As many of you probably know, question 4 on the AP chem test free response question is always a list of reactions you have to write out. Well, my chem teacher tells us the only way to know how to do it is doing it lots of times...except i don't know where to start. How would you approach question 4? Are there any rules on how to write the equations?

The AP people can only ask you about a limited number of reaction (theoretically; how could it be infinite?).  The thing is, you can learn how to write the net ionic equations for most of the reactions they give you.  One really easy kind is (organic molecule) + (burned in oxygen) $\longrightarrow$ $H_2 O$ + $O_2$; by organic molecules i mean the ones with carbon and hydrogen exclusively.  There are rules for the reactions on the exam, so just google "reaction prediction rules ap chem" or something like that and see what you come up with.

I highly recommend getting your hands on a copy of an AP review book (specifically, I've found the Princeton Review one to be much more helpful than Kaplan or other random ones I've tried).  They all have a list of general rules you can use.

Another classic one is lewis acid/base:

Boron hydride (Borane, $\text{BH}_3$) is added to phosphorous trifluoride ($\text{PF}_3$):

$\text{BH}_3+\text{PF}_3\longrightarrow\text{BH}_3\text{PF}_3$

and similar variants (usually with boron and nitrogen/phosphorous).

[quote](organic molecule) + (burned in oxygen)$\longrightarrow H_2 O + O_2$[/quote]

What? Isn't it (organic molecule) + (burned in oxygen)$\longrightarrow H_2 O + CO_2$

I don't have a clue about what the AP chem test is (I'm English...) but your reaction just looked wrong to me. Sorry if I just misinterpreted you.

Nice catch.  Yes, it's $\text{C}_x\text{H}_y +\text{O}_2\longrightarrow\text{H}_2\text{O}+\text{CO}_2$.


calculus
derivative
function
calculus computations

Guys , help out of this freaking S**t:

A right circular tank(pointy end down) is being filled with liquid at a constant rate of 2m^3/s. The tank has a height of 12m and its circular face(at the top) has a radius of 5m. At what rate is the radius of the upper surface of the liquid increasing when the tank is half full?

given formula: V=3/Pi*r^2*h

I used an example in class yesterday that was identical to this except for numbers.

Several points:
- That's not exactly the right formula for the volume of a cone.
- Since all of your information talks about height and volume, you don't want an extra radius variable. Eliminate it.
- The general principle of related rates problems: find an equation which is always true involving the quantities you care about, and differentiate that equation. The result should be easy to solve for the derivative you care about using the "right now" information.

Typo...

V= Pi/3 *r^2 *h

I used (0,0) & (5,12) found a function respect to h, to eliminate height,then h=12/5r,

then V=4/5Pi*r^3 , I derivatived both sides,d/dt(V)=4/5Pi*3r^2*(dr/dt), but how can I use the constant 2?

i know that dv/dt=half V of the tank

[quote="bsbandnick"]i know that dv/dt=half V of the tank[/quote]
You're misreading it. $ \frac{dV}{dt}$ is $ 2\,\frac{m^{3}}{s}$.


Problem: Prove $ (\log_{a}b) \times (\log_{c}d) \equal{} (\log_{c}b) \times (\log_{a}d)$

[hide="My Attempt"]
Using the inverse identity, we have

$ \frac{1}{\log_{b}a} \cdot \log_{c}d \equal{} \frac{1}{\log_{b}c} \cdot \log_{a}d$

This can be re-written as: $ \log_{b}a \cdot \log_{a}d \equal{} \log_{b}c \cdot \log_{c}d$

From the chain rule for logarithms: $ \log_{b}d \equal{} \log_{b}d$
$ \text{Q.E.D}$

[/hide]

Is this a completely rigorous proof and if not, is there a better way to do it?

Just because a series of manipulations on the statement you have to prove yields a true statement (such as $ A \equal{} A$), it doesn't mean that you have proven it. In order to fix your proof, simply reverse your steps. When you are proving something, you must start from what is known to be true and go through a bunch of implications to lead to the statement you want to prove.

EDIT: Fixed my post.

proofs don't have to go "forwards." a proof is just as valid if you work backwards via a series of equivalent statements.

anyway, i don't know what the "inverse identity" is, but the change of base formula does the job:

\[ \frac{\log_a b}{\log_a d}\equal{}\log_d b \equal{} \frac{\log_c b}{\log_c d}\]

and cross-multiplying gives you the result

The inverse identity is essentially using change of base with the base being $ b$:

$ \log_ab \equal{} \frac {\log_bb}{\log_ba} \equal{} \frac {1}{\log_ba}.$

And yes you're right, I'll edit my previous post.

[quote="pluristiq"]proofs don't have to go "forwards." a proof is just as valid if you work backwards via a series of equivalent statements.
[/quote]

Yes, but only because the equivalent statements imply that you can move 'forwards', from the premises to the conclusion.

[quote="mathemagician1729"]Problem: Prove $ (\log_{a}b) \times (\log_{c}d) \equal{} (\log_{c}b) \times (\log_{a}d)$

[hide="My Attempt"]
Using the inverse identity, we have

$ \frac {1}{\log_{b}a} \cdot \log_{c}d \equal{} \frac {1}{\log_{b}c} \cdot \log_{a}d$

This can be re-written as: $ \log_{b}a \cdot \log_{a}d \equal{} \log_{b}c \cdot \log_{c}d$

From the chain rule for logarithms: $ \log_{b}d \equal{} \log_{b}d$
$ \text{Q.E.D}$

[/hide]

Is this a completely rigorous proof and if not, is there a better way to do it?[/quote]

just use the fact that $ {log_a}b\equal{}\dfrac{log_x b}{log_x a}$.

EDIT: IGNORE

Is this right (i have never done a proof in my life and kinda want to learn):
\[ (\log_{a}{b})(\log_{c}{d}) \equal{} (\log_{c}{b})(\log_{a}{d}) \\
 \\
\frac {\log_{c}{b}}{\log_{c}{a}} \times \log_{c}{d} \equal{} \log_{c}{b} \times \frac {\log_{c}{d}}{\log_{c}{a}} \\
 \\
\log_{c}{(b \plus{} d \minus{} a)} \equal{} \log_{c}{(b \plus{} d \minus{} a)}\]
SO because I end up with two equal things, would the first statement be considered proven?  Or in a proof do you start of with a common law or rule and change it to make the first statement?  (I'm really confused)...

EDIT: IGNORE

EDIT: gah....got my log identies confused...

[quote="007math"]Is this right (i have never done a proof in my life and kinda want to learn):
\[ (\log_{a}{b})(\log_{c}{d}) \equal{} (\log_{c}{b})(\log_{a}{d}) \\
 \\
\frac {\log_{c}{b}}{\log_{c}{a}} \times \log_{c}{d} \equal{} \log_{c}{b} \times \frac {\log_{c}{d}}{\log_{c}{a}} \\
 \\
\log_{c}{(b \plus{} d \minus{} a)} \equal{} \log_{c}{(b \plus{} d \minus{} a)}\]
SO because I end up with two equal things, would the first statement be considered proven?  Or in a proof do you start of with a common law or rule and change it to make the first statement?  (I'm really confused)...[/quote]

among other things, you made a mistake akin to the freshman's dream,

http://en.wikipedia.org/wiki/Freshman's_dream

[quote="007math"]Is this right (i have never done a proof in my life and kinda want to learn):
\[ (\log_{a}{b})(\log_{c}{d}) \equal{} (\log_{c}{b})(\log_{a}{d}) \\
 \\
\frac {\log_{c}{b}}{\log_{c}{a}} \times \log_{c}{d} \equal{} \log_{c}{b} \times \frac {\log_{c}{d}}{\log_{c}{a}} \\
 \\
\log_{c}{(b \plus{} d \minus{} a)} \equal{} \log_{c}{(b \plus{} d \minus{} a)}\]
SO because I end up with two equal things, would the first statement be considered proven?  Or in a proof do you start of with a common law or rule and change it to make the first statement?  (I'm really confused)...[/quote]

No, $ \log_{c}{(\frac {bd}{a}}) \equal{} \log_{c}{b} \plus{} \log_{c}{d} \minus{} \log_{c}{a}$

$ \frac {\log_{c}{b}}{\log_{c}{a}} \times \log_{c}{d} \equal{} \log_{c}{b} \times \frac {\log_{c}{d}}{\log_{c}{a}} \equal{} \frac {\log_{c}{b} \times \log_{c}{d}}{\log_{c}{a}} \ne \log_{c}{(b \plus{} d \minus{} a)}$

====================================================

"SO because I end up with two equal things, would the first statement be considered proven?"

Counterexample:

No.  Suppose we are given:
$ 1 \equal{} 2$.
$ 2 \equal{} 1$ . . . Add these equations.
--------
$ 3 \equal{} 3$

The result is true, but the original statement is false.

my own proof also in that way!  :(  :rotfl:

The proof is actually very easy to complete by simply rewriting the logarithms as exponents. For instance let $ log_{a}{b}$ = x and $ log_{c}{d}$ = y and $ log_{c}{b}$ = m and $ log_{a}{d}$ = n. We wish to show that xy = mn. By exponentiating our logarithms we have $ a^x \equal{} b$, $ c^y \equal{} d$, $ c^m \equal{} b$ and $ a^n \equal{} d$. We now have $ a^x \equal{} c^m$ so $ c^{m/x} \equal{} a$. We also have $ a^n \equal{} c^y$ and $ c^{y/n} \equal{} a$. We then have $ y/n \equal{} m/x$ and $ xy \equal{} mn$, as desired.


abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a 2-group, $ Z(G)$ centre of $ G$ and $ s_i$ the number of elements of order $ 2^i$.   Suppose that $ Z(G)$ is elementary abelian and $ exp(G)\equal{}2^r$.  If  $ G\minus{}Z(G)$ has a element of order 2 and $ s_i\equal{}s_j$ for $ i,j\geq 2$, then whether or not $ r\leq 3$?


[b]Problem has been solved by thread creator.[/b]
[b]There is no need for a reply.[/b]
Hello again.
I'm working on a problem that involves Gaussian Elimination. I don't quite understand the process of it all but I think I'm starting to get it.

Instructions: [u]Use matrices to solve each system of equations. Each system has one solution.[/u]

I'm i'm looking at:

2x + y + 3z = 3
-2x - y + z = 5
4x - 2y - 3z = 10

So I decided to turn it into this after doing some process work:

[img]http://www.mathlinks.ro/Forum/latexrender/pictures/1fc0fffdd6a090f1489089efb14dfdf9.gif[/img]

I'm guessing there is something I have to do to the second row because of the row echlon form. Am I allowed to just divide the second row by 1/4 and imagine the numbers getting knocked down so that:

[img]http://www.mathlinks.ro/Forum/latexrender/pictures/9927f331abc4c367a9f5a6ba101ddf37.gif[/img]

I'm guessing that is the right thing for me to do because I am closer to Row echlon form.

What I can't get though is what to do to the third row. Perhaps I should just divide that by -1/4?

I suppose i'm suppose to use an operation on a row and then somehow manipulate that into the third row, but I don't know where that would be coming from.

What do I do next? What am I doing right&&wrong?

My latex skills suck. Anyone know how i would do that type of matrix latex on here?

Matrices belong in the high school forum (Algebra II and above), especially the Gaussian Elimination.  ;)

[quote="math92"]yeah, i don't know what matrices are :blush:[/quote]

I think it's a way to organize data.  They're helpful in solving multi-variable equations.

i saw one in Cyberman's post.
How do they work?

Read Algebra II  :rotfl:
Someone move this topic.

Well, it can help solving a system of equations : http://tutorial.math.lamar.edu/AllBrowsers/2318/SolvingSystemsOfEqns.asp

[quote="math92"][quote="mathmanman"]Enjoy :p : http://www.ping.be/~ping1339/matr.htm[/quote]
but what do matrixes do and how do they do it?[/quote]

http://en.wikipedia.org/wiki/System_of_linear_equations
http://en.wikipedia.org/wiki/Square_matrix
http://www.purplemath.com/modules/mtrxrows.htm
http://en.wikipedia.org/wiki/Gaussian_elimination
http://en.wikipedia.org/wiki/Determinant
http://www.purplemath.com/modules/cramers.htm

Just some stuff.


Hello, is there an elegant possibility to solve this equation?
$ \sqrt [5] {x^3\plus{}2x}$ = $ \sqrt[3] {x^5\minus{}2x}$

I only have followed the complex way, starting with an equation of 25th degree. That is also possible, to find the solutions 0, $ \pm \sqrt{2}$. But is there a way, that is more easy and elegant?

Let $ X\equal{}(x^5\minus{}2x)^{1/3}$. $ X^5\plus{}X^3\equal{}x^5\plus{}x^3$; moreover the function $ t\rightarrow{t^5\plus{}t^3}$ is one to one; then $ X\equal{}x$. Therefore $ x^3\equal{}x^5\minus{}2x$ and after it is easy.

[quote="loup blanc"]Let $ X \equal{} (x^5 \minus{} 2x)^{1/3}$. $ X^5 \plus{} X^3 \equal{} x^5 \plus{} x^3$; moreover the function $ t\rightarrow{t^5 \plus{} t^3}$ is one to one; then $ X \equal{} x$. Therefore $ x^3 \equal{} x^5 \minus{} 2x$ and after it is easy.[/quote]

[color=darkblue]Thank loup blanc. Equation $ \Longleftrightarrow x\equal{}0$ or $ x^4\minus{}x^2\minus{}2\equal{}0$

$ \Longleftrightarrow x\equal{}0$ or $ x\equal{}\pm \sqrt{2}$[/color]

Bonjour, Loup Blanc, je ne peux pas suivre ton idée jusqu'à la fin.

If you substitute $ X \equal{} \sqrt[3] {x^5 \minus{} 2x}$ then the root on the right side of this equation disappears:
$ \sqrt[5] {x^3 \plus{} 2x} \equal{} X$
$ x^3 \plus{} 2x \equal{} X^5$

So, excuse me please, but I cannot get: $ X^5 \plus{} X^3 \equal{} x^5 \plus{} x^3$
Could you help me? The rest of your argumentation is totally obvious, but how to come to this equation?

$ X^5\equal{}x^3\plus{}2x$ AND $ X^3\equal{}x^5\minus{}2x$.

Ah oui, c'est tout clair maintenant, merci beaucoup.


Find a function from R^+ to R^+ such that f(f(x))=(3x+1)/(x+3)

[hide="a solution"]
Let $ f(x)=\frac{ax+b}{cx+d}$
by putting this in $ f(f(x))=\frac{3x+1}{x+3}$ and after some calculations we'll get that:
$ a=d,b=c,a^{2}+b^{2}=3,2ab=1,a\neq b$
so $ (a+b)^{2}=4\text{ or }a+b=+2\text{ or }-2$
let $ a+b=2\rightarrow b=2-a$ (cuz we only want one solution!)
now by $ a^{2}+(a-2)^{2}=3$ a value for $ a$ is $ a=1+\frac{\sqrt{2}}{2}$ and $ b=1-\frac{\sqrt{2}}{2}$
and a  solution is : $ f(x)=\frac{\left(1+\frac{\sqrt{2}}{2}\right)x+1-\frac{\sqrt{2}}{2}}{\left(1-\frac{\sqrt{2}}{2}\right)x+1+\frac{\sqrt{2}}{2}}$
[/hide]


function
complex analysis
complex analysis unsolved

Let $\varphi(z)$ be an analytic function such that $\varphi ' (z_0) \not = 0$, prove that there exist a neighbourhood of $z_0$ so that
$\varphi(z)$ injective.

isn't it exactly local inversion theorem ?


What is the remainder when f(x) = - 3x^17 + x^9 - x^5 + 2x is divided by x + 1?

NOTE:  For the question above, approximate the positive zero correct to [color=red]TWO[/color] decimal places.

Where are you pulling these questions from?

Hint. $f(-1)$ is the remander

Your hint is not clear to me.

Can you show me how to solve this question or not?

NOTE:  The questions are coming from a precalculus textbook.

[quote="Interval"]Can you show me how to solve this question or not?[/quote]
Nope, no one on aops knows how to solve this problem.
Maybe if you actually thought about the hint instead of expecting a ready made answer you would get better at these types of problems?

One of the first theorems in basic algebra is that when a term $(x-a)$ divides a polynomial, the remainder is $f(a)$

If this isn't in your book, go check out an algebra textbook :wink:


There are $ n$ people and there exist $ 2^{n}\minus{}1$ football teams. In each football team we have to choose menager.
BUT: If team  $ X$ is sum $ X\equal{}Y\cup Z$ then the menager $ X$ is the menager of at least one of teams $ Y, Z$
Find the number of possible selections of menager.

This is problem 7 from 2nd series of 1st round of 59th Polish Mathematical Olympiad. The 2nd series of the competition ends at 12.11.2007. Moderators, please lock the topic.

[quote="Rand0m"]Look. I don't know that this question is placed in polish mathematical olympiad. I am from sweden and I am studying in Englang - i am not in high school, and I an not from Poland. I like such a questions, so don't close the topic ! If it is really from POL math olympiad please don't place any answer here, just tell me on private message. If somebody think that he know the answer just make a conversation with me by private message. What's the problem mate?![/quote]
The discussion can be continued at [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=165235[/url].

[quote="Rand0m"]You are moderator?

Thease are open questions! It is not internet competition.... :mad:[/quote]

You are n00b?

It is not an internet competition, it is a written competition... so what? These questions have to be solved by participants WITHOUT EXTERIOR HELP, it's there to see who has the better brain and creativity and not who knows the better professor.

  dg


geometry
circumcircle
geometric transformation
reflection
projective geometry
geometry proposed

Let $ABCD$is convex quadrilateral.$BC$ isn’t parallel$AD$,$AB$isn’t parallel$CD$.

$E\in{AB},F\in{BC},G\in{CD},H\in{AD}$ and$EG\cap{BC}\cap{AD}=\emptyset$

$O\in{ABCD}$

$\angle BOE=\angle AOE,\angle COG=\angle DOG,\angle BEO=\angle OGD$

$FE=EH,FG=GH$

Prove that $\angle FEH=\angle COD$.
 ;)  ;)

Ph-An I waiting you too,or you can't solve this problem? :P  :D

[color=blue]About this problem:[/color]This problem is my generalizatoin of problem in IMO 2000 short list.$G6$[url=http://www.kalva.demon.co.uk/short/soln/sh00g6.html]www.kalva.demon.co.uk/short/soln/sh00g6.html[/url]
  Ph-An I waiting your solution till tomorrov ivening. :nhl:  :spider:

You did understood nothing. You are a obstinate young. I don't want to prove, is it O.K. ? If you wish this, I recognize that I don't know to prove. I am angry with you. You didn't understood me. I say you good bye ! I hope soon to see you better kindly again.

Comon guys,this isn't very hard problem :P  :rotfl:

Ph-An I cheng my post name for you :rotfl:  :rotfl:

[quote="Tiks"]Let $ABCD$is convex quadrilateral.$BC$ isn’t parallel$AD$,$AB$isn’t parallel$CD$.

$E\in{AB},F\in{BC},G\in{CD},H\in{AD}$

$O\in{ABCD}$

$\angle BOE=\angle AOE,\angle COG=\angle DOG,\angle BEO=\angle OGD$

$FE=EH,FG=GH$

Prove that $\angle FEH=\angle COD$.
 ;)  ;)[/quote]

Let the lines BC, AD intersect at a point P and let the bisector of the angle $\angle APB \equiv \angle DPC$ meet the sides AB, CD at points E, G, respectively. Let (U) be the P-Apollonius circle of the triangle $\triangle APB$ and (V) the P-Apollonius circle of the triangle $\triangle DPC$. Assume that $\angle PAB > \angle PBA$ and $\angle PDC < \angle PCD$. This is certainly possible. Then the center of the circle (U) lies on the ray opposite to AB and the center of the circle (V) lies on the ray opposite to CD. Since the line $PE \equiv PG$ bisects the angle $\angle APB \equiv \angle DPC$, $E \in (U),\ G \in (V)$. Let O be the other intersection of the circles (U), (V) different from P, which can certainly be inside of the quadrilateral ABCD. Then the line OE bisects the angle $\angle AOB$ and the line OG bisects the angle $\angle COD$. Therefore,

$(1)\ \ \angle AOE = \angle BOE,\ \ \angle COG = \angle DOG$.

The angle $\angle EPO \equiv \angle GPO$ spans the arcs EO, GO of the circles (U), (V), respectively. Therefore, the isosceles triangles $\triangle EOU \sim \triangle GOV$ are similar and consequently, the angles $\angle AEO \equiv \angle UEO = \angle VGO \equiv \angle CGO$ are equal and the angles

$(2)\ \ \angle BEO = 180^\circ - \angle AEO = 180^\circ - \angle CGO = \angle DGO$

are also equal. Let Q be an arbitrary point on the segment EG and let a normal to EF at the point Q intersect the lines BC, AD at points F, H. Since the line $PE \equiv PG \equiv EG$ bisects the angle $\angle APB \equiv \angle DPC$ and since $FH \perp EG$, it follows that QF = QH. Thus the quadrilateral EFGH has perpendicular diagonals and the diagonal FH is cut by the diagonal EG in half. This means that the quadrilateral EFGH is a kite and

$(3)\ \ FE = EH,\ \ FG = GH$.

All the problem conditions are satisfied. But it is clear that the point O is unique (the other intersection of the circles (U), (V) different from P), hence the angle $\angle COD$ is fixed. On the other hand, the diagonal intersection Q of the quadrilateral EFGH can move within the segment EG, perhaps within smaller limits in order to keep both the points F, H inside the segments BC, AD, respectively, this quadrilateral remains a kite, but the angle $\angle FEH$ varies. As a result, the angles

$(4)\ \ \angle FEH \neq \angle COD$

are generally not equal and [color=blue]the problem proposition is false[/color]. Do not tell me, Tiks, that this is what you intended, because you did not.

Yetti

:oops:  :oops: Soryyyyyyyyyy.......$yetti$:?
I must say also that line $EG$,$BC$,$AD$not intersecting in same point :(
becous,if it is it,we have infiniti many line $FH$ that$FE=FH,FG=GH$ :(
Now  I thike for any $O$ we have only one pers ($F$,$H$),isn't it?
yetti  sory again :(

Guys try solve this problem know it is true :blush: (I edited it :lol: )

so resulting just one question for you Mr Tiks ... is the solution same with yours ?

Davron

[quote="yetti"]Let's finish this unfinished business. Let the lines AB, CD meet at a point X, WLOG on the opposite side of the line BC than the quadrilateral vertices A, D. Since the angles $\angle BEO = \angle DGO$ are equal, $\angle BEO + \angle CGO = 180^\circ$ and the quadrilateral EXGO is cyclic with a circumcircle (Z).

Let the lines OB, OC meet the circle (Z) again at points Q, R, respectively and consider the cyclic hexagon EQXRGO. Denote F' the intersection of the lines GQ, ER. By Pascal's theorem, the intersections B, C, F' of the line pairs (EX, OQ), (GX, OR) and (GQ, ER) are collinear. It follows that

$\angle BOE \equiv \angle QOE = \angle QGE \equiv \angle F'GE$

$\angle COG \equiv \angle ROG = \angle REG \equiv \angle F'EG$

Let the lines OD, OA meet the circle (Z) again at points S, P, respectively and consider the cyclic hexagon EXGPOS. Denote H' the intersection of the lines GP, ES. By Pascal's theorem, the intersections D, A, H' of the line pairs (GX, OS), (EX, OP) and (ES, GP) are collinear. It follows that

$\angle DOG = 180^\circ - \angle SOG = \angle SEG \equiv \angle H'EG$

$\angle AOE = 180^\circ - \angle POE = \angle PGE \equiv \angle H'GE$

By the problem condition, OE, OG are bisectors of the angles $\angle AOB, \angle COD,$ respectively. Therefore,

$\angle F'GE = \angle BOE = \angle AOE = \angle H'GE$

$\angle F'EG = \angle COG = \angle DOG = \angle H'EG$

As a result, the triangles $\triangle EF'G \cong \triangle EH'G$ with the common side EG are oppositely congruent, i.e., one is a reflection of the other in the line EG. This means that the quadrilateral EF'GH' is a kite and moreover,

$\angle AOB = \angle H'GF',\ \angle COD = \angle F'EH'$

The kite diagonals $EG \perp F'H'$ are perpendicular to each other and the diagonal EG cuts the diagonal F'H' in half. As long as the line EG does not pass through the intersection Y of the lines BC, DA (the condition originally missing), it cannot be identical with the bisector of the angle $\angle AYB \equiv \angle DYC$. This implies that the kite EF'GH' is unique, i.e., identical with the given kite EFGH. :cool:[/quote]
At laste yetti... :huh: ,now you realy think that this problem is hard?(I don't think so!)
Dear Davron my idea is same I am saw that ther are only one pair of points  (F,H) with that conditions and considered them from Pascal's theorem ;) .
Anyway
                                     $score$
            $yetti: 2-Virgil Nicula: 0-Darij Grinberg: 0$ :D
Congratulations yetti,but the match hasn't over yet :lol: ,I will post a new problem for this tree geometrs  soon ;) .


A rectangle has perimeter 26 inches and integer length sides, in
inches. What is the number of square inches in the greatest
possible area?

if the perimeter is 26, then the sum of the two sides will be 13. If the area is to be as large as possible then the two sides that add to 13 must be as close as possible. The closest case is 6,7 so the area is 6*7=42

The length and width of rectangle added together is half the perimeter. So: $ l \plus{} w \equal{} \frac{1}{2} \times 26 \equal{} 13$.

The larger the area, the more square-like the rectangle is. This means that the area needs to be as close to a perfect square as possible.

$ 13 \div 2 \equal{} 6.5$

However, the side lengths are integers, so we do the closest numbers -- in this case, 6 and 7. So the length and width are 6 and 7, with an area of $ \boxed{42}$.


geometry
angle bisector
national olympiad

Can anyone tell me what is the date of this year's Mediterranean MO? I know that the previous years it wasn't held in all the participating countries on the same date, but what's the "interval" in which the MO will be held?
P. S. If anyone knows the date of the competition for Bosnia and Herzegovina, it would be a useful information for me... Thanks.

Oooops... :blush:  The Mediterranean competition this year was a bit early... I saw today the dates... But still, can anyone tell me did Bosnia and Hezegovina participate?

I could tell you useful information about participating of Bosnia and Herzegowina.We have participated last year(i have won bronze  :P ) and the competion were on the end of May.I think(but i am not sure) that we will participate this year too(we wont only if competition will be at same dates as our MO :)) and that is usually held in Sarajevo.I think that the dates should be similar to those from last year.Which date do you have?  :D

[quote="silouan"]The Mediiterrenean contest took place at  1 of April . :lol:[/quote]
 I think you are wrong we have just got information that competition is this Sunday.
 http://www.pmf.unsa.ba/UMBIH/olimpijada.html

Maybe we speak for another contest....
Because I have the problems of the meditterrean contest 2006

We sure speak about different competition.I speak about Medditeran Mathematical Olympiad which is held in almost all mediterran countries on same place at same time usualy at the end of may.So there isnt any chance that that was on 1st april
 btw.i just realised that 1th april is day of jokes so I now figure what you meant with that  :D  :D

[quote="hsiljak"]Hmmm... There's something fishy going on here :) Vedran, take a look at this forum entries from Croatia (look at the date!) [url]http://playmath.skolstvo.t-com.hr/forum/index.php?action=vthread&forum=8&topic=198[/url] For God sakes, what is going on here? Btw, look at this: when was the last year's Mediterranean Olympiad held in Croatia:
[url]http://web.math.hr/hmd/vjesnik/22_8mn.htm[/url]....[/quote]
     hmm this is really interesting  :P maybe the problems on mediterrian competition in some countries arent same?I cant se any logic of distances of one month?

Niether do I...
I just hope that we'll do well on Sunday...

So how did you guys do? How did you find the problems? Do you have any ideas on how many points are needed for first, second, third prize?

Problems 1, 2, 3 were OK, but I wasn't able to solve the 4th one. I still hope that I'll get a couple of points on that  one.   :fool:

The fourth one was grose! i think noone did it:) I don't have a clue about how many points will be needed for the medals... my score isn't something I'm proud of:)

Can anyone tell me where can I find tis year's problems? Or can anybody please tell them to me?

Thanks

Here are the problems .
2 April 2006
[color=brown]1[/color] Every point in the plane is colored with two colors ,blue and red .There exist at least one blue point .Is it possible every circle of ray r=1cm contain exactly
ι)one blue point
ιι)two blue points

[color=green]2[/color] Let $P$ an interior point of a triangle ABC and let $A_1B_2$,  $B_1C_2$,  $C_1A_2$  the parallels through  P to the sides AB,BC,CA respectively with $A_1,A_2\in BC$ , $B_1,B_2\in AC$ και $C_1,C_2\in AB$ .Prove that  $(A_1A_2B_1B_2C_1C_2)\geq\frac{2}{3}(ABC)$ where  (...) means area

  [color=darkblue]3[/color] Let a triangle  ABC where a,b,c the sidelenghts are positive integers and GCD(a,b,c)=1 .The interior bisector of the angle Α intersects ΒC at D . Prove that if the triangle ΑΒD is similar to ΑΒC then c is a perfect square .

[color=orange] 4[/color]  Let  $m,n$ positive integers and let  $x_{i,j}\in\left[ 0,1\right]$  for i=1,2,....,m and
j=1,2,....,n

Prove that $\prod_{j=1}^n (1-\prod x_{i,j})+\prod_{i=1}^m (1-\prod_{j=1}^n (1-x_{i,j}))\geq 1$

Solution for (3):
Let $\angle {BAD}$ be $\alpha$. then because triangles $ABD$ and $ABC$ are similar, corresponding angles are equal, then because $\angle {BAD}$ is not equal to $\angle {BAC}$, have $\angle {BCA}=2\alpha$, $\angle {BDA}=\alpha$. So:
$\frac{BA}{BC}=\frac{BD}{AB}$.   (*)
 Also, $AD$ is angle bisector hence $AD=\frac{ac}{c+e}, AB=c,BC=a,AC=b$. Subst this into (*) to get:
$\frac{c}{a}=\frac{ac}{(c+e)c}$
$\frac{c}{a}=\frac{a}{c+e}$
$c^2+ec=a^2$. Because $c|LS$, $c|RS$ ie $c|a^2$. But $gcd(a,c)=1$ hence $c$ is $1$ which is a perfect square.
This is probably wrong because there is just one value for $c$. :?

Yes, you are wrong:
gcd(a,b,c)=1<=>"There does not exist a prime p that divides a, b and c at the same time".

However, each pair of the triple could have a common divisor >1.

I solved a) using the fact that if the product x.y is a perfect square  where x and y are coprime then each of them is a perfect square as well. (I can't figure out what the expressions standing for x and y were, but I also used that gcd(x,y)/y-x).

As for problem 1, I can show one of the possible constructions for (b).
First, colour the whole plain  red. Secondly, construct the family of blue lines y=2.k for every whole k (negative, positive and zero). This way, every circumference of R=1 either intersects exactly one line of the family at exactly two blue points or is "locked" between two neighbouring lines, having one common point in blue with each of them. Hence, the answer is "YES".


abstract algebra
advanced fields
advanced fields unsolved

I have the following problem:

How to prove directly from the definitions that if $ U$ is a direct limit of a direct system $ ( U_{\alpha})_{ \alpha \in \Lambda }$ of sets, then

$ U \equal{} \cup_{ \alpha \in \Lambda } Image( U_{\alpha} \to U)$.

Thank you!

[hide="Hint"]Check that $ U$ fulfils the universal property of the direct limit of the directed family $ \{U_\alpha\}$.[/hide]

I think the hint means that, given the existence of $ U$, we have a canonical inclusion $ V \to U$, where $ V$ is the union of the images, and the universal property shows that this is an isomorphism. by a completely similar argument, the direct sum of modules is generated by the summands, and the tensor product is generated by the factors, etc..... is there any possibilty of generalizing this result? somehow, colimits are always generated by the images of the objects.

sad but true: in general, to verify conveniant properties of universal objects (beyond the universal property), you have to specify concrete models. in this case, take the disjoint union of the $ U_\alpha$ with the obviuos identifications: $ x \in U_\alpha$ and $ y \in U_\beta$ are equivalent iff there is some $ \gamma \geq \alpha,\beta$ such that $ x$ and $ y$ are mapped to the same element in $ U_\gamma$. in every category of algebraic structures directed colimits may be constructed in this way.


It's the first time i notice this odd thing to happen and i don't know why.

tan([b][color=blue]44[/color][/b] / 100 000) = 0.000[b][color=blue]44[/color][/b]000002839466886554982564200603508890882882161731....

tan([b][color=red]432[/color][/b] / 1 000 000) = 0.000[b][color=red]432[/color][/b]00002687385800612275239667366167219232692411299....

tan([b][color=green]1022[/color][/b] / 1 000 000) = 0.00[b][color=green]1022[/color][/b]0003558210313264170396372031593302261567955645....

Can anyone generalize this property and also [b]provide the proof[/b]?
I can't find the proof and it is getting on my nerves. :wallbash_red:  :wallbash_red:

tan(x) is the tangent of x, where x is a real number(that is we express tangent in radians and not degrees)

[quote="t0rajir0u"]This is a consequence of the limit $ \lim_{x \to 0} \frac {\tan x}{x} \equal{} 1$, which is in turn a consequence of the limit $ \lim_{x \to 0} \frac {\sin x}{x} \equal{} 1$.  One way to prove it is a geometric argument to show that $ \sin x < x < \tan x$ for small angles $ x$ and the [url=http://en.wikipedia.org/wiki/Squeeze_theorem]squeeze theorem[/url].[/quote]
I see, but the problem is that we don't have an equality of the form:

tan([b][color=red]432[/color][/b] / 1 000 000) = 0.000[b][color=red]432[/color][/b]

but only an equality of the form:

tan([b][color=red]432[/color][/b] / 1 000 000) = 0.000[b][color=red]432[/color][/b]00002687385800612275239667366167219232692411299....

 That means that the [b][color=red]432[/color][/b] appears inside the number in a specific form [b]and it is followed[/b] by other "random" digits. So we can't use what you have said to prove it. (or can we? :maybe: )

(The four dots "...." after the numbers mean that the numbers are irrational and their digits continue forever.)

If you want a precise bound on the error, that's what [url=http://en.wikipedia.org/wiki/Taylor%27s_theorem]Taylor's theorem[/url] is for.  But for a rough estimate, the first two terms of the Taylor series are

$ \tan x \equal{} x \plus{} \frac{x^3}{3} \plus{} O(x^5)$

which means that for small $ x$ the error in the approximation $ x \approx \tan x$ is cubic.  Including the first term in the approximation we obtain, for example,

$ \tan 0.000432 \approx 0.000432 \plus{} \frac{0.000432^3}{3} \equal{} 0.000432000026873856.$

And you can see that many more of the digits match.  Adding more terms, we obtain even more of the other digits.  (In fact, this is how your calculator actually computes tangents.)

[quote="t0rajir0u"]If you want a precise bound on the error, that's what [url=http://en.wikipedia.org/wiki/Taylor%27s_theorem]Taylor's theorem[/url] is for.  But for a rough estimate, the first two terms of the Taylor series are

$ \tan x \equal{} x \plus{} \frac {x^3}{3} \plus{} O(x^5)$

which means that for small $ x$ the error in the approximation $ x \approx \tan x$ is cubic.  Including the first term in the approximation we obtain, for example,

$ \tan 0.000432 \approx 0.000432 \plus{} \frac {0.000432^3}{3} \equal{} 0.000432000026873856.$

And you can see that many more of the digits match.  Adding more terms, we obtain even more of the other digits.  (In fact, this is how your calculator actually computes tangents.)[/quote]

Oh the mighty Taylor series. It did it again. How did i forget about that??  :wallbash_red:
Thank you! :)


geometry
cyclic quadrilateral
power of a point
radical axis
geometry unsolved

We are given a cyclic quadrilateral $ABCD \quad AB\not=CD$. Quadrilaterals $AKDL$ and $CMBN$ are rhombuses with equal sides. Prove, that $KLMN$ is cyclic

Let $(A)$ be the circle with radius $AK$
Let $(B)$ be the circle with radius $BM$
We have $AK=BM$ so $O$ is on the radical axis of $(A)$ and $(B)$.
This means $OK*OL=OM*ON$ so $K,L,M,N$ are cyclic

Let $ I$,$ J$ be midpoints of  $ AD$,$ BC$
$ KLMN$ cyclic  $ \Leftrightarrow$ $ OM.OK\equal{}OL.ON \Leftrightarrow OI^2\minus{}MI^2\equal{}OJ^2\minus{}NJ^2 \Leftrightarrow OI^2\minus{}(a^2\minus{}AI^2)\equal{}OJ^2\minus{}(a^2\minus{}BJ^2) \Leftrightarrow OA^2\equal{}OB^2\equal{}R^2$
Right!

Let $O$ be the circumcenter of $ABCD$; it suffices to show that $OM \cdot ON - OK \cdot OL$. Consider the circle at $B$ with radius $\overline{BM}$; by power of a point, $$OM \cdot ON = BM^2 - OB^2 = OK \cdot OL$$ as the side lengths are equal.


Euler
function
number theory
number theory unsolved

[quote="jin"]As we have solved:There are infinite positive integers $n$ such that
(1)$p(n)<p(m+1)<p(n+2)$,$p(n)$ is the largest prime factor of $n$.
(2)$w(n)<w(n+1)<w(n+2)$,$w(n)$ is the number of prime factor of $n$.
Does there exist infinite positive integer $n$ such that $f(n)<f(n+1)<f(n+2)$
For (1)$f(n)=d(n)$ which means the number of factors of $n$;
(2)$f(n)=\sigma(n)$ which means the sum of all factors of $n$.
(3)$f(n)=\phi(n)$ which means the Euler fuction.
(4)$f(n)=\pi(n)$ which means the number of primes that less than $n$.
And maybe other number theory functions.[/quote]
4)It is obviosly, that $\pi (n)=\pi(n+1) \ \ or \ \ \pi(n+1)=\pi(n+2)$ for n>2.

[quote="jin"]As we have solved:There are infinite positive integers $n$ such that
(1)$p(n)<p(m+1)<p(n+2)$,$p(n)$ is the largest prime factor of $n$.
(2)$w(n)<w(n+1)<w(n+2)$,$w(n)$ is the number of prime factor of $n$.
[/quote]
How did you solve it? Please write a link or something...
Thanks in advance.

Both already on ML:
http://www.mathlinks.ro/Forum/viewtopic.php?t=2005
http://www.mathlinks.ro/Forum/viewtopic.php?t=82146

ZetaX: thanx for links.
We can define another f:  f(n)|n and (a|n, a>f(n))=>a=n.

It give $f(n)=\frac{n}{p_0(n)}$, were $p_0(n)$ is smallest prime divisor n. This function give very easy jin's problem.

yes,I think so.If $n$ is even,$p_0(n)=p_0(n+2)=2,p_0(n+1)>2$ only finite.
otherwise,$p_0(n)>2,p_0(n+1)=2$ the same.

I knew it wasn't much chalinging. For phi(n) numbers 2^i-2, 2^i-1 and 2^i work,
ill write detailed solution as soon as I have time (hope itl be in the evening CET, so in some five hours).

If f(n) is Eiler function, it is easy to prove that exist infinetely mani i, suth that $f(2^i-2)<f(2^i-1)$, and infinetely many j, suth that $f(2^j-1)<f(2^j)=2^{j-1}$. But it not solved jin's problem.
If $i=12p$, then $f(2^{12p}-1)<f(2^{12p})=2^{12p-1}$. Therefore if $f(2^{12p}-2)=f(2^{12p-1}-1)<f(2^{12p}-1)$ then p satisfyed jin's condition.

What about (1) (2)?Does anyone has some ideas?

[quote="Rust"]... Therefore if $ f(2^{12p} \minus{} 2) \equal{} f(2^{12p \minus{} 1} \minus{} 1) < f(2^{12p} \minus{} 1)$ then p satisfyed jin's condition.[/quote]
Rust you are right, but you have not proved that there exist infinitely many such p  :maybe:


On the Olympiad Resources, AMC 10 2000, how can I find the images?

My guess is that people just haven't gotten around to making them yet..


algebra
polynomial
ratio
geometry unsolved
geometry

Let $H_1$ be an n-sided polygon. Construct the sequence $H_1$, $H_2$, ..., $H_n$ of polygons as follows. Having constructed the polygon $H_k$, $H_{k+1}$ is obtained by reflecting each vertex of $H_k$ through its k-th neighbour in the counterclockwise direction. Prove that if n is a prime, then the polygons $H_1$ and $H_n$ are similar.

[i]Kömal, problem A. 27, November, 2001[/i]

Specially for Pigfly :)

Let $a_0^k$, $a_1^k$, ..., $a_{n-1}^k$ be vertices of $H_k$. Then $a_i^{k+1}=2a_{i+k}^k-a_i^k$, so $a_i^k$ is a linear combination of $a_0^1$, ..., $a_{n-1}^1$, where we assume $a_{i+n}^k=a_i^k$. Let's forget for some time that $a_{i+n}^k=a_i^k$ and associate each $a_i^k$ with a polynomial $P_{ik}(x)=\sum b_jx^j$ iff $a_i^k=\sum b_ja_j^0$.
It is clear that $P_{ik}(x)=x^i P_{0k}(x)$, moreover from $a_i^{k+1}=2a_{i+k}^k-a_i^k$ we conclude $P_{i,k+1}(x)=(2x^k-1)P_{ik}(x)$. Thus \[P_{in}(x)=(2x-1)(2x^2-1)...(2x^{n-1}-1)x^i.\]
Therefore $P_{0n}(x)\in \mathbb Z[x]$. Let $z=\exp(2\pi i/p)$, then $z^{i+p}=z^i$ and $z^1$, ..., $z^{n-1}$ for prime $m$ are lineary independent over $\mathbb{Q}$. It follows that it sufficies to prove $P_{i0}(z)\in \mathbb{Z}$ to complete the proof, since $\sum_{j=1}^{n-1} b_jz^j\in \mathbb Z$ ($b_j\in\mathbb Z$) iff $b_1=b_2=...=b_{n-1}$.
But $P_{0n}(z)=(-2)^{n-1}\prod (1/2-z^i)=(-2)^{n-1}(1+x+x^2+...+x^{p-1})\Big|_{x=1/2}=2^{n}-1$ ($n\geq 3$).

P.S. From the proof we see that ratio will be $2^{n}-1$.

P.P.S. Don't blame me for my extra short explanations.


limit
integration
calculus
real analysis
rate problems
calculus computations

\[ \mathop {\lim }\limits_{n \to \infty } \frac{{\sqrt 1  \plus{} \sqrt 2  \plus{} ..... \plus{} \sqrt {n \minus{} 1} }}
{{n\sqrt n }}
\]

$ \lim_{n\to\infty}\frac {{\sqrt 1 \plus{} \sqrt 2 \plus{} ..... \plus{} \sqrt {n \minus{} 1}}}{{n\sqrt n }} \equal{} \int_0^1 \sqrt {x}dx \equal{} \frac {2}{3}x^{3/2}\bigg|_0^1 \equal{} \frac {2}{3}$

[hide]or using Stolz - Cesaro theorem  :)
for $ x_n \equal{} \sum^{n \minus{} 1}_{i \equal{} 1} \sqrt {i}$ and $ y_n \equal{} n\sqrt {n}$ [/hide]

Hello facis, could you explain how you converted the limit to $ \int_0^1\sqrt{x}\,dx$?

It effectively reads $ \lim_{n\to \infty} \frac1n \sum\limits_{i\equal{}0}^{n\minus{}1} \sqrt{\frac{i}{n}}$.

Mesh size $ \frac1n$, evenly spaced partition of $ [0,1]$ by the points $ 0,  \frac1n, \frac2n$ etc.

Tyl has the right idea.  The limit is a left Riemann sum over the interval $ [0,1]$. It is hard (if not impossible) to take this limit directly as there is no formula for $ \sum_{i\equal{}0}^n \sqrt{i}$.  Recognizing this as an integral, we can then compute it easily by either the fundamental theorem of calculus or with a Riemann sum with varying interval widths.

With the second method, using the points $ 0, 1/n^2, 2/n^2 \ldots (n\minus{}1)^2/n^2$ suffices.  Our sum is then
$ \lim_{n \to \infty} \sum_{i\equal{}0}^{n\minus{}1} \sqrt{\frac{i^2}{n^2}}\frac{2i\minus{}1}{n^2}$
$ \equal{} \lim_{n \to \infty} \sum_{i\equal{}0}^{n\minus{}1} \frac{2i^2\minus{}i}{n^3}$
$ \equal{} \lim_{n \to \infty} \frac{(n\minus{}1)n(2n\minus{}1)}{3n^3}\minus{}\frac{(n\minus{}1)n}{2n^3}$
$ \equal{} \lim_{n \to \infty} \frac{2n^2\minus{}3n\plus{}1}{3n^2}\minus{}\frac{1}{2n}\plus{} \frac{1}{2n^2}$
$ \equal{} \lim_{n \to \infty} \frac{2}{3} \minus{}\frac{3}{2n}\plus{} \frac{5}{6n^2}$
$ \equal{} \frac{2}{3}$

which is the same as before.

btw, what's Stolz-Cesaro theorem?

Stolz-Cesaro is like l'Hopital's for sequences.

If you have two sequences $ (a_n)$ and $ (b_n)$, and $ (b_n)$ is strictly increasing and unbounded and if $ \lim\limits_{n\to\infty} \frac{a_{n\plus{}1} \minus{} a_n}{b_{n\plus{}1} \minus{} b_n} \equal{} L$ finite,

Then $ \lim\limits_{n \to \infty} \frac{a_n}{b_n} \equal{} L$.

[hide]Sort of using HTA's hint, I would choose $ a_n \equal{} \sum\limits_{i\equal{}1}^{n\minus{}1} \sqrt{\frac{i}{n}}$ and $ b_n \equal{} n$.

Then $ \lim\limits_{n\to\infty} \frac{a_{n\plus{}1} \minus{} a_n}{b_{n\plus{}1} \minus{} b_n} \equal{} \lim\limits_{n\to\infty} \frac{1}{1} \equal{} 1$.

So I guess I made a mistake. But you get the idea.[/hide]

Theorem is also valid if $ L\equal{}\infty$.

[quote="Tyl"]Stolz-Cesaro is like l'Hopital's for sequences.

If you have two sequences $ (a_n)$ and $ (b_n)$, and $ (b_n)$ is strictly increasing and unbounded and if $ \lim\limits_{n\to\infty} \frac {a_{n \plus{} 1} \minus{} a_n}{b_{n \plus{} 1} \minus{} b_n} \equal{} L$ finite,

Then $ \lim\limits_{n \to \infty} \frac {a_n}{b_n} \equal{} L$.

[hide]Sort of using HTA's hint, I would choose $ a_n \equal{} \sum\limits_{i \equal{} 1}^{n \minus{} 1} \sqrt {\frac {i}{n}}$ and $ b_n \equal{} n$.

Then $ \lim\limits_{n\to\infty} \frac {a_{n \plus{} 1} \minus{} a_n}{b_{n \plus{} 1} \minus{} b_n} \equal{} \lim\limits_{n\to\infty} \frac {1}{1} \equal{} 1$.

So I guess I made a mistake. But you get the idea.[/hide][/quote]

Your error is that $ a_{n \plus{} 1} \equal{} \sum_{i \equal{} 1}^{n} \sqrt {\frac {i}{n \plus{} 1}}$.  Because of the different denominator in the square root, none of the square roots will cancel.

Choosing $ a_n \equal{} \sum_{i \equal{} 0}^{n \minus{} 1} \sqrt {i}$ and $ b_n \equal{} n\sqrt {n}$, we have

$ \lim_{n \to \infty} \frac {\sqrt {n}}{(n \plus{} 1)\sqrt {n \plus{} 1} \minus{} n \sqrt {n}}$

The top obviously goes to infinity, and the bottom is the average rate of change of $ f(x) \equal{} x^{3/2}$ on $ [n,n \plus{} 1]$, and by mean value theorem there is always a number $ c$ between $ n$ and $ n \plus{} 1$ such that $ f'(c) \equal{} f(n \plus{} 1) \minus{} f(n)$, and $ \lim_{n \to \infty} f'(c) \equal{} \infty$.  Thus we can apply l'Hopital:

$ \lim_{n \to \infty} \frac {1/2 n^{ \minus{} 1/2}}{3/2 \sqrt {n \plus{} 1} \minus{} 3/2 \sqrt {n}}$
$ \equal{} \lim_{n \to \infty} \frac {1}{3 \sqrt {n(n \plus{} 1)} \minus{} 3n}$
$ \equal{} \frac {1}{3} \lim_{n \to \infty} \frac {\sqrt {n(n \plus{} 1)} \plus{} n}{n(n \plus{} 1) \minus{} n^2}$
$ \equal{} \frac {1}{3} \lim_{n \to \infty} \frac {\sqrt {n(n \plus{} 1)} \plus{} n}{n}$
$ \equal{} \frac {1}{3} \lim_{n \to \infty} (\sqrt {\frac {n \plus{} 1}{n}} \plus{} 1)$
$ \equal{} \frac {2}{3}$

as desired.  :lol:


analytic geometry
geometry proposed
geometry

A(-2,3,1) B(2,3,4) C(-8,11,1) find the outside bisector of angle A.

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=303974[/url]


Fie cercurile $C_{1},C_{2}$ tangente interior cercului $C$ in punctele $A$, respectiv $B$, si punctele $D \in C_{1}, E \in C_{2}$ pentru care dreapta $DE$ este tangenta comuna exterioara cercurilor $C_{1},C_{2}$. Sa se arate ca punctul $F \in AD \cap BE$ apartine cercului $C$.

[hide="hint"] punctul cautat este intersectia dintre cercul $\mathcal{C}$ si axa radicala a cercurilor $\mathcal{C}_{1}$ si $\mathcal{C}_{2}$.
[/hide]

[hide="solutie"]fie $P$ punctul de intersectie al lui $\mathcal{C}$ cu axa radicala a cercurilor $\mathcal{C}_{1}$ si $\mathcal{C}_{2}$, fie $C$ intersectia lui $PA$ cu $\mathcal{C}_{1}$ si $D$ intersectia lui $PB$ cu $\mathcal{C}_{2}$. fie $O,O_{1},O_{2}$ centrele cercurilor.
voi arata ca $CD$ e tangenta comuna a cercurilor $\mathcal{C}_{1}$ si $\mathcal{C}_{2}$.

din cauza ca $PC\cdot PA=PD\cdot PB$, rezulta ca $\triangle PCD$ e asemenea cu $\triangle PBA$, si deci $\angle PDC=\angle PAB$.
apoi, cum triunghiurile $OPB$ si $O_{2}DB$ sunt isoscele, rezulta ca $\angle OBP=\angle O_{2}DB=\angle OPB=b$.
ideea ar fi sa arat ca $b+\angle PAB=\frac{\pi}{2}$.
$b=\angle PBA+\angle OBA=\angle PBA+\frac{\pi-\angle AOB}{2}$.
$b+\angle PAB=\angle PBA+\angle PAB+\frac{\pi-\widehat{AB}}{2}=\frac{\widehat{AB}+\pi-\widehat{AB}}{2}=\frac{\pi}{2}$.
deci $\angle O_{2}DB+\angle CDP=\frac{\pi}{2}$, asadar $\angle O_{2}DC=\frac{\pi}{2}$, si deci $CD$ este tangenta la $\mathcal{C}_{2}$.
analog se arata ca $CD$ este tangenta la $\mathcal{C}_{1}$.
[/hide]


function
abstract algebra
superior algebra
superior algebra solved

Can someone define what a Free Group is?
my author does a lousy job, as if he was lazy to define it.

by "define" i mean define what it means a group is free (i know what a free group is).

If $X$ is a subset of a group $F$, then $F$ is a free group with basis $X$ if, for every group $G$ and every function $f: X\to G$, there exists a unique homomorphism $\phi : F\to G$ with $\phi(x)=f(x)$ for all $x\in X$. I think this is easiest to see with a diagram, so I'd try to write one out if I were you (It would be really hard to do on here).

[quote="ZetaX"]A group is free iff it is isomorphic to a free group. ;)[/quote]
but i thought that a free group is a group which is free on some non-empty subset.

And the above is a definition of "being free", so where's the problem.

A free group has either the definition of HilbertThm90, or as "reduced words" over some set, oth are eqivalent.

My author mentions the following:
".... since $G$ is generated by $A$ and it is free means that $G=F[A]$...."
But this is the problem, it IS NOT equal to it. It is isomorphic. Why does the author write "equal to". And he does it even a second time suggesting this is not an error.

Okay, G need not to be equal zu F_A, but it is canonically isomorphic, so F_A and G can be identified.

1)Why is it that no two different products of reduced words is the same?

2)I think I am starting to see why it is called "free" because it is similar to the case of free abelian groups. The generating set that generates G gives unique representations of the finite products in the generating set, because otherwise if that was not the case (and assuming G is iso- to F[A]) then it will means that F[A] violates #1 which cant be the case. Have I got the right idea?

1) What do you mean by different products¿ Two different reduced words are not the same by definition.

2) I think yes (had problems to understand everything).

[quote="ZetaX"]1) What do you mean by different products¿ Two different reduced words are not the same by definition.[/quote]
If $x,y$ are two reduced words. And $a,b$ are two DIFFERENT reduced words. Can it be that $xy=ab$

[quote="ZetaX"]$(a)(b)=ab=()(ab)$ for example ($()$ the empty word).[/quote]

After I posted it I realized that,
$xx^{-1}=1 \cdot 1$
Works also.

But, then why does the book say that? Here:
[quote]
.... no two different formal products in $F[A]$ are equal...[/quote]
What does that mean? Am I misunderstanding the author? Maybe its based on meaning of "formal product". You know what he is talking about?

Probably this means just the reduced words. But I don't know the exact context...


geometry
geometric transformation
rotation
symmetry
3D geometry

I started thinking about binary representations one day and came up with the following problem that I still don't know the answer to.  Obviously the number of distinct binary strings of length n is 2^n.  How many distinct binary "rings" are there? (a ring is just a string that wraps around).  For example, for n =3, we have four distinct ones: 000, 111, 001, 110.

This is a job for Burnside's Lemma (or Polya-Redfield, whatever u want to call it)

We have n possible rotations (identity, shift by one, shift by two, etc).

Every element is fixed under a shift by 0  (identity)

for all other 0<i<n, there are 2i elements fixed under rotation iff i | n.

otherwise, there are 2 elements fixed under a rotation by i. (00.000 and 11...111)

Therefore, the number of binary "rings" is

[tex]\displaystyle\frac{2n + \sum_{i|n} (2^i - 2)}{n}[/tex]

It is a theorem in algebra/group theory which is often used to count the number of different arrangements of some sort when arrangements which are related by a member of a group of symmetry transformations (in your case, rotations of the string) are counted as identical. Cube colouring problems are a common application.

See [url]http://en.wikipedia.org/wiki/Burnside%27s_lemma[/url] or [url]http://www.maths.uwa.edu.au/~schultz/3P5.2000/3P5.12Actions.html[/url]

This thread reminds me of a 1960's puzzle called "Kwazy Quilt" which applied this concept.  The game contained a cardboard frame to hold 13 colorful plastic pieces.  [The plastic pieces were invertible.]

One copy of every possible distinct piece (6-bit rings) was present, with one exception...  The circular piece was replaced by a second piece with one point.

Sometimes, I use this puzzle when I hold a "game day" with my middle school students.  Most students struggle to find a solution.  After analyzing the puzzle, we find the solutions are plentiful and easy.


calculus
derivative
function
limit
real analysis
real analysis unsolved

Let f: (a, b) -> R be a function which is convex (hence continuous, by earlier work I did), but is not assumed to be differentiable.

Prove that the set pf points in (a, b) at which f is not differentiable is at most countable.

Look at the left and right derivatives of $f$: $f^{-}(x)=\lim_{h\to 0^{+}}\frac{f(x)-f(x-h)}{h}$ and $f^{+}(x)=\lim_{h\to 0^{+}}\frac{f(x+h)-f(x)}{h}$. Both exist everywhere and are increasing functions, and $f^{-}(x)\le f^{+}(x)\le f^{-}(y)$ if $x<y$. $f$ is differentiable at $x$ if and only if $f^{+}(x)=f^{-}(x)$.
Wher this fails, we must have a jump discontinuity in $f^{-}$ and $f^{+}$. There can only be countably many such points.
[Edit- Fixed that]

That should, of course, say '...there can only be countably many such points.'


Suppose that for positive integers $a,b,c$ and $x,y,z,$ the equations $a^2+b^2 = c^2$ and $x^2+y^2 = z^2$ are satisfied. Prove that

\[(a+x)^2 + (b+y)^2 \leq (c+z)^2,\]

and determine when the equality holds.

[hide="My solution"]

First, we can expand the given inequality:

$a^2+2ax+x^2+b^2+2by+y^2 \leq c^2+2cz+z^2$

By our two equations given on problem, we can simplify this into:

$(a^2+b^2) + (x^2+y^2) + 2(ax+by) \leq c^2+z^2+2cz\\
c^2+z^2+2(ax+by) \leq c^2+z^2+2cz\\
2(ax+by) \leq 2cz\\
ax+by \leq cz$

Since $a,b,c,x,y,z$ are all positive integers, squaring the inequality will not change the sign.

$(ax+by)(ax+by) \leq c^2z^2$

We can rewrite the right and expand the both sides:

$a^2x^2+2abxy+b^2y^2 \leq (a^2+b^2)(x^2+y^2)\\
a^2x^2+b^2y^2 +2abxy \leq a^2x^2+b^2y^2+a^2y^2+b^2x^2\\
2abxy \leq a^2y^2+b^2x^2\\
abxy \leq \frac{a^2y^2+b^2x^2}{2z}$

This inequality is true by Arithmetic Mean-Geometric Mean Inequality. So, the given inequality holds true as well.

For equality, we can look at the inequality before we squared.

$ax+by \leq cz$

We want:

$ax+by = cz$

This will hold true when $a=x, b=y$ and $c=z$ because either way, we get $a^2+b^2 = c^2$ or $x^2+y^2 = z^2$ which is true by the problem. [/hide]

Is this correct?  :)

I have a different one, after getting $ax+by \leq cz$, then:
$ax+by \leq \sqrt{(a^2+b^2)(x^2+y^2)} = cz$. (by cauchy).

Edit: Silverfalcon actually proved the two variable cauchy.

[quote="Magnara"][hide]It would seem you missed the equality case.  Equality in AM-GM holds when all the variables are equal, which in this case, means that (ay)^2=(bx)^2, that is, ay=bx.[/hide][/quote]

Oh..

If I didn't recognize this on say, USAMTS or USAMO, do you think I'll get points deducted for that reason? I think it's minor but just wondering. :)


inequalities
function
inequalities proposed

Use Convexity Of function as a x or y or Z variable and then discuss about Maximum of function in [0,1] at boundary points.

What about using the method of Lagrange Coefficients, that could help too!

another approach is the following one:
$ f(x)=x^{2}(3b^{2}+3c^{2}-2bc)+x(-2b^{2}c-2bc^{2})+(3b^{2}c^{2})$ where $ b,c \in [0,1]$
now we want to show that for $ x \in [0,1]$ we have $ f(x)\leq 3$
now note that first of all the coefficient of $ x^{2}$ is positive,so the maximum value of $ f$ in the interval $ [0,1]$ happens at its boundaries,i.e $ f_{\max}(x)=\max \{f(0),f(1)\}$ so its sufficient to check the cases where $ a=0,a=1$...


Find all the periodic points of $f(x)=|x-1|$.

This is (fairly)easy.

Since f(x) is always real, all periodic points will be real.

We need to find all x such that \[ \pm(\pm(\pm(...\pm(x-1)-1)-1)...-1)=x  \]

Let m be the number of these plus signs and n be the number of these negative signs.
So, this can be rewritten as \[ (-1)^nx+(n-m)=x  \]
The solution to this is \[ x=\frac{n-m}{1+(-1)^{n+1}}  \] for all non-negative integers m and n (the period of that solution is m+n).

EDIT: (However, they cannot both be 0, otherwise the period would be 0.)
Note that sometimes, division by zero will result.  Also, sometimes solutions will be negative.  So, the actual solution is \[ x=\frac{n-m}{2}  \] for all non-negative integers m and all odd integers n greater than m.

EDIT2: I take back everything.  I didn't remember that the signs of some of the 1's might be compounded.

The correct solution should be: $x \in [0..1]$

Prove:
a) they clearly work (easy to check), more exactly $0.5$ is a fixpoint, all other have order $2$.
b) only those work:
let $x>1$, then $f(x)=x-1$. If $f(x) \leq 1$, then we reach a loop, otherwise the same happens again and again and so on and finally we get $f^{[x]}(x)=x-[x]$ and case a) occurs. Now when $x<0$, then $f(x) >1$ and the previous happens.

It is very easy to extend that proof and generalize all periodic points of \[ f(x)=|x-a|  \] is all x such that \[ 0 \le x \le a  \] iff a is a nonegative real number.  Otherwise f has no periodic points.

Here's a harder one:
What are all periodic points of \[ f(x)=|ax-b|  \]


A boy buys oranges at $ 3$ for $ 10$ cents. He will sell them at $ 5$ for $ 20$ cents. In order to make a profit of $ \$ 1.00$, he must sell:
$ \textbf{(A)}\ 67 \text{ oranges} \qquad\textbf{(B)}\ 150 \text{ oranges} \qquad\textbf{(C)}\ 200 \text{ oranges} \\
\textbf{(D)}\ \text{an infinite number of oranges} \qquad\textbf{(E)}\ \text{none of these}$

Since the unit price at which he bought the oranges is $ \frac{10}{3}$ cents (because unit price is total price divided by quantity), and the unit price at which he sold the oranges is $ \frac{20}{5}\equal{}4$ cents. The net profit he makes per orange is $ \frac{2}{3}$ of a cent. Thus, the number of oranges he must sell to make a profit of $ \$1.00$, or $ 100$ cents is $ \frac{100}{\frac{2}{3}}$, or $ 150$ oranges, which is choice $ \fbox{B}$.

If he buys 15K oranges, he will make 60K-50K=10K cents. Thus, K=10, and the answer is 150.


Αν για τους μη αρνητικούς πραγματικούς ισχύει  $ \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \equal{} 2\sqrt {2}$
να βρεθεί το ελάχιστο τηs    $ \sqrt[6]{a} \plus{}\sqrt[6]{b} \plus{}\sqrt[6] {c}$

[quote="xp"]Αν για τους μη αρνητικούς πραγματικούς ισχύει  $ \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \equal{} 2\sqrt {2}$
να βρεθεί το ελάχιστο τηs    $ \sqrt [6]{a} \plus{} \sqrt [6]{b} \plus{} \sqrt [6] {c}$[/quote]

Ισχύει ότι

$ \left(a \plus{} b \plus{} c\right)^{3} \equal{} a^{3} \plus{} b^{3} \plus{} c^{3} \plus{} 3\left(a \plus{} b\right)\left(b \plus{} c\right)\left(c \plus{} a\right)\geq a^{3} \plus{} b^{3} \plus{} c^{3}\Longleftrightarrow$
$ a \plus{} b \plus{} c\geq \sqrt [3]{a^{3} \plus{} b^{3} \plus{} c^{3}}$

Άρα

$ a^{\frac {1}{6}} \plus{} b^{\frac {1}{6}} \plus{} c^{\frac {1}{6}}\geq \sqrt [3]{a^{\frac {1}{2}} \plus{} b^{\frac {1}{2}} \plus{} c^{\frac {1}{2}}} \equal{} \sqrt {2}$.
Ισότητα όταν $ (0,0,8)$ και κυκλικά...

[quote="Durandal"]Ummm...[hide]$ \sqrt {2}$ στα άκρα ή έχω χάσει κάτι?[/hide]

Cheerio,

Durandal 1707

Edit: hehe, ninja'ed για 3 λεπτά :ninja:[/quote]

Huh :?:  :|

[quote="Durandal"]Edit: hehe, ninja'ed για 3 λεπτά :ninja:[/quote]

"Ninja'ed" είναι internet slang (σαν να λέμε troll, spam κλπ) για όταν απαντάς σχεδόν ταυτόχρονα με κάποιον άλλον... ενώ δηλαδή πας να κάνεις post την απάντησή σου, η προηγούμενη ξεφύτρωσε από το πουθενά χωρίς να τη δείς... και καλά ninja :P

Εξ'ού και το icon :ninja:

Useless information of the day :rotfl:

Cheers,

Pan


I don't get others'; others don't get mine. *sigh* Sometimes I feel like sarcasm should not be used. It makes people feel uncomfortable.

Once my English teacher gave us this article to "analyze". I did my best, and I thought I did well, and then we got it back and I got a low score because the article was actually in a sarcastic tone.

EDIT: I just read over my post and realized it sounded sarcastic. It is not, if you were wondering.

Sarcasm is best picked up/expressed verbally and not written down. For example, take the following sentense and tell if I'm sarcastic or not.

"Wow, you're smart."

If you're concerned with sarcasm on the Net, the usage of "..." is becoming more and more common to hint a sarcastic manner. Or you can be even more obvious and just put "[/sarcasm]" at the end of your sentence.

I <3 sarcasm, but my parents do not  :no:

Use sarcasm tags

<sarcasm> </sarcasm>

I do actually do this occasionally.

In real life, I don't like sarcasm unless it is obvious.  Otherwise it can appear obnoxious.

[quote="archimedes1"]If you're not sure if someone is being sarcastic, then just ask them, "Are you trying to be sarcastic?" or "Is that supposed to be sarcasm?"
See what happens.

[quote="leoxnlin"]EDIT: I just read over my post and realized it sounded sarcastic. It is not, if you were wondering.[/quote]
Sure.[/quote]
I still remember last year whenever someone asked me to stop being sarcastic, I would say, "[sarcasm]Okay, I'm not sarcastic now.[/sarcasm]" and then continue.

Also, I agree that sarcasm is very hard to express on paper. One of the main hints in sarcasm is the tone of voice.

Haha...yeah...sarcasm...

I swear, like at least 25% of my sentences are sarcastic in some way. Obviously, you can't interpret sarcasm on paper...so that article you read shouldn't have been necessarily meant to be interpreted that way...

Sarcasm can be hard to pick up at times. I, personally, when speaking, hardly ever use a sarcastic tone. I feel it's redundant. So, unless you're quick at picking those kinds of things up or know me (and even so, it's "hard"), you couldn't pick up when I'm being sarcastic. I'm pretty quick when picking up sarcasm when conversing in person--and only sometimes in print/type.

So, are you asking for help with sarcasm or what?

Sarcasm should be used. It really should.

[quote="7h3.D3m0n.117"]
If you're concerned with sarcasm on the Net, the usage of "..." is becoming more and more common to hint a sarcastic manner.[/quote]

Hmm...
Sometimes I use it to indicate a lapse of time thinking or a pause of some sort. (Like an awkward pause.)
I don't think I've ever used it sarcastically. At least, not intentionally.

yeah.

i'm always like "WWWOOOOWWWWWWWWW.... that was really smart. To break the (insert object here)"

it's obvious when you see the change in tone of my voice.
but voices aren't possible.

sarcasm is fun. its just that once i nearly got a black eye from it.

if someone doesn't know and still looks at you in that funny cocking-the-head way, then tell them "that was sarcastic btw".

do something like that.
they'll understand.

my class is full of sarcasm. even our teachers are all (except for one) sarcastic.

I'm highly (and yes, [b]highly[/b]) sarcastic in real life, but rarely on the internet. It's pointless to try...

Haha, my fifth grade math teacher always got mad at us for something, and it didn't seem so bad because he used a [b]lot[/b] of sarcasm. Even though his volume was high it was pretty OK. Our reading teacher though...gave us speeches when he got mad, and when he was mad at one particular person he yelled super loud, which made the Cottage shake. Which is scary.

I prefer sarcasm when you are angry...That way it lessens the blow and makes it funny at the same time.

<3 sarcasm <3

I tend to use a lot of sarcasm in my speech, especially in my humor, but it generally runs in the undercurrents. In parties, public gatherings, polite company, etc, the type of sarcasm I use is the best indicator of my mood.

I think the use of good sarcasm is itself an art. And the British have mastered it -- a lot of British people I know can hardly speak a sentence without some sarcasm in it xD

Yes, I find that if people use sarcasm in the wrong way, it makes them look pretty foolish, and rude. I use sarcasm a lot, but I never use it when I'm on the internet.

I was just discussing sarcasm today  :P

Me and my brother use sarcasm alot.  But we constantly have to tell each other "I was being sarcastic... smart one." <-- sarcasm.  :rotfl:

One of my teachers has mastered the art of sarcasm.  He even has that poster:  "National Sarcasm Society--like we need your support."

[hide="My method of detecting verbal sarcasm"]
Here's a good guide for verbal sarcasm.  You can judge roughly by the pitch and volume of the voice.  Let's assume you wanted to say "Oh really?" to somebody.  To sound serious, you need to start with a low pitch and then end up with a raised inflection so you sound interoggative. i.e. "Oh really???" To sound sarcastic, you need to start with a high pitch and then end up with a low, certain tone so you sound declarative when you're really asking a question.  i.e. "Oh REALLY." To sound indifferent (slightly different from sarcastic), have an overall dull tone so it sounds like you really don't care. Example: "Oh really."  Master sarcasmists can sound interrogative and sarcastic at the same time.  :D [/hide]

[hide="My method of detecting written/internet sarcasm..."]
Written:  People will often use written sarcasm by putting a lot of emphasis on the not-so-great aspects of the subject, i.e. "Toothbrushes are amazing.  It has a soft rubber grip for easy use, a necessity that absolutely everybody must have."

Internet: Usually people will stress internet sarcasm with a lot of capital letters followed by periods, italicization or bolded emphasis.  Most commonly I see stuff like "WOW... That was [b]amazing[/b].  It totally blew me away."  I don't ever see people use exclamation points because that is very misleading.  Often they'll use really emotional and extravagant phrases that way overamplifies the situation.  The most obvious form of sarcasm comes in the form <\sarcasm>.  :D
[/hide]

sarcasm in text is just really about reading the context clues. like your original post doesn't seem sarcastic at all, and your disclaimer about it not being sarcastic kinda puzzles me. maybe the way you were taught english was subtly different from that of the rest of us?

like unless like a majority of people in your classroom got a very low grade because they misinterpreted it, it's very likely that the sarcasm in the article was pretty obvious. i would hazard a guess and say that it's simply because you're not being cued in by the diametrically opposing stances and tones of voice the author is switching between

sarcasm is pretty commonplace in the real world. it's probably not used nearly as much in a professional setting, but casually, it's used a lot. so i mean if you have trouble deciphering others' sarcasm, and vice versa, that's probably a problem. like i implied before, it's something that i guess is developed along with other social skills?

[quote="MysticTerminator"]hahahhahaha

epic fail, dude[/quote]

you can't even type hahahahaha right

epic fail :D

that's actually what I always do

it indicates that I WAS LAUGHING SO HARD at your EPIC FAILURE that alternating perfectly between the h and a keystrokes was no longer relevant

[quote="Meta_Knight"]I use sarcasm to annoy people :P

@Lazarus: The sentence in your signature is [b]supposed[/b] to be "There are [b]10[/b] types of people in the world. Those who understand binary and those that don't.[/quote]

LOLWUT
That's exactly what I have. Copied much?

Has anyone ever realized that sarcasm doesnt work on chinese parents. I use sarcasm on my parents alot and they dont pick up on it which drives me crazy.

O and btw i thought it was there are 12 kinds of people in this world, those who understand binary and those who dont.

Then again, with the whole binary thing, the 2 could work, because then you don't understand binary. You used a 2 in Binary (which isn't allowed)


Prove that:
 \sum iaib/ab(a+b) \geq 27/8(a+b+c) in a triangle, where ia is the lenght of the bisector of angle A.  :D  :D  :D

ciprian usually the length of the interior angle bisector of angle A in a triangle ABC is denote by la. :)

I think I have a solution:
la*lb/ab(a+b)=2bc/(b+c) * 2ca/(c+a) *  \sqrt (p(p-a)/bc) *  \sqrt (p(p-b)/ca)=2pc \sqrt ((p-a)(p-b))/(a+b)(b+c)(c+a)
The same with the others and then we add the relations and after some computations we finally get:
8 \sum c \sqrt ((p-a)(p-b)) \geq 27(a+b)(b+c)(c+a)
but  \sqrt ((p-a)(p-b)) \geq 2(p-a)(p-b)/c
And after some more computations we get:
8(a+b+c)^3 \geq 27(a+b)(b+c)(c+a) true

Hope the solution is corect! :)
Cheers :D

Are you sure  that
[tex] \frac{l_al_b}{ab(a+b)} = \frac{2pc\sqrt{(p-a)(p-b)}}{(a+b)(b+c)(c+a)} [/tex]  ??


We call a number a mountain number if its middle digit is larger than any other digit.  For example, 284 is a mountain number.  How many 3-digit mountain numbers are there?

If the middle number is $ 9$, there are $ 8$ possibilities for the first number and $ 8$ for the second.  This makes $ 64$ possible numbers.

If the middle number is $ 8$, there are $ 7$ possibilities for the first number and $ 7$ for the second.  This makes $ 49$ possible numbers.

If the middle number is $ 7$, there are $ 6$ possibilities for the first number and $ 6$ for the second.  THis makes $ 36$ possible numbers.

See a pattern?

The answer should be the sum of the first $ 8$ squares, but that would be counting $ 010$, $ 020$, $ 021$, $ 030$, $ 031$, $ 032$, etc.

Thus, we must subtract the $ 9$-th triangle number, so the answer is $ 204\minus{}45$ or $ \boxed{159}$.

I don't think that's correct.

Note that if we randomly choose three distinct digits, there will be exactly two mountain numbers (with a few exceptions).

If all digits are distinct, then there are $ 2\dbinom{10}3\equal{}240$ mountain numbers. However, we have counted numbers of the form $ \overline{0ab}$, where $ a>b\geq1$, which gives us $ \dbinom92\equal{}36$. The number for this case is $ 240\minus{}36\equal{}204$.

If the numbers are of the form $ \overline{aba}$, where $ b>a\geq1$, then are are $ \dbinom92\equal{}36$ mountain numbers.

Finally, our answer is $ 204\plus{}36\equal{}\boxed{240}$.

It is not correct...for example, for $ b\equal{}2$, where is $ 120$?

Plus, my calculator program says $ 240$. :wink:

Oh thanks  :blush: !

So that means the answer should be 1*2+2*3+3*4+...+8*9=[b][u]240[/u][/b], which is what math154 got

If the middle digit is 9, there are 8 (1-8) choices for the hundreds digit and 9 (0-8) for the units digit.

If the middle digit is 8, there are 7 choices for the hundreds digit and 8 for the units digit.

etc.

8x9 +
7x8 +
6x7 +
...
1x2  (the middle digit cannot be a 1)
=240

The case study is a good approach but far from intuitive. Here is a much better solution in my opinion.
Assuming the middle digit is 2,3,4,5,6, ... We have in each case, 2^2-2, 3^2 -3, ..., 9^2-9 mountain
numbers. This is easy to see since we have two open positions for n digits when the middle one is n,
zero included. So, we have n^2 possiblities. But zero can not be the leading number. So, we have n^2-n
in each case. - Please post this as an alternative solution.

For every 3 digits, there are 2 ways to mak.e a mountain number. So it is C(10,3)*2=240.

We see that with 1 as the middle digit, we have $0\cdot 1$ possibilities for the first and last digit.  For 2 in the middle, we have $1\cdot 2$, for 3 we have $2\cdot 3$.  We see a pattern, and find the answer to be

\[0+2+6+12+20+30+42+56+72=\boxed{240}\]

[quote=policecap]Let the number be abc.

When b=2, there is 1 choice : 121
When b=3, there are 4 choices: 131;132;231;232
When b=4, there are 9.
...
So our final answer is 1+2^2+3^2+...+8^2=[b][u]204[/u][/b][/quote]

i did the same way as you did. But i made a table which make it easier to look at.

I don't get where 9C2 comes from. Can someone explain?

semi-nice problem actually
[hide=sol]for every middle number x, there are x(x-1) possible numbers that can be formed by the given condition. thus, the answer is 1*2+2*3+3*4+...+8*9=2(2c2+3c2+4c2+...+9c2)=2(10c3) (by hockey stick identity) which is then 2*120=240.[/hide]

why doesn't this solution work?
[hide=wrong solution?]
notice that for each middle digit, there are $(n-1)^2$ corresponding mountain numbers. so the solution is $1^2+2^2...+6^2+7^2+8^2=204$
[/hide]

[quote=pieMax2713]units digit can also be 0[/quote]

isn't that already included

[hide=Alternative Solution] For all groups of $3$ distinct integers, there will be $2$ way to arrange it such that the largest number will appear in the middle (the 2nd largest could be on the right or left). Due to this, you can solve it as $2 \cdot \binom{10}{3}$, which would be calculated as $240$.

[b]Questions to be answered:[/b]
Why does $0xy$ never appear?


i am very interested in calculus, but i have not been able to find any good book, i wish AoPS had one from their authors, but anyway i was wondering if someone could tell me some GOOD bibliography on calculus, thanks a lot!!!

You should have posted this in the calculus forum, shouldn't you? :)  Anyway, I would recommend to you the [i]Calculus[/i] books of Tom Apostol and Michael Spivak. Look for them! :P


calculus
integration
function
trigonometry
calculus computations

Do you have any reason for thinking that there's anything interesting about this? You realize that if you just slap random functions together, the most likely result is a function whose antiderivative is not an elementary function.

I was told on the lecture that sine integral is not an elementary function, but it wasn't proved...

could you tell how do you know which functions (defined as integrals) aren't elementary function ?


search
superior algebra
superior algebra solved

Mornin' everybody.
(Sorry for my English)
Is anyone able to help me finding the proof that every set can be a group? I thnk I 've made a proof and want to compare the method. (I've heard that it's equivalence with the AC).
Thanks


It's easy. Write $1+...+k$ as $\frac{k(k+1)}{2}$ and next use the equation $\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}$

can u plz solve it for me and tell me the final answer.
thnx in advance

He definitely meant $\frac{4012}{2007}$  :) The sum will approach 2.


Hi, I'm brand new to the forum, and it's been MANY years since I messed with any of this so I'm not even sure which forum is the most appropriate for my question.  I don't remember dealing with this situation in any of my college or grad school math courses, but have now run into this situation at work.

A relationship that can be described as follows:

\sum\frac{1}{i} (a_i * b_i) = c

If I know, say, the quanties c and \sum\frac{1}{i}(b_i), is there a way to solve for a?

Thanks,

Mark

You'll need to be more clear.  Do you mean that you know

$ \sum_{i \equal{} 1}^{n} a_i b_i \equal{} c$

and $ c$ and $ \sum_{i \equal{} 1}^{n} b_i$?  Then you've got one equation but a heck of a lot more than one unknown, so no, there's no unique solution.  Even if you know all of the $ b_i$, you've still got $ n$ unknowns.

But do you mean that you know

$ \sum_{i \equal{} 1}^{n} a_i b_i \equal{} c_n$

for some sequence $ c_n$?  This is much easier; just take differences.  You can also solve if the relationship is

$ \sum_{i \equal{} 0}^{n} a_i b_{n \minus{} i} \equal{} c_n$

because this type of summation has a really nice structure.  Is this what you mean?  You're probably better off posting the original question.  I've found that oftentimes when you're confused enough, trying to "abstract" the problem to explain your difficulty to someone else makes it harder, not easier, to get your question answered.

Thanks for your reply, t0rajir0u.

In reviewing my original post, I see that I actually misstated my question.  So, taking your advise, let me provide an example:

a     b    a*b
---------------
2    10    20
3    10    30
2    15    30
5    12    60
4    15    60
---------------
16   62   200 = c

So, if I know that c=200 and the sum of b = 62, is there a way to determine the [u]sum of a[/u] (16 in this case)?  I don't think there is, but thought I would ask.

Not even close.  That's two equations and five unknowns.


[quote="Samsa"]Find all positive integers $ n$ such that $ 6n^{2}+5n-4$ is prime.[/quote]

$ 6n^{2}+5n-4 = (3n+4)(2n-1)$
$ n = 1$

[quote="nayel"]Don't you know the basics of factoring? It's just a quadratic. So one way is to solve the equation. Another approach is noticing that $ 6\times (-4)=-24$. So express $ -24$ as the product of two numbers whose sum is $ 5$, which are in our case $ 8$ and $ -3$.

Now the given expression can be written as
$ 6n^{2}+8n-3n-4$
$ =2n(3n+4)-1(3n+4)$
$ =(3n+4)(2n-1)$.[/quote]

Continuing this reasoning, since we want the number to be prime, we either need $ 3n+4=1$ or $ 2n-1=1$. In the first case, $ n=-1$, and in the second case, $ n=1$. Since you wanted positive integers, we exclude $ n=-1$, leaving $ \boxed{1}$ as our answer.


Using Gauss's formula, the sum is
$ \frac{1001(1002)}{2}=501501$

what xpmath said. or you could rewrite it so its
$ 1+2+3+......1000+1001$
$ 1001+1000......+3+2+1$
add the first terms of both sums you get 1002 the second sums also get you 1002. you end up with 1001 1002's because theres 1001 terms in both sums. since you wrote the sum twice, you need to divide it by 2 geting 501501.
(its no much good if you just know the formula. you will only be able to work on these types of equations. some problems relate so it will help if you know how you get the formula)

the answer is 501501. i got this by doing 1001*1002/2

like everyone else stated, Gausses formula to find the sum of numbers $ 1$ to $ n$ where it increases by one is
$ \frac{(n+1)n}{2}$. Plug in the numbers and it should equal $ 501501$.
The reason is because we rewrite the addition but reversely.

$ 1+2+3.....+n=x$
$ n+(n-1)....+1=x$

add the two equal equations. It's $ (n+1)n$ ($ n$ numbers of the $ n+1$). But since we added two same equations, divide it by 2 giving you:

$ \frac{(n+1)n}{2}$


[quote="delegat"]If we know that:

 $ USA \equal{} 0$ , $ PAKISTAN \equal{} 3$ $ BOSNIA \equal{} 1$ $ DENMARK \equal{} 2$ and $ RUSSIA \equal{} 0$ and if $ HOLLAND \equal{} ?$    then find value of $ ?$ Please give an explanation!

NOTE: Countries geographical position, it's development level.. are not important.
Important things are letters that appear in names of countries.

Good luck![/quote]

Does the fact that a comma is missing from between 3 and Bosnia and between 1 and Denmark relate to the answer, or was that just a typo?

[quote="delegat"]If we know that:

 $ USA \equal{} 0$ , $ PAKISTAN \equal{} 3$ $ BOSNIA \equal{} 1$ $ DENMARK \equal{} 2$ and $ RUSSIA \equal{} 0$ and if $ HOLLAND \equal{} ?$    then find value of $ ?$ Please give an explanation!

NOTE: Countries geographical position, it's development level.. are not important.
Important things are letters that appear in names of countries.

Good luck![/quote]
[hide="solution"]
The at least one of the factor(s) that $ USA$ has that $ PAKISTAN$ doesn't have has to equal $ 0$. The only one that $ USA$  has that $ PAKISTAN$ doesn't have is $ U$. so $ U$ is $ 0$.
$ \frac {PAKISTAN}{BOSNIA} \equal{} \frac {3}{1}$, so $ \frac {PAKT}{BO} \equal{} 3$, so $ PAKT \equal{} 3BO$
$ \frac {DENMARK}{BOSNIA} \equal{} \frac {2}{1}$, so $ \frac {DEMRK}{BOSI} \equal{} 2$
This is not working out well...

I'll assume that the letters have to be nonnegative integers in value, otherwise there are too many variables to solve for.
The only nonnegative integers that multiply to $ 1$ are sets containing only $ 1$, so $ B$, $ O$, $ S$, $ N$, $ I$, and $ A$ are all $ 1$.
$ PAKISTAN \equal{} 3$ can now be simplified to $ PKT \equal{} 3$
$ DENMARK \equal{} 2$ can now be simplified to $ DEMRK \equal{} 2$
since $ K$ is common in both $ PKT$ and $ DEMRK$, but $ 3$ and $ 2$ are relatively prime, $ K \equal{} 1$
$ PT \equal{} 3$
$ DEMR \equal{} 2$
we'll assume $ D$ is $ 1$, $ H$ is $ 1$, and $ L$ is $ 1$ because i felt like it, just like my assumption up there.
Then $ HOLLAND$ multiplies together to $ 1$.
$ \boxed{? \equal{} 1}$

I could also try this stuff with addition instead of multiplication, and I'll also find $ H$ and $ L$ hard to figure out.
[/hide]

[hide]
All i could figure out is.
$ PAKISTAN\equal{}KAP,IS,TAN$
$ BOSNIA\equal{}A,BISON$
$ DENMARK\equal{}MEN,DARK$
$ RUSSIA\equal{}IURASS$
Thus,
$ HOLLAND\equal{}LOL,HAND$

So it must be $ \boxed 2$.
[\hide][/hide]

Nice ideas about question but unfortunately I made big mistake becouse I wrote names of countries with capital letters but all names should be written by lower case letters.

For ex. It should be $ usa$, $ denmark$ etc. Then one simple solution can be found.

[quote="delegat"]Since there is no solution yet i am giving some additional informations:

$ austria \equal{} 1$,  $ canada \equal{} 1$, $ romania \equal{} 0$, $ spain \equal{} 1$, $ finland \equal{} 4$ $ albania \equal{} 2$...[/quote]
Are you sure that [i][b]finland = 4 [/b][/i]?  :o

According to this that I wrote, I mean $ finland$ of course it is 4 but according to your one, I mean, [b]finland[/b] it is not 4.

Oh I think that you understand trick [b]AiW[/b]

[hide="solution"]
The values are the amount of times the lowercase letters have things sticking up or down, not including the dot for the $ i$.
for $ holland$, it's $ 4$.
$ ?\equal{}4$
[/hide]

Good work [b]ProtestanT[/b]

Your solution is absolutely correct. :)


Prove that if $ 1\le m\le n$: \[ n\left({n\choose 0}\minus{}{n\choose 1}\plus{}{n\choose 2}\minus{}\dots \plus{}\left(\minus{}1\right)^{m\minus{}1}{n\choose m\minus{}1}\right)\equiv 0 \mod m\]

$ n\left({n\choose 0} \minus{} {n\choose 1} \plus{} {n\choose 2} \minus{} \dots \plus{} \left( \minus{} 1\right)^{m \minus{} 1}{n\choose m \minus{} 1}\right)\equal{}n(\minus{}1)^{m\minus{}1}{n\minus{}1\choose m\minus{}1}$

beacuse $ LHS$ is equal to n times the coefficient of $ x^{m\minus{}1}$ in $ \frac {(1\minus{}x)^n}{1\minus{}x}\equal{}(1\minus{}{n\choose 1}x\plus{}{n\choose 2}x^2\minus{}...)(1\plus{}x\plus{}x^2\plus{}...)$

So
$ LHS\equiv n(\minus{}1)^{m\minus{}1}{n\minus{}1\choose m\minus{}1}\equiv m{n\choose m} \equiv 0 (mod m)$


AMC
AMC 10
AMC 12
AMC 12 B
AwesomeMath
summer program
factorial

[b]Mock AMC 10A by #H34N1
Grand Winners[/b]

[hide="Second Place"]Second place had the following contenders, all tied with 138 points:

 :winner_second: budi713
 :winner_second: stupidityismygam
 :winner_second: Brut3Forc3
 :winner_second: xela1231030
 :winner_second: perfect628
 :winner_second: redcomet46[/hide]

[hide="First Place"]First place, with a narrow lead of 1.5 points, we have:
 :winner_first: Yongyi781
[/hide]

Congratulations to the winners and thank you for participating.  :clap2:

Congrtulations to you all who did much better than me. It was very close though I noticed.

didn't everyone who came 2nd get a 138 not a 132. and first place didn't get a 133.5, he/she got a 139.5.

i completely failed that mock amc10
what grade r u pplz in?
i'm in 7th and i wanna see if i failed mainly bcause of my grade or bcause i really need to learn more
also, do u guys go 2 any math camps or anything?

I am going to 9th grade and i when to mathworks before 7th and 8th (second time was useless) and i will be going to awesomemath this year.

this year i am going into 9th grade also.... my sister is going into 10th grade and when she was in 9th grade instead of going to awesomemath, she had the sets sent to her and now i am working on them for next year. It's really good.

ok
so i'm guessing the camps really helped?
(i really should go to one...)

Woot, second place!
(Actually, I missed that problem about the circle and lines intersection by picking the wrong letter on accident...  :mad: )

I missed the factorial square question. I must have been going to fast.

7th, going in 8th, made problems 18 and 24. I'll do this next time.

I got that narrow lead because I left #25 blank because I had NO IDEA what it meant... :D

[quote="Urc"][quote="budi713"]i did better on this than the amc 10a

mock amc10> amc12b > amc 10a[/quote]

That doesn't make much sense, considering the 12B would be harder than the 10A..but then again, people are different I guess.[/quote]

lolz dont worry budi is a beast that way xD

for me

AMC 12B (114) >AMC 12A(117) >AMC 10A(121.5) >10B (126)

in terms of difficulty not scores.

Will there ever be another Mock AMC exam?

Well it does take up a considerable amount of time to write one with good problems and ones that aren't just copied and pasted from old competitions. cognos559 is writing a mock amc 12 and I'm busy with olympiad level mathematics, so in a short answer, I won't be writing one for some time.

right, my sister and i are making one, and we should be done in 2 weeks or so, so the test will be given in maybe 3 weeks since we need people to test it out for us. we already have the hard problems done (10-17, 19-25).

@cognos is there any way u could inbox the awesomemath sets to me?
@H34n1 could i get a copy of the test?

congratulations to the winners

maththemonster, i would send you the awesomemath sets, but i can't because it is illegal to copy and/or distribute them..... :( sorry... if it was legal i would try do it (although i don't see how i would do that without scanning each page separately and that would take forever because each set has like 30 pages and there are 8 sets so that is 240 pages...)

thanks anyway  :)
@ h34n1 how many hours have u been doing a day?

studying olympiad math, also where can i get a copy of your test? ive been doing about 3 hours of AIME- beginning olympiad (AOPS v2) per day.


can anyone thoroughly explain the Fundamental Theorem of Algebra?

Not sure what you want Mango 070 but how about

http://www.clarku.edu/~djoyce/complex/fta.html

or

http://www.cut-the-knot.org/do_you_know/fundamental2.shtml

or

http://www.cut-the-knot.org/do_you_know/fundamental.shtml

Explain what it says, or explain how to prove it?

The latter is much more difficult to explain than the former.

I presumed that Mango 070 meant the former for the very reason you mention.

yea an explanation of what it says

---thanks for the links AndrewTom

"It states that every polynomial equation over the field of complex numbers of degree higher than 1 has a complex solution."
---from http://www.cut-the-knot.org/do_you_know/fundamental2.shtml

why can't the degree be 1?
e.g. x+1=2, x=1, which is complex

if the degree is zero you would just get something like 5=5 or 7=7, I guess in general, complex number=the same complex number
---so isn't the solution to this infinity? - is infinity a complex number?

also, why can't the degree be negative
e.g. x^-2=4, x = -1/2 and 1/2

does the theorem not include degrees 1 and below because there is an equation, in which the degree is 1 or less that does not have a complex solution? also on the linked website it said complex numbers are algebraically closed, so couldn't n be any complex number and the polynomial would still have at least one complex solution? - except what is the degree of polynomials defined to be? positive integers? if so everything that i just said doesn't matter except for the argument about the degree being 1

[b][u]Addressing some of your concerns and confusions:[/u][/b]

0. The FTA is talking about roots of polynomials. A root of a polynomial $ P(x)$ is a number $ r$ such that $ P(r) = 0$. The FTA claims that, with multiplicity, a polynomial of degree $ n$ has exactly $ n$ complex roots.

1. Conventionally, when you talk about polynomials, you have terms ranging from degree 0 (constant) to some integer degree, where all terms have integer degrees. As for specific cases:

2. Degree one: I'm not sure why they excluded this case, but it is certainly true.

3. Degree zero: The polynomial $ P(x) = 2$ does [i]not[/i] have a root, because it is 2 everywhere (and thus never 0)!

4. Negative degrees: The equation $ P(x) = x^{ - 1} = \frac {1}{x}$ does [i]not[/i] have a root, because 0 is not the reciprocal of any complex number.

[b][u]Important General Fact:[/u][/b] A statement $ p \implies q$ does not say anything about what happens if $ p$ is false. Don't worry about what the FTA is [i]not[/i] talking about -- surely there are equations that aren't polynomials of positive degree, which have a root!

[b][u]The FTA in a nutshell:[/u][/b] One can invoke the FTA and the factor theorem for each step to fully factor a general polynomial, as below:

\begin{eqnarray*} (\text{polynomial of degree }n) & = & (x - r_n) (\text{degree }n - 1) \\
& = & (x - r_n) (x - r_{n - 1}) (\text{degree }n - 2) \\
& = & (x - r_n) (x - r_{n - 1}) (x - r_{n - 2}) (\text{degree }n - 3) \\
& \ldots & \\
& = & (x - r_n)(x - r_{n - 1}) \ldots (x - r_2) (cx - cr_1) \end{eqnarray*}

where at each step, I used the fact that the remaining polynomial has at least one root, and factored out the corresponding monomial.

Of course, the procedure as shown only divides out more than 3 factors if the degree is more than 3 -- otherwise, the procedure will terminate in fewer lines. But you get the idea.

ok, thanks - that definitely clears things up


function
calculus
integration
real analysis
real analysis solved

Here's a problem which I had as a homework:

Find all functions f defined on (-Pi/2,Pi/2) wirh real values which have a primitive F s.t. f(x)+cosx*F(x)=sin(2x)/(1+sinx)^2.

Hi grobber,

Multiply the equation by e^sinx
e^(sinx)f(x)+cosx.e^(sinx)F(x)=
(e^(sinx)F(x))' = 2cosx.e^(sinx).sinx/(1+sinx)
(e^(sinx)F(x))' = 2cosx.e^(sinx)/(1+sinx) - 2e^(sinx).cosx/(1+sinx)
(e^(sinx)F(x))' = 2(e^sinx/(1+sinx))' =>
F(x)= 2/(1+sinx) + A.e^(-sinx)
f(x)=F '(x)= -2cosx/(1+sinx) - A.cosx.e^(-sinx)

Remark:  the idea to multiply by e^(sinx) is not magic, if you start with
equation by derivation and regrouping the terms we have
a differential equation of order one
f'(x) + (cosx+sinx/cosx)f(x)=2cosx/(1+sinx) - 4sinx.cosx/(1+sinx)^3
when you solve without the RHS
f'(x)+(cosx+sinx/cosx)f(x)=0 =>f(x)=B.cosx.e^(-sinx) you see e^(sinx)
http://mathworld.wolfram.com/FirstOrderOrdinaryDifferentialEquation.html

Wich grade this problem ?

This problem is for the 12th grade students!
From now on please post the grade for which the problem was given for. (especially for those in superior maths)

cheers! :D :D

it is hard to post the "grade" of the problem, because most countries have this kind of problem in the undergradutate curriculum!

it's easy to check the equivalent romanian grade: matrices and limits are XI-th, as rings, groups, fields, integrals and differential equations are XII-th grade.


Is there a periodic table song in order by atomic number somewhere on the internet? Just wondering before I go lurk moar...

Definitely not in order - but who can forget [i]The Elements[/i] by [url=http://members.aol.com/quentncree/lehrer/elements.htm]Tom Lehrer[/url]? And mentioning Lehrer is definitely appropriate on this site, for reasons that you can probably figure out.


real analysis
abstract algebra
calculus
integration
function
real analysis unsolved

A quite difficult problem:
   Suppose that $f$ has bounded variation, then the fourier series converges unformly to $f$.

Introduction to Harmonic Analysis, II, (2.8) [Katznelson].
P.S: don't you need continuity ?

Oh, yes, continuity is required. Yet, I would like more to see a solution, not only the title of a book. In particular, are there different approaches?

That's about the last criterion I learned in Fourier Analysis, using Féjer theorem and the tauberian theorem of Hardy ( the one with $f^{*}(n)=O(\frac{1}{n})$, where $f^{*}$ is the nth Fourier coefficient ). Unfortunately my harmonic analysis is a bit rusty and since I did not do any exercise when I learned it, I forgot most of it. That's why I only mentioned the book, because I know this result is essential.

I know of two main ways to say this.

The first is the way Julien suggests. (His memory is pretty good, and Katznelson is a good place to look, although you could go back to the source, Zygmund's [i]Trigonometric Series[/i].) The sequence of Cesaro means $\sigma_n(f)$ of the Fourier series for $f$ is given by convolution of $f$ with the Féjer kernel:

$K_n(x)=\sum_{k=-n}^n\left(1-\frac{|k|}{n+1}\right)e^{ikx}$

When you explictly calculate it, you see that $K_n$ is nonnegative, integrates to 1 around the unit circle (assuming you divide by $2\pi$ in each of your integrals), and it tends to zero away from the origin. I won't write the full formula here (although its derivation is straigtforward) but the essential behavior of $K_n(x)$ around $0$ is something like $\frac{1-\cos(nx)}{n^2x^2}.$

That makes the Féjer kernel a good "approximation of the identity." We can then prove  the following

(a) If $f\in L^p(\mathbb{T})$ for $1\le p<\infty$, then $\|\sigma_n(f)-f\|_p\to0$ as $n\to\infty$

(b) If $f\in C(\mathbb{T})$, then $\sigma _n(f)\to f$ uniformly on $\mathbb{T}.$

It's (b) that we want. We start with a continuous $f$ of bounded variation. The Fourier series is Cesaro summable to $f$ everywhere.

What does bounded variation tell us? By a Stieltjes integration by parts, we get that the $k$th Fourier coeffiecient of $f$ is $O(k^{-1})$ as $k\to\infty.$

Now we invoke that Tauberian theorem of Hardy: if a series is Cesaro summable and the terms are $O(k^{-1}),$ then the series converges.

If we strengthened the condition very slightly to $f$ being absolutely continuous, then we would have that the Fourier coefficients are $o(k^{-1})$ and we could use a much simpler Tauberian theorem.

-----

As for a second approach, see my post #4 on [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=69329]this thread.[/url]

Roughly, the idea is this. Write the partial sums $S_n(f)$ as convolution of $f$ with the Dirichlet kernel $D_n.$ Now, $D_n$ is not a proper or controllable "approximate identity," but you can proceed as shown in that other thread, integrating by parts, differentiating $f$ and integrating $D_n.$ This has to be a Stieljes integration by parts. Push it through, and we get the theorem that the Fourier series of a function of bounded variation converges everywhere to the average of the right and left hand limits.

And then we're mostly done; if $f$ is continuous, the average of the right and left hand limits is the value of the function. What is not so clear to me is how to argue that the convergence is uniform.


geometry
circumcircle
incenter
power of a point
radical axis
geometry unsolved

1.$ {\odot I }$ is inscribed in $ {\triangle ABC }$ the circle touches the side $ {AB,BC,AC}$ at $ {F,D,E}$  and $ {FD}$ meet $ {AC}$ at $ {P}$ ;$ {ED}$ meet $ {AB}$ at $ {Q}$, point $ {M,N}$ are the midpoints of segment $ {EP}$ and $ {FQ}$ ;$ {O}$ is the circumcenter of $ {\triangle ABC}$ prove that $ { IO }$ ⊥$ { MN}$.
2.  $ {ABCD }$ is inscribed in a circle$ {\odot O}$ and $ {AC \cap BD \equal{} P}$ . circle C1 which pass points $ {A,P}$ meet ciecle C2 which pass points $ {P,B}$ at point $ {Q}$ ; and $ {F}$ is the intersection of $ {C1}$ and $ {\odot O}$ ; point $ {E}$ is the intersection of $ {C2}$ and $ {\odot O }$ prove that : the three segments $ {CE,DF,PQ}$ are parallel or meet at one point .
3.given a triangle ABC with three altitude as shown . $ {P \in EF}$ and $ {Q \in FD}$ with $ {\angle QAP \equal{} \angle DAC}$ then prove that      $ {\angle FQA \equal{} \angle PQA}$.
[b]my 100th post  :)
looking forward to your solutions , dear mathlinkers[/b]
[color=red]Really sorry for my bad typo! now pro1 has been corrected M,N are mid points of segments [b]EP,FQ[/b][/color]

I think problem 1 has a mistake. Please check that!
[b]Problem 2:[/b]
Let $ M\equal{}DF\cap (O_1), N\equal{}EN\cap (O_2), I\equal{}DF\cap CE$.
$ \angle DCE\equal{}\angle DBE\equal{}\angle PNE$ then $ DC//PN$
Similarly, $ DC//PM$ therefore $ MN//DC$
$ DCEF$ is cyclic thus $ MNEF$ is cyclic $ \Rightarrow IF.IM\equal{}IE.IN \Rightarrow I$ lies on the radical axis of two circles.
[b]Problem 3:[/b]
Let $ J\equal{}AQ\cap FC$ then $ \Delta AJH\sim \Delta APE$
$ \Rightarrow \Delta AJP\sim \Delta AHE$
$ \Rightarrow \angle QJP\equal{}DHE\equal{}90^o\plus{}\angle DFE/2$
So $ J$ is the incenter of triangle $ FPQ$ and we are done.

Problem1 and 3 seems hard,. do you have any ideas about them? Dear Mathlinkers?

for Problem 1:
i'll use this lemma(it should be posted or maybe u know it,if not tell me to prove it! :) ):
Let $ S,T$ be points on $ CA,BC$ s.t. $ BT\equal{}CS\equal{}BC$.then $ OI \perp ST$.
by constructing $ ST$ as mentioned,we only need to prove that $ ST \parallel{} MN$.which is easy!(use Thale's theorem)
sorry for my bad english!


linear algebra
matrix
integration
vector
induction
advanced fields
advanced fields unsolved

Find the set of all solutions of $ \dot{x} \equal{} Ax\plus{}t^nv$    where $ A$ is a fixed square matrix and $ v$ is a fixed vector.

We can write the solution as
\[ x(t)\equal{}\int_0^ts^ne^{A(t\minus{}s)}v\,ds\plus{}e^{At}x(0)\]
If you want more concrete details of how that is going to work out, we do need to know the relationship between the matrix $ A$ and the vector $ v.$ Is $ v$ an eigenvector of $ A?$ More broadly, is $ (A\minus{}\lambda I)^kv\equal{}0$ for some $ \lambda$ and some $ k?$

There's also something to be said for solving this with Laplace transforms.

I forgot to mention that $ A$ is invertible.

[quote="Kent Merryfield"]That does simplify things a little, since it means that $ 0$ is not an eigenvalue of $ A$ and hence $ t^kv$ is not a solution of the corresponding homogeneous equation for any $ k.$[/quote]

Can you be more explicit?   I know how to compute the case $ n\equal{}0$ and I was trying to use induction but I don't even know how to relate the case $ n\equal{}1$ with $ n\equal{}0$

This is a linear constant coefficient non-homogeneous ODE. Try the method of undetermined coefficients to find a particular solution. That is, assume there is a solution of the form
\[ y\equal{}t^nu_n\plus{}t^{n\minus{}1}u_{n\minus{}1}\plus{}\cdot\plus{}tu_1\plus{}u_0\]
for unknown constant vectors $ u_n,u_{n\minus{}1},\dots,u_0.$

Plug that into the original differential equation and see if you can solve for the $ u_k.$


inequalities
trigonometry
geometry unsolved
geometry

Prove that in any acute triangle have :
$ sin\frac{A}{2}\plus{}sin\frac{B}{2}\plus{}sin\frac{C}{2} \geq \frac{5}{4}\plus{}\frac{r}{2R}$

Call p the Semiperimeter, by briggs formulas we have:

$ 1 \blacktriangleright \sin{\frac {\alpha}{2}} \cdot \sin{\frac {\beta}{2}} \cdot \sin{\frac {\gamma}{2}} \equal{} \frac {(p \minus{} a)(p \minus{} b)(p \minus{} c)}{abc} \equal{} \frac {A^2}{p \cdot abc} \equal{} \frac {r}{4R}$

$ 2 \blacktriangleright R \ge 2r$  'cause $ OI^2 \equal{} R(R \minus{} 2r)$ so $ R \minus{} 2r \ge 0$

$ 3 \blacktriangleright \sin{\frac {\alpha}{2}} \cdot \sin{\frac {\beta}{2}} \cdot \sin{\frac {\gamma}{2}} \le \frac {1}{8}$

edited... :oops:

[quote="¬[ƒ(Gabriel)³²¹º]¼"]$ \sin{\frac {\alpha}{2}} \plus{} \sin{\frac {\beta}{2}} \plus{} \sin{\frac {\gamma}{2}} \ge \frac {5}{4} \plus{} \frac {r}{2R}\ \Longleftrightarrow \ \frac {3}{2} \ge \frac {5}{4} \plus{} \frac {1}{4}$[/quote]
you are wrong,for example:

we want to show that $ 5>\frac 54\plus{}2$,then with your argument,we can say that $ 5>1$ and $ 2>1$ hence:

$ 5>\frac 54\plus{}2\iff 1>\frac 54\plus{}1$ :huh:

and also you have another mistake here:

[quote="¬[ƒ(Gabriel)³²¹º]¼"]
$ 4\blacktriangleright\sin\frac{\alpha}2\plus{}\sin\frac{\beta}2\plus{}\sin\frac{\gamma}2\geq\frac 32$[/quote]

this is wrong to,the reverse one holds:

$ \sin\frac{A}2\plus{}\sin\frac{B}2\plus{}\sin\frac{C}2\leq\frac 32$  :wink: