AMC
AIME
search
geometry
3D geometry
sphere
function

Some of you have already seen a lot of these questions; I post them in the Intermediate Forum to see what responses are but now I want to see if everybody can do it.  Most of you should be able to get a 15; I can get a 15 and I'm only 11....

Note that the questions are in a RANDOM ORDER; they do not get harder as they go along.

PM your answers like Ans1, Ans2, Ans3, ..., Ans15.  Leave all unknown answers blank (say "Unknown").

Please do not search the Intermediate Forum for the answers.

Here it is:
[hide="Problems"]
$1.$  A hollow cone is held vertex down.  A sphere of radius $2$ is dropped in so that it is tangent to the surface of the cone.  A sphere of radius $3$ is dropped in so that it is tangent to the surface of the cone and the first sphere.  A third sphere of radius $5$ is dropped in, and a final sphere of radius $1$ is dropped in so that it is resting on the radius-$5$ sphere and a single point on the surface of the cone.  If the distance from that point to the vertex of the cone is $\sqrt{a}+\sqrt{b}$, compute $a+b$

$2.$  If $a = \sqrt[8]{5+2\sqrt{3}+6\sqrt{4+2\sqrt{3}}+4\sqrt[4]{4+2\sqrt{3}}+4\sqrt[4]{208+120\sqrt{3}}}$ and $f(x) = (x+1)(x-1)$, compute $f^{6}(a)$.

$3.$  Let $Squish(n)$ be a discrete function such that $Squish(n) = n$ for all $n$ with no digits of $0$ in them and for $n$ with $1$ or more digits of $0$, $Squish(n) = n$ with all the digits of $0$ removed.  For example, $Squish(108) = 18$ but $Squish(118) = 118$.  Compute the number of positive integers $n$, $n < 1000$ such that $n | Squish(n)$ but $n \ne Squish(n)$.

$4.$  There exists a $6$-digit number $n$ such that $1n, 2n, 3n, 4n, 5n, \text{and }6n$ are all rearrangements of the digits of $n$.  Compute the remainder when $7n$ is divided by $1000$.

$5.$  Bob lives at $(0, 0)$ and works at $(6, 6)$.  To get to work, he walks $12$ blocks, each either increasing his $x-$ or $y-$ coordinate.  In how many ways can he get to work if $(2, 1)$ and $(2, 4)$?

$6.$  Compute $\left\lfloor{\frac{2006^{2007}(\text{mod }2008)}{10}\right\rfloor}$

$7.$  Me and my brother each pick $3$ $0-9$ digits (not necessarily distinct) and whisper them to our sister.  She rights all the permutations of my $3$ digits next to all the permutations of his $3$ digits.  She sees that none of my permutations share a factor with any of my brother's.  She also sees that the closest two permutations have a difference of one.  Finally, she observes that the farthest apart permutations have a difference is $k$.  Find the smallest possible $k$.

$8.$  Let an integer quadrilateral be a quadrilateral whose sides are all of integer length, whose diagonals form $4$ segments of integer length and whose area is also an integer.  Find the area of an integer quadrilateral with at least $3$ sides of length $13$.

$9.$  Joe leaves his house every morning at $7: 00$ to get to school on time.  If he travels at $40mph$, he is $3$ minutes late.  If he travels at $60mph$, he is $5$ minutes early.  If he travels at $\frac{m}{n}mph$, he is perfectly on time.  Compute $m+n$.

$10.$  An arithmetic sequence with $10$ terms and a common difference of $1$ has the property that the $n^{th}$ term is a multiple of $11-n$.  If the smallest possible first term is $a$, compute the remainder when $a$ is divided by $1000$.

$11.$  $137$ is a "very prime" number because $1, 3, 7, 13, 37, \text{and }137$ are all prime.  Compute the remainder when the largest "very prime" number is divided by $1000$. Consider 1 to be a prime for the purpose of this question.

$12.$  $M(n)$ is defined as follows:
$d_{1}$ is the greatest multiple of $1 \le n$.
$d_{2}$ is the greatest multiple of $2 \le d_{1}$.
$d_{3}$ is the greatest multiple of $3 \le d_{2}$.
.
.
.
$M(n)$ is $k$ such that $d_{k}= k$.  Compute the sum of all positive integers $n$ for which $M(n) = 10$.

$13.$  There exists a set of $k$ distinct integers whose sum is  $2007$, whose mean is $m$, whose maximum is $2007m$, and whose minimum is $k$.  Compute $|\text{The largest possible }k|$.

$14.$  Quadrilateral $ABCD$ has $AB = BC = 120$, $CD = 64\sqrt{10}$, $DA = 160$, $BCD$ is acute, and $DAB$ is a right angle.  Compute the length of the altitude from $D$ to $BC$.

$15.$  $P_{1}P_{2}P_{3}...P_{2}007$ is a regular $2007$-gon with $P_{1}$ at $(1, 0)$, and the farthest point on the circumference of the polygon from $P_{1}$ is at $(3, 0)$.  Let $x_{n}$ and $y_{n}$ be the $x$ and $y$ coordinates of $P_{n}$, respectively.  Let $z_{n}= x_{n}+iy_{n}$.  Compute the remainder when $|z_{1}z_{2}z_{3}\cdots z_{2007}|$ is divided by $1000$.
[/hide]

Hope you like the problems!

uhh.is number 5 unfinished?

and of course you can get a 15, you created the questions :lol:

[quote="Art of Owna"]uhh.is number 5 unfinished?[/quote]

I think the word "if" in question 5 should be "through" (or, alternatively, "if he passes through")

Have you got / Could you write a pdf for this test, please?  :roll:

For number 3, do you mean $Squish(n)|n$? Because $Squish(n)\le n$ for all n, so there aren't any solutions.

For number 5, when you posted this in the forum, it said that you [b]couldn't[/b] pass through (2,1) or (2,4), so I'm assuming that's what you mean now.

For number 11, 1 isn't prime ...

For number 13, we know that the minimum $k$ must be a positive integer, so $2007m<2007$, but $2007m>2007k\ge 2007$, so this is a contradiction ... I don't think that there are any solutions to this problem.

It wasn't such a big deal, just copied and pasted his post in to Texnic and changed a couple things (like add a title and fix the numbering). It probably took less than 10 mts :P

Sorry, there are a couple of typos; I can't edit now.  It is supposed to be $Squish(n) | n$ and "can not pass through (2, 1) and (2, 4)".

[b]@samath[/b]
For the prime question, call 1 a prime.
For the 2007 question, it does not require positive integers.
[size=200]
[b][i][u]Can a mod please edit the original post for me?[/u][/i][/b][/size]

But if $k$ is the minimum and the number of integers, it must be positive

unless there's some way to have a negative number of numbers ...?

It asks for the absolute value of the smallest element.

I didn't find the "easy" AIME that easy.

It's a bit easier than the regular one >.>

[quote="samath"][quote="miyomiyo"]
$13.$  There exists a set of $k$ distinct integers whose sum is  $2007$, whose mean is $m$, whose maximum is $2007m$, and whose minimum is $k$.  Compute $|\text{The largest possible }k|$.[/quote]

I know it does, but it still says there are $k$ integers and the minimum is $k$, so it's impossible.[/quote]  Whoops, I named both the number of integers in the set and the minimum of the set $k$.   :blush:   They're supposed to be different.

[hide="Problems"]
$1.$  A hollow cone is held vertex down.  A sphere of radius $2$ is dropped in so that it is tangent to the surface of the cone.  A sphere of radius $3$ is dropped in so that it is tangent to the surface of the cone and the first sphere.  A third sphere of radius $5$ is dropped in, and a final sphere of radius $1$ is dropped in so that it is resting on the radius-$5$ sphere and a single point on the surface of the cone.  If the distance from that point to the vertex of the cone is $\sqrt{a}+\sqrt{b}$, compute $a+b$

$2.$  If $a = \sqrt[8]{5+2\sqrt{3}+6\sqrt{4+2\sqrt{3}}+4\sqrt[4]{4+2\sqrt{3}}+4\sqrt[4]{208+120\sqrt{3}}}$ and $f(x) = (x+1)(x-1)$, compute $f^{6}(a)$.

$3.$  Let $Squish(n)$ be a discrete function such that $Squish(n) = n$ for all $n$ with no digits of $0$ in them and for $n$ with $1$ or more digits of $0$, $Squish(n) = n$ with all the digits of $0$ removed.  For example, $Squish(108) = 18$ but $Squish(118) = 118$.  Compute the number of positive integers $n$, $n < 1000$ such that $Squish(n) | n$ but $n \ne Squish(n)$.

$4.$  There exists a $6$-digit number $n$ such that $1n, 2n, 3n, 4n, 5n, \text{and }6n$ are all rearrangements of the digits of $n$.  Compute the remainder when $7n$ is divided by $1000$.

$5.$  Bob lives at $(0, 0)$ and works at $(6, 6)$.  To get to work, he walks $12$ blocks, each either increasing his $x-$ or $y-$ coordinate.  In how many ways can he get to work if he can't go through $(2, 1)$ and $(2, 4)$?

$6.$  Compute $\left\lfloor{\frac{2006^{2007}(\text{mod }2008)}{10}\right\rfloor}$

$7.$  Me and my brother each pick $3$ $0-9$ digits (not necessarily distinct) and whisper them to our sister.  She rights all the permutations of my $3$ digits next to all the permutations of his $3$ digits.  She sees that none of my permutations share a factor with any of my brother's.  She also sees that the closest two permutations have a difference of one.  Finally, she observes that the farthest apart permutations have a difference is $k$.  Find the smallest possible $k$.

$8.$  Let an integer quadrilateral be a quadrilateral whose sides are all of integer length, whose diagonals form $4$ segments of integer length and whose area is also an integer.  Find the area of an integer quadrilateral with at least $3$ sides of length $13$.

$9.$  Joe leaves his house every morning at $7: 00$ to get to school on time.  If he travels at $40mph$, he is $3$ minutes late.  If he travels at $60mph$, he is $5$ minutes early.  If he travels at $\frac{m}{n}mph$, he is perfectly on time.  Compute $m+n$.

$10.$  An arithmetic sequence with $10$ terms and a common difference of $1$ has the property that the $n^{th}$ term is a multiple of $11-n$.  If the smallest possible first term is $a$, compute the remainder when $a$ is divided by $1000$.

$11.$  $137$ is a "very prime" number because $1, 3, 7, 13, 37, \text{and }137$ are all prime.  Compute the remainder when the largest "very prime" number is divided by $1000$. Consider 1 to be a prime for the purpose of this question.

$12.$  $M(n)$ is defined as follows:
$d_{1}$ is the greatest multiple of $1 \le n$.
$d_{2}$ is the greatest multiple of $2 \le d_{1}$.
$d_{3}$ is the greatest multiple of $3 \le d_{2}$.
.
.
.
$M(n)$ is $k$ such that $d_{k}= k$.  Compute the sum of all positive integers $n$ for which $M(n) = 10$.

$13.$  There exists a set of $k$ distinct integers whose sum is  $2007$, whose mean is $m$, whose maximum is $2007m$, and whose minimum is $k$.  Compute $|\text{The largest possible }k|$.

$14.$  Quadrilateral $ABCD$ has $AB = BC = 120$, $CD = 64\sqrt{10}$, $DA = 160$, $BCD$ is acute, and $DAB$ is a right angle.  Compute the length of the altitude from $D$ to $BC$.

$15.$  $P_{1}P_{2}P_{3}...P_{2}007$ is a regular $2007$-gon with $P_{1}$ at $(1, 0)$, and the farthest point on the circumference of the polygon from $P_{1}$ is at $(3, 0)$.  Let $x_{n}$ and $y_{n}$ be the $x$ and $y$ coordinates of $P_{n}$, respectively.  Let $z_{n}= x_{n}+iy_{n}$.  Compute the remainder when $|z_{1}z_{2}z_{3}\cdots z_{2007}|$ is divided by $1000$.
[/hide]

I think that these are the correct problems....

Should we start compliing a solutions thread here?


The numbers 1 through 25 are written on 25 cards with one number on each card. Sara picks one of the 25 cards at random. What is the probability that the number on her card will be a multiple of 2 or 5?
Express your answer as a common fraction.

[hide] There are $ 12$ multiples of $ 2$ in the stack, and $ 5$ multiples of $ 5$ in the stack. However there is a slight overcount (since $ 10$ and $ 20$ are both divisible by $ 2$ and $ 5$). So we must subtract these out. Since our denominator is $ 25$, we have $ \frac{12\plus{}5\minus{}2}{25}\equal{}\boxed{\frac35}$.[/hide]


lingomaniac88 is making an ImAoPSAble quiz. It is a parody on these quizzes:

http://www.addictinggames.com/theimpossiblequiz.html
http://www.addictinggames.com/theimpossiblequiz2.html

He has gotten 100 questions out of 340. He will release the first edition soon, but he needs more questions. Post yours here.

The other one was locked.

The ImAoPSible Quiz has 100 complete questions, but I want to make some things run more smoothly.  Therefore I am not going to release it yet.  It should be ready in a few days.

Also, there's something I should say:

The ImAoPSible Quiz is an executable that runs on Windows.  It doesn't work on Macs, but if you run Windows through a Mac *coughBootCampcough*, it might work.  If you don't have access to Windows, then I hope you know somebody that does.  Or you could put the quiz on a flash drive and take it to a computer store and play it there. :P

Sorry, but the software I'm using to create this only makes games for Windows.  I believe a version for the Mac is being made, so if that comes out, I can do all the necessary conversions.

I would have made it with Flash, except I don't have it.  I don't really want to shell out $ \$700$ for it, nor do I want to get it illegally.  (The sad part is that my computer has Photoshop, which is about the same price, which I never even use.)

If anybody knows any free alternatives to Flash, let me know.

Is there no easy conversion into QuickTime?
 :(

37: What is the answer to this question?

A: WTF
B: skip
C: go back one space
D: le

[hide="Answer:"]D, le+37=1337.[/hide]

[quote="asterday1"]102.
if 1=5, 2=10, 3=15, and 4=20, what does 10 equal?

a.50
b.$ \pi$
c.14
d.i don't know what b is!
[hide="answer"]2(in ? number), because 2=10.[/hide][/quote]

so we assume '=' is symmetric? interesting...

@lingomaniac88: why don't you want to get it illegally?

[quote="bubka"]@lingomaniac88: why don't you want to get it illegally?[/quote]
I just don't feel completely comfortable doing stuff like that.

Number is irrelevant, but make the "find x" problem again (the one with the right triangle and you click x), but make it so you have to actually do it. Make the x clickable, but make it the wrong answer.

Oo, I got an idea. So make the screen say at the top MATH TEST.
Put instructions right underneath MATH TEST (don't make it very conspicuous, though) that says "write in all caps"
Then put on the top corner a place that says NAME_____ and DATE (d/m/y, for ex 1/1/08)_____
Then put a few random easy linear equations and a place to type answers.
Finally, put a button at the very bottom that says TURN IN

-1 life if:
they did not label their variables (like they put 1 instead of x=1)
they did not type anything in the "name" line
they did not type the correct date (I'm assuming it's possible to do this, since one of the problems you put had the time)
they miss one or more of the problems

Also, if you're really evil, make it "skippable," and when they skip, it says, "Wait, you didn't turn in your math test yet!" And your skip is spent.

Compute: 1/n(sin x)
1. 6
2. $ \dfrac{\pi}{2}$
3.360/n
4.cos n

[hide="Answer"]6, because the n's cancel out, leaving si x[/hide]

Better may be $ \frac {\sin{x}}{n}$...It's the same thing but cleaner.


Point P is 9 units from the center of a circle of radius 15.  How many different chords of the circle contain P and have integer side lengths?

[quote="Ihatepie"]
[hide]

1.find y and x intercepts, they are both 2. So it is a quarter-circle with radius 2.

pi*2^2/4 so 4pi/4 or pi total possible is 32 so 32/pi

2. The ways that the 2nd best team didn't get 2nd would be if they were on the same bracket side as the 1st place one. so 3/7.

3. longest chord possible is 30 (diameter) shortest is 24. So 24, 25,26,27,28,29,30. But there can be two of each not including biggest and smallest. So 24,25,25,26,26,27,27,28,28,29,29,30, or 12.[/hide][/quote]


[quote="unknown_pattern18"]find the nth term for the sequence below
1, 2, 10, 67, 467, 3268, 22876[/quote]

my darling,

as i saw, u post it in many communities,

u dould only post it in one of the math forums. ;)

[quote="unknown_pattern18"]because no one can answer it,its useless to answer 10000 hard questions and to be unable to answer an easy question.no one can make an nth term for that.
can you ? then what is the point to study higher levels when someone cant answer a basic thing?[/quote]

u mean i cant solve it?

why do u say that , when u dont know wether i can solve it or not?

and another thing. is it a good reason that,

first a person should learn and solve all basically questions and then

learn the higher levels.

if it is, then if u got $10^5$ years old, steel u wont solve and learn them.

and the thired one is the other communities arent a place to post ure

foolish questions there, did u get it?

or i should say some bad things to u?

Every mathematician knows there is no unique answer to such questions. If you have to post it on this math site, try the round table.

[quote="unknown_pattern18"]aghay asheqh englise to khob nist man dorost motavajeh nashodam chi neveshtii,shoma aval baiad spellingeto dorost bekoni .oon still e  na steel,aslan sentencessss to mani nadare,that was too persianesh,not english
shoma manzore mano nagerefti,man manzoram shoma nabodi ke ,ba hama people form goftam,sanian khob nist (foolish questions) begi, medoni man chand salema ? college dars mekhonam,ya university or daneshgah ya tamam kardam? khob nemidoni, medoni?,man ke soal bade nakardam,faghat goftam metonin eno hal konin.va degar khaheshan be kasi ke namidoni kia (my darling)nagi.fekr mekonan to gay asti,khob? gharbi ha mesl ma  ha nistan,
age narahatet kardam bebakhsh,manzore bade nadashtam khodaesh.
if you know the answer for that question then please write back to me the answer for it.
thank you.[/quote]

be darak bezar fekr konan gay am, aslan gay am.

moshkeli dari?

aslan ham mohem nist keh ki hasti, va chand saleteh.

hamoon tor ham keh fer mikoni englisham khoob nist.

farsie to ham vaghean khoob nist.

boro ye fekry be halesh bokon.

aslan hoseleye jaro bahs ba toro nadram.


I was bored so I decided to make a problem....  :lol:

Let $ a,b,c,d$ be real numbers all distinct and satisfy the equation

$ (a \plus{} b \plus{} c \plus{} d)^2 \plus{} (a \minus{} b \minus{} c \minus{} d)^2 \equal{} 0$

If $ \left|\frac {a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2}{bc \plus{} bd \plus{} cd}\right| \equal{} m$, find $ m$.

My bad, I was in hurry and forgot to finish it. There it is now.

[hide]Foiling the two expressions and combining like terms yields $ 2a^2 + 2b^2 + 2c^2 + 2d^2 + 4bc + 4cd + 4bd = 0$  Isolating yields $ a^2 + b^2 + c^2 + d^2 = - 2bc - 2bd - 2cd$.  The expression now becomes $ \left|{\frac { - 2(bc + cd + bd)}{bc + cd + bd}\right|}$, which is $ 2$.[/hide]

Nicely done.

[hide="Even Easier Way"]

Of course, what azn6021023 is the intended method. But let's point out an important key. The problem only said that $ a,b,c,d$ to be real and distinct and we're looking for ratio.  Ratio stays constant regardless of the values of $ a,b,c,d$. So, for convenience, let's set:

$ a \equal{} 0, b \equal{} \minus{}3, c \equal{} 1, d \equal{} 2$

then:

$ \frac{a^2\plus{}b^2\plus{}c^2\plus{}d^2}{bc\plus{}bd\plus{}cd} \equal{} \frac{0 \plus{} 9 \plus{} 1 \plus{} 4}{\minus{}3 \minus{} 6 \plus{} 2} \equal{} \frac{14}{\minus{}7} \equal{} \minus{}2$

And the answer is still 2. [/hide]

Ratio is always nice because to certain extent, you can cut down the unnecessary algebra work.

[hide="Solution"]

Clearly, $ a\plus{}(b\plus{}c\plus{}d)\equal{}a\minus{}(b\plus{}c\plus{}d)\equal{}0 \implies a\equal{}0, \; \; b\plus{}c\plus{}d\equal{}0$, since if two squares add to zero, they must both be zero.

We can write the fraction as $ \frac{b^2\plus{}c^2\plus{}d^2}{bc\plus{}cd\plus{}db} \equal{} \frac{(b\plus{}c\plus{}d)^2\minus{}2(bc\plus{}cd\plus{}db)}{bc\plus{}cd\plus{}db} \equal{} \frac{\minus{}2(bc\plus{}cd\plus{}db)}{bc\plus{}cd\plus{}db}\equal{}\minus{}2$, so the answer is $ \boxed{2}$

[/hide]

[hide="A Bit Different"]Using the same reasoning that Zuton I've found $ a\equal{}0$ and $ b\plus{}c\plus{}d\equal{}0$. Note that $ c\plus{}d\equal{}\minus{}b$, and $ b^2\equal{}\minus{}b(\minus{}b)\equal{}\minus{}b(c\plus{}d)$. Similarly, $ c^2\equal{}\minus{}c(b\plus{}d)$ and $ d^2\equal{}\minus{}d(b\plus{}c)$.

Substituting yields $ \frac{b^2\plus{}c^2\plus{}d^2}{bc\plus{}bd\plus{}cd}\equal{}\frac{\minus{}2(bc\plus{}bd\plus{}cd)}{bc\plus{}bd\plus{}cd}\equal{}\minus{}2$, and so $ m\equal{}2$[/hide]

[hide="Question"][quote]The problem only said that $ a,b,c,d$ to be real and distinct and we're looking for ratio. Ratio stays constant regardless of the values of $ a,b,c,d$.[/quote]

Sorry, I didn't understand this part. Can someone explain/prove this?[/hide]

[quote="Sashsiam_2"][quote]The problem only said that $ a,b,c,d$ to be real and distinct and we're looking for ratio. Ratio stays constant regardless of the values of $ a,b,c,d$.[/quote]

Sorry, I didn't understand this part. Can someone explain/prove this?[/quote]

What I mean is a strategy that works sometimes, not always. But usually with algebra, it is OK to do it, especially if you're running low on the time.

Whenever the question asks for ratio, this ratio always has to be same in all scenarios. Let's "assume" that the original problem has answer to be $ A,B,C,D$ where each correspond to value of $ a,b,c,d$. Now, there are infinite number of ways to choose four numbers such that $ \left|\frac {a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2}{bc \plus{} bd \plus{} cd}\right| \equal{} 2$ because we can "shift" the numbers.

The new solutions could be in the form $ A\plus{}k, B\plus{}k, C\plus{}k, D\plus{}k$ where $ k$ is a real constant. So, even though we're changing the value of the variables, the ratio should stay same because they cancel out (it's like saying if your answer was 5/2, 10/4, and 15/6, they all are same thing because they cancel out even though their variables have different values).

This is often used in easy geometry question as well. Consider the problem where you have to find ratio of one triangle's area over another triangle's area. If the question doesn't specify the length of sides and that there isn't a very strong restriction, we can assume the triangle to be equilateral and solve. Of course, this is not really "solving" the problem but it is a good way to check it. Another way to look at is by asking, "If it works for all triangles, it should work for equilateral triangles too, right?" This, however, is unacceptable to proof questions as you might guess.


My brother can memorize up to 120th digits.. :D

[hide="hint"]divide[/hide]

[quote="DarkKnight"][hide]
$\frac{5005}{55}=91$
He recites 91 digits of pi a minute.  :o [/hide][/quote]Not that unbelievable. Thats actualy slow. 144 a minute is about average

[hide]5005/55 or 1001/11 words a minute. 91 words per minute[/hide]

[quote="MCrawford"]In March 2000, Jake Euget, a high school senior in Fargo, North Dakota, won his school's annual $\pi$ memorization contest by reciting 5,005 digits of $\pi$.  He recited the digits in 55 minutes.  On average, how many digits did he recite each minute?[/quote]

[hide]5005/55=91

91 numbers per minute[/hide]


In circle $ B$, the measure of $ \angle ABC$ is $ 135^\circ$. What is the ratio of the area of the larger sector to the area of the smaller sector? Express your answer as a common fraction.
[asy]size(80);
draw((0,1)..(1,0)..(0,-1)..(-1,0)..cycle);
draw((0,1)--(0,0)--(.707,-.707));
label("$A$",(0,1),N);
label("$B$",(0,0),W);
label("$C$",(.707,-.707),SE);[/asy]

the areas of the sectors are directly proportional to the angles to which they are subtended

therefore, becasue the two angles are $ 135$ and $ 225$, the ratio of the areas of the sectors is:

$ 225/135$ = $ 5/3$


Say you've got a list of frequencies and they are mixed up (i.e., unsorted). It's straightforward to sort that list in ascending or descending order using quicksort and others...but [b]How [/b]do you sort it so that it looks like a bell-shape? (i.e., ascending from first to the middle (where the the list has highest value), then descending till the end of the list).

find the median (linear time), cut the data into $ 2$ equal parts (linear time), and sort each part ($ n \ln(n)$) :maybe:


uh from the looks of mathfriend it's just a bunch of sites strung together. give me an example of like which article you want to download and maybe i can help you

Oh, I also want to download the useful books and articles from it.But I can't understand Veitmese :blush: I only know a few words...
Can you also help me? I want some English books..Thanks. My E-mail
is hawktiger89@yahoo.com

Hey,can you list some books that avilable there for download?As you can guess I don't know Vietnames too :blush: .

You have to login if you are member of mathnfriend.net. If you aren't member of mathnfriend, you have to register.
Then, You can download files in mathnfriend.
Have fun!

what if we don't understand anything there? :D

[quote="navid"]I want to download articles that exist in the :http://www.mathfriend .net
but i cant.
Can any one link them there for me .thanks :gathering:  :!:[/quote]
Now, It's [url]http://www.mathnfriend.org[/url]


[quote="Twins"]$ x,y,z\geq 0$ and $ x^{2}+y^{2}+z^{2}=1$. Prove that: $ x+y+z-2xyz\leq \sqrt{2}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32399


trigonometry
inequalities proposed
inequalities

In a triangle ABC, prove that:
$\cos \frac{A}{2} +\cos \frac{B}{2}+\cos \frac{C}{2} > \cos^2 \frac{A}{2} +\cos^2 \frac{B}{2}+\cos^2 \frac{C}{2}$

:huh:  Maybe I'm wrong, but $0 < \cos \frac{A}{2} < 1$ then $\cos \frac{A}{2} > cos^2 \frac{A}{2}$ (the same for the other angles) and the result follows.


inequalities
logarithms
function
inequalities proposed

If $0\le a\le b\le c\le d$ then the following inequalities hold:
(1)$a^2b+b^2c+c^2d+d^2a\ge abc+bcd+cda+dab$
(2)$(1+a^2b)(1+b^2c)(1+c^2d)(1+d^2a)\ge (1+abc)(1+bcd)(1+cda)(1+dab)$

.

(1)can just sloved by The Karamata inequality   ?

Karamata is the general of Muirhead,I think :D

[quote="arccosinus"]Is Karamata the same as Muirhead? In that case I agree on the first inequality, because $(2,1,0)$ majorizes $(1,1,1)$.

I guess the second one can also be killed by muirhead, but I do not have the patience to do so now...[/quote]

Muirhead only works on symmetric sums.

First one is equivalent to:
$b(b-a)(c-a)+(c-b)(cd-ab)+a(d-c)(d-b)\geq 0$
which is true by $d\geq c\geq b\geq a\geq 0$.

And I think Karamata inequality can be used. Denoting:
$x=\ln d, y=\ln c, z=\ln b, w=\ln a$
We the have: $x\geq y\geq z\geq w$
and we know that array $2w+z,2z+y,2y+x,2x+w$ majorize array $x+y+z,y+z+w,z+w+x,w+x+y$.
Then (1) is Karamata inequality on function $e^x$ and (2) on $\ln (1+e^x)$.


Find all positive integers $n$ such that $2^n \equiv 1$ (mod n).

Variant $n=1$ is evident, and I'll prove, that for all positive integers $n>1$ $2^n\not\equiv1\mod n$. Let's assume, there is such $n$.
Let $p$ be the minimal divisor of $n$, which $>1$. It is, evidently, prime. Let $q$ be the less positive integer, such that $2^q\equiv1\mod p$. Then we have $2^n\equiv1\mod p$(since $2^n\equiv1\mod n$, which divisible by $p$), $2^{p-1}\equiv1\mod p$ by Fermat's Little Theorem, therefore $q|n$ and $q|p-1$  :arrow: $1<q<p$($2^1\not\equiv1\mod n$, since $n\not=1$) and $q|n$ - contradiction with the choise of $p$.


I never did USAMTS but if I were grading, I would prefer PDF.  DVI won't always work, while PDF likely will.

[quote="roadnottaken"]Do the USAMTS graders prefer .pdf format, or is .dvi also acceptable?[/quote]

If in doubt, you could turn in both. You just have to build another file, which takes maybe 10 seconds.


let g(x)=x^5+x^4+x^3+X^2+X+1.  wHAT'S THE remainder when the polynomial g(x^12) is divided by the polynomial g(x)?

[hide="Hint"]$ x^6 \equiv 1 \pmod{g(x)}$[/hide]

[hide="Hint"]Or consider the 5th roots of unity.[/hide]

The two approaches are equivalent.  See the discussion of MellowMelon's approach [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=125861]here[/url].  This can also be solved by repeated application of (32) and (33) [url=http://mathworld.wolfram.com/CyclotomicPolynomial.html]here[/url] (which perform the more difficult task of factoring $ g(x^{12})$ completely).


Does anyone know how to evaluate the sum
$ \sum_{i\equal{}0}^{n}(\minus{}1)^i i^m \binom{n}{i}$ when $ m\geq n$

The [url=http://www.research.att.com/~njas/sequences]OEIS[/url] knows ;)  -- it's sequence [url=http://www.research.att.com/~njas/sequences/A019538]A019538[/url].  (You have to do a little change of variables to see exactly how it corresponds, I think.)

By the way, congratulations on making the IMO team!


geometry
trigonometry
analytic geometry
invariant
modular arithmetic
geometric transformation
dilation

Is anyone going?

[edit: problems here:]
[quote="serialk11r"]3. Find all possible averages of 2 natural numbers a and b such that a^2 and b^2 are congruent mod 100.

4. Anyone want to put the diagram? I'm sure it's not worth the time though, seeing that it has appeared on mathcounts before...its a 2 stepper angle chase.

5. Call a set magic if for any 2 elements i and j in the set, (i+j)/GCD(i,j) is also in the set.

6. Six frogs sit at the vertices of a regular hexagon in the plane. Frogs can move in 2 ways:
Rule 1: Frog A jumps over Frob B, and lands on the line AB at a distance 2AB on the opposite side of B as A.
Rule 2: Frog A jumps over Frog B, and lands on the line AB at a distance AB/2 on the opposite side of B as A.

If we can jump any frogs over each other, and use either rule, is it possible to have a frog reach the center of the original hexagon in a finite number of moves?
(screw part 1, its the same thing)

7. Prove that if a,b,c are angles of an acute triangle, then:
$ \frac {\cos \alpha}{\cos (\beta \minus{} \gamma)} \plus{} \frac {cos \beta}{\cos (\alpha \minus{} \gamma)} \plus{} \frac {\cos \gamma}{\cos (\alpha \minus{} \beta)} \geq \frac32$
[/quote]

Not 'going'... It's coming to us  :ninja:
So hopefully this year's geometry problem won't rely on inversion?

Taking BAMO 8 at SJMC today

There were no geometry problems on BAMO 8 last year :)

Hope it's the same this year.

[quote="AIME15"]There were no geometry problems on BAMO 8 last year :)
[/quote]

Did you typo that smiley or something?

[quote="serialk11r"]GAHHHHH I FAILED.
So like, I semi brute forced number 3, except I got lazy so I probably lots tons of points.
Number 4 I did not see the angle chase, so I trig bashed it (and failed).
Number 5 I'm pretty sure I got, but my proof seems shaky and nonrigorous.

How did everyone do?[/quote]

Please post problems.

I left after an hour! I actually had to do my Spanish project though as we are filming today. I guess I could have stayed but I'd probably only get another 0.5 problems max since I suck.
I got 1.5 problems which is not good for this forum.
#3 and Half of #5 (I did some cases).

I'm going to reword some of them for simplicity's sake.
3. Let a and b be distinct integers with 0<a,b<50. Find all possible (a+b)/2 such that a^2=b^2 mod 100.
4. Geometry, don't want to draw diagram.
5. A set is said to be "magic" if for every 2 elements i,j, (i+j)/GCD(i,j) is also in the set. Find all magic sets.
6. 6 frogs are at the vertices of a regular hexagon. Frogs can jump over each other according to 2 rules:
Rule 1: Frog1----distance d----Frog2
Frog 1 jumps over Frog2 so that:
Frog2---distance 2d---Frog1

Rule 2: Frog1---distance d----Frog2
Frog1 jumps over Frog2 so that:
Frog2--distance 1/2 d---Frog1

a. Using only rule 1 and jumping any frogs over any other frogs, is it possible to reach the center of the hexagon in a finite number of jumps?
b. Using both rules, (you are allowed to pick which frogs jump over which, and which rules to use), can you have one frog reach the center of the hexagon in a finite number of moves?
7. didn't try, don't remember

For number 6...
[hide]
Didn't know how to solve but...
I thought of considering the distance a frog was to the center in terms of the sides of the hexagon. Define the distance's complexity as the least n for which 2^-n for which the distance to the center needs to be described with.
[/hide]
Did anyone solve it?

hey guys #5 was totally on the Bay Area Marathon

This is why we need to keep the marathon going!

I got #7 quite easily, got half of #4(well, I also used trig bash but can't set up the system of equation for it :( ). I proved a ton of "useless" information though(as 2 segments are parallel to AB, inscribed quadrilateral,etc but can't find out the angle E. That problem just matters how to find that angle. So, like in the last 15', I wrote:"Assume that the angle AEB is a, so we see that the number of sides of the regular polygon=360/a because the regular polygon is always inscribed in a circle center E)" Hopefully I got partial credit for this statement(and other things). So SAD! I though I got #3 right but somehow I conclude a^2+b^2 must be divisible by 10, which screw up the rest of my proof. Hopefully they "sympathize" a little bit by thinking I'm considering a and b ends at 5(probably not). On #5, I think it's exactly the same as APMO problem, but don't remember how to do it though. Probably it relates to picking one maximum elements of S and go on. In general, FAILURE!!!!!!
Here's #7:
Prove that in an acute triangle we have:
$ cosA/cos(B\minus{}C) \plus{}cosB/cos(C\minus{}A) \plus{} cosC/cos(A\minus{}B)\geq 3/2$
This year seems to be harder than last year, doesn't it? Or probably because I'm too stupid not to check my work on #3 and couldn't find a proof on #4. :oops:

Heh, #6 seems interesting... well, you can prove that it can only get points with rational triangular coordinates in a certain triangle, but the center also has rational triangular coordinates in that triangle...

EDIT: Solved 6a... you can turn my previous argument into points reachable must have all integer coordinates, and exactly one number must be odd.

Aren't some people still taking the test?

Well anyway, I took BAMO 8...hoping for full marks on 1-3, but I misread 4 so I BS-ed it...

Heres a solution for 6a:

[hide]let the hexagon abcdef have side length 1. Consider lines through a parallel to ce, through c parallel to ae, through e parallel to ac. These form a triangle. It isn't that hard to prove that all points reachable must have integer triangular coordinates with exactly one number odd. QED.
[/hide]

Yes, that was more of a solution outline than a solution, but I'm pretty sure it works.

EDIT: Solved 6b. Nice problem. Do same thing, but consider mod 3. (remember, 1/2 is -1 mod 3).

[quote="archimedes1"][quote="ghjk"]Okay! Here's the #5, which is also the APMO 04 #1:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=474988#474988
Laziness gets "reward" :blush:[/quote]
Not wuite:  The BAMO version has the condition that i,j are distinct.[/quote]
It just means that there doesn't exist such a set S!
By the way, how did you do, archimedes1?? Do you know the cutoff for last year for HM? I'm pretty sure we need to get more than 1 problems completely correct to deserve a prize though. Still mad at myself how I infer a^2+b^2 is divisible by 10 instead of infer that a+b and a-b must be divisible by 5. Then find the lower bound for a-b, then a+b>a-b>that bound. Argh! Such an easy problem. Stupid mistake!

There are infinitely many sets, for example, take $ S \equal{} \{ k, k(k \minus{} 1)\}$.
3 - not very interesting, but forgot to put 10 as a possible average in the end
4 - quick angle chase?
5 was fine until I realized a flaw in my solution towards the end...
6 was a lot of fun--I used some invariant relating to the distance of the points to the origin.  Didn't look at 7 much, although I probably should have.

no time to talk now, but im pretty sure i got 3, 4, 5, and 6 full points. also i was at SJMC, did i see you AIME15?

outline of solution to 6?

[hide](Note: all the variables here without subscripts are vectors.)

Let $ i \equal{} (1,0)$ and $ j \equal{} (\minus{}1/2, \sqrt{3}/2)$. Place the frogs at $ i, j, i\plus{}j, \minus{}i, \minus{}j, \minus{}i\minus{}j$.

Define $ f(v) \equal{} v_x \plus{} v_y$, where $ v \equal{} v_xi \plus{} v_yj$.

If a frog jumps following rule 1 from $ A$ to $ A'$, then $ f(a) \equiv f(a') \pmod{3}$. So f(v) is invariant mod 3.

The six frogs start at locations $ v$ such that $ f(v) \equiv 1 \text{ or } 2 \pmod{3}$ so it is impossible to reach $ 0$ because $ f(0) \equal{} 0$.

For part (b), dilate the plane by a factor of 4 after every application of rule 2. $ f(v)$ remains invariant mod 3.
 [/hide]

I did better than I thought I would :D

Got #3 mostly; for some reason I misread where the problem said $ a$ and $ b$ were positive (so I had 0 down as an answer) and for some reason I thought 45 was attainable.
#4, chased angles but for some reason when I set the things equal that were supposed to be set equal they simplified to 180=180...
#5, proved that if $ \text{gcd}(i,j)\equal{}1$, the series goes infinitely, but I missed the infinitely many 2-element sets (I said there were no solutions).
#6 I didn't feel like reading.
#7 was interesting. I used the fact that $ \alpha\plus{}\beta\plus{}\gamma\equal{}\pi$ and some weird cosine manipulations, then just forced it through AMGM to simplify it a little, but didn't really finish it.

So hoping for lots of partial credit  :wink:

[quote="pythag011"]Hmm... apparently nobody used my solution to 6...

Hmm... can somebody please post #7?[/quote]
I already posted above! Read my post, pythag011! :P

[quote="grn_trtle"]

#4, chased angles but for some reason when I set the things equal that were supposed to be set equal they simplified to 180=180...
#5, proved that if $ \text{gcd}(i,j) \equal{} 1$, the series goes infinitely, but I missed the infinitely many 2-element sets (I said there were no solutions).[/quote]
#4: It is exact what happens to my trig bash(I calculate the area of S(EAB) in two different ways, then simplify, then have the condition: B_1+3B_2/2=90 and also the equality relates these two angles under trig form. But, then after replacing one angle by the other, I get sinB=sinB :mad: .
#5: I also proved that statement by using the fact that since S is finite, then there exists the maximum value a_n. Then it's easy to show that when gcd(i,j)=1, there exists a bigger number, which is contradictory! Unfortunately, I didn't have time to write it down.... Hmm... The case when gcd(i,j)>1, I just got stuck with it immediately! :(
Well, apparently I seem to did worst in this forum, but the only one who got #7??...

[quote="ghjk"][quote="grn_trtle"]
Well, apparently I seem to did worst in this forum, but the only one who got #7??...[/quote][/quote]

How did you do #7? It was a much much harder geometry problem than #4...

Also, Quatto got #7.

[quote="Ubemaya"][quote="ghjk"][quote="grn_trtle"]
Well, apparently I seem to did worst in this forum, but the only one who got #7??...[/quote]

How did you do #7? It was a much much harder geometry problem than #4...

Also, Quatto got #7.[/quote][/quote]
So, he got all?? :o That's pretty bad for the cutoff though I'm probably out, even for Honorable Mention!
On #7, consider the fact that tanA+tanB+tanC=tanA.tanB.tanC. Now transforming the LHS into tanA, tanB and tan C. Then the rest is just brute-forcing(basically, clearing all denominator)+a bit AM-GM. That's it! If I won't be under the time pressure, I would find a much nicer way to do this final step instead of clearing denominator. But I need a solution as quick as possible, so I decided to put it down.
How did you do #4?? I proved a lot of things as: 2 inscribed quadrilateral(I meant quadrilaterals lie on the circle but forget the terminology). I also did prove that 2 lines are parallel to AB and also got the identity that seems to be clueless to bash. A lot of things to prove on that figure, except the final result :( . After you find out the angle AEB, did you basically get the number of sides by 360/degree of AEB?? It's a nice problem though.

[quote="ghjk"][/quote][quote="Ubemaya"][quote="ghjk"][quote="grn_trtle"]
Well, apparently I seem to did worst in this forum, but the only one who got #7??...[/quote]

How did you do #7? It was a much much harder geometry problem than #4...

Also, Quatto got #7.[/quote]
So, he got all?? :o That's pretty bad for the cutoff though I'm probably out, even for Honorable Mention!
On #7, consider the fact that tanA+tanB+tanC=tanA.tanB.tanC. Now transforming the LHS into tanA, tanB and tan C. Then the rest is just brute-forcing(basically, clearing all denominator)+a bit AM-GM. That's it! If I won't be under the time pressure, I would find a much nicer way to do this final step instead of clearing denominator. But I need a solution as quick as possible, so I decided to put it down.
How did you do #4?? I proved a lot of things as: 2 inscribed quadrilateral(I meant quadrilaterals lie on the circle but forget the terminology). I also did prove that 2 lines are parallel to AB and also got the identity that seems to be clueless to bash. A lot of things to prove on that figure, except the final result :( . After you find out the angle AEB, did you basically get the number of sides by 360/degree of AEB?? It's a nice problem though.[/quote][quote="ghjk"][/quote]

Hmm... that solution eventually turns the problem into cyclic sum of x+y/x+y+2z is greater than or equal to 3/2, which is just normalization and cauchy-schwarz lemma. Thats a nicer way.

Hmm... 5 you can just consider the smallest two elements and prove that, since they can't be relatively prime, the second smallest has to be a multiple of the smallest, and furthermore, the second smallest is k^2-k. Similarly, you can get that third smallest is also a multiple of smallest, and that its k(k^2-k-1), but this gives a contradiction if you consider the gcd of the second smallest and the third smallest.

My main problem is that I didn't think of choosing the elements to prove that S exists. Basically, after done with gcd(i,j) must be greater than 1, I didn't know how to continue! Now I ask myself:" Why I didn't think of finding the most simple set, with two elements only, to prove that when gcd(i,j)>1, then there exists such a set like that?" I think of n-element set, which basically kill me! Your proof on #5 on gcd(i,j) is more complicated than mine, Pythag011!

so, not too bad, although quatto got everything??
#3, yeah, mods, w/e
#4, how did everyone miss this? i thought it was really easy... just angle chase and a bunch of isosceles triangles if i'm not mistaken
#5, basically right but got a little error because i didn't prove the two-element sets didn't have any more elements. also, did anyone else include all one-element sets and the null set?
#6, coloring according to Im(z) mod 3 => easiest way, i think
#7, wow geometry and inequalities, tried and failed to do anything on it.

Oh snap I think like no one got the null set for number 5...lol.
Bleh proving that you couldn't have more than 2 elements in S was the toughest part...so not a small mistake? whatever..

[quote="bowei"]
#4, how did everyone miss this? i thought it was really easy... just angle chase and a bunch of isosceles triangles if i'm not mistaken
[/quote]
I was wondering about this as well... I think many people drew in all the lines at once and got carried away?  The problem becomes a lot simpler if you just draw in points E, F, G, A, B, for instance, and then the angles of triangle EAB can be found very quickly.

Solved 3, 4, 6. Got the answer to #5 but accomplished nothing of substance otherwise. Showed the equality case for #7. Fail, fail, fail.

(Though my amount of failure on a certain more recent test far surpasses my failure on the BAMO...)

[quote="ghjk"]How did you do #4?? [/quote]
[hide="STUFF"]
Okay, so say you have triangle $ ABC$ with $ D$ and $ E$ on $ AB$ and $ AC$ respectively with $ AB \equal{} AC$ so that $ BC \equal{} CD \equal{} DE \equal{} AE$.  Let $ \angle ACB \equal{} \alpha$.  Then, $ \angle CDB \equal{} \alpha$, so $ \angle DCB \equal{} 180 \minus{} 2\alpha$, so $ \angle ACD \equal{} 3\alpha \minus{} 180$, so $ \angle CED \equal{} 3\alpha \minus{} 180$, so $ \angle EAD \equal{} \frac {\angle CED}{2} \equal{} \frac {3\alpha}{2} \minus{} 90$.  Now, $ \angle ABC \plus{} \angle ACB \plus{} \angle BAC \equal{} 180$, so $ \alpha \equal{} \frac {540}{7}$, so $ \angle CAB \equal{} \frac {180}{7}$, so the polygon has $ \frac {360}{\angle CAB} \equal{} 14$ sides. [/hide]

Also, here is a weird solution to number 7?
[hide="Solution"]
Let $ \triangle ABC$ be a triangle so that $ \angle BAC \equal{} \alpha$, $ \angle ABC \equal{} \beta$, and $ \angle ACB \equal{} \gamma$.  Let it have circumcenter $ O$, circumradius $ R$, and let the parallel to $ BC$ through $ A$ intersect the circle at $ A'$.  Let the midpoints of $ BC$ and $ AA'$ be $ M$ and $ N$ respectively and let the foot of the altitude from $ A$ to $ BC$ be $ D$.  Let $ AO$ meet $ BC$ at $ X$, $ BO$ meet $ AC$ at $ Y$, and $ CO$ meet $ AB$ at $ Z$.  Notice that $ \angle OBC \equal{} 90 \minus{} \alpha$, so $ OM \equal{} R\cos \alpha$.  Since $ AA'CB$ is an isosceles trapezoid, $ \angle ABA' \equal{} \angle ABC \minus{} \angle ACB \equal{} \beta \minus{} \gamma$.  Thus, $ ON \equal{} R\cos (\beta \minus{} \gamma)$, so $ \frac {OX}{AO} \equal{} \frac {OM}{ON} \equal{} \frac {\cos \alpha}{\cos (\beta \minus{} \gamma)}$ from the similar triangles $ OMX$ and $ ONA$.  Now, notice that\[ \sum \frac {AO}{AX} \equal{} \sum \frac {[ABOC]}{[ABC]} \equal{} 2\]since we count $ [ABC]$ twice when adding the areas of $ ABOC$.  Thus, by Cauchy,\[ 2\sum \frac {AX}{AO} \equal{} \sum \frac {AX}{AO}\sum \frac {AO}{AX}\ge (1 \plus{} 1 \plus{} 1)^2 \equal{} 9\]so\[ \sum \frac {AX}{AO}\ge \frac {9}{2}\implies \sum \frac {\cos \alpha}{\cos (\beta \minus{} \gamma)} \equal{} \sum \frac{OX}{AO}\ge \frac {3}{2}\]
and we are done. [/hide]

[quote="ghjk"][quote="grn_trtle"]
Well, apparently I seem to did worst in this forum, but the only one who got #7??...[/quote]

How did you do #7? It was a much much harder geometry problem than #4...
[/quote]

I'm almost certain I didn't write that.

[quote="bowei"]
#6, coloring according to Im(z) mod 3 => easiest way, i think
[/quote]

That's pretty cool! Anyone want to explain how that works?

[quote="bookaholic"]
(Though my amount of failure on a certain more recent test far surpasses my failure on the BAMO...)
[/quote]

Agree.

#7:Lagrange Bash with partial derivatives. Calculus for the win.

#6: Contraction Mappings in two dimensions with invariants. I like this problem the best.

I found number 4 the easiest one out of the 4 problems I actually managed to do (BAMO8, I hate writing solutions).

Suppose you are the perfect-est (not a word, but w/e) writer in the world. In BAMO, if you get a solution and write it up in your perfect way, but the solution is not elegant, what's the lowest score you can get on hte problem?

I think given that the solution is factually accurate but inelegant, it should get a full score.

However if the solution is EXTREMELY INELEGANT (read: at least 100 cases, bruteforce to near-calculator extremities) points may be deducted...

[quote="SplashD"]Hmm...so like somebody should post the problems in plain sight.[/quote]
Blah I'll type them again:

3. Find all possible averages of 2 natural numbers a and b such that a^2 and b^2 are congruent mod 100.

4. Anyone want to put the diagram? I'm sure it's not worth the time though, seeing that it has appeared on mathcounts before...its a 2 stepper angle chase.

5. Call a set magic if for any 2 elements i and j in the set, (i+j)/GCD(i,j) is also in the set.

6. Six frogs sit at the vertices of a regular hexagon in the plane. Frogs can move in 2 ways:
Rule 1: Frog A jumps over Frob B, and lands on the line AB at a distance 2AB on the opposite side of B as A.
Rule 2: Frog A jumps over Frog B, and lands on the line AB at a distance AB/2 on the opposite side of B as A.

If we can jump any frogs over each other, and use either rule, is it possible to have a frog reach the center of the original hexagon in a finite number of moves?
(screw part 1, its the same thing)

7. Prove that if a,b,c are angles of an acute triangle, then:
$ \frac {\cos \alpha}{\cos (\beta \minus{} \gamma)} \plus{} \frac {cos \beta}{\cos (\alpha \minus{} \gamma)} \plus{} \frac {\cos \gamma}{\cos (\alpha \minus{} \beta)} \geq \frac32$


quadratics
induction
number theory unsolved
number theory

For which numbers $ n$ is there a positive integer $ k$ with the following property: The sum of digits for $ k$ is $ n$ and the number $ k^2$ has sum of digits $ n^2.$

Consider \[ k\equal{}\sum_{i\equal{}1}^n g^{2^i}\] (g - base of system, for example 10). Obviosly $ s(k)\equal{}n$, $ k^2\equal{}\sum_{i,j}g^{2^i\plus{}2^j}$ and $ s(k^2)\equal{}n^2$ if $ g>2$.

Are there any other $ k$?
Btw, how did you come up with that?

It's easy to verify that all of the digits of $ k$ are $ \leq 3$. If $ k$ is satisfying the given condition then $ 10^sk$ as well. So we can assume that $ 10\nmid k$. We make induction with base $ l$-the number of the digits of $ k$. $ l \equal{} 1$ $ \Longrightarrow$ $ k \equal{} 1,2,3$, $ l \equal{} 2$ $ \Longrightarrow$ $ k \equal{} 11,12,13,21,22,31$. For $ l \equal{} 3$ we have that $ k \equal{} 100p \plus{} q$ where $ p\in \{1,2,3\}$, $ q\in \{1,2,3,11,12,13,21,22,31\}$. $ k^2 \equal{} 10000p^2 \plus{} 200pq \plus{} q^2$. If $ p \equal{} 3$ $ \Longleftrightarrow$ $ k^2 \equal{} 90000 \plus{} 600q \plus{} q^2$ $ \Longrightarrow$ the digits of $ q\leq 1$ $ \Longrightarrow$ $ q \equal{} 1,11$. If $ p \equal{} 2$ $ \Longleftrightarrow$ $ k^2 \equal{} 40000 \plus{} 400q \plus{} q^2$ $ \Longrightarrow$ $ q \equal{} 1,2,11,12,21$. If $ p \equal{} 1$ $ \Longleftrightarrow$ $ k^2 \equal{} 10000 \plus{} 200q \plus{} q^2$ $ \Longrightarrow$ $ q \equal{} 1,2,3,11,12,13,21,22$. We can proceed analogically.


Show that the graph of an equation of the form
Cy^2 + Dx + Ey + F = 0, where C does not = zero is:

(a) two horizontal lines if D = 0 and E^2 - 4CF > 0.

(b) Contains no points if D = 0 and E^2 - 4CF < 0.

Following my answer from this topic: [url]http://www.mathlinks.ro/viewtopic.php?p=956740[/url]

if $ E^2\minus{}4CF>0$, there are two different values for $ y$, so they are two horizontal lines; and

if $ E^2\minus{}4CF<0$, $ y$ is not a real number, or in other words, $ Cy^2\plus{}Ey\plus{}F\neq0$ for any real number $ y$, and so it contains no points.


Let $ ABC$ be a triangle and $ PQRS$ a square with $ P$ on $ AB$, $ Q$ on $ AC$, and $ R$ and $ S$ on $ BC$. Let $ H$ on $ BC$ such that $ AH$ is the altitude of the triangle from $ A$ to base $ BC$. Prove that:

(a) $ \frac{1}{AH} \plus{}\frac{1}{BC}\equal{}\frac{1}{PQ}$
(b) the area of $ ABC$ is twice the area of $ PQRS$ iff $ AH\equal{}BC$

a)we have $ \frac {PQ}{AH} \plus{} \frac {PQ}{BC} \equal{} \frac {PS}{AH} \plus{} \frac {PA}{AB} \equal{} \frac {PB}{AB} \plus{} \frac {PA}{AB} \equal{} 1$
b)if $ AH \equal{} BC$ then $ area(ABC) \equal{} 1/2(AH.BC)$ and $ area(PQRS) \equal{} PQ^2 \equal{} AH^2/4$ (by a) $ \equal{} 1/2 area(ABC)$.
if $ area(PQRS) \equal{} 1/2 area(ABC)$ then from : $ (\frac{1}{AH} \plus{} \frac {1}{BC})^2 \equal{} \frac {1}{PQ^2}$ we find$ (\frac{1}{AH} \minus{} \frac {1}{BC})^2 \equal{} 0$ so $ BC \equal{} AH$.

Construct the square on BC, get  PQ/a=h/(a+h), which gives both relations, a being the size of BC, while h - the altitude's from A in the  given triangle.

Best regards,
sunken rock


Hey, I'm having some trouble with a problem right now:

Suppose that the coef. of the equ. $x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}^{x}+a_{0}= 0$ are real and satisfy $0<a_{0}\le a_{1}\le \cdots \le a_{n-1}\le 1$.  Let $z$ be a complex root of the equ. with $|z| \ge 1$. Show that $z^{n+1}= 1$.

I've noticed that $\frac{z^{n+1}-1}{z-1}= z^{n}+z^{n}-1+z^{n}-2+\cdots+1$. After substituting and multiplying, you get that $z^{n+1}=1$.  If this is indeed the only case, then how do I prove that all of the coef. must equal one in order for the abs. value of $z$ to equal $1$?

Are there any cases where it's greater than $1$?

Excuse me!
Has the question  been solved ?

Try the search function. This is an old Chinese Mathematical Olympiad problem (1986, I believe) and has been posted many many times.

Thank you.

Actually, it's hard to use the search function to find the old threads, so why don't people provide links (if they have them) or solutions. This is a problem from AoPS 2.


check it out--atleast 4 SA stream-total amount has been more than doubled!!!!
that's a cool way 2 attract people 4 MSC-when no talent is interested-well,what's ur opinion

how do u know
i don't see any change there the amount is still Rs2000 pm in their site

Edited :rotfl:  :rotfl:  :rotfl:

[quote="Euclidean Geometer"]Par have u forgotten a zero at the end  :?:[/quote]
 :rotfl:  :rotfl:

Rs.4000 pm and Rs.16000 annual contingency!!!!

then SB must be Rs 8000 pm and Rs 20,000 annual contingency :D  :D  :)  :)  :)
great and have they written anything about the added perks like what institutions u have direct admissions to

only 1 --IISER
no other changes
but i think 4 1st time everybody wud have summer prog. at banglore

great that was very nice last time at kolkatta loads of fun food and enjoyment :D ...
u got the letter from DST(deptt. od Science and technology)

When did y'all get this letter? (If you don't mind me asking) When did you get your last instalment?
Also, weren't we supposed to get renewal application in March?

no one from last years scholarship has got the letters i suppose :maybe:
rituraj upadhyaya is thsi years scholarship winner and he is talking about the amount he (SA Stream) will be getting

Yes our scholarship expires thsi march and we here have got our installments very recently(may be 20 days back) of about Rs 5800

General Question. Is the EA scholarship open to all or only previous KVPY Awardees?

no they have to submit a project only SB strem fellows are directly absorbed from SA

Oh then wat do u think will happen to SX stream. any1 here waiting for interviews in may ? Me and alter ego are!

I am tempted....
Slurp...slurp...ummmm.....lots of money....

I cannot wait to lay my hands on the crisp cheque... :rotfl:
(greedy as I am...) :rotfl:

[quote="rituraj007"]is SX and SB+2 same??????[/quote]

Yes they are.

[quote="VARUNARASU"][quote="rituraj007"]to lay ur hands on loads of crisp money
Go to ur nearest bank at night ,confuse guards with ur maths and get it!!!!!!!!![/quote]
ppl if u really want to make money, i dunno wat you're doing in aops rite now :o . dump alll your books and do something constructive :) . likke playing cricket. that'll surely set your pockets jingling( atleast in this country.) :P[/quote]

More specifically playing in the Indian Premier League. I think they should rename it as KBC - Kaun Banega Crorepati.


harmonic polynomial Support calculus integration inequalities

I don't know how to solve the general case, but here's an idea if we assume something stronger: $|u(x)|$ is dominated by a polynomial. Then $u$ defines an element of the dual of Schwartz's space (n french it's called distribution temperee, I guess it's tempered distribution in english) and since $u$ is harmonic, the Fourier transform of $u$ will be a distribution whose support is 0, so by a classical result it implies that the fourier transform is of the form $<v,\phi> =\sum{a_{j}\phi^{(j)}(0)}$. Now, use Fourier inversion theorem to deduce that $u$ is polynomial. I guess however that the general case is much more difficult.

A version of this problem (for tempered distributions) was on a final for me this quarter.

Here's another approach, based on the averaging property: for any radially symmetric distribution $\phi$ of integral 1, $\phi*u=u$. Specializing this, choose $\phi_{\sigma}(x)=\frac1{(\sqrt{2\pi}\sigma)^{n}}e^{-\|x\|^{2}/(2\sigma^{2})}$ for $\sigma>0$- it's a Gaussian with standrd deviation $\sigma$ in each direction.
Now, let $D^{\alpha}$ be a partial differential operator $\frac{\partial^{\alpha_{1}}}{\partial x_{1}^{\alpha_{1}}}\frac{\partial^{\alpha_{2}}}{\partial x_{2}^{\alpha_{2}}}\cdots\frac{\partial^{\alpha_{n}}}{\partial x_{n}^{\alpha_{n}}}$ of degree $|\alpha|=\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}$. $D^{\alpha}(\phi_{\sigma}*u)(x)=(D^{\alpha}(\phi_{\sigma})*u)(x)= \int_{\mathbb{R}^{n}}q_{\alpha}\left(\frac{y}{\sigma}\right)\frac{\sigma^{-|\alpha|}}{(\sqrt{2\pi}\sigma)^{n}}e^{-\frac{\|y\|^{2}}{2\sigma^{2}}}u(x-y)\,dy$, where $q_{\alpha}$ is a fixed polynomial of degree $|\alpha|$. The leading term of $q_{\alpha}(x)$ is $(-1)^{\alpha}x^{\alpha}$.
Substitute $y=\sigma z$; the integral becomes $\int_{\mathbb{R}^{n}}q_{\alpha}(z)\sigma^{-|\alpha|}u(x-\sigma z)\phi_{1}(z)\,dz$

Now, fix $x$ and suppose $\sigma>\|x\|$; $u(x-\sigma z)\le C(1+\|x-\sigma z\|^{d})\le C\left(1+\sigma^{d}(\|z\|+1)^{d}\right)$. If in addition each $\alpha_{i}$ is even, $q_{\alpha}$

If this were a bound on the absolute value as well, the finish would be easy: $|D^{\alpha}u(x)|\le \int_{\mathbb{R}^{n}}\sigma^{d-|\alpha|}C|q_\alpha(z)|(1+\|z\|)^{d}\phi_{1}(z)\,dz\le C'\sigma^{d-|\alpha|}$, which goes to zero as $\sigma\to\infty$. As it is, it's not that simple.

[quote="grobber"]Let $u: \mathbb R^{n}\to\mathbb R$ be a harmonic function such that $u(x)\le C(1+\|x\|^{d}),\ \forall x\in\mathbb R^{n}$ ($d$ is a positive integer, and $\|x\|$ is the Euclidean norm in $\mathbb R^{n}$). Prove that $u$ is a polynomial of degree $\le d$.

P.S.

Notice that the general version of Liouville's Theorem stating that a positive harmonic function on $\mathbb R^{n}$ is constant is a particular case of this result.[/quote]

Dieudonne's book "fundations of mordern analysis"  take this form as "Liouville Thm" (9.11.1)
in the chapter of "analytic functions". using same symbols as his book:

$f$ is entire function in $C^{p}$, with values in a complex Banach space $E$. Suppose there is $a>0, N>0$
such that $\parallel f(z)\parallel \leq a(1+(supz_{j})^{N})$ in $C^{p}$. Then $f(z)$ is the sum of a finite number of "monomials"
$c_{n1n2...np}z_{1}^{n1}...z_{p}^{np}$ with $c_{n1n2...np}\in E$ and $n1+n2+...+np\leq N$.

Proof:
Let $f(z)=\sum_{\nu}c_{\nu}z^{\nu}$ in $C^{p}$, the power series being everywhere abs convergent. The Cauchy inequalities (9.9.5) applied to polydisk $|z_{j}|\leq R,1\leq j\leq p$ yields, for any $\nu=(n1,...,np)$,

$\parallel c_{n1n2...np}\parallel \leq a(1+R)^{N}R^{-(n1+n2+...+np)}$.

Let $R\to \infty$, we see that $c_{n1n2...np}=0$ unless $n1+n2+...+np\leq N$.

For details, please see his book: very good book.

For harmonic functions it is pretty much the same whether you have a bound on $u$ or on $|u|$. Indeed, if $u<M$ in a ball of radius $R$, then, since $u(0)$ is the average of $u$ over the sphere of radius $R$, we must have the average of $u_{-}=\max(0,-u)$ over that sphere not greater than $|u(0)|+M$. Hence, the average of $|u|$ over the sphere of radius $R$ does not exceed $2M+|u(0)|$ whence the maximum of $|u|$ over the ball of radius $R/2$ does not exceed $C(M+|u(0)|)$.

[quote="Jacobson"][quote]
Dieudonne's book "fundations of mordern analysis"  take this form as "Liouville Thm" (9.11.1)
in the chapter of "analytic functions". using same symbols as his book:
[/quote]

Is Dieudonne's book analysis for human beings or an overload of abstractness and theorems? Are there even exercices in his book? Is it French? [I was planning to  ask the librarian to buy that book a while ago]

moderators may may feel free to delete this reply if they find it inappropriate :)[/quote]

 :lol: what do you think of Bourbaki? Many think they are very good, crystal clear. And Exercises are many and difficult.

Dieudonne's  "Treatise on Analysis" series is well organized and is condensed version of Bourbaki: almost all topics will be used later. So one often find himself go back to review previous materials. (Many important topics are not used later so was put in Excercises: there is tons of excercises, often very difficult. But one don't have to do Ex because it is not needed later).

The weakness in both what harazi said and what jmerry said (which they both acknowledged) is that the hypotheses on $u$ do not directly show that it is tempered, because we haven't immediately ruled out $u$ tending rapidly to $-\infty$ in some directions. In fact, what evidence do we have that the convolution with the Gaussian is defined?

[quote="harazi"]... an element of the dual of Schwartz's space (n french it's called distribution temperee, I guess it's tempered distribution in english)[/quote]
Yes, that is precisely the English translation. In this particular case, I think the English vocabulary comes from French-language sources and probably in particular Schwartz's book.

But how, using the resources available in the $\LaTeX$ on this site, does one write the letter "S" in a way that looks right for it being the Schwartz space? (Or "D" for compactly supported infinitely smooth functions.)

Well, fedja's argument patches everything up- it leads to a bound of the same degree for the absolute value.

The convolution with the Gaussian does actually work- either it exists, or it diverges to $-\infty$. That divergence would violate the averaging property, so the integral converges.
I could have used a compactly supported symmetric bump in place of the Gaussian (the derivative scaling still works right), but the Gaussian is a lot nicer.


function
complex numbers
complex analysis
complex analysis unsolved

Hello.. I would like to ask something...

Let f be a function defined by:

f(z) = 0, if z = 0 and

= [Re(z^2)]^2 / |z|^2 , when z is not 0...

further, Re(z) is "The real part of z"...

Show that f is continuous at z = 0.

I am clueless on how to get the limit of f as z -> 0.... I expressed z as x + yi.  But as z -> 0, (x,y) -> (0,0) and my
expression resulted to an indeterminate form... what should I do?

That should read, "Show that $f$ is discontinuous at $0.$"

Let's look at this function on the two axes:

On the real axis, $f(x+0i)=\frac{\text{Re}(x^2)}{|x|^2}=\frac{x^2}{x^2}=1.$

On the imaginary axis, $f(0+iy)=\frac{\text{Re}(-y^2}{|iy|^2}=\frac{-y^2}{y^2}=-1.$

Along these two different paths, the function tends to two different limits. And other limits are possible; if you let $x=(1+i)t$ for real $t,$ the limit at the origin is $0.$

If you want to express things in terms of $x$ and $y$ we have (except at the origin) that

$f(x+iy)=\frac{x^2-y^2}{x^2+y^2}.$

After that, the argument is the same anyway. Approach the origin along different rays.

Thank you for your help... Now I understand it...

But may I comment something?  I notice that you only consider Re(z^2) instead of [Re(z^2)]^2 as your numerator that's why it became discontinuous at z = 0.   :lol:

$0\leqq f(z)=\frac{x^4}{x^2+y^2}\leqq{x^2} \rightarrow 0$ as $|z|\rightarrow 0$.
Am I wrong?


search
geometry
rectangle
projective geometry
geometry unsolved

[color=darkblue]Let $ (O;R)$ is a circle. Let $ P$ is a point lie out $ (O;R)$, $ (PO > R)$. Let $ (d_1)$ and $ (d_2)$ are two lines pass point $ P$ and tangent with circle $ (O;R)$ at $ A$, $ B$ repectively. Let $ BO$ meet circle at $ C$, let $ H$ is a point lie on $ BC$ and $ AH \perp BC$. Let $ PC$ meet $ AH$ at $ I$. Prove that $ I$ is midpoint of $ AH$.[/color]

[size=150][color=darkred]Hi every One. I have done this problem but I don't like that way. So I post this problem at this forum and I want you give me some nice other solutions[/color][/size].

It is an old problem and have some ways to prove this statement .My solution is only one approach .
$ CP$ cuts the circle at $ J$ then $ CAJB$ is a harmonic  quadrilateral . Consider the projection center C to line $ AH$ .Then it is easy to check that :
$ C \to \infty$
$ A \to A$
$ B\to H$
$ J\to I$
Thus $ (\infty,I,A,H)$ is a harmonic division . So :
$ \frac {\overline{IA}}{\overline{IH}} \equal{} \minus{} 1$ .It gives that I is the midpoint of $ AH$

[quote="silouan"]I have posted here http://www.mathlinks.ro/viewtopic.php?search_id=1543654079&t=86552 a very short solution (but it uses some theory) . But in the link there are also many other nice solutions (like Virgil's one , or shobber's one )[/quote]

Hi silouan. Thank you very much. I see your solution, it's nice, but I think we can done by other way.  :P

Let $ CA$ intersect $ BP$ at $ K$ so $ \angle KAB\equal{}90^o$ and $ \angle PAB\equal{}\angle PBA$ so $ K$ is midpoint of $ KB$
$ AH$ parallel with $ KB$ so $ I$ is midpoint of $ AH$

Is know that in a rectangle trangle the symmedian point is the midpoint of the altitude from the right angle. Here PC and AH pass trought the symmedian point af ABC so I is the midpoint of AH.

edit: sorry I see only now the other solution of silouan on the other topic that is the same of mine  :blush:


linear algebra
matrix
vector
algebra
polynomial
abstract algebra
linear algebra unsolved

If $ M$ is the set containing the unit matrix only, the claim fails to be true - even after dropping the assumption on the odd trace.

[quote="Cristi Sav"]Let $ M$ be a finite set of matrixes with integer elements, each of them having odd trace. Knowing that $ M$ is stable under multiplication operation (if $ A,B \in M$ then $ AB\in M$), prove that $ \sum\limits_{A\in M} A$ is a singular matrix.[/quote]
The matrixes are $ 2 \times 2$, sorry I forgot to say this. It is clear now that the unit matrix doesn't belong to $ M$.

OK, then here is something sketchy:

The potential eigenvalues of each $ A\in M$ are $ 0$ or primitive third or sixth roots of unity, i.e. roots of $ X^2\pm X \plus{} 1$: For $ A\in M$ let $ \lambda$ be an eigenvalue of $ A$ with eigen vector $ x\in (\mathbb Z^*)^2$ where $ \mathbb Z^*$ denotes the ring of algebraic integers: $ Ax \equal{} \lambda x$. Then $ \{A^nx\}_n$ is finite, hence $ \lambda^nx \equal{} \lambda^mx$ for some $ m < n$. This implies that $ \lambda \equal{} 0$ or a root of unity. If $ A$ has 0 as eigen value, then 0 is a simple eigen value, since the trace of $ A$ is odd. Then the other eigen value is a rational root of unity, i.e. $ \pm 1$. If 0 is not an eigen value of $ A$, together with $ \lambda$ its conjugate $ \overline{\lambda}$ is an eigen values of $ A$. Then $ 2 Re(\lambda) \equal{} \lambda \plus{} \overline{\lambda} \equal{} Tr(A)\equiv 1 \mod 2$. Using that $ \lambda$ is a root of unity yields that $ \lambda$ is either a third or sixth primitive root of unity, hence $ \lambda$ and $ \overline{\lambda}$ fulfils the polynomial $ X^2\pm X \plus{} 1$. Since $ A$ is diagonalizable, so does $ A$.

Define $ A^* \equal{} \sum_{A\in M}A$. If there is a regular $ A\in M$, then $ AA^* \equal{} A^* \equal{} A^*A$. Multiplying $ A^2\pm A \plus{} 1 \equal{} 0$ with $ A^*$ yields $ A^* \equal{} 0$ which is singular. ;) Alternatively assume that all matrices in $ M$ be singular. Then they all have rank 1, by the above reasoning, such that we have WLOG $ Ae_2 \equal{} \alpha_AAe_1$ with some integer $ \alpha_A$. It holds that $ \alpha_A$ is independent of $ A$, say $ \alpha$. Then we have $ A^*e_2 \equal{} \alpha A^*e_1$, in particular $ A^*$ is singular.

Hi third edition,
I am skeptic about your proof.
1) In the first case you find $ A^*\equal{}0$. But certain matrices $ A\in{M}$ satisfy $ X^2\plus{}X\plus{}I\equal{}0$ and other satisfy $ X^2\minus{}X\plus{}I\equal{}0$ !!!
2) In the second case you must prove that
a) there exists $ e_1$ s.t. for all $ A\in{M}$ $ Ae_1\not\equal{}0$.
b) $ \alpha_A$ is independent of $ A$.

Some ideas:
1) Let $ A\in{M}$. $ M$ is finite then there exist $ l<k$ s.t. $ A^k\equal{}A^l$. $ A$ is not invertible else $ I\in{M}$. Thus $ A^2\equal{}\alpha{A}$ (Cayley Hamilton) (where $ \alpha$ is a non zero integer else $ tr(A)\equal{}0$). For all $ k, A^k\in{M}$ (finite) then $ \alpha\equal{}\pm{1}$.
Conclusion: $ spectrum(A)\equal{}\{0,\pm{1}\}$.
2) We may assume $ U\equal{}\begin{pmatrix}1&0\\0&0\end{pmatrix}\in{M}$.
Proof: Let $ A\in{M}$; if $ A^2\equal{}\minus{}A$ then we take $ A^2$. Thus we may assume $ A^2\equal{}A$. $ A\equal{}\begin{pmatrix}a&b\\c&1\minus{}a\end{pmatrix}$ with $ a(1\minus{}a)\equal{}bc$. Thus $ b$ divides $ a$ or $ 1\minus{}a$. We prove that there exists a matrix of $ A$-eigenvectors $ P$ with $ det(P)\equal{}\pm{1}$.
Case 1: $ b\not\equal{}0$.
An eigenvector associated to $ 0$ is $ [\minus{}b,a]^T$.
An eigenvector associated to $ 1$ is $ [\minus{}b,a\minus{}1]^T$.
We are done because $ b$ divides one of the preceding eigenvectors.
Case 2: $ b\equal{}0$.
An eigenvector associated to $ 0$ is $ [0,1]^T$.
An eigenvector associated to $ 1$ is $ [1,c]^T$.
3) Let $ A\equal{}\begin{pmatrix}a&b\\c&d\end{pmatrix}\in{M}$  s.t. $ A\not\equal{}\pm{U}$. $ UA\equal{}\begin{pmatrix}a&b\\0&0\end{pmatrix}$; then $ a\equal{}\pm{1}$.
$ AUA\equal{}\begin{pmatrix}1&ab\\ca&cb\end{pmatrix}$ with $ tr(AUA)\equal{}\pm{1}$.
Case 1: $ cb\equal{}0$ then $ A$ or $ A^T\equal{}\begin{pmatrix}\pm{1}&0\\c&0\end{pmatrix}$ where $ c\not\equal{}0$.
Case 2: $ cb\equal{}\minus{}2$ then $ a\equal{}1,d\equal{}\minus{}2$ or $ a\equal{}\minus{}1,d\equal{}2$. We obtain $ 8$ candidates.

In fact I think that $ M$ satisfies 1) or 2).
1) $ M\subset\{\begin{pmatrix}\pm{1}&b\\0&0\end{pmatrix},b\in\mathbb{Z}\}$ or
 $ M\subset\{\begin{pmatrix}\pm{1}&0\\b&0\end{pmatrix},b\in\mathbb{Z}\}$
2) $ M\equal{}[U,W]$ where $ W$ is one of the $ 8$ candidates, $ card(M)\equal{}8$ , $ A^*\equal{}0$.

Please observe that if $ A_{k} \equal{}\begin{pmatrix} k & k\\ 1\minus{}k & 1\minus{}k \end{pmatrix}$, then for all finite parts $ K$ of $ N$ we have $ M_{K} \equal{} \{A_{k} | k  \in K \}$ a proper set. I hope it helps.

I prove the results announced in my preceding post.
In particular I give the form of $ M$ but I think there exists an easier way to solve the initial problem.
Let us recall that we may assume $ U\in{M}$.
Case 1: none among the $ 8$ candidates are in $ M$. Then, except $ \pm{U}$, the elements of $ M$ are in the form:$ A\equal{}\begin{pmatrix}\pm{1}&0\\c&0\end{pmatrix}$ or $ B\equal{}\begin{pmatrix}\pm{1}&b\\0&0\end{pmatrix}$ where $ b\not\equal{}0,c\not\equal{}0$. $ AB\notin{M}$ then $ M$ contains only matrices in form $ A$ or only matrices in form $ B$.
Conclusion: $ M\subset\{\begin{pmatrix}\pm{1}&b\\0&0\end{pmatrix},b\in\mathbb{Z}\}$ or $ M\subset\{\begin{pmatrix}\pm{1}&0\\b&0\end{pmatrix},b\in\mathbb{Z}\}$.
Then $ A^*$ or $ A^{*T}$ is in the form  $ \begin{pmatrix}u&0\\b&0\end{pmatrix}$ and we are done.
Case 2: One of the $ 8$ following candidates is in $ M$:  $ A\equal{}\begin{pmatrix}1&\minus{}2\\1&\minus{}2\end{pmatrix},B\equal{}\begin{pmatrix}1&2\\\minus{}1&\minus{}2\end{pmatrix},C\equal{}\begin{pmatrix}1&\minus{}1\\2&\minus{}2\end{pmatrix},D\equal{}\begin{pmatrix}1&1\\\minus{}2&\minus{}2\end{pmatrix},\minus{}A,\minus{}B,\minus{}C,\minus{}D$.
$ A.B,A.C,A.D,B.C,B.D,C.D\notin{M}$. Then $ M$ contains at most one element among $ A,B,C,D$ and at most one element among $ \minus{}A,\minus{}B,\minus{}C,\minus{}D$.
Assume for example $ A\in{M}$. Then $ \{\pm{U},\pm{A},\pm{T},\pm{V} \}\subset{M}$
Where $ T\equal{}\begin{pmatrix}1&\minus{}2\\0&0\end{pmatrix},V\equal{}\begin{pmatrix}1&0\\1&0\end{pmatrix}$ . Suppose there exists in $ M$ one more element $ W$. Then $ W$ is in the form $ W_1\equal{}\begin{pmatrix}1&0\\c&0\end{pmatrix}$ where $ c\not\equal{}0,c\not\equal{}1$   or $ W_2\equal{}\begin{pmatrix}\minus{}1&0\\c&0\end{pmatrix}$ where $ c\not\equal{}0,c\not\equal{}\minus{}1$ or $ W_3\equal{}\begin{pmatrix}1&b\\0&0\end{pmatrix}$ where $ b\not\equal{}0,b\not\equal{}\minus{}2$ or $ W_4\equal{}\begin{pmatrix}\minus{}1&b\\0&0\end{pmatrix}$ where $ b\not\equal{}0,b\not\equal{}2$.
$ W_1.T,W_2.T,W_3.V,W_4.V\notin{M}$.
Conclusion: $ M$ has $ 8$ elements and $ A^*\equal{}0$.

[quote="Cristi Sav"]Please observe that if $ A_{k} \equal{} \begin{pmatrix} k & k \\
1 \minus{} k & 1 \minus{} k \end{pmatrix}$, then for all finite parts $ K$ of $ N$ we have $ M_{K} \equal{} \{A_{k} | k \in K \}$ a proper set. I hope it helps.[/quote]
A proper set ? Do you want to say a proper subset of $ M$?
You can take $ M\equal{}A_k$ because $ A_k^2\equal{}A$.

Loup blanc, thanks a lot for proof reading !!! Apologies for the late reply. Rushing through your solution above I came across the 8 candidates as well. Then it really gets a tedious calculation. Some short remarks on yours:
[quote="loup blanc"]Hi third edition,
I am skeptic about your proof.
1) In the first case you find $ A^* \equal{} 0$. But certain matrices $ A\in{M}$ satisfy $ X^2 \plus{} X \plus{} I \equal{} 0$ and other satisfy $ X^2 \minus{} X \plus{} I \equal{} 0$ !!![/quote]I had the following argument in mind: If there is $ A\in M$ with, say, $ A^2\plus{}A\plus{}I\equal{}0$, then $ A^2A^*\plus{}AA^*\plus{}A^*\equal{}0$ which implies $ A^*\equal{}0$ using $ AA^*\equal{}A^*$. But it is a lot easier: we get $ A^6\equal{}I$ such that $ I\in M$ which is not the case due to the trace condition. The benefit of this route is - as you have outlined already - that every matrix in $ M$ is singular. This implies that they all have rank 1.
[quote]2) In the second case you must prove that
a) there exists $ e_1$ s.t. for all $ A\in{M}$ $ Ae_1\not \equal{} 0$.[/quote]
This is a very crucial point as I simply assumed that $ e_1$ is the first unit vector. To prove that there is some $ x_0\in\mathbb Z^2$ such that $ Ax_0\neq 0$ for all $ A\in M$. Otherwise we have $ \mathbb Z^2\equal{}\cup_{A\in M} ker(A)$, i.e. $ \mathbb Z^2$ is the [b]finite[/b] union of free sub-modules of rank 1. This implies that $ ker(A)\equal{}\mathbb Z^2$ which is absurd (work over $ \mathbb Q$ or use some prime element argument in $ \mathbb Z$).
[quote]b) $ \alpha_A$ is independent of $ A$.[/quote]Here you're breaking my leg. ;) ... and obviously it cannot be proven without any additional assumption.


function
calculus
integration
complex analysis
complex analysis unsolved

Let $ \mathfrak{F}$ be the family of functions $ f$ analytic on the unit disk $ \{ z: |z|<1 \}$ with $ |f(z)| \leq 1$. How do I show that $ \mathfrak{F}$ is an equicontinuous family of functions?

[hide="Hint"]Apply Cauchy's integral formula to $ f(z)\minus{}f(z_0)$ on a circle line around $ z_0$ in the unit disk, then use $ \frac{f(\zeta)}{\zeta\minus{}z}\minus{}\frac{f(\zeta)}{\zeta\minus{}z_0}\equal{}(z\minus{}z_0)\frac{f(\zeta)}{(\zeta\minus{}z)(\zeta\minus{}z_0)}$.[/hide]


[quote="s.m.a"]positive numbers $x,y$ satisfy $x^{2} + y^{3} \geq \ x^{3} + y^{4}.$ prove that $x^{3} + y^{3} \leq \ 2.$[/quote]

This inequality is from the final step of Russian Math Olimpiad :)

[quote="GHAVR"][quote="zhaobin"]We can get by $x^{2} + y^{3} \geq \ x^{3} + y^{4}$:
\[ x^{3} + y^{4} \le x^{2} + y^{3} \le x+y^2 \le 1+y  \]
[/quote]

Could you explain the second sign $\displaystyle \geq$
:read:  :?:[/quote]
Using Cauchy-Shwarz and condition of problem we get $(x+y^2)(x^3+y^4)\geq (x^2+y^3)^2\geq (x^2+y^3)(x^3+y^4)$, hence $x+y^2\geq x^2+y^3$.


algebra
polynomial
trigonometry
quadratics
algebra proposed

Let $ \gamma$ be a fixed angle in $ (0, \pi)$ and consider the polynomial

$ P_n(x) \equal{} x^n \sin \gamma \minus{} x \sin n \gamma \plus{} \sin (n \minus{} 1) \gamma$

where $ n \geq 3$ is a positive integer.

1) How many quadratic polynomials $ Q(x)$ exist such that $ Q(x)$ divides $ P_n(x),$
for all positive integers $ n \geq 3?$
2) How many linear polynomials $ R(x)$ exist such that $ R(x)$ divides $ P_n(x),$
for all positive integers $ n \geq 3?$

[quote="Arne"]Let \gamma be a fixed angle in (0, \pi) and consider the polynomial

Pn(x) = xn sin \gamma - x sin n\gamma + sin (n - 1)\gamma

where n \geq 3 is a positive integer.

1) How many quadratic polynomials Q(x) exist such that Q(x) divides Pn(x),
for all positive integers n \geq 3?
2) How many linear polynomials R(x) exist such that R(x) divides Pn(x),
for all positive integers n \geq 3?[/quote]
[hide]with $ n \geq 3$ we have:
$ P_{n\plus{}1}(x)\equal{}xP_n(x)\plus{}(x^2\minus{}2xcos \gamma\plus{}1)sin(n \gamma)$[/hide]


geometry
ratio
analytic geometry
trigonometry
number theory
relatively prime

In triangle $ABC$, $AB=13,$ $BC=15$ and $CA=17.$  Point $D$ is on $\overline{AB},$ $E$ is on $\overline{BC},$ and $F$ is on $\overline{CA}.$  Let $AD=p\cdot AB,$ $BE=q\cdot BC,$ and $CF=r\cdot CA,$ where $p,$ $q,$ and $r$ are positive and satisfy $p+q+r=2/3$ and $p^2+q^2+r^2=2/5.$  The ratio of the area of triangle $DEF$ to the area of triangle $ABC$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers.  Find $m+n.$

My answer is $61$

Well, I don't know if there is a specific way for solving this kind of problems

[hide="But here is my solution"]
$(p+q+r)^2 = p^2+q^2+r^2 + 2(pq+qr+rp)$

$2(pq+qr+rp) = (p+q+r)^2 - (p^2+q^2+r^2)$

$2(pq+qr+rp) = (\frac{2}{3})^2 - \frac{2}{5} = \frac{4}{9} - \frac{2}{5} = \frac{2}{45}\Rightarrow$

$pq+qr+rp = \frac{1}{45}$

$(ABC) = (BDE)+(CEF)+(ADF)+(DEF)$

$ \begin{array}{c} (BDE) = \frac{1}{2} \cdot BD \cdot BE \cdot sinB \\ \\ (ABC) = \frac{1}{2} \cdot AB \cdot BC \cdot sinB \end{array} \}\Rightarrow$

$\frac{(BDE)}{(ABC)} = \frac{BD}{AB} \cdot \frac{BE}{BC} = (1-p)q$

similarly

$\frac{(CEF)}{(ABC)} = (1-q)r$

$\frac{(ADF)}{(ABC)} = (1-r)p$

so we have

$\frac{(BDE)+(CEF)+(ADF)}{(ABC)} = (1-p)q +(1-q)r +(1-r)p =$

$=(p+q+r) -(pq+qr+rp) = \frac{2}{3} - \frac{1}{45} = \frac{29}{45} \Rightarrow$

$\frac{(DEF)}{(ABC)} = 1-\frac{29}{45} = \frac{16}{45}$

$16$ and $45$ are relatively primes so the sum is $16+45=61$[/hide]

Cool I had the same solution/answer.  Because the triangles share the same "side," you can take those ratios.

Great solution. I liked your reasoning. :)

[hide="Simple Solution"]The barycentric coordinates of D, E, and F are $(1-p,p,0)$, $(0,1-q,q)$, and $(r,0,1-r)$ respectively. We then use the theorem
\[[DEF] = \left| \begin{array}{ccc}d_{1}&d_{2}&d_{3}\\ e_{1}&e_{2}&e_{3}\\ f_{1}&f_{2}&f_{3}\\ \end{array}\right| [ABC] \]
or
\[\frac{[DEF]}{[ABC]}= \left| \begin{array}{ccc}1-p&p&0\\ 0&1-q&q\\ r&0&1-r\\ \end{array}\right|= (1-p)(1-q)(1-r)+pqr=1-p-q-r+pq+qr+pr \]
, then use symmetric sums to find the answer.[/hide]

[quote=droid347]Can we do this without trig or bary?[/quote]

Sorry for the bump, but curious to see if anyone had a synthetic solution? Personally, I used trigonometry (and first tried coordinate bash).

Trigonometry? Bad idea - this is a purely affine problem. The side lengths we're given are completely irrelevant, and only the ratios $p,q,r$ matter. Coordinates aren't a bash here - they're the nicest way to do things. (OK, barycentric coordinates are the nicest way to do things. Throwing in extra pointless variables for a rectangular coordinate system is a waste.)

This is a cute  ratio problem . No need to mess up things with barycentric and stuff . Just  keep using the ratio of bases to ratio of triangles idea ,and the answer comes (as everyone has correctly pointed out ) to be $61$

[b]Solution:[/b] Notice that $\frac{[ADF]}{[ABC]} + \frac{[CEF]}{[ABC]} + \frac{[BED]}{[ABC]} = r(1-q) + q(1-p) + p(1-r) = p + q + r - pq - rq - pr$. We know that $(p+q+r)^2 = \frac{4}{9} \Rightarrow p^2 + q^2 + r^2 + 2(pq + qr + pr) = \frac{2}{5} + 2(pq + qr + pr) = \frac{4}{9} \Rightarrow pq + qr + pr = \frac{1}{45}$. So, $p + q + r - pq - rq - pr = \frac{2}{3} - \frac{1}{45} = \frac{29}{45}$. Therefore, $\frac{[DEF]}{[ABC]} = \frac{16}{45} \Rightarrow \boxed{061}$.


In a graph $ G$, every vertex has degree at least $ k$. Prove that $ G$ has a simple cycle of length at least $ k \plus{} 1$.

Contradiction is the way to go.  If the largest cycle has length $ k$, what can you say about the vertices in it?

t0rajir0u, I don't really understand your approach. My hint would be this:
[hide="Hint"]
Construct a sequence of vertices $ \left(v_0,v_1,v_2,...\right)$ such that for every $ i$, the vertex $ v_i$ is a neighbour of $ v_{i\minus{}1}$ and not among the $ k$ vertices $ v_{i\minus{}1}$, $ v_{i\minus{}2}$, $ v_{i\minus{}3}$, ..., $ v_{i\minus{}k}$. This is possible due to $ \deg v_i\geq k$. Now, the graph has only finitely many vertices, so our sequence cannot go on and on without running into some vertex twice...
[/hide]

  darij

[quote="t0rajir0u"]My apologies; the approach I was thinking of is not in fact any easier.  If the largest cycle $ C$ has length $ k$ then each vertex in it has at least one neighbor not in the cycle.  By assumption, none of these neighbors coincide - in fact, there cannot be a path between any two of these $ k$ neighbors that doesn't pass through $ C$ or else we can enlarge the cycle.  But since each of these neighbors themselves have $ k$ neighbors, which themselves have $ k$ neighbors, etc., this contradicts the graph being finite.[/quote]

I still don't see how it should work. So we have a cycle of length $ k$, and each vertex $ v$ of this cycle has an "ear" attached to it (ear = the set of all vertices $ w$ such that any path connecting $ w$ with the cycle must pass through $ v$). These ears are disjoint. But why are they infinite? There may be smaller cycles inside the ears which are disjoint with our cycle of length $ k$, hence cannot be conjoined to form a $ > k$-cycle. So in your argument "each of these neighbors themselves have $ k$ neighbors, which themselves have $ k$ neighbors, etc.,", some of these $ k \plus{} k^2 \plus{} k^3 \plus{} ...$ neighbours may coincide, leaving a finite amount of neighbours in the end. Am I seeing this right?

  darij

If the ears themselves are smaller cycles, then they too have "ears."  In any case, your argument is much cleaner.


calculus
advanced fields
advanced fields unsolved

Hi there, I am trying to solve the following for T(t), however the 4th power is throwing me off:

A* dT(t)/dt= -B*T(t)^4 +C

Any help would be much appeciated!
Thanks :)
Anika

This has been answered in the Calculus forum, where it belongs. Don't post the same thing multiple times.


prove that $ 19^{n} \minus{} 97\equiv0\mod2^n$ or in other words prove that $ 2^n|19^{n} \minus{} 97$

First, this is false.  (Try $ n \equal{} 3$.)  Second, your misunderstanding of a different (correct) question should never have migrated out of that thread.


[b]This quiz is said to have been written by Einstein in the 19th century, although there is some question as to whether this is true or not. He is said to have stated that 98% of the people in the world cannot solve the quiz.

I can solve it. Can you?[/b]

Facts:

1: There are 5 houses in 5 different colours.
2: In each house lives a person with a different nationality.
3: These 5 owners drink a certain beverage, smoke a certain brand of cigar and keep a certain pet.
4: No owners have the same pet, smoke the same brand of cigar or drink the same drink.

Hints: from the following facts, can you determine WHO KEEPS FISH?

1: The Brit lives in a red house.

2: The Swede keeps dogs as pets.

3: The Dane drinks tea.

4: The green house is on the left of the white house.

5: The green house owner drinks coffee.

6: The person who smokes Pall Mall rears birds.

7: The owner of the yellow house smokes Dunhill.

8: The man living in the house right in the centre drinks milk.

9: The Norwegian lives in the first house.

10: The man who smokes Blend lives next to the one who keeps cats.

11: The man who keeps horses lives next to the man who smokes Dunhill.

12: The owner who smokes Blue Master drinks beer.

13: The German smokes Prince.

14: The Norwegian lives next to the blue house.

15: The man who smokes Blend has a neighbour who drinks water.

[quote="jb05"]That was HARD! It took me about 15 minutes, but after painful toil, I have the answer!!! (I will not put out the entire solution due to its length):

[hide]
yellow; norwegian; water; dunhill; cats
blue; dane; tea; blend; horses
red; brit; milk; pall mall; birds
green; german; coffee; prince; ???
white; swede; beer; blue master; dogs

Therefore, the answer is GERMAN.
[/hide][/quote]

Thats what i got also

[quote="jb05"]That was HARD! It took me about 15 minutes, but after painful toil, I have the answer!!! (I will not put out the entire solution due to its length):

[hide]
yellow; norwegian; water; dunhill; cats
blue; dane; tea; blend; horses
red; brit; milk; pall mall; birds
green; german; coffee; prince; ???
white; swede; beer; blue master; dogs

Therefore, the answer is GERMAN.
[/hide][/quote]

Yes, that is correct.

Would you like me to post Einstein's solution?

I suppose he found something extremely obvious? Well, go ahead. Post Einstein's solution.

[hide="Einstein Solution"]

A, Norwegian, yellow, Dunhill, water, cats
B, Dane, blue, Blend, tea, horses
C, Brit, red, Pall Mall, milk, birds
D, German, green, Prince, coffee, FISH
E, Swede, white, Blue Master, beer, dogs

There are five houses which we will call A, B, C, D, and E (intro), and we can assume they are all in a row since Clue 8 says the man living in it drinks milk. So the man who lives in C drinks milk. (8). The Norwegian lives in the first house (9) so lives in A, and the man whose house is blue lives next to the Norwegian (14), so the blue house is B. The green house is on the left of the white house, so isn’t A [Norwegian], or B [blue], and one in green house drinks coffee (5) so not in C; thus the green house is D [coffee (5)], and the white house E. The Brit lives in the red house (1), so not A [Norwegian], B [blue], D [green], or E [white]; thus the Brit lives in house C and drinks milk, and by elimination, house A where the Norwegian lives is yellow, and he smokes Dunhill (7), and man who has horses lives in blue house at B (11). The Dane drinks tea (3) so isn’t in house A [Norwegian], C [milk], or D [coffee]; the Dane lives in B or E. One who smokes Blue Master drinks beer (12), so isn’t in house A [Dunhill], C [milk], or D [coffee], so also lives in B or E; thus, the Norwegian in A drinks water, which leaves the one in house B [blue, horses] as one who smokes Blend (15). The Swede keeps dogs so not in B and the German smokes Prince so not in B [B, Blend, horses (above)]; thus, the Dane lives in house B, and he drinks tea (3) so by elimination, man in white house at D drinks beer, and smokes Blue Master (12). The German smokes Prince, so not man in house E; he lives in D and the Swede must live in E; so the German in D smokes Prince and the Swede in E keeps dogs (2), and by elimination, the Brit in red house in C who drinks milk is one who smokes Pall Malls and rears birds (6). One who smokes Blend [B, blue, tea, horses] lives next to man who raises cats (10) who isn’t in C [birds]; thus, the Norwegian in A [yellow, Dunhill, water] raises cats, and by elimination, the German in D raises Fish. [/hide]

If I had the time,  I would try to solve it, but for now, the problem looks hard enough.

[hide="my answer"]
German keeps fish[/hide]
I don't think it is a very difficult problem. I just draw a picture and put the conditions under each house and then get the answer.
But it's a pretty good one, really interesting.

had this for a bonus once....never knew it was by Einstein though

[quote="stupidkid"]Are you sure 98% of the population can't solve it?
I think that's a bit too high.[/quote]
Do you yourself the research about it. Try to take a population as heterogeneous as you can, and you'll see that most of the people will give up soon, and a very large number them!

Fortunatelly, in that regard, AoPS and MathLinks is not that heterogeneous, most of the people here is, I would say, very very smart  :D

Mensa = top 2%, so the rest is 98%.

I wonder if any mensa member cannot solve this ;)

[quote="polymorphic"]I don't understand...I found that puzzle extremely exhausting.  After about 15 minutes of getting nowhere, I decided to literally trial and error my way through.  And I got an answer...but apparently it's not the right one.  what is wrong with the following:
[hide](left)
Norwegian: green, blender, coffee, FISH
German:      blue, Prince, water, cats
Swede:       yellow, Durhill, milk, dogs
Brit:            red, Blue Master, beer, horses
Dane:         white, Pall Mall, tea, birds
(right)[/hide][/quote]
I think you misunderstood first.  By first he means left most.

I put the Norwegian in the first house as required.

O sorry, i misread. Ahh...i see
[quote]4: The green house is on the left of the white house.[/quote]
Means that the green house is on the left of the white house with no houses in between.


function
geometry
3D geometry
sphere
topology
induction
advanced fields

Prove that every continuous function $ f: B^n\to B^n$ of a closed $ n$-dimensional ball to itself has a fixed point.

Brouwer's theorem is easy if some homology groups and their properties are used (otherwise the theorem is very hard I think):

Let $S^{n-1}$ be the $n-1$ dimensional sphere, so the border of $B^{n}$, the  $n$ dimensional ball.

A retraction from $A$ to $B \subset A$ is a continuous mapping $r: A \to B$ such that $r(b)=b$ for all $b \in B$.
$B$ is called a retract of $A$ if such a retraction exists.

I will use now without proof that the (reduced) homology groups $H_{n}( \cdot )$ satisfy $H_{k}(B^{m})\cong \{0\}$ (the trivial group) and $H_{k}(S^{m})\cong \{0\}$ iff $k \neq m$ and $\cong \mathbb{Z}$ iff $m=k$.

Lemma: $S^{n-1}$ is not a retract of $B^{n}$.
Proof:
Assume $r: B^{n}\to S^{n-1}$ to be some retraction.
Let $i: S^{n-1}\to B^{n}$ be the canonical injection.
Then $r \circ i$ is the identity on $S^{n-1}$. Now these two mapping give maps
$i_{*}: H_{n-1}(S^{n-1}) \to H_{n-1}(B^{n})$
and
$r_{*}: H_{n-1}(B^{n}) \to H_{n-1}(S^{n-1})$
with $r_{*}\circ i_{*}$ the identity on $H_{n-1}(S^{n-1})$.

But $H_{n-1}(B^{n}) \cong \{0\}$ and $H_{n-1}(S^{n-1}) \cong \mathbb{Z}$, leading to a chain $\mathbb{Z}\to \{0\}\to \mathbb{Z}$ that is supposed to be the identity, clearly impossible. Thus such a retraction cannot exist.

Theorem (Brouwer): Every map $f: B^{n}\to B^{n}$ has a fixed point, so a point $p \in B^{n}$ with $f(p)=p$.
Proof:
Assume again the contrary, so let $f$ be a map without a fixpoint.
Thus for any $p \in B^{n}$ the ray starting at $f(p)$ through $p$ is well defined sicne these points are different. Let $h(p)$ be the intersection of this ray with $S^{n-1}$. It's then easy to see that $h$ is continuous, and that $h$ is the dientity on $S^{n-1}$, in other words: $h$ would be a retraction of $B^{n}$ to $S^{n-1}$, which cannot exist, proving the theorem.

For $n=2$, there is a nice proof using the first homotopy group, in other words the group of homotopic loops, not using that much (but still a bit).

Yes, there is one.
And it is presented in [u]Proofs from the book[/u], by M. Aigner and G. M. Ziegler, which I happen to have in my hands for a few days. :) (It's in the chapter "Pigeon-hole and double counting".)
So, if you're interested in it, and you cannot get a hand on this book, just ask, and I'll post it here.

i have  friend that has this book...

going to his house today and take  it away while he is not paying attention  :wink:

Here is the [url=http://www.cut-the-knot.org/do_you_know/poincare.shtml#brouwertheorem]proof[/url] with Sperner's Lemma

There is a proof using stokes theorem as well: http://www.math.duke.edu/~wka/math204/fixed.pdf

[quote="ZetaX"]

For $ n \equal{} 2$, there is a nice proof using the first homotopy group, in other words the group of homotopic loops, not using that much (but still a bit).[/quote]
Can anyone post this proof?

-Amir.S

See ZetaX's post, the proof is exactly the same..

Yes, in principle it is: replace $ H_1$ by the fundamental group $ \pi_1$.
Then the first part can be done a bit more elementary (without functorial properties). I will repeat the essential steps and change some parts:

A retraction from $ A$ to $ B \subset A$ is a continuous mapping $ r: A \to B$ such that $ r(b) \equal{} b$ for all $ b \in B$.
$ B$ is called a retract of $ A$ if such a retraction exists.

Lemma: $ S^{1}$ is not a retract of $ B^{2}$.
Assume you have a retraction $ r: B^2 \to S^1$. Let $ \alpha : [0,1] \to S^1$ be a loop one times around $ S^1$ (e.g. $ \alpha(t) \equal{} e^ {2\pi t}$). Then we know (from standard fundamental group facts) that $ \alpha$ is not homotopic to a constant loop/map. But we now reach a contradiction: $ h: [0,1] \times [0,1] \to S^1$ defined by $ h(x,t) \equal{} r\left( (1\minus{}t) \cdot \alpha(x) \right)$ is a homotopy from $ \alpha$ to a constant map (as $ h(\cdot,0)\equal{}\alpha, h(\cdot,1)\equal{}r(0)$).

Theorem (Brouwer): Every map $ f: B^{2}\to B^{2}$ has a fixed point, so a point $ P \in B^{2}$ with $ f(p) \equal{} p$.
Proof:
Assume again the contrary, so let $ f$ be a map without a fixpoint.
Thus for any $ p \in B^{2}$ the ray starting at $ f(p)$ through $ p$ is well defined sicne these points are different. Let $ h(p)$ be the intersection of this ray with $ S^{1}$. It's then easy to see that $ h$ is continuous, and that $ h$ is the identity on $ S^{1}$, in other words: $ h$ would be a retraction of $ B^{2}$ to $ S^{1}$, which cannot exist, proving the theorem.

Is it equavelent in brouwer theorem that we prove it for $ f: I^n\rightarrow I^n$ instead of $ f: B^n\rightarrow B^n$?

Yes, because there exists a homeomorphism $ h: I^n\longrightarrow B^n$.

If $ f: I^n\rightarrow I^n$ is a continuous map, then $ f: I^n\rightarrow I^n$ has a fixed point, if and only the continuous map $ hfh^{ \minus{} 1}: B^n\rightarrow B^n$ has fixed point.

You can use induction on $ n$ and construct a homeomorphism from $ B^n$ to $ I^n$.

There is an interesting proof of this theorem in the two-dimensional case using the fact that every game of hex has a winner.

Read this if interested: http://www.math.pitt.edu/~gartside/hex_Browuer.pdf

Here's a combinatorial proof http://amathew.wordpress.com/2010/09/19/the-combinatorial-approach-to-brouwers-fixed-point-theorem/


inequalities induction calculus derivative

Suppose

${a^{2}\geq a}$

Using finite Induction
If this is rigth for $a$, need to be rigth for $a+1$

${(a+1)^{2}\geq a+1}$
${a^{2}+2 a+1\geq a+1}$
${a^{2}+2 a\geq a}$
${a^{2}+a\geq 0}$
${a^{2}\geq-a}$

So

${a^{2}\geq a}$
${a^{2}\geq-a}$

Or

${a^{4}\geq a^{2}}$

If we repeat this spets a lot of times, we get

${a^{2 n}\geq a^{n}}$

That is a more difficult question!!

Take the derivative

${\frac{\partial a^{2 n}}{\partial a}\geq \frac{\partial a^{n}}{\partial a}}$
${2 a^{2 n-1}n\geq a^{n-1}n}$

Since $n \neq 0$ and multiplying both side by $a$

${2 a^{2 n}\geq a^{n}}$

where did this $2$ came!? from hell?!
This is a magic $2$?!?!!? This is the proof that god exist or someting else?!\\
Is $2 = e$ !?

Someone know what the meaning of this $2$?!?!
Please, Tell me more about this $2$

Your post makes very little sense.  The "induction," for example, means nothing.

[quote="Thales418"]${a^{2 n}\geq a^{n}}$[/quote]

If $a^{2}\ge a$ then either $a \ge 1$ in which case this result is obvious or $a \le 0$ in which case this result is false for $-1 < a < 0$ and $n$ even.

[quote="Thales418"]Take the derivative

${\frac{\partial a^{2 n}}{\partial a}\geq \frac{\partial a^{n}}{\partial a}}$[/quote]

Er, in general it is not true that $f(x) \ge g(x) \implies f'(x) \ge g'(x)$.

errr
Forum index » Other Topics » Games & Fun Factory

did you read?

its a joke....

lullaby

so do I

when i read the 'i lose faith in humanity' post by ignite in the other topic, i ignored it, but now i really sympathize with him.

why you mathematicians are too cold!?

is because you dont have friends and a girlfriend?!

the world is good. please. dont kill yourself

http://www.senhorcarteiro.blogspot.com

for other information

I do believe you terminally fail at life.

I'm happy with my existant social life, whatever, you can go talk to your internet friends through Runescape if you want to.