complex analysis
algebra
polynomial
geometry
geometric transformation
reflection
function

Let $ p_n(z)$ be a holomorphic polynomial of degree $ n$.  Assume that
$ p_n(z)$ maps the line $ x \equal{}y \to x \equal{}y$.  If $ p_n(1) \equal{} 0$ show $ p_n(i) \equal{} 0$

This is the reflection principle. $ p(i\overline{z})\equal{}i\overline{p(z)}$ everywhere, since that identity is true on the diagonal.
This is true for any entire function satisfying the condition. For polynomials, we don't even need the analysis.


induction
number theory proposed
number theory

Let $ k \in Z^+ $ and $ (a_n) : a_1=k+1 ; a_{n+1}=a_n^2-ka_n+k $ , for all $ n \geq 1 $
  Prove that : $ \forall m $ is different from $ n $ , we have : $ gcd(a_m,a_n)=1 $

We can verify by induction that $a_{n+1}=k+\prod_{i=1}^na_i$, and that $a_n\equiv1\mod{k}$ for all $n$, and the rest is history...


Well, this one is from Mindsport section of Times Of India (one of India's leading national dailies)

There are n unique balls numbered from 1 to n in an urn. Some number of balls are taken to play a game and placed back after the game is finished. Now some number of balls are withdrawn from the urn. What is the probability that we still can play the same game taking the earlier selected balls from the urn? (Assume at least one ball is required to play a game and minimum number of balls that can be withdrawn is 0.)

I definitely don't understand this problem...  It seems like you need to know something more about the game that is being played...  Maybe I'm missing something obvious. :?

[quote="K81o7"]I think that he's saying...
What is the chance that you drew, in your set of balls just now drawn, each and every ball which was originally drawn?
Let's say that you originally chose x balls.
I'm assuming that the chance to choose a given number of balls is the the same for any number( so the chance to choose 0 is the same as the chance to choose 1, 2, 3, 4 or whatever number.)  In order to play to same game, you have to
1. Have at least chosen the same number of balls
and
2. Have at least chosen each of the original balls

The probability for the first statement to be true is (n-x+1)/(n+1).  The probability for the second to be true varies based on the number which you choose.  For instance, if you choose, x balls, the chance is x!(x-n)!/n!.  The chance if you chose x+1 is (n-x)[(x+1)!][(n-x-1)!]/n!.  So you'd come out with a large sum with a denominator of n!(n+1)!.  The numerator comes out to
([1/nCx]*n!)+(1/nC[x+1])([n-x]C1)+(1/nC[x+2])([n-x]C2)+...+n!
Not exactly sure how to simplifiy that, but i'll come back to it when i'm not so lazy.[/quote]

K81o7's explanation to the problem is correct...well there's a cool solution to this problem with an elegant answer :-)


a,b,c>0, a+b+c=1.Prove
1.$7(a^2+b^2+c^2)+17(a^2b^2+b^2c^2+c^2a^2)+27a^2b^2c^2\ge3$
2.$a^2+b^2+c^2+2\sqrt{3abc}\leq 1$
These two ineq isn't hard.I solved they but my solution isn't nice.If someone has a nice solution, please post here

[quote="Nameless"]a,b,c>0, a+b+c=1.Prove
2.$a^2+b^2+c^2+2\sqrt{3abc}\leq 1$
[/quote]
we just need to prove:
$(ab+bc+ca)^2\geq 3abc(a+b+c)$
use Am-gm to prove it.

[quote="Nameless"]The first ineq is very easy, I have known your solution for a long time.But the second one is harder and I need a nice solution[/quote]
did you mistake 1 for 2?

[quote="zhaobin"]
did you mistake 1 for 2?[/quote]
Yes, I'm sorry

[quote="nttu"]For the first one , just homogenize it :
 $7(a^2+b^2+c^2)(a+b+c)^4+17(a{}^b{}^2+b^2+c{}^2a{}^2)(a+b+c)^2+27a{}^2b{}^2c{}^2 -3(a+b+c)^6 \geq 0$
 I used maple to check it . It's Ok .[/quote]
Thank you,nttu.But if we can prove
$x,y,z>0: 2\sum x^4 +7\sum y{}^2z{}^2\ge 3\sum(y^3z+z^3y)+3\sum x^2yz$
I'll have a nicer solution

[quote="silouan"]If the sums are symmetrics then it is obviously true by Muirhead[/quote]
  No , it's not obvious . Muirhead dosen't work here.

see the generalised Muirhead at the Inequalities Open questions ;)

[quote="spanferkel"]Well, the "generalised Muirhead" is not yet elaborated enough to become a theorem :(. But here we have directly $\sum(x-y)^4\geq0\ \Leftrightarrow 2\sum x^4 +6\sum y^2z^2\geq4\sum(y^3z+z^3y)$.
Add this to $\sum y^2z^2+\sum(y^3z+z^3y)\geq3\sum x^2yz$ (Muirhead) and we obtain
$2\sum x^4 +7\sum y^2z^2\ge 3\sum(y^3z+z^3y)+3\sum x^2yz$ as required.

So Nameless, come on with your beautiful solution! :)[/quote]
Thank you, spanferkel.
With very very useful help from my friend (I don't know how he found it, but he is very intelligent) we have:
$7(a^2+b^2+c^2)+17(a^2b^2+b^2c^2+c^2a^2)+27a^2b^2c^2\ge3 \leftrightarrow\sum\frac{1}{1+a^2}\leq \frac{27}{10} \leftrightarrow\sum\frac{(a+b+c)^2}{a^2+(a+b+c)^2}\leq \frac{27}{10} \leftrightarrow\sum\frac{a^2}{a^2+(a+b+c)^2}\geq \frac{3}{10}$
And by Cauchy we have:$\sum\frac{a^2}{a^2+(a+b+c)^2}\ge\frac{(a^2+b^2+c^2)^2}{a^4+b^4+c^4+(a^2+b^2+c^2)(a+b+c)^2}$We must prove that$\frac{(a^2+b^2+c^2)^2}{a^4+b^4+c^4+(a^2+b^2+c^2)(a+b+c)^2}\ge\frac{3}{10} \leftrightarrow 10(a^2+b^2+c^2)^2\ge 3(a^4+b^4+c^4+(a^2+b^2+c^2)(a+b+c)^2) \leftrightarrow 2\sum a^4 +7\sum b^2c^2\ge 3\sum(b^3c+c^3b)+3\sum a^2bc$
and the problem is solved :)

Another way is

$7(a^2+b^2+c^2)+17(a^2b^2+b^2c^2+c^2a^2)+27a^2b^2c^2\ge3 \leftrightarrow\sum\frac{1}{1+a^2}\leq \frac{27}{10}$

But this is well-known and comes from the ineq $\frac{1}{1+a^2}\leq\frac{27}{50}(2-a)$.

Writing the similars for $b,c$ and adding them we have the desired


algebra function domain calculus Gauss polynomial

Can someone please provide information on how range can easily be determined of a relation. The domain is usually the easy one.
(please provide examples)

It really depends on the specific case.  There is no general idea except to find the min and max and look for gaps.

Range is just the y-values that a relation has.  I can't tell you a way that it can easily be determined for all cases.  There would be one that involve calculus.  There are ways to determine it for other situations.  It's easy to find if you are looking at a graph or a table.  If it s a table with x and y it is just the y values.  A graph it's still just the y values(that is the values it has vertically).

Oh... I don't know calculus but I just know that you might need calculus to find stuff like that.  Maybe JBL, gauss, or someone else who knows calculus could help you out.

Also I m not really sure that there is a calculus method that could find the range for every kind of function.  But this is way out of my league so I'll leave this for someone else to address.

Think about the graph.  For example, any polynomial function with odd degree has range all real numbers.  However, w/rational functions it gets tricky.  Think to yourself "what could the values of y (ie f(x) ) never be?"  Here are some practice problems.

Find range:
(1) f(x)=x^8-3x^4+x^2
(2) f(x)= :sqrt: x+4) +2
(3) f(x)=e^(x+1)
(4) f(x)=1/(x^2+4)
(5) x^2 + (y-3)^2=9

I think that is the chapter I am learning now in calc so I'll be able to tell you in a day or two if nobody answers sooner.

Sorry, that is correct, stupid mistake.  I am way too tired to be answering questions.

[quote="joml88"]Does y=x :^2:  have range all real numbers?   All polynomials have domain over all real numbers but I believe that only odd ones have range over all the reals.[/quote]

No, the range of [tex]y=x^2[/tex] is [tex][0,\infty)[/tex]. Do you see why?

Sigh... I can provide a basic overview of the calculus "method" if you want to call it that, that will work in most cases...

take the limit as y approaches both oo and -oo, to test for the extreme values, check for any relative extrema, and then compare that to the lmits you did first. Check to make sure the curve is continuous... obviously, the domain will affect the range. There is no hard cut "method", these are just steps that will help you.

The full-fledged way to tell what the range of a function is is that it's the set of all points c such that there exists a bin the domain of f(x) with f(x) = c. There aren't any "this-will-always-work-for-any-function" ways to solve this; but there are some calculus methods for differentiable functions, and continuity arguments for continuous fuctions, etc.

But if you can explicitly find the inverse of a function, the domain of the inverse is just the range of the function.

For example: f(x) = x2 on x :ge: 0 has an inverse of f-1(x) = :sqrt:x. The domain of :sqrt:(x) is [0, :inf:) - so that's the range of f(x).

Now there are some one-to-one-ness issues to worry about when finding the inverse of a function; but for ones that have explicit inverses that are easy to find, this is a quick and dirty way to see what the range is.


geometry
circumcircle
geometric transformation
reflection
geometry proposed

Let $ H$ the orthocenter of a triangle $ ABC$ with acute angles $ B,C$   which is inscribed in a circle with center $ K$ and  radius $ R \equal{} 1$. If lines $ HK$ and $ BC$ meet at point $ S$ and it holds $ SK (SK\minus{}SH)\equal{}1$,  determine  the area of  $ ABHCA$.

For a figere see the attachment in the first post :

   http://www.mathlinks.ro/viewtopic.php?t=190370

 where also a solution by a greek contestant is to be seen.

$ R \equal{} 1$ is the triangle circumradius, the orthocenter $ H$ lies on the reflection $ (K_a)$ of the circumcircle $ (K)$ in $ BC.$ $ \overline{SK} \cdot (\overline{SK} \minus{} \overline{SH}) \equal{} R^2$ $ \Longrightarrow$ $ \overline{SK} \cdot \overline{HK} \equal{} \overline{KS} \cdot \overline {KH} \equal{} R^2$ $ \Longrightarrow$ $ S$ is image of $ H$ in inversion with center $ K$ and power $ R^2$ $ \Longrightarrow$ $ H$ lies on the circumcircle of the $ \triangle KBC$ $ \Longrightarrow$ either $ \angle A \equal{} 60^\circ,$ or $ \angle A \equal{} 120^\circ,$ or $ H \equiv S \equiv B,$ or $ H \equiv S \equiv C.$ The last two cases are not acceptable, because then either $ \angle B \equal{} 90^\circ,$ or $ \angle C \equal{} 90^\circ$ are not acute.
$ |S(ABHCA)| \equal{} |S(\triangle ABC) \minus{} S(\triangle HBC)| \equal{} \frac {_1}{^2}\ |BC| \cdot |AH| \equal{} \frac {_1}{^2} \cdot \sqrt 3 \cdot R^2 \equal{} \frac {\sqrt {3}}{2}.$


group theory
abstract algebra
superior algebra
superior algebra solved

Let $p$ be a prime,and let $H$ be a subgroup of $S_{p}$ that contains a transposition and a p-cycle.Show that $H=S_{p}$

Hi,

I once got stuck on this too, and I found this solution myself.  I will gladly give it, please tell me if you understand.

First some things we need to know :

1) in $S_{n}$ the transpositions generate the entire group

2) if a subgroup of $S_{n}$ contains the transpositions (ab) and (ac) it also contains (ab)(a c) (a b) =(b c)

3) if you put all elements 1,2,..., n in a graph, and i give you some transpositions, and you connect all those elements together in a transposition, your generated group will contain a transposition ( a b) if and only if a and b are in the same connected set

4)in particular! if the graph is connected, your generated group is the entire group  :D

4) in $S_{p}$ with p prime, all powers, except the trivial one, of a p cycle are also p cycles

Okay suppose in $S_{p}$ our transposition is simply $( 1 2 )$
now there will be a (nontrivial) power of your p cycle sending 1 to 2
this power will (as said above) be a p cycle itself, let us call it c

c will look like $c= (1 2 a b \cdots p )$

now consider $c^{-1}(1 2) c =(2 a )$

now consider $c^{-1}(2 a ) c =(a b)$

etc...

you will you get a bunch of transpositions in this way, in a loop
well your graph you make with these transpositions is a cycle
it is connected thus
thus the entire group $S_{p}$ is generated

[quote="fredbel6"]

c will look like $c= (1 2 a b \cdots p )$

now consider $c^{-1}(1 2) c =(2 a )$

now consider $c^{-1}(2 a ) c =(a b)$

[/quote]
Thanks, :) Seems the same appoach to the another problem
But I haven't notice that :(
Now,It is clear now,Thanks again.(I think it also should read from right to left :lol: )


The usual proof of Borel-Cantelli lemma proceeds by bounding $ \Pr( \bigcup_{k > n} E_k )$.

But how about this proof?

If $ \sum \Pr(E_n) < \infty$, then the expectation of the $ [0,\infty]$-valued random variable $ \sum 1_{E_n}$ is finite because $ E(\sum 1_{E_n}) \equal{} \sum E( 1_{E_n}) \equal{} \sum \Pr(E_n) < \infty$ and so $ \sum 1_{E_n}$ takes finite value almost everywhere. ■

I've never seen a book on probability that include the above proof, which must mean either that the above proof is wrong and I am not seeing some error or just that I've never seen such a book.

Yes it's a valid proof, but it requires a simple version of monotone convergence theorem. The usual proof does not require the definition of E.


Let's find the function $f(k)$ such that $k\cdot k!=f(k+1)-f(k).$ Then telescoping method will wait for you.

[quote="Altheman"][hide]$k*k!=(k+1)!-k!$[/hide][/quote]
This relationship always makes me smile  :)

Can anyone prove to me why this relationship is true(without using induction)??

$(k+1)! - k! = (k+1) k! - k! = k \times k!$

(I realize this is just a reverse of Altheman's logic, but the logic seems more natural this way...?)

Your explanation is standard, Altheman.

Another example.

$\sum_{k=0}^n \frac{k-1}{k!}$

You are right, but this problem can be used  [b]telescoping method[/b] as well.

[quote="maokid7"]for what kunny posted, would you just split it into 2 sigmas and relate them each to $e$?[/quote]

Of course not.  We're working with finite sums here.

In this case,

$\frac{k-1}{k!}=\frac{1}{(k-1)!}-\frac{1}{k!}$

Oh, sorry for that it was finite sum. :blush:


geometry
trigonometry
trig identities
Law of Sines

[color=darkred]Let be given a triangle $ABC$ and a point $O$ inside it. Let $M,\,N,\,P$ be the projections of $O$ on $BC,\,CA,\,AB,$ respectively. Prove that: $OA\cdot BC=OB\cdot CA=OC\cdot AB$ if and only if $\triangle MNP$ is an equilateral triangle.[/color]

My dear friend April,

thank you for this problem! My solution is very short:

Clearly, $OPAN$ is cyclic, so we can apply the extended law of sines in $\Delta PAN$ to obtain that:

\[\left|OA\right|=\frac{\left|NP\right|}{\sin{A}}\]
Repeat this for $\left|OB\right|$ and $\left|OC\right|$ and proceed like this:

By the law of sines in $\Delta ABC$:

\[{{\left|NP\right|}.\frac{\left|BC\right|}{\sin{A}}=\left|PM\right|.\frac{\left|AC\right|}{\sin{B}}=\left|NM\right|.\frac{\left|AB\right|}{\sin{C}}\iff \left|NP\right|}=\left|PM\right|=\left|NM\right|\]

Which is what the problem asked.


geometry
geometric transformation
rotation
geometry unsolved

$ \bigtriangleup ABC$. $ \angle B \equal{} 60^{o}$. O is the incircle center. $ F\in AC$ such that $ \frac{FA}{FC} \equal{}  \frac{1}{2}$. $ OE\parallel BC(E \in AB)$. Prove that $ \angle AEF \equal{} \frac{1}{2}\angle BAC$

I'll try to post my son's solution, he's too busy for this, sorry if anything coming wrong!
$ AI$ will intersect the arc $ BC$ of the circle $ (ABC)$ at $ N$, $ CI$ the arc $ AB$ at $ M$, $ MN$ will intersect $ AB$ at $ E$ and $ BC$ at $ D$.
Next take $ P$ as symmetrical of $ C$ about $ D$ and $ Q$ the symmetrical of $ A$ about $ E$.
$ M$ and $ N$ being the midpoints of the arcs $ AB$ and $ BC$ respectively, it's easy to see that a $ 60^\circ$ rotation around $ I$ will map $ E$ to $ D$ and $ A$ to $ M$, so $ DM \equal{} AE$, similarly $ NE \equal{} CD$ (actually, the triangles $ MAI$ and $ CNI$ are equilateral).
It's also easy to see that $ MEIA$ and $ NCID$ are cyclic and, as previously mentioned, since the two triangles are equilateral, we shall easily get $ AE \equal{} ME \plus{} EI$ and $ CD \equal{} ND \plus{} DI$. We have also $ BD\equal{}DI\equal{}IE\equal{}BE\equal{}DE$. From all these we get $ ND\equal{}BP$ ( 1 ) and $ ME\equal{}BQ$ ( 2 ).
From the two cyclic quads we shall get that $ \triangle DNI \sim  \triangle EIM$, and from their similarity: $ \frac {DI}{ND} \equal{} \frac {ME}{EI}$ ( 3 ).
From (1), (2) and (3) we get $ \frac {BP}{BD} \equal{} \frac {BE}{BQ}$, i.e. $ PE \parallel DQ$. If $ F' \equal{} PE \cap AC$ and $ G \equal{} QD \cap AC$, from $ E$ - midpoint of $ AQ$ and $ D$ - midpoint of $ CP$ we get $ AF' \equal{} GF' \equal{} CG$, hence $ F' \equiv F$.
Last, seeing, for instance, that $ \triangle DNI \equiv \triangle BPE$ (s.a.s.), we get the required angle equality as well.

[b][color=blue]Remark (own):[/color] [/b] [i] [color=red]Even easier using the above proof, one can show the following: $ \frac {1}{CD} \plus{} \frac {1}{AE} \equal{} \frac {1}{DE}$[/color] [/i]

Best regards,
sunken rock

[u]See[/u] [b]PP17[/b] from [url=http://www.artofproblemsolving.com/blog/56429][color=red][b]here[/b][/color][/url].


1. How many prime factors does 403827 have?

2. How many divisors of 23426 are not prime? (I just need to find the number of prime factors then I can solve)?

1) I will give you a hint: $ 134609$ is a prime number.

2) $ 23426 \equal{} 2\times13\times17\times53$. You can figure out the rest.

OMG LOL....I hope nobody ever sees this thread because they'll probably think I fail...I don't use my brain a lot.

Check out the math jam transcript from May 27.  It has some useful number theory problems similar to yours.


The expression typed below is also located on page 150 of aops vol 2.
V is a vector with elements v1 and v2, and W is a vector with elements w1 and w2. i is the commonly known as i vector (1 0) and j is the commonly known as j vector (0 1). A is a transformation matrix

(v1Ai + v2Aj) x (w1Ai + w2Aj).

Somehow this becomes,
(v1w2 - w1v2)(Ai x Aj) + v1w1(Ai x Ai) + v2w2(Aj x Aj).

I dont see how the former expression can become to second expression, but rather, when I computed out the former expression, I got

A(v1w2 - v2w1)k, where k is the vector (0 0 1). This expression is mentioned a paragraph down on the page as being the more simplified version. For the purpose of knowledge and understanding, can someone please show me how the first expression I typed is equal to the second one?

[quote="jelyman"]
(v1Ai + v2Aj) x (w1Ai + w2Aj).

Somehow this becomes,
(v1w2 - w1v2)(Ai x Aj) + v1w1(Ai x Ai) + v2w2(Aj x Aj).

A(v1w2 - v2w1)k, where k is the vector (0 0 1). [/quote]

(v1Ai + v2Aj) x (w1Ai + w2Aj)=v1Ai x w1Ai + v1Ai x w2Aj + v2Aj x w1Ai + v2Aj x w2Aj. For the third term, we remember that b x a= -a x b, so now we collect like terms to get the second equation. Now, we can evaluate the second equation to get what you got later on by using the determinant rule. just remember that for any vector a, a x a = 0, and that i x j = k.


trigonometry
vector
floor function
absolute value
complex numbers
inequalities open
inequalities

Find the minimum value of $f(x)=|\cos{x}|+|\cos{2x}|+|\cos{3x}|+\cdots+|\cos{nx}|$.

Hmm.. I wonder if we can tackle it using complex numbers.. :)

Looking at it graphically in the complex plane, we have a complex vector $z = e^{ix}$ and we are rotating it and summing the absolute value of the real part of each $z^n$ - that is, we want $\sum_{k=1}^{n} | \Re(z^k) |$.

This is almost definitely minimized by $x = \frac{\pi}{2}$, which gives us (I think) a minimum of $f(x) = \lfloor \frac{n}{2} \rfloor$ because choosing any other value of $x$ will give us too many values of $| \cos kx |$ that aren't equal to zero.


algebra
polynomial
linear algebra
matrix
quadratics

what is the minimal polynomial and characteristic polynomial of an $ n\, by\, n$ matrix all of whose entries are equal to 1.

Note: problems like this should be posted in the linear algebra forum, not the algebra forum. It fits over here, so I moved it.

This matrix has a conventional name: $ J$

I essentially asked this question on the exam I gave in a sophomore-level course last week. (I did specify $ n\equal{}5$ but that's not particularly important.) Here is the exact wording of my question, and there's some redundancy built in - you can get through it without answering everything:

2. Let $ \mathbf{J}$ be the $ 5\times 5$ matrix for which all $ 25$ of its entries are $ 1.$

(a) Compute $ \mathbf{J}^2.$ Can you find a quadratic equation that $ \mathbf{J}$ satisfies?

(b) Compute $ \text{trace}\,(\mathbf{J}).$

(c) Find $ \text{rank}\,(\mathbf{J}).$ What does that tell you about an eigenvalue, and the multiplicity of that eigenvalue?

(d) Find, with multiplicities, all eigenvalues of $ \mathbf{J}.$ Hint: nearly everything before this in problem 1 or problem 2 is applicable.

(Problem 1 included showing, in a specific case, that if $ A$ satisfies a certain polynomial, then the eigenvalues of $ A$ must be roots of that polynomial.)

OK, there's one more unstated assumption there: we're in a field of characteristic zero. Some of the details about multiplicities are different in one case, although the minimal and characteristic polynomials still work out the same.

Note also that if $ J_n$ denotes such a matrix, and if $ x^T \equal{} [1 \,\, 1 \,\, ... \,\, 1]_{1 \times n}$, then $ J_n \equal{} xx^T$. From this you can say immediatly something about the rank of $ J_n$, and it also shows that $ J_n$ is positive semidefinite (why?).

[quote="jmerry"]OK, there's one more unstated assumption there: we're in a field of characteristic zero. [/quote]
When I said it was for an exam in a sophomore-level course, that does carry with it the presumption that the field is $ \mathbb{R}.$ I wouldn't bring finite characteristic into that course - they're confused enough already.

[quote="Carcul"]Note also that if $ J_n$ denotes such a matrix, and if $ x^T \equal{} [1 \,\, 1 \,\, ... \,\, 1]_{1 \times n}$, then $ J_n \equal{} xx^T$[/quote]
This has a nice followup. For this particular $ x,$ $ Jx\equal{}(xx^T)x\equal{}x(x^Tx)\equal{}nx.$ So $ J$ has the eigenvalue $ n,$ with this eigenvector.

You can do something like that for any rank-1 square matrix.


geometry
3D geometry
combinatorics proposed
combinatorics

Is it possible to construct a $2005\times2005\times2005$ cube, consisting of only $2\times2\times2,\ 3\times3\times3,\ and\ 7\times7\times7$ cubes?

I guess the "Hungarian mathematics journal" you are talking about is the famous Komal magazine??

Obviously :) there is only one.
But I have a hard time here typing the 2 points on the o... and I guess most people know it :)


[hide="Solution"]Notice that the product $x_1x_2=(x_1)\frac{2}{x_1}$, the product $x_3x_4=x_3\frac{4}{x_3}$, and etc, such that we're only left with the product of the even numbers. The answer is thus $2\cdot4\cdot6\cdot8=\boxed{384}$. [/hide]

we dont even need $x_1=97$ .... haha .... :P

Looking for clever manipulations, we see that $x_2 = \frac{2}{x_1}$ such that $x_2*x_1 = 2$. Like wise, $x_{2n}*x_{2n-1} = 2n$. We split the terms of the product into pairs like so: $(x_1x_2)*(x_3x_4)*(x_5x_6)*(x_7x_8)$ which equals $2*4*6*8 = \boxed{384}$.

This just expands to:
$$\implies 97\cdot\frac{2}{97}\cdot\frac{3\cdot97}{2}\cdots\frac{8\cdot6\cdot4\cdot2}{7\cdot5\cdot3\cdot97} = 8\cdot6\cdot4\cdot2 = \boxed{384}$$

A bit easy for an AIME don't you guys think? Even though its 40 years old

Wasn't this like the first or second question of the AIME? (also don't double post)

[quote=4everwise]Let $x_1 = 97$, and for $n > 1$ let $x_n = \frac{n}{x_{n - 1}}$.  Calculate the product $x_1 x_2 \dotsm x_8$.[/quote]

so its 97 * 2/97 + 3 * 97/2 * 4 * ....
and then 97 stuff all cancel out
so its just 2*4*6*8=384

[hide=solution]

Note that $x_{2n} \cdot x_{2n-1}=2n$, pairing up we get that, this is just $2 \cdot 4 \cdot 6 \cdot 8=\boxed{384}.$

[/hide]


limit
logarithms
calculus
calculus computations

Find $lim_{x\to 0_+}\frac{2^{2^x}}{x^x}$. Also the limit to infinity of the same.

Is there some easy way to do this that I'm missing?

Well, in 0 it's trivially 2? It is obvious that $2^{2^x} \to 2$ since $2^{2^x}$ is continous in 0. And we have that $x^x = e^{x\ln x} \to e^0 = 1$ by standard limit of $x\ln x$ and $e^x$ being a continous function.

We can rewrite $\frac{2^{2^x}}{x^x} = e^{2^x\ln 2 - x\ln x} = e^{2^x(\ln 2 - \frac{x\ln x}{2^x})}$ which obviously gets arbitrarily large when x gets large.


whats a great way to distinguishing between them.

for example, if i get a question that says i draw cards 5 times, would that be permutation?  I thought it should be but my teacher said since its cards its combination.  but I dont really understand because it says draw 5 times (meaning seperate)

If it was draw all together, wouldnt that be combination?

Or am I totally wrong

Combinations are the number of ways you can select members out of a set (order doesn't matter). Using your example, choosing 5 cards from a deck of 52 cards is combinations, because you are calculating how many such sets of 5 cards you can get.

Permutations are the number of ways you can arrange a certain set (order matters). Again using your example, the number of ways those 5 cards can be arranged (ABCDE, ABCED, ABDCE, etc. ) is permutations.

More info:
[url=http://mathworld.wolfram.com/Permutation.html]Permutations[/url]
[url=http://mathworld.wolfram.com/Combination.html]Combinations[/url]

Here's another example that will help.
The number of ways to choose two letters from {a,b,c} where the order doesn't matter (combination)  is 3:
{a,b},{a.c},{b,c}.
The number of ways to choose two letters from {a,b,c} where the order [i]does[/i] matter (permutation) is 6:
{a,b},{b,a},{a,c},{c,a},{b,c},{c,b}.

In your example of drawing cards the order doesn't matter.  You will still have the same group of cards regardless of what order they were chosen in.  You can think of a combination as the number of ways to choose a group and a permutation as the number of ways to choose AND arrange a group.


I have read and heard a lot about law schools... Is it true that Harvard law school is the best one? and why?

According to who?  US News?  If you go to Harvard Law, then it is probably the best one. If you go to NYU Law, then it is probably the best one.  It depends on what you want in a law school.  Perhaps you should speak to students/graduates at different schools and then decide which is best for you.

[quote="JRav"]If you go to Harvard Law, then it is probably the best one. If you go to NYU Law, then it is probably the best one.[/quote]

honestly, this makes no sense at all.  do you have the slightest clue as to the  merits of various top law schools or are you posting just for the sake of posting?  on a related note, have you ever considered not posting when you have nothing of value to say? (hint: this doesn't just apply to the college forum!)

What I am saying is that there is no official ranking as to which law school is "best".  Does it make a difference if one goes to Harvard, Yale, NYU, Stanford, Chicago, Duke, Berkeley,... in the long run?  Probably not.  Yes, there is a distinction between Harvard and College XYZ but at the top level, there is little to no distinction.  Read between the lines next time.

Harvard, Yale, and Stanford are universally acknowledged as the top three law schools with Yale a clear #1.  Employers know this too and graduates from these schools have a clear advantage over other 'top' schools for the best jobs (of which there are relatively few in law as far as I know, so it is particularly meaningful to talk about which schools are best).

I don't know what you mean about 'in the long run'; one's career earnings tend to increase at a rate faster than linear, so getting as many advantages as one can early in one's career is usually particularly desirable.

e: moreover, whether one wants to quibble over whether x university is better than y university is all fine and dandy, but it doesn't really matter if JRav thinks that all sufficiently high ranked law schools are essentially the same if there are employers who are strongly biased towards top 3 schools.

as well, the aforementioned top 3 schools end up getting a disproportionate amount of the top students for the same reason.  it's like comparing a school like Duke with Harvard, they are both great schools with great undergraduate curricula, but Harvard ends up getting a lot more of the best students, and graduating from such an institution indicates that not only have you received a top-notch education, but a (better) indicator of top-notch ability.

[quote="blahblahblah"]e: moreover, whether one wants to quibble over whether x university is better than y university is all fine and dandy, but it doesn't really matter if JRav thinks that all sufficiently high ranked law schools are essentially the same if there are employers who are strongly biased towards top 3 schools.[/quote]

By your definition then, have the vast majority of lawyers have been turned down a job because they could not afford/did not go to Harvard or Yale? I doubt that an aspiring lawyer will be turned town by firm after firm simply because he did not go to Yale.

[quote="Scenario"]
Wow, your resume looks impressive, and you're definately qualified for the job...but I'm sorry. Your degree from NYU Law isn't up to our standards.
[/quote]

College is where you learn and get a degree. With this degree you can get a job. Now, by going to an Ivy League might help you get a job faster/easier, but it doesn't mean that if you don't go to an Ivy League, then you're going to fail at life.

nice strawman argument, '7h3.D3m0n.117'!  clearly the best way to read my post is to get out of it that 'aspiring lawyers will be turned [sic] town by firm after firm simply because [they] did not go to Yale' and that 'if [one] [doesn't] go to an Ivy League, then [they are] going to fail at life'.

of course most graduates of top 15 law schools are going to be successful.  they just are not generally going to be as successful as the subset of top 3 law school graduates.  this is (partially) because that subset is more qualified and generally more capable (also probably better trained, but i digress).  this is also why that subset is positively selected for by law firms, and if you have a very small number of positions open and hundreds of graduating students each year, you'll see that (especially if your position is a particularly desirable one) it's not really worth looking at schools ranking below a certain threshold.  the opportunity costs of different law schools don't vary as much as between different universities (i.e., it's expensive for everyone and in basically equal amounts), so its safer for firms to assume that students tend to go to their best option and hence safer to make assumptions about students [i]based on the quality of their institution[/i].

[quote="blahblahblah"]...and if you have a very small number of positions open and hundreds of graduating students each year, you'll see that (especially if your position is a particularly desirable one) it's not really worth looking at schools ranking below a certain threshold.[/quote]

You make a very good point, blahblahblah. However, you might also agree that colleges aren't just what some firms (in this case) look for. An experienced lawyer who has worked for 10 years [from a "lower" tiered school] might the same possibility of being hired as a new graduate from say, Harvard. Either one could be hired, so would you say the college of choice mattered?

For people seeking a first career job out of law school, the law school attended matters a great deal, and even the undergraduate college attended before law school can matter. Law is a particularly good occupation to which to apply the "go to the best school you can" principle, because so much of early career performance is hard to evaluate, and thus proxies for performance potential such as where one went to school matter a great deal.


Show that if $x,y,z >0$ then we have:

$ \sum \frac{x}{(y+z)(y+z-x)} \geq \frac{9}{2(x+y+z)} $.

cheers! :D :D

By Cauchy we have

 \sum x(y+z)(y+z-x) \sum x/(y+z)(y+z-x) \geq ( \sum x)^2

so our inequality is equivalent to

2( \sum x)^3 \geq 9 \sum x(x+y)(y+z-x)

but RHS is simply 6xyz and we have to prove

2 \sum x^3 +6 \sum (x+y)xy  \sum 42xyz

obvious by AM-GM.


summer program
PROMYS
Ross Mathematics Program
ARML

Don't know if it's the right place to put this :? , but who here is going to PROMYS this summer?

I'm not sure im going but im applying (Even if i get in i dont know ill go)

I might apply, though I haven't even really done much in terms of working on the application  :?

[quote="t0rajir0u"]I'll go if I get in.  I should probably finish up my application sometime soon...  :blush:[/quote]

you should definitely go.... if you go, then ill decide on PROMYS!!!

[quote="t0rajir0u"]I'll go if I get in.  I should probably finish up my application sometime soon...  :blush:[/quote]

They put a lot of emphasis on the pset, so good luck with that! Also, since PROMYS admits on a rolling basis, it's probably to your benefit to finish it ASAP. But from what I've seen of your work, you shouldn't have too much trouble...

Good luck! PROMYS is a lot of fun, and the NT is absolutely fantastic!

Is it better to apply early?
I'm applying, and probably going if i get in.  Mostly done with the problem set...

If it's a rolling deadline...then, yes, I would say to turn them in as soon as you are comfortable that you won't be making your application look much better.

[quote="K81o7"]how many problems, roughly, do we need to solve to get in?[/quote]

i don't know, i'm guessing around 8, which isn't too bad. I think they're most interested in your approaches and your work rather than the answer though...and aren't the problems just for the scholarship?

I think most people can get in, but only a few get the scholarship. idk though...

[quote="Adunakhor"][quote="paladin8"][quote="K81o7"]how many problems, roughly, do we need to solve to get in?[/quote]

I would suggest solving all of them, or at least all but one. PROMYS is decently competitive and they want to know you've tried on all of them at least (so show anything you've done/tried).

I went to PROMYS in 2004; it's a great program, you'll have fun there for sure.[/quote]

Do you mean full solutions to all of them? Or simply attempt all of them and record what progress you made on those you didn't fully solve? Because I only have 7 full solutions (as far as I can tell, which isn't saying much :blush:), the rest I only answered perhaps half of the question (if it asked for different things).[/quote]

Back when I applied, I think there were only 8 problems. I'm pretty sure had full solutions to 7 and half of the last problem. I haven't looked at the problem set this year yet, but they reuse quite a few problems so I would recommend solving a lot. 7 might be enough if you have a decent amount of work for the ones you didn't completely solve. Of course I can't say for sure because things might have changed so I wouldn't bet on anything I've said... :P

A little story on why I think the problems are important... when I applied the first time I had solutions to I think 6/8 and the other two I didn't think I had enough work to make it worthwhile to write something up. I got rejected that time. I emailed them back seeing if they happened to have open spots later on and they told me to finish the problems. I did one more and half of the last and eventually got in and went. So problems = important. It'd be a lot easier to get accepted the first time around than working your way up from rejection.

I will probably go if I get accepted. I solved all the problems except for like 2 right away, they're not that hard... :)

[quote="pieterminate"]Is PROMYS number theory oriented? i.e. like ross[/quote]

Yes. Most first-year students do purely NT there.

[quote="pieterminate"]Hmmm, supposing one already knows NT well, do they still have to go to first year classes?[/quote]

i think how well you know nt will show on your solutions :P , i think classes will be chosen from that

[quote="pieterminate"]Oof, ouch.  Those problems(are they NT?) stomped me. :oops:[/quote]

most of em are, cept for number 10, 1, and 4 i would think

[quote="pieterminate"]Naw, I'll say PROMYS is too hard for me then.  :oops:  :roll:[/quote]

PROMYS is about persistence. In the last few years, I think there have been two students in all who have not needed to struggle with the program. Everyone else worked their hinies off.

One thing you'll learn quickly is that math is about attitude. And programs like PROMYS make a difference because they empower you to change certain attitudes about yourself.

[quote="paladin8"][/quote][quote="Adunakhor"][quote="paladin8"][quote="K81o7"]how many problems, roughly, do we need to solve to get in?[/quote]

I would suggest solving all of them, or at least all but one. PROMYS is decently competitive and they want to know you've tried on all of them at least (so show anything you've done/tried).

I went to PROMYS in 2004; it's a great program, you'll have fun there for sure.[/quote]

Do you mean full solutions to all of them? Or simply attempt all of them and record what progress you made on those you didn't fully solve? Because I only have 7 full solutions (as far as I can tell, which isn't saying much :blush:), the rest I only answered perhaps half of the question (if it asked for different things).[/quote][quote="paladin8"]

Back when I applied, I think there were only 8 problems. I'm pretty sure had full solutions to 7 and half of the last problem. I haven't looked at the problem set this year yet, but they reuse quite a few problems so I would recommend solving a lot. 7 might be enough if you have a decent amount of work for the ones you didn't completely solve. Of course I can't say for sure because things might have changed so I wouldn't bet on anything I've said... :P

A little story on why I think the problems are important... when I applied the first time I had solutions to I think 6/8 and the other two I didn't think I had enough work to make it worthwhile to write something up. I got rejected that time. I emailed them back seeing if they happened to have open spots later on and they told me to finish the problems. I did one more and half of the last and eventually got in and went. So problems = important. It'd be a lot easier to get accepted the first time around than working your way up from rejection.[/quote]

Thanks for the advice, I guess I'll keep working on the problems a bit longer, see if I can somehow answer a bit more  :roll:

Looks like Promys doesn't let you miss the first two weeks of it...darn...might not be able to go then :(  (not to mention I wasted time on those 10 problems)

Well, it's not wasted time if you learned some math by doing the problems.

What is your schedule conflict with the first two weeks?

[quote="tokenadult"]
What is your schedule conflict with the first two weeks?[/quote]

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=83676

I went to PROMYS last year, and it was awesome!  Don't worry too much about getting "full solutions".  I had never written a proof before PROMYS, so needless to say, my solutions couldn't have possibly been full -- I just sent like ten pages of semi-random work that I thought explained the solutions to most of the problems.  Anyway, if you get in (and if you're on this forum, that shouldn't be a problem) you should definitely go!  Some advice I'd give: think about and prepare for the talent show ahead of time -- don't be like me and shuffle cards for someone else's card trick, heh.

Sorry if I made it sound like getting in was impossible or something; it may be difficult, but I would definitely suggest trying. Even if you can't get a lot of the problems, submit an application. If you get in, you won't regret it at all. PROMYS is really an amazing program.

I agree, it is difficult...though lots of people joked about a 100% acceptance rate, I highly doubt it.  Counselors say that they take ~80/~200 kids, so it's pretty competitive.   That's not to say you have to be a super-genius to get in.  Just do as many problems as you can, look stuff up, and actually cite your sources.  At least, that's what I did last year :)

Ahhh, I miss PROMYS now.  Everything was great, except there aren't too many social activities planned, so YOU will have to organize trips/activities.  Boston/Cambridge is easily accessible thanks to the T...or you can just walk everywhere.

On their site it says they usually have 60 kids at the camp....so I think that was intended to be 80 out of 200 instead of 80% of 200.

Sorry, I meant 80 students out of 200 applicants.  They had around 80-90 students last year, well over the 60 they said they accepted, so the acceptance rate is under 50% (that is, unless the counselor who told me this wasn't telling the truth :roll: ).

How much financial aid do you usually get?

I attended PROMYS last summer, and it was such an amazing experience! You get to learn so much about NT and possibly other topics (like combinatorics), and you're actually in a math-friendly environment (who gets more of that during the school year than through math club meetings and ARML team practives?).

Most of the first-year students will do entirely NT, but the advanced courses are open to first year students, provided they can do the NT problems, and I was one of many students to be in a advanced course. I suggest this: explore an advanced course or two in the beginning, when the NT problems are relatively easy. Then, if the workload gets to be too much, you can drop. Otherwise, you can stay in an advanced course (or theoretically more, but I couldn't even do this).

And finally... YOU MUST THINK OF A TALENT SHOW ACT, AND ONE THAT IS ACTUALLY GOOD!!! OR ELSE... :evilgrin:

Do you have to do a talent show thing? :D

Is 9 problems+conjecture and random stuff (and partial answers/solutions) for a 10th problem enough if I send it by monday?

[quote="scorpius119"]I attended PROMYS last summer, and it was such an amazing experience! [/quote]

Really? Corina and I tried to identify AoPSers, and I don't think we ever IDed you -- a hint, perhaps?

[quote="paladin8"][quote="scorpius119"]
And finally... YOU MUST THINK OF A TALENT SHOW ACT, AND ONE THAT IS ACTUALLY GOOD!!! OR ELSE... :evilgrin:[/quote]

Do they still do the Win a Date with Fergie thing (if he went)? Haha. I won that :P.[/quote]

No, no. Fergie was a JC last year, so he decided that Xiao would be the prize instead. Besides, he cleaned up some, so 1st-years aren't threatened by him anymore.

pst...So YOU'RE the one in the legend?  :rotfl:


What is the cube of the square of the second smallest prime number?

Second smallest prime=3
3^2=9
9^3=729

ANSWER:[b]729[/b]


[quote="symmetry"]The curve y = x/(1 + x^2) is called a serpentine.  Find an equation of the tangent line to this curve at the point (3, 0.3).[/quote]

y' = (1-x^2)/(1+x^2)^2
y' (3) = -0.08

t:y=mx+b
m = y'

P(3,0.3)
0.3 = -0.08.(3)+b <=> b=0.54

t:y=-0.08x+0.54


induction
combinatorics proposed
combinatorics

Let $m$ and $n$ be positive integers and $S$ be a subset with $(2^m-1)n+1$ elements of the set $\{1,2,3,\ldots, 2^mn\}$. Prove that $S$ contains $m+1$ distinct numbers $a_0,a_1,\ldots, a_m$ such that $a_{k-1} \mid a_{k}$ for all $k=1,2,\ldots, m$.

WOW, very nice problem:

By Minsky theorem, if there is no chain of lenght $m+1$ in the division poset, then it is the union of $m$ antichains.

Now let $C_1,C_2...C_m$ be this antichains, For every elemento of a chain $a$, map it to $[2^{m-1}n+1,2^mn]$ in the following way: choose the integer $t$ such that $2^ta$ belongs to that interval.

Now, numbers in an antichain go to different numbers in this map, and there can be overlaps between numbers of different antichains. Suppose the max power of 2 dividing $k$ in $[2^{m-1}n+1,2^mn]$ is $t$, then we can get to $k$ from $t+1$ numbers, or at most $m$, because, there cannot be two numbers in the same antichain with same image.

So we have at most: $2^{m-2}n+2^{m-3}2n+2^{m-4}3n...2^{m-k}(k-1)n...2(m-2)n+2(m-1)n+mn=(2^m-1)n$ possible images counting repetitions, so if we have our Poset partitioned in $m$ antichains, then it has at most $(2^{m}-1)n$ elements, contradiction.

Minsky, or Mirsky theorem is a kind of dilworth dual theorem for Posets.

this is a nice generalization of the erdos problem which is the case $m=1$.

Great Problem

i think it can be done using induction after $m$
$m=1$ well known
now
let
$U=\left\{1,2,\ldots,2^{m}n\right\}$ and $V=\left\{2^{m}n+1,\ldots,2^{m+1}n\right\}$.
and $S\subseteq(U\cup V)$ our subset. we make a partition of $S$ :
$S_{u}\subseteq U$
$S_{v}\subseteq V$.
$S_{u}\cap S_{v}=\emptyset$.
Now, $|V|=2^{m}n$, so let $|S_{v}|=2^{m}n-k$. (because $|S_{v}|\leq|V|$).
Then $|S_{u}|=|S|-|S_{v}|=(2^{m+1}-1)n+1-2^{m}n+k=(2^{m}-1)n+1+k$. (*)
I will call "end of p-chain" a number which is the end of a chain of type asked by the problem, of length $p$. For example, in the chain $1|2, 2|4, 4|8$, $8$ is a "end of a 4-chain".
Using the induction hypothesis, and (*) above, it follows that $S_{u}$ has $k+1$ "ends of (m+1)-chains". This is because, we find the first "end", let it be $e_{1}$, then we look at the set $S_{u}-e_{1}$, etc..
Let the $k+1$ "ends of (m+1)-chains" be $e_{1},e_{2},\ldots,e_{k+1}$.
We take the set $Z=\left\{e_{1},e_{2},\ldots,e_{k+1}\right\}\cup S_{v}$. Also let $t=2^{m}n$. The set $Z$ has exactly $t+1$ elements, and all elements are no bigger than $2t$. Then, it is the well known problem (case $m=1$), so there must be two which divide each other in $Z$. Let this numbers be $a$ and $b$, and $a|b$. If $a\in S_{v}$, then $2a>2t$, and because $a|b$, we get $b\geq 2a>2t$, contradiction. So $a$ must be a "end of (m+1)-chain". we take this chain and complete it with $b$, and there is the $m+2$ chain. this finishes the induction

By Mirsky theorem, we have to prove that the minimal number of anti chains which has union as  $S$ is $\ge m+1$

 We will show that even $m$ anti chains can not cover $S$. There are total $2^{m-1} n$ odd numbers in the set $\left \{ 1, \ldots, 2^m \cdot n \right \} $.Write each number as $2^k \cdot b$, where $b$ is odd. All anti chain will contain distinct odd numbers. (otherwise one will divide the other). We can fill all the anti chains with all odd numbers. So there are total $2^{m-1} n \cdot m$ numbers .(not distinct). Now we remove the numbers that are not distinct.  An odd number from the set $\left \{ 2^{m-k}n + 1, 2^{m-k+1}n - 1 \right \}$ can contribute in at most $ k $ distinct antichains (here $1 \le k \le m-1$ ) and there are $2^{m-k-1}n$ odd numbers in that interval.So there are $(m-k) 2^{m-k-1}n$ entries which are same. (has already appeared). We remove them. So number of distinct entries is $2^{m-1}mn - \sum_{k=1}^{m-1}n (m-k)2^{m-k-1} = (2^m-1)n $. Which is a contradiction since $|S| = (2^m-1)n + 1$. QED

Use induction on M with this Statement:
A: $\{1,2,3,\ldots, 2^mn\}$
If we delet K elements from A (k<n)
then we have N-K subset of A :(B1,B2.....Bn-k)
such that each of them has M+1 elements like a1,a2,....,am+1 such that $a_{k-1} \mid a_{k}$
and if F(Bi)=greatest member of Bi
then all of F(B1),F(B2),....,F(Bn-k) are different.


geometry
trigonometry
arithmetic sequence
trig identities
Law of Sines
area of a triangle
Heron\u0027s formula

A side of triangle $ ABC$ measures $ a$ . The values of the sides form an arithmetic progression and the values of the angles form another. Calculate the value of the area of the triangle.

[hide="what I found"] I used law of sines twice and found that the triangle is necessarily equilateral. Is this correct??[/hide]

I don't think so.

[hide=" click to reveal hidden content"]

I called the sides $ a - d, a, a + d$ and the angles $ \alpha - \delta, \alpha, \alpha + \delta$ and used Heron's formula and $ \frac {1}{2}ab \sin C$ for the area. Equating the two expressions for the area, I got  $ \ \frac {\sqrt {3}a^{2}(\sqrt {6} - 1)}{8}$.

The sum of the angles is $ 180$, so $ \alpha = 60$ degrees.

Heron's formula gives ${ \frac {a}{2}\sqrt {{3}(\frac {a^{2}}{4} - d^{2})}}$ and $ \frac {1}{2}ab \sin C$ gives $ \frac {1}{2}(a^{2} - d^{2})\frac {\sqrt {3}}{2}$.

[/hide]

I have rushed though this and should check the calculations, but I hope it is helpful.

I've checked it again: $ (a^{2} - d^{2})$ should be $ \frac {(a^{2} - d^{2})}{2}$ and I now get $ d^{2} = - 2a^{2}$, so either something is horribly wrong or no such triangle exists. I'll look at it again (but I usually find that when I make a mistake, someone is quick to correct it, so hopefully you'll get the correct answer soon).

Perhaps there is no such triangle! Unless, of course, $ d = 0$ and the triangle is equilateral, as you think.

I'm posting again as I think I can make it clearer.

Using the cosine rule, we get $ a^{2} \equal{} (a \minus{} d)^{2} \plus{} (a \plus{} d)^{2} \minus{} 2(a^{2} \minus{} d^{2})\cos 60$, so $ 2d^{2} \equal{} 0$ and therefore $ d \equal{} 0$, that is the triangle must be equilateral.

Hope this helps.


Compute the following without a calculator: $ \frac{9!}{6!3!}$

Please help. I don't know how to do this without a calculator. The answer should be $ 84$.

[quote="isabella2296"]Compute the following without a calculator: $ \frac {9!}{6!3!}$

Please help. I don't know how to do this without a calculator. The answer should be $ 84$.[/quote]

\[ {\frac{9!}{6!3!}\equal{}\frac{9 \cdot 8 \cdot 7 \cdot 6\cdot5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 }{ (6\cdot5 \cdot 4 \cdot 3 \cdot 2 \cdot 1) \: \times\: (3\cdot2\cdot1)} \equal{} \frac{9\cdot8\cdot7}{3\cdot2\cdot1} \equal{} \frac{(3\cdot3)\cdot (2\cdot 4) \cdot7}{3\cdot2} \equal{} 3 \cdot 4 \cdot 7 \equal{} 84}\]

Sergey Markelov
markelov@mccme.ru

Yeah, because the numbers cancel out. Like Markelov showed, right?


Let k a positive integer. Isn't it true that there exists $N_k$ such that : for any $x_1,...,x_k$ real numbers , if some $N_k$ consecutive terms of the sequence $f(n)=\sum_{i_1+...+i_k=n}{x_1^{i_1}...x_k^{i_k}}$ are integers then all terms are integer. It seems that the method I used to solve Pascual's problem is not very efficient here. Can anyone solve this problem?

Just one question: is $i_j\geq 0$ or $i_j>0$?

Also, could you please link to the problem by Pascual 2005 you are refering to?

All indices are greater than 0. The problem posted was that if 4 consecutive terms of $\frac{x^n-y^n}{x-y}$ are integers, all are integers.


Find all natural numbers $n$ such that there exist $ x_1,x_2,\ldots,x_n,y\in\mathbb{Z}$ where $x_1,x_2,\ldots,x_n,y\neq 0$ satisfying:
\[x_1 \plus{} x_2 \plus{} \ldots \plus{} x_n \equal{} 0\] \[ny^2 \equal{} x_1^2 \plus{} x_2^2 \plus{} \ldots \plus{} x_n^2\]

[quote="Erken"]Find all natural numbers $ n$,such that there exist $ x_1,x_2\dots,x_n,y\in\mathbb{Z}$ and
$ x_1 \plus{} x_2 \plus{} \dots \plus{} x_n \equal{} 0$
$ ny^2 \equal{} x_1^2 \plus{} x_2^2 \plus{} \dots \plus{} x_n^2$.[/quote]

I think you must add $ x_k\neq 0$, else it is trivial for any $ n>0$

Why?I add that $ y\neq 0$.Is it correct now?

Obviosly if $ n$ is solution, then $ n\plus{}2$ is solution too ($ x_{n\plus{}1}\equal{}y\equal{}\minus{}x_{n\plus{}2}$ work).
n=2 solution, therefore any even n solution. It is not hard to prove, that n=1 and n=3 had not non trivial solution.
I think n=5 had solution.

Could you show a solution to $ n\equal{}5$.

[quote="Erken"]Could you show a solution to $ n \equal{} 5$.[/quote]

$ 1\plus{}1\plus{}1\plus{}1\plus{}(\minus{}4)\equal{}0$
$ 1^2\plus{}1^2\plus{}1^2\plus{}1^2\plus{}(\minus{}4)^2\equal{}5\times 2^2$


quadratics
induction
modular arithmetic
number theory unsolved
number theory

Let $ q \equiv 1 (mod 4)$ and $ r \equiv 3 (mod 4)$ denote primes. Then $ n$ is a sum of two squares if and only if

                 $ n\equal{} {2^\alpha}\mu\nu^2$,

where $ \mu$ is a product of primes $ q$, and $ \nu$ product of primes $ r$.

Thanks!

There's a nice infinite descent method for the fact that a prime $ \equiv 1 \pmod 4$ is a sum of squares. I hardly see it on the net, so here it is-
Take $ n < p, x, y$ such that $ pn \equal{} x^2 \plus{} y^2$ (it is equivalent to the existence of a non-trivial solution to $ x^2 \plus{} y^2 \equiv 0 \pmod p$, which follows by the fact that $ \minus{} 1$ is a quadratic residue. Reduce $ x,y$ mod $ p$, etc., to get $ n<p$)
Now, if $ n > 1$, replace $ x,y$ with $ x', y'$ such that
\[ x \equiv x' \pmod n, y \equiv y' \pmod n, |x'|, |y'| \le \frac {n}{2}
\]
Then $ x'^2 \plus{} y'^2 \equiv x^2 \plus{} y^2 \equiv 0 \pmod n$, so there's $ r < n$ (the core of the proof, verify this) such that:
\[ rn \equal{} x'^2 \plus{} y'^2
\]
Multiply by the first equation to get
\[ n^2rp \equal{} (xy' \minus{} x'y)^2 \plus{} (yy' \plus{} xx')^2
\]
Notice that $ xy' \minus{} x'y \equiv xy \minus{} xy \equiv 0 \pmod n, yy' \plus{} xx' \equiv y^2 \plus{} x^2 \equal{} np \equiv 0 \pmod n$, so we can divide both sides by $ n^2$ to get
\[ rp \equal{} (\frac {xy' \minus{} x'y}{n})^2 \plus{} (\frac {yy' \plus{} xx'}{n})^2
\]
With $ r < p$. Repeat the process until $ r \equal{} 1$ (or begin with the minimal $ n$).

Same method will give you that every number is representable as a sum of 4 squares - for a prime $ p$, take $ n,x,y$ such that $ np \equal{} x^2 \plus{} y^2 \plus{} 1$ (an existence of a solution to this congruence follows from a nice counting (of squares) argument). Then, you need an identity similar to that of Vworldv, just (much) larger...

Thank you guys so much!

[quote="Vworldv"]You want us to prove Fermat's two-square theorem? A bit of a sketch:

First note that (considering the situation modulo $ 4$ with quadratic residues $ 0,1$), we can never express numbers of the form $ 4n \plus{} 3$ as a sum of two squares. But we can represent even powers of such primes as squares (as the prime power itself is a square and $ 0$ is a square).

Next note that we can represent $ 2 \equal{} 1^2 \plus{} 1^2$.

Finally, we can use the following beautiful method due to Zagier: consider the finite set $ x^2 \plus{} 4yz \equal{} p$ for $ p \equal{} 4k \plus{} 1$ prime and $ x,y,z$ integers greater than zero. One involution of the set is $ (x,y,z) \to(x,z,y)$, whose fixed points correspond to representations of $ p$ as a sum of squares, and another is:

$ (x,y,z)\to\begin{cases} (x \plus{} 2z,z,y \minus{} x \minus{} z) & x < y \minus{} z \\
(2y \minus{} x,y,x \minus{} y \plus{} z) & y \minus{} z < x < 2y \\
(x \minus{} 2y,x \minus{} y \plus{} z,y) & x > 2y\end{cases}$

The sole fixed point of this last involution (check that it is!) is $ (1,1,k)$. Now, because the number of fixed points of any involution of a finite set is congruent modulo $ 2$ to the cardinality of the set, we note that the cardinality of the given set is odd and therefore the first involution has an odd number of fixed points as well. In particular, it has a positive number of fixed points, meaning that it is representable as a sum of squares! (Note that we can define the latter involution in a simple form, although doing so might sacrifice the positivity of $ x$.)

This seems somewhat hard to motivate, but it is [i]much[/i] more beautiful than an uglier "routine" proof.

Finally, we note the identity $ (x^2 \plus{} y^2)(z^2 \plus{} w^2) \equal{} x^2z^2 \plus{} x^2w^2 \plus{} y^2z^2 \plus{} y^2w^2 \equal{} (xz \plus{} yw)^2 \plus{} (xw \minus{} yz)^2$, so by induction on "representable" numbers, the desired result follows.  :lol:

EDIT: Hmm, there also are two interesting and simple proofs using Gaussian integers but they use some ring theory.[/quote]

Would you like to tell me another two proofs of this problem? Thanks a lot!

Question: Why can we never express numbers of the form $ 4m\plus{}3$ as a sum of two squares?

[quote="foggy"]Question: Why can we never express numbers of the form $ 4m \plus{} 3$ as a sum of two squares?[/quote]Because all squares are of the form $ 4k$ or $ 4k\plus{}1$. So all sum of two squares are of the form $ 4k,4k\plus{}1,4k\plus{}2$.


inequalities function quadratics induction search algebra articles

I shared it from my book, about one of the most useful method solving Inequality

[b][size=167]Section 2. Squares Analysis Method S.O.S[/size][/b]

[b][size=125]Part 2.1. The Begining Problems[/size][/b]

  Generally, if we have an usual inequalities, the ways for us to solve them are neither trying to fumble from well-known inequalities nor finding a mixing-variable method, the best choise is often bringing them back to squaring forms. That is based on the most elementary probperty of a real number : $x^2 \ge 0 ~\forall x \in R$. There are alot of problems, though you are  indeliberate or not, almost used this property in proving. However, what you will read hereafter is likely make you truthly suprising.

  We will start with the inequality $AM-GM$, which can be see as the most well-known inequality of all comom inqualities. But we only consider it in some simply case of $n$. For example, if $n=2$ we have
[b]Problem 1.[/b]
Prove that for all $a,b \ge 0$ we have the inequality
\[ a^2+b^2 \ge 2ab  \]

 There are not many thing to mention in this one, even this is the first time you see a inequality, the solution is very easy. The inequality is equal to $(a-b)^2 \ge 0$, obviously. Now, consider it if $n=3$, we have

[b]Problem 2.[/b]Prove that for all $a,b,c \ge 0$ we have
\[ a^3+b^3+c^3 \ge 3abc  \]

 If asked about one demonstration for it, we fell a bit puzzled. Of course, it's not difficult, the solution is only in one-line

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

 And definitally, this is the sharpest key, because we don't need to use any intermediate stages. Both examples can be proved by Squares analysis method, but in confined meaning. The special advantage of this method is using so little advance knowledge, even you don't need to know any inequality theorems. Moreover, this is still a natural way with our thought.

 If you tried to read the problems in previous chapter carefully, you won't find problems using this method rarely. Now, by general the way using and find the essence of a extremely useful method.

 An important problem which we consider cautious is a very nice and famous, that was introduced in previous chapter, Iran 96 inequality

[b]Problem 3.[ Iran 98 ]  [/b]
Prove that for all $a,b,c \ge 0$ we have
\[ \frac{1}{(a+b)^2}+\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2} \ge \frac{9}{4(ab+bc+ca)}  \]

 This is a problem which exists in a very simply and nice form. In addition, it is also very difficult if you haven't ever seen before. But the first, we will be concerned with the inequality appeared as problem A3 in the International Math Olympiad and find a really natural proving for it.

[b]Problem 4. [ IMO 2005 Pro. A3 ][/b]
Suppose that $x,y,z$ are reals number and $xyz \ge 1$. Prove the following inequality
\[ \frac{x^5-x^2}{x^5+y^2+z^2}+\frac{y^5-y^2}{y^5+z^2+x^2}+\frac{z^5-z^2}{z^5+y^2+x^2} \geq 0  \]

[i][b]Solution[/b][/i]
First, we rewrite it in standard- equal degree form

\[ \frac{x^5-x^2}{x^5+y^2+z^2} \ge \frac{x^5-x^2.xyz}{x^5+(y^2+z^2)xyz} \ge \frac{x^4-x^2yz}{x^4+yz(y^2+z^2)}  \]
\[ \frac{x^4-x^2yz}{x^4+yz(y^2+z^2)}\ge \frac{2x^4-x^2(y^2+z^2)}{2x^4+(y^2+z^2)}  \]

Let $a=x^2,b=y^2,c=z^2$, we need to prove

\[ \sum_{a,b,c} \frac{2c^2-a(b+c)}{2a^2+(b+c)^2} \ge 0\\ \Leftrightarrow \sum_{a,b,c} (a-b)\frac{a}{2a^2+(b+c)^2}-\frac{b}{2b^2+(a+c)^2}) \ge 0  \]
\[ \Leftrightarrow \sum_{a,b,c} (a-b)^2\frac{c^2+c(a+b)+a^2-ab+b^2}{(2a^2+(b+c)^2)(2b^2+(a+c)^2)} \ge 0  \]
Which is obvious true. The equal holds if $a=b=c=1$.

This is not the uequal solution for the inequality, maybe there're some nicer, but the most importance of that is giving us a very natual mothod with 3-variable inequality. Generally, if we have an abiraty inequality, try to rewrite it to the form
\[ S_c(a-b)^2+S_b(a-c)^2+S_a(b-c)^2 \ge 0  \]

  Rewriting the inequality to this base-expression is the first step in the way using S.O.S. If you quitely familier with inequality then construction this expression is relatively simple and easy, we only need some identity and transform. If not, I will explain for you in section "Basic-form of S.O.S method and some analysis techniques.

  Of course, if in that basic-form, all confficinents $S_a,S_b,S_c$ are non-negative, we are done. For a long time, this still only case for us but extremelly, it's only the first simply and easiet application of Squaring analysis thorem. More important, S.O.S help us to solve problems which we treated not to use yet : some of $S_a,S_b,S_c$ are negative.

  Generally, in some sysmetric we can assume that $a \ge b \ge c$. For the cyclic problem, we need to consider  an extra case $c \ge b \ge a$. In case $a \ge b \ge c$ we have a comment

  If $S_b \ge 0$,  because $(a-c)^2 \ge (a-b)^2+(b-c)^2$, so we have
\[ S_c(a-b)^2+S_b(a-c)^2+S_a(b-c)^2 \ge (S_c+S_b)(a-b)^2+(S_b+S_a)(b-c)^2  \]

  And the rest is proving $S_a+S_b \ge 0$, $S_c+S_b \ge 0$. But commonly, two those inequalities can be proved quite easy, because they haven't any square-expression $(a-b)^2,(b-c)^2,(c-a)^2$ yet.

   If $S_b \le 0$,  because $(a-c)^2 \le 2(a-b)^2+2(b-c)^2$, so we have
\[ S_c(a-b)^2+S_b(a-c)^2+S_a(b-c)^2 \ge (S_c+2S_b)(a-b)^2+(2S_b+S_a)(b-c)^2  \]

Also, proving $S_c+2S_b \ge 0$ and $2S_b+S_a \ge 0$ is more simple.}

  In addition, we need some stronger assession. The usual assession is
\[ \frac{a-c}{b-c} \ge \frac{a}{b} (a \ge b \ge c)  \]

From that, if $S_b , S_c \ge 0$ then
\[ S_b(a-c)^2+S_a(b-c)^2=(b-c)^2\bigm(S_b(\frac{a-c}{b-c})^2+S_a\bigm) \ge (b-c)^2(\frac{a^2S_b}{b^2}+S_a)  \]
And we are done if we prove $a^2S_b+b^2S_a \ge 0$ completely.

We combinate all result in one theorem as follow

[b][size=125]Theorem [S.O.S Theorem (Squaring Analysis Theorem)][/size][/b]Consider the expression
\[ S=f(a,b,c)=S_a(b-c)^2+S_b(a-c)^2+S_c(a-b)^2  \]
When $S_a,S_b,S_c$ are functions of $a,b,c$.

[i]1.If $S_a,S_b,S_c \ge 0$ then $S \ge 0$.

2.If $a \ge b \ge c$ and $S_b , S_b+S_c , S_b+S_a \ge 0$ then $S \ge 0$.

3. If $a \ge b \ge c$ and $S_a , S_c , S_a+2S_b , S_c+2S_b \ge 0$ then $S \ge 0$.

4. If $a \ge b \ge c$ and $S_b , S_c \ge 0 , a^2S_b+b^2S_a \ge 0$ then $S \ge 0$.

5. If $S_a+S_b+S_c \ge 0$ and $S_aS_b+S_bS_c+S_cS_a \ge 0$ then $S \ge 0$.[/i]

  The application 5th can be easily prove by quaratic creation.

  Moreover, if $S \ge 0$ for all $a,b,c$, we must have $S_a+S_b |_{a=b} \ge 0$ , $S_b+S_c |_{b=c} \ge 0$ , $S_c+S_a |_{c=a} \ge 0$ ($S_a+S_b|_{a=b}$ mean that we consider the expression $S_a+S_b$ when $a=b$. For sysmetric inequalities, we have $S_a=S_b$ if $a=b$, so $S_a$ must be non-negative. The simply comment is very important for problem which we have to find the best constant satisfy.

  The theorem seems to be so important and if we say that It has a very strong effect for almost 3-variable inequality, It's really incredible. But in fact, S.O.S has done this work and this's very supricing.

  One question is gived that for what expression, we can tranform it to basic S.O.S form? The answer is that we can do it for all sysmetric or cyclic function $f(a,b,c)$ satisfy $f(a,a,a)=0$, $f$ can involve fractions, roots... See the proving in next section.

The complete PDF File here, but there're some misktakes in typo because I translated it in a hurry,

Thank you very much this are really nice...

Davron

[quote="hungkhtn"][quote="pvthuan"]Do you have a better coining for the S.O.S? I do not think it is a method, it is just a technique or an approach at most. In Vietnam some claimed that they invented this  technique. Who should receive this credit for such powerful art of inequalities solving? Please cite any reference in the literature.[/quote]

This's because you don't understand what S.O.S can do and what's a truthly method. With your, what can be called a method? Some examples, of course.

Ceitainly, a so strong method to solve almost problems with 3 variable or more is expectively called a greated method solving inequality.[/quote]

As I said, it is a powerful approach. Aproach or method are interchangeable words sometimes, I prefer the word approach here. I think that your mathematical genuinity is recognized but your English is NOT that good. Read carefully what I say before you can raise your voice.

I ask everyone (not just you) if they have a better term for SOS. That is common when we need to have a proper name for a new thing in science.

Of course, I see Math and Ineq in diffirent way with you, PVThuan. So I can't write as author in your book.

About SOS, there's nothing to say what's its name. Everyone uses and understands it always know that It's very nice and useful method. Nothing else to tell you.

Comeon guys thats enought now continue to discuss the METHOD or APPROACH i dont know Just the message is impotant thanks ...

Davron

[quote="hungkhtn"]Yes, there're many many problem can be solved by this method. I want to say that "Almost problems 3-variables can be solved by this way". There's many exam, you can check in this forum.[/quote]

Is there any known way of generalizing SOS? The usual SOS decomposes an inequality to $\sum S_a (b-c)^2$ form. Is it possible to replace $b-c$ by $f(b,c)$, so that a similar theorem holds? (Some conditions can be imposed to f, I am just very curious to know that)

I wait anyone give at least one solution for three above problem.

In fact the reason I asked for generalization is to see whether there is a systematic way to solve inequality like vasc's famous one ($(\sum x^2)^2 \ge 3(\sum x^3y)$), where a mysterious person Stefan V. gave an incredible solution by directly decomposing it into three squares.

[quote="Flakky"]No! We assume that $ x \plus{} y \minus{} z, y \plus{} z \minus{} x, z \plus{} x \minus{} y$ are possitive. That is the problem. I can't understand why is this possible![/quote]
Dude!
The Iran ineq is:
$ \frac{1}{(x\plus{}y)^2} \plus{} \frac{1}{(y\plus{}z)^2} \plus{} \frac{1}{(z\plus{}x)^2} \ge \frac{9}{4(xy\plus{}yz\plus{}zx)}$

The substitution used is: $ a \equal{} x\plus{}y, b \equal{} y\plus{}z, c \equal{} z\plus{}x$. So we don't have to make sure that $ x\plus{}y\minus{}z$ is positve...

yes,I think it is very helpful for me,thank you~

I don't understand your solution to Iran 96 either

4.if  $ a\ge b\ge c$ and $ S_{b}, S_{c}\ge 0 , a^{2}S_{b}\plus{}b^{2}S_{a}\ge 0$then $ S\ge 0$

but in  this problem ,you write
 $ a\ge b\ge c  \,\,\,S_a\ge 0 \,\,\,(in \,\,fact  \,\,S_b\ge 0)\\
b^2 S_b\plus{}c^2 S_c\ge 0$

How can you use fourth criterion directly?

this inequality "iran 96" can be sloved by "mixing variable" method :lol:  i like it

Hi! I read the article but i have a lot of trouble putting it into practice because i dont know how to factor the inequality. I am not very good at factoring and
Where can i learn how to do that?
You said in the "Basic-form of S.O.S method and some analysis techniques" section but there isnt one and i really want to learn how to factorise! so i can use the SOS method.
thanks

In fact, almost everybody finds them difficulty.

[quote="Ronald Widjojo"]Maybe siily, but is there any short way or method to find $ S_a,S_b$ and $ S_c$ ?

because I find it difficult..  :oops:[/quote]
Solve a lot of problems
Solve a lot of problems
Solve a lot of problems
 after that it seems easy , sure.

Dear hungkhtn !
What about Iran 96 by this way ?

Can we prove Vasc's inequality by using S. O. S. Theorem?

[quote="hungkhtn"]In addition, we need some stronger assession. The usual assession is
\[ \frac{a-c}{b-c} \ge \frac{a}{b} (a \ge b \ge c)  \]

From that, if $S_b , S_c \ge 0$ then
\[ S_b(a-c)^2+S_a(b-c)^2=(b-c)^2\bigm(S_b(\frac{a-c}{b-c})^2+S_a\bigm) \ge (b-c)^2(\frac{a^2S_b}{b^2}+S_a)  \]
And we are done if we prove $a^2S_b+b^2S_a \ge 0$ completely.

...

4. If $a \ge b \ge c$ and $S_b , S_c \ge 0 , a^2S_b+b^2S_a \ge 0$ then $S \ge 0$.
[/quote] Does this still work in the order $a \le b \le c$? It seems like $\frac{c-a}{c-b} \ge \frac ab$ still holds in that case.

[quote="v_Enhance"]
Does this still work in the order $a \le b \le c$? It seems like $\frac{c-a}{c-b} \ge \frac ab$ still holds in that case.[/quote]
Id est, the same: in the ordering $a>b>c$ we obtain $\frac{a-c}{a-b}\geq\frac{c}{b}$.

[quote="Leon"]The exact formulation of the problem proposed in Macedonia MO 1999 is
[color=blue]Let $a$, $b$, $c$ be real numbers such that  $(a^{2}+b^{2}+c^{2}=1, \; a, b, c>0 )$. Show that \[a+b+c+\frac{1}{abc}\geq 4 \sqrt{3}\] [/color]
My question is if is it possible to prove this inequality by the S.O.S. method.  I have a solution in which I make use of Schur inequality.[/quote]
$a^{2}+b^{2}+c^{2}=1\implies abc\leq \frac{\sqrt{3}}{9}$,
$a+b+c+\frac{1}{abc}=a+b+c+\frac{1}{9abc}+\frac{8}{9abc}\geq \frac{4}{ \sqrt{3}}+\frac{8}{9abc}$
$\geq \frac{4}{ \sqrt{3}}+\frac{8}{ \sqrt{3}}=4 \sqrt{3}.$

[quote=hungkhtn]Ok, but I think the method easy using is better. I don't think a generation too difficult in use is good, isn't it? It's my thought :D, although maybe there's some other forms. There're some hardest problem with 3-variable

1.  Prove that for all $a,b,c \ge 0$

\[ (a+\frac{b^2}{c})^2+(b+\frac{c^2}{a})^2+(c+\frac{a^2}{b})^2 \ge \frac{12(a^3+b^3+c^3)}{a+b+c} \]

2. Prove that for all $a,b,c \ge 0$

\[ \frac{a^4}{a^3+b^3}+\frac{b^4}{b^3+c^3}+\frac{c^4}{c^3+a^3} \ge \frac{a+b+c}{2} \]

3. Prove that for all $a,b,c \ge 0$

\[ \frac{a^3}{2a^2+b^2}+\frac{b^3}{2b^2+c^2}+\frac{c^3}{2c^2+a^2} \ge \frac{a+b+c}{3} \][/quote]

Anyone?


$ 30! \equal{} 2^{26}\cdot 3^{14} \cdot 5^{7} \cdot 7^{4} \cdot 11^{2} \cdot 13^{2} \cdot 17 \cdot 19 \cdot 23 \cdot 29$.

Proceed to majorate now every factor by a convenient power of $ 2.$

first of all we prove this n!<(n/2)^n(1), n is a integer.
from (1) we have 30!<15^30 and 15^30 has 36 digits([30log15]+1)
we have 2^124 has 38 digits( [124log2]+1) so 2^124>15^30>30!


What does $2 min(a,b)$ mean?

I just started reading Problem Solving Strategies by Arthur Engel...

Thanks!

$\min{(a,b)}$ means the smaller of the two values. For example, $\min{(1,2)}= 1$ and $\min{(3,2)}= 2$.

ohhh ok thanks! i'm new to some notations..and number theory..


Danny Henry made a waffle on his six-inch-diameter circular griddle using batter containing a half a cup of flour. Using the same batter, and knowing that all waffles have the same thickness, how many cups of flour would Paul Bunyan need for his 24-foot-diameter circular griddle?

The ratio of diameters is $ \frac{1}{48}$.  Thus, the ratio of the areas must be $ (\frac{1}{48})^2$, or $ \frac{1}{2304}$.  So, the answer is $ 2304 \times \frac12$, or $ \boxed{1152}$.

To prove ernie's point, let me illustrate:
6" diameter = 3" radius = 9"pi area
24' diameter = 12' radius = 144" radius  = 20736" area
9/20736 = 2304. But remember, Danny used 1/2 cup of flour. so, 2304/2 = 1152. Much more complicated, but it illustrates that Ernie's way is technically correct. Nice idea, Ernie.  :D

:blush: ...just a random typo I guess. 9/20736 does not equal 2304...20736/9 does  :lol:


[quote="MagnusDahlquist"]Hi there,

I've been trying to figure out this math problem forever. I was hoping you guys can help...=)

Month 1
- 100 sign-ups

Month 2
- 100 sign-ups
- 80 people from month 1 are still members of the club

Month 3
- 100 sign-ups
- 80 people from month 2 are still members of the club
- 40 people from month 1 are still members of the club

Month 4
- 100 sign-ups
- 80 people from month 3 are still members of the club
- 40 people from month 2 are still members of the club
- 2 people from month 1 are still members of the club

What is the average stay?

Thanks a million!

//magnus[/quote]

The average stay is the total number of months stayed by everyone divided by the number of people.

The total number of people is just 400.

Then split it into cases...

Month 1:
20 people stay for 1 month, 40 people stay for 2 months, 38 people stay for 3 months and 2 people stay for 4 months.  so $ 20\cdot 1 \plus{} 40\cdot 2 \plus{} 38\cdot 3\plus{} 2\cdot 4\equal{}222$

Month 2:
20 people stay for 1 moht, 40 people stay for 2 moths, 40 people stay for 3 moths so $ 20\cdot 1 \plus{} 40\cdot 2 \plus{} 40\cdot 3\equal{}220$ months

Month 3:
20 people stay for 1 month, 80 people stay for 2 so $ 20\cdot 1 \plus{} 80\cdot 2\equal{}180$ months

Month 4:
100 people stay for a month so 100 months.

then the total # of months is  722. then the average is 722/400= 1.805


Suppose that $x,$ $y,$ and $z$ are three positive numbers that satisfy the equations $xyz=1,$ $x+\frac{1}{z}=5,$ and $y+\frac{1}{x}=29.$ Then $z+\frac{1}{y}=\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

[hide]
Let $r=z+\frac{1}{y}$. We have the identity
\[\left( x+\frac{1}{z}\right) \left( y+\frac{1}{x}\right) \left( z+\frac{1}{y}\right) =xyz+x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{xyz}\]
Substituting into our given equations (along with the one involving $r$):
\[29\cdot 5r=1+5+29+r+1=36+r\Rightarrow 144r=36\Rightarrow r=\frac{1}{4}\]
so $m+n=\boxed{5}$.
[/hide]

[hide]
We multiply the givens together to get:
\[\left( x+\frac{1}{z}\right) \left( y+\frac{1}{x}\right) \left( z+\frac{1}{y}\right) =xyz+x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{xyz}\]
Therefore,
\[x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{145m}{n}-2 \]
We also know that $x+\frac{1}{z}+y+\frac{1}{x}+z+\frac{1}{y}=34+\frac{m}{n}$, so:
\[34+\frac{m}{n}=\frac{145m}{n}-1\Rightarrow \frac{144m}{n}=36\Rightarrow \frac{4m}{n}=1 \]
Therefore, $n=4m$, so $m+n=5m$.  Since they are relatively prime, $m=1\Rightarrow m+n=\boxed{5}$.
[/hide]

Hint[/hide] [hide]Multiply out each of the eaquations, by the variable that's in the denominator of each equation, respectively. Get everything in terms of x and plug it into xyz=1. Then you find the value of x, and the rest follows.[/hide]

elephantmaster, why would you post essentially the same solution as somebody who beat you to it by [i]two months[/i]?

and Kamil Witek, why would you give a hint to a problem that's been solved [i]three times[/i]?

[quote="Kamil Witek"]SO that I could have more posts? :blush:[/quote]

Dude, you're cool.

[hide=Solution]We are motivated to preserve symmetry by using the unknown expression along with the $5$ and $29$. Since product and sum are the easiest ways to combine multiple numbers, we try those first. Their sum is $x+\frac{1}{z}+y+\frac{1}{x}+z+\frac{1}{y}=x+y+z+\frac{xy+yz+zx}{xyz}=x+y+z+xy+yz+zx$. Their product is $(x+\frac{1}{z})(y+\frac{1}{x})(z+\frac{1}{y})=(xyz+y)(xyz+z)(xyz+x)=(x+1)(y+1)(z+1)=xyz+xy+yz+zx+x+y+z+1=\text{sum}+2$. The second step is by multiplying the first part by $yz$, the second by $zx$, and the third by $xy$, which altogether multiplies by $x^2y^2z^2=(xyz)^2=1$. Thus, letting the third expression be equal to $k$, we have that $145k=k+34+2\implies k=\frac{1}{4}\implies m+n=\boxed{5}$.[/hide]

Answer is 5,multiply all the three equations

This is basically the same as #4 of the Mathcounts Minis

[b]Solution:[/b] Multiplying them together, we get $xyz + x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{xyz} = 1 + (5) + (29) + (a) + \frac{1}{1} = 36 + a = (5)(29)(a) = 145a \Rightarrow 144a = 36 \Rightarrow a = \frac{1}{4} \Rightarrow \boxed{005}$.


I have a word problem that I can't seem to figure out

At the market, a man buys six avocados for $13.35. All produce is 30% off today only. How much will two avocados cost you tomorrow at full price?
A)
$4.00

B) $4.50
C) $5.00

D) $5.50
E) $6.00

The answer on the website I got the question on says $5.00 but I can't figure out how they got that answer. Any help for be great.

[quote="Namingishard"]I have a word problem that I can't seem to figure out

At the market, a man buys six avocados for 13.35. All produce is 30% off today only. How much will two avocados cost you tomorrow at full price?

A) 4.00

B) 4.50

C) 5.00

D) 5.50

E) 6.00
e answer on the website I got the question on says 5.00 but I can't figure out how they got that answer. Any help for be great.[/quote] If he has bought six avocados today then he got 30% off, is how I would interpret it.

So if 6A= 13.35, then A = 2.225 and this is equal to 70% (i.e. 0.7).

If you work out what 100% is, you might get your answer.

EDIT: Although doing it this way doesn't agree with your answer. :maybe:

Are you sure that your starting money is correct?
Because if you were to work backwards from any of the answers...
A: $ \$4.00$ meaning that one avocado is $ \$2.00$ meaning that if they were on sale they would be $ \$1.40$ meaning that six on sale would cost $ \$8.40$
B: $ \$4.50$ meaning that one avocado is $ \$2.25$ meaning that if they were on sale they would be $ \$1.575$ meaning that six on sale would cost $ \$9.45$
C: $ \$5.00$ meaning that one avocado is $ \$2.50$ meaning that if they were on sale they would be $ \$1.75$ meaning that six on sale would cost $ \$10.50$
D: $ \$5.50$ meaning that one avocado is $ \$2.75$ meaning that if they were on sale they would be $ \$1.925$ meaning that six on sale would cost $ \$11.55$
E: $ \$6.00$ meaning that one avocado is $ \$3.00$ meaning that if they were on sale they would be $ \$2.10$ meaning that six on sale would cost $ \$12.60$
So something is wrong with the problem

Thanks for the replies.
I knew something was wrong with the problem, Just wanted to make sure.


What product is expected to obtain when pent-4-en-1-ol is treated with bromine?

It is simply SNi leading to cyclic ether formation

That's right. Are you able to write a correct mechanism for this reaction?

I am directly giving the final stages of the mechanism....

Very good. Don't forget, when writing mechanisms on exams, to include all the non-bonding electrons. You are also using the arrows correctly: an arrow always begins on a pair of electrons and ends on an atom.

Now, what is the precise reason for the oxygen to attack carbon 5 and not carbon 4?

Hello
six member ring is more stable than five member.That's why the oxygen  attacked  carbon 5 and not carbon 4.

Oh, is that so? Then what happens when the first of the following compounds is slightly heated and the second is treated with $ \ce{Bu3SnH}$?

[quote="Carcul"]
Now, what is the precise reason for the oxygen to attack carbon 5 and not carbon 4?[/quote]

Steric hindrance?

And that because the mecanism of oxygen attack is...

I was thinking in SN2...

Now try to solve the problems I posted above (post #7).

What is this  Bu3SnH ?

No idea about first one...


trigonometry
algebra
function
domain
inequalities
quadratics
trig identities

In $\triangle{ABC}$, we have $\lg{\tan{A}}+\lg{\tan{C}}=2\lg{\tan{B}}$
Prove : $\frac{\pi}{3} \leq B < \frac{\pi}{2}$

Condensing logs and then taking ten to power of both sides to cancel, we get $\tan A \tan C = \tan ^2 B$

Then I believe you could assume the range of B is true, and do some case checking of $0 < B < 60$ and $ 60 \le B < 90$ and $90 < B < 180$... Possibly.  I can't think of any trig identities to prove it algebraicly though.

This is a possible way...though I haven't tried yet.
And I have found a straight proof, though it is quite long.

Note: You need not check $B>90^{o}$ since the domain of $y=\lg{x}$ is $x>0$.

Obviously A,B,C<90 , or else you cant take log
by using AM-GM ,

(tanB)^2=(tanA)(tanC)=< {(tanA+tanC)/2}^2

2tanB=<tanA+tanC=tan(A+C) . (1-tanAtanC)= - tanB ( 1- (tanB)^2 )
         = tanB ( (tanB)^2 - 1 )

Hence simplfy and get  (tanB)^2 >= 3  and then get the desired inequality

[quote="BornForMath"]Is $\lg$ log?[/quote]

yes it is  :P

[quote="shyong"]Obviously A,B,C<90 , or else you cant take log
by using AM-GM ,

$\tan^2{B}=\tan{A} \cdot \tan{C} \leq (\frac{\tan{A}+\tan{C}}{2})^2$

$2\tan{B} \leq \tan{A}+\tan{C} = \tan{(A+C)} \cdot (1-\tan{A}\tan{C})= - \tan{B} ( 1- \tan^2{B})
         = \tan{B} ( \tan^2{B}- 1 )$

Hence simplfy and get  $\tan^2{B} \geq 3$  and then get the desired inequality[/quote]
LaTexed by Shobber.

Nice solution [b]shyong[/b]! Applying AM-GM make this problem very easy! My solution is quite long but only used quadratic equations. I shall post it later.


In acute triangle $ABC$, $w$ is the circumcircle and $O$ the circumcenter. $w_{1}$ is the circumcircle of triangle $AOC$, and $OQ$ is the diameter of $w_{1}$. Let $M,N$ be on $AQ,AC$ respectively such that $AMBN$ is a parallelogram. Prove that $MN,BQ$ intersect on $w_{1}$.

Let $L$ be the midpoint of $AB$ and $P$ be the second intersection of $\omega_1$ with $BQ.$ Then $\angle APQ \equal{} \angle BNA \equal{} \angle ABC.$ Thus if $R \equiv BQ \cap AC$ and $D \equiv BN \cap AP,$ then $ PDNR$ is cyclic. But notice that $PQ$ bisects $\angle APC$ since $Q$ is the midpoint of  the arc $AC$ of $\omega_1.$ As a resut, $\angle BPC \equal{} 180^{\circ} \minus{} \angle ABC \equal{} \angle BNC$ $\Longrightarrow$ $BPNC$ is cyclic $\Longrightarrow$ $\angle NPR \equal{} \angle BCA,$ but since $PDNR$ is cyclic, we obtain $\angle NPR \equal{} \angle NDR \equal{} \angle BCA \equal{} \angle ABN$ $\Longrightarrow$ $DR \parallel BA.$ Therefore, the cevian $NP$ of $\triangle BNA$ goes trough the midpoint $L$ of  $AB$ $\Longrightarrow$  $P \equiv BQ \cap MN \in \omega_1.$


ratio
circumcircle
geometry
similar triangles
AMC

In triangle ABC we have AB=7, AC=8, and BC=9. Point D is on the circumscribed circle so that line segment AD bisects angle BAC. What is the value of AD/CD?

I searched briefly for this problem on the forum. It isn't on aops wiki, I think. [hide=" My Difficulties"]I understand each step of the solution, which used similar triangles, but I don't know how I would have thought of it! You can find lots of congruent angles by connecting points and using circle arc angle formulas but how would you determine which triangles to use to find ratios to use in the problem?

In general, when you read a solution and you don't know how you would have thought of it, what is the best approach to learning the concepts and logic behind the problem? Is there anything more to it than just memorizing the methods in the solution and trying them on similar problems?
[/hide]

[hide="quick"]If you don't like working with similar triangles, use [[Ptolemy's Theorem]] for this problem and all (most of) the work is done for you! :) [/hide]
I'm not sure of any good methods of gaining "similar triangle intuition" other than practice.


calculus
derivative
calculus computations

[hide]
1)
$\frac{dy}{dx} e^x=e^x %Error. "longright" is a bad command.
arrow y^{(10)}=\frac{e^x}e$  (You can factor out 1/e before differentiating)

2)
$\frac{dy}{dx} x^n = nx^{n-1}$
Take a series of derivatives:
\[ y=\sqrt{1-x}=(1-x)^{1/2} \]
\[ y'=-\frac12 (1-x)^{-1/2} \]
\[ y''=-\frac14(1-x)^{-3/2} \]
...
Every time you differentiate, another $-1$ term comes from chain rule [d/dx (1-x) = -1]

\[ y^{(5)} = \left(\left(\frac12\right)\left(\frac{-1}2\right)\left(\frac{-3}2\right)\left(\frac{-5}2\right)\left(\frac{-7}2\right)\right)(-1)^5(1-x)^{-9/2}=\frac{-105}{32(1-x)^\frac92} \][/hide]


calculus
number theory open
number theory

This is one of the problem of $250$  $Calculus$ $Problem$ by $Vatslav$ $Sierpinski$ the book has'nt a complete solution for  it so i think this should be hard one.
[color=green]Problem:[/color]
let $f(n)$ be $n$th member of  fibonacci sequence   prove that for $n>12$ ,$f(n)$ cant be show like a power of a natural number.

[quote="payman_pm"]
i think this should be hard one.
[/quote]
Indeed, I think it's still an [b]open[/b] question.

[b]Solved[/b] :
The only square Fibonacci numbers are 1 and 144.
The only cubic Fibonacci numbers are 1 and 8.
The only triangular Fibonacci numbers are 1,3,21 and 55.


prove that the polynomial $4x^{8}-2x^{7}+x^{6}-3x^{4}+x^{2}-x+1$ has no real root.


Do any of you listen to rap music?  For those of you who dont know what rap is:

rap 4  (rp)

noun

1. Slang A talk, conversation, or discussion.

2. a. A form of popular music developed especially in African-American urban communities and characterized by spoken or chanted rhyming lyrics with a syncopated, repetitive rhythmic accompaniment.

b. A composition or performance of such music.

  And no Beethoven's Symphony does not fit under the rap genre.

what's the point of this topic? we all know how to use a dictionary.

i dont think you've read the question i posed......"Do any of you listen to rap music?"

Yeah, once in a while.

And it sucks compared to other genres of music.

I think rap is OK but it sure gets boring after a while.

[quote="bleumoose"]there are times in my life when i need a pulsing, rhythmic beat... then i listen to rap. i try not to listen to the words as much because most of the time they are disgusting or hateful, but i cannot deny that the beat moves me.  :blush:[/quote]

i agree with everything you said.  there is alot of music that you can't dance too, because its not meant to be danced to.  if you need some party music, rap (or disco :rotfl: ) is good for that.

I think rap's intricate balance of rhythmic pulsing and repition makes it, (by far) the most catchy stle of music to listen, or sing along too.  However, the explicit nature of the lyrics, sometimes makes it difficult too.....

i listen to rap as well. it's cool.
:cleaning:

Sure, I listen to rap. And for those of you who complain of all the hateful things in the music, do you put your hands over your ears in movies during any other form of entertainment?

[quote="xxreddevilzxx"][quote="Mariag"]i listen to rap by force because my brother- who shares my room and does not believe in headphones- likes it. i, however, don't like it... i prefer soft music[/quote]

you still share rooms with your brother.... that must not be too good. My sister said she hated it when we did (for like 3 months), because i would listen to too much metallica.[/quote]

Haha... Metallica. I remember when I listened to them last year on the bus from my friends CD player.

And I listen to a little bit of rap.

the catchy phrases r ok but i usually like the ones with less cuss words

Bad raps are [c]rap.

Candy shop = magic stick... wow, there's no creativity. Almost identical beats with lyrics that talked about either women, parties, how rich/poor/wasted you are

i try not to listen to lyrics. however, the instrumental seems fine. It conveys the new type of music. However, since rap industry is mostly consisted of black musicians, many rock-and-roll such as Aaron Kravitz worry that they have forgotten the RnR. Thats what i read somewhere. but still, rap is great except the lyric.

i love all kinds of music :D
i like Ja Rule, Nelly, and Eminem
i don't mind the lyrics or the hate, we need more of those things :D  :D  :D  but i don't like the dirty sex lyrics

yes those are especially annoying especially when its a video :mad: but if the topic is meaningful then i probably will respect the lyrics although the beats are more intresting


$ \sqrt {x} \plus{} \sqrt {2x \plus{} 1} \equal{} \sqrt {x \minus{} 1}$
if one of the roots of equation is [b]a[/b] then find:
$ a^2 \plus{} \frac {1}{a^2}$

$ \sqrt x  \plus{} \sqrt {2x \plus{} 1}  \equal{} \sqrt {x \minus{} 1}  \leftrightarrow \sqrt x  \minus{} \sqrt {x \minus{} 1}  \equal{}  \minus{} \sqrt {2x \plus{} 1}$

But $ \sqrt x  \minus{} \sqrt {x \minus{} 1}$.

Do we solvein complex number?

hey we know a so it could be solved like it!
$ a^2  \plus{}  b^2  \equal{}  a  \plus{}  1/a  \plus{}  2$

[quote="stevenmeow"]$ \sqrt {x} + \sqrt {2x + 1} = \sqrt {x - 1}$
$ x + 2\sqrt {2x^2 + x} + 2x + 1 = x - 1$
$ 2\sqrt {2x^2 + x} = - 2x - 2$
$ \sqrt {2x^2 + x} = - x - 1$
$ 2x^2 + x = x^2 + 2x + 1$
$ x^2 - x - 1 = 0$
${ x - 1 - \frac[1}{x} = 0$
${ x - \frac[1}{x} = 1$
$ x^2 - 2 + \frac {1}{x^2} = 1$
$ x^2 + \frac {1}{x^2} = 3$
blah[/quote]
I think this solution should many condition, but I don't think that's wrong. I am going to think again! Can you see your solution again?  :wink:


geometry
inradius
incenter
geometry proposed

given a triangle ABC, let r be its inradius (its incircle touches BC, CA, AB at D, E, F respectively)
Let the inradii of triangles AFE, BDF, CDE, DEF be r1, r2, r3, r4 .
Prove that  r1 + r2 + r3 + r4 = 2r .

$ J_a,J_b,J_c$ are incenters of $ \triangle AFE,\triangle BFD,\triangle CED.$ Quadrilaterals $DFJ_aE,$  $EDJ_bF$ and $ FEJ_cD$ are cyclic, since

$ \angle DFE \plus{} \angle FJ_aE \equal{} 90^{\circ} \plus{} \frac {_1}{^2}\angle A \plus{} 90^{\circ} \minus{} \frac {_1}{^2}\angle A \equal{} 180^{\circ},$ etc.

Therefore, $ J_a,J_b,J_c$ are  midpoints of the arcs $ EF,FD,DE$ of the incircle $(I).$ If $M,N,L$ denote the midpoints of $ EF,FD,ED,$ by Carnot's theorem for $\triangle DEF,$ it follows that $ IM \plus{} IN \plus{} IL \equal{} r \plus{} r_4.$ But $ IM \equal{} r \minus{} r_1,$  $ IN \equal{} r \minus{} r_2$ and $ IL \equal{} r \minus{} r_3$ $\Longrightarrow$ $r_1 \plus{} r_2 \plus{} r_3 \plus{} r_4 \equal{} 2r.$


geometry
circumcircle
analytic geometry
geometry unsolved

Given a non-isosceles triangle $ABC$, let the medians issuing from $A,B,C$ intersect the circumcircle of the triangle at $M,N,P$.Prove that if $MN=MP$ then $2a^2=b^2+c^2$

[b]Lemma 1:[/b] The centroid G lies on the A-Apollonius circle of the triangle $\triangle ABC$ iff $2a^2=b^2+c^2$.

$m_b^2 = \frac 1 4 (2a^2 + 2c^2 - b^2) = \frac 3 4 c^2$

$m_c^2 = \frac 1 4 (2a^2 + 2b^2 - c^2) = \frac 3 4 b^2$

$\frac{m_c^2}{m_b^2} = \frac{b^2}{c^2}$

$\frac{CG}{BG} = \frac{m_c}{m_b} = \frac{b}{c} = \frac{CA}{BA}$

This means that the pedal triangle $\triangle QRS$ of the centroid G, $Q \in BC, R \in CA, S \in AB$ is isosceles with QR = QS, iff $2a^2=b^2+c^2$.

[b]Lemma 2:[/b] For any point X not on the circumcircle of the triangle $\triangle ABC$, (not necessarily in the triangle interior), the pedal triangle $\triangle QRS$ of X, $Q \in BC, R \in CA, S \in AB$, is similar to the circumcevian triangle $\triangle MNP$ of X, $M \in AX, N \in BX, P \in CX$.

For example, assuming the point X is in the triangle interior,

$\angle AMN = \angle ABN = \angle SQX$

because the first 2 angles span the same arc AN of the triangle circumcircle and the quadrilateral AQXS is cyclic, having right angles at the vertices Q, S, etc.

[b]Lemma 3:[/b] For any point X not on the circumcircle of the triangle $\triangle ABC$, (not necessarily in the triangle interior), the circumcevian triangle $\triangle MNP$ of X, $M \in AX, N \in BX, P \in CX$, is similar to the triangle $\triangle A'B'C'$ obtained from the triangle $\triangle ABC$ by an inversion with the center X.

The cevians AX, BX, CX also intersect the circumcirle at the points M, N, P. The power of the point X to the circumcircle is

$AX \cdot MX = BX \cdot NX = CX \cdot PX$

while the power of an inversion with the center X is

$AX \cdot A'X = BX \cdot B'X = CX \cdot C'X$

Consequently,

$\frac{MX}{A'X} = \frac{NX}{B'X} = \frac{PX}{C'X}$

As a result, the triangles $\triangle QRS \sim \triangle A'B'C' \sim \triangle MNP$ are similar, which means that they are all isosceles with QR = QS resp. A'B = A'C' resp. MN = MP, iff $2a^2=b^2+c^2$.

[color=blue]A nice and easy proposed problem ![/color] Let $ABC$ be a certain triangle.

$\triangle GPM\sim \triangle GAC\Longrightarrow \frac{MP}{b}$$=\frac{GM}{GC}\ ;\ \triangle GNM\sim \triangle GAB\Longrightarrow \frac{MN}{c}=\frac{GM}{GB}\ .$

$\boxed {\ MP=MN\ }\Longleftrightarrow \frac bc=\frac{GC}{GB}\Longleftrightarrow bm_b=cm_c\Longleftrightarrow$

$b^2\left[2\left(a^2+c^2\right)-b^2\right]=c^2\left[2\left(a^2+b^2\right)-c^2\right]\Longleftrightarrow$

$2a^2(b^2-c^2)=b^4-c^4\Longleftrightarrow \boxed {\ b=c\ \ \vee\ \ b^2+c^2=2a^2\ }\ .$

[b]Generalization.[/b] Given a triangle $ABC$ and a interior point $L(x,y,z)$ (with barycentrical coordinates, $x+y+z=1$). The rays issuing from the vertices of the triangle $ABC$ intersect again the circumcircle in the points $M$, $N$, $P$. Prove that $MN=MP$ iff $p(L)(b^2-c^2)+a^2(zb^2-yc^2)=0$, where $p(L)=-(yza^2+zxb^2+xyc^2)$ is the power of the point $L$ w.r.t the circumcircle.

[u]Indication:[/u] $MN=MP\Longleftrightarrow b\cdot LB=c\cdot LC\ ;$
$LB^2=xc^2+za^2+p(L)\ ;$ $LC^2=ya^2+xb^2+p(L)\ ;$ a.s.o.


I need to answer true false questions and don't have the solutions

A is a set, non empty and bounded
with inf the infimum

$ 1.inf \ f(A)=f(inf\ A)\\
2.f(inf(f^{-1} (A))=inf A$
$ 3.f(A)\subset f(B) \ implies \ A\subset B$
$ 4.f^{-1}(A)\subset f^{-1}(B) \ implies \ A\subset B$

Thanks for the help

1. $ f(x)\equal{}\minus{}x$ and $ A\equal{}[0,a]$, $ a>0$, then $ f(A)\equal{}[\minus{}a,0]$ so $ inf~f(A)\equal{}\minus{}a\neq f(inf~A)\equal{}f(0)\equal{}0$. False.

They're all false except #4.

Of course, this is mathematical true/false: it doesn't count unless you have a proof or a counterexample. Can you produce proofs/counterexamples for these?

And #4 is false too (think of a mapping that sends everything to just one point; then the pre-image of every set containing that point is the entire space...)

Ah, yes. I should have paid attention to the difference between the range and the codomain.


$ : ninja: \left( a\minus{}1\plus{}\frac 1b \right) \left( b\minus{}1\plus{}\frac 1c \right) \left( c\minus{}1\plus{}\frac 1a \right) \leq 1.$

http://www.mathlinks.ro/Forum/viewtopic.php?p=354109#354109
LOok there..and please speak english

Daniel