jefferson, and Gunn seem to be enjoying mostly monopolies in terms of USAICO qualification, and at least in our case on high school ACM competitions and TopCoder High School).[/quote]

I think this has more to do with how most people take math for at least three years during high school, so many people are familiar with it.  However, many schools don't even have computer
programming classes, so there are lots of people out there without the tools to do well in USACO contests and stuff.

But there are certainly more than 3 schools that teach computer science in the entire United States, so although that might contribute, I don't think it's the only cause.

I think harry potter is right, or along the right line of thinking.  Math has been around for a loooooot longer than compsci.  Many schools don't even have a teacher that specializes in computer science.  Computer science is certainly not required at most high schools.  I don't think my school had a programming club or anything and I imagine many other clubs don't have it either.  On the other hand, it's a lot easier to get interested in math.  You take it starting in first grade.  If you find that you love it, there's AoPS for you to come to and get really good independent of your school.

I'm going under the assumption here that TJ, Exeter etc have good compsci classes offered at their classes.  At the very least, you're around people who are programming all the time and it'll catch on.  At public schools, that's very rare.  You might see someone writing programs on your calculator, but that's about it.

The only person you need to compete with is yourself.

Competition between schools is generally friendly competition, although partially to draw more good people to a school if it isn't a public school.

[quote="Koda"]I wish maths was a little less competitive ...  :([/quote]
well you are right people i know hate maths since it is so competitive but it isn't really maths it is them :lol:

[quote="chess64"]Meh, I've taught myself all the programming I know :D

For some reason I started learning HTML when I was 7...

I made a Pokemon website :ninja:

I guess it's because my dad works with computers. I don't see any other way I could have become interested...[/quote]

ugh I taught like HTML to myself since my dad was no help (even thou he was really good). I tought myself BASIC through trial and error (don't laugh it's not easy) because there wasn't a good guide I could use at the time. now i'm stuck  :D. i guess the next programming language i learned was LaTeX.

[quote="maokid7"]I agree that AoPS has made many math competitions more competitive. However, I think that having a good coach still plays a role, even though it has been diminished significantly.
about the compsci, my high school had a programming class, but both the teacher and book were horrible. student either had to learn by reading some sort of online help manual or by having an older student teach them. I think its that most teachers in high school are not familiar with what is needed to teach such a class, and the people who do tend to have such skills often end up perusing more "rewarding" (at least financially) jobs.[/quote]
I had to do the same thing in my geometry class.  The teacher was VERY smart but he couldnt communicate well.  the book didnt explain things very well either.


The number of pairs of integers $ (m,n)$ satisfying the equation

\[ m^3 \plus{} 6m^2 \plus{} 5m \equal{} 27n^3 \plus{} 9n^2 \plus{} 9n \plus{} 1\]

is

$ \text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ \text{Infinitely many}$

[hide="Solution"]
\[ m(m \plus{} 1)(m \plus{} 5) \equal{} 3(9n^3 \plus{} 3n^2 \plus{} 3n) \plus{} 1
\]Since $ m$ and $ n$ are integers, look at this equation $ \mod{3}$,
\[ m(m \plus{} 1)(m \plus{} 2) \equiv 1 \pmod{3}
\]However, amongst three consecutive integers one must be divisible by $ 3$, so $ m(m \plus{} 1)(m \plus{} 2) \equiv 0 \pmod{3}$. Thus there are $ 0\ \mathrm{(A)}$ solutions.
[/hide]

Does this solution work?

[hide="Mod 9"]
This isn't as accurate as azjps solution but:

LHS = $ m(m\plus{}1)(m\plus{}5)$ and RHS = $ 27n^3 \plus{} 9n^2\plus{}9n\plus{}1$

Note that RHS is divisible by 9 exception of 1. So, regardless of $ n$, RHS is always 1 on mod 9.

By considering $ m \equal{} 1$ through $ m \equal{} 9,$ we see that LHS cannot be 1 in mod 9 (3,6,6,0,3,3,6,0,0 --> pattern) so answer is A. I tested 1 through 9 because I thought that the pattern will repeat.. [/hide]

 :maybe:

[quote="azjps"][quote="Silverfalcon"]Does this solution work?

[hide="Mod 9"]
This isn't as accurate as azjps solution but:

LHS = $ m(m \plus{} 1)(m \plus{} 5)$ and RHS = $ 27n^3 \plus{} 9n^2 \plus{} 9n \plus{} 1$

Note that RHS is divisible by 9 exception of 1. So, regardless of $ n$, RHS is always 1 on mod 9.

By considering $ m \equal{} 1$ through $ m \equal{} 9,$ we see that LHS cannot be 1 in mod 9 (3,6,6,0,3,3,6,0,0 --> pattern) so answer is A. I tested 1 through 9 because I thought that the pattern will repeat.. [/hide]

 :maybe:[/quote]

It certainly does work  :)  (the values for the LHS listed are all divisible by $ 3$, which suggests that $ \mod{3}$ might work slightly quicker).[/quote]

Thanks, I thought it would but I wasn't quite sure.  :D


linear algebra
matrix
vector
algebra
polynomial

Can someone explain to me how to take the nth root of a matrix?
Thanks.

You have to find the eigenvalues $\lambda_j,\ j = 1, 2, ..., m$ of the given $m \times m$ matrix $A$, then the eigenvectors $\vec{x_j},\ j = 1, 2, ..., m$ of the matrix. These are the values and vectors, such that $\vec{x} \neq 0$ and

$A \cdot  \vec{x} = \lambda \vec{x}$

The eigenvalues are found by solving the equation (where $I$ is the $m \times m$ unit matrix)

$\text{det} (A - \lambda I) = 0$

which is an m-degree polynomial equation for $\lambda$ and the eigenvalues are the roots. An $m \times m$ matrix has m eigenvalues (not necessarily different from each other) and m linearly independent eigenvectors. For a matrix with real elements $a_{jk}$, the eigenvalues are all real, if the matrix is symmetric $(a_{jk} = a_{kj})$. The components $x_{j1}, x_{j2}, ..., x_{jn}$ of the j-th eigenvector $\vec{x_j}$ are found in the following way: Let $(A - \lambda_j I)_{kk}$ be the matrix $A - \lambda_j I$, take away the k-th row and the k-th column. Trying $k = 1, 2, ..., m$, we find the 1st $k$, such that $\text{det} (A - \lambda_j I)_{kk} \neq 0$. Then we can put $x_{jk} = 1$, strip the matrix $(A - \lambda_j I)$ of the k-th row and k-th column and get the remaining $m - 1$ components $x_{jl},\ l \neq k$ of the j-th eigenvector $\vec{x_j}$ by solving a set of $m - 1$ linear equations

$(A - \lambda_j I)_{kk} \cdot \vec x_j' = -d_k$

where $x_j' = (x_{j1}, x_{j2}, ..., x_{j,k-1}, x_{j,k+1}, ..., x_{jm})$ are the remaining components and $d_k = (a_{k1}, a_{k2}, ..., a_{k,k-1}, a_{k,k+1}, ..., {a_{km})}$ is the k-th column of the matrix $A$ without the element $a_{kk}$. The eigenvectors are determined up to an arbitrary multiple by a real number (if $\vec{x}$ is an eigenvector of the matrix $A$, then $k \vec{x}$ is also an eigenvector of this matrix). Once you have all the eigenvectors, you can form another matrix $M$, having the eigenvectors as columns, which diagonalizes the given matrix $A$. The matrix

$D = M^{-1} \cdot A \cdot M$

has non-zero elements (the eigenvalues $\lambda_j$) only on the main diagonal. Then you take the n-th root of the diagonalized matrix $D$ by taking n-th roots of the diagonal elements (if possible) and finally, you transform the matrix $^n\sqrt{D}$ back

$^n\sqrt{A} = M \cdot\ ^n\sqrt{D} \cdot M^{-1}$

The n-th power of this matrix is really equal to the original matrix $A$:

$\left(^n\sqrt{A}\right)^n = \left(M \cdot\ ^n\sqrt{D} \cdot M^{-1}\right)^n =$

$= M \cdot\ ^n\sqrt{D} \cdot M^{-1} \cdot M \cdot\ ^n\sqrt{D} \cdot M^{-1} \cdot ... \cdot M \cdot\ ^n\sqrt{D} \cdot M^{-1} =$

$= M \cdot \left(^n\sqrt{D} \right)^n \cdot M^{-1} = M \cdot D \cdot M^{-1} = M \cdot M^{-1} \cdot A \cdot M^{-1} \cdot M = A$

Simple, but not exactly high school math.

[quote="jcrow60"]Could you please do an actual example. What about the nth power.
I learn better with actual examples.[/quote]

I could, but finding the eigenvectors of a $2 \times 2$ matrix does not illustrate the process well and for a larger matrix, the post would be excessively long. Besides, I do not see much interest at this level in diagonalizing a matrix and thus no reason for the royal we.


limit
integration
real analysis
real analysis unsolved

Calculate:
$ \displaystyle\lim_{n\to\  \plus{} \infty}\int\limits_0^1\sqrt {n} e^{inx^2}\,dx$

Sorry, Just I've solved this problem. The answer is:

$ \frac{\sqrt{2\pi}}{4}\bigl(1\plus{}i\bigr)$.


If $A\subset [a,b]$ is a countable set, prove that there exists a strictly increasing function $f: [a,b]\longrightarrow \mathbb{R}$ s.t. the set of discontinuity points of $f$ is $A$.

We can write the points of discontinuoty of our function in a sequence {x_n}.Let c_n>0 for all n and the series c_n converges .(For example,c_i=1/2^i.)
Let f(x)=sum{c_i : x_i<x}. It is easy to see,that such function has the set of points of discontinuoty all points of {x_n} and only these points.Also this function is monotonic but not strictly.


Everyone please post the names of exams u have cleared
congrats to everyone  from everyone
This is serious stuff :mad:
Leave compliments,discussions etc. for METATHREAD
just post ur scores in following format
[hide]
[dark red]

Name(N)--
Class(C)--
Location(L)--
Exams Cleared(EC)------(comments)

[/dark red]
[/hide]

[b]Remember[/b] Keep editting it as u get more results
[b]P.S.[/b] Even if u haven't cleared (i m pretty sure that won't b the case for anyone here) u can write ur marks in boards

[dark red]

Name(N)-- Vignesh.G
Class(C)-- XII
Location(L)-- Chennai, Tamil Nadu, India
Exams Cleared(EC)------

INMO 2006
AISSE 2006 - 461/500
NTSE 2006
KVPY 2006
(I've mentioned only the national level exams. Dont bother to mention stuff like NCO.)
Top 1% in NSEP (i.e. the top 358)
I jus don't remember much of the other stuff right now. Tell me Shre, if you remember any.

[/dark red]

Why do you want this? It's quite pointless isn't it? Besides, judging is a very subjective term in itself and asking if something judges students better than other things is a silly question.

I wanted this so that those who are giving the exams can use it
Judging means which exam will be able to assess children's skills & knowledge better
My own details
[hide]
N--Rituraj
C--XI
L--Delhi
EC--NTSE 2007
      KVPY 2007
      RMO 2007
      AISSE 2007--467/500
      NSTSE 2008--AIR 1
  (i didn't write about the local exams like SOMT (1))
[/hide]

What is the purpose of this thread ritu???

from the look of it, it's for different people to blow their own trumpet  :dry:

You're right. but when we're asked very nicely we condescend to give our details :). :)

What for has this post been made by rituraj?
Absolutely, it makes no sense.

[quote="Cyber-Shreyas"][quote="SOUMYASHANT NAYAK"][quote]from the look of it, it's for different people to blow their own trumpet[/quote]
you absolutely wrong... :P
I have mentioned few of my modest achievements below.

Achievements:
1) Mr. Grasim International for past 10 years...ever since I learnt to walk(the ramp)...
2) "This guy is surely a cool dude. Imagine doing something with a finger which i manage by two hands and a shoulder."-
Atlas on me in the Mount Olympus Times commenting on my record breaking feat of lifting the sky for a month with my little finger.
3) Awarded Param Vir Chakra posthumuously... for killing my own self and thus saving the Indian army (no pun intended)[/quote][/quote]

You forgot

4)  that you were the one who put his hand into Hanuman's mouth and removed the sun, which he swallowed thinking it to be a fruit.

But then again, you are atheistic arem't you?

5) "Best Terminator" award for terminating the dodo, some 8 species of tiger and also terminating the terminator series.[/quote]

And soumya,you forgot this too..
6) You proved all the unsolved problems of the millenium in 4 hours past 12 on 1-1-2000.. But you din reveal the solutions coz you din want to spoil the thrill other mathematicians had..

7) You were instrumental in taming TR into retirement, and finally ended one of the most illustrious careers in Tamil cinema..

8) & finally, you broke Chuck Norris' hand!!! Wow! That's something great...

Not to mention screaming at Mt. Vesuvius so much that it shut up and didn't erupt. You saved Pompeii and Herculaneum!!

9) I haven't broken Chuck Norris' hand yet. But I hold the Guinness record for being left with maximum number of teeth after fighting him....

LOL! Looking at the poll question, above.

1) Boards - Obviously NOT!

2) IIT-JEE: Well some years ago ppl might have said yes. But the current pattern, while it does test your skill upto a certain level it is not a complete judgement.

3) I did vote for olympiads.... However, while MO does test your talent in maths, again it is incomplete in certain aspects. (Read this by soumya, wow, i've never seen him so serious. :D:D http://www.mathlinks.ro/viewtopic.php?p=830801#830801)

Physics Olympiad, In my opinion does a good job of it.

ha ha. that was brilliant soumya  :rotfl:
you're as eccentric as i am :D

PS oscar wilde FTW!

@hash_include -- if someone wants to blow his trumpet he can do it better by posting a separate topic

@others--its not a futile topic by any means
    one of my friends(who is in 9th class) asked me how do u study for some exam????????
    I told him to join this forum & i will post a topic where
    u can check out who cleared what and then ask them how did they do it as  they will be able to tell u much better than me

ps-he is still not up here!!!!!!!!!!!!

Where did Shridar blow his trumpet?? That was unfair of u ritu, and rude too...

[quote]Physics Olympiad, In my opinion does a good job of it.[/quote]

It actually does a worse or atleast as bad a job...
Having gone for two NIUS(2005,2007) I can tell you that real physics nowhere resembles the physics olympiad kind of problems...
You face [b]real[/b] problematic situations which you don't care about in an olympiad...
You never care where you can go wrong...

Even a simple question may take you a fortnight or so...

Like 'modelling the ionosphere',
In PhO you would normally just use hydrostatic equilibrium, adiabatic assumption, Thomson scattering by electrons,Lambertian atmosphere,solve a trivial differential equation... some other concepts and get an answer and voila you have a good score...
But actually, more often than not your theory will not match your observations...
You will have to correct it ...think creatively of weird ideas that may seem out-of-place and not [b]conventional[/b]...
(like why during sunset and sunrise there are some weird peaks in the graph...from collected data...)
Explain the collected data....make predictions and match them with more data...

Thats how it works....

[/quote]

These posts are the few posts where I write seriously  :rotfl:

But it makes me feel bad when people start yelling you are a genius or the next einstein or sth like that just because you cleared a competitive exam ...

No way can a 4-5 hr thing (competition) mean talent in real research coz the real world is not for idealists where as the olympiad world is in some sense ideal...

Study 24X7X365 and you are through... Start thinking and next moment you are nowhere... I mean think-out-of-the-box...

As such MO has some originality in the probs but rest of the olympiads have limited or nil originality....

scan through few books and you are bound to get all the questions....

But the best thing about olympiad is the exposure you get if you are selected.....

I voted for olympiads coz they judge the students in some sense better than other exams... But had you kept an option like ISEF (Intel Science and Engineering Fair) I would have voted for it...

But therein lies the problem. ISEF is virtually non existent, here in India..... Ask people if they know what FIRST means, and they would say first rank or something like that. But they probably are the best indicators of true talent..... And in a very practical manner too

Yes i too agree wid soumya, one cant determine true talent within 4-5 hrs (even though the best can easily make thro that also)  :)  But what else can people do with constraints?
[hide]
Well they can write constraint eqns  :P   :rotfl:
[/hide]

Ah constraint equations.... You really are trying too hard you know?  :rotfl:

[quote="shruthibaskaran"]Ah constraint equations.... You really are trying too hard you know?  [/quote]

Yes fer iit jee  :P   :rotfl:

[quote="rituraj007"]@hash_include -- if someone wants to blow his trumpet he can do it better by posting a separate topic
[/quote]
not when u give it to them on a platter like you just have  :wink:

and besides, i don't think there's anything wrong with boasting. as long as that 9th  class kid doesn't ask questions like how people manage to win Mr. *cough* grasim international 10 years straight :D

and to prove soumya's and shruthi's points, i haven't heard of ISEF or FIRST till now  :maybe:
i just looked it up on google and it seems to be pretty neat though..

[quote="shruthibaskaran"]As I often say, it is Google and not guys, that were God's gift to the world ;) Pretty neat is underestimating its prowess Shrids. Its pretty much phenomenal there and a matter of great honour to emerge victorious in one of the afforementioned two..... Sigh...[/quote]

Wikipedia pwns Google anyday. W00t!

Anyways, most of today's search engines now adopt the same Page Rank technique as Google so the difference between Google, Yahoo and Live is very little if you do any common search. Nevertheless the future lies in the 2.0 Search Engines like Wikia search. http://search.wikia.com/wiki/Search_Wikia

Ah well.. I gotta agree with this one :D Wiki does pwn google

Don't mind me asking, but wat do u mean wen u say wiki "pwns" google????

See, this goes way back to a game....Old mythical game. Think it was WoW.... Whenever a player got defeated the creator wanted to put up a message saying you got 'owned' but by mistake he typed it as pwned... And its become a phenomenon....

So wiki rocks over google

i dont know the reality but
i have heard thetif someone gets mad and starts editting details wiki will turn to rubble!!!!!!!!!

yes i think neone can edit Wiki but maybe they have an inherent system to try and avoid trashing  :D

hey soumya thats correct coz i am participating this time in the intel international science and engineering fair!
u seee!
i am in the team category
any of u guys or girls there!
 :)

a deoderiser to remove foul gases from the refrigerator!


for what value of k will the following system be dependant?

x+4y+3z=-1
3x-2y-z=5
2x+y+z=k

[hide="Hint"]Set it up as a matrix...[/hide]

[hide="Different Way"]Cramer's rule...[/hide]


inequalities
function
logarithms
calculus
derivative
inequalities proposed

[quote="hungkhtn"]Find the best constant $k>0$ for this inequality is true
$(\frac{a}{b+c+d})^k+(\frac{b}{c+d+a})^k+(\frac{c}{d+a+b})^k+(\frac{d}{a+b+c})^k \ge \frac{4}{3^k}$
$a,b,c,d$ are posituve reals.

The open question's find the minimum of the left expression for each value of $k$, But I can't complete it as soon.[/quote]
hungkhtn, what do you mean the best constant $k>0$ in this case?
Maybe It's true (both of the equality) for many values of $k$...

I mean that the best Real Constant $k$. For Some small valuse of $k$, this one isn't true.

well for all real $k \geq 1$ it holds by jensen as the function $f(x)=(\frac{x}{S-x})^k$ will be convex for $0 \le x \le S$
(here $S$ represents $a+b+c+d$)

[quote="hungkhtn"]Yes. It's not difficult if $k \ge 1$. But if you see the problem with 3 number, You'll see it's difficult. The best value of $k$ for 3 number is $\frac{\ln 3}{2(\ln 3-\ln 2)}$[/quote]
I think you mean half-convex, concave theorem?

No, I didn't use any theorem. Use mixing variable only.

[quote="hungkhtn"]Yes. It's not difficult if $k \ge 1$. But if you see the problem with 3 number, You'll see it's difficult. The best value of $k$ for 3 number is $\frac{\ln 3}{2(\ln 3-\ln 2)}$[/quote]
Im sorry but isnt $\frac{\ln 3}{2(\ln 3-\ln 2)}>1$
for three variables i got the value $\frac{\ln 3-\ln 2}{\ln 2}$
and for four variables i got $\frac{2\ln 2-\ln3}{\ln 3-\ln2}$

Please post the solution for it if you can.

well if one of the variables goes to 0
then $x^k+\frac{1}{x^k}$ should be greater than $\frac{3}{2^k}$ so $2\geq \frac{3}{2^k} \rightarrow k\geq \frac{\ln 3-\ln 2}{\ln 2}$ then by the half convex half concave theorem for the function $f(x)=(\frac{x}{1-x})^k$ for $k\le1$ we need only prove that
for $a \geq k_0$ that the following inequality holds :
$2(\frac{a}{1-a})^k+(\frac{1-2a}{2a})^k \geq \frac{3}{2^k}$ which can be showed by taking the derivative with respect to a
$k_0$ is the inflection point for $f(x)$.
in the same manner for four variables.

I don't know half-concave, so I only use mixing variable to solve it. But for $4$ number, this one is very difficult.

Vasc, can you show any idea?


inequalities
trigonometry
geometry unsolved
geometry

In any triangle $ ABC$ prove that :

\[ 2\left(\sin \frac{A}{2}\sin \frac{B}{2}\plus{}\sin \frac{B}{2}\sin \frac{C}{2}\plus{}\sin \frac{C}{2}\sin \frac{A}{2}\right)\le \sin \frac{A}{2}\plus{}\sin \frac{B}{2}\plus{}\sin \frac{C}{2}\]

\[ 2(\sin\frac{A}{2}\sin\frac{B}{2}+\sin\frac{B}{2}\sin\frac{C}{2}+\sin\frac{C}{2}\sin\frac{A}{2})\leq \frac {2(sin\frac {A}{2}+sin\frac {B}{2}+sin\frac {C}{2})^2}{3}\leq \sin\frac {A}{2}+sin\frac {B}{2}+sin\frac {C}{2}\]


Let p(n) be the greatest  prime divisor  of n .
Prove that  there exist  infinitely many  n  such that:

 p(n)< p(n+1)< p(n+2).


:rotfl: What is the sum of the pattern:

1/2 + 1/4 + 1/8 + 1/16.......infinity  :play_ball:

[hide]1-the last term in the sequence[/hide]

[quote="X_wing1redleader"]If you keep adding half of what is left to reach one, how will you ever reach one?...[/quote]Good question. I rarely post in Middle School but I'll answer your question. You never reach one but you will get so close to 1 that you cannot possibly 'see' how close the sum gets to 1. Can you see infinity? No, because no matter what number you can think of, I can think of a larger number than yours. What am I trying to say? First, the sum is an idea that it will get so close to 1 that we think it's 1. Because the sum goes on indefinitely, we say the sum has a limit and that limit is 1. It cannot go beyond 1 of course as you can see. It reaches a limit.

[quote="ccy"]Think about it like this:

You want to find $S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...$.

Divide that by 2 to get $\frac{S}{2}=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...$.
Subtract that from $S$ to get $\frac{S}{2}=\frac{1}{2}$.
Double that to get $S=1$.

Fun. :)[/quote]

I know that 1/4+1/8+1/16...=1/2, but prove that. I'm not sure why.

Well, technically it will never reach $1$.
It is always going to be off by a really really really small fraction.
The same thing goes for the $\frac{1}{2}$
 :? I think :?

It will be off by a very very small small fraction.
But after all, who looks at the 100th or 1000000th digit after the decimal.  Just call it one.

[quote="soxrox"]It will be Infinity. If you add infinity, It will be infinity.[/quote]

We've already stated that it will never reach 1. Therefore, it will never go up to infinity.

It will reach $1$, by the definition of infinity. If you don't think it will, you need to learn more about infinity.

infinity is a concept, not a number
therefore, there is no answer

*Sigh* That's why someone thought up of a formula, which kyyuanmathcount has given to us.

[quote="kyyuanmathcount"]This is an example of an infinite geometric series.  Because the ratio, r, in the formula $\frac{a(1-r^n)}{1-r}$ for geometric series is less than 1 and n, the number of terms, goes to infinity, $r^n$ is considered zero.

So, you have the formula for infinite geometric series, which is $\frac{a}{1-r}$.[/quote]

What do the $a$ and $r$ stand for?

[quote="sugarwergoingdown"]even if it is infinity, it will never reach one. if it can reach one, then it can go beyond one too. so, the answer can never be definite. and also, it will be always less than 9.9 repeating, which equals one! :) you can think of it that way[/quote]

Just because it can reach one doesn't mean it can go beyond one. Have you ever even taken a basic algebra class? :huh:

There formula comes out to be 1, and they go by the formua in MATHCOUNTS so drop the topic.


Celsius and Fahrenheit temperatures are related by the formula $ C\equal{}\frac59(F\minus{}32)$. If the temperature rises by 5 degrees Celsius, by how many degrees does the equivalent Fahrenheit reading rise?

Simply plug in (because we can cheat and assume that we are going from $ 0^\circ C$ to $ 5^\circ C$, and just find the equivalent $ ^\circ F$ temperature): $ 5^\circ \equal{} \frac59(F \minus{} 32)\implies F \equal{} \boxed{41^\circ}$.

It helps to know the inverse of this: $ F \equal{} \frac95C \plus{} 32$.

Also, memorizing some common temperatures might help (i.e. body, room, freezing/melting, boiling) because they might pop up in questions...(you never know :P ).

Edit: Yeah, sorry.

However, it wants the difference, so it goes from 32-41 so 9 degrees.

Alternatively, using math154's second formula, it is simply 9/5x5=9.

Actually, it's much simpler to just think about the change as the slope. The slope is 5/9, so with a 5-degree Celsius change, there's a $ \fbox{9}$ degree Fahrenheit change.

I also believe ^^ this comment is the best method. Takes less than 3 seconds.


inequalities
function
inequalities proposed

Let $ a_1,\ldots,a_n$ be positive reals so that $ a_1\plus{}a_2\plus{}\ldots\plus{}a_n\le\frac n2$. Find the minimal value of
$ \sqrt{a_1^2\plus{}\frac1{a_2^2}}\plus{}\sqrt{a_2^2\plus{}\frac1{a_3^2}}\plus{}\ldots\plus{}\sqrt{a_n^2\plus{}\frac1{a_1^2}}$.

$ \frac {4\sqrt {2}}{n}$ ??? if it's true i write my sollution  :wink:

Nice little problem :)
We shall prove that the minimum value of the expression is $ n\cdot\frac{\sqrt{17}}{2}$, with equality iff all the numbers are equal.

By the inequality $ x_{1}^{2}+x_{2}^{2}+\cdots+x_{m}^{2}\geq\frac{(x_{1}+x_{2}+\cdots+x_{m})^{2}}{m}$, applied for $ m=17$, $ x_{1}=a_{1}$, and $ x_{2}=x_{3}=\dots=x_{17}=\frac{1}{4a_{2}}$, we obtain $ \displaystyle a_{1}^{2}+16\cdot\frac{1}{16a_{2}^{2}}\geq\frac{(a_{1}+\frac{4}{a_{2}})^{2}}{17},$ hence we have obtained

$ \sqrt{a_{1}^{2}+\frac{1}{a_{2}^{2}}}\geq\frac{1}{\sqrt{17}}\cdot\left(a_{1}+\frac{4}{a_{2}}\right)$. Similarily we have:

$ \sqrt{a_{i}^{2}+\frac{1}{a_{i+1}^{2}}}\geq \frac{1}{\sqrt{17}}\cdot\left(a_{i}+\frac{4}{a_{i+1}}\right)$

and thus by summing all these relations we obtain $ LHS\geq\frac{1}{\sqrt{17}}\left(a_{1}+a_{2}+\cdots+a_{n}+\frac{4}{a_{1}}+\frac{4}{a_{2}}+\cdots+\frac{4}{a_{n}}\right)$    $ (\star)$.

But $ a_{i}+\frac{1}{4a_{1}}\geq 1$, for all $ i$, and since $ \frac{15}{4}\cdot\left(\frac{1}{a_1}+\cdots+\frac{1}{a_n}\right)\geq\frac{15n^{2}}{4(a_{1}+\cdots+a_{n})}\geq\frac{15n}{2},$by summing we obtain

$ a_{1}+a_{2}+\cdots+a_{n}+\frac{4}{a_{1}}+\frac{4}{a_{2}}+\cdots+\frac{4}{a_{n}}\geq n\cdot\frac{17}{2}$, therefore, from the relation $ (\star)$, we have $ LHS\geq n\cdot\frac{\sqrt{17}}{2}$, which is indeed what we wanted to prove.

This is indeed the minimum, as the equality occurs for all the numbers equal to $ \frac{1}{2}$.

[quote="freemind"]Let $ a_1,\ldots,a_n$ be positive reals so that $ a_1 \plus{} a_2 \plus{} \ldots \plus{} a_n\le\frac n2$. Find the minimal value of
$ A\equal{}\sqrt {a_1^2 \plus{} \frac1{a_2^2}} \plus{} \sqrt {a_2^2 \plus{} \frac1{a_3^2}} \plus{} \ldots \plus{} \sqrt {a_n^2 \plus{} \frac1{a_1^2}}$.[/quote]
Using Minkowski and Cauchy-Schwarz inequalities we get
$ A \ge \sqrt{(a_1\plus{}a_2\plus{}\ldots\plus{}a_n)^2\plus{}\left(\frac{1}{a_1}\plus{}\frac{1}{a_2}\ldots\plus{} \frac{1}{n}\right)^2} \\ \ge \sqrt{(a_1\plus{}a_2\plus{}\ldots\plus{}a_n)^2\plus{}\frac{n^4}{(a_1\plus{}a_2\plus{}\ldots\plus{}a_n)^2}}$
By AM-GM inequality:
$ (a_1\plus{}a_2\plus{}\ldots\plus{}a_n)^2\plus{}\frac{\left(\frac{n}{2}\right)^4}{(a_1\plus{}a_2\plus{}\ldots\plus{}a_n)^2}\ge  \frac{n^2}{2}$
Because $ a_1 \plus{} a_2 \plus{} \ldots \plus{} a_n\le\frac n2$ so  $ \frac{\frac{15n^4}{16}}{(a_1\plus{}a_2\plus{}\ldots\plus{}a_n)^2} \ge \frac{15n^2}{4}$
We obtain
$ A \ge \sqrt{\frac{n^2}{2}\plus{}\frac{15n^2}{4}}\equal{}\frac{\sqrt{17}n}{2}$

My solution is similar to to others but it has no trick  :wink:
Using Minkowski and Cauchy-Schwarz inequalities we get
$ A \ge \sqrt {(a_1 \plus{} a_2 \plus{} \ldots \plus{} a_n)^2 \plus{} \left(\frac {1}{a_1} \plus{} \frac {1}{a_2}\ldots \plus{} \frac {1}{n}\right)^2} \\
\ge \sqrt {(a_1 \plus{} a_2 \plus{} \ldots \plus{} a_n)^2 \plus{} \frac {n^4}{(a_1 \plus{} a_2 \plus{} \ldots \plus{} a_n)^2}}$

Let $ a_1\plus{}...\plus{}a_n\equal{}s$ . Consider the function $ f(s)\equal{}s^2\plus{}\frac{n^4}{s^2}$

This function is decreasing for $ s\in \left(0,\frac{n}{2}\right]$

so it attains its minimum at $ s\equal{}\frac{n}{2}$ and we are done ... :)

[quote="freemind"]Let $ a_1,\ldots,a_n$ be positive reals so that $ a_1 \plus{} a_2 \plus{} \ldots \plus{} a_n\le\frac n2$. Find the minimal value of
$ A \equal{} \sqrt {a_1^2 \plus{} \frac1{a_2^2}} \plus{} \sqrt {a_2^2 \plus{} \frac1{a_3^2}} \plus{} \ldots \plus{} \sqrt {a_n^2 \plus{} \frac1{a_1^2}}$.[/quote]

AM-GM: $ a_1^2 \plus{} \frac1{16a_2^2} \plus{} ... \plus{} \frac1{16a_2^2}\ge 17\sqrt [17]{\frac {a_1^2}{(16a_2^2)^{16}}}$

continues ~>done! :)

I think it is a very weak problem for a TST.
Because it is very easy to prove that:
$ \dfrac{1}{a_1}\plus{}...\plus{}\dfrac{1}{a_n}\geq2n$ and
$ \sqrt{a_1^2\plus{}\dfrac{1}{a_2^2}}\geq\dfrac{(a_1\plus{}\dfrac{4}{a_2})}{\sqrt{17}}$

Let $ a_1,a_2,\ldots,a_n$ be positive reals such that $ a_1\plus{}a_2\plus{}\ldots\plus{}a_n\le\frac n2$, and $ b_1,b_2,\ldots,b_n$ be a permutation of $ a_1,a_2,\ldots,a_n$. Prove that\[ \sqrt{a_1^2\plus{}\frac1{b_1^2}}\plus{}\sqrt{a_2^2\plus{}\frac1{b_2^2}}\plus{}\ldots\plus{}\sqrt{a_n^2\plus{}\frac1{b_n^2}}\geq \frac{\sqrt{17}n}{2}.\]


AMC
AIME
calculus
trigonometry
number theory
relatively prime
analytic geometry

There is a way. Working on getting the solution up as we speak.

We can using simple algebra make the equation:

(x-15)^2 + (y-20)^2 = 7^2

Max y: 20+7 = 27
Max x: 15+7 = 22

My answer is 49. I have no idea how you got your solution unless I am reading incorrectly.

You have to maximize $\frac{y}{x}$, you maximized y then x (separately). You have to find the best "trade off" between x being small and y being large :)

Is that method really that bad? That's what I always use on these types of problems.

[quote="JavaMan"]From Mildorf's mock AIME series:
2. If $x^{2}+y^{2}-30x-40y+24^{2}= 0$, then the largest possible value of $\frac{y}{x}$ can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime, positive integers. Determine $m+n$.

I got the solution $\frac{117}{44}$ by drawing the circle on a cartesian plane then finding the upper tangent $y = kx$ on the circle. Quite tedious without a calculator. Is there a simpler solution?[/quote]

You can plug in $y=kx$ into the equation.

Then you can use the discrimant and set it to 0, so that you can solve for k, and the larger value is the answer.

[quote="Iversonfan2007"][quote="JavaMan"]

I got the solution $\frac{117}{44}$ by drawing the circle on a cartesian plane then finding the upper tangent $y = kx$ on the circle. Quite tedious without a calculator. Is there a simpler solution?[/quote]

That is the simplest solution, in my opinion  :wink:

And I don't understand how it's tedious when the $\tan$ values work out nicely.[/quote]

I didn't use tan, could you show me what you did?


Let a,b,c are three length sides of a triangle such that $a^{2}+b^{2}+c^{2}=3$.Prove that:
$\frac{3}{(a+b+c)^{2}-6}+abc\geq 2\sqrt{\frac{a+b+c}{3}}$


A point $ P$ is chosen at random inside a square $ QRST$. What is the probability that angle $ RPQ$ is acute?

[hide]
Imagine a circle with diameter $ QR$. If $ P$ were on the circle, then angle $ RPQ$ would be a right angle. Note that any points inside the intersection of the circle and square would cause angle $ RPQ$ to be obtuse, so if we let $ QR\equal{}x$, then our probability is equal to:
$ \frac{x^2\minus{}\pi(.5)^2}{x^2}$

$ \frac{x^2(1\minus{}.25\pi)}{x^2}$

$ 1\minus{}.25\pi$
It's nice how the probability does not depend on the size of the square.
[/hide]

Thanks.

The area of the semi-circle is $ \frac{1}{2}\pi (\frac{x}{2})^{2} \equal{} \frac{\pi x^{2}}{8}$,

so the required probability is $ \frac{x^{2}\minus{} \frac{1}{8} \pi x^{2}}{x^{2}} \equal{} 1 \minus{} \frac{\pi}{8}$.


let $ z_{1},z_{2}....z_{n}$ be complexe numbers.prove that ther exixt a subset $ J$$ \subset${1,2...n} such that:
$ \left|\sum_{j\in J}z_{j}\right|\geq\frac {1}{4\sqrt{2}}\left|\sum_{1\leq i\leq n}z_{i}\right|$

I think the inequality is much to weak.
Suppose it doesn't happen for both $ J_{1}\equal{}\{1\}$ and $ J_{2}\equal{}\{2,3,\hdots ,n\}$. This means $ |z_{1}| <\frac{1}{4\sqrt{2}}|z_{1}\plus{}z_{2}\plus{}\hdots\plus{}z_{n}|$ and $ |z_{2}\plus{}z_{3}\plus{}\hdots\plus{}z_{n}| <\frac{1}{4\sqrt{2}}|z_{1}\plus{}z_{2}\plus{}\hdots\plus{}z_{n}|$.
But $ |z_{1}\plus{}z_{2}\plus{}\hdots\plus{}z_{n}|\leq |z_{1}|\plus{}|z_{2}\plus{}z_{3}\plus{}\hdots\plus{}z_{n}| <\frac{1}{2\sqrt{2}}|z_{1}\plus{}z_{2}\plus{}\hdots\plus{}z_{n}|$, which is absurd.
So, for at least for one of the sets $ J_{1}$ and $ J_{2}$ the inequality holds. As I said, a much better constant would be $ \frac{1}{2}$.

Actually $ \frac{n\minus{}1}{n}$ is the sharpest bound in the inequality for generic z's

1) The case $ z_{i}\equal{}1$ shows that the sharpest bound cannot be greater than $ \frac{n\minus{}1}{n}$

2) Assume $ \exists z_{1},\ldots,z_{n}$ such that for all proper subsets $ J\subset\{1,\ldots,n\}:\left|\sum_{j\in J}z_{j}\right|<\frac{n\minus{}1}{n}|S|$ with $ S\equal{}\sum_{i\equal{}1}^{n}z_{i}$, then we have the contradiction

$ (n\minus{}1)|S|\equal{}\left|\sum_{i\equal{}1}^{n}(S\minus{}z_{i})\right|\leq\sum_{i\equal{}1}^{n}\left|S\minus{}z_{i}\right|<n\frac{n\minus{}1}{n}|S|\equal{}(n\minus{}1)|S|$


Does anyone know how to say "for better days" in latin?
I want to put it at the end of my play I wrote for english, make it look impressive.

It's a tragedy and in the end, a toast to better days is made.

Maybe "pro diebus melioribus".  But I am not a Latin expert.

That's ok, my language arts teacher probably isn't either. Thanks, now my crappy play has a good ending.
Oh and how many syllables is that?

It's nine syllables, "pro dee-ay-bus mel-ee-or-i-bus" at least according to some pronounciations.


[quote="sri340"]the volume of a rectangle prism is represented by the expression 6x^3-7x^2-x-2 . the length is
2x-1 find expressions to represent the width and height of the prism

a)   x and 3x-2    b)  3x-2 and x-1      c) x and 3x+2       d) 3x+2 and x+1[/quote]
Since the volume is $ 6x^3\minus{}7x^2\minus{}x\minus{}2$ and the length is $ 2x\minus{}1$, we can use either long division or synthetic division to find the quotient when dividing these two polynomials. Whichever you choose....wait $ 2x\minus{}1$ doesn't evenly divide $ 6x^3\minus{}7x^2\minus{}x\minus{}2$...is there a typo or am I doing something wrong?  :S

hello, i also think there is a typo because the term $ 6x^3\minus{}7x^2\minus{}x\minus{}2$ can't be factorized.
Sonnhard.


One fair die has faces $ 1$, $ 1$, $ 2$, $ 2$, $ 3$, $ 3$ and another has faces $ 4$, $ 4$, $ 5$, $ 5$, $ 6$, $ 6$. The dice are rolled and the numbers on the top faces are added. What is the probability that the sum will be odd?

$ \textbf{(A)}\ \frac{1}{3}\qquad
\textbf{(B)}\ \frac{4}{9}\qquad
\textbf{(C)}\ \frac{1}{2}\qquad
\textbf{(D)}\ \frac{5}{9}\qquad
\textbf{(E)}\ \frac{2}{3}$

[hide="Solution"]
We can make odd number by two ways:
1) Odd # from 1st die, even # from 2nd die
2) Even # from 1st die, odd # from 2nd die

1st way: $ \frac{4}{6}*\frac{4}{6}\equal{}\frac{4}{9}$
2nd way: $ \frac{2}{6}*\frac{2}{6}\equal{}\frac{1}{9}$

Therefore, the total probability is $ \frac{5}{9}$.

Answer: D[/hide]