ratio
geometry
3D geometry
prism
rectangle
LaTeX

[quote="grizzland"](Note that your answer is an algebraic expression involving D).
A circular swimming pool has a depth D and a radius that equals five times its depth. A rectangular pool has the same depth as the circular pool. Its width is four feet more than three times its depth and its length is two feet less than six times its depth. The ratio of the rectangular pool's volume to the circular pool's volume is ________________[/quote]
[hide]So the radius is 5D.
The height of the rectangle pool is also D.
The width is 3D+4.
The length is 6D-2.
The volume of the circular swimming pool is $25D^{3}\pi$.
The volume of the rectangular pool is $18D^{3}-2D^{2}-8D$.
We get as our final answer $\frac{18D^{2}-2D-8}{25D^{2}\pi}$.
[/hide]
Edit: Can someone tell me what I did wrong with this latex problem?
Edit2:thnx rwang2

[quote="grizzland"](Note that your answer is an algebraic expression involving D).
A circular swimming pool has a depth D and a radius that equals five times its depth. A rectangular pool has the same depth as the circular pool. Its width is four feet more than three times its depth and its length is two feet less than six times its depth. The ratio of the rectangular pool's volume to the circular pool's volume is ________________[/quote]

[hide="shows my laziness to write latex..."]
the circular pool (call its volume CV):

the radius (r) is 5 times the depth (D). therefore, r = 5D.

CV = D(25(pi)D^2) = 25(pi)D^3

the rectangular pool (call its volume RV):

the width (w) is 4+3D, the length is 6D-2

RV = D(4+3D)(6D-2) = 18D^3+18D^2-8D

RV/CV $={\frac{18D^{3}+18D^{2}-8D}{25D^{3}}}$

[/hide]

[quote="ckck"]Edit: Can someone tell me what I did wrong with this latex problem?[/quote]
You forgot the closing bracket.

[quote="Art of Owna"][quote="superwizard188"][hide]
D = D
r = 5d
L = 6D - 2
W = 3D + 4

V = Circular pool (cylinder) = D*(5D)^2 = 25D^3
V = Rectangular pool (rectangular prism) = D(6D - 2)(3D + 4) = (18D^3 - 18D^2 - 8D)

(18D^3-18D^3 - 8D)/25D^3 = [b](2(9D^2-9D-4)/25D^2[/b] I think.[/hide][/quote]

Arent you forgetting pi in your answer?  And I think the area of the rectangular one should be 18D^3[b]+[/b]18D^2-8D[/quote]

1. yeah they did forget $pi$

2. no, the [b]volume[/b] of the rectangular pool should be $18D^{3}+18D^{2}-8D$

so after all that the answer is

[hide]((18d^3)+(18d^2)-(8d))/((25d^3)(pi))[/hide]


Can somebody help me with this:

\[ \left|\begin{array}{cccccc}1 & 0 & 0 &\ldots & 0 & x\\ 1 &\binom{2}{1}& 0 &\ldots & 0 & x^{2}\\ 1 &\binom{3}{1}&\binom{3}{2}&\ldots & 0 & x^{3}\\ \ldots &\ldots &\ldots &\ldots &\ldots &\ldots\\ 1 &\binom{n}{1}&\binom{n}{2}&\ldots &\binom{n}{n\minus{}1}& x^{n}\\ 1 &\binom{n\plus{}1}{1}&\binom{n\plus{}1}{2}&\ldots &\binom{n\plus{}1}{n\minus{}1}& x^{n\plus{}1}\\ \end{array}\right|\]

tnx

I don't really see what the general element of the matrix is but I think it can be solved by using multiples of the first row to clear out everything in the last column except for its first entry. Then expand along the first row and you should have something much simpler.

[quote="§outh§tar"]I don't really see what the general element of the matrix is...[/quote]

In row $ k$ first $ k$ numbers are binomial coefficient $ \binom{k}{0},\binom{k}{1},...,\binom{k}{k\minus{}1}$ , then are $ (n\minus{}k\minus{}1)$ zeros and finally $ x^{k}$.

[quote="§outh§tar"]multiples of the first row to clear out everything in the last column except for its first entry. Then expand along the first row and you should have something much simpler.[/quote]

I don't see how can this help? In this way we get "ugly" determinant.

Do you know how to compute a determinant of this form?
$ {\begin{array}{*{20}c}. & . & 0 & 0 & 0\\ . & . & . & 0 & 0\\ . & . & . & . & 0\\ . & . & . & . & .\\ . & . & . & . & .\\ \end{array}}$

I have not explicitly solved it but I see that it is reducible to that form (times a constant), in which case it is easily solved. If you haven't gotten it by the time I'm back I will see to getting an answer.


Prove that in any convex 2x-gon,there is a diagonal not parallel to any side.  I know there is a solution, I can't figure it out tho? :blush:

[hide="Hmm.  Well... "]The sides in a $2n$-gon have at most $n$ distinct slopes, but looking at the diagonals from any given point, that's at least $2n-3$ different slopes.  And $n = 3$ should not be a hard special case. [/hide]


Is there a way to make a text size that is smaller than tiny (e.g., explicitly declare the font size) for a string of characters?

You could put the framebox round a parbox whose height is adjustable so \fbox{\parbox[t][3cm][c]{2cm}{text}} looks like this
$ \fbox{\parbox[b][3cm][c]{2cm}{text}}$
This extended version of \parbox has syntax: (the parameters in [ ] are optional)
\parbox[pos][height][inner_po]{width}{text} where
pos = t or b to align with the top or bottom edge of the box
height = height of the box
inner_pos = t,b,c,s whch put the text at the top, bottom, vertical centre or stretch
width = width of the box
text = text in the box

I edited my post; I have a different question now.


algebra
polynomial
limit
arithmetic sequence
algebra unsolved

If x>0, define by <x> the smaller of {x}, {1-x} where {z} denotes the fractional part of z. For any positive x, prove we can make <n^2*x> arbitrarily small (i.e. for any y>0, we can choose n s.t. <n^2*x> is less than y).

I think there's a theorem (due to Weyl) which states that if $P$ is a polynomial with at least an irrational coefficient then the terms of the sequence $\{P(n)\}$ are evenly distributed in $(0,1)$, i.e. if we denote with $S_n(a,b)$ the number of indexes $m$ in the set $\{1,2,\ldots,n\}$ for which $\{P(m)\}\in (a,b)$ then $\lim_{n\to \infty} \frac{S_n(a,b)}n=b-a,\ \forall\ a,b\in (0,1)$. This solves the problem, but I'm not sure I want to use something as strong as this..

Apparently, the density of $\{n^2x\}$ is pretty hard to prove (in your problem, only the case $x\in \mathbb{R\setminus Q}$ is interesting).

However, I found this proof for your problem in some old papers of mine (probably some preparation camp):

We call the distance between $a,b$ the length of the smallest arc connecting them (or rather their fractional parts) on the circle of length $1$. Assume we partition this circle in segments $S_i$ of length $<\frac {\varepsilon}2$. We also partition the set of naturals in $A_i$ s.t. $n\in A_i\iff \{\frac{n^2x}2\}\in S_i$. According to Van der Waerden's thm there is a set $A_i$ which contains three numbers in arithmetic progression, i.e. there are $m>n$ s.t. the distance between any two of $\frac {(m-n)^2x}2, \frac {m^2x}2, \frac {(m+n)^2x}2$ is $<\frac {\varepsilon}2$. This means that the distance between $\frac {(m+n)^2x}2-\frac {m^2x}2, \frac {m^2x}2-\frac {(m-n)^2x}2$ is $<\varepsilon$, but $(\frac {(m+n)^2x}2-\frac {m^2x}2)-(\frac {m^2x}2-\frac {(m-n)^2x}2)=n^2x$, so the distance between $n^2x$ and the origin is $<\varepsilon$, Q.E.D.

There might be some mistakes, but this is the main idea of the solution.

Sorry to be the ignorant one  :D , but could you explain or offer a link explaining exactly what "density" is? thanks!

[quote="grobber"]I think there's a theorem (due to Weyl) which states that if $P$ is a polynomial with at least an irrational coefficient then the terms of the sequence $\{P(n)\}$ are evenly distributed in $(0,1)$, i.e. if we denote with $S_{n}(a,b)$ the number of indexes $m$ in the set $\{1,2,\ldots,n\}$ for which $\{P(m)\}\in (a,b)$ then $\lim_{n\to \infty}\frac{S_{n}(a,b)}n=b-a,\ \forall\ a,b\in (0,1)$. This solves the problem, but I'm not sure I want to use something as strong as this..[/quote]
I think it refers to our beloved van der Corput's Theorem :)
See it on http://mathworld.wolfram.com/vanderCorputsTheorem.html


geometry
3D geometry
sphere
analytic geometry
graphing lines
slope
parameterization

Let A be the set of all point on the [b]2-sphere[/b] (the circumference) with both rational coordinates.
What is the closure of A?

Somehow I feel that it is the whole circumference but I can't guess how to prove that A is dense in it.[/b]

It suffices to consider the upper half of the unit circle. You can project this upper half bijectively onto the axis $ {\mathbb R}\times\{1\}$, and rational points on the circle correspond to rational points on $ {\mathbb R}\times\{1\}$. As the latter lie dense in $ {\mathbb R}\times\{1\}$, the rational points on the upper half of the unit circle lie dense in the circle, too.

The essential fact behind that: a line of rational slope starting from a rational point on the circle intersects the circle again in another rational point.

That projection doesn't project a semicircle to the whole line; it projects the whole circle except for one point (the center we project from) to the line.

[quote="mavropnevma"]Take the parametrization yielding all rational-coordinate points on the unit circle:
$ x = \frac {\lambda^2 - 1} {\lambda^2 + 1}$, ${ y = \frac {2\lambda} {\lambda^2 + 1}}$, $ \lambda \in \mathbb{Q}$.
Clearly it establishes a dense set on $ S^1$.[/quote]

Thank you!

Similarly, take the parametrization
$ x \equal{} \frac {\lambda^2 \plus{} \mu^2 \minus{} 1} {\lambda^2 \plus{} \mu^2 \plus{} 1}$, $ y \equal{} \frac {2\lambda} {\lambda^2 \plus{} \mu^2 \plus{} 1}$, $ z \equal{} \frac {2\mu} {\lambda^2 \plus{} \mu^2 \plus{} 1}$, with $ \lambda, \mu \in \mathbb{Q}$.
You can extend this to any dimension $ d > 3$!


find all real numbers $p,q$ such that $|\sqrt{1-x^{2}}-px-q|\leq\frac{\sqrt{2}-1}{3}\forall x\in[0;1]$

I don't think this is correct but:

we must find p and q so that lim (x->1/2) f(x)=q , this is equivalent with: sqrt(3)-p=2q

[hide="semisolution which is easy to complete"]Substitute $x=\sin t$ and examine min and max values of the function
$f(t) = p \sin t+q-\cos t$ on $t \in [0, \frac{\pi}{2}]$
For $p\le 0$ we have that $f(t)$ decrease hence it's min and max-values  lies on the ends $0$ and $\frac{\pi}{2}$
 ...
For $p > 0$ max-value is at point $t=arctan p$ and min-value is at the one of the ends of the segment.
 ... [/hide]


[quote="tranthanhnam"]find greast value and least value of expression $P =a(b-c)^{3}+b(c-a)^{3}+c(a-b)^{3}$ (  with $a,b,c \geq 0$and$a+b+c=1$)[/quote]
Notice that $P=(a-b)(b-c)(c-a)(a+b+c)= (a-b)(b-c)(c-a)$. It was discussed already at [url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=positive+lt&t=107821[/url].  Maximum is $\frac{\sqrt{3}}{18}$ and minimum is $-\frac{\sqrt{3}}{18}$. :D


Please, does anybody know which player has winning strategy in popular game "xo" on infinite square table?
Or on table $n\times n$ if one should put $k$ signs on the line?

Hi,

I recall the answer is that on an infnite board is that player 1 can win for k \le 5, and player 2 can force a draw for larger k. This implies that player 1 can win for large enough boards for k \le 5. I do not have a reference for this.

I would guess that if player 1 can win for a board, he can also win for any larger board, but I don't immediately see a proof. Finding the maximal n for which the game is a draw for k=4 or k=5 seems hard.

omer


What is the next term of this sequence?
1, 2, 3, 4, 5, 8, 7, ...


please pm me ur old avatar when u don't want it anymore.


calculus integration trigonometry logarithms algebra function domain

evaluate $ \int_{ \minus{} \pi}^{\pi} \ln((a^2 \plus{} b^2 \minus{} 2ab\cos x))\,dx$
where a, b are reals.
(It is equivalent to the original formulation)

For $ |a| < |b|$,
We have:
$ I \equal{} \frac {1}{2} \int \log (a^2 \plus{} b^2 \minus{} 2ab \cos x) dx$
$ \equal{} x\log a \plus{} \frac {1}{2} \int \log \left(1 \plus{} ( \minus{} \frac {b}{a})^2 \plus{} 2( \minus{} \frac {b}{a}) \cos x\right) dx$
$ \equal{} x\log a \plus{} \frac {1}{2} \int \log (1 \plus{} ( \minus{} \frac {b}{a})e^{ix}(1 \plus{} ( \minus{} \frac {b}{a})e^{ \minus{} ix}dx$
$ \equal{} x\log a \minus{} \sum_{n \equal{} 1}^{\infty} (\frac {b}{a})^{n}\frac {\sin xn}{n^2} \plus{} C$

The basic idea is to expand $ \ln (1 \minus{} a \cos x)$ as a power series [where the condition a < 1 justifies the validity of the power series] and then you integrate term by term.

Our integral:

\[ L=\int_{-\pi}^{\pi}\ln(a^2+b^2-2ab\cos x)\, dx = \int_{-\pi}^{\pi}\ln(a^2+b^2)\,dx + \int_{-\pi}^{\pi}\ln\left(1-\frac{2ab}{a^2+b^2}\cos x\right)\, dx\]

put

\[ \lambda=\frac{2ab}{a^2+b^2}\]

Then we have:

\[ =2\pi\ln(a^2+b^2) +\int_{-\pi}^{\pi}\ln(1-\lambda\cos x)\,dx\]

Now define:

\[ I(\lambda)=\int_{-\pi}^{\pi}\ln(1-\lambda\cos x)\,dx\]

Now differentiating under the integral sign (of course we can do it..., anytime.. :roll: ):

\[ \frac{d I(\lambda)}{d\lambda}=-\int_{-\pi}^{\pi}\frac{\cos x}{1-\lambda \cos x}\,dx\]

This integral can easily be calculated to be:

\[ \frac{d I(\lambda)}{d\lambda}=\frac{\pi}{\lambda}\left(1-\frac{1}{\sqrt{1-\lambda^2}}\right)\]

Integrating this easily:

\[ I(\lambda)=2pi\ln(1+\sqrt{1-\lambda^2})+C\]

Notice that $ I(1)=C$. So to find the constant:

\[ I(1)=\int_{-\pi}^{\pi}\ln(1-\cos x)\,dx=-\pi\ln4\]

So we have:

\[ I(\lambda)=2pi\ln\frac{1+\sqrt{1-\lambda^2}}{2}\]

hence the original integral:

\[ L=2\pi\ln(a^2+b^2)+2pi\ln\frac{1+\sqrt{1-\lambda^2}}{2}=2pi\ln\frac{1+\sqrt{1-\lambda^2}}{2}(a^2+b^2)\]

Writing $ \lambda$ back:

\[ L=2\pi\ln\frac{a^2+b^2+|a^2-b^2|}{2}\]

Now I didnt check anything if the integration under the integral sign is ok, or anything else with the domain of $ a,b$..., but hey I'm a physics student.... :D

In yeahnigga's post, in the line before "Integrating this easily:", there is a 2 missing in the numerator.

In evaluating I(1), using 1 - cos x = (sin (x/2))^2 which is symmetric about the origin, and the substitution y = x/2,
I get -4 pi ln2 = -2 pi ln4 instead of -pi ln4, as stated.

instead of I(1), eavaluate I(0) that is 0


function
algebra
functional equation
algebra proposed

For a function $ g: \mathbb{R}\rightarrow\mathbb{R}$, there exists a function $ f: \mathbb{R}\rightarrow\mathbb{R}$ such that for all real number $ x$, $ f(f(x)) \equal{} g(x)$ holds. Does there exist $ g$ such that there is only one $ f$ satisfying the condition?

$ \mbox{If } g(x)\equal{}x \mbox{ and }f(x)\equal{}a\minus{}x, \mbox{ then: }\\
 f(f(x))\equal{}f(a\minus{}x)\equal{}a\minus{}(a\minus{}x)\equal{}x\equal{}g(x); (\forall) x\in\mathbb{R}.$

[quote="mihai miculita"]$ \mbox{If } g(x) \equal{} x \mbox{ and }f(x) \equal{} a \minus{} x, \mbox{ then: } \\
f(f(x)) \equal{} f(a \minus{} x) \equal{} a \minus{} (a \minus{} x) \equal{} x \equal{} g(x); (\forall) x\in\mathbb{R}.$[/quote]

Sure, so $ \minus{}x$, $ 1\minus{}x$, $ 2\minus{}x$ are such that $ f(f(x))\equal{}g(x)$ and so $ g(x)\equal{}x$ is not the answer to the question.

The question is : does it exist a function $ g(x)$ such that the functional equation $ f(f(x))\equal{}g(x)$ has [u][b]exactly [/b][/u]one solution ?


geometry
3D geometry
tetrahedron
geometry proposed

Given reagular tetrahedron $ABCD,P$ is an arebitrary point inside it, let $P_{a},P_{b},P_{c},P_{d}$ be projection of $P$ to face $BCD,CDA,DAB,ABC$ reps $A'$ is point such that $\vec{PA'}=3\vec{PP_{a}}$ similar we get $B',C',D'$ prove that $ABCD,A'B'C'D'$ have the same centroid .

[hide="hint"]Same idea as following problem: Let $M$ be a point lies in equilateral triangle $ABC$ and $MA_{1}\perp BC,MB_{1}\perp CA, MC_{1}\perp AB$. Prove that three points : $M$, centroid of $ABC$, centroid of $A_{1}B_{1}C_{1}$ are lies same line.[/hide]

Can you show clearly? If talk as you all problems of solid geoemtry can be solved by plane geometry ? :D

No, But some problem maybe solved by used ideas of solution of problem similary in plane geometry. And this is the example!

After all what's your idea in the problem of plane ? why can't you say all in a topic, you want to prolong all your post !

Let $M$ be a point lies in equilateral triangle $ABC$ and $MA_{1}\perp BC,MB_{1}\perp CA, MC_{1}\perp AB$. Prove that three points : $M$, centroid of $ABC$, centroid of $A_{1}B_{1}C_{1}$ are lies same line.
Solution. Put $A_{2}B_{3}\parallel AB,B_{2}C_{3}\parallel BC,C_{2}A_{3}\parallel CA$ and $M=A_{2}B_{3}\cap B_{2}C_{3}\cap C_{2}A_{3}$. Then triangles $MA_{2}A_{3},MB_{2}B_{3},MC_{2}C_{3}$ are equilateral, and $3\overrightarrow{MG}=\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}=\overrightarrow{MC_{2}}+\overrightarrow{MB_{3}}+\overrightarrow{MB_{2}}+\overrightarrow{MA_{3}}+\overrightarrow{MA_{2}}+\overrightarrow{MC_{3}}=2(\overrightarrow{MA_{1}}+\overrightarrow{MB_{1}}+\overrightarrow{MC_{1}})=6\overrightarrow{MG_{1}}$, here $G$ is centroid of $ABC$ ,$G_{1}$ is centroid of $A_{1}B_{1}C_{1}$. Done!

Note: $G\equiv G_{2}$ if $G_{2}$ is centroid of $A_{2}B_{2}C_{2}$, here $\overrightarrow{MA_{2}}=2\overrightarrow{MA_{1}},\overrightarrow{MB_{2}}=2\overrightarrow{MB_{1}},\overrightarrow{MC_{2}}=2\overrightarrow{MC_{1}}$.

P.S: Please don't said like that, Now your problem is solved by same idea. I will not post it now, maybe someone trying find it.

I think you can't do for tetrahedron with the same idea, follow me your same idea is  draw throught $P$ four planes parallel four faces of tetrahedron assume them base a tetrahedron $PB_{a}C{}_{a}{}_{D}{}_{a}$ with face $(BCD)$ how do you prove $PB_{a}C_{a}D_{a}$ is regular ?

In my solution of your problem, $PB_{a}C_{a}D_{a}$ is similary for $ABCD$.

Oh I think there isn't concept of similar tetraherdon, or can you define two tetrahedron are similar, and moreover I think if you have that concept you can't prove they are similar easily in a line as you wrote, please understand here is a geometry problem, all thing must be clear.

I am sure that I have got a solution of this problem. But ''maybe someone trying find it'', wait two days.

P.S: I NOT want spam :P . But wait two days :D

Hey my good friend, now is four days from since your last post, what's your solution  :D

Given reagular tetrahedron $ABCD,P$ is an arebitrary point inside it, let $A_{1},B_{1},C_{1},D_{1}$ be projection of $P$ to face $BCD,CDA,DAB,ABC$ reps $A'$ is point such that $\vec{PA'}=3\vec{PP_{a}}$ similar we get $B',C',D'$ prove that $ABCD,A'B'C'D'$ have the same centroid .

Solution. Denote $(AB),(AC),(AD)$ lies on face $BCD$ such that $P(AB)\parallel AB,P(AC)\parallel AC,P(AD)\parallel AD$, etc. It is easy see that $P(AB)(AC)(AD)$ is equilateral tetrahedron and $A_{1}$ is centroid of triangle $(AB)(AC)(AD)$, etc.

By above we have
$\overrightarrow{P(AB)}+\overrightarrow{P(AC)}+\overrightarrow{P(AD)}=3\overrightarrow{PA_{1}}\\ \overrightarrow{P(BC)}+\overrightarrow{P(BD)}+\overrightarrow{P(BA)}=3\overrightarrow{PB_{1}}\\ \overrightarrow{P(CA)}+\overrightarrow{P(CB)}+\overrightarrow{P(CD)}=3\overrightarrow{PC_{1}}\\ \overrightarrow{P(DA)}+\overrightarrow{P(DB)}+\overrightarrow{P(DC)}=3\overrightarrow{PD_{1}}$.

Now note that $\overrightarrow{P(AB)}+\overrightarrow{P(CB)}+\overrightarrow{P(DB)}=\overrightarrow{PB}$, etc. We're done!


Whenever I do math, its very difficult for me to sit down with a book and work through it. I can't focus on one topic for too long. I think this is why I like to do practice contests, because there's different types of questions.

Instead, I've decided to try and diversify my learning through AoPS Volume 2 (which I really want to finish). I start on a chapter, and work through it a bit before I get too bored.
Then I move onto another important chapter in the book, and work through that.
And so on. I come back to the uncompleted chapters often.

I think this is a good idea, because by learning the stuff over a period of time, it also helps you retain the information and make more connections.

I was just wondering, does anyone else do this? Do you think this is a good idea?

I haven't had much experience with math yet.

When I did AoPS 2, I skipped around a bit, but usually sat through the whole chapter before moving on (though I've yet to finish everything)

I did make sure to interject some number theory between geometry chapters, etc, through, which did prevent things from getting boring.


geometry
3D geometry
sphere
calculus
calculus computations

If a liquid has density $ 3$ and velocity field $ v\equal{} \minus{}y{i}\plus{}x{j}\plus{}2z{k}$
How do you find the rate of flow outward through the sphere $ x^2\plus{}y^2\plus{}z^2\equal{}9$

[quote="qwesl"]If a liquid has density $ 3$ and velocity field $ v \equal{} \minus{} y{i} \plus{} x{j} \plus{} 2z{k}$
How do you find the rate of flow outward through the sphere $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 9$[/quote]

If you want us to do your homework you should pay... er, well, donate to the site or something.


Find the biggest positive integrer $n$ such that $4^{19}+4^{98}+4^{n}$ is a perfect square

Rust, [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=20038]computer_nuke[/url] don't understand why $176$ is the biggest solution. Can you explain with detail please???. Thanks.

[quote="BoesFX"]Let me explain this (may I?)

Assume there is solution for $n$ that is bigger than $176$, then since $4^{19}+4^{98}+4^{n}= (2^{19})^{2}(1+4^{7}9+4^{n-19})$ is a square, so is $1+4^{7}9+4^{n-19}$

Since $n-18>2(79)$, then
$(2^{n-19})^{2}< 4^{n-19}+4^{7}9+1 < (2^{n-19})^{2}+2^{n-18}+1=(2^{n-19}+1)^{2}$

Since $4^{n-19}+4^{7}9+1$ is between two consecutive squares, $4^{n-19}+4^{7}9+1$ is not a square.

So there are no solution for $n$ that is bigger than $176$

Since $n=176$ is a solution, then $n=176$ is the biggest solution

[Sorry, if it is too long][/quote]

That should be $4^{79}$, Rust made a typo in his solution. :wink:


search
inequalities unsolved
inequalities

Let $ ABC$ is a triangle. Prove that: $ \boxed{p^2\le2R^2 \plus{} 10Rr \minus{} r^2 \plus{} 2(R \minus{} 2r)\sqrt {R(R \minus{} 2r)}}$


hi,
   i need a proof of theorem 1.5 using only the field axioms and theorems 1.1 through 1.4, as is required by the excercise.... i have a proof using theorem 1.6 so no need to provide tht.... in case you dont have the apostol book i'm attaching a pic of the page of the book containing the the axioms and theorems to be used as well as the theorem to be proved i.e. theorem 1.5 on the pic.....

$ a(b\minus{}c)\equal{}a(b\plus{}(\minus{}c))\equal{}ab\plus{}(a\cdot \minus{}c)\equal{}ab\minus{}ac$, making use the of the distributive axiom and theorem 1.3.

but we have not yet proven that,
          $ a\cdot(\minus{}c) \equal{} \minus{}a\cdot c$

i have proof using theorem 1.6.... if i had the proof without it, i wouldn't be here asking you guys would i......???? although there is no point in writing tht proof here, since you have asked....

assuming theorem 1.6 holds

let
$ x \equal{} a\cdot ( \minus{} c)$

and let
$ y \equal{} \minus{} a\cdot c$

we have to prove $ x \equal{} y$, from which the argument of the previous poster completes without any holes

$ ac \plus{} x \equal{} ac \plus{} a( \minus{} c)$
$ \; \equal{} a (c \plus{} ( \minus{} c))$----------------by Distributive Law
$ \hspace{1in} \equal{} a (c \minus{} c)$--------------------by Theorem 1.3
$ \hspace{1in} \equal{} a.0$---------------------------by definition of -c

$ ac \plus{} y \equal{} ac \minus{} ac \equal{} 0$--------------by definition of -ac

now since we assumed 1.6 holds implying,

     $ a\cdot 0 \equal{} 0\cdot a \equal{} 0$

which implies,

$ ac \plus{} x \equal{} ac \plus{} y$
$ \Rightarrow x \equal{} y$--------------by Theorem 1.1

along with this JRav's argument holds

I glanced through it and it seems fine.

I never said I gave a complete proof.  As ZetaX said, you should provide some of the details.  :wink:

but dude i already had this proof using 1.6, remember!!!!
[quote="in my first post i"]....i have a proof using theorem 1.6 so no need to provide tht[/quote]

...... i'm still waiting....... after 2 whole days.....

The proof of theorem 1.6 does not hinge on theorem 1.5, so there's no reason not to use it.


For $a=b=\frac{1}{\sqrt{3}}$ it is not valid.

You are right, Ramanujan.

[quote="conceive"]Prove or disprove ($a,b>0;a^2+ab+b^2=1$):

$a^3+b^3\ge\frac{2}{3}$[/quote]

What about this ones:

($a,b>0;a^2+ab+b^2=1$):

$a^3+b^3\ge\frac{2}{3\sqrt{3}}$
$a^4+b^4\ge\frac{2}{9}$


Suppose $a_1,a_2,\cdots,a_8$ are eight distinct integers from $\{1,2,\cdots,16,17\}$.  Show that there is an integer $k > 0$ such that the equation $a_i - a_j = k$ has at least three different solutions.
Also, find a specific set of 7 distinct integers from $\{1,2,\ldots,16,17\}$ such that the equation $a_i - a_j = k$ does not have three distinct solutions for any $k > 0$.

Let me solve this old problem:

Let $a_{1}<a_{2}< ... <a_{8}$. Consider the values of $a_{2}-a_{1}$, $a_{3}-a_{2}$ ... $a_{8}-a_{7}$, and the values of $a_{3}-a_{1}$, ... $a_{8}-a_{6}$.

The sum of these $13$ numbers is $(2a_{8}+a_{7})-(2a_{1}+a_{2})$. The greatest value that this can take is $34+16-2-2=46$. However, if the $13$ values above did not repeat a third time anywhere, they would have sum $\geq 47 = 2(1+2+3+4+5+6)+7$.

For the second part, take $1$, $2$, $3$, $5$, $8$, $12$, $17$

Nice proof!

It also works to $ 1, 2, 4, 9, 14, 16, 17$

WLOG let $a_1\le a_2\dots\le a_8$ and let $b_i=a_{i+1}-a_i$ for $i=1,$ $2,$ $\dots,$ $7$. Suppose FTSOC that $a_i-a_j=k$ always has at most two solutions. Then, \[b_1+\dots+b_7\ge 1+1+2+2+3+3+4=16\] Notice we must have equality so $b_i$ must be $(1,1,2,2,3,3,4)$ in some order. Notice that $1$ and $2$ cannot be next to each other because that would yield a third solution to $a_i-a_j=3$. Then, the ones would be forced to be at the end because having two occurrences of a $1$ next to a $3$ would lead to $a_i-a_j=4$ to have at least three solutions. Hence, we need $b_i$ to be $(1,4,2,3,2,3,1)$ or reversed. Then, $a_3-a_1=a_5-a_3=a_7-a_5=5$, a contradiction. For the second part, $(a_1,\dots,a_7)=(1,2,3,5,8,12,17)$ works. $\square$

For part (a), sort the $a_i$ in increasing order, and look at the seven differences $d_i = a_{i+1} - a_i$ for $1 \leq i \leq 7$. Assume for the sake of contradiction that a counterexample existed. Then, $d_i = k$ at most twice for each $k$. Thus, the only possible tuple of $d_i$'s is $(d_1, d_2, d_3, d_4, d_5, d_6, d_7) = (1, 1, 2, 2, 3, 3, 4)$. But this obviously still does not satisfy the condition.

For part (b), we can use $\{1, 2, 3, 5, 8, 12, 17\}$.

[hide=solution]
Assume WLOG that $a_1\geq a_2\geq \dots\geq a_8.$ Now, suppose for the sake of contradiction. Clearly, we have $$(a_8-a_7)+(a_7-a_6)+\dots+(a_2-a_1) = (a_8-a_1)\geq 1+1+2+2+3+3+4 = 16.$$ In other words, we have $a_8 = 17,a_1 = 1,$ and $((a_8-a_7),\dots,(a_2-a_1))$ is a permutation of $1,1,2,2,3,3,4.$ Note that the sum of adjacent terms is equal to differences between two $a_i.$ The only possible numbers adjacent $1$ is $3$ and $4,$ because for $1$ and $2,$ that makes three difference of $2,3,$ which contradicts with our claim. Now, $3$ can't be adjacent twice, as that would yield $3$ differences of $4.$ Thus, $a_8-a_7 = a_2-a_1 = 1.$ Assume WLOG that $a_7-a_6 = 4,a_3-a_2 = 3.$ Thus, the only possible values of $)(a_8-a_7),(a_7-a_6),\dots,(a_2-a_1))$ are $(1,4,2,2,3,3,1),(1,4,2,3,2,3,1),$ and $(1,4,3,2,2,3,1).$ However, the first forms a contradiction with the difference $7,$ formed from $(1,4,2),(2,2,3),$ and $(3,3,1)$. The second and third both form contradictions with the difference $5,$ formed from $(1,4),(2,3),(2,3)$ and $(1,4),(3,2),(2,3),$ respectively. In other words, there always exists an integer $k>0$ such that the equation $a_i-a_j=k$ has at least three different solutions. When $8$ is replaced $7,$ a counterexample would be $\{a_1,a_2,\dots,a_8\} = \{16,13,10,8,6,2,1\}.$
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A sequence of positive integers $ \{a_n\}$ is defined by $ a_1\equal{}1,a_2\equal{}3,$ and $ a_n\equal{}(n\plus{}1)a_{n\minus{}1}\minus{}na_{n\minus{}2}$ (for $ n\ge3$). Find all positive integers $ n$ such that $ 11|a_n$.


If $ x$,$ y$ and $ z$ are positive real numbers prove that:
$ \frac {x^{2}}{x^{2} \plus{} xy \plus{} y^{2}} \plus{} \frac {y^{2}}{y^{2} \plus{} yz \plus{} z^{2}} \plus{} \frac {z^{2}}{z^{2} \plus{} zx \plus{} x^{2}}\geq1$

hello, your inequality is equivalent with
$ x^4y^2\plus{}x^2z^4\plus{}y^4z^2 \geq x^2y^3z\plus{}x^3yz^2\plus{}xy^2z^3$, this is true, because $ a^2\plus{}b^2\plus{}c^2\geq ab\plus{}ac\plus{}bc$ holds for all real $ a,b,c$, when we
put $ a\equal{}x^2y,b\equal{}xz^2,c\equal{}y^2z$.
Sonnhard.

Nice!  :D
Let $ \frac {y}{x} \equal{} a$, $ \frac {z}{y} \equal{} b$, $ \frac {x}{z} \equal{} c$. Then $ abc \equal{} 1$. Substituting, we have to show
$ \frac {1}{a^2 \plus{} a \plus{} 1} \plus{} \frac {1}{b^2 \plus{} b \plus{} 1} \plus{} \frac {1}{c^2 \plus{} c \plus{} 1} \ge 1$
Substitue $ c \equal{} \frac {1}{ab}$, it simplifies to
$ \frac {1}{a^2 \plus{} a \plus{} 1} \plus{} \frac {1}{b^2 \plus{} b \plus{} 1} \ge \frac {ab \plus{} 1}{a^2b^2 \plus{} ab \plus{} 1}$
Expanding, this is equivalent to
$ a^4b^2 \plus{} a^2b^4 \plus{} 1 \ge a^3b^3 \plus{} ab^2 \plus{} a^2b$*
By Muirhead, $ a^4b^2 \plus{} a^2b^4 \ge 2a^3b^3$ and $ ab^2 \plus{} a^2b \le a^3 \plus{} b^3$. So this simplifies to
$ a^3b^3 \plus{} 1 \ge a^3 \plus{} b^3$
$ (a^3 \minus{} 1)(b^3 \minus{} 1) \ge 0$
However, if the middle number of a,b,c is greater than 1, we can let a and b be the higher two of the three, but if the middle number is less than 1, we can let a and b be the smaller two of the three. Then the last inequality is true.
Of course, expanding right away with Muirhead works also.
Note that * must also be true for any a,b. We can prove this by letting $ a^2b \equal{} m$, $ ab^2 \equal{} n$, and $ 1 \equal{} p$, and then using $ m^2 \plus{} n^2 \plus{} p^2 \ge mn \plus{} np \plus{} pm$, similar to what Sonnhard did.


α)Να δειξεται οτι η εξισωση n(n-1)(n-2).....(n-k+1)=x^2  δεν εχει λυσεις για χ>0 (οι n,k,y ειναι θετικοι ακεραιοι.(θα ηθελα κυριοι εαν καποιος την λυσει να ψαξει και για δευτερο τροπο.)
β)Να ερευνησεται οσο περισσοτερο μπορειται την εξισωση:
n(n-1)(n-2).....(n-k+1)=k!x^2 (αν οι n,k,x θετικοι ακεραιοι η εξισωση εχει απειρες λυσεις για καθε k ? ). :wink:

το α ερωτημα ειναι η γενικευση της ασκησης οτι το ν! δεν ειναι τελειο τετραγωνο. αρα ειναι προφανες πως λυνεται...

Ναι Κωστα εχεις δικιο αλλα 2 διαφορετικες λυσεις για το α ερωτημα ειναι δυσκολο να βρεθουν.Παντος  θα ηθελα να κοιταξεται ιδιαιτερα το β ερωτημα. :)

$\\ \beta )$ Xρησιμοποιώντας το θεώρημα του [b]Bertrand[/b]   :D  βρήκα κάτι...

Αν $n=2m$ και $k =m$ τότε $x^{2}= \frac{2m(2m-1)...(m+1)}{m!}$

Όμως υπάρχει πρώτος $p$ με $2m >p > m$ άρα η εξίσωση δεν έχει λύση .

Αν $n =2m+1\,\, , k=m+1$ τότε $x^{2}= \frac{(2m+1)2m...(m+2)}{m!}$.

Σε αυτή την περίπτωση υπάρχει πρώτος $p$ με $2(m+1) > p > m+1$.

 :)

μια αλλη προσεγγιση ειναι η εξης:το γινομενο στην α) γραφεται σαν το συμβολο του Pochhammer. για καποιο αριθμο(εδω το ν-κ+1)αυτο ειναι ισο με το Γ(x+1) προς Γ(x+1-n), οπου Γ ειναι η συναρτηση Γ.,αλλα ειναι πολυπλο το ολοκληρωμα.( χρησημοποιουμε τη συναρτηση Β).μπορει κανενας να βοηθησει? :oops:

Δύσκολα θα βγάλουμε κάτι φαίνεται δύσκολη προσέγγιση... :maybe:

μαλλον εχεις δικιο Ραμανατζαν.καταλειγω οτι πρεπει το  B(X,N)/N πρεπει να ειναι τελειο τετραγωνο.(κατι που τελικα δεν ειναι) αλλα μαλλον δεν εχει προοπτικες....ωραια η λυση σου παντως  :)


Into how many parts at most is a plane cut by n lines?

I used the recursion.
But the book used other approach that I can't understand.
It derived #of parts=$\left(\frac{n}{0}\right)+\left(\frac{n}{1}\right)+\left(\frac{n}{2}\right)$
BTW, above is the combinations not fractions..(i don't know how to write Comb. in Latex)
 :blush:
Can someone derive this?? Thanks,

[hide]
We start with one plane. Each line that is drawn contributes another piece. Also, each time one line intersects another as it is being drawn, another piece of the plane will be cut in two. So the number of pieces of the plane is $1+\text{number of lines}+\text{number of intersections}$.

There are $n$ lines, and since two lines intersect in at most one point, the number of intersections is at most $\binom{n}{2}$. So there are at most $1+n+\binom{n}{2}$ pieces (this is attainable when no two lines are parallel and no three are concurrent).
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[quote="t0rajir0u"]This can of course be generalized to other dimensions.  For $m$ lines in $m$ dimensions it gives $2^m$.  :)[/quote]

A line in three dimension cannot separate the space.

And I think in higher dimension, the subject matter for similiar

situation should be hyperplane?  :)

Yes, yes, that's what I meant.  I should say "cuts," since in three dimensions a cut constitutes a plane, etc.

[quote="kimby_102"]i don't know how to write Comb. in Latex[/quote] \binom{m}{n}


superior algebra
superior algebra solved

Could any one help me prove this question:

V = {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}
V' = {(1 2), (3 4), (1 2)(3 4), e}
where "e" is the identity.

Prove that the permutation groups V and V' are isomorphic.

Thanks a lot!

there are six isomorphisms

just map e to e
and then you can choose ! just make sure you map every element on a different element!

Also, note that these groups are actually isomorphic to the Klein group $\{e,a,b,c\}$ (where $a$, $b$ and $c$ satisfy $a^2 = b^2 = c^2 = e$ and $ab = c,\ bc = a,\ ca = b$).

true arne, there are only two groups of order four : cyclic and klein group

as soon as you find two involutions in your group ->klein group


$ABCD$ is a parallelogram, and $P$ is a point on $CD$ such that $CP: PD=3: 2$. Draw a perpendicular line from $A$ to $BP$, and let $H$ be its intersection. If $AD=AH=HB=BC=3$, find the area of parallelogram $ABCD$.

We know AHB = 90, so the angles of the parallelogram are 45 and 135, and since AH = BH = 3, the sides are 3 and $3 \sqrt 2$.  Then the height is $\frac{3 \sqrt 2 } 2$, and the area, which we note is the product of the longer side and the height, since the triangles at the ends are 45-45-90 when drawing the height in, so the area is $9$.

That's not correct...  :(
The angles of the parallelogram are not $45^\circ$ and $135^\circ$...

Yeah, I thought it seemed too easy... I don't see how I got that, though, but it is wrong.

click and enlarge for better quality
[hide="thanks for being patient with the picture"]
i drew in a few extra points
$BC = AD = BH = AH = CK = DK = 3\\ \bigtriangleup BHA \sim \bigtriangleup CKD \sim \bigtriangleup BRQ \sim \bigtriangleup QIA \sim \bigtriangleup PJC \sim \bigtriangleup DSP$
and they're all right isosceles
thus $BA = CD = 3\sqrt{2}$
because of the 3:2 ratio, $CP = QA = \frac{9\sqrt{2}}{5}$ and $PD = BQ = \frac{6\sqrt{2}}{5}$
thus $BR = RQ = JK = HI = PS = SD = \frac{6}{5}$
and $CJ = BH = JP = QI = KS = IA = \frac{9}{5}$
by some pythagoras, $JH = KI = \frac{3}{5}$
so i think the total area comes out to be
$\begin{eqnarray*} \frac{21}{5}(\frac{9}{5})+\frac{21}{5}(\frac{6}{5})&=&\frac{21}{5}(\frac{15}{5})\\ &=&\frac{63}{5}$
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geometry
parallelogram
geometric transformation
reflection
combinatorial geometry
combinatorics proposed
combinatorics

Find all finite sets $S$ of points in the plane with the following property: for any three distinct points $A,B,$ and $C$ in $S,$ there is a fourth point $D$ in $S$ such that  $A,B,C,$ and $D$ are the vertices of a parallelogram (in some order).

Here's my solution, hope it's correct - it's very easy to make mistakes in combinatorial geometry.
Consider the convex hull of the points. For one side $ A_{1}A_{2}$ (where these two points are the endpoints of that edge), consider the point $ A_{3}$ that maximizes the area of $ A_{1}A_{2}A_{3}$. Now consider the point $ D$ that completes the parallelogram of these 3 points. Since $ D$ must lie on, or inside the convex hull, it is clear that $ DA_{3}$ is parallel to $ A_{1}A_{2}$- so for each side in the convex hull polygon, there is a side parallel to it.
For any edge $ A_{1}A_{2}$ consider the parallel edge $ B_{1}B_{2}$. Suppose they are not of equal length, Wlog $ A_{1}A_{2}$ is longer. Consider the point that completes the parallelogram $ A_{1}A_{2}B_{1}$ - this point must lie outside the segment $ B_{1}B_{2}$, which is a contradiction. Thus for any side, it must have a side parallel to it of equal length.
Suppose on edge $ A_{1}A_{2}$, there exists another point lying on that segment, $ X$. Consider the edge parallel to $ A_{1}A_{2}$, $ B_{1}B_{2}$. Any point that completes the parallelogram $ B_{1}B_{2}X$ must lie outside the convex hull, our outside the segment $ A_{1}A_{2}$, contradiction. So no edge in the convex hull polygon contains 3, or more, vertices.
Label the vertices $ A_{1}A_{2} ... A_{n}$, in cyclic order. Let $ A_{k}A_{k\plus{}1}$ be the parallel edge to $ A_{1}A_{2}$. Suppose $ A_{k\plus{}1} A_{k\plus{}2}$ is not parallel to $ A_{2}A_{3}$. Suppose instead, $ A_{t}A_{t\plus{}1}$ is parallel to $ A_{k\plus{}1}A_{k\plus{}2}$, with $ t \ge 3$. Consider the edge parallel to $ A_{2}A_{3}$. By drawing a diagram, we can see such an edge cannot be part of the convex hull - contradiction. Thus, we can group the vertices into pairs $ (A_{i}, B_{i})$, so that $ A_{i}A_{i\plus{}1}$ is parallel to $ B_{i}B_{i\plus{}1}$.
Consider such a grouping. We can prove that the midpoint of $ A_{i}B_{i}$ is independent $ i$: Let $ O$ be the 'center' of parellelogram $ A_{1}A_{2}B_{1}B_{2}$. Since $ OA_{2} \equal{} OB_{2}$ and $ A_{2}A_{3}$ is parallel to, and equal in length, with, $ B_{2}B_{3}$. Thus $ O$ is also the 'centre' of parellelogram $ A_{2}A_{3}B_{2}B_{3}$. Proceed in the same fashion around the convex hull polygon.
Now suppose the convex hull has $ >4$ points. Consider 3 points $ A_{1}A_{3}B_{2}$. It's easy to see any point completing the parellelogram of these $ 3$ points lies outside the convex hull, a contradiction.
Thus the convex hull has $ 2$ points, in which case there are no other points in the set, (since these 2 points constitute the convex hull); OR there are $ 4$ points in the convex hull, constituting a parallelogram. If there is a point $ T$ inside this convex hull, consider the 3 points $ T, A_{1}, A_{2}$ - the point completing parallelogram must lie outside the convex hull, a contradiction. So we have our second solution set - $ 4$ points constituting a parellelogram.
But we are not done yet - so far we have assumed not all the points are collinear. So we have the case when our points all lie on a line. Let the endpoints be $ A_{1}$ and $ B_{1}$. Let $ O$ be the midpoint of $ A_{1}B_{1}$. For any point $ X$ on the line, by considering the 3 points, $ X, A_{1}, B_{1}$, it's clear that the reflection of $ X$ about $ A_{1}B_{1}$ also lies in our set. Let $ A_{i}$ be the points left of $ O$, and $ B_{i}$ right of $ O$, so that $ A_{i}O \equal{} B_{i}O$. Suppose that $ O$ does not lie in our set - if it does, the following method with slight changes will work.
Consider points $ A_{1}A_{i}B_{2}$ and $ A_{1}B_{i}B_{2}$, for $ i \ge 3$. Since $ A_{i}X > B_{1}B_{2}$, for each $ X\equal{}A_{3}, A_{4} ... B_{4}, B_{3}$. Let $ O'$ be the midpoint of $ A_{1}B_{2}$. For each point in the set $ A_{3} ... B_{3}$, the reflection of that point about $ O'$ also lies in the set. So we can pair up the points. It's clear that if say, $ Y$ lies to the left of $ X$, the reflection, $ Y'$ of $ Y$, lies to the right of the reflection, $ X'$, of $ X$. $ B_{3}$ can't be reflected to $ A_{3}$ or any point to the right of it, since $ A_{1}A_{3} > B{2}B{3}$; thus $ B_{3}$ is reflected to $ A_{2}$. Let $ x_{i} \equal{} A_{i}A_{i\plus{}1}$. The fact that $ B_{3}B_{2}\equal{}A_{1}A_{2}$, gives us $ x_{1}\equal{}x_{2}$. Next, $ B_{4}$ cannot go to $ A_{4}$ or any point to the right of it, since $ B_{2}B_{4} < A_{1}A_{4}$, thus $ B_{4}$ goes to $ A_{3}$, giving us $ x_{2}\equal{}x_{3}$. Continuing in this process, we have $ B_{k\plus{}1}$ goes to $ A_{k}$ for $ k\equal{}2,3...n\minus{}1$, and also $ A_{n}$ goes to itself. The equalities that arise from these facts, give us $ x_{1}\equal{}x_{2} \equal{} ... \equal{}x_{n\minus{}1}\equal{}A_{n}B_{n}$, so we have proven that all $ 2n$ points are equally spaced.
If $ O$ is also part of the set, then we can prove in the same method that all $ 2n\plus{}1$ points are equally spaced. Thus we have a new solution set: any set $ n$ equally spaced points, for any integer $ n$. However, in this proof (of the part when points are collinear), we have assumed that there are at least $ 5$ points - if there are $ 4$ collinear points, they don't have to be equally spaced, as long as $ A_{1}A_{2}\equal{}A_{3}A_{4}$ where the points are $ A_{i}$ going from left to right.
In conclusion solution sets: a parallelogram, a degenerate parellogram that is a line, and any set of equally spaced points.

VERY clever solution due to Ubemaya:
We claim all sets S are 4 points that are vertices of a parallelogram, and sets with less than 3 points.
Suppose there was a set with more than 4 points that would satisfy the conditions.
Take the triangle of maximal area, and let it be ABC. A quadrilateral ABCD contains 4 triangles congruent to ABC, so it must be the convex hull, as a point outside ABCD would result in a triangle of greater area than ABC. There must be some 5th point E inside ABCD because there are more than 4 points. There must be another point F such that AEBF is a parallelogram. However F cannot be inside ABCD, which is the convex hull, so we have a contradiction, and so there cannot be a set with more than 4 points that satisfies the property.

The above post is too clever! here's a different, less slick way, similar to [b]epitomy01[/b].

Let $|S| = n$. Let the convex hull of all $n$ points be $C$. Take two adjacent, distinct points on $C$, say $A$ and $B$. Let the point on $C$ with the maximal perpendicular distance to $\overline{AB}$ (where this means the whole line, not just the line segment) be $P$, and let this perpendicular distance be $d$. Let the slope of $\overline{AB} = \theta$, and the line with slope $\theta$ passing through $C$ be $l$. Observe that $l$ either passes through no points in $S$, or they pass through points in $S$ that have a perpendicular distance of $d$ to $\overline{AB}$, ie points in $S$ on $l$. This is because, if there's a point with perpendicular distance to $\overline{AB}$ greater than $d$, then it lies outside the convex hull by definition of $P$. To further rigorize, you can take a parallel line to $l$, say $l'$ with perpendicular distance to $\overline{AB}$ greater than $d$, then all the points on $l'$ are outside $C$, contradiction.

Suppose $P$ does not lie on $\overline{AB}$, and that there exist no points in $S$ on $l$ besides $P$. Observe that for any three distinct points $M, N, O$ in plane, the only points with $M, N, O$ as vertices of a parallelogram are $M$ reflected over the midpoint of $NO$, $N$ reflected by the midpoint of $OM$, and $O$ reflected over midpoint of $MN$. Then if $P$ is reflected over $AB$ it lies outside $C$, and same for $B$ over $AP$, and $A$ over $BP$. But this means is a contradiction! Therefore, at least one more point lies on $l$.

Case 1: no $3$ points collinear
Let's change our notation. Let the first $m$ points in counterclockwise order be $A_1, A_2, \cdots, A_m$ and the next $p$ points in counterclockwise order be $B_1, B_2, \cdots, B_p$. By our work above, we have that $C$ is a convex polygon with $A_jA_{j+1} \parallel B_jB_{j+1}$ for all $j \le \min(m, n)$. WLOG, $\min{m, p} = p$. The definition of a parallelogram yields the midpoint of $A_iB_i$, say $O$, is fixed for all $i \le n$. Suppose for the sake of contradiction that $p \ge 3$. If $O$ lies outside $\triangle A_1A_3B_2$, this contradicts the counterclockwise ordering of the $A$s then $B$s, so $O$ lies inside $\triangle A_1A_3B_2$. Then $B_2O = A_2O$, so $B_2$ reflected over $A_1A_3$ lies outside $C$. Similarly, we have $A_1$ reflected over $A_3B_2$ and $A_3$ reflected over $A_1B_2$ lies outside $C$. Both of these cases are contradictions, so $p \le 2$. Hence $n \le 4$. The solution here is a parallelogram with $4$ points, any $2$ points, any $1$ point, or the empty set.

Case 2: there exist $3$ collinear points.
Let the points be $X_1, X_2, X_3$ on line $l_1$. If there exists no other points on $l_1$, then it is not hard to see that the solution set is an arbitrary number of equally spaced points on $l$.

Else, there must exist a point not on $l_1$. Then there exists a point $Y_2$ such that $X_1X_2Y_1Y_2$ is a parallelogram. Let $\overline{Y_1Y_2} = l_2$. There must exist a similar point $Y_3$ on $l_2$. Suppose there are $k$ points on $l$. By a straightforward argument, then there exists $k$ points on $l_2$. Continue in this fashion, so we have $n$ lines $l_1, \cdots, l_n$ with $k$ points on each of those lines. Suppose that there are only $kn$ points in $S$. But then we can easily see that there is one more point in $S$ by the problem condition, so there are an infinite number of points in $S$, contradiction. Thus there are no solutions in this case.

Therefore the solutions are an arbitrary number of equally spaced, distinct, collinear points, a parallelogram with $4$ points, any $2$ points, any $1$ point, or the empty set and we are done.

Solution with [b]yunseo[/b].

Suppose $S$ has $n+1$ points for $n\ge 3$ and not all the points are collinear. Then, we have that no three $A$, $B$, $C$ are collinear, since there is no fourth $D$ to make a parallelogram. Thus, no three points are collinear. We now essentially reduce to and solve the one dimensional version of this problem.

Clearly we may pick a line $\ell$ such that the projections of the points in $S$ onto $\ell$ are all distinct (there are only $\binom{n+1}{2}$ bad directions for $\ell$). Imposing a coordinate system on $\ell$, we may assume that the projections are \[S=\{0=x_0<x_1<x_2<\cdots<x_{n-1}<x_n\}.\] Suppose the point corresponding to $x_i$ is $A_i$. For any $1\le i\ne j\le n$, we have some $k\ne 0,i,j$ such that $A_0A_iA_jA_k$ is a parallelogram in some order. Thus, we have some $k\ne 0,i,j$ such that \[x_k\in\{x_i+x_j,x_i-x_j,x_j-x_i\}.\] This is actually enough to show that $n\le 3$.

Note that $x_n+x_i>x_n$ and $x_i-x_n<0$, so we have $x_n-x_i\in S$. Thus, \[x_n-x_1>x_n-x_2>\cdots>x_n-x_{n-1}\] are all in $S$. Since $x_n-x_1>x_0=0$ and $x_n-x_1<x_n$, these must be \[x_{n-1}>x_{n-2}>\cdots>x_1,\] so we have $\boxed{x_i+x_{n-i}=x_n}$ for all $1\le i\le n-1$.

We now do a similar more careful analysis with $x_{n-1}$. Note that $x_{n-1}+x_i>x_{n-1}+x_1=x_n$ for $i\ge 2$, so again, we have \[x_{n-1}-x_2>x_{n-1}-x_3>\cdots>x_{n-1}-x_{n-2}\] all in $S$. Note that $x_{n-1}-x_2<x_n-x_2=x_{n-2}$, so we have that these are exactly \[x_{n-3}>x_{n-4}>\cdots>x_1,\] so $\boxed{x_i+x_{n-1-i}=x_{n-1}}$ for all $1\le i\le n-2$. Subtracting the two boxed equations, we get  \[x_{n-i}-x_{n-1-i}=x_n-x_{n-1}\] for all $1\le i\le n-2$ (assuming $n\ge 4$). This implies that $x_i$ is an arithmetic sequence, so we may write $x_i=di$ for some $d$.

If $n$ is even, then we have $0,dn/2,dn\in S$. We have $dn+dn/2>dn$, so we have $dn-dn/2\in S$. But we want $x_k\in\{x_i+x_j,x_i-x_j,x_j-x_i\}$ for $k\ne i,j$, so this is not possible.

Similarly, if $n\ge 5$ is odd, then $0,d(n-1)/2,d(n-1)$ are in $S$, and a similar argument gives a contradiction. Thus, we have $n\le 3$.

So we have at most $4$ points, so the only allowed configurations are a parallelogram, and any configuration of at most $2$ points. We also have the configuration where all the points are collinear, but this doesn't work for three or more points. So the answer is any set with at most two points, along with a parallelogram.

We claim that $S$ can have atmost 4 points. The construction is easy.  For the proof first , pick the Triangle $ABC$ which has maximal area. Now consider $D$ such that $ABCD$ is a parallelogram. Note that if there is a point $P$ outside $ABCD$ ,then atleast one of the areas of  the triangles $ABP$ , $BCP$ ,$CDP$ ,$DAP$ must be larger than the area of $ABC$ (Contradiction!) .Similarly if P lies inside $ABCD$ ,then consider $P_1$ , $P_2$ , $P_3$ ,$P_4$ such that $APBP_1$ ,$BPCP_2$ , $CPDP_3$ ,$DPAP_4$ . It is clear that atleast one of the $P_i$ lie outside $ABCD$ ,and we arrive at a contradiction


A1=x and A2=x-1 and An=A(n-1)-A(n-2) when n>2. Find A15.

hello, if i have not made a mistake on my sheet of paper so $ a_{15}$ should be $ \minus{}1$.
Sonnhard.


limit
LaTeX
logarithms
real analysis
real analysis unsolved

The limit you are looking for is nothing else
but the famous Glaisher-Kinkelin constant
which it is denoted in the mathematical literature by $A$.

[quote="Cezanne123"]For each integer
\[n \ge 1 \]
set
\[x_{n}= 1^{1}\cdot 2^{2}\cdot 3^{3}\cdot \ldots \cdot n^{n}\]
Prove that
\[\mathop{\lim }_{n \to \infty }\frac{{x_{n}}}{{(e^{-{\textstyle{{n^{2}}\over 4}}}\cdot n^{{\textstyle{1 \over 2}}n^{2}+{\textstyle{1 \over 2}}n+{\textstyle{1 \over{12}}}})}}\]
exists and is a non-zero real number.[/quote]
lim=1
very very simpl! I can prove that $1*2^{2}...*n^{n}$~$((n^{1/2}/e^{1/4})^{n^{2}}*n^{{n/2}+1/12}$ and after is simpl :lol:

[quote="JBL"]Your latex is all messed up; what you should have written (using "\ln" for $\ln$ in place of just "ln," use "\cdot" for $\cdot$ in place of "*," and fixing your bracketing) in the final line was

$x_{n}\sim e^{(1/2)(n^{2}+n+\frac{1}{6})\ln n-n^{2}/4+C_{1}+o(1)}= n^{(n^{2}+n+1/6)/2}e^{-n^{2}/4+C_{1}-o(1)}$ where $C_{1}$ is constant

which is obviously not good enough to get your claimed result that the limit is 1 (which is false) since it gives that the limit is $e^{c}$ for some constant $c$.[/quote]
yes my latex is bad. but you are not right when rewrite my sulution because $x_{n}$ is equal this $e^{(1/2)(n^{2}+n+\frac{1}{6})\ln n-n^{2}/4+C_{1}+o(1)}$ when you write $\sim$ it's not correct, because answer include $o(1)$, and still I don't understand Why you write $-o(1)$ instead $o(1)$??? yes answer is equal $e^{C_{1}}$ but I think $C_{1}$=0
by the stolts theorem we can say that there exist this constant  $C_{1}$ but we cant' say is it equal 0?? I think it is. let's prove it. :lol:

[quote="JBL"]$o(1) =-o(1)$.  It was a product of mechanically rewriting your formula and had no intended meaning.

I can't understand whether you are making assertations or asking questions, but we can't prove it because it's false: the value of this limit, as didilica wrote, is the [url=http://mathworld.wolfram.com/Glaisher-KinkelinConstant.html]Glaisher-Kinkelin Constant[/url].[/quote]
Sorry  :huh: .

[size=200]Set[/size]              [img]http://www.mathlinks.ro/Forum/latexrender/pictures/daf12d3041e4ec52b521567b0e871ddb.gif[/img]    [size=200]=L[/size]

[img]http://bbs.pep.com.cn/attachments/vKvP3mEy_l3BoYDrUqkr9.gif[/img]

kg2100 you mean $C'=0$, but Glaisher-Kinkelin Constant.? Glaisher-Kinkelin said $C'=1.2824271291006226369$ and by mathematica v.4.0 calculate it.

to JBL
[quote="Extremal"][quote="§outh§tar"]Extremal, please prove it for us since you know it[/quote]
OK it's not problem  :lol: .
Easy find by Stolts theorem this
$\sum_{1\leq k\leq n}klnk=\frac{1}{2}(n^{2}+n+\frac{1}{6})lnn-\frac{1}{4}n^{2}+C_{1}+o(1)$ therefore
$e^{\sum_{1\leq k\leq n}klnk}=e^{\frac{1}{2}(n^{2}+n+\frac{1}{6})lnn-\frac{1}{4}n^{2}+C_{1}+o(1)}$
where
$e^{\sum_{1\leq k\leq n}klnk}$=$\prod_{1\leq k\leq n}k^{k}$
and
$e^{(1/2)(n^{2}+n+\frac{1}{6})lnn-(1/4)n^{2}+C_{1}+o(1)}$=$(n^{(1/2)(n^{2}+n+(1/6)))(e^{(1/4)n^{2}-C_{1}+o(1)})^{-1}}$ where $C_{1}$ is constant
I have more good result than this limit. :wink:[/quote]
then you has written
[quote="JBL"] which is obviously not good enough to get your claimed result that the limit is 1 (which is false) since it gives that the limit is $e^{c}$ for some constant $c$.[/quote]
but in my seccond post here I has corrected my answer
 when I written [quote="Extremal"]. where $C_{1}$ is constant
I have more good result than this limit. :wink:[/quote] this already mean $lim=e^{c}$


trigonometry
inequalities unsolved
inequalities

Let $C \leq B \leq A \leq \frac{\pi}{2}$ are angles of triange $ABC$.Find minimum of $cos(\frac{A-B}{2}).sin\frac{A}{2}.sin\frac{B}{2}$.

$cos\frac{A-B}{2}(cos\frac{A-B}{2} - cos\frac{A+B}{2})$

Fix C.  This fixes A+B.  A-B ranges from 0 to $\pi /2 - 3C$.  We obviously take the latter.

So it remains to find the minimum of $\cos \frac{\pi/2 - 3C}{2}$.

This is $\frac{1}{\sqrt{2}}(cos\frac{3C}{2} + sin\frac{3C}{2})$.

Let x = 3C/2 for convenience.  We want to minimize $cos x + sin x$ in the range $0 \le x \le \pi /4
$.  This is $\sqrt{1 + 2cosxsinx} \ge 1$.  So the minimum occurs when $x = 0$.

It is easy to see that $A = \pi /2$ and $B=C=0$ now.  The minimum is zero.

[quote="Singular"] $A = \pi /2$ and $B=C=0$ [/quote]

$A,B,C$ are angles of triange?You are wrong :D

oops, didnt see that condition.

i dont think its fatal.  We just need to find new restriction on A-B.  This restriction is $A-B \in [0, \min \{\pi/2 - C, \pi - 3C\}]$.

Again, we want to minimize $cos \frac{A-B}{2}$, which happens when $\frac{A-B}{2}$ is largest.

So we have two cases: $A-B =  \pi/2 - C \le \pi - 3C$, or $A-B =  \pi - 3C \le \pi/2 - C $.

=============================================

First case:  $C \le \pi /4$.

We minimize $cos ( \pi /4 - C/2) = \frac{1}{\sqrt{2}}(cos\frac{C}{2} + sin\frac{C}{2})$.

Let $x = C/2$; we want to minimize $cos x + sin x = \sqrt{1 + 2cosxsinx} \ge 1$, since $0 \le x \le \pi /8$.

This corresponds to 90, 90, 0 as the angles.
=============================================

Second case:  $C \ge \pi /4$.

We minimize $\cos \frac{\pi/2 - 3C}{2}$.

This is $\frac{1}{\sqrt{2}}(cos\frac{3C}{2} + sin\frac{3C}{2})$.

Let $x = 3C/2$ for convenience.  We want to minimize $cos x + sin x$ in the range $3\pi /8  \le x \le \pi /2
$.  I am stuck here.

This should be easy for someone else to finish though.


geometry
ratio
cyclic quadrilateral
geometry solved

let  ABC  be a triangle  whose angles are in proportion 4:2:1 in that order. prove that.
1/a + 1/b= 1/c
i coulndt seem to get this in my pre tst yesterday it was the only one i didnt solve. yetti i think this one is for you  :P

Let's be a bit more precise about which angles of triangle ABC are in which ratio:

[color=blue][b]Problem.[/b] Let ABC be a triangle such that A : B : C = 4 : 2 : 1. Prove that $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$.[/color]

[i]Solution.[/i] The proportion A : B : C = 4 : 2 : 1 and the equation $A+B+C=\pi$ (sum of angles in triangle ABC) together yield $A=\frac{4\pi}{7}$, $B=\frac{2\pi}{7}$, $C=\frac{\pi}{7}$.

Now, consider a regular 7-gon $A_1A_2A_3A_4A_5A_6A_7$ with its center O. Then, $\measuredangle A_1OA_2=\frac{2\pi}{7}$, but since O is the center of the circle through the points $A_1$, $A_2$, $A_3$, $A_4$, $A_5$, $A_6$, $A_7$, the central angle theorem yields $\measuredangle A_1OA_2=2\cdot\measuredangle A_1A_4A_2$. Thus, $2\cdot\measuredangle A_1A_4A_2=\frac{2\pi}{7}$, so that $\measuredangle A_1A_4A_2=\frac{\pi}{7}$. Similarly, $\measuredangle A_4A_1A_3=\frac{\pi}{7}$ and $\measuredangle A_3A_1A_2=\frac{\pi}{7}$, so that $\measuredangle A_4A_1A_2=\measuredangle A_4A_1A_3+\measuredangle A_3A_1A_2=\frac{\pi}{7}+\frac{\pi}{7}=\frac{2\pi}{7}$.

Now, comparing $\measuredangle A_1A_4A_2=\frac{\pi}{7}$ and $\measuredangle A_4A_1A_2=\frac{2\pi}{7}$ with $C=\frac{\pi}{7}$ and $B=\frac{2\pi}{7}$, we see that $C=\measuredangle A_1A_4A_2$ and $B=\measuredangle A_4A_1A_2$. In other words, $\measuredangle BCA=\measuredangle A_1A_4A_2$ and $\measuredangle CBA=\measuredangle A_4A_1A_2$. Hence, the triangles ABC and $A_2A_1A_4$ are similar. Thus, $BC:CA:AB=A_1A_4:A_4A_2:A_2A_1$. In other words, $a:b:c=A_1A_4:A_4A_2:A_2A_1$. Hence, instead of proving the equation $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$, it will be enough to show that $\frac{1}{A_1A_4}+\frac{1}{A_4A_2}=\frac{1}{A_2A_1}$. This is equivalent to $A_4A_2\cdot A_2A_1+A_2A_1\cdot A_1A_4=A_1A_4\cdot A_4A_2$. But from the symmetries of a regular 7-gon, $A_2A_1=A_1A_7$, $A_1A_4=A_4A_7$ and $A_4A_2=A_2A_7$, so that this equation is equivalent to $A_4A_2\cdot A_1A_7+A_2A_1\cdot A_4A_7=A_1A_4\cdot A_2A_7$. But this equation is clearly true, since it follows from the Ptolemy theorem, applied to the cyclic quadrilateral $A_1A_2A_4A_7$ (cyclic because the 7-gon $A_1A_2A_3A_4A_5A_6A_7$ is regular).

So the problem is solved. By the way, it is a really classical problem (I remember seeing it in a number of books).

  darij

Here's another proof: take $A'$ on $AB$ and $C'$ on $A'C$ such that $\angle ACB =\angle BCA'$ and triangle $BCC'$ is isosceles (see attached figure). Now, it is easy to see that triangles $ABC$ and $AA'C$ are similar, while triangles $ABC$ and $C'BA'$ are equal.
It follows that $\frac {A'C}{AC}=\frac {BC}{AB}$, that is, $\frac {a+b}{b}=\frac{a}{c}$, which is equivalent to $\frac1a +\frac 1b =\frac 1c$.

See alike problem at http://www.mathlinks.ro/Forum/viewtopic.php?t=45515


the given equation is nothing but d/dx{f(x)*f '(x)}............

Dei ranga i thing u read the qsn wrongly. it's '-' not '+'

anyways,
[hide]
heres an idea. i couldn't proceed much..  :maybe:

multiply and divide by $ (f(x))^2$
the resulting polynomial will be $ (f(x))^2 * \frac{d(f'(x)/f(x))}{dx}$
since $ (f(x))^2$ is always +ve, the question reduces to finding where
$ \frac{d(f'(x)/f(x))}{dx}$ is positive
i.e. to find where some function $ g(x) \equal{}  \frac{f'(x)}{f(x)}$ is increasing.

after this, i'm a bit lost :(

[/hide]

Ranga, so what if it were that???

Shrids, yes your g(x) is the right way to proceed. But why dont you think of ways of expressing f(x).


function
inequalities
trigonometry
algebra unsolved
algebra

let $ {\angle A ,\angle B ,   \angle C}$ be three angles of $ {\triangle ABC}$
function :$ {y\equal{}cotA\plus{}\frac{2sinA}{cosA\plus{}cos(B\minus{}C)}}$,then find the mininum value of this function.

Can anybody solve this problem,Dear mathlinkers?  :)

here is my solution for your problem
$ y\equal{}cotA\plus{}\frac{sinA}{sinBsinC}\equal{}\frac{cosAsinBsinC\plus{}(sinA)^2}{sinAsinBsinC}\equal{}\frac{(sinA)^2\plus{}(sinB)^2\plus{}(sinC)^2}{2sinAsinBsinC}$
then use the inequality $ sinAsinBsinC\le\frac{3\sqrt{3}}{8}$

[quote="CCMath1"]let $ {\angle A ,\angle B , \angle C}$ be three angles of $ {\triangle ABC}$
function :$ {y \equal{} cotA \plus{} \frac {2sinA}{cosA \plus{} cos(B \minus{} C)}}$,then find the mininum value of this function.[/quote]
$ \cos(B\minus{}C) \le 1$.
$ \cot A \equal{} \frac{\cot^2 \frac{A}{2} \minus{} 1}{2\cot \frac{A}{2}}$,   $ \sin A \equal{} \frac{2\cot \frac{A}{2}}{\cot^2 \frac{A}{2} \plus{} 1}$,   $ \cos A \equal{} \frac{\cot^2 \frac{A}{2} \minus{} 1}{\cot^2 \frac{A}{2} \plus{} 1}$
Letting $ t \equal{} \cot \frac{A}{2} \in (0;\plus{}\infty)$ we have:
$ y \ge \frac{t^2\minus{}1}{2t} \plus{} \frac{4}{2t} \equal{} \frac{t^2\plus{}3}{2t}$
Using AM-GM we get $ t^2\plus{}3 \ge 2\sqrt{3}t$ and hence $ \frac{t^2\plus{}3}{2t} \ge \sqrt{3}$.
So $ y \ge \sqrt{3}$. Equality whenever $ t\equal{}\sqrt{3}$, $ B\equal{}C$. $ t \equal{} \sqrt{3} \iff \cot \frac{A}{2} \equal{} \sqrt{3} \iff A \equal{} B \equal{} C \equal{} 60^\circ$

Very nice  solutions, :yup:  Thanks a lot Mathias_DK,hoangvn.fix. :)


Let $C$ be a (probably infinite) family of subsets of $\mathbb{N}$ such that for every chain $C_{1}\subset C_{2}\subset \ldots$ of members of $C$, there is a member of $C$ containing all of them. Show that there is a member of $C$ such that no other member of $C$ contains it!

The conclusion is clearly false, for example let $C$ be any finite set without a maximum element.  I'm guessing that we're being asked to prove Zorn's Lemma for subsets of $\mathbb{N}$, which says that there must be a _maximal_ element.

That's easy enough:  We will construct a maximal set $S$ by going through the natural numbers in order, and adding $k$ to $S$ so long as $S$ will still be contained in some set in $C$.  The chain condition insures that our final set $S$ will be contained in $C$.  And if there is a $T \in C$ such that $S \subsetneq T$, then take $k = \min \{T \backslash S\}$.  When we got to $k$ in the construction of $S$, we would have added $k$, as $S$ would still be in $T$, so $k \in S$, a contradiction. So $S$ is maximal in $C$.

[quote="Deedlit"]The conclusion is clearly false, for example let $C$ be any finite set without a maximum element.  I'm guessing that we're being asked to prove Zorn's Lemma for subsets of $\mathbb{N}$, which says that there must be a _maximal_ element. [/quote]

Thanks for the correction :)


calculus
integration
function
limit
real analysis
inequalities
real analysis unsolved

This is from our midsems paper of today. I am looking for a simple solution.

Suppose a function $f$ is nonnegative and continuous on $[a,b]$ with $\sup_{x \in [a,b]}f(x) = M$.
Show that $\lim_{n \rightarrow \infty} \left(\int_{[a,b]}f^{n}\right)^\frac1n = M$.

Here's a proof of the analogous result for Lebesgue integrals.

Namely, what we are trying to show is that 'If $f\in L^{p}\cap L^{\infty}$ for some $p<\infty$, so that $f\in L^{q}$ for $q>p$, then $\|f\|_{\infty}= \lim_{q\to\infty}\|f\|_{q}$'

Take $t\in [0,\|f\|_{\infty})$.  If we set $A_{t}=\{x\in X: |f(x)|<t\}$, then clearly $m(A_{t})>0$.  By Chebyshev's inequality, we get
\[\|f\|_{q}\geq tm(A_{t})^{\frac{1}{q}}\]
If $m(A_{t})<\infty$, then $\lim_{q}m(A_{t})^{\frac{1}{q}}= 1$, and if $m(A_{t})=\infty$, then $m(A_{t})^{\frac{1}{q}}=\infty$, so that
\[\liminf_{q\to\infty}\|f\|_{q}\geq t \]
and since $t$ is arbitrary in $[0,\|f\|_{\infty})$, we get
\[\liminf_{q\to\infty}\|f\|_{q}\geq \|f\|_{\infty}\]
For the other direction, take $p$ such that $f\in L^{p}$.  Then for all $q\in (p,\infty)$, we have
\[\|f\|_{q}\leq \|f\|^{p/q}_{p}\|f\|^{1-p/q}_{\infty}\]
Letting $q\to\infty$ (so that $\|f\|^{p/q}_{p}\to 1$ and $\|f\|^{1-p/q}_{\infty}\to \|f\|_{\infty}$), we see that
\[\limsup_{q\to\infty}\|f\|_{q}\leq \|f\|_{\infty}\]
whence
\[\lim_{q\to\infty}\|f\|_{q}=\|f\|_{\infty}\]
as desired.

For Riemann integrals, the essential idea is to split the integral into two pieces, one of which has $f(x)\ge M-\epsilon$. That piece has positive length $L$, so the integral of $f^{p}$ on it is at least $L(M-\epsilon)^{p}$. Taking $p$th roots, the limit is at least $M-\epsilon$. Repeat for all $\epsilon>0$, and the limit is at least $M$.


[quote="236factorial"]The sum of Sally's age and the square of Melissa's age is 209. THe sum of Melissa's age and the square of Sally's age is 183. What is the sum of the ages?

I remembered this one the best I could.[/quote]

[hide]  We know that among M and S, one of them is odd, while the other is even.  We can also conclude M is larger than S.  Finally, by inspection M and S should be relatively close together.

Hence, we can re-write M as $S + 2n-1$

  Since we have already stated S and M aren't very far apart, we first try $2n-1=1$

  Solving the equation and simplifying yields: $S=13$ and $M=14$

  Checking this we find: $196 + 13 = 209$ and $169 + 14 = 183$  which of course works.
[/hide]

  This isn't a very good solution, but its all I can think of right now..

[quote]The sum of Sally's age and the square of Melissa's age is 209. THe sum of Melissa's age and the square of Sally's age is 183. What is the sum of the ages?
[/quote]

[hide]Knowing that $S + M^2$ and $M + S^2$ are 26 apart, we know that S and M are within 1, 2, or 3. If both $S + M^2$ and $M + S^2$ are odd, one of the numbers must be odd. Therefore, a difference of 2 is not posssible.

Try 10 and 13. The answers would be 113 and 179. We see that, for one thing, the numbers are too small, and for another thing, the difference is too great.

We can see that a difference of 1 is the only difference possible.

Try 12 and 13. The answers are 157 and 181. This is a good difference, but the numbers are still a bit too small.
Trying 13 and 14, the answers come out 183 and 209.
That means the answer is $13 + 14 = \boxed {27}.$[/hide]


logarithms
calculus
calculus computations

Find the derivation:
$y = \log _a x$  ;)
$y^n = ?$

$y = \frac{1}{{1 - x}}$
$y^n = ?$


find all primes $p,q$
such that $2^{p}=q^{q}+q+2$

Suppose that $p>2$ and $q>3$. We have:
$2(2^{p-1}-1)=q(q^{q-1}+1)$
Since $p$ is odd $2^{p-1}-1$ is divisible by 3. But $q^{q-1}+1 \equiv 2 \mod{3}$ since $q$ is also odd. Contradiction.
Let $p=2$. Then it's clear that there are no solutions. Suppose that $q=2$. Then $p=3$. Finally, suppose that $q=3$. Then $p=5$.

So: $(p,q) \in \{(3,2), (5,3)\}$


limit
trigonometry
function
integration
calculus
real analysis
complex numbers

Can someone help me evaluate this:

$\mathop\lim_{n\to\infty}\sum_{k=0}^{n/2}\frac{Sin[k\pi/n]}{n}$

Mathematica says it's $1/\pi$ but I don't know how to determine this.  Just a hint would be ok with me if you like.

Thanks,

Let $n=2l$, then your limit equals:

$\lim_{l\rightarrow \infty}\frac{1}{2l}\sum_{k=0}^{l}\sin \frac{k\pi}{2l}$

The trig. sum, denoted by $S_{l}$, can be calculated exactly, it equals:

$S_{l}=\frac{\sin \frac{\pi}{4}\cos \left(\frac{\pi}{2l}+\frac{\pi(l-1)}{4l}\right)}{\sin \frac{\pi}{4l}}$, hence, via the well-known limit $\frac{\sin x}{x}\rightarrow 1$, you get that

$\frac{S_{l}}{2l}\rightarrow \frac{1}{\pi}$.

Let $m=n/2$. The sum becomes $\frac{1}{\pi}\sum_{k=0}^{m}\sin\left(\frac{k \pi}{2m}\right)\frac{\pi}{2m}$, which is $(1/\pi)$ times a Riemann sum [hide]for the function $f(x)=\sin x$ on the interval $[0,\pi/2]$.[/hide]

Thanks guys.  Nice!  In case anyone's interested, this stems from this problem:

$\mathop\lim_{n\to\infty}\int_{0}^{\infty}\frac{dx}{x^{n}+1}$

which can be evaulated via residues.  One limit to evaluate in this analysis is:

$\mathop\lim_{n\to\infty}\frac{1}{n}e^{\pi i/n}\sum_{k=0}^{n/2}e^{(2k\pi i)/n}$

which is $i/\pi$.  I reduced that limit to an evaluation of the limit I asked about.

I know, what am I doing messin' with this if I can't even do the real part . . .

For calculating that limit you can use this trick:

The substitution $x=y^{1/n}$ transforms your integral to:

$\frac{1}{n}\int_{0}^{\infty}\frac{y^{1-\frac{1}{n}}}{1+y}dy=\frac{1}{n}\beta(\frac{1}{n},1-\frac{1}{n})=\frac{\pi}{n\sin \pi/n}\rightarrow 1$.

Thanks Didi.  Very good! Then perhaps we can use your results and mine to conclude:

$\frac{1}{1-e^{(n-1)\pi i/n}}\frac{-2\pi i}{n}e^{\pi i/n}\sum_{k=0}^{(n-3)/2}e^{2k\pi i/n}= \frac{\pi}{n Sin[\pi/n]}\quad n>2\;\text{and odd}$

Or just formulate it for all n.  May work on it . . .

There is a much easier contour to chose:  An arc in the first quadrant with an argument of $\frac{2\pi}{n}$.  Then only one singularity lies within the contour and this result can then be generalized to all $n\in\mathbb{R}, \ n >1$. Also, the sum would have probably been easier to evaluate if you treated it as a geometric sum.


Hello, I would like to inform everyone that there is a facebook group about IMO 2007 on facebook.

I think facebook is good to stay in touch and to some extent chatting. But for sharing videos youtube is more appropriate. And facebook redudes the resolution of pictures. I rather recommend using the ML/AoPS photo album.

Orl you beat me to it :P I also suggest we use the forums, and/or the Picure Gallery (and Youtube through youtube tags) to share information.


mikham ye soale ghadimi bedam vali omidvaram ke rahe hal haye ghashang va jadidi era'e bedid.
ba farz $ x,y > 0$ va $ x \plus{} y \equal{} 1$ sabet konid:
  $ \frac {x}{\sqrt {1 \minus{} x}} \plus{} \frac {y}{\sqrt {1 \minus{} y}} \geq \sqrt {2}$

in karo man ghablan kardam :lol:

http://www.mathlinks.ro/viewtopic.php?t=205994 :)

vali kollan kheyli soale ghashangie :roll:

There exists a similar topic. Do not do spamming as before jensen. :diablo:

[color=red][Topic Locked by moderator][/color]


A 2 x 2 x 12 rectangular hole in a wall is to be filled with twenty-four 1 x 1 x 2 bricks. In how many different ways can this be done if the bricks are identical?

Find the partitions of 12 numbers 1 and 2.

12 = 12 1's
       10 1's and one 2...
       6 twos

Hence, there are such 7 such partitions.

Or would it be 7x2x2=28?

I'm going with 7 actually since the bricks.

7 ways.

[quote="nilknarf"]it can't be 7 ways, there are more than 7 ways to do this even if the hole was 2x2x2[/quote]

Then it is 28. 7 partitions of 12, 2 of 2, 2 of 2.

2x2x7 = 28 ways!

its not 28 either, i can't describe why, but i know the answer should be in the thousands or millions.

[quote="neelnal"][quote="i_like_pie"]I set the minimum at 30 million (yes, I'm sure). :ninja:[/quote]

 Maybe 28!
 Can we get a solution if there is one? Or actually give me some more time.[/quote]
$28!=3.04888345\cdot10^{29}$

That's 304 octillion.

[quote="nilknarf"]uhhh how did u get that?![/quote]
$28!=1\cdot2\cdot3\cdots28=304888344611713860501504000000$

[quote="nilknarf"]umm...why would it be 28![/quote]
Neelnal suggested $28!$

ugh... I couldn't get this problem on an ARML homework set, only it was a 2x2x17 hole. [hide] If I recall correctly, the solution they gave used some sort of recursion.[/hide]

i found the solution by googling, can any1 tell me if the solution is correct and explain it to me in a way a normal kid can understand
[hide]
http://www.mathpropress.com/archive/iams/vol8.ascii
[quote]Problem:

A 2x2x12 hole in a wall is to be filled with 24 1x1x2 bricks.  In how
many different ways can this be done if the bricks are
indistinguishable?

From: a_rubin@dsg4.dse.beckman.com

Let a_n     be the number of ways of filling a 2x2xn hole with 1x1x2
            bricks.
Let b_{n+1} be the number of ways of filling a 2x2xn hole with a 1x2x1
            distiguished brick alreadyh placed on an end.

Clearly, by considering the posible placement of bricks on the distiguished
end,

a_n = 2 a_{n-1} + a_{n-2} + 4 b_{n-1}
b_n = a_{n-1} + b_{n-1}

a_0 = 1, a_{-1} = 0, b_0 = 0

Formal solution produces:

a_n = ( (2+sqrt(3))^(n+1) + (2-sqrt(3))^(n+1)) / 6 + (-1)^n / 3
    = nearest integer to ((2+sqrt(3))^(n+1)/6

Using this, or evaluating the individual terms, we get:

a_12 = 4,541,161[/quote][/hide]

any1 got this? btw what did u mean by recursive, working my way from 2x2x1 till 2x2x12?


I'm good at all kinds of problems except counting and probability. For example, AMC 12A 2007 #25. I understand the explanation. But at test, I just don't know what to do.

To sum up, my question is when you see counting problem, what do you do first?

[quote="Treething"]You count?[/quote]

of course you count, that's why it's called counting

but counting in a quick way.

Usually, the trick is dividing the problem into a few main, logical cases.

For each case, some things to take a look at are:
- can you solve it directly? (straightforward combinatorics)
- can you draw a bijection to a simpler situation? (such as bit strings or sets)
- can you write it in terms of previous cases? (recursion)
- do you see any meaningful patterns?
- if you can't work with the cases holistically, can you break them into even smaller cases?

Really, you need to approach each problem individually. Looking at past problems will help give you ideas as to possible ways of approaching new ones, because not every counting problem will be 100% original.

[quote="car202"]AMC 12A 2007 #25.[/quote]

This particular question has a really simple recursion that I think is a lot easier than the combinatorial approach.

Recursion in some form is applicable to a lot of hard combinatorial questions, especially ones where it seems like the answer for $n = 12$ (for example) is directly related to the answers for $n = 11, 10, ...$.

yea intermediate counting is almost 99% of the time either straight up counting, balls n urns, or recursion


On December $ 17$, $ 1903$, at Kitty Hawk, N.C., the $ 1903$ Wright Flyer became the first powered, heavier-than-air machine to achieve controlled, sustained flight with a pilot aboard.

\begin{tabular}[t]{|l|c|c|c|}
\multicolumn{4}{c}{\textbf{December 17, 1903 Flights}}\\\hline
 &\textbf{Pilot}&\textbf{Time in Flight}&\textbf{Distance}\\\hline
\textbf{First Flight}&Orville Wright&$ 12$ seconds&$ 37$ meters\\\hline
\textbf{Longest Flight}&Wilbur Wright&$ 59$ seconds&$ 260$ meters\\\hline
\end{tabular}

The average speed for the first flight was $ x$ meters per second.  The average speed for the longest flight was $ y$ meters per second.  What is the average of $ x$ and $ y$? Express your answer as a decimal to the nearest tenth.

[quote="GameBot"]On December $ 17$, $ 1903$, at Kitty Hawk, N.C., the $ 1903$ Wright Flyer became the first powered, heavier-than-air machine to achieve controlled, sustained flight with a pilot aboard.

\begin{tabular}[t]{|l|c|c|c|} \multicolumn{4}{c}{\textbf{December 17, 1903 Flights}} \\
\hline & \textbf{Pilot} & \textbf{Time in Flight} & \textbf{Distance} \\
\hline \textbf{First Flight} & Orville Wright & 12 seconds & 37 meters \\
\hline \textbf{Longest Flight} & Wilbur Wright & 59 seconds & 260 meters \\
\hline \end{tabular}

The average speed for the first flight was $ x$ meters per second.  The average speed for the longest flight was $ y$ meters per second.  What is the average of $ x$ and $ y$? Express your answer as a decimal to the nearest tenth.[/quote]
I think it should look a little something like this....  I think it does show up in the game, but the "review" messes up the $ \text{\LaTeX}$ of the table.
Don't have time to give a solution though, sorry...

Average speed for the first flight: 37/12 (approximately 3.08333)

Average speed for the second flight: 260/59 (approximately 4.4067)

Average of those: 3.74505...

so the answer is $ \boxed{3.74}$ meters per second.

AIME15: Your explanation is correct, but your answer is wrong.
The answer should be to the nearest TENTH: so it's [b]3.7[/b]


Do you mean a frame around the page? Add \usepackage{fancybox} then after \begin{document} add the line \fancypage{\fbox}{}
There's lots more options in the [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/fancybox/fancybox.pdf]fancybox documentation[/url].

[color=red]Please remove the LaTeX from your signature[/color] - see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=810417]Do not use plain TEXT in LaTeX tags![/url] and a number of other posts.


solve equation
(x-3)	\surd(10-x^2)=(x+3)(x-4)

Did you mean: $ (x\minus{}3) \sqrt{(10\minus{}x^2)}\equal{}(x\plus{}3)(x\minus{}4)$

yes I complete my solution but It does not nice


If $a_i\in R$ and the polynomial $P(x)=x^n+a_1x^{n-1}+...+a_n$ is written as a product of first degree factors(with real coeficients) for all $i=1,2,...,n$ prove that
$(n-1)a_1^2\geq 2na_2$

Since the polynomial P(x) has the leading coefficient 1 and is the product of first degree factors with real coefficients, it can be written in the form

$P\left(x\right)=\left(x-x_1\right)\left(x-x_2\right)...\left(x-x_n\right)$

with real $x_1$, $x_2$, ..., $x_n$. Opening the brackets, we obtain

$P\left(x\right)=x^n-\sum_{i=1}^n x_i\cdot x^{n-1}+\sum_{1\leq i<j\leq n} x_ix_j\cdot x^{n-2}+\underbrace{...}_{\text{lower powers of }x}$.

On the other hand, we know that

$P\left(x\right)=x^n+a_1x^{n-1}+a_2x^{n-2}+\underbrace{...}_{\text{lower powers of }x}$.

Hence, comparing coefficients, we get

$a_1=-\sum_{i=1}^n x_i$; $a_2=\sum_{1\leq i<j\leq n} x_ix_j$

Now,

$\left(n-1\right)a_1^2-2na_2=\left(n-1\right)\left(-\sum_{i=1}^n x_i\right)^2-2n\sum_{1\leq i<j\leq n} x_ix_j$
$=\left(n-1\right)\left(\sum_{i=1}^n x_i\right)^2-2n\sum_{1\leq i<j\leq n} x_ix_j$
$=\left(n-1\right)\left(\sum_{i=1}^n x_i^2+2\sum_{1\leq i<j\leq n} x_ix_j\right)-2n\sum_{1\leq i<j\leq n} x_ix_j$
$=\left(n-1\right)\sum_{i=1}^n x_i^2-2\sum_{1\leq i<j\leq n} x_ix_j$.

Now, $\left(n-1\right)\sum_{i=1}^n x_i^2=\sum_{1\leq i<j\leq n}\left(x_i^2+x_j^2\right)$ (in fact, both of these sums count every $x_i^2$ exactly n - 1 times); hence, this becomes

$\left(n-1\right)a_1^2-2na_2=\sum_{1\leq i<j\leq n}\left(x_i^2+x_j^2\right)-2\sum_{1\leq i<j\leq n} x_ix_j=\sum_{1\leq i<j\leq n}\left(x_i^2+x_j^2-2x_ix_j\right)$
$=\sum_{1\leq i<j\leq n}\left(x_i-x_j\right)^2\geq 0$,

since squares are always $\geq 0$. Thus, $\left(n-1\right)a_1^2\geq 2na_2$, and we are done.

  darij

Dariij I admire you for another time. :first:


trigonometry
calculus
number theory
\/closed

Sorry, I was unsure which forum would be best for this question.

I've been thinking about buying one of the Art of Problem Solving books. I see Amazon has two: Basics and And Beyond. I was wondering, from someone who's read these, which would be appropriate for someone of my skill level. I've taken high school algebra, trig and calculus (AP, BC), all of which I did very well in, and my score on AMC is usually right on the cutoff for AIME.

Thank you very much.

I would definately recommend volume 2 in your case, because there are some topics in volume 2 not covered in standard school curriculums, but there is no basic explanation in some topics (if you didn't understand number theory then they won't bother to go over the basics in volume 2; they assume you already know it and are ready to move on).

Take a look at the Table of Contents of [url=http://www.artofproblemsolving.com/Books/Vol1/toc.pdf]volume 1[/url] and [url=http://www.artofproblemsolving.com/Books/Vol2/toc.pdf]volume 2[/url] to see what topics you might not have an understanding of.


Question:

Work out the upper bound and the lower bound of these following
question correct 1 decimal place: x = 4.8, y = 2.4

a) xy  b)x/y c)x+y d) x-y

Scott cycles at a steady speed of 10 metres per second correct to the
nearest metre per second.

a) work out the lower bound of the time Scott takes to cycle exactly
100 meters
b) work out the upper bound of the time Scott takes to cycle exactly
100 metres

[hide]
With Scott, his speed could be anywhere between 9.5 and 10.5 meters per second

To go 100 meters, it would take between $\frac{100}{9.5}$ and $\frac{100}{10.5}$ seconds
Those are the bounds

With your first question, using similar logic, the bounds are

$(4.75)(2.35)$ and $(4.85)(2.45)$

round your final answers as necessary
[/hide]

Thanks for your answer indeed, but can you show me the step(s)?


inequalities
induction
inequalities unsolved

Let a_1,..,a_n be real numbers > 1 such that |a_i - a_(i+1)|<1 for i=1,..,n-1. Prove that

   a_1/a_2 + a_2/a_3 +..+a_n/a_1 <2n-1

This is a really challenging problem. Anyone any ideas? I don't have, or I had and were stupid.

hi, will this help? I dunno.... :(
Let b_i=a_i-a_(i+1)
Rewrite the inequality as
b1/a2+b2/a3+...+b_(n-1)/an-(b1+b2+...+b_(n-1))/a1 < n-1..(*)

How to prove (*)?

[quote="manlio"]Let $a_{1},..,a_{n}$ be real numbers > 1 such that $|a_{i}-a_{i+1}|<1$ for $i=1,..,n-1$. Prove that \[\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+...+\frac{a_{n}}{a_{1}}<2n-1\] [/quote] I think this is not from [url=http://www.kalva.demon.co.uk/aus-pol/ap93.html]Austria-Poland Olympiads 1993[/url]?

Anyway, I will prove this by induction: trivial for $n\le2$, and for induction step we have to show that $\frac{a_{n}}{a_{1}}+2 > \frac{a_{n}}{a_{n+1}}+\frac{a_{n+1}}{a_{1}}$. Here we have 2 cases:

* $a_{n+1}\le a_{n}$: then $\frac{a_{n}}{a_{1}}\ge \frac{a_{n+1}}{a_{1}}$ and $2 > \frac{a_{n}}{a_{n+1}}$ because of $|a_{n}-{a_{n+1}|<1}$ and $a_{n},a_{n+1}>1$.

* $a_{n+1}> a_{n}$: then $1> \frac{a_{n}}{a_{n+1}}$ and since $|a_{n}-{a_{n+1}|<1}$ and $a_{1}>1$ we also have $1 > \frac{a_{n+1}-a_{n}}{a_{1}}$ and we are done.


geometry
Spiral Similarity
ratio
geometric transformation

Let $ABCD$ be a quadrilateral, and let $E$ and $F$ be points on sides $AD$ and $BC$, respectively, such that $\frac{AE}{ED} = \frac{BF}{FC}$. Ray $FE$ meets rays $BA$ and $CD$ at $S$ and $T$, respectively. Prove that the circumcircles of triangles $SAE$, $SBF$, $TCF$, and $TDE$ pass through a common point.

Let $P$ be the second intersection point of the circles $(SAE),(SBF)$. We have $\angle APE=\angle ASE=\angle BPF$, and $\angle PAE=\angle PSE=\angle PBF$. This shows that the triangles $PAE,PBF$ are directly similar, and hence the figures $PAED,PBFC$ are also similar. It follows from here that $PAB,PDC$ are similar, meaning that $\angle PDT=\angle PAS=\angle PET$, i.e. $P$ lies on the circle $(TDE)$. In the same way we prove that it lies on $(TCF)$.

Ok, this is g01ng 2 be teh pwn4ge, you asked for it o_0 lolololol :D

Nice problem, though it doesn't really kick 4ss imho. We are working with directed angles modulo 180° (cuz all teh other angles are l8m). Let $X_{1337}$ be the point of intersection of the lines AD and BC. Let the circumcircles of triangles $X_{1337}AB$ and $X_{1337}CD$ intersect at a point $\Lambda$ apart from $X_{1337}$ (why $\Lambda$? cuz teh letter r00lzorz :P ). Then, $\measuredangle\Lambda BX_{1337}=\measuredangle\Lambda AX_{1337}$, what is equivalent to $\measuredangle\Lambda BC=\measuredangle\Lambda AD$. Similarly, $\measuredangle\Lambda CB=\measuredangle\Lambda DA$. Hence, the triangles $\Lambda BC$ and $\Lambda AD$ are directly similar. The points F and E are corresponding points in these similar triangles, since they lie on the respective sides BC and AD and divide them in the same ratio $\frac{BF}{FC}=\frac{AE}{ED}$. As corresponding points in directly similar triangles form equal angles, this entails $\measuredangle\Lambda FB=\measuredangle\Lambda EA$. This is equivalent to $\measuredangle\Lambda FX_{1337}=\measuredangle\Lambda EX_{1337}$. Thus, the point $\Lambda$ lies on the circumcircle of triangle $X_{1337}EF$.

The rest is trivial using Miquel points. If you are not familiar with Miquel stuff (then j00 are teh n00b :D ), you can do it by angle chasing: Since the point $\Lambda$ lies on the circumcircle of triangle $X_{1337}AB$, we have $\measuredangle\Lambda BA=\measuredangle\Lambda X_{1337}A$. Obviously, $\measuredangle\Lambda X_{1337}A=\measuredangle\Lambda X_{1337}E$. Since the point $\Lambda$ lies on the circumcircle of triangle $X_{1337}EF$, we have $\measuredangle\Lambda X_{1337}E=\measuredangle\Lambda FE$. Thus, $\measuredangle\Lambda BA=\measuredangle\Lambda FE$. In other words, $\measuredangle\Lambda BS=\measuredangle\Lambda FS$. Thus, the point $\Lambda$ lies on the circumcircle of triangle SBF. Similarly, we can prove that the same point $\Lambda$ lies on the circumcircles of triangles SAE, TCF and TDE. Thus, these four circumcircles pass through a common point, what pwnz the level... ehm, problem.

PMFG i sux at writing 1337... [edit: and at being quick  :D ]

  darij

[quote="mathmanman"][quote="darij grinberg"]
...
PMFG i sux at writing 1337... [edit: and at being quick  :D ]
...
[/quote]

And this should be "[b]O[/b]MFG".
Apart from that, 17 d035 0wn j00r m0m!  :D[/quote]

101010101 u just g0t PWN3D by my ub4r1337sp54k!!!1111133337777

$\text{OWNED}\to\text{PWNED}$
$\text{OMFG}\to\text{PMFG}$

  darij

lawlers!

I see, kind of a "slide on the keyboard", eh? :P

*cries* 1 g07 0wn3d..  :surrender:

Nevermind, I'll stop this spam now.

What is scarier is that beneath all the L33tsp33k, there's some really clever maths going on. Wonder what the IMO markers would make of a perfect solution written in L33tsp33k?  :rotfl:

:D
That'd be awesome to try!
At our next TST, I'll do it for one of the exercises, we'll see...  :P

Now it becomes absolutely obvious how much a proper notation is important in maths. :D  :D

[quote="Centy"]Wonder what the IMO markers would make of a perfect solution written in L33tsp33k?  :rotfl:[/quote]

I think it would basicly depend on how readable it is. If it is still mathematically correct and understandable, it shouldn't be much of a problem. Not long ago, I ended a solution to a problem with a quote from Munchkin, what somewhat irritated the correcter, but it still was 7/7. At the most recent TST, I have defined an operation on graphs called "haxxoring" (really, I had no time and no better word came into my mind :D ) and enclosed a part of my solution which I found unnecessary in a <spam> </spam> tag. Participants should have the right to have some fun on exams. Of course, it starts suxx0ring when all the solution is written in 1337, what makes it hard to read normal words and distinguish between letters and numb3rz.

[quote="Centy"]What is scarier is that beneath all the L33tsp33k, there's some really clever maths going on.[/quote]

Well, may be, but this clever maths is over 100 years old ;) . Actually, it's an application of Miquel points. The basic fact is that four lines in the plane enclose four triangles, and that the circumcircles of these triangles have a common point; this point is called the [i]Miquel point[/i] of the four lines. Now, the whole problem comes down to proving that the Miquel point of the lines AD, BC, AB, EF coincides with the Miquel point of the lines AD, BC, CD, EF; in other words, we have to prove that the circumcircles of triangles $X_{1337}AD$, $X_{1337}BC$ and $X_{1337}EF$ have a common point. This is not only a simpler problem (we can forget about S and T now, the point $X_{1337}$ is much easier to handle), but also known: it is a restatement of http://www.mathlinks.ro/Forum/viewtopic.php?t=68157 .

  Darij

[quote="Hikaru79"][quote="darij grinberg"]Not long ago, I ended a solution to a problem with a quote from Munchkin, what somewhat irritated the correcter, but it still was 7/7. [/quote]
You can't leave us hanging like that now :P I think we're all pretty curious![/quote]

Ok, it was an inequality from the ISL 2005, so I won't post it here. I spent about three hours trying to bash it to death by the standard ugly methods (again I won't get more precise now because it is an ISL problem), finally I found a rather nice solution. I was about to end the writeup with a big "PWNED", but I thought it would be too boring (basicly because I would be not the only one to do that - some guys write that under each of their solutions), so I wrote "... thus the problem is dead, dead, dead, any questions left?" (quote from the Krakzilla - I believe this is called Squidzilla in English - card in Munchkin I; this was translated back from the German translation, so I guess it's not quite authentic).

  darij

[quote]That'd be awesome to try! [/quote]

May I suggest that you teach the basis of L33tsp33k to your deputy leader and leader ?  :lol:  :P

Don't forget, when it comes to marking your copy, he's kind of your best ally !  :rotfl:

If  AD // BC, easy to prove
If AD isn’t parallel to BC, call the intersection of AD and BC is X
By a well know lemma we have
(XAB), (XEF), (SAE), (SBF) have a common point
(XCD), (XEF), (TED), (TFC) have a common point
To finish the problem, we will prove that:
(XAB), (XEF), (XCD) have a common point
Because: AE/ED = BF/FC (frac{AE}{ED}=frac{BF}{FC}, similar problem 4 in CANADA MO 2003, we get the result
( Denote that (XYZ) is circumcenter of triangle XYZ)

PS: I think the problem in [b]CANADA MO 2003 [/b]is strong, and beautiful. I can solve many problem with this lemma. For example: [b]China [/b][b]TST 2006[/b]( problem 1 in fouth day)
I’sh that someone can solve problem 5 in [b]VietNam TST 2008 [/b]by this lemma, very thanks

If $ AD // BC$, easy to prove
If $ AD$ isn’t parallel to $ BC$, call the intersection of $ AD$ and $ BC$ is $ X$
By a well know lemma we have
$ (XAB), (XEF), (SAE), (SBF)$ have a common point
$ (XCD), (XEF), (TED), (TFC)$ have a common point
To finish the problem, we will prove that:
$ (XAB), (XEF), (XCD)$ have a common point
Because:  $ \frac{AE}{ED} \equal{} \frac{BF}{FC}$, similar problem 4 in CANADA MO 2003, we get the result
( Denote that $ (XYZ)$ is circumcircle of triangle $ XYZ$)

PS: I think the problem in CANADA MO 2003 is strong, and beautiful. I can solve many problem with this lemma. For example: China TST 2006( problem 1 in fouth day)
I’sh that someone can solve problem 5 in VietNam TST 2008 by this lemma, very thanks

1 minute....
Let $L$ the miquel point of $ABCD$ then there is a spiral similarity sending $BC \to AD$ centered at $L$ and becuase of the ratios, there is a spiral similarity sending $BF \to AE$ centered at $L$ which means that $LSBF, LSEA$ are cyclic. Now we also have that there is a spiral similarity sending $CF \to DE$ centered at $L$ which means that $TCFL, TCDL$ are cyclic. Hence $(SAE), (SBF), (TCF), (TDE)$ share $L$ as common point thus we are done :D

lol remeber when aops was actually productive

[hide=solution]We will show that the Miquel Points of $AEFB,$ $DEFC,$ and $ABCD$ coincide.

There exists a spiral similarity sending $AD$ to $BC.$ However, because $\frac{AE}{ED}=\frac{BF}{FC},$ this spiral similarity also sends $E$ to $F.$ Thus this spiral similarity sends $A$ to $B,$ $E$ to $F,$ and $D$ to $C,$ so the Miquel Points of $AEFB,$ $DEFC,$ and $ABCD$ coincide at the center of this spiral similarity.

The circumcircles of triangles $SAE$ and $SBF$ both go through the Miquel Point of $AEFB.$ The circumcircles of triangles $TCF$ and $TDE$ both go through the Miquel point of $DEFC,$ which coincides with the Miquel Point of $AEFB.$

We are done.[/hide]

The center of spiral sim sending AD to BC, O, is the shared point, which can easily be shown since it is a property of spiral sims that points that break segments in equal ratios are also mapped by the spiral sim, therefore O maps E to F. It is another well known property (someone confirm can I state this in proofs?) that the center of spiral sim is the other shared point by circumcircles of two triangles, one that contains the intersection of the segment with image of segment and the old vertices, and the same intersection with the images of the vertices as the second triangle. Then O is cyclic on all of those circles.

what.

The concurrency point is the Miquel point of $ABCD$. It's the center of the spiral similarity sending $\overline{AD}$ to $\overline{BC}$, which also sends $E$ to $F$, hence it is $(SAE) \cap (SBF)$ and also $(TED) \cap (TFC)$. $\blacksquare$

Extend $AD$ and $BC$ to intersect at $P$. Now, construct the Miquel point of quadrilateral $ABCD$, and call it $M$.
By definition, $(PAB)$ passes through $M$, which means that $MABP$ is a cyclic quadrilateral. Using the fact that $\frac{AE}{ED}=\frac{BF}{FC}$, $\measuredangle MAB = \measuredangle MEF$ by spiral similarity ($M$ is the spiral center).
This means that $MEFP$ is a cyclic quadrilateral, and $(PEF)$ passes through $M$. Since $(PAB)\cap(PEF)=M$, $M$ is the Miquel point of quadrilateral $AEFB$. This means that $(SAE)\cap(SBF)=M$
Now, consider the Miquel point of $EDCF$. This point is defined as $(PDC)\cap(PEF)$, which we already know is $M$. This means that the Miquel point of $EDCF$ is $M$, and by effect, $(TCF)\cap(TDE)=M$
Thus, $(SAE)\cap(SBF)\cap(TCF)\cap(TDE)=M$, as desired. $\blacksquare$

Let $M$ be the Miquel point of $ABCD$. Then $M$ is the center of the spiral similarity taking $\overline{AD}$ to $\overline{BC}$. However, the length condition implies that this spiral similarity also takes $\overline{AE}$ to $\overline{BF}$ and $\overline{ED}$ to $\overline{FC}$, so $M$ is also the Miquel point of $ABFE$ and $CDEF$, which implies the conclusion.

It is well known that the center of spiral similarity mapping $AE \rightarrow BF$ is $(SAE) \cap (SBF)$, and the center of spiral similarity mapping $ED \rightarrow FC$ is $(TED) \cap (TFC)$.

However our given ratio condition implies the two spiral similarities are in fact the same, meaning they have a common center. $\blacksquare$

One liner.
Note that there exists a spiral center that sends $AED$ to $BFC$, so it is the Miquel Point of $AEFB$ and $EDCF$, implying the result.

The spiral similarity taking $AD$ to $BC$ takes $E$ to $F,$ so $ABCD,ABFE,EFCD$ share a Miquel point and thus the circles concur.


induction
number theory solved
number theory

hey ,here's a rather nice problem *:

let $f(x)=\frac 1{1+2\cdot [x]-x}$
let $u_0=0$ and $u_{n+1}=f(u_n)$
prove that for every fraction $\frac ab$ there exists a unique $n$ so that $u_n=\frac ab$.

This problem doesn't get all the attention it needs.
Haven't you noticed what it asks for?
It asks to prove that u_n is a sequencial writing of Q<sub>+</sub>.
I assumed b>0, a>=0.
This not "rather nice", it's extraordinary, so get ot work!

1.Observe that f is injective. ( $f(x)=f(y) \Rightarrow [x]+\{y\}=[y]+\{x\}$)

2.We will work only with pairs of coprime integers.
$f(\frac a{a+b})=\frac {a+b}b$ and $f(\frac{aq+r}a)=\frac a{a+aq-r}$ where $r<a$
so $f(\frac ab)=\frac cd$ implies that $b=c$.
It is easy to se now that $f(x)\neq x$ for any $x\in \mathbb{Q}$.

From 1 and 2 we get that $u_i\neq u_j$ for $i\neq j$ (and because $u_i\neq 0$ for $i>0$)
So we are left to prove that $u$ is surjective in $\mathbb{Q}$.

And now it begins. Basically I want to prove that integers are taken by the sequence in an asscending order, and I have to know that numbers are taken between two consecutive integers. This will be shown using $g$:

3.Let $g:A\to\mathbb{N}$ where A is the set of pairs of coprime integers, such that:
$g(a,b)=g(b,a)$
$g(1,a)=a$
$g(a,a+b)=1+g(a,b)$

4.Lemma
If $\frac cd=f(\frac ab) \Rightarrow \left\{\begin{array}{cc} g(c,d)=g(a,b)&\mbox{,if } b>1\\g(c,d)=1+g(a,b)&\mbox{,if } b=1\end{array}\right.$
[b]Proof:[/b]
if $b>a$ then $b=a+e$ and $f(\frac a{a+e})=\frac {a+e}e$
$g(a,a+e)=1+g(a,e)=1+g(e,a)=g(e,a+e)$

if $a>b$ and $b\neq 1$ then $a=b\cdot q+r$ and $0<r<b$
$f(\frac{bq+r}b)=\frac b{b+bq-r}$ we have to show that
$g(b,bq+r)=g(b,b+bq-r) \Leftrightarrow g(b,b(q-1)+r)=g(b,bq-r) \Leftrightarrow g(b,b(q-2)+r)=g(b,b(q-1)-r) \Leftrightarrow\ldots\Leftrightarrow g(b,r)=g(b,b-r) \Leftrightarrow g(r,b)=g(b-r,b) \Leftrightarrow g(r,b-r)=g(b-r,r)$

if b=1 f(a)=1/(a+1) and
g(1,a+1)=1+g(1,a) so we are done with the lemma.

5.Let M={q in Q: q=u_i for some i in N}. We have to prove that M=Q
Induction by the value of g
Let a0/b0 in Q such that g(a0,b0)=n. Suppose that c/d is in M if g(c,d)<n
g(1,n-1)=n-1 so n-1 is in M (this is the only thing I need from induction)
Let u_i=n-1 Then u_(i+1)=1/n. Let j=max{k:u_k=a/b and g(a,b)=n}

So u_j=a/b and g(a,b)=n and u_(j+1)=c/d and g(c,d)=n+1
This leads to u_j=n and u_(j+1)=1/(n+1)

Remember that g(a0,b0)=n. Apply f to a0/b0 to get a1/b1 then a2/b2 and so on. While b_i<>1 we have g(a_i,b_i)=g(a_(i+1),b_(i+1))=n. This must end so there is k such that b_k=1 resulting a_k=n. So for some m we have f^m(a0/b0)=n=u_j. Recall that f is injective so f^(m-1)(a0/b0)=u_(j-1) and so on to get a0/b0=u_(j-m). So a0/b0 is in M.

There is a thing I didn't prove: that g^(-1)(n) is finite for any n. I am sure is obvious, but I don't have a complete proof.


algebra
polynomial
linear algebra
matrix
linear algebra unsolved

Let $ A$, $ B$ be $ n\times n$-matrices with $ AB \equal{} BA$. Consider the polynomial $ P_{A,B}(x,y) \equal{} \det (yA \minus{} xB)$.

Prove that $ P_{A,B}(A,B) \equal{} 0$.

Well, you didn't quite understand the problem, though this may be due to my bad explanation. Look at the example of $n=2$ and $B=Id$. Let $A=\begin{pmatrix} a&b\\ c&d\end{pmatrix}$. then we have

$P_{A,B}(x,y)=\det(yA-xB)=\det \begin{pmatrix} ya-x&yb\\ yc&yd-x\end{pmatrix}$$= (ya-x)(yd-x) - ybyc = y^2(ad-bc) - xy(a+d) + x^2$.

Then it is our task to show that

$(ad-bc)\begin{pmatrix} 1&0\\ 0&1\end{pmatrix}^2 - (a+d)\begin{pmatrix} 1&0\\ 0&1\end{pmatrix}\begin{pmatrix} a&b\\ c&d\end{pmatrix} + \begin{pmatrix} a&b\\ c&d\end{pmatrix}^2=0$.

This is not just implied by $\det(AI-IA)=0$. Indeed, there are examples of $A,B$ which satisfy $\det(AB-BA)=0$ but not the statement of the theorem.

I wanted to add that I found some even much more far-reaching generalization:

Let $A_1,...,A_m$ be $n$x$n$-matrizes which are pairwisely commutative. Let $B_1,...,B_m$ be $n$x$n$-matrices such that $\sum_{i=1}^{m} B_iA_i=0$. Consider the polynomial $P(x_1,...,x_m)=\det \left(\sum_{i=1}^{m} B_ix_i\right)$.

Prove that $P(A_1,...,A_m)=0$.

[I hope that this is correct; it could be there's some stupid mistake in it.]

[quote="spix"]$P_{A,B}(A,B)=\det(BA-AB)=0$ because $AB=BA$
How can this be a generalisation?[/quote]
No,spix, you can't do so. When you define $P_{A,B}(x,y)=det(yA-xB)$ it is the polynomial in $x,y$ and you can't write $P_{A,B}(A,B)$ as $\det(BA-AB)$.

The statement is clearly true if both $A, B$ are diagonal(isable), and the diagonisable matrices with distinct eigenvalues are dense in $\mathrm{Mat}(n\times n, \mathbf{C})$. So it holds for all such matrices, and because the transcendence degree of $\mathbb{C}$ is $\geq \aleph_0$, we can transfer it to matrices over arbitrary commutative rings.

Is this correct?

[quote="Peter Scholze"]Well, I don't think your proof is correct(besides its brutality - I want nice proofs for nice theorems ;) ). The problem is that you need the same matrix $S$ such that $S^{-1}AS$ are $S^{-1}BS$ are diagonal, but this is clearly only possible for a very small set of pairs $A,B$ (e.g. they need to have the same trace which is a closed condition).[/quote]If $A,B$ commute, they are simultaneously (tri)diagonisable if they are (tri)diagonisable.

And $A,B$ don't need to have the same trace?

Sorry for posting nonsense ;). I was a bit confused as I didn't want such a proof ;). Btw, it looks like your argument does not work for my second generalization( :) ) (at least I don't see how to prove it simply for diagonal matrices).

Suppose that $A_i$ are diagonalisable (we can do so because these matrices are dense). Then there is a basis in which the matrices are simultaneously diagonalisable, so we prove the theorem just for diagonal matrices. Let $b_{i,k}=A_k[i,i]$. From the condition $\sum_{i=1}^n A_iB_i=0$, by looking at the entry $(i,j)$ we see that $\sum_{k=1}^n B_k[i,j] b_{j,k}=0$(*). Expand now the determinant $det(\sum_{i=1}^n B_ix_i)=\sum_{\sigma \in S_n} \prod_{i=1}^n \sum_{i=1}^n B_ix_i[i,\sigma(i)]=\sum_{\sigma \in S_n} \prod_{i=1}^n (\sum_{k=1}^m B_k[i,\sigma(i)]x_k)$(**).

Let's prove now that if we replace $x_k$ by $A_k$ every product in (*) will be the zero matrix. Indeed, clearly $\sum_{k=1}^m B_k[i,\sigma(i)]A_k$ is clearly a diagonal matrix and because of (*) the entry $(\sigma(i),\sigma(i))$ of this matrix is zero. Since $\sigma(i)$ runs through all the values from $1$ to $n$ we deduce that we have a product of diagonal matrices but every entry $(k,k)$ will be zero in some matrix, implying our claim. QED

My solution as well(I guess even for all commutative rings) - and it does not use any normal forms of a matrix like diagonalization(it does not choose a basis to be precise). Indeed, after I proved Cayley-Hamilton that way the results posted here where just corollaries of the proof.

[quote="harazi"]I remember Moubinool posting this result some time ago and as far as I remember my solution worked for any field.[/quote]
This is the link which harazi probably mentioned http://www.mathlinks.ro/Forum/viewtopic.php?highlight=harazi&t=35679#


Show that the product of every $k$ consecutive members of the Fibonacci sequence is divisible by $f_1f_2\ldots f_k$ (where $f_0=0$ and $f_1=1$).

This is a classical problem. A week ago in Rostock, Peter Scholze and me generalized this as follows:

A [i]Mersenne sequence[/i] of numbers (or, more generally, elements of an Euclidean ring) is a sequence $a_1$, $a_2$, ... such that $\gcd\left(a_i;\;a_j\right)=a_{\gcd\left(i;\;j\right)}$ for any two positive integers i and j.

Prove that for any Mersenne sequence $a_1$, $a_2$, ... and for any two positive integers i and j, we have $b_ib_j\mid b_{i+j}$, where $b_n=\prod_{k=1}^n a_k$ for any n.

  Darij


${f=\sqrt{z+\sqrt{y+\sqrt{x}}}+\sqrt{y+\sqrt{z+\sqrt{x}}}+\sqrt{z+\sqrt{x+\sqrt{y}}}+\sqrt{x+\sqrt{z+\sqrt{y}}}+\sqrt{y+\sqrt{x+\sqrt{z}}}+\sqrt{x+\sqrt{y+\sqrt{z}}}}$

${f\leq 6 \sqrt{\frac{\theta }{3}+\sqrt{\frac{\theta }{3}+\frac{\sqrt{\theta }}{\sqrt{3}}}}}$
${x+y+z=\theta }$
${(\theta >0)\land (0<x<\theta )\land (0<y<\theta-x)}$


Let be given three positive integers $a,b,x$ such that: $x^{a+b}=a^{b}.b$.Prove that: $a=x,b=x^{x}$