A circle of radius 50 has a chord of length 50. Another circle passes through the center of the first circle and the endpoints of the chord. Find the area of the intersection.

[hide]Is the radius of the circle ${\frac{50\sqrt{3}}{3}}$? If it is then I think the answer is ${\frac{8570\pi-1250\sqrt{3}}{3}}$[/hide]

I think you're close. Do you have a solution?

[hide]
Start by drawing the radii to the endpoints of the chord. You then have an equilateral triangle with sides 50, and, of course, the center of the circle will be the same as the center of the triangle. You can apply (or derive then apply) the equation for the distance between the center of an equilateral triangle and any of the vertices: $\frac{2s\sqrt{3}}{3}$, (you can derive this using the median concurrence theorem, btw). You get $\frac{50\sqrt{3}}{3}$.

Since this is also the radius of the circle, you find the area of the circle is $\frac{2500\pi}{3}$.

The last step is to find out if all of the smaller circle is contained in the larger circle. We know that since the radius is $\frac{50\sqrt{3}}{3}$, the diameter is twice that. We find that this exceeds 50, and because one of the points on the smaller circle is the center, we know that part of the smaller circle will be outside the larger one.

Could someone please continue from here? I'm having trouble thinking of a simple way of determining the difference...[/hide]

[hide]$\frac{8570\pi-7500\sqrt3}{9}$  Which is my final answer. I have my work attached.[/hide]

[hide]Actually, I think the answer is $\frac{8750\pi-3750\sqrt{3}}{9}$[/hide] Please tell me if I am right or wrong.


trigonometry
logarithms
function
inequalities
geometry unsolved
geometry

Prove that for every triangle ABC,hold:
$(\sqrt{3}-\sin A)(\sqrt{3}-\sin B)(\sqrt{3}-\sin C) \geq \sin A\sin B\sin C$

$(\sqrt{3}-\sin A)(\sqrt{3}-\sin B)(\sqrt{3}-\sin C) \geq \sin A\sin B\sin C$

$\Leftrightarrow \ln \frac{\sqrt{3}-\sin A}{\sin A}+\ln \frac{\sqrt{3}-\sin B}{\sin B}+\ln \frac{\sqrt{3}-\sin C}{\sin C}\geq 0$

Consider the function $f(x)=\ln \frac{\sqrt{3}-\sin x}{\sin x}$, $x\in (0,\pi)$

$f\"(x)=\frac{\sqrt{3}(\sqrt{3}-\sin x-\sin x\cos^2 x)}{\sin^2 x(\sqrt{3}-\sin x)^2}\geq 0$, so $f(x)$ is convex.
(This is because the solution to $\sqrt{3}-k-k(1-k^2)<0$ is $k<-\sqrt{3}$. ;) )
By Jensen Inequality,

$\ln \frac{\sqrt{3}-\sin A}{\sin A}+\ln \frac{\sqrt{3}-\sin B}{\sin B}+\ln \frac{\sqrt{3}-\sin C}{\sin C}$

$\geq 3\ln\frac{\sqrt{3}-\sin (\frac{A+B+C}{3})}{\sin (\frac{A+B+C}{3})}$
$=0$ :D

Equality holds iff $A=B=C=\frac{\pi}{3}$.

P.S. This problem is more 'inequality-like'. :)

Thank you,leepakhin!
But I still think that there is an easier solution for this problem.

We have: \[\prod \sin A\le\prod\left(\sqrt 3-\sin A\right)\\ \Longleftrightarrow \prod\frac{2\tan\frac{A}{2}}{1+\tan^{2}\frac{A}{2}}\le\prod \left(\sqrt 3-\frac{2\tan\frac{A}{2}}{1+\tan^{2}\frac{A}{2}}\right)\\ \Longleftrightarrow 8\prod\tan\frac{A}{2}\le\prod\left(\sqrt 3+\sqrt 3\tan^{2}\frac{A}{2}-2\tan\frac{A}{2}\right)\]
What is truly because: \[\sqrt 3+\sqrt 3\tan^{2}\frac{A}{2}-2\tan\frac{A}{2}=\frac{2}{\sqrt 3}+\frac{1}{\sqrt 3}+\sqrt 3\tan^{2}\frac{A}{2}-2\tan\frac{A}{2}\\ \ge\frac{2}{\sqrt 3}+2\tan\frac{A}{2}-2\tan\frac{A}{2}=\frac{2}{\sqrt 3}\\ \Longrightarrow \prod\left(\sqrt 3+\sqrt 3\tan^{2}\frac{A}{2}-2\tan\frac{A}{2}\right)\ge \frac{8}{3\sqrt 3}\ge 8\prod\tan\frac{A}{2}\]


induction
combinatorics proposed
combinatorics

$ k$ is a positive integer. Prove that there is a graph with all vertices' degrees of $ 3$, that doesn't contain any cycle of lenght lesser than $ k$.

Ok, I think the following construction works. (It is a little hard to follow)

Define $ C_k$ to be a cycle with length $ k$ (so it looks like a regular k-gon)

Now, start with $ C_{6n}$ where $ n$ is some positive integer with $ n > k$

On each of the edges of $ C_{6n}$, draw another $ C_{6n}$ so that it shares an edge and two vertices with the original $ C_{6n}$ such that each pair of adjacent $ C_{6n}$ shares an edge and two vertices.

It is now clear that in the original $ C_{6n}$, all vertices have degree $ 3$ and every vertex incident to any of those vertices has degree $ 3$ and the rest of the degrees are $ 2$. Now, we see that on the boundary, there are are $ 6n \minus{} 4$ vertices that have degree 3 for each $ C_{6n}$, this gives us $ 6n(6n \minus{} 4)$ vertices with degree $ 3$. Label these vertices in order from $ 1$ to $ 6n(6n \minus{} 4)$ and take each of these numbers in modulus $ 2n(6n \minus{} 4)$ so that we have $ 3$ of the same numbered vertex for each residue. For each of these residues, draw a vertex and connect it to each of the residue-labelled vertex. We will result in a 3-regular graph. It is clear that the smallest cycle found will have length $ 3n(6n \minus{} 4) \plus{} 1 > n > k$. Hence, this graph suffices.

This construction is not true. There is always a $ C_6$ in this graph, look at the cycle $ 1 \minus{} A \minus{} 2n(6n\minus{}4)\plus{}1 \minus{} 2n(6n\minus{}4)\plus{}2 \minus{} B \minus{} 2 \minus{} 1$ (since we know that there is such $ A$ and $ B$)

The solution I know is really nice and complicated. We are gonna do induction on $ k$. Small cases are true. Assume that we have a $ 3\minus{}regular$ graph that doesn't consist any cycles of lenght smaller than $ n$. Now assign a different power of $ 2$ and a direction to each of the edges and let $ M$ be a natural number greater than the sum of all these. Let all the vertices divided to $ M$ vertices (that is, think of every vertice a group of $ M$ vertices) and for every vertice $ A$ label them with $ A_1, A_2, ..., A_M$. Connect $ A_i$ and $ B_j$ if and only if 1) $ A$ and $ B$ have an edge between them, let $ 2^k$ be the number assigned to it  2) If the direction of the edge is from $ A$ to $ B$, $ i\minus{}j\equal{}2^k$, otherwise $ j\minus{}i\equal{}2^k$. It is easy to check that this graph is $ 3\minus{}regular$ and doesn't consist any cycles of lenght lesser than or equal to $ n$


When the natural numbers P and P', with P>P', are divided by the natural number D, the remainders are R and R', respectively. When PP' and RR' are divided by D, the remainders are r and r', respectively. FInd a relationship between r and r'.

P :equiv: R (mod D)
P' :equiv: R' (mod D)

Multiplying, we have PP' :equiv: RR' (mod D). Since

PP' :equiv: r (mod D)
RR' :equiv: r' (mod D)

we have r :equiv: r' (mod D).

Actaully, fierytycoon, the relationship goes even more than that...try to make the relationship even closer..

[hide]Since r and r' are both between 0 and D-1, and they're congruent (mod D), r=r'.[/hide]


can someone explain why

1+2+3...+n = n(n+1)/2

Thanks

Pair 1+2+3....+(n-2)+(n-1)+n into groups:
1+n 2+(n-1) 3+(n-2)......the sum of each of these groups is n+1.
Since we have n terms, and we are grouping them by 2's, the sum is n(n+1)/2

Thanks

I have another problem. Maybe you can help me too ?

Pat squared two different positive integers and added
the results. Although Ali did the same thing and got
the same sum, S, Ali's integers were different from Pat's.
What is the least possible value of S ?

[hide]S=65[/hide]

I don't have an elegant solution  :(
I just used guess and check kinda

[quote="SoccerBrainy40"]Going along with this, I have another question.

Say you were trying to find the sum of an arithmetic sequence, but it wasn't counting by ones. What's the formula for that?

For example (something I'm just making up):

2+4+6...+38

or

3+7+11...+27

Thank you![/quote]

The formula for the sum of an arithmetic series where a is the first term and b is the last term, is (b+a)n/2, where n is the number of terms.

here,

try this

if we are adding up the number to 100, we get 5050...

the formula works...

it works because we are actually flipping over the numbers and dividing them in half

like this...

1   +  2 + 3 +....
100+99+98+...

101 + 101 + 101 .... we have 100 of them...
so we multiply them and divide them by 2

extension...

1 + 3 + 5 +...+99

solve this~!

extension...

1 + 3 + 5 +...+99

[hide]
Okay, so as I've learned the formula is n(n+1)/2.

I know that the total number of terms is 50. I have my own way of doing this, erm, let me try to explain...

So I have the pattern 1+3+5+...+99. I try to make a sort of "sequence" to follow to make it so that after applying the sequence to a number, I can get the term number of that number.

For example, I need a sequence to apply to 1 so that after applying it, the answer will be 1 (because it's the first term). And the same sequence needs to be applyed to 3 to get 2. The same sequence needs to be applyed to 5 to get 3. And so on.

I figured out for this arithmetic sequence that the "sequence" is plus one, divide by 2. Try it...(1+1)/2 is 1, (3+1)/2 is 2, and so on. (99+1)/2 is 50, therefore 99 is the 50th and last term.

Applying the formula, (50)(51)/2 = (25)(51) = 1275.

Is that right? :)
[/hide]

OOOOO,,, no it's not right....

thin this way....

look at the picture....

we have 3 by 3 balls and we counted like 1 + 3 + 5
we get
3^2 = 1+3+5...

now try it again....

[hide="the formula you want"]2n - 1 is the last number.... and you just square the n..that's what you want...[/hide]

the sum of the first n odd integers is n^2, but the general trick also works here:

we have an arithmetic sequence with 50 terms, first term 1 last term 99 so the sum is (99+1)*50/2=50^2

Okay, I was getting formulas messed up.

Sum of an arithmetic sequence, adding by 1s, where n is the last term: [n(n+1)]/2

Sum of first n odd numbers: n2

Sum of any arithmetic sequence, where a is the first term, b is the last term, and n is the number of terms: [(a+b)n]/2

So the problem in which I used the wrong formula, the answer should be:
[(1+99)50]/2
=[(100)50]/2
=5000/2
=2500

Gotcha! ;)

Dang... If i had known these before i could do tons of problems like 20 times faster! Gar, guess thats why you go to AoPS.


calculus
integration
trigonometry
calculus computations

The following expression is true: $\int\frac{du}{\sqrt{a^2-u^2}}=\sin^{-1}\left(\frac{u}{a}\right)+C$

but I want to prove that  $\frac{d}{du}\left[\sin^{-1}\left(\frac{u}{a}\right)+C\right]$ is $\frac{1}{\sqrt{a^2-u^2}}$

$let \ y\ =\ \frac{u}{a}$, then $\frac{d}{dy}[\sin^{-1}y+C] = \frac{1}{\sqrt{1-y^2}}$  substituting back $y$ I get: $\frac{1}{\sqrt{\frac{a^2-u^2}{a^2}}}$ which is not same as above.

What is wrong with my solution? :!:

Chain rule! $\frac{d}{du}=\frac{d}{dy}\cdot \frac{dy}{du}$


calculus
integration
logarithms
calculus computations

Other than using integrating by parts, how to use other techniques to solve the integration: S(lnx)^2dx?

From my point of view, you will always have to use integration by parts.

I agree with Carcul. I usually present this in class by starting with the substitution $ t\equal{}\ln x,$ rewritten as $ x\equal{}e^{t},\,dx\equal{}e^{t}\,dt.$ This rearranges the problem to

$ \int(\ln x)^{2}\,dx\equal{}\int t^{2}e^{t}\,dt.$

In this form, the integration by parts is then more obviously presented.


function
abstract algebra
group theory
linear algebra
matrix
linear algebra unsolved

This is a nice problem similiar to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=85457]this one[/url] but the solution is somehow different and also more difficult!

Suppose $f: \mathcal M_n(\mathbb F) \longrightarrow \mathcal M_m(\mathbb F)$ is an injective function which for each $X,Y\in \mathcal M_n(\mathbb F)$ we have $f(X)f(Y)=f(XY)$. Prove that $m\geq n$

It follows from the fact that the maximal cardinality of an abelian subgroup of symetries matrices in $M_n(F)$ is $2^n$.

Here is my solution at the exam (It's not detailed.)
Consider the matrix \[ A=\left( \begin{array}{cccccc} 0&\dots & & & 0& \\ 1 & 0 &\dots & & \BIG {0}\\ \ &1 & \\ & & \ddots \\ \BIG{0} & &\dots&1& 0& \end{array} \right)  \]. You see $A^n=0\neq A^{n-1}$. Also you can prove that $\mbox{Rank}(f(I))>\mbox{Rank}(f(A))>\dots >\mbox{Rank}(f(A^n))$ (We use the injectivity of function here). So $m\geq n$

Here is my proof and is more detailed.
Consider the matrix $A$ above. It's obvious that $\mbox{Im}(f(I))\supset\mbox{Im}(f(A))\supset\dots\supset\mbox{Im}(f(A^n))$. So $\mbox{Rank}(f(I)) \geq\mbox{Rank}(f(A))\geq\dots\geq\mbox{Rank}(f(A^n))$. Suppose that for $i<n, \mbox{Rank}(f(A^i))=\mbox{Rank}(f(A^{i+1}))$. So $\displaystyle f_{|\mbox{Im}(f(A^i))}: \mbox{Im}(f(A^i))\longrightarrow\mbox{Im}(f(A^i))$ is a surjective function. Therefore $\mbox{Rank}(f(A^i))=\mbox{Rank}(f(A^{i+1}))=\dots=\mbox{Rank}(f(A^n))$. You see that $A^n=A^{n+1}=0$. So $f(A^n)=f(A)^n=f(A)^{n+1}$. We conclude that $Id_{|\mbox{Im}(f(A^n))}=f_{|\mbox{Im}(f(A^n))}$. But $\mbox{Im}(f(A^n))=\mbox{Im}(f(A^{n-1}))$ and $Id_{|\mbox{Im}(f(A^{n-1}))}=f_{|\mbox{Im}(f(A^{n-1}))}$. So $f(A^{n-1})=f(A)^{n-1}=f(A)^n=f(A^n)$. Using the fact that $f$ is injective $A^n=A^{n-1}$. But $A^n=0\neq A^{n-1}$. Contradiction proves that $\mbox{Rank}(f(I))>\mbox{Rank}(f(A))>\dots >\mbox{Rank}(f(A^n))$. So $m\geq n$


calculus
integration
complex analysis
calculus computations

show :D

$ \int_{0}^{2\pi}\;e^{e^{\imath\theta}}\;\textbf d\theta\;=\;\boxed{2\pi}$

Use complex analysis:

Let $ u=e^{i \theta}$ which implies $ du=i u d\theta$.  We then have a contour integral over the unit circle:

$ \oint \frac{e^{u}}{i u}du = 2 \pi i \frac{1}{i}$ Residue$ (f(u),0)=2\pi \times 1 =2\pi$


function
limit
calculus
calculus computations

The function $ f(t)$ is defined, for $ t \ne 0$, by

$ f(t) \equal{} \frac{t}{e^{t} \minus{}1}$.

(i) By expanding $ e^{t}$, show that $ \lim_{t\to 0} f(t)\equal{}1$. Find $ f'(t)$ and evaluate $ \lim_{t \to 0} f'(t)$.

(ii) Show that $ f(t) \plus{} \frac{1}{2}t$ is an even function.

(iii) Show with the aid of a sketch that $ e^{t}(1\minus{}t) \le 1$ and deduce that $ f'(t) \ne 0$ for $ t \ne 0$.

Sketch the graph of $ f(t)$.

hello
1.
$ e^{x} \equal{} 1 \plus{} \frac {t}{1!} \plus{} \frac {x^2}{2!}....$
$ \lim_{t\to 0}f(t) \equal{} \frac {1}{1 \plus{} \frac {t}{2!} \plus{} \frac {t^2}{3!}...} \equal{} 1$
$ f'(t) \equal{} \frac {e^{t}(1 \minus{} t) \minus{} 1}{(e^{t} \minus{} 1)^2}$
$ \lim_{x\to 0}f'(t) \equal{} \frac { \minus{} 1}{2}$
2. $ g(t) \equal{} f(t) \plus{} \frac {t}{2} \equal{} \frac {t(1 \plus{} e^{t})}{2(e^{t} \minus{} 1)}$
$ g( \minus{} t) \equal{} \minus{} t\frac {(e^{ \minus{} t}\plus{} 1)}{2(e^{ \minus{} t} \minus{} 1)} \equal{} g(t)$
3. Sketching f(t) ::

$ \lim_{x\to\infty}f(t) \equal{} \plus{} \infty$.
see $ f'(t)\le 0$ as $ e^(t)(1 \minus{} t)\le 1$ for every value of t it is valid .  so $ f(t)$ is decreasing  function.
thank u


Prove that $ \mathbb{R}^d_4$ is a normed linear space

I'm not sure I understand your notation; what is $ \mathbb{R}_4$?

sorry, should have explained..

$ \mathbb{R}^d_p \equal{} \{x \equal{} (\xi_1,\xi_2,...,\xi_d)| (\sum_{i \equal{} 1}^d|\xi_i|^p)^{1/p} < \infty\}$


In triangle $ABC$, $D$ is a point on $AB$ and $E$ is a point on $AC$ such that $BE$ and $CD$ are bisectors of $\angle B$ and $\angle C$ respectively. Let $Q,M$ and $N$ be the feet of perpendiculars from the midpoint $P$ of $DE$ onto $BC,AB$ and $AC$, respectively. Prove that $PQ=PM+PN$.

DL=DK=X   SE=ET=Y  PQ=Z  without loss of generic DP<PE
PE/DE=PN/DK  DP/DE=PM/SE
DP/DE=(Z-X)/(Y-X) PE/DE=(Y-Z)/(Y-X)
PN=X*(Y-Z)/(Y-X)  PM=Y(Z-X)/(Y-X)  PN+PM=(XY-XZ+YZ-XY)/(Y-X)=Z=PQ
So PN+PM= PQ

$ In \ barycentric \ coordonate \ have: \ D(a;b;0), E(a;0;c) \Rightarrow  DE \ equation \ is: \\ \frac{x}{a}\equal{}\frac{y}{b}\plus{}\frac{z}{c} \Leftrightarrow
  \frac{S_{PBC}}{a}\equal{}\frac{S_{PAC}}{b}\plus{}\frac{S_{PAC}}{c}  \Leftrightarrow |PQ|\equal{}|PM|\plus{}|PN| \ (if \ P\in[DE]).$

Dear Mathlinkers,
I want to give a synthetic response to the problem.
The notations are the same as these of "Plane Geometry".
2.PQ = LD + TE
2.PQ = DK + ES
2.PQ = 2.PN + 2.PM
and we are done
Sincerely
Jean-Louis

Construct K,L,M,J be projection of D,E on side ABC. PH//KE and PG//DJ and P is midpoint then 2(PG+PH)=DJ+KE. DEML is an trapezoid so midline PF=1/2(DL+EM). BE,CD is bisector then EK=EM, DJ=DL. From all above, we have PF=PH+PG

Define $f(P)=|PM|+|PN|-|PQ|$. This is linear function $g(x_P,y_P)$ in $x_P$ and in $y_P$ as distance of point to a line is such. So it is sufficient to prove that for two distinct points $P$ we have $f(P)=0$. Of course $f(D)=f(E)=0$ QED.

[quote=jayme]
The notations are the same as these of "Plane Geometry".
2.PQ = LD + TE
[/quote]
I guess this notation is different than in post #2. Otherwise the equality is true iff $DL=ET\vee DP=PE$

[quote=mihai miculita]$ In \ my \ solution \ |PQ|\equal{}|PM|\plus{}|PN| \ is \ true \  (\forall) P\in [DE].$[/quote]

Ummm... I am pretty sure this cannot be true? The lengths are clearly monotonic whatever direction you move in(and its not hard to use similar triangles to literally find the length)

Always, its a nice problem

Let the feet of the altitudes from $D,E$ to $BC$ be $G_1, G_2$ and the foot of perpendicular from $D$ to $AC$ be $G_3$ and the foot of the perpendicular from $E$ to $AB$ be $G_4$. Notice that because $BE$ and $CD$ are angle bisectors of $\angle B$ and $\angle C,$ respectively, we have that $\triangle CG_3D\cong \triangle CG_1D$ and similarly for the other, so this gives that $DG_1=DG_3$ and $EG_4=EG_2$. From here, it is trivial to finish, just note that $PQ=\frac{G_4E+G_3D}{2},$ which gives the desired.

Note: Although the idea was simple, I felt this problem was pretty difficult because it is not at all obvious to draw the altitudes.


If x, y, z > 1 such that 1/x + 1/y + 1/z = 2, prove that

(x + y + z)^(1/2) >= (x - 1)^(1/2) + (y - 1)^(1/2) + (z - 1)^(1/2)

I think I provided you with enough inequalities now !
Have fun solving them, they are all interesting.
You can learn lots of useful methods solving them.

I think I have sth that might work, but I used this lemma I'm not going to prove here:
if m, n, p are positive reals satisfying m^2+n^2+p^2+2mnp=1 then there is a triangle ABC s.t. m=cosA, n=cosB, p=cosC. The way to prove this is by seeing that 0<m, n, p<1, so there are acute angles A and B s.t. m=cosA and n=cosB and then we solve the eqn m^2+n^2+p^2+2mnp=1 having p as the unknown quantity, after replacing m and n. We will find that A+B+C=Pi, where C is the angle s.t. p=cosC. Now to the problem:

Let x=a+1, y=b+1, z=c+1, where a, b, c>0. Then the condition sum (1/x) =2 translates, after some calculations, to ab+bc+ca+2abc=1. Now we apply the lemma for m=(bc)^1/2 and the others, so we get (bc)^1/2=cosA, (ac)^1/2=cosB, (ab)^1/2=cosC, ABC being an acute angled triangle. What we have to prove is sum (a^1/2) <= (a+b+c+3)^1/2. We square this to get 2sum ((bc)^1/2) <= 3 iff 2sum (cosA) <= 3 iff sum (cosA)<=3/2, well known.

Or you can just apply the Cauchy - Scharz inequality :D :

(x + y + z)((x - 1)/x + (y - 1)/y + (z - 1)/z) >=
((x - 1)^(1/2) + (y - 1)^(1/2) + (z - 1)^(1/2))

Now note that (x - 1)/x + (y - 1)/y + (z - 1)/z = 1.
Cancelling this factor and taking square roots we get the result. :P

CAUCHY RULES ! 8-)

Should I post some more inequalities or do you think it's enough now ?

Yeah post some more please. I'm not very good at math but I seem to like inequalities. You can post some more on http://www.mathlinks.ro too.

Cheers :lol:

Why do you say you're not so good at math ?
I think that's not true Maverick ! :?

Well I don't have any remarcable performances at math competitions, you on the other hand have! I hope to do better the following 3 years. :)

And you? I don't remember you posting your age anywhere. If you did I'm sorry I didn't notice.

I'm also 16. But I've got only one year to go at high school. (I did IMO this year and normally next year, but I certainly won't do it in 2005 since I'll be at university then.) But you're saying you are 16 years old and you've still got three years to go ? Well, those three years can change a lot ! You know, I didn't know anything about math a year ago. (Yeah of course I knew the school stuff but I never heard of the AM - GM inequality a year ago for example.) It all began when I won second prize at our flemish olympiad - unexpected - without preparation :) . Then I started doing olympiad problems and you can see I did the IMO only one year later ! :D

SO START WORKING and everything will become fine ;-)

Thanks for the advice. I'll do that. Oh and good luck for the next year. Hope you will do very well at the IMO. :)


function
symmetry
inequalities
topology
triangle inequality
real analysis
real analysis unsolved

Let $ (X,d)$ be a metric space and $ f: \mathbb{R}\to\mathbb{R}$ a function with $ f(0)\equal{}0,f'(t)>0,f''(t)\leq 0$ for $ t>0$. Prove that $ f\circ d$ is a metric on $ X$.

The function f is increasing and concave down.

Let $ d' \equal{} fd$

1. $ d'(x,y) \equal{} 0 \Rightarrow d(x,y) \equal{} 0$ (because f is increasing), so $ x \equal{} y$. Of course, $ d'(x,y) > 0$ for $ x \neq y$ (because f is increasing) and $ d'(x,x) \equal{} 0$ because $ f(0) \equal{} 0$

2. symmetry is easy because d is so.

3. $ d'(x,z) \equal{} f(d(x,z)) \leq f(d(x,y)\plus{}d(y,z))$ (because f is increasing) $ \leq f(d(x,y)) \plus{} f(d(y,z))$ (because f(0) = 0 and f is concave down), so triangle inequality is verified.

[quote="Soarer"]$ \leq f(d(x,y)) \plus{} f(d(y,z))$ (because f(0) = 0 and f is concave down)[/quote]Can you elaborate on this step please?

In general, if f is a concave function on I, $ x,y \in I$, $ 0 \leq t \leq 1$, then
$ f(x) \plus{} f(y) \ge f(tx\plus{}(1\minus{}t)y) \plus{} f(ty\plus{}(1\minus{}t)x)$.
In our case, set x = a+b, y =0, $ t \equal{} \frac{1}{a\plus{}b}$

[quote="Soarer"]In general, if f is a concave function on I, $ x,y \in I$, $ 0 \leq t \leq 1$, then
$ f(x) \plus{} f(y) \ge f(tx \plus{} (1 \minus{} t)y) \plus{} f(ty \plus{} (1 \minus{} t)x)$.
[/quote]How does this follow from $ f(tx\plus{}(1\minus{}t)y)\geq t f(x)\plus{}(1\minus{}t)f(y)$?

I'm sorry that I made a typo in my third post - the sign of inequality is reversed.

Maybe let me rewrite it again.

You want to prove $ f(a\plus{}b) \leq f(a) \plus{} f(b)$ (then you can substitute a = d(x,y), b = d(y,z))

By concavity, $ f(a) \ge \frac{a}{a\plus{}b} f(a\plus{}b) \plus{} \frac{b}{a\plus{}b} f(0)$
$ f(b) \ge \frac{b}{a\plus{}b} f(a\plus{}b) \plus{} \frac{a}{a\plus{}b}f(0)$
Sum them up, you are done.


[quote="me@home"] \[x,y,z\in \mathbb{R}. \text{Prove: }xyz(x+2)(y+2)(z+2)\leq(1+\frac{2(xy+yz+zx)}{3})^{3}\] [/quote]
It's not true: $x=y=-1$ and $z=2.$ :wink:

here is a simple solution if the variables are nonnegative, which I thinnk is the case

Let $xy+yz+zx=a$ then we have the following simple inequalities:
$xyz\le \left(\frac{a}{3}\right)^{\frac{3}{2}}$
$xyz(x+y+z)\le \frac{a^{2}}{3}$
so using the above:
$xyz(x+2)(y+2)(z+2)\le \frac{a^{3}}{27}+8\left(\frac{a}{3}\right)^{\frac{3}{2}}+2a\left(\frac{a}{3}\right)^{\frac{3}{2}}+\frac{4a^{2}}{3}$
now we only need to prove:
$\left(\frac{2a}{3}+1\right)^{3}\geq \frac{a^{3}}{27}+8\left(\frac{a}{3}\right)^{\frac{3}{2}}+2a\left(\frac{a}{3}\right)^{\frac{3}{2}}+\frac{4a^{2}}{3}$
or $1+\frac{7a^{3}}{27}+2a\geq 8\left(\frac{a}{3}\right)^{\frac{3}{2}}+2a\left(\frac{a}{3}\right)^{\frac{3}{2}}$
which follows by just summing up:
$\frac{5a^{3}}{54}+1+\frac{3a}{2}\geq 8\left(\frac{a}{3}\right)^{\frac{3}{2}}$ and
$\frac{a^{3}}{6}+\frac{a}{2}\geq 2a\left(\frac{a}{3}\right)^{\frac{3}{2}}$ :wink:

I initially thought it should work from Bernouli - but for my surprise it doesn't! (or am i that sleepy?).

You know, the "way" of the inequality, the $(1+\frac{x}{n})^{n}$ way etc. everything here says "use Bernoulli"

I mean i thought that $LHS \leq 1+x \leq (1+\frac{x}{3})^{3}$ But my intuition then said: it won't work, i checked it for some random numbers and it doesn't work ;<

PS. Eagle you did it in 10 minutes? :D

What inspired you to find a such solution,life_in_dream?

seems like the official solution. Either way it's very easy.


Is there a way to figure out how many successful games I need to reach a certain precentage?

Ex:

I have 165 successful games and 40 unsuccessful, which is 80% success rate[color=red], rounded to no decimal places[/color].

How may successful games do I need to reach 82%?

[quote="stergiu"][quote="Jeff82"]Is there a way to figure out how many successful games I need to reach a certain precentage?

Ex:

I have [color=red]160[/color] successful games and 40 unsuccessful, which is 80% success rate.

How may successful games do I need to reach 82%?[/quote]

Would you like a simple equation like :

 $ \frac {160 \plus{} x}{200 \plus{} x} \equal{} \frac {82}{100}$

 or something simpler ?[/quote]
That's looks good.  Thanks!

Eh.
There are two scenarios:
[list]Either the given information the $ 165$ and $ 205$ numbers are incorrect since we're assuming $ 80\%$ is an [b]exact value[/b].
[b]OR[/b]
It's because the given percentage is just an [b]approximation[/b] and throwing our procedure off (since approximations generally don't [u]exactly[/u] agree with exact values): $ 0.804878...\approx80\%$.[/list]

I think it's more appropriate to know if the $ 80\%$ given is an [b]approximation[/b] or an [b]exact value[/b] before proceeding (generally most percentages aren't exact, but we don't know if this assumption is always true).

Assuming it's an approximation:
[hide="Hint"]
$ \frac {165 \plus{} x}{205 \plus{} x} \equal{} \frac {82}{100} \implies 82(205 \plus{} x) \equal{} 100(165 \plus{} x)$ See what you can do with this (remember $ x$ [b]must[/b] be an [url=http://www.artofproblemsolving.com/Wiki/index.php/Integer]integer[/url], so just round it after getting the exact value of $ x$).[/hide]

[quote="stergiu"][quote="Jeff82"][quote="alanchou"]165 successfull games and 40 unsuccessful games is not an 80% success rate  :wink:[/quote]
165/(165 + 40) doesn't equal .80487... ? :help:[/quote]

 Yes !
 But this does not change the question. As you can see on the above post , I have pointed it out and corrected the error with red.[/quote]
Does my correction in my first post make it correct?


kengdie a , the link is invalid~~you can try click it,yes.


analytic geometry
integration
trigonometry
calculus
geometry
3D geometry
calculus computations

I am looking at a worksheet and it asks to sketch a solid whose volume is given by:

$ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{\arctan \sqrt{2}}\int_{0}^{2\sec \phi}{\rho }^{2}sin \phi d\rho d\phi d\theta$ $ +$ $ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{\arctan \sqrt{2}}^{\frac{\pi}{2}}\int_{0}^{4 \cot \phi \csc \phi}{\rho }^{2}sin \phi d\rho d\phi d\theta$

I don't know how to go about picturing this. I know that rho represents the distance from the center, but do not know how to interpret a boundary of $ s\sec \phi$ for it. I feel somewhat overwhelmed because it doesn't make any sense and I'm not making any headway after 20 minutes of sketching in my notebook.

Edit: I also have maple, but don't know how to use it to help me sketch this. I've had Maple 11 for a week, and am still learning how to use it.

Well this solid is much more easily visualized in cylindrical coordinates, but you can work it out by playing with spherical equations.

$ \rho = 2 \sec{\phi}$ means $ \rho\cos{\phi}= 2$.  And $ \rho\cos{\phi}= z$.  So you can imagine it will start at the origin and go up until it hits the plane $ z=2$.  Combining this with the $ \theta$ and $ \phi$ limits, you see that the first integral comes out to be [hide] an upside down half cone lying in the positive $ x$ and $ z$ region with base parallel to the $ xy-$ plane and with height $ 2$ and base radius $ 2\sqrt2$. (if you still don't get, I can expand)[/hide]

Try the other one similarly.  Convert $ \rho = 4\cot{\phi}\csc{\phi}$ to rectangular coordinates.


1) Since we know that hot air rises , why it is cooler at the top of a mountain than near the sea level ?

2)The internal energy of a body consist of kinectic energy and potential energy of its particles . How could we distinguish between the internal energy of a body and its temperature ?

3)What justification is there in neglecting the change in gravitational potential energy of molecules in a gas ?

Good morning everyone(I know it's not that early but i just got up now :D )
Here are my guesses......

[hide]1)I have done this question before but i can't remember anything :( Anyway what i'm thinking right now is that the air molecules do not reach that high due to gravitational pull of earth......[/hide]
[hide]2)I suppose the average kinetic energy of the molecules gives the temperature of a substance,right?SO I'm not really able to understand what ur question asks....[/hide]

1) The air that is close to the atmosphere / outer space boundary (of course there's no actual boundary) is cold because the outer space average temperature is about 1 K (I think) and the air here radiates away all of its heat. On the other hand, air clos to the surface is warmed up from the surface (and Earth's surface is being warmed by the Sun). It is true that the warmer air wants to go up and the colder air down but they can't do it fast enough so they lose / gain energy on the way and end up having the right temperature.

3) I'm not sure in what situation is the potential energy neglected but just generally - if you take a 20 °C oxygen and lift it by 1 km, the potential energy gained by each molecule is less than 1/10 of its kinetic energy, so that's probably the reason.

[quote="kubus"]1) I think the loss due to radiation is neglible. Of course there is some, but I'd think it has almost no impact on the temperature differences, especially if we compare temperature at the bottom and the top of a mountain. Imagine that there is a bunch of warm air in the valley and it starts rising uphill. What happens? The pressure decreases! Yeah, it's *something* like shadysaysurspammed said, but said in a statistical point of view. When the pressure decreases, the tepmerature does too, if we assume the change adiabatic or near-adiabatic.
If you go from about 500m elevation (950Mbar) with 20C air to about 2000m (800)Mbar adiabatically (large amount of air rising uphill can be pretty adiabatic), you inveitably get about 5C.[/quote]

I think both my answers were cheap :D

Kubus,when pressuere decreases,doesn't temperature increase...or am i getting confused in the pressure cooker thing which is completely different from the atmospherical concept :|  :|
Well i can'y remember well but last year i thuink we had an experinment to show that the pressure is inversely prop to temperature...i'm not too sure.Is the pressure up in the mountains really less.If so,then kubus is right :D (of course) :blush:

Using the equation of state of an ideal gas:

$pV = nRT$

you see that not changing the volume, lowering the pressure will decrease the temperature proportionally. The problem is that the chunk of air blown upwards has no reason to stay in the same volume. However, if you assume that there is no heat transfer between this chunk of air and the surroundings, you know that $pV^\kappa$ is constant, where $\kappa$ is about 1.4 for air. You can figure out the temperature from this additional assumption.

there's also a lot [i]less[/i] air up there... because its at a higher elevation- so total energy would go down, too

Yes, you're right kubus. I didn't realize that this happens when the air goes up / down... Nevertheless, the reason why air in high altitudes is cold is still that it has to be in equilibrium with the outer space. If it didn't radiate away its energy, the atmosphere would keep getting warmer and warmer and warmer.... what you explained is why the temperature gradient is so steep at low altitudes - and it sounds like you have it right :). Anyway, you're the one who understands this better.

Well ... my opinion (note that I'm calling this an opinion;-) is that even if the temperaure of the outer space was 20C and the whole atmosphere would be set up to have this temperature, after a while (a few pertubations) a gradient would form and it would be about as steep as it is now. Anything that would go up would have its temperature decreased and vice versa. With no heat inflow/outflow, the temperature at the bottom would rise and at the top it would fall, developing exactly the temperature profile that you can calculate, except that it would be a LITTLE less steep, since there is some heat transfer (by radiation also;).

Also, concerning the radiation transfer, I think the air is transparent to the most of the radiation we are to consider here. I don't think radiation as a very important factor in creating the temperature difference.

Please, respond and say what you think.

Aaahh, I finally see what you're saying. Yes, I think you were right again :roll:... But anyway, how does the heat transfer at the topmost layers of atmospehere work? I know that the air doesn't radiate too much energy but there are also some space particles (Hydrogen and stuff :)) that are flying around and have pretty low temperatures and those could cool down the atmospehere a bit - but that's probably not too much, is it? I don't know, I've just always had this feeling that things in space are very very cold and I still can't help this idea...

I didn't say air does not radiate too much energy it surely does radiate much more than outer space, but what I said was that all the air radiates about the same amount ... I guess.

I'm not too sure about what to respond to he second part of your post, but here's what I imagine that happens there - the problem is not with cold molecules hitting the atmosphere, but rather NOT hitting it, therefore letting it expand (as much as gravity allows), and that is the cooling-by-conduction, it you dare to call it that way. Another thing is the radiation, which is really THE way that the Earth (and I guess the atmosphere also) cools itself. Being in space, you feel cold because you radiade a lot of energy and don't receive any (except from the Sun, which is quite a lot around the Earth and so). Hm?

[quote="kubus"]1) I think the loss due to radiation is neglible. Of course there is some, but I'd think it has almost no impact on the temperature differences, especially if we compare temperature at the bottom and the top of a mountain. Imagine that there is a bunch of warm air in the valley and it starts rising uphill. What happens? The pressure decreases! Yeah, it's *something* like shadysaysurspammed said, but said in a statistical point of view. When the pressure decreases, the tepmerature does too, if we assume the change adiabatic or near-adiabatic.
If you go from about 500m elevation (950Mbar) with 20C air to about 2000m (800)Mbar adiabatically (large amount of air rising uphill can be pretty adiabatic), you inveitably get about 5C.[/quote]
When you say that the pressure decreases as we go up shouldnt it increase? cos the density of the gas increases so pressure would increase right? :?  :(
I see the question this way as the warm air rises due to bouyancy it gives away heat by colliding with other particles and staying in contact, so it should be cold at the top because by the time the air reaches there it has already lost all its energy? Is this right?

Riddler that is what bus has been saying(I think)

About the p[ressure thing,i'm still not sure but when you go high up,the pressure does decrease...and ridler remember that the atmosphere becomes thinner as we move up.

so pressure decrease but why does temperature decrease????

As kubus have said, pV = nRT. If p goes down, so does T.

PV = nRT is only for a contained gas at constant Gibbs Free Energy, so it doesn't apply here.

1)  This is a physics problem based on energy flows to and from the earth.  First, there is significant radiation loss from the atomosphere to space.  All gases heavier than Helium escape.  The rest are too heavy at the temperatures to exceed the escape velocity of the earth.  We have bound gases, without a solid barier.

About half of the energy reaching the surface of the earth comes from inside the earth. (Radioactive decay primarly)  The other half comes from the sun.  Most of the energy reaches the earth in a form that passes through the atomosphere, hits the ground and is transferred to infrared, that is absorbed by the greenhouse gases, primarly water.  So while the sun's energy comes from, well the sun, as far as the heat transfer to the atomosphere is concerned, it effectively is delivered at the surface of the earth.  So, given that both halfs of the heat the earth receives effectively come from the surface and the heat is lost from the top, it is not surprising that generally the atomosphere cools off as one travels upward through it.

The reason that the tendency of hot (less dense) materials to rise doesn't dominate the heat flow and move the hot (less dense) material higher is that the heat is being lost at too great a rate, given the turbulance in the atomosphere.  The same is true for most boiling liquids.  Take a thermometer and measure the temperature of heating water (don't touch the bottom of the pan), if you want experiemential proof.  While the same is true for boiling water, it has phase changes that dominate the heat transfer.

2) The kenitic energy is the temperature.  The potential energy of a piece of paper would include the chemical bonds.  When you burn the paper, some of this potential energy is released and there is light and a jump in temperature.  Is this sufficient?

3)  This question is more difficult because of the way it is stated.  The more general question is, "What justification is there in neglecting the change in gravitational potential energy an object at the surface of the earth?"  The earth's diameter is what in miles, the distance over which we are assuming the change in gravitational potential energy is what?  Alternatively, calculate the gravitational potential energy at the surface of the earth (for air or yourself), then calculate what it would be at the top of the atomosphere (say 10 miles up where the planes fly or 100 miles up where objects can orbit).

wizard2378: 1) That's what I was thinking too but then kubus made me believe that the heat isn't really lost from "the top" - do you think that the top layers of atmosphere radiate more energy than the surface of Earth? Although using $pV=nRT$ might not be completely correct, the point is still that as gas expands, it loses temperature. And as the air rises to higher temperatures, it does expand because we all know that the air pressure decreases with distance from the Earth's surface.

bus -- The major loss of heat from the planet's surface is radiation in the infrared range.  The atomosphere is opaque to this frequency.  (Remember all the pointless discussion, laws and treaties about CO2 emissions causing climate change.  This is the same as glass or plastic in a greenhouse.)

bus -- PV = nRT still doesn't work.  Your statement assumes that the volume of the gas remains constant.  Let's imagine that we have a tensionless bag containing a gas and we start at sea level and move it up in the atomosphere.  What happens to the volume of the bag?  It increases as the pressure decreases, since the amount of gas (n) is constant (that's the point of the bag), R is constant and we are working on T.  [While I admit that non-ideal gases will give off temperature as the molecules as moved apart, PV = nRT is the ideal gas law that assumes these effects don't occur.]  Back to the problem.  We move the bag up, P decreases then V increases.  Now we allow the bag to transmit heat (holding the bag in place for a while, after a rapid increase in height will suffice) the T to equalize with the rest of the atomosphere at the height of the bag.  T goes down and V decreases, but T doesn't go down because P goes down.

bus -- There happens to be experiemential support of this as well.  Look up the temperature profile of the atomosphere in a book or online somewhere.  You'll notice that the temperature goes down to below 40 below (F and C -- it's the same at that one point) then increases.  If PV = nRT was the driving force, the decreasing temperature would continue.  The increasing T at highest altitudes is due to radiation from the sun that interacts at the highest levels (breaking oxygen bonds forming monotomic oxygen and the ozone (O3) that provides protection against ultraviolet radiation (by absorbing the radiation and raising the tempertaure of the ozone).

Almost all of the water vapor in the planet's atmosphere is concentrated at the lowest levels of the atmosphere.  Because it's a great greenhouse gas, we can see why the temperature falls with altitude(until we reach the stratosphere)

[quote="wizard2378"]The major loss of heat from the planet's surface is radiation in the infrared range. The atomosphere is opaque to this frequency.[/quote]Hm, that sounds like a problem... :) You're probably right.

And yes, I forgot that both pressure and volume change... but if you assume adiabatic expansion, the air will cool down anyway. I think that this assumption is reasonable.

bus -- You're right again about the location of the water.  The main reason for this is that the water carrying ability of air is very temperature dependent.  At temperatures below about 40 below, there is a very reduced ability for air to carry water.  (That's why so little snow falls an Antarctica.)

It is also fortunate, as the radiation at the outside of the atomosphere would break down the oxygen/hydrogen bond, which would allow the hydrogen to escape, leaving us with much less hydrogen.

However, the green house gases main fuction is to block the heat from the surface, not the sun.  (Look at a distribution of energy from the sun based on frequency.)

[quote="wizard2378"]bus -- You're right again about the location of the water. [/quote]Thanks but it was daveed who said that :).

Anyway, I think that the fact that most of the Earth's heat is radiated by the top layers of the atmosphere and also the adiabatic expansion that kubus mentioned both affect the temperature distribution in our atmosphere. One would probably have to do some calculations to find out which of these effects is dominant at what altitudes. My guess is radiation in high altitudes and kubus' thermodynamics in lower but I'm not sure so don't say I'm wrong :). And that's about as much as I can add to this discussion...


Find the natural numbers $x,y,z$ satisfying the following conditions:
i)$x^3+y^3=2z^3$
ii)$x+y+z$ is a prime number

The first equation has just the trivial solutions $x=y=z$ and $z=0, x=-y$, thus there is not much left to prove (was posted before).

[quote="ZetaX"]$z=0, x=-y$[/quote]
A natural is a positive integer

Did I say something else¿
And I'm the last person you have to tell that  :roll:

Hahaha..  :rotfl:
Everyone knows what you mean, [i]ZetaX[/i].
[i]pavel kozlov[/i] was just trying to be precise. After all, you can't deny the fact that what he said was valid.  :P


geometry
linear algebra
matrix
3D geometry
tetrahedron
vector
trigonometry

Prove that the area of a triangle with vertices $ (x_1,y_2),(x_2,y_2),(x_3,y_3)$ is given by $ \frac{1}{2} \begin{vmatrix}x_1 & x_2 & x_3 \\
y_1 & y_2 & y_3 \\
1 & 1 & 1\end{vmatrix}$.

You should probably indicate that you want to take the absolute value there  :)

In general, I find it cleanest to write $ \frac {1}{2} \cdot \text{abs} \; \det M$ for situations like that, where $ M$ is the matrix...

[hide="Hint"]The value $ \text{abs} \left( \frac {1}{3!} \begin{vmatrix}x_1 & x_2 & x_3 \\
y_1 & y_2 & y_3 \\
z_1 & z_2 & z_3 \end{vmatrix} \right)$

gives the volume of the tetrahedron defined by the points $ (x_i, y_i, z_i), \; i \in \{1,2,3 \}$ and the origin.

If you've never worked with determinants, [url=http://en.wikipedia.org/wiki/Determinant#Determinants_of_3-by-3_matrices]here's a visualization of what the determinant does[/url].
[/hide]

The alternate formula that I know of is 1/2 of the magnitude of the cross product of two of the sides which are treated as vector.  In that case the formula should look like
$ \frac {1}{2} \begin{vmatrix}i & j & k \\
x_1 \minus{} x_2 & y_1 \minus{} y_2 & 0 \\
x_1 \minus{} x_3 & y_1 \minus{} y_3 & 0\end{vmatrix}$

and then take the magnitude of the cross product.

However, how do we show that these two determinants are actually equal (I think we can take the transpose and row/column reduce the determinant in the first post to get two of the three rows to be something like $ x_1 \minus{} x_2, y_1 \minus{} y_2,0$ and $ x_1 \minus{} x_3,y_1 \minus{} y_3,0$ and the row that we have to fix should look something like $ x_1, y_1,1$)?  But how do we transform this row to be a row of $ 1$'s (as we can't get unit vectors i,j,k but replacing them by 1 so that it is a (scalar) determinant instead of a vector resulted from the cross product which we have to take the magnitude from)?

Edit: If you take a look at [url]http://en.wikipedia.org/wiki/Cross_product#Geometric_meaning[/url], you would see that we can also calculate the volume of the tretahedron using $ a \times b \cdot c \equal{} (|a\parallel{}b|\sin \phi)|c|\cos\theta$
 where $ |a \times b|$ is the area of the base.  Can we also show that these two are equal?  :)

Yeah, I think I understand what you said.  It is still surprising that the two determinants (we can't really call that for the formula with cross product) differ only in a row.  However, do you know how was $ \text{abs} \left( \frac {1}{3!} \begin{vmatrix}x_1 & x_2 & x_3 \\
y_1 & y_2 & y_3 \\
z_1 & z_2 & z_3 \end{vmatrix} \right)$ derived?  Does it also follow from the cross product (and/or dot product), or does the derivation follow a totally different logic?

not familiar with the determinant thing though, but expanding that gives,
$ \frac{1}{2}\left[x_1(y_2\minus{}y_3)\plus{}x_2(y_3\minus{}y_1)\plus{}x_3(y_1\minus{}y_2)\right]$
let, $ ABC$ be the triangle, then draw perpendiculars $ AP, BQ, CR$ to $ x\minus{}axis$. we get three trapezia. from fig.
$ ar(ABC)\equal{}ar(ABQP)\plus{}ar(APRC)\minus{}ar(BQRC)$. Put the coordinates you,ll get
$ ar(ABC)\equal{}\frac{1}{2}(y_2\plus{}y_1)(x_1\minus{}x_2)\plus{}\frac{1}{2}(y_1\plus{}y_3)(x_3\minus{}x_1)\minus{}\frac{1}{2}(y_2\plus{}y_3)(x_3\minus{}x_2)$
solving we get the above equation.

The determinant formula in question for the triangle is equivalent to a much larger class of area formulas.

Consider the second way of computing a the determinant of a 3x3 matrix in [url=http://en.wikipedia.org/wiki/Determinant#Determinants_of_3-by-3_matrices]this article[/url]. Does this look familiar?

[hide="Hint"]Try ignoring the 1's that you're multipying in.

Does this look anything like the Shoelace Algorithm?[/hide]

[quote="TZF"][b]To MWD's first question:[/b] You cant. The cross product "determinant" formula is really an abuse of notation; it's not actually a determinant. $ i,j,k$ can't be elements of a matrix because they are vectors, not numbers. The only reason they use that notation is that it's compact and easy to remember.

[b]To MWD's second question:[/b] You probably could, but it would be a roundabout approach, and it would be gross.[/quote]

Why can't the entries of a matrix be vectors?

This area was derived by dropping perpendiculars to the horizontal axis, thus forming a trapezoid. Then, to obtain the area of the triangle, get the difference of the area of the trapezoid and the other smaller trapezoids formed.

[quote="Altheman"]Why can't the entries of a matrix be vectors?[/quote]

You're right, that was poorly articulated on my part.

I meant to say that they can't be used in a matrix that you take the determinant of.


vector
linear algebra
matrix
linear algebra unsolved

Suppose $ A$ and $ B$ are two equally-sized symmetric square matrices with real entries.

Suppose that there exist two matrices $ X$,$ Y$ such that $ AX \plus{} BY$ has full rank.

Prove that $ A^2 \plus{} B^2$ has full rank

$ A^2,B^2,A^2\plus{}B^2$ are non negative symmetric matrices;if $ A^2\plus{}B^2$ isn't invertible then there exist $ z\not\equal{}0$ s.t. $ z^{T}(A^2\plus{}B^2)z\equal{}0$ thus $ z^{T}A^2z\equal{}z^{T}B^2z\equal{}0$ or $ \parallel{}Az\parallel{}^2\equal{}\parallel{}Bz\parallel{}^2\equal{}0$. We can deduce that $ z^{T}A\equal{}z^{T}B\equal{}0$ and $ z^{T}(AX\plus{}BY)\equal{}0$ and $ AX\plus{}BY$ hasn't full rank that is a contradiction.    QED

This problem is very nice problem

let $ x\in ker(A^2 \plus{} B^2)$ where $ x$ a column vector of $ R^n$

$ A^2x \plus{} B^2x \equal{} 0$ multiply left side by  the transpose of $ x$

$ \parallel{}Ax\parallel{}^2 \plus{} \parallel{}Bx\parallel{}^2 \equal{} 0$

we deduce $ Ax \equal{} Bx \equal{} 0$

$ ker(A^2 \plus{} B^2)$ is included in $ Ker(A)$ and $ ker(B)$

Now we use the lemma of "factorisation " (i don't know how to translate in english)
_________________________________________________________________________
lemma of factorisation: E,F,G be K-vectors spaces, $ f \in L(E,F)$ and $ g \in L(E,G)$
Then ( $ ker f$ is included in $ ker g$ ) is equivalent ( there exist $ h \in L(F,G)$ with $ g \equal{} hof$ )
____________________________________________________________________________

Use this lemma twice

there exist $ M,N$ matrices of $ M(n,R)$ such that

$ A \equal{} M(A^2 \plus{} B^2)$

$ B \equal{} N(A^2 \plus{} B^2)$

and

$ X^{T}.A \plus{} Y^{T}.B \equal{} (X^{T}M \plus{} Y^{T}N)(A^2 \plus{} B^2)$   (*)

take the determinant

$ det(X^{T}.A \plus{} Y^{T}.B) \equal{} det(AX \plus{} BY)$ different zero

 take determinant of (*), RHS  is also different to zero

$ A^2 \plus{} B^2$ is invertible

////////////////////////////////////////////////////////////////////////////
 I wonder if there exist a solution using that every real symetric
is diagonalisable with an orthogonal matrix
////////////////////////////////////////////////////////////////////////////


takım seçme sınavının tam tarihi hakkında bilgisi olan var mı?
ayrıca eğer geçen yıllardaki gibi olursa sanırım cahit arf matematik yarışmasının 2. aşaması da o tarihlere denk geliyor. bu yüzden yarışmaya kayıt yaptırıp yaptırmamak konusunda kararsız kalıyorum. en kısa zamanda beni bilgilendirirseniz sevinirim.

Arkadaşlar merhaba..
Yabancı uyruklu öğrenciler katılabılirmi acaba???
Ben çok araştırdım da her hangi bir şey bulamadım.
2003te  katılmıştım IMOya 9. sınıftayken.
Ondan sonra bazı nedenler yüzünden katılamadım. // bu arada söyleyeyim ben azeriyim, Ankarada üni. okuyorum

Yardımcı olursanız sevinirim..

bu bahsettiğim yarışmaların ikisi de lise öğrencileri için ayrıca sanırım türkiye çapında düzenlenen bir üniversitelerarası matematik yarışması yok.


I dont understand number 4 on round 2. I dont know if its against to rules to help understand the problem.

Please do not post anything on the forums.  If you think you have a question that can be answered, PM it to one of the admins, but they more than likely will not be able to answer your questions.

(Click [url=http://www.artofproblemsolving.com/Forum/privmsg.php?mode=post&u=1163]here[/url] to Private Message an administrator with your question.)

(Mods: Lock please.)

We cannot offer help or clarification on a problem while the round is in progress.


What is the $ y$-coordinate of the point where the line whose equation is $ 3x\minus{}5y \equal{} \minus{}15$ intersects the $ x$-axis?

When any line intersects the $ x$-axis, the $ y$-coordinate is $ \fbox{0}$.


conics
ellipse
geometry
geometric transformation
reflection
geometry unsolved

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14642[/img]

 :(

$ a, b$ are directions of the 2 given directions of conjugate diameters, intersecting at the ellipse center $ C$ (otherwise you could not draw the ellipse). When a circle is projected to an ellipse by a parallel projection, any 2 perpendicular diameters of the circle project to 2 conjugate diameters of the ellipse. It follows that any ellipse chord parallel to an ellipse diameter is cut in half by its conjugate diameter. Draw parallels $ a', b'$ to $ a, b$ through $ P.$ $ M$ is intersection of $ a, b'$ and and $ N$ is intersection of $ b, a'.$ Reflect $ P$ in $ M, N, C$ into $ X, Y, Z$ and these 3 points also lie on the ellipse, so we already have 5 points: $ P, Q, X, Y, Z.$ We could do a similar thing with the point $ Q$ to get 3 more points, but it is not necessary. We need the ellipse tangent at $ P$ (or at some other point).

Consider the pentagon $ PXZYQ$ inscribed in the ellipse as a degenerate hexagon $ PPXZYQ.$ By Pascal theorem, the intersections $ C \equiv PZ \cap XY,$ $ S \equiv PX \cap ZQ,$ $ T \equiv PP \cap YQ$ are collinear. The collinearity line $ CS$ can be constructed and $ YQ$ cuts $ CS$ at $ T,$ such the $ p \equiv PP \equiv PT$ is the ellipse tangent at $ P.$ Let this tangent cut $ a$ at $ K.$ Then $ KX$ is the ellipse tangent at $ X.$ If the ellipse was projected to a circle by a parallel projection, the $ KX$ image would be a reflection of the $ KP$ image in the $ a$ image. $ K, M, C \in a$ are collinear. Let the ray $ CK$ cut the unknown ellipse at $ A.$ Again, if the ellipse was projected to a circle, the $ CA$ image would be the circle radious and the $ M$ image would be inversion of the $ K$ image in this circle, so that $ CA^2 \equal{} CM \cdot CK.$ To construct the geometric mean of $ CK, CM,$ reflect $ M$ in $ C$ into $ M',$ draw a circle with diameter $ KM',$ erect perpendicular to $ KM'$ at $ C,$ intersecting this circle at $ D,$ draw a circle with center $ C$ and radius $ CD,$ intersecting the line $ a \equiv M'CMK$ at $ A. A',$ the endpoint of the conjugate diameter $ a.$ [hide="In exactly the same way, construct the endpoints of the the other conjugate diameter."] The ellipse tanget $ p$ at $ P$ cuts $ b$ at $ L.$ Then $ LY$ is the ellipse tangent at $ Y$ and $ L, N, C \in b$ are collinear. Let the ray $ CL$ cut the unknown ellipse at $ B,$ so that $ CB^2 \equal{} CN \cdot CL.$ Reflect $ N$ in $ C$ into $ N',$ draw a circle with diameter $ LN',$ erect perpendicular to $ LN'$ at $ C,$ intersecting this circle at $ E,$ draw a circle with center $ C$ and radius $ CE,$ intersecting the line $ b \equiv N'CNL$ at $ B. B',$ the endpoint of the conjugate diameter $ b.$[/hide]
Given the ellipse conjugate diameters $ AA', BB'$ with their endpoints, erect perpendicular to $ BB'$ at $ C.$ Circle with center $ C$ and radius $ CA$ cuts this perpendicular at $ D$ (we already have $ D$), such that the angle $ \angle BCD$ is acute. Take the midpoint $ O$ of $ BD$ and draw a circle $ (O)$ with center $ O$ and radius $ OC.$ Take its diameter $ UV$ through $ BD,$ so that $ U, B, O, D, V$ follow on the line in this order. Then $ CU, CV$ are the directions of the ellipse major / minor axes and $ UD \equal{} VB, UB \equal{} VD$ are the major / minor axes lenghts. See [url]http://www.mathlinks.ro/viewtopic.php?t=203843[/url] for a proof (the hidden part). Transfer the major axes lengths on the rays $ CU, CV$ into $ CA_1, CB_1$ and also on the opposite rays into $ CA_2, CB_2,$ yielding the ellipse major / minor axes $ A_1A_2, B_1B_2.$ Circle $ (B_1)$ with center $ B_1$ and radius equal to $ CA_1$ cuts the major axis $ A_1A_2$ at the ellipse foci $ F_1, F_2.$

When you say "Construct ellipse ...", that means not only constructing 3 more points, but constructing the major / minor axes and foci. See the attached figure.


Harvard
college
calculus
function
University of Chicago
number theory

My parents have offered to buy me one.  I have very, very, very mild experience with Maple (I've used it maybe 3 or 4 times), so I'm leaning towards Maple.  I understand that Mathematica is more popular, but that doesn't particularly concern me.  Which should I get?

Well either one is very expensive.  People will tell you that Mathematica is the industry standard.  However, it is most important that you <b>buy the same one that the college you are going to uses.</b>  That's because you will have CAS problems, and you have to print out the transcript of what you did.  These will relate to what you are studying in class.  While either are cool to have, you want to get the most use out of it.  I have only used Maple b/c that is what URI uses.  I think both are equally good, but I have never used Mathematica, so I would not be able to make a fair judgement.  Also, most likely the computers in your college will have the CAS installed, or (and I don't do this) you can d/l it for free from Kazaa Lite or something like that (again, that is illegal).

I'm going to get the student edition, which is <$200.

Edit: I'm partial to University of Chicago (I'm going to be a senior in high school this year), and they seem to be partial to Mathematica.

Still a lot of money in my opinion  :)  If you are not going to college any time soon, then I think other people can help you out.  Simon or Joel have probably used both.

I heard today that Maple is better than Mathematica at number theory.  Anyone know about the validity of this statement?

A lot of Maple versus Mathematica arguments are much like Windows versus Mac arguments. I decided on Mathematica because of the HUGE number of third-party books on Mathematica, some of which I had seen in bookstores before I got the student edition. There are whole books on using Mathematica for number theory, and I don' think you will exhaust its capabilities any time soon.

Well, now that we've merged, let's see if we can get a few more responses...eh?

Think of them like a really powerful graphing calculator for the computer.  Like a TI-89, only better, and can do much nicer things because it has the whole monitor to spread out over (nice graphs, etc).

Yeah,  what she said.  They are math programs for the computer.  They can do just about anything that you want them to do(or so I am told).  I don't have either but I want to get Mathematica.  The biggest reason I haven't gotten it yet is it's price.  I do however have Mupad which is free although I hardly ever use it.

[quote="joml88"]I don't have either but I want to get Mathematica.  The biggest reason I haven't gotten it yet is it's price.[/quote]

The student price for Mathematica is hardly higher than list price for a good graphing calculator. It is a steal of a deal: of course the company hopes you become addicted and someday "need" to buy the professional edition. Anyway, I thought the student edition of Mathematica was a necessary expense to make sure my children are prepared for study of higher math.

I think some releases of Nucalc (also known as [url=http://www.pacifict.com/]Graphing Calculator[/url]) are free. Certainly list price for that program is not high, and my son likes it a lot.

Take into account that some colleges will provide you with free access to one or the other.

Personally, I really like Mathematica, but I have no Maple experience.

I actually have both CASs installed and can access both through the university through putty though its like command prompt. Some "not so decent" people I know have downloaded Mathematica 5 and Maple 7 and cracked it". Maple 9.5 is very "new". And the university here uses Maple. So I like it. If u r cheap, or like excellent german products, Mupad 2.5 is available for free. And it is a pretty nice CAS. I suggest you download it.

Harvard (and many other colleges, as well) has a mass licence to Mathematica along with about 10,000 other potentially useful programs.  You can download it free onto your computer for the duration of your time at Harvard.  You might try looking for the Chicago computer services (or whatever they call it) webpage to see what they have; you might already need a password, though.

I unfortunately do not have any experience with Maple.  And mostly I just use Mathematica for fun.  So I don't have great advice for choosing between the two.

[quote="Magnara"][quote="Alison"]Take into account that some colleges will provide you with free access to one or the other.[/quote]

How free?  Like, I have to go to the math lab free, or I can get it from my computer free?  Because I am pretty lazy.[/quote]

Download the trial from Mathematica and then download a keygen or crack is the easiest way. (Note I am not suggesting that you do this. There are alot of bad websites out there with trojans.)

I think he was talking about how free it is at college.  I don't think he was asking how to illegally obtain it because that is what is seems like you are explaining how to do.

[quote="JS1527"]However, it is most important that you <b>buy the same one that the college you are going to uses.</b>[/quote]

In my opinion Maple is better at graphing and visual outputs, while Mathematica is simpler and uses less computer resources. There might be other differences in more complex features which I have not noticed yet.

Otherewise follow JS1527's advice. Not only will you have CAS problems but it also gets confusing trying to remember the inputs for both software (everytime I remember an input for one program I tend to forget the equivalent in the other in a few days :P )

Yea, some colleges use it a lot, and some you will rarely see it.  URI uses Maple a lot in intro calculus classes, but I have a friend who goes to Harvard and took multivariable calculus but has never used it.

I now a few minutes ago, finnished installing Mathematica 9.5. Its amazing. I love the improved interface.
Maple vs. Mathematica.
Here are some features:
Mathematica is obvious more stable and accurate. It takes up less memory and is faster most of the time.
Maple has more packages. These include from Precalculus to LInear Algebra. It was fun using them for Calc 3 this summer. There are alot of tutorials which you can download and view. But still, Mathworld is more organized. There is also a dictionary.

here is zdnet's reviews
mathematica
http://www.eweek.com/article2/0,1759,1498046,00.asp
maple
http://www.eweek.com/article2/0%2C1759%2C1269844%2C00.asp

[quote="churchilljrhigh"]I now a few minutes ago, finnished installing Mathematica 9.5. Its amazing. I love the improved interface.
[/quote]
mathematica 9.5? maybe Maple 9.5 or Mathematica 5?

I'm always using mathematica. I don't know much about Maple.
Mathematica 5 made some radical changes to its Integrate function, some bugs are fixed and some new bugs are produced.
What about Maple?

[quote=mathpro12345]\bumpity \bumpity \bump
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\bumpity \bumpity \bump
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\bumppppp!!!!

(16-year bump lol!!!)[/quote]
[quote=Taco12]Please don't bump really old threads[/quote]

[hide=@#22][url=aops.com/community/user/586791][b]mathpro12345[/b][/url] · 25 minutes ago [url=aops.com/community/p21022092](view)[/url][color=transparent]helo[/color]
\bumpity \bumpity \bump
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\bumpity \bumpity \bump
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\bumppppp!!!!

(16-year bump lol!!!)

-----------
[color=#5b7083][aops]x[/aops] 22[color=transparent]hellloolo[/color] [aops]Y[/aops] 0 [color=transparent]hellloolo[/color] [/hide]
Any changes from 2004 to now?


Here are a few more problems:

1) In the following addition procedure, each letter represents a distinct digit:
$ MQM\ \plus{}\ NQ\ \equal{}\ QHHH$. What is the value of $ N$?

2) A number is chosen at random from the set $ {1,\ 2,\ 3,\ ...,\ 7!}$. What is the probability that the number is a factor of $ 7!$? Express your answer as a common fraction.

The Challenge:
Try to find a way to do these without guess and check.

Please show your work.

Enjoy! :P  :P  :P  :P

[quote="LHM"]Here are a few more problems:

1) In the following addition procedure, each letter represents a distinct digit:
$ MQM\ \plus{}\ NQ\ \equal{}\ QHHH$. What is the value of $ N$?

2) A number is chosen at random from the set $ {1,\ 2,\ 3,\ ...,\ 7!}$. What is the probability that the number is a factor of $ 7!$? Express your answer as a common fraction.

3) What is the greatest factor of $ 11!$ that is one greater thatn a multiple of $ 6$?

The Challenge:
Try to find a way to do these without guess and check.

Please show your work.

Enjoy! :P  :P  :P  :P[/quote]
[hide="1"] Q must be one, since M must be single-digit, and can only carry one over. $ M \equal{} 9$ because $ Q\plus{}N$ can at most be in the teens. Now, solve for H and N, and you get $ N \equal{}\boxed{8}$?
[/hide]
[hide="2"]
$ 7!$ has 60 factors, =$ \frac{60}{7!}\equal{}\boxed{\frac{1}{84}}$? [/hide]
[hide="3"] It cannot be divisible be any multiple of 3 or 2, so try $ 11*7*5*5$, but that doesn't work, so $ 11*7*5 \equal{} /boxed{385}$? I'm not sure about this method. :maybe: [/hide]

wheres number 3?

[hide="EDIT:must hide answers"]1. M must be 9 because a 3 digit number plus a 2 digit numberis always 3 digits unless the hundreds digit is 9. also, Q must be 1. since 9+1=0, the sum must be 1000. then we do 1000-919=81, so n must be 8

2. $ 7! \equal{} 1*2*3*4*5*6*7 \equal{} 2^{4}*3^{2}*5*7$. it has 5*3*2*2=60 factors. $ \frac{60}{7!}\equal{}\frac{1}{84}$[/hide]

[quote="LHM"]The correct answer:

1. 8

2. 1/84[/quote]
So what happened to number 3?

[hide="2"]
$ 7! \equal{} 5040 \equal{} 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2 \equal{} 2^{4}\cdot 2^{2}\cdot 5\cdot 7$
So, the number of factors is $ 5\cdot 3\cdot 2\cdot 2 \equal{} 60\Rightarrow\frac{60}{5040}\equal{}\boxed{\frac{1}{84}}$[/hide]

@ pascal - You messed up a little, and posting the answer to a problem for the 4th time isn't really necessary.


Find the smallest possible two-digit whole number that when divided by each of the numbers 3, 4, and 5 leaves a remainder of 2.

We see that our number has to be two more than a multiple of 3, 4, and 5.  The smallest number that is a multiple of 3, 4, and 5 is $ \text{LCM}(3,4,5)\equal{}3\cdot4\cdot5\equal{}60$.  Two more than 60 is $ \boxed{62}$


Find all real solutions to the equation: $ x^{5}\minus{}15x^{4}\plus{}10x^{3}\minus{}30x^{2}\plus{}5x\minus{}3 \equal{} 0$

:-) As a whole 5-th degree equations is proven that have no common formulae for their solution. I suppose there is a trick. I'm sending a link with a hint(s) to the solution... If you have troubles (for example you don't speak russian I may translate), please ask me.

http://www.umsolver.com/cgi-bin/concur.pl?l=rus&s=math&n=1

Hope this is interesting to you.

A complete solution.

Our equation is equivalent to:
1. $ (x\minus{}1)^{5}\minus{}10x^{4}\minus{}20x^{2}\minus{}2 \equal{} 0$
2. $ (x\plus{}1)^{5}\minus{}20x^{4}\minus{}40x^{2}\minus{}4 \equal{} 0$
3. $ (x\plus{}1)^{5}\equal{} 2(x\minus{}1)^{5}$
4. $ x\plus{}1 \equal{} 2^{\frac{1}{5}}x\minus{}2^{\frac{1}{5}}$

So the answer is: $ x \equal{}\frac{2^{\frac{1}{5}}\plus{}1}{2^{\frac{1}{5}}\minus{}1}$

If it is possible to you by using similar or different approach (but not Cardano), please, try to solve the equation:
$ 2x^{3}\minus{}3x^{2}\plus{}4x\plus{}1 \equal{} 0$


Regarding the problem 1 Quiz 5, I found this result:

Fix $ k\ge 2$. Any squarefree integer $ n>1$ has a divisor $ d$ with $ d\le\sqrt[k]{n}$ such that $ \tau(n)\le (2\tau(d))^{k}$.


integration
limit
function
real analysis
real analysis unsolved

Hello, got 3 problems here and I'll be very thankful is someone can help. Here they are:

1) Prove that closure of $ l_1$ in the metric of $ l_{\infty}$ coincides with $ c_0$. ($ c_0$ is a space of numerical sequences converging to zero)

I am trying to prove that $ \overline{l_1} \subseteq c_0$ and $ c_0 \subseteq \overline{l_1}$. Inclusion $ c_0 \subseteq \overline{l_1}$ seems to be obvious because if we have some converging sequence $ (x_1,x_2,...,x_n,...) \in \overline{l_1}$, we just can choose some $ k \in \mathbb{N}$ and put $ x_n = 0$ for every $ n\geqslant k$ (аctually we can approach each element from $ c_0$ by "rejecting tails"). But how can we prove inclusion  $ \overline{l_1} \subseteq c_0$?(I read that this fact is a consequence of uniform convergence, but it's still not clear enought for me).

2) Two norms $ \parallel{}.\parallel{}_1$,  $ \parallel{}.\parallel{}_2$ in a normed space $ X$ are said to be equivalent if there exist constants $ a,b >0$ such that $ a\parallel{}x\parallel{}_1 \leqslant \parallel{}x\parallel{}_2 \leqslant b\parallel{}x\parallel{}_1$ for all $ x \in X$. Prove that $ 2$ norms $ \parallel{}.\parallel{}_1$,  $ \parallel{}.\parallel{}_2$ on a normed space $ X$ are equivalent if and only if for every sequence from its convergence in norm $ \parallel{}.\parallel{}_1$ implies its convergence in norm $ \parallel{}.\parallel{}_2$, and 	 vice versa.

Well, if we have 2 equivalent norms, then we can just write $ a\parallel{}x_n-x\parallel{}_1 \leqslant \parallel{}x_n-x\parallel{}_2 \leqslant b\parallel{}x_n-x\parallel{}_1$ for all $ x \in X$, so then we have $ \parallel{}x_n-x\parallel{}_1 \to 0$ if  $ \parallel{}x_n-x\parallel{}_2 \to 0$, and vice versa. But how to prove sufficiency (it seems to be convergence' definition consequence, but how to write it - shall we use the fact that converging sequence is bounded and just put $ a = sup\parallel{}x_n\parallel{}_1, b = sup\parallel{}x_n\parallel{}_2$)?

3) Let $ \alpha \in L_{\infty}(a,b)$. Prove, that is $ \alpha (x) \geqslant \alpha_0 > 0$ at $ (a,b)$, then norms $ \parallel{}f\parallel{}_{L_2 (a,b)}$ and $ \parallel{}f\parallel{} = (\int_{a}^{b} \alpha(x)|f(x)|^2 dx)^{1/2}$ are equivalent in $ L_2 (a,b)$. Will be this statement correct if condition $ \alpha (x) \geqslant \alpha_0 > 0$ we replace with $ \alpha (x) > 0$?

Okay, first part (norms equivalence) is proved. How to prove that if  $ \alpha (x) > 0$ then those norms arent equivalent? No ideas..

Thanks to all for answers.

1)  First, clearly $ l_1 \subset c_0$.  Second, taking the uniform limit allows you to interchange the order of limits.

3)  See if you can pick $ \alpha(x)$ and $ f$ such that one norm is finite and the other isn't.  It might help to set $ (a, b) \equal{} (0, \infty)$.

t0rajir0u, thank you. Now I'll try to find an example for 3). And can you please explain 1) - I still dont understand how can this fact help here (the order of limits interchanging)  :(

If a sequence of sequences $ (x_n)_m \in l^1$ converges uniformly to a sequence $ y_n$, then $ \lim_{m \to \infty} \lim_{n \to \infty} (x_n)_m \equal{} 0 \equal{} \lim_{n \to \infty} \lim_{m \to \infty} (x_n)_m \equal{} \lim_{n \to \infty} y_n$.  Hence $ y_n \in c_0$.

Okay, thanks. Now about 3) - I tried different variants for $ \alpha(x)$ and $ f(x)$, but the main problem here is that $ \alpha(x)$ is bounded a.e. and is positive. For example, I tried combinations like $ \alpha(x) \equal{} \left\{\begin{array}{l} |\sin(x)|, x \neq \pi k, k \in \mathbb{Z}, \\
0.1, x \equal{} \pi k , \end{array} \right.$
at $ (0; \plus{} \infty)$
, or $ \alpha(x) \equal{} \left\{\begin{array}{l} \frac {|\sin(x)|} x, x \neq \pi k, k \in \mathbb{Z}, \\
0.1, x \equal{} \pi k , \end{array} \right.$
 at $ (1; \plus{} \infty)$, but I still can't find such $ f(x)$ to let one norm to be finite and another to be infinite. It seems we have too strict conditions for $ \alpha(x)$, and I can't see and variants for it except some combinations of finite trigonometrical functions like $ \sin(x), \cos(x)$ with their absolute values, but most of those combinations arent integrated in elementary functions, so we usually have no limit (finite or infinite) when calculating integral. Any ideas?

Scratch that, actually; it might be a lot easier to just find a sequence that converges in one norm but not the other.  (I suggest something like letting $ \alpha$ approach zero at an endpoint and letting $ f_n$ get large at the same endpoint.)

Okay, I'll think about that now.
edit : problem 2) solved.

Well, tried this, but its still not clear - the main problem I find here is that fact $ \alpha(x)$ is bounded a.e., so if we have finite interval $ (a,b)$ I can't find such $ f(x)$ (or sequence $ (f_n(x)$), and if we put $ b\equal{}\infty$, we get a strict restriction for $ \alpha(x)$. If someone can give a good example, I'll be very thankful - this is last problem I cant' solve, but need solution today  :(


group theory
combinatorics unsolved
combinatorics

[quote="Dr. No"][quote="Jrb599"][img]http://i220.photobucket.com/albums/dd160/jrb599/untitled.jpg[/img][/quote]

A Gray code of order n, is a list of binary words such that the length of each word is n bits. The words in a Gray code only differ from their nearest neighbors by one bit. For a group, such as a Gray code, to be cyclic means that after some fixed number of operations the group begins to repeat itself. For example, if you had a Gray code of order 3 then the highest number you could represent in said code would be 7. Attempting to represent numbers greater than 7 means that you would have to start reusing Gray code words that had already been assigned, i.e. the code would be cyclic.

And yes, the code given appears to be a non-cyclic Gray code of order 3.[/quote]

Ehm... Wrong.

At least according to http://en.wikipedia.org/wiki/Gray_code (or better http://planetmath.org/encyclopedia/GrayCode.html ), a Gray code is called cyclic if the first line differs from the last line by one digit only. A Gray code corresponds to a Hamiltonian path on the $ n$-dimensional hypercube (more precisely, on the graph consisting of its $ 2^n$ vertices and its edges), while a cyclic Gray code is a Hamiltonial cycle on this hypercube (cf. http://www.mathlinks.ro/viewtopic.php?t=15416 ). So yes, the Gray code given in the first post is cyclic.

  Darij

[quote="darij grinberg"]
Ehm... Wrong.

At least according to http://en.wikipedia.org/wiki/Gray_code (or better http://planetmath.org/encyclopedia/GrayCode.html ), a Gray code is called cyclic if the first line differs from the last line by one digit only. A Gray code corresponds to a Hamiltonian path on the $ n$-dimensional hypercube (more precisely, on the graph consisting of its $ 2^n$ vertices and its edges), while a cyclic Gray code is a Hamiltonial cycle on this hypercube (cf. http://www.mathlinks.ro/viewtopic.php?t=15416 ). So yes, the Gray code given in the first post is cyclic.

  Darij[/quote]

I've got no problem with that stipulation. A cyclic group is one that repeats itself after a fixed number operations -- a Gray code wherein the first word and the last word are adjacent certainly qualifies as a cyclic group.


You are given a triangle $ ABC$ in which $ AB>AC$ and $ D$ on $ BC$ so that angle $ BAD \equal{}$ angle $ CAD$. $ E$ is on $ AB$ (betweeen $ A$ and $ B$) so that $ AE \equal{} AC$. Show that $ F$ can be found on $ AD$ so that $ CDEF$ is a rhombus. Deduce that $ BD: DC\equal{}BA: AC$.

[geogebra]d2b090ffbb593ab1045f83c93bdcad9f7a4895dd[/geogebra]
u can easily see that AF_!_EC nd EF = EC now take point G on AD such that GF=FD thus u can prove that CDEG is a rhombus ( here i have mistaken G by F u should clear it urself)
now extend AD to H such that BH parallel AC
u can easily see that tri BDH similar to  tri  ADC
then tri BHA is isosceles using tha side ratios u can prove the latter one also

I apologize: I can't "easily see that AF_|_EC".

Please explain.

consider the two triangles EAF & CAF
AE=AC
<EAF = <CAF
AF =AF (common)
hence the tris. r congruent nd <AFE=<AFC= 90 degree


abstract algebra
topology
geometry
3D geometry
sphere

If $ \pi_1(X)$ is abelian, is there necessarily a space $ Y$ and a map $ \Omega Y \to X$ that is a weak homotopy equivalence?

Mh, what about the spheres? The only spheres which are parallelizable/H-spaces are $ {\mathbb S}^n$ for $ n\equal{}1,3,7$. The other spheres cannot be given the structure of an $ H$-space, and in particular they cannot be pointed homotopy equivalent to a loop space.  :maybe:


Let $ABC$ be a triangle. Let $D$, $E$ and $F$ be the midpoints of $BC$, $CA$ and $AB$ respectively.

Show that $\angle DAC = \angle ABE \iff \angle AFC = \angle ADB$.

Let the centroid be $G$, then
$\angle{ABE}=\angle{GAE} \Longrightarrow \triangle{EAG} \sim \triangle{EBA}$
$\Longrightarrow AE^2=EG \cdot EB \Longrightarrow \triangle{CEG} \sim \triangle{CBE}$
Therefore $\triangle{CEG} \sim \triangle{BEC}$. Thus $\angle{ACF}=\angle{EBC}$. Since $\angle{A}=\angle{BGD}$, so we have $\angle{ADB}=\angle{AFC}$.

[quote="Arne"]Can you also prove that both $\angle DAC = \angle ABE$ and $\angle AFC = \angle ADB$ are equivalent to $2AC^2 = AB^2 + BC^2$?
(It's not hard either.)[/quote]
Oh that's a nicer result.
Since $4BE^2=2AB^2+2BC^2-AC^2$ and $GE=\frac{1}{3}BE$, so $AE^2=EG \cdot EB \Longleftrightarrow AB^2+BC^2=2AC^2$.

$\angle{AFG}=\angle{AGB} \Longleftrightarrow AG \cdot AD = AF \cdot AB$. Use the formula of the length of the median again, we get $AG \cdot AD = AF \cdot AB \Longleftrightarrow AB^2+BC^2=2AC^2$.

Indeed. It's simple, but I think it's an interesting situation :)

[hide="Here is an equivalent enunciation."]
Arne, and I like the your proposed problem (easy , but nice) !

[color=darkred][b]An equivalent enunciation.[/b] Let $L$ be a point wich belongs to the interior of the triangle $ABC$. Denote the intersections $D$, $E$, $F$ of the rays $(AD$, $(BE$, $(CF$ with the opposite sides respectively. Prove that if two  form the following fourth statements are truly, then and others two are truly :
$1.\blacktriangleright \widehat {CAD}\equiv \widehat {ABE}\ .$
$2.\blacktriangleright$ The quadrilateral $BDLF$ is cyclically .
$3.\blacktriangleright 2\cdot AC^2=AB^2+BC^2\ .$
$4.\blacktriangleright$ The point $L$ is the centroid of the triangle $ABC\ .$[/color]

[u]Remark.[/u] The proof $i.\ \wedge\ j.\Longrightarrow 4.$ for $\{i,j\}\subset \{1,2,3\}$, $i\ne j$ are more difficulty ![/hide]

A different way to prove that $2b^2=c^2+a^2$[hide]
As $BDGF$ is cyclic by the power of point theoreme we have $AF*AB=AG*AD$ or $\frac{c}{2}*c=\frac{2m_a}{3}*m_a \Rightarrow 3c^2=4m_a^2=2c^2+2b^2-a^2 \Rightarrow 2b^2=c^2+a^2$ [/hide]
I think  Mr. Nicula means $1.\blacktriangleright \widehat {CAD}\equiv \widehat {ABE}\ .$  :?

More follow up questions.

I can't remember if all these are true...but if we have $2b^2 = c^2 + a^2$, show that:

1. $A, C, G, H$ are concyclic. ($G$ is the centroid, $H$ is the orthocentre.)
2. The Euler line is perpendicular to $BM$. ($M$ is the midpoint of $AC$.)
3. The Euler line, $AC$ and the line joining the feet of the altitudes from $A$ and $C$ to the other side of the triangle are concurrent.
4. $GK$ is parallel to $AC$. ($K$ is the symmedian point.)


I would like to use Asymptote with WinEdt. Could someone teach me how to set up WinEdt so that I can use Asymptote? I couldn't find any documentation online about integrating Asymptote into WinEdt.

I too would be very interested in learning this.

If you have a cable or dsl connection, it may be easier to just download TeXnicCenter and use that instead. The AoPS wiki has [i]lots[/i] of documentation on getting TeXnicCenter and using it to set up asymptote.


algebra
polynomial
calculus
integration
function
domain
abstract algebra

show the ideal (x) is prime in $ %Error. "display" is a bad command.
\mathbb{z}[x]$.

Consider the degree of the polynomials and that $ \mathbb Z$ is an integral domain.

Let $ f(x) \equal{} f_0 \plus{} f_1x \plus{} \cdots \plus{} f_lx^l\in\mathbb{Z}[x]$ and $ g(x) \equal{} g_0 \plus{} g_1x \plus{} \cdots \plus{} f_mx^m\in\mathbb{Z}[x]$ be such that $ f(x)g(x)\in\langle x\rangle$. Then $ f(x)g(x) \equal{} xh(x)$ for some $ h(x) \equal{} h_0 \plus{} h_1x \plus{} \cdots \plus{} h_nx^n\in \mathbb{Z}[X]$. Now
\[ \left (\sum_{i \equal{} 0}^l f_ix^i\right )\left (\sum_{i \equal{} 0}^m g_ix^i\right ) \equal{} x\left (\sum_{i \equal{} 0}^n h_ix^i\right ).
\]
This implies $ f_0g_0 \equal{} 0$ and because $ f_0,g_0\in\mathbb{Z}$, either $ f_0 \equal{} 0$ or $ g_0 \equal{} 0$. If $ f_0 \equal{} 0$ then $ f(x) \equal{} f_1x \plus{} \cdots \plus{} f_lx^l \equal{} x(f_1 \plus{} \cdots f_lx^{l \minus{} 1})\in\langle x\rangle$. If $ g_0 \equal{} 0$ then $ g(x) \equal{} g_1x \plus{} \cdots \plus{} g_mx^m \equal{} x(g_1 \plus{} \cdots g_mx^{m \minus{} 1})\in\langle x\rangle$. Hence $ \langle x\rangle$ is a prime ideal.

$ \mathbb Z[X]\to\mathbb Z,\ p\mapsto p(0)$ is a ring homomorphism into an integral domain, therefore its kernel, $ (x)$, is a prime ideal. owk


I'm planning to take the Advanced Standing Exam for Chemistry at MIT in a few weeks. Does anyone know of the difficulty of this exam relative to the AP test or National Chemistry Olympiad?

It seems an interesting exam, according to the information [url=http://web.mit.edu/chemistry/www/academic/advancedstanding.html]here[/url]. Have you participated in any Chemistry Olympiad? If yes, could you provide some relevant link? I would like to compare the nature and difficulty of the problems.

Does anyone have the book which contains the specified problems? http://acswebcontent.acs.org/education/student/LS_OL07.pdf here is a link of the local exam from last year.

Thanks for the link. Interesting exam. Good luck in your studies.


function
number theory
totient function
number theory proposed

Hello!

For an integer $n>0$ we denote by $\varphi(n)$ (respevtively $d(n)$) the Euler's totient function (respectively the number of divisors of $n$). Prove that if $n\geq 31$ then $\varphi(n)>d(n)$. For which $n$ do we have $\varphi(n)=d(n)$ and $\varphi(n)<d(n)$?

If $n=p_{1}^{e(p_{1})}\cdot\cdot\cdot p_{k}^{e(p_{k})}$ then $\varphi (n)=\prod_{i=1}^{k}(p_{i}^{e(p_{i})}-p_{i}^{e(p_{i})-1})$ and $d(n)=\prod_{i=1}^{k}(1+e(p_{i}))$.

If $n$ is odd then $p_{i}\geq 3$ for all $i$, therefore $p_{i}^{e(p_{i})}-p_{i}^{e(p_{i})-1}\geq 2\cdot 3^{e(p_{i})-1}\geq e(p_{i})+1\;\forall i$, so $\varphi (n)\geq d(n)$ and we have $\varphi (n)> d(n)$ because $n>31$.

If $n$ is even, similary! I think so.


Solve in natural the equation  a)$ n^2\plus{}8\equal{}m!$ b)$ n^2\plus{}2\equal{}m!$ c)$ n^2\plus{}2\equal{}m!$   :)

Are you sure that you wrote with out errors the equation? Because if [b]n^2+8=m![/b] and [b]n^2+2=m![/b] then 8=2 :o

[quote="Pirkuliyev Rovsen"]Solve in natural the equation  a)$ n^2 \plus{} 8 \equal{} m!$ b)$ n^2 \plus{} 2 \equal{} m!$ c)$ n^2 \plus{} 2 \equal{} m!$   :)[/quote]
[hide="a"]If m is greater than 4, then the units digit is going to be 0, and there aren't any squares with a unit digits of 2, therefore we can conclude that m is either 1,2,3 or 4. Testing each one we see that $ (m,n)\equal{}(4,4)$[/hide]

[hide="b"]Same case as last time, if m is greater then 4, units digit is 0, and there aren't any squares with a unit digits of 8, so once again m is either 1,2,3 or 4. Testing each one we get that $ (m,n)\equal{}(3,2)$[/hide]

Isn't c the same as b?


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

and

$ \frac{a^n}{b^{n\minus{}1}}\plus{}\frac{b^n}{c^{n\minus{}1}}\plus{}\plus{}\frac{c^n}{a^{n\minus{}1}} \geq a\plus{}b\plus{}c$ for all positive integer $ n$.

Just AM-GM :

$ \frac{a^3}{b^2}\plus{}b\plus{}b\ge 3a$

Suming the similars is done .

The other inequality can be solved in the same way .

Sorry guys but can this get solved by Murihead?  :huh:
$ (3,\minus{}2,0) \succ (1,0,0)$

[quote="hasan4444"]Sorry guys but can this get solved by Murihead?  :huh:
$ (3, \minus{} 2,0) \succ (1,0,0)$[/quote]
NO hasan......
Pay attention on these:
Muirihead works ONLY FOR SYMMETRIC INEQUALITIES :wink:

To use Muirhead, you have to have terms like $ \frac{a^3}{c^2}$ in the inequality, as $ b$ and $ c$ must be equal respective to $ a$ (I know, it sounds weird, but for example, if there is $ ab$ in inequality, you must have also $ ac$ the same number of times, to make it symmetric and Muirhead doable)

[quote="geniusbliss"]this looks better-
$ \frac {a^{n}}{b^{n \minus{} 1}} \plus{} \frac {b^{n}}{c^{n \minus{} 1}} \plus{} \plus{} \frac {c^{n}}{a^{n \minus{} 1}}\geq \frac {a^{n \minus{} 1}}{b^{n \minus{} 2}} \plus{} \frac {b^{n \minus{} 1}}{c^{n \minus{} 2}} \plus{} \frac {c^{n \minus{} 1}}{a^{n \minus{} 2}}$[/quote][hide="This is also easy to prove."]
It is easily demonstrable through AM-GM that $ \frac {a^{n}}{b^{n \minus{} 1}} \plus{} \frac {b^{n}}{c^{n \minus{} 1}} \plus{} \frac {c^{n}}{a^{n \minus{} 1}} \geq a \plus{} b \plus{} c.$ Then,

$ (n \minus{} 1)(\frac {a^{n}}{b^{n \minus{} 1}} \plus{} \frac {b^{n}}{c^{n \minus{} 1}} \plus{} \frac {c^{n}}{a^{n \minus{} 1}}) \geq (n \minus{} 2)(\frac {a^{n}}{b^{n \minus{} 1}} \plus{} \frac {b^{n}}{c^{n \minus{} 1}} \plus{} \frac {c^{n}}{a^{n \minus{} 1}}) \plus{} a \plus{} b \plus{} c \geq (n \minus{} 1)(\frac {a^{n \minus{} 1}}{b^{n \minus{} 2}} \plus{} \frac {b^{n \minus{} 1}}{c^{n \minus{} 2}} \plus{} \frac {c^{n \minus{} 1}}{a^{n \minus{} 2}}).$[/hide]

very nice elixir10! better than me using holders'

[quote="Tomekk"]Hello, cant we just write it in form

a^5*c^2 + b^5*a^2 + c^5*b^2 >= a^3*b^2*c^2 + b^3*c^2*a^2 + c^3*a^2*b^2

So (5,2,0)>(3,2,2) by Muirhead. Or I am mistaken?[/quote]Again, Muirhead only works for symmetric inequalities, not cyclic ones.

$ a^n, b^n, c^n$ and $ \frac{1}{a^{n\minus{}1}},\frac{1}{b^{n\minus{}1}},\frac{1}{c^{n\minus{}1}}$ are oppositely sorted. The result follows by rearrangement.


[quote="1234567890"]Given that a family has two kids, one is a son, and one kid is named William, what is the probability that the other child is also a son?[/quote]

This problem is weird.

I think the answer is
[hide="this"]1/2[/hide]

but my argument tells me it is
[hide="this"]2/3[/hide]

[hide="Here is my argument"]One child is William. That is a son.

Two cases:
1. The child referred to being a son was William
2. The child referred to being a son was not William

In the first case, there are two possible ways
William and son
William and daughter

In the second case, there is one way
William and son

So it is 2 out of 3 possible outcomes, so [b]2/3[/b], but 1/2 appears to be the intuitive answer.[/hide]

[quote="vishalarul"][hide="Here is my argument"]One child is William. That is a son.

Two cases:
1. The child referred to being a son was William
2. The child referred to being a son was not William

In the first case, there are two possible ways
William and son
William and daughter

In the second case, there is one way
William and son

So it is 2 out of 3 possible outcomes, so [b]2/3[/b], but 1/2 appears to be the intuitive answer.[/hide][/quote]
[hide]I agree with everything up to the last sentence. Since you're making a "tree" of outcomes, the first two major cases each have a probability of 1/2. Then, the two sub-cases of the first major case each have a probability or 1/4, and the only sub-case of the second major case has a probability of 1/2. Therefore, I think the answer should be 1/2 + 1/4 = [b]3/4[/b].[/hide]

vishalarul, there are 4 cases not 3.

1. son named william, another son
2. son named william, girl
3. son, william-son
4. son, william-son

the 3rd and 4th case has a higher chance of appearing because its conditional

(1/2)(1/2)+(1/2).  1/2 chance first son is william, 1/2 chance you have another son.  1/2 chance the first son isnt william, the second must be william and thus be a son.
You have to assume william is a boys name and girls/boys are 50/50 at birth.

EDIT: i guess i like pie beat me to it..


Find all (x, y, z, w)∈N such that (2^x)(3^y)-(5^z)(7^w)=1.


If $ f(x) \equal{} 3x \plus{} 1$, what is $ f(3)$?


[u]$ Question 1$[/u]
A mathematician used to boast that he was $ x$ years old in the year $ x^{2}$. Another mathematician, wishing to improve on this, told Henry E. Dudeney that he was $ a^{2}\plus{}b^{2}$ years old in the year $ a^{4}\plus{}b^{4}$, $ 2m$ years old in the year $ 2m^{2}$, and $ 3n$ years old in the year $ 3n^{4}$. In which years were the two mathematicians born? Find integers $ a, b, m, n$ and tell me, if you would, the names of the two famous mathematicians.

[u]$ Question 2$[/u]
The lengths of two of altitudes of a triangle are $ 29$ and $ 9$.
The length of the third altitude is also a natural number.
What is the probability that it is a prime number?


geometry
incenter
circumcircle
geometric transformation
reflection
Euler
parallelogram

[quote="Nbach"]Let I be the incenter of triangle ABC and (I) tagents BC,CA,AB at points D,E,F. Let any point P in the
plane of triangle ABC. Lines through P and perpendicular to BC,CA,AB intersect them at X,Y,Z.
X',Y',Z' are centroids of triangles XEF,YDE,ZDF. Prove that: 3.R(X'Y'Z')=IP where R(X'Y'Z') is radius
of the circumcenter of triangle X'Y'Z'.[/quote]

[b]Changing notation:[/b] Let $ U,\ V,\ W$ be centroids of the $ \triangle XEF,\ \triangle YFD,\ \triangle ZDE.$ Then $ 6\ R(UVW) \equal{} IP.$
______________________________________________________________________

Let $ O$ be the circumcenter of the $ \triangle ABC.$ $ I$ is the circumcenter of its contact $ \triangle DEF,$ let $ J,\ L$ be its orthocenter and 9-point circle center. Let $ I_D,\ I_E,\ I_F$ be reflections of $ I$ in $ EF,\ FD,\ DE.$ The 9-point circle $ (L)$ is centrally similar to circle $ (I_D) \cong (I)$ with similarity center $ D$ and coefficient 1/2, therefore $ L \in IJ$ (Euler line of the $ \triangle DEF$). Similarly, $ L$ is the midpoint of $ EI_E,\ FI_F.$ The circumcevian triangle $ \triangle J_DJ_EJ_F$ of the $ \triangle DEF$ with respect to the orthocenter $ J$ has incenter $ J.$ It is centrally similar to its tangential $ \triangle ABC$ of the $ \triangle DEF$ (parallel sides perpendicular to $ ID,\ IE,\ IF$), therefore $ O \in ILJ$ and $ \frac {IJ}{OI} \equal{} \frac {r}{R}.$ As a result, $ IDJI_D,\ IEJI_F,\ IFJI_F$ are all parallelograms with the common diagonal IJ, their diagonals cutting each other in half.
_______________________________________________________________________

Assume that the $ \triangle ABC$ is reduced to the point $ I$ by contraction. Let $ a,\ b,\ c$ be parallels to $ BC,\ CA,\ AB$ through $ I$ and let $ X_0,\ Y_0,\ Z_0$ be feet fo normals from the arbitrary point $ P$ to $ a,\ b,\ c.$ Because of the right angles, $ X_0,\ Y_0,\ Z_0$ are on a circle $ \mathcal Q$ with diameter $ IP.$ The centroids $ U_0,\ V_0,\ W_0$ of the degenerated $ \triangle X_0E_0F_0 \equiv \triangle X_0II,$ $ \triangle Y_0F_0D_0 \equiv \triangle Y_0II,$ $ \triangle Z_0D_0E_0 \equiv \triangle Z_0II,$ divide the segments $ IX_0,\ IY_0,\ IZ_0$ in the ratio $ \frac {IU_0}{IX_0} \equal{} \frac {IV_0}{IY_0} \equal{} \frac {IW_0}{IZ_0} \equal{} \frac {1}{3}$ and they are on a circle $ \mathcal S_0$ tangent to $ \mathcal Q$ at $ I$ and with diameter $ \frac {IP}{3},$ i.e., $ 6\ R(U_0V_0W_0) \equal{} IP.$ For the centroid $ U$ of the $ \triangle XEF,$ we have

$ 3\ \vec{IU} \equal{} \vec{IX} \plus{} (\vec{IE} \plus{} \vec{IF}) \equal{} (\vec{IX_0} \plus{} \vec{X_0X}) \plus{} \vec{II_D} \equal{} \vec{IX_0} \plus{} (\vec{ID} \plus{} \vec{II_D})$

By the above lemma, $ \vec{ID} \plus{} \vec{II_D} \equal{} \vec{IJ},$ hence $ 3\ \vec{IU} \equal{} \vec{IX_0} \plus{} \vec{IJ}$ and similarly for the centroids $ V,\ W$ of the $ \triangle YFD,\ \triangle ZDE.$ It follows that

$ \vec{IU} \minus{} \vec{IU_0} \equal{} \vec{IV} \minus{} \vec{IV_0} \equal{} \vec{IW} \minus{} \vec{IW_0} \equal{} \frac {\vec{IJ}}{3}$

The $ \triangle UVW$ is just a translation of the $ \triangle U_0V_0W_0$ by $ \frac {\vec{IJ},}{3},$ i.e., by $ \frac {IJ}{3} \equal{} \frac {r \cdot OI}{3R}$ along the diacentral line $ OI$ of the $ \triangle ABC.$ Their circumcircles $ \mathcal S \cong \mathcal S_0$ are congruent and $ 6\ R(XYZ) \equal{} 6\ R(X_0Y_0Z_0) \equal{} IP.$

[quote="Nbach"]Let I be the incenter of triangle ABC and (I) tagents BC,CA,AB at points D,E,F. Let any point P in the
plane of triangle ABC. Lines through P and perpendicular to BC,CA,AB intersect them at X,Y,Z.
X',Y',Z' are centroids of triangles XEF,YDE,ZDF. Prove that: 3.R(X'Y'Z')=IP where R(X'Y'Z') is radius
of the circumcenter of triangle X'Y'Z'.[/quote]
$ I_1,I_2,I_3$ are the orthogonal projections of $ I$ on $ PX,PY,PZ$. We have $ 3\overrightarrow{Y'Z'} \equal{} \overrightarrow{YE} \plus{} \overrightarrow{FZ} \equal{} \minus{} \overrightarrow{II_2} \plus{} \overrightarrow{II_3} \equal{} \overrightarrow{I_2I_3}$
Similarly, we have $ 3\overrightarrow{Z'X'} \equal{} \overrightarrow{I_3I_1}, 3 \overrightarrow{X'Y'} \equal{} \overrightarrow{I_1I_2}$. It gives us $ 3R_{(X'Y'Z')} \equal{} R_{(I_1I_2I_3)}$
Note that $ I_1,I_2,I_3$ lie on a circle which has $ IP$ to diameter, so we have $ 3R_{(X'Y'Z')} \equal{} R_{(I_1I_2I_3)} \equal{} \frac {1}{2} IP$, as desired.


in latex?
[code]$x^2$[/code]
put it in between dollar signs

for many details just use  [url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php[/url]


[quote="GameBot"]On a given day, the probability of rain is $ 60\%$. What are the odds against rain on that day? Express your answer as $ a: b$ in simplest form.[/quote]

If the probability of rain 60%(which is 60 out of 100), then the odds against rain is 40 : 60, which simplifies to 2:3.


[b]IF[/b] $ m\in [0,3]$ and $ a,b,c$ are positive numbers with  $ abc\equal{}1$ , [b]THEN[/b]  $ \sum\frac{a\plus{}1}{\left[(3\minus{}m)\plus{}mbc\right]\cdot\left[m \plus{}(3\minus{}m) bc\right]}\ge \frac 23 .$

$ \frac {a \plus{} 1}{\left[(3 \minus{} m) \plus{} mbc\right]\cdot\left[m \plus{} (3 \minus{} m) bc\right]} \plus{} \frac {b \plus{} 1}{\left[(3 \minus{} m) \plus{} mca\right]\cdot\left[m \plus{} (3 \minus{} m) ca\right]} \plus{} \frac {c \plus{} 1}{\left[(3 \minus{} m) \plus{} mab\right]\cdot\left[m \plus{} (3 \minus{} m) ab\right]}$

$ \equal{} \frac {a^2(a \plus{} 1)}{\left[(3 \minus{} m)a \plus{} m\right]\cdot\left[ma \plus{} (3 \minus{} m) \right]} \plus{} \frac {b^2(b \plus{} 1)}{\left[(3 \minus{} m)b \plus{} m\right]\cdot\left[mb \plus{} (3 \minus{} m) \right]} \plus{} \frac {c^2(c \plus{} 1)}{\left[(3 \minus{} m)c \plus{} m\right]\cdot\left[mc \plus{} (3 \minus{} m) ab\right]}$

$ \geq \frac {4a^2(a \plus{} 1)}{[(3 \minus{} m)a \plus{} m \plus{} ma \plus{} 3 \minus{} m]^2} \plus{} \frac {4b^2(b \plus{} 1)}{[(3 \minus{} m)b \plus{} m \plus{} mb \plus{} 3 \minus{} m]^2} \plus{} \frac {4c^2(c \plus{} 1)}{[(3 \minus{} m)c \plus{} m \plus{} mc \plus{} 3 \minus{} m]^2}$

$ \equal{} \frac {4}{9}\left(\frac {a^2}{a \plus{} 1} \plus{} \frac {b^2}{b \plus{} 1} \plus{} \frac {c^2}{c \plus{} 1}\right)$.

Since $ \frac {a^2}{a \plus{} 1}\geq \frac {3}{4}a \minus{} \frac {1}{4}\Longleftrightarrow (a \minus{} 1)^2\geq 0$ for $ a > 0$ etc. , we have

$ \frac {a \plus{} 1}{\left[(3 \minus{} m) \plus{} mbc\right]\cdot\left[m \plus{} (3 \minus{} m) bc\right]} \plus{} \frac {b \plus{} 1}{\left[(3 \minus{} m) \plus{} mca\right]\cdot\left[m \plus{} (3 \minus{} m) ca\right]} \plus{} \frac {c \plus{} 1}{\left[(3 \minus{} m) \plus{} mab\right]\cdot\left[m \plus{} (3 \minus{} m) ab\right]}$

$ \geq \frac {4}{9}\left\{\frac {3}{4}(a \plus{} b \plus{} c) \minus{} \frac {3}{4}\right\}$

$ \equal{} \frac {1}{3}(a \plus{} b \plus{} c \minus{} 1)\geq \frac {1}{3}(3\sqrt [3]{abc} \minus{} 1) \equal{} \frac {2}{3}$. Q.E.D.


geometry
perimeter
inequalities open
inequalities

let a,b,c>0 such that a+b+c=1  prove that
$ \sum\sqrt{1\minus{}3ab} \geq \sqrt{6}$

I have two solutions for this one ;)  .
$ \bigstar 1$:
[b]Lemma[/b]:
$ a,b,c,d \ge 0$;$ a\plus{}b\equal{}c\plus{}d$ =>$ \sqrt{a}\plus{}\sqrt{b} \le \sqrt{c}\plus{}\sqrt{d} \leftrightarrow |a\minus{}b| \ge |c\minus{}d|$

Assume :$ c\equal{}min(a,b,c)$.By this lemma ,we can prove :
$ LHS \ge \sqrt{ 1\minus{} \frac{3}{4}(1\minus{}c)^2}\plus{}\sqrt{ \frac{4}{3}\minus{} 3(1\minus{}c)c}\plus{}\sqrt{\frac{2}{3}} \le \sqrt{6}$ ;)
$ \bigstar 2$ . Contractdition ;)

[quote="tranquoc"]let a,b,c>0 such that a+b+c=1  prove that
$ \sum\sqrt {1 \minus{} 3ab} \geq \sqrt {6}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157939
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=156002

Let a,b,c be lengths of sides of a triangle which perimeter 3 Prove that
$ \sum\sqrt{a\plus{}(1\minus{}b)^2}  \geq 3$


can you prove :D

$ \prod_{n\equal{}1}^{\infty}\;\;\frac{\textbf p_{n}\,^{2}\;\plus{}\;1}{\textbf p_{n}\,^{2}\;\minus{}\;1}\;\;\equal{}\;\;\boxed{\frac{5}{2}}$

where $ \textbf p_{n}$ is the $ n^{th}$ prime number


Receives Mexican eyes.

Inserts Lithuanian ears.

UMMMM I MEANT TO REPLY TO THE VENDING MACHINE 2 GAME, BUT I THINK I ACCIDENTALLY CLICKED "NEW TOPIC" WITHOUT THINKING.  ME PHAILZURE BADLY.

Is there any way someone can delete this topic from the forum?  I can see that this


Let f(x): (1,+ \infty )->(- \infty , + \infty) s.t:
   f(x)-f(y)=(y-x)f(xy), for all x,y>1
   Find f(x)?

[quote="Moubinool"]k/x[/quote]
[quote="A1lqdSchool"]Let f(x): (1, \infty)->(-\infty,+\infty) s.t: [/quote]

:? ya, but I wonder how you'll get infinity with k/x if x > 1...

Does anyone say f must be surjective? I think not!

f(x)=k/x is right sol!
I don't think problem said f(x) get infinity.
Find f(x)=k/x is easy but how is solution? :)

oh, don't they HAVE to attain all from -oo to +oo ??

my bad, that's why I never get those function equitations solved :(:(:(


Let $ a,b,c$ be real positive numbers such that $ abc\equal{}1$.Prove that
$ \frac{b\plus{}c}{\sqrt{a}}\plus{}\frac{b\plus{}c}{\sqrt{b}}\plus{}\frac{b\plus{}a}{\sqrt{c}}\ge\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}\plus{}3$

we have by AM-GM

$ \frac {b}{\sqrt {a}} + \sqrt {a}\geq2\sqrt {b}$

$ \implies$  $ \sum{\frac {b + c}{\sqrt {a}}\geq 2(\sqrt {a} + \sqrt {b} + \sqrt {c})\geq \sqrt {a} + \sqrt {b} + \sqrt {c} + 3}$

$ (\sum\sqrt {a}\geq3)$ by AM-GM

:lol:  :oops:

[hide="My solution"]
$ \sum\frac {\sqrt {a}(b \plus{} c)}{a}\ge3\plus{}\sum\sqrt {a}$
Now applying tchebyecheff we get
$ \frac {(\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c})}{3}(\sum\frac {b \plus{} c}{a})\ge\sum\sqrt {a}\plus{}3$

$ (\sum\frac {b \plus{} c}{a})\ge6$ by $ A.M\ge G.M$
and at last $ (\sum\sqrt {a}\geq3)$ which is true

[/hide]

$ \sum \frac {b \plus{} c}{\sqrt a} \ge 2 \sum \sqrt {\frac {bc}{a} } \equal{} 2 \sum \frac {1}{a}$

Now, $ \sum \frac {1}{a} \ge \sum \frac {1}{\sqrt{bc}} \equal{} \sum \sqrt a$ and $ \sum \frac {1}{a} \ge \frac {3}{\sqrt [3] {abc} } \equal{} 3$

Adding up the two inequalities gives the result