Now that the contest is "officially" over, we can discuss the problems.  My solutions can be found at http://www.tjhsst.edu/~tmildorf/math/PracticeAIME3Solutions.pdf.  Based on the solutions, I think this contest was more difficult than the other two. Who keeps on voting "easy"?

Answers (In Spoiler):[hide]

1. 196

2. 435

3. 601

4. 071

5. 936

6. 121

7. 049

8. 472

9. 225

10. 058

11. 035

12. 164

13. 115

14. 468

15. 041

[/hide]

Winners:

1st: 9 MysticTerminator [hide](1, 2, 3, 4, 5, 6, 7, 9, 10)[/hide]

2nd: 6 Beta [hide](1, 2, 3, 4, 7, 9)[/hide]

T-3rd: 5 tetrahedr0n [hide](1, 2, 3, 4, 6) [/hide]

T-3rd: 5 mathfanatic [hide](1, 2, 3, 4, 6)[/hide]

General Statistics for 11 participants:

Mean Score: 3.364

Median Score: 3

Mode(s): {1, 2, 3, 5}

Range: 0-9

Standard Deviation: 2.327

Easiest Problem: #1

Hardest Problem(s): #8, #11, #12, #13, #14, #15

Number of Solvers by Problem:

1. **********

2. ****

3. ******

4. *****

5. **

6. *****

7. **

8.

9. **

10. *

11.

12.

13.

14.

15.

Score Distribution:

15.

14.

13.

12.

11.

10.

9. *

8.

7.

6. *

5. **

4.

3. **

2. **

1. **

0. *

Unfortunately the participation was down to just 11. Does anyone know how this could be improved? Maybe a longer period between the announcing of the contest date and the contest itself?

this Mock AIME seems easier than the last two........ I got a 6

I didn't get my score..... only solved 3. I thought it was 10% easier than last year's AIME.

1. Used Heron's formula. Got 196.
2. Brute forced it (made a table and performed sums in my head). Got 370.
8. Used mods to solve 1*2*4*8*16*32*64*128. Got 456.
Those were the only ones I could do. I tried doing 3 with the methods in Aops V.1 Ch.6(7?) but couldn't get that one.  I also got the answer for 10 in terms of the summation formulas for n 2  and n, but the number was too big to evaluate.  #4- Knew this was related to combanitorics, but couldn't solve correctly. The others I had no idea on. Could someone help me with #3,#4, & #8?

Edit: Sorry, I didn't see the solutions.

my approach on #9

Draw altitude CH of ABC, let CH intersect DB at F. triangle EDA similar to FDC. Let CD=x, DA=16-x, x/(16-x)=2/DF, DF=2x/(16-x), FB=15-2x/(16-x). Note that CF  is the angle bisector of DCB, so by angle bisector theorem,
x/16=DF/FB=2x/(240-17x). x=208/17

I really liked the geometry problems, because I got all geometry problems except #11 and #14 :)  !

I don't understand the explanation of 2. What are strings? Yeah, I made dumb mistakes on 2,3,4, and 8: should've gotten a 3, maybe even a 5. I only got 2 last year that's why I voted easy, somehow I only got 1, though.

*headdesk* on #7.  I found that the ratio of the other two sides was 1:3/2, but didn't realize that their sum was 5.  Bah.

For #2, I did it in a much more complicated fashion: let f(x,y) be the number of y-digit integers that are increasing (as defined in the problem) and end in x.  For example, f(1,7) = 1 and f(2,7) = 7.
We find that f(3,7) = f(2,7) + f(2,6) + f(2,5) + ... + f(2,2) + f(2,1) = $\binom{7}{1} + \binom{6}{1} + \binom{5}{1} + \binom{4}{1} + \binom{3}{1} + \binom{2}{1} + \binom{1}{1} = \binom{8}{2}$ by the hockey-stick theorem.

Similarly, f(4,7) = f(3,7) + f(3,6) + ... + f(3,1) = $\binom{9}{3}$  The same holds true for f(5,7) and on.

Hence, the desired sum f(1,7) + ... f(9,7) = $\binom{6}{0} + \binom{7}{1} + \ldots + \binom{14}{8} = \binom{15}{8}$ again by the hockey-stick theorem.

For #3, I let the three given variables be the roots of a cubic $a_3x^3 + a_2x^2 + a_1x + a_0$, where $a_3 = 1$, and used Newton's sums to find that $a_2, a_1, a_0 = -1$.  Then $s_4$ (the 4th Newton sum, i.e. $C_1^4 + C_2^4 + C_3^4$) is equal to the sum of the three before it, using the equation $a_3s_4 + a_2s_3 + a_1s_2 + a_0s_1 = 0$ and substituting.  The same is true on to infinity, so $s_4 = 11, s_5 = 21, s_6 = 39, s_7 = 71$.

[quote="1234567890"]We find that f(3,7) = f(2,7) + f(2,6) + f(2,5) + ... + f(2,2) + f(2,1) = $\binom{7}{1} + \binom{6}{1} + \binom{5}{1} + \binom{4}{1} + \binom{3}{1} + \binom{2}{1} + \binom{1}{1} = \binom{8}{2}$ by the hockey-stick theorem.

Hence, the desired sum f(1,7) + ... f(9,7) = $\binom{6}{0} + \binom{7}{1} + \ldots + \binom{14}{8} = \binom{15}{8}$ again by the hockey-stick theorem.

How did you find those 2 sums?[/quote]

I made the notation quite confusing, but essentially the idea goes something like this: say you have a string of 7 1's, like 1111111.  If I want the final number to be 1, there's only one way to do it - the way it is.  If I want the final number to be 2, I can change a 1 to a 2 at any of the seven positions and change all the 1's to 2's after that (so that it's still increasing).  Thus there are 7 choices.

But if I want the final number to be 3, it gets a bit more complicated.  I can change the first number to a 2 (and thus all the numbers after it as well), and then the problem reduces to the previous one, since going from 2222222 to a string with the last number as 3 is the same as going from 1111111 to a string with the last problem as 2.   Thus there are 7 numbers for this case. If I change the second number to a 2 (and all the numbers after it...I'm going to omit saying this from now on), then the problem is the same as the previous one, except now I only have 6 choices, so there are 6.  If I change the third number, now there are only 5.  And so on, up until if I change seventh number, now I only have one choice.  Thus if I want the final number to be 3, there are 7 + 6 + ... + 1 = 28 choices.  However, to make the underlying combinatorics more clear, I expressed this as $\binom{7}{1} + \ldots + \binom{1}{1}$, which equals $\binom{8}{2}$ by the hockey-stick theorem (google for this).

If I want the final number to be 4, I do essentially the same thing.  If I change the first number to a 2, then the problem becomes the previous one, with answer $\binom{8}{2}$.  If I change the second number to a 2, then the problem is the same as before, except with one fewer number that can be changed, so there are $\binom{7}{2}$.  If I change the third number to a 2, then there are $\binom{6}{2}$.  And so on.  In all, there are $\binom{8}{2} + \binom{7}{2} + \ldots + \binom{2}{2} = \binom{9}{3}$ by the hockey-stick theorem yet again.

If you don't care to do all the combinatorial arguments, just recognize the pattern.  If I want the final number to be 5, the pattern suggestst that there should be $\binom{10}{4}$.  As I'm growing fond of saying, and so on.

We wanted to find the total number of ways, which is equal to (# of ways with final number 1) + (# of ways with final number 2) + ... + (# of ways with final number 9) $=\binom{6}{1} + \binom{7}{2} + \ldots + \binom{14}{8}$, which by the hockey-stick theorem for the last time, is equal to $\binom{15}{8}$.

For number 2, just note that the for any choice of 10 numbers from the set {1,2,3...9} you can arrange them into exactly one increasing number, so we count the number of choices of 10 elements, with repetition allowed, from {1,2,3...9}. Its well known that this is (10+9-1)C(9-1) = 18C8

[quote="Mildorf"][quote="Rep123max"]I think it went down because people are simply to busy, but, another good time for any mock contest would be during Winter Break and I forget, is Spring Break before or after AIME?[/quote]

* Very well. If I can come up with enough questions between now and then there will be a Mock AIME 4 during winter break.  If someone wants to referee  that would be welcome also.

* Spring break is usually during April, after the real AIMEs.[/quote]

and your Mock AIMEs warrant at least 5 hours ... on this one, I don't think I got to any of the last 4 or 5.

I was just going through the practice aimes when i found this one.

I really like the Zuminglish word problems. I solved using recursion.

Let t(n) be the number of n-letter zuminglish words. Consider a sequence of n letters
there are 2 cases
1) first letter is a consonant (2 choices) . then no restrictions gives 2t(n-1)
2) 1st letter is vowel. means both 2nd and 3rd are consonants (4 choices) . then we have 4t(n-3)

So, t(n) = 2t(n-1) + 4t(n-3)
we can easily find that t(0) = 1, t(1) = 3, t(2) = 9-1 = 8
so t(3) = 2*8 + 4*1 = 20
t(4) = 2*20 + 4*3 = 52
t(5) = 2*52 + 4*8 = 136
t(6) = 2*136 + 4*20 = 352
t(7) = 2*352 + 4*52 = 912
t(8) = 2*912 + 4*136 = 2368
t(9) = 2*2368 + 4*352 = 6144 and finally
t(10) = 2*6144 + 4*912 = 15936 as our answer

[quote="yif man12"][quote="Mildorf"]* Spring break is usually during April, after the real AIMEs.[/quote]

Really? Because my school has spring break March 2 to March 21 this year.  :?[/quote]

wOOt!

I'm sorry; I didn't realize this was an old thread. I tend to only look at the last post. My b...

[quote="Silverfalcon"]PLEASE NEVER REVEAL OLD THREADS

This message goes not only to alpha123 and hockeyadi23, but to any other people who might post after I said this.[/quote]

Hmm what if you find a really cool way to do a problem and you want to share it?

Im glad he revived this thread... I learned a new method.


A chain with two links is 13 cm long. A chain made from three
links of the same type is 18 cm long. How many centimeters are in the length of a chain made from 25 such links?

clearly, each link is 5 cm.
the equation would then be 3+5x, x=#of links
25*5=125+3=128

[hide]not "clearly" if a person had to ask it on here

so, basically what you have when you connect 2 chains is something like this

[code]
  ############  ##############
#                       ##                           #
#                        ##                          #
 ############     #############[/code]

the middle part is an overlap, so we hafta find out the length of the overlap.
call it x
call the length of a chain y
2y-x=13
3y-2x=18
solving gives you
x=3
y=8

25*8-24*3(24 links in a 25 link chain)=128

Jongao, corrrect answer, wrong thought process[/hide]


inequalities
geometry
parallelogram
area of a triangle

If a,b,c, A,B,C are positive numbers satisfying the equation
a+A=b+B=c+C=k where k is a constant.............Prove that aB+bC+cA<k :^2:

hint:[hide]Try looking at the maximum and minimum values for k[/hide]

[hide]:Sigma: a +  :Sigma: A = 2k so  :Sigma: a <= k or  :Sigma: A <= k

Wolog  :Sigma: a <= k then  :Sigma: aA = k( :Sigma: a) - ( :Sigma: ab) < k  [/hide]

Belenos -- if by :Sigma:a you mean a + b + c, then you're wrong -- :Sigma:a + :Sigma:A = 3k, not 2k.  Otherwise, help me understand your notation -- it's not very clear at all, especially since there aren't any capital letters in the last line, which I find ... distressing.

I should really learn TeX.  Oh well.

[hide]I'm going to make it a,b,c,d,e,f just cause I hate capital lettters, with the final statement being ab+cd+ef<k :^2: .

lemma: a^2+b^2 :ge: 2ab

Proof: (a-b) :^2:  :ge: 0, follows directly from the expansion

Problem proof:

a+b=c+d=e+f=k

a :^2: +2ab+b :^2: = c :^2: +2cd+c :^2: =e :^2: +2ef+f :^2:=k^2

By the lemma, we can say that 2ab :le: (a^2+2ab+b^2)/2, so

2ab=2cd=2ef :le: (k :^2: )/2

2ab+2cd+2ef :le: 3/2 * k :^2:

ab+cd+ef :le: 3/4 * k :^2: < k^2

[/hide]

I feel like I probably showed a few too many steps, but better to be safe than sorry.  Sorry about my sporadic switching between  :^2: and ^2.

Isn't your final expression aA + bB + cC, not aB + bC + cA?

How is it I make so many stupid little mistakes when I'm on this webpage?  I should really leave my computer to solve stuff....

Hmm, I did the same thing.  :Thinks harder.:

[hide]aB+bC+cA=a(k-b)+b(k-c)+c(k-a)

 =(a+b+c)k-(ab+bc+ac)

we have to show that

(a+b+c)k-(ab+bc+ac)<k :^2:

k :^2: -(a+b+c)k+(ab+bc+ac)>0

multiply k on both side

k :^3: -(a+b+c)k :^2: +(ab+bc+ac)k>0

subtract abc on both side we have

k :^3: -(a+b+c)k :^2: +(ab+bc+ac)k-abc >-abc

(k-a)(k-b)(k-c)> -abc, which is absolutely true since k>a,b,c

QED[/hide]

This is probably one of the hardest Intemediate problem I have seen

try drawing an equilateral triangle........with it sides k.........each side can be expressed as a+A, b+B, c+C........then by using the area formula of triangle..........u can get 1/2*aBsinC + 1/2*bCsinA + 1/2cAsinB < 1/2 k*k sin A.......since A=B=C=60 degrees.........so we can have aB+bC+cA<k^2

Am I right though? and what do you mean by area formula of triangle? Is it Heron's Formula?

Slick.  It's the area of a triangle given two sides and their included angle.  I guess it's kinda like a cross product in physics that gives the area of the parallelogram made by the two vectors... but a triangle is only half of a parallelogram, so it's 1/2 absinC.


Let $ a;b;c$ be positive real numbers satisfying $ xy+yx+xz=3$
Prove that:
:$ \frac{3+a^{2}}{3-a}+\frac{3+b^{2}}{3-b}+\frac{3+c^{2}}{3-c}\geq 2\sqrt{3(a^{2}+b^{2}+c^{2})}$

I don't understand,maybe you mean $ ab\plus{}bc\plus{}ca\equal{}3$ :maybe: ?


i am stuck up at this one
prove:[b](x^2)[/b]sin[b]x[/b]+[b]x[/b]cos[b]x[/b]+[b]x^2[/b]+1/2>0
i am new to inequalities so plz be as elementary as u can be.

Your problem is easy. I'm lazy and sedning you link with the solution.
http://www.mathlinks.ro/Forum/viewtopic.php?p=355477#355477


[quote="anirudh"]What is the arctan function?[/quote]
It's the inverse tangent. If you input $x$, you get the value of $\theta$ such that $x=\tan\theta$.

[quote="anirudh"]Oh, so is $\arctan x=\tan^{-1}x$?[/quote] Yes. :D :D

Arctan, in terms of a right triangle is the angle itself.

oh. so $\arctan{\tan{x}}=x$ where $0<x<180$?

[quote="anirudh"]oh. so $\arctan{\tan{x}}=x$[/quote]
Well, almost, but not necessarily. Values for $\arctan$ (if I remember correctly) are defined to be between $0^\circ$ and $180^\circ$. This means that, for example, $\arctan{\tan{200^\circ}}=20^\circ$.
However, $\tan{\arctan{x}}=x$ always holds.

Well, yes, I meant x to be between 0-180

(Edited)

Yes. I believe so:

We want a value tan(x) = a/b, so

a/b = (-a)/ (-b) = a/b

So yes, lotrgreengrapes7926 is correct.

-90 to 90 inclusive (in degrees).[/quote]

[quote="moogra"]you know how tan45=1
arctan1=45
for when x is bet 0 and 90[/quote]
It is between $-90^\circ$ and $90^\circ$, as your range would only allow for nonnegatives as the domain.


Well, in fact, in the 3rd one, solutions are directly given.. :)
And, since I've never practiced this method, I wouldn't solve them this way..

I finally realized how to do these after my algebra II textbook came  :D

[hide="1"]$x^2+4x-3=0$
$x^2+4x=3$

To complete the square, add $(\frac{1}{2}b)^2$ to both sides. b is 4, ${(\frac{1}{2}\cdot4})^2=4$. Add that to both sides.

$x^2+4x+4=3+4$

$(x+2)^2=7$

$x+2=\pm\sqrt7$

$x=\pm\sqrt7-2$
[/hide]

[hide="2"]Rewrite as $x^2+3x=4$

Complete the square by adding ${(\frac{1}{2}\cdot3})^2=2.25$ to both sides.

$x^2+3x+2.25=6.25$

$(x+1.5)^2=6.25$

$x+1.5=\pm2.5$

$x=1, -4$
[/hide]

[hide="3"]
$3x^2-x=0$
To complete the square, the coefficient of $x^2$ must be 1. Divide by 3 to get

$x^2-\frac{1}{3}x=0$

complete the square by adding $(\frac{1}{2}\cdot\frac{1}{3})^2=\frac{1}{36}$ to both sides.

$x^2-\frac{1}{3}x+\frac{1}{36}=\frac{1}{36}$

$(x-\frac{1}{6})^2=\frac{1}{36}$. Note that I used subtraction this time.

$x-\frac{1}{6}=\pm\frac{1}{6}$

$x=0, \frac{1}{3}$.
[/hide]

THANK YOU SO MUCH 236!
YOU SAVED MY LIFE!!!!!!!!!!!
 :D


Let $F_1,\ldots,F_n$ be subspaces of a vectorial space $E$. We have to prove that both of the afirmations are equivalent

(a)For each $x\in F_1\cup \ldots F_n$ it's a unique form to be written as $x=x_1+\ldots+x_n$ where $x_i\in F_i$

(b)$\forall j$$F_j\cap (F_1+\ldots+F_{j-1}+F_{j+1}+\ldots+F_n)=\{0\}$.

I have some intuitions but I'm not able to traslate them to mathematics. Anyone could help?

thanks

(a) implies (b):
If $p\in F_j$, we have $p=0+0+\cdots+0+p+0+\cdots+0$, and this representation is unique. If $x_j=x_1+x_2+\cdots+x_{j-1}+x_{j+1}+\cdots+x_n$ with $x_k\in F_k$, we have $p=x_1+x_2+\cdots+x_{j-1}+(p-x_j)+x_{j+1}+\cdots+x_n$. By uniqueness, $x_j=0$ and $F_j$ has a trivial intersection with the sum of the other spaces.

(b) implies (a):
If we reverse as much as possible of the previous argument, we find that we can show the desired result if the sum $\sum_{i\neq j}F_j$ is a direct sum (If $0=\sum x_i$ with $x_i\in F_i$, $x_i=0$ for all $i$). Consider a sum $\sum x_i=0$, and assume some $x_j$ is nonzero. Then $\sum_{i\neq j}x_i=-x_j$, which contradicts (b), so in any such sum, all terms are zero.

We've actually shown that both conditions are equivalent to (c):
$F_1+\cdots+F_n=F_1\oplus\cdots\oplus F_n$


analytic geometry
geometry
perpendicular bisector

Engineers want to connect two parts of an unfinished road. The two parts are parallel, but end at points A and B respectively. We draw line segments AC and BC are perpendicular at point C with AC perpendicular to the two unfinished roads (BC is parallel). The engineers join the two parts of the road with two congruent (the same) circular curves. Each curve is tangent to both one part of the road and the other curve (at the midpoint of line segment AB). Given that AC is 1200 meters and BC is 900 meters, what is the radius of the two curves in meters?

Can someone please give a solution to this question?
I've included a diagram now, see attachment

[quote="SplashD"][quote="lotrgreengrapes7926"][hide] Let's graph this on a coordinate system with $C(0,0)$, $A(1200,0)$, and $B(0,900)$. The circular curve through point $B$ has a center of $(r,900)$ and goes through $(600,450)$, the midpoint of $AB$. So, $(600-r)^{2}+(450-900)^{2}=r^{2}$ from the formula for a circle. Solve to get $r=\frac{1875}{4}$. [/hide][/quote]

The answer is an integer.

I'm also not sure about the claim that the center of the curve through B is $(r,900)$[/quote]
The center [i]is[/i] at $(r,\,900),$ and I don't think it's an integer.

[hide] If a circle must intersect 2 points, the center must be on the perpendicular bisector of the segment joining the 2 points. Through simple algebra work, we get $y=\frac{4}{3}x+275$ and $y=\frac{4}{3}x-975$ as the perpendicular bisectors. The two circles must have centers at $(x,\frac{4}{3}x+275)$ and $(x,\frac{4}{3}x-975)$ and must intersect $(0,900)$ and $(1200,0)$ respectively.
$x^{2}+(\frac{4}{3}x-625)^{2}=(x-1200)^{2}+(\frac{4}{3}x-975)^{2}$
Solve to get $x=600$. Thus the upper circle's center is at $(600,1075)$ and passes through $(0,900)$. By the distance formula, $\sqrt{600^{2}+175^{2}}=\boxed{625}$. [/hide]

Yeah, lotrgreengrapes's answer is correct and his solution looks right too.

I really don't think you need coordinate geometry though.


The powers 2^n and 5^n start with the same digit d. What is this digit?

[hide="Assuming n is a positive integer"]Suppose $ 2^{n}$ has $ x$ digits and $ 5^{n}$ has $ y$ digits. Then $ d \cdot 10^{x}\leq 2^{n}< (d+1) \cdot 10^{x}$ and $ d \cdot 10^{y}\leq 5^{n}< (d+1) \cdot 10^{y}$. Since all terms in these inequalities are positive, we can also say that $ d^{2}\cdot 10^{x+y}\leq 10^{n}\leq (d+1)^{2}\cdot 10^{x+y}$. Quick case analysis shows that the only time this works is when $ \fbox{d = 3}$, since the only other case where a number between the two bounds has leading digits $ 10\ldots$ is $ d = 1$, and that can only work when $ 2^{n}= 10^{x}, 5^{n}= 10^{y}$, both impossible for positive $ n$.[/hide]


geometry
geometric transformation
reflection
circumcircle
Euler
incenter
geometry theorems

I read in Coxeter's Book "Geometry revisited" (NML 19) that the orthocentric quadrangles have many properties, but the author gives only two of them. So, could anyone (perhaps Darij) tell what are the main properties of those quadrangles ?

Hier my little knowledge :
An orthocentric quadrangle is a quadrangle which has the following property : one its vertices is the orthocenter of the triangle formed by the other three.

$1$ It follows from the definition : $ABCD$ is an orthocentric quadrangle $\Leftrightarrow$ $A$ is the orthocenter of triangle $BCD$ $\Leftrightarrow$ $B$ is the orthocenter of triangle $CDA$ $\Leftrightarrow$ $C$ is the orthocenter of triangle $DAB$ $\Leftrightarrow$ $D$ is the orthocenter of triangle $ABC$.

Let $ABCD$ an orthocentric quadrangle.

$2$ Since the reflections of $A$ in the lines $BC$, $CD$, $DA$ lie on the circumcircle of $BCD$, it follows that $ABC$, $BCD$, $CDA$ and $DAB$ have the same circumradius.

$3$ Let $M$, $N$, $P$ be the feet of the altitudes of of triangle $ABC$. Then one can easily see that $M$, $N$, $P$ are the feet of the altitudes of of $BCD$, $CDA$ and $DAB$ as well. It follows that these three triangles have the same Euler-Circle.

Benjamin

Honestly, I seldomly use the notion of orthocentric quadrangles and normally speak of triangles with orthocenters, as most properties of orthocentric quadrangles are, in some way, unfair towards one of the four vertices. To your list, I can contribute:

Let $ABCD$ an orthocentric quadrangle.

[b]4.[/b] The centroid of the quadrilateral $ABCD$ is the common nine-point center of triangles $BCD$, $CDA$, $DAB$, $ABC$.

[b]5.[/b] The circumcenters of triangles $BCD$, $CDA$, $DAB$, $ABC$ are the reflections of the points $A$, $B$, $C$, $D$ in this nine-point center.

Some of the properties of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=109112]Poncelet points[/url] can be seen as constructions based upon an orthocentric quadrangle and a point in the plane (see, e. g., Theorem 3).

Also, since the incenter and the three excenters of a triangle ABC form an orthocentric quadrangle, you can consider properties of incenters and excenters as facts about orthocentric quadrangles.

  Darij


Let $ a_1, a_2, ..., a_{300}$ be nonnegative real numbers, with $ \sum_{i\equal{}1}^{300} a_i \equal{} 1$.

Find the maximum possible value of $ \sum_{i \neq j, i|j} a_ia_j$.

Call 2 of the numbers $ a_i$ and $ a_j$ friendly if either $ i|j$ or $ j|i$ (with $ i \ne j$).

Similarly, we call 2 numbers $ a_i, a_j$ unfriendly if they are not friendly, and $ i \ne j$.

Now, if we take any 2 unfriendly numbers $ a_i, a_j$ each with non-zero values, we can increase the sum by either replacing $ (a_i,a_j)$ with $ (a_i \plus{} a_j,0)$ or $ (0,a_i \plus{} a_j)$. We repeat the process until it eventually halts. Then all of the non-zero numbers must be friendly towards each other, else we could apply the process one more time.

Now, if $ A$ is the set of the $ i$ such that $ a_i$ s non-zero, then we now have

$ \sum_{i \in A} a_i \equal{} 1$

and we need to maximize $ \sum_{i,j \in A, i < j}a_ia_j$.

But this is less than $ (|A| choose 2/|A|^2)(\sum_{i \in A} a_i)^2 \equal{} (|A| choose 2/|A|^2) \equal{} \frac {1}{2}(1 \minus{} \frac {1}{|A|})$. So we need to maximize the size of $ |A|$. Clearly this is can be no larger than when $ A \equal{} {1,2,4,8,16,32,64,128,256}$, which gives $ A$ 9 elements. Hence the max. is $ \frac {1}{2}(1 \minus{} \frac {1}{9}) \equal{} \frac {4}{9}$. We can attain equality when $ a_1 \equal{} a_2 \equal{} a_4 \equal{} a_8 \equal{} a_{16} \equal{} a_{32} \equal{} a_{64} \equal{} a_{128} \equal{} a_{256} \equal{} \frac {1}{9}$ and the rest of the $ a_i \equal{} 0$.

[quote="rofler"]

Now, if we take any 2 unfriendly numbers $ a_i, a_j$ each with non-zero values, we can increase the sum by either replacing $ (a_i,a_j)$ with $ (a_i \plus{} a_j,0)$ or $ (0,a_i \plus{} a_j)$.
[/quote]

I really don't understand this step. Could you explain it again for me?

We are replacing the 2 numbers with one of those 2 pairs. One of them will increase the number, this is easy to see (I could write out the proof, but it is only one or 2 lines).

actually,for a number which contains a prime divisor greater than $3$,we can porve by adjustment that it must be zer0.

Well, we could just take that all two integers $ i,j $ from $ 2^n $ to $ 2^{n+1}-1 $ has "$ i $ doesn't divide $ j $".
So, just $ \sum_{i \neq j, i|j} a_ia_j \leq \prod_{i=0}^{8}(a_{2^i}+...+a_{2^{i+1}-1}) \leq \frac{4}{9}$.

(Well, actually it's the same idea as above. ;) )

[quote=toto1234567890][quote=smy2012]See also Korean 2015[/quote]

Yeah, I took that exam :P (It was really funny to see same problem appearing from an old exam. )
And actually, it's Final 2013. ;) https://www.artofproblemsolving.com/community/c6h526260p2983427[/quote]

It's very old actually. :-D
In my knowledge, it first appear in [url=https://cms.math.ca/openaccess/cjm/v17/cjm1965v17.0533-0540.pdf]here[/url]

The key observation is that if $i$ and $j$ do not divide each other, and $a_i$ and $a_j$ are both nonzero, then replacing either $(a_i, a_j)$ with $(a_i+a_j, 0)$, $(0, a_i+a_j)$ will not decrease the sum, but will increase the number of zero terms.
So apply this operation until it's no longer possible. Let $I = \{ i : a_i \neq 0 \}$. If $I = \{ i_1 < \dots < i_k \}$ we have $i_1 \mid \dots \mid i_k$.
Now the sum in question (by Cauchy or anything else) is at most $\frac{1}{2}(1-k^{-1})$. So we want to maximize $k$, which occurs at $k=9$. This gives the upper bound $4/9$. Equality occurs of course at $a_1 = a_2 = a_4 = \dots = a_{256} = 1/9$.

Really nice smoothing problem.

Notice that for any $i, j$ that don't divide each other, the expression is linear in $a_i$ and $a_j$, thus the smoothing $(a_i, a_j) \to (a_i+a_j, 0)$ or to $(0, a_i+a_j)$ will increase the sum.

In particular, repeating this operation a maximal number of times, we are left with some number of $k \leq 9$ terms, all of which divide each other. The upper bound can easily be seen to be $\frac 49$ now by Cauchy.

Our answer is $\boxed{\frac 49}$, attained with $a_i=\tfrac 19$ if $i=2^j$ and $a_i=0$ otherwise.

Consider two indices $m$, $n$ such that $m \nmid n$, $n \nmid m$. The change in the desired expression by switching $(a_m,a_n) \rightarrow (a_m+a_n,0)$ has opposite sign as the change in $(a_m,a_n) \rightarrow (0,a_m+a_n)$, so we can always (weakly) increase our sum using this algorithm. Further, note $\sum a_i$ remains invariant at 1.

Our algorithm terminates when all nonzero terms have indices which are multiples/divisors of each other, meaning there can be at most 9 nonzero terms (the powers of 2). If the remaining numbers are $b_i, \ldots, b_k$, our expression is
\[\sum b_ib_j = \frac 12 \left(\left(\sum b_i\right)^2 - \sum b_i^2\right) \leq \frac 12 \left(1 - \frac 1k\right) \leq \boxed{\frac 49}. \quad \blacksquare\]


Can someone help me out with the following question?

"Determine the Laurent Expansion of (z^2 + 2z) / (z^3 - 1) in the annulus 0 < lz-1l < R about z = 1. Find the large R one can use."

So far I reach the step after factorizing the denominator, that's:

(z^2 + 2z) / (z^3 - 1) = 1/z-1 - (1/6j)/(z+1/2-3j) + (1/6j)/(z+1/2+3j)

How should I deal with the term 1/z-1? I think I need to use the rule
1/1-z = 1 + z + z^2 + z^3 + ... , but what region should I consider?? (anything to do with the annulus centered at z=1?)

A billion thanks!!

$\frac{1}{z-1}$ will be one of the terms in your Laurent expansion. It's the other two fractionas that require further work: \[\frac{1}{z+1/2-3i}=\frac{1}{(z-1)+3/2-3i}=\frac{1}{3/2-3i}\frac{1}{1+(z-1)/(3/2-3i)}\] Now use $1/(1-w)=1+w+w^{2}+w^{3}+\dots$ with $w=-\frac{z-1}{3/2-3i}$. This series converges when $|w|<1$, that is, when $|z-1|<|3/2-3i|$.

but for the term 1/z-1, I do need to expand it into laurent series also, don't I?
Then how should I deal with the region?
Should I consider the region lzl<1 where 1/z-1 = - 1/1-z = -1 - z - z^2 -...
, or lxl>1 where 1/z-1 = 1 / [z(1 - 1/z)] = 1/z(1 + 1/z + 1/z^2 + ...) = 1/z + 1/z^2 + 1/z^3 + ...
But the two methods above are only for C centered at z=0, aren't they? How about this case that the annulus is centered at z=1??? :huh:  :huh:

The term $\frac{1}{z-1}$ is its own Laurent series. It's a series $\sum_{k}c_{k}(z-1)^{k}$, where $c_{k}=0$ unless $k=-1$, and $c_{-1}=1$.


I'm not sure if this has been asked already, but how long will it take to indicate on the web site that you've received my entrance form? I sent it in about a week ago, and it hasn't shown up yet under "my profile."

If you sent it by fax, then it should have showed up by now.  If by snail mail, it will likely be a few more days as we've been flooded with entries and will be all week.


The chart below gives the air distance in miles between selected
world cities. What is the distance from Honolulu to Cape
Town?
\begin{tabular}{|l|c|c|c|c|}  \hline
& Bangkok & Cape Town & Honolulu & London   \\ \hline
Bangkok & X & 6300 & 6609 & 5944   \\ \hline
Cape Town & 6300 & X & 11,535 & 5989  \\ \hline
Honolulu & 6609 & 11,535 & X & 7240  \\ \hline
London & 5944 & 5989 & 7240 & X  \\ \hline
\end{tabular}

$ \begin{tabular}{|l|c|c|c|c|} \hline
& Bangkok & Cape Town & Honolulu & London \\ \hline
Bangkok & X & 6300 & 6609 & 5944 \\ \hline
Cape Town & 6300 & X & 11,535 & 5989 \\ \hline
Honolulu & 6609 & 11,535 & X & 7240 \\ \hline
London & 5944 & 5989 & 7240 & X \\ \hline
\end{tabular}$

here's the real chart
so looking at it, you see the answer is $ 11535$
very simple
very pointless
shouldn't be classified as math

Find Honolulu on one side and Cape Town on the other.
Find where they meet.
It should be [b]11,535[/b].


trigonometry
calculus
integration
complex analysis
calculus computations

I don't know how to solve this differential equation with initial conditions.

${y^{\prime\prime}}+10y=0, y(0)=\pi, y^{\prime}={\pi}^{2}$

Thanks for any hints. :)

This isn't complex analysis- it goes in the calculus/real analysis section.

I assume the second initial condition should be $y'(0)=\pi^2$?

The standard method:
Write down a general solution $y(x)=af(x)+bg(x)$.
Find $f(0),f'(0),g(0),g'(0)$.
Now $y(0)=af(0)+bg(0)$ and $y'(0)=af'(0)+bg'(0)$.
This is a system of linear equations in $a$ and $b$; solve it.

It'll probably be an easy system to solve- the natural choice of $f$ and $g$ makes some of these values zero.

This is the Simple Harmonic Oscillation-Typed differential equation.

Generally speaking for the positive constant number $\omega$ the general solution of $y''=-\omega ^ 2 y$ is $y=A\cos \omega x+B\sin \omega x.$ or $y=A\sin (\omega x+\theta).$ where $A,B$ are integral constant numbers and $\theta$ is a constant numer.

Another approach: Multiply $y'$ by both sides of $y''=-10y$ we have $y'y''=-10yy'\Longleftrightarrow \left(\frac{1}{2}y'^2\right)'=-10\cdot \left(\frac{1}{2}y^2\right)'.$ Integrate both sides with respect to $x,$ yielding $y'\ ^2+(\sqrt{10y})^2=C^2,$ where $C$ is positive constant numbers.


A little friendly problem from the old National exams of iran.
Let $\mathbb{X}$ be a polynomail with integer coefficients and there exist numbers like $a,b$ such that

$\mathbb{X}(a)\mathbb{X}(b)=-(a-b)^2$

prove that

$\mathbb{X}(a)+\mathbb{X}(b)=0$

I will use $P(x)$ instead of $\mathbb {X}(x)$ to save time...

First, if $a=b$ we easily deduce that $P(a)=P(b)=0$, so we assume that $a \neq b.$
Since $P$ has integer coefficients, we deduce that $a-b$ divides $P(a)-P(b).$

Let $q$ be a prime which divides $a-b.$
Since $q$ divides $P(a)P(b)$ and $P(a)-P(b)$, we deduce that it divides both $P(a)$ and $P(b).$
Let $x,y$ be the exponents of $q$ in the prime decompositions of $P(a)$ and $P(b)$ respectively.
Then, using $x+y$ is the exponent of $q$ in the prime decomposition of $(a-b)^2$, which proves that $x+y$ is even, so that $x$ and $y$ have the same parity, and that $\frac {x+y} 2$ is the exponent of $q$ in the prime decomposition of $a-b.$

Assume that $x<y$. Thus, from above $y \geq x+2$ and $x+1 \leq \frac {x+y} 2$ and $x+1 < y.$
Therefore $q^{x+1}$ divides $a-b$ and $P(b)$, and then it divides $P(a)$. A contradiction.
Thus $x \geq y$, and by a similar reasoning we finally deduce that $x=y.$
It follows that $x$ is the exponent of $q$ in the prime decomposition of $a-b.$

Then, we have proved that for each prime $q$ which divides $a-b$ the exponents of $q$ in the prime decompositions of $a-b,P(a)$ and $P(b)$ are equals. It follows that $a-b$ divides both $P(a)$ and $P(b)$.
Lat $P(a) = m(a-b)$ and $P(b)=n(a-b)$ where $m,n$ are integers.
From $P(a)P(b) = -(a-b)^2$ we deduce that $mn=-1$ so that $m+n=0.$
This easily leads to $P(a)+P(b)=0.$

Pierre.

How do you deduce that $a$ and $b$ are integers?

Because otherwise the result does not hold  :P
Let $P(x) = 1-x - 3x^2.$
We have $P(0)=1$ and $P(\frac 1 2 ) = -\frac 1 4$
so that $P(0)P(\frac 1 2) = - (0- \frac 1 2)^2$
But $P(0)+P(\frac 1 2 ) \neq 0.$

Thus, I think $a,b$ are supposed to be integers in the statement.

Pierre.


algebra
polynomial
trigonometry
arithmetic sequence
real analysis
real analysis unsolved

Could anyone please solve the following difference equations?
1)  y(n+2)-yn=1  ,  if y1=y2=1
2) y(n+2)+y(n)=n, if y1=y2=0

1) $ y_1,\ y_3,\ y_5,\ \cdots$ forms arithmetic sequence with common difference 1 and $ y_2,\ y_4,\ y_6,\ \cdots$ forms arithmetic sequence with common difference 1, that is to say, for some positive integers $ k,$ For $ n\equal{}2k\minus{}1,$ we have $ y_{2k\minus{}1}\equal{}y_1\plus{}1\cdot (k\minus{}1),$ yielding $ y_{2k\minus{}1}\equal{}k,$ let $ 2k\minus{}1\equal{}n\Longleftrightarrow k\equal{}\frac{n\plus{}1}{2},$ we have $ y_n\equal{}\frac{n\plus{}1}{2}\ (n\equal{}1,\ 3,\ 5,\ \cdots),$ similarly by $ y_{2k}\equal{}k\plus{}1,$ we have $ y_n\equal{}\frac{n\plus{}2}{2}\ (n\equal{}2,\ 4,\ 6,\ \cdots).$

These are linear differential equations, so the solutions are of the form:

$ y_n\equal{}h_n\plus{}p_n$

where $ h_n$ is the solution to the homogeneous equation, and $ p_n$ is the particular solution. So, we have:

1)

$ h_{n\plus{}2}\minus{}h_n\equal{}0$

characteristic equation is:

$ r^2\minus{}1\equal{}0$

$ r_{1,2}\equal{}\pm 1$

so:

$ h_n\equal{}A\plus{}B(\minus{}1)^n$

Now, to the particular solution. If the RHS equals $ 1$, the $ p_n$ should be a polynomial. A polynomial of a degree $ 0$, gives us nothing, but with degree $ 1$ we have:

$ a(n\plus{}2)\plus{}b\minus{}a n\minus{}b\equal{}1$

$ 2a\equal{}1$

$ a\equal{}\frac{1}{2}$

so:

$ p_n\equal{}\frac{n}{2}$

also:

$ y_n\equal{}h_n\plus{}p_n\equal{}A\plus{}B(\minus{}1)^n\plus{}\frac{n}{2}$

now:

$ y_1\equal{}1\equal{}a\minus{}b\plus{}\frac{1}{2}$

$ y_2\equal{}1\equal{}a\plus{}b\plus{}1$

so:

$ a\equal{}\frac{1}{4}$

$ b\equal{}\minus{}\frac{1}{4}$

or:

$ y_n\equal{}\frac{1}{4}\left(1\minus{}(\minus{}1)^n\right)\plus{}\frac{n}{2}$

2)

Similar as before, the characteristic equation gives us roots $ r_{1,2}\equal{}\pm i$ so the solution to the homogeneous equation is of the form:

$ h_n\equal{}Ai^n\plus{}B(\minus{}i)^n$

but this can be rewritten as:

$ h_n\equal{}a \cos\left(\frac{n \pi}{2}\right)\plus{}i b \sin\left(\frac{n \pi}{2}\right)$

now, for the particular solution, the same thing as before, it should be of the form $ p_n\equal{}Cn\plus{}D$ (because RHS is a polynomial or degree 1), and:

$ C(n\plus{}2)\plus{}D\plus{}Cn\plus{}D\equal{}n$

$ 2Cn\plus{}2(D\plus{}C)\equal{}n$

from this we have:

$ 2C\equal{}1 \Rightarrow C\equal{}\frac{1}{2}$

$ 2\left(D\plus{}\frac{1}{2}\right)\equal{}0 \Rightarrow D\equal{}\minus{}\frac{1}{2}$

So:

$ y_n\equal{}a \cos\left(\frac{n \pi}{2}\right)\plus{}i b \sin\left(\frac{n \pi}{2}\right)\plus{}\frac{1}{2}(n\minus{}1)$

Also, we have:

$ y_1\equal{}0\equal{}i b \Rightarrow b\equal{}0$

$ y_2\equal{}0\equal{}\frac{1}{2}\minus{}a \Rightarrow a \equal{}\frac{1}{2}$

so:

$ y_n\equal{}\frac{1}{2}\left(\cos\left(\frac{n \pi}{2}\right)\plus{}n\minus{}1\right)$

Thank you very much milin! Can you please explain how I can find a root for the non homogenious case

There is no root for the non homogeneous case. This is the point:

If you have an difference equation of the form:

$ c_0 y(n) \plus{} c_1 y(n \plus{} 1) \plus{} ... \plus{} c_m y(n \plus{} m) \equal{} f_1(n) \plus{} f_2(n) \plus{} ... \plus{} f_k(n)$

then you can define the solution to $ y_n$ as:

$ y_n \equal{} y_1(n) \plus{} y_2(n) \plus{} ... \plus{} y_m(n) \plus{} ... \plus{} p_1(n) \plus{} p_2(n) \plus{} ... \plus{} p_k(n)$

where:

$ y_i(n) \equal{} \sum^{h_i}_{j \equal{} 0}a_{i,j} n^j$

where $ h$ is the i-th root's number of appearances, and $ a_{i,j}$ are some constants. As for $ h$, if you have $ (r \minus{} 2)^2 \equal{} 0$ then $ r$ appears two times so it's $ h \equal{} 2$. Also, $ p_k(n)$ are partial solutions of the difference equation corresponding to $ f_k(n)$.


Well if they want me to help them with something then I am a genius. If they don't want me to help them then I am a nerd. I prefer meaning called geek than nerd, because nerd means someone without social skills and geek means a smart person but I don't really like being called either at all though.

I'm not going to put everyone in my school into one group. A lot of them are nice, and some aren't.

[quote="tennis123"]At your school, what are the kids like? I mean, do they call you a nerd or a genius? Are they into math? Were you ever called a nerd? And also, do you think being called a nerd is derogatory? Just wondering[/quote]

Well, I had left school (high school) 4 years ago. The kids at my school were pretty congenial. People never called me either a nerd or a genius. (I'm neither) Some of the kids at my school were very adept at solving math problems.

Personally, I don't think being called a nerd is derogatory. I opine it's relative-- some people like when others call or address them as a nerd, while some people may loathe being called a nerd. I had seen some nerds who're pretty extroverted. Probably, I will concoct a term for it: Nerdy Extrovert. :D  (Sounds like an oxymoron, ain't it?)

Hmm, in my school, smart people have a smart time getting accepted, and this is a school where 70% (and probably more) is asian.  Although smart fobs get their own group, because they can relate, I guess, the intellectuals (math and science team people) have trouble getting members for the clubs/teams even if the people in charge of those aren't the most socially outcast on the campus.

I think the people at my school are very smart  (quite a few seem to have published research papers), but don't desire to sacrifice their public image.  those who do are uberly motivated.

Hmmm...

The non math boys like to bother people and hang out in groups. In fact, they do really stupid things. The sheriff had to come and do a talk about sexual battery because they were committing  illegal actions.  :o

The non math girls like to attract attention to themselves by being rude and loud and like to expose their bodies. :(

This in Santa Clara Valley, so i think it would be much worse in places like LA


If $a,b,c$ a positive numbers such that $ab+bc+ca=3$ and $p \ge 3+\sqrt{6}$, then

$\frac 1{pa+b}+\frac 1{pb+c}+\frac 1{pc+a}\ge \frac 3{p+1}$.

So, Hungkhtn's Inequality
$\frac 1{(pa+b)^{2}}+\frac 1{(pb+c)^{2}}+\frac 1{(pc+a)^{2}}\ge \frac 9{(p+1)^{2}(ab+bc+ca)}$
is solved for $p \ge 3+\sqrt{6}$.

Dear VASC,

Just want to say to you that accidently, I created this problem (and told to toanhocmuonmau, too) some months ago and already added it to my book. However, my problem is more general

If $a,b,c$ are non-negative real numbers such that $ab+bc+ca=3$ then
\[\frac{1}{pa+b}+\frac{1}{pb+c}+\frac{1}{pc+a}\ge \min (\min f(x,y), \frac{3}{p+1}) \]
in which $xy=3$ and
\[f(x,y)=\frac{1}{px+y}+\frac{1}{py}+\frac{1}{x}. \]
This fact shows that it suffices to consider the minimum of the expression $\frac{1}{pa+b}+\frac{1}{pb+c}+\frac{1}{pc+a}$ just in case $\min(a,b,c)=0$.

My solution is the same as (even much easier) what I have done to prove
\[\frac{1}{(2a+b)^{2}}+\frac{1}{(2b+c)^{2}}+\frac{1}{(2c+a)^{2}}\ge \frac{1}{ab+bc+ca}. \]
This method is Entirely mixing variable (but improved), which is once published in Mathematics Reflection.

I dont know your solution, but I believe that my solution is very nice and short. It it no longer than six lines  :lol:

[quote="hungkhtn"]\[\frac{1}{(2a+b)^{2}}+\frac{1}{(2b+c)^{2}}+\frac{1}{(2c+a)^{2}}\ge \frac{1}{ab+bc+ca}. \]It it no longer than six lines  :lol:[/quote]
Ugly solution , but less than 6 lines  :D  :lol: !

Assume that : $x=min(x,y,z)$ so that $\exists a,b \geq\ 0$ and $y=x+a$ and $z=x+b$ . After clearing denominators  :blush:  we have :
$81(a^{2}-ab+b^{2})x^{4}+108(a^{3}-a^{2}b+ab^{2}+b^{3})x^{3}+6(8a^{4}-7a^{3}b-3a^{2}b^{2}+29ab^{3}+8b^{4})x^{2}+4(2a^{5}+a^{4}b-13a^{3}b^{2}+17a^{2}b^{3}+13ab^{4}+2b^{5})x+ab(4a^{4}-12a^{3}b+5a^{2}b^{2}+12ab^{3}+4b^{4}) \geq\ 0$
Which is true because all off coefficients are non-negative and $x \geq\ 0$ !

[quote="hungkhtn"]Dear VASC,
If $a,b,c$ are non-negative real numbers such that $ab+bc+ca=3$ then
\[\frac{1}{pa+b}+\frac{1}{pb+c}+\frac{1}{pc+a}\ge \min (\min f(x,y), \frac{3}{p+1}) \]
in which $xy=3$ and
\[f(x,y)=\frac{1}{px+y}+\frac{1}{py}+\frac{1}{x}. \]
I dont know your solution, but I believe that my solution is very nice and short. It it no longer than six lines  :lol:[/quote]

I means the solution to this general problem of mine is no longer than six line, tanpham.

[quote="hungkhtn"]I means the solution to this general problem of mine is no longer than six line, tanpham.[/quote]
Oh , sorry  :blush: ! I will try it  :D !

For $p\ge 3+\sqrt{6}$ there is a nice elementary solution.

On the other hand, I found that the inequality

$\frac 1{\frac{4a}{b+c}+\frac b{c}}+\frac 1{\frac{4b}{c+a}+\frac c{a}}+\frac 1{\frac{4c}{a+b}+\frac a{b}}\ge 1$

is stronger that

$\frac 1{(2a+b)^{2}}+\frac 1{(2b+c)^{2}}+\frac 1{(2c+a)^{2}}\ge \frac 1{ab+bc+ca}$

because $\frac{a(b+c)+bc}{(2a+b)^{2}}\ge \frac 1{\frac{4a}{b+c}+\frac b{c}}$.

[quote="Vasc"]For $p\ge 3+\sqrt{6}$ there is a nice elementary solution.

On the other hand, I found that the inequality

$\frac 1{\frac{4a}{b+c}+\frac b{c}}+\frac 1{\frac{4b}{c+a}+\frac c{a}}+\frac 1{\frac{4c}{a+b}+\frac a{b}}\ge 1$
[/quote]

How to prove this one, VASC? I jsut want to know if it is a classical method or advance method or expanding?

About my generalization, it is elementary, too, VASC.

[quote="hungkhtn"]How to prove this one, VASC? I jsut want to know if it is a classical method or advance method or expanding?
[/quote]
I have not a solution to this very sharp inequality, yet. :D

[quote="hungkhtn"][quote="Vasc"]For $p\ge 3+\sqrt{6}$ there is a nice elementary solution.

On the other hand, I found that the inequality
$\frac 1{\frac{4a}{b+c}+\frac b{c}}+\frac 1{\frac{4b}{c+a}+\frac c{a}}+\frac 1{\frac{4c}{a+b}+\frac a{b}}\ge 1$
[/quote]
I jsut want to know if it is an expanding?  :lol:  [/quote]
Base on your idea Hungkhtn !
We assume that $x=min(x,y,z)$ , so that we can put $y=x+a$ and $z=x+b$ after expanding we have :
$16(a^{2}-ab+b^{2})x^{4}+2(10a^{3}-13a^{2}b+15ab^{2}+10b^{3})x^{3}+2(5a^{4}-9a^{3}b+3a^{2}b^{2}+19ab^{3}+5b^{4})x^{2}+(4a^{5}-9a^{4}b-4a^{3}b^{2}+16a^{2}b^{3}+9ab^{4}+4b^{5})x+a(a+b)(a-2b)^{2}b^{2}\geq\ 0$
About your generalization , it's very hard  :o  !


I was wondering how long will the transcripts last before it disappears? I plan to take some notes or print it out before it's gone. Thanks.

Approx. 1 month, but you should have them saved within 1-2 weeks just to be safe.


integration
calculus
calculus computations

The rate of growth in a population of bacteria is given by:
$ \frac{dp}{dt} \equal{} 3t^2\minus{}t$
Where t is the number of hours and p(t) is the population.  If the initial population was 1000 bacteria, find the number present after 24 hours.


Which one do you think will do better in math competitions this year?

the latter option on the poll, but...

honestly

why does this even matter?

NOT NICE CALVIN....

Definitely Ligon...  :)  :)  :)

i vote ligon... though its gonna be close this year and even closer next year (in two years, carnage is gonna pwn)
and calvin, just because you're not there anymore means ligon sucks, right?

@professordad,

Do you go to Carnage? IF SO, LIGON WILL WIN!!  :)

this is kind of off topic, but how well does NC do at the national level?

For MathCounts, NC has been placing near the bottom of the top 10 (enough to get recognition, but not much else).

For ARML, NC has been placing in the top 7 a lot recently, with a first place finish and also a third place finish.

At PUMAC, NC won last year.

Overall, NC is a competitive state.

"None of them will do good"? Will neither do good to society? Will both do evil? How does that work?

On a slightly more serious note, I doubt either will be doing awesomely this year... but we'll see.

Also NC is interesting; we don't have like 1000000000 math people like, say, California, but our good people are 15-20 deep (I hope, at least ;) )

Dude NC is better in the sense that your math people...ACTUALLY DO MATH. I thought my freshmen army was going to be pretty good for random contests, but they are not doing math. They tell me things like "yea i'm doing math, if school math counts." [weep]


Να αποδειξετε οτι για n φυσικο

$ 2(3n \minus{} 1)^n \geq (3n \plus{} 1)^n$

Μια συντομη λυση .
Για $ n > 9$ με λογαριθμιση ισοδυναμα παιρνουμε $ n\ln\left( \frac {3n \plus{} 1}{3n \minus{} 1}\right)\leq{ln2}$ .

Αλλα $ n\ln\left( \frac {3n \plus{} 1}{3n \minus{} 1}\right)\leq n\left(\frac {3n \plus{} 1}{3n \minus{} 1} \minus{} 1\right) \equal{} \frac {2n}{3n \minus{} 1} < ln2$ .
Για $ 1\leq{n}\leq{9}$ επιβεβαιωνεται αριθμητικα...


Um, how is the fact that there are an infinite number of primes related to the divergence of their reciprocals?

There is an infinite number of naturals and perfect squares, but the reciprocals of the naturals diverge and the reciprocals of the perfect squares converge.


[quote="amir2"]Sabet konid hich addade morabae kameli ba addad $0,1$ vojod nadard[/quote]
mesle adade$100$ ;)  :rotfl:

[quote="Omid Hatami"]In masalae ra man ghablan ferestadam va hich kas natavaneste hallesh kone (ba search karadan ehtemalan betoonid peidash konid).[/quote]
raas migi kheyli sakhte :?

bebakhshid mishe soal ro ye kami tozeeh bedid?
man dagheeghan nemifahmam ke chi bayad sabet beshe :oops:


geometry
3D geometry
sphere
symmetry
geometry proposed

A point moves so that the sum of the squares of its distances from the six faces of a cube is constant.Find the locus of the point?

Assume that the cube has vertices $ (\pm1,\pm1,\pm1)$.  Then, the distance from point $ (x,y,z)$ to the parallel faces $ x\equal{}1$ and $ x\equal{}\minus{}1$ of the cube is $ (x \minus{} 1)^2 \plus{} (x \plus{} 1)^2 \equal{} 2x^2 \plus{} 2$, and similarly for the other two pairs of faces, yielding $ x^2 \plus{} y^2 \plus{} z^2 \plus{} 6 \equal{} k$, hence sphere centered at the center $ (0,0,0)$ of the cube.


A billiard ball is hit at a 45-degree angle from a corner of a 4 foot by 7 foot billiards table. How many times will the ball rebound off an edge of the table before landing in a corner?

4?

Just drew a pool table and traced what I thought a ball would do when hit. Didn't use any math.

[quote="isabella2296"]How did you get the $ 45^\circ$ angle right? Did you use a protractor? In MathCounts, I don't think you're allowed to use those... :maybe:[/quote]

Just drew a quick diagram and guessed. What I would like to know is what math could be used to figure out this problem, as my "solution," like you say, is open to many different errors. Also I was not clear from the question which corner the ball starts at, or even whether that would make a difference in the answer.

it would not matter which corner you start at

and i'm pretty sure the answer to this was 5
i'm not good at drawing on this, so i can't really explain

ernie: your diagram is correct but note that the tenth rebound does [b]not[/b] hit the wall, thats why the answer is 9.

are you sure the answer is 9?
i thought it was 5...

To clarify with my memming prowess, 5 is when it just says it is hit at an angle and doesnt give any dimensions. THis is 9.

[quote=riki.math.96]The correct answer is definately nine...
I wonder if I can incorparate math into this...
360 divided by 45 is 8, plus one is nine, but are they related?

please reply![/quote]

that's an interesting thought, but i'm not sure


Let $H_{n}= \sum_{i=1}^{n}\frac{1}{i}$.  Evaluate

$\lim_{n \to \infty}H_{kn}-H_{n}$

[hide]Answer: $\ln k.$ But, I simply used the asymptotic formula for the harmonic series. Not very rigorous I guess.[/hide]

You can use Riemann Sums with infinitesimally smaller intervals..

[hide]Integrals![/hide]

[quote="FieryHydra"]Cool, [b]pkothari13[/b]! I shoud have thought of that earlier. Using integrals is the answer!

Couldn't we also use this result to show the harmonic series diverges :?:[/quote]

Yup... that's essentially putting the Integral Test to work, a basic test for series convergence/divergence.

Thanks, I remember now. Had forgotten the Integral test completely!


function
real analysis
calculus
calculus computations

1. Let f(x) be such that f(x)=1 if x belongs to Q and f(x)=0 if x not belongs to Q. f(x) is not integrable on [0,1]

2. f(x) = 0 for all but finitely many points x in [a,b]. f is integrable on [a,b].

3. define f : [0,1] -> R as follows. Let f(0)=f(1)=1. For x in (0,1) with x not belongs to Q, let f(x)=0. For x in (0,1) with x belongs to Q, write x =a/b where 0<a,b in Z with gcd(a,b)=1, and the let f(x) = 1/b. f is integrable in [0,1]

1 is like 2 and 3 but why is 1 not integrable?

To show that a function is not Riemann integrable, it suffices to demonstrate two sequences of partitions with mesh size approaching $ 0$ (i.e. length of the longest interval approaching $ 0$) such that the corresponding Riemann sums do not approach the same limit.  In question 1, choose one uniform mesh with points sampled at irrational numbers, and another mesh with points sampled at rational numbers -- the first is always equal to $ 0$, while the second is always $ 1$, so the function is not Riemann integrable.

Try to think about why a similar approach to constructing differing Riemann sums would not work for the other two cases, and turn this intuition into a proof.  Hint: For 2, let $ M \equal{} \max \{|f(x)|: x\in [a,b]\}$, and $ N \equal{} |x\in [a,b]: f(x)\neq 0\}$, and show that for any mesh of size $ h$ the Riemann sum is at most $ |MNh|$.  Question 3 is in essence a simple generalization of question 2 -- note that the set $ \{x\in [0,1]: |f(x)| > \epsilon\}$ is finite for each fixed $ \epsilon > 0$, and make some estimates similar to question 2.


inequalities
geometry
trigonometry
area of a triangle

[hide="Solution"] If $ a, b, c$ are not the sides of a triangle then the RHS is negative.  Otherwise, the RHS is $ 16K^2$, where $ K$ is the area of a triangle with side lengths $ a, b, c$.  It remains to prove that

$ \frac {a^2 \plus{} b^2}{4} \ge K$.

But $ \frac {a^2 \plus{} b^2}{4} \ge \frac {ab}{2} \ge \frac {ab \sin C}{2} \equal{} K$ as desired.  We see that equality occurs when $ a \equal{} b$ and $ C \equal{} 90^{\circ} \implies c \equal{} a \sqrt{2}$. [/hide]

[quote="Silverfalcon"]For positive real numbers $ a,b,c,$ prove that

$ (a^2 \plus{} b^2)^2 \geq (a \plus{} b \plus{} c)(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b).$[/quote]
Using the familiar inequality
$ xy\le\frac{(x\plus{}y)^2}{4}, \forall x,y \in R$
we have:
$ (a \plus{} b \plus{} c)(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)\equal{}\left((a\plus{}b)^2\minus{}c^2\right)\left(c^2\minus{}(a\minus{}b)^2\right) \le \frac{\left((a\plus{}b)^2\minus{}c^2\plus{}c^2\minus{}(a\minus{}b)^2\right)^2}{4}\equal{}4a^2b^2\le \frac{(a^2\plus{}b^2)^2}{4}$

[quote="Silverfalcon"]For positive real numbers $ a,b,c,$ prove that
$ (a^2 \plus{} b^2)^2 \geq (a\plus{}b\plus{}c)(a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b).$[/quote]
Britain 2006?


Can someone link me to Mildorf's MOP Notes? Thanks.

The official page is something like http://www.tjhsst.edu/~tmildorf/ . I have a mirror in case tj takes it down one day, so pm me if that doesn't work.

That link does not work. The page I have is http://www.tjhsst.edu/~tmildorf/math/Adv.htm

However, I don't think these are MOP notes.

As discussed before, MOP notes aren't really published online anywhere. But he does have an inequalities packet and stuff. There's the link there, and also

http://web.mit.edu/~tmildorf/www/

[quote="b-flat"][quote="Phelpedo"]That link does not work. [/quote]

It works fine for me  :maybe:[/quote]

It was fixed (notice how the post was edited  :wink: )

It was just a period okay

it's not so hard to delete a period

Hmmm...I remember in August or something I saw MOP notes by him, it might have been directly from AoPS...maybe it was a dream.

Maybe you're confusing him with Reid Barton, who has MOP notes (though they're mostly just problems).


There are five people in a room. Each person is either a knight, who always tells the truth, or a liar, who always lies. Each person is asked the following question: How many liars are among you? The answers are: "one", "two", "three", "four", "five".
How many liars are in the room?

sorry i should explain why

[hide="explanation"]  here are all the possible scenarios:

0-liars
1-liar
2-liars
3-liars
4-liars
5-liars

  If there were 0 liars then they would all be telling the truth, which is impossible since they all gave different numbers.  So we know the number of liars has to be either 1, 2, 3 ,4 or ,5.  Which means one and only one of them are telling the truth.  Thus, there are 4 liars and 1 knight.
[/hide]

[quote="colts18"]Yes I agree with G-Unit. Nice.[/quote]
then write your answer, spammer.


LaTeX
inequalities
Cauchy Inequality
inequalities proposed

if you want to use LaTeX,you must just put dollar signs and between them write formula.For more information click on 'LaTeX Help'.
Here is your problems solutions:
$ \frac{a^{3}+b^{3}+c^{3}}{a+b+c}+\frac{a^{3}+b^{3}+d^{3}}{a+b+d}+\frac{a^{3}+d^{3}+c^{3}}{a+d+c}+\frac{d^{3}+b^{3}+c^{3}}{d+b+c}\geq \frac{(a+b+c+d)^{2}}{4}$
(1)$ \frac{a^{3}+b^{3}+c^{3}}{a+b+c}\geq \frac{a^{2}+b^{2}+c^{2}}{3}$
It is true because:
(1)$ \Longleftrightarrow 2(a^{3}+b^{3}+c^{3})\geq (a^{2}b+a^{2}c+c^{2}a+c^{2}b+b^{2}a+b^{2}c)$which is true because $ a^{3}+b^{3}\geq a^{2}b+b^{2}a$.
So we must just prove that (from(1)) $ 4(a^{2}+b^{2}+c^{2}+d^{2})\geq(a+b+c+d)^{2}$.It is well-known Cauchy-Shwartz inequality.So we are done.
1) problem is well-known Cauchy-Shwartz inequality.
2) is equivalent to $ \frac{a}{b}+\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{c}{b}\geq 2(a+b+c)$ but $ \frac{a}{b}+\frac{a}{c}+1\geq 3a$ (by Cauchy) so we must just prove $ a+b+c\geq 3$ It is Cauchy inequality too.

Thanks!
Is Cauchy this one?
$ \sum_{k=1}^{k}a_{k}b_{k}\geq\sum_{k=1}^{k}a_{k}\sum_{k=1}^{k}b_{k}$
Can you show how to do no.1)?

In no.2) why does $ \frac{a}{b}+\frac{a}{c}+1\geq\ 3a$ and why does $ a+b+c\geq\ 3$

Just use that $ abc=1$.by Cauchy: $ \frac{a}{b}+\frac{a}{c}+1\geq 3\sqrt[3]{\frac{a^{2}}{bc}}$ but  $ \frac{a^{2}}{bc}=a^{3}$.
And $ a+b+c\geq 3\sqrt[3]{abc}=3$

I solved first as I read it but I didn't think about the others, Here is the solution for the first one:
By Cauchy-Shwarz, [(a/sqrt(a+b))^{2}+(b/sqrt(b+c))^{2}+(c/(sqrt(c+d))^{2}+(d/sqrt(d+a)^{2})]*[(sqrt(a+b))^{2}+
(sqrt(b+c))^{2}+(sqrt(c+d))^{2}+(sqrt(d+a))^{2}]>=(a+b+c+d)^{2}=1^{2}=1, or
[a^{2}/(a+b)+b^{2}/(b+c)+c^{2}/(c+d)+d^{2}/(d+a)]*2>=1, from which we get what was needed.

Lasha you wrote the same things as i did.And use LaTeX,please,because it will be easier to read your posts.

Ok but I don't know how to do it. Can't you tell me how to do it?

I wrote about it in the beginning of this topic(2-nd post).But i can write it again,If you want to use LaTeX ,just put dollar signs and between them write formula or something else,

ii) Let $ a\equal{}\frac{x}{y}$, and $ b\equal{}\frac{y}{z}$, then $ c\equal{}\frac{z}{x}$. The inequality becomes [all sums cyclic]:

$ \prod 1\plus{}\frac{xz}{y^{2}}\ge 2(1\plus{}\sum\frac{x}{y})$.

$ \sum x^{3}y^{3}\plus{}x^{4}yz\ge 2\sum x^{3}yz^{2}$

$ \sum x^{3}y^{3}\ge\sum x^{3}yz^{2}$ follows by rearrangement on $ xy$, $ yz$, and $ zx$ [we can say those values have order...$ x\ge y\ge z$ or $ x\ge z\ge y$].

$ \sum x^{4}yz\ge\sum x^{3}yz^{2}\iff\sum x^{3}\ge\sum x^{2}z$ which also follows by rearrangement on $ x$, $ y$, and $ z$.

ii) it's similar to APMO 1998; $ (1\plus{}\frac{a}{b})(1\plus{}\frac{b}{c})(1\plus{}\frac{c}{a}) \geq 2(1\plus{}\frac{a\plus{}b\plus{}c}{\sqrt[3]{abc}})$

[quote="apollo"]ii) it's similar to APMO 1998; $ (1 \plus{} \frac {a}{b})(1 \plus{} \frac {b}{c})(1 \plus{} \frac {c}{a}) \geq 2(1 \plus{} \frac {a \plus{} b \plus{} c}{\sqrt [3]{abc}})$[/quote]
Actually,not only similar,it is the same.I think that condition($ abc\equal{}1$) was given to make the problem bit harder.But anyway it is very easy. :wink:

[quote="snowyowl"]i)Let $ a,b,c,d > 0$ and $ a \plus{} b \plus{} c \plus{} d \equal{} 1$ Prove that $ \frac {a^{2}}{a \plus{} b} \plus{} \frac {b^{2}}{b \plus{} c} \plus{} \frac {c^{2}}{c \plus{} d} \plus{} \frac {d^{2}}{d \plus{} a}\geq\frac {1}{2}$
...[/quote]
We have:
$ \sum_{cyc} \frac{a^2}{a\plus{}b} \geq \frac{(a\plus{}b\plus{}c\plus{}d)^2}{2(a\plus{}b\plus{}c\plus{}d)}\equal{}\frac{a\plus{}b\plus{}c\plus{}d}{2}\equal{}\frac{1}{2}$...

[quote="chien than"][quote="snowyowl"]i)Let $ a,b,c,d > 0$ and $ a \plus{} b \plus{} c \plus{} d \equal{} 1$ Prove that $ \frac {a^{2}}{a \plus{} b} \plus{} \frac {b^{2}}{b \plus{} c} \plus{} \frac {c^{2}}{c \plus{} d} \plus{} \frac {d^{2}}{d \plus{} a}\geq\frac {1}{2}$
...[/quote]
We have:
$ \sum_{cyc} \frac {a^2}{a \plus{} b} \geq \frac {(a \plus{} b \plus{} c \plus{} d)^2}{2(a \plus{} b \plus{} c \plus{} d)} \equal{} \frac {a \plus{} b \plus{} c \plus{} d}{2} \equal{} \frac {1}{2}$...[/quote]
It's same to Erken solution.
[quote="Erken"]No, Cauchy-Shwartz:
$ (\sum_{i \equal{} 1}^{n}a_{i}^{2})(\sum_{i \equal{} 1}^{n}b_{i}^{2})\geq(\sum_{i \equal{} 0}^{n}a_{i}b_{i})^{2}$
Cauchy:
$ \frac {\sum_{i \equal{} 1}^{n}a_{i}}{n}\geq \sqrt [n]{\prod_{i \equal{} 1}^{n}a_{i}}$
1 problem :
by Cauchy-Shwartz:
$ (\frac {a^{2}}{a \plus{} b} \plus{} \frac {b^{2}}{c \plus{} b} \plus{} \frac {c^{2}}{c \plus{} d} \plus{} \frac {d^{2}}{a \plus{} d})((a \plus{} b) \plus{} (c \plus{} b) \plus{} (c \plus{} d) \plus{} (a \plus{} d))\geq (a \plus{} b \plus{} c \plus{} d)^{2}$, so we are done.[/quote]

[quote="snowyowl"]i)Let $ a,b,c,d > 0$ and $ a \plus{} b \plus{} c \plus{} d \equal{} 1$ Prove that $ \frac {a^{2}}{a \plus{} b} \plus{} \frac {b^{2}}{b \plus{} c} \plus{} \frac {c^{2}}{c \plus{} d} \plus{} \frac {d^{2}}{d \plus{} a}\geq\frac {1}{2}$
[/quote]
For the first one just use Cauchy-Shwartz in this form
$ \frac{x_{1}^{2}}{a_{1}}\plus{}\frac{x_{2}^{2}}{a_{2}}\plus{}...\plus{}\frac{x_{n}^{2}}{a_{n}}\geq\frac{(x_{1}\plus{}x_{2}\plus{}...\plus{}x_{n})^{2}}{a_{1}\plus{}a_{2}\plus{}...\plus{}a_{n}}$
where $ x_{i}\in\mathbb{R}$ and $ a_{i}\geq0$  :)

[quote="Erken"][quote="galois01"][quote="snowyowl"]i)Let $ a,b,c,d > 0$ and $ a \plus{} b \plus{} c \plus{} d \equal{} 1$ Prove that $ \frac {a^{2}}{a \plus{} b} \plus{} \frac {b^{2}}{b \plus{} c} \plus{} \frac {c^{2}}{c \plus{} d} \plus{} \frac {d^{2}}{d \plus{} a}\geq\frac {1}{2}$
[/quote]
For the first one just use Cauchy-Shwartz in this form
$ \frac {x_{1}^{2}}{a_{1}} \plus{} \frac {x_{2}^{2}}{a_{2}} \plus{} ... \plus{} \frac {x_{n}^{2}}{a_{n}}\geq\frac {(x_{1} \plus{} x_{2} \plus{} ... \plus{} x_{n})^{2}}{a_{1} \plus{} a_{2} \plus{} ... \plus{} a_{n}}$
where $ x_{i}\in\mathbb{R}$ and $ a_{i}\geq0$  :)[/quote]
I would like to ask mod. to lock this topic,because all of the problems was solved about month ago,and some guys just post my solutuons to them.[/quote]
Erken I didn't read all the solutions so I am sorry if my  solution is the same as yours  :wink:  :blush:

It is OK,but i don't see any reason to continue this topic.


limit
logarithms
function
absolute value
calculus
calculus computations

Define a sequence $ (x_n)_{n\ge 0}$, which satisfies the relation $ x_{n\plus{}1}\equal{}x_n\plus{}e^{\minus{}x_n}$, for any $ n \in \mathbb N$ and $ x_0\in \mathbb R$. Prove that $ \lim_{n\to \infty}{\dfrac{x_n}{\ln {n}}}\equal{}1$

According to the Stolz theorem, since $ \ln{n}$ is monotonous and $ \lim_{n\to\infty}\ln{n} \equal{} \infty$, then

$ \lim_{n\to\infty}\frac {x_n}{\ln{n}} \equal{} \lim_{n\to\infty}\frac {x_{n \plus{} 1} \minus{} x_{n}}{\ln(n \plus{} 1) \minus{} \ln{n}} \equal{}$ $ \lim_{n\to\infty}\frac {1}{e^{x_{n}}\ln\frac {n \plus{} 1}{n}} \equal{} \lim_{n\to\infty}\frac {n}{e^{x_{n}}}$.

On the other hand, we also have $ \lim_{n\to\infty}e^{\ln{n} \minus{} x_n} \equal{} \lim_{n\to\infty}\frac {n}{e^{x_{n}}} \equal{} \lim_{n\to\infty}\frac {x_n}{\ln{n}}$.

Now, notice that if $ \lim_{n\to\infty}\frac {x_n}{\ln{n}} > 1$, then $ \lim_{n\to\infty}(\ln{n} \minus{} x_n) < 0$ so $ \lim_{n\to\infty}e^{\ln{n} \minus{} x_n} < 1$. Similarly for $ \lim_{n\to\infty}\frac {x_n}{\ln{n}} < 1$. Therefore, it remains only $ \lim_{n\to\infty}\frac {x_n}{\ln{n}} \equal{} 1$.

I hope my solution is comprehensible.  :oops:

[quote="Karamata"]Therefore, it remains only $ \lim_{n\to\infty}\frac {x_n}{\ln{n}} \equal{} 1$.[/quote]  Not necessarily: the limit could just fail to exist.

Unrelatedly, a minor point of mathematical English: "[url=http://www.merriam-webster.com/dictionary/monotonous]monotonous[/url]" and "[url=http://en.wikipedia.org/wiki/Monotonic_function]monotonic[/url]" are different words ;)

Since the questions like that appear rather regularly, it may be of interest to consider the general problem $ x_{n\plus{}1}\equal{}x_n\plus{}f(x_n)$ where $ f$ is a positive decreasing continuous function on $ (0,\plus{}\infty)$. Intuitively, it is clear that we can just replace the recursion by the differential equation $ X'(t)\equal{}f(X(t))$ with the same initial condition and get $ x_n\approx X(n)$. The questions that arise are

1) Why is it true?
2) How accurate is this approximation for large $ n$?

The first question can be easily answered if we notice that the difference $ X(n)\minus{}x_n$ is self-correcting: if this difference is positive, then
\[ X(n\plus{}1)\equal{}X(n)\plus{}X'(\xi)\equal{}X(n)\plus{}f(X(\xi))\le X(n)\plus{}f(x_n)\]
with some $ \xi>n$ and $ X(\xi)>X(n)>x_n$, so we add less to $ X(n)$ than to $ x_n$ and the difference decreases at the next step. On the other hand, if it is negative, then it increases unless $ X(n\plus{}1)>x_n$, which would certainly bring it under $ f(x_n)$ in absolute value at the next step. Thus, the difference stays bounded (if it is outside $ [\minus{}\max f,0]$, it'll move towards this interval and if it is inside this interval, it stays there), which, together with the obvious observation that $ x_n\to\plus{}\infty$ as $ n\to\plus{}\infty$, is more than enough to justify that $ \lim_{n\to\plus{}\infty}\frac{x_n}{X(n)}\equal{}1$.

The second question is subtler. To simplify the matters, let us assume that $ \lim_{x\to\plus{}\infty} f(x)\equal{}0$. Then the considerations similar to those above guarantee that the limit of $ X(n)\minus{}x_n$ exists as $ n\to\infty$. The question I'd like to ask now is

[i]Is this limit always $ 0$ under this extra assumption?[/i]. ;)

[quote="JBL"]Unrelatedly, a minor point of mathematical English: "[url=http://www.merriam-webster.com/dictionary/monotonous]monotonous[/url]" and "[url=http://en.wikipedia.org/wiki/Monotonic_function]monotonic[/url]" are different words ;)[/quote]

I observe that before 1950 or so you find "monontonous" used for this type of function.


Suppose medians $m_{a}$ and $m_{b}$ of a triangle are orthogonal. Prove that
a)The medians of that triangle correspond to the sides of a right-triangle.
b)Inequality \[5(a^{2}+b^{2}-c^{2})\geq 8ab\] is valid.

[quote]="N.T.TUAN"]Suppose medians $m_{a}$ and $m_{b}$ of a triangle are orthogonal. Prove that
a)The medians of that triangle correspond to the sides of a right-triangle.
[/quote]

this rezult is very diduced:roll:

[quote]b)Inequality
\[5(a^{2}+b^{2}-c^{2})\geq 8ab \]
is valid.[/quote]

It's very easy becauce if $m_{a}$ and $m_{b}$ of a triangle are orthogonal then $a^{2}+b^{2}=5c^{2}$, therefore $5(a^{2}+b^{2}-c^{2})=5\left(a^{2}+b^{2}-\frac{a^{2}+b^{2}}{5}\right)=4(a^{2}+b^{2})\geq 8ab$, we are done


In the group of ten persons, each person is asked to write the sum of the ages of all the other nine persons. Of all ten sums form the nine-element set $\{ 82, 83,84,85,87,89,90,91,92 \}$, find the individual ages of the persons, assuming them to be whole numbers.

Sum of the smallest nine is 82, largest nine is 92.  So the difference between the smallest and largest is 10. (*)

5+6+7+8+9+10+11+12+13 = 81, so that the least number is atleast 5.

16+6+7+8+9+10+11+12+13 = 92, so that the largest number is at most 16.

From (*) we have either 16,6 or 15,5 as the largest and smallest.  It clear that 16,6 doesnt work (because the smallest is over 82)

So we have 5,....15 with one missing.  The sum 5 + ..  + 15 = 110.  The largest 9 in the set of 10 we want has sum 92; so we remove the smallest which is forced in the set of 10, leaving us 13 to spare, so we remove 13.

The final count is 5,6,7,...,12,14,15.

let $a_1,a_2,a_3...a_{10}$ be the ages of the people. we know that each age is used 9 times so we have
$a_1+a_2+...a_9=82$
$a_2+a_3+...a_{10}=83$
$a_1+a_2+a_3+...a_8+a_{10}=84$
$a_1+a_2+a_3+...a_7+a_9+a_{10}=85$

 and so on (we just leave out one $a_n$ term in each sum.)
we then add them together and find that
$9(a_1+a_2+a_3+...a_{10})=783$
or
$a_1+a_2+a_3+...a_{10}=87$. we then go back to the previous equations and find the value of each $a_n$

[hide]there is one thing that you are missing...you have 9 of the 10 sums, and also using your method, there would be a negative age...

there is a repeat in the set, so we have the sum of 9 of them is 783, and that is divisible by 9, so if we need to add another of the elements to have all 10 sums, the whole sum must be divisible by 9, and adding 90 is the only option to keep it divisible by 9, so we have the sum of the ages is (783+90)/9=97, then using the other sums, we have the ages to be

5,6,7,7,8,10,12,13,14,15[/hide]

[hide="I get multiple answers"]Let $a_1,a_2,\dots,a_{10}$ be the ages of the ten persons.
If $\mathbf{a}$ is a 10-fold vector,
\[ \left(\begin{array}{cccccc} 0&1&1&\cdots&1&1\\ 1&0&1&\cdots&1&1\\ 1&1&0&\cdots&1&1\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 1&1&1&\cdots&0&1\\ 1&1&1&\cdots&1&0 \end{array}\right)\mathbf{a}=\left( \begin{array}{c} 82\\ 83\\ 84\\ \vdots\\ 92\\ k \end{array}\right) \]
where $k$ is the missing sum from the nine-element set. Now add
all the equations together to obtain
\[ \begin{aligned} 9\sum_{i=1}^{10}a_i&=82+83+\dots+92+k=9\cdot87+k\\ \sum_{i=1}^{10}a_i&=87+\frac{k}9 \end{aligned} \]
Assuming the ages are whole numbers, let $k=9m$, for some positive
integer $m$. So we find that
\[ \mathbf{a}=(m+5,m+4,m+3,m+2,m,m-2,m-3,m-4,m-5,87-8m)^\mathrm{T} \]
Since they must be positive, $m-5>0$ and $87-8m>0$, thus obtaining
$10\ge m\ge6$. For each of these values of $m$ we obtain the
following:
\[ \begin{aligned} m=6&\Rightarrow\mathbf{a}=(11,10,9,8,6,4,3,2,1,39)\\ m=7&\Rightarrow\mathbf{a}=(12,11,10,9,7,5,4,3,2,31)\\ m=8&\Rightarrow\mathbf{a}=(13,12,11,10,8,6,5,4,3,23)\\ m=9&\Rightarrow\mathbf{a}=(14,13,12,11,9,7,6,5,4,15)\\ m=10&\Rightarrow\mathbf{a}=(15,14,13,12,10,8,7,6,5,7) \end{aligned} \][/hide]

Here is an easy solution to all RMO aspirants:          Let ages of 10 persons be x1,x2,x3,......,x10  S=x1+x2+x3+......+x10                                                    As x1,x2,x3,......,x10 are whole numbers, hence S is also a whole number.                                               Hence sums written by two persons are(S−x1),(S−x2),(S−x3),.....,(S−x10) which are also whole numbers.   By adding these, we get the sum(S−x1)+(S−x2)+(S−x3)+.....+(S−x10)=10S-(x1+x2+x3+......+x10)=   10S−S =9S                                                            Therefore the sum of all ten sums written by individual persons is a multiple of 9.Since the sums S−x1,S−x2,....S−x10 form a nine−element set, hence two of them must be identical, say S−x1 and S−x2 which show that x1 and x2 are identical.The total of all the nine sums given is 82 + 83 + 84 + ….. + 92 = 783, which it itself a multiple of 9. Therefore the tenth sum, which must be one of the nine given sums, must be a  multiple of 9 is 90, therefore the tenth sum must be 90. Consequently the sum of all  the ten sums = 783 + 90 = 873. The equation 9S = 873 gives S = 97. Subtracting the  numbers 82, 83, …., 89, 90, 91, 92 from 97, we get the individual ages to be
1. 97-82= 15 years
2. 97-83= 14 years
3. 97-84= 13 years
4. 97-85= 12 years
5. 97-87= 10 years
6. 97-89= 08 years
7. 97-90= 07 years
8. 97-90= 07 years ( repeated element )
9. 97-91= 06 years
10. 97-92 = 05 years.
All the best for your preparation.


algebra
function
domain
calculus
integration
Ring Theory
superior algebra

This is very similar to my previous post "Localization."

Let $ R$ be a domain with fraction field $ Q$.  If $ a$ and $ b$ are nonzero elements of $ R$ such that $ R\equal{}(a,b),$ then $ R[1/a]\cap R[1/b]\equal{}R$.

I solved this in a completely inelegant way by showing $ (a,b^n)\equal{}(a^m,b)\equal{}(a^m,b^n)\equal{}R$ for some nonnegative integers $ m,n$.

It seems to me there should be an elementary proof similar to the one in "Localization."

Any ideas?

I think you've chosen the right way ...

lemma1: if $ I$,$ J$ are coprime ideals, then $ I^n, J^m$ are coprime, too (where $ n,m \in \mathbb{N})$).
proof: find $ i \in I$ such that $ i \equiv 1$ mod $ I$. then $ i^n \equiv 1^n \equal{} 1$ mod $ I$, i.e. $ I^n \plus{} J \equal{} R$. repeating this, we get $ I^n \plus{} J^m \equal{} R$.

lemma2: if $ a,b$ are coprime elements of an integral domain, then $ R[1/a] \cap R[1/b] \equal{} R$.
proof: if $ x$ is in the intersection, we find $ n,m \in \mathbb{N}$ such that $ a^n x , b^m x \in R$. lemma1 yields $ u,v \in R$ such that $ u a^n \plus{} v b^m \equal{} 1$, thus $ x \equal{} u a^n x \plus{} v b^m x \in R$.

$ \rm HINT \quad (a^m,b^n) < 1 \implies  max \  M \supset (a^m,b^n) \implies M \supset (a,b) \equal{} 1 \ \Rightarrow\Leftarrow$


MATHCOUNTS AMC USA(J)MO USAMO AIME geometry probability

i got sick the night before States...
My headache distracted me...
I missed the easiest target question... a mistake u would classify with 20+40=50 (otherwise i would have made it)
^ these are just personal bad things that happened and im sure MANY MANY people only made  (or not made) nats because of luck
Im just curious about what people think

Edit
[b]Please read my 2nd post in the edit section.  It summarizes my point of view[/b]

you need decent skill, but luck is definitely a considerable facor.

Yep.

@Fanatic: Exactly! You prepared and you *still* got sick, so you got unlucky.

Uhh, but I rested and got over it.

Mental strength... with a bit of physical strength mixed in.

MATHCOUNTS does test student's mathematical ability in many ways, and it is very different from that of AIME. Apparently, I am not good in MATHCOUNTS style  :oops:

However, there are lucks involved; for example, discussion about "easy states" and "hard states" in state competitions... There are also differences between states: they have different dates! (I think it should be at least on the same day in all the U.S.) One's condition on the day is also a luck - I was nearly "out of my mind" last year :(

So I voted for "Half and half" option :)

[quote="Fanatic"]Are you kidding me? Making dumb errors counts as luck? Uh, dumb errors are part of the game. I make dumb errors all the time -- it's lack of skill, not lack of luck. It's the ability to accurately perform math under the pressure of a timed test.

Getting sick? Deal with it. Mentally strong people will get over it.  Part of the game is also preparing for the competition. i.e. getting enough sleep, staying healthy, washing your hands, etc. I take all of that into consideration. And I still got sick the week of  :blush: Ask anybody on the indiana team, I had an uber cough/runny nose the day of and had just gotten over my massive headache the day before [it lingered slightly, but admittedly not enough to distract me]. And hey, I got off with first place[/quote]

You're saying that Andrew Kaizer is better than Lyndon? Luck [size=200][i][b]is[/b][/i][/size] a factor. Some days you think better. or maybe the problems are just the problems you focused on, or the opposite. there are many ways in which luck is involved.

EDIT: lol that sounded a bit like my SAT essay.

[quote="machack"]Definately, Mathcounts is UNFAIR in the state level because of the fact that as mentioned above, there are "easy states" and "hard states" where non nationals maker in hard states would easily do MUCH MUCH better  than 1st place winners of easy states. Heck, the 30th place scorer in a hard state might do better than the Guam winner (no offense whatsoever to guam. just picking on guam because its small)

edit: just a comment for fanatic
does your 1st place in Indiana logic tell you that sleeping good is mathematical?  How about dumb errors? Is that truely a mathematical fault? Would u consider urself someone who makes a lot of mathematical mistakes if you had a bad sleep and a headache and someone was noisy next to you, so that you couldn't concentrate?  Luck is just a set of circumstances that affected you.  These little circumstances can affect the results greatly, and all im saying is mathcounts is vulnerable to lucky/unlucky incidents because its a set of "easy" questions that everyone can make mistakes on and its luck that determines who makes those mistakes.  Some people would argue that making mistakes is a part of your skills. Yes, it would be if you made a mistake about [i]how[/i] something is done but if u make an arithmetic mistake because of the time contraint and etc, i would consider it unlucky because everyone has the same chances at making such mistakes, and you are one of those who did.[/quote]

Ok, are you saying USAMO, the most prestigious mathematics competition in the USA is fair?  You get 6 problems.  You are a person that can usually just solve numbers 1 and 4, because 2,3,5, and 6 are written to be quite difficult.  However, you blank on #4 because you hadn't studied the subject much and your proof for #1 had a few flaws.

By your logic, this, of course, would be luck.  However, if you had practiced proofs more, would you have gotten points off on #1?  What if you had spent the effort to seek out your weaknesses in the subjects covered by #4, or even strengthened your mathematical abilities enough to get partial to full credit on another problem?

Luck is of course a factor, but if one really wants to be "Awesome" at math competitions, he/she must learn to be able to do well even if luck isn't on your side.  Applying this to MATHCOUNTS, you have to make the problems so trivial that you can fall asleep halfway through the sprint round and still have finished all of the problems, maybe making only a few careless errors.

I do believe it is slightly unfair for somebody in California or a hard state, but that just makes you better, doesn't it  :D ?

Luck is a very minor factor if any. Machak, you're coming across as very sour. Here's my sob story from last year (tops anything you say...)

I had 5th place last year. I had 2nd place with one set of targets to go. I missed both of the targets. On #8, I panicked and erased my answer just as time was called. I panicked and wrote my old answer + 1. I missed it. Had I gotten it, I would've landed in 4th place, making Nationals.

Hey I literally have a "worst second of my life". I can pinpoint that one second and say "that second screwed me over. That is my worst second ever'.

Despite this, I wouldn't say I got "unlucky". I panicked under pressure. I wasn't mentally strong enough to win. I didn't deserve it, period.

I don't think Mathcounts really cares about blah blah blah that can end up post top 12. If you sucked at state, TOO BAD, YOUR FAULT. You weren't going to do amazing at nationals anyways [chances are], so they don't really care that you didn't make it. Some states are easier, some states are harder. That doesn't make it "unfair". If you were truly good, then you wouldn't have to depend on other people sucking. Just yourself doing well.

Mathcounts is Mathcounts. If you're so great, then start preparing for the AIME, and DOMINATE. In Mathcounts, stupid mistakes are real mistakes. Bottom line.

The fact that you treat stupid mistakes like "stupid" mistakes rather than real mistakes just tells me you made a mistake in your preparation for state. You didn't have the right mindset coming in.

Sure there is a certain degree of luck. If you don't know how to do exactly thirty problems, and someone else knows how to do exactly thirty, you can get your butt whipped. However, it does a good job of sorting out the truly good and the not so good. There is never a "total fluke" that waltzes in there and makes top 12.

I had a pretty bad day at my state competition, probably due to the lack of sleep the night before, and finished with a dissapointing 30th place :( . I must admit though, that great skills can probably overcome bad luck. Even if I had a great day at the states I probably still won't be placed into the top 4, unless I was just so lucky that the person graded my test accidently mark everything right! :P

[quote="Fanatic"]Sure there is a certain degree of luck. If you don't know how to do exactly thirty problems, and someone else knows how to do exactly thirty, you can get your butt whipped. However, it does a good job of sorting out the truly good and the not so good. There is never a "total fluke" that waltzes in there and makes top 12.[/quote]
isn't that what your brother used to say? :P

Last year in chapters, I suddenly got very sick the day before the competition even though I tried to stay healthy.  I got 4th written.  Then I lost CD and was bumped to 5th.  Had this been states and my last year, I would have had a better story than yours, I think.  But this year's my second chance, at least.

Actually that was my 7th grade year. And no offense, but i think mine's a better sob story  :lol:.

Anyhow, yuqing, I'm surprised you remember that. Yeah my brother said that. But I understand it completely.

EDIT: I didn't even use it in the same context he used it in. When he used it, he said that you have to make sure you know how to do EVERY problem (extremely quickly), or else anyone can beat you. I just used it as an example of how major luck could totally come into play with a major exaggeration.

I don't consider mistakes as luck. the test is the same for everyone  and whoever makes less mistakes is more consistent. If you consider the target/sprint point difference luck then I have to disagree. The tests are the same for everyone and math is about applying logic creatively and consistently. If you can't do that then that isn't bad luck.

By the way, most of you guys could probably beat me, and if you did I'd shake your hand and say good job.

[quote="Fanatic"]Yes, he was totally new to contest math in 7th grade, and was forced to totally outwork everyone.

/off topic.

Anyhow, overall, I'd say there's 10% luck invovled, but skill overcomes luck anyday. But don't worry about it. As bpms said, life has a lot of luck involved.[/quote]

That is a fantastic way to end the decision.

My dad always tells me about how he grew up in a village, and was expected to become just another village kid - even though luck was against him, his skill brought him to where he is right now. IN the end, skill is what sets you apart, not luck.

Last year the only reason Preet Patel and Sam Lim made nats in FL is because two people who should've made it (one of whom made it last year) made a ton of careless mistakes. Those two didn't make as many. So, they got to go to nats instead. However, both of these people performed extremely well (both made top 35).
So when someone says
[quote]OMFG I SHOULD'VE MADE NATS I MADE 3 CARELESS MISTAKES ON THE SPRINT, AND 3 ON THE TARGET. IF I HAD GOTTEN THESE I WOULD'VE BEEN FIRST[/quote]
how do they know that the people at the top didn't make the same number of careless mistakes, they just knew how to do more problems?

"luck" is just an excuse for doing poorly. like if you do teribbly on a test you say luck was not with me. for me i usually get the score i deserve. even if i happen to guess a random answer on some problem, there is always some other careless mistake you have made *cough* *cough*, my 2006 chapter :( , i guessed a random # for # 8 and got it right but i made a careless mistake on some other target problem

[quote="machack"][quote="Ubemaya"]@Fanatic (or anyone who believes everything is nothing but skill): Can you [i]please[/i] explain to me why I was suddenly completely brain-dead on Chapters? I [i]really[/i] want to know, it's been bugging me for ages. I know it's a "state of mind", but what causes this sudden fluctuation in my "state of mind"?

Just so everyone knows, I personally think that skill far surpasses luck; the only reason I've been supporting luck is because people are giving it absolutely no credit and I was trying to point out some events that I believe is way more luck than skill.

In fact, pretty much everything can be tied down to skill. But most of it is too abstract to be seriously considered "skill." For example, is knowing how to time your sugar high really that skillful? Or am I just really incredibly stupid? The latter is probably true, but I personally don't believe that knowing what a sugar high is or knowing how to prepare your "state of mind" against situations like mine is really pure skill.

Oh yeah, and...

[quote="Fanatic"]Some tests you do well and some tests you don't... that's more of a solid argument.[/quote]

I'm expected to believe that that's skill?

[quote="Fanatic"]However, if you're truly good, then you can't get knocked out of the top four [although perfect may beg to differ].As my brothe said, if you're truly good, then you should be able to dominate even with stupid mistakes.[/quote]

I would know that, I placed 2nd in one of the toughest chapters (where stupid mistakes are all the most deadly) in the nation when I was practically sleepwalking :wink: .[/quote]

this is almost exactly what ive been trying to say, with maybe a few bad wordings (as in my bad wordings, not ubeyama's).  in my state, 2 ppl from last year's nationals didn't make nats and they were 1 and 2 points away from 4th place (i think 3rd and 4th was a tie also) Their skill landed them in the top 10 but maybe the small role from Luck factor booted them down 1/2 scores.  Is the 1 or 2 point difference REALLY a difference in skills? It sure is in the rankings.
Maybe i worded the whole topic badly.  Luck isn't exactly what i meant.  My ultimate reason for this topic was whether u think mc really tests your skills and you are rewarded correctly for your skills levels.  to me, mathcounts (when comparing top 6 ppl perhaps) is just like giving the test to those ppl with 9/10 chacne of getting a question right, whereas in a hard test, its like ppl having 1/2 chance of getting a question right, in which case the results will be more definate.  this is comparable to flipping a coin 5 times compared to flipping a coin 500 times.  Lets say u got 4 heads and a tail and conclude that u have 4/5 chance of getting a heads, but u try 495 times more and get 264 heads and 236 tailes. This is more closer to the calculated probability of 1/2. (do u see my point?)
Yes, i totally and absolutely agree that skills would dominate over luck, but when ppl of similar skills are compared, (maybe the top 6 in a (non existent) state where the scores were 45, 45, 44, 43, 42,42) isn't it more of the "luck factor" that determines the EXACT ranking? Prove be wrong if u can, but i believe the 1st place got first lace not because of skills.  How can it be? Are you saying the 2nd place, who got the same score, lost because s/he had less skills?
As for Countdown, it is still almost all skills except when comparing ppl with similar skills  but there is still the role that luck plays.

Fanatic is almost completely denying the luck factor.
Maybe you have a cocky attitude?[/quote]

In my opinion, outcome of MATHCOUNTS should not be considered as a perfect measurement of our knowledge/skills. This is just the beginning of a series of yet to come interesting and more challenging math contests. As rcv explained in one of the recent postings in reply to life1, this is how MATHCOUNTS is structured and we should all just respect the rules and enjoy our learning as much as we can. More fun is yet to come, i.e: AMC10/12, AIME, USAMO etc, where we will have opportunitity to demonstrate our talents over a longer period of time.

There may be a luck factor. Regardless we should just ignore it and focus on sharpening our skills. I think this is what Fanatic was trying to emphasize.

[quote="mthiyaga"][/quote][quote="machack"][quote="Ubemaya"]@Fanatic (or anyone who believes everything is nothing but skill): Can you [i]please[/i] explain to me why I was suddenly completely brain-dead on Chapters? I [i]really[/i] want to know, it's been bugging me for ages. I know it's a "state of mind", but what causes this sudden fluctuation in my "state of mind"?

Just so everyone knows, I personally think that skill far surpasses luck; the only reason I've been supporting luck is because people are giving it absolutely no credit and I was trying to point out some events that I believe is way more luck than skill.

In fact, pretty much everything can be tied down to skill. But most of it is too abstract to be seriously considered "skill." For example, is knowing how to time your sugar high really that skillful? Or am I just really incredibly stupid? The latter is probably true, but I personally don't believe that knowing what a sugar high is or knowing how to prepare your "state of mind" against situations like mine is really pure skill.

Oh yeah, and...

[quote="Fanatic"]Some tests you do well and some tests you don't... that's more of a solid argument.[/quote]

I'm expected to believe that that's skill?

[quote="Fanatic"]However, if you're truly good, then you can't get knocked out of the top four [although perfect may beg to differ].As my brothe said, if you're truly good, then you should be able to dominate even with stupid mistakes.[/quote]

I would know that, I placed 2nd in one of the toughest chapters (where stupid mistakes are all the most deadly) in the nation when I was practically sleepwalking :wink: .[/quote]

this is almost exactly what ive been trying to say, with maybe a few bad wordings (as in my bad wordings, not ubeyama's).  in my state, 2 ppl from last year's nationals didn't make nats and they were 1 and 2 points away from 4th place (i think 3rd and 4th was a tie also) Their skill landed them in the top 10 but maybe the small role from Luck factor booted them down 1/2 scores.  Is the 1 or 2 point difference REALLY a difference in skills? It sure is in the rankings.
Maybe i worded the whole topic badly.  Luck isn't exactly what i meant.  My ultimate reason for this topic was whether u think mc really tests your skills and you are rewarded correctly for your skills levels.  to me, mathcounts (when comparing top 6 ppl perhaps) is just like giving the test to those ppl with 9/10 chacne of getting a question right, whereas in a hard test, its like ppl having 1/2 chance of getting a question right, in which case the results will be more definate.  this is comparable to flipping a coin 5 times compared to flipping a coin 500 times.  Lets say u got 4 heads and a tail and conclude that u have 4/5 chance of getting a heads, but u try 495 times more and get 264 heads and 236 tailes. This is more closer to the calculated probability of 1/2. (do u see my point?)
Yes, i totally and absolutely agree that skills would dominate over luck, but when ppl of similar skills are compared, (maybe the top 6 in a (non existent) state where the scores were 45, 45, 44, 43, 42,42) isn't it more of the "luck factor" that determines the EXACT ranking? Prove be wrong if u can, but i believe the 1st place got first lace not because of skills.  How can it be? Are you saying the 2nd place, who got the same score, lost because s/he had less skills?
As for Countdown, it is still almost all skills except when comparing ppl with similar skills  but there is still the role that luck plays.

Fanatic is almost completely denying the luck factor.
Maybe you have a cocky attitude?[/quote][quote="mthiyaga"]

In my opinion, outcome of MATHCOUNTS should not be considered as a perfect measurement of our knowledge/skills. This is just the beginning of a series of yet to come interesting and more challenging math contests. As rcv explained in one of the recent postings in reply to life1, this is how MATHCOUNTS is structured and we should all just respect the rules and enjoy our learning as much as we can. More fun is yet to come, i.e: AMC10/12, AIME, USAMO etc, where we will have opportunitity to demonstrate our talents over a longer period of time.

There may be a luck factor. Regardless we should just ignore it and focus on sharpening our skills. I think this is what Fanatic was trying to emphasize.[/quote]

You're Absolutely right but i don't think that's what Fanatic was emphasizing on. i thought he was saying something similar to "Hey You losers i got 1st in Idiana!! and i did that ONly because of SKills!! No luck involved!!! YAY! Im sure that it was ALL skills because "SKILLS > LUCK ANYDAY")

I still retain my opinion of Fanatic being cocky (possibly from the ALL-SKILL win at the states)

[quote="machack]
Fanatic is almost completely denying the luck factor.
Maybe you have a cocky attitude?[/quote]

Ok, I tried to stay out of this, but I have to comment.

The way I see it people who didn't do well call it luck, those who do call it skill.  Let me ask this, if Fanatic's coach just lucky because his teams have won like 22 of 25 years?  Have I been JUST lucky?  I would say of course there is some luck.  But I say Fanatic is taking the best stance.

He is not going lay blame on "bad luck" if he does poorly or be falsley modest and say he was just lucky if he does well.

If you go into a competition you believe is all luck, then you have no argument if you lose based on luck.  If you believe it is skill and you lose then you lost to the better man/woman.

I personally work as hard as I can to not leave things up to luck and generally I find they work out.  I understand it is hard to accept it when you don't reach your goal, and I am sure Fanatic knows that as well as anyone after his experience(and countless posts on this site :lol: ) last year after placing 5th!

It is time to move on guys.  You have nothing to be ashamed of if you finished 8th by just a couple points.  I would say you have something to be ashamed of though if you keep placing blame outside your control.  There shouldn't be blame at all. You did well.  Don't lessen the experience by constantly criticizing the contest.

Is MATHCOUNTS the absolute best test of mathematical ability?  I don't know, but I know if you are taking Calculus and think you automatically should make it because you are "smarter".  Since MATHCOUNTS is designed to test middle school curriculum your higher level classes are probably hurting you in MATHCOUNTS since you are working on other things.  Especially if you haven't absolutely mastered the concepts of middle school.  Maybe you shouldn't have done mathcounts but done just the AMC10/12 and AIME to prove your higher level knowledge.

In my mind MATHCOUNTS is the best competitive program because it reaches the largest number of people and is accessible to the widest range of abilities.  It is not trying to figure out who is "smartest" it is trying to get more kids doing problem solving.  MATHCOUNTS probably got you started in competitions and started you on your way to where you are now.  By insulting MATHCOUNTS you are "biting the hand that feeds you".  Why not just say thanks for the experience and move on!

Sorry that was so long, but this argument has gotten pretty old pretty fast.

Frost

PS Fanatic, I look forward to seeing you at Nationals and testing your skills against ours in friendly competition :lol:  May those with the most skill win and those relying on luck have fun trading pins :D   I'll be rooming with your coach again, so I'll see you for sure.

There will be no luck to you machack. You post forum and ask people to trade the test with you. You ask me to send the test first which I did. and you won't send your test to me. You won't get any luck by doing that!

I guess ur right.
LoL  :blush:  :maybe:  :(  :o  :mad:
Im just sore from not making the nats though i had it in my grasp !! now that i think about it, i don't mind this that much because in fact mathcounts is a mere middle school contest and is not comparable to other higher math competitions.  BUT Im just sore about all the scholarship given to the top 4 and the fun trip to texas! which was lost to me by literally doing 3*4=16

Ok Fanatic, Peace?

(to big m- i told you that my parents couldn't go to a place to get it scanned, which they promised they would.. i apologized several times through pm... and that's the reason why after i was told i can't get my tests scanned, i turned down all other trading offers.)

Can a mod lock this please?

Are you kidding me machack?

Did you not read my story about how I got 5th place last year, and had that happened to you, you would surely be complaining about bad luck. I've only scanned the posts in this thread, as I have little time, but you have GOT to be kidding me if you think I'm saying "look at me I won".

I really don't care. Go whine.

Ubemaya, you've got to be kidding me. 44 on chapter. Wow that's what I got.

You going braindead isn't bad luck. It's you not being ready.

I'll stop posting in this thread now. You people have got to be kidding me. Go whine, please.

And I when I made my closing statement

[quote] Anyhow, overall, I'd say there's 10% luck invovled, but skill overcomes luck anyday. But don't worry about it. As bpms said, life has a lot of luck involved.[/quote]

I did say there was a decent amount of luck.

[quote]You people have got to be kidding me. Go whine, please.
[/quote]
Go, whine? You're seriously saying "Go whine"?
That is downright arrogance.
Why don't you slap yourself out of your arrogance and see how you would be in the real world with it?

There definately is at least a bit of LUCK. Like if you don't know how to do a problem and you just guessed a random answer and it turned out to be correct, that's luck. And if you saw a really hard problem that you have already done before, that is luck. There is almost no such thing as a 100% skill competition/game. But MATHCOUNTS is a very skillful competition which doesn't actually have much luck, but it certainly has SOME luck.

Locked for flaming and the fact that I'm lazy and don't feel good enough to delete the flaming posts.


trigonometry
real analysis
real analysis unsolved

this may be hard...  :D

evaluate
\[ \prod _{k \equal{} 1} ^\infty \; \left\{ 1 \; \minus{} \; \dfrac {1}{k^2 \; \plus{} \; \dfrac {1}{k^2}} \right\}
\]
PS: answer involves $ \cos$ and $ \cosh$

$ 1\minus{}\frac{1}{k^2\plus{}\frac{1}{k^2}}\equal{}1\minus{}\frac{k^2}{k^4\plus{}1}$ $ \equal{}\frac{k^4\minus{}k^2\plus{}1}{k^4\plus{}1}\equal{}\frac{k^6\plus{}1}{(k^2\plus{}1)(k^4\plus{}1)}\equal{}\frac{ 1\plus{}\frac{1}{k^6}}{\left( 1\plus{}\frac{1}{k^2}\right) \left( 1\plus{}\frac{1}{k^4}\right)}$

For expressions of the form $ \prod_{k\equal{}1}^\infty \left(1\plus{}\frac{1}{k^{2p}}\right)$ you can use [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=166267]this[/url] formula.

Eventually your product can be written as $ \frac12 \, \frac{\cosh(\pi)\minus{}\cos\left(\sqrt{3}\,\pi\right)}{\cosh^2 \left(\frac{\pi}{\sqrt{2}}\right)\minus{}\cos^2\left(\frac{\pi}{\sqrt{2}}\right)}$ .

[quote="Dr. Doe"]$ 1 \minus{} \frac {1}{k^2 \plus{} \frac {1}{k^2}} \equal{} 1 \minus{} \frac {k^2}{k^4 \plus{} 1}$ $ \equal{} \frac {k^4 \minus{} k^2 \plus{} 1}{k^4 \plus{} 1} \equal{} \frac {k^6 \plus{} 1}{(k^2 \plus{} 1)(k^4 \plus{} 1)} \equal{} \frac { 1 \plus{} \frac {1}{k^6}}{\left( 1 \plus{} \frac {1}{k^2}\right) \left( 1 \plus{} \frac {1}{k^4}\right)}$

For expressions of the form $ \prod_{k \equal{} 1}^\infty \left(1 \plus{} \frac {1}{k^{2p}}\right)$ you can use [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=166267]this[/url] formula.

Eventually your product can be written as $ \frac12 \, \frac {\cosh(\pi) \minus{} \cos\left(\sqrt {3}\,\pi\right)}{\cosh^2 \left(\frac {\pi}{\sqrt {2}}\right) \minus{} \cos^2\left(\frac {\pi}{\sqrt {2}}\right)}$ .[/quote]

OK, cool....  :D  :clap2: however, your numerator is correct, denominator is not.....  :maybe:

the answer is
\[ \boxed {\dfrac {\cosh(\pi) \minus{} \cos\left(\sqrt {3}\,\pi\right)}{\cosh \left(\sqrt {2}\, \pi\right) \minus{} \cos \left(\sqrt {2}\,\pi \right)} }
\]