The state competition has posted selected problems as a problem of the week, which pretty much implys that we are allowed to talk now.

What did you get on the competition and how hard was it?

Personally, I thought it was hard, and got a 39 off 4 reading errors, thankfully it didn't cost me.

We had 7th go up to 1st, I was 3rd written.  The second and first place people got 42 and 43 respectively.

Our cutoff was around 35.

I am from Michigan.
I got the number one score before the official countdown round with a score of 43.  :lol:
Then the countdown came and I pretty much got owned.  :(

43 with a 29 on Sprint and 14 on the Target. :lol:

The competition wasn't that hard, and I only made careless mistakes so I could've gotten a perfect.
There was a huge gap between second and third place scores, 3 points I believe.

I am happy about my score since I made an improvement from
40 from last year 28 on sprint and 12 on target to 43 this year.

I bombed Sprint 21/14
I got 21/8 last year and considering how hard the sprint was last year, I didn't improve much :( Still enough to barely go to Nats in my state because I won a tiebreaker to get to third.

Ignite, what state are you from

BTW, We are now allowed to discuss questions from STATE.

Mathcounts has posted it on their website.

CT, it's much harder than Delaware considering a 29 last year with 21/14 and 6/10 right on the last 10 questions of the sprint (Basically winning every tiebreaker) was only enough for #8 while in Delaware, 24 qualified for Nats and 30 was first.

3rd and 4th (me) both got a 38 and 22/16. This is in Massachusetts.

Approximate list of all the dumb mistakes I made:

#22.  Forgot to read different.

#30. My brain "shortcircuited."

Target #4.  Didn't see "percent"

#6. Didn't see cube, and didn't know he had tape.

The other one was definitely stupid.

So, those dropped me from a 46 to a 39.

I should be thankful I wasn't in California or something.

I missed 6 problems (3 targets) By one unit because of careless counting mistakes.  If I had gotten those all right, I would have gotten the top score in CA.  One of the problems was:  How many numbers divisible by 4 can you make with the digits 1,2,3,4.  I found all of them except for 4.  I missed #7 because I counted a point not on a line of points, and number 8 by double counting.  I guess I was just nervous :(

I got a 39 (25/14) on the test, and was the top scorer in Maine (before and after countdown).  I thought the sprint round was easier than last year and that the target was also a little easier.

I actually don't know my score yet :( except for target which was 7 :( because on #1 I missed one letter ONE LETTER "f"

on sprint, I know of 3 I got wrong, so probably 27.
Total (hopefully) was 41.

I thought the test was SO MUCH EASIER than last year, and thats not only because of major score improvement. I actually finished all but the last two sprints within 20 minutes

I left a whole page blank. I blame it on my school :( We were in the middle of State Mastery tests the week before the MathCounts competition. The questions are all easy (Think of the 6 * 1/7 question) but the teachers get mad if I finish too early and force me to check my work as I go along. So I was still sorta in that mode during the state Sprint and kept checking my work as I went along giving me the exact same Sprint score as last year :(

I'm going to Nationals though because I got 3 more questions right on the target this time around.

Oh, I got a 14/16 on the Target because of bad handwriting on #7 I think.

I don't know how I didn't make it!!!! I got 6th, yet I think I got like 40 points.   There were only 3 questions I know I got wrong.  2 sprint one target.  Everone else in VA must have gotten like 42's and 43's.  I'm really envious of you guys who get to go with only a 38.

I got 32 and got 7th written. I bombed sprint, 18 :oops: , 14 on target. 10th written got 29 I think, 2nd written got 34. I was so freaking out when I saw that I got to countdown, I was so sure I got 20+ something.
I had 18 last year, and 13 year before that, so I made some good improvement (and I still get to go to nationals!).

Our team got 2nd with 41.25, 1st got 41.5 (noooo!), because we did so badly on team, 12/20, (1st got 16/20).

[quote="kyyuanmathcount"]Well, if I hadn't talked, I would think I would have had 43...[/quote]

Well, I would have gotten a full score, if I read #7 on target correctly and for the last question on the spring I did 7C3=35 instead of 6C3=20( the answer I put), and I would have gotten a perfect.

[quote="Xantos C. Guin"][quote="mathgeniuse^ln(x)"]We got on our team round a 9 because we put 432 for #9 instead of 240, oh well, we still won.[/quote]

I made the same mistake when I did the team round unofficially at home earlier today. I corrected it with 5 seconds to spare.
BTW: Who plays with square tiles?[/quote]

I think I put 432 as well.

I put 432 on the team round as well.. and I was so proud of myself for realizing that you need 20 and not ten (or whatever the trick was) too...

Wasn't it like the smallest number with 20 factors?
We would have missed that, but a teammate not knowledgable of the number of factors formula worked something out, which was wrong but helped me get the right answer.
I had put something too large, 432, probably, but he made an incorrect guess which led me to realize something.

Blah, we missed number 10. The answer I got was either exactly 2x the right answer, or 1/2 x the right answer. Boo on small mistakes.

Wow so many people got #9 Team wrong.

No one on my team knew how to do it except for me, but I let them down.  I got it wrong and guess what I put? 432.

Stupid 2^2 *5.  :mad:

If it was 20 factors, was the number 2^4*3*5? seems kinda easy, but I remember that we got it wrong

Stupid me, read it wrong, and yes, 240 is right.
(Read it as 3^5, not 3*5)

we got #9 right, we just missed #8.

[quote]do you live in california or new york or texas or something?
Oklahoma's population is 800,000.
theirs is 30,000,000+.
their population is 37.5 times greater!
it's kind of odd that the cities of dallas or houston or san antonio have a greater population than a state that's so much huger than delaware.[/quote]

well, I live in Washington and it doesn't have [i]that[/i] many people.  I think there are 10 or 11 million, but I'm not sure. I could be way off.

[quote="Zmastr"]Our team:

me: 37 (really got 40 or so, still working out with the test people)
27
24
19

7 on the team round

and we still got 4th in VA with a 40.25[/quote]

Zack, you didn't "really get 40", you really got a 37.

Arghh I was looking at the team round answers-we got like [b]2[/b] right.

[quote="solafidefarms"]Politics does not belong in a discussion of state scores. Keep bashing/arguing in regard to Mr. Bush in one of the Other Topics forums.[/quote]Ummm i wasn't insulting bush. I would have voted for him. I'm just saying we can't count in florida.

Missed four sprints and a target.  No big hopes at nationals...unless Neal Wu and the entire Texas/Cali teams get sick on may 12...[[u]b]jkjk!!![/b][/u]


omg each natural number is represented once by unique prime factorization, so that is equivalent to the harmonic series, which is divergent

[quote="lingomaniac88"]Altheman's Shift key is broken.  But yes, that is right.  Hide it, please.[/quote]

WHAT?

lol

this is obvious though...and there is no real "figuring this out;" you know it or you don't

Wait. Al, your shift key isn't broken?
Whoa... :huh:

Up to this post... I also could'a sworn.

[quote="white_horse_king88"]Wait. Al, your shift key isn't broken?
Whoa...

Up to this post... I also could'a sworn.[/quote]

:lol:

it is easier to type without capitalizing....YEA THATS RIGHT -


vector
real analysis
real analysis unsolved

Suppose that $ V$ be a vector space with a norm . Prove that its closed unit ball is compact if and only if the space is finite-dimensional.

It is easy using the fact that any closed subspace of the image of a compact operator is finite dimensional: let $ I: X\to X$ be the identity operator. The unit ball is compact iff $ I$ is compact. But the image of $ I$ is $ X$.


inequalities
number theory proposed
number theory

The inequality is equilvalent to :
$ (a_{1} \minus{} (n \minus{} 1))(a_{1} \minus{} n) \plus{} (a_{2} \minus{} (n \minus{} 1))(a_{2} \minus{} n) \plus{} \cdots \plus{} (a_{n} \minus{} (n \minus{} 1))(a_{n} \minus{} n) \leq 0$ but $ a_{i}$ is integer for all $ i$ so for each $ k$ we must have$ a_{k} \equal{} n$ or $ a_{k} \equal{} n \minus{} 1$.
a)we prove that $ a_{i} \equal{} n$ or $ a_{i} \equal{} n \minus{} 1$ so if $ n > 1$ then $ a_{i} > 0$
b)we have :$ n^{2} < n^{2} \plus{} 1 \leq a_{1} \plus{} a_{2} \plus{} \cdots \plus{} a_{n} \plus{} (n \plus{} 1) \geq n^{2} \plus{} n \plus{} 1 < n^{2} \plus{} 2n \plus{} 1 \equal{} (n \plus{} 1)^2$
so this number can't be perfect square .


[quote="NikolayKaz"]$ x,y \in Q \\
(\sqrt {30} \minus{} \sqrt {18})(3\sqrt {x} \plus{} \sqrt {y}) \equal{} 12$

I divided by $ (\sqrt {30} \minus{} \sqrt {18})$ and came up with
$ \sqrt {30} \plus{} 3\sqrt {2} \equal{} \sqrt {y} \plus{} 3\sqrt {x}$

so we have x=2 and y=30

Are there any other solutions, how do i include/exclude that[/quote]

Squaring your last equality, you get $ 9x \plus{} y \plus{} 6\sqrt {xy} \equal{} 48 \plus{} 6\sqrt {60}$ and so $ \sqrt {xy} \equal{} p \plus{} q\sqrt {60}$ with $ p \equal{} 8 \minus{} \frac {9x \plus{} y}6$ and $ q \equal{} 1$

Squaring again, you get $ xy \equal{} p^2 \plus{} 60q^2 \plus{} 2pq\sqrt {60}$ and so $ pq \equal{} 0$ since $ \sqrt {60}$ is irrational while all the other terms are rational.

Since $ q \equal{} 1$, we get $ p \equal{} 0$ and so $ 9x \plus{} y \equal{} 48$ and as a consequence $ xy \equal{} 60$ and so we have the system :

$ 9x \plus{} y \equal{} 48$
$ 9xy \equal{} 540$

So $ 9x$ and $ y$ are the roots of $ X^2 \minus{} 48X \plus{} 540$, that's to say $ 18$ and $ 30$

Hence two solutions :
$ (9x,y) \equal{} (18,30)$ and we get $ (x,y) \equal{} (2,30)$
$ (9x,y) \equal{} (30,18)$ and we get $ (x,y) \equal{} (\frac {10}3,18)$

And plugging back in the original equation, we easily check that these two solutions fit the initial requirements.

[quote="NikolayKaz"]$ x,y \in Q \\
(\sqrt {30} \minus{} \sqrt {18})(3\sqrt {x} \plus{} \sqrt {y}) \equal{} 12$

I divided by $ (\sqrt {30} \minus{} \sqrt {18})$ and came up with
$ \sqrt {30} \plus{} 3\sqrt {2} \equal{} \sqrt {y} \plus{} 3\sqrt {x}$

so we have x=2 and y=30

Are there any other solutions, how do i include/exclude that[/quote]
Generally:
If we have to find $ n,m \in \mathbb{Q}$ given $ a,b \in \mathbb{Q}, \sqrt{a}, \sqrt{b}, \sqrt{ab} \notin \mathbb{Q}$ and the equation
$ \sqrt{n} \plus{} \sqrt{m} \equal{} \sqrt{a} \plus{} \sqrt{b}$ then the only solutions are $ (n,m)\equal{}(a,b)$ and $ (n,m)\equal{}(b,a)$.
Proof:
$ \iff n \plus{} m \plus{} 2\sqrt{nm} \equal{} a\plus{}b\plus{}2\sqrt{ab} \Rightarrow \sqrt{nm}\minus{}\sqrt{ab} \in \mathbb{Q} \Rightarrow \sqrt{nmab} \in \mathbb{Q}$.
And $ \sqrt{nm} \minus{} \sqrt{ab} \in \mathbb{Q}, \sqrt{ab} \notin \mathbb{Q} \Rightarrow \sqrt{ab}(\sqrt{nm}\minus{}\sqrt{ab}) \notin \mathbb{Q}$ or $ nm\equal{}ab$. Since $ \sqrt{ab}(\sqrt{nm}\minus{}\sqrt{ab}) \equal{} \sqrt{abnm}\minus{}ab \in \mathbb{Q}$ we see $ nm \equal{} ab$.
$ n \plus{} m \plus{} 2\sqrt{nm} \equal{} a\plus{}b \plus{} 2\sqrt{ab} \iff n\plus{}m\equal{}a\plus{}b$. So $ n\plus{}m\equal{}a\plus{}b$ and $ nm\equal{}ab$. And hence $ (t\minus{}n)(t\minus{}m) \equal{} (t\minus{}a)(t\minus{}b)$ which gives $ (n,m)\equal{}(a,b)$ or $ (n,m)\equal{}(b,a)$. (By inserting it is obvious that there are indeed solutions)

Here we have $ a \equal{} 30, b \equal{} 18, n \equal{} 9x, m \equal{} y$, which gives the solutions $ (x,y) \equal{} (2,30)$ and $ (x,y) \equal{} \left ( \frac{10}{3}, 18 \right )$


algorithm
modular arithmetic
number theory
Euclidean algorithm
number theory open

Dear arXiv-editor,arXiv-moderator,
Since I am a junior of Mathematics Department, Zhengzhou University, this manuscript is a fresh try as ''a note on Fermatls last theorem'. I have tried my best to have got as many feedbacks as possible so as to make it readable and as strict as I could. There is five theorems as results in this manuscript. Thank you!
Tao Liang

\section{Abel Lemma $ 1$ On Solution Of Fermat's Last Theorem}
\begin{theorem}Theorem: Suppose $ (x, y, z)$ is one solution of Fermat
Equation $ x^n \plus{} y^n \equal{} z^n.$ Under condition of $ (x, y) \equal{} (y, z) \equal{}
(z, x) \equal{} 1, n$ is odd prime, if $ (x, n) \equal{} 1,$ then we get $ x \equal{} ab, z
\minus{} y \equal{} a^n, (a, b) \equal{} 1.$\end{theorem} Proof: Since $ x^n \equal{} z^n \minus{} y^n \equal{}
(z \minus{} y)\frac{x^n}{z \minus{} y}, (x, y) \equal{} (y, z) \equal{} (z, x) \equal{} (x, n) \equal{} 1,$
hence $ (z \minus{} y, nzy) \equal{} 1.$ According to Euclidean algorithm, we get
$ (z \minus{} y, \frac{x^n}{z \minus{} y}) \equal{} 1,$ then we get $ x \equal{} ab, z \minus{} y \equal{} a^n,
(a, b) \equal{} 1.$

\section{Abel Lemma $ 2$ On Solution Of Fermat's Last Theorem}
\begin{theorem}Theorem: Suppose $ (x, y, z)$ is one solution of Fermat
Equation $ x^n \plus{} y^n \equal{} z^n.$ Under condition of $ (x, y) \equal{} (y, z) \equal{}
(z, x) \equal{} 1, n$ is odd prime, if $ (z, n) \equal{} 1,$ then we get $ z \equal{} ab, x
\plus{} y \equal{} a^n, (a, b) \equal{} 1.$\end{theorem} Proof: Since $ z^n \equal{} x^n \plus{} y^n \equal{}
(x \plus{} y)\frac{z^n}{x \plus{} y}, (x, y) \equal{} (y, z) \equal{} (z, x) \equal{} (x, n) \equal{} 1,$
hence $ (x \plus{} y, nxy) \equal{} 1.$ According to Euclidean algorithm, we get
$ (x \plus{} y, \frac{z^n}{x \plus{} y}) \equal{} 1,$ then we get $ z \equal{} ab, x \plus{} y \equal{} a^n,
(a, b) \equal{} 1.$

\section{Lemma $ 1$ On Solution Of Fermat's Last Theorem}
\begin{theorem}Theorem: Suppose $ (x, y, z)$ is one solution of Fermat
Equation $ x^n \plus{} y^n \equal{} z^n.$ Under condition of $ (x, y) \equal{} (y, z) \equal{}
(z, x) \equal{} 1, n$ is odd prime, if $ (x, n) \equal{} 1,$ then we get $ x\equiv1,
\pmod n.$\end{theorem} Proof: For any prime $ p$ of $ \{p_i|x, 0 < i <
j\}, (x, n) \equal{} 1$ which allows $ p_{i_1} \equal{} p_{i_2}.$ Let $ y^{p \minus{} 1}
\equiv z^{p \minus{} 1}\equiv 1, \pmod p.$ In case of $ n\nmid p \minus{} 1:$ Since
we have $ |x|> |z \minus{} y|, (z, y) \equal{} 1,$ according to Euclidean
algorithm, if $ (m, n) \equal{} 1,$ we get $ (z^m \minus{} y^m,z^n \minus{} y^n) \equal{} z \minus{} y.$
Then $ p|(z^{p \minus{} 1} \minus{} y^{p \minus{} 1},z^n \minus{} y^n)\equal{}z \minus{} y.$ It follows that
$ p|(x, z \minus{} y) \equal{} a.$ In fact according to Abel Lemma we get $ x \equal{} ab,
z \minus{} y \equal{} a^n, (a, b) \equal{} 1.$ If $ p_i|a,$ then $ p_i|a^n \equal{} z \minus{} y.$ $ x \equal{}
\prod^j_{i \equal{} 1}p_i,$ yields a contradiction of that $ x|z \minus{} y.$ Then
$ n|p \minus{} 1.$ Hence $ x\equiv1, \pmod n.$

\section{Lemma $ 2$ On Solution Of Fermat's Last Theorem}
\begin{theorem}Theorem: Suppose $ (x, y, z)$ is one solution of Fermat
Equation $ x^n \plus{} y^n \equal{} z^n.$ Under condition of $ (x, y) \equal{} (y, z) \equal{}
(z, x) \equal{} 1, n$ is odd prime, if $ (z, n) \equal{} 1,$ then we get $ z\equiv1,
\pmod n.$\end{theorem} Proof: For any prime $ p$ of $ \{p_i|z, 0 < i <
j\}, (z, n) \equal{} 1$ which allows $ p_{i_1} \equal{} p_{i_2}.$ Let $ x^{p \minus{} 1}
\equiv y^{p \minus{} 1}\equiv 1, \pmod p.$ In case of $ n\nmid p \minus{} 1:$ Since
we have $ |2z|> |x \plus{} y|>|z|, (x, y) \equal{} 1,$ according to Euclidean
algorithm, if $ (m, n) \equal{} 1,$ we get $ (x^m \minus{} (\minus{}1)^my^m, x^n \plus{} y^n) \equal{} x
\plus{} y.$ Then $ p|(x^{p \minus{} 1} \minus{} y^{p \minus{} 1},x^n \plus{} y^n)\equal{}x \plus{} y.$ It follows
that $ p|(z, x \plus{} y) \equal{} a.$ In fact according to Abel Lemma we get $ z \equal{}
ab, x \plus{} y \equal{} a^n, (a, b) \equal{} 1.$ If $ p_i|a,$ then $ p_i|a^n \equal{} x \plus{} y.$ $ z
\equal{} \prod^j_{i \equal{} 1}p_i,$ yields a contradiction of that $ z|x \plus{} y.$
Then $ n|p \minus{} 1.$ Hence $ z\equiv1, \pmod n.$

\section{Odd Prime Case Of Fermat's Last Theorem}
\begin{theorem}Theorem: If $ n$ is odd prime, then $ x^n \plus{}
y^n \equal{} z^n$ have no solution $ x, y, z$ such that $ (x, y) \equal{} (y, z) \equal{}
(z, x) \equal{} 1.$\end{theorem} Proof: Let $ (xy, n) \equal{} 1.$ According to
Lemma $ 1$ Fermat little theorem, we have $ z\equiv x \plus{} y \equiv 1 \plus{} 1
\equiv 2, \pmod n.$ According to Lemma $ 2$ we have $ z\equiv1, \pmod
n.$ This is contradictory.

[quote="Mathias_DK"]I couldn't find any mistakes. But what about the case where $ (n,xyz) \neq 1$ ?[/quote]

In either case where $ n|xyz$ or case where $ (n, xyz) \equal{} 1$, there is $ (n, xy) \equal{} 1$.

If $ u^n \plus{} v^n \equal{} w^n, (n, uvw) \equal{} 1,$ then let $ x \equal{} u, y \equal{} v, z \equal{} w$ which yields $ (n, xy) \equal{} 1.$

In fact if $ (n,xyz) \neq 1,$ that $ n$ is odd yields $ u^n \plus{} ( \minus{} w)^n \equal{} ( \minus{} v)^n$ from $ u^n \plus{} v^n \equal{} w^n$ where $ n|v, (n, uw) \equal{} 1.$ Let $ x \equal{} u, y \equal{} \minus{} w, z \equal{} \minus{} v,$ and we either have $ (n, xy) \equal{} 1.$


trigonometry
quadratics
geometry
similar triangles

Two flag poles of height $h$ and $k$ are situated $2a$ units apart on a level surface. Find the set of all points on the surface which are so situated that the angles of elevation of the tops of the poles are equal.

Takers? I smell trig!

[hide="A very small start"]
If you draw the diagram, the two similar triangles instantly pop up and:
Let the angle that has to be same be $ \theta$.

$ \frac {2a \minus{} x}{x} \equal{} \frac {h}{k}$
Where x is the distance from k to the point.
$ x \equal{} \tan{\theta}*k$

$ \frac {2a \minus{} \tan{\theta}*k}{\tan{\theta}*k} \equal{} \frac {h}{k}$

Still, you have three variables and one equation, so you're still pretty much screwed.
[/hide]

No one? Message is too small. Please make the message longer before submitting.

I get the answer to be:

$ a\pm\sqrt{a^2\minus{}hk}$

@ poincare : Your first equation can be reduced to a quadratic,  then solved. :)

Don't you get $x=\frac{2ah}{h+k}$ if you solve for the first equation (cross multiplying)
And also, I think it's asking for all the points on the surface, not just between them

The problem becomes significantly easier when we realize that the triangles formed are similar triangles, so instead of trig, just use ratios.  ;)


geometry
circumcircle
incenter
trigonometry
geometry proposed

Let $ ABCD$ be a quadrilateral,such that $ AB\equal{}AC\equal{}BD$.Let $ P\in AC\cap BD$.
Prove that $ OI\perp CD$,where $ O,I$ are the circumcenter and incenter of $ \triangle APB$,respectively.
[hide]Comment:
I solved this problem today.with using trigonometry,but i would like to see more elegant proofs,
Thank you. :)
[/hide]

note that:

$ OI\perp CD\iff DI^2 - CI^2 = DO^2 - CO^2$

let the incircle of $ \triangle APB$ with radius $ r$ touch the sides $ PB,PA$ at $ D',C'$ respectively,now according to the power of points $ D,C$ wrt circle $ (I)$ we have:

$ DD'^2 = DI^2 - r^2$ and $ CC'^2 = CI^2 - r^2$ so we have:

$ DI^2 - CI^2 = DD'^2 - CC'^2$ (*)

now let $ AB = p,BP = a,PA = b$ and so $ AB = AC = BD = p$ now according to some well-known theorems we have:

$ DD' = DB - D'B = p - \left(\frac {a + p - b}2\right) = \frac {p + b - a}2$

$ CC' = CA = C'A = p - \left(\frac {b + p - a}2\right) = \frac {p + a - b}2$

$ \Rightarrow DD'^2 - CC'^2 = p(b - a)$

so by (*) we get that:

$ \boxed{DI^2 - CI^2 = p(b - a)} (I)$

now let $ R$ be the radius of circle $ (O)$ so we have:

$ DP.DB = DO^2 - R^2$ and $ CP.CA = CO^2 - R^2$ so we have:

\begin{eqnarray*}DO^2 - CO^2 = DP.DB - CP.CA = \left(DB - PB\right).DB - \left(CA - PA\right).CA = \left(p - a\right)p - \left(p - b\right)p\end{eqnarray*}

$ \boxed{\Rightarrow DO^2 - CO^2 = p(b - a)} (II)$

now from (I),(II) we get that:

$ DI^2 - CI^2 = DO^2 - CO^2$

as wanted...

Thank you,my idea is the same,but with trigonometry.
P.S:
You are very fast,did you know this problem before?

yes,it's a famous problem,it was given in several olympiads...
there is also a similar problem as follow:

[b]let $ D,E$ be points on sides $ AC,BC$ of $ \triangle ABC$ respectively,s.t $ AB\equal{}AD\equal{}BE$ let $ O,I$ be the incenter and cricumcenter of $ \triangle ABC$ respecitively,prove that:

$ i)DE\perp OI$

$ ii)$the circumradius of $ \triangle CDE$ is equal to $ OI$.[/b]

OK,thank you.
Our teacher told us,that there exist solution with using inversion,maybe you know it?
In which olympiads exactly?

[quote="BaBaK Ghalebi"]Belarus 2000 I think...[/quote]
You are right,thank you.


In yeki az soalate olampiade basije ke emruz bargozar shod:
tedade tavabe"e $ f: R\to R$ ke dar rabeteye \[ 2(f(1\minus{}x))^2\plus{}(f(x))^2\equal{}2\] sedgh mikonad chandtast?

Bebin age gofte tedade kolle tavabe'e injoori javabesh mishe binahayat, ta !! vali age bege tedad tavabe'e peyvasteye injoori javabesh mishe $ 2$ ta ! dalilesh ine ke age $ x\longrightarrow 1 \minus{} x$ dar in soorat natije mishe :
$ f(x)^{2} \equal{} \frac {2}{3}$ . pas baraye har $ x$ : $ f(x) \equal{} \frac {\sqrt 2}{\sqrt 3}$ ya $ f(x) \equal{} \minus{} \frac {\sqrt2}{\sqrt3}$
khob maloome ke age tabe'e morede nazar peyvaste nabashe mishe masalan injoori dar nazar gereftesh :
$ f(0) \equal{} \frac {\sqrt 2}{\sqrt 3}$ , $ f(1) \equal{} \minus{} \frac {\sqrt2}{\sqrt3}$ . yani baraye har ozve $ R$ $ 2$ ta entekhab baraye $ f$ e oon adad darim ... khob maloome ke binahayat ta mishe ... vali age gofte bood peyvaste faghat tabeye sabet javab bood ...


witam!!

macie może (znacie, albo macie linka albo odsyłacz do książki) dowód twierdzenia Lejeune-Dirichleta, że w każdym ciągu arytmetycznym $ax+b$, gdzie $nwd(a,b)=1$ istnieje nieskończenie wiele liczb pierwszych

To już bardzo poważna teoria liczb.

Książki :
Władysław Narkiewicz "Teoria liczb"
Tom M. Apostol "Introduction to analytic number theory"
Hua Loo Keng "Introduction to number theory"

link do pliku w internecie :
http://modular.math.washington.edu/129/projects/weissman/project.pdf


Determine the minimum of $|5x^{2}+4xy-3y^{2}|$ such that $(x,y)\in\mathbb{Z}^{2}-\{(0,0)\}$.

Masoud Zargar

If $5x^{2}+4xy-3y^{2}=0$,
Let $t=\frac{x}{y}$ then $t$ is a rational number

$5t^{2}+4t-3=0$. The discrimininant is $4^{2}+4\cdot 15=76$. But since t is rational the discriminant must be a perfect square. Contradiction.

So, since $|5x^{2}+4xy-3y^{2}| \in\mathbb{Z}$ and $5x^{2}+4xy-3y^{2}\neq 0$,  $|5x^{2}+4xy-3y^{2}| \geq 1$

If $x=1$ and $y=2$, $|5x^{2}+4xy-3y^{2}|=1$

So the minimum is $1$


In the 2nd qn , A is quite obvious but why at all to talk about B , i mean i don't think whatever B is , it doesn't seem to have aromaticity ,and hence it should be formed very less (ya certainly i would would be happy to be proved wrong cos i don't know what B is in the first place :D )

[hide]btw also tell me what role does KHS04 perform (which others can't ) in dehydration of glycerol to acrolein[/hide]

isn't the 6 membered ring more stable than 7 membered :?:

[quote="hell_ever"]In first question,

the first product given below is minor, say 5% and the second product is major, say um, 95%? :)  :huh:  :rotfl:[/quote]

Why are you even bothered abt 2 products?  :huh:

btw, what software do you use for drawing diagrams?

for second queswtion, the two products are,

@hell , thanks :)

@da vinci /e

do u think i don't know the answer ?? huh ?
arey , give me the mechanism !!
and also tell that will it be of major significance ?

[hide]btw i am revising first year physics ,so have had some interesting qns , if u all want i can post them [/hide]

lets not forget that the first question is an example of [url=http://en.wikipedia.org/wiki/Tiffeneau-Demjanov_rearrangement]Tiffeneau-Demjanov rearrangement[/url] :D

And the second reaction is RTR!  :huh:
you need the mechanism for [url=http://en.wikipedia.org/wiki/Image:Reimer-Tiemann_Reaction_Mechanism.png]RTR[/url]
 ??  :huh:

Or do you need the mechanism for the formation of second product?  :|

Yay! i just proposed a mechanism for a named reaction. !  :jump:

i gave 2 answers, because the epoxide is also formed no matter in how small amount! During rearrangement, the intermediate carbocations always give products, no matter in how small amount.

@ ritu, the carbocation in second case is stabilised by resonsnce.  :)

sorry da! i'll edit this post and post later OK? gtg  :D

The carbene is formed due to acidic nature of the H of chloroform which reacts with the base, and consequently, a chloride ion leaves forming the carbene.

[quote="hell_ever"]
@ ritu, the carbocation in second case is stabilised by resonsnce.[/quote]
not clear
where's the resonance :(  :(  :(

@hell
I ask again! what software do you use for drawing diagrams?  :huh:

@ritu
back bonding from oxygen!!  :roll:

so the reason is same as for ring contraction
cool  :)


Consider a permutation $ \{a_1,a_2,\dots a_n\}$ of $ \{1,2,\dots ,n\}$. We start piling them in order, into different columns, in the following fashion: For every $ a_i$
(i) We can place $ a_i$ on the top of a pile if it is lesser than the topmost element of the pile.
(ii) A new pile can be initiated.
Lets call a set of piles as different from other if one cannot be obtained from the other by moving the piles. What can we say about the enumeration of sets of piles fro a given permutation?

example: Consider $ 3,5,1,2,4$. Two different sets of piling are $ \{\{1,3\}\{4,5\}\{2\}\}$ and $ \{\{2,3\},\{1,5\},\{4\}\}$. Step wise construct of the first set is as follows:

3
_______

3 5

________

1
3 5

________

1
3 5 2

________

1 4
3 5 2
_________

One way to look at this problem is as a counting problem related to the Clique Cover Problem for a special class of graphs.  The graph problem is as follows: given a graph $ G$ with vertex set $ V \equal{} \{1, \ldots, n\}$, in how many ways can we partition $ V$ so that each part of the partition is a clique in $ G$?  (The Clique Cover Problem asks for a graph $ G$ and an integer $ k$ whether it is possible to partition the vertex set into $ k$ or fewer cliques.)

Now, the relationship: to a permutation $ w \equal{} w_1 \cdots w_n$, associate a graph $ G(w)$ where $ \{w_i, w_j\}$ is an edge in $ G$ if and only it is an inversion in $ w$.  Then cliques of $ G(w)$ are exactly sets of entries of $ w$ that could conceivably end up in the same pile, and a decomposition of $ G(w)$ into cliques is exactly a decomposition of $ w$ into piles.

Another way of phrasing your question is to ask how many ways we can write our given permutation as a union of (disjoint) decreasing subsequences.

Unfortunately, it doesn't seem likely to me that this question is likely to have a nice answer.  Basically, my intuition here is that the set of values should interpolate between 1, for the identity permutation, and the Bell number $ B_n$ for the reversed identity permutation, and I'm somewhat skeptical that a good interpolation exists.  Happy to be proved wrong, though!


calculus
integration
function
probability
analytic geometry
real analysis
real analysis unsolved

http://integrals.wolfram.com/index.jsp

It has the error function in the result: http://mathworld.wolfram.com/Erf.html.

[quote="1/(ln x)"]How would someone symbolical integrate this:

$\frac{1}{\sqrt{2\pi}}\int e^{{\frac{-x^{2}}{2}}}dx$?[/quote]

Since this is the normal probability distribution, the integral over $(-\infty ,\infty)$ is 1 (for this is the statement that the probabilty of something--anything at all--will happen is 1), it can be proven symbolically in a number of ways, an easy one is this:

write $J=\int_{-\infty}^{\infty}e^{{-\frac{x^{2}}{2}}}dx =\int_{-\infty}^{\infty}e^{{-\frac{y^{2}}{2}}}dy$ so that

\[J^{2}=\left( \int_{-\infty}^{\infty}e^{{-\frac{x^{2}}{2}}}dx\right) \left(\int_{-\infty}^{\infty}e^{{-\frac{y^{2}}{2}}}dy\right) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{{-\frac{x^{2}}{2}}}e^{{-\frac{y^{2}}{2}}}dydx = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{\frac{-1}{2}(x^{2}+y^{2})}dydx\]

now change to polar coordinates to get

\[J^{2}=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-\frac{1}{2}(x^{2}+y^{2})}dydx = \int_{0}^{2\pi}\int_{0}^{\infty}e^{-\frac{r^{2}}{2}}r dr d\theta\]

let $u=\frac1{2}r^{2}$ so that $du=rdr$ to get

\[J^{2}= \int_{0}^{2\pi}\int_{0}^{\infty}e^{-u}du d\theta = \int_{0}^{2\pi}\left[-e^{-u}\right]_{u=0}^{\infty}d\theta = \int_{0}^{2\pi}\left[0+1\right] d\theta = 2\pi\]

and hence we have that $J=\int_{-\infty}^{\infty}e^{{-\frac{x^{2}}{2}}}dx=\sqrt{2\pi}$ so it follows that the given integral is 1.


[quote="ridgers"]Show that for each n element of natural numbers, n>=2
(3 to the power of n)-(2 to the power of n)/n is not a integer[/quote]
[hide="Solution"]
Assume the contrary.  Notice that $ 3^n \minus{} 2^n \equiv 0\bmod n$ means that $ 3^n \equiv 2^n \bmod n$, so $ (\frac{3}{2})^n \equiv 1\bmod n$.  Letting $ p$ be the smallest prime factor of $ n$, we have that $ (\frac{3}{2})^n \equiv 1\bmod p$, so if $ k$ is the order of $ \frac{3}{2}\bmod p$, then $ k|\gcd (p \minus{} 1, n)$.  Yet, $ p\minus{}1$ has all prime divisors lower than $ p$ and $ n$ has all prime divisors greater than or equal to $ p$, so $ k|1$, so $ k\equal{}1$, so $ \frac{3}{2}\equiv 1\bmod p$, so $ 3\equiv 2\bmod p$, which is impossible.  Thus, $ \frac{3^n\minus{}2^n}{n}$ is not an integer. [/hide]


function
LaTeX
logarithms
real analysis
real analysis unsolved

is this statement true or false?-give argument

for x element of R,  arcsinh x=ln(((x^2)+1)^(1/2)) -x)

i'm sorry ,i dont know how to use latex

\[ \operatorname{arcsinh}x\mathop  \equal{} \limits^{???} \ln \left( {\sqrt {x^2  \plus{} 1}  \minus{} x} \right),\forall x \in \mathbb{R}.
\]

No, it's actually:

$ \text{arcsinh}\,x\equal{}\ln\,(x\plus{}\sqrt{1\plus{}x^2})$

@bookworm_vn
yes youre right,that is the problem's equation

@milin, we have the same answer, can you show me the proof please?

By definition, $ \sinh x \equal{} \frac{e^x \minus{} e^{\minus{}x}}{2}$. Now let $ y \equal{} \sinh x$, and let's solve the equation for x:

$ y \equal{} \frac{e^x \minus{} e^{\minus{}x}}{2} \equal{}> e^x \minus{} e^{\minus{}x} \equal{} 2y \equal{}> e^{2x} \minus{} 1 \equal{} 2ye^x \equal{}> e^{2x} \minus{} 2ye^x \minus{} 1 \equal{} 0$

and now solving for e^x we get $ e^x \equal{} \frac{2y \pm \sqrt{4y^2\plus{}4}}{2} \equal{}> e^x \equal{} y \plus{} \sqrt{1\plus{}y^2}$

(we take the + sign because the exponential is always positive). Therefore we may conclude that

$ x \equal{} argsinh(y) \equal{} \ln(y \plus{} \sqrt{1\plus{}y^2})$.

Since
\[ \sqrt {x^2  \plus{} 1}  \minus{} x \equal{} \frac{{\left( {\sqrt {x^2  \plus{} 1}  \minus{} x} \right)\left( {\sqrt {x^2  \plus{} 1}  \plus{} x} \right)}}
{{\sqrt {x^2  \plus{} 1}  \plus{} x}} \equal{} \frac{1}
{{\sqrt {x^2  \plus{} 1}  \plus{} x}}
\]
then
\[ \ln \left( {\sqrt {x^2  \plus{} 1}  \minus{} x} \right) \equal{} \ln \left( {\frac{1}
{{\sqrt {x^2  \plus{} 1}  \plus{} x}}} \right) \equal{}  \minus{} \ln \left( {\sqrt {x^2  \plus{} 1}  \plus{} x} \right).
\]
Thus
\[ \operatorname{arcsinh}x \equal{}  \minus{} \ln \left( {\sqrt {x^2  \plus{} 1}  \plus{} x} \right).
\]

1) garry_guitarist, you had -x under the logarithm, and I have +x

2) bookworm_vn, we've concluded that:

$ \text{arcsinh}\,x \equal{} \ln\,(\sqrt {1 \plus{} x^2} \plus{} x)$

so the only thing that could be concluded is that:

$ \text{arcsinh}\,x \equal{} \minus{} \ln\,(\sqrt {1 \plus{} x^2} \minus{} x)$

Your equation:

$ \text{arcsinh}\,x \equal{} \minus{} \ln\,(\sqrt {1 \plus{} x^2} \plus{} x)$

can not be true because we know that $ \forall x > 0$, $ \sinh(x) > 0$. Therefore, $ \forall x > 0$, $ \text{arcsinh}\,(x) > 0$ and that isn't the case with the formula you gave!


Are you sure a solution exists? I looked at them for 10 minutes (that's a lot for me) and haven't gotten it yet. Unless, of course, I'm missing a very simple solution :maybe:

Well, for the second one, if you can use exponents, it's:

$ (4\cdot6)\cdot1^3$

Although I don't know if I can use exponents... :|

If we are allowed to use only three of the numbers:

$ 3*7\plus{}3$

[quote="Dillinger"]Using 4 numbers, adding, subtracting, multiplying, and dividing are ways to find the solution. The numbers are 3, 3, 7, 7
The other set is
1, 3, 4, 6[/quote]

[url=http://www.personal.psu.edu/staff/r/j/rjg5/Math24.htm]According to this[/url], the solutions are $ 7 \times \left(\frac{3}{7} \plus{} 3\right)$ (and some more, but I'm only listing one) for the first one, and for the second, the only one is $ \frac{6}{1\minus{}\frac{3}{4}}$.

Wow those were pretty obscure...good problems though.

yeah...I guess the only way to solve these is to use fractions.

hey, fractions are actually division problems. :wink:

Oops! Forgot to say you can't use exponents.

Thank you isabella2296 for telling me the solutions to the probelms.  :)

@ Math Champion, u can post this in the 24 marathon in the other prblem slving topics, math/strategy games

You didn't find it without a computer. I'm suprised!

although ...
I didn't do that myself

[quote="Math Champion"]Here's another good 24:

3,3,8,8[/quote]
http://www.davidwees.com/24solver says there are no solutions.

EDIT: Apparently that is wrong. I tried myself for a while, with no luck.

how about using your brain instead of a website ;)?

$ \frac{8}{3\minus{}\frac83}\equal{}\frac {8}{\frac93\minus{}\frac83}\equal{}\frac{8}{\frac13}\equal{}8\cdot3\equal{}\boxed{24}$. yay.

lol, and yes, that's the correct answer. And the website i use says that there is a solution PowerofPi.

wow... isabella2296 got it solved. I couldn't even find it out until i saw how she solved it.


limit
trigonometry
algebra
polynomial
calculus
calculus computations

Find limite
\[\huge \begin{array}{l}\mathop{\lim }_{n \to \infty }\frac{{\sin n.\sqrt[3]{{n+1}}}}{n}\\ \mathop{\lim }_{x \to 0}\frac{{1-c{\rm{osx}}{\rm{.cos2x}}{\rm{.cos3x}}........{\rm{cosnx}}}}{{x^{2}}}\\ \end{array}\]
:|  :(

First limit equal to $0$ because $\frac{|\sin{n}|\sqrt[3]{n+1}}{n}\leq \frac{\sqrt{n}\cdot\sqrt[3]{n+1}}{n}\forall n\geq 1$ .

The second one is a nice application of mathematical induction.

Why so hard??
$sinn\leq 1$ therefore our limit is less then $\frac{(1+n)^{1/3}}{n}\to 0$done.
in seccond.instead $cosmx$ we write $1-(xm)^{2}/2+o((x)^{2})$ and easy seen that our limit is equal $1/2(1^{2}+...+n^{2})$ and...
OK!. If you understand all this then try find this limit (it's not hard)
$\frac{1-\prod_{i=1}^{oo}(1-x)^{\frac{1}{i^{2}}}}{x}$ where $x\to 0$

[quote="maokid7"]for the second one i was going so suggest that you use the Taylor polynomial of $\cos(mx)$ and then use L'Hopitol's rule.[/quote]
You don't need L.Hopitol's.


abstract algebra
superior algebra
superior algebra unsolved

Let R be a PID. M is a subset of R^n  a submodule. Prove there exits a complement to M (i.e  M+N= R^n for some subodule N)

iff R^n / M is free.
Give an example where this does not hold.

apply the splitting lemma to

$0 \to M \to R^{n}\to R^{n}/ M \to 0$

and keep in mind that projective modules coincide with free modules over PIDs.


The chart below gives the air distance in miles between selected world cities. If two different cities from the chart are chosen at random, what is the probability that the distance between them is less than 7000 miles? Express your answer as a common fraction.

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

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

There are 8 distances < 7000 and there are 12 distances in total, so the probability is $ \frac{8}{12} \equal{} \boxed{\frac{2}{3}}$.


Prove the inequality ln((x+1)/x) > 1/(x+1)

Condition $ x\in ( \minus{} \infty; \minus{} 1)U(0; \plus{} \infty)$

Let $ y \equal{} \ln {\frac {x \plus{} 1}{x}} \minus{} \frac {1}{x \plus{} 1}$

We get $ y' \equal{} \frac { \minus{} 1}{x(x \plus{} 1)^2}$

If $ x\in ( \minus{} \infty; \minus{} 1) \Longrightarrow y' > 0 \Longrightarrow \mathop {\min y}\limits_{( \minus{} \infty ; \minus{} 1)} > \mathop {\lim }\limits_{x \to \minus{} \infty } y \equal{} 0$

If $ x\in (0; \plus{} \infty) \Longrightarrow y' < 0 \Longrightarrow \mathop {\min y}\limits_{(0; \plus{} \infty )} > \mathop {\lim }\limits_{x \to \plus{} \infty } y \equal{} 0$

Hene $ y>0, \forall x \in (\minus{}\infty; \minus{}1)U(0; \plus{}\infty)$

Then OK


So we had a one hour debate about this with lots of songs being listened to and we still haven't settled on the answer.

IS JOURNEY GAY?
WHICH BAND IS GAYER, JOURNEY OR COLDPLAY?

Coldplay

Journey has nostalgia factor, one of their songs is alright

Coldplay just flat out is lame

Coldplay is like two old men nailing each other in the ***. And I'm not talking about softcore, I'm talking about flat out nailing each other in the ***. Real tight stuff.

(17:35:06) Jathan84: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=720809#720809
(17:35:09) Jathan84: THIS TOPIC NEEDS A BUMP

I HEARTILY ENDORSE THE PREVIOUS STATEMENT


search
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advanced fields unsolved

This is one of my favorite "puzzles".

1. Represent ${\mathbb R}^{3}$ as a disjoint union of circles. (Lines and points are not circles.)
2. Is ${\mathbb R}^{3}$ a disjoint union of circles of radius $1$?

[quote="grobber"][url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=transfinite&t=123504[/url]

If that's correct, then a much stronger assertion than the positive answer to (2) is proven there (using the Axiom of Choice).[/quote]
Nice :D

By the way, the first problem has a constructive solution.

The first problem has been discussed before, separately, and a constructive solution was given (or at least sketched). I'll let you do that search :).


function
vector
abstract algebra
real analysis
invariant
Functional Analysis
real analysis unsolved

There is a version of the Hahn-Banach Theorem as follows:
Let $ p$ be a real-valued function on the vector space $ X$ satisfying $ p(x\plus{}y) \leq p(x)\plus{}p(y)$ and $ p(\alpha x)\equal{}\alpha p(x)$ for $ \alpha \geq 0$. Suppose $ f$ is a linear functional defined on a subspace $ S$ such that $ f \leq p$ on $ S$. Let $ G$ be an Abelian semigroup of linear operators on $ X$ such that for every $ A \in G$ we have $ p(Ax) \leq p(x)$ for all $ x \in X$ while for each $ s \in S$ we have $ As \in S$ and $ f(As)\equal{}f(s)$. Then there is an extension of $ f$ to a linear functional $ F$ on $ X$ such that $ F(x) \leq p(x)$ and $ F(Ax)\equal{}F(x)$ for all $ x \in X$.

What I would like to ask is to use this result to show that there is a set function $ \mu$ defined for all bounded subsets of $ \mathbb{R}$ with the following properties:
(i) If $ A \cap B\equal{}\emptyset$, then $ \mu(A \cup B)\equal{}\mu A\plus{}\mu B$.
(ii) $ \mu(A\plus{}t)\equal{}\mu A$.
(iii) If $ A \subset B$, then $ \mu A \leq \mu B$.
(iv) If $ A$ is Lebesgue measurable, then $ \mu A$ is the Lebesgue measure of $ A$.

Do it on the unit circle first, rotationally invariant, and then use this to define the measure you want on $ [0,2\pi]$ and then just extend it (important of course that the original measure is rotationally invariant).

Let $ X$ be all bounded real-valued functions on unit circle. Let $ S$ be bounded Lebesgue measurable functions on unit circle. And let $ f(s)$ be the lebesge integral. Let $ p(x) \equal{} \inf f(s)$, $ x \leq s$. Let $ G$ consist of the rotations.

Now for a subset $ P$ of the unit circle, let $ \mu(P) \equal{} F(\chi_P)$


There is a circuit containing R,L and C.$V_{i}n=V_{0}e^{i\omega t}$.Find the inductance $L$,if the absolute value of the denominator of $V_{o}ut$ is minimum.

Is it something like \[L = \frac{1+\sqrt{1+4\omega^{2}R^{2}C^{2}}}{2\omega^{2}C}?\]

What do you mean by "the denominator of $V_{out}$ is minimum? :maybe:

It means that you must find an equation for $V_{out}$ and then find the condition for the minimum value of denominator. :)

Oh, I haven't seen that the denominator should be minimal, I minimized the entire thing. But that doesn't make much sense, does it? It would be $L = 1/\omega^{2}C$ then, no?


geometry
circumcircle
trigonometry
Pythagorean Theorem
trig identities
Law of Sines

A triangle has sides of length $a,b,c$ and its circumcircle has diameter $d$
Prove that the triangle is right-angled if and only if
\[d = \sqrt{\frac{a^{2}+b^{2}+c^{2}}{2}}\]

sorry but whats $right$ in frensh :oops:

[b]Solution.[/b] Let $c$ be the hypotenuse. The triangle is right angled if and only if $c$ passes through the circumcentre, that is $c=d$.  Using the Pythagorean theorem,
\[\sqrt{\frac{a^{2}+b^{2}+c^{2}}{2}}=\sqrt{c^{2}}=c=d,\]
as required.

[quote="boxedexe"][b]Solution.[/b] Let $c$ be the hypotenuse. The triangle is right angled if and only if $c$ passes through the circumcentre, that is $c=d$.  Using the Pythagorean theorem,
\[\sqrt{\frac{a^{2}+b^{2}+c^{2}}{2}}=\sqrt{c^{2}}=c=d, \]
as required.[/quote]

i don't see how you proved the reverse direction...[or if you intended to...]

I misunderstood the problem.  I thought it was asking to prove that if it is right angled then the identity follows.

Actually it's a problem from the British MO, and it isn't hard at all (First round!)
I changed the question a bit to make it less obvious.

Arkhammedos: A right angled triangle est un angle droit en français, donc il s'agit d'un triangle rectangle.  :wink:

[hide]
\[d=\sqrt{\frac{a^{2}+b^{2}+c^{2}}{2}}\iff \sin^{2}A+\sin^{2}B+\sin^{2}C=2\]
(by extended law of sines, $a=d\sin A$ etc.)
\[\iff \cos^{2}A+\cos^{2}B+\cos^{2}C=1\iff \cos A\cos B\cos C=0\]
(from the formula $\cos^{2}A+\cos^{2}B+\cos^{2}C+2\cos A\cos B\cos C=1$)
But clearly $\cos A\cos B\cos C$ is 0 if and only if one of the cosines is 0, so this is equivalent to $\triangle ABC$ being a right triangle.
[/hide]

[quote]sorry but whats right in french [/quote]

Un triangle rectangle.

EDIT: didn't realize somebody else posted this  :P  woops.

Yes! I like your solution, scorpius, it's clever and elegant.

The original problem asked to prove the equivalence between the right angle and the identity \[8R^{2}=a^{2}+b^{2}+c^{2}\]

In that case, it's a very clear hint to \[OH^{2}= 9R^{2}-a^{2}-b^{2}-c^{2}\]
and you can solve the problem quickly.


Ean gia ta toksa $x_{1},x_{2},\ldots, x_{2007}$ tou diastimatos $\left[0,\frac{\pi}{2}\right]$ isxiei oti:

$x_{1}+x_{2}+\cdots+x_{2007}=\frac{\pi}{2}$, na vrethei to elaxisto tou athroismatos

$\cos{x_{1}}+\cos{x_{2}}+\cdots+\cos{x_{2007}}$

Alexandros

Μπορούμε να βρούμε και το μέγιστο (πώς ? :wink: )
Για το ελάχιστο [hide]$cos\ x\geq 1-\frac{2x}{\pi}$[/hide]

Poli wraia Silouan. Akrivws auti einai i katallili ipodeiksi.
As tin sinexisei kapoios gia na iparxei oloklirwmeni edw kathws episis na graftei pote isxiei i isotita... E

Einai poli voliki auti i anisotita pou erapse o Silouan.
Gia to megisto kserw oti $\sin{x}\leq x$ (panta sto diastima $\left[0,\frac{\pi}{2}\right]$. Genika isxiei $|\sin{x}|\leq |x|$. ) kai thetontas opou $x$ to $\frac{\pi}{2}-x$ pairnoume $\cos{x}\leq \frac{\pi}{2}-x$ tin opoia kai xrisimopoioume diadoxika.

Alexandros

[quote="cretanman"]
Gia to megisto kserw oti $\sin{x}\leq x$ (panta sto diastima $\left[0,\frac{\pi}{2}\right]$. Genika isxiei $|\sin{x}|\leq |x|$. ) kai thetontas opou $x$ to $\frac{\pi}{2}-x$ pairnoume $\cos{x}\leq \frac{\pi}{2}-x$ tin opoia kai xrisimopoioume diadoxika.
Alexandros[/quote]

Σωστά αλλά δεν χρειάζεται τόσος κόπος  :)  .
[hide]Jensen για την $f(x)=cos\ x$ [/hide]

xmmm... Nomizw oti mono i anisotita Jensen arkei. To allo pou egrapsa pio panw einai apla ena anw fragma alla oxi to megisto (epireasmenos apo tin parapanw lisi)!

Alexandros


Let $ (C_a)$ be the circle with diameter $ DD'$ with $ M_a$ be its center. Then it is followed that $ M_a$ is the midpoint of $ DD'$. But it is very well-known that $ (DD'BC) \equal{} \minus{} 1$ $ \Longrightarrow M_aD^2 \equal{} \overline {M_aB}.\overline {M_aC}$. Or in other word $ (C_a)\perp (O)$, where $ (O)$ is the circumcircle of $ \triangle ABC$. Therefore, $ \mathcal {P}_{O/(C_a)} \equal{} R^2$, where $ R$ is the radii of $ (O)$. With the same argument we also have $ \mathcal {P}_{O/(C_b)} \equal{} \mathcal {P}_{O/(C_c)} \equal{} R^2$. Let $ H$ be the orthocenter of $ \triangle DEF$ and $ D'',E'',F''$ repsectively be the projection of $ D,E,F$ onto $ EF,DE,DF$. Then it is also well-known that $ \overline {HD}.\overline {HD''} \equal{} \overline {HE}.\overline {HE''} \equal{} \overline {HD}.\overline {HD''}$ $ \Longrightarrow$ $ \mathcal {P}_{H/(C_b)} \equal{} \mathcal {P}_{H/(C_c)} \equal{} \mathcal {P}_{O/(C_a)}$. Thus, $ HO$ will be the radical axes of $ (C_a),(C_b),(C_c)$, which implies $ M_a,M_b,M_c$ are collinear $ \square$

Dear Mathlinkers,
1. D', E' and F' are collinear
2. A more synthetic proof can be obtained by using the Newton or Gauss's line to the complete quadrilateral D'EDE'...
Sincerely
Jean-Louis

[quote="nayel"]$ ABC$ is a triangle and $ D,E,F$ are points on $ BC,CA,AB$ respectively, such that $ AD,BE$ and $ CF$ concur. Lines $ EF$ and $ BC$ meet at $ D'$, $ DF$ and $ AC$ meet at $ E'$, $ DE$ and $ AB$ meet at $ F'$. Prove that the midpoints of $ DD',EE',FF'$ lie on a straight line.[/quote]

Midpoint of $ DD'$ is $ A'$ :

$ \frac{A'B}{A'C}\equal{}(\frac{DB}{DC})^2$

With menelaus's theorem we end the proof


I like to swim, but I never bothered to go to lessons or compete, although i am a very strong and quick swimmer just from my experience with ppl in a less organized environment, but especially i love playing waterpolo, even if it is violent, or maybe becuase it is. Unfortunately, i only get to play during the summer holidays cuz no1 around here plays

Swimming is my favorite sport :)
I'm in a high school swimming team.

i'm going to modified swimming, expected to break or come very close to the ultimate school record, so I like to swim.  :P

I used to love swimming, I joined a swim team during the summer and all of the sudden swimming wasnt that fun anymore. I still like to swim for fun though!

swimming is ok but i only like it in a "less organized group" but its really messy afterwards which kind of makes me not want to bother

I can swim but I cant tread.. soo... no i dont swim.  :huh:

umm...you cant tread but can swim? how is that possible :huh:

i dont like to swim unless it allows me to [hide="procasinate"] [b][i][u]and nothing else lets me do that (ex. computer, TV, book, etc.)[/b][/i][/u][/u][/hide]

Swimming is awesome!  Hehehe.  Though I only really like to swim for fun...three cheers for waterslides!!!

cuz.. i cant stay on my back, and i can swim on my stomach, but cant stay still or ill drown.. odd i know.

[quote="whoknowscolinlu"]yeah i like to swim too and i am very very close to beating our schools modified record for the 100 backstoke. what strokes do u guys do and what r ur times??[/quote]

I don't think that 9 seconds is very close. I like the butterfly the best and I got 31.04 in the 50 yard butterfly last winter season, but I'm much weaker in t he longer 100 fly race (1:14.56).

[quote="frt"][quote="whoknowscolinlu"]yeah i like to swim too and i am very very close to beating our schools modified record for the 100 backstoke. what strokes do u guys do and what r ur times??[/quote]
Backstroke is my favorite, too :) (My time is much slower than school records though :lol: )

50 yard backstroke: 32.08
100 yard backstroke: 1.10.97[/quote]

Yeah, I'm pretty far too.

50 Backstroke: 36.03 (but this was long ago)
100 Backstroke: 1:12.28
200 backstroke: 2:39.65

I do modified and I swim competetively. :D  :D

im on a swim team, but i dont like competitive swiming. swimming for fun is fun

I swim competitively it is my favorite sport.  :surf:

[quote=aznmonkey1992]any swimmers 'round here?
im taking lessons[/quote]

check out the swim hub!!!


Let $A$ be a subset of $\{1,2,3,...,100\}$ with $16$ elements.
Prove that we can find 4 distinct numbers $a,b,c,d \in A$ such that $a+b=c+d$.
The existence is simple,we take all the differences(pozitive ones of course) which are 120 and they take values from 1 to 99,but how do i prove we can find 4 [b]distinct[/b] numbers with that property??

have a look at [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=134104]here[/url]. as i said there i have invented a general formula to find this number. and for your problem my formula gives exactly $16$. but i won't post the solution if someone doesn't reply my problem first. :)

edit: but i may change my mind.. :wink:


Who, in your opinion, is the best artist of all time?

I think this question is too broad. First off, are you inquiring about visual artists or musical artists? Or both?

And yeah, it is a matter of personal preference, I guess but I have a hard time choosing between the two.

But if it ever came down to it, I'd pick Mozart. Any day. Too great for this world.

If it is visual, I would say that my favorite is Claude Monet

I like Jan Vermeer of Delft as a visual artist, and Murray Perahia as a performer of other people's piano compositions.

i think the best artist of all time is me! OK I was JK
I think  Monet too.

Mental High Five Funcia!

Musically, I like a whole lot of artists, but some of my favorites include Roger Waters from Pink Floyd, Jim Morrison from the doors, and Jimi Hendrix.

O yeah ! My musical artist=pink floyd! OMG thanks mustafa i forgot about music!

Musical artist, I would say Metallica and AC/DC, I love the bands.

well  before stating ur fav. artist,state which genre he belongs to bcos u can't say ac/dc or pink floyd is da greatest of ALLL the artists nor is mozart teh greatest of all genres,only his operah or whatever he was in.

i'm into rocknroll only,so my choices would be led zeppelin(jimmy page),pink floyd n  gunsnroses(axl rose)(ya u wont agree with gnr but i love them so hehe).Jimi hendrix was great as well but pink floyds music is considered path breaking.

MONEEEEEEYYYY OHH MONEYYYYYYYY..............
"welcome to the machine" is an AoPS 1 chapter did anyone notice that?????
well i guess you are right but this might have to do also with favorite genre apart from stating it or i should say

classical:beethoven
impressionism: monet
rock?:pink floyd
rap: missy elliot
sounding like a girl: justin timberlake
opera: wagner
modern art: whoever made that paper thing in the princeton art museum
songs about hating math: jimmy buffet
truffles: lindt
singing the crazy song: that guy who says " i thnk im crrrrraaaaaazzzzzyyyyy!"
....................and i could go on

the craaazzzyyy guy would be gnarles barkley lol

yeah i just figured that out on itunes well that's hard to remember so if i say crazy guy remember i don't mean the old lead singer of pink floyd(i forgot his name so this is what i call him) or john nash

it's so nice to have pink floyd fans here (:
But David Gilmour pwns Roger Waters. HIs guitars... make me swoon.
Yes, classical rock guitarists are the best. Jimmy Page, Jimi Hendrix, etc. etc. Progressive rock was the best thing to happen to the music world (:

Lasairfhíona Ní Chonaola!!

she is beautiful and sing celtic music! woow... she is the best

Fashion: Marc Jacobs all the way!
architectural: Antoni Gaudi
Music: Panic at the Disco
Best at making people feel extremely awkward when singing: korammath
Painter:idk...hard to say at this point

taylor swift!!!! for pop/country
for dancing to...ke$ha
for hiphop-taio cruz(well actually i dont like him...im just addicted to "break your heart" its my fave song right now)
classical-bethooven?
rock-boys like girls all da way
art?-easy...either me, my friend young, or my friend(who broke her collar bone!) madaline

I can't believe it, almost everyone has forgotten Vincent Van Gogh! high five hi pi :lol:
He's probably my favorite visual artist.

Visual Artist: Picasso. Da Vinci.
Music Artist: That's pretty hard. I'd divide them up in categories:
Baroque: JS Bach
Classical: Mozart
Romantic: Schumann
Techno: Yukana

Composers:
Baroque: J. S. Bach
Classical: Beethoven
Romantic: Chopin (or maybe Liszt)
It's difficult to point one or two best pianists, but I like Horowitz, Argerich, Pollini,  Gulda, Barenboim, and so on.

Classical: Beethoven no doubt. He had amazing solos with beautiful blending with the orchestra.

Guitar: Jimi Hendrix, Van Helsen
Classical: Rachmaninoff, Bach
Visual: Michelangelo
A little bit more contemporary: Dali
I can't think atm -.- will post later.

Classical: Shostakovich, Brahms, and Schubert are among my favorites.
Non-Classical: Belle & Sebastian, anyone?

Dali, O' Keefe, Van Gogh for visual.
Bach for Baroque. I mean, he almost invented the period.
Haydn for Baroque. Hands down.
Mendelssohn for Romantic. His style is unique and beautiful.

I don't know if this counts, but I like the architectural designs by Frank Lloyd Wright because of it's preciseness.
For Baroque, I like J. S. Bach because of his tones that he uses and the patterns he puts in his music.
For Classical, I like Beethoven because there's a nice variety of different pieces to play.
For Romantic, I like Chopin for the same reason as above, and that his pieces are so complex and developed.
For Contemporary, I don't really have a preference; I like all the pieces.

HVE NONE OF YOU EVER HEARD THE BEST ALTERNATIVE ROCK BAND OF ALL TIME!!!
wowow.....
ARCTIC MONKEYS!!!

oh, for piano, definitely DEBUSSY

and for artist, DA VINCI!!!

Artists...
Visual: Pablo Picasso
Composer: Beethoven no doubt... love his violin concertos
Performing: Itzhak Perlman

GUNS and roses $\Huge\text{=}$ The beatles

Guns and rosess are a $\tiny\text{little}$ better


Mhpws kapoios mporei na ekshghsei  to " limma tou jordan"????

http://mathworld.wolfram.com/JordansLemma.html

http://www.math.utah.edu/~milicic/Math_6220/jordan.pdf

http://en.wikipedia.org/wiki/Jordan's_lemma

thanks για την βοήθεια

[quote="giannis"]Για να μην ανοίγω άλλο topic μπορεί κανείς να με βοηθήσει να βρω το

$ \lim_{x \to 0 }x e^{\frac{1}{x}}$[/quote]

Μήπως εννοείς $ x\to 0^{+}$ ? Διαφορετικά δεν υπάρχει στο $ 0$.

Προφανώς $ \lim_{x \to 0^{-}}x e^{\frac{1}{x}}= 0\cdot 0 =0$

ενώ

$ \lim_{x \to 0^{+}}x e^{\frac{1}{x}}=\lim_{x \to 0^{+}}\frac{e^{\frac{1}{x}}}{\frac{1}{x}}\stackrel{L'H\left(\frac{+\infty}{+\infty}\right)}{=}\lim_{x \to 0^{+}}\frac{e^{\frac{1}{x}}\cdot \left(-\frac{1}{x^{2}}\right)}{-\frac{1}{x^{2}}}= \lim_{x \to 0^{+}}e^{\frac{1}{x}}=+\infty$.

Αλέξανδρος

Ευχαριστώ Αλέξανδρε για την βοήθεια , το όριο είναι έτσι , αλλά μπορείς να δεις αν βγαίνε χωρίς De L' Hopsital?
Ευχαριστώ!

Γιάννη ένας άλλος τρόπος  για να βρεις το $ \lim_{x \to 0^{+}}x e^{\frac{1}{x}}$ είναι

$ \lim_{x \to 0^{+}}x e^{\frac{1}{x}}=\lim_{u \to+\infty}\frac{e^{u}}{u}$  (Θέτω $ \frac{1}{x}=u$)

δηλαδή θέλεις να υπολογίσεις το τελευταίο όριο.

Επόμενο βήμα είναι να δείξεις (με τη βοήθεια παραγώγων) ότι για κάθε $ x\geq 0$ ισχύει

$ e^{x}\geq 1+x+\frac{x^{2}}{2}$ άρα για $ x>0$ έχουμε $ \frac{e^{x}}{x}\geq\frac{1}{x}+1+\frac{x}{2}$

Όμως το όριο του δεξιού μέλους , όταν $ x\to+\infty$ είναι $ +\infty$, άρα και το όριο του αριστερού μέλους είναι $ +\infty$ (λόγω της ανισότητας που δείξαμε) και τελειώσαμε.

Αλέξανδρος

Μπορεί να με βοηθήσει κανείς να βρω το

$ \lim_{x \to 0}{\frac{x}{1+e^{\frac{1}{x}}}}$

Καταρχήν $ \lim_{x\to 0^{-}}e^{\frac{1}{x}}= 0$ και $ \lim_{x\to 0^{+}}e^{\frac{1}{x}}=+\infty$

Έχουμε λοιπόν  $ \lim_{x\to 0^{-}}\frac{x}{1+e^{\frac{1}{x}}}= \frac{0}{1+0}=0$ και

$ \lim_{x\to 0^{+}}\frac{x}{1+e^{\frac{1}{x}}}= \lim_{x\to 0^{+}}( x \cdot \frac{1}{1+e^{\frac{1}{x}}})$.

Όμως $ \lim_{x\to 0^{+}}\frac{1}{1+e^{\frac{1}{x}}}) = 0$, άρα το παραπάνω όριο $ \lim_{x\to 0^{+}}\frac{x}{1+e^{\frac{1}{x}}}$, υπάρχει και είναι ίσο με $ \lim_{x\to 0^{+}}(x) \cdot \lim_{x\to 0^{+}}(\frac{1}{1+e^{\frac{1}{x}}})) = 0\cdot 0 =0$

Άρα τελικά υπάρχει και το  $ \lim_{x \to 0}{\frac{x}{1+e^{\frac{1}{x}}}}$ και ισούται με $ 0$.
[u][b]
Σημείωση:[/b][/u] Όρια που με το "μάτι" βγάζεις ότι είναι της μορφής $ \frac{0}{\pm\infty}$ ή $ \frac{\pm\infty}{0}$, όπως το $ \lim_{x\to 0^{+}}\frac{x}{1+e^{\frac{1}{x}}}$, πρέπει να τα σπάσεις αντίστοιχα σε ένα γινόμενο της μορφής $ 0\cdot \frac{1}{\pm\infty}= 0\cdot 0 =0$ και $ \pm\infty\cdot\frac{1}{0}=$ [b](διακρίνεις περιπτώσεις για τον παρονομαστή αν πηγαίνει στο $ 0^{+}$ ή στο $ 0^{-}) =+\infty$ ή $ -\infty$ ή δεν υπάρχει ανάλογα [/b]

Αλέξανδρος

Episis mporeis na to deis kai ws eksis ... Gia kathe x diaforo tou midenos : 0=<1/1+e^1/x <1 sinepagete 0=<|x|/1+e^1/x<|x| kai afou x--->0 exoume 0=<|x|/1+e^1/x=<0 ara (apo Sandwich 'h allios Kritirio paremvolis)|x|/1+e^1/x--->0 kathos x--->0 opote  x/1+e^x-->0 kathos x-->0 (elpizw na einai katanoito etsi opws to grafw giati den exw to Latex)

* =< einai to mikrotero 'h iso kai >= to megaliteri 'h iso...


geometry
incenter
circumcircle
angle bisector
perpendicular bisector
geometry solved

I don't think the solution will have all the necessary details, but here are main ideas:

Let $I^*$ be the projection of $I$ (the incenter) on $AC$. Try to show that $HQP$ is similar to $BCA$ (I don't know if this is necessary, bu it gives some insight), and that $I^*$ is the incenter of $HQP$. Show that $M$ is equidistant from $Q,P\Rightarrow M$ is the point where the angle bisector of $\angle QHP$ cuts the circumcircle of $HQP\Rightarrow MI^*=MQ(=MP)$. If $B^*$ is the point where the bisector of $\angle B$ cuts $AC$, then the fact that $MQHP$ is cyclic is equivalent to $MI^*^2=MB^*\cdot MH$, and this really is well-known (it's also easy to prove).

I haven't got the time to read your solution right now, but I think I have a simpler one.

If the line CQ meets AB at R, then the triangles RQB and CQB are congruent (since BQ = BQ, < RBQ = < CBQ, < RQB = < CQB = $90^{\circ}$); hence, RQ = CQ, and Q is the midpoint of the segment CR. On the other hand, M is the midpoint of the segment CA. Hence MQ || RA, or, in other words, MQ || AB. Thus, if the angle bisector of the angle ABC meets the line CA at X, we have XM / XQ = XA / XB. But on the other hand, since < AHB = < APB = $90^{\circ}$, the points A, B, H and P lie on one circle, and thus by the intersecting chords theorem we have $XH \cdot XA = XP \cdot XB$, thus XA / XB = XP / XH. Together with the former equality XM / XQ = XA / XB, this entails XM / XQ = XP / XH, so that $XM \cdot XH = XP \cdot XQ$. Now, by the converse of the intersecting chords theorem, the points M, H, P and Q are concyclic. OED (maybe you know I'm in Greece right now).

  darij

Nice Proof! This can be done by just chasing angles.

[quote="darij grinberg"]... in other words, MQ || AB. ...[/quote]

After noting that MQ || AB, we could go like this:

 :ang: MQP =   :ang: ABP =  :ang: AHP ( :ang: MHP) since A,B,H,P are concyclic. Hence M,Q,H,P are concyclic.

It's interesting to note that:

The triangle PMQ is isosceles.

The quadrilateral PHQM is similar to  the quadrilateral  ABCD where D is the intersection of the  bisector of :ang: ABC and the circumcircle of ABC.

Another Proof:

Refering to the diagram above. MD is the perpendicular bisector of AC (An easy fact being used for #1 of IMO2004, what a coincidence :) )

 :ang: AMD= :ang: APD=90 :^o: ==> A,M,P,D are concyclic ==>  :ang: PMH= :ang: ADB

 :ang: BQC= :ang: BHC=90 :^o: ==> B,Q,H,C are concyclic ==>  :ang: PQH= :ang: ACB

Hence :ang: PMH= :ang: PQH since :ang: ADB= :ang: ACB.

Therefore P,M,Q,H are concyclic.

QED.

Let $ CQ \cap AB \equal{} {R}$.Then BRC is isosceles triangle.From this we have Q is midpoint of CR.It follows that $ QM$  || $ AB$.After that is just angle-chasing.

Dear Mathlinkers,
this nice circle is from Calabre.
A proof can be see on my website :  http://perso.orange.fr/jl.ayme   vol. 1  Le théorème de Feuerbach  Exemple 1.  (in French)
or in spain :   http://www.oei.es/oim/revistaoim/numero26.htm     el theorema de Feuerbach
Sincerely
Jean-Louis

[quote="Maverick"]Let M be the midpoint of the side AC of a triangle ABC and let H be the footpoint of the altitude from B. Let P and Q be orthogonal projections of A and C on the bisector of the angle B. Prove that the four points H,P,M and Q lie on the same circle[/quote]
Furthermore, the center of this circle lies on the 9 point circle of ABC.

Assume WLOG H is between A and M.
Extend AP to meet BC at R, extend CQ to meet AB at S. Then $ \triangle ARB$ and $ \triangle SBC$ are isoceles, so $ AS \equal{} RC \equal{} a \minus{} c$ and since M is the midpoint of AC, P the midpoint of AR, $ PM \equal{} \frac {a \minus{} c}{2}$, similarly $ QM \equal{} \frac {a \minus{} c}{2}$. So $ \triangle$ PMQ is isoceles, so PHQM is cyclic if $ \angle PHM \equal{} \angle PQM$, or $ \angle PHM \equal{} \angle XPM$, or $ \triangle MPX \sim \triangle MPH$, or $ MX \cdot MH \equal{} MP^2$. But $ MH \equal{} MA \minus{} HA \equal{} \frac {b}{2} \minus{} c \cdot cos A \equal{} \frac {b}{2} \minus{} \frac {b^2 \plus{} c^2 \minus{} a^2}{2b}$. Also $ MX \equal{} MA \minus{} XA \equal{} \frac {b}{2} \minus{} \frac {bc}{a \plus{} c}$. So we have to show
$ (\frac {b}{2} \minus{} \frac {b^2 \plus{} c^2 \minus{} a^2}{2b})(\frac {b}{2} \minus{} \frac {bc}{a \plus{} c}) \equal{} (\frac {a \minus{} c}{2})^2$
Which is easy to verify.


P1,P2,...P11 are 11 distinct points on a line s,t PiPj \leq 1 prove that the sum of all PiPj (distance betweenPi and Pj) \leq 30

11*10/2 = 55 distances to be summed

if P1P2 = 1 (maximal)
then the sum of P1P3 and P2P3 is smaller than 1 (Pn is between P1 and P2)
combine the above: P1P4 + P2P4 + P3P4 is smaller than 2
Adding the N-th number adds the following distance:
(1-x) * the number of the points on the opposite side and x*the number on the own side, x varying for every point.

Therfor it's easy to see that the maximum is thus always attained if half of the points are at one side, and half at another side!

Since there are 11 points, they can at best be divided 5-5 and one variable point, for which it does not matter where it is placed.

The length is therefor smaller then (since they are distinct):
5*5 + 5*(1-x) + 5*x = 30


trigonometry
logarithms
inequalities unsolved
inequalities

Prove that
$ 2^{|sin x|}\plus{}2^{|cos x|} \geq 3 (\forall x)$

Let, $ f(x)\equal{}2^{|\sin x|}\plus{}2^{|\cos x|}$

$ \begin{array}{lll}
\cdot \ \ f \left(x\plus{}\frac{\pi}{2}\right) & \equal{} & 2^{|\sin \left(x\plus{}\frac{\pi}{2}\right)|}\plus{}2^{|\cos \left(x\plus{}\frac{\pi}{2}\right)|} \\
 & \equal{} & 2^{|\cos x|} \plus{} 2^{|\minus{}\sin x|} \\
 & \equal{} & 2^{|\sin x|}\plus{}2^{|\cos x|} \\
 & \equal{} & f(x)
\end{array}$

$ \begin{array}{lll}
\cdot \ \ f \left(\frac{\pi}{2}\minus{}x\right) & \equal{} & 2^{|\sin \left(\frac{\pi}{2}\minus{}x\right)|}\plus{}2^{|\cos \left(\frac{\pi}{2}\minus{}x\right)|} \\
 & \equal{} & 2^{|\cos x|} \plus{} 2^{|\sin x|} \\
 & \equal{} & f(x)
\end{array}$

$ \cdot \ \ f'(x) \equal{} \log 2(2^{\sin x}\cos x \minus{} 2^{\cos x}\sin x) \ \ (x \in [0,\frac{\pi}{2}])$
So, $ f'(x)\equal{}0 \Rightarrow x\equal{}\frac{\pi}{4}$

Therefore,
\[ \begin{array}{|c\parallel{}c|c|c|c|c|} \hline
x & 0 & \cdots & \frac{\pi}{4} & \cdots & \frac{\pi}{2} \\ \hline
f'(x) & \plus{} & \plus{} & 0 & \minus{} & \minus{} \\ \hline
f()x & 3 & \nearrow & 2\cdot2^{\frac{1}{\sqrt{2}}} & \searrow & 3 \\ \hline
\end{array}\]

Thus, $ 2^{|\sin x|}\plus{}2^{|\cos x|} \ge 3 \ \ (\forall x \in \mathbb{R})$

Is there any other solution easier ? :huh:

[quote="No Reason"]Prove that $ 2^{|sin x|} \plus{} 2^{|cos x|} \geq 3\ ,\ (\forall )\ x\in \mathcal R$ .[/quote]
[color=darkblue]The problem $ \boxed {t\in [0,1]\implies 2^{\sqrt t} \plus{} 2^{\sqrt {1 \minus{} t}}\ge 3}$ with equality iff $ t\in\{0,1\}$ is well-known.

For $ t: \equal{} \sin^2x$ obtain the proposed problem. I"ll come back soon with the proof of it.[/color]


function
topology
advanced fields
advanced fields unsolved

Let $X$ be a metric space. How to prove that the characteristic function of open set $A\subset X$ islower semicontinuous in $X$?


Show that $ S_k \equal{} \sum_{n \geq 0} \left(\sum_{i \equal{} 1}^k \frac {1}{2nk \plus{} 2i \minus{} 1} \right) \minus{} \frac {1}{2n \plus{} 2}$ converges and compute its value.  For example, $ S_1 \equal{} \frac {1}{1} \minus{} \frac {1}{2} \plus{} \frac {1}{3} \minus{} \frac {1}{4} \plus{} \ldots$ and $ S_2 \equal{} \frac {1}{1} \plus{} \frac {1}{3} \minus{} \frac {1}{2} \plus{} \frac {1}{5} \plus{} \frac {1}{7} \minus{} \frac {1}{4} \plus{} \frac {1}{9} \plus{} \frac {1}{11} \minus{} \frac {1}{6} \plus{} \ldots$.

[Edited to fix typo caught by jmerry.]

That should be $ \sum_{n\ge 0}\left(\left(\sum_{i\equal{}1}^k \frac{1}{2kn\plus{}2i\minus{}1}\right)\minus{}\frac1{2n\plus{}2}\right)$, to match your examples. As written, the positive terms generally appear in several different copies of the inner sum.

Yes, sorry, I knew something looked too simple :)

Incidentally, I do know how to throw this expression into a computer algebra system.  Therefore, potential future commenters, if you have nothing better to do than plug into a CAS and report the result, please withhold and let people who actually have some of their own thoughts on this do the commenting.


geometry
geometric transformation
reflection
angle bisector
geometry proposed

Let ABCD be a convex quadrilateral. Show that < ADB + < BAC + < CBD + < DCA = 180° holds if and only if either $AB\cdot CD=BC\cdot DA$ or the quadrilateral ABCD is cyclic.

This is from a recent publication: [url=http://forumgeom.fau.edu/FG2006volume6/FG200619index.html]Eisso J. Atzema, [i]A theorem by Giusto Bellavitis on a class of quadrilaterals[/i], Forum Geometricorum, 6 (2006) p. 181-185[/url], where this fact is proven using trigonometry. Bellavitis himself is said to have used complex numbers. How would you prove it?

  Darij

I hope it's correct:

I will prove:
 $ABCD\ \text{is cyclic or}\ AB.CD=AD.BC \longrightarrow \angle ADB+\angle CAB+\angle DBC+\angle ACD=180^{\circ}$
(next one is easier.)

if $ABCD$ be cyclic problem is obviuos if not:
We have $\frac{AD}{CD}=\frac{AB}{BC}$ therefore internal and external angle bisector of angle $D\ ,\ B$ meet $AC$ at points $E,F$ hence $(EF,AC)=-1$

let $\omega$ be a circle with diameter $EF$ and $\omega$ passes through $B,D$ too.

if $\angle DCA=\angle ACB$ it's easy to see that $\angle ADB+\angle CAB+\angle DBC+\angle ACD=180^{\circ}$

if not we can consider that $\angle ACB>\angle DCA$,let $D',G'$ be reflection of $D,G$ wrt $EF$ hence intersection point of $DG',D'G(A')$ is on $EF$ and we have $(EF,A'C)=-1$therefore $A'\equiv A$
like this we have that $AB$ pass through reflection of $H$ wrt $EF$ so we have:

$\angle ADB=\angle G'DB=\angle G'DH+\angle HDB=\angle G'DH+\angle ABD-\angle BDD'=\angle ABD+\angle HCG'$

hence $\angle ADB+\angle DCA=\angle ABD+\angle ACB=180^{\circ}-\angle CAB-\angle DBC$

therefore We have:$\angle ADB+\angle CAB+\angle DBC+\angle ACD=180^{\circ}$

A nice proof! (One typo: $\frac{AB}{CD}=\frac{AB}{BC}$ should be $\frac{AD}{CD}=\frac{AB}{BC}$.)

I had something rather different in mind - in fact, the source in my post above gives a hint to my solution ;) .

  Darij

Hi my dears...

it is proved that: if $ABCD$ is cyclic, or $AB.CD$=$AD.BC$, then:

$\angle ADB+\angle BAC+\angle CBD+\angle DCA = 180$

now i want to prove:  if $\angle ADB+\angle BAC+\angle CBD+\angle DCA = 180$, then  $ABCD$ is cyclic, or

$AB.CD$=$AD.BC$.

i want to prove this fact by $INVERSION$, but i dont invert all of my shape.i only want to figure out, some facts

, which i cant find out them without inversion.

ok..............................................................

let me do an inversion with center $A$ and an arbitrary radius.

and suppose, points $B,C,D$, changes to points $B',C',D',$ respectively.

$\angle ADB=\angle AB'D'$

$\angle BAC=\angle B'AC'$

$\angle DCA =\angle AD'C'$

$\angle DBC=\angle ABC- \angle ABD=\angle AC'B'-\angle AD'B'$

if we some them we should have:

$\angle AB'D'+\angle B'AC'+\angle AD'C'+\angle AC'B'-\angle AD'B'=180$

which is equal to:

$\angle AB'D'+\angle B'AC'+\angle AD'C'+\angle AC'B'-\angle AD'B'$+$(\angle D'B'C'-\angle D'B'C')=180$   ,$(1)$

but in triangle $AB'C'$, we have: $\angle AB'D'+\angle AC'B'+\angle B'AC'+\angle D'B'C'=180$      , $(2)$

then, according to $(1)$and$(2)$ equality, we should have:

$(\angle AB'D'+\angle AC'B'+\angle B'AC'+\angle D'B'C')+\angle AD'C'+-\angle AD'B'-\angle D'B'C'=180$

then:$180+\angle AD'C'+-\angle AD'B'-\angle D'B'C'=180$

and: $\angle AD'C'=\angle AD'B'+\angle D'B'C'$

but,$\angle AD'C'=\angle AD'B'+\angle B'D'C'$

then:  $\angle AD'B'+\angle B'D'C'=\angle AD'B'+\angle D'B'C'$

and we conclude that: $\angle B'D'C'=\angle D'B'C'$

now if: $\angle B'D'C'=\angle D'B'C'=0$, then $B',D',C'$ are colinear, and it shows us that before inversion,

points $A,B,C,D,$ were cyclic.

and if: $\angle B'D'C'=\angle D'B'C$ isnt equal to $0$,

then in triangle$B'D'C'$, $B'C'=C'D'$
.........................................................................................................................................................................

lemma: suppose $AB$ is a segment and $A'B'$ is its inversive segment, with center $O$ and radius $K$.

we have: $A'B'=\frac{K.AB}{OA.OB}$

.........................................................................................................................................................................

then according to lemma:

$B'C'=\frac{K.BC}{AB.AC}$

$C'D'=\frac{K.CD}{AC.AD}$

$B'C'=C'D'$, then :  $\frac{K.BC}{AB.AC}=\frac{K.CD}{AC.AD}$

multipple at $\frac{AC}{K}$:

$\frac{BC}{AB}=\frac{CD}{AD}$

and: $AB.CD=AD.BC$.

and every thing will be ok.

i enjoyed alot with this problem ;) .

[quote="darij grinberg"]Exactly this was my solution ;) . The hint was to [i]invert[/i] (as the source was the inverted title of the thread).

Actually your solution proves not only the direction that if $\angle ADB+\angle BAC+\angle CBD+\angle DCA = 180^{\circ}$, then $ABCD$ is cyclic, but it also proves the converse.

  Darij[/quote]

really?

u made me glad.  ;)

because i thought that only I use inversion to solve problems here :rotfl:

this theorem is really nice.

i tried to find it for pentagon and hexagon and ...

but i got nothing.

may be,this problem hold, only for quadrilateralls. ;)


inequalities
limit
induction
Miscellaneous Problems

Let $\alpha(n)$ be the number of digits equal to one in the dyadic representation of a positive integer $n$. Prove that [list=a] [*] the inequality $\alpha(n^2 ) \le \frac{1}{2} \alpha(n) (1+\alpha(n))$ holds, [*] equality is attained for infinitely $n\in\mathbb{N}$, [*] there exists a sequence $\{n_i\}$ such that $\lim_{i \to \infty} \frac{ \alpha({n_{i}}^2 )}{ \alpha(n_{i}) } = 0$.[/list]

I assume "dyadic" here means "binary".

It is easy to prove by induction on the binary length of numbers that
\[ a(x\plus{}y)\leq a(x)\plus{}a(y).\]

Let $ n\equal{}2^{b_1} \plus{} 2^{b_2} \plus{} \dots \plus{} 2^{b_m}$ be binary representation of an integer $ n$, where $ b_1<b_2<\dots<b_m$. Then $ a(n)\equal{}m$ and
\[ n^2 \equal{} \sum_{i\equal{}1}^m 2^{2b_i} \plus{} \sum_{i\equal{}1}^m\sum_{j\equal{}i\plus{}1}^m 2^{b_i\plus{}b_j\plus{}1},\]
implying that a proof for the problem a.:
\[ a(n^2)\leq \sum_{i\equal{}1}^m 1 \plus{} \sum_{i\equal{}1}^m\sum_{j\equal{}i\plus{}1}^m 1\equal{}\frac{m(m\plus{}1)}{2}.\]

To establish b., it is enough to show that there is infinitely many sets of numbers $ b_1<\dots<b_m$ for which $ |\{ 2b_i, b_i\plus{}b_j\plus{}1 \}|\equal{}\frac{m(m\plus{}1)}{2}$. It is clear that this requirement holds for any sequence $ b_1<\dots<b_m$ that grows fast enough. For example, any sequence for which $ b_{i\plus{}1}>2b_i$ for every $ i\equal{}1,\dots,m\minus{}1$ works.

To prove c., it is enough to take
\[ n_i \equal{} 2^{2^i\minus{}1} \minus{} \sum_{j\equal{}1}^i 2^{2^i\minus{}2^j}.\]


use mod100.

$9^{\phi(100)} = 9^{40} \equiv 1\mod100$
then $9^47 \equiv 9^7 \mod100$
Then work from there.

[quote="xxreddevilzxx"][quote="Altheman"]9
81
729
6561

notice the pattern... 9,1,9,1...

for a more rigirous proof, you need to know modular arithmetic[/quote]

it said tens digit, not the first digit.[/quote]

you beat me to fixing my post...lol

[quote="Altheman"]do you know mods? also you can use binomial expansion[/quote]

I have been trying to figure out mods for the past *insert insane amount of time*. I just can't get it. Can you do it the multiplying-out-and-finding-a-pattern way?

How would you use binomial expansion?

[quote="xxreddevilzxx"][hide="hint"]
$9=10-1$
$9^{47} = \left(10-1\right)^{47}$
[/hide][/quote]

hmm. I'm not sure how that would help. I usually associate that kind of form with finding the coefficients of things (and even then I'm pretty rusty on that idea).

[quote="amcavoy"]Every term in $\left(10-1\right)^{47}$ will have a power of $10$ except for the last, which is just $\left(-1\right)^{47}=-1$.  Therefore it will end in a $-1$ which is otherwise known as $\boxed{9}$ (we're working mod 10 here).[/quote]

tens digit, not the first digit. but thats the binomial way to solve this problem.

Sorry about that.  In that case, it will just be $10\cdot\binom{47}{46}-1=469\equiv 69\pmod{100}$, therefore your answer is $\boxed{6}$.

[quote="xxreddevilzxx"]well.... i dont want to give it away.... so heres another hint:

[hide="hint"]
Well Do you really need to expand all that???? I mean what is the problem asking, and then think again[/hide][/quote]

I think you want the second to last term because you would have one ten...and you want tens place. Thinking (a+b)^x-wise, the second to last term would be ab^x-1. So...for this problem, that would mean ab^46. 47C1 = 47 because there are 47 ways (yeah, thinking aloud) to choose 1 from 47 things. Let a = 10, b= -1. ab^46 = 10. Since the "coefficient" is 47 -- this doesn't make any sense. Am I totally wrong??

[quote="xxreddevilzxx"]well.... ill just give away, but its basically what amcavoy said.

Consider $9^{47}$.

It is $\left(10-1\right)^{47}$ and the expansion is as follows:

$10^{47} + 47 \cdot 10^{46} \cdot (-1)^1 + ... + 47 \cdot 10^1 (-1)^{46} + (-1)^{47}$

We dont need to consider anything but the last two terms (Thats what i meant by do we need to expand all on the second hint) since all the first 46 terms have 0 as the tens digit. Now, the coefficient of the second the last term is $47$, meaning the number is $470$, and the last term is $-1$, so $470-1=469$, meaning the last two digits have to be $69$. thus the tens digit is $6$[/quote]

well, i got half of it. x.X Thanks. :D

If you get used to the idea of using the binomial series, these types of problems become very easy (most of them, at least).


Stie cineva adresele de e-mail ale
-Revista Arhimede
-Revista de matematica si informatica din Constanta
-Revista Octogon?
   Imi trebuiesc ca sa fac si eu abonamente la ele.

pentru revista Arhimede incearca http://www.arhimede.ro si pentru revista din Constanta rmicta@yahoo.com


Compute the numerical value of the infinite sum $\sum_{n=0}^{\infty}(\log_{2000}2)^{n}(\log_{2000}5^{4^{n}}).$

I thought this problem was pretty cool. Here's what I came up with
[hide="Solution"]
When we see infinite sums, we usually think of geometric series or telescoping. Since a product of logs is difficult to telescope, we'll attempt to simplify the sum to a geometric series. Note that
\begin{eqnarray*}(\log_{2000}2)^{n}(\log_{2000}5^{4^{n}}) &=& 4^{n}(\log_{2000}2)^{n}(\log_{2000}5) \\ &=& (\log_{2000}5)(4\log_{2000}2)^{n}= (\log_{2000}5)(\log_{2000}16)^{n}. \end{eqnarray*}
Hence, the sum simplifies to
\begin{eqnarray*}\sum_{n=0}^{\infty}(\log_{2000}2)^{n}(\log_{2000}5^{4^{n}}) &=& \sum_{n=0}^{\infty}(\log_{2000}5)(\log_{2000}16)^{n}\\ &=& (\log_{2000}5)\left(\frac{1}{1-\log_{2000}16}\right) \\ &=& (\log_{2000}5)\left(\frac{1}{\log_{2000}2000-\log_{2000}16}\right) \\ &=& (\log_{2000}5)\left(\frac{1}{\log_{2000}125}\right) =1/3. \end{eqnarray*}
[/hide]

Out of curiosity, does anyone know if there is a solution by [hide]telescoping? I haven't seen any interesting log telescoping problems before.  :D [/hide]

i ono...but this seems kind of trivial...it is obviously a geometric series  :|

obviously, you can telescope a geometric series with the following:

$ar^{n}=a\cdot \frac{1-r^{n+1}}{1-r}-a\cdot \frac{1-r^{n}}{1-r}$

--

obviously product log_k (k+1), etc and sum log[(k+1)/k], etc are common log, telescoping problems


[quote="AIME15"]
[asy]dot((6,0));
dot((4,3));
dot(origin);
draw((6,0)--(4,3)--origin--cycle);
label("A", (6,0), SE);
label("B", (4,3), N);
label("C", origin, SW);[/asy]

What's wrong with my code!?!?[/quote]

I tried debugging, and it seems like the entire dot() command doesn't work.

Read http://www.mathlinks.ro/viewtopic.php?p=1342109#1342109 there is nothing wrong with your code, as you can see now


geometry
3D geometry
number theory unsolved
number theory

Hello.

I have the following problem:

Find the minimum $ n > 0$ such that $ \frac {n}{2}$ is a square of some natural number and $ \frac {n}{3}$ is a 3rd power of some natural number.

Now so far I have:

$ \frac {n}{2} \equal{} x^2 , n \equal{} 2x^2$ and $ \frac {n}{3} \equal{} y^3, n \equal{} 3y^3$ for naturals x and y.

thus we have:

$ 2x^2 \equal{} 3y^3$

and

$ 2x^2 \minus{} 3y^3 \equal{} 0$ .

for mod 2:

$ \minus{} 3y^3 \equal{} 0 (mod 2)$ so $ y^3 \equal{} 0 (mod 2)$  and $ y \equal{} 0(mod 2)$.

for mod 3:

$ 2x^2 \equal{} 0(mod3)$ so $ x^2 \equal{} 0(mod 3)$ and $ x \equal{} 0(mod3)$.

So according to the chinese remainder theorem our solution Z=0 (mod 6).
Now im stuck.

Am I right up to this point? How do I continue from here? Was my direction even correct?

Thanks alot.

Let $ n\equal{}2^a 3^b k$, where $ 2, 3 \nmid k$ and $ a, b$ are positive integers.
It's easy to see that $ k$ is the sixth power of some natural number , so to obtain the minimum value , $ k\equal{}1$
Now $ 2^{a\minus{}1} 3^b$ is a square , $ 2^a 3^{b\minus{}1}$ is a cube
$ \Leftrightarrow a\equal{}1(mod \;\; 2), 0(mod \;\; 3) ;b\equal{}0(mod 2), 1(mod 3)$
To attain the min. value, $ a\equal{}3, b\equal{}4$, i.e. $ n\equal{}2^3 \times 3^4\equal{}648$

[quote="mszew"]$ 2x^2 \equal{} 3y^3$ then mod $ 2$ and $ 3$
$ x \equal{} 3a$ and $ y \equal{} 2b$ replacing $ 18 a^2 \equal{} 24b^3 \Rightarrow 3a^2 \equal{} 4b^3$ again mod $ 2$ and $ 3$
$ a \equal{} 2c$ and $ b \equal{} 3d$ replacing $ 12c^2 \equal{} 108d^3 \Rightarrow c^2 \equal{} 9d^3$ then $ (\frac {c}{3})^2 \equal{} d^3$
with minimum solution for $ d \equal{} 1$ and $ c \equal{} 3 \Rightarrow b \equal{} 3$ and $ a \equal{} 6 \Rightarrow y \equal{} 6$ and $ x \equal{} 18$
$ \boxed{n \equal{} 648 \equal{} 2\times18^2 \equal{} 3\times6^3}$[/quote]

What does it mean that $ 2x^2 \equal{} 3y^3$ then mod $ 2$ and $ 3$ ? You mean the congruence holds both for mod $ 2$ and mod $ 3$ seperately?

I didnt understand how did you decide to place $ x\equal{}3a$ and $ y\equal{}2b$ ?

stephencheng I couldnt understand right at the begining how to choose $ n$ of that particular form...its way too advanced for me to see such things :(

[quote="RealTime"][quote="mszew"]$ 2x^2 \equal{} 3y^3$ then mod $ 2$ and $ 3$
$ x \equal{} 3a$ and $ y \equal{} 2b$ replacing $ 18 a^2 \equal{} 24b^3 \Rightarrow 3a^2 \equal{} 4b^3$ again mod $ 2$ and $ 3$
$ a \equal{} 2c$ and $ b \equal{} 3d$ replacing $ 12c^2 \equal{} 108d^3 \Rightarrow c^2 \equal{} 9d^3$ then $ (\frac {c}{3})^2 \equal{} d^3$
with minimum solution for $ d \equal{} 1$ and $ c \equal{} 3 \Rightarrow b \equal{} 3$ and $ a \equal{} 6 \Rightarrow y \equal{} 6$ and $ x \equal{} 18$
$ \boxed{n \equal{} 648 \equal{} 2\times18^2 \equal{} 3\times6^3}$[/quote]

What does it mean that $ 2x^2 \equal{} 3y^3$ then mod $ 2$ and $ 3$ ? You mean the congruence holds both for mod $ 2$ and mod $ 3$ seperately?

I didnt understand how did you decide to place $ x \equal{} 3a$ and $ y \equal{} 2b$ ?

stephencheng I couldnt understand right at the begining how to choose $ n$ of that particular form...its way too advanced for me to see such things :([/quote]

We have $ n$ can be represent like these : $ 2^{a_1} 3^{a_2} (p_3^{a_3} p_4^{a_4} ...)$, where $ 3<p_3<p_4<p_5<...$ and $ p_3, p_4, p_5, ...$ are distinct primes . So $ n$ is a cube $ \Leftrightarrow 3| a_1, a_2, a_3, ... \Leftrightarrow 2^{a_1} 3^{a_2}$ is a cube and $ (p_3^{a_3} p_4^{a_4} ...)$ (which is defined as $ k$ in my post before) is a cube , similar for square.

For mszew's post , $ 3|x$ because $ 3|R.H.S.\equal{}L.H.S.$

I just don't know which part you don't understand as this is very basic .

P.S.  You should post this topic in High School Basic.

Let $ (x,y)\equal{}d$, $ x\equal{}dx_1, y\equal{}dy_1$ then $ 2x_1^2\equal{}3dy_1^3$ implying $ y_1^3$ divides $ 2$, $ y_1\equal{}1$ hence $ x_1^2\equal{}3d/2$ and $ d\equal{}6u^2$

Then we get $ n\equal{}648u^6$

[quote="RealTime"][What does it mean that $ 2x^2 \equal{} 3y^3$ then mod $ 2$ and $ 3$ ? You mean the congruence holds both for mod $ 2$ and mod $ 3$ seperately?

I didnt understand how did you decide to place $ x \equal{} 3a$ and $ y \equal{} 2b$ ?
[/quote]

Taking mod $ 2$ implies that $ 3y^3 \equiv 0 (2)$ then $ y \equiv 0(2)$ then $ y\equal{}2b$
similarly to get $ x\equal{}3a$


Find real numbers like $m,b$ such that $\sum_{j=1}^{50}(j^2+jm+b)^2$
be minimal.

Expanding gives $\sum_{j=1}^{50}j^4+2mj^3+(m^2+2b)j^2+2mbj+b^2$.

Plugging in values for $\sum_{j=1}^{50}j^k$, we get $65666665 + 3251250m + 42925m^2 + 2550mb + 85850b + 50b^2 =$ $50(b+25.5m+858.5)^2+10412.5(m+51)^2+1732640\ge1732640$ with equality iff $m=-51$, $b=442$.

Ahh, the power of brute force...

I'm sure there's a more elegant proof based on the grouping $j^2+mj+b=\left(j+\frac{m}{2}\right)^2+\left(b-\frac{m^2}{4}\right)$.  In fact, if we minimise the average value of the square term here, and then minimise the average value of the entire expression, we get the same $m$ and $b$. This would need a more analytic justification of course...


superior algebra
superior algebra unsolved

Prove that the only groups such that any two elements of the same order are conjugates are S_3, C_2, and the trivial group.

I am not sure how hard this problem is or how well known.

Your statement is only true for finite groups (where the trivial group should also be included).  For infinite groups, the HNN construction produces an infinite group all of whose non-identity elements are conjugate.

The following three papers use the classification of finite simple groups to handle the finite case:

Fitzpatrick, Patrick. "Order conjugacy in finite groups."
Proc. Roy. Irish Acad. Sect. A 85 (1985), no. 1, 53-58.
http://www.ams.org/mathscinet-getitem?mr=821422

Feit, Walter; Seitz, Gary M. "On finite rational groups and related topics."
Illinois J. Math. 33 (1989), no. 1, 103-131.
http://www.ams.org/mathscinet-getitem?mr=974014

van der Waall, Robert W.; Bensaïd, Adaouia. "On finite groups whose elements of equal order are conjugate."
Simon Stevin 65 (1991), no. 3-4, 361-374.
http://www.ams.org/mathscinet-getitem?mr=1162505


For which positive integers n is it possible to find a sequence of n positive integers whose sum is equal to
the square root of their product?

[quote="john lai"]that pretty easy : davis poul liao :rotfl:[/quote]

wtf

BUMP anyone :ninja:

Eh are you sure the problem is right? I got $n=1$ as the only answer, by AMGM

Ummm... seems to be all n or almost all n. I didn't bother to check, but just take a, b to be solutions to the equation (a-2)(b-2) = n, and then pick the numbers a, a, b, b, 1, 1, 1, ... and it should work. That's if n >= 4. n = 2 is trivly possible, and n = 3 I don't know but it probably isn't that hard to figure out.

Oops, i see what i did. And did badly, for that matter


function
inequalities
geometry
circumcircle
rectangle
linear algebra
matrix

Day 1:

1. http://www.mathlinks.ro/Forum/viewtopic.php?t=24078

Circle $\Gamma$ and line $d$ don't have common points. Line containing the diameter $[AB]$, is perpendicular to $d$, so that $B$ lies between $A$ and $d$. Let $C$ be an arbitrary point on the circle, different from $A$ and $B$, $D$ the intersection point of $AC$ and $d$. Line $DE$ is tangent to $\Gamma$ at $E$ so, that $B$ and $E$ are in the same halfplane defined by $AC$. $F$ is the intersection point of $BE$ and $d$, $AF$ intersects $\Gamma$ at $G$, different from $A$. Prove that the simmetric of $G$ with respect to $AB$ lies on $CF$.

2. http://www.mathlinks.ro/Forum/viewtopic.php?p=190511#p190511

$a^4+b^4+c^4=3$, $a,b,c>0$. Prove that $\sum\frac1{4-ab}\leq1$.

3. http://www.mathlinks.ro/Forum/viewtopic.php?p=190516#190516

\[ A=3\sum_{m=1}^{n^2}(\frac12-\{\sqrt{m}\}) \]
where $n$ is an positive integer. Find the largest $k$ such that $n^k$ divides $[A]$.

4. http://www.mathlinks.ro/Forum/viewtopic.php?p=190523#190523

$n$ is a positive integer, $K$ the set of polynoms of real variables $x_1,x_2,...,x_{n+1}$ and $y_1,y_2,...,y_{n+1}$, function $f: K\rightarrow K$ satisfies
\[ f(p+q)=f(p)+f(q),\quad f(pq)=f(p)q+pf(q),\quad (\forall)p,q\in K. \]
If $f(x_i)=(n-1)x_i+y_i,\quad f(y_i)=2ny_i$ for all $i=1,2,...,n+1$ and
\[ \prod_{i=1}^{n+1}(tx_i+y_i)=\sum_{i=0}^{n+1}p_it^{n+1-i} \]
for any real $t$, prove, that for all $k=1,...,n+1$
\[ f(p_{k-1})=kp_k+(n+1)(n+k-2)p_{k-1} \]

Day 2:

1. http://www.mathlinks.ro/Forum/viewtopic.php?p=204870#p204870

In triangle $ABC$, $M\in(Bc)$, $\frac{BM}{BC}=\alpha$, $N\in(CA)$, $\frac{CN}{CA}=\beta$, $P\in(AB)$, $\frac{AP}{AB}=\gamma$.
Let $AM\cap BN=\{D\}$, $BN\cap CP=\{E\}$, $CP\cap AM=\{F\}$. Prove that
$S_{DEF}=S_{BMD}+S_{CNE}+S_{APF}$ iff $\alpha+\beta+\gamma=1$.

2. http://www.mathlinks.ro/Forum/viewtopic.php?p=204872#204872

Let $m\in N$ and $\displaystyle E(x,y,m)=(\frac{72}x)^m+(\frac{72}y)^m-x^m-y^m$, where $x$ and $y$ are positive divisors of 72.
a) Prove that there exist infinitely many natural numbers $m$ so, that 2005 divides $E(3,12,m)$ and $E(9,6,m)$.
b) Find the smallest positive integer number $m_0$ so, that 2005 divides $E(3,12,m_0)$ and $E(9,6,m_0)$.

3. http://www.mathlinks.ro/Forum/viewtopic.php?p=204875#204875

Does there exist such a configuration of 22 circles and 22 point, that any circle contains at leats 7 points and any point belongs at least to 7 circles.

4. http://www.mathlinks.ro/Forum/viewtopic.php?p=204876#p204876

Find the smallest positive $p$ ($p>1$) such, that $\displaystyle\forall a,b,c\in[\frac1p,p]$ the following inequality takes place
\[ 9(ab+bc+ca)(a^2+b^2+c^2)\geq(a+b+c)^4 \]

Day 3:

1. http://www.mathlinks.ro/Forum/viewtopic.php?p=205736#205736
Let $ABC$ and $A_1B_1C_1$ be 2 triangles. Prove that
\[ \frac a{a_1}+\frac b{b_1}+\frac c{c_1}\leq\frac{3R}{2r_1} \]

2. http://www.mathlinks.ro/Forum/viewtopic.php?p=205737#p205737
In an acute-angled triangle $ABC$, $\angle ABC<\angle ACB$, $O$ is the circumcenter, $D=AO\cap BC$. $E$ and $F$ are the circumcenters of $ABD$ and $ACD$ respectively. $A\in[BG]$, $AG=AC$, $A\in[CH]$, $AH=AB$. Prove that $EFGH$ is a rectangle iff $\angle ACB-\angle ABC=60 <degrees>$

3. http://www.mathlinks.ro/Forum/viewtopic.php?p=205740#p205740
Let $A$ be a square matrix of order $n$, $X_i$ the set of numbers from the $i$th row, $Y_i$ the set of numbers from the $j$th row, $1\le i,j\le n$. Matrice $A$ is called "golden", if the sets $X_1,...,X_n,Y_1,...,Y_n$ are different. Find the smallest natural $n$ such, that there exists a golden matrix of the order 2005 with elements from $\{1,...,n\}$.

4. http://www.mathlinks.ro/Forum/viewtopic.php?t=33056
Given functions $f,g: N*\rightarrow N*$, $g$ is surjective and $2f(n)^2=n^2+g(n)^2$, $\forall n>0$. Prove that if $|f(n)-n|\le2005\sqrt n$, $\forall n>0$, then $f(n)=n$ for infinitely many $n$.

I think condition of the $4$ problem need to be changed.
I wrote right condition in it's topic.

Also how can i find the Moldova 2004? and Ukrine 2004 ?
if u have 'em sent me to my email address.

Rustam Moldova 2004 ni  so'rab nima qilarding. Baribir undan tushmadiyu ahir. :P  :P  :P  :P


While playing around with a problem centered around primes the other day, I happened to discover some surprising properties of the first four primes, 2, 3, 5, and 7. If we take the product of any two of these primes and add it to the product of the other two, the sum is prime. Furthermore, if we take the product of three of these primes and add it to the fourth, the sum is also prime:

2*7 + 3*5 = 29
2*5 + 3*7 = 31
2*3 + 5*7 = 41
2*3*5 + 7 = 37
2*3*7 + 5 = 47
2*5*7 + 3 = 73
3*5*7 + 2 = 107

See if you can find other groups of primes that have this property.

right off the bat, one of the primes is 2; otherwise, some numbers wouldnt be odd
2, 3, 7, 11 works

Since you are creating new integers which must be relatively prime with all the small primes you are using, they chances that they will be prime is very high.

That's a very cool piece of number theoretic exploration.


geometry
geometric transformation
rotation
geometry unsolved

What is the area of an equilateral triangle with the distances from one point to the internal vertices of the triangle equal to 7, 8 and 9.

I found the height of the triangle--> h = 7+8+9....But I think it is not worth

In equilateral triangle ABC of side length d, if P is an internal point with PA = a, PB = b, and PC = c, the following pleasingly symmetrical relationship holds:

$ 3(a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4) \equal{} (a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)^2$

http://www.qbyte.org/puzzles/p117s.html

As suggested, let a, b and c be the distances from P to the vertices A, B and C of the equilateral triangle ABC. A 60 deg rotation around B will bring A to C and P to K, another rotation of 60 deg around C will map B to A and P to L, while a 60 deg rotation around A will map C to B and P to M. See that triangles AMP, BPK and CLP are equilateral of sides a, b and c respectively, the triangles APL, BPM and CPK are congruent, having the sides of a, b and c and the hexagon AMBKCL has its area twice the area of the equilateral triangle ABC (by the way of construction).

Best regards,
sunken rock

Please, what is [b]internal[/b] vertices of a triangle?


linear algebra
matrix
inequalities
linear algebra unsolved

[color=green][b] Let  $A$ be a symmetric $5\times 5$ real  matrix such that   $P(x): = \det(xI - A)= x^5- x^4-0.5x^3+a_3x^2+ a_4x +a_5$  has  the roots (eigenvalues) $r_1\le r_2\le r_3\le r_4\le r_5$ .  Assuming   $\; r_5-r_1 \ge \sqrt{3.6}\; ,\;$  find $\det{(A)}\;$ . [/b] [/color]

[quote="Moubinool"]...Use AM-GM .....[/quote]
Hmm ? Not clear ![color=blue] [b]I appreciate that it's possible to find the exact value of $\det(A)$ .[/b][/color]

[quote="Moubinool"]Your question is find the exact value of det(A) ?[/quote]
If it's possible to compute $\det(A)$ , then  is of interest !.
[size=75][color=green] [b] By the way, a question  : AM-GM inequality (for five numbers)  is valid also for negative numbers ? [/b][/color][/size]

[color=green] [b] By the way, a question  : AM-GM inequality (for five numbers)  is valid also for negative numbers ? [/b][/color][/size][/quote]

We need positive number $a_k\geq 0$ then

${(a_1a_2...a_n)^{\frac{1}{n}}\leq \frac{a_1+...+a_n}{n}}$

I see what you mean


find the value of the term independent of $ a$ in $ (a^2\minus{}\frac{4}{a})^{\frac{9}{2}}$

The extended binomial theorem tells us that $ (x\plus{}y)^r\equal{}\sum_{k\equal{}0}^\infty {r \choose k} x^{r\minus{}k} y^k$, where $ \dbinom{r}{k}\equal{}\frac{r(r\minus{}1)(r\minus{}2)\cdots(r\minus{}(k\minus{}1))}{k!}$.
The constant term occurs when $ k\equal{}3$.
Plugging this in, we get $ \dbinom{\frac92}{3}a^3\cdot\minus{}\left(\frac{64}{a^3}\right)\equal{}\minus{}64\dbinom{\frac92}{3}\equal{}\minus{}\frac{64\cdot\frac92\cdot\frac72\cdot\frac52}{6}\equal{}\boxed{\minus{}420}$.


inequalities
function
integration
calculus
real analysis
real analysis unsolved

Let f(x) be a nonnegetive concave function on [a,b], 0<r<s, to prove that
$ \left( {\frac{{s \plus{} 1}}{{b \minus{} a}}\int_a^b {f^s \left( x \right)dx} } \right)^{\frac{1}{s}}  \le \left( {\frac{{r \plus{} 1}}{{b \minus{} a}}\int_a^b {f^r \left( x \right)dx} } \right)^{\frac{1}{r}}$

That kind of reminds me of Holder's Inequality.

I think this one is quite contradiction stuff. Exchange the role of $ r$ and $ s$, we'll get the same formula again. I guess unless we assume something like $ 0<k<r<s$, there is certainly no point in this inequality.

Sorry for forgetting the conditions. I think you can just separate the part $ (\frac{y+1}{b-a})^{\frac{1}{y}}$ with the part of the integral. Both of them should be progressive with $ y$ by elementary calculus. Then you can easily prove this.

you can easily prove it????  I don't think so !!!!!!!!!!!!!!!!!!!!!!!

still no answrer about the problem??????????????????????

How many times should I repeat that [color=red]this is not a hotline[/color]? :roll:

Anyway, here is a hint. First of all, we can assume that $ [a,b]\equal{}[0,1]$. Second, we can replace $ f$ by its increasing rearrangement (which is still concave). Third, we can check the following general lemma:

[i]If $ f$, $ g$ are two increasing functions on $ [0,1]$ such that $ f\ge g$ on $ [0,c]$ and $ f\le g$ on $ [c,1]$ for some $ c\in[0,1]$, and $ r>0$, then $ \int_0^1 f^r\equal{}\int_0^1 g^r$ implies $ \int_0^1 f^s\le \int_0^1 g^s$ for all $ s>r$[/i]

Fourth, we can use this lemma with our $ f$ and $ g(x)\equal{}ax$ with appropriate $ a>0$ (depending on $ r$).

dear fedja. would you give some detailed explanation about the third the firth points.

Sure. Sorry for the delay: I had very limited access to computers for the last couple of weeks.

The inequality in Step 3 is really quite simple (though I should say that I forgot to mention explicitly that it is assumed that $ f,g\ge 0$ :blush:). Let $ F\equal{}f^r$, $ G\equal{}g^r$, $ p\equal{}s/r>1$. Then $ \int_0^1 (G^p\minus{}F^p)\equal{}\int_0^1 pH^{p\minus{}1}(G\minus{}F)$ where $ H$ is some function squeezed between $ F$ and $ G$ (the mean value theorem). Now let $ F(c)\equal{}G(c)\equal{}Q\ge 0$. Then $ H\le Q$ on $ [0,c]$ and $ H\ge Q$ on $ [c,1]$, whence $ \int_0^c pH^{p\minus{}1}(G\minus{}F)\ge Q\int_0^c(G\minus{}F)$ and $ \int_c^1 pH^{p\minus{}1}(G\minus{}F)\ge Q\int_0^c(G\minus{}F)$ (in the first integral $ G\minus{}F\le 0$, so the smaller positive factor gives larger product and in the second integral the situation is the opposite). Adding these inequalities up, we get $ \int_0^1 pH^{p\minus{}1}(G\minus{}F)\ge Q\int_0^1(G\minus{}F)\equal{}0$.

As to the fourth step, it is exactly as written: fix $ r>0$, find $ a>0$ such that $ g(x)\equal{}ax$ satisfies $ \int_0^1 f^r\equal{}\int_0^1 g^r$. Note that $ f$ and $ g$ satisfy the condition of the lemma (here is where the concavity of $ f$ is used) and draw the conclusion that
\[ \begin{aligned}
\left[(s\plus{}1)\int_0^1 f^s\right]^{1/s}&\le
\left[(s\plus{}1)\int_0^1 g^s\right]^{1/s}\equal{}a
\\
&\equal{}
\left[(r\plus{}1)\int_0^1 g^r\right]^{1/r}\equal{}
\left[(r\plus{}1)\int_0^1 f^r\right]^{1/r}
\end{aligned}\]

the third can be proved by use of majorization theory. thank you very much!!!!!!!!!!

dear fedja, can we prove the result without use of  incresing rearrangment which is exacly what i am especially intersted in??????????

What's wrong with the increasing rearrangement? :? I mean, we can formally avoid it, but it'll really be still present there in some hidden form. Are you really after some generalization of the problem where my solution doesn't work? If so, just state the new question and I'll look at what can be done with it. Otherwise I see no reason to tie my hands with any such restrictions. :)


Este vorba despre problema:
Fie ABCD un patrulater convex şi P un punct exterior patrulaterului. Notăm cu Sab, Sbc, Scd, Sda, Sac, Sbd ariile triunghiurilor PAB, PBC, PCD, PDA, PAC şi respectiv PBD. Să se demonstreze următorul rezultat analog cu teorema lui PTOLEMEU: Sab.Scd+Sad.Sbc=Sac.Sbd.
In attachment veti gasi figura si solutia mea!

[color=darkblue]Aceasta frumoasa [b]relatie metrica intr-un patrulater[/b] ilustreaza ideea ca [b]tratarea vectoriala este un instrument foarte puternic[/b].

Indeosebi prin faptul ca nu tine seama de multitudinea cazurilor posibile precum o tratare sintetica.

Concluzia se desprinde de la sine : [b]aprofundati Geometria vectoriala[/b]. Frumoasa solutie oferita va va indemna spre studiul ei.

O tratare sintetica a acestei probleme este simpla de fapt, insa [b]ne "omoara" multitudinea cazurilor posibile ![/b]

In fiecare din cazuri relatia metrica se va reduce la o banala identitate trigonometrica, insa de fiecare data alta.

[b]Exemplu.[/b] In cazul $ \{\begin{array}{c}BC\cap [DP]=\emptyset\\\ CD\cap (AP)\ne\emptyset\\\ DA\cap [CP]=\emptyset\end{array}$, notand $ \{\begin{array}{c}m(\widehat{BPC})=x\\\ m(\widehat{APB})=y\\\ m(\widehat{DPA})=z\end{array}$, relatia metrica

$ [PAB]\cdot [PCD]+[PAD]\cdot [PBC]=[PAC]\cdot [PBD]$ este echivalenta cu identitatea

$ \sin y\sin (x+y+z)+\sin x\sin z=\sin (x+y)\sin (y+z)$, adica

$ cos (x+z)-\cos (x+2y+z)+\cos (x-z)-\cos (x+z)=\cos (x-z)-\cos (x+2y+z)$.

[b]P.S.[/b] Imi voi permite sa va multumesc, Domnule profesor [b]Mihai Miculita[/b], in numele userilor acestui Site ![/color]