Hi, here's a very nice problem:

[color=blue]Let K and L be the midpoints of the diagonals AC and BD of a convex quadrilateral ABCD. The line KL cuts the lines DA and BC at the points X and Y, respectively. Let the circumcircle of triangle AKX intersect the line AB at a point M. Prove that the circumcircle of triangle BLY also passes through the point M.[/color]

This is actually SPbMO 2004, Elimination Round, 10th Class, problem 6.

Elimination Round, 11th Class, problem 6 asks to prove that [color=blue]the circumcircles of triangles AKX, BLY, CKY and DLX have a common point[/color].

[color=red][Moderator edit: Replaced the problem by the corrected one and added source and additional problem.][/color]

[quote="hardsoul"][color=blue]Let K and L be the midpoints of the diagonals AC and BD of a convex quadrilateral ABCD. The line KL cuts the lines DA and BC at the points X and Y, respectively. Let the circumcircle of triangle AKX intersect the line AB at a point M. Prove that the circumcircle of triangle BLY also passes through the point M.[/color][/quote]

Here is a very nice proof of one of my friends  Andrei (also a Mathlinks member):

If $Z$ is the midpoint of $AB$ then $\angle{DAB}=\angle{LZB}=\angle{MKL}$ hence $KLMZ $ is cyclic.
So we obtain : $\angle{CBA}=\angle{KZA}=\angle{KLZ}$ and the conclusion follows.

Thank you Sailor for this beautiful solution. KLMZ is really cyclic but , <LZB=180-<MKL.

Dear hardsoul , can you post all problems of SPb MO 2004 ?

Unfortunatel for now I don't have LATEX.
If you know Russian you may try http://www.pdmi.ras.ru/~olymp/.

I have really not much time now, so I will have to be brief.

Here are my solutions to both of the problems:

We will use directed angles modulo 180°.

We start with the first problem:

[quote="hardsoul"][color=blue]Let K and L be the midpoints of the diagonals AC and BD of a convex quadrilateral ABCD. The line KL cuts the lines DA and BC at the points X and Y, respectively. Let the circumcircle of triangle AKX intersect the line AB at a point M. Prove that the circumcircle of triangle BLY also passes through the point M.[/color][/quote]

Instead of solving this problem in the given form, I will prove a stronger assertion:

[color=blue]Let Z be the midpoint of the segment AB. Let the circumcircle of triangle KLZ intersect the line AB at a point U (apart from the point Z). Then, the circumcircles of triangles AKX and BLY also pass through this point U.[/color]

Of course, once this assertion will be proven, it will follow that the point U is the point of intersection of the circumcircle of triangle AKX with the line AB (apart from the point A, of course); but this is exactly the definition of the point M. Thus, it will follow that M = U, and since the circumcircle of triangle BLY passes through the point U, it will follow that the circumcircle of triangle BLY passes through the point M, so that the problem will be solved.

So it remains to verify the assertion that the circumcircles of triangles AKX and BLY pass through the point U.

But verifying this assertion is very simple: Since the points K and Z are the midpoints of the segments AC and AB, respectively, the line KZ is a midparallel in triangle ABC, and thus parallel to the line BC. Hence, < (KZ; KL) = < (BC; KL). In other words, < ZKL = < BYL. But since the point U lies on the circumcircle of triangle KLZ, we have < ZUL = < ZKL. Thus, < ZUL = < BYL. In other words, < BUL = < BYL. Thus, the point U lies on the circumcircle of triangle BLY. Similarly, the point U lies on the circumcircle of triangle AKX. Thus, our assertion is proven, and the first problem is solved.

[Of course, this solution is almost the same as Andrei's solution posted by Sailor. The only difference is that I explicitely "re-defined" the point M in a symmetric way, namely as the point of intersection U of the circumcircle of triangle KLZ with the line AB different from the point Z.]

Now to the second problem:

[quote="hardsoul"]prove that [color=blue]the circumcircles of triangles AKX, BLY, CKY and DLX have a common point[/color].[/quote]

Solving this problem requires just simple angle chasing:

From the solution of the first problem, we know that the circumcircles of triangles AKX and BLY both pass through the point U, which lies on the line AB. (Of course, if you want, you can call the point M instead of U - we have shown that M = U.)

Let T be the point of intersection of the circumcircles of triangles AKX and BLY different from the point U. If we now succeed to show that this point T also lies on the circumcircles of triangles CKY and DLX, then it will follow that the circumcircles of triangles AKX, BLY, CKY and DLX have a common point, namely the point T, and thus, the problem will be solved.

Hence, in order to solve the problem, it remains to show that the point T also lies on the circumcircles of triangles CKY and DLX.

Since the points U and T both lie on the circumcircle of triangle BLY, we have < UTY = < UBY. In other words, < UTY = < (AB; BC). Since the points U and T both lie on the circumcircle of triangle AKX, we have < UTK = < UAK. In other words, < UTK = < (AB; AC). Thus,

  < KTY = < UTY - < UTK = < (AB; BC) - < (AB; AC) = < (AC; BC) = < KCY.

Thus, the point T lies on the circumcircle of triangle CKY. Similarly, the point T lies on the circumcircle of triangle DLX. And thus, the problem is solved.

  darij

Here are just 9'th form.Hope that translated them correctly.
Thanks Darij for your solution I'll read it today :) .


A whole set of Hong Kong currency is compiled of coins in: 10 cents, 20 cents,
50 cents, 1 dollar, 2 dollars, 5 dollars ,and also bank notes in: 10 dollars, 20 dollars, 50 dollars, 100 dollars, 500 dollars and 1000 dollars. The problem is that, how many different ways are there to form a sum of 2000 dollars?
e.g. $\$2000=\$1000+\$1000=\$1000+\$500+\$500,...$ and so forth.
Remark:$\$$1 = 100 cents.

See here http://www.mathlinks.ro/Forum/viewtopic.php?t=74494

But the solution given is not so nice. I think there is a method using only combinatorics and maybe this can be solved by the combinations with repetition

Thanks a lot for the information.
In fact, I have ever thought of using generating functions.
The generating function for our problem should be
$(1+x+x^2+...)(1+x^2++x^4...)(1+x^5+x^{10}+...)(1+x^{10}+x^{20}+...)(1+x^{20}+x^{40}+...)(1+x^{50}+x^{100}+...)(1+x^{100}+x^{200}+...)(1+x^{200}+x^{400}+...)(1+x^{500}+x^{1000}+...)(1+x^{1000}+x^{2000}+...)(1+x^{2000}+x^{4000}+...)(1+x^{5000}+x^{10000}+...)(1+x^{10000}+x^{20000}+...)$
Then the answer of our problem should be the coefficient of $x^{20000}$ in the series expansion of this generating function.
But I don't know how to handle this horrible generating function.

A computer algorithm could handle it pretty easily

Do you mean handling the generating function or the original problem?
If computer programming is allowed, a nested-for-loop will  generate all the solutions easily, we don't need to care about generating functions.


I wonder if the AIME tests in the Kalva homepage were the tests in previous years since the answers in those tests are not the integers between 0 and 1000.
   PS Only a week left. Good luck everybody.

They are the questions from the old tests in the exact same order, just worded differently for copyright purposes.  Kalva also changes the format of the answer for copyright reasons as well.  Say on the real AIME the answer to the problem is 1/4.  This isn't an integer of course, so the AIME will ask you to find m+n where the answer to the problem is m/n.  So you bubble in 1+4 = 5

Kalva just asks for the answer, so he'll occasionally have fractions and square roots and multiple answers.


i) The polynomial 9x2 + 30x + 25 is also a perfect square trinomial. Find its factors. Sketch a generic rectangle to verify your factors.

Does anyone know what the answer to this is.. I am stuck on the 9x2 thing

(9x Squared not 9x2)

Well, we know the final form is $(ax+b)^{2}$, and $a^{2}=9\rightarrow a=\pm 3$ and $b^{2}=25\rightarrow b=\pm 5$  We also have $2ab=30$, so a and b are either both negative or both positive.  Since we are drawing a rectangle, I'm assuming they are positive, so the equation factors to $(3x+5)^{2}$.


How many positive integers less than 100 have reciprocals with terminating decimal representations?

THe decimal representation terminates only if:

 - The number is a perfect power of 2
 - The number is a perfect power of 5
 - The number is is a number whose factors are only 2's and 5's

There are 7 perfect powers of 2 (including 1).
There are 2 perfect powers of 5 (excluding 1).
The numbers consisting of 2's and 5's only are:
$ 2\cdot 5, 2^2\cdot 5, 2\cdot 5^2$
(there are 3)

Thus, there are a total of 12 numbers whose reciprocal have terminating decimals. (terminating is non-repeating right?)


1. Ramon said that the set of integers is [i]not [/i]closed for one of the basic operations (addition, subtraction, multiplication, or division). You want to show his statement is correct.

For the operation for which the set of integers is not closed, write an example using:
   - a positive even integer and a zero
   - a positive and a negative even integer
   - two negative even integers

Be sure to explain why each of your examples illustates that the set of integers is not closed for that operation.

[color=brown]<I don't know how to do it, because i don't know [b]which two numbers should add, subtract, multiply and divide[/b]. For an example: The first one" A positive even integer and a zero". Does it mean an operation between a postive even integer and zero? e.g.  2+0=2, 2-0=2, 2*0=0, and 0/2=0?
The second one: a positive and a negative even integer. Does it mean an operation between a positive and a negative even integer? e.g. 2,3 (positive even integers), -2,-3 (negative even integers).  2+(-2)=0 .... 3+ (-3)=0 ....  OR 2+(-3)=-1, 3+(-2)=1 ..... , it means whatever positive and negative even integers can do the operation together.  The same problem as the third one. >[/color]

2. Which set is closed under division?  (1) {1}  (2) counting numbers  (3) integers   (4) whole numbers

[color=brown]<Same problem: The operation between Which two numbers? e.g.  the first one: {1}, it means 1+1=2, 1-1=0, 1*1=1,1/1=1?  THe same thing as any other choices?> [/color]

THanks for replies....!

[quote="MIA6"]For an example: The first one" A positive even integer and a zero". Does it mean an operation between a postive even integer and zero? [/quote]

Yes.  Certainly not closed under division.

[quote="MIA6"]The second one: a positive and a negative even integer. Does it mean an operation between a positive and a negative even integer? [/quote]

Yes.  I don't really understand why the condition is stated that way, but consider what happens if the positive integer is odd.

[quote="MIA6"]Which set is closed under division?  (1) {1}  (2) counting numbers  (3) integers   (4) whole numbers

[color=brown]<Same problem: The operation between Which two numbers? e.g.  the first one: {1}, it means 1+1=2, 1-1=0, 1*1=1,1/1=1?  THe same thing as any other choices?> [/color][/quote]

In this case, "closed under division" means that we can take [b]any two numbers[/b] in the sets listed and find their quotient.

None of those sets are closed under division.  What a strange question.


$ABC$ is a triangle with incircle $\gamma$. The angle bisectors of the triangle intersect $\gamma$ first at $O_A$, $O_B$ and $O_C$. Let $\gamma_A$ be the circle with centre $O_A$ which is tangent to $AB$ and $AC$, and define $\gamma_B$ and $\gamma_C$ similarly. Let $t_A$ be the external common tangent of $\gamma_B$ and $\gamma_C$ (other than $BC$) and define $t_B$ and $t_C$ similarly.

Show that $t_A$, $t_B$ and $t_C$ are concurrent.

This has already been posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=46240 , guess by whom...

  darij


logarithms
number theory proposed
number theory

For an odd prime $p$, let $\mathrm{ord}_{p}(2)$ denote the least positive integer $k$ such that $p$ divides $2^{k}-1$. Prove or disprove: $\sum_{p}\frac{1}{p \, \mathrm{ord}_{p}(2)}< \infty$ is finite.

It converges by comparison to $\sum \frac1{p\log p}$ and then to $\frac1{n\log^{2}n}$.
Obviously, $\text{ord}_{p}2>\log_{2}p$, so $\frac1{p\log p}>\frac1{p\text{ord}_{p}2}$. The $n$th prime is close to $n\log n$, giving us the second comparison.

Ah... Thank you  :blush:  I hadn't noticed this, and had a long and messy proof...  :blush: I needed this fact as an auxiliary lemma:

Suppose in addition that $\sum_{p^{2}\mid 2^{p-1}-1}\frac{1}{\mathrm{ord}_{p}(2)}< \infty$ (this is believed to be the case...) Conclude, using the above convergence, that $\sum_{p}\frac{1}{\mathrm{ord}_{p^{2}}(2)}$ is finite (under this hypothesis); conclude that this implies that $2^{n}-1$ is squarefree for a set of values $n$ of positive density  :!:


geometry
circumcircle
incenter
geometry theorems

[quote="behzad"]Does anyone knows how to calculate the distance between circumcenter and incenter unitage radii of them in a bicenteric pentagon .(without eliptic modulus and other unfamiliar ideas...)[/quote]

Well, you will have to solve a cubic equation at least... See my post #4 in http://www.mathlinks.ro/Forum/viewtopic.php?t=34411 :

[quote="darij grinberg"]Well, if two circles o and i have radii R and r, respectively, and the distance between their centers is d, then there exists a pentagon with circumcircle o and incircle i if and only if $a^3+b^3+c^3=\left(b+c\right)\left(c+a\right)\left(a+b\right)$, where $a=\frac{1}{R+d}$, $b=\frac{1}{R-d}$, $c=-\frac{1}{r}$ (and if this condition is satisfied, then, by Poncelet's porism, there exist infinitely many such pentagons).[/quote]

  Darij


Given a positive integer $ n$, find all functions $ F: \mathbb{N} \rightarrow \mathbb{R}$ such that $ F(x\plus{}y)\equal{}F(xy\minus{}n)$ whenever $ x,y \in \mathbb{N}$ satisfy $ xy>n$.

[quote="moldovan"]Given a positive integer $ n$, find all functions $ F: \mathbb{N} \rightarrow \mathbb{R}$ such that $ F(x \plus{} y) \equal{} F(xy \minus{} n)$ whenever $ x,y \in \mathbb{N}$ satisfy $ xy > n$.[/quote]

Put $ y\equal{}1$ we get $ F(x\plus{}1)\equal{}F(x\minus{}n) \Rightarrow F(x)$ is periodic with period $ n\plus{}1$

So $ F(x\plus{}y)\equal{}F(xy\minus{}n)\equal{}F(xy\plus{}1)$

Now put $ y\equal{}n\plus{}1$ we get $ F(x\plus{}n\plus{}1)\equal{}F((n\plus{}1)x\plus{}1)$
$ \Leftrightarrow F(x)\equal{}F(1)$ for all natural number $ x$

So $ F(x)\equal{}c$, where $ c$ is any real constant, for all $ x$, is the only solution.


Can you make the problem statement more clear?

Third attempt to post something from the pocketpc.

This problem is an exercise in Altshiller-Court on the application of Ceva. By the sine law in triangles APE and CPE, you obtain CE / AE = CP / AP * sin CPE / sin APE. Similarly obtain 5 other equations. Now apply Ceva and notice that some angles have equal sines because of corresponding edges being parallel.

Sorry for brevity. I will go into details when I get back home.

  darij
[quote="April"]Let $ P$ and $ P'$ be two points in interior of the given triangle $ ABC$. Let $ D$, $ E$, $ F$ be the traces on sides $ BC$, $ CA$, $ AB$, respectively with the lines through $ P$ and parallel to $ P'A$, $ P'B$, $ P'C$. Let $ D'$, $ E'$, $ F'$ be the traces on sides $ BC$, $ CA$, $ AB$, respectively with the lines through $ P'$ and parallel to $ PA$, $ PB$, $ PC$.
Prove that $ AD$, $ BE$, $ CF$ are concurrent if and only if $ AD'$, $ BE'$, $ CF'$ are concurrent.[/quote]


The fine for an overdue library book is 8 cents per day. Pokey
is 8 days late with each of three books. How much is Pokey's
fine in cents?

There are 3 books so we might as well think of it as being 24 cents per day.  Then $ 24\times 8\equal{}192$ cents.

[hide]For 8 days, we can can just multiply 8(as in cents) by the number of days(8) which is 64. Then, we know that for just 1 book, not 3, so we just multiply by 3:

$64$
$\times$
$3$
--------

$\boxed{192}$[/hide]


Let $f: R^{+}\rightarrow R^{+}$ be a function, strictly increasing. Let $k\in R^{+}$. Show that $\lim_{x\rightarrow k^{-}}$ exists.

[quote="tjhance"]When you say $\lim_{x\rightarrow k^{-}}$ what does the negative sign next to the k mean?  :blush:[/quote]

It means you are approaching the limit from the left hand side.

[quote="fairyofwind"]See Rudin (1976), p. 95.[/quote]

I don't have it!

Please show me the proof.  :)


In an analysis book i found this proof for the uniqueness of a limit of a function.

"It is easy to show that when a Limit of a function f(z) exists at a point a,it is unique.To do this ,we suppose that

lim f(z) =l and lim f(z)=m as z----> a (z goes to a).

Then,for any positive number ε,there are positive numbers r,δ such that

|f(z)-l |< ε whenever 0< |z-a |<r

and

|f(z)-m |<ε whenever 0< |z-a |<δ

So if 0< |z-a |<θ ,where θ denotes the smaller of the two Nos r and δ,we find that

|l-m |= |(f(z)-m)-(f(z)-l) |=< |f(z)-l | + |f(z)-m | < ε +ε=2ε

But |l-m | is a nonnegative constant, and ε can be chosen arbitrarily small.
Hence

l-m =0 , or l=m."

is that proof correct??

If it's in the book, why are you doubting it?  Choose $ \varepsilon\equal{}\frac{\left|l\minus{}m\right|}{2}$, so you get that $ \left|l\minus{}m\right|<\left|l\minus{}m\right|$, contradiction.

[quote="JRav"]If it's in the book, why are you doubting it?  Choose $ \varepsilon \equal{} \frac {\left|l \minus{} m\right|}{2}$, so you get that $ \left|l \minus{} m\right| < \left|l \minus{} m\right|$, contradiction.[/quote].

If i say that i think the proof is wrong what would you say to that?

[quote="JRav"]Why do you think the proof is wrong?[/quote]

Please go to the Australian forum,and lets discuss the mater there ,since as you said is no good posting in different forums

Thank you

Just post it here or in the other thread, I'll find it.  You've yet to tell me why you think the proof is wrong.


inequalities
vector
AMC
AIME
triangle inequality

what exactly are these generalization of Cauchy? I've seen conflicting definitions on Mathworld, Kedlaya's ineq packet, and the site in the last page of "Useful Inequalities" on this forum

Let x = (x1, ..., xn) be a vector of n real numbers.  Let p :ge: 1.  Then the p-norm of x is defined by the formula
[tex] \displaystyle
||x||_p = \big( \sum_{k=1}^n |x_k|^p \big)^{1/p}.
[/tex]
Note that the case p = 2 is the usual Euclidean norm.

Let x and y be vectors of n real numbers.  Let p :ge: 1.  Then Minkowski's inequality says that
||x + y||p :le: ||x||p + ||y||p.
In other words, Minkowski is just the triangle inequality for the p-norm.

Let p :ge: 1 and q :ge: 1 be such that 1/p + 1/q = 1.  (We say that p and q are conjugate norms.)  Let x and y be vectors of n real numbers.  Then Holder's inequality says that the dot product of x and y is at most ||x||p ||y||q.  Note that the Cauchy-Schwarz inequality is the special case of Holder when p = q = 2.

How do these statements of the inequalities compare with the ones you cited?

I am fond of Holder's inequality because I ended up using it to prove two research theorems.

Wow, those two inequalities are both on my homework this week :)

btw, just the sum inequalities please.

Thanks for your response; it agrees with the resource cited in the "Useful Inequalities" thread and what Simon quoted to me. Kedlaya's packet give the following: (let me try my hand at TeX here ;)):

holder: Let [tex]\displaystyle \mathrm{w_i}[/tex] be positive numbers that sum to one. Then for any positive numbers aij, [tex]\displaystyle \prod_{i=1}^n (\sum_{j=1}^m a_{ij})^{w_i} \ge \sum_{j=1}^m (\prod_{i=1}^n a_{ij}^{w_i})[/tex]

minkowski: let r>s be nonzero real numbers. Then for any positive numbers aij , [tex]\displaystyle \sum_{j=1}^m (\sum_{i=1}^n a_{ij}^r)^{\frac{s}{r}})^{\frac{1}{s}} \ge (\sum_{i=1}^n (\sum_{j=1}^m a_{ij}^s)^{\frac{r}{s}})^{\frac{1}{r}}[/tex]

hm, I suppose these are just generalizations. They look so different though, it threw me off.

bah. saw someone using it and thought it was a convention like \displaystyle. I'll edit - it doesn't look too nice.

I don't mind giving away a few of the ideas that we will cover in the inequalities course in a few weeks.

If you can special case to Cauchy, Holder often does the trick for the more general problem.

Likewise, if you can special case to the triangle inequality, Minkowski often does the trick.

These ideas help me when I'm trying to wrap my brain around an inequality.

USAMO... my AMC + AIME score wasnt high enough but I get to write it for the canadian side unofficially for our IMO selection anywayz  =.=;;    makes little difference to me.

Aw fudge, I didn't get invited by Mr. Barbeau... Blah. Can't say I'm surprised, since I've had a dismal year in math contests. I'm not looking forward to learning about my Euclid score, either...

In any case, good solving.

only 5 students got over 90 in euclid...  jacob with 97, john with 96... I personally got 87  (this is all before remark)

You got all this from Mr. White, of course?

Pathetic attempt to bring thread back on topic: I'll put those inequalities up on the thread, I promise.

yeah of course...  if our school doesnt win euclid this time for a three-in-a-row, it'll all be my fault :(  we sit at 96+88+87=271     I hate it when I mess things up for myself, but its even more unforgivable to do it to the whole school...

what inequalities are you talking about Osiris?  ps I've got a list of inequalities that I've made along with some other notes (okay it sucks but still I made it myself) but I dont have a place to get it uploaded... and I am too lazy to do it now using latex...  guess it will be useless after usamo...

[quote="bugzpodder"]i tried attach it here, but it is 109K which is above the maximum capacity (100k) after i zip it... sorry guys
try here: http://www.geocities.com/bugz_podder/mathnotes.html

just dont make fun of it too much... its only my first attempt[/quote]

I read through it, nice job!

Wow Osiris you didn't make it? From what i've read of your posts you definitely seem good enough to at least qualify for USAMO.

I'm in Canada and didn't get to write AMC or AIME. Teachers wouldn't budge. ~_~


Find natural numbers n iff  1!+2!+...+n! is a perfect square.

[hide="Solution"]For $n\geqslant 7$ we have $\sum_{k=1}^{n}k!\equiv 1!+2!+3!+4!+5!+6!\pmod{7}$ and $1!+2!+3!+4!+5!+6!=873\equiv 5\pmod{7}$, but $5$ isn't a quadratic residue modulo $7$. Checking $1\leqslant n\leqslant6$ by hand, we find $n\in\{1,3\}$.[/hide]

[hide="easy"]For $n\ge 4$ this sum has $3$ as last digit.[/hide]


AMC
AIME
AMC 12
AMC 10
MATHCOUNTS
email
geometry

Is anybody interested in a AMC 12/ AIME competition?
EDIT: This competition will held after the AMC 10/ AIME competition held by AwesomeToad. I just wanted to make that clear.

I guess I'll help.

[hide="participant 0"]
[/hide]

[hide="helpers 1"]
Lord.of.AMC
[/hide]

This is for when AwesomeToad's competition is over. This thread was just to see if anybody was interested in these tests. As I said earlier, the AMC 12 test will be released in late January and in the AIME early March. Also, were doing AMC 12, not 10.

If you are not comfortable with us starting to write the tests now or opening this thread,I'll ask a moderator to lock this thread.

I don't mind joining, although I'll fail
/JOIN

Wait, I thought you were only doing AMC 10 and AIME. Anyway, we need some testwriters too. There is no way my friend and I can write problems past #10 on the AIME.

Also, please put the 'participants' and 'helpers' list in hide tags so when this thread gets larger we don't have to rummage through a ton of pages to find people's usernames.

[hide="Helpers 1"]
Lord.of.AMC
[/hide]

[hide="Participants 2"]
athunder
TheMan1998
[/hide]

I'll join  :lol:
Hmm, I'm in 3 mock competitions right now.

i'll be a helper(although i will just end up writing bunch of combinatoric problems)
[hide="Helpers 2"]
Lord.of.AMC
15!!!
[/hide]

[hide="Participants 3"]
athunder
TheMan1998
AIME15USAMO
[/hide]

[hide="Helpers 2"]
Lord.of.AMC
15!!!
[/hide]

[hide="Participants 4"]
athunder
TheMan1998
AIME15USAMO
PowerOfPi
[/hide]

[quote="MathTwo"]Alright. Please PM me with your problems[/quote]
can we put problems on google doc(just like AwesomToad did). it's more easier that way.

also, we are making AMC12 problems, right? not AMC10

What's "google doc"?

Anyway, yes, we are writing AMC 12 problems.

[hide="Helpers 2"]
Lord.of.AMC
15!!![/hide]

[hide="Participants 5"]
athunder
TheMan1998
AIME15USAMO
PowerOfPi
gaussintraining[/hide]

[hide="Helpers 2"]
Lord.of.AMC
15!!![/hide]

[hide="Participants 6"]
athunder
TheMan1998
AIME15USAMO
PowerOfPi
gaussintraining
bokagadha[/hide]

[quote="bokagadha"][hide="Helpers 2"]
Lord.of.AMC
15!!![/hide]

[hide="Participants 6"]
athunder
TheMan1998
AIME15USAMO
PowerOfPi
gaussintraining
bokagadha
Ihatepie[/hide][/quote]

I throw my hat into the ring.

[hide="Helpers 2"]
Lord.of.AMC
15!!![/hide]

[hide="Participants 6"]
athunder
TheMan1998
AIME15USAMO
PowerOfPi
gaussintraining
bokagadha
stargroup[/hide]

Just like in the other thread, I will note that I can also make problems if the leader so desires. I'll just withdraw from participating and help out with problems.

You might be. It seems that I am second (according to Lord.of.AMC). And I got a 126.

Current top 3:
Brut3Forc3 - 130.5
Jason27603 - 126
simof - 120

lol, I never realized that.
I take WOOT, but I didn't make the problems #24 and #25.
The problems are a bit different, but they are indeed almost the same.
I'm surprised that I never knew that  :( (stupid me)

[quote="varunrocks"]Can I just have the test even though I am not participating
Please. Just p.m it to me. :)[/quote]
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=328510

[quote="Jason27603"]Yay! I'm second only to Brut3Forc3. My goals are accomplished :) Btw, are any of you question makers taking WOOT, because numbers 24 and 25 are almost straight off of WOOT practice AIMEs.[/quote]

Yeah, I wrote #24 and i submitted it to WOOT to be put on the practice AIME because I was too lazy to write new problems. :(

Can anybody tell me how this Mock AMC is related to the real AMC, score-wise?
Like, if you get maybe 100 on the Mock AMC, what would be a predicted score on the real AMC? It doesn't have to be specific, but I just want to compare them in terms of difficulty.

[quote="athunder"]How am i not first place?[/quote]
what did you get?(are you being serious or sarcastic?)

Ok people 3 days left. Good luck to all...

[size=25]@athunder: you are at the bottom of the list with a 30[/size]

The test results:
Mean: 102.6
Standard Deviation: 23.5889135422822

Top Scores:
1. ksun48 - 132
2. Brut3Forc3 - 130.5
3. Jason27603 - 126
4. simof - 120
5. zapi2007 - 118.5
6. edisonchew240 - 117
7. bokagadha - 112.5

I call shenanigans.  It's impossible to beat Brut3forc3  :P, unless of course, you're pythag.

Hmm, I skipped some problems because I thought none of the correct answer choices were on there.


limit
function
calculus
derivative
calculus computations

$\boxed{1}$ Kokushikan University 2005

Find the limit $\lim_{x\rightarrow-4}\frac{x^{3}+4x^{2}+2x+8}{x^{2}+x-12}.$

$\boxed{2}$ Kwansai University 2005

For a function $f(x)$, express $\lim_{h\rightarrow 0}\frac{f(a+3h)-f(a)}{h}$  in terms of $f'(a).$

$\boxed{3}$ Yamagata University/agriculture 1987

Find the values of $a,\ b$ for which both of two curves $y=x^{3}+ax$ and $y=bx^{2}+c$ pass through the point $(-1,\ 0)$ and have a common tangent line at this point.

[hide="(1)"]\begin{eqnarray*}\lim_{x\to-4}\frac{x^{3}+4x^{2}+2x+8}{x^{2}+x-12}&=& \lim_{x\to-4}\frac{x^{2}(x+4)+2(x+4)}{x^{2}+4x-3x-12}\\ &=& \lim_{x\to-4}\frac{(x+4)(x^{2}+2)}{x(x+4)-3(x+4)}\\ &=& \lim_{x\to-4}\frac{(x+4)(x^{2}+2)}{(x+4)(x-3)}\\ &=& \lim_{x\to-4}\frac{x^{2}+2}{x-3}\\ &=&-{18\over 7}\end{eqnarray*}

Or by L'Hospital:

\begin{eqnarray*}\lim_{x\to-4}\frac{x^{3}+4x^{2}+2x+8}{x^{2}+x-12}&=& \lim_{x\to-4}\frac{3x^{2}+8x+2}{2x+1}\\ &=&-{18\over 7}\end{eqnarray*}
[/hide]
[hide="(2)"]$\lim_{h\to 0}\frac{f(a+3h)-f(a)}{h}=3\lim_{h\to 0}\frac{f(a+3h)-f(a)}{3h}=3f'(a)$[/hide]

3. Please, review the text of the problem. You say the curves should "both pass through" the given point and have "common point" at that point - that's one and the same thing. Should it be "they have to have a common tangent" at that point?

[quote="Farenhajt"][hide="(3)"]$f(x)=x^{3}+ax, g(x)=bx^{2}+c$
$f'(x)=3x^{2}+a, g'(x)=2bx$

$f(-1)=0, g(-1)=0, f'(-1)=g'(-1)$

Those give

$\begin{cases}-1-a=0\\ b+c=0\\ 3+a=-2b\end{cases}\iff a=-1,b=-1,c=1$[/hide][/quote]That's a nice way to do it.  I done it slightly differently, but it's essentially the same thing.

[hide="my way"]Let the first curve be f(x) and the second one be g(x).

$f(x)=x^{3}+ax \\ f(-1) =-1-a = 0 \\ a=-1$

$g(x)=bx^{2}+c \\ g(-1)= b+c = 0 \\ b=-c$

$f'(x) = 3x^{2}+a \\ f'(x) = 3x^{2}-1 \\ f'(-1) = 3-1 = 2$

We are trying to find an identical line that is a tangent to g(x) at the same point.  They both go through the point (-1, 0) so we just need to set the derivatives equal to find a tangent with the same gradient at that point, that is

$g'(x) = 2bx\\ g'(-1) = f'(-1) =-2b = 2 \\ b =-1$

And we know that b=-c so c=1, giving the solutions $a=-1, b=-1, c=1$[/hide]

Here is a solution without using differentiation.

We can write that $x^{3}+ax-bx^{2}-c=(x+1)^{2}(x-c).$ :D

[quote="kunny"][quote="ZetaX"]Read http://www.mathlinks.ro/Forum/viewtopic.php?t=26438 ;)[/quote]

I have fogotten what you said. I'm sorry to take troubles.
I see to it that next time.

kunny[/quote]

[i]Moved from Pre-Olympiad to Calculus Computations and Tutorials.[/i]


[quote="Slizzel"]If $M$ is the coeficent of $x^2$ from the following exeresion $E(x)=(1+x)^3+(1+x)^4+...+(1+x)^{11}$, find $M$.

p.s. [i]can anybody give some methods in order to solve this kind of problems.[/i][/quote]

i think it should be 3C2 +4C2 +5C2 +...+11C2  :P

[hide]the method is to find out how many ways you can choose $x$ two times out of $(x+1)^n$, which, by binomial expansion is $\binom{n}{2}x^21^{n-2}$ so the coefficient is what shyong said.  the sum, by the hockeystick theorem is $\binom{12}{3}-\binom{2}{2}=\boxed{219}$[/hide]


A net force of 3,000N accelerates a car from rest to 36.0 km/h in 5.00s. What is the mass of the car?  What is the weight of the car?      Not a clue as how to work this.

There is a constant force acting on the car, so the acceleration it perceives is also constant.  We know the car's velocity at $ t\equal{}5s$ : $ 36km/h\equal{}\frac{10m}{s}$.
We have that $ a(t)\equal{}a_0$.
$ v(t)\equal{}a_0t\plus{}v_0\equal{}a_0t$.
But we know that when $ t\equal{}5s$, $ v(t)\equal{}v(5s)\equal{}a_0(5s)\equal{}\frac{10m}{s} \Leftrightarrow a_0\equal{}\frac{2m}{s^2}$.
Why did I bother to find the acceleration of the car?  Because you know the second law of Newton?  It states $ \sum F\equal{}ma$.  We know $ \sum F$ which is worth $ 3000N$ and we know $ a$.  So work out $ m$, which is the mass of the car.
To get the weight, just multiply the mass of the car by the acceleration of the gravitational force (not the acceleration of the car!), that is by $ \frac{9.8m}{s^2}$.
Good luck.

hey u could do like this:
 [geogebra]fb1b96fa4e71760824c18c5dcb7a938599920a9f[/geogebra]
conversion of the speed into $ m/s$. we get $ 36  *  5  /  18  \equal{}  10 m/s$
after that
use the formula $ a  \equal{} v\minus{}u /t$
we get $ 10/5  \equal{} 2  m/s^2$
now we use the formula $ F \equal{}MA$
f=3000 n
m=?
a=2 m/s^2
m= 1500 kg
and weight = $ m*g$
so weight = 1500*9.8 n


Here's a cute one. I found it in the problem set in Moubinool's last post in the "Math Links" section:

Prove that there are no rings A (which have 1) s.t. A has exactly 5 units (invertible elements).

U(A) has the order 5, -1 in U(A) so (-1)5=1=>-1=1=>1+1=0

For every a, b that commute we have (a+b)2=a2+2ab+b2=a2+b2

U(A) has order 5 so it is cyclic. Let y be one generator of U(A). Clearly y <>1

Then (1+y+y2)2=1+y2+y4
So (1+y+y2)3=(1+y2+y4)(1+y+y2)=...=y3 (check it! :) )

The equation x3=y3 has only one solution x=y because x has to be invertible and we check the values 1, y, y2, y3, y4 (try them!  :) )

But 1+y+y2=y yields y2=1 and also y5=1 so y=1, which is absurd.

So there is no such ring. :)

Thanks for the solution! It's been here for quite some time, hasn't it?:)

related:
[url]https://artofproblemsolving.com/community/c7h61995[/url]
[url]https://artofproblemsolving.com/community/c7h276902[/url]
[url]https://artofproblemsolving.com/community/c7h292862[/url] (26 units)
[url]https://artofproblemsolving.com/community/c7h2010746[/url]
[url]https://artofproblemsolving.com/community/c7h75559[/url] (commutative ring)
[url]https://artofproblemsolving.com/community/c7h1354[/url]
[url]https://math.stackexchange.com/questions/493741/characteristic-of-a-finite-ring-with-34-units[/url] (34 units)
[url]https://artofproblemsolving.com/community/c7h204191[/url] (regular elements instead of units)
[url]https://math.stackexchange.com/questions/384362/group-of-invertible-elements-of-a-ring-has-never-order-5[/url]


Usually I am an "only math" person, but when I'm not doing math, I do chess and ping pong.

[quote="joml88"]Hmm... no baseball(as yifman said) or football.  They are predominantly American sports so I m not too shocked that they aren't on here.  I guess my favorite sport is cross country (running).  It used to be baseball but stopped playing that once I became a runner.[/quote]

I know that its an american sport, but (no offense meant towards anyone who plays it) ultimate frisbee isn't exactly an international sport either  :lol:

chess and soccer, i didn't see chess because i didn't think it is a sport.

[quote="MCrawford"][quote="BHorseMath"]Horseback Riding.  If you call it a sport, I do.  Other wise Crosscountry running or distance running.[/quote]

Have you ever considered training for the pentathlon?  I have a friend who does that, though she hasn't yet qualified for the major international competitions.[/quote]

I have considered training for it, but when I tried it the stadium show jumping was really low and I jump higher even now, and the other events particularly the shooting event did not intrest me that much so I decided to stick to 3-day event horseback riding which consists of dressage on day one, show jumping of day 2, and crosscountry jumping on day 3.  All of those being equestrian events that I am fairly good at.

What about [b][i]REAL[/i][/b] tennis?  (No offense to table tennis intended, even though I like the name "ping-pong" more.  It seems more fitting somehow)

Actually I think that Ping-pong and table tennis are two slightly different things.  I believe that the rules are a little different.  Other than that they are identical.  If I am bored enough I will look into that later.

[quote="yif man12"]I know that its an american sport, but (no offense meant towards anyone who plays it) ultimate frisbee isn't exactly an international sport either  :lol:[/quote] I really must disagree ... me and my friends play frisbee all the time in Romania. The facts that we cannot find disks for sale here :) doesn't mean the sport is not played world-wide :)

[quote="yif man12"]I know that its an american sport, but (no offense meant towards anyone who plays it) ultimate frisbee isn't exactly an international sport either  :lol:[/quote]

Quite the [url=http://www.upa.org/]contrary[/url]...

Didn't they consider checkers as a sport?

Now since the level of skill of chess > checkers, chess should definately count as a sport.

I find it hard to believe that Ihatepie has a free throw average of 100%, when Yao Ming has only around 75% I think.

[quote="serialk11r"]I find it hard to believe that Ihatepie has a free throw average of 100%.[/quote]

With all due respect to Ihatepie's skill, you have to take in consideration how many free throw's he actually attempted  :wink: .

1. I meant Nearly 100%. (Around 80)

2. I do it on my own hoop which is about 8 ft tall, not 10.

3. The spot I marked for free throw is a bit closer than usual.

4. This isn't my free throw percentage in games, so I don't have much pressure on me. (In games it's more like 65.)

Frisbee and soccer.  It's just so much fun to throw a disc around...and soccer is awesome and more fast-paced (usually)

I used to run track, but I can't understand how people could enjoy running.  Sure, it was sometimes fun afterwards, but there's no way I could pass up frisbee or soccer for running.

80% isn't that pro with no one watching. From where I practice at my house (nearly 10 feet), I can do a good 70%, maybe more if I aim.

I play basketball but we just started Ultimate Frisbee in P.E. and my team made this awesome strategy so we won every game.  We just started it today when we were losing 4-1 with 3 minutes left.  We won 5-4.  We have 5 players, but technically we have 3 because the two girls hate Ultimate Frisbee.  One of them is afraid of moving objects and the other doesn't really like sports (or it looks like it).

Oh! I like soccer!

My favourite is Manchester United!

Are you fan of MU?  :D  :D

Lol when I saw MU I automatically thought of Missouri University. I hate soccer but I like Missouri. Go TIGERS!

BTW...this thread has been brought back to life like 6 times after it was considered "dead"

BASKETBALL FTW!

it's the awesome guy asian sport :]

Rowing, anyone?

I feel like a loner on this one :maybe: I wonder how many rowers there are on AoPS

Why. Just why. You necro'd a 6 year dead thread.

Where the heck is baseball? (Just saying since it's been bumped). I voted ping pong (lol)

But baseball is not primarily an American sport. Well it is, I guess, but it's not JUST an American sport, it's played all over the world!


inequalities
calculus
integration
inequalities solved

1. Let $0 \leq b_k \leq 1$ for all $k$ and
$a_1 \geq a_2 \geq \dots \geq a_{n+1} = 0$.
Prove an inequality
\[ \sum_{k=1}^n a_k b_k \leq
  \sum \limits_{k=1}^{\left[ \sum_{m=1}^n b_k \right] + 1} a_k .\]

2. Let $p > 1$, $a_1 \geq a_2 \geq \dots \geq a_{n+1} = 0$, $b_k \geq 0$ and
$b_1^q + b_2^q + \dots + b_n^q \leq 1$ ($1/p+1/q=1$).
Prove an inequality
\[ \left( \sum_{k=1}^n a_k b_k \right)^p \leq \sum_{k=1}^m a_k^p ,\]
with $m = \left[ \left( \sum_{k=1}^n b_k \right)^p \right] + 1$.

The first inequality I gave at Saint Petersburg MO, 1996 and I know many
solutions of this problems. The second is a generalization of the first,
but I can prove integral analog only.

These are indeed beautiful inequalities.
I have read the 2 nice solutions in "Петербургские математические Олимпиады". However, that was after solving the problem myself.

My solution:
We keep $\sum_{k=1}^{n}b_{k}$ constant, that is, RHS - constant. We'll "redistribute" $b_{k}$ to obtain $\sum_{k=1}^{n}a_{k}b_{k}$ maximal.
It is easy to see that we'll have to set all the coefficients of $a_{k}$ equal to 1, as much as possible. That is - for $k$ from 1 to $\left[\sum_{k=1}^{n}b_{k}\right]$, because $a_{k}$ is decreasing. $a_{[\sum_{k=1}^{n}b_{k}]+1}$ will get "what is left": $0\leq\{\sum_{k=1}^{n}b_{k}\}<1$, while other $a_{i}$'s [smaller ones] get 0.
So $\sum_{k=1}^{n}a_{k}b_{k}\leq\sum_{k=1}^{[\sum_{m=1}^{n}b_{m}]}a_{k}+\left({[\sum_{k=1}^{n}b_{k}]\}\right)a_{(1+[\sum_{k=1}^{n}b_{k}])}\leq\sum_{k=1}^{}a_{k}}$.


Hello everyone,

I am having trouble with the following problem. I have written some of my work and thoughts on this question out.

Thank you very much!
---

1. A polynomial F, when divided by $ (x \minus{} 1)$, gives the remainder $ R_1 \equal{} 2$. The same polynomial F, when divided by $ (x \plus{} 1)$ gives the remainder $ R_2 \equal{} 4$. Find the remainder when F is divided by $ (x^2 \minus{} 1)$. (Hint: Write the division statement when F is divided by $ (x^2 \minus{} 1)$ )

[u]My work:[/u]

The division statement: $ Q(x) \equal{} F(x^2 \minus{} 1)$
$ F/(x \minus{} 1) \equal{} Q(x) \plus{} 2$
$ F/(x \plus{} 1) \equal{} Q(x) \plus{} 4$

[hide]$ f(x) \equal{} (x \minus{} 1)Q(x) \plus{} 2$
$ f(x) \equal{} (x \plus{} 1)P(x) \plus{} 4$
$ f(x) \equal{} (x \minus{} 1)(x \plus{} 1)R(x) \plus{} ax \plus{} b$

plugging in for x=1, -1

$ a \plus{} b \equal{} 2$
$ \minus{} a \plus{} b \equal{} 4$

$ a \equal{} \minus{} 1, b \equal{} 3$

the remainder is $ \boxed{\minus{}x \plus{} 3}$[/hide]

[quote="unimpossible"][hide]$ f(x) \equal{} (x \minus{} 1)Q(x) \plus{} 2$
$ f(x) \equal{} (x \plus{} 1)P(x) \plus{} 4$
$ f(x) \equal{} (x \minus{} 1)(x \plus{} 1)R(x) \plus{} ax \plus{} b$

plugging in for x=1, -1

$ a \plus{} b \equal{} 2$
$ \minus{} a \plus{} b \equal{} 4$

$ a \equal{} \minus{} 1, b \equal{} 3$

the remainder is $ \boxed{ \minus{} x \plus{} 3}$[/hide][/quote]

Thank you very much for your reply.

It's probably since it is night time here, but could you please explain how you got from the three division statements to the system of two equations?

How did the system of two equations come about after you plugged in x values into the division statements?

Thanks again!

[quote="unimpossible"][hide]$ f(x) \equal{} (x \minus{} 1)Q(x) \plus{} 2$
$ f(x) \equal{} (x \plus{} 1)P(x) \plus{} 4$
$ f(x) \equal{} (x \minus{} 1)(x \plus{} 1)R(x) \plus{} ax \plus{} b$

plugging in for x=1, -1

$ a \plus{} b \equal{} 2$
$ \minus{} a \plus{} b \equal{} 4$

$ a \equal{} \minus{} 1, b \equal{} 3$

the remainder is $ \boxed{ \minus{} x \plus{} 3}$[/hide][/quote]

Now that he has three, he plugs in $ x\equal{}1$ and results in the following:

$ f(1)\equal{}2$
$ f(1)\equal{}2P(1)\plus{}4$
$ f(1)\equal{}(2)(0)R(1)(a(1)\plus{}b)$

So, this becomes: $ a\plus{}b\equal{}2$, we ignore the second equation, it doesn't tell us anything useful.

Then, doing the same with $ x\equal{}\minus{}1$
$ f(\minus{}1)\equal{}\minus{}2Q(\minus{}1)\plus{}2$
$ f(\minus{}1)\equal{}(0)P(\minus{}1)\plus{}4$
$ f(\minus{}1)\equal{}(0)(\minus{}2)R(\minus{}1)\plus{}a(\minus{}1)\plus{}b$

This becomes $ b\minus{}a\equal{}4$

Now we just solve for the two systems.

[quote="catalan"]could you please explain how you got from the three division statements to the system of two equations?

How did the system of two equations come about after you plugged in x values into the division statements?

Thanks again![/quote]

i got both from the last division statement. because when either x=1 or -1 the R(x) is muliplied by 0, So you're just left with the ax+b part.

well, also you'll need to notice that f(x)=f(x)   :wink:


MIT
college
geometry
geometric transformation
pigeonhole principle
combinatorics unsolved
combinatorics

To $100$ monkeys we give $1600$ bananas , but perhaps some of them will eat none.Prove that at least  $4$ monkeys will eat the same number of bananas.

 [i]I please Darij - or german Mlinkers for a better translation. I can understand the german text but I can not transfer it exactly into english![/i]

 [i]<< Auf 100 Affen werden 1600 Kokosnüsse verteilt, wobei einige Affen auch leer ausgehen können.
Man beweise, dass es – ganz gleich, wie die Verteilung erfolgt – stets mindestens vier Affen mit derselben Anzahl von Kokosnüssen gibt. >>[/i]
[u]Babis[/u]

Correct translation (but yours is understandable and gives the same problem):

We give $1600$ coconuts (not bananas ;) ) to $100$ monkeys. It's possible that some monkeys get none. Prove that there are $4$ monkeys that get the same amount of coconuts.

Ok, now ZetaX has done the hard part (translation). We can move this thread to pre-olympiad then  :P

i think this problem is easy but very useful. :D

Thanks to Zetax.Of course Banana was not the word for  << Kokosnu''sse >> but I ought to find something like it!!!

   a) Everybody ( - 1)  finds the problem easy but I have not yet  seen a solution !

   b) It is better to write a solution to an easy problem , than not to write a solution to a difficult problem !
   c) Of course , good hummor is acceptable in this forum , solutions (and) to easy problems  is more acceptable  , as well !

    Well , I ' m still waiting.

[u]Babis[/u]

Here is my silly idea:
if there no 4 monkeys eat the same number of banana,there would be at least 34 monkeys which eat different number of banana,or else by pigeonhole principle,there will be 4 monkeys eat the same number of banana, contradiction.
To minimize the total number of bananas eaten,let those "number" be 0,1,2,3,...,32,33
So the minimum number of banana $\geq (0+1+2+...+32)*3+33=1617>1600$,contradiction
Therefore,at least 4 monkeys will eat the same number of bananas.

We try to make at most $3$ monkeys having same number of coconuts . First we group three monkeys with the same amount of coconut in one group . That left one monkey with without group . Then for each group , the following coconuts are given , $0,1,2,3...,32$ hence the total coconuts given are $3\cdot \frac{32\cdot 33}{2}=1584$ . This means the last monkey can only recieve $16$ which makes it $4$ monkeys with same number of coconuts . Now , if we withdraw some coconuts from the $k th$ group's monkeys , then one of the three monkeys must have less than $k-1$ coconuts (this will make the number of coconuts of that monkey same as any  $i th$group where $0\leq i<k$ . This still gives us at least $4$ monkeys with same number of coconuts .


inequalities
number theory solved
number theory

well , well.... I enjoy your problems(VERY VERY SIMPLE)...

here is a solution:

The equation can be written like so:
y(x-5)=-3(x+1) <=> y=-3* (x+1)/(x-5)=-3 - 18/(x-5) so x-5 | 18.
we take all the divizors of 18 and we have the solutions.. :D

this one was really simple! You should try harder and then post it here if you really don't know... :D

THE END.

You think they are pretty simple, don't u?!  Well Lagrangia if u are so smart solve this one:

Solve in natural numbers the equation:
7^x+x^4+47=y^2.

 :D  :D  :D  :D

well... well.. A math duel ... wait a couple of minutes and I hope I will solve it.....   :D

I hope I will solve it....  :)

well... EVRIKA!

if x=2k , k is natural <>0. For k>=4 we have:

(7^k)^2<7^(2k)+(2k)^4+47<(7^k+1)^2 . The two inequalities are easy to be proven the first one is obvious and the second one it can be proven using induction. So we  have three cases to check when k is 1 or 2 or 3. Checking them we get natural solutions for k=2 with the solutions: x=4 and y=52.

as for x=2k+1 k natural we have no natural solutions. why? because then we have 7^x+x^4+47==3(mod 4) but there are no squares like this ... so no solutions....

so the only natural solution is : x=4 and y=52...
hope I am right......

:)


[quote="Treething"]Two cubes of equal volume intersect so that their common area is a third cube. This third cube's side length is three times as short as the side length of one of the two larger cubes, and this third cube has a volume of $x$.

Find the total volume of the figure expressed in terms of $x$.[/quote]
[hide]
the third cube has 1/27 the volume of the larger cubes. one larger cube has a volume of $27x$. two larger cubes minus one third cube is $54x-x$, so the answer is $53x$
[/hide]

[quote="sonny"][quote="Quevvy"][quote="Treething"]Two cubes of equal volume intersect so that their common area is a third cube. This third cube's side length is three times as short as the side length of one of the two larger cubes, and this third cube has a volume of $x$.

Find the total volume of the figure expressed in terms of $x$.[/quote]
[hide]Since the area if $x$ then each side is $\sqrt[3]{x}$ Each side of the original cube is three times that length so then each side is $3\sqrt[3]{x}$.

The volume is then $(3\sqrt[3]{x})^{3}=\boxed{27x}$[/hide][/quote]

The question asks for the volume of the whole figure.[/quote]
Ah, yes......
[hide]So then the 2 cubes would be 54x but then you subtract the overlap which is x to get 53x.[/hide]

[quote="Treething"]Two cubes of equal volume intersect so that their common area is a third cube. This third cube's side length is three times as short as the side length of one of the two larger cubes, and this third cube has a volume of $x$.

Find the total volume of the figure expressed in terms of $x$.[/quote]
[hide="Answer"]
The larger cube's side is three times the smaller one's side. That would make the larger cube's volume 27x. By PIE, you have to subtract x since you are counting it twice at the intersection.
27x + 27x - x = [b]53x[/b][/hide]

to use latex, you put a dollar sign around a certain string of numbers or text or anything.
like this:
[code]
$53x$
[/code]
shows up as $53x$

[code]
$23^{34}$
[/code]
shows up as $23^{34}$

[hide="Solution"]
Say that the side of the smaller square is $y$. We then know that the big cubes' side length would be $3y$. The total area of the big squrare would be $27y^{3}$. Therefore, the area of two squares would be $54y^{3}$. However, we have to subtract the intersection area, or $y$. Since the side of the smaller square is $y$ and the volume is $x$, $x=y^{3}$.

Therefore, the answer would be $54y^{3}-y^{3}=53y^{3}\implies\boxed{53x}$.[/hide]


geometry
conics
ellipse
videos
search
geometric transformation
AMC

Wind pummeled the plane, which swerved. Sweating, the pilot yelled into the radio, "We're going to crash! There's too much turbulence!"

"Hang in there," came the reply. "Are there any bodies of water nearby?"

The pilot anxiously answered, "Just the Pacific, and that's not close enough."

Suddenly, the plane flipped over, and the passengers screamed in terror. The aircraft dived towards the jungle. "No!" The pilot's final cry seemed to echo in the cabin, and then all was darkness.

Welcome to the Race to Survival! 15 Challengers will face a series of challenges they must perform. Not all Challengers will have the same ones, especially as the game progresses. The rules are simple: you may talk outside the thread, but you may not post if you die.

Sign-ups begin now.

1. AIME15
2. r15s11z55y89w21
3. Garyzx
4. bubka
5. budi713
6. funtwo
7. eggylv999
8. 1=2
9. Bugi
10. BOGTRO
11. nt811581147
12. mz94
13. hedgehog12
14. ernie
15. scai1

Lol, sounds childish, but to relive the old AoPS times, I'll join!

May as well, although I'll likely die in the first challenge.

[quote="dragon96"]So what does the pilot have to do with this?

I be assistant, but if you don't need help, I'll join.[/quote]

An assistant would be awesome. :)

[quote="bubka"][...]sounds childish[...][/quote]

Childish, eh?  :D

i'm in! and since i know you in real life i can easily persuade you to let me win

:P

jk

I join.
Random words to make message larger.

A dog, Apart lull? Ew! A well ultra pagoda!

Here, but you can't call me alive because I'm a zombie. :D

im here. however...due to my current presence in china accompanied by the WORST comp in the history of existing...i will do my best to participate i guess.

I am alive, (surprised myself!) I actually finished 6th grade...WHEW...last day of school today...:) I'm going on a cruise starting on June 28th and ending July 4th-5th I don't know, so...yeah.

Srrry.. im alive, but i was at a residential camp. i forgot to tell u guys. srry again.  :wink:

not rlly. there are a lot more ppl that have not replied.

I would love to say yes, but I really do wonder which game you're referring to? :P This one died, because only 4-5 people wanted to play at the end.  :(

Aww...   :(  Oh well.   Thanks...  :P   I read some of the thread, and it looked pretty cool, but if it's dead....  well....   :ninja:


$ a,b,c$ are sides of a triangle such that $ a \plus{} b \plus{} c \equal{} 2$. Find all values of the variable $ x$ such that
                        $ (ab)^x \plus{} (bc)^x \plus{} (ca)^x \leq 1$


calculus
integration
limit
trigonometry
calculus computations

Evaluate

\[\lim_{n\to\infty} \int_0^{\pi} x^2 |\sin nx| dx\]

$\lim_{n\to\infty}\int_0^\pi x^2|\sin(nx)| \ dx$

$= \lim_{n\to\infty} \sum_{j=0}^{n-1} \int_{j\left(\frac{\pi}{n}\right)}^{(j+1)\left(\frac{\pi}{n}\right)}(-1)^jx^2\sin(nx) \ dx$

$= \lim_{n\to\infty}\frac{2\pi^2n^2+\pi^2-12}{3n^2}$

$=\frac{2\pi^2}{3}$

Edit:  I changed the $i$s to $j$s since I was just using it as a counter variable and not $\sqrt{-1}$.

Jimmy, can you show me how you got to step two and your reasoning? Id like to understand your methodology.

To get to the summation, I looked at a graph at broke the graph up into $n$ pieces to eliminate the absolute value.

Easy example:

$\int_0^\pi x^2|\sin(2x)| \ dx = \int_0^{\frac{\pi}{2}} x^2\sin(2x) \ dx - \int_{\frac{\pi}{2}}^{\pi} x^2\sin(2x) \ dx$

thanks jimmy, that helped to clarify it for me

See this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=247669#247669]new topic[/url].

Kunny, do you have a "nicer" solution because mine is pretty ugly.

Yes, but the way of thinking is the same as yours, Jimmy.

Here is my solution.

$\lim_{n\to\infty} \int_0^{\pi} x^2 |\sin nx|dx$ (Let $nx=t$)

$=\lim_{n\to\infty}\frac{1}{n^3} \lim_{n\to\infty} \int_0^{n\pi} t^2 |\sin t|dt$

$=\lim_{n\to\infty}\frac{1}{n^3} \sum_{k=0}^{n-1} \int_{k\pi}^{(k+1)\pi} t^2|\sin t|dt$ (Let $t-k\pi=u$)

$=\lim_{n\to\infty}\frac{1}{n^3} \sum_{k=0}^{n-1} \int_0^{\pi} (u+k\pi)^2 |(-1)^k \sin u|du$

$=\lim_{n\to\infty}\frac{1}{n^3} \sum_{k=0}^{n-1} \int_0^{\pi} (u+k\pi)^2 \sin u\ du$

$=\lim_{n\to\infty}\frac{1}{n^3} \sum_{k=0}^{n-1} \left[-(u+k\pi)^2 \cos u +2(u+k\pi)\sin u +2\cos u\right]_0^{\pi}$

$=\lim_{n\to\infty}\frac{1}{n^3} \sum_{k=0}^{n-1} (2\pi ^2 k^2 +2\pi ^ 2 k+\pi ^ 2-4)$

$=\lim_{n\to\infty}\left[\frac{2\pi ^ 2}{n^3} \sum_{k=0}^{n-1} k(k+1)+\frac{1}{n^3}(\pi ^ 2 -4)\cdot n\right]$

$=\lim_{n\to\infty}\left[\frac{2\pi ^ 2}{n^3} \frac{(n-1)n(n+1)}{3}+\frac{1}{n^2}(\pi ^ 2 -4)\right]$

$=\frac{2}{3} \pi ^ 2$

P.S.

The posting by Kent Merryfield is very interesting, Math Links' everyone.

The key word is the average of the ....

kunny


Let ABCD be a quadrilateral inscribed in a circle with center O. Let lines AB and DC intersect at a point P. Line PO intersects AC and BD at E and F respectively let Q be a point such that QE is perpendicular to AC and QF is perpendicular to BD. Prove that triangle QAC and QBD have equal areas.

let $G$ be the point that $AD\parallel CG$ and $DG\parallel AC$
let the intersection of $AC$ and $BD$ be $R$

1. let line $AO$ intersect the circle$O$ at $L$ , line $DO$ intersect the circle $O$ at $K$, so $KL\parallel=AD$.
   $M$ is the intersection of  $BK$ and $CL$
   consider the six points $A,B,K,L,C,D$ we use $Pascal$ theorem to show that $P,O,M$ collinear.

2. let $T$ be the point that $TB\bot DB$ and $TG\bot DG$,   $M$ be the point that $MB\bot DB$ and $MC\bot AC$.
   it is clear that $B,K,M,T$ are collinear and $M,L,C$ are collinear
   let line $KL$ intersect line $TG$ at $N$
   because $LN\parallel AD\parallel CG$ and $TG\parallel MC$, $LN=CG=AD=KL$. therefore $M$ is the midpoint of $KT$.
   consider $\triangle KTD$  $M,O$ are the midpoints of $KT$ and $KD$, so $TD\parallel MO\parallel PO$.

3. it is easy to show that $Q, F, E, R$ cyclic, and $D,B,T,G$ cyclic.
   from $TD\parallel PO$ , we get $\angle BDT=\angle EFR$
   so $\angle BGD=\angle BTD=\pi/2-\angle BDT=\pi/2-\angle EFR=\angle QFE$,
   use almost the same argument, we get $\angle GBD=\angle QEF$.
   hence we have $\triangle QEF\sim\triangle DBG$
   then $\displaystyle \frac{QE}{QF}=\frac{BD}{DG}=\frac{BD}{AC} \Rightarrow {AC}\cdot{QE}={BD}\cdot{QF}$
   therefore the area of $\triangle AQC$ = the area of $\triangle BQD$
[img]sol.JPG[/img]


[quote="manlio"]
\[ \ \frac {1}{a^2 \plus{} 2bc} \plus{} \frac {1}{b^2 \plus{} 2ca} \plus{} \frac {1}{c^2 \plus{} 2ab} \geq \frac {ab \plus{} bc \plus{} ca}{a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2}\]

\[ \sum_{cyc}\frac1{a^2 \plus{} bc}\ge\sum_{cyc}\frac1{a^2 \plus{} 2bc} \plus{} \frac {ab \plus{} bc \plus{} ca}{2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)}\]
P.S. I would like to see a simple Cauchy-Schwarz solution I cannot find.  :blush:
      Thank you very much[/quote]

$ \frac {1}{a^2 \plus{} 2bc} \plus{} \frac {1}{b^2 \plus{} 2ca} \plus{} \frac {1}{c^2 \plus{} 2ab} \geq \frac {ab \plus{} bc \plus{} ca}{a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2}$

is equivalent to

$ (a\plus{}b\plus{}c)(2(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2\plus{}3\sum{b^2c^2(a\minus{}b)(a\minus{}c)})\geq 0$

easy to prove   $ \sum{b^2c^2(a\minus{}b)(a\minus{}c)}\geq 0$  by schur

Dear hedeng123,

your proof is by full expansion. I would like to see a proof by only AM-GM, SHUR and BCS.

However thank you very much for your reply. :)

[quote="manlio"]
\[ \ \frac {1}{a^2 \plus{} 2bc} \plus{} \frac {1}{b^2 \plus{} 2ca} \plus{} \frac {1}{c^2 \plus{} 2ab} \geq \frac {ab \plus{} bc \plus{} ca}{a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2}\]

[/quote]

By AM-GM we have : $ a^2b^2\plus{}c^2a^2\plus{}4b^2c^2 \geq 2a^2bc\plus{}4b^2c^2 \equal{}2bc(a^2\plus{}2bc)$

$ \Leftrightarrow \sum \frac{1}{a^2\plus{}2bc} \geq \sum \frac{2bc}{a^2b^2\plus{}4b^2c^2\plus{}c^2a^2}$

Setting : $ x\equal{}ab \ ; \ y \equal{} bc \ ; \ z\equal{}ca$

We have to prove that :

$ \sum \frac{2x}{x^2\plus{}4y^2\plus{}z^2} \geq \frac{x\plus{}y\plus{}z}{x^2\plus{}y^2\plus{}z^2}$

By Cauchy Schwarz we have :

$ \sum \frac{2x}{x^2\plus{}4y^2\plus{}z^2} \geq \frac{2(x\plus{}y\plus{}z)^2}{\sum x^3\plus{}4xy^2\plus{}z^2x}$

We have to prove that :

$ \frac{2(x\plus{}y\plus{}z)^2}{\sum x^3\plus{}4xy^2\plus{}z^2x} \geq \frac{x\plus{}y\plus{}z}{x^2\plus{}y^2\plus{}z^2}$

It is equivalent to :

$ x^3\plus{}y^3\plus{}z^3\plus{}x^2y\plus{}y^2z\plus{}z^2x \geq 2xy^2\plus{}2yz^2\plus{}2zx^2$

It is true and we have done ! :)

Very nice your proof, Materazzi. :)

Have you also a proof for the second inequality?

Thank you very much.

[quote="dduclam"][quote="manlio"]Dear dduclam,

did you prove these very nice inequalities by SOS or have you a neat proof by Cauchy-Schwarz?

Thank you very much. :)
\[ \ \frac {1}{a^2 \plus{} 2bc} \plus{} \frac {1}{b^2 \plus{} 2ca} \plus{} \frac {1}{c^2 \plus{} 2ab} \geq \frac {ab \plus{} bc \plus{} ca}{a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2}\]
:)
\[ \sum_{cyc}\frac1{a^2 \plus{} bc}\ge\sum_{cyc}\frac1{a^2 \plus{} 2bc} \plus{} \frac {ab \plus{} bc \plus{} ca}{2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)}\]
For the first inequality I found this solution on VIMF:
http://ddbdt.co.cc/forum/showthread.php?t=238

It is nice but it uses SOS method, so I would like to know if you have a solution by only BCS.

Thank you very much[/quote]

Hint for you, manlio:
\[ 2\sum_{cyc}\frac {a^2(b^2 \minus{} bc \plus{} c^2)}{a^2 \plus{} bc}\ge\sum_{cyc}\frac {a^2(2b^2 \minus{} bc \plus{} 2c^2)}{a^2 \plus{} 2bc}\ge ab \plus{} bc \plus{} ca\]
[/quote]

[quote="Materazzi"][quote="manlio"]
\[ \ \frac {1}{a^2 \plus{} 2bc} \plus{} \frac {1}{b^2 \plus{} 2ca} \plus{} \frac {1}{c^2 \plus{} 2ab} \geq \frac {ab \plus{} bc \plus{} ca}{a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2}\]
[/quote]

By AM-GM we have : $ a^2b^2 \plus{} c^2a^2 \plus{} 4b^2c^2 \geq 2a^2bc \plus{} 4b^2c^2 \equal{} 2bc(a^2 \plus{} 2bc)$

$ \Leftrightarrow \sum \frac {1}{a^2 \plus{} 2bc} \geq \sum \frac {2bc}{a^2b^2 \plus{} 4b^2c^2 \plus{} c^2a^2}$

Setting : $ x \equal{} ab \ ; \ y \equal{} bc \ ; \ z \equal{} ca$

We have to prove that :

$ \sum \frac {2x}{x^2 \plus{} 4y^2 \plus{} z^2} \geq \frac {x \plus{} y \plus{} z}{x^2 \plus{} y^2 \plus{} z^2}$

By Cauchy Schwarz we have :

$ \sum \frac {2x}{x^2 \plus{} 4y^2 \plus{} z^2} \geq \frac {2(x \plus{} y \plus{} z)^2}{\sum x^3 \plus{} 4xy^2 \plus{} z^2x}$

We have to prove that :

$ \frac {2(x \plus{} y \plus{} z)^2}{\sum x^3 \plus{} 4xy^2 \plus{} z^2x} \geq \frac {x \plus{} y \plus{} z}{x^2 \plus{} y^2 \plus{} z^2}$

It is equivalent to :

$ x^3 \plus{} y^3 \plus{} z^3 \plus{} x^2y \plus{} y^2z \plus{} z^2x \geq 2xy^2 \plus{} 2yz^2 \plus{} 2zx^2$

It is true and we have done ! :)[/quote]

Unfortunately there is a serious error. It must be $ y$ instead of $ x$ in the third line, it is symmetrical.

We would have to prove that :

$ \sum \frac {2y}{x^2 \plus{} 4y^2 \plus{} z^2} \geq \frac {x \plus{} y \plus{} z}{x^2 \plus{} y^2 \plus{} z^2}$

but that is false. :(

You are right, spanferkel :(

I was so happy to see a so simple solution I thought impossible that I didn't see the mistake. Very sorry. :(  :blush:

[quote="manlio"]
\[ \ \frac {1}{a^2 \plus{} 2bc} \plus{} \frac {1}{b^2 \plus{} 2ca} \plus{} \frac {1}{c^2 \plus{} 2ab} \geq \frac {ab \plus{} bc \plus{} ca}{a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2}\]
[/quote]
I think this inequality does not belong to Dduclam. I have seen Vasc posted it a long time ago. And here is my proof: http://canhang2007.wordpress.com/2009/11/24/inequality-78-v-cirtoaje/

[quote="manlio"]Very nice your proof, my dear friend. :)

I like it much more than SOS solution posted on VIMF:)

I would like to ask for you, please, if you have understood how to use dduclam's hint
\[ 2\sum_{cyc}\frac {a^2(b^2 \minus{} bc \plus{} c^2)}{a^2 \plus{} bc}\ge\sum_{cyc}\frac {a^2(2b^2 \minus{} bc \plus{} 2c^2)}{a^2 \plus{} 2bc}\ge ab \plus{} bc \plus{} ca\]
to prove
\[ \sum_{cyc}\frac1{a^2 \plus{} bc}\ge\sum_{cyc}\frac1{a^2 \plus{} 2bc} \plus{} \frac {ab \plus{} bc \plus{} ca}{2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)}\]
I really cannot how to prove dduclam's hint and how to use it, sorry :blush:
Thank you very much :)[/quote]
Sorry, but in my opinion, this inequality is not nice. And I haven't given it a try yet. I think it will be better for you to ask DDucLam or some other ones. Really sorry.

[quote="spanferkel"][quote="can_hang2007"]
Sorry, but in my opinion, this inequality is not nice. And I haven't given it a try yet.[/quote]
 :lol: Isn't that interesting that unlike others, some ineqs seem "nice" in a [i]quite [/i]objective sens - like, I would guess, all those of a form $ \sum f\ge \sum g$ where $ f$ and $ g$ are relatively simple cyclic or symmetric expressions?! E.g. the first one in manlio's original post.[/quote]
Sorry, but I don't understand what you meant. I think it is better to post a solution. See it here: http://canhang2007.wordpress.com/2009/11/26/inequality-84-d-d-lam/

My dear friend can_hang2007,

you are so clever to solve inequalities that I am really without words. :blush:

I would like to know which is ,in your opinion, the hardest inequality you have ever solved. I mean the inequality has required your greater mental effort. :maybe:

Thank you very much. :)


I dunno whether this has been posted before, but:

Find a perfect square that the numbers -5 and +5 relative to it are also perfect squares.

(ex. say 9 is a perfect square. 9-5 is 4, which is also a perfect square. but 9+5=14, which isnt.)

A perfect square x is defined to be one for which there is some integer n which, when squared, equals x. The square of an integer is always an integer. I think that JesusFreak is correct.

[quote="ToBe"]well... i saw the problem somewhere and the guy says theres an answer. i supose then he means that where $n^2$ (perfect square) $(n^2)-5/+5=(x/y)^2$
its supposed to be a fraction.[/quote]

A fraction  :huh: ? THat's going to take forever unless there's a way to solve this using equations.

This is a very well known problem, try it!  The  original problem is find a real number  $a$ such  that $\sqrt a , \sqrt{a+5}, \sqrt{a-5}$ all are rational.

Good luck.

[quote]i saw the problem somewhere and the guy says theres an answer.[/quote]

Guys are lucky .. becuase if instead of "$5$"  if it was, say 1 (or 2 or 3 or 4 :) ) there are no solutions ...

(There are techniques which  are fun, and can be used  to solve these kind of problems ... But see my  last line comment (abut moving this problem to  more advanced forum)

Now if you want to impress your "guys" say :

[hide][hide=""Of course, One of the answer is:" "]   $\frac{11183412793921^2}{2234116132416^2}$[/hide]

You can check it (may need a calculator :)

[hide](You need not  tell them that there are answers with smaller numbers  like  $x=\frac{41^2}{12^2}$
 note that $x+5 = (49^2/12^2)$ and $x-5 = (31^2/12^2)$ :) .... :) :) [/hide]
Legend has it that Frederick asked this question to Fibonacci, and Fibonacci  found these numbers..
For more discussion , please move this to "Intermediate Form"  -  This problem, IMO, belongs there.[/hide]

Moved to Intermediate by request of ToBe.

What are the tricks to solving this kind of problem?

With logic very similar to find Pythogorian triplets,  take the following three  numbers:

$(a^2-b^2-2ab)^2$
$(a^2+b^2)^2$
$(a^2-b^2+2ab)^2$

They are in airthmatic progression, with difference $(d) = 4ab(a^2-b^2)$

So all you have to  find $a,b$ such that   $d$ is  $5k^2$ for example, if we take $a=5,b=4$
we get the following numbers $31^2,41^2,49^2$ with $d=720 = 5 \cdot 12^2$ so we divide by $12^2$  and get the required result.

The other answer I posted, was when $a=41^2,b=720$  We can find as many more as we want.

I have a number theory book, it goes into this in quite a lot of detail... like finding all numers k such that $\sqrt{x}, \sqrt{x-k}, \sqrt{x+k}$ has solutions in the rationals.

Just number theory, i think. Ill get some details for you if you like.

Yes, i'd appreciate if you gave the name of the Author etc!!

FWIW - Just to add   - I hope, what I was pointing out was clear..
 For every $\sqrt{x-k} , \sqrt x , \sqrt{x+k}$ you have a  corrosponing  pythogorian triplet,

So if the pythogorian triplet   is $(a,b,c)$ (a^2+b^2=c^2)  it produces the sequence  $(a-b),c,(a+b)$ such that their  squares are in Airthmatic progression.

$(3,4,5)$ gives $1,5,7$ (and you indeed see that (1,25,49) has common difference of 12.

The  smaller answer to the above problem was from the pythogorian triplet $9,40,41$  and we get the answer ((40-9),41,(40+9))

So that common difference of $31^2, 41^2, 49^2$ is  $5*12^2$

Hope that helps

Number Theory and its history, by Oystein.

wow
such a beautiful problem  it has no integral solutions
the simplest solution according to me is let
a^2=(n-5) and b^2 as (n+5) subtract both the equations to get a^2-b^2 =-10 and hence forth by looking at the factors you see induvidual equations in two variables
all of them computed are rational fractions and not whole numbers as desired
[code] :roll: [/code]

$\pmod{4}$ kills this. Let the square be $m^2$. Since quadratic residues mod 4 are $0$ and $1$ we consider these cases:
$m^2 \equiv 0 \pmod{4}$
Then $m^2-5 \equiv 3 \pmod{4}$, which is impossible.

$m^2 \equiv 1 \pmod{4}$
Then $m^2+5 \equiv 2 \pmod{4}$, which is impossible.

Edit: Wait maybe not because the problem is kinda strange—idk because the guy who said it was talking about how it didn't have to be a whole number, but that's what a perfect square is?


geometry
3D geometry
tetrahedron
combinatorics proposed
combinatorics

$ n \minus{} 1$ planes are drawn parallel to one face of a regular tetrahedron so that they divide the altitude to that face into $ n$ equal segments, and the same is done for the other three faces.  Into how many regions do these planes divide the interior of the tetrahedron?

translated by JBL

That's not quite comprehensible; does it mean the following?

$ n \minus{} 1$ planes are drawn parallel to one face of a regular tetrahedron so that they divide the altitude to that face into $ n$ equal segments, and the same is done for the other three faces.  Into how many regions do these planes divide the interior of the tetrahedron?

If so, the first simplification I would make is to note the regularity of the tetrahedron is irrelevant: any affine transformation will preserve all the relevant features of the problem, so we might as well assume that our tetrahedron has vertices at $ (0, 0, 0), (0,0,n),(0,n,0)$ and $ (n,0,0)$.

Sorry for my poor English, thank you for translating the context, JBL
Japanese high school students don't usually study affin transformation except for IMO contestant.


In 1986,the greatest prime $2^{216091}-1$ was found.
Can you find the last four digits og that number?

[b]Solution.[/b] Let $x\equiv 2^{216091}-1$ and $y=x+1$.  Then we need to find $y\bmod 625$ and $y\bmod 16$.  $y\equiv 0\pmod{16}$.  Since $\varphi(625)=500$, $y\equiv 2^{216091\bmod 500}\equiv 2^{91}\equiv (2^{13})^{7}\equiv 67^{7}\equiv 114^{3}\cdot 67\equiv 323\pmod{625}$.  By the Chinese Remainder Theorem, $y\equiv 8448\pmod{10000}\implies x\equiv 8447\pmod{10000}$.  Therefore, the last four digits of $2^{216091}-1$ are $8447$.$\Box$

[quote="i_like_pie"][quote="sylow_theory"]Work modulo 1000.

Meaning,

$2^{216091}-1 \equiv x (\mod 1000)$

Where $0\leq x\leq 999$[/quote]
Modulo 1000 only gives 3 digits; the problem asks for 4.[/quote]

Whatever same idea.

You just transformed the problem into symbols.  :wink:

Apparently, the prime

$2^{32582657}-1$

Was discovered in $2006$.  Can we find the last four digits of this?  (It is harder.)  Or five?


algebra
polynomial
binomial theorem
combinatorics unsolved
combinatorics

I am afraid it is false. Put $y=-x^{2}$, then the identity becomes $\sum_{k=0}^{n}\binom{n-1+k}{k}y^{k}= (1+y)^{n}$ for arbitrary $y<0$. Assume that it is true. Then $\sum_{k=0}^{n}\binom{n-1+k}{k}y^{k}= \sum_{k=0}^{n}\binom{n}{k}y^{k}$. Both sides are polynomials of degree $n$ in the variable $y$ and they are equal for any $y<0$, so they must have equal coefficients. But that means $\binom{n-1+k}{k}= \binom{n}{k}$ which is obviously wrong.

thank you

but now i'm wondering if its true still

say i am looking for$x^{4}$ coefficient in $\ (1-x^{2}\ )^{3}$ problem

according to the formula it would be $\binom{n}{k}(-1)^{k}(x)^{2}k$

so $\binom{3}{4}(-1)^{4}x^{2*4}$ am i wrong ?


\[ \begin{array}{l} n \in N \\
a_n \equal{} n^2 \plus{} 20;d_n \equal{} GCD(a_n ,a_{n \plus{} 1} ) \\
\end{array}
\]
prove $ d_n |81$

$ HAPPY$ $ NEW$ $ YEAR!$


In the circle below, $ \overline{AB} \| \overline{CD}$. $ \overline{AD}$ is a diameter of the circle, and $ AD = 36^{\prime \prime}$. What is the number of inches in the length of $ \widehat{AB}$? Express your answer in terms of $ \pi$.
[asy]import olympiad; size(150); import geometry_dev; import graph; defaultpen(linewidth(0.8));
draw(Circle(origin,36));
pair A = 36*dir(130);
pair B = 36*dir(50);
pair D = 36*dir(-50);
pair C = 36*dir(-130);
draw(B--A--D--C);
label("$A$",A,W); label("$B$",B,E); label("$C$",C,W); label("$D$",D,E);
markangle("$50^{\circ}$",D,A,B,radius=10);[/asy]

Draw $ \overline{BC}$. Let the center of the circle be $ O$. Then $ \angle AOB\equal{}180\minus{}50\minus{}50\equal{}80$, and our answer is $ 36\pi\cdot\frac{80}{360}\equal{}8\pi$


modular arithmetic
number theory
greatest common divisor

What is the remainder when $ 5^{30}$ is divided by 7?

I know I've answered this exact one before, but...
$ 5^{30}\equiv(\minus{}2)^{30}\equiv(\minus{}8)^{10}\equiv(\minus{}1)^{10}\equiv \boxed{1} \pmod 7$

Wait, I didn't follow that. Could you please explain it more? I don't really get modular arithmetic. Like, I know what it is, but not how to use it...

@isabella2296 i'm pretty sure a number mod x is just the remainder when the number is divided by x. hope i helped

[hide]DON'T REVIVE OLD THREADS.
[/hide]

hope this one wasn't harsh

blabla, I believe reviving old threads is ok when a solution is too advanced or incomplete, especially in fora like this one.

I'm assuming the steps that are hardest to understand are the ones using the forumla $a^n\equiv b^n\pmod7$ if $a\equiv b \pmod 7$

The proof for this is based on the binomial expansion of $(b)^n$
If you write $b = a + 7k$ with $k$ being an integer, and list out the first few terms, you'll see that you can simplify it:
$(a+7k)^n= a^n+\binom{n}{1}a^{n-1}(7k)^1 + \binom{n}{2}a^{n-2}(7k)^2 + \cdots + \binom{n}{n-1}a (7k)^{n-1} + (7k)^n \\(a+7k)^n= a^n + 7k\left(\binom{n}{1}a^{n-1} + \cdots\right)\\(a+7k)^n \equiv a^n \pmod7$
Of course, this can be generalized for integers other than 7.

if you find a pattern you get this
$ 5^{2} = 4 $
$ 5^{3} = 6 $
$ 5^{4} = 2 $
$ 5^{5} = 3 $
$ 5^{6} = 1 $
$ 5^{7} = 5 $
$ 5^{8} = 4 $
ans so on
so $ (30)/ {6}=5 $ and the fifth number on the list is $\boxed{1}$

this is how i did it. please correct me if i am wrong

I agree with Brut3Forc3. The best thing to do is to use Fermat's Little Theorem which states that $n^{p-1}\equiv 1\pmod{p}$ Where $p$ is a prime number and the $\text{gcd}\, (n, p) = 1$.

Is fermat's thereon or the proof shown above in the intro to number theory book?


There will be scores posted with their names for the questions I posted.
So, please AIM me with your answer, thanks!

Go to here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12234&start=0

Can we PM you, and can non-getting-started people participate for practice?

Yes.  But only if you promise not to tell the answers to any other people in this level

Score:

Topper: 4 correct

I can't give any credits to the question if they'ren't expanded, meaning formally finished.  So, if you just end in 5!, it's not correct answer because you didn't end the computation by 5*4*3*2*1 and get 120.  That's why I can't give you credit for number 5.

Score:

 jobro_gold: 7 correct

It was really nice job.  Only thing I might say is that in number 8, the number is increasing, so you can't have negative value for that one.  And I hope everyone realize that number 5 is this:

[tex](n/2+1)^7[/tex]


inequalities
factorial
induction
algebra
difference of squares
special factorizations

[quote="Phelpedo"]Which is bigger, $2000!$ or $1000^{2000}$?

Which is bigger, $1999^{1999}$ or $2000^{1999}$?[/quote]
Um, isn't the second kinda obvious? [hide="my answer to the second"]The second[/hide]

[hide="1"]
Using Sterling's approximation,
$2000! \approx \sqrt{4000 \pi}\cdot \left( \frac{2000}{e}\right)^{2000}\approx 120 \cdot (740)^{2000}<<< 1000^{2000}$

(Note: the error for Sterling's approximation is tiny after about $100!$ or so, I think, so it should be fine here.)

Also, note that the factors of $2000!$ are only greater than $1000$ for half of the time, and afterwards only give at most an extra factor of two, which will not balance the quite smaller factors in the comparison of the factorial and $1000^{2000}$. [/hide]

For #2, did you mean $1999^{2000}$?

[hide=" If so..."]
A good thing to know, since it comes up alot in problems like these, is that $k^{k+1}> (k+1)^{k}\ \forall k \geq 3$. I think that this can be proven in many ways, including with some sort of induction and with the standard ineqalities. And also, it seems like common sense; the extra one in the RHS won't balance the extra $k$ multiplied to the LHS.[/hide]

Fixed the typo in #1. These are solvable by VERY elementary means and some crude "massaging," you definitely don't need anything like sterling's approximation.

Surely not elegant, but...

$n^{n+1}>(n+1)^{n}$ implies $n\cdot n^{n}>(n+1)^{n}$, that is $(\frac{n}{n+1})^{n}>\frac{1}{n}$.
I'll proceed by induction on n. Of course, with $n=3$ the inequality holds. Supposing it true and multiplying by $\frac{n}{n+1}$ both sides, we get $(\frac{n}{n+1})^{n+1}>\frac{1}{n+1}$.
We still need to prove that $(\frac{n+1}{n+2})^{n+1}>(\frac{n}{n+1})^{n+1}$, which is true, since it leads to $(n+1)^{2}>n(n+2)$ and, trivially, $n^{2}+2n+1>n^{2}+2n$, q.e.d.

Bye!  :)

P.S.: By the way, I'm not able to display high parentheses... writing \displaystyle doesn't help!...

[quote="Ponnamperuma"]P.S.: By the way, I'm not able to display high parentheses... writing \displaystyle doesn't help!...[/quote]

Try [code] \left( [/code] and [code] \right) [/code]. It also works with things like absolute values.

You get a little tighter inequality if you use the first 1999 integers instead of 2000, but it's still got a 2 as a constant.

eh, my way is incredibly similar to pianoforte's (#1), but I'll go:

You can group terms like (1)(1999), (2)(1998), ... , (999)(1001) for 2000!. This is basically x(2000 - x) where x ranges from 1 to 1000, with all the terms multiplied together, then multiplied by 2. Note that x(2000 - x) is always positive when x ranges from 1 to 1000.

Now, x(2000 - x) is [b]always[/b] less or equal to $1000^{2}$. The first term, (1)(1999) is definitely less than $1000^{2}$ even when multiplying it by 2, eliminating the hanging factor of 2 and showing that $\boxed{1000^{2000}> 2000!}$.

edit: bah, that guy with the Pikachu icon pretty much explained my solution before I got to post.

Or, for a sexier solution, divide both sides by $1000^{2000}$. It simplifies to proving that $2[(0.001)(0.002)\cdots(1.999)] < 1$. Group terms and revel in the simpleness.

You know that's the same solution right? You're working with the same numbers, just off by a factor of 1000.


modular arithmetic
number theory unsolved
number theory

We have the equality: $ 1\plus{}10^{a_1}\plus{}10^{a_2}\plus{}\ldots\plus{}10^{a_n}\equal{}z^2$, where $ 1 \leqslant a_1<a_2<\ldots<a_n$ and z is positive integer. Is the equality possible?

I have, that if the z satisfy this eqaution, then $ z$ is odd, so $ z^2\minus{}1 \equiv 0 (mod 8)$, futhermore: $ a_1 \geqslant 3$. Thanks for any further idea  :maybe:

Start the other way around.  (Look at $ a_n$ first.  Can it be odd?  Can you find an upper bound on $ a_{n\minus{}1}$?)

Can you post the whole solution?  :blush:

If $ n \geq 1$,
Both sides modulo $ 10^{a_1}$, we can get $ z^2 \equiv 1 (mod 10^{a_1})$
So $ z\equiv \pm 1 (mod 10^{a_1})$
Let $ z\equal{}10^{a_1} m \pm 1$, m is a integer
Now both sides of the original equation modulo $ 10^{a_1\plus{}1}$
Since $ a_2>a_1$, so $ LHS \equiv 1\plus{}10^{a_1} (mod 10^{a_1\plus{}1})$,
$ RHS \equal{} (10^{a_1} m \pm 1)^2 \equal{} 10^{2a_1} m^2 \pm 2\cdot 10^{a_1} m \plus{}1 \equiv \pm 2 \cdot 10^{a_1} m \plus{}1 (mod 10^{a_1\plus{}1})$
So $ 1\plus{}10^{a_1} \equiv \pm 2 \cdot 10^{a_1} m \plus{}1 (mod 10^{a_1\plus{}1})$
which results in $ 2m \equiv \pm 1 (mod 10^{a_1\plus{}1})$
which is a contradiction.

So $ n < 1$, i.e. $ n\equal{}0$, so $ z\equal{}\pm 1$.

[quote="brianchung11"]If $ n \geq 1$,
Both sides modulo $ 10^{a_1}$, we can get $ z^2 \equiv 1 (mod 10^{a_1})$
So $ z\equiv \pm 1 (mod 10^{a_1})$
[/quote]

I think this step is not valid... $ z^2 \equiv 1 \pmod{10^{a_1}}$ doesn't imply that $ z \equiv \pm 1 \pmod{10^{a_1}}$. For example, say $ a_1\equal{}2$, we have $ 49^2\equal{}2401\equiv 1 \pmod{100}$. Is there a way to fix this detail?

[quote="Jutaro"][quote="brianchung11"]If $ n \geq 1$,
Both sides modulo $ 10^{a_1}$, we can get $ z^2 \equiv 1 (mod 10^{a_1})$
So $ z\equiv \pm 1 (mod 10^{a_1})$
[/quote]

I think this step is not valid...[/quote]I agree. Try again :P


Thank you for your reply,Rafal.

Your answer is correct. :)

kunny


Inside a fixed sphere of radius $ R$ and uniform density $ \rho$, there is spherical cavity of radius $ \frac {R}{2}$ such that surface of the cavity passes through the centre of the sphere. A particle of mass $ m$ is released from rest at centre B of the cavity. Calculate velocity with which particle strikes the centre A of the sphere. Neglect earth's gravity. Initially sphere and particle are at rest.

Thats too easy... :wink:

Now modify the problem to a cavity of radius $ d<R$.

[hide="hint"]antigravity... :roll: [/hide]

[hide="Soln"]the cavity can be considered as an superposition of $ \plus{} \rho$ and $ \minus{} \rho$.Thus we have two spheres one of density $ \rho$ and another of density $ \minus{} \rho$ thus the field at any point inside the spheres is the vector sum of these both fields.
the field at any point inside the sphere due to a sphere of charge density $ \rho$ is $ \frac {GMr}{R^{3}} \equal{} \frac {4\pi G \rho r}{3}$ towards centre so $ \vec{E} \equal{} \frac {4\pi G \rho \vec{r}}{3}$ so net field at any point P inside the cavity is $ \frac {4\pi G \rho}{3}(\vec{ AP} \minus{} \vec{OBP}) \equal{} \frac {4\pi G \rho \vec{AB}}{3}$.
so accelaration of the particle is $ \vec{a} \equal{} \frac {4 \pi G \rho \vec{AB}}{3}$ thus the speed of the particle when it hists the centre is
$ v \equal{} \sqrt {2|\vec{a}|AB} \equal{} \sqrt {\frac {8 \pi G\rho \vec{AB}}{3AB}}$[/hide]

The answer is:
[hide]$ \sqrt{\frac{2}{3}\pi G\rho R^2}$[/hide]
So obviously we went wrong somewhere down there.

[quote="pardesi"] thus the speed of the particle when it hists the centre is
$ v \equal{} \sqrt {2|\vec{a}|AB} \equal{} \sqrt {\frac {8 \pi G\rho \vec{AB}}{3AB}}$[/quote]
The second formulae is wrong instead of multiplying by AB i divided by AB in the second formulae.u multiply by AB u get the rite answer


hello, does anybody know how to prove this inequality:
\[
(b\gamma-c\beta)^2+(c\alpha-a\gamma)^2+(a\beta-b\alpha)^2 < \left(\frac{\pi}{2}\cdot(a+b+c)\right)^2.
\]
Regards,Sonnhard

Maybe you forgot to include some conditions on $\alpha,\,\beta$ and $\gamma$ ?  ;)

yes ,$\alpha, \beta,\gamma$
are angles in some triangle and $a,b,c$ are the side length.
Regards,Sonnhard.

Actually, this inequality can be generalized: The only thing important about the sides a, b, c and the angles $\alpha$, $\beta$, $\gamma$ of a triangle is that all sides and all angles are positive numbers, and that if one side is longer than another one, then the angle opposite to the first side is greater than the angle opposite to the second one. Hence, if we WLOG assume that $a \leq b \leq c$, then we have $\alpha \leq \beta \leq \gamma$. And finally, the sum of the angles of a triangle is $\alpha + \beta + \gamma = \pi$. And our inequality turns out to be a special case of the following one:

[i]If a, b, c, x, y, z are six positive numbers such that $a \leq b \leq c$ and $x \leq y \leq z$, then

  $\displaystyle \left( bz-cy\right) ^{2}+\left( cx-az\right) ^{2}+\left( ay-bx\right) ^{2}<\left( \frac{\left( a+b+c\right) \left( x+y+z\right) }{2}\right) ^{2}$. [/i]

This one I have proved a follows: Since $a \leq b \leq c$, we can define A = b - a and B = c - b, so that $A\geq 0$ and $B\geq 0$ and b = a + A and c = b + B. Similarly, with X = y - x and Y = z - y, we have $X\geq 0$ and $Y\geq 0$ and y = x + X and z = y + Y. Now, a Maple calculation gives

$\displaystyle \left( \frac{\left( x+y+z\right) \left( a+b+c\right) }{2}\right) ^{2}-\left( \left( bz-cy\right) ^{2}+\left( cx-az\right) ^{2}+\left( ay-bx\right) ^{2}\right) $
$\displaystyle =\frac{81}{4}x^{2}a^{2}+\frac{3}{2}Y^{2}aB+Y^{2}aA+7XYa^{2}+6X^{2}aB+12X^{2}aA$
$\displaystyle +\frac{3}{2}xYB^{2}+6xYA^{2}+\frac{27}{2}xYa^{2}+xXB^{2}+12xXA^{2}+27xXa^{2}$
$\displaystyle +7x^{2}AB+\frac{27}{2}x^{2}aB+27x^{2}aA+\frac{1}{4}Y^{2}a^{2}+12xXAB+12XYaA$
$\displaystyle +8xYAB+\frac{1}{4}x^{2}B^{2}+7X^{2}a^{2}+7x^{2}A^{2}+4A^{2}XY+4ABX^{2}$
$\displaystyle +ABY^{2}+B^{2}XY+4A^{2}X^{2}+\frac{1}{4}B^{2}Y^{2}+40xAaX+20xAaY$
$\displaystyle +20xBaX+13xBaY+6ABXY+8XYaB$.

But this is > 0, since all terms containing one of the numbers A, B, X, Y are (at least) nonnegative, and the term $\displaystyle \frac{81}{4}x^{2}a^{2}$ is even strictly positive. This proves the general inequality.

Of course, I am by no means satisfied by the above brain-dead proof. But at least I know now that the inequality is true in its general form.

  Darij

hello Darij, thank you for your nice proof.I will have a look at it. I have worked out with that problem for many days, but i could not solve it.
Sincerely, Sonnhard.


geometry
trigonometry
integration
calculus
calculus computations

Was on my calc 3 exam today.. wondering if I got it right:

A hyperboloid is given by $ x^2 \plus{} y^2 \minus{} z^2 \equal{} 1$ between the planes $ z \equal{} \minus{} 2$ and $ z \equal{} 2$.

$ \bold{(a.)}$ Parameterize the surface.
$ \bold{(b.)}$ Find the stretching factor.
$ \bold{(c.)}$ Find the surface area (just set up the integral).

I think my answers were:
[hide]$ \bold{(a.)}$ $ x \equal{} \sqrt {z^2 \plus{} 1} \cdot \cos \theta$, $ y \equal{} \sqrt {z^2 \plus{} 1} \cdot \sin \theta$, $ z \equal{} z$ where $ 0 \le \theta \le 2 \pi$ and $ \minus{} 2 \le z \le 2$.

$ \bold{(b.)}$ $ |\overrightarrow{\bold{r_{\theta}}} \times \overrightarrow{\bold{r_z}}| \equal{} \sqrt {2z^2 \plus{} 1}$

$ \bold{(c.)}$ $ SA \equal{} \int_0^{2 \pi} \int_{ \minus{} 2}^2 \sqrt {2z^2 \plus{} 1} ~dz ~d \theta$[/hide]

Looks good. One way to evaluate the last integral is to use substitution $ z\equal{}2^{\minus{}1/2}\sinh t$, which (I think) is the reason why $ \sinh$ is called the hyperbolic sine.


$a_1=a_2=1,a_3=p$,$p$ is  a  prime number.$a_{n+1}=\frac{k+a_n a_{n-1}}{a_{n-2}}$.
prove that,if $k=rp-1$,$r$is a integer number,then$a_i$ are all positive integer numbers

thank you very much.it's quiet right. :)


I followed the AoPS user guide for LaTeX (the updated one) and downloaded/installed MiXTeX 2.7 and TeXnic Center.  Everything was fine.
I then continued to follow the guide and wrote my first program, which is saved as 'Hello.tex'.

Then, I tried compiling it, which didn't work.  First, Adobe Reader 8 opens up and says that it couldn't find the file and thus can't open the document.  After I hit 'Close' and exit out of Adobe Reader 8, TeXnic Center says that the line "[DocOpen("%bm.pdf")][FileOpen("%bm.pdf")]" causes an error.

I've done many things to troubleshoot (and spent over an hour and a half doing so):
I changed the DDE as many internet sources tell me to; I've replaced the error line with "[DocOpen("%bm.pdf")][DocOpen("%bm.pdf")]", but that doesn't work either.
Under Build and Define Output Profiles, I selected the viewer tab (which is where the error line is as a command).  I checked the box for "Command Line Argument" instead of "DDE Command", and also hit the arrow button right next to the 'Command:' box and selected "Main file's full path".  This doesn't work.
I made sure that LaTeX => PDF is actually directed towards Adobe Reader 8.0.  This doesn't work as well (because it was directed correctly).
I've checked for spelling errors on my part many, many, many times.  Found none.

Can someone please help me?  Stupid Adobe Reader...

When you compiled hello.tex was hello.pdf, hello.log and other similar files created in the same directory? If hello.pdf was created can you open it Adobe?

No.  That's precisely [u]why[/u] I've posted this: because the compiling process won't work and only an error message shows.  Help would be appreciated.

That is why Adobe complains. There is no pdf file to open so it can't open it. Hence the problem has nothing at all to do with Adobe and so you shouldn't be looking there.
The first thing to do is look at hello.log. If it isn't there then latex hasn't been found and you need to set up TexnicCenter to find it - [[LaTeX:Help#I.27ve_upgraded_MiKTeX_and_now_TeXnicCenter_won.27t_work]] shows you how.

If hello.log does exist then there is an error in the tex file and the log file will tell you what the problem is. Post both hello.tex and hello.log here for further advice.

Now the problem has been resolved.  I see what has been happening (after getting help from a friend): the AoPS guide is flawed.  In fact, the updated version is the one that is flawed, not the old version.
It says to perform one of the three methods to 'View Output' (F5).  This is wrong, because THERE IS no output.  First, I have to build the file, then view it.  The guide never mentions anything about building the file (actually, as I said earlier, the old guide does, but not the updated guide).

Thanks for your help, all you who posted.

[quote="MathAndKnowledge"] First, I have to build the file, then view it.  The guide never mentions anything about building the file (actually, as I said earlier, the old guide does, but not the updated guide).[/quote]
Oops, we'll correct that if it is so (have to wait until the Wiki is up again though :().


You ask a friend to guess a number from 1-10.
Each time the friend answers, you say higher if the friend's number is too low, lower if the friend's number is too high, or you win, if the number is correct. The friend is given 3 times to attempt to win. What is the probability your friend wins?

Maybe my friend guesses the correct number, but I do not want her to win. She will not know which number she has to guess, and nobody will know it. so I can say that the answer is not correct, even when it will be :)

I'll assume the friend wants to guess the correct number and he is clever enough.

[hide="Probability of correct guess"]$ \frac{7}{10}$[/hide]

[hide="Solution."]
For some optimum strategy, we can choose to always pick $ 1$ number first. After acquiring either of two responses, we can choose to pick from $ 2$ other numbers. For each of these possibilities, we can map the responses to $ 4$ other numbers.

Thus in our strategy, we have mapped $ 1 \plus{} 2 \plus{} 4 \equal{} 7$ numbers in our strategy. Assuming the number is chosen randomly from the $ 10$ numbers and such an optimum strategy is taken, there is a $ \boxed{\frac {7}{10}}$ chance that the randomly chosen number is an element of the $ 7$ numbers mapped in our strategy.
[/hide]


I know how to solve a basic equation with the CRT however, I heard from a friend that there is a wierd exception when x in a problem is raised to a power
x^28  is congruent to two seperate numbers in deifferent mods. Do you know what this exception is, he said it had something to do with adding the power and the mod time n where n is any integer, to find solutions.

Well, unfortunately what you ask is a little imprecise :). Do you recall more info about what you are asking?

Andrei

in china,we call the theory "sunzi",which is the name of a general in ancient china.
he used the theory to check the number of his soldiers. :D

No, they are just being more spcific :P

NEway, I think realaec was asking about the restrictions of the CRT (exceptions, I think)

[quote="georgew_bush"][quote="fredbel6"]are you saying that the chinese don't call the chinese remainder theorem chinese? :D[/quote]
I mean we call it another name,"sunzi" theorem.He is said to be the first person to use this theorem(about 300 BC) :D[/quote]
that person who use this theorem for his soldiers is "hanxin"...


$ a,b,c,d>0$ and $ \sum{a}\equal{}4$

$ \sum\frac {a}{b^2 c\plus{}1} \geq 2$


For a,b,c real numbers   0<a,b,c<1 prove that

a^c*b^(1-c)+ (1-a)^c*(1-b)^(1-c) \leq 1

[i]Taken from Mitrinovic's Book  "Elementary inequalities"[/i]

Yes, it's not very hard ! I  can easily generalize :
If c,a(k),b(k) are positive numbers such that 0 < c <1 and   Sum{a(k)}=Sum{b(k)} = s , then Sum{(a(k)^c)*(b(k)^(1-c))} =< s^(3/2)

Can you prove it ?


N students take a test.  Then, the teacher takes their tests and randomly matches up one student with one test.  (I know, it actually doesn't happen this way in the classroom.)  The student grades the test that he was assigned.  However, due to fear of cheating, the students aren't allowed to grade their own tests.  If a student ends up with his own test, he must switch with another student so neither students are grading their own tests.  What's the formula for n that represents the probability that at least one student needs to switch tests?

[hide]
For one person to have his own test, there are ${n\choose 1}$ ways to choose that person, and $(n-1)!$ ways to arrange the tests of the other kids.

This double counts the cases where $2$ kids have their own tests. The number of those cases is equal to ${n\choose 2} \times (n-2)!$.

Using PIE to correct the counting of the rest of the cases, we get: ${n\choose 1}(n-1)! - {n\choose 2}(n-2)! + ... + (-1)^n$

There are a total of $n!$ ways to distribute the tests, so the probability is then:

$\displaystyle \frac{{n\choose 1}(n-1)! - {n\choose 2}(n-2)! + ... + (-1)^n}{n!}$
[/hide]

Oh, didn't see that. Looks simpler that way  :)


What is the coefficient of $ x^2$ when $ (...(((x\minus{}2)^2\minus{}2)^2\minus{}2)^2)...\minus{}2)^2$ is expanded and like terms are combined?


$ a_1, a_2, ...$is a sequence of real numbers where $ a_n$ is the arithmetic mean of the previous $ n \minus{} 1$ terms for $ n > 3$ and $ a_{2004} \equal{} 7$. $ b_1, b_2,...$ is a sequence of real numbers in which $ b_n$ is the geometric mean of the previous $ n \minus{} 1$ terms for $ n > 3$ and $ b_{2005} \equal{} 6$. If $ a_i \equal{} b_i$ for $ i \equal{} 1, 2, 3$
and $ a_1 \equal{} 3$, then compute the value of $ (a_2)^2 \plus{} (a_3)^2$

[hide="Hint"]
Look at what changes between $ a_n$ and $ a_{n\plus{}1}$, and between $ b_n$ and $ b_{n\plus{}1}$.  It's not much.
[/hide]


Prove that if $\alpha+\beta+\mu=\pi$ then

$\tan(\alpha)\tan(\beta)\tan(\mu)=\tan(\alpha)+\tan(\beta)+\tan(\mu)$

A typo at the beginning - $\tan(\alpha)$ should be $\tan(\beta)$

[quote="FieryHydra"][quote="Altheman"][quote="M4RI0"]$\tan(\alpha)=tan(\pi-\alpha-\mu)=-tan(\alpha+\mu)=-\frac{tan(\alpha)+tan(\mu)}{1-tan(\alpha)tan(\mu)}$
$\Rightarrow \boxed{tan(\alpha)+tan(\beta)+tan(\mu)=tan(\alpha) tan(\beta) tan(\mu)}$[/quote]

What if $1-\tan\alpha\cdot\tan\beta=0$?[/quote]
In that case, $\alpha+\beta = \frac{\pi}{2}\Rightarrow \mu = \frac{\pi}{2}$, and the original equation still holds.[/quote]

I don't think you can take $\tan\frac{\pi}{2}$.

Of course, I cannot do that but look at it in a "limiting" sense.

[quote="arkhammedos"]i think that its verry easy .or not[/quote]

it has been solved, but the problem statement is wrong since you cannot have a right angle, (so it only holds for $\alpha,\beta,\gamma \ne \frac{\pi}{2}$)


What perfect-square integer is closest to 273?


Let $a,b,c$ be three positive reals numbers such that $abc=1$. Prove that $(a-1+\frac{1}{b})(b-1+\frac{1}{c})(c-1+\frac{1}{a})\le 1$

let $a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$
then prove $xyz\ge(x+y-z)(y+z-x)(z+x-y)$

By Schur's inequality we have:
$\Rightarrow x(x-y)(x-z)+y(y-z)(y-x)+z(z-x)(z-y) \geq 0$
$\Rightarrow xyz \geq (x+y-z)(y+z-x)(z+x-y)$
$\Rightarrow 1 \geq (\frac{x}{y}+\frac{y}{z}-1)(\frac{y}{x}+\frac{z}{x}-1)(\frac{z}{y}+\frac{x}{y}-1)$

Replacing: $\frac{x}{y}=a$;$\frac{y}{z}=b$;$\frac{z}{x}=c$

We get: $1 \geq (\frac{1}{c}+b-1)(\frac{1}{a}+c-1)(\frac{1}{b}+a-1)$  $\mathbb{QED}$


superior algebra
superior algebra solved

Give a countexample to the follwing statement: if R is a ring, n is a positive interger, and na=0 for some a in R then nr = o for all r in R.

Thanks

$R=\mathbb Z/6\mathbb Z$
For $n=3$, $3\cdot\hat 2=\hat 0$, but $3\cdot\hat3\neq0$.


Hello

I never participated in any of the AMC/AIME contests in high school. But I want to try and do decent on some college math competition (Putnam or any other one). I have not really had any experience in writing proofs, aside for those in Euclidean Geometry. Would going through a text like "Discrete Mathematics with Applications" by Kenneth Rosen be good way to learn how to write proofs? Also, would the Art of Problem Solving Books Vols. 1 and 2 also help in preparing for college competitions?

Any comments and suggestions are appreciated.

Thanks

AoPS volume 2 would help for some college math competitions. (There are regional competitions besides the well known Putnam competition.) Paul Zeitz's book The Art and Craft of Problem Solving, which is now in its second edition, has a lot of content related to college level math competitions. There are also books about old Putnam exams that are very useful.


The question says itall..

wht is your favorite tv show?

[quote="joml88"]How could you leave Family Guy off? :?[/quote]

I second this. I think that you can't have a poll for something as broad as this, or else other will probably get about 90%.

Maybe he dislikes the crudeness of family guy and refuses to accept the existance of the show  :D

Simpsons, of course.

I really like 7th Heaven, Everwood, and The OC.

haha simpsons is winning i voted for the simpsons!!! and i just happend 2 finish watching it. good episode. check my topic about it!!!

ARRESTED DEVELOPMENT!!!

I'm so going to buy the first season DVD and watch every tid bit during Fall Break.

Favorite show ever: Mythbusters, on the Discovery Channel.

lol wow.......... many shows i have never even HEARD abuot

[quote="theone853"]I dwell on a lot of old television series...
Anyone here ever heard of "Family Matters"?[/quote]

Heard of it.  I thought it was pretty good, but I don't watch it anymore.

Eh, CSI is nice, but it can't compare to the bona fide enormous explosions, and the unbridled carnage loosed upon Buster the Crash Test Dummy.

o yes.... ive seen that [u]once[/u] it was pretty cool

Hey, the hosts Adam and Jamie were on the special effects teams that did Star Wars Episodes I and II, and also did the effects for Terminator 3, among many other distinguished credits.  You can't compete with that.


Can you find numbers such that their factorials end in EXACTLY $23$ zeroes? How many such numbers can you find?

[hide]we know 100! ends in 20+4=24 zeros. therefore 99! ends with 22 zeros. so there are no numbers such that their factorials with in 23 zeros.[/hide]

Which numbers $x$ have no values of $k$ such that $k!$ ends in $x$ zeroes?

I believe numbers of the form $6n-1$, but I'm not completely sure on this one.


[quote="luimichael"]A positive integer N, when divided by 10,9,8,7,6,5,4,3, and 2, leaves remainders 9,8,7,6,5,4,3,2 and 1 respectively. Find the least value of N.[/quote]
[hide="Hint"]
What happens when you add one to the number?
[/hide]

[hide="Solution"]
We add one to the number.  Now, notice that our number plus one is divisible by 10, 9, 8, 7, 6, 5, 4, 3, and 2.  We can find $ n \plus{} 1 \equal{} lcm [2, 3, 4, 5, 6, 7, 8, 9, 10]$ when $ n$ is the number we seek.  $ lcm [2, 3, 4, 5, 6, 7, 8, 9, 10] \equal{} 2520$.  Thus, $ n \plus{} 1 \equal{} 2520\implies n \equal{} \boxed{2519}$. [/hide]


calculus
integration
function
real analysis
real analysis unsolved

Find all $f$:$R+\rightarrow R+$, piecewise continous, decreasing
integrable $\int_{0}^{+\infty}f(t)dt<+\infty$
$f(0)=1$
for any $x \in R+$, $\int_{x}^{+\infty}f(t)dt\leq f(x)$

Hmm... What kind of answer do you expect here? It is pretty much the same as to ask to describe all decreasing to $0$ convex functions $F(x)$ (the antiderivatives of $-f$) such that $F'(x)\le -F(x)$ for all $x$ and $F'(0)=-1$ but it is not clear how one can reformulate these properties in more "basic" terms or what those "more basic terms" may possibly be.  :?


Hi,
I want to show that if A(nxn) is symmetric then it is orthognally equvalent to a symmetric matrix wit equal elements in the diagonal.For example for 2x2 if A=[a b;b c] there exist a Q  O(2) (Q is orthognal matrix)such thet Q 'AQ=[d e;e d][/quote]

I don't think it's on the right place here in pre-olympiad, please post problems containing matrices in the superior algebra forum.
And please also do not double post ( http://www.mathlinks.ro/Forum/viewtopic.php?p=218676&highlight=#218676 )


topology
real analysis
real analysis unsolved

Describe the weak and weak* topologies of $ \ell^{1}$ [b]considered as the dual of $ c_{0}$[/b].

I think that since in $ l^1$, if $ x_n \rightharpoonup x$ then $ x_n \to x$. So that weak and strong topologies of $ l^1$ are the same  :oops: .

That's not true. Consider the sequence $ (e^{(m)})$ where $ e_k^{(m)}$ equals $ 1$ for $ k\equal{}m$ and zero otherwise.

$ e^{(m)}$ tends to zero in the weak* topology (considered as the dual of $ c_0$) but does not tend to zero in the norm topology of $ \ell^1.$

As for the weak topology on $ \ell^1,$ that involves testing against elements of $ (\ell^1)^*\equal{}\ell^{\infty}.$ Give me a little while to think of an example here.