geometry
3D geometry
percent
tetrahedron
ratio
rectangle
function

Please give solutions... Enjoy!

Sprints:

1) given the equation 1/m +1/n = 2/15, where m and n are positive integers, find the value of m+n such that m>n and m is not a multiple of n

2) The dimensions of a rectangular block are 30 cm, 16 cm, and 12 cm.  It is made from cubes which are each 1 cm :^3: in volume.  If the outside of the block is painted red, what percent of the block is made of cubes with exactly 2 sides painted red? Express your answer to the nearest tenth.

3) A sculpture for a math classroom is a figure composed of a 1" cube stacked upon a 3" cube which is in turn stacked upon a 5" cube.  The entire sculpture is painted. Find the number of square inches of the surface that will be painted.

4) Suppose a coin purse contains 30 coins which are either nickels, dimes, and/or quarters.  How many different combinations of these coins are there whose value is $5?

5) In the diagram (below), RECT and LONG are rectangles.  How many square units are in the area of LONG?

6) How many different 4-digit numbers can be formed using the digits 2,4,5,6, & 7 such that no digits repeat and the number is divisible by 4?

7) The first 6 rows of Pascal's triangle are shown below, beginning with row zero.  Except for the 1 at each end, row 4 consists of only even numbers, as does row 2. How many of the first 20 rows of Pascal's triangle have this property?

                   1
                1    1
             1    2    1
           1   3    3   1
         1   4   6    4  1
        1  5  10 10  5  1

8) The piece (below), with all indicated lengths of 5 units, can be folded to form a regular tetrahedron.  What is the volume of the tetrahedron? Express your answer as a fraction in simplest radical form.

9) How many distinct triangles can be constructed by connecting 3 vertices of a cube?

10) In the diagram (below), circle Q is congruent to circle W, and both are tangent to circle O and to each other.  Circle S and circle T are congruent and are tangent to circle O, to circle Q, and to circle W.  Find the ratio of the area of the smallest circle to the largest circle.

Targets:

11) There are positive integers k,n,m such that 19/20< 1/k+1/m+1/n< 1. What is the least possible value of k+m+n?

12) Angle BAC and the exterior angle CBD of triangle ABC are each bisected.  If the angle formed by these bisectors when they meet is 50 degrees, find the number of degrees in angle C of triangle ABC.

13) How many rectangles are in the diagram (below)? All angles shown are right.

14) The function f(x) is called a nested function, and is defined as: f(x)= x-2 for x> 5, and x-f(f(x+3)) for x :le: 5. Find f(1).

Diagrams:

Sorry about these sizes-I have to work fast....Here's #13's picture...

6) How many different 4-digit numbers can be formed using the digits 2,4,5,6, & 7 such that no digits repeat and the number is divisible by 4?

(Answer in spoiler)

[hide]36

A number is divisible by 4 if its last two digits are divisible by 4.  Using the five digits given, the six numbers 24, 64, 72, 76, 52, and 56 are divisible by 4.  For each of these, the first two digits will be from the remaining three, and C(3,2)=3.  For each choice of two digits a and b for the first two digits, they can be arranged as either ab or ba.  So, there are 6*3*2=36 numbers satisfying the conditions.[/hide]

That was fun.. I miss mathcounts =/

7) The first 6 rows of Pascal's triangle are shown below, beginning with row zero. Except for the 1 at each end, row 4 consists of only even numbers, as does row 2. How many of the first 20 rows of Pascal's triangle have this property?

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

Answer in spoiler...

[hide]4

As I recently discovered, you can find the number of odd numbers in a given row n of pascal's triangle by the following process: write n in base two.  Count the number of 1s.  Raise 2 to that power.  That is the number of odd members of row n.  Clearly, if n is a power of 2, there will be one 1 in its binary expansion (10000..).  So, there are two odd numbers in that row; namely the 1s at the beginning and end.  Hence, the rows with only even numbers (besides the 1s) will be rows with a row number that is a power of 2.  There are 4 powers of 2 less than 20 (2, 4, 8, 16), so among the first 20 rows, 4 have the property stated in the question.[/hide][/hide]

12) in spoiler

[hide]Drawing it out, we will call the intersection of BC and the bisector of A the point E, and the intersection of the bisectors F.  We will now look at the triangle BAE.  The angle ABC has measure 180-DBC, the angle BAE has measure BAC/2, and angle BEA has measure DBC-BAE.  Next, we know that EBF has measure DBC/2 angle BFE has measure 50, and angle BEF has measure 180 +. BAE - DBC.  Performing some algebra gives BAC to be 15, DBC to be 115, and so ABC is 65, and BCA is 100[/hide]

2) The dimensions of a rectangular block are 30 cm, 16 cm, and 12 cm. It is made from cubes which are each 1 cm  in volume. If the outside of the block is painted red, what percent of the block is made of cubes with exactly 2 sides painted red? Express your answer to the nearest tenth.

spoiler..

[hide]3.6%

The cubes with two sides painted red will be those along the edges, but not on the corners.  There are four edges in each direction, and the number of cubes on a given edge that is not on the corner is the length of that edge minus 2 (2 corners).  So, there are 4(28)+4(14)+4(10) = 208 cubes with exactly two sides painted red.  There are 30*16*12=5760 cubes in the total figure.  208/5760 = .03611, or 3.6%.[/hide]

3) A sculpture for a math classroom is a figure composed of a 1" cube stacked upon a 3" cube which is in turn stacked upon a 5" cube. The entire sculpture is painted. Find the number of square inches of the surface that will be painted. In spoiler

[hide]

The surface area of the 5" cube visible is 6*25-9=141.  The surface area of the 1" cube visible is 5*1=5.  The surface area of the 3" cube visible is 9*5-1=44.  Thus you paint 190 sq. in.

This might be wrong because I included the very bottom.[/hide]

Caroline, isn't it past midnight for you?

3) A sculpture for a math classroom is a figure composed of a 1" cube stacked upon a 3" cube which is in turn stacked upon a 5" cube. The entire sculpture is painted. Find the number of square inches of the surface that will be painted.

[hide]190

Disregarding overlap, the total surface area is 6(25+9+1)=210.  However, we're counting the 9 square inches of overlap between the 3" and 5" twice when it shouldn't be counted at all, so we have 210-18=192.  Similarly, we're counting the 1 square inch of overlap between the 1" and the 3" twice, so 192-2=190.[/hide]

[quote="Combinatoricsman"]
Caroline, isn't it past midnight for you?[/quote]

Yes, that's why I'm doing mathcounts problems =)

Didn't realize you'd done number three already, sorry..

8) The piece (below), with all indicated lengths of 5 units, can be folded to form a regular tetrahedron. What is the volume of the tetrahedron? Express your answer as a fraction in simplest radical form. In spoiler.

[hide]When this tetrahedron is folded up, it has 5 units for each edge.  Thus, we can just use the formula of 1/12  :rt2: r :^3: and get 125 :rt2: /12[/hide]

9) How many distinct triangles can be constructed by connecting 3 vertices of a cube? In spoiler.

[hide]We can just say that the side of the cube is 1.  Thus, there are 3 distinct trangles, the one with sides 1,  :rt2: , :rt3: , the one with sides 1, 1,  :rt2: , and the one with all sides  :rt2: .  I know that these are the only ones because I found every triangle for a related problem a long time ago.[/hide]

13) In spoiler.

[hide]46.  I counted them systematically, and I don't see any other way.[/hide]

[quote] Didn't realize you'd done number three already, sorry...[/quote]

By all means, do problems someone else has done already!

For #1:

[hide]Let m = x + y and n = x - y with x and y positive integers.  After some simple algebra, we get 2x/(x^2 - y^2) = 2/15.  Cross-multiply and simplify, and you get 15x = x^2 - y^2.  You then easily get (x-15)x = y^2, and it's not too hard to find two squares 15 apart (16 and 1).  Note that 49 and 64 also work, but it results in m | n.  Other values will not work (for the sake of completeness, (a+1)^2 - a^2 and (a+3)^2 - a^2 have already been used, and (a+5)^2 - a^2 = 10a + 25 > 16).  With 16 and 1, we get x = 16 and y = 4.  Plugging in, we see that indeed 1/12 + 1/20 = 2/15.

I think there's an easier way (e.g., guess and check) but this works... [/hide]

For #4:

[hide]Divide everything by 5 and the problem is how many ways to use 30 total coins of 1, 2, and 5 to achieve 100.  Starting with twenty 5's, and then add 0.8 coins for each 1 you use and 0.6 coins for each 2 you use.  This may seem very strange and illogical, but I don't know how to describe it clearly here.  We have 0.8x + 0.6y = 10 (the # of coins we need to add).  We can have pairs (x, y) as (11, 2), (8, 6), (5, 10), and (2, 14).  That gives us 4 pairs and therefore 4 combinations.[/hide]

For #5:

[hide] :tri: CON  :sim:  :tri: LOT, so OC/7 = 3/1, and from that, OC = 21.  (OC+TO)(CN+NE) = (21+1)(7+3) = 220 is the area of RECT (notice that LOT  :cong: GEN).  We subtract the triangles: 220 - (21*7/2 + 21*7/2 + 3*1/2 + 3*1/2) = 70.[/hide]

For #10:

[hide]Let the radius of Q be x, and let the radius of T be y.  x^2 + (2x-y)^2 = (x+y)^2 in the triangle created from joining the centers of circles Q, T, and O, simplifying to 4x^2 - 6xy = 0.  Thus, y/x = 4/6 = 2/3.[/hide]

For #11:

[hide]Obviously k (or m or n; it doesn't matter) must be 2, because if it were 3, then 1/3 + 1/3 + 1/4 is not sufficiently large.  Now, we have 9/20 < 1/m + 1/n < 1/2.  Again, we get that m (or n; it doesn't matter) must be 3 because 1/4 + 1/5 = 9/20.  Finally, we get 7/60 < 1/n < 10/60.  7n < 60 < 10n, so we see that n = 7 at best.  We get k=2, m=3, and n=7.[/hide]

These solutions are mostly way too rigorous (though still not completely rigorous) for MathCounts, but it was at least fun coming up with them.

It's still easier with guess and check ;-)

A really cool way for doing #1:

1/x+1/y = 2/15, x<y

15(x+y) = 2xy

2xy-15(x+y) = 0

xy-7.5(x+y) = 0

(x-7.5)(y-7.5) = 7.5 :^2:

(2x-15)(2y-15) = 15 :^2: = 225

So if a and b are integers such that ab = 225 and a<b, then x = (a+15)/2 and y = (b+15)/2.  If x and y are such that x|y can't be possible, then (b+15)/(a+15) cannot be an integer.  The possible values of (a,b) are (1,225), (3,75), (5,45), (9,25).  Testing those pairs, (25+15)/(9+15) = 5/3 which isn't an integer.  So a=9 and b=25.  Then x=12 and y=20.

1/12+1/20=2/15.

Btw, that was a really nice sol'n for #10.  [b]REALLY[/b] nice.

Cool!
But please, do not bump threads for little purpose.


[i]Find all pair of functions $ f; g$  : $ f: \mathbb{R} \ \to \ \mathbb{R} \ ,\ g: \mathbb{R} \ \to \ \mathbb{R}$ such that :

 $ f(x\plus{}g(y)) \equal{} xf(y) \minus{} yf(x) \plus{} g(x) \ \ \forall \ x;y \ \in  \mathbb{R}$[/i]

[quote="nguyenvuthanhha"][i]Find all pair of functions $ f; g$  : $ f: \mathbb{R} \ \to \ \mathbb{R} \ ,\ g: \mathbb{R} \ \to \ \mathbb{R}$ such that :

 $ f(x \plus{} g(y)) \equal{} xf(y) \minus{} yf(x) \plus{} g(x) \ \ \forall \ x;y \ \in \mathbb{R}$[/i][/quote]

Already posted : see http://www.mathlinks.ro/Forum/viewtopic.php?t=81122


Is computer science a subfield of electrical engineering or is it considered a separate field (like the relationship between math and physics)?

I think nowadays they are considered separate fields. EE usually is hardware-oriented, while CS is software/theory. But naturally there is a lot of overlap; I believe in the past the distinction was more ambiguous.

Some programs, e.g. Berkeley, put the two majors together (EECS -- although I'm not positive how it works there), but from what I know CS majors often do not need to know much EE and vice versa.

In the beginning, CS used to be subfield of Math... but that dates back to times when computers virtually didn't exist :D. In my school we have applied CS and theoretical CS - the first one is more EE oriented and the second one is more pure math oriented. CS is a broad term.

MIT lumps EE and CS together under one major primarily, as I understand it, because that's the way it's always been done.  These days, (theoretical) CS is more or less a branch of math, whereas EE is engineering, which is why we also have an option where EECS majors can also be part math (although I don't know the details).


Χθες ψάχνοντας σε κάτι σημειώσεις βρήκα άλυτη την εψής άσκηση που με παίδεψε χθες αλλά δυστυχώς χωρίς επιτυχία .

Βρείτε τους ακεραίους $x,y,z$  έτσι ώστε ο $x$ να μην εχει διαιρέτες της μορφής 6κ+1 ώστε να ικανοποιείται η σχέση
$xy^{2}=z^{3}+1$

[quote="perfect_radio"]It has been posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=hard&t=28590]here[/url] by harazi.

Albanian Eagle will post his solution any minute now...[/quote]

Thanks  :)  I was searching for it


inequalities
quadratics
geometry
trigonometry
AMC
AIME
summer program

Hi guys,

I'm thinking of running a thread like what towersfreak2006 did with inequailty. For those who don't know what I'm talking about, he did a thread on topic "inequailty" and posted almost all useful inequality theorems that are known (starting from trivia to Schur).

I like to do another one here on topic [b]you guys[/b] choose. I made a poll on following topics so please cast your vote.

1. [b]Algebra[/b]: Starting from basic quadratic equations, we'll attack some hard substitution problems and use of binomial theorem. The problems will range from easy AMC-10 to difficult AMC-12 and easy AIMEs. No proof will be in here.

2. [b]Geometry[/b]: We'll attack Euclidean geometry (school geometry) and trig geometry. The problems WILL include proofs but the very basic ones. Also, there will be many well-known theorems from Pythagorean to Ptolemy's Theorem. Theorem like Ceva won't be used in here since that's above Getting Started. Problems will range from easy AMC-10 to difficult AMC-12 and early AIMEs.

3. [b]Counting[/b]: There won't be much of a "talk" in this topic. It will be basically set of problems from AMC-10 level to AMC-12. No AIME level (even easy) counting problems will be posted. This topic won't go heavy as other topics since this is a topic that is NOT taught in school.

4. [b]Number Theory[/b]: Similar to Counting, this will be most of problem set. But, it will include some very useful theorems (Like Fermat's Little Theorem --> Proof of this won't be posted here) and problems that have clever method using number theory. Problems will range from AMC-10 to AMC-12. No AIME level problems.

The poll will last from today, 9/18/05, to next Sunday. The topic that has most people supporting will be the topic of the thread I'll be running. More information will be available but you are free to ask questions here!

[quote="JesusFreak197"]I like both Algebra and Number Theory, but voted for Number Theory.[/quote]

Just like me, because i like to learn about primes  :D

I had enough algebra after reading algebra 2 about 5 times.  :roll:

My MathPath counselors say I like number theory; therefore I do. Q.E.D. Tombstone.

333rd post!!!!!

[quote="chess64"]I like number theory, but there's no point of doing it on something I'm already good at...so I pick Geometry! :D (my weakest subject)[/quote]
Contrary. I don't think Getting Started geometry will do any good for me, but I know nothing about Number Theory. Therefore I voted for it.

This year at Mathcounts Georgia's team cheer was:

If Number Theory makes you weary
Never fear, Georgia's here!

I like Number Theory!

[quote="mathnerd314"]solafidefarms, are you from Georgia?

For some other team: "Math and alcohol don't mix. Don't drink and derive."

I've never been too good at counting either. I've never understood which parts to overcount and by which factor, and whether to subtract or divide. Links would be appreciated.[/quote]

  INTRODUCTION TO COUNTING AND PROBABILITY

lol i agree. i was really bad at probability before i took that course :blush:

Interesting.

No one topic is dominating (excluding counting since it has only two people now but who knows what will hapen in next couple days).

I really don't mind which topic gets picked. All 4 are of them are my favorites and to be honest, they are all topics that will help you in no matter what level of math competitions you in (from Mathcounts to USAMO --> Maybe not getting a good grade in USAMO since I never been there yet  :roll: but it will give you solid base, hopefully).

im really bad at geometry so i picked it

[quote="chess64"]if its geometry (which it is right now :)) then can the topic go in intermediate?[/quote]

How far do you mean by "Intermediate"?

You have to know that geometry can get very hard from Getting Started to Intermediate. I think there isn't much big increase from Intermediate to Pre-Olympiad compared to Getting Started to Intermediate because Intermediate is where you learn all those theorems (Ceva, Stewart, etc) while Pre-Olympiad is just testing your knowledge and extra not-well known theorem area.

Pick Number Theory!! I need to work on number theory.  ;)

How come there wasn't an intermediate number theory course?

[quote="chess64"]if it's not going to be intermediate-forum level, then i don't really benefit from it...[/quote]

Some problems may be intermediate. I'm betting that geometry and algebra will have most problems since it's mostly used. I'll try add basic problems to very difficult problems that will be more like Intermediate.

Only thing that's not intermediate is "lecture." While learning theorems like Ceva and Menealus are important, only theorems you'll see in first 5-6 AIME (pretty sure) are basic trig combined with geometry and/or geometry with algebra using Pythagorean Theorem, Ptolemy's Theorem, and Law of Cosine, Sine. You have to understand that there had been people in past who ONLY finished AoPS volume 1 and made to USAMO (not necessarily doing well but at least they made it). That's because they covered the AoPS volume 1 geometry thoroughly.

I have two geometry texts other than AoPS volume 1. Challenging Problems in Geometry and Geometry Revisited. Geometry Revisited WILL NOT be used in this because even I don't understand the stuffs in there. Challenging Problems in Geometry may be used as lecture-tool and sample problems.

Most problems will from real math competitions.

i picked algebra because i'm not very good at it.

[quote="236factorial"]Careful about the difficulty in case it gets moved.... I hate it when that happens. I think the one in intermediate by towersfreak2006 was moved to pre-olympiad.[/quote]

Yes. I'm very aware of that.

That's why I say problems only go for early AIME. If not, it will become too difficult for Getting Started.

Since nobody seem to be voting anymore, I'll stop the poll now.

The topic is therefore.. (drumroll).... GEOMETRY!

There will be separate thread for this. I'm hoping to post it pretty soon (not necessarily today but pretty soon) so thanks to all people who participated in this poll!


SMO 2007 Open, #3

Let $A$ be the set of any 20 points in a circle. Joining any two points in $A$ produces one chord of a circle. Suppose every three such chords are not concurrent. Find the number of regions within the circle which are divided by all these chords.

Comment: I didn't quite understand the question, didn't attempt this one.

[hide]Let $f(x)$ be the number of regions if you have $x$ points on the circle with this property.  We want $f(20)$.  We can express $f(x+1)$ in terms of $f(x)$.  Think about adding a vertex.  Now create the $x$ new chords.  The number of new regions created by each is 1 more than the number of chords it intersects.  And the number of chords each of these chords intersects is the number of points on one side times the number of points on the other.  Thus, the total number of intersections is $0(x-1)+1(x-2)+...+(x-2)1+(x-1)0=\frac{1*2}{2}+\frac{2*3}{2}+...+\frac{(x-2)(x-1)}{2}=\binom{x}{3}$.  Thus, $f(x+1)=f(x)+\binom{x}{3}+x$.  Because $f(3)=4$, we have $f(x)=4+(4+5+...+[x-1])+\binom{3}{3}+\binom{4}{3}+...+\binom{x-1}{3}$
$=-2+\frac{x(x-1)}{2}+\binom{x}{4}$.

Setting $x=20$, we get $-2+190+4845=\boxed{5033}$[/hide]


Let $ F_n$ denote the Fibonacci sequence, so that $ F_0\equal{}F_1\equal{}1$ and $ F_{n\plus{}2}\equal{}F_{n\plus{}1}\plus{}F_n$ for $ n \ge 0$. Prove that:
$ (i)$ The statement "$ F_{n\plus{}k}\minus{}F_n$ is divisible by $ 10$ for all positive integers $ n$" is true if $ k\equal{}60$ and false for any positive integer $ k<60$;
$ (ii)$ The statement "$ F_{n\plus{}t}\minus{}F_n$ is divisible by $ 100$ for all positive integers $ n$" is true if $ t\equal{}300$ and false for any positive integer $ t<300$.

Find first $ n_1>1$, suth that $ a|F_{n_1}$. For $ a\equal{}10$ $ n_1\equal{}15$. For all k, suth that $ a|F_k$ we had $ n_1|k$ (else $ n_1$ is not minimal). Then find minimal  $ n_0\equal{}n_1k$, suth that $ F_{n_0\plus{}1}\equal{}1\mod a$.
It give $ F_{n_0\plus{}n}\equiv \mod a \forall n$ and $ F{k\plus{}n}\equiv F_n \forall n$ only if $ n_0|k$.
For $ a\equal{}10$ $ n_0\equal{}60$, for $ a\equal{}100$ $ n_0\equal{}300$.


algebra
system of equations
number theory proposed
number theory

Prove that: \[x^4+y^4=z^4\]
\[xyz\neq 0\] has no integer solutions.

[hide]Show that \[x^4+y^4=z^2\] has no solutions [hide]By the well ordering principle, every non-empty subset of $\mathbb{N}$ has a smallest element,  let $z_0$, be the smallest solution for $z$ of the equation $x^4+y^4=z^2$ , then show that there exists a smaller one leading to a contradiction.[/hide][/hide]

BTW, how do i write a system of equations?

Fermat's Principle of infinite descent!!!

This can e proved using the pYhtagorean triad formula. WLOG assume x,y,z are pairwise coprime and proceed from there.

BOmb


for all iit-jee aspirants, i would suggest 'MODERN CHEMICAL CALCULATONS' by R.C.MUKHERJEE

well actually i prepared for medical,i.e. a bit lower level
and i found the book easy

u shud use that book only if u lack consistency  :)

and also if u are extremely gud at calculations

calc are lethal in the book  :rotfl:

in that case,u shud refer S.N.MANDAL's 'IIT-CHEMISTRY'
A VERY NICE BOOK FOR IIT-AIEEE aspirants.......

@Rituraj007
Which all books do you suggest for JEE/IIT Physics and Chem?

i didnt nething properly excpt my modules(that also medical ones)

 just study chem NCERT well.esp inorganic ...it'll really help :)

the following books r generally used by students in mumbai:
1.h.c.v.
2.halliday & krane
3.university physics
4.galaxy physics-ashish arora
5.arihant physics books-d.c.pandey
6.i.e.irodov

I think ,dinesh publications' IIT-PHYSICSIS A NICE BOOK FOR PRACTICE.
It contains a lot of innovative problems..........

jee now focuses on ncert syllabus,so that the dependance of students on petty coaching classesis the least :idea:  :roll:

this yrs paper really surprised

there was no high level stuff..just normal concepts :D  :D


If the integers $a$, $b$, and $c$, are all powers of $2$, and if $a^{3}+b^{4}=c^{5}$, what is the least possible value of $a+b+c$?

[hide]It's pretty easy to see that the sum of two powers of two is at most the next power of two and that this condition is only satisfied if both of them are equal.

Thus $a^3=b^4=2^{12k}$ and $a^3+b^4=2^{12k+1}=c^5$. The smallest $k$ that has a solution is $k=2$, from which we get $a=2^8, b=2^6, c=2^5$ and $a+b+c=256+64+32=352$.[/hide]


What is the most rating you have lost during a losing streak in FTW?
Remember, it doesn't have to be in one day, but it can be.
An example of a losing streak might be after you did really bad in a competition, and just didn't feel like doing math anymore, or it could just be a time where the ppl you pick to play against are just the wrong ppl.
Mine is 115.
 :(  :(  :(

You should do a poll on this.
I think I lost 54 pts once during a losing streak.  :wallbash_red:

Ah, about 200 I think. <- mostly to k00l and pianogirl because of 2 problem 5 seconds games. I stopped about about 20~. LOL FAIL

I win.  :winner_first:

Edit: Forget the wining part.....

600?!? WOW!
but you are now at 2000-ish, so it doesnt matter anymore

[quote="AIME15"]If bombing counts, ~600.[/quote]

If it counts, ~800. :P

+50ish

 :rotfl:

cuz I played mem game with people +400 me multiple times.  I'd lose, but still gain  rating (most of the time). :D

ummmmm, most [b]LOSING[/b] streak, +50 isnt losing  :P

But I didn't win a single one then... so it was a losing streak.

[quote="$ LaTeX$"][quote="jjfun1"][quote="k00lperson"]wait, rating or number of games? Oh..it seems like rating..hm how do you lose 600?[/quote]
Bomb rating. Keep losing. It's not hard.[/quote]

I would like to second this. Don't leave out 2 problem 5 second games either.[/quote]
Especially if you keep tying with a person that has lower rating.

I barely have a losing streak; I usually win a game and then lose a game.

Mine was around 200 points (1200 to 1000).  Mostly because I played against the wrong people (I started off by playing a bunch of games against ilovepink).

I'm still currently in my streak of about 200


Well, there is a topic about the Penn State Site, but unfortunately, NC will not be going to Penn State.

How many teams will be going to Georgia?

[quote="Anaxerzia"]SC repppin'!!

Assuming they dont kick me out  :P[/quote]

I don't think that's very likely Ravi  :roll:

And NC is coming to Georgia, I'm pretty sure.

NC is quote definitely going to Georgia!
And Carowinds afterwards...

Excellent! We'll have Georgia, Kentucky, SC, NC, and Colombia here for the party... anyone else? :)

I would suspect that Florida and Alabama will also come to the Georgia site if they have teams, but I could be wrong. And does anyone know if Tennessee has a team?

Alabama has a team and will come to the Georgia site.  I don't know about Texas.  I think Florida has a team; if they do have a team I'm certain they're coming to Georgia.

[quote="jb05"]Is Texas going???[/quote]
texas => iowa

[quote="Anaxerzia"][quote="solafidefarms"]Excellent! We'll have Georgia, Kentucky, [b]SC[/b], NC, and [b]Colombia[/b] here for the party... anyone else? :)[/quote]

Umm.... ?[/quote]

As in the nation of ...

Does anyone know more specifically who the Colombians are?  (I have a Colombian friend who is curious -- he assumes it's a single school, but we don't know which one.)

Why would Colombia, of all countries, come to Georgia?

Because Georgia is the closest of the four sites to Colombia, and Atlanta is a major travel hub?  (By the way, for those curious: the trip is actually organized by the Colombian olympiad organization, not by any individual schools.  I don't know how they select their team, but it probably includes the Colombian IMO team.)

This is the second appearance for the Colombian team. They came to ARML 2007 at Penn State, and finished in a tie (with NYC J) for 87th place overall. What I can't tell from the results was whether they were a full 15-member team or whether they arrived short-handed.

Last year's international guest teams were Colombia, two teams from Taiwan, and one team from the Philippines. The Taiwan and Philippines teams have been coming to the western site for quite a few years now. They'll be back this year, joined by another international guest team at the western site - a team from Hong Kong.

I've heard that there might be some off-site international guest teams, but I don't know the details of that.

International teams are not considered to belong to either Division A or Division B, although both Taiwan teams would qualify for Division A on the basis of their record. And individuals on international teams are not eligible for the prizes that attach to performance on the tiebreaker. However, last year, several individuals from Taiwan did participate in the tiebreaker (with questions hastily translated into Chinese), and a sponsor stepped forward to match the prize money.

[quote="Kent Merryfield"]I've heard that there might be some off-site international guest teams, but I don't know the details of that. [/quote]

One of the ARML site directors/coordinators said that those are the teams that couldn't get plane tickets over. I know that Turkey is one of them, but I'm not sure about anyone else.


Given that a,b, and c are greater than 0, prove:

$\frac{c}{a} + \frac{a}{b+c} + \frac{b}{c} \ge 2$

[hide]

$\displaystyle\frac{c}{a} + \frac{a}{b+c} + \frac{b}{c} \ge 2$

$\displaystyle\frac{c}{a} + \frac{a}{b+c} + \frac{b}{c} + 1\ge 3$

$\displaystyle\frac{c}{a} + \frac{a}{b+c} + \frac{b+c}{c} \ge 3$

which is true by AM-GM

[/hide]


integration logarithms trigonometry calculus function inequalities algebra

I've decided that my 100th topic should be special. And I hope you will enjoy the following problem:
  It is very simple to prove that the sequence x_n=|sin1*sin2*...*sinn| has limit 0. What about the series x_1+x_2+...+? Is it convergent? Maybe divergent?

I think this solusion is acceptable.
   Lemma: there is at least one of the numbers n,n+1,n+2,called x, satisfies |sin(x)|<=1/2.

   By using this lemma we have: |sin(1).sin(2)...sin(3k)|<1/2^k
then |sin1|+|sin1sin2|+(|sin1sin2sin3|+...)<2+3(1/2+1/2^2+...)=5.

But what about the series ax_1+a^2x_2+...+a^nx_n+...?, where a is a positive real?

I think that the lemma is valid for 4 conecutive numbers, not 3, but the solution is indeed very nice. :)

My idea for the last problem is a similar lemma:

For every t>0 there exists one positive integer k s.t. for every n there is one x in n, n+1, n+2, ..., n+k-1 s.t. |sin(x)|<t.

Then a|sin(1)| + a2|sin(1)sin(2)|+... \leq a+a2+...+ak-1 + (1+a+a2+...+ak-1)(tak+(tak)2+...) so for a less than 1/t1/k the serie converges.

I only need to show that t1/k->0, but I don't even know if this is true. Any ideas? :)

let s_k= \sum_1^n a^k*x_k. i will treat this sequence as a series. with logaritmic criterion we obtain:

lim_{n->00}ln(1/s_n)/ln(n)=
-lim_{n->00}(n*ln(a))+ln(x_n)/ln(n)   (**)
=(-00)lim_{n->00}(ln)a)+ln|sin(n)|)=00, if a<1 so the series is convergent. i show that, if a>1, series diverge.
conform to Kroneker's thm. ther exists k_n\in N, k_n-->00,
such that |k_n*\pi-[k_n*\pi]|-->\pi/2, so
|sin(k_n*pi)|-->1==> ln|sin(k_n*pi)|-->0, then by (**), and a>1 we obtain
lim_{n->00}ln(1/s_n)/ln(n)=-00, so series diverge.
that demonstration works in case a=1 too. :)

Since the last post in this old thread (the one which is assumed to solve the problem) is incorrect and since I liked the problem a lot, I want to bring it back to life. So

For what $a>0$ does the series $\sum_{n\geq 1}a^n|\sin1\cdot\sin2\cdot...\cdot\sin n|$ converge?

My conjecture is $a\leq 2$.

Hmm... Obviously caugust's proof is wrong, since trongcanh's solution shows that series converges for all $a<\sqrt[3]2$, doesn't it?

[quote="harazi"]Very very nice, indeed. I didn't think about this, but this is a very neat solution.[/quote]

Yeah, perfect scorers from the IMO sometimes invent nice solutions.  ;)

Indeed, it is nice solution, but it was very clear for this problem. ;)

[quote="Myth"]I have a strong supposition that answer is something like $a<exp\left(-\int_0^{\pi/2}\ln\sin t\,dt\right)\approx 2.971$. Unfortunately, I don't know whether we are able to calculate this integral?[/quote]

Mith, it is rather $\exp\left(-\frac2\pi\int_0^{\pi/2}\ln\sin t\,dt\right)=2$
(the average value, not the integral). And, indeed, one can show that $a<2$ is fine. But how about $a=2$?

Oh, you are right! I forgot this factor $2/\pi$.

Just to confirm fedja's observation, I will show that when $a < 2$, the series converges.  In a later post I will try to show that when $a > 2$, the series diverges.  I don't know what the answer is when $a = 2$.

Suppose that $a < 2$.  Let $\epsilon$ be a positive number such that
\[\int_{\epsilon}^{\pi - \epsilon} \ln | \sin x| dx < -\frac{\pi}{2} \ln(2a).\]
Define f(x) to be $\ln |\sin x|$ when $\epsilon \le x \bmod \pi \le \pi - \epsilon$, and $\ln |\sin \epsilon|$ otherwise.  Then f is a continuous function with period $\pi$.  It is known that the integers $\bmod \pi$ are equidistributed on $[0, \pi]$.  In particular,
\[\frac{1}{n} \sum_{k=1}^n f(k) \to \frac{1}{\pi} \int_0^{\pi} f(x) dx < \frac{1}{\pi} \int_{\epsilon}^{\pi-\epsilon} \ln|\sin x| dx < -\frac{1}{2} \ln(2a).\]

Thus for sufficiently large n, we have $\sum_{k=1}^n \ln |\sin k| \le -\frac{1}{2} \ln(2a) n$.  Rearranging gives $a^n \prod_{k=1}^n |\sin k| \le (a/2)^{n/2}$.  Thus our series is bounded by a convergent geometric series.

[quote="Myth"]Yes, I obtained my supposition in the similar way. But usually I don't like such argument, so I called it "supposition".

BTW. Fedja! I am [b]Myth[/b], not Mith.  :([/quote]

I apologize, [b]Myth[/b]... :blush:

I also used a truncation of $\log|\sin x|$. There is another approach though.
Let $\zeta=e^{2i}$. Then $\prod_{k=1}^n (2|\sin k|)=|p_n(\zeta)|$ where $p_n(z)=(z-1)(z^2-1)\cdot...\cdot(z^n-1)$. This is quite a famous polynomial and, if my memory does not betray me, there is some combinatorial argument I don't remember that shows that all its coefficients are quite small. I wonder if anybody can finish the proof along these lines... On the other hand, what's so wrong with good old Weyl's lemma that nobody seems to like it?

[quote="Ravi B"]Just to confirm fedja's observation, I will show that when $a < 2$, the series converges.  In a later post I will try to show that when $a > 2$, the series diverges.  I don't know what the answer is when $a = 2$.[/quote]
Well I did promise to prove divergence when $a > 2$, and fedja asked me to post my proof, so here goes.

Suppose $a > 2$.  Choose $\epsilon = (\frac{1}{100}\ln(a/2))^8$.  Let the function f be the same truncation of $\ln |\sin x|$ that I defined in my previous post.  Then we have
\begin{eqnarray*}
 \sum_{k=1}^n \ln |\sin k|
    & = & \sum_{k=1}^n f(k)
             - \sum_{\substack{k=1 \\ |\sin k| < |\sin \epsilon|}}^n                          \ln \frac{|\sin \epsilon|}{|\sin k|} \\
   & = & \frac{n}{\pi} \int_0^{\pi} f(x) \, dx - o(n) -
       \sum_{\substack{k=1 \\ |\sin k| < |\sin \epsilon|}}^n
            \ln \frac{|\sin \epsilon|}{|\sin k|} \\
   & \ge & \frac{n}{\pi} \int_0^{\pi} \ln |\sin x| \, dx - o(n)
   -   \sum_{\substack{k=1 \\ |\sin k| < |\sin \epsilon|}}^n
            \ln \frac{|\sin \epsilon|}{|\sin k|} \\
   & = & -n \ln 2 - o(n) -
             \sum_{\substack{k=1 \\ |\sin k| < |\sin \epsilon|}}^n
                \ln \frac{|\sin \epsilon|}{|\sin k|}.
\end{eqnarray*}

So we just need to show that the rightmost summation is small.  Let $t$ be an arbitrary positive number.  By using Hata's theorem in a similar way that I did in a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22093]Tough infinite series[/url], we can show that integer solutions for k of the inequality $|\sin k| \le t$ must differ from each other by at least $\frac{1}{10} (1/t)^{1/8}$.  In particular, there are at most $10 n t^{1/8}$ integer solutions from 1 to $n$.  Thus we have
\begin{eqnarray*}
\sum_{\substack{k=1 \\ |\sin k| < |\sin \epsilon|}}^n
                \ln \frac{|\sin \epsilon|}{|\sin k|}
  & = & \sum_{\substack{k=1 \\ |\sin k| < |\sin \epsilon|}}^n
              \int_{|\sin k|}^{|\sin \epsilon|} \frac{1}{t} \, dt \\
  & = & \int_{0}^{|\sin \epsilon|}
               \frac{| \{ k \in 1..n : |\sin k| \le t \} |}{t} \, dt \\
  & \le & \int_{0}^{|\sin \epsilon|}
                \frac{10 n t^{1/8}}{t} \, dt \\
  & = & 10n \int_{0}^{|\sin \epsilon|}
                        \frac{1}{t^{7/8}} \, dt \\
  & = & 80n |\sin \epsilon|^{1/8} \\
  & \le & 80n |\epsilon|^{1/8} \\
  & \le & \frac{4}{5} \ln(\frac{a}{2}) n.
\end{eqnarray*}

Plugging back in to the original equation shows that
\[\sum_{k=1}^n \ln |\sin k| \geq -n \ln a\]
for sufficiently large n.  Thus $a^n \prod_{k=1}^n |\sin k| \ge 1$.  Thus the terms of the series are bounded away from 0, and in particular the series diverges.

I don't know how to solve the case $a = 2$.  That seems to be a much more delicate question than I know how to solve.  If I had to guess, I would say the series diverges.  I would even guess that the terms of the series don't converge to 0.  But that's just a guess.

Very nice piece of analysis, Ravi! Your proof actually shows that, for $a>2$, the general term of the series tends to $+\infty$. Now about $a=2$. You are right, the series will diverge (so my original conjecture was wrong). Here is a hint. I mentioned earlier that
\[
\prod_{k=1}^n |2\sin k|=\left|\prod_{k=1}^n\left(1-e^{2\pi i \frac1\pi k} \right) \right|
\]
One useful trick is to think of irrational numbers as of irreducible fractions with arbitrarily large denominators. So assume that $\frac 1\pi =\frac pq$ with $(p,q)=1$ where $q$ is large. Then the product becomes $\prod_{k=1}^n\left(1-e^{2\pi i \frac  pq k} \right) $. Is it possible now to find some large $n$ (depending on $q$) for which this product is easily computable and large?  :)

The answer to the hint question is $n=q-1$. This will give a product of the differences between $1$ and all $q$-th roots of unity different from $1$, which is just $q$. Now, unfortunately, we cannot write $\frac 1\pi=\frac pq$. What we have to use instead is that $\left|\frac 1\pi-\frac pq\right|\leq \frac c{q^2}$ for infinitely many $q$ with $c=\frac 1{\sqrt 5}<\frac 12$ (this is true for every irrational number in place of $\frac 1\pi$, the proof can be found in any textbook on continuous fractions). Thus, instead of the exact $q$-th roots of unity $e^{2\pi i \frac mq}$, $m=1,\dots,q-1$, we shall have the perturbed roots of unity $e^{2\pi i (\frac mq+\delta_m)}$ with $|\delta_m|<\frac cq$. It is an elementary exercise to show that such a perturbation can introduce a factor that is bounded from below by $\prod_{1\leq m\leq q/2}\left(1-\frac c m\right)^2\geq \text{const }q^{-2c}$. Therefore, the $q-1$-th term of the series will be bounded from below by $q^{1-2c}\to +\infty$. Note that in this proof one only controls a subsequence of terms in the series, so it gives less information than the proof by Ravi B for $a>2$.

Cool proof.  An amusing contrast: Even though we are both proving divergence, my proof uses the fact that $\pi$ cannot be approximated too closely by rationals, whereas your proof uses the fact that $\pi$ can be approximated closely by rationals!


[quote="236factorial"]A point is randomly selected inside a rectangle with sides 2' and 3'. What is the probability that the point is more than 1' from every vertex of the rectangle? Use 3.14 for pi, and round your answer as a decimal to the nearest hundredth.[/quote]

[hide]For them to be all at least one foot away from each vertex, quarter-circles are drawn on each vertex. The area of the rectangle is $6$, and the area of all the quarter-circles is $\pi$. The probability is $\frac{6-\pi}{6}\approx\boxed{.48}.$[/hide]

[hide="answer"]47.67%[/hide]
When you say round your answer, I'm assuming it's supposed to be a percent?

[quote="The question:"]and round your answer as a decimal to the nearest hundredth.[/quote]

 :)

[quote="X_wing1redleader"][hide]make a quarter circle at each vertex
These quarters make up a total circle of area pi
6-pi/(2*#)=.476666666666666666666666666[/hide][/quote]

You're supposed to round it.

But yeah, you get a quarter circle at each vertex, put them together to get a circle with radius 1,$1^2*\pi=\pi. \frac{6-\pi}{6}\approx 47.67$%


Would it be acceptable for a Physics Teacher to write this rec? or a Chemistry teacher?  I just ask because these classes aren't completely outside the world of mathematics and therefore would slightly intersect with the math recommendation.

They are fine. My science teacher wrote my non-math teacher recommendation and I got in.


Hello, Mathlinkers!

When I try to type document using my language (Russian) it does not
show the letters, except stops. If I want to see them, what must I do? :)
Maybe there exists some package?

The first result I got on a search for "cyrillic LaTeX" was [url=http://www.cromwell-intl.com/russian/latex.html]this[/url], which appears to be one solution.

Thanks a lot, though it seems that that article is written for linux...
(Perhaps, the only reason it doesn't work is that I'm to silly  :oops: )

Nevertheless, I'll try to reveal the secret of cyrillic LaTeX  :roll:

What happens if you use
[code]\usepackage[T1,T2A]{fontenc}
\usepackage[koi8-ru]{inputenc}
\usepackage[english,russian]{babel}[/code]in the preamble? See page 33 of [url=http://tug.ctan.org/tex-archive/info/lshort/english/lshort.pdf]The Not So Short Introduction to LaTeX2e[/url] which is aso available in Russian on [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=lshort]CTAN[/url].
See also [url=http://tug.ctan.org/tex-archive/macros/latex/doc/cyrguide.pdf]Cyrillic languages support in LaTeX[/url].


$1$. Daca $a$,$b$,$c$,$d$ sunt reale si $a^{2}+b^{2}+c^{2}+d^{2}\leq 4$, atunci $a^{3}+b^{3}+c^{3}+d^{3}\leq 8$.

$2$.Tom a primit o pusculita in care are $5$ lei, dupa care isi propune sa economiseasca saptamanal o suma de bani astfel: $5$ lei intr-o saptamana impara si $2$ lei intr-o saptamana para.Se cere sa se afle:
a)Ce suma va avea peste $7$ saptamani?
b)Daca va avea vreodata in pusculita o suma de bani patrat perfect.

$3$.Paralelogramul $ABCD$ si patratul $BDEF$ sunt asezate in plane perpendiculare. Fie $M$ si $N$ proiectiile punctelor $E$ si $F$ pe $AC$. Daca $AB=8$, $AD=5$ si $m(DAB)=60$ aflati:
a)lungimea proiectiei segmentului $AC$ pe $EF$.
b)Distanta dintre planele $(EDM)$ si $(FNB)$.

$4$.Fie $ABCA'B'C'$ o prisma regulata. Sa se demonstreze ca $AA'^{2}/AB^{2}=3/4$ daca si numai daca $(A'BC)\perp(AB'C')$.

[quote="luke999"]1)Cum patratele o sa fie intotdeauna pozitive si cuburile pot fi negative vom demonstra doar pentru nr pozitive(dak sunt negative le facem pozitive si suma cuburilor pozitive o sa fie mai mare decat suma cuburilor negative) .Pt ca inegalitatea este omogena putem pres ca $a \le b \le c \le d$. Dar cum
$a^{2}+b^{2}+c^{2}+d^{2}\le 4$=>$d \le 2$
Cum $a^{3}+b^{3}+c^{3}+d^{3}\le da^{2}+db^{2}+dc^{2}+d^{3}\le 8$[/quote]

cat m-am gandit la o solutie de asta... pan' la urma am facut-o cu cauchy.

[quote="luke999"]2 a)Mi-e lene sa calculez...
   b)Facem mod 7....(ink n-am calculat da asa pare)[/quote]

asta a fost si ideea mea

Notez AC' int CA' cu O si AB' int A'B cu O'. Cum ACA'C' dreptunghi si ABB'A' dreptunghi=>O' mijloc AB' si 0 mijloc AC'=>AO=AO' si A'O=A'O'=>Triunghiurile astea sunt isoscele
Fie M mijloc OO'(OO' e si lm in tr. A'BC)=>AM inaltime in tr. AOO' si A'M inaltime in tr A'OO'. Deci unghiul dintre plane o sa fie egal cu m(<AMA').
Notez AC=a si AA'=b=>
$OO'=\frac{a}{2}si A'O'=AO=A'O=AO'=\frac{sqrt{a^{2}+b^{2}}}{2}\Rightarrow AM=A'M=\sqrt{\frac{9a^{2}}{16}+\frac{b^{2}}{2}}$

si pan la urma o sa dea ca AMA' dreptunghic<=>$4AA'^{2}=3=AC^{2}$

Dreapta de intersectie a planelor este OO'. Conform "celebrei" Teorema acoperisului ( chiar asa am scris in teza), adica lema de paralelism (suna mai bine asa) este paralela cu BC. Apoi m-am cam complicat, am luat M pe d, unde d e dreapta paralela cu OO' si trece prin A astfel incat m(MOO')=90. Analog pt un punct N apartinand lui BC. Proiectam A si B pe OO', apoi am demonstrat ca proiectiile coincid folosindu-ma de celelalte puncte...dar apoi am continuat ca tine.

Mai si eu am auzit de Th. Acoperisului...da' am uitat exact km era

[b]Teorema acoperisului [/b]spune ca daca [u]doua drepte paralele [/u]date sunt continute in [u]doua planele secante[/u], atunci [u]dreapta lor de intersectie este paralela cu dreptele date[/u].


I think there's a media thing which doesn't allow us to say who's on it?

Anyway, you guys know who you are. And you deserve a [b]huge[/b] congratulations.

So, yeah, well done, and congrats. And of course, good luck for July. ;)

[quote="vulgarfraction"]I think there's a media thing which doesn't allow us to say who's on it? [/quote]

Is it allowed to say ML user names, assumed there are some registered qualified members? Otherwise your post is pretty useless.  ;)

At least a few have usernames, but you can easily find real names from them. They know who they are anyway, and it'd be announced.... sometime..... You just wait and see.  :)

hahahaha

nobody cares - seriously, its not that big. our principal told the whole school who was in the teams just after april last year, the whole official thing is meaningless.

the maths team i think is graham, vinoth, matt? who've i missed...

How many of them are from the James Ruse Agriculture High School?


Prove that all separable, infinite-dimensional Hilbert spaces are isometric

This is equivalent to saying that every such space has a countable orthonormal basis, and hence is isometrically isomorphic to the sequence space $ \ell^{2}.$

On the one hand, it can be shown that $ \ell^{2}$ is separable and hence anything isometric to it must be separable.

On the other hand, start with a separable Hilbert space. It has a countable dense subset. Write that subset as a sequence. Apply the Gram-Schmidt process to that sequence. (Along the way, many elements of the sequence may turn out to be in the span of the preceding elements; we discard those). We arrive at an orthonormal sequence. The span of this sequence includes all of the elements of our original countable dense subset and is thus dense. An orthonormal set with a dense span is a "complete" orthonormal set, or orthonormal basis. And if the original space was infinite-dimensional, no finite set could have spanned it, so our orthonormal basis is infinite.


MATHCOUNTS
blogs
geometry
FTW
3D geometry
pyramid
AMC

what was your mathcounts state score??? poll is above.

Come on, why did you have to seperate it as 35 and 36?

I got 35, 21/14. Screwed up sprint.

I got 21/12=33. Dumb mistake on target and I was just recovering from being sick  :( so I went really slow on sprint.

I bet that whoever said they got a 46 was lieing. (unless it is stevenmeow)

yeah, it was a really hard test this year comparably to other years. It seems impossible for some one to have gotten it. (Maybe it seemed hard because I was sick.)

I got a 13.... Go ahead laugh im an idiot

[quote="cognos599"][quote="ra5249"]Don't worry, stevenmeow did NOT get a 46.[/quote]

what were the texas state scores?[/quote]
BLAH BLAH BLAH IGNORE THIS.

[quote="cognos599"][quote="ra5249"]Don't worry, stevenmeow did NOT get a 46.[/quote]

what were the texas state scores?[/quote]

hm
well if i told you the rest of the texas team will mob me
but i can tell you that there was a three-way tie for first place (me, bshen, kli)
and we all had dang close to a perfect

edit: OKAY DING NOW WE ARE GOING TO MOB YOU

and he was first written in texas right?

XPMATH PLEASE EDIT YOUR POST and delete my scores thnaks

why are you guys being so secretive about the mathcounts scores? It's done and over with.

i don't think texas wants rival states to know how good they and stuff cuz it might influence their preparation i guess. i know indiana is into the whole secrecy thing, only if i gave away the indiana scores, i would be mobbed by the top 4 and the coach so i am keeping my mouth closed. if you really want to know all the the scores were in the interval [-$ \infty$,$ \infty$].  :)

I messed up the Sprint round...19/14. I said 502 for #8 on the Target and it was 503  :oops_sign:

I had a base program that made it really easy, but then I forgot that if 502 works, so will 503...

How can base programming on graphing calculators be permitted if it only gives an unfair advantage?

Wow,....Then that would be an advantageous thing to do....along with your brain, to win :lol:

[quote="Yongyi781"][quote="leoxnlin"]They just are. You can program anything you want on your calculator. Yes, some people even program their calculators to find the answer to the nth problem of the sprint or target test of the mth year at levels 1, 2, 3, or 4, where 1=school, 2=chapter, 3=state, 4=nationals.[/quote]

I have to comment on this :rotfl:

You can't use calculators on the sprint... so what good would programming the answers to the sprint problems do for you?[/quote]

Because it's possible that on the team round, one of the problems was on a sprint round 15 years ago.

No, I'm talking about Target #8 from this year's state contest.

[quote="leoxnlin"]They just are. You can program anything you want on your calculator. Yes, some people even program their calculators to find the answer to the nth problem of the sprint or target test of the mth year at levels 1, 2, 3, or 4, where 1=school, 2=chapter, 3=state, 4=nationals.[/quote]

but what's the point?  If someone just does that instead of doing the problems, then nobody would find failing the test a surprise for that person.

Funny thing is, my nats score was two points higher than my state score. That's how much I screwed up at states.

lol, my huge rank change in comparison with others in states showed how i screwed at chapter :P

Delaware did horrible. 4th place in written had a freaking 20 lol! Idk about the other 3 though.

I did pathetic. 13/6=19. Maybe I actually should have practiced =O


[quote="vietnam2007"]I want to know if a participant can smoke during the exam.[/quote]

I think it is not officially forbidden: http://www.imo2007.edu.vn/index.php?module=ViewNewDetailType2&NewId=11

But I think it is common sense that you should not bother your fellow contestants and thus you are not supposed to do that.

I am still waiting for a reply from  someone who had participated last years in the IMO.

I have, and I'm pretty sure that they would not allow it (even if it's not explicitely stated in the rules).
But in fact rather impudent to annoy other contestants this way. If you want to smoke, you can ask if you can do this at the toilets, but doing in the room cannot be tolerated in my oppinion (in general smoking at places non-smokers are is a very bad habit of some smokers; it's ok if they agree only to smoke at some special places non-smokers can avoid).

I totally agree with you ZETAX it's forbidden for me and for all contestants to smoke in the room but it will be so fine if  smoking at the toilets is not forbidden.But I don't think that during the last teen years for example no contestant smoked  and if not how they did ?

I can only say that I never say anyone smoking during exam, even not at toilets.
I rarely saw some contestents smoking at other times.
Well, to be completely sure, you probably have to ask the organizers.

It does sound unlikely that they will allow you to smoke. In Sweden you could bring "snus". I don't know how to translate, but it is stuffed under your lip and is quite potent nicotine-wise. However, this is a scandinavian thing, so failing that I would suggest bringing nicotine patches or gum or some other quit-smoking stuff to get your nicotine during the exam.

One English word for that is "snuff," which must be a related word. I suppose the etymological question is which direction the word migrated. Of course, the person using that has to occasionally spit - it's really quite a disgusting habit. I doubt that spittoons would be provided.

Well most snus sold comes in small bags, so there is no need to spit.

come on vietnam2007, you cannot stop smoking during 4 hours ! If so, maybe it is your chance to try to stop your addiction  :D

I can't understand such a question has come, isn't limited that IMO partipants to high school or junior highschool students, less than or equal to 18 years old?
If Japanese high school students would do smoke, they are forced to receive dismissaling from school.
I wonder, if possible, can university students participate in IMO except Japan? If so, it's unfair.

I reassure you kunny I am not a university student.In morocco I had too many problems to smoke during the exams but they had understood me after a deep conversation with the reponsibles.I inform you that I have 18.2 years old :P

University students don't participate the IMO.
You shouldn't be surprised because In some countries it is allowed to smoke even if under age.

i have to agree though that for smokers IMO is as frustrating as it would have been for nonsmokers to have someone smoking near them.

even if you could smoke on the toilet it would be annoy the people sitting next to you... so i wouldn't understand if someone would allow you to smoke during the contests... and i wouldn't be really pleased if i had to sit next to you...
@albanian eagle: in germany e.g. from the age 16...

Naphthalin

i am asking too if someone can go out during the IMO exam in order to relax?

Very likely not as every single person would need an observer then...

Hey people this is a mathematical competition and not a "let's see what I'm allowed to do"-contest.
If you have reasons why something should not be forbidden, ask them (I know they forbid for some strange reasons things like coke...), but anything that annoys others or clearly can be abused to cheat a lot is forbidden for very good reasons.

[quote="ZetaX"]Hey people this is a mathematical competition and not a "let's see what I'm allowed to do"-contest.[/quote] :rotfl:  yes i agree totaly withe u  :|

You can smoke, just not inside the contest hall, and not in the dining lounge. Everywhere else, anytime, as long as you don't bother anyone you can smoke.
That was our guides' first questions last year - "does anyone smoke??? You can do it over there -->>  "


quadratics
modular arithmetic
number theory proposed
number theory

Let $p$ be an odd prime, and denote by $r(p)$ the least quadratic residue modulo $p$ in $\{1,2,\ldots,p-1\}$. Prove that $r(p) < \sqrt{p} + \frac{1}{2}$. Enjoy :). There is a cute 1-line solution.

--Vesselin

P.S. There are better bounds available, but I am not aware of an elementary better bound. In contrast, the above admits a completely elementary, 1-line solution.

[Edit: Sorry for the typo; yes, I wanted to write "least" in place of "last". It is now corrected. Thanks, ZetaX :) ]

What does "the last" mean?The smallest or the biggest or some other meaning?

last -> least
(and everything else makes no sense)

[quote="vess"]denote by $r(p)$ the least quadratic residue modulo $p$[/quote]

Also, you must have meant "the least quadratic [b]non[/b]residue". Anyway, the title suggests what the real problem is, so don't worry :). This is not a one-line solution, unfortunately. I'm sure that deep within your heart, you'll find the power to forgive me for this :).

Let $t$ be the smallest positive integer larger than $\sqrt p+\frac 12$, and let $a$ be the largest positive integer which is $<\sqrt p+\frac 12$. Since $at>p$, it's easy to see that as we compute $t,2t,\ldots,(a-1)t,at\pmod p$, at some point we'll reach a residue modulo $p$ which lies in the interval $[1,\sqrt p+\frac 12]$. This means that $t$ is expressible as a quotient of two residues in this interval, and if all of them were quadratic residues, $t$ would be one as well. We now repeat the argument for $t+1$, and show that it is expressible as a quotient of two residues modulo $p$ in the interval $[1,t]$, and so on. The conclusion is that all residues modulo $p$ would be quadratic, contradiction.


Please, I need other ideas to prove this ineq and it's my first time to post here my own problem.
Let $ x,y,z$ be positive real numbers such that $ x \plus{} y \plus{} z \equal{} 1$ Prove that $ 4(xy \plus{} yz \plus{} zx)\leqslant3\sqrt{(x^6 \plus{} \frac {1}{1 \minus{} z})^2 \plus{} (y^6 \plus{} \frac {1}{1 \minus{} x})^2 \plus{} (z^6 \plus{} \frac {1}{1 \minus{} y})^2} \plus{} 1$
Thank you for your nice proof. :clap:

Hello
we have $ 1\geq 3(xy\plus{}yz\plus{}zx)$
and $ \sqrt{3(\frac{1}{(x\plus{}y)^{2}}\plus{}\frac{1}{(y\plus{}z)^{2}}\plus{}\frac{1}{(z\plus{}x)^{2}})}\geq \frac{9}{2(x\plus{}y\plus{}z)}$
so $ \sqrt{\frac{1}{(x\plus{}y)^{2}}\plus{}\frac{1}{(y\plus{}z)^{2}}\plus{}\frac{1}{(z\plus{}x)^{2}}}\geq \sqrt{27/4}\geq 9\sqrt{3}/2(xy\plus{}yz\plus{}zx)$
if it's correct your inequality  :blush: can't keep its beauty . and i don't hope so.

Sorry, but I don't know how to applied your ineq to my problem. Could you explain to me please?  :lol:

By the way RHS is more stronger that LHS,  Now I see it clearly that the good ineq shouldn't use many ideas to crack it  :(

Thank you for your nice proof.


conics hyperbola geometry circumcircle algebra polynomial Vieta

Points A, B and C lie on the hyperbola xy=1. The circumcircle of triangle ABC intersects the hyperbola in D. Prove that D is the mirror image, in the origin, of H (the orthocentre of the triangle).

I have a feeling that [hide="a magic wand"] complex numbers [/hide] will work here.

Consider the system

$xy=1$

$(x-p)^{2}+(y-q)^{2}=r^{2}$

If we eliminate $y$ in the second equation and multiply by $x^{2}$, we get a polynomial in $x$ for which the coefficient of the $x^{4}$ and $x^{0}$ terms both $1$. So if $A$, $B$, and $C$ are $\left(a,\frac{1}{a}\right)$, $\left(b,\frac{1}{b}\right)$, $\left(c,\frac{1}{c}\right)$, respectively, then $D$ is: $\left(\frac{1}{abc},abc\right)$ by vieta's sums.

It suffices to show that $-D$ is the orthocenter of $ABC$...i.e. $(A+D)\cdot (B-C)=0$. A short computation reveals that this is true.

$\left(a+\frac{1}{abc},\frac{1}{a}+abc\right)\cdot \left(b-c,\frac{1}{b}-\frac{1}{c}\right)=0$.

Comment: We can determine $(p,q)$ by using the fact that $3\vec{OG}=\vec{OH}$. Consequently, we can determine $r^{2}$ by plugging a variable into the equation for the circle.


probability
counting
derangement
function
complex analysis
combinatorics proposed
combinatorics

Here's a challenging problem; it took me a long time to solve and I'd be interested in different solutions:

$n$ people participate in Secret Santa one year (they line up in a single file, then randomly select someone else's name out of a box and gives that person small gifts for a week or so; they cannot choose their own names: assume that that never happens, or that if it does, some mechanism is in place to respond to it, e.g. the person puts it back, and if the last person has to pick his or her name, everything starts over).  What is the probability that at least one two-cycle (where a person selects the same person who selects him or her) exists?

Sorry if that explanation's too long, but here's an example: let's say Jack, Jane, Joe and John participate in Secret Santa.  Here's one possible outcome (Jack picks first, Joe second, John third, and Jane fourth):
Jack picks Jane
Joe picks John
John picks Joe
Jane picks Jack
so there are two two-cycles.

What is the answer for $n=4$?  How about $n=5$?  Can you generalize?

How about counting derangements and derangements with at least one two-cycle?  For example, for $n=4$, there are 9 derangements, and 3 derangements with at least one two-cycle.  Is the probability therefore $\frac{3}{9}=\frac{1}{3}$?

The answer is one minus the quotient of the $n^{th}$ coefficient of $\frac 1 {1-t} e^{-t-t^2/2}$ and the $n^{th}$ coefficient of $\frac 1 {1-t} e^{-t}$ which are the generating functions of the permutations with no 1 and 2- cycles and no 1-cycles, respectively.

The inverse symbolic calculator suggests that the quotient converges very rapidly against $\frac 1 {\sqrt e}\sim 0.60653065971263342360$ which one probably gets from singularity analysis as the quotient $\frac{e^{-1-\frac 12}}{e^{-1}}$
of the (complex analysis) residues at $t=1$.

(In particular, the probability that 2-cycles happen is about $0.4$ for $n \ge 6$.)

Edited to remove superfluous *don't* in the last sentence.

Wow, pretty good, although $0.4$, according to your argument, should be the approximate probability that there [i]is[/i] at least one two-cycle (not the other way around).

There are closed form definitions for the number of permutations with no one-cycles and the number of permutations with no one-cycles but at least one two-cycle, by the way, but I guess they aren't necessary to derive.

The only real problem with your argument is the assumption that every derangement is equally likely.

It is true that I assumed this, but since your mechanism of avoiding 1-cycles was labelled "e.g.", I decided to assume the mechanism "if someone picks their own name, start over".

Is the probability distribution affected by your mechanism?

My mechanism aside, just the process of selection by a single file affects the probability distribution, somewhat surprisingly.

Let's just assume that if someone picks their own name, everyone starts over (which, probability wise, is the same thing as my "e.g. mechanism").  What is the probability then?

[quote="kokosai"]My mechanism aside, just the process of selection by a single file affects the probability distribution, somewhat surprisingly.

Let's just assume that if someone picks their own name, everyone starts over (which, probability wise, is the same thing as my "e.g. mechanism"). [/quote]

I don't think so. Whether we start over immediately when someone picks their own name or after picking has finished doesn't affect the probability distribution.

Let's say A, B, C, and D participate in Secret Santa one year, and that A->B->C->D->A  ("->" means "chooses").  There are then 24 (4!) arrangements (orders) for A, B, C, and D to pick in.  For example, let's say A picks first, B picks second, C picks third and D picks fourth:
A->B (Probability: 1/4 that A picked first, 1/3 that A picked B)
B->C (Probability: 1/3 that B picked second, 1/3 that B picked D)
C->D (Probability: 1/2 that C picked third, 1 that C picked D)
D->A (Probability: 1 that D picked fourth, 1 that D picked A)
Note that I'm using conditional probabilities, since it's a single file; for example, there is a 1/3 probability after A picked first that B would then pick second.  Also, no individual can choose him or herself, so C must choose D, because otherwise D would choose him or herself (and everything would restart).  The probability of ths particular arrangement is therefore 1/216.  If you go through all the arrangements, you'll find that the average probability of all the arrangements of A->B->C->D->A is $\frac{13}{108}$, and since there are six permutations that are equivalent to A->B->C->D->A, the probability of a four-cycle is $6*\frac{13}{108}=\frac{13}{18}$.  With four people participating in Secret Santa, there will either be two two-cycles or one four-cycle, so the probability of a two-cycle is 5/18, not 1/3.

If you want to do this on your own, note that since you're dealing with a single four-cycle, you can just assume that A picks first.  You'll then only get six arrangements, but if you did the twenty-four arrangements, each of the six would be repeated four times.

But the different probability distribution comes from the fact that you go on picking after you hit a fixed point, not from the fact that you pick in single file. For example, if you just mix up the name and give them to everybody and only the fixed-point people return them and mix again, this will also affect the distribution. And if you pick in single-file and restart after hitting a fixed point, you will get equidistribution.

The computer says that the probs for your permuation generation gives:

n=4  .7222
n=5  .56597
n=6  .6212
n=7  .628
n=8  .616
n=9  .6139
n=10  0.613414

$\frac1 {\sqrt e}=.60653...$

So, the  probability for large $n$ is very close to the one for equidistribution and your best bet for a proof is to show that the distribution is not affected too much after all.

You're right, it's not just because of the single file, it's because people pick another name after they pick their own name, and the second-to-last-to-choose prevents the last-to-choose from choosing him or herself.  However, this "replace" mechanism isn't just some arbitrary, artificial construct; I think that's how most real-live Secret Santas take place, since it's much more efficient than other selecton methods.  Anyway, thanks for pointing out that some single file selection methods will be probabilisticaly equidistributional; that was a real oversight on my part.

I also noticed that the probability distribution for the replace method approaches equidistribution, but I couldn't find a proof.  It's not difficult (with a horrendous nested sum) to find the exact probability of a derangement, but I couldn't determine rigorously the effect of the replace method on the distribution as $n$ goes to infinity.  The distribution can be quite complicated; for example, for $n=6$, a derangement with two three-cycles is less likely than a derangement with one six-cycle.  However, intuitively and numerically, it seems that the distribution should approach equidistribution relatively quickly.  We need to show that the probability of any derangement is asymptotic to $\frac{1}{d_n}$, where $d_n$ is the number of derangements of $n$ objects.

Do you have any ideas for a rigorous proof, kunterbunt?


First, prove that if $u_1,u_2,u_3...u_n=1$ then $u_{n+1}=1$.
Since $u_1=1$, use the principle of mathematical induction to show that $u_1,u_2,u_3...=1$.  Hence, all terms are integers.

[quote="darij grinberg"][quote="maskman"]Let $(u_n )$ be the sequence defined by :
$u_1 = 1$ and $u_{n + 1} = \frac{{1 + \sum\limits_{k = 1}^n {u_k^2 } }}{n}$
Are all the terms of $(u_n )$ integers?[/quote]

Pwned!

http://www.research.att.com/~njas/sequences/A003504
http://mathworld.wolfram.com/GoebelsSequence.html

  Darij[/quote]
Darij are ther the solution of this problem? :?
If yes,I can't find it  :( .

It says there that first nonintegral $u$ is $u_{43}.$ (Read carefully.)

[quote="ZetaX"]Since $u_2=2$, this obviously doesn't work.[/quote]
Sorry, I thought that the denominator was $n+1$.


limit
logarithms
algebra unsolved
algebra

[quote="Nga"]Find $\lim_{n\to +\infty}n\cdot (a^{\frac{1}{n}}-1)$,where $n \in \mathbb N^*$[/quote]
This is well-known.
$a^{\frac{1}{n}}-1=e^{\frac{\ln{a}}{n}}-1{\sim}\frac{\ln{a}}{n}$,so $\lim_{n\to +\infty}n\cdot (a^{\frac{1}{n}}-1)=\ln{a}$


Yo everybody!!!!!!!I want to become a moderator. How do I? Oh yeah and since I named this topic New Definition of Spam: This is spam: spamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspamspa
 :spam:  :spam:

You really should get off. Thene-thene pleaseee...don't shut all the windows I had up....  :mad:
you already(or was it cbc) shut my chess game!


modular arithmetic
number theory proposed
number theory

p is a prime and a, b, c, are distinct positive integers less than p such that a^3 = b^3 = c^3 mod p. Show that a^2 + b^2 + c^2 is divisible by a + b + c.

We clearly have $p\ne 2$.

Since such $a,b,c$ exist, it means that $p=M_3+1$, and $b\equiv ta,\ c\equiv t^2a\pmod p$, where $t\ne 1$ is a root of the equation $x^3-1=0$ in $\mathbb Z_p$. We have $a+b+c\equiv a(1+t+t^2)\equiv 0\pmod p$, and since $a+b+c<3p$, we have $a+b+c\in\{p,2p\}$. The conclusion is now clear: $a^2+b^2+c^2\equiv a^2(1+t^2+t)\equiv 0\pmod p$, and, additionally, if $a+b+c=2p$, then $a+b+c$ is even, and so is $a^2+b^2+c^2$, so $2p|a^2+b^2+c^2$.

Excellent Solution, When i first solve it i didnt use any field thing, but 2 days ago when I saw the problem i could solve it as you did. Thanks for this solutions

It means "number of the form $3k+1$". The $M_3$ stands for "multiple of $3$".


trigonometry
calculus
integration
real analysis
real analysis unsolved

Does the serie : $ \sum\limits_{n \equal{} 1}^{\infty} \sin(\pi(2 \plus{} \sqrt {3})^{n})$ converge

:blush:
good question , therefore I give you only the hint:
Compare $ (2 \plus{} \sqrt {3})^{n}$ with integer $ (2 \plus{} \sqrt {3})^{n} \plus{} (2 \minus{} \sqrt {3})^{n}$
generalization of this problem :
Consider :
$ \sum \sin (\pi (\sqrt {m} \plus{} m \minus{} 1)^{n} )$
 :roll:

Thank for your hints ,Extremal :) i could do it.and this is another serie with the same question  :D
$ \sum\limits_{n\equal{}3}^{\infty} \frac{1}{n^{a}(lnn)^{b}}(a,b>0)$
Could you show me the way to do it? :blush:

$ a < 1$:  divergence.  Compare to $ \frac {1}{n^{a \plus{} \epsilon}}$.

$ a \equal{} 1, b \le 1$:  divergence.  Integral test.

$ a \equal{} 1, b > 1$:  convergence.  Integral test.

$ a > 1:$  convergence.  $ p$-series.


I just started with this program.

How do write a multi-line text (actually it's a matrix) in my window?

(kind of cout<<blabla<<endl<<blablabla<<endl, but in windows)

What is borland c 6.0 - is that some version of Borland C++Builder? Where do you want to write your text / what do you want to write it into? You could pop a message box, write it directly into your form, create some kind of label control... (I'm not familiar with C++Builder so I'm not sure how are all the controls called in it.)


Let a and b be positive integers and $K=\sqrt{\frac{a^2++b^2}{2}}$, $A=\frac{a+b}{2}$. If $\frac{K}{A}$is a positive integer, prove that $a=b$

You are asking for $M = \frac{2(a^2+b^2)} {(a+b)^2}$ to be the square of a positive integer.
Note that $M = 2 - \frac{4ab} {(a+b)^2}$ and that $0 < 4ab \leq (a+b)^2$ with quality iff $a=b$. Since want that $\frac{4ab} {(a+b)^2}$ to be an integer, it forces the equality case above, and we are done.

Pierre.


inequalities function vector calculus integration

Hey! I have a brainless solution for this problem, but it's very ugly so I'm looking for a nice one. Please post it if you find one.

For $x,y,u,v>0$ prove that
\[\frac{xy+xv+uy+uv}{x+y+u+v}\geq \frac{xy}{x+y}+\frac{uv}{u+v}\]

Have fun!  :lol:

Maybe this will work... let $f(a,b) = \frac{ab}{a+b}$. Since $f$ is increasing and concave in both $a$ and $b$, we have

$f(x+u,y+v) = 2f\left(\frac{x+u}{2}, \frac{y+v}{2}\right) \ge f(x,y)+f(u,v)$

by Jensen's, which is exactly our inequality.

[quote="Farenhajt"]How do you apply Jensen on two-argument functions?[/quote]

There is a statement [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=jensen+multivariable&t=31173]here[/url]. Basically, just use vectors for your variables, so if $f$ is convex then

$f\left(\sum a_{i}X_{i}\right) \ge \sum a_{i}f(X_{i})$

where $\sum a_{i}= 1$ and $X_{i}$ are vectors. In this case,

$a_{1}= a_{2}= \frac{1}{2}$ and $X_{1}= (x,y)$, $X_{2}= (u,v)$

and $f$ is concave.

Hmm, now I'm not sure whether

$f_{aa}\ge 0$ and $f_{bb}\ge 0$

are enough to call $f$ a convex function.

Possibly is, but you can circumvent that thus:

$f(a,b)=af\left(1,{b\over a}\right)$, so put $g(r) : =f(1,r)={r\over 1+r}$, and that function is definitely concave. Then, if $y=px, v=qu$, we have

\begin{eqnarray*}f(x+u,y+v) &=& f(x+u,px+qu)\\ &=& (x+u)g\left({px+qu\over x+u}\right)\\ &\geqslant& (x+u)\left[{x\over x+u}g(p)+{u\over x+u}g(q)\right]\\ &=& xg(p)+ug(q)\\ &=& xf\left(1,{y\over x}\right)+uf\left(1,{v\over u}\right)\\ &=& f(x,y)+f(u,v)\end{eqnarray*}

Ah, very nice. I always thought multivariable Jensen's was quite slick, though :wink:.

Wow Farenhajt, thank you for the [u]very very cool[/u] proof!   :thumbup:

I can hardly find anything on the net on multivariable Jensen, except for integral multivariable Jensen. If it would be slick, it would become a very powerfull tool!


if Pat is 100/3% taller than Joel, than Joel is what percent shorter than Pat. How would you do this problem?

[hide]The ratio of Joel's height to Pat's height is 3:4, so Joel is 25% shorter. [/hide]


What's being randomly chosen here?  Are we choosing the coefficients randomly out of $\mathbb{R}$?  Or the roots?  Or what?

[quote="akcagil"]P(x) is a third degree polynomial. What is the probability that
P(x) have two extreme points?[/quote]

maybe instead of extreme point, you mean critical point, i.e. a point $(x_{0},P(x_{0}))$ such that $P'(x_{0})=0$

and i am not clear how the probability is distributed, do you mean that each coefficient has an equal probability of being any real number?

let $P(x)=ax^{3}+bx^{2}+cx+d$

$P'(x)=3ax^{2}+2bx+c$

real roots using the disriminant:

$4b^{2}-4(3a)(c)\ge 0$

$b^{2}\ge 3ac$

then you would graph this in a cube with $(\pm1,\pm1,\pm1)$ as all verticies, and take the ratio of area that satisfies divided by the whole area

We are choosing the coefficients randomly out of R .  So every choosing  coefficients  give us a polynomial. But some of these plynomials have two extreme points.

[quote="akcagil"]We are choosing the coefficients randomly out of R .[/quote]

"Chosen randomly out of $\bf R$" unfortunately doesn't mean anything -- you need to give a distribution.

Altheman's solution sounds basically correct, where the distribution is "uniformly distributed over [-1, 1] and zero elsewhere."  Except in the last line, he said "area" instead of "volume."


integration
trigonometry
function
calculus
calculus computations

Is $f:\mathbb {R}^2\to\mathbb R$
\[ f(x,y)={\begin{cases}
\displaystyle\int_{\frac{\pi}6}^{\frac{\pi}3}\dfrac{2xy}{x^2\sin^2t+y^2\cos^2t}\,dt\qquad\text{if}\quad(x,y)\not=(0,0)\\
\quad\\
0\qquad\qquad\qquad\qquad\qquad\qquad\quad\text{if}\quad(x,y)=(0,0)
\end{cases}}\]
continuous in $\mathbb{R}^2$?

The function is discontinuous at $(0,0).$

One way to see this: for $x\ne0,\,f(x,0)=0$ but $f(x,x)=\frac{\pi}3.$

In general:

Suppose $f$ is positively homogeneous of degree zero, which means that for all $\lambda>0$ and for all $(x,y)\ne(0,0),$

$f(\lambda x,\lambda y)=f(x,y).$

Then $f$ is continuous at $(0,0)$ if and only if $f$ is constant.

As always O.K. Kent. I worked out the exercice so:
\[{{{{{ f(x,y)=\dfrac{2x}y\int_{\pi/6}^{\pi/3}\dfrac{dt}{\cos^2t(\tan^2t+
\dfrac{x^2}{y^2}}})}\\=
2(\arctan\dfrac{x\sqrt{3}}y-\arctan\dfrac{x}{y\sqrt{3}})}}\\=
2(\arctan\dfrac{y\sqrt{3}}x-\arctan\dfrac{y}{x\sqrt{3}})}\]
since $\forall u\in\mathbb{R}_0$ we have $\arctan|u|+\arctan\frac1{|u|}
=\frac{\pi}2$
The function is continuous in all points (x,y) $\in\mathbb{R}_0\times
\mathbb{R}_0$
Then
if $ x_0\ne 0 $ and $\forall \;(x,y)\;\in\mathbb{R}_0\times\mathbb R:$
\[ |f(x,y)-f(x_0,0)|\le 2|\frac yx|(\sqrt{3}-\dfrac1{\sqrt{3}})\]
The function is continuous in all points $(x_0,0)$
if $ y_0\ne 0 $ and $\forall \;(x,y)\;\in\mathbb R\times\mathbb{R}_0:$
\[|f(x,y)-f(0,y_0)|\le 2|\frac xy|(\sqrt{3}-\dfrac1{\sqrt{3}})\]
The function is continuous in all points $(0,y_0)$


SirErnest, the old mod, never logs on anymore and my membership of the Indiana "usergroup" is still pending. so, i would like to be included in this indiana group. SHould we get a new mod? if so, how?

how would someone become a mod? not like i want to or anything, but i'm just curious :lol: ...

How would you apply to get into this "usergroup" ?

to apply for membership in a usergroup, there is a button under your Private Messages button (or somewhere near it) that says "Usergroups". Click on that and it will give you a little scroll link where you can choose which usergroup you want to join. Pick indiana and u can probably figure out the rest from there...


function
combinatorics unsolved
combinatorics

Let positive integer $ n$.
Find the number of pair $ (x,y,z,t)$ such that: $ 2^x3^y5^z7^t\leq n$

What makes you think that there is a nice closed form formula for that? :? The number of integer points inside a dilated "irrational" simplex is a very ugly function even in dimension 2.

I agree with fedja.
I never think that there would be such a nice closed form.


Hi, I'm Ezbakhe Yassin, from Morocco.

Glad to learn from you all. THANKS! :)

By the way, are there other people from Morocco?

hi,

why didn't you prost here:www.mathlinks.ro/Forum/topic-51533.html

you'll find what you are looking for :)

Oh,  :oops: you are right.

Nice to meet you, by the way!


In an equilateral triangle ABC ,suppose P,R are points on AB,BC such that PA+AQ=RC+CS=AC.Prove that PS,QR intersect at an angle of 60 degree.

Oops sorry for the wrong question. :oops:
In an equilateral triangle ABC,suppose P is a point on AB,R is a point on BC and Q,S are points on AC such that PA+AQ=RC+CS=AC. Prove that PS ,QR intersect at an angle of 60 degree


$abcd$ is a 4 digit integer such that $3,000<abcd<4,000$ and $abcd=a^a+b^b+c^c+d^d$. Find $abcd$.

is it supposed to have different digits or can a, b, c, and d all represent the same digits? i'm assuming they can't be the same.

[hide]
You can use either $5^5=3125$, $4^4=256$, $3^3=27$, $2^2=4$, and $1^1=1$.
Clearly the first digit is 3.
You also have to have a 5 to get the sum in the 3000 range.
So right now you have 27+3125=3152
From here you just have to play around with adding 256s, 27s, 4s, and 1s.
You will eventually get 3435, which is $3^3+4^4+3^3+5^5=3435$[/hide]


function
limit
topology
calculus
calculus computations

Let $ (f_n)$ be a sequence of real-valued continuous functions on a metric space $ X$. Suppose that $ (f_n)$ converges uniformly to a function $ f$. Prove that if all the terms of the sequence $ (f_n)$ are bounded, then $ f$ is bounded.

So $ |f_{n}(x)| \leq M_{n}$ for all $ x \in X$ and $ n \in \mathbb{N}$. Now for every $ \epsilon > 0$ there exists $ N \in \mathbb{N}$ such that if $ n > N$ and $ x \in X$, $ d(f(x), f_{n}(x)) < \epsilon$.  Thus $ d(f(x), f_{n}(x)) \leq d(M_n, f)$ for all $ n \in \mathbb{N}$.Thus $ |f| < \sum_{i \equal{} 1}^{n} M_n$?

Now what?

Just note that $ \left|f(x)\right|\equal{}\lim_{n\rightarrow\infty}\left|f_{n}(x)\right|\leq\sup_{n}\left\|f_{n}\right\|_{\infty}\equal{}C$ and $ \left\|f\right\|_{\infty}\leq C$, since $ \left\|f\right\|_{\infty}\leq\left\|f\minus{}f_{n}\right\|_{\infty}\plus{}\left\|f_{n}\right\|_{\infty}<\varepsilon\plus{}\left\|f_{n}\right\|_{\infty}$.


After simplify the equation we find that 7^3*3=1029.


I'm currently working on a smaller text based task i hope someone can help me with.

"A copper plate on 0.5 m^3 is o be transformed into a cylinder shaped copper wire with a lenght of 200km. How big will the copper wire's diameter be? (The volume of the cylinder is V = pi x r^2 x h)."

Now, i set 'h = 200,000'. If i increase h, V will just become bigger, and that doesn't make much sence since it shouldn't give a bigger diameter when you increase the lenght. Can anyone help? Thx in advance.

The moving of the letters was the hardest work, but thx, it helped a lot. However, i get way different results than what you get. I've tried giving the calculator the whole calculation and did provide it with a lot of brackets that i usally don't have any problems in placing the right place, and i tried calculating it step by step, but i end up with:
d=0.000892
Can you agree on this results? Or what about this equation:
d=2*sqrt((0.125)/(pi*200,000))

Does this seem right in your eyes or did i do something wrong?

Plz, can you tell me what went wrong since we got different results?

NVM, i found out what went wrong, so np now.


linear algebra
matrix
linear algebra unsolved

show that a n by n matrixx A is symmetric and positive definite
if and only if it can be written as A = GG ^ t where G is a lower triangular matrix
with positive diagonal elements. This representation is known as the Cholesky
factorization of A.

can anyone please help with this


calculus
integration
trigonometry
function
calculus computations

can someone help me in inegration of the following by parts

INTEGRATE  x arctan[sqrt[x]]

I am not able to get past the first integration by parable to get to first integration by parts ..can anyone help [/img][/code]

Let $\tan^{-1}\sqrt{x}=u,$

$\int x\tan^{-1}\sqrt{x}\ dx=\int \tan^{2}u\cdot \frac{2\tan u}{\cos^{2}u}\ du$

$=2\int u\cdot \tan^{3}u\cdot (\tan u)'\ du$

$=2\int u\left(\frac{1}{4}\tan^{4}u\right)'\ du.$

Alternately, substitute $t=\sqrt{x}$ first. If you integrate by parts without this substitution, you'll need to do something like it later anyway.

The problem is reasonable in terms of either trig functions or inverse trig functions, but some substitution is essential. If you leave it in inverse trig form, you should differentiate the $\arctan$ and integrate whatever's left over.

Integrator shows the result as shown in the image ..not sure if I get the result if I use arctan sqrt(x) = u

Hey Thanks Jmerry, t= sqrt (x) substitution worked , it made it esier to integerate down th e line. I was getting struck at integral

integral (X)^3/2 /(1+x) after the first integration by parts.


Let $ k$ and $ d$ be integers such that $ k>1$ and $ 0 \le d < 9$.    Prove that there exists a positive integer $ n$ such that the decimal representation of the number $ 2^n$ has digit $ d$ in the position $ k$ (from the right)

for any k digits $ d_k,d_{k \minus{} 1},...,d_1$ suth that $ 2^k|d_1 \plus{} d_2*10 \plus{} ... \plus{} d_k*10^{k \minus{} 1}$ exist infinetely many n, suth that
$ 2^n \minus{} d_1 \minus{} d_2*10 \minus{} ... \minus{} d_k*10^{k \minus{} 1}\equiv 0\mod 10^k$, because 2 is primitive root mod $ 5^k$ with period $ 4*5^{k\minus{}1}$.

I don't really get it. The primitive root gives us $ 4\cdot 5^{k\minus{}1}$ numbers, coprime to $ 10^{k}$. But we need something like $ 9^{k\minus{}2}\cdot 4$ nubers to be sure every digit appears on the $ k^{\mathrm{th}}$ position (there can be $ 9^{k\minus{}3}\cdot 4$ with $ 0$ on the begining, with $ 1$, with $ 2$...). So there must be other (maybe trivial) condition for those numbers to fulfil, which I don't see.
Could you explain it more formally?


Hi , i'm not god on mathematics sow please help me.I'm working on software who must calculate an algorithm :
i have 16 rows with 3 values and i have to calculate the maximum of options for this algo.
ex: 1. a1 a2 a3
2. b1 b2 b3
3. c1 c2 c3
....................
....................
16. q1 q2 q3
the algo is like the game for socker .you must choice 1 value for 1 string : a1
or a2 or a3 for first string so on until the end.So , my problem is what algo to use ?Please help me bigs..!!!

wait a minute "soccer"? Could you explain yourself more? From what I get you should get some software to try each row and not to keep the value you have if it is lower than that of the last highest number and store tha in a vector but I am not sure.. :blush:

:D funcia...my english is like my german..very bad :)
  the game is like this :
row 1:   1 or 2 or 3
      2:   1 or 2 or 3
      3:   1 or 2 or 3
      etc..
     16.   1 or 2 or 3
  from a row you can choice just a option.How many option are ?
and what is the algo ?

from the above it seems the algorithm (?) is [b]3^16[/b], giving the number 43046721.


Prove that if $a,b,c\geq 0$, then $a^5+b^5+c^5\geq 5abc(b^2-ca)$.

it is equal to
$a^5+b^5+c^5+5a^2bc^2 \ge 5ab^3c$
let $x^{10}=a^5c^5$
so we should only to prove
$2x^5+b^5+5bx^4 \ge 5x^2b^3$
let $y=\frac{b}{x}$
so we need to prove:
$y^5-5y^3+5y+2 \ge 0$
let $f(y)=y^5-5y^3+5y+2 \ge 0$
$f'(y)=5y^4-15y^2+5$
i think the rest is easy...

Have a look at :
  [url]http://www.mathlinks.ro/Forum/topic-4367.html[/url]


algebra polynomial function inequalities calculus derivative vector

Let $P(x)$ be a polynomial of degree $n$ with real zeros $a_{1},\cdots,a_{n}$, and let $P^{\prime}(x)$ have zeros $b_{1},\cdots,b_{n-1}$. Let $f$ be a convex function defined on the real line. Prove that
\[\frac{f(a_{1})+\cdots+f(a_{n})}{n}\geq \frac{f(b_{1})+\cdots+f(b_{n-1})}{n-1}. \]

[quote="arqady"]
Then since $f$ is convex function the following inequalities holds:
$f(a_{1})+f(a_{n})\geq f(b_{1})+f(b_{n-1}),$ $f(a_{1})+f(a_{n-1})\geq f(b_{1})+f(b_{n-2}),...,$ $f(a_{1})+f(a_{2})\geq2f(b_{1}),$
$f(a_{2})+f(a_{n})\geq f(b_{2})+f(b_{n-1}),$ $f(a_{2})+f(a_{n-1})\geq f(b_{2})+f(b_{n-2}),...,$ $f(a_{2})+f(a_{3})\geq2f(b_{2}),$

$...,$

$f(a_{n-1})+f(a_{n})\geq 2f(b_{n-1}).$[/quote]

Why? Karamata? then, please, show that left side majorizes right side because it's not obvious for me. I suspect your proof is wrong.

[quote="Megus"]I suspect your proof is wrong.[/quote]
Why, Przemyslaw? $f(a_{1})+f(a_{n})\geq f(b_{1})+f(b_{n-1})$ is true or wrong? Thank you. :)

The sentence "I suspect your proof is wrong" was separated from prior considerations by a dot, hence (in this case) it's a new thought independent of what was before  :wink:

Anyway, can you show why $f(a_{1})+f(a_{n})\geq f(b_{1})+f(b_{n-1})$ holds? I really don't see it.  :maybe:

[quote="Megus"]The sentence "I suspect your proof is wrong" was separated from prior considerations by a dot, hence (in this case) it's a new thought independent of what was before[/quote]
You  realy  philosopher! :D
[quote="Megus"]
Anyway, can you show why $f(a_{1})+f(a_{n})\geq f(b_{1})+f(b_{n-1})$ holds? I really don't see it.  :maybe:[/quote]
Oops. It can be false. Thank you, Megus!

If it's true then it's very nice problem.

Hans Rullgård's proof, appeared in the solutions of Nordic mathematical team contest 2007:

Every convex function can be approximated by piecewise linear function. Therefore WLOG we may assume that $f$ is linear between the roots of $P$. The case $f$ is linear is easy to check. Furthermore, if the inequality hold for $f(x)=\max (0,x-A)$, it holds for a positive linear combinations of such functions.

Let $Q(x)$ be the polynomial with roots $c_{j}=\max(A,a_{j})$. and let $Q'(x)$ have roots $d_{1},\cdots,d_{n-1}$. Then
\[\frac{\sum_{i=1}^{n}f(a_{i})}{n}=\frac{\sum_{i=1}^{n}f(c_{i})}{n}=\frac{\sum_{i=1}^{n-1}f(d_{i})}{n-1}. \]
Assuming the zeros are written in increasing order, the proof will be completed by showing $b_{j}\leq d_{j}$ for $j=1,\cdots,n-1$.

Let $a_{j}(t)=a_{j}+t(c_{j}-a_{j})$ for $0\leq t\leq 1$, let $P_{t}(x)$ be the polynomial with zeroes $a_{j}(t)$, and let $b_{j}(t)$ be the zeroes of $P_{t}^\prime(x)$. Since
\[\frac{{P_{t}^'}(x)}{P_{t}(x)}=\sum_{i=1}^{n}\frac{1}{x-a_{i}(t)}\]
it follows by substituting $x=b_{j}(t)$ that
\[0=\sum_{i=1}^{n}\frac{1}{b_{j}(t)-a_{i}(t)}. \]
Differentiating this with respect to $t$ gives
\[\frac{db_{j}(t)}{dt}\left (\sum_{i=1}^{n}\frac{1}{(b_{j}(t)-a_{i}(t))^{2}}\right )=\sum_{i=1}^{n}\frac{c_{1}-a_{1}}{(b_{j}(t)-a_{i}(t))^{2}}\]
showing that $db_{j}(t)/dt\geq 0$ and completing the proof.

I also tried a proof with Jensen inequality but I was unable to find suitable coefficients. Inequality holds also if the zeros of $P$ are complex numbers (De Bruijn-Springer conjecture), but the proof is more difficult and I have not seen it. I think, however, that it can be found on

N.G. de Bruijn, T.A. Springer, On the zeros of a polynomial with its derivative II, Indag. Math. 9 (1947) 264-270.

Very nice theorem!
Note that by majorization theory, the statement is equivalent to

$(\ \underbrace{a_{1},...,a_{1}}_{n-1},...,\underbrace{a_{n},...,a_{n}}_{n-1}\ )\succ (\ \underbrace{b_{1},...,b_{1}}_{n},...,\underbrace{b_{n-1},...,b_{n-1}}_{n}\ )$.

(Note I have inversed the order of the $a_{k}$'s etc. but that doesn't really matter)
For $k=1,...,n-1$ let $e_{k}: =\frac{n-k}na_{k}+\frac{k}{n}a_{k+1}$. Then it's easy to show by comparing the partial sums that $(e_{1},...,e_{n-1})$ is maximal (in the sense of $\succ$) as a vector satisfying
$(\underbrace{a_{1},...,a_{1}}_{n-1},...,\underbrace{a_{n},...,a_{n}}_{n-1})\succ (\underbrace{e_{1},...,e_{1}}_{n},...,\underbrace{e_{n-1},...,e_{n-1}}_{n})$.

Even though $\succ$ is only a partial order, does that allow us to conclude for the zeros of $P^\prime(x)$ that $b_{k}\le e_{k}$? At first glance, that seems obvious. Is it really?  :maybe:

I cannot edit this anymore.  :mad: [quote="spanferkel"]
does that allow us to conclude for the zeros of $P^\prime(x)$ that $b_{k}\le e_{k}$? [/quote]
Of course that was nonsense. Reverting to the original notation, $a_{1}\le b_{1}\le a_{2}\le...\le b_{n-1}\le a_{n}$, the only thing we can obviously infer from the majorization is that
$e_{1}\le b_{1}\le a_{2}$  and $a_{n-1}\le b_{n-1}\le e_{n-1}$. Is it possible to find similar restrictions for others of the $b_{k}$'s?


Prove: Starting at a point on a side of a triangle (not a vertex) and constructing a parallel to another side intersecting the third side at a second point, and repeating this five more times, the seventh point is the starting point.

This problem is related to [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=347821#p347821[/url]


How many natural numbers $ n$ will make $ \frac{36}n$ a natural number?

factors of 36. $ 36\equal{}2^2\cdot3^2$, which has $ (2\plus{}1)(2\plus{}1)\equal{}\boxed9$ factors.


ok...thanks to all i've managed to grab something out of inequalities now...now i face a couple of problems...

1) (5x-6)/(x+6) < 1

a. -6 < x < 6
b. -6 < x < 0
c. -6 < x < 3
d. None of the Above

I solve it to get

x<3, x> -6 (or) x > 3, x< -6

This is answer choice c... but the given solution is d. none of the above....Why? How?

[quote="coffeym"]$\frac{5x-6}{x+6}<1$

Now, $x$ may be positive or negative, so to ensure positivity (i.e so that we can keep the inequality sign as it is) we must multiply through by $(x+6)^{2}$. So

$(5x-6)(x+6)-(x+6)^{2}<0$

$(x+6)(4x-12)<0$

Which means $-6<x<3$

Oh dear...that's not answer $d$  :lol: Maybe the book is wrong  :blush:[/quote]

Lol I think the book is wrong too  :D

[quote="coffeym"]$\frac{5x-6}{x+6}<1$

Now, $x$ may be positive or negative, so to ensure positivity (i.e so that we can keep the inequality sign as it is) we must multiply through by $(x+6)^{2}$. So

$(5x-6)(x+6)-(x+6)^{2}<0$

$(x+6)(4x-12)<0$

Which means $-6<x<3$

Oh dear...that's not answer $d$  :lol: Maybe the book is wrong  :blush:[/quote]

Another way is to subtract $\frac{x+6}{x+6}$ from the LHS and $1$ from the RHS.

$\frac{4x-12}{x+6}<0\Longrightarrow\frac{x-3}{x+6}<0\Longrightarrow-6<x<3$


calculus
integration
combinatorics unsolved
combinatorics

Let's think about it ! This is a very nice  problem from my friend

well, $C_n^{k}$ is $n \choose{k}$.
anyway it seems quite simple.We know: $\sum_{k=0}^{n} {n \choose{k}}y^k=(1+y)^n$.so,$\sum_{k=1}^{n} {n \choose{k}} (x-1)^{k}=(1+(x-1))^{n}-1=x^n-1$;so
 $\sum_{k=1}^{n} {n \choose{k}} (x-1)^{k-1}=1+x+x^2+...+x^{n-1}$;integrating both sides we get
$\sum_{k=1}^{n} \frac{{n\choose{k}}(x-1)^k}{k}=\sum_{k=1}^{n}\frac{x^k}{k} +C$ for some constant C.so the given summation is simply the constant C.To get C ,simply plug in $x=0$ so that all that remains is evaluation of $\sum_{k=1}^{n} \frac{(-1)^{k+1} {n \choose{k}}}{k} $but that again comes from the preceding procedure during the integration-simply plug $x=1$ in the integral to get $C=-(1+1/2+....1/n)$ and since finally we need -C, we get the result.


I found some interesting things when working with factorials:

$ 10\times9\times8\equal{}6!$
$ 10\times9\times8\times7\equal{}7!$
$ 3!\times5!\times7!\equal{}10!$

Can anybody explain this? Does a similar thing work with different numbers? Are these just coincedinces?

I'm not sure if this happens with other numbers but $ 10*9*8$ IS $ 6!$, only the some numbers have been re-arranged:
$ 10*9*8\equal{}5*2*3*3*4*2$
If you re-arrange these numbers, you find: $ 6*5*4*3*2$ which is $ 6!$, the same is with the other cases.

Well I think that these are just coincidences. Cause I think that these don't occur much with higher factorials.

I just realized, it is kind of the same thing with Pythagorean Triples. I agree that they are coincedinces.

apparently, if u pay attention... u can factor the factorials into bigger #s...
 :)

i think this has something to do with the factors

[quote="isabella2296"]Isn't it just because you chose numbers that when multiplied together, have the same prime factors?[/quote]
i'm pretty sure that's the reason.

[quote="batteredbutnotdefeated"]This is a bit unrelated, but you should have noticed that all factorials beyond 5 end in 0. If you know this then you can answer certain EASY factorial questions.[/quote]
Obviously. Starting at 5!, all have a factor of 5, and multiple factors of 2, leading to at least one 10.

[hide="Mod Edit: Stuff too advanced for forum"]
This is very interesting indeed. Actually, for each $ n \in \mathbb{N}$, $ n \geq 2$ there are infinite number of solutions for the equation
\[ a_1!\cdot a_2!\cdot \ldots \cdot a_{n \minus{} 1}! \equal{} a_n!
\]
To show this, just look at this example:
\[ n! \cdot (n! \minus{} 1)! \cdot (n!! \minus{} 1)! \cdot \ldots \cdot (n \underbrace{!!\ldots !}_{k \minus{} 1 \text{ signs}} \minus{} 1)! \equal{} n \underbrace{!!\ldots !}_{k \text{ signs} }
\]
or prettier
\[ n!\cdot \prod_{i \equal{} 1}^{k \minus{} 1}\left( \left( n \underbrace{!!\ldots !}_{i \text{ signs}} \minus{} 1 \right)! \right) \equal{} n \underbrace{!!\ldots !}_{k \text{ signs} }
\]
This is true because
$ n! \cdot (n! \minus{} 1)! \equal{} n!!$,
$ n!! \cdot (n!! \minus{} 1)! \equal{} n!!!$
and so on.  :)[/hide]

i agree that these are just coincidences -they have the same factors

for 10*9*8=6!
you're  just saying that 2*5*3*3*2*2*2=3*2*5*2*2*3*2*1
you can see that this is true by crossing out factors

for 10*9*8*7=7!
you're just multiplying both sides of the last one by 7

for 3!*5!*7!=10!
this one is a bit more interesting, but works for the same reason
1*2*3*1*2*3*2*2*5*1*2*3*2*2*5*2*3*7=1*2*3*2*2*5*2*3*7*2*2*2*3*3*2*5

umm..
my previous post (before the one about the hard stuff) said exactly that
just a little more vaguely :ninja:

@nick42: they aren't "special", they're just coincidences.

There are some more, like:

$ 3!\times2!\times2! \equal{} 4!$
$ 3!\times5! \equal{} 6!$

I think there are infinitely many of these. For example, you take a factorial, like $ 5!$, and multiply it by the factorial of one less than that value. In this example, it would become $ 5!\times119! \equal{} 120!$.

Some are coincidences, but others follow a pattern.

EDIT: I just noticed that you can take any factorials, like $ 3!$ and $ 4!$, and multiply them to get $ 144$. Multiply this by $ 143!$ and get $ 3!\times4!\times143!\equal{}144!$.

There are lots of these, but some are just harder to find, like $ 3!\times5!\times7!\equal{}10!$. This is because $ 3!\times5!\equal{}720\equal{}10\times9\times8\equal{}6!$.

[quote="PowerOfPi"]There are some more, like:

$ 3!\times2!\times2! \equal{} 4!$
$ 3!\times5! \equal{} 6!$

I think there are infinitely many of these. For example, you take a factorial, like $ 5!$, and multiply it by the factorial of one less than that value. In this example, it would become $ 5!\times119! \equal{} 120!$.

Some are coincidences, but others follow a pattern.

EDIT: I just noticed that you can take any factorials, like $ 3!$ and $ 4!$, and multiply them to get $ 144$. Multiply this by $ 143!$ and get $ 3!\times4!\times143! \equal{} 144!$.

There are lots of these, but some are just harder to find, like $ 3!\times5!\times7! \equal{} 10!$. This is because $ 3!\times5! \equal{} 720 \equal{} 10\times9\times8 \equal{} 6!$.[/quote]

You just find numbers with the right factors which are coincidences.
The patterned ones like powerofpi said are obviously not coincidences, like $ 3!\times4!\times143! \equal{} 144!$ or $ 2!\times4!\times5!\times5759! \equal{} 5760!$ which are easy to find and follow a pattern.

[quote="PowerOfPi"]$ 10\times9\times8 = 6!$
$ 10\times9\times8\times7 = 7!$[/quote]

They are the same.. but, what about:
"Find all (n,m) in N such that (m+n)! = (n!)^2$ ?.
We can do better. "Find all (a,b,n) in N with n>1 such that a!=b^n$".

Just look modulo 3 from a certain point.. :wink:


function
induction
inequalities
algebra
polynomial
algebra unsolved

The functions $ T_{n}(x)$, for $ n\equal{}0, 1, 2, 3, ...$, satisfy the recurrence relation

$ T_{n\plus{}1}(x) \minus{}2xT_{n}(x) \plus{} T_{n\minus{}1}(x) \equal{} 0$  $ (n\geq 1)$   $ (*)$.

Show by induction that

$ (T_{n}(x))^{2} \minus{} T_{n\minus{}1}(x)T_{n\plus{}1}(x) \equal{} f(x)$,

where $ f(x) \equal{} (T_{1}(x))^{2} \minus{} T_{0}(x)T_{2}(x)$.

In the case $ f(x) \equiv 0$, detemine (with proof) an expression for $ T_{n}(x)$ in terms of $ T_{0}(x)$ (assumed to be non-zero) and $ r(x)$, where $ r(x) \equal{} \frac{T_{1}(x)}{T_{0}(x)}$. Find two possible expression for $ r(x)$ in terms of $ x$.


find all integers $k$ such that there exist an integer $n>1$ and relatively prime integer $x,y$ such that
$x^n+y^n=3^k$

I'm quite sure to have solved this one here before... ;)

Pierre.

[quote="kai07"]Well if $n=2k$ for some $k$ then 3 is a prime of the form $4k+3$ and divides a sum of two squares. Then $=>3|gcd(x,y)=>$ contradiction. Now we conclude $n$ is odd. Then $x+y|x^{n}+y^{n}$ and $x+y>1$. Then both of $x^{n}+y^{n}$ and $x+y$ are powers of 3. But By Zsigmondy's theorem we get that for $n>3$ there exists a prime $p$ s.t. $p|x^{n}+y^{n}$ and $p$ does not divide $x+y$.
So for $n>3$ the equality is impossible. So $n=3$ and this is the only solution.[/quote]

If $n=3=>(x+y)(x^{2}-xy+y^{2})=3^{k}$. We have $x+y>1=>3|x+y$. Now we use LTE:
$v_{3}(x^{3}+y^{3})=v_{3}(x+y)+1=>v_{3}(x^{2}-xy+y^{2})=1=>(1) x^{2}-xy+y^{2}=3$. After solving equation $(1)$ the only solutions are $(x;y)=(2;1);(1;2)$. There is one possible value of $k$ and it is $k=2$.


Find all values of $m,n,p$ such that $m,n$ are positive integers and $p$ is a prime number that satisfy:

$p^{n}+144=m^{2}$.

Yeah, $s = 0$ and $r = 2$ gives $p = 5$. Nice solution, though.


An o-Pod MP3 player stores and plays entire songs. Celeste has 10 songs stored on her o-Pod. The time length of each song is different. When the songs are ordered by length, the shortest song is only 30 seconds long and each subsequent song is 30 seconds longer than the previous song. Her favorite song is 3 minutes, 30 seconds long. The o-Pod will play all the songs in random order before repeating any song. What is the probability that she hears the first 4 minutes, 30 seconds of music - there are no pauses between songs - without hearing every second of her favorite song? Express your answer as a common fraction.

We use complementary probability:

Case 1: Her song players first.

This is $ \frac{1}{10}$.

Case 2: 30 second song, then her song.

This is $ \frac{1}{10} \times \frac{1}{9}\equal{}\frac{1}{90}$.

Case 3: 60 second song, then her song.

This is $ \frac{1}{10} \times \frac{1}{9}\equal{}\frac{1}{90}$.

The probability that she hears all of her favorite song is $ \frac{1}{10}\plus{}\frac{1}{90}\plus{}\frac{1}{90}\equal{}\frac{11}{90}$.  Thus, the probability that she doesn't hear her entire song is $ 1\minus{}\frac{11}{90}\equal{}\boxed{\frac{79}{90}}$.


[quote="rnwang2"]Can I just attach my $\LaTeX$ code and PDF document to the email I send to solutions@usamts.org? Or should I copy all my code into the email text?[/quote]
You can send the files as attachments. It's clarified on the http://www.usamts.org page