geometry
circumcircle
LaTeX
ratio
parallelogram
Euler
power of a point

Figure, [b]K[/b] is the symmedian point of [b]ΔABC[/b], [b]O[/b] is the circumcenter; [b]BK,CK[/b] intersect  [b]AC,AB[/b] at [b]D,E[/b], and [b]FP⊥BC [/b] .
（1）：If [b]OK||DE[/b], then prove：[b]ΔPED~ΔABC[/b] .
（2）：If [b]PE/PD=AB/AC[/b], then prove：[b]OK||DE [/b].

（*）：Let [b]K[/b] be [b]I[/b], then [b]OI||DE <=> ∠BAC=60[/b]
Answer for （*）：[url]http://www.aoshoo.com/bbs1/dispbbs.asp?boardid=43&Id=16362[/url]（2nd floor）

Using ${ \color[rgb]{0.7,0,0}}\text{\LaTeX}$ is far easier than what you do. If you typed:
[code]Figure, $K$ is the symmedian point of $\triangle ABC$, $O$ is the circumcenter; $BK,CK$ intersect  $AC,AB$ at $D,E$, $AK$ interscets $DE$ at $F$ and $FP \perp BC$.
(1): If $OK \parallel DE$, then prove: $\triangle PED \sim \triangle ABC$.
(2): If $\frac{PE}{PD}=\frac{AB}{AC}$, then prove: $OK \parallel DE$.[/code]
then your post would look like this:
[quote="lym"]Figure, $ K$ is the symmedian point of $ \triangle ABC$, $ O$ is the circumcenter; $ BK,CK$ intersect  $ AC,AB$ at $ D, E$, $ AK$ intersects $ DE$ at $ F$ and $ FP \perp BC$.
(1): If $ OK \parallel DE$, then prove: $ \triangle PED \sim \triangle ABC$ .
(2): If $ \frac {PE}{PD} = \frac {AB}{AC}$, then prove: $ OK \parallel DE$.[/quote]

Tangents of $ (O)$ at $ A, B, C$ cut $ BC, CA, AB$ at $ X, Y, Z$ (centers of the coaxal Apollonius circles). Brocard axis $ KO$ is their common radical axis $ \Longrightarrow$ $ KO \perp XYZ.$ Since $ KO \parallel DE$ $ \Longrightarrow$ $ DE \perp XYZ$ as well. Let $ AK$ cut $ BC$ at $ Q.$ $ DE, EQ, QD$ go through $ X, Y, Z.$ Cross ratio $ (Z, E, A, B) = - 1$ is harmonic $ \Longrightarrow$ pencil $ X(Z, E, A, B)$ is harmonic. Since $ XE \perp XZ,$ it follows that $ XE, XZ$ bisect the angle $ \angle AXB$ internally and externally. This means that $ OK \parallel XDE$ are both perpendicular to the internal bisector $ AU$ of the $ \angle A$ (where $ AU \in BC$). Consequently, $ AD = AE$ or

$ \frac {b^2c}{a^2 + b^2} = \frac {bc^2}{c^2 + a^2} \Longleftrightarrow a^2(b - c) = bc (b - c)$

Isosceles $ \triangle ABC$ with $ b = c$ would have $ KO \perp DE,$ hence $ b \neq c$ and $ a^2 = bc.$ Using this,

$ \frac {CU}{CD} = \frac {ab}{b + c} \cdot \frac {c^2 + a^2}{a^2b} = \frac {c(c + b)}{a(b + c)} = \frac {c}{a} = \frac {a}{b} = \frac {CB}{CA}$

$ \Longrightarrow$ $ DU \parallel AB$ and similarly, $ EU \parallel AC.$ As a result, $ ADUE$ is parallelogram with perpendicular diagonals $ DE \perp AU,$ a rhombus. Regardless of this, the cross ratio $ (X, F, D, E) = (X, Q, C, B) = - 1$ is harmonic $ \Longrightarrow$ pencil $ P(X, F, D, E)$ is harmonic and $ PF \perp (BC \equiv PX)$ $ \Longrightarrow$ $ PF, PX$ bisect the angle $ \angle DPE$ internally and externally. Perpendicular bisector of $ DE$ and the external angle bisector $ PX$ meet on the circumcircle $ \odot (PDE).$ In our case, $ AU$ is the perpendicular bisector of $ DE$ $ \Longrightarrow$ $ PDEU$ is cyclic. But then $ \angle EPD = \angle DAP = \angle CAB,$ $ \angle PDE = \angle PUE = \angle BCA \mod 180^\circ$ and $ \angle DEP = \angle DUP = \angle ABC \mod 180^\circ,$ so that $ \triangle PDE \sim \triangle ABC$ are similar.

$ \frac {AB}{AC} = \frac {PE}{PD}$ itself is insufficient for $ KO \parallel DE.$
[hide]In any $ \triangle ABC,$ it is easy to calculate $ \frac {PE}{PD} = - \frac {\overline{FE}}{\overline{FD}} = \frac {c^2 + a^2}{a^2 + b^2}.$ The given condition is therefore equivalent to $ \frac {a^2 + b^2}{a^2 + c^2} = \frac {b}{c}$ $ \Longleftrightarrow$ $ a^2(b - c) = bc(b - c)$ $ \Longrightarrow$ $ a^2 = bc$ or $ b = c.$ But $ KO \parallel DE$ does not hold for an isosceles $ \triangle ABC$ with $ b = c,$ which was not excluded. On the other hand, $ a^2 = bc$ with $ b \neq c$ immediately implies $ AD = AE$ and $ X$ finite $ \Longrightarrow$ $ DE \perp AU$ and $ X$ finite $ \Longrightarrow$ $ DE \perp XYZ$ $ \Longrightarrow$ $ DE \parallel KO.$[/hide]

 __________________________________________________________________________________________
The configuration $ KO \parallel DE$ has tons of other nice properties, which can be used for creating geometry problems.

Thank you, Yetti！
My way is similar with yours, and it also has anther way to do. This problem I have give a prompt in a chinese website, but nobody prove, Yetti, you are really good！
But You said $ \frac {PE}{PD} \equal{} \frac {AB}{AC}$ was insufficient for $ OK \parallel DE$, I think you are wrong：

Because we can have：$ \frac {PE}{PD} \equal{} \frac {AB}{AC} \Longrightarrow \frac {EF}{DF} \equal{} \frac {AB}{AC}$ .

Yes, you are right, add a condition：$ AB<AC$ for （2）.

Thanks. But it takes more to create a very nice problem with a solution, than just solve it.

[quote="lym"] $ I$ is the incenter of $ \triangle ABC$, $ O$ is the circumcenter; $ BI,CI$ intersect  $ AC,AB$ at $ D, E$.
(*): Prove: $ OI \parallel DE$ $ \Longleftrightarrow$ $ \angle A \equal{} 60^\circ$[/quote] (edited)

External bisectors of $ \angle A, \angle B, \angle C$ cut $ BC, CA, AB$ at $ U, V, W.$ $ (O)$ is 9-point circle and $ I$ orthocenter of the excentral triangle, $ OI$ its Euler line. Let $ (O')$ be its circumcircle, $ O' \in OI.$ $ UVW \perp O'OI$ as radical axis of $ (O'), (O).$ $ OI \parallel DE$ $ \Longleftrightarrow$ $ UVW \perp DE.$ $ DE$ goes through $ U.$ Cross ratio $ G(V, E, A, B) \equal{} \minus{} 1$ is harmonic $ \Longrightarrow$ pencil $ U(V, E, A, B)$ is harmonic and  $ UE \perp UV$ $ \Longrightarrow$ $ UE, UV$ bisect $ \angle AUB \equiv \angle AUC$ internally and externally. But then $ \angle DUB \equal{} \angle EUC \equal{} \frac {\angle AUB}{2} \equal{} \frac {|\angle B \minus{} \angle C|}{4}$ and from $ \triangle BUD, \triangle CED,$ $ \angle IDE \equal{} \frac {\angle B \plus{} \angle C}{4} \equal{} \angle IED$ $ \Longrightarrow$ $ ID \equal{} IE.$ Since $ AI$ bisects $ \angle DAE,$ $ ADIE$ is cyclic, $ \angle DAE \plus{} \angle EID \equal{} \angle A \plus{} \frac {180^\circ \plus{} \angle A}{2} \equal{} 180^\circ$ $ \Longrightarrow$ $ \angle A \equal{} 60^\circ.$ The proof is reversible.


calculus
vector
abstract algebra
superior algebra
superior algebra theorems

This is trivial and silly, but it might be worth a couple of minutes for someone who is staring at the axioms for a Lie algebra for the first time:

(1) Consider $\mathbb{R}^{3}$ together with the cross product (as normally defined in a multivariable calculus course). Treating $u\times v$ as if it were the $[u,v]$ called for in the axioms, show that this satisfies the axioms for a Lie algebra.

(2) Consider the set of $2\times 2$ real matrices with trace zero. Define the "usual" bracket operation: $[A,B]=AB-BA.$ Show that this satisfies the axioms for a Lie algebra.

(3) How are (1) and (2) related?

I've been doing some Googlin' but I still don't get it.

(1) One can prove the Jacobi identity using the formula $u \times \left( v \times w \right) = \left( u \cdot w \right) v-\left( u \cdot v \right) w.$
Can we prove this formula without using brute force?

(2) Kind of obvious.

(3) Are they supposed to be isomorphic? I'm not getting anywhere close to proving that.
Nonetheless, I found with the help of Google that (1) is isomorphic to the group of skew-symmetric matrices of order $3$.

On (1), there needs to be at least some brute force somewhere.

It turns out that all non-commutative 3-dimensional Lie algebras are isomorphic, so yes, (1) and (2) are supposed to be isomorphic. Constructing the isomorphism is the challenge.

Vocabulary note - what is "non-cummutative Lie algebra" supposed to mean anyway? The answer is that a commutative Lie algebra is one for which $[a,b]=0$ for all $a$ and $b$ - that is, just any vector space with a trivial bracket operation. The only Lie algebras worth studying under that title are non-commutative.

(1) $u\times(v\times w)+v\times(w\times u)+w\times(u\times v)$ is 3-multilinear and alternating, hence equals $\det(u,v,w)\cdot x$ for some fixed vector $x$. By rotational invariance, $x=0$.

As to "non-commutative" Lie algebras: typically, the word "(non-)abelian" is preferred. Abelian Lie algebras come up as Cartan subalgebras in complex semisimple Lie algebras, and the fact that you typically do not hear much about them probably has to do with the difficulties in their representation theory: even in the simplest case of the one-dimensional (automatically abelian) Lie algebra, representations correspond to endomorphisms of vector spaces, and their classification amounts to the Jordan normal form.

owk

Multilinearity pops up when I least expect it :w00tb:

What about that isomorphism? Maybe a small push in the right direction?

You should be warned that the Lie algebras are not isomorphic over $\mathbb R$, you need to switch to complex coefficients. Just writing down an isomorphism is not very enlightening, ideally one should talk about Lie groups as well. But I'll try anyway. So let us call (1) $\mathfrak{so}(3)$ and (2) $\mathfrak{sl}_{2}(\mathbb R)$. Then, $\mathfrak{so}(3)$ has the obvious basis $e_{1},e_{2},e_{3}$ satisfying the relations $[e_{i},e_{j}]=\epsilon_{ijk}e_{k}$ (i.e., $[e_{1},e_{2}]=e_{3}$ and cyclic permutations). These relations are satisfied by the Pauli matrices (up to a factor $2i$ which you can get rid of by considering $\sigma_\nu/2i$), and they generate $\mathfrak{sl}_{2}(\mathbb C)$. owk


calculus derivative function trigonometry inequalities

I guess this is semi-Round Table but...

When I heard that the Poincare conjecture was pretty much solved by some mathematicians (http://en.wikipedia.org/wiki/Poincare_conjecture) I realized that the name "Poincare" is pronounced really weirdly.

How do you pronounce it and the names of the other people who came up with the other six insanely hard problems? (Well, I guess "P vs NP is pretty obvious but...")

Just a question.

Hodge is like HOJJ
Reimann is RAI-MANN
Yang Mills is obvious
so are the rest
Birch is BURCH
Only Poincare is odd in fact.

Most people I know pronounce Riemann as Ree mahn

I thought it was "Pwon-cah-RAY" (as opposed to "Poyn - cah -RAY") but I was never really sure.

I always say "REE-man" too.

From dictionary:  Poincaré: pwan-ka-rey
Riemann: ree-mahn.
[color=yellow]
[size=59]Of course, the Riemann Hypothesis is actually the Ramen Hypothesis given by the FSM considering the strip of Ramen between 0 and one half of the derivative of the zeta function[/size].[/color]  :)

I've always wondered how do I pronounce "Cauchy"? :maybe:

[quote="bubka"]Well my version is much more stylish and fun pronouncing. So there.[/quote]
RAI-MANN is sort of like rayman which is a dumb game, and not stylish or fun.So RAI-MANN is not stylish or fun.
Contradiction.
therefore ree-mahn must be the right pronounciation.
QED

[quote="mathnerd314"][quote="Xevarion"]AWW-guss-teen LOO-eese KOH-shee[/quote]

I always thought it was "loo-WEE"...[/quote]
Actually that's probably right. I never studied French. Don't trust me.

http://nsm1.nsm.iup.edu/gsstoudt/history/ma350/pronounce.html

A pronunciation guide to mathematicians. :D

cauchy = co-she

kinda like cosine except not really

Riemann should be pronounced Ree-man since it's a german's name, hence we should use the German pronounciation. Lawless flogic,  isn't it?

Cauchy rhymes with Yoshi.

How do you pronounce Jensen?


The number of real number solutions (x,y,z,w) to the following equations is what?

2y=x+17/x

2z=y+17/y

2w=z+17/z

2x=w+17/w

Ugghhh.  I've done this problem, but the solution is sort of long and I have to study for a midterm tomorrow so I'll post it later.  However, here is a hint

[hide="Hint"]Use AM-GM[/hide]


Is countdown official in your state? It is in FL.

The 16 person Countdown isn't official in Connecticut.

Yes, it is in Maine. (unfortunately for me.  :( )
I don't think such a small state like Maine should have an official countdown, since we don't have a very high chance of getting anyone in the national countdown.

official in OK. lucky me, i'm good at it. :)

not in IN tho... i would have gotten a higher place too...

[quote="pianoforte"]Yes, it is in Kansas (thankfully for me).[/quote]

same with me for tx

Not in california (lucky for me i suck at countdown).

[quote="Ignite168"][quote="dogseatcheese"]Really? Its not official in CA? But you get a trophy and stuff. And would if people in the top ten have the same score?[/quote]
Yeah but their competition is split  up into two state competitions so it's kinda impossible to have a Countdown round.[/quote]

Yea.....I guess that its because my state is too big.

its not in TN. it wouldnt really matter for me. i woulda made it both ways. they give prizes to top 4 in countdown. kinda pointless because there is only 8 people in countdown.

CO has an official ladder format countdown round with 10 people. Good for me last year, didn't matter this year.

[quote="binonunquineist"]The countdown round in Texas is official, but it is in a ladder format, so even if you lose the first round you'll only go down in your ranking one place.[/quote]

Every state it's official in it's ladder, only Nats has an official bracket.

Unofficial in Missouri, which allows them to pair up 10th-place with 1st-place. It's always amusing when 10th-place wins... (like they did last year)

[quote="dogseatcheese"]Really? Its not official in CA? But you get a trophy and stuff. And would if people in the top ten have the same score?[/quote]

If people have the same score they use that stupid thing called tiebreakers :mad:


symmetry
vector
linear algebra
matrix
linear algebra unsolved

This sounds like homework.  Tell us what you've done so far.

Remember each inner product induces a norm, so my guess is find all the norms on $ \mathbb{C}$. And all the norms has the form $ \alpha|z|$ for some $ \alpha > 0$

Please correct me if I am wrong . Thanks

To be an inner product it has to satisfy

1. Linearity       <ax+y, z>= a<x,z> + <y,z>   for x,y,z bleonging to V, a belonging to F
This is true for V=C

2. Symmetry: is not satisfied on V=C because of complex conjugation

3. <x,x> greater than 0 is not satisfied

So I'm guessing the only inner product space is the subset R (real numbers)

Is this correct?

That's the definition of a real inner product, not a complex inner product, and it's not what the question is asking.  Depending on convention, an inner product on a complex vector spaces satisfies linearity in the first variable and [b]conjugate symmetry[/b], which says that

$ \left< x,y \right> \equal{} \overline{ \left<y, x \right> }.$

where $ \bar{z}$ denotes the complex conjugate.  This implies that $ \left< x, x \right>$ is real; we also require that it's greater than or equal to zero.  The standard inner product on $ \mathbb{C}$ is

$ \left< x, y \right> \equal{} x \cdot \bar{y}.$

Can you think of any others?

[quote="t0rajir0u"]That's the definition of a real inner product, not a complex inner product, and it's not what the question is asking.  Depending on convention, an inner product on a complex vector spaces satisfies linearity in the first variable and [b]conjugate symmetry[/b], which says that

$ \left < x,y \right > \equal{} \overline{\left < y, x \right > }.$

where $ \bar{z}$ denotes the complex conjugate.  This implies that $ \left < x, x \right >$ is real; we also require that it's greater than or equal to zero.  The standard inner product on $ \mathbb{C}$ is

$ \left < x, y \right > \equal{} x \cdot \bar{y}.$

Can you think of any others?[/quote]

I'm thinking of $ \left < x, y \right > \equal{} \overline{y} Q x$ where Q is a hermitian and positive definite matrix but I'm not sure.

A Hermitian positive definite $ 1\times 1$ matrix would have to be what?

The only number I can think of is 1 but that's just the standard inner product...

Actually it could be any positive real number. For $ 1\times 1$ matrices, "Hermitian" implies real and "positive definite" implies positive.


Hello friends,

                       Let us denote the k-th prime by P(k) (where k, of course, is a natural number). Can we find 2 primes P(j) & P(j+1) (i.e two succesive primes) such that
P(j+1) > P(j)^2.

Any help received will be appreciated.

Thank You

I don't think so (there's always a prime between $a$ and $2a$)

Why should a prime always exist between a & 2a? Please do reply because this thing really sounds curious.

Thanks in advance


algebra
function
domain
real analysis
real analysis unsolved

Ok, given a function $ f: R^n \minus{}> R^m$ that's 1-1 and continuous on $ R^n$. If A is open and disconnected in $ R^n$, prove f(A) is open and disconnected in f($ R^n$). Now, the solution says that the exercise is wrong, and that there is a counterexample.

so the counterexample is this: Let f: R -> R^2.

f = $ (cos(\frac{2 \pi x}{1 \plus{} x} \minus{} \frac{\pi}{2}), 1 \minus{} sin(\frac{2 \pi x}{1 \plus{} x} \minus{} \frac{\pi}{2}))$ if $ x \geq 0$
$ (cos(\frac{2 \pi x}{1 \minus{} x} \minus{} \frac{\pi}{2}), \minus{}1 \plus{} sin(\frac{2 \pi x}{1 \minus{} x} \minus{} \frac{\pi}{2}))$ if $ x < 0$

So the counterexample is basically the graph of two circles, one going counterclockwise from (0, -2) to (0, 0) to (0, -2) as x goes from $ \minus{} \infty$ to 0, and then jumping to a clockwise function going clockwise from (0, 2) to (0,0) back to (0,2) as x goes from 0 to $ \infty$. So the range is connected at (0,0). The thing is - I don't understand how the domain is disconnected. x encompasses all values of x from -infinity to infinity. And the $ (\frac{2 \pi x}{1 \minus{} x} \minus{} \frac{\pi}{2})$ to $ (\frac{2 \pi x}{1 \plus{} x} \minus{} \frac{\pi}{2})$ function is continuous at x = 0. So where is the disconnectedness in the domain?? Thanks!

You can disconnect the domain (by removing a point from it) without disconnecting the range.


you have mistake 2 with $ \sqrt{2}$.
Because when $ a\equal{}b\equal{}c\equal{}1$ then $ \sum_{cyc}\frac{a\sqrt{b\plus{}c}}{b\plus{}c\plus{}1}\equal{}\sqrt{2}$

[quote="hophinhan"]Let $ a,b,c > 0$ such at $ abc \equal{} 1$.

Prove : $ \sum_{cyc} \frac {a\sqrt {b \plus{} c}}{b \plus{} c \plus{} 1} \ge \sqrt 2$[/quote]
Nhan, Holder Inequality works here.  :maybe:

Assume that:$ a\geq b\geq c$. So $ b\plus{}a\geq c\plus{}a\geq b\plus{}c$.
Let: $ f(x)\equal{}\frac{\sqrt{x}}{x\plus{}1}$ on $ R^{\plus{}}$.We have $ f'(x)<0$ so: $ f(b\plus{}c)\geq f(a\plus{}c)\geq f(a\plus{}b)$.
Using ineq Chebyshev;we have:
$ 3\sum_{cyc}\frac{a\sqrt{b\plus{}c}}{b\plus{}c\plus{}1}\ge (\sum a)(\sum\frac{\sqrt{b\plus{}c}}{b\plus{}c\plus{}1})$
Using ineq Cauchy_Schwar We have:$ (2\sum a\plus{}3)(\sum\frac{\sqrt{b\plus{}c}}{b\plus{}c\plus{}1})\geq (\sum\sqrt[4]{b\plus{}c})^2$
So: $ (2\sum a\plus{}3)(\sum\frac{\sqrt{b\plus{}c}}{b\plus{}c\plus{}1})\geq 9\sqrt{2}$.Because $ (\sum\sqrt[4]{b\plus{}c})^2\geq 3\sqrt[4]{2}$ (Using AM-GM).But $ 2\sum a\plus{}3\leq 3\sum a$.So QED.

[quote="hophinhan"]Let $ a,b,c > 0$ such at $ abc \equal{} 1$.

Prove : $ \sum_{cyc} \frac {a\sqrt {b \plus{} c}}{b \plus{} c \plus{} 1} \ge \sqrt{2}$[/quote]

It ' s problem 2 in this file :)

I knew it is  problem in that ebook but the solution not very nice

Thanks Materazzi, tronghieu ...and math10-your solution is what i'm fiding

Another solution.
Applying AM-GM Inequality, we have: $ 1 \le \frac{{\sqrt a (b \plus{} c)}}{2}.$
Thus, we only need to prove that: $ \sum\limits_{sym} {\frac{a}{{\sqrt {b \plus{} c} (\sqrt a  \plus{} 2)}} \ge \frac{{\sqrt 2 }}{2}}$
using CS inequality, we have:
$ \sum\limits_{sym} {\frac{a}{{\sqrt {b \plus{} c} (\sqrt a  \plus{} 2)}}}  \ge \frac{{(a \plus{} b \plus{} c)^2 }}{{2\sum {a\sqrt {b \plus{} c} }  \plus{} \sum {a\sqrt {a(b \plus{} c)} } }} \ge \frac{{(a \plus{} b \plus{} c)^2 }}{{2\sqrt {2(\sum {a)(\sum {ab)} } }  \plus{} \sqrt {(\sum {a)(\sum {ab(a \plus{} b))} } } }}.$
Setting $ t^2\equal{}a\plus{}b\plus{}c;u^2\equal{}ab\plus{}bc\plus{}ca$. Rewrite inequality, we need to prove:$ \sqrt 2 t^3  \minus{} 2\sqrt 2 u \ge \sqrt {t^2 u^2  \minus{} 3}$. By AM-GM ineq, $ u^2<\equal{}t^2/3.$ Thus,
$ (\sqrt 2 t^3  \minus{} 2\sqrt 2 u)^2  \minus{} t^2 u^2  \plus{} 3 \ge \frac{5}{3}t^6  \minus{} \frac{{8\sqrt 3 }}{3}t^5  \plus{} \frac{8}{3}t^4  \plus{} 3\ge 0$(Using AM-GM: $ \frac{5}{3}t^6  \plus{} \frac{8}{3}t^4  \plus{} 3 \ge 24\sqrt[{24}]{{\frac{{t^{122} }}{{3^{37} }}}} \ge \frac{{8\sqrt 3 }}{3}t^5 (t \ge \sqrt 3 )$)


inequalities
number theory unsolved
number theory

Let $n=\overline{a_1a_2...a_n}$ and denote $S(n)=a_1+a_2+\cdots +a_n$
Find all $10\leq n\leq 99$ such that:
 \[170 \mid S(10^n-n)\]

probably not what you'd expect, a little brute force, but here goes:

we obviously have $10^n - n = \overline{\underbrace{9...9}_{n-2}xy}$, so $S(10^n - n) = 9(n-2) + S(100-n)$. it is also clear that $S(99-n) = 18 - S(n)$. therefore $S(10^n - n) = 9n - S(n-1)$ which is smaller than $9 \cdot 99 = 891$. hence if $S(10^n - n) = 170k$, than $k \in \{1,...,5\}$
above all $S(n-1) \leqslant 17$. therefore $k$ must satisfy $9n > 170k \geqslant 9n-17$ for $n$ that we're looking for.
setting $k=1$ to these inequalities gives us $n=20$, which satisfies the condition $170k = 9n - S(n-1)$. similarily, for $k \in \{2,...5\}$ we get $n \in \{39,58,77,96\}$

if anyone has a solution leading direclty to the fact that $n$ is of the form $19k+1$ i'd like to see it, as mine is not very elegant


Hexagon ABCDEF has the following properties:
i) diagonals AC,CE and EA are all the same length
ii) angles ABC and CDE are both 90 degrees
iii) all the sides of the hexagon have lengths whcih are different integers

a) What is the minimum perimeter of ABCDEF if AC =square root of 85
b) What is the smallest length of AC for which ABCDEF has all these properties? WHat is the minimum perimeter in this case?

I think i have completed part a), and i got an answer of 27.66 units by multiplying the square root of 85 by 3, however could some one please verify my answer.

Also i have some difficulty in deriving part b)

Help is much appreciated

Infinity

please ban this topic. answers shouldnt be given. ;)

ruins the integrity of the maths challenge event runnign in Australia.


geometry
circumcircle
geometric transformation
reflection
geometry proposed

Let $ABC$ be a triangle and let $A'B'C'$ be its orthic triangle, $H$ its orthocenter. Let $M$ be the midpoint of $AH$ and $X,Y$ be the feet of perpendiculars from $A'$ to $MB', MC'$. Prove that $B,C,X,Y$ are concylic.

[i]Note.[/i] The [i]orthic triangle[/i] of $\triangle ABC$ is defined as the triangle formed by the feet of the altitudes of triangle $ABC$.

My proof is based on Steiner's theorem which states that the Simson line of a point $P$ (lying on the circumcircle of $ABC$) with respect to $ABC$ will bisect the segment $PH$ where $H$ is the orthocenter of $ABC$.

Consider a special case when $ABC$ is isosceles at $A$. Let $A'$ be the antipode of $A$. Then the orthocenter $H$ is the reflection of $A'$ through $BC$ and the Simson line of $P$ wrt $ABC$ will bisect $PH$.  Hence if we call $T$ the intersection of $PA'$ and $BC$, then the Simson line of $P$ wrt $ABC$ will bisect $PT$.

Back to the problem, let $T$ be the intersection of $B'C'$ and $BC$. Consider the isosceles triangle $B'MC'$ at $M$ and $XY$ the Simson line of the point $A'$ wrt $B'MC'$. Then, based on what we have developed above, we obtain $XY$ passes through the midpoint $N$ of $A'T$.

Since $(TA'BC) = -1$, it follows that $NA'^2 = NB.NC$. On the other hand, since the center of the circumcircle of triangle $A'XY$ is the midpoint of $MA'$, that circle has to touch $BC$ at $A'$. Thus, $NA'^2 = NX.NY$. Therefore, $NX.NY = NB.NC$,or equivalently $B, C, X, Y$ are concyclic


calculus
function
algebra solved
algebra

The maximum value is 2. This follows easily from the fact that the expression is convex in each variable, thus it attains its maximum for a,b,c in {0;1}. It leads to the conclusion

Pierre.

i think he was asking for a solution without using calculus (weierstrass' theorem) :)

which weierstrass theorem do you mean here?
Many theorems refer to him...

that theorem that states that a continuous convex function reaches its maximum in extremes. :)

i've never heard of this theorem.
where can i find it? i searched for the web but got no result.

it's a classical calculus result. with this you can solve most of the problems which are simillar with this one. I forgot to mention that f must be defined on a compact set.

[0,1]x[0,1]x[0,1] is obviously a compact set. :D

[quote="liyi"]find the max value of

a/(bc+1) + b/(ac+1) + c/(ab+1)

where 0\leq a,b,c\leq 1.

(The max value is 2, when a=b=1 and c=0)[/quote]
========================
WLOG we can assume a  \leq b  \leq c
Then
a/(bc+1)+b/(ac+1) + c/(ab+1)  \leq (a + b + c)/(ab+1) (1)

But 0  \leq (a-1)(b-1) => a + b  \leq ab+1 (2)
and  c  \leq 1  \leq ab + 1 (3)

From (1), (2), (3) we can conclude that LHS of (1)  \leq 2. The equality holds, for example, when a = 0, b = c = 1.


[quote="azn6021023"]Logarithm properties

$ \log{ab} \equal{} \log{a} \plus{} \log{b}$
$ \log{\dfrac{a}{b}} \equal{} \log{a} \minus{} \log{b}$
$ \log{(a^n)} \equal{} n\log{a}$[/quote]

So wouldn't $ \log{(1/12)}\equal{}\log{1}\minus{}\log{12}$?

Of course it would. But that doesn't mean that the first representation isn't correct. It just means that you can transform logs in many ways depending on what you need in particular problem. For instance, it could have been also $ \log \frac{1}{4}\plus{}\log\frac{1}{3}$ etc.

Btw, notice that $ \log 1\equal{}0$, so you get $ \log \frac{1}{12}\equal{}\minus{}\log 12$, which is the third property in azn's post.

[quote="elcric"]Why does $ \log{(1/12)} \equal{} \log{(1/6)} \plus{} \log{(1/2)}$?

Thanks![/quote]
Using the properties from azn6021023's post,
$ \log{\frac {1}{12}} \equal{} \log{\frac {1}{(6)(2)}} \equal{} \log{\left[\left(\frac {1}{6}\right)\left(\frac {1}{2}\right)\right]} \equal{} \log{\frac {1}{6}} \plus{} \log{\frac {1}{2}}$

erm.. I guess this sorta shows why the first two properties are what they are:

$ \log{ab}\equal{}\log{a}\plus{}\log{b}$

If $ x\equal{}\log{a}$, then $ a\equal{}10^x$
If $ y\equal{}\log{b}$, then $ b\equal{}10^y$

Plug in those into the LHS to get

$ \log{10^{x\plus{}y}}\equal{}\log{a}\plus{}\log{b}$
$ x\plus{}y\equal{}\log{a}\plus{}\log{b}$, since $ \log_{\text{dur}}\text{dur}^\text{mer}\equal{}\text{mer}$ :D
$ \log{a}\plus{}\log{b}\equal{}\log{a}\plus{}\log{b}$

papaya.

You might want to see the [url=http://www.artofproblemsolving.com/Wiki/index.php/Logarithms#Logarithmic_Properties]wiki article[/url] on this.[/url]


Steve will write the list of consecutive integers, starting with 1, at the speed of 5 digits per second. What will be the sum of the final two digits he writes on his list at the end of 10 seconds?

Well 5 digits per second means 50 after 10 seconds, so the last two are 49 and 50, so the sum is 99.

[hide]5 per second is 50 DIGITS, not numbers. Therefore, he only gets to 3 in 30. The last two digits he writes are the 9 in 29 and the 3 in 30.
9 + 3 = 12[/hide]

[hide=Solution]We start with $[1,9]$ in $\frac{9}{5}$ seconds, then $[10,28]$ in $\frac{38}{5}$ seconds. We only can write three digits more ($\dfrac{3}{5}$ seconds to spare). We write the $2$ in $\textbf{29}$, then the $9$ in $\textbf{29}$, then the $3$ in $\textbf{30}$. We then are done.

The last two digits we wrote were the $9$ in $29$ and the $3$ in $30$. \[9+3 = 12\] And the answer is $\boxed{12}$.

[/hide]


What is the common fraction equivalent to $ 0.0\overline{45}$?


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Let $p\in\mathcal l N^*$.Find the matrix $A\in\mathcal l M_n(C)$ such that :
  $\begin{enumerate} \bullet tr(A)=n\\\bullet A^p=I_n \end{enumerate}$.

From the condition $A^p=I_n$ it follows that all the eigenvalues of $A$  are of the form $e^{i\alpha_k},k=1,...,n$.Thus $|tr(A)| \leq \sum|{e^{i\alpha_k}}|\leq n$, which means that all the eigenvalues are $1$.The characteristic polynomial of $A$ is of the form $P_A(x)=(X-1)^n$, hence the minimal polynomial $m_A(x)=(X-1)^k$ for some $k \leq n$. Since the polynomial $f(x)=x^p-1$ annulate $A$, $m_A(x)=(X-1)^k$ divide $f(x)=X^p-1$ and since the last one has only simple roots in $\mathbb{C}$ we have that $p=k=1$ and $A=I$.

The eigenvalues are egual to $1$ I don`t understand what you did.

Yes, i misunderstood the problem condition

if we prove that the eigs are all equal to 1, we can prove ,by assuming the contrary, that the minimal polynomyal of $A$  is $X-1$

all the eigs are 1, and we prove by assuming the contrary that the minimal polynomial is X-I...   :ninja:

all the eigenvalues are equal to $1$. Moreover the matrix is diagonalizable. So it is similar to the identity:  it is the identity.


It looks like the URL for this "comprehensive exam" is too long to fit into the source field, so here it is

http://www.math.unl.edu/image/graduate/PhDComps/958Jun1995.pdf

This appears to be set at a higher level than the other exams I have posted; does it look that way to you?

[quote]Question 1. State and prove Turan's Theorem concerning graphs not containing Kr.[/quote]

is that really problem solving? :?

Maybe this set of problems belongs in College Playground. Part of what I'm doing here is trying to figure out the level of the higher-level forums here on the new merged site. I don't have moderator powers here, so I don't think I can move the thread myself, but I would be happy to have it moved.

it's not because it's too hard. It's just: "state and prove the theorem of ..."

that's nothing but schoolish study work. I just don't get the meaning of this on a problem solving site :)


Well, I'm not sure if this is a silly question, since I know nothing about number theory yet, but...

I was reading a proof about the probability of picking 2 random (integer) numbers that were relatively prime... And at the proof it was stated the following:

[quote]Let $ m$ and $ n$ be the given random numbers and let  $ a$ be any prime.
[b]Now $ 1/a$ of all the integers are divisible by $ a$, because $ a$ divides every $ a$'th one.[/b][/quote]

I verified this proposition... It was indeed true, but could someone, please, prove it?

Maybe, I am just too dumb to do it, but I was quite troubled because my possible (ugly) solution was the following:

- All numbers can be written in a essentially unique combination of prime factors (OK), but how does someone finds the probability of the numbers in a given range being divisible by a given (prime) factor?

I mean for {1..20}, we have a x number of primes and in {1..100} we have y number of primes, is is possible to know the number of primes in a given interval?

And we kinda need this value to compute the probability, or not?

This has nothing to do with the distribution of the primes.  The numbers divisible by $ a$ are $ a, 2a, 3a, 4a, \ldots$, i.e., every $ a$th number.  Thus (in some sense that can be made precise) exactly $ \frac{1}{a}$th of the positive integers are divisible by $ a$.

Yeah, I just figured (I just woke up) that one thing was not linked to another...

Thanks for answering.

But I still didn't quite get it =/ (sorry for being dumb)

I am probably missing something really elementary or am just too amazed to believe it, but what does allow someone to state that:
that $ 1/a$ th of the positive integers are divisible by a? For example for 2, we have $ 1/2$ and for 3 we have $ 1/3$.

Why is 1/2 > 1/3 being both primes? Why is that there are more (in a finite range) $ 2$th numbers then $ 3$th numbers?

For instance:
$ 1$
$ 2$
$ 3$
$ 2^2$
$ 5$
$ 2 . 3$
$ 7$
$ 2^3$
$ 3^2$
$ 2 . 5$

Is it something like that?
Okay, a term is the last one plus 1.
2 appears every 2 of them.
3 appears every 3 of them. - if this 3th is also a multiple of 2 (6th) it is also divisible by 2
5 appears every 5 of them. - the same occurs as above
..
an so on.
Are these the so called $ a$th numbers?

A simple argument for this is by using modular arithmetic. Consider a set of $ n$ arbitrary integers, and create $ n$ buckets, labeling each bucket with an integer from $ 0$ to $ n\minus{}1$. Place each integer in the bucket corresponding to its remainder when divided by $ n$. Assume one bucket has two elements, their difference must be divisible by $ n$, yet since the largest possible difference is $ n\minus{}1$, no bucket can have more than one element.

Therefore, among every $ n$ consecutive integers, one integer is divisible by $ n$.

If you consider the integers from $ 1$ to $ an$, you get that all the integers of the form $ kn$ for $ 1 \le k \le a$ are divisible by $ n$, and none else.

Thanks a lot! Both of you helped me a lot.

$ a$th, as in, 4th, 5th, 6th, 7th, 8th, etc.  (Also, this basic property does not rely on $ a$ being prime: in the sense we're talking about, $ \frac{1}{4}$th of the integers are divisible by 4, $ \frac{1}{6}$th are divisible by 6, and so on.


Let $ x, y, z > 0$. Prove that the following inequality holds:
\[ \frac {x^2 \plus{} y^2 \plus{} z^2}{xy \plus{} yz \plus{} zx}\geq 1 \plus{} \left(\frac {x \plus{} y}{z} \plus{} \frac {z \plus{} x}{y} \plus{} \frac {y \plus{} z}{x}\right) \minus{} 4\left(\frac {x}{y \plus{} z} \plus{} \frac {y}{z \plus{} x} \plus{} \frac {z}{x \plus{} y}\right).
\]

the inequality is not true...

it is equivalent to $ \sum\dfrac{(x \minus{} y)^2}{2(xy \plus{} xz \plus{} yz)}\geq \sum\dfrac{z(x \minus{} y)^2}{xy(x \plus{} y)}$

if we let $ y \equal{} z$ we are left to prove that $ \dfrac{(x \minus{} y)^2}{2xy \plus{} y^2}\geq 2\dfrac{(x \minus{} y)^2}{x(x \plus{} y)}$, which is not always true...


[quote="vishalarul"]Prove that $n!<\frac{(n+1)^{n}}{2^{n}}$ for all positive intgegers $n>1$.[/quote][hide]AM-GM on the first positive $n>1$ integers: \[\frac{1+2+3+\ldots+n}{n}>\sqrt[n]{1\cdot2\cdot3\cdots n}.\] Hence, \[\frac{(n)(n+1)}{2n}>\sqrt[n]{n!},\]  \[\frac{n+1}{2}>\sqrt[n]{n!}.\] Raising both sides to the $n$th power gives \[n!<\frac{(n+1)^{n}}{2^{n}},\] as desired.[/hide]

Already posted:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=103273


Assume that it is wrong, i.e. $ \min\left\{\left(a-b\right)^{2},\left(b-c\right)^{2},\left(c-a\right)^{2}\right\}>\frac{1}{2}\left(a^{2}+b^{2}+c^{2}\right)$. Then $ \left(a-b\right)^{2}+\left(c-a\right)^{2}>a^{2}+b^{2}+c^{2}$ which is equivalent to $ 3a^{2}>2a\left(a+b+c\right)$. If $ a=0$ it is a contradiction, if $ a\neq 0$, it gives $ \frac{a+b+c}{a}<\frac{3}{2}$ or $ \frac{b+c}{a}<\frac{1}{2}$. Adding up two analogously produced inequalities we get $ \sum \frac{b+c}{a}<\frac{3}{2}$ which is clearly false since $ \sum\frac{b+c}{a}= \sum\frac{b}{a}+\sum\frac{a}{b}\geqslant 6$.


conics
ellipse
hyperbola
geometry proposed
geometry

let A,B be the foci of an ellipse
let P and Q be two points on that ellipse
let R and S be the intersection of PA, QB and PB, QA respectively
prove R and S are also on an ellipse (different) with the same foci A and B

Dude, I'm, looking at it on my piece of paper and it doesn't look like an ellipse. Are you sure R and S aren't on a hyperbola of foci A and B?

I know this related problem (I'll use the same notations):

The following conditions are equivalent (PRQS is a quadrilateral, B is the int of PS and RQ; A is the int of PR and SQ):

a) PRQS is a circum,scribed quadrilateral;
b) PA+PB=QA+AB;
c) RA-RB=SA-SB.

This problem you posted is equivalent to proving b)<=>c), but this would prove R and S on a hyperbola of foci A and B.

I think I could post a soln of that prb I posted.

I knew I missed something!! sorry.  P and Q are suppose to be on the *opposite* side of the major axis

I don't really have time for a complete soln right now, but the problem is quite similar to the one I posted on this very topic 2 posts ago ( :D ). We use the fact that the quadrilateral AQBP has an exinscribed circle (which isn't, I think, hard to prove, although I haven't actually written a soln), meaning that there is a circle tangent to all the lines AP, PB, BQ, QA, only that this circle is on the outside of the quadrilateral. Then we just use the fact that the tangents from a pt to a circle are equal (we use this a couple of times).


Καλησπερα, χτες εγινε στο Πολυτεχνειο διαγωνισμος μαθηματικων μεταξυ των σχολων ΣΕΜΦΕ και Πολιτικων Μηχανικων, ενα απο τα προβληματα που μπηκαν ηταν το παρακατω:

Να βρεθει το οριο:

$ \lim_{n \rightarrow \plus{}\infty}{tan(\frac{\pi}{2k\plus{}1}(\sqrt{n^2 \plus{} m}))}$ οπου k ειναι θετικος ακεραιος και m θετικος πραγματικος αριθμος

Μονο μια λυση μπορεσα να σκεφτω αλλα δεν γνωριζω εαν ειναι σωστη. Η λυση που εδωσα εκανε χρηση υπακολουθιων και εδειξα οτι το οριο δεν υπαρχει (και πραγματι δεν υπαρχει). Το σημειο για το οποιο δεν ειμαι σιγουρος ειναι το εξης:
Πηρα την υπακολουθια για $ n \equal{} a(2k\plus{}1)$ και αφου πηρα το $ n^2$ κοινο παραγοντα και το εβγαλα εξω απο τη ριζα ειπα οτι κοντα στο απειρο η παρασταση ειναι ασυμπτωτικα ιση με $ tan(a\pi)$ το οποιο ειναι 0.

Η Ιδια μου ηρθε οταν θυμηθηκα τη λυση που ειχε δωσει o Durandal εδω:
http://www.mathlinks.ro/Forum/viewtopic.php?t=281271

Θα ηθελα να μου πειτε αν χανει καπου η λυση και αν πραγματι χανει θα ηθελα να δω τις δικιες σας λυσεις.

[quote="Durandal"][quote="NickNafplio"]Μονο μια λυση μπορεσα να σκεφτω αλλα δεν γνωριζω εαν ειναι σωστη. Η λυση που εδωσα εκανε χρηση υπακολουθιων και εδειξα οτι το οριο δεν υπαρχει (και πραγματι δεν υπαρχει). Το σημειο για το οποιο δεν ειμαι σιγουρος ειναι το εξης:
Πηρα την υπακολουθια για $ n \equal{} a(2k \plus{} 1)$ και αφου πηρα το $ n^2$ κοινο παραγοντα και το εβγαλα εξω απο τη ριζα ειπα οτι κοντα στο απειρο η παρασταση ειναι ασυμπτωτικα ιση με $ tan(a\pi)$ το οποιο ειναι 0.[/quote]

Huuuh? Σε χάνω Nick... :huh:

Το όριο αυτό καθαυτό είναι μάλλον τετριμμένο... αφού $ \sqrt {n^2 \plus{} m}\sim n$ για μεγάλα $ n$, το όριο προφανώς δεν υπάρχει (επειδή το όριο της $ \tan x$ δεν υπάρχει... αυτό εννοείς ότι σου θύμισε αυτό που είχα γράψει?). Τώρα, αν αυτό δεν αρκεί σε αυτούς που διορθώνουν, too bad ;)

Cheerio,

Durandal 1707[/quote]

Αυτο ακριβως εδωσα ως λυση, απλα με λιγο διαφορετικο τροπο (βγαζοντας πρωτα το n εξω απο τη ριζα), μετα απλα εδειξα με υπακολουθιες οτι το οριο της εφαπτομενης του νπ δεν υπαρχει... την ιδια ασκηση εψαξα και τη βρηκα μετα το διαγωνισμο στο βιβλιο του καθηγητη που το διοργανωσε, αλλα πισω στις λυσεις το μονο που δινει ειναι ενα hint οτι το limsup και το liminf δεν ειναι ισα, πραγμα το οποιο χρησημοποιουμε για να δειξουμε οτι το οριο tanνπ δεν υπαρχει. Ας ελπισουμε οτι θα το παρουν για σωστο :P


I had to leave early for a piano festival, so can anyone tell me the results?  I epically phailed my ciphering and possibly my written, so I won't be too surprised when I don't place. :)

Anyways, I thought the ciphering was pretty difficult and the written was all right (though they had calc on the test...how dare they?!!!) apart from that parking question...  Grr...  Everyone knows I'm too short to even reach the gas pedal let alone parallel park a car and know the math behind it...

This is a good example of why you shouldn't moan about how terribly you do on every test, because it's really inconsiderate of other people when you do better than they do.  Maybe it seems like humility to you, but it is really not nice.  Try to think more of other people.  You got fifth; I was first.  I think Claus was second, Maja was third, and Allen was fourth.  Northview won varsity and JV, and I think Walton was second in varsity.

I think the ciphering problems would not have been too hard as written test problems, but it was difficult to complete them in the short time limits.

There seemed to be more people at the tournament than there were two and three years ago---a good thing, of course, but it means that it was harder for teams and individuals to place.

I have to completely agree with Miles on this one. Maybe you DIDN'T perform to your expectation, but there are a lot of people who will NEVER even get to your poor level. Talking about how you (cough, cough) FAILED, when you beat almost everyone else is just rude.

Congratulations to everyone who did well.
Tom

Sitan, you know that means I'm coming after you.

I remember two years ago, when I was in tenth grade,  Norcross went to its first real out of school tourney and I placed third in JV at Lassiter.  And I wasn't at awards because of Chinese school, so the people said that I lost to second place by like 3 points and by first place at 30 points.  I still haven't forgotten who got 1st place, Mr. I only got $ \frac{236}{250}$.

And I'll need to talk to Norcross about not even hearing about this tourney.

Well, almost all of the open math tournaments in Georgia are posted at [url=http://www.gctm.org]GCTM website[/url].  Currently, the URL is http://www.gctm.org/Default.aspx?pageId=130271 .  (I think the tournaments that aren't posted are mostly meets that are run by and open only to the schools of a particular county.)

Varsity Individuals
1. Miles [I didn't hear the score]
2. Claus 150
3. Maja [I didn't hear the score]
4. Allen 130
5. Sitan 130 [Broken by written score]

Varsity Teams
1. Northview 465 [People tell me it was 460, but I heard 465]
2. Walton 420
3. Alpharetta 400
4. Pace Academy 315
5. Rockdale 300

I didn't scrawl down JV rankings, but Northview won JV team.

I would pitch in on the lashback against Sitan, but I think there's already been enough <_<.

Congratulations to Miles and Northview!

Um, maybe it's just me, but aren't you guys misinterpreting my post a bit?  I said I "epically phailed" out of humility, not because I was so inconsiderate as to denounce the achievements of others who placed behind me.  Perhaps you guys don't really know me that well,  but when I said I "epically phailed ciphering," I meant that my ciphering score of 35 was significantly lower than any ciphering score I've ever received.  True, others may have scored lower, but how would I know that?  I appreciate all of the responses you guys had, but I don't think a misunderstanding of my post deserves that much criticism of my character.  I'll try to be more positive next time, but please don't add to the handful of comments denouncing my "rudeness" and "inconsiderateness."  I'll stop using the word "fail" and any variations of it.  I hope I'm up to the task... :(

Perhaps I just don't understand how people should respond to their performance at a competitive event, in which case I apologize.  So yeah, don't hate me for my misspeaking.  I'm sorry for my behavior and I'll try to improve my outlook on life next time.  And no, that wasn't sarcasm.

I don't think it's as much as you complaining about the difficulty of the ciphering in that time limit, but it's more because you constantly complain about phailing, even before you learn your placing and while your placing is perfectly good. I still remember a time when there was a max score of 8, but a very good score was actually a score of 2 or 3(Eric's practice ARML problems). How you did worse numerically to other tournaments is no measure of how you actually did.

Too many people assume they "fail" and do horribly and don't realize their true potential. This isn't just you; there are several others I know who "fail" constantly while doing incredibly for their age. Frankly, I wish they would stop that.

Sorry for being harsh, Sitan.  I don't doubt that you were trying to be humble, and [b]you are by no means the only offender[/b].  But when you are interacting with other people, the effect that you actually produce is more important than the effect you intend to produce.  Another person looking at what you post may or may not see you trying to be humble.  But whether or not you see that, it is discouraging to see somebody else complaining about doing better than you.  It's like the difference between trying to let somebody else know that you're trying to be polite and actually making the person feel better.  Does that make sense?

I'm glad you're changing your ways. :)

What they all said. I wasn't trying to bash on you Sitan. I agree that there are some that "fail/phail" way to often! Maybe this thread should be printed and given to some of those offenders!

See you folks soon!

Hey, just saw this thread.

Just wanted to add, we stayed for a bit after the tournament, and Santosh's number got messed up or something. So, he actually got 2nd I think. o.o

The whole series of posts were supposed to be NICE to Sitan, Heng.

[quote="FrisbeeBaller425"]Sitan, you fail.[/quote]

Um...okay...  You fail even more and have no life or friends and should go suck on something?...

Maybe we should stop.

Can someone lock this or something? This might be like a potential flame thread or w/e...>.<

Um, calm down, this is a local community forum.  No admin cares enough to lock this, and no mods exist for this forum. :wink:  :wink:

Which doesn't really mean that we shouldn't stop.

[quote]Sitan, you fail.[/quote]<== Unneeded, provoking post. To be ignored.

[quote]Um...okay... You fail even more and have no life or friends and should go suck on something?...[/quote]<== Not ignoring.

Of course, I acknowledge that I also should be ignoring this. I apologize for my misbehavior.

[quote="Arvind_sn"], and Santosh's number got messed up or something. So, he actually got 2nd I think. o.o[/quote]
I tried to register as an individual, but the tournament wasn't set up for individual scoring. So, Mr. Ogelsby was kind enough to let me he Mount de Sales team and not pay anything :). Since I registered before the coach from Mount de Sales did, they gave me the number 181 for ciphering. However, since they knew that two students from Mount de Sales would show up, they gave me the scantron with 183, and I thus assumed that I was also 183 for ciphering, and thus was recorded as having 0 points on the ciphering round.

Fortunately, this was all straightened out, and I got a score of 105 + 45 = 150. I'm not sure what place that was.

Anyway, while I felt that the tournament was highly computational and the ride up there was boring, it was free for me. Also, lunch and a few* free cupcakes made it well worth competing.

Contrary to the discussion, I will say that I failed ... at getting more than one person from my school to attend the competition. But we were aware of its existence, which is an improvement from last year. Maybe next year, we could shoot for 2 students to come to this competition.

I believe I am noticing a trend. (I might be wrong though) For every tournament this school year that exactly two of Miles, Eric, and I attend:

If Miles and I attend, Miles wins.
If Eric and I attend, I win.
If Miles and Eric attend, Eric wins.

Anyone else notice this, or have a counterexample from this year?

Hm, well, tournaments that exactly two of us attend . . . don't happen very much, I think.  In particular, I think I've only been to Cobb Invitational, Rockdale, UGA, North Springs, and this this year (aside from the Cobb County meets).

By the way, we do have a moderator (Howard).  And administrators do care about the local forums, but I don't think they read most of them very often.  I'm pretty sure we've only had to lock a thread once, and that was highly amusing.


Let $H$ an interior point of a triangle $ABC$. Lines $AH , BH , CH$ meet the sides of the triangle at points  $D, E, F$  respectively. If $H$ is the incenter of the triangle $DEF$ , prove that $H$ is the orthocenter of triangle $ABC$.
[u] Babis [/u]

 [i]This a variation of an unsolved problem that I found in an old issue of Gazeta Matematica in 1991.I hope you will find a better solution - mere elementary -  than me [/i]

[color=darkblue][b]Hello, Stergiu ![/b] Here is an easy and well-known property : " $AD\perp BC$ $\Longleftrightarrow$ $\widehat{EDA}\equiv\widehat{FDA}$ ".
[b]Proof.[/b] I"ll with the orientate segments. Let $d$ be the line for which $A\in d$ and $d\parallel BC$. Denote the intersections $M\in d\cap DE$ and $N\in d\cap DF$.
$d\parallel BC$ $\Longrightarrow$ $\frac{EC}{EA}=\frac{DC}{MA}\ ,\ \frac{FA}{FB}=\frac{DB}{NA}$. Apply the [u]Ceva's theorem[/u] to the point $H$ for the triangle $ABC$ :
$\frac{DB}{DC}\cdot\frac{EC}{EA}\cdot \frac{FA}{FB}=-1$ $\Longrightarrow$ $MA=AN$. Therefore, $AD\perp BC$ $\Longleftrightarrow$ $DA\perp MN$ $\Longleftrightarrow$ $\widehat{MDA}\equiv\widehat{NDA}$.

[b]Remark.[/b] You prove immediately this problem with the harmonical division.
Denote the intersections $X\in AD\cap EF$ and $Y\in EF\cap BC$.
Thus, $(B,C;D,Y)$- h.d. $\Longrightarrow$ $(E,F;X,Y)$- h.d. and $d\parallel BC$ $\Longrightarrow$ $NA=AM$.
Therefore, $AD\perp BC$ $\Longleftrightarrow$ $XD\perp BC$ $\Longleftrightarrow$ $\widehat{EDX}\equiv\widehat{FDX}$ $\Longleftrightarrow$ $\widehat{MDA}\equiv\widehat{NDA}$. [/color]

hallo Virgil !

 I'm glad to see you here again for the new school -and olympiad - year !!!

  I have also used harmonic ratios.I liked also the solution by Ceva.
  [u]Babis[/u]


Consider a triangle with vertices $A(0,0,0),B(1-a,0,0),C(0,1,a)\ (0<a<1)$.Find the value of $a$ for which the volume of solid body which is generated by rotating the triangle $ABC$ around the $z$-axis is maximized, then the maximum value of the volume.


We have $2005$ points in the plane such that any three are not collinear. Prove that there is a circle which contains in its interior $1002$ and in its exterior another $1002$ points.

[i]Petre Bătrăneţu[/i]

Nobody wants to take a shot at this question? I couldn't solve it  :blush:

Specifically, what does it mean by "Interior" and "Exterior"? I'm not understanding it. Does it mean inside the points so no points are left out of the circle is considered "Exterior" and if there are points outside of the circle, it is "Interior"? Please someone clarify.

A circle of center $O$ and radius $r$ is the locus of the points $A$ such that $OA = r$.

The interior of a circle of center $O$ and radius $r$ is the locus of the points $A$ such that $OA < r$.

The exterior of a circle of center $O$ and radius $r$ is the locus of the points $A$ such that $OA > r$.

stancioiu: for "mediator," use "perpendicular bisector."

Alternatively, it's easy to find a line which has the desired property -- passes through some point P of the set and has 1002 other points on each side.  Say that, aside from P, the next nearest point in the set to the line is at distance d.  It is easy to find a (very very large) circle which is tangent to the line at P and which is never at distance greater than d/2 from the line inside the convex hull of the points.

A similar method: take a point not on any line joining two points in the set.  Invert in some circle around this point.  Find a line in the inverted space with the desired properties, and invert again.  The line becomes a circle which works, I believe.

[quote="stancioiu sorin"]Yes, you're right.Neither me I can solve it. :( But I post Marius Ghergu's solution.

  We can take all the mediators of the segemnts formed by the $\ 2005$ given points.They are $\ C_{2005}^{2}$.We take a point $\ O$ which isn't on these mediators.Let's note the points begining whith the closest, and the we have $\ OA_{1}<OA_{2}<...<OA_{2005}$.We observe that $\ OA_{i}$ isn't equal whith $\ OA_{j}$, because $\ O$ cannot be on the mediator of the $\ (A_{i}A_{j})$.The cirqle whith the radius $\ OA_{1003}$ will contain the points $\ A_{1},A_{2},...,A_{1002}$ in the intherior(there are $\ 1002$ points) and the points $\ A_{1004},...,A_{2005}$(there are $\ 1002$ points)..sorry for my english :blush:
[hide="ps"]mai puteai folosi inversiunea de pol[/hide][/quote]
waht is mediatror?
and there need to be one point on the circumference of the circle right?

[quote="srulikbd"]and there need to be one point on the circumference of the circle right?[/quote]
Yes, that would be $A_{1003}$.

Well, there are a couple of solutions for this nice problem.
First solution: Note that if by some construction we can make the points sorted in ascending order according to distance from a center, we are done. And such a point can't be on a perpendicular bisector of any line created by the pairs of points. So, consider a point $O$ that's not any perpendicular bisector of all possible pair of lines. Then $OP_1<OP_2<...<OP_{2005}$ in some order. Therefore, draw a circle with radius $OP_{1003}$ and we are done.

Second solution: Sort them based on angles created with X(or Y) axis and the left-most point of the set($P$). Then join all points with $P$ and we get $2014$ lines. Consider the $1003$th line. It has exactly two point on it, $1001$ on one side and $1003$ on the other. Now, draw a circle that touches this line at $P_{1003}$. Translate its center until it covers $1002$ points exactly.


geometry inequalities rectangle calculus function algebra polynomial

Here is a good problem I found

Let ABCD be a rectangle with BC=10. Let E be a point on CD. If you fold B to E, the crease is GH with G lies on BC and H lies on AD. Find EC when area of triangle GCE is maximized.

BTW it's a challenge.

Its easy to do if you can use calculus or a calculator. let GC = x and CE = y. We are trying to maximize xy/2. GE = BG because BG is folded onto GE, therefore GE = 10 - x. Now we can find y as a function of x, y = sqrt((10-x)^2 - x^2).  xy/2 has the same maximum as x^2*y^2, which is x^2*(100-20x). Maximizing this whatever way you like, you get that x = 10/3, y=10/sqrt(3), xy/2 = 50/(3*sqrt(3))

The only way I can do it is to find the maximum value of a 3rd degree polynomial within certain bounds.  Are you sure this can be done with < 3rd degree?

Edit: I therefore second the person before me (he must have posted while I was writing).

[quote="schulmannerism"] Maximizing this whatever way you like, you get that x = 10/3, y=10/sqrt(3), xy/2 = 50/(3*sqrt(3))[/quote]

How did you maximize it? and please, no calculator or calculus

probability1.01 wrote:The only way I can do it is to find the maximum value of a 3rd degree polynomial within certain bounds.  Are you sure this can be done with < 3rd degree?

I am not quite sure what you are doing but it's a lot simpler than that.

hint: using schulmannerism's labeling and inequality( maximizing (x)^2(5-x))   [hide]Use AM-GM inequality[/hide],

bigger hint: [hide]how can you change (x)^2(5-x)) such that the changed product can be expressed as product of 3 numbers that the add up to a constant [/hide]

that is a 3rd degree polynomial (similar to mine, but different definitions were used).  I don't see how [hide]AM-GM[/hide] helps.

Practically give it away:

 [hide]x^2(x-5)/4=(x/2)(x/2)(5-x). (x/2)+(x/2)+(5-x)=5. Now use AM-GM.[/hide] Do you see it now?

oops! x can't be negative or :ge: 5. duh! Sorry I hadn't been paying enough attention to the problem.


inequalities
vector
real analysis
real analysis unsolved

Exercise is from analisys and linear algebra.

Let $\{x_{1},..., x_{m}\}$ be a linear independent set of vectors in the normed space $X$. Prove that
$\exists c>0 \forall \alpha_{1},..., \alpha_{m}\in K \parallel \alpha_{1}x_{1}+...+\alpha_{m}x_{m}\parallel \geq c(| \alpha_{1}|+...+|\alpha_{m}| )$.

I have been thinking over this exercice for a few hours and I haven't reached the aim.  Thank you for help.

do you know that on a finite dimensional normed vector space, all norms are equivalent ? This is a consequence.

Yes, I know that in every finite dimensional spaces two norms are equivalent.

However I have misprinted the exercise, now it is written correct. At the right side of the inequality there is the sum of absolute values (not the norm of alpha's). Sorry for the mistake. Now it's not so clear.

Same thing: you can define another norm on the linear span of $x_{i}$s by $\parallel \sum \alpha_{i}x_{i}\parallel _{1}: =\sum|\alpha_{i}|$. This norm must be equivalent to $\parallel \cdot \parallel $, hence the conclusion.


calculus
integration
logarithms
calculus computations

I have two integrals containing $\ln(x)$. I did one of them with the exception that i am EXACTLY 1 one from the result. I don't know how to approach the second task though.

$1)$

$\int_e^{e^2}\frac{1}{x\cdot \ln(x)}dx=\int_e^{2e^2}\frac{1}{t}dt$

$\int_e^{e^2}\frac{1}{x\cdot \ln(x)}dx=[\ln(t)]_e^{2e^2}$

$\int_e^{e^2}\frac{1}{x\cdot \ln(x)}dx=1,6931$

...which is exactly 1 above the result.

$2)$

$\int_1^e\frac{\ln(x)}{x}dx$

If you could give me a clue i should be able to do the rest of it myself. This was actually the first of the two tasks, and for that reason it ought be to easier than the above stated task.

For 1, what substitution are you using?

For 2, you have in the integral a product of two things, one of which is easy to derivate and the other of which is easy to integrate.  This should suggest something to you.

$1)$

$\int_a^bf(g(x))\cdot g'(x)dx=\int_{g(a)}^{g(b)}f(t)dt$

In this case:

$f(x)=\frac{1}{t}$, $g(x)=ln(x)$ and $g'(x)=\frac{1}{x}$

This is how my book describes it. I don't know if you are american, in that case you might be using different words.

EDIT: Nevermind, now i can see what is wrong here. It is my new bounds that are not right. The new bounds are $\int_1^2$, right?

$2)$

$\int_1^e\frac{\ln(x)}{x}=[\ln(x)\cdot \ln(x)]_a^b-\int_a^b\frac{1}{x}\cdot \ln(x)$

So you get a new integral that is equal to the first one, right?

Yes and yes.  On the second one, once you've reached that step you can solve for the value of the integral directly (Xevarion would write that you have I = A - I) and you just have to evaluate that expression A and you're done.

These are examples with nice indefinite integrals as well, which the same methods will reveal.

OK, this was the second time i was told how to do handle such an integral - now i should be able to handle them again. I just forgot that point, sorry. Thx for your help ;)

You could also do 2) directly by substitution. Let $u = \ln x$ and $du = 1/x \ dx$

Then your integral becomes,

$\int_1^e \frac{\ln x}{x} dx = \int_0^1 u \ du$

OK, i thought i couldn't use the substitution method as i had only a $u$ and a $du$. But i will take that into consideration for another problem. Thx...


Evaluate $ (22 \times 6) \plus{} (6 \times 38)$.

Instead of multiplying out, we notice that there is a factor of $ 6$ in each term. Factoring out makes it much easier: $ 6(22\plus{}38)\equal{}6(60)\equal{}\boxed{360}$.


Does this have anything with hexdecimal system? If it does, I'm too lazy to convert.

Different base, but that's not the main cypher.

is the plain text:
[hide]canyoudecodethismessage[/hide]?


I study Computer Science in Italy, and, after the 3-year graduation (maybe I'll also do the 2-year Specialization here), I hope to have the chance to study in some American college (Master or PhD).

Which are the requirements in order to have a chance to be accepted? I mean, I don't think it's enough to have the maximum note in the graduation.

A recent thread: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=139142]Graduate school - application process[/url]
Computer science is not my field. I have a (possibly wrong) impression that a significant number of CS graduate students are supported by a professor's research grant or startup fund. If this is the case, having a prior contact with someone at the department could be very important.

The admissions requirements vary widely depending on the university that you want to attend.  The quality of the programs vary quite a lot as well.  If you have a good mathematics background (at least a few Calculus courses and Linear Algebra,) good marks in a decent CS program, good programming skills, money, and passing scores on the Test of English as a Foreign Language, then you can probably find a program in the US that will accept you.

The real question for you is what do you really hope to accomplish with your degree and stay in the US.  Schools with the more restrictive admission standards are often those that can most easily help you achieve your goals.

There are many graduate programs in CS at regional and local universities in the US that need students.  Enrollments fell dramatically after the dot-com bust and are just starting to rebound a bit.  Foreign nationals often make up large portions of technical graduate programs in the US, especially in Computer Science.  There are other computer-related programs in the US including Information Technology and Management Information Systems.  They tend to be less technical and more business-related than CS.

The better programs tend to have funds to support graduate students with assistantships (teaching or research.)  Weaker programs may not be able to cover your tuition costs with funding.

Thanks to everybody for the complete replies.

I'm definitely interested in pure CS and not in other CS related programs. My hope is to find some good programs which could support me with their funds, since I don't think I'll have enough of it to live and study in the US. I do have a Calculus and Linear Algebra background (and still have time to improve it, since I'm only in the first year of graduation), programming skills, and some Cambridge certificates for English Tests, so I see that I have some chances.

Regarding the quality level of an eventual college where I'd like to study, I sure I want to try to some of the finest and most famous ones, but I'll always be ready to try something else.


An argument against evolution, "Because evolution does not fully explain everything , ID is right/a resonable alternative", is Pyrrhonic.

This brings it really close to being a logical fallency, because there are some underlying assumptions by evolutionists that are basic biology facts, and are better explained by a biologist than someone arguing about evolution. Questioning the basics ad infinum is idiotic once you reach a point.

Also, I have been unable to get some kind of theory as to how the earth/universe was created, why it happened, when, where, by what methods, etc.

It might just be me, but ID seems more like saying "Evolution is wrong, and because it is wrong, our ideas are right, and if you question our ideas, you are wrong because evolution is wrong."

ID is creationism. In order to be presented in a scientific guise, it is dressed up in more "scientific" terms like "designer."

ID is consitent with everything - there is nothing that can imply that a designer can exist, and nothing that can imply one doesn't. ID is completely consistent with evolution.

In order to attempt to argue with evolution as a theory, ID supporters falsely assume that ID and evolution are mutually exclusive and cover all possibilities (ie, it's either one or the other, not neither nor both.) This is pretty obviously false - neither is evolution the sole possiblity not ID exclusive with evolution.

[quote="chess64"][quote="PenguinIntegral"][b]It might just be me[/b], but ID seems more like saying "Evolution is wrong, and because it is wrong, our ideas are right, and if you question our ideas, you are wrong because evolution is wrong."[/quote]

It [i]is[/i] just you. ;)[/quote]
And me.

If you look at the ID/evolution thread, you will notice that there have been several posts asking for evidence for ID rather than evidence against evolution, and none has been given.

[quote="PenguinIntegral"][quote="chess64"][quote="PenguinIntegral"][b]It might just be me[/b], but ID seems more like saying "Evolution is wrong, and because it is wrong, our ideas are right, and if you question our ideas, you are wrong because evolution is wrong."[/quote]

It [i]is[/i] just you. ;)[/quote]
Ok, so ID would involve how the world was created? Because then it wouldn't be arguing about evolution, we would be into the big bang, and that is COMPLETLY different.[/quote]

Ah, but if ID involves how the world was created, ID would be religious, wouldn't it?

Locked because there are far too many evolution/intelligent design threads. Please keep it to a reasonable number and as of now, stop creating more.

You folks shoudl also read my comments on the (now locked) "how to argue..." thread.


inequalities
function
inequalities unsolved
BPSQ
india

If $ a$ and $ b$ are positive and $ a \plus{} b \equal{} 1$, prove that
\[ \left(a\plus{}\frac{1}{a}\right)^2\plus{}\left(b\plus{}\frac{1}{b}\right)^2 \geq \frac{25}{2}\]

Jensen kills it! This inequality often comes in Japanese university entrance exam, few students would notice Jensen, but most students, if they can solve it, they would use AM-GM.

$ \[ \left(a\plus{}\frac{1}{a}\right)^{2}\plus{}\left(b\plus{}\frac{1}{b}\right)^{2}\ge 2\left(a\plus{}\frac 1a\right)\left(b\plus{}\frac 1b\right)\equal{}X$

$ X\equal{}\frac{2(a^2\plus{}1)(b^2\plus{}1)}{ab}\equal{}2ab\plus{}\frac{2a^2\plus{}2b^2\plus{}2}{ab}\equal{}2ab\minus{}4\plus{}\frac{4}{ab}$

Let $ ab\equal{}y$

$ X\equal{}2y\minus{}4\plus{}\frac{4}{y}$, we have to prove $ 2y\plus{}\frac{4}{y}\ge \frac{33}{2}$, when $ 0\le y\le \frac{1}{4}$

Let $ f(y)\equal{}2y\plus{}\frac{4}{y}$

For the given interval, if $ p>q$, then $ f(p)<f(q)$ with easy check.
Then $ f(y)$ is smallest for $ y\equal{}\frac 14$, when it is exactly $ \frac{33}2$

[quote="Leon"]If $ a$ and $ b$ are positive and $ a \plus{} b \equal{} 1$, prove that
\[ \left(a \plus{} \frac {1}{a}\right)^2 \plus{} \left(b \plus{} \frac {1}{b}\right)^2 \geq \frac {25}{2}\]
[/quote]
Applying $ x^2\plus{}y^2 \ge \frac{1}{2}(x\plus{}y)^2$ with $ x\equal{}a\plus{}\frac{1}{a},y\equal{}b\plus{}\frac{1}{b}$ we just have to prove:
$ \frac{\frac{1}{a} \plus{} \frac{1}{b}}{2} \ge \frac{2}{a\plus{}b}$ which is AM-HM for two variables.

You can prove $ 2y\plus{}\frac{4}{y}\ge\frac{33}{2}$ also with AM-GM
$ (2y\plus{}\frac{1}{8y})\plus{}\frac{31}{8y}\ge 2\sqrt{\frac{1}{4}}\plus{}\frac{31}{2}\equal{}\frac{33}{2}$

I have just found my solution [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=310929#310929]here[/url] :lol:

As you mentioned Jensen I thought of $ f(x)\equal{}(x\plus{}\frac{1}{x})^2$, which solves it even quicker:
$ f(a)\plus{}f(b) \ge 2f(\frac{a\plus{}b}{2})\equal{}\frac{25}{2}$

Generalization:

Let $ a_k>0\ (k\equal{}1,\ 2,\ \cdots \ , n)$ with $ \sum_{k\equal{}1}^n a_k\equal{}1$.

Prove that $ \sum_{k\equal{}1}^n \left(a_k\plus{}\frac{1}{a_k}\right)^2 \geq \frac{(n^2\plus{}1)^2}{n}.$

This follows again with Jensen and the function $ f(x) \equal{} (x \plus{} \frac {1}{x})^{2}$
But I'm wondering, are there other techniques than Jensen to solve this generalization almost as effectiv?

Edit: found another one

[hide="Power mean makes it 1 liner"]
$ \frac{(a\plus{}\frac{1}{a})^{2}\plus{}(b\plus{}\frac{1}{b})^{2}}{2}\geq \frac{(a\plus{}b\plus{}\frac{1}{a}\plus{}\frac{1}{b})^{2}}{4}$

[/hide]

The following also holds.

If $ a_1 ,a_2 ,...,a_n$ are positive real numbers such that $ a_1 \plus{} a_2 \plus{} ...\plus{}a_n\equal{}1$ ,then

$ \left( a_1 \plus{} \frac{1}{a_1} \right)^2  \plus{} \left(a_2 \plus{} \frac{1}{a_2} \right)^2 \plus{} ...\plus{}\left(a_n \plus{} \frac{1}{a_n} \right)^2 \geq \frac{(n^2 \plus{}1)^2}{n}$

[hide="solution"]

Multiplying by $ n$ ,and applying Cauchy -Schwartz inequality we have

$ LHS \geq \left(a_1 \plus{}a_2 \plus{} ...\plus{}a_n \plus{} \frac{1}{a_1} \plus{}\frac{1}{a_2} \plus{}...\plus{}\frac{1}{a_n} \right)^2 \equal{}(1 \plus{} \sum { \frac{1}{a_i}})^2 \geq (1\plus{}n^2)^2$

 $ \sum {\frac{1}{a_i}} \geq n^2$ is true by Cauchy-Schwartz ,$ \sum{\frac{1}{a_i}} \left(a_1 \plus{}a_2 \plus{}...\plus{}a_n \right) \geq n^2$
[/hide]

[quote="Maths Mechanic"][hide="Power mean makes it 1 liner"]
$ \frac {(a \plus{} \frac {1}{a})^{2} \plus{} (b \plus{} \frac {1}{b})^{2}}{2}\geq \frac {(a \plus{} b \plus{} \frac {1}{a} \plus{} \frac {1}{b})^{2}}{4}$

[/hide][/quote]

Excuse me, but how is this a one liner?
What you did is all right....but the RHS does not simply be equal to $ 25/2$ or be reachable trivially......

BTW Jensen [b]IS[/b] the best way to do this....

Also ENDBOX I think kunny had already posted your question.....so why repost it??

[hide="one liner"]$ (a\plus{}\frac1a)^2\plus{}(b\plus{}\frac1b)^2\ge\frac{(a\plus{}b\plus{}\frac1a\plus{}\frac1b)^2}2\ge\frac{(a\plus{}b\plus{}\frac4{a\plus{}b})^2}2\equal{}\frac{25}2$

(the first one is power mean, the second is cauchy schwarz)[/hide]

shouldnt this be in the inequality forum?
i think i saw the same sum(with solution) in some pre-oly marathon ..

If $ a$ and $ b$ are positive and $a\text{ +}b=1$, prove that
	\[{{\left( a+\frac{1}{a} \right)}^{2}}+{{\left( b+\frac{1}{b} \right)}^{2}}\ge \frac{{{\left( a+\frac{1}{a}+b+\frac{1}{b} \right)}^{2}}}{2}\ge 2\left( a+\frac{1}{a} \right)\left( b+\frac{1}{b} \right)\ge 2\left( \frac{a+b}{2}+\frac{2}{a+b} \right)^2=\frac{25}{2}.\]

If $ a$ and $ b$ are positive and $a+b=2$, prove that
	\[{{\left( a+\frac{1}{a} \right)}^{2}}+{{\left( b+\frac{1}{b} \right)}^{2}}\ge \frac{{{\left( a+\frac{1}{a}+b+\frac{1}{b} \right)}^{2}}}{2}\ge 2\left( a+\frac{1}{a} \right)\left( b+\frac{1}{b} \right)\ge 2\left( \frac{a+b}{2}+\frac{2}{a+b} \right)^2=8.\]

If $ a$ and $ b$ are positive and $a+b\leq2$, prove that
	\[{{\left( a+\frac{1}{a} \right)}^{2}}+{{\left( b+\frac{1}{b} \right)}^{2}}\ge \frac{{{\left( a+\frac{1}{a}+b+\frac{1}{b} \right)}^{2}}}{2}\ge 2\left( a+\frac{1}{a} \right)\left( b+\frac{1}{b} \right)\ge 2\left( \frac{a+b}{2}+\frac{2}{a+b} \right)^2\ge8.\]

Inequality is equivalent to
\[
\left(\frac{(a+b)^2}{a}+a\right)^2+\left(\frac{(a+b)^2}{b}+b\right)^2-\frac{25}{2} (a+b)^2 \geq0
\]

After bringing all terms to common denominator it is enough to prove that $f(a,b)\geq0$ where
\[
f(a,b)=2 a^6+8 a^5 b-a^4 b^2-18 a^3 b^3-a^2 b^4+8 a b^5+2 b^6
\]
Inequality is symmetric and we may take $a\geqq$. That implies the existence of non-negative $x$ such that $a=b+x$. Then
\[
f(a,b)=f(b+x,b)=49 b^4 x^2+98 b^3 x^3+69 b^2 x^4+20 b x^5+2 x^6 \geq0
\]

[quote="sqing"]If $ a$ and $ b$ are positive ,  then\[\left(\frac{(a+b)^2}{a}+a\right)^2+\left(\frac{(a+b)^2}{b}+b\right)^2\geq (4b+a)^2+(4a+b)^2\geq\frac{25}{2} (a+b)^2 .\][/quote]
Nice result.
Both LHS and RHS can be proved by method that I have used for original inequality.
Also, after expanding RHS is equivalent to  $a^{2}+b^{2}\geq2ab$ that is well known.
Another proof for RHS is to directly apply well known $x^{2}+y^{2}\geq\left(\frac{x+y}{2}\right)^2$

observe that
$ ( a + 1/a + b + 1/b) = ( a + b + (a+b)/a + (a+b)/b) >= 5 $ since $ a/b + b/a >=2 $ for a,b>0 by trivial ineq.

now by cauchy
$ \[ \left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2 >={ ( a + 1/a + b + 1/b)^2}/2 = 25/2 $
hence proved

[quote=sqing]If $ a$ and $ b$ are positive and $ a \plus{} b \equal{} 1$, prove that
\[ \left(a\plus{}\frac{1}{a}\right)^2\plus{}\left(b\plus{}\frac{1}{b}\right)^2 \geq 2\left(a\plus{}\frac{1}{a}\right)\left(b\plus{}\frac{1}{b}\right)\geq \frac{25}{2}.\][/quote]

If $ a$ and $ b$ are positive and $ a \plus{} b \equal{} 1$, prove that
\[ \left(a+\frac{1}{b}\right)\left(b+\frac{1}{a}\right) \geq \frac{25}{4}\]
\[ \left(a+\frac{1}{a}+1\right)\left(b+\frac{1}{b}+1\right) \geq \frac{49}{4}\]\[ \left(a+\frac{1}{b}+1\right)\left(b+\frac{1}{a}+1\right) \geq \frac{49}{4}\]

[quote=sqing]

If $ a$ and $ b$ are positive and $ a \plus{} b \equal{} 1$, prove that
\[ \left(a+\frac{1}{b}\right)\left(b+\frac{1}{a}\right) \geq \frac{25}{4}\]
[hide=*]\[ \left(a+\frac{1}{a}+1\right)\left(b+\frac{1}{b}+1\right) \geq \frac{49}{4}\]\[ \left(a+\frac{1}{b}+1\right)\left(b+\frac{1}{a}+1\right) \geq \frac{49}{4}\][/hide][/quote]
$$ \frac{2a^2-2a+2}{3} \ge \frac{1}{2}\ge \sqrt{a(1-a)}$$
$$\frac{2a^2-2a+2}{3} \ge \sqrt {a(1-a)}\iff a^2+b^2+a^2b^2\geq \frac{9}{4}ab\iff  \left(a+\frac{1}{b}\right)\left(b+\frac{1}{a}\right) \geq \frac{25}{4}$$
Thank [b]electrovector[/b].


geometry
calculus
rectangle
real analysis
real analysis unsolved

If $ S$ and $ T$ are both bounded sets, and we use notation: $ A^ \plus{} \left(S\right)$ denote the Jordan outer area of the set $ S$

show that: $ A^ \plus{} \left(S\bigcup T\right) \plus{} A^ \plus{} \left(S\bigcap T\right) \leq A^ \plus{} \left(S\right) \plus{} A^ \plus{} \left(T\right)$

The above exercise is taken from "Introduction to Calculus and Analysis Volume 2 Chapter 4 Exercise 4.1-3 "  Author : Richard Courant and Fritz John

Cover $ S$ with (essentially disjoint) rectangles $ R_j$ and cover $ T$ with (essentially disjoint) rectangles $ P_k$. Take all possible intersections $ R_j\cap P_k$ to get a bunch of rectangles such that any two of them are either identical or essentially disjoint. The repeated rectangles in the bunch cover $ S\cap T$; the rest cover $ S\cup T$.


function
limit
topology
calculus
calculus computations

Let $ f(x) \equal{} 0$ if $ x$ is irrational and let $ f(\frac {p}{q}) \equal{} \frac {1}{q}$ if $ \frac {p}{q}$ is in lowest terms. What is the function $ g$ defined by $ g(x) \equal{} \lim_{y\to\ x}f(y)$?

I'm completely lost on this one. If anybody could even explain this problem to me, that would be helpful.

Thank you.

I think there is a slight trouble. If $ g(x)\equiv 0$ for all $ x$, that means the limit of $ f(x)$ exists at all points (the limit equals $ 0$). Let's see whether this is so.
First of all, note that in the vicinity of any rational point on the number line, we can find infinitely many irrational points and vice versa.
Now, consider an arbitrary point $ x\equal{}a$ and consider a sequence $ \{x_n\}$ of irrational numbers that has a limit $ a$. Since, the set of irrational numbers is very dense and so we can always find such a sequence. Then the sequence $ \{f(x_n)\}$ has a limit zero, since $ f(x_n)\equal{}0$ for all indices $ n$.

Next, consider another sequence $ \{z_n\}\to a$, this time, of rational numbers. Then, the corresponding sequence of the function values $ \{f(z_n)\}$  does not approaches zero (because $ f(z_n)\neq 0$ for any $ n$). Accordingly, using the Cauchy's definition for the limit, we see that the function $ f$ has no limit for any $ a$.

Thus, I assert that no such $ g(x)$ exists.

[quote="kaymant"]Next, consider another sequence $ \{z_n\}\to a$, this time, of rational numbers. Then, the corresponding sequence of the function values $ \{f(z_n)\}$  does not approaches zero (because $ f(z_n)\neq 0$ for any $ n$). Accordingly, using the Cauchy's definition for the limit, we see that the function $ f$ has no limit for any $ a$.[/quote]

If $ f$ were a [url=http://en.wikipedia.org/wiki/Dirichlet_function]Dirichlet function[/url] or something similar to this, your argument would be true. But this function is designed so that $ f(z_n)$ does approach to zero. If $ x_0 \in \mathbb{R}$ and $ r$ is a rational number, then denominator of $ r$ grows as $ r$ approaches to $ x_0$ so that the size of $ f(r)$ decreases to zero.

In fact, this function is a famous and typical example which deserves attention for its counter-intuitive behavior of discontinuity. You can find more on this function [url=http://en.wikipedia.org/wiki/Thomae%27s_function]here[/url].

what do you mean by?
[quote="iamhe"]What is the function $ g$ defined by $ g(x) \equal{} \lim_{y\to\ x}f(y)$?[/quote]
Regardless it can be shown that $ g$ is continuous at all irrational $ x$ and discontinuous at non-zero rational $ x$. I'm not sure about $ x\equal{}0$ but I think it might be continuous there.

The $ g(x)\equiv 0$ statement seems to make sense since the rationals form a set with measure 0.

[quote="maokid7"]what do you mean by?
[quote="iamhe"]What is the function $ g$ defined by $ g(x) \equal{} \lim_{y\to\ x}f(y)$?[/quote]
Regardless it can be shown that $ g$ is continuous at all irrational $ x$ and discontinuous at non-zero rational $ x$. I'm not sure about $ x \equal{} 0$ but I think it might be continuous there.

The $ g(x)\equiv 0$ statement seems to make sense since the rationals form a set with measure 0.[/quote]

Basically, find the limit mentioned above.

[quote="sos440"]If $ r$ is a real number and $ \varepsilon$ is a positive number, then there are at most finite $ x$ near $ r$ (but distinct from $ r$ itself) such that $ |f(x)|$ exceeds $ \varepsilon$. This indicates that $ f(x)$ can be made arbitrarily small for $ x$ sufficiently near $ x$, or in other words, $ f(x) \to 0$ as $ x \to r$. Therefore, $ g \equiv 0$. You may polish this idea to give a formal proof like this:

[hide="Proof"]Fix any $ r \in \mathbb{R}$ and pick $ \varepsilon > 0$. Choose $ n$ such that $ \varepsilon > \tfrac{1}{n}$ and put $ \delta \equal{} \min \{ | r \minus{} \tfrac{k}{n!} | \, : \, k \in \mathbb{Z} \}$. Clearly $ \delta > 0$. Also, if $ 0 < |x \minus{} r| < \delta$, then either $ x$ is irrational or $ x$ is a rational number such that if $ x \equal{} \tfrac{p}{q}$ is in lowest term, then $ |p| > n$. This implies that $ |f(x)| < \frac {1}{n} < \varepsilon$. So $ \lim_{x \to r} f(x) \equal{} 0$.[/hide][/quote]

How would you justify that  $ \delta>0$?

[quote="limsup"]How would you justify that  $ \delta > 0$?[/quote]

If you know some basic topology, then it immediately follow from that $ \tfrac{1}{n!}\mathbb{Z} \equal{} \{\tfrac{k}{n!} \, | \, k \in \mathbb{Z} \}$ is discrete, hence closed. But of course we can argue it directly.

If $ r$ is irrational, then $ | r \minus{} \tfrac{k}{n!}|$ never vanishes. Also, if we put $ M \equal{} |r \minus{} \tfrac{k_0}{n!}|$ for some $ k_0 \in \mathbb{Z}$, then for any $ |k| > n!(|r| \plus{} M)$, we have $ |r| < |\tfrac{k}{n!}| \minus{} M \leq |r| \plus{} |\tfrac{k}{n!} \minus{} r| \minus{} M$, so we have $ M < |\tfrac{k}{n!} \minus{} r|$. This shows that, in fact, $ \delta$ reduces to the minimum among certain finite nonzero numbers. Therefore $ \delta$ cannot be zero.

Well done sos440! Can you please justify the following also?
[quote="sos440"] if $ x \equal{} \tfrac{p}{q}$ is in lowest term, then $ |p| > n$. [/quote]
Moreover, what is the significance of using $ n!$ in your definition of $ \delta$?

[quote="limsup"]Well done sos440! Can you please justify the following also?[/quote]

Sorry, it was $ q$, not $ p$. That is, we have $ |q| > n$.

Ya I too realized just now.Moreover, what is the significance of using $ n!$ in your definition of $ \delta$. I think it should be $ n$ instead of $ n!$ there.

Moreover,  what would be your $ \delta$ for rational $ r$?


TPBM is a Teletubby :D

TPAM is a teletubby too!

TPBM is confused.

TPAM is confused.

wait..is TPBM the person before me or the person below me?

and TPAM is the person above me or the person after me?

strange it works either way...

Wow, I really am confused...

TP above M is confusing.
TP below M's new nickname is Bob.

I'm never confused. :D

TPBM likes the Teletubbies :D

TPaboveM has a strange fascination with teletubbies O_o
TPbelowM has a pet monkey =D

(am I doing this correctly?!)

Pet monkey?

The Person above ime s wrong.

The person below me has a TI-89 graphing calculator.  :D

yes

TPBM also had agraphing calculator that is a TI-89

TPAM is partly correct.
"had" is incorrect. TPAMBM HAS a TI-89.

TPBM has an N-spire

yes.

TPBM also went to the 2008 National Mathcounts Competition

Unfortunately, no.

TPBM has an addiction to chocolate.

eh... not really

my parents accuse me of being skinny.

TPBM knows my gender.

Yes. You're a boy :D

TPBM likes my blog.

... you scare me. I thought you were beating up John Sullivan.

TPAM is insane, like ra and stevenmeow.

yes, because we're teletubbies who have no life and keep pressing f5 on each other's blogs :D

TPBM likes stevenmeow. (I hope he does!)

erm.... ok? no preference.

TPBM <3's Voldemort.

ERR: SYNTAX

TPAM should sprekinzy the same lingitty as me.

How old are you?

TPBM thinks that the Telettubies sucks.

TPBM is neither gauss1181 nor tinytim.

TPAM did not make IMO this year.

TPBM did not make MOSP this year.

:|

TPAM loves the Teletubbies

TPBM thinks that pie is cool.

tpam is currently eating a coconut in his helicopter

nope :mad:

TPAM is gonna have a cocoanut in his/her head
TPBM is neither gauss1181 nor ra5249

Wrong! I'm gauss1181. :D

TPBM is either tinytim, ra5249, or stevenmeow :D

Wrong! I'm pacman2812 :mad:

TPBM plays the Mafia Game 35

Nope. I don't play Mafia.

TPBM likes monkeys. :|

not so fond of monkeys...

TPBM can type

ummm....gee i wonder....

TPBM is wondering about perfect628's recent inactivity

(if in Mole 2 or Mafia 34)

[quote="dannyhamtx"]Yes. But I am wondering why you are inactive to my PM.

TPBM should have a TI 83 plus silver edition calculator[/quote]

you never pmed me...

no i dont

TPBM got >5 on AIME

I didnt take it xD
TPBM likes the show SpongeBob

Uh, there's another thread just like this. (And it's not the one that's old).

Locked.


[quote]Section 3.1: Arrange in Order
Example 2: Given $2n+2$ points in the plane, no three collinear, prove that two of them determine a line that separates $n$ of the points from the other $n$.[/quote]
It's not [i]quite[/i] the same problem, but I used their same method to solve #5: work with the west-most point, label it red, and draw balancing lines in order from the least-clockwise blue point (starting from North) to the furthest-clockwise blue point.  At least one of those lines works.


function
geometric series
combinatorics unsolved
combinatorics

Hello,
How can we find a closed form expression for the generating function for the number of ways to make n cents out of nickels, dimes and quarters?

Thanks,

The nickel generating function is $ (1 \plus{} x^5 \plus{} x^{10} \plus{} ...)$, the dime generating function is $ (1 \plus{} x^{10} \plus{} x^{20} \plus{} ...)$, and the quarter generating function is $ (1 \plus{} x^{25} \plus{} x^{50} \plus{} ...)$.  These are all geometric series and their product is

$ \frac{1}{(1 \minus{} x^5)(1 \minus{} x^{10})(1 \minus{} x^{25})}$.


function
induction
real analysis
real analysis unsolved

X={x in R; x>0}
Show that there is no function f: X-> X that satisfies
 f(x+y)> f(x)(1+yf(x)) for all x,y in X

This problem is still without solution!!  :o  .. Any way..
Let we suppose that there is a such function $ f(x \plus{} y) \geq f(x)(1 \plus{} yf(x))$ .  First of all, manifestly $ f$ is an increasing function.
I denote $ g_1(x) \equal{} x \plus{} \frac 1{f(x)}, g_2(x) \equal{} g_1(x) \plus{} \frac 1{f(g_1(x))},...,$$ g_{n \plus{} 1}(x) \equal{} g_1(g_n(x)) \equal{} g_n(x) \plus{} \frac 1{f(g_n(x))},...$
By induction we can see that : $ f(g_1(x))\geq 2f(x),...,f(g_{n \plus{} 1}(x)) \equal{} f(g_1(g_n(x)))\geq 2f(g_n(x))\geq f(x)2^{n \plus{} 1},...$
We can write $ g_{n \plus{} 1}(x) \equal{} x \plus{} \frac 1{f(x)} \plus{} \sum_{k \equal{} 1}^n\frac 1{f(g_k(x))} \leq x \plus{} \frac 1{f(x)}\sum_{k \equal{} 0}^n \frac 1{2^k} < x \plus{} \frac 2{f(x)}$
But $ f$ is an increasing function, so $ f(x \plus{} \frac 2{f(x)}) \geq f(g_{n \plus{} 1}(x)) \geq f(x)2^{n \plus{} 1}$ for all $ n$ and that's impossible because $ f(x)>0$ .
Finally a  such function doesn’t exist.   :cool:


There is 6 points on a plane.Every 2 point is connected by a segment.From 6 points perpendiculars are drawen to this line segments.Find the maximum value of
intersection points of these perpendiculars.(perpendiculars dont have length they are unlimited from both points)

can you post whole solution i think there are less than 1500 points

still nobody solved :S  its not hard question
hint[hide]3 heights in a triangle intersects on same point[/hide]


vector
linear algebra
matrix
linear algebra unsolved

Can someone give an inductive proof for this theorem?
Let $ x_{1},\dots ,x_{n}$ and $ y_{1},\dots ,y_{n}$ be two bases. Then $ k$ of the vectors $ y_{1},\dots ,y_{n}$ can be swapped with the vectors $ x_{1},\dots ,x_{k}$ so that we get again two bases. (The original proof is easy but a clever one.)

Consider these vectors in terms of the basis $ \left( x_{1},\ldots, x_{n}\right)$.

Let $ X \equal{} I_{n}$ be the matrix whose columns are $ x_{1},\ldots, x_{n}$ and define $ Y$ likewise.

Suppose the result isn't true.

Now swap $ x_{1},\ldots, x_{k}$ with, say, $ y_{1},\ldots, y_{k}$.*

The new matrices will be $ X^{\prime}\equal{}\begin{pmatrix}y_{1}&\ldots & y_{k}& e_{k\plus{}1}&\ldots & e_{n}\end{pmatrix}$ and $ Y^{\prime}\equal{}\begin{pmatrix}e_{1}&\ldots & e_{k}& y_{k\plus{}1}&\ldots & y_{n}\end{pmatrix}$.

By assumption, $ \det X^{\prime}\equal{} 0$ or $ \det Y^{\prime}\equal{} 0$. That is, either the $ k\times k$ upper principal minor or the lower $ (n\minus{}k)\times (n\minus{}k)$ lower principal is $ 0$.

If we return to *, we see that could have chosen any other vectors instead of $ y_{1},\ldots, y_{k}$. The result is that, given a $ k\times k$ minor contained in the first $ k$ rows of $ Y$, either that minor is $ 0$, or its "complement" is $ 0$.

But by Laplace we have that $ \det Y \equal{}\sum_{T\text{ is a }k\times k\text{ minor}}\text{sgn}\left( T\right)\cdot\det\left( T\right)\cdot\det\left(\text{complement of }T\right) \equal{} 0$, contradiction.

[i]Note.[/i] This is not exactly an inductive proof. But since it took me quite a while to come up with this argument, I thought I'd post it.

Your idea is the same as the one used in Prasolov's book. But I wonder if there exists an inductive proof for it. (The case $ k\equal{}1$ is assumed to be trivial!)


[size=150]That la buon qua!

Doi tuyen TOAN than yeu cua chung ta da bi out khoi top 5 roi. Chang biet co so nhu the nao ma nen noi nay? Nam nay chung ta da thua xa lac NAM TRIEU TIEN (SOUTH KOREA), doi ma gan nhu luc nao cung theo duoi chung ta ca.

Chung ta da khong giu duoc thanh tich lam ca the gioi phai nga mu kinh phuc: 20 nam troi luon trong top ten cua the gioi. Dac biet trong 5-7 nam gan day luon o top five.

Chang le hao kiet ca nuoc nam nay khong con nua hay sao? Bo giao duc phai tim cho ra duoc nguyen nhan!

Chi con co cach la TRO LAI THAT AN TUONG trong nam 2006 toi ma thoi truoc khi chung ta chu nha: 4 huy chuong vang va 2 huy chuong bac, voi 2 thi sinh diem tuyet doi 42/42!

Phai lam cho the gioi biet duoc rang: PHONG DO CUA CHUNG TA CO THE DOI LUC BI THAT BAI NHUNG DANG CAP CUA TRI TUE TOAN HOC VIET NAM LA KHONG HE THAY DOI! Cho cua doi tuyen nam toi la o tren cao![/size]

What goes up must come down ! :D

Vice versa! What goes down must come up  :roll:  :roll:  :roll:

[color=blue]Đúng vậy! Có xuống rồi sẽ có lên lại thôi. Thế nào rồi trong kì thi [b]IMO2006[/b] này chúng ta sẽ trở lại $top$$five$ thôi[/color]

Theo toi kq thi chi la mot phan .Quan trong nhat la chat luong nen toan hoc nuoc nha!

[color=blue]Nước ta vẫn có các nhà toán học nổi tiếng như Giáo sư [b]Ngô Bảo Châu[/b] hay là [b] Lê Văn Thiêm[/b] đó thôi.[/color]

Cũng đơn giản thôi mà!!!tại toán học và cả nền giáo dục nước người ta được đầu tư nhiều hơn mình,còn mình thì sao chứ?? 1 lớp 50 HS,cơ sở hạ tầng xuống cấp,thiết bị dạy học cũ xì,mấy cái thiết bị mới cho cải cách toàn là xài hàng trung quốc,đụng tới là gãy,đổ,bể,vỡ,làm thí nghiệm mà em mún nổi khùng luôn!!nước mình nhân tài thì nhiều,nhưng mà không được ưu đãi cho lắm nên mới như vậy đóa!!! :(  :(  :(


search
number theory unsolved
number theory

For each $ m,n>2$ prove that there exists a sequence $ a_0,a_1,...,a_k$ such that $ a_0\equal{}m,a_k\equal{}n$, and $ a_i\plus{}a_{i\plus{}1}|a_ia_{i\plus{}1}\plus{}1$ for all $ i\equal{}0,1,...,k\minus{}1$.


limit
search
real analysis
real analysis unsolved

A hard problem. :) $ f(x)\in C(0,\infty)$,and for any $ x_{0}>0$,$ \lim_{n\rightarrow\infty}f(nx_{0})\equal{}0$,show that $ \lim_{x\rightarrow\infty}f(x)\equal{}0$.

Some hints:the book says we can prove an indirect result:there exists an closed interval $ [a,b]$ with $ b>a>0$ s.t.for any $ \epsilon>0$,existing $ N\in \mbox {N}$ s.t $ \forall n\geq N,|f(nx)|<\epsilon,\forall x\in [a,b]$.But how to prove the above? :(

This problem has been discussed before, but it's hard to search for. If I could find a link, I'd post it.

[Edit: See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=137954]here[/url] or [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=70016]here[/url]. Thanks to mathmanman.]

I'm pretty sure this was given on some Putnam. Maybe that can narrow down the search  :)


How many parts can we get when dividing a plane with $n$ lines?

at least $n$, at most $\frac{n^{2}+n+2}{2}$

you not forgett anything?


show that:if x and y are positive real numbers and x^3+y^3=x-y then x^2+4y^2 less than 1

Actually we can prove that
$ x^2\plus{}4y^2\plus{}\frac{(2\sqrt{5}\minus{}4)y^2x}{x\minus{}y}\le1$ which is stronger than original problem

[quote="hoangvn.fix"]Actually we can prove that
$ x^2 \plus{} 4y^2 \plus{} \frac {(2\sqrt {5} \minus{} 4)y^2x}{x \minus{} y}\le1$ which is stronger than original problem[/quote]

How?

we have$ \frac{y^3\plus{}y}{x}\equal{}1\minus{}x^2$ and $ 1\minus{}x^2\minus{}4y^2\equal{}\frac{y(1\plus{}y^2\minus{}4xy)}{x}$($ x>y$)
$ x^3\plus{}y^3\plus{}y^2(x\minus{}y)\ge (2\sqrt{5}\minus{}4)x^2y\plus{}4xy(x\minus{}y)$
$ \Leftrightarrow1\plus{}y^2\minus{}4xy\ge\frac{(2\sqrt{5}\minus{}4)x^2y}{x\minus{}y}$ so we have
$ 1\minus{}x^2\minus{}4y^2\ge\frac{(2\sqrt{5}\minus{}4)y^2x}{x\minus{}y}$
equality happen with $ x\equal{}\sqrt{\frac{5\sqrt{5}\minus{}5}{5\sqrt{5}\plus{}1}}$ and $ y\equal{}\sqrt{\frac{\sqrt{5}\minus{}1}{5\sqrt{5}\plus{}1}}$

[quote="hoangvn.fix"]we have$ \frac {y^3 \plus{} y}{x} \equal{} 1 \minus{} x^2$ and $ 1 \minus{} x^2 \minus{} 4y^2 \equal{} \frac {y(1 \plus{} y^2 \minus{} 4xy)}{x}$($ x > y$)
$ x^3 \plus{} y^3 \plus{} y^2(x \minus{} y)\ge (2\sqrt {5} \minus{} 4)x^2y \plus{} 4xy(x \minus{} y)$
$ \Leftrightarrow1 \plus{} y^2 \minus{} 4xy\ge\frac {(2\sqrt {5} \minus{} 4)x^2y}{x \minus{} y}$ so we have
$ 1 \minus{} x^2 \minus{} 4y^2\ge\frac {(2\sqrt {5} \minus{} 4)y^2x}{x \minus{} y}$
equality happen with $ x \equal{} \sqrt {\frac {5\sqrt {5} \minus{} 5}{5\sqrt {5} \plus{} 1}}$ and $ y \equal{} \sqrt {\frac {\sqrt {5} \minus{} 1}{5\sqrt {5} \plus{} 1}}$[/quote]

nice!


Show that if $a_{1}+a_{2}+a_{3}+\ldots$ is a convergent series of real numbers and $v_{1}, v_{2}, v_{3}, \ldots$ is a subsequence of the sequence $1, 2, 3, \ldots$, then $(a_{1}+a_{2}+\ldots+a_{v_{1}})+(a_{v_{1}+1}+\ldots+a_{v_{2}})+(a_{v_{2}+1}+\ldots+a_{v_{3}})+\ldots = \sum_{n=1}^{\infty}a_{n}$.

Can anyone help me with this problem? THanks.

The partial sums of your new sequence are a subsequence of the old partial sums.


algebra
polynomial
parameterization
quadratics
algebra proposed

hi
solve in $ \mathbb{R}$
\[ x^{4} \minus{} (2n \plus{} 1)x^{3} \plus{} (n \minus{} 1)x^{2} \plus{} (2n^{2} \plus{} 1)x \plus{} n \equal{} 0,
\]
where $ n$ is real.

The standard trick in these problems is to regard the polynomial as a polynomial in the parameter $ n$ rather than in $ x$, since it's quadratic in $ n$ and if there is a factorization it will be much easier to spot this way.


geometry
perimeter
combinatorics solved
combinatorics

10 vertices of a regular 100-gon are coloured red and ten other (distinct) vertices are coloured blue. Prove that there is at least one connection edge (segment) of two red which is as long as the connection edge of two blue points.

[hide="Hint"]Possible approaches are pigeon hole principle, proof by contradiction, consider turns (bijective congruent mappings) which maps red in blue points.
[/hide]

Let's refer to the [i]minimum[/i] number of edges between two nodes along the perimeter of the 100-gon as the 'distance' between the nodes. Obviously the edge-length between two nodes is proportional to the distance between those nodes so we may speak of the distance between nodes and the length of the edges between those nodes synonymously.
Suppose, on the contrary, that we have a placement of the red vertices such that the maximum distance between two of them is less than the minimum distance between any two blue vertices. Obviously the minimum distance between any two blue vertices <= 100/10 = 10. This means that the maximum distance between any two red vertices must be <=9 - i.e. all 10 red vertices must be on sequential nodes of the 100-gon. This means that all 10 blue vertices have a sequential, 90-node path  to be placed on, each at a distance of 10 or more from the previous one, an obvious contradiction.

Here's what I did (hope it's correct):

We consider the 100 distances along the circle, in a counter-clockwise direction, between a blue point and a red point (the number of segments between them). It's obvious that if 2 such distances are equal, then the problem is solved. Let's assume they're all distinct. We thus have 100 distinct distances along the circle, in a counter-clockwise direction, but this is impossible, since there are only 99 such distances: from 1 up to 99.

We are not asked to prove that there are any lengths that are equal, but that there is one edge between two red veritices that is [b]as long  as[/b]  the edge between two blue vertices. We also are concerned with the distance between red nodes compared to the distance between blue nodes, not the distance between red and blue nodes.  Also, there would C(10,2) =45 such distances between the blue and 45 edges between the red nodes yielding 90 distances total.

I think you should reconsider that:

Assume B1 and B2 are blue vertices and R1 and R2 are red vertices. It can quickly be seen that if the distance (measured on the circle in a counter-clockwise direction) B1R1 is the same as B2R2 then B1B2=R1R2: assume the points are arranged on the circle in the order (measured in a counter-clockwise direction) B1 B2 R1 R2. Then B1R2=B1R1+R1R2=B1B2+B2R2, so from B1R1=B2R2 we get B1B2=R1R2, which is what we want . Something similar happens if the order of the points is, for example, B1 R1 B2 R2.

Is it clear now? Sorry if my initial solution wasn't as clear as it should have been.

Beautiful! As soon as I read your post I assumed that you didn't understand the problem so I didn't give your argument any real thought.  You actually come at it from a different angle, that there is an R1R2 distance that is [b]the same as [/b]a B1B2 distance. This may actually be what orl wanted anyway (maybe[b] I[/b] didn't undestand the problem!) although 'as long as' implies to me that there exists an RiRj >= BxBy.  Either way, I like your approach better.


function
superior algebra
superior algebra theorems

Im rather new to modern algebra and i was just wondering what techniques you use to show two groups with either one or two operations are isomorphic besides looking at there tables and also say that you are not a given a function that maps group A bijectively to B do you just generate one and what are the best ways to do that?


[quote="Ulanbek_Kyzylorda KTL"]|sinx|^|cosx|=|cosx|^|sinx|
Find x[/quote]

Let study the equation in $ [0,2\pi]$

1) $ x \in [0,\frac\pi2]$

The equation is   $ {\sin x}^{\cos x}\equal{}{\cos x}^{\sin x}$

$ 0 \le LHS \le 1$ and it is monotonically increasing, $ 1 \ge RHS \ge 0$ and it is monotonically decreasing. So the solution is one and it is when $ {\sin x}\equal{}{\cos x} \rightarrow x\equal{}\frac \pi4$

2) $ x \in [\frac\pi2, \pi]$

The equation is   $ {\sin x}^{\minus{}\cos x}\equal{}{\minus{}\cos x}^{\sin x}$

$ 1 \ge LHS \ge 0$ and it is monotonically decreasing, $ 0 \le RHS \le 1$ and it is monotonically increasing. So the solution is one and it is when $ {\sin x}\equal{}{\minus{}\cos x} \rightarrow x\equal{}\frac {3\pi}4$

3) $ x \in [\pi,\frac{3\pi}2]$

The equation is   $ {\minus{}\sin x}^{\minus{}\cos x}\equal{}{\minus{}\cos x}^{\minus{}\sin x}$

$ 0 \le LHS \le 1$ and it is monotonically increasing, $ 1 \ge RHS \ge 0$ and it is monotonically decreasing. So the solution is one and it is when $ {\minus{}\sin x}\equal{}{\minus{}\cos x} \rightarrow x\equal{}\frac {5\pi}4$

4) $ x \in [\frac{3\pi}2, 2\pi]$

The equation is   $ {\minus{}\sin x}^{\cos x}\equal{}{\cos x}^{\minus{}\sin x}$

$ 1 \ge LHS \ge 0$ and it is monotonically decreasing, $ 0 \le RHS \le 1$ and it is monotonically increasing. So the solution is one and it is when $ {\minus{}\sin x}\equal{}{\cos x} \rightarrow x\equal{}\frac {7\pi}4$

In conclusion all the solutions are $ x\equal{}\frac \pi4 \plus{} \frac{k\pi}2$ where $ k \in \mathbb{Z}$


A motor boat went 14 kilometres downstream in 30 minutes. The return trip upstream took 6 minutes longer. Find the rate of the boat in still water and the rate of the current.
How would you solve it using the concept of linear systems?

set the rate of the boat in still water as x, and the rate of the current as y
you have:
(x+y)/2=14
(x-y)*3/5=14

then you solve with a system..

hope that helps


geometry
angle bisector
geometry unsolved

Let $ ABCD$ be a cyclic quadrilateral.Points $ M,N,P$ and $ Q$ are chosen on the sides $ AB,BC,CD$ and $ DA$,respectively,such that $ \frac{MA}{MB}\equal{}\frac{PD}{PC}\equal{}\frac{AD}{BC}$ and $ \frac{QA}{QD}\equal{}\frac{NB}{NC}\equal{}\frac{AB}{CD}$ Prove that $ MP$ is perpendicular to $ NQ$

$ E: AB \cap CD \ \ F: BC \cap DA$. Is know that the angle bisector of $ \angle BEC$ is perpendicolar to the angle bisector of $ \angle CFD$, then appling [url=http://www.mathlinks.ro/viewtopic.php?t=116208]this lemma[/url] we have the result.

Call E the common point of AC and BD; since the triangles AED and BEC are similar we get AD/BC=AE/BE. Consequently, EM is the internal bisector of the angle <AEB, etc.

Best regards,
sunken rock


probability
number theory
relatively prime
advanced fields
advanced fields unsolved

We know that the probability for 2 integers to be relatively prime is 1/zeta(2) ; can we generalize this result and say that the probability for n numbers to be relatively prime in their set is 1/zeta(n)?! , it seems logical since 1/zeta(n) increases and tends to 1 when n tends to infinity...!!?

See http://www.mathlinks.ro/Forum/viewtopic.php?t=31028 , also the generalisation is mentioned there.

A caveat: There is no such thing as a uniform probability distribution on the integers. A correct statement:
Choose an integer at random (uniformly) between $-M$ and $M$, and another random integer between $-N$ and $N$. As $M$ and $N$ tend to $\infty$, the probability that the chosen integers are relatively prime approaches $\frac1{\zeta(2)}$.

You've made the right guess for the general version. The probability that a prime $p$ divides all $n$ integers approaches $\frac1{p^{n}}$, and these events tend toward independence for different primes. This should give a limiting probability of $\prod_{p\text{ is prime}}1-\frac1{p^{n}}=\frac1{\zeta(n)}$.

The convergence does need proof, and I don't want to go through that proof right now.


topology
function
real analysis
real analysis unsolved

How do we prove that $f: U_{N}\to \mathbb{R}^{2}$ where $U_{N}= S^{2}-{N}, \,\, (N=(1,0,0) )$

and $f(x,y,z) = \left( \frac{y}{1-x}, \frac{z}{1-x}\right)$ is a homeomorphism ?

Please explain  :)

Sorry if it's trivial or if it is in the wrong place.

$f$ is continuous because both components are continuous functions of $x,y,z$. It remains to prove that the inverse of $f$ is also continuous. Solve the system
$\begin{cases}u=\frac{y}{1-x}\\ v=\frac{z}{1-x}\\ x^{2}+y^{2}+z^{2}=1\end{cases}$
to obtain
$\begin{cases}x=\frac{u^{2}+v^{2}-1}{u^{2}+v^{2}+1}\\ y=\frac{2u}{u^{2}+v^{2}+1}\\ z=\frac{2v}{u^{2}+v^{2}+1}\end{cases}$
Clearly, $x,y,z$ are continuous functions of $u$ and $v$.


Prove that
$\ \ \prod_{k=1}^{n-1}\sin\left(\frac{k\pi }{n}\right)=\frac{n}{2^{n-1}}$