LaTeX
linear algebra
matrix
logarithms
probability
vector
trigonometry

Has anyone heard about the ICTM competition that happened in February?
Did anyone qualify for State, aha?

=D.
[url]http://www.ictm.org/links/contest.html[/url]
That's the link for it.

I had never heard about it before now.

Is it fun? =D

Basically all high schools do it. Most of the problems are either trivial or esoteric. It's only fun if your high school just messes around and plays ultimate in the UIUC quad.
I'm going to state.

[url]http://www.mathlinks.ro/viewtopic.php?t=260081[/url]

I have time and want to get better at $ \text{\LaTeX}$, so I'll type them up. These are the regional questions for the Pre-Calc individual. So, the second-hardest set of written questions it offers. The 2-person team has harder questions (sometimes involving number theory! \sarcasm) and the orals can be fun, but they never choose good topics.

ICTM Regional 2009 AA (you are allowed a calculator)

[hide="Question 1"]
$ \begin{bmatrix}
3 & 2 \\
4 & 6 \\
\end{bmatrix}
\begin{bmatrix}
5 \\ -1 \end{bmatrix} = \begin{bmatrix} k \\ w \end{bmatrix}$ Find the value of $ (2k + 3w)$
[/hide]

[hide="Question 2"]
Given the set: $ \{ \log_2(16), \ \log_3(16), \ \log_4(16), \ \log_5(16) \}$. If one of the 4 members of the set is drawn at random, find the probability it could represent a positive integer.
[/hide]

[hide="Question 3"]
Find the ordered pair that represents the sum of the following two vectors: $ \langle -5, 6 \rangle$ and $ \langle 17, 7 \rangle$
[/hide]

[hide="Question 4"]
Let $ n$ be a positive integer. For $ n \geq 1,\ x_{n+2} = x_n + x_{n+1}$. If $ x_1 = 3$ and ${ x_2 = 1}$, find the value of $ x_{10}$
[/hide]

[hide="Question 5"]
Matrix A contains 23 rows and 13 columns. Matrix B contains 13 rows and 92 columns. Find the number of rows in the product matrix AB
[/hide]

[hide="Question 6"]
In $ \triangle ABC$ as with $ \overline{AB} = 20, \ \overline{BC} = 80, \ \angle ABC = 120^{\circ}$. Find the exact length of $ \overline{AC}$
[/hide]

[hide="Question 7"]
A woman is standing in the stands behind a football field. In front of her are two players 100 yards apart. One player is at an angle of depression of $ 15 ^{\circ}$ and the other is at an angle of $ 8^{\circ}$. Assuming the field is horizontal, find the number of feet in the vertical height of the woman's eye above the plane of the football field.
[/hide]

[hide="Question 8"]
If three real geometric means are inserted between $ 15$ and $ \frac{5}{27}$, find the value of the middle term.
[/hide]

[hide="Question 9"]
Consider $ \triangle ABC$ with $ \angle CAB = 90^{\circ}$ Point D is on $ \overline{BC}$ such that $ \overline{AD} \perp \overline{BC}$. Let the lengths of $ \triangle CAB$ be $ a, \ b, \ c$ with opposite angles $ A, \ B, \ C$. If $ b \sin C (b \cos C + c \cos B) = 41$, and if the length of $ \overline{AD}$ is an integer, find the largest possible value of $ \overline AD$.
[/hide]

[hide="Question 10"]
The roots of $ x^3 + kx^2 - 59x + 420 = 0$ are $ 5, \ -7,  \ 12$ Find $ k$
[/hide]

[hide="Question 11"]
When $ (2x - 3)^8$ is expanded and simplified, find the coefficient of $ x^5$
[/hide]

[hide="Question 12"]
The first term of an infinite geometric sequence with real terms is $ 225$ and the fourth term is $ 14.4$. Find the sum of the terms of this infinite sequence.[/hide]

[hide="Question 13"]
The sum of the terms of an arithmetic sequence with 13 terms and common difference of 3 is found to be 286. Find the thirteenth term.
[/hide]

[hide="Question 14"]
Let there be $ k$ objects, each of weight $ w$. When these objects are weighed in pairs, the sum of the weights of all possible pairs is $ 770$. When these objects are weighed in groups of four, the sum of the weights of all possible groups is $ 9240$. Find $ (k + w)$
[/hide]

[hide="Question 15"]
Find $ |\sqrt{6} + i \sqrt{8}|$
[/hide]

[hide="Question 16"]
Tom is a gambler with fair dice. He selects two numbers from 1...6. He rolls the dice. If both numbers match his selections, he wins 108. If exactly one matches, he comes even. If neither, he loses 54. Find his expected earnings.
[/hide]

[hide="Question 17"]
Let $ D$ be the point $ \frac{1}{3}$ of the way from $ B$ to $ A$ in equilateral $ \triangle ABC$. Let $ P$ be a point on $ \overline{BC}$ such that $ AP + PD$ is minimum. If the perimeter of $ \triangle ABC = 63$, find the exact value of $ (AP + PD)$
[/hide]

[hide="Question 18"]
$ k, \ w \in \mathbb{R}$. If $ (k + wi)(4 + i) = 33 - 13i$ then find $ k + w$
[/hide]

[hide="Question 19"]
Find the acute angle between the normal vectors of the following two planes:
$ 3x + 2y + 5z + 4 = 0 \text{ and } -2x - 3y + 6z - 2 = 0$
[/hide]

[hide="Question 20"]
Points $ A, B, C, D$ are the vertices of a rectangle. Point $ E$ is in the interior of this rectangle such that $ AE = 8, \ EC = 21$. The lengths of $ \overline{DE}, \overline{BE}, \overline{BC}$ are integers. If $ DE < BE$, find the largest possible value of $ \overline{BC}$
[/hide]

ICTM is in general really easy; although there are a few tricky/annoying/weird/long problems on there.  And yes, my school is Naperville North and we're going to win state this year :P  Good luck to all (except Naperville Central and IMSA, I hope they write all of their answers one line down xD) anyways

I was the only one who qualified individually in my school and then there's the 8-person team from our school. I don't know if it was just me, but I thought the 8-person team questions for frosh-soph were hard compared to previous years.

And yea, there are some plain annoying problems on there.

I agree.
One of the Algebra 1 problems was something like 2x+3=x-8 or something.
I was just wondering, aha.
This is my first year doing any of these math contests.

I would have to agree with ManHatTaNatiVe94. We are going to win! :-D

I was in Naperville North's chapter. We tied them for Fresh-Soph 8-person :o , and our oralist beat them. I bet IMSA will win at State. I did Geometry, and some problems were ridiculously easy, while one I didn't know how to do.

Well obviously everybody solves all the easy problems...The people who place well are the ones who can solve the few hard ones.

ALEX!

btw central 2p >>>>> north 2p  :wink:

NO SCREW YOU NORTH IMSA IS GOING TO WIN!!!!!!!!!! :first: :first:
Unless I decide to do all the problems wrong because I'm epically pissed that they put me on fresh soph relays instead of a junior senior event.  :mad:
But I won't because I want some medals/trophies!  :winner_first:

All of us AOPSers should go meet up at states and chill. We can play frisbee. =)

this reminds me a lot of the failed attempt to do the same at HMMT
also I have the feeling a lot of us won't have much time due to multiple events  :P

North>IMSA (at least until next year when I come to IMSA :D)

I apologize, but I believe IMSA>North

=P I was going to go to north, but I went to IMSA instead! haha

Guys, guys, stop bickering over 2nd and 3rd place.

[quote="laughinghead505"]IMSA>>>>>>>>>>North
who are you, ManHatTanNatiVe94? I think we'll have a really strong math team next year, lots of good incoming sophomores :D but I don't know about freshmen... hopefully gauss will get in :P[/quote]

Who am I?  That's irrelevant.  But who are you?

And I'm sorry Khu, but even though Central has a lot of "good" people like you, you can't make up for incompetence in other areas of your math team. =P

Ya, all you have is like you and Kate. Not gonna be enough :P
I'm a girl that made USAMO that's not Kate and goes to IMSA. That should be more than enough. :P

Yes, our incompetence that got us a score greater than yours or IMSA's at regionals :P

Central has a pretty beast fr soph 2 person team ... two state mathcounts winners =0 My money is on central winning it all.

but seriously you guys we have to play frisbee at state. or kevin might prefer to call it fribee.

Dude, you beat us by three points, that's nothing.

Regionals doesn't matter though.  =P
And wow laughinghead, you're such a pro.  I need to become more pro like you are.

Wow you guys are all n00bs. Clearly Glenbrook South is the best. Geez.

Riiggghht Winston no comment lol
I'm not a pro, imona epically fail usamo D:
Oh you were talking about regionals? Those don't count at all, all the smart IMSA people were at HMMT. :P

Dude my dad got a perfect on all levels.

it's not hard.

[quote="Aryth"]oh and josh is not on central's frosh/soph 2p...[/quote]

There better be a good reason for this  :o

Hm.  Who won state again?  Can't remember, help me out.

Wow, the results were just like in NSML. Same first, second, and third.

Two of our most important people had bad days. It happens.

Dude but our individuals were not very good at reporting scores? So we had our hopes up...
like, 134 - 118 Freshmen is not "edging out"
124 - 114 Sophomores is not "a huge victory margin"
132 - 118 Juniors... well that one was pretty accurate...
134 - 118 Seniors was also pretty accurate

calculator... was okay actually (according to expectations)
Frosh/Soph 8, meh. We did not do as well as we could have, but it wasn't terrible.
Junior/Senior 8, yay.

Relays and orals were decent, I thought.

My favorite, though:

Frosh/Soph 2 Person

10th - 21 pts
9th - 23 pts
8th - 23 pts
7th - 24 pts
6th - 25 pts
5th - 26 pts

4th - 30 pts
3rd - 30 pts
2nd - 30 pts

1st - 44 pts

I couldn't believe we were in a tie for second with 30 when you guys got 44...

It's cuz khu haxes everything.
I'm disappointed in you central people, I thought it would be close.  You guys were looking pretty threatening after nearly beating us in DVC and winning in Regionals.

I was pretty disappointed too. We could've done a lot better, but things happen. Always next year? :D

I was pretty disappointed at the results too, but mostly particularly at the high school that's located in my hometown...

Which I won't say. :P

hmm well i'm even more disappointed to see that one guy from that high school didn't make USAMO this year, and he did so last year. :(

[quote="gauss1181"]

hmm well i'm even more disappointed to see that one guy [/quote]

waaaait. Individual results are up?

ehh I don't think so. I was just mentioning this random guy from my local high school and how my local high school phails.

[quote="stickfigure"]eck. I don't know what's with ICTM and individual scores. My school has a copy of all of last year's individual scores but the ICTM people never seem to make it available to the general public.... I'm pretty sure its not a matter of privacy as they do announce everybody's scores as they go up.[/quote]
If you know people by their school and their initials...
http://www.ictm.org/links/contest.html (results have been posted for this year)

Lol wow now that I look at this I tied so many people with 30 on Geometry (2 reading errors -___- shoulda gotten at least 34) but I think I won all the tiebreakers. :D

haha on number 20 it was like find ratio of [ABC] : [DEFG] or whatever and i found [ABC] : [AFG].  :roll:

On that sphere in a cone one, I did
base angle of isosceles = 180 - 2*(vertex angle)
...

And on 20 I put k/w instead of k+w.

Stupid mistakes suck in ICTM. I messed up two problems pretty badly (for the 30 40 50 triangle I thought it asked for circumradius....) and ended up with a 15.

Then I got lucky cause the rest of the 3AA geometry kids didn't do all that well  :roll:


group theory
superior algebra
superior algebra unsolved

Our teacher told us that Burnside proved that the order of any simple group (finite) has at least three prime divisors, but also that the proof is very difficult. Does someone have any reference about this result?

That's what I wanted to know. Thanks a lot, we are doing some lectures on linear representations of finite groups. I hope in the end of the semester I will be able to understand a proof.

[quote="Gockel"]What exactly do you want to know? How this can be proved? All known proofs make afaik use of the representation theory of groups.[/quote]

In Kurzweil's book you can find an "elementary" proof of this $p^{a}q^{b}$ theorem.


function
induction
real analysis
number theory
prime factorization

Let $n=p_{1}^{e_{1}}\ldots p_{k}^{e_{k}}$ be a prime factorization of $n$. If $d(x)$ represents the number of positive divisors of the positive integer $x$, then prove that

$\sum_{j|n}d(\frac{n}{j})d(j) = \binom{e_{1}+3}{3}\ldots \binom{e_{k}+3}{3}$.

[quote="Elemennop"]Are you sure it holds? For $n=3^{4}$, it's

$3^{4}+3^{3}(2)+3^{2}(3)+3(4)+5 = 179 \neq 35 = \binom{4+3}{3}$.[/quote]

yes i had to edit it. it's fixed now.

I edited the formula now. I think that this is the nicest way to represent it.

[hide]

The formula holds . To show this, we need only show that . But the sum on the left is equivalent to , which after some computation boils down to , as desired.

Now, since  is a multiplicative function, we know that

and by induction on the number of prime factors it must hold for all . Then the sum for any particular  is simply the product of the sum for each particular prime power of , but those are simply , so the product of those is .

[/hide]

[quote="t0rajir0u"]For those who don't know, the strange operation on the left is known as [url=http://en.wikipedia.org/wiki/Dirichlet_convolution]Dirichlet convolution[/url].[/quote]

interesting how i started looking at it after considering the positive integer solutions to $xyzw=n$. i never knew that this form was so important with multiplicative functions in general.


come on...can anybody say something??? :blush:

[quote="luke999"]If a,b,c positive numbers and a+b+c=6 proove that
\[ \sum 4 \frac{3(3-a)(3-b)(3-c)}{a^2}\leq \frac{7(a^2+b^2+c^2)}{2}  \][/quote]

If we write inequality without cyclic sum, we have:
$12(3-a)(3-b)(3-c)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\leq\frac{7(a^2+b^2+c^2)}{2}$

$3=\frac{a+b+c}{2}$ so LHS is equal to:
$\frac{3}{2}(a+b-c)(b+c-a)(c+a-b)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)$

If $a>b+c$ when LHS is negative and RHS is positive and inequality is true. So we continue with $a<b+c$ and we can say that $a, b, c$ are sides of a triangle.

Make a substitution $a=x+y, b=y+z, c=z+x$ and $x+y+z=3$.

Inequality now is:
$12xyz\left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}\right)\leq\frac{7((x+y)^2+(y+z)^2+(z+x)^2)}{2}$

$mn\leq\left(\frac{m+n}{2}\right)^2$, so $\frac{xyz}{(x+y)^2}\leq\frac{z\left(\frac{x+y}{2}\right)^2}{(x+y)^2}=\frac{z}{4}$

It follows that:
$12xyz\left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}\right)\leq12\left(\frac{x}{4}+\frac{y}{4}+\frac{z}{4}\right)=9$

Now RHS is:
$\frac{7((x+y)^2+(y+z)^2+(z+x)^2}{2}=\frac{7((x+y+z)^2+x^2+y^2+z^2}{2}=\frac{7(9+x^2+y^2+z^2}{2}=\frac{7(6+x^2+1+y^2+1+z^2+1}{2}=\frac{7(6+2x+2y+2z}{2}=\frac{84}{2}=42$

And we are done.

I think it is correct unless I made a mistake in my first step :)

Good solution...look at http://www.mathlinks.ro/Forum/viewtopic.php?t=85945 to...it's one of my other geo inequality (as I call it...)

Again we can use triangle and the ineq becomes:
$\frac{S^2}{a^2} \leq \frac 78 (a^2+b^2+c^2)$.And from this form I think we can express everything using $sin$ and $R$ .I`m not sure this can be finnalysed as a complete proof,but I`ll think about it.

Something Like that Andrei...You proove that the left member is the sum of
\[ h_a^2+h_b^2+h_c^2 \]
if a,b,c are   numbers that can be triangle lengths...The right part is the sum
\[ m_a^2+m_b^2+m_c^2 \]
(you can proove this by using papus)
And app andrei:"Bravo ma baiatule!I-ai luat la Valcea!"

[quote="andyciup"]:lol:  :lol:[/quote]

   For Andrei: Andrei, in ce clasa esti?Si, in al doilea rand, la ce concurs se referea Luca?Am fost si eu la 3 concursuri la Rm-Valcea.

Cls a8a,I have been to the Pitagora contest this saturday.That is the contest Luca is talking about.

Ok, what place did you take?At <proba colectiva> I took the third place and at <proba individuala> I took <mentiune>.I'll post the subjects at seventh grade on the Romanian Forum.

I got first place at both individual and team!! :)  :)

For Andrei:mai, dar chiar ca i-ai <spart> la Valcea.Oricum, pentru mine rezultatul este multumitor,... :blush: Felicitari!

:lol:  :lol: Bv andrei & sorin...Firstly for Valcea ... Secondly for posting a lot of messages for this problem...


Let $ a,b,c$ be nonzero real numbers for which there exist $ \alpha, \beta, \gamma \in \{ \minus{}1,1 \}$ with: $ \alpha a\plus{}\beta b\plus{}\gamma c\equal{}0$. What is the smallest possible value of: $ \left( \frac{a^3\plus{}b^3\plus{}c^3}{abc} \right)^2?$

[quote="Rust"]0, $ a \equal{} b,c \equal{} \minus{} a\sqrt [3] 2$.[/quote]

Wow ! Great! and could you, please, give us the values of $ \alpha,\beta$ and $ \gamma\in\{\minus{}1,\plus{}1\}$ which matches your solution ?

Thanks in advance.

Assume that $ a,b,c$ all have the same sign.

Then $ |a^3 \plus{} b^3 \plus{} c^3| > |a^3 \plus{} b^3 \minus{} c^3|$.

Thus, the minimal value has two of $ a,b,c$ having the same sign.

WLOG, we can assume that two are positive, and one is negative.

So, $ a > 0$, $ b > 0$, $ c < 0$

So either $ c \equal{} \minus{} (a \plus{} b)$, $ c \equal{} \minus{} (a \minus{} b)$ [when $ a > b$], or $ c \equal{} \minus{} (b \minus{} a)$ [when $ b > a$]

In case 1:

$ \left(\frac {a^{3} \plus{} b^{3} \minus{} (a \plus{} b)^{3}}{ab(a \plus{} b)}\right)^{2}$
$ \equal{} \left(\frac {(3a^2b \plus{} 3ab^2)}{ab(a \plus{} b)}\right)^{2}$
$ \equal{} \left(\frac {(3ab(a \plus{} b)}{ab(a \plus{} b)}\right)^{2}$
$ \equal{} 3^2$
$ \equal{} 9$

In case 2:

$ \left(\frac {a^{3} \plus{} b^{3} \minus{} (b \minus{} a)^{3}}{ab(b \minus{} a)}\right)^{2}$
$ \equal{} \left(\frac {a^{3} \plus{} b^{3} \minus{} (b^3 \minus{} 3a^2b \plus{} 3ab^2 \minus{} a^3)}{ab(b \minus{} a)}\right)^{2}$
$ \equal{} \left(\frac {2a^3 \plus{} 3a^2b \minus{} 3ab^2}{ab(b \minus{} a)}\right)^{2}$
$ \equal{} \left(\frac {2a^2 \plus{} 3ab \minus{} 3b^2}{b^2 \minus{} ab)}\right)^{2}$
$ \equal{} \left(\frac {3b^2 \minus{} 3ab \minus{} 2a^2}{b^2 \minus{} ab)}\right)^{2}$
$ \equal{} \left(3 \minus{} \frac {2a^2}{b^2 \minus{} ab}\right)^{2}$
But $ a > b\Rightarrow ab > b^2\Rightarrow b^2 \minus{} ab < 0$
$ \equal{} \left(3 \plus{} \frac {2a^2}{ab \minus{} b^2)}\right)^{2} > \left(\frac {\pm (3a^2b \plus{} 3ab^2)}{ab(a \plus{} b)}\right)^{2} \equal{} 9$

Case three gives the same result as case 2, but with $ a$ and $ b$ switched.

So in summation, the minimum is $ 9$ for $ a$ and $ b$ having the same sign, and $ c \equal{} \minus{} (a \plus{} b)$.

I am sorry, I think $ \alpha, \beta, \gamma \in [\minus{}1,1]$.
 Your question is find min, when $ c\equal{}\pm a \pm b$.
If all of them had one sign, then it is maximum. Obviosly we can change signs, suth that 2 of them positive (for example a,b), and c negative.
Obviosly min, when c=-a-b, then value of expression is 9.


Consider the plane $ 2x \plus{} 3y \minus{} z \equal{} 100$.

Given an example of a point $ q$ not on the given plane, how do you know it does not line on the given plane?

Does this point $ q$ lie on the same side of the given plane? Why or why not?

Find any point $ (x,y,z)$ such that $ 2x\plus{}3y\minus{}z\ne 100$. The origin will do.

I'm not sure what you mean by a point on the same side of the plane. Do you mean that given two points not on the plane and you want to determine whether they are on the same side or opposite side with respect to the plane?


geometry
parallelogram
analytic geometry
graphing lines
slope
geometric transformation
reflection

[b][size=100][color=DarkBlue]A triangle $ \bigtriangleup ABC$ is given and let $ P$ be, a arbitrary point on $ BC,$ between $ B,\ C.$ We denote the points $ D,\ E,\ F,$ on $ BC,\ AC,\ AB$ respectively, such that $ EF\parallel BC$ and $ DF\parallel AP.$ Prove that $ BQ\parallel MN,$ where $ M,\ N,$ are the midpoints of the segments $ BC,\ AP$ respectively and $ Q\equiv AD\cap PE.$[/color][/size][/b]

Kostas Vittas.

Let a line pass over $ Q$ paralell with $ CA$ cut $ BC$ at $ R$. We have:
$ \frac {BD}{DP} \equal{} \frac {BF}{FA} \equal{} \frac {CE}{EA} \equal{} \frac {CP}{PD}.\frac {DQ}{QA} \equal{} \frac {CP}{PD}.\frac {DR}{RC}$
(Apply Menelaus we have $ \frac {AE}{EC}.\frac {CP}{PD}.\frac {DQ}{QA} \equal{} 1$)
It mean $ \frac {BD}{DR} \equal{} \frac {CP}{RC}\Rightarrow \frac {BR}{DR} \equal{} \frac {RP}{RC} \equal{} \frac { \minus{} BR \plus{} RP}{ \minus{} DR \plus{} RC} \equal{} \frac {BP}{CD}\Rightarrow \frac {BR}{BP} \equal{} \frac {DR}{CD} \equal{} \frac {RQ}{AC} \equal{} k$
Now $ 2\overrightarrow{MN} \equal{} \overrightarrow{BP} \plus{} \overrightarrow{CA}$ and $ \overrightarrow{QB} \equal{} \overrightarrow{QR} \plus{} \overrightarrow{RB} \equal{} \frac{2}{k}.\overrightarrow{MN}$ then OK!

Thank you dear [b][size=100]tdl[/size][/b] for your interest and your elementary solution, much more simple that the one I have in mind.

Alternatively, from the relatiοn $ \frac {BR}{BP} \equal{} \frac {RQ}{AC}$ you mentioned, we can also so say that $ \frac {BR}{P'C} \equal{} \frac {RQ}{AC}$ $ ,(1)$ where $ P'$ is the point on $ BC,$ such that $ MP' \equal{} MP.$

Then, form $ (1)$ we conclude that the triangles $ \bigtriangleup BQR,\ \bigtriangleup P'AC$ are similar and so we have $ \angle QBR \equal{} \angle AP'C$ $ ,(2)$

From $ (2)$ $ \Longrightarrow$ $ BQ\parallel AP'\parallel MN$ and we are done without using vectors.

Best regards, Kostas Vittas.

PS. I would like to preserve this nice proof in the future dear [b][size=100]tdl[/size][/b] and if it is possible, tell me please your name and where do you come from. Thank you in advance.

[size=117][color=darkblue][b]Very nice problem ! Thank you, [u]Vittasko[/u].[/b][/color][/size]

[quote="Kostas Vittas"][color=darkred]Let $ ABC$ be a triangle and let $ P$ be a fixed point of the sideline $ BC$ . Denote the reflection $ R$ of $ P$ w.r.t. the

midpoint of $ [BC]$ . For a mobile point $ D$ of the sideline $ BC$ define the point $ E\in AC$ for which $ \frac {\overline {EA}}{\overline{EC}} \equal{} \frac {\overline{DP}}{\overline{DB}}$ .

Prove that the geometrical locus of the point $ Q\in AD\cap EP$ is a line $ d$ so that  $ B\in d$ and $ d\parallel AR$ .[/color][/quote]

[color=darkblue][b][u]Proof 1[/u].[/b] I'll prove only in the particular case $ D\in (BC)$ of the proposed problem. Suppose w.l.o.g. that $ D\in (AP)$ . We can prove analogously in generally.

Denote $ S\in AR$ for which $ QS\parallel BC$ . Apply the [u][b]Menelaus[/b]'s theorem[/u] to the transversal $ \overline {QPE}$ in the triangle $ ACD$ : $ \frac {QD}{QA}\cdot\frac {EA}{EC}\cdot\frac {PC}{PD} \equal{} 1\Longleftrightarrow$

$ \frac {QD}{QA}\cdot\frac {DP}{DB}\cdot\frac {PC}{PD} \equal{} 1$ $ \Longleftrightarrow$ $ \frac {QA}{QD} \equal{} \frac {PC}{BD}$ $ \Longleftrightarrow \frac {SA}{SR} \equal{} \frac {RB}{BD}$ $ \Longleftrightarrow$ $ \frac {SA}{AR} \equal{} \frac {RB}{RD}$ $ \Longleftrightarrow$ $ \frac {QS}{RD} \equal{} \frac {RB}{RD}$ $ \Longleftrightarrow$  $ QS \equal{} RB$ $ \Longleftrightarrow$ $ BRSQ$ is a parallelogram $ \Longleftrightarrow$

$ BQ\parallel RS\equiv AR$ . Since the points $ R$ and $ B$ are fixed obtain that the locus of the point $ Q$ is a line $ d$ for which $ B\in d$ and $ d\parallel AR$ .

[b][u]Proof 2 (analytic)[/u].[/b] Choose $ A(0,a)$ , $ B(b,0)$ , $ C(c,0)$ , $ P(k,0)$ , $ R(b \plus{} c \minus{} k,0)$ , $ D(m,0)$ , where $ b < c$ , $ a > 0$  and $ b\ne m\ne k\ne b \plus{} c$ .

For the point $ E\left(\alpha , \beta\right)$ we have  $ \frac {\overline{DB}}{\overline{DC}} \equal{} \frac {\overline{EC}}{\overline{EA}}$ $ \Longleftrightarrow$ $ \frac {b \minus{} m}{k \minus{} m} \equal{} \frac {c \minus{} \alpha}{0 \minus{} \alpha} \equal{} \frac {0 \minus{} \beta}{a \minus{} \beta}$ $ \Longleftrightarrow$ $ E\left(\frac {c(k \minus{} m)}{k \minus{} b}\ ,\ \frac {a(m \minus{} b)}{k \minus{} b}\right)$ .

Equation of $ EP\ \ ;\ \ \left|\begin{array}{ccc} x & y & 1 \\
\ k & 0 & 1 \\
\ c(k \minus{} m) & a (m \minus{} b) & k \minus{} b\end{array}\right| \equal{} 0\Longleftrightarrow a(m \minus{} b)\cdot x \plus{}$ $ \left(k^2 \plus{} cm \minus{} kb \minus{} kc\right)\cdot y \minus{} ka(m \minus{} b) \equal{} 0$ .

Equation of $ AD\ \ ;\ \ a\cdot x \plus{} m\cdot y \equal{} am$ . Denote $ Q(u,v)\in AD\cap EP$ and the slope $ p \equal{} \frac {v}{u \minus{} b}$ of the line $ BQ$ , i.e.

$ pu \minus{} v \equal{} pb$ . Therefore, $ \left|\begin{array}{ccc} a(m \minus{} b) & k^2 \plus{} cm \minus{} kb \minus{} kc & ka(m \minus{} b) \\
\ a & m & am \\
\ p & \minus{} 1 & pb\end{array}\right| \equal{} 0$ $ \Longleftrightarrow$

$ p\left[am\left(k^2 \plus{} cm \minus{} kb \minus{} kc\right) \minus{} kam(m \minus{} b) \plus{} bma(m \minus{} b) \minus{} ab\left(k^2 \plus{} cm \minus{} kb \minus{} kc\right)\right] \plus{}$ $ \left[a^2m(m \minus{} b) \minus{} a^2k(m \minus{} b)\right] \equal{} 0$  $ \Longleftrightarrow$

$ p\left(k^2 \plus{} cm \minus{} kb \minus{} kc \minus{} km \plus{} bm\right) \plus{} a(m \minus{} k) \equal{} 0$ $ \Longleftrightarrow$ $ p(k \minus{} m)(k \minus{} c \minus{} b) \plus{} a(m \minus{} k) \equal{} 0$ $ \Longleftrightarrow$  $ \boxed {\ p \equal{} \frac {a}{k \minus{} c \minus{} b}\ }$ .

Observe that the slope $ p$ of the line $ BQ$ doesn't depend by $ m$ and the line $ BQ\parallel AR$ because the slope of $ AR$ is $ \frac {a \minus{} 0}{0 \minus{} (b \plus{} c \minus{} k)}$ .

[b][u]Remark[/u].[/b] The particular case $ k \equal{} b \plus{} c\Longleftrightarrow AR\perp BC$ $ \Longleftrightarrow$ the fixed point $ P$ is the reflection of the  foot of $ A$-altitude

 w.r.t.  the midpoint of $ [BC]$ . Here is this nice and interesting particular particular case in a synthetical form :[/color]

[quote][color=darkred]Let $ ABC$ be a triangle and let $ P$ be the reflection of the foot of the $ A$-altitude w.r.t. the midpoint

of $ [BC]$ . For a mobile $ D$ of the sideline $ BC$ define  $ E\in AC$ for which $ \frac {\overline{EA}}{\overline{EC}} \equal{} \frac {\overline{DP}}{\overline{DB}}$ . Prove that

the geometrical locus of the point $ Q\in AD\cap EP$ is a perpendicular line on $ BC$ in a point $ B$ .[/color][/quote]

[color=darkblue]Here are another two  particular cases :[/color]

[quote][color=darkred]Let $ ABC$ be an $ A$-isosceles  triangle and let $ P$ be the midpoint of the side $ [BC]$ .

For a mobill  $ D$ of the sideline $ BC$ define $ E\in AC$ for which $ \frac {\overline{EA}}{\overline{EC}} \equal{} \frac {\overline{DP}}{\overline{DB}}$ . Prove that

the geometrical locus of the point $ Q\in AD\cap EP$ is a perpendicular line on $ BC$ from $ B$ .[/color][/quote]

[quote][color=darkred]Let $ ABC$ be a triangle and let $ P$ , $ R$ be the tangent points of the incircle and $ A$-exincircle respectively with

the sideline $ BC$ . For a mobile $ D$ of the sideline $ BC$ define  $ E\in AC$ for which $ \frac {\overline{EA}}{\overline{EC}} \equal{} \frac {\overline{DP}}{\overline {DB}}$ . Prove that

the geometrical locus of the point $ Q\in AD\cap EP$ is a parallel line with $ AR$ from the point $ B$ .[/color][/quote]

Thank you very much dear [b][size=100]Virgil[/size][/b] for your interest and nice solutions.

All your remarks are useful and let me to say another interesting result, which is true in the configuration of the proposed problem.

( Please see at the schema [b][size=100]t=273768[/size][/b] ) [b][size=100][color=DarkBlue]The  line segment $ MN,$ always bisects the segment $ DE.$[/color][/size][/b]

I have in mind an elementary proof of this result and I will post here next time more details.

Best regards, Kostas Vittas.


calculus
derivative
function
search
calculus computations

Suppose $ f$ is a real function on $ (\minus{}\infty , \infty)$, and $ A$ be a constant such that $ A<1$. Show that if $ |f'(t)|\le{A}$ for all real $ t$ then $ f$ has a fixed point.

I have already shown that $ f$ can only have at most 1 fixed point, but don't know how to show that there is a fixed point.

If $ |f^\prime (t)|\leq A$, then $ f(0)\minus{}At\leq f(t)\leq f(0)\plus{}At$ for $ t\geq 0$ and $ f(0)\plus{}At\leq f(t)\leq f(0)\minus{}At$ for $ t\leq 0$. Use this to find $ u$ such that $ f(u)\leq u$ if $ f(0)\geq 0$ or $ f(u)\geq u$ if $ f(0)\leq 0$.

The Banach fixed point theorem applies. There must be a number of topics here that address it. I put a sketch of the proof in the last post of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=2100597974&t=107979]this topic.[/url]


calculus
integration
trigonometry
real analysis
real analysis unsolved

It's so-called Fresnel integral. I don't think it has a specific value for limits like that.
So the answer is
$ \int_0^{\frac{\pi}{2}} \sin x^2 \, \textbf d x \equal{} \boxed{\textbf S \! \left( \frac {\pi}{2} \right)}$

how to prove that $ \int_{0}^{\infty} sin(x^2) \equal{} \sqrt{2\pi}/4$?

[quote="yangzhe1990"]how to prove that $ \int_{0}^{\infty} sin(x^2) \equal{} \sqrt {2\pi}/4$?[/quote]

it has been shown several times before:  $ \int_{0}^{\infty} \sin(x^k) \;dx \equal{} \sin \left(\frac{\pi}{2k}\right) \cdot \Gamma \left(\frac{1}{k}\plus{}1\right)$

take $ e^{\imath\,  z^k}$ , pick $ \gamma$ as a contour starting at $ 0$ going on the $ \plus{}ve$ real axis to $ R$, then go  counterclockwise from angle $ 0$ to $ \frac{\pi}{2k}$ , then come back to $ 0$, and finally let  $ R\to\infty$

See this topic http://www.mathlinks.ro/viewtopic.php?t=202045


algebra
polynomial
superior algebra
superior algebra unsolved

Is $ \{a \plus{} b\sqrt {2}: a,b\in \mathbb{F}_{11}\}$ a field ?

 ( if it is then it is a field of $ 11^2$ elements )

Is there any general condition like:

 if there is no square root of $ 2$ in $ \mathbb{F}_p$ then $ \{a \plus{} b\sqrt {2}: a,b\in \mathbb{F}_{p}\}$ is a field of $ p^2$ elements ?

Yes, in fact $ \{a \plus{} b\sqrt {2}: a,b\in \mathbb{F}_{p}\} \equal{} \mathbb{F}[x]/(x^2\minus{}2)$, the factorization of the ring of polynomials with coefficients in $ \mathbb{F}_p$ by the ideal generated by the irreducible polynomial $ x^2\minus{}2$.

[quote="MathJimmy"]Is there any general condition like:

if there is no square root of $ 2$ in $ \mathbb{F}_p$ then $ \{a \plus{} b\sqrt {2}: a,b\in \mathbb{F}_{p}\}$ is a field of $ p^2$ elements ?[/quote]

The general statement is as follows: If $ \mathbb{F}_p$ is a finite field of order $ p$ (with $ p$ prime) and $ g(x)$ is a monic irreducible polynomial of degree $ n$ in $ \mathbb{F}_p[x]$, then $ \mathbb{F}_p[x]/\langle g(x)\rangle$ is a field of order $ p^n$.


AMC
AIME
geometry
rotation
geometric transformation
analytic geometry
trigonometry

Just like in other marathons, post your solution and a new problem as well.

[size=150]Problem 1[/size]

AIME 1988 #15

In an office, at various times during the day, the boss gives the secretary a letter to type, each time putting the letter on top of the pile in the secretary's in-box. When there is time, the secretary takes the top letter off the pile and types it. There are nine letters to be typed during the day, and the boss delivers them in the oder 1, 2, 3, 4, 5, 6, 7, 8, 9.
While leaving for lunch, the secretary tells a colleague that letter 8 has already been typed, but says nothing else about the morning's typing. The colleague wonders which of the nine letters remains to be typed after lunch and in what order they will be typed. Based on the above information, how many such after-lunch typing orders are possible? (That there are no letters left to be typed is one of the possibilities.)

(posted by alan)

[quote="breez"][hide="Maybe solution"]The typing comes in this order:

$9$
$8$
$7$
$6$
$5$
$4$
$3$
$2$
$1$

Each letter can have one of $2$ states: Typed or Not Typed.
$9$ could have arrived while $8$ was being typed. So in that case, there are $8$ letters we don't know if she typed. The number of arrangements of typed/untyped states is thus $2^8$ or $256$ typing arrangements after lunch.

$\boxed{256}$[/hide][/quote]
Your answer is simply the result of the number of elements in the power set

$\left|\mathcal{P}\left(S\right)\right|=2^{8}$

where $\left|S\right|=8$ because there are 8 papers left.  However, there is something wrong with this answer because it doesn't ask only for the amount of subsets, but also the different combinations of each subset.  I'd have to look over it again, but I'm almost certain there are quite a few more than 256.

[hide="Solution"]Since letter 8 has been typed, letters 1-7 are already in the tray, and possibly 9 is.  All the remaining letters must be typed in decreasing order, so all that matters is which letters (besides for letter 8) have already been typed.

So we have letters 1, 2, 3, 4, 5, 6, and 7 that can be there.

If 0 of those have been typed: $\binom{7}{0} \cdot 9$, the 9 is for the possible "places" that letter 9 can be added: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | or not at all (where | represents where the 9 goes).

If 1 of those have been typed: $\binom{7}{1} \cdot 8$ (use above reasoning)

So we want
$S=\binom{7}{0} \cdot 9 + \binom{7}{1} \cdot 8 +...+ \binom{7}{7} \cdot 2$
which also equals:
$S=\binom{7}{0} \cdot 2 + \binom{7}{2} \cdot 2 + ...+ \binom{7}{7} \cdot 9$

adding the two equations:
$2S = 11\left(\sum_{k=0}^{7} \binom{7}{k}\right)=11\cdot 2^{7}$

So $S=11\cdot 2^{6}=\boxed{704}$.

[/hide]

[size=150]Problem 2[/size]

AIME 1994 #8

Find $ab$ if $(0, 0)$, $(a, 11)$, $(b, 37)$ is an equilateral triangle.

[hide="wrong ans i guess"]1138/4?[/hide]

Just wondering that do all the problems have to be AIME ones or any other problem is allowed as well of the same level?(If in any case my dumb ans is right,someone else can post a question)

[quote="eryaman"][size=150]Problem 2[/size]

AIME 1994 #8

Find $ab$ if $(0, 0)$, $(a, 11)$, $(b, 37)$ is an equilateral triangle.[/quote]

[hide="Solution"]We have two equations:

$a^2+11^2=b^2+37^2$
$(a-b)^2+26^2=a^2+11^2$

or

$a^2-b^2=1248\qquad (1)$
$2ab-b^2=555\qquad (2)$

From $(2)$ we have

$a=\frac{555+b^2}{2b}\qquad (3)$

Substituting $(3)$ into $(1)$ we get

$\left(\frac{555+b^2}{2b}\right)^2-b^2=1248$, or after simplifying

$b^4+1294b^2-102675=0$

This is a biquadratic equation and the only positive solution is $b^2=\frac{-1294+\sqrt{1294^2+4\cdot 102675}}{2}=75$

From $(3)$ we get $ab=\frac{555+b^2}{2}=\boxed{315}$

PS. This is a really excrutiating problem for a competition where the calculators aren't allowed...[/hide]

AIME 1994 #8

Find ab if (0, 0), (a, 11), (b, 37) is an equilateral triangle.

$\\(a+i11)e^{i\frac{\pi}{3}}=b+i37\\ (a+i11)(\frac{1}{2}+i\frac{\sqrt{3}}{2})=b+i37\\ a/2+ia\frac{\sqrt{3}}{2}+i11/2-11\frac{\sqrt{3}}{2}=b+i37\\ a/2-11\frac{\sqrt{3}}{2}=b\\ a\frac{\sqrt{3}}{2}+11/2=37\\ a=2\frac{37-11/2}{\sqrt{3}}=\frac{63}{\sqrt{3}}\\ b=\frac{37-11/2}{\sqrt{3}}-11\frac{\sqrt{3}}{2}=\frac{30}{2\sqrt{3}}\\ ab=(\frac{63}{\sqrt{3}})(\frac{30}{2\sqrt{3}})=\boxed{315}$

:)
Do I post the next problem?

EDIT: I posted later. :(
EDIT2: Since it's AIME, we are not supposed to use calculators, if you used them.

Masoud Zargar

And why I am oh sooo stupid not to think of complex numbers?  :?  :rotfl:

Note: You should just mention the case when $a,b$ are negative, but the answer is still the same.

Post the next problem, boxedexe, my solution is yuck.

Should it be from an AIME or at an AIME level?

EDIT: Here's a problem meanwhile with unknown source.

[size=150]Problem 3:[/size]

Let there be $m+1$ equally spaced parallel lines, intersecting at right angles by $k+1$ equally spaced parallel lines. If $m=4$ and $k=5$, what is the total number of squares in the network.  It is kinda made-up. :P

Masoud Zargar

I don't have a problem handy, so feel free to post one if you have it.

I'm not really much into AIME (would need to review the problems from the past years, which I haven't done), so I can't give you an answer.

[quote="boxedexe"]That's correct.  I suppose you can post the next problem.  By the way, what woudl you rank that problem at an AIME level out of 15?

Masoud Zargar[/quote]

I'd think that'd be one of the first five problems, at most

[quote="Altheman"][hide="very quick solution to 1994 #8"]Let u and v be the other points, we have $|u|=|v|$ and $u*v=|u|^2*\frac{1}{2}$. So $ab+37*11=\frac{1}{2}*(a^2+11^2)=\frac{1}{2}*(b^2+37^2)$, rearrange, multiply and let $x=ab$:
$(2x+11*63)(2x-37*15)=x^2 \implies x=315$ (one may use RRT to help solve)[/hide][/quote]

that's kind of what I did, but I used shoelace theorem to find out the area of the triangle and then equated it to te area using $a$ and $b$ using the advantage that it was equilateral.

I'm not 100% on this solution, but:

[hide="Solution"]By simplifying a bit, I get

$\binom{200}{100}=\frac{2^{50}\left(199\cdot 197\cdots 101\right)}{50!}.$

From this, the greatest two-digit prime number that divides this must also be in the set $\left\{101,103,\dots ,199\right\}$ when multiplied by 3.  The greatest two-digit prime to satisfy this is $\frac{183}{3}=\boxed{61}$.[/hide]

[size=150]Problem 60[/size]
AIME 1984, Problem 14

What is the largest even integer that cannot be written as the sum of two odd composite positive integers?

[hide="Solution"]
Consider the process of subtracting the integers $3 \times 5, 5 \times 5, 7 \times 5$ from an even integer.  These three numbers are distinct $\bmod 3$, so one of them must be divisible by $3$ and will be composite unless it is $3$ - therefore, we have written that even integer as a sum of two composite odd integers.

The largest even integer for which this is impossible is found by considering if subtracting $7 \times 5$ gives $3$ itself, that is, $\boxed{38}$.
[/hide]

[size=150]Problem 61[/size]
AIME 1999, Problem 9

A function $f$ is defined on the complex numbers by $f(z)=(a+bi)z,$ where $a$ and $b$ are positive numbers.  This function has the property that the image of each point in the complex plane is equidistant from that point and the origin.  Given that $|a+bi|=8$ and that $b^2=m/n,$ where $m$ and $n$ are relatively prime positive integers.  Find $m+n.$

(Easy problem, but I have a neat solution and I want to see if others can find it.  :) )

Well we have:

$|{z-(a+bi)z|=|{(a+bi)z}|}$.  Simplifying yields:

$1-a+bi=a+bi$ or $1-a+bi=-a-bi$.  Only the first has solutions in real numbers with
$a=\frac{1}{2}$

Now we use the fact that $a^2+b^2=64$

This gives us that $b=\sqrt{\frac{255}{4}}$

Therefore the answer is 259.

Heh, that's much simpler than how I remember doing that problem  :blush: Next problem?

Check your work on that last step.  :)

[hide="My marginally neater solution"]
The points $0 + 0i, z, (a+bi)z$ form an isosceles triangle with side lengths $8|z|, 8|z|, |z|$.  $b |z|$ is the altitude of this triangle, which, by Pythagorean theorem, is equal to $\sqrt{ \frac{255}{4} } |z|$.  Hence $b^2 = \frac{255}{4}$ and the answer is $\boxed{259}$.
[/hide]

[size=150]Problem 62[/size]
AIME 1983, Problem 11

The solid shown has a square base of side length $s$. The upper edge is parallel to the base and has length $2s$. All edges have length $s$. Given that $s=6\sqrt{2}$, what is the volume of the solid?
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=399[/img]

(I hope this hasn't been done before on this marathon; also a problem with a very neat solution)

(Edit:  Has been posted, but not with the neatest solution. :) )

[hide] We realize that putting two solids together along the square base yields a regular tetrahedron with side length $12\sqrt{2}$. The volume of the tetrahedron is $\displaystyle\frac{(12\sqrt{2})^3 \sqrt{2}}{12} = 576$, so the volume of the solid is $288$.[/hide]

[size=150]Problem 63[/size]

(AIME 1986 #7, taken from Kalva)

The sequence 1, 3, 4, 9, 10, 12, 13, 27, ... includes all numbers which are a sum of one or more distinct powers of 3. What is the 100th term?

[hide="Solution"]
The terms of the sequence are precisely the base $2$ numerals interpreted in base $3$ (since those are the numbers that use only up to one of each power of $3$).  $100_{10} = 1100100_2$, so our desired number is $1100100_3 = \boxed{981}$.
[/hide]

Someone else can post the next problem.

[size=150]Problem 64[/size]

(AIME 2004 I, Problem 8) A regular $n$-star is the union of $n$ equal line segments $P_1P_2, P_2P_3, \ldots , P_nP_1$ in the plane such that the angles at $P_i$ are all equal and the path $P_1P_2\cdots P_nP_1$ turns counterclockwise through an angle less than $180^\circ$ at each vertex. There are no regular 3-stars, 4-stars or 6-stars, but there are two non-similar regular 7-stars. How many non-similar regular 1000-stars are there?

[size=75][color=green]mod edit: changed problem, which was already posted before[/color][/size]

[url=http://www.upal.ca/adeel/marathon.pdf]First 64 problems of this marathon[/url]

No solutions yet; that will take a long time... :)

[quote=thinker123]\bump. let's revive this thread! anyone have a new problem?[/quote]

u literally revived a 15 year old thread wao

[size=150]Problem 65[/size]

(AIME 2010 I, Problem 2) Find the remainder when $9 \times 99 \times 999 \times \cdots \times \underbrace{99\cdots9}_{\text{999 9s}}$ is divided by $1000$.

[hide]$9 \times 99 \times 999 \times \cdots \times \underbrace{99\cdots9}_{\text{999 9s}} \equiv 9 \times 99 \times (-1)^{997} \equiv -9 \times 99 \equiv \boxed{109} \pmod{1000}.$ [/hide]


This isn't just some random spammy advertisement, this is something I actually think a lot of you guys will like. Basically, those of us in the 2009 Research Science Institute with math projects got together and decided to start a group blog at http://deltaepsilons.wordpress.com. We're all high-schoolers, we all like math (no actually we secretly hate math but we're pretending otherwise to keep our mentors happy...NO NOT REALLY), and we all have a high-school perspective on math (okay, that's actually partly a lie, but...whatever, go see for yourself). So we tend to say lots of things you may find interesting about research, contests, and other stuff, and therefore everyone here should go check out our blog. DO IT.

For the sake of fairness, anyone else here can also advertise their blog or anything else on this thread (and only this thread) provided that the subject of the advertisement is mathy and the advertisement is worded with reasonable intelligence.

Seconded, contingent on more people who aren't Akeel posting ;)

Five of us have made posts now, and four of them aren't Akhil :P I'm sure things will balance out a lot more once RSI is over and we all get a lot more free time.


x,y,z positive
$ x\plus{}y\plus{}z\equal{}\sqrt{xyz}$
Prove that
$ xy\plus{}zy\plus{}xz \ge 9(x\plus{}y\plus{}z)$

$ xy\plus{}yz\plus{}zx\ge9(x\plus{}y\plus{}z)$

$ \Leftrightarrow (x\plus{}y\plus{}z)(xy\plus{}yz\plus{}zx)\ge9(x\plus{}y\plus{}z)^2$

$ \Leftrightarrow 3xyz\plus{}\sum_{sym}x^2y\ge9xyz$

$ \Leftrightarrow \sum_{sym}x^2y\ge6xyz$

This is AM-GM

[quote="NikolayKaz"]x,y,z positive
$ x \plus{} y \plus{} z \equal{} \sqrt {xyz}$
Prove that
$ xy \plus{} zy \plus{} xz \ge 9(x \plus{} y \plus{} z)$[/quote]

By Am-Gm $ x\plus{}y\plus{}z \geq 3\sqrt[3]{xyz}$ ,and from $ x\plus{}y\plus{}z\equal{}\sqrt{xyz}  \implies xyz \geq 3^6$

$ xy\plus{}zy\plus{}xz \geq 9(x\plus{}y\plus{}z) \iff xy\plus{}zy\plus{}xz \geq 9\sqrt{xyz}$

Again by  Am-Gm $ xy \plus{}zy \plus{}xz \geq 3\sqrt[3]{(xyz)^2} \geq 9\sqrt{xyz} \iff xyz \geq 3^6$,we're done


[quote="13375P34K43V312"]When does round 2 come out?

And since it's past the deadline, can we discuss the problems freely now?[/quote]

i think they both happen after the math jam on thursday

Correct.  Do not discuss the problems until after the Math Jam, and the Round 2 problems will be available tomorrow night.

At about what time today will round 2 come out? I am going out of town tommorow early in the morning and I could get a lot of work done on the airplaine.
Are the problems going to out within $\frac{1}{2}$ hour?
THanks,
BPMS.


inequalities
trigonometry
geometry
inequalities proposed

Let $ a$, $ b$, $ c$ and $ R$ be the sides and circumradi of a non obtuse triangle. Then

\[ 8R^2\le a^2\plus{}b^2\plus{}c^2\]

:)

[quote="ElChapin"]Let $ a$, $ b$, $ c$ and $ R$ be the sides and circumradi of a non obtuse triangle. Then
\[ 8R^2\le a^2 \plus{} b^2 \plus{} c^2
\]
:)[/quote]

Just remember

$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2(s^2 \minus{} r^2 \minus{} 4Rr)$

and you get for the inequality under consideration

$ 2R \plus{} r < s$

The result thus follows.

equality holds iff the triangle is right

Also, it is equivalent to

\[ (a^2\plus{}b^2\minus{}c^2)(b^2\plus{}c^2\minus{}a^2)(c^2\plus{}a^2\minus{}b^2)\ge 0\]

:)

this is equivalent to $ \sum \sin^2A\geq 2$, which is equivalent to $ 1\minus{}2\cos A\cos B\cos C\equal{}\sum\cos^2A\leq 1$, which is true  :D

another proof...

... if you get the equivalent form $ \sum{cos^2(A)}\le 1$

Remember that $ OH^2\equal{}AH^2\plus{}BH^2\plus{}CH^2\minus{}3R^2\equal{}4R^2\sum{(cos^2(A))}\minus{}3R^2$ where $ H$ is orthocenter and $ O$ is circumcenter.

Also note that in a non obtuse triangle $ OH^2\le R^2$ and we are done.

:)


So, should I be moving this to the Self-Pity forum?  We can get a big collection -- all the people who got a 144 on the AMC, mis-bubbled an answer on AIME, were so close to making their IMO team, can get together and bemoan the unfairness of it all.  Actually, maybe I'll go put this on the wish-list right now.

I like the idea...it can facilitate community commiseration. Dang, I'm out of lemonade...

JBL, no need to be sarcastic about this. I remember that in my senior year, after the first 2 TSTs I was on the 6th (qualifying position) for the Balkan Math Olympiad team, and at the last TST I managed to get on the 7th position, at 0,5 points and 1 point (out of 84) from places 6 and 5 respectively. Most people have no idea how tough that can be like  :(

[quote="JellyWorld"]Our TSTs for the IMO team was held today and yesterday, but since I'm not officially  in the "national team", which is sort of like the shortlist for IMO selection, I'm not considered for the IMO team selection.  But I attended the TSTs anyway, just for the fun of it, and in the end I came out [b]6th[/b].  But I won't be in the IMO team because I didnt make the cut for the shortlist last year, even though I'm somehow managed to smoke my way into the top 6 for the TSTs. :(

Oh well, I'm still in 9th grade, so there's still 3 more years to try :([/quote]

Dont worry you are very young . Future is yours  :)  :)

IMO is not the only thing in your life... When I got this idea, I started practising just for fun, not for the obsessive and degrading idea to go to an IMO.....

[quote="joml88"]Well, the question is:  should people be encouraged to vent their frustration and disappointment of matters such as USAMO qualification, IMO qualification, etc?

Also, where should they vent their emotions...round table, amc, gaff, their blog, etc?

Now, my opinion...I don't think we should discourage all of the posts of this type.  But round table is definitely not the place for it.  I think blogs are probably the most appropriate.[/quote]

agreed. It's pointless to complain about these things. I mean, come on, you were really close to making IMO, how would people who didn't even make AIME feel (whether they genuinely didn't perform well or misbubbled) about this after reading. They'd feel worse about themselves, trust me ;)

Back in 1994, I placed 7th on the USAMO so I barely missed the cutoff, and my cumulative score at MOP ranked 6th so theoretically one could argue that I "deserved" to be on the team.  But, truth be told, I'm very happy I [b]wasn't[/b] on the team because the actual team proceeded to make history with the first ever perfect score on the IMO, and I'm pretty sure if I were on the team I would have ruined that!  :lol:

[quote="yif man12"]If you actually feel like crap about something like this, you don't post it on some online forum. The people who post sobs about not making something are just looking for pity. If you're truly disappointed about something, you'd be down for a while, and then come back with a vengeance the next time around, not go wallowing in some forum about your sob story and how God is out to get you.[/quote]

amen


modular arithmetic
number theory proposed
number theory

Problem. Let $b_{n,k} := \binom{n}{k} \pmod{2}$ be the parity of the binomial coefficient $\binom{n}{k}$ (i.e. $0$ or $1$, accordingly to whether it is even or odd). Let
\[
a_n := \sum_{0 \leq j \leq n} b_{n,j}2^j.
\]
Show that, for any $n$, $a_{2n+1} = 3a_{2n}$.

I think this can be put in a slightly more general setting:

Given $a_n$, show that $a_{n+1}$ is obtained by performing the $\text{XOR}$ operation (or addition in $\mathbb Z_2$, if you prefer) on $2a_n$ and $a_n$, and then show that this implies the result stated by vess. In order to prove this last implication we can use the fact that $\binom n i$ is odd iff in binary representation, $n$ has all the digits of $i$. From here we can easily conclude that we can't have two consecutive $t$ for which $\binom {2n} t$ is odd, so we can't have two consecutive $1$s in $a_{2n}$, meaning that addition in $\mathbb Z_2$ is the same as the regular addition (we don't run into any $1+1$ stuff) for $2a_{2n}$ and $a_{2n}$.


function algebra domain Asymptote calculus polynomial conics

im not quite familar with the method as to how to graph rational functions like this:

$ (2x^5 \minus{} x^3 \plus{} 2) / (x^3 \minus{} 1)$

any help is appreciated. thanks.

I think basic steps of analyzing graph of a rational function include:

1) Find domain of rational function ----> (-infinity, 1) U (1, infinity)
2) Write function in lowest terms -----> i believe it simplifies into 2x^2 - 1 + [(2x^2 + 1)/(x^3 - 1)]
3) Locate x and y intercepts ------> (0, -2); x-ints can't be found w/o calc
4) Test for x-axis, y-axis, and origin symetry ---> in this case, there are none
5) Locate vertical asymptotes ----> x=1
6) Locate horizontal or oblique asymptotes, and determine if the graph intersects them at all ----> there are                              none since the power of the numerator (5) is 2 more than power of denominator (3)
7) Graph ----> now we should pick sample points (apart from those already determined in previous steps) and try and find some pattern to the graph

These are basic precalc methods at analyzing rational functions. In calculus, you will learn more about curve sketching with topics such as concavity and increasing and decreasing functions. I hope this helps :).

Using polynomial long division, write the rational function in the form $ P(x) \plus{} \frac{Q(x)}{R(x)}$, where $ P,Q,R$ are polynomials and $ \text{deg} Q < \text{deg} R$. You can further simplify the $ Q/R$ term (if you want) using the [url=http://en.wikipedia.org/wiki/Partial_fractions]method of partial fractions[/url].

$ \frac{2x^{5}\minus{}x^{3}\plus{}2}{x^{3}\minus{}1} \equal{} (2 x^2\minus{}1)\plus{} \left(\frac{x}{x^2\plus{}x\plus{}1}\plus{}\frac{1}{x\minus{}1} \right)$

Note that because of the degree conditions on Q and R, the terms that sum to Q/R go to zero, and the graph of P+Q/R looks like that of P when you zoom out, with some disturbances in a limited interval. In this case, when you zoom out, the graph will look like $ 2x^2\minus{}1$ when the sub-linear terms go to zero, so draw $ 2x^2\minus{}1$ as a sort of parabolic 'asymptote' to work from.

Now let's look at the remaining terms separately. The term $ \frac{x}{x^2\plus{}x\plus{}1}$ can be written as $ \frac{1}{1\plus{}x\plus{}\frac{1}{x}}$, from where you should see that the extreme values occur at $ x\equal{}\pm 1$. You should be able to figure out roughly what it looks like from here.

The term $ \frac{1}{x\minus{}1}$ is your usual rectangular hyperbola with a vertical asymptote at $ x\equal{}1$, and decaying fairly quickly for $ x$ far from $ 1$.

Now just draw a curve near your 'parabolic asymptote' that takes these perturbations into account. It obviously doesn't hurt to plot a few points to help you.


Given segment $XY$ and line $m$ that is parallel to $XY$, find the midpoint of $XY$ using only a pencil and a straightedge ([i]NO compass[/i]).


function
inequalities
limit
algebra unsolved
algebra

Find the maximum value of $f(2006)-f(2005)$where f is a real valued function that satisfies the inequality

$|f(x)-f(y)| \leq  (x-y)^{2}$

for all real numbers x and y

It's $0$.

$\left|{f(x+h)-f(x)\over h}\right|\leq |h|$, so $f'(x)=\lim_{h\rightarrow 0}{f(x+h)-f(x)\over h}=0$ exists.
And so $f$ is differentiable with $f'\equiv 0$, and so $f$ is constant by the mean value theorem.


geometry
perpendicular bisector
geometry proposed

ABC is a non-isosceles triangle. E and F are points outside triangle ABC such that $ \angle ECA \equal{}
\angle EAC \equal{} \angle FAB \equal{} \angle FBA \equal{}\theta$. The line through A perpendicular to EF intersects the
perpendicular bisector of BC at D. Determine $ \angle BDC$.

Construct point $ D'$ on perpendicular bisector of $ BC$ outside $ \triangle ABC$ such that $ BD'C\equal{}2\theta$. Consider the complex plane. Denote $ A(a), B(b), C(c), F(f), E(e), D'(d)$. Hence $ (f\minus{}b)e^{(\pi \minus{}2\theta)i}\equal{}f\minus{}a$. We obtain $ f\equal{}\frac{b.e^{(\pi \minus{}2\theta)i}\minus{}a}{e^{(\pi \minus{}2\theta)i}\minus{}1}$ but $ e^{(\pi \minus{}2\theta)i}\equal{}\frac{\minus{}1}{e^{2\theta i}}$ hence $ f\equal{}\frac{a.e^{2\theta i}\plus{}b}{{e^{2\theta i}}\plus{}1}$. Analogically $ e\equal{}\frac{c.e^{2\theta i}\plus{}a}{e^{2\theta i}\plus{}1}$ $ , d\equal{}\frac{c.e^{2\theta i}\minus{}b}{e^{2\theta i}\minus{}1}$  $ \Rightarrow$ $ f\minus{}e\equal{}\frac{(a\minus{}c)e^{2\theta i}\plus{}b\minus{}a}{e^{2\theta i}\plus{}1}$ and $ a\minus{}d\equal{}\frac{(a\minus{}c)e^{2\theta i}\plus{}b\minus{}a}{e^{2\theta i}\minus{}1}$. Hence $ \frac{f\minus{}e}{a\minus{}d}\equal{}\frac{e^{2\theta i}\minus{}1}{e^{2\theta i}\plus{}1}\equal{}\frac{ \minus{}1\plus{}cos2\theta\plus{}isin2\theta}{1\plus{}cos2\theta \plus{}isin2\theta}\equal{}\frac{i.2sin2\theta}{(1\plus{}cos2\theta)^2\plus{}sin\theta^2}\equal{}k.i$ where $ k\in \mathbb{R}$. So $ AD'\perp FE$ but $ AD\perp FE$ $ \Rightarrow D\equiv D'\Rightarrow \angle BDC\equal{}2\theta$


How many positive integers less than 100 have an odd number of
distinct factors?

This is equivalent to finding perfect squares. We have $ 1^2$ to $ 9^2$ for a total of $ \boxed9$.

extension:
Here is a method for finding the number of factors of a number. Say i want to find the number of prime factors of 42.
Take the prime factorization:
\[ 2 \cdot 3 \cdot 7
\]
But there is an understood exponent of 1 on each of the bases. (Make sure you put an exponent of 1 on all bases with an understood exponent of 1)
\[ 2^1 \cdot 3^1 \cdot 7^1
\]
Now take the exponents ... $ 1,1,1$ (call this step 1)
and add 1 to each exponent: $ 2,2,2$ (call this step 2)
The last step is to multiply all these numbers.
[hide="why it works"]
Suppose a number has prime factors $ p_1, p_2, p_3 \ldots$ with corresponding exponents $ e_1, e_2, e_3 \ldots$ (with prime factorization $ p_1^{e_1}p_2^{e_2}p_3^{e_3} \ldots$)
Then each factor can be written in the form
\[ p_1^{k_1}p_2^{k_2}p_3^{k_3} \ldots
\]
Where $ k_1$ ranges from 0 to $ e_1$, $ k_2$ ranges from 0 to $ e_2$, and in general, $ k_n$ ranges from 0 to $ e_n$
Therefore, there are $ e_n \plus{} 1$ ways to choose the value for each $ k_n$, and since each of the $ k_n$ are independent, we can use the counting principle and multiply the possibilities for each individual $ k_n$ to find the total number of possibilities for all the $ k_n$.
So, the number of ways to assign values to the $ k_n$ (or, equivalently, the number of factors of the original number) is $ (e_1 \plus{} 1)(e_2 \plus{} 1)(e_3 \plus{} 1) \ldots$
[/hide]
For the given problem, we want on odd number of factors.
This means all of the numbers in step 2 are odd, and all of the numbers in step 1 are even. If the prime factorization of a number has only even exponents, then it is a square number.


trigonometry
geometry
circumcircle
Pythagorean Theorem

We have an acute triangle ABC.  Altitude AY is drawn to side BC, and altitude BX is drawn to AC.  BY = 48 and AY=55.  What is the sine of AXY?

Simple enough...
[hide="one solution"]By the Pythagorean Theorem, $AB=\sqrt{48^2+55^2}=73$.  $\triangle{AYB}$ and $\triangle{AXB}$ have a common hypotenuse, and are right triangles.  Therefore, both of their circumcircles have this hypotenuse as a diameter, and thus are the same circumcircle.  Thus, this is the circumcircle of $\triangle{AXY}$, so $\sin{AXY}=\frac{AY}{AB}=\frac{55}{73}$[/hide]


modular arithmetic
number theory unsolved
number theory

Find all pair integer numbers $ (x,y)$ x,y>0 satisfy equation: $ \frac {x^{29} \minus{} 1}{x \minus{} 1} \equal{} y^{12} \minus{} 1$

Thank.

[hide="hint"]
use the fact that if $ q | \frac{x^p\minus{}1}{x\minus{}1}$ then we have either $ q\equal{}p$ or $ q \equiv 1 \pmod {p}$ where $ q,p$ are primes.
[/hide]
that's like $ \frac{x^7\minus{}1}{x\minus{}1}\equal{}y^5\minus{}1$ (shortlist)

[quote="shoki"][hide="hint"]
use the fact that if $ q | \frac {x^p \minus{} 1}{x \minus{} 1}$ then we have either $ q \equal{} p$ or $ q \equiv 1 \pmod {p}$ where $ q,p$ are primes.
[/hide]
that's like $ \frac {x^7 \minus{} 1}{x \minus{} 1} \equal{} y^5 \minus{} 1$ (shortlist)[/quote]

But unfortunately, 12 isn't a prime. :(

here is the solution:
we have two cases :
1) $ 29 \nmid \frac{x^{29}\minus{}1}{x\minus{}1}$
so by the lemma we know that each prime dividing $ (y\minus{}1)$ and $ (y^{11}\plus{}...\plus{}y\plus{}1)$ is congruent to $ 1$ modulo $ 29$
which means that $ y\minus{}1 \equiv 1 \pmod{29}$ or $ y \equiv 2 \pmod{29}$ which implies finally :
$ y^{11}\plus{}...\plus{}y\plus{}1 \equiv 2^{11}\plus{}...\plus{}2\plus{}1\equal{}2^{12}\minus{}1 \equiv 6 \pmod {29}$
so there must exist a prime like $ p$ which divides $ y^{11}\plus{}...\plus{}y\plus{}1$ and which isn't equivalent to $ 1$ modulo $ 29$.
so we r done!
2)$ 29 \parallel{} \frac{x^{29}\minus{}1}{x\minus{}1}$ which implies that $ 29 \parallel{} (y^4\minus{}1)(y^8\plus{}y^4\plus{}1)$
but we know by the lemma that each prime divisor of $ \frac{(y^4)^3\minus{}1}{y^4\minus{}1}$ is either $ 3$ or equivalent to $ 1$ modulo $ 3$ and since $ 29 \equiv \minus{}1 \pmod{3}$ so we must have $ 29 | y^4\minus{}1$ which implies that $ y^4 \equiv 1 \pmod {29}$
and finally it follows that $ y^8\plus{}y^4\plus{}1 \equiv 3 \pmod {29}$ so we must have a prime which divides $ y^8\plus{}y^4\plus{}1 \equiv 3 \pmod {29}$ and isn't equivalent to $ 1$ modulo $ 29$ so we get another contradiction and we r done!


Assuming that the birth of a boy or a girl is equally likely, what
is the probability that the three children in a family include at
least one boy and one girl? Express your answer as a common
fraction.

First, we need to list the different favorable outcomes.

B = boy

G = girl

BGB --> 3 ways
BGG --> 3 ways

For the first way, $ \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times 3 \equal{} \frac{3}{8}$.

For the second way, we get the same result, $ \frac{3}{8}$. We add them together and get $ \boxed{\frac{3}{4}}$.

I'm not convinced my answer is correct. If I'm wrong, someone please show me where I messed up and the correct solution. Thanks.

The only way that it is not possible would be if all of the children are boys or girls. There are 8 possible ways to have 3 children. Thus, the probability of having at least one boy and girl would be $ \frac{8}{8}\minus{}\frac{2}{8}\equal{}\frac{6}{8}\equal{}\frac{3}{4}$.

We can use complementary counting.  There are two cases that do not satisfy the requirement that there will be at least one boy and one girl:  if all three kids are boys or all three are girls.  The probability that this will occur is $\frac{2}{8}=\frac{1}{4}$.  The desired probability is $1-\frac{1}{4}=\boxed{\frac{3}{4}}$.


[quote="bpms"][quote="nebula42"]A sequence of numbers $a_{n}$ is defined as $a_{1}=7$, $a_{2}=6$, and $a_{n}=a_{n-1}-a_{n-2}$. Find the sum of the first 2006 terms of the sequence.

Generalization:
What would the sum of the first 2006 terms be if the sequence were defined as $a_{1}=x$, $a_{2}=y$, and $a_{n}=a_{n-1}-a_{n-2}$? (Express your answer in terms of $x$ and $y$.)[/quote]
[hide]I have a feeling that there is a pattern.
Let me start with the generaliztion.
These are the terms
x, y, y-x, -x, -y, x-y, x, y Note the pattern. Every six terms repeats. x+y+y-x-x-y+x-y=0. 2006 mod 6=2. So x+y is the answer. So for part one it would obviously be 13.[/hide][/quote]
Yes, that's correct.  :)
In fact, for some problems, it's actually easier to solve the generalization first and then apply it to a specific case.

[hide]repeats after 6, all but last 2 cancel out, $13$[/hide]


induction
number theory unsolved
number theory


calculus
integration
function
limit
Support
real analysis
real analysis unsolved

If f is any integrable function on $ \mathbb R$ , then :

1) $ \int f(x) dx \equal{} \int f(x \plus{} t) dx$ .

2) $ \lim_{ t \to 0 } \int\limits_{ \minus{} \infty}^{\infty} | g(x) [f(x) \minus{} f(x \plus{} t) ]|dx \equal{} 0$, where $ g$ is any bounded measurable function.

[quote="Kent Merryfield"]You can work out (1) for yourself.

For (2), what would you have if $ f$ were continuous and compactly supported?

Now: continuous compactly supported functions are dense in $ L^1.$ What can you get from that?[/quote]

$ f$ will be integrable ( since u say they are dense in $ L^1$ ) .

Let me tell u what I'm trying to do :

since $ g$ is bounded then we can assume there is $ M$ such that $ |g(x)|\leq M$.

if $ \int\limits_{ \minus{} \infty}^{\infty} | g(x) [f(x) \minus{} f(x \plus{} t) ]|dx  \leq  \int\limits_{ \minus{} \infty}^{\infty} M |f(x) \minus{} f(x \plus{} t) |dx$

Now , if $ f$ is continuous and vanishes outside a finite interval then it would be integrable .

from problem (a) , we could say $ \int\limits_{ \minus{} \infty}^{\infty} |f(x) \minus{} f(x \plus{} t) |\equal{}0dx$

Does my thinking true ???

Sorry , if u see anything bad in my  mathematical written . :oops:

BTW, I'm not a student

f is uniformly continuous on [a,b] its compact support
$ \forall \epsilon>0, \exists \delta>0$ s.t. $ |x-y|<\delta \to |f(x)-f(y)|<\frac{\epsilon}{(b-a)M}$
then
$ \int_{ - \infty}^{\infty} M|f(x) - f(x + t) |dx = \int_a^b M|f(x) - f(x + t) |dx <\epsilon$
for small enough t, and as we let $ \epsilon \to 0^+$ then we get what you wanted


calculus
integration
logarithms
calculus computations

[hide="Solution"]$\int_{-e}^{-\frac{1}{e}}\left|\ln\left|x\right|\right|\,dx=\int_{-e}^{-1}\ln\left(-x\right)\,dx-\int_{-1}^{-\frac{1}{e}}\ln\left(-x\right)\,dx$

Part integration gives \[\int\ln\left(-x\right)\,dx=x\ln\left(-x\right)-x,\] so the answer is \[1-\ln 1-\frac{2}{e}+\ln 1+1=\boxed{2-\frac{2}{e}}\] [/hide]

\[I\equiv\int_{-e}^{-\frac{1}{e}}\left|\ln\left|x\right|\right|\ dx=\frac{2(e-1)}{e}\ .\] $x: =-x\Longrightarrow \boxed{\ \int_{a}^{b}f(x)\ dx=\int_{-b}^{-a}f(-x)\ dx\ }\Longrightarrow I=\int_{\frac{1}{e}}^{e}\left|\ln x\right|\ dx\ .$

$x: =\frac{1}{x}\Longrightarrow \boxed{\ \int_{\frac{1}{a}}^{a}f(x)\ dx=\int_{\frac{1}{a}}^{a}\frac{1}{x^{2}}\cdot f\left(\frac{1}{x}\right)\ dx\ }\ \ \wedge\ \ \boxed{\ \int_{\frac{1}{a}}^{a}f(x)\ dx=\int_{1}^{a}\left(1+\frac{1}{x^{2}}\right)f\left(\frac{1}{x}\right)\ dx\ }$ $\Longrightarrow$ $I=\int_{1}^{e}\left(1+\frac{1}{x^{2}}\right)\ln x\ dx\Longrightarrow$

$\boxed{\ \begin{array}{ccc}u(x)=\ln x & \Longrightarrow & u'(x)=\frac{1}{x}\\ v'(x)=1+\frac{1}{x^{2}}& \Longrightarrow & v(x)=x-\frac{1}{x}\end{array}\ }$ $\Longrightarrow$ $I=\left[\left(x-\frac{1}{x}\right)\ln x\right]_{1}^{e}-\int_{1}^{e}\left(1-\frac{1}{x^{2}}\right)\ dx=$ $e-\frac{1}{e}-\left[x+\frac{1}{x}\right]_{1}^{e}=$ $e-\frac{1}{e}-e-\frac{1}{e}+2=$ $2-\frac{2}{e}\ .$


algebra
polynomial
calculus
integration
Rational Root Theorem

Let $f(x)=x^{n}+5x^{n-1}+3$ where $n>1$ is an integer. Prove that $f(x)$ cannot be expressed as the product of two polynomials, each of which has all its coefficients integers and degree at least $1$.

since its IMO, you can easily find in Contest Resource page  :wink:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=372292#p372292

There is a generalisation of this problem, which goes as follows:

If $a,m,n$ are positive integers and $p < a-1$ is a prime, then $f(x)=x^{m}(x-a)^{n}+p$ is irreducible.

Is somebody interested in solving this?  :)

[b]Solution (IMO 1993)[/b] By the extended [i]Eisenstein's Irreducibility Criterion[/i], $f(x)$ has an irreducible factor of degree at least $n-1$; $f(x)$ may have another factor $P(X)\in\mathbb{Z}[X],\ \deg(P(X))=1$.
Assume to the contrary that $f(x)$ has a factor of $P(X)\in\mathbb{Z}[X],\ \deg(P(X))=1$.  But by the [i]Rational Root Theorem[/i], if there exists a rational root $\frac{p}{q}$, then $q|1\wedge p|3$.  But these properties do not lead to a root.  This leads to a contradiction. $\blacksquare$

Masoud Zargar

Due to a combination of confusing notation and english, I don't really understand the end of joshua_mex's solution, but isn't it sufficient to say that if $g$ has all coefficients divisible by $3$, then $f$ must be divisible by $3$ at all integral $x$, which it is not?

EDIT: in the book [i] International Mathematical Olympiads 1978-1999 [/i] (Kuczma), the solution says "let $l$ be the least index for which $b_{l}$ is not divisible by $3$..." and then proceeds to find a contradiction. Am I right in thinking that this is not rigorous, because it does not deal with the possibility that no such $l$ exists?

I think I know what he's trying to say.  He just happens to bring in $p(x)$ without introduction and necessity.  He is trying to prove by contradiction that such a polynomial is irreducible in $\mathbb{Z}[x]$.  He attempts to lead to the fact that $g(-5)=\pm 3^{k}$, where $k$ is the number of roots of $g(x)$, but from $f(-5)=g(-5)h(-5)=3$, we see that the maximum degree of a factor of $f(x)$ other than $h(x)$, i.e. $g(x)$, must be 1.  But since there are no integer roots to $f(x)=0$, we have a contradiction.

I am not sure what you are exactly talking about to identify whether it is rigorous or not, so could you please reveal the details?

Masoud Zargar

Well, we can prove that if $f(x)=g(x)h(x)$, then both $g$ and $h$ have degree > 1 (this follows from testing $\pm 1$ and $\pm 3$ as roots). Then we can prove that all the coefficients of $g$ (WLOG) are divisible by $3$. This implies that $f$ is divisible by $3$ at all integer $x$, which can be disproved with a mod argument.

The book I was referring to took the least coefficient of $g$ that was not divisible by $3$, but I pointed out that this might not exist (in fact, it does not exist).

Could you please write out the book's solutions if possible? :)

Assume $f(x)$ is the product of polynomials $F$ and $G$:
\[f(x)=F(x)G(x) \]
And
$F(x) = a_{0}+a_{1}x+...+a_{k}x^{k}$
$G(x) = b_{0}+b_{1}x+...+b_{m}x^{m}$
Then the coefficient of $x^{n}$ in the product $F(x)G(x)$ is $a_{k}b_{m}$. $a_{0}b_{0}=3$ and $a_{k}b_{m}=1$, so $a_{k}=b_{m}=\pm1$ and WLOG, $a_{0}= \pm 1$, $b_{0}= \pm 4$.

Let $l$ be the least index for which $b_{l}$ is not divisible by $3$. Since $b_{0}= \pm 3$ and $b_{m}= \pm 1$, we see that $0 < l \leq m$. The coefficient of $x^{l}$ in $F(x)G(x)$ equals $a_{l}b_{0}+a_{l-1}b_{1}+...+a_{0}b_{l}$. $a_{0}b_{l}$ is not divisible by $3$ and all the other terms are so the sum is not divisible by $3$. Therefore, $l=n-1$. Then $k=1$ and $F$ is linear. Plugging in $pm 1$ into $f$, neither of them are roots so neither fo them is a root of $F$, which is impossible.

My point was that $l$ does not necessarily exist if all the coefficients of $G$ are divisible by $3$. This case can be handled in the manner mentioned in my previous post.


$\displaystyle\sum (a+b-2c)^2 = \sum (a^2+b^2 + 4c^2 - 2(a)(2c) - 2(b)(2c) + 2ab)$
$= \sum a^2 + \sum b^2 + 4\sum c^2 - 4\sum ac - 4\sum bc + 2\sum ab = 6\sum a^2 - 6\sum ab$
$3\sum a^2 - 6\sum ab + 3\sum b^2 = 3\sum (a-b)^2$
$\Rightarrow \displaystyle\sum (a+b-2c)^4 + 2\sum(a+b-2c)^2(b+c-2a)^2$
$= 9\sum (a-b)^4 + 2\sum 9(a-b)^2(b-c)^2$
$\displaystyle\sum 9(a-b)^2(b-c)^2 \geq \sum(a+b-2c)^2(b+c-2a)^2$
$\Rightarrow \displaystyle\sum (a+b-2c)^4 \geq \sum (a+b-2c)^4 + 2\sum(a+b-2c)^2(b+c-2a)^2 - 2\sum 9(a-b)^2(b-c)^2$
$\displaystyle = 9\sum (a-b)^4$

So, all we need do is prove:
$\displaystyle\sum 9(a-b)^2(b-c)^2 \geq \sum(a+b-2c)^2(b+c-2a)^2$
$\Rightarrow \displaystyle\sum (9(a-b)^2(b-c)^2 - (a+b-2c)^2(b+c-2a)^2) \geq 0$

This expands to:
$\displaystyle\sum (8b^4 - 4a^4 - 4c^4 + 12a^2b^2 + 12b^2c^2 - 24c^2a^2 - 16ab^3 +20ac^3 + 20ca^3 - 16b^3c - 12a^2bc - 12abc^2 + 24ab^2c - 4bc^3) \geq 0$

When all cyclic sums are taken and the eq. is divided by 4, it simplifies to:

$\displaystyle \sum ab^3 \geq 0$

Edit: Big Edit:

[quote]This expands to:
$\displaystyle\sum (8b^4 - 4a^4 - 4c^4 + 12a^2b^2 + 12b^2c^2 - 24c^2a^2 - 16ab^3 +20ac^3 + 20ca^3 - 16b^3c - 12a^2bc - 12abc^2 + 24ab^2c - 4bc^3) \geq 0$[/quote]

That is not all true, I left out a $-4ab^3$. When put back in we find $0 \geq 0$ but really $0 = 0$ which then tells us $\displaystyle\sum 9(a-b)^2(b-c)^2 = \sum(a+b-2c)^2(b+c-2a)^2$ which in turn tells us $\displaystyle\sum (a + b - 2c)^4 = \sum (a-b)^4$


A robot is programmed to do the following in a coordinate system: start at the orgin facing in the positive y direction and move 1 unit. For each subsequent move, it is to turn 90 degrees clockwise and go forward 1 unit further than it went in the previous move. What is the sum of the coordinates of the position of the robot after it just moved 18 units in one direction?

[hide]19[/hide]

[hide]
Find a pattern.

When the robot stops moving, it is facing the right, since 18/4=4 remainder 2.

On the first move, it moves to (0,1).  Then (2,1), (2,-2), (-2,-2), (-2,3), (4,3).

Notice that on the second move it is at (2,1), and on the sixth move it is at (4,3).  This pattern continues... so

2--> (2,1)
6--> (4,3)
10--> (6,5)
14--> (8,7)
18--> (10,9)

10+9=[b]19[/b]

[/hide]


\[ \begin{tabular}{|c|c|} \hline x & f(x) \\
\hline \hline 0 & a \\
\hline 1 & 24 \\
\hline 2 & b \\
\hline
\end{tabular}
\]

The table above shows some values for the function $ f$. If $ f$ is a linear function, what is the value of $ a \plus{} b$?

(A) 24
(B) 36
(C) 48
(D) 72
(E) It cannot be determined from the information given.

[hide="Solution"]
Let $ f(x) \equal{} mx \plus{} k$, such that,
$ f(0) \equal{} m\cdot 0 \plus{} k \equal{} a\implies k \equal{} a$

So $ f(x) \equal{} mx \plus{} a$.

also $ f(1) \equal{} x \plus{} a \equal{} 24$ and $ f(2) \equal{} 2x \plus{} a \equal{} b$

adding $ a$ to $ f(2)$ you get:

$ b\plus{}a\equal{}2x\plus{}2a\implies b \plus{} a \equal{} 2(x \plus{} a)\implies b \plus{} a \equal{} 2\cdot 24 \equal{} \boxed{48}$ so the answer is $ \mathbf{(C)}$.
[/hide]

[hide="Solution"]Since $ f$ is a linear function, we have $ \frac{24\minus{}a}{1\minus{}0}\equal{}\frac{b\minus{}24}{2\minus{}1}$.

Hence, we have $ 24\minus{}a\equal{}b\minus{}24 \implies a\plus{}b\equal{}48$.

Answer: (C) 48[/hide]

Could you just make the function $ f(x)$ $ \equal{} 24b$ with $ a$ being zero?

In which case $ f(0) \equal{} 0$ and $ f(2) \equal{} 48$ so $ a \plus{} b$ $ \equal{} 48$.

Lol, yeah. Even better, assume $ f$ is a constant function (so $ f(x)\equal{}24$ for all $ x$). Then obviously $ a\equal{}b\equal{}24$ giving the answer.
That's the beauty of questions that ask for a value dependent on multiple variables... you can make stupid assumptions.

[hide]
For some value $ x$, we have

$ 24\minus{}x \equal{} a$
$ 24\plus{}x \equal{} b$.

From this, we have $ a \plus{} b \equal{} 24 \minus{} x \plus{} 24 \plus{} x \equal{} 48$.
[/hide]

$ b \minus{} 24 \equal{} 24 \minus{} a, a \plus{} b \equal{} 48$


algebra
polynomial
calculus
function
derivative
analytic geometry
graphing lines

How do you find the maximum value of $y$ and $x$ in this graph generally,
$y=ax^n+bx^{n-1}....+z$?
Im asking for HOW, please explain :?
Can you use only $\frac{\delta{y}}{\delta{x}}$ or gradients please as I have only been taught that yet

I calculated the gradient of this curve,
$\frac{\delta{y}}{\delta{x}}=anx^{n-1}+b(n-1)x^{n-2}.....+0$
But now I dont understand what to do,
In class sir says that this line is a curve and at the highest point the gradient is $0$
But you could only work that out by looking at the graph
How can you do it without that? :?

If I understand correctly you want the minumum value of a real polynomial ?

If $n$ is odd, then $min P(x), x \in R = -\infty$
If $n$ is even and $a > 0$, then $min P(x), x \in R$ exists for a $x$ but there are no way to compute the min value exactly
If $n$ is even and $a < 0$, then $min P(x), x \in R = -infty$

But I'm not sure I understand your question  :huh:

Edit : actually you were looking for maximum value, but it's the same reasonning  :lol:

In our question in school
It was....
$y=(20-2x)^2*x$
So you have to find $y$ when it is MAXIMUM,
but in class sir told us that this would be when,
$f'(x)=0$
I dont understand how though.

What you're talking about is a theorem usually attributed to Fermat. Phrased in calculus terminology, if a differentiable function has a local maximum or local minimum, then the derivative at that point must be zero.

Since Fermat did his work a generation before Newton and Leibniz, to use the calculus terminology is a bit anachronistic, but Fermat was doing the algebraic manipulations needed to calculate the derivatives of polynomials, so it's still appropriate.

See this as a picture. Draw a picture of a curve with a local maximum. Now the derivative is the limit of the slope of a secant line. Draw the secant line connecting the maximum point to a point a little to the left. That secant line has a positive slope. The limit of a bunch of positive numbers has to be $\ge 0.$ Repeat this with a secant line connecting the maximum point to a point a little to the right. This secant line has a negative slope. The limit of a bunch of negative numbers has to be $\le 0.$ The slope at the maximum point has to be equal to both limits, and the only way for it to be simultaneously $\ge0$ and $\le0$ is for it to be zero.

tµtµ was bringing in the idea of global or absolute maximum, which is somewhat different than local maximum. Consider the function $f(x)=(20-2x)^2x=4x^3-80x^2+400x.$ If you think of this as a function whose domain is the whole real line, then as $x\to\infty,\,f(x)\to\infty$ and the function does not have a global maximum. That's not very interesting, and likely not what your problem was referring to. Perhaps this came from a physical optimazation problem. Let's see: maximize the volume of the open-topped box you get starting with a $20\times20$ sheet of cardboard, cutting squares out of each corner, and folding it up. Is that it?

In any case, the point is likely to be that this formula doesn't really make sense for all $x$ but only for $x$ in some restricted domain. Perhaps we're only concerned with $0\le x \le 10.$

This function will have a maximum, and you can find it by setting its derivative equal to zero.


[quote="jeez123"]Solve for $ x$ using logs:
$ 9^{x}\equal{} 2e^{x^{2}}$

What I did:

$ \frac{9^{x}}{2}\equal{} e^{x^{2}}$
$ ln(9^{x}/2) \equal{} x^{2}$

then how do you isolate the $ x$?[/quote]

[hide]$ ln 9^{x}\minus{}ln 2 \equal{} x^{2}$

$ 0\equal{}x^{2}\minus{}x\cdot ln 9\plus{}ln 2$

now you can just use the quadratic formula.[/hide]


[hide]$\sin(x)=\cos(\pi/2-x)$ so every possible x from the original equation are used for $(4\sin^{2}(\pi/2-x)-3)(4\cos^{2}(\pi/2-x)-3)$.
Multiplying through, we get $16\sin^{2}(\pi/2-x)\cos^{2}(\pi/2-x)-3$.  SEt this equal to $4\sin^{2}{z}-3$.  One gets $2\sin(\pi/2-x)\cos(\pi/2-x)=\sin(z)$.  This looks like the double angle formula, so $z=\pi-2x=2x$ since taking the sine of either will have the same value.  The only factor that isn't paired is the median, which then becomes the maximum in the new group of factors.  However, the signs keep alternating because $\cos(\pi-2x)=-\cos(2x)$.  Noting that if this entire equation was reduced down to k=2 and the denominator being 4, the product is 3, so it should similarly follow that for k=$2^{1}999$, the product is also 3.[/hide]


[hide="BEST PICTURE EVER"]
[img]http://www.planetavp.com/images/planetavpfeatures/pegleg.gif[/img]
[color=red][size=150]Peg Leg John[/size][/color][/hide]

[i]
Moderator edit: This is getting out of hand, my computer is loading slower. I like your pictures and all, but maybe you could hide them and tell them it's a picture so that they can unhide if they want to see it. That way the page won't load quite so slowly. By the way whats your upload quota?  :lol:  ;)  :D  :P [/i]

[color=darkred]Upload quota? Close to 1%.[/color]

[color=indigo][hide]If he is a human with 2 kinds of socks the maximum number of socks is 3, if he has 3 kinds of the maximum is 4, if he is an alien the maximum is 7.  :lol:[/color][/hide]

[quote="h_s_potter2002"]John is a human who has a whole bunch of socks that he wears.  He will only wear a pair of socks if each one is the same type of sock.  When he goes to chose which socks he will wear for that day, he chooses one sock at a time untill he has a pair of socks from the socks he took.

What is the maximum number of socks that John can pull out before he gets a pair if he has two different type of socks?  Three different types of socks?  What if he is an alien with 3 legs, and has 3 different types of socks?[/quote]

[hide]two kinds- three
three kinds- four
alien 3 legs, 3 kinds- seven[/hide]

[quote="h_s_potter2002"]John is a human who has a whole bunch of socks that he wears.  He will only wear a pair of socks if each one is the same type of sock.  When he goes to chose which socks he will wear for that day, he chooses one sock at a time untill he has a pair of socks from the socks he took.

What is the maximum number of socks that John can pull out before he gets a pair if he has two different type of socks?  Three different types of socks?  What if he is an alien with 3 legs, and has 3 different types of socks?[/quote]

[hide="hint"]pigeonhole[/hide]

If you would like to make the problem harder:

Given a creature with $x$ legs and $y$ sets of $x$ socks, and every 3rd sock he pulls out of a dresser is instantly destroyed, what is the maximum number of socks he can pull out to insure that he has at maximum, one complete pair of socks?

what if the creature had 3 legs and 2 sets of 3 matching socks?
then, he could never get a matching set.
if he had 1 set of red socks and one set of blue socks, he could pick like this:
1 red sock
1 blue sock
1 red sock (destroyed)
1 blue sock
1 red sock
1 blue sock (destroyed)

after going through all the socks, the creature would have 2 red socks and 2 blue socks, no matching sets. if you set it up that way, it doesn't have a solution.


quadratics
linear algebra
matrix
algebra
system of equations
linear algebra theorems

2 weeks ago i've read the Kurosh's book on the Algebra the following:

let $\lambda_0$ be the characteristik root of matrix A(real symmetric) with the multiplisity k.
From the theorem that states that there exists such orthogonal matrix Q s.t. it $Q^{-1}AQ$ is diagonal matrix and this diagonal matrix is  unique it follows that there are k linear independent solutions in the system of equations $(A-\lambda_0E)X=0$.

It was all clear 2 weeks ago, but i am a little bit confused about "[b]it follows that there are k linear independent solutions in the system of equations $(A-\lambda_0E)X=0$.[/b]" and no one here told me why it is so...
please help..  :(

you forgot to mention that your matrix A must be real symmetric

well, that is the point of being diagonisable
the algebraic multiplicity equals the geometric multiplicity
in your diagonal form you will clearly see k different basis eigenvectors of your eigenvalue

Or was that not your question?

Forgive me my insolence, but I must say your questions are not always that easy to understand.

My question is why  [b]there exist k-linear independent vectors(basis) in the subspace X[/b].
In that book it was said that [b]"...exists such orthogonal matrix Q s.t. it transforms the given one to diagonal form and this diagonal form is unique[/b](i know that on the diagonal stand all the eigenvalues).[b]That's why there exist k-linear independent vectors(basis) in the subspace X"[/b]

but i can't still understand it
and what u mean saying [color=darkblue][b]the algebraic multiplicity equals the geometric multiplicity [/b][/color]?

Is it a correct prove?
If not i dunno what book should  study


topology
inequalities
triangle inequality
real analysis
real analysis unsolved

consider the finite  set of symbols  $ \Lambda \equal{} \{ a_{1},...,a_{r}\}$ and consider the set $ \Omega$ of maps $ f: \mathbb{Z} \to \Lambda$.
Let $ d$ be metric on the set $ \Omega$ such that :
$ d(x,y) : \equal{} \inf\{ 1/k\plus{}1 : x(i)\equal{}y(i) \text{for all } i, |i|<k \}$
for all $ x,y \in \Omega$
Prove that $ \Omega$ is compact .

What characterization of "compact" will be applicable to this space?

Extremal, you, of cousre, meant that the distance is $ \frac {1}{(k \plus{} 1)}$, right? Because if this is not the case(if distance is $ \frac {1}{k} \plus{} 1$), then the problem is false, you can just take an open covering by balls with radius $ \frac {1}{2009}$ and get the contradiction :wink:
P.S To prove that our set is compact one can first prove that given space is complete and then using standart arguments that from any  sequence we can choose convergent subsequence.

Yes I meant that distance is $ \frac {1}{k + 1}$
I just wanted to see solution by construction homeomorphism of $ \Omega$ to Dekarts product of two sets Cantor type.
(P.S if distance is ${ \frac{1}{k}+1}$ then $ \Omega$ is not metric space, there is problem near zero hence triangle inequality can't be always true :) )

If you consider something like $ 1\plus{} \frac {1}{k\plus{}1}$ or $ 1\plus{}\frac {1}{k}$ (where $ f : \mathbb {N}\to\omega$) instead, then your space, of course, would be metric, that what I wanted to say in my previous post. :)


trigonometry
complex numbers
complex analysis
complex analysis solved

Let $\alpha,\beta$ be distinct complex numbers such that $\frac{\alpha}{\beta}+\frac{\overline{\alpha}}{\overline{\beta}}=2$ and $\alpha \neq 0,\ \beta \neq 0$.Determine the shape of $\triangle 0\alpha \beta$ in a complex plane.

sorry for my miss typo, Diogene.

I have just edited.

You have forgot something again: what is $\frac{\alpha}{\beta}+\frac{\overline{\alpha}}{\overline{\beta}}=?$.

Thanｋ you for pointing out my miss typo, Myth. :)  :blush:

I have just edited.

kunny

uhm...
let's set $ a = |\alpha|$,$b = |\beta|$, $\gamma = \theta - \phi$, $\theta = \arg {\alpha}$, $\phi = \arg {\beta}$.
the given condition is $ {a \over b}(e^{i\gamma} + e^{-i\gamma}) = 2$, so $ a \cos {\gamma} = b$.
so we have that $\triangle 0\alpha \beta$ has a right angle in $\beta$.

$\frac{\alpha}{\beta}+\overline{\left(\frac{\alpha}{\beta}\right)}=2\Longleftrightarrow Re\left(\frac{\alpha}{\beta}\right)=1\Longleftrightarrow \frac{\alpha}{\beta}-1\in pi\ (p\neq 0,p\in \mathbb{R})$ from $ \alpha\neq \beta \Longleftrightarrow \frac{\alpha-\beta}{\beta}\in pi$,yielding $\angle 0\beta\alpha =\frac{\pi}{2}$.


search
number theory unsolved
number theory

Prove that $ 1 \plus{} \frac {1}{2} \plus{} ... \plus{} \frac {1}{p \minus{} 1} \equal{} \frac {a}{b}$ with $ p^2 \mid a$ if $ p$ is a prime greater than $ 3$

Search the forum for Wolstenholme's theorem.

thanks

 The message is too small. Please make the message longer before submitting.


The volume of oblique prism $ABC-A_1B_1C_1$ is $V$. $P$ on lateral edge $BB_1$ such that $PB=\frac{1}{2}BB_1$. Find the volume of $P-ACC_1A_1$.

[hide]Let $S=[ABC]=[A_1B_1C_1]$ and $H$ be the height of the prism. Then $V=SH$ and $V_{PABC}=V_{PA_1B_1C_1}=S\cdot\frac{H/2}{3}={V\over 6}$, hence $V_{PACC_1A_1}=V-V_{PABC}-V_{PA_1B_1C_1}={2\over 3}V$[/hide]

[hide]This is a triangular prism, so call the height of the triangular faces $h$, and the area of the polygon $ACC_1A_1$ $a$.  Then $V=\frac{ah}{2}$.  Since $P-ACC_1A_1$ is a pyramid with the same height and base as the prism, it's volume is $\frac{ah}{3}$.  Thus, $\frac{\frac{ah}{3}}{\frac{ah}{2}}=\frac{2}{3}$ and the volume of $P-ACC_1A_1$ is $\frac{2}{3}V$.[/hide]

[quote="chesspro"]Please hide your answers. Thanks.  :)[/quote]I hid it for him. :D