If $ 200 \le a \le 400$ and $ 600 \le b \le 1200$, then the largest value of the quotient $ \frac{b}{a}$ is

\[ \textbf{(A)}\ \frac{3}{2} \qquad
\textbf{(B)}\ 3 \qquad
\textbf{(C)}\ 6 \qquad
\textbf{(D)}\ 300 \qquad
\textbf{(E)}\ 600 \qquad
\]

[quote=PlaneGod][quote=Cheese12345]If b is maximum possibility and a is minimum,1200/200=6?[/quote]

For anyone who had doubts about the solution, I can confirm that this is correct.[/quote]

i solved it way too fast that I had doubts

[quote=PlaneGod][quote=Cheese12345]If b is maximum possibility and a is minimum,1200/200=6?[/quote]

For anyone who had doubts about the solution, I can confirm that this is correct.[/quote]

Would you please not bump threads that are more than 11 years old?  When you bump these threads, they clog up the forum. Thanks

You are allowed to bump threads, as long as you have a contribution to make.

why is it the maximum possibility / minimum possibility

a larger numerator is a larger number, and a larger denominator is a smaller number. It's pretty intuitive if you ask me, but it does require some thinking to make complete sense of it.

i told my dad that maximum / minimum is biggest value
he said i was wrong and asked what the maximum of two numbers -1000<a<1000 and 100000<b<10000000000000
i said 10000000000000/-1000 but he say this is not right?????

[quote=mathlete5451006]the minimum is the smallest value, the maximum is the largest value. They are not synonyms[/quote]

what does you mean???
it says max number / smallest number = maximum number
so i did maximum number which is 10000000000000 divided by minimum number which is -1000
my dad say answer is 10000000000000/1
why????

duh, because positive numbers are greater than negative numbers :)
so your dad is right

Applies until you get to negative values. sorry for being a bit vague.


induction
combinatorics proposed
combinatorics

$ a_1,a_2,\dots ,a_n$ be $ n$ real numbers arranged on the circumference of a circle in the same order with $ a_1$ being adjacent to $ a_n$ with the property that $ \exists \quad 1\le i\le n$  so that any consecutive sum (say anticlockwise) of $ i$ real numbers of the sequence is always positive. Prove that there exists an index $ j$ so that every partial sum starting from that index ie,
$ a_j$
$ a_j + a_{j + 1}$
$ a_j + a_{j + 1} + a_{j + 2}$
$ \dots$
$ \underbrace{a_j + \dots_}_{n}$ is positive.

Denote the sum of all numbers by $ \sigma$. By summing all the positive sums of $ i$ consecutive numbers one gets $ i\sigma$, hence $ \sigma$ is positive.

Check now [url]http://www.mathlinks.ro/viewtopic.php?t=153964[/url] post $ \#17$.

For good measure, I include two proofs.
[hide]First, a slight rewording of the  problem (it is equivalent to the original).
Some $ n$ real numbers with zero sum are written on a circle, labeled in clockwise order $ a_i$, $ i \in \mathbb{Z}_n$. Prove there exists $ m$ such that $ \displaystyle \sum_{i=0}^k a_{m+i} \geq 0$ for all $ 0 \leq k \leq n-1$.

Induction proof. Group them in contiguous groups of same sign (if only one, made of non-negatives, trivial), so we have values $ p_i, -m_i$, $ p_i, m_i \geq 0$, $ i \in \mathbb{Z}_t$, $ 2t \leq n$.
If $ 2t < n$, apply induction, then go back to initial values. If $ 2t = n$, then at least once do we have $ p_s \geq m_s$, else $ \sum p_i - \sum m_i < 0$. Group $ p_s - m_s > 0$. Induction yields labeling. Go back to initial values.

Combinatorial proof. Find least sum $ v = \displaystyle \sum_{i=0}^j a_i$; suppose it is negative, else take $ m=0$. Take $ m = j+1$. Then $ \displaystyle \sum_{i=0}^k a_{m+i} \geq 0$ for all $ 0 \leq k \leq n-1$, since for $ \displaystyle \sum_{i=0}^k a_{m+i} < 0$ one has $ \displaystyle \sum_{i=0}^{m+k} a_i < v$, absurd.

All these also works for the famous version with cars on a circular track, where the sum of the petrol quantities $ x_i$ covers the circuit. Take $ a_i = x_i - d_i$, where $ d_i$ is the quantity needed to cover the leg $ X_iX_{i+1}$; then $ \displaystyle \sum_{i=0}^{n-1} a_i = 0$, etc.[/hide]

It was noted above that the sum has to be non negative. Suppose that the claim is false and that for each $ i\in \mathbb{Z}/n\mathbb{Z}$ there exists $ f(i)\in \mathbb{Z}/n\mathbb{Z}$ so that the sum of $ a_j$'s going clockwise from $ i$ to $ f(i)$ is negative. Start from any $ i_1$ and let $ i_{k \plus{} 1} \equal{} f(i_k) \plus{} 1$. At some point you will have $ i_{k} \equal{} i_m$ for some $ k,m \in \mathbb{Z}/n\mathbb{Z}$. Which means that the circle can be partitioned in disjoint intervals with negative sum. This gives the desired contradiction.


Hi everyone,

In this post, http://www.artofproblemsolving.com/Forum/viewtopic.php?t=84425, it is mentioned that Carnegie Mellon University has an Honors program which allows undergrads to get their undergrad degrees and a Master's in 4 years? Can anyone:
1. Tell me a bit more about this program, and
2. Kindly guide me towards other schools which may also have this program?

Your help is greatly appreciated  :)

I graduated from this program in 1992.  What exactly do you want to know?

Well, a website would be helpful -- I've heard of it in passing but I don't know the name of it, etc. How are the job opportunities coming out of this pogram? Finding a place to do a doctoral dissertation? Too accelerated for a student's good?

Thanks so much for your reponse Mr. Patrick!

Their website is [url]http://www.math.cmu.edu/ug/ungrd/index.html[/url], but there's not much info there.  Click on "Special Program" for what little details there are.  Back in the day (20 years ago, when I was in high school, before there were websites, when I had to walk to school in heavy snow, uphill both ways) they used to publish a little booklet about Math Studies and the Honors Program...maybe they still do.

If you click on the link on the above page that says "Life After CMU", it'll show both the grad school and job placements of their recent graduates.  I can't personally speak to the job market, but I can say that I got into every grad school that I applied to, and ended up going to MIT.

The main downside is that the program will eat up most or all of your free electives.  This is not a trivial concern -- if you want to take a significant number of non-math courses, it will be really hard in this program.

If you want more detailed information about undergrad math at CMU, the guy to talk to is Dr. Russ Walker.  Dr. Walker is the Associate Department Head and was for a long time in charge of all matters undergraduate in the math department at CMU.  You can find his contact info on the CMU web site.  (If you get in touch with him, you can tell him that I said that you should talk to him!)

One more thing about CMU -- the world-famous buggy races (OK, maybe they're only Pittsburgh-famous) are going to be [url=http://www.cmu.edu/cmnews/extra/060420_buggy.html]streamed live[/url] April 21 and 22 from 8-Noon ET.


Calculate the following expressions without using calculator.

1. $ 41*271$.

2. $ 7^4\minus{}7^3\minus{}7^2$.

3. $ \frac{111111111}{9}$.

[quote="kunny"]Calculate the following expressions without using calculator.

1. $ 41*271$.

2. $ 7^4 \minus{} 7^3 \minus{} 7^2$.

3. $ \frac {111111111}{9}$.[/quote]
[hide]1. $ (40 \plus{} 1)(270 \plus{} 1) \equal{} 10800 \plus{} 40 \plus{} 270 \plus{} 1 \equal{} 10840 \plus{} 271 \equal{} 11111$
2. $ 49(49 \minus{} 7 \minus{} 1) \equal{} 49(41) \equal{} 2009$
[/hide]
I'm not sure on any of these @_@

$ \frac{111,111,111}{9}\equal{}\frac{1,001,001\cdot 111}{9}\equal{}\frac{3\cdot 333,667 \cdot 3 \cdot 37}{9} \equal{} 333,667 \cdot 37 \equal{} 12,345,679$
Of course, you could have just used good ol' division, too. :P

rts2007, you should see other's solution before posting solution. :wink:


[quote="236factorial"]The tickets for a dance recital cost 5 dollars for adults and 2 dollars for children. If the total number of tickets sold was 295 and the total amount collected was 1220 dollars, how many adult tickets were sold?[/quote]

[hide]

$x+y=295$
$2x+5y=1220$

You finish from there [/hide]

Maybe I'll finish from there.
[hide]$x+y = 295$
$2x+5y = 1220$
Solve the system of equations.
Multiply the equation $[x+y = 295]$ by $-2$ to get
$-2x-2y =-590$
Add it to the second equation to get
$3y = 630$
Divide by 3 to get
$y = 210$
Plug in y and solve for x to get
$x = 85$
So, 210 adult tickets were sold.[/hide]

[hide="complete solutions"]There are a few solutions. They all start with two equations,
$a+c=295$
$5a+2c=1220$

I'll use substitution.
$a=295-c$
$1475-5c+2c=1220$
$1475-3c=1220$
$3c=255$
$c=85$
$a=210$[/hide]

[hide] $x+y=295$
$5x+2y=1220$
We know that $5x+5y=1475\Rightarrow 3y=255\Rightarrow y=85\Rightarrow x=210$[/hide]

[quote="236factorial"]The tickets for a dance recital cost 5 dollars for adults and 2 dollars for children. If the total number of tickets sold was 295 and the total amount collected was 1220 dollars, how many adult tickets were sold?[/quote]

[hide]
5x+2y=1220
x+y=295

5x+5y=1475
3y=255
y=85
x=210[/hide]


function
calculus
derivative
integration
complex analysis
complex analysis unsolved

Let $f$ be an anlytic function in the open unit disk of the complex plane such that $|f(z)|\leq \frac{C}{1-|z|}$ for all $z$ in the unit disk, where $C$ is a positive constant.Prove that $|f(z)'|\leq \frac{4C}{(1-|z|)^2}$

Doing a simple estimate using the Cauchy Integral Formula for the derivative, we see that if $|f|\le M$ everwhere on the closed disk of radius $r$ centered at $z$, then $|f'(z)|\le\frac{M}r.$

For a given $z$, use the disk of radius $r=\frac{1-|z|}2.$ The maximum value of $f$ on that disk is no greater than its estimate at the point closest to the boundary of the unit disk, namely $M=\frac{2C}{1-|z|}.$

The result follows.

The same argument can be used to show that if $f$ is analytic on the unit disk and for some $a\ge0,$ $f(z)=O((1-|z|)^{-a}),$ then $f'(z)=O((1-|z|)^{-a-1}).$


Let ABCDE be a regular pentagon with area 1. Draw lines AC, AD, BD, BE, and CE. These lines form a new pentagon in the middle. What is the area of the new pentagon?
I think the answer is $ (1/4)(1\minus{}\tan18/\tan54)^2$, but it seems too ugly to simplify...

[hide="A simpler form is..."]$ A\equal{}{\cos^2{72^\circ}\over\cos^2{36^\circ}}$, which further simplifies into $ A\equal{}{3\minus{}\sqrt{5}\over 3\plus{}\sqrt{5}}\equal{}{7\minus{}3\sqrt{5}\over 2}$[/hide]

May I know how you arrived at such simpler form?

I'll just describe the procedure, and leave the details to you.

Let $ P \equal{} AD\cap CE$ and $ Q \equal{} BD\cap CE$. By the properties of similarity, the required area is $ \left({PQ\over AB}\right)^2$, but the ratio under the square sign is equal to the ratio of D-altitude of $ \triangle DPQ$ to the D-altitude of $ \triangle DAB$.

Did you arrive at the ratio $ A\equal{}{\cos^{2}{72^\circ}\over\cos^{2}{36^\circ}}$ directly? I got the square of $ \frac{\sin36}{0.5\tan72}$ first and then arrive at the same answer after some simplifications.


algebra
polynomial
algorithm
geometry
geometric transformation
induction
number theory

$\mathbb{N}_{10}$ is generalization of $\mathbb{N}$ that every hypernumber in $\mathbb{N}_{10}$ is something like: $\overline{...a_2a_1a_0}$ with $a_i \in {0,1..9}$
(Notice that $\overline {...000} \in \mathbb{N}_{10}$)
Also we easily have $+,*$ in $\mathbb{N}_{10}$.
first $k$ number of $a*b$= first $k$ nubmer of (first $k$ number of a * first $k$ number of b)
first $k$ number of $a+b$= first $k$ nubmer of (first $k$ number of a + first $k$ number of b)
Fore example $\overline {...999}+ \overline {...0001}= \overline {...000}$
Prove  that every monic polynomial in $\mathbb{N}_{10}[x]$ with degree $d$ has at most $d^2$ roots.

People nobody could solve this this is an excellent problem think on it  :P  :P  :P

Just no one see this nice problem within numerous spam messages... :?

[quote="sam-n"]just generalization:prove that any polynomial with degree d in  $\mathbb{N}_{p_1.p_2...p_k}[x]$has   at most $d^k$[/quote]
This sentence immediately reveals how can we prove statement: by induction. We are able to show that $a_1$, $a_2$, ... can be step by step uniquely restored from $a_0$. Do you need further explanations?
It is really excellent problem.

yes just u need a isomomorphism and because $N{}_p{}_i[x]$ is fieldand all polynomial in it has at most d roots.

Please explain more because I was the only peson solved this problem at the exam and my solution was different from the one you say.
Well if you want me I'll send my solution that is very beautiful.

Omid Hatami! Whom are you asking? Sam-n? Me?

Well each of you that send your solution I'd be thankful.Also if you want me I'll send my solution.

I've been waiting here for a few days and you haven't sent your solution.

Well please send your solution. I'm in a hurry.

I am not under the right circumstances now to think about olympiad problems, especially about hard number theory problems, but I just wanted to note that a particular case of the problem, namely to show that the equation $x^2=x$ has only four solutions: x = 0, x = 1 and two numbers x = ...90625 and x = ...09376 which can be constructed by a special algorithm, is solved in

Е. Б. Дынкин, В. А. Успенский, [i]Математические беседы[/i], Moscow 1952, section 2, chapter II, 2. (Maybe this book has an English translation - it certainly has a German one: E. B. Dynkin, W. A. Uspenski, [i]Mathematische Unterhaltungen[/i].)

The next paragraph, 3, tells something about certain p-adic numbers, but I fear these are not the "real" p-adic numbers considered in number theory. Anyway I am posting the above since it may be a bit helpful.

Myth, you are a real master of one-line messages!

  Darij

Hi Omid Hatami!
I found small difficulty in my proof yesterday. And today I have realized that your statement is wrong.

There are exist non-zero number $a\in\mathbb{N}_{10}$ s.t. first $k$ digits of $a$ are divisible by $2^k$ ($a=...2112$). I am sure that you are able to prove it ;)
Analogously, there are exist non-zero number $b\in\mathbb{N}_{10}$  s.t. first $k$ digits of $b$ are divisible by $5^k$ ($b=...3125$)

Consider equation $ax=0$. It has one trivial root $x=...000$. Moreover, it has one nontrivial root $b$ !!!

Contradiction, since $d=\deg(ax)=1$ and $d^2=1$.

What is wrong in my constuction? I still remember your fault on "golden sets" ;)

I can proof your statement if polynomial has coefficients from $\mathbb{Z}_+$ or, more precisely, not all of its coefficients are divisible by "$2^\infty$" (as number $a$) and $5^\infty$.

Well I'm so sorry. This is my second fault.
The polynomial is monic. :blush:  :blush:  :blush:  :blush:

Well I also did this and I used divisibility by $2^{\infty}$ and $5^{\infty}$.
Of course my solution was induction on $d$ using the equation $d^2=(d-1)^2+2(d-1)+1$
But where is the solution using isomorphism on the field $\mathbb{N}{}_p{}_i[x]$
I also apologize becuase of my fault and I promise be careful before posting something.

Hi Omid Hatami!
Maybe your fault will be good lesson for you for next time.

I have changed my mind. We don't need induction.

Here my solution. It is not fully detailed, but I hope everyone can complete it.

Consider $\mathbb{N}_p$.
We can well define number $-a$: first $k$ digits of $-a$ are first $k$ digits of $p^k-a$.

Suppose $f\in\mathbb{N}_p[x]$ and $a$ is a root of $f$. Then we can write $f(x)=(x-a)\cdot g(x)$, where $g\in\mathbb{N}_p[x]$, $\deg(g)=\deg(f)-1$. Indeed, we obtain coefficients of $g$ from Horner's method. It is easy to show that product of non-zero numbers in $\mathbb{N}_p[x]$ isn't equal to zero (...00). So we conclude $f$ have at most $d=\deg(f)$ roots in $\mathbb{N}_p[x]$.

Consider $\mathbb{N}_{p_1p_2...p_k}$ and $f\in\mathbb{N}_{p_1p_2...p_k}[x]$. Suppose $a$ is a root of $f$. Define number $a^{(p_i)}\in\mathbb{N}_{p_i}$:
first $k$ digits of $a^{(p_i)}$ are (first $k$ digits of $a$ as $m$-ary number)$\mod p_i^k$. We state that $a^{(p_i)}$ is well defined. Ok?

Finally, we have $a^{(p_i)}$ is a root of $f^{(p_i)}(x)$ (we replace coefficients of $f$ with their $...^{(p_i)}$-form). Since $f$ is monic polynomial $f^{(p_i)}(x)$ is monic too, hence each of $f^{(p_i)}(x)$ ($i=1..k$) has at most $d$ roots. Thus we conclude that each root $a$ of $f(x)$ defines a set of $k$ roots of $f^{(p_1)}(x)$, ..., $f^{(p_k)}(x)$. Moreover, due to Chinese residue theorem each set of roots of $f^{(p_1)}(x)$, ..., $f^{(p_k)}(x)$ defines unique root of $f$. It follows that number of roots of $f$ isn't exceeded $d^k$.

hello Omid,  I think the solution you were asking about is like this.
Actually the problem is almost completely obvious if you know two
relevant facts:

1. The Chinese remainder theorem, which in this case says
    that N10 (a ring) is the direct product N2 x N5.  In other
    words, a solution x in N10 of   P(x)=0  is an ordered pair
    of solutions   (x2, x5)  where P(x2)=0 in N2 and P(x5)=0 in N5.

2. N2 is the ring of 2-adic integers, which lives inside the field
    Q2 of 2-adic rational numbers.  Therefore any polynomial of
    degree d  has at most d  solutions in N2.  Same story for N5.

I think the hard part if you don't know these things is to prove
the equivalent of fact #2  by bare hands, it would be a hard
problem (and congratulations  if you solved it during an exam).


If 64 is divided into three parts proportional to 2, 4 and 6, the smallest part is:
\[ \text{(A)}\ 5\frac{1}{3} \qquad \text{(B)}\ 11 \qquad \text{(C)}\ 10\frac{2}{3} \qquad \text{(D)}\ 5 \qquad \text{(E) none of these answers} \]

4everwise has posted twice in multiple forums, i don't get it, you only need to post once.

mabye hes trying to increase his rank (the 5 starts below you name) quickly and not to mention unfairly


Hello,

Given is $yln(y)dx+(x-ln(y))dy=0$ then multiply by $\frac{1}{dx}$

$yln(y)+(x-ln(y))\frac{dy}{dx}=0$ then multiply by $\frac{1}{x-ln(y)}$

and we get $\frac{yln(y)}{x-ln(y)}+\frac{dy}{dx}=0$ we can rewrite and then we get $\frac{dy}{dx}+\frac{ln(y)}{x-ln(y)}y=0$

but now I don't know who go further if the x in the second term was not there it could not a problem.

Who can help me? Greets.

Hmm... You could try to find an integrating factor to make the equation exact.

I will try this later, but for now try to find some function $\mu$ so that for $Pdx+Qdy=0$, you have

\[ \frac{\partial \mu P}{\partial y}=\frac{\partial \mu Q}{\partial x}. \]

Sometimes it's hard to do so try it out and see what you get.  Possibilities are $\mu =f\left(x\right)$, $\mu =f\left(y\right)$, and $\mu=x^{\alpha}y^{\beta}$.


modular arithmetic
number theory unsolved
number theory

prove that 2^20 is congruent to 37 modulo 83.

We know that $ 2^{7}\equal{}128\equiv 45\pmod{83}$. So $ 2^{14}\equiv 45^{2}\equiv 33 \pmod{83}$ by squaring both sides. Note that $ 2^{6}\equiv 64\pmod{83}$. So $ 2^{20}\equiv 2112\equiv 37\pmod{83}$.

Or just $ 2^20\equal{}1048576\equiv 37 (mod 83)$
Of course, don't use this for big numbers.

Aadil, u can post these problems in Pre olympiad section as the people here [not me] are extremely expert looking out for challenging stuff.


vector
linear algebra
matrix
linear algebra unsolved

Let $ \mathcal{F}$ the set of finite order elements in $ \mbox{GL}_n( \mathbb{Z})$, and $ \mathcal{P}\equal{}\mathcal{F} \cap \mathcal{M}_n(\mathbb{N})$.
1. Describe $ \mathcal{P}$.
2. Show that there is no $ S$, $ \mathbb{N} \subsetneq S \subset \mathbb{Z}$, such that $ \mathcal{P}\equal{}\mathcal{F} \cap \mathcal{M}_n(S)$.

For 1.
1) Lemma: Let $ X\in{GL}_n(\mathbb{R})$ s.t. $ X$ and $ X^{\minus{}1}$ have non negative coefficients. Then in each $ X$-row and in each $ X$-column there exists only one non zero coefficient.
2) Let $ A\in\mathcal{P}$; then $ A^k\equal{}I$ and $ A^{\minus{}1}\equal{}A^{k\minus{}1}\in\mathcal{M}_n(\mathbb{N})$; according to the lemma, in the line $ i$, there exists only one non zero coefficient $ a_i$.  $ det(A)\equal{}\pm{a_1}\cdots{a_n}\equal{}\pm{1}$; thus $ a_i\equal{}1$ and $ A$ is a permutation matrix. The converse is obvious.

Nice lemma loup blanc, that makes a neat proof for 1.
Question 2. is even simpler than 1 :wink:

Hi JC_math.
Let $ S\equal{}\mathbb{N}\cup\{\minus{}2\}$, $ n\equal{}2$ and $ A\in\mathcal{F}\cap\mathcal{M}_2(S)\setminus\mathcal{P}$. We may assume that $ a_{11}\equal{}\minus{}2$ or $ a_{21}\equal{}\minus{}2$ (switch the basis vectors or change $ A$ into $ A^T$). Moreover $ det(A)\equal{}\pm{1}$.
Case 1. $ a_{11}\equal{}\minus{}2$: $ A$ can take essentially 3 values (eventually change $ A$ into $ A^T$): $ A\equal{}\begin{pmatrix}\minus{}2&1\\1&0\end{pmatrix},\begin{pmatrix}\minus{}2&3\\1&\minus{}2\end{pmatrix},\begin{pmatrix}\minus{}2&5\\1&\minus{}2\end{pmatrix}$. These matrices haven't finite order.
Case 2. $ a_{21}\equal{}\minus{}2$: by similar way $ A\equal{}\begin{pmatrix}1&0\\\minus{}2&1\end{pmatrix}$; this matrix hasn't finite order.
  My result is contradictory with your; who is right ?

loup blanc,

You are right, and I didn't express properly my thoughts for 2).
In fact for $ S$, I had in mind a particlar type of subsets, in the form $ \{m, m\plus{}1,\ldots\}$, with $ m\in \mathbb{Z}\setminus \mathbb{N}$.


geometry
parallelogram
geometric transformation
reflection
geometry unsolved

Is it possible to construct a convex quadrilateral which is not a parallelogram, in which a pair of opposite angles and a pair of opposite sides are equal? Prove or disprove.

[quote="izzystar"]Easy.
We first construct an equilateral triangle, then cut it into two triangles,
each having an angle of degree 60.  The final step is to adjust the position
of one of the two triangles so that they could form a convex quadrilateral.[/quote]

there are two problems over here at what angle you are going to cut it.More over how you are going to re-orientate it so that they form a quadrilateral but not parallelogram.Note one pair of opposite sides should also be equal. :|
please show the construction.

Ok, I will give more details.

Let $ \triangle ABC'$ be an equilateral triangle.  We pick a point $ C$
on the side $ BC'$,  which is not the midpoint of $ BC'$.  Construct the
line $ l$, such that $ A$ is the reflection of $ C$ about $ l$.  Finally, we
construct the point $ D$, which is the reflection of $ C'$ about $ l$.
Then the quadrilateral $ ABCD$ is the desired one, because
$ \angle ABC\equal{}\angle ADC\equal{}60^\circ$ and $ AB\equal{}CD$.
[/img]

[quote="izzystar"]Ok, I will give more details.

Let $ \triangle ABC'$ be an equilateral triangle.  We pick a point $ C$
on the side $ BC'$,  which is not the midpoint of $ BC'$.  Construct the
line $ l$, such that $ A$ is the reflection of $ C$ about $ l$.  Finally, we
construct the point $ D$, which is the reflection of $ C'$ about $ l$.
Then the quadrilateral $ ABCD$ is the desired one, because
$ \angle ABC \equal{} \angle ADC \equal{} 60^\circ$ and $ AB \equal{} CD$.
[/img][/quote]

can u attach your construction
it will help me a lot to understand your point of view  :blush:
More over is there any general idea to draw such Quadrilateral for e.g. one pair of opposite angle is '$ \theta$' and one pair of opposite side of lengths '$ a$'

[quote="izzystar"][quote="makar"]
can u attach your construction
it will help me a lot to understand your point of view  :blush:
More over is there any general idea to draw such Quadrilateral for e.g. one pair of opposite angle is '$ \theta$' and one pair of opposite side of lengths '$ a$'
[/quote]

Please see here.[/quote]

thanks a lot.


For the Power Test, if I find useful information on the internet, can I simply cite the source I used and then assume that that fact holds true?

Also, must all answers be justified with proof, or only when the test asks for it?

All answers have to be proven to a certain extent using only the properties that have been proven before it; therefore, you can't use something you prove in part $ 3$ to solve a question in part $ 2$.  And no, according to my math coach, you don't need to cite your sources.

Every answer must include a proof, but if the problem is trivial (i.e. purely computational) a simple explanation of the logic used is enough.

[quote="Z-coli"]All answers have to be proven to a certain extent using only the properties that have been proven before it; therefore, you can't use something you prove in part $ 3$ to solve a question in part $ 2$.[/quote]

Well can't a proof simply be copied from a future question into an earlier section? If a proof solves a problem, it solves the problem.

I think it's more like: If problem 3 asks you "prove x", but you can't figure it out, you can still use x in problem 4, but you can't in problem 2.
If you can prove x, though, then of course it can be used in problem 2.

On justification, it's like what cyberspace said, if it's simple, a short explanation would still be nice.

However, the problems are "Prove or Disprove" so that doesn't help.


It is known that $a^{2005} + b^{2005}$ can be expressed as the polynomial of $a + b$ and $ab$. Find the coefficients' sum of this polynomial.

If $a^{2005}+b^{2005}=P(a+b,ab)$ then the requested sum equals $P(1,1)$. Therefore, take $a=\frac{1+i\sqrt{3}}{2}$ and $b=\frac{1-i\sqrt{3}}{2}$. We obtain $P(1,1)=1.$

I think have a solution (don't use complex number)

There is an official solution which use the period of g(x)=coeff sum of a^x+b^x expressed as the polynomial of a+b,ab , but i think it use the idea of complex number, which can produce odd period function.

(i used complex number in the competition.i think it should be accepted.)

I only   find the coefficients' sum of this polynomial$P(x)$ which $P(x+\frac{1}{x})=x^{2005}+\frac{1}{x^{2005}}$.This Polynomial is unique.
 defined by $P_{0}(x)=2,P_{1}(x)=x,P_{n+2}(x)=xP_{n+1}(x)-P_{n}(x)$


inequalities
algebra
polynomial
trigonometry
inequalities proposed

Find for $ k$ a value as large as possible such that the inequalities below hold for any positive real numbers $ a,b,c$ with $ abc\equal{}1$:

$ (1) \ \ \ \ \ a\plus{}b\plus{}c\minus{}3\ge k(\frac 1{a\plus{}1}\plus{}\frac 1{b\plus{}1}\plus{}\frac 1{c\plus{}1}\minus{}\frac 3{2})$;

$ (2) \ \ \ \ \ a\plus{}b\plus{}c\minus{}3\plus{}k(\frac 1{a\plus{}1}\plus{}\frac 1{b\plus{}1}\plus{}\frac 1{c\plus{}1}\minus{}\frac 3{2}) \ge 0$.

[b]For ineq#1:[/b]
I was checking with values, and found that $ k$ can take a very high value indeed! The maximum $ k$ I found is $ k\leq 134.014977$, but I think $ k$ can take even bigger values! I am eagerly waiting to see the solutions!
 :)

But arqady, I got larger values of $ k$ by means of simply testing.
Take $ a\equal{}32, b\equal{}\frac{1}{8}, c\equal{}\frac{1}{4}$ :maybe:

Sunkern_sunflora, for your value of $ k$ the inequality is wrong. Try $ a\equal{}4$ and $ b\equal{}c\equal{}\frac{1}{2}.$ :wink:

Yes I have noticed it...I was compuitng on the wrong track. Thanks arqady.

[b]To Vasc:[/b]
Post your solution please.
:)

[quote="Vasc"]For $ abc\equal{}1$:

$ (1) \ \ \ \ \ a \plus{} b \plus{} c \minus{} 3\ge 30(\frac 1{a \plus{} 1} \plus{} \frac 1{b \plus{} 1} \plus{} \frac 1{c \plus{} 1} \minus{} \frac 3{2})$;

$ (2) \ \ \ \ \ a \plus{} b \plus{} c \minus{} 3 \plus{} 50(\frac 1{a \plus{} 1} \plus{} \frac 1{b \plus{} 1} \plus{} \frac 1{c \plus{} 1} \minus{} \frac 3{2}) \ge 0$.[/quote]

Using RLCF-Theorem or mixing variables (arqady), it suffices to consider $ b\equal{}c$.
Then, the inequalities becomes

$ (b\minus{}1)^2[2b^2(b\minus{}3)^2\plus{}3b\plus{}1) \ge 0$,

$ (b\minus{}1)^2(2b^4\plus{}28b^3\minus{}22b^2\plus{}3b\plus{}1) \ge 0$,

respectively. Can someone to finish the second proof ? (there is nice solution)

[quote="Vasc"]I found nice solutions for the following:

$ (1) \ \ \ \ \ a \plus{} b \plus{} c \minus{} 3\ge 30(\frac 1{a \plus{} 1} \plus{} \frac 1{b \plus{} 1} \plus{} \frac 1{c \plus{} 1} \minus{} \frac 3{2})$;

$ (2) \ \ \ \ \ a \plus{} b \plus{} c \minus{} 3 \plus{} 50(\frac 1{a \plus{} 1} \plus{} \frac 1{b \plus{} 1} \plus{} \frac 1{c \plus{} 1} \minus{} \frac 3{2}) \ge 0$.[/quote]

Dear Vasc  :)
for (1):
WlOG $ 1\geq ab$ then $ \frac {2}{\sqrt {ab} \plus{} 1} \geq \frac {1}{a \plus{} 1} \plus{} \frac {1}{b\plus{} 1}$
but  $ LHS \equal{} a \plus{} b \plus{} c \minus{} 3 \equal{} \frac {1}{bc} \plus{} b \plus{} c \minus{} 3 \geq 2\frac {1}{\sqrt {c}} \plus{} c \minus{} 3$
so our inequality is equivalent to : $ \frac {(x \minus{} 1)^2(x^4 \plus{} 3x^3 \minus{} 12x^2 \plus{} 18x \plus{} 2)}{x(x \plus{} 1)(x^2 \plus{} 1)} \geq 0$ , with $ x^2 \equal{} c$ , for the second I think we can do the same thing , but I'am not sure , I will try later  :wink:

[quote="anas"]
WlOG $ 1\geq ab$ then $ \frac {2}{\sqrt {ab} \plus{} 1} \geq \frac {1}{a^2 \plus{} 1} \plus{} \frac {1}{b^2 \plus{} 1}$
[/quote]
It can be $ \frac {2}{\sqrt {ab} \plus{} 1} \geq \frac {1}{a \plus{} 1} \plus{} \frac {1}{b \plus{} 1}$
Cool, anas! Your method gives the maximal estimation! How it can be? :maybe:

[quote="Vasc"][quote="Vasc"]For $ abc \equal{} 1$:

$ (1) \ \ \ \ \ a \plus{} b \plus{} c \minus{} 3\ge 30(\frac 1{a \plus{} 1} \plus{} \frac 1{b \plus{} 1} \plus{} \frac 1{c \plus{} 1} \minus{} \frac 3{2})$;

$ (2) \ \ \ \ \ a \plus{} b \plus{} c \minus{} 3 \plus{} 50(\frac 1{a \plus{} 1} \plus{} \frac 1{b \plus{} 1} \plus{} \frac 1{c \plus{} 1} \minus{} \frac 3{2}) \ge 0$.[/quote]

Using RLCF-Theorem or mixing variables (arqady), it suffices to consider $ b \equal{} c$.
Then, the inequalities becomes

$ (b \minus{} 1)^2[2b^2(b \minus{} 3)^2 \plus{} 3b \plus{} 1) \ge 0$,

$ (b \minus{} 1)^2(2b^4 \plus{} 28b^3 \minus{} 22b^2 \plus{} 3b \plus{} 1) \ge 0$,

respectively. [/quote]
There are some mistakes here. Actually,there are nice solutions to

$ (1) \ \ \ \ \ a \plus{} b \plus{} c \minus{} 3\ge 59(\frac 1{a \plus{} 1} \plus{} \frac 1{b \plus{} 1} \plus{} \frac 1{c \plus{} 1} \minus{} \frac 3{2})$;

$ (2) \ \ \ \ \ a \plus{} b \plus{} c \minus{} 3 \plus{} 30(\frac 1{a \plus{} 1} \plus{} \frac 1{b \plus{} 1} \plus{} \frac 1{c \plus{} 1} \minus{} \frac 3{2}) \ge 0$

Using RLCF-Theorem, or mixing variables (arqady), or above method (anas), it suffices to consider $ b \equal{} c$. Then, the inequalities becomes

$ (b \minus{} 1)^2[4b^4 \plus{} 65b^3 \minus{} 53b^2 \plus{} 6b \plus{} 2) \ge 0$,

$ (b \minus{} 1)^2(2b^4 \minus{} 12b^3 \plus{} 18b^2 \plus{} 3b \plus{} 1) \ge 0$,

respectively. The second is easy. Can someone prove the first, that is $ 4b^4 \plus{} 65b^3 \minus{} 53b^2 \plus{} 6b \plus{} 2) \ge 0$ ?

[quote="Vasc"][quote="Vasc"][quote="Vasc"]For $ abc \equal{} 1$:

$ (1) \ \ \ \ \ a \plus{} b \plus{} c \minus{} 3\ge 30(\frac 1{a \plus{} 1} \plus{} \frac 1{b \plus{} 1} \plus{} \frac 1{c \plus{} 1} \minus{} \frac 3{2})$;

$ (2) \ \ \ \ \ a \plus{} b \plus{} c \minus{} 3 \plus{} 50(\frac 1{a \plus{} 1} \plus{} \frac 1{b \plus{} 1} \plus{} \frac 1{c \plus{} 1} \minus{} \frac 3{2}) \ge 0$.[/quote]

Using RLCF-Theorem or mixing variables (arqady), it suffices to consider $ b \equal{} c$.
Then, the inequalities becomes

$ (b \minus{} 1)^2[2b^2(b \minus{} 3)^2 \plus{} 3b \plus{} 1) \ge 0$,

$ (b \minus{} 1)^2(2b^4 \plus{} 28b^3 \minus{} 22b^2 \plus{} 3b \plus{} 1) \ge 0$,

respectively. [/quote]
There are some mistakes here. Actually,there are nice solutions to

$ (1) \ \ \ \ \ a \plus{} b \plus{} c \minus{} 3\ge 59(\frac 1{a \plus{} 1} \plus{} \frac 1{b \plus{} 1} \plus{} \frac 1{c \plus{} 1} \minus{} \frac 3{2})$;

$ (2) \ \ \ \ \ a \plus{} b \plus{} c \minus{} 3 \plus{} 30(\frac 1{a \plus{} 1} \plus{} \frac 1{b \plus{} 1} \plus{} \frac 1{c \plus{} 1} \minus{} \frac 3{2}) \ge 0$

Using RLCF-Theorem, or mixing variables (arqady), or above method (anas), it suffices to consider $ b \equal{} c$. Then, the inequalities becomes

$ (b \minus{} 1)^2[4b^4 \plus{} 65b^3 \minus{} 53b^2 \plus{} 6b \plus{} 2) \ge 0$,

$ (b \minus{} 1)^2(2b^4 \minus{} 12b^3 \plus{} 18b^2 \plus{} 3b \plus{} 1) \ge 0$,

respectively. The second is easy. Can someone prove the first, that is $ 4b^4 \plus{} 65b^3 \minus{} 53b^2 \plus{} 6b \plus{} 2\ge 0$ ?[/quote]
We need to prove that
\[ 4b^4 \plus{} 65b^3 \minus{} 53b^2 \plus{} 6b \plus{} 2\ge 0\]
\[ \Leftrightarrow (b^2\minus{}2b\plus{}2)(2b\minus{}1)^2 \plus{}b(77b^2\minus{}70b\plus{}16) \ge 0\]
which is true since the polynomial $ 77b^2\minus{}70b\plus{}16$ has a discriminant
\[ \Delta^{\prime} \equal{}35^2 \minus{}77\cdot 16 \equal{}1225\minus{}1232 \equal{}\minus{}7<0\]
:)
And we have proved it. :)

[quote="can_hang2007"]
We need to prove that
\[ 4b^4 \plus{} 65b^3 \minus{} 53b^2 \plus{} 6b \plus{} 2\ge 0
\]

\[ \Leftrightarrow (b^2 \minus{} 2b \plus{} 2)(2b \minus{} 1)^2 \plus{} b(77b^2 \minus{} 70b \plus{} 16) \ge 0
\]
which is true since the polynomial $ 77b^2 \minus{} 70b \plus{} 16$ has a discriminant
\[ \Delta^{\prime} \equal{} 35^2 \minus{} 77\cdot 16 \equal{} 1225 \minus{} 1232 \equal{} \minus{} 7 < 0
\]
[/quote]
Yes, can_hang. A little simpler is $ (b^2 \plus{} 2)(2b \minus{} 1)^2 \plus{} b(69b^2 \minus{} 62b \plus{} 14) \ge 0$. :P

[quote="arqady"][quote="anas"]
WlOG $ 1\geq ab$ then $ \frac {2}{\sqrt {ab} \plus{} 1} \geq \frac {1}{a^2 \plus{} 1} \plus{} \frac {1}{b^2 \plus{} 1}$
[/quote]
It can be $ \frac {2}{\sqrt {ab} \plus{} 1} \geq \frac {1}{a \plus{} 1} \plus{} \frac {1}{b \plus{} 1}$
[/quote]
you are right arqady , I made a typo , sorry about that  :blush:

[quote="Vasc"]An extension to 4 variables is the following:

If $ a,b,c,d$ are positive real numbers such that $ abcd \equal{} 1$, then

$ a \plus{} b \plus{} c \plus{} d \minus{} 4 \ge 17(\frac a{a \plus{} 1} \plus{} \frac b{b \plus{} 1} \plus{} \frac c{c \plus{} 1} \plus{} \frac d{d \plus{} 1} \minus{} 2)$.[/quote]
The hint is to use Mathlinks contest 2008
"If $ a \ge b \ge c \ge d>0$ such that $ abcd\equal{}1$, then
\[ \frac{1}{a\plus{}1}\plus{}\frac{1}{b\plus{}1}\plus{}\frac{1}{c\plus{}1} \ge \frac{3}{\sqrt[3]{abc}\plus{}1}\]
Then, it suffices to consider the original inequality in the case $ a\equal{}b\equal{}c$. :)

[quote="Vasc"][quote="can_hang2007"][quote="Vasc"]An extension to 4 variables is the following:

If $ a,b,c,d$ are positive real numbers such that $ abcd \equal{} 1$, then

$ a \plus{} b \plus{} c \plus{} d \minus{} 4 \ge 17(\frac a{a \plus{} 1} \plus{} \frac b{b \plus{} 1} \plus{} \frac c{c \plus{} 1} \plus{} \frac d{d \plus{} 1} \minus{} 2)$.[/quote]
The hint is to use Mathlinks contest 2008
"If $ a \ge b \ge c \ge d > 0$ such that $ abcd \equal{} 1$, then
\[ \frac {1}{a \plus{} 1} \plus{} \frac {1}{b \plus{} 1} \plus{} \frac {1}{c \plus{} 1} \ge \frac {3}{\sqrt [3]{abc} \plus{} 1}
\]
Then, it suffices to consider the original inequality in the case $ a \equal{} b \equal{} c$. :)[/quote]
Very nice hint.
The following statement is also valid.

If $ a,b,c,d$ are positive real numbers such that $ abcd \equal{} 1$, then

$ a \plus{} b \plus{} c \plus{} d \minus{} 4 \ge 32(\frac 1{a \plus{} 1} \plus{} \frac 1{b \plus{} 1} \plus{} \frac 1{c \plus{} 1} \plus{} \frac 1{d \plus{} 1} \minus{} 2)$.[/quote]
It came from the following statement
"If $ 0 \le a \le b \le c \le d$ such that $ abcd\equal{}1$, then
\[ \frac{1}{1\plus{}a}\plus{}\frac{1}{1\plus{}b}\plus{}\frac{1}{1\plus{}c} \le \frac{3}{1\plus{}\sqrt[3]{abc}}\]

[quote="can_hang2007"] [quote="Vasc"]

If $ a,b,c,d$ are positive real numbers such that $ abcd \equal{} 1$, then

$ a \plus{} b \plus{} c \plus{} d \minus{} 4 \ge 32(\frac 1{a \plus{} 1} \plus{} \frac 1{b \plus{} 1} \plus{} \frac 1{c \plus{} 1} \plus{} \frac 1{d \plus{} 1} \minus{} 2)$.[/quote]
It came from the following statement
"If $ 0 \le a \le b \le c \le d$ such that $ abcd \equal{} 1$, then
\[ \frac {1}{1 \plus{} a} \plus{} \frac {1}{1 \plus{} b} \plus{} \frac {1}{1 \plus{} c} \le \frac {3}{1 \plus{} \sqrt [3]{abc}}
\]
[/quote]
Indeed, Can_hang2007. However, the following can't be proved by this way.

If $ a,b,c,d,e,f$ are positive real numbers such that $ abcdef\equal{} 1$, then

$ a \plus{} b \plus{} c \plus{} d \plus{}e\plus{}f\minus{} 6 \ge 30(\frac 1{a \plus{} 1} \plus{} \frac 1{b \plus{} 1} \plus{} \frac 1{c \plus{} 1} \plus{} \frac 1{d \plus{} 1}\plus{} \frac 1{e \plus{} 1} \plus{} \frac 1{f \plus{} 1} \minus{} 3)$.

For any positive real numbers $ a,b,c$ with $ abc \equal{} 1$, the inequaliy

$ a \plus{} b \plus{} c \minus{} 3\geq k\left(\frac 1{a \plus{} 1} \plus{} \frac 1{b \plus{} 1} \plus{} \frac 1{c \plus{} 1} \minus{} \frac 3{2}\right)$

holds if and only if

$ \minus{} 31.0133833965\cdots \equal{} \frac {28}{3} \minus{} \frac {8}{3}\sqrt {397}\sin\left[\frac {\pi}{6} \plus{} \frac {1}{3}\arccos\left(\frac {16663}{630436}\sqrt {397}\right)\right]$

$ \leq k\leq \frac {28}{3} \plus{} \frac {8}{3}\sqrt {397}\cos\left[\frac {1}{3}\arccos\left(\frac {16663}{630436}\sqrt {397}\right)\right] \equal{} 59.4473024163\cdots$.

If $ a,b,c,d$ are positive real numbers such that $ abcd \equal{} 1$, then

$ a \plus{} b \plus{} c \plus{} d \minus{} 4 \geq k\left(\frac 1{a \plus{} 1} \plus{} \frac 1{b \plus{} 1} \plus{} \frac 1{c \plus{} 1} \plus{} \frac 1{d \plus{} 1} \minus{} 2\right)$

holds if and only if $ \minus{}17.474140574\cdots\equal{}k_1\leq k \leq k_2\equal{} 46.149534123\cdots$,

where $ k_1$ and $ k_2$ are two roots of the irreducible polynomial $ k^4\minus{}28k^3\minus{}824k^2\minus{}596k\minus{}1444.$

[quote="arqady"][quote="Vasc"]Find for $ k$ a value as large as possible such that the inequalities below hold for any positive real numbers $ a,b,c$ with $ abc \equal{} 1$:

$ (1) \ \ \ \ \ a \plus{} b \plus{} c \minus{} 3\ge k(\frac 1{a \plus{} 1} \plus{} \frac 1{b \plus{} 1} \plus{} \frac 1{c \plus{} 1} \minus{} \frac 3{2})$;
[/quote]
Mixing variables kills it.
Here $ \max k \equal{} \min_{(0,1)}\frac {2(2b^4 \plus{} 3b^3 \plus{} 3b^2 \plus{} 3b \plus{} 1)}{b^2 \minus{} b^3} \equal{} 59.44730242...,$
where $ b_{min} \equal{} 0.460811...$ is a middle root of the following equation:
$ x^3 \minus{} 3x^2 \minus{} x \plus{} 1 \equal{} 0.$[/quote]
How do you find it?  :maybe: I don't know why

[quote="lenhamquy"][quote="arqady"][quote="Vasc"]Find for $ k$ a value as large as possible such that the inequalities below hold for any positive real numbers $ a,b,c$ with $ abc = 1$:

$ (1) \ \ \ \ \ a + b + c - 3\ge k(\frac 1{a + 1} + \frac 1{b + 1} + \frac 1{c + 1} - \frac 3{2})$;
[/quote]
Mixing variables kills it.
Here $ \max k = \min_{(0,1)}\frac {2(2b^4 + 3b^3 + 3b^2 + 3b + 1)}{b^2 - b^3} = 59.44730242...,$
where $ b_{min} = 0.460811...$ is a middle root of the following equation:
$ x^3 - 3x^2 - x + 1 = 0.$[/quote]
How do you find it?  :maybe: I don't know why[/quote]
Let $ f(a,b,c) = a + b + c - 3 - k(\frac {1}{a + 1} + \frac {1}{b + 1} + \frac {1}{c + 1} - \frac {3}{2})$ and $ a = \max\{a,b,c\}.$
Then $ bc\leq1$ and we can assume $ k\geq0.$ :wink:
$ f(a,b,c) - f(a,\sqrt {bc},\sqrt {bc}) = b + c - 2\sqrt {bc} - k(\frac {1}{b + 1} + \frac {1}{c + 1} - \frac {2}{\sqrt {bc} + 1}) =$
$ = (\sqrt b - \sqrt c)^2(1 + \frac {k(1 - \sqrt {bc})}{(b + 1)(c + 1)(\sqrt {bc} + 1)})\geq0.$
Id est, it remains to understand when $ f(a,b,b)\geq0,$ where $ a = \frac {1}{b^2}$ and $ b\in(0,1).$
We obtain: $ f(a,b,b)\geq0\Leftrightarrow\frac {1}{b^2} + 2b - 3\geq k(\frac {b^2}{b^2 + 1} + \frac {2}{b + 1} - \frac {3}{2})\Leftrightarrow$
$ \Leftrightarrow\frac {(b - 1)^2(b + 2)}{b^2}\geq\frac {k(1 - b)^3}{2(b + 1)(b^2 + 1)}\Leftrightarrow k\leq\frac {2(2b + 1)(b + 1)(b^2 + 1)}{b^2(1 - b)}.$
Let $ g(b) = \frac {2(2b + 1)(b + 1)(b^2 + 1)}{b^2(1 - b)}.$
Hence, $ g'(b) = - \frac {4(b^2 + b + 1)(b^3 - 3b^2 - b + 1)}{b^3(1 - b)^2}.$
Id est, $ \max k = \min_{(0,1)}g = g(b_{min}) = 59.4473024...,$
where $ b_{min}$ is a middle root of the equation $ b^3 - 3b^2 - b + 1 = 0.$


[quote="orangefronted"]Find the number of 3-digit numbers ABC, such that 2A+3B=5C.[/quote]

[hide]
The quantity 5C has to end in a zero or a five.
If the first digit is allowed to be a zero, I have:

EDIT:

000
053
111
164
222
275
333
386
397
444
497
502
555
613
666
724
777
835
884
946
999

21 numbers . . . EDIT

[/hide]


1. At a dance party a group of boys and girls exchange dances as follows: one boy dances with 5 girls, a second boy dances with 6 girls, and so on, the last boy dancing with all the girls. If b represents the number of boys and g the number of girls, then find b in terms of g

2.The tens digit of a two-digit number exceeds its units digit by 4. The number exceeds twice the number obtained by reversing the digits of the orinial number by 10. What is the original number?

3.George Washington was born 11 years before Thomas Jefferson. In 1770 Washington's age was 3 more than 7 times the age of Jefferson in 1748. What was the sum of the two men's ages in 1750?

4. MIke and Joey bought identical loaves or bread nad packages of bologna. Mike made sandwiches with 5 slices of bologna and had 4 slices of bread left when he ran out of meat. Joey made sandwiches with 4 slices of bologna and had 4 slices of meat when he ran out of bread. How many slices of bread were in each loaf?

The first problem did not mention different girls, so $ \boxed{b \equal{} g\minus{}4}$.

Sorry, I dont have time for the others. :P

[hide="1"]
$ b \equal{} g \minus{} 4$
[/hide]
[hide="2"]
Let the number be $ ab$.
Then, $ a \equal{} b \plus{} 4$ and $ 10a \plus{} b \equal{} 2(10b \plus{} a) \plus{} 10$. Solving, we get $ a \equal{} 6, b \equal{} 2$, so the number is $ \boxed{62}$.
[/hide]
[hide="3"]
Let $ x$ be the birth year of Washington and $ y$ be the birth year of Jefferson.
Then, $ x \equal{} y \minus{} 11$ and $ (1770 \minus{} x) \equal{} 3 \plus{} 7(1748 \minus{} y)$. Solving, we get $ x \equal{} 1732, y \equal{} 1743$. Thus, the sum of the ages in $ 1750$ is $ (1750 \minus{} 1732) \plus{} (1750 \minus{} 1743) \equal{} \boxed{25}$
[/hide]
[hide="4"]
Let there be $ m$ slices of bologna and $ n$ slices of bread.
Then, $ n \minus{} \frac {2m}{5} \equal{} 4$ and $ m \minus{} \frac {4n}{2} \equal{} 4$ (assuming a sandwich uses $ 2$ slices of bread). Solving, $ m \equal{} 60$ and $ n \equal{} 28$, so there are $ \boxed{28}$ slices of bread per loaf.[/hide]

how did you get 2/5 and 4/2 and all that?
The solution really confuses me


Hewas almost 20 years old when he was taking his AMC12 A, why did the AMC officials led him to take the test?

I think if you pay, they'll let you take the test (they want the money) but I doubt they would let him go to USAMO or MOSP should he qualify.

The answer sheet has a bubble for students who are 19. If he was [u]almost[/u] 20, he was certainly not older than 19.

Plus, why would he take it if he couldn't qualify for anything? If you just want to take a test for fun, you can always find it on AoPS or Kalva  :P

No, he is already 20 years old!!!!!!!  He celebrated his 20th birthday last year October or November.

This forum is not an appropriate venue for discussion of alleged rules violations.

Please see the Teachers' Manual, page 6, Section VI:  Policy Statements,
http://www.unl.edu/amc/d-publication/d1-pubarchive/2007-8pub/1012TM/2008AMC1012A-TM.pdf
in particular in the second column: "...the Contest Manager... must report all the facts of the situation to the AMC Director, who in conjunction with the Chair of the CAMC will determine what to do."

It appears that you have misunderstood the important role of the Contest Manager.  With over 150,000 students taking the A-date contest in all 50 states and over 20 countries outside of the USA, the "AMC officials" are in no practical position to enforce rules.  Instead, as we have done since the beginnings of the contest, we rely on the mathematical interest, integrity and honesty of the Contest Managers, that is, the teachers in the schools who register, purchase the contests and take the time to administer the contest to you in your schools.

In your message, you provide no practical means of investigating the alleged situation.  If you have a complaint, then you should first speak to your teacher or Contest Manager.  If failing satisfaction in that, then you can contact  the AMC office by any of the means listed in all of our publications.

Steve Dunbar
MAA Director, American Mathematics Competitions


If we start with any whole number $ n$ and reverse its digits to get $ m$, is $ 3$ always a factor of $ n \minus{} m$?

[hide="Explanation"]
Let $ n\equal{}10a\plus{}b$ where $ a$ and $ b$ are digits. Then $ m\equal{}10b\plus{}a,$ and $ n\minus{}m\equal{}(10a\minus{}a)\plus{}(b\minus{}10b)\equal{}9(a\minus{}b).$ This is a multiple of $ 9$ and is thus divisible by $ 3.$[/hide]

[hide="Fullest generalization  :) "]Let $ n\equal{}10^k  a_1\plus{}10^{k\minus{}1} a_2 \plus{} ... \plus{} a_{k\plus{}1}$

Reversing the digits,

$ m\equal{}10^k a_{k\plus{}1}\plus{}10^{k} a_2 \plus{} ... \plus{} a_{1}$

From this, $ n\minus{}m$ is

$ n\minus{}m\equal{}(10^k\minus{}1) a_1\plus{}(10^{k\minus{}1}\minus{}10) a_2 \plus{} ... \plus{} (1\minus{}10^k) a_{k\plus{}1}$

When a power of 10 is subtracted by another power of 10, it is always divisible by 9. (Even 10^a-10^b where a>b>0, since you factor out whatever power of 10 and arrive to 10^c-1, which is 9999... for c-1 digits).

Thus, n-m is always divisible by 3.[/hide]

[hide="A simpler explanation"]Let the original number $ n$ leave remainder $ x$ upon division by $ 9$. Since this remainder is determined by the sum of the digits, then $ m$, which contains the same digits, also leaves the same remainder. Then we can write:

$ n \equiv x \bmod{9}$
$ m \equiv x \bmod{9}$
$ n \minus{} m \equiv x\minus{}x \equiv 0 \bmod{9}$[/hide]


I'm working on a physics problem, and i got stuck on an integral.  the entire question is as follows:

----------------------------
Magnetostatics treats the "source current" (the one that sets up the field) and
the "recipient current" (the one that experiences the force) so asymmetrically that it is by no means obvious that the magnetic force between two current loops is consistent with Newton"s third law. Show, starting with the Biot-Savart law and the Lorentz force law, that the force on loop 2 due to loop 1 can be written as

$F2 = -\frac{\mu_o}{4\pi}I_1 I_2 \oint_1 \oint_2 \frac{\widehat{r}}{r^2} d\overrightarrow{l_1} \cdot d\overrightarrow{l_2}$

In this form it is clear that F2 = — F1, since i changes direction when the roles of 1 and 2 are interchanged. (If you seem to be getting an "extra" term, it will help to note that $d\overrightarrow{l_2} \cdot \widehat{r} = dr$.)
--------------------------------------

I'm okay with most of the proof - just a specific part I'm stuck on.  i got to the part where i have
$F2 = \frac{\mu_o}{4\pi}I_1 I_2 [\oint_1 \oint_2 \frac{d\overrightarrow{l_1} dr}{r^2} - \oint_1 \oint_2 \frac{\widehat{r}}{r^2} d\overrightarrow{l_1} \cdot d\overrightarrow{l_2}]$

i want to prove that the first integral is zero, so i need to show that this integral along a closed loop:

$\oint_2 (1/r^2) dr$

is equal to zero.  but i'm not sure how to prove it.  i was wondering if someone can show me a rigid proof for this.  Note that the integral is along loop 2.  Note also that the $\overrightarrow{r}$ is the distance from a point on loop 1 to a point on loop 2 - that is, $\overrightarrow {r} = (x2-x1, y2-y1, z2-z1)$, so in the integral above, where r is a scalar, then $r = [(x2-x1, y2-y1, z2-z1)]^{1/2}$.  i'm not sure how to rewrite dr in terms of x2, since r is a scaler.  should it be dr = dxdydz?  I think i'm missing something here because I'm not really that familiar with loop integrals.

Before I state my proof, I'd like to make sure that your expression $\vec a \cdot \vec b$ actually signifies a vector product of vectors $\vec a$ and $\vec b$ (the common notation for the vector product is $\vec a \times \vec b$, while the dot signifies a scalar product -- at least this is my experience). In the following proof I will use $\times$ to denote the vector product.

The magnetic field induced by loop 1 at an infinitesimally small length of loop 2 can be found by Biot-Savart's law: \[ \text d\vec B = \frac{\mu_0I_1}{4\pi}\frac{\text d\vec l_1 \times \vec r}{r^3} \Rightarrow \vec B = \frac{\mu_0I_1}{4\pi}\oint_1\frac{\text d\vec l_1 \times \vec r}{r^3}.  \] The force acting on this part of loop 2 can be obtained by Ampere's law (which easily follows from the expression for the Lorentz force): \[ \text d\vec F = I_2 (\text d \vec l_2 \times \vec B) = \frac{\mu_0 I_1I_2}{4\pi} \left(\text d \vec l_2 \times \oint_1\frac{\text d\vec l_1 \times \vec r}{r^3}\right) \Rightarrow \] \[ \Rightarrow \vec F = \frac{\mu_0 I_1I_2}{4\pi} \oint_2\oint_1 \text d \vec l_2 \times \frac{\text d\vec l_1 \times \vec r}{r^3}.  \] Now, since $\vec r/r = \hat r$ and $\text d\vec l_1 \times \text d\vec l_2 = -(\text d\vec l_2\times \text d\vec l_1)$, we get \[ \vec F = -\frac{\mu_0}{4\pi}I_1I_2\oint_1\oint_2\frac{\hat r}{r^2}\times (\text d\vec l_1 \times \text d\vec l_2).  \] Note that we've switched the integrals in the last step.

I hope this helps!


algebra
polynomial
quadratics
superior algebra
superior algebra unsolved

Determine the irreducible polynomial for $ \alpha$ = $ \sqrt {3} \plus{} \sqrt {5}$ over each of the following fields.

(c) Q($ \sqrt {10}$) (d) Q($ \sqrt {15}$)

So I'm having trouble with both c and d. I can try $ (x \minus{} \sqrt {3} \minus{} \sqrt {5})^2$ or $ x^2 \minus{} (\sqrt {3} \plus{} \sqrt {5})^2$ but it's quite difficult to find irreducible polynomials for those fields this way (I can do it for $ \sqrt {3}$ easily.

For (c), the conjugates are $ \sqrt{3}\plus{}\sqrt{5}$, $ \sqrt{3}\minus{}\sqrt{5}$, $ \minus{}(\sqrt{3}\plus{}\sqrt{5})$ and $ \minus{}(\sqrt{3}\minus{}\sqrt{5})$. That gives the irr. polynomial.

For (d), the conjugates are $ \sqrt{3}\plus{}\sqrt{5}$ and $ \minus{}(\sqrt{3}\plus{}\sqrt{5})$.

For (d), squaring $ \alpha$ immediately gives a quadratic in $ \mathbb{Q}(\sqrt{15})$ with $ \alpha$ as a root, so it's enough to show $ \alpha \not\in \mathbb{Q}(\sqrt{15})$.

For (c), square $ \alpha$, isolate the $ \sqrt{15}$ part, and square again to get a quartic in $ \alpha$ with leading coefficient 1 and coefficients in $ \mathbb{Z}[\sqrt{10}]$ (which has field of fractions equal to $ \mathbb{Q}(\sqrt{10})$. To show it's irreducible, it has no linear factors (show that $ \alpha \not\in \mathbb{Q}(\sqrt{10})$), so any nontrivial factorization in $ \mathbb{Q}(\sqrt{10})$ would imply a factorization into two quadratics with leading coefficients 1 and coefficients in $ \mathbb{Z}[\sqrt{10}]$. This makes the factorization of the form $ (\alpha^2\plus{}p\alpha \plus{}q)(\alpha^2\plus{}r\alpha \plus{}s)$ where $ p,q,r,s\in \mathbb{Z}[\sqrt{10}]$, set coefficients equal to get a bunch of equations and arrive at a contradiction.

Okay, thanks for the replies. Another question: are ALL integers (and perhaps ALL rational numbers) elements of ANY field $ Z[\sqrt{d}]$


real analysis
calculus
calculus computations

Determine if exists the cesaro $ (C,1)$-mean for the following series:

i)$ \displaystyle \sum_{n\equal{}1}^\infty i^{n\minus{}1}$

ii)$ \displaystyle \sum_{n\equal{}1}^\infty (\minus{}1)^{n\minus{}1}n$

In the first case:

$ \sum_{k\equal{}1}^ni^{k\minus{}1}\equal{}\frac{1\minus{}i^n}{1\minus{}i}$, so

$ \frac1N\sum_{n\equal{}1}^{N}\sum_{k\equal{}1}^ni^{k\minus{}1}\equal{}\frac1N\sum_{n\equal{}1}^{N}\frac{1\minus{}i^n}{1\minus{}i}$

$ \equal{}\frac1N\left(\frac{N}{1\minus{}i}\minus{}\frac{i\minus{}i^{N\plus{}1}}{(1\minus{}i)^2}\right).$

In the limit as $ N\to\infty,$ this tends to $ \frac1{1\minus{}i}.$ The sum is Cesaro summable despite not being convergent.

We also have $ \sum_{n\equal{}1}^{\infty}i^{n\minus{}1}r^{n}\equal{}\frac{r}{1\minus{}ir}.$

In the limit as $ r\to1^\minus{},$ this tends to $ \frac1{1\minus{}i}.$ Hence, this is also Abel summable, and to the same limit.

------

For the second problem, let's write the sum out:

$ 1\minus{}2\plus{}3\minus{}4\plus{}5\minus{}6\plus{}7\minus{}8\plus{}\cdots.$

Now write the sequence of partial sums:

$ 1,\minus{}1,2,\minus{}2,3,\minus{}3,4,\minus{}4,\dots.$

Now find the partial sums of these partial sums:

$ 1,0,2,0,3,0,4,0,\dots.$

And divide by $ N$ to get the sequence of Cesaro means:

$ 1,0,\frac23,0,\frac35,0,\frac47,\dots.$

This sequence of Cesaro means has a limsup of $ \frac12$ and a liminf of $ 0$ and hence fails to converge. The series is not Cesaro summable.

However, $ \sum_{n\equal{}1}^{\infty}(\minus{}1)^{n\minus{}1}nr^n\equal{}\frac{r}{(1\plus{}r)^2}$. The limit of this as $ r\to1^\minus{}$ is $ \frac14.$ Hence the series is Abel summable to $ \frac14$ despite not being Cesaro summable.

In fact, if $ \sum_{n\equal{}1}^{\infty} a_n \equal{} \infty$ as usual summation, then its Cesaro summation and even Abel summation also diverge to $ \infty$.

sos440: that does not apply to either of these problems.

Ah, I misunderstood the first problem. :blush:  Clearly, it does not apply to both problems.


You assume that all the x_i add up to 1. Then you use that the right side is less than k(sum x_i)^2 with k the constant for the case x_1=...=x_n. Thn you use Jensen for (1-x)^3/x which is convex because x<=1.

Misha


MATHCOUNTS AMC AIME USA(J)MO USAMO geometry probability

Hello everyone!

Many of us have taken the AMC 8 last Tuesday, and a few of us have yet to take it. MathCounts is right around the corner, and what better way to prepare than...AoPS Jeopardy!

A lot of tournaments take some time, and I know most of us are bogged down with school work and other activities. But AoPS Jeopardy is fun, helps you get ready, and doesn't take too much of your time!

Here's how it works: there are 9 people in total, three in each group. The groups will be called Group A, Group B, and Group C. (Sorry, you can't make your own names. Too much bickering and it wastes time. Plus, A, B, and C are easier to type.) Each of these groups will have a Member 1, a Member 2, and of course, a Member 3.

There are three stages, each with 25 rounds. (Don't worry, they're really short.) In the first stage and round, Member 1 of Group A will compete against the other Member 1's. He/She will select a category and dollar price from the game board. The Member 1's will be given two days to solve the problem. Details and rules concerning each stage and round will be posted at the appropriate time. If the person gets a problem wrong, and the problem was worth x dollars, he/she will lose x dollars from his/her total amount. If the person got it right, he/she will gain that amount of dollars. (Yes, it's possible to get negative amounts of money.) Then, in the next round, the Member 2's compete, with Group B's Member 2 selecting the category and question. The next round will allow Group C's Member 3 to choose the categories and questions.

Each member of the group has their own individual amount of money. If Member 1 loses 400 dollars, it doesn't affect Member 2 or 3. At the end of each stage, the person in each group with the lowest amount of money is sent home, and the other two members go on.

In the third and final stage, the ultimate champion is declared. The Ultimate Champion will, as a prize, receive the True Math Geek Badge.

If that didn't really make much sense, don't worry. It'll become clearer once we begin.

If you have any questions, [i][b]please private message me.[/b][/i] To sign up, also private message me. If you sign up in the thread, I'll ignore it. (I'm doing this to keep the thread clean.)

Hurry and sign up before all the slots are taken!

[color=blue]Group A:[/color]
Member 1 -
Member 2 - AIME_is_hard
Member 3 - AIME15

[color=red]Group B:[/color]
Member 1 - BOGTRO
Member 2 - kookamunga
Member 3 -

[color=green]Group C[/color]
Member 1 - 27r
Member 2 - ernie
Member 3 - Ttocs45

Note: The 100 dollar questions will be AMC 8 level, and the 500 dollar questions will be somewhere between MathCounts States and AMC 10.

I will be problem writer!

It's not that I can't write mathcoutns level problems, its that I find USAMO problems more fun. But I can also write mathcounts level problems.

Sure, you can contribute problems. The Categories for Stage 1 are Number Theory, Geometry, Algebra, Counting, and Expected Value. (100 through 500 dollar prizes.)

One whole category for expected value?!?!?!?!?

I think the categories should be:

Number Theory
Algebra
Counting
Geometry
Probability
Miscellaneous

[quote="izzy"]If you have any questions, [i][b]please private message me.[/b][/i] To sign up, also private message me. If you sign up in the thread, I'll ignore it. (I'm doing this to keep the thread clean.)[/quote]

Sorry guys, but I said that you had to pm me and sign-ups in thread would be ignored.

Here's the official list of people participating:

Group A:
1. nikeballa96
2. AIME_is_hard
3. AIME15

Group B:
1. BOGTRO
2. kookamunga
3. FantasyLover

Group C:
1. 27r
2. ernie
3. Ttocs45

We will be beginning shortly. I will send out reminders for the three contestants participating in this round.

[size=200]Stage 1, Round 1[/size]

Good afternoon everyone, and welcome to AoPS Jeopardy! Today we will be starting Round 1, Stage 1. Our three contestants are: nikeballa96, BOGTRO, and 27r. Hi, guys. How are you? Ready for some competition? If so, great!

May I please direct your attention to the game board shown below.

$ \begin{tabular}{|l\parallel{}c|c|c|c|c|}\hline & Number Theory & Geometry & Algebra & Counting & Expected Value \\
\hline 100 Dollars & AoPS Jeopardy! & AoPS Jeopardy! & AoPS Jeopardy! & AoPS Jeopardy! & AoPS Jeopardy! \\
\hline 200 Dollars & AoPS Jeopardy! & AoPS Jeopardy! & AoPS Jeopardy! & AoPS Jeopardy! & AoPS Jeopardy! \\
\hline 300 Dollars & AoPS Jeopardy! & AoPS Jeopardy! & AoPS Jeopardy! & AoPS Jeopardy! & AoPS Jeopardy! \\
\hline 400 Dollars & AoPS Jeopardy! & AoPS Jeopardy! & AoPS Jeopardy! & AoPS Jeopardy! & AoPS Jeopardy! \\
\hline 500 Dollars & AoPS Jeopardy! & AoPS Jeopardy! & AoPS Jeopardy! & AoPS Jeopardy! & AoPS Jeopardy! \\
\hline \end{tabular}$

nikeballa96, please select a category and dollar price. Write your choice in a post in this thread. For example, if you were to choose 300 dollars for Geometry, write:

"I choose Geometry, 300 dollars."

[quote="isabella2296"]
If that didn't really make much sense, don't worry. It'll become clearer once we begin.
[/quote]

Okay it started now...

If all three contestants get the question right, then they all get the money right?

not playing, but how will you decide who buzzes in the answer first?

Can I join even though I think some of the few problems I submitted are in here?

Patience, patience. Some of us have school, you know. :)

dragon96 replaces 27r, Twin Prime Conjecture replaces xpmath, thanks to both of you. Nikey, sorry but no.

Algebra for 500:

Izzy's favorite number $ i$ is a three-digit number. Nikey's favorite number $ n$ is also a three digit number. In addition, $ n < i$ and when $ i$'s units digit is swapped with its hundreds digit, $ n$ is the result. If $ i \minus{} n$ is a perfect cube, how many different numbers can $ i$ be?

can i join?

and [i][b]i[/b][/i] is not a real number :P  [size=25]i know what it really is[/size]

Waiting for Twin Prime Conjecture, who has 24 hours from now.

So far AIME15 has incorrect, Ttocs has it correct.

Twin Prime Conjecture has submitted the correct answer! It was 0.

Stats:

Team A:

nikeballa96: 1500
AIME15: 1600

Team B:

BOGTRO: 1800
Twin Prime Conjecture: 2200

Team C:

dragon96: 2800
Ttocs45: 1200
LHM: cheerleader!!!
Dojo: cheerleader!!!

This time we have nikeballa96, BOGTRO, and dragon96! Someone please select a question.

$ \begin{tabular}{|l\parallel{}c|c|c|c|c|}\hline & Geometry & Algebra & Polyominoes & Probability & Proofs \\
\hline 100 Dollars & AoPS Jeopardy! & AoPS Jeopardy! & AoPS Jeopardy! & AoPS Jeopardy! & AoPS Jeopardy! \\
\hline 200 Dollars & AoPS Jeopardy! & AoPS Jeopardy! & AoPS Jeopardy! & AoPS Jeopardy! & AoPS Jeopardy! \\
\hline 300 Dollars & AoPS Jeopardy! & AoPS Jeopardy! & AoPS Jeopardy! & AoPS Jeopardy! & AoPS Jeopardy! \\
\hline 400 Dollars & AoPS Jeopardy! & AoPS Jeopardy! & AoPS Jeopardy! & AoPS Jeopardy! & AoPS Jeopardy! \\
\hline 500 Dollars & AoPS Jeopardy! & & AoPS Jeopardy! & AoPS Jeopardy! & \\
\hline \end{tabular}$

@everyone who sent me problems - I lost them because my inbox died. Could you please resend them? Thanks very much!

PROBABILITY FOR 500:

How often do you have to roll two dice simultaneously such that the probability of obtaining at least one double 6 is larger than 1/2?

Wait so if the answer is 42 pairs of dice, then does that mean you roll all 84 dice at once? And when I submit, do I have to post why it's 42 pairs of dice?

Everyone must submit in [b]24 hours.[/b]


The quadrangle $ABCD$ circumscribe $(I)$. $(I)$ touches $AB$ at $M$, $BC$ at $N$, $CD$ at $P$ and $DA$ at $Q$.
The outer bisector from $A$ cut one from $B$ at $M'$. Similary we have $N',P',Q'$.
Proof that $MM',NN',PP',QQ'$ concurrent.

Ouch ! I'm sorry , my mistake. $ABCD$ is also concyclic  :D

This is from grade 10, ARO 2004.
Anyway, the essential part is to note that MNPQ and M'N'P'Q' are homothetic, which is easy to justify.

See also http://www.mathlinks.ro/Forum/viewtopic.php?p=17850 .

  darij


The probability that Rory finishes a race is 0.55 and The probability that Maclaurin finishes the same race is 0.6 . Because of team spirit, there is an 80% chance that Maclaurin will finish the race if Rory finishes the race.
Find the probability that
a) both will finish the race
b) Rory finish the race given that Maclaurin finishes.

R= Rory finishes the race
M=Maclaurin finishes the race
M|R=Maclaurin finishes the race if Rory finishes the race

P(R)=0.55  ;
P(M)=0.6   ;
P(M|R)=0.8;

MR=both finish the race;

a.) P(MR)=P(M|R).P(R)=0.8*0.55;

R|M=Rory finishes the race given that Maclaurin finishes

b.) P(R|M)=P(RM)/P(M) = P(MR)/P(M)= 0.8*0.55/0.6 ;

[hide]$P(R) = \frac{11}{20}$
$P(M) = \frac{3}{5}$

If this means R wins 11 times out of 20 and M wins 3 times out of 5, then the total number of possible outcomes is $2*5*20$.

This is the sample space.

$P{MR} = \frac{11 * 4}{200} + \frac{3 * 11}{200} = \frac{77}{200}$

I think b would be $\frac{77}{104}$ [/hide]


probability
number theory
relatively prime

John rolls a pair of standard 6-sided dice. What is the probability that the two numbers he rolls are relatively prime? Express your answer as a common fraction.

$ \begin{array}{|c\parallel{}c|c|c|c|c|c|}\hline
& 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \hline
1 & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \bigcirc \\ \hline
2 & \bigcirc & \times & \bigcirc & \times & \bigcirc & \times \\ \hline
3 & \bigcirc & \bigcirc & \times & \bigcirc & \bigcirc & \times \\ \hline
4 & \bigcirc & \times & \bigcirc & \times & \bigcirc & \times \\ \hline
5 & \bigcirc & \bigcirc & \bigcirc & \bigcirc & \times & \bigcirc \\ \hline
6 & \bigcirc & \times & \times & \times & \bigcirc & \times \\ \hline
\end{array}$

Thus $ \boxed{\frac{23}{36}}$.

Why would a roll of 1,1 (snake eyes) be considered relatively prime?

The definition of relatively prime is"numbers not having any common divisors other than 1." 1 has only one divisor, namely 1. So 1 and any positive integer $ x$ are always relatively prime. This thus applies to (1,1).

Thank you for the solution!!!!  There his not one provided by alcumus(or it was slow to load, probably this reason)


[quote="Drunken_Math"]Can you make it so that there's a button to put dollar signs around highlighted text? So it'll be much easier to have to $\LaTeX$?[/quote]That would be simple to do, but not as simple as actually typing in the dollar :) In plus it would create mess as people tend to forget tags open when using the buttons and that would screw up both that person´s post and our texer on the server (resulting in delays in loading LaTeX).

How do you put a dollar sign in Latex? how do you disable it?

[quote="chess64"][code]\[/code]

$\$\$\$ $\Rightarrow$ $\$

In other words, dollar sign, \, dollar sign. :D[/quote]
I didn't know that. So cool :lol:


How do you divide polynomials?
What is $\frac{2x^2 + 19x + 35}{x+7}$?

How to say polynomial:

What did the keeper say to the parrot who needed to go on a diet?

[hide]Polly-no-meal  :rotfl:

From a textbook!!!!!!!!!!!!!!!!!
[/hide]

You can use long or synthetic division. Here's synthetic:

$\begin{tabular}[b]{c|ccc}
-7 & 2 & 19 & 35 \\
\ \ & \  & -14 & -35  \\
\ \ & 2  & 5 &  0\\
\end{tabular}$

That's the best I can do on the computer  :P So the answer is $2x+5$. Search on the internet for a better explanation.

[quote="math92"]I vaguely remember this from a textbook.  It was kinda complicated, i don't know how to explain it.It is just like long division though.[/quote]

:huuh: no it isn't! It's completely different, and much faster. I'll try to explain now.

To divide $ax^2+bx+c$ by $x+d$, for example,

\begin{tabular}[b]{c|ccc}
-d & a & b & c \\
\ \ & \ & -ad & (-bd +ad^2)  \\
\ \ & a & b-ad & c-bd+ad^2\\
\end{tabular}

So the answer is $ax+b-ad$ with a remainder of $c-bd+ad^2$. It's much easier with real numbers :D

That's just complicated, chess.

Take it like this (best I can do on a computer):

$\frac{2x^2+19x}{x+7}=2x$ with remainder $5x$, because $2x(x+7)=2x^2+14x$, which you then subtract from the top. Then, you have $\frac{5x+35}{x+7}=5$, so the answer is $2x+5$. In this case, it's easy because the top factors to $(2x+5)(x+7)$, but in some cases you'll be left with a remainder, or it's not so easy to factor, if, for example, it's a third- or fourth-degree polynomial, if not higher. In those cases, it's very useful to use synthetic division, especially if you know what term to divide by.

Sorry :blush:

[hide="Let me try again."]You can only use synthetic division when dividing by something of the form $x+c$.

Say we want to find $\frac{2x^2 + 19x + 35}{x+7}.$

Step 1:

Write down $-c$, in this case $-7$ on the left side. Write the coefficients of the dividend on the right.

$\begin{tabular}[b]{c|ccc}
-7 & 2 & 19 & 35 \\
\ \ & \ \ & \ \ & \ \  \\
\ \ & \ \  & \ \  & \ \ \\
\end{tabular} $

Step 2:

Bring down the number in the 1st column (the 1st coefficient).

$\begin{tabular}[b]{c|ccc}
-7 & 2 & 19 & 35 \\
\ \ & \ \ & \ \ & \ \  \\
\ \ & 2  & \ \  & \ \ \\
\end{tabular} $

Step 3:

Multiply the number on the left by the number on the 3rd row. Write it under the 2nd coefficient.

$\begin{tabular}[b]{c|ccc}
-7 & 2 & 19 & 35 \\
\ \ & \ \ &-14 & \ \  \\
\ \ & 2  & \ \  & \ \ \\
\end{tabular} $

Step 4:

Add the numbers in the 2nd column, and write the result below. Repeat steps 3 and 4 until there is a number in the last column. The degree of the result is 1 less than before dividing, so the number in the last column is the remainder:

$\begin{tabular}[b]{c|ccc}
-7 & 2 & 19 & 35 \\
\ \ & \ \ &-14 & -35  \\
\ \ & 2  & 5  & 0 \\
\end{tabular} $

Thus the answer is $2x+5$ with a remainder of 0. [/hide]

Is that better?

I don't understand the notation fully, but it looks much cleaner. I suppose your previous one works as a formula as well, but if he doesn't get it in the first place, I was thinking an example would help more. ;)

While that was part of his question, he also asked how to divide polynomials in general; notice that those of us who have answered addressed both questions.

thank you everyone. I've just happened to read about dividing polynomials using factorization, long, and synthetic division in an Algebra II review book just yesterday. Synthetic division is hard to show on the computer, but it's very simple actually.

Chess64's explanation makes a lot of sense to me, but that's probably only because I've already learned it!

By the way, synthetic division is only possible when dividing polynomials in the form
x-n?

No, it's possible when dividing any polynomial by any other polynomial smaller than itself, but it becomes more difficult the larger the divisor(?) is. (I always forget the names of the terms; it's the smaller one.)

[quote="JesusFreak197"](I always forget the names of the terms; it's the smaller one.)[/quote]

I think that's the divisor.
It doesn't always have to be smaller, what about the fraction 3/5? divisor is bigger.

True, I guess you can end up with a fraction rather than a polynomial after the division... I know that with standard fractions the divisor can be larger, but I was specifically referring to polynomial division.

it is just like normal division, at the very beginning u try to get rid of the x^2 term so u multiply by 2x which leaves u with 5x+35 and then u divide again getting rid of the x term so u multiply by 5 and which is 5x+35 so u have no remainder so it would be (2x+5)(x+7)=2x^2+19x+35


Let a triangle ABC .A circle (O)  meet BC,CA,AB at M,Q,N,R,P,S,respectivily.the circumcircles of triangles ANP,BPM,CMN at I.The circumcircles of triangles ARS,BSQ,CQR at J.
a) Show that OI=OJ with O is center of (O)
b)Suppose that AM,BN,CP meet at U
Show that AS,BR,CQ meet at V.
Show that UV parralel IJ

Anybody?Is it difficult?I haven't any solution to it.Help me!


The problem states: Prove that for [tex]a > b^2[/tex]
[tex]a - b \sqrt{a + b \sqrt{a - b \sqrt{a + \cdots}}} = a - \frac{3}{4} b^2 - \frac{1}{2}b.[/tex]

I have found a different answer, so I am skeptical.

The RHS must be $\sqrt {a - \frac 3 4 b^2} - \frac 1 2 b$.

Pierre.

Mmm, I do not get this either.  Could you check this?

Ooops...I read the given expression with a big square root for more.
Thus the result is the square of what I've said.

Pierre.

That works, thank you.

There's one another thing which is bugging me: The way to solve this problem is to set $f(x) = a - b \sqrt{a + b \sqrt{x}}$ and solve the equation $f(x) = x$, which produces some answer $x = r$.  However, this argument assumes that the underlying sequence converges.

More explicitly, let $x_0$ be some real number, and let $x_{n + 1} = f(x_n)$.  Then the expression is $\lim_{n \rightarrow \infty} x_n$.  However, for certain values of $a$ and $b$, this sequence does not converge, except when $x_0 = r$.

We can calculate that
\[|f'(r)| = \frac{b^2}{4(a - b^2)}.\]
For example, if we set $a = 4$ and $b = 1.9$, then $|f'(r)| = 2.314$, which is greater than 1.  By chaos theory, the fixed point $r$ is repelling, which means that unless you start with $x_0 = r$, the sequence diverges.  But of course, if you didn't know the answer, why would you start with this value?

To summarize: For certain values of $a$ and $b$, the given expression, seen as a limit, does not converge, which means that the problem is not well-defined.  What do other people think?

I agree with you. I had the same problem with 'my' expression...

Pierre.

that's probably why the problem didn't make it too the shortlist

except for that thingie it's quite a nice problem. :)


a wire of legth 10cm is given.
find the bounded figure of maximum area,that can be made by that wire.

please explain elaborately....
 :) thank you sir....

Its petty hard to prove, but givan any lenght of wire the greater area occurs when its shaped in a circle.

yah, i know the answer that it will be a circle.
but cant prove it :(
can anybody help me out?
thank you.

http://en.wikipedia.org/wiki/Isoperimetric

is the best i can do.

Alternatively, prove that among among the n-gons with fixed perimeter the one with biggest area is the regular one. And then that the one with n+1 sides would have more area than the one with n sides with same perimeter. then limit it as n tends to infinity. Not quite rigerous, but as close as your going to realistically expect.

It's actually not hard at all to prove, provided your definition of a circle is the right one.  Suppose you have a curve that you claim is maximal.  Choose two points on the curve, $ A$ and $ B$, which bisect it, so half the length of the curve is in arc $ AB$ and half is in arc $ BA$.  Note that we only really need to deal with one half of the curve at a time (if maximal shapes exist, then maximal shapes exist with mirror symmetry across the line $ AB$).  Choose any other point $ C$ on the curve, and draw lines $ AC$ and $ BC$.  Leaving arcs $ AC$ and $ CB$ rigid, consider the point $ C$ as a hinge.  The area determined by arc $ AB$ and the segment $ AB$ (which is [i]not[/i] held rigid) is equal to the area $ X$ determined by arc $ AC$ together with segment $ AC$, the area $ Y$ determined by arc $ BC$ together with segment $ BC$, and the area of triangle $ ABC$.  Since we're holding $ AC$ and $ BC$ rigid while we hinge, areas $ X$ and $ Y$ don't change.  Thus maximizing the area in question is reduced to maximizing the area of triangle $ ABC$.  This occurs when angle $ C$ is $ 90^{\circ}$, and the result follows.


Q:  $ n\in \mathbb N$, is not greater than $ 1800$ .If $ n$ is coprime with $ 1800,$
How many possible  solutions to such $ n$?

Soln:$ 1800\equal{}2^3.3^2.5^2 \implies 1800\times\left(1\minus{}\frac{1}{2}\right)\left(1\minus{}\frac{1}{3}\right)\left(1\minus{}\frac{1}{5}\right)\equal{}480$

I read it somewhere and wonder whether  it is true or not, anybody can explain it.Thanks. :lol:

[quote="hsiljak"]1. It is true.

2. http://mathworld.wolfram.com/TotientFunction.html

3. It's discussed all over ML, just do a search on "totient function".[/quote]

Thanks.

I am just a beginner in this field. :D  :blush:


calculus
integration
limit
logarithms
probability and stats

How can we show that : almoste everywhere $ \int_{0}^{u} \frac{B_u}{u} dt  < \plus{}oo$  ?

:)
can you show that $ B_t$ is a.e. finite?

more interesting:
\[ \int_0^u \frac{B_t}{t} \, dt\]
Hint: $ \lim_{t \to 0} \frac{B_t}{ \sqrt{t} } \equal{} 0$ or you can use the law of the iterated logarithm


inequalities
calculus
derivative
inequalities proposed

Well,This is not a real problem.It is an exercise but i find it interesting,so i am posting it:

Let us consider $f(a,b)=\frac{a+b}{ab}$ for the $a ,b$ which are in $\mathbb{R}^+$
Then prove :

$\frac{1}{f(a,c)}+\frac{1}{f(b,d)}\le \frac{2}{f(\frac{a+b}{2},\frac{c+d}{2})}$
I hope you enjoy solving it.

In another word :

$\frac{ac}{a+c}+\frac{bd}{b+d}\le \frac{(a+b)(c+d)}{a+b+c+d}$

There were some typos in the problem.I have edited it right know.

Sorry my opinion is WRONG

This inequality is wrong

a=c=30,b=d=1[

Hmm.I think your counterexample is wrong.

Some Brute Force
 (bc-ad)^2>=0

hence equality holds if and only if bc=ad

[i]Dear Sir Arsene Lupin[/i], this little exercise is a very nice inequality. I note $f:(0,\infty )\rightarrow R,$ where $f(x)=\frac{cx}{x+c}+\frac{bd}{b+d}-\frac{(x+b)(c+d)}{x+b+c+d}$.
Because $f'(x).s.s.(bc-dx)$ and $f'\left( \frac{bc}{d}\right) =0$, results $f'(x)\le 0,\ \forall x>0$ and $f(x)=0\Longleftrightarrow x=\frac {bc}{d}$. From this results the proposed inequality
$(x.s.s.y\Longleftrightarrow x=y=0\ or\ xy>0$).
Have you a solution without derivatives ? I don't. I don't know what is happened with this Latex. Can you to explain me ? Thanks you.

My solution is not too interesting .
I used brute force as Ghang Hwan.
Equality occures when $\frac{a}{c}=\frac{b}{d}$

The proposed inequality is equivalently with the inequality

$c\left( 1-\frac{c}{a+c}\right) +d\left( 1-\frac{d}{b+d}\right) \le (c+d)\left[ 1-\frac{c+d}{a+b+c+d}\right] \Longleftrightarrow$

$\frac{c^2}{a+c}+\frac{d^2}{b+d}\ge \frac{(c+d)^2}{a+b+c+d}$ $\Longleftrightarrow c^2 (b+d)^2 +d^2 (a+c)^2\ge 2cd(a+c)(b+d)\Longleftrightarrow$

 $[c(b+d)-d(a+c)]^2\ge 0$. We have equality iff c(b+d)=d(a+c) iff bc=ad.


Prove that, if for some $a,b\in Z$ number $a\cdot 2^n + b$ is a perfect square for every $n\in N$, then $a=0$.

Some time I have also posted this problem: :D  :D
See:[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=2%5En%2Bb&t=41463]www.artofproblemsolving.com/Forum/viewtopic.php?highlight=2%5En%2Bb&t=41463[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=2%5En%2Bb&t=34569]www.artofproblemsolving.com/Forum/viewtopic.php?highlight=2%5En%2Bb&t=34569[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=2%5En%2Bb&t=4399]www.artofproblemsolving.com/Forum/viewtopic.php?highlight=2%5En%2Bb&t=4399[/url]
Enjoy!!!!! :D

tnx... I thought this problem has much nicer solution... oh well...

bye


I was reading chapter 27 in campbells about prokaryotes. It stated that the average diameter of a prokaryote was approx 1-5 microns, but then it said that they travel 50 microns a second... [i]about 100 times their size[/i] .... I have a problem with that fact... shouldnt it be 10 times since the are only 1-5 microns? Or did i misread...

50 microns per second is anywhere from 10-50 times their length per second, not 100.


[quote="Nora.91"]Prove that for any positive integer [b]n[/b], $ 1^n \plus{} 2^n \plus{} 3^n \plus{} 4^n$ is divisible by five, if and only if, [b]n[/b] is not divisible by 4.[/quote]

For $ n$ odd, $ (1\plus{}4)|(1^n\plus{}4^n)$ and $ (2\plus{}3)|(2^n\plus{}3^n)$ gives $ 5|1^n \plus{} 2^n \plus{} 3^n \plus{} 4^n$

For $ n\equal{}2 \pmod{4}$, $ 1^n\plus{}2^n\plus{}3^n\plus{}4^n \equiv 1^2\plus{}2^2\plus{}3^2\plus{}4^2 \equiv 0 \pmod{5}$ by Fermat's little theorem.

For $ 4|n$, $ 1^n\plus{}2^n\plus{}3^n\plus{}4^n \equiv 1\plus{}1\plus{}1\plus{}1 \equiv 4 \pmod{5}$, done.


$ f(n)\equal{}9^{n\minus{}1}...$ baghiye be ohdeye shoma  :D

chera too topice khodesh ([url]http://www.mathlinks.ro/viewtopic.php?t=247476[/url]) naneveshti? :maybe:


calculus
integration
function
limit
complex analysis

Let $ C_r$ be the closed path $ z\equal{}re^{i\theta} \ \ 0\leq \theta \leq 2\pi$ and $ f$ a continuous function at the origin of $ \mathbf{C}$. Show that\[ \lim_{r \to 0}\oint_{C_r}\frac{f(z)}{z}\, {\rm d}z\equal{}2\pi i \cdot f(0) \]

oh i remember learning it in complex analysis.  oh that takes me back.  i dont remember the solution though.  there was supposed to be some formula for this kind of thing.

I am not sure but I try by this way :
in your case your closed path $ C_{r}$ is homotopical to point zero. Know Let's use theorem that integral like this is same at homotopical way. and finnaly use fact from complex analysis Cauchy formula for complex function about representation of complex function by integral

We can't use the Cauchy integral formula because we weren't told that $ f$ was analytic - merely that it was continuous at $ 0.$

A preliminary computation based on explicitly parameterizing the curve:

$ \oint_{C_r}\frac1{z}\,dz\equal{}\int_0^{2\pi}\frac{ire^{it}}{re^{it}}\,dt\equal{}2\pi i.$

Then $ \left|\oint_{C_r}\frac{f(z)}{z}\,dz\minus{}2\pi if(0)\right|\equal{}
\left|\oint_{C_r}\frac{f(z)\minus{}f(0)}{z}\,dz\right|$

$ \le\oint_{C_r}\frac{|f(z)\minus{}f(0)|}{r}\,|dz|$

$ \le \max_{z\in C_r}\,|f(z)\minus{}f(0)|\frac{1}{r}\oint_{C_r}|dz|\equal{}2\pi\max_{z\in C_r}\,|f(z)\minus{}f(0)|$

Since $ f$ is continuous at $ 0,$ we can make $ \max_{z\in C_r}\,|f(z)\minus{}f(0)|$ arbitrarily small by making $ r$ small enough.


The sequence {a[size=75]n[/size]} of positive integers is defined by
a[size=59]0[/size]=m and a[size=75]n+1[/size]=a[size=75]n[/size]^5+487 , for all n>=0
Determine all values of m for which the sequence contains as many square numbers as possible. :roll:

[quote="Armenia"]The sequence ${a_{n}}$ of positive integers is defined by
$a_{0}=m$ and $a_{n+1}=a_{n}^{5}+487$ , for all $n\geq 0$
Determine all values of $m$ for which the sequence contains as many square numbers as possible. :roll:[/quote]

Aren't you going to learn using $LaTeX$?

I don't know how to use LaTex.
Can you give me some lessons? :lol:
Ara mekel mi ban el
Karas Jivaja Geometrian tas mi dzev??? :wink:

the best way is to move your cursor over a latex output wait see and learn :wink:

First, the link suggested above does work and you are probably doing something wrong. But if you wish, go to the http://www.artofproblemsolving.com and see there (in the left) the link $LaTeX$.

Second, it is not polite to use your own language in this sub-forum! :P

And third, I don't have the pograme you were asking about! :wink:

[hide="Only for Mr Hakobyan"] "Ara" is the number of your car :rotfl: (if you have one).[/hide]

try put your text between two dollar signs :lol:


topology
function
algebra
polynomial
number theory
real analysis
real analysis unsolved

Let $A,B \subset \mathbb{R}$ be two dense, countable subsets of the reals. Does there necessarily exist a bijection $f: A \to B$ that is continuous under the subspace topology, but such that $f^{-1}$ is discontinuous at every point?

Since any two subspaces of $\mathbb R$ obtained from a countable dense real set are homeomorphic, we may as well assume that, for example, $A=\mathbb Q,\ B=f(\mathbb Q)$ where $f:\mathbb R\to\mathbb R,\ f(x)=x^3-5x,\ \forall x\in\mathbb R$, and the function from $A$ to $B$ that we are looking for is the restriction of $f$ to the rationals.

1 - Is it so trivial that $f$ is injective on the rationals (this seems to be a nice problem for the number theory forum, using congruences mod 5)?

2 - I don't understand why $f^{-1}(x)$ is not continuous for large values of $x$.

Oh, I forgot about large values of $x$ :). As for the other question, it's not trivial, but it's not hard either.

I think the example can easily be repaired (we should define the function by gluing together graphs of polynomials similar to this one) by using this idea, but I don't have the time right now.

Edit:

Simpler yet, you use this function as it is. It provides a continuous function from the set of small enough rationals to the set of small enough rationals of the form $k^3-5k$ with rational $k$, and we take these sets to be $A,B$, while replacing $\mathbb R$ with a small enough open interval. Everything is the same as with the "large version" (:)) of $\mathbb R$ up to homeomorphism, so there really is no difference, topologically.


I need to find the inverse of a 6x6 matrix

 _                       _
|2     2       4     -2     8      18|
|3     4      11    -5    18     26|
|-1   -1     -1      0    -6      -5 |
|2     2      4      -1    14     21|
|2     2      4      -5    -9      7 |
|-1   -6   -43     27   -2      -67|

So first. If anyone wants to do this that would be GREAT. But i doubt not

So if not, about how long is this going to take?

Thanks

Oh lol.

Thanks. I have to do it by hand

:(

Oh and by the way, just saw your location

I go to baylor ( live in waco :)

Did the assignment really say "Find the inverse of this matrix by hand"? That would be very strange, even for a place like Baylor.  :D Maybe there is a way to find what you're looking for without inverting the matrix.

[quote="Kent Merryfield"]For matrices in which every entry is nonzero, I generally put $ 4\times 4$ as the usual practical limit for hand computation, although under some circumstances, I might be willing to attempt a $ 5\times 5.$ I'd allow about half an hour for that $ 5\times 5,$ and I'm a child of the pre-calculator era with long-practiced arithmetic skills.

I think the operations count for inverting an $ n\times n$ (using Gaussian elimination, of course) is about $ \frac43\,n^3$ multiplications/divisions and about the same number of additions/subtractions. I might have the coefficient wrong in one of those cases, but I'm sure of the exponent.

Note that $ 6^3 \equal{} 216$ and $ \frac43\,6^3 \equal{} 288.$ So we'll need nearly 300 multiplications/divisions and nearly 300 additions/subtractions (some of which will probably be the adding of fractions.)

The real problem with the hand calculation is how do do that many computations without making an arithmetical error.

The same considerations apply to the machine computations, except that a few hundred arithmetic operations take a truly miniscule amount of time, and there's no concern with making mistakes (although there is a concern with roundoff error in floating-point arithmetic.)

One numerical analyst's attitude: who needs an inverse? An LU factorization is just as good for nearly all purposes, and the operation count for finding the LU factorization is $ \frac13\,n^3,$ which is a fourth of the time it takes to find the inverse.[/quote]

I see.... lol Awesome... So this is going to take a  while!

And yes it says " Find the inverse of the matrix using row reduction"

And he said it would be several pages of hand written work.

ugh

well i'll be up for the next couple hours or so doing this

[quote="Kent Merryfield"]As long as we're talking Texas ... I was an undergraduate at Rice (1971-1975) and taught at UT Austin (1981-1984).[/quote]

Awesome lol. I used to live in Long Beach (90810)

:)

Ok, so i dont even know how to star this. What numbers am I using for the other side of the matrix? like for a 3x3 it would be
[1 0 0
 0 1 0
 0 0 1]

but what about for a 6x6??

Same thing: ones on the diagonal, zeroes elsewhere. It's called the identity matrix.

I would use a computer or calculator (whatever can subtract vectors and multiply them by a scalar) to perform row operations, and then record the results of each step on paper.

which method would i be using if i found the determinant of M, or |M|and then found the adjugate of M, and then scalar divided adj(M) by |M|?

Would that be using row-reduction?

And if not

which method is associated w/ row-reduction?

Thanks!

No, inversion with determinants is not row reduction. You can read about row reduction [url=http://www.sparknotes.com/math/algebra2/matrices/section4.rhtml]here[/url].

[url=http://www.math.ncsu.edu/ma114/tools/row_ops.html]Row reduction applet[/url] This ought to help.  :)

gotcha, thanks for the app link

Haha I am so confused though. I have my identity matrix now.

But i dont have a clue where to start.

What the heck do i even do? I have been all over the net trying to figure out where to start.

I even have an example of a 3x3 solved.... but I don't know how to relate what they are doing to a larger scale matrix.

Like the first step for a 3x3 would be to swap row 1 and row 2

then the next step would be to make row 2  equal to  -Row 1 + row 2, and Row 3 equal to  -2 x Row 1 + row 3.

But how does this apply to a 6x6?

[quote]Like the first step for a 3x3 would be to swap row 1 and row 2
then the next step would be to make row 2 equal to -Row 1 + row 2, and Row 3 equal to -2 x Row 1 + row 3. [/quote]
Wrong. Wrong. Wrong.
Wrong. Wrong. Wrong.
Wrong. Wrong. Wrong.

It all depends on what the entries are. You are trying to creates zeroes in the matrix by cancelling some of the terms by others. Which rows to swap and which to add, and by what number to multiply - it all depends on the numbers that are in those rows. Read the SparkNotes page linked above. Then work out a 3x3 matrix or two. And only then consider doing something about 6x6.

Oh my god.

So is there a TON of trial and error? when trying to inverse a 6x6 by hand?

How is one (and I'm talking about normal ppl like me, and excluding genius's like you ppl on this board) supposed to know where even to star?

It's not exactly trial and error, since you know that you'll need 1's along the diagonal and 0's everywhere else. I'll do a 2x2 example: If we wanted to invert the matrix $ \left(
\begin{array}{cc}
5&7\\
1&3\\
\end{array}
\right)$ using row reduction, we would set it up like this:
\[ \left( \begin{array}{cc|cc}
5&7&1&0\\
1&3&0&1\\
\end{array}
\right)\]
Now we need to make the left matrix the identity matrix. First make the upper left corner 1. This can be done in two ways: either:

-multiply the first row by a constant (in this case, 1/5), or
- subtract a multiple of the second row from the first (in this case, we would subtract a number of 1's from 5 to get 1. This would be 4.)

I'll do the second one, since I want to deal with fractions as little as possible:
\[ \left( \begin{array}{cc|cc}
1&\minus{}5&1&\minus{}4\\
1&3&0&1\\
\end{array}
\right)\]
(We subtract four times the second row from the first row.) Next, the lower left corner must be a 0. You can't multiply a row by the constant 0, so this requires subtracting a multiple of the first row from the second row. In this case, it's easy since the entries in the first column are the same: just subtracting one multiple of the first row will be needed to get a zero in the lower left corner:
\[ \left( \begin{array}{cc|cc}
1&\minus{}5&1&\minus{}4\\
0&8&\minus{}1&5\\
\end{array}
\right)\]
The left column is finished with. Now we need to make the upper right corner 0 and the lower right corner 1.

Now we're stuck having to deal with fractions, since in the right column we have an 8 and a 5. If we multiply the second row by a constant to make the lower right corner 1, that constant must be 1/8. If we subtract $ k$ times the first row from the second, we'll need $ 8\minus{}(\minus{}5)k\equal{}1$, or $ k\equal{}\minus{}5/7$. That would mean [i]adding[/i] 5/7 times the first row to the second.

In this case, I'll choose to divide the second row by 8.
\[ \left( \begin{array}{cc|cc}
1&\minus{}5&1&\minus{}4\\
0&1&\minus{}1/8&5/8\\
\end{array}
\right)\]
Finally, make the upper right corner 0. Since the lower right corner is 1, we can just add 5 times the second row to the first.
\[ \left( \begin{array}{cc|cc}
1&0&3/8&\minus{}7/8\\
0&1&\minus{}1/8&5/8\\
\end{array}
\right)\]
We're done. The inverse is now $ \left(
\begin{array}{cc}
3/8&\minus{}7/8\\
\minus{}1/8&5/8\\
\end{array}
\right)$.

Yes, row reduction sucks, and it gets worse quickly. One thing is that you'll more choices to make when deciding how to make an entry 1 or 0. In my opinion, the best rule is to avoid fractions as long as possible. (Had I divided by 5 in the beginning, I would have fractions the whole time, and the 5's in the denominator would just go away in the end.) But as long as you make sure to make the entries 1's or 0's, you will know what to do (apart from messing up in the computation).

How do you know where to start? It's so simple, we can tell a computer how to do it. You only [b]need[/b] to swap rows if you can't divide by your pivot element (because it's zero). If you're working by hand, swaps that make the pivot element $ \pm 1$ are also a good idea- but that only makes sense if it was rigged to come out nicely anyway.

There's a fairly natural row swap to make in the first step here- if you do it, you'll only have to divide a row by something other than $ \pm 1$ once. (I tested the computations in a spreadsheet)

Row reduction is a much better method than Cramer's rule for finding inverses of reasonably large matrices, because finding a determinant is itself best done by row reduction to a triangular matrix. That's the $ \frac13n^3$ multiplications/divisions Kent mentioned- and if you need $ n^2\plus{}1$ determinants, that's $ \frac13n^5$.

[quote="scorpius119"][u]Long quote removed by a moderator[/u][/quote]

Thanks! That does help

I guess all I am saying is since it is a larger matrix.. I have to make sure that for each "thing" i do to it (swap, multiply, add subtract) I have to make sure it doesnt mess up what I already have done.

So is the best way to work from left to right?

Thanks everyone really

finished it guys!

Thanks for all the help

only took 2hrs and 50 min!


Edit: [hide="click to see my wrong solution"]
\[ (\frac13)^{2x} \plus{} (\frac13)^{\frac1x} > (\frac13)^{ \minus{} 1 \minus{} \log_34}\]
[color=red]This next step doesn't work.[/color]
\[ 2x \plus{} \frac1x > \minus{} 1 \minus{} 2\log_32\]
To find endpoints/beginning points:
\[ 2x^2 \plus{} 2x\log_32 \plus{} 2 \equal{} 0\]

\[ x^2 \plus{} x\log_32 \plus{} 1 \equal{} 0\]
Solve to get: $ \frac { \minus{} \log_32\pm\sqrt {2\log_32 \minus{} 4}}2$
The solutions are on the outside of these two point. I don't have time to TeX that up right now.[/hide]

hello, your calculation is wrong, i thionk this inequality can only solved by a numerical way.
Sonnhard.

The inequality is not interesting for $ x > 0$, since $ (1/3)^{\text{positive}} < (1/3)^0 \equal{} 1$.

For $ x \leq 0$, substitute $ z \equal{} \minus{} x$, so that $ z$ is positive. The inequality then reads:
$ 3^{2z} \plus{} 3^{1/z} > 12$

[b]Strategy:[/b] Look for simple solutions to the associated equation (where both sides are equal). Since the left side is continuous for positive $ z$, these solutions will divide the positive numbers into intervals, each of which is either a solution to the inequality or not a solution.

Using simple common inequalities[url=http://en.wikipedia.org/wiki/AM_GM_inequality],[/url] find the minimum value of the left-side expression, and find which interval it corresponds to. Then, use extreme values of $ z$ to locate intervals that clearly satisfy or clearly don't satisfy the inequality. From there, you should be able to figure it out.


Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers with $a_{1}\le a_{2}\le \cdots \le a_{n}$ such that the arithmetic mean of $a_{1}^{2}, a_{2}^{2}, \cdots, a_{n}^{2}$ is $1.$ If the arithmetic mean of $a_{1}, a_{2}, \cdots, a_{n}$ is $m$ and if $a_{i}\le m$ for some $1 \le i \le m,$ prove that $n-i \ge n(m-a_{i})^{2}.$


Does anyone know whether matrices or vectors are (or have ever been) on the AMC tests?

Fierytycoon

I don't remember having seen it. If it's on them, it's very rare.

I don't think those topics are required to solve any of the problems on any AMC tests. But they can sometimes be useful in solving some problems in quicker or easier ways.

Yes.  Certain solid geometry problems can possibly be solved faster using vectors/matrices.  This can be seen on problem 12 of the 1989 AIME.

-interesting_move


group theory
abstract algebra
superior algebra
superior algebra unsolved

Suppose there is an isomorphism from a group G to a subgroup of another group H & there is another isomorphism from H to a subgroup of G.Then are G & H isomorphic??

I asked this question some time ago. I can't remember the examples, but take, for instance, the free group on $2$ generators and the free group on $3$ generators.

I dont understand will they be isomorphic or not?

No, they are not isomorphic, but each one can be thought of as a subgroup of the other one. It's clear that the free group on $2$ generators can be regarded as a subgroup of the one on $3$ generators, and there's a theorem stating that the reverse is also possible.

So the idea is that no, such two groups need not be isomorphic.

Hmm thanx for the examples.Well I am sorry that the case are not very interesting after all :(


linear algebra
matrix
vector
linear algebra unsolved

we know:

Assume $ A \in M_{n}(\mathbb{R}).$  If $ A^TA\equal{}I_{n}.$ and $ \lambda_1, \lambda_2$ are eigenvalues of $ A$   where $ \lambda_1 \not\equal{} \lambda_2$

If there exists $ \alpha_1, \alpha_2$

satisfy: $ A\alpha_1\equal{}\lambda_1\alpha_1.$ $ A\alpha_2\equal{}\lambda_2\alpha_2$

Obviously,  $ (\alpha_1,\alpha_2)\equal{}0.$

[color=red]My problem is:[/color]

If $ \lambda$ is a eigenvalue of $ A$, and it has more than one linear independent vectors, are these vectors pairwise orthogonality?

in other words: if $ \alpha_1, \alpha_2,...\alpha_k$ are $ \lambda's$ linear independent vectors, do we have: $ (\alpha_i, \alpha_j)\equal{}0$,when $ i \not\equal{}j$

It seems false :oops:

Why would they be? Linear independence is not a strong condition at all; $ \alpha_1,\alpha_2$ are independent iff $ \alpha_1,(\alpha_1\plus{}\alpha_2)$ are independent, after all... :)

But you have a choice about how to represent a basis for a non-simple eigenspace. If you [i]want[/i] orthogonal vectors, you can take the independent set you have and perform the Gram-Schmidt orthogonalization on it.

Now, you do realize that much of the time, many of the eigenvalues of a real orthogonal matrix are complex and not real?


[hide]6 ways to choose 2 points (4c2). Each set of 2 points can make two circles, one pointing one way and the other pointing the other way. 6*2 is 12, and i think thats the answer.[/hide]

[hide]
Well, you can choose the 2 points that are on the circle in $\binom 42=6$ ways. Each of those combos has 2 ways for the circle; each in another direction.
So $6\cdot2=12$[/hide]

[hide]
12, there are 4C2=6 and two sides for each one.
I remember this problem.[/hide]

For a short explanation of why a circle can pass through any points...
A circle can be constructed through any 3 points that are not collinear. For it to have 1 radius, we must find a point on the perpendicular bisector of the two points that is one point apart from each of them and define a circle. Now, for that 3rd point to be collinear with the first 2 and on the perpendicular bisector, the points must be 2 units apart. Since the farthest away 2 points are in that square (a diagonal) is $\sqrt{2}$, you can make a circle through any 2 points in that square.


There are nine unit triangles in the three rows of this
triangle. If you add three more rows to the bottom of this
triangle so that each row has 2 more unit triangles than
the previous row, how many unit triangles will there be?

[asy]draw((0,0)--(15,26)--(30,0)--cycle);
draw((5,8.7)--(10,0)--(20,17.3));
draw((20,0)--(25,8.7)--(5,8.7));
draw((20,0)--(10,17.3)--(20,17.3));[/asy]

Ok, the next three rows will have 7,9, and 11 units. So ad all the units together and you get [u][b]36[/b][/u]

There is another quite interesting solution. First notice that with one row there is 1 unit triangle, 4 (1+3) with the addition of the second row, 9 (1+3+5) with the addition and the third row and so on. One may thus notice that there for n rows there are n^2 individual triangles. Thus, in the 6th row, there will be [b]36[/b] triangles.

I find this a much more satisfying approach because it employs patterns without the encumbering addition and, just a point of interest, artofproblemsolving.com has an applet showing that the sum of the first n natural odd numbers starting with one is n^2, so this triangle problem is a variation of this applet. Therefore, artofproblemsolving.com may wish to add this applet directly to the problem solution in order to expand the solution quality.


[quote="arqady"][quote="quykhtn-qa1"]Prove that in any triangle ABC, the following inequality holds:
                       $ cos^2Acos^2B \plus{} cos^2Bcos^2C \plus{} cos^2Ccos^2A \geq\ 2cosAcosBcosC(cos^2A \plus{} cos^2B \plus{} cos^2C)$[/quote]
We can assume that $ \Delta ABC$ is an acute angle triangle and after a substitution
$ x \equal{} b^2 \plus{} c^2 \minus{} a^2,$ $ y \equal{} a^2 \plus{} c^2 \minus{} b^2,$ $ z \equal{} a^2 \plus{} b^2 \minus{} c^2$ we obtain  the Schur's inequality.[/quote]
with$ \Delta ABC$ is an acute angle triangle.Let
$ \frac{cosAcosB}{cosC}\equal{}\frac{z}{2},\frac{cosBcosC}{cosA}\equal{}\frac{x}{2},
\frac{cosCcosA}{cosB}\equal{}\frac{y}{2}$
We have :$ xy\plus{}yz\plus{}zx\plus{}xyz\equal{}4$.must prove that:$ x\plus{}y\plus{}z\geq\ xy\plus{}yz\plus{}zx$
It very nice and easy :)