Imma crash your cello at school :rotfl: jk :P

I used to play clarinet in our middle school band....was first chair, but I dropped out of band so I could play soccer and be in choir.  :(

[quote="Tunio"]Yay~~~ well I play French Horn in band...[/quote]
she just said she dropped out of band... why are you happy? thats mean
 she even had a  :( are you a      sadist??
-_-;; tut tut tut

NONO I meant I was happy that someone else besides me plays clarinet you Mafia jk

I played clarinet for a year, but then dropped it for the sax.

[quote="spang"]Clarinets are the best  :lol:! Songs I like and have played include Concerto by Mozart; Concertino and Concerto by Weber; Saint Saens' Sonata; Brahms' Sonata; and the Rhapsodies. I've been playing almost 7 yrs now and it has always been a great time. For sure, Clarinets UNITE![/quote]
Absolutely right! I haven't played Brahms yet but I have played Saint Saens, Weber, and Mozart, but my favourites are usually contemporary, such as Bernstein Sonata and Muczynski Time Pieces. This is my 7th year as well. I've attached a list of clarinet "standards" that maybe you would be interested to look at. I can't wait to get my A clarinet  :lol:  :lol:

That's a very comprehensive list of songs! Are the ones that you have played marked with checks? Also, did you take the "上海考級本"? (Just wondering because your flag says Taiwan). I have an A clarinet but I personally like my Bb better I think it has a nicer tone and is easier to play with. However, playing on (almost) any A would beat transposing in Orchestra  :) .

Yap I have taken the Shanghai ones... depends which one you're talking about, the auburn colour or the green colour. I'm taking the 音樂學院十級 this november and ABRSM G8... Oh and the checks are the ones I own, not play, but I basically touched on all of them. I just put it there cause people borrow my music regularly. Why do you like your Bb better? Is it because it's a better model? I use Buffet Festival series... I'm already content with that one... and I'm hoping my dad would say yes to the basset clarinet :P

Cool so I'm guessing you are in Shanghai? What school do you go to? Hmm well I like my Bb better just because its a better model => it seems to play better; sound is more focused; articulations seem to come a tad easier etc.  :) . I'm prolly going to do a lot of Baermann over the summer hehe should be fun I guess.

Yeap I go to Shanghai American School. And I don't like Baermann. Eww. I'm just going to concentrate on my exams.

Shanghai American High School? Do you know Jean Xu? She went to my middle school.

Colts18: care to elaborate?  :) .

Alright I played violin... that got boring so I played clarinet. We don't play many good songs because there from a "workbook" . But sometimes we play other types of genres other than classical like jazz. It stinks for me because i'm in 6th and only the higher grades get to play real good music. The best song I ever played was Ronde by Tielman Susiato. :oops: Yeah i know it stinks.

I just started playing the clarinet, and my teacher didn't really show me how to

tongue. He said that since your tongue is usually on the reed anyways, you lift your

tongue to emphasize a note. Help!

I play clarinet! I have a Buffet E11.

Trying out for Sr. Districts next year... :)

Me too!  :P

[quote="SoccerBrainy40"]
I play clarinet! I have a Buffet E11.

Trying out for Sr. Districts next year... Smile
[/quote]

HEY! Surprise, surprise.

I play clarinet, and it is my biggest obsession (next to math, of course). I have been playing since the 4th grade and have played in the Southern California and All State honor bands and orchestras. My favorite pieces are the Saint-Saens Sonate and the Poulenc Sonata.

I've been playing clarinet for 7 years.  I've played the mozart concerto, Saint-Saens, Poulenc, and all that good stuff.  Clarinet is definitely where it's at.  No matter what the song, it CAN be played on the clarinet because of range and ease of articulation.  It always gets melody, and it's amazingly light on the marching field.  Plus, if you're clarinet section leader during marching band, there are always a lot of people in your section to boss around. . . I mean "lead".  Yes, clarinet is definitely the best.  I love it.

I play the clarinet, and unlike most Asians I'm just a typical band dude, middle chair. Been playing since 4th grade. No private lessons.

Edit: I also love the range of keys on the clarinet, in agreeance with the above post.

I played clarinet for a year but i sucked at it so i noobed down to Bass Clarinet. I even played the BBb Contrabass Clarinet in my high school Wind Ensemble. They begged me to play it because apparently i was the only Bass Clarinetist who could play loud.

the only time you see "Bass Clarinet" and "loud" in the same sentence is if the word "isn't" is in the middle. they only gave me a bass clarinet because nobody could hear it.

lol.
I play violin and clarinet.
No private lessons for clarinet, but still been 1st chair for over 2 year... and I don't even practice.
I suck compared to you guys though. The highest I ever got was District Honor Band in eighth grade, b/c I accidentally bombed my all-state audition... long story.


ceiling function
number theory proposed
number theory

Let $ n$ be a positive integer congruent to 1 modulo 6. Show that  $ \frac {3}{n}$  can be expressed as :
$ \frac{1}{a_{1}} \plus{}\frac{1}{a_{2}}\plus{}...\plus{} \frac{1}{a_{k}}$ for some distinct positive integers $ a_{1}$, $ a_{2}$, ..., $ a_{k}$, and also find minimal value of $ k$.

A constructive solution that should work
[hide]Set $ n \equal{} 6h \plus{} 1$, with $ h > 0$, and let $ l$ be the smallest positive integer such that $ \frac{1}{6l} < \frac{3}{6h \plus{} 1}$. Setting $ a_{1} \equal{} 6l$ we have to find $ b_{2}, ..., b_{k}$ such that
\[ \frac{1}{6h \plus{} 1}\left(\frac{1}{b_{2}} \plus{} ... \plus{} \frac{1}{b_{k}}\right) \equal{} \frac{3}{6h \plus{} 1} \minus{} \frac{1}{6l},\] which is equivalent to finding $ b_{2}, ..., b_{k}$ such that
\[ \frac{1}{b_{2}} \plus{} ... \plus{} \frac{1}{b_{k}} \equal{} 3 \minus{} \frac{6h \plus{} 1}{6l}\]
[b]Lemma:[/b] $ 18l \minus{} (6h \plus{} 1)$ can be written as sum of distinct divisors of $ 6l$.
Since $ l \equal{} \left\lceil\frac{6h \plus{} 1}{18}\right\rceil$, \[ 18l \minus{} (6h \plus{} 1) < 18\left(\frac{6h \plus{} 1}{18} \plus{} 1\right) \minus{} (6h \plus{} 1) \equal{} 18.\]
But the residues of $ 18l \minus{} (6h \plus{} 1)$ modulo $ 18$ are $ 17, 11, 5$, according to $ h$ congruent to $ 0, 1, 2$ modulo $ 3$, respectively. And so the lemma holds since for example $ 5 \equal{} 2 \plus{} 3$, $ 11 \equal{} 2 \plus{} 3 \plus{} 6$ and $ 17 \equal{} 2 \plus{} 3 \plus{} 12$.
From here it's easy to finish. Then it remains to prove it when $ n \equal{} 1$.[/hide]


1.How manny solutions  has the sistem?$ x,y,z$ are integers
\[ x \plus{} y \plus{} z \equal{} 0
\]

\[ x^3 \plus{} y^3 \plus{} z^3 \equal{} 18
\]
2.How many pairs of positive real numbers verify the system
\[ \sqrt {x} \plus{} \sqrt {y} \equal{} 1
\]

\[ \sqrt {x \plus{} 2} \plus{} \sqrt {y \plus{} 2} \in N
\]
3.How many pairs of real numbers verify the sistem?
\[ |x \plus{} 1| \plus{} |y \plus{} 1| \equal{} 8
\]

\[ |x \minus{} 1| \plus{} |y \minus{} 1| \equal{} 4
\]

[hide="2"]We have $ \sqrt{x}\plus{}\sqrt{y}\equal{}1\ (1)\Rightarrow 0\leq x,y\leq 1.$
$ \Rightarrow 2\sqrt{2}<\sqrt{x\plus{}2}\plus{}\sqrt{y\plus{}2}<2\sqrt{3}$
$ \Rightarrow \sqrt{x\plus{}2}\plus{}\sqrt{y\plus{}2}\equal{}3\ (2)$
$ (1)\Rightarrow x^2\plus{}y^2\minus{}2xy\minus{}2x\minus{}2y\plus{}1\equal{}0$
$ (2)\Rightarrow x^2\plus{}y^2\minus{}2xy\minus{}18x\minus{}18y\plus{}9\equal{}0$
$ \Rightarrow x\plus{}y\equal{}\frac{1}{2}$
$ \Rightarrow xy\equal{}\frac{1}{16}$
$ \Rightarrow x\equal{}y\equal{}\frac{1}{4}$[/hide]

[quote="skand"][hide="1"]
we have,
$ a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc \equal{} (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca)$
so,if $ a \plus{} b \plus{} c \equal{} 0$,then,$ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3abc$

using this formula,we can find,$ xyz \equal{} 6$
so $ (x,y,z) \equal{} (1,1,18),(1,2,9),(1,3,6),(2,3,3)$ and all other permutations[/hide][/quote]
I have forgoten to say....$ x,y,z$ are integers....sorry...therefore your solutions don't verify the system


inequalities
function
calculus
algebra unsolved
algebra

[quote="scarface"]If $ |x^3 \plus{} 3x^2 \minus{} 33x \minus{} 3|\geq2x^2$ and $ |x|\geq n$, find the smallest possible value of $ n$ positive integer.

NOTE: I found a solution with graphs of functions  $ y \equal{} |x^3 \plus{} 3x^2 \minus{} 33x \minus{} 3|$ and $ y \equal{} 2x^2$. Please don't use a computer program in your solution. Answer is 9.[/quote]

This is a crazy problem and I think that if somebody gave it to you for olympiad training, better to move to another training team.

Obviously, for $ x$ great enough $ x^3 \plus{} 3x^2 \minus{} 33x \minus{} 3\geq 2x^2$,or also $ x^3 \plus{} x^2 \minus{} 33x \minus{} 3\geq 0$
Obviously, for $ \minus{} x$ great enough, $ \minus{} x^3 \minus{} 3x^2 \plus{} 33x \plus{} 3\geq 2x^2$, or also, using $ \minus{} x$ instead of $ x$ : $ x^3 \minus{} 5x^2 \minus{} 33x \plus{} 3\geq 0$

And you just have some basic calculus :
For $ x^3 \plus{} x^2 \minus{} 33x \minus{} 3$, seems clever to test around $ x^3 \equal{} 33x$ : for $ x \equal{} 5,6,7$, $ x^3 \plus{} x^2 \minus{} 33x \minus{} 3 \equal{} \minus{} 18,51,158$
For $ x^3 \minus{} 5x^2 \minus{} 33x \plus{} 3$, seems clever to test around $ x^2 \equal{} 5x \plus{} 33$ : for $ x \equal{} 8,9,10$, $ x^3 \minus{} 5x^2 \minus{} 33x \plus{} 3 \equal{} \minus{} 69,30,173$

So, suspected value is 9 and it remains to prove that :
$ x^3 \plus{} x^2 \minus{} 33x \minus{} 3\geq 0$ $ \forall x\geq 9$, which is obvious since $ x^3 > 33x$ and $ x^2 > 1$
$ x^3 \minus{} 5x^2 \minus{} 33x \plus{} 3\geq 0$ $ \forall x\geq 9$, which is obvious since $ x^2 > 5x \plus{} 33$ and so $ x^3 > 5x^2 \plus{} 33x$

Very very poor problem in an olympiad forum, IMHO.

[quote="scarface"]tnx. your solution is enought for me, the problem can be easy or not but keep your luminous opinion yourself.[/quote]
Sorry to have bothered you.

Keep training as you want, sure. Could you just avoid to put non olympiad problem in olympiad fora ? There exist some other fora better adapted to this level.
Thanks in advance.
And sorry again.


algebra
polynomial
quadratics
quadratic formula

One of the roots of ax^2+bx+c=0 is 2-3[i]i[/i]. Find  b^2-4ac when a=1.
A) -9
B) -13
C) -24
D) -30
E) -36

[u]Two notes:[/u]
1. You should add that b and c are real.
2. You mean to say "discriminant," not "determinant"  :wink:

[u]Solution:[/u]

If 2-3i is a root, then 2+3i is also a root. Hence, the polynomial is (x-2-3i)(x-2+3i) = x^2 - 4x + 13

The discriminant is (-4)^2 - 4(1)(13) = -36, so the answer is [b]E[/b]

[hide="Discussion"] You don't need to actually calculate $ a, b, c$.  Let $ \Delta$ be the discriminant.  You know that the complex roots are $ \frac{\minus{}b}{2} \pm \frac{ \sqrt{\Delta} }{2}$, where the second term is the imaginary part.  It follows that $ \frac{ \sqrt{\Delta} }{2} \equal{} 3i \implies \Delta \equal{} \minus{}36$. [/hide]

t0rajir0u,
how do you get the (-b/2)+((sqrt (discriminant))/2) and (-b/2)-((sqrt (discriminant))/2)?
Is that a specific formula or something?

And yes, sorry for all the typos in the problem. That was my bad.

That is the [url=http://en.wikipedia.org/wiki/Quadratic_formula]Quadratic Formula[/url], a general solution to the quadratic equation.

We know that the roots are $ 2 \pm 3i$, so the $ \pm \frac {\sqrt{b^2 \minus{} 4ac}}{2a} \equal{} \pm 3i$.

We multiply by 2a to get $ \pm \sqrt{b^2 \minus{} 4ac} \equal{} \pm 6ai \equal{} \pm 6i$, and square to get $ b^2 \minus{} 4ac \equal{} \minus{} 36$, regardless of the signs.


Solve the system of equations:
$ x(x\plus{}y)^2\equal{}9$
$ x(y^3\minus{}x^3)\equal{}7$

Help me

hello, solving your first equation for $ y$ we get $ y\equal{}\minus{}\frac{x^{3/2}\minus{}3}{\sqrt{x}}$ or $ y\equal{}\minus{}\frac{x^{3/2}\plus{}3}{\sqrt{x}}$ setting the first in the second equation and simplifying this we get
$ \minus{}2x^{11/2}\plus{}9x^4\minus{}27x^{5/2}\plus{}27x\minus{}7x^{3/2}\equal{}0$, this equation has only $ x\equal{}1$ as the solution, setting the second equation from above in the second one we get
$ 2x^{11/2}\plus{}9x^4\plus{}27x^{5/2}\plus{}27x\plus{}7x^{3/2}\equal{}0$ with no real solution.( we have $ x>0$.)
With $ x\equal{}1$ we get $ y\equal{}2$.
Sonnhard.


For $ a,b,c \in \mathbb{R^ \plus{} }$ and $ abc \equal{} 1$. Prove that,
\[ \dfrac{a^2 \plus{} b^2 \plus{} c^2}{2}\ge \dfrac{1}{a \plus{} b} \plus{} \dfrac{1}{a \plus{} c} \plus{} \dfrac{1}{b \plus{} c}\]

Have a Happy Thanksgiving!

Hello.

By $ AM\minus{}GM$we have:

$ \frac{1}{2 \sqrt[]{ab}} \plus{} \frac{1}{2\sqrt[]{bc}} \plus{}\frac{1}{2\sqrt[]{ca}} \ge \frac{1}{a\plus{}b} \plus{} \frac{1}{b\plus{}c} \plus{} \frac{1}{c\plus{}a}$

$ \frac{1}{2 \sqrt[]{ab}} \plus{} \frac{1}{2\sqrt[]{bc}} \plus{}\frac{1}{2\sqrt[]{ca}}\Leftrightarrow \frac{\sqrt[]{a}\plus{}\sqrt[]{b}\plus{}\sqrt[]{c}}{2}$

From there we need to show that $ a^2 \plus{} b^2 \plus{} c^2 \ge \sqrt[]{a} \plus{} \sqrt[]{b} \plus{}\sqrt[]{c}$

$ a^2 \plus{} b^2 \plus{} c^2 \ge \frac{1}{3}(a\plus{}b\plus{}c)^2 \ge \frac{1}{3}3\sqrt[3]{abc}(a\plus{}b\plus{}c)\equal{}a\plus{}b\plus{}c$. Hence, proved.

you need to do that one more time to show that
a+b+c>sqrt(a)+sqrt(b)+sqrt(c)

Yes I know, but I assumed that giving solution like this would be understandable for everyone who is reading this post.  :)

+I was to lazy to copy/paste and change it abit  :P

Good solution, other than that last step that was missing but still understandable,

[hide="My Solution"]
We know that,
\[ \dfrac{a^2\plus{}b^2\plus{}c^2}{2}\ge \dfrac{ab\plus{}ac\plus{}bc}{2}\equal{}\dfrac{1}{2c}\plus{}\dfrac{1}{2b}\plus{}\dfrac{1}{2a}\equal{}\dfrac{1}{2}\left(\dfrac{1}{a}\plus{}\dfrac{1}{b}\plus{}\dfrac{1}{c}\right)\]

Therefore, it would suffice to show that,
\[ \dfrac{1}{2}\left(\dfrac{1}{a}\plus{}\dfrac{1}{b}\plus{}\dfrac{1}{c}\right)\ge \dfrac{1}{a\plus{}b}\plus{}\dfrac{1}{a\plus{}c}\plus{}\dfrac{1}{b\plus{}c}\]
I have two solutions for this,
[hide="First one"]
We know by the Cauchy-Schwarz inequality that,
\[ (a\plus{}b)\left(\dfrac{1}{a}\plus{}\dfrac{1}{b}\right)\ge 4\implies \dfrac{1}{a}\plus{}\dfrac{1}{b}\ge \dfrac{4}{a\plus{}b}\]
Therefore, $ \dfrac{1}{a}\plus{}\dfrac{1}{c}\ge \dfrac{4}{a\plus{}c}$ and $ \dfrac{1}{b}\plus{}\dfrac{1}{c}\ge \dfrac{4}{b\plus{}c}$.

Adding these up we have,
\[ \dfrac{2}{a}\plus{}\dfrac{2}{b}\plus{}\dfrac{2}{c}\ge \dfrac{4}{a\plus{}b}\plus{}\dfrac{4}{a\plus{}c}\plus{}\dfrac{4}{b\plus{}c}\implies \dfrac{1}{2}\left(\dfrac{1}{a}\plus{}\dfrac{1}{b}\plus{}\dfrac{1}{c}\right)\ge \dfrac{1}{a\plus{}b}\plus{}\dfrac{1}{a\plus{}c}\plus{}\dfrac{1}{b\plus{}c}\]
[/hide]
[hide="Second one"]
We know that $ x^2\plus{}y^2\plus{}z^2\ge xy\plus{}xz\plus{}yz$, now let $ x\equal{}\dfrac{1}{\sqrt{a}}, y\equal{}\dfrac{1}{\sqrt{b}}, z\equal{}\dfrac{1}{\sqrt{c}}$

So that we have,
\[ \dfrac{1}{2}\left(\dfrac{1}{a}\plus{}\dfrac{1}{b}\plus{}\dfrac{1}{c}\right)\ge \dfrac{1}{2}\left(\dfrac{1}{\sqrt{ab}}\plus{}\dfrac{1}{\sqrt{ac}}\plus{}\dfrac{1}{\sqrt{bc}}\right)\]
Also we know that $ \dfrac{1}{\sqrt{ab}}\ge \dfrac{2}{a\plus{}b}$, now just apply it cyclically and add it up and your done,
\[ \dfrac{1}{2}\left(\dfrac{1}{a}\plus{}\dfrac{1}{b}\plus{}\dfrac{1}{c}\right)\ge \dfrac{1}{a\plus{}b}\plus{}\dfrac{1}{a\plus{}c}\plus{}\dfrac{1}{b\plus{}c}\] :ninja:
[/hide]
[/hide]

Nice solutions tonypr.  :)

I especially liked "First One".


Jack and Jill take a total of $ \$80$ to an amusement park. Including admission, Jack spends $ \$24$ and Jill spends $ \$20$. At the end of the day, Jack has twice as much money as Jill. How many dollars did Jill have at the end of the day?

Jack brings $ x$ dollars
Jill brings $ y$ dollars.

So we know that $ x\plus{}y\equal{}80$
$ x\minus{}24\equal{}2(y\minus{}20)$
Thus, $ x\equal{}2y\minus{}16$
$ 3y\minus{}16\equal{}80 \implies y\equal{}32 \implies x\equal{}48$

At the end of the day, Jill has $ 32\minus{}20\equal{}12$ dollars.


geometry
circumcircle
rotation
geometric transformation
homothety
perpendicular bisector
geometry unsolved

Let $P_0$, $P_1$, $P_2$ be three points on the circumference of a circle with radius $1$, where $P_1P_2 = t < 2$. For each $i \ge 3$, define $P_i$ to be the centre of the circumcircle of $\triangle P_{i-1} P_{i-2} P_{i-3}$.

(1)  Prove that the points $P_1, P_5, P_9, P_{13},\cdots$ are collinear.

(2)  Let $x$ be the distance from $P_1$ to $P_{1001}$, and let $y$ be the distance from $P_{1001}$ to $P_{2001}$. Determine all values of $t$ for which $\sqrt[500]{ \frac xy}$ is an integer.

[color=blue][b](a)[/b][/color] $P_3$ is the circumcenter of the triangle $\triangle P_0P_1P_2$, hence the triangle $\triangle P_1P_2P_3$ is isosceles with $P_1P_3 = P_2P_3$. Let $\phi = \angle P_1P_2P_3 = \angle P_2P_1P_3$ be its base angle measure. $P_3, P_4$ both lie on the perpendicular bisector of the segment $P_1P_2$, hence $P_1P_2 \perp P_3P_4$. The triangle $\triangle P_2P_3P_4$ is also isosceles with $P_2P_4 = P_3P_4$, with the angle $\angle P_2P_4P_3 = 2 \angle P_2P_1P_3 = 2 \phi$ and the with base angles $\angle P_2P_3P_4 = \angle P_3P_2P_4 = 90^\circ - \phi$. By the similar reasoning, the isosceles triangle $\triangle P_3P_4P_5$ has the base angle $\angle P_3P_4P_5 = \angle P_4P_3P_5 = 90^\circ - (90^\circ - \phi) = \phi$. Thus the isosceles triangles $\triangle P_1P_2P_3 \sim \triangle P_3P_4P_5$ are similar and the base $P_3P_4$ of the later triangle is rotated $90^\circ$ with respect to the base $P_1P_2$ of the former triangle. Again, by similar reasoning, the isosceles triangles $\triangle P_3P_4P_5 \sim \triangle P_5P_6P_7$ are similar and the base $P_5P_6$ of the later triangle is rotated $90^\circ$ with respect to the base $P_3P_4$ of the former triangle. It follows that the isosceles triangles $\triangle P_1P_2P_3 \sim \triangle P_5P_6P_7$ have the corresponding sides parallel, hence, they are centrally similar and the lines $P_1P_5,\ P_2P_6,\ P_3P_7$ connecting their corresponding vertices concur at their homothety center H. Likewise, starting with the triangle $\triangle P_1P_2P_3$ instead of the triangle $\triangle P_0P_1P_2$, we can show that the lines $P_2P_6,\ P_3P_7,\ P_4P_8$ also concur, so that their concurrency point is identical with H. In general, the lines $P_{n}P_{n+4}$ concur at H for n > 0. This means that the points $P_n, P_{n+4}$ are collinear with H for any n > 0, in particular, the points $H, P_1, P_5, P_9, P_{13}, ...$ are all collinear.

[b]bump[/b]

Is there a solution to part (b)?

The official solutions can be found here

https://cms.math.ca/Competitions/OMC/


algebra polynomial calculus integration induction

(1) show that if f(x) is a polynomial of degree n with real coefficients, which takes integral values on a certain set of n+1 consecutive integers, then f(x) is integer-valued.
(2)show that if f(x) is an integer-valued polynomial of degree n, then n!f(x) is a polynomial with integral coefficients.
(3)suppose that f(x) is an integer-valued polynomial of degree n and that g=g.c.d(f(0),f(1),,,,,,,,,f(n)). show that g divides f(k) for all integers k.

[hide="The relevant lemma"] [b]Lemma:[/b]  $ f(x)$ is an integer-valued polynomial of degree $ n$ if and only if it can be written in the form

$ f(x) \equal{} a_n {x \choose n} \plus{} a_{n\minus{}1} {x \choose n\minus{}1} \plus{} ... \plus{} a_0 {x \choose 0}$

for integers $ a_n, a_{n\minus{}1}, .... a_0$. [/hide]
[hide="Hint to the proof"] The left direction is easy.  The right direction proceeds by induction; specifically, $ f(0) \in \mathbb{Z} \implies a_0 \in \mathbb{Z}$, and we then induct on each coefficient in turn. [/hide]


prove that all positive integers except the powers of 2
can be written as the sum of at least two consecutive integers

[hide]All symbols in the sequel are denoting integer numbers; moreover, we look for consecutive $ {\bf positive}$ integers.
Let $ n \equal{} 2^{a}b$, $ a \geq 0$, $ b \geq 1$, $ b$ odd. We want to have $ n \equal{} (m\plus{}1) \plus{} (m\plus{}2) \plus{} \cdots \plus{} (m\plus{}k)$, with $ m \geq 0$ and $ k \geq 2$, hence $ k(2m \plus{} k \plus{} 1) \equal{} 2^{a\plus{}1}b$.
If $ b \equal{} 1$, it follows $ k \equal{} 2^{\alpha}$, with $ 1 \leq \alpha \leq a\plus{}1$, but then $ 2m \plus{} k \plus{} 1 > 1$ is odd, so there are no solutions (notice that if we consider just consecutive integers, not necessarily positive, we do have a solution, namely $ k \equal{} 2n$ and $ m \equal{} \minus{}n$).
If $ b \geq 2^{a\plus{}1}\plus{}1$, then take $ m \equal{} \frac {1} {2} (b \minus{} 2^{a\plus{}1} \minus{} 1)$ and $ k \equal{} 2^{a\plus{}1}$.
If $ b \leq 2^{a\plus{}1}\minus{}1$, then take $ m \equal{} \frac {1} {2} (2^{a\plus{}1} \minus{} 1 \minus{} b)$ and $ k \equal{} b$.[/hide]


Given that $\frac{1}{4}$ of $\frac{1}{3}$ of $\frac{1}{2}$ of a number is 48, what is the number?

[hide]i think this is what you do:
Multiply 48 by the denominators. $48 \cdot 2 \cdot 3 \cdot 4=\boxed{1152}$[/hide]

its basically $\frac{1}{24}x = 48$ and thus you get that $x = 1152$

[hide]You multiply 48 by 4 then by 3 then by 2 and you get 1152.[/hide]

[hide]48 is just (1/4)(1/3)(1/2) of x or 1/24 of x. Solving for x gets x=[b]1152[/b].[/hide]


algebra
polynomial
linear algebra
matrix
linear algebra unsolved

If A, B, C, D are matrices in Mn(C) s.t. det(A) ≠ 0, det(C) ≠ 0 and A^k*B = C^k*D for each integer k ≥ 1 show that B = D

[hide]Prove that there exists a polinomial such that $P(A)=P(C)=I_n$ and $P(0)=0$.[hide] $P=-(\frac 1 {\det A}C_A-1)(\frac 1 {\det C}C_C-1)$ where $C_A$ is the characteristic polinomial of matrix $A$.[/hide] [/hide]

OK, here is the solution.

If P is the characteristic polynom of A then P(X) = det(A-X*I) and using the Hamilton-Cayley theorem A verifies the ecuation P(X) = 0,
so: $A^n - a_{n-1}A^{n-1} + ... + (-1)^na_0I_n = O_n$ where $a_0 = det(A) \neq 0$.
the same: $C^n - c_{n-1}C^{n-1} + ... + (-1)^nc_0I_n = O_n$ where $c_0 = det(C) \neq 0$.
We consider the polynom :
$Q = \frac{1}{(-1)^na_0(-1)^nc_0} * (X^n - a_{n-1}X^{n-1} + ... + (-1)^na_0) * (X^n - c_{n-1}X^{n-1} + ... + (-1)^nc_0) = 1 + Xb_1 + ... + X^{2n}b_{2n}$.
Of course $Q(A) = Q(C) = 0$, so $I_n + b_1A + ... b_{2n}A^{2n} = O_n$ and $I_n + b_1C + ... b_{2n}C^{2n} = O_n$.
Now multiply each by B and D, substract and get B = D.


Five numbers from a list of nine integers are $7,8,3,5,$ and $9$. The largest possible value of the median of all nine numbers in this list is

$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 7\qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9$

[hide]
The answer is E.
You make all the other numbers higher than nine to get the highest median and the median will be 9.[/hide]

It's actually 5 numbers, not 6 :P
[hide]
E, because you have 3,5,7,8,9.
There are 9 numbers, so add 4 numbers to the right side and get
3 5 7 8 9 10 11 12 13 15 17, orsomething like that.[/hide]

The original problem included 5 in the list, so the answer is D.

Sorry to revive this topic but there's no answer choice E for this thread or the contest resource.

[quote="pascal_1623"]It's actually 5 numbers, not 6 :P
[/quote]

I agree... The answer would turn out different if it was 6 numbers. since its five, it would be E (right?)

Got my AHSME 1996 out of storage.  The six number list given in the problem is:
7, 8, 3, 5, 9, and 5

The choices of answer are:
(A) 5  (B) 6  (C) 7  (D) 8  (E) 9

Then the answer would be D because if the rest of the numbers are bigger than 9 then there are 4 numbers above and below 9 so it would be D,8.

@apranav Pls dont revive old/thread threads to say unnecessary information such as then the answer is 8, when @calc rulz said that 8 years ago.


function
integration
real analysis
real analysis unsolved

For any integrable function $ f: \left[ a,b \right]\to \mathbb{R}: \int_{a}^{b}{f(x)e^{f(x)}\,dx}\ge \frac {1}{b \minus{} a}\left\{\int_{a}^{b}{f(x)\,dx}\cdot \int_{a}^{b}{e^{f(x)}\,dx} \right\}.$

Suppose a=0 and  b=1

Suppose $ f$ is continuous

For any $ x,y \in [0;1]$  we have $ f(x)e^{f(x)}\plus{}f(y)e^{f(y)}\minus{}f(x)e^{f(y)}\minus{}f(y)e^{f(x)}\geq 0$

this function in two variables is continous on $ K\equal{}[0;1]^2$

integrate on $ K$

With the theorem of my great friend Fubini

You will get the inequality.

[quote="Moubinool"] ...
Suppose $ f$ is continuous
...[/quote]
But the statement says "integrable function " ... :D
 :cool:

[quote="Kalle"]Where does Moubinool even use continuity? I don't see it ...[/quote]
He don't use it , but he says it ...  :rotfl:
 :cool:

[quote="Moubinool"]
...   my great friend Fubini  ...

[/quote]
 :rotfl:


Firstly i would like to say "Hi mathlinks!Hi, everyone!".
I don't know to much maths in English (i am Romanian), but  i have a question:Is it true that a perfect square's form is either 4k or 4k+1?
Thanks!
P.S:could you prove it?

Yes, it's true.   :lol:

[hide]Consider the 4 residues class $\bmod 4$.
If $a \equiv 1 \bmod 4 \Longrightarrow a^{2}\equiv 1$
If $a \equiv 2 \bmod 4 \Longrightarrow a^{2}\equiv 0$
If $a \equiv 3 \bmod 4 \Longrightarrow a^{2}\equiv 1$
If $a \equiv 0 \bmod 4 \Longrightarrow a^{2}\equiv 0$.[/hide]

Without as much "residues mod 4" stuff:

[hide]
Let $x^{2}$ be a perfect square, where x is an integer.

Consider the case when $x$ is even. Then $x$ can be expressed as two times another integer $n$, or $2n$.
Then $x^{2}=(2n)^{2}=4n^{2}$.
Let $k=n^{2}$; $k$ is also an integer. Then $x=4k$.

Now consider the case when $x$ is odd. Then $x$ can be expressed as twice an integer $n$ plus one, or $2n+1$.
Then $x^{2}=(2n+1)^{2}=4n^{2}+4n+1$.
Let $k=n^{2}+n$; again, $k$ is an integer. Then $x=4k+1$.

QED
[/hide]

[quote="ashwinrk_jain"]Even Squares are always in the form of $4k$ and $8k+1$. Now prove it.I have the solution but u should try.[/quote] No, it's never $8k+1$. That implies it's odd.  :huh:
They're always $4k$.

Though, odd squares are always $8k+1$, never $8k+5$.

[quote="ashwinrk_jain"]Even Squares are always in the form of $4k$ and $8k+1$. Now prove it.I have the solution but u should try.[/quote]
I believe you are refering to $n^{2}\equiv 1\pmod 8\ \forall n\equiv 1\pmod 2$.  The proof is very elementary.

Masoud Zargar

in want to say that every odd's square is always in the form of $8k+1$ now prove it.

[hide="Method 1"] Simply work $\in \mathbb{Z}_{8}$. [/hide]
[hide="Method 2"] Square $(2k+1)$. [/hide]
[hide="Method 3"] Reason from Euler's Totient Theorem. [/hide]

yes the solution given by t0rajir0u is exactly true.


If $ a,b,c >0$ prove $ \dfrac{a^2}{a^2\plus{}2bc}\plus{}\dfrac{b^2}{b^2\plus{}2ca}\plus{}\dfrac{c^2}{c^2\plus{}2ab} \ge 1 \ge \dfrac{bc}{a^2\plus{}2bc}\plus{}\dfrac{ca}{b^2\plus{}2ca}\plus{}\dfrac{ab}{c^2\plus{}2ab}$.

[quote="Raja Oktovin"]If $ a,b,c > 0$ prove $ \dfrac{a^2}{a^2 \plus{} 2bc} \plus{} \dfrac{b^2}{b^2 \plus{} 2ca} \plus{} \dfrac{c^2}{c^2 \plus{} 2ab} \ge 1 \ge \dfrac{bc}{a^2 \plus{} 2bc} \plus{} \dfrac{ca}{b^2 \plus{} 2ca} \plus{} \dfrac{ab}{c^2 \plus{} 2ab}$.[/quote]
[hide]
The left ineq is a simple application of andreescu inequality.

The right is actually the left but written in a different way:
$ \sum\frac {bc}{a^2 \plus{} 2bc} \le 1 \Longleftrightarrow \sum\frac {2bc}{a^2 \plus{} 2bc} \le 2 \Longleftrightarrow \sum\frac {2bc}{a^2 \plus{} 2bc} \minus{} 1 \le \minus{} 1 \Longleftrightarrow \sum\frac {a^2}{a^2 \plus{} 2bc} \ge 1$.[/hide]

[quote="R.Maths"]It's too easy , chebychev works[/quote][hide]Cauchy kills it, too. :)[/hide]

[quote="Raja Oktovin"]If $ a,b,c > 0$ prove $ \dfrac{a^2}{a^2 + 2bc} + \dfrac{b^2}{b^2 + 2ca} + \dfrac{c^2}{c^2 + 2ab} \ge 1 \ge \dfrac{bc}{a^2 + 2bc} + \dfrac{ca}{b^2 + 2ca} + \dfrac{ab}{c^2 + 2ab}$.[/quote]

In fact ,both left and right inequality are equivalent :
[hide="solution"]
First we prove the left :

By the Cauchy Schwartz inequality ,we have :
\[ LHS \geq \frac {(a + b + c)^2}{a^2 + b^2 + c^2 + 2(ab + bc + ca)} = \frac {(a + b + c)^2}{(a + b + c)^2} = 1\]

Right :

Observe that $ \frac {2bc}{a^2 + 2bc} = 1 - \frac {a^2}{a^2 +2bc}$.We need to prove that $ 2 \geq \sum_{cyc}{\frac {2bc}{a^2 + 2bc}}$ , it is equivalent with :
\[ 1 - \frac {a^2}{a^2 + 2bc } + 1 - \frac {b^2}{b^2 + 2ca } + 1 - \frac {c^2}{c^2 + 2ab} \leq 2 \iff \sum_{cyc}{\frac {a^2}{a^2 + 2bc } \geq 1}\]
[/hide]

[quote="enndb0x"][quote="Raja Oktovin"]If $ a,b,c > 0$ prove $ \dfrac{a^2}{a^2 \plus{} 2bc} \plus{} \dfrac{b^2}{b^2 \plus{} 2ca} \plus{} \dfrac{c^2}{c^2 \plus{} 2ab} \ge 1 \ge \dfrac{bc}{a^2 \plus{} 2bc} \plus{} \dfrac{ca}{b^2 \plus{} 2ca} \plus{} \dfrac{ab}{c^2 \plus{} 2ab}$.[/quote]

In fact ,both left and right inequality are equivalent :
[/quote]

Look at my solution enndb0x :wink:

Very sorry Dimitris   , I didn't read it   :(  :oops:


Three tokens $A$, $B$, $C$ are, each one in a vertex of an equilateral triangle of side $n$. Its divided on equilateral triangles of side 1, such as it is shown in the figure for the case $n=3$

Initially, all the lines of the figure are painted blue. The tokens are moving along the lines painting them of red, following the next two rules:

[b](1) [/b]First $A$ moves, after that $B$ moves, and then $C$, by turns. On each turn, the token moves over exactly one line of one of the little triangles, form one side to the other.
[b](2)[/b] Non token moves over a line that is already painted red, but it can rest on one endpoint of a side that is already red, even if there is another token there waiting its turn.

Show that for every positive integer $n$ it is possible to paint red all the sides of the little triangles.

number the vertices of the small triangles from A, B, and C separately in the same sense,i.e., from A towards B vertex is numbered 2 then that towards C is numbered 3 and so on , then for B first towards C and from C first towards A...
then write the moves for one point only say from A and for each move note down the move in all the three forms example for a move from 6,2,8 to 9,3,4 as the moves for A 6-9 for B 2-3 and for C 8-4 so you get three sequence of moves make sure no move is common in the sequences
then move only to vertices not previously visited and donot move to the end points in the middle of the sequence
so there are three choices at each vertex at worst only two choices have already been cancelled as they have occured in the  sequence of moves according to numbering from B or C
so anyway there is one choice remaining at each point
continue making moves until you have made n(n+1)/2 moves
finally repeat the sequence of moves generated from A's numbering and repeat it from point B and C
so no moves overlap for the way the sequence has been chosen so no edge is coloured twice
(for if we end up in the same vertex again after a sequence of moves then it must have one choice of going out if it doesnot have 1 means its been visited by at least one of the other tokens but if any token has come in then it must have gone out so two choices gone and this current token has also been there before so more two choices have gone so there is no way left to go to the vertex again, contradiction)
and as there were n(n+1)/2 moves from each starting point all the edges [3(n(n+1)/2] have been coloured red
so it is possible to colour all the edges for any number n


The positive inetegrs are colored into 2 colors.
Prove there are even integers a_1, a_2.. s.t.
a_1, (a_1+a_2)/2, a_2, (a_2+a_3)/2... are the same color

I forgot to say don't use van der Waerden!

Let the colors be r (red) and b (blue).

Let's prove a slightly differrent thing, which can be used to show this. I intend to show that we can find $\{a_i\}_{i\ge 1}$ with the same parity s.t. $a_i$ and $\frac {a_i+a_{i+1}}2$ all have the same color.

Assume no such sequence exists. Then each sequence ends with an "endpoint", i.e. a sequence of red points ends with $x_r$ (the index shows it's red) s.t. for all $y_r>x_r$ we have $(2y_r-x_r)_b$, There are red odd points and red even points, as well as blue odd and even points, or the property we want to show would follow trivially. If we start a sequence with an odd (even) point, then the endpoint would be odd (even), so we have red&blue, odd&even endpoints, in all possible combinations.

Let $M_r,\ M_b$ be the sets of red and blue points respectively, each chosen from a convenient point upwards (not necessarily from the beginning of the naturals), and let $x_r,\ y_b$ be a red and a blue endpoint respectively, having the same parity. In this case $2M_r-x_r\subseteq M_b,\ 2M_b-y_b\subseteq M_r$. Then there's no pairs of points $x'_r,\ y'_b$ s.t. $2x'r-x_r=2y'_b-y_b$, so we can't have $x'_r-y'_b=t=\frac {x_r-y_b}2$. Now take a progression with common difference $t$. If it contains a blue point, then all the points following it are blue, because a red point can't follow a blue point, and we're done, and if it contains only red points (starting from a threshold), then we are done again.

I have to say again that here we don't work with the actual sets, but the sets $\setminus [0,n_0]$ for some $n_0\in \mathbb N$

Now if we do this sort of thing with the even numbers instead of the naturals, replace the words "odd" and "even" with "$M_4+2$" and "$M_4$", we find what we want ($M_4$ means multiple of $4$).

Since I'm pretty sure there must be something wrong with the whole thing, I'm waiting for your reply, iura :).

Well, Iura, Van der Waerden theorem does not hold for monochromatic sequences with infinite length.

Pierre.

But maybe there's some other way of using it..


hm. i thinks its e.... whats the answer? can u pm me if you dont want to post it?

oh I see. I missed the inside X so that's why I couldn't get any formula to work. Great problem.

No it is E since you don’t know that the angle is right

Can't you just say that there are three segments, one with length 8 and two with sum 8, and same thing with 6 to get 16+12 = 28?
Or does that not work because of the fact that we don't know the angle?

It is E since we can't assume that the angles are right. And what question is this from what test?

[quote=PlaneGod]Alternatively, (in AIME15's diagram) if you shifted XZ to the right so it lines up with BY and you shifted XY down so it lined up with CZ, you would get a rectangle with perimeter 28.[/quote]

12 year bump? Really?

You can't assume that $X$ halfway between lines $AB$ and $CZ$

[quote=WyattB]What is it with these bumps lately :mad:
EEeeEEEEE[/quote]

how do they even find these old posts????????????

It is in the search function. I would assume they search AMC 8 problems, and then do the oldest posts first, so yeah.

[quote=PlaneGod]Alternatively, (in AIME15's diagram) if you shifted XZ to the right so it lines up with BY and you shifted XY down so it lined up with CZ, you would get a rectangle with perimeter 28.[/quote]

wow this is really smart

[quote=sky1983]$$\text{perimeter} = 8 + 6 + 8 + 6 = \boxed{28}$$[/quote]

hello, that is not true because we don't know if the lines are perpendicular

[quote=AIME15][asy]draw((0,0)--(0,6)--(8,6)--(8,3)--(4,3)--(4,0)--cycle);
label("6",(0,3),W);
label("8",(4,6),N);[/asy]

Given that all angles shown are marked, the perimeter of the polygon shown is

\[ \textbf{(A)}\ 14 \qquad
\textbf{(B)}\ 20 \qquad
\textbf{(C)}\ 28 \qquad
\textbf{(D)}\ 48 \qquad
\textbf{(E)}\ \text{cannot be determined from the information given} \qquad
\][/quote]

I'm not sure what the "Given that all angles shown are marked" means but yeah assuming right angles it's D if not it's E

it's possible that by marked they meant they are all the same, like how when sides of a shape are the same there are dashes through them.

[quote=benchkodakshark][quote=hh99754539]Because we do not know any of the angles, we cannot assume it is right.[/quote]

the angles look right to me idk[/quote]

But it isn't marked. It could be 90.0000000000000000000000000001°. Then, it is obtuse. Since the angle is 0.0000000000000000000000000001° greater than a right angle, it looks like a right angle, but it is greater than a right angle, so it is obtuse.

I think it's safe to assume that this is [url=https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_13]1986 AJHSME Problem #13[/url] just diagrammed differently, due to the answer choices and lengths in the diagram being the same, in which case, the angles are all right angles, and the problem was just transcribed incorrectly. This is technically not provable, but it is by far the most likely theory.

[quote=AIME15][hide="Not quite..."]
[asy]draw((0,0)--(0,6)--(8,6)--(8,3)--(4,3)--(4,0)--cycle);
label("6",(0,3),W);
label("8",(4,6),N);
label("A",(0,6),NW);
label("B",(8,6),NE);
label("C",(0,0),SW);
label("X",(4,3),NW);
label("Y",(8,3),E);
label("Z",(4,0),S);[/asy]

Note that $ BY\plus{}XZ\equal{}AC$, and $ CZ\plus{}XY\equal{}AB$.  Thus, the perimeter of polygon $ ABYXZC$ is $ 2(8\plus{}6)\equal{}28$, or $ \boxed{\textbf{(C)}}$.
[/hide][/quote]

Also the original poster of the problem confirmed that the answer they were looking for was not E.

bruh if you go to wiki they're marked as right angles


3, 6, 9, 12... are some multiples of three. How many muiltiples of three are there between 10 and 226? Don't tell me, give me a hint.

[hide=" hint"]try dividing 226 by 3 and then........[hide="click if you have to"] subtracting how many multiples of 3 are there from 1 to 10[/hide][/hide]

[hide="hint"]Try finding the smallest multiple greater than 10 and the largest multiple less than 226. [/hide]

[hide="hint"]
get the numbers between 10 and 226, then divide the quotient by 3.[/hide]


AMC
USA(J)MO
USAMO
AIME
probability
AMC 10
AMC 12

The first selection for the USAMO is from AMC 12A and AMC 12B. From those they determine the floor value. However, it states that they then take top contestants who took meet the floor value the AMC 10A or 10B. What about 9th and 10th graders taking the AMC 12A or B? Do they have to make it through the first selection process if take the AMC 12?

I don't really understand that. For the second selection do they ONLY look at AMC 10 contestants?!?!

For one thing, (as just about everyone who's been involved in USAMO coaching will tell you) the USAMO selection process is not set in stone. Not only that, but it's likely that there'll be changes of some sort every year.

Well I am in 10th grade and I am wondering which AMC test I should take based upon which one I would have a better chance in qualifying to USAMO with since I am hoping on qualifying using the floor value system.

There are no strategic disadvantages to taking the AMC 12, except that you have to take a harder test.  It would be nice if the AMC would post their descriptions in nice, simple terms, but my first sentence is all you need to know.

Thank you, this is what I wanted to know, since I didn't wanna not be eligible for the 2nd selection for the USAMO if I took the AMC 12. :)

1.  I just looked again at my diary of the selection process from 2004.  I can confirm that  _all_  9th and 10th grade students with an AIME score greater than or equal to 8 were invited to the USAMO.  Some had qualified for the AIME through the AMC 12 and some had qualified for the AIME from the AMC 10.

2.  "It would be nice if the AMC would post their descriptions in nice, simple terms"     I agree.  However, I can count at least 18 different paths that lead to the USAMO, and simultaneously we are trying to satisfy several selection criteria with slightly different goals.  Before 2002, the process was demonstrably biased against 11th and 12th graders.  The present rule attempts to provide a fair balance among all 18 paths and the selection criteria.
If you have an alternative wording of the rules, I am willing to present it to the Committee on the American Mathematics Competitions  for consideration.  Rules changes won't be implemented before 2006.

3.  I believe the evidence points strongly in favor of qualifying for the USAMO by taking the AMC 12.  Last year, more 9th and 10th graders qualified for the USAMO by taking the AMC 12 than did 9th and 10th graders taking the AMC 10.   My records indicate 126 students at 10th grade and below who had a an 8 or above on the AIME.   Of these, 40 qualified via the AMC 10 and 86 qualified via the AMC 12.

Steve Dunbar
AMC Director

[quote="AMC web site"]The second selection of USAMO participants will be from the highest USAMO indices among students who took the AMC 10A or AMC 10B and the AIME, and got an AIME score at least as high as the floor value.[/quote]
I would recommend changing "who took the AMC 10A or AMC 10B and the AIME" to "who are in 10th grade or below".

I take it back -- the wording itself is generally not confusing (although the statement Ravi points to is factually incorrect, which should be changed).  It does seem to me, however, that you could re-arrange your materials in a more helpful way.  In particular, a FAQ page would make a lot of the information more accessible.  I think this forum and the questions you've answered here over and over again would be a good start for questions to be in the FAQ.  In particular, a page that laid out (correctly) all the routes that one can take to get to USAMO would be extremely helpful.  As it is, a student would have to look at both the AIME and USAMO pages and would be left with some false information (as well as not be informed about "alternative" methods of qualification, such as via USAMTS).

[quote]3. I believe the evidence points strongly in favor of qualifying for the USAMO by taking the AMC 12. Last year, more 9th and 10th graders qualified for the USAMO by taking the AMC 12 than did 9th and 10th graders taking the AMC 10. My records indicate 126 students at 10th grade and below who had a an 8 or above on the AIME. Of these, 40 qualified via the AMC 10 and 86 qualified via the AMC 12. [/quote]

Perhaps I am stating  what you may already know, but this  does not  suggest, i would think,  that a given student has more probablity of  of qualifying for  USAMO if he takes AMC-12 (vs AMC 10).  From the students I know/knew,   a   very good student  is more likely to take AMC-12, (just for the fun of it.) ... (That is, if you offer a choice  to  the shudents who are likely to qualify for USAMO  many may choose to take AMC-12, while those who are not   would more likely to prefer  to stay with AMC-10....Am I missing something?

In any case, FWIW i'd say, who cares .. take the  exam which you think would be more fun.

[quote="AMCDirector"]1.  I just looked again at my diary of the selection process from 2004.  I can confirm that  _all_  9th and 10th grade students with an AIME score greater than or equal to 8 were invited to the USAMO.  Some had qualified for the AIME through the AMC 12 and some had qualified for the AIME from the AMC 10.
[/quote]

Wait, wasn't the floor a 7 last year?

or take both. thats what im doing. but im taking amc10 just for the chance at a perfect score and amc12 for real practice/experience. either way, like its been said, 10th graders and below just have to make the aime floor to qualify for usamo, regardless of the amc they took.

Should I take both 10 and 12 JUST in case there's going to be a change this year?

i'm pretty sure the current rules will apply to this year's usamo selection. they can't change it on such short notice, cant they? if you aren't shooting for a perfect amc10, there really isn't an advantage to taking the amc10 (other than an easier, less torturing test) so you could just take the amc12 to be safe, maybe? i dont really know.

The advice I give to the mathletes on my team is to take both the AMC 10 and the AMC 12 during the years when they are eligible for both. (That is, to begin taking the AMC 12 during the same school year when they learn precalculus.) Correspondingly, my advice for 11th and 12th graders is to take the AMC 12 on both the "A" date and the "B" date.


Quadrilateral ABCD is circumscribed about a circle. Three sides of the quadrilateral are 9, 17, and 12. Find the length of side AD.

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

pair A,B,C,D;

draw(unitcircle);

B=(-1.8,-1);
C=(1.18,-1);
D=(0.9,0.7);
A=(-0.4,1.25);

draw (A--B--C--D--A);

label("A",A,NNW);
label("B",B,WSW);
label("C",C,ESE);
label("D",D,NE);

label("$12$",midpoint(A--B),NW);
label("$17$",midpoint(B--C),S);
label("$9$",midpoint(C--D),NE);[/asy]

For these types of quadrilaterals (called cyclic, right?), the sum of two opposite sides is equal to the sum of the other two sides.

AD+17=9+12
AD=4

sorry to revive but is there a proof for that assertion?


conics
parabola
vector
real analysis
real analysis unsolved

I signed up here because I am somewhat inexperienced with differential equations and have an apparently simple problem:

You have a parabola described by the equation y=ax^2.

Now, find the curve with constant distance d to this parabola.

The distance is orthogonal to the tangents of both curves.

Good luck finding the solution!

Let $\mathcal{C}$ be a curve and $\mathcal{\gamma}$ be the curve with constant distance $d$ to $\mathcal{C}$.
Let $M$ be a point on $\mathcal{C}$ and $\overrightarrow{n}$ the unitary normal vector in $M$. Let $\epsilon\in\{\pm1\}$. Then $P$ s.t. $\overrightarrow{OP}=\overrightarrow{OM}+\epsilon{d\overrightarrow{n}}$ is a point on $\mathcal{\gamma}$.  The tangents in $P$ and in $M$ have the same direction: $\frac{\overrightarrow{dP}}{ds}=(1-\epsilon\frac{d}{R})\overrightarrow{t}$ where $s$ is the elementary length of $\mathcal{C}$, $\overrightarrow{t}$ its unitary tangent vector in $M$ and $R$ the radius of curvature in $M$ ($qed$ except if $d=\epsilon{R}$ or $R=0$).
Here $\overrightarrow{OM}=[u,au^{2}]$ therefore $\overrightarrow{t}=[\frac{1}{\sqrt(1+4a^{2}u^{2}},\frac{2au}{\sqrt(1+4a^{2}u^{2}}]$ and $\overrightarrow{n}=[\frac{-2au}{\sqrt(1+4a^{2}u^{2}},\frac{1}{\sqrt(1+4a^{2}u^{2}}]$. We deduce easily $\overrightarrow{OP}$.


superior algebra
superior algebra solved

i got this question from a recent test and dun know how to approach it, hope someone could kindly point me to the rite direction.

Prove that there is no line in the plane such that every point on the line is a constructible point.

thanks.

I will move this to superior algebra because it fits better there.

One approach: any line has a continuum as cardinality, but the set of constructible points is countable.
More constructive: if $a,b$ are constructible points on the line, then e.g. $\sqrt[3]{2}a+(1-\sqrt[3]{2})b$ is also on that line, but if it would be constructible, then $\sqrt[3]{2}$ would be.


What's wrong with a little GPML marathon?  It's just so depressing to see such a fun contest series' forum so deserted!  I'm sure we all know the rules of a marathon, if you solve the problem, post a new one.  For this one, individual questions (sprint and target, in other words) will be preferred, but if you have a nice team round or perhaps even relay question, go ahead and post away!

I'll start us off witha pretty easy one.

What is the remainder when $2^{6581}$ is divided by 7?

(Sprint round GPML-10401, 2003-2004 Season).

[quote="ashwinrk_jain"]remainder is 4.[/quote]

YOu should really post a solution on how you got that answer, and post the next problem as well.

Problem 2:

What is the remainder when $6^{343}+8^{343}$ is divided by $49$?

[quote="ashwinrk_jain"]please post complete solution.[/quote]

you're supposed to post a solution.

[hide="Problem 1 Solution"]
We want to find $2^{6581}\pmod{7}$

We know it will repeat every six terms.  How do we know this?  Well, the mod is 7.  So to find the number of terms it takes to repeat, we do $\phi{7}$.

$\phi{7}=6$ so it repeats every six terms.

Knowing that, we can subtract 6 from the exponent an unlimited number of times, to get:
$2^{5}\pmod{7}$
$32 \pmod{7}$
Answer: $\boxed{4}$

[/hide]

[quote="ashwinrk_jain"]remainder is $0$[/quote]

Post your solution. Stop telling people to post THEIR solutions. In a marathon and any other place for that matter, you are expected to show how you got your answer. Also, in a marathon you should post the next question after you show your solution.

Stop it Ashwin. I see you going around different forums and posting the answers only. Who cares if you are right or wrong, just post the solution. Oh, and hide your work if possible.

[hide="hint for 2"]Try writing $6$ as $7-1$ and $8$ as $7+1$[/hide]

sorry for not posting the whole solution because the whole solution is large. but ok i will submit it.
[hide]we can write $6^{3}43+8^{3}43$ as $-(1-7)^{3}43+(1+7)^{3}43$
Now we use binomial expansion $-(^{343}C_{0}-^{343}C_{1}7^{1}...(Multiple of 49))+(^{343}C_{0}+^{343}C_{1}7^{1}... Multiples of 49)$and also $^{343}C_{1}7$ is multiple of 49. Hence we can say that the remailnder is $-^{343}C_{0}+^{343}C_{0}$ and hence it is $0$.
[/hide]
My next question is
 In a triangle ABC, AD is an altitude and H is the orthocenter, Prove that
 $AH: DH = (tanB+tanC): tan A$

When you use ^a number in latex, if the number has more than 1 digit, use 6^{343} for $6^{343}$

Same for subscript.

For choose, you can do $_{123}C_{123}$ or $\binom{123}{123}$


Let (O) be the circumcircle of a triangle $\triangle ABC$. Let the line AO meet the circumcircle (O) at another point D different from A and let the tangent to the circumcircle at the point D meet the (extended) line BC at a point E. Let the line EO connecting the point E with the triangle circumcenter O meet the remaining 2 (extended) triangle sides AB, CA at points P, Q, respectively. Show that the segments PO = QO are equal.

A generalization:

[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=719079#719079[/url] :wink:


Hi guys. I have a question. What are Newton's Sums and what are they used for? I'm reading Volume 2 but I'm having some difficulty comprehending the topic. I'd appreciate it if you could help me out. Thanks.

I [i]know[/i] we've discussed this before -- try using the search feature (found in the upper-right part of the screen) to look for older discussions on it.  And if you still have questions, feel free to ask again.


if you could change your name to anything what would it be?

I guess my name would be like Peyton or someting.


Ρε παιδιά  ποιό είναι το max της παράστασης
a*cosθ+b*sinθ όπου a,b σταθερά :?:

[quote="r_boris"]Είναι τo $ \sqrt {a^2 \plus{} b^2}$[/quote]
Ναι σωστά   :wallbash:
(από την Bcs) τόση ώρα δεν το έβλεπα

$ a\cos \theta \plus{} b\sin \theta \equal{} \sqrt {\left(a\cos \theta \plus{} b\sin \theta\right)^2} \leq \sqrt {a^2 \plus{} b^2}\cdot\sqrt {\cos^2\theta \plus{} \sin^2\theta} \equal{} \sqrt {a^2 \plus{} b^2}$

Η ισότητα ισχύει όταν $ \frac{a}{\cos\theta} \equal{} \frac{b}{\sin\theta}\Leftrightarrow \frac{b}{a} \equal{} \tan\theta$

Ανάποδα η ανισότητα. :D  Θα πιάσει λόξυγγας τον Cauchy. :D  :D

Έλεγχος της ισότητας?  :maybe:

Έκανα edit το topic.

(Υ.Γ: Σιλουανέ με πρόλαβες!)

$ (a\cos \theta \plus{}b\sin \theta)^2\plus{}(a\sin \theta \minus{}b\cos \theta)^2\equal{}a^2\plus{}b^2$. Since $ (a\sin \theta \minus{}b\cos \theta)^2\geq 0$, we have $ (a\cos \theta \plus{}b\sin \theta)^2\leq a^2\plus{}b^2\Longleftrightarrow |a\cos \theta \plus{}b\sin \theta |\leq \sqrt{a^2\plus{}b^2}$.

$ a\cos \theta \plus{} b\sin \theta\equal{}\sqrt{a^2\plus{}b^2}\cdot\sin(x\plus{}\theta)$

όπου $ \tan{x}\equal{}\frac{a}{b}$

QED

Geometric solution:

$ a \cdot cos(\theta) \plus{} b \cdot \sin(\theta) \le |a| \cdot |cos(\theta)| \plus{} |b| \cdot |\sin(\theta)|$

so it all comes down to finding the max value of:

$ a \cdot cos(\theta) \plus{} b \cdot \sin(\theta)$ for $ a,b \ge 0$ and $ 0 \le \theta \le \frac {\pi}{2}$

let $ A(a,0), B(0,b), O(0,0)$ then $ ABO$ is a right trinagle (with $ AB \equal{} \sqrt {a^2 \plus{} b^2}$)

now it holds that:

$ a \cdot cos(\theta) \plus{} b \cdot sin(\theta) \equal{} a\frac {b}{AB} \plus{} b\frac {a}{AB} \equal{}$

$ \equal{} 2\frac {ab}{AB} \equal{} 2\frac {ab}{\sqrt {a^2 \plus{} b^2}} \le$ (from AM-GM) $ 2\frac {\frac {a^2 \plus{} b^2}{2}}{\sqrt {a^2 \plus{} b^2}} \equal{} \sqrt {a^2 \plus{} b^2}$

Let $ \overrightarrow {p}\equal{}(a,\ b),\ \overrightarrow {q}\equal{}(\cos \theta ,\ \sin \theta)$, we have $ a\cos \theta \plus{}b\sin \theta \equal{}|\overrightarrow {p}\parallel{}\overrightarrow {q}|\cos \alpha$
$ \leq\sqrt{a^2\plus{}b^2}$. The equality holds when two vectors are pararell in same direction.

[quote="Orfikoss"]$ a\cos \theta \plus{} b\sin \theta \equal{} \sqrt {\left(a\cos \theta \plus{} b\sin \theta\right)^2} \leq \sqrt {a^2 \plus{} b^2}\cdot\sqrt {\cos^2\theta \plus{} \sin^2\theta} \equal{} \sqrt {a^2 \plus{} b^2}$

Η ισότητα ισχύει όταν $ \frac {a}{\cos\theta} \equal{} \frac {b}{\sin\theta}\Leftrightarrow \frac {b}{a} \equal{} \tan\theta$[/quote]

Vasika i prwti isotita pou egrapses einai lathos.....Den ksereis an a*cost+b*sint einai thetiko 'h arnitiko , kai fusika otan grafeis afto to athrisma san tetragono is tin tetragoniki riza , exeis apoliti timi otan ta vgazeis

Το ειπε και ο kunny οτι $ |a\cos \theta \plus{} b\sin \theta |\leq \sqrt {a^2 \plus{} b^2}$ το οποιο ομως γραφεται

$ \minus{}\sqrt {a^2 \plus{} b^2} \leq a\cos \theta \plus{} b\sin \theta \leq \sqrt {a^2 \plus{} b^2}$ οποτε παλι το max ειναι το $ \sqrt {a^2 \plus{} b^2}$


The area of a triangle is 176 $ \text{cm}^2$. What is the number of centimeters in the length of the base given that the height is 22 cm?

$ \frac{bh}{2}\equal{}11b\equal{}176 \implies b\equal{}16$

$ A \equal{} \frac{bh}{2}$

$ 176 \equal{} \frac{22b}{2}$

$ 352 \equal{} 22b$

$ \boxed{16} \equal{} b$

edit: beaten

yes this is a mistake. i reported an error.


algebra
polynomial
number theory unsolved
number theory

How many solutions are there to the congruence $ x^{p \minus{} 1} \equal{} 1(\mod p^2)$, where $ x\in \{0,1,...,p^2\minus{}1\}$? Can you specify which intervals they are in?

I think, that p - is prime.
If $ p|x$ then $ x^p\not \equal{} 1.$
If $ p\not |x$, then it equavalent to $ x^p \equal{} x\mod p$. But $ (x \plus{} yp)^p \minus{} x \minus{} yp \equal{} x^p \minus{} x \minus{} yp\mod p^2$.
Therefore for any $ z \equal{} 1,2,...,p \minus{} 1$ we get solution $ x \equal{} z \plus{} yp$, were $ y \equal{} \frac {z^{p} \minus{}z}{p}\mod p$.
It give $ p \minus{} 1$ solutions.

It is special case of Polynomial congruence .
If $ P(x)\equiv 0 (\mod p)$ has k solution $ x_1,..,x_k$  and
$ \gcd (P'(x_i,p) \equal{} 1,\forall i \equal{} 1,...,k$ then the equation
$ P(x)\equiv 0 (mod p^m)$ has exactly k solution for all $ m\in N$

[quote="Rust"] $ x \equal{} z \plus{} yp$, were $ y \equal{} \frac {z^{p \minus{} 1} \minus{} 1}{p}\mod p$.
[/quote]

Then $ x\equal{}z\plus{}z^{p\minus{}1}\minus{}1$? I think it does not satisfy!

$ x\equal{}z^p\mod p^2,z\equal{}1,2,...,p\minus{}1.$
Really $ x^p\equal{}(z\plus{}(z^p\minus{}z))^p\equal{}z^p\mod p^2\equal{}x.$

[quote="nguyentrang"]Can you specify which intervals they are in?[/quote]

It should be a deep hard question.

Can it happen that $ a^{p\minus{}1}\equal{}1(\mod p^2)$ and $ (a\plus{}1)^{p\minus{}1}\equal{}1(\mod p^2)$ at the same time for some $ a$?

Yes. Let $ a\equal{}\frac{p^2\minus{}1}{2}.$ and $ 2^{p\minus{}1}\equal{}1\mod p^2$, then $ 2^pa^p\equal{}(p^2\minus{}1)^p\equal{}\minus{}1\mod p^2$, $ 2^p(a\plus{}1)^p\equal{}(p\plus{}1)^p\equal{}1\mod p^2.$ It give $ a^{p\minus{}1}\equal{}1\mod p^2$ and $ (a\plus{}1)^{p\minus{}1}\equal{}1\mod p^2.$


logarithms
geometric sequence
algebra unsolved
algebra

Prove that a finite number of geometric sequences which have only natural terms don't include all the natual numbers.

Thanks for help. :(

The first of all we show that if a geometric sequence has only natural number then it multiple is a natural .
Suppose that $ x_{n + 1} = dx_n$
So $ x_n = x_1.d^{n - 1}$ .
If d is a rational and not a integer then : $ d = \frac {p}{q}$ where $ \gcd(p,q) = 1$ and $ q > 1$
From the condition we must have $ q^n|x_1,\forall n\in N$  this contradition  if we chose n large enough.
Now consider your problem :
Call finite sequence is ${ \{c_1d_1^n\},...,\{c_kd_k^n}$
Because has only finite term so $ p (c_{1}d_{1}...c_{k}.d_{k})$ has only finite term where $ p(n)$ is the set of prime divisor of n  .
It mean that has a prime p such that $ p\not \in p(c_1d_1...c_kd_k)$
Easy to check has no term in the union of these sequence divisible by p so p is not in each of this sequence.
Problem claim .

Or just note that a finite union of geometric sequences contains only $ O(\log x)$ elements below $ x$.


AMC
AIME
search
AMC 12
AMC 10
AMC 8
USA(J)MO

Can the schools request another AMC school winner pin?
'cause i accidentally lost it, not really lost, i mean, it just disappeared.

Anyway guys, can I ask my math teacher to request another one from the MAA? It's really important to me, since i had a goal to have a pin each year to add into my collection.

I'm so dead!!!!!   :(

I have no idea but...
Hrm, they've been sent out to schools already? I better go and ask my teacher about it. I'll have an AMC 8, AMC 10, and an AMC 12 pin...yay.

Yes, certainly your teacher may request a replacement pin from us, we fulfill this kind of request frequently.  Please have your teacher contact the AMC office directly, with all the relevant information, for instance, contest manager name, school address including city and state and a contact phone number or email.  Having the CEEB number is helpful, it's how we record information relevant to the school registration, but it is not essential, we can find everything from the information above.

Yes, the school winner pins (and medals, and certificates, everything) went with the school report mailed about two or three weeks after the contest dates.  If you got your AIME, everything was in the same envelope, so you know it had to have arrived.

Steve Dunbar
Director, American Mathematics Competitions

hmmm i was my school winner and i havn't got a pin yet.

indianamath-

Either have I (for this year)

But when I was at your school last year, Mrs. Arnold gave me one at the end of the school year..at like the ceremonies..

So close to a pin in Amc 8... :o
To clarify, it is for the highest score in the school, not a section?

Yeah, it's for the school. Just got my pin today. (collected all 3 pins, woot.)
So it's in the same evenlope as the AIME...so I should have gotten a certificate too. I'll tell my math teacher to search for it.

Yeah...it's pretty sweet.
Now I should get all three medals too. Gold, Silver, and Bronze. To continue my domination in math contests at my school.  :D

Hmm.... I remember reading about pins somewhere...
Is it too late to get pins for winning in 05, 06 and 07? :)
I guess my teacher never bothered to distrubute them, or never got them, or something.
Also, are there separate pins for the 12A and 12B?

@davidyko: I got second on the AMC 8, but I got a certificate and not a medal, so I'm not certain that we will be getting medals this year...

does anyone know how much the pin replacement cost and how long does it take to arrive?

[quote="kstan013"][quote="kstan013"]if i was the only one in my school to take it, do I get a school winner pin?[/quote]Anyone know????[/quote]
yeah same here...
i won in amc12 amc10 amc8 aime and will win in usamo
but thats cause im only person who took... XD

Read indianamath's post. Or maybe you weren't asking. I dunno.

then why haven't i gotten one yet? does it come in the mail?

It should've come with the AIME test.
And you don't get pins for school winner for the AIME, or the USAMO, if I recall correctly.


inequalities
search
real analysis
real analysis unsolved

let $ a_{1}, a_{2}, a_{3}, a_{4}, a_{5},\ldots$ be a sequence of positive real numbers $ \mathbb R^{\plus{}}$

prove that :D

$ \sum_{n\equal{}1}^{\infty}\left(a_{1}\cdot a_{2}\cdot a_{3}\cdot a_{4}\cdot a_{5}\;\cdots\;a_{n}\right)^{\frac{1}{n}}\;\le\; e\sum_{n\equal{}1}^{\infty}a_{n}$

[ [b]note:[/b] it suffices to say that the inequality will [u]not[/u] hold if
$ e$ is replaced by a smaller number... :| ( here, $ e$ is $ 2.718281828\ldots$ ) ]

Carlman inegality it was dicussed on the forum

Actually, it's Carl[b]e[/b]man: http://www.mathlinks.ro/viewtopic.php?search_id=1014824041&t=5016


An ellipsoid - basically a stretched sphere.

[quote="binomial_4eva"]what does the equation $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1$ describe?[/quote]

well $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}$ describes an ellipse so adding z, then it would be the 3-D version of an ellipse (which thanks to The Zuton Force i learned that it is called an ellipsoid)


Let $C_1,...,C_n$ some closed convex plane sets, each having diameter at most 1 and such that for all $1\leq i<j\leq n$ the union of the sets $C_i,C_{i+1},...,C_j$ is convex. Is it true that the union of these sets cannot contain a disk of diameter greater than $\sqrt{3}$?

Hmm... How about $C_j$ being the sectors of the disk of radius $1$ of opening $\frac{2\pi}n$ ($n\geq 6$) enumerated counterclockwise? Seems like they satisfy the conditions of the problem and their union is the unit disk of diameter $2>\sqrt 3$. This makes me wonder whether there is some upper bound on the diameter of the disk contained in the union at all...

The union of all sectors except 1 is not convex.

Oops!  :blush: You're right. I should be more attentive...


If $ x,y,z>0$ show that :

$ \frac{1}{x^2\plus{}yz}\plus{}\frac{1}{y^2\plus{}zx}\plus{}\frac{1}{z^2\plus{}xy}\le \frac{1}{2}\left(\frac{1}{xy}\plus{}\frac{1}{yz}\plus{}\frac{1}{zx}\right)$

$ \sum (\frac{1}{x^2\plus{}yz}\minus{} \frac{1}{2yz}) \ge 0$
$ \sum (\frac{(yz\minus{}x^2)(y\plus{}z)}{yz(x^2\plus{}yz)(y\plus{}z)} \ge 0$
We can assume $ a \ge y \ge z$
And $ (yz\minus{}x^2)(y\plus{}z) \le (zx\minus{}y^2)(z\plus{}x) \le (xy\minus{}z^2)(x\plus{}y)$
and $ \frac{1}{yz(x^2\plus{}yz)(y\plus{}z)} \le ...$
By chebyshev,we complete.
Equality occur if and only a=b=c.

[quote="secrets"]$ \sum (\frac {1}{x^2 \plus{} yz} \minus{} \frac {1}{2yz}) \ge 0$
$ \sum (\frac {(yz \minus{} x^2)(y \plus{} z)}{yz(x^2 \plus{} yz)(y \plus{} z)} \ge 0$
We can assume $ a \ge y \ge z$
And $ (yz \minus{} x^2)(y \plus{} z) \le (zx \minus{} y^2)(z \plus{} x) \le (xy \minus{} z^2)(x \plus{} y)$
and $ \frac {1}{yz(x^2 \plus{} yz)(y \plus{} z)} \le ...$
By chebyshev,we complete.
Equality occur if and only a=b=c.[/quote]

Nice . My solution use only $ MA\minus{}MG$ .

This problem is old and very easy,my friend
$ \frac {1}{x^2 \plus{} yz} \le \frac {1}{2\sqrt {xy.xz}} \le \frac {1}{4}(\frac {1}{xy} \plus{} \frac {1}{xz})$
and similar,we get
$ \frac {1}{x^2 \plus{} yz} \plus{} \frac {1}{y^2 \plus{} zx} \plus{} \frac {1}{z^2 \plus{} xy} \le \frac {1}{2}( \frac {1}{xy} \plus{} \frac {1}{yz} \plus{} \frac {1}{zy} )$
Equality occur if and only a=b=c.


Let $A$ be a set with $2n$ elements, and let $A_1, A_2...,A_m$ be subsets of $A$e ach one with n elements. Find the greatest possible m, such that it is possible to select these $m$ subsets in such a way that the intersection of any 3 of them has at most one element.

Call the elements $1, 2, ..., 2n$. Suppose two elements $i \neq j$ both belong to $A_1, A_2$. Then they can't belong (together) to any of the other subsets.

Note that the $m$ subsets have $mn$ elements together, possibly some repeated. Each element appears $M_1, ..., M_{2n}$ times, with $M_1+...+M_{2n}=mn$. Now element $i$ appears in 3 subsets in $ \frac {M_i(M_i-1)(M_i-2)}{6}$ ways. There are $ \frac {m(m-1)(m-2)}{6}$ ways to appear in 3 sets in total, so we must clearly have

\[M_1(M_1-1)(M_1-2)+...+M_{2n}(M_{2n}-1)(M_{2n}-2) \leq m(m-1)(m-2).\]

It is clear that if $x>y+1$, given $x(x-1)(x-2)+y(y-1)(y-2)$, we can transform it into $(x-1)(x-2)(x-3)+(y+1)y(y-1)=x(x-1)(x-2) +y(y-1)(y-2) -3(x-1)(x-2) +3y(y-1)$ which reduces the sum. So the LHS of the above equation is minimized when the $M_i$ differ by at most 1. So we can take the euclidean representation of $mn$,
\[2nk+r=mn \\
n(2k-m)=r \]
So $r$ will be $0$ or $n$ depending on whether $m$ is even or odd, respectively. This yields, for $m=2a$, $mn=2an, M_i=a$, so

\[2na(a-1)(a-2) \leq 2a(2a-1)(2a-2) \\
na-2n \leq 4a-2 \\
a(n-4) \leq 2n-2 \\
a \leq \frac {2n-2}{n-4} \\
m \leq 4 \frac {n-1}{n-4}.\]
For $m=2a+1$, we have $mn=(2a+1)n, M_1, ..., M_n=a, M_{n+1}, ..., M_{2n}=a+1.$ This yields
\[n(a(a-1)(a-2)+(a+1)a(a-1)) \leq (2a+1)2a(2a-1) \\
na(a-1)(2a-1) \leq (2a+1)2a(2a-1) \\
n(a-1) \leq (2a+1)2 \\
a \leq \frac {n+2}{n-4} \\
m \leq \frac {3n-2}{n-4}.\]

Well, with these one can find that for $n>16$, the limit is $4$, which is easily achievable taking $(1, 2, ..., n), (n+1, ..., 2n), (1, 3, ..., 2n-1), (2, 4, ..., 2n)$. For smaller $n$ I'd have to check by hand if $5$ or $6$ works...

For $n=3,4,5$ the value $m=6$ works... :?

one can also get a different bound that might help for smaller $n$ as follows:
the condition given is also equivalent to saying that for any pair of elements the numer of sets containing that pair is atmost $2$.so again counting triples $(a,b,A)$ where $a\neq b,a,b \in A$ is on the one hand $mn(n-1)$;on the other hand counting the same number over all the pairs we get that this number is also $\displaystyle\sum_{a\neq b} n_{a,b}$ where $n_{a,b}$ is the number of sets containing $a,b$ and by assumption $n_{a,b}\leq 2$ so that this sum is atmost $2(2n)(2n-1)$,so $m\leq \frac{2(2n)(2n-1)}{n(n-1)}=8+\frac{4}{n-1}$ for smaller n($n \geq 6$) the bound is uniformly $8$.moreover in case of $n=3$,the bound specified here is $10$ and this can be achieved as follows:
 $\{1,2,3\},\{1,3,4\},\{1,4,5\},\{1,5,6\},\{1,2,6\},\{2,3,5\},\{2,4,6\},\{2,4,5\},\{3,4,6\},\{3,5,6\}$.
in addition,for $n=6$ i got an example with $m=6$:
 $\{1,2,3,4,5,6\},\{5,6,7,8,9,0\},\{9,0,a,b,1,2\},\{3,4,7,8,a,b\},\{1,3,5,7,9,a\},\{2,4,6,8,0,b\}$.

i think that $n\geq 7$ the best bound is $4$.

I see - I was just too lazy to do the small $n$ cases by hand... :P

In fact for $n$ 11 or above, $m$ can be narrowed down to $4$... so there's 10, 9, 8, 7 to check.


Does anyone know of any sinusoidal functions that model real world phenomena? I searched it on Google but I couldn't find anything. Listing a site would help a lot. Thank you.

You may have heard of [url=http://en.wikipedia.org/wiki/Simple_harmonic_motion]simple harmonic motion[/url] -- this can be used to model phenomena such as:
- The oscillations of a mass on a spring
- The motion of pendulum
- The vibration of a plucked string
- The stored energy of a capacitor (and the current in) an LC circuit
- (etc.)

Other phenomena that involve sinusoids:
- The potential energy of a planet in orbit, as a function of angle
- Electromagnetic waves


Forgive me if this is not in the right spot. I have a few questions that I've answered and want to make sure I've done them properly. Thank you!!

1. The highest recorded shorthand speed was 300 words per minute for 5 min with 99.64% accuracy. How many errors were made? (answer to the nearest whole number)

.9964 x 300(5)= x
 .9964x 1500=x
1494.6=x

5 errors

2.An accountant and her boss have to travel to a nearby town. The accountant catches a train traveling at 50 mi/hr while the boss leaves 1 hour later ina car traveling at 60mi/hr. They have decided to meet at the trqin station and, strangely enough they get there at the exact same time. If the train and the car traveled in straight lines on parallel paths, how far is it from one town to the other?

60 T=50 (T+1)
60 T=50 T+51
60 T-50 T=51
10T/10 =51/10
T=5.1

51 miles apart

3.The total amount spent in corporate travel and entertainment increased by 21% in a 2 year period. If the amount spent reached $ \$$115billion at the end of the period, how much was spent at the begining? I came up with $ \$$24billion

[b]Problem #1[/b]
Your solution for #1 is correct:

$ 0.9964 \cdot (300 \cdot 5) \equal{} 1494.6$, so there were 1495 correct words, showing there were $ \boxed{5}$ errors.

[b]Problem #2[/b]
Your solution for #2 has a mistake. You wrote:
$ 60T \equal{} 50(T \plus{} 1)$
$ 60T \equal{} 50T \plus{} 51$

The last step should have been $ 60T \equal{} 50T \plus{} 50$. After that step, you solved for $ T$ correctly, but since $ d \equal{} rt$ ($ distance \equal{} rate \cdot time$), and the boss traveled at 60 mph for 5.1 hours, the distance should have been $ d \equal{} rt \equal{} 60 mph \cdot 5.1 hours \equal{} 306 miles$. (However, the value $ T \equal{} 5.1$ used in this calculation is incorrect, so this is not the right answer either.)

One way to solve it is:
Let $ d$ be the distance between the towns, let $ t_a$ be the time it took the accountant travel, and let $ t_b$ be the time it took the boss to travel. We know that:
$ d \equal{} 50 t_a$
$ d \equal{} 60 t_b$
$ t_b \equal{} t_a \minus{} 1$

By substitution:
$ d \equal{} 60 \cdot \left( t_a \minus{} 1 \right) \equal{} 60 t_a \minus{} 60$.
By substituting for $ d$, we obtain
$ 60 t_a \minus{} 60 \equal{} 50 t_a$
$ t_a \equal{} 6$

Now we have the time it took the accountant to travel, but we want the distance $ d$. $ d \equal{} 50 t_a$, so $ d \equal{} 50 \cdot 6 \equal{} 300$, so the answer is $ \boxed{300 miles}$.

[b]Problem #3[/b]
Let $ s$ be the amount spent at the beginning. 121% of $ s$ is 115 billion dollars, so $ 1.21s \equal{} \$115billion$, which gives $ s \equal{} \frac{\$115billion}{1.21} \equal{} \boxed{\$95billion}$ to the nearest billion dollars.

[quote="Arrange your tan EDIT"]

1. The highest recorded shorthand speed was 300 words per minute for 5 min
 with 99.64% accuracy. How many errors were made? (answer to the nearest
 whole number)

.9964 X 300(5) = N
.9964 X 1500 = N
1494.6 = N
N rounds to 1495

5 errors

[/quote]

Kristin010306,

see, in my edit, you don't want to use a times sign and
a variable together, such as x, that are identical looking.
I used a capital ex, and N, to differentiate the multiplication
going on on the left-hand side and a different looking variable
that you're solving for on the right-hand side, respectively.

Again, good work, process, interpretation of final result and
answer for the first problem.


Find $ x, y, z \in R$ of this system of equation: $ \begin {cases} xy\equal{}x\plus{}3z\\yz\equal{}2(2y\plus{}z) \\ x\plus{}y^2\plus{}z^2\equal{}11\sqrt[11]{\frac{2.8^5}{3^9.9^6}} \end {cases}$

Has anyone solved this? I tried for a long time but nothing... Only thing I could say was that;

$ x \equal{} a\sqrt [11]{c}$
$ y \equal{} b\sqrt [22]{c}$
$ z \equal{} d\sqrt [22]{c}$

Where $ a,b,d \in\mathbb{R}$

could be of the form like that. Also, if they are in the form given above then $ a \plus{} b^2 \plus{} d^2 \equal{} 11$ is also something we can say.

Where $ c$ is $ \frac {2.8^5}{3^9.9^6}$

There are also several stuff I have found, mainly the values of $ x$ and $ y$ in terms of $ z$, being;

$ x \equal{} \frac {3z} {y \minus{} 1}$

$ y \equal{} \frac {2z} {z \minus{} 4} \implies x \equal{} \frac {3z^2 \minus{} 12z} {z \plus{} 4}$

BUT, that move just makes things EVEN worse, which is making you desperate and you have to guess the simplest form of them, which could be what I've written above, WHICH makes things A LOT worse... I'm sure the trick is at the 3rd equation given by the question, but what...

I think that there's must be something with AM-GM, because I see $ x\plus{}y^2\plus{}z^2\equal{}11\sqrt[11]{\frac{2.8^5}{3^9.9^6}}$, that $ 11$ need to think about, I think we will try to expand $ x\plus{}y^2\plus{}z^2$ into something with 11 terms.

hello, from our first equation we get $ z=\frac{1}{3}(-x+xy)$, inserting this in equation two and three we have $ \frac{1}{3}y(-x+xy)=2(2y+\frac{1}{3}(-x+xy))$ and $ x+y^2+\frac{1}{9}(-x+xy)^2=\frac{22\cdot 2^{\frac{5}{11}}}{3\cdot3^{\frac{10}{11}}}$. From our second equation above we get
$ x=\frac{12y}{2-3y+y^2}$, inserting this in our last equation and simplifying we have
${ \displaystyle{88\cdot2^{\frac{5}{11}}\cdot3^{\frac{1}{11}}-216y-1762^{\frac{5}{11}}\cdot3^{\frac{1}{11}}y-72y^2+110\cdot2^{\frac{5}{11}}}\cdot3^{\frac{1}{11}}y^2+216y^3-22\cdot2^{\frac{5}{11}}\cdot3^{\frac{1}{11}}y^3-45y^4+9y^5=0}$.
This polynomial has three real solutions.
$ y_1 \approx -1.605909$, $ y_2\approx 0.304221640$ and $ y_3\approx 1.36117840695$.
Sonnhard.

Well... I did the same thing but in terms of $ z$, but never really found the solutions you have. Did you use some kind of a calculator? (In fact, I couldn't find anything from what I had anyway...  :rotfl: )


problem19: {a_n},n=1,2,.. and sum(k=0 to n) a_k/(k+1)^2 =0. prove that a_0+a_1x+...+a_nx^n has root in (0,1).

Assume that equation hasn't root on $(0,1)$, WLOG assume that $a_{0}+a_{1}x+...+a_{n}x^{n}>0\forall x\in (0,1)$.

Put $g(x)=a_{0}+\frac{a_{1}x}{2}+...+\frac{a_{n}x^{n}}{n+1}$ and $f(x)=xg(x).$ Then $f'(x)>0\forall x\in (0,1)$ therefore $f(x)$ is increasing on $[0,1]$, so $f(x)>f(0)=0\forall x\in (0,1)$ and therefore $g(x)>0\forall x\in (0,1)$.

Now denote $h(x)=\frac{a_{0}x}{1^{2}}+\frac{a_{1}x^{2}}{2^{2}}+...+\frac{a_{n}x^{n+1}}{(n+1)^{2}}$, then $h'(x)=g(x)>0\forall x\in (0,1)$ therefore $h(x)$ is increasing on $[0,1]$, contradiction by from hypothesis we have $h(1)=h(0)=0$.


inequalities
AMC
AIME
algebra
polynomial
quadratics
number theory

In addition to the other thread on Inequalities, I'd like to discuss ways to study Algebra and Number theory, assuming one has a basic grip of AIME level material.

1) I know in Number Theory that many college level results are used. Being the idiot I am, could anyone reccomend one(1) book on number theory to read after AoPS 2 to get me up to IMO level, at least knowledge-wise? Many people in the Olympiad board are using many theorems unframiliar to me.

2) This is a long shot, but I need lots of problems that are strictly instructive Algebra and Number Theory problems. I know 102 Combinatorial Problems helped me greatly, is there a .pdf or book like this? I don't count 102 Algebra problems because it's out of stock every place I look, and it isn't getting good reviews... :ninja:

Just an aside, but am I the only the that loves problems like  AIME 2006 II 15, AIME 2005 II 7, or AIME 2003 II 15? They are my favorite kind...

You need look no further than your own friendly AoPS Resources section:

http://www.artofproblemsolving.com/Resources/Papers/SatoNT.pdf

I think he says at the beginning that it is only supposed to be used as a reference. In any case, I'd prefer something with a little more range of problems, more proofs, and more explanations...

[quote="Phelpedo"]An Introduction to the Theory of Numbers (Hardy and Wright).[/quote]

Much of that isn't directly relevant to Olympiads? But I guess it's good nevertheless. I'd suggest just using the aforementioned packet and posting here if the explanations in there aren't enough. You usually learn more by figuring stuff out on your own though.

Everything you need to know for <= AIME is pretty much all in AoPS... I'd say. Also, it looks like Titu/Zuming/Dorin are coming out with two new books (104 Number Theory Problems, Number Theory: A Problem-Solving Approach) so maybe look into that in a month or so.

Ok, any help on number 2 than? I've long given up on finding any kind of instruction but maybe there is a .pdf collection I have missed? Or if there are some algebra and factoring tricks, tips, or secrets anyone would like to share?

A few of mine would be:
1) Know your basic factoring formulas plus the Sophie Germain one like the back of your hand. Know how to factor a^K +- b^k.

2) Geometric series or roots of unity in polynomial problems are sometimes useful

3) Look for quadratics in terms of different quantities

4) Look for clever substitutions, geometrical interpretations, or telescopes

I linked to Hojoo Lee's site in the Inequalities topic. Check out his Problems in Elementary Number Theory! (I think it's ~ a good 100 pages of just problems) For the tough problems, I suggest either looking up the sources (most of which are included) or posting in the Olympiad forum.

[quote="Phelpedo"]Number Theory (Andrews) is also supposed to be quite good. I'm not sure if it's tailored to a problem-solving audience, and it has a huge emphasis on partitions, but overall it does have pretty much everything you need to know.[/quote]

I have that book. It's good, but isn't written for olympiad preparation (not like 102 combinatorial problems or engel is). I think it's meant to be an undergrad book, but has lots of problems and instruction like you're looking for.


[quote="nktp"]Let $ a,b,c > 0$ satisfy $ ab\plus{}bc\plus{}ca\leq 3abc$. Prove that:
$ \sum\sqrt{ma^{2}\plus{}n}\geq\sum\sqrt{pa^{2}\plus{}q}$, in which $ m,n,p,q > 0,m\plus{}n \equal{} p\plus{}q,m > p$[/quote]
Since, $ 3\minus{}\frac{1}{a}\minus{}\frac{1}{b}\minus{}\frac{1}{c}\geq0$ enough to prove that
$ \sum_{cyc}\left(\sqrt{ma^{2}\plus{}n}\minus{}\sqrt{pa^{2}\plus{}q}\right)\geq\frac{m\minus{}p}{\sqrt{m\plus{}n}}\cdot\left(3\minus{}\frac{1}{a}\minus{}\frac{1}{b}\minus{}\frac{1}{c}\right).$
But $ \sum_{cyc}\left(\sqrt{ma^{2}\plus{}n}\minus{}\sqrt{pa^{2}\plus{}q}\right)\geq\frac{m\minus{}p}{\sqrt{m\plus{}n}}\cdot\left(3\minus{}\frac{1}{a}\minus{}\frac{1}{b}\minus{}\frac{1}{c}\right)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\sqrt{ma^{2}\plus{}n}\minus{}\sqrt{pa^{2}\plus{}q}\minus{}\frac{(m\minus{}p)(a\minus{}1)}{a\sqrt{m\plus{}n}}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(m\minus{}p)(a\minus{}1)\left(\frac{a\plus{}1}{\sqrt{ma^{2}\plus{}n}\plus{}\sqrt{pa^{2}\plus{}q}}\minus{}\frac{1}{a\sqrt{m\plus{}n}}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{(a\minus{}1)\left(a^{2}\sqrt{m\plus{}n}\minus{}\sqrt{ma^{2}\plus{}n}\plus{}a\sqrt{p\plus{}q}\minus{}\sqrt{pa^{2}\plus{}q}\right)}{a\left(\sqrt{ma^{2}\plus{}n}\plus{}\sqrt{pa^{2}\plus{}q}\right)}\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{(a\minus{}1)^{2}\left(\frac{ma^{2}(a\plus{}1)\plus{}n(a^{3}\plus{}a^{2}\plus{}a\plus{}1)}{\sqrt{ma^{2}\plus{}n}\plus{}a^{2}\sqrt{m\plus{}n}}\plus{}\frac{q(a\plus{}1)}{\sqrt{pa^{2}\plus{}q}\plus{}a\sqrt{p\plus{}q}}\right)}{a\left(\sqrt{ma^{2}\plus{}n}\plus{}\sqrt{pa^{2}\plus{}q}\right)}\geq0.$ :)

[quote="arqady"]Since, $ 3\minus{}\frac{1}{a}\minus{}\frac{1}{b}\minus{}\frac{1}{c}\geq0$ enough to prove that
$ \sum_{cyc}\left(\sqrt{ma^{2}\plus{}n}\minus{}\sqrt{pa^{2}\plus{}q}\right)\geq\frac{m\minus{}p}{\sqrt{m\plus{}n}}\cdot\left(3\minus{}\frac{1}{a}\minus{}\frac{1}{b}\minus{}\frac{1}{c}\right).$[/quote]
[quote="arqady"]$ \Leftrightarrow\sum_{cyc}\left(\sqrt{ma^{2}\plus{}n}\minus{}\sqrt{pa^{2}\plus{}q}\minus{}\frac{(m\minus{}p)(a\minus{}1)}{a\sqrt{m\plus{}n}}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(m\minus{}p)(a\minus{}1)\left(\frac{a\plus{}1}{\sqrt{ma^{2}\plus{}n}\plus{}\sqrt{pa^{2}\plus{}q}}\minus{}\frac{1}{a\sqrt{m\plus{}n}}\right)\geq0\Leftrightarrow$
[/quote]

Man,how can you think of these??? :D Great solution. :)

Here is my solution:
$ \sum (\sqrt{ma^{2}\plus{}n}\minus{}\sqrt{pa^{2}\plus{}q}) \equal{}\sum\frac{(m\minus{}p)(a^{2}\minus{}1)}{\sqrt{ma^{2}\plus{}n}\plus{}\sqrt{pa^{2}\plus{}q}}$
$ \equal{}\sum\frac{(m\minus{}p)(a\minus{}\frac{1}{a})}{\sqrt{m\plus{}\frac{n}{a^{2}}}\plus{}\sqrt{p\plus{}\frac{q}{a^{2}}}}$
$ \geq\frac{m\minus{}p}{3}\sum (a\minus{}\frac{1}{a}).\sum\frac{1}{\sqrt{m\plus{}\frac{n}{a^{2}}}\plus{}\sqrt{p\plus{}\frac{q}{a^{2}}}}$(By Chebysev inequality)
Let $ \sum (a\minus{}\frac{1}{a}) \equal{}\sum a\minus{}\sum\frac{1}{a}\geq 0$ because: $ \sum a\geq 3\geq\sum\frac{1}{a}$


logarithms
trigonometry
function
algebra
binomial theorem
complex analysis
complex numbers

As always, a question like this is not entirely well defined - but this is a first quadrant angle, and there's a most obvious choice for the argument. If we make that choice,

$\log(1+i)=\frac12\ln2+i\frac{\pi}4.$

Then $i\log(1+i)=-\frac{\pi}4+\frac{i}2\ln 2$ and thus

$(1+i)^{i}=e^{-\pi/4}e^{i\ln 2/2}= e^{-\pi/4}\left(\cos\frac{\ln2}2+i\sin\frac{\ln 2}2\right).$

I really don't understand how this fits as an exam question. There's absolutely nothing to it?

[quote="kunny"]Find $(1+i)^{i}$.[/quote]
Use the binomial theorem  :rotfl:  (joking).

As Kent points out this isn't well defined.

I will do this problem a little bit different. In complex analysis there is something called "multi-valued function".

[b]Definition:[/b] A "multi-valued function (complex)" is a function mapping a set of set of complex numbers into a power set of $\mathbb{C}$.

I will use the notation $\mbox{Ln}(z)$ to represent the multi-valued function $\ln z = w$ as $e^{w}= z$ (for $z\not =0$).

Then,
$(1+i)^{i}= \exp (\mbox{Ln}(1+i)i)$

Now the rule if if you found one logarithm then you found them all, i.e. just keep on adding $2\pi$.

The principal logarithm is,
$\ln (1+i) = \ln |1+i|+i\arg (1+i) = \ln \sqrt{2}+(\pi/4)i$
So,
$\mbox{Ln}(1+i) = \ln \sqrt{2}+((\pi/4)+2\pi n)i$

So, continuing from previous lines:
$\exp ( \ln \sqrt{2}i-(\pi/4+2\pi n)) = e^{-\pi/4-2\pi n}\cdot e^{\ln \sqrt{2}i}= e^{-\pi/4}e^{2\pi n}\cdot \left( \cos \frac{\ln 2}{2}+i\sin \frac{\ln 2}{2}\right)$
(I renamed $n$ to $-n$ in last lines).

Now if you chose $n=0$ (principal logarithm) you get what Kent got.
But if you are looking for multi-values then the above shows it.


number theory
prime numbers
number theory proposed

I need to find all prime numbers p, q, r that suit this equation -> p(p+1)+q(q+1)=r(r+1)
Any suggestions on how this could be done?  :?

You could start by subtracting $q(q+1)$ from both sides and factoring the RHS, maybe.

Solved by grobber in this post:
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=519]www.mathlinks.ro/Forum/viewtopic.php?t=519[/url]

Thanks a lot! Couldn't find it anywhere...  :D


function
real analysis
real analysis unsolved

Let $ (a_n)$ be a sequence of nonzero real numbers.  Prove that the sequence of functions $ f_n(x) \equal{} \frac {1}{a_n}\sin(a_nx) \plus{} \cos(x \plus{} a_n)$ has a subsequence that (uniformly) converges to a continuous function.

Obviously false. A uniformly convergent sequence of bounded functions has a bounded limit, and the sequence is in fact uniformly bounded.
If $ a_n\equal{}\frac1n$, the $ n$th function takes values near $ n$, and no sequence is uniformly bounded.

So, what's the fix? Maybe this should be about uniform convergence on compact sets. This is a normal family.


A chord is perpendicular to a diameter of a circle at a point which divides the diameter into segments with lengths in the ratio 1:3. In what ratio does the chord divide the circumference?

A. 1:2
B. 1:3
C. 1:4
D. 2:3
E. 3:4

[hide]
B - 1:3

Assuming that the diameter is 4 (1+3), the radius would be 2, and the 2 triangles formed in the diagram are 30 - 60 - 90 triangles. Thus, the angle of the minor arc is 120 degrees, 1/3 of the circumference. Hence the ratio is 1:3.

[/hide]


A sequence of numbers satisfies the recursion $ x_0\equal{}0$ $ nx_n\equal{}(n\minus{}2)x_{n\minus{}1}\plus{}1$ Find a closed form expression for $ x_n$

Trying out a couple of numbers, I noticed that $ x_n\equal{} 1/2$ for $ 2$ or greater but how do I prove this rigorously?

[hide]You could try induction with base case $ 2$.
You can solve for $ x_1$ and $ x_2$ and get that $ x_1\equal{}1$ and $ x_2\equal{}\frac{1}{2}$ so we have proved it for the base case. Now assume that $ x_n\equal{}\frac{1}{2}$ for $ n\geq 2$. Then we want to prove it for $ x_{n\plus{}1}$. We have that $ (n\plus{}1)x_{n\plus{}1}\equal{}(n\minus{}1)x_{n}\plus{}1$. Substituting we have $ (n\plus{}1)x_{n\plus{}1}\equal{}\frac{1}{2}(n\minus{}1)\plus{}1$ and this is equivalent to $ 2(n\plus{}1)x_{n\plus{}1}\equal{}n\plus{}1$ and since $ n\geq 2$ dividing both sides by $ n\plus{}1$ results in $ 2x_{n\plus{}1}\equal{}1$ so $ x_{n\plus{}1}\equal{}\frac{1}{2}$ which proves that for every integer $ n\geq 2$ we have that $ x_{n}\equal{}\frac{1}{2}$.[/hide]

how does multiplying by $ (n\minus{}1)$ help

Let $ y_n\equal{}(n\minus{}1)nx_n$, you have $ y_n\minus{}y_{n\minus{}1}\equal{}n\minus{}1\ (n\geq 1).$

The answer is $ x_0\equal{}0,\ x_1\equal{}1,\ x_n\equal{}\frac 12\ (n\geq 2)$.


Let $a,b,c$ be real numbers such that $a\geqq 1,b\geqq 1,c\geqq 1$.

Prove that $\frac{a}{a^5+a^4+1}+\frac{b}{b^5+b^4+1}+\frac{c}{c^5+c^4+1}\leqq 1+\sqrt{\frac{a+b+c}{3}-1}$.

This looks funny.
\[
\begin{gathered}
  \sum {\frac{a}
{{a^5  + a^4  + 1}}}  \hfill \\
   \leqslant \sum {\frac{a}
{{3a^3 }}}  \hfill \\
   = \frac{1}
{3}\sum {\frac{1}
{{a^2 }}}  \hfill \\
   \leqslant 1 \hfill \\
   \leqslant 1 + \sqrt {\frac{{a + b + c}}
{3} - 1}  \hfill \\
\end{gathered}
\]


topology
analytic geometry
real analysis
real analysis unsolved

Let $ M$ and $ N$ be non-empty metric spaces.
(a)If $ M$ and $ N$ are connected prove that $ M\times N$ is connected.
(b)What about the converse?
(c)Answer the queston again for path-connectness.

(a) here you don't need the assumption of being metric spaces. pick some point $ (x,y)\in M\times N$. note that $ M \times \{y\}$ is connected because it's homeomorphic to $ M$, and note that $ \{u\} \times N$ is connected for any $ \{u\}\in M$ for the same reason. the space $ S_u \equal{} M\times \{y\} \cup \{u\} \times N$ is then connected as the union of connected spaces with the point $ (u,y)$ in common. now of course $ \bigcup_{u\in M} S_u \equal{} M\times N$, and it is connected because all the $ S_u$ have the point $ (x,y)$ in common.

(c) let $ M$ and $ N$ be path connected and let $ (x_0,y_0)$ and $ (x_1,y_1)$ be points in $ M \times N$. since $ M$ is path connected, there exists continuous $ f: [0,1]\to M$ such that $ f(0) \equal{} x_0, f(1) \equal{} x_1$, and there exists continuous $ g: [0,1]\to N$ such that $ g(0) \equal{} y_0, g(1) \equal{} y_1$. then $ h: [0,1]\to M\times N$ defined by $ h(t) \equal{} (f(t),g(t))$ is continuous since the coordinates are continuous and provides the needed path.

In the converse direction: $ M$ and $ N$ are images of $ M\times N$ under continuous maps, namely coordinate projections.


AMC
USA(J)MO
USAMO
geometry
number theory
relatively prime

[quote="mathking123"]The USAMO was this week, on tuesday and wednesday, right? If so, can anyone post a link to the problems or tell me when they will be available?

Thanks in advance.[/quote]

No, it's next week, three days from now.

But the test page problems posted are pretty funny:

1. Find $ \displaystyle\sum_{i \equal{} 1}^{2009} 1$.

2. Prove that $ \sqrt {2}$ is rational.  You may not use proof by contradiction

3. Give three proofs of the Pythagorean Theorem.

Plus, the date on the paper is April 1, 2009.  (April fools day)

[quote="tenniskidperson3"]
2. Prove that $ \sqrt {2}$ is rational.  You may not use proof by contradiction
[/quote]

False?  :P

[quote="tenniskidperson3"]But the test page problems posted are pretty funny:

1. Find $ \displaystyle\sum_{i \equal{} 1}^{2009} 1$.

2. Prove that $ \sqrt {2}$ is rational.  You may not use proof by contradiction.

3. Give three proofs of the Pythagorean Theorem.

Plus, the date on the paper is April 1, 2009.  (April fools day)[/quote]

1. The answer is 0, because it states that $ i \equal{} 1$. 'Nuff said.
1. (Answer 2) The answer is 0 because there is no set to sum up. The numbers on top, below, and to the right of the sigma have no meaning, and are only there for decoration.

2. Because $ \sqrt {2}$ = $ \frac {\sqrt {8}}{2}$, and we all know that $ \sqrt {8}$ is an integer. (Proof by false premise)
2. (Answer 2) Assume that $ \sqrt {2} \equal{} \frac {p}{q}$ for two relatively prime integers $ p$ and $ q$. Therefore, $ p^2 \equal{} 2q^2$. But only if the example exists can we make this assumption. We have made the assumption, therefore an example must exist. (Proof by assumption)

3. Draw out squares on the sides of the triangle with sidelengths equal to the side it branches off of. Then, stick the two smaller squares together, and squish the smaller square to deform it, so that you can apply it as a sort of coating onto the medium-sized square. After you apply the coating as evenly as possible, and make it into a square, it kinda looks like it has the same area as the largest square, thus completing the proof. (Proof by estimation)
3. (Answer 2) The theorem is false. It is possible to cut the smallest square up into infinitely many small pieces, and rearrange these pieces into the larger square, meaning that $ a^2 \equal{} c^2$, with no need for the $ b^2$. (Proof by Banach-Tarski)
3. (Answer 3) The theorem is false because it involves division by zero. (Proof by irrelevance)

[quote="math154"]They will be available after the USAMO, not before. :P  :wink:[/quote]

Oops...i thought they were this week, on the 21st and 22nd.

.......... :oops:  :blush:  :oops:

Also, for problem 2, here is an alternative solution:

sqrt(2)=1.414213562373095048801  (try squaring it on your calculator!), and since any number with a terminating decimal expansion is rational, it follows that sqrt(2) is rational.

And the reason for that is rather interesting.
[url]http://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox#The_von_Neumann_paradox_in_the_Euclidean_plane[/url]

Well, while we're on the subject of silly proofs (This does not include Banach-Tarski) I have one I will use in case I have no time left but still a problem to answer:

Let the statement to be proven be statement $ X$.  Consider the statement $ Y: Y\Leftrightarrow X$.  The only time these two are consistent is when $ X$ is true, as seen by the truth table:

\begin{tabular}{c|c|c|c|c}\\
Y=&(X&\Leftrightarrow&Y)&?\\
0&(0&1&0)&0\\
0&(1&0&0)&1\\
1&(0&0&1)&0\\
1&(1&1&1)&1\end{tabular}

Also, the proof: If it were not true, then it would not be on the contest, contradiction.  So it must be true. :P

I got those problems too. I was confused the first time I saw it xD.

Anyways,
This statement contradicts itself. Prove that the preceding statement does not contradict itself.

Proof by contradiction: Assuming that the statement does contradict itself, it stating that it contradicts itself means that it does not contradict itself, clearly contradicting itself. Therefore, the statement must not contradict itself, and the proof is done.

Good luck on USAMO everyone!

[quote="MellowMelon"][quote="sdkudrgn88"]3. (Answer 2) The theorem is false. It is possible to cut the smallest square up into infinitely many small pieces, and rearrange these pieces into the larger square, meaning that $ a^2 \equal{} c^2$, with no need for the $ b^2$. (Proof by Banach-Tarski)[/quote]
I believe Banach-Tarski requires three dimensions. Two doesn't work.[/quote]

Alright, by pseudo-Banach-Tarski, then.


geometry
vector
calculus
3D geometry
sphere
geometry unsolved

Problem 1
Find the maximum volume of a cylinder contained in the intersection of
a sphere with center O and radius R and a cone with vertex O meeting
the sphere in the circles of radius r, having the same axis as the cone.
----------------------------------------------------------------------------------
Problem 2
Find the maximum area of the orthogonal projection of a unit cube
onto a plane
---------------------------------------------------------------------------------
Problem 3
Given a triangle ABC, let a,b,c denote the lenghts of its sides and
m,n,p the lenghts of its medians. For every positive real q, let
f(q) be the real number satisfaying

a^q + b^q + c^q = (f(q))^q.(m^q+n^q+p^q)

i) Compute f(2)
ii) Determine the limit of f(q) as q tends to 0
iii) For wich triangle ABC is f(q) independent of q ?

For prb 3:

i) We know that ma^2 (ma is the median corresponding to side a) = (b^2+c^2)/2-a^2/4. By adding all 3 relations of this type we get a^2+b^2+c^2 = 4/3*(ma^2+mb^2+mc^2) = f(2)^2*(ma^2+mb^2+mc^2), so f(2)= 2/(3^1/2)

ii) This limit is well known: lim q->0 ((a^q+b^q+c^q)/3)^1/q= (abc)^1/3 for positive reals a, b, c. f(q)= ((a^q+b^q+c^q)/(m^q+n^q+p^q))1/q. As q->0 this expression tends to ((abc)^1/3)/((mnp)^1/3).

iii) If f(q) is independent of q this means that f(2)=lim q->0 f(q), so 2/(3^1/2)= ((abc)^1/3)/((mnp)^1/3), so (abc)^2/(mnp)^2= 64/27, so (abc)^2/product (2a^2+2b^2-c^2) = 1/27. If we expand that product (let?s call it P) and we apply AM-GM we can see that P >= (abc)^2, with equality iff a=b=c, so the triangles we are looking for must be equilateral. Now we must check whether the equilateral triangle verifies this condition. It does, since for the equilateral triangle a=2/(3^1/2)*ma=f(q)*ma. If we take these to power q and add them up we get a^q+b^q+c^q=(f(q))^q*(ma^q+mb^q+mc^q).

the area of projection onto the plane equals the abs value of the dot product of the unit vectors peprpendicular to the face and the plane..let x,y,z be the dot products,then max area is maxima of x+y+z under the condition x 2 +y 2 +z 2 =1.
use cauchy and u r done.the answer is  \sqrt 3
the 1st problem is pure drudgery ,calculus and all
cheers


for your index rank, you will probably have to wait until they release the usamo qualifying scores/index on http://www.unl.edu/amc/index.html

for the amc, i don't think you can see your rank in the country yet, but you can see how well you did in your state by comparing yourself to the statistics here: http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2006-12a/06AMC1012dir-pict.html

since i didn't qualify for aime, thats about as much as i know for the rankings; i'm sure there are others who can tell you a lot more

[quote="kevin gong"]for your index rank, you will probably have to wait until they release the usamo qualifying scores/index on http://www.unl.edu/amc/index.html[/quote]

USAMO index rankings are not published, from what I know. They give you a list of USAMO qualifiers by state and grade on the AMC website. The summary of awards published every year lists AIME and AMC scores of USAMO qualifiers (along with everyone else). So although in theory you could calculate the index of every single qualifier and see where you placed, that really doesn't sound like the way I like spending my Friday nights.

Kevin's right about the other part.

[quote="MithsApprentice"]So although in theory you could calculate the index of every single qualifier and see where you placed, that really doesn't sound like the way I like spending my Friday nights.[/quote]

And that's what computer programs are for  :D

[quote="Go Beyond"]After I get my AIME score, or index score (AMC+AIME), can I know my index score ranking in the country, especially for those people who make USAMO?
For example, somebody gets 230, which is ranking about 180 or something like that.
Thank you! :P[/quote]

Why would you want to know your "ranking" anyway since you already made USAMO?

[quote="probability1.01"]You have to have some pretty advanced technology to be able to actually read the physical results packet and calculate the rankings :P.[/quote]

If you are desperate, pay 20 bucks to your little brother or sister to waste their friday night typing datas into the computer, and then you use the program :rotfl:

[quote="probability1.01"]You have to have some pretty advanced technology to be able to actually read the physical results packet and calculate the rankings :P.[/quote]

Actually a lot of your "all in one" centers like HP printers can scan your documents and "read" the text off them. The only thing is you have to then program it to read from a PDF document and calculate all that stuff....... and you would be a :fool: to waste your time doing that. Unless you REALLY like your AMCs. Oh don't they send an e-mail to your school?

[quote="beta"][quote="t0rajir0u"][quote="probability1.01"]You have to have some pretty advanced technology to be able to actually read the physical results packet and calculate the rankings :P.[/quote]

I was going to, uh, use the online statistics once they come out... :)[/quote]

Well you can't use online statistics because they use separate statistics for AMC and AIME and they don't have USAMO index statistics. Yeah the only way to calculate the ranking is to use the Result/Award book and calculate everyone's USAMO index.[/quote]

Or simply beg to the admins to make a new section for indexes :)

Somehow I don't think one person begging the CAMC is enough leverage to get a publication of the USAMO indices.

[quote="filletwho"]Start an online petition.[/quote]

Yep aops assemble!


How many different six-digit numbers can be formed using three
5's, two 4's, and one 6?

There are $ 6!$ different ways of arranging $ 6$ different letters.  Since we do not have different letters, we must divide by $ 3! \cdot 2$:

$ \frac{6!}{3! \cdot 2} \equal{} \frac{6 \cdot 5 \cdot 4}{2} \equal{} 6 \cdot 5 \cdot 2 \equal{} \boxed{60}$ different numbers.


48 AoPSers' , Billy Bob, and his evil twin Billy Bobo are entered in a lottery in which there are 50 prizes. After a person is drawn, his/her name is immediately put back. There is no limit on the number of prizes a person can get. What is the probability that:

a) Billy Bob gets NO prizes?
b) Billy Bobo gets all the prizes?
c) Billy Bob gets exactly one prize ?
d) [color=white]akalra1[/color] gets 5 prizes?[color=white]**[/color]

First ten people to answer correctly get a[size=75]*[/size] meatspin sandwich!  :D

[size=59]*[color=yellow]virtual. Offer only valid while supplies last[/color].[/size]
[size=59][color=white]**pm me if you want your name here[/color][/size]

[hide]a) The probability that Billy Bob doesn't get a prize the first time is 49/50. So [b](49/50)^50?[/b]

b) The probability that Billy Bob gets a prize the first time is 1/50. So [b](1/50)^50[/b]

c) The probability that Billy Bob gets a prize the first time is 1/50. Then the rest are 49/50 because he only gets 1 prize.  But his prize can be on the first, second, etc etc. draw, so multiply that by 50. (1/50)*(49/50)^49 * 50 = [b](49/50)^49[/b]

d)The probability that AoPSer X gets a prize on the first draw is 1/50. The prob he doesn't get a prize is 49/50. Therefore, we have (1/50)^5 * (49/50)^45. But the 5 prizes can be on any draw. So (1/50)^5 * (49/50)^45 * 50C5, or (50 * 49 * 48 * 47 * 46)/(5*4*3*2*1) = [b]2,118,760 * (1/50^5) * (49/50)^45[/b]

Do I need to simplify them...?[/hide]

Bonus:
Due to the success of the aformentioned lottery, a new one has been formed. This time, however, there are an infinite number of participants and a matching number of prizes. [i]Now[/i] what's the probability that Billy Bob wins:

a) 1 prize
b) No prizes
c) In this lottery, find a general formula for the probability of Billy Bob winning $n$ prizes