inequalities
real analysis
integration
complex numbers
real analysis solved

Let $f: [0,1] \to \mathbb{C}$ -Lebesgue integrable. Prove that there exists a measurable subset $A$ of $[0,1]$ such that $|\int_A f(x)dx | \geq \frac{1}{\pi} \int_0^1 |f(x)|dx$.

for $\theta \in [0;2 \pi]$ define $F_{\theta}=f^{-1}(A_{\theta})$ where $A_{\theta}$ is the set of complex number $z$ satisfying $\Re(ze^{-i\theta})>0$. (draw a picture, it is a half plane). Define also $g(x)=arg(f(x)) \in [0; 2 \pi[$

Then $H(\theta)=| \int_{F_{\theta}} f(x)dx | = | \int_{F_{\theta}} f(x) e^{-i \theta}dx |$ satisfies $H(\theta) \geq \int_{F_{\theta}} |f(x)| cos(a(x)- \theta)dx = \int_{[0;1]} |f(x)| cos^+(a(x)- \theta)dx$ (decompose real and imaginary part to see this) where $cos^+(x) = \max(cos(x),0)$.
Hence
$\int_{0}^{2 \pi} H(\theta) d \theta \geq \int_{0}^{2 \pi} (\int_{[0;1]} |f(x)| cos^+(a(x)- \theta)dx) d \theta$.
But Fubinni theorem shows that $\int_{0}^{2 \pi} (\int_{[0;1]} |f(x)| cos^+(a(x)- \theta)dx) d \theta=2 \int_{[0;1]} |f(x)| dx$. The conclusion then follows because:

$2 \pi \max_{\theta} H(\theta) \geq \int_{0}^{2 \pi} H(\theta) d \theta \geq 2 \int_{[0;1]} |f(x)| dx$.

thanks, Alekk, my solution was the same. Btw, the particular case of this inequality was also discussed and the idea was the same http://www.mathlinks.ro/Forum/viewtopic.php?highlight=subset&t=36584


geometry
geometric transformation
reflection
perimeter
power of a point
radical axis
geometry unsolved

Let $ C_{b}$ and $ C_{c}$ be the excircles opposite to the vertices $ B$ and $ C$ of a triangle $ ABC$; let $ B_{0}$ and $ C_{0}$ be the mid points of $ AC$ and $ AB$ resp. Let $ C_{b}'$ and $ C_{c}'$ be the circles symmetric respectively to $ C_{b}$ and $ C_{c}$ obtained by reflection in the points $ B_{0}$ and $ C_{0}$. Prove that the radical axis of $ C_{b}'$ and $ C_{c}'$ halves the perimeter of $ ABC$.


Welcome to Mafia Game 36: Mt. Chimney!

[b][u]Story:[/u][/b]
The Hoenn Intelligence Agency, or HIA, has received evidence that Team Magma has developed a plan to make the climate of the Hoenn region much hotter, causing ocean levels to drop severely.  At the same time, Team Aqua has developed a plan to raise the sea level, submerging most of the Hoenn region in water.  Both of their plans require the use of Mt. Chimney, a dormant volcano.  To stop them, the HIA has sent a team of Pokemon trainers to accompany and sabotage these devious schemes.  Unfortunately, members of Team Magma and Team Aqua showed up at the HIA briefing as well, and the HIA's secretary misplaced some important paperwork.  Therefore, 24 trainers have set up camp on Mt. Chimney:

BOGTRO
(^_^)
Brut3Forc3
pacman2812
chenhsi
math_explorer
jackdillon
Bicameral
xpmath
shaggy75
mnmath
zapi2007
iLord
Peter
#H34N1
mathelete9999
mz94
jnub
142857
1=2
crazychinesemaniac
erabel
Sunny
kimmystar94

After setting up camp, you all notice that the HIA gave you a walkie-talkie.  The HIA chief comes in clearly, saying:
"Attention, agents.  If you haven't realized by now, Team Magma and Team Aqua are among you, which makes your job much more difficult.  I order you to set up a lynching system based on voting.  As soon as someone achieves a majority of votes, he or she must be immediately lynched.  Since time is of the essence, you must lynch someone every day.  Good luck."

[b][u]Notes:[/u][/b]
All power roles will specify whether the trainer or the Pokemon performs a night action.  If the trainer performs the action, keep in mind that you do not need your Pokemon to perform it.  Also keep in mind that all night roles are optional, if you choose not to act.
As you know, Pokemon are not perfect, so their attacks sometimes fail.  If you submit a night action and you notice that its effect did not take place the next day, this may be the reason why.

Role PMs are being sent out now.  Do not post anything until I post again with the general rules.

[b][u]General Rules[/u][/b]:[list][*]Votes must be in [b]bold[/b].  If you do not bold your vote, it will not be counted.
[*]Please be attentive and [b]unvote[/b], if necessary, before casting a new vote.  This is not [i]required[/i], but I’d appreciate it.
[*]Lynches will require a simple majority of votes.  Once a player has reached the necessary majority, his pleas are useless and any attempts to unvote will be unheeded. As specified in the introduction, you must lynch someone every day.  A vote of [b]No lynch[/b] will not be counted.
[*]The game is not to be discussed outside of the thread unless your role specifically states that you may do so. [b]If your role PM tells you the alignment of other players, you may privately talk to them.[/b]
[*]Once your death scene has been posted, you’re dead.  Stop typing.
[*]Don’t edit/delete previously submitted posts.
[*]Don’t quote any PMs from me.
[*]If you have a night choice to make, it is due by the posted deadline.  I will not wait for you – if you do not submit a choice to me, too bad.
[*][b]If you anticipate being unavailable for more than a few days, please post a notice to that effect in the thread.[/b]  Treat this game as a commitment.  Be considerate – don’t leave us hanging.
[*]Remember, it’s a game.  It’s supposed to played.  It’s supposed to be [i]fun[/i].  Do your part.[/list]

The "townie" PM looks like this, where the blanks contain a Pokemon and the Pokemon's type respectively:
[quote="Townie PM"]You are an [b]HIA trainer[/b], and your Pokemon is [b]___[/b], a [b]___[/b] type.  Unfortunately, it doesn’t have any abilities that empower you in your situation, so it will be relatively useless to you.
You win when all scum are dead.[/quote]

If you did not receive a role PM, or if you have any questions about your role, please PM me.

[b]It is now day 1.[/b]
[i]With 24 people alive, it takes 13 to lynch.[/i]

GO!

[b]Vote:zapi2007[/b] for being the first to post besides lingomaniac.
Woot, pokemon! Actually, I was very  :huh: upon reading the role PM.

[b]Vote: Ash[/b] for having a cheesy last name.

cool...

i love my TI

[b]Random Vote: Bicameral[/b]

and oh yess we should have a Digimon Mafia

(whats Digimon? ;])

AH hahaha I LOVE the pokemon theme.  I was so addicted to that show!

[quote="zapi2007"]Here!
Pokemon, haha.
Maybe someone should do Harry Potter someday or something.
or Digimon   :lol:[/quote]

At MafiaScum a Harry Potter game was played, I don't remember where.

At MafiaScum almost every imaginable game has been played.

[b] vote: pacman2812 [/b] for being silly and not starting his Mish-Mash game!

[b]Vote: chenhsi[/b] because he is prodding me to make something related to a game OUT of AoPS.

I haven't started it because I want more people to sign-up.

[quote="iLord"][b]Vote mz94[/b] for using a calculator!

Real men use abaci  :roll:

Theme is cool.

There's a Harry Potter game on Sally right now.[/quote]

alright. please suggest a way i can implement a random integer function on an abacus, then we talk ;]

Simple. Shake the abacus and mess up the beads. There is your random number!

[b]vote: crazychinesemaniac[/b] because he has the longest username. :D

[b]Vote: (^_^)[/b] for having an underscore.

I suspect abacus-shaking is biased, so you should flip it repeatedly to get binary digits. It's a lot slower, noisy, and potentially violent, but it's not biased (as much).

Far as I can see, we should leave the teams alone so one of them destroys the other, thus cutting our job in half.

Actually, we should just move. I'm getting all the regions mixed up, ugh.

Sorry for losing the game for us, guys.  Well done, jnub.

Jnub beasted. What made brut3 think he was a vig?

I just realized, the setup of this game and Game 34 are eerily similar. 2 scum groups, 1 sk, and almost all people have power roles. Out of curiosity lingo, what was your setup inspiration?

[quote="zapi2007"]actually, i typed in "brain" in google and that was one of the first images that popped up[/quote]

Yep, it's third.  Scary.

[quote="mewto55555"]Out of curiosity lingo, what was your setup inspiration?[/quote]
I felt it would be appropriate to have two mafia groups (Magma and Aqua).  Then I threw in some of the basic roles, trying to think up appropriate trainers and Pokemon for them.  I threw in a serial killer and Team Rocket for some extra twists.

imo, someone could make a harry potter themed one.
or maybe some other fantasy series  (i guess pokemon would count as one)

I just past post $ 2^9!!!!!!!$

Im making a star wars one. although I am tempted to make it HSM.... :rotfl:

please PM this argument or move it to the moderator thread.
Klebian, please lock this.

[quote="crazychinesemaniac"][quote="zapi2007"]I can't say anything but good game guys!

Thanks to lingo for modding the game (the guy with the mario ambassador (?) avatar)  :)[/quote]

acctually thats off the simpsons. Just shows you how much about the simpsons i know :D[/quote]
Wrong. That mario is from Futurama

that didn't seem like it was from the simpsons
for one thing, he wasn't yellow

lastpost (this is used to preserve order of locked topic)


Duke
college
geometry
Pythagorean Theorem

Let $ \triangle ABC$ be a triangle where $ AB\equal{}25$ and $ AC\equal{}29$.  $ C_1$ is a circle that has $ AB$ as a diameter and $ C_2$ is a circle that has $ BC$ as a diameter.  $ D$ is a point on $ C_1$ so that $ BD\equal{}15$ and $ CD\equal{}21$.  $ C_1$ and $ C_2$ clearly intersect at $ B$; let $ E$ be the other point where $ C_1$ and $ C_2$ intersect.  Find all possible values of $ ED$.

This would be difficult to do in ten minutes.

[hide="Duke Individual 2007 #8"]
 [geogebra]6f131f82f28fa0bcc60b3a731d779ed6561c83ec[/geogebra]

Point $ E$ is on side $ AC$ and point $ D$ is on side $ BC$.  In addition, each define an altitude of triangle $ ABC$:

$ E$:  The centers of the circles $ C_1$ and $ C_2$ are midpoints of sides $ AB$ and $ BC$.  The centers and points $ B$ and $ E$ create a kite, meaning that diaganol $ BE$ is bisected by $ C_1C_2$, landing $ E$ on side $ AC$.  Because $ AB$ is a diameter of circle $ C_1$ and because $ E$ is on $ C_1$, $ \angle{AEB}$ is a right angle.

$ D$:  $ D$ is on $ C_1$, so $ \angle{ADB}$ is a right angle.  By the Pythagorean theorem, $ AD$ has a length of 20.  Also shown by the Pythagorean theorem, $ \angle{ADC}$ is a right angle because $ 20^2 + 21^2 = 29^2$.  Thus, $ D$ is on side $ BC$.

[geogebra]e843ce81ee6225a7737eeee3f8aecae6105cd466[/geogebra]

$ A$, $ B$, $ D$, and $ E$ are all on circle $ C_1$, defining a cyclic qudrilateral.  This defines that $ \triangle{AEO}\sim \triangle{BDO}$ and $ \triangle{ABO}\sim \triangle{EDO}$  Let $ AO = ax$, $ OD = by$, $ BO = bx$, and $ OE = by$, with $ \frac {a}{b} = \frac {AE}{BD}$ and $ \frac {x}{y} = \frac {AB}{ED}$.

Using the Pythagorean theorem on $ \triangle{AEB}$ and $ \triangle{CEB}$, and given $ AC = 29$:
\begin{eqnarray*} AB^2 - AE^2 & = & BC^2 - EC^2 \\
AE^2 - EC^2 & = & AB^2 - BC^2 \\
29(AE - EC) & = & 25^2 - 36^2 \\
AE - EC & = & - 671/29 \\
AE + EC & = & 29 = 841/29 \\
AE & = & 85/29 \end{eqnarray*}

Thus $ \frac {a}{b} = \frac {85/29}{15} = \frac {17}{87}$.  Let $ a = 17$ and $ b = 87$

From the common area:
\begin{eqnarray*} AD*BC & = & AC*BE \\
BE & = & \frac {AD*BC}{AC} \\
& = & 720/29 \end{eqnarray*}

Equating the diaganols to the sum of their parts yields:
\begin{eqnarray*} BO + OE = BE & \text{ and } & AO + OD = AD \\
bx + ay = \frac {720}{29} & \text{ and } & ax + by = 20 \\
\end{eqnarray*}

With $ a$ and $ b$ defined:
\begin{eqnarray*} 17x + 87y & = & 20 \\
87x + 17y & = & \frac {720}{29} \end{eqnarray*}

Which yields $ x = 29/116$ and $ y = 21/116$.
$ \frac {x}{y} = \frac {AB}{ED}$
$ ED = \frac {y*AB}{x} = \frac {21*25}{29} = \frac {525}{29}$
[/hide]

That is an excellent solution!  I am glad you kept working on it.  I wonder if there is a shorter way of doing the problem.  Also, I wonder if anyone answered that question correctly at Duke.

Last year, apparantly nobody got it. (Look at the distributions in the Duke Math Meet 07 thread)


function
geometry
rectangle
limit
real analysis
real analysis unsolved

Let $ f(x,y)$ be a function defined on the unit square $ 0 \leq x \leq 1$, $ 0 \leq y \leq 1$ which is continuous in each variable separately. Show that $ f$ is a measurable function of $ (x,y)$.

Any ideas? Thank you.

Slice the rectangle into horizontal strips of width $ \frac1n.$ Build the following function:

$ g_n(x,y)\equal{}f\left(x,\frac {k}n\right)$ for $ \frac kn \le y<\frac{k\plus{}1}n,$ $ 0\le k\le n.$

Each $ g_n$ is measurable - that's not too hard to show.

For each fixed $ x,$ $ f(x,y)$ is uniformly continuous as a function of $ y$. From that we can show that $ \lim g_n(x,y)\equal{}f(x,y)$ on a pointwise basis.

The pointwise limit of measurable functions is measurable.

By the way, the [i]et al.[/i] in your subtitle is Antoni Zygmund. I had this course from Zygmund himself when I was a first-year grad student; he used his own notes, which were a precursor to this book.

We will be using his book for the graduate course on measure theory next semester at my school.  Kent, how do you feel he did with it?  Just in my personal opinion, from what I've glanced at, the book seems okay but rather dry and not particularly motivating.

Prof. Kent
The function $ g_{n}$ are so well-defined, could you please explain why you know to define such functions $ g_{n}$?


calculus
integration
Ring Theory
abstract algebra
vector
superior algebra
superior algebra unsolved

maybe it is classical but cute:
assume that $S$ is ring which is integral over it subring$R$,and also finitly generated over it.if $p\in Spec(R)$ then prove that there exit only finitly many prime $q$ in $S$ over $p$.

The conditions mean that $S$ is finite as $R$-module, so is $S_{p}\otimes_{R}R_{p}/pR_{p}$ as $R_{p}/pR_{p}$-module, i.e. it is a finite dimensional vector space over $R_{p}/pR_{p}$, so the fibre at $p$ is finite.

would you mind elaboraing more i didnt get your soln

I'm just using that the fibre of a morphism $R \to S$ at $p$ is obtained by tensoring with the residue field.


Post here if you're looking for teammates, but please try finding people at your school to be on your team first.

If you don't find people by December, please email us. If you want to compete, we'll do our best to make sure you can.

I am looking for a couple of teammates. I paid full $90, but have only 4 members and 1 alternate. Could you help me find a teammate by Nov. 13th? Thank you!!

We've switched to matching people up on the day of the contest. Just show up with as many people as you have, and if there's a group that brings seven people, for example, we'll add the seventh person to your team.

Yasha


What's the difference between \frac and \cfrac and \dfrac?

I was using Aurora (http://elevatorlady.ca/) to make Latex images on Microsoft Word and I have trouble making fractions consistent-looking. Thanks!

\frac (=\tfrac) is for text mode where you want to include a fraction like this $ \tfrac{1}{2}$ in a sentence whereas \dfrac gives  larger fraction like this suitabe for display mode.
\[ \frac{1}{2}\]
\cfrac is for continuous fractions like
\[ 1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\dotsb}}}\]
Using \displaystyle at the start of the formula automatically turns all \frac into \dfrac so typing
\displaystyle \frac{1}{2} + \frac{3}{4}
gives the same result as
\dfrac{1}{2} + \dfrac{3}{4}

[size=75]Edited: removed error pointed out by roadnottaken[/size]

I believe that last line should be \dfrac{1}{2} + \dfrac{3}{4}.  Additionally, the surest way to make fractions consistant-looking is to actually use LaTeX, rather than just sticking LaTeX images in a traditional word processor.


geometry
geometric transformation
reflection
circumcircle
perpendicular bisector

Find the area of a cyclic hexagon with sides 1, 1, 1, 2, 2, 2.

I know I've seen this problem on the boards here before, but I can't find it. Here's my solution.

[hide]
Let's say our hexagon is $ ABCDEF$ with $ AB\equal{}BC\equal{}CD\equal{}1$ and $ DE\equal{}EF\equal{}FA\equal{}2$. Reflect D across the perpendicular bisector of $ CE$ to $ D'$. We have $ CD\equal{}ED'\equal{}1$ and $ CD'\equal{}DE\equal{}2$. Since $ CE$ doesn't change, the areas of $ CDE$ and $ CD'E$ are equal, so the areas of $ ABCDEF$ and $ ABCD'EF$ are equal. Also, $ D'$ lies on the circumcircle of $ ABCDEF$. Now reflect $ A$ across the perpendicular bisector of $ BF$ to $ A'$. Just like above, we get $ A'F\equal{}1$ and $ A'B\equal{}2$ with $ BF$ unchanging, so the areas of $ ABCD'EF$ and $ A'BCD'EF$ are equal. Also, $ A'$ on the circumcircle of $ ABCDEF$. We now have a cyclic hexagon with side lengths 1, 2, 1, 2, 1, 2, which can easily be seen as an equilateral triangle of side length 4 with three equilateral triangles of side length 1 cut off. Thus the area of $ A'BCD'EF$ (which is also the area of $ ABCDEF$, what we are looking for), is $ 4\sqrt{3}\minus{}3(\frac{\sqrt{3}}{4})\equal{}\frac{13\sqrt3}{4}$.[/hide]


Prove that for any $ n \in \mathbb{N}^*$ the number $ A\equal{}\frac{4^n\plus{}5^n\minus{}9^n}{5}$ is an integer.

$ 9^n \equal{} (4 \plus{} 5)^n \equal{} \sum_{r \equal{} 0}^n \binom{n}{r}4^{n\minus{}r}5^{r}$
So, $ 4^n \plus{} 5^n \minus{} 9^n \equal{} 4^n \plus{} 5^n \minus{} (4^n \plus{} \sum_{r \equal{} 1}^{n \minus{} 1} \binom{n}{r}4^{n\minus{}r}5^{r} \plus{} 5^n) \equal{} \minus{} \sum_{r \equal{} 1}^{n \minus{} 1} \binom{n}{r}4^{n\minus{}r}5^{r}$.
Hence, $ 5| \minus{} \sum_{r \equal{} 1}^{n \minus{} 1} \binom{n}{r}4^{n\minus{}r}5^{r}$.

Thus, $ A$ is an integer.

[hide="induction"]
For $ 1$ it's true. Assume it's true for $ n$. Now for $ n \plus{} 1$ we have:

$ 4^{n \plus{} 1} \plus{} 5^{n \plus{} 1} \minus{} 9^{n \plus{} 1} \equal{} 4(4^n \plus{} 5^n \minus{} 9^n) \plus{} 5^n \minus{} 5\cdot 9^n \quad \blacksquare$[/hide]

$ 9^n\equiv(\minus{}1)^n\pmod5\\*
4^n\equiv(\minus{}1)^n\pmod5\\*9^n\minus{}4^n\equiv(\minus{}1)^n\minus{}(\minus{}1)^n\equal{}0\pmod5$.
$ \minus{}(9^n\minus{}4^n)\equal{}4^n\minus{}9^n$ is divisible by 5 and $ 5^n$ is also divisible by 5 and thus $ A\equal{}\frac{4^{n}\plus{}5^{n}\minus{}9^{n}}{5}$ is an integer

sorry, the same solution, i didn't see it :oops:


inequalities
function
inequalities unsolved

Let $ a,b,c,x,y,z>0$ and $ x\geq y\geq z>0$ such that:
$ a\geq x$
$ a^2\plus{}b^2\geq x^2\plus{}y^2$
$ a^3\plus{}b^3\plus{}c^3\geq x^3\plus{}y^3\plus{}z^3$
Prove that $ a^6\plus{}b^6\plus{}c^6\geq x^6\plus{}y^6\plus{}z^6$

I derived a generalization of Karamata's inequality which can be used to solve
this type of problems easily in [url]http://btw.blobtrox.net/~btw/karamata.pdf[/url].

The theorem is as follows:

Let $ h_0, h_1, \dots, h_n$ be increasing convex functions on an interval $ I \subset \mathbb{R}$.
Define $ H_k \equal{} h_k \circ h_{k\minus{}1} \circ \dots \circ h_0$. Suppose that $ a_1, \dots, a_n$ and
$ b_1, \dots, b_n$ are increasing sequences of numbers in $ I$. Assume that the inequality

$ \sum_{k\equal{}1}^{m\plus{}1} H_m(a_k) \ge \sum_{k\equal{}1}^{m\plus{}1} H_m(b_k)$

holds for every $ m \in \{0,1,\dots,n\minus{}1\}$. Then we also have the inequality:

$ \sum_{k\equal{}1}^{n} H_{n}(a_k) \ge \sum_{k\equal{}1}^{n} H_{n}(b_k)$.

A proof of the theorem (which I would like someone to check) is given in the paper above.


LaTeX
algorithm
modular arithmetic
number theory
relatively prime
Euclidean algorithm

Let $a$ and $b$ be integers. Prove that $a$ and $b$ are relatively prime if and only if there exists $x$ and $y$ such that $ax+by=1$.

[hide="solution"]
Assume $a$ and $b$ have some common factor, $k>1$.  Then $ka'x+kb'y=1$.  Dividing both sides by $k$ yeilds $a'x+b'y=1/k$, which is a contradiction because the left hand side is an integer, but the right hand side is non-integer.
[/hide]

That is the easy direction.  Can you do the other direction?

calculuslover800 proved that if $ax+by = 1$ for some integers x and y, then a and b are relatively prime. The proof of the converse is more complex. It would require a lot of latex, so I won't explicitly write out the equation for every step.
If a and b are relatively prime, then by repeatedly applying the Euclidian Algorithm, we will eventually get 1. The Euclidean Algorithm will generate for us a series of linear equations, going
$a = c_{1}b+r_{1}$
$b = c_{2}r_{1}+r_{2}$
...
$r_{n}= c_{n+2}r_{n+1}+1$
Now we can reverse these equations, solving for $r_{1}, r_{2}, ..., 1$. We can then substitute the first equation into the second and rearrange the terms to get $r_{2}= x_{1}a+y_{1}b$, where $x_{1}, y_{1}$ are integers.
We can keep on substituting down the series of equations until at last we get
$ax+by = 1$
which is what we were trying to prove.

The question doesn't state that $x$ and $y$ are integers= gameover.

[quote="t0rajir0u"][hide="Shortest possible proof"] Let $S = \{ s \in \mathbb{N}| \exists x, y \in \mathbb{Z}: s = ax+by \}$.  It is non-empty since $a^{2}+b^{2}\in S$, so consider its minimal element $d$.

Consider the remainders $r, s$ when $a, b$ are divided by $d$.  We have

$a = dp+r, 0 \le r < d$
$b = dt+s, 0 \le s < d$

So $r = a-dp, s = b-dt$ are also linear combinations of $a, b$.  But we assumed $d$ was minimal, so $r, s \not \in S$.  It follows that $r, s \not \in \mathbb{N}$, so they equal $0$.

Then $d | a, d | b, gcd(a, b) = 1 \Rightarrow d = 1$.  QED. [/hide][/quote]

That must be the best proof I've ever seen for Bezout Theorem! Thank you for this rigorous proof!

[hide="Another short proof"]First, let's assume that $a$ and $b$ aren't coprime. Then there exists $d>1$ such that $a=da_{1}$ and $b=db_{1}$. But then $1=ax+by=d(a_{1}x+b_{1}y)$, which means that $d|1$, and that's impossible.

Now, let's assume that $a$ and $b$ are coprime. We'll prove that the numbers $0, b, 2b,\dots,(a-1)b$ give pairwise distinct remainders modulo $a$. Assume the opposite, that there are distinct $k,l\in\{0,1,\dots,a-1\}$ such that $kb\equiv lb\pmod{a}$. Then $a|(k-l)b$, but as no prime factor of $a$ is a prime factor of $b$, it means that $a|k-l$. Since $|k-l|<a$, that implies $k=l$, contrary to the presumption.

Therefore, $a$ different numbers give $a$ different remainders modulo $a$, so one of those remainders must be $1$. Hence there exists $k_{0}<a$ such that $k_{0}b\equiv 1\pmod{a}$, which means that there exists $l_{0}\in\mathbb{Z}$ such that $k_{0}b-1=l_{0}a$, so we can take $x=-l_{0}$ and $y=k_{0}$. QED.[/hide]

Those two proofs posted above by t0rajir0u and Farenhajt were really neat, I have never seen them before.

[quote="Kurt Gödel"]That's nice too  :)  Every proof I've seen so far used the Euclidean algorithm, which isn't really rigorous...[/quote]
Why does use of the Euclidean Algorithm make an algorithm non-rigorous? For the proof I posted, a set of steps was given, using the Euclidean Algorithm, that finds integers x and y. Because we can use these steps to find x and y for any a and b (as long as $(a, b) = 1$), and because there is always a finite number of steps, x and y must exist. Perfectly rigorous, as far as I can see.

There's nothing wrong with using the Euclidean Algorithm, but a proof that does it rigorously has to prove that the extended algorithm terminates, which is tedious.


Hi
Please give me some source about application of graph in game theory I think you can't explain it here beacuse this topic is little prof.  so give me some source about it.

Thank  :roll:

look for combinatorial game theory in the web


induction inequalities algebra polynomial

We prove $ x_{n\plus{}1}\equal{}4x_n\minus{}x_{n\minus{}1}$ by induction on n. Suppose that $ x_{n\plus{}1}\equal{}4x_n\minus{}x_{n\minus{}1}$, we have
\[ {x_{n \plus{} 2}} \equal{} \frac{{{x_{n \plus{} 1}}^2 \plus{} 3}}{{{x_n}}} \equal{} \frac{{{{\left( {4{x_n} \minus{} {x_{n \minus{} 1}}} \right)}^2} \plus{} 3}}{{{x_n}}} \equal{} \frac{{16{x_n}^2 \minus{} 8{x_n}{x_{n \minus{} 1}} \plus{} {x_{n \minus{} 1}}^2 \plus{} 3}}{{{x_n}}} \equal{} \frac{{16{x_n}^2 \minus{} 8{x_n}{x_{n \minus{} 1}} \plus{} {x_n}{x_{n \minus{} 2}}}}{{{x_n}}} \equal{} 16{x_n} \minus{} 8{x_{n \minus{} 1}} \plus{} {x_{n \minus{} 2}} \equal{} 4\left( {4{x_n} \minus{} {x_{n \minus{} 1}}} \right) \minus{} \left( {4{x_{n \minus{} 1}} \minus{} {x_{n \minus{} 2}}} \right) \equal{} 4{x_{n \plus{} 1}} \minus{} {x_n}\]
Therefore it suffices to find $ x_n$ satisfies that $ \left\{ \begin{array}{l}
 {x_1} \equal{} 1,{x_2} \equal{} 2 \\
 {x_{n \plus{} 1}} \equal{} 4{x_n} \minus{} {x_{n \minus{} 1}} \\
 \end{array} \right.$, which is easy

[quote="dvtvd"]$ \left\{\begin{array}{l} {x_1} \equal{} 1,{x_2} \equal{} 2 \\
{x_{n \plus{} 1}} \equal{} 4{x_n} \minus{} {x_{n \minus{} 1}} \\
\end{array} \right.$[/quote]

$ x_n$ is Chebyshev Polynomial of the First Kind evaluated at $ x\equal{}2$.
Chebyshev Polynomial of the First Kind : $ T_n(x)\equal{}\left\{\begin{array}{l} T_0(x) \equal{} 1, T_1(x) \equal{} x \\
T_{n\plus{}1}(x) \equal{} 2x T_n(x) \minus{} T_{n\minus{}1}(x) \\
\end{array} \right.$

[quote="Kouichi Nakagawa"][quote="dvtvd"]$ \left\{\begin{array}{l} {x_1} \equal{} 1,{x_2} \equal{} 2 \\
{x_{n \plus{} 1}} \equal{} 4{x_n} \minus{} {x_{n \minus{} 1}} \\
\end{array} \right.$[/quote]

$ x_n$ is Chebyshev Polynomial of the First Kind evaluated at $ x \equal{} 2$.
Chebyshev Polynomial of the First Kind : $ T_n(x) \equal{} \left\{\begin{array}{l} T_0(x) \equal{} 1, T_1(x) \equal{} x \\
T_{n \plus{} 1}(x) \equal{} 2x T_n(x) \minus{} T_{n \minus{} 1}(x) \\
\end{array} \right.$[/quote]

[size=150][b]Thanks!But how could you found that $ \left\{\begin{array}{l} {x_1} \equal{} 1,{x_2} \equal{} 2 \\
{x_{n \plus{} 1}} \equal{} 4{x_n} \minus{} {x_{n \minus{} 1}} \\
\end{array} \right.$
and this problem is related to "Chebyshev Polynomial"?[/b][/size]

[quote="admire9898"]Thanks!But how could you found that $ \left\{\begin{array}{l} {x_1} \equal{} 1,{x_2} \equal{} 2 \\
{x_{n \plus{} 1}} \equal{} 4{x_n} \minus{} {x_{n \minus{} 1}} \\
\end{array} \right.$[/quote]
I supplement dvtvd's post.
So you had better refer to dvtvd's post.

[quote="admire9898"]and this problem is related to "Chebyshev Polynomial"?[/quote]
Yes.
$ x_n$ is Chebyshev Polynomial of the First Kind evaluated at $ x\equal{}2$.


I've been trying to fax in my USAMTS solutions all evening today and the line was busy. Is it possible that they would allow me to fax them tomorrow? (if I don't succeed within the next hour)

They are due at midnight tonight (Pacific Time).


There are $ n$ sets having $ 4$ elements each. The difference set of any two of the sets is equal to one of the $ n$ sets. $ n$ can be at most ? (A difference set of $A$ and $B$ is $ (A\setminus B)\cup(B\setminus A) $)

$\textbf{(A)}\ 3  \qquad\textbf{(B)}\ 5  \qquad\textbf{(C)}\ 7  \qquad\textbf{(D)}\ 15  \qquad\textbf{(E)}\ \text{None}$

What exactly is "the difference set"? Is it $ A\setminus B$, or perhaps $ (A\setminus B)\cup(B\setminus A)$, or something else?

either $ A \setminus B$ or $ B \setminus A$. (It is wrong). Follow second post after this post.

Then I don't understand the problem correctly, since you can have as many disjoint sets as you like, and then $ A\setminus B\equal{}A$ for all ordered pairs $ (A,B)$.

most probably it is $ (A\setminus B)\cup(B\setminus A)$. You're right the number of elements must be $ 4$. I've looked the original text, i think it is $ (A\setminus B)\cup(B\setminus A)$.

[hide="If that is so, then..."]Let $ A\triangle B\equal{}C$

Then $ A\triangle C\equal{}A\triangle(A\triangle B)\equal{}(A\triangle A)\triangle B\equal{}\emptyset\triangle B\equal{}B$

and $ B\triangle C\equal{}B\triangle(A\triangle B)\equal{}(B\triangle B)\triangle A\equal{}\emptyset\triangle A\equal{}A$

Hence we can't "move out" of three sets.

An example for three sets is easily constructed: $ A\equal{}\{a,b,c,d\}, B\equal{}\{a,b,e,f\},C\equal{}\{c,d,e,f\}$

Hence $ n\equal{}3$ and the answer is $ A$.

Note: the above properties of operation "$ \triangle$" (symmetric difference of sets) are analogous to the properties of the logic operation "exclusive or".[/hide]

answer key says different
[url=http://www.mathlinks.ro/viewtopic.php?t=274048#]second post[/url]

Then there's some problem in the wording and/or translation.

No, you just haven't proven what you think you've proven.  When did you rule out the addition of the set $ \{a, c, e, g\}$ for some $ g$?

Good point - there can be additional "triplets" of sets.

[hide="Correction (not too strong towards the end, but could be improved)"]We must start from $ A, B$ and $ A\triangle B$. Let's add another set $ C$. Then we must have $ A\triangle C$ and $ B\triangle C$ as well, which in turn means we must have $ A\triangle B\triangle C$ too.

Now $ (A\triangle B)\triangle(A\triangle C) \equal{} B\triangle C$ and $ (A\triangle B)\triangle(B\triangle C) \equal{} A\triangle C$ yield no new possibilites.

Since $ |U\triangle V| \equal{} |U| \plus{} |V| \minus{} 2|U\cap V|$, it means that every two sets must have exactly two elements in common (given $ |U\triangle V| \equal{} |U| \equal{} |V| \equal{} 4$ for all $ U,V$). But as $ \binom {4}{2} \equal{} 6$, there's only $ 6$ ways to "distribute" two elements from the initial set to the other sets, hence there can be no more than $ 7$ in total.[/hide]

Okay, now what if we replace "4" with "6," or "$ 2n$"?  (This seems like it should relate to some known combinatorial object, but I don't recognize it.)

For sets of size 2, every maximal collection looks like (12, 23, 13).
For sets of size 4, we've seen that every maximal collection looks like (1234, 1256, 3456, 1357, 2457, 2367, 1467).
Actually, after a moment's thought, it seems that perhaps we should consider $ 2^n$ instead of $ 2n$.  In particular, maximal solutions for sets of size 6 look like (123456, 123789, 456789), i.e., they are just "blown up" copies of the solution for sets of size 2.  And I think one can argue that the same thing should happen for sets of size $ 2^m \cdot k$ where $ k$ is odd, i.e., every solution will just be a "blown up" copy of a solution for sets of size $ 2^m$.

Edit: well, here's a construction that always gives $ 2^{n \plus{} 1} \minus{} 1$ sets of size $ 2^n$ on a universal set of size $ 2^{n \plus{} 1} \minus{} 1$: for $ n \equal{} 0$, take the set $ \{1\}$.  Now build inductively: to generate the sets of size $ 2^n$, first look at your solutions of size $ 2^{n \minus{} 1}$.  Suppose one of these is $ \{a_1, \ldots, a_{2^{n \minus{} 1}}\}$.  Then include as one of your sets of size $ 2^n$ the set $ \{2a_1 \minus{} 1, 2a_1, \ldots, 2a_{2^{n \minus{} 1}} \minus{} 1, 2a_{2^{n \minus{} 1}}\}$.  Also include the set $ \{1, 3, \ldots, 2^{n \plus{} 1} \minus{} 1\}$, and its symmetric differences with each of the previously included sets.  The resulting collection has the desired properties.  Now, how do we prove that it is maximal?  (This is actually pretty clear, in that if you try to build a solution by hand you essentially are forced to go through the construction I just described, but this isn't exactly a proof.)

If we take $ 2^n$ elements per set, similarly as above we prove that $ |A\cap B| \equal{} 2^{n \minus{} 1}$ for every pair $ (A,B)$. Another detail which was omitted from my previous post is this: no three sets can be "concurrent" (is there a term for this in the theory of sets?), meaning there can be no triplet $ A,B,C$ satisfying $ A\cap B \cap C \neq\emptyset$, because then $ A\triangle B\triangle C$ would have at least $ 3\cdot 2^{n \minus{} 1}$ elements.

Hence, starting from the initial set $ A$, every other set is uniquely determined by the exact $ 2^{n \minus{} 1}$ elements which are "redistributed" into it from $ A$, so it seems the number of sets can't exceed $ \binom{2^n}{2^{n \minus{} 1}} \plus{} 1$. Though this still must be looked into, but now I have to go...

Edit on my previous post: The $ A\cap B\cap C\neq\emptyset$ should actually be $ A\cap B = A\cap C = B\cap C$

Possible construction would go like this.

Take $ 2^n$ elements and put them into the first set $ S_0$. Now form all the possible combinations of the length $ 2^{n - 1}$ out of those elements and assign them as "roots" of the remaining sets. Among those "root sets", $ A,B$ will be called complementary if $ A\cap B = \emptyset\land A\cup B = S_0$. Now take another ${ 2^n - 1}$ elements (i.e. such that none is already contained in $ S_0$) and form all the possible combinations of the length $ 2^{n - 1}$ out of them. A particular combination of the second collection of elements will be united with two arbitrary chosen complementary "root sets", thus forming two of the desired sets.

It is easy to show that $ \binom{2^{n}}{2^{n - 1}} = 2\cdot\binom{2^n - 1}{2^{n - 1}}$, hence such an algorithm can always be performed.

Your suggested construction doesn't work.  Here are several reasons why:
1) When I said (in the parenthetical in my previous post) that "This is actually pretty clear, in that if you try to build a solution by hand you essentially are forced to go through the construction I just described," I was right, and if you actually try to construct these sets by hand you'll quickly see that this is the case.
2) A collection of sets closed under taking symmetric differences must have order $ 2^m$ for some $ m$: such a set is actually an abelian group under $ \Delta$ (the identity is the empty set) such that every element is self-inverse.  But such a group is of the form $ \left({\bf Z}/2{\bf Z}\right)^m$ for some $ m$.  In our case, we are excluding the empty set, so the answer must be $ 2^m \minus{} 1$ for some $ m$.
3) Suppose we begin with the set $ A \equal{} \{1, 2, 3, 4, 5, 6, 7, 8\}$.  Then the second set we add is generically of the form $ B \equal{} \{1, 2, 3, 4 ,9, 10, 11, 12\}$.  Taking symmetric differences, we must include the set $ C \equal{} \{5, 6, 7, 8, 9, 10, 11, 12\}$.  Now you claim that I can include some set that intersects $ A$ in the set $ \{1, 2, 3, 5\}$.  This set must not contain any of 4, 6, 7 or 8.  Then its intersection with $ B$ must contain exactly one of 9, 10, 11 and 12.  But also its intersection with $ C$ must contain exactly three of 9, 10, 11 and 12.  This is obviously a contradiction.

To rescue oneself from the contradiction we came across in (3), one must follow the construction I suggested in my previous post.  And this is what I mean by (1).


${2 (3 n-1)^{n}\geq (3 n+1)^{n}}$

${\left(\frac{3 n-1}{3 n+1}\right)^{n}\geq \frac{1}{2}}$
${\left(1-\frac{2}{3 n+1}\right)^{n}\geq \frac{1}{2}}$

${f(n)=\left(1-\frac{2}{3 n+1}\right)^{n}}$

This function is slowly incressing for $n \geq 1$. Remember Euler.
So it have a minimum on $n = 1$

${f(1)=\frac{1}{2}}$

Analagous we can see that

$e^{-{\frac{2}{3}}}\geq f(n) \geq 1$

I have a way without calculus, but i think it is a horrible solution, someone should look for a better one.
[hide]
[b]Lemma:[/b] For each positive integer $n$ and nonnegative real number $x$, we have
\[x^{n}\geq 1+n(x-1)+\binom{n}{2}(x-1)^{2}+\binom{n}{3}(x-1)^{3}\]
[b]Proof:[/b] If $n\leq 3$, we have equality. For $n\geq 4$, factor $x^{n}-1$ into $(x-1)(1+x+x^{2}+\cdots+x^{n-1})$. Also, we have
\[(1+x+x^{2}+\cdots+x^{n-1})-n=(x-1)\sum_{k=0}^{n-2}(n-1-k)x^{k}\]
\[\left( \sum_{k=0}^{n-2}(n-1-k)x^{k}\right)-\binom{n}{2}=(x-1)\sum_{k=0}^{n-3}\binom{n-1-k}{2}x^{k}\]
\[\left( \sum_{k=0}^{n-3}\binom{n-1-k}{2}x^{k}\right)-\binom{n}{3}=(x-1)\sum_{k=0}^{n-4}\binom{n-1-k}{3}x^{k}\]
(These all follow from hockey stick.) So basically, this gives us
\[x^{n}-\left[ 1+n(x-1)+\binom{n}{2}(x-1)^{2}+\binom{n}{3}(x-1)^{3}\right]\]
\[=(x-1)^{4}\sum_{k=0}^{n-4}\binom{n-1-k}{3}x^{k}\geq 0\]

Now we can deal with the problem by applying the lemma to $x=1-\frac{2}{3n+1}$, so we get
\[\left( 1-\frac{2}{3n+1}\right)^{n}\geq 1-\frac{2n}{3n+1}+\frac{2n(n-1)}{(3n+1)^{2}}-\frac{4n(n-1)(n-2)}{3(3n+1)^{3}}\]
The right hand side is at least 1/2, after multiplying out, it reduces to $(n-1)(n^{2}+10n-3)\geq 0$.
[/hide]


Find the explicit formula for the nth term of the sequence: 5,8,10,12,14,16,18,....

(remember: explicit not recursive and the sequence starts with +3 and then continues with +2).

What do you mean by your remember statement... surely it needs punctuation or something, because I can't figure out what you meant.

I would use a piecewise formula but you can always use a_n=2n+4-floor(1/n), where floor(x) is the greatest integer less than or equal to x.


$ r \plus{} s \equal{} 44 \minus{} r \times s$
$ s \plus{} t \equal{} 12 \minus{} s \times t$
$ t \plus{} r \equal{} 74 \minus{} t \times r$

solve for negative real values of $ r, s, t$.

Hint: Let r' = 1+r, s' = 1+s and t' = 1+t

hello, from the first equation we have $ s \equal{} \minus{} \frac {r \minus{} 44}{1 \plus{} r}$, $ r \equal{} \minus{} 1$ is impossible. Inserting this in your second equation we have $ \minus{} \frac {13r \minus{} 32 \minus{} 45t}{1 \plus{} r} \equal{} 0$, from here we get $ t \equal{} \frac {13}{45}r \minus{} \frac {32}{45}$, inserting this in your last equation we have to solve $ \frac {13}{45}r^2 \minus{} \frac {3362}{45} \plus{} \frac {26}{45}r \equal{} 0$, with $ r_1 \equal{} \minus{} 1 \plus{} \frac {15}{13}\sqrt {195}$,$ r_2 \equal{} \minus{} 1 \minus{} \frac {15}{13}\sqrt {195}$.
Sonnhard.

[hide="Answer"]

$ r \plus{} s \equal{} 44 \minus{} r \times s$

follows

$ (r \plus{} 1) \times (s \plus{} 1) \equal{} 45$

and by same way we get

$ (s \plus{} 1) \times (t \plus{} 1) \equal{} 13$

and

$ (t \plus{} 1) \times (r \plus{} 1) \equal{} 75.$

Multiplying them follows

$ \{(r \plus{} 1) \times (s \plus{} 1) \times (t \plus{} 1) \}^2 \equal{} 45 \times 13 \times 75.$

So $ (r \plus{} 1) \times (s \plus{} 1) \times (t \plus{} 1) \equal{} \pm \sqrt {45} \times \sqrt {13} \times \sqrt {75}$

and so

$ r \plus{} 1 \equal{} \pm \frac {\sqrt {45} \times \sqrt {75}}{\sqrt {13}}$

$ s \plus{} 1 \equal{} \pm \frac {\sqrt {45} \times \sqrt {13}}{\sqrt {75}}$

$ t \plus{} 1 \equal{} \pm \frac {\sqrt {13} \times \sqrt {75}}{\sqrt {45}}$

$ r \equal{} \minus{} 1 \pm \frac {15 \sqrt {15}}{\sqrt {13}}$

$ s \equal{} \minus{} 1 \pm \frac {\sqrt {39}}{\sqrt {5}}$

$ t \equal{} \minus{} 1 \pm \frac {\sqrt {65}}{\sqrt {3}}$

[/hide]


function
algebra
domain
algebra proposed

Find all functions $ f: \mathbb N \rightarrow \mathbb N$ satisfying the condition:
\[ f\left(f\left(n\right) \minus{} n\right) \plus{} f\left(f\left(n\right) \plus{} n\right) \equal{} 8n
\]
for all $ n\in\mathbb N$.

This is only a hint cause I still stuck with it.

Let a = f(n) + n
     b = f(n) - n

So, f(a) + f(b) = 4a - 4b
  with a, b

f(a) + f(a) = 2f(a) = 4a - 4a = 0.

So, f(x) = 0 ??????

I don't think you can set $a=b$ since that would imply $f(n)+n = f(n)-n$, which is clearly absurd.

Hint
As $f$ is defined from $N$ to itself then $f(n)>=n$
This gives $4n>=f(n)>=n$ then $f(0)=0$

But its seems for me that not only $f(n)=2n$ satisfy :?
Starting for example whit $f(1)=1$ i found no contradiction!
But If $f$is increasing :?: then contradiction!

i think that the $f(n)=2n$ is the only solution for this problem!
i have no more moves sorry!  :mad:

Anybody can find correct solution of this??

I was trying to think of this as lattice points. I've got the same conclusion, since we cannot send values to negative ones, because we're restricting the domain to naturals, we have

$4n\geq f(n)$

Well, as we can see by lattice points, we know that $f(0)=0$, but only if you consider 0 as a natural number. For $4\geq f(1)$, there aren't many possibilities...try a construction of this functions by the lattice points...maybe we can have a not monotonic one...i'll think more about

very nice.
somebody has one complete solution for this problem?  :)


trigonometry
complex numbers
absolute value

There are $2n$ complex numbers that satisfy both $z^{28}-z^{8}-1=0$ and $|z|=1$. These numbers have the form $z_{m}=\cos\theta_{m}+i\sin\theta_{m}$, where $0\leq\theta_{1}<\theta_{2}< \dots <\theta_{2n}<360$ and angles are measured in degrees. Find the value of $\theta_{2}+\theta_{4}+\dots+\theta_{2n}$.

[hide="Hint"] Problem just screams roots of unity.  [/hide]

[hide="If I am lucky"]
...and I'm probably not, but here goes.

by demoivres, get
cos28x+isin28x-1=cos8x+isin8x

notice that the imaginary parts on both sides must be equal, so sin28x=sin8x.

this means cos28x-1=cos8x
putting that in the unit circle, we get 8x=+/- 120, 28x=+/-60

setting up coterminal angles (8x=360k+/-120, 28x=360j+/-60), we get
k=(2j-2)/7

plugging in a few values of j that make k an integer, we get 105, 195, 285, and 15. by the conjugate root theorem, we then get 75, 345, 255, and 165. we add up the ones we need, and get 840.

is that right? the solution isn't great...[/hide]

[hide="not sure what I am missing"]
we know $\sin8\theta=\sin28\theta$ and $\cos8\theta+1=\cos28\theta$
this only works if $8\theta$ is coterminal with $120^\circ$ and $28\theta$ is coterminal with $60^\circ$, or $8\theta$ is coterminal with $-120^\circ$ and $28\theta$ is coterminal with $-60^\circ$

so I set up this system:
$8\theta=360k+120$
$28\theta=360j+60$
$20\theta=360(j-k)-60$
$\theta=18(j-k)-3$
$8\theta=144(j-k)-24=360k+120$
$144(j-1)=504k$
$2(j-1)=7k$
obviously, $j$ and $k$ must be integers.
the first few ordered pairs, $(8,2),(15,4),(22,6),(29,8),(36,10)$ yield $\theta$ values of $105^\circ,195^\circ,285^\circ,15^\circ,105^\circ$, respectively, so I have hit a repeat. By the conjugate root theorem, $255^\circ,165^\circ,75^\circ,345^\circ$ must be solutions also.

To check the other side, where $28\theta$ is coterminal with $-60^\circ$, I set up a similar system:

$8\theta=360k-120$
$28\theta=360j-60$
$20\theta=360(j-k)+60$
$\theta=18(j-k)+3$
$8\theta=144(j-k)+24=360k-120$
$144(j+1)=504k$
$2(j+1)=7k$
this gives these ordered pairs: $(6,2),(16,4),(20,6),(27,8),(34,10)$, giving $\theta$ values that have already been discovered, so these must be all solutions.

to answer the question, I get $75+165+255+345=\boxed{840}$
[/hide]

I actually think I was right, because Kalva agrees with me. But I'm not sure if Altheman was saying I was wrong or not...

i was just suggesting another method for solving the problem

$z^{8}(z^{20}-1)=1$, $|z|^{8}|z^{20}-1|=1$, $|z^{20}-1|=1$

consider on the complex plane, $z^{20}$ is the intersection of two unit circles, one centered at 0 [since $|z^{20}|=1$], the other centered at 1 [since $|z^{20}-1|=1$], making two equilateral triangles, or w/e, basically, $z^{20}=\text{cis}\, \pm 60$. So the solution set is a subset of: $\text{cis}\, \pm 3+18k$ for $k\in\mathbb{Z}$

I guess we have to check these using demoivre's theorem...

$\pm 144+144k=360j$
$2(\pm 1+k)=5j$
for +, k=4,9,14,19...
for -, k=1,6,11,16

so our answer is $3*4+18(4+9+14+19)=840$

The solution looked better before I noticed that I added a lot of solutions.

[quote="mdk"]Can someone please explain why the solution set is a subset of cis+-3+18k and everything after that ?[/quote]

z^20=cis 60, or cis-60

so z=cis 3+18k, or cis -3+18k
[this should be easy]

so i have to check the solutions back in the original equation...because that is 40 solutions...and it is degree 28 initially

so i plugged that back in
note z^20-1=cis +- 60
we know that
z^8=+-24+144k

so we multiply those, that means adding the angles
cis +-144+144k=1
so that means +-144+144k=360j for some integer j

then reduce or w/e...easy from here to find pairs for k,j

[quote="Altheman"]$ z^{8}(z^{20} - 1) = 1$, $ |z|^{8}|z^{20} - 1| = 1$, $ |z^{20} - 1| = 1$

consider on the complex plane, $ z^{20}$ is the intersection of two unit circles, one centered at 0 [since $ |z^{20}| = 1$], the other centered at 1 [since $ |z^{20} - 1| = 1$], making two equilateral triangles, or w/e, basically, $ z^{20} = \text{cis}\, \pm 60$. So the solution set is a subset of: $ \text{cis}\, \pm 3 + 18k$ for $ k\in\mathbb{Z}$

I guess we have to check these using demoivre's theorem...

$ \pm 144 + 144k = 360j$
$ 2(\pm 1 + k) = 5j$
for +, k=4,9,14,19...
for -, k=1,6,11,16

so our answer is $ 3*4 + 18(4 + 9 + 14 + 19) = 840$

The solution looked better before I noticed that I added a lot of solutions.[/quote]

How did you get this first step?
$ z^{8}(z^{20} - 1) = 1$, $ |z|^{8}|z^{20} - 1| = 1$, $ |z^{20} - 1| = 1$
how do you know that z^8 is equal to 1?

He didn't use that. He used the fact that $ |ab| \equal{} |a| \cdot |b|$

He took the absolute value of both sides of the equation. The absolute value of $ 1$ is just $ 1$.

The absolute value of $ z^8$ is $ |z^8| \equal{} |z| ^8$, using that identity at the top of the post. Because it is given that $ |z| \equal{} 1$, we know that $ |z|^8$ is also equal to $ 1$. That doesn't mean that $ z^8$ is equal to $ 1$, though.

And then, the absolute value of $ z^{20}\minus{}1$ was left unsimplified.

[quote="TZF"]He didn't use that. He used the fact that $ |ab| \equal{} |a| \cdot |b|$

He took the absolute value of both sides of the equation. The absolute value of $ 1$ is just $ 1$.

The absolute value of $ z^8$ is $ |z^8| \equal{} |z| ^8$, using that identity at the top of the post. Because it is given that $ |z| \equal{} 1$, we know that $ |z|^8$ is also equal to $ 1$. That doesn't mean that $ z^8$ is equal to $ 1$, though.

And then, the absolute value of $ z^{20} \minus{} 1$ was left unsimplified.[/quote]

Thank you, but how do we know |z^8|=|z|^8?

Generally, modulus of a product is a product of modulii: answer to your question follows immediately.

[quote]we know $\sin8\theta=\sin28\theta$ and $\cos8\theta+1=\cos28\theta$
 this only works if $8\theta$ is coterminal with $120^\circ$ and $28\theta$ is coterminal with $60^\circ$, or $8\theta$ is coterminal with $-120^\circ$ and $28\theta$ is coterminal with $-60^\circ$
[/quote]
Why is this true? Can you prove
Thanks

the first line is derived from demoivre's and matching up the real and imaginary parts
if $8\theta = a$ and $28\theta = b$ then we have $\sin{a}=\sin{b}$ and $\cos{a}+1=\cos{b}$, meaning that $a=28\theta$ and $b=8\theta$ are not coterminal, since their sine values are equal but their cosine values aren't.
from there, you square both equations and add them, to get $\sin^{2}{a}+\cos^{2}{a}+2\cos{a}+1=\sin^{2}{b}+\cos^{2}{b}\rightarrow 2+2\cos{a}=1\rightarrow \cos{a}=-\frac{1}{2}$

wait what's wrong with this i thought i had it down but it looks like im slightly off (p10):

Note that since $|z^8|=1$, $z^8$ lies on the unit circle, so $z^8+1$ lies on the circle shifted one to the right. Specifically, the equation of the unit circle is $x^2+y^2 = 1$, and the equation of the circle shifted one to the right is $(x-1)^2+y^2 = 1$. But $|z^{28}| = |z^8+1| = 1$, so $z^8$ must be either of the intersections of these two circles, which is at $\text{cis} \pm 60$, but this is apparently wrong according to [url=https://artofproblemsolving.com/wiki/index.php/2001_AIME_II_Problems/Problem_14]official sol??[/url] it seems like the circle should be $(x+1)^2+y^2=1$ but why would it be $x+1$ instead of $x-1$?? agh

bumping i'm still not sure what's wrong here

[quote=OlympusHero] it seems like the circle should be $(x+1)^2+y^2=1$ but why would it be $x+1$ instead of $x-1$?? agh[/quote]

I think $z^8+1$ is supposed to be one of the intersection points of $x^2+y^2=1$ and $(x-1)^2+y^2=1$, not $z^8$; that might be where your confusion is arising from.

@above Thanks!

Note that since $|z^8|=1$, $z^8$ lies on the unit circle, so $z^8+1$ lies on the circle shifted one to the right. Specifically, the equation of the unit circle is $x^2+y^2 = 1$, and the equation of the circle shifted one to the right is $(x-1)^2+y^2 = 1$. But $|z^{28}| = |z^8+1| = 1$, so $z^8+1$ must be either of the intersections of these two circles, which is at $\text{cis} \pm 60$. Therefore, $z^8 = -\frac{1}{2} + i \frac{\sqrt3}{2}, -\frac{1}{2} - i \frac{\sqrt3}{2}$, or $z^8 = \text{\cis 120}, \text{\cis 240}$. The solutions to $8x \equiv 120 \pmod {360}$ and $8x \equiv 240 \pmod {360}$ are $x \equiv 15 \pmod {45}$ and $x \equiv 30 \pmod {45}$ respectively. Therefore, the angles $15, 30, 60, 75, 105 \ldots$ work... but that's wrong. What happened?

I used de moivre's. Nice complex number problem


geometry
inradius
angle bisector
geometry unsolved

[quote="primoz2"]Let ABC be a triangle and I the center of its incircle. Let D be the touching point of AB with the incircle. Let T be on the line CD, such that CT=TD. Let  L be the intersecsion point of TI and AB. Prove that,  AL=LB[/quote]
This is most likely not the best way to solve it, but here is my solution:
[hide="Solution"]
The problem is equivalent to showing that $ T$, $ I$, and $ M$ are collinear, where $ M$ is the midpoint of $ BC$.  Draw $ AI$ and let it meed $ BC$ at $ K$.  Now, let $ AB \equal{} x \plus{} y$, $ BC \equal{} y \plus{} z$, and $ AC \equal{} x \plus{} z$ where $ BD \equal{} y$ and $ DC \equal{} z$.  Drop altitude $ AX$ to $ BC$.  Notice that $ \triangle KID\sim \triangle KAX\implies \frac {AK}{IK} \equal{} \frac {AX}{ID} \equal{} \frac {AX}{r}$, where $ r$ is the inradius of $ \triangle ABC$.  Also, notice that $ BC\cdot AX \equal{} 2[ABC] \equal{} r(AB \plus{} BC \plus{} AC)$.  Thus, $ \frac {AX}{r} \equal{} \frac {AB \plus{} BC \plus{} AC}{BC}$.  Hence, we have that $ \frac {AI}{IK} \equal{} \frac {AK}{IK} \minus{} 1 \equal{} \frac {AX}{r} \minus{} 1 \equal{} \frac {AB \plus{} AC}{BC} \equal{} \frac {2x \plus{} y \plus{} z}{y \plus{} z}$.

By the Angle Bisector Theorem, we have that $ \frac {BK}{CK} \equal{} \frac {AB}{BC}$ and $ BK \plus{} KC \equal{} BC$, so $ BK \equal{} \frac {BC\cdot AB}{AB \plus{} AC}$ and $ CK \equal{} \frac {AC\cdot BC}{AB \plus{} AC} \equal{} \frac {(x \plus{} z)(y \plus{} z)}{2x \plus{} y \plus{} z}$.  Hence $ KM \equal{} CK \minus{} CM \equal{} \frac {(x \plus{} y)(y \plus{} z)}{2x \plus{} y \plus{} z} \minus{} \frac {y \plus{} z}{2} \equal{} \frac {(y \plus{} z)(z \minus{} y)}{2(2x \plus{} y \plus{} z)}$ and $ MD \equal{} MB \minus{} BD \equal{} \frac {y \plus{} z}{2} \minus{} y \equal{} \frac {z \minus{} y}{2}$.  Therefore, $ \frac {KM}{DM} \equal{} \frac {\frac {(y \plus{} z)(z \minus{} y)}{2(2x \plus{} y \plus{} z)}}{\frac {z \minus{} y}{2}} \equal{} \frac {y \plus{} z}{2x \plus{} y \plus{} z}$.

Finally, we can finish.  Notice that $ \frac {AI}{IK}\cdot \frac {KM}{DM}\cdot \frac {DT}{AT} \equal{} \frac {2x \plus{} y \plus{} z}{y \plus{} z}\cdot \frac {y \plus{} z}{2x \plus{} y \plus{} z}\cdot 1 \equal{} 1$ and by Menalus's Theorem, we have that $ T$, $ I$, and $ M$ are collinear, as desired. [/hide]

Let $ D'$ be, the tangency point of $ AB,$ to the $ C$-excircle. So we have $ LD \equal{} LD',$ where $ L$ is the midpoint of the side-segment $ AB.$

It is well known (and easy to prove) that the line segment $ LI,$ is parallel to the line segment $ CD'$ $ ($ see at the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=181220]Parallel segments - Incircle[/url] $ ).$

So, the line segment $ LI,$ passes through the midpoint $ T$ of the segment $ CD$ and the proof is completed.

Kostas Vittas.

The problem is the same as the one posted in http://www.mathlinks.ro/Forum/viewtopic.php?t=184582 .

  darij


abstract algebra
algebra
function
domain
Ring Theory
superior algebra
superior algebra solved

Hi, I have this problem: show that if a commutative ring A is noetherian and of dimension greater than 1, then Spec(A) is infinite. Don't have any idea!! Thanks! :)

[quote="-oo-"]clearly wrong (take some discrete valuation ring, such as $ \mathbb{Z}_p$).[/quote]

DVRs are of dimension 1 aren't they? I think greater there means really greater and not greater or equal.

Anyway. Because R is noetherian, it has finitely many minimal primes and because of correspondences between Spec(R) and Spec($ R_p$) and Spec(R/p), we can assume that R is a local domain of dimension 2 and we have to prove that there are infinitely many primes of height 1.
Assume for a contradiction there are finitely many. By the prime avoidance lemma, we can then pick for each prime ideal of height 1 an element that is in this ideal and not in the rest of them. Sum all these elements. This sum is not in any of the ideals of height 1 so in particular it is not 0. But then Krull's Haupt... gives the contradiction.


[hide]Add $ \minus{}\cos A\cos B\cos C\plus{}\cos A\cos B\cos C$ to get

$ \sin^2A\plus{}\cos A(\sin B\sin C\minus{}\cos B\cos C)\plus{}\cos A\cos B\cos C$

$ \equal{}\sin^2A\plus{}\cos A(\minus{}\cos (B\plus{}C))\plus{}\cos A\cos B\cos C$

$ \equal{}\sin^2A\plus{}\cos^2 A\plus{}\cos A\cos B\cos C$

$ \equal{}1\plus{}\cos A\cos B\cos C$

In order to maximize, we assume $ \cos A,\cos B,\cos C>0$.  Thus, by AM-GM and Jensens, we have

$ \cos A\cos B\cos C\le\left(\frac{\cos A \plus{}\cos B\plus{}\cos C}{3}\right)^3\le\cos^3 \frac{\pi}{3}$

Thus, the maximum value is $ \frac{9}{8}$.[/hide]

[quote="WindingFunction"]If $ A$, $ B$, and $ C$ denote the angles of a triangle, determine the maximum value of
\[ \sin^2A \plus{} \sin B\sin C\cos A.
\]
[/quote]

\[ \sin^2A \plus{} \sin B\sin C\cos A\equal{}\frac{a^2}{4R^2}\plus{}\frac{b}{2R}\cdot\frac{c}{2R}\cdot\frac{b^2\plus{}c^2\minus{}a^2}{2bc}\equal{}\frac{a^2\plus{}b^2\plus{}c^2}{8R^2}\equal{}\frac{9R^2\minus{}OH^2}{8R^2}\le \frac{9R^2}{8R^2}\equal{}\frac{9}{8}\]
Equality occurs when $ OH\equal{}0$... The triangle is equilateral.

May I ask the source of this problem?


Hi every1,

Help frnds. This problem is killing me. Plzzz kill it---

there is a 6-digit number, m, with the following property--

2m is permutation of digits in m.

so is 3m

so is 4m

so is 5m

so is 6m....

Find m.

The only advancement (if it can be called advancement) i made was 9|m.

Plz solve it. thanx in advance

Well, we have this property for the number 142857 which is the period of 1/7.
Obviously the nuber cannot contain two identical digits (no two multiples can begin with the same digit) and can't contain 0.
We have 6m<=999999 so the first digit is 1 and the second is at most 6.
Suppose it doesn't contain 2. So 2m must begin with 3 and the second digit in m is 5 or 6.
For m=15**** we have 450000<3m<480000 and 600000<4m<640000, so no multiple of m begins with 5 - contradiction.
For m=16**** we have 320000<2m<340000, 2m can't contain 2 and can't begin with 33, so contradiction.
So m contains 2 and  the second digit is at most 4.
Suppose m contains 9. Then we need m to be more than 900000/6=150000, contradiction.
If m=12****, we have  360000<3m<390000. But 6m<840000, so 3m can't contain 8 and must begin with either 36 or 37. If it begins with 36, we have 2m begining with 24, so some multiple must begin with 4. It's obviusly 4m so m<125000, contradiction. If 3m begins with 37, we have 4m begining with either 49 or 50, both contradiction.
Suppose m contains 3. The number 2m begins with 2, so the number begining with 3 must be 3m. Because m can't begin with 12, we have m=13****, then 3m=39****, contradiction.
If m contained 6, 5m would begin with 6 (because 4m<4.150000 and 6m>6000000). So we would have m begining with 12 or 13, both contradictions.
So m begins with 14, the other digits are 2, 5, 7, 8. It's easy to show that no two multiples may end with the same digit, so 2 and 8 can't be at the end (6m ends the same as m then), as well as 5, because 2m can't end with 0. So the number ends with 7. So we have m=14***7, wher the asterisks can be replaced by 2,5 and 8 only. The six cases are very easy to finish...
I know my solution is messy, but i don't have much time right now.

Other numbers exhibit this property - for example, the first six digits of the decimal expansion of $\frac{1}{13}$ exhibit this property for [b]odd[/b] multiples.

Can anyone generalize?  :)

[hide="Hint 1"]Find a $16$-digit number with this property ;) [/hide]

[hide="Hint 2 (well, its not much of a hint, if this made sense to you youd already know how to solve the problem)"] Residue classes ;) [/hide]

Let p be a prime, k=ord(10) mod p. Then the first k digits of 1/p have the property we want when taking some (p-1)/k multiples (for 13 these arent the odd ones, but the residues of $10^i$ mod 13, which are 1, 3, 4, 9, 10, 12).  This is quite obvious and the answer for hint 1 follows...
But in these eneral cases we have to include the zero's in the begining.  For 13 we have just to take m=076923 and for 17 m=0588235294117647.

But how would you show that this solution for 17 (or other prime>7) is the only one? And is there some way how to find all the primes mod which 10 is primitive root?


modular arithmetic
number theory proposed
number theory

Let $ n$ be a natural number, such that $ (n,2(2^{1386}\minus{}1))\equal{}1$. Let $ \{a_{1},a_{2},\dots,a_{\varphi(n)}\}$ be a reduced residue system for $ n$. Prove that:\[ n|a_{1}^{1386}\plus{}a_{2}^{1386}\plus{}\dots\plus{}a_{\varphi(n)}^{1386}\]

You can easily prove that $ \{2a_{1},2a_{2},\dots.2a_{\varphi(n)}\}$ is a reduced residue system for $ n$ and then
 $ a_{1}^{1386}\plus{}a_{2}^{1386}\plus{}\dots\plus{}a_{\varphi(n)}^{1386}\equiv (2a_{1})^{1386}\plus{}(2a_{2})^{1386}\plus{}\dots\plus{}(2a_{\varphi(n)})^{1386}(mod n)$ and we get result .

This problem feels somewhat weird.

Notice that $\{2a_1, 2a_2, \cdots, 2a_{\phi(n)}\}$ is also a complete reduced residue system mod $n$ because $n$ is odd. As a result, we have the congruence $$\sum_{i=1}^{\phi(n)} (2a_i)^{1386}  = 2^{1386} \sum_{i=1}^{\phi(n)} a_i \equiv (2^{1386} - 1) \sum_{i=1}^{\phi(n)} a_i \pmod n.$$ So $n$ divides their difference, which is the result.


algebra polynomial function calculus integration quadratics

Just draw some fourth degree polynomials with leading coefficients of 1 for #12 and see what happens. The shape of them will be -infinity at the left, increasing to some value a, decreasing down to a value b, a>b, and then increasing to +infinity.

18) Log(1)=0 Log(0) approaches -infinity, therefore cos(log(x)) on 0<x<1 is the same as cos(y) on -infinity<y<0 so there are infinitely many zeros.

if you have P(x) and Q(x), both fourth degree polynomials with leading coefficient 1, [hide]the degree of P(x)-Q(x) is at most 3, meaning there are at most 3 roots of P(x)-Q(x) = 0 or P(x)=Q(x)[/hide]

Theres a standard way for no. 17:

17. Let P(x) be a polynomial such that when P(x) is divided by x - 19, the remainder is 99, and when P(x) is divided by x - 99, the remainder is 19. What is the remainder when P(x) is divided by (x - 19)(x - 99) ?

[hide]

Write P(x) = Q(x)(x-99)(x-19) + ax + b, and sub in x = 19 and x = 99.

[/hide]

[quote="MagnusRex"]I need help understanding these:

12. What is the maximum number of points of intersection of the graphs of two different fourth-degree polynomial functions y=P(x) and y=q(x) each with different leading coefficient 1?

17. Let P(x) be a polynomial such that when P(x) is divided by x - 19, the remainder is 99, and when P(x) is divided by x - 99, the remainder is 19. What is the remainder when P(x) is divided by (x - 19)(x - 99) ?

18. How many zeroes does f(x) = cos(log(x)) have on the interval 0<x<1?[/quote]

12. p(x)-q(x) is a third degree polynomial and we're trying to find the maximum number of roots it has. hmmm...i'm guessing 3

17. p(x)=q(x)(x-19)(x-99)+ax+b. plug in x=19 and x=99 => 99=19a+b, 19=99a+b, a=-1, b=118. ax+b is 118-x.

18. cos(log(x))=0 => log x=pi/2+k*pi (integral k), x=10^(pi/2+kpi). We want to find how many of these are between 0 and 1. We can find an upper bound for k easily (though I don't feel like it). However, there is no lower bound for k, and thus k can range all through the integers less than the upper bound  = infinite roots.

note to 17. q(x) is the quotient polynomial, btw. As we are dividing by (x-19)(x-99), which is a quadratic, the remainder will be of degree less than that, i.e. first-degree. So we write this as ax+b. That is the remainder.  a and b are just the coefficients of the remainder that we are trying to find. And ... um...this is just a standard method. Remember it.

Mystic, I don't like your way for #18 much. It's a bit more verbose than necessary.

blah. i was trying to be nice and polite and helpful. blah.

[quote="MagnusRex"]12 and 18: I now understand very well, thanks Nukular and Confuted.

17. TripleM: Ok but how did you get that/(why does it work), and what are a and b?[/quote]]

You are dividing a polynomial P(x) by a quadratic.. so you must get a remainder that is linear. Call that ax+b, and you P(x) = something x the quadratic + remainder..

its in the polynomial chapter in aops v2 btw


trigonometry
analytic geometry
function
geometry
geometric transformation
logarithms
trig identities

[quote="A+MATH"]No i dont do trig yet finishi algebra 2 this year. :)[/quote]
Well then don't reply to this.

[quote="TheSouthpaw"]Yeah, I already know the very basics of trig, but not enough for any high-school level math competitions. What I'm looking for is a book that takes you through all the necessary trig identities (even some of the obscure ones) with derivations and excercises....does anyone have suggestions?[/quote]

In my humble opinion, the trigonometry book by my avatar, I. M. Gelfand, is one of the best. Here is my FAQ file on the whole Gelfand series:

My son has taken an algebra 1 course by correspondence and will take a classroom-based accelerated algebra I & II course in the coming school year. Although he has already had exposure to other algebra materials, I will be going through algebra-level materials with him from the Gelfand Correspondence Program in Mathematics

http://gcpm.rutgers.edu/

such as I. M. Gel'fand and A. Shen's book Algebra, which I bought on-line from Singaporemath.com

http://www.singaporemath.com/

You only get "coverage" of all the usual topics of beginning algebra courses in the United States if you use that with Gelfand's The Method of Coordinates and his Functions and Graphs, but those are also very good books. The books were originally designed for a correspondence course so they are largely self-teaching. They are a good bit more challenging and deep than anything about algebra presented to most English-speaking students, while still being very engaging and approachable. It's reassuring to me that Gelfand is a world-famous mathematician and math educator, who formerly lived in the Soviet Union and now lives in the United States. Professor Hung-hsi Wu of Berkeley's math department has an interesting review of the Gelfand texts at his Web site

http://math.berkeley.edu/~wu/

which appears to refer to the first printing of the Algebra book, now available in a corrected second printing. As I recall, I discovered the Gelfand-Shen textbook through Professor Richard Askey's review of that book.

http://www.aft.org/publications/american_educator/fall99/amed1.pdf

Askey's review is actually a review of Ma Liping's book Knowing and Teaching Elementary Mathematics that quotes a passage of Gelfand's book in a sidebar. My son enjoys both the THINKING and the humor in the Gelfand books, so they are a good relief from more tedious math programs written by lesser mathematicians. Gelfand's attitude toward teaching mathematics is summed up in a statement he makes in reaction to American math teachers: "I would like to make one comment here. Some of my American colleagues have explained to me that American students are not really accustomed to thinking and working hard, and for this reason we must make the material as attractive as possible. Permit me to not completely agree with this opinion. From my long experience with young students all over the world, I know that they are curious and inquisitive and I believe that if they have some clear material presented in a simple form, they will prefer this to all artificial means of attracting their attention--much as one buys books for their content and not for their dazzling jacket designs that engage only for the moment. The most important thing a student can get from the study of mathematics is the attainment of a higher intellectual level." Gelfand is optimistic--based on long experience--that many learners can attain a higher intellectual level in mathematics, perhaps because of his flexibility as a teacher: "Students have no shortcomings, they have only peculiarities. The job of a teacher is to turn these peculiarities into advantages."

My son was eight when we started using Gelfand's algebra. We mostly used Singapore the next couple of years, and later the EPGY correspondence course

http://epgy.stanford.edu/

for math through algebra 1. But we dip into Gelfand's Algebra every once in a while for a change of pace. The first several problems are easy for kids who have used Miquon Math, and then the Gelfand problems just keep relentlessly getting more challenging, and remain very interesting. I'll try to finish up Gelfand's Algebra with my son before his twelfth birthday. In the early going, when my son was eight, he would ASK FOR more problems from the Gelfand book, so I have to give Miquon Math credit for giving him a good attitude about math and Gelfand's book credit for setting some very interesting problems.

Amazon.com links (with prices from a few months ago):

http://www.amazon.com/exec/obidos/tg/detail/-/0817636773/

(Algebra $19.95)
http://www.amazon.com/exec/obidos/tg/detail/-/0486425649/
(Functions and Graphs $7.95)

http://www.amazon.com/exec/obidos/tg/detail/-/0486425657/

(Method of Coordinates $6.95)
http://www.amazon.com/exec/obidos/tg/detail/-/0817639144/
(Trigonometry $19.95)

The book republished by Dover called Sequences, Combinations, Limits and edited by S. I. Gelfand (who I think is I. M. Gelfand's son) appears to be part of the same series.

http://www.amazon.com/exec/obidos/tg/detail/-/0486425665/

(Sequences, Combinations, Limits $8.95)

Hope this helps!

u might want to look into that Southpaw, i heard that this gelfand guy has some pretty decent material out...

Here's a good one:

[url]http://www.amazon.com/exec/obidos/tg/detail/-/0534390501/qid=1064031874/sr=8-1/ref=sr_8_1/103-1466144-0993436?v=glance&s=books&n=507846[/url]

It's titled [u]Algebra and Trigonometry with Analytic Geometry[/u].

Hope it helps; I liked it.

-interesting_move (not posting much...)

[i]Mathematical Olympiad Challenges[/i] is good for trigonometry, but it's a rather advanced. Titu Andreescu is a big fan of trigonometry, so his problem books (or lecture notes, if you can get them) are excellent.

hm. I have that book (though I'm not going to attempt looking at it for another couple of months), but where can you get these lecture note things?

[quote]but where can you get these lecture note things?[/quote]

From a MOPper.

[quote="Tare"]You could always use the AoPS books, but I strongly recommend Japanese books. Even if you can't read Japanese, they're still much better than average textbooks. [b]That[/b]'s  how good they are.[/quote]

Tare, where would someone here in the Midwest of the United States obtain Japanese-language math books? I can read a little bit of Japanese (probably enough to follow a secondary math book) because I can read Chinese readily and have studied some Japanese. Thanks for any ordering sources for Japanese-language math books you can suggest.

P.S. I am already aware of the translations of Japanese secondary school mathematics textbooks from the University of Chicago School Mathematics Project, and have seen the seventh grade volume in that series by interlibrary loan. It is quite good--years ahead of American seventh grade books.

Haha I can get some Japanese math texts!

i think ill pass on those japanese texts. whats the symbol for sin in japan? some of u moppers should upload your lecture notes so the rest of us can gain some insight.

I speak Japanese fluently, it's quite understandable since all the mathematical notations that I know of are the same (ie all the Trig notations, operations, functions, etc.)

Correct, almost every single country in the world uses the same notations, and I think the trigs (sin, cos, tan, cosec, sec, cot) and logarithm too.

Try [url=http://www.amazon.co.jp]amazon[/url] first. If it's not there, then you could always contact individual publishers, but I'm not sure if you know Japanese enough to do that...

There is no way I'm going to scan in several hundred pages of MOP notes, even if they weren't 300 miles away.


Please could somebody just clear this up for me. Im ok with the first part of the sum but having probs with the last digit. This is calculating magnitude of acceleration.
10.6*10^3/5.3*10^3=    is it.........
2m/s^2
2*10^3m/s^2
or
something else??

If sombody could help me out that would be great. Thanks.

That depends on the units of the two numbers you divide. Normally it should be in m/s^2. (That is if the rest of the numbers are in m, m/s, s, etc.)


Let $T = \{9^k : k \ \text{is an integer}, 0 \le k \le 4000\}$.  Given that $9^{4000}$ has 3817 digits and that its first (leftmost) digit is 9, how many elements of $T$ have 9 as their leftmost digit?

[hide]First, let's examine how a number $9^n$ can have its leftmost digit as $9$. For this to be true, it must also be true that $9^{n-1}$ has a leftmost digit starts with $1$ and has the same number of digits.

With a similar arguement, we see that if $9^n$ starts with anyting but $9$, it has one more digit than $9^{n-1}$. We know that $9^{4000}$ has $3816$ more digits than $9^1$, so there are $3816$ cases in which the first digit of $9^n$ was something other than $9$. That leaves $\boxed{184}$ cases where the first digit of $9^n$ is $9$.[/hide]

Interesting part here

The  answer (About 183 or 184) would be  approximately .. same even if you connsider  a  set of $8^k$  [hide="or"](or in fact ANY number other than a power of 10.) [/hide]..instead of $9^k$

In other words    if $a$ is not a(integer)  power of 10,  about  4.6% of the numbers of the type $a^k$ would start with a 9.

Can anyone prove that?

I'm almost sure there is a theorem about that, I just don't remember the name  :(  pretty cool theorem!
it's something with logarithm.
and how do you prove the fact that if $9^{n}$ starts with anyting but $9$, it has one more digit than $9^{n-1}$?
thanks!
remembered  :)  http://en.wikipedia.org/wiki/Benford%27s_law

[quote="srulikbd"]How do you prove the fact that if $9^{n}$ starts with anyting but $9$, it has one more digit than $9^{n-1}$?[/quote]I'll use a leading digit of $8$ as an example. If $9^{n}$ has an 8 as the first digit, then it can be written in the form $8.\text{something}\times 10^{k}.$ for an integer $k.$

It follows that $9^{n-1}=\frac{8.\text{something}}{9}\times 10^{k}< 1 \times 10^{k}$ and we must decrease the power of $10$ by $1$ to compensate for this.

Uh...4everwise, Kalva says the answer is 184

[hide]The number on the left is 9 if and only if 9^(n-1) does not increase by one digits when multiplied by 9.

10^4000 has 4001 digits so you subtract 3817 from 4001 and get 184[/hide]

I think this logic seems right...

[quote]The number on the left is 9 if and only if 9^(n-1) does not increase by one digits when multiplied by 9.

$10^{4}000 has 4001$ digits so you subtract 3817 from 4001 and get 184 [/quote]

I dont really get your logic. Can you explain your solution in more detail. Thanx :wink:

I'm pretty sure my Ti-89 can perform that operation in a few seconds.

[quote="lingomaniac88"][quote="nsato"]...Given that $9^{4000}$ has 3817 digits and that its first (leftmost) digit is 9...[/quote]
Wow, apparently the people at AIME have a LOT of time on their hands, figuring something like that out.  Oh, boy...[/quote]

Sorry for bring back an old thread, first of all.
And also I was referring to this comment.

[quote="lingomaniac88"][quote="nsato"]...Given that $9^{4000}$ has 3817 digits and that its first (leftmost) digit is 9...[/quote]
Wow, apparently the people at AIME have a LOT of time on their hands, figuring something like that out.  Oh, boy...[/quote]

They could just use logs :wink: .

Yea, heh. Titaniums can do 9^4000 in a matter of seconds. In addition, finding the number of digits isn't that hard... just use logs.

Sorry for the bump, but I noticed that on the wiki for this problem (1990 AIME #13), the answer is 184, but on the contest page, the official solution gives an answer of 185 even though the correct answer is 184.

This may cause confusion for some people.

[quote=Phelpedo]Just a reminder that the "official" solution still has an answer of 185, which is wrong. Here's how I think about it:

[hide] As noted, $ 9^{n}$ has one more digit than $ 9^{n\minus{}1}$ except when $ 9^n$ starts with $ 9$. Starting with $ 9^0 \equal{} 1$, then we expect to add $ 1$ digit each time we multiply by $ 9$, except when the number starts with $ 9$. We do this $ 4000$ times and add $ 3816$ digits, so the number of times we got a first digit of $ 9$ was $ 4000\minus{}3816 \equal{} 184$. I do not think that the fact that $ 9^{4000}$ starts with $ 9$ is significant. [/hide][/quote]

I am pretty sure the fact that $9^{4000}$ ends in $9$ is not significant, what is significant is that $9^0=1,$ or else we could "lose" the $1$ because we are decrementing for $n-1$.

\(\lfloor (x+1)\log_{10}(9) \rfloor=\lfloor (x\log_{10}(9)) \rfloor\) implies that the left most digit is same where x varies from \({1,2,3....,3999}\) and now because \(9^{4000}\) has \(3817\) digits we can claim that the number of times 9 must have come on the front is \(4000-3817+1\)


just got out and am sitting in econ class, supposedly working on my presentation

so how hard do u think it was?

I'm eating lunch at home(my school is 600 steps, 5 minutes away). I was planning to make this thread when I turned on my computer, hoping that no one hasn't came home yet. :D
I voted for "a joke". First hour was all I needed. Unless I made a stupid mistake(hopefully not), I think I got them all.

lightrhee, you made IMO didn't you? Unless I'm mixing you up with someone else. Anyway, it [i]should[/i] be really easy for you! ;)

now, i'm not any good at math, but i thought it was easy (or, easier than normal).  i just love to make stupid mistakes on every other question.  I never liked the comc anyway, aime is where it's at.  Of course if you're IMO level, that speaks for itself.

Thought it was tougher than in recent years (didnt vote on poll though.)

ps. David Rhee, how can you solve all in 1 hour?  wow

Tougher.... I expect about the same score as last year but I made fewer stupid mistakes this time around (argh for solving for the wrong angle in A7 though)

I think that's easy, since it's not tougher than I expected.
I got everything correct, but I hope there's no writing mistakes(since my english always get "sick" when I'm impatient, my grammar become really funny, even started to use different [b]tenses[/b] in writing answers...) :P  :lol:

Probably easier, but that doesn't mean I did well.

For me the most annoying thing was that, the first second opening the test, I found the print was a lot smaller than the past exams I downloaded from the website. :blush:


Hey everyone!

I have a few questions that have been bothering me for quite a bit of time.

#1) How much in depth do I usually have to go for Bio SAT II and AP Bio? Do I need to know stuff like NAD is really nicotinamide adenine dinucleotide? Should I know the nine steps of glycolysis? What about some special molecules of the electron transport chain?

#2) I've always wondered why they don't teach chemistry before biology. All the energy transfer stuff would be so much easier with a foundation in chemistry. For instance, I know that the electron transport chain pumps hydrogen ions with the energy from the electrons, but what exactly is the chemical mechanism behind that? I don't feel entirely comfortable with just knowing that somehow it happens since it doesn't feel lodged in my brain.

#3) Kind of relating to #2. This may be a bit difficult, and I definitely thank the person who takes the time to type it up and explain this to me, but say, for example, glycolysis or the Kreb Cycle. How do the molecules interact to form intermediaries and release ATP or whatnot. How does CoA help acetyl-coA get into the Kreb Cycle? I'm more interested in a deeper understanding so it stays in my brain.

#4) For these tests and such, would I need to know some obscure names (or maybe they're only obscure to me  :oops: ), such as Hans Kreb (you notice all my stuff is focused on cellular respiration since I'm rereading that at the moment)? I know people like Lamarck and Darwin are probably necessary but what about people like Hook?

#5) This is a specific question. Is there a reason why we never see AMP? We only see that ATP is energized state and ADP is not. What happens to AMP?

There may be a bit more in the future, but thanks for anyone that helps me with these :)

Okay:

#1. For AP Biology and SAT II (Don't trip, I got a 5 and an 800 respectively of these tests) you don't need the specifics, but, for example, you will need to know [i]what[/i] glycolysis and the electron transport chain are and what they're good for. You might need to know what $\text{NAD}^{+}$ does but not what it stands for. Most generalized biology tests do not ask for extremely specific questions, just knowledge of themes. I mean, I will guarantee you that you wont get a question concerning specifics about phosphofructokinase or phosphoenolpyruvate or something really specific.

#2. Biology is NOTHING, I repeat, NOTHING without chemistry. I would agree with anyone who says that chemistry should be taught first. There is a chemical mechanism for ALL that stuff. Even ethology has a chemical basis.....everything does and everyone needs to know chemistry whether it's fun or not. I personally hate doing chemistry but you have to know it and know it well in order to have complete understanding of some biological topics. Study all sciences- math, physics, chemistry, biology, everything so you can realize the awesomeness of the intertwining threads of Nature.

#3. IM me at "tyvolkoff" for more on metabolism but essentially, glycolysis and the Citric Acid Cycle follow all the rules of regular chemistry. Aerobic respiration is simply a catabolic process done efficiently in steps to harness chemical energy. Remember, subtrates are held weakly in active sites by hydrogen, ionic, or other weak bonds by thr R-groups of some amino acids in the active site. Remember that chemical reactions take place by CONTACT between molecules. Remember how chemical gradients work and how like charges repel each other; all of this stuff contributes to formation of ATP  in aerobic respiration. If you need more specific information, IM me.

#4. No, they won't ask you about names. They [i]might[/i] discuss differences between "Lamarck"-ian and Darwinian evolutionary theories though.

#5. We see AMP all the time. Books only seem to mention it a little bit but it's EXTREMELY important acting as an allosteric activator in some processes and hanging around for conversion into ADP or ATP. ATP is a high energy molecule for obvious reasons and AMP is a lower energy molecule for obvious reasons. "Clamp" together three extremely negatively charged phosphates $\text{PO}_{4}^{3-}$ and see what happens when you use it for energy. Just bond one to adenoside and it's not such a big deal. AMP is always hanging around.

Thanks! As you can see, I'm not too great at biology yet  :blush:. I haven't learned chemistry so some of my foundations are not as well as they should be, but with the help of people that are far more knowledgeable than me, I should be able to get by  :P

hello zyloch,
I was the same as you are. :mad:  i hate learning chemistry but later i found that some of the topics need a bit of chemistry even if you are a dealing with biology. As adidasty mentioned in biology they will not expect too much of specific information, here most of the times it is generalised. ;)  but if the topic is related with biochemistry like this one you need to know some basic terminologies which help in the understanding the changes e.g. carboxylation, oxidation, reduction, phosphorylation,etc. these terms once you know you can make out what kind of change is taking place.

yeah zyloch, for some biochemistry stuff you'll want to know the names of certain processes like appasha mentioned just so you can label what's going on

Hey adidasty, I was going through Campbell's and I feel extremely overwhelmed. Is this helpful for USABO prep, though?

Also, what is the best way to self-study for AP Bio, in your opinion? Please be detailed.

AMP also serves as a secondary messenger in cell signal/transduction pathways.

As for independent studying in preparation for the AP Biology exam, I think the best way would be to read/skim through a textbook like Campbell's.  Not necessarily read every chapter in depth, but all the major ones.  Then really read/study a prep book like Cliffsnotes in order to make sure you covered all the important topics.  However, reading a dry textbook requires a good deal of self-motivation and may not be the best method for everyone.

[quote="adidasty"]yeah if you want to be nearly 100% sure of getting a 5 on ap biology, read campbells completely.[/quote]

or at least most of it  :D
some of the animal stuff gets a little too indepth. cliff notes was good enough for the plants/animals/invertables.


wats ur favorite type of literature and why?

Action becuase all other types are boring...occasionally i like an adventure but action mostly

true, i like action, but a mixture is much better  :lol:

fantasy is best.(not king arthur merlin fantasy) A mix of all types of book (except romance) with some imagination. (it's been proven that fantasy readers are the smartest in the class)

Manga! :lol:
Actually I like fantasy, adventure, action, and math books

Manga, Adventure, and Humor :D
And Fantasy and Mathbooks

I would definitely want a book with a mix of it. Otherwise, you'll get a bit too bored, and it will be too easy to predict what will happen next. :wink:

I prefer mystery because there are lots of suspense.

Romance+Mathematics  :D (note, they do exist).

Where is the option "Math books"
I am currently reading Introduction to Algebra - AoPS
G-R-E-A-T B-O-O-K!

Can you pick 2 things on a pole? Because I like both Adventure and Mystery.(They are pretty much alike, but still..)

I enjoy action and adventure but every good book has action!!!!!!

Adventure, Action, Mystery, Drama, Romance, Humor, and when you blend finely, you get a big batch of goodness!

ROMANCE!!!!! :love:

[size=200]JKJKJKJKJKJKJKJKJKJKJKJKJKJKJKJKJKJKJKJKJK!!!!!![/size]

Humor and Action

I like a mixture of adventure action and mystery. Sometimes romance, like in Pride and Prejudice or Wuthering Heights.

Usually  fantasy, but I read a bit of everything though only romance if it's at least a hundred years old.

[quote="dbkarp"](it's been proven that fantasy readers are the smartest in the class)[/quote]

Wow, that must be true. I read too much fantasy...Living with Robin Hood, being a Knight of the Round Table, being one of the Baker Street Irregulars...Where to begin? I am the smartest in the class though...(in the school, besides the adults), so maybe that'll support that reason.

I like philosophy books...action, romance,adventure seems to be booring

MATH BOOKS!!! I am currently on my hypnotized-by-math phase that occurs about every 2 months but have recently be occuring more and more often... and then after that, I like....

LITERARY NONSENSE!!!!!!! (look it up.)

[hide="my favorite nonsense poetry"]

'You are old, Father William', the young man said,
   'And your hair has become very white;
And yet you incessantly stand on your head --
   Do you think, at your age, it is right?'

'In my youth', Father William replied to his son,
   'I feared it might injure the brain;
But, now that I'm perfectly sure I have none,
   Why, I do it again and again.'

'You are old', said the youth, 'as I mentioned before,
   And have grown most uncommonly fat;
Yet you turned a back-somersault in at the door --
   Pray, what is the reason of that?'

'In my youth', said the sage, as he shook his grey locks,
   'I kept all my limbs very supple
By the use of this ointment - one shilling the box -
   Allow me to sell you a couple?'

'You are old', said the youth, 'and your jaws are too weak
   For anything tougher than suet;
Yet you finished the goose, with the bones and the beak -
   Pray, how did you manage to do it?'

'In my youth', said his father, 'I took to the law,
   And argued each case with my wife;
And the muscular strength, which it gave to my jaw,
   Has lasted the rest of my life.'

'You are old', said the youth, 'one would hardly suppose
   That your eye was as steady as ever;
Yet you balanced an eel on the end of your nose -
   What made you so awfully clever?'

'I have answered three questions, and that is enough,'
   Said his father, 'don't give yourself airs!
Do you think I can listen all day to such stuff?
   Be off, or I'll kick you downstairs!'

'Twas brillig, and the slithy toves
Did gyre and gimble in the wabe:
All mimsy were the borogoves,
And the mome raths outgrabe.

"Beware the Jabberwock, my son!
The jaws that bite, the claws that catch!
Beware the Jubjub bird, and shun
The frumious Bandersnatch!"

He took his vorpal sword in hand:
Long time the manxome foe he sought --
So rested he by the Tumtum tree,
And stood awhile in thought

And as in uffish thought he stood,
The Jabberwock, with eyes of flame,
Came whiffling through the tulgey wood,
And burbled as it came!

One, two! One, two! and through and through
The vorpal blade went snicker-snack!
He left it dead, and with its head
He went galumphing back.

"And has thou slain the Jabberwock?
Come to my arms, my beamish boy!
O frabjous day! Callooh! Callay!"
He chortled in his joy.

'Twas brillig, and the slithy toves
Did gyre and gimble in the wabe:
All mimsy were the borogoves,
And the mome raths outgrabe.

They dined on mince, and slices of quince
  Which they ate with a runcible spoon;
And hand in hand, on the edge of the sand,
  They danced by the light of the moon,
   The moon,
   The moon,
They danced by the light of the moon.

Hey diddle diddle,
The cat and the fiddle,
The cow jumped over the moon
The little dog laughed to see such sport,
And the dish ran away with the spoon.

 :rotfl:  :lol:  How does this stuff ever get published????
[/hide]

Your literary nonsense that was posted was quite intriguing and deep  :) What I thought to be was juvenile (and funny) was the cat and the fiddle part. Wonders.

I like reading mysteries, classics, and other types of books, like actions. Comics are fun too

[quote="moogra"]Comics are fun too[/quote]

It's been proven that comics have helped enhance young readers' reading skills. I read that in a TFK article.  :lol:

i luv manga, and its about the only thing i ever read, so i'll base my opinion on genres of manga:

a lot of the mangas that i read r romance, but thats because usually those hav a lot of action/fighting in them, like inuyasha and rurouni kenshin. my fav type is plain old action without the romance, like shaman king, although those arent that easy to come by if u want a good one. even tho i'm a girl, i usually dont read girl's manga. usually i like boy's mangas since they hav more action in them. basically, i like books with fighting in them. (i'm sorta a tomboy when it comes to manga.)

A mixture of of fantasy, a bit of science ..etc.Something like "the da vinci code." Or the foundation series of Eisaak asimove.

I like Sci-Fi, but not those cheesy ones, like "The Blob". 1984 was good, and so was Dr. Franklin's Island. I also like Classics - Charles Dickens(esp. Great Expectations), etc. (But I hated Little Women.) Oh, and of course, math books.  :D

I like Sci-Fi, but then again, I also spend a lot of time in the non-fiction section of the library picking up random books. (Mainly on math, space, and philiosophy)

[quote="Cronus"]Eisaak asimove.[/quote]

... Isaac Asimov.

Who likes the Inheritance series?

Asimov is one of the best Sci-fi writers :D  cool :lol:

I think he's the best.(better than Arthur conan doyle,joule varne..and others that i've read anyway)


How do I verify that $z=f\left(\frac{y}{x}\right)$ satisfies ${x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=0}$

Without knowing $y$ or $x$?

Calculate using the chain rule: if $f$ is differentiable and $x\neq0$,
$\frac{\partial z}{\partial x}=\frac{-y}{x^2}\cdot f'\left(\frac yx\right)$
$\frac{\partial z}{\partial y}=\frac{1}{x}\cdot f'\left(\frac yx\right)$.
The rest is obvious.


\[ \begin{array}{l}
 x,y \in R^ \plus{}  ;m,n \in Z^ \plus{}   \\
 prove \\
 \left( {\frac{{x \plus{} y}}{{m \plus{} n}}} \right)^{m \plus{} n}  \ge \left( {\frac{x}{m}} \right)^m \left( {\frac{y}{n}} \right)^n  \\
 \end{array}
\]

[quote="apratimdefermat"]i think induction will help[/hide][/quote]

whydon't u post a soln :mad:

:blush: Unfortunately i don't have one

i just think induction will help. :blush:

so don't waste any more space spamming, give the others a chance! :mad:

[quote="anorithdialga"]
\[ \begin{array}{l} x,y \in R^ + ;m,n \in Z^ + \\
prove \\
( {\frac {{x + y}}{{m + n}}} )^{m + n} \ge ( {\frac {x}{m}} )^m ( {\frac {y}{n}} )^n \\
\end{array}
\]
[/quote]

$ (\frac{x+y}{m+n})^{m+n}=(\frac{m\frac{x}{m}+n\frac{x}{n}}{m+n})^{m+n}\ge  ( {\frac {x}{m}}) ^m ( {\frac {y}{n}})^n $ last by weight AM-GM


imo 1976
geometry
trigonometry
inequalities
area of a triangle
geometric inequality
IMO

In a convex quadrilateral (in the plane) with the area of $32 \text{ cm}^{2}$ the sum of two opposite sides and a diagonal is $16 \text{ cm}$. Determine all the possible values that the other diagonal can have.

Let's introduce notations:

[color=blue][b]Problem.[/b] Let ABCD be a convex quadrilateral with the area 32, and assume that AB + BD + CD = 16. Find AC.[/color]

[i]Solution.[/i] Let S and S' be the areas of triangles ABD and CDB, respectively. Then, the area of the quadrilateral ABCD is obviously the sum of these two areas S and S'; since we know that the area of the quadrilateral ABCD is 32, we thus get 32 = S + S'.

Since the area of a triangle equals half the product of two sides multiplied with the sine of the angle between them, the area S of triangle ABD must be $S=\frac12\cdot AB\cdot BD\cdot\sin\measuredangle ABD$. Similarly, $S^{\prime}=\frac12\cdot CD\cdot BD\cdot\sin\measuredangle CDB$. Hence,

$32=S+S^{\prime}=\frac12\cdot AB\cdot BD\cdot\sin\measuredangle ABD+\frac12\cdot CD\cdot BD\cdot\sin\measuredangle CDB$.

Now, we have $\sin\measuredangle ABD\leq 1$ with equality if and only if < ABD = 90°, and we have $\sin\measuredangle CDB\leq 1$ with equality if and only if < CDB = 90°. Also, by the AM-GM inequality, applied to the numbers AB + CD and BD, we get

$\left(AB+CD\right)\cdot BD\leq\left(\frac{\left(AB+CD\right)+BD}{2}\right)^2=\left(\frac{AB+BD+CD}{2}\right)^2$
$=\left(\frac{16}{2}\right)^2=64$.

Hence,

$32=\frac12\cdot AB\cdot BD\cdot\sin\measuredangle ABD+\frac12\cdot CD\cdot BD\cdot\sin\measuredangle CDB$
$\leq\frac12\cdot AB\cdot BD\cdot 1+\frac12\cdot CD\cdot BD\cdot 1=\frac12\cdot\left(AB+CD\right)\cdot BD\leq\frac12\cdot 64=32$.

Thus, we have a chain of inequalities with the very left and the very right hand side being equal. This means that all the inequalities inbetween must be equalities. Now, the inequality $\sin\measuredangle ABD\leq 1$ becomes an equality if and only if < ABD = 90°, the inequality $\sin\measuredangle CDB\leq 1$ becomes an equality if and only if < CDB = 90°, and the inequality $\left(AB+CD\right)\cdot BD\leq 64$, being derived from the AM-GM inequality applied to the numbers AB + CD and BD, becomes an equality if and only if AB + CD = BD. Hence, we must have < ABD = 90°, < CDB = 90° and AB + CD = BD.

Now, < ABD = 90° yields $AB\perp BD$, and < CDB = 90° yields $CD\perp BD$; hence, AB || CD. Thus, by Thales, $\frac{AB}{CD}=\frac{BR}{DR}$, where R is the point of intersection of the segments AC and BD. Hence, $\frac{BD}{CD}=\frac{AB+CD}{CD}=\frac{AB}{CD}+1=\frac{BR}{DR}+1=\frac{BR+DR}{DR}=\frac{BD}{DR}$. Hence, CD = DR, so the triangle CDR is an isosceles right triangle (right since < CDR = < CDB = 90°). Hence, its hypotenuse CR equals $CR=\sqrt2\cdot DR$. Similarly, $AR=\sqrt2\cdot BR$. Now, since AB + CD = BD, we have $16=AB+BD+CD=\left(AB+CD\right)+BD=BD+BD=2\cdot BD$, so that $BD=\frac{16}{2}=8$. Thus,

$AC=AR+CR=\sqrt2\cdot BR+\sqrt2\cdot DR=\sqrt2\cdot\left(BR+DR\right)=\sqrt2\cdot BD=\sqrt2\cdot 8$.

  Darij

I found a very short solution,

Notation: Let $ABCD$ be the quadrilateral with the area 32 with AB + BD + CD = 16, let $h_a$ and $h_c$ the length of the altitudes from $A,C$ to $BD$.

$$
h_a\cdot BD+ h_c\cdot BD=64
$$
$$
BD^2+h_a\cdot BD+ h_c\cdot BD=BD^2+64
$$
$$
BD(BD+h_a+h_c)=BD^2+64
$$

Now, $AB\geq h_a$ and $CD\geq h_c$, then $16\cdot BD\geq BD^2+64$, this implies $BD=8$ and $h_a=AB, h_c=CD$, the rest is easy, $AC$ is a diagonal of a 8-side square then $AC=\sqrt{2}\cdot 8$.


algebra
polynomial
function
calculus
derivative
real analysis
real analysis unsolved

Suppose that $ p(x)$ is the interpolating polynomial of degree less than or equal to $ n$ to ($ n \plus{} 1$)-continuously differentiable function $ f(x)$ at distinct points $ x_i$, $ 0\le i\le n$. Prove that for any $ k$, $ 0\le k\le n$, there are points $ y_0,\ldots,y_{n \minus{} k}$ so that for any $ x$ there exists a corresponding $ \xi_x$ satisfying,
\[ f^{(k)}(x) \minus{} p^{(k)}(x) \equal{} \frac {f^{(n \plus{} 1)}(\xi_x)}{(n \minus{} k \plus{} 1)!}\prod_{i \equal{} 0}^{n \minus{} k}(x \minus{} y_i),
\]
where $ f^{(k)}$ and $ p^{(k)}$ denote the $ k$th derivative of $ f$ and $ p$ respectively.


ok,my ugly solution:
Solution:Use Caughy-Schwarz inequality we have:
\[ \frac{a(b+c)}{b^2+bc+c^2}+\frac{b(c+a)}{c^2+ca+a^2}+\frac{c(a+b)}{a^2+ab+b^2}=\sum_{cyc}\frac{a^2}{a(b^2+bc+c^2)/(b+c)}  \]
\[ \ge \frac{(a+b+c)^2}{\sum_{cyc}a(b^2+bc+c^2)/(b+c)}  \]
then we should prove:
\[ \frac{(a+b+c)^2}{\sum_{cyc}a(b^2+bc+c^2)/(b+c)}\ge 2  \]
Expand it ;) .It is equal to:
\[ \sum_{sym}a^4b \ge \sum_{cyc}a^3b^2  \] :D


Let $ a \in \mathbb{C}$ be a fixed number . Solve the equation in $ x \in \mathbb{C}$ , where $ |x| \equal{} 1$ also :
\[ x^2 \plus{} a^2 \minus{} xa \minus{} x \minus{} a \plus{} 1 \equal{} 0 .
\]


[hide]
Answer = 0
Consider the various intervals of x and find the appr. values in the intervals
[/hide]

[quote="madness"]
Answer = 0
Consider the various intervals of x and find the appr. values in the intervals
[/quote]

I think you should reconsider your opinion :

$ f(0)\equal{}2000!>1$ and $ f(1)\equal{}0$, so at least one solution in $ (0,1)$
$ f(2001)\equal{}2000!>1$ and $ f(2000)\equal{}0$, so at least one solution in $ (2000,2001)$

...

In fact there are at least two solutions in each interval $ (n,n\plus{}1)$ for any $ n\in\{1,2,\ldots,1998,1999\}$

so then is the answer

[hide]4000?[/hide]

But so far we know there are only "at least two solutions" in any given interval $ [n, n\plus{}1]$. What if $ |f(x)|$ takes on a value of 1 more than twice in a given interval $ [n ,n\plus{}1]$?

[b]We expect there to be one "hump" of maximum height greater than one between each pair of consecutive roots.[/b]

[u]Proof of hump uniqueness:[/u]
- First, note that iIf there are n factors of (x-a) in a polynomial, then there are (n-1) factors of (x-a) in its derivative, by the chain rule: $ ((x \minus{} a)^n g(x)) ' \equal{} n(x \minus{} a)^{n \minus{} 1} g(x) \plus{} (x \minus{} a)^{n} g'(x)$.
- Applying this to each of the 2000 unique roots means that all but 2000 roots (counting multiplicity) of f(x) are roots of f'(x) - that is, one root for each unique root value.
- We also know that f'(x) has one less root than f(x) since its degree is one lower. Hence, there must be 1999 other roots of f'(x). - We know by the mean value theorem that there must be an extremum between each pair of integer roots.
- There are 1999 such pairs, and this exhausts the remaining 1999 roots of f'(x), and thus f'(x) has no other roots.
- Thus, there can therefore be no other local extrema of the polynomial except for one hump between each pair of consecutive roots.

[u]Proof of hump height:[/u] Let's start with the least favorable case, and put as many factors less than 1 in as possible. In particular, let's consider the points between x=1999 and x=2000. Now consider the point x=1999.5. The values less than 1 are contributed by $ (x \minus{} 1999)^{1999}(x \minus{} 2000)^{2000}$, which contributes $ 2^{3999}$ in the denominator. However, note that the $ (x \minus{} 1995)^{1995}$ term alone contributes $ 4.5 ^ {1995} > 4 ^ {1995} \equal{} 2^{3990}$ in the numerator. This leaves $ 2^9 \equal{} 512$ to go, which is easily covered by the $ (x \minus{} 9)^9 \equal{} 1990.5^9 > 2^9$ term. So clearly, all the hump heights are greater than 1.

[b]Also, since the end behavior of polynomials tends to increase or decrease indefinitely in both x directions, there will be one root left of the left-most root, and one root to the right of the right-most root.[/b]

[u]Proof of uniqueness:[/u]
Again, because all the roots of f'(x) have been exhausted, we know we cannot have any more extrema, and so there can only be one root on each side of the interval [1,2000].

There are thus [b]2(2000-1) + (1)(2) = 4000[/b] unique real solutions to the equation $ |f(x)| \equal{} 1$

For a problem like this, I would assume that if the relative extrema of $ f(x)$ occur at times when $ |f(x)|>1$.  It seems rather obvious, since you'd think that other terms in $ f(x)$ would compensate for a decimal, and TZF has shown that this is the case.

[quote="aidan"]Let $ f(x) \equal{} (x \minus{} 1)^{1} (x \minus{} 2)^{2} (x \minus{} 3)^{3} \cdots (x \minus{} 2000)^{2000}$. Determine the number of solutions to $ |f(x)| \equal{} 1$.[/quote]

[hide="answer"]4002000[/hide]

... Do you mean real roots?