Let $ (x_n)_{n \in \mathbb{N}}$ be a sequence of positive reals, $ \sigma : \mathbb{N} \rightarrow \mathbb{N}$ a bijection, and $ (u_n)_{n \in \mathbb{N}}$ and $ (v_n)_{n \in \mathbb{N}}$ two sequences such that:

$ u_n=x_1+x_2+...+x_n, n \in \mathbb{N}$

$ v_n=\sqrt{x_1 x_{\sigma(1)}}+\sqrt{x_2 x_{\sigma(2)}}+...+\sqrt{x_n x_{\sigma(n)}}, n \in \mathbb{N}$

Prove that if the sequence $ (u_n)$ is convergent, then the sequence $ (v_n)$ is also convergent.

$ u_n \; \mathrm{convergent} \Rightarrow$ for each $ \epsilon>0$ there exists $ n_0$, such that $ \left|x_{n_0}+x_{n_0+1}+\ldots\right|<\frac{\epsilon}{2}$   (1)

Further on there exists $ n_1$, such that $ \left\{1, 2, \ldots, n_0-1\right\} \subset \left\{\sigma(1),\sigma(2),\ldots,\sigma(n_1)\right\}.$

Now we just show, that $ \left|\sqrt{x_{n_1}x_{\sigma(n_1)}}+\sqrt{x_{n_1+1}x_{\sigma(n_1+1)}}+\ldots \right|<\epsilon.$  (2)

We assume in the following way:

Each $ n\geq n_{1}$ appears in (2) at most twice: as $ \sqrt{x_{n}x_{\sigma(n)}}$ and maybe as $ \sqrt{x_{n^{\prime}}x_{\sigma(n^{\prime})}}$, where $ x_{\sigma(n^{\prime})}=x_{n}$.

We know that $ \sqrt{x_{n}x_{\sigma(n)}}+\sqrt{x_{n^{\prime}}x_{\sigma(n^{\prime})}} < 2x_{n} + x_{\sigma(n)}+ x_{n^{\prime}}$. But since we will go over all $ n\geq n_1$, we will take $ x_{\sigma(n)}$ and $ x_{n^{\prime}}$ for $ n_2=\sigma(n)$ and $ n_3=n^{\prime}$. Therefore we can estimate:

$ \left|\sqrt{x_{n_1}x_{\sigma(n_1)}}+\sqrt{x_{n_1+1}x_{\sigma(n_1+1)}}+\ldots \right|<2\left|x_{n_1}+x_{n_1+1}+\ldots\right|<\epsilon$  (We have used (1) here.)

Therefore $ \{v_n\}$ is convergent.


[quote="pbornsztein"]I must be blind, but where is the problem?
$2004^{2002!} > 2004^{2002} = 2004 \cdot 2004 \cdots 2004 > 2003\cdot 2002 \cdots 2 = 2003!.$

Pierre.[/quote]
Pierre,you are right. Thank you for your reply. Everyone will think this problem is very easy and self -evident.What I want to say is how this problem is made.The answer is  tangent line.

I agree that this problem may probably be solved by different ways, but since the desired estimate is very far from best, I don't think that some exotic approachs are really surprising and useful here. ;)

Pierre.


Find all positive integer $ n\ge2$ and $ a_1 ,a_2 ,...,a_n  \in \{ 0,1,...,9\}$ in decimal such that
$ \sqrt {\overline {a_1 a_2 ...a_n } }  \minus{} \sqrt {\overline {a_1 a_2 ...a_{n \minus{} 1} } }  \equal{} a_n$

Let $ x\equal{}\overline {a_1 a_2 ...a_{n \minus{} 1} },y \equal{} a_n$, then $ 10x\plus{}y\equal{}x\plus{}y^2\plus{}2y\sqrt x$, therefore $ x\equal{}k^2$ and $ 9k^2\minus{}2yk\equal{}y(y\minus{}1)$ and $ k\equal{}\frac{y\pm \sqrt{10y^2\minus{}9y}}{9}$. Chek $ y\equal{}0,x\equal{}0$, $ y\equal{}1,x\equal{}0$, $ y\equal{}9,x\equal{}16.$


I doubt if the following really is true, and can anybody help me to prove this? I've already stuck in this question for a whole week. :(

$ A_{2}$ is a subspace of $ A_{1}$. $ A_{i}$ are all linear spaces.
Prove: $ A_{1}\cap (A_{2} \plus{} A_{3}) \equal{} A_{2} \plus{} (A_{1}\cap A_{3})$.

It's false as written, but may be a typo. Note that the left hand side is a subspace of $ A_1,$ while the right hand side need not be a subspace of $ A_2.$

How about this: Let $ \mathbf{e}_1,\mathbf{e_2},$ and $ \mathbf{e_3}$ be the standard basis of $ \mathbb{F}^3.$ (Any field $ \mathbb{F}$ - it doesn't matter.)

Let $ A_k\equal{}\text{span}(\mathbf{e}_k).$

Then $ A_1\cap (A_2\plus{}A_3)\equal{}\{0\}$ and $ A_2\plus{}(A_1\cap A_3)\equal{}A_2.$

Now, did you actually mean this:

$ A_1\cap (A_2\plus{}A_3)\equal{}(A_1\cap A_2)\plus{}(A_1\cap A_3)\ ?$

[quote="Kent Merryfield"]Let $ A_k \equal{} \text{span}(\mathbf{e}_k).$

Then $ A_1\cap (A_2 \plus{} A_3) \equal{} \{0\}$ and $ A_2 \plus{} (A_1\cap A_3) \equal{} A_2.$

Now, did you actually mean this:

$ A_1\cap (A_2 \plus{} A_3) \equal{} (A_1\cap A_2) \plus{} (A_1\cap A_3)\ ?$[/quote]

Indeed it was $ A_1\cap (A_2 \plus{} A_3) \equal{} (A_1\cap A_2) \plus{} (A_1\cap A_3)$, but my book was a simplified version, with $ A_{2}\subset A_{1}$, so I edited the expression when posting. Indeed, without the condition $ A_{2}\subset A_{1}$, is $ A_1\cap (A_2 \plus{} A_3) \equal{} (A_1\cap A_2) \plus{} (A_1\cap A_3)$ true? Thanks.

Any element in $ (A_1\cap A_2)\plus{}(A_1\cap A_3)$ is of the form $ u\plus{}v$, where $ u$ is in both $ A_1$ and $ A_2$ and $ v$ is in both $ A_1$ and $ A_3$. So $ u\plus{}v$ is both in $ A_1$ (as $ u,v$ both are) and $ A_2\plus{}A_3$ (as $ u\in A_2,v\in A_3$).

This proves $ (A_1\cap A_2)\plus{}(A_1\cap A_3)\subseteq A_1\cap (A_2\plus{}A_3)$. The reverse containment isn't always true: in $ \mathbb{F}^2$ where $ \mathbb{F}$ is any field, let $ A_1,A_2,A_3$ be the subspaces generated by $ (1,1),(1,0),(0,1)$ respectively. Then $ A_1$ has a trivial intersection with both $ A_2$ and $ A_3$, but not their sum.

For the original question, where $ A_2\leq A_1$, it remains to show that $ A_1\cap (A_2\plus{}A_3)\subseteq A_2\plus{}(A_1\cap A_3)$. To prove this, any element in $ A_1\cap (A_2\plus{}A_3)$ is of the form $ u\plus{}v$, where $ u\in A_2,v\in A_3$ (being in $ A_2\plus{}A_3$).

But also $ u\in A_1$ and $ u\plus{}v\in A_1$, so we must have $ v\in A_1$. So $ u\in A_2,v\in A_1\cap A_3$, meaning our element $ u\plus{}v$ is in fact an element of $ A_2\plus{}(A_1\cap A_3)$. So in this case, we actually have $ A_1\cap (A_2\plus{}A_3)\subseteq A_2\plus{}(A_1\cap A_3)$. Containment in both directions means the two spaces are equal.


Findall nonconstant polynomials p so that p(x^2)=P(x)P(x-1) for all x

Suppose P is a polymnomial of degree 3n so

P(0)=P(3)=...=P(3n)=2
P(1)=P(4)=...=P(3n-2)=1
P(2)=P(5)=...=P(3n-1)=0

Find n if P(3n+1)=730.

The second problem has been posted before.  There was a very nice, but somewhat complex, solution involving a direct Lagrange Interpolation, if I recall correctly.

The URL please

BTW problem 2 is USAMO 1980 i think


modular arithmetic
algebra unsolved
algebra

Show that 19^93 - 13^99 is divisible by 162...


A car has an engine which delivers a constant power. It accelerates from rest at time t=0, and at t=$ t_0$ its acceleration is $ a_0$. What is its acceleration at t=$ 2t_0$?

Tons of results through the search function
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=244769
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=190300


[u]Problem [/u]Find all  $f: \mathbb{N}\to \mathbb{N}$ such that$f(f(n)+f(n+1)=n+2$with every ${n}\in{N}$

Can you post solution of problem or link in here ? :D

Algebra-Unsolved problems- 30 October - tmbd

sorry but I can't find that topic  :(
Can you post [b]link[/b] in here ? Please :)


trigonometry
algebra
system of equations

A block of mass $m$ resting on the incline of $\theta$ is attached to a spring parallel to the incline.  The spring has a constant of $k$ and the coeff of static friction between the block and the incline is $\mu_{s}$.  The spring is pulled from the high end until the block starts to move.  Find the necessary $\mu_{k}$ such that the block comes to rest just as the spring is returned to equilibrium.

[hide]
One may write the following system of equations:

\begin{eqnarray}f+mg\sin\theta-kx &=& 0 \\ N-mg\cos\theta &=& 0 \\ f &=&\mu_{k}N \end{eqnarray}

Solving the system simultaneously

$\mu_{k}= \frac{kx}{mg}\sec\theta-\tan\theta$
[/hide]

I hope that I understood the problem correctly
$\frac{1}{2}kx^{2}=m_{k}mgxcosa+mgxsina$
  $m_{k}mgcosa=kx$
  $x=\frac{m_{k}mg}{k}$
 We put the value of x in first equation

 $\frac{1}{2}k(\frac{m_{k}mg}{k})^{2}=\frac{m_{k}m^{2}g^{2}}{k}(sina+m_{k}cosa)$
 $m_{k}=2sina+2m_{k}cosa$
 $m_{k}(1-2cosa)=2sina$
 $m_{k}=\frac{2sina}{(1-2cosa)}$

At the streched state,
$kx=\mu_{s}mgcosa+mgsina$ ,as for the block to start moving it has to first overcome the static friction.
also,
$\frac{1}{2}kx^{2}=\mu_{k}mgxcosa+mgsina$
solving the two,gives
$\mu_{k}=\frac{\mu_{s}cosa-sina}{2cosa}$


In the AMC 12 problem series, will the problems be from the recent AMC 12s, or the AHSME's?


IN THE NAME OF GOD
HELLO, MY BROTHERS

There are 12 people in a room. 6 people are wearing socks and 4 people are wearing shoes, 3 people are wearing both.

How many people are in bare feet?

[hide="Solution"]

$ 6 \plus{} 4 \minus{} 3 \equal{} 7$ are wearing shoes or socks or both. Therefore $ 12 \minus{} 7 \equal{} 5$ are wearing neither, that is, $ 5$ are barefoot.

Hope this helps; you could also draw a Venn diagram.

[/hide]

So if 6 people are wearing socks and 3 people are wearing socks+shoes, then 3 people are wearing only socks. The same goes for the 4 who are wearing shoes, so only 1 person has just shoes.
3 people wearing only socks
+1 person wearing only shoes
+3 people wearing both
= 7 people wearing footwear.
Thus, 12 - 7 = $ \boxed{5}$ people in bare feet.

thanks
for
AndrewTom & goldenboy1.618

Knowledge is light and the ignorance is darkness

You're welcome, salahcool. Anymore questions?

oh also instead of saying unuseful stuff, just add it to your signiture panel.


Electrically charged drops of mercury fall from an altitude h into a spherical metal vessel of radius R in the upper part of which there is a small opening. The mass of each drops is m and the charge on each drop is q. What will be the number n of the last drop that can still enter the sphere?

[hide="Ans"]$ \frac{4 \pi \epsilon_o mg(h\minus{}R)R}{q^2}$
[/hide]

I am getting a different ans  :(

can somebody help me  :?:

Answer this question: What will be the distribution assumed by the charged mercury drops inside the sphere?

Looking at the answer I think the drops collect at the centre of the sphere.

Use energy conservation.

0 = $ mg(h \minus{} r)$ + Work done by electric force.

Now work done by electric force is $ \minus{} Knq^2(\frac {1}{R} \minus{} \frac {1}{H})$. Pluggin this is in, I am also getting different answer :(

Almost...

As the vessel is metallic the charge will be distributed  all over the surface (supposing a very small volume of mercury) as a spherical uniform charged shell.

So
   $ \;\;mg(h\minus{}2R)\;\equal{}\;{{n q^2} \over{4 \pi \epsilon_o}}\;({1\over R} \minus{}{1\over{h \minus{} R}})\;\;$

"As the vessel is metallic the charge will be distributed all over the surface (supposing a very small volume of mercury) as a spherical uniform charged shell. "

That's precisely the point.

All mass will be concentrated at the centre of the sphere , ie at a height R.
Hence loss of PE = nmg(h-R)

Now this energy is used in building interaction energy = QV
[img]http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/f/5/3/f535ee080c2238ad853d370a8dddb11643a126e8.gif[/img][/img]

I think what u want to ask in better terms is how can i say that it is the last drop only and that there cannot be any more drops coming in

If it is so
[hide]then see , if it is the last drop then only the centre of mass will be at the centre else the centre of mass will be shifted from the centre of sphere :) [/hide]

else
[hide]sorry i can't get ur qn  :rotfl:  :rotfl:[/hide]

Immanuel Bonfils approach seems better and is more concrete.I dont know where it goes wrong?

Sorry, but it's not going wrong.

I just left people do the last calculation, isolate n, to get the same answer as anandghegde's...

Didn't any one notice it?  I'm sure Carcul did...

[quote="golduck"][quote="slogger"][quote="Immanuel Bonfils"]Almost...

As the vessel is metallic the charge will be distributed  all over the surface (supposing a very small volume of mercury) as a spherical uniform charged shell.

So
   $ \;\;mg(h \minus{} 2R)\; \equal{} \;{{n q^2} \over{4 \pi \epsilon_o}}\;({1\over R} \minus{} {1\over{h \minus{} R}})\;\;$[/quote]

I don't think this is giving the right answer[/quote]
actually, it is giving the right answer :)[/quote]

Sorry really sorry  :oops_sign:

I think i was drunk  :rotfl:


USAMTS
geometry
AMC
AIME
trapezoid
HMMT
AIME II

Hi everyone,

I just noticed that PEN problem K5 is identical to this round's #4:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=150748

Shouldn't this be mentioned in the official solutions?

Also, #5 was posted on aops before; I can't find the exact post, and the numbers might have been different, but it was the same problem.

Hmm... yes, I saw something that looked astonishingly like #5 somewhere in the olympiad geometry forum just the other day, but now I can't find the link.

Aren't these problems written by a committee of some kind? Pretty big coincidence.

[quote="worthawholebean"]Last year's 19/4/3 was in Engel essentially verbatim.[/quote]

Round 4 Problem 3 or Round 3 Problem 4?

this seems pretty unfair. so i guess usamts is olympiad level sometimes.

Er - it was 19/4/4. See Engel Chapter 7 Problem 22.

20/2/2 was quite similar to 2004 AIME II #12.

Let ABCD be an isosceles trapezoid, whose dimensions are and AB = 6, BC = 5 = DA, and CD = 4. Draw circles of radius 3 centered at A and B and circles of radius 2 centered at C and D. A circle contained within the trapezoid is tangent to all four of these circles. Its radius is $ \frac { \minus{} k \plus{} m\sqrt {n}}{p}$ where k,m,n, and p are positive integers, n is not divisible by the square of any prime, and k and p are relatively prime. Find k + m + n + p. (AIME II 2004 #12)

Let C1, C2, C3, and C4 be four circles with radii 1, 1, 3, and 3, respectively, such that circles C1 and C2, C2 and C3, C3 and C4, and C4 and C1 are externally tangent. A fifth circle C is smaller than the four circles and is externally tangent to each of them. Find the radius of C. (USAMTS 20/2/2)

Just a substitution of one number but the problems are essentially the same.

Also, several of the geometry problems this year are strangely familiar to to the old AIMEs... like 1983 or 1984. I forget where exactly.

[quote="123456789"]Wow, that's interesting.

This was actually almost identical to one of the problems on the 1988 IMO Shortlist.[/quote]

The shortlist problem is the same except asks for a specific value of the function. I just did the shortlist problem a few days ago to prep functional equations :)

i used the exact same solution as found in the IMO compendium :D

[quote="grn_trtle"]Aren't these problems written by a committee of some kind? Pretty big coincidence.[/quote]
The problem sets are written by a single person. The first writer was George Berzsenyi, the second was Richard Rusczyk, and the current writer is David Patrick.

These writers do not invent twenty problems a year on their own, however. George Berzsenyi often credited the sources of his problems in the official solutions: he invented about eight a year himself, and the other twelve were suggested by a network of mathematicians he knew or copied from a foreign mathematics contest of a prior decade.

[quote="123456789"][quote="rrusczyk"]There was also a problem on the AIME last year that had been on the USAMTS either earlier that year or the year before.  These things happen . . .[/quote]

Do you mean the arctan one?[/quote]

I think hes refering to the trapezoid problem. It was essentially the same, but the numbers were different.

Indeed, there was a problem on this year's HMMT that appeared on the 1990 AIME, which wasn't discovered until after the test.  When a problem is really good, it's likely that problem writers will write ones very similar to it on later tests.  When things like this happen, I think it is more of a tribute to how good a problem is than anything else.


Hi,

does somebody know how many summands is contained in the multinomial formula (that is, for general $ n,m$ with the following convention  http://en.wikipedia.org/wiki/Multinomial_theorem )? For the binomial theorem ($ m\equal{}2$) it is easy...

thanks
michael

I see what you don't prove by yourself you don't have...   :cleaning:
anyway, the answer is $ \binom{m\plus{}n\minus{}1}{n}$


[hide]exponents same, so divide 66666 by 22222 and get 3. Then we have 3^4 or 81. [/hide]

[hide]
66,666 and the 22,222 cancel out to make $\frac{3^4}{1}$
which is $\boxed{\boxed{\boxed{\boxed{81}}}}$[/hide]


Let $M$ be a point inside the equilateral triangle $ABC$ and let $A^{\prime },$ $B^{\prime },$ and $C^{\prime }$ be its projections on the sides $BC,$ $CA,$ and $AB,$ respectively. Denote by $ r_{1},r_{2},r_{3},r_{1}^{\prime },r_{2}^{\prime },$ and $r_{3}^{\prime }$ the inradii of the triangles $MAC^{\prime },$ $MBA^{\prime },$ $MCB^{\prime },$ $MAB^{\prime },$ $MBC^{\prime},$ and $MCA^{\prime }.$ Prove that

\[
r_{1}+r_{2}+r_{3}=r_{1}^{\prime }+r_{2}^{\prime }+r_{3}^{\prime }.
\]

This is very easy.

First we prove a lemma:

[b]Lemma 1.[/b] If ABC is an arbitrary right-angled triangle with the right angle at C, then the inradius r of triangle ABC equals $r=\frac{a+b-c}{2}$, where, of course, a = BC, b = CA, c = AB.

[i]Proof.[/i] Let the incircle of triangle ABC have center I and touch the sides BC and CA at the points X and Y, respectively. Then, < IXC = 90 and < IYC = 90. On the other hand, since the triangle ABC is right-angled with right angle at C, we have < BCA = 90, so < XCY = 90. Hence, the quadrilateral IXCY has three right angles (namely, the ones at its vertices X, Y and C); therefore, this quadrilateral is a rectangle. This yields IY = CX. But IY is clearly the inradius r of triangle ABC, and a well-known formula gives $CX = \frac{a+b-c}{2}$. Thus, we get $r=\frac{a+b-c}{2}$, and Lemma 1 is proven.

Now to the solution of the problem:

Applying Lemma 1 to the right-angled triangles MBA' and MCA' with the right angles at the vertex A', we get for their inradii

$r_2 = \frac{BA^{\prime}+MA^{\prime}-BM}{2}$ and $r_3^{\prime} = \frac{CA^{\prime}+MA^{\prime}-CM}{2}$.

Thus,

$r_2-r_3^{\prime} = \frac{BA^{\prime}+MA^{\prime}-BM}{2} - \frac{CA^{\prime}+MA^{\prime}-CM}{2} = \frac{BA^{\prime}-CA^{\prime}}{2} - \frac{BM-CM}{2}$.

Now, if s = BC = CA = AB is the sidelength of the equilateral triangle ABC, then $BA^{\prime}+CA^{\prime} = BC = s$ and thus

$BA^{\prime}-CA^{\prime} = \frac{BA^{\prime}\ ^2 - CA^{\prime}\ ^2}{BA^{\prime}+CA^{\prime}} = \frac{BA^{\prime}\ ^2 - CA^{\prime}\ ^2}{s}$.

But by the Pythagoras theorem in the right-angled triangles MBA' and MCA', we have $BA^{\prime}\ ^2 = BM^2 - MA^{\prime}\ ^2$ and $CA^{\prime}\ ^2 = CM^2 - MA^{\prime}\ ^2$, so that

$BA^{\prime}\ ^2 - CA^{\prime}\ ^2 = \left(BM^2 - MA^{\prime}\ ^2\right)-\left(CM^2 - MA^{\prime}\ ^2\right) = BM^2 - CM^2$.

Thus,

$BA^{\prime}-CA^{\prime} = \frac{BA^{\prime}\ ^2 - CA^{\prime}\ ^2}{s} = \frac{BM^2-CM^2}{s}$,

and

$r_2-r_3^{\prime} = \frac{BA^{\prime}-CA^{\prime}}{2} - \frac{BM-CM}{2} = \frac{BM^2-CM^2}{2s} - \frac{BM-CM}{2}$.

Similarly,

$r_3-r_1^{\prime} = \frac{CM^2-AM^2}{2s} - \frac{CM-AM}{2}$;
$r_1-r_2^{\prime} = \frac{AM^2-BM^2}{2s} - \frac{AM-BM}{2}$.

Summing up, we get

$\left(r_1+r_2+r_3\right)-\left(r_1^{\prime}+r_2^{\prime}+r_3^{\prime}\right) = \left(r_2-r_3^{\prime}\right)+\left(r_3-r_1^{\prime}\right)+\left(r_1-r_2^{\prime}\right)$
$= \left(\frac{BM^2-CM^2}{2s} - \frac{BM-CM}{2}\right) + \left(\frac{CM^2-AM^2}{2s} - \frac{CM-AM}{2}\right) + \left(\frac{AM^2-BM^2}{2s} - \frac{AM-BM}{2}\right)$
$= \left(\frac{BM^2-CM^2}{2s}+\frac{CM^2-AM^2}{2s}+\frac{AM^2-BM^2}{2s}\right) - \left(\frac{BM-CM}{2}+\frac{CM-AM}{2}+\frac{AM-BM}{2}\right)$
$=0-0=0$.

Thus, $r_1+r_2+r_3=r_1^{\prime}+r_2^{\prime}+r_3^{\prime}$, and we are done.

I also have solutions for some of the other BMO problems (they are all very nice - thanks Enescu), but I have no time now since I'm trying to install a LAN internet connection and it just wouldn't work.

  Darij


[quote="anuj kumar"]if $a,b,c$ are positive reals such that $abc=1$.then prove that
\[\sqrt{\frac{a}{b+3}}+\sqrt{\frac{b}{c+3}}+\sqrt{\frac{c}{a+3}}\ge\frac{3}{2}\]
[/quote]
My ugly proof:
Let $a=\frac{x}{y},$ $b=\frac{z}{x}$ and $c=\frac{y}{z},$ where $x,$ $y$ and $z$ are positive numbers.
Hence, $\sqrt{\frac{a}{b+3}}+\sqrt{\frac{b}{c+3}}+\sqrt{\frac{c}{a+3}}\ge\frac{3}{2}\Leftrightarrow\sum_{cyc}\frac{x}{\sqrt{yz+3xy}}\geq\frac{3}{2}.$
Let $x+y+z=1$ and $x=\min\{x,y,z\}.$
Thus, $\sum_{cyc}\frac{x}{\sqrt{yz+3xy}}\geq\frac{1}{\sqrt{3xyz+3(x^{2}y+y^{2}z+z^{2}x)}}.$
But $\frac{1}{\sqrt{3xyz+3(x^{2}y+y^{2}z+z^{2}x)}}\geq\frac{3}{2}\Leftrightarrow$
$\Leftrightarrow4(x+y+z)^{3}\geq27(x^{2}y+y^{2}z+z^{2}x+xyz)\Leftrightarrow$
$\Leftrightarrow9x(x^{2}+y^{2}+z^{2}-xy-xz-yz)+(4y+z-5x)(x+y-2z)^{2}\geq0.$ :)

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=139744]The sharper
[/url]
:)

nice proof i liked it .
mine is different but long

$\sum_{cyc}\sqrt{\frac{a}{b+k}}\geq\frac{3}{\sqrt{1+k}}$ is true for all $k\geq8$ and for all $0\leq k\leq3.$ :wink:


Columbia
geometry
3D geometry
sphere
geometry unsolved

This problem was posted before in another section. There wasn't any solution so I am posting it here.

Is it possible to divide a sphere in 15 regions of equal volume using 4 planes?

Does anyone know a solution for this problem???

Construct reqular tethrahedron with center of inscribed and circumscribed sphere same as the center of the 'big' sphere and volume $ \frac{1}{15}$ from 'big' sphere's volume. Four planes to be the sides of this tetrahedron. Is this the solution of the problem?

borislav,
I don't think this a solution for the problem.
I think that it is impossible to divide the sphere in such way.
Anyway I don't know how to prove this.

If you think it is impossible - you should take a look if there any combinations places of the planes that divide sphere in 15 parts with the same volumes. If there any (If I'm not wrong - it is the only such case) you should prove that the volumes of the 15 parts are not all equal.
You are right it is not solution at all but you may use something like this. I believe that someone else will find a better solutions. It is only a draft. It was the first thing that came in my mind. I wanted to believe  :) this composition divide the sphere in 15 parts. And then to prove it is the only possible situation for the planes, then to prove or disprove ... the parts have same volumes (it is not very easy but it is not impossible).

P.S. I think the problem is not easy but is not so difficult as you think. You just should spend more time by solving it ... You may separate this problem to more simpler cases for the planes.


geometry
incenter
ratio
geometry proposed

By George Lee:

Let ABC be a triangle with incenter I and let the incircle touch sides BC; CA; AB
at D;E; F. Let lines AI and EF intersect at P; let K be the intersection of BP and
DF,  and L that of CP and DE. Show that KL is parallel to EF.

PS
Actually a little bit more general result holds.

By me:

P could be a generic point on AI and still KL is parallel to EF.

With P being the point of intersection of the lines AI and EF, you get the same problem as posted by al.M.V. in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=111382]the thread "Dizzy construction"[/url] and solved by me. The proof I gave there can be directly used to show the more general assertion for a general point P on the line AI.

  Darij

If R=BI^DF and S=CI^DE then, as <ISD+<IRD=90+90=180, ISDR is ciclyc. From here is easy to see that c(ISDR) is homotetic with the incircle c(DEF) then SR//EF.

By the invariance wrt the projection, it holds that the cross ratio of {F,K,R,D} is equal to the cross ratio of {A,P,I,Q} where Q=AI^BC, i.e. (RK/RD)/(FK/FD) = (IP/IQ)/(AP/AQ).

Similarly we have that (SL/SD)/(EL/ED) = (IP/IQ)/(AP/AQ).

From SR//EF then FD/RD = ED/SD. Therefore combining all this relation it hols that SL/EL = RK/FK. From which by Thales follows the thesis.

If, under the conditions of the initial problem, $ A'$ is the middle of $ EF$, $ { K } \equal{} A'B \cap FD$, $ { L } \equal{} A'C \cap DE$, similarly getting $ MN$ and $ PQ$, then $ KL, MN$ and $ PQ$ are concurrent.

Best regards,
sunken rock

$ \boxed 1$ Let $ \triangle ABC$ with incircle $ C(I)$ . Denote the midpoint $ D$ of the side $ [BC]$ . Consider  $ E\in (AC)$ , $ F\in (AB)$

for which $ BF \equal{} CE$ . For any point $ P\in AI$ denote $ K\in DF\cap BP$ , $ L\in DE\cap CP$ . Show that $ KL\parallel EF$ .

$ \boxed 2$ Let $ \triangle ABC$ with incircle $ C(I)$ . Denote the midpoints $ D$ , $ E$ of the sides $ [BC]$ , $ [AC]$ respectively. Suppose

$ AC \equal{} 2\cdot AB$ . For any point $ P\in AI$ denote $ K\in AD\cap BP$ , $ L\in DE\cap CP$ . Show that $ KL\parallel AC$ .

$ \boxed 3$ Let $ \triangle ABC$ with incircle $ C(I)$ . Consider $ D\in (BC)$ , $ E\in (CA$ , $ F\in (BA$ so that $ BD \equal{} BF$

and  $ CD \equal{} CE$ . For any point $ P\in AI$ denote $ K\in DF\cap BP$ , $ L\in DE\cap CP$ . Show that $ KL\parallel EF$ .

****************

[b]Remark.[/b] I'll come back soon with a common extension for which these four problems are simple

and particular cases. Maybe you will find it using eventually the metrical relation from [url=http://www.mathlinks.ro/viewtopic.php?p=1694355#1694355][color=red][b]here ![/b][/color][/url]

[color=darkblue][b]Here is the promised extension ![/b][/color]

[quote="Virgil Nicula"][color=darkred]Let $ ABC$ be a triangle with the incircle $ w \equal{} C(I)$ . Consider the points $ D\in (BC)$ , $ E\in (CA$ , $ F\in (BA$

for which $ \frac {DB}{DC} \equal{} \frac {BF}{CE}$ . For a point $ P\in AI$ denote $ K\in DF\cap BP$ , $ L\in DE\cap CP$ . Show that $ KL\parallel EF$ .[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] Denote $ M\in AI\cap BC$ . Using my [/color] [url=http://www.mathlinks.ro/viewtopic.php?p=1694355#1694355][color=red][b]relation[/b][/color][/url] [color=darkblue]obtain $ \left\|\begin{array}{c} \frac {KF}{KD} \equal{} \frac {PA}{PM}\cdot\frac {BF}{BD}\cdot\frac {BM}{BA} \\
 \\
\frac {LE}{LD} \equal{} \frac {PA}{PM}\cdot \frac {CE}{CD}\cdot\frac {CM}{CA}\end{array}\right\|$

from which results $ \frac {KF}{KD} \equal{} \frac {LE}{LD}$ , i.e. $ KL\parallel EF$ because $ \frac {BF}{BD} \equal{} \frac {CE}{CD}$ and $ \frac {BM}{AB} \equal{} \frac {CM}{CA}$ .[/color]


$a, b, c$ are non-zero reals, and $x$ is real and satisfies $\frac{[bx + c(1-x)]}{a} = \frac{[cx + a(1-x)]}{b} = \frac{[ax + b(1-x)]}{b}$. Show that $a = b = c.$

This is unclear for me :
Does $[...]$ denote the integer part?
Is $x$ a given number or do the equations hold for all real $x$?

Pierre.

The result is not entire true.

See problem 319 [url=http://www.cms.math.ca/Competitions/MOCP/2004/sol_jul.html]here[/url] for the solution.


I m stucked on this problem....

It looks complicated but maybe its easy

If a cat and half eat a fish and a half in a day and a half, how many  days will take 14 cats eat 14 fishes?

 :lol:  ;)

got a solution i can look at?

i really need it

The correlation is 1.5 cats:1.5 fishes in 1.5 days. It follows, then, that 1 cat:1 fish in 1.5 days, 2:2, etc.

err my brain hurts....

u know y it ratio is not 1:1:1?

.....

this is getting kinda hard

These types of problems often get very confusing!  Don't stress too much.

maybe the guy made this problem didnt give enough info.. :rotfl:

LOL!!!!! Maybe the guy that made the problem....is.....you!  :rotfl:
Thanks for the laughs, i needed it lol

[quote="kidwithshirt"]i still got like 20 sheet of reviews my teacher gave me havent done yet...

that fishy one was one of them[/quote]

Post some more! Post some more!  The more practice the better!  Thanks!  ANd if you really really want to make it to nationals, I suggest you get started on those sheets of review *chuckles* hehe

[quote="kidwithshirt"]o they are just plain work up from mathcounts... if u like some i ll post a new thread when i get more[/quote]
Post them! Post them!Post them!Post them!Post them!Post them!Post them!
I wana see em!

Okay lol thanks but I gota log off now.  See ya later!  Bye!

About that cat and a half problem... just think about like this. Ignore the days... because they'll just make you confused. You realize that one cat and a half eats a fish and a half, and you have fourteen cats eating fourteen fish. Those proportions are the SAME, so the answer is a day and a half. Don't let all the information confuse you or bog you down... it's easy only if you can sort out all the info.

I hate these kinds of problems when I have to make it harder than it really is.
For example:

There are 32 socks in a drawer, 8 blue, 8 brown, 8 yellow, and 8 green. If you choose 5 socks, what is the probability that at least 2 socks will be the same?

I was about half way into multiplying fractions until I saw that the problem didn't ask for a common fraction or a decimal. When I looked close ly at the problem, I saw that the probability was 1, because there are only 4 types of socks, so if you choose 5 socks, there has to be one that is the same as one of the others.

Yes, the answer is a-day-and-a-half to this cat and fish problem.


1)let f be a real polynomial of degree n and suppose constants c_0,c_1,c_2,...c_n satisfy
  \sum {j=0-->n} c _j *f[(2^j)*x] = 0
show that c_j=0 for all j

2)if f be a real polynomial of degree n,show that there exists a polynomial g of degree n+1 such that \sum {i=1 to k} f(i) = g(k)
for every positive integer k

Maybe Lagrange's formula works for the 2'nd one: g has degree n+1, so it's defined by n+2 values. In this case,let Sk=\sum{i=1->k}f(i). Then we have g(x)=\sum{i=1->n+2}Si*[\pi{t=1->n+2}(x-t)/(x-i)]/[\pi{j=1->n+2;j=/=i}(i-j)].

This ensures that g(k)=Sk for k from 1 to n+2, but I don't know if it's easy to show that it works for all positive integers k.

Maybe it's useless, but maybe it isn't :D.

Let S(p,k)=1^p+2^p+...+k^p. Then it is well-known that this is a polynomial of degree p+1 in k. Let f(x)=a_0+a_1X+...+z_n*X^n. Then f(1)+...+f(k)=ka_0+S(1,k)a_1+...+S(n,k)a_n and clearly this will be a polinomial of degree n+1 in k.

The first one is wrong. Take f(X)=X^n and take non-zero c_0,...,c_n such that c_0+c_1*2^n_+c_2*4^n+...+c_n*2^n^2=0.


[quote="mz94"]can anyone make htem? theyre pretty hard especially with ks and all. any tips?[/quote]

Don't try to make each letter correspond to another letter. Try to make compromises along the way, if it doesn't turn out absolutely correctly. For example, not each letter must be capitalized or decapitalized.

Use a lot of extra lines. That's all I can really describe..I made decent one with my name though.


Prove that $ (a_n)_{n\ge 1} , a_n>0$ is arithmetic progression $ \Longleftrightarrow (\forall)n\ge 2\ ,\ \sum_{k\equal{}2}^n\frac{1}{\sqrt{a_{k\minus{}1}}\plus{}\sqrt{a_k}}\equal{}\frac{n\minus{}1}{\sqrt{a_1}\minus{}\sqrt{a_n}}$

...not the right forum.

mods/admins, move.

nike...you go through all the forums saying...
"too hard move to another forum"
BUt u NEVER solve a question!!!
@Pain rinnegan...nike is always like that...

It is too hard for this forum though. Nike is right. As was yours in the Mc forum.

I think that would be best in Intermediate or Pre-Olympiad, but don't take my word for it as I don't know much on that topic.

I'll try to prove the "if" part of the "if and only if" (which I'm assuming the arrow means). Rationalizing, $ \frac{1}{\sqrt{a_{k-1}} + \sqrt{a_k}} = - \frac{\sqrt{a_{k-1}} - \sqrt{a_k}}{d}$ where $ d$ is the common difference between the terms. Summing these, we find that
\begin{align*}
\sum_{k=2}^{n}\frac{1}{\sqrt{a_{k-1}}+\sqrt{a_{k}}} &=-\frac{\sqrt{a_1} - \sqrt{a_n}}{d} \\
&= \frac{\sqrt{a_n} - \sqrt{a_1}}{\frac{a_n - a_1}{n-1}} \\
&= \frac{n-1}{\frac{\sqrt{a_n} - \sqrt{a_1}}{a_n - a_1}} \\
&= \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}}.
\end{align*}
Pain rinnegan, did I make a sign mistake in that last step? :maybe:

[quote="ThinkFlow"]
Pain rinnegan, did I make a sign mistake in that last step? :maybe:[/quote]

Probably he made a typo .

"$ \Leftarrow$"

We have for $ n\minus{}1$ :

$ \sum_{k\equal{}1}^{n\minus{}1}\frac{1}{\sqrt{a_k}\plus{}\sqrt{a_{k\minus{}1}}}\equal{}\frac{n\minus{}2}{\sqrt{a_1}\plus{}\sqrt{a_{n\minus{}1}}}\Leftrightarrow \minus{}\sum_{k\equal{}1}^{n\minus{}1}\frac{1}{\sqrt{a_k}\plus{}\sqrt{a_{k\minus{}1}}}\equal{}\minus{}\frac{n\minus{}2}{\sqrt{a_1}\plus{}\sqrt{a_{n\minus{}1}}}$

Summing this with sum for $ n$ we get :

$ \frac{1}{\sqrt{a_n}\plus{}\sqrt{a_{n\minus{}1}}}\equal{}\frac{n\minus{}1}{\sqrt{a_1}\plus{}\sqrt{a_n}}\minus{}\frac{n\minus{}2}{\sqrt{a_1}\plus{}\sqrt{a_{n\minus{}1}}}$

which is equivalent with :

$ a_1\equal{}(n\minus{}1)a_{n\minus{}1}\plus{}(n\minus{}2)a_n\ ,\ (\forall)n\ge 2$ .

Setting $ n\rightarrow n\plus{}1$ we have :

$ a_1\equal{}na_n\plus{}(n\minus{}1)a_{n\plus{}1}$

$ \Rightarrow 2a_n\equal{}a_{n\minus{}1}\plus{}a_{n\plus{}1}\ ,\ (\forall)n\ge 2$


Let $ABC$ a acute triangle.

(a) Find the locus of all the points $P$ such that, calling $O_{a}, O_{b}, O_{c}$ the circumcenters of $PBC$, $PAC$, $PAB$:
\[\frac{ O_{a}O_{b}}{AB}= \frac{ O_{b}O_{c}}{BC}=\frac{ O_{c}O_{a}}{CA}\]

(b) For all points $P$ of the locus in (a), show that the lines $AO_{a}$, $BO_{b}$ , $CO_{c}$ are cuncurrent (in $X$);

(c) Show that the power of $X$ wrt the circumcircle of $ABC$ is:
\[-\frac{ a^{2}+b^{2}+c^{2}-5R^{2}}4\]
Where $a=BC$ , $b=AC$ and $c=AB$.

I made the locus the single point $ H$, which seems wrong (and renders the other two parts very easy).

a) Call M the midpoint of AP, N of PB and L of PC. So we want to prove that $ \triangle O_aO_bO_c$ and $ \triangle ABC$ are similar and then $ \angle O_bO_cO_a \equal{} \angle ACB$ and so on. $ NPMO_c$ is cyclic and so $ \angle O_bO_cO_a \equal{} 180 \minus{} APB$ and others. Then P stay on the symmetric of the circumcircle of ABC w.r.t. AB, w.r.t BC and w.r.t. AC. But this 3 circle have the common point in the orthocenter H of ABC, so the locus is H.
b)cause the orthocenter of $ \triangle ABC$ is the circumcenter of $ \triangle O_aO_bO_c$, then the circle that pass true M,L,N is the Feuerbach circle of $ \triangle ABC$ and of $ \triangle O_aO_bO_c$, so the 2 triangle are equal and then they have the homothetic center on the center of the Feuerbach circle.
c) $ \displaystyle pow_\Gamma(P) \equal{} R^2 \minus{} PO^2 \equal{} R^2 \minus{} \frac {9R^2 \minus{} a^2 \minus{} b^2 \minus{} c^2}{4} \equal{} \frac {a^2 \plus{} b^2 \plus{} c^2 \minus{} 5R^2}{4}$

a) Locus of $ P$ is $ H$, with $ H$ is the orthorcenter of  $ \delta ABC$
b) easy to proof that each of lines $ AO_{a}, BO_{b}, C_O{c}$ pass through the midpoint of $ (OP)$
c) $ OX \equal{} \frac{OP}{2}$, $ OP$ can calculate from $ a, b, c$ and $ R$


Prove that the number 11...1(there are 91 numbers 1) is a prime number

That will be rather difficult since it's composite.
See http://www.research.att.com/~njas/sequences/A004023

[quote="Solar Plexsus"]The natural number $N$ consisting of 91 ones (i.e. the digit 1) is not a prime since

$N \;=\; \frac{10^{91}\:-\: 1}{9}\;=\; \frac{10^{7}\:-\: 1}{9}\: \sum_{k=0}^{12}10^{7k}$.

Hence 1111111 is a divisor of $N.$[/quote]
Generally: if $n$ is composite, then $\sum_{0\leqslant i\leqslant n-1}10^{i}$ is composite.


I have a few tables with a  border line in the middle. I attach an example.

My questions:
1. Is there any  Latex code to display the middle border line in the table?
or
2. I have to save the table as a figure and then includes it as an .eps figure?

3. How to write Latex for the first cell with a line in the middle of PI and TL.

4. I use Paint shop to convert the figure into .eps. But the table is not fully displayed afterwards.

5. BTW, ETD requires .eps for figures.

Thank you.

It's quite difficult to do this. But it is possible with a combination of things:
1. \setlength{\doublerulesep}{\arrayrulewidth} means that \hline\hline or \hline\cline gives a bolder horizontal rule;
2. the array package together with !{\vrule width 0.5pt}c|} gives a bolder vertical rule;
3. \multicolumn allows different cells to have different vertical rules;
4. the slashbox package gives a diagonal line in a cell with \backslashbox.

Here is an example to see how it works:
[code]\documentclass{article}
\usepackage{array}
\usepackage{slashbox}

\begin{document}
% optionally uncommenting the next line makes the lines thicker
%\setlength{\arrayrulewidth}{1pt}
\setlength{\doublerulesep}{\arrayrulewidth}

\begin{tabular}{|c|c|c|c|} \hline
 \backslashbox{A}{B} & 1 & 2 & 3 \\ \hline
O & \multicolumn{1}{!{\vrule width 0.5pt}c|}{A}  & B  & C \\ \hline\cline{2-2}
V & X  &  \multicolumn{1}{!{\vrule width 0.5pt}c|}{Y} & Z \\ \hline\cline{3-3}
1000 & 100 & 10 & \multicolumn{1}{!{\vrule width 0.5pt}c|}{1}  \\ \hline \cline{4-4}
\end{tabular}

\end{document}[/code]

Stevem helped me with the bold border lines ..
and I have footnotes and want the table centered and widen to \textwidth..so I have the following codes..
but the bold border lines are still in the original place..how to fix this problem

\documentclass{article}
\usepackage{array}
\usepackage{slashbox}
\usepackage{threeparttable}
\begin{document}
% optionally uncommenting the next line makes the lines thicker
%\setlength{\arrayrulewidth}{1pt}
\setlength{\doublerulesep}{\arrayrulewidth}

 \begin{threeparttable}
\begin{tabular*}{\textwidth}{@{}@{\extracolsep{\fill}} |c|c|c|c|c|c| @{}} \hline
 \backslashbox{A\tnote{a}}{B\tnote{b}} & 1 & 2 & 3&5&6 \\ \hline
O & \multicolumn{1}{!{\vrule width 0.5pt}c|}{A}  & B &D&E & C \\ \hline\cline{2-2}
V & X  &  \multicolumn{1}{!{\vrule width 0.5pt}c|}{Y\tnote{a}} & &e&QZ \\ \hline\cline{3-3}
1000 & 100 & 10 & \multicolumn{1}{!{\vrule width 0.5pt}c|}{1} &Q&5 \\ \hline \cline{4-4}
\end{tabular}
\begin{tablnotes}
\item[a] aaaaa
\item[b] bbbb
\end{tablenotes}
\end{threeparttable}
\end{document}

The problem is that you've used @{\extracolsep{\fill} which just increases the intercolumn spacing NOT the width of the columns. This is shown by \cline{2-2} which only goes across the column not the spacing between them.
Use tabularx instead of tabular* as this calculates the column widths and makes them equal width without adding to the intercolumn spacing.
[code]\usepackage{tabularx} % automatically loads the array package
...
\begin{tabularx}{\textwidth}{|X|X|X|X|X|X|}  \hline
...
\end{tabularx} [/code]See the [url=http://tug.ctan.org/get/macros/latex/required/tools/tabularx.pdf]tabularx documentation[/url] if you wish to customise it further.

[i]Note: I have deleted your repeated post.[/i]

how to make all the numbers in the table centered??

some of them are left-justified and some are centered?

I got all centered..but the method is an  painstaking..any easier one ?

I did say look at the documentation if you wish to customise it which tells you all about \tabularxcolumn. X means p{[i]width[/i]} so all you need to do is add a centering command to all of them using \tabularxcolumn. Since using \centering in the last column means that \\ is a new line in a cell you need to use \tabularnewline instead.
[code]\renewcommand{\tabularxcolumn}[1]{>{\centering}p{#1}}
\begin{threeparttable}
\begin{tabularx}{\textwidth}{|X|X|X|X|X|X|}  \hline
\backslashbox{A\tnote{a}}{B\tnote{b}} & 1 & 2 & 3&5&6 \tabularnewline \hline
O & \multicolumn{1}{!{\vrule width 0.5pt}c|}{A} & B &D&E & C \tabularnewline \hline\cline{2-2}
V & X & \multicolumn{1}{!{\vrule width 0.5pt}c|}{Y\tnote{a}} & &e&QZ \tabularnewline \hline\cline{3-3}
1000 & 100 & 10 & \multicolumn{1}{!{\vrule width 0.5pt}c|}{1} &Q&5 \tabularnewline \hline \cline{4-4}
\end{tabularx}
...[/code]


In the expression $ \frac{x \plus{} 1}{x \minus{} 1}$ each $ x$ is replaced by $ \frac{x \plus{} 1}{x \minus{} 1}$. The resulting expression, evaluated for $ x \equal{} \frac{1}{2}$, equals:

$ \textbf{(A)}\ 3\qquad
\textbf{(B)}\ \minus{}3\qquad
\textbf{(C)}\ 1\qquad
\textbf{(D)}\ \minus{}1\qquad
\textbf{(E)}\ \text{none of these}$

[hide="Solution"]
When $ x\equal{}\dfrac{1}{2}$, $ \dfrac{x\plus{}1}{x\minus{}1}\equal{}\minus{}3$, substituting $ \minus{}3$ for $ x$ in the original equation we get:

$ \dfrac{\minus{}3\plus{}1}{\minus{}3\minus{}1}\equal{}\dfrac{\minus{}2}{\minus{}4}\equal{}\dfrac{1}{2}\implies \text{ Ans. } \mathbf{(E)}$
[/hide]


consider the reaction
$ \ce{Et}^{.} \plus{} \ce{HBr} \rightarrow \ce{EtH} \plus{} \ce{Br}^{.}$.both the forward and backward reactions are bimolecular
the forward reaction has an $ E_{act,f} \equal{} \minus{} 4.2Kj/mol$ and pre-exponential factor $ A_{f} \equal{} 10^{9} L \minus{} mol^{ \minus{} 1} s^{ \minus{} 1}$
for reverse reaction $ E_{act,b} \equal{} 53.3kJ/mol$ and rate constant $ k_{b} \equal{} 1.4 \times 10^{11}L\minus{}mol^{\minus{}1}s^{\minus{}1}$
Find the standard enthalpy and gibbs free energy of formation of ethyl free radical using these data
all species are in gaseous state

Calc. T  by using the data given abt kb and Ea. then calc. kf using Arrhenius eq. kf/kb gives Kc which can be converted to Kp and then dG=-2.303logkp. dunno if this is rite, coz I suck at phy chem. And also coz I wrote this on the spur of the moment.

u cant's calculate the temp. with Eact backward and kb u need the pre exponential factor also .
but anyways assume the temperature to be 298K
also u have to find the $ \Delta G$ for formation of ethyl free radical and [b]not [/b]the reaction

Maybe we could start by calculating $ \Delta H$ as $ E_{a_{f}} \minus{} E_{a_{b}}$  :maybe:

again....the question is [b]not[/b] to find the $ \Delta H$ of the reaction but for the formation of ethyl free radical


The triangle $ABC$, where $AB < AC$, has circumcircle S. The perpendicular from $A$ to $BC$ meets $S$ again at $P$. The point $X$ lies on the line segment $AC$, and $BX$ meets $S$ again at $Q$.
Show that $BX=CX$ if and only if $PQ$ is a diameter of $S$.

my solution[hide]Suppose first that $XB=XC$ then $\widehat{XBC}=\widehat{XCB}=a$;we have also $\widehat{PAC}=\widehat{PBC}=b$.But $a+b=90$(in a triangle rectangle...) then $\widehat{PBQ}=a+b= 90$ so $PQ$ is diameter of $S$.
The reciprocal is obvious. :wink: [/hide]


geometry
parallelogram
modular arithmetic
inequalities
number theory
relatively prime
superior algebra

Prove that $ \frac{\mathbb{Z}[i]}{a\plus{}ib},a^2\plus{}b^2>0$ has $ a^2\plus{}b^2$ elements.
More generally $ \frac{\mathbb{Z}[\sqrt{d}]}{a\plus{}\sqrt{d}},a^2\plus{}b^2>0$ with $ d$ free of square has $ |a^2\minus{}db^2|$ elements.

The fundamental parallelogram of the lattice spanned by $ a \plus{} bi$ is a square with side length $ \sqrt{a^2 \plus{} b^2}$.  :)

Let $ R$ be ring of integers of a number field. It also can be proved that $ \left|\frac {R}{(a)} \right| \equal{} N(a)$ which $ N$ is the norm of $ a$ in the number field.

[hide]
Let $ g\equal{}\gcd (a,b)$ with $ a\equal{}gx$ and $ b\equal{}gy$, and $ m\equal{}x^2\minus{}dy^2$. Then $ y$ and $ m$ are relatively prime, so there exists $ y_0$ such that $ yy_0\equiv 1\pmod{m}$. So in $ \mathbb{Z}[\sqrt{d}]/(a\plus{}b\sqrt{d})$, we get $ g\sqrt{d}\equal{}\minus{}gxy_0$. (This is so because $ \sqrt{d}\equiv \minus{}xy_0\pmod{x\plus{}y\sqrt{d}}$.) Also, $ gm\equiv 0$ in this ring. So any element of $ \mathbb{Z}[\sqrt{d}]$ can be reduced modulo $ a\plus{}b\sqrt{d}$ to a number of the form $ j\plus{}k\sqrt{d}$ where $ 0\leq j<gm,\ 0\leq k<g$.

We now need to show our set of representatives are distinct. Working modulo $ a\plus{}b\sqrt{d}$, if $ j_1\plus{}k_1\sqrt{d}\equiv j_2\plus{}k_2\sqrt{d}$ where the $ j$'s and $ k$'s satisfy the above inequalities, then $ (j_1\minus{}j_2)\plus{}(k_1\minus{}k_2)\sqrt{d}\equiv 0$. This difference element must be a multiple of $ g$, so $ j_1\minus{}j_2$ and $ k_1\minus{}k_2$ both are. $ |k_1\minus{}k_2|<g$, so $ k_1\equal{}k_2$.

To show $ j_1\equal{}j_2$, we have $ \frac{j_1\minus{}j_2}{g}$ a multiple of $ x\plus{}y\sqrt{d}$. Then it can be shown that $ \frac{j_1\minus{}j_2}{g}$ is a multiple of $ m$, but it is less than $ m$ in absolute value. So $ j_1\equal{}j_2$.

So the ring $ \mathbb{Z}[\sqrt{d}]/(a\plus{}b\sqrt{d})$ has it elements uniquely represented by an element of the form $ j\plus{}k\sqrt{d}$ where $ 0\leq j<gm$ and $ 0\leq k<g$. So the number of elements is $ g^2m\equal{}a^2\minus{}db^2$, as desired.
[/hide]


inequalities USAMTS AMC AIME USA(J)MO USAMO function

I've noticed a lot of solutions are derived using Jensen's Inequality, can someone please tell me what it is?

Recall that the definition of convexity is that f((x+y)/2) <= (f(x)+f(y))/2. What Jensen's inequality says is that we can put more variables in and it remains true. That is, if f is convex, then f((x1+x2+...+xn)/n) <= (f(x1)+f(x2)+...+f(xn))/n. The inequality is reversed for concave functions.

As an example of the power of Jensen's Inequality, let's derive AM-GM as an instant corollary. Let f(x)=-log(x), which is convex. -log((x1+x2+...+xn)/n) <= -1/n*log(x1x2...xn), and then we can multiply by -1 and switch the inequality sign and then take the antilogarithm, and there's AM-GM.

a convex equation simple refers to a curve that is "concave up" in calc lingo?

which, now that I look at it, makes perfect sense, lol.

[quote="Spoon"]That is really cool.. matter of fact, all of this inequality stuff is cool. I wish I had known about it four years ago... Tare, your lucky!!!

I hope a lot of these inequalities are taught in AoPS, I should be getting Vol 1+2 either today or tomorrow!!! :), cant wait[/quote]
Not really. I haven't know about MC until this year (which is my last year to qualify) and soon I'll be disqualified from all American competitions such as, but not limited to, AMC, USAMTS, AIME, USAMO, and Mandelbrot because I'll need to go back to my country. My time here is running out and I still have to do homework...I'll have to ditch school to make it to the AIME/NATS, that's how desperate I am, especially with MC and AMC 10 disqualification happening in the same year...I mean, I haven't even read half of my AoPS 1! I'm doomed...*stressed*

...sorry, got carried away...I'd like to say something relevant but... :wink:

[quote]take the antilogarithm[/quote]

Hmm, I can probably guess what the antilogarithm is, but could you explain a bit further on when it can be used and why it can be used?

The antilog for the common log would be 10^x, it is jus the inverse function... for ln(x), the antilog is e^x, at the end of that proof, you end up with log(....) :ge: log(,,,,), so when you take the anitlog (or the inverse function), both log's disappear and you just have .... :ge: ,,,,

[quote]you end up with log(....)  log(,,,,), so when you take the anitlog (or the inverse function), both log's disappear and you just have ....  ,,,,
[/quote]

Yeah, I understand the concept, but how can you show that the logs can cancel algebraically?

Using the property that f(f_inv(x))=x is how I know it.... I will leave it to the pro's on this forum to prove it another way.

Hmm..that's true... Thanks for the insight.

[quote="Spoon"]
I hope a lot of these inequalities are taught in AoPS, I should be getting Vol 1+2 either today or tomorrow!!! :), cant wait[/quote]

Not really.  AoPS does have a chapter on inequalities, which covers mostly just AM-GM and Cauchy-Schwarz.  Kiran's A<B, which is on the AMC website is the best resouce I know of for olympiad inequalities.

Hehe, he probably knows that by now, he posted that about 4 months ago :D

haha! I saw my name off to the side and didn't recall making a post. Thanks for the tip!

[quote]Kiran's A<B, which is on the AMC website is the best resouce I know of for olympiad inequalities.[/quote]

if you know some french, then http://www.animath.fr/cours/inegalites.pdf
is a very good resource.

[quote="Spoon"]
I hope a lot of these inequalities are taught in AoPS, I should be getting Vol 1+2 either today or tomorrow!!! :), cant wait[/quote]

We are now offering an inequalities class here at AoPS.  It starts in mid-March and focuses very much on developing the problem solving concepts that should help students take the subject to a higher level on their own.

[quote]Wow that's really good! It gives me an opportunity to work on Inequalities and brush up on French at the same time.[/quote]

yea, the page http://www.animath.fr/tutorat.html has some [b]really good[/b] olympiad resources in french.


integration
trigonometry
real analysis
real analysis unsolved

Let $ f\in C(\mathbb{R})$ be such that $ f(t)\equal{}0, |t|\ge\pi,$ $ c_{k}: \equal{}\int_{\minus{}\pi}^{\pi}{f(t)e^{ikt}dt}.$

à) Prove that if Fourier transform of $ f$ belongs to $ L_{p}(\mathbb{R})\ (1\le p<\infty)$ then $ \sum_{k\equal{}\minus{}\infty}^{\infty}{|c_{k}|^{p}}<\infty.$

á) Prove that if $ f(t)\equal{}0$ holds for $ |t|>\pi\minus{}\varepsilon\ (\varepsilon>0)$ and $ \sum_{k\equal{}\minus{}\infty}^{\infty}{|c_{k}|^{p}}<\infty$ then Fourier transform of $ f$ belongs to $ L_{p}(\mathbb{R}).$

I was getting a little annoyed with the notation, so I'd like to redefine

$ c_{k}: \equal{}\frac{1}{2\pi}\int_{\minus{}\pi}^{\pi}f(t)e^{\minus{}ikt}\,dt$

Compared to the original, that flips $ k$ and $ \minus{}k$ around and divides by $ 2\pi.$ The essence of the problem remains the same.

Now, let the Fourier transform be $ \widehat{f}(\omega)\equal{}\frac{1}{2\pi}\int_{\minus{}\infty}^{\infty}f(t)e^{\minus{}i\omega t}\,dt.$

Here's a formula: for sufficiently smooth $ f,$

$ \widehat{f}(\omega)\equal{}\sum_{k\equal{}\minus{}\infty}^{\infty}c_{k}\,\frac{\sin [\pi (\omega\minus{}k)]}{\pi(\omega\minus{}k)}.$

When $ \omega$ is an integer, this simplifies to $ \widehat{f}(k)\equal{}c_{k}.$

For other $ \omega$ - that's not a very nice formula, since $ \sum_{k\equal{}\minus{}\infty}^{\infty}\frac{\sin [\pi (\omega\minus{}k)]}{\pi(\omega\minus{}k)}$ doesn't converge absolutely.

The path ahead is not clear.


Let a,b,c be 3 sides of a triangle.Knowing that $ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\equal{}\frac{5}{2}$,prove that:
$ a^2\plus{}b^2\plus{}c^2\plus{}3(ab\plus{}bc\plus{}ca)$≥$ 4$

[quote="SUPERMAN2"]Let a,b,c be 3 sides of a triangle.Knowing that $ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \equal{} \frac {5}{2}$,prove that:
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 3(ab \plus{} bc \plus{} ca)$≥$ 4$[/quote]

$ x\equal{}\frac {1}{a}...$
We must prove:
$ \sum\frac{1}{x^2}\plus{}3\sum\frac{1}{xy}\ge 4$
$ \sum x^2y^2\plus{}3\sum x^2yz \ge 4x^2y^2z^2$
$ (x\plus{}y\plus{}z)^2(\sum x^2y^2\plus{}3\sum x^2yz)\ge 25 x^2y^2z^2$
but
$ (x\plus{}y\plus{}z)^2(\sum x^2y^2\plus{}3\sum x^2yz)\ge 108x^2y^2z^2$ by AM-GM hmmm  :huh:

mmmmmm,
by cauchy,
$ a\plus{}b\plus{}c\geq \frac{9}{\sum \frac{1}{a}}\equal{}\frac{18}{5}$
we have also $ ab\plus{}bc\plus{}ca\geq \frac{1}{2}(a^2\plus{}b^2\plus{}c^2)\geq \frac{1}{6}(a\plus{}b\plus{}c)^2\geq \frac{54}{25}$
then $ a^2\plus{}b^2\plus{}c^2\plus{}3(ab\plus{}bc\plus{}ca)\equal{}(a\plus{}b\plus{}c)^2\plus{}ab\plus{}bc\plus{}ca\geq \frac{324}{25}\plus{}\frac{54}{25}\equal{} \frac{378}{25}>4$
 :?:  :?:

Here is my solution(may be ugly than yours)
Since a,b,c are 3 sides of a triangle,we can let
$ a\equal{}y\plus{}z,b\equal{}x\plus{}z,c\equal{}x\plus{}y$
=>$ \frac{1}{a\plus{}b}\plus{}\frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a}\equal{}\frac{5}{2}$
=>$ ab\plus{}bc\plus{}ca$≥$ 1$(This is easy to prove)
=>Q.E.D


I know there are many problems in mathlinks, but how can I find them?

You can always go to the Contest Section.

http://www.artofproblemsolving.com/Forum/resources.php

Or use the search function.  Or simply browse the forums.

[quote="BanishedTraitor"]You can always go to the Contest Section.
[/quote]

Yes, there have many problems which Resourse seems to miss, like Chinese Math Olympiads.


function
real analysis
real analysis unsolved

Fix r>0, let B(x) denote the open ball {y: |x-y|<r}.
Define g(x)=sup{f(y): y in B(x)}. It is easy to show that g(x) is lower semi-continuous ( {g>a} is open for all a).

Now if B(x) is closed ball, can we find a counterexmaple? That is, find f(x) such that g(x) is not LSC?

Thanks.

Let $f(0)=1$ and $f(x)=0$ otherwise. Then $g(x)=1$ on $[-r,r]$ and $0$ otherwise - not lsc.

Of course. Let $f$ be 1 at a point and zero elsewhere. Counterexamples are extremely easy to find in most spaces.


A right triangle has leg lengths of 7 and 24 units. A second triangle is similar to the first triangle and has a hypotenuse of 100 units. What is the length, in units, of the shorter leg of the second triangle?

If you know the Pythagorean triples, you know that the length of the missing side is 25.
The second triangle is 4 times bigger, so multiply $ 7\cdot4\equal{}28$


limit
trigonometry
calculus
calculus computations

[b]1. The problem statement, all variables and given/known data[/b]
Evaluate limit in terms of the number $ \alpha\equal{}\lim_{x\rightarrow 0}\frac{sinx}{x}$

$ \lim_{x\rightarrow 0}\frac{tan^2x\plus{}2x}{(x\plus{}x^2)}$

It is equal to $ \lim_{x\to 0}\left(\frac {\sin^2x}{x\cdot\cos^2x} \plus{} 2\right) \equal{} 2 \plus{} \lim_{x\to 0}\frac {\sin^2x}{x} \equal{} 2 \plus{} \lim_{x\to 0}\left[x\cdot\left(\frac {\sin x}{x}\right)^2\right] \equal{} 2$.

$ \lim_{x\rightarrow 0}\frac{tan^2x\plus{}2x}{x}\minus{}\lim_{x\rightarrow 0}\frac{tan^2x\plus{}2x}{(1\plus{}x)}$

Why it's not equal to $ \lim_{x\rightarrow 0}\frac{tan^2x}{x(x\plus{}1)}\plus{}2$

[quote="azatkgz"]$ \lim_{x\rightarrow 0}\frac {tan^2x \plus{} 2x}{x} \minus{} \lim_{x\rightarrow 0}\frac {tan^2x \plus{} 2x}{(1 \plus{} x)}$

Why it's not equal to $ \lim_{x\rightarrow 0}\frac {tan^2x}{x(x \plus{} 1)} \plus{} 2$[/quote]
I don't understand.  :maybe:

And,actually,can we say that $ \lim_{x\rightarrow 0}sinx \equal{} 0$ and $ \lim_{x\rightarrow 0}cosx \equal{} 1$?

Thanks,I understood how he got this answer.


1.A Baller's freethrow percentage is 70%. What is the probability that he will get exactly 7 freethrow points out of 10 attempts during a given game? Answer to the nearest percent. (No, its not 100%)

[hide]
$ \binom{10}{7} (.7)^7 \cdot (.3)^3$

$ \equal{} 120\cdot .0823543 \cdot .027$

$ \approx .267$

$ \approx \boxed{27}$

Nearest percent, as math154 pointed out (:
[/hide]

[quote="DaReaper"]why did you pick the powers 7 and 3? does it matter what the power is as long as it adds up to 10?[/quote]

Yes, it does. The probability of making a particular free throw is $ \frac{7}{10}$. You want him to make $ 7$ and miss $ 3$, so the probability of this occurring is $ \left(\frac{7}{10}\right)^7$ (since you're making seven free throws, each with a probability of $ \frac{7}{10}$). Similarly, the probability of missing a free throw is $ \frac{3}{10}$, and since you want to miss three of them you raise it to the third power. Then you multiply by $ \binom{10}{7}$ because there are that many different ways to choose seven of the ten shots to be the ones that go in.


algebra
function
domain
polynomial
superior algebra
superior algebra unsolved

Suppose we have finite field extensions $ B$ and $ K$ of $ F$ of respective degrees $ m$ and $ n$, all contained in some larger field $ E$. Prove that a domain which is a finite dimentional algebra over a field must be a field, so $ B \otimes K$ will be a field if it has no zero divisors.

Pick an element $ a$. The infinite set $ 1,a,a^2,\cdots$ is linearly dependent, so $ a$ has a minimal polynomial $ p_a(x)\equal{}x^n\plus{}c_{a,n\minus{}1}x^{n\minus{}1}\plus{}\cdots\plus{}c_{a,0}$ over $ F$. This $ p_a$ must be irreducible; if $ p_a \equal{}q_1\cdot q_2$, $ q_1(a)\cdot q(2)\equal{}0$, and either $ q_1(a)$ or $ q_2(a)$ must be zero since we're in a domain.
Now, if $ c_{a,0}\equal{}0$, $ p_a(x)$ has $ x$ as a factor and must be $ x$, making $ a\equal{}0$. Otherwise, we can write $ a^{\minus{}1}\equal{}\frac{\minus{}1}{c_{a,0}}\cdot\left(a^{n\minus{}1}\plus{}c_{a,n\minus{}1}x^{n\minus{}2}\plus{}\cdots\plus{}c_{a,1}\right)$, from the fact that $ p_a(a)\equal{}0$.

Everything that's nonzero is invertible, making this a division ring. Is it a field? Not necessarily; the real quaternions are a counterexample. Your original question avoids this by forcing commutativity through inclusion in some large field.


trigonometry
logarithms
LaTeX
trig identities

These are some problems from my honors trig class, I need help with them mostly because I'm not seeing the process on some of them and on others it has just been a while since I've worked with log and ln. Please use mostly trig identities and give fairly detailed solutions as i don't want to have to ask for help very often.

1.)$\log 10^{\tan{x}}=\tan{x}$

2.)$10^{\log{\mid\sin{t}\mid}}=\mid\sin{t}\mid$

3.)$-\ln \mid \sec{\theta}-\tan{\theta}\mid=\ln\mid\sec{\theta}+\tan{\theta}\mid$

4.)$\ln \mid \csc{x}-\cot{x}\mid=-\ln\mid\csc{x}+\cot{x}\mid$

5.)\[\frac{\sin{\alpha}\cos{\beta}+\cos{\alpha}\sin{\beta}}{\cos{\alpha}\cos{\beta}-\sin{\alpha}\sin{\beta}}=\frac{\tan{\alpha}+\tan{\beta}}{1-\tan{\alpha}\tan{\beta}}\]
First time using LaTex for real so there might be errors

[quote="daermon"][quote="Jamminjake"]
1.)$\log 10^{\tan{x}}=\tan{x}$
[/quote]
do you mean $\log_{10}\tan{x}=\tan{x}$?[/quote]

no what he said is true.

what you said isnt.

btw, for that problem, bring down tan x because its the exponent and log 10 = 1.

Thanks for clarifying that for him. To whom it may concern, sorry about posting this in the intermediate forum I meant to put it in basics.

[hide="5"]
So first you start with:

$\frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\cos\alpha\cos\beta-\sin\alpha\sin\beta}$
We divide the numerator and the denominator of the expression by $\cos\alpha\cos\beta$ and use the fact that $\tan\theta=\frac{\sin\theta}{\cos\theta}$ to get
$\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$ as desired.
[/hide]

thanks but i posted this under basics forum instead so u can stop posting

[quote="Jamminjake"]
5.)
\[\frac{\sin{\alpha}\cos{\beta}+\cos{\alpha}\sin{\beta}}{\cos{\alpha}\cos{\beta}-\sin{\alpha}\sin{\beta}}=\frac{\tan{\alpha}+\tan{\beta}}{1-\tan{\alpha}\tan{\beta}}\][/quote]

What identity might this be?

[hide="hint"] $\sin{(\alpha+\beta)}= \sin{\alpha}\cos{\beta}+\cos{\alpha}\sin{\beta}$[/hide]

I know that, I've figured out #5 i just haven't worked with logs in a while so those are the one]s i mainly need help with.

[quote="Leao"][quote="anirudh"][hide="For #5"]

You take the original equation.

$\frac{\sin{\alpha}\cos{\beta}+\cos{\alpha}\sin{\beta}}{\cos{\alpha}\cos{\beta}-\sin{\alpha}\sin{\beta}}=\frac{\tan{\alpha}+\tan{\beta}}{1-\tan{\alpha}\tan{\beta}}$

$\sin\alpha\cos\beta+\cos\alpha\sin\beta=\sin(\alpha+\beta)$
$\cos\alpha\cos\beta-\sin\alpha\sin\beta=\cos(\alpha+\beta)$

$\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\tan(\alpha+\beta)$.

This is true, because $\frac{\sin\varphi}{\cos\varphi}=\tan{\varphi}$.

Therefore, I have established this identity.
[/hide]

EDIT: Fixed a couple of typos[/quote]
I think he wants a proof of

$\frac{\sin{\alpha}\cos{\beta}+\cos{\alpha}\sin{\beta}}{\cos{\alpha}\cos{\beta}-\sin{\alpha}\sin{\beta}}=\frac{\tan{\alpha}+\tan{\beta}}{1-\tan{\alpha}\tan{\beta}}$
so by assuming its truth in the first line, the proof is reversed.  I think you should work your way up from the last line, as in reverse the steps.:maybe:[/quote]

It is proved. I am unclear. Does he want a proof of the addition formulae?

[quote="aahmadi"]Can anyone enlighten me on the identity(ies) that explain 3 and 4? Neither one is ringing a bell in my head. Thanks.[/quote]

$\frac{1}{\sec{\theta}-\tan{\theta}}= \frac{\sec{\theta}+\tan{\theta}}{\sec^{2}{\theta}-\tan^{2}{\theta}}$.

Do you recognize the denominator?


$ a,b,k,n>0$ and $ ab>ka\plus{}nb$  . prove:

$ a\plus{}b>(\sqrt{k}\plus{}\sqrt{n})^{2}$

Yes it's easy just using cauchy-shwarts
$ (\sqrt{k}\plus{}\sqrt{n})^2\equal{}(\sqrt{ka/a}\plus{}\sqrt{nb/b})\equal{}<(ak\plus{}nb)*(1/a\plus{}1/b)<ab*((a\plus{}b)/ab)\equal{}a\plus{}b$
and we are done.Yes it is easy.


I don't know where to post this but I think this is best fit.

Is the number $ \frac{{11^{100}  \minus{} 7^{100} }}{{11^4  \minus{} 7^4 }}$ odd or even?

[quote="hasan4444"]I don't know where to post this but I think this is best fit.

Is the number $ \frac {{11^{100} \minus{} 7^{100} }}{{11^4 \minus{} 7^4 }}$ odd or even?[/quote]

of course odd:

Let $ x\equal{}11^4$ and $ y\equal{}7^4$ then we have

$ \frac {x^{25}\minus{}y^{25}}{x\minus{}y}\equal{}x^{24}\plus{}....\plus{}y^{24}$

sum of odd number of odds is odd, so sum is odd...................

Yes. This is Number Theory.

For more information about this subject the Introduction to Number Theory helps give you the basics.

[quote="hasan4444"]Seems you get angry from the question  :huh:

But is this a part of what they call it "Number Theory"?[/quote]

NO i didn't get angry  :)
i just thought it's a little bit easy....so i said "of course"
anyway don't mind =)


Va rog frumos sa-mi explicati si mie ceva...pt ca am o nelamurire :roll: la aceasta problema:
Se considera $ m\in R, m\geq 1$ si functia$ f: [m, \infty) \to [1,\infty), f(x)\equal{}x^2\minus{}2mx\plus{}m^2\plus{}1$
a)aratati ca functia f este o bijectie
b)rezolvati ecuatia : $ x^2\minus{}2mx\plus{}m^2\plus{}1\equal{}m\plus{}\sqrt{x\minus{}1}$

Am gasit baremul pentru problema aceasta, dar nu inteleg ceva: la subpunctul b), dupa ce se gasesc radacinile $ x_1\equal{}\frac{2m\plus{}1\plus{}\sqrt{4m\minus{}3}}{2}$ si $ x_2\equal{}\frac{2m\plus{}1\minus{}\sqrt{4m\minus{}3}}{2}$, ei spun " se verifica usor ca sunt $ \geq1"$. Dar nu trebuia verificat daca sunt $ \geq m$? Pentru ca f este definita pe  $ [m, \infty)$, cu $ m\geq1$. Aici este baremul, clasa a X-a, subiectul 2
http://mategl.com/atasamente%20pentru%20site/UNIREA%202009.rar
Ati putea sa ma lamuriti si pe mine? Va multumesc! :wink:

Da, asa e, este o greseala in solutia oficiala. Trebuie pusa conditia $ \ge m$. Mai mult, vezi [url]http://mateforum.ro/viewtopic.php?t=3336[/url]


Prove that for positive reals x,y,z xyz=1

$ \frac{1}{\sqrt{1\plus{}8x}}\plus{}\frac{1}{\sqrt{1\plus{}8y}}\plus{}\frac{1}{\sqrt{1\plus{}8z}}\geq 1$

See hear:  http://www.mathlinks.ro/viewtopic.php?t=182763     :wink:

oke!!!
we have a,b,c >0 . a,b,c satisfy ::
x=ab/c² y=bc/a² z=ac/b² ;
oke!! we obtain:
a/√(a²+8bc) + c/√(c²+8ba)+b/√(b²+8ac) ≥ 1
oke!!!
it's a problem in IMO2001
I think it's easy now!!!

Posted: Thu Mar 06, 2008 5:58 am

ProfilePMEmail QuoteEdit


What is the sum of the smallest two positive real numbers whose digits are all 1s
when written in decimal form and whose product and sum are the same? Express
your answer as a decimal to the nearest tenth.

Thank you!

[hide] Let the real numbers be $ x$ and $ y$. Then, we have

$ xy\equal{}x\plus{}y$

$ \implies xy\minus{}x\minus{}y\equal{}0 \implies xy\minus{}x\minus{}y\plus{}1\equal{}1 \implies (x\minus{}1)(y\minus{}1)\equal{}1$

Therefore, $ x\minus{}1$ and $ y\minus{}1$ are reciprocals, and hence one of $ x$ and $ y$ is between $ 1$ and $ 2$ (we can rule out other cases for various reasons). Rememebering that all the digits are $ 1$, after some trial and error, we get that $ 1.1$ and $ 11$ are the smallest solutions and hence their sum is $ 1.1\plus{}11\equal{}\boxed{12.1}$.[/hide]


calculus
integration
trigonometry
real analysis
real analysis unsolved

Let $ I = \int_{0}^{\infty}\cos(x^{2}+ax+a^{2}) \; dx$, $ J = \int_{0}^{\infty}\sin(x^{2}+ax+a^{2}) \; dx$.

Then $ y = I+iJ$ satisfies the differential equation $ y'+\frac{1}{2}e^{ia^{2}}-\frac{3ai}{2}y = 0$ with the initial condition $ y(0) = \frac{\sqrt{\pi}}{2}e^{\frac{\pi i}{4}}$.

But I think the solution for this equation is not elementary... my solution involves Fresnel integral.

@sos440: I agree I dont think you can do it.  :maybe:
@misan: Maybe the limits should be $ -1/2$ to $ \infty$ which is doable.

sos440 is right.
The answer is $ y(1)=\sqrt{\pi/8}(cos(3/4)+sin(3/4)-Im[(1+I)(cos(3/4)+Isin(3/4))erf(\frac{1-I}{2\sqrt{2}})])\approx{0.5166}$, where for all complex $ x$: $ erf(x)=\frac{2}{\sqrt{\pi}}\int_{0}^{x}{e^{-t^{2}}}dt$; this last integral is well defined because $ e^{-t^{2}}$ is an entire function.


the formula:

v1/1 + (v1-v2)/j1  = 6 angle 0

reduces to

v1(1-j) + jv2 = 6 angle 0

where v1 and v2 are voltages and j is complex.

how does this reduce? i cant figure it out. thanks for the help

[hide=""Click to reveal a property of the complex j""]$\frac{1}{j}=-j$[/hide]
If you still don't know what I am talking about, maybe the problem you are doing is too hard for you.

Does 6 angle 0 = 6 degrees? In any case,

Multiply the numerator and the denomenator of the second fraction by $j$ (remember that $j^2 = -1$):

$\frac{v_1}{1} + \frac{v_1 - v_2}{j}$
$= v_1 + \frac{(v_1 - v_2)j}{j^2}$
$= v_1 + \frac{v_1j - v_2j}{-1}$
$= v_1 + (-v_1j + v_2j)$
$= v_1(1-j) + v_2j$


Gauss
number theory proposed
number theory

Let $ a,b,c$ be positive integers and $ a\geqq b\geqq c$.

For all integers $ k > 2$, there exist just two solutions in positive integer $ n$ satisfying the equation:

$ n \minus{} \left[ \frac {n}{a} \right] \minus{} \left[ \frac {n}{b} \right] \minus{} \left[ \frac {n}{c} \right] \equal{} k$

Determine all ($ a,b,c$).

The definition of Gauss sign, from the condition we have $ 1\minus{}\frac{k}{n}\leq \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}<1\minus{}\frac{k}{n}\plus{}3$.


Please help me to solve these type of matrix.

Here is one of the example of this type of matrix

You are told a 2 by 2 matrix A has eigenvaues 1 and 0 corresponding to eigenvector $ \dbinom{1}{1} and \dbinom{1}{\minus{}1}$

How to sketch the set $ x^TAx$ = 1?

First step: Consider traveling outward from the origin in the direction of one of the eigenvectors. How far will you go before you cross this curve - or will you ever reach the curve at all? Do that for each eigenvector.

The problem is I don't know how far the length of semi-axes is. When I take $ \dfrac{1}{\sqrt[2]{lamda}}$where the second eigenvalues is 0 , this is undefine! So I stuck. I think I go in the wrong direction.

Could you please work out this problem, so I can learn from it?
Thanks

If $ x \equal{} cv,$ where $ v \equal{} \begin{bmatrix}1 \\
\minus{} 1\end{bmatrix},$ which is to say that $ v$ is the eigenvalue associated to the eigenvector $ 0,$ then $ x^TAx \equal{} c^2v^TAv \equal{} c^2\cdot 0 \equal{} 0.$ Which means it never equals $ 1,$ which means that if you travel in that direction, you never reach the curve in question.

So the curve is not an ellipse, and it does not enclose the origin.

Let the two eigenvectors be $ v$ and $ w.$ Let $ x \equal{} (av \plus{} bw).$ Suppose $ x^TAx \equal{} 1.$ What does that tell you about the coefficients $ a$ and $ b?$ (I would suggest that you go through and distribute out that product, using $ Av \equal{} 0,Aw \equal{} 1,$ and $ v^Tw \equal{} 0$ where needed.)

[hide="By the way, ..."]
You've told us exactly what the matrix $ A$ is.

$ A\equal{}\frac12\begin{bmatrix}1&1\\1&\minus{}1\end{bmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}1&1\\1&\minus{}1\end{bmatrix}\equal{}\frac12\begin{bmatrix}1&1\\1&1\end{bmatrix}.$

Which means that $ x^TAx\equal{}\frac12(x_1^2\plus{}2x_1x_2\plus{}x_2^2)\equal{}\frac12(x_1\plus{}x_2)^2.$[/hide]


It would be nice if the following is true: for any sets of indexes $k_0 < k_1 < \dots < k_n$ and $l_0 < l_1 < \dots < l_m$, from equality:
\[ \sum_{0 \leq i \leq n, 0 \leq j \leq m} 2^{k_i + 2 l_j} = \sum_{0 \leq i \leq n, 0 \leq j \leq m} 2^{2k_i + l_j} \]
we get $k_0 = l_0, k_1 = l_1, \dots, k_n = l_m$.
Thanks

[quote="alexilic"] $k_0 = l_0, k_1 = l_1, \dots, k_n = l_m$.
[/quote]
What do you mean??? We have $n$ k-s and $m$ l-s. Thanks ;)

and $n = m$ i.e. from those equality we get equality of sequences $k$ and $l$.
Thx

Sorry for all those non Sebian people, if this is correct and if
somebody is interested I can write it again in English...
\[ \sum_{0\leq i \leq n,0\leq j \leq m} 2^{k_i+2l_j}=\sum_{0\leq i \leq n,0\leq j \leq m} 2^{2k_i+l_j} \]
 Dakle, indukcija... Najveci stepen dvojke koji deli
levu stranu ocigledno je $k_0+2l_0$ a desnu $2k_0+l_0$ pa odatle
sledi da je $k_0=l_0$.

E sad pretpostavimo da je za neko $x$
ispunjeno $k_0=l_0, k_1=l_1,...,k_x=l_x$, onda se podsume leve sume
$\sum_{0\leq i \leq x,0\leq j \leq x} 2^{k_i+2l_j}$ i desne sume
$\sum_{0\leq i \leq x,0\leq j \leq x} 2^{2k_i+l_j}$ potiru, pa
najveci stepen dvojke koji deli levu stranu sada je
$min\{k_0+2l_{x+1}, k_{x+1}+2l_0\}$ a desnu $min \{l_0+2k_{x+1}, l_{x+1}+2k_0\}$ i naravno oni mora da budu jednaki.

Pretpostavimo recimo da je $k_{x+1}<l_{x+1}$, onda je $min\{k_0+2l_{x+1}, k_{x+1}+2l_0\}=k_{x+1}+2l_0$ a kako je on manji od $l_0+2k_{x+1}$
sledi da je $min \{l_0+2k_{x+1}, l_{x+1}+2k_0\}=l_{x+1}+2k_0$. Dakle
sad imamo $k_{x+1}+2l_0=l_{x+1}+2k_0 \Rightarrow k_{x+1}=l_{x+1}$,
dakle kontradikcija. Isto je i ako pretpostavimo da je
$k_{x+1}>l_{x+1}$ dakle, sledi da je $k_{x+1}=l_{x+1}$, i ovo ti je
sad kraj dokaza jer ako pretpostavimo da je $n<m$ onda dobijamo da
je $2^{k_0+2l_{n+1}}=2^{l_{n+1}+2k_0}$ a to nije moguce.

Najveci stepeni dvojke ne moraju da budu jednaki... Jer mogu da ti se sabiraju neki clanovi (za takvo pokazivanje treba da pokazes da postoji jedinstven najveci stepen dvojke koji deli obe strane)!
Pozdrav, Aca

Can someone solve this little problem?
Greetings!

Ali jeste, vidi, prvo za $k_0$ i $l_0$ ono su NAJMANJI brojevi na levoj i desnoj strani jer vaze stroge nejednakosti izmedju brojeva $k$ i $l$. Dalje kod indukcijskog koraka: zamisli da napises sve one sume, onda sve lepo pokratis sto je jednako na levoj i desnoj strani.
Sada, najmanji brojevi koji ti ostanu, i koji su [b]strogo manji[/b] od svih ostalih brojeva su bas oni koje sam smestio u $min\{...\}$, i posto su strogo manji od svih ostalih samo njih razmatras. Pretpostavis da je$k_{x+1}>l_{x+1}$ pa vidis sta se dobije, isto i za obrnuto i lepo zakljucis.

Ja stvarno ne vidim problem ovde....


A skew-Hermitian matrix, $ \mathsf{W}$, is one such that $ \mathsf{W}^{\dagger} \equal{} \minus{} \mathsf{W}$. What can be said about the eigenvalues of a skew-Hermitian matrix?

If $ \mathsf{S}$ is real symmetric and $ \mathsf{T}$ is real skew-symmetric, show that $ \mathsf{T} \pm i\mathsf{S}$ is skew-Hermitian. State a property of the eigenvalues of $ \mathsf{T} \plus{} i\mathsf{S}$ and hence, or otherwise, show that
\[ \det (\mathsf{T} \plus{} i\mathsf{S} \minus{} \mathsf{I}) \neq 0.\]Show that the matrix
\[ \mathsf{U} \equal{} (\mathsf{I} \plus{} \mathsf{T} \plus{} i\mathsf{S})(\mathsf{I} \minus{} \mathsf{T} \minus{} i\mathsf{S})^{\minus{}1}\]is unitary. For
\[ \mathsf{S} \equal{} \begin{pmatrix} 1 & 1 \\ 1 & 1\end{pmatrix}, \qquad \mathsf{T} \equal{} \begin{pmatrix} 0 & 1 \\ \minus{}1 & 0 \end{pmatrix},\]show that the eigenvalues of $ \mathsf{U}$ are $ \frac{\pm(1 \minus{} i)}{2}.$

$ Wx \equal{} \lambda x \equal{}> x^*W^* \equal{} \minus{}x^*W \equal{} \bar{\lambda} x^* \equal{}> x^*W \equal{} \minus{} \bar{\lambda} x^*$. So

$ x^*Wx \equal{} \minus{} \bar{\lambda} x^*x \equal{}> x^* \lambda x \equal{} \minus{} \bar{\lambda} x^*x \equal{}> \lambda \equal{} \minus{} \bar{\lambda}$,

which means $ \lambda$ is purely imaginary.

Ok, and to prove $ \det (\mathsf{T}\plus{}i\mathsf{S}\minus{}\mathsf{I})\neq 0$ we just use the fact that $ (\mathsf{T}\plus{}i\mathsf{S}\minus{}\mathsf{I})$ is still skew-Hermitian?

Also, $ \mathsf{U}^{\dagger} \equal{} (\mathsf{I}\minus{}\mathsf{T}\minus{}i\mathsf{S})^{\minus{}1}(\mathsf{I}\plus{}\mathsf{T}\plus{}i\mathsf{S})$, but I can't find a way to manipulate it to obtain $ \mathsf{UU}^{\dagger}\equal{}\mathsf{U}^{\dagger}\mathsf{U} \equal{} \mathsf{I}$

If T + iS is skew-hermitian, than its eigenvalues are purely imaginary. Therefore, 1 is not an eigenvalue of such matrix, and so $ \det(T \plus{} iS \minus{} I) \neq 0$.

If $ U \equal{} (I \plus{} T \plus{} iS)(I \minus{} T \minus{} iS)^{ \minus{} 1}$ then $ U^* \equal{} (I \plus{} T \plus{} iS)^{ \minus{} 1} (I \minus{} T \minus{} iS)$. Moreover, the matrices $ I \plus{} T \plus{} iS$ and $ I \minus{} T \minus{} iS$ commute. Hence,

$ U^*U \equal{} (I \plus{} T \plus{} iS)^{ \minus{} 1} (I \minus{} T \minus{} iS)(I \plus{} T \plus{} iS)(I \minus{} T \minus{} iS)^{ \minus{} 1}$

$ \equal{} (I \plus{} T \plus{} iS)^{ \minus{} 1} (I \plus{} T \plus{} iS) (I \minus{} T \minus{} iS)(I \minus{} T \minus{} iS)^{ \minus{} 1} \equal{} I$,

which of course also implies that $ UU^* \equal{} I$ and $ U$ is unitary.

[quote="Carcul"]If $ U \equal{} (I \plus{} T \plus{} iS)(I \minus{} T \minus{} iS)^{ \minus{} 1}$ then $ U^* \equal{} (I \plus{} T \plus{} iS)^{ \minus{} 1} (I \minus{} T \minus{} iS)$. Moreover, the matrices $ I \plus{} T \plus{} iS$ and $ I \minus{} T \minus{} iS$ commute.[/quote]

Oh I calculated my Hermitian transposes wrongly  :blush: Thank you


function
functional equation
algebra
IMO Shortlist

The function $f$ is defined on the set $\mathbb{Q}$ of all rational numbers and has values in $\mathbb{Q}$. It satisfies the conditions $f(1) = 2$ and $f(xy) = f(x)f(y) - f(x+y) + 1$ for all $x,y \in \mathbb{Q}$. Determine $f$.

A short solution to a hard problem:
   Take y=1 and find that $ f(x+1)=f(x)+1$. Then replace y by y+1 and use again the condition given and the fact that $ f(x+1)=f(x)+1$. You will find that $ f(xy+x)=f(xy)+f(x)-1$. Thus $ g(x)=f(x)-1$ is aditive and then it's easy.

You may study this problem in different interesting cases.
Solve it for :
a) $f:N \rightarrow N$
b) $f:Q \rightarrow Q$
c) $f:R \rightarrow R$

Pierre.

My solution:
$f(xy)=f(x)f(y)-f(x+y)+1$
Put $y=1$ to get $f(x+1)=f(x)+1$  .......$*$
An easy induction on $n$ yields two important results:
$1)$for $n$=integer , we have $f(n)=n+1$.
$2)$ for $n$=integer and $x$=rational we have $f(x+n)=f(x)+n$
Now putting $x=q$ and $y=\frac{p}{q}$ for some integers $p,q$ we have
$f\left(\frac{q.p}{q}\right)=f(p)=f(q)f\left(\frac{p}{q}\right)-f\left(q+\frac{p}{q}\right)+1$. Then using the two results we have :
$f\left(\frac{p}{q}\right)=1+\frac{p}{q}$. Hence we have
$f(r)=1+r$ for all rational $r$.

Easy one:
Let $P(x,y)$ be the assertion.
$P(x,1)\implies f(x)+f(x+1)=f(x)f(1)+1\implies f(x+1)=f(x)+1$ with $f(1)=2,$ we can easily get $f(x)=x+1\forall x\in\mathbb{Z}$
An easy induction gives $f(x+n)=f(x)+n\forall x,n\in\mathbb{Z}$
$f\left(\frac{p}{q}+q\right)=f\left(\frac{p}{q}\right)+q$
$P\left(\frac{p}{q},q\right)\implies f\left(\frac{p}{q}\cdot q\right)+f\left(\frac{p}{q}+q\right)=f\left(\frac{p}{q}\right)f(p)+1$
$\implies f(p)+f\left(\frac{p}{q}\right)+q=f\left(\frac{p}{q}\right)f(p)+1$
$\implies p+q=qf\left(\frac{p}{q}\right)\implies \boxed{f\left(\frac{p}{q}\right)=\frac{p}{q}+1}$

It can be solved for Q even if the f(1)=2 wasn’t given. Then the answer becomes
1)f(x)=x+1
2)f(x)=1
It gradually reaches f(1)=2 or 0. The 0 part gets ugly.

$P(x,0)\Rightarrow f(0)=1$
$P(x,1)\Rightarrow f(x+1)=f(x)+1\Rightarrow f(x+n)=f(x)+n\Rightarrow f(n)=n+1\forall n\in\mathbb Z$
$P\left(\frac p{q+1},q+1\right)\Rightarrow f\left(\frac pq\right)=\frac pq+1\Rightarrow\boxed{f(x)=x+1}$ which works.

$P(x,0): f(0)=f(x)f(0)-f(x)+1$.

$P(x,1): f(x)=2f(x)-f(x+1)+1\implies f(x+1)=f(x)+1$. So $f(x)=x+1\forall x\in \mathbb{Z}$.

For rational $\frac{p}{q}$, where $p$ and $q$ are integers,

$P\left(\frac{p}{q},q\right): f(p)=f\left(\frac{p}{q}\right)f(q)-f\left(\frac{p}{q}+q\right)+1$.

Thus, $p+1=f\left(\frac{p}{q}\right)(q+1)-f\left(\frac{p}{q}\right)-q+1$. So we have $qf\left(\frac{p}{q}\right)=p+q\implies f\left(\frac{p}{q}\right)=\frac{p}{q}+1$.

Thus, $\boxed{f(x)=x+1}$, which works.


Let f from the integers to integers be surjective funcion. Prove there are surjective functions g,h from integers to integers s.t. f(x)=g(x)h(x) for all integers x.

Bomb

It's pretty straightforward, I think. I also think it was used in a recent Romanian TST.

Find first $a_0=b_0=x$ such that $f(x)=0$, and set $g(x)=h(x)=0$. Then take $a_1>a_0,b_0$ and put $g(a_1)=1,\ h(a_1)=\frac{f(a_1)}{g(a_1)}$. After that, take $b_1>a_0,b_0,a_1$ and put $h(b_1)=1,\ g(b_1)=\frac{f(b_1)}{h(b_1)}$. I think it's clear what we're doing here. Once we've defined $a_0=b_0<a_1<b_1<a_{-1}<b_{-1}<a_2<\ldots<a_n<b_n$, find $a_{-n}>b_n$ such that $-n|f(a_{-n})$ and set $g(a_{-n})=-n,\ h(a_{-n})=\frac{f(a_{-n})}{g(a_{-n})}$, then choose $b_{-n}>a_{-n}$ such that $-n|f(b_{-n})$ and put $h(b_{-n})=-n$, after which we find $a_{n+1},b_{n+1}$, and so on. On integer values other than $a_i,b_j$ define $g$ to be $1$ and $h$ to be $f$.

Another approach, possibly simpler.

If we can find surjective $G,H$ such that $G(n)H(n)=n$, then $g(n)=G(f(n))$ and $h(n)=H(f(n))$ will work.
For each $n\ge0$, let $G(n^2)=H(n^2)=n$.
For each $n\ge1$, let $G(n^2+n)=-n$ and $H(n^2+n)=-n-1$.
Let $G(-1)=1$, $H(-1)=-1$.
For all other $n$, $G(n)$ and $H(n)$ can be any integers satisfying $G(n)H(n)=n$.

[quote="grobber"]It's pretty straightforward, I think. I also think it was used in a recent Romanian TST.
[/quote]

Indeed, it was used in a Romanian TST for IMO 2001.


For $a,b,c \ge 0$, prove that: \[\sqrt[3]{a \over b+c}+\sqrt[3]{b \over c+a}+\sqrt[3]{c \over a+b}\ge 2.\]

This is an very old inequality and has been post many times (I've post it once :D ). Can you show me your solution?


Let $ x,y,z,k >0$. Prove that

[list]$ (x^2\plus{}y^2\plus{}z^2\plus{}k^2) \left(\frac{1}{x\plus{}k}\plus{}\frac{1}{y\plus{}k}\plus{}\frac{1}{z\plus{}k} \right) \geq 2(x\plus{}y\plus{}z).$[/list]

[quote="4865550150"]Let $ x,y,z,k > 0$. Prove that

[list]$ (x^2 \plus{} y^2 \plus{} z^2 \plus{} k^2) \left(\frac {1}{x \plus{} k} \plus{} \frac {1}{y \plus{} k} \plus{} \frac {1}{z \plus{} k} \right) \geq 2(x \plus{} y \plus{} z).$[/list][/quote]

By Cauchy Schwarz we have :

$ LHS \geq \frac{9(x\plus{}y\plus{}z\plus{}k)^2}{4(x\plus{}y\plus{}z\plus{}3k)}$

We have to prove that :

$ \frac{9(x\plus{}y\plus{}z\plus{}k)^2}{4(x\plus{}y\plus{}z\plus{}3k)} \geq 2(x\plus{}y\plus{}z) \Leftrightarrow (x\plus{}y\plus{}z\minus{}3k)^2 \geq 0$

We have done :)


geometry
inequalities proposed
inequalities

[quote="xzlbq"]Let ABC be a triangle,prove that:
$ \frac {w_a^2}{(w_a \plus{} r_b)^2} \plus{} \frac {w_b^2}{(w_b \plus{} r_c)^2} \plus{} \frac {w_c^2}{(w_c \plus{} r_a)^2} \geq \frac {3}{4}$[/quote]
Sorry, but what do you mean by $ w_a$? Are they the lengths of the angle-bisectors? :oops:

Are they the lengths of the angle-bisectors?
yes!
BQ

*************************************************8
a, b, c,            lengths of sides of the triangle``    s,                  s:=(a+b+c)/2, half the perimeter`

`    R,                  circumradius`
`    r,                  radius of the incircle`
`    ra, rb, rc,         radii of excircles, respectively`
`    ha, hb, hc,         altitudes respectively`
`    ma, mb, mc,         lengths of medians, respectively`
`    wa, wb, wc,         bisectors of angles,
`    S,                  Area of the triangle`

`    A, B, C,            angles of the triangle, respectively`

**************************************************


geometry
3D geometry
Pythagorean Theorem

The diagonal of a face of a cube is $7\sqrt2$ cm. What is the volume of the cube?

[hide="i forgot again"]
root 98 is hte diagonal. therefore, each side is 7. 7^3=343 cubic centimeters
[/hide]

[hide]
You use the Pythagorean Theorem to get the side lenght.
$2a^2 = 7\sqrt{2}^2 = 49 * 2$
$a^2 = 49$
$a = 7$
$7^3 = \framebox{343}$centimeters[/hide]

[hide="click if you want to"]
since the diagonal is $7 \sqrt{2}$ then one side of the cube is 7. $7^3=\boxed{343}$[/hide]

[quote="236factorial"]Thanks for hiding your answer, pascal  :roll:[/quote]

That was useless. Next time, pm me about all the hides i missed. Let's forget all the hides for today, and start tomorrow. Next time don't spam.  :spam:


[quote="rustam"]Find the $\lim_{n\to\infty} \sum_{i=n^2}^{(n+1)^2} \frac{1}{\sqrt(i)}$[/quote]
I'm not so sure. Do you mean here $\lim_{n\to\infty} \sum_{i=n^2}^{(n+1)^2} \frac{1}{\sqrt{(i)}}$, or else?
If so, the result is 2:

\[ \lim_{n\to\infty} \sum_{i=n^2}^{(n+1)^2} \frac{1}{\sqrt{(i)}}\leq\lim_{n\to\infty} \sum_{i=n^2}^{(n+1)^2} \frac{1}{n}=\lim_{n\to\infty}\frac{2n+1}{n}=2 \]

Another way,

\[ \lim_{n\to\infty} \sum_{i=n^2}^{(n+1)^2} \frac{1}{\sqrt{(i)}}\geq\lim_{n\to\infty} \sum_{i=n^2}^{(n+1)^2} \frac{1}{n+1}=\lim_{n\to\infty}\frac{2n+1}{n+1}=2 \]

:P  I am a beginner in Univercity (or as u say college) education(i am freshman) and i  have a lot of questions . But this one i have solved  two days after i announced it here.  i rarely visit Internet(it is expensive here). i do solved it as xufrog did . I am so silly . but what i should do to get bettre at limits? I have done the Demidovich and i am trying to solve the problems from the book Sadovnichiy" Math olympiads for students " . what you recommend to me ?


calculus
integration
geometry
calculus computations

Find the area bounded by $ y\equal{}x^2\minus{}|x^2\minus{}1|\plus{}|2|x|\minus{}2|\plus{}2|x|\minus{}7$ and the $ x$ axis.

Let $ f(x)\equal{}x^{2}\minus{}|x^{2}\minus{}1|\plus{}|2|x|\minus{}2|\plus{}2|x|\minus{}7$

Hence, $ f(\minus{}x)\equal{}(\minus{}x)^{2}\minus{}|(\minus{}x)^{2}\minus{}1|\plus{}|2|(\minus{}x)|\minus{}2|\plus{}2|(\minus{}x)|\minus{}7\equal{}f(x)$

From that we also have $ Area_{x<0}\equal{}Area_{x>0}$.

[color=blue]Case 1: $ 0<x\leq 1$[/color]

$ f(x)\equal{}x^{2}\plus{}(x^{2}\minus{}1)\plus{}(2\minus{}2x)\plus{}2x\minus{}7\equal{}2x^{2}\minus{}6$

$ \therefore Area_{0<x\leq 1}\equal{}\int_{0}^{1}\left|2x^{2}\minus{}6\right|dx\equal{}\frac{16}{3}$

[color=blue]Case 2: $ x>1$[/color]

$ f(x)\equal{}x^{2}\minus{}(x^{2}\minus{}1)\plus{}(2x\minus{}2)\plus{}2x\minus{}7\equal{}4x\minus{}8\rightarrow f(2)\equal{}0$

$ \therefore Area_{x>1}\equal{}\int_{1}^{2}\left|4x\minus{}8\right|dx\equal{}2$

Finally, $ Area_{x\in\mathbb{R}}\equal{}2Area_{x>0}\equal{}2\left[Area_{0<x\leq 1}\plus{}Area_{x>1}\right]\equal{}\frac{44}{3}$ :)


group theory
abstract algebra
function
superior algebra
superior algebra unsolved

[quote="exp"]Consider the following Bilinear Map:

e: g X g maps to G where g and G are two groups of some prime

e(aK,bR) = e(K,R)^ab.

where K, R are elements of group g and a,b are integers.

I thought that if K is a generator of g then e(K, K) is a generator of G. I was told that I am right. However, how do I prove this please?[/quote]

Why have you used the additive notation with the group $ g$ and the multiplicative notation with $ G$?
Anyway I write it in a more comfortable way (at least for me).

$ e: G \times G \to G'$

Where $ G \cong \mathbb{Z}_p$ and $ G' \cong \mathbb{Z}_q$ where $ p,q$ are two, not necessary distinct, prime.

For any $ s, t \in G$ and $ a, b \in \mathbb{Z}$:

$ e(s^a, t^b) \equal{} e(s,t)^{ab}$

Let's be $ g' \in Im(e)$, then there exist two elements $ s, t \in G$ such that $ e^{\minus{}1}(g) \equal{} (s, t)$.

Now, $ G \equal{} <g>$ and so $ s\equal{}g^n$ and $ t\equal{}g^m$ and so $ e(s,t)\equal{}e(g^n,g^m)\equal{}e(g,g)^{nm}$. Then $ \forall g' \in Im(e),\ \exists k \in \mathbb{Z}$ such that $ g'\equal{}e(g,g)^{k}$.

Now, $ Im(e) \le G'$?

$ G'$ has only two subgroup $ \{1_{G'}\}$ and $ G'$. $ Im(e)\equal{}\{1_{G'}\}$ iff $ e(g,g)\equal{}1_{G'}$

Let's consider the function $ f: G \to G'$ defined as $ g^m \mapsto e(g^m,g)\equal{}e(g,g)^m$. It's an homomorphism, $ Im(f) \subseteq Im(e) \subseteq G'$, and $ Im(f) \le G'$.

An homomorphism from $ G \cong \mathbb{Z}_p$ to $ G \cong \mathbb{Z}_q$ must be trivial or it's an isomorphism. It's trivial if $ p\ne q$ and it's an isomorphism if $ p\equal{}q$.

Let's suppose $ p\ne q$, than $ Im(f)\equal{}\{1_{G'}\}$ and $ e(g,g)\equal{}1_{G'}$ and so $ Im(e)\equal{}\{1_{G'}\}$.

If $ p\equal{}q$ then $ Im(f)\equal{}G'$ and so $ G' \equal{} <e(g,g)>$.

[quote="aadil"]what is a generator?[/quote]

A generator is an element of a group such that the subgroup generated by this element is the group itself. In other word is an element $ g \in G$ such that $ \forall a \in G, \ a \equal{} g^n$ for some $ n \in \mathbb{Z}$.

[quote="vict85"][Moderator says: There is effectively no reason to quote a full solution just to say "Thanks"][/quote]

Thanks!


inequalities
search
inequalities proposed

If $ x,y,z$ are three reals with $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} 1$ and $ x,y,z >1$

Prove that :

$ \frac {3}{2}\sqrt {x \plus{} y \plus{} z \minus{} 1} \ge \sqrt {x \minus{} 1} \plus{} \sqrt {y \minus{} 1} \plus{} \sqrt {z \minus{} 1}$

$ \sum_{x,y,z}\sqrt{x\minus{}1}\equal{}\sum_{x,y,z}\frac{\sqrt{x\minus{}1}}{\sqrt{x}}\cdot\sqrt{x}\leq\sqrt{\left(\sum_{x,y,z}\frac{x\minus{}1}{x}\right)\cdot\left(\sum_{x,y,z} x\right)}\equal{}\sqrt{2(x\plus{}y\plus{}z)}$
Let $ x\plus{}y\plus{}z\equal{}A$,so we need to prove that:
$ \sqrt{2A}\leq\frac{3}{2}\sqrt{A\minus{}1}$,or
$ A\geq 9$,which is obvious,because as we know from CBS inequality:
$ A\equal{}A(\sum_{x,y,z}\frac{1}{x})\geq 9$

Another solution, please ?

I've two lines solution  :D

Your condition has a little problem, if x, y,z be reals , let's try $ x\equal{}\minus{}1, y\equal{}1, z\equal{}1$ .

Hmm..
I'm sorry, it must be $ x,y,z>1$
I'll edit it..

If $ x,y,z$ are three reals with $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} 2$ and $ x,y,z > 1$

Prove that :

$ \sqrt {x \plus{} y \plus{} z } \ge \sqrt {x \minus{} 1} \plus{} \sqrt {y \minus{} 1} \plus{} \sqrt {z \minus{} 1}$

My Solution :

Because $ \frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\equal{}1$
then $ xy\plus{}yz\plus{}xz\equal{}xyz$

We may set :
$ x\equal{}\text{tan} A \text{tan} B$
$ y\equal{}\text{tan} B \text{tan} C$
$ z\equal{}\text{tan} C \text{tan} A$
with $ A\plus{}B\plus{}C\equal{}\pi$

And then our Inequality 'moved' to

$ cos A\plus{}cos B\plus{}cos C\le \frac{3}{2}$

Which is well known..  :P

[quote="FOURRIER"][quote="Ronald Widjojo"][quote="lambruscokid"]Just do the QM-AM and AM-HM[/quote]

Can you post your solution ??

I don't understand how to use QM-AM and AM-HM in this problem..

I think it doesn't work..  :maybe:  :huh:[/quote]

for you  :P
http://www.mathlinks.ro/viewtopic.php?search_id=1411496829&t=144[/quote]

My problem is different..

This is not Iran 1998..  :P


Let $ t$, $ u$ and $ v$ be the roots of the polynomial $ x^3\plus{}ax^2\plus{}bx\plus{}c$. If $ t^3$, $ u^3$ and $ v^3$ are the roots of $ x^3\plus{}a^3x^2\plus{}b^3x\plus{}c^3$, find all possible values of $ a$, $ b$ and $ c$.

from the first equation we have that $ \minus{} a \equal{} t \plus{} u \plus{} v$ and from the second that $ \minus{} a^3 \equal{} t^3 \plus{} u^3 \plus{} v^3$. cubing the first relation and substracting the second one it follows that $ 3(t \plus{} u)(u \plus{} v)(v \plus{} t) \equal{} 0$.

suppose without loss of generality that $ u \plus{} v \equal{} 0$. then $ a \equal{} \minus{} t, b \equal{} \minus{} u^2, c \equal{} tu^2$. it's easy to verify from this that $ t^3,u^3,v^3$ are the solutions for the second equation. then, all posible values of $ (a,b,c)$ are of the form $ ( \minus{} t, \minus{} u^2, tu^2)$ and we're done.

hope i haven't made a mistake  :P


[quote="ch1n353ch3s54a1l"]Grandview High School is a four-year high school. Among the 560 upperclassmen (sophomores, juniors, and seniors) at Grandview H.S., the ratio fo boys to girls is 3 to 4. Among the freshmen, there are 110 boys and 90 girls. Overall, what percent of the students are boys? Express your answer to the nearest whole number.[/quote]

That's 200 freshmen + 560 upperclassmen= 760 total.
110 freshman boys+ 240 upperclassmen boys= 350 total boys.
$\frac{350 boys}{760 total}$=about 46%

[hide]110+260
----------- = 46 percent
560+200[/hide]

This could be a spring round problem... When did they make target round problems so easy?


Given quadrilateral ABCD with A$ (3,7)$, B$ (3,5)$, C$ (\minus{}1,1)$, and D$ (\minus{}1,3)$, find the number of square units in the area of the quadrilateral.

Two of the x coordinates are the same, and so are the other two. This means we have a trapezoid. Now loom at the y coordinates. Repsective pairs are two units apart, and thus we have a parallelogram. Use the area formula A = bh.

base = 7-5 = 3-1 = 2

height = 3--1 = 4

$ 2 \cdot 4 \equal{} 8$

you can also use the shoelace method
this is very helpful when you have more coordinates


geometry
circumcircle
trapezoid
power of a point
radical axis
angle bisector
geometry solved

Assume $\mathcal G$ cuts $AB$ again in $N$. Then $D$ is the midpoint of the arc $MN$, so $MN\|BC$. This means that $\angle APM=\angle ANM=\angle ABC$. This is equivalent to $BC$ being tangent to the circumcircle of $APB$. We have thus found that $BC$ is a common tangent for the circles $\mathcal G$ and $(APB)$.

Let $T=AP\cap BC$. $T$ lies on the radical axis of the two circles $\mathcal G$ and $(APB)$, so $TB^2=TD^2$.

[quote="pestich"]Circle G is tangent to the circumcircle of ABC at A because tangent point D is the
    point of intersection of angle bisector AD and BC.  Rays AB and AC intersect circles
    at the ends of parallel segments.  In trapezoid CMNB  line thru A connects
    midpoints of parallel sides and diagonals intersection. The same happens in 2
    trapezoids  made by AD cutting CMNB.[/quote]

Okay. Now let's talk about the problem...

  darij

i was trying to approach the solution using the tangent-secant product theorem,
i am just wondering if there is an solution using such theorem
because i couldnt understand what grobber has said

[color=blue][b]Initial problem.[/b] Let ABC be a triangle, and let the angle bisector of the angle BAC meet the side BC at a point D. Let G be a circle which passes through the point A and is tangent to the line BC at the point D. Let this circle G meet the side AC of triangle ABC at a point M (apart from the point A), and let the line BM meet the circle G at a point P (apart from the point M). Prove that the line AP is a median of triangle ABD.[/color]

[i]Solution.[/i] Here is Grobber's solution with more details for everybody who has difficulties with understanding it:

Let the circle G meet the side AB of triangle ABC at a point N (apart from the point A).

Since the circle G is tangent to the line BC at the point D, we have < BDN = < DAN by the secant-tangent angle theorem. On the other hand, since the point D lies on the angle bisector of the angle BAC, we have < DAN = < DAM. Finally, < DAM = < DNM, since the points D, M, A and N lie on one circle (namely, on the circle G). Thus, < BDN = < DAN = < DAM = < DNM, what yields BC || MN. Thus, < ABC = < ANM. On the other hand, since the points A, M, N and P lie on one circle (namely, on the circle G), we have < ANM = < APM. Thus, < ABC = < APM. Thus, 180° - < ABC = 180° - < APM. In other words, 180° - < ABC = < APB. By the converse of the secant-tangent angle theorem, this shows that the line BC is the tangent to the circumcircle of triangle APB at the point B. Therefore, by the intersecting secant and tangent theorem, we have $TB^{2}=TA\cdot TP$, where T is the point of intersection of the lines AP and BC. On the other hand, since the line BC is tangent to the circle G at the point D, the intersecting secant and tangent theorem yields $TD^{2}=TA\cdot TP$. Consequently, $TB^{2}=TD^{2}$, so that TB = TD, and the point T must be the midpoint of the segment BD. Thus, the line AP, passing through this point T, must be a median of triangle ABD.

  darij

Just wanted to say thanks to Darij Grinberg. That explanation helpd a lot. Thanks!

typerintegra, could you post the problem again, since you edited your original posting, and now I dont know what problem are we treating with. thanks

@typerintegra: Be careful. You can get in serious troubles with moderators if you delete your posts. Many people tend to ask questions and then to delete them for a particular reason: the questions are from current olympiads or from homework. If you delete a question of yours, you automatically get suspect of such behavior.

What was the original problem? Well, let me try to rewrite it:

[color=blue][b]Initial problem.[/b] Let ABC be a triangle, and let the angle bisector of the angle BAC meet the side BC at a point D. Let G be a circle which passes through the point A and is tangent to the line BC at the point D. Let this circle G meet the side AC of triangle ABC at a point M (apart from the point A), and let the line BM meet the circle G at a point P (apart from the point M). Prove that the line AP is a median of triangle ABD.[/color]

@ Pestich:

[quote="pestich"]    Let me try again:
    Circle G is mapped onto line BC  from A, MN || BC,  X= BM x CN  on a line from A
    thru the midpoints of  MN and BC.   Corresponding trapezoids have their
    diagonals intersect on the line thru X parallel to BC  (Pappus ? ).
    I do not think a proof is needed here that line thru intersection of diagonals
   and non-parallel sides divides trapezoid bases in half.

    As you can see, an obscure remark can be a challenge, but I can not think of
    a solution simpler than this.[/quote]

And I cannot think of a reason why this is a solution...

  darij

I am quite fed up with the offtopic discussion here, so let me just put things clear:

The problem asks to show that the line AP is a median of triangle ABD, i. e. that the point P lies on the A-median of triangle ABD. What you showed is that the line joining the point A with the point of intersection of the diagonals of one of your three trapezoids is the A-median of triangle ABD. The two points are different, and I don't see a simple way to show that they are collinear with A. Maybe you see one, but then please share it with us. So, if you want your "solution" to be a solution, please complete the missing steps. That's all.

[b]EDIT:[/b] Okay, Pestich, now I have found out how to complete your solution. One [i]has to add an important observation.[/i] Still I'm quite sure that you didn't do this observation, as you didn't mention a magic word anywhere in your posts. The word is "Pascal".

Here is, for all other readers of this topic, the completed solution:

We begin, like in Grobber's solution, with showing that BC || MN.

Let $W=AD\cap MN$ and $V=DN\cap BW$. Then, the relation BC || MN becomes BD || NW. Thus, the quadrilateral BDWN is a trapezoid. It is well-known that, in a trapezoid, the line joining the point of intersection of the (extended) legs with the point of intersection of the diagonals bisects the bases. Thus, in the trapezoid BDWN, the line joining the point of intersection A of the (extended) legs BN and DW with the point of intersection V of the diagonals DN and BW bisects the bases BD and NW. So we obtain that the line AV bisects the segment BD. In other words, the line AV is the A-median in triangle ABD.

The problem asks us to show that the line AP is the A-median in triangle ABD. Hence, the problem will be solved when it will be showed that the lines AV and AP coincide, i. e. that the points A, V and P are collinear.

(And now, the argument comes that is completely missing in your posts:)

The hexagon NDDAPM is cyclic, since all vertices of this hexagon lie on the circle G. Thus, we can apply the Pascal theorem to this hexagon, and conclude that the points $ND\cap AP$, $DD\cap PM$ and $DA\cap MN$ are collinear; hereby, the line DD is the tangent to the circle G at the point D. But since the circle G is tangent to the line BC at the point D, the line DD, being the tangent to the circle G at the point D, must coincide with the line BC, and thus $DD\cap PM=BC\cap PM=B$. Also, trivially, $DA\cap MN=W$. Hence, the points $ND\cap AP$, B and W are collinear. In other words, the point $ND\cap AP$ lies on the line BW. Thus, the lines ND, AP and BW concur, so that the point V, being the point of intersection of the lines ND and BW, must lie on the line AP. In other words, the points A, V and P are collinear. Proof complete.

Thanks, Pestich, for the idea of this solution. But still, some important steps had to be done until it became a solution!

  darij

Another approuch i found, was to invert with respect a circle with center A and radius AD  :D


MIT
college
geometry
topology
number theory

I am a high school Senior and I'm right now highly considering the University of Michigan in Ann Arbor. I was wondering if anyone could give me their opinions of the university as a whole, the math department, how it compares to... in my case Mit, Dartmouth, or Cornell? Or just say anything you can think of relating to University of Michigan, lol.

University of Michigan is a great place. I was accepted there for graduate school but ultimately decided not to go. I know very little about the undergraduate program there, but apparently it's a fine place to go if you like algebraic geometry. (I'm interested in algebraic number theory, and unfortunately the number theory program there is falling/has fallen apart, but they're trying to rebuild it. A few years ago, Michigan was considered by many, including Wiles and Sarnak, the best place in the world to study number theory.)

Reformed calculus.
Heavy teaching load on graduate students and postdocs.
High cost of living in Ann Arbor.
But one of the very best math departments, in particular for algebraic geometry, geometric topology, several complex variables, and applied stuff.

[quote="mlok"]High cost of living in Ann Arbor. [/quote]
High relative to what?

Hey thanks alot for all of your input- I'm not very well versed in the different branches of mathematics past high school, but I tend to shy away from Number Theory and love Geometry. Plus I can always go somewhere else for grad school if i decide to that.

Also In terms of cost of living, I looked up my lovely Connecticut hometown on the site mlok used and it has an index of [b]163.9[/b]! Anyway, I have a pretty sweet scholarship package at UM.

I'm considering Michigan. I don't have the time or money to visit unfortunately.

I didn't get any scholarships. :( You think there's anything I can do? Do you know what in particular you got it for?

[quote="mlok"][url=http://www.bestplaces.net/city/Ann_Arbor_MI-2603000000.aspx]To the U.S. average.[/url] If $100$ is the national average, then Ann Arbor, MI is $111$.
And Bryan, TX is $74$. Yes, I'm cheap.[/quote]

Hrm, well, fair enough.  In my defense, though, my current zipcode rates a 268, my (likely) future zipcode rates a 170, and my previous zipcode rates a 172.

Michigan is a very nice place -- I also considered attending graduate school there but decided against it.  Overall, it is less prestigious than any of the other three schools you mentioned, although Dartmouth doesn't have much of a reputation for mathematics, as far as I know.


pigeonhole principle
arithmetic sequence
number theory
relatively prime

Let $ p$ be a prime,$ a_1,a_2,....a_p$ is a growing arithmetic sequence whose common difference is $ d$,and each $ a_i(1<\equal{}i<\equal{}p)$ is a prime,$ a_1>p$.Prove that $ p|d$.

[hide]Suppose not then d is congruent to something nonzero mod p and since p is prime then d is relatively prime to it and there is a multiple of d, qd q<=p congruent to every residue mod p, including -a_1[/hide]

[hide="Solution"]Consider the sequence $ \mod p$. Since $ a_1 > p$ and $ p$ is prime, we have that none of the $ a_i\equiv0\mod p$. Now, by pigeonhole, since there are only $ p \minus{} 1$ residues $ \mod p$, we must have some $ a_i\equiv a_j\mod p$. So for some positive integers $ m$ and $ n$ such that $ n < m < p$, we have that $ a_1 \plus{} md\equiv a_1 \plus{} nd\mod p$, or $ (m \minus{} n)d \equal{} kp$. Since $ m \minus{} n < p$ and $ p$ is prime, we have that $ m \minus{} n \equal{} b$ and $ d \equal{} cp$ for some positive integer $ b$ and $ c$ such that $ bc \equal{} k$. Therefore, $ p|d$. $ \boxed{\text{QED}}$.
[/hide]