Well, heres a hint. Try to split it up into parts. First each team by themselves. Then touchdown with extra point, field goal, touchdown with no extra point, safety, etc.

Well, others will I'm sure jump in if I'm wrong, but I don't think there's any "quick" way of solving a problem like this.  The approach that I've seen suggested most often for problems like this is "Make an organized list".  In a team competition, it would probably make sense to have one very careful person work on this one while the rest of the team went on to other problems.

The "organized" part of the strategy is key to making sure you get all solutions.  If I am making an organized list for a problem like this, I always start with the greatest possible number of the highest point score, then the greatest number of the next highest point score that will still "fit" within the total, and so on down the line, so
 [hide="for 20 points"]

The possible scores are 7 (touchdown plus extra point -- keep together as a single unit), 6, 3 and 2.  Keep  in mind that you must have an even number of odd scores to get an even total...

2x7 + 1x6
2x7 + 2x3
2x7 + 3x2
1x7 + 1x6 + 1x3 + 1x2
1x7 + 3x3 + 1x2
3x6 + 1x2
2x6 + 2x3 + 1x2
2x6 + 4x2
1x6 + 4x3 + 1x2
1x6 + 2x3 + 4x2
1x6 + 7x2
6x3 + 1x2
4x3 + 4x2
2x3 + 7x2
10x2

(fwiw, that took me about 2 minutes to enumerate -- you can see patterns unfolding as you go)
[/hide]

You would obviously have to do the same for 17 points.   If someone else has a faster way, I'd be happy to see it, but I think organized brute force may be the best way to go here.

In a professional football it is possible to score 2 points (for a safety) , 3 points (for a field goal), or 6 points ( for a touchdown). If a touchdown is scored it is possible to score one additional point. The score of a football game is 21 to 17. Find the total number of possible combinations of points to achieve this final score.

[b]16[/b] ways to acheive the score of 21:

[hide]6T+3F+2S+1E=21

  T    F     S    E

  3    1     0    0
  3    0     2    1
  3    0     0    3
  2    3     0    0
  2    2     1    1
  2    1     3    0
  2    1     2    2
  1    5     0    0
  1    4     1    1
  1    3     3    0
  1    2     4    1
  1    1     6    0
  0    7     0    0
  0    5     3    0
  0    3     5    0
  0    1     9    0[/hide]

[b]9[/b] ways to acheive the score of 17:

[hide]6T+3F+2S+1E=17

  T    F     S    E
  2    1    1    0
  2    1    0    2
  2    0    2    1
  1    3    1    0
  1    2    2    1
  1    1    4    0
  0    5    1    0
  0    3    4    0
  0    1    7    0[/hide]

[hide]9+16=25[/hide]

[quote]
You would obviously have to do the same for 17 points.
[/quote]

... and 21 since that was the actual score listed in the problem, not 20.  That was just an example to give you the idea.  Yeah, that's the ticket  :blush:

The answer is 180. Did not know how to get this number. I got more than that.

Jli, wouldn 't you have to multiply those instead of add?

[hide="18 ways to score 21"]
3x7
2x7 +           1x3 + 2x2
1x7 + 2x6 +           1x2
1x7 + 1x6 + 2x3 + 1x2
1x7 + 1x6 +           4x2 -- this is one that JLI missed (2 0 4 1)
1x7 +           4x3 + 1x2
1x7 +           2x3 + 4x2
1x7 +                     7x2 --  this is one that JLI missed (1 0 7 1)
          3x6 + 1x3
          2x6 + 3x3
          2x6 + 1x3 + 3x2
          1x6 + 5x3
          1x6 + 3x3 + 3x2
          1x6 + 1x3 + 6x2
                    7x3
                    5x3 + 3x2
                    3x3 + 6x2
                    1x3 + 9x2
[/hide]

[hide=" 10 ways to score 17"]
2x7 +           1x3
1x7 + 1x6 +        + 2x2
1x7 +           2x3 + 2x2
1x7 +                     5x2 -- JLI missed this one (1 0 5 1)
          2x6 + 1x3 + 1x2
          1x6 + 3x3 + 1x2
          1x6 + 1x3 + 4x2
                    5x3 + 1x2
                    3x3 + 4x2
                    1x3 + 7x2
[/hide]

Then as eryaman says, you do have to multiply rather than add, which does indeed give the desired 180 answer.

For the benefit of those still trying to grasp the system, allow me to expand on Mathmoms solution, which illustrated the exact sequences of scenarios for 21 points and 17 points in solving the problem systematically.

A table should be used (sorry I did not spend time to draw it nicer, see below) during a real test to make it visually easier, which can help to avoid careless omission or incorrect placement.

A key to solving this problem is to recognize that it is the same as solving 2 coin-combination problems (quarters, dimes, nickels and pennies) then multiply the results.

A good system to solve this type of combination problems involves constructing all the possible scenarios starting with the maximum number of the highest valued item to the minimum number of the same item, then move on to the next item.

The method is verbose to describe, but in practice the table is populated rather rapidly.

In this case we must first determine the items (in coin problem 25-10-5-1 are assumed common knowledge), we have

TD+1 = 7pts
TD+0 = 6pts
Field Goal = 3pts
Safety = 2pts

Here are the scenarios for 17 points (same as Mathmoms):

S	7	6	3	2	Tot
1	2	0	1	0	17
2	1	1	0	2	17
3	1	0	2	2	17
4	1	0	0	5	17
5	0	2	1	1	17
6	0	1	3	1	17
7	0	1	1	4	17
8	0	0	5	1	17
9	0	0	3	4	17
10	0	0	1	7	17

We start with the highest point item, 7pts:

Scenario 1.
-	2  7pts (maximum allowed), leaving 17-14=3
-	0  6pts is possible
-	1  3pts makes the total 2x7 + 1x3=17
-	0  2pts needed

Scenario 2.
-	1  7pts (no more 2x7pts possible), leaving 17-7=10
-	1  6pts is possible, leaving 10-6=4
-	0  3pts is possible, as 4-3=1 ,and 1 is not an item
-	2  2pts makes the total 7 + 6 + 4 = 14

and so on.

(edited)

In Mathmoms solution, 1x7 + 1x6 +  2x2  does not work.

In MathChauffeur's solution, 10  0  0  1   8   19  does not sum to 17.

If I use the method you used here, I would not have time for the other 9 problems.

[quote="cj"]In Mathmoms solution, 1x7 + 1x6 +  2x2  does not work.[/quote]

Sure it does.  7 + 6 + 4 = 17

[quote]In MathChauffeur's solution, 10  0  0  1   8   19  does not sum to 17.[/quote]

Yeah, he meant 10 0 0 1 7 ;-)

[quote]If I use the method you used here, I would not have time for the other 9 problems.[/quote]

It's a team round problem.  Have one or two (to double-check one another for missed items) members of your team work through the solutions carefully while the others start on the other problems.  It doesn't take [i]that[/i] long to enumerate all of the solutions if you use a methodical approach.  Also, once you've briefly discussed the approach among  yourselves, you don't really need your strongest team members to work on it, just someone with a bit of a perfectionist streak ;-)

[quote="cj"]I know that  7 + 6 + 4 = 17 . But you can't have a 7 (1x7) and a 6 (1x6) at the same time. That is why it does not work.[/quote]

I don't see why not.  You could have two touchdowns, but only score one additional point on one of them.

[quote="cj"]In MathChauffeur's solution, 10  0  0  1   8   19  does not sum to 17.

If I use the method you used here, I would not have time for the other 9 problems.[/quote]

Sorry I had to run before I had a chance to check my math ...

It was supposed to be 10 0 0 1 7 17 ...

Also, this and the "coin combination" problems are usually are among the last problems to be solved in a given round/set as they tend to take a long time.

Is there a faster way?

Well, if you look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15314]here[/url] on post 10, there's another way to list all of the ways.

Quoting from the post Mr. Potter referenced:

 [quote]nways(N,K) = nway(N, K-1) + nway(N-coin[K], K);

"the number of ways to make N cents with the first k
coins is the number of ways to make N cents using the first K-1
coins (i.e., without using the Kth coin) plus the number of
ways to make N-coin[K] cents using the first K coins. For the
second set of ways, we then add the Kth coin to arrive at a
total of N cents. [/quote]

We can use that to say that:

nways(17, {7,6,3,2})
  = nways(17, {6,3,2} ) + nways(10, {7,6,3,2})
   = nways (17, {3,2}) + nways (11, {6,3,2}) + nways (10, {6,3,2} + nways(4, {3,2})
    =... and so on

This is great for writing a computer program, but IMO harder than just enumerating the cases for a human.

You could probably stop at the point where you were down to the {2,3} coin examples and do those by inspection, but still, it's at least as messy as just listing the combinations.

[quote="dengmi"][hide]  There is no quick way[/quote]

I got a quick way.


24. The first four terms in an arithmetic sequence are x+y, x-y, xy, and x/y, in that order. What is the fifth term?
(a) -15/8 (b)-6/5 (c) 0 (d) 27/20 (e) 123/40

25. how many distinct four digit numbers are divisible by 3 and have 23 as their last two digits?
(a) 27 (b) 30 (c)33 (d) 81 (e) 90

I want to know how to do it . please show ur solution

You can always find solutions by going to the Contests/Resources section (look in the top toolbar with Profile/Private Messages). Find the problem you're looking for and click its number - this will bring you to the original thread.

Merged to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1370128]#24[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1370131]#25[/url] and locked.


Source: Bay Area Math Meet 2002

Let obtuse triangle $ABC$ satisfy $AB \cdot BC \cdot CA = 3\sqrt 3 \sin A \sin B \sin C$.  Find the upper bound of the area of $ABC$.

By the Sine Law we have

$3\sqrt{3}=\frac{AB}{\sin C}\cdot\frac{BC}{\sin A}\cdot\frac{CA}{\sin B}=(2R)^3$

$(2R)^3=3^{3/2}\implies R=\frac{\sqrt 3}{2}$

Since the triangle is obtuse, the upper bound of its area is the area of the equilateral right triangle with the hypotenuse $2R$, and that's $R^2={3\over 4}$. Hence $A<{3\over 4}$


I was trying to help a friend of mine in her geometry class and came across this problem.  Not quite sure how to prove it:

Given circle O, and an arbitrary point P (can be outside, on, or inside the circle).  X, X', X''... are points on the circle.
Prove that the locus of the midpoints of PX, PX', PX''... form another circle with half the radius of O.

Anyone?

Let $Y_1, Y_2, Y_3, ...$ be the midpoints of the segments $PX_1, PX_2, PX_3, ...$, where $X_1, X_2, X_3, ...$ are arbitrary points of the circle $(C)$. Denote $(D)$ the locus of points $Y_i$. For arbitrary indices $m \neq n$, the triangles $\triangle PY_mY_n \sim \triangle PX_mX_n$ are similar by SAS, because the angle $\measuredangle Y_mPY_n \equiv \measuredangle X_mPX_n$ is common and $\frac{PY_m}{PY_n} = \frac{\frac 1 2 PX_m}{\frac 1 2 PX_n} = \frac {PX_m}{PX_n}$,  with coefficient similarity $\frac{PY_i}{PX_i} = \frac 1 2$. It follows that $\measuredangle PY_mY_n = \measuredangle PX_mX_n$,  $\measuredangle PY_nY_m = \measuredangle PX_nX_m$,  $Y_mY_n = \frac 1 2 X_mX_n$ and $Y_mY_n \parallel X_mX_n$. Since this is true for all possible pairs of the corresponding points $(Y_m, X_m)$ and $(Y_n, X_n)$, it followes that the figures $(D)$ and $(C)$ themselves are similar with coefficient of similarity $\frac 1 2$, i.e., the locus $(D)$ is a circle with half the radius of the circle $(C)$.

This does not have much to do with circles and chords. For any figure $(C)$, say a picture of cat, the locus $(D)$ of midpoints $Y_i$ of the segments $PX_i$, where $X_i \in (C)$, is a picture of cat half the size of the cat $(C)$. The point $P$ is called the center of similarity.


integration
trigonometry
calculus
real analysis
real analysis solved

[quote="skylian"]As $\displaystyle f \in C^{2}[0, \pi], f(0)=1$. define $f(\pi)$ so that $\int_{0}^{\pi}(f(x)+f''(x))sin x dx=0$[/quote]
Not so hard.

Note that $\sin(\pi) = \sin 0 = 0, d\sin x = \cos x\,dx, d\cos x = -\sin x\,dx, \cos \pi  = - \cos 0 = -1$ and applying integral by part, we have:

$\int_{0}^{\pi}(f(x)+f''(x))sinx\,dx = \\ \int f(x) \sin x\,dx + \int \sin x\, df'(x) = \\\int f(x) \sin x +0 - \int \cos x f'(x)
=\\ \int f(x) \sin x - \int \cos x df(x) =\\ \int f(x)\sin x + f(\pi) - f(0) - \int f(x)\sin x = \\f(\pi) $


let $ABCD$ be a convex qualilareral  ,  AC meet BD at $M$ .
                     the lines through $M$ and perpendicular $AB,BC,CD,DA$ at $E,F,G,H$ repectively.
              prove that :$S_{EFGH} \leq 1/2 S_{ABCD}$
          (let S be area of ABCD and EFGH)


algebra
polynomial
LaTeX
superior algebra
superior algebra solved

How do I prove that for any prime $p$, $\overline{\mathbb{F}_p} = \bigcup_{i \in \mathbb{N}} \mathbb{F}_{p^i}$?

i think you need to show both inclusions:
- $\displaystyle\overline{\mathbb{F}}_p \subset \bigcup_{i \in \mathbb{N}} \mathbb{F}_{p^i}$: the splitting field $\displaystyle\mathbb{S}_{f,\mathbb{F}_p}$ of any polynomial $f \in \mathbb{F}_p[x]$ of degree $\partial f = d$ is $\displaystyle\mathbb{S}_{f,\mathbb{F}_p} \subset \mathbb{F}_{p^{d!}} \subset \bigcup_{i \in \mathbb{N}} \mathbb{F}_{p^i}$.
- $\displaystyle\bigcup_{i \in \mathbb{N}} \mathbb{F}_{p^i} \subset \overline{\mathbb{F}}_p$: take a polynomial in $\displaystyle \sum_{k=0}^{\partial g} a_kx^k = g\in \displaystyle\left(\bigcup_{i \in \mathbb{N}} \mathbb{F}_{p^i}\right)[x]$: consider the maximum $m = \max_{0\le l \le \partial g} \min \{k|a_l \in \mathbb{F}_{p^k}\}$. you get that the splitting field of $g$ is contained in $\overline{\mathbb{F}}_p$, since it's an algebraic extension of an algebraic extension of $\mathbb{F}_p$.

i hope i've been clear enough.. it's been tiring to write everything in LaTeX! :D

The only thing I couldn't do is proving that $\overline{\mathbb{F}_p}$ is algebraically closed.

So, I don't follow your last argument :blush: Am I missing something?

uhm, right...
$g \in \mathbb{F}_{p^m}[x]$, and $\mathbb{F}_{p^m}$ is obviously a finite extension of $\mathbb{F}_p$. its splitting field is a finite extension $\mathbb{K}$ of $\mathbb{F}_{p^m}$, and so you have that $[\mathbb{K}: \mathbb{F}_p]<+\infty$ and so $\mathbb{K} = \mathbb{F}_{p^n}$ for some $n$ (which i suppose has to be $n\le m\cdot (\partial g)!$), and so $\mathbb{K}\subset \bigcup_{i \in \mathbb{N}} \mathbb{F}_{p^i}$...
is it ok? :) hope so


floor function
blogs
inequalities
number theory
greatest common divisor
modular arithmetic

1)Let $ x,y$ be integers greater than $ 1$ such that $ 2x\minus{}1$ is dividible by $ y$ and $ 4y\minus{}1$ is dividible by $ x$.Find all pairs $ (x,y)$.

2)Find all prime $ p$ for which $ (p^2\minus{}1)!$ is dividible by $ p^4$ but not by $ p^5$.

3)Let $ a,b,c$ are integers such that $ 0<a<b<c$ and $ a\plus{}b\minus{}1$ is dividible by $ c$,$ b\plus{}c\minus{}1$ is dividible by $ a$,and $ c\plus{}a\minus{}1$ is dividible by $ b$.Find all $ (a,b,c)$.

4)$ a>1,b>1$ are integers and relatively prime.
Express $ \sum_{k\equal{}1}^{a\minus{}1}\lfloor \frac{kb}{a}\rfloor\plus{}\sum_{k\equal{}1}^{b\minus{}1}\lfloor \frac{ka}{b}\rfloor$ in terms of $ a,b$.

[hide="2"]
Motivated by: [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=141316]This Blog[/url]

$ v_p ( (p^2\minus{}1)! ) \equal{} p\minus{}1 \equal{} 4$

Therefore $ p\equal{}5$
[/hide]

[hide="Hint for #1"]
Neither $ x$ or $ y$ can be even because the expressions $ 2x \minus{} 1$ and $ 4y \minus{} 1$ are always odd.[/hide]

[hide="Number 1"]
$ \frac {4y \minus{} 1}{x}, \frac {2x \minus{} 1}{y} \in \mathbb{N}$ $ \Longrightarrow \frac {4y \minus{} 1}{x} \cdot \frac {2x \minus{} 1}{y} \equal{} \left(4 \minus{} \frac 1y\right)\left(2 \minus{} \frac 1x\right) \equal{} 8 \minus{} \frac {2x \plus{} 4y \minus{} 1}{xy} \in \mathbb{N}$. Thus $ \frac {2x \plus{} 4y \minus{} 1}{xy} \in \mathbb{Z}$, and since $ x,y > 1$, the quantity is positive, and we have $ \frac {2x \plus{} 4y \minus{} 1}{xy} \ge 1 \Longrightarrow 7 \ge (x \minus{} 4)(y \minus{} 2)$.

Note that $ 4y \minus{} 1, 2x \minus{} 1$ are odd, so $ x,y$ cannot be even. For $ x \equal{} 3$, we have $ y|5 \Longrightarrow y \equal{} 5$, which does not work. Otherwise, the inequality above implies only one of $ x \minus{} 4 \ge 3$ or $ y \minus{} 2 \ge 3$ may hold, so either $ x \equal{} 5$ or $ y \equal{} 3$. For $ x \equal{} 5$, we have $ y|9 \Longrightarrow y \equal{} 3,9$, and $ (x,y) \equal{} \boxed{(5,9)}$ works. For $ y \equal{} 3$, $ x|11 \Longrightarrow x \equal{} 11$, and $ \boxed{(11,3)}$ works. [/hide]

[hide="Number 3"]
$ c|a \plus{} b \minus{} 1 < 2c \minus{} 1 \Longrightarrow c \equal{} a \plus{} b \minus{} 1$. Then $ b | c \plus{} a \minus{} 1 \equal{} 2a \plus{} b \minus{} 2 < 3b \minus{} 2$, so either $ b \equal{} c \plus{} a \minus{} 1$ or $ 2b \equal{} c \plus{} a \minus{} 1$. In the former case, we have $ a \minus{} 1 \equal{} b \minus{} c \equal{} c \minus{} b \Longrightarrow b \equal{} c$, contradiction. Thus,\[ \begin{cases}c \minus{} b & \equal{} a \minus{} 1 \\
2b \minus{} c & \equal{} a \minus{} 1 \end{cases} \Longrightarrow \begin{cases}b & \equal{} 2a \minus{} 2 \\
c & \equal{} 3a \minus{} 3 \end{cases}
\]It follows that $ a | b \plus{} c \minus{} 1 \equal{} 5a \minus{} 6$ or $ a | 6$, and $ a \equal{} 1,2,3,6$. Substituting, we have contradictions for $ a \equal{} 1,2$, but $ (a,b,c) \equal{} \boxed{(3,4,6)(6,10,15)}$ work.
[/hide]

[hide="Number 4"]
$ \text{gcd}\,(a,b) \equal{} 1$ implies that $ ka \pmod{b}$ is distinct for $ 1 \le k \le b\minus{}1$. Let $ \{x\} \equal{} x \minus{} \lfloor x \rfloor$ denote the fractional part of $ x$. Thus, $ 0 < \left\{\frac{ka}{b} \right\} < 1$ is distinct for $ 1 \le k \le b\minus{}1$, with sum (in some order) $ \frac{1}{b} \plus{} \frac{2}{b} \plus{} \cdots \plus{} \frac{b\minus{}1}{b} \equal{} \frac{(b\minus{}1)b/2}{b} \equal{} \frac{b\minus{}1}{2}$.  Hence, \[ \sum_{k\equal{}1}^{b\minus{}1}\left\lfloor \frac{ka}{b}\right\rfloor \equal{} \sum_{k\equal{}1}^{b\minus{}1} \frac{ka}{b} \minus{} \sum_{k\equal{}1}^{b\minus{}1} \left\{\frac{ka}{b} \right\} \equal{} a\frac{b\minus{}1}2 \minus{} \frac{b\minus{}1}{2} \equal{} \frac{(a\minus{}1)(b\minus{}1)}{2}.\] The desired sum is then $ (a\minus{}1)(b\minus{}1)$.
[/hide]


superior algebra
superior algebra solved

Let t in C\Q be s.t. the set A={a+bt|a,b in Z} is a ring with respect to the addition and multiplication in C. Prove that if A has exactly 4 invertible elements, then A is Z.

[i]Romanian NMO/final round/Suceava 1997/problem 1/12'th grade/proposed by Marcel Tena.[/i]

Not that tough, but I thought it was a nice result.

I solved this problem on another topic. Look for it in the Proposed Problems Section. It is a big topic on the 1997 NMO posted by grobber.


MATHCOUNTS
AMC
AMC 8
ARML
USA(J)MO
USAMO
geometry

Ligon sent the top three individuals, the top countdown performer, and of course fielded the #1 team.  The written winner was Nick Tobey and the Countdown winner was Calvin Deng.

I am curious to know how many of the top individuals were younger students.  A 7th grader named Yu Wang from Phillips middle school was 9th.  One 6th grader made the top 10 from Hanes middle school, but I've forgotten the name.

(P.S. The Phillips team came in third!)

Oh yeah, the other nationals qualifier from Ligon was Allen Yang. Fourth place, also qualifying for nats, was a girl called Connie (Chen?) from Metrolina .

[quote="n^4+4"]One 6th grader made the top 10 from Hanes middle school, but I've forgotten the name.[/quote]

The 6th graders name is Sammy Luo and he placed 7th.

[quote="n^4+4"]A 7th grader named Yu Wang from Phillips middle school was 9th.  [/quote]

Dude... this guy is [i]pretty cool[/i]

He was in my dad's math class at Chinese school

as were like... probably many pretty good people

Damien I actually try to coach Yu actively.  He's over at my house a lot doing problems and that's why I'm wondering what sort of competition he'll have next year   :D

Also, didn't Alex Kelly from Phillips do well?

So do the top AMC 8 scorers and teams usually dominate Mathcounts too? Or is Mathcounts more speed-intensive?

Eh I think that top AMC8 scorers tend to do well, but it's not guaranteed that they win. Of course, AMC8 is pretty speedy as well.

Alex Kelly was 39th at states but his raw score was tied with Yu for 6th/7th at chapters.  Phillips beat Smith at chapters and Smith did not qualify for states, so apparently the AMC 8 isn't worth everything.  (I'd also like to think my coaching helped some)

Nobody knows of any other young students who did well?

um yeah he's in 7th grade and got 3rd (?)

And, Jeffrey An, 8th place, Connie Chen, 4th, both 7th grade

Hmm I know a Connie Chen who is a senior who goes to my school.

Anyway, congrats everyone! I'm excited to see the future of NC ARML and USAMO (and IMO?) participation, even if mathcounts in itself is kind of dumb.

It's Connie Zhou....

She was sitting across from me at the competition....

O well

I think these were the top 10 or so

1.Nick T. Ligon
2.Allen Y. Ligon
3. Calvin D. Ligon
4. Connie Z. Metrolina
5. Tom L. Lincoln
6. Lisa Z. Martin
7. Sammy Luo Hanes
8. Jeffrey A. West Cary
9. Yu Wang Phillips
10. Some wierd-named girl. The last name was like Vezuba or something

Now lets take out the 8th graders =)

3. Calvin D. Ligon
4. Connie Z. Metrolina
7. Sammy Luo Hanes
8. Jeffrey A. West Cary
9. Yu Wang Phillips
10. Some wierd-named girl. The last name was like Vezuba or something

My friends, the next National mathcounts team

3. Calvin D. Ligon
4. Connie Z. Metrolina
7. Sammy Luo Hanes
8. Jeffrey A. West Cary

[quote="not_trig"]err dude this is highly inaccurate

for example

I was like 100th in 7th grade

and 6th in 8th

dude anyway I sucked at states[/quote]

But I don't have the top 100 people

my point exactly?

making Nats predictions doesn't always work :wink:

[quote="greeneggs"]Ligon had the 2nd highest AMC8 school total, while Smith in Chapel Hill took first.  Did Smith come in 2nd at Mathcounts?[/quote]

Smith did not participate at state Mathcounts.  I'm not sure if they competed at the chapter level but they were in one of the toughest districts in NC.

Smith got 6th in chapters and lost to Phillips for state qualifying by one team round question

[quote="n^4+4"]Note that Phillips went on to get 3rd at states even though they barely got 5th at chapters  :D

As far as nats predictions, I can only say that I've been extremely impressed with Yu.  He seems to understand concepts very quickly and has a lot of potential.  The people who make it far in math contests are the ones that work the hardest though, regardless of potential...[/quote]

oversimplification

Perhaps, but it's the sort of oversimplification that encourages people to work... and certain people are reading this that need to work...

By the way, the "weird-named" girl
who got 10th place was Velina.
She goes to Community House. :]

Btw, connie is short for constance. She was rather talkative at our table.... Well, it was mainly her in Lisa. I was busy stealing James' calculator =)

[quote="istos"]... yeaa.
That's pretty inaccurate...
You can't really assume stuff.
I beat Connie at chapter. and
she owned me in state. :[[/quote]

The old official guy told me that she got perfect, so how did you beat her?

[quote="MysticTerminator"]PURE WILLPOWER[/quote]

lol

very efficient, I assure you.

He's old, which means he's almost blind.  Senility does that to you.  At any rate, he probably just needs a stronger prescription for glasses.

[quote="greeneggs"][quote="bisbmard9"]How is that possible? They were the highest AMC8 average in the state, weren't they?[/quote] As a follow-up to my own question, I've compared the Raleigh regional Geometry results with the AMC 8 scores of most of those participants. Lists are published at the State Math Contest website and that of the NC AMC Homepage.

     Numbers 1-20 are the order of the top scorers on the geometry test.

      Name        AMC score

1. Deng           23
2. Yang           24
3. Yang           23
4. Zheng          21
5. Li                23
6. Chen           20
7. Westerhold   17
8. Lee             18
9. Greer           21
9.  Small          21
10. Cao           22
11. Haravu       21
12. Delosreyes  21
13. Palpant
14. Wu            22
15. Joy            19
16. Kim
17. Lu             25
18. Wang
19. Han
20. On             19

I suppose that those who score a 23 or better almost always finish near the top of a subject test. A blank AMC score probably means no test was taken but could also be the result of a score below the level of the published honor roll.[/quote]

James lost to lisa (by 2 points)

[quote="dnkywin"]How is this related to state mathcounts? You should have put that post in state math.[/quote]

An analysis of his question with a few assumptions:

calvin is either one of the following, a alternator (1lie then 1 truth the 1 lie then...), or a knave (all lies.)

Therefore.. if 'How is this related to state mathcounts?' is true.........(its not related)... then you should've put it here... good so far

Since the first statement is TRUE.... this is the right spot to put it

 :lol:

Well, originally I was interested in the corellation between performance on AMC 8 and Mathcounts.
I then ran a comparison with the posted results from the state math regional contest, and followed up with a post on the same thread.
 I see your point, though.

[quote="eshen"][quote="istos"]No, Connie didn't get perfect in chapter,
or else I wouldn't have beaten her....
Elizabeth Shen got a perfect score.
He must have gotten
the asians mixed up...  :D[/quote]

Hmmm... who is this? (Sara???)

Er... I'm Eliz. Shen.[/quote]

who's sara?
me?.
nope.


Find all integers $n,m$ that:
\[m|n^2+1\]
\[n|m^3+1\]

nothing useful in this thread :D :
http://www.mathlinks.ro/Forum/viewtopic.php?t=13875

Nobody is interested in this problem????  :)  :)

Omid, have you solved it? I tried a lot long time before and I couldn"t find anything useful. By the way, where is this problem from? Is it surely solved?

Well I see that it was in on of the exams of Iran team for being ready for IMO 2004.


Yeah, same here.  I'm one of the IMSA kids who got an 18  :? .  Congrats on getting #19 right  :) .

Oh, that was entirely unfair.  This year, the school got a subscription to "Math Horizons" for the upper math kids.  In the problems section in one issue, there was a problem "Prove $\frac{1}{sin\frac{\pi}{7}}=\frac{1}{sin\frac{2\pi}{7}}+\frac{1}{sin\frac{3\pi}{7}}$.  I thought that the identity was really cool.  I saw the problem and remembered the identity.  I figured it had to be the answer.  I didn't actually solve it :)

Hah, my school gets those too... I just never read them... lol.  Actually, after some brute-forcing the problem reduces to $\cos{\frac{7x}{2}}\cos{\frac{3x}{2}}=\cos{\frac{7x}{2}}\cos{\frac{x}{2}}$, so yeah it wasn't that bad.

[edit] Oh wait, they gave us solutions... heh, I kinda missed that.  :)

IMSA had quite a nice turnout seeing as how no seniors ever go (our graduation is same day as ARML). Five of us got into AB team, including me and ThAzN1. Now about that homework...

I've actually done most of it already.  It's really easy to focus on math when I have a 5 page research paper due on Tuesday....

I made the AB team too. I'm a freshman from GBS.

And the homework seems daunting (a little) at first, but it's mostly pretty easy... I mean, the solutions all fit in the space provided. (I only have two problems left. That last one was fun.  :D )

I thought the last one was the most boring.  Only because I've seen it a thousand times and have two solutions at my fingertips.

congrats winluo2 for making it as a freshman, thats quite an accomplishment.  I'm a sophomore at naperville north.  Lowest ABC qualifier with a 12.. personally.. i think I should be on D  :?  Ahh well, unless... was that crossing the desert thing a typical power question or whatever they're called?  cuz i didn't like it... at all... hm, we'll see.

djshowdown: Thanks. Although I disagree about that desert question... I thought it was pretty fun. I've heard the problem before, but never thought about it mathematically. Just as one of those brainteasers.


How many ten letter words can one make using the letters $ A,B$ and $ C$ such that no two consecutive letters are the same?

Well first choice I can choose A,B, or C.
Second I can only choose 2 of the 3 (not repeating)
Third, I can only choose 2 of the 3 (not reapeating from the second one)

and 2 choices so on

so 10 choices is $ 3*2^{9}=1536$.


f(x) = x^2+bx+c
M is the maximum of |f(x)| (-1≤x≤1)
find the minimum of M

the answer is 0.5
I want to know the solution.


Let $p>2$ be a prime number such that $3 \ | \ {p-2}$. Let
$\{S={{y^2}-{x^2}-1} | {x} \ and \ y \ are \ integers \ , 0\le{x,y}\le {p-1}\}$.
Prove that at most $p-1$ elements of $S$ are divisible by $p$.


Let $ K$ is a compact set, $ F$ is subset closed of metric space $ (X,d)$.
Prove: $ K\cap F\ne \varnothing \Leftrightarrow \inf\{d(x,y): x\in K, y\in F\}\equal{}0$

I apologize ahead of time for backseat moderating, but it'd probably be better to have a more descriptive topic title than "Prove!!!!" Otherwise it clutters the forum.

Clearly the implication $ (\implies)$ is true, taking $ x \equal{} y \in K \cap F$.

In the reverse direction, the function $ f: X \to \mathbb{R}$ defined
by  $ f(x) \equal{} \mathrm{inf}\{d(x,y) : y \in F\}$ is continuous. Therefore
it attains a minimum value on $ K$. Therefore there is a $ x \in K$ where
$ f(x) \equal{} 0$, so that $ d(x,F) \equal{} 0$. This means that $ x$ is a limit point of $ F$;
consequently, $ x \in F$ since $ F$ is closed. Therefore $ x \in F \cap K$.


Is there a time limit on the re-entry of MathJam from its starting point? :(

I do not believe so. Some math jams have gone over the 1 hour limit. But the Admins choose to end it.

I must have not been clear. I mean is there a timelimit to the re-entry into the mathjam (can join a mathjam at any time during the "jam") not a timelimit for the total time of the mathjam?

Yes.  After 1 hour (typically) you can no longer enter a Math Jam.


quadratics
algorithm
algebra
difference of squares
special factorizations

Two problems from IMTS

Year 11 Problem 1

Express $\frac {19}{94}$ in form of $\frac {1}{m} + \frac {1}{n}$, where $m$ and $n$ are positive integers.

Year 17 Problem 2

Find all pairs of positive integers $(m,n)$ for which $m^2-n^2 = 1995$.

If this is too hard (I don't think they are even though they're from IMTS), feel free to remove this to appropriate forum.

When answering these, please show your work! :)

Do you mean 19/84 for number one?  And what is IMTS?

for the second question:

factor the difference of squares into (m+n)(m-n)=1995=3*5*7*19. therefore u can divide this problem into different cases.

E.g. case 1: m-n=3, m+n=665, adding these two, we get (334, 331). Repeat for all other factors. Altogether we get 7 possible pairs of (m,n)

(334, 331), (146, 139), (202, 197), (46,11), (74, 59), (62, 55), (58,37)

I second that question: are u sure it's not 19/84? I'm going to answer it on the assumption it is.

1/m+1/n=(m+n)/mn=19/84
thus: m+n=19, mn=84
Using the properties of quadratic equations, this can be seen as the coeff of the equation x^2-19x+84=0 where the roots to this eq is m and n.
Factoring, we get (x-7)(x-12)=0. There are two choices: either m+n=7 or m+n=12. Trying both we find only m+n=7 works. This gives us two solutions to (m,n):

(3,4) and (4,3)

However, if the question isn't 84 but is 94...then I'm stuck, because if u use the same method, you get a quadratic with imaginary roots. however since both m and n are integers, it doesn't work. so i'm hoping it's 84.

Nope. It is 19/94.

Here is how I did. This was pure luck.

[hide="Sol"]

Since 19/94 = 0.2021276596.., and 1/5 = 0.2, I decided to subtract 1/5 from 19/94. The result is 0.0021276596... And luckily, this corresponds to 1/470. Therefore, m and n are 5 and 470.[/hide]

I'm not sure whether there's a better way (I'm pretty sure there is) but I'm not confident on what that way is.

For the second one, what about {998,997}? :lol:

[quote]Factoring, we get (x-7)(x-12)=0. There are two choices: either m+n=7 or m+n=12. Trying both we find only m+n=7 works. This gives us two solutions to (m,n):

(3,4) and (4,3)[/quote]
I agree with all your math but wouldn't you have the final combinations of (m,n) as (12,7) and (7,12) not (3,4) and (4,3)?

The way I initally tried to do the first one was by writing (1/m)+(1/n) as
(m+n)/(mn).  Therefore m+n=19 and mn=94.  But Silverfalcon's answer doesn't work for this even though it works when you plug it in.  What is wrong with my logic?

[quote="eryaman"]The way I initally tried to do the first one was by writing (1/m)+(1/n) as
(m+n)/(mn).  Therefore m+n=19 and mn=94.  But Silverfalcon's answer doesn't work for this even though it works when you plug it in wrong.  What is wrong with my logic?[/quote]

How do you know that m+n is 19 and mn is 94...can't they be, say, 19a and 94a, where a is some constant?

number 1 is simple with greedy algorithm for what are known as egyptian fractions, i.e. fractions made up of unit fractions ($\frac{1}{k}$ for some positive integer k). for problems like this, greedy algorithm usually works; that is, take out the largest unit fraction possible. From what remains, take out the largest unit fraction. Repeat.

in fact, the greedy algorithm always works.  if i'm not wrong that is.

you just need to show that after every step in the greedy algorithm, the numerator of the resulting fraction is strictly decreasing, so at some point it'll be equal to 1.

let your fraction be p/q where (p,q)=1 here p, q positive integers
find integer m s.t. 1/(m-1) > p/q  >= 1/m.
then p/q - 1/m = (pm-q)/mq.
if pm-q >= p
then p(m-1) >= q, p/q >= 1/(m-1) contradiction.
thus the numerator always decreases.

thus every rational number can be expressed as the sum of distinct unit fractions.

[quote]In fact, the greedy algorithm always works. if i'm not wrong that is. [/quote]

I meant that the greedy algorithm will not always generate the most efficient solution. But you are correct too, kennyworm.

For the second question, I think mandy_pal just made a small calculation error. It should be (62,43) :P


Let a positive integer $ n\ge 2$ and $ q$ is the smallest prime divisor of $ n$. Prove that all prime divisors of $ 2^n\minus{}1$ are greater than $ q$.

[hide="Solution"]
Assume that $ 2^n \minus{} 1\equiv 0\bmod p$, where $ p$ is a prime less than or equal to $ q$.  Then, we have that $ 2^n \equiv 1\bmod p$, so letting the order of $ 2\bmod p$ be $ k$, we have that $ k|n$ and $ k|(p \minus{} 1)$, so $ k|\gcd (n, p \minus{} 1)$.  Since $ n$ has all of its divisors other than one greater than or equal to $ q > p \minus{} 1$ and $ p \minus{} 1$ has all of its divisors less than or equal to $ p \minus{} 1$, we have that $ \gcd (n, p \minus{} 1) \equal{} 1\implies k \equal{} 1$, so $ 2\equiv 1\bmod p$, which is impossible, so $ p > q$. [/hide]


linear algebra
matrix
inequalities
triangle inequality
number theory
linear algebra theorems

I want to find these two things: $ \|L\|$ and $ \|L\|_{tr}$ if $ L$ is a linear mapping from $ \mathbb{R}^2$ to $ \mathbb{R}^2$ and $ L$ has matrix

$ [L] \equal{} \begin{bmatrix} 1 & 1 \\
0 & 1 \end{bmatrix}$

 :huh: Is $ \|L\|$ just the det. here?

$ \|L\|$ can mean many different things. Determinant is not likely. (Since the determinant doesn't scale linearly, it doesn't much resemble a norm.) I'm afraid we can't answer this without knowing what the question is.

The norm $ \|\cdot\|$ is just the determinant. But how do you calculate the trace norm?

Thanks in advance,

Los

That's nonstandard terminology, so we can't be sure. Of the norms I know, the most likely is the Frobenius norm $ \|A\|\equal{}\sqrt{\text{tr}(A^TA)}$. This is associated to the inner product $ \langle A,B\rangle\equal{}\text{tr}(A^TB)$.

If the field is $ \mathbb{C}$, replace transpose by Hermitian adjoint $ (\overline{A})^T$.

This norm scales linearly; its most undesirable property is that it isn't 1 at the identity, so equality in the inequality $ \|AB\|\le \|A\|\|B\|$ is rare.

In standard language, the determinant is not a matrix norm. It blatantly fails the triangle inequality, and isn't even positive. The only context in which I've seen a determinant referred to as a norm is number theory, where the multplicative property is essential and we don't care about scaling as long as it's homogeneous.

Trace norm is nonstandard?

Wikipedia has the following definition:

The '''trace norm''' is defined as
$ \|A\|_{tr}
\equal{}\operatorname{trace}(\sqrt{A^*A})\equal{}\sum_{i\equal{}1}^{\min\{m,\,n\}} \sigma_{i}.$

I'm not sure how to apply this!

Los  :oops:

The two definitions you just gave contradict each other. One is the norm I mentioned, and the other is the 2-norm $ \max_{x\neq 0}\frac {\|Ax\|_2}{\|x\|_2}$.

The first is $ \sqrt {n}$ at the identity, and the second is $ 1$.

Actually, in retrospect, the first version you gave is almost certainly a mistake. Taking the square root before the trace is just weird. It does have a unique meaning (the positive definite square root), but I don't think it's intended.


How about can complex nos. be primes,imaginary are ilke 6i,so....

I think he is inquirying about objects called 'gaussian primes'. Go google it or ask the question in a more advanced forum. This is only Getting Started forum.

Should I move it to Intermeditate topics, or some other forum?

What is a prime number?  :)

Asking yourself that question gets you into some very complicated stuff.  Let me try to explain it without getting bogged down in the complications.

The normal primes that you see are primes like $7$.  The reason they're prime is because we can't find two integers $a$ and $b$ such that $a \cdot b = 7$ without having one of those integers be $1$ or $-1$.  Now, $1$ and $-1$ are called [b]units[/b] because of this special property.

The set of integers (we certainly can't talk about whether, say, $\pi$ is a prime!) is called $\mathbb{Z}$, and $1$ and $-1$ are the units of $\mathbb{Z}$.

In complex numbers (more precisely, in the set of complex numbers with integer coefficients, or $\mathbb{Z}[i]$, or the Gaussian integers), on the other hand, there are four units:  $1, -1, i, -i$.

So, we can define a "complex prime," or a "Gaussian prime," $a + bi$ where $a, b$ are integers, such that we can't find two other complex numbers $c + di, e + fi$ such that $(c + di)(e + fi) = a + bi$ without one or the other of the complex numbers being a unit.

For example, $1 + 2i$ is a Gaussian prime, but $1 + 3i$ is not because

$(1 - i)(-1 + 2i) = 1 + 3i$

This is a very confusing subject.  :P


I have an another Algebra test on Friday.... Any tips for not getting overly nervous...? I hate them...

Any particular part of algebra? It's pretty extensive  :roll:

Don't study the day before the test..and eat a good meal...then study before the test.. :lol:

It's only a school algebra test I think.

That method usually works. But verridith $ \neq$ hunter34. So I don't know.


If $f(x)=5x^2+3x+4$, what is the value of $f(-2)$?

[hide]

$\displaystyle \displaystyle f(-2) = 5(-2)^2+3(-2)+4$
$\displaystyle \displaystyle f(-2) = 5 * 4 + (-6) + 4$

So:

$\framebox{ f(-2) = 18}$

[/hide]


How to carry out the following conversion?  :maybe:

A possible strategy is to begin with an assymetric epoxidation (or just epoxidate and then separate enantiomers). Another path is simply to react with two equivalents of lithium dimethylcuprate, and then separate the stereoisomers (you should get mainly two).

How do you add methyl to the epoxide formed?

You have to protect the carbonyl group, because addition to it is faster than addition to epoxide.

Wouldnt the gillman reagent give 1,4 product in majority due to weak nucleophilic character?

No. With the cuprate reagent there is first conjugate addition, which yields an enolate anion, which can then react with a second equivalent of the Gilman reagent to give the desired product (as a mixture).

How does the enolate ion react with another carbanion? What is the site of attack?

:fool: Oops, my mistake. After Gilman reagent and obtaining an enolate anion, we must obviously add an electrophile, in this case, mehtyl iodide for example.


yea probably a dumb question but how would you find the roots to

$x^4+x^3+x^2+x+1$ without using a calculator?


Please help me with this cyclic inequality. I have stuck at this problem for a long time.

let a,b,c>0, abc=1
prove that 1/b(a+b)  +  1/c(b+c)  + 1/a(c+a)  >=3/2

Substituting $ a \equal{} \frac {x}{y}$,  $ b \equal{} \frac {y}{z}$, $ c \equal{} \frac {z}{x}$, we find that the inequality is equivalent to

$ \sum_{cyc} \frac {z^2}{y^2 \plus{} zx} \geq \frac {3}{2}$ for positive real numbers $ x,y,z$.

By Cauchy-Schwartz, we obtain

$ LHS \equal{} \sum_{cyc} \frac {z^4}{y^2 z^2 \plus{} z^3 x} \geq \frac {(x^2 \plus{} y^2 \plus{} z^2 )^2 }{\sum_{cyc} (x^2 y^2 \plus{} xz^3 )}$

So, it suffices to show that $ 2(x^2 \plus{} y^2 \plus{} z^2 )^2 \geq 3(x^2 y^2 \plus{} y^2 z^2 \plus{} z^2 x^2 \plus{} x^3 y \plus{} y^3 z \plus{} z^3 x)$

which can be proved by adding the following inequalities:

$ x^4 \plus{} y^4 \plus{} z^4 \geq x^3 y \plus{} y^3 z \plus{} z^3 x$ (AM-GM)
$ (x^4 \plus{} x^2 y^2 ) \plus{} (y^4 \plus{} y^2 z^2 ) \plus{} (z^4 \plus{} z^2 x^2 ) \geq 2(x^3 y \plus{} y^3 z \plus{} z^3 x)$ (AM-GM)

It is also equivalent to

$ \frac {1}{b(a\plus{}b)} \plus{} \frac {1}{c(c\plus{}a)}\plus{} \frac {1}{a(c\plus{}a)} \geq \frac {3}{1\plus{}abc}$
which is well known


Hi,
I have just been "messing" around with LaTeX and tried to compile a document with the following information:

[code] \documentclass[12pt,a4paper]{article}
 \usepackage[dvips]{color}
 \usepackage{multicol}
 \pagestyle{empty}
 \setlength{\oddsidemargin}{0pt}
 \setlength{\textwidth}{16cm}
 \setlength{\textheight}{22cm}
 \setlength{\parindent}{0pt}
 \setlength{\parskip}{0pt}

 \begin{document}
 \renewcommand*{\DefineNamedColor}[4]{%
    \textcolor[named]{#2}{\rule{7mm}{7mm}}\quad
    \texttt{#2}\strut\\}

 \begin{center}\Large Named colours in \texttt{dvipsnam.def}
 \end{center}
 \begin{multicols}{3}
 \input{dvipsnam.def}
 \end{multicols}
 \end{document}  [/code]

You're meant to then get this: http://tex.loria.fr/graph-pack/grf/gr2.htm , however the colours on my document are all black! Is there any way in which I can resolve this?

Many thanks for any help.

[quote="stevem"]Make sure you compile using LaTeX => PS => PDF which will use dvips (ie DVI->PS) to compile.[/quote] Hi Steve,

I only have the following three options:

LaTeX =>PS
LaTeX => PDF
LaTeX => DVI

I chose the PS one but it says "cannot excute". I think the problem might be with my output profiles because I've never used the other two, but no error message appears. :maybe:

You need to set up the LaTeX => PS => PDF profile which means that you must also have Ghostscript installed (which does the PS to PDF bit). The easiest way is to download a profile. Get the file latex-ps-pdf.zip in the second post at http://www.latex-community.org/viewtopic.php?f=31&t=302 Unzip it and import it into TexnicCenter via Build, Define Output Profiles, Import.
Then you must change the paths in LaTeX, Postprocessor and Viewer tabs.

Hi Steve,

Thanks a lot it worked! I once again didn't read your whole reply and was going to post "It doesn't work" again, but thankfully I then spotted your last line. (Which was a tedious task indeed!)

Cheers.   :)


Let $x_1,x_2,...,x_{2005}\geq 0$ with $x_1+x_2+...+x_{2005}=1$.
Find the maximum value of
$S=\sum_{k=1}^{2005}\sqrt{x_k^2+x_k}$.

(any solution is welcome, but I'm especially interested in elementary solutions)

But relation
$\sqrt{x_k^2+x_k}\leq \frac{x_k+(x_k+1)}{2}$
becomes an equality only if $x_k=x_k+1$, which is impossible ([b]this[/b] was the trap, btw  :P  ).

And for $x_k=\frac 1{2005}$ we don't get $S=\frac{2007}{2}$  :!:

Sorry but,  Yes, I noticed it. Let's start the discussion about this problem.

Great! :)

We could have also used Jensen inequality for $f(x)=\sqrt{x^2+x}$ (since f is concave on $[0.\infty)$, we have    $f(\frac{x_1+x_2+...+x_n}{n})\geq \frac{f(x_1)+...+f(x_n)}{n}$ and equality holds iff $x_1=x_2=...=x_n$).


You have two pieces of rope. Each, once lit on fire, will completely burn in exactly one hour.

The ropes are not of uniform width you so you cannot be sure that, for example, a rope burnt halfway has been burning for 30 minutes. However, how do you measure exactly 45 minutes of time?

[quote="randomdragoon"]You have two pieces of rope. Both, once lit on fire, will completely burn in exactly one hour.

The ropes are not of uniform width you so you cannot be sure that, for example, a rope burnt halfway has been burning for 30 minutes. However, how do you measure exactly 45 minutes of time?[/quote]

get a stopwatch lol

[hide]dunno[/hide]

You must mean "each" rope burns for 1 hour, and not "both" which would often get married with "together".

solution

[hide]take rope A and burn from both sides.
simaltenously burn B from one side.
when A is burnt completely, 30 min have passed.
now set Bs other end on fire as well. now another 15 mins pass before B is finished.30+15=45.[/hide]

rishabh,

without giving it away, your solution would be clearer/more accurate if you changed "sides" and "side," respectively, to
"ends" and "end," respectively.


1)The product of two positive integers x and y is equal to 3x+4y. What is the largest possible value of x?

Honestly I don't get how you finished this problem either (I mean, the algebra is obvous of course, but then ?) However you did it, thats not the right answer

Now, thales got (15,4) which doesn't make sense.

Heres how I got my answer:

I'm with thales up to x = 4 + 12/(y - 3).
The denominator should obviously be as small as possible, so y = 4.  then you get x as 16.

x = 16
y = 4

sorry

i type  wrong

${f(2)=-8\land f(4)=16}$

[hide="much much much simpler"]
$xy=3x+4y$
$(x-4)(y-3)=12$
x is maxed out at 16. ((16,4))[/hide]

Thales, please don't post advanced methods such as limits, they aren't included in the MATHCOUNTS curriculum; using them is ok and can be helpful to your problem solving skills.  Bpms' method is much more traditional and intended.


calculus
integration
function
real analysis
real analysis unsolved

Let $ f: [a,b] \rightarrow \mathbb R$ to be a bounded variation function with variation $ V_{f}(a,b)$. Show that:

\[ \int_{[a,b]}{|f'|} \leq V_{f}(a,b)\]

where the integral is the lebesgue integral.

This is what i´ve done by now:

Let $ \varphi$ any symple function with $ 0 \leq \varphi \leq |f'|$. Then we can write,
\[ \varphi \equal{} \sum_{i \equal{} 1}^{n}{a_j (x_j \minus{} x_{j \minus{} 1})}\]
, where $ a \equal{} x_0 < x_1 < ... < x_n \equal{} b$ and $ a_j \leq |f'(x)|$ for all $ x \in [x_{j \minus{} 1},x_j]$ in which |f'| is defined.

Now, I´m in doubt if we can use the mean valeu theorem to $ f$ because $ f$ is not differentiable, but, differetiable almost everywhere.

:(

You wrote some strange things there. Are you familiar with the fact that $ \int_a^b f' \leq f(b) \minus{} f(a)$ if $ f$ is increasing?

First prove that $ |f'(x)| \le V_f ' (a,x)$ for a.e. $ x$ (hint: it isn't hard!). Coupled it with the fact mentioned in WWW's message, you're almost done.


t(x)=p(x)/q(x) is a rational function.  Fill in the blanks:
t(x) has a bounce point at -4.  Therefore _____ has a factor of _____.
(would this be  __[u]p[/u]__ has a factor of __[u]4[/u]__?
t(x) has a vertical asymptote at x=3.  Therefore _____ has a factor of_____.
t(x) has a slant asymptote.  What can you say about p(x) and q(x)?

I'm not sure what a bounce point is, but as for the others:

q(x) has a factor of 3 [q(3) = 0, which makes a vertical asymptote at x=3.
and p(x)'s degree is greater than q(x)'s


[9:02] * MysticTerminator HAS SEXY LEGS

[8:22] mewto55555: 40 is th only number

[11:30] MysticTerminator: YOU CAN BE MY HONEY
[11:30] hxvlnmt321: sure

[9:12] isabella2296: [asy]size(500);void kn(int n) {pair[] points;for(int i=0; i<n; ++i) {points[i]=dir(360.0/n*i);}for(int i=0; i<n; ++i) {for(int j=i+1; j<n; ++j) {draw(points[i]--points[j]);}}}kn(45);[/asy]

Got more?

[9:11]xoangieexo: * MysticTerminator HAS SEXY LEGS
[8:22] xoangieexo: I AM KIRBY!

i was copying and pasting :P
and the other one
i said it cuz meta knight was there :D

[9:20] mewto55555: now where did i put my dentures

[9:24] * Welcome to AoPS For The Win! Alpha Edition. Please review terms of service regularly. Searching for ratings bug - expect game server to occasionally go down.

[6:59] xoangieexo: MYSTIC JUST SAID SOMETHING
[6:59] MysticTerminator: I DID?
[6:59] MysticTerminator: DOESN'T THAT HAPPEN PRETTY OFTEN

[9:58] MysticTerminator: YOU KNOW WHAT I DON'T GET
[9:58] MysticTerminator: HOMOSEXUALITY
[9:58] MysticTerminator: I THINK THAT IS NOT A TOPIC TO DISCUSS IN ROUND TABLE THOUGH

[9:01] Meta Knight: I Want to learn how to tackle Richard Rusczyk and Mathew Crawford

[7:04] MysticTerminator: DANG YO
[7:04] MysticTerminator: I COULD TOTALLY SHOW YOU
[7:04] MysticTerminator: I HAVE TACKLED A LOT OF PEOPLE

Still, mewto55555's "Where did I put my dentures?" was funny.

[10:30] mathblitz: i was a toga dressed monkey....

[10:27] jjx1: i got a weird story to tell you angie
[10:27] xoangieexo: OH
[10:28] xoangieexo: OK
[10:28] xoangieexo: WHAT
[10:28] jjx1: it was on saturday night
[10:28] 27r: involving ruth
[10:28] jjx1: okay
[10:28] jjx1: so i woke up around like 2:30
[10:28] xoangieexo: UH HUH
[10:28] jjx1: and i looked around
[10:28] mathblitz: and...
[10:28] [27r]: on the bed
[10:28] mathblitz: yes...
[10:28] jjx1: and iwas in a weird room
[10:28] xoangieexo: WITH RUTH
[10:28] 27r: maybe RUTH'S ROOM?
[10:28] mathblitz: with...
[10:28] jjx1: i looked to the right
[10:28] mathblitz: nah
[10:28] xoangieexo: UH HUH
[10:28] mathblitz: hotel
[10:28] jjx1: and guess what i saw
[10:28] xoangieexo: RUTH
[10:28] 27r: RUTH
[10:28] mathblitz: but ruth there
[10:29] xoangieexo: CLOWN
[10:29] jjx1: no
[10:29] xoangieexo: CLOWN
[10:29] mathblitz: lol
[10:29] 27r: YO MOMMA
[10:29] jjx1: I SAW THE AMAZON RIVER
[10:29] xoangieexo: UH
[10:29] jjx1: and then i woke up
[10:29] Neji0313: yo mammas mamma
[10:29] mathblitz: thats nothing
[10:29] jjx1: it was a dream
[10:29] 27r: THATS... FASCINATING
[10:29] xoangieexo: UHH...

[13:00] splatyango: angie is a dictator
OHHHHHHHHHHHH WHAT NOW?

lol. niiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiice, guys.  :rotfl:  :rotfl:  :rotfl:   btw, who's ruth? gf? bff?

his crush lol

[16:07] [-->billybobjoe95]: all guys LOVE to shop[16:07] [billybobjoe95]: yah thats what i do alot

ahahah you just had to say the wrong thing at the wrong time... rotfl

[quote="xoangieexo"] [13:00] splatyango: angie is a dictator
OHHHHHHHHHHHH WHAT NOW?[/quote]

WHAT NOW? WE OVERTHROW YOU. REVOLT!!!!  :D

mathblitz: jjx1...you are now the charman
mz94: hi charman(der)!
mathblitz: *chairman
xoangieexo: CHARMANDER! LOL

xoangieexo: CHARMANDER!
jjx1: what?

xDDDD
and now you know the origin of jjx's new nickname  :P

ha i feel so bad for not_trig:

[18:53] not_trig: umm
[18:53] not_trig: some people call me "trig"
[18:53] not_trig: but my name clearly implies
[18:53] not_trig: otherwise
[18:53] not_trig: XD
[18:53] xoangieexo: lol
[18:53] isabella2296: hi pplz!
[18:53] xoangieexo: wait... you dislike trig?
[18:53] xoangieexo: hi izzy
[18:53] isabella2296: hi angie!!
[18:53] not_trig: what
[18:53] isabella2296: hi hurdy!!
[18:53] not_trig: I am just bad at trig
[18:54] isabella2296: hi trig!
[18:54] xoangieexo: NOOOOOOOOOO
[18:54] not_trig: however I am a lot better than I used to
[18:54] not_trig: HAHAHA
[18:54] not_trig: OH MAN
[18:54] not_trig: SEE

later...

[18:56] xoangieexo: aww bye not_trig
[18:56] xoangieexo: ha izzy say bye
[18:56] isabella2296: 9% of a possible score *sniff*
[18:56] isabella2296: aww bye tirg!
[18:56] xoangieexo: i wonder waht she would say
[18:56] isabella2296: *trig
[18:56] xoangieexo: HAHAHAHHA
[18:56] xoangieexo: AHHAHAHAHAHAHHAHA
[18:56] isabella2296: tirg lol
[18:56] xoangieexo: HAHAHAHAHAHHAHAHAHAHAHHAHAHAH
[18:56] xoangieexo: dude i feel so bad for you
[18:57] xoangieexo: ahhahahahahahahhahahahahah

ha izzy has no idea...

lol, i think the goal of this thread is to mute ppl

 :rotfl:

[18:53] mz94: mew is officially a genius

[18:54] FantasyLover: uhh you gave me this??
[18:54] mz94: as did my virginian friends

[19:54] pianoforte: WHY ARE WE TALKING ABOUT TAPPING?
[19:54] mz94: WHY IS CHESS64 RICKROLLING
[19:54] chess64: IIIIIIIII JUST WANNA TEEEELL YOU HOW I'M FEEEEEELIN
[19:55] pianoforte: WHY IS CHESS64 HOTT HOTT HOTT?

[10:05] xoangieexo: KIRBY DOESNT SUCK
[10:05] xoangieexo: HE SWALLOWS
[10:05] nikeballa96: whats not nice?
[10:05] xoangieexo: oh dam
[10:05] pianoforte: FINE
[10:05] nikeballa96: i didnt say anthing
[10:05] xoangieexo: that sounded so - wrong

crazychinesemaniac: my anus is bleeding!
xoangieexo: YAY!

 :rotfl:  :rotfl:

angie was talking to me btw

[11:13] pianoforte: I HAVE A GOOD PARTNER TOO
[11:13] not_trig: BECAUSE T IS NOT VERY HELPFUL AT 2V2
[11:13] pianoforte: BUT NOT FOR STARCRAFT
[11:13] not_trig: UMMM SURE PIANO

[11:15] * You are muted for violation of the Terms of Service

 :rotfl:  :rotfl:  :rotfl:

[quote="mz94"][quote="johnnybeard"][10:15] xoangieexo: woah
[10:15] Dojo: whoa
[10:15] smellydonut: I BEAT SUPER
[10:15] isabella2296: SUPER
[10:15] mz94: o_o
[10:15] smellydonut: I BEAT SUPER!
[10:15] isabella2296: OMG SUPER
[10:15] superllamel: meh
[10:15] superllamel: lol
[10:15] smellydonut: OMG I BEAT SUPER!!!!!!!!!!!!!
[10:15] Dojo: oh no super

YEAH THATS RIGHT!!! I PWNED SUPERLLAMEL!!!!!!!!!!!!!!!! OH YEAH!!![/quote]

YOURE SO SMART DUDE... :wink:[/quote]

HAHAHAHAHAHAHAHAHHAHAHAHAH

CAUGHT YOU!

WRONG ACCOUNT!!!

AHAHAHAHAHHAHAHAHAHAHAHA

JOHNNYBEARD WASN'T ON FTW AT THAT TIME

HA!

good job johnnybeard  :D u did a great job giving urself away

you have exactly 0 posts

do you know how much that sucks

wow i just got banned

do you know how much that sucks

OMG johnnybeards really realllly stupid.

did he really beat superllamel though??? thats scary...still super good to beat someone like that!


Determine all integers greater than 1 such that $\frac{(2^{n}+1)}{n^{2}}$ is an integer.

Is this not a little bit hard for pre-olympiad?

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=366466#p366466]IMO 1990/3[/url]

Seems pretty easy, actually...
[hide]
If $n$ is even, the statement is impossible, because we have odd divided by even.  Thus, assume $n$ is odd. \[2^{n}\equiv-1\mod{n^{2}}\iff 2^{2n}\equiv 1\mod{n^{2}}\]  \[\iff \phi(n^{2})|2n\] But $\phi(n^{2})=n\phi(n)$, by a completment argument (take all the naturals not relatively prime to $n$ less than n, and add $0n, n, 2n, ..., (n-1)n$ to each to get the number of naturals less than $n^{2}$ not relatively prime to $n^{2}$: namely, $n$ times as many). \[\phi(n^{2})|2n \iff \phi(n)|2 \iff \phi(n)=1, 2\] But \[\phi(n)=n\prod_{p|n}{(1-\frac{1}{p})}\] Thus, $\phi(n)=1 \iff n=1, 2$, $\phi(n)=2 \iff n=2^{2}, 3, 2*3$.
We already discarded even $n$, so that leaves our two solutions, $\boxed{n=1, n=3}$...[/hide]

EDIT: Thanks to Arnav for the correction...there is an error here...

[quote="K81o7"]Seems about right for Pre-Olympiad, actually...
[hide]
If $n$ is even, the statement is impossible, because we have odd divided by even.  Thus, assume $n$ is odd. \[2^{n}\equiv-1\mod{n^{2}}\iff 2^{2n}\equiv 1\mod{n^{2}}\]  \[\iff \phi(n^{2})|2n\] But $\phi(n^{2})=n\phi(n)$, by a completment argument (take all the naturals not relatively prime to $n$ less than n, and add $0n, n, 2n, ..., (n-1)n$ to each to get the number of naturals less than $n^{2}$ not relatively prime to $n^{2}$: namely, $n$ times as many). \[\phi(n^{2})|2n \iff \phi(n)|2 \iff \phi(n)=1, 2\] But \[\phi(n)=n\prod_{p|n}{(1-\frac{1}{p})}\] Thus, $\phi(n)=1 \iff n=1, 2$, $\phi(n)=2 \iff n=2^{2}, 3, 2*3$.
We already discarded even $n$, so that leaves our two solutions, $\boxed{n=1, n=3}$...[/hide][/quote]

Sorry, why does phi(n^2) divide 2n?

Hm, I did miss something then?

Oh...apparently I did...you're right

I assumed 2 had order $\phi(n^{2})$, when really we only know $2^{\phi(n^{2})}\equiv 1\mod{\phi(n^{2})}$

Sorry for bumping this, but is the blog still available? Can someone share the link?


A natural number $ n$ is said to be blue if the sum of its digits are equal to the sum of the digits of $ 3n \plus{} 11$. Prove that there is an infinite amount of blue numbers.

$ 8 \times 10^{k} (k \geq 2)$ are blue numbers.


integration
limit
real analysis
real analysis unsolved

Suppose $ \mu$ is a positive measure on $ X$ ,$ \mu (X) < \infty$, $ f \in L^\infty (\mu )$, $ \| f \|_\infty > 0$, and $ \alpha_n \equal{}  \int_X |f|^n d \mu \quad (n \equal{} 1,2,3,\ldots )$. Prove that $ \lim \frac{\alpha_{n\plus{}1}}{\alpha_n} \equal{} \| f \|_\infty$ .

we have $ a_{n+1}\leq a_n\parallel{}f\parallel{}_{\infty}$, then $ \limsup\limits_{n\to\infty}\frac{a_{n+1}}{a_n}\leq\parallel{}f\parallel{}_{\infty}$.
for all $ 0<\epsilon<\parallel{}f\parallel{}_{\infty}$, there is a subset $ A_{\epsilon}\subset X$ such that $ f(x)\geq \parallel{}f\parallel{}_{\infty}-\epsilon$ $ \forall x\in A_{\epsilon}$. we put $ b_n=\frac{1}{\mu(A_{\epsilon})}\int\limits_{A_{\epsilon}}f^n(x)d\mu\geq (\parallel{}f\parallel{}_{\infty}-\epsilon)^n$
We have
$ a_{n+1}\geq (\parallel{}f\parallel{}_{\infty}-\epsilon)\mu(A_{\epsilon})b_n$
$ a_n=\int\limits_{A_{\epsilon}^c}f^n(x)d\mu+\int\limits_{A_{\epsilon}}f^n(x)d\mu=\int\limits_{A_{\epsilon}^c}f^n(x)d\mu+\mu(A_{\epsilon})b_n$
Therefore
\[ \frac{a_{n+1}}{a_n}\geq \frac{\parallel{}f\parallel{}_{\infty}-\epsilon}{1+\frac{1}{\mu(A_{\epsilon})}\int\limits_{A_{\epsilon}^c}\frac{f^n(x)}{b_n}d\mu\]
Let $ n\to\infty$ we obtain
$ \liminf\limits_{n\to\infty}\frac{a_{n+1}}{a_n}\geq \parallel{}f\parallel{}_{\infty}-\epsilon$ $ \forall 0<\epsilon<\parallel{}f\parallel{}_{\infty}$ thus $ \liminf\limits_{n\to\infty}\frac{a_{n+1}}{a_n}\geq \parallel{}f\parallel{}_{\infty}$.
Because $ \frac{f^n(x)}{b_n}\leq (\frac{f(x)}{\parallel{}f\parallel{}_{\infty}-\epsilon})^n\to 0$ when $ n\to\infty$ by $ f(x)< \parallel{}f\parallel{}_{\infty}-\epsilon$ in $ A_{\epsilon}^c$.
by the Lebesgue's dominated convergence theorem, we have
\[ \int\limits_{A_{\epsilon}^c}\frac{f^n(x)}{b_n}d\mu\to 0\]
Two results above finished our proof.


I have to design and build a device using 5 sheets of regular paper and some tape to hold an egg and make the egg survive a 2 meter drop without cracking. Any advice on how i should make it?

I know that type of problem.
The best is as good as always to build some kind of parachute (if you can manage to get this out of what you have, e.g. the sheets of paper should be big enough). This methode has also the advantage that it will probably work from any height.

The general idea is that you want the egg to hit the floor with as low a velocity as possible.

When dropping an object, air resistance will cause it to accelerate to a maximal velocity, known as the "terminal velocity," which is usually a function of the surface area and other geometric properties of the object.  We want to make the force of air resistance as large as possible, and usually the easiest way to do that is by the expedient of a parachute.

The reason this technique will work from any height is that the terminal velocity is independent of height.  Once an object reaches its terminal velocity, it will not accelerate faster unless its geometry changes.

Please discuss this in the physics board. (Moving from RT)

Though the parachute solution seems feasible in theory; you might find it infeasible in practice.

The concept of a parachute is that when deployed the drag force on the large area of the canopy reduces the terminal velocity of the parachutist to a level such that the parachuist's impact with the ground is essentially harmless.

Nonetheless, there is a shock when that impact with the ground occurs that an eggshell might not withstand.

The momentum of the egg as it falls is given by: $\vec{p}= m\vec{v}_{term}.$

As the egg impacts the ground its velocity changes: $\vec{v}_{term}\to 0.$

It follows then that $\Delta \vec{v}= 0-\vec{v}_{term}=-\vec{v}_{term}.$

The change in momentum resulting from this impact is: $\Delta \vec{p}= m\Delta \vec{v}=m\frac{\Delta \vec{v}}{\Delta t}\Delta t.$

$m\frac{\Delta \vec{v}}{\Delta t}= \vec{F}$

$\Rightarrow \Delta \vec{p}= m\Delta \vec{v}=-m\vec{v}_{term}= \vec{F}\Delta t.$

As $\Delta t \to \epsilon : \vec{F}=-\frac{m\vec{v}_{term}}{\epsilon}.$

The smaller the time it takes the egg to come to a stop, the greater the shock force on the shell, regardless of how small the terminal velocity might be. (This is why parachuists "crumple" and fall down when they land -- so the shock of impact isn't transmitted to their bones.) Furthermore, the breaking strength of an eggshell is miniscule to begin with. With that in mind, my advice would be to take some of your paper and make a shock absorbing cushion with it.

I also have to do that for my physics class too, except that my egg seems to lounge toward the hard floor faster than everybody else


algebra
polynomial
Vieta
algebra unsolved

Let $ r,s,$ and $ t$ be the roots of the cubic polynomial: $ p(x)\equal{}x^3\minus{}2007x\plus{}2002.$

Determine the value of: $ \frac{r\minus{}1}{r\plus{}1}\plus{}\frac{s\minus{}1}{s\plus{}1}\plus{}\frac{t\minus{}1}{t\plus{}1}$.

Isn't this problem just a computational matter?

Find $ \sigma_1 \equal{} s\plus{}r\plus{}t$, $ \sigma_2 \equal{} sr\plus{}rs\plus{}st$ and $ \sigma_3 \equal{} rst$ using Vieta's formulas. Then the sought fraction is $ \frac{3\sigma_3 \plus{} \sigma_2 \minus{} \sigma_1 \minus{} 3}{\sigma_3 \plus{} \sigma_2 \plus{} \sigma_1 \plus{} 1}$.

It's easier to let $ g(x) \equal{} \frac{x \minus{} 1}{x \plus{} 1}$ and compute the coefficients of $ p(g^{\minus{}1}(x))$.

oh!!!so its already a depressed cubic..well just apply formula due  to cardona...efficient n straight forward  n den divide p(x) by (x-r) to get s n t...n then simple computation[/quote]

The given expression may be written as $ 3 \minus{} 2 \left[\frac{1}{r\plus{}1}\plus{}\frac{1}{s\plus{}1}\plus{}\frac{1}{t\plus{}1} \right]$

Now, if (r,s, t) are roots of $ x^3 \minus{} 2007x \plus{} 2002 \equal{} 0$ then (r+1,s+1,t+1) are roots of

$ (y\minus{}1)^3 \plus{} 2007 (y\minus{}1) \plus{} 2002 \equal{} 0$ which is equivalent to $ y^3\minus{}3y^2 \plus{} 2010y \minus{}6 \equal{} 0$

Further, $ \left(\frac{1}{r\plus{}1},\frac{1}{s\plus{}1},\frac{1}{t\plus{}1} \right)$ are roots of the reciprocal equation to the last one i.e. $ 6z^3 \minus{} 2010 z^2 \plus{} 3z^2 \minus{}1 \equal{} 0$

Hence, $ \left[\frac{1}{r\plus{}1}\plus{}\frac{1}{s\plus{}1}\plus{}\frac{1}{t\plus{}1} \right] \equal{} \frac{2010}{6}$ and

hence the expression evaluates to 338.

can some one plz solve this?
8 sin^3x -6 sinx +1=0

Assume the roots of the equation $x^3-2007x+2002=0$ are $\alpha,\beta,\gamma$ By TOR we get, $$\frac{\alpha-1}{\alpha+1}=t\implies\boxed{\alpha=\frac{t+1}{1-t}}$$ putting thus value on given equation we get, $$\left(\frac{t+1}{1-t}\right)^3-2007\left(\frac{t+1}{1-t}\right)+2002=0$$ $$\implies(t+1)^3-2007(1-t)^2(t+1)+2002=0\implies\boxed{2006t^3-2010t^2-2010t+4=0}$$ Therefore we get, $$\sum_{cyc}{}\left(\frac{\alpha-1}{\alpha+1}\right)=\frac{1005}{1003}$$
Vieta's Relation!!!! :love:


linear algebra
calculus
calculus computations

Let $ \alpha ,\ \beta$ be positive real numbers such that $ \alpha ^ 2 \plus{} \beta ^2 \equal{} \alpha ^3 \plus{} \beta ^3$. Find the range of $ \alpha \plus{} \beta$.

[hide]
The minimum value must be $ 0$, and it occurs when $ a\equal{}b\equal{}0$
It is known that if $ a$ and $ b$ are two positive real numbers, then :
$ {\frac {a^2\plus{}b^2}{2}}^{\frac {1}{2}} \leq {\frac {a^3\plus{}b^3}{2}}^{\frac {1}{3}}$
Because $ a^2\plus{}b^2\equal{}a^3\plus{}b^3$, then :
$ {\frac {a^3\plus{}b^3}{2}}^{\frac {1}{2}} \leq {\frac {a^3\plus{}b^3}{2}}^{\frac {1}{3}}$. Write $ \frac {a^3\plus{}b^3}{2}\equal{}m$, then :
$ {m}^{\frac {1}{2}} \leq {m}^{\frac {1}{3}}$, $ m \leq 1$
It is also known that :
$ \frac {a\plus{}b}{2} \leq {m}^{\frac {1}{2}} \leq 1$, then $ a\plus{}b \leq 2$. The equality holds when $ a\equal{}b\equal{}1$.
Sorry if I am wrong.
[/hide]

You have almost solved the problem, but $ \alpha >0,\ \beta >0$.

[code]$\left(\frac{B}{A}\right)$[/code]

You will see $ \left(\frac{B}{A}\right)$.

If $ \alpha^2 \plus{} \beta^2 \equal{} \alpha^3 \plus{} \beta^3$ can you use linear algebra to find $ \alpha, \beta$? Because it its one equation with two unknowns. And the range of $ \alpha \plus{} \beta$ is just found by plugging in $ \alpha$ and $ \beta$?

[quote="rofler"]Let $ a + b = p$, $ ab = q$

$ p^2 - 2q = p^3 - 3pq$

$ q(3p - 2) = p^3 - p^2$

If $ p = \frac {2}{3}$ then this clearly leads to a fallacy, so assume otherwise.

$ q = \frac {p^3 - p^2}{3p - 2} = \frac {p^2(p - 1)}{3p - 2}$. This is positive only when $ p < \frac {2}{3}$ and $ p > 1$. From now on, we assume that p lies in one of $ (0,\frac {2}{3})$ or $ [1,inf)$

Note that we have $ 4q\le p^2$, so $ \frac {4p^2(p - 1)}{3p - 2}\lep^2$

$ \frac {4(p - 1)}{3p - 2}\le1$

$ 4p - 4\le 3p - 2$

$ p\le 2$

Therefore, p has to lie in one of $ (0,\frac {2}{3})$ or $ [1,2]$.

For every p in one of these ranges, we can find a q, since reversing the steps just shown, we get a $ q$ s.t. $ 4q \le p^2$ and $ p^2 - 2q = p^3 - 3pq$. Hence the union of these 2 ranges is our desired range.[/quote]

That's right. The answer is $ 1<x\leq 2$.


If $ \gcd(n, k) \equal{} 1$, prove that $ n | {n \choose k}$.

[hide="Hint"] You can either make this tedious or trivial.  What is $ k {n \choose k}$?  Alternately, try to sort the subsets of size $ k$ into groups each with $ n$ subsets. [/hide]

[hide="Yet Another Hint"]You can also look at $ {n\minus{}1 \choose k}$[/hide]

=)

Since $ \gcd(n,k)\equal{}1$ ,

if $ n | k\binom{n}{k}$ then $ n|\binom{n}{k}$

it is easy to see

$ k \binom{n}{k}\equal{} n \binom{n\minus{}1}{k\minus{}1}$

$ \blacksquare$

Also could we say $ \binom{n}{k} \equal{} \frac{n!}{k!\left(n\minus{}k\right)!}$ which is an integer. Since $ \gcd(n,k) \equal{} 1$, $ n|\binom{n}{k}$

Actually this also seems to have the property that for $ p \equal{} \text{prime}$, $ p|\binom{p}{k}$ for $ k \in \mathbb{Z}$.

This is not true when $ k \equal{} 0$ or $ k \equal{} p$.  (You do need the condition $ (p, k) \equal{} 1$, and then it's a special case.)  And your argument, as written, gives no explanation for the last step.

if (n,k) = 1 - meaning that they don't have a common integer divisor that will leave a interger as a result of dividing either number by it (other than 1) -
nCk  = (n!)/(k!(n-k)!) if divided by n will be an integer, for any integer n that is coprime to k and all integer numbers less than k and greater than one - because n won't be simplified by its divisors (in nCk)

(n doesnt have to be prime)


Given an integer n decide if it's possible to represent it a sum of two squares of integer.

All I found: A natural number n is the sum of two squares of integers if and only if the factorization of n into prime factors does not contain any prime of the form 4k+3 that has an odd exponent.

But if n is big (10^12 for example) how to check efficiently is n a sum of two squres of integers or not??

This question is a classic result -- try searching the forums, as I'm sure it's been discussed here before.

I found that:
a)Those numbers n whose prime divisors of the form 4*m+3 divide n an even number of times can be written as the sum of two squares.
b)Jacobi's Two Square Theorem: The number of representations of a positive integer as the sum of two squares is equal to four times the difference of the numbers of divisors congruent to 1 and 3 modulo 4.

But again - when n is big (10^10 or 10^12) it's hard to check all divisors or find all prime divisors of the form 4*m+3.

Can anyone give me a hint how to efficently check whether n can be represented as the sum of two squares or not??


Hello !

Just wondering which skin is the most popular or most used.
Please participate!  :lol:

idc but i just keep aops cause it's already here.

I first found it as AoPS, so I just keep it as AoPS.

AoPS FTW!!!

No, FTW is not the opposite of WTF

AoPS.

[size=50]Stupid size limitations.  This message is just as large as it need be.  Hmph.[/size]

I think the AoPS skin is kind of ugly but I still use it because it has all of the relevant links (for me) on the side, and I'm not doing all of the MathLinks Olympiad stuff.  I think Mathlinks is more of Olympiad stuff.
However, on the MathLinks skin, every other post has a blue background. So you can read the sneaky white text posted by very clever people, and without painting over it with the mouse like you would have to do on AoPS.[color=white][b]...............nyah,nyah!!!!!  :P [/b][/color]
But I discovered AoPS on the AoPS skin so I'm more used to it.

I think the Mathlinks skin looks better, especially the hexagons.

Nevertheless, I use the AoPS URL because it looks better in the title bar.

I was an AOPSER but then i switched to the magic of mathlinks.

[quote="mathnerd314"][quote="probability1.01"][quote="mathnerd314"]Nevertheless, I use the AoPS URL because it looks better in the title bar.[/quote]
wtf? That is... a pretty good reason...[/quote]

Plus, it's more fun to type artofproblemsolving than mathlinks. (Of course, I have AoPS as one of my home pages, so it doesn't really matter.)[/quote]
wtf dude
skin has nothing to do with url wtf

Precisely.

I use the http://www.artofproblemsolving.com URL because
1) It's more fun to type
2) It looks cooler in the address bar

I use the http://www.mathlinks.ro skin because
1) It looks prettier

Any questions?

Odd preferences  :)

but I agree with you that mathlinks is prettier. Its more "fresh", not too dark
and feels a bit more new.

 thanks for posting anyway!  :D


let graph $G$ has $n$ vertexs .Find $f(n)$ is smallest s.t  $deg$ of each vertex $=f(n)$ and two vertex $A,B$ didn't cornected then there $C$ so $A,C$ and $B,C$ is cornected


LaTeX
abstract algebra
advanced fields
advanced fields unsolved

Let $ R$ be a commutative ring with unity. Thinking of the torus $ T$ as a two closed annuli $ A$ and $ B$ with their boundaries identified, let $ A \cap B \equal{} C_1 \sqcup C_2$ where $ C_1$ and $ C_2$ are circles. Then $ (T, A, B)$ is an exact triad. A portion of its M-V sequence is

$ 0 \longrightarrow H_2(T) \stackrel{\partial_2}{\longrightarrow} H_1(C_1 \sqcup C_2) \stackrel{\alpha}{\longrightarrow} H_1(A) \oplus H_1(B)$ $ \stackrel{\beta}{\longrightarrow} H_1(T) \stackrel{\partial_1}{\longrightarrow} H_0^{\#}(C_1 \sqcup C_2) \longrightarrow 0.$

Examining $ \alpha$, we have $ \ker \alpha \cong R$ and $ \operatorname{im} \alpha \cong R$, thus we have $ H_2(T) \cong R$.

But, we also have $ \ker \beta \cong R$ and $ \operatorname{im} \partial_1 \equal{} H_0^{\#}(C_1 \sqcup C_2) \cong R$. From this, how can we conclude that $ H_1(T) \cong R \oplus R$?

[quote="iks i grek"]Let $ \phi$ be the map $ H_0^{\#}(C_1 \sqcup C_2) \longrightarrow 0$. By the isomorphism theorem $ \operatorname{Im} \partial_1 \cong H_1(T)/ \operatorname{Ker} \partial_1$. Since $ \operatorname{Ker} \partial_1 \cong R$ (because $ \operatorname{Im} \beta \cong R$) and $ \operatorname{Im} \partial_1 \cong R$ (because $ \operatorname{Ker} \phi \cong R$) we get $ H_1(T) \cong \operatorname{Im} \partial_1 \oplus \operatorname{Ker} \partial_1 \cong R \oplus R$.[/quote]

Thanks for replying, but what I wanted to know is that how can we know $ 0 \longrightarrow \operatorname{im} \beta \longrightarrow H_1(T) \longrightarrow H_0^{\#}(C_1 \sqcup C_2) \longrightarrow 0$ splits. But now I know this is possible because $ R$ is a projective $ R$-module.

Looks like there are latex errors everywhere; might want to check in preview mode before submitting in the future!

The posts are error-free. It is a current site $ \text{\LaTeX}$ problem with the \operatorname command.


IF ANYONE IS ALIVE PLEASE MAKE YOURSELF KNOWN

[quote="ra5249"]greetings, i am dead[/quote]

whoa a texan in my (and all other dead peoples) forum I am honored!

I suppose to make this useful:

Prove $ \sin^4{x} \minus{} \cos^4{x} \equal{} \minus{}\cos{2x}$

[quote="lifesapain"]Why bother?[/quote]

Because..... idk.  Anyway long time no see lifesapain!

Why make it useful (I cant even do that... :(   )? We have no mods.....

[quote="hunter34"]I suppose to make this useful:

Prove $ \sin^4{x} \minus{} \cos^4{x} \equal{} \minus{} \cos{2x}$[/quote]

i see no one bothered to prove this so i will go ahead and do it
by diff of squares we have (sin^2(x)-cos^2(x))=-(cos2x) okay done

I lived in Alabama.....wonder if that counts.

im here, but ive been in summer camp for a looooooong time so i can't go 2 the AL forum. sorry dynamo :(

lol chestnut its ok. I cannot reserve the right to be mad at you because you beat me at nats. Ive just been bored for like 4 days.

now i was away from a computer for another week bc of summer camp. and this week im going 2 miami

ok its fine.... and... I got an email from your mom  :rotfl: .

[hide]ME AND OTHER COOL PEOPLE OKAY WHY ARE YOU POSTING HERE ANYWAYS?[/hide]

im a cool person who hasnt been 2 aops in a loooooonnnnnnggggggg time

[quote="dynamo729"] We have no mods.....[/quote]

What???

I'm still around... occasionally... and I'm actually in California now.

REVIVE!!
REVIVE!!
I revive ye from the dead!!!


trigonometry
geometry
angle bisector
trig identities
Law of Cosines

Point $D$ is on side $CB$ of triangle $ABC$.  If \[ \angle{CAD} = \angle{DAB} = 60^\circ,\quad AC = 3\quad\mbox{ and }\quad AB = 6, \] then the length of $AD$ is
$\text{(A)} \ 2 \qquad \text{(B)} \ 2.5 \qquad \text{(C)} \ 3 \qquad \text{(D)} \ 3.5 \qquad \text{(E)} \ 4$

You can use Law of Cosines and/or the property of angle bisector, but here's a cool approach.
[hide]Take point $E$ on the extension of $BA$ such that $E$, $A$, $B$ are collinear in this order and $EA=3$.
Then since $\angle EAC=60^\circ$ and $AC=AE=3$, $\triangle EAC$ is an equilateral triangle, which means $\angle AEC=60^\circ$.

$\angle BAD=\angle BEC=60^\circ$, so $AD\parallel EC$.
Hence $\frac{BA}{BE}=\frac{AD}{EC}$.
$\therefore AD=EC\times\frac{BA}{BC}=3\cdot\frac{2}{3}=2$

$\boxed{\text{(A)}}$

[/hide]

[hide="solution"]Draw a line $L$ through $B$ parallel to $AC$, let $AD \cap L=E$, we get $\angle AEB=60$, making $\triangle ABE$ equilateral. That makes $AE=6=AD+DE$ (1), and since $\angle ABC=\angle EDB$, and and the 60 degree angles, $\triangle ADC \sim \triangle EDB$, $\frac{AD}{DE}=\frac{1}{2}$ (2), using (1) and (2), $AD=2\implies \boxed{A}$[/hide]

[hide]Draw a line through $ D$ parallel to $ AC$.  Then let $ s$ be the side of the equilateral triangle formed.  $ AD \equal{} s$, and $ \frac {s}{6 \minus{} s} \equal{} \frac {1}{2}$.[/hide]


inequalities induction function trigonometry logarithms calculus derivative

In the same spirit as http://www.mathlinks.ro/Forum/viewtopic.php?t=40857.

Let's go (1) review of RMS-AM-GM-HM, (2) review of Cauchy, (3) Jensen, (4) Weighted Power Mean, (5) Holder, (6) Rearrangement, (7) Chebyshev, (8) Schur, (9) Newton, (10) Maclaurin, (11) Majorization, (12) Popoviciu, (13) Bernoulli, (14) Muirhead, (15) Karamata; Mildorf's standard dozen + Karamata.

[size=150]1 RMS-AM-GM-HM[/size]

[color=darkblue]For positive reals $x_{1},x_{2},\ldots,x_{n}$, we have
\[\sqrt{\frac{x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}}{n}}\ge \frac{x_{1}+x_{2}+\cdots+x_{n}}{n}\ge \sqrt[n]{x_{1}x_{2}\cdots x_{n}}\ge \frac{n}{\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{n}}}, \]
with equality iff $x_{1}=\cdots=x_{n}$.[/color]

[b]1.1[/b] Prove RMS-AM-GM-HM.

[size=150]2 Cauchy-Schwartz[/size]

[color=darkblue]For real numbers $x_{1},x_{2},\ldots,x_{n}$ and $y_{1},y_{2},\ldots,y_{n}$, we have
\[(x_{1}y_{1}+x_{2}y_{2}+\cdots+x_{n}y_{n})^{2}\le (x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2})(y_{1}^{2}+y_{2}^{2}+\cdots+y_{n}^{2}), \]
with equality iff $\frac{x_{1}}{y_{1}}= \cdots = \frac{x_{n}}{y_{n}}$.[/color]

[size=117]2.1 Engel Form[/size]

[color=darkblue]For real numbers $x_{1},x_{2},\ldots,x_{n}$ and $y_{1},y_{2},\ldots,y_{n}$ with $y_{1}+y_{2}+\cdots+y_{n}>0$, we have
\[\frac{x_{1}^{2}}{y_{1}}+\frac{x_{2}^{2}}{y_{2}}+\cdots+\frac{x_{n}^{2}}{y_{n}}\ge \frac{(x_{1}+\cdots+x_{n})^{2}}{y_{1}+\cdots+y_{n}}. \] If $y_{1}+y_{2}+\cdots+y_{n}< 0$, the inequality is reversed.
[/color]

[b]2.1.1[/b] Prove Cauchy's Engel form (try induction on $n=2$).

[b]2.1.2[/b] Prove Cauchy using Engel form.

[b]2.1.3[/b] (Mathematical Olympiad Treasures) Let $a$ and $b$ be positive real numbers. Prove that
\[8(a^{4}+b^{4}) \ge (a+b)^{4}. \]
[b]2.1.4[/b] (Mathematical Olympiad Treasures) Let $x,y,z>0$. Prove that
\[\frac{2}{x+y}+\frac{2}{y+z}+\frac{2}{z+x}\ge \frac{9}{x+y+z}. \]
[b]2.1.5[/b] (Mathematical Olympiad Treasures) Let $a,b,x,y,z$ be positive real numbers. Prove that
\[\frac{x}{ay+bz}+\frac{y}{az+bx}+\frac{z}{ax+by}\ge \frac{3}{a+b}. \]
[b]2.1.6[/b] (IMO 1995) Let $a,b,c$ be positive numbers such that $abc=1$. Prove that
\[\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(a+c)}+\frac{1}{c^{3}(a+b)}\ge \frac{3}{2}. \]

[hide="2.1.1"]
We proceed by induction. For the base case $n=2$, we find that the claim is equivalent to $(x_{1}y_{2}-x_{2}y_{1})^{2}\ge 0$. Now suppose the claim is true for $n=k$, i.e.
\[\frac{x_{1}^{2}}{y_{1}}+\cdots+\frac{x_{k}^{2}}{y_{k}}\ge \frac{(x_{1}+\cdots+x_{k})^{2}}{y_{1}+\cdots+y_{k}}. \]
Adding $\frac{x_{k+1}^{2}}{y_{k+1}}$ to both sides, we find that
\[\frac{x_{1}^{2}}{y_{1}}+\cdots+\frac{x_{k}^{2}}{y_{k}}+\frac{x_{k+1}^{2}}{y_{k+1}}\ge \frac{(x_{1}+\cdots+x_{k})^{2}}{y_{1}+\cdots+y_{k}}+\frac{x_{k+1}^{2}}{y_{k+1}}\ge \frac{(x_{1}+\cdots+x_{k+1})^{2}}{y_{1}+\cdots+y_{k+1}}\]
since we know the claim is true for $n=2$. Thus the claim is true for $n=k+1$ as well. Therefore, the claim is true for all $n\ge 2$ and it is trivial for $n=1$, so it is true for all positive integers $n$ and all reals $x_{1},\ldots,x_{n}$ and $y_{1},\ldots,y_{n}$. $\blacksquare$
[/hide]

[hide="2.1.1"]
\[\frac{x_{1}^{2}}{y_{1}}+\frac{x_{2}^{2}}{y_{2}}+\cdots+\frac{x_{n}^{2}}{y_{n}}\ge \frac{(x_{1}+\cdots+x_{n})^{2}}{y_{1}+\cdots+y_{n}}\]
Multiply the left hand side by $y_{1}+\hdots+y_{n}$, and so the conclusion follows by Cauchy-Schwartz Inequality.[/hide]
[hide="2.1.2"]  Quite ridiculous, but okay (your problem statement is unclear). Assuming that Cauchy in Engel Form in true, multiply both sides by $y_{1}+\hdots+y_{n}$, you obtain Cauchy-Schwartz Inequality in standard form.  Just substitute![/hide]
[hide="2.1.3"] By power means inequality (generalization of AM-GM), we have
\[\sqrt[4]{\frac{a^{4}+b^{4}}{2}}\geq\frac{a+b}{2}\implies 8(a^{4}+b^{4})\geq (a+b)^{4}, \]
as required.[/hide]
[hide="2.1.4"]
\[\left(\sum_{cyc}(x+y)\right)\left(\sum_{cyc}\frac{2}{x+y}\right)\geq (3\sqrt{2})^{2}=18, \]
as required.[/hide]
[hide="2.1.5?"]By Cauchy-Schwartz Inequality,
\[[(ay+bz)+(az+bx)+(ax+by)]\left(\frac{x}{ay+bz}+\frac{y}{az+bx}+\frac{z}{ax+by}\right)\geq (\sqrt{x}+\sqrt{y}+\sqrt{z})^{2}. \]
If we prove that $\frac{(\sqrt{x}+\sqrt{y}+\sqrt{z})^{2}}{x+y+z}\geq 3$, we will be done.  This follows by the Am-QM Inequality.[/hide]
[hide="2.1.6"] You can either multiply everythign out and use Muirhead, or substitute $a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$.[/hide]

EDIT: I was not connected to mathlinks for several minutes, so my post messed up.

2 people both quoted the wrong problem; 2.1.4 isn't up yet.
[hide="2.1.4 (really)"]
Here's the Cauchy:
\[\sum \frac{x}{ay+bz}=\sum \frac{x^{2}}{axy+bxz}\geq \frac{(\sum x)^{2}}{(a+b)(\sum xy)}\geq \frac{3}{a+b}\]
noting that $(\sum x)^{2}\geq 3\sum xy \iff \sum x^{2}\geq xy$ which is provable a bazillion ways.

Here's an interesting rearrangement:Sorry never mind the rearrangement I can't salvage it :(
[/hide]

Isn't Minkowski just a general case of one of the others, and hence we should put it a bit later?

[quote="me@home"]2 people both quoted the wrong problem; 2.1.4 isn't up yet.
[hide="2.1.4 (really)"]
Here's the Cauchy:
\[\sum \frac{x}{ay+bz}=\sum \frac{x^{2}}{axy+bxz}\geq \frac{(\sum x)^{2}}{(a+b)(\sum xy)}\geq \frac{3}{a+b}\]
noting that $(\sum x)^{2}\geq 3\sum xy \iff \sum x^{2}\geq xy$ which is provable a bazillion ways.

Here's an interesting rearrangement:
\[\sum \frac{x(ax+bx)}{(ay+bz)(ax+bx)}\geq \sum \frac{x(ay+bz)}{(ax+bx)(ay+bz)}\]
which is $6$-expression rearrangement on the sets:
\[A=(\frac{xa}{(ax+bx)(ay+bz)},\frac{xb}{(ax+bx)(ay+bz)},\frac{ya}{(ay+by)(az+bx)},\frac{yb}{(ay+by)(az+bx)},\frac{za}{(az+bz)(ax+by)},\frac{zb}{(az+bz)(ax+by)}) \]

\[B=(x,x,y,y,z,z) \]
Now,
\[\sum \frac{x(ax+bx)}{(ay+bz)(ax+bx)}\geq \sum \frac{x(ay+bz)}{(ax+bx)(ay+bz)}\iff \sum \frac{x}{ay+bz}\geq \sum \frac{x}{ax+bx}=\frac{3}{a+b}\]
which proves the problem :rotfl:
[/hide][/quote]
What do you mean I quoted the wrong problem?  There seems to be two 2.1.4's up there, so I thought it should be 2.1.5.

Masoud, you used Cauchy to prove Engel's form, which you then used to prove Cauchy! :D

[quote="paladin8"][quote="PenguinIntegral"]I heard Holder was a special case of Minkowski, do you think we should do that first?[/quote]

I don't think this statement is true.[/quote]

[quote="Boy Soprano II"]Hölder is pretty neat, though, as well as elegant and fairly widely applicable for a sophisticated inequality (and it's a special case of Minkowski, which is almost as neat).  [/quote]

I'm confused...

OK, I'll just do it the [url=http://mit.edu/~tmildorf/www/Inequalities.pdf]Mildorf[/url] way then.

It seems that the reverse is true. Holder can be used to prove Minkowski according to [url=http://en.wikipedia.org/wiki/Minkowski_inequality]Wikipedia[/url].

[quote="chess64"]
[b]2.1.6[/b] (IMO 1995) Let $a,b,c$ be positive numbers such that $abc=1$. Prove that
\[\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(a+c)}+\frac{1}{c^{3}(a+b)}\ge \frac{3}{2}. \]
[/quote]

This inequality is equivalent to proving

$\Sigma \frac{a^{2}b^{2}}{c(a+b)}\geq \frac{3}{2}$

By Cauchy's Inequality

$(\Sigma \frac{a^{2}b^{2}}{c(a+b)})(c(a+b)) \geq (ab+bc+ca)^{2}$

Thus $LHS \geq \frac{(ab+bc+ca)^{2}}{a(b+c)+b(c+a)+c(a+b)}= \frac{ab+bc+ca}{2}\geq \frac{3}{2}$

Nitpick:  Engel form requires the condition $\sum y_{k}> 0$, otherwise the inequality is reversed.

[quote="13375P34K43V312"]isn't that equivalent to having its first derivative increasing over a given interval?[/quote]

Yes, but that explanation requires a calculus background to understand.  You don't need calculus to understand convexity.

Here's another way to define it, also entirely without using any calculus:  A function is convex if any secant line drawn on its graph is "above" the graph.  In other words,

$f \left( tx+(1-t)y \right) \le tf(x)+(1-t) f(y)$

Where the RHS is a parameterization of the $y$-values of the secant line from $(x, f(x))$ to $(y, f(y))$ and the LHS is the corresponding point below the secant line on the graph.  The simplest case is when $t = \frac{1}{2}$, so that we have

$f \left( \frac{x+y}{2}\right) \le \frac{f(x)+f(y)}{2}$

Edit:  This is, by the way, the simplest statement of Jensen's inequality.

[b]Schur's Inequality[/b]

Let $t \geq 1$ be a real number. Then for all non-negative real numbers $x,y,z$

$x^{t}(x-y)(x-z)+y^{t}(y-x)(y-z)+z^{t}(z-x)(z-y) \geq 0$

where the equality occurs only when $x=y=z$ not equal to $0$ or some two of $x,y,z$ are equal and the third is zero.

A simple example:

Prove that for any positive $a,b,c$

$a^{3}+b^{3}+c^{3}+3abc \geq a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)$.

For positive numbers $a,$ $b$ and $c$ prove that
$a^{a}(a-b)(a-c)+b^{b}(b-a)(b-c)+c^{c}(c-a)(c-b)\geq0.$

[hide="Solution"]
Let $a\ge b \ge c \implies a^{a}\ge b^{b}\ge c^{c}$. The signs of the expression: $(+)(+)(+)+(+)(-)(+)+(+)(-)(-) \ge 0$
\begin{eqnarray*}&\iff& a^{a}(a-b)(a-c)+c^{c}(a-c)(b-c)\ge b^{b}(a-b)(b-c) \\ &\iff&(a-b+b-c)a^{a}(a-b)+c^{c}(a-c)(b-c)\ge b^{b}(a-b)(b-c)\\ &\iff& a^{a}(a-b)^{2}+a^{a}(a-b)(b-c)+c^{c}(a-c)(b-c)\ge b^{b}(a-b)(b-c)\\ &\iff& a^{a}(a-b)^{2}+c^{c}(a-c)(b-c)+(a^{a}-b^{b})(a-b)(b-c)\ge 0 \end{eqnarray*}
Thus all are true since the last is a sum of positive terms.
Equality must have $a=b$, and once of $c^{c},a-c$ must be zero. Since $a,b,c$ are positive (non-zero) we must have $a=b=c$ as the only equality case.[/hide]

[quote="me@home"]
Let $a\ge b \ge c \implies a^{a}\ge b^{b}\ge c^{c}$. [/quote]
It's wrong. :wink:

yes, but we [i]do[/i] have
\[a^{a}+c^{c}\geq b^{b}\]
and that's enough to solve it (just turn $(a-c)(b-c)$ into $(b-c)^{2}+(a-b)(b-c)$). I wonder if there's a way to prove $a^{a}+c^{c}\geq b^{b}$ without calculus, though...

I knew the inequality seemed too weak...

about the calculus... I think I can find a slick way with Jensen ideas... but I'm busy right now :(

[quote="I"]I wonder if there's a way to prove $a^{a}+c^{c}\geq b^{b}$ without calculus, though...[/quote]

i found something!
[hide]
If $b\geq 1$, then it's simple: $a^{a}+c^{c}>a^{a}\geq a^{b}\geq b^{b}$. Otherwise, $0<b<1$ so $b^{b}<1$. Then if we can prove
\[x^{x}>\frac{1}{2}\]
for positive real $x$, we will be done. Let $y=\frac{1}{x}$ (this makes sense because $x$ will be small when $x^{x}$ approaches its minimum), this is equivalent to
\[2^{y}>y\]
for positive real numbers $y$. But for this, we can just induct on $\lfloor y\rfloor$.

The base cases, $0<y<1$ and $1\leq y<2$, are clearly true; for the first case, $2^{y}>1>y$, and for the second case, $2^{y}\geq 2>y$.

The inductive step (prove for $y+1$ given true for $y\geq 1$) follows from $2^{y+1}>2y\geq y+1$. Therefore, the statement is true for all positive $y$.
[/hide]

[quote="Svejk"]We may consider the function $f: [0,\infty)\to \mathbb R$ $f(x)=x^{x}$ $f ''(x)>0$ $f$ is strict convex.So we have $\sum f(a)(a-b)(a-c)\geq 0$ which is Vornicu-Schur[/quote]

$f$ isn't standardly defined at $0$, so should the interval not be $(0, \infty)$?

[quote="arqady"]I think, Vornicu-Schur not apply for following inequality:
let $a,$ $b$ and $c$ are positive numbers. Prove that:
$\sum_{cyc}\frac{a^{3}}{(2a^{2}+b^{2})(2a^{2}+c^{2})}\leq\frac{(a+b+c)}{3(a^{2}+b^{2}+c^{2})}.$ :wink:[/quote]

[hide="a start... maybe?"]
Multiply by $3(a^{2}+b^{2}+c^{2})$, then RHS-LHS is equal to
\[\sum_\text{cyc}a-\frac{3a^{3}(a^{2}+b^{2}+c^{2})}{(2a^{2}+b^{2})(2a^{2}+c^{2})}=\sum_\text{cyc}\frac{a(a+b)(a+c)}{(2a^{2}+b^{2})(2a^{2}+c^{2})}(a-b)(a-c) \]
hmm... it appears that if $a\geq b\geq c$, then we can prove
\[\frac{a(a+b)(a+c)}{(2a^{2}+b^{2})(2a^{2}+c^{2})}+\frac{c(c+a)(c+b)}{(2c^{2}+a^{2})(2c^{2}+b^{2})}\geq \frac{b(b+a)(b+c)}{(2b^{2}+a^{2})(2b^{2}+c^{2})}\]
Unfortunately, multiplying this all out would be a bit of a mess, and no single term on the left hand side will always dominate the right hand side (so the more complicated-looking statement is needed).
[/hide]

[quote="Svejk"]We may consider the function $f: [0,\infty)\to \mathbb R$ $f(x)=x^{x}$ $f ''(x)>0$ $f$ is strict convex.So we have $\sum f(a)(a-b)(a-c)\geq 0$ which is Vornicu-Schur[/quote]
Yes I was exactly thinking of Vornicu Schur's when I tried to give my "proof" but you actually pulled it through...


In 1950, the minimum wage for non-farm workers was $ \$0.75$ per hour. In 1993, it was $ \$4.25$ per hour. What was the percent increase expressed as a mixed number?

$ 4.25 \equal{} \frac {17}{3}\times 0.75$, so $ \frac {17}{3} \minus{} 1 \equal{} 4 \frac {2}{3}$ and the percent increase is $ 466 \frac {2}{3}$.

EDIT: hmm?  :D

Percentage increase: Change/Original Value x 100:
\[ \frac {4.25 \minus{} 0.75}{0.75} \cdot 100 \implies \frac {1400}{3} \implies 466\frac {2}{3}
\]

AIME15 that's incorrect  :P .


prove that for any positive numbers x1,x2,....xk (k>=4)
$\frac{x_{1}}{x_{k}+x_{2}}+\frac{x_{2}}{x_{1}+x_{3}}+...+\frac{x_{k}}{x_{k-1}+x_{1}}\geq 2$

prove if you can replace 2 by a greater number.

[b][b]I think it easy to solve the problem.but I am from china .I can not write it in English. :maybe:  :maybe: [/b][/b]

We'll use induction. We may assume that $x_{k}$ is the smallest of the numbers $x_{1},x_{2},\ldots,x_{k}$. We need to prove that \[\frac{x_{1}}{x_{k}+x_{2}}+\frac{x_{k-1}}{x_{k-2}+x_{k}}+\frac{x_{k}}{x_{k-1}+x_{1}}\geq \frac{x_{1}}{x_{k-1}+x_{2}}+\frac{x_{k-1}}{x_{k-2}+x_{1}},\] which is true because $\frac{x_{1}}{x_{k}+x_{2}}\geq \frac{x_{1}}{x_{k-1}+x_{2}}$ and $\frac{x_{k-1}}{x_{k-2}+x_{k}}\geq \frac{x_{k-1}}{x_{1}+x_{k-2}}$. To prove that we can't replace 2 with greater number, take $x_{1}=x_{k}$ and $\frac{x_{1}}{x_{2}}=\frac{x_{2}}{x_{3}}=\ldots=\frac{x_{k-2}}{x_{k-1}}\rightarrow+\infty$.


I hope this is the correct forum for this, not sure if this should be in the grade school section.

i wish i could find a consistent way to solve these.

How many gallons of a 5% acid solution should be mixed with 20 gallons of a 10% acid solution to obtain a 8% acid solution.

what i get is .05x+2=.08 which is not even close to the correct answer. x is not 56 lol.

$ 20\cdot \frac{1}{10}\equal{}2$ gallons of acid.
$ 2\plus{}.05x$ gallons of acid total.
$ \frac{2\plus{}.05x}{20\plus{}x}$ is percentage acid.

Solving it is giving me $ x\equal{}13.33$?

back of the book says its 40/3.. which is 13.3333....and is also .4/.03..

i like how you solve them. thanks again.

The problem can be written in the form of the equation:

0.05x + 2 = 0.08 (20 + x)
0.05x + 2 = 1.6 + 0.08x
0.05x - 0.08x = 1.6 - 2
-0.03x = - 0.4
0.03x = 0.4
x = 0.4/0.03
x= 13.3333...

Answer: x = 13.3 i.e. we'll need 13.3 gallons of the 5% acid solution.

To learn more you may click the links below and go through video lessons on 'Percent mixture problems':
http://www.winpossible.com/lessons/Algebra_I_Percent_mixture_problems.html


find all functions , that f (x + y + f(y))=f(x) + 2y ; f : Z->Z; x,y : Z


It seems that JGeneson has gone on a lengthy vacation to somewhere far away.

so i'll come to you guys for help :D

Given a, b, c, d all real numbers such that  [tex]a^2+b^2+c^2+d^2=36[/tex], what is the maximum value of [tex]2a+3b+6c+24d?[/tex]

btw- how does one type coherent sentences using tex here?  it seems that tex here is automatically stuck in math mode.

[quote="Nosepuppy"]btw- how does one type coherent sentences using tex here?  it seems that tex here is automatically stuck in math mode.[/quote]

You need surround all non-math (i.e. text) with the \textrm{} command while in LaTeX mode.  For example,

$\textrm{The solutions to the equation } x^2 - 3x + 2 = 0 \textrm{ are } 2 \textrm{ and } 1.$

[code]$\textrm{The solutions to the equation } x^2 - 3x + 2 = 0 \textrm{ are } 2 \textrm{ and } 1.$[/code]

I'd been wondering how to do that for a while. Thanks!

(nitpick: in your sample equation, I think you meant 1 and 2 for the solutions)


modular arithmetic
algebra
polynomial
Vieta

For how many primes $ p$, there exits unique integers $ r$ and $ s$ such that for every integer $ x$ $ x^{3} \minus{} x \plus{} 2\equiv \left(x \minus{} r\right)^{2} \left(x \minus{} s\right)\pmod p$?

$\textbf{(A)}\ 0  \qquad\textbf{(B)}\ 1  \qquad\textbf{(C)}\ 2  \qquad\textbf{(D)}\ 3  \qquad\textbf{(E)}\ \text{None}$

[hide="trivialness?"]Assume (r,s) is a solution. Then (r+p,s+p) is a solution. Thus for no primes are there a unique solution. [b]A[/b][/hide]

$ p \equal{} 2, r \equal{} 1, s \equal{} 0$ is a solution.

Do you mean "distinct" instead of "unique"?

For $ p \equal{} 2$, $ LHS$ is $ 0$. When we take $ r \equal{} 1$ and $ s \equal{} 0$ or vice versa, we get $ 0 \equal{} (x)^2(x\minus{}1) \pmod 2$. This is true for all $ x$.
For example, let $ p \equal{} 3$.  Thus $ x^{3}\minus{}x\plus{}2\equiv 2 \equiv\left(x\minus{}r\right)^{2}\left(x\minus{}s\right)\pmod 3$. For $ x \equal{} r$ or $ x \equal{} s$, $ 2 \not\equiv 0 \pmod 3$. So for prime number $ p \equal{} 3$, there is no $ (r,s)$ which satisfies $ x^{3}\minus{}x\plus{}2\equiv \left(x\minus{}r\right)^{2}\left(x\minus{}s\right)\pmod 3$ for all $ x$.

I have yet no idea for others.

I have an idea for $ p \equal{} 13$.
$ x^3 \minus{} x \plus{} 2 \equiv x^3 \minus{} 27x \plus{} 54 \equal{} (x \minus{} 3)^2(x \minus{} 6) \pmod {13}$. So for $ p \equal{} 13$, there exists unique $ (r,s) \equal{} (3,6)$ which satisfies the congruence for all $ x$.

So far $ p \equal{} 2$ and $ p \equal{} 13$ are desired primes.

Shall I ask the problem in olympiad section?

I have learnt a formula by that [url=http://www.mathlinks.ro/viewtopic.php?p=1478013#1478013]topic[/url].
The formula is
$ \partial_{3} \equal{} G^{2} \plus{} 4H^{3}$
where $ G \equal{} a^2d \minus{} 3abc \plus{} 2b^3$ and $ H \equal{} ac \minus{} b^2$ where $ P\left( x\right) \equal{}  ax^3 \plus{} 3bx^2 \plus{} 3cx \plus{} d$.
If $ \partial_3 \equal{} 0$, then $ P\left(x\right)$ has multiple roots.

For the polynomial $ P\left(x\right) \equal{} x^3 \minus{} x \plus{} 2$, $ G \equal{} 2$ and $ H \equal{} \frac {\minus{}1}{3}$. Thus $ \partial_3 \equal{} 4 \plus{} \frac{\minus{}4}{27} \equal{} 4\cdot \frac {26}{27}$.
For the prime numner $ p$, $ 4\cdot \frac {26}{27} \equiv 0 \pmod p$. Thus  $ p \mid 4 \cdot 26 \Rightarrow p \equal{} 2$ or $ p \equal{} 13$.
In fact, $ (p,r,s) \equal{} (2,1,0)$ and $ (13, 3, 6)$ are the solutions.

But is there a simpler way to solve the problem without knowing $ \partial_3$?

[quote="t0rajir0u"][b]Lemma:[/b]  Let $ f(x)$ be a polynomial over $ \mathbb{F}_p$.  $ f$ has repeated roots if and only if $ \gcd(f, f')$ has degree greater than zero.[/quote]
Hey, that looks pretty useful!

[hide="Solution"] If $ p \equal{} 3$, then $ x^3 \minus{} x \plus{} 2 \equiv 2 \bmod 3$ for every $ x$, so this is impossible.  Otherwise, $ (x^3 \minus{} x \plus{} 2)' \equal{} 3x^2 \minus{} 1$ and

$ \gcd(3x^3 \minus{} 3x \plus{} 6, 3x^2 \minus{} 1) \equal{} \gcd(\minus{}2x \plus{} 6, 3x^2 \minus{} 1)$

so $ r \equal{} 3$, which gives $ 26 \equiv 0 \bmod p$ and $ p \equal{} 2, 13$. [/hide]

$ x^{3} - x + 2\equiv \left(x - r\right)^{2} \left(x - s\right)\pmod p$

$f(r) = r^{3} - r + 2\equiv 0 \pmod p $  $..*$

$f'(x) = 2(x-r)(x-s)+(x-r)^2 \Rightarrow f'(r) = 3r^2-1 \equiv 0 \pmod p $  $..**$

$r^{3} - r + 2\equiv 3r^2-1 \equiv  0 \pmod p $

Since  $p=3 \Rightarrow -1 \not\equiv 0 \pmod3$,

$3r^{3} - 3r + 6\equiv (3r^2-1)r -2r+6 \equiv  0 \pmod p $. $..***$

Merge $..**$ and $..***$.

$-2r+6 \equiv  0 \pmod p \Rightarrow $ $r \equiv 3 \pmod p $ or $p=2$.

If $r \equiv 3 \pmod p $, then $26 \equiv 0 \pmod p $, then $p=2$ or $p=13$.

In both cases we have $(p,r) = (13,3)$ and $(p,r) = (2,3) \equiv (2,1)$.

From Vieta formula we have sum of the roots are $0 \pmod p$. So $(p,q,r) = (13,3,7)$ or $(p,q,r) = (2,1,0)$.


Let be $ m,n\in \mathbb{Z}_\plus{}$ . Show that :

$ |n\sqrt{2005}\minus{}m|>\frac{1}{90n}$

[hide="Hint"]Start with $ |2005n^2\minus{}m^2|\geqslant 1$ - why it holds?[/hide]


function
real analysis
real analysis unsolved

Let n > 1 be an integer. Show that there exists a function f : R -> R such
that,
(a)  f(x) + f(2x) + · · · + f(nx) = 0, for all real x.
(b)  f(x) = 0 iff x = 0.

There is a simple example,do you mean another ?
  f:R----> R   f(x) = 0

That second condition is that $ f(x) \equal{} 0$ if and [b]only[/b] if $ x \equal{} 0$.

The real numbers are irrelevant here; the rational numbers would make it essentially the same problem, as only being able to divide by integers matters.

The Problem is,

Let $ n \geq 2$ be an integer. Show that there exists a function $ f : R \rightarrow R$
such that,

\[ (a)\ \ \ \ \ \ \ \forall x \in R,         f(x) \plus{} f(2x) \plus{} \dots \plus{} f(nx) \equal{} 0; \]
\[ (b)\ \ \ \ \ \ \ \forall x \in R, f(x) \equal{} 0 \Longleftrightarrow x \equal{} 0. \]

now if $ n \equal{} 2$, I can define such $ f : R \rightarrow R$.

if $ x > 0$, we know there exists a unique $ q \in Z$ such that $ 2^q \leq x < 2^{q\plus{}1}$. if $ q$ is Odd, $ f(x) \equal{} 1$, and if $ q$ is even $ f(x) \equal{} \minus{}1$.

if $ x < 0$,  $ f(x) \equal{} f(\minus{}x)$.

and $ f(0) \equal{} 0$.


calculus
integration
algebra
function
domain
polynomial
Ring Theory

Let A be an integral domain, and a,b are elements in A. Let B=A[x]/(ax+b), where (ax+b) is ideal generated by ax+b. Suppose $ (a)\cap(b) \equal{} (ab)$, show that B is an integral domain.

I think it suffices to show that (ax+b) is prime under the assumption, and hence need to show that if f(x)g(x) is in (ax+b), then either f(x) or g(x) is in (ax+b).

thank you for help!

> Let D be a domain, a,b in D\0 such that  a,b|e => ab|e
> for all e in D. Show that  f = ax+b  is prime in D[x].

HINT Use McCoy's theorem [1] to pullback f prime in K[x] to D[x],
K = fraction field of D. See my sci.math post [2] for details.

[1] sci.math, 4 May 2004, problem with rings of polynomials
http://google.com/group/alt.algebra.help/msg/d3179d6f17afa48c
http://google.com/groups?selm=y8z3c6fwpme.fsf%40nestle.csail.mit.edu

[2] sci.math, 7 Sep 2009, linear polynomial is prime iff superprimitive
http://groups.google.com/group/sci.math/msg/ecf77ef1d635ea41
http://google.com/groups?selm=y8ztyzemc4a.fsf%40nestle.csail.mit.edu