analytic geometry
Euler
topology
geometry
3D geometry
sphere
invariant

1.  Start with the unit disc $ \mathbb{D}$ and identify points on the boundary if their angles, thought of in polar coordinates, differ by a multiple of $ \frac {\pi}{2}$. Let $ X$ be the resulting space, use the Van Kampen's theorem to compute $ \pi_{1}(X,*)$

2. Let $ X \equal{} S^{2}/\{p_{1} \equal{} \ldots \equal{} p_{k}\}$  be the topological space obtained by identifying $ k$ distinct points $ p_{k}$  on it ( $ k\geq 2$) :
a) find the fundamental group of $ X$
b) Find the Euler characteristics of $ X$
c) find the homology groups of $ X$

I am still a beginner  of learning AT, I am learning it by myself so  it would be great for me if you could give me as details as possible.  Thanks  so much .

By the way, I have a question : is there any problem - solution book in Algebraic topology ? if there is, would you please give me some names?

1. Cover $ X$ with open sets $ A$ and $ B$, where $ A$ is the disc centered at the origin of radius $ 3/4$ and $ B$ is the complement of the closed disc centered at the origin with radius $ 1/2$. Then $ A$ is contractible, so by the van kampen theorem $ \pi_1(X) \equal{} \pi_1(B) / \pi_1(A \cap B)$, where the quotient is induced by the inclusion $ \pi_1(A \cap B) \to \pi_1(B)$.

$ B$ deformation retracts by a map $ r$ onto the boundary of the disc, consisting of four copies of a loop $ \alpha$ which traverses precisely $ 1/4$ of the boundary of the disc. $ A \cap B$ is a thin ribbon, which itself deformation retracts to a circle, which is carried by $ r$ to the concatenation $ \alpha^4$. Therefore $ \pi_1(X) \equal{} \langle \alpha \rangle / \langle \alpha^4 \rangle \equal{} \mathbb{Z}/4$. (This is to be expected, since $ \pi_1(\mathbb{R}P^2) \equal{} \mathbb{Z}/2$ by a similar argument.)

thanks  for your solution for #1 Nukular , it is great

let's try and break down number 2 also...

so..
for point a) we can notice that our space deformation retracts on the wedge of a 2-sphere and a bouquet $ k\minus{}1$ circles, so fundemental group, by van kampen, is free over $ k\minus{}1$ generators.
for point b) we can triangulate the sphere in such a way that the points we identify are vertices, so euler characteristic just decreases by $ k\minus{}1$, and then it's $ 3\minus{}k$.
for point c), we have three ways: either we use point a) and b) and notice that the space is connected, so we can deduce rational homologies (and they have rank $ 1,k\minus{}1,1$ respectively over $ \mathbb{Q}$), or we use mayer-vietoris sequence with the decomposition we already used in point a), or we use another one, which is seeing identification as "summing" with a segment having their endpoints on distinct points on the sphere and then using mayer-vietoris again.

since you're a beginner, i'll leave you the details :)

b) and c) should not require triangulation; that's too much work;

that deformation retract is all you need. It tells you directly that $ H_1(X) = H_1(S^2) \times \prod_{i = 1}^{k - 1} H_1(S^1) = \mathbb{Z}^{k - 1}$. Furthermore $ H_2(X) = \mathbb{Z}$, and $ H_0(X) = \mathbb{Z}$. Therefore the Euler characteristic is ${ \mathrm{Rank}(H_2(X)) - \mathrm{Rank}(H_1(X)} + \mathrm{Rank}(H_0(X)) = 3 - k$.

ma_go, is there an easy way to see that deformation retract? Seems plausible, but it's a bit weird lookin.

Hi all. One way to see the homotopy equivalence to $ {\mathbb S}^2\vee({\mathbb S}^1)^{\vee (k\minus{}1)}$ formally is as follows: for $ 1\leq i<k$, attach a line segment $ s_i$ connecting $ p_i$ and $ p_{i\minus{}1}$ to $ {\mathbb S}^2$. Call the resulting space $ X$. The union $ S$ of the $ s_i$ is contractible, and since we are working with "nice" spaces, the map $ X\to X/S$ contracting this segment to a point is a homotopy equivalence. Thus we can concentrate on the computation of the invariants of $ X$: Next (for $ 1\leq i<k$) consider arcs $ c_i$ on $ {\mathbb S}^2$ connecting $ p_i$ and $ p_{i\plus{}1}$. Again, the union $ C$ of these arcs is contractible, and thus $ X\to X/C$ is a homotopy equivalence. But $ X/C\cong{\mathbb S}^2\vee({\mathbb S}^1)^{\vee (k\minus{}1)}$.

[quote="Nukular"]b) and c) should not require triangulation; that's too much work;[/quote]
right, but for b) it has the advantage of not requiring to compute any homology group, so i thought it was nice :)
[quote="Nukular"]ma_go, is there an easy way to see that deformation retract? Seems plausible, but it's a bit weird lookin.[/quote]
quite pointless after hanno's post, but you can also picture it in another way. join the "center" of the sphere with the $ k$ points on the surface with $ k$ segments and "pull the strings" towards the center: this tells you that this sort of spider web is homotopy equivalent to our quotient space. then you can just move your $ k$ points to a single point without changing homotopy type, and you see your wedge there.
(the center was actually useless, but it was nice seeing it that way).


[hide]6 must be greater than or equal to x, so, the only non-negative integers which work are 1,2,3,4,5, and 6.[/hide]

i edited the last post. you are very selective in your answers.(just kidding) :D

[quote="236factorial"]How many positive integers x satisfy:

$x+6\ge2x$[/quote]
[hide]
I did this problem by chosing a large number and narrowing the numbers down. 8 is too big, so is 7.
6 works, so there are 6 possible integers, since 0 isn't positive.[/hide]

[quote="236factorial"]How many positive integers x satisfy:

$x+6\ge2x$[/quote]
[hide]
6+6=2x6
so 6 is the highest

so 6 numbers satisfy this condition

[/hide]


Hi, I have a problem with the bibliography of my document. I'm writing the document in a language different  from english and I'm using BibText to create de bibliography. My problem is that when I have a reference with many authors, in the pdf it joins the  last 2 authors with "and" instead of translating "and" to the language of the document. Is this normal???  :huh:  or do I have to do something to specify the language of the bibliography??
thanks in advance!

Are you using the Babel package? Try putting \usepackage[nameoflanguage]{babel} at the top of your document and see if that works.

yes, I'm using the Babel package and the \usepackage[nameoflanguage]{babel}, but it's not working....

thanks!!


Look at it this way.
In 1 hour, 1 bacteria will make 1 new bacteria, but won't die.

[quote="sunnykim"]Bacteria have the following biological rhythm for spreading: each one lives an hour and in each half of hour of its life produces a new one. Starting from one, how many bacteria are after 6 hours?

[hide]I know the answer:$ 377$ but I can't get it :( [/hide][/quote]

let's do every 30 minutes

0 -> 1
30 -> 1
60 -> 2
90 -> 3
120 -> 5
150 -> 8

Notice that if we let $ B_{n}$ be the number of bacteria after $ n$ minutes, then $ B_{30k}$ seems to be equal to $ F_{k\plus{}1}$.  In fact, there is a reason.

Hint: Try thinking about the bacteria in having two phases, phase 1 is the first 30 min, then phase 2 is the second 30 min,
In phase 1, they make a new bacteria
In phase 2, nothing happens.
Repeat.
Each phase takes 30min
And then you have two types of bacteria, type 1 has phase 1 on the hour, and phase 2 on the half hours.
Type 2 does the opposite.

um, did you read at all or did you copy?

[quote="alanchou"]then the answer should be $ 610$, unless I can't count to 14 correctly[/quote]

the 14th fibonacci number is 377.
Your reasoning is correct.

In fact this problem is the same problem that led fibonacci to the discovery of the pattern.

The only difference in his problem was he had bunnies.

Yeah the question was like the bunnies reproduce every month, and die in two months.


Do there exist a convex polygon which can be partitioned into non-convex quadrilaterals?

I would say no, if I did not miss something in the following :

Assume that $P$ is a convex $n \geq 3$-gon for which such a dissection does exists. Let $m$ be the number of non-convex quadrilaterals used. The vertices of these quadrilaterals are :
- vertices of $P$
- or other points on the boundary of $P$, say that there are $r$ such vertices,
- or interior points in $P$ which are between two vertices of some of the non-convex quadrilaterals (they are on the boundary of a quadrilateral but not a vertex of it), say that there are $q$ such points,
- or interior points in $P$ which are not between two vertices of some of the non-convex quadrilaterals, say that there are $p$ such points.

We will compute the sum $S$ of the interior angles of all the quadrilaterals in two ways.

Because of the $r$points, $P$ has to be considered as a $(n+r)$ convex polygon, so that the sume of its interior angles is $(n+r-2) \pi.$ Moreover, each of the $p+q$ interior points contributes for $2 \pi$. Thus, $S = (n+r-2) \pi + 2(p+q) \pi.$ (1)

In another hand, each of the $q$ points defined above contributes for $\pi$ in the sum of the interior angles in exactly one of the quadrilaterals (which in that case has to be considered as a degenerated $t$-gon, where $ t >4$). It means that if $Q$ is one of the quadrilaterals with $\alpha$ of the $q$ points on its boundary (but not as one of its verteices), the total sum of the interior angles of $Q$ is $(4+ \alpha - 2) \pi$. Thus, summing over $Q$, it leads to $S = q \pi + 2m \pi.$ (2)

From (1) and (2), we deduce that :
$p = \frac {2m-q+2-n-r} 2 \leq \frac {2m- n +2} 2 < m.$ (3)

But, since each of the quadrilaterals is non-convex it has exactly one interior angle greater than $\pi$. This angle is necessarly at some interior point of $P$ (since $P$ is convex) and more precisely at some of the $p$ points defined above (since each of the $q$ others used an angle of $\pi$). Obviously, any of the $p$ points can be used in that way by only one of the quadrilaterals. Therefore, we must have $m \leq p$, which contradicts (3), and we are done.

Pierre.


Greetings,

We have been gradually building up our [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Contests.php]contest directory[/url] and would like more information about Canadian Math Competitions.  I know there are tests called Pascal, Fermat, etc., but know little about them (grade level, what they are used for, are they national?).  I would appreciate any information about those and any other Canadian competitions we could add to our directory.

Thank you in advance!

Mathew,

Have you seen this?

http://cemc.uwaterloo.ca/english/contests/index.shtml

They are all national.  The PCF contests (pascal cayley fermat) are all pretty basic.  Good students don't have any problem scoring 140+ on then (e.g, students who can consistently score over 120 on the AMC).  The Euclid and COMC are both written contests.  They are the most important ones for better students (and obviously the CMO).  The COMC is a feeder contest for the CMO (top 50 or so make it)  It is mostly easy, although the last few problems are hard.

I have seen that site in passing.  Thanks for the nice summary.

Hello, info about Contests in Canada:

Grades 4-6
http://www.mathematica.ca/eng/home.htm

Grades 7-8
Gauss Contest - http://www.cemc.uwaterloo.ca/english/contests/gauss.shtml

Grades 9
Pascal Contest - http://www.cemc.uwaterloo.ca/english/contests/pascal.shtml
Fryer Contest - http://www.cemc.uwaterloo.ca/english/contests/fryer.shtml

Grades 10
Cayley Contest - http://www.cemc.uwaterloo.ca/english/contests/cayley.shtml
Galois Contest - http://www.cemc.uwaterloo.ca/english/contests/galois.shtml

Grades 11
Fermat Contest - http://www.cemc.uwaterloo.ca/english/contests/fermat.shtml
Hypatia Contest - http://www.cemc.uwaterloo.ca/english/contests/hypatia.shtml

Grades 12
Euclid Contest - http://www.cemc.uwaterloo.ca/english/contests/euclid.shtml

Additional Contests –
Lakehead University: http://math.lakeheadu.ca/math_competition.htm
Nova Scotia: http://www.mathstat.dal.ca/~mathleague/problems.html
BC Math Contest- http://www.sci.ouc.bc.ca/instructors/clee/bcchsmc/BCCHSMC.htm
AB Math Contest- http://www.math.ualberta.ca/~ahsmc/
Calgary Math Contest- http://www.math.ucalgary.ca/education/JrMath/
Maritime Math Contest  - http://ace.acadiau.ca/math/MMC/mmc.htm
Problem of the Month (WLU) - http://www.wlu.ca/page.php?grp_id=43&s_id=71
Newfoundland Math Contest - http://www.math.mun.ca/index.php3?content=includes/outreach/blundon.html
Mathleague- For grades 4 – 12 http://www.mathleague.com/

Olympiads-
COMC/CMO/APMO/MOCP - http://camel.math.ca/CMS/Olympiads/

Hmm, if you're including all the little local contests, then there might be more.

Like the Edmonton Junior High Math Competition

[Edit]  Heya lightrhee, do you have the site for the edmonton contest?  I sure don't

Thanks for all of the great information!  We have added them to [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Contests.php]our list[/url].

If you have more to suggest (with weblinks), please post them here.

[quote="Sunny"]
Heya lightrhee, do you have the site for the edmonton contest?  I sure don't[/quote]
No idea. I'm not even sure whether there is one.

In British Columbia, about two years ago, the MATHCOUNTs group terminated their association with MATHCOUNTS and renamed themselves MATHChallengers. I believe the licencing and fees charged by MATHCOUNTS became too great, so the organizers decided to start their own competition pretty much from scratch. There was also talk about the Saskatchewan group joining MATHChallengers, but I don't know if that plan ever went anywhere.
FYI: Here is the link to the MATHChallengers site

http://www.apeg.bc.ca/mathchallengers/index.htm

Regards,

Hey, is there anyone here in British Columbia? I heard that if you have high mark for Euclid, you will be accepted to the math honour class of UBC right? Do you know exactly how many percent do they need?


If $a,b,c$ are the sidelengths of a triangle, then \[ \frac{b}{c+1} + \frac{c}{b+1} > \frac{2a}{a+2} .  \]

[i]Victor Tudor[/i]


[b]A circle is drawn with center at the origin and radius $ 3$. Find the coordinates of all intersections of the circle with an origin-centered square of side length 4 whose sides are parallel to the coordinate axes[/b]

[quote="Prime factorization"]This sounds pretty simple. We have circle

$ x^{2}+y^{2}= 3^{2}$ and square

$ y=-2$ $ y = 2$ $ x=-2$ $ x=2$ (the intersection of those 4 lines forms a square). Now just plug in those values into the circle equation to find our vertices.

For instance,

$ (-2)^{2}+y^{2}= 9$
$ y^{2}= 5$
$ y = \pm \sqrt{5}$

Upon similar calculations, our intersection points are:

$ (-2, \;-\sqrt{5})$
$ (-2, \; \sqrt{5})$
$ (2, \;-\sqrt{5})$
$ (2, \; \sqrt{5})$

Or, simply written, $ (\pm 2, \pm \sqrt{5})$ and $ (\mp 2, \pm \sqrt{5})$

Hopefully I didn't overlook anything  :blush:[/quote]

that's a very reasonable answer and i believe its correct..its just i have some doubts about something...since the square is centered at the origin and has side length 4, does that mean that the coordinates of its vertices are (2,2), (2,-2), (-2,-2), (-2,2)..wouldn't that mean that the square intersects the circle at its vertices..im probably wrong.

[quote="mihail911"

that's a very reasonable answer and i believe its correct..its just i have some doubts about something...since the square is centered at the origin and has side length 4, does that mean that the coordinates of its vertices are (2,2), (2,-2), (-2,-2), (-2,2)..wouldn't that mean that the square intersects the circle at its vertices..im probably wrong.[/quote]

Are any of those points on the circle?  :wink:

yeah, I don't think they intersect, since the length of the diagonals of the square= $ 4\sqrt{2}$, which is $ <$ the diameter of the circle = $ 9$

Yes, but they intersect in 8 places, not 4 as mihail thought.

Also, the diameter of the circle is not 9. :P

oops, I somehow got confused with the numbers somewhere in there.  :blush:
Ok, I agree with Prime Factorization now.


Prove: If $ f(x)$ is differentiable at $ x\equal{}a$ then $ f(x)$ is continuous at $ x\equal{}a$.

See [url=http://www-math.mit.edu/~djk/18_01/chapter02/proof04.html]here[/url].  Note that the converse is false.


Let $ A$ be a $ n\times n$ matrix with $ a_{ij}\equal{}i\in\mathbb{N}$  $ \forall i\equal{}j$ and $ a_{ij}\equal{}k\in\mathbb{R}$  $ \forall i\neq j$.Find $ \det A$.

Let $ X$ be the $ n\times n$ matrix with every entry equal to $ k$.  Let $ Y$ be the diagonal matrix with diagonal entries $ 1\minus{}k$, $ 2\minus{}k$, ..., $ n\minus{}k$.  Then the given matrix $ A$ is equal to $ X\plus{}Y$.  To compute the determinant of a sum, we will use the formula developed in the post
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=591700#591700[/url].

We need to handle only the cases when 0 or 1 rows are taken from $ X$.  As long as $ k$ is not equal to 1, 2, ..., and $ n$, we get the answer
\[ \prod_{i \equal{} 1}^{n}(i\minus{}k)\left[ 1\plus{}k\sum_{i \equal{} 1}^{n}\frac{1}{i\minus{}k}\right ].\]
If $ k$ happens to equal 1, 2, ..., or $ n$, we get the answer $ (\minus{}1)^{k\minus{}1}k! (n\minus{}k)!$.


abstract algebra
function
geometry
geometric transformation
superior algebra
superior algebra unsolved

we say that R is a euclidean ring,if R be abelian and unity.and exist a function f: R_{0}   $ %Error. "display" is a bad command.
\to$(positive integer numbers) so that:1)if a and b be nonzero in the R and a$ %Error. "display" is a bad command.
\mid$    b then f(a)   $ %Error. "display" is a bad command.
\le$f(b)   2)for every a and b   $ %Error. "display" is a bad command.
\neq$    0,exist q and r so that a=bq+r that r=0 or f(r)<f(b).now suppose R={a+b$ %Error. "display" is a bad command.
\sqrt {3} i$ $ %Error. "display" is a bad command.
\mid$  a&b$ %Error. "display" is a bad command.
\in$ Z or 2a&2b be odd integer numbers},show it is a euclidean ring.

[quote="m_jackson"]we say that R is a oghlidosy ring,if R be abelian and unity.and exist a function f: R_{0}   $ %Error. "display" is a bad command.
\to$(positive integer numbers) so that:1)if a and b be nonzero in the R and a$ %Error. "display" is a bad command.
\mid$    b then f(a)   $ %Error. "display" is a bad command.
\le$f(b)   2)for every a and b   $ %Error. "display" is a bad command.
\neq$    0,exist q and r so that a=bq+r that r=0 or f(r)<f(b).now suppose R={a+b$ %Error. "display" is a bad command.
\sqrt {3} i$ $ %Error. "display" is a bad command.
\mid$  a&b$ %Error. "display" is a bad command.
\in$ Z or 2a&2b be odd integer numbers},show it is a oghlidosy ring.[/quote]

Hello.
You can take the function $ f: s + ti \mapsto N(s + ti) = s^{2} + 3 t^{2}$.
If a is in $ R$ and $ b$ in $ R_{0}$, put $ a/b = x + y \sqrt {3} i$, with $ x$ and $ y$ rational numbers. Let $ \beta$ denote the closest integer to $ 2y$ and $ \alpha$ the closest integer to $ 2x$ which has the same parity as $ \beta$. Put $ q = (\alpha + \beta \sqrt {3} i)/2$ and $ r = a - bq$.
Well, I must confess that I copy this in the French book "Théorie des Nombres" of Daniel Duverney, 2007, p. 55, and that I didn't check it carefully.
If there remain problems, we can have a closer look.

Friendly,
P.

"oghlidosy ring" sounds like a failed translation of "euclidean ring" ;).


Could someone please explain angle chasing to me and give me a basic example?  Thanks.

angle chasing is basically when you have a diagram and you're given certain angles and you use angle properties in order to find another angle.

There is a chapter in AoPS volume 1 called angle chasing, which lists all of the basic angle chasing techniques and gives lots of nice problems :wink:

angle chasing is basically finding the measures of angles with various "tools" that nclude:

sum of angles in a triangle
sum of angles in a quadrilateral
angles intercepting arcs
angles with together form a line
angles in a right traingle
inscribed angles subtending the same arc angle bisectors
the various kinds of lines, angles, and triangles

Thisis what I understand about angle chasing.

As I understand it, angle chasing can be used in any problem when you want to find the measure of an angle.  Many times you need to find the measures of other angles first before you find the measure of the angle you are looking for.

2006 ARML Individual #3 is a good example of angle chasing.  You can find more ARML problems and solutions at [url=http://www.arml.com]the ARML website[/url].

Thanks, guys.  BTW, I went to ARML this past June.


1- a,b,c ar positive real numbers and a+b+c=1 prove:
5(a^2+b^2+c^2)<=6(a^3+b^3+c^3)+1

2- for a real x we say <x> i the distance of x from the closest number to it
prove for any irrational a and e>0 prove there exists natural n that
<n^2a> is smaller than e.

3-find all f from natural numbers to natural numbers that
2n+2001<=f(f(n))+f(n)<=2

4-does exist  a {bi} of positive numbers that for any positive m
b(m)+b(2m)+b(3m)+...+1/m

5-find all f from Q to Q that
2f(f(x))=f(2x)           f(x)+f(1/x)=1

6-find all functions on R-{0} that
xf(x+1/y)+yf(y)+y/x=yf(y+1/x0+xf(x)+x/y

7- We call a polynomial good if and only if p(x)^2+c=p(x^2+c) for a constant complex number c
a)if c is ratio and not integer find all such polynomials
b)if c= a+bi (i^2=-1) and a^2+B^2>4  :)

hi!

nice of you to supply these questions, but please put each problem in a separate topic into its proper subfolder :) also specify the source (country, olympiad type, year -> there is a topic that explains how to do that :) ).


search
function
modular arithmetic
quadratics

Source: Polish MO 2006
1. Determine all nonnegative integers $n$ for which $2^n+105$ is a perfect square.

[/hide]

It's been posted before.  If the search function were working properly...  :|

Anyway, it's a simple problem, cause we see that $n$ must be even (looking $\mod 3$), so

$2^{2k}+105=m^2 \rightarrow (m-2^k)(m+2^k)=105$

but u haven't fiinshed yet...does for every even n exist integer m?

We can solve it considering that $105=3 \cdot 5 \cdot 7$. Since $2^k-m$ and $2^k+m$ are integers and factors of 105, the easiest way is to try all possible "combinations" of factors, for example

$2^k-m=35$
$2^k+m=3$

or

$2^k-m=105$
$2^k+m=1$

and check which numbers $n$ work for the problem.
I didn't write it because it's a "standard way" to solve this kind of problems, and it's not really interesting

Assume there is an integer m such that

$(1) \; 2^n + 105 = m^2.$

If $n$ is an odd integer, then

$m^2 = 2^n + 105 \equiv 2 \pmod{3}$

which is impossible because 2 is quadratic nonresidue of 3. Hence $n$ is even, i.e. $n = 2p$ where $p$ is a non-negative integer. So

$(m - 2^p)(m + 2^p) = 105$

by (1), thus

$(2) \; m - 2^p = d$

$(3) \; m + 2^p = 105/d,$

where $d$ is an positive divisor of 105 and $d < 105/d$. In other words, $d \leq 10$ and $d\,|\,105 = 3 \cdot 5 \cdot 7$. The only possibilities are $d = 1, \, 3, \, 5, \, 7$. Subtracting (2) from (3) we get

$(4) \; 2^{p+1} = 105/d - d$.

Inserting $d = 1, \, 3, \, 5, \, 7$ in (4), the results are $2^{p+1} = 104, \, 32 = 2^5, \, 16 = 2^4, \, 8 = 2^3$. Consequently $p = 2, \, 3, \, 4$, which implies that (1) has 3 solutions, namely

(m,n) = (11,4), (13,6), (19,8).


geometry
3D geometry
analytic geometry
modular arithmetic

I don't where to post this problem , so I am posting it right here , so please help me in solving this problem...

There is a tenstory building having 10×10 number of rooms on each floor (consider the building
like a 10×10×10 lattice with each 1×1×1 cube as a room).The electrician has made such an
arrangement for lighting that if you switch on alight in a room all the rooms above and below,all
the rooms to its right and left,and all the rooms in its front and back are also lighted. So as to say
if there is a building with 3 ×3× 3 number of rooms and if welight the center room all but the
diagonally neighboring rooms are lighted.To optimize on the cost of electricity bill find out the
least number of rooms required to light the whole building.

[hide="Hint"] Note that the number of other rooms that a single light can light is independent of the location of the light.  That gets you a lower bound. [/hide]

I do not have time, but the answer is
[hide="Answer"] 50 [/hide]

[b]Edit:[/b] Thanks diophantient, yeah, I don't know what I was on

[quote="The Zuton Force"]Mmm?

You have 10 in the column, plus 8 others around it on the same floor, so that makes 18 rooms lit per room powered, and 50 = 1000/20 < 1000/18...

Also, in response to the hint by t0rajir0u, the hint you gave isn't completely true:
[hide="Clarifier"] It's true as long as the room isn't on a side-face of the building, or an edge (vertical); in these cases it gets cut down a bit[/hide][/quote]
Doesn't each room light up 27 (9 in the row, 9 in the column, 9 in the pillar)?

[hide]50 lights is possible.  Consider splitting up the building into eight $ 5\times 5\times 5$ cubes.  Then we choose two small cubes opposite each other and light up 25 rooms in each cube such that no two of the 25 rooms lie on a line parallel to one of the edges of the cube.  If we can do this, then each of the 25 vertical lines throught the small cube we chose contain exactly one light, so all the rooms in the vertically adjacent small cube are lit.  Similarly all the rooms in the rest of the eight small cubes will be lit since each small cube is adjacent to one of the small cubes we chose initially (or is one of the ones we chose initially).

To prove this, give the individual cubes inside a $ 5\times 5\times 5$ cube coordinates from $ (0,0,0)$ to $ (4,4,4)$.  We want to pick 25 points such that no two points share more than one coordinate.  Consider the set of points $ \{(x,y,z)|x\plus{}y\plus{}z\equiv 0\pmod{5}\}$.  This set contains one point for each choice of $ x$ and $ y$, so it contains 25 points.  Since we can't change one variable without changing another, it also satisfies the condition that no two points share more than one coordinate.  Therefore we can light the whole building with 50 lights.

I think you can prove that 49 lights is impossible by considering a floor with only 4 lights, but it's a little messy.[/hide]


So apparently Mathcounts is coming up (along with the somewhat less important to me this year AMCs).

Are there still any middle schoolers at [i]all[/i] who come on this forum? Bleh I apparently have chapter on Febuary 7th, three days before the AMC As...meh.

I was also going to create an AMC topic, but there's still a bit of time.

lol, I have chapter Mathcounts on feb. 7 as well. :)

Argh dude
this means that the AMC is Feb 10?

Okay I'm going to an Academic challenge tourney this Saturday dang
I'll take next Monday and prep for those dudes..

hmm
I should not make 3-2=2 errors

MY school does the 10B I think, so it's the 25th for me. According to my teacher...

Hm my school apparently signed up for the 10A on the deadline, so I'm taking it there. Yay.

Possibly someone should create an AMC thread.

hm..Lol.
AMC thread
EDIT: sorry. revived a dead thread...


In $\triangle ABC$, an angle bisector $AD$ is drawn. Write $BD$ in term of side lengths $a,b,c$.

We have :
$\frac{c}{b}=\frac{x}{y}$
$\frac{c}{b}=\frac{x}{a-x}$
$(a-x)c=bx$
$x(b+c)=ac$
$x=\frac{ac}{b+c}$

Click on the picture, then it'll be more clear.


I forgot how to do completing the square.  Can someone show me how to do it?

Completing the square is when you have an equation like $x^2+bx$ and you add $(\frac{1}{2}b)^2$ to the equation.

For example, in the equation $x^2+2x$, you add $(\frac{1}{2}(2))^2=1$ to get $x^2+2x+1$. Completing the square makes it a perfect square. MAke sure the coefficient of $x^2$ is 1!

thanks, i got my math packet yesterday.  It has 72 problems and its due next wednesday.
it has about 10 completing the square problems.  Thanks 236!


A sequence $ \{a_n\}$ is said to be [i]of bounded variation[/i] if the series
\[ \sum |a_n\minus{}a_{n\plus{}1}|\]
is convergent.
Prove that if $ \{ b_n\}$ is of bounded variation, and the partial sums of the series $ \sum a_n$ are bounded, then the series $ \sum a_n\cdot b_n$ is convergent.

Use summation by parts (also often called Abel's formula). Difference the $ b_n$ and sum the $ a_n$ in your summation by parts; the hypotheses look custom-fitted to make that work.

Except there's a potential issue involving the boundary terms. Note that the hypothesis that $ b_n$ is of bounded variation does not necessarily imply that $ b_n\to 0.$ It does imply that $ b_n\to b$ for some fixed $ b.$ The statement is actually false unless we supply an additional hypothesis.

Example: Let $ a_n\equal{}(\minus{}1)^n,$ and $ b_n\equal{}1$ for all $ n.$ Then $ \{b_n\}n$ is of bounded variation, and $ \sum_{n\equal{}1}^Na_n$ is bounded but $ \sum (\minus{}1)^n\cdot 1$ is divergent.

It looks like we must add the hypothesis that $ b_n\to 0$ in order to make this work.


How many ways are there to label 999 points arranged around in a circle, using labels A,B, and C, in such a way that between any two occurrences (clockwise and anti-clockwise) of one label the number of labels of the other two types is even (so between any two A's the sum of B's and C's is even).

[hide="Uh"]$ 0$, right? Consider all points marked $ A$ or $ B$. Between each consecutive pair is an even number of $ C$s. Then the total number of $ C$s must be even. Similarly for $ A,B$. But there are an odd number of points total. Contradiction.[/hide]

The wording is a little tricky and you've misread: between each consecutive pair of A, the [i]total number[/i] of B and C must be even.  You also missed three trivial colorings (all A, all B and all C).

Welll when I first approached this problem I thought there was only three possible solutions like JBL said. However suppose we had something like
 AAAAAAAAA.........AAAAABCAAAAAA....AAAAA So there would be 997 A's which satifies the property. So I ask some teachers and they recommended me to try smaller example... however I still have not made much progress.


Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc\equal{}2.$ Prove that:
\[ a^2b\plus{}b^2c\plus{}c^2a\geq a\sqrt{b\plus{}c}\plus{}b\sqrt{c\plus{}a}\plus{}c\sqrt{a\plus{}b}\]

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc \equal{} 2.$ Prove that:
\[ a^2b \plus{} b^2c \plus{} c^2a\geq a\sqrt {b \plus{} c} \plus{} b\sqrt {c \plus{} a} \plus{} c\sqrt {a \plus{} b}
\]
[/quote]

Cauchy Schwarz works
\[ a\sqrt {b \plus{} c} \plus{} b\sqrt {c \plus{} a} \plus{} c\sqrt {a \plus{} b} \le \sqrt {\left( {a^2 b \plus{} b^2 c \plus{} c^2 a} \right)\left( {\frac {{b \plus{} c}}{b} \plus{} \frac {{c \plus{} a}}{c} \plus{} \frac {{a \plus{} b}}{a}} \right)}
\]

\[ \left( {a^2 b \plus{} b^2 c \plus{} c^2 a} \right)\left( {\frac {{b \plus{} c}}{b} \plus{} \frac {{c \plus{} a}}{c} \plus{} \frac {{a \plus{} b}}{a}} \right) \le \left( {a^2 b \plus{} b^2 c \plus{} c^2 a} \right)^2
\]

\[ \frac {{b \plus{} c}}{b} \plus{} \frac {{c \plus{} a}}{c} \plus{} \frac {{a \plus{} b}}{a} \le a^2 b \plus{} b^2 c \plus{} c^2 a
\]

\[ 6 \le a^2 b \plus{} b^2 c \plus{} c^2 a
\]
Nice inequality.

Prove http://www.mathlinks.ro/viewtopic.php?t=248093 with the same.

I think your solution is not correct... You haven't proved $ a^2b\plus{}b^2c\plus{}c^2a \ge 6 \ge \frac{a\plus{}b}{c}\plus{}\frac{b\plus{}c}{a}\plus{}\frac{c\plus{}a}{b}$ correctly, while this inequality is obviously wrong since $ RHS \ge 6$.

[quote="Honey_S"]
\[ \frac {{b \plus{} c}}{b} \plus{} \frac {{c \plus{} a}}{c} \plus{} \frac {{a \plus{} b}}{a} \le a^2 b \plus{} b^2 c \plus{} c^2 a
\]
[/quote]
I think it is true because it was equivalent to \[ 3 \le \frac{a}{c} \plus{} \frac{b}{a} \plus{} \frac{c}{b}
\]

I thinks that he just haven't proof that : sigma(b+c)/b<=sigma(a²b)
wich is equivalent to : 3abc+sigma(a²b)<=2sigma(a²b) (multiplying by : abc)
<=> sigma(a²b)>=3abc wich is true by AM-GM


analytic geometry
induction
combinatorics proposed
combinatorics

Let $n$ be a positive integer and $\mathcal S$ be the set of points $(x, y)$ with $x, y \in \{1, 2, \ldots , n\}$. Let $\mathcal T$ be the set of all squares with vertices in the set $\mathcal S$. We denote by $a_k$ ($k \geq 0$) the number of (unordered) pairs of points for which there are exactly $k$ squares in $\mathcal T$ having these two points as vertices. Prove that $a_0 = a_2 + 2a_3$.

[i]Yugoslavia[/i]

We shall first prove that $ |\mathcal{T}| \equal{} \frac {n^2(n^2 \minus{} 1)}{12}$

Let's first count the number of squares that have their sides parallel to coordinate axis. It is not to difficult to see that if a square has a side of length $ i$ there are exactly $ (n \minus{} i)^2$ sqaures of such size. So the number of such squares is equal to $ \sum_{i \equal{} 1}^{n \minus{} 1}i^2$.
For any other square draw parallels to coordinate axis through the vertecies of the square-they intersect at latice points- therefore any other square that doesn't have sides parallel to the axis is inscriebed in some(exactly one) square that has sides parallel to coordinate axis. We can inscribe $ i \minus{} 1$ squres into a sqaure with side length of $ i$. Combining that with previous result we get that the total number of squares is $ \sum_{i \equal{} 1}^{n \minus{} 1}i^2 \plus{} \sum_{i \equal{} 1}^{n \minus{} 1}i^2(n \minus{} i \minus{} 1) \equal{} \sum_{i \equal{} 1}^{n \minus{} 1}i^2(n \minus{} i)$. It can be proven by the means of induction that this equals $ \frac {n^2(n^2 \minus{} 1)}{12}$.

We continue our proof with observation that $ a_k \equal{} 0$ for every $ k \ge 4$. So $ a_0 \plus{} a_1 \plus{} a_2 \plus{} a_3 \equal{} \frac {n^2(n^2 \minus{} 1)}{2}$, since every pair of points gets counted in one of the first four terms.

We can also note that $ |\mathcal{T}| \equal{} \frac {a_1 \plus{} 2a_2 \plus{} 3a_3}{6}$.(Each square correspondes to six pairs of points each gets counted the appropriate number of times in the right term).

We see that $ 6|\mathcal{T}| \equal{} a_0 \plus{} a_1 \plus{} a_2 \plus{} a_3$ (combinig first two results). By putting in the new value for $ |\mathcal{T}|$ we get. $ a_1 \plus{} 2a_2 \plus{} 3a_3 \equal{} a_0 \plus{} a_1 \plus{} a_2 \plus{} a_3$ form where the desiered result $ a_2 \plus{} 2a_3 \equal{} a_0$ follows.

In a "20 years after" fashion, check Problem 4 at this Balkan MO 2014 (see Contests section).


I am sorry for asking a very basic question about the Math Main Exam of
IIT-JEE. In main exam, there are too many questions for 120 minutes.
I cannot believe that we should sove as many as 15 to 20 questions
in only 120 minutes.

  I wonder how many questions in the IIT-JEE Math Main exam we should
solve in this very limited 120 minites.

Did you come out through a time machine from the stone age?

There are no mains no more. :wink:

I am ashamed of asking a silly question. Actually, I am not Indian but Japanese
(I come from not a stone age but from abroad)
and I bought a book on IIT-JEE (from 1992-2005) during my stay in India,
and knew about the IIT-JEE. (until 2005, I remeber that there WAS a Main Exam).

Comparing with the Japanese math entrance exam (for example,
in Tokyo University, there are 6 questions within 150 minutes)
and INMO (6 or 7 questions within 3 hours?),
I was surprised at the vast amount of questions in IIT-JEE.
And I wonder whether the candidates were expected to solve so many
questions when there was a main exam.
(or was it practically selective in that it was enough to solve 7 or 8 questions out of 15 to 20).

as far as I remember the papers were always solvable within the time (or at least most of it, even without rigorous time training, something I'm dead against).

Thank you for your reply. I learned that the IIT-JEE is different
from INMO or Japanese entrance exams of mathematics.
(A question of Tokyo University of typical level is, for example,
For -1<x<1, x \neq 0, prove (1-x)^{1-1/x} < (1+x)^{1/x}.
Hence or otherwise, prove 0.9999^{101}<0.99<0.9999^{100},
which we are expected to solve within 150/6=25 minutes.)

That involves a lot of "pen work", besides "brain work". The JEE however, is more of "brain work" and [b]less[/b] of "pen work", in [b]most[/b] questions.

(those words are not made bold for the fun of it, I really mean to stress on those)


geometry
analytic geometry
graphing lines
slope
trigonometry
rectangle

$\triangle ABC$ is right-angled at $B$. Point $M$ is located on side $AB$ so that $AM=BC$, and point $N$ is on side $BC$ so that $CN = MB$. Find the angle between $AN$ and $CM$.

When two lines intersect there are two different (supplementary) angles you can measure, which one are you referring to?

Loads of analytical geometry says 45 degrees.

if you do the slopes (make one of them a, and the other b), and the tangent subtracting formula, you get 45

Well , I tried for a while for a geometrical proof but I failed. I send an analytical proof.

We put $NA = x$ , $BN = y$ , so $BM = x$   and $MC = x + y$ .Consider a Cartesian system at  $B$ with axis $BC , BA$. Then the slope of  $CN$  is  $- \frac{y }{2x+y}$ and the slope of $AM$  is $- \frac{x + y }{x}$ .

 If $\phi$ is the desired angle ,  we find that $\tan\phi$ = 1 and the angle is 45 .

 [i]Now I see that this is a hint of Altheman , too![/i]

Babis

We consider the rectangle $ABCD$ .Take a point $F$ on $CD$ with  $CF = AM$. Then $\triangle FNA$ is an isosceles and right triangle. But  $CM // FA$and $\angle FAN = 45$.

 see also the solution of M@re  in

 http://www.mathlinks.ro/Forum/viewtopic.php?t=57468

 and my observations in the topic.

 [i]Babis[/i]


prove:

$ 1/1^2 \plus{} 1/2^2 \plus{} ... \plus{} 1/n^2 \equal{} 1^2 \plus{} 2^2 \plus{} ... \plus{} n^2$  $ mod(n \plus{} 1)$

sorry if this is too easy for this section, i was not sure..

edit: n+1 is prime

The divisions on the LHS are clearly not all valid unless $ n \plus{} 1$ is prime.

In the book i am reading it says that $ 1/1 \plus{} 1/2 \plus{} ... \plus{} 1/(p \minus{} 1) \equal{} 1 \plus{} 2 \plus{} ... \plus{} p \minus{} 1$ $ mod(p)$ where p is prime
the proof is that each non-zero residue mod(p) is equal to the reciprocal of one and only one non-zero residue mod(p)

it then states the equality given in the OP. in this equality we can see that the numbers on the RHS do not cover all the non-zero residues of mod(p) so how do we know the LHS is a permutation of exactly those residues on the RHS? this is probably real simple but i can't see it atm so please show me?

Let $ a_1, a_2 \ldots a_{p \minus{} 1}$ be a permutation of $ 1, 2, \ldots p \minus{} 1$ so that:

$ 1 \cdot a_1 \equiv 1$
$ 2 \cdot a_2 \equiv 1$
$ 3 \cdot a_3 \equiv 1$
$ \ldots$
$ (p \minus{} 1)a_{p \minus{} 1} \equiv 1$

We have $ m \equiv \frac {1}{a_m}$ from here, which was what was used to solve your example proof. We can just square this to get $ m^2 \equiv \frac {1}{a_m ^2}$

In other words, the same permutation that solves your example question will solve your original problem.

ok thanks, i suck.

btw can you tell me what a generator of z/pz means?


Prove that the angles of a triangle add up to 180 degrees.

This is easy.

Draw a triangle ABC.

Then, draw a line, XY, containing A that is parallel to BC with points Y and C to the right of A and the other two points to the left.
Then, extend BC segment into a line.

Then, using corresponding angles, <YAC = <ACB and <XAB = <ABC.

Since <YAC+<XAB+<BAC = 180,

the three angles of ABC also add up to 180.

[hide]Exterior angles are the sums of two remote angles, so $\angle 1=\angle B+\angle C;\,\angle 2=\angle A+\angle C;\,\angle 3=\angle A+\angle B.$

We can combine that into $\angle 1+\angle 2+\angle 3=2\angle A+2\angle B+2\angle C.$

The sum of the exterior angles of a polygon are 360, so $\angle 1+\angle 2+\angle 3=360.$

Substituting, we get $2\angle A+2\angle B+2\angle C=360.$

Simplifying gives us $\angle A+\angle B+\angle C=180.$

[img]http://img308.imageshack.us/img308/4405/trianglenf8.jpg[/img][/hide]

[quote="junggi"][quote="lotrgreengrapes7926"][quote="i_like_pie"]The sum of the exterior angles of a polygon are 360[/quote] That seems like a more advanced theorem than what we are trying to prove.  :wink:[/quote]
well don't we get that fact from what we're trying to prove
so its a bit circular reasoning[/quote]

consider a walk around a polygon, at each vertex, you turn precisely the exterior angle, once you return to your starting vertex, you have spun around 360 degrees

[quote="Altheman"][/quote][quote="junggi"][quote="lotrgreengrapes7926"][quote="i_like_pie"]The sum of the exterior angles of a polygon are 360[/quote] That seems like a more advanced theorem than what we are trying to prove.  :wink:[/quote]
well don't we get that fact from what we're trying to prove
so its a bit circular reasoning[/quote][quote="Altheman"]

consider a walk around a polygon, at each vertex, you turn precisely the exterior angle, once you return to your starting vertex, you have spun around 360 degrees[/quote]
ok that's cool


number theory
greatest common divisor
number theory unsolved

let:
 \[ \frac{1\plus{}xy}{x\minus{}y}\equal{}k, \frac{1\plus{}xz}{z\minus{}x})\equal{}m,\frac{1\plus{}yz}{y\minus{}z})\equal{}l\]
prove that :
\[ gcd(m.l)\equal{}gcd(m,k)\equal{}gcd(l,k)\equal{}1\]
($ x,y,z \in R$)
($ m,l,k \in Z$)

$ km\plus{}ml\plus{}lk\equal{}1$ $ \Longrightarrow gcd(m,l)\equal{}gcd(l,k)\equal{}gcd(k,m)\equal{}1$


geometry
combinatorics unsolved
combinatorics

Some $1\times 2$ dominoes, each covering two adjacent unit squares, are placed on a board of size $n\times n$ such that no two of them touch (not even at a corner). Given that the total area covered by the dominoes is $2008$, find the least possible value of $n$.

Consider a set of dominoes placed on an $ n \times n$ board, no touching allowed. Extend the board to $ (n\plus{}1) \times (n\plus{}1)$ by adding an empty column on the right and an empty row at the top. For each domino consider it's "shadow" which are all the squares lying directly above, directly to the right, or directly above and to the right of a cell covered by the domino.

In the diagram below, the dominoes are the XX cells, and the shadows are designated by the dots.

[code]
...
XX.

..
X.
X.
[/code]

It is clear that each domino has precisely $ 4$ shadow cells, and that different dominoes have disjoint shadows. The point of extending the board is that now even dominoes that were placed along the original right or top borders of the board have their full shadows.

If there are $ 1004$ dominoes, then they cover $ 6 \times 1004$ cells (including their shadows), so the extended $ (n\plus{}1) \times (n\plus{}1)$ board must have area at least $ 6024 > 77^2$. Therefore $ n \geq 77$.

If $ n\equal{}77$, we can achieve the required result by placing all the dominoes horizontally on the odd-numbered rows, having a single cell separating each domino from the next one. We can pack $ 26$ dominoes per row since $ 26 \times 3 \minus{} 1 \equal{} 77$, and they lie in $ 39$ rows, so overall we have $ 39 \times 26 \equal{} 1014$ dominoes, a bit more than necessary.


geometry
radical axis
power of a point
geometry solved

Given a triangle $ABC$ and $AA_0,BB_0,CC_0$ are converge at M.
The circle with diameter $BC$ intersect the circle with diameter $AA_0$ at $A_1$ and $A_2$. Similar we get $B_1,B_2$ and $C_1,C_2$. Prove that six points $A_1,A_2,B_1,B_2,C_1,C_2$ are lie on a circle

This was posted before at http://www.mathlinks.ro/Forum/viewtopic.php?t=3535 , but there is no correct solution yet.

  darij

Let $A', B', C'$ be the midpoints of the sides $BC, CA, AB$ and $D, E, F$ the feet of altitudes $AD, BE, CF$ from the vertices $A, B, C$. The circles $(A'), (B'), (C')$ with diameters $BC, CA, AB$ obviously pairwise meet at the vertices $A = (B') \cap (C')$, $B = (C') \cap (A')$, $C = (A') \cap (B')$ and also at the feet of the altitudes $D = (B') \cap (C')$, $E = (C') \cap (A')$, $F = (A') \cap (B')$. Hence, the altitudes $AD, BE, CF$ are the pairwise radical axes of the circle pairs $(B') - (C')$, $(C') - (A')$, $(A') - (B')$ and the orthocenter $H$ is their radical center. Let $P, Q, R$ be the midpoints of the cevians $AA_0, BB_0, CC_0$. Similar claim can be made for the circles $(P), (Q), (R)$ with diameters $AA_0, BB_0, CC_0$ - they pass through the feet of the altitudes and their radical center is the triangle orthocenter $H$. Actually, the 3 cevians $AA_0, BB_0, CC_0$ do not even have to be concurrent at all. For example, the circle $(P)$ intersects the side $BC$ at the point $A_0$ and at one other point $D'$. Since $AA_0$ is the circle diameter, the angle $\measuredangle AD'A_0 = 90^o$ is right and consequently, $AD'$ is the altitude from the vertex $A$ and the points $D' \equiv D$ are identical. Similarly, the circles $(Q), (R)$ with diameters $BB_0, CC_0$ pass through the feet $E, F$ of the altitudes $BE, CF$. Consequently the power of the ortocenter to the circles $(P), (Q), (R)$ is $HA \cdot HD = HB \cdot HE = HC \cdot HF$, respectively, is the same and the orthocenter $H$ therefore must be the radical center of these 3 circles. Of course, the power of the orthocenter to the circles $(A'), (B'), (C')$ is exactly the same as for the circles $(P), (Q), (R)$ and the orthocenter is thus the common radical center of all 6 circles $(A'), (B'), (C'), (P), (Q), (R)$, regardless of whether the cevians concur at a point $M$ or not. But then the radical axis of any pair from these 6 circles passes through the orthocenter and the power of the orthocenter to all 6 circles is exactly the same, in particular,

$HA_1 \cdot HA_2 = HB_1 \cdot HB_2 = HC_1 \cdot HC_2$

As a result, [color=blue]the 6 points $A_1, A_2, B_1, B_2, C_1, C_2$ lie on a single circle, regardles of whether the cevians $AA_0, BB_0, CC_0$ are concurrent or not[/color]. In the original thread [url=http://www.mathlinks.ro/Forum/topic-3535.html]Geo4 [the points A1, A2, B1, B2, C1, C2 lie on a circle][/url], grobber arrived to the same conclusion. Gotcha, Darij.  :D

[quote="yetti"]in particular,

$HA_1 \cdot HA_2 = HB_1 \cdot HB_2 = HC_1 \cdot HC_2$

As a result, [color=blue]the 6 points $A_1, A_2, B_1, B_2, C_1, C_2$ lie on a single circle, regardles of whether the cevians $AA_0, BB_0, CC_0$ are concurrent or not[/color].[/quote]

Sorry, Yetti - same error as Grobber did...

  darij

[quote="darij grinberg"]
Sorry, Yetti - same error as Grobber did...

  darij[/quote]

All right, I see that I am wrong. Darn.

Yetti

Phew!!! Thanks everyone , I think I finally got it.  :D
Let $X,Y,Z$ and $U,V,T$ be the midpoint of $BC,CA,AB$ and $AA_0,BB_0,CC_0$ respectively.
So we easily find that $XU,YV,ZT$ are concurent at point K.
Hence $K$ is also the center of $(A_1A_2B_1B_2)$ and $(B_1B_2C_1C_2)$ and $(C_1C_2A_1A_2)$ and we get the result

[quote="Stun"]Phew!!! Thanks everyone , I think I finally got it.  :D... [/quote]

After a good sleep, I got it as well. The centers of the circles $(P), (Q), (R)$ with the diameters $AA_0, BB_0, CC_0$, are the midpoints of these cevians, hence, they lie on the sides $B'C', C'A', A'B'$ of the medial triangle $\triangle A'B'C'$. From similarity of the triangle $\triangle ABC$ and its medial triangle,

$\frac{C'P}{BA_0} = \frac{B'P}{CA_0},\ \frac{A'Q}{CB_0} = \frac{C'Q}{AB_0},\ \frac{B'R}{AC_0} = \frac{A'R}{BC_0}$

Consequently,

$\frac{C'P}{B'P} \cdot  \frac{A'Q}{C'Q} \cdot \frac{B'R}{A'R} = \frac{BA_0}{CA_0} \cdot  \frac{CB_0}{AB_0} \cdot \frac{AC_0}{BC_0}$

i.e., the cevians $A'P, B'Q, C'R$ of the medial triangle $\triangle A'B'C'$ are concurrent iff the cevians $AA_0, BB_0, CC_0$ of the original triangle $\triangle ABC$ are concurrent. Due to

$HA_1 \cdot HA_2 = HB_1 \cdot HB_2 = HC_1 \cdot HC_2$

the point quadruples $(B_1, B_2, C_1, C_2)$, $(C_1, C_2, A_1, A_2)$, $(A_1, A_2, B_1, B_2)$ are concyclic, and the centers of these 3 circles are $A_3 \equiv B'Q \cap C'R$, $B_3 \equiv C'R \cap A'P$, $C_3 \equiv A'P \cap B'Q$, respectively. Iff the cevians $A'P, B'Q, C'R$ concur, the centers $A_3, B_3, C_3$ are identical and because the 3 circles pairwise intersect in at least 2 points, they cannot be concentric, hence, they are identical.


induction
advanced fields
advanced fields unsolved

The statement is here http://eridian.xhost.ro/autom.jpg
What do the hints mean, exactly?
Is it possible to apply induction in order to solve the problem?
Thanks!

[quote="tomsofts"]The statement is here ...[/quote]

No, it isn't. owk

[img]http://eridian.xhost.ro/autom.jpg[/img]
Hope it works now


inequalities
search
AMC
USA(J)MO
USAMO
graph theory
inequalities theorems

[quote="rem"]Hello,

I heard that one has to prove Muirhead on a competition, and you can always site AM-GM and other trivial inequalities. So could some one post here which inequalities can be just sited on a contest, and which have to be proved?

Thank you.[/quote]

Hey rem, I think it mostly depends on the contest.For example I think in the IMO you can easily use Muirhead and other rather famous ineq's without any proof. But if you use that kind of inequalities in let's say a regional olympiad  you  will probably have hard time with jury :lol: .

It does depend in the contest, but I think that if you can attribute a theorem to a particular person (i.e. [b]Schur's[/b] inequality, [b]Holder[/b] or [b]Muirhead[/b]) then you should always be able to quote them. I see no reason why a jury in a regional olympiad would have a problem with someone quoting Muirhead- it's not a particularly obscure theorem.

[quote="tim1234133"]I see no reason why a jury in a regional olympiad would have a problem with someone quoting Muirhead- it's not a particularly obscure theorem.[/quote]

One reason is that if the jury don't have enough qualification you can easily get missunderstood and lose important points. Also the problems in that kind of olympiads are usualy easy and you can avoid using complex theorems and formuls. Believe me that I am telling you this form my own experience that it is better to solve problems as simple as possible especialy in the regional or other similar olympiads. :wink:

I recently read this somewhere (a couple days ago?) that a person in the 2002 IMO was describing their experiences as part of the committee and later grader, and a student sited an obscure graph theory theorem that trivialized (maybe not trivialized but helped a lot) on a problem. No one had heard of it, and for days the committee tried to prove it, and no one could construct a counterexample. Finally, a counterexample was found and a 0 was given on the problem; however, according to the writer, he would have recieved a 7 if someone could prove it. I think this was on an IMO compendium or something, really sorry I can't remember where I found this but try a google search :wink:

Half-way down the (very long) first post of this thread: http://www.mathlinks.ro/Forum/viewtopic.php?t=15623

I think Tiks' point is that in some smaller olympiads, (not IMO) there will be much less leeway as to what is considered 'well-known'.

[quote="jb05"]Does anybody have specific information about Muirhead (especially on the USAMO)? Some people have told me it's okay, some people that you have to prove it. Since the proof is NASTY and it is extremely well known, I would like some more info.[/quote]

I am sure you don't need to prove it in IMO.Although,I am not sure about USAMO(never participated :blush: ), but I think that you don't need to prove it there, too.


function
search
topology
real analysis
real analysis unsolved

let $ X$ be a compact metric space and $ f: X\to X$ be a function  that

$ d(f(x),f(y))\geq d(x,y)$  for all $ x\neq y$ , show that $ f$ is isometric.


Place the integers $ 1,2,3,...,n^2$ (without duplication) in any order onto an $ n$ x $ n$ chessboard, with one integer per square. Show that there exist two adjacent entries whose difference is at least $ n\plus{}1$. Adjacent means horizontally or vertically or diagonally adjacent.

I think it's just because 1, 1+n, 1+2n, ... and 1+(n-1)n < n^2 for n > 2.

So starting from 1, you need the shortest path (counting diagonals, horizontal, vertical) from 1 to n^2 to be at least n, but in n by n board, the shortest distance from a square block to another is at most n-1.

And for n = 2, it's clear.

[hide="Proof"]A path can be traced from square $ 1$ to square $ n^2$ in such a way that the minimal amount of squares is covered along this path. The length of this path, in squares, is at most $ n\minus{}2$. This means the number of differences between square $ 1$ and square $ n^2$ is at most $ n\minus{}1$.

The difference between $ n^2$ and $ 1$ is $ n^2\minus{}1$, or $ (n\plus{}1)(n\minus{}1)$.

On average, the difference between any two adjacent squares on this path is $ \frac{(n\plus{}1)(n\minus{}1)}{n\minus{}1}\equal{}n\plus{}1$.

Therefore, at least one pair of adjacent squares exists such that their difference is $ n\plus{}1$ or greater.[/hide]


geometry
circumcircle
projective geometry
conics
hyperbola
geometry proposed

Let H and O be the orthocenter and the circumcenter of a triangle ABC, respectively, and let D, E, F be the orthocenters of triangles AHO, BHO, CHO. Prove that the midpoints of the segments AD, BE, CF are collinear.

Ah yes, and I've almost forgotten to say that the line through these midpoints touches the nine-point circle of triangle ABC at the point X(115) (the center of the Kiepert hyperbola). This wasn't part of the olympiad problem, but it helps in the solution.

  Darij

I proved the following LemmaRL041104 from which a more general results of that claimed follows.
Actually O no needs to be the circumcenters and H be the orthocenter and again the midpoints of the segments AD, BE, CF remain collinear. Where D, E and F need to be redefined in a little bit more general way.
I don't have time now to translate in english my items: I'll do it later and will post here.

LemmaRL041104

Let O be a oint on the plane of triangle  ABC and let a,b and c the parallel lines from O to BC, CA and AB respectively. If d is a line through A wich intersect BC at D and l is a parallel to d from B, let R=b^l, T=a^l and S=c^l, then ST/TR=CD/DB.

Proof.

As RTO~ADC then RT/OT = AD/CD. As OTS~BDA then OT/TS = BD/DA. From this the thesis.

Now one can prove the following LemmaRL041104a

Let O be a oint on the plane of triangle  ABC and let a,b and c the parallel lines from O to BC, CA and AB respectively and let a', b' and c' three line through A,B and C which are parallel eachother. Let a'^a=A1, a'^b=A2 and a'^c=A3, and Bi and Ci are defined similarly (see fig. 1 attached).
If X=AB^A1C3 and Y=AC^A1B2 then XY//CC3.

Proof
If C'3=c'^B2A1 then C'3C=B3B. Indeed B2B1/C1C'3 = B1A1/A1C1 = B2B1/B1B3 which leads to C1C3'=B1B3 from which, as CC1=BB1, C'3C=B3B=A3A=C3C' where C'=c'^AB.
Now, from various similarity, it results XC3/C3A1 = C'C3/A3A1 = AA3/A3A1 = YC'3/C'3A1 which, by Thales implies the thesis.

From the latter lemma, appropiately appling Desargues, one can deduce  the fact that midpoints of AA1, BB2, CC3 are collinear.

I will give later the details.

The points referred are those you can see in the fig. 2 attacched. I renamed the point to match as much is possible the original problem.

Lets have a look at the triabgles BRE and FSC. From lemma, it holds that BE^FC = oo, BR^FS = X ans CS^ER = Y are collinear. Then, by Desargues, BF, EC and RS are concurrent, lets say, at T (*).
Let, now, have a look at the triangles ABc and DEF. Since AD//BE//FC are concurrents then, by Desargues, R = AB^DE, S = AC^DF and U = BC^EF are collinear.
It is a well known fact that UT bisects BE and all chords of triangle UBE parallel to BE, in particular UT bisects FC. As ASd~CSF, UT bisects  AD  too.

Now what is missing is to prove that the original problem is a corollary of this fact, but this is quite easy.

(*) This fact, by Brianchon, iplies that B,C,D,E,T,F stay on a conic. Perhaps this is the hyperbola mentioned by Darij  :maybe: .

Sorry for bumping a thirteen year old topic, but this problem seems nice.

Note that the midpoint of $AD$ is the antipode of the midpoint of $OH$ in the nine point circle of $AOH$, so all we need to show is that the nine point centers of $AOH,BOH,COH$ are collinear. This in turn is equivalent to showing that they are coaxial, since they share the midpoint of $OH$ asa common point. But the center of the rectangular circumhyperbola through $O$ is on the nine point circle of all the three triangles. (Since $H$ is always on this hyperbola). So they are coaxial and we are done.

Just for fun, here's a solution using Cartesian coordinates.
Suppose that hyperbola is $xy=1$ and the point $X$ on it has a parameter $t$ when $X=(t,\frac 1t)$. Suppose the parameters of the points are denoted by the smaller letter. Then $habc=-1$ (well known). We also have $daho=-1$, so the midpoint of $AD$ is $\frac{1}{2} \left( a + \frac{bc}{o}, \frac 1a + \frac{o}{bc} \right)$. So the slope of the line joining the midpoints of $AD, BE$ is $\frac{o}{abc}$. Thus we are done.

Note that both the proofs show the fact that $O$ doesn't need to be the circumcenter of $ABC$ and the problem is true for any point replacing $O$.


Prove that the square of any odd multiple of $ 3$ is the difference of two triangular numbers.

The difference between the nth and n+1th triangular numbers is n+1, so every number can be expressed as the difference of 2 triangular numbers.


There are 2n+1 pirates.The distance between each one of them is different.
Everyone has only one bullet left in their guns.Now each one is shooting bullet to that pirate who is standing nearest to him.

Prove that there would be no one who will be shot by more than 5 pirate's bullets.


Hello!

I know different constructions of real numbers, one with Dedekind's cuts, another one with Cauchy's sequences...

I understand them and they are "pretty", BUT these constructions are not natural for me, I mean they are not intuitive...

I have heard that a construction of real numbers by decimal expansions is possible, and it seems very intuitive then...

But the problem is that arithmetical operations (+,x) are difficult to define formally, and show that R is a field too...

Could everyone give me links or complete proofs with construction of real numbers by decimal expansions please?

Thank you in advance, this subject is wonderful and thinking about constructions of "basic" sets is very interesting :)

Best regards.

PS: I am french, and my english is poor!

Try Spivak's book Calculus. There's a chapter at the end which constructs the real numbers (by Dedekind cuts) and then has a pair of long exercises leading you through alternate constructions (Cauchy sequences and decimal expansions).

In any case, the important theorem is that there is only one complete ordered field up to isomorphism. The real numbers are unique any way we construct them, and the construction merely proves existence.

We don't normally use the decimal expansion construction because it's harder in every way. The ambiguity of decimal representations complicates everything, and arithmetic is hard because we can't just inherit it from the rationals.

I know all this, but what I am looking for is a way to construct real numbers that is intuitive and natural!

And this is because construct them with decimal expansions is a difficult job that I asked this question here: how define arithmetical operations (+,x) on decimal expansions and how prove that one gets a field...

The great advantage of this method is that with decimal expansions, the fact that every nonempty set of real numbers with an upper bound has a least upper bound is "obvious" !

Thanks for the link, I will try to find the book

But I still need your help if someone want to help me :)

If someone can confirm or scan the exercice about construction of real numbers by decimal expansions in Spivak book, it would be so great! Thanks !

Intuitively, what makes us believe that every decimal expansion corresponds to a real number?  Doesn't this depend on already having accepted (on intuitive grounds) the completeness axiom?  We have an idea of what real numbers are and how they behave before we know anything about decimal expansions.  I don't find it especially natural to start with decimal expansions as a foundation for understanding what the real numbers are.

In my mind, I start with an intuitive understanding of what real numbers are supposed to be... an intuitive understanding which is captured by the axioms for the real numbers.  In my mind, the only reason to construct the real numbers at all, using Dedekind cuts or any other method, is just to prove that the axioms for the real numbers are consistent.  (Of course, we expected they would be, but it doesn't hurt to check.)

Kind of philosophical and vague, but that's my two cents.

Is the axiom of upper bound not actually an axiom? By axiom of upper bound I mean that every non-empty set of real numbers bounded upwards has a least upper bound. I always learned this as an axiom in school. Is it not true?

There is always one completeness axiom. There are many equivalent choices for this axiom- least upper bounds are simply the most common choice (and generally easy to work with).


geometry
3D geometry
sphere
vector
abstract algebra
trigonometry
rectangle

Ok, this is a classical idea and you probably know it, but here it is anyway:

A sphere of radius $R$ is charged with the surface charge density given by $\sigma = \vec a \cdot \hat r$, where $\vec a$ is a constant vector, and $\hat r$ is the unit radius-vector at the given point. Find the intensity of the electric field in the center of the sphere.

I'm not sure that I understood the problem well :( .But according to what I understood,I got the answer $E=\frac{aR}{e_{0}}$
What is the answer?

I think it mustn't be empty, because $sigma$ is changing with radius $\sigma=ar$

The sphere is empty, yes. And it's not changing with the radius, but with the direction of the radius vector ($\hat r$ is a unit vector, and $\vec a\cdot \hat r$ is its dot product with the constant $\vec a$).

If r with a hat is r vector/r than is the answer E=4a/3epsilon ?  a is the the module of the vector

thats the right answer :D .now for extra points find it for a solid sphere. also show that the resulting charge distribution foe a hollow sphere can be represented as the sum of two solid charged spheres with a small distance between their centres.

You've mentioned the neat trick :) But the answer isn't correct, it doesn't have the 4...

If it has the three and not the four then could you post the sollution?

By examining the dot product we can notice that the charge density is $\sigma (\phi) = a \cos\phi$, so the sphere can be considered to be made up of rings of various radii and charges. We have: \[E(\phi) = \frac{kq(\phi) z}{R^{3}},\] which is the well-known formula for the el. field of a charged ring; a ring corresponding to the angle $\phi$ has a charge $q(\phi) = 2a\pi R^{2}\sin\phi\cos\phi\text d\phi$, and is at height $z = R\cos\phi$.

Now, plugging these back into $E(\phi)$ and integrating over $(-\pi/2, \pi/2)$ while taking care of the changing of the sign of $\sigma$, we get \[E = \frac{a}{3\varepsilon_{0}}.\]

No, the dimensions would be wrong then (for a start...)

than the surface on which q(phi) is is 2 pi sqr(R ) sin phi d(phi). how come?

The surface is a rectangle; one side is $R \text d\phi$, and the other one is $2r\pi$, where $r = R\sin\phi$.

I hope so :) Personally, I think I've explained horribly... but if you have any questions do ask!


Just out of curiosity, are we allowed to use pen in the AMC 10?

Well the forms are scantrons, so I would think, no.

I'm sure that you would be allowed to use a pen to solve the problems, however, you must use a #2 pencil to bubble in your answers.


We have
$ 8\sum{\frac{a^2\plus{}b^2}{a\plus{}b}}\plus{}16\sum_{cyc}{\frac{ab}{a\plus{}2b\plus{}c}}\leq 16(a\plus{}b\plus{}c)$
It is equivalent to
$ \sum{[3a^3bc\plus{}21a^2bc(b\plus{}c)\plus{}77ab^2c^2\plus{}2b^2c^2(b\plus{}c)](a\minus{}b)(a\minus{}c)}$
$ \plus{}\sum{a^2b^2(a\plus{}b\plus{}3c)(b\minus{}c)^2}$
$ \plus{}128(abc)^2(a\plus{}b\plus{}c)$
$ \geq0$.

Could you write a full detailed solution,
such that everyone would understand all your steps?
Thank you very much.

After expanding using software, we can see this is not a symmetric polynomial(it is cyclic, but not symmetric),this is
quite surprising, so can't be solved by uvw method. BTW,seems that change RHS ‘25’to '24', still holds.
[quote=Inequalities Master]Let $ a,b,c\geq 0$, no two of which are zero. Prove that:
$ 8\sum{\frac {a^2 \plus{} b^2}{a \plus{} b}} \plus{} 64\sum_{cyc}{\frac {ab}{a \plus{} 2b \plus{} c}}\leq 25(a \plus{} b \plus{} c)$[/quote]

this solution is wrong, after full expanding ,they are not equivalent inequality.
[quote=fjwxcsl]$ 8\sum{\frac {a^2 \plus{} b^2}{a \plus{} b}} \plus{} 64\sum_{cyc}{\frac {ab}{a \plus{} 2b \plus{} c}}\leq 25(a \plus{} b \plus{} c) (*)$

is equivalent to

      $ \sum{[18a^4(b\plus{}c)\plus{}180a^3bc\plus{}210a^2bc(b\plus{}c)\plus{}572ab^2c^2\plus{}20b^2c^2(b\plus{}c)](a\minus{}b)(a\minus{}c)}$

     $ \plus{}21(a\plus{}b\plus{}c)(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2\plus{}64\sum{a^2b^2(a\plus{}b\plus{}3c)(b\minus{}c)^2}\geq0$,

the (*) holds![/quote]


calculus
integration
calculus computations

[quote="vijaymenon"]if [size=150][b]f'(X)[/b] [/size] [b]=[size=150] f(X) + integral 0 to 1 of f(X) dx then integral f(X) dx = [/size] :?: [/b]

please give me a solution..[/quote]

The question is $ f'(x)\equal{}f(x)\plus{}\int_{0}^{1} f(x) dx.....................(1)$ then $ \int f(x) dx\equal{}?$

[b]Solution:[/b]
Let $ k\equal{}\int_{0}^{1} f(x) dx$ as a constant.

Differentiate both sides of [b](1)[/b] with respect to $ x$ , then we get $ f''(x)\equal{}f'(x)$

$ \frac{d}{dx}(e^{\minus{}x}f'(x))\equal{}e^{\minus{}x}(f''(x)\minus{}f'(x))\equal{}0$

So $ f'(x)\equal{}ce^{x}$, where $ c$ is any constant, which gives $ f(x)\equal{}ce^{x}\minus{}k$ from $ (1)$

Now $ k\equal{}\int_{0}^{1} f(x) dx\equal{}\int_{0}^{1} (ce^{x}\minus{}k) dx\equal{}ce\minus{}k\minus{}c$

$ \Rightarrow k\equal{}\frac{c(e\minus{}1)}{2}$
$ \Rightarrow f(x)\equal{}ce^{x}\minus{}\frac{c(e\minus{}1)}{2}$

So $ \int f(x) dx\equal{}ce^x\minus{}\frac{c(e\minus{}1)}{2}x\plus{}C$, where $ c$ and $ C$ are any constants.

[quote="vijaymenon"]gr8 solution but how did u get the idea of differentiating [img]http://alt2.mathlinks.ro/latexrender/pictures/e/3/8/e3829e5a2157817b62803e68c845c0f974c7a119.gif[/img] this.please help[/quote]

thats  something  which we come across quite often .

Hi :) !!!

i think that the result of [b]stephencheng[/b] is not enough and is correct if $ f$ is the class $ C^2$ ( because we just know that $ f$ are one time derivable) and thanks !!

my answer :

$ f(x)\equal{}0$ is one solution obvious...

the other:

$ f'(x) \equal{} f(x) \plus{} a$ with $ a\equal{} \int_0^1 f(x)dx$

the soulution of differentielle equation : $ y'\minus{}y \equal{} a$ are the fuctions: $ x \rightarrow Ae^x \minus{} a$ with $ A$ is a real constante.

than $ f(x) \equal{} Ae^x \minus{} a$

and $ a\equal{}  \int_0^1 f(x) dx \equal{} Ae \minus{} a \minus{}A \Rightarrow  a \equal{} \frac{A(e\minus{}1)}{2}$

than $ f(x) \equal{} Ae^x \minus{} \frac{A(e\minus{}1)}{2}$

the rest are easy ...

PS: i didn't talk by the second derivate ....


the sequence (an) is difined  by a1>0 and a_{n+1}=c.an²+an where c is a costant.prove that:
a) an >=sqrt(c^{n-1}.n^n.a1^{n+1})
b)a1+a2......+an >n(n.a1-1/c)   ,where n is a naturel number


modular arithmetic
quadratics
superior algebra
superior algebra unsolved

Prove that the following propositions (i) and (ii) are equivalent

$ (i)$ $ char\mathbb{F}_q\equal{}2$ , there exist a prime number $ p$ and a primitive $ p\minus{}$ root of unity say $ \zeta$ such that $ \deg Irr\left(\zeta\plus{}\zeta^{\minus{}1},\mathbb{F}_q\right)\equal{}\frac{p\minus{}1}{2}$ and $ \mathbb{F}_{q^n}\equal{}\mathbb{F}_q\left(\zeta\plus{}\zeta^{\minus{}1}\right)$

$ (ii)$ There exist a prime $ p$ such that $ n\equal{}\frac{p\minus{}1}{2}$ and at least one of the following holds:

   $ (a)$ 2 is primitive element of $ \mathbb{Z}_{2n\plus{}1}^{\star}$ or

    $ (b)$ $ 2n\plus{}1\equiv 3 \pmod{4}$ and $ 2$ is a generator of the quadratic residues of $ \mathbb{Z}_{2n\plus{}1}$ .

Please give all the details of the proof.

Can anyone post a proof for this proposition? It seems not very difficult.

Thanks a lot,

Alexandros


Hmm...is it just me, or does the "Log me on automatically each visit" function not work on AoPS right now?

its probably just you, try deleting the artofproblemsolving cookie and that should fix your problem.

Yep, good call, everything works now. Thanks.

Actually, that problem is happening to me too. I'd get logged out randomly. Let me check something...

I found the cookie, but it said nothing conclusive. I can venture a guess that somehow the AOPS forum cookie expiration setting was made too low - so it'd expire after half a day or so, and you have to relogin - but this is only a guess.

No, since I checked the forum right after I had clicked on the option and then closed the browser.

But everything works now. :)


search
number theory unsolved
number theory

Find all $n\in N$ such that

$x+y+z+w = n\sqrt{xyzw}$

has a solution $x,y,z,w\in N$

Search the forum for Harazi's solution (I think).

I am sure it use the same idea as I have written in http://www.mathlinks.ro/Forum/viewtopic.php?t=113734


complex numbers
calculus
calculus computations

or $ x_{n}\equal{}\sqrt{5 \plus{}\sqrt{13\plus{}x_{n\minus{}1}}}$ and $ x_{0}\equal{}0$
at first you need to prove that there exist limit.
and after you suppose that $ |x_{n}\minus{}x_{n\minus{}1}| \to 0$ as $ n \to \infty$
and your last step will be to solve this equation:
$ x\equal{}\sqrt{5 \plus{}\sqrt{13\plus{}x}}$
[hide="answer"]
3
[/hide]

I make $ x\equal{}\sqrt{5\plus{}\sqrt{13\plus{}x}}$
$ (x^2\minus{}5)^2\minus{}13\minus{}x\equal{}0$
$ x^4\minus{}10x^2\minus{}x\plus{}12\equal{}0$
$ Xn<1\plus{}\sqrt{5\plus{}\sqrt{13}}$
$ Xn\plus{}1\equal{}\sqrt{5\plus{}\sqrt{13\plus{}Xn}}<1\plus{}\sqrt{5\plus{}\sqrt{13}}$
I am stuck :roll:

Well, Mathematica gives some ugly solutions. One is integer and equals 3, two are suspiciously complex (one with negative real part and one with positive but smaller than $ \sqrt {5 \plus{} \sqrt {13}}$) and one is real but negative. And by "suspiciously complex" I mean that numerically, their imaginary parts are of the order $ 10^{\minus{}6}$. Therefore, the answer should be $ x \equal{} 3$ (and that is backed up with some numerical approximation).


ΑΣΚΗΣΗ 1
Να βρεθούν όλα τα πολυόνυνα με πραγματικούς συντελεστές ώστε να ικανοποιούν την παρακάτω σχέση για όλα τα $x \in R$.
$P(2x^2-1)=\frac{(P(x))^2}{2} -1 $.

Νομίζω ότι αυτή είναι άλυτη πάνω από μια βδομάδα. Θα μπορούσες να δώσεις ένα hint Στέργιο?

ok, tha doso kai ligi akoma boitheia.... to p(1) einai ritos i arritos ??

Afu to $P(x)$ einai polionimo eukola pairnume apo ti dosmeni sxesi oti to $P$ eite einai artio eite einai peritto. An einai peritto, dld $P(x)=-P(-x)$ tote $P(0)=0$.
$x=0$ stin arxiki $\rightarrow P(-1)=-1$
$x=-1$ stin arxiki $\rightarrow P(1)=-\dfrac{1}{2}$
pu erxetai se antithesi me tin "peritotita" tu $P$.
Ara $P$ artio kai $P(1)=P(-1)$.
$x=-1$ stin arxiki $\rightarrow P(1)=1+\sqrt{3}=P(-1)$ h $P(1)=1-\sqrt{3}=P(-1)$.
Theoro tin proti periptosi pu einai akrivos idia me ti deuteri.
$x=0$ stin arxiki $\rightarrow (P(0))^2=(1+\sqrt{3})^2$
Paragogizo tin arxiki sxesi kai pairno $P'(2x^2-1)4x=P(x)P'(x)$(1), ap' opu $P'(0)=0$. Theoro akoluthia $a_n$
me $a_1=0$ kai $a_{n+1}=\sqrt{\dfrac{1+a_n}{2}}$. Einai apli epagogi na deiksei kaneis oti $a_n=\cos(\dfrac{\pi}{2^n})$ . Epomenos i $a_n$ einai gnisios monotoni kai diafori tu 0. Akoma apo tin (1) exume:
$P'(a_n)4a_{n+1}=P(a_{n+1})P'(a_{n+1})$. An deiksume oti $P(a_{n+1})$ diaforo tu 0 exume teleiosei afu tote exume $P'(a_n)=0$ gia ola ta $n$, pu simainei oti $P'(x)=0$ dld $P(x)$ stathero.
Esto oti $P(a_n)=0$ gia kapoio $n$. Apo tin arxiki sxesi exume oti $P(a_n)=\dfrac{(P(a_{n+1}))^2}{2}-1$. Tote $P(a_{n-1})=-1$, $P(a_{n-2})=-\dfrac{1}{2}$ kai profanos oso proxorame pros ta piso tha pairnume ritus, pragma pu simainei oti $P(a_1)=P(0)$ ritos,atopo afu $(P(0))^2=(1+\sqrt{3})^2$

Telika $P$ stathero kai $P(x)=1+\sqrt{3}$ h $P(x)=1-\sqrt{3}$

[quote="Anto"]Geia sas, tora mporesa na mpo sto internet proti fora. Signomi gia tin kathisterisi alla den briskomai spiti.

Tha doso tin apantisi kai prospathiste ti lisi...

To p(x) einai stathero...[/quote]

me prolaves prin kano post tin lisi!

Oraia lisi  ;)  ksero alles 2 alla kai i dio diaforetikes, sas grafo tin diki mou.
Esto oti den einai stathero.  :o
Os polionimo to P einai paragogisimo. 'Esto oti $P'(x)=(x-1)^r \cdot q(x)$, opou r fisikos mporei kai miden,alla q(1) pote miden. Tote paragogizontas kai antikathistontas, exo oti : $4x(2x^2-2)^r q(2x^2-1)= P(x)(x-1)^r q(x)$. Alla apo tin ipothesi to q(x) den einai to mideniko kai epeidi milame gia polionima kai i pano sxesi isxuei gia apeira x. Pairno oti $2^{r+2}x(x+1)^r q(2x^2 -1)=P(x)q(x)$ gia ola ta x. Tote omos to P(1) einai akeraios arithmos (apla balte stin teleutaia sxesi apou x to 1).
Antifasi dioti stin arxi sxesi an baloume opou x to 1, pairnoume oti to x einai arritos.

opos sinithizeis teleutaia, pio apli lisi apo mena!
paromoio stil :

$P$ polionimo tetoio oste $P(x^2-1)=(P(x))^2-1$. Na vrethei to $P$.

Τέθηκε στη Σερβία στο TST φέτος. Πολύ καλή άσκηση . Υπάρχει λύση στο φόρουμ αλλά καλό θα είναι να δούμε εδώ διαφορετικές.

[quote="galeos"]tin allaksa ligo tora![/quote]

Θα προτιμούσα να μην την άλλαζες. Να βλέπαμε μια ΕΛΛΗΝΙΚΗ λύση στο ατόφιο πρόβλημα

[quote="silouan"]

Θα προτιμούσα να μην την άλλαζες. Να βλέπαμε μια ΕΛΛΗΝΙΚΗ λύση στο ατόφιο πρόβλημα[/quote]
to arxiko provlima  periorize tin idia sxesi se polionima peritu vathmu. k efoson exei idi lithei sto forum to ekana na exei ligo perisotero endiaferon(oso endiaferon mporei na exei meta ti lisi tis proigumenis ;) :roll:  )

Geia sas,
(den exo poli xrono na parakoloutho to forum, giauto na me sinxoreite pou kathistero)
Loipon, oso afora to problima pou these o galeos. Den to brika katholou eukolo kai den to exo lisi.
Grafo parakato ti proodo pou exo kanei :
An o bathmos tou P einai perritos tote einai "sxetika" eukolo na deiksoume oti $P(x) =x$.(to afino auto os askisi gia ta ipolipa meli)
An o bathmos tou P einai artios arithmos, tote an einai 0 exoume statheres liseis.
An einai dinami tou 2, tote mporoume na deiksoume oti iparxei akrivos mia lisi, i $f(f(...f(x)...))$ opou $f(x)=x^2-1$.
Kai genika mporoume na deiksoume oti gia kathe bathmo iparxei to poli mia lisi.
Mia sigkrisi sinteleston se 6 bathmou polinomio, odigei se antifasi. Pragma pou me odigise na ipotheso (eikaso) oti an o bathmos tou P einai artios tote prepei na einai dinami tou 2.

Sizitisa to problima auto kai me allous lites kai den mporesan oute auti na to lisoun.

Antrea, exeis lisi se auto to problima ? i to egrapses apo periergia ?

Xereto,
Stergios

[quote="Anto"]
Antrea, exeis lisi se auto to problima ? i to egrapses apo periergia ?
[/quote]
distixos to deutero  :yinyang: Stin arxi nomiza oti den eixe tpt idiaitero, alla argotera "eida" oti mporume na parume apeires liseis k sineiditopoihsa oti den einai kai toso trivial. :blush:

Loipon, istera apo ola auta pou eixa grapsei stamatisa na to skeftomai...
alla simera to proi, enas filos edo, nomizo oti einai grammenos sto site me to nickname Fiachra, brike mia lisi sto problima. Kai pragmati i eikasia mou itan sosti. An to n einai artio , kai oxi miden, tote einai dinami tou 2. Kai an perittos tote P(x)=x. Einai oles i liseis. I lisi tou, pou einai teleios steixeiodeis, apoteleitai apo ta parakato bimata, prota deixnei oti i sintelestes tou P(x) einai akeraioi arithmoi. Meta xrisimopoiei mia morfi epagogis kai apodeikniei oti an $2^a | n$ kai $n\geq 2^{a+1} +1$ tote $2^{a+1} | n$. Entiposiastika poli dioti i lisi tou , otan tin diavazeis, fainete poli apli na tin skefteis kai poli profaneis... anyway an thelete olokliri tin lisi mporo na tin grapso, alla den to brisko katholou endiaferon, apla ergapsa tin idea. Opoios endiaferete parapano tha mporouse na to apodeiksi apo monos tou, dioti apotelei mia poli kali askisi.

Stergios


let XOY be a fixed angle .and (C) is a fixed circle in the interior of it.

draw a circle wich is tangent to OX,OY,and (C).

oh... yes dear Arne.but that was because of seeing my solution.

now im waiting for u nice solutions. ;)

Is this problem open ? If yes we shall cry for humanity for not being able to solve this problem ,but i can assure you that this problem is not Open ,therefore it should not be in this section .:

As for the problem ,this is one of the  problems which ,i have personally disccused it with you lots of times ,so are you playing with these guys here man , time is valuable .

You will find nice conversations here ,mate:

http://www.artofproblemsolving.com/Forum/topic-46945.html

shhhhhhhhhhhhhhh,

i dont need u informations :mad:


A certain kind of tree increases in height by 25 percent each year. It was 2.5 feet tall when it was planted in January 2000. Approximately how tall will the tree be in December 2008?

[hide=" click to reveal hidden content"]

$ 2.5 \cdot(1.25)^{8} \equal{} 14.90116119...$, or qapproximately $ 14.9$ metres.

[/hide]

[quote="AndrewTom"]

$ 2.5 \cdot(1.25)^{8} \equal{} 14.90116119...$, or qapproximately $ 14.9$ metres.

[/quote]

Closer to $ 9$ years pass, so shouldn't we have $ 2.5 \cdot(1.25)^{9}$.

[quote="PhireKaLk6781"]True, but it is not yet $ 9$ years, and the question does not specify on sub-annual growth amounts. It could suggest continuous growth, but I doubt that we're dealing with tropical trees here, as that isn't the tree the normal person, or at least the normal math-question writer, would think of.[/quote]

The question asks for an approximation.


Let $ P(x) \equal{} (x \minus{} 1)(x \minus{} 2)(x \minus{} 3)$. For how many polynomials $ Q(x)$ does there exist a polynomial $ R(x)$ of degree 3 such that $ P(Q(x)) \equal{} P(x) \cdot R(x)$?

$ \textbf{(A)}\ 19\qquad
\textbf{(B)}\ 22\qquad
\textbf{(C)}\ 24\qquad
\textbf{(D)}\ 27\qquad
\textbf{(E)}\ 32$

Check out the AMC 12A math jam from last year, it goes over the last five problems on the AMC 12, so this one is obvious included.

[quote="10000th User"]Considering this is a AMC test question, Hamster1800 perhaps you should try another method...[/quote]

I would....but I don't see the other method......

[hide="trying something else"]
I'm going to put the polynomials in the form of (x-a)(x-b) and (x-c)(x-d)(x-e) in hopes of getting an easier system.
so....

$((x-a)(x-b)-1)((x-a)(x-b)-2)((x-a)(x-b)-3) = (x-1)(x-2)(x-3)(x-c)(x-d)(x-e)$

OH!!! dur!!!

ok...say $y=Q(x)$

$(y-1)(y-2)(y-3) = (x-1)(x-2)(x-3)(x-c)(x-d)(x-e)$
ok. Now each factor on the left is the product of two on the right.

I think I'm still missing something....The answer that I got is 90....but that's way to high. Am I at least on the right path now though?

Oh...ok I see something else.

It only works when the the two a values for the two (x-a) factors add up to the same number. Then the products have to be 3 consecutive numbers.

So....I know how to do it but I don't know how to quickly just eliminate all the ones that won't work.
[/hide]
[/hide]

[hide]
First we notice that $Q(x)$ is degree two . Then we know that $Q(1),Q(2),Q(3)$ can only be $1,2,3$ since it must satisfy $P(Q(1))=P(Q(2))=P(Q(3))=0$ and $P(x)$ is degree 3 . So now , if $Q(1)=1$ , then $Q(2)=Q(3)=2$ or $=3$ because $Q(x)$ is quadratic function and the most is only two same roots. So this happens to be one of the way $Q(x)$ can be formed.Similarly we can work out such permutation and get $3\times 3P2=18$ . But we also have another case which is all $Q(1),Q(2),Q(3)$ is different which has $3!$ method . So total is $\boxed{24}$ ways  ;)
[/hide]

The answer is actually B) 22.
[hide]Since $R(x)$ has degree three, then $P(x)\cdot R(x)$ has degree six. Thus, $P(Q(x))$ has degree six, so $Q(x)$ must have degree two, since $P(x)$ has degree three. \[ P(Q(1))=(Q(1)-1)(Q(1)-2)(Q(1)-3)=P(1)\cdot R(1)=0, \] \[ P(Q(2))=(Q(2)-1)(Q(2)-2)(Q(2)-3)=P(2)\cdot R(2)=0, \] \[ P(Q(3))=(Q(3)-1)(Q(3)-2)(Q(3)-3)=P(3)\cdot R(3)=0. \] Hence, we conclude $Q(1)$, $Q(2)$, and $Q(3)$ must each be $1$, $2$, or $3$. Since a quadratic is uniquely determined by three points, there can be $3*3*3=27$ different quadratics $Q(x)$ after each of the values of $Q(1)$, $Q(2)$, and $Q(3)$ are chosen.

However, we have included $Q(x)$ which are not quadratics. Namely, \[ Q(1)=Q(2)=Q(3)=1 \Rightarrow Q(x)=1, \] \[ Q(1)=Q(2)=Q(3)=2 \Rightarrow Q(x)=2, \] \[ Q(1)=Q(2)=Q(3)=3 \Rightarrow Q(x)=3, \] \[ Q(1)=1, Q(2)=2, Q(3)=3 \Rightarrow Q(x)=x, \] \[ Q(1)=3, Q(2)=2, Q(3)=1 \Rightarrow Q(x)=4-x. \] Clearly, we could not have included any other constant functions. For any linear function, we have $2\cdot Q(2)=Q(1)+Q(3)$. Again, it is pretty obvious that we have not included any other linear functions. Therefore, the desired answer is $27-5=\boxed{22}$.[/hide]

[quote="Smartguy"][quote="towersfreak2006"]
[hide]Since $ R(x)$ has degree three, then $ P(x)\cdot R(x)$ has degree six. Thus, $ P(Q(x))$ has degree six, so $ Q(x)$ must have degree two, since $ P(x)$ has degree three.
\[ P(Q(1)) \equal{} (Q(1) \minus{} 1)(Q(1) \minus{} 2)(Q(1) \minus{} 3) \equal{} P(1)\cdot R(1) \equal{} 0,
\]

\[ P(Q(2)) \equal{} (Q(2) \minus{} 1)(Q(2) \minus{} 2)(Q(2) \minus{} 3) \equal{} P(2)\cdot R(2) \equal{} 0,
\]

\[ P(Q(3)) \equal{} (Q(3) \minus{} 1)(Q(3) \minus{} 2)(Q(3) \minus{} 3) \equal{} P(3)\cdot R(3) \equal{} 0.
\]
Hence, we conclude $ Q(1)$, $ Q(2)$, and $ Q(3)$ must each be $ 1$, $ 2$, or $ 3$.[/hide][/quote]

Why must Q(x) be 1,2,3?[/quote]

They don't have to be, that's just one of the possible options which doesn't give you a quadratic polynomial for $ Q(x)$. Can you specify which part of his solution you are unclear about? There are multiple parts of the solution you're question can be asking about.

[quote="sunehra"]
There are multiple parts of the solution you're question can be asking about.[/quote]

the part i quoted

Why he used P(Q(1)), and P(Q(2)) etc.

[quote="Smartguy"]OK, why did he plug in x=1,2,3??
If you do Q(1,2,3) what will happen?[/quote]If you read carefully, $ x\equal{}1,2,3$ are zeroes of $ P(x)$. Your objective is to look for polynomials $ Q(x)$ satisfying $ P(Q(x))\equal{}P(x)\cdot R(x)$, and now do you see why he plugged in $ x\equal{}1,2,3$? (Obvious hint: They are also zeroes of $ P(Q(x))$!!!  :ninja:

thanks, i get it now. I got a little confused over $ 3^3$ but i see how Q(1), Q(2), Q(3) each can be 1, 2 or 3.

Thanks everyone

$P(Q(x)) = P(x) \cdot R(x) \implies $ $(Q(x)-1)(Q(x)-2)(Q(x)-3)=$ $(x-1)(x-2)(x-3)R(x)$. From here, we can conclude that $Q(1)$ is either $1,2,3$, $Q(2)$ is either $1,2,3$, and $Q(3)$ is either $1,2,3$, yielding $3^3=27$ possibilities but there's a catch! Because the degree of $R$ is $3$, the degree of $1$ must be $2$, so if $Q(1)=Q(2)=Q(3)=1$, $2$, or $3$, then $Q$ is not a quadratic, but a horizontal line. Similarly, if $Q(1)=1,Q(2)=2,$ and $Q(3)=3$, then we have the line $y=x$ or if $Q(1)=3,Q(2)=2,Q(3)=1$, we have the line $y=-x+4$. In essence, so long as $Q$ satisfies that $Q(1),Q(2),Q(3) \in \{ 1,2,3 \}$ and not one of the five cases highlighted above, then it satisfies the given equation. Hence, there are $27-5 = \boxed{22}$ possibilities for $Q$.

[hide=another sol?]Let $Q(x)=ax^2+bx+c$ as it has a degree of at most $2$ since $P(x) \cdot R(x)$ has a degree of $6$. Thus, we have $(ax^2+bx+c-1)(ax^2+bx+c-2)(ax^2+bx+c-3)=(x-1)(x-2)(x-3)(R(x))$. Plugging in $x=1,2,3$, we get the equations $(a+b+c-1)(a+b+c-2)(a+b+c-3)=0, (4a+2b+c-1)(4a+2b+c-2)(4a+2b+c-3)=0, (9a+3b+c-1)(9a+3b+c-2)(9a+3b+c-3)=0$. Choosing one condition from each, we get that there are $27$ possible conditions. However, we cannot have $a=0$, because this will force the degree of $R(x)$ to not be $3$. The cases where this happens is with $a+b+c=4a+2b+c=9a+3b+c=1, a+b+c=4a+2b+c=9a+3b+c=2, a+b+c=4a+2b+c=9a+3b+c=3, a+b+c=1; 4a+2b+c=2; 9a+3b+c=3$, and $a+b+c=3; 4a+2b+c=2; 9a+3b+c=3$. To get these cases, we do: [hide]Basically, let $a+b+c=t_1, 4a+2b+c=t_2, 9a+3b+c=t_3$. With the first two equations, we have $3a+b=t_2-t_1$ and with the last two equations, we have $5a+b=t_3-t_2$. Note that if $t_2-t_1=t_3-t_2$, then this will force $a$ to be $0$. The only cases where this happens is $(t_1,t_2,t_3)=(1,1,1), (2,2,2), (3,3,3), (1,2,3), (3,2,1)$.[/hide]These make up $5$ cases, so we have $\boxed{22}$. I have a feeling that this solution might be flawed in some way. Apologies if I missed something or messed something up.[/hide]

Clearly $Q(x)$ is a quadratic polynomial we have $P(Q(1))=P(Q(2))=P(Q(3))=0$ so $Q(1), Q(2), Q(3) \in \{1,2,3\}$ but also $Q(1) =Q(2)=Q(3)$ is not permissible since $Q(x)$ is a quadratic polynomial.\\

A quadratic is uniquely determined by three points so $Q(1),Q(2),Q(3)$ can be from  $\{1,2,3\}$ only now $3 \times 3 \times 3=27$ ways are there to place numbers

But exclude cases where all are same that is $3$\\

And there are just two cases where it is linear\\

$Q(1)=1,Q(2)=2,Q(3)=3 ,Q(x)=x$ and $Q(1)=3, Q(2)=2 ,Q(3)=1 , Q(x)=4-x$

So $27-3-2=\boxed{22}$ such polynomials are there


given dat,if a+b+c=0,then a3+b3+c3=3abc.show dat (2+rt.5)^1/3  +(2-rt.5)^1/3 is a rational number.

[quote="williams"]given dat,if a+b+c=0,then a3+b3+c3=3abc.show dat (2+rt.5)^{1/3 } +(2-rt.5)^{1/3} is a rational number.[/quote]

The question is that:

given that,if $ a\plus{}b\plus{}c\equal{}0,$ then $ a^3\plus{}b^3\plus{}c^3\equal{}3abc$.show that $ (2\plus{}\sqrt 5)^{1/3}  \plus{}(2\minus{}\sqrt 5)^{1/3}$ is a rational number.

Is it right?

[hide="The first one"]
$ a\plus{}b\plus{}c\equal{}0 \implies a\plus{}b\equal{}\minus{}c \implies (a\plus{}b)^3\equal{}\minus{}c^3$
$ \implies a^3\plus{}b^3\plus{}c^3\equal{}\minus{}3ab(a\plus{}b)\equal{}3abc$
since $ a\plus{}b\equal{}\minus{}c$
[/hide]

[quote="vinskman"][quote="williams"]given dat,if a+b+c=0,then a3+b3+c3=3abc.show dat (2+rt.5)^{1/3 } +(2-rt.5)^{1/3} is a rational number.[/quote][/quote]

The question is that:

given that,if $ a \plus{} b \plus{} c \equal{} 0,$ then $ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3abc$.show that $ (2 \plus{} \sqrt 5)^{1/3} \plus{} (2 \minus{} \sqrt 5)^{1/3}$ is a rational number.

Is it right?

[hide="The first one"]
$ a \plus{} b \plus{} c \equal{} 0 \implies a \plus{} b \equal{} \minus{} c \implies (a \plus{} b)^3 \equal{} \minus{} c^3$
$ \implies a^3 \plus{} b^3 \plus{} c^3 \equal{} \minus{} 3ab(a \plus{} b) \equal{} 3abc$
since $ a \plus{} b \equal{} \minus{} c$
[/hide]

show that $ (2 \plus{} \sqrt 5)^{1/3} \plus{} (2 \minus{} \sqrt 5)^{1/3}$ is a rational number.

[hide=" The second one "]

Let $ a\equal{}(2 \plus{} \sqrt 5)^{1/3},b\equal{}(2 \minus{} \sqrt 5)^{1/3}$

Since
$ a^3\plus{}b^3\equal{} 2 \plus{} \sqrt 5\plus{}2 \minus{} \sqrt 5\equal{}4$

$ ab\equal{}(2 \plus{} \sqrt 5)^{1/3} (2 \minus{} \sqrt 5)^{1/3}\equal{}(\minus{}1)^{1/3}\equal{}\minus{}1$

Then $ (a\plus{}b)^3\equal{} a^3\plus{}b^3\plus{}3ab(a\plus{}b)\equal{}4\minus{}3(a\plus{}b)$

Suppose that $ x\equal{}a\plus{}b$

$ \therefore x^3\plus{}3x\minus{}4\equal{}0$

$ \implies (x\minus{}1)(x^2\plus{}x\plus{}4)\equal{}0$

$ \delta_x\equal{}1^2\minus{}4\cdot1\cdot4<0 \implies x^2\plus{}x\plus{}4\equal{}0$ no real roos.

$ \therefore x\equal{}1$ only.

$ \implies a\plus{}b\equal{}1 \quad$  so that the sum of two cubic roots is rational.

[/hide]

[hide="Alt solution"]

for the first one, simplification gives

$ a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc \equal{} (a\plus{}b\plus{}c)(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ac)$

$ a \plus{} b \plus{} c \equal{} 0 \ \Rightarrow \ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3abc$

next part, take $ a \equal{} (2\plus{}\sqrt{5})^{1/3} , b \equal{} (2\minus{}\sqrt{5})^{1/3}, c \equal{} \minus{}1$

Observe that $ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3abc$

and $ a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} ac \minus{} bc > 0 \ if \ a \neq b \neq c$ (from A.M > G.M)

so a + b = -c = rational number.

[b]Proved.[/b]

[/hide]

[quote="Nora.91"][hide="Alt solution"]

for the first one, simplification gives

$ a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc \equal{} (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ac)$

$ a \plus{} b \plus{} c \equal{} 0 \ \Rightarrow \ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3abc$

next part, take $ a \equal{} (2 \plus{} \sqrt {5})^{1/3} , b \equal{} (2 \minus{} \sqrt {5})^{1/3}, c \equal{} \minus{} 1$

Observe that $ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3abc$

and $ a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} ac \minus{} bc > 0 \ if \ a \neq b \neq c$ (from A.M > G.M)

so a + b = -c = rational number.

[b]Proved.[/b]

[/hide][/quote]

good, very nice solution :P

[color=green][b]Thank You vinskman[/b][/color]  :D

[quote="kunny"]Let $ x \equal{} \sqrt [3]{2 \plus{} \sqrt {5}} \plus{} \sqrt [3]{2 \minus{} \sqrt {5}}\Longrightarrow x^3 \minus{} (2 \plus{} \sqrt {5}) \minus{} (2 \minus{} \sqrt {5}) \equal{} 3x\sqrt [3]{2 \plus{} \sqrt {5}}\cdot \sqrt [3]{2 \minus{} \sqrt {5}}$

$ \therefore x^3 \plus{} 3x \minus{} 4 \equal{} 0\Longleftrightarrow (x \minus{} 1)(x^2 \plus{} x \plus{} 4) \equal{} 0$, yielding $ x\in\mathbb{R},\ x \equal{} 1.$[/quote]

haha , very funny

yours is just as the same as mine. :P  :lol:

[quote="kunny"]No. My solution is $ x \equal{} a \plus{} b\Longleftrightarrow x \plus{} ( \minus{} a) \plus{} ( \minus{} b) \equal{} 0\Longrightarrow x^3 \plus{} ( \minus{} a)^3 \plus{} ( \minus{} b)^3 \equal{} 3x( \minus{} a)( \minus{} b).$

Your solution is $ x \equal{} a \plus{} b\Longleftrightarrow x^3 \equal{} (a \plus{} b)^3\Longleftrightarrow x^3 \equal{} a^3 \plus{} b^3 \plus{} 3ab(a \plus{} b).$[/quote]

Sorry , mistaken  :D

No problem, but both of us finally obtained $ x^3\plus{}3x\minus{}4\equal{}0.$ :lol:

A simple way for number 2:

$ \left(\dfrac{1\plus{}\sqrt{5}}{2}\right)^3 \equal{} 2 \plus{} \sqrt{5}$

$ \left(\dfrac{1\minus{}\sqrt{5}}{2}\right)^3 \equal{} 2 \minus{} \sqrt{5}$

Then the question becomes: is $ \dfrac{1\plus{}\sqrt{5}}{2}\plus{}\dfrac{1\minus{}\sqrt{5}}{2}$ a rational number?

$ \dfrac{1\plus{}\sqrt{5}}{2}\plus{}\dfrac{1\minus{}\sqrt{5}}{2}\equal{}\dfrac{2}{2}\equal{}1$ so the answer is yes.

[quote="tonypr"]A simple way for number 2:

$ \left(\dfrac{1 \plus{} \sqrt {5}}{2}\right)^3 \equal{} 2 \plus{} \sqrt {5}$

$ \left(\dfrac{1 \minus{} \sqrt {5}}{2}\right)^3 \equal{} 2 \minus{} \sqrt {5}$

Then the question becomes: is $ \dfrac{1 \plus{} \sqrt {5}}{2} \plus{} \dfrac{1 \minus{} \sqrt {5}}{2}$ a rational number?

$ \dfrac{1 \plus{} \sqrt {5}}{2} \plus{} \dfrac{1 \minus{} \sqrt {5}}{2} \equal{} \dfrac{2}{2} \equal{} 1$ so the answer is yes.[/quote]

[b] Good One !!  [/b]   :coolspeak:

Hmm is there a fast way to simplify $ \sqrt[3]{2\plus{}\sqrt5}$ to $ \frac{1\plus{}\sqrt5}2$?
Of course it's easier to verify than to derive...


Do the items in the online school Archive ever expire?  If so, when do they expire?

We don't anticipate removing them, but we're reluctant to ever say forever.

Is there a way to save every question with its solution using a script as a PDF file that we can save on our computers


Let $ p$ be a prime number from $ 4\mathbb{Z}\plus{}3$

then prove $ x^2 \equal{}\minus{}1 [p]$ impossible

thanks

Use Euler's criterion http://en.wikipedia.org/wiki/Euler's_criterion
Or if you dont want to quote it, learn its proof and apply to this situation!

[quote="FOURRIER"]Let $ p$ be a prime number from $ 4\mathbb{Z} \plus{} 3$

then prove $ x^2 \equal{} \minus{} 1 [p]$ impossible

thanks[/quote]

$ 1\equiv x^{p\minus{}1}\equiv (x^2)^{\frac{p\minus{}1}{2}}\equiv (\minus{}1)^{\frac{p\minus{}1}{2}}\equiv \minus{}1$ impossible


Prove that every point in the Cantor Set is a cluster point of its complement.

the cantor set is the set of numbers in $ [0,1]$ that don't require the digit $ 1$ for their base-3 representations. let $ x$ be a member of the cantor set. we need to show that there are numbers arbitrarily close to $ x$ which require $ 1$'s in their base 3 representation. by definition, we may assume $ x\equal{}0.d_1d_2d_3...$, where each $ d_i$ is 0 or 2. if infinitely many $ d$'s are 0, then the set of numbers obtained by replacing exactly one of those 0's by 1 is a set of numbers all requiring 1 in their base 3 rep's and containing elements arbitrarily close to $ x$. if only finitely many digits are 0, the digits are all 2 from some point on. this representation of $ x$ may then be replaced with one having 0's from some point on. then, as in the first case, the set of numbers obtained by replacing exactly one of these 0's by a 1 does the job.


i was solving a problem,and then i found this inequality.i think it is nice but i don't know it is hard or not,because i didn't solve it yet. :D


[quote="mdk"]Find the only integer n>32 that satisfies                               (n-23)!(23)!=(n-32)!(32)![/quote]

[hide]
Divide both sides by $23!$ to get, $(n-23)!=(n-32)!(32*31*30*...*25*24)$. Divide both sides by $(n-32)!$ to get, $(n-23)(n-24)(n-25)...(n-30)(n-31)=32*31*30*...*25*24$ So, since both of these are products of 9 consecutive integers, the terms have to be equivilent, so $n-23=32, n-24=31,...,n-31=24$ So, $n=55$[/hide]

[quote="mathgeniuse^ln(x)"]Right from Math League #2, huh.

And yeah, that is the best way of approaching it, or notice the symmetry involved, and there you have it.[/quote]

yep

[quote="Altheman"]take reciprocals, and multiply by n!, the equation is then $\binom{n}{23}=\binom{n}{32}$, which has the obvious connection to $\binom{n}{k}=\binom{n}{n-k}$, giving $(n-k)+k=23+32=55$[/quote]

even prettier than the first response.

[quote="1=2"]prettiest:

n-23=32

n-32=23

Both give n=55[/quote]

nice observation. Your answer is nice.


Prove that if $ p$ is a prime number greater than $ 3$, then $ 24$ divides $ p^{2}-1$.

[hide]
Since all primes greater than three are in the the form $ 6k \pm 1$, then
$ (6k \pm 1 )^{2}-1 = 36k \pm 12k+1-1 = 48k , 24k \Rightarrow \boxed{24 | 48k \ , \ 24 | 24 k }\ \ \ \mathbb{Q.E.D.}$
[/hide]

[hide="alternate solution"]
$ p^{2}-1 = (p+1)(p-1)$, and p is prime, so p mod 4 = 1 mod 4 or 3 mod 4. When p = 3mod4, then p+1 = 0 mod 4, and p-1= 2mod4, so the product is divisible by 8. When p=1 mod 4, then (p+1)(p-1) = 2mod4 * 0 mod 4, so it is also divisible by eight. Finally, it must also be divisible by 3, to have 24 divide it, and p must be 1 mod 3 or 2 mod 3. By same process as for 4, you get that it is divisible by 3. [/hide]

btw, how did you know that all primes after 3 are $ 6k\pm1$? is there a proof or something? :huh:

[hide="proof for the prime form"]
looking at $ \mathbb{Z}$ in mod 6

$ 0 \mod{6}$ is divisible by 6
$ 2 \mod{6}$ is divisible by 2
$ 3 \mod{6}$ is divisible by 3
$ 4 \mod{6}$ is divisible by 2

which leaves us with $ 1,5 \equiv \pm{1}\mod{6}$ or $ 6k\pm1$
[/hide]

What do you mean? Prove what? Both of the proofs posted above are valid proofs.

[quote="The great one"]So how do you prove?[/quote]
Look at AstroPhys's proof (first reply).


calculus
MIT
college
videos
vector
calculus computations

I recently finished Calculus 2 and I'm looking forward to setting out a journey to Multivariable Calculus. Can anyone recommend me any good resources or books to study so that I can acquire the most out of it?
I'm also planning on brushing up my skills on calc 1 a little bit more at the same time. That will be really nice if anyone knows how to study them at once.

Are you doing this alone or do you have a teacher? What book did you use for Calculus 2? Are you interested in science and/or engineering?

I'm very interested in science and engineering. I would like to learn chemistry more in depth and hopefully astronomy and all other sorts of fields. My teacher for calc 2 used Stewart single variable calc book which I didn't think so self-contained and also Robin Gottlieb's calc textbook, which in my opinion, better than Stewart, but somewhat lacked in explanation. I'm going to learn Multivariable on my own, so I will be happy if the textbook or other kinds of resources could explain really well.

A good resource for you might be the MIT OpenCourseWare site for [url=http://ocw.mit.edu/OcwWeb/Mathematics/18-02Fall-2007/CourseHome/]18.02 Multivariable Calculus[/url]. They have video lectures, (sparse) notes, problem sets, and exams.

You also might find some of the resources from http://teachers.sduhsd.k12.ca.us/abrown/classes.htm (under Calculus D) and http://multivariablecalculus.wordpress.com/ useful. There are lots of homepages for multivariable calculus courses out there with course notes and practice tests.

As for books, one of my favorite multivariable calculus books (especially for aspiring physicists/engineers) is [u]Vector Calculus[/u] by Marsden and Tromba. It has good application problems, but it is not an easy book without a teacher. It's not easy to read on your own and, though most of the problems are computational, they are generally harder than the ones in Stewart.

On the other hand, a book with good explanations and with pretty decent problems and exercises is [u]Calculus Early Transcendentals[/u] by Varberg and Purcell. It isn't very advanced though. Just bare bones multivariable calculus computation skills.

For a book that may be in the middle - try [u]Thomas' Calculus[/u] (10th ed or later) by Thomas, Weir, Hass, and Giordano. It comes in various formats, but they're all basically the same. Of the "basic" calculus books that cover both single and multivariable calculus, I think it has the best exercises for people interested in science/engineering. It's very similar to Stewart in style and difficulty, though Stewart has more concept questions and less application questions. There's also a solutions manual for the 11th edition floating around on the internet that's not hard to find.


Suppose f:[a,b]--> is bounded and f^2 is Riemann integrable on [a,b]. Must f also be Riemann integrable on [a, b]? Either prove that the answer is "Yes" or give an example showing that it is "No".

Good Luck :)

This is very much not the correct forum for this question.  On the other hand, I do not think it is too hard.  Take a non-integrable function $g: [a, b] \to \{1, -1\}$ and any integrable function $h: [a, b] \to \bf R$ which is within our bound.  Then $f = g\cdot h$ satisfies $f^2 = h^2$ is integrable while $f$ is not.  (If the bound is of a more general nature than $|f| < R$ for some $R > 0$, this method obviously doesn't work.)