analytic geometry
trigonometry
conics
parabola
graphing lines
slope

A ball is thrown from the top of a hemispheric hill of radius R at an initial speed V. Find, in terms of R, V and g = 9.81ms^-2, the minimum distance between the ball's point of landing and the base of the hill.

Picture the hill as an isolated rock on a plain, and remember that the ball must not crash into the hill before it lands on the plain. Neglect air resistance.

At first I thought that this was a very hard question, but after some thinking and drawing I'm not so sure... I'll write my solution:

First, let's introduce the cartesian coordinates so that the center of the hill is in $(0,0)$ and the $x$-axis is parallel to the horizontal direction. The innitial coordinates of the ball are $(0, R)$. Let $\phi$ denote the angle between the velocity $\vec v$ and the $x$-axis, so that the projections of $\vec v$ are $\vec v_{\rm x} = v \cos\phi$ and $\vec v_{\rm y} = v \sin\phi$. The coordinates of the ball after time $t$ are then given by $x = vt\cos\phi$ and $y = R + vt\sin\phi - \frac{gt^2}{2}$. After substituting $t$ we get the equation of the parabola described by the ball: \[ y = R + x\tan\phi - \frac{g}{2v^2\cos^2\phi}x^2.  \] Note that we have assumed that $\phi\neq 90^\circ$ -- a justified assumption because if we hurl the ball directly upwards it will land on the hill.

This is where I got confused a bit. Clearly, the problem was to find the parabola so that $y = 0$ gives the minimal possible $x$. An additional constraint was that this parabola wasn't supposed to intersect with the circle given by the equation $x^2 + y^2 = R^2$. However, the parabola and the circle will intersect if and only if, at the point where the parabola intersects the $x$-axis, the $x$ coordinate is not greater than $R$! A plausible (I hope) proof is that the slope $\theta$ of the parabola (its tangent given by $\tan\theta = \frac{dy}{dx}$) is always lesser than the slope of the circle, so that the parabola will always 'run away' from the circle until it reaches its maximum and meats the circle again (in point $(0, R)$). I'm not sure that I've made myself clear, and I'm sure that I haven't given the correct mathematical proof, but I really don't have the nerve to go around squaring roots and making complicated expressions ;) Anyway, this means that if the ball can reach the base of the hill (point $(R,0)$) then it will not touch the hill during its fall and $R$ will be the required minimal distance. Furthermore, the ball will reach the maximal horizontal distance if thrown under a certain angle and if this distance is lesser than $R$ the ball will not be able to fall down without touching the hill.

I hope that there is sense in what I just said :)

Your solution looks good, except i disagree with the fact that the maximum distance will be attained when the angle $\phi = 45^\circ$. This is true for the motion of a projectile which lands on the same level it begins on, but not for one which lands on a different level.You could reason this out by intuiting that the particle spends more time in the air when it is landing on a different level from the one it started on, and therefore it can "spend" more of its velocity on the horizontal component and less on its vertical component. Other than that, I believe the ball can indeed fall at a point very, very, very close to $(0, R)$, which is essentially falling at the base of the hill. It can never actually fall onto the base of the hill, since a parabola opening downwards can never be tangent to a circle at the point where the circle's slope reaches infinity (or undefined). I think I made a correct analysis of the problem, but the more advanced max/min problems always seem to thwart me when I try to intuit the answer.

What has gotten into me? Every problem I try to do these days turns out wrong :?

Ok, but you must agree that there will be a maximal $x$ the ball can reach. After some boring trigonometric deriving and equation-solving it can be obtained... I'll change my post accordingly.

Well Djole, you got it that the gist of the problem is to find the conditions such that the parabola will not intersect the hemisphere (x^2 + y^2 = R^2 ). And this is where the hard part is.
Obviously the ball will crash into the hill if it is launched too steeply or gently, or simply just too slowly. On the other hand, it may be launched too fast for the minimal distance condition to be satisfied at any angle. Notice that the ball can never land exactly at the base of the hill, for this will require a zero horizontal velocity (or infinite vertical velocity).

Good luck  :)


(From the Polya Competition, 1996, Team Round Part 2)

4.1  Josh and Athina play a match of pong.  The first player to have BOTH:

(a) at least 10 points and
(b) at least 2 more points than their opponent

wins the game.  The first player to win BOTH

(a) at least 5 games
(b) at least 2 more points than their opponent

wins the set; the first player to win two sets wins the match.  How small a fraction of games can Josh win and still win the match?  Express your answer as a fraction in lowest terms.

4.2 Let r be the numerator of the answer to the previous question.  Two circles with radius x are tangent to each other.  A third circle of radius r is tangent to both circles.  All three circles are disjoint and the three circles share a common tangent line.  Find x.

4.3 Let x be the answer to the previous question.  The numbers a^2, b^2, and c^2 form an arithmetic progression.  If a + b = -120 and c - b = x, find (c-a)/(b-a).

4.4 Let a be the answer to the previous question.  A piece of paper that is 4a cm wide and 5a + 1 cm long is folded so that one pair of opposite vertices coincide.  Find the length of the crease.  (That is, find the length of the line the ofld makes in the paper.)

4.5 Let q be the answer to the previous question.  If the roots of x^2 + bx + c = 0 are r1 and r2, and r1 + r2 = 25 and r12 + r22 = q, find c.

Enjoy.  Please post your solutions if at all possible.

1. [hide] If Athina wins 4 games and Josh wins 5, Josh could have two points more than his opponent. So, 5/9 per set, and 5/9 per match (if the scores start anew after each set). [/hide]


MIT college Harvard Princeton Stanford CMU

I'm studying in Hong Kong, which practices the British 6-5-2-4 System, and I've just now finished Form 5 (Grade 11, if you want to make it in the American way). I'm now considering to attend college / university in the US, after I finish my Form 6.

This is my brief resume.

International Student

Competitions:
Member and chairman-to-be, leading the team, and myself, getting various top 1-3 awards in my city, and went to China for several competition (China West Math Oly x 2, China Southeast Math Oly x 1), scoring mostly silver medals
IMO 2009 Team member
HKPhO (you can say that it's a national PhO) top scorer of 2009, aiming (and hopefully) to get into IPhO team in 2010

Academic:
Top score in all Chemistry, Biology and Physics in same year (last year) + Rank 1
Top score in both Economics and Math this year + Rank 2
All my non-language subjects are in top 10, the whole form / grade consists of 200 students, while my school being the top 3 school in my city / region
Likely getting 9A in HKCEE (equivalently as the IGCSE, though lots more difficult, and does not award A*)
Took GCE A Level Pure math in Form 4 (Grade 10), got A
Took GCE O Level math in Primary 6 (Grade 6), got A
Actively participating gifted programmes offered in a local university.

Plans:
Take SAT Reasoning Test and SAT II Math II and Physics
Universities likely applying: MIT, Caltech, Harvard, Princeton, Stanford

Worries:
I don't have much extra-curricular activities mentions little science, also, my languages (Chinese and English), are just fair, though I'll have to brush my English for SAT Reasoning Test

Please give me some advice, on what to improve, (if you are in the above universities, is there anything that I particularly lack :P), are there other good universities that will suits me?

[quote="calc rulz"]I'm not an expert on admission for foreign students, but if you are a member of the Chinese IMO team, then I expect that you will be admitted to most if not all of those universities.[/quote]

Well. I'm not a Chinese IMO team member, the IMO team I refer to, is only Hong Kong (a Chinese city which participate individually in IMO).

[quote="Darmani"]MIT international admissions can be insane. I heard there was a perfect-scorer on the Chinese IMO team who was rejected.[/quote]

Then how about the other universities?

Also, I guess if in that case, I should keep my focus on some extra-curricular and extra-MO activities. Would this help?

[b]@ KenHungKK[/b]

The title of this thread ought to be:[b] "Advice on College Needed" [/b] and not [b] "Advise on College Needed". [/b]

(Advice is a noun, and advise is a verb, so, in this particular context, the word should be "advice")

I should have some confidence in myself, mistaken it by copying that from the title of another thread. Thanks, gen.

do you have results from your applications?

Look at Case Western, it is a good Math and Science school. Also you will easily make it in to MIT. :)

Sorry to have you guys waiting, I've actually forgotten this post :P

Caltech - Accepted and planned to enroll
CMU - Accepted
MIT, H, P - Rejected

Congrats! Are you going to study math at CalTech?


probability
algorithm
induction
conditional probability
probability and stats

A standard deck of 52 cards (26 red and 26 black) is shuffled so that all orderings are equally likely. We then play the following game: I begin turning the cards over one at a time so that you can see them. At some point (before I have turned over all 52 cards) you say “I’m ready!” At this point I turn over the next card and if the card is red, you receive one dollar; otherwise you receive nothing. You would like to design a strategy to maximize the probability that you will receive the dollar. How should you decide when to say “I’m ready”?

Play hunches. Or develop an algorithm based on the numbers in your last fortune cookie. It doesn't matter - every strategy is as good or bad as every other strategy.

One rule, of course: if there's only one card left and it's red, then you will say "I'm ready" in that circumstance.

Here's the claim: if there are $ q$ cards left to be played, of which $ p$ are red, then your chances of winning are $ \frac pq,$ no matter what your strategy is.

We prove this by induction on $ q.$

There are two $ q\equal{}1$ cases: state $ (p,q)\equal{}(1,1)$ and state $ (p,q)\equal{}(0,1).$ In the first case, by the one rule I mention above, you will say "I'm ready," and you will win. In the other case, you've already lost and it doesn't matter what you say. In both cases, that agrees with what we're trying to claim by induction.

Now suppose we have proved my result for $ q\minus{}1$ and you find yourselves in state $ (p,q).$ You have two strategic choices.

You could say "I'm ready," and the game will be resolved immediately. Your probability of winning is $ \frac pq.$

Or you could pass to the next turn. Two things can then happen:

With probability $ \frac pq,$ a red card is drawn, leaving you in state $ (p\minus{}1,q\minus{}1).$ By induction your conditional probability of winning is $ \frac{p\minus{}1}{q\minus{}1}.$

With probability $ \frac{q\minus{}p}{q},$ a black card is drawn, leaving you in state $ (p,q\minus{}1).$ By induction your conditional probability of winning is $ \frac{p}{q\minus{}1}.$

Combine those and your probability of winning (under the strategy of passing the turn) becomes \[ \frac pq\cdot\frac{p\minus{}1}{q\minus{}1}\plus{}\frac{q\minus{}p}{q}\cdot\frac p{q\minus{}1}\equal{}\frac{p^2\minus{}p\plus{}pq\minus{}p^2}{q(q\minus{}1)}\equal{}\frac{p}{q}.\]

So the the probability of winning is $ \frac pq$ no matter which strategy you employ. And by induction, this holds for all possible positions.

Since our strategy has no effect on your chances of winning, you're then free to decide whether you'd rather have a long game or a short game.

The game show "Deal or No Deal" is a somewhat different beast; any speculation about that should probably be in another topic. But consider this variation: Each time after a card is dealt, the dealer makes you an offer of $ \frac pq$ dollars to end the game before any more cards are dealt.

thanks, it was great.
If you would like to discuss "Deal or No Deal", start a new thread, and I'd be happy to read, even though I am not familiar with that show.
thanks again


Let under  equation \[ az^3\plus{}bz^2\plus{}cz\plus{}d\equal{}0\]
We knew that $ \alpha,\beta,\gamma$ are three solution of this equation
find the value $ \alpha\beta\gamma$
Let you try doing.

[quote="minhphuc.v"]Let under  equation
\[ az^3 \plus{} bz^2 \plus{} cz \plus{} d \equal{} 0\]
We knew that $ \alpha,\beta,\gamma$ are three solution of this equation
find the value $ \alpha\beta\gamma$
Let you try doing.[/quote]

This is difficult. Maybe few people get full marks.
Anyway, the problem should be moved to appropriate forum.

[quote="Joao Pedro Santos"][quote="kunny"]This is difficult. Maybe few people get full marks.[/quote]
I don't agree... Using Viète's Formulas, we find out the result is $ \minus{} d/a$.
Unless at least two of them are equal and there are other solution(s)...
But if they are all the solutions the problem is very easy.[/quote]

I think either there's something missing, or he was being sarcastic :D

[quote="Joao Pedro Santos"][quote="kunny"]This is difficult. Maybe few people get full marks.[/quote]
I don't agree... Using Viète's Formulas, we find out the result is $ \minus{} d/a$.[/quote]

Sorry, you get 0 point. :(  Why is $ a\neq 0?$

I assumed $ a\ne0$... :D
Anyway, $ \alpha$, $ \beta$ and $ \gamma$ are distinct, right?

No they are any solution. Joao Pedro Santos is all right .This problem's solution is $ \minus{}\frac{d}{a}$.
But how do you do it?

It's easy too see that:
$ az^3\plus{}bz^2\plus{}cz\plus{}d\equal{}a(z\minus{}\alpha)(z\minus{}\beta)(z\minus{}\gamma)\equal{}$
$ \equal{}az^3\minus{}a(\alpha\plus{}\beta\plus{}\gamma)z^2\plus{}a(\alpha\beta\plus{}\alpha\gamma\plus{}\beta\gamma)z\minus{}a(\alpha\beta\gamma)$
And the conclusion comes from there.
But if they are not necessarily different, the solution must be incorrect, even though the result is correct.

[quote="Joao Pedro Santos"]
But if they are not necessarily different, the solution must be incorrect, even though the result is correct.[/quote]

No, being make a distinct solution of $ \alpha ,\ \beta ,\ \gamma$, Vieta's formula is valid.

If there are 3 distinct solutions, you need not mention in the problem that $ a\ne0$ because
It would follow immediately from the Fundamental Theorem of Algebra that indeed $ a\ne0$, and so by using Vieta's Formulas, the product of the roots is $ \minus{}\frac{d}{a}$ as stated.


I get $ \boxed{192}$ using the path counting method mentioned [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=115126610&t=246549]here[/url].

Edit: Wow, I'm being so stupid. You're correct.

My apologies; you're correct. I forgot to draw the last column on my paper because I did it in a hurry. :oops:

That will surely take less than a minute huh??

[quote=Kouichi Nakagawa]$ \binom{7\plus{}5}{5} \minus{} \left(\binom{2\plus{}2}{2} \cdot \binom{5\plus{}3}{3}\right) \minus{} \left(\binom{2\plus{}3}{3} \cdot \binom{5\plus{}2}{2}\right) \plus{} \left(\binom{2\plus{}2}{2} \cdot \binom{0\plus{}1}{1} \cdot \binom{5\plus{}2}{2}\right)$
$ \equal{} \binom{12}{5} \minus{} \left(\binom{4}{2} \cdot \binom{8}{3}\right) \minus{} \left(\binom{5}{3} \cdot \binom{7}{2}\right) \plus{} \left(\binom{4}{2} \cdot \binom{1}{1} \cdot \binom{7}{2}\right)$
$ \equal{} 792 \minus{} 6 \cdot 56 \minus{} 10 \cdot 21 \plus{} 6 \cdot 1 \cdot 21$
$ \equal{} 792 \minus{} 336 \minus{} 210 \plus{} 126$
$ \equal{} \boxed{372}$[/quote]

Explain?

You get to exactly one of the four red points. Then you just sum the products of the binomial coefficients.

Using the method above and some combinatorics, you should get $\boxed{372}.$


number theory
greatest common divisor
Euler
number theory unsolved

Prove,with $ k \in N,k\geq\ 2$ There is infinite positive integers $ n$ such that
$ {2^2}^n \plus{} k$ is not prime

Let $ t$ = order of $ 2 ,mod \frac {\phi (k \plus{} 4)}{2}$

CLAIM: For $ n \equal{} tr \plus{} 1$ ,$ {2^2}^n \plus{} k$ is composite $ \forall r \in N$

Proof:
t is order of 2 mod $ \frac {\phi (k \plus{} 4)}{2}$

Hence,
$ 2^{tr} \equal{} (2^t)^r \equiv 1(mod \frac {\phi (k \plus{} 4)}{2})$
Let $ 2^{tr} \minus{} 1 \equal{} u [\frac {\phi (k \plus{} 4)}{2}]$
Hence,
$ 2^{2^n \minus{} 2} \equal{} 2^{2^{tr \plus{} 1} \minus{} 2} \equal{} 2^{\phi(k \plus{} 4)u} \equiv 1 (mod (k \plus{} 4))$(by Euler's Theorem)
[We may assume k to be odd and thus $ gcd (k\plus{}4,2)\equal{}1$]

Hence,
$ 2^{2^n} \equiv 4 \equiv \minus{} k (mod (k \plus{} 4))$
$ 2^{2^n} \plus{} k \equiv 0 (mod (k \plus{} 4))$

Hence $ 2^{2^n} \plus{} k$ is composite.

Thanks,a nice solution,i have just had a proof,but it is hard thank you

[quote="aakansh92"]Let $ t$ = order of $ 2 ,mod \frac {\phi (k \plus{} 4)}{2}$
[/quote]
Hey, I have found a flaw in this statement:

I think Order -t will not necessarily exist as it is possible that Gcd($ 2 ,mod \frac {\phi (k \plus{} 4)}{2}$) is NOT =1

Am I right?

AND

can you post your proof please???

Iy means your proof is wrong,sorry because I did not check carefully
Here is My proof( can you complete it,my English is bad)
first $ k\equal{}2^s\minus{}1$ there $ s$ is odd.With $ a\in N$ choose $ m$ such that $ 2^{2^m}>a\minus{}k$
Then,with $ l$ is national number,$ l\geq\ s,l\geq\ m$
If $ 2^{2^l}\plus{}k$ is not prime.we finish our proof
in other case set $ p\equal{} 2^{2^l}\plus{}k$ so $ p\minus{}l\equal{}2^sh_1$ there $ h_1$ is odd
Use Euler theorem $ 2^{s\plus{}\phi(h_1)}\equiv 2^s (mod(p\minus{}1))\equiv 2^l$
so ưe can easy to see that $ 2^{2^{l\plus{}\phi(h_1)}} \plus{}k$ is not prime( use Fecmat treorem)
My proof is completed
Thank your proof


symmetry
group theory
abstract algebra
geometry
geometric transformation
reflection
superior algebra

Let $K=\{I,H\}$ be a subgroup of the symmetry group $D4$. What are the conjugates $g^{-1}Kg$, $g\in D_{4}$? What is $[K]$?

[quote="rustam"]What is I,H and what you mean by $[K]$? :maybe:[/quote]

I assume since K is a subgroup I is the identity and H is an element of order 2.

I'm almost certain $[K]$ is meant to be the equivalence class of conjugate subgroups.

Sorry, I was so tired yesterday.

$I$ is the identity map and $H$ is the map which maps a square $ABCD$ to $CDBE$ i.e. a reflection with respect to the line through the vertices. $[K]$ is indeed the equivalence class of conjugate subgroups.

I solved the problem today.


I am only able to prove one part: $x^y+y^x <2$.
Since $x<1$ and $y<1$, so $x^y<1$ and $y^x<1$, hence $x^y+y^x < 2$.

[quote="leepakhin"]Use Bernouli Inequality for the left part.[/quote]

can you show it ?  :roll:

See here for example : http://www.mathlinks.ro/Forum/viewtopic.php?t=484 :)


given a.b.c is natural numbers satisfying a^2+b^2=c^2.
  Prove that:a*b/12=k with k is natural number.

Exit $ a$ or $ b$ is $ \vdots 3$ because if $ a,b$ isn't $ \vdots 3$ then $ a^2\equiv 1(mod3), b^2\equiv 1(mod3)$ then $ a^2 \plus{} b^2\equiv 2(mod3)$, contradiction with $ c^2\equiv 0,1(mod3)$.
Exit $ a$ or $ b$ is even because if $ a,b$ are odd then $ a^2 \plus{} b^2\equiv 2(mod4)$, contradiction with $ c^2\equiv 0,1(mod4)$.
If $ a \equal{} 2a',b \equal{} 2b'$ then OK!
If $ a \equal{} 2a',b \equal{} 2b' \plus{} 1$ then $ c \equal{} 2c' \plus{} 1$, now:
$ 4a'^2 \equal{} (2c' \plus{} 1)^2 \minus{} (2b' \plus{} 1)^2 \equal{} (2c' \minus{} 2b')(2c' \plus{} 2b' \plus{} 2) \equal{} 4(c' \minus{} b')(c' \plus{} b' \plus{} 1)$
Because $ (c' \minus{} b') \plus{} (c' \plus{} b' \plus{} 1) \equal{} 2c' \plus{} 1$ then exit one number is even, then $ (c' \minus{} b')(c' \plus{} b' \plus{} 1)\vdots 2$
Then $ 4a'^2\vdots 8$, it mean $ a'$ is even, $ a\vdots 4$
It is OK!

PS: We can use general type of root: $ a \equal{} m^2 \minus{} n^2,b \equal{} 2mn,c \equal{} m^2 \plus{} n^2$.

[hide="solution"]WLOG

$ a=m^2-n^2$

$ b=2mn$

b is even, YAY.

Now we check all residues mod 3 of m and n:

If one of m or n is divisible by 3, then b is divisible by 3.

If m and n are not divisible by 3, they are either one or 2 mod 3. The difference of their squares is therefore 0 mod 3, and a is divisible by 3. Therefore, ab is sure to be divisible by 3.

Now we check m and n's parity:

If m and n are of the same parity, then a is sure to be even. If m and n are of different parity, then mn is even. Therefore, ab is sure to be divisible by 4.

Therefore, etc.

$ \mathtt{QED}$[/hide]


How do I change the following code to add the word "Chapter" to the table of contents?  w5.tex contains the \tableofcontents command.  Thank you.

\documentclass[12pt]{report}

\usepackage{amsmath,amsfonts,amsbsy,amssymb}
\usepackage{amstext,amsthm,amscd}
\usepackage{latexsym}
\usepackage{amsgen,amsopn,amsxtra,enumerate,verbatim}
\usepackage[mathscr]{eucal}
\usepackage{setspace}

\everymath{\displaystyle}

\makeatletter
\renewcommand\contentsname{Table of Contents}
\renewcommand*\l@chapter[2]{%
  \ifnum \c@tocdepth >\m@ne
    \addpenalty{-\@highpenalty}%
    \vskip 1.0em \@plus\p@
    \setlength\@tempdima{1.5em}
    \begingroup
      \leftskip \z@ \rightskip \@tocrmarg \parfillskip -\rightskip
      \parindent \z@
      \leavevmode \bfseries
      \advance\leftskip\@tempdima
      \hskip -\leftskip
      #1\nobreak
      \leaders\hbox{$ \m@th%Error. "mkern" is a bad command.
\@dotsep mu\fbox{.}%Error. "mkern" is a bad command.
\@dotsep mu$}
      \hfil \nobreak\hbox to\@pnumwidth{\hss #2}\par
      \penalty\@highpenalty
    \endgroup
  \fi}
\makeatother

\begin{document}

\include{w0}

\pagenumbering{roman}

\include{w1}
\include{w2}
\include{w3}
\include{w4}
\include{w5}

\pagenumbering{arabic}

\include{w6}
\include{w7}
\include{w8}
\include{w9}
\include{w10}

\end{document}

Use the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=tocloft]tocloft package[/url]
[code]\usepackage{tocloft}
\setlength\cftchapnumwidth{7em}
\renewcommand\cftchappresnum{Chapter~} [/code]

It does work but you've overriden it with your \makeatletter ... \makeatother code. If the purpose of that codeis to add dots to chapter entries then use
[code]\renewcommand{\cftchapdotsep}{\cftdotsep}[/code]instead.
[code]\renewcommand\contentsname{Table of Contents}[/code] can be left in though.
For any further customisation see the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=tocloft]tocloft documentation[/url]


a, b, c are positive, real numbers satisfying ab+bc+ac=3

Prove that [img]E:\mathtype\1.gif[/img][/list][/b][/code]

We won't be able to solve it either, if you don't type it in LaTeX.


vector
analytic geometry
geometry unsolved
geometry

[color=darkblue]Given are two points $A$, $B$ and four points $X_{k}\ ,\ k\in\overline{1,4}$ for which we know $X_{k}=x_{k,a}\cdot A+x_{k,b}\cdot B\ ,\ k\in\overline{1,4}$. Ascertain the typical condition $R(x_{k,a}\ ,\ x_{k,b}\ |\ k\in \overline{1,4})=0$ for which the points $X_{k}\ ,\ k\in\overline{1,4}$ are conciclycally.

[b]I don't know. [u]Maybe interesting for all ![/u][/b][/color]

We can see this problem depend the choice of center vectors. If we can choose center $O$ of vectors that triangle $AOB$ is right isoceles(right at $O$) then $(x_{k,a},x_{k,b})$ are only coordinate of $X_{k}$ in a Catersian coordinate, then they are concyclic when the coordinate satify a equation of a cricle example $x^{2}+y^{2}+Ax+By+C=0$
but if $O$ is given point then we have $OAB$ is a affine coordinate and $(x_{k,a},x_{k,b})$ is affine coordinate of $X$ maybe it is too hard!  :)


I am not an expert in the conformation of organic molecules, especially when it comes to cyclic molecules. This is something I picked up when browsing a blog online:

[quote]From an organic chemist’s perspective, ask yourself “what’s wrong with these molecules.”  I’m surprised a referee didn’t pick up the mistake.[/quote]

Any comments?

Hey what has this got to do with conformations? I think these are basically zwitterions . but they have been written in the non-ionic form . I can't see anythign related to conformation s.do correct if wrong .


OK, the question is really just asking what value of $k$ results in there being only 1 distinct solution. Since it's a quadratic, the solution needs to be in the form $(x+a)^{2}=0$, so we're trying to find a value of $k$ such that $x^{2}-kx+4=(x+a)^{2}$. Through whatever method, you find that, if $k$ is 4, then the equation works.

lol..elaboarting on that...

$x^{2}-kx+4=(x-a)^{2}$
$x^{2}-kx+4=x^{2}-2ax+a^{2}$
$kx-4=2ax-a^{2}$

Comparing Coefficients we see..
$k=2a$
$a^{2}=4\implies{a=\pm{2}}$

Therefore $k=2a=\boxed{\pm{4}}$

repeat solutions -> discriminant=0, $k^{2}-4(4)=0\implies k=\pm 4$

I thank you all for showing me different methods of
finding k = 4.

[quote="Interval"]Find k such that the equation x^2 -kx + 4 = 0 has a repeated real solution.

Is there another way to say the same thing as the above question?

Also, what is the value of k?

I need steps NOT JUST the answer.[/quote]

[hide]

$x^{2}-kx+4 = 0$

To restate the question in different words, we're looking for a value of k that results in only one real solution (one value of x).

The discriminant must be zero.

$b^{2}-4ac = 0$

Plug and chug.

$k^{2}-4(1)(4) = 0$

$k^{2}= 16$

$k = \pm 4$

Let's check them out.

$x^{2}-4x+4 = 0$

$(x-2)^{2}= 0$

This one is true.

$x^{2}+4x+4 = 0$

$(x+2)^{2}= 0$

This one is also true.

So, the values of k are $\pm 4$.
[/hide]

erik-the-red,

Your steps are simply divine.

Thanks.

[quote="nutz_for2.718281828"][hide="yet ANOTHER way:"]

The product of the roots is 4, so the roots are either both 2 or -2. k is the sum of the roots, so k is an element of (2+2,-2-2), so k is either 4 or -4.[/hide]

[hide="some useful formulas that you might want to know"]We have the roots of a quadratic equation

$\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. We sum them:

$\frac{-b+\sqrt{b^{2}-4ac}}{2a}+\frac{-b-\sqrt{b^{2}-4ac}}{2a}=\frac{-2b}{2a}=\frac{-b}{a}$.

We take the product of the roots:

$\frac{-b+\sqrt{b^{2}-4ac}}{2a}*\frac{-b-\sqrt{b^{2}-4ac}}{2a}=\frac{b^{2}-b^{2}+4ac}{4a^{2}}=\frac{c}{a}$[/hide][/quote]
Instead of proving it that way, see what happens when you expand $(x-a)(x-b)$ which has roots a and b.


Find the least positive integer $ n$ so that the polynomial $ P(X)\equal{}\sqrt3\cdot X^{n\plus{}1}\minus{}X^n\minus{}1$ has at least one root of modulus $ 1$.

Let $ X \equal{} exp(2\pi \phi i)$, then we get $ \phi n \equal{} \pm \frac 16 \mod 1, \ \phi (n \plus{} 1) \equal{} \frac {1}{12} \mod 1$.
It give $ n \equal{} 10\mod 12$ and exactly 2 roots $ X \equal{} exp(\frac {\pm \pi i}{6})$.


geometry
Pythagorean Theorem
geometry unsolved

Let $ABC$ be a triangle with $\hat{A}=90^{o}$.Let M $\in$ AB ,N $\in$ AC such that MN= a ,CM = a$\sqrt{3}$ , BN = a$\sqrt{2}$ ,$3AB= BC$.Caculate $P=AB+BC+CA$ and areas of triangle $ABC$

[hide="Hint :lol: "]Let $AB=x,AM=u, AN=v$.
Then $BC=3x$ and $AC=2\sqrt{2}x$.
By Pythagorean theorem, we have
$a^{2}=u^{2}+v^{2}$
$3a^{2}=u^{2}+8x^{2}$
$2a^{2}=v^{2}+x^{2}$
By adding the last 2 ones and subtracting for the first one, we deduce $9x^{2}=4a^{2}$. From this, it's easy to solve the problem.[/hide]


algebra polynomial search counting distinguishability

I just have a quick question.  Is there a formula that says how many terms will be in $\left(\sum_{k=1}^{n}a_{k}\right)^{m}$?  After trying a few values, it seems as though there would be $\binom{n+m-1}{m}$ terms, but I'm not sure.  Also, if this is the correct formula, does anyone have a proof?

Well, $\sum_{k=1}^{n}a_k$ will always give you a polynomial of $n$ terms, so you just need to find the number of terms in the expansion of that to the $m$ power...

There is a [url=http://mathworld.wolfram.com/MultinomialSeries.html]Multinomial Theorem[/url], but it's kind of on the long side...  I guess it's better than multiplying it out though...

I have seen that, but I just wanted to know about the number of terms in total (not really with regard to the values of coefficients).  Hmm...

Well, for a polynomial of the form $a_nx^n+a_{n-1}x^{n-1}+...+a_2x^2+a_1x+a_0$ raised to the $m$ power, the number of terms is $mn+1$, but if we're dealing with more than one variable it gets a lot harder...

consider each term:

it will have $a_1^{e_1}*a_2^{e_2}...$

where the exponents $e_1+e_2+...e_n=m$

and $e_i$ are nonegative integers

this is a very common problem, often refered to as the ball and urn problem...i suggets you search, but the general idea is that you have m dots, and separate them by n-1 walls, and between a set of walls is a value for $e_i$, and all permutations is your total (considering a dot or wall to be indistinguishable)

Wow, I actually know about the ball and urn problem and the formula for that, but I didn't connect that to this at all...


The mean of a collection of five positive integers is 7.  If the collection has a unique mode, what is the greatest possible member of the collection?

The answer says 31, but couldn't it be 1,1,1,1,45?


induction logarithms inequalities trigonometry

What is the biggest number you have already thought?!

 I have spent a lot of time in this wonderfull game and i think i have won

the biggest is a Random Number
for more information:

[url=http://senhorcarteiro.blogspot.com/2006/10/nmeros-de-random.html]This is a Blue Link (for more informationRead the Book (The Blue Link And Wonderfulland)[/url]

the biggest number ever seen is of course [b]1million and three!!!!!![/b]

Thales I'm really beginning hate you and your Mexican links.

[quote="Thales418"]yes mister Tret

you looosee

dont cry

read some calc book and post again here[/quote]
errr...his name isnt Tret its Treething and how will reading a calc book help us learn spanish?

The biggest number is -0.2

proof by induction

-0.2>x means that
-0.2>x-1
so -0.2 is greater than all numbers

QED

hey mister Pie

Suppose

$x = x+1$ for a very large $1$

So, using the fast fourier transform together with newtons method

$x = Pie$

Hm... Delicious!!

lets Eat

bon apetit

[b]BIGGEST NUMBER:[/b] ${\log_{b}(\pi )}$

[b]Log Rhythm[/b]

[i]What rolls down stairs, alone or in pairs, rolls over your neighbor's dog?
It fits on your back, it's great for a snack- it's log! log! log!
Loo-og, loo-og, it's big, it's heavy, it's wood.
Loo-og, loo-og, it's better than bad, it's good!
Everyone wants a log. Come on and get your log. You're gonna love your log
It's [b]log! log! log! [/b][/i]

the biggest number possible is $4873$ here i willl show you.

assuming the theory is true:
suppose $x$ and $y$ are bigger than $4873$ and $x>y$. therefore $x$ and $y$ are bigger than the biggest possible number. Since $x>y$, $x$ is already bigger than the biggest possible number, which is not possible.

assuming the theory is false:
suppose $x$ and $y$ are bigger than $4873$ and $x>y$. since $y>4873$, $y$ could be the biggest possible number. since it could be the biggest possible number, $x$ could be bigger than the biggest possible number. that itself is a contradiction, so $x>4873$, $y>4873$, and $x>y$ can't always work.

so in either case no values, not even $x$ and $y$, can ever be bigger than $4873$.

here's how $1000003$ (one million and three) is less than $4873$:

in the text below, replace every $x$ you see with $1000003$.

assuming the theory is true:
suppose $x$ and $y$ are bigger than $4873$ and $x>y$. therefore $x$ and $y$ are bigger than the biggest possible number. Since $x>y$, $x$ is already bigger than the biggest possible number, which is not possible.

assuming the theory is false:
suppose $x$ and $y$ are bigger than $4873$ and $x>y$. since $y>4873$, $y$ could be the biggest possible number. since it could be the biggest possible number, $x$ could be bigger than the biggest possible number. that itself is a contradiction, so $x>4873$, $y>4873$, and $x>y$ can't always work.

the same can be done for replacing $x$ with $-0.2$ and $Pie$.

Suppose $x$ is the biggest number of all the numbers. so we have some easy relations

${x>x-1}$

This is easy to see. Since $x$ is the biggest number, the number before it need to be the smallest. Lets prove this by contradiction.

Suppose the number before $x$ is not the smallest, so it need to be the biggest, since it is not the smallest, but the biggest is $x$, subsituing

${x>x}$

What is a contradiction!

So $x$ is the biggest and $x-1$ the smallest

Since the smallest number is one. because this number is the more thin. Solving

${x-1=1}$
${x=0}$

Lets another proof

Suppose

${x>x-1}$

For very larges $x$. So it is true for

${x+1>x}$

That is the same as

${x<x+1}$

Combining this $2$ inequalitys

${x-1>x+1}$
${x>x+2}$

This proof that $x$ is the biggest number. It is more big than $2$ numbers after it.

Thats why we have the minimum and maximum when

${f'(x)=0}$

${0=\tan \left(\frac{\pi }{2}\right)}$

Since  $x=0$

${x=\tan \left(\frac{\pi }{2}\right)}$
${f(x)=f\left(\tan \left(\frac{\pi }{2}\right)\right)}$
${f(x)=f(0)}$

That is every True

${Q.E.D}$

I dont get it...  :huh:
Is this funny? Sorry, maybe im missing something

The induction case for n+1 instead of just n-1 is trivial, so I didn't bother to put it in.

this is not funny

but you guys doesnt understand and bullying me that is cool

huauhauhuauh

I wasn't bullying you, just asking a question.
If that sounded like a bullying remark, sorry. It wasn't meant to be.

bullying its just another joke

hauhuauha

you never understand

huahua

so funny


My name looks like SOUMYASHANTnp&**NAYAK but its still SOUMYASHANT NAYAK in profile....

How come??

&nbsp is a HTML code for space ... i tHink!!!

yes i too think that is so... :)

official roll call
(copy and paste the above statement...)

#5yes i too think that is so...

official roll call
(copy and paste the above statement...)

you all got it wrong it's because of
[size=200]
[color=blue]CHUMMA JUJUBI[/color][/size] :rotfl:  :rotfl:  :rotfl:  :P

[quote="madness"]and what's this new thing abt numbering the posts?? :D :D[/quote]
that's also
[size=200]
[color=blue]CHUMMA JUJUBI[/color][/size] :rotfl:  :rotfl:  :rotfl:  :P

How dare you take [size=150][color=red]JUJUBI[/color][/size]'s name in your foul mouth....

why yes .....   is exactly the HTML code 4 space

NBSP - No(N) Blank Space.

I found the reason da. There was a recent complaint in the Support forum that long usernames render the page unreadable so Valentin has edited the forum settings to make LONG USERNAMES automatically wrap. And since "SOUMYASHANT" is enough letters for one line. It tries to wrap it to the next line, and your name is stored as I think Soumyashant&nbspNayak, and it gives the line break after the nb. (See the source code of the page)

See this: http://www.mathlinks.ro/viewtopic.php?t=192228

cool man--that's too good an explanation

HEY PT STUDENTS I JUS HEARD  THAT U ALL HAVE TO READ 2 HEAVY BOOKS FOR UR MONDAYS EXAM

HEAVY books 4 pt/pe?
i thought that was a freak's work!!!!


One diagonal of a rhombus measures 48 inches, and one side of the rhombus measures 26 inches. How many square inches are in the area of the rhombus?

the other diagonal is 20, so 480. (t becomes a 10-24-26, which is a blown p 5-12-123)

[hide="Really long, but hopefully helpful solution"]Well, try to draw a rhombus with the two diagonals. It sort of looks like this: [url=http://www.mathplanet.com/media/44260/area_rhombus_499x300.jpg]Click this[/url]

Note that the two diagonals make 4 right angles. They also split the diagonals in half.
So if we are given a side length of ${26}$, and a diagonals length of ${24}$, we can find the other diagonal.

Using the Pythagorean theorem here is what we get. We have leg; x, and 24 (half of the diagonal), and we have hypotenuse 26.

We get ${x}^{2} + {24}^{2} = {26}^{2}$

Simplifying:

${x}^{2} + {576} = {676}$

${x}^{2} = {676} - {576}$

${x}^{2} = {100}$

${\sqrt{{x}^{2}} = \sqrt{100}}$

${x} = {10}$

So we have DIAGONALS ${20}$ (${10} \cdot {2}$) and ${48}$

A formula given to find the area of a rhombus with diagonals is:

$\frac{d_1 \cdot d_2}{2}$

So we plug in $\frac{48 \cdot 20}{2}$

And finally get the area of $\boxed{480}$

[/hide]


inequalities
algebra
polynomial
inequalities unsolved

Sorry to say Ji Chne but one should prove that inequality, not just type some random 16th degree polynomial. Proving and guessing are completely different things.


[quote="xxxyyyy"]$ a^b \plus{} b^a \equal{} c^a$

$ a$, $ b$ and $ c$ are positive integers.

What maximum $ 10$ digit value of $ c$ with all different digits ($ 0$, $ 1$, $ 2$ $ \cdots$, $ 9$) is possible which satisfies the above equation and what are the corresponding values of $ a$ and $ b$?[/quote]

$ a\equal{}1$
$ b\equal{}9876543209$
$ c\equal{}9876543210$

[quote="pco"][quote="xxxyyyy"]$ a^b \plus{} b^a \equal{} c^a$

$ a$, $ b$ and $ c$ are positive integers.

What maximum $ 10$ digit value of $ c$ with all different digits ($ 0$, $ 1$, $ 2$ $ \cdots$, $ 9$) is possible which satisfies the above equation and what are the corresponding values of $ a$ and $ b$?[/quote]

$ a \equal{} 1$
$ b \equal{} 9876543209$
$ c \equal{} 9876543210$[/quote]

Very clever and nice. :thumbup:


Find A, B, C, D.  :maybe:
The product D is cycloalkane carboxylic acid.

[hide="Answer"]A is sodium ethyl acetoacetate.

B is 2-(5-bromo-1-pentyl)acetoacetic esther.

C is 1-acetyl-1-ethoxicarbonylcyclohexane.

D is acetylcyclohexane.[/hide]

Oh yes of course, did not notice the active methylene group!!  :o

This is the old Acetoacetic Esther Synthesis.


I have a quesiton :what is the meaning of finite field,or Garois field?Please help me .

I gave a short overview over algebraic structures in http://www.mathlinks.ro/Forum/viewtopic.php?p=584714&#584714.

Then A finite field / Galois field is just a field with a finite number of elements.
We also know that there is (up to isomorphy) exactly one field with $q$ elements iff $q=p^{k}$ is a power of a prime $p$, and there are no fields with $q$ elements otherwise.
If $q$ is a prime power, one often writes $\mathbb F_{q}$ or $GL(q)$ (the latter vanishing a bit in modern times) for the field with $q$ elements.
One can find the $p$ from above as the characteristic of the field, so as the number of $1$'s you have to add untill you reach $0$ again.
The field with $q=p$=prime elements is the one $\mod p$, and all others can be seen as extensions of those.

"Thank you very much,Zetax.I'try to prove some of the "??( I forgot how to spell :blush: )" in your post that you haven't proved.


Let $a,b,c,d$ be digits such that the prime $q$ divides $\overline{abcd},\overline{badc},\overline{cdab},\overline{dcba}$.

Prove that at least one of the numbers $a+b+c+d,a+b-c-d,a-b+c-d,a-b-c+d$ is divisible by $q$.

If $q \not | a+b+c+d$ then $q | 1111 =11 \cdot 101$ (to see this just notice that $q | \overline{abcd}+\overline{badc}+\overline{cdab}+\overline{dcba}$). So $q = 11$ or $q=101$ and we are left with easy casework.

Denote $A=\overline{abcd}, B=\overline{bcda}, C=\overline{cdab}, D=\overline{dabc}$. Now it's easy to see that

\begin{eqnarray*}A+B+C+D &=& 11\cdot 101(a+b+c+d)\\ A+B-C-D &=& 3^2\cdot 11^2(a+b-c-d)\\ A-B+C-D &=& 3^2\cdot 101(a-b+c-d)\\A-B-C+D &=& 3^4\cdot 11(a-b-c+d)\end{eqnarray*}

All of these are divisible by $q$. If $q=3$, the parentheses in the first expression must be divisible by $q$; if $q=11$, the parentheses in the third are divisible by $q$; if $q=101$, the parentheses in both second and fourth are divisible by $q.$ Otherwise all $4$ are divisible by $q$.


A pyramid of cubes is constructed as follows: the top layer has 1 cube, the second layer has 3 cubes, and the third layer has 6 cubes, the 4th layer has 10 cubes, and so on. If we build a pyramid with 50 layers, what is the total number of cubes in the pyramid?

[hide="Solution"]The sum of the first $ n$th triangular numbers is $ \displaystyle\frac{n(n\plus{}1)(n\plus{}2)}{6}$. Proof by induction.

Base case: $ n\equal{}1$. $ 1\times2\times3/6 \equal{} 1$, so the base case is true.

We assume the claim is true for some integer $ k$.

Consider the sum of the first $ k\plus{}1$ triangular numbers. This is equal to $ \displaystyle\frac{k(k\plus{}1)(k\plus{}2)}{6} \plus{} \displaystyle\frac{(k\plus{}1)(k\plus{}2)}{2} \equal{} \displaystyle\frac{k(k\plus{}1)(k\plus{}2) \plus{} 3(k\plus{}1)(k\plus{}2)}{6} \equal{} \displaystyle\frac{(k\plus{}1)(k\plus{}2)(k\plus{}3)}{6}$.

Therefore, by induction, the sum of the first $ n$th triangular numbers is $ \displaystyle\frac{n(n\plus{}1)(n\plus{}2)}{6}$.

Let $ n \equal{} 50$. $ 50 \times 51 \times 52 / 6 \equal{} 22100$.[/hide]

[quote="itzmiko"]or because the rows are 1, 3, 6, 10 ...

1 = 1 = 2C2
2 = 3 = 3C2
3 = 6 = 4C2
4 = 10 = 5C2
50 = X = 51C2

X = 1275[/quote]

if we do it this way, we can't stop there
we would have to add up all the combinations, since the problem asks for the total number of cubes

adding up 50 combinations doesn't seem very appealing to me

although you can do it this way, it would be easier to do it darkdieuguerre's way

[quote="itzmiko"]or because the rows are 1, 3, 6, 10 ...

1 = 1 = 2C2
2 = 3 = 3C2
3 = 6 = 4C2
4 = 10 = 5C2
50 = X = 51C2

X = 1275[/quote]

Building off this, we can now us the Hockey Stick Identity so all of those added together gives $ \binom{52}{3}$ which is equal to $ 22100$.


There are 2000 white balls in a box, and unlimited supply of white, green and red balls, initially outside the box. in each turn, we may replace: 2 white with 1 green, 2 reds with green, 2 greens with white and red, a white and green with a red, or a green and red with a white.
(a) if we end up with just three balls, prove at least one ball is green.
(b) can we end up with just 1 ball left?

a) Let be $ n$ the number off balls inside the box and $ g$ the number of green balls inside the box.
We will show that $ n\equiv g \mod 2$, before the first turn, this is true

So we have 3 different cases
i)when we  replace (2 white with 1 green or 2 reds with 1 green)
then $ n\rightarrow(n\minus{}1)$ and $ g\rightarrow(g\plus{}1)$
therefore $ n\minus{}1\equiv (g\plus{}1) \mod 2$

ii) When we replace 2 greens with white and red
$ n\rightarrow n$ and $ g\rightarrow(g\plus{}2)$
$ n\equiv (g\plus{}2) \mod 2$
iii) When we replace (a white and green with a red, or a green and red with a white)
$ n\rightarrow (n\minus{}1)$ and $ g\rightarrow(g\minus{}1)$
then $ (n\minus{}1)\equiv (g\minus{}1) \mod 2$

This means that if $ n$ is odd, g is also, so when $ n\equal{}3$ $ g\equal{}2k\plus{}1$ for $ k\geq 0$ and at least one ball is green.

b) Yes. just prove that of $ 2j\plus{}1\equal{}n$, then we can reach $ 2j\minus{}1\equal{}n$ using that $ n\equiv g \mod 2$ and just focusing on 3 balls

[quote="streeler"]a) Let be $ n$ the number off balls inside the box and $ g$ the number of green balls inside the box.
We will show that $ n\equiv g \mod 2$, before the first turn, this is true

So we have 3 different cases
i)when we  replace (2 white with 1 green or 2 reds with 1 green)
then $ n\rightarrow(n \minus{} 1)$ and $ g\rightarrow(g \plus{} 1)$
therefore $ n \minus{} 1\equiv (g \plus{} 1) \mod 2$

ii) When we replace 2 greens with white and red
$ n\rightarrow n$ and $ g\rightarrow(g \plus{} 2)$
$ n\equiv (g \plus{} 2) \mod 2$
iii) When we replace (a white and green with a red, or a green and red with a white)
$ n\rightarrow (n \minus{} 1)$ and $ g\rightarrow(g \minus{} 1)$
then $ (n \minus{} 1)\equiv (g \minus{} 1) \mod 2$

This means that if $ n$ is odd, g is also, so when $ n \equal{} 3$ $ g \equal{} 2k \plus{} 1$ for $ k\geq 0$ and at least one ball is green.

b) Yes. just prove that of $ 2j \plus{} 1 \equal{} n$, then we can reach $ 2j \minus{} 1 \equal{} n$ using that $ n\equiv g \mod 2$ and just focusing on 3 balls[/quote]

Sorry,but I think the answer for (b) is NO :blush:
Because if we regard $ g$ as $ \minus{}1$,$ w$ as $ i$,$ r$ as $ \minus{}i$ where $ i^2\equal{}\minus{}1$,and we focus on the product of all numbers which never changes :!:
But if there was still one ball left,it would be $ i^{2000}\equal{}1$,but no ball satisfies it :?:


HEY THIS TOPIC IS A LITTLE MEAN TO ummmm.....uhhhhh........ hehehehehehehehehehehehe SHORT people  :P
how is it possible that NOBODY here (so far) is less than 5 feet??
im not going to vote :mad:

[quote="junggi"]this is one of the best topics in this forum[/quote]

Glad you appreciate it.

Dude I'm more than 3 miles tall. There should be a section for over 3 miles tall.

Dude like I tottaly agree.
I just grew a mile overnight.

[quote="MysticTerminator"]DANG DUDE

I DON'T KNOW WHAT YOU HEARD ABOUT ME

BUT IN TURNS OUT THAT IN FACT MOST PEOPLE ARE UNABLE TO PROCURE DOLLARS FROM ME

IT IS ALSO TRUE THAT I HAVE NEITHER A CADILLAC NOR A PERM[/quote]
dude what
Arnav wants a mile long perm and a Cadillac?
whoa

[quote="james4"]i don't get the point of this topic

convert, select poll option, comment, done

that simple[/quote]and then brag about your 1337 height.

[quote="DiscreetFourierTransform"]What's considered 1337 height?
I'm *kinda* tall..
Not really THAT tall compared to an NBA basketball team, though.[/quote]

Just because this is a spam forum and i want to spam more, you can't compare height to a basketball team.


function
logarithms
inequalities unsolved
inequalities

Let $ a,b,c$ are non-negative real numbers.Prove that:$ (a^2\plus{}b^2\plus{}c^2)^2(ab\plus{}bc\plus{}ca)\geq (a^2b^2\plus{}b^2c^2\plus{}c^2a^2)(a\plus{}b\plus{}c)^2$

[quote="toanIneq"]Let $ a,b,c$ are non-negative real numbers.Prove that:$ (a^2 \plus{} b^2 \plus{} c^2)^2(ab \plus{} bc \plus{} ca)\geq (a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)(a \plus{} b \plus{} c)^2$[/quote]
The general problem is the following:
Let $ a,b,c$ be positive reals and $ p,q$ are real numbers. Then
\[ \frac{(a^p\plus{}b^p\plus{}c^p)^2}{a^pb^p\plus{}b^pc^p\plus{}c^pa^p} \ge \frac{(a^q\plus{}b^q\plus{}c^q)^2}{a^qb^q\plus{}b^qc^q\plus{}c^qa^q}
\]
for all $ p \ge q \ge 0$ and
\[ \frac{(a^p\plus{}b^p\plus{}c^p)^2}{a^pb^p\plus{}b^pc^p\plus{}c^pa^p} \ge \frac{(a^q\plus{}b^q\plus{}c^q)^2}{a^qb^q\plus{}b^qc^q\plus{}c^qa^q}
\]
for all $ q \le p \le 0$.
We can prove it considering the function
\[ f(p)\equal{}\frac{(a^p\plus{}b^p\plus{}c^p)^2}{a^pb^p\plus{}b^pc^p\plus{}c^pa^p}
\]
$ f'(p)$ has the same sign as $ \sum a^pb^p(a^p\minus{}b^p)(\ln a \minus{} \ln b)$.


what does something like
$ \binom 6i$
mean? :huh:  :help:

It's called a binomial coefficient or a combination. Take a look at this page from the AoPS Wiki: [url]http://www.artofproblemsolving.com/Wiki/index.php/Binomial_coefficient[/url].

Ooh, are you referring to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1272624#1272624]this[/url] post?

The $ \binom{a}{b}$ symbol simply denotes $ a$ choose $ b$. It is equivalent to $ _aC_b$ or $ C(a, b)$ or $ nCr(a, b)$.

If you [i]were[/i] referring to that post, let me just tell you that the $ \binom 6i$ is not important here. The important point is that the expression is a sum of 7 terms, each term corresponding to when $ i \equal{} 0$ to $ i \equal{} 6$.


function
calculus
integration
irrational number
real analysis
real analysis unsolved

Let f: [2,5] -> R be the function

f(x) = 0             if x is an irrational number       or
f(x) = 1/q          if x is a rational number, expressed as p/q
                       in "lowest terms."

Prove that the integral of f(x)dx from 2 to 5 exists, and evaluate this integral.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=128991]Just two days ago.[/url]


Claire Jogs 7 miles per hour and Megan jogs 6 miles per hour. They start together at their campsite and jog to an outpost 28 miles away. When Claire gets to the outpost, she immediately turns around and heads back towards Megan. How many miles will they be from the outpost when they meet? Express your answer as common fraction.

[hide]ans is 28/13[/hide]

Well, when claire finishes, megan is at 24 miles. The question is now, how much time would it take for them to meet up. They would have to jog the same amount of time after that, and the distance they cover would equal 4, so you have 7x + 6x = 4, so x = 4/13.... You multiply that by 7, because Claire will be running away from the outpost, so you get [b]28/13[/b]


Six distinct integers are picked at random from {1,2,3,4,5,6,7,8,9,10}. What is the probability that among those selected, the second smallest is 3?

[hide="Answer"]
There are $ 2$ ways to pick the smallest number, the second smallest number is fixed, and so there are $ {7\choose 4}$ ways to pick the remaining four numbers from the last 7 numbers.

In total, there are $ {10\choose 6}$ total ways to pick 6 elements, so the probability is $ \frac{2 \cdot {7\choose 4}}{{10 \choose 6}} \equal{} \boxed{\frac 13}$.
[/hide]

You were there also at KU yesterday April 5??  Me too!!  BTW: I think this years questions were easy (or atleast the 7-9 division was).  Anyway, for this problem you basically find the denominator and since two ways are available to pick the smallest number, there are 2*4C7, which is the numerator (or is it 7C4, i forgot how to write it) and get 1/3.

[quote="pianoforte"]dude, pm me who you are

And yes, I was (you might as well post yours in this forum as well) - I know I scored a 4, but I doubt if anyone else solved #5 so it's all good. A lot of people have been saying the junior high was easy, might as well hit them in their wallet  :wink: .[/quote]

Junior High was like the easiest.  I got a 5/5.  At least (i think) i'll get honorable mentions!  Or maybe a bunch of bills in my pocket.

Related problem: Which has a greater chance? To choose 6 integers and have 3 as the second smallest or to choose 3 integers and have 3 as the second smallest?

OMG! I found a loophole in the original problem. Because the set was{1,2,3,4,5,6,7,8,9,10}, and not {0,1,2,3,4,5,6,7,8,9}, there are only nine integers. Using this, can anyone now find the solution?


I checked the search, and where it says #6, #8 is actually posted.

Lupe went to the store and paid for her purchase with a 10 dollar bill. She found that the digits making the amount of her purchase could be rearranged to make the amount she received back in change. If her purchase amount and her change amount were different and each amount was at least 1 dollar, how many possible amounts of change could she ahve received.

[quote="Fanatic"]I checked the search, and where it says #6, #8 is actually posted.

Lupe went to the store and paid for her purchase with a 10 dollar bill. She found that the digits making the amount of her purchase could be rearranged to make the amount she received back in change. If her purchase amount and her change amount were different and each amount was at least 1 dollar, how many possible amounts of change could she have received.[/quote]

The dollar signs messes with Latex so I changed it for you

[hide]If the purchase price is A.BC then you have five possibilities

1.)A.BC+A.CB=10.00 Doesn't make sense, the .BC+.CB have to equal 1.00 therefore B+C=10 therefore B+C=9 therefore B+C doesn't=B+C

2.)A.BC+B.CA=10.00 A+C=10 C+B=9 A+B=9
 C+B=A+B C=A
A+C=10
 A=5 and C=5
 5+B=9, B=4 therefore 5.45+4.55=10.00

3.)A.BC+B.AC=10.00 C+C=10 A+B=9 and B+A=9  or C+C=0 A+B=10 and A+B=9 9=10 therefore C+C=10

C=5 A+B=9 A>0 B>0 which leaves you with 8 answers

4.)A.BC+C.BA=10.00 B+B=9 not possible

5.)A.BC+C.AB=10.00 C+B=10 A+B=9 A+C=9 therefore C=B and both equal 5
5+A=9 A=4

4.55+5.45=10.00

for a total of nine answers when you take out redundancy
1.85+8.15
2.75+7.25
3.65+6.35
4.55+5.45
5.45+4.55
6.35+3.65
7.25+2.75
8.15+1.85
[/hide]

[quote="zserf"][quote="Fanatic"]I checked the search, and where it says #6, #8 is actually posted.

Lupe went to the store and paid for her purchase with a 10 dollar bill. She found that the digits making the amount of her purchase could be rearranged to make the amount she received back in change. If her purchase amount and her change amount were different and each amount was at least 1 dollar, how many possible amounts of change could she have received.[/quote]

The dollar signs messes with Latex so I changed it for you

[hide]If the purchase price is A.BC then you have six possibilities

1.)A.BC+A.BC=10.00 A.BC=5.00

2.)A.BC+A.CB=10.00 Doesn't make sense, the .BC+.CB have to equal 1.00 therefore B+C=10 therefore B+C=9 therefore B+C doesn't=B+C

3.)A.BC+B.CA=10.00 A+C=10 C+B=9 A+B=9
 C+B=A+B C=A
A+C=10
 A=5 and C=5
 5+B=9, B=4 therefore 5.45+4.55=10.00

4.)A.BC+B.AC=10.00 C+C=10 A+B=9 and B+A=9  or C+C=0 A+B=10 and A+B=9 9=10 therefore C+C=10

C=5 A+B=9 A>0 B>0 which leaves you with 8 answers

5.)A.BC+C.BA=10.00 B+B=9 not possible

6.)A.BC+C.AB=10.00 C+B=10 A+B=9 A+C=9 therefore C=B and both equal 5
5+A=9 A=4

4.55+5.45=10.00

for a total of nine answers when you take out redundancy
1.85+8.15
2.75+7.25
3.65+6.35
4.55+5.45
5.00+5.00
5.45+4.55
6.35+3.65
7.25+2.75
8.15+1.85
[/hide][/quote]

Uh I think there are only 5 cases not 6 (a.bc+a.bc is excluded), so the answer should be 8 because 5.00 and 5.00 mean change and purchase are equal.  Your solution is exactly what I did, but it took like 5+ mins to caculate everything..

You're right, I always overlook a tiny detail.

During the target I thought that the answer was 4 because I failed to realize that she can switch the cost and change.
:roll:

Here's an easier solution.

If you do 1000-abc where abc is a three digit number, you get the new number with digits in order, 9-a, 9-b, 10-c.

9-a= either b or c. Can't be a or else a =4.5
If 9-a=c then 10-c =10-(9-a)= a+1 which must equal b since it can't be a.  9-b must then be a but that contradicts 9-a=c.

So 9-a can't be c and must be b.  Then 9-b=a and 10-c=c, so c =5.  a+b=9, and you just solve for pairs of a and b.
[/u]


Goran has a standard deck of 52 cards.  He considers an ace to have a value of 1, a jack has a value of 11, a queen has a value of 12, a king has a value of 13, and all other cards are their face value.  What is the probability that a randomly selected card will have an even value?  Express your answer as a common fraction.

there's total 6 even card, and there's total 4 kinds of cards.

so its 6*4=24.

it becomes 24/52, which is 6/13.

answer ; 6/13


How many paths from $(0,0)$ to $(n,n)$ of length $2n$ are there with exactly $k$ steps. A step is an occurence of the pair $EN$ in the path

[quote="Rushil"]A step is an occurence of the pair $EN$ in the path[/quote]

What do you mean?

i think it means East and then North. Though I;m not sure!!


$x,y,z$ are distinct positive integers such that:

$x$ is a divisor of $y+z+yz$
$y$ is a divisor of $z+x+xz$
$z$ is a divisor of $x+y+xy$

Show that $x,y,z$ can't all be primes.

And what next? Sorry I don't quite get it. :blush:

hehe.this is quite easy if one knows imo 1992.from $x | (y+1)(z+1)-1 , y | (x+1)(z+1)-1 ,z | (y+1)(x+1)-1$ we easily get that $x,y,z | (x+1)(y+1)(z+1)-1$.assuming that $x,y,z$ are all prime numbers(and using the fact that they are distinct) we clearly have that $xyz | (x+1)(y+1)(z+1)-1\Rightarrow  (x,y,z)=(2,4,14)$or$(1,3,7)$(it is kind of easy to prove it,but if you can't do it yourself look on the forum :wink: )and so our problem is solved because $4,14$ are not primes in the first solution and $1$ isn't in the second.

See, I was thinking about it but couldn't justify the step $x|(z+1)(y+1)-1\implies x|(x+1)(y+1)(z+1)-1$
but now i see that it is really easy:
We are given: $(z+1)(y+1)-1\equiv 0 \text{ mod x}$ $\implies (z+1)(y+1)\equiv 1 \implies (x+1)(y+1)(z+1)\equiv (x+1)$ the last step is valid because $gcd(x+1,x)=1$, $(x+1)(y+1)(z+1)\equiv (x+1)$ $\implies (x+1)(y+1)(z+1)\equiv 1$ $\implies (x+1)(y+1)(z+1)-1\equiv 0$ so we are done...[quote="ehsan2004"]Find all integers $\,a,b,c\,$ with $\,1<a<b<c\,$ such that \[(a-1)(b-1)(c-1)\] is a divisor of $abc-1.$[/quote]
Hmmm that's the imo problem you were talking about, it is a completely different problem!

Just an suggestion. If this was a past national olympiad, couldn't you just find the solution from google or the national olympiad's website or kalva.

i guess it is still unclear how do i get those solutions isn't it?well as i said after you get $xyz | (x+1)(y+1)(z+1)-1$ this is just imo 1992 problem 1.look for it on the forum.i remember there is a solution for it and even if it isn't it is not hard at all to do it yourself.

Let $x,y,z$ be primes and $x>y>z$.

Then the above three devisions

$x \ | \ y+z+yz$

$...$

are equivalent to the following one:

$xyz \ | \ x+y+z+xy+yz+zx=R$

Now it need nothing but brute nonintellectual force to obtain an absurd:

1) If $z=2$ then $xyz$ is even, but $R$ is odd.

2) If $x>y>z=3$ then evidently $3+8x<10x \le 2xy$ hence $R<3+8x+xy<3xy=xyz$

3) If $x>y>z \ge 5$ then evidently $x+y+z<3x<2xy$ and $xy+yz+zx < 3xy$ hence $R<2xy+3xy=5xy \le xyz$

P.S. A strange thing - still yesterday there was no post here but the author's ones ...

P.P.S. My own problem appears to be more complicated:
 I saw the topic yesterday [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=yz+divisor&t=109963]there[/url] but found it through the author's profile in the morning today  here.  :D


Dear everyone.

Please think of the following.
There are three circles s, t, u so that t, u are tangent to s. Three lines m, n, o are common tangents of t and u. Then AB is parallel to the line o.

Thank you for reading. :)

I'm surprised this problem went unnoticed as it is nothing but an extraversion of the so called "Parallel tangent theorem". See the threads [url]http://www.artofproblemsolving.com/community/c6h15945[/url], [url]http://www.artofproblemsolving.com/community/c6h430441[/url] for proofs.


I have two questions I am having trouble on:

40) Which combinations of equal volumes of solution will result in buffer solutions?

I. 0.1 M HCl and 0.1 M NH3
II. 0.1 M HNO2 and 0.05 M NaOH
III. 0.05 M HNO2 and 0.05 NH3

(A) I only
(B) II only
(C) I and III only
(D) II and III only

[hide]The answer key claims the answer is C.  However, I think this might be an error (I found these questions/answers at http://dbhs.wvusd.k12.ca.us/webdocs/NChO/NChO-Menu.html--not an official site, and have noticed a few typos in the problems, so the answer key could indeed be wrong).  I don't think that the first part will give a buffer solution.  II definitely should.  I'm not sure about the third part, are you allowed to make buffers with mixed weak acid-weak bases or only conjugate pairs?[/hide]

60) Which formula(s) could represent more than one compound?

I. CH2Cl2
II. Pt(NH3)2Cl2

(A) I only
(B) II only
(C) both I and II
(D) neither I nor II

[hide] The correct answer is B.  What are the possibilities for the compounds this could be?[/hide]

Thanks for any ideas you can provide.

I was also confused by the first question. I thought only II would make a good buffer, and that the other two choices would create either no buffer or very bad ones. Also, for question two, Pt 2+ has 8 electrons in its 5d orbital, and would have a square planar structure. The NH3's and Cl-'s can be arranged in a total of 2 different ways.[/hide]

Oh, right that makes sense.  Thanks a lot.  Is there a way in general to tell the 3D shape of a transition metal complex?

Same VSEPR methods used for ordinary molecules, remembering that usually you're dealing with coordinate covalent bonds.

As per the first question - choice I would not give a very useful buffer solution for adding acids, but I think it would be a fair buffer for the addition of base... choice II is a fine buffer, and choice III should also act as a buffer, since the undissociated HNO2 will neutralize added base and the undissociated NH3 will neutralize added acid.  So D is probably the best choice.


Show
\[ S\equal{}\sum_{k\equal{}1} ^n (\minus{}1)^k{n\choose k}k^l \equal{} 0
\]
for $ l\equal{}1, 2, ..., n\minus{}1$.

[hide="First of all..."] The result is much more easily proven if stated in the following equivalent form:  $ \sum_{k \equal{} 0}^{n} ( \minus{} 1)^k {n \choose k} {k \choose r} \equal{} 0, r \equal{} 0, 1, 2, ... n \minus{} 1$.  (The convention here is that $ {0 \choose r} \equal{} 0$.  Note that it is equally valid to start the sum at $ k \equal{} r$.) [/hide]
[hide="Solution 1"] $ (x \plus{} y \plus{} z)^n \equal{} \sum_{k \equal{} 0}^{n} {n \choose k} (x \plus{} y)^k z^{n \minus{} k} \equal{} \sum_{k \equal{} 0}^{n} {n \choose k} z^{n \minus{} k} \sum_{r \equal{} 0}^{k} {k \choose r} x^r y^{k \minus{} r}$.

The result then follows by setting $ z \equal{} \minus{} 1, y \equal{} 1$ and comparing coefficients of $ x$.  As a bonus, we conclude that when $ l \equal{} n$ the original sum evaluates to $ (\minus{}1)^n n!$. [/hide]
Edit:  [hide="Solution 2"] The combinatorial interpretation for what we did above is as follows:  among $ n$ people, we choose $ k$ to be in a committee and $ r$ to be a board of directors of that committee.  For fixed $ k$ and $ r$ this is done in $ a_k \equal{} {n \choose k} {k \choose r}$ ways.  For fixed $ r$ we can ask how many ways we can do this so that everybody is in the board of directors.

For $ r < n$ there are clearly $ 0$ ways.  But we can start with the number of ways to put everyone in the committee, which is $ a_n$, and subtract the number of ways to put $ n\minus{}1$ people in the committee (since clearly there are people not in the board of directors this way), which is $ a_{n\minus{}1}$.  But we have overcounted; we have to add back the number of ways to put $ n\minus{}2$ people in the committee, and so forth.  This is, of course, [url=http://en.wikipedia.org/wiki/Inclusion-exclusion_principle]PIE[/url], and the process continues until we put $ r$ people in the committee. [/hide]

[quote="t0rajir0u"][hide="First of all..."] The result is much more easily proven if stated in the following equivalent form:  $ \sum_{k = 0}^{n} ( - 1)^k {n \choose k} {k \choose r} = 0, r = 0, 1, 2, ... n - 1$.  (The convention here is that $ {0 \choose r} = 0$.  Note that it is equally valid to start the sum at $ k = r$.) [/hide]
[hide="Solution 1"] $ (x + y + z)^n = \sum_{k = 0}^{n} {n \choose k} (x + y)^k z^{n - k} = \sum_{k = 0}^{n} {n \choose k} z^{n - k} \sum_{r = 0}^{k} {k \choose r} x^r y^{k - r}$.

The result then follows by setting $ z = - 1, y = 1$ and comparing coefficients of $ x$.  As a bonus, we conclude that when $ l = n$ the original sum evaluates to $ ( - 1)^n n!$. [/hide]
Edit:  [hide="Solution 2"] The combinatorial interpretation for what we did above is as follows:  among $ n$ people, we choose $ k$ to be in a committee and $ r$ to be a board of directors of that committee.  For fixed $ k$ and $ r$ this is done in $ a_k = {n \choose k} {k \choose r}$ ways.  For fixed $ r$ we can ask how many ways we can do this so that everybody is in the board of directors.

For $ r < n$ there are clearly $ 0$ ways.  But we can start with the number of ways to put everyone in the committee, which is $ a_n$, and subtract the number of ways to put $ n - 1$ people in the committee (since clearly there are people not in the board of directors this way), which is $ a_{n - 1}$.  But we have overcounted; we have to add back the number of ways to put $ n - 2$ people in the committee, and so forth.  This is, of course, [url=http://en.wikipedia.org/wiki/Inclusion-exclusion_principle]PIE[/url], and the process continues until we put $ r$ people in the committee. [/hide][/quote]

Nice solutions! I'm not sure how you got the equivalent form
\[ \sum_{k = 0}^{n} ( - 1)^k {n \choose k} {k \choose r} = 0, r = 0, 1, 2, ... n - 1$ .  (The convention here is that${0 \choose r} = 0
\]

Indeed.  The key point in the above argument is that $ S$ is linear.  See also [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=178381]this blog post[/url].

[quote="JBL"][quote="jli"]I'm not sure how you got the equivalent form
\[ \sum_{k \equal{} 0}^{n} ( \minus{} 1)^k {n \choose k} {k \choose r} \equal{} 0, r \equal{} 0, 1, 2, ... n \minus{} 1.
\]
[/quote]  Please look over your posts after you've made them to avoid the LaTeX and other errors.  Just as every polynomial of degree $ r$ or less in the variable $ x$ can be written uniquely as a sum of multiples of $ x^0, x^1, \ldots, x^r$, so every polynomial of degree $ r$ or less can be written uniquely as a sum of multiples of $ \binom{x}{0}, \binom{x}{1}, \ldots, \binom{x}{r}$, where $ \binom{x}{k} \equal{} \frac {x(x \minus{} 1)\cdots(x \minus{} k \plus{} 1)}{k!}$.  (More generally, a set of polynomials $ P_0, P_1, \ldots, P_n$ has the property that the polynomials of degree $ r$ or less can be written uniquely as a sum of the $ P_i$ if and only if $ n \equal{} r$ and $ \{\deg P_0, \deg P_1, \ldots, \deg P_r\} \equal{} \{0, 1, \ldots, r\}$.)

You have an operator $ S(P) \equal{} \sum_{k \equal{} 1}^n ( \minus{} 1)^k \binom{n}{k} P(k)$, and your claim is that it gives 0 on every polynomial $ P(x) \equal{} x^l$ for $ l \leq n \minus{} 1$.  Since every polynomial can be written as a combination of the $ x^l$, this is equivalent to the claim that $ S(P) \equal{} 0$ for every polynomial of degree at most $ n \minus{} 1$.  Then this implies that $ S\left(\binom{x}{r}\right) \equal{} 0$ for every $ r \leq n \minus{} 1$.  However, since every polynomial of degree less than $ n$ can be written as a sum of the $ \binom{x}{r}$, this implication is actually an equivalence.

(I hope I haven't said anything stupidly wrong in any of this.)

The short version of this post: $ S$ is a linear transformation from the set of polynomials of degree less than $ n$ to the reals.  The claim is that $ S$ is 0 on some basis of this space.  This is equivalent to $ S$ being 0 on every basis of this space.[/quote]

Thanks for the answer.  so basically...

we want to show that $ S(P)\equal{}0$ for $ P\equal{}x^l$ so if we prove it for all $ P$ we are done.
I'm not sure why it is implied that $ S({x\choose r})\equal{}0$.  are'nt we trying to show it for all $ P$?
why didnt we just say from the start that since all polynomials can be written as the sum of $ {x\choose r}$, we need to show $ \sum_{k \equal{} 0}^{n} ( \minus{} 1)^k {n \choose k} {k \choose r} \equal{} 0, r \equal{} 0, 1, 2, ... n \minus{} 1$

thanks so much

Again, the key is that $ S$ is linear, which makes the following three statements [i]equivalent:[/i]

- $ S(P) \equal{} 0$ for $ P \equal{} x^l, l \equal{} 0, 1, 2, ... n\minus{}1$
- $ S(P) \equal{} 0$ for $ P \equal{} {x \choose l}, l \equal{} 0, 1, 2, ... n\minus{}1$
- $ S(P) \equal{} 0$ for $ P$ of degree $ l$ or less.


Friday, May 16.  Darryl Wu is scheduled to appear on [i]Live with Regis and Kelly[/i].  In most markets this show starts at 9AM or 10AM.  See [url=http://bventertainment.go.com/tv/buenavista/regisandkelly/local_listings.html]Local Listings.[/url]  See [url=http://regisandkelly.go.com/guest_guide.html?xml=guest_guide20080512.xml&load_day=5]Guest Guide.[/url]

Live With Regis and Kelly is a talk show with "celebrities", broadcasted live every weekday. Like Kevin Chen, Darryl Wu gets to appear on the show, since he has become a sort of "celebrity" be winning the national competition.

Sarcasm not intended, in case you were wondering.

I heard Darryl countdowned one of the hosts

[quote="serialk11r"]Hmmm one thing I never understood was why Mathcounts gets so much more publicity than any other math contest.[/quote]

Probably in part because of all of its big corporate sponsors. Also, things like the countdown round (and the spelling bee...) are in general more impressive than hardcore problem solving to an audience of average Americans, unfortunately.

Well, there was that one year that the US IMO team was on some magazine (was it TIME?).....


Prove that if  a, b and c are  the lengths of the sides of a triangle. Then 1/a+b    1/b+c  1/c+a    are  the lengths of the sides of a triangle   :?:  :?:

[size=150][color=red]mathematicien@naseej.com[/color][/size]

We only need to prove that \[\displaystyle \frac{1}{a+c} + \frac {1}{b+c} > \frac{1}{a+b}.\] The left-hand side is equal to \[\displaystyle \frac{a+b+2c}{(a+b)c+ab+c^2}.\] Noticing that $a+b>\frac{ab}{a+b}$ and $c^2 <(a+b)c$, we have that \[\displaystyle \frac{a+b+2c}{(a+b)c+ab+c^2}>\cfrac{\cfrac{ab}{a+b}+2c}{2(a+b)c+ab}=\frac{1}{a+b},\] which is what we wanted to prove.

good solution [size=150][color=red]k2c901_1[/color][/size]

can anyone give other solution :rotfl:


I'm having a little trouble with this problem...

You have a square box (x by x by 1) of Ping-Pong [b][u]balls[/u][/b] and arrange [b][u]them[/u][/b] so [b][u]they[/u][/b]form a square pyramid whose side faces are [i][b]equilateral triangles[/b][/i]. What is the fewest number (actually the only number that works) of [b][u]balls[/u][/b] needed to form such a pyramid?

[hide=" A Hint!"]No the answer isn't one- note the plural forms used![/hide]

The only reason why balls was under lined because isn't it impossible for some spherical object to make "perfect" lines? But if they could, would the answer be five if there were four ping-pong balls for the base, and then one ball on top of all of them? What I really want to know is how to get to the answer.

The answer has to be $ n^2 \equal{} \frac {x(x \plus{} 1)(2x \plus{} 1)}{6}$ where $ n$ is an integer and $ x$ is an integer.$ n^2$ would be the answer.  For a square pyramid with equilateral sides, the bottom will be a square with side length $ x$, the level above it will be a square with side length $ x \minus{} 1$ and so on.  You will have to find the first number that is a sum of the first $ x$ square numbers, and that number has to be a square too.

Brute Force reveals the answer is $ \boxed{4900}$.  $ n \equal{} 70$, and $ x \equal{} 24$.  I don't know an easier way...I just brute forced it.

[quote="joyofpi"]Wait, I'm a little confused, could you put "let statements" (I know, I sound like a teacher)? I don't quite get it  :blush:  :|[/quote]

I probably did a horrible job of explaining.  Let $ n$ and $ x$ be an integer, where as $ n^2$ is the answer you are seeking.

In case you didn't know, this is the formula for finding the sum of the first $ x$ squares; $ \frac {x(x \plus{} 1)(2x \plus{} 1)}{6}$
And if the base of the pyramid is $ 9$, and there are $ 4$ on top of of that square, and $ 1$ at the very top, so basically the total number of ping pong balls is $ 9 \plus{} 4 \plus{} 1$, or $ 3^2 \plus{} 2^2 \plus{} 1^2$, which is $ 14$. Plugging three into the formula above, you should get $ 14$.

So we have;

$ n^2 \equal{} \frac {x(x \plus{} 1)(2x \plus{} 1)}{6}$
Multiplying by $ 6$ we get;

$ 6n^2 \equal{} x(x \plus{} 1)(2x \plus{} 1)$
Then you solve from there, but if $ 4900$ really is the value of $ n^2$, that means $ n \equal{} 70$, so guess and check is not going to work, so you probably have to do something different.


Simplify $ \frac{3^4\plus{}3^2}{3^3\minus{}3}$  . Express your answer as a common fraction.

We can factor and therefore cancel out a $ 3$ from both the denominator and the numerator. Our expression thus becomes $ \frac{30}{8}$. $ \frac{30}{8} \equal{} \boxed{\frac{15}{4}}$.


Find the sum of the distinct solutions to
$ (x^{2}\minus{}x\minus{}13)^{x^{2}\plus{}x\minus{}20}$
[hide="My solution"]
So anything raised to the zero power is one, so setting the top to 0 yields a sum of the roots to be $ \minus{}1$. Also, the base can be 1, and 1 to the anything also equals 1. So setting the base to 1, yields $ x^{2}\minus{}x\minus{}14$. So the sum of the roots will be 1. So the sum of the roots will be 0. [/hide]

[hide="Messed up"]
Even though I get this answer, the answer key says that the answer to this problem is -3. Is this problem flawed?[/hide]

I don't understand how their can be solutions with out an equality. Do they mean "distinct" solutions? (As in values the function can take?)

I assume that you mean the solutions to "that expression equals one." Anyways,
[hide="hint"]Notice that $ \minus{} 1$ raised to any even power is also equal to $ 1$.
Also, the solution $ x \equal{} 4$ makes the top equal to zero and the bottom equal to negative one so be sure not to count it twice.[/hide]


Peter's notebook has 100 pages. Each page of the notebook contains a curious statement. The first page "says": This notebook has exactly one wrong statement. And the $i$-th, $1 \leq i \leq 100$ "says": This notebook has exactly $i$ wrong statements. Which page(s) contain wrong and which correct statements ?

The 99th page is right and the other 99 pages are wrong.

Yes. Perhaps with a little mathematics involved: (is rather getting started level though)

The 100th must be false since it claims itself false. So we only consider the first 99 statements.

for every i in [1,100] we have: if i was true, then next to 100 there would be i-1 false statements. There are i-1 statements before i that cannot co-exist with statement i, so they must all be false. That means all statements in the interval [i,99] must be true. However, none of those can co-exist, which means the size of the interval is 1. So the only i for which it can be true is page 99.

Woah, thats a bit complicated isn't it? You can never have two of the statements true since they contradict each other and one must be true, so 99 must be false..

Indeed. Still I have no clue why Orl didn't post this in the getting started forum :P

Yeah, maybe somebody will it move to intermediate...  :)

Besides orl, how did you know my notebook has 100 pages? :D


Problem from a book: Let m, n positive integers with sqrt(m!)=n!*sqrt(n+1). Show that n+1 is prime. I think you can use Bertrand's postulate.

That's equivalent to $ m!\equal{}n!(n\plus{}1)!$, which I believe the only solution is $ m\equal{}10,n\equal{}6$.

You mean $ m!\equal{}n!^2(n\plus{}1)!$?  :D

:P [hide="No"]
$ \begin{eqnarray*} \sqrt {m!} & = & n! \sqrt {n + 1} \\
m! & = & \left(n! \right)^{2} \left( n + 1 \right) \\
& = & n! \left( \left( n + 1 \right) n! \right) \\
& = & n! \left( n + 1 \right)!$.
[/hide]

:wallbash:
... Now announcing my withdrawal from math forever.

[quote="Brut3Forc3"]Now announcing my withdrawal from math forever.[/quote]

*falls on knees and dramatically shouts with dramatic music in the background:* [size=200]NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO!!!!!!!!!!!!!!!!!!!![/size]

Where could I find Bertrand's Postulate?

Bertrand's Postulate is the same thing as Chebyshev's Theorem.


Suppose $ x,y,z$ are positive integers satisfying the equation
\[ \frac{1}{x} \minus{} \frac{1}{y} \equal{} \frac{1}{z},
\] and let $ h$ be the highest common factor of $ x,y,z$.

Prove that $ hxyz$ is a perfect square.

Prove also that $ h(y\minus{}x)$ is a perfect square.

[hide="An Idea?"]
Let $ x,y,z \equal{} hx',hy',hz'$.

Then $ hxyz\equal{}h^4x'y'z'$, which is a perfect square when $ x'y'z'$ is a perfect square.

Similarly, $ h(y\minus{}x) \equal{} h^2(y'\minus{}x')$ which is a perfect square when $ y'\minus{}x'$ is a perfect square.

Then $ 1/x \minus{} 1/y \equal{} 1/z \iff \frac{y'\minus{}x'}{x'y'z'} \equal{} \left( \frac{1}{z'} \right) ^2$.

Now what?[/hide]

[hide="Solution"]Rewrite the equation as $ (z \minus{} x)(z \plus{} y) \equal{} z^2$. Thus we conclude that $ z \minus{} x \equal{} m^2k, z \plus{} y \equal{} n^2k,z \equal{} mnk$ for some integers $ m,n,k$ such that $ (m,n) \equal{} 1$ ([url=http://www.mathlinks.ro/weblog_entry.php?t=201834]proof[/url]). That means $ (x,y,z) \equal{} (mk(n \minus{} m),nk(n \minus{} m),mnk)$. Since $ (m,n) \equal{} (m,n \minus{} m) \equal{} (n,n \minus{} m) \equal{} 1$, we conclude that $ k \equal{} h$, and that $ hxyz \equal{} (mnk^2(n \minus{} m))^2$, $ h(y \minus{} x) \equal{} (k(n \minus{} m))^2$, which concludes the proof.[/hide]


trigonometry
symmetry
advanced fields
advanced fields theorems

Prove that $\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\dots=\frac{{\pi}^4}{90}$.

i know Euler's proof for, $\frac{1}{1^2}+\frac{1}{2^2}+\dots=\frac{{\pi}^2}{6}$.
and was wondering if this could be used in any way to solve the above problem.
also this might seem immature, but why are we not able to find the value $\zeta(x)$ when $x$ is an odd integer?

any help/proof would be greatly appreciated  :) .

Sure. Euler's method involves using the Viete relations on the infinite product $\sin \pi x=\pi x(1-\frac{x^2}{1^2})(1-\frac{x^2}{2^2})(1-\frac{x^2}{3^2})\cdots$
This gets us the infinite sums $\sum_n \frac1{n^2}$, $\sum_{m< n}\frac1{m^2n^2}$, etc. How do we get $\sum_n\frac1{n^4}$? Square $\sum_n \frac1{n^2}$ and subtract twice $\sum_{m< n}\frac1{m^2n^2}$. This can be extended to any even $n$, although it grows more complicated very rapidly.

For this method, you can't get $\zeta(n)$ for odd $n$ because the method started with even powers. Other simple methods all fail for simple reasons at odd $n$; you might get a tautological $0=0$ rather than an interesting equality involving the sum you care about, or something else equally obvious.
If you look at the analytic continuation, we can calculate $\zeta(n)$ for all negative integers $n$. There's a symmetry equation relating $\zeta(s)$ and $\zeta(1-s)$; when $s$ is an odd positive integer, it merely says that $\zeta(1-s)=0\cdot\zeta(s)$ and we have nothing.

ok but how do u find $\sum_{m<n}\frac1{m^2n^2}$?

Look at the coefficient of $x^5$ in the taylor expansion of $\sin(\pi x)$ and in the product.

But I dislike this way (comparing coefficients of products and series) since to show it (and the used expansions) rigorously is not that easy as it seems (why sould they fulfill that equality of coefficients¿).

Also, this theme already occured sometimes, e.g. see http://www.mathlinks.ro/Forum/viewtopic.php?t=52359 .


Flights are arranged between 13 countries. For $ k\ge 2$, the sequence $ A_{1} ,A_{2} ,\ldots A_{k}$ is said to a cycle if there exist a flight from $ A_{1}$ to $ A_{2}$, from $ A_{2}$ to $ A_{3}$, $ \ldots$, from $ A_{k \minus{} 1}$ to $ A_{k}$, and from $ A_{k}$ to $ A_{1}$. What is the smallest possible number of flights such that how the flights are arranged, there exist a cycle?

$\textbf{(A)}\ 14  \qquad\textbf{(B)}\ 53  \qquad\textbf{(C)}\ 66  \qquad\textbf{(D)}\ 79  \qquad\textbf{(E)}\ 156$

[quote="xeroxia"][url=http://www.mathlinks.ro/viewtopic.php?t=272312]Turkey NMO First Round - 1999[/url]

Flights are arranged between 13 countries. For $ k\ge 2$, the sequence $ A_{1} ,A_{2} ,\ldots A_{k}$ is said to a cycle if there exist a flight from $ A_{1}$ to $ A_{2}$, from $ A_{2}$ to $ A_{3}$, $ \ldots$, from $ A_{k \minus{} 1}$ to $ A_{k}$, and from $ A_{k}$ to $ A_{1}$. What is the smallest possible number of flights such that how the flights are arranged, there exist a cycle?

A) 14  B) 53  C) 66  D) 79  E) 156[/quote]

The typo is corrected.

I think the answer is D. One note is that the flights are directed.

Label the cities $ C_1, C_2,...,C_{13}$. If a flight goes from $ C_i \to C_j$ and another goes from $ C_j \to C_i$ then we have a loop. There are $ \binom{13}{2} \equal{} 78$ ways to pick two countries which can have at most two flights between them (going in different directions). Therefore by the box principle if there are $ \boxed{79}$ flights we must get a two country loop. Now we show that this is the minimum.

Construct the flights such that if $ i > j$ then a flight goes from $ C_i \to C_j$. Now any sequence $ A_k$ has a strictly increasing $ k$ and we can never get back to the start - hence there will be no loop. So the answer is $ \boxed{D}$

Okay.  What if a loop only counts if it involves three or more vertices?

How do you get that "Yang Mills Theory" or "Rieman ..." under your avatar.

Just keep on posting. You'll see it change at 25, 60, 120, 250, 500, 1000, 2000 I think.


trigonometry
arithmetic sequence
complex numbers

How can one prove the formula for the sum of the sines of an arithmetic progression? That is,
\[ \displaystyle \sin{\varphi} \plus{} \sin{(\varphi \plus{} \alpha)} \plus{} \sin{(\varphi \plus{} 2\alpha)} \plus{}
\cdots \plus{} \sin{(\varphi \plus{} n\alpha)}\equal{}\frac{\sin{\left(\frac{(n\plus{}1) \alpha}{2}\right)} \cdot \sin{(\varphi \plus{} \frac{n \alpha}{2})}}{\sin{\frac{\alpha}{2}}}.\]

Hey, If you know Proof by Mathematical Induction. Use that

Yes, I know I can do that, but I guess I meant I want to see it derived.

Something keeps telling me double angle.

[quote="123456789"]How can one prove the formula for the sum of the sines of an arithmetic progression? That is,
\[ \displaystyle \sin{\varphi} \plus{} \sin{(\varphi \plus{} \alpha)} \plus{} \sin{(\varphi \plus{} 2\alpha)} \plus{} \cdots \plus{} \sin{(\varphi \plus{} n\alpha)} \equal{} \frac {\sin{\left(\frac {(n \plus{} 1) \alpha}{2}\right)} \cdot \sin{(\varphi \plus{} \frac {n \alpha}{2})}}{\sin{\frac {\alpha}{2}}}.
\]
[/quote]
[hide="This proof is used in Complex numbers"]
The original question $ \equal{} \text{Im } \left \{cis \varphi[1 \plus{} cis\alpha \plus{} cis2\alpha \plus{} .... \plus{} cisn\alpha]\right\}$

$ \equal{} \text{Im } \left \{cis \varphi[1 \plus{} cis\alpha \plus{} (cis\alpha)^2 \plus{} .... \plus{} (cis\alpha)^n]\right\}$

consider $ cis \varphi[1 \plus{} cis\alpha \plus{} cis2\alpha \plus{} .... \plus{} cisn\alpha]$

$ \equal{} cis \varphi.\frac {cis^{n \plus{} 1}\alpha \minus{} 1}{cis\alpha \minus{} 1}$

$ \equal{} cis \varphi.\frac {cis(n \plus{} 1)\alpha \minus{} 1}{cis\alpha \minus{} 1}$

$ \equal{} cis \varphi.\frac {cis(n \plus{} 1)\alpha \minus{} 1}{cis\alpha \minus{} 1}.\frac {cis( \minus{} \alpha) \minus{} 1}{cis( \minus{} \alpha) \minus{} 1}$

Denominator $ \equal{} (cis\alpha \minus{} 1)[cis( \minus{} \alpha) \minus{} 1]$

$ \equal{} cis0 \plus{} 1 \minus{} cis\alpha \minus{} cis( \minus{} \alpha)$

$ \equal{} 2 \minus{} [cis\alpha \plus{} cis( \minus{} \alpha)] \equal{} 2 \minus{} 2\cos\alpha \equal{} 4\sin^2\frac {\alpha}{2}$

Nominator $ \equal{} cis \varphi[cisn\alpha \minus{} cis( \minus{} \alpha) \minus{} cis(n \plus{} 1)\alpha \plus{} 1]$

$ \equal{} cis(n\alpha \plus{} \varphi) \minus{} cis(\varphi \minus{} \alpha) \minus{} cis[(n \plus{} 1)\alpha \plus{} \varphi] \plus{} cis\varphi$

Its imginary part $ \text{Im'} \equal{} \sin(n\alpha \plus{} \varphi) \minus{} \sin(\varphi \minus{} \alpha) \minus{} \sin[(n \plus{} 1)\alpha \plus{} \varphi] \plus{} \sin\varphi$

$ \equal{} \sin(n\alpha \plus{} \varphi) \minus{} \sin[(n \plus{} 1)\alpha \plus{} \varphi]) \minus{} \sin(\varphi \minus{} \alpha) \plus{} \sin\varphi$

$ \equal{} 2\cos (\frac {2n \plus{} 1}{2}\alpha \plus{} \varphi)\sin\frac { \minus{} \alpha}{2} \minus{} 2\cos(\varphi \minus{} \frac {\alpha}{2})\sin\frac { \minus{} \alpha}{2}$

When we cancel the common factor $ 2\sin\frac {\alpha}{2}$ from denominator and nominator, thus

Denominator $ 4\sin^2\frac {\alpha}{2}$ becomes $ 2\sin\frac {\alpha}{2}$

Nominator becomes
$ \equal{} \minus{} \cos (\frac {2n \plus{} 1}{2}\alpha \plus{} \varphi) \plus{} \cos(\varphi \minus{} \frac {\alpha}{2})$

$ \equal{} 2\sin(\varphi \plus{} \frac {n\alpha}{2})\sin\frac {n \plus{} 1}{2}\alpha$

After $ 2$ is cancelled from denominator and nominator , then the result got.

 [/hide]

 :)  :P

Let, $ S=\sum_{k=1}^{n+1} \sin (\varphi + (k-1)\alpha)$.
If, $ \alpha \neq 2m\pi$,
$ 2\sin (\varphi + (k-1)\alpha) \sin \frac{\alpha}{2} = \cos\left( \varphi + \frac{2k-3}{2}\alpha \right) - \cos\left( \varphi + \frac{2k-1}{2}\alpha \right)$.
So,
\begin{eqnarray*}
2S \sin \frac{\alpha}{2}
& = & \sum_{k=1}^{n+1} 2\sin (\varphi + (k-1)\alpha) \sin \frac{\alpha}{2} \\
& = & \cos\left( \varphi - \frac{\alpha}{2} \right) - \cos\left( \varphi + \frac{2n+1}{2}\alpha \right) \\
& = & 2 \sin \left( \varphi + \frac{n}{2}\alpha \right) \sin \frac{n+1}{2} \alpha \\
\end{eqnarray*}
Thus, $ S = \frac{\sin \left( \varphi + \frac{n}{2}\alpha \right) \sin \frac{n+1}{2} \alpha}{\sin \frac{\alpha}{2}}$.

[quote="Kouichi Nakagawa"]Let, $ S = \sum_{k = 1}^{n + 1} \sin (\varphi + (k - 1)\alpha)$.
If, $ \alpha \neq 2m\pi$,
$ 2\sin (\varphi + (k - 1)\alpha) \sin \frac {\alpha}{2} = \cos\left( \varphi + \frac {2k - 3}{2}\alpha \right) - \cos\left( \varphi + \frac {2k - 1}{2}\alpha \right)$.
So,
\begin{eqnarray*} 2S \sin \frac {\alpha}{2} & = & \sum_{k = 1}^{n + 1} 2\sin (\varphi + (k - 1)\alpha) \sin \frac {\alpha}{2} \\
& = & \cos\left( \varphi - \frac {\alpha}{2} \right) - \cos\left( \varphi + \frac {2n + 1}{2}\alpha \right) \\
& = & 2 \sin \left( \varphi + \frac {n}{2}\alpha \right) \sin \frac {n + 1}{2} \alpha \\
\end{eqnarray*}
Thus, $ S = \frac {\sin \left( \varphi + \frac {n}{2}\alpha \right) \sin \frac {n + 1}{2} \alpha}{\sin \frac {\alpha}{2}}$.[/quote]
Yes , you can do the same for Cosine as well.
Just break it down into a difference by taking something common , so that you left with the nth term and the 1st term.
Now use the above method to derive a relation for Cosine , and see if you can extend it to tan and see the difference in it.


analytic geometry
quadratics
geometry
power of a point
radical axis
angle bisector
geometry proposed

Given four  collinear point A,M,N,B in this order. Find a couple of points X and Y s.t. (X,Y)  is harmonic with both (A,B) and (M,N).

PS
Perhaps (as I suspect) this is a well known classic result.

Let l be the line through the points A, B, M and N. Let c be the circle with diameter AB and d the circle with diameter MN. Then, a well-known result states that two points X and Y lying on the line l are harmonic with respect to the segment AB if and only if these two points X and Y are inverses of each other with respect to the circle c with diameter AB. Similarly, the points X and Y are harmonic with respect to the segment MN if and only if these two points X and Y are inverses of each other with respect to the circle d. Hence, we have to find two points X and Y on the line l which are inverses of each other with respect to both circles c and d.

Since the points M and N lie between the points A and B, the circle d with diameter MN lies inside the circle c with diameter AB, and hence the radical axis of the circles c and d completely lies outside of the circles c and d. Hence, for every point on this radical axis, there exists a (real) circle centered at this point and orthogonal to both of the circles c and d. For instance, let the radical axis of the circles c and d intersect the line l at a point P; then, there exists a circle p which has center P and is orthogonal to both circles c and d. Let this circle p meet the line l at the two points X and Y (these two points always exist, since the line l is a diameter of the circle p, and a circle always intersects a diameter in two points). Since the circle p is orthogonal to both circles c and d, it is easy to see that these points X and Y are inverses of each other with respect to both circles c and d. And we are done.

  Darij

[quote="grobber"]The problem is: can we construct the points $X,Y$ with a straightedge alone? This would solve that problem I posted a long time ago, about constructing the midpoint of $BC$ with a straightedge having collinear $A,B,C,D$ s.t. $AB=CD$.[/quote]

Yes this is the real problem,   also for me in general ;)

But in the case I have now in mind we can use the compass.  My idea was that the problem can be proved  be equivalent to that which requires to construct a triangle of wich are given the median, the angle bisector and  the altitude all relative to the same side.

You know, in case you're interested, I posted that problem a long time ago on MathForum, and I got some replies, but no actual proof for the general case. [url=http://mathforum.org/epigone/geometry-college/droxsningdend]Here[/url]'s the link.

[quote="grobber"]The problem is: can we construct the points $X,Y$ with a straightedge alone?[/quote]

Generally no. The coordinates of the points A, B, M, N on the line l can be rational, while those of X and Y can be irrational at the same time (roots of a quadratic equation).

[quote="grobber"]This would solve that problem I posted a long time ago, about constructing the midpoint of $BC$ with a straightedge having collinear $A,B,C,D$ s.t. $AB=CD$.[/quote]

That's a $hell$ of a problem... I had tried for a week, now I've given it up... Sorry...

  Darij

if c is the circle which have MN as diameter, let P(*) be a point on c then trace the pencil P{A,M,N,B}. From a point L (*) on PA trace the parallel p and the ortogonal line o to PB. If H and K are the intersection of o with PN and PM let I and J be the intersection of c(PHK) with p. The lines (PI, PJ) are harmonic wrt both (PA,PB) and (PM,PN).

(*) if P is chosen s.t. <APM is equal to the (given) angle between median and bisectrix and L is distant from AB the given altitude we have the way to construct a triangle of which are given median, altitude and bisectrix of the same vertex.


geometry
3D geometry
tetrahedron
geometry unsolved

[quote="perfect_radio"]Solid geometry seems to be unattractive on the forum...[/quote]
Probably, you are right :(

[quote="perfect_radio"]We have $6$ points on the edges of a tetrahedron. (each point on different edge) the three lines passing through points which lie on opposite edges are collinear iff three of the faces of the tetrahedron have this property:

$(P)$ Three points on the edges of a face of the tetrahedron are feet of three concuring cevians. (in the plane of that face)

I'm sure that Ceva's Theorem is implied here, but I don't know how to use it to get something useful  :blush:[/quote]

Let's denote the points on the edges as $X_{AB},X_{BC},...$ on $AB, BC...$ respectively.

Let's prove that if the lines are concurrent (H is the point of intersection) $\Rightarrow (P)$ :
We can see that points $H, X_{AC}, X_{DB}, X_{BC}, X_{AD}$ lies in one plane $\alpha$
a)If $\alpha \parallel AB \Rightarrow X_{AC}X_{BC} \parallel X_{AD}X_{DB} \parallel AB \Rightarrow$
   $\frac{AX_{AD}*DX_{DB}}{X_{AD}D*X_{BD}B}=1=\frac{AX_{AC}*CX_{BC}}{X_{AC}C*X_{BC}B}$
b)If $\alpha \nparallel AB \Rightarrow \alpha \cap AB = T \Rightarrow$

   Under Menelaus' Theorem ( ;) ) we have

   $\frac{AX_{AD}*DX_{DB}}{X_{AD}D*X_{BD}B}=\frac{BT}{TA}=\frac{AX_{AC}*CX_{BC}}{X_{AC}C*X_{BC}B}$

In any case we have $\frac{AX_{AD}*DX_{DB}}{X_{AD}D*X_{BD}B}=\frac{AX_{AC}*CX_{BC}}{X_{AC}C*X_{BC}B}$

Similarly we have $\frac{DX_{BD}*BX_{AB}}{X_{BD}B*X_{AB}A}=\frac{AX_{DC}*CX_{CD}}{X_{DC}C*X_{AC}A}$
And $\frac{BX_{AB}*AX_{AD}}{X_{AB}A*X_{AD}D}=\frac{BX_{BC}*CX_{DC}}{X_{BC}C*X_{CD}D}$

After multiplying this 3 expressions ("left part to left part")
We get $\left(\frac{AX_{AD}*DX_{DB}*BX_{AB}}{X_{AD}D*X_{BD}B*X_{AB}A}\right)^2=1 \Rightarrow$

$\frac{AX_{AD}*DX_{DB}*BX_{AB}}{X_{AD}D*X_{BD}B*X_{AB}A}=1$
And this three points lie exactly on the segments $AB, BD, DA \Rightarrow$
Under the Ceva's Theorem ( :!:  :) ) $AX_{BD}, DX_{AB}, BX_{AD}$ are cevians.
Similarly we have such fact for other faces of tetrahedron.

In the other hand we have:
If $(P) \Rightarrow$ in the similar manner we get $X_{AD}, X_{BD}, X_{BC}, X_{AC} \in \alpha_1$
$X_{AD}, X_{DC}, X_{BC}, X_{AB} \in \alpha_2$
$X_{AB}, X_{BD}, X_{DC}, X_{AC} \in \alpha_3$
Then we can see that there exist point $T=X_{AD} X_{BC} \cap X_{BD}X_{AC} \cap X_{DC}X_{AB}$

It finishes the solution.
V:)


With one skein of wool yarn, Sharon can knit three inches of a scarf that is nine inches wide. What is the least whole number of skeins with which she can knit a scarf that is six inches wide and two feet long?

with one skein, Sharon can make 9*3=27 square inches

if she needs to make a 6 in * 2 ft scarf, she needs 6 * 24 or 144 square inches of skein

divide 144/27=5.33333333333333333333333333333333

she must use a whole number, so, the answer is [b]6[/b]

This marks the end of my 60th post.

idky i put that.

my rating is currently 1068. :harhar:


I'm very sorry for making a mistake and taking troubles to you,everyone.

There is a mistake in my problem.I will edit right now.

Best regards.

kunny

Well...

cos 2x = cos^2 (x) - sin^2 (x), so...

Cute problem!!

The answer is of course, [hide]2005

[hide]You can use cos(2x) = (1-t^2)/(1+t^2) where t = tan x ..
[hide]Or  for a really unfair short-cut - use the physicist way of assuming   one angle is close to zero, while the other is close to pi/2 . or  one angle = pi/4  etc..  Anyway[/hide] [/hide]
[/hide]

Let $\tan{\alpha}=1$, $\tan{\beta}=\frac{1}{\sqrt{2005}}$.
So $\cos{2\alpha}=0$, $\cos{2\beta}=\frac{1-\frac{1}{2005}}{1+\frac{1}{2005}}=\frac{1002}{1003}$.
Plug this in we get $2005$.

Kunny, please post your official solution!  :)

[quote="kunny"][quote="shobber"]Let $\tan{\alpha}=1$, $\tan{\beta}=\frac{1}{\sqrt{2005}}$. [/quote]
Why can you set like this, shobber? :?[/quote]
Since $\tan{\alpha} \cdot \tan{\beta} = 1 \cdot \frac{1}{\sqrt{2005}}=\frac{1}{\sqrt{2005}}$.
The two given values satisfy the equation you gave, and moreover, $\alpha=45^o$, so it will become much easier.

[quote="kunny"]You should consider all $\alpha,\beta$ satisfying  $\tan \alpha\cdot \tan \beta =\frac{1}{\sqrt{2005}}$ , shobber :)[/quote]
Yep...that is the official solution I asked you to post, kunny.
I am having a little trouble with this problem, I haven't yet found an easy way to solve it...


Me and my friend were playing 24 with only +, -, / and *. He said: 3, 3, 8, 8. I can't solve this 24 and there IS an answer. Can somebody help?

If you tell me what 24 is, I'll be able to help you

I guess he want to take the number $24$ from the numbers $3,3,8,8$
using $+,-,/,*$

I believe it is $\frac{8}{3-\frac83}$

... sorry didn't see post...

Making 24 with 2 3's and 2 8's was actually discussed here before:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=91738


trigonometry quadratics function parameterization calculus integration algebra

Solve the following differential equation :)

y''+y'+y=e^(-x)/(4+x^2)+Cosx

Is the question as follows: $ y'' \plus{} y' \plus{} y \equal{} \frac {e^{ \minus{} x}}{4 \plus{} x^2} \plus{} \cos x$?

To find the homogenous(?) solution, you have to solve $ y'' \plus{} y' \plus{} y \equal{}0$. You do this by solving the characteristic equation, which is of the form, $ \lambda^2 \plus{} \lambda \plus{} 1 \equal{} 0$, whose solution can be found using the quadratic formula.

How far have you got with this?

For the non-homogeneous part, you'll have to do a lot of algebra. Laplace transforms should be useful here.

If $ y''\plus{}2y'\plus{}y\equal{}\frac{e^{\minus{}x}}{4\plus{}x^2}\plus{}\cos x$, then we can easily solve it.

Unfortunately the equation is as given :(
I'm having trouble in finding a partial solution of y''+y'+y=e^(-x)/(4+x^2), i' ve used the method of variation of parameters.
I don't think that the Laplace transformation is going to be useful.

I have never come across Laplace transforms before, and have only solved naive differential equations. However given that the homogeneous solution is of the form:
$ y \equal{} \exp\{\frac { \minus{} x}{2}\} (Z\cos (\frac {3x}{2}) \plus{} Y\sin (\frac {3x}{2}))$,

Can we not say that our ansatz for the particular solution is of the form: $ y \equal{} (Ax^2 \plus{} Bx \plus{} C) \plus{} De^{ \minus{} x} \plus{} E\cos(x)$ and play around with this? OR does it have to be: $ y \equal{} \frac{De^{\minus{}x}}{Ax^2 \plus{} Bx \plus{} C}\plus{} E\cos(x)$

 (I am using the method of undetermined coefficients).

The fraction on the right hand side is causing me to scratch my head!

[hide="My working out! (no paper nearby...)"] $ \lambda \equal{} \frac { \minus{} 1 \pm \sqrt {1 \minus{} 4}}{2} \equal{} \frac { \minus{} 1 \pm 3i}{2}$ so
a^2 + a +1 = 0[/hide]

At this point, I agree with kunny- I see no reason to believe that the solutions are elementary functions, and it looks like a mistake. The $ \frac{e^{\minus{}x}}{4\plus{}x^2}$ term is completely intractable- unless $ \minus{}1$ is the only root of the characteristic equation.

[quote="Andre@s"]hmm... The method of undetermined coefficients does not work when the nonhomogeneous term is a rational function ( i am refering to the fraction ). So another way should be followed. Any other ideas are welcomed ![/quote] Thanks. I hadn't previously been aware of that.

[quote="jmerry"] I see no reason to believe that the solutions are elementary functions, and it looks like a mistake. [/quote] Well, you can still solve the corresponding linear system in the method of variation of parameters and write the solution as a combination of antiderivatives of certain functions. The fact that the antiderivatives are not elementary functions should not bother you much: after all it just means that you do not need to do that boring integration by parts here :). I remember that when I taught the course in ODE, I gave my students a few problems like that (though I was even more lazy and instead of inventing some functions you cannot integrate explicitly, I just put $ f(x)$ on the right hand side and asked for an ansver in terms of integrals involving $ f$).


$a,b,c,t$ are antural numbers and $k=c^{t}$ and $n=a^{k}-b^{k}$.
a) Prove that if $k$ has at least $q$ different prime divisors, then $n$ has at least $qt$ different prime divisors.
b)Prove that $\varphi(n)$ id divisible by $2^{\frac{t}{2}}$

sorry :( , but I can't understand your solution, could u explain it more?
[/hide][/quote]

It's not hard to solve this problem after using [b] Zsigmondy's Theorem. [/b]

Let $c$ have the prime divisors $p_1, p_2, \cdots, p_q.$ Then, consider the numbers $a^{p_i^j} - b^{p_i^j}$ as $i, j$ vary in $[q] \times [t].$ Note that they are increasing as $p_i^j$ increases. Call these numbers [b] tasty. [/b]

By Zsigmondy's Theorem, each of the tasty numbers has a prime divisor which does not divide any smaller tasty numbers, because $p_i^j \neq 6$ and this statement is trivial when $p_i^j = 2.$

Now, taking these primes gives us $qt$ prime numbers that divide $a^{c^t} - b^{c^t}.$

$\square$

As for $b$, any number with at least $qt$ prime divisors is divisible by $2^{qt - 1}$ because at least $qt-1$ of them are odd, so the problem easily follows.

$\square$