AMC
USA(J)MO
USAMO
geometry
greatest common divisor
trigonometry
Vieta

I'm trying to get a perfect AMC 10 and 135 AMC 12 this year.
I'm sure others would like to improve there AMC 10/12 score, so please post problems based on one of the following levels:

$ \star$=Easy AMC 10/12 (1-5)
$ \star\star$=Middle Easy AMC 10/12 (5-13)
$ \star\star\star$=Middle AMC 10/12 (14-18)
$ \star\star\star\star$=Middle Hard AMC 10/12 (19-22)
$ \star\star\star\star\star$=Hard AMC 10/12 (23-25)

Don't worry too much about the exact level.

Here is the first problem: $ \star\star$
Compute $ (1000\minus{}998)\plus{}(998\minus{}996)\plus{}(996\minus{}994)\plus{}\cdots\plus{}(330\minus{}328)\plus{}(328\minus{}326)$

[quote]I'm trying to get a perfect AMC 10 and 135 AMC 12 this year. [/quote]

Weird, it won't let me edit my previous post.  Anyways, I guess I will just seperate this since its a totally different topic, if you are an 8th grader, I advise you not to take the AMC 12 but the AMC 10 2 times.  They have the junior math olympiad this year, so its an advantage to take the AMC 10.

Also, a 135 on the AMC 12 is probably as hard as gettign 5+ on USAMO, so its an extremely lofty goal, especially if you have not taken pre-calc (way harder than 150 on AMC 10).  Of course, I am not saying that you shouldn't shoot for your goals, but I am just cautioning you.  I know a person last year that got like 102 or so on the AMC 12 but missed blue MOP by 1 or 2 points on USAMO.

[quote="the future"]Hi, AwesomeToad.  I used to visit this website a lot, but I am kind of busy right now (I'm a senior).  However, once I am done with my college apps and my research, I will be posting here a lot.

I want to do well this year on the AMC/AIME/USAMO (since its my last year), so I will definitely working hard to get a good score.  I don't know who else still visits this website, but I think I will try to make some mockAMC/AIME problems like starting in December through March.

As for your first problem, the problem is essentially 2+2+2...x amount of times.
To find x, we have (1000-328)/(2)+1 = 337  because the difference between each term is 2.

So it becomes 2(337) = 674.

Next problem (AMC 12 1-5): Individually, it takes Alex 10 hours to mow the lawn and it takes Bob 5 hours to mow the lawn.  What is the fewest amount of hours that it will take for them to mow the lawn together, if they start together?[/quote]

lol, the futuer, another way to solve that problem is to see that it just telescopes so the answer is just 1000-326 = 674.  :) . As for your problem

1/(1/5 + 1/10) = 10/3 hours.

Harmonic mean.  Eh.  All I want is 100 AMC 10 and I'd be happy.......AMC 12 is very hard :o

(2 star difficulty[AMC 10'ish])

A number x is three larger than it's reciprocal.  What is the value of $ (x\minus{}1/x)^4$?

[hide="New Problem"]Suppose a square is inscribed in a circle. The radius of the circle is 5. Find the area of the square[/hide]$ \star\star$

Okay, so our diameter is $ 10$, which is the same as the diagonal of the square.  Thus, our answer is $ \frac{10^2}{2}$ or $ \boxed{50}$.

[hide="New Problem (shamelessly stolen from Ihatepie)"]The roots of $ ax^2\plus{}bx\plus{}c\equal{}0$ are irrational, but their calculator approximations are $ \minus{}1.780776407$ and $ .2807764065$. If $ a$, $ b$, and $ c$ are integer whose $ \text{GCF}$ is $ 1$, and $ a>0$, what is the ordered triple $ (a,b,c)$?[/hide]

[hide="Solution"]The sum of roots is $ \minus{}1.5$ and product $ \minus{}0.5$ so $ (a,b,c)\equal{}(2,\minus{}3,\minus{}1)$[/hide]
[hide="NP"]What is the least positive integer $ \phi$ so that $ \sin{3\phi}\equal{}\cos{7\phi}$[/hide]

[quote="cognos599"]9[/quote]

Uh, you should probably explain.

Although that is correct, using $ \sin{\theta}\equal{}\cos{90\minus{}\theta}$ so $ \sin{3\phi}\equal{}\cos{90\minus{}3\phi}$ and $ 90\minus{}3\phi\equal{}7\phi$ so $ \phi\equal{}9$

[hide="New Problem"]Let $ r$ and $ s$ be roots of $ 3x^2\minus{}2x\plus{}12$ Find $ (r\minus{}1)(s\minus{}1)$[/hide]

Hint: Expand the given product. Use the formulas for sum/product of roots.

[quote="AwesomeToad"][quote="cognos599"]9[/quote]

Uh, you should probably explain.

Although that is correct, using $ \sin{\theta} \equal{} \cos{90 \minus{} \theta}$ so $ \sin{3\phi} \equal{} \cos{90 \minus{} 3\phi}$ and $ 90 \minus{} 3\phi \equal{} 7\phi$ so $ \phi \equal{} 9$

[hide="New Problem"]Let $ r$ and $ s$ be roots of $ 3x^2 \minus{} 2x \plus{} 12$ Find $ (r \minus{} 1)(s \minus{} 1)$[/hide][/quote]

[hide="Solution"]$ (r\minus{}1)(s\minus{}1)\equal{}rs\minus{}r\minus{}s\plus{}1$

By Vieta's formulas, 12 is the product of the roots, and -2 is the sum of the roots times minus one.  In other words, $ rs\equal{}12$, $ \minus{}(r\plus{}s)\equal{}\minus{}2.$  So, by substitution, $ 12\minus{}2\plus{}1\equal{}\fbox{11.}$[/hide]

[hide="NP"]

Let $ k\equal{}2008^2\plus{}2^{2008}.$  What is the units digit of $ k^2\plus{}2^k$?

[hide="Answer Choices"]

$ (A)0$ $ (B)2$ $ (C)4$ $ (D)6$ $ (E)8$[/hide][/hide]

I'm bored.

[hide="Solution"]

Hmm.... Units digit of $ k$ is $ 2008^2 \plus{}2^{2008} \equiv 4\plus{}6 \equiv 0 \mod 10$ (check that $ 2^k$ is periodic mod 10).

Also easy to see that $ k$ is divisible by $ 4$ and hence $ 2^k \equiv 6 \mod 10$ while $ k^2 \equiv 0 \mod 10$ so our answer is $ D$.

[/hide]

[u]New problem[/u]

There exists a point $ P$ inside an equilateral triangle  $ ABC$ such that $ PA \equal{} 3, PB \equal{} 4,$ and $ PC \equal{} 5.$ Find the length of $ AB$.

[hide]Find the sum of all 4 digit integers that can be made from the 5 integers 6, 7, 8, 9, and 0, each used once.[/hide]

why would you revive an obviously dead thread, especially when one for the same purpose has already been [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=468965]created more recently[/url]???

I PMed AwesomeToad, and he/she said it was okay :)

Wait, for the problem, can you only use each digit once, or can you use each as many times as possible?

I don't think this is right...
[hide="Solution"]
Another attempt:
There are $4*4*3*2=96$ possible numbers.
If you have one number in the thousands digit, there are $4*3*2=24$numbers with that number in the thousands digit.
Since you can't have 0 in the thousands digit, the sum is:
$24,000*(6+7+8+9)+1,800*(6+7+8+9)+180*(6+7+8+9)+18*(6+7+8+9)=\boxed{779,940}$
I know, this is my horrible attempt at this problem.
[/hide]

[quote="Nitzuga"]I [i]do[/i] think this is right...
[hide="Solution"]You have $4*3*2$ numbers with a $6, 7, 8,$ or $9$ For the thousands digit. So you have $24*(6+7+8+9)=720$ for the thousands digit, so its $720,000$ for the thousands digit. For the hundreds digit, you have $3*3*2$ for the numbers $6, 7, 8,$ and $9.$ (0 won't add anything.) This is $18*(6+7+8+9)=540.$ The same applies for the 10's and one's digit. So that's $111*540=59940.$  $59940+720000=779940$ [b]Answer[/b][/hide][/quote]
What if 0 is one of the numbers?
EDIT:I thought of this wrong. (plus edited my solution)

[hide="NP"]
Pat wants to buy $4$ donuts from an ample supply of $3$ types of donuts: glazed, chocolate, and powdered. How many different selections are possible?
[/hide]

[hide]There are 4 donuts and 2 dividers. This makes 6 objects. There are $\binom{6}{2}=\boxed{15} $ways to put the 2 dividers in.[/hide]
[hide]How many triangles can be found when connecting the 9 diagonals of a hexagon?[/hide]

[hide]
I tried a casework counting attempt:
Outside triangles = $18+12+6+12+6+6=60$
Inside triangles = $12+6+6+2=26$
$60+26=\boxed{86}$
[/hide]
If that is correct...
[hide]
If the exterior angle of a polygon is 1, how many sides does it have?
[/hide]

[hide]360, the exterior angles add up to 360.[/hide]
[hide]Joe and JoAnn each bought 12 ounces of coffee in a 16-ounce cup. Joe drank 2 ounces of his coffee and then added 2 ounces of cream. JoAnn added 2 ounces of cream, stirred the coffee well, and then drank 2 ounces. What is the resulting ratio of the amount of cream in Joe's coffee to that in JoAnn's coffee? Source: AMC
 [/hide]

[hide]
Joe has 2 ounces of cream. JoAnn has 12/7 ounces of cream.
The ratio is $\boxed{\frac{7}{6}}$.
[/hide]
[hide]
Three distinct vertices of a cube are chosen at random. What is the probability that the plane determined by these three vertices contains points inside the cube?
[/hide]

[hide]Doesn't it... have to? So 100%?[/hide]
[hide]The Fibonacci sequence is defined as $F_n=F_{n-1}+F_{n-2}$ for all $F>2$, where $F(1)=1$ and $F(2)=1$. What is the remainder when the $F_{15}$th Fibonacci number is divided by 3?[/hide]

I should have said strictly inside. You can have 3 points on once face of the cube.

[quote="MCL"]@minimario, answer is [hide="Don't want to spoil it."]1[/hide][/quote]
Ummm... you might want to post a solution.

Oh sorry,

[hide]
Look at the remainders when divided by 3.

F1 = 1
F2 = 1
F3 = 2
F4 = 0
F5 = 2
F6 = 2
F7 = 1
F8 = 0
F9 = 1
F10 = 1
F11 = 2
F12 = 0
F13 = 2
F14 = 2
F15 = 1.[/hide]

Wasn't that a problem from a MathCounts State test? I remember solving it...

[hide="Solution to Cube Problem"]
First, find the total number of ways to choose three points. This is $\frac{8*7*6}{3*2*1}=56$
Then, you need to see how many of those sets of three don't work. These are the sets of points that for a plane on the side of a triangle. There are $4$ of these per side, so there's a total of $6*4=24$
$\frac{56-24}{56}=\frac{32}{56}=\frac{4}{7}[b] Answer[/b][/hide]

I don't know many problems, so can someone post another problem for me?

Eh, please try to use original problems.

You don't have to post a new problem every time.

Sorry for interrupting the marathon; carry on. :)

[hide]
You might see a pattern with the number of intersecting points:
2, 6, 12, ...
Anyways, the answer is the 3rd once, which is $\boxed{12}$.
Besides the pattern, I think you can look at it this way.
Choose 2 circles to intersect. They intersect twice. So:
general formula = $\binom{n}{2}*2$
[/hide]
[hide="NP"]
What is the value of $2^{-1}-2^{-2}+2^{-3}-2^{-4}+...+2^{-9}-2^{-10}$?
[/hide]

[hide="Solution"]I don't like negative powers so i'll just write it as 1/2 - 1/4 + 1/8 ... + 1/512 - 1/1024. Next, use the geometric sequence formula with A = 1/2 and r = -1/2. Our formula is A (1-r^n)/(1-r) = 1/2 * (1-1/2048)/(1+1/2) = 2047/3.[/hide]


trigonometry
algebra
polynomial
inequalities
geometry
rectangle
absolute value

[quote="darij (can't login again)"]Great work, Orl!

Little remark: Problem 4 = Problem 3...

  Darij[/quote]

Thank you.  :)  But I don't see that problems 3 and 4 are the same...

But they are... I would write both of them in the following form:

Given a triangle ABC. Prove that the equation $\sin A \sin B \leq \sin^2 \frac{C}{2}$ holds if and only if there exists a point D on the segment AB such that CD is the geometric mean of AD and BD.

  Darij

[b][size=150]IMO 1974 ShortListed Problems: (Posted in Forums)[/size][/b]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=358062&#p358062]1.(USA 4)[/url] [b]IMO1[/b] Three players $A,B$ and $C$ play a game with three cards and on each of these $3$ cards it is written a positive integer, all $3$ numbers are different. A game consists of shuffling the cards, giving each player a card and each player is attributed a number of points equal to the number written on the card and then they give the cards back. After a number $(\geq 2)$ of games we find out that A has $20$ points, $B$ has $10$ points and $C$ has $9$ points. We also know that in the last game B had the card with the biggest number. Who had in the first game the card with the second value (this means the middle card concerning its value).

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2026232#p2026232]2.  (POL 1)[/url] Prove that the squares with sides $\frac{1}{1}, \frac{1}{2}, \frac{1}{3},\ldots$ may be put into the square with side $\frac{3}{2} $ in such a way that no two of them have any interior point in common.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=358045&#p358045]3.  (SWE 3)[/url] [b]IMO6[/b] Let $P(x)$ be a polynomial with integer coefficients. We denote $\deg(P)$ its degree which is $\geq 1.$ Let $n(P)$ be the number of all the integers $k$ for which we have $(P(k))^{2}=1.$ Prove that $n(P)- \deg(P) \leq 2.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2026233#p2026233]4. (USS 4)[/url] The sum of the squares of five real numbers $a_1, a_2, a_3, a_4, a_5$ equals $1$. Prove that the least of the numbers $(a_i - a_j)^2$, where $i, j = 1, 2, 3, 4,5$ and $i \neq j$, does not exceed $\frac{1}{10}.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2026234#p2026234]5. (GBR 3)[/url] Let $A_r,B_r, C_r$ be points on the circumference of a given circle $S$. From the triangle $A_rB_rC_r$, called $\Delta_r$, the triangle $\Delta_{r+1}$ is obtained by constructing the points $A_{r+1},B_{r+1}, C_{r+1} $on $S$ such that $A_{r+1}A_r$ is parallel to $B_rC_r$, $B_{r+1}B_r$ is parallel to $C_rA_r$, and $C_{r+1}C_r$ is parallel to $A_rB_r$. Each angle of $\Delta_1$ is an integer number of degrees and those integers are not multiples of $45$. Prove that at least two of the triangles $\Delta_1,\Delta_2, \ldots ,\Delta_{15}$  are congruent.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=358034&#p358034]6. (ROM 4)[/url] [b]IMO3[/b] Prove that for any n natural, the number \[ \sum \limits_{k=0}^{n} \binom{2n+1}{2k+1} 2^{3k}  \]
cannot be divided by $5$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2026236#p2026236]7. (POL 2)[/url] Let $a_i, b_i$ be coprime positive integers for $i = 1, 2, \ldots , k$, and $m$ the least common multiple of $b_1, \ldots , b_k$. Prove that the greatest common divisor of $a_1 \frac{m}{b_1} , \ldots, a_k \frac{m}{b_k}$ equals the greatest common divisor of $a_1, \ldots , a_k.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=358019&#p358019]8. (NET 3)[/url] [b]IMO5[/b] The variables $a,b,c,d,$ traverse, independently from each other, the set of positive real values. What are the values which the expression \[ S= \frac{a}{a+b+d} + \frac{b}{a+b+c} + \frac{c}{b+c+d} + \frac{d}{a+c+d} \] takes?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2026238#p2026238]9. (CUB 3)[/url] Let $x, y, z$ be real numbers each of whose absolute value is different from $\frac{1}{\sqrt 3}$ such that $x + y + z = xyz$. Prove that
\[\frac{3x - x^3}{1-3x^2} + \frac{3y - y^3}{1-3y^2} + \frac{3z -z^3}{1-3z^2} = \frac{3x - x^3}{1-3x^2} \cdot \frac{3y - y^3}{1-3y^2} \cdot \frac{3z - z^3}{1-3z^2}\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=357983&#p357983]10.(FIN 3)[/url] [b]IMO2[/b] Let $ABC$ be a triangle. Prove that there exists a point $D$ on the side $AB$ of the triangle $ABC$, such that $CD$ is the geometric mean of $AD$ and $DB$, iff the triangle $ABC$ satisfies the inequality $\sin A\sin B\le\sin^2\frac{C}{2}$.

[hide="Comment"][i]Alternative formulation, from IMO ShortList 1974, Finland 2:[/i] We consider a triangle $ABC$. Prove that: $\sin(A) \sin(B) \leq \sin^2 \left( \frac{C}{2} \right)$ is a necessary and sufficient condition for the existence of a point $D$ on the segment $AB$ so that $CD$ is the geometrical mean of $AD$ and $BD$.[/hide]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2026241#p2026241]11.(BUL 1)[/url][b]IMO4[/b] We consider the division of a chess board $8 \times 8$ in p disjoint rectangles which satisfy the conditions:

a) every rectangle is formed from a number of full squares (not partial) from the 64 and the number of white squares is equal to the number of black squares.

b) the numbers $\ a_{1}, \ldots, a_{p}$ of white squares from $p$ rectangles satisfy $a_1, , \ldots, a_p.$ Find the greatest value of $p$ for which there exists such a division and then for that value of $p,$ all the sequences $a_{1}, \ldots, a_{p}$ for which we can have such a division.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2026242#p2026242]12.(USS 1)[/url] In a certain language words are formed using an alphabet of three letters. Some words of two or more letters are not allowed, and any two such distinct words are of different lengths. Prove that one can form a word of arbitrary length that does not contain any non-allowed word.


floor function
number theory unsolved
number theory

$ n,k$ are positive integers and $ n \ge2 ,k \le \lfloor \frac {n}{2} \rfloor$
Prove there exists a prime $ p$ with $ p > k$ and $ p \mid \binom{n}{k}$
Someone told me it is Sylvester Theorem,is it true?Could somebody offer a prooF?

[quote="pbornsztein"]Sylvester's theorem states that the product of $ k$ consecutive integers greater than or equal to $ k$ has a prime divisor greater than $ k$.
There are "elementary" proofs but they are quite long.

Pierre.[/quote]
Thank you so much :lol:

pbornsztein,can we use this theorem to prove $ a,d,n$ are positive integers $ (a,d)\equal{}1$
so $ a(a \plus{} d)\cdots (a \plus{} id \minus{} d)(a \plus{} id \plus{} d)\cdots (a \plus{} nd)$ has a prime factor not smaller than $ n$? :)


[quote="towbomb"]Does anyone have any interesting strategies for solving problems?

For instance, the "assume integers" strategy: Once, at a meet, I was working on a problem that was essentially the "British Flag" theorem, but at the time, I didn't know of it.  So, intead of setting up a complex system of equations, I assumed that the measurements were going to be integers, and I easliy got the solution in seconds![/quote]

Well, I don't really consider this a strategy for solving problems, but when you're working with triangles:

All triangles are pythagorean, equilateral, or right isosceles.


quadratics function MATHCOUNTS AMC USA(J)MO USAMO algebra

It does have text storage capability, but not a keypad like a Palm Pilot or a BLackberry. Is it legal?

Billy

Yes: since it's legal for the SAT, it's legal for mathcounts. Any calc with a QWERTY (typewriter) keyboard though is illegal, like the TI Voyage 200

[quote="kidwithshirt"]it doesnt really make whole lot of sense cuz u still can record some key formulas or threoms with a ti-84 silve+....

So its kinda same with a QWERTY pad[/quote]
Just let it slip and use the 84s and the 89s... I DONT WANT THEM TO BAN THE 89 TITANIUM!!!

well QWERTY basicly means a typing keyboard, a typing keyboard inputs alphabets, althought Ti graphing stuff doesnt really have a QWERTY but it input things, if u can input things that basicly means u can record datas... which basicly means u can cheat... which basicly means.... :blush:

True.
As long as they allow the graphing ones, I'm fine
Lets hope the lawyers don't get to MATHCOUNTS.
Otherwise...
I don't think they'll ban the 89T, but you never know...

[quote="solafidefarms"]I have programmed my calulator, and quite enjoy the ease of the Quadratic Formula and finding all divisors of a number.

Why store the theorems whille you can make a program to do the same thing faster?[/quote]

I've got like 21 programs that I use to speed find things.  :D

[quote="solafidefarms"]Mind posting some of the code?  :lol:  :lol: I'll give you mine! ALLDIVISors, FIBONACCi, QUADRATIc formula, and POLygonAREA.[/quote]

Can you post the alldivisors, ficonacci and polygonarea programs pls?

AllDivisors:

{1}STO>$L_{4}$
Input "NUMBER? ",N
N STO> M
While iPart(N/2)=N/2
2 STO> $L_{4}$(dim $L_{4}$+1)
N/2 STO> N
End
For(A,3,iPart($\sqrt$ M)+2,2)
If iPart(N/A)=N/A
Then
While iPart(N/A)=N/A
A STO>$L_{4}$(dim $L_{4}$+1)
N/A STO>N
End
End
End
Disp $L_{4}$
Stop

FIbonacci numbers:

Input "NUMBER OF FIBBS? ",N
N-2 STO>N
round(N/7) STO>N
1 STO>A
1 STO>B
{1,1} STO> $L_{4}$
Disp A,B
Pause
For(K,1,N)
For(M,1,7)
A+B STO>C
B STO>A
C STO>B
B STO> $L_{4}$(dim $L_{4}$)
Disp B
End
Pause
End
Disp $L_{4}$
Stop

Tell me what calculator you are using for the Quadratic program, it has several different versions.

POLYAREA:

Input "SIDES? ",N
Input "CENTER-TO-VERTICE? ",B
Disp NB2sin (pi/N) cos (pi/N)
Stop

The ALLDIVIS program will not always display all divisors, especially when the divisors become greater than about 100. Also, on large numbers, it will take up to a minute, especially if you are using a slower calculator.

Do ya'll have any?

Thank you, Billy

Thanks for the programs...I have a TI-89 Titanium.. and I got tons of programs (aka systems solver, pynomial root solver {including greater than 3!!!}) from [url]education.ti.com[/url] Do your programs work on 89s or do they only work on the 83/84 series?

Wow! I think if it has a [PRGM] key and List capability it will work on any TI. I know they all work on a 82 also, if anyone has one of those.

[2nd][4]. The 4 key should have a L4 above it. If it doesn't, too bad. Going to have to do without. I'll adapt it not to need list if necessary, it's just not so pretty.

[quote="shake9991"]
the only good thing about living in Texas: TI donates free calculators... every competitor got an 83+
they're giving 89 T's at nationals...quote]

Shake, did they tell you this at States?  Usually you don't know what you're getting until you get it:-)

I need a TI-89 Titanium for my high school next year.  And luckily, I'm going to nationals this year! :lol:  :lol:  :lol:  :lol:  :lol: :lol:  :lol:

[quote="shake9991"]
by the way: using your programs in mathcounts in illegal... they explicitly say that the use of any programs which are not included with the calculator is prohibited..[/quote]
Where does it say that????

I didn't know that, though I didn't use any of them.

Hey, how about the Apps that come with the TI-84+ Silver? They come with the calculator?!

Billy

hold on, WHAT???!!!!

But, but thats half the reason we use them in the competittion!

This is all MATHCOUNTS says about Calculators.  Everya year I've been to Nationals, the coaches discuss the program issue and the answer is always the same.  there is no way to regulate programs as it can not be expected of coaches to know how to clear the memory of every eligible calculator.  The only way that "no programs" rule could be used would be for MC to provide Calculators at every level of competition.  that's never going to happen, and you wouldn't want a calc you weren't familiar with for the comp anyway.

[quote]Calculators are not permitted in the Sprint or Countdown rounds, but they are permitted in the Target, Team and Masters rounds. Where calculators are permitted, students may use any calculator that does not contain a QWERTY (i.e., typewriter-like) keypad. Calculators that have the ability to enter letters of the alphabet but do not have a keypad in a standard typewriter arrangement are acceptable. Personal digital assistants (e.g., Palm Pilots) are not considered to be calculators and may not be used during competitions. Students may not use calculators to exchange information with another person or device during the competition.

Coaches are responsible for ensuring that their students use acceptable calculators, and students are responsible for providing their own calculators. Coordinators are not responsible for providing Mathletes with calculators, AC outlets or batteries before or during MATHCOUNTS competitions. Coaches are strongly advised to bring backup calculators and spare batteries to the competition for their team members in case of a malfunctioning calculator or weak/dead batteries. Neither the MATHCOUNTS Foundation nor coordinators shall be responsible for the consequences of a calculator’s malfunctioning.
[/quote]

Whew...you had me worried there for a second.  I have about 40 programs.

It is legal!  I used it myself and my calculater was checked and they said that it was fine to use :roll:

[quote="shake9991"]you just wasted your time... they have that in the Aps for the 83+... its under Prob. Simulator
the only good thing about living in Texas: TI donates free calculators... every competitor got an 83+
they're giving 89 T's at nationals... if only I hadn't screwed up at state...
by the way: using your programs in mathcounts in illegal... they explicitly say that the use of any programs which are not included with the calculator is prohibited..[/quote]

About writing that simulation program, I didn't waste my time.  It's all about learning new things for programming and gaining experience.

[quote="tarquin"]Yep I got an 83 + and the top ten I think all got 89 Titaniums[/quote]

No. Top ten and coaches got 84 silvers at the competition. I think top four get 89 Titaniums. Thats what Mark said, but we haven't gotten them yet.

[quote="solafidefarms"]AllDivisors:

{1}STO>$L_{4}$
Input "NUMBER? ",N
N STO> M
While iPart(N/2)=N/2
2 STO> $L_{4}$(dim $L_{4}$+1)
N/2 STO> N
End
For(A,3,iPart($\sqrt$ M)+2,2)
If iPart(N/A)=N/A
Then
While iPart(N/A)=N/A
A STO>$L_{4}$(dim $L_{4}$+1)
N/A STO>N
End
End
End
Disp $L_{4}$
Stop

[/quote]

This will cause mistakes when you factor... For example, if you were to factor 52, it would display {1 2 2}

TI-84 Silver editions are almost the exact same as TI-83 plus calculators.  There are a few differences - Processor speed, Archive memory, and a few extra commands.  I don't see why it wouldn't be allowed.  If it were a TI-84 calculator (not silver), then the only differences are processor speed and archive memory.  Anyhow, I hope I helped.  I didn't read the other posts (I'm in a hurry so I just scanned over them), so I hope I didn't cross-post anybody and I hope I'm not making myself look stupid. :?

Frost13, does anyone take the Masters round anymore? I thought it disappeared.

yeah.  my state used today.  it is certainly interesting.  It s optional for states to use.  My state always does, but mostly because in NH we fear change :)   We still use the 1980s version of the countdown round.  It wil be used at nationals for sure.

When we watched the nationals competition at state, and they kept on referring to how well they did in the masters round


[b]Let $ABC$ be an acute-angled triangle and $P$ a variable point in its plane. Let  $C_{1}$ be a point in the exterior of $ABC$ such that
 $ABC_{1}$ is equilateral. Prove that \[\parallel CC_{1}\parallel \le \parallel PA\parallel+\parallel PB\parallel+\parallel PC\parallel \; .\] Find the equality case.
[/b]

[size=75][b]
See: [i] Problem 305 [/i], Matematicki Vesnik 10(25)(1973),p.289, also 11(26)(1974),p.265.
[/b] [/size]

$PA+PB+PC$ takes minimum if and only if $P$ is the fermat point of $\triangle{ABC}$. That is, $PA+PB+PC \geq FA+FB+FC$. But since we have that $FA+FB+FC=CC_{1}$, so $CC_{1}\leq PA+PB+PC$.

[quote="shobber"]$PA+PB+PC$ takes minimum if and only if $P$ is the Fermat point of $\triangle{ABC}$. That is, $PA+PB+PC \geq FA+FB+FC$.
But since we have that $FA+FB+FC=CC_{1}$, so $CC_{1}\leq PA+PB+PC$.[/quote]
Nice solution,proposer


1) How many ordered triple $(a, b, c)$ satisfy the equation $(a^2-1)^2+(b^2-4)^2+(c^2-9)^2=0$ ?

We know that for any real x  $x^2\geq0$.  We have the sum of three squares equaling 0.  All three of them must equal zero.  So:
\[\begin{eqnarray*}
a^2-1&=&0\\
b^2-4&=&0\\
c^2-9&=&0\\
\\
a^2&=&1\Rightarrow a=\pm1\\
b^2&=&4\Rightarrow b=\pm2\\
c^2&=&9\Rightarrow c=\pm3
\end{eqnarray}\]

So there are 2 choices for a, 2 for b and 2 for c.  Thus there are 2*2*2=8 ordered pairs.

To answer your question:
(1,2,3)
(-1,2,3)
(1,-2,3)
(1,2,-3)
(-1,-2,3)
(-1,2,-3)
(1,-2,-3)
(-1,-2,-3)

For the record, just because it was #1 doesn't mean that it was the easiest question on the test - that would depend on who you asked.

Also, most #1s on real AMC 12s will be easier than this one. It seems as if all the mock tests were more difficult than a normal AMC 12.


Enjoy, [url=http://www.tkcs-collins.com/truman/monopoly/monopoly.shtml]it's really interesting.[/url]

Cool. Now I can just memorize that list and win every time! ;)

That person must've been either bored or really desperate to win to be able to do all that research. Too bad my monopoly game probably has spider webs on it.

i wonder exactly how many people know about that site and used it before in their games........ :D

Now I could try a monopoly money-winning competition and WIN!!!  :winner_first:

my sister is in love with monopoly and she made this whole stategy that doesn't even work. When she sees this she'll be like:  :o  :wow:

[quote="NoSoupForYou"]Maybe this should be moved to the Math & Strategy Games forum.

Can monopoly be considered a strategy game?[/quote]

Definitely - you have to make many decisions. Like whether to actually buy the property or not, whether to put houses, and if so, how many houses? Hotel? Should you bail yourself out of jail or pay a fee? etc. etc.

Quite interesting. Sadly, I think it's more fun to play this time of game without any form of 'strategy' in mind (unless it's like a competition or something), but most of the time it's with family or friends.


At the very top of every page, there is an "a:0:{}}}');" ;)

In the AOPS view, it doesn't show up- except in the "Topic review" from the posting screen.

[Edit- on the "topic review", it has the ?> as Klebian noted.]

Interestingly, mine shows

a:0:{}}}'); ?>

On the AoPS domain, it shows Klebian. On the ML domain, it shows what ZetaX has.

Now I see a:0:{}}}'); ?> , too :D
I'm interested in an explaination why this thing occurs... (the one in the old thread doesn't satisfy me)

Check the source code.  For some reason which we don't know, it's written into the webpage.

It's the end of a php statement that somehow got cut off...

Heh, I posted a topic on this a few seconds after you: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=169051

EDIT: when I do egosearch, it shows a white page with only the a thing, and then it logs me out.  :o

From what I think it is is that its a bug onto one of the header template files. Its reading something extra toward the end

[code]
?>
[/code]

and interpreting the > as its ascii counterpart.

[quote="jmerry"]I'm not seeing it now- is it fixed for anyone else?[/quote]
Me too!  :lol:

[quote="minicon"]Yep, I was getting that last night too.   :huh:  Maybe Valentin can fix the code.  :lol:[/quote]I fixed it last night I forgot to mention it here though.

@ZetaX: the explanation remains the same as before: corrupt cache file. I have to regenerate it when the error appears which fixes it instantly. Problem is finding out how the file gets corrupted in the first place so I can correct it for good.

So people whenever you see the error please report the first page you see it on, or whatever blank pages you get (you should never get blank pages) so that I can backtrace the errors (and hopefully find&destroy them).

I got that when I logged in. (I use MathLinks board)
It stuck on every AoPS page I went to.  Thanks for fixing it!  :D


probability geometry conics parabola inequalities

I clicked the first stupid too!!! What are the odds of that...

Also--better on teh math SAT than MAP math??? :wink: sketchy..

:rotfl:  :rotfl:  :rotfl:  FIRST 11 TOPICS ARE SPAM

Tinytim, the probability of you picking the same as math154 is $ \frac{1}{2}$

The probability of you both picking the first one is $ \frac{1}{4}$

These probabilities of course assume that each stupid is equally likely.

More interestingly, psychologically, it would be around 70%-85% that a person picks the 1st one. For example, right now, we have $ \frac {5}{7}\approx71\%$

Gosh, the Bay area forum is only a [i]little[/i] spammier than ours! Soon, MO will be the sapmmiest...

So actually, the probability of 2 people picking the same stupid is $ \frac{29}{49}$

An actual math question:

Prove that given 2 options, with probabilities of $ p$ and $ q$ (obviously, $ p\equal{}1\minus{}q$) that the probability of two people picking [b]different[/b] options is greatest when $ p\equal{}q\equal{}\frac{1}{2}$

For fun. EDIT-This message is too short!

w/e

So the probability that they are different is $ p(1 \minus{} p)\equal{}\minus{}p^2\plus{}p.$ The vertex of the parabola with equation $ y \equal{} \minus{} x^2 \plus{} x$ is when $ x \equal{} \frac {1}{2}$ thus, $ p(1 \minus{} p)$ is maximized when $ p \equal{} \frac {1}{2}$

[gasp]

Another stupid proof:

We know that when $ p \equal{} 1 \minus{} p$, the probability is $ p^2 \equal{} \frac {1}{4}$. Now we let $ p\neq{1 \minus{} p}$. Suppose $ p$ is now $ \frac {1}{2} \minus{} x$ for any real x. Then the probability of $ not(p)$ is $ \frac {1}{2} \plus{} x$. Thus the probability of $ p(1 \minus{} p)$ is $ \frac {1}{4} \minus{} x^2$. Since $ x$ is real, and $ x^2\geq0$ by the Trivial Inequality, we know that no matter what value $ x$ takes on, it will still be less than $ \frac {1}{4}$. Hence the probability of $ p\cdot{not(p)}$ is greatest when $ p \equal{} \frac {1}{2}$.

The map test? In kansas we take the map test for reading and math but dont you get a 3 digit number as your score?


calculus
derivative
calculus computations

The spread of measles in a certain school is given by $P(t) = \frac{200}{1+e^{5-t}}$, where $t$ is the number of days since measles first appeared, and $P(t)$ is the total number of students who have caught the measles to date.
...
(c) When will the rate of spread of measles be greatest?  What is this rate?

Ok, so I know I'll need to find the second derivative and set it equal to $0$, but i just can't get it.  I got close, but I had a random $ln(200)$ still in my equation when it shouldn't have been.  If it's not too much trouble, a thorough explanation is much appreciated.

The solution, if you want to check your work:[hide]After 5 days, when then rate is 50 students/day[/hide]


Me and a friend of mine saw a weird animal today, and thought it was pretty strange...From what I hae found out it doesnt fall into insecta....I was wondering if anyone here could help me find out what it was

It was like a ball with wings...literally
It looked like it was onyl a ball about 2 times as big as your thumbnail...It had 2 wings overlapping and extended maybe an inch out after his head/thorax thing

It had no legs, and when we touched it, it just barely got away from us by flapping its wings against the ground and had no control on where he was going...From what we saw, he could not fly..

It didnt look like it had a thorax or abdomen...very strange thing

It looked like it had an exoskeleton from what I saw, and 2 black eyes

Just wondeing if any of you could help me find out what this weird animal was...

thanks

I'm not an entomologist, but I would utilize an insect identification guide, in this case.

The following I quoted from [url]http://www.amentsoc.org/insects/what-bug-is-this/adult-key.html#a2[/url]

Start at Question 1 and follow the links until you've identified your insect. Some of the questions recquire greater detail of the insect you may have.

You seem to have posted this three years ago, while I just joined the AOPS community, some months back, sorry. If you still remember some of the details of the bug from that long ago, I hope this helps.

-arb95.

[quote][list]1. Insect has wings? Go to 2
Insect wingless or with poorly developed wings. Go to 29

2. One pair of wings. Go to 3
Two pairs of wings. Go to 7

3. Body grasshopper-like, with enlarged hind legs and pronotum extending back over abdomen = Orthoptera
Insects not like this. Go to 4

4. Abdomen with 'tails'. Go to 5
Abdomen without 'tails'. Go to 6

5. Insects <5mm long, with relatively long antennae: wing with only one forked vein = Hemiptera
Larger insects with short antennae and many wing veins: tails long = Ephemeroptera

6. Forewings forming club-shaped halteres = Strepsiptera
Hind wings forming halteres (may be hidden) = Diptera

7. Forewings hard or leathery. Go to 8
All wings membranous. Go to 13

8. Forewings tough apart from membranous tip = Hemiptera
Forewings of uniform texture throughout. Go to 9

9. Forewings (elytra) hard and veinless, meeting in centre line. Go to 10
Forewings with many veins, overlapping at least a little and often held roofwise over the body. Go to 11

10. Abdomen ending in a pair of forceps : elytra always short = Dermaptera
Abdomen without forceps: elytra commonly cover whole abdomen = Coleoptera

11. Insects with piercing and sucking beaks = Hemiptera
Insects with chewing mouthparts: cerci ('tails') usually present. Go to 12

12. Hind legs modified for jumping = Orthoptera
Hind legs not modified for jumping. Go to 49

13. Tiny insects covered with white powder. Go to 14
Insects not like this. Go to 15

14. Wings held flat at rest: mouth-parts adapted for piercing and sucking = Hemiptera
Wings held roofwise over body at rest: biting mouthparts = Neuroptera

15. Small, slender insects with narrow, hair-fringed wings: often found in flowers = Thysanoptera
Insects not like this. Go to 16

16. Head extending downwards into a beak = Mecoptera
No such beak. Go to 17

17. Wings more or less covered with scales: coiled proboscis (tongue) usually present = Lepidoptera
Wings usually transparent although often hairy. Go to 18

18. Wings with a network of veins, including many cross veins. Go to 19
Wings with relatively few cross veins. Go to 23

19. Abdomen with long terminal threads. Go to 20
Terminal appendages short or absent. Go to 21

20. Forewings much larger than hind wings: wings held vertically over body at rest: 2 or 3 terminal threads = Ephemeroptera
Wings more of less equal in size or hind wings larger: wings folded close to body at rest: 2 terminal appendages = Plecoptera

21. Antennae very short: body at least 25mm long = Odonata
Antennae longer: greater than width of head. Go to 22

22. Tarsi 3-segmented = Plecoptera
Tarsi 5-segmented = Neuroptera

23. Wings noticeably hairy. Go to 24
Wings not noticeably hairy. Go to 25

24. All wings more or less alike: front tarsi swollen = Embioptera
Hind wings usually broader than forewings: front tarsi not swollen = Trichoptera

25. Tarsi with 4 or 5 segments. Go to 26
Tarsi with 1 - 3 segments. Go to 27

26. All wings alike = Isoptera
Hind wings much smaller than forewings = Hymenoptera

27. Hind wings similar to or larger than forewings: abdomen with cerci = Plecoptera
Hindwings smaller than forewings: no cerci. Go to 28

28. Tiny insects with at least 12 antennal segments = Psocoptera
Never more than 10 antennal segments: piercing and sucking beak present = Hemiptera

29. Insects with slender, twig like body = Phasmatodea
Insects not like this. Go to 30

30. Insects with grasshopper-like body and long back legs = Orthoptera
Insects not like this. Go to 31

31. Small, soft-bodied insects living on plants, often under protective shield or scale = Hemiptera
Insects not like this. Go to 32

32. Minute soil-living insects, <2mm long without antennae = Protura
Insects not like this. Go to 33

33. Insects with cerci or other abdominal appendages. Go to 34
Insects with other appendages. Go to 41

34. Abdominal appendages long and conspicuous. Go to 35
Abdominal appendages short or hidden under body. Go to 38

35. Abdominal appendages forming pincers. Go to 36
Abdominal appendages not forming pincers. Go to 37

36. Tarsi 3-segmented = Dermaptera
Tarsi 1-segmented = Diplura

37. Abdomen with 3 long terminal appendages = Thysanura
Abdomen with only 2 terminal appendages = Diplura

38. Tiny jumping insects, head points downwards forming a beak = Mecoptera
No sign of beak. Go to 39

39. Small or minute insects with a forked springing organ under rear of abdomen: generally found in soil or decaying vegetation = Collembola
Insects not like this. Go to 40

40. Tarsi usually 4-segmented = Isoptera
Tarsi 3-segmented: front tarsi swollen = Embioptera

41. Parasites in fur or feathers: insects generally flattened side-to-side or dorso-ventrally. Go to 42
Insects not parasitic and not usually flattened. Go to 46

42. Jumping insects flattened from side-to-side = Siphonaptera
Insects not flattened from side-to-side. Go to 43

43. Insects of moderate size: head partly withdrawn into thorax. Go to 44
Small minute insects: head not withdrawn into thorax. Go to 45

44. Antennae very short: very 'leggy' insects with strong claws well suited to clinging to a host mammal = Diptera
Antennae long: body somewhat circular, with less prominant legs and claws = Hemiptera

45. Prothorax distinct: biting mouthparts = Mallophaga
Thoracic segments fused into one unit: sucking mouthparts = Anoplura

46. Abdomen with pronounced 'waist': antennae often elbowed = Hymenoptera
No such features. Go to 47

47. Body >5mm long, clothed with flattened hairs and scales: vestigial wings present = Lepidoptera
Body usually <5mm long, bald or occasionally scaly: vestigial wings rarely present. Go to 48

48. Head a wide or nearly as wide as body: biting mouthparts: insects often found among dried materials = Psocoptera
Head narrower than body: sucking mouthparts: abdomen often with a pair of tubular outgrowths near hind end: insects found on growing plants = Hemiptera

49. First pair of legs used to grasp prey and held close to the body at rest = Praying Mantids
Front legs not like this and body flattened = Cockroaches[/list][/quote]

Again, I am not quite sure, for I'm not an expert. [b]Does it have a greenish-bluish tinge?[/b]
If so, I consulted with my grandfather, recently, and together, we assumed it to be a Euglossine bee of genus Euglossa.

The one you could have saw could have been injured; curling itself into a ball to protect itself. An injury would prove the your statement of "it just barely got away from us by flapping its wings against the ground and had no control on where he was going."

The following is an image of the (Euglossa Bee.) It may be a more ball-shaped Euglossa than this one, of the same genus.
-arb95
[size=59][img]http://www.orchidspecies.com/orphotdir/mormodebadia.jpg[/img][/size]


AMC
USA(J)MO
USAMO
Harvard
college
Ross Mathematics Program
geometry

A few of you have tossed around mods in some solutions to problems on this board.

I'm curious - when did most of you learn modular arithmetic?  I was kind of expecting someone to see those solutions and say 'what's this mod stuff'.  I didn't see it until I was preparing for the USAMO in 10th grade - when I first saw a solution with mods it looked like encrypted Sanskrit to me.  That was 15+ years ago, maybe more students are being taught it now...

I learned mods mainly from a freshman at harvard who was usamo in 11th and 12th grade only this year (im a freshman in high school). But I was able to finetune my mod skills with aop volume 1. I still have to learn the division rules (havent started vol 2 yet, busy with vol 1 and nationals practice).

I leraned at Ross before my freshman year...

I'm not entirely sure, I think it was when I went to my first ever (New Zealand team selection) camp.. but they're one of my favourite parts of problem solving now. Basically because even if you have no idea how to solve the problem you can play around and play around until you may get a better idea of what to do.

I learned it from AoPS, of course! Actually, I had learned the very basics in a great course taught at the high school. I think it was Precal. However, I only really got to appreciate the use of modular arithmetic through AoPS.

I learned it last year in a cryptography course at Johns Hopkins CTY...

-Isaac

This is probably extremely unusual, but my second grade teacher introduced the concept of a mod right along with remainders when we were doing our unit on division.  In eighth grade I think I saw them again, but I cant remember the context.  I guess it wasnt until around ninth grade that I ever saw them being used to help solve problems.  My math class back then wasnt great so its very likely that my first real introduction to problem solving with mods came from AoPS in the summer going into that year.  There arent too many data points, but perhaps this trend toward learning about modular arithmetic earlier (in the US) has a lot to do with an awesome discussion of it early in Volume I.  The placement of that section was brilliant.  =)

I don't remember exactly when I learned mods. I know I understood them when I was in junior high school and possibly even in elementary school, but I don't remember ever being taught them. I don't think I really ever had any difficulty using mods for simple problems on competitions even without being shown how to do them. Perhaps I have some intuition for number theory. Then again, I was taught how to program in Pascal in junior high school, and I had studied Pascal programming on my own before then, and I'm pretty sure I used mods for some of my early programs. What I do know is that when I got to high school and started doing competitions, I knew most of the theorems needed, but I often would not be able to use them properly to solve the problems. I think I'm getting better now after four years of contests.

I'd seen mods previously while preparing for competitions in Singapore - I moved to the Bay Area about 2 years ago - and also learnt about them from SJSU math circles...besides that I studied a bit of number theory over my 8th grade summer.

Um...I'm feeling pretty stupid, but what are mods?  I might actually have used them and not have known the name.  A breif summary please.

A mod is simply a remainder function....for example....22 divided by 7 leaves a remainder of 1...therefore 22mod7=1

other examples:
16mod4=0
98mod8=2
etc...its easy...

mods are used in number theory and are VERY helpful in many proofs...

And things that you can do with mods:

Add + subtract (if the mod is the same):  4 (mod 6) + 2 (mod 6) == 0 (mod 6).
Multiply:  4 (mod 6) x 2 (mod 6) == 8 (mod 6) == 2 (mod 6).

You <cannot> divide however there is a cancellation rule: If ab == ac (mod x) and a and x have no common factos, then b == c (mod x). (For example if 4a == 6b (mod 5) then 2a == 3b (mod 5). But if 4a == 6b (mod 22) then you cannot cancel the two.)

Also a famous theorem is if p is prime then a^p == a (mod p), and using the above cancellation rule if a and p have no common factors, ie if a is not a multiple of p, then a^(p-1) == 1 (mod p).

Eg any non multiple of 17 to the power of 16 is one more than a multiple of 17.

It should be an equal sign but with 3 lines instead of 2. I guess he used == instead because you can't get that symbol easily on the keyboard.

My dad showed me a little about them, but forgot about it. I learned the definition at a computer programming camp (%), and played around with them to get what I know now.

i learned last year i think, so like beginning of 7th grade. twas from aops book. then some help from me broski (brother)

All these people who have learned modular arithmetic... I feel like such an IDIOT.  I claim I have never learned modular arithmetic... I never heard of such thing until about 4 weeks ago when I was doing counting.  I still can't use it in a problem though... I only know the basic idea.

Feeling extremely stupid,

Taylor

Don't feel stupid - you'll know it by the end of the summer.  You'll be 2 years ahead of where I was 18 years ago...

Even though I know what mod is, and know when to use it, but I probably only know about 25% of all there is about mods, cause there's always new stuff, and then you learn the new stuff, and then there's new stuff, and then when you learn those, still new stuff. AAAAAAAAAAAAAAAAAA (new means stuff I didn't find before, not like new new. Oh yeah, are they still making new theories in math??? (They probably are, but just wondering)

If you ever know 25% of what there is to know about anything, that's incredibly good. (Most people never get to know anything like 25% of what there is to know about anything.) It seems to me that the more I learn the less I know. I always find more problems that I can't solve, more theorems I don't know, etc. when I read books. Yes, they're certainly coming up with new theorems and branches of mathematics all the time. When people come up with more questions, more mathematics comes out of them.

The very little I learned of mods was when I was in about 6th grade and decided to learn C, I used them a lot more when I made a casino program in 8th grade and I had to use a randomizer function and mod arithmetic for it.., I don't quite know how to use mod's for problem solving(yet) but I will once my book comes  :D

I learned mods in 6th grade, to help limit complicated problems.  Mods are also the the unit digit of the same base as the mod of a number.

Example:
12 mod 5 = 2
12 base 5 = 22 (5*2+1*2)

It was when I got my first PC (486 66 MHz) in 1994 (I'm born 1985), when I did some C++ programming with my dads books.

Not that anyone cares but here's my two rupees,

I've heard the word mod before but my laziness prevented me from learning what it was. It was when I needed to actually prepare for the USAMO and solve some USAMTS problems that I learned to use mods. Actually, it was because of AoPS I & II that really helped me with mods.

I should bust out a testimonial for AoPS!

I used to be afraid of mods and that weird sign that's not really an equal sign but is just bleh. *Wipes tears from eyes* Now I can understand mods just like normal people, I can actually live a normal life now!

Yeah, the american system failed me.. I mean I was a kid who had to use baseball cards to do decimal division!! And once you've gone that far you can forget about mods...

Cheers,
IW.OAV


funny problem! It seems almost more combinatorical than related to ineqalities.

First note the well-known ineq $\frac1{(1+a)^{2}}+\frac1{(1+b)^{2}}\ge\frac1{1+ab}$ for $a,b>0$.
Next note (easy to see) that for $x,y,z>0$, the ineq $\sum\frac x{x+y}\ge\frac32$ holds iff $x\ge y\ge z$ or cyclical. Putting $u=\frac{y}{x}$ etc., this means that for $uvw=1$, the ineq $\sum\frac 1{1+u}\ge\frac32$ holds if (at least) two of $u,v,w$ are $\le1$.

So the problem is solved if we can show that there are four different indices $i,j,k,l$ such that $x_{i}x_{j}\le1$ and $x_{k}x_{l}\le1$. Let's go!  :blush:

Assume wlog $x_{1}\ge\cdots\ge x_{6}$.

If $x_{3}x_{5}\le1,$ we have $x_{4}x_{6}\le x_{3}x_{5}\le1,$ done.
If $x_{3}x_{5}>1$, we have $x_{2}x_{4}\ge x_{3}x_{5}>1$ so we must have $x_{1}x_{6}<1$ because of $\prod x_{i}=1$.
Now the five terms $x_{i}x_{6}<1$contribute negative factors to $P=\prod_{i<j}(x_{i}x_{j}-1)$, all others (except $x_{4}x_{5}$) are $\ge x_{3}x_{5}>1$ and thus contribute 9 positive factors. As $P\ge0$, we must have $x_{4}x_{5}\le1$, and we are done.  :)   :D

I wonder if there is a more symmetrical solution, but I doubt it.

wow! very nice solution :thumbup:
Is the generalisation true?
$n\geq 3$ and $x_{1}x_{2}\cdots x_{n}=1$ $\prod (1-x_{i}x_{j})\geq 0$
$\sum \frac{1}{(1+x_{i})^{2}}\geq \frac{n}{4}$

(I have done a typo writing $P$,  :blush:  but doesn't affect my proof)
the condition is $\prod (x_{i}x_{j}-1)\geq 0$ not $\prod (1-x_{i}x_{j})\geq 0$

It is well possible this holds for all $n$.
But we would need a different method. Maybe mine can be extended for even $n$, not tried yet.

Maybe even the stronger one holds $\sum_{i<j}\frac{1}{1+x_{i}x_{j}}\geq \frac{n(n-1)}{4}$. (Easy for $n=3,4$.)
I'm not sure at all for this latter one if $n>4$. If it is wrong for $n=5$ (too many 'bad' terms on the LHS?), i guess it is wrong for bigger $n$ too.

$\frac1{(1+a)^{2}}+\frac1{(1+b)^{2}}\ge\frac1{1+ab}$
probably trivial, but how do you get this?

well the stronger one holds:
$\frac{1}{1+a^{2}}+\frac{1}{1+b^{2}}\geq \frac{2}{1+ab}$ (try expanding)

well for the generalisation I relised that $\binom{n}{2}$ should be odd or $n=4$ otherwise we can find a counter example.

[quote="Albanian Eagle"] I realised that $\binom{n}{2}$ should be odd or $n=4$ otherwise we can find a counter example.[/quote]Wasn't that just the matter of having falsely written $\prod (1-x_{i}x_{j})\geq 0$ instead of $\prod (x_{i}x_{j}-1)\geq 0$ (which makes a difference iff $\binom{n}{2}$ is odd)? :?:


A two-digit integer is divisible by $ n$ and its last digit is $ n$. What is the greatest possible value of $ n$?

99 is divisible by 9, and its last digit is 9.

obviously, 9 is the greatest possible digit.

answer : 9


function
combinatorics solved
combinatorics

How many sequences of 0's, 1's, 2's, and 3's are there, each with a total of n symbols, consisting of an even number of 0's, an odd numberof 1's, and number of 2's and 3's.

its the coefficient of x^n/n! in the generating function:
g(x)=1/16(e^x+e^-x)(e^x-e^-x)^3

hence answer= 2^(2n+1) -2^(n+2) for odd n.
and  0 for even n.


For a convex polygon $P$ in the plane let $P'$ denote the convex polygon with vertices at the midpoints of the sides of $P.$ Given an integer $n \geq 3,$ determine sharp bounds for the ratio

\[ \frac{\text{area } P'}{\text{area } P}, \] over all convex $n$-gons $P.$


Please post votes here! The candidates:

Ernie : 12 votes
AIME15: 0 votes
Myyellowducky82: 2 votes

People who need to vote:

Skaldjfhrulez
Violin321
(^_^)
Mathpassion101
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[quote="mewto55555"]Please post votes here! The candidates:

Ernie : 10 votes
AIME15: 0 votes
Myyellowducky82: 0 votes

People who need to vote:

Skaldjfhrulez
Violin321
Kl2836
(^_^)
Mathpassion101
Lotsofmath
Ernie
Yoh2241
Vallon22
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Jongao
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Vamathletes
Guineapiggies
Pinkpiglet
Jain_harshi
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Mathblitz
Weird888
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Kookamunga
Nikeballa96
Pi guy[/quote]

I vote myself.  And please tell me why you made another thread.

THis is just for the elction, the other one was littered with spam.

ELECTION UPDATE:

ernie (11): needs 8 to clinch VP <--- yay! thanks to all the people that voted for me!
myyellowducky82 (0): needs 19 to clinch VP
AIME15 (0): needs 19 to clinch VP

i vote ernie 4 my tourneys!  :rotfl:  :rotfl:

I VOTE FOR AIME15!! EVERYBODY VOTE AIME!

I'm sorry johnnybeard, but you were not a registered member of CTC when the election started (and you still aren't!)

ernie (12)- 7 to clinch <--- yay!
myyellowducky82 (3)- 16 to clinch
AIME15 (0)- 19 to clinch

Voters Left: 21

[quote] All the letters in your topic title are uppercase (A, B, C, ...). Please make some of them lowercase (a, b, c, ...)[/quote]

GAH!

Please post votes here! The candidates:

Ernie : 14 votes
AIME15: 0 votes
Myyellowducky82: 3 votes

People who need to vote:

Skaldjfhrulez
Violin321
(^_^)
Mathpassion101
Lotsofmath
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Myyellowducky82
Vamathletes
Jain_harshi
Splatyango
Weird888
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Kookamunga
Pi guy

i'll vote....

[hide="my vote"][hide="i vote for...."][hide="this person"]ernie[/hide][/hide][/hide]

Please post votes here! The candidates:

Ernie : 17 votes
AIME15: 1 votes
Myyellowducky82: 3 votes

People who need to vote:

Skaldjfhrulez
Violin321
(^_^)
Mathpassion101
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Vamathletes
Jain_harshi
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I vote for AIME15 cuz he has no votes!! :P

Eletion update: 1 more vote for ernie or 2 more votes for AIME15 and ernie clinches! The following people still must vote:

Skaldjfhrulez
Violin321
(^_^)
Mathpassion101
Lotsofmath
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AIME15
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Vamathletes
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Weird888
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Dannyhamtx

[quote="mewto55555"]Please post votes here! The candidates:

Ernie : 19 votes
AIME15: 1 votes
Myyellowducky82: 3 votes

People who need to vote:

Skaldjfhrulez
Violin321
(^_^)
Mathpassion101
Yoh2241
AIME15
Jongao
Budi713
Myyellowducky82
Vamathletes
Jain_harshi
Weird888
Dannyhamtx

CONGRATULATIONS VICE PRESIDENT ERNIE!


Let $ F: M\to N$ be an injective Immersion, which is proper.show that $ F$ is an embeding,and $ F(M)$ is a closed regular submanifold,and conversly.

(Picks up manifolds textbook. Looks up "regular submanifold" in the index. Goes to the one mention of it: "Embedded submanifolds are also called [i]regular submanifolds[/i] by some authors" in the section where embedded submanifolds are first introduced.)
Now, there are several bits to this question.
- The image of a smooth embedding $ F$ is an embedded submanifold (Appears as a theorem, shortly after introducing the definition of an embedded submanifold). It's straightforward, building slice charts from the constant rank of $ F$. It's also essential to all sorts of things.
- A injective, continuous, closed map is an embedding, and proper maps are closed. Easy point-set topology, proved in an appendix.
For the converse, we have
- An embedded submanifold is closed (if and) only if the inclusion map is proper. This one appears as a problem, so I'll prove it. Note that since the map from $ M$ to its image is a homeomorphism, $ F$ is proper if and only if the inclusion map from the image to $ N$ is proper.

Let $ X$ be embedded in $ Y$, and let $ X$ be closed. The intersection of a compact set $ K$ with $ Y$ is closed in $ K$, hence compact both as a subset of $ X$ and as a subset of $ Y$. That intersection is the preimage of $ K$ under inclusion, and thus the preimages of compact sets are compact, making the inclusion proper.


Let  a right  triangle $ABC$ with $\angle A = 90$.On side $BC$  we take point $D$ , such that
  $BD = AC$ . Prove that the median $AM$  , the bisector $CE$ and the altitude  $DH$ of $\triangle DAC$  have a common point (are concurrent )

[u]Babis[/u]

i dont know what kind of sketch u got, but in mine they definitly dont have a common point, even more: entering it into a geometry sketcher it seems it never has a common point ...

(i might have misunderstood u, bisector is the angle bisector isnt it?)

$M\in (DC),\ MD=MC\ ;$
$E\in (AD),\ \frac{EA}{ED}=\frac{b}{a-c}\ ;$
$\ H\in (AC),\ \frac{HC}{HA}=\frac{a-c}{b}\ .$
======================================
$\frac{MD}{MC}\cdot \frac{HC}{HA}\cdot \frac{EA}{ED}=1\Longrightarrow AM\cap DH\cap CE \ne\emptyset.$

\[ \underline {\overline {\left| \ \blacksquare\ \emptyset\emptyset\emptyset\ \blacksquare\ \right| }} \]\[ (\ empty set\ ) \]

[color=blue][b]To peeta.[/b][/color]

 I think that the figure is simple.Take M midle of  $DC , CE$ is angle bisector of $\angle C$. I think , PH -AN , has helped with his nice and simple  solution.

[u] Babis[/u]

[b]Note: [/b]try to find another solution without CEVA.Consider the common point P of the median and the bisector and try to prove that DP is altitude(!)

i think its wrong...

i thought that we should say $DH$ is the altitude of triangle $DAB$.and i tried to prove it.

suppose $DH$ is the altitude of triagle $DAB$.then we should have:

$DH$is paralell to $AC$,and because of the triangle $AMC$ is isoceles,we should have:

$MD: DC$=$MK: KA$ and $AD$should be the bisector ofangle $CAM$.

then we have: $AM: AC$=$MD: DC$

$AM$=$a: 2$

$AC$=$b$

$MD$=$-a: 2-b$

$DC$=$a-b$

then we have:$a^2: 2$+$b^2$=$0$

wichisnt trough.

Please see the figure in the pdf file. I could not imagine that you will have problem with it. Sorry.

[u]Babis[/u]

i guess i was confused by the median and thought it was of $BC$ and not of $DC$ ...
i got it now, but i want to say that $\overline{DC}=a-b$ not $a-c$ as Ph-An is saying ...

If someone does not know Ceva , can asume that the median and the bisector meet at $R$ and then prove means Thales that

   $DR // AB$. Of course we '' ll use ( and begin with) the bisector theorem on triangle $CDA$ for $CR$.
This could be another formulation for the same problem.

 [u]Babis[/u]


hello, we have
$ \frac {(a \minus{} b)^2}{c^2} \plus{} \frac {(b \minus{} c)^2}{a^2} \plus{} \frac {(c \minus{} a)^2}{b^2} \minus{} 18 \equal{} \frac {(a \minus{} b)^2}{( \minus{} a \minus{} b)^2} \plus{} \frac {( \minus{} 2a \minus{} b)^2}{b^2} \plus{} \frac {(a \plus{} 2b)^2}{a^2}\minus{}18 \equal{} \frac {(a \minus{} b)^2(2a \plus{} b)^2(a \plus{} 2b)^2}{(abc)^2}\geq0$.
Sonnhard.

Dr Sonnhard, do you have any software for factorising algenraic expressions? How are you able to do it very first?

[quote="nguoivn"][quote="Dr Sonnhard Graubner"]hello, we have
$ \frac {(a \minus{} b)^2}{c^2} \plus{} \frac {(b \minus{} c)^2}{a^2} \plus{} \frac {(c \minus{} a)^2}{b^2} \minus{} 18 \equal{} \frac {(a \minus{} b)^2}{( \minus{} a \minus{} b)^2} \plus{} \frac {( \minus{} 2a \minus{} b)^2}{b^2} \plus{} \frac {(a \plus{} 2b)^2}{a^2} \equal{} \frac {(a \minus{} b)^2(2a \plus{} b)^2(a \plus{} 2b)^2}{(abc)^2}\geq0$.
Sonnhard.[/quote]

Your proof isn't true. Remember that $ a \plus{} b \plus{} c \equal{} 0$, so, equality doesn't occurs when $ a \equal{} b \equal{} c$
@Luat: Your inequality is trues and nice (althought easy).
PS: But you shouldn't post it here  :mad:[/quote]
Lol... Read his proof once more....   :roll:

@Aravind Srinivas L: I will provide an explanation on how one could have guessed the factorisation :)

Substitute $ c \equal{} \minus{}a\minus{}b$: $ \left ( \frac{a\minus{}b}{a\plus{}b} \right ) ^2 \plus{} \left ( \frac{a\plus{}2b}{a} \right ) ^2 \plus{} \left ( \frac{2a\plus{}b}{b} \right ) ^2$. Then it's is easy to see that there are equality when $ a\equal{}b$. Always! Then it is natural to assume that there are equality when $ a\equal{}b$ or $ b\equal{}c$ or $ c\equal{}a$.
Since $ \left ( \frac{a\minus{}b}{c} \right ) ^2 \plus{} \left ( \frac{b\minus{}c}{a} \right ) ^2 \plus{} \left ( \frac{c\minus{}a}{b} \right ) ^2 \minus{} 18 \equal{} \frac{\sum_{cyc} (a\minus{}b)^2a^2b^2 \plus{} 18a^2b^2c^2}{a^2b^2c^2}$ we can probably figure out that $ \sum_{cyc} (a\minus{}b)^2a^2b^2 \plus{} 18a^2b^2c^2 \equal{} k \cdot (a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2$ for some $ k$. Playing around with the numbers gives $ k\equal{}1$! And we are finished with the proof :)

Actually, I obtained
$ (a\minus{}b)^2\left((ab)^2\plus{}(a\plus{}b)^2ab\plus{}(a\plus{}b)^4 \right)\ge 0$
which is obviously true.

[quote="Dr Sonnhard Graubner"]hello, why is my proof not true? Inserting $ c \equal{} \minus{} a \minus{} b$ we got
$ \frac {(a \minus{} b)^2}{( \minus{} a \minus{} b)^2} \plus{} \frac {( \minus{} 2a \minus{} b)^2}{b^2} \plus{} \frac {(a \plus{} 2b)^2}{a^2} \minus{} 18$ and this is equivalent with
$ \frac {(a \minus{} b)^2(2a \plus{} b)^2(a \plus{} 2b)^2}{(abc)^2}$
Note that $ 2a \plus{} b \equal{} a \minus{} c$ and $ a \plus{} 2b \equal{} b \minus{} c$.
Sonnhard.[/quote]

Oh, I'm sorry. Today I'm so busy, so I see your proof not exactly. Very sorry  :(
We can prove easily the general problem:
$ \sum\ (\frac{a\plus{}b\minus{}kc}{a\minus{}b})^2 \geq\ 2(k\plus{}1)^2$
<=> $ (\sum\ \frac{a\plus{}b\minus{}kc}{a\minus{}b})^2 \geq\ 0$

[quote="great math"]Actually, I obtained
$ (a \minus{} b)^2\left((ab)^2 \plus{} (a \plus{} b)^2ab \plus{} (a \plus{} b)^4 \right)\ge 0$
which is obviously true.[/quote]
You have an error in your calculations. There should be equality when $ a\equal{}\minus{}2$ and $ b\equal{}1$, but then:
$ (a \minus{} b)^2\left((ab)^2 \plus{} (a \plus{} b)^2ab \plus{} (a \plus{} b)^4 \right) \equal{} 3^2(2^2 \plus{} (\minus{}1)^2 \cdot \minus{}2 \plus{} (\minus{}1)^4) \equal{} 27$..... Please check it again ;)

[quote="Mathias_DK"][quote="great math"]Actually, I obtained
$ (a \minus{} b)^2\left((ab)^2 \plus{} (a \plus{} b)^2ab \plus{} (a \plus{} b)^4 \right)\ge 0$
which is obviously true.[/quote]
You have an error in your calculations. There should be equality when $ a \equal{} \minus{} 2$ and $ b \equal{} 1$, but then:
$ (a \minus{} b)^2\left((ab)^2 \plus{} (a \plus{} b)^2ab \plus{} (a \plus{} b)^4 \right) \equal{} 3^2(2^2 \plus{} ( \minus{} 1)^2 \cdot \minus{} 2 \plus{} ( \minus{} 1)^4) \equal{} 27$..... Please check it again ;)[/quote]
It does not seem that!
One thing should be looked at is $ a\equal{}\minus{}2, b\equal{}1$ and $ c\equal{}1$
That's what I have shown. The equality holds if we have at least 2 or 3 variables equal.

[quote="great math"][quote="Mathias_DK"][quote="great math"]Actually, I obtained
$ (a \minus{} b)^2\left((ab)^2 \plus{} (a \plus{} b)^2ab \plus{} (a \plus{} b)^4 \right)\ge 0$
which is obviously true.[/quote]
You have an error in your calculations. There should be equality when $ a \equal{} \minus{} 2$ and $ b \equal{} 1$, but then:
$ (a \minus{} b)^2\left((ab)^2 \plus{} (a \plus{} b)^2ab \plus{} (a \plus{} b)^4 \right) \equal{} 3^2(2^2 \plus{} ( \minus{} 1)^2 \cdot \minus{} 2 \plus{} ( \minus{} 1)^4) \equal{} 27$..... Please check it again ;)[/quote]
It does not seem that!
One thing should be looked at is $ a \equal{} \minus{} 2, b \equal{} 1$ and $ c \equal{} 1$
That's what I have shown. The equality holds if we have at least 2 or 3 variables equal.[/quote]
Please rephrase yourself. You grammar is so bad, that I do not understand you.

[quote="Mathias_DK"][/quote][quote="great math"][quote="Mathias_DK"][quote="great math"]Actually, I obtained
$ (a - b)^2\left((ab)^2 + (a + b)^2ab + (a + b)^4 \right)\ge 0$
which is obviously true.[/quote]
You have an error in your calculations. There should be equality when $ a = - 2$ and $ b = 1$, but then:
$ (a - b)^2\left((ab)^2 + (a + b)^2ab + (a + b)^4 \right) = 3^2(2^2 + ( - 1)^2 \cdot - 2 + ( - 1)^4) = 27$..... Please check it again ;)[/quote]
It does not seem that!
One thing should be looked at is $ a = - 2, b = 1$ and $ c = 1$
That's what I have shown. The equality holds if we have at least 2 or 3 variables equal.[/quote][quote="Mathias_DK"]
Please rephrase yourself. You grammar is so bad, that I do not understand you.[/quote]
I mean my aim was letting a equals to b. By the way, we can interchange the role of c and a in my proof as you suggested.
You pointed out that a=2 and b=-1. Now, because a, b, c are cyclic so you can position the value of a into c. Does this explanation satisfy you?
Well, I admit that my English is bad so if you get into some trouble with the understanding, I will recorrect it!

[quote=pi_quadrat_sechstel][quote=tranquoc]Let abc$ \ne 0$ but a+b+c=0 .Prove that:

$ (\frac{a\minus{}b}{c})^2\plus{}(\frac{b\minus{}c}{a})^2\plus{}(\frac{c\minus{}a}{b})^2 \geq 18$[/quote]

$$
\sum_{cyc}\frac{(a-b)^2}{c^2}-18=\frac{(a-b)^2(b-c)^2(c-a)^2+2abc(a+b+c)(a^2+b^2+c^2-2ab-2bc-2ca)}{a^2b^2c^2}\geq0
$$[/quote]
[url]https://artofproblemsolving.com/community/c6h2796100p24624418[/url]

[quote=pi_quadrat_sechstel][quote=tranquoc]Let abc$ \ne 0$ but a+b+c=0 .Prove that:

$ (\frac{a\minus{}b}{c})^2\plus{}(\frac{b\minus{}c}{a})^2\plus{}(\frac{c\minus{}a}{b})^2 \geq 18$[/quote]

$$
\sum_{cyc}\frac{(a-b)^2}{c^2}-18=\frac{(a-b)^2(b-c)^2(c-a)^2+2abc(a+b+c)(a^2+b^2+c^2-2ab-2bc-2ca)}{a^2b^2c^2}\geq0
$$[/quote]

bravo


Hi all,

I've just come back from the Chinese IMO TST.
The six lucky contestants are:
Zhu Qingsan from Attended school of South China Teachers' College;
Lin Yuncheng from Shanghai Middle School;
Huang Zhiyi from Attended school of South China Teachers' College;
Yang Shiwu from Hubei Huanggang Middle School;
Li Xianyin from Attended school of Hunan Teachers' College;
Peng Minyu from Yingtan No.1 Middle School.

The camp lasted two weeks from 16 March to 2 April.
Eight tests with 24 problems are held during that time.

I'll post the problems of the Final TST soon.

Regards,

Wang Chen (wpolly)

Did you participate in the tests wpolly?

That would be great if wpolly can bring us the chinese problems since I can't translate chinese to english  :D

[quote="iura"]Can you please say how namy questions one needed to solve to pass the
tests?[/quote]
The problems of the first 6 tests are worth 7 points while the final TST Problem worth 21 points, so the full mark is 252.
My estimate is the cutting-off of IMO team is roughly 215-225 points.(the results is not public so I can only make a estimate.)
I've got about 180 points in those tests(which make me about 16th-21st in all the 30 contestants.)

that's a nice result wpolly congrats! sorry though you won't make it to Greece this year :(

how many selections can you take now? (that is in what grade are you?)

"Attended School of South China Teachers' College" always gets good mark in Math Olympiad contests in China.

wpolly, why is it 'attended', i think it should be 'affiliated'.

Could u please bring these problems quicker ????
:))))))))
LOL:))))))))))

[quote="wpolly"]I'm at grade 11 now so I have another chance for IMO 2005 in Mexico.[/quote]

I am younger than you . :lol:

can you sent me problems of contest CMO round 2


function
algebra
domain
linear algebra
matrix

Could anyone please help me with this functional proof:

Suppose $f: A \longrightarrow A$ and$ g: A \longrightarrow A$ are both bijections.

Prove ${(g \circ  f)}^{-1} = {{f}^{-1}  \circ  {g}^{-1}}$

What's really confusing me is the inverse function compositions. I know to prove two function are equal, we must show their domains are equal and for all elements in their domains, both functions give identical outputs. But i don't know how to show the two functions have equal domains for a start.

Could anyone help? Thanks.

I don't know if we can say this seeing I'm bad enough at linear algebra when I keep my mouth shut...

Let $M$ be the matrix representation of $f$ and $N$ be the matrix representation of $g$. Then $(g\circ f)^{-1}=(NM)^{-1}$. But $(NM)^{-1}=M^{-1}N^{-1}=f^{-1}\circ g^{-1}$.

From where did matrices come????

I was thinking of picking arbitary elements of sets and then solving. Is it right?


Do you guys know what software to use to draw nice pictures for geometry? (ie. triangle ABC inscribe in a circle c1, bla bla bla)

windows paint can do the job but it is not fast enough.
can anybody tell me a link or a site to download this kind of software?

thx :D

or you can try google " geometer's sketchpad ". I think the latest version is 4.06?

[quote="Iversonfan2005"][quote="BornForMath"]Geometer Sketchpad.
But i dont' know any free websites to download that.[/quote]

download.com, it's free and legal[/quote]
Ive tried that and I couldnt find it  :?  :(


algebra
polynomial
number theory
prime factorization

I cannot prove the following problems, can you help me?
1. PT 1280000401 is composite  :blush: (sorry but I had typed an incorrect no)
2. if m ends with the digit 5, PT (12^m+9^m+8^m+6^m) is divisible by 1991
(however, thanks for your sol of prob 2)

Some hints :D
[hide]Easy,
1)128000401 is divisible by 41
2)Last digit of m is 5 so m=10k+5
$ 12^m\plus{}9^m\plus{}8^m\plus{}6^m\equal{}(2^m\plus{}3^m)(3^m\plus{}4^m)$
Then use modulo we have solution[/hide]

according to my calculator, the prime factorization of 1280000401 is 421*3040381.
Working backwards, i still can't figure out how to arrive at that answer.

Yeah I have made mistake for pro.1
For pro.2 try to prove that
$ (2^m\plus{}3^m)$ is divisible by 11 and $ (3^m\plus{}4^m)$ is divisible by 181 :lol:

[quote="MellowMelon"][hide="1"]$ P(x) \equal{} x^7 \plus{} x^2 \plus{} 1 \equal{} (x^2 \plus{} x \plus{} 1)Q(x)$ for some $ 5$th degree polynomial $ Q$, which can be seen by plugging in a third root of unity. $ P(20) \equal{} 1280000401 \equal{} 421 Q(20)$, so the number is composite.[/hide][/quote]
How did you get $ P(x)$ (from sketch)?

[hide="Answer to mathwizarddude"]The $ 128, 4, 1$ make you think of powers of $ 2$, and then you realize they're at just the right places for the number to be a sum of powers of $ 20$. The number ends up being $ 20^7 \plus{} 20^2 \plus{} 20^0$, and that's when you try to factor $ x^7 \plus{} x^2 \plus{} 1$.[/hide]


floor function
number theory proposed
number theory

Prove that of each positive integer $n\geq 2$ there exist a positive integer $N=a^{2}+b^{2}$ where $a,b$ are natural numbers
such that $n\leq N<n+3\sqrt[4]{n}$

I have a solution for all  $n$ .
But your solution is also near to the idea I used .

Interesting...
Therefore $\forall b>5\in N : \exists A\in R : \forall n\in N: \lfloor A^{b^{n}}\rfloor=x^{2}+y^{2}$

silouan could you post your solution

P.S Could you tell me which TST was it

It was a TST in Greece .
I will post my solution in two days .

[quote="Αρχιμήδης 6"]Juna are you sure?As i know there is real nymber A such as [A^b^n]=p. and not [A^b^n]=x^2+y^2.......[/quote]

http://www.cs.uwaterloo.ca/journals/JIS/VOL8/Caldwell/caldwell78.pdf

It follows from theorem 7

[b]silouan[/b], I'm wait impatiently for your solution. :)

Moreover, I have so interesting question in this connection: is estimation $c\sqrt[4]{n}$ a best for gaps between contiguous numbers which are sums of two squares?

Namely, let $s_{n}$ is a sequence of numbers which are sums of two squares. Is there a subsequence $s_{n_{k}}$ such that $s_{n_{k}+1}-s_{n_{k}}>cs_{n_{k}}^{\frac{1}{4}}$ for some positive constant $c>0$ and for all $k\in\mathbb{N}$?

[quote="Αρχιμήδης 6"]How?Iread it fast but i don't understand how.....[/quote]

Theorem 7. Let $S = \{a_{n}\}$ be any sequence of integers satisfying the following property: there exist real numbers $x_{0}$ and $w$ with $0 < w < 1$, for which the open interval $(x, x+x^{w})$ contains an element of $S$ for all real numbers $x > x_{0}$. Then for every real number $c > \min\left(\frac{1}{1-w},2\right)$ there is a number $A$ for which $[A^{c^{n}}]$ is a subsequence of $S$.

Let $S$ be a sequence of numbers which are a sums of two squares and $w=\frac14+\varepsilon$. In this case theorem asserts that for every real number $c>\frac43+\varepsilon_{1}$ there is a number $A$ for which $[A^{c^{n}}]=a^{2}+b^{2}$ for some $a,b\in\mathbb{N}$. For example, it is true for $c = 2$.

Interesting get $A$.
It is possible that this constant is a algebraic number for several numbers $c$

[b]silouan[/b], I'm still wait impatiently for your solution... :(

Ok here is my solution
Let $n=k^{2}+m$  where $0\leq m\leq 2k$ and $k=\left\lfloor \sqrt{n}\right\rfloor$  with $m\leq 2\sqrt{n}$

Now take $m=l^{2}+t$  where $0\leq t\leq 2l$
so $l=\left\lfloor \sqrt{m}\right\rfloor \leq \sqrt{m}\leq \sqrt{2}\sqrt[4]{n}$

Now if t=0 it is obvious .
If $t\geq 1$ we consider the number $N=k^{2}+(l+1)^{2}$
In this case
$0\leq N-n=2l+1-t\leq 2l\leq 2\sqrt{2}\sqrt[4]{n}<3\sqrt[4]{n}$

and we are done


When written in base 76 or base 133, the expansion of 1/n terminates. If
n is a positive integer greater than 1, then what is the least possible value
of n?(Answer in base 10)

Isn't it just 2?

$ 2_{10} \equal{} 2_{76 \text{ and }133} \equal{} \frac12 \equal{} 0.5 \equal{} \text{ terminate}$  So, probably, yes.

EDIT: Oh dang, it has to terminate in that base.  SO it's not 2  :P (directed towards me)

Nope. What's $ 0.\overline1_3$?

In general, the decimal expansion of $ \left(\frac1m\right)_n$ terminates if and only if the base $ n \equal{} p_1^{k_1}p_2^{k_2}\cdots$ contains all prime factors of $ m$.
[b]
Edit:[/b] Fixed a typo.

So are we just trying to find the least common factor between the two?

Yes it's the least common factor other than 1.

I guess it's an intimidating problem with a simple 5-second solution.

[quote="math154"]I $ n \equal{} p_1^{k_1}p_2^{k_2}\cdots$
.[/quote]

What's that mean?

Sorry if I wasn't clear; it's the prime factorization of $ n$.

Math154's long speech in a few words:

If the union of the prime factors of $ n$ and the prime factors of $ m$ is the set of prime factors of $ m$.


symmetry
geometry
geometric transformation
rotation
vector

Hi everyone,

A changing magnetic field in the $ z$ direction can induce electric field lines following circles in the $ xy$ plane.  The book's explanation for this is that it follows by symmetry (two books give that explanation actually).  How does the observation follow by symmetry?  By symmetry, couldn't the field lines point radially outwards from the circles?

-Johnny

The symmetry argument alone is indeed deficient. This is exactly what happens when you try to explain electrodynamics without Maxwell's Equations. Here are the two things you need to know:

Changing magnetic field induces a rotational component to the electric field.
Electric charge creates the radial component to the electric field.

In this case, there is no net charge density anywhere, so there is no radial component of any kind. Because there is a changing magnetic field, there will be a purely rotational electric field.

[i]Now[/i] you can talk about symmetry to learn more about [i]how[/i] the rotational component looks like.

To get a better feel for what I mean by "radial" and "rotation", read about the [url=http://en.wikipedia.org/wiki/Divergence]divergence[/url] and [url=http://en.wikipedia.org/wiki/Curl_(mathematics)]curl[/url] of a vector field. Look at the pictures and examples to get an intuitive feel of these if you don't want to deal with the vector calculus.

Now, look up Maxwell's Equations in differential form in terms of free charges and currents (you should see E's and B's instead of D's and H's). Try to make sense of them.


inequalities trigonometry limit calculus derivative conics ellipse

how to prove

$ sinx < x$ for all $ x>0$

4 Unit Question today :wink:  How did you go?

Let $ f(x) \equal{} x \minus{} \sin{x}$ and hence $ f'(x) \equal{} 1 \minus{} \cos{x}$.

Since $ \cos{x} \leq 1$ for $ x \in \mathbb{R}^{ \plus{} }$ (in fact, this is always true). Hence $ f'(x) > 0$ for all $ x >0$ and thus $ f(x)$ is an increasing function.

Therefore $ f(x) > f(0) \equal{} 0$ for $ x > 0$

Thus $ x > \sin{x}$.

Of course, you can always argue from a graphical perspective (though you have to be very accurate in your curve sketching).

Geometric:

Clearly if x>1 the inequality holds as sinx < 1

Now if you have a unit circle, consider a central angle with measure x starting at the x-axis and going up counter-clockwise (like usual).  Now sinx is the length of the perpendicular down from where the angle meets the circle, and x is the length of the arc.  But we know that the shortest distance from a point to a line is the perpendicular, so sinx<x

[hide="Overkill"] $ \frac {\sin x}{x} \equal{} \cos \frac {x}{2} \cos \frac {x}{4} \cos \frac {x}{8} ... \cos \frac {x}{2^n} ... < 1$

:D

Strictly speaking, this is not a proof, since the proof of this identity relies on the result that $ \lim_{x \to 0} \frac {\sin x}{x} \equal{} 1$, which in turn relies on the desired result.  But this is a nice way to see, in a concrete algebraic manner (and without resorting to the boring Taylor series) that the result is true.  :) [/hide]

I like [b]nateharman1234[/b]'s proof, because I thisk the proof using differentiation seems to be circular reasoninig.

Genellaly speaking, $ x < \sin x < \tan x$ for $ 0 < x < \frac {\pi}{2}.$

[quote="BanishedTraitor"]4 Unit Question today :wink:  How did you go?

Let $ f(x) \equal{} x \minus{} \sin{x}$ and hence $ f'(x) \equal{} 1 \minus{} \cos{x}$.

Since $ \cos{x} \leq 1$ for $ x \in \mathbb{R}^{ \plus{} }$ (in fact, this is always true). Hence $ f'(x) > 0$ for all $ x > 0$ and thus $ f(x)$ is an increasing function.

Therefore $ f(x) > f(0) \equal{} 0$ for $ x > 0$

Thus $ x > \sin{x}$.

Of course, you can always argue from a graphical perspective (though you have to be very accurate in your curve sketching).[/quote]
Hi thanks for solution.

Not too bad. question 1-6 Maybe about 85% however I left out 5(d) the pentagon question.

qn 7 left this qn and the parts which accompany it i left out.

Ellipse I struggled and harder 3 unit only could do part (b)

Hopefully above average.

How about you?

Lets kill 3 unit now :)

goodluck with the rest of your exams pal

and thanks everyone for your soltuions

Got quite a bit done (screwed up a bit as well though). But never mind. Good luck Bos1234.

Just curiousity, How many points can you get for the proof using differentiation?

In Japan, we will get 3/10.

For that part, I would guess full marks because there really aren't any restrictions on how to prove things. I am quite certain that this question was set to test the students' understanding of what:

$ f'(x) > 0$ means.


What is the sum of all positive integers less than 100 that are squares of perfect squares?

We are looking for fourth powers. $ 1^4\plus{}2^4\plus{}3^4\equal{}1\plus{}16\plus{}81\equal{}\boxed{98}$.


What is the only pair of consecutive integers whose square roots are also consecutive integers?

That could be a school countdown problem maybe

Can anyone prove that 0, 1 are the only two integers that work? It seems obvious, but some quote says that that's not allowed in mathematics.  :lol:

I know that proofs are generally above the level of MATHCOUNTS, so remove this if you wish.

sure,

Lets say the numbers are $x$ and $x + 1$.  There square roots are $x^2$ and $x^2 + 2x+ 1$.  But, there square roots must be consecutive too meaning:

$x^2 + 1$ = $x^2 + 2x + 1$

  simplifying this we get 0 =2x, which means x has to be 0.  And our numbers are x, x+1  or 0 and  1


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?

[hide]Obviously we must have quarters because nickles and dimes alone won't be enough.  20 quarters=$5.00 so we must have less than 18 quarters.  The minimum is 14 because any less would never add up to $5.  If we had 14 quarters, we need  $1.50.  That's 14 dimes and 2 nickles.  But since 2 nickles=1 dime and we're only allowed 30 coins, we can't substitute 2 nickles for a dime because that would exceed the limit.  We find that this is also true for 15 quarters which could only be paid with 10 dimes and 5 nickles.  16, where we need to make a dollar with 14 coins also has only one solution of 6 dimes and 8 nickles and 17 also has one solution of 2 dimes and 11 nickles.  All in all, that's four possible solutions.[/hide]

I know this seems a bit off topic, but how do you usually do these kinds of problems? As far as i see it, you need both good number skills and have to be a bit systematic, the latter of which I am not (i like to use shortcuts, etc.)


I think this was mentioned but not proven in another post, but I'm running into a brick wall.  I have two connected compact $ n$-manifolds $ M_1$ and $ M_2$.  Why is $ H_{n \minus{} 1}(M_1 \# M_2) \equal{} H_{n \minus{} 1}(M_1) \oplus H_{n \minus{} 1}(M_2)$?

More specifically, I suspect the inclusion $ M \minus{} \{p\} \to M$ (where $ p \in M$) induces an isomorphism on the $ n \minus{} 1$-th homology, but can't prove it.

Use excision: $ H_m(M,M\minus{}\{p\}) \equal{} H_m(D^n,D^n\minus{}\{p\}) \equal{} H_{m\minus{}1}(D^n\minus{}\{p\}) \equal{} H_{m\minus{}1}(S^{n\minus{}1})$.

So when $ m \equal{} n\minus{}1$, the relative homology is zero. However, when $ m \equal{} n$, the relative homology is $ \mathbb{Z}$. So, the inclusion
does not induce an isomorphism on $ n\minus{}1$ homology necessarily; but it's certainly a surjection.

I figured this out earlier.  When the manifold is orientable, the inclusion actually induces an isomorphism on the $ n\minus{}1$ homologies.  We get an exact sequence

$ H_n(M) \to H_n(M, M \minus{} x) \to H_{n \minus{} 1}(M \minus{} x) \to H_{n \minus{} 1}(M) \to H_{n \minus{} 1}(M, M \minus{} x) \equal{} 0$.

But the first map is actually an isomorphism (orientations are the same as local orientations), which makes the second map injective, which makes the third map an isomorphism!

Rather it makes the second map a zero map :).


AMC
USA(J)MO
USAMO
function
quadratics
number theory
modular arithmetic
\/closed

Is this class a counterpart to your Intro. Number Threory Class in September?

What I mean is: are the materials covered essentially the same to the last Intro. class? A while ago, I heard you talking about teaching some interesting theorems like McNugget Theorem (if I remembered correctly) and similar things. I'm more interested in the 'other' theorems. So can you also please clarify the topic in specifics?

This is the same as the Intro to Number Theory - same topics covered.  What specific topics are you interested in?

If you are not well-versed in using mods, then you should strongly consider taking this class.

What sort of other theorems did you have in mind?

How deep is this class going to get? And also, isn't there going to be an intermediate number theory class near the time of this one?

At this point it is uncertain when an intermediate number theory course will be available.  I will write one sometime, but I have another class to prepare first so an intermediate class will be at least a few months away.

Topics covered in intro:
Divisibility -- This goes deeper than many students might think.  There are many AMC/AIME and even a few USAMO problems that use relatively simple math of primes, composites, divisibility, etc., but do so in a way that requires more creative depths of thinking.
Divisors -- Ways to identify and count them and to count various kinds of them (even, multiples of 20, perfect squares, etc.).
Diophantine Equations -- Along with some applications.
Modular Arithmetic -- This topic is covered thoroughly with many AMC/AIME level problems discussed in great detail.

The intermediate course is not prepared, but I suspect it will revolve at least partially around the following:
Divisibility Revisited -- Going up to some Olympiad level problems
Euler's Phi Function
Euler's Theorem and Fermat's Little Theorem -- This might be a significant part of the course.  I could certainly find several days worth of problems to discuss.
Wilson's Theorem
Quadratic Reciprocity

If there are topics that you want to see covered that are not mentioned here, please ask about them.

The Chicken McNugget Theorem is part of the intro course and we go through a proof using Diophantine equations.

Partitions are part of the Intermediate Counting class.

[quote="answerseeker95"]What about Chinese Remainder Theorem?[/quote]

We walk up through an understanding of why it works and how to apply it in the intro course.


function
number theory
prime numbers
combinatorics unsolved
combinatorics

Let $S=\{1,2,...,100\}$ . Find number of functions $f: S\to S$ satisfying the following conditions
a)$f(1)=1$
b)$f$ is bijective
c)$f(n)=f(g(n))f(h(n))\forall n\in S$, where $g(n),h(n)$ are positive integer numbers such that $g(n)\leq h(n),n=g(n)h(n)$ that minimize $h(n)-g(n)$.

Trivially, one such function is determined from the values it takes on prime numbers.

Suppose $f(n)$ is a prime number. Then, either $f(g(n))$ or $f(h(n))$ equals $1$, but from injectivity we get either $g(n)$ or $h(n)$ equals 1 $\Rightarrow$ $n$ is a prime number. Since $f$ is bijective, this implies $f(p)$ is a prime for all primes $p$.

Now, for each prime number $p$, let $\alpha(p)$ be the number of primes $q$ such that $pq \leq 100$. For example, $\alpha(2) = 15$, because $2q \leq 100 \Rightarrow q=2,3,5,7,11,13,17,19,23,29,31,37,41,43,47$.

If $p,q$ are prime numbers such that $pq \leq 100$, then $f(pq) = f(p)f(q) \leq 100$. Thus, $\alpha(f(p)) \geq \alpha(p)$. Let us compute the values of $\alpha$:

$\alpha(2) = 15$
$\alpha(3) = 11$
$\alpha(5) = 8$
$\alpha(7) = 6$
$\alpha(11) = \alpha(13) = 4$
$\alpha(17) = \alpha(19) = 3$
$\alpha(23) = \alpha(29) = \alpha(31) = 2$
$\alpha(37) = \alpha(41) = \alpha(43) = \alpha(47) = 1$
For all the others: $\alpha(p) = 0$.

From these we get $f(2) = 2$, $f(3) = 3$, $f(5) = 5$, $f(7) = 7$. Now, $f(11) = 11$ or $f(11) = 13$, but from $f(99) = f(9)f(11) = 9f(11)$ we get $f(11) = 11$, hence also $f(13) = 13$.
Going on, $f(17)$ and $f(19)$ are $17$ and $19$ in some order, and the same thing happens with $23, 29$ and $31$. But, from $f(92) = f(4)f(23) = 4f(23)$ we get $f(23) = 23$, so now we know $f(29)$ and $f(31)$ are $29$ and $31$ in some order.
Same thing happens with $37,41,43,47$ and with the set of all primes between $50$ and $100$.

Until now, we have $2! \cdot 2! \cdot 4! \cdot 10!$ ways to define the function on the prime numbers. It is not hard to see that each of these gives a function that works. So the answer is $\boxed{96 \cdot 10!}$.


[quote="Phelpedo"]Pardon the ignorance of a first-time USAMT-er, but how exactly do I submit the entry and solution forms?[/quote]

You can send in your entry form along with your 1st round solution form.

There are various ways you can send in your solutions including $\LaTeX$, check the website.

Yeah but do I send them by email or mail, or what?

How you can submit solutions: http://www.usamts.org/Problems/U_ProblemsSoln.php

FAQ:
http://www.usamts.org/TipsFAQ/U_FAQ.php

[quote="Phelpedo"]Ah, thanks, I've got this figured out now. Out of curiosity, what's the point of including the usamts.tex file in the folder?[/quote]

It's a template for the solutions.tex. without it, you get errors when you build

[quote="DPatrick"][quote="Phelpedo"]Ah, thanks, I've got this figured out now. Out of curiosity, what's the point of including the usamts.tex file in the folder?[/quote]The solutions.tex file won't compile without the usamts.tex file.[/quote]

so will we have to send the usamts.tex file with our solution files?

[quote="Phelpedo"]Ah, thanks, I've got this figured out now. Out of curiosity, what's the point of including the usamts.tex file in the folder?[/quote]

To put it simply:

It defines new "commands" which you can think of as "functions."  Normally, when you write \realname{Sean} you would get an error.  But the USAMTS.tex file defines that command to output the text "Real Name: Sean" or whatever it says.  This way, LaTeX beginners have a simple document, instead of having to go through complex code and change certain parts.  The USAMTS.tex definitions also produce all the fancy boxes at the top.

Also, to interpret your question differently, I guess it is not neccessary to send the usamts.tex file in your email.  USAMTS staff could always just add that file in and compile.  But sending it probably helps them a lot, so they don't have to find it and drag it into your folder.

You do not need to send the usamts.tex file.  In fact, please don't.  We have a copy. :)

You only need to send your solutions.tex file and the output PDF (or DVI) file, plus any diagram files that we might need.

Umm...when I build the file, it says there are 75 errors, but when I open the PDF, nothing is wrong.

Will that be a problem? :maybe:

Hm, what are the errors?  There are two buttons next error and previous error, I believe, and you can use these to figure out what the errors are.  75 is a pretty big number so you can just name a few :P

All of them seem to be the same 3 errors:

! Missing = inserted for \ifdim.
  <to be read again>

! Missing number, treated as zero.
  <to be read again>

! Illegal unit of measure (pt inserted).
  <to be read again>

I can't seem to find any pattern for what creates an error.

Some are on \end{solution}, which I don't understand. And some errors are on my $\LaTeX$, when I can't find anything wrong. And some errors just occur on blank lines.  :maybe:

Illegal unit of measure:

I can help you with that.

I bet you used a command such as \vskip 0.5.

You need to replace this with \vskip 0.5 cm.  You are missing a unit of measurement.  That will kill about 30 or 40 errors right off the bat.

Actually, I didn't use any command like that.  :maybe:

For example, it says "Illegal unit of measure" for "\end{solution}"

Normally, I would say send me your LaTeX, but that obviously can't happen.  Can you post a list of commands that caused this error?

Also: (correct me if I am wrong) an illegal unit of measure is an error, but it won't affect your output.  So you can ignore that error, if you need to.  I bet you can ignore the other two as well (becasue of <to be read again> ) but I am not familiar with those two, so you definitely need a second opinion.

MAKE SURE ALL OF YOUR {'S ARE CLOSED!!!!!

This causes the vast majority of all LaTeX errors.

Hm, I don't know.  Trying a couple of things I was unable to recreate those errors.  If the output looks OK, then just don't worry about it.  Though, if you figure it out let us know what caused those.  I'm curious. :P


function
search
algebra
functional equation
algebra unsolved

Find all functions $ f: \mathbb{N}_{0}\rightarrow\mathbb{N}_{0}$ such that
\[f(m \plus{} f(n)) \equal{} f(f(m)) \plus{} f(n)\]
for all $m,n\in\mathbb{N}_{0}$.

I'm not very good with functional equations, so please check if I took wrong/too many steps.  :D

let $ P(m,n)$ denote the assertion $ f[m \plus{} f(n)] \equal{} f[f(m)] \plus{} f(n).$

$ P(0,0) \Rightarrow f(0) \equal{} 0.$
$ P(n,0) \Rightarrow f(m) \equal{} f(f(m)).$

suppose for some $ m$, $ f(m) \equal{} k \Rightarrow f(k) \equal{} k.$
then, $ P(k,m) \Rightarrow f(2k) \equal{} f(k) \plus{} f(k).$
This is the Cauchy functional equation, and since $ f(k) \equal{} k$, therefore $ f(x) \equal{} x$ for the given. :nhl:

i think we get $ f(n.k) \equal{} nk$  for $ \forall n\in\mathbb{N}_{0}$
could you explain better ?

[quote="cnyd"]$ f: \mathbb{N}_{0}\rightarrow\mathbb{N}_{0}$
$ f[m \plus{} f(n)] \equal{} f[f(m)] \plus{} f(n)$ for $ \forall m,n\in\mathbb{N}_{0}$[/quote]

Let $ P(x,y)$ be the assertion $ f(x \plus{} f(y)) \equal{} f(f(x)) \plus{} f(y)$

$ f(x) \equal{} 0$ $ \forall x$ and $ f(x) \equal{} x$ $ \forall x$  are two trivial solutions and we'll now search for other solutions :

Let $ A \equal{} f(\mathbb N_0)$
Let $ u \equal{} \gcd($all non zero elements of $ A)$
$ P(0,0)$ $ \implies$ $ f(0) \equal{} 0$
$ P(0,x)$ $ \implies$ $ f(f(x)) \equal{} f(x)$ and so $ f(x) \equal{} x$ $ \forall x\in A$ and so $ f(x \plus{} y) \equal{} x \plus{} y$ $ \forall x,y\in A$ and so $ f(mx \plus{} ny) \equal{} mx \plus{} ny$ ** edited ** : "$ \forall x,y,m,n\in\mathbb N_0$" ==> $ \forall x,y\in A$, $ \forall m,n\in\mathbb N_0$

We know that $ \exists x_0,y_0\in A^*$ such that $ \gcd(x,y) \equal{} u$.
So $ \exists m,n,m',n'\in\mathbb N_0$ such that $ m'x_0 \plus{} n'y_0 \equal{} mx_0 \plus{} ny_0 \plus{} u$

Then $ P(u,mx_0 \plus{} nx_0)$ $ \implies$ $ f(u) \equal{} u$ and so $ A \equal{} \{nu, n\in\mathbb N_0\}$

So, up to now, we know that $ f(m)$ is always a multiple of $ u$ and that $ f(m) \equal{} m$ for any multiple of $ u$.
We also know that $ f(m \plus{} ku) \equal{} f(m) \plus{} ku$ and so it remains to define $ f(m)$ for $ m\in[1,u \minus{} 1]$ in order to have a full definition for $ f$.
But we'll now show that $ f$ may take any values (multiple of $ u$) for $ m\in[1,u \minus{} 1]$ :

Let $ g(m)$ any function from $ [0,u \minus{} 1]\to\mathbb N_0$ such that $ g(0) \equal{} 0$ and $ f(n)$ defined as :

$ \forall a\in[0,u \minus{} 1],\forall b$ : $ f(a \plus{} ub) \equal{} u(g(a) \plus{} b))$

Then let $ m \equal{} a \plus{} ub$ and $ n \equal{} c \plus{} ud$ :
$ f(n) \equal{} u(g(c) \plus{} d)$
$ m \plus{} f(n) \equal{} a \plus{} u(b \plus{} g(c) \plus{} d)$ and so $ f(m \plus{} f(n)) \equal{} u(g(a) \plus{} b \plus{} g(c) \plus{} d)$
$ f(m) \equal{} u(g(a) \plus{} b)$ and so $ f(f(m)) \equal{} u(g(a) \plus{} b)$ and so $ f(f(m)) \plus{} f(n) \equal{} y(g(a) \plus{} b \plus{} g(c) \plus{} d) \equal{} f(m \plus{} f(n))$

Hence the results :
$ f(x) \equal{} 0$ $ \forall x$
$ f(x) \equal{} u\cdot g(u\{\frac xu\}) \plus{} u[\frac xu]$ where $ u$ is any positive integer and $ g(m)$ any function from $ [0,u \minus{} 1]\to\mathbb N_0$ such that $ g(0) \equal{} 0$.
(Notice that $ u \equal{} 1$ gives the solution $ f(x) \equal{} x$)

[quote="earldbest"]... please check if I took wrong/too many steps.  ...[/quote]

First, you can get $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$, but only for $ x,y\in f(\mathbb N_0)$ and not for any $ x,y$
Second, Cauchy equation implies $ f(x)\equal{}ax$ only if it is true at least over $ \mathbb Q$, so you could not use this conclusion here.

[quote="pco"]

First, you can get $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$, but only for $ x,y\in f(\mathbb N_0)$ and not for any $ x,y$
Second, Cauchy equation implies $ f(x) \equal{} ax$ only if it is true at least over $ \mathbb Q$, so you could not use this conclusion here.[/quote]

oh, ok.  :D

Cuneyt what is the source of this problem?


can anyone solve this question please,

$ evaluate$    $ \sqrt{1\plus{}2\sqrt{1\plus{}3\sqrt{1\plus{}4\sqrt{1\plus{}5\sqrt{\cdots}}}}}$

i know the answer is $ 3$

but i could not get it

thanks in advance

Firstly, please don't double [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=310883]post[/url]

Secondly, the problem posted is not at the high school basics level [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32837]Read BEFORE Posting Problems[/url]

Finally, this problem comes from [url=http://en.wikipedia.org/wiki/Srinivasa_Ramanujan#Attention_from_mathematicians]Srinivasa Ramanujan[/url] although the solution is not there, the solution has something to do with the equation:
\[ x + n + a = \sqrt {ax + (n + a)^2 + x\sqrt {a(x + n) + (n + a)^2 + (x + n)\sqrt {\cdots}}}\]
In particular, what you are asking is the case where $ a = 0,n = 1$ and $ x = 2$, which indeed gives $ 3$.

Hope that helps, and remember to wait some time for an answer, don't just post it again.


Let $ABC$ be a triangle. $AA_{1}, BB_{1}, CC_{1}$ concurrent cevians with $A_{1}$ on segment $BC$, $B_{1}$ on $CA$ and $C_{1}$ on $AB$. Let $P$ be the concurrence point. Let $X$ be a point on $AA_{1}$, in the same semiplane with $A$ respect to $BC$, such that $|XA_{1}|=3|PA_{1}|$, the same $Y$ for $B$ on $BB_{1}$ and $Z$ for $C$ on $CC_{1}$.
Prove that:

                     $\frac{AX}{AA_{1}}+\frac{BY}{BB_{1}}+\frac{CZ}{CC_{1}}=0$

 :)  :lol:  :!:

(I think this is what I want to mean)

I gather you want to argue that the only solution is when $X = A$, $Y = B$, $C = Z$ and the triangle is equilateral, or something?  It's false: try isosceles triangle $ABC$ with $m\angle A = 120^\circ$.  Take $P$ to be the midpoint of altitude $AA'$.

sorry you're right; but no, it is for all triangles and not necesarly with X=Y=Z, in fact I think it is also for exterior cevians.
I only need to rewrite the not exterior part in a clearer way  :D Now you can see the correct problem.


geometry
3D geometry
sphere
vector
superior algebra
superior algebra solved

Show that M=S^n X R   (The Cartesian Product of n-sphere and the real line) is parallelizable for all n.

May I first know what is M actually??

$ S^1\times \mathbb R$ is an infinite cylinder. But you can think of it as a finite cylinder without boundary, because $ \mathbb R$ is diffeomorphic to $ (0,1)$. You can visualize $ S^n\times \mathbb R$ as an open spherical shell $ \{x\in\mathbb R^{n\plus{}1}\colon 1<|x|<2\}$.

this is nice: the normal bundle of $ S^n\subset \mathbb{R}^{n \plus{} 1}$ is isomorphic to $ \varepsilon^1_n$ (the trivial vector bundle of rank $ 1$ over $ S^n$). $ TS^n\oplus \varepsilon^1_n$ is then isomorphic to the tangent bundle of $ S^n$ as a submanifold of $ \mathbb{R}^{n \plus{} 1}$, so it's a trivial bundle of rank $ n \plus{} 1$.
the fact that $ M$ is parallelizzable follows from the fact that $ M$ retracts onto $ S^n$ and $ \varepsilon^{n\plus{}1}_n$ is the pullback of $ TM$ through the inclusion $ S^n \simeq S^n\times\{0\} \subset M$.


function
integration
real analysis
real analysis theorems

(i)Let f_n be a sequence of real valued functions in L1(R,m).Suppose that Sum 1..oo int_R |f_n-f}dm <oo
prove that fn->f m a.e.

(ii)Let t_n be a sequence in the interval [0,1] and an be a sequence in [0,oo) s.t. Sum 1..oo an <oo
Prove that Sum 1..oo an/sqrt(|s-t_n|) converges for almost all s in [0,1]

$\sum \int |f_n-f| dm < \infty$, hence $\int |f_n-f| dm \to 0$, whence $f_n$ converges to $f$ in measure. (I think you know $L^1$-convergences implies convergence in measure)

In order to show $\sum \frac{a_n}{\sqrt{|s-t_n|}}$ converges for almost $s\in [0,1]$, we shall show that $\int_0^1 \left(\sum \frac{a_n}{\sqrt{|s-t_n|}}\right) ds < \infty$. Since $\frac{a_n}{\sqrt{|s-t_n|}}\geq 0$, we can integrate by terms here. So $\int_0^1 \left(\sum \frac{a_n}{\sqrt{|s-t_n|}}\right) ds = \sum \int_0^1 \frac{a_n}{\sqrt{|s-t_n|}}ds = \sum (2a_n(\sqrt t_n + \sqrt{1-t_n})) \leq \sum 2a_n\cdot \sqrt 2 < \infty$

p.s. I don't think this post should be in Theorems and Formulas