symmetry
superior algebra
superior algebra unsolved

hello, iv'e picked up an algebra book and started to work through it, i am curious to see another solution to this exercise if possible.
[u]a first course in abstract algebra 6ed John B. Fraleigh chapter 1.3 ex 38[/u]
prove that a nonempty set $ G$, together with an associative binary operation $ *$ on $ G$ such that $ a*x\equal{}b$ and $ y*a\equal{}b$ have solutions in $ G$ for all $ a,b \in G$

here's my solution:
counting all the possible equations $ a*b\equal{}x$ we have $ |G|^2$ such unique equations. counting all the possible equations of the form $ a*x\equal{}b$ we have at least $ |G|^2$ so we conclude there is one and only one solution to each equation of the form $ a*x\equal{}b$ similarly for $ y*a\equal{}b$ which are permutations of the other set of equations.

now let $ e*a\equal{}a$ and $ a*a'\equal{}e$
$ e*a\equal{}a$
$ e*a*a'\equal{}a*a'$
$ e*e\equal{}e$
let $ z*b\equal{}b$ and $ b*b'\equal{}e$
$ z*b\equal{}b$
...
$ z*e\equal{}e$  now $ z\equal{}e$ since there is only one solution to equation $ y*e\equal{}e$ so $ e$ is a left identity element.
repeat the same procedure but on the right, ie use $ a*i\equal{}a$ and $ a"*a\equal{}i$ we get $ i*i\equal{}i$ but $ e*i\equal{}i$ also. so $ e\equal{}i$. $ e$ is a left & right identity.

now for each $ a \in G$ there is a unique $ b$ that is the solution to $ a*x\equal{}e$. looking at the equation $ y*b\equal{}e$, $ a$ must be the unique solution here or else we have two solutions for $ y*b\equal{}e$ similarly if $ c*b\equal{}e$ then $ c\equal{}a$. so each $ a$ has a unique right inverse $ b$ which means the set of inverses are permutations of the elements and each element has a left inverse. therefore $ G$ with $ *$ form a group.

i was wondering if there is a proof of this without the argument in the beginning that the equations have a unique solution? something more algebraic..

Note that your proof only works if $ G$ is finite. And you never used that $ G$ is associative although maybe you did implicitly. I'll have to think some more though on another solution.

actually now im a little confused by the problem. what are we trying to show? you seem to be trying to prove G is in fact a group?

opps sorry, i missed the last 3 words   prove...[b]"is a group"[/b] oh and i didn't prove left and right inverses are the same...  but i can add that now,
let $ a*b \equal{} e$ and $ c*a \equal{} e$ then
$ a*b \equal{} e$
$ c*a*b \equal{} c*e$
$ e*b \equal{} c*e$
$ b \equal{} c$

as for $ G$ being finite, the problem never mentioned it.. the counting argument fails on infinte sets yes?
this is a good reason to look for another proof now.

Yeah it does fail. The hard part is finding a universal identity element. Once we have that based on our hypothesis we can find inverses and thus $ G$ will be a group.

[quote="question?"]hello, iv'e picked up an algebra book and started to work through it, i am curious to see another solution to this exercise if possible.
[u]a first course in abstract algebra 6ed John B. Fraleigh chapter 1.3 ex 38[/u]
prove that a nonempty set $ G$, together with an associative binary operation $ *$ on $ G$ such that $ a*x \equal{} b$ and $ y*a \equal{} b$ have solutions in $ G$ for all $ a,b \in G$

here's my solution:
counting all the possible equations $ a*b \equal{} x$ we have $ |G|^2$ such unique equations. counting all the possible equations of the form $ a*x \equal{} b$ we have at least $ |G|^2$ so we conclude there is one and only one solution to each equation of the form $ a*x \equal{} b$ similarly for $ y*a \equal{} b$ which are permutations of the other set of equations.

now let $ e*a \equal{} a$ and $ a*a' \equal{} e$
$ e*a \equal{} a$
$ e*a*a' \equal{} a*a'$
$ e*e \equal{} e$
let $ z*b \equal{} b$ and $ b*b' \equal{} e$
$ z*b \equal{} b$
...
$ z*e \equal{} e$  now $ z \equal{} e$ since there is only one solution to equation $ y*e \equal{} e$ so $ e$ is a left identity element.
repeat the same procedure but on the right, ie use $ a*i \equal{} a$ and $ a"*a \equal{} i$ we get $ i*i \equal{} i$ but $ e*i \equal{} i$ also. so $ e \equal{} i$. $ e$ is a left & right identity.

now for each $ a \in G$ there is a unique $ b$ that is the solution to $ a*x \equal{} e$. looking at the equation $ y*b \equal{} e$, $ a$ must be the unique solution here or else we have two solutions for $ y*b \equal{} e$ similarly if $ c*b \equal{} e$ then $ c \equal{} a$. so each $ a$ has a unique right inverse $ b$ which means the set of inverses are permutations of the elements and each element has a left inverse. therefore $ G$ with $ *$ form a group.

i was wondering if there is a proof of this without the argument in the beginning that the equations have a unique solution? something more algebraic..[/quote]

Let $ x,y\in{G}$ By assumption there is a solution to $ x*e_x\equal{}x$ and then also to $ x*x'\equal{}e_x$. Similarly with y. Using your proof (which doesn't depend on your counting argument)  we see that $ e_x*e_x\equal{}e_x$

Now consider the equation $ e_x*z\equal{}e_y$. Then $ e_y\equal{}e_x*z\equal{}(e_x*e_x)*z\equal{}e_x*(e_x*z)\equal{}e_x*e_y$
So we have shown $ e_y\equal{}e_x*e_y$

A similar result shows that $ e_x\equal{}e_y*e_x$

Now look at the equation $ w*e_x\equal{}e_y$ Then similar to the proof above we get that $ e_y\equal{}e_y*e_x$ and from above this shows that $ e_x\equal{}e_y$

But $ x$ and $ y$ were arbritary so we have actually shown there is a universal right identity. The proof for the left identity is similar. Thus using hypothesis we also have that inverses exist for both left and right identities. (left and right identities are always the same but for this problem it seems easier to separate the cases, inverses are also unique as you probably know) $ \Box$

ah ok thxs for that.
i find that it is helpful to review problems where i fiddled around to get the solution. writing out a logical path to the solution provides a clearer approach. so here's one for this problem:

0. the inverse equations require an identity, so we must start with finding an identity.
1. proving there is a left identity and left inverse is enough to solve the problem (this was actually previous exercise)
2. show there is a left identity for any element (simply a solution to $ y*a\equal{}a$ say $ e$ for $ a$)
3. we want to show the left identity is the same for all elements so create $ i*b\equal{}b$ (another left identity)
4. to show $ e\equal{}i$ we will need intermediate expressions, what simple expression can be equal to the identity? $ e*e,i*i,e*i,i*e$ at least one of these much feature before the last line, unless you plan to collapse 3 symbols into one, unlikely.(intuitively $ e\equal{}e*i\equal{}i$ looks better then $ e\equal{}e*e\equal{}i$)
5. so you want to get from $ e$ to $ i$ this will involve somewhere in the calculation an expression with at least 3 symbols, so somewhere there will be an expansion and a reduction. what can $ e$ and $ i$ expand to? using the eqns we can expand $ e$ into $ i*x$ or $ y*i$ also into $ e*x$ and $ y*e$ but all these equation introduce a new symbol. we need to use this symbol and get rid of it. so what purpose does it serve? associativity. so we introduce something like $ e*x$ and we want to use what is to the left of $ e$ here (_$ *e$) which is $ i*e$ or $ e*e$. both are yet un-derived.
6. we go to derive $ e*e\equal{}e$ as the other has a symmetry that suggests if we can derive $ e*i\equal{}i$ then we can do $ e\equal{}e*i$ aswell and this is our final solution. so this is our last step.
7. put the pieces together.

establish $ e*e\equal{}e$ and $ i*i\equal{}i$ for some $ e,i \in G$
then let $ a$ be the solution to $ y*i\equal{}e$
$ a*i\equal{}e$
$ a*(i*i)\equal{}e$
$ (a*i)*i\equal{}e$
$ e*i\equal{}e$
now we want to show $ e*i\equal{}i$
let $ b$ be the solution to $ e*x\equal{}i$
$ e*b\equal{}i$
$ (e*e)*b\equal{}i$
$ e*(e*b)\equal{}i$
$ e*i\equal{}i$


inequalities
calculus
inequalities unsolved

[i]35-th Austrian Mathematical Olympiad 2004
Final Round

Part 2 – May 26–27
First Day
[/i]
[i]Prove without using advanced (differential) calculus:
(a) For any real numbers $a, b, c, d$ it holds that $a^6+b^6+c^6+d^6-6abcd\ge -2$. When does equality hold?
(b) For which natural numbers $k$ does some inequality of the form $a^k + b^k + c^k + d^k -kabcd \ge M_k$ hold for all real $a, b, c, d$? For each such
$k$, find the greatest possible value of $M_k$ and determine the cases of
equality.[/i]

$\dots$ the only case of equality is when $a = b = c = d$

 :D .


trigonometry
real analysis
real analysis unsolved

Solve $ y'\plus{}\tan y\equal{}\frac{x}{\cos y}.$

[hide="approach"] Let $ v \equal{} \sin y$ and the ODE becomes a simple nonhomogeneous linear one.  The general solution is
\[ y \equal{} \arcsin \left( x \minus{} 1 \plus{} ce^{ \minus{} x}\right)
\]
[/hide]


[quote="Treething"]Phil has six socks in one drawer and six socks in the other. If he has at least one pair with both socks in one drawer AND at least one pair with both socks in different drawers, what is the largest number of different colors of socks he can have?[/quote]
Do we assume these twelve socks make 6 matching pairs?


Each year, starting with 1999, five new state quarter designs were introduced. How many different state quarter designs were introduced by the end of 2005?

There are a total of $ 2005\minus{}1999\plus{}1\equal{}7$ years. The answer is simply $ 7\cdot5\equal{}\boxed{35}$. (Notice how we add one when we take how many numbers are between $ a$ and $ b$ [b]inclusive[/b])


geometry
circumcircle
geometric transformation
reflection
geometry proposed

The point $ P$ is a fixed point on a circle and $ Q$ is a fixed point on a line. The point $ R$ is a variable point on the circle such that $ P,Q,$ and $ R$ are not collinear. The circle through $ P,Q,$ and $ R$ meets the line again at $ V$. Show that the line $ VR$ passes through a fixed point.

Let $ l$ be the fixed line and $ \mathcal K$ the fixed circle. Inversion with center $ P$ and arbitraty power takes $ l$ into a fixed circle $ \mathcal L'$ through $ P$ and $ \mathcal K$ through $ P$ into a fixed line $ k'.$ The fixed point $ Q \in l$ goes to a fixed point $ Q' \in \mathcal L'$ and the variable point $ R \in \mathcal K$ goes to a variable point $ R' \in k'.$  Circumcircle $ \odot(PQR)$ goes to the line $ Q'R'$ cutting $ \mathcal L'$ again at the image $ V'$ of $ V,$ and line $ VR$ goes to the circumcircle $ \odot (PV'R').$ Let $ S$ be its center. $ \angle PSR' \equal{} 2 \angle PV'R' \equal{} 2 \angle PV'Q' \equal{} \text{const}$ $ \Longrightarrow$ the isosceles triangle $ \triangle PSR'$ with the fixed vertex $ P$ and $ R' \in k'$ remains similar for all $ R' \in k'$ $ \Longrightarrow$ locus of $ S$ is a fixed line $ s$ $ \Longrightarrow$ circles $ \odot (PV'R')$ intersecting at $ P$ form a pencil, they also intersect at the reflection $ W'$ of $ P$ in $ s$ $ \Longrightarrow$ lines $ VR$ go through the fixed image $ W$ of $ W'.$

Denote $ k$ the fixed circle and $ k'$ the variable circle $ (PQR).$ Let $ M$ be the second intersection of $ PQ$ with $ k.$ The parallel line through $ M$ to the fixed line cuts $ k$ again in the fixed point $ S.$ By Reim's theorem applied to $ k,k'$ with common chord $ PR,$ the points $ R,S,V$ are collinear. The line $ VR$ goes through the fixed point $ S$


Let $ n\geq 1$ be a natural number. Let $ a_{n}$ be the arithmetic mean of the positive divisors of $ n$,  let $ b_{n}$ be the geometric mean of the positive divisors of $ n$, and let $ \alpha$ and $ \beta$ be positive quantities.

Study the convergence of the following two series:

$ \sum_{n=1}^{\infty}\frac{a_{n}}{n^{\alpha}}$

and

$ \sum_{n=1}^{\infty}\frac{b_{n}}{n^{\beta}}$.

$ b_{n}= \sqrt{n}$ for all $ n$, so the latter converges if and only if $ \beta > \frac{3}{2}$.

If $ \alpha \leq 2$ then $ \sum_{p \textrm{ prime}}\frac{a_{p}}{p^\alpha}= \sum_{p \textrm{ prime}}\frac{p+1}{2p^\alpha}$ which diverges because $ \sum_{p \textrm{ prime}}\frac{1}{p}$ diverges.  However, $ a_{n}< n$ for all $ n$, so $ \alpha > 2$ is sufficient for convergence.

Can we say anything for general power-means?  E.g. if $ c_{n}$ is the harmonic mean of the divisors of $ n$, study the convergence of $ \sum_{n \geq 1}\frac{c_{n}}{n^\gamma}$ for positive $ \gamma$.

Positive power means are easy: the $ r$ mean of 1 and $ p$ is between $ 2^{-1/r}p$ and $ p$ for $ p$ prime, so the series converges for $ \gamma>2$ and diverges for $ \gamma<2$. For negative $ r$, the primes lead to smaller terms than non-primes; since the $ r$ mean of $ 1$ and $ p$ is asymptotically constant, this gives divergence for $ \gamma\le1$.
More generally, the $ r$ mean of the divisors of $ n$ is less than $ d(n)^{-1/r}$ for $ r<0$, where $ d(n)$ is the number of divisors. Since $ d(n)$ grows slower than any power of $ n$, we have convergence for $ \gamma>1$.


Prove that, for $k \ge 1, \frac{9^k - 1}{8}$ is a triangular number (that is, that it can be expressed in the form $T_n = \frac{n(n+1)}{2}$ for some integer $n$.

[hide]
$\frac{n^2+n}{2}=\frac{9^k-1}{8}$
$n^2+n=\frac{9^k-1}{4}$
In order for $\frac{9^k-1}{8}$ to be a triangular number $n$ has to have a positive integral solution.
$n^2+n-\frac{3^k-1}{2}*\frac{3^k+1}{2}=0$
$(n-\frac{3^k-1}{2})(n+\frac{3^k+1}{2})=0$
It is easily seen that $\frac{3^k-1}{2}$ is an integer for $k\geq 1$, therefore, we are done.[/hide]

[hide]
$\frac{9^k-1}{8}=\frac{(3^k+1)(3^k-1)}{8}$
let $a=3^k-1$ so $\frac{(a+2)(a)}{8}$
a is even by inspection so let a = 2b
Now you have $\frac{(2b+2)(2b)}{8}$
factor 2's out on the top and cancel then you have
$\frac{(b+1)(b)}{2}$ $QED$
[/hide]


[quote="sexynsmartjenny"]how do you find a locally lip**** function?[/quote]

i don't get it. I can't figure out what goes in the 4 *'s, even with all the nasty things I thought of :rotfl:

[quote="miyomiyo"][url=http://mathworld.wolfram.com/LipschitzFunction.html]Lipschitz[/url] Function.[/quote]

A function f such that
|f(x)-f(y)|<=C|x-y|

for all x and y, where C is a constant independent of x and y, is called a Lipschitz function. For example, any function with a bounded first derivative must be Lipschitz.


[b]Five points $ O, A, B, C, D$ are taken in order on a straight line with distances $ OA\equal{}a$, $ OB\equal{}b$, $ OC\equal{}c$, and $ OD\equal{}d$. $ P$ is a point on the line between $ B$ and $ C$ and such that $ AP: PD\equal{}BP: PC$. Find $ OP$ in terms of $ a, b, c, d.$[/b]

[hide="solution"]
$ \frac{AP}{PD}\equal{}\frac{BP}{PC}\Rightarrow\frac{AP}{AD}\equal{}\frac{BP}{BC}\Rightarrow$
$ \frac{AB\plus{}BP}{d\minus{}a}\equal{}\frac{BP}{c\minus{}b}\Rightarrow BP\equal{}\frac{(a\minus{}b)(c\minus{}b)}{a\minus{}b\plus{}c\minus{}d}$
Then $  OP\equal{}OB\plus{}BP\equal{}\frac{ac\minus{}bd}{a\minus{}b\plus{}c\minus{}d}$
[/hide]


Which of the following cannot be the last digit of the sum of the squares of seven consecutive numbers?

(A) 3   (B) 5    (C) 6    (D) 7    (E)  8

[quote="Dr Sonnhard Graubner"]hello, the right answer is $ \text{D} 7$.
Sonnhard.[/quote]

Hello, how did you get this?

We can take mod 4, as we know all squares are 0,1 mod 4.

I will consider that their last digits follow this pattern (starting from 1):
1,4,9,6,5,6,9,4,1,0, 1,4,9,6,5,6,9,4,1,0... Thus, we shall consider when the digit of the first number is:

First half of pattern:
1: last digit of consecutive: 0
4: 3
9: 0
6: 2
5: 0

Second half pattern:
6: 5
9: 8
4: 5
1: 6
0: 1

Thus, it can be 0, 1, 2, 3, 5, 6, or 8, but not 4, $ \boxed{7}, or 9$.
P.S. Sonnhard uses Maple whenever he answers problems.

The units digits of the squares of numbers with a unit digit from 0-9 are;

0 - 0
1 - 1
2 - 4
3 - 9
4 - 6
5 - 5
6 - 6
7 - 9
8 - 4
9 - 1
The sum of the numbers from 0-6 have a units digit of one.
The sum of the numbers from 1-7 have a units digit of zero.
The sum of the numbers from 2-8 have a units digit of three.
The sum of the numbers from 3-9 have a units digit of zero.
The sum of the numbers from 4-0 have a units digit of one.
The sum of the numbers from 5-1 have a units digit of six.
The sum of the numbers from 6-2 have a units digit of five.
The sum of the numbers from 7-3 have a units digit of eight.
So three, five, six, and eight are all possibilities so seven is the answer.

I am curious as to why Sonnhard never posts solutions, just answers.


If $ p,q$ are distinct odd primes, and $ n \equal{} pq, m \equal{} (p\minus{}1)(q\minus{}1)$
Suppose there is a number $ a$ such that $ a^m \equiv 1 \mod n$
Prove that there is a smaller number $ b < m$ such that $ a^b \equiv 1 \mod n$

Because $ pq$ has no primitive root .
All primitive root have form $ 2,4,p^a ,2p^a$
But has some other ways to prove your problem .
We only chose $ b=\text{lcm} (p-1,q-1)=\frac{m}{\gcd(p-1,q-1)}$ because p,q odd so
$ \gcd(p-1,q-1)>1$ so $ b<m$


How does one calculate the energy change of a gas whose temperature rises from $T_{0}$ to $T$, Pressure rises from $P_{0}$ to $P$, and Volume rises from $V_{0}$ to $V$.  The $C$ of the gas changes for changes in temperature, pressure, and volume, so how could one do this mathematically in terms of the function $C(P,V, T)$

You'd only need two variables, not three, because the third can be determined...

The change in internal kinetic energy would be $\frac{3}{2}Nk_{B}(T-T_{0})$, and the change in potential energy (ie work done) would be $\int_{P_{0}}^{P}{VdP}$.


function
integration
probability and stats

If $ \phi_n(t)$ is a sequence of characteristic functions such that $ \phi_n(t)\to 1$, $ \forall t \in ( \minus{} a,a)$, then $ \phi_n(t)\to 1$, $ \forall t \in \mathbb{R}$.

Let $ \mu_n$ be the associated sequence of measures.  Then for $ t<a$ we have the estimate

\[ \mu_n(\{|x|\geq 2/t\} \leq \frac{1}{t}\int_{[\minus{}t,t]} (1\minus{}\varphi_n(s))ds\to 0
\]

as $ n\to\infty$ by the dominated convergence theorem.  So the $ \mu_n$'s are tight (why?).  So we can pass to weakly convergent subsequences, but if $ Y_{n_k}\implies Y$, then this means that $ \varphi_Y\equal{}1$ on $ (a,a)$.  But if $ \varphi_Y(t)\equal{}1\plus{}o(t^2)$ at $ 0$, then it is everywhere equal to $ 1$, and hence $ Y\equal{}0$.

[quote="Mr.Doe"]Aren't you missing something here
[quote="blahblahblah"]
\[ \mu_n(\{|x|\geq 2/t\} \leq \frac {1}{t}\int_{[ \minus{} t,t]} (1 \minus{} \textrm{Re }\varphi_n(s))ds\to 0
\]
[/quote]
since $ \varphi_n$ can take value in $ \mathbb{C}$
[quote="blahblahblah"]
So the $ \mu_n$'s are tight (why?).
[/quote]
Indeed, why is that?

[quote="blahblahblah"]But if $ \varphi_Y(t) \equal{} 1 \plus{} o(t^2)$ at $ 0$, then it is everywhere equal to $ 1$ [/quote]
Can you explain this please?[/quote]

For the first part, the dominated convergence theorem applies perfectly well to complex valued functions, the second follows pretty much immediately from stating the definition of tightness in terms of the $ \mu_n$'s instead of their distribution functions, and the third one follows from the fact that the behavior of the characteristic function near $ 0$ encodes behavior about the moments of $ Y$.  To be more specific, there's a result that says something like if $ (\phi_Y(t)\minus{}1)/t^2\to c$ as $ t\downarrow 0$, then $ EY\equal{}0$ and $ E|Y^2|\equal{}\minus{}2c$, from which the result follows immediately.

Ok, for a given $ \varepsilon>0$, you claim that you can find a compact set $ K$ such that $ \mu_n (\mathbb{R}\setminus K)\leq \varepsilon, \forall n\in\mathbb{N}$. But as I see it, that set depends on $ n\in \mathbb{N}$ in this case. Or am I missing something?


A set $A$ of distinct positive integers is such that when any of its elements is removed, the remaining ones can be divided into two subsets with equal sums. What is the maximum possible cardinality of $A$?

n=7. Example {1,3,5,7,9,11,13}. It is not hard to prove that this works.
Obviously all of the numbers have same parity. If all numbers are even then we divide all of them by 2 and we get other example. We do it while all numbers are even.
So we may think that all of numbers are odd.
Let s be the sum of all numbers. Then s-x is even for arbitrary number x(from the set). From here we have that s is odd and so is n.
Since n=1, and n = 3 obviously aren't solutions the remaining case is n=5.
From here it is simple case checking.

Okay, but I think your solution is for the minimum possible cardinality, not the maximum..

n can be as big as you want because from:

Lemma 1
If A can be divided into two subsets with equal sums then A-k and A+k too if k and minA-k are naturals.

Lemma 2
B={1,3,5,7} can be divided into two subsets with equal sums  {1,7} and {3,5}

Lemma 3
Baronovs example for n=7

then for n=7+4k
A= {1,3,5,7,9,11,13, ...} also works
because splitting in three sets with 4i, 7, and 4(k-i) elements and applying lemma 1,2 and 3.

is it correct?


Let ABC be a triangle such that 0<A,B,C<pi/2.
Prove that :
sin(2A)tanB+sin(2B)tanC+sin(2C)tanA \geq sin(2A) \sqrt (tanBtanC)+
sin(2B) \sqrt (tanAtanC)+sin(2C) \sqrt (tanAtanB)

It's too easy . Please give me your solution . Thanks.

hi i will give you the solution soon
but ivery bad at english so wait

[quote="Master"]hi i will give you the solution soon
but i am very bad at english [/quote]
 may be i am too supid because i am only know one way to solve this
the inquelity
2(sin2AtanB+sin2BtanC+sin2CtanA)  \geq 2(sin2A \sqrt tanBtanC+....)
so we use Cauchy inquelity
2sin2A \sqrt tanBtanC \leq sin2A(tanB+tanC)
then we will prove that
2( \sum sin2AtanB)= \sum sin2A(tanB+tanC)
so is it right?

Yes , you are right .
It's also my solution .


real analysis
advanced fields
advanced fields unsolved

Do there exists a partition of $\mathbb{R}=A\cup B$ such that $A\cap B=\emptyset$, $A$ and $B$ are uncountable, and $A$ and $B$ are dense subsets of $\mathbb{R}$?

Yes, but I want to ask something stronger: can we show it without the axiom of choice?

Well, you can find an uncountable subset of measure zero (e.g, the cantor set), which is nowhere dense.  Then glue it together with the rationals to make your set $A$, and let $B$ be the rest of $\mathbb{R}$.

I found a even simpler form:

$A$ contains the nonnegative rationals and negative irrationals.
$B$ contains the negative rationals and positive irrationals.


quadratics
algebra
quadratic formula
algebra unsolved

Find the positive real numbers ${x,y}$ if

$\frac{x+y}{2}, \sqrt{xy}, \sqrt{\frac{x^{2} + y^{2}}{2}}, \frac{2xy}{x+y}$

are four integer numbers which sum is 66.

Am I supposed to just find [u]any[/u] solutions to this?

In that case, I would try to set [i]x[/i] equal to [i]y[/i].  In this case, every one of the above expressions is an integer when [i]x[/i] (and [i]y[/i]) are integers.  Now, I will string them together in an equation and solve using algebra.

Before, solving for [i]x[/i] (or [i]y[/i]) was practically impossible, but now every one of those expressions simplfies to [i]x[/i] (or [i]y[/i]).

However, the solution to x+x+x+x=66 yield x=33/2, which is not an integer and causes our expressions to be nonintegers.

I thought I had something going there, didn't you?

There aren't any integer solutions for x,y to this.
I tried every integer combination of x and y less than 133 (any higher and the left expression isn't an integer).

draperp, the problem stipulates $x, y$ real, not integer :)

Edit:

[hide="Really long solution:  Do not read"]We know that $x \neq y$ because $x = y$ does not produce a solution.  By AM-GM-HM-QM we then have

$\frac{2xy}{x+y} < \sqrt{xy} < \frac{x+y}{2} < \sqrt{ \frac{x^2 + y^2}{2} }$ and all of these are integers that add up to $66$, so the harmonic mean must be $\le 15$.

Note that $\frac{x+y}{2} \times \frac{2xy}{x + y} = xy = ( \sqrt{xy} )^2$, so we can write $\frac{x+y}{2} = a, \frac{2xy}{x+y} = b, \sqrt{ab} = \sqrt{xy}$.

Next $a^2 = \frac{x^2 + 2xy + y^2}{4} \implies 4a^2 = x^2 + y^2 + 2xy$, so after some algebra, we get $\sqrt{ \frac{x^2 + y^2}{2} } = \sqrt{ 2a^2 - ab}$ for the integers $a, b$ that we have defined.  We then have as our condition:

 $a + b + \sqrt{ab} + \sqrt{2a^2 - ab} = 66$.

Dividing the last term by the second-to-last term gives us $\sqrt{ \frac{2a - b}{b} }$ is rational, which means it must be a perfect square.  We have a few cases.

Case 1:  $b$ and $2a-b$ are both perfect squares.  We have $b < 16$ so $b = 1, 4, 9$.  Setting $b = 1$ means we want $a$ such that $\sqrt{a}$ is an integer and $\sqrt{2a-1}$ is also an integer.  This gives us $a = 25$, so we try it:

$25 + 1 + 5 + \sqrt{ 25(50 - 1) } = 66$

So our first solution is $a = 25, b = 1$.  Solving for $x$ and $y$, we have $x + y = 50, xy = 25$ so we want the solutions to the quadratic equation $u^2 - 50u + 25$, and quadratic formula gives us...[/hide]

[hide="The answer"] $x, y = \frac{ 50 \pm \sqrt{2500 - 100} }{2} = \boxed{25 \pm 10 \sqrt{6}}$. [/hide]

There's probably a shorter way  :|

Yes, I realize that x,y aren't necessarily integers, but I was just trying out possiblities that [u]might[/u] be solutions.
(It turns out that they weren't.)

By the way, I liked your proof very much.
But what is "AM-GM-HM-QM", and what does the hamonic mean being equal to or greater than 15 have to do with the proof.

[quote="111222333444555"]
$\frac{x+y}{2}$[/quote]
AM (Arithmetic Mean)

[quote="111222333444555"]
$\sqrt{xy}$[/quote]
GM (Geometric Mean)

[quote="111222333444555"]
$\sqrt{\frac{x^{2} + y^{2}}{2}}$[/quote]
QM (Quadratic Mean)

[quote="111222333444555"]
$\frac{2xy}{x+y}$[/quote]
HM (Harmonic Mean)

I think that answers your question draperp ;)


search
number theory proposed
number theory

Determine all positive number non bigger than 1 so that for any couple of different factors a and b of n, the number a + b-1 is also a factor of n.

non bigger?????

however forall n prime.
in fact, if n is composite then (a \neq 1 \neq n \neq b): a|n, b|n, a+b-1|n, 2a+b-2|n, ...ka+b-k|n forall k positive integer. but for large k: ka+b-k> n, absurde.

I think you made a mistake.

    why if a/n, b/n and a+b-1/n it means that 2a +b -2 divides n????

    And there are many n that work like 63 or 5 to the 5 power times 3124.

   Or 81 times 79. Or some others

       anybody????

so, i don't undestand the problem: can you repeat clearly the answer? :|

Sorry, i was completely wrong

Find all odd positive integers n bigger than 1 that satisfy that for any couple of different [b] prime [/b]factors of n, a and b, the number a+b-1 also divides n.

I dont have the solution but this is what i've got:

This n fulfills

      a=3b=7              n=63       3+7-1=9 and divides n.

    a=3 b=79             n=3^4*79        3+79-1 dividesn

  I found some other cases that do fullfill the conditions, but cant find any n with 3 prime factors.

also  b=a^x -a +1         and b doesnt divide a+b-1 if b>a and no other prime divisors.

also clearly b-1 is congruent to 0 mod a.

hope you can help :(

[quote="chingonkan999"]This n fulfills

      a=3b=7              n=63       3+7-1=9 and divides n.

    a=3 b=79             n=3^4*79        3+79-1 dividesn

[/quote]

don't forget $ n\equal{}12$ :wink:

this is from russia 2001:

[url]http://www.mathlinks.ro/viewtopic.php?search_id=1773834756&t=74746[/url]
[url]http://www.mathlinks.ro/viewtopic.php?search_id=1796151311&t=112155[/url]
[url]http://www.mathlinks.ro/viewtopic.php?search_id=1796151311&t=163365[/url]


AMC
USA(J)MO
USAMO
Olimpiada de matematicas

Indicar los pasos para trisecar usando regla y compás un ángulo de $27^\circ$

Para hacer eso no hay que inscribirle un ángulo recto primero??

[color=blue]Trácese un ángulo de $90^\circ$ en cualquier parte.
Luego transcríbase tres veces el ángulo de $27^\circ$ sobre el anterior, de forma que estos tres ángulos sean adyacentes, y uno tenga un lado común con el ángulo de $90^\circ$.
Sobrará una sección del ángulo, que medirá $90^\circ-3 \times 27^\circ = 9^\circ$.
Es fácil, por último, transcribir este último ángulo obtenido en el dado por el enunciado.[/color]

Muy bien, si saben de otros ángulos que se puedan trisecar porfavor posteen.

[color=blue]Aplicando el método análogo (casi idéntico, en realidad), bien puede trisecarse ángulos de $9^\circ;18^\circ;36^\circ;45^\circ;54^\circ;63^\circ;72^\circ;81^\circ$. Ha de haber más ángulos, pero la lista es bastante extensa (en efecto, no la terminé, pues corresponden a ángulos múltiplos de $9$).[/color]

ahora q me acuerdo hay un problema relacionado con todo esto, de la usamo 1981 si no me equivoco, que dice asi:

Si $n$ no es múltiplo de 3, probar que el ángulo $\frac{\pi}{n}$ puede ser trisecado con regla y compás.

espero sea de su interes :)

[quote="Jutaro"]ahora q me acuerdo hay un problema relacionado con todo esto, de la usamo 1981 si no me equivoco, que dice asi:

Si $ n$ no es múltiplo de 3, probar que el ángulo $ \frac {\pi}{n}$ puede ser trisecado con regla y compás.

espero sea de su interes :)[/quote]No todos los ángulos con medida $ \dfrac{\pi}n$ pueden ser construídos con regla y compás (por ejemplo, $ \dfrac{\pi}7$), entonces es necesario suponer que el ángulo con medida $ \dfrac{\pi}n$ ya está construído.

Es dado un ángulo $ \angle AOB$, cuya medida es $ \dfrac{\pi}n$. Construya un ángulo $ \angle A'O'B'$, cuya medida es $ \dfrac{\pi}3$. Existen números enteros positivos: a,b, tales que an-3b=1, determinados de la siguiente manera:[list][*]Si n=3k+1, entonces a=1, b=k
[*]Si n=3k+2, entonces a=2, b=2k+1[/list]Son construídos ángulos con medidas $ \dfrac{a\pi}3$ y $ \dfrac{b\pi}n$. Finalmente, es construído un ángulo cuya medida sea igual a la diferencia: $ \dfrac{a\pi}3\minus{}\dfrac{b\pi}n\equal{}\dfrac{an \minus{} 3b}{3n} \equal{} \dfrac1{3n}$. Copiando dos veces esta medida en el ángulo original, éste será trisecado.

En general trisecar un ángulo no es posible con regla y compas, salvo ciertos ángulos particulares. La operación regla-compas  equivale analiticamente a la resolución de ecuaciones lineales, cuadraticas y hasta algunas bicuadraticas. La trisección del ángulo en general involucra la resolución de una ecuacion cubica irreducible, por lo tanto no es en general posible. Una tecnica no-convencional para la trisección gráfica del ángulo es la llamada ''operación'' deslizar la regla y el uso de ciertas curvas algebraicas como la trisectriz de Maclaurin, la  trisectriz de Newton, la concoide de Nicomedes, entre otras.


I think this problem is pretty cool.

A three-man jury has two members each of whom independently has probability $p$ of making the correct decision and a third member who flips a coin for each decision (majority rules).  A one-man jury has probability $p$ of making the correct decision.   Which jury has the better probability of making the correct decision?

[hide="Um"]The probability that all three jurors make the right decision is $\frac{p^2}{2}$. The probability that the two jurors with probability p make the right decision and the coin-flipper doesn't is also $\frac{p^2}{2}$. The probability that one of the p-jurors and the coin-flipper make the right decision but the other p-juror doesn't is $\frac{p(1-p)}{2}$. The probability that the other p-juror and the coin-flipper make the right decision but the first p-juror doesn't is also $\frac{p(1-p)}{2}$. Adding all that together gives $p^2+p(1-p)=p(p+1-p)=p\cdot 1=p$. So the 3-person and the 1-person juries have the same probability of making the right decision.[/hide]

Here's what I got, but I'm not sure if it's right.

[hide]Let's label each person in the jury $a$, $b$, and $c$, with $a$ and $b$ having a $p$ chance of making the correct decision, and $c$ has a $\frac{1}{2}$ chance.  There a four cases in which the jury makes the correct decision -

If $a$ and $b$ are right, but $c$ is wrong.  This has a $\frac{p^2}{2}$ chance of happenning.

If $a$ and $c$ are right, but $b$ is wrong.  This has a $\frac{p(1-p)}{2}$ chance of happenning.

If $b$ and $c$ are right, but $a$ is wrong.  This has a $\frac{p(1-p)}{2}$ chance of happenning.

If $a$, $b$, and $c$ are all right.  This has a $\frac{p^2}{2}$ chance of hapenning.

When you ass them all up, you get $p$, which is the same chance as the one man jury.  That means that they both have the same chance of making the right decision.[/hide]

call the jurors on the three man jury: x,y,z.... (z flips the coin)
now we look at different cases:
[hide]I. x,y, and z are correct:
probability is: p 2 /2
II. x,y are correct, z is wrong:
probability is: p 2 /2
III. y and z are correct, x is wrong:
probability is: (1-p)(p)/2
IV. x and z are correct, y is wrong:
probability is: (1-p)(p)/2
so lets take the sum of these probabilirties:
p 2 + p -p 2 =[b]p[/b][/hide].
SO the probability that the one man jury is also [b]p[/b].
therefore, it is equally likely to get correct decision out of a one and three man jury[/hide]

yeah, um [hide]just briefly looking, i'd say that the calculations amount to p, so they have an equal probability![/hide]


MATHCOUNTS
articles
geometry
Pythagorean Theorem

The chapter tests are available on the Mathcounts website, at least my coach/mom has access now, so I would guess we all do.  I redid the questions I missed and without any trouble got 4/5 of them right.  I know it's way different at home, with no time contraints, but I missed some SUPER easy problems.

I would appreciate ideas on how to avoid making dumb mistakes.  In other words, how best to succeed in answering correctly questions that one should get correct...

[quote="topper"]I would appreciate ideas on how to avoid making dumb mistakes.  In other words, how best to succeed in answering correctly questions that one should get correct...[/quote]

Have you seen our site owner's article "[url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php]Stop Making Stupid Mistakes[/url]"? It's quite helpful. It seems to be the kind of article worth reviewing after every major math test.

cool thanks! lets see how many i get this year after a bad sprint in chapter last year. it cost me a 3rd place trophy  :?  :(
i'm going to do it under timed conditions tomorrow.

do you know where it is on the website? i'm just wondering about the questions. i'm still wondering what kind of stupid thing i probably did on 11 and 21

Go to the Mathcounts website, then go to "contents", scroll down a little and you'll find it.

I'm glad I'm not the only one to miss #11, but I bet nobody else missed #9!

Thanks guys. Of course, now that i see the problem again i forgot what i put for my answer :?
Oh well

Edit:  :-x  I didn't see that you didn't have to have a quarter! ARG!

how do you guys know what you got wrong? still trying to figure out which 4 i missed =[

They give the coaches a data sheet with scores and which ones each person on the team missed and other stuffs.  Usually the coach makes copy for the members.

I did really bad than I did on the chapter, thank god... That made me to go state, though... Anyway, since it's over, how do ya do question like 18, 29, 30.... I got others wrong but rest of them I think wouldn't be that hard to solve... I did really bad and only got 18 right. -.-;;;; I think I got 24 on chapter but... Anyway, so will anyone help me through those three questions?

18-[hide]you can just count blocks, using 4 blocks each time and plotting different points[/hide]

30-[hide]if the height is 30, and the area is 63960, then the base must be 63960/30*3, which is 6396. That must be a square, so each side is 2 :sqrt: 1599. Then, you do the pythagorean theorum with AB :^2: =1599+6396, getting AB :^2: = :sqrt: 7955 and then you can round that.

I might've made a mistake there, and also, I'm in a rush so I just skipped 29, which is just finding the average[/hide]

Yifster, are you sure about #30? I got [hide]50[/hide], here's how:

[hide]Let the side of the base be b. Then:

1/3 * b :^2: * h = 63960

1/3 * b :^2: * 30 = 63960

So b :^2: = 6396.

Now, you can see that AB is the hypotenuse of a right triangle with legs h and b/2. So use the Pythagorean Theorem:

(b/2) :^2: + h :^2:  = AB :^2:

(1/4) * b :^2: + 900 = AB :^2:

(1/4) * 6396 + 900 = AB :^2:

2499 = AB :^2:

 :sqrt: 2499 is much closer to 50 than it is to 49, so the answer is 50.

[/hide]

oops. yeah, jeffy's rite. I multiplied by the wrong thing.

Can somebody plz still do question 29? I'm apparently missing something, because I'm getting an answer that's about 5 times the answer they give.

Thanx.


MATHCOUNTS
geometry
3D geometry
pyramid
frustum
trapezoid

Can you please help me in this one, I am more confused on this???

A plane parallel to the base of a right, square pyramid divides the pyramid into a smaller pyramid and a frustum that have equal volumes. If the height of the smaller pyramid is 6 cm, what is the height, in centimeters, of the original pyramid? Express your answer in simplest radical form.

It's basically saying you have a square pyramid and you chop it horizontally so you have a small pyramid and a thing that has 4 trapezoids connecting 2 parallel bases, called a frustum. The small pyramid has a height of 6, and it's area is equal to the frustum.

Equation for V of Pyramid: $ V= \frac 13 hB$

Equation for V of Frustum: $ V = \frac 13 h(B_1 + B_2 + \sqrt{B_1 \cdot B_2})$

The base of the smaller pyramid is square, and is equal to the upper base of the frustum, $ B_2$. $ B_1$ and $ B_2$ are both squares. Let's put in what we know.

$ \frac 13 \cdot 6B_2 = \frac 13 h(B_1 + B_2 + \sqrt{B_1 \cdot B_2})$
$ 6B_2 = {B_1 + B_2 + \sqrt{B_1 \cdot B_2}}$
$ 5B_2 - B_1 = \sqrt{B_1 \cdot B_2}$

Sorry, I have to go right now, I'll come back if I can and finish it, but work with this, alright?
[/hide]


induction
logarithms
number theory proposed
number theory

Find all positive intengers n such that $\frac{3^{n}-1 }{2^{n}}$ is an intenger.

By induction we can prove that $v_{2}(3^{n}-1)=v_{2}(n)+2$ for all $n>1$ ($v_{2}$ being the $2$-adic valuation, thus the number of factors $2$).
We need $\log_{2}(n)+2 \geq v_{2}(n)+2=v_{2}(3^{n}-1) \geq n$, which gives us just small cases to check.

We need $\log_{2}(n)+2 \geq v_{2}(n)+2$

Hi.I didn't understand that!

$n>0$ can be only divisible by $2^{k}$ if $n \geq 2^{k}$ or equivalent:
$\log_{2}(n) \geq k$.
Choosing $k=v_{2}(n)$, the result follows.

why we need to have that $n$ is divisible by $2^{k}$?

hahahahaha--- :rotfl:  :rotfl:

Sorry, i am so idiot today, i don't believe i did that question!!!!Thanks for your pacient  :rotfl:

see also the equivalent problem from Romanian TST2005: (same actually...)
http://www.mathlinks.ro/Forum/viewtopic.php?p=198299#p198299


LaTeX articles parameterization geometry

So can anyone tell me how to put something in columns?  I need to do my science project in the correct format, with 4 columns, equally spaced.

And the paper HAS to be landscape, not portrait, and that's a hard one too... further, I need a way to force LaTeX to jump to the next column even though I didn't fill up the previous.  This is very important, it's due on Thursday and I prefer to start moving today instead of procrastinating.

I hope stevem comes and answers this  :)

EDIT: FYI I have MiKTeX 2.7 beta 5.  The latest version! YAY.

[quote="mgao"]And the paper HAS to be landscape, not portrait, and that's a hard one too...[/quote]That's the easy bit - just add landscape as the documrntclass option eg \documentclass[landscape]{article}.

The [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=multicol]multicol[/url] package seems to be made for what you need. If you read the [url=http://tug.ctan.org/tex-archive/macros/latex/required/tools/multicol.pdf]documentation[/url] you'll see it has a \columnbreak command which will allow you to change columns wherever you want. All you need in the preamble is \usepackage{multicol} then you can use \begin{multicols}{4}...\end{multicols}.

thanks!  I thought I could extend

\documentclass[landscape,twocolumn,letterpaper]{article}

to

\documentclass[landscape,fourcolumn,letterpaper]{article}

but the multicol package seems to be way better.

And I thought it would be stevem and not anyone else who would respond.. I made a prediction!

 :rotfl:

wait... the multicol package does not show up in the package downloader.... I'll try to search...

It's part of the tools package which is there and may already have been installed if you downloaded the larger version of MikTeX.

Tools?  Isn't MiKTeX Tools for Linux?

Sorry if I sound pretty stupid...

and no, I didn't download the larger version.  I dual boot Vista with XP, and that's why my Vista drive, the one I installed on has only 30GB of space and I don't want that to get any lower...

nvm i found it

OK, now it seems that the landscape parameter and multicol are incompatible.  The result was a PORTRAIT paper.
What my document's LaTeX looks like right now:

\documentclass[landscape,letterpaper,10pt]{article}
\usepackage{multicol}
\begin{document}
\begin{multicols}{4}

<insert text here>

\columnbreak

<insert text here>

\columnbreak

<insert text here>

\columnbreak

<insert text here>

\end{multicols}
\end{document}

This is, I think, a perfectly fine LaTeX document.  What's bugging it?

It does work but seems to redefine the margins. So try using the lscape package which has a very novel way of doing things as you'll see when you compile it and works well presuming you want to print it out,
[code]\documentclass[letterpaper,10pt]{article}
\usepackage{multicol}
\usepackage{lscape}
\begin{document}
 \begin{landscape}
  \begin{multicols}{4}
   Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident, sunt in culpa qui officia deserunt mollit anim id est laborum
   \columnbreak
Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident, sunt in culpa qui officia deserunt mollit anim id est laborum
   \columnbreak
Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident, sunt in culpa qui officia deserunt mollit anim id est laborum
   \columnbreak
Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident, sunt in culpa qui officia deserunt mollit anim id est laborum  \end{multicols}
 \end{landscape}
\end{document}[/code]

An alternative method is to get the geometry package to set the correct paper size for landscape:
[code]\documentclass[letterpaper,10pt]{article}
\usepackage{multicol}
\usepackage[landscape]{geometry}
\begin{document}
  \begin{multicols}{4}
   Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident, sunt in culpa qui officia deserunt mollit anim id est laborum
   \columnbreak
Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident, sunt in culpa qui officia deserunt mollit anim id est laborum
   \columnbreak
Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident, sunt in culpa qui officia deserunt mollit anim id est laborum
   \columnbreak
Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident, sunt in culpa qui officia deserunt mollit anim id est laborum
\end{multicols}
\end{document}[/code]

You know that lorem ipsum dolor sit amet thing? COOL!

WAIT.

I tried, and the horizontal columns only go about halfway across the 8.5" side of the page as if they were dealing with a portrait paper instead...


algorithm
algebra
polynomial
analytic geometry
Galois Theory
geometry open

The Fermat Weber Problem is the problem of finding the point such that the sum of its distances from $n$ given points in a plane is minimum.
I Have some questions regarding this.

1) Darij and Grobber had pointed in earlier posts that there exist algorithms for finding P. Are you sure there are no geometical methods(not for actual construction -- just prooving)?? I think I have a proper proof though I will post it later. What's the difference between this and Steiner's Problem?

2) What is the solution to this problem in a general 3D Space?? Can you quote any article?

3) Would writing an article on the geometric proof I have and on 3D spaces be considered 'research'???
[hide](If you have seen my other recent posts, my ideas for the proof are present there!)[/hide]
Thanks!!!

Steiner's problem allows connecting the points by an arbitrary union of 1-dimensional curves.  As lines are geodesics, it is enough to consider line segments, and removing redundant line segments one arrives at the following problem:

given some points in the plane, add some finite number (possibly zero) of additional points, and draw a spanning tree consisting of segments in the plane between these points.   Minimize, over all choices of the extra points and the spanning trees, the length of the spanning tree.

Using Fermat's principle for 3 points one sees that the vertices in the spanning tree have degree at most 3 and where three edges meet, the angles between them are 120 degrees.   This reduces the calculation to a finite combinatorial problem, but I believe that it is intractable (NP-complete) and can be found under the name EUCLIDEAN STEINER TREE in the complexity theory compendia such as Garey and Johnson.

I do not know the complexity of the minimum distance sum problem, and would not be surprised if it too is intractable, as there are $n$ square roots (a field extension potentially of degree $2^n$) and the methods for solving polynomial systems, such as Groebner bases, can reach the complexity level of PSPACE which is even worse than NP-complete.

Thanks for replying!!! Although, you ave prsented a mathematically true version of the general problem and even given a 'computing' tinge to it, I was looking for a more geometrical (at least in 2D and 3D) proof and solution of the FW Problem. DO you think that is possible to solve fully? Are you aware of any such research? Think you can answer my other questions???

The sum of distances problem has several easy-to-solve geometrical cases in one, two and (I think) three dimensions.  Other than that, it is the solution of a computational problem that is potentially quite intractable; other problems of the same type involving Euclidean distances are known to be NP-hard.  The existence of a minimum is easy to prove but for actually computing it, there are difficulties with the finite-accuracy arithmetic of square roots, with the potential sea of false minima (which of course would defeat the "physics" solution of suspending masses on strings), and the instability of these minima with respect to the data in the problem.

[quote="Rushil"]

1) Darij and Grobber had pointed in earlier posts that there exist algorithms for finding P. Are you sure there are no geometical methods(not for actual construction -- just prooving)?? I think I have a proper proof though I will post it later. What's the difference between this and Steiner's Problem?

There are many solutions. Please see,

I. M. Yaglom: Geometric Transformations I., Random House, 1962
E. Molnar (ed): Elemi matematika II., Tankonyvkiado, Budapest, 1967
G. Jalal, J. Krarup: Geometrical solution to the Fermat problem with arbitrary weights, Annals of Operations Research, vol. 123, pp. 67-104, 2003

2) What is the solution to this problem in a general 3D Space?? Can you quote any article?

E. Weiszfeld: Sur le point pour lequel la somme des distances de n points donn´es est minimum, Tˆohoku Mathematical Journal, vol. 43, pp. 355-386, 1937

This approach can be used also (as far as I know).

Bye.

Weizfield has presented an algorith for the general case. I am searching for a purelu geometric solution.

What is a "geometric" algorithm?  It is unlikely that the minimum is ruler-and-compass constructible, and this should be demonstrable by computing Galois groups of randomly chosen instances.  On the other hand, the coordinates of the solution satisfy a system of two  polynomials in $x$ and $y$ of degree $2^n$ that can be computed from the given points, so is an intersection point of two algebraic curves, if that is considered "geometric".  Other than that, finding the solution would require exhibiting it as a limit of an infinite sequence of points, and one can ask whether there is some "geometric" construction to produce the next point from the known ones, but Weiszfeld's iteration already does that.   What exactly are you looking for that isn't already done?

Yes, it has already been proved ( I have the paper) that the Fremat Point CANNOT be constructed. But is that the same as saying that we CANNOT determine its properties. I'm not asking for a construction - merely a characterisation. For e.g. you might say that for a Pentagon (actually special ones) - , the FP is the point where all sides subtend equal angles. Now you know you cant construct this using ruler and compass or intersection og curves etc - Galois theory eliminates everything - but you still know the properties of that point!

If the Galois group is large (which is very likely), then a characterization such as you describe cannot work, as it would compute the point as intersection of a fixed number of low-degree curves.   It is worth specifying clearly what is a "characterization" for purposes of this problem.

You mentioned that you are interested in this as a project for a science competition. Determining whether or not the Galois group of this problem is the maximum possible  would be an excellent result (easily a winner in the contest), and finding the Galois groups rigorously for a given finite number of points such as $n=9$ is already a significant task.


Do you know any good 3D picture-compilation programs?

[url]http://en.wikipedia.org/wiki/POV-Ray[/url]

POV-Ray.

I think it can still be downloaded for free at
[url]http://www.povray.org/[/url].

I did it and... I have to admit I can't work with it, it's too complicated for me. :oops:


[quote="mathgeniuse^ln(x)"]I don't think so, but I know this is approaching $\frac{\pi^{2}}{6}$[/quote]
Well, yes, when I first looked at this I thought that perhaps I could evaluate $ \sum^{\infty}_{n=k}\frac{1}{n^{2}}$, but that didn't work out. Maybe it can be solved with calculus...

[hide="Well..."] Let the generating function of the sequence be

$g(x) = \sum_{k=1}^{n}\frac{x^{k}}{k^{2}}$

Then

$g'(x) = \sum_{k=1}^{n}\frac{x^{k-1}}{k}$
$x g'(x) = \sum_{k=1}^{n}\frac{x^{k}}{k}$
$\frac{d}{dx}\left( x g'(x) \right) = \sum_{k=1}^{n}x^{k-1}= \frac{x^{n}-1}{x-1}$

So

$g(x) = \int_{0}^{x}\frac{f(y)}{y}\, dy+C_{1}$

Where

$f(y) = \int_{0}^{y}\frac{t^{n}-1}{t-1}\, dt+C_{2}$

And the sum in which we are interested is $g(1)$.

Since $f(1)$ will give us the partial sums of the harmonic series, which I do not think has a nice form, I do not think these integrals are trivial to evaluate. [/hide]

But you can (with not too much difficulty) expand it for large $n$  in terms of  power of $(1/n)$

For example it is easy to prove that the value is between $\frac{\pi^{2}}{6}-\frac{1}{n}$ and  $\frac{\pi^{2}}{6}-\frac{1}{n+1}$

The expansion may look like $\frac{\pi^{2}}{6}-\frac{1}{n}+\frac{1}{2n^{2}}-\frac{1}{6n^{3}}-\frac{1}{30n^{5}}$ which may give very accurate values of large  $n$
(In practice this can be very useful, For example, to sum first 100 terms, you get 10 figure accuracy by going up to (1/n^5)  without having to add all  100 the terms)

How about combining to make a common denominator to get
\[\frac{\frac{(n)(n+1)(2n+1)}{6}}{(n!)^{2}}=\frac{(n)(n+1)(2n+1)}{6(n!)^{2}}\]
or do things with factorials not count as closed forms?

[quote="NeverOddOrEven"]How about combining to make a common denominator to get
\[\frac{\frac{(n)(n+1)(2n+1)}{6}}{(n!)^{2}}=\frac{(n)(n+1)(2n+1)}{6(n!)^{2}}\]
[/quote]

Except that the numerator is not $1^{2}+2^{2}+\cdots+n^{2}$.

[quote="bpms"]Is there a formula for $\frac{1}{1^{2}}+\frac{1}{2^{2}}+...+\frac{1}{(n-1)^{2}}+\frac{1}{n^{2}}$
?[/quote]

Well, it's equal to the sum of the squares of the numbers 1-n taken n-1 at a time all over (n!)^2


geometry
3D geometry
tetrahedron
inequalities
geometry proposed

Given a tetrahedron $ T$. Valentin wants to find two its edges $ a,b$ with no common vertices so that $ T$ is covered by balls with diameters $ a,b$. Can he always find such a pair?

[i]A. Zaslavsky[/i]

I think we can always find such pair, unless there is some special
construction with a particular inequalities between faces.

Let $BC=a,AC=b,AB=c,AD=d,BD=e,CD=f$.  WLOG let
$a^{2}+d^{2}= \max\{a^{2}+d^{2}, b^{2}+e^{2}, c^{2}+f^{2}\}$.  We claim that
balls with diameters $a,d$ with centers at the midpoints of edges should cover the tetrahedron.
To prove that it suffices to show that they cover four faces.

Consider arbitrary face $ABC$. Let
$L$ be the projection of $D$ onto $(ABC)$ and let $H$ be the orthocenter of triangle $ABC$.
Denote by $X,Y,Z$ the projections of $H$ onto triangle sides.
Also denote by $X',Y', Z'$  the projections of $L$ onto triangle sides.
The balls with diameters $(a,d)$ in the plane $(ABC)$ are circles $(BXYC)$ and $(AX'Y')$.
Observe that $AX' \geq AX, AY' \geq AY$, because
$AX'^{2}-BX'^{2}=AD^{2}-BD^{2}=d^{2}-e^{2}\geq b^{2}-a^{2}=AC^{2}-BC^{2}=AX^{2}-BX^{2}$.
$AY'^{2}-CY'^{2}=AD^{2}-CD^{2}=d^{2}-f^{2}\geq c^{2}-a^{2}=AB^{2}-BC^{2}=AY^{2}-BY^{2}$.
Thus the circle $(AX'Y')$ contains the region not covered by the circle $(BXYC)$ and we are done.

The only thing that disturbs me are obtuse traingles, but it is still should not change the essence, unless I am wrong.


geometry
3D geometry
tetrahedron
combinatorics proposed
combinatorics

$18$ points in the space are given such that no $4$ of them lie in the same plane. Prove that between tetrahedrons they define (as vertices) there is at least one of them such that each of its edges is the greatest edge of some other tetrahedron.

Well, actually it is pretty easy, so I will just give a hint: 18 is one of the very well-known numbers (they have a name, there are infinitely many of them and there are only finitely of them known till now).

It is known that $R(4,4)=18$, now for every tetrahedron color its biggest side in blue. So there is a blue tetrehedron or an uncolored tetrahedron, but there is no uncolored tetrahedron since we colored at least one edge from each.

Very nice problem! however it would have been very hard without the clue! Ramsey is very hidden in the problem

A nitpick...

Doesn't Pascual's proof show that there is some tetrahedron, each edge of which is the greatest edge of *some* tetrahedron. The problem asks to prove that it is the greatest edge of some _other_ tetrahedron.

It can be fixed: during the coloring procces if a tetredron has already an edge colored, we dont color more of its edges....this way when we get the final monocrhomatic tetrahedron, all its edeges are colored becuase they are the longest of other tetrahedrons...

[quote="Pascual2005"]It can be fixed: during the coloring procces if a tetredron has already an edge colored, we dont color more of its edges [/quote]
That I don't understand. How are we going to get a fully colored tetrahedron then?  :? Or, if we understand this phrase as "we do not specially color the longest edge of a tetrahedron if some its edge is already colored", what if the longest edge of the monochromatic tetrahedron was colored first and is not the longest edge of any other tetrahedron?  :?
I do not say that a fix is impossible or that the idea of the fix is wrong, I only say that it should be presented more accurately  ;)

ok, here is another idea:

If a side of the obatined tetraedron, is only the biggest side of that tetraedron, then uncolor it, then we have that still every tetraedron has at least one colored edge, so there is another monocromatic tetraedron, keep doing this, this procces cannot go forever, since in every step we are removing an edge, so there must be a time when we find a good monocromatic tetraedron. ;)

It works! My question was just a nitpick :-)

An elegant solution to a nice problem... Well done.


algorithm
combinatorics theorems
combinatorics

Today I got an assignment in the informatics class which sounded like: determine whether a graph is bipartite. I first observed that if we have an odd cycle, then the graph is not bipartite. In the other cases, the graphs were bipartite, so I "concluded" that a graph is bipartite iff it has no odd cycles, but the teacher said I must prove it first.

It turns out that this is in fact true:

[b]A graph is bipartite if and only if its cycles (if any) are even.[/b]

But how can we prove such a fact? Well, intuitively, it's true, but I can't find a rigorous argument... Can you help me?

PS: What's the optimal (and elementary) algorithm for determining whether a graph is bipartite? What about an algorithm which finds the way to divide the graph?

[Edit] Later on, I found out that this was discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=15325 .

it is rather easy to see this. if the graph is bipartite then obviously it cannot have odd cycles(since then there would have to be an internal edge in one of the two partitions and that is impossible). conversely, suppose that a graph has no odd cycles. start with a vertex and start two coloring the vertices in the following way: color the first vertex $1$ and then color all its neighbors $2$. Extend this same coloring to those vertices not already colored. The only way a vertex is labelled ambiguously is if there are two paths to the vretex each leading to a different color. But since all cycles are even, any two paths from our initial vertex to this later vertex must have the same parity and so this gives a proper two-coloring of the grpah, thereby a bi-partition.
in fact this can be used as an efficient algorithm to check for bipartiteness.

i thought also about this algorithm but i didn't believe it can be considered a "proof". seems i was wrong


x+y+z+6, -2x+y+2z+0, and -2y-3z=-7 solve for the value of Y
i dont remember how to do this

Assuming you meant
x+y+z=6
-2x+y+2z=0
-2y-3z=-7
[hide]First you would subtract the second equation from the first, which gives an equation with just x and z, no y's. Then multiply the second equation by 2 and add it to the third equation, making a second equation with just x and z. Then you have a two-variable system of equations you can easily solve and substitute the values for x and z into any of the equations to find y.[/hide]

x+y+z=6
-2x+y+2z=0
-2y-3z=-7
[hide="Full Solution"]First we eliminate x from the last two equations using the first. Happily, the third equation already contains no x's. So we multiply the first equation by two: 2x+2y+2z=12. We then add that to the second equation to get 3y+4z=12. Now we have:
3y+4z=12
-2y-3z=-7
We multiply the top equation by 2 and the last by 3:
6y+8z=24
-6y-9z=-21
Adding them:
-z=3
So z=-3. Then y=8 and x=1, so {x,y,z}={1,8,-3}[/hide]

When you put x + y +z + 6, did you mean x + y +z + 6 = 0? If so, then the three equations would be

x + y + z = -6
-2x + y +2z = 0
-2y - 3z = -7

[hide]Using Cramer's rule x, y, and z are  13, -64, and 45 respectively[/hide]

[quote="dangvh"]
Using Cramer's rule x, y, and z are  13, -64, and 45 respectivel[/quote]
what is the $\text{Cramer's rule}$? :?

[quote="shobber"]
what is the Cramer's rule? :?[/quote]

Cramer's rule is a fast way solve for one variable without having to solve the whole system. [url=http://www.purplemath.com/modules/cramers.htm]For more details, visit this page.[/url]

But like the first three posters demonstrated--all you need to do is use basic algebra to add/subtract systems of equations so that you can eliminate variables and solve for the one you need.

By the way...this smells like a homework question.  :?


[quote="Illos"]Prove that an $ n$ digit integer is divisible by $ 2^n$ if its last $ n$ digits are divisible by $ 2^n$.[/quote]
[hide="Solution"]
Since the last $ n$ digits of the number are divisible by $ 2^n$, let the number be $ 10^n \plus{} j\cdot 2^n \equal{} 2^n\cdot j \plus{} 2^n\cdot 5^n \equal{} 2^n(j \plus{} 5^n)$, so $ 2^n$ is a factor of the number. [/hide]


function
algebra
domain
vector
topology
calculus
calculus computations

Let $ f$ be a real-valued function with domain $ E$. Prove that $ F$ is continuous if and only if for each open set $ O$ there is an open set $ U$ such that $ f^{-1}/[O]=E\cap U$.

[quote="queue_zniv16"]Let $ f$ be a real-valued function with domain $ E$. Prove that $ F$ is continuous if and only if for each open set $ O$ there is an open set $ U$ such that $ f^{-1}/[O] = E\cap U$.[/quote]

What type of function is $ f$? Is it a vector function $ f:\mathbb{R}^{n}\mapsto\mathbb{R}^{m}$.

If so follow this proof below. (I'm not proving for the more general case when the domain is $ E\subseteq\mathbb{R}^{n}$ but rather when $ E=\mathbb{R}^{n}$. It will be your job to generalize this proof).

[b]Theorem:[/b] Let $ f:\mathbb{R}^{n}\mapsto\mathbb{R}^{m}$ be a continous function. Let $ O\subseteq\mathbb{R}^{m}$ be an open set then the pre-image $ f^{-1}[ O ] = U =\{x\in\mathbb{R}^{n}| f(x)\in O\}$ is an open set.

[b]Proof:[/b] Let $ a\in U$ we will show that $ a$ is an interior point which will complete the proof that that $ U$ is open. Now $ f(a)\in O$, which is open, hence $ |y-f(a)|<\epsilon$ in $ O$ for any $ \epsilon$ sufficiently small. But since $ f$ is continous it means $ |x-a|<\delta\implies |f(x)-f(a)|<\epsilon$. Thus, $ |x-a|<\delta$ all lie in $ U$. So $ a$ is an interior point. Q.E.D.

[b]Theorem:[/b] Let $ f:\mathbb{R}^{n}\mapsto\mathbb{R}^{m}$ be a map so that the pre-image of any open set $ O$ is an open set $ U$, then $ f$ is a continous map.

[b]Proof:[/b] Let $ a\in\mathbb{R}^{n}$ we will show $ f$ is continous at $ a$ to complete the proof. Now define $ N(f(a),\epsilon)$ the open ball around $ f(a)$ with radius $ \epsilon$. We can think of this ball as a set of point $ |y-f(a)|<\epsilon$ for $ y\in\mathbb{R}^{m}$. Now let $ U$ be the pre-image of $ S$. By hypothesis this is open. And $ a\in U$ because $ |f(a)-f(a)|=0<\epsilon$. That means by definition that $ |x-a|<\delta$ is fully contained in $ U$ which implies $ |f(x)-f(a)|<\epsilon$. Q.E.D.


Using the least upper bound axiom, prove that there is a real number $ x$ satisfying $ x^3 \equal{} 2$.

This isn't a number theory problem.

Consider the set $ S: \equal{} \{x \in \mathbb{R} | x^3 < 2\}$; show it is bounded above and take it's least upper bound.

Got it - the course I'm taking is called Numbers and Sets, so I wasn't sure where to post this either  :)


Hehe my AoPS vol 1 is experiencing some wear and tear after being transported too and from school every other day. A hardback would certainly help  :) .


Find the remainder when 45^2001 is divided by 41.

(MAML Olympiad 2001)

[hide] First,  see that 4 :equiv: 4 mod 41.  Then it's equivalent to finding the remainder of 4^2001 / 41.

Fermat's little theorem says 4^40  :equiv: 1 mod 41.  So 4^2000 :equiv: 1 mod 41.  Then 4^2000*4  :equiv: 4 mod 41.

[/hide]


function
logarithms
algebra
domain
real analysis
real analysis unsolved

I want the following conditions for a function $F(n,k)$:

(i) $F(n,k)>0$ for any positive integers $n$ and $k$

(ii) $F(n,k)-F(n,k-1)\ge 1/k+k/n$ for any integers $n>k>0$

(iii) $F(n,k)-F(n-k,k)\ge 1/k+k/n$ for any integers $n>k>0$

(iv) for given $c>0$, $F(n,k)=o(k)$ provided $c<k^2/n<1/c$.

May all these conditions be satisfied? At fact, even if (ii) and (iii) are satisfied for $k<\sqrt{n}\ln n$, it would be good.

In a domain $c<k^2/n<1/c$ function $M(\ln n+\ln k)$ satisfies these conditions, but this domain is not stable after transforms $(n,k)\rightarrow (n,k-1),\, (n-k,k)$.

From (i) and (ii) for $n=1$, we get $F(1,k)\geqslant \frac 12 k^2$ ($k>1$). From (iii), we conclude that $F(k^2+1,k)>F(1,k)\geqslant \frac12 k^2\neq o(k)$. Am I missing something?  :?

No, you don't. Everything you wrote is clear.

I am very sorry. It arises in analysis of partitions $p(n,k)$ of $n$ onto summands $\le k$, so I did not think about $k>n$ at all. I just edited the question, but I'm afraid it is not much better then before.

Hmm, still something is bad: (iii) used with $k=1$ implies $F(n,1)\geqslant n-1$ and then (ii) gives $F(n,k)\geqslant F(n,1)\geqslant n-1$ for all $n,k$. Are some more corrections to the formulation due? Maybe there is no $\frac 1k$ in (iii)?   :?

Indeed. $1/k$ in (iii) is essential. Never mind, forget this topic, or better please delete it.


geometry
topology
real analysis
advanced fields
advanced fields theorems

So, one of the central ideas in geometric measure theory is the projection properties of sets, and Besicovitch (as with everything else in GMT) made remarkable progress in this area in the early years. One thing he proved, which I will state in the 2-dimensional case only although the result (and my question) generalize to $\mathbb{R}^{n}$.

Suppose that $E \subset\mathbb{R}^{2}$ has exact Hausdorff dimension 1. That is, $0 < \mathcal{H}^{1}(E) < \infty$. Then:

(a) If $E$ is regular (that is, its density is 1 at almost every point in the set; analogous to Lebesgue density theorem), then for all $\theta$ (except possibly one value) the orthogonal projection $\pi_\theta(E)$ onto the line $L_\theta$ through the origin at angle $\theta$ has exact Hausdorff dimension 1.
(b) If $E$ is irregular (i.e., not regular), then for almost all $\theta$, $\mathcal{H}^{1}(\pi_\theta(E)) = 0.$

So, in view of the characterization of regular 1-sets in the plane as subsets of countable unions of rectifiable curves (plus a set of measure 0), (a) makes a lot of sense. Basically the only thing that can go wrong is if $E$ is a subset of a straight line segment (or a bunch of parallel ones). However, I don't really understand the reason for (b). I have looked at the proof, but it is a really ingenious geometric proof due to Besicovitch, and I haven't understood it completely yet. I have a feeling that it's somehow connected with the fact that irregular sets have tangents almost nowhere, but I just can't see it.

Could somebody help me understand the intuitive reason for this result?

The word "regular" is dangerously ambiguous: these days it often means that $C^{-1}r\le \mathcal{H}^{1}(E\cap B(x,r))\le Cr$ for all $x\in E$ and $r>0$ (with an upper bound for $r$ if $E$ is bounded).

I never read a proof of that theorem. Here is a stupid idea: suppose that the projection of $E$ onto $x$-axis has positive measure. Define $f(x)=\sup\{y\colon (x,y)\in E\}$ for $x$ in the projection. If $f$ is measurable (which it probably is), then it is approximately continuous almost everywhere, which means that $E$ has some structure resembling tangent lines... But honestly, the best way to get an intuition is to work through an example. Perhaps the simplest example of a purely unrectifiable set is the square of $(1/2)$-dimensional Cantor set [obtained by removing the middle $1/2$ of each segment).


Determine whether there exists a function $ f: \mathbb{N}\longrightarrow \mathbb{N}$ such that
$ f(n)\equal{}f(f(n\minus{}1))\plus{}f(f(n\plus{}1))$ for all natural numbers $ n\ge 2$.

We will prove that if $ f(n)\ge{k}$ ($ k\ge{2}$) for all naturals $ n$ ($ n\ge{2}$), then the equality $ f(n)\ge{k \plus{} 1}$ is also true for all naturals $ n$.

$ n \plus{} 1\ge{3} > 2$, so $ f(n \plus{} 1)\ge{k}\ge{2}$, so $ f(f(n\plus{}1))\ge{k}$,
$ f(n) \equal{} f(f(n \minus{} 1)) \plus{} f(f(n \plus{} 1))\ge1 \plus{} f(f(n \plus{} 1))\ge{k\plus{}1}$ - proved.

$ f(n) \equal{} f(f(n \minus{} 1)) \plus{} f(f(n \plus{} 1))\ge{1 \plus{} 1 \equal{} 2}$
From the statement above we have $ f(n)\ge{2 \plus{} 1 \equal{} 3 > 2}$, that means $ f(n)\ge{3 \plus{} 1 \equal{} 4 > 3}$, that means $ f(n)\ge{4 \plus{} 1} \equal{} 5 > 4...$

By mathematical induction $ f(n) > {m}$, where m is any natural number, so there is no such function.

[quote="Bachukas"]Wait! is $ \infty$ a natural number ?[/quote]
No.  Look up the definition.

Eh, doesn't $ f(n) \equal{} 0$ for all $ n\in \mathbb{N}$ work?

EDIT: Blah, fail. :oops:

The symbol $ \bf N$ is ambiguous -- in my field of work, it most commonly refers to $ {\bf Z}_{\geq 0}$, and some other symbol is used for $ {\bf Z}_{> 0}$.  However, from context it's clear that the function we're looking for is from the positive integers to the positive integers, since otherwise the given equation should have started at $ n \geq 1$.

I take $ \mathbb{N}$ as >0, and $ \mathbb{N}_0$ as >=0, of course integers.


geometry
parallelogram
circumcircle
perpendicular bisector

[color=red]I'm sure that you will enjoy this problem![/color]

Let a parallelogram $ ABCD$. The bisector of angle $ A$ meets side $ CD$ at $ Z$ and line $ BC$ at point $ E$. If $ O$ is

 the circumcenter of triangle $ CEZ$ and $ M$ is the midpoint of $ BD$ , prove that

 $ OM \perp BD$

 [u]Babis[/u]

Yes, nice problem Mr. Babis. But what a pity, it's known.
Look at Huyen Vu's solution: http://www.mathlinks.ro/viewtopic.php?t=54065

Yes ! This problem is from Latvian olympid 1997 I think.The inverse is problem 2 at IMO 2007. But , as stated in the subtitel , I tried to reformulate it to a more elementary form.I put it only to share with you this different form. In the morning I found the same solution as Huen Vu. If you have time , please write again this hint, so we 'll have problem and solution together!

 Babis

[quote="April"][color=red]O.K. I'll rewrite [b]Huyen Vu[/b]'s solution here[/color] :D
[quote="stergiu"] Let a parallelogram $ ABCD.$ The bisector of angle $ A$ meets side $ CD$ at $ Z$ and line $ BC$ at point $ E.$ If $ O$ is the circumcenter of triangle $ CEZ$ and $ M$ is the midpoint of $ BD,$ prove that $ OM \perp BD$

 [u]Babis[/u][/quote]
[b]Huyen Vu's solution.[/b] We have $ \angle BAE=\angle CZE\quad (\textrm{because}\; AB\parallel CD)$ and $ \angle DAE=\angle CEZ\quad (\textrm{because}\; AD\parallel BC)$
So $ CE=CZ.$ We also have $ OE=OZ,$ this implies that $ OC$ is perpendicular bisector of $ EZ,$
thus $ \angle OCE=\angle OCZ\implies \angle OZD=\angle OCZ=\angle OCB\quad (1)$
On the other hand, we have $ \angle DAZ=\angle BAZ=\angle DZA\implies ZD=DA=CB\quad (2)$

Consider two triangles $ OZD$ and $ OCB,$ we have $ \left\{\begin{array}{c}OZ=OC\\ \angle OZD=\angle OCB\\ ZD=CB\end{array}\right\|\Rightarrow\triangle OZD=\triangle OCB\Rightarrow OD=OB\Rightarrow OM\perp BD$[/quote]

          Thank you April ! I apraciate your help.

 Babis

[color=darkblue]This [b]nice and easy[/b] problem becomes [b]very easily[/b] if we'll state equivalently thus : [/color]
[quote][color=purple][b]Let $ ABC$ be an $ A$- isosceles triangle with the circumcircle $ w=\mathcal C(O,R)$. Let the points

$ \{\begin{array}{c}M\in AB\ ,\ A\in (BM)\\\ N\in AC\ ,\ C\in (AN)\end{array}$ so that $ AM=CN$. Prove that $ OM=ON$.[/b][/color][/quote]
[color=darkblue][b]Proof I.[/b] $ \{\begin{array}{c}OA=OC\ ,\ AM=CN\\\\ m(\widehat{OAM})=m(\widehat{OCN})=90^{\circ}+C\end{array}\|$ $ \implies\over{S.A.S.}$ $ \triangle \ OAM\equiv\triangle\ OCN$ $ \implies$ $ OM=ON$.

[b]Proof II[/b] ([/color][color=darkred][b]Huyen Vu's[/b][/color][color=darkblue]). $ \{\begin{array}{c}OB=OA\ ,\ BM=AN\\\\ m(\widehat{OBM})=m(\widehat{OAN})=\frac{A}{2}\end{array}\|$ $ \implies\over{S.A.S.}$ $ \triangle\ OBM\equiv\triangle\ OAN$ $ \implies$ $ OM=ON$.

[b]Proof III.[/b] Observe that $ BM=AN$ $ \implies$ $ MA\cdot MB=NA\cdot NC$, i.e. the points $ M$, $ N$ have the same power w.r.t. the circle $ w$ $ \implies$ $ OM=ON$.

[b]Remark.[/b] The quadrilateral $ OAMN$ is cyclically and $ OM=ON=R\cdot\frac{MN}{AB}$ !.[/color]


AMC AIME MIT college USA(J)MO USAMO function

I know that MIT likes the applicants to report AMC score since SAT SUB MATH 2 is easy and lots of people got 800.

AIME certification probably helps. What else???

Like No.1 on in a state or top 3 in a state in either index or AMC or AIME?
Or qualified or USAMO?

Any seniors here used AMC or AIME to help them got into colleges?
Tell me how please. Thank you

Of course it helps. It's not going to get you in by itself, though. College should not be your primary reason for taking contests; in fact, it probably shouldn't be your primary reason for doing anything. You can put your scores and achievements on your resume or write about them in an essay.

Obviously it does help, but the question is how much.  I still think that your high school grades play a much larger role.  There is currently a long debate on College Confidential regarding the "hook" value of USAMO qualification.  You can check there for more information.

USAMO qualification is a valuable hook, however, it will not make or break the deal.  Metaphorically speaking, USAMO qualification is often one of the gems in the crowns of superstar college applicants.

math camps, research, internships, and higher level classes are all equally valuable in my opinion. im still going to try and make USAMO tho  :P

[quote="chinesemather"]Any seniors here used AMC or AIME to help them got into colleges?
Tell me how please. Thank you[/quote]
It's simple.  Instead of bubbling in the correct answers, you bubble in the cheat code:  065, 077, 067, 065, 077, 067, up, up, down, down, ...

In all seriousness, there are a few things you should know.

1.  Your username suggests that you are probably an international applicant, or that you have citizenship elsewhere but go to school here.  If that's the case, international application at top colleges in the United States is very difficult and anything that stands out helps, but there is nothing that will guarantee acceptance, not even an IMO gold medal.

2.  No applicant is told what parts of his or her application helped and what parts didn't.  To find out that kind of information you need a friend in the admissions office.  In other words, no particular applicant will be able to tell you if his or her AMC/AIME scores were a deciding factor or even particularly important to them getting accepted to a particular school.

3.  Whether colleges care about the AMC at all is a delicate function of the college.  If you're applying to a school with a strong math program, you can bet that they'll pay attention to such things, but some colleges may not even know what the AMC is.

4.  What do you plan on doing with this information?  If you decide how much time to spend studying for contests because of college admissions, you've got it all wrong.  College admissions is not a process where you prove that you're good enough to a college - it is a process where you [b]find[/b] a college that [b]fits your interests.[/b]  Part of that requires you to be honest in your application about the distribution of your time - that is, decide how you want to spend your time based on [b]what interests you.[/b]  If that happens to be studying for the AMC, great!  Write about how much you enjoy math in your essay, and that will probably make [b]much[/b] more of an impact than whether you got 116.5 or 120 on the AMC 12 in sophomore year.

5.  People don't like to admit this, but college admissions is much more a function of what kind of a day your college admissions officer has been having than a function of small deviations in test scores (that is, once you've gotten past the first few rounds).  If you're going to do something to stand out in your application, a number in one small corner isn't it.

It's difficult to say, but personally I doubt that most competitions carry any weight: admissions officers probably give negligible consideration to math league, AIME qualification, or any sort of local contest. USAMO qualifiers generally do quite well in terms of colleges, but then those are probably students talented enough to excel in other areas as well. According to my counselor, the most important things are recommendations, transcript, and application essays.

As t0rajir0u said, it depends on the school. Places like MIT and Yale which ask for scores probably do pay some degree of attention to them, especially if they are really good.

While that is quite true (there is a reason four of of the five senior IMO 2008 team members are here), I don't think it translates to "work harder at Olympiads in order to get into MIT."  I think it translates to "if you are not passionate about [things such as] Olympiads, then [i]perhaps you shouldn't go to MIT[/i]."

It's probably not that if you do well in the USAMO, you'll make MIT. It's more like, if you are good enough to do well on the USAMO, you'll make MIT whether or not you had taken it.

Yes, exactly!  Doing well on the USAMO correlates to a lot of other things which will also show up on an application and any sufficiently large subset of those things will be good enough.  As Matt McGann is fond of telling [url=http://www.cee.org/programs/rsi]Rickoids[/url], going to RSI doesn't guarantee acceptance to MIT - it's just that going to RSI correlates to having done things that give you a strong enough application to get accepted anyway.

[quote="mathemonster"][quote="DataBox"]As far as MIT is concerned the following is some good information:

"As far as how much of an edge a high AMC or AIME score provides, the only thing I know from meetings with adcoms as an EC (educational counselor) , is that an AIME score greater than 7 makes the candidate a "super candidate". USAMO, Siemens or Intel finalists get even a greater boost and MIT tends to attract and aggressively recruit a disproportionate number of them. A USAMO finalist or Intel finalist is much more attractive to MIT than a 2400 SAT scorer. MIT participates in a lot of collegiate math competitions such as the Putnam as well as many science competitions and wants to perform well at them. MIT may not recruit olympic caliber athletes but it does recruit math and science olympians."[/quote]
Wow, I'm definately going to work hard and get atleast 7 on the AIME next year. That's not even that hard..[/quote]

No, that's not what any student's goal should be. Work hard to get a 7 because you WANT to, not because MIT finds that a "super candidate." The people who get into schools like MIT or other top colleges are there because they are passionate about their interests, not simply MIT interests.

[quote="mathemonster"]Wow, I'm definately going to work hard and get atleast 7 on the AIME next year. That's not even that hard..[/quote]
There's a reason schools don't release information like this to the public.  Metrics like that are shortcuts to help a school measure something else correlated to, say, getting a 7 on AIME.  If you game the system, the metric stops measuring what it's supposed to measure, so if enough people game the system they'll just [i]change metrics.[/i]

(Apologies where appropriate to fans of metric spaces and measure theory.)


A long piece of paper 5 cm wide is made into a roll for cash registers by wrapping it 600 times around a cardboard tube of diameter 2 cm, forming a roll 10 cm in diameter.  Approximate the length of the paper in meters (in terms of pi).  (Pretend the paper forms 600 concentric circles with diameters evenly spaced from 2 cm to 10 cm)

Hmmm...this is a tough one. Anyway, you know that the cardboard tube is a cylinder which starts off with radius of 1 and increases its radius by 4. So for every 150 circles, the radius increases by 1 inch. Hmmmm...........oh wait, I think I got it!

Allright, just look at the figure from an aerial view. You have a cylinder in a cylinder. So the area of the tube would be 1^2pi x 5 and so it would be 5pi. The area of the larger cylinder (the paper) would be (5^2 x 5pi)-5pi=[b]120pi[/b].

[quote="Palytoxin"][quote="Iversonfan2005"]Am I right?   :w00tb:  :?:[/quote]
sorry but i dont think question is asking for area, its asking for approximate length of paper...[/quote]

Oh you're right.... :wallbash: Okay, that makes it a lot harder, hmmm

[quote="Iversonfan2005"]oh wait then, since I already have the area of 120pi, all I have to do is 5x=120pi, so the length would then be [b]24pi.[/b] If that's not right then I'm very frustrated :furious:[/quote]
i got something different but ill need to double check first before posting it

What did you get, I think I got the right answer but I'm waiting for other people's responses :read:

here is my solution
[hide="my solution"]
let $r$ be the radius of empty roll
$R$ be the radius of full roll
$\delta$ be the content of paper, the paper wraps, the width of the annulus
$t$ be the thickness of paper
$n$ be the number of wraps around tube

so we have $R-r=\delta$ and $\delta=nt$, so $t=\frac{\delta}{n}=\frac{R-r}{n}$

now lets find the length of paper used for each full wrap around tube (circumference)

for first wrap $r_1=r+t$ and this wrap is $c_1=2\pi r_1=2\pi(r+t)$

for second wrap $r_2=r_1+t=r+2t$ and this wrap is $c_2=2\pi r_2=2\pi(r+2t)$

for third wrap $r_3=r_2+t=r+3t$ and this wrap is $c_3=2\pi r_3=2\pi(r+3t)$

and so on...

for nth wrap $r_n=r+nt$ and this wrap is $c_n=2\pi r_n=2\pi(r+nt)$

total length of paper is
\[c_1+c_2+c_3+\cdots+c_n=2\pi (nr+(1+2+3+\cdots+n)t)=2\pi (nr+\frac{tn(n+1)}2)\\
\sum_{i=1}^n c_i=2\pi (nr+\frac{(R-r)(n+1)}2)\]

we know $r=1\,\text{cm},R=5\,\text{cm},n=600$

so we finally get answer
\[\sum_{i=1}^n c_i=2\pi (nr+\frac{(R-r)(n+1)}2)=2\pi (600\cdot 1 +\frac{(5-1)(600+1)}2)=\boxed{3604\pi\,\text{cm}}\]
or $\boxed{\frac{901}{25}\pi\,\text{m}}$ because he asked in meters  :D [/hide]

Is that the correct answer snowstorm :(  :wallbash:  :wallbash_red:
Hopefully mine is though  :?


Give ellipse $ (E)$ : $ \frac{x^2}{a^2} + \frac{y^2}{b^2} =1$ (a>b>0)
 A;B;C $ \in$ $ (E)$. find $ \triangle ABC$ such that $ S_{ABC}$ is maximal.

This is very well known. Try to show that any triangle having the center of the ellipse as its center satisfies this.

well,i had tried but nothing became more clear. can you give me more details

Apply an affine transformation that converts the ellipse into, say, the unit circle. It is clear that the area of any triangle with vertices on the ellipse has its area multiplied by a fixed constant to obtain the area of its image on the unit circle.  But the triangles in the unit circle of maximal area are equilateral.  Undoing the affine transformation, we find the class of maximal area triangles (none of which are equilateral) with the origin as their common centroid. Moreover, a simple calculation shows their area to be $\frac{3ab\sqrt{3}}{4}.$

Usually I would agree, Mildorf, but in this particular case all you need is to project a circle on the plane so that it becomes the ellipse. Mentioning the affine transformation (even though the projection is, itself, an affine transformation :)) is overkill here :).


floor function
combinatorics unsolved
combinatorics

For any positive integer $n$, evaluate
\[ \sum_{i=0}^{\lfloor\frac{n+1}{2}\rfloor} \mbox{\small \left(\!\!\begin{array}{c} n-i+1\\ i\end{array}\!\!\right) } \ , \]
where  ${\mbox{ \left(\!\!\begin{array}{c} m\\ k\end{array}\!\!\right) } = \frac{m!}{k!(m-k)!}}$ and $\lfloor\frac{n+1}{2}\rfloor$ is the greatest integer less than or equal to $\frac{n+1}{2}$.

Let $s_n=\displaystyle\sum_{i=0}^{\lfloor\frac{n+1}{2}\rfloor}{n+1-i\choose i}$.  Then we have

$s_n=\displaystyle\sum_{i=0}^{\lfloor\frac{n+1}{2}\rfloor}{n+1-i\choose i}$

$=\displaystyle\sum_{i=0}^{\lfloor\frac{n+1}{2}\rfloor}\displaystyle\left [{n-i\choose i-1}+{n-i\choose i}\right ]$

$=\displaystyle\sum_{i=0}^{\lfloor\frac{n+1}{2}\rfloor -1}{n-1-i\choose i}+\displaystyle\sum_{i=0}^{\lfloor\frac{n+1}{2}\rfloor}{n-i\choose i}$

$=\displaystyle\sum_{i=0}^{\lfloor\frac{(n-2)+1}{2}\rfloor}{(n-2)+1-i\choose i}+\displaystyle\sum_{i=0}^{\lfloor\frac{n+1}{2}\rfloor}{(n-1)+1-i\choose i}$

All we have is that pesky limit on the second sigma notation and we have a recursion.  Now, notice that for even $n$, we have $\displaystyle\left\lfloor\frac{n+1}{2}\right\rfloor=\frac{n}{2}=\left\lfloor\frac{n}{2}\right\rfloor$.  So for even $n$ we have what we want.

And for odd $n$, we have $\displaystyle\left\lfloor\frac{n+1}{2}\right\rfloor=\left\lfloor\frac{n}{2}\right\rfloor +1$.  Then we have

$\displaystyle\sum_{i=0}^{\lfloor\frac{n+1}{2}\rfloor}{(n-1)+1-i\choose i}$

$=\displaystyle\sum_{i=0}^{\lfloor\frac{n+1}{2}\rfloor -1}{(n-1)+1-i\choose i}+ {n-\lfloor\frac{n+1}{2}\rfloor\choose\lfloor\frac{n+1}{2}\rfloor}$

$=\displaystyle\sum_{i=0}^{\lfloor\frac{(n-1)+1}{2}\rfloor}{(n-1)+1-i\choose i} +{n-\lfloor\frac{n+1}{2}\rfloor\choose\lfloor\frac{n+1}{2}\rfloor}$

$=\displaystyle\sum_{i=0}^{\lfloor\frac{(n-1)+1}{2}\rfloor}{(n-1)+1-i\choose i}$

So we finally have that for all $n\ge 3$ that $s_n=s_{n-1}+s_{n-2}$.

Then we can just solve this recursion with $s_1=2$ and $s_2=3$, and we end with

$s_n=\displaystyle\left (\frac{5+3\sqrt{5}}{10}\right )\displaystyle\left (\frac{1+\sqrt{5}}{2}\right )^n+\displaystyle\left (\frac{5-3\sqrt{5}}{10}\right )\displaystyle\left (\frac{1-\sqrt{5}}{2}\right )^n$

Another way of seeing this recursion is to note that a given $i$ in the summand counts the number of ways of placing $i$ $2 \times 1$ bricks and $n-2i+1$ $1 \times 1$ bricks in a line of length $n+1$, so the total sum counts the number ways of tiling a $1 \times n+1$ strip with $1 \times 1$and $2 \times 1$ dominoes.  Now the recursion follows by considering whether the first domino in the strip is $1 \times 1$ or $2 \times 1$

That is $F_{n+2}$, the (n+2)-th Fibonacci number. The formula is somewhat classic, see formula (61) at http://mathworld.wolfram.com/FibonacciNumber.html


Find all functions $f: \mathbb{N}\rightarrow\mathbb{N}$ such that \[ f(m-n)f(m+n)=f(m^2) \] for $m,n\in\mathbb{N},\;m>n.$

Also at:

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=f%28m-n%29f%28m%2Bn%29&t=65708

but no complete solution given there though

The question is : can we consider $0 \in \mathbb{N}$ ?

I don't think we can.  Instead of m >= n, the condition is explicitly m > n, which prevents the possibility of f[0] anyways.


inequalities
function
inequalities proposed

This problem is dedicated to E.lopes and Silouan, who shown a very nice solution to my preceding problem. Hope that they will find a nice solution to this problem, too. Welcome all Mathlinkers.

Let $a,b,c$ be non-negative real numbers. Prove that
\[\frac{a}{11a+9b+c}+\frac{b}{11b+9c+a}+\frac{c}{11c+9a+b}\le \frac{1}{7}. \]

I am sorry, E.Lopes. Just a mistyping only, and I did edited.

Thank you for the nice gift Hungkthn
 we have to prove that
$9(a^{3}+b^{3}+c^{3})+75(b^{2}a+c^{2}b+a^{2})-93abc-53(a^{2}b+b^{2}c+c^{2}a)\geq 0$

Now this is linear as a variable $(abc)$ so by ABC method of Anh-Cuong we have to examine two cases .
1st when two variables are equal the inequality is equivalent to
$160a^{3}-284a^{2}c+88ac^{2}+36c^{3}\geq 0$ or $4(a-c)^{2}(40a+9c)\geq 0$

2nd when one variable is 0 we have to prove that
$36a^{2}-24ac+4c^{2}\geq 0$ or $4(3a-c)^{2}\geq 0$ which is true

so we are done .

can you please explain this method in detail or a link.

see this link, anuj_kumar:

http://reflections.awesomemath.org/2006_4/2006_4_proving_inequalities.pdf

(solving inequalities with linear functions)

Nice solution, Silouan! (if it's correct!)
I have one ugly way to prove the open version of the inequality, nothing nice!  :(

I think your solution is not true, silouan.

what's your proof, hungkhtn?
with brute force, we can 'easily' prove the open version, but i can't find one nice solution! ;\

Silouan is not true because the inequality is not symmetric. ABC is false in applying this case. However, you are good to realize that the equality can hold for $(a,b,c)~ (3,1,0).$

@ E. Lopes. My solution is very simple and easy (for a very general problem). But I still wait for a nice solution.

[quote="toanhocmuonmau"][quote="hungkhtn"][quote="e.lopes"]I'm trying in two ways, at the moment!

i) opening, we have to prove that $9(a^{3}+b^{3}+c^{3})+75(b^{2}a+c^{2}b+a^{2}c) \geq 93abc+53(a^{2}b+b^{2}c+c^{2}a)$ (ugly!)

ii) using cauchy to find one nice expression for the minimum of the expressions of form (11a+9b+c)!

If this two ideas don't kill the problem, i will try with Smothing!

If i solve the problem, i will post my solution!

Again, thanks for this nice problem, hungkhtn!  :)[/quote]

Expanding is a natural way. According to the case of equality, can you solve this problem, E.lopes? Toanhocmuonmau said that he can find a solution, too. So I wait for him,[/quote]
Okie, Here is my solution
Note that
\[\sum a^{3}-3abc=(a+b+c)((a-b)^{2}+(a-c)(b-c)) \]

\[\sum ab^{2}-3abc =c(a-b)^{2}+b(a-c)(b-c) \]

\[\sum a^{2}b-\sum ab^{2}=(a-b)(b-c)(a-c) \]
Then the inequality can be rewritten as
\[9(\sum a^{3}-3abc)+22(\sum ab^{2}-3abc) \ge 53(a-b)(b-c)(a-c) \]
or equivalently
\[9(a+b+c)((a-b)^{2}+(a-c)(b-c))+22c(a-b)^{2}+22b(a-c)(b-c) \ge 25(a-b)(b-c)(a-c) \]
Now, suppose that$c=\min\{a,b,c\}$, setting $x=a-c,y=b-c$, the inequality is equivalent to
\[9(3c+x+y)((x-y)^{2}+xy)+22c(x-y)^{2}+22xy(c+y) \ge 25xy(x-y) \]
It's obvious that this is an increasing function in terms of $c$, hence it suffices to consider the case $c=0$, then the inequality becomes
\[9xy(x+y)+22xy^{2}\ge 25xy(x-y) \]
It's easy to check this inequality. We are done. :)[/quote]

Good work of EMV  :lol: Thank you.

Prove that if $a,b,c,d\geq0$ then
\[\frac a{8a+6b+c+d}+\frac b{8b+6c+d+a}+\frac c{8c+6d+a+b}+\frac d{8c+6d+a+b}\leq\frac14.\]


number theory
prime numbers
algebra proposed
algebra

let $ k_1,\dots,k_n$ be a sequence of mutually prime numbers, $ n>1$.

is it true that for every posotive integers $ a_1,\dots,a_n$ the sum
\[ \sqrt[a_1]{k_1}\plus{}\dots \plus{} \sqrt[a_n]{k_n}\]
is irrational ?

We need the condition $ a_1 a_2 ... a_n > 1$.

Yes, of course it should be
I forgot to write this assumption but I want it :wink:

And then the sum will be irrational ?
Could you show some approach to the proof of it ?

Thanks

In fact, we can show that the degree of the extension $ \mathbb{Q}[ \sqrt[a_1]{k_1} \plus{} ... ] : \mathbb{Q}$ is $ a_1 a_2 ... a_n$.  Unfortunately, I'm not familiar with the proof.


[quote="13375P34K43V312"]If 554 is the base $b$ representation of the square of the number whose base $b$ representation is 24, then what is the value of $b$ when written in base 10?[/quote]
[hide]$5*b^{2}+5b+4=(2b+4)^{2}$
$b^{2}-11b-12=0$
$b=12$[/hide]
Great problem :)

I just added some content to the number base entry on the AoPSWiki.  Someone had previously posted some very hard problems for the article and I wanted to create some simple beginner ones.

If anyone woud like to solve the problems I wrote, go here:
http://www.artofproblemsolving.com/Wiki/index.php/Base_numbers/Conversion#Examples

Then you can edit the examples section to include your solutions.

It would be great to post some more problems like the one just discussed on this thread. The AoPSWiki is developing into a great resource!


How many numbers of the form $\overline{abcd}$, $a \neq 5$, have the property that $N = a+4b+3c-d$ is divisible by $13$?

[i]Nicolae Stăniloiu[/i]

You could find the general solution to $a+4b+3c-d \equiv 0 \pmod{13}$ and then look at all cases where values of all a, b, c, d are between 0 and 9. This would get pretty tiresome though...

is $a, b, c, d$ un-distinct??? I think so...
[hide="solution"]
we take the number to be $1000a + 100b + 10c + d$, mod it by 13 yields $-a -4b -3c + d$, the exact inverse of $N$!!!.  Since $N$ is divisible by $13$, the number is divisible by 13.  Therefore, there are $\lfloor{9999/13} - %Error. "flor" is a bad command.
{1000/13} = 693$ 4 digit numbers that's divisble by $13$.  But $a != 5$, so we have to subtract $\lfloor{5999/13} - \lfloor{5000/13} = 77$.  The answer is $616$.
[/hide]

nice problem!!!


Solve in positive integers : $ x(x+1)=y^{2007}$

[hide="Er..."] $ gcd(x, x+1) = 1$, so both must be perfect $ 2007^{th}$ powers; clearly impossible. [/hide]

How about the following problems?

Find all solutions in positive integers:

1. $ x(x+y)=y^{2007}$

2. $ x^{2007}(y-x)=y^{2007}$

Have fun!


Prove that the following equation: $ A.a^x\plus{}B.b^x\equal{}A\plus{}B$ (where $ a>1,0<b<1,A,B \in R$) has at most two solutions

One solution is x=0.
Consuder $ f(x)\equal{}\frac{a^x\minus{}1}{1\minus{}b^x}$. For $ x\not \equal{}0$ equation equavalent to $ f(x)\equal{}\frac{B}{A}$.
Let $ g(x)\equal{}f'(x)\frac{(1\minus{}b^x)^2}{(ab)^x}\equal{}b^{\minus{}x}ln(a)\minus{}a^{\minus{}x}ln(b)\minus{}ln(a)\plus{}ln(b), g(0)\equal{}0.$
$ g'(x)a^xb^x\equal{}ln(a)ln(b)(b^x\minus{}a^x)>0$ if $ x>0$ and less than 0, if $ x<0$. Therefore $ g(x)>0$ if $ x\not \equal{}0$ or f(x) increasing function. Because $ f(\infty)\equal{}\infty, \ f(\minus{}\infty )\equal{}0$ for any $ c>0$ we get one solution $ f(x)\equal{}c$.
But if $ \frac{B}{A}\equal{}\frac{ln(a)}{\minus{}ln(b)}$ then these solution is $ x\equal{}0$ and your statement is not true.

For example $ A\equal{}B\equal{}1,a\equal{}e,b\equal{}e^{\minus{}1}$, then it equavalent to $ sh(x)\equal{}1$ had only one solution $ x\equal{}0$.

[quote="every"][quote="Rust"] your statement is not true.[/quote]
I don't think so.I think your solution is not true.Do you check your solution?[/quote]
rust's solution is true   :wink:


[quote="IndoChina"]Is the sign in the middle really negative? Thx.[/quote]

yes, of course.

Great, then it will be straight forward...

Regrouping $ \left( 2x^{3}\plus{}3xy^{2}\minus{}7x\right) dx\minus{}\left( 8y\minus{}3x^{2}y\minus{}2y^{3}\right) dy\equal{}0$

$ 2x^{3}\minus{}7x\plus{}\frac{3}{2}\left( 2xy^{2}\plus{}2x^{2}y\frac{dy}{dx}\right) \plus{}\left( 2y^{3}\minus{}8y\right) \frac{dy}{dx}\equal{}0$

Integrating, $ x^{4}\minus{}7x^{2}\plus{}3x^{2}y^{2}\plus{}y^{4}\minus{}\allowbreak 8y^{2}\equal{}C$


Q.1 : The speeds of three cars are in the ratio $3:4:5$.What is the ratio of the time they take to cover a given distance ?

Q.2 : Find the mean proportional to 9/16 and 16/9.

P.S: Since this is in the Middle school forum , give an explanation along with your answer.

[quote="karthik_k"]Q.2 : Find the mean proportional to 9/16 and 16/9.[/quote]
The term "mean proportional" is undefined to most middle school students.  In modern middle school language, this question should be phrased:

Q2:  Find the geometric mean of 9/16 and 16/9.

:oops: Oops....Sorry for that.Anyway , if any of the moderators feel that the above questions are too much for the middle schoolers , I request them to move this topic to the high school section... :)

Is question 2:

[hide]
$\frac{9}{16}x(x) = \frac{16}{9}$
$\frac{9}{16}x^2 = \frac{16}{9}$
$x^2 = \frac{16^2}{9^2}$
$x = \frac{16}{9}$

so, $\frac{9}{16}$ times $\frac{16}{9}$ is 1, the geometric mean.

Well, I'm not sure what geometric mean really is...
[/hide]

Well , 236factorial , your answer is right.
As for the first question , I'm waiting for replies.To make it simpler , let me give you four options for the first question -
$A ) 6:5:4$
$B ) 4:5:6$
$C )15:12:10$
$D ) None of these.$

Obviously B can't be the answer because as the speed decreases, the time to cover a distance increases.
Let's pick a distance like 60. the times are 20, 15, and 12 for speeds of 3, 4, and 5.
There's no such ratio in the choices, so the answer is D.
Hmm... Why does that look wrong?

For number 1 i get[hide]c.  B can't be the answer because the answer is increasing, and A just doesn't look right.  i don't think its none of these, so i guess [b]c[/b][/hide]

Time is inverse of speed
So its
1/3:1/4:1/5    x60=
20:15:12

[quote="karthik_k"][quote="236factorial"]
Hmm... Why does that look wrong?[/quote]

It just looks wrong .... The answer is right.[/quote]
you're both right.  He is right even though it might look wrong


Hi! I am an engineering student who is studying mathematics on his own. I was wondering if anyone could help me find books in the main topics of undergraduate mathematics (especially abstract and linear algebra) that would be suitable for self-study. I don't want the books to be easy; however I am looking for books that at least explain the proof of the main theorems accurately and completely. Thanks!

I didn't think engineers needed to learn abstract algebra...

For linear algebra, I used Friedberg, Insel, and Spence, which is very proof-based.  I think it's a pretty good book.  The book by Bretscher is pretty widely used, and is more application-based.


Stoicism, Epicurism, and Skepticism. the online definitions are hard to understand

can someone sum it up as to what each is ?  :lol:

thanks!

I'll try.

Stoicism: The cosmic soul (or any equivalent term, such as god or nature) is everything. One should live peacefully and rationally and try to achieve happiness through virtue. People should be ambivalent towards their destiny and should respect each other.

Epicureanism: Matter is built from atoms which move about randomly. The final purpose of life is to achieve happiness through rational thinking; one should always strive for intellectual happiness.

Skepticism: What the name says; don't trust anything you witness if it can't be proven; especially don't believe in dogmatic knowledge. Now there are types of skepticism, depending on the object of one's doubt, but this idea mainly covers all of them.