$ABCD$ is an isosceles trapezoid with parallel sides $AB$ and $DC$. $AB=8$, $CD=12$, and the two sides are each $6$. Point $P$ is on $BC$ such that $DP$ divides the trapezoid into two regions of equal area. What is the length of $BP$?

[hide]$\left[ABCD\right] = 40\sqrt{2}$

So $\left[DPC\right] = 20\sqrt{2} = \frac{1}{2} \cdot 12 \cdot (6 - x) \cdot \frac{4\sqrt{2}}{6}$.

Hence $x = BP = 1$.[/hide]

Yes, that's correct, but maybe you should have briefly shown how you got the area of the trapezoid ;)

For those who're interested: You can solve this problem without finding the area of the trapezoid.

another way to solve without using area, but just proportionality
[hide]
extend AD and BC so that they meet at a point E.
then EB = 12 and EC = 18.
$\bigtriangleup AEB$ has an area that is $\frac{4}{9}$ of $\bigtriangleup DEC$ because they're similar and their sides have a ratio of 2:3
assign F to be the point between E and C where EF is $\frac{4}{9}$ of the way from E to C.
$[\bigtriangleup DEF] = \frac{4}{9} [\bigtriangleup DEC]$
$[\bigtriangleup DEF] = [\bigtriangleup AEB]$
now assign point P between F and C so that $FP = PC$
since $FC = 6 + 12 - \frac{2}{3}EC = 10$
$FP = PC = 5$
$[\bigtriangleup DFP] = [\bigtriangleup DPC]$ because they have the same altitude
$BC - PC = 1$
[/hide]


calculus integration function derivative algebra polynomial quadratics

[quote="xirti"]Let $\alpha$ be a real number and
\[ \mathcal F=\{f\in C^2[0,1];f(0)=0,f(1)=0,f'(0)=\alpha\}  \]
Find $\inf_{f\in\mathcal F}\int_0^1(f''(x))^2dx$ and the function for which the equality holds.[/quote]
The minimizing function must have vanishing fourth derivative, hence it is a cubic polynomial of the form $f(x) = \alpha x + \beta x^2 + \gamma x^3$. Also, $\beta + \gamma = -\alpha$, since $f(1) = 0$. Now compute $\int_0^1(f''(x))^2dx$ in terms of $\beta$ and $\gamma$ and minimize this quadratic form subject to the constraint $\beta + \gamma = -\alpha$. The infimum should be $3 \alpha^2$.

hi,I also can't understand why "The minimizing function must have vanishing fourth derivative."
Anyway,here is my approach(Haven't finished,and by hint of hpe.)
Set $f'(1)=\beta.$
Use Cauchy-Schwarz's inequality,we have:
\[ \int_0^1(f''(x))^2dx \int_0^1 (1-ax)^2 dx \ge \left(\int_0^1 f\"(x)(1-ax) dx \right)^2 \]
and we have:\[ \int_0^1 f\"(x)(1-ax) dx=\int_0^1f\"(x)dx-a \int_0^1 xf\"(x)dx \]
\[ =\beta-\alpha-a\left(xf'(x)|_0^1-\int_0^1 f'(x) dx\right)=(1-a)\beta -\alpha \]
and \[ \int_0^1 (1-ax)^2 dx =\frac 1 3(a^2-3a+3) \]
thus we get:\[ \int_0^1(f''(x))^2dx \ge 3\frac{\left((1-a)\beta -\alpha\right)^2}{a^2-3a+3} \]
because  the above inequality is ture for any $a$.then fixed $\beta$,we calculate the maximum of $\frac{\left((1-a)\beta -\alpha\right)^2}{a^2-3a+3}$ about $\beta$.define it is $t(\beta)$.
then calculate the minimum of  $t(\beta)$,we get the desired inequality.I think it is inf of $\int_0^1(f''(x))^2dx$.(Because hpe said it is a cubic polynomial,then the inequality can be hold.)
but I haven't finished it... :blush:

[quote="xirti"][quote="hpe"]
The minimizing function must have vanishing fourth derivative.[/quote]

This is the big ideea in your proof. But why is this true? We don't know even if the function $f$ is differentiable four times..  :?[/quote]It's not a big idea if you know a little calculus of variations :)

Here's an elementary argument why a cubic polynomial with these properties attains the infimum: Let $f_o(x) = \alpha x + \beta x^2 + \gamma x^3$ be the [b]cubic polynomial[/b] that minimizes the integral in question. If you compute this function, you will find that $f_0''(1) = 0$.

Now let $g$ be any other $C^2$ function that satisifies $g(0) = g(1) = 0, \, g'(0) = \alpha$. Set $\psi(x) = g(x) - f_0(x)$, then $\psi(0) = \psi(1) = \psi'(0) = 0$. Consider $\Phi(t) = \int_0^1 \left( f_0''(x) + t\psi''(x)\right)^2 dx$. This is quadratic in $t$. The first derivative at $t=0$ equals $\Phi'(0) = 2\int_0^1 f_0''(x)\psi''(x) dx = 2\int_0^1 f_0''''(x)\psi(x) dx = 0$ (two integrations by parts, use $f_0(0) = 0 = \psi(0) = \psi'(0) = \psi(1) = f_0''(1)$), and the second derivative at $t=0$ equals $\Phi''(0) = 2\int_0^1 (\psi''(x))^2 dx > 0$. So for any $C^2$ function $g \neq f_0$, $\Phi(1) = \int_0^1(g''(x))^2dx > \Phi(0) = \int_0^1(f_0''(x))^2 dx$. Done.  :lol:

[quote="hpe"][quote="xirti"][quote="hpe"]
The minimizing function must have vanishing fourth derivative.[/quote]

This is the big ideea in your proof. But why is this true? We don't know even if the function $f$ is differentiable four times..  :?[/quote]It's not a big idea if you know a little calculus of variations :)
[/quote]

What calculus of variations means? I haven't heard about this.. :?  Can someone show me some usefull methods use in this kind of problems?  :?

[quote="xirti"]What calculus of variations means? I haven't heard about this.. :?  Can someone show me some usefull methods use in this kind of problems?  :?[/quote]
From [url=http://en.wikipedia.org/wiki/Calculus_of_variations]Wikipedia[/url]:[color=blue] Calculus of variations is a field of mathematics which deals with functions of functions, as opposed to ordinary calculus which deals with functions of numbers.  [/color]

This is a whole field of mathematics that was invented by the Bernoullis around 1700 (brachystochrone problem). One of Hilbert's 23 problems from 1900 adresses a basic question in this area ("are minimizers always regular", e.g. how come a minimizer of an integral that involves functions in $C^2$ is automatically in $C^4$?).  Question such as "what is the equilibrium shape of a loaded elastic body", "what is the shape of a soap film", "what is the optimal trajectory for a spacecraft from Earth to Jupiter" all can be addressed (if not always solved) with methods from this theory.  :)

Hpe's proof is completely new to me,Thanks to make me learn more :) .
I find in my appoach,you just let $a=1$ then you will get the result:
which means just use:\[ \int_0^1(f''(x))^2dx \int_0^1 (1-x)^2 dx \ge \left(\int_0^1 f\"(x)(1-x) dx \right)^2  \]
and  $\int_0^1 (1-x)^2=\frac 1 3,\int_0^1 f\"(x)(1-x) dx=\alpha$
the function when the equality hold is:$\alpha x^3-3\alpha x^2+2 \alpha x$
but I find my solution is not good,because I know the answer first,then come out the solution.

[quote="hpe"][quote="xirti"]What calculus of variations means? I haven't heard about this.. :?  Can someone show me some usefull methods use in this kind of problems?  :?[/quote]
From [url=http://en.wikipedia.org/wiki/Calculus_of_variations]Wikipedia[/url]:[color=blue] Calculus of variations is a field of mathematics which deals with functions of functions, as opposed to ordinary calculus which deals with functions of numbers.  [/color]

[/quote]
I think that your link does not work. At least it not works for me. CAn u solve the following problem?

Find $\lambda$ such that the following ineq holds for any continuous $f$ with $f'\le 1$ and $f(0)=0$such that

\[ \left(\int_0^1f(x)dx\right)^2\le\lambda(\int_0^1f^3(x)dx)  \]/

Did you mean this? [quote="xirti"]
I think that your link does not work. At least it not works for me. CAn u solve the following problem?

Find $\lambda$ such that the following ineq holds for any continuous $f$ with $f'\le 1$ and $f(0)=0$such that

\[ \left(\int_0^1f(x)dx\right)^2\le\lambda(\int_0^1f^3(x)dx)  \][/quote]
Set $f(x) = \epsilon x$ with $0 < \epsilon \le 1$, then the left hand side is $\frac{\epsilon^2}{4}$ and the right hand side is $\lambda\frac{\epsilon^3}{4}$. This would require an infinite $\lambda$.

The link is to Wikipedia.

I don't know why my previous post looks like that :?  I meant $\ge$ instead of $\le$. sorry :blush:

It's a typo. \lamba does bad things; you want \lambda.

[quote="hpe]  how come a minimizer of an integral that involves functions in $C^2$ is automatically in $C^4$? [/quote]

Do you know where can i get a proof for this fact?

[quote="xirti"]How come a minimizer of an integral that involves functions in $C^2$ is automatically in $C^4$?
Do you know where can i get a proof for this fact?[/quote]You need some additional assumptions such that an integral of the form $\int_a^bF(x,y(x),y'(x),y''(x))dx$ has a minimizer that is automatically in $C^4$ (or let's say has inetgrable fourth derivative). I guess a good introduction is the text by Gelfand and Fomin,  Calculus of Variations, available from Dover.


Let $ n\geq 2$ be a positive integer. Find the [u]maximum[/u] number of segments with lenghts greater than $ 1,$ determined by $ n$ points which lie on a closed disc with radius $ 1.$

WLOG the points are on a circle; just project all points onto the boundary. A segment is good iff the angle between the two points from the center of the disk is more than $ \pi/3$. Taking the graph of $ n$ vertices with each edge corresponding to a good segment, this leads to there being no $ K_6$. Now apply Turan's theorem.


induction
number theory
relatively prime
number theory unsolved

Here is an interesting induction proof.

How may we prove, by induction, that any integer > 31 can be expressed as multiples of 5 and 9?.

For instance:

$ 9\cdot{3}\plus{}5\cdot{1}\equal{}32$

$ 9\cdot{2}\plus{}5\cdot{3}\equal{}33$

$ 9\cdot{1}\plus{}5\cdot{5}\equal{}34$

$ 9\cdot{0}\plus{}5\cdot{7}\equal{}35$

$ 9\cdot{4}\plus{}5\cdot{0}\equal{}36$

and on and on. I see there is a definite pattern here in the multiples of 5 and 9.

This is kind of tough.  Any insight is valued. Thank you.

I'm gonna prove this for all $ n > 36$.

Suppose that the statement holds for every integer $ m$ with $ 31 < m < n$.
Suppose that $ n \minus{} 5$ is expressed as $ 5a\plus{}9b$.
Now $ n \equal{} 5 \plus{} (n \minus{} 5) \equal{} 5(a\plus{}1)\plus{}9b$.

in general, if $ a$ and $ b$ are positive integers with $ \gcd(a,b)\equal{}1$, the largest positive integer *not* expressible in the form $ ax\plus{}by$, with $ x,y\geq 0$, is $ (a\minus{}1)(b\minus{}1)\minus{}1$.

in the following, by "non-negative solution" to $ ax\plus{}by\equal{}N$ we mean a solution with $ x,y\geq 0$.

Proof: let $ N$ be a positive integer. since $ a$ and $ b$ are relatively prime, there exist integers $ u,v$ (not necessarily both positive) such that $ au\plus{}bv\equal{}N$. all solutions to $ ax\plus{}by\equal{}N$ are then given by

\[ (x,y)\equal{}(u\minus{}bt,v\plus{}at) \mbox{ (*) }
\]

choose $ t_0$ so that $ u\minus{}bt_0\in [0,b\minus{}1]$ (note that $ t_0$ is unique), and let $ (x_0,y_0)$ be the solution corresponding to $ t_0$. now note that there will be a non-negative solution $ (x,y)$ to the equation $ ax\plus{}by\equal{}N$ if and only if $ (x_0,y_0)$ is a non-negative solution. to see this, note that if we had $ y_0<0$ then making $ y_0\geq 0$ would require increasing $ t_0$, which would mean decreasing $ x_0$, hence making it negative, in which case there would be no non-negative solutions.

it follows that the largest integer $ N$ for which there is no non-negative solution to $ ax\plus{}by\equal{}N$ is $ a(b\minus{}1)\plus{}b(\minus{}1)\equal{}ab\minus{}a\minus{}b\equal{}(a\minus{}1)(b\minus{}1)\minus{}1$.

Thank you very much for the replies and help.

well if $ 32$ to $ 36$ are possible than any other larger integer is basically one of these $ \plus{}5k$

I seen a pattern in the sums of factors of 5 and 9 and it's not that bad.

$ (1)(5) \plus{} (3)(9) \equal{} 32$

$ (3)(5) \plus{} (2)(9) \equal{} 33$

$ (5)(5) \plus{} (1)(9) \equal{} 34$

$ (7)(5) \plus{} (0)(9) \equal{} 35$

$ (0)(5) \plus{} (4)(9) \equal{} 36$

$ (2)(5) \plus{} (3)(9) \equal{} 37$

$ (4)(5) \plus{} (2)(9) \equal{} 38$

$ (6)(5) \plus{} (1)(9) \equal{} 39$

and so on.

$ N \equal{} 5p \plus{} 9q$

If q>0, then we can subtract a 9 and add 2-5's

$ N \plus{} 1 \equal{} 5(p \plus{} 2) \plus{} 9(q \minus{} 1)$

If q=0, then we subtract 7-5's and add 4-9's.

And

$ N \plus{} 1 \equal{} 5(p \minus{} 7) \plus{} 9(q \plus{} 4)$

Because $ 5(p \minus{} 7) \plus{} 9(q \plus{} 4) \equal{}5(p\plus{}2)\plus{}9(q\minus{}1)\equal{} \underbrace{5p \plus{} 9q}_{\text{N}} \plus{} 1 \equal{} N \plus{} 1$

And we have the induction.


analytic geometry
vector
calculus
trigonometry
calculus computations

One of the questions in my Calculus textbook says:

Using spherical coordinates, show that

$ \phi \equal{}\cos^{\minus{}1}\left(\frac{\vec{u}\cdot\hat{k}}{\parallel \vec{u}\parallel }\right)$

where $ \vec{u}\equal{} x\hat{i}\plus{}y\hat{j}\plus{}z\hat{k}$.

Does this simply amount to showing that $ \vec{u}\cdot\hat{k}\equal{} z$ and $ \parallel \vec{u}\parallel \equal{}\sqrt{x^{2}\plus{}y^{2}\plus{}z^{2}}$?

The definition given in my book:

The spherical coordinates of $ (x, y, z)$ is the triple $ (\rho,\theta,\phi)$, defined as follows:

$ x \equal{}\rho\sin{\phi}\cos{\theta}$
$ y \equal{}\rho\sin{\phi}\sin{\theta}$
$ z \equal{}\rho\cos{\phi}$

where

$ \rho\geq 0$
$ 0\leq\theta < 2\pi$
$ 0\leq\phi\leq\pi$

Very easy, since $ z\equal{}\| u\|\cos\phi$ then $ \cos\phi \equal{}\frac{z}{\|u\|}\equal{}\frac{\overrightarrow u\overrightarrow k }{\|u\|}$ which gives the answer. OK?

If that's the part of the calculus book you're in, you might be interested in the exercise in my last post on [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=124599]this topic.[/url]


[hide="solution"]6!/(3!3!) = 20 ways[/hide]

[hide]There are 6 books total. There are 3 identical algebra books and 3 identical geometry books so its 6!/(3!3!)= 20 ways[/hide]

[hide]Split it into six spaces for the books.  _ _ _ _ _ _  Obviously the total number of ways they can be arranged is the number of ways 3 books can be placed out of those six, since the remaining books are identical and can only go in one way.  So the answer is $\displaystyle\binom{6}{3}=20$.[/hide]

[hide]
6C3=20 because there are 6 spaces:
= = = = = =
and you need to choose 3 for the algebra books:
x=x==x(example)
and the geometry books fill in.
[/hide]

6 C 3, so that's (6X5X4)/(3X2X1)=20. I think that should be right.

[quote="MCrawford"]In how many ways can 3 identical algebra books and 3 identical geometry books be arranged on a shelf?[/quote]

[hide="Solution"]
6*5*4*3*2*1/3/2/1/3/2/1

[b]20[/b].
[/hide]


Prove that if and only if natural number n can be written in form
$ n \equal{} a^2 \plus{} b^2 \plus{} c^2$ where a,b,c are also natural,then 8 not divide n - 7

You better learn to check your posts  :wink:
In this case, repairing is probably irrelevant as the three-squares-theorem was discussed often enough -> take a search.


The values of $ k$ for which the equation $ 2x^2\minus{}kx\plus{}x\plus{}8\equal{}0$ will have real and equal roots are:

$ \textbf{(A)}\ 9 \text{ and }\minus{}7 \qquad
\textbf{(B)}\ \text{only }\minus{}7 \qquad
\textbf{(C)}\ \text{9 and 7} \\
\textbf{(D)}\ \minus{}9 \text{ and }\minus{}7 \qquad
\textbf{(E)}\ \text{only 9}$

[hide="Click for solution"]
$ 2x^2\minus{}x(k\minus{}1)\plus{}8\equal{}0$.  For real, equal roots, the discriminant $ (k\minus{}1)^2\minus{}64\equal{}0$ and $ k\minus{}1\equal{}\pm 8 \implies k\equal{}9$ or $ \minus{}7$ so the answer is $ \boxed{\textbf{(A)}}$.
[/hide]


Put the code in when it asks you, when you go to order something.  That's all I recall doing.

[quote="calc rulz"]Well I got this gift certificate for an AoPS class or book, so how exactly do I use it?[/quote]

Congrats.  There's a chance for you to enter it when you check out.


- Jthan's unlimited spamming
- Team forums (at least ours got used a lot)
- My post count
- The sort of tight knit community thing with the board
- Captain descriptions (as well as descriptions like "Sir Spamsalot")
- The old color scheme
- The old user names (wtf is up with all these new names?  I took me 2 years just to get to know everyone!)
- THE WHAT IF GAME!!! sniff.  I'm not sad to see word association go though...
- Spam threats from Yeeland
- Ghetto equations (sqrt(x+(1/x))=15/(4x^2))
- Many other things...
Only post to this thread if you have something to add.

To raise my post count, I won't edit
- Custom word filters (STFU=flirt with me; Wayne=can't even remember anymore, but it was funny)

Thou shalt BOW to the word association game.... :P

word association game

we can always start that again... ;)

I set up a team forum for us offsite.  Would you like me to host one for you?

this place is ok though i liked the old forum better

[quote="towersfreak2006"]I think a private forum defeats the purpose of spamming, thoobik.  :P

And I missed my post count too. I was closing in on 1337. =([/quote]

I didn't know spamming actually had a purpose  :D .

Lol, ghetto equations. But what old forum?  :?


Solve the ordinary differential equation: $ x'''\minus{}7x''\plus{}15x'\minus{}9x\equal{}2e^{3t}$ with initial condition $ x(0)\equal{}1,\ x'(0)\equal{}0, x''(0)\equal{}1$.

[quote="kunny"]Solve the ordinary differential equation: $ x''' \minus{} 7x'' \plus{} 15x' \minus{} 9x \equal{} 2e^{3t}$ with initial condition $ x(0) \equal{} 1,\ x'(0) \equal{} 0, x''(0) \equal{} 1$.[/quote]
first we have to solve $ x''' \minus{} 7x'' \plus{} 15x' \minus{} 9x \equal{} 0$!
we know the answer is something in the form $ x\equal{}e^{at}$ where $ a$ is a constant.
$ \rightarrow t^3\minus{}7t^2\plus{}15t\minus{}9\equal{}0\rightarrow \left(t\minus{}1\right)\left(t\minus{}3\right)^2\equal{}0$

so the homogenues answer is $ x_{c}\equal{}c_{1}e^t\plus{}c_{2}e^{3t}\plus{}c_{3}te^{3t}$

by replacing the initial conditions, $ c_{1},c_{2}, c_{3}$ will be found!

hello, i have found
$ x(t)\equal{}\frac{1}{2}t^2e^{3t}\plus{}\frac{9}{4}e^t\minus{}\frac{5}{4}e^{3t}\plus{}\frac{3}{2}e^{3t}t$.
Sonnhard.

[quote="V.V."]$ x(t) \equal{} \frac {9e^t}{4} \plus{} e^{3t}\left(\frac {t^2}{2} \plus{} \frac {3t}{2} \minus{} \frac {5}{4}\right)$.[/quote]

[quote="Dr Sonnhard Graubner"]hello, i have found
$ x(t) \equal{} \frac {1}{2}t^2e^{3t} \plus{} \frac {9}{4}e^t \minus{} \frac {5}{4}e^{3t} \plus{} \frac {3}{2}e^{3t}t$.
Sonnhard.[/quote]

Yes, your answers are correct. :)

hello kunny, i have the same answer.
Sonnhard.


AMC
AMC 12
calculus
AIME
geometry
MIT
college

Umm, I guess it depends on what level you are at...I would think so, although my advice would be to work as many AHSME's and the AMC12s as you can get your hands on...

It also depends on what your goal is for the exam.

[quote="mit_dream"]is that possible to prepare for AMC12 in 1 month? (i am taking pre-calculus currently)...[/quote]

If you're studying math in Taiwan, which is what I guess from the location in your member profile, you are already way ahead of most of the people who will be taking the AMC12 this year. But, definitely, try to obtain some of the old exams to practice on according to the time limits so you have an idea how to pace yourself. Then you are likely to do fine when you do the real thing.

May I ask what school you are studying at? I used to live in Taiwan; my third son was born there. (My wife is from there originally; we met over there two decades ago.) What materials are you using to study precalculus? My oldest son gets his precalculus course next year. The course you are in right now ought to get you most of the math background you need for the AMC12.

Good luck. Welcome to the forum.

I thought the courses in Taiwan are rather rated by grades, not topics. Therefore, is precalculus equivalent to grade 10 math in Taiwan (just a thought)? Also, is AIME also organized in Taiwan?

I think that AMC 12 topics are relatively easy for Taiwanese students since they have such a strong algebratic background. However, they're probably not as experience in number theories, advanced countings, or various geometry techinques. Try to study these topics and you will notice a significant increasement in your score.

Anything is possible if you put in enough time.  For instance, there's a crazy guy on this forum who wants to do 8-10 hours of math a day (twice as much time as he plans to spend sleeping) in order to make MOP.

I suppose an hour or three each day would suffice, though.  I personally would advise you to take some old AMC tests, in order to get used to the format, difficulty level, etc.

yeah, that guy whos doing that mop preparation is really messed up

you can do it! just work hard, and im sure you can already do well at the moment but you just dont know it yet

am i missing something, or are you talking about mathfanatic?

perhaps, but you can accomplish many a thing with work, so go for it mit_dream

[quote="mit_dream"]I used to attend the CKHS in Taipei (the best high school in there)... and I just moved here about 1 month ago...[/quote]

If you used to attend Chien-kuo, and you have the level of English that is evident in your online postings, you should do fine. Do be sure to get some old AMC12 tests to practice with.

[quote="Tare"]Um I just had this thought: How come Taiwanese people participate in the U.S. IMO Team selection process? Why not another country?[/quote]

As far as I understood, they can take the exam, but only for fun and for a high score (and perhaps it is used as a form of selection for the Taiwanese IMO team, if it exists; I'm not sure on that one).  High School students in Taiwan cannot join the USA IMO team, or qualify for USAMO, for that matter.

[quote="mathfanatic"][quote="Tare"]How come Taiwanese people participate in the U.S. IMO Team selection process?[/quote]

As far as I understood, they can take the exam, but only for fun and for a high score (and perhaps it is used as a form of selection for the Taiwanese IMO team.[/quote]

There is definitely a Taiwan IMO team, and the coach of that team has put together a very good Amazon.com Listmania [url=http://www.amazon.com/exec/obidos/tg/listmania/list-browse/-/1F6PA8P28D7HY/]list of math problem-solving books[/url] (in English). The core of my collection of math problem-solving books consists of books, both English and Chinese, which I bought before I left Taiwan in July 2001. The bookstores there are evidently advised by the IMO coach about what books to keep in stock, so it is EASY to get all the classic training titles in Taiwan.

I am quite sure that Mathfanatic is correct that the MAIN reason young people in Taiwan take AMC8, AMC10, etc. is simply to participate in world-class mathematical culture. I think this is wonderful, as it shows Americans how pathetically low the academic aspirations for math are in the United States. That there are so many young people in Taiwan who can score well on a challenging math test that is given, after all, in a second language shows that American young people are hardly reaching the top level of human potential, on average. There was a good article by Mark Saul in one of the AMS or MAA publications a few years ago about a visit to Taiwan as these competition programs began to be established there. Saul was quite impressed by the attitude and ability of the young people he saw in Taiwan, as I always am.

Um, interesting. As opposed to you, mit_dream, I'm actually moving back to Taipei (I used to live there) in a few months. This is why I'm quite interested. Do people in CKHS do other math contests as well?


quadratics
algebra
number theory open
number theory

From the description we have $ x^3 \minus{} 6x \plus{} 4$ and $ x^2 \plus{} 2x \minus{} 2$ are both rational. Since $ x^3 \minus{} 6x \plus{} 4 \equal{} (x \minus{} 2)(x^2 \plus{} 2x \minus{} 2)$ and $ x \minus{} 2$ is irrational, we must have $ x^2 \plus{} 2x \minus{} 2 \equal{} 0\Longrightarrow x^3 \minus{} 6x \equal{} 2x \minus{} 2x^2 \minus{} 6x \equal{} \minus{} 4$. This implies that both irrational roots $ \minus{}1\pm\sqrt {3}$ of the quadratic equation $ x^2 \plus{} 2x \minus{} 2 \equal{} 0$ are the only solutions of the problem.

[quote="Ulanbek_Kyzylorda KTL"]x is irrational number with conditions x^2+2x is rational number and x^3-6x is rational too. Find x.[/quote]
Let $ a \equal{} x^2\plus{}2x, b \equal{} x^3\minus{}6x$, so $ a,b \in \mathbb{Q}$. Let $ p(t) \equal{} t^2 \plus{} 2t \minus{} a$ and $ q(t) \equal{} t^3 \minus{} 6t \minus{} b$. Then $ p(x) \equal{} q(x) \equal{} 0$. Thus $ q(x) \minus{} xp(x) \plus{} 2p(x) \equal{} 0$ - but then $ q(x) \minus{} xp(x) \plus{} 2p(x) \equal{} \minus{}(2\minus{}a)x \minus{} 2a\minus{}b \equal{} 0$. If $ a \neq 2$ we have $ x \equal{} \frac{2a\plus{}b}{2\minus{}a} \in \mathbb{Q}$, which is not possible, so $ a\equal{}2$ (and from the equation $ b\equal{}\minus{}4$). Thus $ x^2\plus{}2x\minus{}2\equal{}0 \iff x \equal{} \minus{}1 \pm \sqrt{3}$. Inserting gives $ x^2\plus{}2x \equal{} 2$, and $ x^3\minus{}6x \equal{} \minus{}4$, so they are both solutions, and by the argumentation the only solutions :) QED


calculus
derivative
calculus computations

Decide if the given value of x is a critical number for f, and if so, decide wheter the point for x on f is a relative minimum, relative maximum or neither.
f(x) = 2(x)^3 - 3(x)^2 - 12x + 18  .............x = 2
f'(x) = 6(x)^2 - 6x - 12
f'(x) =6(x^2 - x - 2)
f'(x) =6(x - 2)(x + 1)

critical x = 2   and x = -1
Can someone explain how to find relative minima , relative maxima.
Thank You all

One way is to check the value of the second derivative of $ f$ at the critical point.

Given that $ x \equal{} a$ is a critical point of $ f$, that point is a relative maximum if $ f''(a) < 0$ or a relative minimum if $ f''(a) > 0$.  However, if $ f''(a) \equal{} 0$, this "second derivative test" fails; another method must be used to determine even if that critical point is also a relative extremum.  (For example, $ f(x) \equal{} x^4$ very clearly has a relative minimum at $ x \equal{} 0$, even though $ f''(0) \equal{} 0$.)

The other method is to find all you critical points and use a table of signs. Take an interval $ [a,b]$ containing that critical point but no others. If $ f'(a) < 0$ and $ f'(b) > 0$ then you have a minimum, otherwise you have a maximum.

There is the rare case where you'll have a straddle point; an example is $ x^{3}$ at $ x \equal{} 0$. The derivative gives $ 3x^{2}$ but since the derivative is $ > 0$ for all reals, it fails the first derivative test.


I understand the process for solving congruences of two terms, but what if we have three terms, all congruent to each other?

For example, $ 11k \equiv 21k \equiv 31k \pmod{100}$ (I don't know if this has a solution, I just picked some numbers to show what I meant)

You can just solve for congruences two at a time, and see if there are overlaps.


graph theory
combinatorics solved
combinatorics

Prove that a finite graph is bipartite if and only if all of its cycles are even cycles.

[i]Alternative formulation.[/i] Prove that a finite graph is bipartite if and only if it contains no cycle of odd length.

"only if" part : suppose $G$ is a bipartite graph $G=(X,Y)$. If there is an odd cycle with starting vertex $x$ and ending point $y$. Since it is odd, $x,y$ are in the same vertices set, hence it is not a cycle...

Suppose that $G$ has no odd cycles, we will prove that it is bipartite that is we may find a partition of the vertex set into two parts, say $A,B$ such that each edge joins a vertex from $A$ to a vertex from $B$.
Clearly, we may assume that $G$ is connected, otherwise we use the following for each connected component.

Let $M$ be any arbitrary vertex, and for any vertex $P$ let $d(P)$ be the length of a minimal path from $M$ to $P$ (so that $d(M) = 0$).
Let $A$ be the set of the vertices $P$ such that $d(P)$ is even, and $B$ be the set of vertices $P$ such that $d(P)$ is odd.
If there is an edge between two vertices, say $Q,P$ from the same subset, then using a minimal path from $M$ to $P$ the edge $PQ$ and a minimal path from $Q$ to $M$ (which can be a path of length 0), we construct a cycle with length $d(P)+d(Q)+1$ which is odd, since $d(P),d(Q)$ have the same parity. A contradiction.

Pierre.

Oh....I forgot to say that the reference of this result is König, 1936.

Pierre.

Yeah, it turns out to be a truly elementary problem, but it wasn't clear to someone that has had virtually no experience in Graph Theory... :(

Thanks.


This is in reference Problem number 1.218 of Irodov. I encountered 2 solutions to this problem giving 2 different answers.

Two satellites move around the earth in a common plane with radii of 7000 and 6930. What time interval separates the periodic approach of the satellites to each other over the minimum distance.

Just consider the case where they are moving in the same sense

Solution 1::
[hide]$ \frac{GMm}{r^2} \equal{} mrw^2$
$ w\equal{}\sqrt{ \frac{GM}{r^3}}$

Moving in the same sense, they will be closest after a time $ t$ such that
$ w_1t \minus{} w_2t \equal{} 2\Pi$
w1 = 1.09 x 10^(-3)
w2 = 1.08 x 10^-3

So substituting value of t = 7.27 days.[/hide]

Solution 2
[hide]$ w1\equal{}\sqrt{ \frac{GM}{r^3}}$ (Outer Orbit)
$ w2\equal{}\sqrt{ \frac{GM}{(r\minus{}\delta{r})^3}}$ (Inner Orbit)

Time between closest approach
$ \frac{2 \Pi}{w1\minus{}w2}$
But $ \delta w \equal{} 3/2 \frac{ \delta r}{r}$
Now solving we get answer = 4.5 days which is Irodov's answer
[/hide]

Which do you think is right? Why?

[quote="Cyber-Shreyas"]
But $ \delta w \equal{} 3/2 \frac {\delta r}{r}$
[/quote]
You certainly meant $ \frac{\delta w}w \equal{} 3/2 \frac {\delta r}{r}$. See if this correction reconciles the answers. If not, increase the precision in your first approach computations.

yes fedja's reply makes sense....is that what u got wrong...i am in no mood to check the calculations

in your first approach --

a lot of difference is made
by taking

w1 = 1.09 x 10^(-3) and not = 1.097 x 10^(-3)

now using w1 as this and w2 as what u took only

the answer comes as 4.27 days

so as we will increase the decimal places(precision), the answer will come nearer to perfect

 :)


Let $x\in\left[\frac{\pi}2,\pi\right]$ satisfy \[ \sin^4 x+\cos^4 x=\frac59 \]
Find $\sin2x$.

[hide]
This problem is intermediate level; no trig in this forum.

Anyway, the solution:
$(\cos^2{x}+\sin^2{x})^2=(\cos^4{x}+2\cos^2{x}\sin^2{x}+\sin^4{x}$
$2\cos^2{x}\sin^2{x}=sin^2{2x}=8/9$
$\sin{2x}=\frac{-2\sqrt{2}}{3}$

[/hide]

hmm...i guess there is no trig allowed so i will move it


inequalities
trigonometry
inequalities proposed

Prove that any triangle with the standard notations the following inequality takes place:

                    $ m_a+m_b+m_c\leq 4R+r $ ;)

Here is a inferior boundary: \[\frac{a^2+b^2+c^2}{2R}\ \le m_a+m_b+m_c\]Direction: can use the inequality: $4Rm_a\ge b^2+c^2$ with equality iff b=c or a=90 a.s.o.

Ineq. $m_a+m_b+m_c\le 4R+r\ $ is true only in acute triangle: $m_a\le R+R\cos A, \ \ m_a\le 2R\cos^2\frac {A}{2},\ \ 2rm_a\le a(p-a)\ and\ \sum a(p-a)=2r(4R+r)\ a.s.o.$

I was mistaken. It is true and in a obtuse triangle. Indeed, I suppose A>90. Thus,
$-\cos A+\cos B+\cos C =\frac{r_a}{R}-1\ \ and\ \  m_a+m_b+m_c\le R(3+\frac{r_a}{R}-1)=
R+r_a.\  But\  2R+r_a\le 4R+r\ \leftrightarrow\ 2R\ge r_a-r\ \leftrightarrow\ 2sR\le ar_a\ \leftrightarrow\ \tan \frac{A}{2}.\sin A\ge 1\ \leftrightarrow\ 2sin^2\frac{A}{2}\ge 1\ \leftrightarrow\ \cos A\le 0,\  it\  is\ true.$
[b]Generally[/b], The distance from O to BC is R|cosA| and $m_a\le R(1+|\cos A|).\ \ \ \ Ave,\ Cezar\ !$

The last inequality hold in every acute triangle.Actually is equivalent with $sin^2 A/2 \leq\frac{a^2}{b^2 +c^2}$ :roll:

[quote="cezar lupu"]Actually it's not $b^2+c^2$ it is $ 2(b^2+c^2) $.It is a particular case of Ballieu's inequality. ;)[/quote]

what's that? i tried googling it, but came up empty

I'm talking about the inequality with $sinA/2$ ;)

[quote="cezar lupu"]The last inequality hold in every acute triangle.Actually is equivalent with $2sin^2 A/2 \leq\frac{a^2}{b^2 +c^2}$ :roll:[/quote]

 what about the inequality above. ;)


linear algebra
matrix
linear algebra solved

$n\geq 3$ integer
For any $X\in M_n(R)$, $(A+X)^{*}=A^{*}+X^{*}$

Prove that  $A=0_{M_n(R)}$

$M^{*}$ is the matrix such that $M.M^{*}=det(M)I_n$

Since the elements of $M^*$ are the minors corresponding to the elements of $M^t$, multiplied by $\pm 1$, we can apply that problem with $\det(A+X)=\det A+\det X$ to the submatrices obtained from $A$ by deleting a line and a column. Is this wrong? :?

$M^{*}=(Com(M)^{t}$ we work with $Com(M)$ comatrix

$X=A$ gives
$Com(2A)=2Com(A)$
$Com(2I_n)Com(A)=2Com(A)$  since $Com(MN)=Com(M)Com(N)$
$2^{n-1}Com(A)=2Com(A)$ since $n>2$
we have $Com(A)=0$

The equation reduce to (1) $Com(A+X)=Com(X)$

Let $r=rank(A)$ suppose that $0<r$

$A=P^{-1}J_rQ^{-1}$ choose $X=P^{-1}Q^{-1}$ in (1), after simplification

$Com(J_r+I_n)=I_n^$

$diag(2^{r-1},...,2^{r-1},2^r,...,2^r)=diag(1,......,1)$ contradiction

So $r=rank(A)=0$


[quote="Phelpedo"]Prove that there does not exist an integer which is doubled when the intitial digits is transferred to the end.[/quote]

Do you mean initial digit?

Let the original number be $10^{n}a+b$, where a, b, and n are integers, $n \geq 1$, $1 \leq a \leq 9$, and $b<10^{n}$. Then,

\begin{eqnarray*}2(10^{n}a+b) &=& 10b+a \\ 2 \cdot 10^{n}a+2b &=& 10b+a \\ (2 \cdot 10^{n}-1)a &=& 8b \end{eqnarray*}

Since $2 \cdot 10^{n}-1$ is not divisible by 2, a must be a multiple of 8. However, a cannot be more than 9, so $a=8$. That means that $b=2 \cdot 10^{n}-1$. However, b can be at most $10^{n}-1$, and so there is no integeral solution to this equation. Therefore, no such integer exists. QED


superior algebra
superior algebra theorems

In my algebra class, this theorem came up. I don't quite understand how this works.

Say we have a permutation S = (1 2 3).
So S(1) = 2, S(2) = 3, and S(3) = 4

According to some websites I came across,
S = (1 2 3) = (1 3)(2 3)(1 2)(1 2)

I'm interpreting this as: 1>2>1>3 .. This is clearly not true, as S(1) = 2 .. Please explain how to read these. I'm lost.

Thanks

Your interpretation is correct. First thing to note: $(12)(12)$ does nothing at the end since the second one just undoes the first one. I don't think that website was reliable.

It is true that you can break it into $2$-cycles though. It should be $S=(13)(12)$.

P.S. I know it is just a typo by how you worded the post. $S(3)=1$ not $4$. :wink:

Thank you very much. I understand now. I guess I can't always trust a random math professor's page, heh


What is the fewest number of colors that can be used to color the 19 hexagonal pieces so that no two of the hexagonal pieces sharing a side are the same color?

[asy]for(int i = 0; i<6; ++i)
{
	draw(dir(60i)--dir(60(i+1)));
}

for(int j = 0; j<6; ++j)
{
	for(int i = 0; i<6; ++i)
	{
		draw((dir(60i)+sqrt(3)*dir(30+60j))--(dir(60(i+1))+sqrt(3)*dir(30+60j)));
			draw((dir(60i)+3*dir(60j))--(dir(60(i+1))+3*dir(60j)));

draw((dir(60i)+2sqrt(3)*dir(30+60j))--(dir(60(i+1))+2sqrt(3)*dir(30+60j)));
	}
}[/asy]

I think I'm wrong, but maybe 3? Because there are $ \boxed{3}$ hexagons that share each vertex. Please correct me if I'm wrong.

I am not sure, but I think izzy's right, because of the vertex sharing.

Yes- 3 is right- because if you are given 2 hexagons that are different, then the ones that fit in the intersection must be a different color from either of them.


algorithm
real analysis
real analysis unsolved

$a_n$ sequence of $R+$ such that $\sum a_n$ converge and $\forall n$ $a_n\leq \sum_{k\geq n+1}a_k$

Let $S=\sum_{k\geq 0}a_k$

Prove that for any $x\in ]0,S]$ there exist a subsequence $a_{n_{k}}$ of $a_n$
such that $x=\sum_{k\geq 0}a_{n_{k}}$

Are we allowed to write $x$ as a finite number of terms of the sequence too?

Assuming that we are indeed allowed to do that, so that no problems arise, it shouldn't be too difficult to see that the obvious greedy algorithm works:

First of all, we can assume the sequence $(a_n)_n$ is non-increasing, since given $a_n,a_{n+1}$ with $a_n<a_{n+1}$, permuting the two terms won't affect the property given in the hypothesis. Next, given some $x\in(0,S]$, at each step we add the first $a_n$ which is less than or equal to the difference between $x$ and sum of the terms we have so far.


logarithms
calculus
calculus computations

It's questionable whether $x^{-x}$ is defined for $x<0$. But the one-sided limit $x\to 0+$ exists and is equal to $1$. Indeed, the logarithm of $x^{-x}$ is $-x\log x$, and this converges to $0$ by the L'Hospital rule.


function
limit
calculus
calculus computations

Hello. I blocked on this problem:

Prove that if $f(x) = 2f(x)$   $\forall x$  then $f(x) = Cte$

I can imagine why this is true graphically, but I need a more rigorous solution.

Thank you for any suggestion.

:huh:  Check the condition!
If not $f=0$ :D

Ops, I mistyped it: if $f(x) = f(2x)$

Sorry :blush:  effectively, the answer f = 0 is immediate for my first formula.

I'm supposing you don't know about f's continuity.

Consider the indicator function $\mathbf{1}_\mathbb{Q}$. It satisfies your requirements, but it's not constant.

Thank you for the remark, I'm looking for a function continued in any point of $\mathbb{R}$

I'm wondering if this works:

$f'(x) = \lim_{2x-x \to 0}\frac{f(2x)-f(x)}{2x-x}$

$f'(x) = \lim_{x \to 0}\frac{f(2x)-f(x)}{x}$

$f'(x) = \lim_{x \to 0}\frac{0}{x}$

If actually $f$ is continued everywhere, including in 0, then the division is defined, and we have $f'(x) = 0$


Find the best number $ \alpha, \beta$ so that $ (1\plus{}\frac{1}{x})^{x\plus{}\alpha}<e<(1\plus{}\frac{1}{x})^{x\plus{}\beta}$ for all $ x>0$.

[quote="doicu"]are $ \alpha, \beta$ integers ?[/quote]
No, $ \alpha,\beta$ are real!

[quote="doicu"]what do you mean with "best" numbers ?[/quote]
It mean $ \alpha$ max and $ \beta$ min!


geometry
geometric transformation
reflection
combinatorics unsolved
combinatorics

Let $ n\geq 2$ be a positive integer. There are $ 2n$ points in the plane, and $ n^{2} \plus{} 1$ segments joining them. Prove there there are $ 2$ triangles sharing an edge.

http://www.mathlinks.ro/Forum/viewtopic.php?t=37430

Ivan Borsenco proved a generalization of this in [url=http://reflections.awesomemath.org/2008_1/K_k%20versus%20K_%7Bk+1%7D-%7Be%7D.pdf][i]K_k versus K_{k+1}\{e}[/i], Mathematical Reflections 2008/1[/url].

  darij


Sequence
number theory
recurrence relation
IMO Shortlist

Find the greatest natural number $n$ such there exist natural numbers $x_{1}, x_{2}, \ldots, x_{n}$ and natural $a_{1}< a_{2}< \ldots < a_{n-1}$ satisfying the following equations for $i =1,2,\ldots,n-1$: \[x_{1}x_{2}\ldots x_{n}= 1980 \quad \text{and}\quad x_{i}+\frac{1980}{x_{i}}= a_{i}.\]

The maximum is six, obtained by $\{11, 6, 5, 3, 2, 1 \}$.  To see that this is maximal, note that $1980$ is a product of six primes;  we must divide these primes among the $x_{i}$, and we cannot split them all up, since $2$ appears twice.  So the primes are split among at most five $x_{i}$, and one more $x_{i}$ can be $1$, for a maximum of six.


function
algebra
polynomial
complex analysis
absolute value
complex numbers
complex analysis unsolved

If an entire function maps every unbounded sequence into an unbounded sequence, does it follow that this function is a polynomial ?

It does.

A holomorphic function maps every neighborhood of an essential singularity onto a dense subset of the plane. If $\infty$ were to be an essential singularity of the entire function $f$, then we would be able to find arbitrarily large (in absolute value, of course) complex numbers mapped by $f$ onto complex numbers which are close to $1$, contradicting the hypothesis. Hence $\infty$ is only a pole, i.e. $f$ is a polynomial.


7 distinct primes can be written as a, b, c, a + b + c, a + b - c, a - b + c, -a + b + c, where two of
a, b, c sum to 800. Let d be the difference between the largest and the smallest of the 7 primes. What
is the largest possible value of d?

I remember this problem from our APMO camp last year. I post the first half of my solution: (the following will be taken modulo 3)
Checking the parity, we can easy deduce that $a,b,c \neq 2$; furthermore, suppose that none of the primes equal 3. Then if $a \equiv b \equiv c$, $a+b+c \equiv 0$ is not prime; then WLOG we can assume that $a \equiv b$; but then $a + b -c \equiv 0$ is not prime. Therefore one of the primes must equal 3, and $d=a+b+c-3$.


Prove that $ |z \plus{} 1|\ge \frac {1}{\sqrt {2}}$ or $ |1 \plus{} z^2|\ge 1$ for any complex number $ z$.

[quote="Pain rinnegan"]Prove that $ |z \plus{} 1|\ge \frac {1}{\sqrt {2}}$ or $ |1 \plus{} z^2|\ge 2$ for any complex number $ z$.[/quote]

Wrong : choose $ z\equal{}\minus{}\frac 12$ and  then : $ |z \plus{} 1|\equal{}\frac 12< \frac {1}{\sqrt {2}}$ and  $ |1 \plus{} z^2|\equal{}\frac 54< 2$

[quote="pco"][quote="Pain rinnegan"]Prove that $ |z \plus{} 1|\ge \frac {1}{\sqrt {2}}$ or $ |1 \plus{} z^2|\ge 2$ for any complex number $ z$.[/quote]

Wrong : choose $ z \equal{} \minus{} \frac 12$ and  then : $ |z \plus{} 1| \equal{} \frac 12 < \frac {1}{\sqrt {2}}$ and  $ |1 \plus{} z^2| \equal{} \frac 54 < 2$[/quote]

Sorry i meant $ |1\plus{}z^2|\ge 1$. Check it now.

I have checked your edited problem, sorry, but what is your question?

[quote="Pain rinnegan"]$ \boxed{\text{Prove that}}$ $ |z \plus{} 1|\ge \frac {1}{\sqrt {2}}$ or $ |1 \plus{} z^2|\ge 1$ for any complex number $ z$.[/quote]

that's

[quote="pco"][quote="kunny"]I do know it.

Can anyone expalin what his problem is?[/quote]

He asks for a proof of : "$ \forall z\in\mathbb C$ : either $ |z \plus{} 1|\ge \frac {1}{\sqrt {2}}$, either $ |1 \plus{} z^2|\ge 1$ (either both)"[/quote]

Yes, that's it. The most probably way is by contradiction, but i couldn't find it yet.

[quote="pco"][quote="kunny"]I do know it.

Can anyone expalin what his problem is?[/quote]

He asks for a proof of : "$ \forall z\in\mathbb C$ : either $ |z \plus{} 1|\ge \frac {1}{\sqrt {2}}$, either $ |1 \plus{} z^2|\ge 1$ (either both)"[/quote]

Thank you.


What happens if you lose your financial aid award letter. Is it possible to call your school and ask them to send you a copy?


irrational number
approximation
number theory

Prove the following statement: If $r_1$ and $r_2$ are real numbers whose quotient is irrational, then any real number $x$ can be approximated arbitrarily well by the numbers of the form $\ z_{k_1,k_2} = k_1r_1 + k_2r_2$ integers, i.e. for every number $x$ and every positive real number $p$ two integers $k_1$ and $k_2$ can be found so that $|x - (k_1r_1 + k_2r_2)| < p$ holds.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

Okay...this solution may not be correct, please tell me where I'm wrong.
We are given that  r1/r2 is irrational, which implies that r1 and r2 are both irrational.

Now, put z=k1r1 +k2r2.

Choosing r1, and r2 as a base, we can construct a coordinate system so that any number can be written in the form of z.

So, any number p can be written as the absolute value of the difference of two numbers. That number can be as small as we please. Hence the proof. Is this correct?! :)

Sorry for the bad presentation....how does this latex thing work?


limit
logarithms
calculus
calculus computations

Define $a_{n}$ as satisfying: $\left(1+\frac{1}{n}\right)^{n+a_{n}}=e$.
Find $\lim_{n\rightarrow\infty}a_{n}$.

I believe the answer is just zero, is it not?
Since we have $(1+1/n)^{n}= e$

Or am I missing something?

No. It's not 0. That's what I believed and tried to prove during the contest :(
If you want, I can post the solution...

edit: Ohps, read the problem wrong.

Well, this is certainly not more difficult the way it is phrased. Taking logarithms we get $a_{n}= \frac{1}{\ln\left(1+1/n\right)}-n$ and by taylor expansions we get $\lim_{n}a_{n}= 1/2$

[quote="prowler"]Also the idea that $\lim_{n \to \infty}a_{n}= \frac{1}{2}$ can
be observed from the Stirling formula:

(1) $n! \equiv \sqrt{2\pi}\cdot \frac{n^{n+\frac{1}{2}}}{e^{n}}$

$(n+1)! \equiv \sqrt{2\pi}\cdot \frac{(n+1)^{n+1+\frac{1}{2}}}{e^{n+1}}$ or

(2) $n! \equiv \sqrt{2\pi}\cdot \frac{(n+1)^{n+\frac{1}{2}}}{e^{n+1}}$

Comparing them $e\equiv \left (\frac{n+1}{n}\right)^{n+\frac{1}{2}}$[/quote]
Yeah.. prowler :) Do you think I don't who proposed this problem? :)
Good luck!

I don't think the limit equal $ \frac{1}{2}$ becaese we have $ a_n<0,\forall n$.
Indeed $ \frac{1}{\ln\left(1\plus{}\frac{1}{n}\right)}\minus{}n<0$ is equavalent to $ 1<\ln\left(1\plus{}\frac{1}{n}\right)^n$.
But that is clear because we have $ 2<\left(1\plus{}\frac{1}{n}\right)^n<3$.
The anwser is 0

I'm afraid I can't agree. $ \left(1\plus{}\frac{1}{n}\right)^n$ is monotone increasing with limit $ e$, hence $ \left(1\plus{}\frac{1}{n}\right)^n<e$, so $ \ln\left(1\plus{}\frac{1}{n}\right)^n<1$.


The convex hexagon ABCDEF is inscribed in a circle. Prove that the diagonals AD, BE and CF are concurrent if and only if $ (AB)(CD)(EF)\equal{}(BC)(DE)(FA)$.

I have proved the "only if" part using similar tringles and ratios and whatnot, but I'm struggling with the "if" implication. If someone could help then that would be great.

You proved the important part.

Granted it is true that when the diagonals are concurrent then the ratios are as shown, then you just prove that a diagonal not passing through the point of concurrency produces a different ratio.  I guess that falls into the category of a contradiction.  It's best to think of the equation as $ (\frac{AB}{BC})(\frac{CD}{AF})(\frac{EF}{ED})\equal{}1$.

[hide] Pick a diagonal, (I'll take BE).  Move point E so that diagonal BE is no longer concurrent with AD and FC.  Notice that the ratio of $ \frac{FE}{ED}$ varies continuously from 0 to infinity as E moves from F to D along the circle.  So there exists only one point E such that the ratio is what you need it to be. [/hide]

Thanks. I think I have a nice solution now.

Let $ ABCDEF$ be an inscribed hexagon with the required ratio of sides.

Draw $ AD$ and $ BE$ to meet at a point $ X$. Draw $ CX$ to meet the circle at G. Since the lines $ AB$, $ BE$ and $ CG$ all concur, $ ABCDEG$ is another hexagon with the required ratio. We now have:

$ \frac{(AB)(CD)(EF)}{(BC)(DE)(FA)} \equal{} \frac{(AB)(CD)(EG)}{(BC)(DE)(GA)}$

$ \frac{EF}{FA} \equal{} \frac {EG}{GA}$

$ \frac{EF}{EG} \equal{} \frac{FA}{GA}$

Angles $ EFA$ and $ EGA$ are equal by angles in the same segment, so by the side-angle-side version of similarity, triangles $ EFA$ and $ EGA$ are similar.

However, $ \frac{AE}{AE} \equal{} 1$ so the ratio of sides between the two triangles is $ 1$ meaning that they are in fact congruent. This implies that $ F$ and $ G$ are the same point, and that the line $ CX$ actually meets the circle at $ F$

Therefore, if $ ABCDEF$ is an inscribed hexagon with the required ratio of sides, then $ AD$, $ BE$ and $ CF$ concur.


Who's coming to the national champ's ARML practice? :D

Yeah, go NC!!

I hope I make A team... AMC10 score won't expedite that...

I'll be there.  Other possibilities from East include:
Oliver, Thomas, Jerry, Arnav, Varun, Arthur, Jane and Alex

Jane and Varun are both not bad contenders at least for B team.  I'm not sure who actually will come though.

Oliver? Thomas? Jerry? Varun? Alex?

Huh?

Please confirm whether or not Noah is going to ARML either. Thanks.

and Jane... she can probably contend for A team (she [i]was[/i] on the B team last year)

Just to make this very clear: Noah will not participate in any high school math competition ever again  :(

I do not fully comprehend why 138 will not expedite being on the ARML A team... I mean it's a team of FIFTEEN people for goodness sake! And 138 is very good, even in NC. Personally, I believe people can vent about high scores in forums as long as they don't insist not only that they are unhappy, but that it is a bad score. I imagine you'll probably make the A team.

I won't be there :(.

Okay, 8 people from East who you think could make ARML. 1 from Hoggard, and definitely no more. This is unarguable proof that East is a lot better than what you make of it.

lol, everybody's confused about Noah. The only reason I was at first was that Arnav told me he was still there...

jb05, as I said only Varun and Jane have any real chance at making either team.  I certainly wish there were more interest at both of our schools.  Noah left to go to college.  I no longer have anyone to room with  :(, and I anticipate getting very lost at ARML without him.

How did it go?  Unforetunately, I had MathCounts chapters today, so I was unable to attend the practice session  :(

I did terribly, as I went immediately to urgent care afterwards for pain...

but I did pretty well on team/relay, which doesn't count for much.

what's the national champ's arml practice?

ask arnav... or jeremy...

NC=nat'l ARML champs

NC ARML practice=nat'l champ's ARML practice


Three girls A, B, and C, and nine boys are to be lined up in a row. Let $ n$ be the number of
ways this can be done if B must lie between A and C, and A, B must be separated by exactly 4
boys. Determine $ \displaystyle {\left \lfloor \frac {n}{7!} \right \rfloor}$.

Floor function: least integer greater than or equal to $ \frac {n}{7!}$  :)

[hide="Solution, I think"]Call b a boy. We have that A and B must be placed as AbbbbB. Consider this one unit, X. We must now place X and C in that order among 5 boys. There are $ \binom{7}{2}$ ways to do this, and $ 9!$ ways to order the boys, so
\[ \left\lfloor\frac {n}{7!}\right\rfloor \equal{} (9)(8)\binom{7}{2} \equal{} \boxed{1512}
\]
[/hide]

[quote="aidan"]I have [hide]3024[/hide], the girls could either be arranged ABC or CBA?[/quote]Yeah, that's right, missed that.


inequalities
trigonometry
inequalities unsolved

let $ a$ , $ b$ , $ c$ $ \in \mathbb R$ , $ \minus{} 1 < x < 1$  , and , $ ax^2 \plus{} bx \plus{} c \le \frac {1}{\sqrt {1 \minus{} x^2}}$ :
find maximum of :  $ \frac {a}{2} \plus{} c$

Is the inequality above true with all x belongs to interval $ [\minus{}1,1]$

$ \boxed{x<\tan x<\frac{x}{\sqrt{1\minus{}x^2}}\ \left(0<x<\frac{\pi}{2}\right)}$

Putting $ x\equal{}\frac{1}{\sqrt{2}}$ and $ x\equal{}\frac{\minus{}1}{\sqrt{2}}$ we have $ \frac{a}{2}\plus{}c\leq \frac{2\minus{}b}{\sqrt{2}}$ and $ \frac{a}{2}\plus{}c\leq \frac{2\plus{}b}{\sqrt{2}}$ so $ \frac{a}{2}\plus{}c\leq \sqrt{2}$. We will prove that equality holds.

$ a\equal{}\sqrt{2}$ , $ b\equal{}0$ , $ c\equal{}\frac{1}{\sqrt{2}}$ $ \Longrightarrow$ By AM-GM $ (2\sqrt{2}\minus{}2\sqrt{2}x^2)(\sqrt{2}x^2\plus{}\frac{1}{\sqrt{2}})(\sqrt{2}x^2\plus{}\frac{1}{\sqrt{2}})\leq2\sqrt{2}$ hence $ \sqrt{2}x^2\plus{}\frac{1}{\sqrt{2}}\leq \frac{1}{\sqrt{1\minus{}x^2}}$ for all $ \minus{}1< x<1$.


Inspired by the success of gfour4's and Caelestor's Mock AMC12, I have decided to write a mock MATHCOUNTS Nationals Sprint Round. Anyone else, feel free to write the rest of the rounds! I expect it to come out by mid-October if I work on it at a steady pace, however, I don't know the exact dates.

I would like 1 other person to help me administor the test.

Also like the mock AMC12, there will be a short survey at the end.

To sign up, post your name on this thread.

Enjoy!

If we sign up, what do you mean by help "administer" the test?

Help me write problems is what I mean. And PMing it.

And @athunder, its coming out mid-October, early nov so ill post in thanksgiving or something like that

join? my name is ... i forgot please help me...  :(

Vallon22 will be helping me administer the test. Let's all sign up!

I'll join.

But I'd prefer a mock Mathcounts School Round :P
:rofl:

Hopefully it won't be too hard.

Ill join.

But by november Ill probably need to be PMed a reminder. Also, you will probably get more participants if you put this is the MST forum.

I suggest this be discussed on a blog or through PMs, but not in HSB. I'll lock this.


Paul and Jenny both have a whole number of pounds.
He says to her: "If you give me 3 pounds, I will have n times as much as you."
She says to him: "If you give me n pounds, I will have 3 times as much as you."

Given that all these statements are true and that n is a positive integer, what are the possible values of n?

oops i meant [b]british[/b] maths olympiad 1. There is no way of changing the source now is there?[/b]

[quote="Myth"]Yes :) I understood it is BRITISH MO :D
Tyr to edit your post.[/quote]

oh i didn't think it was possible oops (= ah okay its edited.

[b]towersfreak2006[/b] is right.

Solving the system we get $p=\frac{4n^{2}+3n+3}{3n-1},j=\frac{13n+9}{3n-1}$. Rewrite $j$ as $j=4+\frac{n+13}{3n-1}$. Since $j$ is integer, we must have $\frac{n+13}{3n-1}\geqslant 1\iff n\leqslant 7$. Manual checking shows that only $n=1,2,3,7$ give integer values for $j$, and all those values also give integer values for $p$. Hence the solution is $n\in\{1,2,3,7\}$

[quote="towersfreak2006"]I think the second equation should be \[j+n=3(p-n).\] [/quote]
Ah, I confused her with him... :blush:


Let $p>3$ be a prime number. For each nonempty subset $T$ of $\{0,1,2,3,...,p-1\}$, let $E(T)$ be the set of all $(p-1)$-tuples $(x_1,...,x_{p-1})$ where each $x_i \in T$ and $x_1 + 2x_2 +...+(p-1)x_{p-1}$ is divisible by p and let $\vert E(T)\vert$ denote the number of elements in $E(T)$.
Prove that $\vert E(\{0, 1, 3\})\vert \geq \vert E(\{0,1,2\})\vert$ with equality if and only if $p=5$.

I have a solution. it's not too nice. any alternates would be appreciated.

oh, it is, is it.  very good, congratulations

Ekh... No one wants to solve problems, but just only read ISLs or something else...

Sorry, I don't have a pen and a paper to check my words, but here's an idea:
Consider $P(t,x_1\ldots , x_{p-1})=t^{x_1+2x_2+\ldots (p-1)x_{p-1}}$. If we consider $S(t)$ to be the sum of $P$ over all possible values of $x_i$ in the set $|T|={a_1,\ldots, a_t}$, then I think $S=F(t^1)F(t^2)\ldots F(t^{p-1})$, where $F(t)=t^{a_1} + \ldots + t^{a_t}$
Now, we are interested in $\sum_{i=1}^{p-1}S(\epsilon ^i)$, where $\epsilon$ is a $p^{th}$ root of  unity!
I think/hope that for this particular sets, we could be able to find this values......
Please give any feedback : wrong :(, something :?, very good :D
[a small edit] :blush:


quadratics
algebra
polynomial
conics
parabola
symmetry
algebra proposed

Do there exist quadratic polynomials f(x), g(x), h(x) such that the equation

f(g(h(x))) = 0

has exactly the numbers 1, 2, 3, 4, 5, 6, 7, 8 as its solutions?

Well///since ((x 2 ) 2 ) 2  = x 8 , I think your argument is not correct....

Pierre.

In another hand, your conclusion is correct : there are no such polynomials.

Suppose, for a contradiction, that such a polynomials do exist.
Then h(1), h(2),..., h(8) are the roots of the 4-th degree polynomial f(g(x)). But since h is a quadratic polynomial, its graph is symmetric with respect to an axis (it is a parabola), and h(x) = h(y) iff x = y or x,y are symmetric with espect to this axis.
It follows that h(1) = h(8), h(2) = h(7), h(3) = h(6) and h(4) = h(5), and the axis of symmetry is x = 9/2.
Moroever either h(1) < h(2) < h(3) < h(4) or h(1) > h(2) > h(3) > h(4).

Since g(h(1)), g(h(2)), g(h(3)) and g(h(4)) are the roots of the quadratic f(x), we deduce in a similar way than above that g(h(1)) = g(h(4)) and g(h(2)) = g(h(3)), from which it follows that h(1) + h(4) = h(2) + h(3) (since g is a quadratic). But for h(x) = ax 2  + bx + c, this forces a = 0, a contradiction.

Pierre.

Pierre,
Why do the parabolas have to be necessarily symmetrical about the y-axis? What if they are not in the form of ax2+c?

I agree with you about general parabolas, but theses ones are graphs of quadratic polynomials so that they have respective equations of the form y = ax 2  + bx + c.

Pierre.

Oh, sorry, i misread your solution. When you said axis, I thought you meant one of the x/y-axes.

But you can do it with 4 quadratics - for example f(g(h(j(x)))) with

f(x) = x^2 - 5760x
g(x) = x^2 - 72x
h(x) = x^2 + 14x
j(x) = x^2 - 9x

has roots 1,2,3,4,5,6,7,8 (plus another 8 roots)


The sequence $\{a_{n}\}_{n \ge 1}$ is defined by \[a_{n}= 1+2^{2}+3^{3}+\cdots+n^{n}.\] Prove that there are infinitely many $n$ such that $a_{n}$ is composite.

We has $ a_{2n \plus{} 1}\equiv 0(\mod 2)$and $ a_{2n \plus{} 1} > 2,\forall n\in N$
So $ a_{2n \plus{} 1}$ is composite.

Actually, only if $ n\equiv0,3\pmod4$, not for odd $ n$ (for example, $ a_5$ is odd - and even prime).

But the problem is still way too easy. I don't think this is the correct statement.

The following is the correct statement:
[i]Prove that the sequence [/i]$ a_n\equal{}1\plus{}2^2\plus{}3^3\plus{}\ldots\plus{}n^n$ [i] contains infinitely many odd composite numbers.[/i]
It was also given at All-Soviet MO in 1988, as 5th (or 1st in the second day) problem for Grade 9.

[quote="freemind"]The following is the correct statement:
[i]Prove that the sequence [/i]$ a_n \equal{} 1 \plus{} 2^2 \plus{} 3^3 \plus{} \ldots \plus{} n^n$ [i] contains infinitely many odd composite numbers.[/i][/quote]That looks a bit better indeed. :) However don't we have the same kind of trivial solution with mod 3 then?

[quote="freemind"]It was also given at All-Soviet MO in 1988, as 5th (or 1st in the second day) problem for Grade 9.[/quote]Interesting, do you have a reference for that?

[quote="ZetaX"]Let $ p, q$ be coprime integers. Prove that the sum get's divisible by $ p$ and coprime to $ q$ (both at the same time) infinitely often.[/quote]Exactly my thoughts. :) Maybe I should put the extended version in PEN. Do you have a nice and elementary solution, ZeTaX?

$ (k \plus{} mp(p \minus{} 1))^{k \plus{} mp(p \minus{} 1)}\equiv k^k\mod p$, therefore $ p|a_{mp^2(p \minus{} 1)} \ \forall m\in N$.
Exactly, if $ a_n \equal{} \sum_{k \equal{} 1}^n f(k)$ and $ f(k)\mod p$ had period $ T_p$, then $ p|a_{mpT_p} \equal{} \sum_{k \equal{} 1}^{T_p}\sum_{l \equal{} 0}^{mp \minus{} 1} f(k \plus{} lT_p)\equiv \sum_{k \equal{} 1}^{T_p}f(k)mp\equiv 0\mod p$ for any $ m\in N$.

I think problem very easy
obviously $a_{4k+3}\equiv 0(mod 2)$
Then there are infinitely many $n$ such that $a_n$ is composite.


Are you sure the problem is correctly stated? I found the following:

Assume that $ |A_{1}\cup A_{2}\cup ....\cup A_{910}|\leq 100$.

If an element $ x$ belongs to $ >10$ different sets among $ A_1, A_2, ... , A_{910}$, say $ A_1, A_2, ..., A_{11}$, then because of 2) we have that $ |A_1 \cup A_2 \cup ... \cup A_{11}| \geq 1 \plus{} 9 \times 11 \equal{} 100$, which contradicts the assumption. [b](B)[/b]

Now choose a set $ A_k$ among $ A_1, A_2, ... , A_{910}$ and consider its elements $ x_1, x_2, ..., x_{10}$. Throw away all sets $ A_j, j \neq k$, such that $ A_j$ contains any of $ x_1, x_2, ... , x_{10}$. Then because of (B) you have thrown away $ \leq 10 \times 9 \equal{} 90$ different sets. Now throw away also $ A_k$. Of the remaining sets $ A_i$, none can contain any of $ x_1, x_2, ..., x_{10}$.

This procedure can be repeated at least $ 10$ times (you throw away at most $ 91$ sets among $ A_1, A_2, ... , A_{910}$ each time), and for every time the procedure is repeated you find 10 new unique elemets $ x$.

Therefore $ |A_1 \cup A_2 \cup ... \cup A_{910}| \geq 100$.

hmm is it ok? :
$ |A_1 \cup A_2 \cup ....\cup A_{910}| \equal{} n$
$ 910 \cdot {10 \choose 2} \le {n \choose 2} < \frac {n^2}{2}$
so $ 100 < 286 < n$
it looks strange but I don't see any mistakes

isn't having 100 sets quite enough?

apparently I am mistaken somewhere [hide="but"]

suppose that the union of all sets contains less than 100 elements, so if you consider the multiset $ A$ containing the elements of $ A_1,A_2,\dots,A_{100}$, it would be possible to write down the set as $ A\equal{}\{x_1,x_1,\dots,x_i,x_i,\dots,x_{99},x_{99}\}$, where
$ x_i$ appears $ t_i$ times and $ t_i$ is a non-negative integer, besides, $ t_1\plus{}t_2\plus{}\dots\plus{}t_{99}\equal{}1000$. Observe that $ t_i<11$, indeed, if $ t_i>11$, it'd mean that $ x_i$ belongs to at least $ 11$ sets which might not be true.

But then $ t_1\plus{}t_2\plus{}\dots\plus{}t_{99}\leq 10\cdot 99 <1000$, contradiction.[/hide]

Let $ m \equal{} | A_1 \cup A_2 \cup \cdots \cup A_{2009} |$.

WLOG we can assume { $ 1,2,\cdots,m$ } $ \equal{} A_1 \cup A_2 \cup \cdots \cup A_{2009}$.

Then let $ x_i$ as number of sets which contains i. ($ i \equal{} 1,2,\cdots\,m$)

We know $ \sum x_i \equal{} 2009 \times 10$.

Now, let $ S \equal{}${$ (a,b,c)|a \in A_b , A_c$}, and let's find |S|.

If we selected arbitrary a first, there are $ x_a$ sets which contain a. Therefore, $ |S| \equal{} \sum \binom{x_a}{2}$. (1)

If we selected arbitrary b,c first, there are at most 1 number that $ A_b ,A_c$ commonly contain. Therefore, $ |S| \le \binom {2009}{2}$. (2)

Because of (1),(2), $ \frac {1}{2} \sum {(x_a^2 \minus{} x_a )} \equal{} \sum \binom {x_a}{2} \le \binom {2009}{2} \equal{} 2009 \times 1004$.

Now, substitute $ \sum {x_a} \equal{} 2009 \times 10$, then we can get $ \sum {x_a^2 } \le 2009 \times 2018$.

And, because of cauchy, $ \sum {x_a^2 } \ge \frac {1}{m} ( \sum {x_a} )^2 \equal{} \frac {1}{m} ( 2009 \times 10 )^2$.

$ \therefore m \ge \frac {(2009 \times 10)^2}{2009 \times 2018} \equal{} \frac {200900}{2018} \ge 99.5 > 99$

QED.


probability
logarithms
combinatorics proposed
combinatorics

What is an easy way to compute the entropy of a binomial?

Picking an event out of $ e_0, \ldots, e_n$ where the probability of $ e_k$ is $ p_k\equal{}\binom{n}{k}/2^n$ gives us (on average) $ H\equal{}\minus{}\sum_k p_k \lg p_k$ bits of information (I use $ \lg$ for $ \log_2$). [url=http://en.wikipedia.org/wiki/Binomial_distribution]Wikipedia says[/url] that $ H\equal{}\frac{1}{2} \ln\frac{\pi n e}{2} \plus{} O(1/n)$. My question is: [i]How[/i] to derive this formula (as easy as possible)? The tricky part is, of course, computing $ \sum_k \binom{n}{k} \lg \binom{n}{k}$.

doesn't it follow from Stirling approximation?


calculus
integration
LaTeX
calculus computations

How do you find F'(x) if F(x)=the integral from 0 to x^2 of squ(1-t) dt?

If F(x)=the integral from x^3 to 1 of e^(-t^2)) dt?

${F(x)=\int_{h(x)}^{g(x)}M(t) \, dt}$

${F'(x)=\frac{d\left(\int_{h(x)}^{g(x)}M(t) \, dt\right)}{dx}=M(g(x)) g'(x)-M(h(x)) h'(x)}$

By calculus Fundamental Theorem.

First of all, learn to use LaTeX and your posts will be much more readable.  (There is a primer on this site -- ignore all the parts about downloading things; all you need to know is the basic codes, then slap them between dollar signs, and everything will turn out pretty.)
Second of all, use the Fundamental Theorem of Calculus and the chain rule and all will be well.  (This is just what Zarark wrote.)


calculus
function
logarithms
number theory open
number theory

It's my night shift, so I'm sitting in front of the telly trying to kill the time till 7.00am, watching Discovery on channel 11. Then I suddenly noticed that
2^3+3=3^2+2
So I thought:
maybe, for two integers a,b>1 satisfying a>b, the following equation:

a^b+b=b^a+a

might be solvable.

So, is it? I've been working on it for like 4 hours and have used most of my weapons for Diophantine equations and haven't found a way to prove it with elemntary number theory. I'd appreciate any advice/help/hint/relevant fact, but please don't say you used Calculus or Catalan's conjecture to deal with the problem :)

The pair $(2, 3)$ is the only solution. Indeed, since $a > b$, we have $b^{a}= a^{b}+b-a < a^{b}$. Hence, $b^{1/b}< a^{1/a}$. But the function $x^{1/x}$ is decreasing for $x > e$ (do we use Calculus? :blush:). Therefore, $b$ must be less than $e$: $1< b < e$. This means $b=2$. The only value of $a$ s.t $a^{1/a}>\sqrt{2}$ is 3.

10x for the immediate reaction :)

I'll try to find another solution in the next two days, last night my brain was far from its best :)

I think it's a pretty nice fact, although, as you've showed, there's a solution that can prove it straightaway. :oops:

[quote="manuel"][quote="Yury"] But the function $x^{1/x}$ is decreasing for $x > e$ [/quote]
sorry, but can you prove this :blush:[/quote] \[\left(x^{1/x}\right)' = \left(e^{\frac{\ln x}{x}}\right)' = \frac{1-\ln x}{x^{2}}e^{\frac{\ln x}{x}}< 0\text{ for }x > e.\]


The length of the shortest trip from $ A$ to $ B$ along the
edges of the cube shown is the length of 3 edges. How
many different 3-edge trips are there from $ A$ to $ B$?

[asy]size(4cm,4cm);

pair a1, b1, c1, d1;

a1=(1,1);
b1=(0,1);
c1=(1.6,1.4);
d1=(1,0);

pair e1, f1, g1, h1;

e1=(0,0);
f1=c1-(a1-d1);
g1=b1+(c1-a1);
h1=e1+(g1-b1);

draw(a1--d1--e1--b1--a1);
draw(b1--g1--c1--a1);
draw(c1--f1--d1);
draw(g1--h1--e1,dotted+1pt);
draw(h1--f1,dotted+1pt);

label("$A$",e1,SW);
label("$B$",c1,NE);[/asy]

Notice that at the start, there are $ 3$ possible paths to take to depart vertex $ A$. For each of these cases, there are $ 2$ ways to continue, after which there is only $ 1$ way to immediately reach vertex $ B$. Thus, there are $ 3\times2\times1\equal{}3!\equal{}\fbox{6}$ ways to reach vertex $ B$ from vertex $ A$ by traveling three edge lengths.


Circle O has a radius of $x$.

Another circle O has radius of $\frac{x}{2}$

Yet another circle O has radius of $\frac{x}{4}$

And another circle O has raduis of $\frac{x}{8}$

Find the sum of the areas of all these circles, in terms of $x$

If you don't understand what I mean, it's just 4 circles, each with the same center, but they have smaller and smaller radiuses.

[quote="Drunken_Math"]
Find the sum of the areas of all these triangles, in terms of $x$
quote]

You mean Circles right?

[hide]
$x+\frac{x}{2}+\frac{x}{4}+\frac{x}{8}=y$
$8x+4x+2x+x=8y$
$15x=8y$
$y=\frac{15x}{8}$
[/hide]

EDIT: I forgot it sayed areas. :oops:

[hide]
$z=x^2\pi$
$z+\frac{z}{2}+\frac{z}{4}+\frac{z}{8}=y$
$8z+4z+2z+z=8y$
$15z=8y$
$y=\frac{15z}{8}$
$y=\frac{15x^2\pi}{8}?$
[/hide]

EDIT: It's still wrong, isn't it :?

[hide="answer"]The area of the circles are $x^2\pi,\frac{x^2\pi}{4},\frac{x^2\pi}{16},\frac{x^2\pi}{64}$.  Find a common denometer and add them together to find the combined area of the circles.  $\frac{64x^2\pi+16x^2\pi+4x^2\pi+x^2\pi}{64}=\frac{(64+16+4+1)x^2\pi}{64}=\boxed{\frac{85x^2\pi}{64}}$[/hide]


x,y,z are arbitrary reals,
x+y+z=a
xy+yz+zx=b
xyz=c
prove:10(a^2-2b)^3/2+12ab+27c>=6a^3

actually, i've worked it out...
not hard indeed.i'm waiting for someone to post a nice solution,


geometry
3D geometry
arithmetic sequence
algebra
factorization
sum of cubes
number theory proposed

Find an infinite non-constant arithmetic progression of natural numbers such that each term is neither a sum of two squares, nor a sum of two cubes (of natural numbers).

Recall that squares are congruent to $ 0, 1 \bmod 4$ and that cubes are congruent to $ 0, 1, \minus{}1 \bmod 9$.  No sum of squares is congruent to $ 3 \bmod 4$ and no sum of cubes is congruent to $ 4 \bmod 9$, so we can take the positive integers congruent to $ 31 \bmod 36$.


Can anyone provide more information on this program? There's not too much on the website. Specifically, how hard is it/what kind of math abilities do you need to get in? MOP, graduate classes, etc.?

I know nothing about the level of the participants or
the selection procedure, but I can mention that the
instructors and guest lecturers were very high-level.
As I recall there were groups led by Pavel Etingof
and Richard Stanley, who are rather famous
mathematicians.

[quote="RenaudianOntology"]what kind of math abilities do you need to get in? MOP, graduate classes, etc.?[/quote]

Yes, stuff like MOP, and grad classes. The group representation group last year had abstract algebra as a prerequisite. I know one of the kids well who went last year, and know who most of the others are. It's a fabulous program, and costs nothing to apply, or attend. If you can get in, it will be a wonderful experience. But don't get your hopes up unless you walk on water mathematically. Very selective (they only take 12). Not even all Moppers can get selected. Lots of IMO types. Not intended for "mere mortals".

I was at the Clay math program in its first year, when the applications were actually invitations.  I imagine the application process is very competitive, but there's no harm in applying.  I enjoyed my time there very much -- met some cool people, did some very cool math.  I'm not sure I can really shed more light on what it's like now, but if you had specific questions about the experience I'd do my best to answer them.

Is someone froim the forum going to this year academy?

I am!! I just found out yesterday. I will be in the combinatorics group.

[quote="Pascual2005"]great, so I guess we will be working together![/quote]Wow you are going Pascual? Great! :D

I am also going. I will be in the combinatorics group as well.

I'm going in the combinatorics group  :)

[quote="answerseeker95"]oh, Tom Bellulovich and Hyun-Soo Kim from AAST are going as well(I think).[/quote]

Yup, we are. I'm doing combinatorics and hsk is doing representation.

But its Belulovich, not Bellulovich :)


ratio
geometry
trigonometry
geometry unsolved

Given is a circle $ O$ and point $ A$ outside it. The two tangents from $ A$ to the circle touch the circles at $ D,E$. Let $ B,C$ be on rays $ AD,AE$ such that $ BC$ is tangent to the circle as well. Show that the ratio $ \frac{AB\cdot AC}{BD\cdot CE}$ is independent of the selection of $ B,C$.

equivalent to: on an triangle ABC we have the semiperimeter p fixed and the angle $ \angle BAC$ fixed. The incircle meet BC on D. Then $ \frac{bc}{\overline{DB} \cdot \overline{DC}}$ is fixed.

Let $ p\minus{}a\equal{}x$, $ p\minus{}b\equal{}y$ and $ p\minus{}c\equal{}z$ so $ \frac{bc}{\overline{DB} \cdot \overline{DC}}\equal{}\frac{(x\plus{}y)(x\plus{}z)}{yz}\equal{}\frac{xp}{yz}\plus{}1 \equal{} p^2 \tan^2{\frac{A}{2}}\plus{}1$ that is fixed.

Well, is there a nice synthetic solution?

We draw the line through the center $ O$ of the given circle $ (O)$ and perpendicular to $ AO,$ which intersects the rays $ AD,\ AE,$ at points $ B',\ C',$ respectively.

We denote the points $ X\equiv DE\cap B'C,\ Y\equiv DE\cap BC'$ and we will prove that $ BX\parallel AE$ and $ CY\parallel AD.$

$ \bullet$ We denote as $ D',$ the point on $ (O),$ the other that $ D,$ such that the line segment $ B'D',$ tangents to $ (O)$ and so, we have that $ B'D'\parallel AE,$ from $ \angle D'B'C' \equal{} \angle A'B'C' \equal{} \angle AC'B'.$

Hence, in the configuration of the circumscribed quadrilateral $ BB'\infty C,$ based on the [b][size=100]Newton theorem[/size][/b], we conclude that its ''diagonal'' $ B\infty\parallel B'D'\parallel CE,$ passes through the point $ X\equiv B'C\cap DE$ $ ($ the line segment $ A'D',$ where $ A'$ is the tangency point of $ (O),$ with the line segment $ BC,$ also passes through the point $ X$ $ ).$

That is, we have $ BX\parallel AE$ $ ,(1)$ and similarly $ CY\parallel AD$ $ ,(2)$

$ \bullet$ From $ (1)$ $ \Longrightarrow$ $ \frac {AB}{BD} \equal{} \frac {XE}{XD}$ $ ,(3)$

From $ (2)$ $ \Longrightarrow$ $ \frac {AC}{CE} \equal{} \frac {YD}{YE}$ $ ,(4)$

From $ (3),$ $ (4)$ $ \Longrightarrow$ $ \frac {AB}{BD}\cdot \frac {AC}{CE} \equal{} \frac {XE}{XD}\cdot \frac {YD}{YE}$ $ ,(5)$

Applying the [b][size=100]Menelaus theorem[/size][/b], in the triangle $ \bigtriangleup ADE,$ with transversal $ B'XC,$ we have that $ \frac {XE}{XD}\cdot \frac {B'D}{B'A}\cdot \frac {CA}{CE} \equal{} 1$ $ \Longrightarrow$ $ \frac {XE}{XD} \equal{} \frac {B'A}{B'D}\cdot \frac {CE}{CA}$ $ ,(6)$

Similarly, in the triangle $ \bigtriangleup ADE,$ with transversal $ BYC',$ we have that $ \frac {YD}{YE}\cdot \frac {C'E}{C'A}\cdot \frac {BA}{BD} \equal{} 1$ $ \Longrightarrow$ $ \frac {YD}{YE} \equal{} \frac {C'A}{C'E}\cdot \frac {BD}{BA}$ $ ,(7)$

From $ (5),$ $ (6),$ $ (7)$ $ \Longrightarrow$ $ \frac {AB}{BD}\cdot \frac {AC}{CE} \equal{} \frac {B'A}{B'D}\cdot \frac {CE}{CA}\cdot \frac {C'A}{C'E}\cdot \frac {BD}{BA}$ $ ,(8)$

From $ (8)$ $ \Longrightarrow$ $ \frac {AB}{BD}\cdot \frac {AC}{CE} \equal{} \frac {B'A}{B'D}$ $ ,(9)$ $ ($ because of $ B'A \equal{} C'A$ and $ B'D \equal{} C'E$ and  here, may be completed the solution of the proposed problem $ ).$

$ \bullet$ From the similar right triangles $ \bigtriangleup OAB'$ and $ \bigtriangleup DOB',$ we have that $ \frac {B'A}{OB'} \equal{} \frac {OB'}{B'D}$ $ \Longrightarrow$ $ B'D \equal{} \frac {(OB')^{2}}{B'A}$ $ ,(10)$

From $ (9),$ $ (10)$ $ \Longrightarrow$ $ \frac {AB}{BD}\cdot \frac {AC}{CE} \equal{} \frac {(B'A)^{2}}{(OB')^{2}} \equal{} \frac {(AO)^{2} \plus{} (OB')^{2}}{(OB')^{2}} \equal{} 1 \plus{} \frac {(AO)^{2}}{(OB')^{2}}$ $ ,(11)$

But, from the similar right triangles $ \bigtriangleup DAO$ and $ \bigtriangleup DOB',$ we have that $ \frac {AO}{O'B} \equal{} \frac {AD}{OD}$ $ ,(12)$

From $ (11),$ $ (12)$ $ \Longrightarrow$ $ \frac {AB}{BD}\cdot \frac {AC}{CE} \equal{} 1 \plus{} \frac {(AD)^{2}}{(OD)^{2}} \equal{} 1 \plus{} \frac {(AO)^{2} \minus{} (OD)^{2}}{(OD)^{2}} \equal{} \frac {(AO)^{2}}{(OD)^{2}}$

Hence, we conclude that $ \frac {AB}{BD}\cdot \frac {AC}{CE} \equal{} \frac {(AO)^{2}}{(OD)^{2}}$ and the solution is completed.

Kostas Vittas.

Let radii of circle $ O$ is $ R$.
$ i)$ $ B\in[AD]$ and $ C\in[AE]$ $ \Longrightarrow$ Let $ BC$ touches the circle $ O$ at $ K$. $ m(KDB) \equal{} \beta$ and $ m(DAE) \equal{} \alpha$.Hence $ |BD| \equal{} Rtan\beta$,$ |AB| \equal{} R(tan\alpha \minus{} tan\beta)$,$ |EC| \equal{} Rtan(\alpha \minus{} \beta)$,$ |AC| \equal{} R(tan\alpha \minus{} tan(\alpha \minus{} \beta))$ and $ \frac {|AB\parallel{}AC|}{|BD\parallel{}EC|} \equal{} \frac {1}{cos^2\alpha}$ is fixed.
$ ii)$ $ D\in[AB]$ and $ E\in[AC]$ $ \Longrightarrow$  Let $ BC$ touches the circle $ O$ at $ L$. $ m(IBL) \equal{} \beta$ , ($ I$ is center of the circle $ O$) and $ m(BAC) \equal{} 180 \minus{} 2\alpha$. Hence $ |BD| \equal{} Rcot\beta$,$ |AB| \equal{} R(tan\alpha \plus{} cot\beta)$,$ |EC| \equal{} Rcot(\alpha \minus{} \beta)$,$ |AC| \equal{} R(tan\alpha \plus{} cot(\alpha \minus{} \beta))$ so $ \frac {|AB\parallel{}AC|}{|BD\parallel{}EC|} \equal{} \frac {1}{cos^2\alpha}$ is fixed.The proof is ended.

In the triangle $ ABC$ (with the standard notations) we have $ \frac {AB}{BD}\cdot\frac {AC}{CE} \equal{} \frac {c}{p \minus{} b}\cdot\frac {b}{p \minus{} c} \equal{} \frac {bc}{(p \minus{} b)(p \minus{} c)} \equal{} \frac {1}{\sin^2\frac A2}$ (constant).

[b][u]Observation[/u]. Discredit[/b], why wish you a synthetical proof for this easy problem (one row - upper proof) ?!

If you want keenly it, then you can find it by yourself. My dear [b]Vittasko[/b] is a kindman.


geometry
calculus
analytic geometry
trigonometry
parameterization
integration
graphing lines

In the XY plane, A = (0,1), B = (0,0), and C = (1,0). Let S be the set of line segments XY of length 1 unit, such that X lies on AB and Y lies on BC. Find the area of S.

Hint- Don't assume it's a cutout of a circle!

Well I think I can do it with calculus, but it seems hard otherwise...

Let $\phi = \angle XCB$. The points $X, Y$ then have coordinates $X = (\sin \phi, 0),\ Y = (0, \cos \phi)$ and their distance is

$XY = \sqrt{\sin^2{\phi} + \cos^2{\phi}} = 1$

as required. The line $XY$ has an equation

$F(x, y, \phi) = \frac{x}{\sin \phi} + \frac{y}{\cos \phi} - 1 = 0$

This is a family of lines depending on the parameter $\phi$. The enveope of this family of lines is obtained by eliminating the parameter $\phi$ from 2 equations: the equation of the lines and

$\frac{\partial F}{\partial \phi} = \frac{x \cos \phi}{\sin^2{\phi}} - \frac{y \sin \phi}{\cos^2{\phi}} = 0$

Dividing this by $\sin \phi \cos \phi$ and taking the 3rd root, we get

$\frac{x}{\sin^3{\phi}} - \frac{y}{\cos^3{\phi}} = 0,\ \ \frac{x^{\frac 1 3}}{\sin \phi} = \frac{y^{\frac 1 3}}{\cos \phi}$

Substituting this to the line equation yields

$\frac{x}{\sin \phi} + \frac{y}{\cos \phi} = \frac{x^{\frac 2 3}y^{\frac 1 3}}{\cos \phi} + \frac{y}{\cos \phi} = \frac{y^{\frac 1 3}}{\cos \phi} (x^{\frac 2 3} + y^{\frac 2 3}) = 1$

$\frac{x}{\sin \phi} + \frac{y}{\cos \phi} = \frac{x}{\sin \phi} + \frac{y^{\frac 2 3}x^{\frac 1 3}}{\sin \phi} = \frac{x^{\frac 1 3}}{\sin \phi} (x^{\frac 2 3} + y^{\frac 2 3}) = 1$

On the other hand, squaring the previous equation yields

$\frac{x^{\frac 2 3}}{\sin^2{\phi}} = \frac{y^{\frac 2 3}}{\cos^2{\phi}},\ \ \frac{x^{\frac 2 3}}{\sin^2{\phi}} = \frac{y^{\frac 2 3}}{1 - \sin^2{\phi}},\ \ \frac{x^{\frac 2 3}}{1 - \cos^2{\phi}} = \frac{y^{\frac 2 3}}{\cos^2{\phi}}$

${x^{\frac 2 3} (1 - \sin^2{\phi}) = y^{\frac 2 3} \sin^2{\phi}},\ \ x^{\frac 2 3} \cos^2{\phi} = y^{\frac 2 3} (1 - \cos^2{\phi})$

$\frac{\sin^2{\phi}}{x^{\frac 2 3}} (x^{\frac 2 3} + y^{\frac 2 3}) = 1,\ \ \frac{\cos^2{\phi}}{y^{\frac 2 3}} (x^{\frac 2 3} + y^{\frac 2 3}) = 1$

It follows that the envelope equation is $x^{\frac 2 3} + y^{\frac 2 3} = 1$, which can be parametrized as $x = \sin^3{\phi},\ y = \cos^3{\phi}$, because $\frac{x^{\frac 1 3}}{\sin \phi} = \frac{y^{\frac 1 3}}{\cos \phi} = 1$. This is equation of an astroid with the semi-major axis $a = 1$. The area under the curve $y(x)$ in the 1st quadrant is given by

$S = \int_0^1 {y(x) \text dx} = \int_0^{\frac \pi 2} {y(\phi) \frac{\text dx}{\text d\phi}\ \text d\phi} = \int_0^{\frac \pi 2} {\cos^3{\phi} \cdot 3\sin^2{\phi} \cos \phi\ \text d\phi} =$

$= 3 \int_0^{\frac \pi 2} {\cos^4{\phi} \sin^2{\phi}\ \text d\phi} = \frac 3 8 \int_0^{\frac \pi 2} {(1 + \cos{2\phi})^2 (1 - \cos{2\phi})\ \text d\phi} =$

${= \frac{3}{16} \int_0^{\pi} {\sin^2{\theta} (1 + \cos \theta)\ \text d\theta} = \frac{3}{16} (\int_0^{\pi} \frac{1 - \cos{2 \theta}}{2}\ \text d\theta} + \int_0^0{z^2\ \text dz}) = $

$= \frac{3}{32}(\pi - \frac 1 2 \int_0^{2\pi}{\cos \omega\ \text d\omega}) = \frac{3\pi}{32}$

Hexaditidom, could you give us a hint as to how to do this in a non-calculus way?

No clue. I can dish it out, but I sure can't solve it. :blush:
That was a very nice solution, though. I'm sure there is a simpler way to do it... that's the fun of problem solving.

I was thinking something like this: The midpoints of all segments XY define the curve (probably, I could be wrong). Then calculus could be used to find the area under it, which I'm sure you just did.

Nah, the midpoint only meets the curve when the slope of the line segment is -1.  If you trace the midpoint of the curve, you'll actually get one quarter of a circle with radius 1/2, or so it seems from the diagram I made on The Geometer's Sketchpad.

The right angle triangles $\triangle XBY$ all have hypotenuse $XY = 1$ of the same length. The midpoints $M$ of the segments $XY$ are their circumcenters and their circumradii are $R = \frac{XY}{2} = \frac 1 2$. The segments $BM = R = \frac 1 2$ are their medians of constant length, so the locus of midpoints $M$ is indeed a circle centered at the point $B$ and with radius $BM = \frac 1 2$.


http://www.cnn.com/2008/WORLD/europe/08/28/russia.georgia.cold.war/index.html

I'm curious -- what's your take on this? It personally wouldn't surprise me at all, especially since there are some major unanswered questions as to why the Georgian leader was so confident of Western support that he felt it prudent to suddenly move into the region.

It's not a bad angle, it's reasonable to think that the threat of a new cold war would play into McCain's campaign.  But I don't really buy it.  For one thing, if the administration was setting up the conflict in Georgia, they would have known that Russia would respond with force (Russia openly stated this position), and would have known that they would have no good counter except a statement of condemnation.  That seems like a bad situation to willingly put yourself into.  But even if they had, then at least one would expect they would actually [i]have[/i] a strong statement of condemnation prepared for immediate release.  They didn't; the US response was cautious until days later when they stepped up the rhetoric.

Besides, the way the administration has been talking and acting on other foreign policy issues -- Iran in particular -- suggests they're trying desperately to avoid any hint of the cowboy politics that marked the first half.  For those reasons I do believe the state department really was urging Georgia to avoid conflict, just as they've been saying they had been.  Especially since it was no secret that Russia was looking for an excuse to avenge Kosovo, the administration would almost certainly realize the risk of emerging the political loser in the event of a conflict.  Would GW tarnish his legacy more than it already is to boost McCain a couple points? Nah.

The connection to McCain's campaign is much more direct than that- his top foreign policy adviser is a lobbyist for Georgia. McCain had the extremely aggressive response immediately.

The escalating rhetoric came when it appeared that Russia might be occupying substantial parts of Georgia.

Right, but my response was to the assertion being made by Putin that the conflict was instigated in order to help one of the Presidential candidates (which I presume to be McCain) in the election.  Furthermore he seems to point the finger specifically at the US government, and even Bush in particular, not merely at some individual connected to one of the campaigns.  At least that's what I gather from these quotes:
[quote]"The suspicion arises that someone in the United States especially created this conflict with the aim of making the situation more tense and creating a competitive advantage for one of the candidates fighting for the post of US president."[/quote]
[quote]"It should be admitted that they would do so only following direct orders from their leaders. Therefore, they were acting in implementing those orders, doing as they were ordered, and the only one who can give such orders is their leader."[/quote]
So while it may or not be true that McCain has interests in Georgia, Putin's suggestion seems to be that the situation was 'created' by the administration specifically for the purpose of improving one of the candidate's campaign position.

But anyway, how does McCain's adviser being a lobbyist for Georgia give him the means or incentive to help create a situation disadvantageous to Georgia?

It doesn't. I think it was Georgia pushing things all along, and trying to convince the US to support them. That connection simply explains why McCain was so aggressive.

Sometimes the people running countries do very stupid things. There isn't always a rational explanation.

My impression was that this conflict was a very unpleasant surprise to the US government and that they didn't expect at all that the situation would unfold so tragically. In particular, official US government statements were very moderate at first; they asked all parties to cease fire and affirmed Georgia's territorial integrity (without explicitly blaming any party involved).

On the other hand, McCain was very critical of Russia even before this conflict; it was clear he would strongly support Georgia under any circumstances. I am wondering to which extent this conflict will affect the US presidential race. If McCain successfully plays the "rise of the evil empire" card, he can get several points.


vector
geometry
geometric transformation
homothety
linear algebra
matrix
advanced fields

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=654178#654178

This belongs here in the advanced section.

If there's no proof more elementary than yours, that's surely true. But are you sure that's teh case? ;)
Also, it's probably better to post this in the thread of the original problem, or to PM a moderator about it so that they can move the post. :)

Here's a possible different idea: because $\parallel a\times b\parallel ^{2}=\parallel a\parallel ^{2}\parallel b\parallel ^{2}-<a,b>^{2}$, we deduce that $<u,Bu><v,Bv>-<u,Bv>^{2}=\parallel u\parallel ^{2}\parallel v\parallel ^{2}-<u,v>^{2}$, where $B$ is the product between $A$ and its transpose. Now, take $x,y$ two different eigenvalues of $B$ corresponding to vectors $u,v$. Because $B$ is symmetric, $<u,v\geq 0$ and so we find that $xy=1$. We need to prove that all eigenvalues of $B$ are equal to 1 because in this case by diagonalisability of $B$, it will be the identity and so $A$ will be orthogonal. By the previous argument, we know that any two distinct eigenvalues of $B$ are related by $xy=1$. If all eigenvalues are equal, $B$ is a homothety and we are done (quick check). So, exactly two eigenvalues are distinct and $B$ has eigenvalues $x,x,1/x$ for some $x\ne 1$ and $x>0$. Thus $B$ is unitarily similar to the diagonal matrix $x,x,1/x$. Unfortunately, I have no elegant way to continue from here. :maybe:


function
real analysis
real analysis unsolved

Let x=(a_1,a_2,..,a_n) be a real n-tuple .
|x|is defined as
 |x|=sup{ |a_1,|a_2|,…,|a_n| }
 Prove that the function g(x)=|x| does not satisfy
|x+y|^2+|x-y|^2=2 { |x|^2+|y|^2 }

we have to deal with the general case.a_1,..a_n may be any set of real munbers.

Take $\{3, 2\}$ and $\{3,-2\}$. Then we have $36+16 = 52$ on the LHS and $2(9+9) = 36$ on the RHS.

One example of where it fails proves what you wanted, regardless of dimension (add some 0s, if you like).

I hope I didn't make a stupid mistake... :P


inequalities
calculus
derivative
integration
inequalities proposed

Can anyone prove that
 x^a<=ax + (1-a) for x=>0 and 0<a<1 without using calculus but by elementary means?
This is a version of Bernoulli's inequality

I've tried with no success.

Thanks!

[quote="Albanian Eagle"]is weighed AMGM allowed? :D[/quote]

As long as no derivatives, limits, or integrals are used.


Fie $f:R \to R$ injectiva astfel incat:
                 $f(x^3-9x^2-20x+1)=f^3(x)-9f^2(x)-20f(x)+1,x \in R$
Demonstrati ca exista $x_1,x_2,x_3 \in R$ diferite astfel incat $f(x_1)+f(x_2)+f(x_3)=9$.

Indeed, consider three roots of the equation $x^3-9x^2-20x+1=x$. They are all real and distinct (easy task!). Denote them $x_1,x_2,x_3$. Then we have $f(x_i)=f^3(x_i)-9f^2(x_i)-20f(x_i)^2+1$. Since function $f$ is injective we conclude that $f(x_1),f(x_2),f(x_3)$ are three distinct root of the same equation $x^3-9x^2-20x+1=x$, but sum of its roots is 9 by Viet's theorem!


1986 AHSME, #29

Two of the altitudes of the scalene triangle $ ABC$ have length 4 and 12. If the length of the third altitude is also an integer, what is the biggest it can be?

A. 4
B. 5
C. 6
D. 7
E. none of these

[hide="Hint"]
Use variables of the two sides at the base of the known altitudes to determine the range of the third side, then use area to calculate the third altitude length.

[hide="Solution"]
Let, in $ \triangle\text{ABC}$, the altitude of 4 be drawn from A, and the altitude of 12 be drawn from B.  Further, let side $ b\equal{}x$ and let side $ a\equal{}3x$.  Therefore: $ 2x<c<4x$, and $ 6>\text{altitude}>3$.  The largest integer length it can be 5 (Answer B). [/hide][/hide]

I have one question, regarding the concepts behind this problem.  Is the length of the base of an altitude proportional to the length of a base of another altitude.  For instance, if the altitude is 4, and it's corresponding base is 6 and another altitude is 8, will the corresponding base be 12, will the corresponding base be proportional, relating to the altitude length?

No, however they are inversely proportional. So in your case the corresponding base would be 3, not 12. In other words, the base and altitude must always multiple to the same thing.

thank you dgreen, that helped clear some things up.


modular arithmetic
number theory theorems
number theory

Is $1^m+2^m+...+(p-1)^m$ really $0(modp)$ for $p-2\geq  m\geq 0$? I saw this lemma somewhere, and I was wondering about the case when $m$ is even. I don't think the proof is trivial

[quote="sunchips"]Is $1^{m}+2^{m}+...+(p-1)^{m}$ really $0(modp)$ for $p-2\geq m\geq 0$? I saw this lemma somewhere, and I was wondering about the case when $m$ is even. I don't think the proof is trivial[/quote]

Yes the idea is that all prime mods have generators, therefore let $g$ be a generator in $p$, then:

$\{1,2,...,p-1\}= \{g^{0},g^{1},...,g^{p-1}\}$

(not necessarily in that order, of course).

Then the question is if

$(g^{m})^{0}+(g^{m})^{1}+...+(g^{m})^{p-1}\equiv 0 \pmod{p}$

This follows immediately. In particular, one easy way to verify this is through the summation formula for geometric series.

We can use there is a primitive root in modulo p(where p is prime number). And i think you mean with "generator" to "primitive root". Am i wrong? :wink:

"Generator" is just the general term in groups.


AMC
AIME
geometry
3D geometry
dodecahedron
icosahedron
tetrahedron

Whoever solves the problem correctly (with solutions) first should post another problem (credit the source), and we'll go from there. Because of a lack of GPML problems, some AMC and AIME problems will be allowed (although local GPML sprint and target is about the level of the AMC 12).

An easy one:
GPML 2001-2002, Invitational Round 1: Sprint #25 and 26

25. Which of these is not a regular solid?
(A) Dodecahedron (B) Cuboctahedron (C) Icosahedron (D) Tetrahedron (E) None of the above

26. The five regular solids are named after whom?

25. The five platonic solids are the tetrahedron, the cube, the octahedron, the dodecahedron, and the icosahedron. A cubeoctahedron is not one of those, so the answer is B.

26. The platonic (regular) solids are named after Plato.  :lol:

And I think that GPML sprints are like AMC 1-15. They are definately not anything like 20-25 on the AMC.

Anyway, here's the next problem. I don't have any GPML problems on me, so I'll post a different one.

Two perpindicular chords AB and CD, which are inside a circle, intersect at E. If AE = 3, BE = 4, CE = 6, and DE = 2, find the radius of the circle.

Source: AoPS II

... what do people not understand about posting both a new problem and your solution to the problem in marathons.  Is it so hard to read the two posts above yours? :mad:

$\frac{\sqrt{65}}{2}$
[hide="Solution"]Draw $PQ$, the diameter of the circle parallel to $AB$. Then from the center $O$ of the circle, draw radius $AO$. If CD intersects $PQ$ at $F$, $CF=DF=4$. Then a perpendicular segment from $O$ bisecting $AB$ at $M$ will have length $2$, and $AM=\frac{7}{2}$. Use the Pythagorean Theorem to get $AO=\frac{\sqrt{65}}{2}$ (Sorry without a diagram).[/hide]
[u]1991 AIME #1[/u]:
Find $x^{2}+y^{2}$ if $x$ and $y$ are positive integers such that
$xy+x+y=71$ and $x^{2}y+xy^{2}=880$.


[quote="GameBot"]For the quadrilateral shown, how many different whole numbers could be the length of the diagonal represented by the dashed line?

[asy]draw((0,0)--(5,5)--(12,1)--(7,-8)--cycle,linewidth(0.7));
draw((0,0)--(12,1),dashed);
label("8",(2.5,2.5),NW);
label("10",(8.5,3),NE);
label("16",(9.5, -3.5),SE);
label("12",(3.5,-4),SW);[/asy][/quote]

By the triangle inequality,

$ 18 > x$.

Thus, $ 17$.

The answer on the actual FTW says 13. Lol.

[hide="Solution"]By the triangle inequality, we have $8+10>x$ and $12+x>16$, or $18>x>4$. We have the range $[17,5]$ for $x$ which gives us $17-5+1=\boxed{13}$ values.[/hide]


geometry
3D geometry
tetrahedron
inequalities

Consider a tetrahedron DABC which has the following property:$AB^{2}+CD^{2}=AC^{2}+BD^{2}=AD^{2}+BC^{2}$. Show that at least one of the triangles ABD, ABC, BDC, ADC  is acute.

[hide="Solution"]WLOG, let AB be the longest side.  Assume all the triangles are obtuse.  Then we have in triangles ABC and ABD, $AB^2 > BC^2 + CA^2, AB^2 > BD^2 + DA^2$.  Adding these two inequalities we get $2AB^2 > BC^2 + CA^2 + BD^2 + DA^2 = (AC^2 + BD^2) + (AD^2 + BC^2) = 2(AB^2 + CD^2) > 2AB^2$, contradiction.[/hide]


I'm kinda new to LaTeX and not really familiar with the TeXnic center compile output.  Often when I compile my solutions the compiler lists errors but I see no errors in the output file.  Is it acceptable to send in solutions that compile with errors but look fine?

Check the output file very carefully. There may be some minor mistake as a result of a $\text{\TeX}$ error. But sometimes it will come out correctly. Often an extra } will be added automatically if you leave it out.

On the $\LaTeX$ toolbar there are two buttons.  One has a right arrow pointing toward an exclamation mark and the other is just the same thing but backwards.  If you click either of these it will bring you to the error(s).  If you can't find what I am saying you could just press F9 (or Shift + F9) also.

If the PDF file looks the way you want it to when you compile, then you're fine.

Thanks joml88, that tip really helped.  Amazing how good LaTeX is finding bugs now that I use this feature.  Wish they had an easy to use "next error" feature in Visual C++.

Also its good to now that I can submit an error ridden file, just in case.


From "Nicolae Paun" contest, Romania, 2003 ,grade10!

Give an example of function f:[b]R->R[/b] with propriety f(f(x))=-x for every real number x.

well, partitioning the reals into quadruplets of the form {x,y,-x,-y} (this can be done easily, with any bijective function from R to R, which doesn't have a fixed point).

then just take f(x)=y, f(y)=-x, f(-x)=-y, f(-y)=x. You will see that this function verifies.

PS I was thinking about [url=http://www.mathlinks.ro/viewtopic.php?p=12656#12656][THIS][/url].

A possible construction:
  Take f(x)=x+1 if there is a natural number n such that 2n<=x<2n+1 and take f(x)=-x+1 if there is a natural number n such that 2n+1<=x<2n+2. Thus, we defined f on the nonnegative reals. Then put f(-x)=-f(x) and you will find a function defined on the reals such that f(f(x))=-x for all x.

[quote="mr_flirt"]
Hey harazi, there are many functions with this property?>i mean your example is equivalent with Valentin's one???
[/quote]

[quote="Valentin Vornicu"]
(this can be done easily, with [b]any[/b] bijective function from R to R, which doesn't have a fixed point)
[/quote]

Trust me mr_flirt, there are really [i]many[/i] such functions...   ;)


[hide="Solution"]
[b]Solution.[/b] Let $f(n)=a_{n}$.  Then the problem is equivalent to finding all sequences $\{a_{n}\}$ such that $a_{1}=1$.    We want to express $a_{n}$ in the form $a_{n}=\lambda^{n}$.  Hence,
\[a_{n+2}-3a_{n+1}+2a_{n}=0\implies \lambda^{2}-3\lambda+2=0\implies \lambda_{1,2}=1,2.\]
As a result,
\[a_{n}=A(2^{n})+B.\]
But, $1=a_{1}=2A+B\implies B=1-2A$, so
\[a_{n}=2A(2^{n-1}-1)+1.\]
Therefore, the desired set of functions $f$ is
\[f(x)=2c(2^{x-1}-1)+1\]
for any constant $c$.$\Box$[/hide]