Determine all prime p for which the system:

                                     p + 1 = 2.x^2

                                 p^2 + 1 = 2.y^2

has a solution in integers x,y.

This is equivalent to finding solutions to 1+p+p^2+p^3 = n^2, which is discussed in another recent thread.

here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14926

actually as it turns out, this problem has a simpler solution (and should be moved to number theory)

wlog we can assume x,y are non-negative (in fact we can take x,y to be positive since x=0 or y=0 do not work)

Notice that we have x < y < p, and that 2x^2 = 2y^2 mod p.  But p must be odd, so we can cancel the 2 to get x^2 = y^2 mod p or x = :pm: y mod p

Now x <> y, so we must have that x+y=p.  Plugging in we have that:

p^2+1=2(p-x)^2 = 2p^2 - 4px + 2x^2 = 2p^2 - 4px + p + 1.

Which implies that p = 2p-4x+1 or that p = 4x-1.  Then (4x-1)+1=2x^2 or x = 0 or 2.  x = 0 doesn't work, but if x = 2 we have that p = 7 and y = 5

so the only solution (p, x, y) is (7, 2, 5).


ceiling function
geometry
geometric transformation
rotation
group theory

[quote="sylow_theory"][quote="vidyamanohar"]Find the number of ways of arranging three things of same kind and twelve things of other kind around a circle.[/quote]

I think the general formula if $ \frac{n!}{n}= (n-1)!$, does that make sense?

Like sitting 7 people around a table is $ 6!$ not $ 7!$ because we need to divide by the number of seats.[/quote]
Yeah that is true.  But how your answer is going to help solve this problem.  Can you work out and give the solution for this problem?

Would you just list out?
There was a problem in AOPS 1 like it and it just said to list out carefully..

This problem is not easy!
We are to consider whether the arranging is symmetric or not.

Indeed, the problem does not succumb easily to a simple formula you know.  The entry in the OEIS (shifted by a few terms) for the sequence is [url=http://www.research.att.com/~njas/sequences/A007997]here[/url].  The formula (where 12 is replaced by $ n$) is [hide]$ \left\lceil\frac{(n+1)(n+2)}{6}\right\rceil$.[/hide]  "Intelligent listing" in some form or another isn't a terrible way to solve the given instance.

Additional thoughts, for those who've read the answer already:
[hide]The answer is also equivalent to $ \left\lceil\frac{1}{n+3}\binom{n+3}{3}\right\rceil$.  This is interesting because it's exactly what you would expect if you didn't have to worry about the symmetries that Kunny pointed out.  Moreover, investigating the same question when we have $ n$ of one type and 0, 1 or 2 of the other type also gives the same result, i.e. the number of arrangements is in all of these cases $ \left\lceil\frac{1}{m+n}\binom{m+n}{m}\right\rceil$.  So, the question is: is this always the case?  Do there exist $ m$ and $ n$ such that in placing $ m$ things of one color and $ n$ of another around a circle, the number of possible arrangements is [i]not[/i] $ \left\lceil\frac{1}{m+n}\binom{m+n}{m}\right\rceil$?  What about the corresponding question for larger numbers of colors?[/hide]

I can derive what I posted above using some Group Theory and a Combinatorics thing called [url=http://en.wikipedia.org/wiki/Burnside_lemma]Burnside's Lemma[/url]. The reason why I didnt post this approach is because this is not a Superior Algebra forum and I didnt think it would be appropriate. But if anyone wants to see if I will happily post it.

[quote="mdk"]post it  sylow_theory[/quote]
Okay, but dont say I didnt warn you.  :bruce:

Let there be $ n\geq 3$ people (otherwise it makes no sense for less). We want to show the number of circular seatings we can make is $ (n-1)!$.

To remind ourselves what "Burnside's Lemma" says is that given a finite group $ G$ and a set $ X$ a finite G-set, then the [b]number of orbits in $ X$ under $ G$ is:[/b]
$ \frac{1}{|G|}\sum_{g\in G}|X_{g}|$

Let $ X$ deonote the set of all possible ways to seat them (counting order) and there are of course $ n!$ possibilities. We define the [i]action[/i] on $ X$ to be a rotation of all memebers once clockwise. These "actions" for a rotation group $ G$, which is cyclic of order $ n$. Now the only element which leaves any arrangement fixed is $ e$, i.e. the identity element (a non-trivial rotation).
That means $ |X_{g}|=0$ for $ g\not = e$ and $ |X_{g}|=n!$ for $ g=e$. Note, the number of orbits in $ X$ under $ G$ is the number of seatings not counting order. So by Burnside's Counting Lemma:
$ =\frac{1}{n}\cdot n! = (n-1)!$

[hide="Similar Problem"]
Given a 11 different beads. How many necklaces can you make out of those beads? Because we can flip it over, and so there are fewer then $ (11-1)!$.
[/hide]

What are you talking about "warn you"? That was very easy to understand, but nonetheless nice.

For those of us who don't have a necessary background to understand sylow's proof, there is a common sense way of looking at how to find permutations of circular arrangements:
Let's take, for example, 4 distinct objects in a straight line. These objects can be arranged 4! ways. If we are to join the ends of this line (and in this process creating a circular arrangement) 4! is no longer the number of permutations because the arrangement can undergo rotation. To find the number of permutations of the new arrangement we can fix one of the objects and find the number of permutations of all the other objects around it, in this case, 3!. If we generalize this to $ n$ objects and we fix one of them, we are then looking for $ (n-1)!$. I probably made this explanation unnecessarily confusing...

As for sylow's problem, the answer is $ \frac{10!}{2}$, because we'll double count if we "flip over" the necklace.


function
logarithms
algebra unsolved
algebra

hi everyone
is there any method to solve the equation : a^x=x  (a<1)
and this cos x  =  x
(rather than using the calculator    :lol: )

Solution to $a^{x}= x$ can be expressed in terms of [url=http://mathworld.wolfram.com/LambertW-Function.html]Lambert W-function[/url] as $x = \frac{W(-\ln a)}{-\ln a}.$

thanks for that
and anyone know how to solve cosx=x???


geometry
geometric transformation
rotation
calculus
integration
vector

I need some help on this problem

analysis
1. Angular momentum not conserved,
2. Length of rope decreases.

BTW, is it legal for me to post problems from books like Introduction to Mechanics by Kleppner and Kolenkow, so that you guys can help me (am I violating copyright rules).

Any tips are welcome!

You are correct that angular momentum is not conserved about the axis of the cylinder - so what's the problem?  Keep in mind that there have been a few flawed USAPhO problems in previous years, and this is one of them (the official solutions can be flawed, that is).  Don't take the official solutions as holy writ.

I'd like to note that all three problems you've just posted have already been discussed, either in this forum or in the Introductory Physics forum.  Please consider searching for such threads before starting three entirely new ones (although this is understandable since you're a relatively new member).

[hide="HINT"]

We can assume that for a small time dt , the body is doing pure rotation about a fixed radius. Now who is providing centripetal acceleration?[/hide]

And please do post problems from such books/sources, dont worry about copyright and stuff, its no big deal.

[b]EDIT[/b]: Just noticed the first line of your solution. Energy is not conserved, tension is doing positive work.

[quote="integration"]
$ v_t\equal{}v_0\equal{}constant$

[/quote]

How is that?
Will angular momentum not be conserved?

Like:

$ \frac{d}{dt}(I\omega) \equal{} 0
\\
\\
\\
2ml_t\omega_t\frac{dl}{dt} \plus{} ml_t^2\frac{d\omega}{dt} \equal{} 0
\\
\\
\\
2l_t\omega_t^2 \equal{} \minus{}l_t^2\alpha, \;\; \alpha \equal{} \text{Angular acceleration.}
\\
\\
\frac{d\alpha}{dt}\equal{}0, \; \text{ie: \minus{} }\; \alpha \text{ is constant.}$

Can anyone clear this for me?
I'm quite confused... :(

Kinetic Energy [b]is conserved[/b];
Tension [b]is not doing any work[/b] ; it's a centripetal force (read you hint, he he he...)

Thnx.  Apparently, I solved it the right way, using energy.

To Rajiv: Angular momentum is not conserved about the center of the cylinder, because the torque= (T)(R), where R is the radius of the cylinder and T is tension of the chord.  R is non-zero.

If tension is not doing any work, how is the block coming in i.e reducing the radius.

By definition: W= {integral}F . dx, where F is the force and dx is the differential vector of displacement.  In this case, [i]instantaneously[/i] Tension is always perpendicular to the direction of motion.  Thus, F.dx is always zero, and thus, W= 0.  Is that satisfactory?

If tension is doing work, I have no idea how to solve this problem..0

Conjurer, can you derive an expression for the work done by tension?  And if tension is doing work, how much more KE is obtained after a time t, for example?

Yes F.dx is work, why do you think it is always zero, the path of the particle is not complete circular, it cannot be approximated to be perpendicular when considering the whole motion. I think Kleppner has one solved example like this in the starting chapter.Please have a look and I will too.

OK, so if work is done, how can you quantify it?  B/c if you can't quantify it, I don't see how you calculate the KE, and thus velocity of the particle.


limit
logarithms
calculus
calculus computations

[ sorry if i asked this before...   :oops:  ]

show

$ \lim_{m\to1^{\leftarrow}}\;\sum_{n=1}^\infty\;\left( \frac{1}{n^{m}}\;-\;\frac{1}{m^{n}}\right)\;=\;\boxed{\gamma}$


geometry
circumcircle
geometric transformation
reflection
analytic geometry
graphing lines
slope

I need a way to solve the folowing problem using only Euclidean geometry.
[img]http://i34.tinypic.com/v5fh1s.gif[/img]

I think Posted before.
anyway extend BO and CO to cut the circle then use Pascal theorem.

can you show me how exactly?
we did not learn yet the pascal theorem so maby there is a simplear way?

is there a way without Pascal theorem or complex numbers?

Apply this lemma when $ P \equal{} O$. Then the problem is solved easly.
[color=blue]Let $ ABC$ be a triangle, and $ g$ an arbitrary line. The line $ g$ intersects the lines $ BC, CA, AB$ at the points $ X, Y, Z.$ Let $ O$ be the circumcenter of triangle $ ABC,$ and let $ P$ be the orthogonal projection of the point $ O$ on the line $ g.$ Let $ X', Y', Z'$ be the reflections of the points $ X, Y, Z$ in the point $ P.$ Then, the lines $ AX', BY', CZ'$ concur at one point, and this point lies on the circumcircle of triangle $ ABC.$[/color]
for a proof, see Theorem 4 of http://www.cip.ifi.lmu.de/~grinberg/Butterfly.zip

I forced a very ugly proof with coordinates, I don't feel like typing out everything.
We can have O(0,0) A(1,0), B(sinX,cosX),C(sinY,cosY). Let MN be the line y=mx.
We have to show that the angle between
$ (sinX,cosX)$ $ (1,0)$ $ (sinY,cosY)$
In that order, is the same as the angle between
$ ((sinY\plus{}1/(m\minus{}(cosX\minus{}1)/sinX))/2,(cosY\plus{}m/(m\minus{}(cosX\minus{}1)/sinX))/2)$, $ (0,0)$ $ ((sinX\plus{}1/(m\minus{}(cosY\minus{}1)/sinY))/2,(cosX\plus{}m/(m\minus{}(cosY\minus{}1)/sinY))/2)$
This can be done by finding the slopes and using arctans.


Is there any efficient algorithm to count number of spanning tree of a graph?

For example, the graph below (4 vertexes, 5 edges ) has 8 spanning trees:
Edges of the graph

3 4
4 2
2 3
1 2
1 3

Vertexes are numbered from 1 to 4.

maybe the link will help..please check
[url]http://en.wikipedia.org/wiki/Kirchhoff's_theorem[/url]

Yeah! the theorem is easy to understand and O(n^3) algorithm is too, but I am still not clear with O(n^4) algorithm (which doesn't use divisions).
Counting problems interest me!!!

Others.

1) Count the number of distinct substrings of a string.
Ex:
ABABA: number of distinct substrings: 9
len=1 : A,B
len=2 : AB,BA
len=3 : ABA,BAB
len=4 : ABAB,BABA
len=5 : ABABA

2) Count the number of distinct subsets of a set.
Ex:
{1,1,2}:number of distinct subsets:
1 element-set  : {1},{2}.
2 elements-set: {1,1},{1,2},{2,1}
3 elements-set: {1,1,2},{1,2,1},{2,1,1}

the first one is easy if u make a trie .. that would be O(n^2)...
another way would be to consider all suffixes of the string.. and consider the length of the longest common prefix of all such strings... i leave the details to u :)
an 0(n) is possible if u are really willing to use heavy machinery like suffix trees

second one i'll think

[code]Another one.
Count the number of maximum bipartite matching in a graph.

Ex

Given a matrix
  a b c
1 1 1 0
2 1 0 1
3 1 1 0
matrix[i][j]=1 if i-j is an edge.

there are 2  maximum bipartite matching

1)                        2)
1--a                      1--b
2--c                      2--c
3--b                      3--a
[/code]


[hide]
$ \cos6x \minus{} \cos4x \plus{} 4\cos3x \plus{} 4 \equal{} 2\cos^2 3x \minus{} 1 \minus{} \cos4x \plus{} 4\cos3x \plus{} 4 \equal{} 2(\cos 3x \plus{} 1)^2 \plus{} 1 \minus{} \cos4x$

Since $ (\cos3x \plus{} 1)^2 > 0$ except when $ \cos 3x \equal{} \minus{}1$, and $ 1 \minus{} \cos4x > 0$ except when $ \cos 4x \equal{} 1$, the equation

$ \cos6x \minus{} \cos4x \plus{} 4\cos3x \plus{} 4 \equal{} 0$

has solutions when $ \cos3x \equal{} \minus{}1,\cos4x \equal{} 1$, when $ x\equal{}2k\pi\plus{}\pi$.[/hide]

[quote="cosinator"]
$ \cos6x \minus{} \cos4x \plus{} 4\cos3x \plus{} 4 \equal{} 2\cos^2 3x \minus{} 1 \minus{} \cos4x \plus{} 4\cos3x \plus{} 4 \equal{} 2(\cos 3x \plus{} 1)^2 \plus{} 1 \minus{} \cos4x$

Since $ (\cos3x \plus{} 1)^2 > 0$ except when $ \cos 3x \equal{} \minus{} 1$, and $ 1 \minus{} \cos4x > 0$ except when $ \cos 4x \equal{} 1$, the equation

$ \cos6x \minus{} \cos4x \plus{} 4\cos3x \plus{} 4 \equal{} 0$

has solutions when $ \cos3x \equal{} \minus{} 1,\cos4x \equal{} 1$, when $ x \equal{} 2k\pi \plus{} \pi$.[/quote]

Thank you very much.


ARML
geometry
3D geometry
algebra
polynomial
trigonometry
conics

Does anybody have the answers to the 2004 ARML individual round? Problems at http://www.arml.com/2004%20Contest/2004%20ARML%20Individuals.pdf

[hide="Solutions to 4, 5, and 6"]
4.
Well, if you're VishalArul at our ARML practices, you happen to know the cube root of two to 5 decimal places  :lol:

Or, if you're anyone else, you can basically see that it a has to be 1
Then, you can cube the whole thing, use Newton's Method, and work it down so that $\frac{1}{b+\frac{1}{c+1...}}$ is < 0.2599, so B has to be 4 to make the whole thing after it positive.

(I think that's right).

5. Multiply the ugly polynomial by $(x-1)$ This introduces an extraneous root at x = 1, but the resulting polynomial is $x^{8}-1 = 0$, which forms a regular octagon on the complex plane.

By taking away the point x = 1, we form a convex septagon, with six congruent isoceles triangles (with an interior angle of 45 degrees), and one larger triangle (with an interior angle of 90 degrees).
We use the formula for an area of a triangle $\frac{AB\sin \theta}{2}$
and get the area of a small triangle to be $(1)(1)(\sin 45)(0.5)$ = $\frac{\sqrt{2}}{4}$
There are 6 of these for an area of $\frac{3\sqrt{2}}{2}$.
Also, we add the big one in $(1)(1)(\sin 90)(0.5) = \frac{1}{2}$
for a grand total of $\frac{3\sqrt{2}+1}{2}$

6.
If we use the fact that the powers of $i$ repeat mod 4, we can notice that the sequence of powers of the Fibonacci Subtraction dealie repeat mod 6.

Then the sequence is
$(i^{3}+i^{3}+i^{2}+i^{1}+i^{3}+i^{0})$
Because 6 divides 2004 exactly 334 times, it's just this the evaluation of the sequence * 334.
The sequence works out to $-2i$, so multiplied out the answer is $-668i$.
[/hide]

[quote="DiscreetFourierTransform"][hide="Solutions to 4, 5, and 6"]
4.
Well, if you're VishalArul at our ARML practices, you happen to know the cube root of two to 5 decimal places  :lol:

Or, if you're anyone else, you can basically see that it a has to be 1
Then, you can cube the whole thing, use Newton's Method, and work it down so that $\frac{1}{b+\frac{1}{c+1...}}$ is < 0.2599, so B has to be 4 to make the whole thing after it positive.

(I think that's right).
[/hide][/quote]

[hide="#4"]
Or, you can guess the cube root of two. It should be between 1.2 and 1.3, because if you cube them (just multiply by hand), 2 is in between those two. Then, add a digit, and try to guess the closest one. I took my around 2 tries to get the right digit. I guessed 1.25, and I cubed it, and it was slightly under 2. So, I just said it was 1.253 or something. The thousandths digit doesn't matter. We just need to know that it's greater than 1.25.

Anyway, note that $a,b,c,d,\cdots$ have to be positive, so $a$ must equal to $1$. The fractional part must be less than $1$ because $b$ is a positive integer.

Now, $b$ has to equal to the units digit of the reciprocal of $0.253$, and that's $3$ for sure, because it's slightly less than $4$.

P.S. My approximation is off, but it doesn't affect the value.[/hide]

Yeah, I actually typed the wrong number in my solution.  :blush:  It should be $b=3$.

#1

Solution:
[hide]
Given:
$a-b=2$--------->(1)
$c-b=-3$------->(2)
$c-d=4$--------->(3)
$e-d=-5$------->(4)

Then:
$(1)-(2)+(3)-(4)+2e=a+e=14+2e$

Clearly $a+e=14+2e$ is minimized for the smallest value of $e$, which is $e=-10$. Substituting and solving back gives $(a,b,c,d,e)=(4,2,-1,-5,-10)$ which satisfies the distinctiveness condition. Thus the answer is $\boxed{-6}$[/hide]

#2

Solution:
[hide]
From the congruent arcs and congruent small circles it is easy to convince ourselves that the triangle formed by the intersecting tangent lines around the central small circle is equilateral and hence has all 60 degree angles.

Then consider the triangle formed by a vertex of the central triangle, the center of the outer small circle nearest that vertex, and a tangent segment connecting the small outer circle to that vertex. This is a 30-60-90 triangle.

Hence for the small radius $r$ and the big radius $R=1$ we have $5r=R$ and thus the area of any one of the four small circles is $\boxed{\frac{\pi}{25}}$[/hide]

#3

Solution:

[hide]
The two ellipses are congruent but one is displaced $k$ units in the positive direction. Thus it is easy to convince ourselves that for all k, 0<k<6, AB=BC=CD=DA. The quadrilateral will be a square when its diagonals are equal, that is AC=BD.

From the displacement, $BD=6-k$.

If the two ellipses intersect at $(x_{i}, y_{i})$ and $(x_{i},-y_{i})$ then we want $k$ such that $6-k=2y_{i}$.

Hence,

$6-k=2y_{i}=2\sqrt{1-\frac{x_{i}^{2}}{9}}$

Subtracting the two ellipse equations and solving for $x_{i}$ gives $x_{i}=\frac{k}{2}$.

Thus,

$6-k=2\sqrt{1-\frac{\frac{k^{2}}{4}}{9}}$

Manipulating and solving with the quadratic formula yields

$5k^{2}-54k+144=0$

$k=\frac{24}{5},6$. Given the restrictions the answer is $\boxed{\frac{24}{5}}$[/hide]

#7

Solution:

[hide]

The bounds on the inequality can be determined by the equations:

$\frac{nx}{2004}-1<\frac{n}{2004}$ and

$1-\frac{nx}{2004}<\frac{n}{2004}$

These yield, respectively,

$x<\frac{2004}{n}+1$ and $x>\frac{2004}{n}-1$

The difference between these two places on the number line is always 2, so there will always be exactly 2 integer solutions except when $\frac{2004}{n}$ is an integer, in which case there will be only 1 solution.

Thus we desire the number of factors $n$ of 2004 on the given bounds, which is all of them. $2004=2^{2}\times{3}\times{167}$ and since each prime factor is either used or not used to create factors of 2004, there are (3)(2)(2)=12 factors. Thus there are 2004-12=$\boxed{1992}$ values of n that fit the requested criteria.
[/hide]

#8

Solution:
[hide]
I may have missed small details, so this solution should be read critically and I would appreciate affirmation or correction.

I will consider six cases, and the remaining six are symmetric to the ones already under consideration:

---A----B
1) HH--HT
2) HH--TH
3) HH--TT
4) HT--HH
5) HT--TH
6) HT--TT

Clearly the A rolls TH and TT cases will produce the same probabilities as the cases 1-6 described already.

Two deductions come immediately:

-If A's and B's rolls start with the same flip, the probabilities of A and B winning will be equal because the next roll has an equal probability of being correct for A's or B's sequence.

-In cases of complete symmetry between A's and B's rolls, both A and B will have equal probability of winning.

The only cases where equity is not preserved are cases 2 and 6.
Setting up a coin flips diagram for C, we see that:

Compressing the sample space to the six cases described, case 2 has a $\frac{1}{2}\frac{1}{4}$ chance of occuring, and in case 2, P(A wins)=$\frac{1}{4}$. Likewise, case 6 has a $\frac{1}{2}\frac{3}{8}$ chance of occuring, and in that case P(A wins)=$\frac{3}{4}$. The remaining sample space is $1-\frac{1}{2}(\frac{1}{4}+\frac{3}{8})$. Thus the probability A wins is:

$\frac{1}{2}\frac{1}{4}\frac{1}{4}+\frac{1}{2}\frac{3}{8}\frac{3}{4}+\frac{1}{2}(1-\frac{1}{2}(\frac{1}{4}+\frac{3}{8}))=\boxed{\frac{33}{64}}$.
[/hide]

I just did some clever address-hunting and found the 2004 individual solutions:

http://arml.com/2004%20Contest/2004%20ARML%20Individual%20Solutions.pdf

Plus, I am happy to say all my answers were correct, including #8. :D


There is a chess competition between A and B. They have to play 7 matches. If A or B won 4 matches, the game ends and he/she will become a champion.

You are a bettor (meaning that you bet on A or B in order to win the money). Before the game, you don't know how many matches they will play. You can bet on A or B with any amount of money, and you have to bet on every match. However, since you are a mathematician, you want to bet so that no matter who win or loses, you will gain or loss only 100 dollar.

Here is the question: How much money do you have to bet on the first game in order to win or lose 100 dollar after the competition ends?

Please don't double post.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=167960


A 65.0 kg ice skater moving forward with a velocity of 5.00 m/s throws a 0.240 kg snowball forward with a velocity of 30.0 m/s.
(a) What is the velocity of the ice skater after throwing the snowball? Disregard the friction between the skates and the ice.
(b) A second skater initially at rest with a mass of 60.0 kg catches the snowball. What is the velocity of the second skater after catching the snowball in a perfectly inelastic collision?

A tennis player places a 66 kg ball machine on a frictionless surface. The machine fires a 0.057 kg tennis ball horizontally with a velocity of 27 m/s toward the north. What is the final velocity of the machine?

After being struck by a bowling ball, a 1.5 kg bowling pin moving to the right at 6.5 m/s collides head-on with another 1.5 kg bowling pin initially at rest. Find the final velocity of the second pin in the following situations.
(a) The first pin moves to the right after the collision at 0.8 m/s

http://answers.yahoo.com/question/index?qid=20080410141742AARe2so&r=w&pa=AptqD2bwHTHc7.OAYKg.uW1riuFSjRf98tmcr.OWtwuyXQ--&paid=voted

The "best answer" from Yahoo answers seems to be garbage, not even mentioning the idea that you can use conservation of momentum to solve this problem.  If I have time I'll try to answer later, but basically just use conservation of momentum.

Here's an answer to the first question, part a):
Let $ v$ be the skater's velocity after throwing the ball.
Conservation of momentum tells us $ (65.0 \plus{} .240)\cdot 5.00 \equal{} 65.0 \cdot v \plus{} .240 \cdot 30.0$.  You can solve for $ v$.

And here's part b): Let $ w$ be the second skater's velocity after catching the snowball.  Conservation of momentum tells us $ 60.0 \cdot 0 \plus{} .240 \cdot 30.0 \equal{} (60.0 \plus{} .240) \cdot w$.  So you can solve for $ w$.


limit
number theory unsolved
number theory

Find all pairs $ (m, n)$ of positive integers such that $ m^n \minus{} n^m \equal{} 3$.

Through pure speculation, we can see that a solution satisfying $ (m,n)\equal{}(4,1)$. Now we will show that this is our only solution to the case.

Suppose that without loss of generality, we have positive integers $ a,b$ s.t. $ a>b>1$.

Then we can see that

$ \lim_{b\to\infty}b^a\minus{}a^b\equal{}\infty>3$

Once it reaches a point, it just becomes greater than $ 3$.
If we consider a different expression, $ a^b\minus{}b^a$, we can see that this will fail as well for $ a>b>1$:
Similarly, the expression will not be satisfied to equal 3 for small positive integers and as $ b\to\infty$

$ \lim_{b\to\infty}a^b\minus{}b^a\equal{}\minus{}\infty$

Thus, our only solution will be $ m\equal{}4$ and $ n\equal{}1$

Because my EL is qyute bad so i cant post my proof here
My  ideal is proving with 2 case ans use $ (1\plus{}\frac{1}{n})^m<e<n$ if $ 3\leq m<n$
Then we ony check with n=1


geometry
incenter
geometric transformation
homothety
circumcircle
geometry solved

Let $ABCD$ be a quadrilateral. Denote by $I$ the point of intersection of the lines $AC$ and $BD$. Let $I_{1}$, $I_{2}$, $I_{3}$, $I_{4}$ be the centers of the incircles of the triangles $IAB$, $IBC$, $ICD$, $IDA$, respectively. Prove that the quadrilateral $I_{1}I_{2}I_{3}I_{4}$ has a circumscribed circle if and only if the quadrilateral $ABCD$ has an inscribed circle.

This is a hard problem. You can find a solution in

Toshio Seimiya, Peter Y. Woo, [i]Solution of problem 2338[/i], Crux Mathematicorum 4 / 25 (1999) pages 243-245.

See the attached PDF file.

[b]EDIT:[/b] See also http://www.mathlinks.ro/Forum/viewtopic.php?t=21768 .

  Darij

Let $I$ be the incenter of $ABCD$ and $J_1, J_3$ the incenters of the triangles $\triangle ABD, \triangle BCD$. The diagonal $AC$ is the (external) homothety axis of the incircles $(I), (J_1), (J_3)$. Let $K$ be the external homothety center of $(J_1), (J_3)$ on $AC$. Then the line $BK$ is the (external) homothety axis of $(I_1), (J_1), (J_3)$, of $(I_2), (J_1), (J_3)$, of $(I_1), (I_2), (J_1)$ and of $(I_1), I_2), (J_3)$. The external homothety center of $(I_1), (I_2)$ is therefore on $BK$ and on also their their common external tangent $AC$, i.e., at $K$. Similarly, the external homothety center of $(I_3), (I_4)$ is also at $K$. The lines $I_1I_2$ and $I_3I_4$ meet at $K$ on $AC$.

Since $ABCD$ is tangential, the circles $(J_1), (J_3)$ touch at a point $M$ on $BD$ and $KM \perp BD$. Let $T_2, T_3$ be the tangency points of $(I_2), (I_3)$ with $AC$. Denote $s$ the semiperimeter of a triangle, $X$ the diagonal intersection of $AC \cap BD$ and assume that the points $B, M, X, D$ follow on the line $BD$ in this order.

$XT_2 = s(\triangle BCX) - BC$
$XT_3 = s(\triangle DCX) - CD$
$T_2T_3 = s(\triangle BCX) - s(\triangle DCX) - (BC - DC) = \frac{BX - DX}{2} - \frac{BC - DC}{2}$

$BX = BM + XM = s(\triangle BCD) - DC + XM$
$DX = DM - XM = s(\triangle BCD) - BC - XM$
$BX - DX = 2 \cdot XM + BC - DC$

$T_2T_3 = XM$

Suppose we move a point $C'$ on the line $AC$ and draw the external tangents to the circles $(I_2), (I_3)$ (other than $AC$) intersecting $BD$ at points $B', D'$. The incenter $J_3'$ of the triangle $\triangle B'C'D'$ then moves on the normal $KM$ to $BD$, because the distance of the tangency point $M$ form $X$, equal to $T_2T_3$, does not depend of the position of the point $C'$ on the line $AC$. If we move the point $C'$ to coincide with the point $K$, then $C'M \equiv KM$ becomes the bisector of the $\measuredangle B'C'D'$, perpendicular to the side $B'D'$. It follows that the $\triangle B'C'D'$ becomes isosceles and consequently, the lines $KM, KX$ are isogonals of the lines $KI_2, KI_3$. Let $XI_1$ cut $KM \perp BD$ at $U$. From here, $\angle KUI_1 = 90^\circ - \frac{AXB}{2} = \frac{\angle AXD}{2} = \angle KXI_4$ $\Longrightarrow$ $\triangle KI_1N \sim \triangle KXI_4$ by ASA $\Longrightarrow$ angles $\angle XI_1K = \angle XI_4K$ are equal $\Longrightarrow$ the right angle triangles $\triangle XI_1I_2 \sim \triangle XI_3I_4$ are similar and

$\frac{XI_1}{XI_2} = \frac{XI_4}{XI_3}$ and $XI_1 \cdot XI_3 = XI_2 \cdot XI_4$.

In addition, a convex tangential quadrilateral has one concyclic quadruple of triangle incenters and three concyclic quadruples of triangle excenters. Four concyclic quadruples of triangle incenters/excenters exist even in a concave tangential quadrilateral with an incircle or in a tangential quadrilateral (convex or concave) with an excircle. The triangle incenter and excenter pairs just have to be mixed up.

Let $E$ the cut point $MD$ and $I_3I_4$, now $E$ is the insimilicenter of  $I_3$ and $I_4$, further $K,E$ are harmonic conjugates of  $I_3$ and $I_4$. Also $K,M$ are harmonic conjugates of $J_1$ and $J_3$.  Now $I_3I_4$ is transversal wrt triangle $J_1DJ_3$ then $J_3I_4$ cut $J_1I_3$ at $F$ on the altitude $DM$, hence angle $I_4MD$=angle $I_3MD$ by Blanchet´s theorem. Follow $I_1XI_4$ and $I_2XI_3$ are similar, now: $I_1X.I_3X=I_2X.I_4X$, so $I_1I_2I_3I_4$ are cyclic. QED.


linear algebra
matrix
quadratics
combinatorics open
combinatorics

It is an open question to me finding all pairs (n,k) such from that a set with n elements can choose k substes with at least 2 elements such that each pair of elements belongs to exactly one subset.
Of course this problem is equivalent to find $n$ subsets of a set with $k$ elements such that each element is contained in at least $2$ sets and every $2$ sets intersect in exactly one element.
One can easily deduce
(*) $(n,1)$ and $(n, a(n-a)+2)$ are solutions
(**) was proven a long time ago by N>Bourbaki:
If $k>1$ then $k \geq n$
I think I've managed to find more pairs:
(***) $(p^2+1, p^2+p)$ for $p$ prime
(****) $(xy+1, p^2+1)$ for $x,y<p$ and $x+y \geq p+2$ where $p$ prime
(*****) $(p^2+p+1, p^2+p+1)$. This is a special case: there are as many sets as elements, every set has the same cardinality $(p+1)$, each element belongs to $(p+1)$ sets and every two sets intersect in exactly $1$ element.

Try to prove (**)-(*****). They are quite nice,
I believe (***), (****), (*****) hold without the assumption $p$ is prime, but I can't prove it. Can you?
What other pairs can you find?

Thank you for your attention :)

No one wants to try this?
It's quite beautiful!

Really? (**) is in Bourbaki? Have you the reference?
Well, it is also in one of my paper on the use of linear algebra methods into combinatorics :

Let $A_1, \cdots , A_n$ be pairwise distinct subsets of the set $X = \{x_1, \cdots, x_k \}$ such that $|A_i \cap A_j | = 1$ for all $i \neq j$.
Then $n \leq k$ (the condition for all $i$, the element $x_i$ belongs to at least two of the $A_j$'s is useless).

Consider the matrix $M = (( m_{i,j} ))$ with $k$ rows and $n$ columns such that $m_{i,j} = 1$ if $x_i \in A_j$, and $m_{i,j} = 0$ otherwise.
Now the matrix $U=M^tM$ (where $M^t$ is the transposed matrix of $M$) is a square matrix of size $n$, where the general term $u_{i,j}=|A_i \cap A_j |$.
It follows that $u_{i,j} = 1$ if $i \neq j$.
Thus $u_{i,i} = |_i \cap A_i| \geq 1$ for all $i$.
Now there do not exist two distinct subscripts, say $i,j$, such that $u_{i,i} = u_{j,j} = 1$ because in that case, we would have $A_i = A_i \cap A_j = A_j$, a contradiction. Therefore, at most one of the $u_{i,i}$ is equal to $1$.
It follows that $U$ represents the bilinear form associated to the quadratic form :
$f(y_1, \cdots , y_n) = ( \sum_{i=1}^n y_i )^2 + \sum_{i=1}^n (u_{i,i} - 1)y_i^2 $
which is clearly defined positive, and then non degenerated. Thus, $U$ is invertible, and so has rank $n$. It follows that $M$ has rank at least $n$ and then $k \geq n$, as desired.

Pierre.

It's probrem $M5$ from Kvant and there it's author is written N>bourbaki.
The original proof is also quite long.
However I managed to find a short one using a well-known idea used in such problems.
Of course if $m>1$ then no element is not contained in every subset.
Take an element $x$ and a set $A$ not containing $x$.
Let $x$ belong to $d$ subsets and $A$ have $k$ elements.
Every element of $A$ can be contained in exactly one subset containing $x$.
Hence $d \geq k$. So
So $\frac{1}{n-d} \geq \frac{1}{n-k}$.
Summing over all pairs $x,A$ we get $n+(m-n)\sum{\frac{1}{n-d}} \geq m$
If $m<n$ the $\sum{\frac{1}{n-d}} \leq 1$ which is wrong obviously.

I agree Iura, and I think this kind of proofs can be handled here, but I always see them as quite magical. I think it is clearer to understand why the result is what it is from linear algebra which gives a conceptual approach.

Pierre.


function
trigonometry
real analysis
real analysis unsolved

Given the function

$ f(x)\equal{}\begin{cases}
1&\text{if } x \in \mathbb{Q} \\
0&\text{if } x \in \mathbb{R} \minus{} \mathbb{Q} \end{cases}$

So I know this is a classic function in Intro to Analysis classes. Normally the question is whether this function is ever continuous.

My question is, if one were to graph this function, how would it look like? Would it just look like the line $ y \equal{} 1$ and $ y \equal{} 0$ ?

That seems a little weird to me.

Thanks!

Yes, and there's a reason for it: the density of rationals and irrationals in R.

[quote="bachikarn"]Given the function

$ f(x) \equal{} \begin{cases} 1 & \text{if } x \in \mathbb{Q} \\
0 & \text{if } x \in \mathbb{R} \minus{} \mathbb{Q} \end{cases}$

So I know this is a classic function in Intro to Analysis classes. Normally the question is whether this function is ever continuous.

My question is, if one were to graph this function, how would it look like? Would it just look like the line $ y \equal{} 1$ and $ y \equal{} 0$ ?

That seems a little weird to me.

Thanks![/quote]

You can also construct functions in this way that are only continuous at one, a finite or a countable number of points, like

$ f(x) \equal{} \begin{cases} x & \text{if } x \in \mathbb{Q} \\
0 & \text{if } x \in \mathbb{R} \minus{} \mathbb{Q} \end{cases}$

which is continuous at 0. This would look like the line $ y \equal{} x$ and $ y \equal{} 0$.

 Better yet, there is a function which is everywhere continuous and nowhere differentiable.

But such example could actually be obtained in a closed form, and wouldn't require definition over different sets! An example could be:

$ f(x)\equal{}\sum^{\infty}_{n\equal{}0}\frac{1}{3^n}\sin \left(4^n x\right)$

that converges, but

$ f'(x)\equal{}\sum^{\infty}_{n\equal{}0}\left(\frac{4}{3}\right)^n\cos \left(4^n x\right)$

does not.

Also, you could construct an function that isn't integrable in the same manor as:

$ g(x)\equal{}\sum^{\infty}_{n\equal{}0}\frac{1}{3^n}\sin \left(\frac{x}{4^n}\right)$

And a function that isn't either, as the sum of the above two! Weierstrass function is also a good example, and Blancmange function is a good example for a set-defined function.

[quote="milin"]But such example could actually be obtained in a closed form, and wouldn't require definition over different sets! An example could be:

$ f(x) \equal{} \sum^{\infty}_{n \equal{} 0}\frac {1}{3^n}\sin \left(4^n x\right)$

that converges, but

$ f'(x) \equal{} \sum^{\infty}_{n \equal{} 0}\left(\frac {4}{3}\right)^n\cos \left(4^n x\right)$

does not.

Also, you could construct an function that isn't integrable in the same manor as:

$ g(x) \equal{} \sum^{\infty}_{n \equal{} 0}\frac {1}{3^n}\sin \left(\frac {x}{4^n}\right)$

And a function that isn't either, as the sum of the above two! Weierstrass function is also a good example, and Blancmange function is a good example for a set-defined function.[/quote]

Do those functions converge uniformly? I'm not terribly familiar with the concept (yet!) so I couldn't say myself.

A note on etiquette- it's never appropriate to quote the entire post just before yours. That clutters the thread- we assume you're responding to it unless you state otherwise. If you want to focus on a particular part of that post, uote just that part.

The function milin named is a different counterexample, with no particular relation to nowhere continuous functions- it's [url=http://en.wikipedia.org/wiki/Weierstrass_function]continuous everywhere and differentiable nowhere[/url]. For technical reasons, it's easiest to construct something like this as a uniformly convergent series.

The "not integrable" part, however, is nonsense.

[quote="bachikarn"]Back to the original function. What exactly is the definition of a line besides the shortest distance between two points? I'm thinking those two 'lines' aren't actually lines, but then what are they? Just an infinite amount of points really close together?[/quote]

More or less.  To be more precise, the points on each "line" are so close that between any two points there is another point.  Like pic said, this is because the rationals and irrationals are both dense in the reals.


induction
arithmetic series
number theory unsolved
number theory

An arithmetic series is given:
2,5,8.....an

How can we prove the following:
2*3a(n)+3*2a(n+1)
is divisible by 38 for every n

I was thinking maybe we could use 2*33n-1+3*23n+2=38k. n=1 to infty

I added (52*33n+252*22n)/3 to both sides and arrived at

2*33n+2+3*23n+5=38k+4(13*27n+63*4n)/3.

Does this go anywhere?. I don't think so. Induction can be a booger?.  There is, no doubt, many ways to tackle this.

If you use induction, express the $n+1$th term in terms of the $n$th, plus some junk that is obviously divisible by $38$.

Alternately...

You want to show that $f(k)=18\cdot 3^{3k}+96\cdot 2^{3k}$ is divisible by $19$ for every $k$.  When we look at residues $\bmod 19$, we get

$f(k)\equiv -(3^3)^k+(2^3)^k\equiv -8^k+8^k\equiv 0$

so we're done.

Thanks, that sure is a quick way to go about it. Clever.

[quote="blahblahblah"]If you use induction, express the $n+1$th term in terms of the $n$th, plus some junk that is obviously divisible by $38$.

Alternately...

You want to show that $f(k)=18\cdot 3^{3k}+96\cdot 2^{3k}$ is divisible by $19$ for every $k$.  When we look at residues $\bmod 19$, we get

$f(k)\equiv -(3^3)^k+(2^3)^k\equiv -8^k+8^k\equiv 0$

[color=red]Wouldn't this be -27k+8k?. Also, why 19 instead of 38?. I see where you

arrived at 18*33k+96*23k, from 2*33n+2+3*23n+5=38k.

We didn't divide through by 2. Why 19?. I know this is probably a dumb question, but I am not too versed in

modular arithmetic.[/color]

so we're done.[/quote]

A number is divisible by $38$ iff it is divisible by $2$ and $19$.  Divisibility by $2$ is trivial, so we need to prove divisibility by $19$.

By the laws of exponents, we have $3^{3k}=(3^3)^k=27^k$, and we mod the base out by $19$ to get what we want.  Similarly for $2^{3k}$.

Hope that helps.

Yes, thanks. The residue of 18 mod 19 is -1, hence -1(27)k. The residue of 96 mod 19 is 1, hence 8k.

I didn't know we could do the same with -27k mod 19. Is it because -27 mod 19 is -8?. Then we have -8k+8k=0.

That's $-(27)^k$, not $(-27)^k$.  Anyways, you can look at $27^k$ as $27\cdot\27\cdot \ldots \cdot 27$, and you mod out each of those individually by $19$.


We can simplify $ \sqrt{5}\plus{}\sqrt{45}\plus{}\sqrt{125}$ as $ \sqrt{5}\plus{}3\sqrt{5}\plus{}5\sqrt{5}5\equal{}9\sqrt{5}$. $ (9\sqrt{5})^2\equal{}\boxed{405}$.


Define $ \{a_n\}$ as follows:$ a_{n \plus{} 1} \equal{} a_1 \plus{} \frac {n}{a_n}$.
a.Find $ \lim_{n \minus{} > \infty}\frac {a_n}{\sqrt {n}}$
b. Give $ b_n \equal{} a_{n \plus{} 1} \minus{} 2a_n \plus{} a_{n \minus{} 1}$.Show that $ \lim_{n \minus{} > \infty}\frac {b_{n \plus{} 1}}{b_n} \equal{} 1.$

[color=green]His LaTeX code :

\lim_{n->\infty}\frac{a_n}{\sqrtn}

So part a, means : Find $ \lim_{n\to\infty}\dfrac{a_n}{\sqrt{n}}$[/color]

Assuming that the limit exists,
Let $ \lim_{n\to\infty}\frac{a_{n}}{\sqrt{n}}$ be $ x$.
$ \frac{a_{n\plus{}1}}{\sqrt{n}}\equal{}\frac{a_{1}}{\sqrt{n}}\plus{}\frac{\sqrt{n}}{a_{n}}$
Limiting both sides,
$ x\equal{}\frac{1}{x}$
$ x\equal{}\pm 1$

But I do wonder if limit exists.

I think the solution by KenHungKK is incorrect because we can't take $ \lim_{n\to\infty}\frac{a_{n}}{\sqrt{n}} \equal{} \lim_{n\to\infty}\frac{a_{n\plus{}1}}{\sqrt{n}}$ please tell me if i am wrong.

[quote="ashwinrk_jain"]I think the solution by KenHungKK is incorrect because we can't take $ \lim_{n\to\infty}\frac {a_{n}}{\sqrt {n}} \equal{} \lim_{n\to\infty}\frac {a_{n \plus{} 1}}{\sqrt {n}}$ please tell me if i am wrong.[/quote]

I think if exist $ \lim_{n\rightarrow \infty}\frac{a_n}{\sqrt{n}}$ then $ \lim_{n\rightarrow \infty}\frac{a_{n\plus{}1}}{\sqrt{n}}\equal{}\lim_{n\rightarrow \infty}\frac{a_{n\plus{}1}}{\sqrt{n\plus{}1}}\times\lim_{n\rightarrow \infty}\frac{\sqrt{n\plus{}1}}{\sqrt{n}}\equal{}\lim_{n\rightarrow \infty}\frac{a_n}{\sqrt{n}}$

ok...thks then i think your solution is correct. :roll:

[quote="keira_khtn"]Both two limits are $ 1$.Any help??[/quote]

Can anybody help me to reject -1 as a wrong limit?

The problem is that $ a_1$ is not specified. Thus $ \minus{}1$ cannot be ruled out for the following reason:
If for a certain value of $ a_1$ the limit of the sequence is $ 1$, then the sequence obtained
starting from $ a_1' : \equal{} \minus{}a_1$ is the sequence $ \minus{}a_1,\minus{}a_2,\minus{}a_3,\ldots$, which then has the  limit  $ \minus{}1$.

So the statement of the problem doesn't contain enough information to rule out $ \minus{}1$ as a limit.

Oh,I'm sorry :blush: .I forget one impotant condition :a1>0.But how do these limits exist anyway??

In the problem no1, i think that we can use the Stolez's theorem.


This is a very tedious probability question.

Dora the Explorer goes to three new locations everyday and then returns home. At the beginning of each day, there is a $\frac{1}{4}$ in chance she'll need to help of The Map and a $\frac{1}{6}$ chance she'll need it after she visits her first new location, $\frac{1}{8}$ after her second location, and a $0$ chance after her third location. Everytime she visits a new location, there's a $\frac{1}{3}$ chance Swiper the Swiper will be there and will try to swipe something. As every second passes, there is a $\frac{1}{5}$ chance she will notice Swiper the Swiper and start saying "Swiper, no swiping!" It takes Swiper $13$ seconds to swipe an object, but everytime he hears "Swiper, no swiping," there is a $\frac{1}{2}$ chance the swipe will be unsuccessful. What is the probability Dora will return home without using the Map and without anything swiped?

tedious is an understatement. I bet I could do this if I tried, but I'm way too lazy. It would take insane concentration, which I lack.


algorithm
number theory unsolved
number theory

Let gcd(a,b)=1. Proof that gcd(11a+2b, 18a+5b) is equal 1 or 19.

We know that gcd(11a+2b, 7a+3b)=gcd(19b, a-5b). Why this fact implies that gcd(19b, a-5b)=1 or gcd(19b,a-5b)=19? I know that gcd(a,b)=1 but I don't know how to use it.

Assume $c=(19b,a-5b)$. Then it follows that $c|19b$ i.e. $c=1,c=19,$ or $c|b$. But if $c|b$, then because $c|a-5b$, $c|a$. But $(a,b)=1$ so $c=1$.


geometry
USAMTS
ARML
Euler
number theory
linear algebra
modular arithmetic

I posted earlier about the classes I planned on taking at the local University next year. Well one of the classes that I was really looking forward to taking was the Number Theory class since that is one of the topics that I know very little about (a problem that I need to rectify if I am to do well on contests in college). But when I looked a little closer I found out I got the pre reqs wrong apparently you need to take intro to alg I and II before you can take Number Theory and before you take those you need to take a class called Intro to Abstract Math (as far as I can tell this is supposed to help you learn how to write like formal proofs, logic, cardinality etc). I don't understand do you really need all that just to learn number theory it dosen't seem right to me since for combinatorics the only pre req is Linear Algebra I. Not to mention Number Theory and Combinatorics for that matter seems to be known by many on AOPS who do not have that kind of background. Oh and the pre reqs for Intro to Abstract Math is Linear Alg II. I mean idk if I try and take some of this stuff without all the pre reqs and stuff as far as I know they won't like try and stop me or anything I just don't know what the best course of action is and was hoping AOPSers could give me some thoughts on the matter. If you need some really basic course descriptions (they are like two sentences each) or something in order to make an assessment let me know and I can post those.

Number theory comes in very different flavors.  "Elementary number theory" (what you've seen around AoPS, most probably) shouldn't have any prerequisites.  But "algebraic number theory" and "analytic number theory" both have important prerequisites (abstract algebra and analysis, respectively).

There are probably more different prerequisite structures in this area than in most others. In our case, the system goes like this:

Intro to Abstract Math is prerequisite for Elementary Number Theory.

Intro Linear Algebra is prerequisite for Linear Algebra (II).

Either Elem. Number Theory or Linear Algebra II (one or the other, you don't need both) is prerequisite for Abstract Algebra.

To a large extent, this system is set up to serve those of our majors who are planning to be high school teachers. For them, the Number Theory course is required; the second Linear Algebra course isn't.

A Number Theory course with an Abstract Algebra prerequisite is a different creature than the Elementary Number Theory course that I'm talking about, so my advising experiences won't translate all that well.

Prerequisites are also administered by (and can be overridden by) human beings. In many cases, the instructor of the course can let in anyone he or she wants to. Often, there is also someone in the department (a chair or associate chair or undergraduate advisor) whom you can talk to. In the case of our department that would be me. Of course, I do spend a lot of time saying "no". After all, we don't want student taking courses that they're going to fail. (And note that I'm not using the word "fail" in the casual unrealistically high standards mode so common on this site - I'm talking about one of the letters "DFW" appearing on a transcript.)

For someone coming out of a high school contest and problem solving background, in my opinion, the Intro to Abstract Math course might be unnecessary - in particular, if I could see something from the proof-writing side, like USAMTS or participation on an ARML team, that might clinch it. As for the content of the Elementary Number Theory course: someone from the contest background would probably already know about half of it (such as modular arithmetic, Fermat and Euler) but might be missing some systematic stuff and some of the basic proofs that hold the subject together. My counterpart at another institution may not be that familiar with the high school contest world, so don't count on that familiarity.

The strange thing is that there does not seem to be any lower level Number Theory courses although I suppose the material may be covered a pre reqs under a different name.

I don't think ill be allowed to skip anything on the basis of really good contest performance (although I am working on fixing that for the future). I do however really want to take the intro the alg (and there's a geo one too) two course sequence in the Autumn and Winter quarters respectively and of course the Abstract math. The problem is that the Abstract math is a pre req to the Alg and the Geo, and Lin Alg II is a pre req to the Abstract Math again I don't really understand what the deal is here since all of these classes seem to cover completely different things. (Again I can post some descriptions if necessary). I was thinking I might be able to take the Lin Alg, and the Abstract Math, Algebra, maybe Geo stuff concurrently rather than one after the other. Idk if that would work though one friend of mine insists I have no idea what I am dealing with and don't know what I would be getting into. Ya basically he told me that if I did that it would be a terrible idea and that I would crash and burn. I am not sure how right he is though?

I was considering Intro to Analysis (I/II) for the winter and spring quarters respectively but again that has intro to Abstract math as a pre req so if I don't take that in the Autumn I am kinda screwed. Oh and for any who read my other post on potential classes yes ppl's advice and some new info have led to my overhauling it that's why these new candidates have shown up.


analytic geometry
inequalities unsolved
inequalities

This is a linear function. It can only be increasing or decreasing (in the case is increasing), but no extremes.

note that the range of x is -25 to 13. so I think there are maximum and minimum values

maximum occurs at $ x\equal{}\minus{}6\plus{}\frac{19}{\sqrt{2}}$.

minimum occurs at $ x\equal{}\minus{}6\minus{}\frac{19}{\sqrt{2}}$.

Find local maxima of:

$ f(x)\equal{}\sqrt{25\plus{}x}\plus{}\sqrt{13\minus{}x}\plus{}|x|$

hello, i have got for $ y\equal{}\sqrt{25\plus{}x}\plus{}\sqrt{13\minus{}x}\plus{}x$
$ x_{max}\equal{}\minus{}6\minus{}1/8\sqrt{150\plus{}6\sqrt{17}}\plus{}3/8\sqrt{17}\sqrt{150\plus{}6\sqrt{17}}$
$ y_{max}\equal{}\minus{}6\plus{}3/8\sqrt{150\plus{}6\sqrt{17}}\plus{}3/8\sqrt{17}\sqrt{150\plus{}6\sqrt{17}}$
Sonnhard.

hello, the minimum has the coordinates $ x_{min}\equal{}\minus{}25,y_{min}\equal{}\minus{}25\plus{}\sqrt{38}$
Sonnhard.


number theory
greatest common divisor
superior algebra
superior algebra unsolved

Let $ n,m\in \mathbb{N}$. We know that
\[ Hom(\mathbb{Z}_n ,\mathbb{Z}_m ) \equal{} \{ f: \mathbb{Z}_n  \to \mathbb{Z}_m \,|\,f{\text{ is a homomorphism}}\}.
\]
Find $ |Hom(\mathbb{Z}_n ,\mathbb{Z}_m )|.$

Look at where $ 1$ goes- it can go to anything which is zero mod $ m$ when multiplied by $ n$. These are the multiples of $ \frac{m}{\text{gcd}(m,n)}$; there are gcd$ (m,n)$ of them.


LaTeX
number theory unsolved
number theory

Let n£N
We say that n is [b]"Good"[/b] if all digits under n can be the sum of some different n's divizors.

For example : The divizors of 6 are 1  2  3  6
Since we have  [img]http://www.gnux.be/latex/data/50347a3f14aea923e9f8eac867fd3bb1.png[/img] and [img]http://www.gnux.be/latex/data/a08af937649cb606cb4b1c27efecd4c5.png[/img] and [img]http://www.gnux.be/latex/data/693c55cb4cd51db599edeef83763f685.png[/img] and

[img]http://www.gnux.be/latex/data/e90f55eade93e21c6ac56d0919c40f24.png[/img] and [img]http://www.gnux.be/latex/data/ed0d35abfb5a9903e3d8a035fa308969.png[/img] and [img]http://www.gnux.be/latex/data/78d6e1f999f17d23dde068e9be4a9dd3.png[/img]
We say that 6 is good  :lol: .
Prove that the product of two good digits is also good.
Thanks
PS: it wont let me post more than 3 images..

suppose that $a$ and $b$ is good,($X=ab$)
if $k\le X$
1) if $k=nm$ such that $n\le a$, $m\le b$
$\exists (d_{i})_{i\in I},(q_{j})_{j\in J},\ such\ that\ \forall (i,j)\in I\times J,\ d_{i}|a,q_{i}|b,\ and\ \sum_{i\in I}d_{i}=n,\sum_{j\in J}q_{j}=m$
then $k=\sum_{(i,j)\in I\times J}d_{i}q_{j}$ and we have $\forall (i,j)\in I\times J,\ d_{i}q_{j}|k$
2)we must see if $\forall (n,m)\in [|0,a|]\times[|0,b|],\ k\neq nm$
but for exemple, if $ab-1$ is prime,what can we do?

Hi
I didnt understand a lot of things:
First: What is k please?

$X=ab$ is good $\Longleftrightarrow \ \forall k\le X,\ \exists \{d_{1},d_{2},..,d_{s}\}\in \{d>0| \ d|X\}$(it's your definition for an good numbre)
so, $k$ is an unspecified $\le X$ number which I test broke up into a sum of divieurs of X.


vector
search
linear algebra
matrix
combinatorics unsolved
combinatorics

Given are $ 2n\plus{}1$ real numbers. If you choose any one of them, the rest can be divided into two groups with equal sum.

Show that all the numbers are equal.

This is false as stated: see, for example, http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1248082#1248082

The correct statement requires that the two groups have equal size, as well.  That thread contains a solution for the case of integers.  This trivially extends to the case of rational numbers.  To get to the case of real numbers, you can either take a basis of $ \mathbb{R}$ as a $ \mathbb{Q}$ vector space and do some magic, or argue that there exists an appropriate integer $ m$ such that you can multiply all $ 2n \plus{} 1$ numbers by $ m$ and get things that are simultaneously "close" to integers, then round off and use the integer version.

My office mate is reading over my shoulder and claims there's a way to do it with linear algebra over the reals, but I'm not sure I believe him.

Here's another thread in which it was discussed: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=817988558&t=142892

A related problem (linked from the first thread I linked) is http://www.artofproblemsolving.com/Forum/viewtopic.php?t=224623

[quote="JBL"]My office mate is reading over my shoulder and claims there's a way to do it with linear algebra over the reals, but I'm not sure I believe him.[/quote]  Here's the proof (constructed or reconstructed by Yan Zhang and Po-Ru Loh):  it's equivalent to showing that if $ (2n \plus{} 1) \times (2n \plus{} 1)$ matrix $ M$ has 0s on the diagonal and each row has $ n$ 1s and $ n$ -1s, that its null space is one-dimensional.  (Since we already know that vectors with all entries the same are in the null space, this shows that indeed there are no others.)  Equivalently, we have to show that it has rank $ 2n$.  It's clear that its rank is at most $ 2n$.  Let $ M'$ be the matrix with its last row and column removed.  Then $ \text{rank } M' \leq \text{rank } M$, so it suffices to show that $ \text{rank } M' \equal{} 2n$.  Consider $ M'$ modulo 2, and perform the following row operations: first add every other row to the first row.  This makes the first row all-ones.  Now add the first row to each of the other rows.  The result is upper-triangular with 1s on the diagonal, so it has determinant 1.  Thus $ \det M' \equiv 1 \mod 2$ and in particular $ \det M' \neq 0$, so $ M'$ has full rank, and we're done.


geometry
incenter
rhombus
geometry proposed

Given that triangle ABC is inscribed in a circle and angle BAC=60. The tangent to the circle passing through A cuts the extension of CB at P. The bisector of angle APC intersect AB and AC at D and E respectively. Given that AD=15, AC=40. Find BC.

From simple angle chasing $ \angle DEA \equal{} \angle EDA \equal{} \frac{\pi}{3}$, then $ \overline{EA}\equal{}15$ and $ EC\equal{}25$. But $ \frac{\overline{BD}}{\overline{DA}}\equal{}\frac{\overline{BP}}{\overline{PA}}\equal{}\frac{\overline{PA}}{\overline{PC}}\equal{}\frac{\overline{AE}}{\overline{EC}}\equal{}\frac{3}{5}$ then $ \overline{BD}\equal{}9$ and $ \overline{AB}\equal{}24$.

Then for law of cos we get $ \overline{BC}^2 \equal{} 24^2 \plus{} 40^2 \minus{} 24 \cdot 40 \equal{} 1216$ and so $ \overline{BC}\equal{} 8 \sqrt{19}$

[quote="borislav_mirchev"] [color=darkred]Let $ ABC$ be a triangle with $ A = 60^{\circ}$ and the circumcircle $ w$ . Denote $ P\in AA\cap BC$ . The bisector

of  $ \widehat {APC}$ intersects $ AB$ and $ AC$ at $ D$ and $ E$ respectively. If  $ AD = 15$ ,  $ AC = 40$ , then find $ BC$ .[/color] [/quote]

[color=darkblue]If $ AD = x$ , then $ x = \frac {bc}{b + c}$ and $ a^2 = b^2 + c^2 - bc$ $ \implies$ $ \boxed {\ a = \frac {b}{b - x}\cdot\sqrt {b^2 - 3bx + 3 x^2}\ }$ .

In the particular case  $ b = 40$ and $ x = 15$ obtain $ \boxed {a = 8\sqrt {19}}$ . Is it O.K. ?!

I am surely that [b]¬[ƒ(Gabriel)³²¹º]¼[/b] doesnn't miss.[/color]

[hide="Proof."] [color=darkblue] Denote $ F\in AI\cap BC$ , where $ I$ is the incenter of $ \triangle ABC$ . Observe that $ m(\widehat {PAF}) = m(\widehat {PFA}) = C + \frac A2$

$ \implies$ $ PA = PF$ and the quadrilateral $ ADFE$ is a rhombus, where $ x = AD = \frac {bc}{b + c}$ because $ \triangle PAB\sim\triangle PCA$ $ \implies$

$ \frac {AB}{CA} = \frac {PB}{PA}$ $ \implies$ ${ \frac {DA}{DB} = \frac {PA}{PB} = \frac bc}$ $ \implies$ $ \frac {DA}{b} = \frac {DB}{c} = \frac {c}{b + c}$ $ \implies$ $ DA = \frac {bc}{b + c}$ $ \implies$ $ \boxed {\ c = \frac {bx}{b - x}\ }$ .

Thus, in the particular case $ A = 60^{\circ}$ , i.e. $ a^2 = b^2 + c^2 - bc$ obtain $ a^2(b - x)^2 = b^2\left[(b - x)^2 + x^2 - x(b - x)\right]$ , i.e.

$ \boxed {\ a = \frac {b}{b - x}\cdot\sqrt {b^2 - 3bx + 3x^2}\ }$ .[/color][/hide]
[color=darkblue][b][u]Remark.[/u][/b] If $ A = 120^{\circ}$ , then show analogously that $ a^2 = b^2 + c^2 + bc$ $ \implies$ $ \boxed {\ a = \frac {b}{b - x}\cdot\sqrt {b^2 - bx + x^2}\ }$ .

In the particular case $ b = 40$ and $ x = 15$ obtain $ \boxed {a = 56}$ . [b]Borislav Mirchev[/b], it is more nicely ![/color]


group theory
abstract algebra
induction
superior algebra
superior algebra solved

Given a group with finitely many elements of finite order, prove that these elements form a subgroup. Sounds really nice, huh? It's not even that hard (I couldn't solve it, I checked out the solution  :( ).

For a long time I've thought this problem not to be true.

Let $F_0$ be the set of elements of $G$ that has finite order. Let $F_k=F_0^k\setminus F_{k-1}$. This is the set of elements of $G$ than can be written as a product of $k$ elements of $F_0$ and no less. By hypothesis, $F_0$ is finite. Let $n$ be the number of elements of $F_0$.

I will prove that $F_{n+1}$ is void.

Let $f_g:G\to G$ defined by $f_g(x)=g^{-1}xg$. This is an inner automorphism of $G$. If $x^m=1$ then $(g^{-1}xg)^m=g^{-1}x^mg=g^{-1}g=1$. This proves that $f_g$ leaves $F_0$ invariant. Let $\{f_1,f_2,\ldots ,f_n\}=F_0$. Then for all $i,j\in \overline{1,n}$ there exists $k$ such that $f_j^{-1}f_if_j=f_k \Rightarrow f_if_j=f_jf_k$ (*).

Let $x\in F_{n+1}$. Then there exist $i_1,i_2,\ldots ,i_{n+1}\in \overline{1,n}$ such that $x=f_{i_1}f_{i_2}\ldots f_{i_{n+1}}$.
Because we have $n+1$ numbers in a set of $n$ elements, two of them must be equal. Assume $i_1=i_{n+1}$. Then by (*), $x=f_2'f_{i_1}f_{i_3}\ldots f_{i_{n+1}}=f_2'f_3'f_{i_1}\ldots f_{i_{n+1}}=\ldots =f_2'f_3'\ldots f_n' f_{i_1}^2\in F_n$. This proves that $F_{n+1}$ is void.

If $x\in F_0$ then $x^{-1}\in F_0$. Let $F$ be the subgroup of $G$ generated by $F_0$. Then if is easy to see that $F=\bigcup_{k\geq 1}F_0^k=\bigcup_{k\geq 1}F_k=\bigcup_{k=1}^n F_k=\bigcup_{k=1}^n F_0^k$. Because $F_0$ is finite, $F_0^k$ is also finite, so $F$ is finite. But $F$ is a group and finite, so all the elements of $F$ have finite order, so $F=F_0$ and $F_0$ is a group which concludes our problem.

I'm sorry to revive such an incredibly old topic, but I randomly remembered this problem last night, and I found the following proof, which I just had to write down :).

I'll write the statement again:

Let $G$ be a group, and let $H$ be the subset of $G$ consisting of elements of fintie order. If $H$ is finite, then it's a subgroup of $G$.

We prove this by induction on $|H|$ (it's obvious for $|H|=1$), and we assume WLOG that $G$ is actually generated by $H$. Every inner automorphism of $G$ induces a permutation of $H$, so we have a homomorphism from $G$ onto a finite permutation group. The kernel of this homomorphism is the centralizer of $H$, and, since $H$ generates $G$, it's precisely the center $Z=Z(G)$ of $G$. There are now two possibilities:

(a) $Z\cap H\ne\{1\}$

In this situation, we factor everything through $Z\cap H$ (which is obviously a group), and use the induction hypothesis, which works because the set of elements of finite order in $G/(Z\cap H)$ is the set of classes of elements in $H$ modulo $Z\cap H$, and there are fewer than $|H|$ of these.

(b) $Z\cap H=\{1\}$

Let $m$ be the order of $G/Z$. The transfer $V: G\to Z$ (see, for example, the next-to-last message in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=transfer&t=105047]this[/url] topic) is $x\mapsto x^{m}$. The kernel of $V$ is thus contained in $H$. On the other hand, if $h\in H$, then $h^{m}\in Z\cap H=\{1\}$, so $\ker V=H$, which means that $H$ is indeed a subgroup of $G$.


How can I write 'n choose r' in LaTeX? This question has been very frustrating. Thanks guys.

EDIT: changed sorry hehe thanks stevem

[quote="chenwb"]How can I write 'n choose r' in $ %Error. "latex" is a bad command.
$? This question has been very frustrating. Thanks guys.[/quote]
1. This page is taking a long time to load as \latex is not a recognised command. chenwb could you change \latex to \LaTeX and then you'll find it will work correctly?

2. n choose r can rendered as n \choose r  $n \choose r$
Or, if you load amsmath, you can use
\binom{n}{r} $\binom{n}{r}$
or
\tbinom{n}{r} $\tbinom{n}{r}$


What is your (realistic) goal on the AIME.

I hope to get a 9 because I think I need that to be safe on passing.

And good luck to everybody next week!  Ill need it

EDIT: this poll is all messed up for some reason.

uh oh I tried to fix the poll but ended up adding more options. This is messed up indeed.

I need to get 11 correct to have a safe index of 210.5, maybe 12 for a 220.5, but I'll fail again and get an 8 instead. If my adrenaline works negatively like it did for this year's AMC 12, I'll get around a 5.

For some reason, getting a 10+ on the AIME is a lot easier than getting a 120+ on the AMC (atleast for me).

I have never gotten 120+ on the AMC (and only twice during practice), but on the AIMEs, if I dont freak out, I can constantly get 10+.  But Ill probably be so nervous I forget everything on test day.

Haha yeah.. kinda the same for me, since with 3 hours, I'm calmer and consequently make fewer mistakes.

But wow, 10+? I constantly get 7, 7, 7, 8, 7, 7... So I guess I should really be aiming for a 6... But that would be devastating. I'm hoping for a 10 on the WOOT AIME we just had, but I probably made mistakes. I'm pretty sure I bruteforced #15 correctly, though. :)

Ok maybe not constantly...just some wishful thinking.

I tried the 2002 AIMEs and I got 7 on AIME I but 14 on the AIME II.  I also tried the 2004 AIME and got 8 on AIME I and 14 on AIME II...so averaging its pretty high, but the fluctuation is too huge to say a constant score...I just hope that this one is as easy as the 2002 AIME II because I went through that test in 2 hours...number 14 on that one has got to be the easiest number 14 ever.

i'm hoping for at least an 11 or 12, but i'll probly end up getting like a 4...  :(


function
combinatorics unsolved
combinatorics

1. there are 5 distinct gifts. we are to give all of these 5 gifts to 3 men such that each man has at least 1 gift. how many ways can we do this?
2. there are 5 distinct gifts. we are to give these gifts to 3 men such that each man has at least 1 gift (it is not necessary to give all of 5 gifts to these men). how many ways to do this?
please help me, thank you!

1. We can have the following ways:

(1,1,3) --> 3! = 6 ways to order(since we can also have 3,1,3, etc.)

(1,2,2) --> 3! = 6

Thus there are $ 6 \plus{} 6 \equal{} \boxed{12}$ ways.

i think you considered these gifts are the same.but the question asks [u]distinct[/u] gifts.

1.
There are two cases:
Case 1: There is a man received a gift and the others, each man receives two. Suppose the first man receive a gift, the second and last, each man receives two gifts, we have $ C_5^1.C_4^2.C_2^2 \equal{} 30$. But the second and third, each can also receive a gift. So we have $ 30 \plus{} 30 \plus{} 30 \equal{} 90$.
Case 2: There is a man received three gifts and the others, each man receives one. Suppose the first man receive three gifts, the second and last, each man receives one, we have $ C_5^3.C_2^1.C_1^1 \equal{} 20$. But the second and third, each can also receive three gifts. So we have $ 20 \plus{} 20 \plus{} 20 \equal{} 60$.
Thus, we have $ 90 \plus{} 60 \equal{} 150$ (ways).

Another solution that may help some people.  Consider the exponential generating function for this problem:
\[ f(x)\equal{}\left(\frac{x}{1!}\plus{}\frac{x^2}{2!}\plus{}\hdots\right)^3\equal{}(e^x\minus{}1)^3\equal{}\sum_{k\equal{}0}^{\infty}\left(3^k\minus{}3\cdot 2^k\plus{}3\right)\frac{x^k}{k!}\minus{}1\]
Therefore, the answer to 1 is $ 3^5\minus{}3\cdot 2^5\plus{}3\equal{}150$ and for 2 is $ \binom{5}{3}(3^3\minus{}3\cdot 2^3\plus{}3)\plus{}\binom{5}{4}(3^4\minus{}3\cdot 2^4\plus{}3)\plus{}\binom{5}{5}(3^5\minus{}3\cdot 2^5\plus{}3)\equal{}390$.


Prove that we have sin(2)=sin(2*{pi/2}).
where {x}=x-[x];

 Solve the equation:
x={x}/[x].

Solve the equation:
x={x}/[x].

[b]Solution:[/b]

It's kute! :D :D
You write the equation like so:
[x]={x}(1-[x]).
We have from here that we must have [x]*[x]/(-[x]+1)<1 and [x]<=1. But we know that [x] is integer so from the last two inequalities we get that [x]=-1 and x=-1/2.

Prove that we have sin(2)=sin(2*{pi/2}).
where {x}=x-[x];

[b]Solution:[/b]

We see that sin(2*{pi/2})=sin(2(pi/2-[pi/2]))=sin(pi-2[pi/2])=sin(pi-2*1)=sin(2).
We can also haver that sin(2)=2sin(1)*cos(1)
we also see that sin{pi/2}=sin(pi/2-[pi/2])=sin(pi/2-1)=cos(1)
cos{pi/2}=cos(pi-[pi/2])=cos(pi/2-1)=sin(1).
From the last three relations we have that:
sin(2)=2cos{pi/2}*sin{pi/2}=sin(2*{pi/2}).


inequalities
induction
inequalities unsolved

I proposed this inequality in
http://www.mathlinks.ro/Forum/viewtopic.php?p=276980#p276980

But no solution was given.

It can be easily solved with Lagrange multipliers.

thanks VASC.I don't know this is a problem of you :lol:
and one more question:
do you think that your "n-1 principle theorem" can slove it?(I think it can)

Nice Problem, Vasc. It's really hard. I'll try now.  :lol:

About problem of Zhoabin. I use induction to solve it, even I had creat the more general for it in my Ebook. ( About 4 problems for this kind ).

[quote="zhaobin"]and one more question:
do you think that your "n-1 principle theorem" can slove it?(I think it can)[/quote]
Yes. It works. :lol:

Could anyone enlighten me about the [b]n-1 principle theorem[/b]?

Thanks in advance ;)


search
function
inequalities
inequalities unsolved

I think this has been posted before, but I couldn't find it with the Search function:

For $a,b,c>0$, prove

$\frac{a^n}{b+c}+\frac{b^n}{c+a}+\frac{c^n}{a+b} \ge \frac{a^{n-1}+b^{n-1}+c^{n-1}}{2}$.

I'm curious if this solution works:
[hide]
$\sum \frac{a^{2n-2}}{a^{n-2}(b+c)} \ge \frac{(a^{n-1}+b^{n-1}+c^{n-1})^2}{\sum a^{n-2}b}$ by Cauchy.

But since $a^{n-1}+b^{n-1}+c^{n-1} \ge \frac{\sum a^{n-2}b}{2}$ by rearrangement, we have

$\frac{(a^{n-1}+b^{n-1}+c^{n-1})^2}{\sum a^{n-2}b} \ge \frac{(a^{n-1}+b^{n-1}+c^{n-1})\frac{\sum a^{n-2}b}{2}}{\sum a^{n-2}b} \ge \frac{a^{n-1}+b^{n-1}+c^{n-1}}{2}$ as desired.
[/hide]

The question is trivial from rearrangement+nesbitt

We have

$\sum \frac {a^n}{b+c}\geq \frac {1}{3}\left(\sum a^{n-1}\right)\left(\sum \frac {a}{b+c}\right)\geq \frac {\sum a^{n-1}}{2}$

Also, we get

$\sum \frac {a^n}{b+c}\geq \frac {1}{2}\left(\sum \frac {b^{n}+c^{n}}{b+c}\right)$

But $(b^{n-1}+c^{n-1})(b+c)\leq 2(b^n+c^n)$

Thus we have

$\sum \frac {a^n}{b+c}\geq \frac {1}{2}\left(\sum \frac {b^{n}+c^{n}}{b+c}\right)\geq \frac {1}{2}\left(\sum \frac {b^{n-1}+c^{n-1}}{2}\right)=\frac {1}{2}\left(\sum a^{n-1}\right)$

The source is "inequalities class final [exam]".
It is surprising that courses about inequalities exist -- may I ask where?


linear algebra
matrix
linear algebra unsolved

It is like this one
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=152320&highlight=#152320
-----------------------------------------------------------------------------------------------------

$X,Y\in M_{n}(C)$
$XY=YX$
$X^a=Y^b=I_n$ where $a,b$ are both even positives integers

Prove that $X-wY$ is invertible where $w$ are any root of $z^{ab}=-1$ in $C$

$(w^m \neq 1)\& (X^m = Y^m \ invertible)\& (XY=YX) \Longrightarrow  X - wY \  invertible$
:cool:


Let be given a triangle $ABC$. Prove that:
$1.\quad\sum\sin A\ge\sum\sin{2A}$
$2.\quad\sum\cos A+\sum\cos{2A}\ge 0$

[quote="April"]Let be given a triangle $ABC$. Prove that:
$1.\quad\sum\sin A\ge\sum\sin{2A}$
[/quote]

using $\sum\sin{A}=4\prod\cos{\frac{A}{2}}, \sum\sin{2A}=4\prod\sin{A}$ and $\sin{A}.\sin{B}\leq\cos^{2}{\frac{C}{2}}$

We have:
\[1.\quad\sum\sin A\ge\sum\sin 2A\Leftrightarrow\prod\cos\frac{A}{2}\ge\prod\sin A\Leftrightarrow\frac{1}{8}\ge\prod\sin\frac{A}{2}\]
\[2.\quad\sum\cos A+\sum\cos 2A\ge 0\Leftrightarrow\left(1+4\prod\sin\frac{A}{2}\right)-\left(1+4\prod\cos A\right)\Leftrightarrow\prod\sin\frac{A}{2}\ge\prod\cos A \]
But it's simple because: $\cos B.\cos C=\frac{1}{2}\left(\cos(B-C)+\cos(B+C)\right)\le\frac{1}{2}(1-\cos C)=\sin^{2}\frac{A}{2}$


Ring Theory
algebra
function
domain
superior algebra
superior algebra unsolved

Could you give an example of a PID, which is not Euclidian domain.

* I have seen the example $ \mathbb{Z}[\alpha]$, where $ \alpha\equal{}[\frac{1\plus{}\sqrt{\minus{}19}}{2}]$.  Do you know any other one ?

[quote="eugene"]Could you give an example of a PID, which is not Euclidian domain.

* I have seen the example $ \mathbb{Z}[\alpha]$, where $ \alpha \equal{} [\frac {1 \plus{} \sqrt { \minus{} 19}}{2}]$.  Do you know any other one ?[/quote]

Hello.
Daniel Perrin, Cours d'algèbre (Paris, 2004), chapter II, exerc. 5, pp. 62-63, gives many hints in order to prove that the ring $ \mathbb{R}[X,Y]/(X^{2} \plus{} Y^{2} \plus{} 1)$ is principal but is not euclidean.
But I confess that I didn't solve this exercise...

Friendly,
P.


function
trigonometry
calculus
derivative
complex analysis

this a very nice problem:
find all the pairs  of entire functions $f,g $  (holomorfic in all the plane) such  $f+g=1$

We have $f^2(z)+g^2(z)=1$ if and only if there exists an entire function $h$ such that $f(z)=\sin h(z)$ and $g(z)=\cos h(z)$.

First, if $h(z)$ is entire, $\sin h(z)$ and $\cos h(z)$ are entire- we can differentiate by the chain rule. Also, it is obvious that $\sin^2h(z)+\cos^2h(z)=1$.

Conversely, we can construct such a function $h$ from $f$ and $g$. The functions $\sin^{-1}(z)$ and $\cos^{-1}(z)$ are analytic except at their branch points $\pm 1$, so only bad behavior at these singularities could cause problems.
Away from the singularities, we can start by choosing a branch; find $a$ such that $\cos a = \cos(-a) = g(b)$ and then either $\sin a = f(b)$ or $\sin(-a)=f(b)$. With this choice, $\sin^{-1}(f(z))$ and $\cos^{-1}(g(z))$ are analytic at b, and both are identical; their derivatives are $\frac{f'(z)}{\sqrt{1-f^2(z)}}=\frac{f'(z)}{g(z)}$ and $\frac{g'(z)}{\sqrt{1-g^2(z)}}= -\frac{g'(z)}{f(z)}$ respectively, and since $2ff'+2gg'=0, \frac{f'}{g}+\frac{g'}{f}=0$.
Define $H(z)=\frac{f'(z)}{g(z)}= -\frac{g'(z)}{f(z)}$. H is analytic except possibly where f=0 or g=0. Since f and g are never both zero, one of the definitions is always analytic, and $H$ is entire. An analytic function in a simply connected region has an antiderivative, so we can find $h$ with $h'(z)=H(z)$ and $h(b)=a$; then $f(z)=\sin h(z)$ and $g(z)=\cos h(z)$

Nice proof,....

that's what I did: (It cost me like a year)

Well, since $f$ and $g$ are holomorfic in tha hole plane then $f+ig$, $f-ig$ are so, and they hay no zeros (since $f^2+g^2=1$), then must be $u,v$ entires such that
$f+ig=e^u$ and $f-ig=e^v$ then since $1=f^2+g^2=e^{u-v}$ we may take
$u=-v$. Finally $f+ig=e^u$ and $f-ig=e^{-u}\Longrightarrow f=\sin u$ and $g=\cos u$ .


[quote="kyyuanmathcount"]A solar system contains four planets and a sun.  Planet A completes an orbit around the sun in 20 months; planet B completes an orbit in 36 months; planet C completes an orbit in 40 months; and planet D completes an orbit in 48 months.  If the four planets lie along a straight line in May of year 2000, what is the next year that they will again lie along the same line? Assume that the orbits are circular and that a year has exactly 12 months.[/quote]

[hide]You need to find the Least Commom Multiple (LCM) of 20, 36, and 48 which is 720 months. Then you add it to May 2000. 720 divided by 12 (because there are 12 months in a year) which is 60. 2000+60=[size=150]2060[/size].

MY 50TH POST!! WOHOO!!!!  :clap2:  :clap: [/hide]

[hide]The LCM of 20,36,40, and 48 is 720, so the planets will meet again in 720 months. This will occur 60 years later, in the year 2060.[/hide]

You guys are talking about full rotation.

[hide]The planets are aligned when they're halfway (sorry if that didn't make any sense.

Here's an example: Imagine a clock and the center point is the sun.  All of the planets are aligned at 9 o'clock.  The next time they are aligned is 3 o'clock, not 9.

So thus, I believe that the right answer is $2030$[/hide]

Actually it is full rotation... the problem says when is the next year they will lie on the [i]same[/i] line.
[hide]LCM(20,40,36,48) = 2^4*3^2*5 = 720
720/12 = 60
So the next year they will be on the same line is [color=green]2060[/color].[/hide]

Actually, robinhe has the correct answer.  Lying on the same line does not assume that the sun is in the middle.

You can have:

Planet 1    Sun      Planet 2       planet 3      planet 4.

I thought same line meant the line they were on originally... but I guess halfway around is the original line.

I remember this problem!

Yeah, the entire team did 2060 instead of 2030. :|

It's a line, not a segment or a ray. It continues infinitely in either direction. So, in $30\, \text{years}=360\, \text{months}$,  Planets A, B, and C will have done complete orbits and Planet D will be halfway around. You can find this answer without too much work, as well. Notice that the number must be a multiple of $20$ so that Planet C can be either halfway or all the way around in its orbit. Similarly, it must be a multiple of $18$ and $24$. $LCM(18,24)=72$, so we simply multiply until it becomes a multiple of $20$, which gives $360$. This is also how you can be sure that there are no smaller answers.

[hide]
Its 2030.
Half the lcm because they can be on the other side of the sun![/hide]

[hide]
Factoring the months for orbits:
$20: 2^2 \cdot 5$
$36: 2^2 \cdot 3^2$
$40: 2^3 \cdot 5$
$48: 2^4 \cdot 3$

Since it doesn't have to be on one side of the sun, it can be on the other side as well. Since it is half an orbit, you have to divide the months for an orbit by 2.
The least common multiple of:
$10: 2 \cdot 5$
$18: 2 \cdot 3^2$
$20: 2^2 \cdot 5$
$24: 2^3 \cdot 3$

is $2^3 \cdot 3^2 \cdot 5 = 360$
360 months is 30 years.
So, it is 2030.[/hide]


I am not able to convince myself that $ \mathbf{S^{1}} \times \mathbf{S^{1}}$ is a torus.For example its easy to visualize that $ (0,1) \times (0,1)$ is the unit open square.How does one do so about the torus?

A circle is just a unit interval with its endpoints identified.  You can probably visualize this process occurring.  A torus is therefore a unit square with its opposite sides identified.  Visualize one pair of sides being identified (obtaining a cylinder) and then the other (the two ends of the cylinder getting pasted together).

[quote="Soarer"][quote="pluristiq"]well, you get a torus by dragging a circle through space along a circular path, hence the homeomorphism to $ S^1 \times S^1$.[/quote]

Is this a valid argument? What if I use the same idea to visualize mobius band and the cylinder? In this way they sound like the same.[/quote]

how does that sentence describe a cylinder?

my description was [i]merely[/i] intended to help visualize a torus as $ S^1 \times S^1$... every point on the torus can be associated with a unique pair $ (a,b)$, where $ a$ is a point on the first circle (the one being dragged), $ b$ a point on the second circle (the path of the first circle), in the natural way...

What I mean is, both mobius band and cylinder are "dragging a line along a circular path". More generally it seems like what he said would imply any vector bundle of same rank over a manifold would be isomorphic.

Well the reason why it's not totally foolproof is that it doesn't account for the fact that you can twist the fiber. It's still a good visualization though. Thinking of it as the unit square with opposite sides identified also helps.
http://www.youtube.com/watch?v=0H5_h-RB0T8&feature=PlayList&p=21815B0942D07F94&playnext=1&playnext_from=PL&index=12 is a good video illustrating these ideas.


inequalities
trigonometry
inequalities unsolved

1) Let $ \alpha\in(0;\frac{\pi}{2}]$. Prove $ \sqrt{\sin\alpha}\minus{}\sqrt{\alpha}\geq 1\minus{}\sqrt{\frac{\pi}{2}}$

2) Let $ \alpha\leq 6,\beta\leq \minus{}8,\gamma\leq 3$. Prove that
$ x^4\geq\alpha x^2\plus{}\beta x\plus{}\gamma$ with $ x\geq 1$

The first one:We can use Jordan Inequality and monotonicity of quardratic function.

[quote="Algadin"]Let $ \alpha\in\left(0;\frac {\pi}{2}\right]$. Prove $ \sqrt {\sin\alpha} \minus{} \sqrt {\alpha}\geq 1 \minus{} \sqrt {\frac {\pi}{2}}.$[/quote]$ f(\alpha)\equal{}\sqrt {\sin\alpha} \minus{} \sqrt {\alpha}\Longrightarrow$

$ f'(\alpha)\equal{}\frac{1}{2}\sqrt{\frac{\cos\alpha}{\tan\alpha}}\minus{}\frac{1}{2\sqrt{\alpha}}<0$ on $ \left(0,\frac{\pi}{2}\right),$

$ \Longrightarrow f(\alpha)\geq f\left(\frac{\pi}{2}\right)\equal{}1 \minus{} \sqrt {\frac {\pi}{2}}.$

[quote="Algadin"][quote="Cai Ni"]2$ LHS \minus{} RHS\ge(x \minus{} 1)^2*[(x \plus{} 1)^2 \minus{} 4]\ge 0$[/quote]
Why??? :maybe:  :maybe:[/quote]

$ (x \minus{} 1)^2 *[(x \plus{} 1)^2 \minus{} 4] \equal{} (x^2 \minus{} 1)^2 \minus{} 4(x \minus{} 1)^2 \equal{} x^4 \minus{} 6x^2 \plus{} 8x \minus{} 3$
$ LHS \minus{} RHS \equal{} x^4 \minus{} \alpha x^2 \minus{} \beta x \minus{} \gamma$
Therefore $ LHS \minus{} RHS\ge(x \minus{} 1)^2*[(x \plus{} 1)^2 \minus{} 4]$, because $ \alpha\leq 6,\beta\leq \minus{} 8,\gamma\leq 3$.


complex analysis
function
limit
complex numbers

Let $ G$ be an open subset of $ \mathbb C$ such that $ z\in G$ implies $ \bar{z}\in G$. Show that if $ f$ is holomorphic on $ G$, then the function $ g(z)\equal{}\overline{f(\bar{z})}$ is also holomorphic on $ G$.

Also, could we express $ g'(z)$ in terms of $ f'(z)$?

let ${ \bar{a}\in G}$ since $ f$ is analytic at $ \overline{a}$ then we have $ f = \sum_{n\geq 0} A_{n}(z - \bar{a})^{n}$ , $ g(z) = \overline{f(\bar{z})} = \sum_{n\geq 0}\bar{A_{n}}(z - a)^{n}$, hence $ g$ is analytic at $ a$ , therefore $ g$ is analytic on $ G$. It is easy to find $ g'(z)$ in term of $ f'$

Is there an "elementary" way to do it, say, by using the definition of a holomorphic function (a function which is defined and differentiable on an open subset of $ \mathbb C$), and the theorems about Cauchy-Riemann equations?

Since $ f$ is holomorphic on $ G$, we have

$ \lim_{h\to0}\frac{f(z\plus{}h)\minus{}f(z)}h\equal{}f'(z)$ for all $ z\in G.$

To prove $ g(z)\equal{}\overline{f(\overline{z})}$ is holomorphic on $ G$, we have to show that the limit

$ \lim_{h\to0}\frac{g(z\plus{}h)\minus{}g(z)}h$

exists for every $ z\in G$. Now,

$ \lim_{h\to0}\frac{g(z\plus{}h)\minus{}g(z)}h$

$ \equal{}\lim_{h\to0}\frac{\overline{f(\overline{z\plus{}h})}\minus{}\overline{f(\overline{z})}}h$

$ \equal{}\lim_{h\to0}\frac{\overline{f(\overline{z}\plus{}\overline{h})\minus{}f(\overline{z})}}h$

$ \equal{}\lim_{h\to0}\frac{f(\overline{z}\plus{}\overline{h})\minus{}f(\overline{z})}{\overline h}$ [since $ \overline{a/b}\equal{}\bar a/\bar b$]

$ \equal{}f'(\overline z).$

Thus $ g$ is holomorphic on $ G$.

[quote="tmrfea"]Are also $ f(\overline{z})$ and $ \overline{f(z)}$ holomorphic?[/quote]
No, they are not.

I see now:

For $ g(z)=f(\overline{z})$:

\begin{eqnarray*}
\lim_{h\to 0}\frac{g(z+h)-g(z)}h&=&\lim_{h\to0}\frac{f(\overline{z+h})-f(\overline{z})}h=\lim_{h\to0}\frac{f(\overline{z}+\overline{h})-f(\overline{z})}h=\\
&=& f'(\overline z) \quad \mathrm{for} \quad h\in \mathbb{R}\\
&=& -f'(\overline z) \quad \mathrm{for} \quad h\in \mathbb{C}\setminus \mathbb{R}
\end{eqnarray*}

For $ k(z)=\overline{f(z)}$:

\begin{eqnarray*}
\lim_{h\to 0}\frac{k(z+h)-k(z)}h&=&\lim_{h\to0}\frac{\overline{f(z+h)}-\overline{f(z)}}h=\lim_{h\to0}\overline{\frac{f(z+h)-f(z)}{\overline{h}}}=\\
&=& f'(z) \quad \mathrm{for} \quad h\in \mathbb{R}\\
&=& -f'( z) \quad \mathrm{for} \quad h\in \mathbb{C}\setminus \mathbb{R}
\end{eqnarray*}

But for $ l(z)=\overline{f(\overline{z})}$:

\begin{eqnarray*}
\lim_{h\to 0}\frac{l(z+h)-l(z)}h&=&\lim_{h\to0}\frac{\overline{f(\overline{z+h})}-\overline{f(\overline{z})}}h=\lim_{h\to0}\overline{\frac{f(\overline{z+h})-f(\overline{z})}{\overline{h}}}=\lim_{h\to0}\overline{\frac{f(\overline{z}+\overline{h})-f(\overline{z})}{\overline{h}}}\\
&=& \overline{f'(\overline{z})}
\end{eqnarray*}

[quote="tmrfea"]I see now:

For $ g(z) = f(\overline{z})$:

\begin{eqnarray*} \lim_{h\to 0}\frac {g(z + h) - g(z)}h & = & \lim_{h\to0}\frac {f(\overline{z + h}) - f(\overline{z})}h = \lim_{h\to0}\frac {f(\overline{z} + \overline{h}) - f(\overline{z})}h = \\
& = & f'(\overline z) \quad \mathrm{for} \quad h\in \mathbb{R} \\
& = & - f'(\overline z) \quad \mathrm{for} \quad h\in \mathbb{C}\setminus \mathbb{R} \end{eqnarray*}

For $ k(z) = \overline{f(z)}$:

\begin{eqnarray*} \lim_{h\to 0}\frac {k(z + h) - k(z)}h & = & \lim_{h\to0}\frac {\overline{f(z + h)} - \overline{f(z)}}h = \lim_{h\to0}\overline{\frac {f(z + h) - f(z)}{\overline{h}}} = \\
& = & f'(z) \quad \mathrm{for} \quad h\in \mathbb{R} \\
& = & - f'( z) \quad \mathrm{for} \quad h\in \mathbb{C}\setminus \mathbb{R} \end{eqnarray*}
[/quote]
No, it's not correct. The limits do not exist for $ h\in \mathbb{C}\setminus \mathbb{R}$. (Did you mean $ h\in i\mathbb R$?)

In fact, a function $ f$ is holomorphic iff $ \partial f/\partial \overline z = 0$. Which means that if $ f$ is holomorphic, then $ f(\overline z)$ or $ \overline{f(z)}$ can happen to be holomorphic only when $ f$ is constant.

Thank you, zhoraster.

[quote="zhoraster"]No, it's not correct. The limits do not exist for $ h\in \mathbb{C}\setminus \mathbb{R}$. (Did you mean $ h\in i\mathbb R$?) [/quote]

Yes, I meant $ h\in i\mathbb R$.

[quote="tmrfea"]
Yes, I meant $ h\in i\mathbb R$.[/quote]
Anyway, you have other problems there as well :)


geometry
geometric transformation
reflection
trigonometry
circumcircle
symmetry
geometry proposed

Let $ ABC$. $ T$ is Torixeli point. $ AT,BT,CT$ cut $ BC,CA,AB$ at $ A_1,B_1,C_1$
$ T_1,T_2,T_3$ is the symmetric poit of $ T$ by $ BC,CA,AB$. Prove that $ A_1T_1,B_1T_2,C_1T_3$ are concurrent

Denote $ T'$ the isogonal conjugate point of Torixeli point $ T, T'_a$ the reflection of $ T'$ across $ BC$. We will show that $ T'_a\in AT$.
For every point $ X$ which lies on $ AT$. We have $ \frac{S_{AXB}}{S_{AXC}}\equal{}\frac{BA.AT.\sin \angle BAT}{CA.AT.\sin \angle CAT}\equal{}\frac{BT}{CT}$
So to prove $ T'_a\in AT$ we will prove $ \frac{S_{ABT'_a}}{S_{ACT'_a}}\equal{}\frac{BT}{CT}$
Or $ \frac{AB.BT'_a.\sin \angle ABT'_a}{AC.CT'.\sin \angle ACT'_a}\equal{}\frac{BT}{CT} (1)$
Put $ \angle ABT\equal{}y, \angle ACT\equal{}z$ then $ \frac{\sin z}{\sin y}\equal{}\frac{AC}{AB}\equal{}\frac{\sin C}{\sin B} (2)$
$ (1)\Leftrightarrow \frac{\sin C}{\sin B}.\frac{\sin z}{\sin y}.\frac{\sin (B\plus{}y)}{\sin (C\plus{}z)}\equal{}\frac{\sin (C\minus{}z)}{\sin (B\minus{}y)}$
$ \Leftrightarrow \sin^2C.\sin (B\plus{}y).\sin (B\minus{}y)\equal{}\sin^2 B.\sin (C\plus{}z).\sin (C\minus{}z)$
$ \Leftrightarrow \sin^2C.(\cos2B\minus{}\cos 2y)\equal{}\sin^2B(\cos 2C\minus{}\cos 2z)$
$ \Leftrightarrow \sin^2C.(1\minus{}2\sin^2B\minus{}1\plus{}2\sin^2y)\equal{}\sin^2B(1\minus{}2\sin^2C\minus{}1\plus{}2\sin^2z)$
$ \Leftrightarrow \sin^2C.\sin^2y\equal{}\sin^2B.\sin^2z$ (It's true from $ (2)$)
Therefore $ T'$ lies on $ T_1T$. Similarly we are done.

[quote]$\triangle ABC$ is scalene and $ F,F'$ denote its 1st Fermat point and 1st isodynamic point. $ F_a,F_b,F_c$ are the reflections of $ F$ across $ BC,CA,AB.$ Prove that the cevian triangle $ \triangle A''B''C''$ of $ F$ WRT $ \triangle ABC$ is perspective with $ \triangle F_aF_bF_c$ through $ F'.$[/quote]
Since pedal triangle of $F'$ with respect to $ \triangle ABC$ is equilateral, the reflections $ A',B',C'$ of $ F'$ about $ BC,CA,AB$ form another equilateral triangle centrally similar to its pedal triangle $\Longrightarrow$ Perpendiculars from $ A,B,C$ to the sidelines of the pedal triangle of $ F'$ concur then at the isogonal conjugate of $ F'$ WRT $ \triangle ABC.$ Therefore,  $ AX\perp B'C',$ $BY \perp A'C'$ and $ CZ\perp A'B'$ concur at $ F,$ where $ X \in B'C',$ $Y \in A'C'$ and $Z \in A'B'.$

Note  that $ BC,CA,AB$ become perpendicular bisectors of $ F'A',F'B',F'C',$ thus it follows that $ A,B,C$ are circumcenters of $ \triangle F'B'C',\triangle F'A'C'$ and $\triangle F'A'B'$ $ \Longrightarrow F$ is the circumcenter of $ \triangle A'B'C'$ $ \Longrightarrow$ $ A' \in AF, $ $B' \in BF$ and $ C' \in CF.$ Therefore, the reflections $ F_aF',F_bF',F_cF'$ of the lines $ AF,BF,CF$ across $ BC,CA,AB$ pass through $ A'',B'',C''$ $ \Longrightarrow$ $ \triangle F_aF_bF_c$ and $ \triangle A''B''C''$ are perspective through $ F'.$

[quote="Luis González"]$\triangle ABC$ is scalene and $ F,F'$ denote its 1st Fermat point and 1st isodynamic point. $ F_a,F_b,F_c$ are the reflections of $ F$ across $ BC,CA,AB.$ Prove that the cevian triangle $ \triangle A''B''C''$ of $ F$ WRT $ \triangle ABC$ is perspective with $ \triangle F_aF_bF_c$ through $ F'.$[/quote]
Clearly $F_a$ is the isogonal conjugate of $A$ wrt $\triangle F'BC$. Now $AA_1, AF'$ are isogonals.($A_1=AF'\cap BC$). We know that $F'$ lies on A-Apollonius circle. Hence $F'A'', F'A_1$ are isogonal, i.e $F'A'',F'A $ are isogonal $\Longrightarrow F',A'', F_a$ are collinear. Similarly $F',B'',F_b$ are collinear and same thing holds for $C$. So done.

We can prove the stronger result:

Let $ P $ be a point and $ Q $ be the reflection of $ P $ with respect to $ BC $ ,
then $ AP,AQ $ are isogonal line of $ \angle BAC $ iff $ P $ lie on A-Apollonius circle .

Proof:

[b]Lemma :[/b]

Let $ P,P' $ be the isogonal conjugate of triangle $ ABC $ .
Let $ \ell $ be the isogonal conjugate of $ AP $ WRT $ \angle BPC $ .
Let $ \ell ' $ be the isogonal conjugate of $ AP' $ WRT $ \angle BP'C $ .

Then $ \ell, \ell' $ are symmetry with respect to $ BC $.

[b]Proof of the lemma:[/b]

Let $ X $ be the intersection of $ BP' $ and $ CP $ .
Let $ Y $ lie on $ BC $ such that $ XA,XY $ are isogonal conjugate WRT $ \angle BXC $ .

Since $ CA,CY $ are isogonal conjugate WRT $ \angle P'CP $ ,
so $ A,Y $ are isogonal conjugate of triangle $ CXP' $ .
i.e. $ P'Y,P'A $ are isogonal conjugate WRT $ \angle BP'C $ .
Similarly, we can prove $ PY, PA $ are isogonal conjugate WRT $ \angle BPC $ .

Since $ \angle PYB=\angle CYP' $ ,
so we get $ PY,P'Y $ are symmetry with respect to $ BC $ .
____________________________________________________________
[b]Back to the main problem:[/b]

Let $ P' $ be the isogonal conjugate of $ P $ WRT to $ \triangle  ABC $ .

If $ P $ lie on A-Apollonius circle

Since $ PB:PC = \sin \angle PCB : \sin \angle PBC = \sin \angle P'CA : \sin \angle P'BA $ ,
so we get $ \angle AP'B=\angle AP'C \Longrightarrow $ from lemma we get $ AP' $ pass through $ Q $ .

Similarly, we can prove the converse .

By the way,[b]Luis[/b]'s method also work .


function
integration
calculus
calculus computations

We define the function $ g: [0,2\pi]\rightarrow\mathbb{C}$ such that for some  $ k\in\mathbb{C}$ et  $ \forall t\in\ [0,2\pi]$ we have $ |g(t)|\le\ |k|$.

prove that if $ \int_{0}^{2\pi}(k-g(t))dt=0$ so $ \forall t\in\mathbb{C}$ we have $ g(t)=k.$

If $ k \equal{} r e^{i\alpha}$, $ r>0$, multyplying this by $ e^{\minus{}i\alpha}$ and taking real parts, you reduce this problem to real line.


So which book should I buy if I want to practice for the AMC and maybe AIME. What kind of material does the book have? Do you recommend buying the AHSME books from the old years? Which books are better: The Art of problem solving or AHSME books for the AMC

If you can't get at least 130 on AMC12, the Art of Problem Solving books are much better since they teach you strategy. If you can get at least 130 or so, then all you need is practice in which case you should get books of past tests and just go through them.

But of course, the Art of Problem Solving is a very helpful book anyways, and even if you have gotten above 130, you will still almost certainly benefit from reading it.  (I know I did, and I got above 130...)

[color=cyan]I think I'm going to make it a habit of disagreeing with Simon.  I think that you need to take a couple of the more recent exams, and then decide what your bigger problem is -- would you benefit more from more knowledge, or do you get questions wrong/take too long solving questions in areas you know?

Because if you just don't know the math on half of the AMC, then AOPS would be more helpful.  On the other hand, even if you know the material (have taken geometry or algebra 2, say, or even if you've already just learned it from other ocntests or other routes entirely) but just are having difficulty doing it quickly/accurately, then you'd probably benefit more from practice problems.  You don't have to wait until you score very highly to start practicing -- in fact, that hardly makes any sense at all.  If anything, I'd go for both books, and anything else of a similar difficulty you can find.  You might think about the ARML-NYSML books for more problems to do of a similar (slightly harder, in most cases) difficulty level.  They are very well written.  Larry Zimmerman really knows what he's doing, as did the other author whose name escapes me.[/color]

definitely get the aops books. both of em. and make sure to get the solutions.

the first isnt only for mathcounts. it helps of course. but they have many ahsme problems. i dont know how good you are, but test it out for yeself. even if you are advanced for it, you can always pick up lotsa stuff

then i suggest both, youll improve dramatically if you do all the problems and know how to do them afterwards

[quote="hello"]but isn't the first book only for mathcounts. i am going to 12th grade. Why are the book so expensive. With solutions almost 4o dollars?[/quote]

Definitely not.  The Art of Problem Solving is much harder than mathcounts.  For instance, one of the first chapters of AoPS talks about imaginary numbers, which mathcounts never even goes into.  By the end, you're starting on proofs, and throughout the books there are lots of challenging problems.

Consider this: I got a 137 on the AMC, and I find AoPS Vol. 1 challenging and helpful.  I've learned lots of things from it.

And as for the cost?  Really, it's not that much.  Zeitz is $70, I believe, and it's one book with no solutions.  You can get both books and solution manuals for less than $70.  Book 1 is $25.  That's how much you'd pay for an SAT prep book, and believe me, you will learn much more from this than from an SAT prep book.

How about this?  Hypothetical situation: you are a guy who's done math competitions all his life, and you decide to write a book to teach other people how to do math competitions better.  You spend thousands of hours figuring out exactly the right wording that will help your students, buying problem sets and sorting through them to figure out which questions are appropriate for which chapter, writing out nice, concise solutions that will help your readers understand, you spend money out of your own pocket to build a site where your readers can come to hang out, discuss math, etc.  You have to pay to get the book published.  You and another guy split the proceeds.  When it all comes down, you have to realize that Mr. Rusczyk hardly gets any money from selling one book.  And I'm sure I'm underexaggerating this, too.

And after I think about it, you really don't need an AMC problem book as much as you'll need AoPS.  AoPS features something like 1100 problems, with solutions, with lessons - everything you need!

well said mathfanatic
3 cheers for Mr. Rusczyk. hip hip...

[color=cyan]was it really necessary to quote that entire thing?  Not that I disagree, or anything.[/color]

[quote="Syntax Error"]well said mathfanatic[/quote]

Thank you.  We try...

[quote="Syntax Error"]3 cheers for Mr. Rusczyk. hip hip...[/quote]

Yes, indeed.  Thanks for the great book, website, forum, math jams, and everything!  And let's make that 3.14159 cheers....

that was extremely pointless. so is this if i left it at that, but how hard is aops v2?

This is not a flame but if you have time to post that many digits of pi, you honestly need a life.

all he did was copy and paste it. he can find it especially fast cause its linked on his site.


geometry
perimeter
trigonometry
geometry unsolved

In a triangle $ABC$, sides $a, b, c$ and angles $A, B, C$ and perimeter $2p$. Show that $a=p\frac{sin(\frac{A}{2})}{cos(\frac{B}{2})cos(\frac{C}{2})}$


This question came in my exam today.Lots of controversy around it. :p

a1,a2,a3 ....a20 are AMs inserted between the numbers 3 and 77. Find the max. value of a1a2a3a4.....a20.

a)20^20               b)40^20
c)80^20               d) None of these.

My answer was d) None of these because  of 2 reasons:

1)The no.s have a definite value, so product should also be fixed.
2)Equality holds when a1=a2=a3=..=a20, which is possible when d=0 which is surely not the case.

If you dont consider these two facts then answer obviously is b) which most of my friends are saying but I dont agree.What say?

aren't the value of AMs remains fixed????
what is this maxima????

Your answer is wrong. That is because inserting 20 AM's means basically means that the 22 numbers (Including a,b) must be in AP.

But your answer does not follow the rule. Surely, 1,2,2,3. Is not in AP.

aah well, if it was in a school exam, i would've written B, because the only property to be used is that AM if the 20 AM's = AM of 3 and 77 :P
otw, your answer is right.
but as shreyas pointed out, your second reason is wrong.
:)

[quote="Conjurer"]Ye, thats what I have said in my first argument.So what do u think d) right?

@ above : Where did this series come from 1,2,2,3[/quote]

Just giving an example of what your statement implies. Basically you say that 1,2,2,2,2,2,2,3 are in AP which is wrong, isn't it??

So I saw that the two facts must not be considered because they are against the basic definition of AM.

You got me wrong.I am also saying it is not possible.

wudn't the values of 20 AM's remain fixed?

i doubt shreyas is arguing on that point.
i think he's refering to this point of yours
[quote="Conjurer"]
2)Equality holds when a1=a2=a3=..=a20, which is possible when d=0 which is surely not the case.
[/quote]
the thing is, that you can't consider this case itself. because 3, 20 ,20 are NOT in AP..

Oh.OK it is NOT possible but max value will be 40^20 when all the terms are equal na, so it is not possible isnt it.

PLz tell me what is the most appropriate answer?

definitely D only.
But why are u fretting about it so much!? it's not as if it's a jee qsn  :huh:

My coaching institute question.And I had arguments with the "toppers" .Wanna own them hard  :D

even if key says B u can rgue on ur point :D