http://www.handdrawngames.com/DesktopTD/

Enjoy. My best score is 3240. The trick is to make a maze with the guns so to prolong the time that the monsters are being fired upon.

the best would be a really long zig zag straight up and down as it makes them cover the most distance  for flying use frost and the squirt guns  (frost slows them allowing for more hits)

[quote="D3m0n Shad0w"][quote="PenguinIntegral"]Does anyone have any insight as to optimal maze strategy?

I find some diagonal sections work best because then you can stack guns on top of each other like
{}{}{}{}{}{}
-------------
{}{}{}{}{}{}

Versus
{}{}---
{} - - {}
  ---{}{}
{}{}
Also, how to best deal with flying enemies? I don't want to waste money on towers that are only anti air, but I think a few of those in the middle may be the best strategy.[/quote]

i dont know how to deal with the flying ones.....but u can stack em??[/quote]
I ment make them denser, sorry.

Could someone post a screenshot of what they consider to be the perfect set up?

EDIT:  I keep getting killed on level 43.

This game is awesome.

I'll see if I can think of a good strategy.

Let's try to beat normal first:

Try to make the corridors as small as possible. I made the mistake of planning my things with corridors a 1 turret big, and it was a mistake. In general, make a very tight winding maze and plan it out.

Diagonal corridors are the best, because when I when up and down many planes went through

Never use pellet (5 a pop) guns in except in the VERY beginning to make your maze. The bulk of your maze should be white fast shooting guns. Upgrading pellet guns is always a waste, even if you go all the way to sniper.

Put ice guns only at turning points, and only use one or two throughout your maze.

The long range, slow firing guns are a waste; don't use them. Period.

One fully upgraded gun beats is better than many half-upgraded ones.

I currently use 5 different anti-air guns in the middle by the time I get to about 35. I have all of them level 4 or more.

Current high score: 3968

[img]http://i3.tinypic.com/29yjr7n.jpg[/img]

This is what I think is the best starting configuration. From here out, it's all squirt towers.

The best I think I will ever do with my current assumptions. Does anyone wish to submit some other configuration that would produce a higher score?

[img]http://i6.tinypic.com/44ufaqg.jpg[/img]

how do you send the next level early?

well something you should do is place dart towers near the begining of maze, and fast towers near the end. Fast towers are good at finishing off and damaging stuff overall. Dart towers are only effective in the begining of the maze when theres alot of stuff to attack. It is actually VERY effective at the begining of the maze.

Sometimes it can allow you to send 3 levels at once

[quote="ProtestanT"]how do you send the next level early?[/quote]It can only be done on the normal level or harder, not on the easy.

I got 3964+682 by putting 3 fully upgraded dart towers in the upper left corner of your maze.

I'm getting slightly more by just using pellet towers instead of squirt towers to make the maze, and by bringing the diagonal portion in the starting position in to make the corridor smaller.

Ok. I think the best strategy would be to get AS MANY typhoon towers as possible. I have 6 in the attached screen shot, and, as you can see, hardly needed anything else.

Those flying things are a bit tricky to deal with though.

[quote]Could someone post a screenshot of what they consider to be the perfect set up?[/quote]

well ur picture 've shown isnt really good. I sort of make the evil little dudes swerve around like in a maze

i'll show ya guys a perfect pic latr once i get a good score

I have a screenie but it's 3841kb... is it still attachable?

I got to lvl 79 on challenge with a score of 10366... I copied a maze off YouTube :D... and I scored higher than the guy in the vid too!

Wow, I figured out how not to lose lives on the spawn waves. You have to place your towers straight: not diagonally, or else they pop out.

Like this:

[] [] []
[]     []
[][][][]

Not like this:

[]   []   []
 []      []
   []  []
     []

However, I still can't beat hard mode. Any suggestions?

juggle

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

i got to level 48 on hard :ninja:
i got inspiration from youtube but i didnt copy it exactly which is probably why i didnt beat it

yay, finally beat hard!
i think i didnt before cuz i was focusing too much on bashes

would've had a lot of lives left had i not screwed up a flying level about halfway through... :ninja:

Bash towers are cool to put at corners but they cost too much to upgrade.

the best way to do bash towers is to put them in a line next to a line of squirts/darts so that the squirts/darts can shoot at the people while they're stunned

Anyone have a good way to kill the first spawn and flying waves on hard? I keep losing like 10 lives on them.

I dislike those waves too... although I find the spawning much harder. Against both the best strategy is to have tons of pellet towers, so basically get 1 dart of squirt and upgrade once and then spend hte rest of your money building your maze. A long maze should help against the spawns and lots of towers where the flyings go help against the flyings. Also, be prepared to drop pellet towers outside of your maze whenever some make it past your defenses and sell them later.

There seems to be a way to hack into the servers or something because lately tons of bogus scores have been popping up.

And it also seems that people are able to beat the 100 level challenge... which I thought was impossibly because I heard that math people on the forums calculated that in the later flying waves (lvl 80+), even if you spent all your money on swarm towers, you still wouldn't be able to beat them.

Dude does anyone play this anymroe besides me?

I just got past level 79... AND I WAS SO BUSY PAYING ATTENTION TO THE FLYINGS THAT I DIDN'T EVEN THINK ABOUT MY JUGGLING... AND I LET ABOUT 18 FAST BY (I had lost 1 life a lot earlier in like level 7 because the spawns were being g*y and then two lives on the 31k hp lvl 79 flying wave).

I play so much, it aint funny. by the way, I got to level 93 of the 1oo level challange in one try. I'll post a screenie later.

ha... I finally got a good Hard score. A bit mad because my first try I would have gotten over 7.1k except that I messed up a juggle by like half a second and died :( .

7844 on normal! common ubameya, I have no compitition in normal, do you still play?


A rectangular field is 50 feet wide and 80 feet long. Fence posts are placed every 10 feet around the field. How many posts are needed?

If we unravel the fence, it is 260 feet long. Since there are posts every 10 feet, there are $ \boxed{26}$ posts.


Hello, I have recently downloaded the asymptote and am facing problems when it comes to drawing a circle.
 :(
Can anyone please tell me the CODE for drawing a circle? :blush:

Sorry for disturbing you.

[quote="fedja"][b]draw(circle(A,r));[/b] where $ A$ is the center and $ r$ is the radius. In general, read the manual before playing with the program; you can find a lot of useful information there...[/quote]
Yeah, I thik I should read the manual first..... :oops:
Thanks alot for your advice! :)


quadratics
number theory unsolved
number theory

Prove that there exist infinitely many naturals of the form $ 18^m\plus{}7^m\minus{}9^m\minus{}14^m$ which are divisible by $ 59$ , where $ m$ is an odd natural.

[quote="Inequalities Master"]Prove that there exist infinitely many naturals of the form $ 18^m \plus{} 7^m \minus{} 9^m \minus{} 14^m$ which are divisible by $ 59$ , where $ m$ is an odd natural.[/quote]

1. Note that 59 is prime.
2. Recall Fermat's little theorem.
3. Conclude that if $ k$ is not a multiple of 59, then $ k^{58} \equiv 1\bmod 59$.
4. Conclude that if $ k$ is not a multiple of 59, then $ k^{58n} \equiv 1\bmod 59$ for any $ n\ge1$.
5. Apply this observation to $ k\equal{}7,9,14,18$.

[quote="test20"]

1. Note that 59 is prime.
2. Recall Fermat's little theorem.
3. Conclude that if $ k$ is not a multiple of 59, then $ k^{58} \equiv 1\bmod 59$.
4. Conclude that if $ k$ is not a multiple of 59, then $ k^{58n} \equiv 1\bmod 59$ for any $ n\ge1$.
5. Apply this observation to $ k \equal{} 7,9,14,18$.[/quote]

Note that $ m$ needs to be odd, so multiples of 58 won't due. But $ m \equal{} 29$ satisfies the given condition, and so anything of the form $ 29 \plus{} 58x$ for positive integer $ x$ will work and all of those will be odd.

And here's a way to find that 29 works without using a calculator:
9 is obviously a quadratic residue mod 59 and you can use quadratic reciprocity to find that 7 is too. So $ 7^{29}\equiv 1$ and $ 9^{29}\equiv 1$ Then 14 and 18 are quadratic residues iff 2 is, but since 59 isn't $ \pm 1 \bmod 8$ 2 isn't. So we also have $ 14^{29}\equiv \minus{} 1$ and $ 18^{29}\equiv \minus{} 1$


limit
trigonometry
logarithms
real analysis
real analysis unsolved

\[ \lim_{x\rightarrow \frac{\pi}{2}^\minus{}} \cos(x)^{\tan(x)}\]

I take log and then I got -infinity then the limit should be zero, am i right?

What is the question?
Is it:

$ \lim_{x\to\frac{\pi}{2}^\minus{}} (\cos x)^{\tan x}$

or:

$ \lim_{x\to\frac{\pi}{2}^\minus{}} \cos \left(x^{\tan x}\right)$

If it is the first one,

$ \lim_{x\to\frac{\pi}{2}^\minus{}} (\cos x)^{\tan x} \equal{} \lim_{x\to\frac{\pi}{2}^\minus{}} e^{\tan x \ln \cos x}$

As $ x\to\frac{\pi}{2}^\minus{} , \;\; \cos x \to 0^\plus{}$.

[quote="kelvin911"]
\[ \lim_{x\rightarrow \frac {\pi}{2}^ \minus{} } \cos(x)^{\tan(x)}
\]
I take log and then I got -infinity then the limit should be zero, am i right?[/quote]
if it is cos^tan then limit==0.
if it is cos(x^tan(x)) then limit is any value between 1 and -1.

the limit is zero, how do you construct an $ \epsilon$-$ \delta$ proof of it?

[quote="Rajiv"][quote="ith_power"]if it is $ \cos(x^{\tan x})$ then limit is any value between 1 and -1.[/quote]

It'd be $ \cos 1$, not [b]any[/b] value![/quote]
Its equal to
=Cos((pi/2)^(+infinity))
=cos(infinity)
=any val between [1,-1]

Sorry! I calculted $ \tan 0$ instead of $ \tan 90$!


calculus
derivative
geometry
rectangle
calculus computations

A rectangular beam is cut from a cylindrical log of radius 30 cm.  The strength of a beam of width w and height h is proportional to $wh^{2}$.  Find the width and height of the beam of maximum strength.

I'm not sure what to do to set this up... After the setup, It'll be easy.. I'll just take the derivative of the strength equation to maximize it...

All i have so far is $s = kwh^{2}$.  I don't know how to write any other equations to help with writing one variable in terms of another for substitution.

Draw the picture; you're cutting a rectangle of sides $w$ and $h$ out of a circle of radius 30. $w$ and $h$ will be related by the fact that the corners of the rectangle lie on the circle.


[quote="chien than"]Find all  $ n \in Z^{ \plus{} }$ such that:
 $ \frac {7n \minus{} 12}{2^{n}} \plus{} \frac {2n \minus{} 14}{3^{n}} \plus{} \frac {24}{6^{n}} \equal{} 1$[/quote]
From $ \frac{1}{2}\plus{}\frac{1}{3}\plus{}\frac{1}{6}\equal{}1$
$ 7n\minus{}12\geq 2^{n\minus{}1},\forall n\geq 6$
$ 3^{n\minus{}1}\geq 2n\minus{}14,\forall n\in N$
$ 24\geq 6^{n\minus{}1},\forall n\geq 3$
Try to cacurlas with $ n\equal{}1,..,6$ we have it has no solution.


Hi,

Does anybody know the average score of the person winning National MathCounts over the past couple of years? I'm only talking about the written round, so that I know what to shoot for.  :P

-biblioman

I think last year's written winner got a 43 to 45...somewhere in that range. Then again, 2008-2009 was not a very hard year.

OK. Do you know the historical average? I am intending to score well *cough* first *cough* this year, and I need to know what a good score is.

Thanks,
biblioman

Oh. Crud.

I guess I have to go study now...  :P

Thanks, both of you!

First things first, you actually have to GET to nats to do well..........

So you've got a LONG road ahead of you(states-nats is a very hard leap in many states) just like me.

But really you'd rather just make countdown with like a 37-42 and if you're good at that you can win.

[quote="Maybach"]But really you'd rather just make countdown with like a 37-42 and if you're good at that you can win.[/quote]
The top four have an advantage, since they have to win one less matchup.

[quote="athunder"]You cannot say last years test was too much easier then the past years because 2 years ago the written round winner did not win last year.[/quote]

Most of this is pretty much due to luck.  After all, most of the top 12 are smart enough to score perfects, but don't due to stupid mistakes  and such.

They got 10th last year and 9th the year before that, so I would say it is one of the harder states. I think it is about 35-42 for qualifying (that is a VERY inaccurate guess).

Also, what kind of chapter are you in (as in, is it one with major city in it)?

No, I'm in one with Sterling, (i think) Reston, and Fairfax, so it should be not so hard.

OK, I retested myself on the states and need to improve by a few more points.

[quote="biblioman"]Okay, Maybach. One other thing, though I don't know if this belongs in the Virginia forums.

How hard is the VA States - Nats leap?[/quote]

This isn't the Virginia forum...........this is Middle School Contests forum  :wink:


calculus
derivative
function
calculus computations

Water leaks out of a tank at a rate proportional to the square root of the depth of water at that time. The water level starts at 100 inches and drops to 64 inches in 1 hour.
What's the equation which models the water that left in the tank?

[hide="Hints"] The problem seems pretty realistic, since the physical analysis really shows that the rate of the leaking out is proportional to the sqrt of the depth.
Now, as far as the equation is concerned: How can you express the leak rate in terms of calculus(is it a derivative? is it something like velocity?)
On the other hand, you have the proportionality with $ \sqrt h$. So the rate is a product of it and a real constant. How can you find the constant? Can you use the data about the level drop in one hour? Before searching for the constant, you'll have to solve the differential equation you get after expressing the leak rate in appropriate form in function of depth and time.[/hide]


$ \lim_{x \to 0} \frac{1\minus{}x\sqrt{sinx^2}}{1\minus{}cos3x}$

The numerator approaches $ 1$ and the denominator approaches $ 0$.

hello

$ \lim_{x \to 0 } \frac {1 \minus{} x\sqrt {sinx^2}}{1 \minus{} cos3x} \equal{} \plus{} \infty$


The times between $ 7$ and $ 8$ o'clock, correct to the nearest minute, when the hands of a clock will form an angle of $ 84$ degrees are:

$ \textbf{(A)}\ \text{7: 23 and 7: 53} \qquad
\textbf{(B)}\ \text{7: 20 and 7: 50} \qquad
\textbf{(C)}\ \text{7: 22 and 7: 53} \\
\textbf{(D)}\ \text{7: 23 and 7: 52} \qquad
\textbf{(E)}\ \text{7: 21 and 7: 49}$

[hide="Click for solution"]
$ 210\plus{}x\minus{}12x\equal{}84 \implies 11x\equal{}126 \implies x\equal{}\frac{126}{11}$.  Then $ \frac{126 \cdot 60}{11 \cdot 30} \approx 23$ and $ 210\plus{}x\plus{}84\equal{}12x \implies 11x\equal{}294 \implies x\equal{}\frac{294}{11}$.  Then $ \frac{294 \cdot 60}{11 \cdot 30} \approx 53$, so the times are $ 7: 23$ and $ 7: 53$ which is $ \boxed{\textbf{(A)}}$.
[/hide]

instead of spamming intermediate topics with random problems nobody requested in order to raise your post count, try the resources section

I'm posting these for the resources section :wink:

In this case does it mean it's impossible to have exactly 84 degrees since the answers turn out to include a fractional part of a minutes (which in reality the difference of 1 sec usually don't change the angle of the minute and hour hand unless the second hand just passed the minute hand in one of those instances, if my understanding of how the hands work is correct)?


A circle is colored in 3 colors prove that there infinitely many isosceles triangles with vertices on this circle and having the same color.

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=van+der+waerden&t=5434

Pierre.


Has anyone heard about People to People Future Leaders program??

I saw them sitting on the steps when I was in Italy one year. They looked thoroughly bored. Oh, and one girl lost her credit card. But that's totally random, and you probably don't care.

But if you're asking whether that program is highly selective, then the answer is no.


[quote="anirudh"]Use the clues below to determine the value of n.
         The sum of the digits of n is 11, and 50 < n < 100.
         When n is divided by 2, the result is a prime integer.
What is the value of n ?[/quote]

[hide]
25<n/2<50, and n is a prime.
so they can be
29
31
37
41
43
47
x2=
58
62
74
82
86
94

[b]74[/b]
[/hide]

[hide]Since  it leaves an integer when divided by 2, it must be even.
Find the integers with digit sum 11 and even:

56, 74, 92,
56/2=28, not prime.
74/2=37, prime.
92/2=46, not prime.

74. [/hide]

[hide]
here are all the integers between 50 and 100 that digits add to 11
56
65
74
83
92
so when it is divided by two the result is a prime integer.  So that means the number is even
56/2=28, which is not prime
74/2=37, which is prime so then 72 is the answer
[/hide]

[quote="anirudh"]Use the clues below to determine the value of n.
         The sum of the digits of n is 11, and 50 < n < 100.
         When n is divided by 2, the result is a prime integer.
What is the value of n ?[/quote]
[hide="answer"]50<n<100 and n/2 is prime.
the numbers can b: 29*2=58
                                31*2=62
                                37*2=74
                                41*2=82
                                43*2=86
                                47*2=94
the only one that works is [b]74[/b]. :| [/hide]

[hide]
The numbers whose integers add up to 11 are:
56
65
74
83
92
The numbers that are also even are:
56
74
92
The only number that works is 74

[/hide]


vector
linear algebra
linear algebra unsolved

Can someone suggest improvements to how I should be proving this?

$ K,L$ are fields such that $ K\subset L$. If $ V$ is a finite dimensional vector space over $ L$ of dimension $ n$ and $ L$ is  a finite dimensional vector space over $ K$ of dimension $ m$, I need to show that $ V$ is a finite dimensional vector space over $ K$ of dimension $ mn$.

So first, I decided to sum the vectors of $ V$ with coefficients from $ K$(my reasoning being that because $ K$ is a subset of $ V$ I could express the vectors of $ V$ as a linear combination with vectors from $ K$)

$ \sum_{i \equal{} 1}^{m}k_{i}v_{i}$ (1)

Next, I did the same with vectors from $ V$ and coefficients from $ L$

$ \sum_{i \equal{} 1}^{n}l_{i}v_{i}$  (2)

and since we're given that $ L$ is a finite dimensional vector space over $ K$

$ \sum_{i\equal{}1}^{m}l_{i}k_{i}$    (3)

Next, I took (1) and substituted (3) in for $ k_{i}$

$ \sum_{i\equal{}1}^{m}[\sum_{i\equal{}1}^{n}l_{i}k_{i}]v_{i}$  (4)

my rationale being that although (3) is a linear combination between $ l$ and $ k$, these are still vectors in $ K$.

and then, from (4) gives us a dimension of $ mn$

my other question is should i index each of the sums differently( ex. $ j\equal{}1$ instead of $ i \equal{} 1$)

Your very first step either assumes the result, or is nonsense. This is not a proof.

Writing down expressions without setting them equal to anything won't help.
There are three parts to the proof:
(1) Construct the set that will be a basis for $ V$ as an $ K$-vector space.
(2) Show that it spans $ V$.
(3) Show that it's linearly independent.

Also, please do use different dummy variables for nested sums. If you set it up right, the coefficients will depend on both indices.

I can write the set that will be a basis as  {$ l_{i}v_{i}$} for all $ i \equal{} 1,..., n$.

then , i proceed to express the set for any vectors $ v$ in $ V$ as a linear combination of vectors in $ K$, or the sum of the elements in my previous set.

$ \sum_{i\equal{}1}^{n}l_{i}v_{i}$ (1)

Then I set (1) equal to zero to show that it's linear independent. If the sum is equal to zero and we're dealing with nonzero basis elements from $ V$ then this implies that the $ l_{i}$ are equal to zero, and thus linear independence is proved.

If the above is correct, should i proceed with showing that $ L$ can be expressed as a linear combination of vectors in $ K$?

It's still nonsense.

[quote="mstrip"]should i proceed with showing that $ L$ can be expressed as a linear combination of vectors in $ K$?[/quote]
This is also nonsense.

[Note- I also just fixed a mistake in my earlier post.]

I find it a little strange that you haven't used the word "basis" yet, except to repeat what I said. That is what the two assumptions about finite-dimensional vector spaces give us- and you need these bases explicitly to construct the basis for $ V$ over $ K$.

I didn't understand the implication of a finite dimensional vector space, but now I do. I appreciate the help.


geometry
parallelogram
circumcircle
geometric transformation
reflection
perpendicular bisector
geometry proposed

Let two chords $AC$ and $BD$ of a circle $k$ meet at the point $K$, and let $O$ be the center of $k$. Let $M$ and $N$ be the circumcenters of triangles $AKB$ and $CKD$. Show that the quadrilateral $OMKN$ is a parallelogram.

It's well-known that if we construct a line through $K$ which cuts the circles $(AKB)$ and $(CKD)$ in $X,Y$ respectively, then the locus of the midpoint of $XY$ is a circle $\omega$ centered at the midpoint of $MN$ and which passes through $K$ (and, of course, $P$, the other intersection pt of the two circles). We have to show that the midpoints of $KO$ and $MN$ coincide. The lines $AC$ and $BD$ cut $\omega$ at their midpoints. Consider the perpendiculars to these two lines through the midpoints. They must intersect $(a)$ in $O$ and $(b)$ in the point on $\omega$ diametrally opposite to $K$. This means that the center of $\omega$, which is the midpoint of $MN$, is also the midpoint of $KO$, and we're done.

I think I've proved that thing about the locus somewhere else on the forum. It's not at all hard to prove.

Seems like you are making the problem more difficult than it actually is! Here is my solution:

I will use the coolest of all types of angles, i. e. directed angles modulo 180. The points O and M, being the circumcenters of triangles CAB and KAB, lie on the perpendicular bisector of the segment AB, so that we have $OM\perp AB$.

If K' is the reflection of K in N, then KK' is a diameter of the circumcircle of triangle CDK. Hence, after Thales, $\measuredangle KCK^{\prime }=90^{\circ }$. But the sum of the angles of triangle KCK' is

  $\measuredangle K^{\prime }KC+\measuredangle KCK^{\prime }+\measuredangle CK^{\prime }K=0^{\circ }$.

As a chordal angle, $\measuredangle CK^{\prime }K=\measuredangle CDK$, hence $\measuredangle CK^{\prime }K=\measuredangle CDB=\measuredangle CAB$ (again from chordal angles). Moreover, we have $\measuredangle KCK^{\prime }=90^{\circ }$, as already said. Hence,

  $\measuredangle K^{\prime }KC+90^{\circ }+\measuredangle CAB=0^{\circ },$

so that

  $\measuredangle \left( KN;\;AB\right) =\measuredangle \left( KN;\;AC\right) +\measuredangle \left( AC;\;AB\right) =\measuredangle K^{\prime }KC+\measuredangle CAB$
  $=\left( \measuredangle K^{\prime }KC+90^{\circ }+\measuredangle CAB\right) -90^{\circ }=0^{\circ }-90^{\circ }=-90^{\circ }=90^{\circ }$

(remember the angles are directed modulo 180). Hence, $KN\perp AB$. Together with $OM\perp AB$, this yields KN || OM. Similarly, KM || ON; therefore, OMKN is a parallelogram.

  Darij

Well, what you did makes it look much easier, but that's what came to me first, so I posted it :).


linear algebra
matrix
linear algebra unsolved

If p is a prime number and A is the sum of all $X \in M_n(Z_p)$ with $det(X) \neq 0$ then $det(A) = 0$

For any non-singular $B \in M_n(\mathbb{Z_p})$ we have that $AB=A$ which means that $A(B-I)=0$ and taking for example $B=diag(2,2,...,2)$ we have that $det(B-I) \neq 0$ with meaning that $det(A)=0$.

but... :oops: how do you compute B - I ?

I denoted $B=diag(2,2,...,2)$ and $B-I$ is the identical matrix $diag(1,1,...,1)$.

Technically, your argument only works in characteristic >2. We could tweak it to work in any field with more than three elements (in which it shows that $A=0$), but we need more when the field is $\mathbb{Z}_2$.

In particular, we need an invertible matrix $B$ such that $B-I$ is also invertible. Let $p(x)=x^n+b_{n-1}x^{n+1}+\cdots+b_1x+b_0$ be such that $p(1)$ and $p(0)$ are nonzero. We can do this by choosing $p$ irreducible, or by merely choosing $b_{n-1}=-1,b_0=1$ and $b_i=0$ for each other $i$. Now the matrix $B=\begin{bmatrix}0&0&0&\cdots&0&-b_0\\ 1&0&0&\cdots&0&-b_1\\0&1&0&\cdots&0&-b_2\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\cdots&0&-b_{n-2}\\0&0&0&\cdots&1&-b_{n-1}\end{bmatrix}$ has determinant $(-1)^np(0)$, and $B-I$ has determinant $(-1)^np(1)$. In the example above, $p(0)=p(1)=1$.
Now $A(B-I)=0$. Since $B-I$ is invertible, $A=0$.

The above works in any (finite) field.

Another solution, i would like to share it here, hope it is ok.

If $A_1,...,A_m$ are all the matrices X with $det(X) \neq 0$ they are forming a group $M = U(M_n(Z_p))$.
Suppose A is one of this $A_i$.
$A^2 = (A_1 + ... + A_m)*A_1 + ... + (A_1 + ... + A_m)*A_m = (A_1 + ... + A_m) + ... + (A_1 + ... + A_m) = m*A$.
I used $(A_1 + ... + A_m)*A_i = (A_1 + ... + A_m)$, in some other order maybe, because M is group.
But A is invertible, so $A = mI_n$. [X is invertible in $M_n(Z_p)$ if and only if $det(X) \neq 0$, because p is prime].
On the other hand p | m, hence $A = O_n$ which has determinant 0, contradiction.

For p | m, we consider the relation $m = card(U(M_n(Z_p))) = (p^n - 1)(p^n - p)...(p^n - p^{n-1})$ [classical one].
This can be proved like this: an element X in $U(M_n(Z_p))$ can be build by choosing the first column, then the second, and so on.
We need to take care that $det(X) \neq 0$.
The first column can be choose from $Z_p$ in $p^n - 1$ ways [all $p^n$ possible cases without one case, all elements are 0].
The second column can be choose from $Z_p$ in $p^n - p$ ways [all $p^n$ possible cases minus the cardinal of the generated space of the first already filled column].
The third column can be choose from $Z_p$ in $p^n - p^2$ ways [all $p^n$ possible cases minus the cardinal of the generated space of the first and second already filled columns].
And so on through the last column. :)


Let $ a,b,c>\minus{}1$. Prove that

$ \sqrt{\frac{a\plus{}2}{a\plus{}b\plus{}5}}\plus{}\sqrt{\frac{b\plus{}2}{b\plus{}c\plus{}5}}\plus{}\sqrt{\frac{c\plus{}2}{c\plus{}a\plus{}5}}\leq\frac{3\sqrt{2}}{2}$

It's obviously trues because: $ \sum\ \sqrt{\frac{x}{x\plus{}y}} \leq\ \frac{3}{\sqrt{2}}$  :maybe:
With $ x\equal{}a\plus{}1$...

[quote="nguoivn"]It's obviously trues because: $ \sum\ \sqrt {\frac {x}{x \plus{} y}} \leq\ \frac {3}{\sqrt {2}}$  :maybe:
With $ x \equal{} a \plus{} 1$...[/quote]

Thank you very much.

[quote="nguoivn"]It's obviously trues because: $ \sum\ \sqrt {\frac {x}{x \plus{} y}} \leq\ \frac {3}{\sqrt {2}}$  :maybe:
With $ x \equal{} a \plus{} 1$...[/quote]
I think with $ x\equal{}a\plus{}1;y\equal{}b\plus{}1;z\equal{}c\plus{}1$, this ineq

$ \Leftrightarrow \sum \sqrt{\frac{x\plus{}1}{x\plus{}y\plus{}3}}\le \frac{3}{\sqrt{2}}$

[quote="Ligouras"]Let $ a,b,c > \minus{} 1$. Prove that

$ \sqrt {\frac {a \plus{} 2}{a \plus{} b \plus{} 5}} \plus{} \sqrt {\frac {b \plus{} 2}{b \plus{} c \plus{} 5}} \plus{} \sqrt {\frac {c \plus{} 2}{c \plus{} a \plus{} 5}}\leq\frac {3\sqrt {2}}{2}$[/quote]

MAY be nguoivn means $ x\equal{}a\plus{}2$... :wink:

[quote="anas"]MAY be nguoivn means $ x \equal{} a \plus{} 2$... :wink:[/quote]

Even with $ x\equal{}a\plus{}2...$, we have this ineq

$ \Leftrightarrow \sum \sqrt{\frac{x}{x\plus{}y\plus{}1}}\le \frac{3}{\sqrt{2}}$

[quote="gigaman"][quote="anas"]MAY be nguoivn means $ x \equal{} a \plus{} 2$... :wink:[/quote]

Even with $ x \equal{} a \plus{} 2...$, we have this ineq

$ \Leftrightarrow \sum \sqrt {\frac {x}{x \plus{} y \plus{} 1}}\le \frac {3}{\sqrt {2}}$[/quote]

but :  $ \frac{x}{x\plus{}y\plus{}1} \leq  \frac{x}{x\plus{}y}$ :D

[quote="anas"][quote="gigaman"][quote="anas"]MAY be nguoivn means $ x \equal{} a \plus{} 2$... :wink:[/quote]

Even with $ x \equal{} a \plus{} 2...$, we have this ineq

$ \Leftrightarrow \sum \sqrt {\frac {x}{x \plus{} y \plus{} 1}}\le \frac {3}{\sqrt {2}}$[/quote]

but :  $ \frac {x}{x \plus{} y \plus{} 1} \leq \frac {x}{x \plus{} y}$ :D[/quote]

I know, I just say if

[quote="gigaman"][quote="anas"]MAY be nguoivn means $ x \equal{} a \plus{} 2$... :wink:[/quote]

Even with $ x \equal{} a \plus{} 2...$, we have this ineq

$ \Leftrightarrow \sum \sqrt {\frac {x}{x \plus{} y \plus{} 1}}\le \frac {3}{\sqrt {2}}$[/quote]

[quote="gigaman"]
I think with $ x \equal{} a \plus{} 1;y \equal{} b \plus{} 1;z \equal{} c \plus{} 1$, this ineq

$ \Leftrightarrow \sum \sqrt {\frac {x \plus{} 1}{x \plus{} y \plus{} 3}}\le \frac {3}{\sqrt {2}}$

[/quote]
But $ \sum \sqrt {\frac {x \plus{} 1}{x \plus{} y \plus{} 3}}\le \sum \sqrt {\frac {x \plus{} 1}{x \plus{} y \plus{} 2}}$ (obviously trues)  :wink:


The front and back covers of a book are each 3 mm thick, and a stack of 25 sheets of paper is 4 mm thick. Al’s book is 38 mm thick, and all of the pages are numbered front and back starting with page
1. What is the number of the last page facing the back cover?

$ 3$ mm for each of the front and back. So there's $ 32$ mm for the pages, dividing by $ 4$ yields $ 8 \times 25\equal{}200$ sheets, for $ \boxed{400}$ pages.

Actually the answer to the question is the [i]number of the last page facing the back cover[/i].  (Using LaTeX's work.)  After finding that there are 400 pages, you can know that the number of the last page is 400.  This is because the backside of the first sheet is numbered 2, the backside of the second sheet is numbered 4, and so on.


function
real analysis
real analysis unsolved

Up to my knowledge, a semi-convex function ($f(\frac{x+y}{2})\ge \frac{f(x)+f(y)}{2}$) need not to be convex. Can you give me a counterexample.

The question was answered in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=293926#p293926]http://www.mathlinks.ro/Forum/viewtopic.php?t=46490 post #10[/url]. The answer in brief: Take a non-continuous solution of the Cauchy functional equation.

  Darij

Thank you. I didn't think about that functions they have a lot of interesting properties also. For example their graphs are dense in $\mathbb{R}^2$ ;)


function
modular arithmetic
quadratics
number theory
prime numbers
algebra unsolved
algebra

For odd prime numbers $ p$, find all functions $ f: \mathbb{Z}\rightarrow\mathbb{Z}$ satisfying following conditions:

(1) If $ m\equiv n\mod p$ , $ f(m)\equal{}f(n)$.
(2) For all integers $ m$ and $ n$, $ f(mn)\equal{}f(m)f(n)$.

[quote="FantasyLover"]For odd prime numbers $ p$, find all functions $ f: \mathbb{Z}\rightarrow\mathbb{Z}$ satisfying following conditions:

(1) If $ m\equiv n\mod p$ , $ f(m) \equal{} f(n)$.
(2) For all integers $ m$ and $ n$, $ f(mn) \equal{} f(m)f(n)$.[/quote]

$ f(n) \equal{} 1$ $ \forall n$ is a trivial solution.
$ f(n) \equal{} 0$ $ \forall n$ is a trivial solution.
$ f(n) \equal{} 1$ $ \forall n\neq 0\pmod p$ and $ f(n)\equal{}0$ $ \forall n\equal{}0\pmod p$ is a trivial solution.

If $ f(u)\equal{}0$ for some $ u\neq 0\pmod p$, then let $ y\equal{}\frac{v}{u}\pmod p$. then $ v\equal{}uy\pmod p$ and so $ f(v)\equal{}f(u)f(y)\equal{}0$ $ \forall v$

Consider now we have $ m$ such that $ f(m)\notin\{0,1\}$
$ f(0) \equal{} f(m)f(0)$ $ \implies$ $ f(0) \equal{} 0$.
$ f(m*1) \equal{} f(m)f(1)$ $ \implies$ $ f(1) \equal{} 1$
Let $ x\neq 0\pmod p$. Exists $ y$ such that $ xy \equal{} 1\pmod p$ so $ f(xy) \equal{} f(1)$ so $ f(x)f(y) \equal{} 1$ and so $ f(x)\in\{ \minus{} 1,1\}$ $ \forall x\neq0\pmod p$ (since $ f(x)\in\mathbb Z$ and $ f(x)|1$). So $ f(m) \equal{} \minus{} 1$

Let now $ u\neq 0\pmod p$ a quadratic residue mod p. we have $ u \equal{} v^2\pmod p$ and so $ f(u) \equal{} (f(v)^2 \equal{} 1$
So $ m$ is not a quadratic residue mod p.
Let $ w$ a non quadratic residue mod p. $ mw$ is a quadratic residue and so $ f(mw) \equal{} 1$. But $ f(mw) \equal{} f(m)f(w) \equal{} \minus{} f(w)$ and so $ f(w) \equal{} \minus{} 1$

So $ f(x)$ is the Legendre symbol $ \left(\frac {x}{p}\right)$

And all the solutions are

$ f(x) \equal{} 0$

$ f(x) \equal{} 1$

$ f(x) \equal{} 0$ $ \forall x\equal{}0\pmod p$ and $ f(x)\equal{}1$ $ \forall x\neq 0\pmod p$

$ f(x) \equal{} \left(\frac {x}{p}\right)$


calculus
integration
number theory
prime factorization

Find all n from 1-50 such that  the fraction $n!/((a+1)!(n-a)!)$ is an integer for all integers a such that $0< a < n$

[hide]First, we can rewrite the expression as $\frac{{n+1\choose a+1}}{n+1}$.

(defining new variables for lemma)
-------------------------------
We know that ${m \choose n}$ must be divisible by m if m is prime, and 0<n<m. This is because the result is integral, and nothing in the denominator cancels with m.

Also, if m is even, then ${m \choose 2}=\frac{m\cdot(m-1)}{2}$ m-1 must be odd, and so the result will have one less 2 in it's prime factorization than m. Thus ${m \choose n}$ for even m can never be divisible by m for all 0<n<m.

If m is divisible by 3 (but of couse not equal to 3) , then considering the expression ${m \choose 3}$ in a similar fashion tells us that the expression is not divisible by m.

Likewise for 5, 7, and so on.
-----------------------------
(end lemma)

Thus the only n that work for the problem are n such that (n+1) is a prime.

There are 14 such n:

2,4,6,10,12,16,18,22,28,30,36,40,42,46

[/hide]

Well I think you should have said like

[hide="ribbit"]
WHOA WHAT IF $p | m$??????

OH WAIT I GUESS THAT'S PRETTY TRICKY WHEN WE SAY ${ m \choose p }$

cause like we can think hmmm oh no p can't divide anything in the numerator except m! THE TURKEYS HAVE COME!!!
[/hide]

Edit--darn, if you want me to be all

[hide="articulate and all that"]
then suppose there exists a prime $p < m, p | m$.  Then ${ m \choose p }$ is not divisible by m because

$\frac{{ m \choose p }}{m} = \frac{(m-1)\cdots (m-p+1)}{p!}$

and nothing in the numerator of that is divisible by p.
[/hide]

Edit2--whoa I kinda missed that generalizing part of your solution, so your solution is actually right  :blush: . Yeah my excuse is that I really really need to go to sleep now. ;)

Um, are you talking about my solution? Sorry I'm not really sure what you're saying.  :blush:


I'm not really use to using a graphing calculator so if anyone can help me with the steps to this problem it'll be greatly appreciated.

the number of U.S military peronnel M (in thousands) on active duty for the years 1989 through 1996 is shown in the table, where t represents the time (in years), with t=0 corresponding to 1990.

charts
t   -1      0         1         2
M  2130  2044   1986   1807

t    3          4          5        6
M  1705   1611     1518     1472
questions are:
a, use the regression capabilities of a graphing utility to find a cubic model for the data.
b, use a graphing utility and model to create a table of estimated values of M.  compare the estimated values w/ the actual data
c, use the remainder theorem to evaluate the model for the year 2001.  even though the model is relatively accurate for estimating the given data, would you use this model to predict the number of military personnel in the future? explain

Do you have a TI-8x Family Calculator?
(x can be 3, 3+, 4, 4 SE, ...)

If so, here's how:
Open the statistical list editor:
Hit STAT, then the number 1.
This should bring you to a screen that has List Names at the top, and columns below the lists.

Let's make L1 our time list.

In each space, input the time values, and hit enter.
So your editor should look like this
L1     |L2      |L3      |L4      |L5      |L6     |
-1
0
2
3
4
5
6
--------------------------------------------

Now we want to input the M values. Let's use list two.

L1     |L2      |L3      |L4      |L5      |L6     |
-1      2130
0       2044
1       1986
2       1807
3       1705
4       1611
5       1518
6       1472
--------------------------------------------

Ok, we now have enough data for a regression.
Hit 2ND --> Quit
Hit STAT
Hit the right arrow cursor so that the STAT Menu shows CALC in bold.

You should now have a list of items.
Scroll down until you see CubicReg (item #6). This stands for cubic regression.

Hit Enter. This will bring you to the home screen and paste the function CubicReg on it.
This function needs arguments, namely, the lists to run the regression on.

We want to graph L1(time) as X values, and L2(number of ppl) as Y values.
Hit 2ND --> L1
Hit the comma key
Hit 2ND --> L2
Hit the comma key
Hit VARS
Hit Right Cursor
Hit enter over Y-VARS
Hit 1 for FUNCTION
Hit 1 to store the regression in function Y1.

Now hit enter.
Let the calculator run for a bit, it shouldn't take long.

You now have a function stored in the Y1 Space.
Hit the grey Y= button at the top of the calc.
Check to make sure you have a function.
Now, hit the window key.
Set these values:

Xmin = -1
Xmax = 6
Xscl = 1
Ymin = 1400
Ymax = 2200
Yscl = 100
Xres=1 (in general ALWAYS leave this as 1)

We're almost done.
Hit 2ND --> Stat Plot
Mouse over until 1: is bolded.
Hit Enter
Set the Xlist to L1, and the Ylist to L2
Set the mark to the middle one. (or whatever you want)
Set the type to the the upper right most type (scatter plot).
Finally, turn the scatter plot on by mousing over the ON option and hitting enter.

Here's a summary of what you just did:
Inputted data into to lists.
Ran a cubic regression to find a trend line.
Saved the trend line to the Y1 function.
Made a scatter plot with the data you inputted.

We can now hit graph.
This shows you the data points, and the line we made.

That should help with the rest of the problem.


For all numbers [i]p[/i] and [i]q[/i], where [i]p[/i] is not equal to 4. let [i]p[/i]//[i]q[/i] be defined as [i]pq[/i]/[i]p[/i]-4. For what value of [i]p[/i] does [i]p[/i]//7 = 21

This isn't a multiple choice question, but a grid-in. I tried to just plug in numbers, but that is tedious and takes some time. Is there a formula for this?

For all numbers p and q, where p is not equal to 4. let p//q be defined as pq/p-4. For what value of p does p//7 = 21

This isn't a multiple choice question, but a grid-in. I tried to just plug in numbers, but that is tedious and takes some time. Is there a formula for this?

there isn't a formula because it's already defined in the problem.
so in $p$//$7$ remember that 7 is the $q$ so we just plug in 7 to the defined equation. so

$\frac{7p}{p-4}= 21$

im pretty sure you know how to find p from here.
don't plug in numbers.

a high school way of solving the fraction would be
$21(p-4)=7p$
$21p-84=7p$
$-84=-14p$
$p=6$


I'm pretty sure you can do it without case analysis (at least, not very many cases), by partitioning.

If we did not have the requirement that $d\leq12$, there would be $\binom{18}{3}$ solutions. To eliminate the cases where $d>12$, we consider the first partition to be $d$; we see that when $d=12+k$, with $0<k\leq5$, there are $5-k$ remaining units to partition, and thus $\binom{6-k}{2}$ combinations. So we just subtract this, for all possible values of $k$, from the total amount (noting that when $k=5$, then $d=17$ and there's really only one choice for the rest: $a=b=c=0$):
\[\binom{18}{3}-\binom{5}{2}-\binom{4}{2}-\binom{3}{2}-\binom{2}{2}-1 \]
Whatever that evaluates to should be the answer, I think. I mean, yes, there are cases technically, but it is not at all hard or tedious to find $\binom{6-k}{2}$ for $0<k\leq5$..

I have a slight discrepency in my numbers:
[hide]We note that the number of solutions in nonnegative integers to $a+b+c+d=17$ is the number of ways to put $17$ ball into $4$ urns, which is $\binom{20}{3}=1140$. Similarly, the number of solutions in nonnegative integers to $a+b+c=k$ is $\binom{k+2}{2}$. Now we let $d$ range from $0$ to $12$, so $k$ ranges from $17$ to $5$. Then the number of solutions is

$\sum_{k=5}^{17}\binom{k+2}{2}= \sum_{k=7}^{19}\binom{k}{2}$

$=\sum_{k=2}^{19}\binom{k}{2}-\sum_{k=2}^{6}\binom{k}{2}$

$= \binom{20}{3}-\binom{7}{3}$
[/hide]

I ran a quick little Python script:
[code]counter = 0
for a in range(0,18):
	for b in range(0,18):
		for c in range(0,18):
			for d in range(0,13):
				if (a+b+c+d==17):
					counter+=1
print "There are " + str(counter) + " solutions."[/code]
And it returned:[code]adrian-petrescus-computer:~ adrianpetrescu$ python prob.py
There are 1105 solutions.[/code]
So you are right :(, and I am wrong. (Your answer evaluates to 1105, mine to 795)

[quote="Phelpedo"]How many nonnegative solutions are there to $a+b+c+d=17$, provided that $d \leq 12$?

Wondering if there's a solution not involving case analysis.[/quote]

Well, I don't know if there is one without case analysis, but here is mine.

[hide]You know that, the total number is $\binom{20}{3}$.

Next, we assume that for $d=13$, we get $\binom{6}{2}$
$d=14$, $\binom{5}{2}$
...
$d=17$, $\binom{2}{2}$.

So we add up all the disallowed, and we get $\binom{7}{3}$

So the answer is $\binom{20}{3}-\binom{7}{3}=1105$

$1105$[/hide]

There is actually a nice geometric/combinatoric way to do this, consider the tetrahedral numbers (natural extention of the triangular), and if you do it that way there is no casework.

In case anyone wondered, the number of nonnegative solutions to $x_{1}+x_{2}+\cdots+x_{r}=n$ is

[hide="formula"]$\binom{n+r-1}{r-1}$[/hide]

And yes, my solution is exactly mathgeniuse^ln(x)'s.


linear algebra
matrix
linear algebra theorems

Let $A=(a_{ij}| a_{ij}+a_{ji}=0 , a_{ii}=0 )$ be a nxn matrix(I don't know it's name in eglish, but in russian we say "kososimmetricheskaya" matrix).
What can we say about Jordan form of a such matrix?
What can we say about Jordan form of $A^{T}$ if we already know the Jordan form of $A$

In English, I'd say "skew-symmetric" or "antisymmetric".

You didn't say whether $A$ was real. If $A$ is real and skew-symmetric, then $iA$ is Hermitian. It follows then that $A$ is diagonalizable (thus answering your question about Jordan form), and that all of its eigenvalues are pure imaginary.

If $A$ is not real, I don't think we can draw such nice conclusions.


Let $ a,b,c >0$ and $ n$ is an integer greater than 1
Prove that

$ \frac {(a\plus{}b\plus{}c)^2}{a^2\plus{}b^2\plus{}c^2} \leq \sqrt[n] {\frac {2a}{a\plus{}b}} \plus{}\sqrt[n] {\frac {2b}{b\plus{}c}} \plus{}\sqrt[n] {\frac {2c}{c\plus{}a}} \leq 3$

Please help me. :(
I have posted this in IE open Question, but nobody help me.

For right inequality:

if we define $ x \equal{} \frac {2a}{a \plus{} b} ,y \equal{} \frac {2b}{b \plus{} c} ,z \equal{} \frac {2c}{c \plus{} a}$

by power mean inequality we have
\[ (\frac {x^{\frac {1}{n}} \plus{} y^{\frac {1}{n}} \plus{} z^{\frac {1}{n}}}{3})^n\le (\frac {x^{\frac {1}{2}} \plus{} y^{\frac {1}{2}} \plus{} z^{\frac {1}{2}}}{3})^2
\]
so it's enough to prove that
\[ A \equal{} \sqrt {\frac {2a}{a \plus{} b}} \plus{} \sqrt {\frac {2b}{b \plus{} c}} \plus{} \sqrt {\frac {2c}{c \plus{} a}} \leq 3
\]
by cawchy we know
\[ (A)^2\le (2(a \plus{} b \plus{} c))(\sum\frac {2a}{(a \plus{} b)(a \plus{} c)})
\]
so we must prove that
\[ 4((a \plus{} b \plus{} c)(\sum ab \plus{} ac))\le9(\sum a^2b \plus{} a^2c \plus{} 2abc)
\]
or
\[ 6abc\le \sum a^2b \plus{} a^2c
\]
that is AM-GM

What's about the left side?
Maybe it 's harder

The left inequality is wrong if we have a big $ n$

[quote="Dread Phantom"]Why? Can you show me one example? :huh:  :huh:[/quote]

 \[ n\rightarrow\infty , a\equal{}2,b\equal{}1,c\equal{}1\]

Emm , but this example is wrong.
I have tried to find several example but I can't find one.


LaTeX Asymptote AoPSwiki articles integration trigonometry parameterization

How can I have latex text wrap about an asymptote image? Let's say I want text on the left side, and an image to the text's right side.

You would wrap the text around in the same way as you would any image. The [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=wrapfig]wrapfig package[/url] will do this. The documentation is in .../tex/latex/wrapfig/wrapfig.sty at the end.
Adapting the AoPSWiki example at [[Asymptote:_Advanced_Configuration#Using_Asymptote_in_LaTeX]] to put the text next to the yellow smiley:
[code]\documentclass{article}
\usepackage[pdftex]{graphicx}
\usepackage{asymptote}
\usepackage{wrapfig}
\begin{document}
Hello.
I like to make pics with Asymptote like this one. This line carries on over the image:\\
\begin{wrapfigure}{r}{1in}
 \begin{asy}
  include graph;
  size(1inch);
  filldraw(circle((0,0),1),yellow,black);
  fill(circle((-.3,.4),.1),black);
  fill(circle((.3,.4),.1),black);
  draw(arc((0,0),.5,-140,-40));
\end{asy}
\end{wrapfigure}

It makes me happy, since I can still type my normal LaTeX stuff around it:
\(\int_0^{\pi}{\sin{x}}\,dx=2\)

More text that goes along the line but will be wrapped when it reaches the image\\
More text that goes along the line but will be wrapped when it reaches the image\\
More text that goes along the line but will be wrapped when it reaches the image\\
More text that goes along the line but will be wrapped when it reaches the image or carry on underneath it.
\end{document}[/code]
Do read the wrapfig documentation as it has a lot more parameters you can adjust.

How about if I want do this inside a table cell? (Like the the middle column of the table i was describing in the other thread)

Just put the asy code inside the cell as if it were text.

For some reason the formatting got all messed up.

Possibly for that particular cell you want to have
\begin{minipage}
your_cell_text&image_here
\end{minipage}
? Minipages are often useful.


linear algebra
matrix
inequalities
inequalities unsolved

For positive integers $ n,k$ such that $ 1\leq k\leq2n$ prove that:

$ \begin{pmatrix}2n+1 \\ k-1 \end{pmatrix}+\begin{pmatrix}2n+1 \\ k+1 \end{pmatrix}\geq2\frac{n+1}{n+2}\begin{pmatrix}2n+1 \\ k \end{pmatrix}$

Adding in both members $  2\left(\begin{array}{c}2n+1 k\end{array}\right)$ and using the recurrance formula for combinations, the inequality becomes
$  \left(\begin{array}{c}2n+3 k+1\end{array}\right)\geq 2\left(\begin{array}2n+1 k\end{array}\right)\cdot\left(\frac{n+1}{n+2}+1\right)\Leftrightarrow \frac{n+1}{(k+1)(2n-k+2)}\geq\frac{1}{n+2}\Leftrightarrow$
$ \Leftrightarrow (n-k)(n-k+1)\geq 0$
If $ k<n$ both factors are positive.
If $ k=n$ or $ k=n+1$ the product is zero.
If $ n+2\leq k\leq 2n$, both factors are negative, so the product is positive.


I'm going to start a math marathon. Here are the rules.

Rules:
1.Keep difficulty ranging from easy MOEM problems to problems from AMC10.
2.Always add a new problem after you post your solution.

Problem 1: Write 5.12929292929292...... as a common fraction.

I think we have a lot already. Maybe if another one dies, this would be a good idea, but for now I think 3 is good.

[quote="Thunder365"]...but for now I think 3 is good.[/quote]

Indeed.


An IV-bag with 1000 mL of fluid is delivering fluid to a patient at the
rate of 3 mL per minute. After the first three hours, how many milliliters
of fluid will remain in the IV-bag?

Obviously, 3 hours means 180 minutes.
Since the fluid is released at 3 ml a minute, 540 ml is released over 3 hrs. (180*3)
Since there were 1000 ml before, 1000-540(released)=460 ml.


geometry
3D geometry
sphere
probability
tetrahedron
symmetry
geometric transformation

Four points are chosen at random on the surface of a sphere. What is the probability that the center of the sphere lies inside the tetrahedron whose vertices are at the four points?

That does look a bit too small... I'll stay tuned for the proof.

Ummm....Notice that the tetrahedron contains the center of the sphere if the fourth point of the tetrahedron lies on the projection of the other 3 points about the center of the sphere. There are $2^{3}$ projected 3-points with respect to three of the vertices of the tetrahedron. By symmetry, the projections from three of the 3-point selection out of $\binom{4}{3}$ are also contained in the $8$.  The are a total of $32$ projected not necessarily distinct 3-point sets. So the answer is $\frac{4}{32}=\frac{1}{8}$.  Is this correct? By the way, I'm interested to see what you did, Farenhajt. :)

Masoud Zargar

I also got $\frac{1}{8}$.

After choosing two points, the attached image shows a cross section containing $P_{1}$,$P_{2}$, and the centre $O$. The third and fourth must be on different sides of this cross-section (probability = $1/2$). Call the line joining them $P_{3}P_{4}$, and their intersection with said cross-section $Q$. Now, if the tetrahedron is to contain $O$ then $\bigtriangleup P_{1}P_{2}Q$ must do the same. Therefore $Q$ is in the shaded area.

Q's location is evenly distributed across this cross section, so the probability of being in the shaded area is $\frac{\angle P_{1}OP_{2}}{2\pi}$. Note that the average angle between the first two points is $\frac{\pi}{2}$. (Why? consider the 1st point to be the north pole, any 2nd point is equally likely as its reflection across the equator). Thus the chance of crossing the shaded area = $\frac{\frac{\pi}{2}}{2\pi}= \frac{1}{4}$.

Thus the total probability is $(\frac{1}{2})(\frac{1}{4}) = \boxed{\frac{1}{8}}$.

What do you mean by $P_{3}$ and $P_{4}$ must be on different "sides?"

On the other side of the plane containing $P_{1},P_{2}$ and $0$

[quote="Phelpedo"]Four points are chosen at random on the surface of a sphere.[/quote]Sorry, but how is "at random" defined here? You really need to give this extra information for the problem to make sense... Maybe that also explains the different answers.

They are independently chosen relative to a uniform distribution.

This was set as a past year putnam problem. The key is to realize that, given any 3 points on the sphere, the only way for the tetrahedron to contain the center is for the 4th point to lie within the triangle formed by the antipodal points of the original 3 points.

Hence, it reduces to finding the expected area of a triangle (ratio) on the surface of a sphere. Given any 3 points A, B, C, their antipodes A' B' C' would yield 8 triangles that cover the surface of the sphere. Hence, the expected area (ratio) of the triangle would be 1/8.

After placing the three points $P_1,P_2,P_3$, the third point $P_4$ will enclose the center of the sphere if and only if it lies in the spherical triangle $P_1'P_2'P_3',$ such that the center lies on the straight line $QQ'.$ Its obvious that the probability is the area of $P_1P_2P_3$ divided by the area of the sphere. WLOG the area is 1. But since eight triangles, namely $P_1P_2P_3, P_1'P_2P_3,$ et al., with each having equal likelihood, the probability is $\boxed{1/8}.$


calculus
integration
real analysis
real analysis unsolved

Evaluate  $ \int_{0}^{1}\frac{\arctan\frac{x}{x\plus{}1}}{\arctan\frac{1\plus{}2x\minus{}2x^2}{2}}$

_______________________________

Azerbaijan Land of the Fire   :ninja:

You received responses the first time.  If you don't understand them, ask questions there.  Posting the same problem repeatedly is rude.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=309243


AIME
geometry
quadratics
incenter
area of a triangle
Heron\u0027s formula

Triangle  $ABC$ has $BC=20$. The incircle of the triangle evenly trisects the median $AD$. If the area of the triangle is $m \sqrt{n}$ where $m$ and $n$ are integers and $n$ is not divisible by the square of a prime, find $m+n$.

[hide="Solution"]WLOG $AB<AC.$

Say the incircle meets $BC$ at $E, AB$ at $F.$  Then $DE^2 = x(2x)$ by power of a point.  Similarily $DF^2 = 2x^2$.

It's well known we can write $CE, AF = s-c, s-a$, where $s$ is the semiperimeter.  Then $DE = |s-c-10|$, and $AF = s-a$.

So we have $(s-a)^2 = (s-c-10)^2$.  Solving, $b = 10$ or $a = c+10 \Rightarrow c = 10$.  We chose $c = AB < AC = b$, so $c = 10$.

Then $s = \frac{a+b+c}{2} = 15 + \frac{b}{2}$.  From Stewart's theorem, we can easily derive $2x^2 = \frac{b^2 + c^2 - 200}{9}$.

Finally, $(s-a)^2 = \frac{b^2 + c^2 - 200}{9}$.  Putting all variables in terms of $b$, we get the quadratic $b^2 -36b + 260$.  Solving, $b = 10$ or $b = 26$.  Obviously, $b = 26$ is what we want.

Moving to Heron's formula, Area = $\sqrt{(28)(8)(18)(2)} = 24\sqrt{14}$.  The answer is 038.[/hide]

really sorry for reviving old thread.

Singular:  but how would you prove "$ CE \equal{} s \minus{} c$? i think it invovles something with tangents but I wasn't sure. any help would be appreciated.

Notice that CE=CD, BD=BF, AF=AE. Then, a=BD+DC, b=CE+EA, c=AF+FB. Solve.

thanks. i completely overlooked equal tangents :wallbash:

[quote="Singular"][hide="Solution"]WLOG $ AB < AC.$

Say the incircle meets $ BC$ at $ E, AB$ at $ F.$  Then $ DE^2 \equal{} x(2x)$ by power of a point.  Similarily $ DF^2 \equal{} 2x^2$.

It's well known we can write $ CE, AF \equal{} s \minus{} c, s \minus{} a$, where $ s$ is the semiperimeter.  Then $ DE \equal{} |s \minus{} c \minus{} 10|$, and $ AF \equal{} s \minus{} a$.

So we have $ (s \minus{} a)^2 \equal{} (s \minus{} c \minus{} 10)^2$.  Solving, $ b \equal{} 10$ or $ a \equal{} c \plus{} 10 \Rightarrow c \equal{} 10$.  We chose $ c \equal{} AB < AC \equal{} b$, so $ c \equal{} 10$.

Then $ s \equal{} \frac {a \plus{} b \plus{} c}{2} \equal{} 15 \plus{} \frac {b}{2}$.  From Stewart's theorem, we can easily derive $ 2x^2 \equal{} \frac {b^2 \plus{} c^2 \minus{} 200}{9}$.

Finally, $ (s \minus{} a)^2 \equal{} \frac {b^2 \plus{} c^2 \minus{} 200}{9}$.  Putting all variables in terms of $ b$, we get the quadratic $ b^2 \minus{} 36b \plus{} 260$.  Solving, $ b \equal{} 10$ or $ b \equal{} 26$.  Obviously, $ b \equal{} 26$ is what we want.

Moving to Heron's formula, Area = $ \sqrt {(28)(8)(18)(2)} \equal{} 24\sqrt {14}$.  The answer is 038.[/hide][/quote]
I think you're wrong! Since DE^2=x.3x not x.2x although I don't know what x is

[quote="ghjk"][quote="Singular"][hide="Solution"]WLOG $ AB < AC.$

Say the incircle meets $ BC$ at $ E, AB$ at $ F.$  Then $ DE^2 \equal{} x(2x)$ by power of a point.  Similarily $ DF^2 \equal{} 2x^2$.

It's well known we can write $ CE, AF \equal{} s \minus{} c, s \minus{} a$, where $ s$ is the semiperimeter.  Then $ DE \equal{} |s \minus{} c \minus{} 10|$, and $ AF \equal{} s \minus{} a$.

So we have $ (s \minus{} a)^2 \equal{} (s \minus{} c \minus{} 10)^2$.  Solving, $ b \equal{} 10$ or $ a \equal{} c \plus{} 10 \Rightarrow c \equal{} 10$.  We chose $ c \equal{} AB < AC \equal{} b$, so $ c \equal{} 10$.

Then $ s \equal{} \frac {a \plus{} b \plus{} c}{2} \equal{} 15 \plus{} \frac {b}{2}$.  From Stewart's theorem, we can easily derive $ 2x^2 \equal{} \frac {b^2 \plus{} c^2 \minus{} 200}{9}$.

Finally, $ (s \minus{} a)^2 \equal{} \frac {b^2 \plus{} c^2 \minus{} 200}{9}$.  Putting all variables in terms of $ b$, we get the quadratic $ b^2 \minus{} 36b \plus{} 260$.  Solving, $ b \equal{} 10$ or $ b \equal{} 26$.  Obviously, $ b \equal{} 26$ is what we want.

Moving to Heron's formula, Area = $ \sqrt {(28)(8)(18)(2)} \equal{} 24\sqrt {14}$.  The answer is 038.[/hide][/quote]I think you're wrong! Since DE^2=x.3x not x.2x although I don't know what x is[/quote]
[quote="Singular"]Similarily $ DF^2 \equal{} 2x^2$.[/quote]This should be $ AF^2\equal{}2x^2$.[asy]defaultpen(fontsize(8));
pair A=26*expi(acos(61/65)), B=(20,0), C=(0,0), D=(10,0), E, F, I;
I=incenter(A,B,C);E=foot(I,B,C);F=foot(I,A,B);
draw(incircle(A,B,C));draw(A--B--C--A--D);dot(A^^B^^C^^D^^E^^F);
label("A",A,(1,1));label("B",B,(1,-1));label("C",C,(-1,-1));label("D",D,(0,-1));label("E",E,(0,-1));
label("F",F,(1,-1));label("$x$",A/6+D*5/6,(0,1));label("$x$",(A+D)/2,(0.5,-1));label("$x$",A*5/6+D/6,(0,-1));[/asy]

Sorry! I'm wrong! :blush: .
Here is my another solution:
AD^2=2(AB^2+AC^2)-BC^2/4
Now, let F,E,G are intersections of the incircle and AB,BC,AC, respectively. Then proving exactly similar as Singular did I have the following equation:
x^2(124-6x^2)=0 with x is one-third of AD.
Now, we just need to calculate AB,AC and applying Heron! :)
PS: I think it's still too long! Does anybody have a solution with under 10 lines??

I did it differently.  Let the center of the circle be $ O$.  Draw the perpendicular from $ O$ to $ AD$ to point $ G$, seeing that it divides $ AD$ into two equal segments (why?). Drawing $ OA, OD, OF,$ and $ OE$, we get congruent triangles $ AOG$  and $ DOG$.  So, we know that $ DO \equal{} AO$.  Since we also know that $ EO \equal{} OF$ and $ < DEO \equal{} < AFO \equal{} 90^{\circ}$, we know that $ \triangle{DEO}$ is congruent to $ \triangle{AFO}$.  So, $ AF \equal{} DE$, and since $ EB \equal{} FB$, it follows that $ AB \equal{} DB \equal{} 10$.  After that, labelling the point that circle $ O$ meets $ AC$ $ H$, I set $ AH \equal{} AF \equal{} b$, $ FB \equal{} EB \equal{} 10 \minus{} b$, and $ CE \equal{} CH \equal{} 10 \plus{} b$.  Therefore, $ AC \equal{} 2a \plus{} 10$.  Drawing down the altitude from $ A$ to $ BC$, I got an equation for $ b$, which yielded $ b \equal{} 8$.  Hence, $ AC \equal{} 2(8) \plus{} 10 \equal{} 16 \plus{} 10 \equal{} 26$.  Now, looking at isosceles triangle $ ABD$, we see that $ AD \equal{} \frac {3b}{\sqrt {2}} \equal{} \frac {24}{\sqrt {2}}$. Drawing down the altitude from $ B$, we get that it equals $ \sqrt {28}$.  Hence, the area of this triangle is $ \frac {\frac {24}{\sqrt {2}} \bullet \sqrt {28}}{2} \equal{} 12\sqrt {14}$.   Since $ CD \equal{} DB$ and both have the same altitude from $ A$, triangles $ ADB$ and $ ADC$ have the same area. Their combined area is therefore $ 12\sqrt {14}$, and our answer is thus $ 24 \plus{} 14 \equal{} \boxed{38}$.

less than $ 10$ lines!!!  :P

I'm assuming this is revivable since it is a contest thread.

[hide="Faster Solution Outline"]
We let AD = $3a$. Let the incircle be tangent to E on BC, F on AB, and G on AC.

By PoP, $DE = AF = a\sqrt2$. So this means $AG = a \sqrt2, CE = GC = a\sqrt2 + 10$. We also see that $AB = AD = 10$ by tangency theorems.

We can now apply Stewart, get that $a = 4\sqrt2$, and plug it in to get side lengths of 26, 10, and 20.

We finish with Hero to get the area as $24\sqrt{14} \rightarrow \boxed{038}$.

[/hide]

[hide=Easy for 15?]
(diagram from AoPS Wiki)
[asy]
size(300); pointpen=black;pathpen=black+linewidth(0.65); pen s = fontsize(10);
pair A=(0,0),B=(26,0),C=IP(circle(A,10),circle(B,20)),D=(B+C)/2,I=incenter(A,B,C);
path cir = incircle(A,B,C); pair E1=IP(cir,B--C),F=IP(cir,A--C),G=IP(cir,A--B),P=IP(A--D,cir),Q=OP(A--D,cir);
D(MP("A",A,s)--MP("B",B,s)--MP("C",C,N,s)--cycle); D(cir);
D(A--MP("D",D,NE,s)); D(MP("E",E1,NE,s)); D(MP("F",F,NW,s)); D(MP("G",G,s)); D(MP("P",P,SW,s)); D(MP("Q",Q,SE,s));
MP("10",(B+D)/2,NE); MP("10",(C+D)/2,NE);
[/asy]
Let $AP=x$. By Power of a Point, we have $AP\cdot AQ=x\cdot 2x=AF^2\implies AF=x\sqrt{2}$. Similarly, we have $DE=x\sqrt{2}$. By the two-tangent theorem we have $AC=AF+FC=DE+EC=10$, and we also have $AB=AG+GB=AF+BE=10+2x\sqrt{2}$. By LoC on $\triangle{ADC}$, we have \begin{align*}AC^2+CD^2-2\cdot AC\cdot CD\cdot \cos{C}=AD^2 \\ \implies 200-200\cos{C}=9x^2.\end{align*} By LoC on $\triangle{ABC}$, we have \begin{align*}AC^2+BC^2-2\cdot AC\cdot CB\cdot \cos{C}=AB^2 \\ \implies 500-400\cos{C}=100+40x\sqrt{2}+8x^2.\end{align*} Solving the system of equations gives $x=4\sqrt{2}$, so $AB=10+2x\sqrt{2}=26$. Now this is just Heron's: \[K_{ABC}=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{28\cdot 8\cdot 18\cdot 2}=24\sqrt{14}\longrightarrow \boxed{38}.\]
[/hide]

It really bothers me that nobody used coordinates
[hide=bash bash bash]
Let $a$, $b$, $c$, denote the lengths of $BC$, $CA$, and $AB$ respectively. We can obtain that $\triangle ABD$ is isosceles with vertex angle $B$ as in the above solutions, and so $c=CD=10$. Now, scale the diagram down by a scale factor of $5$ to obtain $a'=4$, $b'=2n$, $c'=2$.
----------------------------------------------------------------------------------------------------------------
We note that since the centroid of a triangle divides its medians into $2:1$ ratios, the centroid of $\triangle ABC$ lies on the incircle. We now use barycentric coordinates wrt $\triangle ABC$ ($A=(1,0,0)$, etc.):

denote the centroid as $G=(1:1:1)$, and the intouch points as $D$, $E$, and $F$, which yields that
$P=(x:y:z)\in (DEF)\iff a'^2yz+b'^2zx+c'^2xy=(x+y+z)((s'-a')^2x+(s'-b')^2y+(s'-c')^2z)$. Now, with $s'-a'=n-1$, $s'-b'=3-n$, and $s'-c'=n+1$, we substitute $G$ into the above equation to obtain
$$(4^2)(1)(1)+(2n)^2(1)(1)+(2^2)(1)(1)=(1+1+1)((n-1)^2(1)+(3-n)^2(1)+(n+1)^2(1))$$
$$16+4n^2+4=3((n-1)^2+(3-n)^2+(n+1)^2)$$
$$5n^2-18n+13=0$$
$$n=\frac{13}{5}, \cancel{1}$$
where $1$ is not a possible value of $n$, since it would fail to satisfy the triangle inequality.
--------------------------------------------------------------------------------------------------
Finally, scaling back up again, we have that $a=20$, $b=26$, $c=10$, so Heron's Formula yields that
$$[ABC]=\sqrt{s(s-a)(s-b)(s-c)}$$
$$=\sqrt{(28)(8)(2)(18)}$$
$$=24\sqrt{14}$$
and so the answer is $24+14=\boxed{038}$, and we're done.
[/hide]

Let $\omega$ denote the incircle, $\omega$ touch $AB, BC$ at $Z, X$ respectively, $AD$ meet the incircle at $E$ and $F$ such that $AE < AF$, and $AE = EF = FD = x$. It's clear that $$AZ^2 = Pow_{\omega}(A) = AE \cdot AF = x \cdot 2x$$ $$= DF \cdot DE = Pow_{\omega}(D) = DX^2$$ so $$AB = AZ + BZ = DX + BX = BD = 10.$$ Now, we can compute $$2x^2 = AZ^2 = (s - a)^2 = \left( \frac{b}{2} - 5 \right)^2$$ and the median formula implies $$36x^2 = 2b^2 + 2c^2 - a^2 = 2b^2 - 200.$$ Solving the system and checking the two possible values for $b$ against the Triangle Inequality yields $b = 26$. Now, Heron's gives $$[ABC] = 24 \sqrt{14} \implies \boxed{038.}$$

[b]Remark:[/b] The POP argument is easy to spot. Once we realize that we have an equation relating $b$ and $x$, it becomes natural to utilize the median formula, as it relates $AD = 3x$ to the side lengths of $ABC$.

Nice and simple.

Let $PQR$ be the contact triangle of $ABC$, and WLOG set $AB<AC$ [s]because we are Americans here[/s] oh wait, the triangle is obtuse.

Two power of a point arguments yield
\begin{align*}
AR^2&=(s-a)^2=\frac 29 AD^2 = \frac 1{18}(2b^2+2c^2-a^2) \\
DP^2 &= \left(\frac 12 a-s+b\right)^2 = \frac 29 AD^2 = \frac 1{18}(2b^2+2c^2-a^2).
\end{align*}
Simplifying, $$9(b+c-a)^2=9(b-c)^2=2(2b^2+2c^2-a^2).$$ From the first two equations, we find $a=2c$, and from the second display, $$9(b-c)^2=2(2b^2-2c^2) \iff (b-c)(5b-13c) = 0.$$ So $a=20, b=26$, and $c=10$. Heron yields $24\sqrt{14}$, so extract $\boxed{038}$.

[hide=Killed with PoP and Stewarts. ]

[asy]
size(300); pointpen=black;pathpen=black+linewidth(0.65); pen s = fontsize(10);
pair A=(0,0),B=(26,0),C=IP(circle(A,10),circle(B,20)),D=(B+C)/2,I=incenter(A,B,C);
path cir = incircle(A,B,C); pair E1=IP(cir,B--C),F=IP(cir,A--C),G=IP(cir,A--B),P=IP(A--D,cir),Q=OP(A--D,cir);
D(MP("A",A,s)--MP("B",B,s)--MP("C",C,N,s)--cycle); D(cir);
D(A--MP("D",D,NE,s)); D(MP("T",E1,NE,s)); D(MP("F",F,NW,s)); D(MP("E",G,s)); D(MP("P",P,SW,s)); D(MP("Q",Q,SE,s));
MP("10",(B+D)/2,NE); MP("10",(C+D)/2,NE);
[/asy]
Let $AD=3x$. We quickly see $AE^2=DT^2=AF^2=2x^2$, by PoP. Then $AC=10,$ and $AB=x\sqrt{2}+10$. By Stewarts on $\Delta ABC$, we see $x=4\sqrt{2}$, then we have a $10-20-26$ triangle, and by herons we get $\boxed{038}$

Thanks to [b]Evan Chen[/b] for the walkthrough. Use the diagram in #7, but with E as the unlabeled point and T as the previous point E.

By PoP, we have $AE^2 = 2x^2 = AF^2$, and also $DT^2 = 2x^2$, so $AE = AF = DT$. Thus, $DT + TB = FB + AF = 10 = AB$, and $CE = CT = 10+x\sqrt2, AE = AF = x\sqrt2$. This means that $AC = 10+2x\sqrt2$, and Stewart's Theorem gives $x = 4\sqrt2$. Applying Heron's formula gives $24\sqrt{14} \implies \boxed{38}$.

Finally began doing computational geometry!

First of all define some stuff.

Let $\omega$ be the incircle of $\Delta ABC$.
$P$ is the point of tangency of $\omega$ with $AB$, $Q$ is the point of tangency of $\omega$ with $CA$, $R$ is the point of tangency of $\omega$ with $BC$.
$X$ is the first intersection of $AD$ with $\omega$, and $Y$ is the second intersection of $AD$ with $\omega$.
Let $d=AD$, $b=AC$, $c=AB$.

By Power of a Point,
\[AX\cdot AY=2\left(\frac{d}{3}\right)^2=AQ^2
=\left(\frac{b+c+20}{2}-20\right)^2=\left(\frac{b+c-20}{2}\right)^2.\]
Since $A$ and $D$ have the same power with respect to $\omega$, $AP^2=DR^2$, so this implies that $AB=BD=10$.

By Stewart's Theorem, we may prove that
\[2d^2=b^2-100\]
(simply plug and chug). Since $c=10$, use the displayed PoP expression to get
\[\frac{2d^2}{9}=\left(\frac{b-10}{2}\right)^2.\]
Substitute $b^2-100$ for $2d^2$, solve for $b$ in the resulting quadratic ($b^2-36b+260=0$, simplified) so $b$ is $10$ or $26$.

$10$ cannot work, as sides of $10$, $10$, $20$ violate the Triangle Inequality. So $AC=26$.

Now compute with Heron's Formula, to obtain that the area is $\boxed{24\sqrt{14}}$. $\blacksquare$

[hide=solution]
Let $AD=3x$, and let $\triangle TEF$ be the intouch triangle of $\triangle ABC$. Let $AD$ intersect the incircle at 2 points, $X$ and $Y$ such that $X$ is between $A$ and $Y$. Then $AX=XY=YD=x$.

Note that $AF^2=AE^2=AX\cdot AY=2x^2$ and $DT^2=DY\cdot DX=2x^2$, so $AE=AF=DT=x\sqrt{2}$. Note that \[AF=s-a=s-20=DT=10-(s-b)=10-s+b,\]
so $b=2s-30=20+b+c-30$. Hence, $c=10$.

By Stewart's Theorem, \[2AD^2=2(3x)^2=18x^2=b^2+c^2-\frac{a^2}{2}=b^2-100,\]
and $\frac{1}{2}(b-c)=x\sqrt{2}$, so $2x^2=\frac{1}{4}(b-c)^2$. Hence, plugging this back into the first equation gives \[b^2-36b+260=(b-26)(b-10)=0,\]
so $b=26$. ($b\ne 10$ because that would not satisfy the triangle inequality.)

Therefore, by Heron's formula, the area of $\triangle ABC$ is $24\sqrt{14}$, and our answer is $24+14=\boxed{38}$.
[/hide]

Solved with hint from [b]Trigonometry in the AIME.[/b]
[hide=sketch]The key is power of a point, then Stewart's. Indeed, DE^2=DQ*DP=AP*AQ=AF^2, so DE=AF=AG and from CF=CE=c we have that CA=CD=10. Now BG=BE=20-c and AG=DE=10-c, so AB=30-2c and by Stewart's (or median formula but i dont have it memorized) we get that (3m)^2*20+20*10*10=10^2*10+(30-2c)^2*10. On the other hand, from pop we had that 2m^2=(10-c)^2, and solving we get that c=2 (c=10 makes a 10-10-20 triangle, impossible). Now plugging in c to get our side lengths we get an area of 24sqrt14 for a final answer of $\boxed{038.}$[/hide]

Alternatively, I just noticed it was also in OTIS


geometry
3D geometry
tetrahedron
circumcircle
inequalities
geometry proposed

Let $ ABCD$ be a tetrahedron with the centroid $ G$ and the circumradius $ R$. Prove that
$ GA \plus{} GB \plus{} GC \plus{} GD \ge 4R$

First, you posted incorrectly, $ GA\plus{}GB\plus{}GC\plus{}GD\le 4R$
Second, inequalties in geometry don't usually interest Mathlinkers because they don't really contain the beauty of geometry.  :)


From 11 positive integer scores on a 10-point quiz, the mean is 8, the median is 8, and the mode is 7.  Find the maximum number of perfect scores possible on this test.

[hide]
For every 10, there is at least one 7.
So, no more than five tens.
5*10+6*7=92 and average is more than 8.
So, no more than four tens.
4*10+5*7 and two other numbers. For median to be 8, both must be greater than seven.  However, then the mean would be greater than 8.
4*10+6*7 must be added to 6 for mean to be 8.  Then, median is not 8.
4*10+7*7=89 and mean is greater than 8
So, no more than three tens.
3*10+9*2+8+4*7+4=88. Mean is 8, and so is median.  7 is the mode. So, this fits, and maximum perfects is 3.
[size=150]3[/size]
[/hide]

For this problem, as with all problems of this kind, I usually write out the sum of all the terms, fill in the median, and leave blanks in the empty spaces... you quickly figure out that you cannot have more than 4 10s, and 4 itself doesn't work, leaving your answer at three (the sequence would be 5,7,7,7,7,8,8,8,10,10,10, and everything I did was in numeric order to keep things in order)


superior algebra
superior algebra unsolved

Let $ A$ be a ring, $ P_1,\ldots,P_r$ prime ideals, $ I$ an ideal, and $ x$ an element of $ A$. If $ xA \plus{} I \nsubseteq P_1 \cup \cdots \cup P_r$ then there is a $ y \in I$ such that $ x\plus{}y \notin  P_1 \cup \cdots \cup P_r$.

[quote="yannick"]Let $ A$ be a ring, $ P_1,\ldots,P_r$ prime ideals, $ I$ an ideal, and $ x$ an element of $ A$. If $ xA \plus{} I \nsubseteq P_1 \cup \cdots \cup P_r$ then there is a $ y \in I$ such that $ x \plus{} y \notin P_1 \cup \cdots \cup P_r$.[/quote]I suspect that the assumption is too weak to invoke the implication just looking at the case $ r\equal{}1$. Presumably you meant the following changed version which is a classical exercise in commutative algebra:[quote]If $ x \plus{} I \nsubseteq P_1 \cup \cdots \cup P_r$ then there is a $ y \in I$ such that $ x \plus{} y \notin P_1 \cup \cdots \cup P_r$.[/quote]

What you wrote is more like a classical exercise in set theory  :wink:

In fact I should have written:
[quote]Let $ A$ be a ring, $ P_1,\ldots,P_r$ prime ideals, $ I$ an ideal, and $ x$ an element of $ A$. If $ x \plus{} I \nsubseteq P_i$ for every $ i$, then there is a $ y \in I$ such that $ x \plus{} y \notin P_1 \cup \cdots \cup P_r$.[/quote]What about this problem within the field of commutative algebra?

I checked again and I stated the problem correctly....

[b]Source:[/b] Matsumura: Commutative ring theory, Ex 16.8

[b]Chapter subjects:[/b] Regular sequences, quasi-regular sequences, depth (including the approach with Koszul complexes) and grade.


Is it possible to do long division in Latex?? If so, how? Thanks!

If it's the format that you're referring to , it's possible in a couple of ways - The first way is to click on the "Code" option here in the forum . If you're using texnic center , I think the \begin{verbatim} and the \end{verbatim}
command should do you fine.This forum . however , doesn't seem to accept the \begin{verbatim} command....Anyone has any idea why it's so ?

EDIT : I dont get why you need long division...... :?

If you mean polynomial long division (and/or synthetic division) then try the polynom package. Full details and a demo can be found [url=http://www.ctan.org/tex-archive/help/Catalogue/entries/polynom.html]here[/url]


The following equation was written on the chalkboard   Δ + Δ + ◊ + ¤ – Δ = Ω
If Δ = 4, ¤ = 3 and Ω = 12, what is the value of ◊ ?

[hide]Substitute all of the information in and you get:
4 + 4 + ◊ + 3 - 4 = 12
7 + ◊ = 12
◊ = 5[/hide]

[hide]Substituting, we get $ 4+4+Δ+3-4=12$, then $ 11+Δ=16$ then $ Δ=5$[/hide]


inequalities
geometry
incenter
geometry proposed

[quote="gemath"]Given are triangle $ABC$ with centroid $G$ and incenter $I$ prove that
$\max\{IA,IB,IC\}\ge\min\{GA,GB,GC\}$
 :D[/quote]
I think this inequality is true for any two points in the triangle and it is not hard to prove :wink: .

That's right Tiks  :D , It isn't hard to prove but it is a nice inequality for triangle
can you post your solution?

[quote="gemath"]
Can you post your solution?[/quote]

Sure! :wink:

So,we have two points $X$ and $Y$ in the $ABC$.Let's suppose that point $X$ is in the triangle $AYB$,then
\[2min(AX;BX;CX)\leq AX+BX\leq AY+BY\leq 2max(AY;BY;CY) \]
so :D
\[min(AX;BX;CX)\leq max(AY;BY;CY) \]
:wink:

Nice solution but actually
$max\{YA,YB,YC\}\ge min\{XA,XB,XC\}$ only need $X$ inside triangle , $Y$
don't need inside! :)

[quote="gemath"] actually
$max\{YA,YB,YC\}\ge min\{XA,XB,XC\}$ only need $X$ inside triangle , $Y$
don't need inside! :)[/quote]
The same solution works for that case too :wink: .

[quote="gemath"]$\max\{IA,IB,IC\}\ge\min\{GA,GB,GC\}$.[/quote]
For any triangle $ABC$ with sides $a,b,c$, angle-bisectors $w_{a},w_{b},w_{c}$, incenter $I$, centroid $G$ and area $\Delta$, we have

$3\max\{IA,IB,IC\}\geq3\max\{GA,GB,GC\}$

$\geq2\max\{w_{a},w_{b},w_{c}\}\geq GA+GB+GC\geq IA+IB+IC$

$\geq\sqrt{\frac{a^{2}+b^{2}+c^{2}}{2}+2\sqrt{3}\Delta}\geq3\min\{GA,GB,GC\}$.
[hide="Reference."]
The fourth inequality was proved by Zhen Wang in 1994. see:

Ji Chen, Congjie Chen, Linear inequalities in a triangle (Chinese), Geometric Inequalities in China (Edit by Zun Shan, Ji Chen, Yao Zhang, Shiguo Yang), Jiangsu Education Publishing House, 1996, pp. 87-110.[/hide]


Find, in terms of n, the values of $ x$ and $ y$ so that

$ \binom {x}{y} \equal{} \sum_{k \equal{} 0}^{n \minus{} 2} \binom {n}{k} \binom{n}{k \plus{} 2}$

Hmm, this is interesting.

[hide="Hint"]
Do you know the closed form of $ \sum_{k\equal{}0}^{n}\binom{n}{k}^2$ (or more importantly, how to prove it)? Because it's really the same thing.
[/hide]

[hide="Solution w/ combinatorial argument"]

Consider a pile of $ 2n$ objects. We split them into two piles of $ n$ objects. Now, suppose we choose $ n\minus{}k$ from the first pile and $ k\plus{}2$ from the second pile. We've then chosen $ n\plus{}2$ objects. When we sum over $ k$ from $ 0$ to $ n\minus{}2$, we're counting all the ways in which this can happen.

This happens in $ \sum_{k\equal{}0}^{n\minus{}2}\binom{n}{n\minus{}k}\binom{n}{k\plus{}2}$. Using the identity $ \binom{n}{n\minus{}k}\equal{}\binom nk$, we have $ \sum_{k\equal{}0}^{n\minus{}2}\binom nk\binom{n}{k\plus{}2}$.

However, we know that we can choose $ n\plus{}2$ from $ 2n$ in $ \boxed{\binom{2n}{n\plus{}2}}$ ways.[/hide]

Thanks for the solution. This problem has annoyed me for quite some time.

And no, I don't know how to close $ \sum_{k\equal{}0}^{n} \binom {n}{k}^2$

You're welcome :)

Try finding the closed form of $ \sum_{k\equal{}0}^{n}\binom{n}{k}^2$, after having read the solution I posted above to the other problem. The idea is similar.

[hide="Hint for the first step"]
Work backwards by first using the symmetry identity $ \binom{n}{n\minus{}k}\equal{}\binom nk$ on the sum.
[/hide]

I had already figured, using your combinatorial idea, that it's $ \binom {2n}{n}$.
I just thought there was an algebraic proof or something alike. Was that the only way to prove it?

Yeah, $ \binom{2n}n$ is right.

Here is the algebraic proof of it, I guess:

Consider the polynomial $ (x\plus{}1)^n\cdot(x\plus{}1)^n$.

$ (x\plus{}1)^n$ expands to $ \sum\binom{n}{k}x^k$.

The coefficient of $ x^n$ in this mess of stuff is given by taking all the terms $ x^k$ in the first one and multiplying them by the terms $ x^{n\minus{}k}$ in the second. The coefficient of $ x^k$ in the first is, of course, $ \binom nk$. The coefficient of $ x^{n\minus{}k}$ in the second is $ \binom{n}{n\minus{}k}\equal{}\binom{n}{k}$.

Therefore, the coefficient of $ x^n$ is $ \sum\binom{n}{k}^2$.

However, we also obviously know that the coefficient of $ x^n$ in $ (x\plus{}1)^{2n}$ is $ \binom{2n}{n}$, so the conclusion follows.

(This is also a special case of Vandermonde's identity).

Wow, that is some brilliant proof.
I tried half an hour on induction using $ \binom{n}{k}\plus{}\binom{n}{k\plus{}1}\equal{}\binom{n\plus{}1}{k\plus{}1}$ without much avail

You could also use a block-walking argument. Below, I'm showing the case for $ n \equal{} 2$:
[hide]
Suppose that you wanted to walk from point A to point B using only steps going up and to the right:
  ____B
 |_|_|
 |_|_|
A

You could say that the number of paths from A to B is given by $ \binom{2(2)}{2}$, because there are that many ways to get to order the steps going up and the steps going to the right.
But you could also argue that the number of steps from A to B is the same as the number of steps from A to some arbitrary point C and then from C to B.
We will consider three possibilities:
Case 1. Choose C as the upper left corner. There are $ \binom{2}{0}$ ways to go from A to C and $ \binom{2}{0}$ ways to go from C to B.
Case 2. Choose C as the center of the diagram. There are $ \binom{2}{1}$  to go from A to the new C and $ \binom{2}{1}$ ways to go from C to B.
Case 3. Choose C as the lower right corner. There are $ \binom{2}{2}$ ways to go from A to the new C and $ \binom{2}{2}$  ways to go from C to B.
Because we counted the same thing in two different ways, the two expressions for the number of ways must be the same; that is, $ \binom{2(2)}{2} \equal{} \binom{2}{0}^2 \plus{} \binom{2}{1}^2 \plus{} \binom{2}{2}^2$.
[/hide]

You could tweak the proof to cover a general square path, which would prove the case for all $ n$.


List some of your favorite TV shows.  Here are some of mine:

[list]Whose Line Is It Anyway?
Family Guy
Scrubs[/list]

Saturday Night Live

The Tonight Show

American Idol


trigonometry
vector
parameterization
parametric equation

A man volunteers to test out a new space capsule for NASA. They use an explosive charge to fire him up into the air with an initial velocity of 350 m/s at an angle of 60 degrees above the horizontal. How high up in the air does the man go? How long does it take him to reach this highest point.

I solved the first part using the maximum height formula, but I am having trouble finding the time.

Well, all we need to concern ourselves with is $ v_{y}\equal{}v\sin{\theta}\equal{}350\sin{60}\equal{}175\sqrt{3}$.  So now we have the initial velocity in the y-direction.  Now we know that at its maximum height, $ v_{f}\equal{}0$.  So now we use the formula $ v_{f}\equal{}v_{i}\plus{}at$ and plug in: $ 0\equal{}175\sqrt{3}\plus{}(\minus{}9.8)t$.  Solve for $ t$ and we get $ t\equal{}30.9\mathrm{s}$.

you forgot about air resistance... :wink:

In problems like this, air resistance is commonly neglected, and I am pretty sure that is the purpose of the problem.

In projectile problems like this, always start by writting the parametric equations for the position,

$ x \equal{} v_o \cos \alpha \cdot t$
$ y \equal{} v_o \sin \alpha \cdot t \minus{} \frac{1}{2}gt^2$

and for the velocity,

$ v_x \equal{} v_o \cos \alpha$
$ v_y \equal{} v_o \sin \alpha \minus{}gt$

Now, what happens to the velocity vector when the projectile reaches the maximum height?

[hide="Answer"]At the maximum height, $ v_y \equal{} 0$, and so $ t \equal{} \frac{v_o \sin \alpha}{g}$. To find the maximum height, just plug this expression for time in the parametric equation above for y.[/hide]

I often find that using the range equations is somewhat boring.  Yes their derivations are simple and plugging in is straightforward but it takes the fun out of solving the problem.


symmetry
probability
geometry
geometric transformation
reflection
circumcircle

In a triangle $\Delta ABC$, we know that $\angle A > \angle B > \angle C$. Let $O$ be the midpoint of the circumscribed circle, and $I$ the midpoint of the inscribed circle, and call $D$ the midpoint of $BC$.
Let $M$ be the midpoint of the arc $BC$, not containing $A$. Let $F$ be a point on $ID$, such that $D$ is the midpoint of $IF$. Let $Q$ be the intersection of $MF$ with the circumscribed circle.

(a) Show that $Q$ lies on the arc $AC$, not containing $B$.

(b) Show that $AQ-BQ+CQ =0$.

Hmmm seems to have been pretty tricky! Part a is pretty easy, so I'll just do part b (assuming part a).

Let P be the point on BQ such that IP || MQ. We first show that AIPB is cyclic--this is true since $\angle IPQ = \angle PQM = \angle BAI$. Then let N be the midpoint of arc AB. It is well-known that NA = NB = NI, so we also have NA = NP. Since $\angle NQA = \angle NQP$, we also then have QA = QP. Now pick the point R such that RBQC is a parallelogram. By virtue of the fact that QM bisects $\angle BQC$, we have BP = BR = QC. Hence QB = QP + BP = QA + QC.

[quote="probability1.01"]Now pick the point R such that RBQC is a parallelogram. By virtue of the fact that QM bisects $\angle BQC$, we have BP = BR = QC. Hence QB = QP + BP = QA + QC.[/quote]
Maybe i have a blind spot but didnt get this part :blush:  (particulary where do you use that D is the middle of IF). Will you please clarify?

I forgot to note that R must lie on IP due to all the symmetry with respect to D.

NIce!!

I don't think you needed N though: <QAP = <QPA follows from <QPA = <CQP + <ABI = (<BAC + <CBA)/2 and <AQP = <BCA

Also, it appears that this problem is supposedly very hard  :P

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=42577
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=42568
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15773

I'm very surprised to hear that ThAzN1, I would never guess that this problem comes from a Chinese Olympiad...

Perhaps from one of the first rounds, which would still surprise me.
I've always heared that Chinese problems are extremely hard, but if you would have told me it came from a British Olympiad (the first example that popped into my head) right after I solved it, I would have believed you.  :lol:

Sure silouan. I hope there's no mistake... It looks a lot like probabilty 1.01's proof, but we seem to be solving the problem in different directions  :lol:

(b) Draw a point $J$ on segment $\left[BQ\right]$, with $\left|QJ\right|=\left|AQ\right|$.

Construct the line through $B$, parallel to $QC$, and call $K$ the point of intersection with the line $IJ$.

Because $\Delta QAJ$ has two equal side lengths, we see that $\measuredangle JAQ = \measuredangle QJA = \measuredangle CBI+\measuredangle IAC$

Now, we can easily deduce that

\begin{eqnarray}\measuredangle JAI &=& \measuredangle JAQ+\measuredangle QAC+\measuredangle CAI \\ &=& \measuredangle CBI+\measuredangle IAC+\measuredangle QAC+\measuredangle CAI\\ &=& \measuredangle CBI+\measuredangle QBC \\ &=& \measuredangle QBI \\ &=& \measuredangle JBI \end{eqnarray}

and thus obtaining that $AIJB$ is a cyclic quadrangle. This leads us to the conclusion $\measuredangle BJK = \measuredangle BAI = \measuredangle BQM$, so it's now clear that $IJ \parallel QM$.

Now, call $L$ the intersection of lines $IJ$ and $DM$. It's clear that $\Delta IDL \cong \Delta FDM$, so $L$ is the image of $M$ under the reflection with respect to point $D$. Similarly, $I$ is the image of $F$, and $B$ is the image of $C$.
So the line $IJ$ is the image of line $MF$, and it follows that $K$ is the image of $Q$.

Also note that $\angle BKJ = \angle BJK$, it's easy to see why.

We can now see that \[\left|CQ\right| = \left|BK\right| = \left|BJ \right| = \left|BQ \right|-\left|AQ \right|\]

And so our problem is no longer a problem.  :)


Let be given $32$ positive integers with the sum $120$, none of which is greater than $60.$ Prove that these integers can be divided into two disjoint subsets with the same sum of elements.

We suppose that it's impossible to do the sets.
We suppose that every number is different.
Every number is less than 60, then, they have different congruences (mod 60). If we have a number with congruence $ a$ and another with congruence $ \minus{} a$, the result becomes trivial (we put the two numbers on a set, and all the others in the another one, the both groups have 60 as sum of elements), then we know that the congruence of the numbers are on the set:  $ {0,\pm 1,\pm 2,...,\pm 29, 30}$
But, there are 32 numbers, and there are 31 different congruences, then there are two numbers that have the opposite congruence, and we put that numbers on a set, and the others on the another one.
Then, there are at less two equal numbers.
(to be continued)

If anything is not understood, let me know, i'm sorry for having a poor english.

"We suppose that every number is different."

Then we cannot make 120  :oops:  sum of positive 32 integers.

[quote="Justyo"]...then we know that the congruence of the numbers are on the set:  $ {0,\pm 1,\pm 2,...,\pm 29, 30}$
But, there are 32 numbers, and there are 31 different congruences, then there are two numbers that have the opposite congruence, and we put that numbers on a set, and the others on the another one.
...[quote]
note that 2 numbers can't both have congruence 0, cause those 2 numbers would have to be 60 and 60, which already sum to 120.

I've seen what I post after I posted, and I realize the stupid thing I made.
Later, I found an strategy, but I didn't make a demostration of this... Since you remember me this problem, I will try to remember the strategy for you:
[hide]You call $ a_1 \leq a_2 \leq ... \leq a_{32}$ the numbers. You put $ a_{32}$ in a first subset. In the next step, you put $ a_{31}$ and $ a_{30}$ in a second subset. If $ a_{31}\plus{}a_{30} > a_{32}$ then we put $ a_{29}$ in the subset with $ a_{32}$, if not, we put $ a_{29}$ with $ a_{31}$ and $ a_{30}$. In every step, we put $ a_k$ in the subset with the smaller sum. If the two subsets are equal, we can choose, it's the same thing. When we finish (we put $ a_1$) the two subsets are equal...[/hide]
If I have a good memory, this is the strategy, and it look it works, but I don't really know how to prove it, and I totally forgot the problem after doing this... I post this because no one post nothing, then maybe someone gets inspired by my strategy :p
(And as I said in posts before, excuse my poor english...)


A crew of 80 men can do a job in 24 days. If the contractor increases the work force by 1/2 and productivity is unchanged, how many days will be saved by adding the additional workers?

Answer:
[hide]8.0[/hide]

[hide="Solution"]The $ 80$ men each contribute $ 24$ days of work. Therefore, $ 80\times 24 \equal{} 1920$ days of work are put into the project. An increase in the workforce by $ 1/2$ yields $ 120$ men. If Therefore, since productivity is unchanged, each man works $ 1920/120 \equal{} 16$ days, so the job will be done in $ 16$ days now. $ 24 \minus{} 16 \equal{} \boxed{8}$ days are saved.[/hide]

[hide="A slightly more straightforward solution"]
We have the ratio men:days as 80:24.
We want there to be 1.5 more men, right? So we multiply the mens side by 1.5. Since we have more men to work on the same amount of job, the days are going to decrease, right? So we divide 24 days by 1.5 since the two are proportional. Now if we have 80x1.5=120 men, we need 24/1.5=16 days to work on the project. So 24-16=[b]8 [/b] days were saved [/hide]

If the contractor increases the work force by 1/2 this mean  the men will become 80 + 40 =120

(80)  men can do a job in   (24 )

(120) men can do a job in   (x)

then 120x =1920

therefore, x=16

the contractor will be saved  24-16=8 days

salahcool  :gathering:


How to prove this??

lim sup (an+bn)=<lim sup(an)+lim sup (bn)

$ s\equal{}\lim \sup s_n$ iff

$ (i): \forall \epsilon > 0, \exists N^{(s)}_\epsilon$ s.t. $ n\ge N_\epsilon \to s_n < s\plus{}\epsilon$
$ (ii): \forall \epsilon >0, \forall N, \exists N'>N$ s.t. $ |s_{N'} \minus{} s |< \epsilon$

$ (i) \equal{} \lim \sup s_n \le s,(ii) \equal{} \lim \sup s_n \ge s$

fix $ \epsilon > 0$, then $ \exists N\equal{}\max (N^{(a)}_{\frac{\epsilon}{2}},  N^{(b)}_{\frac{\epsilon}{2}})$

$ n \ge N \to a_n \plus{}b_n < a\plus{} b \plus{} \epsilon \to \lim \sup (a_n \plus{} b_n ) \le a \plus{} b$

if we could prove (ii) for general sequences, we would actually have $ \lim \sup (a_n \plus{}b_n ) \equal{} a \plus{} b$

but we know this isnt true, for instance, $ a_n \equal{} (\minus{}1)^n, b_n \equal{} \frac{1}{2} (\minus{}1)^{n\plus{}1}$

then $ a_n \plus{}b_n \equal{} \minus{}b_n$ which will give us our contradiction


Help my old brother to get this nonlinear autonomous third system or  is called The Lorenz Equation.
$\frac{dx}{dt}=\sigma(-x+y)$
$\frac{dy}{dt}=rx-y-xz$
$\frac{dz}{dt}=-bz+xy$
where $\sigma$ , $r$ , and $b$ is constant.

What do you mean by "get"?

This is a nonlinear autonomous system. Finding closed-form solutions for $x,y,z$ in terms of $t$ would be painful and complicated at best, with a reasonable likelyhood of being impossible.

Is perhaps the question to find and characterize (as stable, unstable, or whatever) all the stationary points of the autonomous system?


I heard the University of California branches and CSULB require a year of visual arts. Is that true? Though I am an American citizen, I live in India. Unfortunately, my schools never had that facility (visual arts). Does that make me ineligible?
Also, I want to know what the universities in California look for while giving scholarships. Actually, my family is really poor(well, at least when compared to people in USA) and cannot afford the fees unless I get plenty of scholarships or federal loans. But then, I do not know whether I am eligible for federal loans since I haven't lived in USA for a year. Though my SAT score is quite good
(2250-Math 800, Reading 760, Writing 690) my GPA is around 2.8.
The only reason I am applying to universities in California is that I will be treated as an in-state student after my first year. I do not have any outstanding achievements except having made it to the IMO selection camp once (didn't do well at the camp). Well, I could attach two articles to my application (I can only wish they were difficult enough to be called research papers).
Any information in this matter will be really appreciated and, to be honest, is badly needed. Also let me know which universities I should apply to.

Regarding your first question (the only one I have enough experience to answer), there is a requirement for one year of visual and performing arts.  As an out of state student, I did not know about this requirement and only took a half year of music theory.  When I tried to submit my application, it warned me that I was missing this requirement and said to explain in the Additional Information section.  I wrote that my school only required a half a year of visual and performing arts and that I played the piano outside of school and was accepted to Berkeley and UCLA.

They wouldn't penalize you for something you didn't have access to, so explain that in the Additional Information section.  Also if you have extracurricular experience with visual and performing arts, it could be worth mentioning there also.


Prove that if $\overline{abc}$ is prime, then $b^{2}-4ac$ is not a perfect square.

[quote="harazi"]By Cohn's theorem, $ax^{2}+bx+c$ is irreducible in $Q[X]$ and so it cannot have rational roots. Thus conclusion.[/quote]
What is Cohn's Theorem?


Anyone here have any programs for a TI-83+ in TI-BASIC that they could show me?

how do you put the programs on the calcultor. :?

Do you have any kind of link cable? If not, then buy one. ;)

yeah, i have a link cable.  What do i do with it?

Which kind is it? They usually come with instructions...

You can also read in the manual how to create basic programs.  The language is very easy to learn and you will be able to make your own programs.

What kinds of programs are you looking for?  Games? Math? ...

i already have created 12 programs on my own, but i was wondering how to get Super Mario, or Snake.

you can download them online if you have a link cable and link it to the computer
or, you can get it off a friend.

Look at the instructions! It depends on what type of cable you have.

The instructions didn't tell me anything!!!

Well it might be a good idea to start by telling us what type of cable you have..... :roll:


linear algebra
matrix
induction
function

Problem Let $ A \in M_n{\mathbb{(C)}}$ , Prove that $ \det(A\overline{A} \plus{} I_n) \geq 0$ Here $ \overline{A}$ is conjugate complex matrix of $ A$
 Proof not complete but easy understand .
 [b]By induction [/b]
 With $ n \equal{} 1$ it is true .
 Suppose that it is correct to $ n \geq 1$, now take $ A \in M_{n \plus{} 1}{\mathbb{(C)}}$
case 1) If $ rank(A) \equal{} r < n \plus{} 1$ then the exist $ P,Q \in GL_{n \plus{} 1}{\mathbb{(C)}}$
such that  $ A \equal{} P\begin{bmatrix}I_r & O \\
O & O\end{bmatrix}Q$
 Put $ Q\overline{P} \equal{} \begin{bmatrix}A_1 & A_2 \\
A_3 & A_4\end{bmatrix}$
 Here $ A_1 \in M_r{\mathbb{(C)}},A_4 \in M_{n \plus{} 1 \minus{} r}{\mathbb{(C)}}$,$ A_2 \in M_{rx(n \plus{} 1 \minus{} r)}{\mathbb{(C)}}$,$ A_3 \in M_{(n \plus{} 1 \minus{} r)xr}{\mathbb{(C)}}$
 Then $ A\overline{A} \equal{} P\begin{bmatrix}I_r & O \\
O & O\end{bmatrix}Q\overline{P}\begin{bmatrix}I_r & O \\
O & O\end{bmatrix}\overline{Q}$=$ P\begin{bmatrix}I_r & O \\
O & O\end{bmatrix}\begin{bmatrix}A_1 & A_2 \\
A_3 & A_4\end{bmatrix}\begin{bmatrix}I_r & O \\
O & O\end{bmatrix}\overline{Q}$=$ P\begin{bmatrix}A_1 & O \\
O & O\end{bmatrix}\overline{Q}$
 Use $ \det(XY \plus{} I) \equal{} \det(YX \plus{} I)$ we have :
 $ \det(A\overline{A} \plus{} I_{n \plus{} 1}) \equal{} \det(\overline{Q}P\begin{bmatrix}A_1 & O \\
O & O\end{bmatrix} \plus{} I_{n \plus{} 1})$=$ \det(\begin{bmatrix}\overline{A_1} & \overline{A_2} \\
\overline{A_3} & \overline{A_4}\end{bmatrix}\begin{bmatrix}A_1 & O \\
O & O\end{bmatrix} \plus{} I_{n \plus{} 1})$=$ \det(\begin{bmatrix}\overline{A_1}A_1 & O \\
\overline{A_3}A_1 & O\end{bmatrix} \plus{} I_{n \plus{} 1}) \equal{} \det(\overline{A_1}A_1 \plus{} I_r) \geq 0$ (by induction $ r < (n \plus{} 1)$)
case 2) [b]Now please make[/b] with $ rank(A) \equal{} n \plus{} 1$ (It is hard ) but It correct to $ k \equal{} 1,2,..,n$ and $ rank(A) < n \plus{} 1$ (I am so bad  :blush: )
Ps : If you have a proof $ \forall A \in GL_n{\mathbb{(C)}}$ that $ \det(A\overline{A} \plus{} I_n) \geq 0$
 then It is so easy to prove that If $ rank(A) < n$ then $ \det(A\overline{A} \plus{} I_n) \geq 0$
[hide=" Answer with"] $ rank(A) \equal{} n \plus{} 1$ If you have a solution please post it  :clap2:  [/hide]

You can see that If $ \overline{P(\lambda)} \equal{} P(\overline{\lambda}) ,\forall \lambda \in \mathbb{C}$ , $ P \in \mathbb{K}[X]$
then $ P(x) \in \mathbb{R}[X]$
 Applied we have $ P(\lambda) \equal{} \det(A\overline{A} \plus{} \lambda I_n) \in \mathbb{R}[\lambda]$ or $ A\overline{A}$ have eigenvalues real $ \lambda_1,...,\lambda_k$ and conjugate complex $ r_1,\overline{r_1},...,r_m,\overline{r_m}$
Now we consider $ f(A\overline{A})$ such that $ f(\lambda_i) \plus{} 1 \geq 0,1 \leq i \leq k$, and $ \overline{f(r_j) \plus{} 1} \equal{} f(\overline{r_j}) \plus{} 1,1 \leq j \leq m$, $ k \plus{} 2m \equal{} n$
Applied for $ f(A\overline{A}) \equal{} e^{A\overline{A}}$,$ f(A\overline{A}) \equal{} sin(A\overline{A})$,$ f(A\overline{A}) \equal{} cos(A\overline{A})$,$ f(A\overline{A}) \equal{} Ch(A\overline{A})$ We have :
 $ \det(e^{A\overline{A}} \plus{} I_n) \geq 0$,$ \det(sin{A\overline{A}} \plus{} I_n) \geq 0$ ,$ \det(cos{A\overline{A}} \plus{} I_n) \geq 0$
$ \det(ch{A\overline{A}}\pm I_n) \geq 0$
 Note :Can have many function $ f$ above $ f \equal{} 1$ are case $ \det(A\overline{A} \plus{} I_{n})\geq 0$ But $ f$ don't such that above
add For $ A_1,A_2,..,A_m \in M_n{\mathbb{(C)}}$  then $ \det(e^{\sum_{k \equal{} 1}^{m}{A_k\overline{A_k}}}) \geq 0$
If $ A_iA_j \equal{} A_jA_i 1 \leq i,j \leq m$ then $ f(A_1\otimes{\overline{A_1}},...,A_m\otimes{\overline{A_m}})$
 Please for more information about function $ f$ Thanks you very much ! :rotfl:
Question : Let $ A,B \in M_n{\mathbb{(C)}}$ such that : $ AB \equal{} BA$ then $ \det(A\overline{A} \plus{} B\overline{B} \plus{} \lambda I_n) \in \mathbb{R}[\lambda]$ ?


When is the next AMC 12 class will be offered after the June session? Thanks!

Probably starting in October 2007, ending in late January or early February 2008 just before the AMC test.


Let $ ABCD$ be a convex quadrilateral such that $ \angle DAB \equal{}\angle ABC \equal{}\angle BCD$. Let $ H$ and $ O$ denote the orthocenter and circumcenter of
the triangle $ ABC$. Prove that $ H,O,D$ are collinear.

I love this combination "Easy" + "Unsolved Problems"...

http://www.mathlinks.ro/Forum/viewtopic.php?t=25859

  darij


$\ a,b,c\ghe0$ .
$\frac{a}{a+\sqrt{(a+b)(a+c)}}+\frac{b}{b+\sqrt{(b+c)(b+a)}}+\frac{c}{c+\sqrt{(c+a)(c+b)}}\leq1$
     Good luck  :D

we have : Ineq's BCS $(a+b)(b+c) \geq (\sqrt{bc}+\sqrt{ba})^{2}$
  hence :$b+(a+b)(b+c) \geq (\sqrt{bc}+\sqrt{ba})^{2}+b = \sqrt{b}.( \sqrt{a}+\sqrt{b}+\sqrt{c}).$
    from this,  we prove pro very easy


Let x, y, z be positive real numbers,
xyz=1
Prove:
2/[(x + 1)^2 + y^2 + 1] + 2/[(y + 1)^2 + z^2 + 1] + 2/[(z + 1)^2 + x^2 + 1 ] <=1

Pls help. Can't do it for months. omg

Well, I don't have a complete solution so I'll post what I predict how it's gonna work  :wink:
First notice that $ (x\plus{}1)^{2}\plus{}y^{2}\plus{}1\equal{}x^{2}\plus{}y^{2}\plus{}2x\plus{}2\geq 2xy\plus{}2x\plus{}2\equal{}2(xy\plus{}y\plus{}1)$
thus:
$ \sum_{cyc}\frac{2}{(x\plus{}1)^{2}\plus{}y^{2}\plus{}1}\leq \sum_{cyc}\frac{1}{xy\plus{}y\plus{}1}$
What is missing in my pseudo-solution is the next part, being specific proving that:
$ \sum_{cyc}\frac{1}{xy\plus{}y\plus{}1}\leq \sum_{cyc}\frac{1}{xy\plus{}z\plus{}1}$
(to be honest I'm not quite sure if it is really true :wink: )
But if that's true than our solution is finished, because $ \forall_{a\in R_{\plus{}}}a\plus{}\frac{1}{a}\geq 2$
So
$ \sum_{cyc}\frac{1}{xy\plus{}z\plus{}1}\equal{}\sum_{cyc}\frac{1}{\frac{1}{z}\plus{}z\plus{}1}\leq \frac{1}{3}\plus{}\frac{1}{3}\plus{}\frac{1}{3}\equal{}1$
It'd be great if someone would tell me if it is a good way to approach that problem :wink:

[quote="polskimisiek"]Well, I don't have a complete solution so I'll post what I predict how it's gonna work  :wink:
First notice that $ (x \plus{} 1)^{2} \plus{} y^{2} \plus{} 1 \equal{} x^{2} \plus{} y^{2} \plus{} 2x \plus{} 2\geq 2xy \plus{} 2x \plus{} 2 \equal{} 2(xy \plus{} y \plus{} 1)$
thus:
$ \sum_{cyc}\frac {2}{(x \plus{} 1)^{2} \plus{} y^{2} \plus{} 1}\leq \sum_{cyc}\frac {1}{xy \plus{} y \plus{} 1}$
What is missing in my pseudo-solution is the next part, being specific proving that:
$ \sum_{cyc}\frac {1}{xy \plus{} y \plus{} 1}\leq \sum_{cyc}\frac {1}{xy \plus{} z \plus{} 1}$
(to be honest I'm not quite sure if it is really true :wink: )
But if that's true than our solution is finished, because $ \forall_{a\in R_{ \plus{} }}a \plus{} \frac {1}{a}\geq 2$
So
$ \sum_{cyc}\frac {1}{xy \plus{} z \plus{} 1} \equal{} \sum_{cyc}\frac {1}{\frac {1}{z} \plus{} z \plus{} 1}\leq \frac {1}{3} \plus{} \frac {1}{3} \plus{} \frac {1}{3} \equal{} 1$
It'd be great if someone would tell me if it is a good way to approach that problem :wink:[/quote]

you don't have to prove that part....if you look more carefully at $ \sum_{cyc}\frac {1}{xy \plus{} y \plus{} 1}$, you will see that
$ \sum_{cyc}\frac {1}{xy \plus{} y \plus{} 1}\equal{}1$ holds...cheers :wink:

[quote="Wolfbrother"][quote="behemont"]if you look more carefully at $ \sum_{cyc}\frac {1}{xy \plus{} y \plus{} 1}$, you will see that
$ \sum_{cyc}\frac {1}{xy \plus{} y \plus{} 1} \equal{} 1$ holds...cheers :wink:[/quote]
Uff...Why is it 1?[/quote]
$ \frac {1}{xy \plus{} y \plus{} 1}\plus{}\frac{1}{yz\plus{}z\plus{}1}\plus{}\frac{1}{xz\plus{}x\plus{}1}\equal{}$
$ \equal{}\frac {1}{xy \plus{} y \plus{} 1}\plus{}\frac{xy}{y\plus{}1\plus{}xy}\plus{}\frac{y}{1\plus{}xy\plus{}y}\equal{}1.$ :wink:

Hehehe, yup right. It turned out that I complicated everything myself :wink: Looking at it now it seems quite obvious that my "part" would even fail. :wink: Anyway also thanks.