integration
calculus
function
real analysis
real analysis unsolved

Solve for $ f\in C'(\mathbb{R})$>

$ f(x^\alpha)=f'(x)x^\alpha$

I do not have a full solution, but I can offer the following for thought:

Clearly, $ f \equiv 0$ and $ f(x)=x$ are solutions.  One can expand on the second solution to include $ f(x)=kx$ where $ k$ satisfies $ k^{\alpha-1}=1$.

I am inclined to think that these are the only solutions, but that is just an opinion.

Alternatively, ( :idea: ) it may be useful to rewrite the equation as follows:

$ f(x)= \lambda \int_{0}^{x^{\frac{1}{\lambda}}}\frac{f(u)}{u^{2-\lambda}}\hspace{1 mm}du$

where $ \lambda=\frac{1}{\alpha}$.

This is an integral equation, and there is established theory for these things.  Unfortunately this one looks to be singular and there is no $ g(x)$ being added to the integral.  So maybe those with more experience can chime in and solve this.

Here is one more idea. Everything would be easy if we were looking for a function $ f\in C^{1}([0,1])$ with this property and if we knew that $ \alpha>0$. Then we could run the following argument.

Obviously, $ f(0)=0$. Let $ M=\max_{[0,1]}f'$. Assume that this maximum is attained at some point $ x>0$. Then we would have $ M=f'(x)=\frac{f(x^\alpha)}{x^\alpha}$. On the other hand, $ f(x^\alpha)=\int_{0}^{x^\alpha}f'(t)\,dt\le M x^\alpha$ and the equality can be attained only if $ f'\equiv M$ on $ [0,x^\alpha]$. Thus, we must have $ M=f'(0)$. Similarly, the minimum of $ f'$ over $ [0,1]$ equals $ f'(0)$. Therefore, $ f'$ is constant on $ [0,1]$, so $ f(x)=kx$ with some real $ k$.

I'm not quite sure what the right formulation of the problem should be ($ x^\alpha$ doesn't make sense for $ x<0$ and non-integer $ \alpha$) but, at least, the question makes sense for $ f\in C^{1}([0,+\infty))$ with $ \alpha>0$ and for $ f\in C^{1}((0,\infty))$ and all $ \alpha\in\mathbb R$. Let us see if somebody can solve any of these two versions ;).

Alternatively, require $ \alpha = \frac{m}{n}$ for integers $ m, n$ with $ n$ odd.


A bag contains 7 white, 9 blue, and 4 red marbles. If the marbles are pulled from the bag, what is the probability that two are blue and one is red?

We have three cases: Blue-Blue-Red, Blue-Red-Blue, and Red-Blue-Blue.

For our first case, the probability is $ \frac{9}{20} \times \frac{8}{19} \times \frac{4}{18} \equal{} \frac{4}{95}$.

For the second case, we have $ \frac{9}{20} \times \frac{4}{19} \times \frac{8}{18} \equal{} \frac{4}{95}$.

And the third case is $ \frac{4}{20} \times \frac{9}{19} \times \frac{8}{18} \equal{} \frac{4}{95}$.

Notice anything? This is because each product has the same denominators and numerators, so we could have just done $ \frac{4}{95} \times 3 \equal{} \frac{12}{95}$, which is the final answer.


Can you explain the pattern of the exponents? The first three terms imply powers of 2, but obviously 200 is not a power of 2.

Perhaps he meant \[ \left(1+\frac{1}{a}\right)\left(1+\frac{1}{a^2}\right)\left(1+\frac{1}{a^4}\right)\cdots\left(1+\frac{1}{a^{2^{100}}}\right)  \]

i think i know where this problem came from.
and it is supposed to be \[ 1-\frac 1{a^{2^{100}}} \]
hope that clears things up.
its really sorta easy to solve just think difference of squares.

[quote="maokid7"]i think i know where this problem came from.[/quote]

I vaguely remember seeing it in the Art and Craft of Problem Solving.  Is that what you were thinking of?

No this isn't from ACOps and please hide your thoughts....no solution yet,yeah maokid gave the idea.


The product of two of the four roots of the quartic equation $ x^4\minus{}18x^3\plus{}kx^2\plus{}200x\minus{}1984\equal{}0$ is $ \minus{}32$. Determine the value of [i]k[/i].

[quote="1=2"]The product of two of the four roots of the quartic equation $ x^4 \minus{} 18x^3 \plus{} kx^2 \plus{} 200x \minus{} 1984 \equal{} 0$ is $ \minus{} 32$. Determine the value of [i]k[/i].[/quote]

[hide="Solution Outline"]Factor into $ (x^2 \plus{} ax \minus{}32)(x^2 \plus{}bx \plus{} 62) \equal{} 0.$ Solve for $ a, b$ and expand.[/hide]

Denote the roots by $ a,b,c,d$ and wlog assume $ ab\equal{}\minus{}32$. From $ abcd\equal{}\minus{}1984$ we conclude $ cd\equal{}62$. Substitution into $ \minus{}200\equal{}abc\plus{}abd\plus{}acd\plus{}bcd\equal{}ab(c\plus{}d)\plus{}(a\plus{}b)cd$ yields $ \minus{}200\equal{}\minus{}32(c\plus{}d)\plus{}62(a\plus{}b)$. Combining with $ 18\equal{}a\plus{}b\plus{}c\plus{}d$ and solving for $ c\plus{}d$ we conclude $ a\plus{}b\equal{}4$ and $ c\plus{}d\equal{}14$. Therefore, the quartic polynomial is given by

$ (x\minus{}a)(x\minus{}b)(x\minus{}c)(x\minus{}d)\equal{}[x^2\minus{}x(a\plus{}b)\plus{}ab][x^2\minus{}x(c\plus{}d)\plus{}cd]\equal{}[x^2\minus{}4x\minus{}32][x^2\minus{}14x\plus{}62]\equal{}\ldots\plus{}[\minus{}32\plus{}62\plus{}(\minus{}4)(\minus{}14)]x^2\plus{}\ldots$

i.e. $ k\equal{}86$


algebra
polynomial
inequalities solved
inequalities

If $x,y,z,t \in [-1, \infty)$ and $x+y+z+t=2$ then prove that we have: $x^3+y^3+z^3+t^3 \geq \frac{1}{2}$.

made by Safta I.

We have (x + 1)(2x - 1)2 \geq 0 and the analogous inequalities.
Hence 4x3 \geq 3x - 1 (and the analogous inequalities).
Adding, 4 \sum x3 \geq (3 \sum x) - 4 = 2, which is what we want.

[quote="3X.lich"]In my experience, the polynomial P(x) = 4x 3 -3x+1 has come in handy relating linear terms with cubic terms. Memorize it :)[/quote]

I don't think that this makes sense ... it works only for a very specific case.
It's better to keep in mind the method itself...


Let $ a,b,c>0$
Prove that $ \sum \frac{\sqrt{a^2\plus{}bc}}{b\plus{}c} \geq \frac{3\sqrt{2}}{2}$

[quote="Toan_VN_LC"][quote="fesha3t"]Let $ a,b,c > 0$
Prove that $ \sum \frac {\sqrt {a^2 \plus{} bc}}{b \plus{} c} \geq \frac {3\sqrt {2}}{2}$[/quote]
it's wrong :lol:[/quote]

no, it's right

Vasile solve it.:)
use $ (x\plus{}y\plus{}z)^2 \ge 3(xy\plus{}yz\plus{}zx)$

Try with the harder: $ \sum\ \frac{\sqrt{a^2\plus{}2bc}}{b\plus{}c} \geq\ \frac{3\sqrt{3}}{2}$
And I have had a similar result:
$ \frac{\sqrt{a^2\plus{}256bc}}{b\plus{}c}\plus{}\frac{\sqrt{b^2\plus{}256ca}}{c\plus{}a}\plus{}\frac{\sqrt{c^2\plus{}256ab}}{a\plus{}b} \geq\ 10$
But my proof is so long. Wait for another proofs  :)

[quote="fesha3t"][quote="secrets"]Vasile solve it.:)
use $ (x \plus{} y \plus{} z)^2 \ge 3(xy \plus{} yz \plus{} zx)$[/quote]

oh, ...
what's Vasile, spam[/quote]
Vasile is "vasile critoaje", namely "Vasc", a famous member at the AoPS. He has written the book "Algebraic inequalities". :blush:
His technique must have solved it (according to secrets). But unfortunately, I do not see how. :(

[quote="fesha3t"]Let $ a,b,c > 0$
Prove that $ \sum \frac {\sqrt {a^2 \plus{} bc}}{b \plus{} c} \geq \frac {3\sqrt {2}}{2}$[/quote]
With this one, use Holder ineq, we need to prove:
$ (\sum\ a^2\plus{} \sum\ ab)^3 \geq\ \frac{9}{2}\sum\ (a^2\plus{}bc)^2(b\plus{}c)^2$  (1)
And pqr works here, of course, not so nice (notice that the equality of (1) occurs when $ a\equal{}b\equal{}c$ or $ a\equal{}b;c\equal{}0$) :wink:

[quote="nguoivn"][quote="fesha3t"]Let $ a,b,c > 0$
Prove that $ \sum \frac {\sqrt {a^2 \plus{} bc}}{b \plus{} c} \geq \frac {3\sqrt {2}}{2}$[/quote]
With this one, use Holder ineq, we need to prove:
$ (\sum\ a^2 \plus{} \sum\ ab)^3 \geq\ \frac {9}{2}\sum\ (a^2 \plus{} bc)^2(b \plus{} c)^2$  (1)
And pqr works here, of course, not so nice (notice that the equality of (1) occurs when $ a \equal{} b \equal{} c$ or $ a \equal{} b;c \equal{} 0$) :wink:[/quote]

I'm  stupid, so, can you continue?!
i'm Sorry

With this first ineq of Vasc, you can see the proof of can_hang for the stronger in PhamVanThuan's book:
$ \sum\ \sqrt{\frac{a^2\plus{}bc}{b^2\plus{}bc\plus{}c^2}} \geq\ \sqrt{6}$
And with the harder:
$ \sum\ \frac{\sqrt{a^{2}\plus{}2bc}}{b\plus{}c}\geq\ \frac{3\sqrt{3}}{2}$
can_hang and me had a nice proof by Am-Gm and CS  :)
But with the last ineq:
$ \frac{\sqrt{a^{2}\plus{}256bc}}{b\plus{}c}\plus{}\frac{\sqrt{b^{2}\plus{}256ca}}{c\plus{}a}\plus{}\frac{\sqrt{c^{2}\plus{}256ab}}{a\plus{}b}\geq\ 10$
I haven't found any nice proof for it.  :oops:

I have 1 problem:
Find K( in Z ) max such that INEQ 's right with a,b,c.
Let$ a,b,c > 0$and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$
$ \sum \frac {a}{\sqrt [k]{b \plus{} c \minus{} a}} \geq 3$


function
limit
integration
geometry
real analysis
real analysis unsolved

Does there exist such function $ f$ that $ f$ is continuous on $ (0, 1]$, positive and satisfies following conditions:
$ i$) $ \displaystyle\lim_{\varepsilon\to \plus{} 0}\displaystyle\int^{1}_\varepsilon f(x)\,dx$ is finite;
$ ii$) For every $ n\in \mathbb{N}$ holds: $ \overline{\displaystyle\lim_{x\to \plus{} 0}} x^n f(x) \equal{} \plus{} \infty$.

Thanks.

Of course. One way to substantiate this claim is to consider a suitably tuned arrangement of impulse-like sharp bumps. Here we demonstrate a specific construction: Let $ w$ be a bump function defined by

$ w(x) \equal{} \begin{cases} e^{\minus{}cx^2 / (1\minus{}x^2)} & |x| < 1 \\ 0 & \text{otherwise} \end{cases}$

where $ c$ is chosen to normalize the total area under $ w$. Choose a strictly decreasing sequence $ x_n$ convergent to 0. Now choose each $ d_n > 0$ sufficiently small that $ x_{n\plus{}1} \plus{} d_{n\plus{}1} < x_n \minus{} d_n$. Then define

$ w_n(x) \equal{} \frac{1}{d_n} w\left( \frac{x\minus{}x_n}{d_n} \right)$.

Intuitively, all we did is make a copy of the original bump, slide it close to the origin, stretch it upward while not changing its total area, and repeat this process using it as a new prototype. Under this construction, the function

$ f(x) \equal{} \sum_{n\equal{}1}^{\infty} a_n w_n (x)$

is continuous on $ (0, \infty)$ for any choice of $ a_n > 0$ by the observation that there are only finite summands which survive on $ (\varepsilon, \infty)$ for any $ \varepsilon > 0$. The followings are also easy consequences of this observation:

(1) $ \lim_{\varepsilon \to \plus{} 0} \int^{1}_{\varepsilon} f(x) \, dx \leq \sum_{n\equal{}1}^{\infty} a_n$

(2) $ \limsup_{n\to\infty} \frac{a_n x_n^k}{d_n} \leq \limsup_{x \to \plus{}0} x^k f(x)$

Now all that remains is to make clever choices to $ x_n, d_n$ and $ a_n$ which meet the conditions. For example, $ x_n \equal{} 2^{\minus{}n}$, $ d_n \equal{} 1/(4n)!$ and $ a_n \equal{} n^{\minus{}2}$ will work.


let $f:R...R $such that for x,y member of R we have
$f(x^3+y^3)=(x+y)(f(x)^2-f(x)f(y)+(f(y))^2)$
prove that $f(1999x)=1999f(x)$

we have $f(nx)=nf(x)$ for all $ n \in\ N \ $  ;)

Well, how did you got that. Please give some strong reason for what you say.


hello, we have $ \frac{6!\minus{}5!}{4!}\equal{}\frac{5!\cdot6\minus{}5!}{4!}\equal{}\frac{5!(6\minus{}1)}{4!}\equal{}\frac{5!\cdot5}{4!}\equal{}\frac{4!\cdot5\cdot5}{4!}\equal{}25$
Sonnhard.

This is $ \frac{720 \minus{} 120}{24}$, which simplifies to $ \frac{600}{24}$, and ultimately, the answer is $ \boxed{25}$.

EDIT: Beaten.

Note how well Sonnhard's solution generalizes.  What if we had $ \dfrac{40!\minus{}39!}{38!}$?

[quote="AIME15"]Note how well Sonnhard's solution generalizes.  What if we had $ \dfrac{40! \minus{} 39!}{38!}$?[/quote]

But we didnt, did we?

But in the future, please use Dr. Sonnhard's way.

hello, by the same procedure as above we have
$ \frac{40!\minus{}39!}{38!}\equal{}\frac{39!\cdot40\minus{}39!}{38!}\equal{}\frac{39!(40\minus{}1)}{38!}\equal{}\frac{39!\cdot39}{38!}\equal{}\frac{38!\cdot39\cdot39}{38!}\equal{}39^2\equal{}1521$.
Sonnhard.


Let $ A$ represent the set of all functions $ f : \mathbb{N} \rightarrow \mathbb{N}$ such that for all $ k \in \overline{1, 2007}$, $ f^{[k]} \neq \mathrm{Id}_{\mathbb{N}}$ and $ f^{[2008]} \equiv \mathrm{Id}_{\mathbb{N}}$.
a) Prove that $ A$ is non-empty.
b) Find, with proof, whether $ A$ is infinite.
c) Prove that all the elements of $ A$ are bijective functions.
(Denote by $ \mathbb{N}$ the set of the nonnegative integers, and by $ f^{[k]}$, the composition of $ f$ with itself $ k$ times.)

Let's consider any subset $ E \equal{} \{x_1, \cdots, x_{2008} \}$ of $ \mathbb{N}$ which contains exactly $ 2008$ elements.
Define $ f(x_i) \equal{} x_{i\plus{}1}$ for $ i\equal{}1, \cdots, 2008$ (where $ x_{2009} \equal{} x_1$), and $ f(x) \equal{} x$ for each $ x \notin E$.
Clearly $ f \in A$.
This proves a) and b).
c) is trivial since $ f^{[2008]}$ is bijective...

Pierre.


For reals [b]x,y,z>0[/b]:

$x^xy^yz^z\geq (xyz)^{\frac{x+y+z}{3}}$

posted many times, see for example [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=47354[/url]

 :)


In $\triangle ABC$, $AC$ and $BC$ are extended, and the circle tangent to these extended lines and $BC$ has radius 4. What is the perimeter of $\triangle ABC$?

A. $4 \sqrt 3$
B. $6 \sqrt 3$
C. $8 \sqrt 3$
D. 12
E. 14

equilateral triangle?

[hide]the identity: $\frac{1}{r_A}+\frac{1}{r_B}+\frac{1}{r_C}=1/r$ where the $r_A$ is the radius of the excircle opposide of angle A.

we know $r_A=r_B=r_C=4$
so $r=\frac{4}{3}$

from there we find the perimeter of $\triangle ABC$ to be $8\sqrt{3}$[/hide]

[quote="Nosepuppy"][hide]from there we find the perimeter of $\triangle ABC$ to be $8\sqrt{3}$[/hide][/quote]

How? :?

I think 4/3 is the inradius of the triangle.

And making a small 30-60-90 triangle, with the radius being the shortest leg and half a side being the longer leg, you can see that half a side is $4\sqrt{3}/3$, so all 3 sides is 6 times that, or $8\sqrt{3}$

Here's how I did it tho:

Draw the perpendiculars to AC and BC from the outer radius. Let the midpoint of BC be D.
Let the center of circle be O. ODC is a 30-60-90 triangle with longer leg 4, and shorter leg which is half of a side. Using the fact that its 30-60-90, that half of side is $4/\sqrt{3}$.

Just an exradius of an equilateral triangle....

Exradii are one of those peculiar things that, when they come in handy in an Olympiad problem, they come in VERY handy. Good intuition and general skills with recognizing and manipulating exradii are pretty helpful.

Exradius? :? *looks it up on MathWorld* :idea: http://mathworld.wolfram.com/Exradius.html :huuh: cool :)


linear algebra
matrix
linear algebra unsolved

First, the easier one : How do we find that whether there exists a characteristic matrix or not which is equivalent to a matrix, say A? Is that way the only one? Or what is the most convenient way to use in order to answer this question?

Second, the harder question : If A is an nxn complex matrix, then there exist a diagonalizable matrix D and a nilpotent matrix N such that  A = D + N , and  DN = ND. How can we prove this? By the way, are these matrices D and N uniquely determined?

Are you familiar with Jordan normal forms?

Yes, i can say that i'm familiar with canonical forms though i'm not that good at these kind of upper-intermediate problems.


algebra
polynomial
Gauss
superior algebra
superior algebra theorems

Any cyclotomic polynomial $P_{n}$ (of degree $\varphi(n)$) is irreducible over $\mathbb{Z}$. If $n$ is prime, there is an easy proof using Eisenstein's criterion. For arbitrary $n$ there is an earlier but harder proof by Gauss, which can be seen in Lang's [i]Algebra[/i] and in various other books, and here:
http://www.mathreference.com/fld-cyc,pirr.html
Is any other proof known? I suspect that Gauss's proof can be improved.

We had this proof in class a two days ago and I found it by far the most impressive proof of the irreducibility of the cyclotomic polynomial. Really, there's nothing to be improved there, it's just perfect! There's another short proof using Dirichlet's theorem about primes in arithmetic progressions, but this one is by far the easiest and most beautiful one.


suppose $ X$ and $ Y$ are two topological space
is there any extra condition on the spaces or function that any function $ f : X \rightarrow Y$ that preserve connectedness and compactness should be countinous?

Continuous functions preserve connectedness and compactness, but (apart from this) I don’t think there’s any other relation between continuousness and connectedness/compactness. Continuous functions can exist between non-connected or non-compact spaces (e.g. a function that maps everything to a single point is always continuous, regardless of what spaces it’s mapping to and from). It is also possible for non-continuous functions to exist between two connected spaces or two compact spaces (i.e. some non-continuous functions can also “preserve” connectedness and compactness).

i think the following could be a counterexample: consider a partition $ A\cup B$ of the set of rational numbers into two dense (dense in the real line, with the euclidean topology) subsets.
the rationals, with the topology inherited as a subspace of the real line with the euclidean topology, are a totally disconnected subspace, so connectedness is preserved by any function from $ \mathbb{Q}$ to any space. on the other hand, defining $ f: \mathbb{Q}\to\mathbb{R}$ as $ 0$ on $ A$ and $ 1$ on $ B$, we have that also compactness is preserved.

but this function is not continuous, since it's not the restriction of any continuous real-valued functions from the reals.

Well, you could also try this example. Let $ S\equal{}\{0,1\}$ and $ X$ and $ Y$ be the following topological spaces: $ X$ is $ S$ with the indiscrete topology $ \{\O,S\},$ and $ Y$ is $ S$ with the topology $ \{\O,\{1\},S\}.$

Then $ X$ and $ Y$ are both connected and compact, so any function between them can be regarded as “preserving” their connectedness and compactness. However the identity mapping on $ S$ is not a continuous function from $ X$ to $ Y.$

you're right. anyway, my counterexamples wins on hypothesis on $ X,Y$ like hausdorff/metrizable or second countable. examples on the same line should show that not even compactness+metrizability+second countable is enough.

i wonder if metrizability+completeness is enough..

you are right.

its easy to find a family of function that are not continous but preserve compactness and connectedness.

but i want to find some condition on spaces and function that preserving  compactness and connectedness
presuade f to be continous.

i think both condition on function and spaces are needed ,for example(a trivial one)if $ Y$ is hausdorff and $ X$ is compact and $ f$ is bijection .

but " preserving the connectedness" is not easy to work(for me)


modular arithmetic
number theory unsolved
number theory

Prove that
\[ 2^{\frac {p \minus{} 1}{2}} mod p \equal{} {( p \minus{} 1)} ^ {\frac {p^2 \minus{} 1}{8} }mod p
\]
where p is a prime.

$ p \not \equal{}2$

We have

$ 2^{\frac{p\minus{}1}{2}}\equiv \bigg( \dfrac{2}{p} \bigg) \pmod p$

and

$ (p\minus{}1)^{\frac{p^2\minus{}1}{8}} \equiv (\minus{}1)^{\frac{p^2\minus{}1}{8}} \equal{} \bigg ( \dfrac{2}{p} \bigg) \pmod p$.

So the relation is true.


I think some people have already created a game similar to this, but I guess I'll start another thread.

[hide="Basic Instructions"]
There will be 15 people in this game. First come First serve. I will start out by saying the word: DOCK. The next person that signed up first will post a new word, but taking out only one letter in the word that I said and changing it into another word. So for example: I say Dock, he says mock. Get it? But if someone cannot think of an answer within 2 days, he is disqualified. If the next person in line does NOT answer within 2 days, he will also be disqualified. (Probably me...)  :P     So it's pretty simple.
[/hide]

[hide="Advanced Instruction"]
If we eliminate 8 people, then there would be 8 people left right? (Counting me.) So then, if I were still alive, then I would pick a 5-letter word to start with, and we begin the game again. If I were not alive, then I would randomly pick someone to choose a 5-letter word. And if we got down to 2 people left, it's the same steps as the 8 people, but this time someone else or me has to choose a 6-letter word! (Poor grammar.)  :ninja:  :huh: [/hide]

[hide="Sign Ups"]
1.Me. Jonathanchou711
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
[/hide]

I'm in!

The message is too small. Please make the message longer before submitting.

Um.. Anyone else?  :maybe:  This is a fail.

bump for jonathonchou......

I'd like to play in lots of different games, but I'll just get dq-ed in all of them because I'll be away from a computer for 1.5 weeks.

Dudes. Does like everyone hate me? Cause I've seen games like this and people sign up in like 1 day. And this? 500 millenniums and counting.

Hm... Ok. So there's

1. yaofan
2. bluecarneal
3. weird888
4. jonathanchou711
5. mathforlife
6. auroraak.

Let's just get 2 more people.


geometry
3D geometry
geometric transformation
reflection
geometry unsolved

A ray of light is issued from a point interior to a given cone. All the internal surface of this cone can reflect the light.
Prove that after a finite number of reflections, the ray will never meet the cone again.

Pierre.

this is a pretty common techinque for this sort of problem...

denote by $ C_1 $ the initial cone and by $ V $ its vertex.  let the ray first hit the cone at point $ P_1 $ (if it does hit the cone at all).  reflect $ C_1 $ in the tangent plane to $ C_1 $ at the point $ P_1 $ to get the cone $ C_2 $.  note that the vertex of $ C_2 $ is also $ V $.  extend the ray outwards from the cone $ C_1 $ to hit the cone $ C_2 $ at $ P_2 $.  now do the same thing at point $ P_2 $: reflect $ C_2 $, etc., to get cone $ C_3 $, whose vertex is also $ V $; and so on.  observe that the points $ P_i $ correspond to points in our original cone, $ C_1 $, where the ray hits it.

now, note that the angles $ P_iVP_{i+1} $ are all equal, and that, placing enough of them successively around the point $ V $ (in this sort of "fan"), at one point the last angle placed will be so that the ray won't intersect its outer "leg"; this corresponds to the ray not hitting our original cone...


For every $0 \leq a, b, c < \frac{1}{\sqrt{3}}$ prove the inequality

$\frac{a+b}{1-ab} + \frac{b+c}{1-bc} + \frac{a+c}{1-ac} \leq 2 \frac{a+b+c-abc}{1- ab-bc-ca}$ .

[quote="KMN"]For every $0 \leq a, b, c < \frac{1}{\sqrt{3}}$ prove the inequality

$\frac{a+b}{1-ab} + \frac{b+c}{1-bc} + \frac{ac}{1-ac} \leq 2 \frac{a+b+c-abc}{1- ab-bc-ca}$ .[/quote]

   I think by putting: a=tanx,b=tany and c= tanz...we can reduce the given inequality to the simple form ...
the condition : $0 \leq a,b,c \leq \frac{1}{ \sqrt{3}}$ gives us: $0 \leq x,y,z \leq \frac{ \pi }{3}$
From my opinion  it should be a simplest way ... :P

Please post full :P  solutions, and not only hints that are confusing


Prove the Cassini's identity  $ ?$

$ \sum_{k\equal{}1}^{n}\binom{n}{k}2^kf_k\equal{}f_{3n}$


[img]http://www2.hkedcity.net/sch_files/a/lpl/lpl-lsk/public_html/cake2.gif[/img]
As shown in the figure, ABCD is a cake in the shape of a square with AB = 10 inches. The cake will be cut into pieces along straight lines  AE, BF, CG and DH. Besides, AH = BE = CF = DG.
If it is required that the area of square PQRS should be 50 square inches, how long should  AH be chosen?

This looks strangely similar to 1985 AIME question 4...

I tried to sovle it  and my answer is [hide="this"]$AH=40-20\sqrt{3}$[/hide]. Tell me if is correct so I can post the entire solution :)

Funny because I got exactly half that.

[hide]Let $SC$ be $b$, $SD$ be $a$, and $CD$ be $c$. Then, $SR$ must be $b - a$
$\triangle FRC$ is similar to $\triangle DSC$ so to find $FR$ we set up proportions

$\frac ab = \frac x{a}$

$x = \frac {a^2}{b}$

Then, solving for $FC$ we have

$\sqrt{\frac {a^4}{b^2} + a^2} = FC$

$\sqrt {\frac {a^2(a^2 + b^2)}{b^2}} = FC$

$\sqrt {\frac {100a^2}{b^2}} = FC$

$FC = \frac {10a}{b}$

Solving for $b$

$a = \frac {25}{b}$

$b^2 + \frac {625}{b^2}=100 \Rightarrow b^4 - 100b^2 +625 = 0$

$b = \sqrt {50 + 25 \sqrt 3}$

Substituting into $a^2 + b^2 = 100$ yields $a = \sqrt {50 - 25 \sqrt 3}$
Plugging into $\frac {10a}{b}$ we have

$\frac {10\sqrt {50 - 25 \sqrt 3}}{\sqrt {50 + 25 \sqrt 3}}$

Rationalizing give us the final answer $20 - 10\sqrt 3$[/hide]

Well I did it [hide="this way"]Like in the picture the area of $ABCD$ is equal to $2A_{AEB}+2A_{APSG}+A_{PQRS}$.  We know that $A_{PQRS}=50$. I noted $AH=x$ so $A_{AEB}=5x$. (The $A$ notation means Area)

From the 2 trapezium $APSG$ and $QECR$ I built an rectangular which has also the notation $PQRS$ but it's lenghts are:$PS=RQ=5\sqrt{2}$ and $RS=PQ=GS+RC=AP+QE$. But from the triangle $ABE$ we have $EB=AH=x$ and $AB=10$ and because it's a right triangle $AE=\sqrt{100+x^2}$.

And $AE=PQ+(AP+QE)\Rightarrow AP+QE=AE-PQ=\sqrt{100+x^2}-5\sqrt{2}=RS$. Now we know the lenghts or the builted rectangular so it's area is $PS\cdot RS=5\sqrt{2}(\sqrt{100+x^2}-5\sqrt{2})=5\sqrt{200+2x^2}-50$

Now returning to the first relation $A_{ABCD}=2A_{AEB}+2A_{APSG}+A_{PQRS}\Longleftrightarrow$
$100=2A_{AEB}+ \text{Area-of-the-builted-rectangular} +50\Longleftrightarrow 50=10x+5\sqrt{200+2x^2}-50\Longleftrightarrow 100-10x=5\sqrt{200+2x^2}$

$20-2x=\sqrt{200+2x^2}\Longleftrightarrow 400-80x+4x^2=200+2x^2$ or else $x^2-40x+100=0$. The solutions are $x_{1}=\frac{40+20\sqrt{3}}{2}$ but it's grater that $AB$ :D and $x_{2}=\frac{40-20\sqrt{3}}{2}=20-10\sqrt{3}$

So tarquin it's right. :D:D. :oops:

Last night maybe I was tired :P. Indeed it's a nice problem :clap: [/hide]


Express as a mixed number the mean of all whole numbers from 1 to 100, inclusive.

we see that every pair of the nth term and the (100-n+1)th term has the same average, because these are consecutive numbers. this average is (101+n-n)/2=101/2=50 1/2


trigonometry function algebra polynomial

Here is some trig problems that look pretty basic but require a bit of work: Enjoy! (Reminder:I am looking for solutions not just answers)

1) $4 \arctan{X}+\arctan{ \frac{1}{239}}= \frac{ \pi }{4}$

Find X (Note this question has stumped a lot of people try at your own risk)

2) Find the value of this fraction at $X =\frac{ \pi }{12}$

 $\frac{ ( \tan{X})^{4}+2 ( \tan{X})^{2}+1 }{ \sec{X}^{2}}$

I believe Farenhajt's solution for number 2 is correct

[quote="Farenhajt"]Two errors: $+$ should be $-$ and $\theta$ should be $\pi$.

And it's posted here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=109619[/url][/quote]

Nice try but sorry I typed them correct; they are from a 1996 Vestavia Hills Mathematics Tournament Comprehensive Written Test. Also the test is no calculators, so keep the computation low.

I actually did consider to put these problems on the Intermediate or Pre-Olympiad level but decided against it considering these questions were designed for Pre-Calc students and anyone with a fair understanding of Trig could solve them.  But if the moderators reading this post feel this problem is too hard feel free to move this post.

they are right hes refering to $pi$ as radians and its pretty basic if you know trigonometry identities

[quote="me@home"]Cmon, somebody plug that into mathematica! :P (I don't own it...)[/quote]

 :D  With Mathematica, the roots are:
$x_{1,2}= \frac{-17 \pm 13 \sqrt{2}}{7}, \ \ x_{3,4}= \frac{7 \pm 13 \sqrt{2}}{17}$.

Though, it doesnt make sense to have four roots, since the $arctan$ function is one-to-one...

(edit: evaluating each, only $x_{1}= \frac{-17+13 \sqrt{2}}{7}$ works. Is there an easier way to do this?)

Can you use mathematica to find the lowest degree polynomial which has a given root? If you can, check to see if $x_{1}= \frac{-17+13 \sqrt{2}}{7}$ satisfies something nicer...
edit: by hand I found that $x_{1},x_{2}$ are the roots of $7x^{2}+34x-7$ (if I didn't mess up...)

[quote="Kolumbus"][quote="lotrgreengrapes7926"]And then you get the quartic $119x^{4}+480x^{3}-714x^{2}-480x+119=0$.  :maybe: [/quote]

Substitute $u: =x-1/x$. Then $x^{2}+1/x^{2}=u^{2}+2$ and so $119x^{2}+480x-714-480x^{-1}+119x^{-2}=0$ gives $0=119(u^{2}+2)+480u-714=119u^{2}+480u-476$ and solving this for $u$ we get $u\in\left\{-\frac{34}7,\frac{14}{17}\right\}$.
From $x-1/x=u=-\frac{34}7$ it follows that $x=\frac{-17\pm13\sqrt{2}}{7}$, whereas from $x-1/x=u=\frac{14}{17}$ it follows that $x=\frac{7\pm13\sqrt{2}}{17}$.
Since $\arctan x=\frac{\frac\pi4-\arctan\frac1{239}}4\in[0,\frac\pi{16}]$and $\frac{-17+13\sqrt{2}}7$ is the only root in $[0,1]$, we must have $x=\frac{-17+13\sqrt{2}}7$.[/quote]

Nice solution!

Is there a way to determine what substitutions will work to reduce a given polynomial, like the one above?

[quote="D.P.L"]Correct me if im wrong but the tan function has a period of $\pi$[/quote]

Yes, but the $\arctan$ (or $\tan^{-1}$) function is one-to-one, so if we have $\arctan{X}=\frac{ \frac{\pi}{4}-\arctan \frac{1}{239}}{4}$, then there's only one $X$ that will satisfy the equation (if the RHS is between $\frac{-\pi}{2}\text{ and }\frac{\pi}{2}$, which it is).

You could try to find the formula for tangent of 4 times an angle and then use angle addition from there.
Also, once again this test was administered to pre-calc students without calculators in 1 hour time frame keep the computation calc-free, low, and efficient.

[hide="hint"]
$\tan{4 \theta}= \frac{4 \tan{ \theta}-4( \tan{ \theta})^{3}}{1-6( \tan{ \theta})^{2}+( \tan{ \theta})^{4}}$ [/hide]

are you sure the first problem is correct?

i think it is $4\tan^{-1}x-\tan^{-1}\frac{1}{239}=\frac{\pi}{4}$

then it works out nicely

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=239&t=68487

Some possible approaches for the first problem.

Luckily, our tournament is multiple choice (however, there is an option for none of the above)

Somehow I think that might have been a misprint or typo and they wanted it to be the famous identity.

On the other hand, this test was most likely written by Mathew Crawford himself, so he might have been sadistic enough to intend the plus sign to be there.

Someone should send a PM off to him and ask.

[quote="Elemennop"]Somehow I think that might have been a misprint or typo and they wanted it to be the famous identity.
[/quote]

Which famous identitiy is it? Does anyone know the official name? Why is it so famous?

It also helps if you know the trig function values at $\frac{\pi}{12}$

Hey guys stoppped all your work! (Like anyone is still troubling over this problem  :lol: )

It turns out that there was a type-o on the test and it is a famous formula by John Machin

It is:
$\frac{ \pi}{4}= 4 \arctan \frac{1}{5}-\arctan \frac{1}{239}$

Thanks for all the work anyway!  :)

Wikipedia source:
[url]http://en.wikipedia.org/wiki/John_Machin[/url]

MathWorld Source:
[url]http://mathworld.wolfram.com/MachinsFormula.html[/url]

Which was pointed at way back in Post #2 :P

Yeah kind makes you feel bad doesn't it?  :blush:

sorry to bring this back but,
i dont get why number 2 is the answer that it is...can someone explan to me why the answer is $4(2-\sqrt{3})$-4th post from top-and there have been no other answers to dispute that but i dont get that answer.

(because the problem is $\sec x^{2}$, not $(\sec x)^{2}$)


induction
strong induction
combinatorics unsolved
combinatorics

Define a sequence $ < a_n > _{n\geq0}$ by $ a_0 \equal{} 0$, $ a_1 \equal{} 1$ and
\[ a_n \equal{} 2a_{n \minus{} 1} \plus{} a_{n \minus{} 2},\]
for $ n\geq2.$

$ (a)$ For every $ m > 0$ and $ 0\leq j\leq m,$ prove that $ 2a_m$ divides
$ a_{m \plus{} j} \plus{} ( \minus{} 1)^ja_{m \minus{} j}$.

$ (b)$ Suppose $ 2^k$ divides $ n$ for some natural numbers $ n$ and $ k$. Prove that $ 2^k$ divides $ a_n.$

$ (a)$

Like $ 2a_{m}\equal{}a_{m\plus{}1}\minus{}a_{m\minus{}1}$, then we can prove by induction that $ 2a_{m}|a_{m\plus{}j}\plus{}(\minus{}1)^ja_{m\minus{}j}$

Suppose that it's true for all $ j \le k$

$ 2a_{m}|2a_{m\plus{}k}\plus{}2(\minus{}1)^ka_{m\minus{}k}\equal{}a_{m\plus{}k\plus{}1}\minus{}a_{m\plus{}k\minus{}1}\plus{}(\minus{}1)^ka_{m\minus{}k\plus{}1}\plus{}(\minus{}1)^{k\plus{}1}a_{m\minus{}k\minus{}1}$

$ 2a_{m}|a_{m\plus{}k\plus{}1}\plus{}(\minus{}1)^{k\plus{}1}a_{m\minus{}k\minus{}1}\minus{}(a_{m\plus{}k\minus{}1}\plus{}(\minus{}1)^{k\minus{}1}a_{m\minus{}k\plus{}1})$

$ \implies 2a_{m}|a_{m\plus{}k\plus{}1}\plus{}(\minus{}1)^{k\plus{}1}a_{m\minus{}k\minus{}1}$

$ (b)$
We can prove by induction that $ a_{d\plus{}e}\equal{}a_{d\minus{}1}a_{e}\plus{}a_{d}a_{e\plus{}1}$

Then we can prove that if $ d|e$ then $ a_d|a_e$.

So if we prove that $ 2^k|a_{2^k}$, then the problem is solved because $ a_{2^k}|a_n$

From $ a_{d\plus{}e}\equal{}a_{d\minus{}1}a_{e}\plus{}a_{d}a_{e\plus{}1}$, for $ d\equal{}e$ we have:
$ a_{2d}\equal{}a_{d}(a_{d\minus{}1}\plus{}a_{d\plus{}1})\equal{}2a_{d}(a_{d\minus{}1}\plus{}a_{d})$

Then $ 2a_d|a_{2d}$, so like $ a_2\equal{}2$ we can prove by another induction that $ 2^k|a_{2^k}$

By the way guys, how many questions do you think one needs to do or the minimum marks required to get through the INMO?

[hide="A trickle of a Solution (Due to antiparticle)"]Part one follows by strong induction on $ j$.

Part two follows by plugging in $ j \equal{} m$ in part one followed by strong induction on $ v_2(n)$.[/hide]

we get from the theory of Difference equations that
\[a_n=\frac{(1+\sqrt{2})^n-(1-\sqrt{2})^n}{2\sqrt{2}}\]
let $n=2^kj$
we want to prove that \[\frac{(1+\sqrt{2})^{2^kj}-(1-\sqrt{2})^{2^kj}}{2^{k+1}\sqrt{2}}\] is an integer
the expression is equal to \[\frac{(1+\sqrt{2})^{2^{k-1}j}+(1-\sqrt{2})^{2^{k-1}j}}{2}.\frac{(1+\sqrt{2})^{2^{k-2}j}+(1-\sqrt{2})^{2^{k-2}j}}{2}....................\frac{(1+\sqrt{2})^j-(1-\sqrt{2})^j}{2\sqrt{2}}\]
\[=\frac{(1+\sqrt{2})^{2^{k-1}j}+(1-\sqrt{2})^{2^{k-1}j}}{2}.\frac{(1+\sqrt{2})^{2^{k-2}j}+(1-\sqrt{2})^{2^{k-2}j}}{2}....................a_j\]
we can expand the terms of the form $\frac{(1+\sqrt{2})^x+(1-\sqrt{2})^x}{2}$.....(i) to get\[\frac{\displaystyle\sum\limits_{i=0}^x\binom{x}{i}2^{\frac{i}{2}}+\displaystyle\sum\limits_{i=0}^x\binom{x}{i}(-1)^i2^{\frac{i}{2}}}{2}\]
note that terms of the form $\binom{x}{2k-1}2^{\frac{2k-1}{2}}$ are the only terms which are not integers and they cancel out,leaving terms divisible by two and $\binom{x}{0}+\binom{x}{0}$.so we conclude that all the terms of the form (i) are integers and their product multiplied by $a_j$ is also an integer.
                                                                                $QED$

[quote=shoki]a)as gilcu3  did, we use induction.
b)...
we have $ a_n\equal{}\frac{1}{2\sqrt{2}} \left((1\plus{}\sqrt{2})^n\minus{}(1\minus{}\sqrt{2})^n \right)$
or in other words $ a_n\equal{}\sum_{i\equal{}1}^{[\frac{n\plus{}1}{2}]} \binom{n}{2i\minus{}1}2^{i\minus{}1}$
we know that $ 2^k \parallel{} n$
and we want to prove that $ 2^k | a_n$.(in fact $ 2^k \parallel{}a_n$, as well).
anyhow, we have $ n\equal{}2^kr$
and so we get for any $ 1 \le t \le [\frac{n\plus{}1}{2}]$
$ v_2(\binom{n}{2t\minus{}1}2^{t\minus{}1})\equal{}v_2(\binom{2^kr}{2t\minus{}1})\plus{}v_2(2^{t\minus{}1})$

$ \equal{}\sum_{i\equal{}1}^{\infty} \left([\frac{2^kr}{2^i}]\minus{}[\frac{2^kr\minus{}(2t\minus{}1)}{2^i}]\minus{}[\frac{2t\minus{}1}{2^{i}}] \right)\plus{}(t\minus{}1)$

$ \ge \sum_{i\equal{}1}^{k} \left([\frac{2^kr}{2^i}]\minus{}[\frac{2^kr\minus{}(2t\minus{}1)}{2^i}]\minus{}[\frac{2t\minus{}1}{2^{i}}] \right)\plus{}(t\minus{}1)$

$ \equal{}\sum_{i\equal{}1}^{k} \left(\minus{}[\frac{\minus{}(2t\minus{}1)}{2^{i}}]\minus{}[\frac{2t\minus{}1}{2^{i}}] \right) \plus{}(t\minus{}1)\equal{}\sum_{i\equal{}1}^{k}( 1 )\plus{} (t\minus{}1)\equal{}k\plus{}(t\minus{}1)$

so for each $ t$ the product $ \binom{n}{2t\minus{}1}2^{t\minus{}1}$ is divisible by $ 2^k$ and if $ t>1$ then it is divisible by $ 2^{k\plus{}t\minus{}1}$ and if $ t\equal{}1$,then obviously it is exactly divisible by $ 2^k$. and so we have $ 2^k \parallel{} a_n$.[/quote]

How did you even get that closed form?

The part one is killed by (strong) induction.
For the part 2, set $m=2^{k}$ and induct on $k$. The base case is true, as $a_{2}=2$. Now assume that it is true for $m=2^{j}$. Then by the part one,
$$2^{j+1}|2a_{m}|a_{m+m}+(-1)^{m}a_{m-m}=a_{2m}=a_{2^{j+1}}$$
and hence the induction is complete.

The second part can also be proved independently.
Firstly, an easy induction shows that the parity of $a_{k}$ is the same as that of $k$ for all $k \in N$.
Next, we claim that $2a_{m}|a_{2m}$ for all $m \in N$. To prove our claim, note that
$$a_{2m}=2a_{2m-1}+a_{2m-2}=5a_{2m-2}+2a_{2m-3}=12a_{2m-3}+5a_{2m-4}=...$$
This recursive process shows that $a_{2m}=a_{j+1}a_{2m-j}+a_{j}a_{2m-j-1}$ for all $1 \le j \le 2m$. Set $j=m$ to get
$$a_{2m}=a_{m+1}a_{m}+a_{m}a_{m-1}=a_{m}(a_{m+1}+a_{m-1})$$
And since $a_{m+1}$ and $a_{m-1}$ have the same parities, their sum is divisible by 2. Thus, the RHS is divisible by $2a_{m}$, and hence $2a_{m}|a_{2m}$, as claimed.
The result now immediately follows  :-D


Yes, I'm wondering if there is a program I can download for free that generates Algebra/Intermediate Algebra problem.

I'm hoping that the program breaks the problems into topics such as Exponents or Graphing. I would also like the problems to be different about 90% of the time. If it has the choice of giving me the answer if I can't figure out the problem, that would be wonderful.

I was wondering if anyone knew of a program that existed.

I notice that when I get done with most of this Intermediate Algebra book that I will need to keep up my skills. To do so, I will need to be tested on problems regulary.

Does anyone know where I can download/get a program like this for free?

such a program seems complex, yet possible. all you need is a program that randomly chooses variables and creates equations. however, the solutions to those equations may get long and ugly. i dont know where you can find a program like this.

You can create a problwm builder by Scientific Workplace, but unfortunately it is not free. By this software you can also type math in WYSIWYG mode (like equation in word) and have the output in Latex. The software also can do algebraic manipulation (they have Maple and Mupad imbedded) and draw graph.

I think you can try the evaluation version for 30 days.

Wasn't the Elias Saab Mathcounts page a generator of Mathcounts problems?

On math.com you can crete algebra problems by entering in guidelines for quad, 2 step, etc.  I think it was under the Algebra section and random problem generator.  You might get some problems twice though.

well, theres alot of problems on AoPS that you could use


geometry
3D geometry
combinatorics proposed
combinatorics

Does there exist an infinite sequence which consists only $ 123$ so that every adjacent blocks are different?(for example:$ 1212$ or $ 11$ are not allowed.

See [url]http://en.wikipedia.org/wiki/Squarefree_word[/url]

These are called square-free words, and one needs an alphabet of at least three letters. With two letters only, The Thue-Morse word is cube-free.

Thank you very much for such a nice link!
I came up with this problem when  dealing with "cubefree" sequence.
But unfortunetly I failed to create a solution.
Now I am wondering if there exists more thinkable construction and if the problem become simpler when replacing $ 123$ by $ 1234$?


What is the smallest possible value of the expression
\[ \sqrt {a^2 \plus{} 9} \plus{} \sqrt {(b \minus{} a)^2 \plus{} 4} \plus{} \sqrt {(8 \minus{} b)^2 \plus{} 16}
\]
for real numbers $ a$ and $ b$?

[hide]$ \sqrt {145}$[/hide]

Thanks in advance!

[EDIT] Should this be in the Intermediate Forum?

Consider each as the hypotenuse of a right triangle. Then the sum is equal to the sum of 3 segments which is minimized when the segments are all on one line (since the distance between the top right point and bottom left point is fixed, apply the triangle inequality a bunch). This happens when the triangles are all similar. Then (see diagram) we have a right triangle with side lengths $ 3 \plus{} 2 \plus{} 4 \equal{} 9$ and $ a \plus{} b \minus{} a \plus{} 8 \minus{} b \equal{} 8$ so the hypotenuse is $ \sqrt {145}$. This can happen, by setting up proportions to represent similarity and solving a linear system we find that $ b \equal{} 40/9$ and $ a \equal{} 8/3$

Sorry for the bad diagram.

EDIT: Here's the same solution except written more algebraically. Let $ x\equal{}a\plus{}3i$, $ y\equal{}(b\minus{}a)\plus{}2i$, $ z\equal{}(8\minus{}b)\plus{}4i$. Then Our sum$ \equal{}|x|\plus{}|y|\plus{}|z|\geq |x\plus{}y\plus{}z|\equal{}|8\plus{}9i|\equal{}\sqrt{145}$. And as above, equality occurs (set the argument equal).

Why is it sqrt145? :?:
we can get a smaller value of this expression by setting a=b=0.

Did you forget to square that 8-b in there or something?

$ (a,b) \equal{} (0,0) \implies f(a,b) \equal{} 5 \plus{} \sqrt {80} > 5\plus{}\sqrt{64} \equal{} \sqrt{169} > \sqrt {145}$


Does anyone have the links to the 2008, 2007, and 2005 National Count Downs? If so, please PM me. I really want to watch those because it features people from my school  :)

I posted a thread with a similar request here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=293968[/url]

But it was fruitless.......

but I'm sure somewhere on YouTube you can find it.

Oh darn...why'd they have to take them off? I really wanted to see that...

Well, Mathcounts.org took it off, but I'm pretty sure you can find it somewhere..............


group theory
number theory unsolved
number theory

Let $ p$ be an odd prime. Let $ g$ be a primitive root modulo $ p$. Find all the values of $ p$ such that the sets $ A \equal{} \{k^2 \plus{} 1: 1\leq k\leq \frac {p \minus{} 1}{2}\}$ and $ b \equal{} \{g^m: 1\leq m\leq \frac {p \minus{} 1}{2}\}$ are equal modulo $ p$.

Σ(k^2+1)-Σ(g^m)=0 mod p
=> Σ(g^2m +1) - Σ(g^m) = 0 mod p
{g^2m} is a cyclic group w.r.t multiplication mod p
=> Σ(g^2m) = 0 mod p
Also Σg^m = 0 mod p, m from 1 to (p-1)/2 because g^m = -g^((p-1)/2+m) mod p

Finally 0+(p-1)/2 - 0 = 0 mod p which is not possible.
So, no solution


number theory
greatest common divisor
quadratics
number theory unsolved

x,y,z are positive integer. z is gcd(x,y) Find x,y,z so that $ x \plus{} y^2 \plus{} z^3 \equal{} xyz$

Let
x = az
y = bz ,  GCD (a,b) = 1

az + (bz)^2 + z^3 = abz^3
a +zb^2  + z^2 = abz^2
a = 0 mod z,  therefore a = cz, GCD (c,b)=1
c + b^2 + z = cbz^2
b^2 - (cz^2)b +c+z = 0
this is a quadratic in positive integers therefore the discriminant must be a square

(cz^2)^2 - 4c - 4z = n^2
but for z >2
(cz^2)^2 > (cz^2)^2 - 4c - 4z > (cz^2 - 2)^2,  (this is trivial to check)
and (cz^2)^2 - 4c - 4z and (cz^2 -1)^2 have opposite parity so z=2,1

z = 1 yields
c^2 - 4c - 4 = n^2
(c-2)^2 -8 = n^2, c = 5

z = 2 yields
16c^2 - 4c - 8 = n^2
64c^2 - 16c - 32 = n^2
(8c-1)^2 - 33 = n^2,  c = 1

The only solutions are
x,y,z = (5,3,1), (5,2,1), (4,6,2), (4,2,2)


I know the blogs are under construction, but this is really making me angry:

I can edit my own posts now, but I can't edit OR delete my contributor's posts.  So they have essentially the same rights as me.

i have the oposite problem, i can post but my guest friends can't!
even if i allow them in the blog options! :!:


[quote="vo thanh van"]Let a,b,c are three length sides of a triangle such that a+b+c=1.Prove that:
$\frac{a^{3}}{b(a+b-c)}+\frac{b^{3}}{c(b+c-a)}+\frac{c^{3}}{a(c+a-b)}\geq 1$[/quote]

a+b-c>0
b+c-a>0
c+a-b>0

LHS is greater or equal then (a+b+c)^3/ 3[b(a+b-c)+c(b+c-a)+a(c+a-b)] etc

Let a=x+y,b=y+z,c=z+x. We have
    x+y+z=1/2. And need to prove
   $\large\frac{(x+y)^{3}}{\2(y+z)y}$+$\large\frac{(y+z)^{3}}{\2(z+x)z}$+$\large\frac{(y+z)^{3}}{\2(z+x)z}\[\geq 1$
  Assume that x>=y>=z
  $(x+y+z)\]\large\frac{(x+y)^{3}}{\(y+z)y}$+$\large\frac{(y+z)^{3}}{\(z+x)z}$+$\large\frac{(y+z)^{3}}{\(z+x)z}\[\geq 1$
  $\large\frac{(x+y)sqrt{x+y}}{sqrt{y+z}}$+$\large\frac{(y+z)sqrt{y+z}}{sqrt{z+x}}$+$\large\frac{(z+x)sqrt{z+x}}{sqrt{x+y}}\]\geq 1$
  By permutation Ineq we have result

I think it's not very hard.
Look http://www.mathlinks.ro/Forum/viewtopic.php?p=815712#815712  :lol:
(post 4) HAVE FUN!

That is my inequality,my solution is very nice :),I use AM-GM,very short ^^,but you can use the Cauchy-Schwarz with SOS.
Have fun!


If an arc of $ 45^\circ$ on circle $ A$ has the same length as an arc of $ 30^\circ$ on circle $ B$, then the ratio of the area of circle $ A$ to the area of circle $ B$ is

$ \textbf{(A)}\ \frac {4}{9} \qquad \textbf{(B)}\ \frac {2}{3} \qquad \textbf{(C)}\ \frac {5}{6} \qquad \textbf{(D)}\ \frac {3}{2} \qquad \textbf{(E)}\ \frac {9}{4}$

If the arc is 45, then it represents $ 1/8$ of the total circumference.  So let's say circle $ A$ has radius 8.  Then its circumference is $ 16\pi$ and so the arc length is $ 2\pi$.  Now the other circle's arc length is $ 1/12$ of its circumference so set up a proportion and we find out that circle $ B$ has a radius of 12.  Then its area is $ 144\pi$ in comparison to $ A$'s $ 64\pi$.  So the ratio is $ 4/9$.


Copycatting from the Illinois forum, I thought we could have a big game, password, INDIANA MC for all Indiana residents to enjoy.  Who wants to play?

Sure, I have nothing better to do  :lol:

Of course, I don't think we'll have very high turnout.

Without asking a mod, you can't guarantee that everyone's from Indiana Maybach.

Oh great username change eh?


Hey,

I'm trying to write that a function goes from $ \mathbb{R}^n \to \mathbb{R}^m$ in TeXnicCenter; however the \mathbb{} command is generating an error. Do I need to download an additional plug-in or package? (I have MikTex basic). If so, does anyone know where I can get this?

Thanks!

~ tZf

Just make sure you have the amsmath and amssymb packages loaded; you should use them for all documents with
\usepackage{amsmath, amssymb}
If you are using theorems, definitions etc then also add amsthm to this list of packages.


let $ABC$ be a triangle.and $a,b,c$ be the sidelength of it. prove that area of triangle $ABC$ is $\frac{1}{2} \times a \times b \times sin(C)$.

ohhhh...

jensen its too hard for basics :mad:

[quote="solafidefarms"]Shouldn't proofs be in Getting Started High School???[/quote]
sorry i though you can prove it.  :blush:

It's actually not even that hard to prove this, but unfortunately, the basics is the basics, and this isn't supposed to be here.

[quote="ashegh"]236factorial...

i said that its hard for the basics not the others... ;)[/quote]

Isn't that what I said?  :huh:

[quote="chess64"][hide="hint"]
compare to the more well-known formula for area
[/hide][/quote]
[hide="Another hint"]
Draw a perpendicular line from $A$ to $BC$, and let $D$ be the intersection. Express $AD$ in terms of $b$ and angle $C$ ;) [/hide]

[hide]Draw a perpendicular line from $A$ to $BC$, cut $BC$ at {H}.

The area of the triangle $A=\frac{1}{2}ah$

In triangle $AHC$, $h=bsinC$

So $A=\frac{1}{2}absinC$[/hide]

[quote="jensen"][quote="Altheman"]you posted this already...

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=56862[/url][/quote]
thanks but i think i know what i have posted before.anyway thanks a lot.
 :D[/quote]

 :D

haha...youre only supposed to post the same topic ONCE

ususally proofs go the intermediatte but this one is a little on the easy side


Pascal\u0027s Triangle
algebra proposed
algebra

Suppose you have convergent alternating series such as

 1/1 - 1/2 + 1/3 - 1/4 + ... (pattern continues)

and

 1/1 - 1/3 + 1/5 - 1/7 + ... (pattern continues)

These are just two I am using for examples, but other convergent alternating series would work as well.

As with series such as these, at a minimum, they can require
thousands of terms to be summed to get a few correct
digits to the right of the decimal point.

I will make use of some Binomial numbers and the 2nd series from above.  The row of numbers below is a cumulative sum working from right to left of the row above it:

....1.....7...21 35 35 21 7 1
128 127 120 99 64 29 8 1

Then the numbers get placed in the formula (I could have used
a different line of Pascal's triangle) like so:

(1/128)(128*x1/x2 - 127*x3/x4 + 120*x5/x6 - 99*x7/x8 +
64*x9/x10 - 29*x11/x12 + 8*x13/x14 - 1*x15/x16)

where the x sub 1 through x sub 16 are positive integers.

I DON"T have to start with 1 in this formula. In fact I get a jump
on the value if I get the exact sum of the first so many terms BEFORE I apply the formula to where I stopped adding the terms.

Case in point:

1/1 - 1/3 + 1/5 - 1/7 ~ 0.7238095238

(1/128)(128*1/9 - 127*1/11 + 120*1/13 - 99*1/15 + 64*1/17 -
29*1/19 + 8*1/21 - 1*1/23) ~ 0.0615890713

These two results sum to about 0.7853985951

Because this series is equal to (pi)/4, then multiply this last
result by 4 to get approximately:

3.14159438

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

Other Binomial numbers can be used that involve shorter calculations, but fewer correct digits will turn out on average.

For example, ......1 3 3 1
...........................8 7 4 1

Leading to  (1/8)(8*x1/x2 - 7*x3/x4 + 4*x5/x6 - 1*x7/x8)

****
I left the first series as an exercise; answers will vary depending
on which formula you use AND how far along in the series you put in the formula.


Anyone have a guess what the Orals topci for next years 1A/2A would be...
I kinda thought it would just be last year's 3AA/4AA topic, but i'm unsure. Can anyone clarify this?
Thanks in advance...

while the 1A/2A topic is often the previous year's 3AA/4AA topic, there have been one or two exceptions in the past few years.

i am fairly certain the topic and source for 1A/2A have been decided, but i am not sure if it has been made public yet.

the topic for 3AA/4AA will be Fair Division, but the source has not yet been announced.

I have official word now that the 1A/2A topic and source are "Secret Codes, Remainder Arithmetic, and Matrices" by Lyman C. Peck (published by NCTM), chapters 1-2 for regionals, 1-3 for state.

hmm, that sounds like an interesting topic,also it could be useful in a cryptography job, too bad the seniors get priority for the oral competition

I'm going to be doing this, but I have no prior experience.  I have started researching the topic and I was wondering if the pressentation is on logic or numbers like the triangles and simplexes or a combination of the two.


conics parabola trigonometry quadratics algebra

If line $ x\cos \alpha\plus{}y\sin \alpha\equal{}p$ touches $ y^{2}\equal{}4a(a\plus{}x)$ then which of the following holds true
a)$ p\cos \alpha\plus{}a\equal{}0$
b)$ a\cos \alpha\plus{}p\equal{}0$
c)$ p\sin \alpha\plus{}a\equal{}0$
d)$ a\sin \alpha\minus{}p\equal{}0$

[hide="Note"] $ x\cos \alpha\plus{}y\sin \alpha\equal{}p$-Normal Form of the line
where $ p$ and $ \alpha$ have their usual meanings

[/hide]

hello, inserting $ y\equal{}\frac{p\minus{}x\cos(\alpha)}{\sin(\alpha)}$ in $ y^2\equal{}4a(a\plus{}x)$ we get
$ \frac{(p\minus{}x\cos(\alpha))^2}{\sin(\alpha)^2}\equal{}4a(a\plus{}x)$ solving this quadratic equation for $ x$ we get
$ x_1\equal{}\frac{\minus{}p\cos(\alpha)\minus{}2\sin(\alpha)^2a\plus{}\sqrt{4\sin(\alpha)^2a(p\cos(\alpha)\plus{}a)}}{\sin(\alpha)^2\minus{}1}$
or
$ x_2\equal{}\minus{}\frac{p\cos(\alpha)\plus{}2\sin(\alpha)^2a\plus{}\sqrt{4\sin(\alpha)^2a(p\cos(\alpha)\plus{}a)}}{\sin(\alpha)^2\minus{}1}$
because the line touches the parabola we get that
$ p\cos(\alpha)\plus{}a\equal{}0$.So the right answer is $ \boxed{a)}$
Sonnhard.

[quote="TZF"][hide="Hint"]The equation of the line basically reads, "the line tangent to the origin-centered circle of radius $ p$ at the point $ (p \cos \alpha, \; p \sin \alpha )$

You'll want to use the reflective property of the parabola, [url=http://upload.wikimedia.org/wikipedia/commons/9/97/Parabola_focus_directrix.svg]shown here[/url].[/hide][/quote]
For some strange reasons I couldn't upload the document so could you please elaborate the desired solution a little further?


A guy walks into a 7-11 store and selects four items to buy.  The clerk at the counter informs the gentleman that the total cost of the four items is $\$$7.11. He was completely surprised that the cost was the same as the name of the store. The clerk informed the man that he simply multiplied the cost of each item and arrived at the total. The customer calmly informed the clerk that the items should be added and not multiplied. The clerk carefully added the items together and informed the customer that the total was still exactly $\$$7.11. (There was no sales tax involved)  Your task is to determine the exact costs of each item?

[size=84][color=green]Use dollor signs like $\$$ \ $\$ \$$[/color][/size]

[hide]
If each item is $x$, then we have $x^4=711$ and $4x=711$.
Thus $x^4=4x$
$x^3=4$
$x=\sqrt[3]{4}$[/hide]

How do you know that all 4 of the items are the same price.....? :?

you dont.
There are infinite solutions to this problem, in the form
$w+x+y+z=wxyz$
I chose the simplest case and solved

maybe he means whole numbers only. IN that case, I don't know if that would work...

well i cant imagine a place other than gas stations where the price of some thing is not a whole cent


[hide]

it was a coincidence  :D

[/hide]

it was a coincidence  :D : [b]No, the word "because" implies causation.[/b]

As with other Lateral Thinking puzzles, yes-no questions are encouraged.

If a pen had been present, would he have used it to write something crucial to his survival?

Could he have accomplished the same task with an otherwise identical object that was shaped exactly like a pen but contained no ink?

[quote="moogra"]did he kil himself?[/quote]
yeah, not having a pen can really get you down... :P

Did another person cause his death?:
[b]Not directly.  He wasn't murdered, if that's what you're thinking.[/b]

If a pen had been present, would he have used it to write something crucial to his survival?: [b]Yes, except that he wouldn't have been the one to write it.[/b]

Could he have accomplished the same task with an otherwise identical object that was shaped exactly like a pen but contained no ink?: [b]No.[/b]

did he kil himself?: [b]No.[/b]

did he need the pen to live? did it contain some sort of medication/poison?:
[b]No, any normal pen (actually, any functional writing utensil) would have been sufficient.[/b]

[hide]Was he dying from something such as an asthmatic attack when someone else came along who could have cured him if he had written instructions on a card?[/hide]

Was he dying from something such as an asthmatic attack when someone else came along who could have cured him if he had written instructions on a card?: [b]No, those cards are preprinted.  But instructions are indeed what would have been written.[/b]

[quote="E^(pi*i)=-1"]A man died because there was no pen in the room.  Why?[/quote]

Got it!

[hide]
not exactly sure but Aman was a fictional work by Tolkien (it was a continent). Since ther was no pen in the room Aman did not exist, therefore it "died".

[/hide]

[hide="Maybe..."]The room was a doctor's office, and the patient was diagnosed with a life-threatening disease that could be cured by a perscription drug.  But since there was no pen in the room, the doctor couldn't prescribe it to the patient, and therefore the patient died from the disease.[/hide]

is he unable to talk?  is the person who would have written stuff if there was a pen unable to talk?

were the instructions that would have been written relating to some medical thing, like he had some disease or was choking or something?

were the instructions like, how to get out of a burning room, and the room happened to be burning?

Was the man who died blind, deaf, or otherwise physically impaired?

[quote="E^(pi*i)=-1"]I urge everyone to ponder this question, that I thought you would raise: why didn't the doctor find the nurse and tell her verbally (he knew where she was)?[/quote]

Was the nurse deaf?

I'm not sure this has been asked before, but I HAVE READ THE WHOLE THREAD since it was started, did the nurse have medical problems?

is this doctor wearing a mask?: [b]No.[/b]

or is asthmatic?: [b]No.[/b]

but i thought he couldn't talk b/c of heart attack...:[b]My attempts at cunning clues are backfiring . . .  Disregard that.[/b]

Oh well, I think I know the elements of the solution but don't quite get it still...:[b]Yes, you do.[/b]

are we missing just some of the minor details, or is there still a major element that we still don't have?: [b]Minor details - so minor that I think that perhaps I should stop being so nitpicky and declare it solved.[/b]

Was the nurse deaf?: [b]No.[/b]

did the nurse have medical problems?: [b]No.[/b]

Is the cause of the heart attack relevant?: [b]No.[/b]

[hide]Did the patient want to play with the pen, then became so bored he wanted one and tried to write the doctor's instructions how to get him one, but then he had a heart attack, and could not ask the doctor for a pen, and bored himself to death[/hide]

What puzzles me is that you said the nurse knew that the doctor was dying. Yet he died of a sudden heart attack. I would guess that the doctor was suffering from some condition that would inevitably kill him (and the nurse knew this), but you've said the the cause of the heart attack is irrelevant.

So, just for clarity, did the nurse know that the doctor was dying before she found him dead (and the "we're all dying" argument doesn't count :P)?

did the doctor die from old age (probably not relevant)?
what about the patient?

-edit0
old age+heart attack

[quote="me@home"]I speaking on my behalf but I think others as well by declaring that you won, E^pi i=1, and should disclose the full solution...[/quote]

Agree, but... something wrong in your Euler Formula? :D

[quote="funcia"][quote="E^(pi*i)=-1"][quote="me@home"]I speaking on my behalf but I think others as well by declaring that you won, E^pi i=1, and should disclose the full solution...[/quote]

Okay, here is the solution I intended, hopefully I didn't give too many wrong or misleading clues:

[hide]
The doctor was one of a small handful in the world who knew how to treat the disease.  He examined the patient and discovered a simple surgical procedure that could save his life, but he was not a surgeon.  He wanted to write it down, but there was no pen in the room.  The nurse was also not there, so he went to the nurses' station to find her.  On the way he had a heart attack.  He shouted, but by the time the nurse arrived he was dead.  By the time another doctor who could treat the disease could arrive, the patient was dead.
[/hide][/quote]
wow minor details[/quote]

Yeah . . . that's why I declared it solved.

That was such a random puzzle, how did you ever come up with it? I'd like to see your thought process :P

How is someone supposed to figure that out!  :D


There are 12 different-colored crayons in a box. How many ways can Karl select four crayons, if the order in which he draws them out does not matter?

He will have $ 12$ choices for the first crayon, $ 11$ for the second, $ 10$ for the

third, and $ 9$ for the fourth. However, Consider the sequence

ABCD, where A,B,C,D are crayons.

ABCD
ABDC
ACBD
ACDB
...

become the same ordering, since  he will have those same crayons!

Therefore, we divide by $ 4!\equal{}24$ to get $ 495$.

Note: The answer to these problems in general, with $ n$ being the total number

of things and $ k$ being the number of things we are taking, is $ \dbinom{n}{k}\equal{}\frac{n!}{k!(n\minus{}k)!}$.


Here is your qindex:
\[\textrm{AMC 12 Score}\quad+\quad \textrm{AIME Score}\quad \times \quad 10 \quad+\quad 2 \quad \times \quad \left(\sum_{n=1}^{3}\textrm{indiv. score on meet n}\right) \]
Let's see what I get:
\[102+10 \times 6+2 \times (3+6+6) = 102+60+30 = 192 \]
Theoretically, the max score is: $150+150+48 = 348$.

Eh? This is different than what I heard, but oh well.
$112.5+10(5)+2(3+5)=178.5$

I didn't go to the first meeting, so it's really not fair.  :P
Anyway, I thought they took your best two scores from the first four meetings, and added it to your index divided by 10?

um i suck
112.5+30+2(3+2+2)=156.5
stupid mistakes are extremely annoying :ninja:
i should have at least 180 something... :(

[quote="junggi"]I thought it was
$\text{AMC INDEX}/10+\text{2 BEST ARML INDIV SCORES}$[/quote]

Yeah, that's what I thought too.

Moi so far = $\frac{162.5}{10}+(3+5)=24.25$.

Mine would be $27.45$
dang i feel proud of myself after that 7 today
woo

I was under the influence that it was AMC index divided by ten, plus top two scores as well.

That gives me $\frac{158.5}{10}+3+2 = 20.85$.

Based on your method, I have
$158.5+2(3+2+1) = 170.5$

(132+50)/10 + (6+7)= 31.2

>_< i used to not make any mistakes on math contests....

Well for the other index mines is $\frac{124.5+80}{10}+(6+7) = 33.45$.

[quote="Ubemaya"]Yeah, but people at least try calculating using the new method to see what you would get with that.

@Vishal, where did you get the qindex formula from?[/quote]

I just made it up...

I wanted to see how well people do based on AMC/AIME, and the 1st 3 practices.


limit
logarithms
calculus
calculus computations

$ \lim_{x\to\infty}\left(1\plus{}\frac{a}{x^b}\right)^{c x^d}$

$ \lim_{n\to\infty}\sum_{i\equal{}1}^{n}\frac{i}{n^2\plus{}n\plus{}i}$

1) is equivalent to $ e^{acx^{d \minus{} b} \minus{} \frac{a^{2}cx^{d \minus{} 2b}}{2}}$ after is simpl.
the seccond somwhere here already was posted

[hide="Number 1"]$ L \equal{} \lim_{x \to \infty}\left[\left(1 \plus{} \frac {a}{x^b}\right)^{x^b}\right]^{\frac {cx^d}{x^b}}$

$ \equal{} (e^a)^{\infty} \equal{} \infty$,if d > b;

$ \equal{} (e^a)^{c} \equal{} e^{ac}$, if d = b;

$ \equal{} (e^a)^{0} \equal{} 1$, if d < b,

for b,d > 0.[/hide]

[quote="Carcul"][hide="Number 1"]$ L \equal{} \lim_{x \to \infty}\left[\left(1 \plus{} \frac {a}{x^b}\right)^{x^b}\right]^{\frac {cx^d}{x^b}}$

$ \equal{} (e^a)^{\infty} \equal{} \infty$,if d > b;

$ \equal{} (e^a)^{c} \equal{} e^{ac}$, if d = b;

$ \equal{} (e^a)^{0} \equal{} 1$, if d < b.[/hide][/quote]
No.
for $ a\equal{}b$ as I  wrote we have :
$ \exp(ac\minus{}\frac{a^{2}c}{2x^{b}}\plus{}o()   )$ this mean that if $ b>0$ then you are right, but if $ b<0$ then limit is equal to $ 0$

Yes, you are right, I didn't thought in the case b < 0.

case if $ b\equal{}d<0$ then
let $ x^{b}\equal{}\frac{1}{y}$ where $ y\minus{}>\plus{}oo$
then
$ (1\plus{}ay)^{\frac{c}{y}}$~$ e^{\frac{\ln y}{y}}\minus{}>1$


For each $ n \in \mathbb{N}$ we have two numbers $ p_n, q_n$ with the following property: For exactly $ n$ distinct integer numbers $ x$ the number \[ x^2 \plus{} p_n \cdot x \plus{} q_n\] is the square of a natural number. (Note the definition of natural numbers includes the zero here.)

Let $p_{n}=0, q_{n}=2^{n+1}$. Then we need to prove that there are exactly $n$ integers $x$ such that
$$x^{2}+2^{n+1}=y^{2}$$
for some integer $y$. This is equivalent to $(y-x)(x+y)=2^{n+1}$. Let $x+y=2^{a}$ for some nonnegative integer $a$. If $a=0$ or $a=n+1$, we obtain a contradiction, because $x+y$ and $y-x$ must have the same parity. For $1\leq a\leq n$, the system
\begin{align*}
x+y&=2^{a}\\
y-x&=2^{n+1-a}
\end{align*}
has the unique integer solution $(2^{a-1}-2^{n-a},2^{a-1}+2^{n-a})$. Thus the equation $x^{2}+2^{n+1}=y^{2}$ has $n$ solutions, as requested.


a boat moves relative to water with velocity which is n=2 times less than the river flow velocity .at what angle to the stream direction must the boat move to minimize drifting?

Unless I am not understanding your English, shouldn't the angle be 180º?

no .the angle is 120 but l dont know the solution

Two questions:

1) Doesn't this problem have a picture?
2) What is the meaning of "minimize drifting"?


let P is a prime .prove that
$\sum\limits_{k=1}^{p-1} k^{m} \equiv \left\{\begin{array}{l} {-1 (mod P) if (p-1)\vdots m}\\{{0 ( mod P) if (p-1) \not\vdots m}}\end{array}\right.$

See http://www.mathlinks.ro/Forum/viewtopic.php?t=40171 (it's clear that the problem can be reduced with $k \mod p-1$).


geometry
parallelogram
geometry proposed

Given triangle $ ABC$ and a point $ P$ lies inside. $ A'B'C'$ is pedal triangle of $ ABC$ wrt $ P$. Denote $ H_a, H_b, H_c$ the orthocenter of triangle $ AB'C', BA'C', CA'B'$. Prove that $ A'H_a, B'H_b, C'H_c$ are concurrent.

Since $ AH_c\perp AC$ and $ PB'\perp AC$ we have $ PB'\parallel{}A'H_c$ and analogously $ PA'\parallel{}B'H_c$ hence $ A'H_cB'P$ is parallelogram. In the same way we prove that $ PB'H_aC'$ is parallelogram hence $ A'H_cH_aC'$ is parallelogram and the intersection point of $ A'H_a$ and $ C'H_c$ is their common midpoint. In the same way we prove that this midpoint lies also on $ B'H_b$ which means that $ A'H_a$, $ B'H_b$ and $ C'H_c$ are concurrent.

They actually intersect at the midpoint of HbB',HaA',HcC'


geometry
3D geometry
blogs
MATHCOUNTS
Mafia
dodecahedron

this could be like, the reverse of "answer a question, ask a question".  the first person posts an answer, then the next person posts a question that has that answer.  (we could get really creative with this...) for example:

Answer: a large furry animal, native to china, that eats bamboo

and so the next person would post something like:

Question: What is your history teacher like?  :lol:

and after you post a question, post another answer for the next person.

i'll start:

Answer: chaotic noise and disorder

Note: spelling is diarrhoea

Question: draw a bacterium.

Answer: a pitchfork

what did ya hit ur sis with?

answer:  a ping pong ball.

Q: what did you put in the cookies to make them taste so good?

A: the pythagorean theorem

What did you get wrong on the test yesterday?

Edison.

Question: who didn't invent the lightbulb?

Answer:  tofu a la soybean

What you found in the garbage can by your teacher (from experiance).

 :idea:

what is it that u don't hav in ur brain?

answer: roger federer

How do you spell Roger Federer?

Answer: rubix cube

Question: What is it that you have solved but can't spell properly? (it's rubik's btw)

Answer: fat

Question: What would you call a man 4ft by 4ft?

Answer : As high as a six foot three inches tree.

What did you write on your essay today?

 :|

Whom are you supporting: Barcelona or Arsenal.

three red elephants.

what caused that big bang last night??!!

answer: geeksta rap

What nose has the ability to smell?

Elmer's Glue and Kleenex.

What did you put into the toilet?

pencil shavings

What do you throw away from the box-thingy in an electric pencil sharpener?

EMOTICONS!

what is $EMOTICONS(EMOTICONS-1)(EMOTICONS-2)...(2)(1)$?

green eggs and ham

What is Sam I Am's favorite food?

Superfluidity

What are you reminded of when Sam I Am speaks in an extempore?

A 15-inch ruler.

What is similar but shorter than a 16-inch ruler?

A dodecahedron

Name a word that you pronounced wrong.

Wrong

Not in base 6, 5, 4, 3, or 2.

So you are wrong.

What is the name of an angle with a measure of 90 degrees?

Two of the three western movies that won the Oscar title of Best Picture

(In case you're wondering where that answer came from, it was the Final Jeopardy clue yesterday. Sorry...)

what was the final jeopardy clue yesterday?

the other western movie that won best picture

What's mathnerd's idea of entertainment?

Argentina 6 Serbia and Montenegro 0


A prism has 15 edges. How many faces does it have?

How would you solve this? Please help. Thanks!

The vertical edges and horizontal edges from the top or the bottom faces must be the same. Therefore, the number of edges on the top is $ 15\div3 \equal{} 5$. It should be a pentagon prism and it has 7 faces totally.


algebra
polynomial
function
algebra proposed

$p(x)$ is a polynomial with real coefficients such that $p(x^2 - 1) = p(x) p(-x)$. What is the largest number of real roots that $p(x)$ can have?

Well, i think i could be wrong, if I am, correct me please.

Suppose that the polynomial has a real root, say $\alpha$; so

$p(\alpha^2-1)=0$

Well, $\alpha^2-1$ is real, so...we take $x=\alpha^2-1$

and we get that $(\alpha^2-1)^2-1$ is also a real root of the polynomial...and so on...it seems like a serie, and even if it convergente, it would have infinit values...

please correct me if I'm wrong...
thanks

You're wrong , aodeath ! This sequence $a , a^2-1 , (a^2-1)^2-1 ,...$ must be periodic .
  I think the answer is 4 and we can find this function

Well, actually, not exactly for all the values this sequence is periodic, for exemple, for $a=3$ it seems that the sequence diverges though. But we're now trying to solve the recurrence relation

$a_{n+1}=a_n^2-1$

wich obviously, depends on the first term. This first term might be essencial for the result; it will probably "decide" whether our sequence is or not periodic.

 ;)

Well, I couldn't solve that recurrence, but investigating more about that polynomial, we get

$p(-1)=p^2(0), p^2(0)p(1)=p(0)$

and so we have 0 as one possible value for real root. The other roots are obtained when we solve:

$x^2-1=\pm x$, which gives us $x=\frac{1\pm \sqrt{5}}{2}$.

Well, its just to prove that those values remais a periodic serie, as solving that recurrence relation.

Well, first of all, we know that the last x particularly solves

$a_0=x \rightarrow a_1=x^2-1=x \rightarrow a_2=x^2-1$...seems like a constant sequence.

For $a_0=0 \rightarrow a_1=-1 \rightarrow a_2=0$...

Are there any other possible roots? I can't see by now =]


Let $ a_0 \equal{} 1/2$ and let $ a_{n \plus{} 1} \equal{} 1 \minus{} (a_n)^2$. Find $ lim_{n\to \infty}a_{2n\plus{}1}$ .

I'm pretty sure this converges to $ 1$.

[hide="Start"]
Since we are only concerned about the limiting behavior of $ a_1,a_3,a_5,\dots$, define a new sequence $ b_n$ by $ b_1\equal{}a_1$ and $ b_n\equal{}a_{2n\minus{}1}$. Then by the definition of $ a_{n\plus{}1}$, we have $ b_{n\plus{}1}\equal{}2b_n^2\minus{}b_n^4\equal{}b_n^2(2\minus{}b_n^2)$. By AM-GM we see that this is less than or equal to $ 1$, so now we have to show that $ 1$ is indeed the least upper bound.
[/hide]

$ a_0 \equal{} \frac{1}{2}$ also if $ 0< a_n < 1$ then $ 0 < 1 \minus{} a_n^2 < 1 \Longrightarrow 0 < a_{n\plus{}1} < 1$
So all terms $ a_i \in (0,1)$

Now as Yongyi suggested we define a sequence $ b_0\equal{}a_1\equal{}\frac{3}{4}, b_{n\plus{}1}\equal{}b_{n}^2(2\minus{}b_{n}^2)$
We show that $ b_{n\plus{}1}>b_n$

$ b_{n}^2(2\minus{}b_{n}^2) > b_n \Longrightarrow b_n(2\minus{}b_n^2) > 1$
$ (1\minus{}b_n)(b_n^2\plus{}b_n\minus{}1) > 0$
Which is true for $ b_n > \frac{\minus{}1\plus{}\sqrt{5}}{2} \left( \frac{3}{4} > \frac{\minus{}1\plus{}\sqrt{5}}{2}\right)$

So we have a strictly increasing set of positive real numbers $ \{b_i\}$, bounded at $ 1$

As $ n\to \infty, a_n\equal{}a_{n\plus{}1}$ therefore $ a_n(1\minus{}a_n)(a_n^2\plus{}a_n\minus{}1) \equal{} 0$

This gives roots $ a_n \equal{} 0, \frac{\minus{}1\pm \sqrt{5}}{2}, 1$ but all $ \{b_i\}$ are greater that the first three roots so the least upper bound is $ 1$.


$n>2$ so $q>p$,hence $q=kp-1,k\in\mathbb N$.

 Denote $n=2m,m\in\mathbb N$.
We have $(p^m+1)\left(\frac {p^m-1} {p-1}\right)=q\left(\frac {q+1} p\right)$

But $p^m+1>\frac {p^m-1} {p-1}$,$gcd(p^m+1,\frac {p^m-1} {p-1})=1$ and $q$ is prime

so:$p^m+1=qt$,and,$\frac {p^m-1} {p-1}=\frac {q+1} t$.

$qt\equiv1(mod p)$,$q\equiv-1(mod p)$ so$t\equiv-1(mod p)$ hence $t\geq{p-1}$

Now$q+1=(qt-2)(\frac {t} {p-1}) \geq q(p-1)-2$,that is,$q(p-2) \leq 3$.We then get $p=2$.

$p=2$ then $q+1=t(qt-2)$ so $t\geq 2$.Thus $q+1 \geq 2(q-2)$ ,$q \leq5$.

We'll get $(p,q,n)=(2,5,4)$

I remeber i solved it long time ago, somewhere in Proposed probs. section! There is a more general problems which i do not know how to do it

$\frac{p^n-1}{p-1}=\frac{q^m-1}{q-1}$

where $p,q$ are unknown primes and $m,n$ are given positive integer numbers.

Hello Nalpaction,

I agree in your final answer, but there are two issues about your solution.

[quote="nalpaction"]We'll show that $p=2$.Assume the contrary
 $n>2$so $q>p$.But  $q(q+1)=p(p^{n-1}+...+p+1)$,so $q=kp-1,k\in\mathbb N$.$p$is odd then $k$'s even
 Thus the even number $k(kp-1)$ equal to $p^{n-1}+...+p+1$,which is odd,contradiction.
[/quote]

Why do you say that $p^{n-1}+...+p+1$ is odd. Remember, $n$ is even, therefore in such a sum you have $n$ odd terms, an even number of odd numbers added, that is even, you still don't have your contradiction (I still haven't found it either)

[quote="nalpaction"] But $t(qt-2)\geq {q-2}$,so $q\leq 5$. We'll get $(p,q,n)=(2,5,4)$.[/quote]

I think that $t(qt-2)\geq {q-2}$ just imply that $t\geq1$, but not $q\leq5$. How do you justify your statment.

Best,

To djimenez:Thank you for your claim.

With question 1:yes,you are right.I've edited my wrong solution.

With question:sorry,I was right.I meant $q+1=t(qt-2)\geq 2(q-2)$,so,$q\leq 5$

Thank you  $nalpaction$  for your proof( I'll read it today),and  $djimenez$  for your corrections :)

[quote="nalpaction"]To djimenez:Thank you for your claim.[/quote]

You're very welcome  ;) . Anyway, it is a nice (and clever) solution.

Best,

My solution goes like this:
$p^n+p^{n-1}+.........+p = q^2+q$
$p(p^n-1)=q(q+1)(p-1)$
thus $p|q+1$  $(*)$
$(p^{\frac{n}{2}}+1)(p^{\frac{n}{2}}-1)= q{\frac{(q+1)(p-1)}{p}}$
$\frac{(q+1)(p-1)}{p}$ is an integer using $(*)$
well,  $(q,\frac{(q+1)(p-1)}{p}) = 1$
Now my idea is to factorize RHS in all possible ways  such that we get a product of two numbers say $\alpha$ and $\beta$ differing by 2.
Thus, we have the relation that $(\alpha,\beta)$ = 1 or 2. and $\alpha\beta=q{\frac{(q+1)(p-1)}{p}}$
Take it case wise,
We can assume $\alpha$ and $\beta$ like two boxes. Since q is a prime it will lie in one box individually
i) if   $\alpha$         has only         $ q$                   then   $\beta$  is   $\frac{(q+1)(p-1)}{p}$
j)  if  $\alpha$  has   $qr  $   then $\beta$   is  $\frac{(q+1)(p-1)}{pr}$  for,$q+1>r>1$ and $r |   \frac{(q+1)(p-1)}{p} $

It can't be taken as $\alpha$ and $\beta$ as $qr-\frac{(q+1)(p-1)}{pr}= \frac{pq(r-1)+q-p+1}{pr} > 2$.
Thus only case i) is possible .
We have two further cases like this,
$(p^{\frac{n}{2}}-1) =q$ and $(p^{\frac{n}{2}}+1)$ = $\frac{(q+1)(p-1)}{p}$
Solving this doesn't give any solution.
now,  $(p^{\frac{n}{2}}+1) = q$   and $(p^{\frac{n}{2}}-1) = \frac{(q+1)(p-1)}{p}$.
$\Rightarrow$ $3p = q+1$
$\Rightarrow$ p =2 and q=5 and thus n = 4  is the only solution.
Correct me if I'm wrong. :)

[quote="rationalist"]
 if   $\alpha$  is $q$  then  $\beta$  is   $\frac{(q+1)(p-1)}{p}$   ($\alpha$ and $\beta$ are supposed symmetric)
 if  $\alpha$  is   $qr$   then $\beta$   is  $\frac{(q+1)(p-1)}{pr}$  for,$r>1$
I didn't understand this part.
2) can't be possible as $\boxed{qr-\frac{(q+1)(p-1)}{pr} > 2\text{why?}}$.
now,  $(p^{\frac{n}{2}}+1)=q$.
I don't understand this too, why?[/quote]
Here is my solution.
We consider $q>p>2$. Now let $n=2k$. Then, $p(p^k+1)(p^k-1)=q(q+1)(p-1)\implies p|q+1$
It is obvious that $p^k+1|(q+1)(p-1)$ but $\gcd(p^k+1,p-1)=2$, so $p^k+1|2(q+1)\implies p(p^k+1)|2(q+1)\implies 2(q+1)\ge p(p^k+1)$
But we have $q|p^k-1\implies p^k-1\ge q\implies 2p^k\ge 2(q+1)\ge p(p^k+1)$ too. This inequality is true  if $(2-p)p^{k-1}$, false.
So, $p=2$. Then $2(2^k-1)|q+1,q|2^k+1\implies 2^k+2\ge q+1\ge2(2^k-1)\implies 3\ge 2^{k-1}$, true for $k=2$ giving $k=2$. Then $n=4,q=5$.
I am afraid if I had missed some trivial calculations.

@mathmdmb I have edited my earlier post. I hope it is clearer now.

[quote="hardsoul"]Find all primes $p,q$ and even $n>2$ such that $p^n+p^{n-1}+...+1=q^2+q+1$.[/quote]

Because $n>2$ then $p<q$ , also it is easy to see that for $q=p+1$ no solutions so $q>p+1$ ,  set $n=2k$

$p(p^{2k-1}+p^{2k-2}+\cdots+p^2+p+1)=q(q+1)$   (****)  so  because $p<q$

$q|p^{2k-1}+p^{2k-2}+\cdots+p^2+p+1$

$q|(p^{k-1}+\cdots+p^2+p+1)(p^k+1)$

Because $q>p+1$ then

$q|\dfrac{p^k-1}{p-1}\dfrac{p^k+1}{p+1}$

 so at any case must

$q\leq(p^k+1)$    (##)

Now from (##) and (****) we get that

$p^{2k}+\cdots+p^2+p\leq(p^k+1)(p^k+2)=p^{2k}+3p^k+3$ so $p=2$ , $k=2$ , $q=5$ , $n=4$

[quote="Αρχιμήδης 6"][quote="aditya21"]my solution =

the given equation is equivalent to
$p(p^n-1) = q(q+1)(p-1) $
now , assuming $p,q$ both odd

we can easily see that $q+1, p-1 , q$ cannot divide $p$ since it is evident that $q>p$ and $gcd(p-1,p)=1$

thus we dont have any solution for both $p,q$ as odd as this case implies $ p^n-1 = q(q+1)(p-1) $ and $p=1$
a contradiction!

CASE 2 = now $p<q$ now assume that $p=2$
than , the given equation becomes
$2(2^n-1) = q(q+1) $

this is $2(2^{n/2} - 1)(2^{n/2}+1) = q(q+1) $

thus as $gcd(2^{n/2}-1 , 2^{n/2}+1) = 1$ thus one of the factors divide $q+1$ and other divide $q$.

considering all such cases it is easy to find the only solution as
$ (p,q,n) = (2,5,4) $ :P :)

please tell me if i am correct[/quote]

The last part is false because if $xy=zw$ and $(x,y)=(z,w)=1$ dont imply that $x=z$ or $x=w$

exmple: $15*7=3*35$ with $(15,7)=(3,35)=1$[/quote]

thanks!!

i shall extend my solution now and somewhat edit it.

edit::

after we get
$2(2^{n/2}-1)(2^{n/2}+1) = q(q+1) $
thus one of the factors of lhs must divide $q$.
now $2$ cannot divide $q$.
so either $2^{n/2}-1 $ or $2^{n/2}+1$ divide $q$.
now considering both cases
we get our solution as $(p,q,n) = (2,5,4) $ :)

[quote=aops29][hide=Solution, please check]
This becomes
\[p(p^n - 1) = q(q+1)(p-1)\]
Now, suppose \(p>2\). Clearly we have \(q>p\). Thus, it is easy to see that \(q(q+1) \mid p^n - 1\), so
\[p(p^n - 1) = q(q+1)(p-1) \leq (p^n - 1) (p-1) \implies p \leq p-1\]

Therefore \(p=2\) and we get the only solution \((p,q,n) = (2,5,4)\)
[/hide]

[color=#f00][u]Remark:[/u][/color]
Me, MathPassionForever and Mathotsav *together* took 1-1.5 hrs to solve this, but now that I look back at it, it wasn't that hard.[/quote]

Where exactly were you solving this *together*?

@below LOL

[quote=Math-wiz]Where exactly were you solving this *together*?[/quote]
Whatsapp ;)

[quote=aops29]
This becomes
\[p(p^n - 1) = q(q+1)(p-1)\]
Now, suppose \(p>2\). Clearly we have \(q>p\). Thus, it is easy to see that \(q(q+1) \mid p^n - 1\), so
\[p(p^n - 1) = q(q+1)(p-1) \leq (p^n - 1) (p-1) \implies p \leq p-1\]

Therefore \(p=2\) and we get the only solution \((p,q,n) = (2,5,4)\)
[/quote]

How do you claim \(q(q+1) \mid p^n - 1\)?
Because if this were to happen, then $p$, which divides $q+1$, also divides \(p^n - 1\) which is false...


calculus
integration
geometry
MIT
college
Ross Mathematics Program
analytic geometry

i have two options at my disposal:
1)introduction to calculus and analysis by richard courant & john fritz.
2) Differential and integral calculus by richard courant.

i am somewhat inclined towards the first option because it has problems also in physics which interest me as well, so what should i choose?
(i must say that i did a big mistake once and took them both, in hope that they will complememt each other it got me nutsy).  :oops:

You can try for

1. Calculus & Analytic Geometry
     By- George B Thomas, Jr.
            MIT
            Ross L. Finney
     ADDISON WESLEY Publishers.

2.  A Problem Book In Mathematical Analysis
     By- GN Berman
     MIR PUBLISHERS, RUSSIA

3. A Problem Book Of Calculus In One Variable
    By- IA Maron
    MIR PUBLISHERS, RUSSIA

Bye.  :lol:

i am pretty much close at courant, so your reply wasn't any of help.
but thank you.

i guess it's all up to me, then i choose the former.

Taking a look at here might help you to make your choice ;) :
http://www.amazon.com/gp/product/0471608408/102-5107576-4784157?v=glance&n=283155
http://www.amazon.com/gp/product/354065058X/102-5107576-4784157?v=glance&n=283155

[quote="fskilnik"]Hi there, Alon!

I prefer "Differential Calculus".  It seems to me that the former is really more "physics inclined" but the second one is (in my opinion) more "mathematically-built".  The sections are (as far as I remember) more complete/rigorous and I do remember there are very-good problems for checking your understanding of the concepts.

The fact that the first book has many good diagrams and important techniques presented as "theory" or exercises (chain-Rules in special cases, for instance) makes me feel it would be a good thing to complement the second book with the first but... I wouldn´t do the other way-round.

Well, it´s just a matter of opinion, for sure. Both books are more-than-respected, that´s a fact.

Best regards and... Happy Chanuka!
fskilnik[/quote]
hi Fskilnik!
thank you for the great advice.


We have triangle ABC such that angle ABC is twice angle ACB.  Now, take incenter I of this triangle.  Given that IC=AB, prove that angle BAC=60.

It is trivial to find $ IC$ in terms of the sides. It probably turns out to be 60 degrees or something.


a rigit 255ml cylinder contains 0.305 mol of N2 at 235K.  A 0.544 mol of H2 is added to the cylinder, where a reaction occurs to produce NH3.  calculate the total pressure in atm in the cylinder at 235 K after the reaction is complete?

1) Partial pressures are directly proportional to concentrations.

2) Write the balanced chemical equation of the reaction that takes place:

$ \ce{N2} \plus{} 3\, \ce{H2} \longrightarrow 2\, \ce{NH3}$

3) Compute the initial partial pressures $ P_{\ce{N2}, o}$ and $ P_{\ce{H2}, o}$

4) According to reaction stoicheometry, in the equilibrium we will have

$ P_{\ce{N2}} \equal{} P_{\ce{N2}, o} \minus{} x$

$ P_{\ce{H2}} \equal{} P_{\ce{H2}, o} \minus{} 3x$

$ P_{\ce{NH3}} \equal{} 2x$

5) It is just me, or do we need $ K_P$ (or $ K_C$)?

I think the reaction goes to completion.  There will be some excess of one of the reactants most likely.  and then just use PV=nRT.  This is just speculation though.


How do you count in bases higher than 2?

I'm confused...

Especially on how you count in a base higher than 10.

You can use numbers larger than one.

so 222 is a base 3 number. Otherwise it works mostly like a base 2 number. 222_3 in base ten is 2*9 + 2*3 + 2 = 26

When you get to bases over 10, you begin to use letters. Like in Base 16, here's the first few numbers:

1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F, 11, 12, 13, 14, 15, ....

in base smaller than 10, there won't be some digits that are in base 10. Base 5:

1, 2, 3, 4, 11, 12, 13, 14, 21, 22, 23, 24, 31, 32, 33, ....


Now that selection is done, several colleges have announced spring travel programs in which admission officers will go to regional information sessions at cities around the country. Harvard's programs are listed on the Harvard Web site

[url=http://www.admissions.college.harvard.edu/utilities/travel_schedule/index.cgi]Harvard in Your Hometown[/url]

and most of those programs are joint programs offered by the Exploring College Options consortium

http://exploringcollegeoptions.org/

of Duke, Georgetown, Harvard, Penn, and now Stanford.

MIT is part of a different joint travel program, announced on its Web site

[url=http://www.mitadmissions.org/topics/mityou/spring_recruitment_travel_schedule/index.shtml]MIT Spring Recruitment Travel[/url]

with Brown, Princeton, Swarthmore, Wellesley, and Yale.

I look forward to attending the Exploring College Options program in my town and hope I'll see you there. I'd be delighted to read your comments about one of those regional information sessions here on AoPS after you have attended the meeting in your town.

tokenadult, your info is very helpful. We will certainly participate in these sessions. Thanks.


function
limit
calculus
calculus computations

how do you prove that some function $f$ is not $C^{1}$?

Find a point $x_{0}$ where one of two things happen: either $f'(x_{0})$ does not exist, or $f'(x_{0})\ne\lim_{x\to x_{0}}f'(x)$.


[quote="tranthanhnam"]3 thi  quoc gia bang B va bai 6 thi quoc te = tieng viet di[/quote]
Ban nen qua diendantoanhoc.net la tot nhat,neu ban muon xem tieng Viet


linear algebra
matrix
algebra
polynomial

1.) If A is an n by n matrix, show that A is diagonalizable if and only if transpose of A is diagonalizable

2.) If A is diagonalizable, show A^k for k>=1 is also diagonalizable

3.) If A is diagonalizable, show p(A), p(x) any polynomial is diagonalizable

Write $A=X^{-1}DX$, where $D$ is diagonal. What happpens when you transpose/take a power of this relation?


I have spent a good hour trying to work this one out to no avail. I am trying to insert a graphic using the graphicx package and am getting the warning:

LaTeX Warning: File `myimage.png' not found on input line 7.
Error:  (file myimage.png): cannot find image file
 ==> Fatal error occurred, the output PDF file is not finished!

the file is definately in the correct directory

I am using the example on http://www.artofproblemsolving.com/LaTeX/AoPS_L_PictHow.php

and am running miktex 2.5 and texnic centre beta 7.01

any help would be greatly appreciated- i have only just started using latex and am seriously tempted to go back to msword :S

The only reason that I can see for the LaTeX Warning: File `myimage.png' not found on input line 7 is that the file isn't there. Make absolutely sure that myimage.png really is in the [b]same[/b] directory as the latex file you are compiling.

You could try moving both files to a different directory without spaces in their names, but I doubt that is the cause of your problem.

I would give a definite thought to making sure there aren't any spaces in directory names. But another thought is Texnic going through dvi or ps to get to pdf? I am pretty sure those formats don't support the raster formats, but instead look at eps, so you might try dropping the extension, then the compilers look for their preferred extension type.


The attached identity is posted in the generating functions math jam (http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=113).

What is the basis for this identity?

You mean binomial theorem (in the general case of Newtons)?

[quote="Phelpedo"]Could you walk me through how the binomial formula is used to derive this identity?[/quote]
It's derived by expanding

$(1+x+x^{2}+...)^{n}$.

You can examine the expansion term by term.

a. There are $\binom{n}{0}= \binom{n}{n}= \binom{n-1}{n-1}$ ways to choose no $x$s and all 1s.

b. There are $\binom{n}{1}= \binom{n}{n-1}$ ways to choose one $x$ and the rest 1s.

c. There are $\binom{n}{2}= \binom{n}{n-2}$ ways to choose two $x$s and the rest 1s, and $\binom{n}{1}= \binom{n}{n-1}$ ways to choose one $x^{2}$ and the rest ones.  This gives $\binom{n}{n-2}+\binom{n}{n-1}= \binom{n+1}{n-1}$ ways to choose an $x^{2}$ term in all.

You see the pattern.  You could prove by induction that for the $x^{k}$ term, you have the coefficient $\binom{n+k-1}{n-1}$ in front of it, because you'll have $\binom{n}{1}+\binom{n}{2}+...+\binom{n}{k}= \binom{n+k-1}{k}= \binom{n+k-1}{n-1}$.  Does this help?

Yes this does help...so the name for this identity is Newton's?


geometry
circumcircle
trigonometry
search
geometry proposed

[quote="bimbo"]1. Let ABC be a triangle and H its orthocenter. Prove:
$\frac{AH+BH+CH}{2}=R+r$,
where R is the circumradius and r the inradius.

2. Let ABC be a triangle. The bisectrices intersect the circumcircle at 6 points: $A, B, C, A', B', C'$. Prove that $(ABC)\leq(A'B'C')$, where $(XYZ)$ is the area of the triangle XYZ.

3. Let $\angle A, \angle B,\angle C$ be the internal angles of a triangle then
\[ 2\cos\angle A\cos\angle B\cos\angle C = 1 - \cos^2\angle A - \cos^2\angle B - \cos^2\angle C.  \]

I hope you will enjoy this problems. If you find mistakes please post a message with the correction.
Good luck,
Juan.[/quote]
[quote][b]Some problems from myself[/b][/quote]
You mine these are your own problems?
How I know these are well know and very old problems :(  :lol: .(and not hard too ;) )

[quote="bimbo"]I know they are easy, but where have it appeared, or where can I find 'em?[/quote]
 I can't say you how to finde this problems in the net,but if you really want I can post solutions here ;)(also you can try to search I am sure you will get results).


A Mersenne Prime is a prime defined by $2^p-1$, where p is a prime.

Prove that $2^{p-1}(2^p-1)$ is a perfect number.

First let us assume  $p = 2^k-1$ is a prime number, and set $n = 2^{k-1}(2^k-1)$.  To show that n is perfect we can  show that  $sigma(n) = 2n$. $Sigma$ is multiplicative and $sigma(p) = p+1 = 2^k$, hence:

$sigma(n) = sigma(2^{k-1}).sigma(p) = (2^k-1)2^k = 2n$ Hence n is perfect, to show the converse:

Suppose n is any even perfect number and write n as $2^{k-1}m$ where m is an odd integer and k greater than one. It follows that:

$sigma(2^{k-1}m) = sigma(2^{k-1}).sigma(m) = (2^k-1).sigma(m).$

Since n is perfect we know: $sigma(n)=2n=2^km$
Hence:
$2^km=(2^k-1)sigma(m)$
So $2^k-1$ divides $2^km$ hence $2^k-1$ divides m. Say $m = (2^k-1)M$
$2^kM=sigma(M)$
$2^kM=sigma(m) \geq m+M = 2^kM$
Hence $sigma(m)=m+M$ Which means its only two divisors are itself m and one M. Thus $m=2^k-1$ is a prime and we have shown that n has the prescribed form.

Prove that an even number is perfect iff it is of the form asrcibed above.

Bomb

To: Bomb

It is enough when $ 2^{k}\minus{}1$ be prime.
See $ frac{\sigma(n)}{n}\equal{}\prod_{i\equal{}1}^{k}(\sum_{j\equal{}0}^{e_{i}}frac{1}{p^{j}})$
Then, You can get easily.


[quote="VIMF"]see here http://ddbdt.co.cc/forum/ba-t-ae-ng-tha-c-sa-ng-ta-o-thpt/43-da-nh-cho-ca-c-ba-n-tha-ch-sos.html[/quote]
But $ \sum_{cyc}\frac{a\plus{}b}{a^2\plus{}b^2}\leq\sqrt{\frac{27}{a^2\plus{}b^2\plus{}c^2}}$ is wrong.
Try $ a\rightarrow\plus{}\infty$ and $ b\equal{}c\equal{}1.$ :wink:

[quote="arqady"][quote="VIMF"]see here http://ddbdt.co.cc/forum/ba-t-ae-ng-tha-c-sa-ng-ta-o-thpt/43-da-nh-cho-ca-c-ba-n-tha-ch-sos.html[/quote]
But $ \sum_{cyc}\frac {a \plus{} b}{a^2 \plus{} b^2}\leq\sqrt {\frac {27}{a^2 \plus{} b^2 \plus{} c^2}}$ is wrong.
Try $ a\rightarrow \plus{} \infty$ and $ b \equal{} c \equal{} 1.$ :wink:[/quote]
sorry because I write in Vietnamese, so you do not understand

Title is: cho a,b,c là 3 cạnh tam giác . chứng minh

[quote]for a, b, c is the length of three sides of a triangle[/quote]

[quote="VIMF"]If $ a,$ $ b,$ $ c$ are the side-lengths of a triangle, then
$ \frac{a\plus{}b}{a^2\plus{}b^2}\plus{}\frac{b\plus{}c}{b^2\plus{}c^2}\plus{}\frac{c\plus{}a}{c^2\plus{}a^2} \le \sqrt{\frac{27}{a^2\plus{}b^2\plus{}c^2}}.$[/quote]
Try $ a\equal{}1.1$ and $ b\equal{}c\equal{}1.$


suposse the cevians AP, BQ, CR of triangle ABC meet at T.Prove that:
$\frac{TP}{AP}+\frac{TQ}{BQ}+\frac{TR}{CR}=1$.

It is a well known simple problem.

(TP / AP) + (TQ / BQ) + (TR / CR) =

= [(BTC) / (ABC)] + [(CTA) + (ABC)] + [(ATB) / (ABC)] = 1

(TP / AP = AD / TD' = (BTC) / (ABC), where AD, TD', the altitudes of ABC, TBC, respectively etc).

Kostas vittas.


inequalities
geometry
incenter
geometry proposed

To start with,we have these identities:
$ cos^{2}A\plus{}cos^{2}B\plus{}cos^{2}C\equal{}1\minus{}2cosAcosBcosC$
and
$ cos(A\minus{}B)\plus{}cos(A\plus{}B)\equal{}2cosAcosB$
then,
$ 3\sum{cosA}\le\sum{cos(A\minus{}B)}\plus{}2\sum{cos^2A}$
$ \iff 0\le\sum{(cos(A\minus{}B)\plus{}cos(A\plus{}B))}\plus{}2(1\minus{}2cosAcosBcosC\minus{}\sum{cosA})$
$ \iff 0\le2\sum{cosAcosB}\plus{}2(1\minus{}2cosAcosBcosC\minus{}\sum{cosA})$
$ \iff cosAcosBcosC\le (1\minus{}cosA)(1\minus{}cosB)(1\minus{}cosC)$
let $ cosA\equal{}\frac{b^2\plus{}c^2\minus{}a^2}{2bc},cosB\equal{}\frac{a^2\plus{}c^2\minus{}b^2}{2ac},cosC\equal{}\frac{b^2\plus{}a^2\minus{}c^2}{2ba}$
and $ a\equal{}x\plus{}y,b\equal{}y\plus{}z,c\equal{}z\plus{}x$
$ cosAcosBcosC\le (1\minus{}cosA)(1\minus{}cosB)(1\minus{}cosC)$
$ \iff \prod(y^2\plus{}xy\plus{}yz\minus{}xz)\le8x^2y^2z^2$
$ \iff 2\sum_{sym}a^3b^2c\plus{}\sum_{sym}a^4bc\le\sum_{sym}a^4b^2\plus{}\sum_{sym}a^2b^2c^2\plus{}\sum_{sym}a^3b^3$

[quote="∫FaiLurE∮"]$ \iff 2\sum_{sym}a^3b^2c \plus{} \sum_{sym}a^4bc\le\sum_{sym}a^4b^2 \plus{} \sum_{sym}a^2b^2c^2 \plus{} \sum_{sym}a^3b^3$[/quote]
By Schur we have:
$ \sum_{sym}a^2b^2c^2 \plus{} \sum_{sym}a^3b^3 \ge 2\sum_{sym}a^3b^2c$
And by muirhead:
$ \sum_{sym}a^4bc \le \sum_{sym}a^4b^2$

(Just in case anyone didn't know)

Mathias, why can't you tell the first also by Muirhead as
$ (\sum_{sym}a^2b^2c^2 \minus{} \sum_{sym}a^3b^2c) \plus{} (\sum_{sym}a^3b^3\minus{} \sum_{sym}a^3b^2c) \ge 0$.

[quote="Agr_94_Math"]Mathias, why can't you tell the first also by Muirhead as
$ (\sum_{sym}a^2b^2c^2 \minus{} \sum_{sym}a^3b^2c) \plus{} (\sum_{sym}a^3b^3 \minus{} \sum_{sym}a^3b^2c) \ge 0$.[/quote]
How can we derive the first from Muirhead?
Note that $ \sum_{sym}a^2b^2c^2 \minus{} \sum_{sym}a^3b^2c \color{red} \le \color{black} 0$

this solution is not from me:
obviously, it is true when the triangle is obtuse or right-angled.
when the triangle is acute
$ cosAcosBcosC\le (1\minus{}cosA)(1\minus{}cosB)(1\minus{}cosC)$
$ \iff \prod (1\minus{}\frac{b^2\plus{}c^2\minus{}a^2}{2bc}) \ge \prod \frac {b^2\plus{}c^2\minus{}a^2}{2bc}$
$ \iff \prod(a\plus{}b\minus{}c)^2\ge\prod(a^2\plus{}b^2\minus{}c^2)$
$ \iff\prod[(a\plus{}b\minus{}c)^2(b\plus{}c\minus{}a)^2]\ge\prod[(a^2\plus{}b^2\minus{}c^2)(b^2\plus{}c^2\minus{}a^2)]$
since $ (a\plus{}b\minus{}c)^2(b\plus{}c\minus{}a)^2\ge(a^2\plus{}b^2\minus{}c^2)(b^2\plus{}c^2\minus{}a^2)$
$ \iff 2(c\minus{}a)^2(c^2\plus{}a^2\minus{}b^2)\ge0$
$ QED$

[quote="SUPERMAN2"]Given a triangle ABC,prove that:
3$ \sum cos A$≤$ \sum cos(A \minus{} B)$+2$ \sum (cos A)^2$[/quote]
$ \left\|\begin{array}{c} \mathrm cos(A) \plus{} cos(B) \plus{} cos(C) \equal{} \frac {R \plus{} r}{r} \\
\mathrm cos(A \minus{} B) \plus{} cos(B \minus{} C) \plus{} cos(C \minus{} A) \equal{} \minus{} 1 \plus{} \frac {r^2}{2R^2} \plus{} \frac {s^2}{2R^2} \plus{} \frac {r}{R} \\
\mathrm cos^2(A) \plus{} cos^2(B) \plus{} cos^2(C) \equal{} 3 \minus{} \frac {s^2 \minus{} r^2 \minus{} 4rR}{2R^2}\end{array}\right\|$

So we must prove :$ \boxed{ s^2 \le 4R^2 \plus{} 4Rr \plus{} 3r^2}$ which is obviously true ( If $ I$ is incenter and $ H$ is orthocenter, then $ IH^2 \equal{} 4R^2 \plus{} 4Rr \plus{} 3r^2 \minus{} s^2$) :)

Sorry Mathias, I never knew the proper conditions of applying Muirhead's Theorem. Sorry.

May be my solution isn't very nice:
We have already known the ineq in triangle:
$ \frac{r}{R}\plus{}\sum \frac{(a\minus{}b)^2}{16R^2}$≤$ \frac{1}{2}$(I created my problem from this one)
=> cosA+cosB+cosC-1+$ \sum \frac{(sinA\minus{}sinB)^2}{4}$≤$ \frac{1}{2}$
=>4(cosA+cosB+cosC)+2($ \sum (sinA)^2\minus{}\sum sinBsinC)$≤6
=> 4(cosA+cosB+cosC)-2$ \sum sinBsinC$≤$ 2(3\minus{}\sum (sinA)^2)$
=> 4(cosA+cosB+cosC)-$ (\sum cosA\plus{}\sum cos(B\minus{}C))$≤$ 2\sum (cosA)^2$
=> $ 3\sum cosA$≤$ \sum cos(B\minus{}C)\plus{}2\sum (cosA)^2$


algebra polynomial trigonometry algorithm Vieta

Please can some one tell me how to solve cubic equations.

I was having problem in the following question: :(
$x^3-3x+1=0$

A help in the sort of an explaination is requested.

You can use some Italian guy's cubic formula, but that turns out real ugly. The thing is, this cubic will no integer or rational solutions, making it especially nasty to factor. So I believe the polynomial is irreducible.
The problem is related to the impossibillity of trisecting a 90 degree angle.

if you follow this carefully, then you can solve this

[url]http://mathworld.wolfram.com/CubicFormula.html[/url]

For $ax^3+bx^2+cx+d=0$, you may use the cubic formula, which is
$x=\sqrt[3]{\left(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\right)+\sqrt{\left(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\right)^2+\left(\frac{c}{3a}-\frac{b^2}{9a^2}\right)^3}}+\sqrt[3]{\left(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\right)-\sqrt{\left(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\right)^2+\left(\frac{c}{3a}-\frac{b^2}{9a^2}\right)^3}}-\frac{b}{3a}$

EDIT: If you think this is not convinient, take a look at the quartic formula...[url=http://planetmath.org/encyclopedia/QuarticFormula.html]planetmath.org/encyclopedia/QuarticFormula.html[/url]
Masoud Zargar

First of all, how would you get the insight to use trig to simplify a polynomial?

[quote="Gyan"]...$2(4cos^3\theta - 3cos \theta) +1 = 2(cos(3\theta))+1$...[/quote]

Also, why is the above step true?

And by the way, to boxedexe, I am amazed you took the time to tex the cubic formula!   :P

matt276eagles- $\cos (3\theta) = 4 \cos^3 \theta -3 \cos \theta$ is fairly well know formula. Also using trig to solve cubic is fairy standard (and simplest) . (See any math book) Basic idea here is,  given a cubic of the form $ax^3+bx+c=0$ just start with $x=p\cos \theta$ and select $p$ such that the equation is reduced to the nice form given above.  BTW try $x = \cos 40^o = 1.53208889..$  in the equation above and you can verify that the answer does work..

[quote="Gyan"]You can also, let $x=2cos\theta$
so you get $8cos^3\theta-6 cos\theta + 1 = 2(4cos^3\theta - 3cos \theta) +1 = 2(cos(3\theta))+1 = 0$
This gives $cos(3\theta) = -(1/2)$ so $3\theta = 120^o \ or \ 240^o \ or \ 480^o$

so $x = 2 cos (40^o)$or $x= 2 cos( 80^o)$ or $x = -2 cos (20^o)$

BTW one of  popular problem is:
To prove $cos(20^o) cos(40^o) cos(80^o) = (1/8)$
This becomes obvious from the above equation, and seeing the products of the threee roots :)[/quote]

I solved this problem as well.This equation is Archimedes ????? equation.What do you call this equation generally? :?

I believe that there is a little something in AoPS I that explains just enough of it.  I had looked at some web-sites, and was discouraged by the mountains of co-efficients and radicals.  However, that little page in AoPS I shows us the way very well, with no mammoth equations at all.  After reading that, I felt confident enough to confront the algorithm for quartics.

All roots can be expressed in the form $p \cos{\theta}$?  That is amazing, it gives me new reason to practice more trig.

[quote="Gyan"]

BTW one of  popular problem is:
To prove $cos(20^o) cos(40^o) cos(80^o) = (1/8)$
This becomes obvious from the above equation, and seeing the products of the threee roots :)[/quote]

you can also do this by this method
$cos(20^o)cos(40^o)cos(80^o)=P$
$sin(20^o)cos(20^o)cos(40^o)cos(80^o)=Psin(20^o)$
and then use double angle over and over, and at the end note that $sin(160^o)=sin(20^o)$
i personally prefer this method.

[quote="breez"]Where can I get a link to a proof for that Cos (3A) form?[/quote]
[quote="seamusoboyle"]Just use Demoivre...[/quote]

Yes, or, if you don't know De Moivre, you can prove it noting that : $cos(3a) = cos(2a+a)$ ;)
So, you can find it using the fact that : $cos(a+b) = cos(a) cos(b) - sin(a) sin(b)$

By the way, it was Vieta that came up with the trig solution.