MATHCOUNTS
probability
geometry
3D geometry
AMC
AIME
LaTeX

Okay, was this week's material, like, way too easy? I think I overcompensated for last week or something. Anyway...do the problems, I think they're pretty interesting. And as before, discuss them here. Last week's class problem answers are 1. 126 2. 468000 3. 210 4. 20 5. 89/91 6. 19 7. 10 8. 495000 9. 121 10. 1124. Week 1 homework answers are 1. 120 2. 40 3. 60 4. 32 5. 1680 6. 28 7. 192 8. 360 9. 120 for part 1, 18 for part 2 10. 560. I just realized that I didn't give a picture in problem 10, despite promising to do so in the problem...sorry. It should still be pretty clear though. Also, everyone please do your best to find nice solutions to #8-10 because they're really important. Hints:
[hide]9. Consider this: ##|#|###|#. The three |s divide the 7 #s into 4 groups. The leftmost group goes into the first lunchbox, the second group goes into the second lunchbox, and so on. How do you count the number of ways that 3 dividers and 7 #s can be arranged? This also applies to 8. (In 8, first pretend the customers are identical.)[/hide]
[hide]10. Once you've picked the times at which the targets in one column are shot, you can only shoot those targets in one order. So you might as well pretend that they're indistinct. This now collapses to problem 5--do you see how?[/hide]

EDIT: okay, due to confusion, I'm posting all of this week's class/HW problems.
Class Problems
1. A fair coin is flipped 4 times in a row. Find the probability that it shows heads every time.
2. Find the probability that the coin in problem 1 displays exactly two heads.
3. A bag contains six white marbles, five red marbles and four blue marbles. Norka will reach into the bag and pull out three marbles at the same time. What is the probability that all three marbles will be the same color? (2005-2006 Warm-up 15 #10)
4. I have three cookies. One is yellow on both sides, one is yellow on one side and brown on the other, and one is brown on both sides. I pick a cookie at random and place it on the table on a random side. The top of the cookie is yellow. What’s the probability that the bottom is yellow?
5. Russell rolls a pair of dice with his eyes closed. His sister tells him that the sum of the dice is greater than or equal to nine. Knowing this information, Russell calculates the probability that he rolled doubles. What is that probability? (2005-2006 Warm-up 13 #7)
6. In a small town of 100 men, 85 are married, 70 have a telephone, 75 own a car, and 80 own their own home. On this basis, what is the smallest possible number of men who are married, have their own telephone, own their own car, and own their own home? (1993 National Sprint #28)
7. Richard is hitchhiking from Decatur, AL, to Amherst, VA. The probability that Richard will see a car within the next 20 minutes is 609/625. What is the probability that he will see a car within the next 5 minutes? Assume that the probability of seeing a car at any moment is uniform (the same) for the entire 20 minutes. (Mandelbrot #2)
8. A bug starts at one vertex of a cube and moves along the edges of the cube according to the following rule. At each vertex the bug will choose to travel along one of the three edges emanating from that vertex. Each edge has equal probability of being chosen, and all choices are independent. What is the probability that after seven moves the bug will have visited every vertex exactly once? (2006 AMC 10A #25)
9. A point P is selected at random from the interior of the pentagon with vertices A=(0,2), B=(4,0), C=(2pi+1,0), D=(2pi+1,4), and E=(0,4). What is the probability that angle APB is obtuse? (2001 AMC 12 #17)
10. A fair coin is to be tossed ten times. Find the probability that heads never occur on consecutive tosses. (1990 AIME #9)

Homework Problems
(Hint: whenever you’re about to panic, remember P(A)=1-P(not A), and try to find P(not A) and subtract from 1 instead of directly calculating P(A). This won’t always work, but it might.)
1. The ace of hearts, the ace of clubs, the ace of diamonds, and the ace of spades are face down on a table. Two different cards are selected at random from the set of four cards. What is the probability that at least one of the cards is a red ace? (MathCounts 1985)
2. A blue urn contains 4 black marbles and 2 blue ones. A black urn contains 4 black marbles and 11 blue ones. One marble is drawn at random from each of the two urns. What is the probability that both of the marbles drawn are blue? (MathCounts 1988)
3. In a school of 100 students, 50 own dogs, 32 own cats, 57 own goldfish, 5 own all three, and 14 own none of them. Find the number of students that own exactly two out of the three types of pets.
4. Two digits between 1 and 9, inclusive, are selected at random. The same digit may be selected twice. What is the probability that their product is a multiple of 3? (Mu Alpha Theta 1987)
5. In a raffle 20 tickets are sold. Two prizes will be given. A student buys 2 tickets. What is the probability that this student wins at least one prize? (MathCounts 1988)
6. Bertie throws two darts at the given target, both of which fall somewhere within the target. What is the probability that Bertie’s total score is 5 or greater?
[img]http://img.photobucket.com/albums/v189/bigsister/math/w2hwp6.gif[/img]
7. Instead of using two standard cubical dice in a board game, three standard cubical dice are used so that the game goes more quickly. In the regular game, doubles are needed to get out of the ‘pit’. In the revised game, doubles or triples will get you out. How many times as likely is it for a player to get out of the ‘pit’ on one toss under the new rules as compared to the old rules? (MathCounts 1987)
8. A box contains exactly five chips, three red and two white. Chips are randomly removed one at a time without replacement until all the red chips are drawn or all the white chips are drawn. What is the probability that the last chip drawn is white? (2001 AMC 10 #23)
9. Two three-letter strings, aaa and bbb, are transmitted electronically. Each string is sent letter by letter. Due to faulty equipment, each of the six letters has a 1/3 chance of being received incorrectly, as an a when it should have been a b, or as a b when it should have been an a. However, whether a given letter is received correctly or incorrectly is independent of the reception of any other letter. Let S_a be the three-letter string received when aaa is transmitted and let S_b be the three-letter string received when bbb is transmitted. Find the probability that S_a comes before S_b in alphabetical order. (1991 AIME #10)
10. A drawer contains a mixture of red socks and blue socks, at most 1991 in all. It so happens that, when two socks are selected randomly without replacement, there is a probability of exactly 1/2 that both are red or both are blue. What is the largest possible number of red socks in the drawer that is consistent with this data? (1991 AIME #13)

itll be nice if you say your name(first) too

whos slt again? i can't think of any initials -_-

[quote="Bluestalk"]OKAY. SLT, you are my twin, correct?

So are we starting a new forum each week?[/quote]
no we're starting a new topic/thread each week. a forum is where you make a bunch of topics/threads.

[quote="Edz504"]Hihihihihihimezisignedupfordaheckofit...[/quote]
um good job spamming

[quote="Edz504"]Hihihihihihimezisignedupfordaheckofit...[/quote]

whos edz?
ed zhou?
ed zhang? wait no

[quote="Bluestalk"]um...How do you quote?

Yes it does if you're in 8th grade.....

Okay then. So next week there's going to be a #3. Wonderful. [/quote][/quote]

there s a quote button

[color=violet]Bookaholic[/color] and [color=violet]ccy[/color]...for number 7 on the Week 1 Class problems...(the one about cubes)...

Bookaholic got 2 different cubes and ccy got 360 cubes..and I got 10 different cubes.

Yeah....

Anybody else get anything different (or same?)

And pardon me, [color=violet]ccy[/color], if you don't like your name in purple. I'll change it if you want me to. BUT I think [color=violet]bookaholic's[/color] favorite color is purple/lavender/violet. Something like that.

Darn; I was being stupid and misread/screwed up the problem.  Here's my new brute-force solution.

[hide="7"]
Obviously, we can have the cube be all white or gray in 1 way each ($2$ ways total).  If we paint 1 white/5 gray or 1 gray/5 white, we still have 1 way each ($2$ ways total).  Now, consider 2 white/4 gray (or 2 gray/4 white).  You can choose the white faces so that they're either opposite each other or adjacent, so we have 2 ways each ($4$ total).  For 3 white/3 gray, you get 2 configurations, either having 3 whites/grays in a sequence or 3 whites/grays all adjacent (hard to explain) to get $2$ more.  So unless we're both being stupid :P it's probably $10$.
[/hide]
Hmm random thought:  they probably didn't put "black" in the problem to be politically correct.

[quote="bookaholic"]Congyue, you have mod status?![/quote]They didn't have any mods in this forum, so I asked and they made me one (Yay!).

[quote]And pardon me, ccy, if you don't like your name in purple.[/quote]
I don't care. :|

Yeah...it is 10. I checked the handbook from last year...

So...bookaholic should change her answer on the very first post.

And...I got something different from ccy on number 8 too (palindromes). Something like 495000? Which I THINK is the same as bookaholic. Oh...and number 6...it's odd integers only and you kinda missed the point of the problem.....

would you care if I wrote your name in [color=hotpink]hot pink[/color]? Or would you still not care? just wondering...

//edit//: bookaholic...you know problem one on the homework problems from week 1? did you say something about if they don't tell you if it's a combination or a permutation that it's usually a combination? (or SOMETHING like that)

i got same answer for #8
[hide="solution idea"]compute average and number of palindromes.[/hide]

PS HOT PINK ROCKS OMG.

What the...that was stupid. I somehow read the words "three gray faces and three white faces" into that problem. Gaaah.

About permutations versus combinations: use common sense. Usually the problem writers will be as unambiguous as possible and will state whether or not the order matters. If they don't, look for wording in the problem that suggests it. "In how many ways can person X choose blah" probably means the order doesn't matter and you want C(n,k), unless you have other reasons to believe otherwise.

Oh yeah, and 9 and 10 in the class problems are both subject to my undying hatred. Especially 10 because if I hadn't been so incredibly stupid about it I'd have made the USAMO. (I counted up to 104! instead of 100!. Um. Yeah.)

[quote="bookaholic"]Oh yeah, and 9 and 10 in the class problems are both subject to my undying hatred. Especially 10 because if I hadn't been so incredibly stupid about it I'd have made the USAMO. (I counted up to 104! instead of 100!. Um. Yeah.)[/quote]
whoa.
in that case, 10 will be subject to my undying love.

so is that good or bad? 100! is a sad number

hey bluestalk, for analysis do we have to do the stupid worksheet too

[quote="junggi"][quote="moogra"]ask ccy? doesn't ur teacher use inclass?[/quote]
are you stupid.
dont you think i already asked him you idiot?

ccy delete these posts[/quote]
wtf is this ccy?

too bad. im posting again

and if you delete, im gonna spam even more

plus give me the week 2 problems. i want to do them

The week 2 problems are in the first post.

What are we doing next Tuesday?

 :jump:

That guy is so bouncy, like a little basketball.

Mr. Berghout has sooo many wrinkles on his forehead. Ami and I counted FOUR.

Next week will be a practice sprint round, so I won't come, but there'll be homework. I believe it's also the due date of USAMTS Round 1, which I'm still procrastinating on.

yes! practice sprint round. but im not going -_-
i'll have ccy gimme the problems. what year is it anyways?
 :first:

cool. oh and i found out that kevin gave up on the terman people. he went and "saw that they were sad, so he didn't do anything YET. he thinks they are in trouble though." so hes not helping...


Angie starts with 8 pieces of paper, and at least one of these pieces is cut into 8 smaller pieces. Then at least one of these smaller pieces is cut into 8 smaller pieces. Then at least one of these smallest pieces is cut into 8 smaller pieces. If $x$ represents the final number of pieces of paper that Angie has, find the sum of all possible distinct values of $x$.

Angie has
$8+7x_{1}$ pieces, where $x_{1}\le 8$ on the 1st step,
$8+7(x_{1}+x_{2})$ pieces, where $x_{2}\le 8+7x_{1}$ on the 2nd step,
....
$8+7(x_{1}+x_{2}+...+x_{k}$ pieces, where $x_{2}\le 8+7(x_{1}+...+x_{k-1})$ on the k-nd step

[quote]If  $x$ represents the [b]final[/b] number of pieces of paper that Angie has, find the [b]sum [/b]of all possible distinct values of $x$.[/quote]
Sorry, I am not sure I have got what you mean.

I guess you have meant the all possible numbers of pieces after some actions. If it is so the answer is evident: they must be $1+7k, \ \ k=1,2,...$ and all this possibilities are achievable.

For some strange reason the answer is 1200375.  :huh:

Well, the problem states that she does it three times.


geometry
trapezoid
AMC
AIME
similar triangles

$ABCD$ is a quadrilateral with $AB = 8, BC = 6, BD = 10$, $\angle{A} = \angle{D}$ and $\angle{ABD} = \angle{C}$. Find $CD$

Draw a line from B to say...X on CD such that BX is parallel to AD forming isoceles trapezoid BZAD. Then AX=BD=10 and <AXD=<ABD=<C. Since <ADX=<BXC (parallel lines), triangles BCX and AXD are similar. Then CX/XD=CB/XA=3/5, so CX=8*3/5=24/5.
Since DX=10, CD=64/5.

....... you got both, have you done these before :huh:  :|  :o   well  cool.

Nope. I haven't solved this problem before, but I remembered part of beta's solution for the problem. I think he actually extended CD and AB to meet, forming an isoceles triangle.

I did the other problem when I was prepping for AIME, but it's been a while.

Extend $BC$ until it meets the line which contains $AD$ at $E$.  Then, with the congruent angles you gave, we have two congruent angles in $ABD$ and $DCE$ and thus those two triangles are similar.  Because angles $BED$ and $BDE$ are congruent, we have $BE=10$.  Then, using similar triangles, $CD=\frac{8*16}{10}=\boxed{\frac{64}{5}}$

[quote="riddler"]Im just curious to know when actually are you allowed to do AIME?[/quote]

Read about it at

http://www.unl.edu/amc


graph theory
combinatorics proposed
combinatorics

Let $n$ and $k$ be positive integers. Each of the edges of the complete graph $K_n$ shall be coloured with one of $k$ colours such that afterwards there is no monochrome circle. Given $k$ what is the maximum $n$.

If I'm not wrong $n_{max} = 2k$, but I would like to know what is folklore in Germany Orlando   ;)

Assume that we may color the edges of $K_n$ with $k$ colors and no cycle.
Then, none of the colors can be used more than $n-1$ times, otherwise we would have at least $n$ edges from one color and the monochrome induced subgraph would have no more than $n$ vertices, so that it would contain a cycle.
Thus, $\frac {n(n-1)} 2 \leq  k(n-1)$ so that $n \leq 2k$ as claimed.

Now we prove that $K_{2k}$ can be colored with $k$ colors with no cycle. It is sufficient to prove that $K_{2k}$ can be decomposed in $k$ spanning trees pairwise edge-disjoint. This will be leads to obvious $k$ classes of colors with no cycle.
Let's label the vertices $A_1, \cdots , A_{2k}$. From now, subscripts are considered modulo $2k$.
For each $i=1, \cdots , k$ let's consider the partition of the set of vertices formed by $E_i = \{A_j / j=2i-1,2i-2, \cdots, 2i-k \}$ and $F_i = \{A_j / j=2i,2i+1, \cdots, 2i+k-1 \}$. Clearly $|E_i| = |F_i| = k.$
Now, we color with color $c_i$ all the edges of the form $A_{2i-1}A_j$ for $A_j \in E_i$, the edges $A_{2i}A_j$ for $A_j \in F_i$ and the edge $A_{2i-1}A_{2i}$.
The induced monochrome subgraph is then a spanning tree of $K_{2k}$ with diameter 3. It is easy to verify that these trees are pairwise edge-disjoint. And we are done.

Pierre.

Just a little remark:
$K_{2k}$ can be partitioned not in trees, but even in chains.
To see this, label the vertices $1,\cdots 2k$ clockwise.
Then for $1\leq i\leq n$ let $C_i$ be the chain $i,i+1,i-1,i+2,i-2,\cdots i+k-1,i-k+1,i+k$.
$C_1,\cdots C_n$ are our partition.


algebra
polynomial
Duke
college
ARML
geometry
3D geometry

I encountered this problem from a regional Duke meet which is quite similar to one of the ARML problems this year, if I recall correctly:

-> Find the sum of the cubes of the roots of

x^10 + x^9 + x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x^1 + 1=0

Please provide an explanation of your solution.  :)

This should be moved to intermediate.

[hide]

multiply the equation by (x-1).[/hide]

I don't think that's an advanced problem, which is suppose to be medium to hard olympiad problems

edit to fix a slight error
Let the roots be [tex]a_1, ...a_{10}[/tex]
SUm of the roots is -1
and [tex]a_1a_2+...a_9a_{10}=1[/tex]
Thus sum of squares of the roots are
[tex](-1)^2-2(1)=-1[/tex]
Let S_1=sum of cubes of roots,[tex] S_2=a_1^2a_2+...a_9^2a_{10}[/tex]
[tex](a_1^2+a_2^2+...a_{10}^2)(a_1+...a_{10})=[/tex]
[tex]S_1+S_2=1[/tex]
[tex](a_1+...a_{10})^3=S_1+3(S_2)+6(a_1a_2a_3+...a_8a_9a_10)[/tex]
[tex]-1=S_1+3S_2-6[/tex]
Solve to get[tex] S_1=-1[/tex]

sorry, I should have posted this in the intermediate forum. (I didn't see the "Olympiad Problems" label".

Alternate solution using Newton's Sums:

We have the following family of equations:

[tex]a_n \cdot S_1 + a_{n-1} = 0[/tex]

[tex]a_n \cdot S_2 + a_{n-1} \cdot S_1 + 2 \cdot a_{n-2} = 0[/tex]

[tex]a_n \cdot S_3 + a_{n-1} \cdot S_2 + a_{n-2} \cdot S_1 + 3 \cdot a_{n-3} = 0[/tex]

...where the [tex]a_k[/tex] denote the coefficients of the polynomial and the [tex]S_k[/tex] denote the sums of the kth powers of the roots.

Since all the coefficients in the given polynomial are 1, we can rewrite the equations as:

[tex]S_1 + 1 = 0[/tex]

[tex]S_2 + S_1 + 2 = 0[/tex]

[tex]S_3 +  S_2 + S_1 + 3 = 0[/tex]

We find successively that [tex]S_1 = -1[/tex], [tex]S_2 = -1[/tex], and [tex]S_3 = -1[/tex].

Thus the sum of the cubes of the roots of the given polynomial is [b]-1[/b].

(I agree that this should be moved to intermediate.)

A third one :
The given polynomial is $P(x)= \frac {x^{11} - 1} {x-1}$ thus its roots are the complex numbers $x_k = e^{ \frac {2ik \pi} {11} }$ for ${k=1,2,...,10}$.
Thus the desired sum is $S = \sum_{k=1}^{10} (e^{ \frac {2ik \pi} {11} } )^3 $.
But :
$1+S= \sum_{k=0}^{10} (e^{ \frac {2ik \pi} {11} } )^3 = \sum_{k=0}^{10} (e^{ \frac {6i \pi} {11} } )^k = 0$ (summing over a geometric serie)
Thus $S = -1$.

Pierre.


function
vector
Functional Analysis
real analysis
real analysis unsolved

$H$ a Hilbert space, $a(.,.): H \times H \to  \mathbb{R}$ a bilinear form satisfying:
1/ there exists $C_1>0$ such that: $|a(u,v)|\leq C_1|u||v|$ for any $u,v \in H$
1/ there exists $C_2>0$ such that: $|a(u,u)|\geq C_2|u|^2$ for any $u \in H$.

Prove that for any continuous function $f: H \to  \mathbb{R}$, there exists a unique vector $u \in H$ such that:
$a(u,v)=f(v)$ for any $v \in H$. Is $v$ depends continuously of $f$ ?

Your Hilbert space is, of course, a real space, and that $f$ actually has to be linear, not merely continuous, right? :). In other words, it should be a linear functional.

For every $x\in H$, the map $y\mapsto a(x,y)$ is a bounded linear functional, so, by the Riesz Representation Theorem, there is a unique $Ax$ such that $a(x,y)=(Ax,y),\ \forall y\in H$. Clearly, the map $A$ sending $x$ to $Ax$ is linear, and since $|(Ax,y)|=|a(x,y)|\le C_1\|x\|\cdot\|y\|,\ \forall x,y\in H$, we get $\|Ax\|^2=(u_x,u_x)\le C_1\|Ax\|\cdot\|x\|,\ \forall x\in H\Rightarrow\|A\|\le C_1$, so $A$ is a bounded linear operator on $H$ satisfying $a(x,y)=(Ax,y),\ \forall x,y\in H$.

For any linear functional $f$, there is a unique $t\in H$ such that $f(v)=(t,v),\ \forall v\in H$ (Riesz's Theorem again), and conversely, to each $t$ there corresponds an $f$. Since $A$ is obviously one-to-one (this follows from the second condition), proving that for any $f$ there is a unique $u$ such that $f(v)=(Au,v),\ \forall v\in H$ is equivalent to proving that $A: H\to H$ is onto. The problem is thus:

Given a bounded linear operator $A: H\to H$ which satisfies $(Ax,x)\ge C\|x\|^2,\ \forall x\in H$ for some $C>0$, prove that $A$ is onto.

On the one hand, since $(*)\ \|Ax\|\ge C\|x\|,\ \forall x$, the image of $A$ is a closed subspace of $H$: if $Ax_n\to y$, then $(Ax_n)_n$ is a Cauchy sequence, and $(*)$ implies that $(x_n)_n$ is also Cauchy. This means that $x_n\to x\Rightarrow Ax_n\to Ax\Rightarrow y=Ax$. On the other hand, if $u\in(\mbox{Im}\ A)^\perp$, then $(Ax,u)=0,\ \forall x\in H\Rightarrow \|u\|^2\le\frac 1C(Au,u)=0\Rightarrow u=0$, so $(\mbox{Im}\ A)^\perp$ is trivial, meaning that $\mbox{Im}\ A$ is dense. Being dense and closed, the image of $A$ must be the entire $H$, and we’re done.

But then
$f(x)+f(y) = a(u,x)+a(u,y) = a(u,x+y) = f(x+y)$. So $f$ - must be linear
 :?:

yes, as Grobber said, I wanted to write "functional". Anyway, nice proof Grobber  :)

Oh, and I forgot to answer the second question. Yes, $u$ depends continuously on $f$, because the map sending $f$ to $t$ such that $f(v)=(t,v),\ \forall v$ is an isometry, $A^{-1}$ is continuous, and the map sending $f$ to $u$ such that $f(v)=(Au,v),\ \forall v$ is the composition of these two.

another approach:
One wants to show that $A$ is surjective. Take a vector $f \in H$. For $r>0$ small enough, the function: $T(u)=u-rA(u)+rf$ is a contraction (easy computation). Banach fixed point theorem allows us to conclude.


The distinct reals $x_1, x_2, ... , x_n$ ,($n > 3$) satisfy $\sum_{i=1}^n x_i = 0$, $\sum_{i=1}^n x_i ^2 = 1$. Show that four of the numbers $a, b, c, d$ must satisfy:
\[ a + b + c + nabc \leq \sum_{i=1}^n x_i ^3 \leq a + b + d + nabd  \].

Does exist someone on this forum who was able do kill this problem?

Let $F(x, y, z) = x + y + z + nxyz$

[i]Lemma:

If there exist two there-element subsets $U = \{x, y, z\}$ and $V = \{t, u, v\}$ of $X = \{x_{1}, x_{2},\ldots,x_{n}\}$ such that

$F(x,y,z) \leq \sum_{i=1}^{n} x_{i}^3 \leq F(t,u,v)$ then there exist four numbers $\{a,b,c,d\}$ as desired.

Proof:

1)  $U \cap V = \emptyset$

Consider 4 numbers

$F(x,y,z) , F(x,y,v), F(x,u,v), F(t,u,v) $

As $F(x,y,z) \leq \sum_{i=1}^{n} x_{i}^3 \leq F(t,u,v)$ , there must exsit two consecutive numbers among the four such that $\sum_{i=1}^{n} x_{i}^3$ belongs to the segment founded by that two numbers. Then we could easily select four numbers satisfying problem conditions

2) $card(U \cap V) = 1$

Without loss of generality we suppose that z = v, then consider the 3 numbers for $F(x,y,z) , F(x,u,z), F(t,u,z) $ .....

3) $card(U \cap V) = 2$

$U \cup V $ gives us 4 desired numbers
[/i]

Back to the main problem, it remains to prove that there exist three-element subsets $U = \{x, y, z\}$ and $V = \{t, u, v\}$ of $X = \{x_{1}, x_{2},\ldots,x_{n}\}$ such that

$F(x,y,z) \leq \sum_{i=1}^{n} x_{i}^3 \leq F(t,u,v)$

Let’s see these things.

Let a be the least of all and b be the largest of all.
Then, let c be the second largest one and let d be the second smallest.
Consider a fact that,
      To any xi,
        (Xi—a)(Xi—b)(Xi—c)<=0
And we can write as,
         Xi^3-(a+b+c)Xi^2+(ab+ac+bc )Xi—abc<=0
If we add all the inequalities( i from 1 to n)
       that’s X1^3+X2^3+……+Xn^3<=a+b+c+n*abc—(ab+bc+ca)*(X1+X2+X3)<=a+b+c+n*abc
Ok , we solve one side,
and the same to the other side(using a,b,d)


integration
real analysis
real analysis unsolved

Let $\phi \in C[0,1]$ and let $T_\phi: C[0,1] \to \mathbb{R}$ given by $T_\phi f = \int_{0}^{1}\hspace{-5pt}f(x)\phi (x) dx$.

I've already proved that $T_\phi$ is a continuous linear functional and that $\parallel T_\phi \parallel \hspace{3pt}\leq \int_{0}^{1}\hspace{-5pt}|\phi (x)|dx$. But I'm asked to prove that equality holds and I couldn't manage... Can someone give me any idea?

Note: $\parallel T_\phi \parallel = \sup_{\parallel f\parallel \leq 1}|T_\phi f|$ where $\parallel f\parallel =\sup_{x\in [0,1]}|f(x)|$.

Let $g(x)=\text{sgn}(\phi(x)).$ That is, $g(x)=\begin{cases}1,&\text{if }\phi(x)>0\\0,&\text{if }\phi(x)=0\\-1,&\text{if }\phi(x)<0\end{cases}.$

We have $T_{\phi}(g)=\int_{0}^{1}|\phi(x)|\,dx.$ Now, if $g$ were continuous, we would be done. But, of course, we may not assume that $g$ is continuous, because $\phi$ may change sign.

The idea: can we find a sequence $f_{n}\in C[0,1]$ with $\|f_{n}\|\le 1$ such that $f_{n}$ tends to $g$ is some sense that you can exploit?


function
inequalities
trigonometry
inequalities proposed

For $ 0<x,\ y,\ z\leq \frac{\pi}{2}$, prove that $ \sin x\plus{}\sin y\plus{}\sin z\plus{}\cos x\plus{}\cos y\plus{}\cos z\geq 3\sin x\sin y\sin z$.

[quote="kunny"]For $ 0 < x,\ y,\ z\leq \frac {\pi}{2}$, prove that $ \sin x \plus{} \sin y \plus{} \sin z \plus{} \cos x \plus{} \cos y \plus{} \cos z\geq 3\sin x\sin y\sin z$.[/quote]

$ \sin x \plus{} \sin y \plus{} \sin z \ge 3\sqrt[3]{{\sin x.\sin y.\sin z}} \ge 3\sin x.\sin y.\sin z$

Oh, sorry, the problem seemed to be easy.  :blush:  My solution isn't same as yours.

I should have posted that $ (\sin x\plus{}\sin y\plus{}\sin z\plus{}\cos x\plus{}\cos y\plus{}\cos z)^2\geq 9\sin x\sin y\sin z\cos x\cos y\cos z$.

[quote="kunny"]Oh, sorry, the problem seemed to be easy.  :blush:  My solution isn't same as yours.

I should have posted that $ (\sin x \plus{} \sin y \plus{} \sin z \plus{} \cos x \plus{} \cos y \plus{} \cos z)^2\geq 9\sin x\sin y\sin z\cos x\cos y\cos z$.[/quote]
this  prob  with  $ 0 \leq x \leq \pi/2$
 It's  easy  ,also is killed   by AM-GM
$ (\sin x \plus{} \sin y \plus{} \sin z \plus{} \cos x \plus{} \cos y \plus{} \cos z)^2 \ge 36 \sqrt [3]{\sin x\sin y\sin z\cos x\cos y\cos z} \ge 9\sin x\sin y\sin z\cos x\cos y\cos z$


Find the number ways to expressing a natural number as the sum of natural numbers less than or equal to it without repetition.
Example: 6 has 4 representations.
                6= 6
                6= 2+4
                6= 1+ 5
                6=  1+2 +3
:)

[size=75]Akash kumar wrote:use falling powers....this is the famous cribbage problem[/size]

To Akash : Would you elaborate a little or lead me to a Url that deals with this problem ?

Thanks and Regards,
Srikanth

A related question: I have n boxes with first box containing k balls, second k+1 and so on up to nth box containing k+n-1 balls. Move balls from i th box to j th box  (with i < j ) such that for any given a >b, a th box contains more balls than b th box. One such movement gives a new arrangement. How many such arrangements are there ?

It is just the number of partitions into distinct parts, [url]http://www.research.att.com/~njas/sequences/A000009[/url].  The second question is unclear to me.  Do you mean, "How many partitions of $ k+(k+1)+\ldots+(k+n-1) = \frac{n}{2}(n+2k-1)$ are there into $ n$ distinct parts?"  If I'm not mistaken, the answer is $ k^{n-1}$.

@JBL:the second problem is the same as the first but from an other opinion:[b]in how many ways can we partition $ n$ similar objects into similar parts s.t the number of members of each part is distinct... :?: [/b]

@ JBL : My second question asks for numbers of distinct partitions of the sum
             k + k+1 + .... + K+n-1
into 'n' distinct parts.
How do you prove it equals to 2^(n-1) ?

I did look at the
	http://www.research.att.com/~njas/sequences/A000009
I had asked this question keeping expansion of product ( 1 + x )^m in mind.But is there a closed form for that sequence ? Is there a combinatorial argument to derive it ?

Regards,
Srikanth


Show that it is possible to color the set of integers \[M=\{ 1, 2, 3, \cdots, 1987 \},\] using four colors, so that no arithmetic progression with $10$ terms has all its members the same color.

We can replace $ 10$ by $ 7$,and consider $ M$ in base $ 7$(this was a well known problem,it must have been discussed somewhere but I just cannot find a link)

I want to solve this question by double counting. It is clear that you can color the set of integers {1,2,..., 1987} is 4^1987 . Let the quantity of arithmetic progressions with 10 terms be A. Now form a table which its rows are all that progressions and its columns are all the conditions of coloring that set. Now write 1 the intersection of the row I and column j if and only if the coloring condition j has just one color for the progression I. Now the sum of the numbers of each row is 4×(4^1977) . Now the sum of table(S) is A×(4^1978). Now the average of sum of the numbers of the table is S÷(4^1978)=(4^-9)×A
Now if we have average < 1 then A< 4^9 and now done.


limit
real analysis
real analysis unsolved

1.  Let f be continuous on R.  Let c be in R with f(x)=c for all x in Q.  Shw that f(x)=c for all x in R.

2.  Let L be a positive real number.  Let (Xn) and (Yn) be sequences of positive real numbers and suppose that Xn/Yn --> L.  If (Yn) --> infinity, then (Xn) --> infinity.

1) Let $x\in\mathbb{R}$. There exist a sequence $\{r_n\}\subset \mathbb{Q}$ such that $r_n\to x$. By continuity of $f$ then we have
\[ f(x)=\lim_{n\to \infty} f(r_n)=\lim_{n\to \infty} c=c  \]

2) Assume $(x_n)$ has a limit. If $x_n\to K$ then
\[ \lim \frac{1}{y_n}=\lim \frac{x_n}{y_n}\cdot x_n= LK  \]
So $\lim y_n=\frac{1}{KL}$ (contradiction!)

That's not a very convincing solution to #2; a sequence that doesn't diverge to $\infty$ doesn't have to converge.

[hide="Hint"]Choose $C>0$, $C<L$. There exists $N$ such that $C<\frac{x_n}{y_n}$ for all $n>N$.[/hide]


In soale male khodame , soale bahalie :
hameye tavabe'e $ f:\mathbb{R}\to\mathbb{R}$ ro peyda konid ke baraye har $ x,y$ haghighi dashte bashim :
$ f(x^{2}\plus{}y^{2})\equal{}(f(x))^{2}\plus{}(f(y))^{2}$

motmaenni in ghabele halle?
bebin age gharar bedi $ y \equal{} 0$ natije mishe:

$ f(x^{2}) \equal{} f(x)^{2}\plus{}f(0)^{2}$ pas $ f(x)^{2}\equal{} f(x^{2})\minus{}f(0)^{2}$

$ \Rightarrow f(x^{2}\plus{}y^{2}) \equal{} f(x^{2})\plus{}f(y)^{2}\minus{}2f(0)^{2}(*)$

khob hala gharar bede $ g(x) \equal{} f(x^{2})\minus{}2f(0)^{2}$ pas avvalan $ g:\mathbb{R^{\plus{}}}\cup\{0\}\to\mathbb{R}$ hamchenin ba tavajoh be rabete bala darim:

$ g(x\plus{}y) \equal{} g(x)\plus{}g(y)$

khob hala bedoone sharte peivastegi nemishe ino hal kard,mishe?  :huh:
yeni man kolli javabe joorvajoor vasat miaram ke too in sedgh kone va dar natije too moadele asli...
pas ya soorate soale to naghese ya inke harfe man ghalate...  :maybe:

pas haddeaghal bayad sharte peivastegi ro bezari oonmoghe chon $ f$ peivastas pas $ g$ peivastas,yeni $ g(x) \equal{} kx$ pas $ f(x^{2}) \equal{} kx\plus{}2f(0)^{2}$

hamchenin deghat kon ke age jaye $ x$ ro ba $ \minus{}x$ avaz konim natije mishe ke $ f(\minus{}x) \equal{}\pm f(x)$...

hala ba gharar dadane $ kx\plus{}2f(0)^{2}$ bejaye $ f(x^{2})$ too rabete $ (*)$ darim:

$ k(x\plus{}y)\plus{}2f(0)^{2}\equal{} kx\plus{}2f(0)^{2}\plus{}ky\plus{}2f(0)^{2}\minus{}2f(0)^{2}$

ke hamishe bargharare,pas fek konam laaghal bayad sharte peivastegio bezari...

[quote="lasha"]
We see, that it's cauchy equation and it's well known, that $ f(x) \equal{} xf(1)$[/quote]
notice that in cauchy equation,$ f$ must be continuous...
so here you cant say that $ f(x)\equal{}xf(1)$...  :)

ye shartesh yadam raft farz konin $ f(0)\equal{}0$ , hala ghabele halle :)

[quote="khashi70"]ye shartesh yadam raft farz konin $ f(0) \equal{} 0$ , hala ghabele halle :)[/quote]
nist,bazam sharte peivastegi lazeme...

Khob hamishe ke tavabe'e Cauchy ba peyvastegi hal nemishan masalan age sabet konin tabe yeknava hastesh ham hokm sabet mishe , too in masale ham sabet konin tabe soo'oodie :)

gharar bedid $ y \equal{}\epsilon$ ke $ \epsilon\to 0$ oon moghe darim:
$ f(x^{2}\plus{}\epsilon) \equal{} f(x^{2})\plus{}f(\epsilon)^{2}\geq f(x^{2})$
$ \Rightarrow f(x^{2}\plus{}\epsilon)\geq f(x^{2})$
faghat to ye zahmati bekesh va rahe halle caushy ro baraye tavabe yeknava benevis age mishe ke rahe halle in masale kamel beshe...

avvalan too rahe hal tavvajoh konin ke shoma tabe ro faghat tooye adade mosbat mitoonin cauchy bezanin pas felan too adade mosbat va gooya darim : $ f(x) \equal{} cx$ . hala shoma adade gonge $ y$ ro dar nazar begirin dar in soorat ye donbale az adade gooya mesle $ a_{n}$ voojood dare ke $ a_{n}\longrightarrow y$ va ye donbale az adade gooya mesle $ b_{n}$ voojood dare ke $ y\longleftarrow b_{n}$ . hala ba tavvajoh be soo'oodi boodane tabe darim :
$ f(a_{i})\geq f(y)\geq f(b_{j})$ hala chon $ f(a_{i}) \equal{} ca_{i}$ va $ f(b_{j}) \equal{} cb_{j}$ va mishe $ a_{i}$ va $ b_{j}$ ro ta hadde delkhah be $ y$ nazdik kard pas lazeme ke $ f(y) \equal{} cy$ .
hala tabe rooye adade mosbat shode $ f(x) \equal{} cx$ . hala age gharar bedin be dast miad $ c \equal{} 0$ ya $ c \equal{} 1$ pas tabe rooye adade mosbat be shekle $ f(x) \equal{} 0$ , ya $ f(x) \equal{} x$ hast .
age $ f(x) \equal{} 0$ az oonjayi ke $ f(x) \equal{}\pm f(\minus{}x)$ pas $ f(\minus{}x) \equal{} 0$ yani tabe rooye adade manfiam sabete sefre .
amma age $ f(x) \equal{} x$ oon moghe baraye har $ x$ e manfi darim $ f(x) \equal{} x$ ya $ f(x) \equal{}\minus{}x$ . be sadegiam mishe barresi kard ke sedgh mikone chon tooye samte rast hatman $ f(x^{2}\plus{}y^{2})$ mosbate va dar tarafe dovvom ham $ f(x)^{2}$ age $ x$ manfi bashe bazam chon be tavane $ 2$ mirese $ x$ ya $ \minus{}x$ hich farghi nemikonan , omidvaram manzooramo fahmide bashin , baraye inke roshan tar she ye mesal mizanam , masalan momkene baraye adade $ 1,\minus{}1,2,\minus{}2$ maghadire tabeye $ f$ injoori beshe : $ f(1) \equal{} 1 , f(\minus{}1) \equal{} 1 , f(2) \equal{} 2 , f(\minus{}2) \equal{}\minus{}2$
baraye ghesmate avval halate noozooli boodanam daghighan hamin sheklie faghat jahate namosavia barmigarde :)
hala babak ye soali oonam inke age man sharte $ f(0)\equal{}0$ ro nemidadam yani too oon yeki halat ke $ f(0)\equal{}\frac{1}{2}$ ham tabdil be cauchy mishod ? age beshe khob mishe in shartam vardasht !

hala ye mesale khoobo sade ke ba hamin soo'oodi boodane hal mishe ro hal konin :
hameye tavabe'e $ f:\mathbb{R}\to\mathbb{R}$ ro peyda konid ke baraye har $ x,y$ haghighi dashte bashim :
$ 1) f(x)\plus{}f(y) \equal{} f(x\plus{}y)$
$ 2)$ agar $ x > 0$ oon moghe $ f(x)\geq 0$

[quote="khashi70"]
hala babak ye soali oonam inke age man sharte $ f(0) \equal{} 0$ ro nemidadam yani too oon yeki halat ke $ f(0) \equal{}\frac{1}{2}$ ham tabdil be cauchy mishod ? age beshe khob mishe in shartam vardasht !
[/quote]
are fek konam,poste avvale mano bekhoon,age ghaar bedi $ g(x^{2})\equal{}f(x^{2})\minus{}2f(0)^{2}$ oon moghe bedast miad:
$ g(x^{2}\plus{}y^{2})\equal{}g(x^{2})\plus{}g(y^{2})$ amma chon $ f$ soudie pas $ g$ ham soudie,pas $ g(x)\equal{}kx$ baraye $ x$haye mosbat ya be ebarati:
$ f(x)\equal{}kx\plus{}2f(0)^{2}$ baraye $ x$haye mosbat ke age jagozari konim be dast miad:
$ f(x^{2}\plus{}y^{2})\equal{}f(x)^{2}\plus{}f(y)^{2}\equal{}f(x^{2})\plus{}f(y^{2})\minus{}2f(0)^{2}$
$ \Rightarrow k(x^{2}\plus{}y^{2})\plus{}2f(0)^{2}\equal{}kx^{2}\plus{}2f(0)^{2}\plus{}ky^{2}\plus{}2f(0)^{2}\minus{}2f(0)^{2}$ ke inam hamishe bargharare,vase manfiasham ke mese halatie ke khodet gofti...
yeni too har halati mishe az caushy estefade kard...

[quote="khashi70"]
hameye tavabe'e $ f:\mathbb{R}\to\mathbb{R}$ ro peyda konid ke baraye har $ x,y$ haghighi dashte bashim :
$ 1) f(x)\plus{}f(y) \equal{} f(x\plus{}y)$
$ 2)$ agar $ x > 0$ oon moghe $ f(x)\geq 0$[/quote]
soudi boodane inam mese ghablie sabet mishe dge...
[hide]
gharar bedi $ y \equal{}\epsilon\to 0^{\plus{}}$ pas avvalan $ f(\epsilon)\geq 0$ pas darim:
$ f(x\plus{}\epsilon) \equal{} f(x)\plus{}f(\epsilon)\geq f(x)$ pas $ f(x\plus{}\epsilon)\geq f(x)$...[/hide]

Ye rahe asoone Dge baraye esbate soo'oodi boodanesh ba tavvajoh be inke $ f(x+y)=f(x)+f(y)$ pas :
$ f(y+(x-y))=f(y)+f(x-y)$ , hala farz konid ke $ x>y$ ba tavajoh be sharte masale natije mishe :
$ f(x)=f(y)+f(x-y)$ amma $ x-y>0$ pas $ f(x-y)\geq0$ va dar natije $ f(x)\geqq(y)$ . ke sabet mishe soodie :)
(albatte fekr konam ta hadde kheyli ziadi hamoon rahe babake :D)

Khob inam ye soale tabeyie Dge marboot be cauchy mishe in yekiam :
Hameye tavabe'e $ f:\mathbb{R}\to\mathbb{R}$ ro peyda konid ke baraye har $ x,y$ e haghighi dashte bashim :
$ f(x^{3}\plus{}y^{3})\equal{}x^{2}f(x)\plus{}yf(y^{2})$ .

[quote="khashi70"]Hameye tavabe'e $ f:\mathbb{R}\to\mathbb{R}$ ro peyda konid ke baraye har $ x,y$ e haghighi dashte bashim :
$ f(x^{3}\plus{}y^{3}) \equal{} x^{2}f(x)\plus{}yf(y^{2})$ .[/quote]
bedoone cauchy ham mishe hallesh kard:

gharar bedid $ x \equal{} y \equal{} 0$ natije mishe:

$ f(0) \equal{} 0$

hala gharar bedid $ y \equal{} 0$ pas $ f(x^{3}) \equal{} x^{2}f(x)$

hala gharar bedid $ x \equal{} 0$ pas $ f(y^{3}) \equal{} yf(y^{2})$

az do tasavie bala natije mishe:

$ f(x^{3}) \equal{} x^{2}f(x) \equal{} xf(x^{2})$ (I)

pas ba jagozari dar rabete asli be dast miad:

$ f(x^{3}\plus{}y^{3}) \equal{} f(x^{3})\plus{}f(y^{3})$

yeni baraye har $ x,y$e haghighi $ f(x\plus{}y) \equal{} f(x)\plus{}f(y)$ (*)

(az inja be bado mishe az caushy estefade kard vali age nakhaim estefade konim intori mishe)

khob avval az hame az (I) natije mishe age $ x\not \equal{} 0$ oon moghe $ f(x^{2}) \equal{} xf(x)$

yeni $ f((x\plus{}1)^{2}) \equal{} (x\plus{}1)f(x\plus{}1)$ faghat bayad havasemoon bashe $ x\plus{}1\not \equal{} 0$ yeni $ x\not \equal{}\minus{}1$

amma tebghe rabete (*) darim $ f(x\plus{}1) \equal{} f(x)\plus{}f(1)$ pas tasavie akhar be shekle zir dar miad:

$ f((x\plus{}1)^{2}) \equal{} (x\plus{}1)(f(x)\plus{}f(1)) \equal{} (x\plus{}1)f(x)\plus{}f(1)\plus{}xf(1)$ (1)

amma az tarafi:

$ f((x\plus{}1)^{2}) \equal{} f(x^{2}\plus{}2x\plus{}1) \equal{} f(x^{2})\plus{}2f(x)\plus{}f(1) \equal{} xf(x)\plus{}2f(x)\plus{}f(1)$

$ \equal{} (x\plus{}1)f(x)\plus{}f(x)\plus{}f(1)$ (2)

hala az (1),(2) natije mishe:

$ f(x) \equal{} xf(1)$ hala gharar bedid $ f(1) \equal{} c$ pas $ f(x) \equal{} cx$ be ezae $ x\not \equal{} 0,\minus{}1$ khob $ x \equal{} 0$ ke avvale esbat barresi shod ke rabete $ f(x) \equal{} cx$ vasash sadeghe,hamcheni vase $ x \equal{}\minus{}1$ darim:
age bejaye $ y$ too (*) gharar bedim $ \minus{}x$ natije mishe:

$ f(0) \equal{} f(x)\plus{}f(\minus{}x)$

pas $ f(\minus{}x) \equal{}\minus{}f(x)$ yeni $ f(\minus{}1) \equal{}\minus{}f(1) \equal{}\minus{}c \equal{} (\minus{}1)c$ pas baraye har $ x\in\mathbb{R}$ darim:

$ f(x) \equal{} cx$


A clock advances to 3 minutes per day. If it is hit at noon, calculate how many days elapsed for it to be right again.

Portuguese question:

Um relógio adianta-se 3 minutos por dia. Se ele for acertado ao meio-dia, calcule quantos dias transcorrerão para que ele esteja certo novamente.

ANSWER: 20 days

How to do ?  :D

The answer looks wrong. After exactly $ 20$ days, when a true clock shows noon again, our clock would have advanced $ 20\cdot 3 \equal{} 60$ minutes, so it shows exactly one PM. The period for a clock to show the true time is $ 12$ hours (for an analogue clock - for a digital one is $ 24$ hours). So our clock will need $ 12\cdot 20 \equal{} 240$ days to advance $ 12$ hours and thus be right again.

I  got a different answer than mavropnevma.
[hide="Solution"]
If every day the clock gains 3 minutes in order for the clock to be correct again we need to find $ \frac{720}{3}$ because 720 is half of the number of minutes in a day. We know that  $ \frac{720}{3} \equal{} 240$ so it would take 240 days for the clock to be correct again.[/hide]

Interested, yeah, i think the answer is wrong too. Thank you very much ;)


abstract algebra
group theory
linear algebra
matrix
superior algebra
superior algebra solved

if |G:Z(G)| show this can't be prime for a group G.

Show that if $G/Z(G)$ is cyclic, then the group is abelian, and the result follows from here.

Suppose $[G:Z(G)]=p$ is a prime. Notice that $Z(G)$ is a subgroup of the centralizer of $x\in G$, $[G:Z(G)]=[G:C(x)][C(x):Z(G)]$, where $C(x)$ is the centralizer of $x$. Thus $[G:C(x)]=1,p$ and $[C(x):Z(G)]=1,p$. If $[C(x):Z(G)]=1$ then $C(x)=Z(G)$. So $x\in Z(G)$ and $gx=xg, \forall g\in G$. It follows that $C(x)=G$ and $Z(G)=G$, which is a contradiction with $[G:Z(G)]=p>1$. Hence $[C(x):Z(G)]=p$ for all $x$. Thus $[G:C(x)]=1$ for all $x$, which implies that $G$ is abelian, thus $Z(G)=G$. We meet a contradiction again.

WOW thanks for hte help. Along the same lines i have another question.

Show G=D4 (dihedral group) is a non-ableian group G such that G/Z(G) is abelian???

I am so lost in this class, do you guys know any good tutorials or anything.....????

Thanks
JAson

One way to exhibit the dihedral group is to consider it to be the set of matrices over $\mathbb{Z}_4$ that are of the form:

$\begin{bmatrix}1&x\\0&1\end{bmatrix}$ or $\begin{bmatrix}1&y\\0&-1\end{bmatrix}$

By direct examination, we find that the center of the group is

$\left\{\begin{bmatrix}1&0\\0&1\end{bmatrix}, \begin{bmatrix}1&2\\0&1\end{bmatrix}\right\}.$

So the order of the center is two. Therefore, the order of the quotient group is 4, and all groups of order 4 are abelian. In this case, it happens to be the Klein 4-group.

It is a special case of the more general statement: for every group $G$ with $|G|=p^3$ and $p$ prime $G/Z(G)$ is abelian.

Proof. If $G$ is abelian $G=Z(G)$ and the statement is trivial. If $G$ is non-abelian then by class equation we must have $|Z(G)|\in\{p,p^2\}$. But $|Z(G)|=p^2$ is impossible since because of the first problem posted here it would follow that $G$ is abelian and $|Z(G)|=p^3$. In the case $|Z(G)|=p$ we get $|G/Z(G)|=p^2$ and each group of order $p^2$ is either isomorphic to $\Bbb Z_p\times\Bbb Z_p$ or $\Bbb Z_{p^2}$ which are both clearly abelian.

Misha


Ali, Bonnie, Carlo and Dianna are going to drive together to
a nearby theme park.  The car they are using has four seats: one
driver's seat, one front passenger seat and two different back seats. Bonnie
and Carlo are the only two who can drive the car. How many
possible seating arrangements are there?

Either Bonnie or Carlo can drive the car, and there are $ 3!$ ways to put everybody else. Thus our answer is $ 2\cdot3!\equal{}2\cdot6\equal{}\boxed{12}$.


geometry
projective geometry
superior algebra
superior algebra unsolved

Hello, sorry if this is the wrong place for projective geometry questions
suppose $Q$ is a quadric in $PG(n,q)$, nonsingular,$n\geq 1$ let us say

let g be the maximal dimension of  subspace on the quadric

prove that there are two subspaces on the quadric with dimension g that contain no common point

It appears as a seemingly simple antidegeneration axiom in the definition of polar spaces, and I am trying to prove you can make a polar spaces with a nonsingular quadric, but for some reason I cannot see it

So uhm has anyone tried this problem
it seems so trivial yet it doesn't work
Is this the right place for geometry?  It is not really algebra


Bottle A contains more soda than Bottle B. Pour from Bottle A into Bottle B as much soda as B already contains. Now pour from
B into A as much soda as A now contains. Then pour from A into
B as much soda as B now contains. Each bottle now contains 64
ounces. How many more ounces of soda were in Bottle A than
Bottle B at the beginning?

Work backwards:

$ 64$ and $ 64$

Half of what $ B$ has was poured from $ A$, so we have:

$ 96$ and $ 32$

Half of what $ A$ has was poured from $ B$, so we have:

$ 48$ and $ 80$

Finally, half of what $ B$ has was poured from $ A$, so we have:

$ 88$ and $ 40$

The answer, thus, is $ 88\minus{}40$, or $ \boxed{48}$.


Έστω $a,b,c>0$ τέτοιοι ώστε $abc=1$  .Δείξτε ότι
$\frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}\geq \frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}$

Θα την χαρακτηριζα μέτριας δυσκολίας..... :)

ωραια....
Θετουμε $a+b+c=x ,ab+ac+bc=y$
οποτε  μετα απο πραξεις παιρνουμε παιρνουμε...
(3+4x+y+x^2)/(2x+y+x^2+xy)<=(12+4x+y)/(9+4x+2y)
Που δινει
$(3x^{2}y-5x^{2}-12x)+(xy^{2}-y^{2}-3x-3y)+(6xy-9x-27)\geq 0$
που ισχυει ,αφου y,x>=3

ΡΑΛΛΗΣ :D

Ωραία Ράλλη ......Νομίζω αυτή είναι η πιο σύντομη λύση.....


Let $a,b,c$ be positive real numbers

Prove that

$a^2 + b^2 + c^2 + 2abc + 3 \geq (1+a)(1+b)(1+c)$

It's a particular case of Turkevici's ineq  ;)

Yes, indeed. I saw it for the first time in RMO shortlist in 2003 proposed by Gil. ;)

because it is posted too many times.

$a^2+b^2+c^2+2abc+1 \ge \sum ab + \sum a$

$a^2+b^2+c^2+2abc+1 \ge a^2+b^2+c^2+ 3(abc)^{2/3} \ge 2(ab+bc+ca)$ schur
$\sum a^2 + 3 = \sum (a^2+1) \ge 2 \sum a$
Sum up and done.

Sorry guys but I totally don't understand at all what Soar did!
  Probably I am dump!
 But the inequality is equivelent to:
If $a,b,c \geq0$ then
$a^2+b^2+c^2+abc+2 \geq (a+b+c)+(ab+bc+ca)$ (1)
  I can only prove that  (1) is corrected in case one of $a,b,c$ is in the interval $[-1..3]$
And what is Turkevici's ineq? Can anyone give me a link?

OOps.... I am sorry but I found the sols in case $a,b,c \geq3$
Ok, I will prove that :
If $a,b,c \geq -1$ then
$a^2+b^2+c^2+abc+2 \geq (a+b+c)+(ab+bc+ca)$ (1)
Suppose $a \geq b \geq c$
Case 1: If $-1 \leq c \leq 3$ we have :
$V(a,b,c)=a^2+b^2+c^2+abc+2 - (a+b+c)-(ab+bc+ca)$
-->$V(a,b,c)=(a-\frac12+ \frac{bc}{2}-\frac{b+c}{2})^2+ \frac14(c+1)(3-c)(b-1)^2+(c-1)^2$
Done !
Case 2: If $c\geq 3$ because $a \geq b \geq c$  so $a \geq b \geq c \geq 3$
We have $a^2+b^2+c^2 \geq ab+bc+ca$
And $abc+2 >a+b+c$ (It is easy to prove because $a \geq b \geq c \geq 3$we can denote $a=3+x$ $b=3+y$$c=3+z$ and $x,y,z \geq 0$ Expand them and you will see
Right :o


AMC
USA(J)MO
USAMO
calculus
invariant
pigeonhole principle
MATHCOUNTS

I think i got into USAMO (AMC:139.5/AIME:not so sure... about 8~11)
But this is my first time taking the actual test
I printed out the questions from the past,
but it seems i just can't solve any of them(except for few odd ones) :oops:
Could please anyone help how i should study for it?

I'm also very inexperienced with this stuff. I looked at past problems and I don't think I can get a single one...Are they really all solvable with pre-calculus and under knowledge?

[quote="Hamster1800"]How long have you been working on each problem? To get anything at all out of them, you have to work on them for half an hour, but that's a bare minimum. To really understand the problem, you should work on it for an hour or more. Only once you've tried absolutely everything you can think of should you read the solution. This way you will understand the solution better - how they came up with it and how you can come up with similar solutions in the future.[/quote]
That's very tedious...an hour for a problem...I don't think I've ever taken that long on a problem that I had a chance of solving.

[quote="Hamster1800"]How long have you been working on each problem? To get anything at all out of them, you have to work on them for half an hour, but that's a bare minimum. To really understand the problem, you should work on it for an hour or more. Only once you've tried absolutely everything you can think of should you read the solution. This way you will understand the solution better - how they came up with it and how you can come up with similar solutions in the future.[/quote]
I agree.  Even though I am pretty n00b at Olympiad problem-solving, I've found that the problems I do solve usually come after 45 to 90 minutes of repeated bashing and hopelessly trying new techniques.

[quote="ProtestanT"]
That's very tedious...an hour for a problem...I don't think I've ever taken that long on a problem that I had a chance of solving.[/quote]

On the actual USAMO, you have 90 minutes for each problem. However, unless you can actually solve 3/6 difficulty problems, you will probably be spending 2 hours on the first two. Practicing for this is valuable.

Speaking from personal experience, I actually have taken an actual USAMO where I didn't know how to do any of the problems but managed to get a solution after 1.5 hours of work. (This was my freshman year btw)

People straight out of MATHCOUNTS/AMC/AIME sometimes think that if they can't get the solution in the first 30 minutes, then they will never get it ... but it's not true. Especially #1 and #4 are designed so that you can keep experimenting around and looking for patterns if you aren't really sure how to proceed.

Also, food helps. We always buy a huge amount of food for every USAMO, hehe

I'm pretty much in the same position... I've found that about 75% of the time or so I can solve the first on a page fairly quickly (10 minutes to "solve" and another 30 or so to put in writable form) and then I can get the second on a page about 25% or less of the time after considerable work. Thirds are just beyond me.

I bought Arthur Engel's book though to help :) (Problem-Solving Strategies)... so we'll see how it goes. Thing is the only way I can get into MOP is through Black though... so yeah :/

Dang you are from IL@ Who are you (you can PM that)? Also, don't expect any of the techniques you learn from Engel to directly help you on the USAMO. If you can do that many problems, you should focus your practice on doing past USAMOs.

Alright, I pm'd you.

As for Engel, I know for a fact that invariant and pigeonhole stuff he has is pretty useful (although he got a bit too detailed into Ramsey numbers in my opinion which aren't relevant at all). Also the NT section is particularly good (which is good since I need a lot of work there :P). I did find the geo one a bit disappointing, admittedly. Overall though I don't think it was a waste.

hey, two IL people. pm me while you're at it :D
Agreed with the Mathcounts comment above. That's me right there -scowl-

Hm but you will understand the techniques much better if you come up with them through working on problems rather than reading them in a book. Some people find books useful though, although personally I don't feel like I've benefited very much from the few I've read, except maybe Geometry Revisited. That could be because I was fortunate enough to know people that already knew a lot of the knowledge base for Olympiads, but like Alex said if you can work through an entire book you might as well do past USAMOs. I'm a bit more perseverant than others, but I often spend over a day on a problem before I give up. Definitely spend at least an hour trying to solve a problem.

I don't really know what general level you're at, but if you can get 8 on the AIME then you should be able to solve a fair amount of USAMO #1's if you actually put in the required thought.

I've found that USAMTS problems are really helpful for the sort of "bash it for a while to get a solution" training as while they are not incredibly difficult problems, they take a while to get in my cases. I'm really glad I did it this year.

Edit: I generally average about 10 on the AIME but can very rarely solve USAMO problems of any caliber. I've been working my way up through the older ones with easier problems. By the way, how do people think the difficulty on the old ones (pre-'96) compares to the current one? Is an old 3 about equal to a current 1?

hmm im not studying.

im thinking more of it as wasting 9 hours skipping school.

since even if i study, i wont improve that many points (if any at all), so ill just hope i get some easy question i can solve.

[quote="JSteinhardt"]Hm but you will understand the techniques much better if you come up with them through working on problems rather than reading them in a book.[/quote]

That's true but have you actually taken a look at Engel? I'd say its approximately 20/80 in text/problems (the latter including solutions)... So it's actually pretty useful in that respect. The problem is that if you can't do 3/6-level problems you can't do them. Simple as that. You have to work your way up via easier problems which is where a book comes in handy I think.

@Firecricket: Thanks for making it easier for me! :D

Both Alex and Jacob have definitely studied Engel and know the style of PSS. Even so, they believe that solving problems from past exams is the best way to go. I think so too, as you'll notice that Engel's problems are not really the type you see on the USAMO.

Food is the answer to all your problems. (not kidding) I just took a olympiad style test and I would have been way off track without food. I don't know why but if you stop every once in a while to grab a handful or crackers or trailmix or something like that you can concentrate for a longer period of time.  Just make sure that its not candy or anything too sugary. :) I hope that food can help you as well. :D

[quote=Pks]I will be giving RMO this year,which is just a bit lower in difficulty compared to USAMO.Could anyone give me advice on how to stay concentrated on a problem for hours?:)[/quote]

Hi there- do check out the India forum, where you'll find loads of discussion on this! As for RMO being USAMO level... it's more of between AMC and AIME, truthfully. USAMO level is probably our TSTS...

Is the stuff about USAMO 1/4 being solvable still true?

[quote=ericlee7979]
Could please anyone help how i should study for it?[/quote]

Is the answer different now that the USAMO is seemingly harder?

[quote=worthawholebean]Do problems. Lots of problems. Work your way up to USAMO level - do contests like the Canadian NMO and old USAMOs. Read Zeitz' book (The Art and Craft of Problem Solving).

At least, that's what I've been doing.[/quote]

Any other good contests? What about the CMO?


suppose we construct an infinite network on the cartesian plane by joining every pair of adjacent lattice points (like (1,1) and (1,2), (0,0) and (1,0), but not (1,1) and (2,2))with a one ohm resistor such that it forms an infinite square grid. so now the question is: what is the resistance between two adjacent lattice point? what is the electrical resistance between two random points?

[url=http://www.geocities.com/frooha/grid/node2.html]this[/url] might help :wink:


Does anyone know how to do this one? I can't figure out how one would find an exact solution to this...

[hide="try"] $x=\pi$
it was the odd+even=odd that made me think -1+0=-1[/hide]


[color=darkblue]Let $ ABC$ is a triangle, known that $ \frac {cosA}{3} \plus{} \frac {cosB}{4} \plus{} \frac {cosC}{5} \equal{} \frac {5}{12}$. Prove that $ ABC$ is right triangle.[/color]

Let's prove the ineq: $ 2yz.cosA + 2zx.cosB + 2xy.cosC\leq x^2 + y^2 + z^2 (1)$  where $ x,y,z$ are real numbers.
We proceed from ineq $ (z - x.cosB - y.cosA)^2 + (x.sinB - y.sinA)^2\geq 0 (2)$
(1) was proved by (2), the eq in (1) occurs when $ \frac {x}{sinA} = \frac {y}{sinB} = \frac {z}{sinC}$
Apply (1) with $ x = 4,y = 5,z = 3$ which gives us $ 40cosA + 30cosB + 24cosC\leq 50$
$ \Rightarrow \frac {cosA}{3} + \frac {cosB}{4} + \frac {cosC}{5}\leq \frac {5}{12} (3)$
The eq in (3) occurs when $ \frac {sinA}{3} = \frac {sinB}{4} = \frac {sinC}{5}%Error. "Rightlefttarrow" is a bad command.
\frac {a}{3} = \frac {b}{4} = \frac {c}{5}$
Hence, triangle ABC is right.

[color=darkblue]Thank you very much.[/color]

Prove ineq $ 2yz.cosA \plus{} 2zx.cosB \plus{} 2xy.cosC\leq x^2 \plus{} y^2 \plus{} z^2$
see at: http://www.mathlinks.ro/viewtopic.php?t=206088


If $ x$ and $ y$ are reals so that $ x \in [2,3]$ and $ x\minus{}5y\plus{}2\equal{}0$, then prove that: $ \sqrt{(x\plus{}2)^2\plus{}2y^2}\plus{}\sqrt{(x\minus{}3)^2\plus{}2(y\minus{}1)^2}$ is constant.

hello, setting $ y\equal{}\frac{x\plus{}2}{5}$ in $ \sqrt{(x\plus{}2)^2\plus{}2y^2}\plus{}\sqrt{(x\minus{}3)^2\plus{}2(y\minus{}1)^2}$ we get
$ \frac{3}{5}\left(\sqrt{(x\plus{}2)^2}\plus{}\sqrt{(x\minus{}3)^2}\right)$ and for $ x$ with $ 2\le x \le3$ i have got
$ 3\sqrt{3}$ and this is really constant.
Sonnhard.


Hi,

I don't know if I have already asked this question in this forum before but I think it's a nice problem...

Show that for every [tex]q \in \mathbb{Q}^+[/tex] exist four positiv integers [tex]a,b,c,d[/tex] such that:

[tex]q=\frac{a^3+b^3}{c^3+d^3}[/tex]

Just for future: http://www.mathlinks.ro/Forum/viewtopic.php?t=19763


real analysis
calculus
calculus computations

Dear all,

When solving the exercise of Real Analysis, I was ask a question: "When is a product of two numbers positive?" Can anyone help me on how to explain this?

Thank you.

Is that the whole question?  If $ a,b\in\mathbb{R}$, then $ ab>0$ if both have the same sign.

I consider it like this:
Since multiplication is just addition, multiplying two numbers of opposite signs is like adding a positive amount of negative numbers.
Eg. $ (5)(\minus{}3) \equal{} \minus{}3\minus{}3\minus{}3\minus{}3\minus{}3 \equal{} \minus{}15$
$ (\minus{}5)(\minus{}3) \equal{} (\minus{}1)(\minus{}1)[3\plus{}3\plus{}3\plus{}3\plus{}3] \equal{} (\minus{}(\minus{}1))[15] \equal{} 15$
$ (5)(3) \equal{} 3\plus{}3\plus{}3\plus{}3\plus{}3 \equal{} 15$.

With basic properties of complex number manipulation, this would be easy to rationalize.

Unfortunately, that doesn't constitute a proof.  To successfully prove it, you need to show that $ \minus{}1\cdot \minus{}1\equal{}1$.

what about assuming the opposite?

If $ a$ and $ b$ have opposite sign, then $ ab < 0$. contradiction

Take a look at this thread.

http://www.mathlinks.ro/viewtopic.php?t=210726

What is the set of axioms that you've used to define $ \mathbb{R}$? As pointed out in the thread above, you'll have to use those axioms for proving the claim.

I am still at the very basic level. I havent learn Axioms or the lower bound stuff. Guess that is a bit advanced eh :roll: . The actual question is as follows:

Fanf all number x for which (x-1)(x-3)>0 (When is a product of two numbers positive?)

I had solve for x. Now is just how to answer this question (When is a product of two numbers positive?)! Its really confusing.

Well, this question doesn't look like a question to me. More of a hint.

(x-1)(x-3)>0 (Hint: when is a product of two numbers positive?)

The hint in the problem helps you solve the problem: hey, iff the numbers have the same sign, their product is positive, so it must be either x-1>0 and x-3>0 or x-1<0 and x-3<0).

No real need for answering the question, IMHO.

Hi. I solve this question by drafting a graph. Then from the question (x-1)(x-3)>0, from the graph, it is obvious the answer is x<1 and x>3.

So, hsiljak; are you sure the answer is x-1>0 and x-3>0 or x-1<0 and x-3<0? I am getting confused now.

Don't worry, your answer is correct, as well as mine (I didn't give my answer in the previous post, just the crucial step in the solution).

The question in the problem: when is the product of two reals positive?

Answer: when those reals have the same sign.

So, if a=x-1, y=x-3, ab>0 if

1. case: a>0, b>0

2. case: a<0, b<0

Now, it's fairly easy:

1. x-1>0, x-3>0, they're both positive if x>3
2. x-1<0, x-3<0, they're both negative if x<1

There you go.

Thank you, hsiljak. You clear up my confusion and doubt of the question.

[quote="JRav"]Unfortunately, that doesn't constitute a proof.  To successfully prove it, you need to show that $ \minus{} 1\cdot \minus{} 1 \equal{} 1$.[/quote]

I did show... it is the same as subtracting $ \minus{} 1$.... by the multiplication/addition style I think of it.

Don't have to prove subtracting negative gives positive. The axiom it would represent is just a logical idea. If you take away 4 cold degrees, you get 4 degrees warmer...

But I am not going about rigorously proving these products. I am just showing how it works.


Six distinct positive integers are randomly chosen between 1 and 2006 inclusive. What is the probability that some pair of these integers has a difference that is a multiple of 5?

The probability is 1. Each number can be either $ \equiv (1, 2, 3, 4, 5) (mod 5)$. When any two numbers are $ \equiv x (mod 5)$ their difference is a multiple of 5. Since there are 6 numbers, there must be at least 2 numbers that are $ \equiv x (mod 5)$ thus some pair must have a difference that is a multiple of 5.

[hide="Pigeonhole"] There are five possible values a number can have in mod 5. From Pigeonhole, we find that atleast two of the six numbers are the same in mod 5, and hence, their difference is divisible by 5. Our answer is 1.[/hide]


Assign eight numbers which sum to one to the vertices of a cube. Furthermore, assign to each edge the product of the numbers at its vertices. What is the maximal value the sum of the numbers assigned to the edge can take?

Nobody got an idea? Well here's a hint:

[hide="Hint"]The answer is $ \frac{1}{4}\equal{}\left(\frac{\sum x_i}{2}\right)^2.$[/hide]


omg...cant put sites, can someone tell me how to put sites on here

Put the desired url in between the "[url]"and"[/url]" things. So, something that looks like this:

[code][url]www.artofproblemsolving.com[/url][/code]

Will look like

[url]http://www.artofproblemsolving.com[/url]


Dear Vasc and Arqady

The following problem is dedicated for you, whom I'm very admire about your good job in Inequality. I hope you will enjoy it.

[b][u]Problem.[/u][/b] Let $a,b,c$ be non-negative real numbers satisfying $ab+bc+ca=1$. Prove that
\[ \frac{1+ab}{kc^2+ab}+\frac{1+bc}{ka^2+bc}+\frac{1+ca}{kb^2+ca} \ge \frac{12}{k+1},  \]
in which $k$ is a real constant and $k \ge 2+\sqrt{3}$ (the best constant).

This one is also for the other if you're intersted in it.

Actually I posed without proof the following inequalities in my book

1) Let $x,y,z$ be non-negative numbers such that $xy+yz+zx=1$, prove that
\[ \frac{1+yz}{x^2+yz}+\frac{1+zx}{y^2+zx}+\frac{1+xy}{z^2+xy}\geq4. \]
2) Let $x,y,z$ be non-negative numbers such that $xy+yz+zx=1$, prove that
\[ \frac{1+yz}{(y+z)^2}+\frac{1+zx}{(z+x)^2}+\frac{1+xy}{(x+y)^2}\geq\frac52. \]

[quote="Vasc"]I posted and proved on this forum the inequality
$\frac{1}{kc^2+ab}+\frac{1}{ka^2+bc}+\frac{1}{kb^2+ca} \geq \frac{9}{k+1}$,
for $k \geq 3+\sqrt 7$ and $a,b,c$ non-negative real numbers satisfying $ab+bc+ca=1$.
I think we can prove your inequality in the same manner.[/quote]

Yes, VASC. Two problem is quite the same, but the solution of mine is diffirent from yours. I used S.O.S method. I think it's nice and short than the solution using expanding which was posted in this forum.

Dear Arquady. Could you solve this problem?  ;)

[quote="arqady"][quote="hungkhtn"]

[b][u]Problem.[/u][/b] Let $a,b,c$ be non-negative real numbers satisfying $ab+bc+ca=1$. Prove that
\[ \frac{1+ab}{kc^2+ab}+\frac{1+bc}{ka^2+bc}+\frac{1+ca}{kb^2+ca} \ge \frac{12}{k+1},  \]
in which $k$ is a real constant and $k \ge 2+\sqrt{3}$

[/quote]
Dear hungkhtn! I have found only here it:
$\frac{1+ab}{kc^2+ab}+\frac{1+bc}{ka^2+bc}+\frac{1+ca}{kb^2+ca} \ge \frac{12}{k+1}\Leftrightarrow$
$\Leftrightarrow(k^2+k)(a-b)^2(a-c)^2(b-c)^2+\sum_{cyc}\left(2k(k^2-4k+1)a^3b^3+2k(3k-2)a^4bc\right)+$
$+\sum_{cyc}\left((k^3+1)(a^3b^2c+a^3c^2b)-6(k^2-1)(2k-1)a^2b^2c^2\right)\geq0,$
which obviously true for $k\geq2+\sqrt3.$[/quote]

Oh... I wonder where did you find out this expression. I only solve inequality without computer, so for me it's too hard work  :oops:  But S.O.S is worked here and I think it's very nice.

[quote="pvthuan"]:o arqady, how could you manipulate with such monster computations? Did you use computer? Usually, I am too lazy for such work.[/quote]
You can be lazy, as at you the head cooks better.:)
 To me nothing remains how to remember, that
$(a-b)^2(a-c)^2(b-c)^2=$
$=\sum_{cyc}\left(a^4b^2+a^4c^2-2a^3b^3-2a^4bc+2a^3b^2c+2a^3c^2b-2a^2b^2c^2\right).$ ;)


When is confirmation of a solution set sent by email generally given? If I sent mine in 36-48 hours ago and have not received a checkmark, should I consider printing a set and mailing them in on the postmark deadline day?

In general, don't send a second solution. I sent mine by email early too, and I haven't gotten anything, so I wouldn't be worried.

There won't be any email confirmations until probably early next week, so don't worry yet.

I sent mine on December 18th because I was going on vacation and was worried I would miss the deadline ... should I be worried if it has not been confirmed, or will those not be posted until next week as well?

Also, should I worry about them losing my solutions since I sent them so long ago?  Sorry if this has been asked too many times, but I don't want to not get credit for them.


inequalities
number theory
relatively prime
number theory proposed

Let $ x_1,x_2,...,x_n,y$ be positive integers such that

a) $ x_1 \leq x_2 \leq ... \leq x_{n\minus{}1} \leq x_{n}$,

b) the numbers $ x_1,x_2,...,x_n$ are relatively prime,

c) $ y^y\equal{}x_1^{x_1}x_2^{x_2}...x_n^{x_n}$.

Prove that $ x_1\equal{}x_2\equal{}...\equal{}x_{n\minus{}1}\equal{}1$.

Let $ n\equal{}4,x_1\equal{}x_2\equal{}x_3\equal{}x_4\equal{}2,y\equal{}4$.  :lol:

So as for $ x_1,x_2,......,x_{n }$ to be relatively primes there is (if there is ) only one $ x_k\equiv0(mod2)$

[quote="Rust"]Let $ n \equal{} 4,x_1 \equal{} x_2 \equal{} x_3 \equal{} x_4 \equal{} 2,y \equal{} 4$.  :lol:[/quote]

They must be relatively prime  :lol:

I don't doubt that the problem is correct...

The case $ n\equal{}2009$ was given during the Armenian Mathematical Olympiad. However, I don't think that there is any considerable difference between the two cases. If you can solve the problem for $ 2009$ variables, you can (probably) do it for any $ n$ variables. Good luck Rust  :wink:

hi
if you raise both sides to the power 1/y , we get

y = x1 ^ (x1/y) * x2 ^ (x2/y) * ..

so y divides x1 , y divides x2 , ....

but x1 , x2 ,x3 .. xn are relatively prime. So , y must be equal to 1
hence each x1 x2 = 1

Here is my solution .

Let p be a prime factor of y then $ p^y/y^y$

but because $ y^y\equal{}{x_1}^{x_1}\cdot{x_2}^{x_2}....{x_n}^{x_n}$ and $ x_1,x_2,...,x_n$ are relatively primes ,we have that p is a prime factor of one of them, lets say $ x_n$ .So we take that $ p^y\equal{}p^{x_n} \Rightarrow y\equal{}x_n$ so we have $ {x_n}^{x_n}\equal{}{x_1}^{x_1}\cdot{x_2}^{x_2}....{x_n}^{x_n} \Rightarrow x_1\equal{}x_2\equal{}...\equal{}x_{n\minus{}1}\equal{}1$ .

[quote="kritikos"]Here is my solution .

Let p be a prime factor of y then $ p^y/y^y$

but because $ y^y \equal{} {x_1}^{x_1}\cdot{x_2}^{x_2}....{x_n}^{x_n}$ and $ x_1,x_2,...,x_n$ are relatively primes ,we have that p is a prime factor of one of them, lets say $ x_n$ .So we take that $ p^y \equal{} p^{x_n} \Rightarrow y \equal{} x_n$ so we have $ {x_n}^{x_n} \equal{} {x_1}^{x_1}\cdot{x_2}^{x_2}....{x_n}^{x_n} \Rightarrow x_1 \equal{} x_2 \equal{} ... \equal{} x_{n \minus{} 1} \equal{} 1$ .[/quote]
If $ p|y$, then exist (only one, btcause they are relatively prime ) $ x_i\equal{}x$. Then we get $ yord_p(y)\equal{}xord_p(x)$.
Because for any i $ y\ge x_i$ we get $ ord_p(y)\le ord_p(x)$. But it prove only $ y\le x_1....x_n.$

If there is a p/y then we have that p/$ {x_n}$ because i said p is a prime factor of y . So because they are relatively primes(the x's) we take that $ p^y = p^{x_n}$ because there is only one $ x_n$  for which p/$ {x_n}$ because they are relatively primes all the x's . So we have that $ p^y = p^{x_n} \Rightarrow y = x_n$ and because $ y = {x_n}$ we take that $ {x_n}^{x_n} = {x_1}^{x_1}....{x_n}^{x_n} \Rightarrow 1 = {x_1}^{x_1}....{x_{n - 1}^{x_{n - 1}} \Rightarrow x_1 = x_2 = .... = x_{n - 1} = 1}$ because $ x_1,x_2,...,x_n$ are relatively primes.

We'll prove that $ x_{n \minus{} 1} \equal{} 1$. Let $ x_{n \minus{} 1} \equal{} {p_1}^{\alpha_1}...{p_k}^{\alpha_k}$ ($ p_i < > p_j$ when $ i < > j$, are prime numbers).

Let it be otherwise: $ x_{n \minus{} 1} < > 1$ $ \equal{} >$ $ k \geq 1$ and we can easily prove the inequality $ {p_1}^{\alpha_1} \geq {\alpha_1}^2$. $ ^{(1)}$

Let us choose a positive integer $ \beta_1$ such that $ y\parallel{}{p_1}^{\beta_1}$ (i.e. $ y$ is divisible by $ {p_1}^{\beta_1}$ but not divisible by $ {p_1}^{\beta_1 \plus{} 1}$). Next, using the equality and the fact that only $ x_{n \minus{} 1}$ is divisible by $ p_1$ (since the numbers $ x_i$ are relatively prime)  we have: $ {p_1}^{y\beta_1} \equal{} {p_1}^{x_{n \minus{} 1}\alpha_1}$ $ ^{(2)}$.

After, using the same way we have: $ {p_i}^{y\beta_i} \equal{} {p_i}^{x_{n \minus{} 1}\alpha_i}$ $ ^{(2)}$ $ \equal{} >$ $ {x_{n \minus{} 1}}^{x_{n \minus{} 1}} \equal{} ({p_1}^{\beta_1}...{p_k}^{\beta_k})^y \equal{} {z_{n \minus{} 1}}^y$, where $ z_{n \minus{} 1} \equal{} {p_1}^{\beta_1}...{p_k}^{\beta_k}$.

With the analogue way (using analogue substitutions) we have $ {x_i}^{x_i} \equal{} {z_{i}}^y$. Note that the number $ z_i$ is relatively prime with $ x_j$ and $ z_j$ if $ i < > j$ $ ^{(3)}$.

Further, using the equality we have: $ y \equal{} z_1...z_n$ $ \equal{} >$ by $ ^{(2)}$ $ {p_1}^{x_{n \minus{} 1}\alpha_1} \equal{} {p_1}^{\beta_1z_1...z_n}$ $ \equal{} >$ $ x_{n \minus{} 1}\alpha_1 \equal{} {\beta_1z_1...z_n}$ $ \equal{} >$ using $ ^{(3)}$ we have that $ \alpha_1$ is divisible by $ z_1...z_{n \minus{} 2}z_n$ $ \equal{} >$ $ \alpha_1 \geq z_1...z_{n \minus{} 2}z_n$.

On the other hand, using the inequality $ ^{(1)}$ and the problem condition $ x_{n} \geq x_{n \minus{} 1}$ (note that $ \equal{} >$ $ z_{n} \geq z_{n \minus{} 1}$) we have:

$ x_{n \minus{} 1}\geq {p_1}^{\alpha_1}\geq {\alpha_1}^2 \geq (z_1...z_{n \minus{} 2}z_n)^2 \geq z_1...z_n \equal{} y$ $ \equal{} >$ $ x_{n \minus{} 1} \geq y$ $ \equal{} >$ $ x_n \equal{} 1$ $ \equal{} >$ $ x_{n \minus{} 1} \equal{} 1$, cotradiction

$ \equal{} >$ (using the problem's condition) $ x_1 \equal{} x_2 \equal{} ... \equal{} x_{n \minus{} 1} \equal{} 1$.

$ Q \minus{} E \minus{} D$

The results of the Olympiad will be known tomorrow. Hope to be in the top 3  :roll:.

Do you have any other solutions/ways ?

P.S.            :D   I'm in the top 3 !!!  :D


I have just started studying functions and know almost nothing about them. I came across this problem:
Prove that the function $f: \mathbb{Z} \rightarrow \mathbb{Z} :$   $f(m+f(f(n)))=-f(f(m+1))-n$ is bijective.
Injective I can do, but how to prove that it's surjective? Please help.

Sorry, I thought this was kind of the basics and everybody would know this. I'll give you the book's definitions:

[b]Surjective:[/b] a function is said to be onto (or surjective) if every point of the codomain is mapped to by some or other point in the domain, that is, the range is all of the codomain.

[b]Injective:[/b] in general, a point in the range could have more than one point in the domain which is mapped to it, but a function is said to be one-to-one (injective) if every point of the range has exactly one point in the domain mapped to it

[b]Bijective:[/b] a function is said to be bijective if it is both surjective and injective.

I suggest you try [url=http://www.cs.odu.edu/~toida/nerzic/content/function/definitions.html]this site :)[/url]

Shobber's from China, so he may not be familiar with the terminology used.

There are quite a few ways of proving a function is surjective.

First, directly assume that you have two values $f(a)=f(b) \implies a=b$

Second, use the contrapositive.

I suggest playing around by plugging in values (probably $n=0$) and taking inverses.  Good luck!

[quote="Melissa"]surjective[/quote]
$f(f(f(-n-f(f(1))))) = n$  :roll:


show if there are integer numbers x,y,z that verify the equation
x^x+y^y=z^z

Assume that such positive integers do exist. Then, clearly, we have $ z > x$ and $ z>y$.
Wlog, we may assume that $ x \geq y$.
Then $ 2x^x \geq x^x \plus{} y^y \equal{} z^z$.
But since $ z \geq x\plus{}1$, we have
$ z^z \geq (x\plus{}1)^{x\plus{}1} \geq x^{x\plus{}1} \plus{} (x\plus{}1)x^x \equal{} (2x\plus{}1)x^x > 2x^x$, a contradiction.

Thus, there are no positive integers satisfying the desired equation.

Pierre.


The figure shown has three circles tangent to each other
with radii of $ x$, $ y$ and $ z$ units. If $ x + y = 12$, $ x + z = 14$ and $ y + z = 22$, what is the product of the three radii?
[asy]import geometry_dev; size(150); defaultpen(linewidth(0.8));
triangle x = triangleabc(12,14,22,80);
draw(x);
draw(Circle(x.point(0),12));
draw(Circle(x.point(2),2));
draw(Circle(x.point(1),10));
pair A = x.point(0);
pair B = x.point(1);
pair C = x.point(2);
label("$z$",A--((12*B + 10*A)/22),E);
label("$y$",B--((2*B + 10*C)/12));
label("$x$",(C.x-2, C.y)--((2*A + 12*C-28)/14),W);[/asy]

Do you multiply the expressions together?

Subtracting the second equation from the first, we see that $ z\minus{}y\equal{}2$. Thus, we have a system of equations together witht he third equation.

$ z\minus{}y\equal{}2$.

$ y\plus{}z\equal{}22$.

Adding, $ 2z\equal{}24 \implies z\equal{}12$. We also get that $ y\equal{}22\minus{}z \implies y\equal{}22\minus{}12 \implies y\equal{}10$

and $ x\plus{}z\equal{}14 \implies x\equal{}14\minus{}z \implies x\equal{}14\minus{}12 \implies x\equal{}2$.]

Thus, $ xyz\equal{}2(10)(12)\equal{}\boxed{240}$.

$ x+y = 12 $ -->1
$ y+z = 22 $ -->2
$ x+z = 14 $ -->3

we can now just substitute

using #1 equation wwe colclude x=12-y-->4

put equation  4 into equation 3 and get 12-y+z=14 and y+z=22 so doing the rest we get

x=2
y=10
z=12

[hide="aNsWEr"]240[/hide]

[size=200][color=#BFBF00][b][i][u]PLEASE RATE MY POST THANK YOU SO MUCH!!![/u][/i][/b][/color][/size]

:jump::jump::jump::jump:

Peoples you are missing a simple solution. Just add all the equations together and divide by 2. you get x+y+z = 24.
Now subtract all the equations like this.
 x+y+z = 24                 x+y+z = 24                  x+y+z = 24
 x+y     = 12                 x+    z = 14                      y+z = 22
______________________________________________

         z = 12                      y    = 10                   x         = 2

Now just multiply all of these together. 12*2 = 24 * 10 = 240.


ratio
geometry
incenter
trigonometry
geometry proposed

Let $ P,Q,R,S$ be the intersection of second internal common tangent to $ AB,BC,BD,CD$, and $ N,M$ be intersection of diagonal $ AR,DQ$ and, $ BS,CR$.And $ E,F,G,H$ are tangency between incircle of $ ABD,BCD$ and $ AB,QR$.$ BC,BR$. Suppose that $ P$ lies on expanded line $ BC$, then by menelaus's theorem on triangle $ BCD$ by cut $ PRS$ , $ \frac {BP}{CP}.\frac {DR}{BR}.\frac {CS}{DS} \equal{} 1$, and on triangle $ BDS$ by cut $ CMR$ ,$ \frac {BR}{BD}.\frac {CM}{RM}.\frac {DS}{CS} \equal{} 1$.
In short form, $ \frac {BP}{CP}.\frac {CM}{RM}.\frac {DR}{BD} \equal{} 1$.
And simmilar way, $ \frac {DR}{BD}.\frac {AN}{RN}.\frac {BQ}{AQ} \equal{} 1$.
Now, Let $ T$ be the intersection of $ AC$ and the second internal common tangent.By menelaus's theorem on $ ABC$ by cut
$ PQT$, $ \frac {BP}{CP}.\frac {AQ}{BQ}.\frac {BT}{AT} \equal{} 1$.
Therefore, $ \frac {BT}{AT} \equal{} \frac {RN}{AN}.\frac {CM}{RM} \equal{} \frac {RG}{AE}.\frac {CF}{RH} \equal{} \frac {r}{AE}.\frac {CF}{r`} \equal{} \frac {tan\frac {A}{2}}{tan\frac {C}{2}}$.

Fang-jh, is your solution the same as Ponclete's?
[quote="Fang-jh"]Let $ ABCD$ be a quadrilateral (convex or concave). Let the internal common tangent (different from $ BD$) of the incircles of triangles $ ABD$ and $ BCD$ cut the diagonal $ AC$ at $ M$. Prove that: $ \frac {AM}{MC} \equal{} \frac {tan\frac {C}{2}}{tan\frac {A}{2}}.$[/quote]

My solution:

Let $ I_1, I_2 $ be the incenter of $ \triangle ABD, \triangle CBD $, respectively .
Let $ \ell $ be the internal common tangent of $ (I_1) $ and $ (I_2) $ (different from $ BD $)
Let $ X=\ell \cap BD, Y=\ell \cap DA, Z=\ell \cap BC $ and $ r_1, r_2 $ be the radius of $ (I_1), (I_2) $, respectively .

Since $ \angle DI_1B=90^{\circ}+\angle YAI_1, \angle BI_2D=90^{\circ}+\angle ZCI_2 $ ,

so $ \frac{AM}{MC}=\frac{AY \cdot \sin \angle MYA}{CZ \cdot \sin \angle MZC} $

$ =\frac{ r_1 \cdot ( \cot \angle YAI_1 + \cot \angle I_1YA) \cdot \sin \angle MYA}{r_2 \cdot ( \cot \angle ZCI_2 + \cot \angle I_2ZC) \cdot \sin \angle MZC } $

$ =\frac{r_1 \cdot ( \tan \angle DI_1I_2- \tan \angle DI_1B) \cdot \sin \angle MYA}{r_2 \cdot ( \tan \angle I_1I_2D - \tan \angle BI_2D) \cdot \sin \angle MZC } $

$=\frac{\text{S}[\triangle I_1BD] \cdot ( \tan \angle DI_1I_2- \tan \angle DI_1B) \cdot \sin \angle MYA}{\text{S}[\triangle I_2BD] \cdot ( \tan \angle I_1I_2D - \tan \angle BI_2D) \cdot \sin \angle MZC } $

$=\frac{( I_1B \cdot I_1D \cdot \sin \angle DI_1B ) \cdot \sin DI_1I_2 \cdot \sin \angle I_2I_1B \cdot \cos \angle BI_2D}{( I_2B \cdot I_2D \cdot \sin \angle BI_2D ) \cdot \sin I_1I_2D \cdot \sin \angle BI_2I_1 \cdot \cos \angle DI_1B } $

$=\frac{( I_1B \cdot I_1D \cdot \sin \angle DI_1B ) \cdot I_2D \cdot I_2B \cdot \cos \angle BI_2D}{( I_2B \cdot I_2D \cdot \sin \angle BI_2D ) \cdot I_1D \cdot I_1B \cdot \cos \angle DI_1B } $

$ =\frac{\cot \angle  BI_2D}{\cot \angle DI_1B}=\frac{\tan \frac{1}{2} \angle BCD}{\tan \frac{1}{2} \angle DAB} $

Q.E.D

This was also discussed in the following threads:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=487427[/url] (Lemma)
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=354536[/url] (Lemma at post #2)


calculus
integration
trigonometry
calculus computations

(1) Calculate the indefinite integral :

$ \int e^{\minus{}x}\sin 2x\ dx$.

(2) Evaluate the definite integral :

$ \int_0^{\pi} e^{\minus{}x}|\sin 2x|\ dx$.

(1) Calculate the indefinite integral :

$ \int e^{ \minus{} x}\sin 2x\ dx$.

One method:

By using integration by parts twice, integrating the exponential part within the integral and differentiating the trigonometric part each time (or vice versa), one can obtain an algebraic expression for the integral in terms of itself... solving this, one obtains:

$ \int_0^{\pi} e^{ \minus{} x}|\sin 2x|\ dx \equal{} \minus{}\frac{2}{5}e^{\minus{}x}\cos 2x \plus{} \frac{1}{5}e^{\minus{}x}\sin 2x$

(2) Evaluate the definite integral :

$ \int_0^{\pi} e^{ \minus{} x}|\sin 2x|\ dx$

The integrand is nonnegative for $ 0 \leq x \leq \frac{\pi}{2}$ and nonpositive for $ \frac{\pi}{2} \leq x \leq \pi$.  Thus:

$ \int_0^{\pi} e^{ \minus{} x}|\sin 2x|\ dx \equal{} \int_0^{\pi/2} e^{ \minus{} x}|\sin 2x|\ dx \minus{} \int_{\pi/2}^{\pi} e^{ \minus{} x}|\sin 2x|\ dx \equal{} \frac{2}{5}e^{\minus{}\pi}(e^{\frac{\pi}{2}} \plus{} 1)^2$

I hope my work is correct.

[quote="kunny"](1) Calculate the indefinite integral :

$ \int e^{ \minus{} x}\sin 2x\ dx$.

(2) Evaluate the definite integral :

$ \int_0^{\pi} e^{ \minus{} x}|\sin 2x|\ dx$.[/quote]
We have:
$ \int e^{ \minus{} x}\sin 2x\ dx \equal{} \minus{} \frac {e^{ \minus{} x}}{5}\left(\sin 2x \plus{} 2\cos 2x \right) \plus{} C$

_____
Sorry, kunny. I've edited my post.

Your have seemed to make a mistake in your calculation. :(

My answer on the definite integral is  $ \frac{2}{5}(e^{\minus{}pi}\plus{}e^{\minus{}pi/2}\plus{}1)$


[hide]There are 3 different choices for the meat patties, and $\binom80 + \binom81 + \binom82 + \dots + \binom88$ which is 256, so the total number is $768 \Rightarrow \boxed{C}$[/hide]

[hide]
It is obvious that there are 3 ways to choose patties, 1, 2, or 3.\
For the condiments, there are 8.
You can choose any number of them, so now we have to sum $\binom 80 +\binom 81 + \binom 82 \cdots \binom 88$.
Remember that formula $2^n$? Well apply it here to get $2^8=256$ ways to choose the meat.
Multiply by 3 for the patties to get $\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{768}}}}}}}$[/hide]

but I like to $\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{BOX}}}}}}}}}}$ stuff  :D

How do you prove the $2^n$ formula thing?

Say that we have a set with $n$ items in it and are trying to find the number of subsets. It would be $\binom n0 + \binom n1 + \binom n2...\binom nn$.
Let's look at each of the $n$ items. It can either be in or out of the subset.
We made $n$ choices, each with 2 options. It therefore is $2^n$
QED

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=406480&highlight=%5C%5Cbinom%7Bn%7D%7B2%7D#406480]This thread[/url] includes discussion and proofs on that topic.

[color=darkred][size=75]Moderator rcv's note:  Following was posted as a new topic, rather than as a reply in this thread.  This is the best I can do to merge the two threads.[/size][/color]

[quote="bryantacademy"][quote]Henry's Hamburger Heaven offers its hamburgers with the following condiments: ketchup, mustard, mayyonaise, tomato, lettuce, pickles, cheese, and onions. A customer can choose one, two, or three meat patties, and any collection of condiments. How many different kinds of hamburgers can be ordered?

(A) 24 (B) 256 (C) 768 (D) 40320 (E) 120960 [/quote]

for the one patties: 8+7+6+5+4+3+2+1=144

it is the same for the other patties,(right?) so the answer is 144*3=432[/quote]


Let $ N\equal{}1*3*5*...*2549\plus{}2*4*6*...*2550$
find the remainder from dividing $ N$ by $ 2551$

This goes in the Intermediate forums.

[hide="Hint"]How can we express the second product in terms of the first product $ \bmod{2551}$?[/hide]

omg I am so proud of my stupid self... I figured it out! :jump: And I guess if I figured it out, it does belong in a lower-level forum.

OK sorry.

[hide]
$ N\equal{}1\cdot3\cdot5\cdot\ldots\cdot2549\plus{}2\cdot4\cdot6\cdot\ldots\cdot2550 \\ \equiv 1\cdot3\cdot5\cdot\ldots\cdot2549\plus{}(\minus{}2549)(\minus{}2547)\ldots(\minus{}3)(\minus{}1) \\ \equiv 1\cdot3\cdot5\cdot\ldots\cdot2549 \minus{} 1\cdot3\cdot5\cdot\ldots\cdot2549 \\ \equiv \boxed{0} \bmod{2551}$.
[/hide]

[quote="1=2"]
[hide="hint"]  How can we express the second sum in terms of the first sum $ \bmod{2551}$?
[/hide]
[/quote]

I think you mean product, not sum.  :wink:


1)  Jake purchases a selection of screwdrivers. His bill is 78 dollars. He buys the same number of 1.50 and 2.50 dollar screwdrivers, and half as many of 4 dollar screwdrivers. The number of 3 dollar screwdrivers is one more than the number of 4 dollar screwdrivers. How many of each did he purchase?

2)  A collection of coins has a value of 64 cents. There are two more nickels than dimes and three times as many pennies as dimes. How many of each kind are there?

[hide="Problem 1"]If he buys $ x$ of the 4 dollar screwdrivers, he buys $ 2x$ of the 1.5 and 2.5 dollar screwdrivers along with $ x\plus{}1$ 3 dollar screwdrivers. So, $ 4x \plus{} 3x \plus{} 5x \plus{} 3x \plus{} 3 \equal{} 78$, or $ 15x \equal{} 75$, or $ x \equal{} 5$. So, he bought:

$ 10$ 1.5 dollar screwdrivers
$ 10$ 2.5 dollar screwdrivers
$ 6$ 3 dollar screwdrivers
$ 5$ 4 dollar screwdrivers

[/hide]

[hide="Problem 2"]

If there are $ x$ dimes, there are $ 3x$ pennies and $ x\plus{}2$ nickels. So, $ 10x \plus{} 3x \plus{} 5x\plus{}10 \equal{} 64$, so $ 18x \equal{} 54$, so $ x \equal{} 3$. Therefore, he has:

$ 9$ pennies.
$ 5$ nickels.
$ 3$ dimes.

[/hide]


Νδο

$ \frac{2n}{2n\minus{}1}<\sqrt[n]{2}<\frac{n}{n\minus{}1}$ για καθε $ n\in N$ με $ n\geq 2$

[hide="Υποδειξη"]ανισοτητα Bernoulli[/hide]

[quote="mailo"]Νδο

$ \frac {2n}{2n \minus{} 1} < \sqrt [n]{2} < \frac {n}{n \minus{} 1}$ για καθε $ n\in N$ με $ n\geq 2$

[hide="Υποδειξη"]ανισοτητα Bernoulli[/hide][/quote]
Μπορείς α στηριχτείς στο ότι η $ (1\plus{}\frac{1}{n\minus{}1})^n$ είναι γνήσια φθίνουσα(κομψή απόδειξη με ανισότητα ΑΜ-ΓΜ) και συγκλίνει στο $ e$ Για το αριστερό μέλος χρησιμοποίησε το δεξί και εμφάνισε μια υπακολουθία της προηγούμενης

Μια πιο ευκολη αποδειξη χρησιμοποιοντας την ανισοτητα Bernoulli

$ \frac {2n}{2n \minus{} 1} < \sqrt [n]{2}\Leftrightarrow\frac {2n \minus{} 1}{2n} > \frac {1}{\sqrt [n]{2}}\Leftrightarrow\left(1 \minus{} \frac {1}{2n}\right)^n > \frac {1}{2}$

Το οποιο ισχυει γιατι για καθε $ n\geq2$ εχουμε απο την ανισοτητα Bernoulli οτι

$ \left(1 \minus{} \frac {1}{2n}\right)^n > 1 \minus{} \frac {n}{2n} \equal{} \frac {1}{2}$

Επισης

$ \sqrt [n]{2} < \frac {n}{n \minus{} 1}\Leftrightarrow\left(1 \plus{} \frac {1}{n \minus{} 1}\right)^n > 2$

Το οποιο επισης ισχυει αφου για καθε $ n\geq2$ εχουμε απο την ανισοτητα Bernoulli οτι

$ \left(1 \plus{} \frac {1}{n \minus{} 1}\right)^n > 1 \plus{} \frac {n}{n \minus{} 1} > 1 \plus{} 1 \equal{} 2$

Ωραια χρηση της ανισοτητας Bernoulli. Ενα typo μονο,στους εκθετες ειναι n και οχι 2

@mailo:Εχεις δει το pm που σου εστειλα?

Αλεξανδρε εχεις δικιο, το διορθωσα.

ΥΓ. σου εστειλα πμ


When time, in which Y, by flowing theory becomes Y+N, the quantity YN becomes (Y+N)N.

Is this true? Or what is the answer

Bonus: By the method of infinite series, what is the result of Y, when this formula is applied to the resulting time answer from above:

YN + noYN-1 + (nn-n)/2 ooYN-2

?

I don't understand what you're trying to say. Please use standard notation.


function
LaTeX
topology
superior algebra
superior algebra theorems

Can anyone tell me something about The Principles of Contraction to functions.

Definition of Contraction Mappings: Given a metric space S with distance measure d(), and $T:S \to S$ mapping S into itself, T is a contraction mapping if for some coefficient $k\in(0,1)$, $d(Tx,Ty) \leq kd(x,y)$ for all x and y in S.

Simpler definition in terms of real numbers: Given the metric (or distance map) $d(x,y)=|x-y|$ defined on R, f is a contraction mapping if for some coefficient $k\in(0,1)$, $d(Tx,Ty)=|f(x)-f(y)| \leq kd(x,y)=k|x-y|$ for all x and y in R.

Contraction principle states (roughly speaking) that if we have a contraction mapping f, then there exists a unique fixed point in f. This is due to Banach.

Definition of Contraction Mappings: Given a metric space S with distance measure d(), and T:S->S mapping S into itself, T is a contraction mapping if for some coefficient $k\in(0,1)$, $d(Tx,Ty) \leq kd(x,y)$ for all x and y in S.

Simpler definition in terms of real numbers: Given the metric (or distance map) $d(x,y)=|x-y|$ defined on R, f is a contraction mapping if for some coefficient $k\in(0,1)$, $d(Tx,Ty)=|f(x)-f(y)| \leq kd(x,y)=k|x-y|$ for all x and y in R.

Contraction principle states (roughly speaking) that if we have a contraction mapping f, then there exists a unique fixed point in f. This is due to Banach.

[i]Note[/i]: The LaTeX code has changed since the previous message was posted, thus making it unreadable by humans; so I've redone it.


I'm not very comfortable with posting diagrams directly, hence the attachment. Sorry for the inconvenience caused. The attachment contains a couple of triangle related problems.

1. See the attached diagram for additional notation.

What needs to be found is $ k : \equal{}{p\over c}$, and then the desired area ratio is equal to $ k^{2}$.

We have

$ {p\over e}\equal{}{BQ\over 8}\land{e\over c}\equal{}{3\over 4}\implies k\equal{}{3BQ\over 32}\implies BQ\equal{}{32k\over 3}$

$ {p\over d}\equal{}{AR\over 8}\land{d\over c}\equal{}{5\over 7}\implies k\equal{}{5AR\over 56}\implies AR\equal{}{56k\over 5}$

$ BQ\plus{}AR\equal{}AB\plus{}QR\equal{}AB\plus{}kAB\equal{}8\plus{}8k$

Therefore we get an equation

$ {32k\over 3}\plus{}{56k\over 5}\equal{}8\plus{}8k\implies k\equal{}{15\over 26}$

Finally, $ {[PQR]\over[ABC]}\equal{}k^{2}\equal{}{225\over 676}$

For the second problem see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=126063


topology
advanced fields
advanced fields unsolved

Once I read about a theorem's proof which seems to use a fact as follows: Let $ A\equal{}\mathbb{R}^4/\Gamma$, where $ \Gamma$ is the group $ \mathbb{Z}\oplus\mathbb{Z}$ with (noncommutative) groups operation $ (j, k)*(j',k')\equal{}(j\plus{}j', j'k\plus{}k')$, and the group action is described as $ (x,y,w,t)\rightarrow (x,y\plus{}i,w\plus{}jt,t)$, then the fundamental group is just $ \Gamma$.

Does anybody know about the reason this holds?  And is there a theorem giving the fundamental groups of the quotient of some contractible space over a group?  Thanks!

If you have a "good" action of a group G on a simply connected space X, then $ \pi_1(X/G)\equal{}G$. Good means continuous, properly discontinuous and free. I hope I didn't skip anything.

I have a problem with your group though. It doesn't seem to have a neutral element.


Hi everyone!

Is here anybody who comes to this year's British-Hungarian math (olympic) camp in December? (and January)
Because I go and it seemed a good idea to me to get acquainted with each other even before the camp :D
And be glad till I'm not speaking, just writing :P

kdano

Hey good idea. I'll be there, but I only joined the UK squad in September so I'm fairly naive in the ways of IMO.

Apparently there was some issue with our discussion of Bath and Hungary camps on the Cambridge-based NRICH message board in the past though, so I don't know how legitimate such a discussion is, but I see no harm in getting to know each other online first.

I'm 'the new guy'. My favourite maths is Number Theory. Outside of maths I am a fairly 'Renaissance man', with a very keen interest in music, theatre and literature (in fact, I won our school poetry reading competition today, so I'm very happy :D).

I'll be there too. (were you there last year, kdano?)

If we're going to go into introductions, then I suppose my favourite maths used to be number theory, but it is fast being replaced by combinatorics. I play a fair amount of sport, particularly cricket. Though I play clarinet, I wouldn't describe myself as musical (and definitely not theatrical).

I think that on Mathlinks this kind of discussion should not be an issue.

[quote="Ilthigore"]Apparently there was some issue with our discussion of Bath and Hungary camps on the Cambridge-based NRICH message board in the past though, so I don't know how legitimate such a discussion is, but I see no harm in getting to know each other online first.[/quote]

I'm sorry if I did something wrong, I'm really a newbie here :P.

[quote="tim1234133"]I'll be there too. (were you there last year, kdano?)[/quote]

Unfortunately not, since my English teacher not supported me enough :D But there were two of my classmates (András Eisenberger and László Lovász, if [i]you[/i] were there :P).

I'm an 11th grade student from our country's best math-educating school. (I hope it means what i want it to mean :) so this school has the best math classes)
I don't have any favourite maths.. Maybe later it will be number theory or graph theory or i dunno.. I play the piano and I like to play football (soccer) as well. And of course I enjoy solving problems in maths and especially in informatics. And in my free time (if i have) I'm one of the few contributors of the Hungarian [i]Wikipédia[/i] :P

kdano

I shall be there too. Oh, Andras and his crazy ties: brings back a lot of memories, mostly about snowmen.
Dominic

[quote="Inquisitor"]Oh, Andras and his crazy ties: brings back a lot of memories, mostly about snowmen.[/quote]

Yeah, he gladdened us with those too :P But I don't know that snowmen-story..  :maybe:

kdano

Photographs.

The best photos are on http://ukhungary0607.blogspot.com/
but Alison's Facebook is also an excellent source of photographs (she has uploaded her entire collection there).


let $ x,y,z$ reals satisfying $ x \leq y \leq z$ and $ xy \plus{} yz \plus{} zx \equal{} 1$
Then prove $ xz < \frac {1}{2}$

This was first part of question that appeared in Mediterranean M.C. 2007
In second part of question it was asked:"Is it possible to change constant $ \frac{1}{2}$ by better one?"

Answer is NO (try to prove that one It is not so difficult) Here is solution for [b]FOURRIER[/b]'s problem:

From $ x\leq y \leq z$ it follows that $ (y\minus{}x)(y\minus{}z)\leq 0$ and consequently:

$ y^{2}\minus{}xy\minus{}yz\plus{}xz\leq 0$ which is equivalent to: $ y^{2}\minus{}(xy\plus{}yz\plus{}zx)\plus{}2xz\leq 0$ which implies $ y^{2}\minus{}1\plus{}2xz \leq 0$.

Thus, $ 1\minus{}2xz\geq y^{2}\geq0$ i.e. $ 1\minus{}2xz\geq 0$ and therefore $ xz\leq \frac{1}{2}$.

If $ xz\equal{}\frac {1}{2}$ then $ 0\geq y^{2}\geq 0$ i.e. y=0 and from $ xy\plus{}yz\plus{}zx\equal{}1$ it follows that $ xz\equal{}1$, contradiction so equality can not occur.

Hence $ xz < \frac{1}{2}$

Thats it
there's another solution with cases

$ x,y,z >0$ etc.


A generator of constant voltage but of variable frequency is connected to a series RLC circuit. Gradually changing the frequency the current reaches a maximum, when the voltage across the inductor is $ U_{1}$. If the frequency is further increased the current decreases, but the voltage across the inductor still increases. Later the voltage across the coil reaches its maximum of $ U_{2}$ and then it decreases.

Determine the quotient of $ \frac{U_{2}}{U_{1}}$ in terms of $ R$, $ L$ and $ C$.

I would like to check with you people my answer which is:

$ \frac{U_{2}}{U_{1}}\equal{}\frac{1}{\sqrt{1\minus{}\frac{R^{2}C}{4L}}}$

Is there an easy way to get U2 ? I did it by minimizing U w.r.t. frequency.