1. There are three points in the plane. All of the points are not on the straight line. How many straight lines we can draw that have same distance from these three points?

2. There are four points in the space. All of the plane that can be made by any three points, don't meet one another. Then, how many planes we can draw that have same distance from these four points?

3. There are four points in the plane with all of the points are not on straight line and either they are not on the same circle. How many circles or lines we can draw that have same distance from those four points?  (But, any three points of them can be on the straight line and same circle)

For 1 :
Let A,B,C be the three given points. Let d be a line equidistant from these 3 points. Clearly, the three points are not on the same side of d, because it would contradicts the hypothesis that they are not collinear.
Thus, there are two points on the same side of d, say A and B, and the third one is on the other side.
In that case, d is clearly the line which is parallel to the line (AB) and is the perpendicular bissector of [CC'] where C' is the orthogonal projection of C on the line (AB). Thus, in that case d is uniquely defined, and conversly this line is a solution.
Thus, there are exactly 3 equidistant lines, according to the pair of points chosen in {A,B,C}.

Pierre.

For 2 : Well, it is a quite similar reasoning as above, since if A,B,C,D are the four given points and P is an equidistant plane, then A,B,C,D are not on the same side of P (if not they would be coplanar, contradicting the hypothesis).

- If three of the points, say A,B,C are on the same side of P, and D is on the other side :
Clearly, A,B,C are not collinear, if not with D they would be colplanar. Thus, they determine an unique plane (ABC), which is parallel to P. And as above P contained the midpoint of [DD'], where D' is the orthogonal projection of D on (ABC). This determine uniquely P, and conversely this plane is an equidistant plane.
Thus, there are 4 such plane in that case.

- If two of the points, say A and B, are on the same side of P and the two others are on the other side :
As above, it is easy to see that the lines (AB) and (CD) are not coplanar, and that P has to be parallel to these two lines. More precisely, P is parallel to the plane P' which is parallel to the line (AB) and wich contains the line (CD), and P is parallel to the plane P'' which is parallel to the line (CD) ad which contains the line (AB). And P is equidistant from P' and P'' (it goes through the midpoint of [AA'], where A' is the orthogonal projection of A on P'). Thus uniquely defined the plane P, and conversely this plane is an equidistant plane.
Since there are exactly 3 ways of partitionning {A,B,C,D} into pairs, it leads to 3 equidistant planes in that case.

Thus, there are 4+3 = 7 equidistant planes.

Pierre.

for 3 :
The reasoning is similar than above, but we have sometimes to distinct between circles and lines :

Let A,B,C,D be the given points.
Let d (resp. w) be an equidistant line (resp. circle).

- If the four points are on the same side of d ( resp. they are all inside w, or all outside w), then they are all on a same parallel to d (resp. on a same circle conentric with w), which contradicts the hypothesis.

- If, three of the points, say A,B,C are on the same side of d (resp. w), and the other on the other side :
According to the fact that A,B,C are colinear or not, it follows that A,B,C are either on a parallel d' to d or on a circle w' concentric with w, and clearly theses cases exclude each other. Moreover d' or w', according to the case, must contain the midpoint of the segment [DD'], where D' is the orthogonal projection of D on d' or w'. Thus, in each case, d' or w' is uniquely defined, and conversely is an equidistant line/circle.
Thus, there are a total of 4 equidistant lines or circles in that case.

- If two of the points, say A and B, are on the same side of d (resP; w), and the two others on the other side :
Let k (resp. k') be the perpendicular bissector of the segment [AB] (resp. CD]). First note that k and k' cannot coincide, since it would mean that the four points are concyclic, contradicting the hypothesis.

a) If k and k' intersects at O :
Then, any equidistant circle must have O for its centre and radius R = (OA  + OC)/2, and then is uniquely determined, and conversely is an equidistant circle.
In the other hand, any equidsitant line has to be parallel to the line (AB), and to the line (CD). but, since k and k' intersect, this cannot be done. Thus, there is no equidistant line in that case.

b) If k and k' are parallel :
As above, we see that there is no equidistant circle, and that an equidistant line has to be a common parallel to the lines (AB) and (CD) (which, in that case, are parallel), and equidistant to these two lines. Thus, there is an unique equistant in that case.

Thus, in each case a) or b), we have exactly one equidistant set (line or circle). It gives a total of 3 equidistant sets, according to the pairing of the  points.

Then, we have a total of 4+3 = 7 equidistant set.

Pierre.


Solve:
$\large 3^{\log_{3}x}+4^{\log_{3}x}=7^{\log_{3}(x^{2}-6)}$
It is intrested
Very good at M..

Bad title http://www.mathlinks.ro/Forum/viewtopic.php?t=144403 Locked!


analytic geometry
rotation
geometry
geometric transformation
reflection

Find the coordinate of the image of the point (-2,1) if it is translated under <6,-5>, relected in y=1, and then rotated 270 degrees about the origin.

(How do u find how to translate, reflect over a certain line,reflect over a certain point, and rotate a point?) (i need a generic concept on how to do these different effects)

Translations are simple; move it "over" by the amount specified. In the x coordinate, positive is to the right, negative to the left. In the y coordinate, positive is up, negative is down. Reflecting over a line isn't so hard either; it's best not to memorize a formula for this, just understand how it works. You can determine the equation of the perpendicular of the line through the point, find the perp. distance from the point to the line, then find the coordinates of the point that's the same distance from the original line on the other side of the line on the perpendicular. There might be a simpler way, but I don't know one. Rotating about the origin is simple if your coordinates are expressed in polar form; just add the the number of degrees or radians to $ \theta$. Otherwise, it's pretty easy to figure out the formula yourself by converting your rectangular coordinates to polar ones and then back after rotation.


Show that, for every positive integer n,
121^n - 25^n + 1900^n - (-4)^n
is divisible by 2000.


AMC
AIME
geometry
trigonometry
perpendicular bisector

i was wondering if anyone has any hints for problems 9, 12, 14, and 15? i can't seem to get the geometry ones...

and if you  have questions about other problems, please post!

thanks,

sara

here are some hints for those problems.  if they're too confusing or obscure, please tell me.

[hide="hint for problem 9"]
notice that $\frac{9x^2\sin^2(x) + 4}{x \sin (x)}=\frac{9(x\sin(x))^2+ 4}{(x \sin (x))}$
how much do you care about the fact that it's $x\sin(x)$, and not, say, $x$?
[/hide]
[hide="hint for problem 12"]
let $AB=10m+n$ (where $m$ and $n$ are the digits), and $DC=10n+m$.
we want $OH=\sqrt{\Big(\frac{AB}{2}\Big)^2-\Big(\frac{DC}{2}\Big)^2}$ to be rational, so that means we want $2OH=\sqrt{(10m+n)^2-(10n+m)^2}$ to be a perfect square.
there ends up only being one solution for $0 \leq m,n \leq 9$
[/hide]
[hide="hint for problem 14"]
I think a nice technique for this problem is to identify what information the circles give you, mark that, and then forget about the circles entirely.  Also, drawing your picture so that $QR$ is horizontal can make the problem seem more clear.
[hide="what information the circles give you"]
try drawing the perpendiculars from $A$ and $B$ to $QR$.  what do you know about where they meet $QR$?  after that you can erase the circles, and the problem is a lot simpler.
[/hide][/hide]
[hide="hint for problem 15"]
first, ignore the circle for a moment.  say you're given a point $A$ and a line $BC$.  what is the set of points $D$ such that $AD$ is bisected by $BC$?
if you can't quite see it, try drawing $A$ and $BC$ and then find a bunch of points $D$ that work and mark them down.  see what they look like, and then it might be clear.
[/hide]

[hide="Hint for 15"]If you draw a circle with diameter AO, that circle must be tangent to BC.[/hide]

[hide="hint 2"] OD is the perpendicular bisector of BC[/hide]
[hide="hint 3"] Draw line AH perpendicular to the expension of BC intersecting BC at H, you can figure out the lengths of just about all sides in the diagram, and they are almost all integers[/hide]

Trig would help a lot here, but I dont know much of it.


calculus integration function logarithms trigonometry limit Euler

Identifying $x!$ as $\Gamma (x+1)$, How can I show this formula?

$\int_{0}^{1}\ln (x!) \, dx = \ln \sqrt{2\pi}-1$

Use this trick:

$I=\int_{0}^{1}\ln \Gamma(x)dx=\int_{0}^{1}\ln \Gamma(1-x)dx=\int_{0}^{1}\ln \frac{\pi}{\Gamma(x)\sin \pi x}dx$

It follows that

$2I=\ln \pi-\int_{0}^{1}\ln \sin \pi x dx$, and the last integral can be calculated by using the standard techniques.


geometry
trapezoid
trigonometry
trig identities
Law of Cosines

Prove that if the sides of trapezoid $ ABCD (AB\parallel{}CD)$ satisfy the condition $ AC^2 \plus{} BD^2 \equal{} (AB \plus{} CD)^2$, then $ ABCD$ is orthodiagonal. It's #44 in page 31 aops 2. Can somone help me?
And I still don't know if I post in the right forum or I need to post it in intermediate.

Quoted from the Solution Manual,

Extend $ AB$ to $ E$ so the $ BECD$ is a parallelogram. Thus, $ AE\equal{} AB \plus{} BE \equal{} AB \plus{} CD$ and applying the law of cosines to $ \bigtriangleup ACE$ gives

$ AE^2 \equal{} (AB \plus{} CD)^2 \equal{} AC^2 \plus{} EC^2 \minus{} 2(AC)(CE)cos\angle ACE$.

Since $ EC\equal{}BD$ and $ (AB \plus{} CD)^2 \minus{} (AC^2\plus{}BD^2) \equal{} 0$, we have $ cos \angle ACE \equal{} 0$. Thus, $ \angle ACE \equal{} 90^\circ$. Since $ BD\parallel EC$, $ BD\perp AC$ and $ ABCD$ is orthodiagonal.

This may be more of intermediate, but there's no harm, as long as it's "reasonably close"  :)

I don't understand where is $ (AB\plus{}CD)^2\minus{}(AC\plus{}BD)^2\equal{}0$ come from? And if we have that, we would have $ AE^2\equal{}AC^2\plus{}BD^2  (CD\equal{}BE, (AB\plus{}CD\equal{}AE), BD\equal{}CE)$ which is true by Pythagorean? So why we need the law of cosine here?

[quote="doabli"]I don't understand where is $ (AB \plus{} CD)^2 \minus{} (AC \plus{} BD)^2 \equal{} 0$ come from? And if we have that, we would have $ AE^2 \equal{} AC^2 \plus{} BD^2 (CD \equal{} BE, (AB \plus{} CD \equal{} AE), BD \equal{} CE)$ which is true by Pythagorean? So why we need the law of cosine here?[/quote]
Umm, aguoman said $ (AB\plus{}CD)^2\minus{}(AC^2\plus{}BD^2)\equal{}0$ (you misquoted him).  This statement is obviously true by the given.

Ok, I see, thank you. So you don't need to use the law of cosine like the solution book  :huh:


A bug is placed on a 2 dimensional grid. Every second, the bug moves 1 unit up, down, left or right, with each direction having an equal chance of being chosen. If the bug is allowed to move for 6 seconds (6 moves) what is the probability that the bug ends up where it started?

There are $ 4^6\equal{}4096$ total possibilities.

In order for the ant to end up at the origin, it must have an equal number of ups and downs, as well as rights and lefts.

If there are three ups, three downs, and no other moves (or three rights, three lefts, and no other moves), then there are $ \dbinom63\equal{}20$ arrangements of the movements.
If there are two ups, two downs, one right, and one left (or two rights, two lefts, one up, and one down), then there are $ \dbinom62\dbinom42\dbinom21\equal{}180$ orders of movements.

Since there are two possibilities for each case, there are $ 400$ ways for the ant to end back at the origin, and the probability is $ \frac{400}{4096}\equal{}\boxed{\frac{25}{256}}$.


1.Given x,y,z, three different integers.
Prove that $((x-y)^{5}+(y-z)^{5}+(z-x)^{5})$ is divisible by $5(x-y)(y-z)(z-x)$.
2.Find all the natural n such that $n!$ is not divisible by $n^{2}$.
3.What is greater, $31^{11}$ or $17^{14}$? (no calculators) :)

[quote="Bictor717"][hide="2"]Find $n$ such that $(n-1)!$ is not divisible by $n$. Wilson's Theorem, $(p-1)!\equiv-1\mod p$ for $p$ prime. Then there's 4. So final answer: 4 and primes.[/hide]
[hide="3"]$17^{14}>16^{14}=2^{56}>2^{55}=32^{11}>31^{11}$[/hide][/quote]

I got same answers for these. For the first one, I just thought it might have something to do with roots of polynomial.

[quote="Bictor717"][hide="2"]Find $n$ such that $(n-1)!$ is not divisible by $n$. Wilson's Theorem, $(p-1)!\equiv-1\mod p$ for $p$ prime. Then there's 4. So final answer: 4 and primes.[/hide]
[hide="3"]$17^{14}>16^{14}=2^{56}>2^{55}=32^{11}>31^{11}$[/hide][/quote]

Can you prove that 4 is the only non-prime answer?

$n!$ is not divisible by $n^{2}$ if and only if $n$ only appears once in $n!$. This occurs if $n$ is prime, because the next occurrance of $n$, because it is prime is $2n$, which is obvioustly not in $n!$. This can also occur if the number of 2 in $n^{2}$ is more than in $n!$, this only occurs for $n=4$. $4^{2}=2^{4}$ and $4!=2^{3}*3$. I'm sure the number of $2$ in $n!$ can be put into a general formula and this can be proved in a more general term, but for right now, this is the best I can do.


logarithms
calculus
calculus computations

I need help with half-life relating to population.

A=100.326e^-.1314*t

In this decay problem, the starting number is 100.326, the multiplication rate is -.1314, and the time is unknown.  How can the half-life be found?

This is a very standard homework or test question; your textbook should have examples.

Hints:

1. Think about exactly what "half-life" means. What equation should you be solving for $t$?

2. Take the logarithm - in this case, the natural logarithm - of your equation.

3. The number 100.326 is irrelevant.

Half-life is a first order reaction, meaning that the change in quantity is proportional to the quantity: $\displaystyle \frac{dQ}{dt} = kQ$. $k$ is just the constant of proportionality.  Seperate the variables to obtain $\displaystyle \frac{dQ}{Q} = k dt$.  Integrating both sides yields $\displaystyle \log_e Q = kt+C \Rightarrow Q=Ce^{kt}$.  Note that $C$ is the initial quantity or $Q_0$.  Thus $Q=Q_0e^{kt}$.  Don't memorize this, but understand it.  So for half-life, set $Q$ to $\frac{1}{2}$ and set $Q_0$ to 1.


rotation
geometry
geometric transformation
reflection

Two circles intersect at A and B. A line through A meets the circles at C and D. M and N are the midpoints of arcs BC and BD, both not containing A. K is the midpoint of CD. Show that $ \angle MKN\equal{}90$.

Let $B'$ be the inverse of $B$ under the inversion with center $A$ and power $\overline{AC} \cdot \overline{AD}.$ Circles $(O),(O')$ go to lines $DB'$ and $CB'$ $\Longrightarrow$ $M' \equiv AM \cap DB'$ and $N' \equiv AN \cap CB'$ are the inverses of $M,N,$ respectively. Let $U \equiv M'N' \cap CD.$ Since the pencil $A(C,B,M,N)$ is harmonic, it follows that pencil $B'(M',N',A,U)$ is also harmonic $\Longrightarrow$ $(C,D,A,U)$ is harmonic $\Longrightarrow$ $\overline{AC} \cdot \overline{AD}=\overline{AK} \cdot \overline{AU},$ i.e. $U$ is the inverse of $A$ under the referred inversion. Since $U,M',N'$ are collinear, then $K,$ $M,$ $N,$ $A$ are concyclic and the conclusion follows.

Take a random point $X$ on $(AB$ produced (to see that $\angle MBX=\angle ACM\ (*)$ and $\angle NBX=\angle ADN\ (**)$).
Rotate triangle $\triangle MBN$ about $M$ of angle $\alpha=\angle BAC$; $B$ will go to $C$ and $N$ to a point $N'$ so that $\angle MCN'=\angle MBN$, implying, from $(*), (**)$ that  $CN"\parallel DN$; with $CN'=ND$ we get that $N'$ is the reflection of $N$ in $K$, done (do not forget, the rotation preserves the distances, hence $MN'=MN$).

Best regards,
sunken rock


A sequence is defined by $ a_{n\plus{}1} \equal{} a_n \plus{} 2n$, where the first term is $ a_1 \equal{} 4$. What is the sum of seventh term and eighth term?

note that the difference between the n and n+1 term is 2n. that means the difference between the first and seventh term is 2(1)+2(2)+2(3)...2(7-1) which becomes $ 2\frac{(6)(7)} {2}\equal{}6*7\equal{}42$ so the seventh term is 46.the 8th term is 2*7=14 more than that. so the 8th term is 60. $ 60\plus{}46\equal{}\boxed{106}$


[quote="mr. math"]If $x^{1/2}= k^{3/2}= y^{5/6}$, what is the value of x in terms of y?[/quote]
[hide]Set them all equal to a. So we have
$x^{\frac{1}{2}}=a$.
$k^{\frac{3}{2}}=a$
$y^{\frac{5}{6}}=a$
So we want x in terms of y.
Simplifying the first equation we have that $x=a^{2}$. If we go to the last equation, and square that one, we have that $y^{\frac{5}{3}}=a^{2}$. So now we can set these two equations together and have our answer. ${x=y^\frac{5}{3}}$[/hide]

What is the purpose of introducing k?...it is completely unneccessary.I would understand if x and y were defined parametrically in terms of k, and one would have to find the constraint equation.

Probably just to confuse you. But as everyone says, all you have to do is square the equation.


analytic geometry
graphing lines
slope
algebra
linear equation

Hello forumers. I have a lot of trouble in math so I posted examples of what I need help in. Thanks in advance.

Getting the equation of the line with the slope and y-intercept given.
eg.
3; (3,4)
-1; (2,-3)
$\frac 12$; (-3,2)

Getting an equation from points already given.
eg.
A(3,5) and B(-1,2)

Getting equations in the $y = mx + b$ form.
eg.
$x - 2y - 4 = 0$
$y = \frac 34x + 2$

Getting equations in the from of $Ax + By + C = 0$.
eg.
$y = 3x - 2$
$y = \frac{-2}{3}x - 5$

Finding slopes and y-intercepts from $Ax + By + C = 0$ equations.
eg.
$6x + 2y - 3 = 0$

Getting coordinates from $Ax + By + C = 0$ equations.
eg.
$2x + y + 4 = 0$

Those are the important ones I could think of. Help on any of them is greatly appreciated.

EDIT - Read the post I made thats a couple down from this one. Itll clear up this post.

[quote="Syrin"]
** - Thats supposed to be negative 2 over 3. I cant get the negative in the fraction. It comes out as a negative sign over 2 with a 3 beside it. $y = \frac -2{3}x - 5$ <-- Like that.[/quote]

y=\frac{-2}{3}-5

Since you are new,you should know one thing that mostly this forum is not for solving homework problems but of course one of the administrators has said that it is not wrong to help people if they have some problems in their homework.So never post excess homework problems and expect people to help you all the time unless you have tried it a lot,searched for it,etc ;)

P.S--I know I'm no moderattor but this is what they usually say,so if I am wrong in any way,please do correct me :)

don't you have a math textbook? These are fairly simple straight forward math. There isn't much to explain. It's just merely knowing how to do these problems, and understanding what they (like the linear equations you have) are. There is no trick to it.

Assumint you have a book, if these were given to you as homework, then i expect that your book explains how to do these types of problems. Have you tried reading it?

Make sure you understand the basic linear equation: $y=ma+b$.

the [i]x[/i] and [i]y[/i] terms are just what they are. the values of y in relation to x.
the [i]b[/i] term is the "y-intercept" (the point where the line of the equation crosses the y-axis)
the [i]m[/i] term is the "slope" it determines the tilt of your line. A greater positive slope means a greater increase in y per increase in x. if it's negative then a greater decrease in y per increase in x.

if you know this, you should be able to do all those problems up there.

Generally school homework is not permitted to be posted in this forum... :ninja:

Ok, sorry for the big list of work then. I know now. And akimatsu, I know the letter meanings to an extent and my textbook does have examples and solutions to them but I dont know how its really done and solved. My book quickly goes over them by showing the questions and the steps to solving it. It dosent give an in-depth explination of the equation. It basically assumes you know everything before it perfectly and have an idea on how the new one works. In my case, I dont understand things you learn in grades 7 and 8. Read my other post and youll see why. It takes me a long time to grasp most things involving math. Some people have the benefit of being good learners and others have trouble in things, like me in math. Its not that I dont want to learn it, its that by the time I understand one problem, Ive got five more that need to figured out. Its just how I am but unfortunately, the math I know is sufficient in the everyday world but its not enough to please the government. So until I understand whats troubling me, Im screwed.

EDIT - Ok, I changed the first post. I dont really need the answers, just how theyre done. Now you cant say I need my homework done for me. Thanks in advance.


advanced fields
advanced fields unsolved

After doing a few problems on Turing machine programming, I couldn't intuitively resolve a dilemma appeared:

Most of the problems are "Given a string of 'ones' on the tape with head scanning the leftmost 'one', do..." (the allowed alphabet is (1,_)). Now, my question is: is the rest of the tape surely blank? I know for sure that the two neighboring fields of my string are blanks, but are all others blanks too, or they can contain both ones and blanks? Intuitively, I assume the rest of the tape is blank by default, but knowing that problems with counter machines require clearing contents of registers, I'm aware of the possibility that the 'memory' of the system does not necessarily contain only the data necessary for the observed procedure.

[EDIT: I should have posted it in [i]Theorems, Formulas and Other Questions[/i], sorry.]

[quote="hsiljak"][EDIT: I should have posted it in [i]Theorems, Formulas and Other Questions[/i], sorry.][/quote]  Or perhaps in Computer Science.  In Sipser's [u]Introduction to the Theory of Computation[/u], the blank symbol is a special symbol that belongs to the tape alphabet but not to the input alphabet.  This seems like a "reasonable" restriction on a TM, although it is likely that it doesn't matter to the ultimate computing power of the machine.


How would you solve for x algebraically??

$x^{3}e^{\frac{-a}{x}}= b$

where a and b are some constants.

Generally, most arbitrary problems do not have closed-form solutions.  This is one of them.

What do you mean by "closed-form" solution?

I mean that you can't write the solution in the form

$x = \text{blah}$

Where $\text{blah}$ is some algebraic expression involving the constants in the problem.

What about this?
$x = \frac{a}{3W(\frac{ab^{-1/3}}{3})}$

where $W(\cdot)$ is the Lambert W function. This is a closed form solution right?

[quote="t0rajir0u"]The Lambert $W$ function is cheating.   :P

(Specifically, it's not algebraic.)[/quote]
What does it matter if it is algebraic or not? I mean logs are non-algebraic and most of us don't consider using logs as "cheating." Lambert W function is pretty much like logs too. Instead of being the inverse of $e^{x}$, W is the inverse of $xe^{x}$.

I see your point.

http://mathworld.wolfram.com/AlgebraicFunction.html

Very well.  :P


Putnam
probability
irrational number
college contests

If a is an irrational number, $0 < a <1$, is there a finite game with an honest coin such that the probability of one player winning the game is a? (An honest coin is one for which the probability of heads and the probability of tails are both 1/2, and a game is finite if with probability 1 it must end in a finite number of moves).

Something tells me e gets involved somewhere...

[quote="justdudxit"]Something tells me e gets involved somewhere...[/quote]

Something was lying to you :)

The key is to consider [hide]the binary representation of $a$, and build your game around it.[/hide]


geometry MIT college Harvard Yale Princeton Stanford

what are my chances of getting in an university:

Gender: Male
Location: india
College Class Year: 2012
High School: Private
High School Type: rarely sends grads to top schools
Will apply for financial aid: Yes

Academics:

GPA - Unweighted: 93.00/100
GPA - Weighted: 92.00/100
Class Rank: top 5 students(usually first or second)
Class Size: 55

Scores:SAT I
740(CR),780(MATH),777

Extracurriculars:

Significant Extracurriculars: partcipated in intel isef
winner of the city level science fair
winner of the state level science fair
winner of the regional science fair
winner of the national science fair

Leadership positions: captain of college debating team(went to france to represent the country)
captain of the college quiz team
captain of the college science team
member of the literary board of the college

Athletic Status - list sport and level:
swimming :college level
track and field :school level
hockey :college level

Honors and Awards: recipient of the micfest award
recipient of the literary fest award

Area:
Importance of cost: Extremely important

Well what type of college are you referring to?  At a top school, I would say you have as good of a chance as anyone else.  No better, no worse.  However, since you are international and money is very important, that may hurt your chances a bit.

I am also the member of my college magazine named "Constantia"
and I also have letters of recommendations from my principal and important staff members who are closely associated with me and some of the education department personnel from the government of India.

college list:
MIT
Harvard
Yale
Princeton
Stanford

please comment on my chances of getting admitted or not:

thank you in advance

First note that I am not an admissions officer.  Second, unfortunately, I have see VERY impressive resumes of international students who have been flat out rejected from those schools.  It's a crapshoot.

@JRav i seriously need your help as i think you know much better than me.
what are my chances ?or are there any chances?

As I said, I am not an admissions officer.  How should I know whether or not you will get in?  As I said, I have seen people with better resumes not get in, and people with worse resumes get in.  It's a total crapshoot.

Your grades, extracurricular activities, and test scores are easily enough to make any admissions officer very comfortable with you. So good, you're halfway there. But we can't put a number on you because you haven't put it together yet.

What you have to do is step back from the lists and scores and develop an image of yourself. Who are you?
- What are the core aspects of your personality?
- What are your interests and goals?
- What do you want out of college?
- What are some experiences or people that have shaped your life or given you inspiration?
- What makes you a unique and interesting person?

Once you have developed an image of yourself, this will become the centerpiece of your application. You have to focus on bringing this message out through every single piece of your application. Your scores and extracurricular activities are probably already aligned with this message. Now you have to align your essays in this direction as well. Now you have to pick people to write letters of recommendation that will corroborate this image.

yippie!that was great TZF
thanks a lot!
i really thank you !
just one more thing do u really feel i can get into MIT or HARVARD (i am an international student and i also need Fin Aid)

please reply frankly!
and also tell some more good colleges where i can apply!

Your resume looks pretty impressive for a freshman. But anything can happen in the next 4 years, so I wouldn't stress it. Try and expand your extracurriculars a little. Join some sports teams (simply saying "I play hockey" won't cut it), or some other college-approved clubs (not science or mathematics related). And see if you can rack up some community service hours.

Also, I wouldn't worry if you don't end up at an Ivy as an undergrad. As my friend said, it is better to be valedictorian of a decent college than an average of an Ivy. It's the diploma of the grad school that counts.

EDIT: rt_08 just posted again.

Honestly, nothing stands out of your resume that screams Ivy league. Your resume is almost like mine with the exception of a slightly better GPA and more awards. (And I know I'm not getting into an Ivy league nor am I planning on it). But the good news is that you have 4 years to rack up some points  :)

thanks

and what about CALTECH or UPenn?

I mean my chance at UPenn or CALTECH(although UPenn is a part of ivy league,i thought admissions might be easier there)

They're both extremely competitive.  Their acceptance rates are comparable, although Caltech gets far fewer applicants.

Is it compulsary to take TOEFL?

by the way thank you all for your support and comments.

any others???????/

so what would be a good score for me(considering my ec's) :blush:  :blush:  :blush:

I can't say, "Get 800's across the board and you're in" because it's not true.  The people who apply to top schools have resumes similar to yours, with the scores so it comes down to intangibles which none of us are qualified to assess.  An important note is cost; you say cost is very important to you.  At some schools, that won't matter but at other schools, it may.

Honestly, there's no point in setting a goal on your score except that it might help motivate you to keep studying and practising and learning. And you can motivate yourself to do that without setting a goal too. There's no way we can tell what would be the threshold between safe scores and unsafe scores. Just try to get as high a score as you can get, and put it in, and know that most likely, it won't be the factor that makes or breaks your application.

thanks JRav and TZF

JRav:i thought top tier colleges never cared much about Fin Aid

It's trickier because you're an international student. A few very rich universities can afford to extend their financial aid program equivalantly to foreign students, but most don't. Many other top tier private schools in the country are only need-blind and full-need for their domestic students.

So, for a particular good school, what happens will be one of the following, depending your application, other peoples' applications, and the particular school:

1. A good school with generous international aid accepts you, and then gives you a great package
2. A good school likes you a lot, and doesn't mind giving you a fair amount of aid to encourage you to come
3. A good school likes you enough to accept you, but not enough to give you a huge financial aid grant. They still give you the acceptance to let you pick and prioritize.
4. A good school is comfortable in accepting you, but not interested in providing financial aid, which you state you need, so it rejects you in favor of an international student who can pay.
5. A good school does not offer you admission from a merit perspective. This is common; there are so many impressive applications to any good school that it must turn down many very qualified people to fit the class size. It does not necessarily mean that your application was deficient.

well now i have to improve on my essays i think

Just to interrupt a bit: For a student speaking English as second language, what will be the SAT score that can eliminate the need to take TOFEL?

read the previous post by JRav (i think it is about 680)

I would check that number.  I'm not completely sure.  It's something like that.

approximately i mean

also what are the minimum good scores requirse for getting admission in yale or caltech

I've answered this question too many times.  There is no minimum score.  Top tier schools look at the whole package.  They could very easily accept someone with 650's across the boards over a 2400 scorer if the first person has a better overall package.

I think you will definately get into atleast  [b]1[/b] top school and thats all it takes. Pinning your hopes on any 1 school isn't a good idea.
My Resume (Currently Sophmore) please comment
Gender: Male
GPA: 4.2
Extracurrics: Currently in AP CALC BC, will take 2 years of math classes at Brown before graduating. Nationally ranked in chess (top 50), captain varsity chess team, Nationally ranked in Squash(#1 at school), Winner of a few poetry contests - nothing amazing but have applied to more prestigious ones, 1st place freshman state math leauge, High score in the state AMC 10 2008 - planning to make USAMO this year, on school debate team, math team, and creative writing club. Did a Quant internship last summer, and I have a few other things... Currently participating in USAMTS, Mandelbrot, State Math Leauge, NEML, AMC's.

What are my prospects? any advice? I want to do PROMYS this summer....

Anyone knows the requirement for SAT score(to not taking Toefl)? I'm really worried since my score is far from 680!! :(

If you're far from it, just take the TOEFL. You are not automatically out just because you know English less well than many native speakers.

At mathemonster, it's too early to tell.  Your EC's look nice but you're only a sophomore.

Don't be scared about taking TOEFL, it's there for if you're not a native speaker.  Of course, you still need to well on TOEFL because if you don't know enough english, you'll have trouble in classes, and as a result you won't be admitted.


The minimal example is this:
z+y+z+a+b
=z+y+x+c

The LHS is too long. So I want to put the Equivalent sign and the RHS into the second line..

How to write the code?

I tried:

\begin{equation}
\begin{split}
z+y+z+a+b\\
=z+y+x+c
\end{split}
\end{equation}

It is wrong, how to correct it?

You don't say what you are trying to achieve. Maybe you should use & which works like a tab so
[code]\begin{equation}
\begin{split}
z+y+z+a+b\\
&=z+y+x+c
\end{split}
\end{equation}[/code]
but it's not clear if that's what you need.
You really must get used to saying [i]exactly[/i] what is required as there's no way to read your mind.

the equation is like this
x+y+z+a+b=o+p+q+r+s

this equation is too long. So I want to split like this
x+y+z+a+b
=0+p+q+r+s

How to deal with it? hope it is clear now this time. Thanks！

I understand that but I don't know how you want it laid out. What is wrong with the two codes, yours and mine?

I indeed tried both codes.
The ouput of my code comes like this
                     x+y+z+a+b
=(o+p)*(q+r+s)                          (1)

The output of yours comes like this:

x+y+z+a+b
               =(o+p)*(q+r+s)            (1)

Since the LHS and RHS of my equation are very long..the output of the code you gave to me still cannot full display the equation.

It looks weird if I split (o+p)*(q+r+s) into two line.  That is why I want it looks like
x+y+z+a+b
=(o+p)*(q+r+s)                           (1)

But the 3 output examples you give are all the same!!
So I am still mystified. You will need  explain better than this with an actual example, and if you prefer, upload a pdf with the incorrect code and what is wrong with it.

Attached is what happens with the code you gave and my suggestion. Comment on these.

I dont understand why the computer shows the three equations with the same layout.

But I really cannot get what I want from both your and my codes.

Your pdf looks decent and looks like my code works in your computer.

But you still haven't explained what you want! Is the picture of your equation not what you want? How would you like it to look? If you can't show it please explain what it is is wrong.

Please post the latex code for your equation so I don't have to recreate it then I can try it out. In fact put it in a complete document as the margins and font size may have an effect. Then post a pdf of what you get and I will see if I get the same.

\begin{split}
\frac{\partial E\left[(V^{\ast}-Z^{\ast}-\pi)U'(W)
\right]}{\partial \sigma} \\
&=\lbrack\frac{1-e^{-\eta rT1}}{1-e^{-rT1}}f(T1,q)e^{-rT1}+\frac{e^{-\eta rT1}}{1-e^{-rT2}}g(T2,q)e^{-rT2}\rbrack E\lbrace U'\epsilon[-\rho L_{f}(V-Z-\pi)+1]\rbrace
\end{split}
\end{equation}

This is the code that produced the equation I just posted..
Removing the sign & before the sign =  (which is my original code) still would not produce the result that I want.  I tried to make the First line and second line both are left-justified..that is, make the the sign= right under the this first line.

I dont know how you can get the decent one from my code in your pdf..

[quote="weng"]\I tried to make the First line and second line both are left-justified..that is, make the the sign= right under the this first line.[/quote]
Aha! Finally you have explained what layout you want.
You need to use the multline environment (note the spelling; it is [b]not[/b] mult[b][color=red]i[/color][/b]line but multline). BUT you haven't told me the margins or font size as I asked which may affect the result.

Try
[code]\begin{multline}
\frac{\partial E\left[(V^{\ast}-Z^{\ast}-\pi)U'(W)
\right]}{\partial \sigma} \\
=\left[\frac{1-e^{-\eta rT1}}{1-e^{-rT1}}f(T1,q)e^{-rT1}+\frac{e^{-\eta rT1}}{1-e^{-rT2}}g(T2,q)e^{-rT2}\right] E\{ U'\epsilon[-\rho L_{f}(V-Z-\pi)+1]\}
\end{multline}[/code]
but if that isn't good enough you can force the second line to the left with \shoveleft{...}
[code]\begin{multline}
\frac{\partial E\left[(V^{\ast}-Z^{\ast}-\pi)U'(W)
\right]}{\partial \sigma} \\[2ex]
\shoveleft{=\left[\frac{1-e^{-\eta rT1}}{1-e^{-rT1}}f(T1,q)e^{-rT1}+\frac{e^{-\eta rT1}}{1-e^{-rT2}}g(T2,q)e^{-rT2}\right] E\{ U'\epsilon[-\rho L_{f}(V-Z-\pi)+1]\}}
\end{multline}[/code]
I have made a few changes:
1. the second line is a bit too close to the first vertically so I have increased the gap with \\[2ex]
2. You don't need \lbrack and \rbrack since they are the same as [ and ]
3.  You don't need \lbrace and \rbrace since they are the same as \{ and \}
4. I have used \left[ ... \right] so that the [...] are resized to cover the fractins on the second line.

You need to do 2 things:
a. Read [url=http://tug.ctan.org/tex-archive/info/math/voss/mathmode/Mathmode.pdf]Math Mode[/url] which explains very well how to type mathematics;
b. When asking questins you must give [b]complete[/b] detail not bits. It took 9 posts for you to explain that you wanted left justification. You still didn't provide a [b]complete[/b] document which shows packages, margins and fonts used as they have a relevance. Read [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]Minimal example please![/url].

\documentclass[a4paper,fleqn]{article}
\usepackage{graphicx}
\usepackage{amsmath}

\usepackage{setspace}

These are what I set for my article.  Is the package or documentclass problem?

Thanks!!! BTW, i did make a mistake when I tried to use \begin{multiline} instead of \begin{multline}..

I will give the complete minimal example next time！
Thanks again!


Wierd geometry questions  :P

http://www.8foxes.com

I couldn't solve #95.  Is 8*1=8 the only solution ?
Can we prove that there is no rectangle bigger than area of 8, that can slide tru the corner?


Note that 1990 can be "turned into a square" by adding a digit on its right, and some digits on its left; i.e., $419904 = 648^2$. Prove that 1991 cannot be turned into a square by the same procedure; i.e., there are no digits $d,x,y,..$ such that $...yx1991d$ is a perfect square.

Suppose $x^2 = 19910 + d$ modulo 100000. for any d = 0 ,1,2,....9.

Modulo 100, x^2 can only be 1 4 9 16 25 36 49 64 81 0.  So x = 4 or 0 mod 10, and d can only be 0 or 6.

Suppose x = 10k   :   Then k^2 = 1991 mod 10000.  It implies k^2 = 91 mod 100; impossible.

Suppose x = 10k+4 :  Then 100k^2 +80k + 16 = 19916 mod 100000.  It implies 100k^2 + 80k = 1990 mod 100000,

so that 10k^2 + 8k = 199 mod 10000.  But the left side is even, so this is impossible.

Contradiction.

[quote="Singular"]Modulo 100, x^2 can only be 1 4 9 16 25 36 49 64 81 0.  So x = 4 or 0 mod 10, and d can only be 0 or 6.[/quote]

i think it is not true......

as, $17^2=289=89($mod $100)$ and infinitely many more!!!!!!!!!

am i mistaking somewhere?

[quote="the_boss"]
i think it is not true......

as, $17^2=289=89($mod $100)$ and infinitely many more!!!!!!!!!

am i mistaking somewhere?[/quote]

In this case, we must have $0 \le d \le 9$.

yeah, in my example, d=9........ i didn't understand my mistake!!!!!!!!


A plane flies from city A to city B, a distance of 150 miles and then turns through an angle of 30 degrees and heads toward city C. If the distance from city A to city C is 400 miles, how far is it from city B to city C?

use cosine equation.
400^2=(150)^2+x^2-2.150.x.cos(120)
it will be a quadratic equation in x. (Suposed x is the distance from city B to city C).


quadratics
linear algebra
matrix
linear algebra unsolved

Suppose $ A \in M_4 (\mathbb{F}_p)$, is $ AA^{t} \equal{} \minus{}I_4$, $ \forall p \equiv 3 (mod 4)$? It seems that when $ p \equiv 1 (mod4)$ the result holds since $ \minus{}1$ is a quadratic residue in this case. I could use some direction on the former case, though. any help would be appreciated.

If you are asking about the existence of such an $ A$ - yes, it exists. Take $ a$ and $ b$ in $ \mathbb{F}_p$ such that $ a^2 \plus{} b^2 \equal{} \minus{} 1$ (such $ a$ and $ b$ exist, since $ \left\{x^2\mid x\in\mathbb{F}_p\right\}\cap\left\{ \minus{} 1 \minus{} x^2\mid x\in\mathbb{F}_p\right\}\neq\emptyset$, what follows from $ \left|\left\{x^2\mid x\in\mathbb{F}_p\right\}\right| > \frac12\left|\mathbb{F}_p\right|$ and $ \left|\left\{ \minus{} 1 \minus{} x^2\mid x\in\mathbb{F}_p\right\}\right| > \frac12\left|\mathbb{F}_p\right|$). Now let

$ A \equal{} \left(\begin{array}{cccc}a & b & 0 & 0 \\
\minus{} b & a & 0 & 0 \\
0 & 0 & a & b \\
0 & 0 & \minus{} b & a\end{array}\right)$.

Similarly you can construct such a matrix in $ \mathrm{M}_{2n}\left(\mathbb{F}_p\right)$ for any integer $ n$ and any prime $ p$.

[b]Generalization.[/b] Let $ \mathbb{F}_q$ be a finite field with $ q$ odd, and $ U$ an invertible symmetric $ n\times n$ matrix over $ \mathbb{F}_q$. Then, there exists a matrix $ M\in\mathrm{GL}_n\left(\mathbb{F}_q\right)$ satisfying $ MM^T \equal{} U$ if and only if $ \det U$ is a square in $ \mathbb{F}_q$.
(Source: L. E. Dickson, [i]Linear Groups[/i], chapter 7, Dover, New York, 1958.)

  darij


geometry
geometric transformation
rotation
rectangle
inequalities unsolved
inequalities

Let $H$ be a rectangle with angle between two diagonal $\leq 45^{0}$. Rotation $H$ around the its center with angle $0^{0}\leq x\leq 360^{0}$ we have rectangle $H_{x}$. Find $x$ such that $[H\cap H_{x}]$ minimum, where $[S]$ is area of $S$.

[img]http://i34.tinypic.com/1o7kuf.jpg[/img]
Let $ H: =ABCD$ and $ H_x: =A'B'C'D'$ (like in the diagram). Then we are looking for the minimum of $ [WXYZ]$. But
\begin{align*}[WXYZ]&=AB\cdot WX\\
&\ge AB\cdot A'B'\\
&= AB^2\end{align*}
(Since $ A'B'$ is the perpendicular distance between the parallels $ A'D'$ and $ B'C'$, $ A'B'\le WX$; and since $ H_x$ is a rotation of $ H$, $ A'B'=AB$.)

Hence equality holds iff $ A'B'\perp AB$, i.e. iff $ x=90^\circ$.

I think we need to consider the case where $ B'$ is between $ A$ and $ B$ which might need in that case we'll need that $ \le 45^0$ condition.


Seven beakers are on the table, all upside down. Each move, we can turn around $4$ of these seven beakers. Can we get all beakers with their correct side up?

[hide]
No, as there will always be an even number of beakers turned up.
[/hide]

I'll try to formalize what rjt correctly sayd...  :)

[hide]I consider the configuration UUUUDDD (U means up, D means down), which is the same as the proposed one, 'cause it's the necessary development of the game. I observe we have 4 beakers up, so an even number.
Now, when I choose a move, I decide to turn a certain number of beakers, which are already up: this number, $k$, can vary between 1 and 4 (even if with 4 it's stupid!). Anyway, after this move, I'll still have $4-k$ beakers up. How many beakers will I consequently turn from down to up? Again this number is $4-k$. So, after a generic move, I'll have $(4-k)+(4-k)=2(4-k)$ beakers up, surely an even number!... q.e.d. [/hide]

P.S.: When will I get the writing "P vs NP" in my profile??  :roll:

[quote="Ponnamperuma"]P.S.: When will I get the writing "P vs NP" in my profile??  :roll:[/quote]

At 20 posts (and Hodge when you get to 50) :)

I see... I thought it was after 10...  still 2 posts...  :)


hello, consider all possible remainders of the given numbers, for example $ a\equiv 0,1 \mod 2$ and $ b\equiv 0,1 \mod 2$ and so on.
Sonnhard.

[quote="Nora.91"]Prove that for any two integers a and b,

$ ab ( a^{60} \minus{} b^{60} )$ is divisible by [b]56786730[/b].

[hide="I know that"]
56786730 = 2 x 3 x 5 x 7 x 11 x 13 x 31 x 61

But how do i proceed further?

is it an application of Euler's function? [/hide][/quote]

From Fermat's little theorem, $ a^{p\minus{}1}\equal{}b^{p\minus{}1} (\mod p)$ if $ p \nmid a,b$ , where $ p$ is any prime.

Now for each of $ p_i\equal{}2,3,5,...,13,31,61$, if $ p_i|a$ or $ b$ then $ p_i|ab ( a^{60} \minus{} b^{60} )$, if not, then from Fermat's little theorem, $ p_i$ divides the number too (because $ p_i\minus{}1\equal{}1,2,4,6,10,12,30,60$ are all divisors of $ 60$).


ok, I took AMC 12A at my school, and 10B at Chapman University. After 10B, I compared my answers with the answer keys on the internet, and found out that i got 145.5 on it, which qualifies for AIME.

However, my school math teacher who is responsible for AMC doesn't have my score. (of course he does have my score for 12A, which i took at school, but no 10B)

Since people are taking AIME today at 8:45 AM, I cannot take it because still doesn't have my score.

So, if I register for alternative AIME, can I still get a chance a qualify for USAMO? and in order to register for alternative AIME, what do I have to do? I'm guessing that i have to contact the contest manager at Chapman University and have her tell my score to my school math teacher?

[quote="davidyko"]This should go in the AMC forum after it's been unlocked.
Yes, I believe you should do what you said at the end of your post.[/quote]

I agree - start by talking to the people who administered the test at Chapman.


let $ x^2 \plus{} y^2 \plus{} 2gx \plus{} 2fy \plus{} c \equal{} 0$ be a circle. Find the locus of the feet of the perpendicular from the origin to a chord subtending right angles at the origin.Find also the locus of the midpoints of the chord.
(Can anyone give a geometric interpretation)

hello
sorry,I just read that perpendiculars are drawn from the center So I gave the hint for that question.
I will post the answer to this question later.

i don't quite get it , better u post the ans coz i think u  misinterpreted the problem.


I have nine school days in March! :D

March 2-3 : teachers' convention
March 9-10 : PD days
March 13 : APMO
March 15-20 : Gathering for Gardener
March 27-31 : spring break

I'm thinking whether I should show up on 14th.  :rotfl:

[quote="jpark"]what's  gathering for gardener?[/quote]
Opps! I have a spelling error. :blush: It's supposed to be 'Gatherings for Gardner'.
I'm too lazy to explain. Read about Martin Gardner [url=http://en.wikipedia.org/wiki/Martin_Gardener]here[/url]. It also has a link to Gatherings for Gardner site.

You serious Elyot?

Mother called from London, and she didn't mention anything about snow.  But man...it's like 20 degrees down here.  So hot... :P

[quote="Elyot"]Actually, we just had a snow day yesterday, and many places west of here are totally iced in today because of lots of snow, followed by freezing rain, followed by rain, followed by a flash freeze (this morning, it was +8 degrees at 4am and went down to windchill of -20 by 7:30am).  Around London-ish, all the highways are closed, because they're just sheets of ice.

Do you often get snow days in Edmonton?  I know it gets very cold there, but do you ever get freaky weather like that?[/quote]
We used to get a lot of snow. However, we couldn't see much snow this year. We usually get a week that goes down to -40(not considering wind chill), but we didn't have such week this year. Edmonton got too warm comparing to the previous winters. :)

[quote="Wolfie"]We(Montreal) just got hit by a goodly amount of freezing rain. I had to walk up to the train station, and got to school bruised and battered. I had hoped that, one day, my school would just declare it unsafe for students to come, but no.

Even during the biggest storm of the year, even though most schools on the Island of Montreal were closed, Loyola still didn't give up. It was [i]exam week[/i], after all. And we wouldn't want to miss that.[/quote]

And that, my friend, would be the storm that just hit us yesterday (yea, Montreal gets K/W weather with a delay of 24 hours)

Actually, I've just read the paper: winds of 110km/h reached us, made a whole big mess on the highway. I hope it doesn't get that bad where you are.

[quote="lightrhee"]
We used to get a lot of snow. However, we couldn't see much snow this year. We usually get a week that goes down to -40(not considering wind chill), but we didn't have such week this year. Edmonton got too warm comparing to the previous winters. :)[/quote]

I just came back from Edmonton, wow it was sad.  It was freezing in Calgary, so we assumed Edmonton was going to be like...-20 degrees worse.  We spent most of the time without our jackets....fun stuff.

Lucky.  March sounds like an awesome month for you.  Personally, I can't wait for April.  Now that's going to be a fun month for me.

Just leave it to Canadians to talk about the weather.  There are very few countries whose population does that as much as we do.

[quote="Wolfie"]Oh, come on, eh. Don't yeh know it affects our lumberjackin' skill, eh? And if it's too warm, the igloos melt, eh. Weather has a lot to do with our lives in Canada, eh. It affects wether we wear our tuques or not. Eh.[/quote]

 :D


Six points A,X,Y,B,D,C are on the same circle in order(clockwise).Prove (AX+BX)/(AY+BY)=(CX+DX)/CY+DY) if and only if AB∥CD.

Apply the theorem of Ptolemy:

$(AX+BX)/(AY+BY)=(CX+DX)/CY+DY) \iff(AX*CY-AY*CX)+(AX*DY-AY*DX)+(BX*CY-BY*CX)+(BX*DY-BY*DX)=0 \iff (-AC*XY)+(-AD*XY)+BC*XY+BD*XY=0 \iff AC+AD=BC+BD$.

Now,we only need prove$AC+AD=BC+BD \iff AB//CD$,but this is very simple,please try it by yourself!


[quote="erdugan"]Prove that the equation  x^3 +y^3=z^4 _ t^2  has infinitely many positive solution x,y,z,t such that gcd(x,y,z,t)=1[/quote]

See http://www.mathlinks.ro/Forum/viewtopic.php?t=288655


number theory
greatest common divisor
floor function
binomial coefficients
combinatorics unsolved
combinatorics

Given positive integer n.
Determine $gcd(\binom{n}{1},\binom{n}{2},...,\binom{n}{n-1})$

And here is the proof if $n\neq p^{k}$ then if we suppose that exhist prime q
which divides $\binom{n}{k}$ for k=1,2...n-1. we must obviously have q/n.
and since $n\neq q^{l}$ base q representation of n must have at least two non-zero digits and let the rightmost non-zero digit be l .From this and Luca's theorem it follows that if we take k has the same digits(in base q) except l-1 instead of l we will have $0<k<n$ and $\binom{n}{k}\neq 0(mod q)$ which is a contradiction.

An example when n=3;
$Gcd(\binom{3}{1},\binom{3}{2})=3$

[quote="Baronov"]I think you mean
$gcd(\binom{n}{1},\binom{n}{2},...,\binom{n}{n-1})$

otherwise the way you have written it all numbers are $0$. Now if $n=p^{k}$ where p is prime answer is
p because all binomial coefficients are divisible by p (follows easily by Lusa's theorem).
If not ($n\neq p^{k}$)i think the answer is 1.[/quote]
I am interested...I can't find anything on Lusa's Theorem-I haven't encountered it before.

[b]7erry [/b], i think you don't know the notation  $\binom{n}{k}$  because $\binom{3}{1}=\binom{3}{2}=3$, oh may be by gcd you don't mean greatest common divisor.

[b]funcia [/b], the Luca's theorem states:
Let $p_{1},p_{2}....p_{l}$ and $q_{1},q_{2}...q_{l}$ be the digits in base p of the numbers n and k respectively then
$\binom{n}{k}\equiv\binom{p_{1}}{q_{1}}\binom{p_{2}}{q_{2}}...\binom{p_{l}}{q_{l}}(mod \ p)$

Oh, sorry, that's my mistake while typing. I've edited.

[quote="Ravi B"][quote="Baronov"]Now if $n=p^{k}$ where p is prime answer is
p because all binomial coefficients are divisible by p (follows easily by Lusa's theorem).[/quote]
Your argument shows that the greatest common divisor (gcd) is a multiple of $p$, but doesn't show that the gcd is exactly $p$.[/quote]

Yup, but I find out that the gcd must be $p$.
Consider $N ={{p^{k}}\choose{p^{k-1}}}= \frac{(p^{k})!}{(p^{k-1})! \cdot (p^{k}-p^{k-1})!}$.
Let $S$ be the highest power of $p$ divides $N$.
$S = \sum_{i=1}^{\infty}\lfloor \frac{p^{k}}{p^{i}}\rfloor-\sum_{i=1}^{\infty}\lfloor \frac{p^{k-1}}{p^{i}}\rfloor-\sum_{i=1}^{\infty}\lfloor \frac{p^{k}-p^{k-1}}{p^{i}}\rfloor$
$= \sum_{i=0}^{k-1}p^{i}-\sum_{i=0}^{k-2}p^{i}-\sum_{i=1}^{k-1}(p^{i}-p^{i-1})$
$= 1$

Thus, $p \parallel N$ and the gcd is equal to $p$.  :)

NB : I think the real name is Lucas' theorem  :D


What is the sum of the measures of the interior angles of a hexagon?

[hide] connect each vertice with every other one to construct 3 triangles.  Thus, the sum of its interior angles must be 180 * 3= [b]540 degrees[/b][/hide]

interesting follow up.  whats the sum of the exterior angles ?  :D

The sum fo the exterior angles of any polygon is $180^\circ$

G-Unit, $540^\circ$ would have been the sum of the interior angles in a pentagon. The sum of the inerior angles in a hexagon is $720^\circ$. I think that you were thinking about drawing a line to 3 of the vertices, which would create four triangles.

uh....The sum of the exterior angles of any polygon is $360^\circ$. This can be shown in a hexagon easily. dividing a regular hexagon into 6 equilateral triangles, you can see that the measure of each angle is 120, and the measure of each exterior angle is 60. $60\cdot6 = 360^\circ$

Yeah, I was thinking that Hamster1800. Where did 180 come from?

Interior: 4*180 or 720
Exterior: 360.


A belt fits tightly around two pulleys with diameters of 24 units and 6 units. If the centers of the pulleys are 18 units apart, how long is the belt?

Hint:

[hide]Draw a sketch, and think about the circumference of the pulleys.[/hide]

Solution:

[hide]Split this up into sections.

24 unit diameter pulley - Covers half of the circumference of the pulley, that is $ 12 \pi$ units
6 unit dimater pulley - Again, covers half of the circumference of the pulley, that is $ 3 \pi$ units
In between the two pulleys - Covers the 18 unit distance between the two pulleys twice, so that's a total of $ 36$ units here.

Adding these up we get $ \boxed{36+15 \pi}$ units.[/hide]

[quote="vikas"]I get a totally different answer, as in $ 18\pi+18\sqrt{3}$

That does not make any sense..
it does not cover half of the circumference..(you have to find angles first)
its not 18..because the belt is slanted..[/quote]Sorry, you're right.  That was very stupid from me  :P

[quote="vikas"]I get a totally different answer, as in $ 18\pi+18\sqrt{3}$

[quote="ad absurdum"]24 unit diameter pulley - Covers half of the circumference of the pulley, that is $ 12 \pi$ units
6 unit dimater pulley - Again, covers half of the circumference of the pulley, that is $ 3 \pi$ units
In between the two pulleys - Covers the 18 unit distance between the two pulleys twice, so that's a total of $ 36$ units here.

Adding these up we get $ \boxed{36+15 \pi}$ units.[/quote]

That does not make any sense..
it does not cover half of the circumference..(you have to find angles first)
its not 18..because the belt is slanted..[/quote]

I'm lost with your solution, how did you find the angles?

ad absurdum made some  mistakes.  The pulleys cover more than half the circumference.  And 18 is  the distance between the [i]centers[/i].

[hide="HINT"]Similar right triangles[/hide]
A very similar problem was solved [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=151478]here.[/url]

[quote="archimedes1"]ad absurdum made some  mistakes.  The pulleys cover more than half the circumference.  And 18 is  the distance between the [i]centers[/i].
[/quote]

Well actually arch, the circumference is the easy part of this problem. That part isn't made to be hard. It's got a word trick (Slanted vs. Unslanted), and on top of that, its a $ \Delta R_{r}= R-r$. I used that to get the slope, then did a tan inverse and some other stuff I can't remember, but I got the same answer as vikas. Call it intuition, call it whatever you want but I almost [i]KNOW[/i] those pulleys go tangent at half the circumference.

A bold statement, iguil. :)   However, your intuition does not prove anything.
All tangents are perpendicular to the diameter.
Thus, the only way exactly half the circles could be covered would be if the circles were of the same size, which they are not.

[hide="BIG HINT"][img]9518[/img] Solve for x using similar right triangles.  Now, what special type of right triangles are these?[/hide]

$ 30^\circ$, $ 60^\circ$  :mad:
[b]
Edit:[/b] I rated a 6.  :D

[quote="sha"]there's this friend of mine who's worked out this question. The answer, apparently, is 2√243 + 18∏. the belt does not cover half the circumferences of the circles. similarity is the right method to do it. :lol:

an expert is someone who knows more and more about less and less.. :)[/quote]

guess what 2√243 + 18∏ is $ 18\sqrt{3}+18\pi$..which we already have gotten..


geometry
Ring Theory
advanced fields
advanced fields unsolved

'If $ X$ is a curve and $ x\in X$ is a smooth point, then the maximal ideal $ m_{x,X}$ itself is principal.'

This part is the 15th line from the bottom on the page 17 of  "Algebraic geometry 1:Complex Projective Varieties(by David Mumford)"

Can you explain this ??

Isn't the local ring $ \mathcal O_{X,x}$ of $ X$ at $ x$ a discrete valuation ring?


Solve the system equation:
$ \{\begin{array}{l} x^2+2yz=x\\y^2+2zx=y\\z^2+2xy=z \end{array}$

Adding the equations up we get $ x\plus{}y\plus{}z\equal{}0$ or $ x\plus{}y\plus{}z\equal{}1$

Subtracting the second equation from the first we get $ (x\minus{}y)(x\plus{}y\minus{}2z\minus{}1)\equal{}0$

Now it's just crunching the cases and solving some easy quadratics, to get

$ \{x,y,z\}\in\left\{\{0,0,0\},\{{1\over 3},{1\over 3},{1\over 3}\},\{0,0,1\},\{\minus{}{1\over 3},\minus{}{1\over 3},{2\over 3}\}\right\}$

This system was a problem to Austria math olympiad between 1986-1989 - I'm not sure. You may solve it by substracting the equations.
I've similar problem:
$ x^{2} \plus{} yz \equal{} 2$
$ y^{2} \plus{} zx \equal{} 2$
$ z^{2} \plus{} xy \equal{} 2$

You may try the following problem composed by me when I was 9-th grade:
$ \sqrt {x^{2} \plus{} y^{2}} \plus{} |z| \equal{} 1$
$ \sqrt {y^{2} \plus{} z^{2}} \plus{} |x| \equal{} 1$
$ \sqrt {z^{2} \plus{} x^{2}} \plus{} |y| \equal{} 1$

[quote="Farenhajt"]Adding the equations up we get $ x \plus{} y \plus{} z \equal{} 0$ or $ x \plus{} y \plus{} z \equal{} 1$

Subtracting the second equation from the first we get $ (x \minus{} y)(x \plus{} y \minus{} 2z \minus{} 1) \equal{} 0$

Now it's just crunching the cases and solving some easy quadratics, to get

$ \{x,y,z\}\in\left\{\{0,0,0\},\{{1\over 3},{1\over 3},{1\over 3}\},\{0,0,1\},\{ \minus{} {1\over 3}, \minus{} {1\over 3},{2\over 3}\}\right\}$[/quote]
OK! It isn't a hard problem! My solution is similarly to you!


There are 6 speakers $A,B,C,D,E,F$ who have to address a press conference. In how many ways can they address the press if it is given that B adresses after A.

I solved this but the answer I got did not match. I am giving my solution, if you all agree then tell.
[hide]
no. of ways they can address press= $\frac{6!}{2!}$

In the denominator 2! is there because, when we have say ABCDEF and BACDEF, then only ABCDEF is to be considered. So by $6!$ we get the required case doubled. So divide by $2!$.
Hope I am correct. :| [/hide]

[hide="solution"]
number the order
123456
the number of ways that A and B can be in the order is $\binom{6}{2}$. For each of these, C,D,E, and F can be in $4!$ orders, so the final answer is $\binom{6}{2}\cdot4! = \boxed{360}$.
[/hide]

Well you have done the same thing what I did. So, I feel I was correct.
You can also try other problem which I have posted.

Thanks,
Varun

Do it by placements: we know that $C,D,E,F$ are not affected by $A$ and $B$ therefore they will always be $4 \cdot 3 \cdot 2 \cdot 1$, but we have two spaces that are left, and to find how many of those two spots can remain for $A$ and $B$ and we must use combinations since we want $A$ before $B$ therefore we have $6C2=15$, now $15 \cdot 4!=\boxed{360}$.

[quote="mathmanman"]Yes varun, you were correct.

"but the answer I got did not match" wtf?[/quote]

Well, mathmanman what do you mean by "wtf"


invariant
algebra
polynomial
Putnam
Rational Root Theorem

[quote="hpanda"]Prove that $ \sqrt {2} \plus{} \sqrt {3} \plus{} \sqrt {5}$ is irrational.

This is pretty straight forward that this is true, but I don't see how I can prove it.

I got to the point that $ \sqrt {2} \plus{} \sqrt {3}$ is irrational, since $ (\sqrt {2} \plus{} \sqrt {3})^{2} \equal{} 5 \plus{} 2\sqrt {6}$, as well as $ \sqrt {2} \plus{} \sqrt {5}$ and $ \sqrt {3} \plus{} \sqrt {5}$.

But from there, I don't know how I prove it.[/quote]

[hide="solution"]
Assume that $ \sqrt{2}\plus{}\sqrt{3}\plus{}\sqrt{5} \in Q$,
then $ (\sqrt{2}\plus{}\sqrt{3}\plus{}\sqrt{5})^2\equal{}(2\plus{}3\plus{}5)\plus{}2(\sqrt{6}\plus{}\sqrt{15}\plus{}\sqrt{10}) \in Q$.
$ \Rightarrow (\sqrt{6}\plus{}\sqrt{15}\plus{}\sqrt{10})^2\equal{}(6\plus{}15\plus{}10)\plus{}2(\sqrt{90}\plus{}\sqrt{150}\plus{}\sqrt{60}) \in Q$
$ \Rightarrow \sqrt{60} \plus{}\sqrt{90}\plus{}\sqrt{150} \in Q$
$ \Rightarrow \frac{\sqrt{60}\plus{}\sqrt{90}\plus{}\sqrt{150}}{\sqrt{2}\plus{}\sqrt{3}\plus{}\sqrt{5}}\equal{}\sqrt{30} \in Q$, contradiction, done.
[/hide]

Ket $ u\equal{}\sqrt{2}\plus{}\sqrt{3}\plus{}\sqrt{5}$.

Consider $ f(x)\equal{}\prod (x\pm\sqrt{2}\pm \sqrt{3}\pm\sqrt{5})$ where we multiply over all 8 possible combinations of the $ \plus{}$s and $ \minus{}$s.

$ f(x)$ is invariant under conjugation with respect of any of the roots (i.e. replacing $ a\plus{}b\sqrt{2}$ with $ a\minus{}b\sqrt{2}$ where a,b have no part of sqrt(2)).

So $ f(x)$ is a polynomial with integer coefficients with lead coefficient of $ 1$. Thus by the rational root theorem, the roots of $ f$ are integers if they are rational. It is easy to see that $ f(u)\equal{}0$. So if $ u$ is rational, then it is an integer. But $ 5<u<6$. So $ u$ is irrational.

[quote="hpanda"]Prove that $ \sqrt {2} \plus{} \sqrt {3} \plus{} \sqrt {5}$ is irrational.
[/quote]

This is the first problem in the book: Putnam and Beyond.


While eating out, Mike and Joe  each tipped their server
\$2. Mike tipped 10\% of his bill and Joe tipped
20\% of his bill. What was the difference, in dollars,  between their bills?

Mike: $ 2\cdot10\equal{}20\$$
Joe: $ 2\cdot5\equal{}10\$$
Difference: 10 dollars


$ \sqrt{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10} \equal{}$ ?

[hide="solution"]$ \sqrt{2\cdot5\cdot10\cdot4\cdot9\cdot8\cdot6\cdot3\cdot7}$
$ 10\cdot2\cdot3\sqrt{2\cdot4\cdot2\cdot3\cdot3\cdot7}$
$ 60\cdot2\cdot2\cdot3\sqrt{7}$
$ \boxed{720\sqrt{7}}$[/hide]

Edit: Gahhh, beat again!

[hide]$ \sqrt{2 \times 3 \times 2 \times 2\times 5 \times 2 \times 3 \times 7 \times 2 \times 2 \times 2 \times 3 \times 3 \times 2 \times 5}$

=2x2x2x2x3x3x5x$ \sqrt{7}$
=$ 720\sqrt{7}$[/hide]

[quote="small_unicorn"]$ \sqrt {1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10} \equal{}$ ?[/quote]
hi all
[url=http://www.servimg.com/image_preview.php?i=427&u=12899977][img]http://i99.servimg.com/u/f99/12/89/99/77/almonk10.gif[/img][/url]


geometry
3D geometry
algebra
polynomial
algebra unsolved

Prove that for any polynomial $f \in \mathbb C[X]$, there are $g,h \in \mathbb C[X]$ such that $f = g^{3}+h^{3}$ if and only if there are $u,v,w \in \mathbb C[X]$ such that $u+v+w=0$ and $f = uvw$.

$\zeta=e^{2\pi i\over 3}={-1+\sqrt{3}i\over 2}$ a primitive $3$-rd root of unity satisfies $\zeta^{2}+\zeta+1=0$ and $x^{3}+1=(x+1)(x+\zeta)(x+\zeta^{2})$.
Then $g^{3}+h^{3}=\zeta^{3}(g+h)(g+\zeta h)(g+\zeta^{2}h)=(\zeta g+\zeta h)(\zeta^{2}g+h)(g+\zeta^{2}h)$.
So we can take $(u,v,w)=(\zeta g+\zeta h,\zeta^{2}g+h,g+\zeta^{2}h)$ and conversely $(g,h)=({\zeta^{2}u+\zeta v+w\over 3},{\zeta^{2}u+v+\zeta w\over 3})$.


[quote="Pinaki_Mishra"]0>a,b,c>1[/quote]

Nice condition :) . Anyway, check problem 6 in http://www.mathlinks.ro/Forum/viewtopic.php?t=5175 , or http://www.mathlinks.ro/Forum/viewtopic.php?t=11484 , http://www.mathlinks.ro/Forum/viewtopic.php?t=54443 .

  darij


i'm not sure if this is in the right forums, but......

Prove that 2^1/2 (sq root of 2) is not a rational number.

this is the right forum but this has been proven [b]countless[/b] times (well maybe not countless but a lot) in here. USE THE SEARCH FEATURE!!!

This is copied from a text book  :P

If $\sqrt{2}$ is a rational number, then there are natural numbers $m$ and $n$ such that $\sqrt{2}=\frac{m}{n}$. Square both sides.
$(\sqrt{2})^2=\big(\frac{m}{n})^2$
$2=\frac{m^2}{n^2}$
$2n^2=m^2$
Since  a perfect square has an even number of prime factors, then $m^2$ has an even number of prime factors and $2n^2$ has an odd number. This is impossible. Therefore, $\sqrt{2}$ cannot be a rational number.


Simplify as
\[ a^2-(2x^2+1)a+x^4+x=0 \]
which is a quadratic with respect to $ a $.  Using the quadratic formula we get either
\[ x^2+x-a=0 \]
or
\[ x^2-x-a+1=0 \]
From here use the quadratic formula again and test to find which solutions work and which are extraneous.

It doesn't 'appeal' to make a quadratic in a constant, but logically, it seems all right!!!


LaTeX
inequalities
inequalities proposed

For all odd integers $ n > 1$, find the greatest positive real number $ k_n$ such that for all $ x_1, x_2, \dots ,x_n \in \mathbb{R}$ with $ x \equal{} \frac {x_1 \plus{} x_2 \plus{} \dots \plus{} x_n}{n}$,

$ \frac {\sum_{i \equal{} 1}^{n} (x_i \minus{} x)^2}{n} \geq k_n (\frac {\sum_{i \equal{} 1}^{n} |x_i \minus{} x|}{n})^2$

Since $ \sum_{i\equal{}1}^{n} a_i^2 \ge \dfrac{1}{n}\left(\sum_{i\equal{}1}^n a_i \right)^2$ for all reals $ a_i$, then we can substitute $ a_i\equal{}|x\minus{}x_i|$ as defined. then it seems that $ latex k_n\equal{}1$ for all $ k$,,, just look at the equality case $ x_1\equal{}x_2\equal{}\dots\equal{}x_n$.
i don't really understand the problem..  :( am i wrong?

When we have the equality case you mentioned, $ x_1 \equal{} x_2 \equal{} \dots \equal{} x_n \equal{} x$. But then the LHS and the RHS of the above inequality are zero, which means that you can increase $ k_n$ without making the inequality false. Thus it is not an equality case.
To prove that $ m \equal{} k_n$ is the greatest positive real number for the inequality, you need to prove that $ \frac{\sum_{i \equal{} 1}^n (x_i\minus{}x)^2}{n}\geq k_n(\frac{\sum_{i \equal{} 1}^n |x_i\minus{}x|}{n})^2$ for all $ x_1, x_2, \dots , x_n \in \mathbb{R}$ and your equality case has to make $ (\frac{\sum_{i \equal{} 1}^n |x_i\minus{}x|}{n})^2$ positive.

ah ya... thank you.. i've understood my misunderstanding. this is really nice!

How do I put up my solution while giving readers of this topic a choice whether to read it or not?

Use the hide button^^
for example: [hide="click here"]here you have to write down your solution[/hide]
I hope that I told you that what you wanted to know…^^

Thanks FelixDD
[hide="Here is my solution:"]
without loss of generality let $ x_1 \geq x_2 \geq \dots \geq x_m \geq x \geq x_{m \plus{} 1} \geq \dots \geq x_n$, where $ 1 \leq m \leq n \minus{} 1$ . The let $ a_i \equal{} x_i \minus{} x$ for all $ 1 \leq i \leq m$, and $ b_i \equal{} x \minus{} x_{m \plus{} i}$ for all $ 1 \leq i \leq n \minus{} m$. Then we have

$ a_i \geq 0$ and $ b_i \geq 0$ for all applicable $ i$,

$ a_1 \plus{} \dots \plus{} a_m \equal{} b_1 \plus{} \dots \plus{} b_{n \minus{} m} \equal{} c$ since $ x \equal{} \frac {x_{1} \plus{} x_{2} \plus{} \dots \plus{} x_{n}}{n}$,

$ \frac {\sum_{i \equal{} 1}^{n}(x_{i} \minus{} x)^{2}}{n} \equal{} \frac {\sum_{i \equal{} 1}^{m} a_{i}^2 \plus{} \sum_{i \equal{} 1}^{n \minus{} m} b_{i}^2}{n}$, and

$ \frac {\sum_{i \equal{} 1}^{n}|x_{i} \minus{} x|}{n} \equal{} \frac {\sum_{i \equal{} 1}^{m} a_{i} \plus{} \sum_{i \equal{} 1}^{n \minus{} m} b_{i}}{n} \equal{} \frac {2c}{n}$.

Then  $ \frac {\sum_{i \equal{} 1}^{m} a_{i}^2 \plus{} \sum_{i \equal{} 1}^{n \minus{} m} b_{i}^2}{n} \geq \frac {\frac {\sum_{i \equal{} 1}^{m} a_{i}}{m} \plus{} \frac {\sum_{i \equal{} 1}^{n \minus{} m} b_{i}}{n \minus{} m}}{n} \equal{} \frac {(n \minus{} m)c^2 \plus{} mc^2}{nm(n \minus{} m)} \equal{} \frac {c^2}{m(n \minus{} m)}$

Now since n is odd, $ m \minus{} \frac {n \minus{} 1}{2}$ and $ m \minus{} \frac {n \plus{} 1}{2}$ are consecutive integers. Therefore the product of the above two terms is nonnegative, so that
$ m(n \minus{} m) \equal{} \frac {n^2 \minus{} 1}{4} \minus{} (m \minus{} \frac {n \minus{} 1}{2})(m \minus{} \frac {n \plus{} 1}{2}) \leq \frac {n^2 \minus{} 1}{4}.$
Therefore $ \frac {c^2}{m(n \minus{} m)} \geq \frac {4c^2}{n^2 \minus{} 1} \equal{} \frac {n^2}{n^2 \minus{} 1}\left(\frac {\sum_{i \equal{} 1}^{m} a_{i} \plus{} \sum_{i \equal{} 1}^{n \minus{} m} b_{i}}{n}\right)^2$.

Since we have equality in all of the above when $ m \equal{} \frac {n \minus{} 1}{2}$, $ a_i \equal{} n \plus{} 1$ and $ b_i \equal{} n \minus{} 1$ for all applicable $ i$ and the term $ \left(\frac {\sum_{i \equal{} 1}^{m} a_{i} \plus{} \sum_{i \equal{} 1}^{n \minus{} m} b_{i}}{n}\right)^2$ is positive the largest possible value of $ k_n$ is $ \frac {n^2}{n^2 \minus{} 1}$.[/hide]


modular arithmetic
number theory proposed
number theory

Let n  be postive number such that n>3 and  n be odd number
Call k,t be smallest number such that kn,tn+1 both square number
Prove of k,t>(n/4) then n be prime number

Writing $n = a \cdot c^{2}$ and $k = b \cdot d^{2}$ where $a$ and $b$ are squarefree numbers, we obtain

$kn = ab \cdot (cd)^{2}$.

To obtain the smallest number $k$ making $kn$ a square number, we set  $c = d = 1$ and $a = b$. Thus $k = n > 4$ is a squarefree odd integer.

Moreover $t$ is defined as the smallest natural number such that $tn+1$ is a square number, i.e.

$(1) \;\; t = \frac{m^{2}-1}{n}$

for a natural number $m$. Hence $t$ is given by the smallest value of $m$ which satisfies the congruence

$(2) \;\; m^{2}\equiv 1 \pmod{n}.$

Let $r$ be the number of primes which is divisible by $n$. The congruence (2) has $2^{r}$ incongruent solutions modulo $n$. Therefore, if $r \geq 2$, (2) has an integer solution $m \in [2, n/2)$, so by (1)

$0 < t < \frac{m^{2}}{n}< \frac{(n/2)^{2}}{n}= n/4$.

In other words, if $t > n/4$, then $n$ is a prime.


inequalities
trigonometry
inequalities unsolved

Let $ n$ and $ N$ be natural number. Prove that for any $ \alpha$, $ 0\le\alpha\le N$, and any real $ x$, it holds that \[{ |\sum_{k=0}^n}\frac{\sin((\alpha+k)x)}{N+k}|\le\min\{(n+1)|x|, \frac{1}{N|\sin\frac{x}{2}|}\}\]


algebra
polynomial
superior algebra
superior algebra unsolved

Let $P(x)$ be complex monic polynomial of degree $n$ .Let $W$ be the set of real $x$ for those $|P(x)| \leq 1$. Prove that the linear measure(measure on $\mathbb{R}$) of $W$ doesn't exceed
a)$2n$.
b)Some contsant, independent of $n$.

[quote="harazi"]To be sincere I do not understand what is the role of n... Should it be the degree of the polynomial?[/quote]
Yes, i've already edited it.

It is clear we are to prove your statement only for real-zeros polynomials.

1. It is true, since $W\in \bigcup [x_k-1,x_k+1]$, where $x_k$ are roots of $P$.

2. Clearly we can even to determine exact maximal measure $|W|$. It is $2^{\frac{2n-1}{n}}$. I hope you will quickly  reconstruct my solution from this hint.


If a water bottle is $\frac{3}{4}$ full, and you drink 80% of the water, what percent of water is left in the bottle?

[hide]Since $\frac{3}{4}$ is the same as 75%it is just $.75*.20=.15$. So it is 15%.[/hide]


[quote="bjorgy"]There is probably an easy way to do this but I don't see it...  There are 2 planes leaving from destination A.  They will both arrive at destination B.  Plane 1 is going 540mph and arrives at B 8 minutes late, Plane 2 is going 480 mph and arrives at B 53 minutes late.  I won't tell you how many people I have given this too that can't give me the answer.  HELP![/quote]
[hide]Using the equation D = RT, you can make 2 equations.  Since Distance is the same for both, you can use the same D.  The first equation is $D=540(T+ \frac{8}{60})$.  Since plane one was 8 minutes late, it takes them 8 extra minutes.  The second equation is $D=480(T + \frac{53}{60})$.  Now since both equations have D =, you can combine them into one equation $540(T+ \frac{8}{60})=480(T + \frac{53}{60})$ .  Now you have $540T+ 72 = 480T +424$ which simplifies to $60T =352$ and $T=\frac{352}{60}$ and finally $T=5 \  hours \ and \ 52 \  minutes$
Now you didn't actually ask a question, but this means plane 1 takes 6 hours and plane 2 takes 6 hours 45 minutes[/hide]

[quote="bjorgy"]oops...the question is what is the distance between A and B[/quote]

[hide]Well if plane A takes 6 hours at 540mph, then it is 3240 miles[/hide]


Solve the system:
$ x^y\equal{}y^x$
$ x^p\equal{}y^q$ ,where $ (x,y>0); p,q\in R;p\neq q$

(1, 1) is a solution.

Now, assume neither x nor 1 are 1. $ x \equal{} y^{\frac{q}{p}}$. Substituting this into the first equation, $ y^{\frac{yq}{p}} \equal{} y^{y^{\frac{p}{q}}}$. Therefore, $ \frac{yq}{p} \equal{} y^{\frac{p}{q}}$, so $ \frac{q}{p} \equal{} y^{\frac{p}{q} \minus{} 1}$. We can easily solve for $ y$ and $ x$ from here.

[quote="Zhero"](1, 1) is a solution.

Now, assume neither x nor 1 are 1. $ x \equal{} y^{\frac {q}{p}}$. Substituting this into the first equation, $ y^{\frac {yq}{p}} \equal{} y^{y^{\frac {p}{q}}}$. Therefore, $ \frac {yq}{p} \equal{} y^{\frac {p}{q}}$, so $ \frac {q}{p} \equal{} y^{\frac {p}{q} \minus{} 1}$. We can easily solve for $ y$ and $ x$ from here.[/quote] And make the assumption that $ p\ne0$. Want to complete your solution when $ p \equal{} 0$?

EDIT: Actually, nevermind. This leads to the obvious $ p\equal{}q$.

Neh, that's reasonable. I felt like I was skipping a few annoying special cases.


so heres how it works: 3 people control a bomb and they a gun. there are x good people and you all have been locked underground because of the explosive. This is a timed game the "mafia" have already set the bomb and it will blow in 10 days. I am in charge of the security system and the people may choose to kill someone that day. when the last mafia dies you will find the deactivation code and be able to unlock it. We have a secret agent who is a third party and is trying to keep everyone from killing the mafia thus blowing the bomb and then he wins because another mafia has decided that if he succeeds before the bomb blows he will disappear.. (how?)

there must be at least 16 people.

Now one more thing: I WIN if the neutral guy and one mafia member remain BEFORE the 10 day deadline. more info will be posted or pmed later. mafia wins if they kill everyone including the agent who can only be lynched

every day i or the controller will reveal a slight bit of information about someone and or an item he has which i will give everyone.

 and i will probably give this game to someone else i can trust to control if for some reason i cant do this... (most likely)

This is EXACTLY mafia, pretty much, with a different theme....

If it doesn't get locked, thoguh, i join :D

Just sign up on the Mafia waiting list. Just because you have no patience you still don't get to skip the line.

ok ive changed my mind ill open this up to newbies and have older mafia players give a kinda tutorial.

new set of rules: 8 players 2 mafia coached by a longer player
4 townies coached by 2 longer players a cop and a doctor with another coach
arrangement is in a jail cell with a hanging place. ill explain more later

Ok this game makes sense. A mafia game in the wrong forum modded by a guy who doesn't even know how to play [b]and[/b] is listing all of the roles.

[quote="mewto55555"]listing all of the roles.[/quote]

So? Open setup...

Hm true chenhsi, but I distinctly recall reading [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=177743]here[/url] that we should stop doing that.

Yeah, but this would be a newbie game, and it makes it easier for them. But maybe instead of saying how many there are of each role, you could just say the roles.

And, if this doesn't get locked, then I would also join.

SuperMathBoy, I used to make the same mistake. It's ok though. Please don't post on a thread if the last post was like a month ago. It is like spamming, and obviously every1 has stopped caring about this game, but it is ok, just don't do it again. (like me)


How many ways are there to put $ n$ balls in $ m$ boxes if the balls are not distinguishable and neither are the boxes?


Problem 1

Each vertex of a regular $1997$-gon is labeled with an integer, such that the sum of the integers is $1$. Starting at some vertex, we write down the labels of the vertices reading counterclockwise around the polygon.
Can we always choose the starting vertex so that the sum of the first $k$ integers written down is positive for $k =1,...,1997$ ?

I doubt that this is a analysis problem.
Anyway, my solution is inspired by an informatics problem.

Replace $1997$ by $n$. Fix a vertex of the regular $n$-gon and then let $x_{k}$ be the integer (the fact that they are integers seems to play no role) placed in the $k$th vertex (called $A_{k}$), starting from the chosen one (the vertex) and going counter-clockwise. Now define $y_{0}= 0$ and $y_{k}= x_{1}+\ldots+x_{k}, \, \forall k \in \overline{1,n}$ (so that $y_{n}= 1$).
It's clear that $x_{i}+x_{i+1}+\ldots+x_{j}= \begin{cases}y_{j}-y_{i-1}, \, \textrm{for}\, \, i \leq j; \\ y_{j}-y_{i-1}+1, \, \textrm{for}\, \, i > j. \end{cases}\, \, \, \, \left( \star \right)$
Note that the indices of the $x_{i}$'s are taken modulo $n$.

The conclusion is equivalent to finding $m$ such that for all $i \in \overline{0,n-1}$ we have that $x_{m}+x_{m+1}+\ldots+x_{m+i}> 0$.
By $\left( \star \right)$, that's equivalent to finding $m$ such that $\begin{cases}y_{m+i}> y_{m-1}, \, \textrm{for}\, \, m \leq m+i; \\ y_{m+i}> y_{m-1}-1, \, \textrm{for}\, \, m > m+i. \end{cases}$

The problem is almost done. Look at the $y_{i}$'s ($i \in \overline{0,n-1}$). Choose the $y_{i}$ with the least value. If this "least value" appears more than one time, take $i$ to be the greatest such $i$. Actually, we can take $i = m-1$, and this $m$ is obviously the sought one.

So, what you think? Is it correct?


function
trigonometry
real analysis
real analysis theorems

I was reading in an analysis book about the oscillation of a function and it was said that: $\omega_{f}(x)$ can be unbounded, even if $f$ was assumed to be bounded, but i couldn't find an example of such function. Can somebody help?

ps. I don't know if the correct english word is [i]oscillation[/i] or [i]variation[/i].

I think [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=variation+bounded&t=39534]this[/url] $\left( x \mapsto \sin \left( \frac1{x}\right) \right)$ will help you.

PS: I assumed that by "oscillation/variation" you meant $\sup \sum \left| f \left( x_{i+1}\right)-f \left( x_{i}\right) \right|$.

[quote="perfect_radio"]
PS: I assumed that by "oscillation/variation" you meant $\sup \sum \left| f \left( x_{i+1}\right)-f \left( x_{i}\right) \right|$.[/quote]

The oscillation/variation was defined like this: $\omega_{f}(x)=\inf\omega_{f}(V\cap E)$ where the $\inf$ is taken after all vecinities of $x$ if the function is $f: E\to\mathbb{R}$. So i don't think that example works.

Nobody ? Can somebody at least tell me if this property is true?

[quote="Thor"]The oscillation/variation was defined like this: $\omega_{f}(x)=\inf\omega_{f}(V\cap E)$[/quote]
Well, what is $\omega_{f}(V\cap E)$ then? Suppose it is what I would call the oscillation of $f$ on the set $V\cap E$, which is $\sup|f(x_{1})-f(x_{2})|$ over all $x_{1},x_{2}\in V\cap E$. Then it must be bounded for bounded $f$, because $|f(x_{1})-f(x_{2})|\le 2\sup_{x\in E}|f(x)|<\infty$.  :huh:

sin(1/x) oscillates a lot near the origin,

take the sequence 1/(n*pi/2) then on this sequence the function assume the values 1, 0, -1 infinitely often and

 sup summation{ f(X i+1) - f(X i) } is greater than when summed over the above sequence which grows like a the harmonic series which diverges.


linear algebra
matrix
limit
linear algebra unsolved

Find the nessesery and sufficient statement for existance $\lim_{n\to\infty}A$ for matrix A

Wrong. See for example the identity. If I'm not mistaken, one must have as condition the fact that all eigenvalues have modulus at most 1 and those that have modulus 1 are simple.

[quote="rustam"][quote="Fedor Petrov"]Fix a misprint, please. It is senseless now.[/quote]

I did not understand too.
I found the problem in Sadovnichiy, Podkolzin "Zadachi studenchiskih olimpiad po matematike"Moskva"Hauka"1978
problem number 360:(in russian)
Nayti neobhodimoe i dostatochnoye usloviye na matritsu A dlya togo chtobi sushestvoval $\lim_{n\to\infty}A^n$.[/quote]

Replace $A$ to $A^n$ in your initial post.

As for the answer, it is as follows: all eigenvalues belong to the set $\{|z|<1\}\cup\{1\}$, and for the eigenvalue 1 corresponding geometric and algebraic multiplicities must concide (i.e. there are no Jordan cells of size more then $1\times 1$, which correspond to the eigenvalue 1).


[color=red][i]Ο Θαλής πλησιάζει. Ας αρχίσουμε με κανα θεματάκι ! [/i][/color]

[b]ΑΣΚΗΣΗ[/b]
 Να  αποδείξετε ότι οι προβολές  κάθε σημείου του περιγεγραμμένου κύκλου ενός τραπεζίου πάνω στις μη παράλληλες πλευρές και στις διαγωνίους του είναι τέσσερα ομοκυκλικλά σημεία.

 Μπάμπης

Ωραία άσκηση κ. Στεργίου!

Δίνω ένα hint γιατί δεν έχω πολύ χρόνο να γράψω όλη τη λύση...

[hide]Χρησιμοποιούμε το θεώρημα του Simson (με τις προβολές του σημείου στις παράλληλες) και μετά δύναμη σημείου! [/hide]


For all $ n$ be positive integer real number. Denote that $ d(n)$be maximum odd divisor of $ n$.
Compute the sum : $ S\equal{}d(503)\plus{}d(504)\plus{}...\plus{}d(2009)$

For the odd numbers, the $ n$ is the largest odd divisor, and for the even numbers, $ \frac {n}{2}$ is the largest odd divisor. From there its just lots of easy but tedious calculations to do.

[quote="Thunder365"]for the even numbers, $ \frac {n}{2}$ is the largest odd divisor. [/quote]
Unless, y'know, it's even.   :P

[hide="outline"] For a number $ a$, if $ 2^n\parallel{}a$, then $ d(a) \equal{} \frac {a}{2^n}$.

Luckily, the highest power of 2 that can divide any of these numbers is $ 10$, so brute forcing is not that bad.

$ n \equal{} 0$: $ (503 \plus{} 505 \plus{} ... \plus{} 2009)$

$ n \equal{} 1$: $ (253 \plus{} 255 \plus{} 257 \plus{} ... \plus{} 1003)$

$ n \equal{} 2$: $ (127 \plus{} 129 \plus{} ... \plus{} 501)$

$ n \equal{} 3$: $ (63 \plus{} 65 \plus{} ... \plus{} 251)$

$ n \equal{} 4$: $ (33 \plus{} 35 \plus{} ... \plus{} 125)$

$ n \equal{} 5$: $ (17 \plus{} 19 \plus{} ... \plus{} 61)$

$ n \equal{} 6$: $ (9 \plus{} 11 \plus{} ... \plus{} 31)$

$ n \equal{} 7$: $ (5 \plus{} 7 \plus{} ... \plus{} 15)$

$ n \equal{} 8$: $ (3 \plus{} 5 \plus{} 7)$

$ n \equal{} 9$: $ (1 \plus{} 3)$

$ n \equal{} 10$: $ (1)$

Now, if you want, you can add this up. I suppose I will try to look for a better way now  :P .[/hide]

EDIT: $ d(n) \plus{} d(n \plus{} 1) \plus{} ... \plus{} d(2n \minus{} 1) \equal{} n^2$, proven by induction. Much faster brute forcing now  :D , as in just a couple calculations, I believe.


trigonometry
real analysis
real analysis unsolved

$ \lambda\in\mathbb{R}$. Find all the solutions $ x: [0,u[\rightarrow\mathbb{R}$ of the ODE $ x''x\minus{}x'^2\equal{}\lambda{e^t}$ s.t. $ x\in{C^2}$ and $ x(0)\equal{}0$.

For $ \lambda > 0$ we would have $ x'(0)^2 < 0$ and hence no real classical solutions.  For $ \lambda = 0$ we find the solutions $ x(t) = c_1e^{c_2t}$ and hence the only solution satisfying the BC is $ x\equiv 0$.

For $ \lambda < 0$, we look for solutions of the form $ x(t) = \sqrt { - l} e^{\frac {t}{2}}y(t)$ which gives rise to the equation $ y'^2 - yy'' = 1$, an equation which upon inspection one can see is satisfied by $ y = \frac {1}{c}\cos (ct + d)$.  Thus, a family of solutions satisfying the original problem is:
\[ x(t) = \pm \frac {\sqrt { - \lambda}}{c}e^{\frac {t}{2}}\cos (ct - \frac {\pi}{2}))
\]
for each $ c\in \mathbb{R}$ and $ \lambda < 0$.  No idea about uniqueness.

There exist other solutions.
And what about the solutions $ x: \mathbb{R}\rightarrow\mathbb{R}$ s.t. $ x\in{C^2},x(0)\equal{}0$ ?


Find all polynomial$P(x)\in Z[x]$such that $P((x+1)^n)=P(x)^n+(x+1)^n-x^n\forall x\in R$where$n$ is given number

[quote="mumath"]This problem was solved in CRUX,but this solution is very difficult.From my solution,I suggested the general problem.
Find all polynomial$P(x)\in R[x]$satisfying the following condition
 $P((x+1)^n)=P(T(x))^n+(x+1)^n-H(x)^n$where $T(x),H(x)$are given polynomials and$degH(x)\leq 1$,n is given positive integer.[/quote]

Can you post your solution?

i think you should think a lot about the problem,because it's not very hard


How do you guys know how to rearrange formulas to solve for certain variables. For example how would you tackle a problem like this:

$1 + 2 / 1 - 2$ = $2 + b / 2 - b$

When you need to take variables and put them on the opposite side how do you know if the variable should be placed either in the numerator or denominator?

wait...why do you have $2/1$.  couldn't you just write 2, or is that a typo.

One easy approach is to multiply both sides by a common number to eliminate any fractions. Then isolate the variable you want to solve for and do the necessary divisions or multiplications.

Well I get something like this:

$-3 * 2 - b$ = $2 + b$

= $-8 / 2$ = $2b / 2$

= $ - 4 = b$

but I know for sure the answer is 4 so, I'm not sure what I'm doing wrong.

all right you really need to use paranthesis in your original problem, or write some how that it would be understandable, because it's so confusing now... ;)

Sorry, let me try again.

Original Problem:

$(1 + 2) / (1 - 2) = ( 2 + b) / ( 2 - b)$

Solving for b:

cross multiplication

$-3 * 2 - b = 2 + b$

move -b to right side

$-6 = 2 + 2b$

isolate variable

$-8 / 2 = 2b / 2$

$-4 = b$

thing is I know b should equal 4 not -4

putting this in fraction helps. To put stuff in fraction:

\frac{numerator}{denominator}

so $\displaystyle (1 + 2) / (1 - 2) = ( 2 + b) / ( 2 - b)$
becomes
$\frac{1 + 2}{1 - 2} = \frac{2 + b}{ 2 - b}$

[quote="laptof89"]Original Problem:

$(1 + 2) / (1 - 2) = ( 2 + b) / ( 2 - b)$

Solving for b:

cross multiplication

$-3 * 2 - b = 2 + b$[/quote]

No, cross multiplication yields:
$\begin{eqnarray*}
-3&=&\frac{2+b}{2-b}\\
-3\left(2-b\right)&=&2+b\\
-6+3b&=&2+b\\
2b&=&8
\end{eqnarray*}\\
\center{\boxed{b=4}}$

I noticed the beautiful symettry that this problem has. What I would do is factor out a 2 from both the numerator and denomenator on the right side so we have:
$(1+2)/(1-2)=2(1+b/2)/2(1-b/2)$.
The two's cancel out and we can clearly see that the equation is the same on both sides if (b/2)=2.  Therefore $b=4$.

yeah, you've made a mistake in your cross multipication, the final asnwer is $b=4$

was my solution simple enough for your sig amirhtlusa

Simply use cross-multiplication and then use division to isolate the variable b, getting you the answer (-4)

um....

there is a flaw in the original problem if the solution is 4

if you plug in 4's for b

you get....

1 = 0

which can't be true

change the mistake :)


$ABC$ is an equilateral triangle. $A_{1}, B_{1}, C_{1}$ are the midpoints of $BC, CA, AB$ respectively. $p$ is an arbitrary line through $A_{1}$. $q$ and $r$ are lines parallel to $p$ through $B_{1}$ and $C_{1}$ respectively. $p$ meets the line $B_{1}C_{1}$ at $A_{2}$. Similarly, $q$ meets $C_{1}A_{1}$ at $B_{2}$, and $r$ meets $A_{1}B_{1}$ at $C_{2}$. Show that the lines $AA_{2}, BB_{2}, CC_{2}$ meet at some point $X$, and that $X$ lies on the circumcircle of $ABC$.

WLOG we can suppose that A2 lie on the segment B1C1.
than it is easy to show that AA2C1 ∽ B2BC1
So ∠(AA2, BB2)= 60
Similary, ∠(BB2, CC2)=60, ∠(CC2, AA2)=120
by this results, the problem is obvious!

We have $\frac{A_{2}C_{1}}{A_{2}B_{1}}=\frac{(p,r)}{(p,q)}$, etc,.. Where $(p,q)$ is algebra dintances between $p$ and $q$. By Ceva's theorem, we have first part. For second part $(AC_{1},AA_{2})=(A_{1}A_{2},A_{1}C_{1})=(C_{1}C_{2},A_{1}C_{1})=(CB,CC_{2})) (\mod \pi)$ , hence $(AB,AX)=(CB,CX)(\mod \pi)$. Done :D


I was wondering where you disappeared to.

tetrahedr0n, who are you?
For some reason you're the only AoPSer with lots of posts at Mathcamp that I can't seem to identify.

How many of you showed up to the Space Battle simulation yesterday? Were any of you perchance in the Deep Freeze ship? It took a good beatin', eh.  :D I was pummeling it with Taiwan.

(The first hour though was rather traumatising - some MCers broke the "Be Excellent to Each Other" rule a few hundred times...)

i think i will let you guess 314159265358979323846. but i do speak the same language as dorin.

[quote="Oasis"]314159265358979323846...you are Jennifer Lin?  Nice to meet you.  Haha =)

Spaceship Combat is pretty fun, I was the Engineer on Archangel.  We disabled and destroyed 3 ships...funnnn, though we started too late so there wasn't a lot of time.[/quote]

Yes, reports about Archangel popped up all over my screen... I think you guys destroyed Canada?

[quote="Nikolas"]Hello everybody... I am not a US citizen so u may excuse me for not knowing what this mathcamp is... I would be gratefull if you could just inform me... Thanks in advance![/quote]

Hi Nikolas: please see [url]http://www.mathcamp.org/[/url].

Hello, Osiris! Thanks for the info... :-)

[quote]Yes, reports about Archangel popped up all over my screen... I think you guys destroyed Canada? [/quote]

Yup, who were you in the game?

Strange an AoPS meeting haven't occured at mathcamp yet.

See you in tomorrow's game!

Yeah. I was (and will be) yuethomas, and I manned Taiwan. Canada was kind of stupid - its helm actually locked weapons on me for a split second.

See you in tomorrow's game. Oh, assuming that you'll partake in the puzzle hunt, don't fret to give me a frantic plea for help should you (and your team) be stuck. AIM SN is MCTomYue. If you need MSN, I'll give it to you over AIM.

[quote]See you in tomorrow's game. Oh, assuming that you'll partake in the puzzle hunt, don't fret to give me a frantic plea for help should you (and your team) be stuck. AIM SN is MCTomYue. If you need MSN, I'll give it to you over AIM.[/quote]

My MSN is chrissu_1226@hotmail.com, I don't have AIM :P

Are you writing/connected to the writing of the hunt?

No; but hunt organisers will tell you (or will have told you, at this point) that you're welcome to use whatever resources available to you. That includes me.  :D

[quote="Nikolas"]Hi again! Osiris, i checked the link u gave me and i found many more, but sadly i just finished school so i missed my last chance to participate in mathcamp and other similar projects... :(  Does anyone knows any other projects dedicated for new math undergraduates? Thanks in advance![/quote]

A lot of these camps look for people to come serve as counselors (teachers, assistants).  If that's up your alley, you may want to contact some and see if that's something you could do for them.

Out of interest...which mathcamp was this?

I was there (Katie, England)...who is this?


analytic geometry
function
MATHCOUNTS
modular arithmetic
number theory
relatively prime

I need to know how to solve these problems in a really quick way!!  :(

1. What is the remainder when $ 2^{2005}$ is divided by 7?

2. On the horizontal line y=5, there are 2 points exactly 5 units from the point (7,1). What is the sum of the x-coordinates of these two points?

hello, to your problem 1), note that $ 2^3\equiv1 \mod 7$ and $ 2^{2005}\equal{}2^{2004}\cdot2\equal{}(2^3)^{668}\cdot2\equiv1\cdot2\equiv2 \mod 7$.
Sonnhard.

how would you do this using Euler's totient function?

Since this is MATHCOUNTS, it is usually best to look for other methods (there will usually be some trick like that illustrated in Dr Sonnhard Graubner's post above).

[hide="Using the Totient Theorem for 1"]
Since $ \phi(7)\equal{}6$, and $ 2$ is relatively prime to $ 7$, $ 2^6\equiv1\pmod7$, and the rest can be solved in a similar fashion to that of Dr Sonnhard Graubner's solution.

Note that this is weaker than the simple condition $ 2^3\equiv1\pmod7$, again showing that it is better to use elementary methods first.
[/hide]
[hide="2"]
Note that the $ x$ coordinates will be $ 7\minus{}5$ and $ 7\plus{}5$, so the sum is $ \boxed{14}$.
[/hide]


8,000,000 is the product of two whole numbers, neither whole number contains the digit “0”.  What is the positive difference between the two whole numbers?

A.15,113  B.25,550	  C.94,200  D.106,117  E. 999,992

We can prime factorize this as 5^5*2^8.  This is 3,125 and 256.  The difference of the two numbers is 2,869.  I'm almost certain this is the correct answer, but it's not an answer choice.

[hide=" click to reveal hidden content"]

Hi Maybach. You've read $ 8000000$ as $ 800000$. Otherwise, your answer is, of course, correct.

$ 8000000 \equal{} 5^{6}\cdot2^{9} \equal{} 15625 \cdot 512$ and $ 15625\minus{}512 \equal{} 15113$, that is, option $ A$.

[/hide]


geometry
circumcircle
search
perpendicular bisector
angle bisector
projective geometry
power of a point

Let $AD,BE,CF$ be the internal angle bisectors of the scalene triangle $ABC$. The perpendicular bisector of $AD$ meets $BC$ at $X$, that of $BE$ meets $CA$ at $Y$ and of $CF$ meets $AB$ at $Z$. Prove that $X,Y$ and $Z$ are collinear.

Call $Q$ the intersection point of the external bisector of $\angle BAC$ and $BC$.
It is clear that $2 \left|XD\right|=\left|QD\right|$.
We deduce from the angle bisector theorem:

\[\frac{b}{c}= \frac{2\left|XD\right|-\left|CD\right|}{2\left|XD\right|+\left|BD\right|}\]
\[\iff \left|XD\right|c-b \left|BD\right| = \left|XD\right|b\]
\[\iff \frac{\left|XD\right|}{\left|BX\right|}=\frac{b}{c}\]
So we can use this relation to find that

\[\frac{2\left|XD\right|-\left|CD\right|}{2\left|XD\right|+\left|BD\right|}= \frac{c\left|CX\right|+b\left|BX\right|}{(b+c)\left|BX\right|}= \frac{b}{c}\]
\[\iff \frac{\left|CX\right|}{\left|BX\right|}= \frac{b^{2}}{c^{2}}\]

We repeat the same calculations for the other angles of the triangle, and Menelaos gives us that:

\[\frac{\left|CX\right|}{\left|BX\right|}.\frac{\left|ZB\right|}{\left|ZA\right|}.\frac{\left|YA\right|}{\left|YC\right|}= 1\]

[img]http://img118.imageshack.us/img118/7824/nayelexternalanglebiseckw6.jpg[/img]

[b][size=100]GENERAL PROBLEM.[/size][/b] - A triangle $\bigtriangleup ABC$ is given and let $P$ be, an arbitrary point inwardly to it, with its cevian triangle $\bigtriangleup DEF.$

On the sideline $BC,$ we define the point $D',$ as the harmonic conjugate of $D,$ with respect to the points $B,$ $C$ and similarly we define the points $E',$ $F',$ on the sidelines $AC,$ $AB,$ respectively.

Prove that the midpoints of the segments $DD',$ $EE',$ $FF',$ are collinear.

[b][size=100]PROOF.[/size][/b] - It is easy to define the points $D',$ $E',$ $F',$ as the intersection points of the sidelines $BC,$ $AC,$ $AB$ respectively, from the segment lines $EF,$ $DF,$ $DE,$ as well known.

These points are collinear ( well known and easy to prove ) and now, based on [b][size=100]Gauss’s theorem[/size][/b], we conclude that the midpoints of the segments $DD',$ $EE',$ $FF',$ as the diagonals of the complete quadrilateral $DED'E'FF',$ are collinear and the proof is completed.

Kostas vittas.

PS. This problem is coming from the past, as well known.

Here is my solution to the general problem:
Let $ H$ be the orthocenter of triangle $ DEF$. Denote$ D_1,E_1,F_1$the projections of$ D,E,F$ onto $ EF,FD,ED$. We call $ X,Y,Z$ are the midpoints of $ DD',EE',FF'$, and$ (X),(Y),(Z)$ are the circles with diameter $ DD',EE',FF'$, respectively. Because $ (D',D,C,B) \equal{} \minus{} 1$, hence, we get, $ XD'^2 \equal{} XD^2 \equal{} \overline {XC}.\overline {XB}$. As the result, $ (X)$ is orthogonal wrt $ (O)$, where $ O$ is the circumcenter of triangle $ ABC$ $ (1)$. Let $ R$ be the radii of $ (ABC)$. From $ (1)$, we get, the power of $ O$ wrt $ (X)$ is equal to $ R^2$. With the same argument, we also have that the power of $ O$ wrt $ (Y),(Z)$ are equal to $ R^2$. Because $ H$ is the orthocenter of triangle $ DEF$, as a very well- known result, we get, $ \overline {HD}.\overline {HD_1} \equal{} \overline {HE}.\overline{HE_1} \equal{} \overline {HF}.\overline {HF_1}$, which means that the powers of $ H$ wrt $ (X),(Y),(Z)$ are the same. Therefore, $ HO$ is the radical axis of $ (X),(Y),(Z)$, which leads to the fact that $ X,Y,Z$ are collinear, further, the line $ d$ through $ X,Y,Z$ is perpendicular to $ OH$.
Our proof is completed :lol: .

Dera mathlinkers,
this result is well known.
See
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1601085176&t=17064
Sincerely
Jean-Louis


[b]Moe uses a mower to cut his rectangular 90x150 ft lawn. The swath he cuts is 28 in. wide, but he overlaps each cut by 4 in to make sure that no grass is missed. He walks at a rate of 5000 ft. per hour while pushing the mower. Which of the following is closest to the number of hours it will take Moe to mow his lawn?[/b]

[b]A clock chimes once at 30 minutes past each hour and chime on the hour according to the hour. For example at 1 pm there is one chime and at noon and midnight there are 12 chimes. Starting at 11:15 on February 26, 2003, on what date will the 2003rd chime occur?[/b]

[hide="Answer to #1"]Since he overlaps 4 in he basically mows 24in each time. The dimensions are 1080 in.*1800in. Says he mows along the width (90 ft). To find how many "stripes" he has to mow, 1800/24=75.
75*90ft=6750
6750/5000=1.35=1 hours 21 min
If you try that he mows along the length you get the same answer.[/hide]

i got 2.7 hours for the 1st one because he overlaps each one, so doesn't he actually mow 2*75=150 "stripes"?
as for the 2nd question, i think its 11:15 pm.

[quote="xpmath"][hide="#1"]So whevever he walks a foot more, he cuts 2 feet (28-4). So he cuts 10000 sq. feet in an hour. I get 1.35.[/hide]
[hide="hint for #2"]Maybe you should find out how many chimes you get in a period of time, say every 24 hours[/hide].

I think #1 was an old AMC 12 problem.[/quote]

both of them were old amc 10s and 1 was old 12. iv taken the 10 and the 12 and both times i get 1 wrong. as for the second one, i did that. i got march 10th, when it should be march 9th.

[quote="mihail911"] i got march 10th, when it should be march 9th.[/quote]

That probably means the time is 11:15 am, not pm.  That would make sense if the test maker was referring to a 24-hour clock.