inequalities
algebra
polynomial
inequalities proposed
122

Let $ a,b,c$ be real numbers in the interval $ [0,1].$ Prove that
$ a(1\minus{}b^2)\plus{}b(1\minus{}c^2)\plus{}c(1\minus{}a^2) \le \frac{5}{4}.$

The left-hand side of the inequality can be written as:

    $ (a \plus{} b \plus{} c) \minus{} (ab^2 \plus{} bc^2 \plus{} ca^2) \leq \frac {5}{4}$

But from $ AM \minus{} GM$, we have that: $ ab^2 \plus{} bc^2 \plus{} ca^2 \geq 3\sqrt [3]{(abc)^3} \equal{} 3abc$

Hence it suffices to prove that:
 $ (a \plus{} b \plus{} c) \minus{} 3abc \leq \frac {5}{4}$ $ (1)$

Set $ a \equal{} 1 \minus{} x,b \equal{} 1 \minus{} y,c \equal{} 1 \minus{} z$ where $ 0 \leq x,y,x \leq 1$.

Then $ (1)$ becomes:

  $ 3 \minus{} (x \plus{} y \plus{} z) \minus{} 3(1 \minus{} x)(1 \minus{} y)(1 \minus{} z) \leq \frac {5}{4}$ or
  $ \minus{} 2(x \plus{} y \plus{} z) \plus{} 3(xy \plus{} yz \plus{} zx) \minus{} 3xyz \geq \minus{} \frac {5}{4}$

Consider the polynomial: $ P(x) \equal{} x^3 \minus{} 2x^2 \plus{} 3x \minus{} 3$. We would like to show that $ P(x) \geq \minus{} \frac {5}{4}$

I couldn't factor this polynomial hence i considered the polynomial now $ Q(x)$ such that:

   $ Q(x) \equal{} 4P(x) \plus{} 5$ and I want to show that $ Q(1) \geq 0$ .

Subsituting x=1 to the polynomial $ Q(x) \equal{} 4P(x) \plus{} 5 \equal{} 4x^3 \minus{} 8x^2 \plus{} 12x \minus{} 12 \plus{} 5$ we get that:

$ Q(1) \equal{} 4 \minus{} 8 \plus{} 12 \minus{} 7 \equal{} 1 > 0$.

I am not sure about the validity of my solution though.It probably contains mistakes.Feel free to comment on it.

Peace

~Kostas

[quote="new_member"]The left-hand side of the inequality can be written as:

    $ (a \plus{} b \plus{} c) \minus{} (ab^2 \plus{} bc^2 \plus{} ca^2) \leq \frac {5}{4}$

But from $ AM \minus{} GM$, we have that: $ ab^2 \plus{} bc^2 \plus{} ca^2 \geq 3\sqrt [3]{(abc)^3} \equal{} 3abc$

Hence it suffices to prove that:
 $ (a \plus{} b \plus{} c) \minus{} 3abc \leq \frac {5}{4}$ $ (1)$
[/quote]
Dear friend, I think you cant use (1) to prove the desired inequality, because (1) is false (for example $ a \equal{} b \equal{} 1,c \equal{} 0$).

[quote="Tourish"][quote="can_hang2007"]Let $ a,b,c$ be real numbers in the interval $ [0,1].$ Prove that
$ a(1 \minus{} b^2) \plus{} b(1 \minus{} c^2) \plus{} c(1 \minus{} a^2) \le \frac {5}{4}.$[/quote]

Oh,I think this problem belongs Ji Chen.Sorry if I am wrong.
I have got a solution that written in Chinese(may be wrong).I don't understand why we can assume that $ 0\leq b\leq c\leq a\leq 1$

http://www.aoshoo.com/bbs1/UploadFile/2005-11/2005112818514556808.pdf[/quote]
Thank you for your information. I got this problem at http://www.irgoc.org/viewtopic.php?f=15&t=9&sid=2d963e2ef316c979e834216cdad6205a , kuing posted that problem and it is the reason why I thought he is the author of the problem.

Actually, I think it does not matter so much about who is the author of the problem. The way we think of solving it is more important.

Oops  :blush:  Apparently,you're right.I am sorry.I will try it again!

Can_hang,I didn't know that the inequality $ [\sum x^2y \le \frac{4}{27}\left( \sum x\right)^3]$ is well-known.
Could you please give a solution for that inequality?
Thanks

[quote="new_member"]Can_hang,I didn't know that the inequality $ [\sum x^2y \le \frac {4}{27}\left( \sum x\right)^3]$ is well-known.
Could you please give a solution for that inequality?
Thanks[/quote]
Dear friend,
For this inequality, by assuming $ x\equal{}\max\{x,y,z\},$ we can show that
$ x^2y\plus{}y^2z\plus{}z^2x \le (x\plus{}z)^2y \le \frac{4}{27}(x\plus{}y\plus{}z)^3.$
By the way, the stronger inequality holds, too
$ x^2y\plus{}y^2z\plus{}z^2x\plus{}xyz \le \frac{4}{27}(x\plus{}y\plus{}z)^3.$
The known solution of this inequality is to assume $ y$ to be the number between $ x$ and $ z$ and then use some suitable estimations. But I think it can also be proved by using Cauchy Schwarz, can someone find it out?

[quote="new_member"]Can_hang,I didn't know that the inequality $ [\sum x^2y \le \frac {4}{27}\left( \sum x\right)^3]$ is well-known.
Could you please give a solution for that inequality?
Thanks[/quote]
As Can said the stronger $ \sum x^2y \plus{} xyz \le \frac {4}{27}( \sum x)^3$ is also true.

The crux move to solve it is by setting $ y \equal{} x \plus{} a$ and $ z \equal{} x \plus{} b$ where of course $ a,b \ge 0$ because we have consider $ x$ as minimum :wink:

[quote="can_hang2007"]Let $ a,b,c$ be real numbers in the interval $ [0,\ 1].$ Prove that

$ a\left(1 \minus{} b^2\right) \plus{} b\left(1 \minus{} c^2\right) \plus{} c\left(1 \minus{} a^2\right) \le \frac {5}{4}.$[/quote]$ \frac {5}{4} \minus{} \sum_{cyc}{a\left(1 \minus{} b^2\right)}$

$ \equal{} \frac {3 \plus{} 4(a \plus{} b \plus{} c)}{108}\left[2(a \plus{} b \plus{} c) \minus{} 3\right]^2 \plus{} \sum_{cyc}{\frac {1 \minus{} a}{27}(3 \plus{} 2a \minus{} b \minus{} 4c)^2}\geq0.$

This inequality is the [b]Theorem 2[/b] in

Ji Chen, The extension and improvement of a test question of the All-Russian Mathematical Olympiad (Chinese), [url=http://www.shuxuetongxun.com/]Bulletin of Mathematics (Wuhan)[/url], 1995, No. 9, pp. 28 - 29.


I dont know where to put this game, so i guess ill place it in the Games Forum.

If it should be in the Fun Factory, Modertaors, Please move this game in the FF.

Ok, on with the game.

There are 4 switches, 1,2,3,4. One sets off a bomb. the other 3 dont. I can handle up to 8 ppl who want to play.

Just say if you wnat to play.

Example: (Pretend the Bomb switch is #2, chosen by random.org)

Bob hits switch #1.

I say:

Nothing happens.

BUT, if

Bo hits switch #2

I say: The bomb blew up. You lose a point.

Score: Bob, -1
          Joe, 0
          Billy, 0
etc.

The game ends when someone gets to -5 points. (VOTE: Do you want the game to end whne someone has -5 points, or -10 points.)

Person with least "dies" wins, after someone gets to -5 (or -10) pts.

I will not cheat, or pick favorites. I will use Random.org for this game.

If you like this game, tell me, and maybe I could create Bombs Away! II.

Signups begin now.

P.S. If you have any questions, or stuff I left out, that you would lke to know, or stuff thats not clear, state your questions in this thread.

Remember! Only 8 ppl! Sign up Fast!!

P.P.S. If it is your turn, and you do not say which switch you flick, after 3 days, the bomb blows up, and you automatically get -1 pt.

Please Join!

i join

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

ONLY ONE SPOT LEFT!!

HURRY HURRY HURRY!

SIGN UP!!!!

(btw, 7 signups in less then 24 hours. wow.)

ok, we start!

ernie, 3 days to pick a switch!

1,2,3, or 4

GO!!

Let everyone pick a switch at the same time, so each round doesn't go 20+ days.

good idea, aime.

actually, lets vote.

a. we play to -5 pts or -10 pts
b. 3 switches or 4 switches?

hold on

pm me your switches.

so there wont be any cheating. :)

please, [b]NO DISCUSSION OUT-OF-THREAD!!!!![/b]

thanks

btw, the timer is ticking!!!

i live in californian time zone, so at 9:00 am (californian), (12:00 pm for u guys in the east, i think) on April 5 is the deadline!

pm me your switches!!

and there can be discussion here, about plans, and stuff, but if i suspect you guys are talking out of thread, i will automatically give you -2 points.

so dont anger me!

lol

Joe. I think we should post numbers on forum. cause if we pm we might pm the same numbers as other people.

that will definently happen, no matter what.

8 ppl, 4 switches (or 3 switches....)

there will definentlay be repeats.

btw,

about one day passed.

only one person pm me.

ernie did

LETS ALL CLAP FOR ERNIE!!!

*Clap! Clap! Clap!*

YAY ERNIE!!

anyway, two more days. PM ME!!!!

btw, add your opinions to the following questions to your pm to me about the switch.

[b]a. do you want 3 switches to choose from, or 4 switches?

b. do we play till someone gets to -5 pts, or -10 pts.[/b]
Ernie, as you were the first one, you may post your descions in the thread. PLEASE EVERYONE GIVE ME YOUR OPININIONS!

darn

i cnt spll.

EIGHT switches. Then you don't say who died till end of round, and peeps post in the thread.

and -5 points...

just giving my opinion as a helpful person.

hmmmm

good idea spidey

do u want to be a replacement?

ok, now, use spiedy's rules.

good idea spiedy.....

post in thread now!

and please hide them, so we wont have all you eight ppl giving me all different switches. then it wont be fun.

wow

one day left, only 3 ppl pm me

HURRY!

what? don't get it.
switch 3 I guess? If thats legal

hmmmm

not alot of ppl sent in their switches.

if you do not want to play, then tell me in this thread.

i dont want to waste my time checking this thread.

sure, ecxept that this thread is dead, unless ppl who are playing, still want to play.......

sorry coolts (for right now)


Ten (indistinguishable) children order ice-cream at a store featuring 31 flavors. How many orders are possible in which at least two children get the same flavor?

Is there an easy way of doing this without partitions.

[hide="Solution?"]We can model this as distributing 10 balls into 31 boxes.

Example:

$|\,\, *\,\, *\,\, |\,\,\,\,\,\, |\,\, *\,\, |\,\, *\,\, *\,\, *\,\, | \,\,\,\,\,\, | \,\,\,\,\,\, |$....and so on.

A box is represented by $|\,\,\, |$ and a ball by $*$. There are 10 $*$'s and 30 $|$'s excluding the first and last $|$. These can appear in any order and the total number of arrangements is the number of ways we could permute 40 objects, 10 being alike of one kind and 30 alike of a second kind.

Total $=\frac{40!}{10!30!}=\dbinom{40}{10}=847660528$.

The number of arrangements where each child chooses a different flavor is

$\dbinom{31}{10}= 44352165$.

So the number of arrangements where at least 2 children choose the same flavor is

$\dbinom{40}{10}-\dbinom{31}{10}= 803308363$.[/hide]

[quote="rnwang2"][hide="Solution?"]We can model this as distributing 10 balls into 31 boxes.

Example:

$|\,\, *\,\, *\,\, |\,\,\,\,\,\, |\,\, *\,\, |\,\, *\,\, *\,\, *\,\, | \,\,\,\,\,\, | \,\,\,\,\,\, |$....and so on.

A box is represented by $|\,\,\, |$ and a ball by $*$. There are 10 $*$'s and 30 $|$'s excluding the first and last $|$. These can appear in any order and the total number of arrangements is the number of ways we could permute 40 objects, 10 being alike of one kind and 30 alike of a second kind.

Total $=\frac{40!}{10!30}=\dbinom{40}{10}=847660528$.

The number of arrangements where each child chooses a different flavor is

$\dbinom{31}{10}= 44352165$.

So the number of arrangements where at least 2 children choose the same flavor is

$\dbinom{40}{10}-\dbinom{31}{10}= 803308363$.[/hide][/quote]

Oh, that is what they were asking.  Well there it is.


function
induction
number theory proposed
number theory

Let $ f$ a function from $ \mathbb{N}$ to $ \{1,2,...,m\}$. Prove that there exist a real $ \alpha$ such that $ [\alpha]^{n}\equiv f(n)(\bmod m)$

We can find $ m\le\alpha < 2m$. By induction $ \beta_{n}<\beta_{n\plus{}1}<\alpha <\gamma_{n}<\gamma_{n\plus{}1}.$


geometry
geometric transformation
reflection

I did #2 recursively did anyone else do it recursively or find another solution besides the casework based one presented in the math jam?

I did it recursively as well. Here's my solution.

[hide]
(a) Going clockwise from 1, the first circle is greater than 1, so that circle must be greater than both its
neighbors. By that logic, the second circle must be less than both its neighbors, the third must be greater, and
so on, until finally the sixth circle must be less than both its neighbors. But it is adjecent to 1, which it must
be greater than, so we have a contradiction. Therefore, there are no possible ways to fill in the circles as described.
This argument can be extended to any even number of circles that have to be filled in (that is, an odd number of
circles total, counting the 1). Going around the circle clockwise, we will always find that the last circle must be
less than both its neighbors, which is not possible because it neighbors 1.

(b) Using the same kind of logic as above, we conclude that we must write a permutation of the numbers 2 through 8
such that every other number is less than both its neighors. That is, the numbers alternate between decreasing and increasing.
Call such permutations Alternating Permutations.
To make the problem simpler, we consider Alternating Permutations of 1 through 7 instead of 2 through 8 (one can be convinced quite easily
that this does not change the final answer). Actually, we will generalize and consider permutations of 1
through $(2n+1)$. Let $F(2n+1)$ be the number of Alternating Permutations
of 1 through $(2n+1)$; thus the final answer
we seek is $F(7)$. We will attempt to define $F(2n+1)$ recursively. To do so, consider the location of the 1 within
an Alternating Permutation. The 1 must be in an even location. Call that location $2k$. The question we
now must answer is this: how many Alternating Permutations of 1 through $(2n+1)$ contain a 1 at the location $2k$? Consider that the 1
partitions the numbers 2 through $(2n+1)$ into two sets, one with $2k-1$ numbers and one with $(2n+1)-2k$ numbers.
This partition can happen in ${2n\choose 2k-1}$ ways. Now, the set of numbers to the left of 1 and the set of numbers
to the right of 1 must both be put in an order. We can see that these two sets must be ordered into two Alternating Permutations.
That is, the set on the left can be ordered in $F(2k-1)$ ways and the set on the right can be ordered in $F([2n+1]-2k)$ ways. Since the 1 at position $2k$ will always be less than both its neighbors, we know that iff the two sets are Alternating Permutations, the entire permutation
will also be an Alternating Permutation. Combining the above results, we find that there are ${2n\choose 2k-1}F(2k-1) F(2n+1-2k)$
Alternating Permutations of 1 through $(2n+1)$ with the 1 at location $2k$. Now
we must simply add up the results for all possible values of $k$. We find that \[F(2n+1) = \sum_{k=1}^{n}{2n\choose 2k-1}F(2k-1) F(2n+1-2k)\] along with the obvious $F(1) = 1$. Using our definition of $F(2n+1)$,
$F(3) ={2\choose 1}F(1)F(1) = 2$
$F(5) ={4\choose 1}F(1)F(3)+{4\choose 3}F(3)F(1) = 16$
$F(7) ={6\choose 1}F(1)F(5)+{6\choose 3}F(3)F(3)+{6\choose 5}F(5)F(1) = 272$

Thus the answer we seek is 272.
[/hide]

I hope it's not hard to follow, but it's definitely nicer than casework, especially the kind given in the math jam solution.

Niiice.  I did it similarly, I defined an "alternating permutation" I called it a "windy sequence."  However I started with $\ R_{1}$ and worked all the way up to $\ R_{7}$  I'll post my solution later.

I tried to do it recursively, but I failed.

I tried looking at smaller cases, and I brute forced 4 and 6 circles. It didn't get me anywhere. I completely solved part a, so how many points do you think I can get?

I did it kind of recursively, but differently. I started with the case of the numbers 1-3, and worked up from there. My solution is:[hide]In this proof, you can assume that when a string of numbers or letters is shown with commas in between (e.g. 1,3,5,2,6,8,7,4) it represents either the ring with the first number at the top, the next number immediately to the right of it, and the rest of the numbers proceeding around the ring counterclockwise, and looping back to the first number, or a partial ring of the sort.

Let us define a “run” in a ring as a place where there are either 3 ascending integers in a row, or three descending integers in a row (for example, 1,6,7). Obviously, this would make a ring not work, as the middle number would have one number higher than it, and one number lower than it. We can also see that the converse of this is true. If a ring is “illegal”, and does not satisfy the given condition, that means that there is some number that is neither higher nor lower than both of the numbers adjacent to it, so it is higher than one, and lower than the other. Therefore, the number and the two next to it are of ascending or descending order, and form a run.

Let us define “fixing” a run to be when we insert a number somewhere in the run so that run no longer exists, so that if there was only one run in the ring and we fix it, the ring will now be a solution.

Lemma 1: There is one and only one way to fix a run by inserting a number higher than any number previously in the ring.

Proof: Let us assume that the run is composed of a,b,c. Obviously, in order to fix the run, we will have to insert the new number, d, between either a and b, or between b and c, or else a, b, and c would all still be next to each other, and there would still be a run. However, it is not obvious why there is exactly one way to fix the run. Let us consider the two possible cases, the run is ascending, or the run is descending.

Case 1: The run is ascending, and a<b<c. When we insert d, it must either be between a and b, or between b and c if we hope to fix the run. If we insert it between a and b, it will fix the run. a will be between two numbers higher than it because if the number to the left of it was lower than a, it would be a different run, and a<d. d will be between two numbers lower than it because it is the highest number in the ring. Since b<c, and b<d, b will be less than both of the numbers next to it. Finally, since c>b and c must be higher than the number to the right of it or there would be two runs, it is between two numbers lower than it. So, putting do between a and b fixes the run. However, putting d between b and c does not fix the run, because we will be left with the sequence a,b,d where a<b<d.

Case 2: The run is descending and a>b>c. We can easily see that this is exactly the same as case 1, because if we flip the entire ring around, we will get an ascending run.

Therefore, with either an ascending run or a descending run, there is exactly one way to fix a run by inserting a number higher than any number previously in the ring.
-End lemma-

Lemma 2: Any place that we insert a number higher than any number already existing in the ring that does not fix a run, adds a run

Proof: There are four different cases of where we could put this higher number, d, in the ring: before an ascending run (which is the same as at the end of a descending run), after an ascending run (the same as the beginning of a descending run), the place in the middle of the run which does not fix the run, and somewhere not touching or in a run.

Case 1: At the beginning of an ascending run, a,b,c. Let us say that before adding d, the number to the immediate left of a was e, and the number to the left of e was f. So we now have the sequence f,e,d,a,b,c, with a<b<c, and d is the biggest of all of the numbers. If e<a, than e,a,b was a run, and adding d fixed it by lemma 1. But we said that d does not fix a run, so e must be greater than a. If f>e, then f,e,a was a run, and d fixed it. But we said that d wouldn’t fix a run, so e>f. Therefore, f,e,d forms a run, and d added a run.

Case 2: at the end of an ascending run, a,b,c. This leaves us with a,b,c,d with a<b<c and d is the biggest of the numbers. Therefore, b,c,d is a run and d added a run.

Case 3: at the place in the run that does not fix the run. We have run ascending run a,b,c, and we know from lemma 1 that inserting d between a and b fixes the run. However, inserting them between b and c leaves the run a,b,d, so d added a run. (Note that in the case of a descending run, we can again flip the entire ring and be left with an ascending run.)

Case 4: not in the middle of a run or next to one. Let us say that the numbers to the right and left of d are a and b respectively, and the numbers to the right of a and to the left of b are c and e respectively. So we have e,b,d,a,c. Obviously, since there are no equal numbers in these rings, either a>b, or b>a. We can assume WLOG that b>a. since e,b,a was not a run, b must also be greater than e. so e,b,d forms a run and d added a run.

Therefore, wherever we put a number higher than any number previously in a ring, it must either fix a run, or add a run.

-End Lemma-

Solution to part A: The problem does not specify whether or not to count reflections as two solutions or not, so we will solve for them as one, and then also determine the answer if they count as two.

With the numbers 1, 2,and 3, there is obviously only one ring: 1,2,3 (since the position of the 1 is fixed, and we are not double counting reflections). Obviously, this has exactly one run in it. Because of lemma 2, when we add 4 to the ring, we either add a run or subtract one. The same holds true when we add 5 to any of these rings of 4. It also holds true when we add 6 to the ring of 5, and add 7 to the ring of 6. It is therefore easy to see that we must end up with an odd number of runs. Therefore, there cannot be 0 runs, and there is no way to complete the circle under the given conditions, with or without reflections. Therefore, the answer to part A is 0.

Solution to part B: Again, we will solve this problem as if reflections counted only once, and then double our answer at the end for if they count as two.

Again, with the numbers 1, 2, and 3, there is only one ring: 1,2,3, which has exactly one run. When we add the number 4, there is one run, so one place we can put four that will fix a run, and both of the other places will add a run. Therefore, there is one ring of four with 2 runs, and 1 ring of four with no runs.

When we add the number five anywhere in the ring with no runs, it will result in a ring with no run, so there are four rings of five with one run. However, we can also fix either of the rings with 2 runs each in two different ways, which leaves us with 2*2=4 more rings of 5 with 1 run, for a total of 8 different rings of five with one run. However, we can also make rings of 5 with 3 runs by simply adding the 5 in any of the 2 remaining places in either of the circles with 2 runs. This leaves us with 4 rings of 5 with 3 runs. Therefore, there are 4 rings of 5 with 3 runs, and 8 rings of 5 with 1 run.

When we add the number six, if we want a ring with 0 runs, we must do that by fixing the run in any of the rings with one run. This leaves us with 8*1=8 of these rings. If we want a ring with 2 runs, we can either add it in any of the 4 remaining places of the rings with one run, or fix any of the three runs in a ring with three runs. This leaves us with 4*8+3*4=44 rings of six with 2 runs. Finally, we can get a ring with four runs, However, since we can only subtract one run each time, these will never give us a ring of 8 with no runs, so we can discard them. So, there are 8 rings of six with 0 runs, and 44 rings of six with 2 runs.

When we add the number seven, if we want a ring with 1 run, we must do that by adding a run to any of the six places in ring with no runs, or by fixing a run in one of the two places of a ring with two runs. This leaves us with 8*6+44*2=136 rings of seven with one run. Again, we can discard the rest of the rings, since we will not be able to fix enough runs before we get to rings of 8.

Finally, when we add the number eight, we can only get 0 runs by fixing the run in a ring of seven with only one run. Therefore, there are exactly 136*1=136 rings of 8 with 0 runs.

Therefore, there are 136 ways to fill the circles under the given conditions if we don’t allow for reflections, and 272 ways if we do. WWWWW
[/hide]

It's long, but I think not too bad to follow

My solution is basically the same as asquared's one.  Did anyone find an explicit formula for the number given n circles?  I spent a good bit of time hung up on the 2 circles --> 1 way, 4 circles --> 2 ways, 6 circles --> 16 ways, so I expected there to be a formula ($2^{something}$), but then 8 gives 272 and 10 gives 7936 (I think), so I'd guess that there isn't an explicit formula, since these numbers seem to factor fairly randomly $(2^{0}, 2^{1}, 2^{4}, 272 = 17*2^{4}, 7936 = 31* 2^{8})$, but that's obviously not a proof.

I did a recursion too.  I think I made it more complicated than it needed to be, but I got the right answer :P


limit
number theory
prime factorization
calculus
calculus computations

Simply find $ \lim_{x\rightarrow 6}\frac{x\minus{}6}{\sqrt{x\plus{}3}\minus{}3}$

Multiply up & down by $ \sqrt{x\plus{}3}\plus{}3$

Did that, and simplified but I get an answer of 0, it should be 6.

Or if you prefer

\begin{eqnarray*}\mathop{\lim }_{x\to 6}\frac{{x-6}}{{\sqrt{x+3}-3}}&=&\mathop{\lim }_{x\to 6}\frac{{x+3-9}}{{\sqrt{x+3}-3}}\\ &=&\mathop{\lim }_{x\to 6}\frac{{\left({\sqrt{x+3}+3}\right)\left({\sqrt{x+3}-3}\right)}}{{\sqrt{x+3}-3}}\\ &=&\mathop{\lim }_{x\to 6}\left({\sqrt{x+3}+3}\right)\end{eqnarray*}

Does that make sense?

Ahh, I didn't spot the diff. of squares. Thanks a bunch! Also, where the heck is my mistake here?

$ \lim_{x\rightarrow 6}\frac{x\minus{}6}{\sqrt{x\plus{}3}\minus{}3}\equal{}\lim_{x\rightarrow 6}\frac{x(1\minus{}\frac{6}{x})}{\sqrt{x(1\plus{}\frac{3}{x})}\minus{}3}$

$ \equal{}\lim_{x\rightarrow 6}\frac{x^{1}(1\minus{}\frac{6}{x})}{x^{\frac{1}{2}}\sqrt{(1\plus{}\frac{3}{x}) }\minus{}3}$

$ \equal{}\lim_{x\rightarrow 6}\frac{x^{\frac{1}{2}}(1\minus{}\frac{6}{x})}{\sqrt{(1\plus{}\frac{3}{x}) }\minus{}3}\times\frac{\sqrt{1\plus{}\frac{3}{x}}\plus{}3}{\sqrt{1\plus{}\frac{3}{x}}\plus{}3}$

$ \equal{}\lim_{x\rightarrow 6}\frac{\Big(x^{\frac{1}{2}}(1\minus{}\frac{6}{x})\;\Big)\left(\sqrt{1\plus{}\frac{3}{x}}\plus{}3\right) }{ (1\plus{}\frac{3}{x})\minus{}9}$

Between lines $ 2$ and $ 3$, you did $ \frac{x}{x^{1/2}}\equal{}x^{1/2}$, but this wasn't allowed due to the fact that the denominator is $ x^{1/2}\sqrt{x\plus{}3}\minus{}3$ and not $ x^{1/2}\left(\sqrt{x\plus{}3}\minus{}3\right)$...very easy to miss due to the square root almost blending in to the fraction line.

[quote="Prime factorization"]Simply find $ L \equal{}\lim_{x\rightarrow 6}\frac{x\minus{}6}{\sqrt{x\plus{}3}\minus{}3}$[/quote]
[b]Proof.[/b] Denote $ \sqrt{x\plus{}3}\equal{} t$, i.e. $ x \equal{} t^{2}\minus{}3$. Thus, $ x\to 6$ means $ t\to 3$ and $ L \equal{}\lim_{\begin{array}{c}t\to 3\\ t\ne 3\end{array}}\frac{t^{2}\minus{}9}{t\minus{}3}\equal{}\lim_{\begin{array}{c}t\to 3\\ t\ne 3\end{array}}(t\plus{}3) \equal{} 6$.

All solution has same method so let
$ t\equal{}\sqrt{x\plus{}3}$ not many diference from above solution.

In this very simple case no, but generally, yes, is a great difference !

Now I have an even trickier problem:

$ \lim_{x\rightarrow 2}\frac{\sqrt{6\minus{}x}\minus{}2}{\sqrt{3\minus{}x}\minus{}1 }\equal{}\lim_{x\rightarrow 2}\frac{\sqrt{6\minus{}x}\minus{}2}{\sqrt{3\minus{}x}\minus{}1 }\times\frac{\sqrt{3\minus{}x}\plus{}1}{\sqrt{3\minus{}x}\plus{}1}$

$ \equal{}\lim_{x\rightarrow 2}\frac{\sqrt{(6\minus{}x)(3\minus{}x)}}{2\minus{}x}\minus{}\frac{2\sqrt{3\minus{}x}}{2\minus{}x}\plus{}\frac{\sqrt{6\minus{}x}}{2\minus{}x}\minus{}\frac{2}{2\minus{}x}$

But then I get division by 0 for every single fraction :o

[quote="Prime factorization"]Ascertain $ L \equal{}\lim_{x\rightarrow 2}\frac{\sqrt{6\minus{}x}\minus{}2}{\sqrt{3\minus{}x}\minus{}1 }$.[/quote]
[b]Proof.[/b] $ L \equal{}\frac{1}{2}\cdot$  $ \lim_{x\to 2}\frac{\left[\left(\frac{6\minus{}x}{4}\right)^{\frac{1}{2}}\minus{}1\right]}{\frac{6\minus{}x}{4}\minus{}1}\cdot\frac{(3\minus{}x)\minus{}1}{(3\minus{}x)^{\frac{1}{2}}\minus{}1}\equal{}\frac{1}{2}$. I used the remarkable limit  $ \lim_{t\to 1}\frac{t^{\alpha}\minus{}1}{t\minus{}1}\equal{}\alpha\ \ (*)$, where $ \alpha\in\mathcal R^{*}$. You can choose $ m \equal{} 3$ !

For [color=darkred][b]TTsphn[/b][/color] and [color=darkred][b]Prime factorization[/b][/color] I"ll present a difficult exercise ([b]without l'Hospital's rules[/b]).

$ \lim_{x\to 0}\frac{1}{x^{2}}\cdot\left(\sqrt [m]{1\plus{}x}\plus{}\sqrt [m]{1\minus{}x}\minus{}2\right)$, where $ m\in N$, $ m\ge 2$. Here the limit $ (*)$ is very usually. You can choose $ m\equal{}3$ !

No problem
$ \sqrt[m]{x\plus{}1}\minus{}1 \equal{}\frac{x}{{\sqrt[m]{x\plus{}1}}^{m\minus{}1}\plus{}....\plus{}1}$

[color=darkblue]Go on, [b]TTsphn[/b] ... , you'll meet again $ \frac{0}{0}$ ?![/color]


greatest common divisor
number theory unsolved
number theory

Two players play the following game. They alternately write divisors of 100! on the blackboard, not repeating any of the numbers written before . The player after whose move the greatest common divisor of the written numbers equals 1, loses the game. Which player has a winnuing strategy?

Suppose that a prime $p$ is written at some point. Then all numbers written (before and after) must be multiples of $p$, and any multiple of $p$ can be written. Then, when the game ends, there are $100!/p$ numbers on the blackboard, which is even. So the first player must then lose.

Now, if the first player writes a prime $p$ on his first move, he loses. If he writes a composite number $n=pk$, with $p$ a prime divisor, then the second player writes $p$, and again the first loses. So the second player has a winning strategy.


MATHCOUNTS Duke college AMC AIME USA(J)MO USAMO

Did you take the SAT (TIP)? How did you do if you did?

No, it doesn't belong in MathCounts. I'm moving it to The Round Table.

Typically, the home of questions like this seems to be the College forum, so it is getting re-forwarded.

I got

730 Math
540 Verbal  :mad:
500 Writing  :blush:

The reason I posted it in Mathcounts is so I could see how people who do mathcounts do in 7th grade...

I'm happy to have posts about SAT I scores (at any age) posted here. Yes, any participant is welcome to apply the "show me yours before I show you mine" rule to discussions of SAT I scores (or, for that matter, to discussions of other test scores).

College Confidential threads on the same subject include

http://talk.collegeconfidential.com/showthread.php?t=40823

and

http://talk.collegeconfidential.com/showthread.php?t=78732

[quote="SplashD"]It's common courtesy to give your score before asking others for theirs[/quote]
SOrry. 800-Math
640- Verbal
600-Writing

I can't remember exactly how I did, but it was something like

800 Math
650 Verbal
750 Writing

Can this be moved back to mathcounts? I was trying to see how people who do well in Mathcounts do on the Sat in 7th grade through TIP...

The MATHCOUNTS moderator doesn't want this thread in the MATHCOUNTS forum, because it doesn't have much to do with MATHCOUNTS.

740 math
720 english
lets say ididn't do as well in writing

760 math
570 CR

i took it 2 years ago, the last old sat, no essay, hehe :P

I think the OP was asking about SAT scores in 7th grade to qualify for Duke TIP.  I'll give you my score from back then, but it was from when we only had 2 sections.

M - 780
V - 650

OK

I took TIP as 7th grader in January 2006.

Math:640
Verbal:490 (ran out of time a lot lol)
Writing:530

I have only done the National Mythology Exam and the intro to latin one and the iowa but i didn't do the SAT though i made the scores so yeah...i might do it this year though!!!1

EDIT: TO PLEASE joml88 and to be nice  :wink:

(a) At least try to make coherent sentences.  Good grammar would be a start...
(b) Yes, those are the lyrics.  Go on youtube.com and search for "white and nerdy".

According to the most recent practice test,

Math: 800
CR: 650
Writing: 730

I really need to make sure I don't zone out during reading comprehension...well I did take it from like 5:30PM to 10PM...

A question about the PSAT...is it so important in RSI admission?

[quote="tokenadult"][quote="K81o7"]A question about the PSAT...is it so important in RSI admission?[/quote]

I hear that it's one thing that is important. I think there are some alumni here who can tell you more.[/quote]
Really?  :huh:
Wow, I never knew that. (Good thing I know now though.  :) )

Also, one thing I've been noticing about my SAT/PSAT scores is that my Writing score is pretty low (I took a practice PSAT a few weeks ago, which was the first time I ever did a test with a Writing section).
Is Writing easy or hard to improve? Also, what is the best way to do so?
And also, my Reading score is pretty good, but how can I get it to about perfect? (There always seem to be a few questions I can't get.  :maybe: )

[quote="tokenadult"][quote="K81o7"]A question about the PSAT...is it so important in RSI admission?[/quote]

I hear that it's one thing that is important. I think there are some alumni here who can tell you more.[/quote]

Not sure alumni would be particularly knowledgeable about it :P, but in the application it says

[quote="RSI Application"] PSAT math scores should be at least 75, and combined math, verbal and writing PSAT scores should be at least 220. ACT minimum math scores should be 29 and reading 27. Lower scores must be offset by strong indications of mathematical, scientific, and academic potential exemplified in recommendations, high school grades, and science activities.[/quote]

Though if I recall correctly it "recommended" PSAT composite of >225 when I applied last year.

[quote="paladin8"]Not sure alumni would be particularly knowledgeable about [admission requirements for Research Science Institute] :P, but in the application it says

[quote="RSI Application"] PSAT math scores should be at least 75, and combined math, verbal and writing PSAT scores should be at least 220. ACT minimum math scores should be 29 and reading 27. Lower scores must be offset by strong indications of mathematical, scientific, and academic potential exemplified in recommendations, high school grades, and science activities.[/quote][/quote]

Thanks. Yeah, every RSI alumnus I've ever heard from, online or in person, doesn't have a definitive answer about what it takes to get in, but good PSAT scores apparently don't hurt.

[quote="Kent Merryfield"]In my own case, I'll have 7 students taking the Putnam. I know who the three members of the team are. I've let it be known who one of the three is. (He's the most experienced member of the group; you probably met him in Las Vegas last June.) I'm leaving the identities of the other two as a mystery. That way, every one of the 7 will come knowing that their score might become part of the school's team score.[/quote]

This is the case at my school as well.  The coordinator has let four students know that they are the team + alternate, in some combination.  As the person with the highest individual score last year and the strongest math background, I am not amused by this, since it's not like there's any doubt that any of the four of us will fail to show up on contest day.

[quote="Iversonfan2007"][quote="RSI Application"]ACT minimum math scores should be 29 and reading 27.[/quote]

wow....do they superscore the ACT or something, a 220 PSAT is much much much much harder to get than a 29 and 27.[/quote]

I'm not sure. As far as I know, this is the first year they have included an ACT "minimum" score, so maybe they don't want to discourage applicants or something.

[quote="K81o7"]The reason I'm asking is as follows...I don't want to apply with a PSAT score below their cutoff and then find out that 220 is a REQUIREMENT as opposed to a recommendation.  It just seems that PSAT scores aren't a very good indicator of ability.  I mean, I will be doing math if I do get in, so will they really pay attention to a lower than expected CR score if I have teacher recommendations, contest scores, stuff from other programs, etc?[/quote]

Well, like it says, as long as you have sufficient other things to back it up, you should go ahead and apply.

wait, does that mean that the ACT is required for RSI this year?

[quote="nebula42"]
Is Writing easy or hard to improve? Also, what is the best way to do so?
[/quote]
Read Strunk and White's [i]The Elements of Style[/i].  It's a very very concise and excellent book on writing style and common errors.  Outside of the essay portion, you can improve multiple-choice by studying the types of grammatical errors they use when writing questions: subject-verb agreement, tense agreement, etc.
[quote="nebula42"]
And also, my Reading score is pretty good, but how can I get it to about perfect? (There always seem to be a few questions I can't get.  :maybe: )[/quote]
Besides just lots of practice, I think learning to read slowly and carefully helped me a lot.  I used to read too fast to get the gist and make silly mistakes (on math problems too.  I used to solve for area instead of altitude a lot...)  There are some common strategies that Princeton Review uses that can be helpful.  Their book is probably quite useful and lot more economical than any class.

Regarding RSI:
I dislike how everyone places so much emphasis on RSI.  It's not the end of the world if you don't get in.  I think it can be beneficial not to go and instead look for an interesting lab at your local university.  It's my opinion that it's more important to be really enthusiastic about your summer research and do a good job because of it rather than feel pressured to work hard just because RSI is a competitive program.[/i]


$ 6$ colors of paint are are available.   Each face of a cube is to be painted a different color.  In how many different ways can this be done if $ 2$ colorings are considered the same when one can be obtained from the other by rotating a cube?

Source: Challenging Math Problems with Elementary Solutions Volume $ 1$

We went through this one in training. If I remember my solution correctly,

[hide]
To make comparing easier, pick one color and put it on top all the time. You have 5 ways to paint the opposite face. The remaining 4 faces can be chosen in $ 3! \equal{} 6$ ways (it's a circular permutation of 4 items). We thus have $ 30$ total possibilities.
[/hide]

Not quite what I did.
I overcounted, and then divided by the necessary means.
I did this:
The first face has $ 6$ possibilities, the second $ 5$, the third $ 4$... giving us $ 6!$ combinations.  Of course, this is overcounting.  For each arrangement, choose $ 1$ color.  Place this on any of the $ 6$ faces.  Notice, now, that the rest of the cube is set, other than the $ 4$ possible twists of it.  Therefore, for each combination, we are overcounting it $ 24$ times.  Therefore, the answer is $ \frac{6!}{24}\equal{}\boxed {30}$

This is a rather special case of [url=http://en.wikipedia.org/wiki/Burnside%27s_lemma]Burnside's lemma[/url] (specifically, the case where every stabilizer looks the same).  Here's a much harder question:  what if the colors don't have to be different?  (You still get at most $ 6$ of them.)

Casework? You have: $ 1, 2, 3, 4, 5,$ and $ 6$ colors to deal with; I would approach each case- can't take TOO long.

Would you like to try it?  :)  I would encourage you to try the Burnside's lemma calculation first.


let $x,y,z \in [1;2]$ .
  find the maximum value  and min value of the expression :
    $P= \dfrac{10a}{bc} +\dfrac{11b}{ac} +\dfrac{12c}{ba}$

For the minimum use AM-GM and then put $a=b=c=2$ so

$P\geq 3 \sqrt[3] {165}$

Are you sure this minimum can be attained?  :D


The problem:

A factory is producing cars, small cars, trucks, buses and oil trucks. The demand for these two products is 10 buses and 20 oil trucks in a three month period, for the other three products demand is unlimited. The capacity of the erecting shop is 100000 working hours, for the carpentry department it is 36000, for the car polishing department it is 45000, for the upholstering department it is 90000, and for the car electricity department the capacity is 24000 working hours. The specific value of the different products is shown in the table below
Departments	  Car 	Small Car 	Truck	Bus	Oil Truck
Erecting Department	      100	40	150	300	100
Carpentry Department	0	0	60	40	0
Upholstering Department	30	10	0	100	15
Car Polishing Department	30	10	5	30	20
Car Electricity Department	10	5	5	20	12
in an hour / vehicle dimension.

This table shows , that the producing of one bus requires 300 working hours in the erecting department, 40 working hours in the carpentry department, and so on.

How many products does the factory have to produce from the different types, if they cannot exceed the capacity of the different departments, they have to satisfy the demand for buses and oil trucks of the country, the number of cars plus the number of small cars produced have to be equal to the number of trucks produced, and the firm should maximize the number of vehicles produced.

Hmmm...I got a maximum of 550 automobiles when I used the minimum for buses and oil trucks. I don't knwo if that's the absolute maximum, though.


function
real analysis
real analysis solved

Consider a continuous function $f(x):[0;1]->R$ and $f(0)=f(1)$.
Prove that $\forall a \in (0;1)$ we have (1) or (2) is true

(1)$\exists x \in [0;a]$ such that $f(x)=f(x+1-a)$
(2)$\exists x \in [0;1-a]$ such that $f(x)=f(x+a)$

Having $g:[0,1] \to R$,
               so that $g(x)=f(x)-f(x+1-a)$.
We have:$g(0)=f(0)-f(1-a),g(a)=f(a)-f(1)$ ,
               so $g(0)g(a)=[f(0)-f(1-a)][f(a)-f(1)]$
Also,$h:[0,1] \to R$,
                so that $ h(x)=f(x)-f(a+x)$
Also we have that:$h(0)=f(0)-f(a),h(1-a)=f(1-a)-f(1)$ ,
                  so $h(0)h(1-a)=[f(0)-f(a)][f(1-a)-f(1)]$
So we have $g(0)g(a)=h(0)h(1-a)$,and than from Darboux,we have the conclusion:
So I belive that $(1) \Leftrightarrow (2)$.

Assume none of those is true.

Assume $f(x+a)>f(x),\ \forall x$ s.t. $f(x+a)$ makes sense (we may assume this WLOG). We then have $f(a)>f(0)=f(1)$, meaning that for $x=a$, we have $f(x+1-a)<f(x)$. This means that we have $f(x+1-a)<f(x),\ \forall x$. Consider now a point $x_0$ where $f$ reaches its minimum on $[0,1]$. Then, at least one of $x_0-a,x_0+1-a$ belongs to $[0,1]$. Suppose it's $x_0-a$. We thus have $f(x_0-a)<f(x_0)\le f(x_0-a)$, which is a contradiction.


Considering the IMO team leaves tomorrow for training, I think its a good time to wish them all luck on the IMO.  Hehehehe make us proud!!!!

Yay, go Canada!

Bonne chance à tous les hommes qui représentent le Canada aux Olympiades internationales de mathématiques. Je suis désolé—je ne peux pas trouver les mots exacts pour «International Mathematics Olympiad». Je viens espérer que je serai dans l'équipe d'OIM quand je suis à la dixième année!  :D

... and I did look up words if you're wondering.  :roll:

Yeah, good luck guys..And don't forget, you guys have by far the coolest logo out there!! :D

[quote="billzhao"]d.lam.86, why are you posting in french ?!?!  :wacko:[/quote]
I don't know. I just felt like writing French because I had no French classes today. I spent the entire afternoon (3½ hours) outside so I missed Science and French classes. [b]Anyone care to add me to their Messenger list?  :D [/b]

[quote="Sunny"][quote="d.lam.86"]
I don't know. I just felt like writing French because I had no French classes today.[/quote]

You strange strange person.   :D[/quote]
D`oh!   :ewpu:

[quote="Elyot"]French eh?

我認為它會有更多道理如果加拿大IMO 隊的官方語言是漢語。

For David:
행운을 빕니다!
For the rest of the team:
祝好運!

Does that work?  You guys need to have Chinese and Korean installed on your computers...

I hope that made at least some sense... in Japanese, I would say GAMBATTE!!![/quote]

:rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl:
Yeah it makes sense alright.....
Even though the proper grammar for the Chinese one should be (I think, my Chinese is rusty)
我认为加拿大IMO队的官方语言因该是汉语。
For Elyot:
Good luck!

[quote="Richard Peng"]:rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl::rotfl:
Yeah it makes sense alright.....
Even though the proper grammar for the Chinese one should be (I think, my Chinese is rusty)
我认为加拿大IMO队的官方语言因该是汉语。
For Elyot:
Good luck![/quote]

Hmmm... can't read half of what you put (I only have traditional Chinese installed on my computer... it would probably show up if I DLed simplified.)

Wow Richard... you've like... almost doubled your post count today!

[quote="Elyot"]Wow Richard... you've like... almost doubled your post count today![/quote]
That's because today is my second day posting on aops. As you can see, I'm quite consistent so far :lol:

Good job... at 4 posts/day, only 5 or 6 months until you catch sarah and yufei... that's assuming they don't post though... :P

[quote="Elyot"]French eh?

我認為它會有更多道理如果加拿大IMO 隊的官方語言是漢語。

For David:

행운을 빕니다!

For the rest of the team:

祝好運!

Does that work?  You guys need to have Chinese and Korean installed on your computers...

I hope that made at least some sense... in Japanese, I would say GAMBATTE!!![/quote]
Oh!!! How did you get the translation? It's perfectly written! :?

[quote="Elyot"]David, another useful website that may be helpful:

[url=http://babelfish.altavista.com]http://babelfish.altavista.com[/url]

Translates anything (not perfectly... but pretty good.)[/quote]
Eloyt is my neighbour in Banff. I saw him typing this. His typing is slower than what I imagined(even though he is faster than me).

[quote="lightrhee"][quote="Elyot"]David, another useful website that may be helpful:

[url=http://babelfish.altavista.com]http://babelfish.altavista.com[/url]

Translates anything (not perfectly... but pretty good.)[/quote]
Eloyt is my neighbour in Banff. I saw him typing this. His typing is slower than what I imagined(even though he is faster).[/quote]

These keyboards are shiz... whenever I try to type fast, I make a mistake.  I've hit the cAPS LOCK KEY SO Many times when trying to type an A.

Hahaha...what's it like?  Is it raining?

[quote="Sunny"]Hahaha...what's it like?  Is it raining?[/quote]
No, it stopped to rain.Clouds are starting to disappear. :)

[quote="Sunny"][quote="lightrhee"]
No, it stopped to rain.Clouds are starting to disappear. :)[/quote]

Lucky.  My mom was an hour late picking me up because of traffic due to flooding...[/quote]
Holy sh!t, there is no rain or flooding in Toronto. Thank goodness.  :?

Hahahaha...but they are scary when mad!!

[quote="lightrhee"]Read this. This is one of the things that were placed on my table in Banff Centre .It's funny.(This is not everything. I just typed interesting ones)
[b]
Elf Calving Season has begun(May 15 - July 1)[/b]
Do  not approach elk. They are all potentially dangerous.
 :moose: =  :diablo:

Stay back 100 metres.
 :moose:  <----100 metres -----> :)
[b]
Signs of Elk Aggression [/b]
Teeth exposed
 :D

Elk looking directly at you with whites of eyes showing
:w00t:

Ears back
:clap:

Some aggressive elk have been eartagged or marked with a paint ball on the hip or shouder.
:rotfl:

Aggressive elk may charge. If charged, get behind a tree or any large object.
 :?:[/quote]
Does this concern me at all? I don't want to be attacked by elks.  :huh:  :(


function
limit
trigonometry
algebra
domain
induction
algebra unsolved

Let f(x) be a real-valued function defined for all positive x, satisfying
$ f(x\plus{}y) \equal{} f(xy)$ for all positive x, y. Prove that f is a constant function.

let $ c,r>0$. then $ x\plus{}\frac{c}{x} \equal{} r\Leftrightarrow x^2\minus{}rx\plus{}c\equal{}0\Leftrightarrow x\equal{}\frac{r\pm \sqrt{r^2\minus{}4c}}{2}$.

so for $ r\geq 2\sqrt{c}$, we can set $ x\equal{}\frac{r\plus{}\sqrt{r^2\minus{}4c}}{2}$ and $ y\equal{}\frac{c}{x}$ to get $ f(r)\equal{}f(c)$ for $ r\geq 2\sqrt{c}$. letting $ c\to 0$ shows that $ f$ must be constant.

I didn't understand why letting $ c\minus{}>0$ then f must be constant.

pleurestique concludes that when $ c$ is arbitrarily small, $ f(r) \equal{} L$ for all $ r \ge 2 \sqrt{c} > 0$ (where $ L \equal{} \lim_{x \to 0^{\plus{}}} f(x)$).  The problem, of course, is that it's not clear that $ L$ exists.

to make this more precise...

we want to show that for any $ x,y>0$, $ f(x)\equal{}f(y)$. we may assume WLOG that $ x<y$. choose $ c>0$ such that $ 2\sqrt{c}<x$. then by the argument i gave above, $ f(x)\equal{}f(c)\equal{}f(y)$.

Question: would it be enough to prove that $ P(2)\equal{}P(1)$ if $ x, y \equal{} 1$, then $ P(n\plus{}1)\equal{}P(n)$, if $ x\equal{}n, y\equal{}1$, which concludes that $ P(1)\equal{}P(2)\equal{}p(3)\equal{}\cdot\cdot\cdot\equal{}P(n)$? This shows that $ P(x)$ is independent of $ x$, right? So $ P(x)$ is a constant?

your argument only shows that $ f$ is constant on the natural numbers. $ f$ is defined on the positive reals.

For example, the function $ P(x) \equal{} \sin (2\pi x)$ is constant on the naturals (that is, has period $ 1$), but of course it is far from constant.

why it is enough to show that $ f(x)\equal{}f(y)$ for all $ x,y>0$ to assure that f must be constant?

that's what it means for $ f$ to be constant

I can set for example f(x)=sin x and f(y)=sin y, if I set $ x\equal{}\pi/2$ , $ y\equal{}\pi/2$ or $ x\equal{}5\pi/2$, $ y\equal{}5\pi/2$, then $ f(x)\equal{}f(y)$ and they are not constant functions.

The result was proved for [i]any[/i] x and y.

You can conclude, for example, that, being c = f(0), f(x) = c for every x.

danilorj: in your post prior to your last one, you yourself stated "for all $ x,y > 0$", then in the "counterexample" you gave you only picked a particular pair $ x,y$. if for ALL $ x,y$ in the domain of $ f$, $ f(x) \equal{} f(y)$ (call this statement (*)), then certainly $ f$ is constant: fix any $ c$ in the domain of $ f$; then $ f(x) \equal{} f(c)$, by statement (*). get it now?

it is still not clear to me why f(x)=f(c)=f(y) when c is arbitrarily small. Could you explain it in more details?

i think you should try to think about it a little more on your own; that would help a lot.

You get that $ f(r)\equal{}f(c)$ for $ r>\equal{}2\sqrt{c}$. But f must be constant for every x and y without this restriction.
Just below this problem there is a hint to prove by induction that f(x+n)=f(x) for every positive real x and positive integer n in order to remove this restriction, but I still don't know how to use it.

[quote="danilorj"]You get that $ f(r) \equal{} f(c)$ for $ r > \equal{} 2\sqrt {c}$. But f must be constant for every x and y without this restriction. [/quote]

Given any $ x$ and $ y$, you can find $ c < \left( \frac{ \text{min}(x, y)}{2} \right)^2$ such that $ f(x) \equal{} f(y) \equal{} f(c)$.  The key is that $ c$ is arbitrary.

[quote="pleurestique"]let $ c,r > 0$. then $ x \plus{} \frac {c}{x} \equal{} r\Leftrightarrow x^2 \minus{} rx \plus{} c \equal{} 0\Leftrightarrow x \equal{} \frac {r\pm \sqrt {r^2 \minus{} 4c}}{2}$.

so for $ r\geq 2\sqrt {c}$, we can set $ x \equal{} \frac {r \plus{} \sqrt {r^2 \minus{} 4c}}{2}$ and $ y \equal{} \frac {c}{x}$ to get $ f(r) \equal{} f(c)$ for $ r\geq 2\sqrt {c}$. letting $ c\to 0$ shows that $ f$ must be constant.[/quote]
A slightly different solution

$ \boxed{f(x\plus{}y) \equal{} f(xy)}$

for any $ x>0$:

for $ y\equal{}1$ we get $ f(x\plus{}1)\equal{}f(x) \ \ (1)$

for $ y\equal{}\frac{1}{x}$ we get $ f(x\plus{}\frac{1}{x}) \equal{} f(1) \ \ (2)$

For any $ w\geq 2$ let $ x \equal{} \frac{w\plus{}\sqrt{w^2\minus{}4}}{4}$

Then $ w\equal{}x\plus{}\frac{1}{x} \Rightarrow f(w)\equal{}f(1), \  \forall w \geq 2$

Now, for any $ a\in (1,2)$, from $ (1),(2)$ we get $ f(a)\equal{}f(a\plus{}1)\equal{}1$

and then, from $ (1)$ we get  $ f(a\minus{}1)\equal{}f(a)\equal{}1$

So finally, for any $ x>0$ it is $ f(x)\equal{}f(1)$


calculus
group theory
abstract algebra
function
linear algebra
matrix
topology

When I Pass exam in Algebra : there was some question at the first establish a bit of the notation which will be used in this question.
Defination: $G$ Group We can write as half direct product of two subgroup  $N$ and $H$ if:

1)$N$ is normal subgroup in $G$.
2)$H$ is subgroup in $G$.
3) intersection of $H$ and $N$ is equal to $e$ where $e$ is identity element
4) $\forall{g}\in{G}$,  $g=hn$ where $h\in{H}$ and $n\in{N}$ only
then we can write $G=N\bullet{H}$.
if $N$ is normal subgroup in $G$ then we write (defination) $N\triangleleft{G}$. if $H$ is subgroup in $G$ then we write (defination): $H\leq G$
Theorem: if $G=N\bullet{H}$ then we have $G/N\cong{H}$.in this case  $\cong$ is isomorphism, this mean biection mapping. and if $f$ is isomorphism $f: G\to G'$ then we have $f(x*y)=f(x)\circ{f(y)}$ where $\circ$ is binnary operation in $G'$ and $*$ is binnary operation in $G$.
defination:$N$ is Normal subgroep in $G$ mean $gN=Ng$ where $\forall{g}\in{G}$.
question:  show some example when we have isomorphism  $G/N\cong{H}$ where $N$ is normal subgroup in $G$ and $H$ is subgroup in $G$, but there is not half direct poduct.  I want exampl where $G$ will be from "real calculus" I mean $G$ is some category of function or functional, or polinomial.. from "real calclulus".

[quote="mlok"]I am trying to think of a non-abelian group that comes up in real calculus... :maybe: Allowing matrix groups would mean no restriction at all.[/quote]
My answer in exam was: $GL(n,k)/\lambda{E}\cong{SL(n,k)}$. Where K is from some good field : I mean if $ab=0$ where $a\in{K},b\in{k}$ then $a=0$ or $b=0$. We easy check this is isomorphism, but $SL(n,k)$ intersection $\lambda{E}$ is equal to $(e)$ and $\forall{g}\in{G}$, $g=a'b'$ where $a'\in{SL(n,k)}$ and $b'\in{\lambda{E}}$. therefore there exist half direct product  :huh: . Then I wanted use real calculus and wanted took example from real calculus, but I cant find nothing.

[quote="Kent Merryfield"]How about the set of strictly increasing continuous functions from $[0,1]$ onto $[0,1]$ with composition of functions as the group operation?

Or, more generally, the set of homeomorphisms from some compact set onto itself?[/quote]
thanks. identity element will be $f(x)=x$. there allways exist return elemnt $g^{-1}(x)$ becauset it is continuous increasing function. Good. and what's about normal subgroup? we must chekc: if $N$ is normal subgropup if and only if when $\forall{f}\in{F}$ we have $f\circ{n}\circ{f^{-1}}\in{N}$ where $n\in{N}$.

A clarification may be needed here. If all you wanted was a group $G$ and two subgroups $N\triangleleft G$ and $H<G$ such that $G/N\cong H$ but $N\cap H\ne\{e\}$, then this is easy to arrange. For example, $G=\mathbb R^{2}$ with $H=N=\mathbb R\times\{0\}\subset \mathbb R^{2}$.
But I think that the problem was to find $G$ and $N\triangleleft G$ so that $G/N$ is isomorphic to a subgroup of $G$, but no such subgroup forms a semidirect product with $N$. You can try

[quote]the Weil group $W_{\mathbb R}$ of ${\mathbb R}$. This group is a non-split extension of the multiplicative group $\mathbb C^{*}$ of $\mathbb C$ by $\mathrm{Gal}(\mathbb C/\mathbb R)=\mathbb Z/2\mathbb Z$. Explicitly, $W_{\mathbb R}={\mathbb C}^{*}\cup\sigma{\mathbb C}^{*}$, where $\sigma^{2}=-1$ and $\sigma z \sigma^{-1}=\overline{z}$ for $z\in\mathbb C^{*}$.[/quote]
[url=http://www.stieltjes.org/archief/rep9899/node16.html]source[/url]

I admit that this does not look like real calculus.

[quote="mlok"]
I admit that this does not look like real calculus.[/quote]
Thanks anyway


Hello everyone,

I would appreciate any help in the following two problems. Thanks!

---

[u]Write the equation of a sine function that has the given characteristics.[/u]

1. Amplitude: 3
Period: $ \pi$

Textbook's answer: [hide]$ 3\sin 2x$[/hide]
My answer: $ 3\sin\pi x$

According to the textbook's answer, the period was calculated with Period = $ \frac{2\pi}{\pi}$ = 2. My question is: Why do we need to find the new period when the question originally gave us $ \pi$ as the period?

2. Amplitude: 3
Period: $ 3\pi$
Phase shift: $ \frac{\minus{}1}{3}$

Textbook's answer: [hide]$ 3\sin\frac{2}{3}(x\plus{}\frac{1}{3})$[/hide]
My answer: $ 3\sin 3\pi(x\minus{}\frac{1}{9\pi})$

Same question here.

Note that the period of $ \sin x$ is $ 2\pi$, so thus, if you have $ \sin kx$, then it's period is $ \frac{2\pi}{k}$.  Changing amplitude is simply a matter of multiplying the entire function by a constant, so you are both correct in that part.

To shift, you simply add the negative of how much you want to go to the right.

Shift before you change the period or amplitude!

I'm not sure where you are getting the extra $ \pi$s from in your answer...  That would make the function have period 2, for #1...


QUESTION 9

As you have found the mathematical model can be expressed in the form H = ab^n.
The relationship between fatling distance(s) and time(t) can be expressed as s = 0.5g^t

g the acceleration due to gravity on Earth is considered to be constant at 9.8ms^-2

Using the above information, calculate the total distance that your ball will travel and how long this will take.

(Remember you need to establish your answer with thorough mathematical reasoning and justification)

Would you clearify the question for us please....

Clearly a homework question, and you need information about the ball anyway. How long it is falling, perhaps?

If your trying to find how much distance does the ball travels after it is done bouncing, it's quite easy to find it using infine geometric series....


See my link in your original thread : http://www.mathlinks.ro/Forum/viewtopic.php?t=81659 .  :)


In Pat's first 30 foul shots at basketball practice, he made 2 baskets for every 3 that he missed. How many shots must he now make consecutively to bring his percentage up to $ 50\%$?

He made 12 and missed 18.  He must make $ \boxed{6}$ more.

*better solution*

He got 12 and the total is 12+18=30. To bring his score up to 50% with $ x$ being the answer, \[ \frac{12\plus{}x}{30\plus{}x}\equal{}\frac{1}{2}\]\[ 24\plus{}2x\equal{}30\plus{}x \implies 24\minus{}30\plus{}2x\equal{}x \Rightarrow 24\minus{}30\equal{}\minus{}x \implies x\equal{} \boxed{6}\]

Or we can note that it is one "unit of 5", so $ \frac{30}{5}\equal{}\boxed{6}$.

No offense, akhil, but AIME's solution actually [i]is[/i] better.

^Sorry for backseat modding, but you didn't need to revive a 3 1/2 year thread just for that post


analytic geometry
search
combinatorial geometry
topology
real analysis
real analysis unsolved

Does there exist a compact subset $K$ of $\mathbb R^n$ such that for any $x\in\mathbb R^n\setminus K$, the minimal distance from $x$ to $K$ is achieved for more than one point of $K$?

Take the point $p$ in $K$ with the largest $x_1$ coordinate and take $x=p+e_1$ where $e_1$ is the first coordinate vector. Then $p$ is the unique closest point to $x$ in $K$. More interesting question: is it always true that the set of the points $x\in \mathbb R^n\setminus K$ such that the closest to $x$ point of $K$ is unique has positive measure?

[quote="fedja"]Is it always true that the set of the points $x\in \mathbb R^n\setminus K$ such that the closest to $x$ point of $K$ is unique has positive measure?[/quote]

I think the answer is 'yes' and that the following (sketch of an) argument can be extended to thegeneral case, but I'm only (almost :)) sure it works for the plane right now.

Let $K$ be the respective compact subset of $\mathbb R^2$, and let $\mathcal C$ be the convex hull of $K$. Let's assume that $\mathcal C$ has non-empty interior. For any $x\not\in\mathcal C$, there is a unique point $p(x)$ on the boundary of $\mathcal C$ such that $|x-p(x)|=\min_{y\in\mathcal C}|x-y|$. Now, if $p(x)\in K,\ x$ is one of the points which reach minimal distance to $K$ in a single point of $K$, and if $p(x)$ is an extremal point of $\mathcal C$, then it belongs to $K$, since all extremal points of $\mathcal C$ are actually points in $K$. This means that it suffices to prove that $p^{-1}(\mbox{Ex}(\mathcal C))$ has positive measure, where $\mbox{Ex}(\mathcal C)$ is the set of extremal points of $\mathcal C$.

Take a point $x_0$ in the interior of $\mathcal C$, and consider the map $f: \mathbb R^2\setminus\mathcal C\to\mathbb R^2$ defined by $f(x)=x-(p(x)-x_0)$. It's easy to argue that $f(\mathbb R^2\setminus\mathcal C)=f(p^{-1}(\mbox{Ex}(\mathcal C)))$ (i.e. any point in the image of $f$ is the image through $f$ of a point which projects through $p$ onto an extremal point of $\mathcal C$), and that $f(\mathbb R^2\setminus\mathcal C)=\mathbb R^2$. Now, since $f$ decreases distances (this is one of the things that I am not sure hold in $\mathbb R^n$, in general) and takes $p^{-1}(\mbox{Ex}(\mathcal C))$ (which is a closed set) onto the entire plane, $p^{-1}(\mbox{Ex}(\mathcal C))$ must have positive measure, and this is what we wanted.

Yes, that's correct and works in all dimensions :) . Now the most difficult question in this line: is it true that the set of points for which the closest point is not unique always has zero measure?  ;)

Thanks for having the patience to read it :). I did manage to prove that that map I was worried about is distance-decreasing in all dimensions, I think, and I wanted to post it, but every time I tried, I got too lazy to write it down :).

[quote="fedja"]is it true that the set of points for which the closest point is not unique always has zero measure?  ;)[/quote]

I think this is indeed the case. I was aware of everything needed in the proof when you asked the question, but it took me an embarrassingly long time to put two and two together :).

It suffices to show that almost all points outside $K$ achieve their distance to $K$ uniquely. As has been [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=349947457&t=40546]discussed before[/url], the map $x\mapsto d(x,K)$ sending a point to its distance from $K$ is differentiable at a point outside $K$ iff that point achieves its distance uniquely. Coupled with the obvious fact that $x\mapsto d(x,K)$ is $1$-Lipschitz and [url=http://planetmath.org/encyclopedia/RademachersTheorem.html]Rademacher's Theorem[/url] on the almost everywhere differentiability of a Lipschitz map between Euclidean spaces, this gives the desired conclusion.

A recent development along these lines (in two dimensions):
[quote="Bishop and Hakobyan"]The central set $C(D)$ of a domain $D$ is the set of centers of maximal discs included in $D$. Fremlin showed that a central set has to have zero area and asked wether it can have Hausdorff dimension strictly larger than $1$. We answer this affirmatively by constructing a domain with central set of Hausdorff dimension $2$. In fact, given any Hausdorff measure function $h(t)$ such that $h(t)/t^{2}$ tends to infinity, $D$ can be chosen so that $\mathcal H_{h}(C(D)) > 0$.[/quote]
[url=http://www.math.sunysb.edu/~hhakob/papers/central.pdf]Link to the preprint.[/url]


analytic geometry
trigonometry
trig identities
Law of Cosines

1. How many liters of a 25% solution do I have to add to 8 liters of a 12.5% solution to make a 20% solution?

2. danb80 and I both start at the same point. I head north at 5 miles per hour while he heads northeast at 10 miles per hour. What is the distance between us after 5 hours?

3. I start at the origin on a coordinate plane. I go east 1 mile, turn left and go north 2 miles, turn left and go west 3 miles and so on. After I go east 13 miles, what is the sum of the x and y coordinates of my position?

4. On a straight path between A and B, a male cat starts at point A and heads for point B at 6 miles per hour. A female cat starts at point B and heads for point A at 4 miles per hour. When the two cats start off at the same time, a fly starts from the nose of the male cat and flies to the female cat. When the fly touches her nose, it flies back and so on... until he is smashed by the two noses (he takes no time in turning around). If the fly flies at a constant rate of 16 mph and flies for 5 hours before he is crushed, what is the distance from A to B?

5. How many liters of a 20% acid and a 40% acid should I mix to get 20 liters of a 36% acid solution?

only one that i could do pretty quickly was #1. i'll work on rest later.

1. [hide] 12 gallons.
kinda plugged and chugged.
first tried 4 gallons, then 8, then 12. all in multiples of 4 so i could calculate the percentage easily.[/hide]

[/hide]

I tried. The picture is for #2(At least what I think it is..).

[hide="Answers"][hide="#1"]1. let $ x$ be what we are trying to get, which is how many liters of 25% solution.

Using the information given, I set up an equation like this: $ \frac{0.25x+1}{8+x}=\frac15$

I cross-multiplied to get $ 1.25x+5=8+x$ and $ x$ comes out to be $ \boxed{12}$, like what spider got.[/hide]
[hide="#2"]2. I think the picture explains the best. I drew a picture like the one in the very bottom.
I drew vertical(imaginary) line perpendicular from location of danb80 after 5 hours to the path that Chinaboy could not travel in 5 hours.

Since the degrees between North and Northeast is $ 45^{\circ}$, I concluded that my right triangle is right isosceles triangle.

So I got the length of my imaginary line, which I believe is $ 25\sqrt{2}$ and the distance Chinaboy have not traveled, which would be $ 25\sqrt{2}-25$.
From here, I used Pythagorean to get the distance between them.
So $ (25\sqrt{2}-25)^2+(25\sqrt{2})^2=d^2$

I think this equation yields the result of $ \boxed{\approx{36.84063296}}$[/hide]
[hide="#3"]3. This one, I played around with it and it seems to have a pattern. I come up with the answer of $ \boxed{1}$, which was derived from $ (7,-6)$, I think..[/hide]
[hide="#4"]4. I have no idea how I did this.. so it's probably wrong. My work shows that letting $ d$ be the distance of $ \overline{\text{AB}}$, the fly travels $ (d-10)+(d-20)+(d-30)+(d-40)+(d-50)$ miles in 5 hours.

We were given that the fly traveled 16 mph for 5 hours, I made these two equations equal each other:$ (d-10)+(d-20)+(d-30)+(d-40)+(d-50)=16\times5$

By solving this, I get $ \boxed{46}$ miles.[/hide]
[hide="#5"]5. I think this is the one that you use system of equations.
Letting $ A$ be the amount of 20% acid, and $ B$ be the amount of 40% acid, we can have:
$ A+B=20$
$ .2A+.4B= 20\times .36=7.2$
Multiply the last equation by 10 and solving, I get $ \boxed{A=4}$ and $ \boxed{B=16}$ [/hide][/hide]

[hide="another solution for 2"]
Although not quite as elementary level

So we take the picture that DK drew, and by law of cosines

$ d^2\equal{}25^2\plus{}50^2\minus{}2(25)(50)\cos 45$
$ d^2\equal{}625\plus{}2500\minus{}2500\cos 45$
Since $ \cos 45\equal{}\frac{\sqrt{2}}{2}$,
$ d^2\equal{}625\plus{}2500\minus{}2500\frac{\sqrt{2}}{2}$
$ d^2\equal{}625\plus{}2500\minus{}1250\sqrt{2}$
$ d^2\equal{}3125\minus{}1250\sqrt{2}$
$ d\equal{}\sqrt{3125\minus{}1250\sqrt{2}}$
$ d\equal{}25\sqrt{5\minus{}2\sqrt{2}}$

Which is DK's answer, except in exact form
[/hide]

Couldn't resist :)

[hide="um, different answer on number 4"]
Imagine it's just one cat running from A to B at 6+4=10 mph (hence the flew goes from the cat to B, and back and forth).  Since the fly flew for 5 hours, the distance the cat ran is $ 5(10)$ or $ \boxed{50}$ miles.
[/hide]

[quote="alanchou"][hide="another solution for 2"]
Although not quite as elementary level

So we take the picture that DK drew, and by law of cosines

$ d^2 \equal{} 25^2 \plus{} 50^2 \minus{} 2(25)(50)\cos 45$
$ d^2 \equal{} 625 \plus{} 2500 \minus{} 2500\cos 45$
Since $ \cos 45 \equal{} \frac {\sqrt {2}}{2}$,
$ d^2 \equal{} 625 \plus{} 2500 \minus{} 2500\frac {\sqrt {2}}{2}$
$ d^2 \equal{} 625 \plus{} 2500 \minus{} 1250\sqrt {2}$
$ d^2 \equal{} 3125 \minus{} 1250\sqrt {2}$
$ d \equal{} \sqrt {3125 \minus{} 1250\sqrt {2}}$
$ d \equal{} 25\sqrt {5 \minus{} 2\sqrt {2}}$

Which is DK's answer, except in exact form
[/hide]

Couldn't resist :)

[hide="um, different answer on number 4"]
Imagine it's just one cat running from A to B at 6+4=10 mph (hence the flew goes from the cat to B, and back and forth).  Since the fly flew for 5 hours, the distance the cat ran is $ 5(10)$ or $ \boxed{50}$ miles.
[/hide][/quote]
Oh, I see now. Thanks.


inequalities
geometry
circumcircle
inequalities unsolved

If $ R$ is the circumradius of triangle $ ABC$, the prove that

$ a b c (b c \plus{} c a \plus{} a b) \leqslant 27 \sqrt{3} R^5$,

with equality if and only if $ ABC$ is equilateral.

[quote="pacoga"]If $ R$ is the circumradius of triangle $ ABC$, the prove that

$ a b c (b c \plus{} c a \plus{} a b) \leqslant 27 \sqrt {3} R^5$,

with equality if and only if $ ABC$ is equilateral.[/quote]
$ abc\equal{} 4Rpr$  ; $ ab\plus{}bc\plus{}ca \equal{} r^2\plus{}p^2\plus{}4rR$ ;  $ p\le \frac{3\sqrt{3}}{2}R$ ; $ 2r \le R$ ; :)


geometry
rectangle
incenter
trapezoid
cyclic quadrilateral
angle bisector
perpendicular bisector

$ABCD$ is cyclic quadrilateral, $E,F,G,H$ are middle point of arcs $AB,BC,CD,DA$. $P,Q,R,S$ are center of incenter of triangle $ABC,BCD,DCA,DBA$ resp.Prove that $PQRD$ is rectangle whose sides are parallel to $EG,FH$.

$R$ is the incenter of triangle $ACD$ so $GR=GC=GD$
$Q$ is the incenter of triangle $BCD$ so $GQ=GC=GD$
Thus $GR=GQ$. Otherwise $GE$ is the angle bisector of $\angle AGB$ of isosceles triangle $RGQ$ so $GE$ is the perpendicular bisector of $QR$
Similarly $GE$ is the perpendicular bisector of $SP$
This means $SPQR$ is a isosceles trapezium ($SP//RQ$) $(1)$
Similarly $SRQP$ is a isosceles trapezium ($SR//PQ$) $(2)$
From $(1)$ and $(2)$ we have $PQRD$ is rectangle whose sides are parallel to $EG,FH$.


Yes I know the solutions were posted, but where can I find the problems? If there is no site that has the problems yet, can someone tell me what number 4 was? I heard its extremely easy.

On the AMC website (http://www.unl.edu/amc/) click the links that say USAMO Day 1 Test (http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2009-ua/09USAMOday1.pdf) and USAMO Day 2 Test (http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2009-ua/09USAMOday2.pdf).

you could also look on the top bar and go to contests


geometry
inequalities
projective geometry
analytic geometry
ratio
trapezoid
geometry unsolved

I'm sorry I wont post it topic (no inequality)

Let O be the intersection point of the diagonals of the convex quadrilateral
ABCD, AO = CO. Points P and Q are marked on the segments AO and
CO, respectively, such that PO = OQ. Let N and K be the intersection
points of the sides AB, CD, and the lines DP and BQ respectively. Prove
that the points N, O, and K are colinear.

[quote="vittasko"]I can only see that the points $ N,\ K,\ L,$ are collinear (easy to prove), where $ L\equiv AD\cap BC.$

Please, clarify your problem.

Kostas Vittas.[/quote]

i think you understood $ AB \cap CD \equal{}N$, $ DP \cap  BQ \equal{} K$.  But in problem given: $ AB \cap DP \equal{}N$ $ CD \cap  BQ \equal{}K$.
can you  show proof of $ AB \cap CD  ,DP \cap  BQ , AD \cap BC$ is colinear. If it is true then by desargues theorem problem will be proved.

[hide="my solution is ugly"]In coordinate system:
$ O(0,0) ,A(\minus{}a,0) , C(0,a) ,B(p,q), D(\minus{}pk, \minus{}qk),P(\minus{}b, 0), Q(b,0), N(x_0, y_0), K(x_1,y_1)$ then $ x_o \equal{} \frac{apk\minus{}bka\minus{}bkp\minus{}ab}{ak\plus{}b}$ , $ y_0 \equal{} \frac{ak(a\minus{}b)}{ak\plus{}b}$, $ x_1 \equal{} \frac{ab(k\plus{}1)\plus{}pk(b\minus{}a)}{bk\plus{}a}$, $ y_1 \equal{} \frac {qk(b\minus{}a)}{bk\plus{}a}$ then $ \frac{x_0}{y_0} \equal{} \frac {x_1}{y_1} \equal{} \frac {ab(k\plus{}1)\plus{}pk(b\minus{}a)}{qk(b\minus{}a)} \rightarrow O \in NK$[/hide]

Sorry dear all my friends because of I was inattentive in the notations of the problem and thank you dear [b][size=100]Little 1000t[/size][/b], for the clarification.

It is a good case for a solution based on the double ratio (= cross ratio) theory.

So, because of the pencil $ D.APOC,$ is intersected from the line segments $ AC,\ AE,$ where $ E\equiv AB\cap CD,$

we have $ (A,N,B,E) \equal{} (A,P,O,C)$ $ ,(1)$

Similarly, because of the pencil $ B.AOQC,$ is intersected from the line segments $ AC,\ CD,$

we have $ (A,O,Q,C) \equal{} (E,D,K,C) \equal{} (K,C,E,D) \equal{} (C,K,D,E)$ $ \Longrightarrow$ $ (A,O,Q,C) \equal{} (C,K,D,E)$ $ ,(2)$

But, because of $ O,$ is the midpoint of the segments $ AC,\ PQ,$ it is easy to show that $ (A,P,O,C) \equal{} (A,O,Q,C)$ $ ,(3)$

From $ (1),$ $ (2),$ $ (3),$ $ \Longrightarrow$ $ (A,N,B,E) \equal{} (C,K,D,E)$ $ ,(4)$

From $ (4),$ we conclude that the points $ N,\ O,\ K,$ are collinear and the proof is completed.

Kostas Vittas.

As [b][size=100]Little 100t[/size][/b] said, we can have another approach, based on [b][size=100]Desarques theorem[/size][/b], as follows:

$ \bullet$ We denote the points $ E\equiv AB\cap CD$ and $ Z\equiv DP\cap BQ$ and $ F\equiv AD\cap BC$ and we will prove that they are collinear.

Because of the line segment $ BD,$ taken $ B$ as the intersection point of the diagonals of the quadrilateral $ ACEF,$ passes through the midpoint of $ AC,$ we conclude that $ ACEF$ is a trapezium with $ EF\parallel AC.$

So, we denote the point $ Z\equiv EF\cap BK$ and it is enough to prove that the line segment $ DPN,$ passes through the point $ Z.$

From $ AC\parallel EF$, by [b][size=100]Thales theorem[/size][/b], we have $ \frac {QC}{FZ} = \frac {BC}{BF} = \frac {AC}{EF} = \frac {DA}{DF}$ $ \Longrightarrow$ $ \frac {DA}{DF} = \frac {AP}{FZ}$ $ ,(1)$ because of $ QC = AP.$

From $ (1)$ we conclude the points $ D,\ P,\ Z,$ are collinear and so, the collinearity of the points $ E,\ Z,\ F,$ it follows.

Hence, by [b][size=100]Desarques theorem[/size][/b], we conclude that the triangles $ %Error. "bigtriangles" is a bad command.
ADN$ and $ \bigtriangleup CBK,$ are perspective $ ($ $ E\equiv AN\cap CK$ and $ Z\equiv DN\cap BK$ and $ F\equiv AD\cap CB$ $ ).$

So, we conclude that the line segments $ AC,\ DB,\ NK,$ are concurrent at one point.

Hence, the line segment $ NK,$ passes through the point $ O,$ as the intersection point of $ AC,\ BD$ and the proof is completed.

Kostas Vittas.

[quote="Virgil Nicula"][color=darkred][b][u]An easy extension.[/u][/b] Let $ ABCD$ be a convex quadrilateral for which $ O\in AC\cap BD$ .

For the points  $ P\in (AO)$ , $ Q\in (CO)$ define the points  $ \begin{array}{c} N\in AB\cap DP \\
\ K\in CD\cap BQ\end{array}$ .

Prove that $ O\in NK$ $ \Longleftrightarrow$ $ \frac {PO}{PA}\cdot \frac {QC}{QO} \equal{} \frac {OC}{OA}$ .[/color][/quote]

[color=darkblue][b][u]Proof I.[/u][/b] Denote $ \begin{array}{c} X_1\in AD\cap ON \\
\ X_2\in AD\cap OK\end{array}$ . Apply the [b]Menelaus' theorem[/b] to the transversals :

$ \left\{\begin{array}{cc} \overline {DPN}\mathrm {\ in\ }\triangle\ AOB\ : & \frac {DO}{DB}\cdot\frac {NB}{NA}\cdot \frac {PA}{PO} \equal{} 1 \\
 \\
\overline {ONX_1}\mathrm {\ in\ }\triangle\ ABD\ : & \frac {X_1D}{X_1A}\cdot \frac {NA}{NB}\cdot\frac {OB}{OD} \equal{} 1\end{array}\right\|\ \bigodot\ \Longrightarrow\ \frac {X_1A}{X_1D} \equal{} \frac {OB}{DB}\cdot\frac {PA}{PO}$

$ \left\{\begin{array}{cc} \overline {BQK}\mathrm {\ in\ }\triangle\ COD\ : & \frac {BO}{BD}\cdot\frac {KD}{KC}\cdot \frac {QC}{QO} \equal{} 1 \\
 \\
\overline {OKX_2}\mathrm {\ in\ }\triangle\ ACD\ : & \frac {X_2D}{X_2A}\cdot \frac {OA}{OC}\cdot\frac {KC}{KD} \equal{} 1\end{array}\right\|\ \bigodot\ \Longrightarrow\ \frac {X_2A}{X_2D} \equal{}$ $ \frac {BO}{BD}\cdot\frac {QC}{QO}\cdot\frac {OA}{OC}$

Therefore, $ O\in NK$ $ \Longleftrightarrow$ $ X_1\equiv X_2$ $ \Longleftrightarrow$ $ \frac {X_1A}{X_1D} \equal{} \frac {X_2A}{X_2D}$ $ \Longleftrightarrow$ $ \frac {PO}{PA}\cdot \frac {QC}{QO} \equal{} \frac {OC}{OA}$ .

[b][u]An interesting remark.[/u][/b] If $ P$ , $ C$ are harmonical conjugate w.r.t. $ A$ , $ O$ , i.e. $ \frac {PO}{PA} \equal{} \frac {CO}{CA}$ and  $ Q$ , $ A$

are harmonical conjugate w.r.t. $ C$ , $ O$ , i.e. $ \frac {QC}{QO} \equal{} \frac {AC}{AO}$ ,  then $ \frac {PO}{PA}\cdot \frac {QC}{QO} \equal{} \frac {OC}{OA}$ , i.e. $ O\in NK$ .

[b][u]Proof II.[/u][/b] Prove easily that (well-known !) $ \boxed {\ O\in NK\ \Longleftrightarrow\ \frac {NB}{NA}\cdot \frac {KC}{KD}\equal{}\frac {OB\cdot OC}{OA\cdot OD}\ }\ \ (*)$ .

Apply the [b]Menelaus' theorem[/b] to the transversals :

$ \left\{\begin{array}{cc}
\overline {DPN}\mathrm {\ in\ }\triangle\ AOB\ : & \frac {DO}{DB}\cdot\frac {NB}{NA}\cdot\frac {PA}{PO}\equal{}1\\\\
\overline {BQK}\mathrm {\ in\ }\triangle COD\ : & \frac {BD}{BO}\cdot\frac {KC}{KD}\cdot\frac {QO}{QC}\equal{}1\end{array}\right\|\ \bigodot\ \Longrightarrow$ $ \frac {NB}{NA}\cdot\frac {KC}{KD}\equal{}\frac {OB}{OD}\cdot\frac {PO}{PA}\cdot\frac {QC}{QO}$ .

Using the relation $ (*)$ obtain $ O\in NK$ $ \Longleftrightarrow$ $ \frac {OB\cdot OC}{OA\cdot OD}\equal{}\frac {OB}{OD}\cdot\frac {PO}{PA}\cdot\frac {QC}{QO}$ $ \Longleftrightarrow$ $ \frac {PO}{PA}\cdot \frac {QC}{QO} \equal{} \frac {OC}{OA}$ .[/color]


What is the units digit of $ 10!\plus{}9!\plus{}8!\plus{}7!\plus{}6!\plus{}5!\plus{}4!\plus{}3!\plus{}2!$ ?

For those that might not know what the $ !$ symbol represents, it is called a factorial and is defined as $ n!\equal{}n(n\minus{}1)(n\minus{}2)....(1)$

Since all factorials $ 5!$ and above have a units digit of $ 0$, they don't matter, and so thus we just find the units digit of $ 4!\plus{}3!\plus{}2!$. $ 4!\plus{}3!\plus{}2!\equal{}2!(4(3)\plus{}3\plus{}1)\equal{}2(12\plus{}4)\equal{}2(16)\equal{}32$. The units digit of the result is therefore $ \fbox{2}$.

I don't think the factoring at the end is nessacary, or saves time, especially since this is a countdown round. Easier is to just do $ 4\plus{}6\plus{}2$, which gives 2.


function
trigonometry
calculus
derivative
real analysis
real analysis unsolved

Find $m\in\mathbb{R}$ such that the function $f(x)=\left\{\begin{array}{ll}\displaystyle|x|^m\sin\frac{1}{x},&x\in\mathbb{R}\backslash\{0\}\\0,&x=0 \end{array}\right.$ is differentiable.

This is a well-known family of examples. The function is clearly differentiable for $x\ne 0$, so the only issue is the differentiability at zero. Write out that difference quotient, and note that $\sin(1/x)$ is bounded by 1 but that no better statement can usefully be made about its behavior as $x\to 0$. The function is differentiable for $m>1$ and not differentiable at zero for $m\le 1$.

Now try the followup questions: for what values of $m$ is the function in $C^1$ (that is, for which its derivative is continuous)? For what values of $m$ (if any) does it have a second derivative? For what values of $m$ is it in $C^2$?


topology
calculus
integration
real analysis
Functional Analysis
\/closed

I don't understand why real analysis and algebra-questions are posted in advanced fields? I don't look in advanced fields because I don't know what a lot of the questions are about, so I miss a lot of questions about things I know about.

I don't think that topology, measure, integration and stuff like that belong in advanced fields.

Where else should topology go¿
Algebra and integration are  just misplaced there...

Maybe I don't look at advanced fields as often as I should, but when I do see obvious misplacements, I move the topics. Topology and functional analysis I usually leave there.

Feel free to point out to me things you think ought to be moved. (I might not always agree.)

A lot depends on the perceived difficulty of the question. I view "Advanced fields" as the place for problems that either do not fit in any other category, or are perceived as much more difficult or requiring more advanced knowledge than normal. As Kent said, feel free to PM any of the College Playground moderators if you think that something should be moved; just be specific in your requests and don't expect us to always agree with your point of view :).


Refer to http://www.artofproblemsolving.com/Forum/viewtopic.php?p=746234#746234 .  The answers should be clear.

jli said:

"just a note, whenever there is an imainary root, its conjugate is also a root."

Question:  Can you prove your statement?

in other words

if you have
$a+bi$
as a root then

$a-bi$ is also a root...unless im mistaken? were gonna do complex conjugates tommorow i think in math so im not sure

[quote="7h3.D3m0n.117"]in other words

if you have
$a+bi$
as a root then

$a-bi$ is also a root...[/quote]

Yes; this is true of any polynomial with [b]real[/b] coefficients.

[quote="Interval"]For each question below, information is given about a complex polynomial f(x) whose coefficients are real numbers.  Find the remaining zeros of f.

[1] Degree 5; zeros: 0, 1, 2, i

[2] Degree 6; zeros: i, 3 - 2i, -2 + i[/quote]
[hide="1"]Using the Complex Conjugate Root Theorem (What other people have also been talking about) The last zero is $-i$.[/hide]
[hide="2"]Same thing, if $a+bi$ is a root, then $a-bi$ is also a root so then the three remaining zeros are $-i, 3+2i,-2-i$.[/hide]


It's wrong. Try $ a\equal{}b\equal{}1$ and $ c\equal{}0.$

[quote="duc 95"]Thank you very much! Can you find max of LHF[/quote]
Not exists. $ a \equal{} b\rightarrow0^ \plus{}$ :wink:
The minimum is $ 2.$

[quote="duc 95"]Thank you very much! Can you find max of LHF[/quote]

$ a\minus{}>0,b\minus{}>0,c\equal{}1$

max ??

[quote="duc 95"]The minimum is easy but max is difficult[/quote]

no max ...... see 4# & 5#... :wink:


congratulations!!

if I were you, I'd lock my door at night and make sure I don't misplace my keys somewhere else. :P

OH NOES!  MY SLEEP-TIME HAEV BEN STOELED!!!!

Well... the guy had just lost 800 bucks online, so I thought letting him win a few play-money games might make him feel a little better. ;)

Actually, I have a lot to learn from Alex... he's very good at NL heads-up hold 'em.

Cheers.

[quote="blahblahblah"][quote="Singular"]any challengers?[/quote]

HORSE H/U? gogogo[/quote]

Probably you'd have better luck against him than me; my HU skills are seriously lacking.

[quote="Elyot"][quote="jpark"]LOL the arrogance!!

Did you guys sleep last night, even?![/quote]

Me?  Sleep?  I took a brief nap in the morning.  Tonight I plan on getting a full night's sleep and hopefully returning to a "normal" sleep cycle, since I'm going to Wonderland tomorrow and starting work next week...[/quote]

Oh wow! That's exactly what I was thinking!!

[quote="Elyot"][quote="jpark"]
Oh wow! That's exactly what I was thinking!![/quote]

Really?  You're going to wonderland too?!  AWESOME!!!  :P  :P  :P[/quote]

*rolls eyes*

[quote="jpark"]

*rolls eyes*[/quote]

Hence the  :P  :P  :P.

Gotta love the last few weeks of summer.  Hardcore boredom at its best.

Where can I play HORSE H/U  (preferably in a HU sng?)

Stars or Full Tilt.

Unfortunately for various reasons I have no moneys on the internet right now (thankfully not BUSTO), and probably won't for another 10 days.  I'll play you then?

i also have no money on the internets (or anywhere else for that matter... completely bustoooooooo..)

i dont think stars has HU horse at any limit (free or real), not sure tho

[quote="Singular"]any challengers?  maybe we can have a tourney with me and some of the other regulars here.

ps i light my cubano cigar with 100 USD bills

pps now i am broke from all the partying[/quote]

Such a noob! Real players use 100 USD bills as toilet paper, and light their cigars with origonal action comics!

yea, if we can organized a tournament that would be great. I don't have real money either, but I think play money will be fine with people on this forum. I'm sure everyone here is very competitive :)


In a trapezoid $ABCD$ with $AB\parallel CD$, $\angle ADB+\angle DBC=180$. Show that $\frac{AB}{AD}=\frac{DC}{BC}$

[quote="radio"][color=darkred]Let $ABCD$ be a quadrilateral with $AB\parallel CD$. Show that
$\frac{AB}{AD}=\frac{CD}{CB}$ $\Longleftrightarrow$ $AD\parallel BC\ \ \vee\ \ m(\widehat{ADB})+m(\widehat{CBD})=180^{\circ}$.[/color][/quote]
[color=darkblue][b][u]Proof.[/u][/b] Denote $m(\widehat{ABD})=m(\widehat{BDC})=z$, $m(\widehat{ADB})=x$ and $m(\widehat{CBD})=y$.
Apply the Sinus' theorem in the triangles $ABD$ and $BCD$ : $\{\begin{array}{c}\frac{AB}{AD}=\frac{\sin x}{\sin z}\\\\ \frac{CD}{CB}=\frac{\sin y}{\sin z}\end{array}$ . Therefore,
$\frac{AB}{AD}=\frac{CD}{CB}$ $\Longleftrightarrow$ $\sin x=\sin y$ $\Longleftrightarrow$ $x=y\ (AD\parallel BC)\ \ \vee\ \ x+y=180^{\circ}$.[/color]

Extend $AD$ and $BC$ to meet at point $E$. Then $\angle{ADB}+\angle{BDE}=180^{\circ}$ and from the first  q.e.drelation the we get that the triangle $EDB$ is isoscelles, with $EB=ED$.
Triangles $EDC$ and $EAB$ are similar, so \[\frac{ED}{DA}=\frac{EC}{CB}\Rightarrow \frac{EB}{DA}=\frac{EC}{CB},\] which means $\frac{EB}{EC}=\frac{DA}{BC}$.  Again from the similarity of the triangles $EDC$ and $EAB$, we get $\frac{EB}{EC}=\frac{BA}{DC}$. Consulting the last two relations, we get $\frac{DA}{BC}=\frac{BA}{CD}\Rightarrow  \frac{AB}{AD}=\frac{CD}{CB}$  q.e.d.

Take $E$ on $(CB$ produced so that $AE\parallel BD$; obviously, $AEBD$ is an isosceles trapezoid, hence $\Delta ABD\cong\triangle EDB$, so $DE=AB, BE=AD\ (\ 1\ )$ and $\widehat{EDB}=\widehat{ABD}$, but $\widehat{ABD}=\widehat{BDC}$ (from $AB\parallel CD$), so $DB$ is the bisector of $\angle CDE$ in $\Delta CDE$, and $\frac{DE}{CD}=\frac{BE}{BC}$ with the equalities $(1)$ we are done.

Best regards,
sunken rock

Let  AE//BC, then:

$\left. \begin{array}{l}
 {\rm{          }}AE//BC \\
  \\
 A\hat DE + D\hat BC = 180^o  \\
 \end{array} \right\} \Rightarrow \mathop {}\limits^{} \left\{ \begin{array}{l}
 AEB \approx CBD \\
  \\
 AE = AD \\
 \end{array} \right.\mathop {}\limits_{}  \Rightarrow \mathop {}\limits^{} \frac{{CB}}{{CD}} = \frac{{AE}}{{AB}} = \frac{{AD}}{{AB}}$


Suppose $n$ coins are available that their mass is unknown. We have a pair of balances and every time we can choose an even number of coins and put half of them on one side of the balance and put another half on the other side, therefore a [i]comparison[/i] will be done. Our aim is determining that the mass of all coins is equal or not. Show that at least $n-1$ [i]comparisons[/i] are required.

Let $x_1,\ldots,x_n$ be the masses of the coins. A comparison tells us whether $x_{a_1}+\ldots+x_{a_k}-(x_{b_1}+\ldots+x_{b_k})=0$ or not for some subset $\{a_i\}\cup\{b_j\}\subset\overline{1,n}$.

Suppose we make $n-2$ comparisons. We want to prove that there can be two different situations, one in which all $x_i$ are equal and one in which they are not, which cannot be distinguished by the $n-2$ comparisons. These $n-2$ comparisons give us a homogeneous linear system with $n$ unknowns and $n-2$ equations, so its solution space has dimension $\ge 2$. This means that in addition to the family of solutions given by $x_1=x_2=\ldots=x_n$, it must have other solutions as well, and this is what we wanted.

[quote="epitomy01"]what i dont understand is:
why cant the exact same argument work to show that we need at least n-1 comparisions? since then we would have a solution space of 1, which is still not good enough since we can still have infinitely many solutions, i.e. if we have 2 equations and 3 unknowns we have infinitely many solutions over the reals.
and what happens if you make one comparison and it turns out they are unequal? it straight away gives you the answer no and it takes 1 comparision .. so i think the correct version of the problem is: suppose a set of n masses are equal but we dont know that yet. at least how many comparisions does it take to confirm that they are equal?[/quote]
At least $n-1$ comparisons means that for the worst case we need $n-1$ comparisons, and it turns out where they are all equal is the worst case.
But the problem statement is correct.

Being equal = infinitely many solutions as you can never pin down the weight of one particular coin, so a 1-dimensional solution space is what you want.


calculus
integration
logarithms
calculus computations

Let

$f(x) = \int_1^x \frac{\log{t}}{1+t} dt$ for $x > 0$.

Compute $f(x)+f\left(\frac{1}{x}\right)$.

Let $g(x)=f(x)+f\left(\frac{1}{x}\right),$ use the fundamental theorem of calculus we have $f'(x)=\frac{\ln x}{1+x},$ so $g'(x)=\frac{\ln x}{x}.$ Thus by $f(1)=0$ integrate $g'(x)$ with respect to  $x,$ yielding $g(x) =\frac{1}{2}(\ln x)^2.$


induction
real analysis
real analysis unsolved

Find all $\lambda\in\mathbb{C}$ such that every sequence $\{a_{n}, n\ge1\}\subset\mathbb{C}$ which satisfy $\mid \lambda a_{n+1}-\lambda^{2}a_{n}\mid <1$ for every $n\ge1$ is bounded.
(A. Prymak)

Any body else try this one ?

After searching I get $|\lambda|\leq 1$ is the condition

uhm.. i've found $|\lambda|<1$...
since it's easy to manage the case $|\lambda|\neq 1$, let's suppose $\lambda = e^{i\theta}$, and call $\lambda a_{n+1}-\lambda^{2}a_{n}= \rho_{n}e^{i\theta_{n}}$, with $\rho_{n}< 1$.
my aim is to find an unbounded sequence $a_{n}$: so suppose $a_{0}= 0$, $\theta_{n}= (n+2)\theta$, and $\rho_{n}= 1/2$. it seems like you get $a_{n}= (n/2)e^{in\theta}$... right?

the case $|\lambda|>1$ is more or less the same as my previous post, just a bit easier, maybe...

then, if $|\lambda|<1$, you just have $|a_{n+1}| < |\lambda|^{-1}+|\lambda| |a_{n}|$.
then, by induction, $|a_{n}| < \frac1{|\lambda|}\frac{1-|\lambda|^{n}}{1-|\lambda|}+|\lambda|^{n}|a_{0}|$. and since the right term converges in $n$, the sequence is bounded.


inequalities
trigonometry
trig identities
Law of Cosines

Prove or disprove the inequality:

[tex]a^{6}+b^{6}+c^{6} \geq 6a^{2}b^{2}c^{2}[/tex] where a,b,c are the sides of an acute triangle.

thank you very much.

[hide]

Case 1: a = b = c

a^6 + a^6 + a^6 >= 6a^(2+2+2)

3a^6 >= 6a^6

3 >= 6

This is false.  Thus the inequality is untrue if a=b=c.

Case 2: a = b, but not equal to c.

We know that a^2 + b^2 > c^2 in an acute triangle.  If a^2 + b^2 was equal to c^2, then our triangle would have the maximum side  lengths.  Thus the maximum of the LHS of the inequality would occur when a^2 + b^2 = 2a^2 = c^2.  We can show that even at the maximum, the left hand side will be smaller, not greater, than the right-hand side:

Sqrt[2a^2] = :rt2:a

2a^2 + (a:rt2:)^6  :ge: 6a^4 (a :rt2: )^2

10a^2   :ge: (6)2a^6

10a^2 :ge: 12a^6

10 :ge: 12

This is obviously false, so the inequality is not true if a=b.

[/hide]

I think if you put the sign the other way the inequality might be provable by multiplying equations obtained using the law of cosines.  I'll get back to this later, but law of cosines looks like a good choice to me.


Does anyone know where I can find values for the solubility of naphthalene in hexane when they are at room temperature?

Thanks!

I would advise you to go to your library and look for handbooks of physical-chemistry, perhaps.

Hmmmm... yeah, I was able to find the mole fractions of mixtures involving them.... But I want to know the solubility in grams of naphthalene per mL of hexane.... hmmmm...

[url=http://srdata.nist.gov/solubility/sol_detail.asp?sys_ID=59_92]Look here[/url]. This might help. And I have just made a google search.

May I ask why you need such a data?


If f(x) = |(x^2-12)(x^2+4)|, how many numbers in the interval -2 <= x <= 3 satisfy the conclusion of the Mean Value Theorem?

A. None
B. One
C. Two
D. Three
E. Four

Note #1: Problem Sets posted this same exact problem earlier today.

Note #2: Problem Sets and croatian_dan use the same IP address.

Quit wasting our time.


Yeah...this is impossible, couldn't he have any finite number of M&M's?

I think that you would need to know how many m&m's he had.

We could, but that would get pretty nasty.  You'd have to be like, since there are 5 colors of M&Ms, there are $5^{n}$ different combinations he could have, and then you would have to do the casework for each of those...

Just Wondering...did you get this on a competition or out of your head?

[quote="nutz_for2.718281828"]In how many ways can Horatio give M&M's to 3 kids? (M&M's come in red, yellow, green, brown, and tan.)

I don't think that it is possible to solve, since we don't know if he has even only 5 M&M's or not.[/quote]

tan?
how many m&m's?

1. It's from AoPS Volume 1.

2. It says that there are 125 ways.

Also, there are no typos.

P.S. I think that there has to be something wrong with the answer.

[quote="nutz_for2.718281828"]In how many ways can Horatio give M&M's to 3 kids? (M&M's come in red, yellow, green, brown, and tan.)

I don't think that it is possible to solve, since we don't know if he has even only 5 M&M's or not.[/quote]
I think we are assuming that he has all of red, yellow, green, brown and tan.
It was an excerise in AoPS, not much of a big deal. It was simply to give an understanding of the concept.

If the answer is $125=5^{3}$, then it's assumed that each kid receives ONE m&m of whatever color.

It ASSUMES that he was giving one to each child. :wink:  :wink:

Generalizing with $x$ M&Ms is rather easier if you use the Hockey Stick Identity.

If you don't know what that is, buy the Introduction to Counting & Probability book.  It helped me a lot  :D

Well obviously Horatio has three or more of each color and each kid gets one M&M of any color...