[quote="mr. math"]A cube with a side length of 5 inches is painted and divided into unit cubes. What is the ratio, expressed as a fraction in lowest terms, between the number of cubes with 2 faces painted to the number of cubes with no faces painted?

Because the question does not say exactly 2 faces painted, should I also include the cubes with 3 faces painted?

I solved the problem including the cubes with 3 faces painted and got the following answer. I don't have the answer for the problem, so it would be great if someone could check it for me

[hide]

(5-2) ^3 = 27 with no faces painted.
 (5-2) * 12 = 36 for two faces painted.
8 = three faces painted.

(36+8)/27 = [b]44/27[/b][/hide][/quote]

I think it's just talking about 2 faces.

Total number of faces: $ 5^3 \equal{} 125$
No faces have color: $ 3^3 \equal{} 27$
Two faces colored: $ 12\times3 \equal{} 36$

Thus, the answer is $ \frac {36}{27} \equal{} \boxed{\frac {4}{3}}$

Wait, but isn't it:

Total number of cubes: $ 5^3 \equal{} 125$

Total number of cubes with no faces colored: $ 3^3 \equal{} 27$

Total number of cubes with only one face colored: $ 3^2\cdot6 \equal{} 54$

Total number of cubes with three faces colored: $ 6$

Total number of cubes with two faces colored: $ 125 \minus{} 27 \minus{} 54 \minus{} 6 \equal{} 38$

Therefore, the answer would be: $ \frac{38}{27}$

[quote="Fibonacci11235"]Wait, but isn't it:

Total number of cubes: $ 5^3 \equal{} 125$

Total number of cubes with no faces colored: $ 3^3 \equal{} 27$

Total number of cubes with only one face colored: $ 3^2\cdot6 \equal{} 54$

[b]Total number of cubes with three faces colored: $ 6$ [/b]

Total number of cubes with two faces colored: $ 125 \minus{} 27 \minus{} 54 \minus{} 6 \equal{} 38$

Therefore, the answer would be: $ \frac {38}{27}$[/quote]

It's 8 not 6

OHH MY BAD. For some reason I thought there were 6 vertices.


logarithms
calculus
calculus computations

How can I solve:

$ \frac{dy}{dx} \equal{} xe^{\minus{}y} \plus{} \frac{2}{x}$

Thanks.

Let $ v(x) \equal{} \frac {e^{y(x)}}{x^2}$, then
\[ v' \equal{} \frac {v}{x}(xy' \minus{} 2) \equal{} \frac {v}{x}\left(\left(x^2e^{ \minus{} y} \plus{} 2\right) \minus{} 2\right) \equal{} \frac {1}{x}
\]
Integrating both sides gives $ v(x) \equal{} c \plus{} \log \vert x\vert$, and hence
\[ y(x) \equal{} \log\left(x^2 \vert v \vert \right) \equal{} \log\left(x^2\left(c \plus{} \log \vert x \vert \right)\right)
\]
for $ \vert x \vert > e^{\minus{}c}$.


Hi all,

in case someone has read Nelson's REPT, I was wondering, what does Nelson mean by "finite"? Does he take it to mean limited/standard? It's horribly ambiguous. Is any element of N - including the nonstandard ones - finite?

Thanks

Philippe

I have read it but it's been a while. Can you give context?

By the way, the book appears to be available online on Nelson's homepage.  You can google for it.


integration
calculus
calculus computations

$ \sum_{k\equal{}1} ^{\infty} \frac{(\minus{}1)^k}{k}\equal{}?$

$ \sum_{k\equal{}1}^{\infty }\frac{(\minus{}1)^kx^k}{k}\equal{}\minus{}\int \frac{dx}{1\plus{}x} \equal{}\minus{}ln(1\plus{}x).$

So $ exp(\sum_{k\equal{}1}^{\infty} \frac{(\minus{}1)^{k\minus{}1}}{k})\equal{}exp(\log2)$ ?

Yes (why did you put that $ exp$ there¿). By the way, surely not number theory, and even more not open...


We have to sort $ x_1, \ldots, x_n$, but not the usual way.

We do not know the values of the $ x_i$'s but instead we know the position $ m_i$ that $ x_i$ would occupy in the sorted array $ x_1, \ldots, x_i$ (or in insertion sort terms, $ m_i$ is the position at which you insert $ x_i$ after you've sorted $ x_1, \ldots, x_{i \minus{} 1}$).

What is best complexity of an algorithm that computes the sorted sequence?

I know how a $ O \left( n \log n \right)$ algorithm for this one looks like (it's very strange)... However, I cannot help it wonder if something better is possible. I have no idea, the hypothesis here seem so strong (but still so weak), but I cannot connect the dots.

Just out of curiosity, does the n logn algorithm sort by m_i's, and then traverse again to sort by x_i?


modular arithmetic
number theory
least common multiple
AMC
AIME

[quote="xxxyyyy"]Find at least one number ( there are many ) which when divided by $9$ leaves a remainder of $8$; when divided by $8$ , leaves $7$ ; by $6$ leaves $5$ ; and so on till $1$ .

Extended version:
if no repetitions are allowed ; is there a number possible ?
If not why not ? No restriction on its size though .[/quote]
What means "and so on"? You wrote that some integer $x$: $x\equiv-1 \left( \bmod 9,8,6\right)$. $9,8,6,...$ isn't a well known sequence, so maybe you meant $9,8,7,6,5,4,3,2$ and maybe $8,6,4,2$. But in both cases the solution for problem:
[color=blue]$x$ gives remainder $n_{i}-1$ modulo $n_{i}$ for $i=1,...,k$, then find all posiible $x$[/color]
is easy: $x\equiv n_{i}-1\pmod{n_{i}}\iff x\equiv-1\pmod{n_{i}}\iff$ $x+1\equiv 0\pmod{n_{i}}\iff n_{i}\mid x+1$, so just $x+1=l\cdot\mathrm{lcm}\left(n_{1},...,n_{k}\right) \iff x=l\cdot\mathrm{lcm}\left(n_{1},...,n_{k}\right)-1$, where $l$ is an arbitrary integer.

My solution gives that $x=l\cdot\mathrm{lcm}\left(2,3,4,5,6,7,8,9\right)-1= l\cdot 2520-1$, so the least positive valu of $x$ is $2519$, the next are $5039,7559,10079,12599,...$, so you were right. But I don't like using computer in such easy problems.

[quote="xxxyyyy"]
Find $n$ such that:
$n \equiv 8 (mod 9)$
$n \equiv 7 (mod 8)$
$n \equiv 6 (mod 7)$
$n \equiv 5 (mod 6)$
$n \equiv 4 (mod 5)$
$n \equiv 3 (mod 4)$
$n \equiv 2 (mod 3)$
$n \equiv 1 (mod 2)$[/quote]

If it's not necessary to find the smallest such $n$ then the answer is very simple.

$n = 9!-1$

:)

In fact, this result is deeper than it seems, as it suggests a method for proving that there exist consecutive primes arbitrarily far apart.

[quote="xxxyyyy"][quote]In fact, this result is deeper than it seems, as it suggests a method for proving that there exist consecutive primes arbitrarily far apart.[/quote]

this is interesting.... didnt thought about this.[/quote]

O yes, you can prove that there are infinite number of consecutive primes and that there are primes n number away from each other.

[quote="computer_nuke"]O yes, you can prove that there are infinite number of consecutive primes and that there are primes n number away from each other.[/quote]

You might want to reconsider this statement.  You can prove that there exist [i]arbitrarily long[/i]  (not infinite) strings of [i]composite[/i] numbers.  I'm not sure how you would use this to find primes of gap exactly $n$, though (especially for odd $n$  :wink: ).

Also, jh, that argument has been given twice already (both times by Rzeszut).

[quote="jhcreinhardt"]um...negative mods anyone?
[hide]
simple:  -1mod(integers 1-10)

first number 2519=2520-1

2520 LCM of 1-10[/hide][/quote]

I'd appreciate it if, in the future, you read over the other posts in a topic to avoid repeating a solution that has already been given.   :wink:

i had almost the same problem as my #2 on my first mock aime  :)


$ x^{2} \plus{}7 \equal{} 2^{n}$

If x and n are naturals, how many solutions exist for this equation

Obviously, $ x \equal{} 2k \plus{} 1$
$ k^2 \plus{} k \equal{} 2^{n \minus{} 2}$
$ k^2 \plus{} k \equal{} a$, $ a$ is always even.
$ a \equal{} 2^{n \minus{} 2}$
$ k^2 \plus{} k \minus{} a \equal{} 0$
$ k \equal{} \frac { \minus{} k \plus{} \sqrt {k^2 \plus{} 4k^2a}}{2}$
$ k \equal{} \frac {k( \minus{} 1 \plus{} \sqrt {1 \plus{} 4a})}2$
$ 2 \equal{} \minus{} 1 \plus{} \sqrt {1 \plus{} 4a}$
$ a \equal{} 2$
$ n \equal{} 3$
$ x \equal{} 1$
One solution? $ (x,n) \equal{} (1,3)$

Just by looking at the problem you are missing the solution x=3, n=4

[quote="Bugi"]Obviously, $ x \equal{} 2k \plus{} 1$
$ k^2 \plus{} k \equal{} 2^{n \minus{} 2}$[/quote]  Try again ;)  (Also, we can solve $ k^2 \plus{} k \equal{} 2^m$ much more quickly: $ k^2 \plus{} k \equal{} k(k \plus{} 1)$ is the product of an odd and an even number, so the odd number must be 1.)

JBL is right.
It's $ k^2\plus{}k\plus{}2\equal{}2^{n\minus{}2}$. Now no idea

Bugi missed a -2 in his solutions.

Just taking a random stab at it:

As Bugi said, obviously x is odd, so x = 2k + 1 where k is a nonnegative integer.

$ (2k \plus{} 1)^2 \plus{} 7 \equal{} 2^n$
$ 4k^2 \plus{} 4k \plus{} 8 \equal{} 2^n$
$ k^2 \plus{} k \equal{} 2^{n \minus{} 2} \minus{} 2$
$ k(k \plus{} 1) \equal{} 2(2^{n \minus{} 3} \minus{} 1)$

Both sides are in terms of the product of an even and odd number now... If my steps are right the answer should follow.

EDIT>
k = 0, x = 1, n = 3
k = 1, x = 3, n = 4
k = 2, x = 5, n = 5

Checking them in my head they seem to be right.

But even number can be product of even number and odd number...

You're missing $ (11, 7)$ and $ (181, 15)$.  This is not an easy problem and has appeared on the forums multiple times; the standard solution is due to Nagell and appears on page 50 [url=http://www.math.sc.edu/~filaseta/gradcourses/TheMath784Notes.pdf]here[/url].


[quote="solafidefarms"]$\sqrt{x+\sqrt{x+\sqrt{x+...}}}=64$. Find x.[/quote]

[hide="Hint"]
So, $\sqrt{x+\sqrt{x+\sqrt{x+...}}}=64$.

Is it OK to make it equal to $\sqrt{x+64}=64$ ? Because since it's infinity here, one less doesn't matter. So, is it OK?

[hide]Yes, it is OK[/hide]

So with that equation, try and solve it NOW.

[/hide]

[quote="Drunken_Math"]
[hide="huh"]How did you get
 $\sqrt{x+64}=64$
from
 $\sqrt{x+\sqrt{x+\sqrt{x+...}}}=64$.

[/hide][/quote]

EDIT: OH I see it now! :)

[quote="nonie"][quote="Drunken_Math"]
[hide="huh"]How did you get
 $\sqrt{x+64}=64$
from
 $\sqrt{x+\sqrt{x+\sqrt{x+...}}}=64$.

[/hide][/quote][/quote]

$64=\sqrt{x+\sqrt{x+\sqrt{x+...}}}$

Ignore the first $\sqrt{x+}$

That leaves you with $\sqrt{x+\sqrt{x+\sqrt{x+...}}}$, the same thing we started out with. And I simply substituted 64 for that value.

[hide="answer"]Using what Drunken Math said, $ \sqrt{x + 64} = 64$.
Squaring both sides, we get $ \sqrt{x + 64}^2 = 64^2$ or $x + 64 = 4096$ Subtract $64$ from each side and get $x = \boxed{4032}$
To solve the generic equation $ \sqrt{x + \sqrt{x + \sqrt{x + ...}}}=N$ The answer in $N^2-N$ as shown above.[/hide]


trigonometry
logarithms
real analysis
real analysis unsolved

[color=darkblue][b]Find: $ lim\displaystyle \left ( \frac {cos^{2}1}{1.2} \plus{} \frac {cos^{2}2}{2.3} \plus{} ... \plus{} \frac {cos^{2}n}{n(n \plus{} 1)}\right)$[/b][/color]

[quote="Doe John"]$ \sum_{k \equal{} 1}^\infty \frac {\cos^2(k)}{k\, (k \plus{} 1)} \equal{} 1 \minus{} \sin^2(1)\cdot\ln\Big(2\sin (1)\Big) \minus{} \frac {\sin(2)}{2} \cdot\left(\frac {\pi}{2} \minus{} 1\right)$[/quote]

:huh: :huh:
How? At least give me a hint  :P

Use the partial fraction decomposition $ \frac {1}{k\, (k \plus{} 1)} \equal{} \frac {1}{k} \minus{} \frac {1}{k \plus{} 1}$

and the Fourier series $ \sum_{k \equal{} 1}^\infty \frac {\cos kx}{k} \equal{} \minus{} \ln\left(2\sin\frac {x}{2}\right) \quad , \quad \sum_{k \equal{} 1}^\infty \frac {\sin kx}{k} \equal{} \frac {\pi \minus{} x}{2}$ .

$ S: \equal{} \sum_{k \equal{} 1}^\infty \frac {\cos 2k}{k\, (k \plus{} 1)} \equal{} \sum_{k \equal{} 1}^\infty \frac {\cos 2k}{k} \minus{} \sum_{k \equal{} 1}^\infty \frac {\cos 2k}{k \plus{} 1}$

$ \equal{} \minus{} \ln\left(2\sin\frac22\right) \minus{} \sum_{k \equal{} 1}^\infty \frac {\cos(2k \minus{} 2)}{k} \plus{} 1$ ,whereat

$ \cos(2k \minus{} 2) \equal{} \cos(2k)\, \cos(2) \plus{} \sin(2k)\,\sin(2)$

Hence $ S \equal{} \minus{} \ln(2\sin 1) \plus{} \cos(2)\, \ln(2\sin 1) \minus{} \sin(2)\, \frac {\pi \minus{} 2}{2} \plus{} 1$

$ \equal{} \minus{} \underbrace{(1 \minus{} \cos 2)}_{2\, \sin^2(1)}\, \ln(2\sin 1) \minus{} \sin(2)\left(\frac {\pi}{2} \minus{} 1\right) \plus{} 1$

Because of $ \cos^2(k) \equal{} \frac12 \left(\cos(2k) \plus{} 1\right)$ and $ \sum_{k \equal{} 1}^\infty \frac {1}{k\, (k \plus{} 1)} \equal{} 1$

$ \sum_{k \equal{} 1}^\infty \frac {\cos^2(k)}{k\, (k \plus{} 1)}$ must be $ 1 \minus{} \sin^2(1)\, \ln\left(2\sin(1)\right) \minus{} \frac {\sin(2)}{2}\, \left(\frac {\pi}{2} \minus{} 1\right)$ .


A sphere with charge $ Q_1$ lies inside a spherical shell with charge $ Q_2$. In what follows, suppose that charges are uniformly distributed in the volume of both sphere and spherical shell.

(a) Find the electric field inside the sphere of radius a.

(b) Find the electric field for r > a.

(c) Find the electric potential in any point of the region a < r < b.

(d) Find the distribution of charges in the sphere of radius a.

(e) Solve the above supposing you were only told that the charges were uniformly distributed in the volume of the spherical shell.

The problem settles charge uniformly distributed in volume; this implies dielectric medium.
Then it asks for the distribution of charges in the sphere???

Charges are uniformly distributed in sphere of radius a, and charges are uniformly distributed in the shell (region b < r < c), not in all space inside the shell. It is understood that for a < r < b there is vacuum.

Still, item d) asks for the charge distribution in the sphere...

For $ r<a$, $ E \equal{} \frac{1}{4\pi\varepsilon_0}\frac{qr}{a^3}$.

[b]Immanuel[/b], I still don't understand where is your problem.

Immanuel's problem is that:

"In what follows, suppose that charges are uniformly distributed in the volume of both sphere [...] Find the distribution of charges in the sphere of radius a"

So the problem statement says that in what follows the charges are uniformly distributed in the spheres, and still you ask in d, to find the distribution of the charges...

I see. By "distribution of charges", it is meant "the expression for the surface or volumetric charge density". Anyway, this is just a problem taken from a Physics exam for Biochemistry students.

Yes, but that is already given "the expression for the surface or volumetric charge density", it is given that the volumetric charge density is uniform, so what is there to calculate?


calculus integration logarithms trigonometry inequalities derivative function

Prove the following inequality.

\[\frac{1}{12}(\pi -6+2\sqrt{3})\leq \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \ln (1+\cos 2x) dx\leq \frac{1}{4}(2-\sqrt{3})\]

[quote="Kent Merryfield"]Hint: for $u>-1,\,u-\frac{u^2}2\le\ln(1+u)\le u.$

(How would you prove this hint?)[/quote]

We can prove inequalities using derivatives and when we have inequalities take $t=cos2u$, and then integrals.

I shouldn't have said $u>-1.$ The logarithm inequality above is only true for $u\ge0.$ If you integrate kunny's inequality, you get that, and you get the opposite inequality for $-1<u<0.$

Fortunately for us in this problem,

$x\in\left[\frac{\pi}6,\frac{\pi}4\right]\implies 2x\in\left[\frac{\pi}3,\frac{\pi}2\right]\implies \cos(2x)\ge0.$

[quote="Kent Merryfield"]I shouldn't have said $u>-1.$ The logarithm inequality above is only true for $u\ge0.$ If you integrate kunny's inequality, you get that, and you get the opposite inequality for $-1<u<0.$[/quote]

for $-1<u<0$ we have $ln(1+u)<u-\frac{u^2}{2}<u$ ,but as you said in this problem it's not important (our lecturer called this way of proving inequalities (using derivatives) 'lemma about two runners' :) (you just check that for functions f,g f=g in one point and then you look at the behavior of derivatives, very simple technique and very effective)

Integrating u - u^2/2 or cos(2x) - ((cos 2x)^2)/2  = cos(2x) - (1+ cos 4x)/4 from pi/6 to pi/4,
I get 1/2 -pi/48 -7sqrt(3)/32 and not (pi - 6 +2sqrt(3))/12. Am I doing something silly?

i am not sure about this but cant we use   CAUCHY-BUNYAKOVSKY   inequality here? :?:

Perhaps. But my problem is that  the posts above seem to indicate that we can substitute cos 2x for u in u - (u^2)/2 <= ln(1+u) <= u and integrate. It works for the right-hand-inequality. Does it work for the left-hand-inequality?

I've just realised that as I've shown that the integral is greater than 0.5566..., then I've shown that it is greater than 0.05047...
= (pi - 6 +2sqrt3)/12, but still haven't managed to get the latter number. It seems like it should be straightforward but I just can't see it. It's so annoying. Any ideas?

I got the same thing as Andrew. Is it ok to just crow bar it in because, in any case, 0.0566 > 0.05047?


geometry
3D geometry
ARML
LaTeX
rotation
factorial
percent

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Hmm... That's always a rule... but w/e

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Rule 12: It is the phrase [b][i][color=blue]AAST PWNS[/color][/i][/b] that must be colored, italicized, and bolded, and if one should fail to do so, it will be in direct violation of this rule.

[hide="Score"]BOGTRO: 21
bluecarneal: 3
james4l: 1
pythag011: 5
isabella2296: 29
Auroraak: 8
FantasyLover: 11[/hide]

Rule 1: All rules require such that if a|b, Rule b follows Rule a.
Rule 2: All rules must have fail written in them.
Rule 3: All rules must not fail the previous rules
Rule 4: No rules shall fail to include a color in them.
Rule 5: All rules whose numbers are not squares shall fail to start with "No rules."
Rule 6: Every rule whose number is perfect shall not fail to include its factors. For example, 1,2,3,6.
Rule 7: Following this rule, each prime rule shall fail to have the phrase rule in it, excluding the initial "rule n:".
Rule 8: Every rule that contains a perfect cube shall not fail to have the root of that cube at the end. 2
Rule 9: No rules shall fail to have a [color=red][b]bold[/b][/color] word in it.
Rule 10: The [color=blue]bolded[/color] words shall not fail to also be [b][i]italicized[/i][/b].
Rule 11: All [i][b][color=blue]comandments[/color][/b][/i] must not fail to include the phrase 'AAST PWNS' in them. Also, the phrase 'AAST PWNS' shall not be neither the very first phrase nor the very last phrase.
Rule 12: It is the phrase [b][i][color=blue]AAST PWNS[/color][/i][/b] that must be colored, italicized, and bolded, and if one should fail to do so, it will be in direct violation of this rule.
Rule 13: Of course, the phrase, [b][i][color=red]AAST PWNS[/color][/i][/b], is a fail phrase as they failed ARML. Now on fail must be used at least x-10 times where x is your directive's number.

[hide="Score"]BOGTRO: 21
bluecarneal: 3
james4l: 14
pythag011: 5
isabella2296: 29
Auroraak: 8
FantasyLover: 11[/hide]

Sorry; off topic, but why does everyone hate AAST?  :(

1. No word may contain two consecutive instances of the same glyph.
2. Unistring.
3. Mispel.
4. Eighchar.
5. [i]Italicks.[/i]

I was going to say "Coloured", to see how many people would let it slide, but nah.

1. No word may contain two consecutive instances of the same glyph.
2. Unistring.
3. Mispel.
4. Eighchar.
5. [i]Italicks.[/i]
6. [i]Espaniol.[/i]
7. [i][color=red]Colorádo[/color].[/i]

The misspelling is the misplaced accent. I was tempted to just put "colorado", but never mind.

1. No word may contain two consecutive instances of the same glyph.
2. Unistring.
3. Mispel.
4. Eighchar.
5. [i]Italicks.[/i]
6. [i]Espaniol.[/i]
7. [i][color=red]Colorádo[/color].[/i]
8. [i][color=red]Unonúmer.[/color][/i]

new thread. it died.

1. Each word in all rules must start with different letter.

1. Each word in all rules must start with different letter.
2. No word may finish using a like symbol either.

i.e. Each word in a rule must end with a different letter as well.

[quote="sdkudrgn88"]1. Each word in all rules must start with different letter.
2. No word may finish using a like symbol either.
3. Initials create English.

First letter of each word forms another word. Here, it is I-C-E.

1. Each word in all rules must start with different letter.
2. No word may finish using a like symbol either.
3. Initials create English.
4. Three words only.
5. [b]Bold, or die.[/b]

Yes, bod is a word.

1. Each word in all rules must start with different letter.
2. No word may finish using a like symbol either.
3. Initials create English.
4. Three words only.
5. [b]Bold, or die.[/b]
6. [b][color=red]Colors are necessary.[/color][/b]
7. [b]$ \text{\LaTeX}$ [color=blue]ordinance passed.[/color][/b]

BTW: I hate color tags. (Chrome handler)

1. Each word in all rules must start with different letter.
2. No word may finish using a like symbol either.
3. Initials create English.
4. Three words only.
5. [b]Bold, or die.[/b]
6. [b][color=red]Colors are necessary.[/color][/b]
7. [b]$ \text{\LaTeX}$ [color=blue]ordinance passed.[/color][/b]
8. [b][color=green]Nix all $ \textbf{primes}$.[/color][/b]

(As in, when numbering the rules, nix all primes.)
(Bold the LaTeX too.)

Screw this...

1. Each rule: spellable with a standard Scrabble set.

(To clarify: blanks can be any letter.)

1. Each rule: spellable with a standard Scrabble set.
2. Craziest rule - one word pts > 18.
3. Obnoxiously difficult rule - two words' points > 17.

(Numerals don't count towards the quota.)


Solve the following equation :
( 3- x  2 ) / (1 - x 2 ) + (3-x 2)(1-3x 2) = (4-4x 2 ) ( x 4  - 6x 2 +1)

--------
Sorry about my mistake! :D

I don't know what 's the role of this equation:Where is the "="???  :(  :D

[quote="A1lqdSchool"]Solve the following equation :
( 3- x  2 ) / (1 - x 2 ) + (3-x 2)(1-3x 2) + (4-4x 2 ) ( x 4  - 6x 2 +1)[/quote]
 :lol: I ate your equality sign A1lqdSchool  :D


Do you know which country proposed this problem?
  I have proposed something like this one for the National Olympiad 2005.


Prove that the number $512^{3} +675^{3}+ 720^{3}$ is composite.

[quote=Python54]Where did the motivation for the factorization of $x^3+y^3+z^3 = x^3+y^3+(-z)^3 -3xy(-z)$ to $(x+y-z)(x^2+zx-xy+zy+z^2+y^2)$ come from?[/quote]

It is well known identity


What age would you rather be?  :huh:

i would like..5 years

i would want to be 7, collect current newspapers which have winning lottery numbers on it, and ...WIN!

Hmmm.........I would probably want to be anywhere near 2 or 3...... No one around can scold you for anything then...

Sixteen, but than I would mean, I want to be in the same situation again as I was back then

Mmmm hmmm, sweet 16 once again.  That'd be fun.  Though it had its down points...it was still fun.

Yeah, of course now I am soo old, but reconsidering : studies and responsibilities were a lot easier for instance.  It did have downsides.  Being a minor was a bit restricting

I would want to be my age (13) forever and win mathcounts when I should be 25 :lol: and then I could play in junior league forever and everyone would say "oh no, that big kid with a beard is pitching" :D so fun

I would like to be 11 because thats sixth grade... i can start all over for mathcounts.

but I would rather be 12 .... and stay 12!

back in elem school or in college or 7th grade

i'd like to be 11.  Last year of elementary school, yeah!

Actually, I rather like my current age (17.)

Barring that, I think 13 would be a nice year to go back to.  First, it was back in the good old days of Mathcounts; second, it was arguably my best year, schoolwise; third, that was before two of my best friends at the time moved away.

10 to have the chance to try to go to imo

I'm ACTUALLY 11!!!!!!!!!!!!!!!!!!!!! :wow: *POP!* *thud*

(My Mom): EEEWWW! Matt's head has popped!

13 so I can possibly win nats again :?

or make that 11 so I have three more chances to win

this problem is good, but why should we discuss a bout a problem,

which isnt trust, at all?

i would like to be 185 again.

and stay there forever.

*sigh*

Hmmm...

I would want to be 5, with the same brain as I do now...

Or I would want to be 20, still young enough to be in good physical condition yet old enough to have independence.


floor function
factorial
linear algebra
matrix
induction
logarithms

$2^0+0\times{5}=1$
$2+0\times{0}\times{5}=2$
$-2+0+0+5=3$
$20\div{(0+5)}=4$
$\sqrt{20+0+5}=5$
$2^0+0+5=6$
$2+0+0+5=7$
$2+0!+0+5=8$
$20-\sqrt{0!+5!}=9$

Using the digits of the year, in order, and the standard symbols $+$, $-$, $\times$, $\div$, $\sqrt$, brackets and $!$, as above, how far can you extend the list? Find the longest unbroken list possible and a long list with as few breaks as possible. Enjoy :)

I don't get it. How does $\sqrt{5!}=11$?

It doesn't :)  $\sqrt{0!+5!}=11$. I guess felipesa just made a little calculation error.

$2^{0!+0!}*7=28$

EDIT: where did i get the 7 from...

[quote="Altheman"]$2^{0!+0!}*7=28$[/quote]
Umm... the 7 should be a 5?

Edit: oh, I see you edited

Isn't anybody going to try and find 35 and 36? Or maybe you can even extend the list further than I could.

Ummm... you may not change the order of 2 0 0 5 ;)

[quote="Melissa"]When I did this problem I got up to 36, but also missed 28, so I think there is no way to do 28.[/quote]
How about $28=|\sqrt{200}| \div .5$ (or $|2 \div \sqrt{.005}|$if you prefer.
Or $28=2+0!+0!+\Gamma 5$ :)

[quote="Melissa"] Using the digits of the year, in order, and the standard symbols $+$, $-$, $\times$, $\div$, $\sqrt$, brackets and $!$...[/quote] :)

[quote="Melissa"] Using the digits of the year, in order, and the standard symbols $+$, $-$, $\times$, $\div$, $\sqrt$, brackets and $!$.. :)[/quote]
Yes, I noticed the absence  of the word "only" as in
[quote] Using the digits of the year, in order, and [b]only [/b] the standard symbols $+$, $-$, $\times$, $\div$, $\sqrt$, brackets and $!$.. :)[/quote]
:) Happy New year!

[quote="10000th User"][quote="Gyan"]$28=|\sqrt{200}| \div .5$ (or $|2 \div \sqrt{.005}|$if you prefer.
Or $28=2+0!+0!+\Gamma 5$ :)[/quote]
28 took me two weeks of failed attempts!

First two are valid I think but there should be the brackets like this:

$\left\lfloor\sqrt{200}\right\rfloor\div .5=28$ or

$\left\lfloor2\div\sqrt{.005}\right\rfloor=28$, if that's what Melissa and everyone is looking for.

Gamma not valid, I think because connective 'and' don't need the word 'only'.  :roll:[/quote]
I'm sorry, but you may not use $\lfloor$ or $\rfloor$. Nor may you use the decimal point.

So I really think 28 is not possible. But nobody got 35 or 36 yet, and I know they are possible, because I got them :D

[quote="Melissa"]I'm sorry, but you may not use $\lfloor$ or $\rfloor$. Nor may you use the decimal point.

So I really think 28 is not possible. But nobody got 35 or 36 yet, and I know they are possible, because I got them :D[/quote]
OK, what are your 'brackets'? Because your 'brackets' can be different from my 'brackets'. By 'brackets', I understood either $\lfloor,\rfloor$, or $\left[,\right]$, or $\{,\}$

By brackets, I meant $(,)$. Sorry for the confusion everybody :oops:

Okay, if you allow $log$ (well you are allowing "power" so its only fair to allow "inverse of power" too! ) and decimal point  (A small "dot" is not going to hurt anyone) then I  am able to write all integers from $1$ to $2005$  ;)  :P  :idea:   :)

[quote="Melissa"][

So I really think 28 is not possible. But nobody got 35 or 36 yet, and I know they are possible, because I got them :D[/quote]

On the contrary:

[quote="moesasji5"]((2+0!)!+0!)*5=35
(5+0!)^(2+0)=36
(5+0!)^2+0!=37[/quote]

Only his 36 and 37 were not in order.

At this point in time, $28$ is unsolved.  $36$ and up are also unsolved by everyone but you, Melissa (she has solved $36$).

How about $28=((2+0!)!)!!!! + 0! +5!!$

[hide](By the way "Double(!!) and  Triple (!!!) factorials (Multifactorial ) are defined as: (Say "answer.com")

A common related notation is to use multiple exclamation points to denote a multifactorial, the product of integers in steps of two (n!!), three (n!!!), or more.

n!! denotes the double factorial of n and is defined recursively by

    $n!!= \{ \begin{matrix} 1,\qquad\quad\ &&\mbox{if }n=0\mbox{ or }n=1; \\ n(n-2)!!&&\mbox{if }n\ge2.\qquad\qquad \end{matrix} \right$

For example, $8!! = 2\cdot 4\cdot 6\cdot 8 = 384$  and $9!! = 1 \cdot 3 \cdot 5 \cdot 7 \cdot 9 = 945$ .

The double factorial is the most commonly used variant, but one can similarly define the triple factorial (n!!!) and so on. In general, the k-th factorial, denoted by n!(k), is defined recursively as

    $n!^{(k)}= \{ \begin{matrix} 1,\qquad\qquad\ &&\mbox{if }0\le n<k; \\ n(n-k)!^{(k)},&&\mbox{if }n\ge k.\quad\ \ \, \end{matrix} $

So $(2+0!)!)!!!! + 0! +5!!= (3!)!!!!+1+5 \cdot 3 = 6!!!! +1+15 = 12+1+15=28$ [/hide]

Or (Added later:  Much simpler is  [hide]   $(2+0+0+5)!!! = 7!!! = 7\cdot4\cdot1=28$
[/hide]

In any case ., I want to see if any one  finds the method I was thinking of to write  ALL numbers from 1 to 2005 using only the symbols suggested above with the addition of "log", "decimal point"

And NO you can not use $i*i$  or $e^{i\pi}$  or other cute thingies... :)

I think this question is more easily solved with a computer program.  It's becoming to outrageous.

[quote="FMako"][quote="Melissa"]On the contrary:

[quote="moesasji5"]((2+0!)!+0!)*5=35
(5+0!)^(2+0)=36
(5+0!)^2+0!=37[/quote]

Only his 36 and 37 were not in order.[/quote][/quote]
Oh sorry I didn't see that. Well, 36 is still up for grabs :D

$(2+0!)!(0!+5)=36$

Melissa: can we use multifactorial like Gyan suggested?

Congrats on finding 36. Actually, according to the original problem you cannot use multifactorial. But since you have already found all possible successive solutions that exist (according to me) I suppose you may do as you like and try to extend the list further.

Hey, I just thought of something...

Since it's almost new year, let's do the same for 2006 - in other words,using the digits of the year 2006, in order, and ONLY the standard symbols $+$, $-$, $\times$, $\div$, $\sqrt$, NORMAL brackets and SINGLE $!$, how far can you extend the list? Find the longest unbroken list possible and a long list with as few breaks as possible.

I'll start:
$2-0!+0\times 6=1$
$2+0\times 0\times 6=2$
$2+0!+0\times 6=3$
$-2+0+0+6=4$
$2\times 0-0!+6=5$
$2\times 0\times 0+6=6$
$2\times 0 +0!+6=7$
$2+0+0+6=8$
$2+0!+0+6=9$
$(2+0)(-0!+6)=10$
...

prototypex24, see this...

[quote="Melissa"]using the digits of the year 2006, in order, and ONLY the standard symbols $+$, $-$, $\times$, $\div$, $\sqrt$, NORMAL brackets and [b]SINGLE $!$[/b][/quote]

Okay, my new year’s resolution is not to be restricted by [i]all [/i]the narrow rules – think outside the box, so to speak and generalize the solution.  Here is a proof  by induction that solution exists :)
$1 = - \ln_2 \ ( \ln_{(0!+0!)}\ (\ \sqrt{\sqrt{\varphi (5)}}))$
$2 = - \ln_2 \ ( \ln_{(0!+0!)}\ (\ \sqrt \sqrt\ \sqrt{\varphi (5)}}}))$
$3 = - \ln_2 \ ( \ln_{(0!+0!)}\ (\ \sqrt{\sqrt \sqrt\ \sqrt{\varphi (5)}}}))$
$4= - \ln_2 \ ( \ln_{(0!+0!)}\ (\ \sqrt{\sqrt{\sqrt \sqrt\ \sqrt{\varphi (5)}}}}))$

etc ....

Of course,  Another one of my  this years resolution is to let go one of the  $\sqrt$ sign , as in:
$1 = - \ln_2 \ ( \ln_{(0!+0!)}\ (\ \sqrt{\varphi (6)}))$
$2 = - \ln_2 \ ( \ln_{(0!+0!)}\ (\ \sqrt{\sqrt{\varphi (6)}}))$
$3 = - \ln_2 \ ( \ln_{(0!+0!)}\ (\ \sqrt \sqrt\ \sqrt{\varphi (6)}}}))$
$4= - \ln_2 \ ( \ln_{(0!+0!)}\ (\ \sqrt{\sqrt \sqrt\ \sqrt{\varphi (6)}}}))$
$5= - \ln_2 \ ( \ln_{(0!+0!)}\ (\ \sqrt{\sqrt{\sqrt \sqrt\ \sqrt{\varphi (6)}}}}))$

etc ....
Happy New year to  you all.

PS-  (I kow I am using "ln" and "phi" functions..)

As usual $\varphi (6) = 2$ as (1,5) are only numbers less than 6 which are prime to 6. ( and  $\varphi (5) = 4$)


A sequence $ \{a_n\}_{n\in N}, a_1\equal{}50 , a_n\equal{}a_{n\minus{}1} \minus{}4n\plus{}21, n\ge2$.

$ a_n\equal{}?$

Find the possible greatest value of $ a_1\plus{}a_2\plus{}\cdots\plus{}a_n$.

 :P

$ \boxed{a_n\equal{}a_1\plus{}\sum_{k\equal{}1}^{n\minus{}1} (a_{k\plus{}1}\minus{}a_k)\ (n\geq 2)}$

[quote="kunny"]$ \boxed{a_n \equal{} a_1 \plus{} \sum_{k \equal{} 1}^{n \minus{} 1} (a_{k \plus{} 1} \minus{} a_k)\ (n\geq 2)}$[/quote]

How to find the possible  greatest value of $ \sum_{k\equal{}1}^n a_k$?

[quote="vinskman"][quote="kunny"]$ \boxed{a_n \equal{} a_1 \plus{} \sum_{k \equal{} 1}^{n \minus{} 1} (a_{k \plus{} 1} \minus{} a_k)\ (n\geq 2)}$[/quote]

How to find the possible  greatest value of $ \sum_{k \equal{} 1}^n a_k$?[/quote]

$ \sum_{k \equal{} 1}^n a_k$ is maximal when we take sum for the terms such that $ a_n>0$.

[quote="kunny"][quote="vinskman"][quote="kunny"]$ \boxed{a_n \equal{} a_1 \plus{} \sum_{k \equal{} 1}^{n \minus{} 1} (a_{k \plus{} 1} \minus{} a_k)\ (n\geq 2)}$[/quote]

How to find the possible  greatest value of $ \sum_{k \equal{} 1}^n a_k$?[/quote]

$ \sum_{k \equal{} 1}^n a_k$ is maximal when we take sum for the terms such that $ a_n > 0$.[/quote]

Thanks, :lol:  :blush: fully  got the points to solve.


How do you do those problems that area like "How many rectangles are in this figure?" fast. The only thing i do is just count them using case work, but i usaully miss a case. [geogebra]0445ac013e88213911c4cf8cd12a78428a89d65f[/geogebra]

I think casework's the only way unless there's a formula.
First count the # of rectangles on the top half.
# of small ones: 4
# of rectangles made of 2 small ones: 3
# of rectangles made of 3 small ones: 2
# of rectangles made of 4 small ones: 1
Total: 10

Since the bottom half is the same, there are also 10 rectangles in the bottom half.
But don't forget the 2 rectangles that are each half of the whole figure on the left and right side and the whole rectangle itself.
10+10+3=23 rectangles. Am I right?

Usually splitting them up by organized cases helps...like in this one, the top rectangle and the bottom rectangle by themselves have $ \binom22\binom52\equal{}10$ each, and then there are $ 3$ which have parts from both the top and bottom, for a total of $ 2\cdot10\plus{}3\equal{}\boxed{23}$.


The sequence of numbers $u_n$ is defined by:
     \[ u_n=\frac 1{n}(1+\frac 12+\frac 13+...+\frac1{n})\]
find the limit of this sequence of numbers.

You apply Cesaro Stolz end get that the limit is 0.

Or use that fact that $ \frac {1} {k+1} \leq ln(k+1) - ln(k)$.
Summing these inequalities, you get the well-known :
$0 < 1 + \frac {1} {2} + ... + \frac {1} {n} \leq 1 + ln(n)$, from which the result follows easily.

Pierre.


:(  I use WinEdt (Miktex) In my computer but It doesn't work  :( . When I try to compile any document this message appear:

"Error starting Latex!
see Winedt's Tex Configuration Wizard
for detailed diagnosis of your Tex System"

Could anybody help me???? It's very urgent!.

thanks

Hi,
Did you follow [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_Downloads.php]these[/url] installation instructions? In your case, instead of TeXnicCenter you would install WinEdt.
SSN.


Let $ x_1, x_2,..., x_n$ be arbitrary real numbers chosen in $ [a,b] (a\leq x_i \leq b)$. Prove that
$ \sum\limits_{i \equal{} 1}^n (x_i \minus{} \frac {a \plus{} b}{2})^2 \geq \sum\limits_{i \equal{} 1}^n (x_i \minus{} \frac {1}{n}\sum\limits_{i \equal{} 1}^n x_i)^2$
When does equality occur?

posted before.
here : [hide]http://www.mathlinks.ro/Forum/viewtopic.php?t=175102[/hide]

Hix, sorry, i have just participed the forum!


[quote="arqady"][quote="Twins"]$a,b,c>0$, $abc=1$, prove that
$(a+b)(b+c)(c+a)\geq 4(a+b+c-1)$[/quote]
Mixing of variables kills it.
$(a+b)(b+c)(c+a)\geq 5(a+b+c)-7$ true too.[/quote]
I'm sorry, but I really can't catch it...
Can you explain it a little more?

[quote="silouan"]Nice Arqady  :wink:  .Here is mine

We have that $(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-1$
So letting $a+b+c=s$ and $ab+bc+ca=q$ we have to prove that $sq-1\geq 4s-4$ . But $q\geq \sqrt{3s}$ so we have to prove that $s\sqrt{3s}\geq 4s-3$ or equivalently
$(s-3)(3s^{2}-7s+3)\geq 0$ which is true  :wink:[/quote]

simple and nice solution :wink:

[quote="silouan"]Nice Arqady  :wink:  .Here is mine

We have that $(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-1$
[/quote]
It's really nice... But I don't know how did you think out the first step :?:

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=6347 .

  darij

[quote="Twins"] It's really nice... But I don't know how did you think out the first step :?:[/quote]

Thank you . I used the well-known identity
$(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc$   :wink:

[quote=OlympusHero]Why did you bump a 13 - year old thread, and how did you even find it? :maybe:[/quote]

Wow sqing has 18.3K posts!

[quote=sqing][quote=OlympusHero]Why did you bump a 13 - year old thread, and how did you even find it? :maybe:[/quote]
[url=https://artofproblemsolving.com/community/c6h2175345p16226988]here[/url][/quote]

thank! :)

[quote=Twins]$a,b,c>0$, $abc=1$, prove that
$(a+b)(b+c)(c+a)\geq 4(a+b+c-1)$[/quote]
For all $a,b,c \ge 0$, [url=https://artofproblemsolving.com/community/c6h2175345p16288374]Prove that[/url]
$$(a+b)(b+c)(c+a) \geq  \frac{8}{9}(a+b+c)(ab+bc+ca)\geq  4(a+b+c-1) \sqrt{abc} $$


calculus
integration
function
logarithms
calculus computations

hello there :D

can you integrate

$ \int_0^{\frac{\pi}{2}}\;\frac{x}{\tanh\,x}\;\;dx$

hello, your integral can't be expressed by the known elementary functions. The value is approximately $ \int_0^{\frac{\pi}{2}}\;\frac{x}{\tanh\,x}\;\;dx \approx 1.9649319757669596454728555871$
Sonnhard.

[quote="Yuriy Solovyov"]$ \int_0^{\frac {\pi}{2}}\;\frac {x}{\tanh\,x}\;\;dx \equal{} \frac {\pi}{2} \plus{} \pi\sum_{n \equal{} 1}^{\infty}\frac {( \minus{} 1)^{n \plus{} 1}}{2n \plus{} 1}\cdot\frac {\zeta (2n)}{2^{2n}}.$[/quote]

May I know the derivation of the above identity?

It's just the power series for the function integrated.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=218894]Cross-reference.[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=218877]Cross-reference.[/url]

I'm just trying to keep the family linked together.

hello, Mathematica gives the following answer:
$ \int_{0}^{\frac {\pi}{2}}\frac {x}{\tanh(x)}dx \equal{} \frac {1}{24}\left( \minus{} 7\pi^2 \plus{} 12\pi\log\left(\exp(\pi) \minus{} 1\right) \minus{} 12PolyLog(2,\exp( \minus{} \pi))\right)$
Sonnhard.

$ x\cdot\frac {e^{2x} \plus{} 1}{e^{2x} \minus{} 1} \equal{} x\cdot (1 \plus{} e^{ \minus{} 2x})\sum_{k \equal{} 0}^{\infty}{e^{ \minus{} 2kx}} \equal{} \sum_{k \equal{} 0}^{\infty}{xe^{ \minus{} 2kx}} \plus{} \sum_{k \equal{} 1}^{\infty}{xe^{ \minus{} 2kx}} \equal{} x \plus{} 2\cdot \sum_{k \equal{} 1}^{\infty}{xe^{ \minus{} 2kx}}$

$ \int_{0}^{\frac {\pi}{2}}x\coth x\;dx \equal{} \frac {\pi^2}{8} \plus{} 2\cdot\sum_{k \equal{} 1}^{\infty}\int_{0}^{\frac {\pi}{2}}{xe^{ \minus{} 2kx}dx}$

$ \int_{0}^{\frac {\pi}{2}}{xe^{ \minus{} 2kx}dx} \equal{} \frac {1}{4k^2} \minus{} e^{ \minus{} \pi k}\cdot \frac {1 \plus{} \pi k}{4k^2}$

$ \int_{0}^{\frac {\pi}{2}}x\coth x\;dx \equal{} \frac {5\pi^2}{24} \minus{} \frac {1}{2}\cdot\sum_{k \equal{} 1}^{\infty}{e^{ \minus{} \pi k}\cdot \frac {1}{k^2}} \minus{} \frac {\pi}{2}\cdot \sum_{k \equal{} 1}^{\infty}{e^{ \minus{} \pi k}\cdot \frac {1}{k}}$

Second sum is very easy: $ \sum_{k \equal{} 1}^{\infty}{e^{ \minus{} \pi k}\cdot \frac {1}{k}} \equal{} \minus{} \ln(1 \minus{} e^{ \minus{} \pi}).$

For the first we get: $ f(x) \equal{} \sum_{k \equal{} 1}^{\infty}{e^{ \minus{} \pi k x}\cdot \frac {1}{k^2}}.$ Now $ f'(x) \equal{} \minus{} \pi \cdot \sum_{k \equal{} 1}^{\infty}{e^{ \minus{} \pi k x}\cdot \frac {1}{k}} \equal{} \pi \ln (1 \minus{} e^{ \minus{} \pi x}).$

So: $ f(1) \equal{} \frac {\pi^2}{6} \plus{} \pi\cdot \int_{0}^{1}\ln (1 \minus{} e^{ \minus{} \pi x})\;dx.$ This integral could not be evaluated $ 100$%.

For simplification use $ \ln (1 \minus{} e^{ \minus{} \pi x}) \equal{} \ln (e^{\pi x} \minus{} 1) \minus{} \pi x.$

Final answer: $ \int_{0}^{\frac {\pi}{2}}x\coth x\;dx \equal{} \minus{}\frac {\pi^2}{8} \plus{} \frac {\pi}{2}\cdot\ln(e^{\pi} \minus{} 1) \minus{} \frac {\pi}{2}\cdot \int_{0}^{1}\ln (e^{\pi x} \minus{} 1)\;dx.$


I don't know if I'm an idiot or not, but I don't see how to rearrange $ xyz \equal{} x \plus{} y \plus{} z \plus{} 2$ and get $ \frac{1}{1\plus{}x} \plus{} \frac{1}{1\plus{}y} \plus{} \frac{1}{1\plus{}z} \equal{} 1$

Could someone help me?
Thanks

Add $ xy + yz + zx + x + y + z + 1$ to both sides to get $ xyz + xy + yz + zx + x + y + z + 1 = xy + yz + zx + 2x + 2y + 2z + 3$. With some nice rearrangement, this is
\begin{align*}(x + 1)(y + 1)(z + 1) & = (xy + x + y + 1) + (yz + y + z + 1) + (zx + z + x + 1) \\
& = (x + 1)(y + 1) + (y + 1)(z + 1) + (z + 1)(x + 1),\end{align*}
which is what we want.


Well I took the time and I've added a new search feature, one that has been requested for a while now. You have the option of searching only through the first post now, so basically searching only through the text of the problems. Have fun with it.

How about another restriction: First post in thread AND problem in resources section?

[quote="Valentin Vornicu"]I've added the little checkbox orl. Note that this means that the post you are looking for must be in the resources section (ie if the keyword is in the same topic as a post in the resources section, then it's useless.)[/quote]

I am a bit confused on what you say, especially the latter part. I thought if a post is part of the resources section and matches the keyword(s) that is what I was searching for. Why is it useless then?

[quote="orl"][quote="Valentin Vornicu"]No I'm saying if you are looking for a keyword that is in the TOPIC of discussion of a post in the resources, that won't work. If the keyword is in the post itself that will work.[/quote]

Does that mean that it will look only in the first post of this thread which commonly contains the problem statement? The resources section also references solutions: Will the solution post(s) of a particular thread be searched too?[/quote]It will basically look in the first post only as of right now.

Can you add the option to find all threads started by a specific user?

Click on 'Memberlist' button above (next to where you click on 'Profile', 'Private Messages', etc.). Search the person you want. Click on their name. Then, click on where it says 'Find all posts by [member]'.

idk, maybe you know the guy who posted but wanna see what he posted...and most of our posts ae replying to them, so uit would be a lot easier to look at first post...

So putting everything together: you (and I) want a search option that just checks the first post, but also considers the name, not only the text.
And this is usefull, e.g. if you prefer problems by someone.

[quote="buzzer11"][quote="007math"]But he wants to know the threads [i]started[/i] by that user.[/quote]

For what purpose?[/quote]
Sometimes I like to see threads about specific members. You can find interesting problems quickly.

Searching the first post with some search term and a user specified works nicely; the keyword "sum" and me combine for two results, since I haven't started many threads. The same search unrestricted finds over 300 results.

Leaving off the search term doesn't work. Searching for me, first post only, with no search term specified, turns up all of the over 4000 threads I have ever posted in.


geometry
similar triangles
geometry proposed

I have solution for this problem I'm not sure about... Anyone can help me finding error in it?

problem states: in quadrilateral $ABCD$ we put point $M$ such that $AM = MC$. Also we have $\angle BAD = \angle BMC = \angle CMD$. Prove that points $A$,$B$,$C$ and $D$ lay on the same circle.

[hide]my proof: We have circle $\omega_{1}$ such that $M$ is centre of it and points $B$ and $D$ lay on it. then we find that $A$ also lays on circle $\omega_{1}$. Then, becouse $AM=MC$ point $C$ also lays on this circle, $QED$
edit:
Ok, heres error: I assumed inproperly that $BM=MD$. but still, problem is worth a look :) and fortunately that's not the only solution I found (the other one is correct) just found this and at first glance couldn't figure out what's wrong about it[/hide]

Well, solution is pretty easier and bases on similar triangles and finding "magical point". $D^{*}$ and honestly I don't get what your hint leads to, can you explain?

(sorry, i'm such a fool when it comes to geometry)

well my hint says exactly that triangles MBC and MCD are similar !


Sorry if I post a mistake solution!
We have : $ e^x\equal{}\sum^{\infty}_{n\equal{}1} \frac{x^n}{n!}$ then:
$ e\equal{}\sum^{\infty}_{n\equal{}1} \frac{1}{n!},\frac {1}{e}\equal{}e^{\minus{}1}\equal{}\sum^{\infty}_{n\equal{}1} \frac{(\minus{}1)^n}{n!}$
Add two equation we have:
$ \sum^{\infty}_{n\equal{}1} \frac{1}{(2n)!}\equal{}2(e\plus{}\frac{1}{e})$

Better problem for one of the Algebra sections


How many 3-digit numbers are divisible by 11?

$ \left\lfloor \frac {999}{11} \right\rfloor \minus{} \left\lfloor \frac {99}{11} \right\rfloor \equal{} 90 \minus{} 9 \equal{} \boxed{81}$.

[hide=Solution]
We get the list $110, 121, \ldots , 990$. Dividing $11$ gives us $10, 11, \ldots , 99$ and subtracting $9$ we get $1, 2, \ldots , 81$. So there are $\boxed{81}$ $3$-digit numbers that are divisible by $11$.
[/hide]
[hide=Alternative Solution]
We see that there are $9$ multiples of $11$ in every hundred. There are $9$ hundreds that are $3$-digit so we have $9\times9=\boxed{81}$ as our answer.
[/hide]


I found this link about fear of math.

Check it out and tell me what most of you think.

Go to:

http://www.petersstudysteps.com/fear_of_math.htm

interval

That is really weird.  I feel sorry for all the poor people who can't have fun with math any more :(  Lol I bet most people here don't have anything like that (You don't run to a math forum to get away from math) :D

A large portion of the problem lies in the ways in which math is taught to students today.  In the US, students are frequently taught the rules of math, but rarely will they learn why a specific approach to a math problems work.  Should students be provided with a foundation of “deeper understanding” of math, it may prevent the development of phobias.

So the teachers are the cause of this. We just need to find teachers who explain everything, and go slow.

It's not surprising.  The American educational system isn't geared towards deep understanding.  It's inherently vocational - you go to learn skills that you apply in adult life.  Curriculums are not predisposed to changing - the administrative costs are too high, so kids continue to learn much the same things and in much the same ways as before.  It's clear, though, that these methods aren't very effective.

As far as this "math anxiety" goes, it's relatively simple classical conditioning.

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

Mathematics becomes associated with unpleasant experiences in poorly-taught math classes, and therefore it produces an anxiety response.

I'm frankly not surprised! Don't even get me started about the lousy math system in schools...
Math teachers in my city are mostly the type to induce such a math-phobia rather than get rid of it. I complain so much because for no reason, my classmates are sweating and shivering and having nervous breakdowns in the exams for one of  my favourite subjects!! :P  :mad:
Rarely do u find a good math teacher in my city who neither complicates all the sums nor gives it all for homework without explaining!!
People HATE math only because they are confused and fed up, and they make ridiculously childish mistakes!
Seriously...if robots replaced teachers....how AMAZING school would be!!  :clap2:


linear algebra
matrix
complex numbers
linear algebra unsolved

Does anyone have a natural solution for the following problem:
  For any square complex matrix A we have $ det(I_n+AB)\geq 0$ where $ b_ij$ are the conjugates of the elements of $A$. I have a monstrous solution and I would really like to see something nice and easy.

I'll try:
It is easy to verify that $det(xI_n-A\bar{A})=(x-a_1)(x-a_2)\cdots(x-a_n)$, where $a_i$ is an eigenvalue of $A\bar A$. Then
$det(I_n+A\bar{A})=(-1)^ndet(-I_n-A\bar{A})=(-1)^n(-1)^n(1+a_1)(1+a_2)\cdots(1+a_n)=(1+a_1)(1+a_2)\cdots(1+a_n)  (1)$.
But $det(A\bar{A})=a_1a_2 \cdots a_n$ and it is real also ($det(A\bar{A})=det(A)\bar{detA}$). So all the $a_i$ came in (1) in conjugate pairs. Since $(1+a_i)(1+\bar{a_i})=1+2Rea_i+(Rea_i)^2+(Ima_i)^2\geq0$ we get our result.

It seems to me right now that i have missed something but i can't find it now.

[quote="eugene"]
But $det(A\bar{A})=a_1a_2 \cdots a_n$  it is real

So all the $a_i$ came in (1) in conjugate pairs.

[/quote]

Some $a_i$ could be real

a teacher told me he never seen a natural solution for this one  :D
It would be very nice if someone could find it ..

[quote="harazi"]Eugene, why if the product of some complex numbers is real, then these complex numbers come in pairs of conjugate numbers?[/quote]
Yes, of course what i said was wrong-i realized it in few minutes after posting but i had to go out.
Maybe one could prove that all real $a_i$ which are negative come in even degrees: i mean that if $a_i<0$ then the matrix $A\bar{A}$ has the eigenvalues $a_i$ of even multiply.
It is just supposions which i couldn't prove

if $v_i$ is an eigenvector of $A$ then:
$Av_i=\lambda_i v_i\Rightarrow \bar{A}v_i=\bar{\lambda_i}v_i\Rightarrow A\bar{A}v_i=\bar{\lambda_i}\lambda_iv_i$
Thus $a_1,...,a_n\geq 0$
I think I'm wrong since I have $det(I_n+A\bar{A})\geq 1$ but where ?

[quote="Rodolphe2005"]if $v_i$ is an eigenvector of $A$ then:
$Av_i=\lambda_i v_i\Rightarrow \bar{A}v_i=\bar{\lambda_i}v_i$[/quote]

The entries of $v_i$ are complex numbers so you have $\bar{A}\bar{v_i}=  \bar{\lambda_i}\bar{v_i}$

Here another direction to explore to find a new proof

Write $A=X+iY$ with $X,Y \in M_n(R)$

Work with $det(I+A\bar{A})=det(I_n+X^2+Y^2+i(YX-XY))=f(X,Y)+ig(X,Y)$

Moubi, is there a solution starting with this observation. When I received the problem, I tried this approach and I couldn't find relevant results.


There are 90 balls numbered from 0 to 89.we have to select three balls without replacement such that sum of the numbers on the balls is 90.how many ways are possible.
_________________

[hide="hint"]
In general $ x_1\plus{}x_2\plus{}...\plus{}x_n\equal{}k$ have $ \binom{k\minus{}1}{n\minus{}1}$ solutions[/hide]

Dimitris X, can you please expand on that. I am not sure how you are getting that.

I think this this problem can be solved by a little tweak of the Hockey Stick Identity. This problem is analogous to the number of ways to distribute $ 90$ pieces of candy among $ 3$ kids with the restriction that every kid must get a different number of candy. (Perhaps this the same thing Dimitris suggested?  :maybe: )

Yes, it is the same thing that Dimitris X suggested. He just put in in equation form, where $ k$ is equal to the number of kids and $ n$ is equal to the number of candies.

Don't you mean where k is the amount of candies and n is the number of kids?

Sunehra was right. Also Dimitris's \binom expression shoudl have been $ \binom{n\minus{}1}{k\minus{}1}$. :wink:

i dont think this will work

as the binomial formula also accounts for solutions like 1, 1, 88
but there is only one ball numbered 1.
Give it a thought

I think you have to subtract out those cases.

we have to find $ i, j, k$ such that
$ i\plus{}j\plus{}k\equal{}90$
$ 1x1x1x1x1x1x1x......1x1$
consider the 90 1's
if we place 2 sticks in the 89 x's, we get the 90 1's divided into 3 parts.
consider the number of $ 1's$ in each of the groups
the first group contains i $ 1's$ and select the ball numbered $ i$
do as above for the remaining 2 balls, selecting from the other 2 groups.......$ 1$

but the ball numbered 0 won't be selected
thus add $ \binom{89}{1}$and subtract 1 from that for $ (45,45)$ also, we would have selected twice every number in $ 1$
$ (i,i,j)$
$ (1,1,88), (2,2,86),...... (44,44,2)$
44 triples
subtract this and add 1 for$ (30,30,30)$
the answer is
$ \dbinom{89}{2}\plus{}89\minus{}1\minus{}44\plus{}1\\*\equal{}\dbinom{89}{2}\plus{}45$

is that correct?

[quote="Ihatepie"]Does the order of picking the balls matter?[/quote]
I don't think so. It doesn't mention specifically in the original problem statement. :maybe:

If order doesn’t matter, you could do it this way. Let $ a$ be the smallest number on the three balls, $ b$ be the number in between and $ c$ be the largest number on the three balls.

If $ a \equal{} 0,$ you can have $ b \equal{} 1,c \equal{} 89;\,b \equal{} 2,c \equal{} 88;\,\ldots;\,b \equal{} 44,c \equal{} 46.$ There are $ 44$ choices for the other two balls.

If $ a \equal{} 1,$ you can have $ b \equal{} 2,c \equal{} 87;\,b \equal{} 3,c \equal{} 86;\,\ldots;\,b \equal{} 44,c \equal{} 45.$ There are $ 43$ choices for the other two balls.

If $ a \equal{} 2,$ you can have $ b \equal{} 3,c \equal{} 85;\,\ldots;\,b \equal{} 43,c \equal{} 45$ so $ 41$ choices for the other two balls.

If $ a \equal{} 3,$ you can have $ b \equal{} 4,c \equal{} 83;\,\ldots;\,b \equal{} 43,c \equal{} 44$ so $ 40$ choices for the other two balls.

[list]$ \vdots$[/list]

$ a \equal{} 2n\ \Rightarrow\ b \equal{} 2n \plus{} 1,c \equal{} 89 \minus{} 4n;\,\ldots;\,b \equal{} 44 \minus{} n,c \equal{} 46 \minus{} n\ \Rightarrow\ 44 \minus{} 3n$ choices for the other two balls.

$ a \equal{} 2n \plus{} 1\ \Rightarrow\ b \equal{} 2n \plus{} 2,c \equal{} 87 \minus{} 4n;\,\ldots;\,b \equal{} 44 \minus{} n,c \equal{} 45 \minus{} n\ \Rightarrow\ 43 \minus{} 3n$ choices for the other two balls.

[list]$ \vdots$[/list]

$ a \equal{} 28\ \Rightarrow\ b \equal{} 29,c \equal{} 33;\,b \equal{} 30,c \equal{} 32\ \Rightarrow\ 2$ choices for the other two balls.

$ a \equal{} 29\ \Rightarrow\ b \equal{} 30,c \equal{} 31 \Rightarrow\ 1$ choice for the other two balls.

Hence the total number of ways of choosing (with order not mattering) is $ \sum_{n\, \equal{} \,0}^{14}\left[(44 \minus{} 3n) \plus{} (43 \minus{} 3n)\right]\ \equal{} \ 675.$

NB: If order does matter, multiply by $ 3!\equal{}6.$ :-D


LaTeX
calculus
derivative
inequalities
calculus computations

I would like to solve the following differential equation:

y''''[x]+2y''[x]+y[x]=-x

Latex doesn't really work, I'm new here so please forgive me   :o
$\frac{d^4y}{dx^4}+2\frac{d^2y}{dx^2}+y=-x$

Can someone help me with it?

Thanks in advance,
Wilb

[quote="ashwinrk_jain"]I am not getting the equation properly. please write it in simplified form.[/quote]

I placed the equation in the first post above the strange latex outcome. I can't get the latex code to work somehow. Note that the first term is the fourth derivative of y.

Wilb

[quote="hydro"][quote="wilb"]I would like to solve the following differential equation:

$\frac{d^4y}{dx^4}+2\frac{d^2y}{dx^2}+\frac{dy}{dx}=-x$

[/quote]

is it that?[/quote]

Finally i made it... my problem is in the first post edited it ;). Took me some time, i hope somebody can help me with it.

Wilb

Now i will get the equation now i try to solve it.

[quote="Soarer"]$y = Axcosx + Bxsinx + Ccosx + Dsinx - x$
This is the inequality forum which is not supposed to have differential equations here.
By the way, find a particular solution first.[/quote]
How do you solve this ,Soarer?
I don't know more about differential equation,so if it is easy,fogive me.


Let $ x,y,z \geq 0 $ and $ x+y+z =1 $ . Prove that :
  $ 0 \leq xy+yz+xz - 3xyz \leq \frac{1}{4} $

We know that $(x+y+z)(xy+yx+xz)\geq 9xyz$ so $(xy+yx+xz)\geq 9xyz$.
The second part goes from:
$ 1 = (\sum x)^3\ = \sum x^3\ + 6xyz + 3\sum x^2y + y^2x $ > $4\sum x^2y + y^2x = 4[(x+y+z)(xy+yx+xz)-3xyz] $
(by Schur Inequality)
So $ (xy+yx+xz)-3xyz $ < $ 1/4 $

Well, but this similar ineq look more difficult:
   $ x,y,z >0$ ; $ x+y+z=3$
Prove that $ xy+yz+zx - xyz \leq 2$

I have another solution :
  Suppose  $ x \geq y \geq z \geq 0 $
  $ ==> z \leq \frac{1}{3}   ,   3xyz \leq xy $
  So $ xy + yz + xz - 3xyz \geq yz + zx \geq 0 $
 Let $ z = \frac{1}{3} - k  === >  x+y = \frac{2}{3} + k $
  By AM-GM :
   $ xy \leq (\frac{1}{3}+\frac{k}{2})^2 $
  So $ xy + yz + xz -3xyz \leq (\frac{1}{3}-k)(\frac{2}{3}+k) + (\frac{1}{3}+\frac{k}{2})^2  . 3k \leq \frac{1}{4} $


A rectangular $ 2n$ polygon wth centre $ O$was painted by $ n$ colors such that:
  1)Each vertice was painted by exactly 1 color.
  2) One color painted exactly two vertices.
  Two way paint seem similar if this way is image of that way pass a $ R_O$.
            How many way to color the polygon.

[hide="My hint"]
The answer is : $ [\frac {2n\minus{}1}{2}].\frac {(2n\minus{}2)!}{{2}^{n\minus{}1}}$
[/hide]


Indicate in which one of the following equations $ y$ is neither directly nor inversely proportional to $ x$:

$ \textbf{(A)}\ x \plus{} y \equal{} 0 \qquad\textbf{(B)}\ 3xy \equal{} 10 \qquad\textbf{(C)}\ x \equal{} 5y \qquad\textbf{(D)}\ 3x \plus{} y \equal{} 10$
$ \textbf{(E)}\ \frac {x}{y} \equal{} \sqrt {3}$

[hide]
For $ A$, we can rewrite it as $ y\equal{}\minus{}x\Rightarrow\frac{y}{x}\equal{}\minus{}1\Rightarrow\frac{x}{y}\equal{}\minus{}1$, a direct proportion.
For $ B$, $ xy\equal{}k$, where $ k$ is a constant, so $ x$ and $ y$ are inversely proportional.
For $ C$, we can rewrite it as $ \frac{x}{y}\equal{}5$, a direct proportion.
For $ E$, $ \frac{x}{y}\equal{}k$, where $ k$ is a constant, so $ x$ and $ y$ are directly proportional.
By process of elimnation, we are left with $ \boxed{D}$ as our answer, and $ 3x\plus{}y\equal{}10$ cannot be written as either $ \frac{x}{y}\equal{}k$ or $ xy\equal{}k$.
[/hide]


[hide]there are no solutions:

the equation is equivalent to:
$(x+y)^2=122-xy$

looking at it modulo $2$, we see, that both $x$ and $y$ need to be even. so $x=2x'$ and $y=2y'$ with alle variables integers.

$4(x'+y')^2=2\cdot 61+4x'y'$

now the RHS is $0\pmod4$ but the LHS is $2\pmod4$ $\implies$ a contradiction.
so there are no integer solutions

[/hide]

[quote="Altheman"]i was always wondering, what does RHS and LHS mean? i see it all the time, but i have no idea what they stand for[/quote]
RHS = right hand side

LHS = left hand side

Bye ;)

[hide]$(x+y)^2+xy=122$
$k+xy = 2 \pmod {4}, k=0 \text { or } 1$

If $xy$ is divisible by 4 it is impossible.

If $k=0$, then $xy \equiv 2 \pmod {4}$, so only one of $x$ or $y$ is even. But if $k=0$, then $(x+y)^2 \equiv 0 \pmod {4} \Rightarrow (\text { odd } + \text { even })^2 \equiv 0 \Rightarrow (\text { odd } )^2 \equiv 0 \pmod {4}$, which is impossible.

If $k=1$, then $xy \equiv 1 \pmod {4}$. Therefore, neither are even. But if $k=1$, then $(x+y)^2 \equiv 1 \pmod {4} \Rightarrow (\text { odd } + \text { odd })^2 \equiv 1 \Rightarrow (\text { even } )^2 \equiv 1 \pmod {4}$, which is also impossible.

Therefore $k \ne 0 \text { or } 1$, which is a contradiction, because all perfect squares must be congruent to either 0 or 1, $\mod {4}$.[/hide]


Let $ a,b,c$ positive real numbers. Prove or disprove that

$ \frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a} \geq 3\sqrt[12]{(\frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca})^7}$

______
 :?:  :?:

You can see the stronger here:
http://www.mathlinks.ro/Forum/viewtopic.php?t=252165
 :)


Dedicated to those who liked the post "A Mathematician's Love Letter."

[color=green]CIRCUIT LOVE

I was all alone and all was dark
Beneath me and above
My life was dull of volts and amps
But not the spark of love
         But now you are here with me
         My heart is over-joyed
         You turn the square of my heart
         Into sinusoid
         You load things from my memory
         Onto my system bus
         My life was just assembly code
         Now its C++
I love the way you solder things
My circuit you can fix
The voltage across your diode is
Much more than just point six
         With your amp and risistors
         You have built my integrator
         I can not survive without you
         You are my function generator
You have changed my life
Increase my gain and made my math discreal
And now I will end my poem here
Control, Alt and Delete.[/color]

This wasn't as great as the last post but....

i like it
 :rotfl:
where do you get these?


Δοθεντων.

1) $ y\geq 0$ yεN (natural Nos)

2) $ y\neq 0$

Να αποδειχθη .........................x<x+y.......................(x,y) εN.............Χωρις να χρησιμοποιησουμε επαγωγη (induction)

Πώς ορίζεις τα < και +;

[quote="Demetres"]Πώς ορίζεις τα < και +;[/quote]

τα συμβολα ειναι ................(=)...........(')............(+)...............(.)...........(0).................

Τα αξιωματα ειναι.

1)$ \forall x\forall y( x'\equal{}y'\rightarrow x\equal{}y)$

2)$ \forall( x'\neq 0)$

3)$ \forall x ( x\plus{}0\equal{}x)$

4)$ \forall x\forall y [ x\plus{}y'\equal{}(x\plus{}y)']$

5 )$ \forall x ( x0\equal{}0 )$

6)$ \forall x\forall y ( xy' \equal{} xy \plus{} x)$.

Τωρα για ανισοτητες υπαρχουν πολλοι ορισμοι χρησιμοποιησε τον καταλληλο

Η ασκηση δεν ειναι δικη μου την ειδα σε ενα αλλλο φορουμ και μου εκανε εντυπωση που καποιος εδωσε μια απαντηση χωρις επαγωγη (Induction)

[quote="τρικλινο"]
Τωρα για ανισοτητες υπαρχουν πολλοι ορισμοι χρησιμοποιησε τον καταλληλο
[/quote]

Τότε διαλέγω να χρησιμοποιήσω $ \forall x \forall z [(x < z) \leftrightarrow \exists y (y \neq 0) \wedge (x\plus{}y\equal{}z)]$  :D

Και εμένα μου κάνει εντύπωση γι'αυτό και ζήτησα τους ορισμούς. Ο ορισμός της πρόσθεσης π.χ. που παρέθεσες, δεν δουλεύει χωρίς επαγωγή.

Αν μπορείς βάλε ένα link στην συζήτηση στο άλλο forum διότι είμαι αρκετά περίεργος να δω πως γίνεται.

[quote="Demetres"][quote="τρικλινο"]
Τωρα για ανισοτητες υπαρχουν πολλοι ορισμοι χρησιμοποιησε τον καταλληλο
[/quote]

Τότε διαλέγω να χρησιμοποιήσω $ \forall x \forall z [(x < z) \leftrightarrow \exists y (y \neq 0) \wedge (x \plus{} y \equal{} z)]$  :D

Και εμένα μου κάνει εντύπωση γι'αυτό και ζήτησα τους ορισμούς. Ο ορισμός της πρόσθεσης π.χ. που παρέθεσες, δεν δουλεύει χωρίς επαγωγή.

Αν μπορείς βάλε ένα link στην συζήτηση στο άλλο forum διότι είμαι αρκετά περίεργος να δω πως γίνεται.[/quote]

Θυμαμαι καλα την αποδειξη.

Ο τυπος λοιπον εχρησιμοποιησε τον ορισμο.

........r<s <=====>υπαρχει ,u ετσι ωστε (r + u)'=s...........................................................................................

...........$ \forall r\forall s[ r<s\Longleftrightarrow\exists u( (r \plus{} u)'\equal{}s)]$.............................................................................

Συμφωνα με αυτο τον ορισμο λεει λοιπον οτι.

 $ y\neq 0\rightarrow y>0$ και συμφωνα με τον ορισμο υπαρχει z ετσι ωστε (z+0)'=y η z'=y συμφωνα με το αξιωμα (3)

Τωρα απο x+y=x+y εχουμε x+z'=x+y.Αλλα συμφωνα με το αξιωμα (4) εχουμε (x+z)'=x+y και ετσι συμφωνα με τον ορισμο ξανα ...............................................x<x+y..........................................................................................................


Let $ ABCD$ be a quadrialteral such that the opposite sides are not parallel, excribed a circle $ (O)$.Let $ M,N,P,Q$ be
midpoint of  $ AB,BC,CD,DA$ respectively.Prove that $ O$ is the intersection of $ MP$ and$ NQ$ iff $ OA.OC\equal{}OB.OD$

The point of intersection of the lines $ MP$ and $ NQ$ is simply the barycenter (= centroid = center of gravity) of the quadrilateral $ ABCD$, so that the problem is http://www.mathlinks.ro/Forum/viewtopic.php?t=35309 .

  darij


Βρείτε όλους τους τριψήφιους αριθμούς που διαιρούνται τέλεια με το 11 και το πηλίκο που αφήνουν ισούται με το άθροισμα των τετραγώνων των ψηφίων τους.

Έχω μια λύση αλλά με πολλές περιπτώσεις.

Νεκτάριε, καλώς ήλθες κι εσύ στην παρέα μας  :)
Ενδιαφέρον πρόβλημα, και πιστεύω ότι άξιζε να ασχοληθεί κανείς.
Αλλά πραγματικά είχε αρκετές περιπτώσεις...
πολύ κακό για το τίποτα, αφού προκύπτουν μόνο δύο λύσεις.

Έστω $a=100x+10y+z = 11b \ \ (x \neq 0)$, και είναι γνωστό ότι $b=x^2+y^2+z^2$

Πρώτα θα προσπαθήσουμε να εκφράσουμε το $z$ συναρτήσει των $x,y$

$a =(99x+x)+(11y-y)+z = (99x+11y)+(x-y+z) \Rightarrow$

$11b = 11(9x+y)+(x-y+z)$

Άρα $11 \mid (x-y+z)$ (αυτό προκύπτει και από το γνωστό "κριτήριο διαιρετότητας με το 11")

Επομένως γράφουμε $x-y+z = 11k\Rightarrow$
$11b = 11(9x+y)+11k \Rightarrow$
$b=9x+y+k$

Το $k$ μπορεί να πάρει μόνο τις τιμές $0,1$. Πράγματι, αφού τα $x,y,z$ είναι ψηφία έχουμε:
$ \begin{array}{l} 1\leq x\leq 9 \\ -9\leq -y\leq 0 \\ 0\leq z\leq 9 \end{array}  \}\Rightarrow -8\leq x-y+z\leq 18$

Τα μόνα πολλαπλάσια του $11$ σ'αυτήν την περιοχή είναι οι αριθμοί $0,11$

$x-y+z=11k\Rightarrow z=11k+y-x$
Για την απλούστευση των παραστάσεων που ακολουθούν θέτω $m=11k$

Έτσι θα γράφουμε $z = m+y-x$

Θα βρούμε πρώτα την εξίσωση με παράμετρο το $m$ και μετά θα δούμε δύο περιπτώσεις: $m=0,m=11$
Θα εκφράσουμε το $b = x^2+y^2+z^2$ συναρτήσει των $x,y$

$z = m+y-x \Rightarrow$
$z^2 = (m+y-x)^2$
$z^2 = m^2+x^2+y^2 +2my-2xy-2mx \Rightarrow$
$x^2+y^2+z^2 = x^2+y^2 + m^2 +x^2+y^2-2xy -2mx+2my \Rightarrow$
$x^2+y^2+z^2 = 2x^2+2y^2 -2xy -2mx+2my+ m^2$

Αλλά $x^2+y^2+z^2 = b = 9x+y+k \Rightarrow$

$2x^2+2y^2 -2xy -2mx +2my + m^2 = 9x+y+k \Rightarrow$
$2x^2+2y^2 -2xy -(2m+9)x +(2m-1)y + m^2 - k = 0 \ \ (1)$

[hide="a) k=0"]$z=11k+y-x \Rightarrow z=y-x$ ,άρα θα είναι $y\geq x$

Επίσης έχουμε $m=11k \Rightarrow m = 0 \Rightarrow$
$-(2m+9)=-9$
$2m-1=-1$
$m^2-k = 0$

Η $(1)$ γράφεται:

$2x^2+2y^2 -2xy -9x -y = 0 \Rightarrow$
$\boxed{2x^2 - (2y+9)x + (2y^2 -y) = 0 }$

Είναι ένα τριώνυμο του $x$. Υπολογίζουμε τη διακρίνουσα:

$D = (2y+9)^2-8 (2y^2-y)$
$D = 4y^2+36y+81-16y^2+8y$
$D = (-12)y^2+44y+81$

Για να έχουμε ακέραιες λύσεις, η διακρίνουσα πρέπει να είναι τέλειο τετράγωνο

Η $D$ είναι και αυτή ένα τριώνυμο. Βρίσκουμε τις ρίζες...

$D_1 = 44^2-4(-12)81 = 5824 = 64 \cdot 91 \Rightarrow \sqrt{D_1} = 8 \sqrt{91}$

Οι δύο ρίζες είναι $y_{1,2} = \frac{-44 \pm 8 \sqrt{91}}{-24} = \frac{11 \pm 2 \sqrt{91}}{6} \Rightarrow$

$y_1 = -1,34646$
$y_2 = 5,013131$

Η $D$ παίρνει θετικές τιμές μόνο μεταξύ των ριζών.
Επομένως, το $y$ πρέπει να βρίσκεται στο διάστημα $[y_1,y_2]$ και ταυτόχρονα να είναι και φυσικός αριθμός.
Άρα $y \in \{ 0,1,2,3,4,5 \}$

Επίσης, η $D$ πρέπει να είναι τέλειο τετράγωνο.

Αφού βρούμε όλες τις διακρίνουσες που είναι τέλεια τετράγωνα, στη συνέχεια θα βρουμε το $x$ από τον τύπο
$x=\frac{(2y+9) \pm \sqrt{D}}{4}$, και πρέπει να είναι φυσικός αριθμός και επιπλέον μικρότερος ή ίσος του $y$.

$\begin{array}{|c|c|c|c|c|} \hline y&D&2y+9&x_1&x_2 \\ \hline 0&81&9&0&4,5 \\ \hline 1&113&&& \\ \hline 2&121&13&0,5&6 \\ \hline 3&105&&& \\ \hline 4&65&&& \\ \hline 5&1&19&4,5&5 \\ \hline \end{array}$

Όπως βλέπουμε στον πίνακα, η  $D$ είναι τέλειο τετράγωνο για $y=0,2,5$
Δεκτή είναι κάθε φορά όποια από τις ρίζες $x_1,x_2$ είναι ακέραια.

Έτσι, για $y=0,2,5$ παίρνουμε αντίστοιχα $x=0,6,5$

Από αυτές δεκτή είναι μόνο η $x=5$, αφού $x \neq 0$ και $x\leq y$

Στην πρώτη λοιπόν περίπτωση $(k=0)$ έχουμε  μοναδική λύση την $(x,y)=(5,5)$

Παίρνουμε $z=5-5=0$

Άρα ο ζητούμενος αριθμός είναι $a=550 \Rightarrow b=\frac{a}{11}=50$
και $x^2+y^2+z^2 = 5^2+5^2+0^2 = 25+25=50$[/hide]

[hide="b) k=1"]$z=11k+y-x \Rightarrow z=11+y-x$

$z\leq 9 \Rightarrow 11+y-x\leq 9 \Rightarrow x\geq y+2$

Επίσης έχουμε $m=11k \Rightarrow m = 11 \Rightarrow$
$-(2m+9)=-31$
$2m-1=21$
$m^2-k = 11^2-1=120$

Η $(1)$ γράφεται:

$2x^2+2y^2 -2xy -(2m+9)x +(2m-1)y + m^2 - k = 0 \ \ (1)$

$2x^2+2y^2 -2xy -31x +2y + 120 = 0 \Rightarrow$

$\boxed{2x^2 -(2y +31)x+2y^2 +2y + 120 = 0}$

$D = (2y+31)^2 -8(2y^2 +2y + 120)$

$D = 4y^2+124y+961 -16y^2 -16y -960$

$D = -12y^2+108y+1$

[hide="Ρίζες της D"]
$D_1 = 108^2+4 \cdot 12 = 11712 = 64 \cdot 183 \Rightarrow \sqrt{D_1}=8\sqrt{183}$

$y_{1,2} = \frac{-108 \pm 8\sqrt{183}}{-24}= \frac{27 \pm 2\sqrt{183}}{6}\Rightarrow$[/hide]

$y_1 = -0,009249753$
$y_2 = 9,009249753$

Θα αναζητήσουμε τα $y$ στους φυσικούς αριθμούς από $0$ ως $9$.

Η $D$ γίνεται τέλειο τετράγωνο μόνο για $y=0,2,7,9$

Το $x$ θα δίνεται σε κάθε περίπτωση από τη σχέση $x_{1,2}=\frac{(2y +31) \pm \sqrt{D}}{4}$

$\begin{array}{|c|c|c|c|c|} \hline y&D&2y+31&x_1&x_2 \\ \hline 0&1&31&7,5&8 \\ \hline 2&169&35&5,5&12 \\ \hline 7&169&45&8&14,5 \\ \hline 9&1&49&12&12,5 \\ \hline \end{array}$

Για $y=0,2,7,9$ οι ακέραιες τιμές για το $x$ είναι αντίστοιχα $8,12,8,12$

Οι τιμές 12 απορρίπτονται γιατί το $x$ παριστάνει ψηφίο
Επίσης, απορρίπτεται το ζεύγος $(7,8)$ γιατί είπαμε ότι πρέπει να είναι $x\geq y+2$

Έχουμε μοναδική λύση και σ'αυτήν την περίπτωση, την $(0,8)$. Τότε $z=11+y-x=11+0-8=3$.
Άρα ο ζητούμενος αριθμός είναι
$a=803\Rightarrow b=\frac{803}{11}=73$

και $8^2+0^2+3^2=64+9=73$[/hide]


The perimeter of a square lot is lined with trees, and there are three yards between the centers of the adjacent trees.  There are eight trees on a side, and a tree is at each corner.  What is the number of yards in the perimeter of the lot?

I can't picture it......

I don't really get the question....

um... this requires drawing a pictures...

I tried to put my picture on here... but it said something about bmp extension not allowed, but anyways:

[hide]the perimeter of the square is equal to 3x, where x is equal to the number of spaces between the trees (the centers of the adjacent trees, the exact word of the problem). and that would be 4*8-4 = 28 because the four corners are counted twice. Or you could count it if you try a picture which is much easier. 3*28 = 84.... [/hide]

I have a slight question about the problem:

[hide]What about the length/width/space the actually trees take up? should the problem tell you that too?[/hide]

.bmp files are very large and would quickly take up too much space on our server.  If you are able to turn the picture into a .gif or .jpeg then the picture will probably post.

[quote="LuCy4EvA"]not really good with these kinda stuff, but o well, i've already delete the picture, thank you[/quote]

Since you said bmp im guessing you used the paint program. To save it in gif or jpeg, press file, save as, and look for save as type and choose gif or jpeg.

thank you, GoBraves, as you can tell, I'm not a computer person. Thanks!

I redid the picture:

[quote="DarkKnight"][hide]
I got 84 but i'm wondering about the space the tree's take up.
[/hide][/quote]

Same here, but i guess we just have to do it like this.

[hide]great question i liked it
28
that picture is correct[/hide]

wat bout the space that the tree takes up!
how do u find how much space that covers?
 :(
 :D
coz i got 84 but dont no how to find the space the trees cover? :blush:

[hide="About the space the tree takes up"]This has already been counted as we have the distance between the [i]centers[/i]of the trees, not the trees themselves.[/hide]

[hide="Solution"]As there are 8 trees on a side, there are 7 spaces between trees on each side. 7*4=28. 28 spaces * 3 yards each = 84 yards in the perimeter.[/hide]

[hide="click text"]Is it 441? Sory that is the area. The perimeter is 84.[/hide]

[hide]When he says "and a tree at each corner", does that mean that there would be 10 trees per side?  If what I said was true, then it would be 108[/hide]

No, it wouldn't, because the eight trees includes the corner trees :P .

[color=white]84[/color] yards of sacrifices for the Treething god.


Factor $ w^4\minus{}16$ as much as possible.

Use differences of squares.

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

hello, you can also write
$ (w\plus{}2)(w\minus{}2)(w\minus{}2i)(w\plus{}2i)$ with $ i^2\equal{}\minus{}1$.
Sonnhard.

Yeah, but FTW doesn't accept those kinds of answers.

umm this is mathcounts...

and i think it should say 'factor over real parts' in the problem if the answer involves i.

[quote="Dr Sonnhard Graubner"]hello, please tell me what's FTW.
Sonnhard.[/quote]

FTW is an online game played with guests and real FTW players.

It is a math game, where you try to answer the problem correctly in the fastest time.


The great Indian mathematical genius '[b]Srinavasa[/b] Ramanujan[b]' said that all numbers are his Friends[u].With this statement,i believe that this poor,yet,indian genius believe that each[b] numbers[/b] in the mathematical world had a distinctive[u] personalities.Can anyone tell me if there are available sources in the internet that explain the '[b]Number[/b] with its distinctive[b] personalities[/b]? I'm working on a particular thesis work related to game[/u] theory[u].[/u]

                                                  Respectfully'
                                                  A.Eddington


induction
number theory unsolved
number theory

$x_{1}$,$x_{2}$,...,$x_{n}$ and R are rational numbers.Also we have this equation:
$\sqrt{x_{1}}+\sqrt{x_{2}}+...+\sqrt{x_{n}}=R$
 Prove that;

$\sqrt{x_{1}}$,$\sqrt{x_{2}}$,...,$\sqrt{x_{n}}$ are rational numbers.

sory to my bad english :blush:

Induction with respect to $n$. $n=1$ is clear. Now if $\sqrt{x_{1}}+\cdots+\sqrt{x_{n}}=R_{1}$ and $\sqrt{x_{1}}+\cdots+\sqrt{x_{n+1}}=R_{2}$ are rational numbers, then $\sqrt{x_{n+1}}=R_{2}-R_{1}$ is a rational number.

Oops. :blush: A silly mistake.

And what if $\sqrt{x_{1}}+...+\sqrt{x_{n}}$ and $\sqrt{x_{n+1}}$ are both irrational?

See http://www.mathlinks.ro/Forum/viewtopic.php?t=27089 for a more general version.

Zetax I think it is different

İsn't it ?


algebra
polynomial
number theory proposed
number theory

$a,b\in\mathbb{N},b>1$ and f is a polynomial with integer coefficient such that
for all $n\in\mathbb{N}$:
$b^{n}|a^{n}+f(n)$
prove that $b|a$and $f(n)=0$for all $n\in\mathbb{N}$

[quote="hamedbakhshandeh"]very easy
find all a,b such as a<powel>-b+b<powel>2=b<powel>a+3[/quote]
what's that ??

find all a,b such as $a^{-b}+b^{2}=b^{a}+3$ ?? :ninja:

[quote="harazi"]Write the relation for $n$ and $n+1$, multiply the one for $n$ by $a$ and then subtract. You find that $b^{n}$ divides $f(n+1)-af(n)$. Because $b^{n}$ grows too fast, the last relation implies $f(n+1)=af(n)$ for sufficiently large $n$. This obviously implies the conclusion if $a>1$ and if $a=1$ everything follows from the growth of $b^{n}$.[/quote]thank harazi,can you sulve this one?: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=662470#662470[/url]

Yes, I can, but it was already dicussed on the forum. By the way, there exists an extremely easy solution if $f$ is not constant!  :wink:


hello guys, can anyone of you recommend me a free ebook of inequalitis thanks  :roll:  :roll:

http://can-hang2007.blogspot.com/
There're many materials of inequality in this blog.

http://www.gbas2010.wordpress.com - George Baskedis
http://www.irgoc.org/index.php?sid=dcebbe6b5f60fc58bf4b24eb3d8335ed -Chinese inequality forum
http://css-mu.blogspot.com/ - Nguyen Huy Tung


Euler
geometry
geometric transformation
vector
topology

let $ M_{g}$ be a closed orientable surface of genus $ g$, $ f: M_{g}\to M_{g}$ be a continuous map homotopic to the identity. For which $ g$ does it follow that $ f$ must have a fixed point?

thanks

the answer lies in lefschetz fixed-point theorem (you can look it up on wikipedia): the lefschetz number of a map homotopic to the identity is the euler characteristic of the surface. hence if the surface is not a torus, it doesn't vanish, and then the map has a fixed point. on the other hand translations are maps on the torus (seen as $ S^1\times S^1$) with no fixed points.

if you ask for the map to be a diffeomorphism, anyway, you don't (hopefully) need to invoke the fixed point theorem, but you can just use the correspondence (that seems at least plausible) between vector fields and diffeomorphisms isotopic to the identity.


[quote="mehdi"]hi all romanian mathematicians
i'm an iranian boy and i really need romanian mathematical and informatics olympiad contests.especially combinatorics problems.but this site doesn't have enough for me.would you say a site that has all the problems(or many of them) from long ago by now?thank you very much.[/quote]There are over 40,000 problmes on this site (mostly mathematics). If that is not enough for you, then nothing is.

Check the [url=http://www.mathlinks.ro/Forum/resources.php]resources section[/url]. Topic closed.


Like the subject says. I've taken it once so far, and will be a senior next year. I was just curious how many times people here have taken it. My parents want me to retake it since many of the kids at my school have taken it at least twice, but I don't really want to :( Is there any benefit to retaking the SAT if you can score higher (how much higher)?

There might be a benefit to retaking if you can score significantly higher. How much higher is "significantly" higher depends on how high you have already scored, in relation to the usual scores at the colleges to which you are applying.

I was wondering the same thing. I got a 2250 (760 V 800 M 690 W) and will probably apply to schools such as MIT and Stanford. If I think I can break 2300 with a little self-preparation (I'm not planning to spend the kind of time that taking a prep-course would require), would it be worth it to retake the SAT in October? It seems a lot of AoPSers have only taken it once, but I dunno, maybe it's because a lot of them got 2300+ scores on their first attempt  :P

From my experience, it's mainly the week before that counts, just so you can get accustomed with the test.

Now, if you went in cold and got a 2250, then with a bit of preparation, you would likely score much higher, and you really should take them again.

If you prepared for the 2250, then I guess it's really up to you. I did worse than you did on the SAT's, but I'm not retaking them in October because I think my time is better invested in other things.

In your case, since you did very well on Math and Verbal (which matter more than the writing), I don't know if it will be worth your time.

That's based on how I work though, it's really up to you though, and how you study and how your studying style affects your performance.


Here is a question about Taylor's approximations that I can't do.

[quote]1. Use a Maclaurin polynomial for the function f(x)=arccos(0.5+x) to estimate arccos(0.52) correct to 6 decimal places.
2. Find the Maclaurin power series expansion for f(x)=(3+x) to the power of -1, its radius of convergence, and the values of x for which it converges.[/quote]

Can anyone help me please? I really don't know where to start, if I did, I can do the problem better.

Hi, can someone help please? Sorry for double posting.


Consider the sequences of n terms in which each terms is -1,0 or 1.  Let Sn denote the number of such sequnces in which noterm of 0 occurs after a term of 1.  FInd a formula expressing Sn as a function of n.

Consider the sequences of n terms in which each term is -1,0 or 1.  Let Sn denote the number of such seqeunces that contain an even number of 1s.  Find a formula expressing Sn as a function of n.

Hints.
[hide="1"] Condition on the total number of $ 0$s and $ 1$s.  (This might not lead to a quick solution.)  [/hide]
[hide="2"] Also consider the number of sequences with an odd number of $ 1$s and write down a system of recursions. [/hide]

but how to you solve a second-order homogeneous linear difference equation with constant coefficients BUT plus a term that involves n such like 2*(3^(n-2))

The same way you solve a differential equation of a similar form:  write down the homogeneous solution, then find a particular solution.  In that particular case, you'd want to find the homogeneous solution and find a particular solution of the form $ c \cdot 3^n$.

This approach runs into difficulties when the recurrence and the extra term have common roots, same as in the differential equation scenario:  then you need to add polynomial terms, i.e. things like $ cn \cdot 3^n$.  Generally, if $ P$ is the characteristic polynomial of the homogeneous recurrence and $ Q$ is the characteristic polynomial of a recurrence that the extra term satisfies (in this case, $ x \minus{} 3$), the entire thing satisfies a larger recurrence with characteristic polynomial $ PQ$.

t0r: I think he's asking about nonautonomous recurrences, not inhomogeneous ones. Those are harder in general, and I don't believe there's a uniform way to solve them.
EDIT: Scratch that. cdymd, you called that a homogeneous recurrence; it isn't. ;)

That said, 1 can be written in the form of a second-order recurrence not dependent on $ n$, so you might want to rethink your method.

2 has a very nice solution without recursions.
[hide="Hint"]$ \minus{} 1$ to $ 0$, $ 0$ to $ 1$, and $ 1$ to $ \minus{} 1$.[/hide]
[hide="Solution Sketch"]Consider the operation that sends $ \minus{} 1$ to $ 0$, $ 0$ to $ 1$, and $ 1$ to $ \minus{} 1$. Take an arbitrary sequence with an even number of $ 1$s. Perform the operation on the first element. If that didn't change the number of $ 1$s, do it on the second element, and so on. Stop as soon as a $ 1$ is added or removed; you now have a sequence with an odd number of $ 1$s. You can show this process is reversible, so we have a bijection. The only time this algorithm fails is on the sequence of all $ \minus{} 1$s. This implies the number of sequences with an odd number of ones is exactly one less than the number of sequences with an even number of ones, and it readily follows the answer is $ \frac {3^n \plus{} 1}{2}$.[/hide]

Is the first one equals to S(n) = 3S(n-1) -S(n-2)?


analytic geometry
AIME
modular arithmetic
algebra
system of equations

In a rectangular array of points, with 5 rows and $N$ columns, the points are numbered consecutively from left to right beginning with the top row.  Thus the top row is numbered 1 through $N,$ the second row is numbered $N+1$ through $2N,$ and so forth.  Five points, $P_1, P_2, P_3, P_4,$ and $P_5,$ are selected so that each $P_i$ is in row $i.$  Let $x_i$ be the number associated with $P_i.$  Now renumber the array consecutively from top to bottom, beginning with the first column.  Let $y_i$ be the number associated with $P_i$ after the renumbering.  It is found that $x_1=y_2,$ $x_2=y_1,$ $x_3=y_4,$ $x_4=y_5,$ and $x_5=y_3.$  Find the smallest possible value of $N.$

anybody else want to provide a concise solution, i have an answer, but it took me a while, and it was messy

[hide]i used the idea $P_i$ has the row of $i$, and the column of $x_i-(i-1)n$

and $y_i$ in terms of the row and column is =5(column-1)+row

then i got a horrible system of equations and found that
$n\equiv 5 \mod 24$ and $365n+51 \equiv 0 \mod{124}$

then that was a pain to solve, but i got that the lowest value is $\boxed{149}$...? i am not sure though[/hide]

Anybody want to post a solution to this one please?

Does anyone have any clue how to do this? A hint, or a nice solution?

I'd probably do this one with coordinates, i.e. assign each point a coordinate corresponding to its row and column. You can then get the value of the point using the coordinates, then solve a system of equations. Seems pretty messy, but it is an AIME #11...

I think I did it the same way as Altheman...

[hide]
We need a coordinate system that we can write both $x_{i}$ and $y_{i}$ in terms of. There might be a better way, but I let $P_{i}= (i,a_{i})$. Then $x_{1}= a_{1}$, $x_{2}= N+a_{2}$, $x_{3}= 2N+a_{3}$, etc. and $y_{1}= 5(a_{1}-1)+1$, $y_{2}= 5(a_{2}-1)+2$, etc. Now we get the system of equations:
\begin{align}a_{1}&= 5(a_{2}-1)+2\\ N+a_{2}&= 5(a_{1}-1)+1\\ 2N+a_{3}&= 5(a_{4}-1)+4\\ 3N+a_{4}&= 5(a_{5}-1)+5\\ 4N+a_{5}&= 5(a_{3}-1)+3 \end{align}
We solve the system to get
\[\left(a_{1},a_{2},a_{3},a_{4},a_{5}\right) = \left(\frac{23+5N}{24},\frac{19+N}{24},\frac{51+117N}{124},\frac{5+73N}{124},\frac{7+89N}{124}\right). \]
$a_{1},a_{2}$ will be integers iff $N\equiv 5\pmod{24}$ and $a_{3},a_{4},a_{5}$ will be integers iff $N\equiv 25\pmod{125}$. Solving these congruences simultaneously by standard methods gives $N\equiv \boxed{149}\pmod{744}$.
[/hide]

Sorry to bring this back up. But how would you solve the three equations for $ a_3 , a_4, a_5 $ ?

chess64 gave a completely correct solution . In fact,  I got the exact variable by variable solution as his .

This was a nasty problem ngl .


1) In $\Delta ABC$ $D$ is the point on $\overline{BC}$ for which $\overline{AD}$ bisects median $\overline{BE}$. Find $BD: DC$

2) The length of a radius of a circle is $3+2\sqrt{3}$.  Three congruent circles are drawn in the interior of the original circle, each internally tangent to the original circle and externally tangent to the others.  Find the length of the radius of one of the three congruent circles.

3) Circles $O$ and $P$ have radii 16 and 9 respectively, and are tangent externally at point $A$.  Point $B$ is chosen on circle $O$ such that $AB=8$, and $BC$ is drawn tangent to circle $P$ at point $C$.  Compute $BC$.

[hide="(1)"]Let $M$ be the midpoint of $BE$.
The parallel line to $MD$ passing through $E$ intersects $BC$ at $H$. Then of course $D$ is the midpoint of $BH$, namely $BD=DH$.

In the triangle $ADC$ we have $EH\parallel AD$ and $E$ is the midpoint of $AC \Rightarrow  H$ is the midpoint of $DC$, that is $DH=DC$

And the ratio is $\frac{BD}{DC}=\frac{BD}{BD+DH+DC}=\frac{1}{3}$[/hide]
[hide="(2)"]Let $A,B,C$ be the centers of the three internal circles with radius $r$
Let $(O,R)$ be the external circle

$AB=BC=CA=2r \Rightarrow  \triangle ABC$ is equilateral

$OA=OB=OC=R-r \Rightarrow  O$ is the circumcenter of $\triangle ABC$

So, $AC = OC\sqrt{3}\Rightarrow  2r = (R-r)\sqrt 3$

$r(2+\sqrt 3) = R\sqrt 3 \Rightarrow  r=\frac{R\sqrt 3}{2+\sqrt 3}=\frac{2\cdot 3+3\sqrt 3}{2+\sqrt 3}= 3$[/hide]
[hide="(3)"]The ray $BA$ intersercts again the circle $P$ at the point  $D$. Then $BC^{2}= BA\cdot BD$.

The isosceles triangles $OBA,PDA$ are similar because $\angle OAB = \angle PAD$

So $\frac{AD}{BA}= \frac{PA}{OA}\Rightarrow \frac{BA+AD}{BA}= \frac{PA+OA}{OA}\Rightarrow  \frac{BD}{BA}= \frac{PA+OA}{OA}\Rightarrow $

$BD\cdot BA = \frac{PA+OA}{OA}\cdot BA^{2}\Rightarrow $

$BC^{2}= \frac{PA+OA}{OA}\cdot BA^{2}= \frac{9+16}{16}\cdot 8^{2}= \frac{5^{2}}{4^{2}}\cdot 8^{2}\Rightarrow $

$BC = \frac{5}{4}\cdot 8 = 10$[/hide]


So how did everyone do in this year's TEAMS competition?

I thought the state competition was relatively harder than last year. Especially because of the transportation and hybrid electric problems.

The national part was about the same difficulty.

TEAMS, it is part of the Junior Engineering Technical Society ([url=http://www.jets.org]JETS[/url]).
What happens during TEAMS is there are several "teams" from each high school that go under the "Junior Varsity" or "Varsity" category.  On one day in March they take a test composed of multiple chose and a seperate part of out-ended questions.  The questions are made up of about 6 or 7 engineering problems.  They provide you with the formulae and explain each of the topics.  They have examples on the [url=http://www.jets.org/teams/index.cfm]TEAMS website[/url].


Hi,

can somebody help me to solve this recurrence eq
\[ x_m\equal{}a_{m\minus{}1}x_{m\minus{}1}\minus{}b_{m\minus{}2}x_{m\minus{}2}\] ?
We suppose that $ a_{m\minus{}1},b_{m\minus{}2}$ are generally different and NON-constant coeffs. Is there a solution for $ x_k$ after a given number of iterations ($ k>2$)? $ x_0,x_1$ are known. If this is too general, can we assume something about $ a_{m\minus{}1},b_{m\minus{}2}$ to have a solution (eg, it must be positive or their ratio $ a_{m\minus{}1}/b_{m\minus{}2}>1$ or so)?

Thanks a lot

If you don't know anything special about the coefficients, there's nothing faster than just iterating the recursion/multiplying the matrices. Your proposed conditions don't do anything interesting, either.

Here is what I did, maybe it's not good...

Let's assume that we can write $ x_k$ in the following form: $ x_k \equal{}q^k$ Where $ q$ is some nonzero real number.

then we have:

$ q^m\equal{}a_{m\minus{}1}q^{m\minus{}1} \minus{} b_{m\minus{}2}q^{m\minus{}2}$

So:

$ q^2 \minus{}\frac{a_{m\minus{}1}}{b_{m\minus{}2}}q \plus{}1 \equal{}0$

From here we have for $ q$:

$ q\equal{}\frac12 \frac{a_{m\minus{}1}}{b_{m\minus{}2}} \pm \sqrt{\frac{a_{m\minus{}1}}{b_{m\minus{}2}}\minus{}4}$

So we only have a solution if $ \frac{a_{m\minus{}1}}{b_{m\minus{}2}} \geq 4$

This is a condition for a solution... :D , but I cant prove or not that anymore types of solutions exist... :huh:

yes but, actually, I'm not interested in the speed of the algorithm but in a general formula to conclude something about the solution
to yeahnigga: I think you cannot always assume a solution in the form you proposed and it is exactly the case.

Given any sequence of positive reals, I can choose $ a_i$ and $ b_i$ (including with the condition $ a_i > b_i$ or whatever) so that it obeys your recursion.  You're going to need much, much more solid information about the $ a_i, b_i$ if you hope to learn anything whatsoever about the sequences that result.  (Information about their absolute (not relative) sizes is the sort of thing that might allow for some concrete statements about rate of growth, etc.)


trigonometry
function
algebra
polynomial
calculus
calculus computations

The question is on the image below:

[url=http://img29.imageshack.us/i/ucf.jpg/][img]http://img29.imageshack.us/img29/3663/ucf.th.jpg[/img][/url]

can someone give me an explanation why it's a yes or no?

Is the right hand side this:
$ 5x\sin^2 x \plus{} 5x \cos^2 x$? or $ 5x \sin (x^2) \plus{} 5x \cos (x^2)$ ?

In order for you to use the method of undetermined coefficients, the function on the right hand side has to be a nice function [i.e Constants, polynomial, trigonometric functions, exponentials, etc] so that you are able to guess what it will look like in general.


geometry
circumcircle
power of a point
radical axis
geometry proposed

Let $ \Omega$ is circle with radius $ R$ and center $ O$. Let $ \omega$ is a circle inside of the $ \Omega$ with center $ I$ radius $ r$. $ X$ is variable point of $ \omega$ and tangent line of $ \omega$ pass through $ X$ intersect the circle $ \Omega$ at points $ A,B$. A line pass through $ X$ perpendicular with $ AI$ intersect $ \omega$ at $ Y$ distinct with $ X$.Let point $ C$ is symmetric to the point $ I$ with respect to the line $ XY$.Find the locus of circumcenter of triangle $ ABC$ when $ X$  varies on $ \omega$

[color=blue][b]Lemma.[/b]denote the power of point $ A$ wrt $ C(O,R)$ by $ \mathcal{P}_O(A)$, and let $ h$ be distance of $ A$ from radical axis of $ C(O,R)\ ,\ C(O',R')$, then we have $ \mathcal{P}_O(A)\minus{}\mathcal{P}_{O'}(A)\equal{}2h.OO'$[/color]
[i]Proof.[/i] To the Readers.

Denote the circumcircle of triangle $ ABC$ by $ \omega$.By applying the lemma we have :$ \mathcal{P}_O(I)\minus{}\mathcal{P}_{\omega}(I)\equal{}2IX.OO'$ (O' is center of $ \omega$) and cause $ \mathcal{P}_{\omega}(I)\equal{}2IX^2$ hence $ OO'$ is constant.


Level: Easy Chapter Mathcounts Problem

The total area of four squares (some may be of the same size), each with whole-number side
measurements is 23 square centimeters. In centimeters, what is the positive difference between
the perimeter of the largest square and the perimeter of the smallest square?

This is pretty easy, I just wanted to know if you can solve it without guess and check.

btw- Has this been posted before?

In 10 seconds, I was able to do guess and check.

[hide="my answer"]8[/hide]

I wish that there was an easier way to do this, but this is the best I know how.

Sorry. Mathgeeks right. I was stupid. Edited.

[hide]The areas of the squares are 9,9,4,1. The perimeters are 12,12,8,4. So the difference is 8.[/hide]

Billy

[hide]I can't think of any other way to do this problem other than guess and check, but finding four square numbers adding up to 23 yields 9, 9, 4, and 1, or squares with sides 3, 3, 2, and 1, or perimeters 12, 12, 8, and 4. 12-4=[b]8[/b][/hide]

[quote="236factorial"]you're not in mathcounts, DarkKnight??

I've done this problem before, and the answer is 8.

the perimeter is not 4 times the area.[/quote]

Haha... I wasn't thinking sry guys. I thought the area was a side for some reason. Sorry for all the confusion. *Whacks self on head* Yeah. I'm not in Mathcounts but i'm hoping to do it next year.


Compute the exact value of
\[ \sqrt\left\frac{1}{2^1} + \sqrt\left\frac{1}{2^1 2^2} + \sqrt\left\frac{1}{2^1 2^2 2^4} + \sqrt\left\frac{1}{2^1 2^2 2^4 2^8} + \cdots\]