ratio
geometry
incenter
circumcircle
geometry proposed

For the triangle ABC ($b\ne c$) specify the position of the (remarkable) point
$P\in IG\cap BC$ if there is one from the following relations:

 $1).\ \ \frac{2}{a}=\frac{1}{b}+\frac{1}{c}\ \ \ \ 2).\ \ \frac{a}{b+c}=\frac{bc}{b^2+c^2}.$

Denote a = BC, b = CA, c = AB the sides of the triangle $\triangle ABC$. Let D be the midpoint of the side a = BC, K and K' the intersection of the internal and external bisectors of the angle $\angle A$ with the line BC, and P the midpoint of the segment KK' (the center of A-Apollonius circle of the triangle $\triangle ABC$). Consider the triangle $\triangle ADK$. The centroid G divides the median AD in the ratio

$\frac{GD}{GA} = \frac 1 2$

The incenter I divides the internal bisector AK of the angle $\angle A$ in the ratio

$\frac{IA}{IK} = \frac{c + b}{a}$

The internal bisector AK of the angle $\angle A$ divides the opposite side BC internally in the ratio of the adjacent sides, while the external bisector AK' of the angle $\angle A$ divides the opposite side BC externally in the ratio of the adjacent sides:

$\frac{KB}{KC} = \frac{K'B}{K'C} = \frac {AB}{CA} = \frac c b$

WLOG, assume that $\angle C > \angle B$, hence, c > b. The points B, K, C, K' then follow on the line BC in this order.

[color=blue][b](1)[/b][/color] Case [color=white].[/color] $\frac 2 a = \frac 1 b + \frac 1 c$.

$KC = \frac{ab}{c + b},\ \ K'C = \frac{ab}{c - b}$

$K'K = K'C + KC = \frac{ab}{c - b} + \frac{ab}{c + b} = \frac{2abc}{c^2 - b^2}$

$K'D = K'C + CD = \frac{ab}{c - b} + \frac a 2 = \frac{a(2b + c - b)}{2(c - b)} = \frac{a(c + b}{2(c - b)}$

$\frac{K'K}{K'D} = \frac{2abc}{c^2 - b^2} \cdot \frac{2(c - b)}{a(c + b)} = \frac{4bc}{(c + b)^2}$

$\frac{K'K}{K'D} \cdot \frac{GD}{GA} \cdot \frac{IA}{IK}= \frac{4bc}{(c + b)^2} \cdot \frac 1 2 \cdot \frac{c + b}{a} = \frac{2bc}{a(c + b)}$

By Menelaus' theorem for the triangle $\triangle ADK$, the points I, G, K' are collinear, i.e., the line IG passes through the point K', iff

$\frac{2bc}{a(c + b)} = 1,\ \ \frac 2 a = \frac{c + b}{bc} = \frac 1 b + \frac 1 c$

[color=blue][b](2)[/b][/color] Case [color=white].[/color] $\frac{a}{b + c} = \frac{bc}{b^2 + c^2}$.

$PK = \frac{K'K}{2} = \frac{abc}{c^2 - b^2}$

$PD = K'D - \frac{K'K}{2} = \frac{a(c + b}{2(c - b)} - \frac{abc}{c^2 - b^2} = \frac{a[(c + b)^2 - 2bc]}{2(c^2 - b^2)} = \frac{a(c^2 + b^2}{2(c^2 - b^2)}$

$\frac{PK}{PD} = \frac{abc}{c^2 - b^2} \cdot \frac{2(c^2 - b^2)}{a(c^2 + b^2)} = \frac{2bc}{c^2 + b^2}$

$\frac{PK}{PD} \cdot \frac{GD}{GA} \cdot \frac{IA}{IK}= \frac{2bc}{c^2 + b^2} \cdot \frac 1 2 \cdot \frac{c + b}{a} = \frac{bc (c + b)}{a(c^2 + b^2)}$

By Menelaus' theorem for the triangle $\triangle ADK$, the points I, G, P are collinear, i.e., the line IG passes through the point P, iff

$\frac{2bc (c + b)}{a(c^2 + b^2)} = 1,\ \ \frac{a}{c + b} = \frac{bc}{c^2 + b^2}$

Therefore, $P\in BC\cap AA$, i.e. $\frac{PB}{PC}=\frac{AB^2}{AC^2}$ (AA is the tangent in A at the circumcircle of ABC).
****************************************************************************
[b]The one's solution.[/b]

Let $X\in BC$ for which $\overline {XB}=m\cdot\overline{XC},m\in R,m\ne 1$. Thus,

$X\in GI\Longleftrightarrow\left|\begin{array}{ccc}1&1&1\\\\a&b&c\\\\0&1&{-m}\end{array}\right|=0\Longleftrightarrow m=\frac{c-a}{a-b}.$

[u]Case 1.[/u].....$a=\frac{2bc}{b+c}\Longrightarrow m=\frac{c-\frac{2bc}{b+c}}{\frac{2bc}{b+c}-b}\Longrightarrow m=\frac{c}{b}.$

[u]Case 2.[/u].....$a=\frac{bc(b+c)}{b^2+c^2}\Longrightarrow m=\frac{c-\frac{bc(b+c)}{b^2+c^2}}{\frac{bc(b+c)}{b^2+c^2}-b}\Longrightarrow m={\left(\frac{c}{b}\right)^2.}$


group theory
abstract algebra
superior algebra
superior algebra unsolved

Prove that $\mathcal{Q}$ is not isomorphic to a subgroup of $S_{n}$ for $n \leq 7$.

The quaternions $Q$ are defined by $i^{2}=j^{2}=k^{2}=ijk=-1$. If $Q$ is isomorphic to a subgroup of $S_{n}$ ($n \leq 7$), then $i$, having order $4$, must be identified either with a $4$-cycle $(a,b,c,d)$ or the disjoint composition of a $4$-cycle and a transposition. Either way, the square of such an element $u$ will be $(ac)(bd)$, where $(abcd)$ is the cycle in its decomposition. Moreover, $-i$ will be identified with $u^{3}$ (*). The key is that $(abcd)$ and $(adcb)$ are the two only $4$-cycles whose squares are $(ac)(bd)$. Therefore, among $i,j,k$, two elements at least are identified with the disjoint composition of the $4$-cycle $(a,b,c,d)$ (or $(a,d,b,c)$) and a transposition. The transpositions of those compositions must be pairwise distinct by (*). But this contradicts the fact that $ijk=-1$, because the composition of $3$ transpositions cannot be the identity (since the identity has signature $1$).


[quote="aleCTY"]There are two movie theaters positioned adjacent to each other, both theaters sell tickets to a two-hour movie. One theater starts selling tickets at 15.00 and the other at 15.15 both theaters play the movie the same amount of times. Which one sells more tickets? Why?[/quote]

Hm. Is this really a math problem?
The cheaper one could sell more tickets because everyone is trying to save money.
The more expensive one might have a better movie, though.
Is that all the information there is?

This is an economics problem. Assuming demand is not perfectly inelastic (i.e. a change in price induces no change in the amount demanded), and watching movies is not a Giffen good (when the price increases, you buy more of it), we should expect the theater that has a lower cost for its movies will sell more tickets.

The tickets cost the same. :D
This has to do with the times.It is a little hard to see how to map this into an equation at first, but it can solved many different ways.

This problem also assumes a knowledge of human behavior in this case: 1. people will come to the two theaters at random times. 2. They will buy a ticket to a time that is closest to their time of arrival (without traveling backwards in time.) This problem also assumes a steady flow of people.

So 15.00 and 15.15 are times, not prices?

Well, the theater that opens at 15:15 should sell more tickets as people that arrive late to the other theater will go like "Oh no!  The movie started.  Wait a second, the other theater's movie hasn't started yet.  Let's go there!"  Also, the time 15:15 is more likeable since its easy for people to remember.

How would you make this into a mathematical equation?  Is it even possible?  My answer is not very realistic  :D

[quote="aleCTY"]The tickets cost the same. :D[/quote]

 :blush: I read royalflower's post and assumed the costs were different.

In theory, if more people come before 15:00 than between 15:00 and 15:15, then the theater that shows the movie earlier sells more tickets, because (from personal experience), we humans are very impatient and we want to see our movies now.

From my personal experiences, humans are always late :D

True, but does that mean the theater will bar them for entering? (Then again, I've never gone to a theater, so I don't know the rules.)

Besides, if you've paid for something, although it would be economically beneficial to buy the later ticket to go see the movie, most people wouldn't buy another ticket.

Generally, people buy the ticket when they come to the theater.
 :o You've never gone to a movie theater :o You should go see Avatar in IMAX 3-D as your first movie :o

[quote="darkdieuguerre"]In theory, [b]if more people come before 15:00 than between 15:00 and 15:15[/b], then the theater that shows the movie earlier sells more tickets, because (from personal experience), we humans are very impatient and we want to see our movies now.[/quote]

I purposefully assume something about the actual (not expected) distribution of number of arrivals with respect to time, so your statement doesn't make sense then.

(Of course, prior knowledge could have something to do with it, but I think, at that point, we're delving a bit too much into specifics.)

[quote="aleCTY"]One theater starts selling tickets at 15.00 and the other at 15.15 both theaters play the movie the same amount of times. Which one sells more tickets? Why?[/quote]

You've told us what time the theaters start selling tickets and that they play the movie (I'm assuming the same movie) the same amount of times (per day), but you haven't told us what times the movies start. Are the movies at each theater also starting 15 minutes apart, or are they starting at the same times?

"selling tickets" simply refers to the show times. 007math, I think your solution has more to do with human behavior than with math. Think of two waves with the same frequency, but out of phase with each other by 15 minutes. darkdieuguerre got the abstract answer correct.

The first one will sell more tickets. Taking into account human behavior -- most people will try to make the earlier movie, using the later movie as a backup in case the first is sold out or they are late.

Sorry, I was misunderstanding the problem.  I thought is said that the movie theaters [i]open[/i] at the given times, whereas it actually says that start selling tickets at those times.

Lets see. People buy tickets around 1 hour before the movie, with maximum occuring in the middle. So let wave 1 be
$ sin(x)$
wave 2 will be
$ sin(x\minus{} \frac {\pi}4)$
with $ pi$ as defined as one hour
Hmm, now to analyze it.
[hide="My perspective"]
Can't find a difference
[/hide]

I don't think price is really an issue here.

For every 2 hour period, the later theater will only sell in a particular 15 minute interval, while the first one will sell in a 105 minute interval. So that makes the first theater sell more?

Or am I not getting something here?

I believe this is pretty much all that was meant but we'd better wait until the OP confirms or denies it. :)

[quote="fedja"]It seems like everybody is misinterpreting the problem (which is no wonder given how the OP made ambiguous [b]everything[/b] he could plus a few things he couldn't when posting it). So, here is what (to the best of my mind-reading abilities) was meant.

There are two theaters showing the same 2-hour movie. The first theater starts the shows at 3PM, 5PM, 7PM, and 9PM sharp. The second one starts its shows at 3:15, 5:15, 7:15, and 9:15. A few people who have nothing better to do in the afternoon arrive to the street at random times and buy tickets to the earliest show they can go to. Which theater will sell more tickets?[/quote]

Well if it is like that, then if a person arrives at X:00-X:15 will go to the second one, and X:15-(X+2):00 will go the the first one.
(where X is an odd number)
The first interval is much greater, so there will be more people there

Yes, that is the answer and the question. Sorry for the misunderstanding.  :oops:


geometry
circumcircle
analytic geometry
Euler
Vietnam

Let $ k$ is a given positive number. Let $ ABC$ be an acute triangle, $ AB\neq BC\neq CA$, let $ O$ be its circumcenter, and $ AD$, $ BE$, $ CF$ be its internal angle bisectors. Points $ L$, $ M$, $ N$ lie on rays $ AD$, $ BE$, $ CF$, respectively such that $ \frac {AL}{AD} \equal{} \frac {BM}{BE} \equal{} \frac {CN}{CF} \equal{} k$. Denote by $ (O_1)$, $ (O_2)$, $ (O_3)$, respectively the circle passes through $ L$ and tangent to $ OA$ at $ A$, passes through $ M$ and tangent to $ OB$ at $ B$, passes through $ N$ and tangent to $ OC$ at $ C$.
$ 1.$ When $ k \equal{} \frac {1}{2}$, prove that circles $ (O_1)$, $ (O_2)$, $ (O_3)$ have two common points and the centroid $ G$ of triangle $ ABC$ lies on line connecting these common points.
$ 2.$ Find all $ k$, for which circles $ (O_1)$, $ (O_2)$, $ (O_3)$ are coaxial.

Tangents to the circumcircle $(O)$ through $A,B,C$ cut $BC,CA,AB$ at the centers of the the A-,B-,C- Apollonian circles $O_A,$ $O_B,$ $O_C.$ Hence, perpendicular bisectors of $AL,BM,CN$ cut $AO_A,BO_B,CO_C$ at $O_1,O_2,O_3,$ respectively. Isosceles $\triangle ALO_1$ and $\triangle ADO_A$ are similar with similarity coefficient $k,$ a.s.o. Thus, $\frac{_{AO_1}}{^{AO_A}}=\frac{_{BO_2}}{^{BO_B}}=\frac{_{CO_3}}{^{CO_C}}=k.$ Now, using barycentric coordinates WRT $\triangle ABC,$ we have

$O_A(0:b^2:-c^2) \ , \ O_1((b^2-c^2)(1-k):b^2k:-c^2k)$

$O_B(-a^2:0:c^2) \ , \ O_2(-a^2k:(c^2-a^2)(1-k):c^2k)$

$O_C(a^2:-b^2:0) \ , \ O_3(a^2k:-b^2k:(a^2-b^2)(1-k))$

$OA=OB=OC$ implies that  $O$ has equal power to $(O_1),(O_2),(O_3).$ Thus, $(O_1),$ $(O_2),$ $(O_3)$ are coaxal $\Longleftrightarrow$ $O_1,O_2,O_3$ are collinear $\Longleftrightarrow$

$\left [\begin {array}{ccc} (b^2-c^2)(1-k) & b^2k& -c^2k\\ -a^2k &  (c^2-a^2)(1-k)& c^2k\\  a^2k& -b^2k& (a^2-b^2)(1-k)\end{array}\right]=0$

$ \Longleftrightarrow (k-1)(2k-1)(b^2-c^2)(c^2-a^2)(a^2-b^2)=0$

Since $\triangle ABC$ is not isosceles, $(O_1),$ $(O_2),$ $(O_3)$ are coaxal $ \Longleftrightarrow$ either $k=1$ or $k=\frac{_1}{^2}.$  If $k=1,$ then $(O_1),$ $(O_2),$ $(O_3)$ coincide with the Apollonian circles $(O_A),$ $(O_B),$ $(O_C),$ whose common radical axis is  the Brocard axis of $\triangle ABC.$ If  $k=\frac{_1}{^2},$ then $O_1,$ $O_2,$ $O_3$ are the midpoints of $AO_A,$ $BO_B,$ $CO_C$ $\Longrightarrow$ $O_1,$ $O_2,$ $O_3$ lie on the orthic axis of $\triangle ABC$ $\Longrightarrow$ common radical axis of $(O_1),(O_2),$ $(O_3)$ is the Euler line of $\triangle ABC.$

You can draw pictures of items for everyone! :D.


linear algebra
matrix
search
linear algebra unsolved

Let $A=[a_{i,j}]\in{GL}_{n}(\mathbb{C})$ be a tridiagonal matrix ($a_{i,j}=0$ if $|i-j|>1)$ and $b\in\mathbb{C}^{n}$; here $n\approx{10^{6}}$. $A,b$ are known and you search $x\in\mathbb{C}^{n}$ s.t. $Ax=b$.
 If I give you $A^{-1}$ (explicitly) what do you do ?

I ignore $A^{-1}$ and solve the system by Gaussian elimination. (I am hoping that it's not ill-conditioned.) How many multiplications/divisions do I need for that? It's something like $3\times 10^{6},$ I think. The problem with $A^{-1}$ is that I'm pretty sure that it's not tridiagonal - that it has nonzero entries in most places. If that's so, then to execute the product $A^{-1}b$ takes about $10^{12}$ multiplications.

Next time, don't give me $A^{-1}.$ Give me $A=LU$ instead.

Well done Kent.
The key is here:
"I'm pretty sure that it's not tridiagonal - that it has nonzero entries in most places. "

Oh, and when I said "ignore $A^{-1},$" that includes that I don't even download the file that contains it. That's a terabyte file - it's going to take up an awful lot of storage space and take an awfully long time to transmit. We can store $A$ in a few megabytes because we don't need to store the zeros.


geometry
geometric transformation
geometry proposed

Let P1,P2,Q three points from the plane s.t. <QP1P2= \pi/4. Now let consider the points P1' and P2' s.t. QPiPi' =\pi /2 and PiQ=PiPi',i=1,2.
Proove that QP1, P1'P2' and the mediator of P1P2 are concurent.
Hope that is correct :)

[i]This doesn't have a source... i was inspired from a problem from Gamow wich has an interesting story... i will post it after u'll give some solutions(hope that are more than one...i found only one).[/i]

don't u have any idea?
i could solve it only with complex numbers
if u want i can post here a solution

This may be stupid but...what is a mediator? :(

it's a direct (mot-a-mot) translation of "segment bisector line" from Romanian :D.


Hi, everybody! I'm new here, and this is my first post.

Getting to the point, I'm having trouble with one of this year's Warm-Up problems. I can't figure out how to get the answer that they got. Is there something I missed? I'd like it if somebody could help me out. Thanks!

The problem is:
A deck of 52 playing cards is printed on a 4-by-13 rectangular grid. If stacking the pieces is allowed after each cut, but no folding is allowed, what is the least number of cuts required to seperate all the cards? (Cuts can be any length, but may go in only one direction per cut.)

The answer is 6, but no matter how many different ways I tried it I always ended up with 7. Thanks in advance!


Hey guys I have a math question.

If, f(x)= 11/ (x-10) and g(x)= 13/ (x+11)
What is the domain of (f/g) (x) ?

Here's what I found:
First, f/g is (11x + 121) / (13x - 130) and its domain is (-infinity, 10) U (10,infinity).

But my book sais the domain is wrong. Can anyone help me out?

What does your book say, out of curiosity? That does seem right; I can't see why they would claim that $ x \equal{} \minus{}11$ is not in the domain because the quotient of the two functions permits $ x\equal{}\minus{}11$ as a valid input... unless there's some convention I'm not aware of which states that the domain of $ (f/g)(x)$ is the intersection of the domains of $ f(x)$ and $ g(x)$.

Nope. You are perfectly correct. The domain is $ ( \minus{} \infty, 10) \cup (10, \infty)$. You can also write it as $ \{ x | x \in R$ \ $ 10 \}$


Find the least positive integer with $ 96$ divisors.

[hide="Use this formula"]If $ x\equal{}p_1^{a_1}\cdot p_2^{a_2}\cdots p_n^{a_n}$ where $ p_i$ are primes,  then the no. of divisors $ x$ has is $ (a_1\plus{}1)(a_2\plus{}1)\cdots(a_n\plus{}1)$[/hide]

[hide="never mind"]I got something else. $ 96 \equal{} 2^5\cdot3$
\[ 2^2\cdot3\cdot5\cdot7\cdot11\cdot13 \equal{} \boxed{60060}\][/hide]

I agree with AndrewTom.

$ 27720 \equal{} 2^3 \times 3^2 \times 5 \times 7 \times 11$.

$ 2^3 < 13$, so we are allowed to make that substitution.


5.  The perimeter of a sector of a circle is the sum of the two sides
formed by the radii and the length of the included arc. A sector
of a particular circle has a perimeter of 28 cm and an area of
49 sq cm. What is the length of the arc of this sector?

6. John noticed that the angle formed by the minute hand and
hour hand on a standard 12-hour clock was 110 degrees when
he left home after 6 p.m.; it was also 110 degrees when he
returned after 6 p.m. but before 7 p.m. that same night. If he
left home for more than five minutes, for how many minutes
was he away?

7.  Let T be a positive integer whose only digits are 0s and 1s. If
X = T/12 and X is an integer, what is the smallest possible
value of X?

8. Using each of the digits {1,2,3,4} exactly once and using
zero or more + signs, how many different totals can be
obtained? No digit can be used as a power, root, etc. Only
addition can be done. When n digits are written next to each
other, they represent an n-digit number and not the product of
n factors. Four of the totals to be included are [u]1234,[/u]
[u]55[/u]= 12+43, [u]127[/u] = 123+4 and [u]37[/u]=31+2+4.

I am at library right now, so I'll do these as soon as I get back. :lol:

EDIT: Mr.dan80, what level is this? Is this NATS level?
This is challenging!

Before I head home, I looked at #7.
I think I have the answer. I'll only post the answer now and when  I get home, I'll post the solution.
EDIT: I am back, so will edit the solution
[hide="#7"]
I tried to solve T first.
T has to be factor of 12, which also has to be factor of 2 and 3.
Note that T only has 1's and 0's for its digits.
0 can cover up 2, so all we have to do is make the sum of the digits multiples of 3.
So my first guess was 1110, but it gave me 92.5 and 92.5 does not qualify as an answer since X= integer.
So $ \ 11100$ should be T, so $ \ 11100\div12 \equal{} \boxed{925}$
What about the other ones?
They wouldn't work because you'll need one or more 0 in the number.

[/hide]

I think I also have the solution for #8 (Is there easier way to do this?)
I am not sure if I got this right.... :(
P:S I had to count them and tried them and find a pattern....

[hide="#8"]It's really hard to explain....so bear with me.. :wink:
First thing I want to tell you is that there is a pattern to this problem.

I divided into 3 sections by amount of usages of +'s

[u]No +'s :[/u] $ 4!\plus{}(4\cdot3\cdot2)\plus{}(4\cdot3)\plus{}4\equal{}64$
[u]1 + :[/u] You have to be picky about this one. They can NOT be these(took me awhile :lol: ):
[hide="Numbers"]1,2,3,4,12,13,14,21,23,24,31,32,34,41,42,43,123,124,132,134,142,143,213,214,231,234,241,243,312,314,321,324,341,342,412,413,421,423,431,432,1234,1243,1324,1342,1423,1432,2134,2143,2314,2341,2413,2431,3124,3142,3214,3241,3412,3421,4123,4132,4213,4231,4312,4321.[/hide]

So either you can count or realize there are 3 numbers in every set.
So $ \ 17\times3\equal{}51$ numbers + $ 5$ numbers$ \rightarrow(46,64,73,37,55) \equal{} 56$

[u]2 +'s : [/u] You can only achieve $ 2 \rightarrow (19,28)$

3 +'s : Only $ 1 \rightarrow (10)$

Now you add them up to get $ \boxed{123}$ sums.[/hide]

Pheww....... Now I think I can sleep and do the rest two. :lol:

I got a solution for #6...I am not sure though..

[hide="#6"]Since hour-hand moves $ \frac {3}{5}^{\circ}$ in a minute and minute hand moves $ 6^{\circ}$ in a minute, we can make an equation like this:

Let $ m$ be minutes after 6'O clock.
$ \frac {3}{5}m \plus{} 6m \equal{} 110$ and $ m$ equals out to be $ 16\frac {2}{3}$ so when he leaves it's [u]6:17 PM.(rounded up)[/u]

You have to get the time the hour-hand and minute hand elapses together,(synchronizes) and the formula for it is $ x \equal{} \text {head start} \plus{} \frac {x}{12}$ and you get [u]6:33 PM(rounded up)[/u]

Then you add 17 minutes again to 6:33 PM to get [u]6:50 PM.[/u]
Since the problem asks you how many minutes he was gone, the answer is $ \boxed{33}$minutes

EDIT: On second thought on this, can't you just get the minutes after 6'O clock when it synchorizes(before 7'O clock)?[/hide]

Now I have 1 more question to go!! :lol:

[quote="DonkeyKong"][quote="Chinaboy"]DK, the only reason why 1110 did not work is because 12 is divisible by not only 2 and 3, but also 4 because 12 = 3*4

For #8, you have to use each number once, so you can't have 123. But you're on the right track of dividing it all into different cases.[/quote]

Oh... okay. Thanks!
But on #7, did I get it right?

On #8, are you saying that you have to use all four once?(i didn't do that..)[/quote]

Yep, you got 7!

And you do have to use all four at once (or "exactly once")

[quote="chinaboy"]And you do have to use all four at once (or "exactly once")
[/quote]

That makes it clearer.. Thank you!

Oh.. Do you know if I got #6 right?

Can anyone give me some hints on #5?

[hide=" Where I am so far"]Perimeter=28
Area=49

[u]Perimeter:[/u]
$ \ 2r\plus{}(\cfrac{\text{central angle}}{180^{\circ}}\cdot \pi \cdot r) \equal{} 28$

[u]Area:[/u]
$ \frac12r^{2}\cdot \theta\equal{}49$    OR     $ \frac12r^{2}\cdot 2\pi \equal{}49$[/hide]

Am I on the right track?

[quote="DonkeyKong"]Can anyone give me some hints on #5?

[hide=" Where I am so far"]Perimeter=28
Area=49

[u]Perimeter:[/u]
$ \ 2r \plus{} (\cfrac{\text{central angle}}{180^{\circ}}\cdot \pi \cdot r) \equal{} 28$

[u]Area:[/u]
$ \frac12r^{2}\cdot \theta \equal{} 49$    OR     $ \frac12r^{2}\cdot 2\pi \equal{} 49$[/hide]

Am I on the right track?[/quote]

For #6, the hour hand does not move .6 degrees. Check you work again.

For #5, I think you're using degrees for the first equation and radians for the second one. Choose one of the two. I think you should go ahead with degrees on this one just because it's more Mathcounts-friendly.

[quote="chinaboy"]For #6, the hour hand does not move .6 degrees. Check you work again.[/quote]

Wait....would it be $ {\frac12}^\circ?$

[hide="If it is.."]I tried it with $ {\frac12}^\circ$ for hour-hand, but still gave me same answer... :wink: [/hide]

I misunderstood #8....I tried to edit on my previous post, but it won't let me..
So I'll post another one.

P:S=Is there easier way than counting??(except dividing into sections..)

[hide="New Solution for #8"]
I started from taking the obvious ones out and divided into sections by the # of usages of +'s.

We can easily perceive that with NO +'s, we get [u]24 possible sum.[/u]
Also on using ALL the +'s would get you 10 no matter what.. So [u]1 possible sum. .[/u]

Then I divided into sections.

One +: [u]16 possible sums.[/u]
[hide=" Possible Sums"]46,55,64,37,73,127,136,325,235,217,316,145,244,415,343,433[/hide]

Two +'s: [u]2 possible sums.[/u]
[hide=" Possible Sums"]19,28[/hide]

So we add the possible sums to get $ \boxed{43}$ possible sums.[/hide]

[quote="Chinaboy"]Talk to me next week about this. I like to draw the picture when I explain clock problems.

I really admire your work ethic, DK! Keep on trucking![/quote]

Okay!

wait..that means I didn't get it right... :(

So far only #7 has been answered correctly.  Maybe our Chinaboy can give some more hints at practice!

Hey, you weren't there at school today! So I couldn't use your microwave to heat my soup.  :(

Hint for 5.
[hide]Come up with two equations (one for perimeter and one for area), each using r for radius and theta for angle (in degrees!).[/hide]

Hint for 6
[hide]Before you do anything, you have to find out how many degrees the hour hand and minute hand move in one minute. (I'll have to go over this with you guys during class because this is VERY important)[/hide]

Hint for 8
[hide]Break it into cases.
Case 1. No + sign
Case 2. 1 + sign
...[/hide]

REVISED:
[hide="#8"]
I started from taking the obvious ones out and divided into sections by the # of usages of +'s.

We can easily perceive that with NO +'s, we get [u]24 possible sum.[/u]
Also on using ALL the +'s would get you 10 no matter what.. So [u]1 possible sum. .[/u]

Then I divided into sections.

One +: [u]17 possible sums.[/u]
[hide=" Possible Sums"]235,244,325,343,424,433,136,145,316,415,127,217,46,55,37,64,73[/hide]

Two +'s: [u]2 possible sums.[/u]
[hide=" Possible Sums"]19,28[/hide]

So we add the possible sums to get $ \boxed{44}$ possible sums.[/hide]

Is it right?? I actually tried all of them..
If it's not right, I might cry... :(

[hide]number 6# is 220/5.5=40 so the answer is [size=200][b][u]40![/u][/b][/size][/hide]

The circumference of a particular circle is 18 cm. In square centimeters, what is the area of the circle? Express your answer as a common fraction in terms of $\pi$.


[quote="mayhem"]Can anybody suggest a proof or web link about well known AM-GM inequality $(a+b+c)^{3}\geq27abc$  in format without words .  I interested proofs without words. Thanks.[/quote]

Wikipedia is your friend.

Okay. Proofs aren't till like Geometry/PreCalc, so um...I believe you should post in Intermediate.

http://en.wikipedia.org/wiki/AM-GM

I think by proofs without words, he means the little picture things (like the ones in the left menu bar on the AoPS website).

The Wikipedia for AM-GM does not have these.

I haven't seen any myself. However,

one possible idea stems from the AM-GM demonstration that uses a semicircle (attached). However, that one contains only two variables - if you use a semi-elipse, you may be able to extend it into three variables somehow, though it would be difficult to get it into the form you want visually.

The other possibility is to use 3-D graphics. However, not only would this be hard to make, it would be difficult to demonstrate that your graphic is a representative case (unlike in the semicircle demonstration of the case of 2, in which case it is "obvious" that it  holds true for other inputs).

I donno... you'll have a tough time, but you might be able to make something out of it.


can you prove :maybe:

$ \prod_{n\equal{}1}^{\infty}\;(1\minus{}x^{2n})\cdot(1\plus{}x^{2n\minus{}1}\cdot y^{2})\cdot\left(1\plus{}\frac{x^{2n\minus{}1}}{y^{2}}\right)\;\;\equal{}\;\;\boxed{\sum_{p\equal{}\minus{}\infty}^{\infty}\;x^{p^{2}}\cdot y^{2p}}$

This identity is called Jacobi Triple Product.

You can find a proof at [url=http://mathworld.wolfram.com/JacobiTripleProduct.html]Mathworld[/url] and at [url=http://en.wikipedia.org/wiki/Jacobi_triple_product]Wikipedia[/url].


This is an Exercises 3.3.4 of Hatcher's Algebraic Topology (p257).

"Given a covering space action of a group G on an orientable manifold M by orientation-preserving homeomorphisms, show that M/G is also orientable."

Some attempts so far,

Since M is an orientable manifold, there exists an orientation $ x \mapsto \mu_x$ such that, for each $ x \in M$, we can assign a local orientatation $ \mu_x \in H_n(M, M \minus{}\{x\})$ satisfying a "local consistency property" with respect to a neighborhood $ U_x \subset R^n$ containing x in M.

By the definition of a covering space action, for a non-identity element g in G, we choose $ U_x$ containing x such that $ gU_x \cap U_x \equal{} \emptyset$.

Then, $ H_n(M, M\minus{}\{x\})\equal{}H_n(U_x, U_x\minus{}\{x\}) \equal{} H_n(gU_x, gU_x\minus{}\{gx\})$.

Is this correct? Then, how do I proceed from here?

Thanks.

Here something in another language which should easily translate to your reference's:

Since $ M$ is orientable, there is a global continuous section $ s: M\to Or(M)$ of the two-fold orientation covering $ \pi_M: Or(M)\to M$. The quotient map $ \psi: M\to M/G$ induces the natural map $ Or(\psi): Or(M)\to Or(M/G)$ with $ \pi_{M/G}\circ Or(\psi)=\psi\circ \pi_M$ with the two-fold covering $ \pi_{M/G}: Or(M/G)\to M/G$. Since $ G$ is orientation preserving, the mapping $ Or(\psi)\circ s: M\to Or(M/G)$ descends to a continuous section $ \overline{s}: M/G\to Or(M/G)$ of the $ \pi_{M/G}$. This is equivalent to that $ M/G$ has an orientation (such that $ \psi$ is orientation preserving).


[quote="IntrepidMath"]What is the ratio of the area of a circle and the area of a square inscribed inside that circle?[/quote]

[hide]  lets say the side of the square is x:

Then the area of the square would be $ x^2 $

And the area of the circle would be $\frac{x\sqrt2}{2} $ * $\frac{x\sqrt2}{2} $  * $\pi$= $\frac{x^2\pi}{2}$

  Thus, there ratios are $x^2$ : $\frac{x^2\pi}{2}$, siplified = $ 2 : \pi $[/hide]

EDIT: $ 2 : \pi$ is reffering to the "square : circle".......

Let's say that the radius of the circle is $2$, then the area of that circle is $4pi$, then the diagonal of the square could be $4$ and the side length would be $4/(2^{1/2})$, so the area of the square would be $8$, so the ratio would be $pi/2$.

[hide]If each side of the square is $x$, then the area of the square would be $x^2$

So, the diameter of the circle would be $x\sqrt{2}$ and the radius would be$ \frac{x\sqrt{2}}{2}$

So the area of the circle would be $\pi \left (\frac{x\sqrt{2}}{2}\right)^2$

Simplified:

$\pi\left(\frac{2{x^2}}{4} \right)$

$\frac{\pi{x^2}}{2}$

The ratio would be $\frac{\frac{\pi{x^2}}{2}}{x^2}$

or

$\left(\frac{\pi{x^2}}{2}\right)*\left(\frac{1}{x^2}\right)$

$\frac{\pi}{2}$

$\pi:2$[/hide]

Area of circle : pi*r^2
Diameter of circle: 2r

So diagonal of square = 2r

Using 1,1, radical2 relationship, side length of square = 2r/radical2 =

2r*radical2
-------------- = r*radical2
2

(r*radical2)^2= area of square = 2r^2

pi*r^2
--------        simplifies to: pi/2 (Answer)
2r^2

it is a generally known fact that the area of a square inscribed in a circle is $2r^2$  just as it is a generally known fact that an inscribed dodecagon has an area of $3r^2$

$2r^2:\pi r^2 = 2:\pi$


geometry
trigonometry
trig identities
Law of Cosines
similar triangles
area of a triangle
Heron\u0027s formula

In triangle $ABC$, $a=11$, $b=13$, and the median to side $c$ is 10.  Compute the area of the triangle.

[hide]Stewart's Theorem on ABC

121(c/2)+169(c/2)=100c+2(c/2)^3

solve for c/2=3 :rt5:

Law of Cosines on BCM, where M is midpt. AB

121+100-220cos :theta: =(3 :rt5:)^2=45

cos :theta:=4/5, sin :theta: =3/5

2[BCM]=[ABC]=11(10)(3/5)=66[/hide]

Here's what I have so far:

[hide]Drawing the three medians, we have the medium to c is split into 10/3 and 20/3.  However, this is all I've got, and I'm not sure where to go from here.  I'm trying similar triangles but not getting anywhere, and I don't think this problem requires the use of trig (which I don't know a whole lot about).  I thought of using Heron's formula, but then I'd need to know side c.  Can anyone offer me small hints?  Thanks.[/hide]


Three Students were asked to find the identity of the metal in a particular sulfate salt. They dissolved a 0.1472-g sample of the salt in water adn treated it with excess barium chloride, resulting in the precipitation of barium sulfate. After teh precipitate had been filtered and dried, it weighed 0.2327 g.
Each student analyzed the data independently and came to different conclusions.
The conclusions were that the metal is sodium, titanium or gallium. What further tests you suggest to determine which student is most likely correct.

I have worked it out, but I find the same solution... it could be either sodium, titanium and gallium.

i have tried using a CRC, but I to no avail.

does anyone know how to approach this problem? Thanks

I've a strong background in chemistry, but can't figure out this college problem.... it's not for a grade, but I'm obsessed to find the answer.

thanks!

What is your opinion about this problem?

What about Atomic Absorption Spectroscopy (A.A.S) or Flame Test?

Why in the first place the metal should be sodium, titanium, or galium?


tetrahedr0n wrote:Find all integral solutions to a^2 + b^2 + c^2 = a^2*b^2.

[hide](a-1)(a+1)(b-1)(b+1)=c:^2:+1. Either the LHS is odd, or 4|LHS, as if 2 divides one factor, it will divide the one that is 2 away from it. However, 4 can never divide the RHS, as c:^2: is never 3 mod 4. So, a, b, and c must all be even. a=2x, b=2y, c=2z => (2x-1)(2x+1)(2y-1)(2y+1)=4z:^2:+1 => 4x:^2:y:^2:=x:^2:+y:^2:+z:^2:. Considering this mod 4, we have 0:equiv:sum of three squares. all of them must be 0, or multiples of 4. Thus, x=4l, y=4m, z=4n. So 64l:^2:m:^2:=l:^2:+m:^2:+n:^2:. We can continue to apply mod 4 analysis, so we have that l, m, and n must all be integers that continually yield integers upon divison by 4. The only such integers are 0. Thus, l=m=n=0, and the only solution is a=b=c=0.[/hide]


[quote="fjwxcsl"]Let $ a,b,c,d\ge \frac {1}{7}$ such that $ a \plus{} b \plus{} c \plus{} d \equal{} 1$. Show that:

$ {\frac {7203}{16}}\ge \frac {1}{abcd} \minus{} \left(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2} \plus{} \frac {1}{d^2}\right)\ge 192;$

$ {\frac {16807}{16}}\ge \frac {2}{abcd} \minus{} \left(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2} \plus{} \frac {1}{d^2}\right)\ge 448.$

Generally,if $ k\ge{\frac {174}{353}},$then

$ {\frac {2401}{16}}(4k \minus{} 1)\ge \frac {k}{abcd} \minus{} \left(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2} \plus{} \frac {1}{d^2}\right)\ge {256k \minus{} 64}.$[/quote]

>f:=subs({a=p+1/7,b=1/7+q,c=1/7+r,d=1/7+s},{p=3/7*x/(x+y+z+w),q=3/7*y/(x+y+z+w),r=3/7*z/(x+y+z+w),s=3/7*w/(x+y+z+w)},k/a/b/c/d-1/a^2-1/b^2-1/c^2-1/d^2-m):
>f:=numer(f):

>factor(subs(x=1,y=1,z=1,w=1,f)):
>MIN:=solve(%,m);

MIN := 256*k-64

>factor(subs(x=0,y=0,z=0,w=1,f)):
>MAX:=solve(%,m);

MAX := 2401/4*k-2401/16

>symprove(numer(subs(w=(x+y+z)*t,m=MIN,k=u+174/353,f)));

                   it is symmetric for . {x, z, y}

                             ITS = [u, t]

                            IS = 1, N = 2

                        The inequality holds.

                             ITS = [u, t]

                            IS = 1, N = 2

                        The inequality holds.

                             ITS = [u, t]

                            IS = 1, N = 1

                        The inequality holds.

                               DEGU = 2

                               DEG = 2

                           ITS = [u, ee, t]

                            IS = 1, N = 5

                        The inequality holds.

                           ITS = [u, ee, t]

                            IS = 1, N = 4

                        The inequality holds.

                           ITS = [u, ee, t]

                            IS = 1, N = 3

                        The inequality holds.

                  The original form is psd!(SYM3(1))

                          它是平凡非负的!

                             TIME = 2.921

>symprove(numer(subs(w=(x+y+z)*t,m=MAX,k=u+174/353,-f)));

                   it is symmetric for . {x, z, y}

                             ITS = [u, t]

                            IS = 1, N = 1

                        The inequality holds.

                             ITS = [u, t]

                            IS = 1, N = 1

                        The inequality holds.

                             ITS = [u, t]

                            IS = 1, N = 1

                        The inequality holds.

                               DEGU = 2

                               DEG = 0

                  The original form is psd!(SYM3(1))

                               它是平凡非负的!

                             TIME = 2.094


Hi all and especially romanian people!!

Does anyone have the problems of the current romanian olympiad(2006),
all rounds (or those that have already taken place) as well as the dates of the contests??

Maybe this can be the thread for the RMO 2006!!

Thanks and good luck  to the participants!

 :)  :)

[quote="perfect_radio"]The romanian olympiad has not yet begun. The first round takes place somewhere in February, in the first week after the vacantion. (I think)

Also, I think Valentin will post all the problems.[/quote]

Thanks, perfect_radio. :)
The first round is the local one??

[quote="perfect_radio"][quote="socrates"]The first round is the local one??[/quote]

Yes. Here are the subjects for the eleventh grade.

Moubinool asked me to post them.[/quote]

Μany thanks perfect radio!
Could you post problems for the other forms, too?/

 :)


Let $ ABCD$ be a convex quadrilateral such that $ AB$ extended and $ CD$ extended cut at a right angle. Prove that $ AC\cdot BD>AD\cdot BC$.

i'm assuming that $ B$ lies between $ A$ and $ E$, where $ E$ is the intersection of $ AB$ and $ CD$... suppose wlog that $ AC\geq BD$... note that angles $ BCD$ and $ ACD$ are obtuse, so $ AD>AC$ and $ BD>BC$, so $ AD\minus{}BC>AC\minus{}BD\geq 0$... since $ AB\perp CD$ we have that $ AD^2\plus{}BC^2\equal{}AC^2\plus{}BD^2$, so the inequality follows after squaring the last inequality obtained and using this equality...  :D


Παιδιά βάζω μια πάρα πολύ δύσκολη κατά τη γνώμη μου άσκηση την οποία δεν μπόρεσα ούτε να προσεγγίσω. Όποιος τη λύσει πραγματικά μπράβο.

Θεωρούμε ένα θετικό ακέραιο κ. Αποδείξτε ότι sqrt(k+1)-sqrt(k) δεν μπορεί να είναι το πραγματικό μέρος μιγαδικού αριθμού ζ τέτοιου ώστε ζ^ν=1 για κάποιο θετικό ακέραιο ν.

Βρήκα μια πάρα πολύ ωραία και έξυπνη λύση για την οποία θα δώσω μια υπόδειξη αύριο αν δεν έχει βρεθεί λύση.

Κοίταξε . Η άσκηση είναι πολύ καλή και η εκφώνησή της δίνει κάτι και είναι ελκυστική .Αλλά γενικά όταν έχει συνημίτονα τα πολυώνυμα Chebychev σκοτώνουν σχεδόν τα πάντα . Να η λύση μου

Η συνθήκη είναι ισοδύναμη με το νδο η εξίσωση
$\cos{m}=\sqrt{k+1}-\sqrt{k}$, όπου  $z=\cos{m}+i\sin{m}$ και
$z^{n}=1$  δεν έχει λύσεις . Τότε από DeMoivre έχουμε
 $\cos{nm}=1$ άρα ο αριθμός  $\sqrt{k+1}-\sqrt{k}$ είναι ρίζα του  $g(x)-1$, όπου  $g(\cos m)=\cos{nm}$ (Θυμίζει πολυώνυμο Chebychev  :wink: ) .
Παρατηρούμε ότι ο αριθμός $A=\sqrt{k+1}-\sqrt{k}$ είναι αλγεβρικός και το ελάχιστό του πολυώνυμο είναι
$f(x)=x^{4}-(4k+2)x^{2}+1$, άρα  έχουμε ότι το $f(x)$ διαιρεί το $g(x)-1$
το οποίο όμως είναι άτοπο αφού το γινόμενο των μέρτων των ριζών του $f(x)$ είναι ίσο με $1$ ενώ του $g$ όχι  , άτοπο

QED  :wink:

Έδωσες μια πιο σύντομη λύση από αυτή που είχα βρει που όμως στηρίζεται στην ίδια λογική. Για να σου πω την αλήθεια δεν μπορώ να τις καταλάβω απόλυτα. Κάπου χάνομαι και στις δύο (στο ίδιο σημείο). Μάλλον μου διαφέυγει κάτι από θεωρία. Σίγουρα δηλαδή... Οπότε εμπιστεύομαι απόλυτα τη λύση σου.

Μπράβο!

Το σημείο που χάνομαι αν μπορείς να μου εξηγήσεις είναι εκεί που λες ότι το g(x)-1 έχει γινόμενο ριζών διάφορο του 1. Έχω δει κάπου μια απόδειξη για τα πολυώνυμα Τσεμπίτσεφ ότι το Τn(x)-1 έχει ρίζες στο {-1, 1}και μάλιστα της μορφής coskπ/n και το k ανάλογα με την κλάση του ν ως προς το δύο (άρτιος, περιττός). Αυτό είναι? Ελπίζω δηλαδή γιατί αλλιώς κάτι δεν καταλαβαίνω καθόλου.

Τις απορίες σου στις απάντησα σε PM .
Μπορείς να γράψεις τη λύση που έχεις υπόψην σου ? Thanks

Σιλουανέ ευχαριστώ σου έστειλα ένα pm. Δε βάζω τη λύση διότι είναι σχεδόν η ίδια. Τη διαφορά στην έγραψα στο pm. :wink:

Επίσης ζητώ συγγνώμη σε όσους είχαν στείλει pm και δεν είχα απαντήσει γιατί νόμιζα ότι σε ενημέρωνε άμα έπαιρνες pm  :blush: και δεν τα είχα δει. :oops:

A και καλό καλοκαίρι σε όλους διότι φεύγω διακοπές. Ελπίζω να έχω σύνδεση internet εκεί που θα πάω για να μπαίνω. Σε διαφορετική περίπτωση να μου συγχωρέσετε μια βδομάδα αποχής.


How did Stanley Miller's classic experiment apply mechanism to the evolution of life?

--

This question is from a workbook related to the Campbell AP Biology Textbook, but I can't seem to find a direct concise answer.  Could someone please help?  Thanks in advance!

actually it didnt provide the mechanism--it was just a proof for the mechanism proposed by oparin and haldane

other scientists added various mix of chemicals to find out how the present organic chems evolved :)


darkquantum wrote:Find the largest integer that is always a divisor of
 13^n+5^n-10^n-8^n

[hide]30. n=0, 1 gives 0, n=2 gives 30, n=3 gives 810, and n=4 gives 503*30. The largest integer possible can be 30. Let us show that 30 does indeed divide 13^n+5^n-10^n-8^n for all positive integer n.

30=2*3*5. 2 obviously divides that (there are 2 odd and 2 even terms). We can write that expression as (13^n-10^n)-(8^n-5^n)=

(13-10)(13^(n-1)+13^(n-2)*10+13^(n-3)*10^2)+...+10^(n-1))-(8-5)(8^(n-1)+8^(n-2)*5+...+5^(n-1).

We can see that both terms are divisible by 3. Finally we write that expression again as (13^n-8^n)-(10^n-5^n). A similar analysis shows that it is divisible by 5. Therefore, 30 divides it for all n.[/hide]

[color=cyan]nicely done.  Darkquantum, do you know where Linda got this problem from?  It's very nice.  An old Mandelbrot, perhps?  Or is it her own creation?[/color]

um, im not sure. She gave a problem similar to it as a lesson problem, so, im not sure.

And mystic, ur correct!
the key to this problem is to kno that (a-b) l (a^n-b^n)

Show the expression divides 5, 3, and then obviously it divides 2 by a simple parity check

I don't know what you're asking, teratomato, but it's divisible by a judicious use of (a-b) | (a^n - b^n) for n >= 1,  and it's divisible by 2 because there are four things being added and two are odd and two are even which makes for an even number.

I don't know what you think I'm asking or why you think I'm asking anything.

I'm not asking. I'm telling.

[quote="darkquantum"]Show the expression divides 5, 3, and then obviously it divides 2 by a simple parity check.[/quote]
The expression doesn't divide 5, 3, or 2. It is divisible by those numbers. That is what I meant. That's why I said:
[quote="teratomato"]What you mean is that it is divisible by 5, 3, and then obviously divisible by 2 by a simple parity check.[/quote]

Sorry.  I thought you wrote, "What do you mean it's divisible . . ."

My fault.

oops
bad diction on my part
thx hooyoung


Can't wait to get my score :( ... You guys are really testing my patience lol  :D

I generally takes up to a month to grade solutions, and though it certainly seems that it's been that long, we still have over a week left to go before we hit that mark, so they might be testing your patience just a little bit longer  :lol:

We hope to have the scores up this week.

Yes! That's great news. I've been waiting for what seems to be an eternity!

be grateful -- last year, it was way slower than this, but that was probably result of not as much people getting to work on grading and stuff.

Ahhh, that was good.  It took me 5 or 10 seconds to catch the pun.  Sorry, man.  If I was a grader, I would have left you hanging.


analytic geometry
graphing lines
slope
calculus
derivative
function
calculus computations

:D :D :ddr:  Find the $ x$ coordinate of the point that the slope of the tangent line is maximum. The equation is:

$ y \equal{} \minus{} x^4 \plus{} 12x^3 \minus{} 54x^2 \plus{} 21341234220445723893240575x \plus{} 13936048234505238907$

What do you notice about the last two coefficients (other than their size)? There is something interesting about them that affects the problem. Just think carefully, and you will notice a pattern.

[quote="t0rajir0u"]What's your answer to the first part?  The "interesting" thing is that the answer doesn't depend on either of the last two coefficients.[/quote]
Exactly, so what is the answer?

Since we know those two coefficients don't matter, how about replacing them with $ 108$ and $ \minus{}81$?

[quote="jmerry"]Since we know those two coefficients don't matter, how about replacing them with $ 108$ and $ \minus{} 81$?[/quote]
In your mind you can, but in the problem, the coefficients stay. :evilgrin: Uh oh! The return of the evil smiley!

That was meant to be a hint to the solution. In that form, I don't even need to do any calculus to see the answer.

then what is the answer? you still have to figure out the answer!

jmerry isn't providing the answer because he's leaving the problem as an exercise for others, especially for you since this is (presumably) your homework.

In any case, here's what he is saying:  if you replace the last two coefficients with the coefficients he gave, then $ y \equal{} \minus{}(x \minus{} 3)^4$ by the binomial theorem.  Now it's obvious what the answer has to be.

[quote="t0rajir0u"]jmerry isn't providing the answer because he's leaving the problem as an exercise for others, especially for you since this is (presumably) your homework.

In any case, here's what he is saying:  if you replace the last two coefficients with the coefficients he gave, then $ y \equal{} \minus{} (x \minus{} 3)^4$ by the binomial theorem.  Now it's obvious what the answer has to be.[/quote]

If this is my homework, then that would be fun! I wish this was actually a pop quiz. I would love that. But I cannot stay in my world of fantasies.

Okay, there is no maximum. This was a trick. However, there is a relative maximum. Do not change the equation now.

$ y \equal{} \minus{} x^4 \plus{} 12x^3 \minus{} 54x^2 \plus{} 21341234220445723893240575x \plus{} 13936048234505238907$

At what $ x$ coordinate is the slope of the tangent line at a relative maximum?

[quote="jmerry"]None. The first derivative is strictly decreasing on the whole real line.[/quote]

I guess you did not get the "relative part".

No, I'm saying that the critical point is not a relative maximum or minimum. It's just a flat spot for the derivative, which decreases on both sides.

Please, please, [i]please[/i] do not assume that the people who are answering your questions know less than you do.  $ f'(x) \equal{} 0$ is not a sufficient condition for either a relative maximum or a relative minimum; for a twice differentiable function, a relative minimum also requires $ f''(x) > 0$, whereas here we have $ f''(x) \equal{} 0$.  Graphically, we do not have either a relative maximum or a relative minimum.

Okay, I was just trying to throw trick questions. Now let us change the problem a little:

$ y \equal{} \minus{}x^4 \plus{} 10x^3 \minus{} 36x^2 \plus{} 21341234220445723893240575x \plus{} 13936048234505238907$

At what x coordinate is the slope of the tangent line at a relative maximum?


Find all continuous functions $ f: \mathbb R\rightarrow\mathbb R$ satisfying $ f(f(x))\equal{}f(x)\plus{}2x$

Please check if the answer below makes sense.
http://www.mathlinks.ro/viewtopic.php?p=1328709#1328709


Compute the value of the infinite product $ \frac{9}{10} \times \frac{99}{100} \times \frac{999}{1000} \times \cdots$.

You won't get anything simpler than what the [url=http://en.wikipedia.org/wiki/Pentagonal_number_theorem]pentagonal number theorem[/url] tells you.

It converges to something close to $ 0.8900100999989988$. I don't think you'll get a nicer answer than that.

Yep; that's what the first few terms $ 1 \minus{} \frac {1}{10} \minus{} \frac {1}{10^2} \plus{} \frac {1}{10^5} \plus{} \frac {1}{10^7} \minus{} \frac {1}{10^{12}} \minus{} \frac {1}{10^{15}}$ get you.  Examining the decimal representation carefully here is actually an interesting way to [b]discover[/b] the pentagonal number theorem, although I don't know if it would be easy for anyone to discover any of the proofs independently, none of which (that I know of) are trivial.


Very easy:
$ x \equal{}\tan{\alpha},y\equal{}\tan{\beta},z\equal{}\tan{\gamma}\Rightarrow\sum\alpha \equal{}\pi$,because if $ \tan{\alpha}\plus{}\tan{\beta}\plus{}\tan{\gamma}\equal{}\tan{\alpha}\tan{\beta}\tan{\gamma}$,then $ \alpha\plus{}\beta\plus{}\gamma\equal{}\pi$, so we have to prove that $ \sum\cos{\alpha}\leq\frac{3}{2}$,which is true by Jensen :)

There is another proof of the last inequality:
$ \sum\cos{\alpha}\leq\frac{3}{2}$,for all $ \alpha,\beta,\gamma\geq 0,\sum\alpha\equal{}\pi$.
Proof(if you don't want to use Jensen):
$ \sum\cos{\alpha}\equal{}\frac{R\plus{}r}{R}\leq\frac{3}{2}$,because $ 2r\leq R$. :wink:

[quote="FOURRIER"][quote="Erken"]$ x \equal{}\tan{\alpha}$[/quote]

You supposed that $ \alpha$£$ [0,\pi/2]$

I'm I right?[/quote]
Yes,you are right $ \alpha\in\[ 0,\frac{\pi}{2}\]$

[quote="FOURRIER"][quote="Erken"]$ \sum\cos{\alpha}\equal{}\frac{R\plus{}r}{R}$[/quote]

Hello  :)

Could you please write the proof of this or give me a link[/quote]
Use $ r\equal{}4R\prod\sin\frac{A}{2}$.

[quote=chien than]Let $ x,y,z$ be positive real numbers satisfying $ x\plus{}y\plus{}z\equal{}xyz$. Prove that:
\[\frac1{\sqrt{1+x^2}}+\frac1{\sqrt{1+y^2}}+\frac1{\sqrt{1+z^2}}\leq\frac{3}{2}\][/quote]
Let $ x,y,z$ be positive real numbers satisfying $ x\plus{}y\plus{}z\equal{}xyz$. Prove that:
\[\frac1{\sqrt{1+x^2}}+\frac1{\sqrt{1+y^2}}+\frac4{\sqrt{1+z^2}}\leq\frac{33}{8}\]

Let $ x,y,z$ be positive real numbers satisfying $ x+y+z=xyz$. Prove that
$$\frac1{\sqrt{1+x^2}}+\frac1{\sqrt{1+y^2}}+\frac k{\sqrt{1+z^2}}\leq\frac{2k^2+1}{2k}$$
Where $k\in N^+.$

[quote=sqing]Let $a,b,c$ be positive real numbers satisfying $ ab+bc+ca=1$. Prove that
$$\frac1{1+a^2}+\frac1{1+b^2}+\frac{k}{1+c^2}  \leq\frac{(2k+1)^2}{4k}$$
Where $k\geq 1.$[/quote]

Thank you for your hard-working, update old problem. Respects!

[quote=sqing][quote=chien than]Let $ x,y,z$ be positive real numbers satisfying $ x\plus{}y\plus{}z\equal{}xyz$. Prove that:
\[\frac1{\sqrt{1+x^2}}+\frac1{\sqrt{1+y^2}}+\frac1{\sqrt{1+z^2}}\leq\frac{3}{2}\][/quote]
[url]https://artofproblemsolving.com/community/c4h2866653p25472821:[/url]
Let $a=\frac{1}{x}, b=\frac{1}{y}, c=\frac{1}{z} \Rightarrow ab+bc+ca=1$.
We have $$\frac{1}{\sqrt{1+x^2}} =\frac{a}{\sqrt{a^2+1}}=\frac{a}{\sqrt{(a+b)(a+c)}} = \sqrt{\frac{a^2}{(a+b)(a+c)}} \leq \frac{1}{2}(\frac{a}{a+b} + \frac{a}{a+c})$$
$$\Rightarrow \frac{1}{\sqrt{1+x^2}} + \frac{1}{\sqrt{1+y^2}} + \frac{1}{\sqrt{1+z^2}} \leq \frac{1}{2}(\frac{a}{a+b} + \frac{a}{a+c}+\frac{b}{b+c} + \frac{b}{b+a} + \frac{c}{c+a} + \frac{c}{c+b}) = \frac{3}{2}$$[/quote]

how we get $\sum{\frac{1}{\sqrt{x^2+1}}}=\sum{\frac{a}{\sqrt{a^2+1}}}$

a = 1/x, so x = 1/a. Substituting this and factorize 1/a^2 from the square root, then put it on the numerator.


A topological space $ X$ is called pathwise connected if for every $ x,y\in X$, there exists a continuous map $ \gamma: [0,1]\to X$ with $ \gamma(0)\equal{}x$ and $ \gamma(1)\equal{}y$. Show that every pathwise connected space is connected and show that the converse is false.

Suppose $ X$ pathwise connected and not connected. Then let $ U\cup V\equal{}X$ be a separation of $ X$. Choose $ x\in U$ and $ y\in V$. Let $ \gamma: [0,1]\to X$ be the path between $ x$ and $ y$. Then $ \gamma^{\minus{}1}(X)\equal{}\gamma^{\minus{}1}(U\cup V)\equal{}\gamma^{\minus{}1}(U)\cup\gamma^{\minus{}1}(V)\equal{}[0,1]$, but also $ \gamma^{\minus{}1}(U)\cap\gamma^{\minus{}1}(V)\equal{}\emptyset$. Thus we have a separation of $ [0,1]$ which is a contradiction since $ [0,1]$ is clearly connected.

Take $ \{(x,y): y\equal{}\sin(1/x) \ 0<x\leq 1\}\cup\{(0,0)\}$ as a subspace of $ \mathbb{R}^2$. There is no path from $ (0,0)$ to any other point on the curve. It is sort of tedious to make that argument rigorous though.


Let $n$ be positive integers and let $\alpha$ be real numbers such that $0<\alpha <1.$ Denote by $N(n)$ the number among $\sqrt{1},\ \sqrt{2},\ \sqrt{3},\ \cdots ,\sqrt{n}$ whose decimal part is less than or equal to $\alpha.$

Prove that $\lim_{n\to\infty} \frac{N(n)}{n}=\alpha.$

Oh! this is my posted one!  :lol:
In fact,I already post an answer on Japanese Communities ,but I have a trouble in translating into English.Sorry...:blush:


6!x7!=10!
 :?: Give the values of m,n and r such that m!xn!=r!.

Well, consider the case where n = m! - 1 and r = m!. Clearly this gives infinite values of m, n, and r, but you can tell that it doesn't give all of them.


LaTeX
inequalities
geometry unsolved
geometry

Find three angles of a triangle if

  $ 27p^{2} \plus{} 7r^{2} \equal{} 108 Rr$
   I can't  solve
  Dear Virgil Cula and anybody can help me? :blush:

i dont know how to use latex,sorry,

there is no triangle with property [27p^2+7r^2=108Rr]
 we always have p^2>4Rr

  (a+b+c)/3>(abc)^(1/3) so p^3>(27/8).abc
R=abc/4s r=s/p
4Rr=4.abc/4s.s/p=abc/p  so p^2>abc/p=4Rr

I think you are wrong , please be careful

i wrote the answer in a clear way  plz download the attachment

Sorry for my terrible mistake
   It shoud be
      $ 27r^{2} \plus{} 7p^{2} \equal{} 108 Rr$
 :oops:
     Please try agian
   Sorry

[size=117][color=darkblue][b]This problem is easily and nicely ![/b][/color][/size]

[hide="Only for nkht-tk14."]
[color=darkblue]With the greatest pleasure !

By the way, my name is [b]Virgil Nicula[/b] and not [u]Virgil Cula[/u], [b]nkht-tk14[/b] ...[/color][/hide]
[quote="nkht-tk14"][color=darkred]Prove that $ 7p^{2} + 27r^2 = 108Rr$ $ \Longleftrightarrow$ $ a = b = c$.[/color]  [/quote]
[color=darkblue][b][u]Proof.[/u][/b]

We observe that if $ a = b = c$ , then $ \{\begin{array}{c} p = 3r\sqrt 3 \\
\ R = 2r\end{array}\|$ and $ 27r^2 + 7p^2 = 108Rr$ $ \Longleftrightarrow$

$ 27r^2 + 7\cdot 27r^2 = 108\cdot 2r^2$ $ \Longleftrightarrow$ $ 27\cdot 8 = 108\cdot 2$ , what is truly.

The well-known inequalities $ \|\begin{array}{ccc} \boxed {p^2 + 5r^2\ge 16Rr} & \bigodot & 7 \\
 \\
\boxed {R\ge 2r} & \bigodot & 4r\end{array}\ \bigoplus$ $ \implies$

$ 7\cdot(p^2 + 5r^2) + 4r\cdot R\ge 7\cdot 16Rr + 4r\cdot 2r$ $ \implies$ $ 7p^2 + 27r^2\ge 108Rr$ .

Therefore, $ 7p^2 + 27r^2 = 108Rr$ $ \Longleftrightarrow$ $ \|\begin{array}{ccc} p^2 + 5r^2 = 16Rr \\
 \\
R = 2r\end{array}\|$ $ \Longleftrightarrow$ $ a = b = c$ .

[b][u]An easy extension.[/u][/b] For any $ m\ge 0$ and $ n\ge 0$ we have

$ mp^2 + (5m - 2n)r^2 = (16m - n)Rr$ $ \Longleftrightarrow$ $ a = b = c$ .

For example, for $ m = 2$ and $ n = 5$ we obtain $ 2p^2 = 27Rr$ $ \Longleftrightarrow$ $ a = b = c$ .[/color]

My proof to the ineq:   $ p^{2} \plus{} 5r^{2} \geq  16 Rr$ is very long
     Can you  please show me a good proof and another ineq in triangle  you usually use , Virgil Nicila?

[quote="nkht-tk14"][color=darkred]Prove that $ p^{2} \plus{} 5r^{2} \ge 16 Rr$ ... , [b]Virgil Nicila[/b] ?[/color] [/quote]
[hide="Again, only for nkht-tk14"]
[color=darkred]By the way, my name is [b]Virgil Nicula[/b] and not [b]Virgil Nicila[/b] .[/color].. [/hide]
[color=darkblue][b][u]Proof.[/u][/b] From the [b]Lagrange's identity[/b] $ \boxed {\sum XA^2 \equal{} 3\cdot XG^2 \plus{} \frac 13\cdot\sum a^2}$ obtain

the inequality $ 3\cdot\sum XA^2\ge \sum a^2$ for any point $ X$, with equality iff $ X: \equal{} G$ .

Using $ IA^2 \equal{} \frac {bc(p \minus{} a)}{p}$ a.s.o., the upper inequality for $ X: \equal{} I$ becomes :

$ 3\sum \frac {bc(p \minus{} a)}{p}\ge \sum a^2$ $ \Longleftrightarrow$ $ 3p\cdot\sum bc\ge 9abc \plus{} p\cdot\sum a^2$ $ \Longleftrightarrow$

$ 3(p^2 \plus{} r^2 \plus{} 4Rr)\ge 36Rr \plus{} 2(p^2 \minus{} r^2 \minus{} 4Rr)$ $ \Longleftrightarrow$ $ \boxed {p^2 \plus{} 5r^2\ge 16Rr}$ ,

with equality if and only if $ I\equiv G$ , i.e. the triangle $ ABC$ is equilateral.

[b][u]Remark.[/u][/b] $ IN^2 \equal{} 9\cdot IG^2 \equal{} p^2 \plus{} 5r^2 \minus{} 16Rr\ge 0$ .[/color]

[size=117][color=red][b]Only so much ![/b][/color][/size]


How many real numbers $ a$ are there such that, for these values of $ a$, the equation $ x^{3}= ax+a+1$ has at least one root $ x_{0}$ satisfying
$ (i)$ $ x_{0}$ is an integer and is even.
$ (ii)$ $ |x_{0}| < 1000.$


\[ {\rm{Prove: }} \& {\rm{cosec}}^6 \theta  - {\rm{cot}}^6 \theta  = 3{\rm{cot}}^2 \theta {\rm{ cosec}}^2 \theta  + 1
\]

$ \Leftrightarrow \frac{1\minus{}cos^6\theta}{sin^6\theta}\equal{}\frac{3cos^2\theta\plus{}sin^4\theta}{sin^4\theta}$
$ \Leftrightarrow 1\minus{}cos^6\theta\equal{}3cos^2\theta sin^2\theta\plus{}sin^6\theta$
$ \Leftrightarrow 1\equal{}(sin^2\theta\plus{}cos^2\theta)^3$


inequalities
geometry
circumcircle
trigonometry
Euler
blogs
geometry proposed

Given are $\triangle ABC$ area $S$ and circumcenter $O$ with any $M$ on plane prove that
$a^{2}MA^{2}+b^{2}MB^{2}+c^{2}MC^{2}\ge 4\sqrt{3}S(R^{2}-OM^{2})$
equal when $ABC$ is euqaliteral and $M\equiv O$
an especial case $M\equiv O\Rightarrow a^{2}+b^{2}+c^{2}\ge 4\sqrt{3}S$
$M\equiv I\Rightarrow \cos^{2}\frac{A}{2}+\cos^{2}\frac{B}{2}+\cos^{2}\frac{C}{2}\ge\frac{\sqrt{3}}{2}(\sin A+\sin B+\sin C)$ :lol:

[color=purple]Let D, E,F are on BC, CA, AB such that $MD\perp BC;ME\perp CA; MF\perp AB$. By Euler theorem we have$S_{DEF}=\frac{S(R^{2}-OM^{2})}{4R^{2}}$ thus$4\sqrt{3}S(R^{2}-OM^{2}=\frac{4S_{DEF}}{R^{2}}$
   So your inequality is equipvalent to
 $MA^{2}sin^{2}A+MB^{2}sin^{2}B+MC^{2}sin^{2}C\geq 4\sqrt{3}S_{DEF}\leftrightarrow EF^{2}+DE^{2}+FD^{2}\geq 4\sqrt{3}S_{DEF}$
  The last inequlity is well- known.And with my solution we can solve the genenral  :D
$MA^{2}xa^{2}+MB^{2}yb^{2}+MC^{2}zc^{2}\geq 4\sqrt{xy+yz+zx}S(R^{2}-OM^{2})$
  Do you have more gemath?  :D  [/color]

[quote="April"]Now, [b]gemath[/b] don't take part in Mathlinks Forum but you can find more Geometric Inequalities unsloved of [b]gemath[/b] at  [url=http://www.mathlinks.ro/Forum/weblog.php?w=659]his blog[/url][/quote]
 [color=purple]    What a pity!  :(  Thank April I saw them.Can you send a message to gemath for me? I have many problem and i want to discuss with gemath [/color]

Why you want to discuss with [b]gemath[/b]? You can post your problems in this Forum and many Mathlinkers can help you  :wink:

That's right you should post your problem in ML and many people better than me can help you.
And
[quote="evarist"]Let D, E,F are on BC, CA, AB such that $MD\perp BC;ME\perp CA; MF\perp AB$. By Euler theorem we have$S_{DEF}=\frac{S(R^{2}-OM^{2})}{4R^{2}}$ thus$4\sqrt{3}S(R^{2}-OM^{2}=\frac{4S_{DEF}}{R^{2}}$
   So your inequality is equipvalent to
 $MA^{2}sin^{2}A+MB^{2}sin^{2}B+MC^{2}sin^{2}C\geq 4\sqrt{3}S_{DEF}\leftrightarrow EF^{2}+DE^{2}+FD^{2}\geq 4\sqrt{3}S_{DEF}$
  The last inequlity is well- known.And with my solution we can solve the genenral  :D
$MA^{2}xa^{2}+MB^{2}yb^{2}+MC^{2}zc^{2}\geq 4\sqrt{xy+yz+zx}S(R^{2}-OM^{2})$
  Do you have more gemath?  :D [/quote]
You have true solution but don't think I don't know "the general problem"  :rotfl: , I don't post it because it don't have any sense.
Let me show a general problem like your "general problem" in thinking of children: They think $4$ is divisilble by $2$, $6$ is divisible by $2$ and $8,10..$ is too... hmm there is a problem here ah I think I can generalize it ok we will have $2n$ is divisible by $2$  :rotfl:

[color=purple]You sneer me ? Please post your solution for this problem and do you get the stronger as me ? It's very funny gemath!  :rotfl: [/color]

Ok the following is the stronger
given are $\triangle ABC$ area $S$ and circumcenter $O$ with any $M$ on plane prove that
$a^{2}MA^{2}+b^{2}MB^{2}+c^{2}MC^{2}\ge 4\sqrt{3}S|R^{2}-OM^{2}|$
equal when $ABC$ is equilateral and $M\equiv O$ hope you like  :rotfl:

Consider the pedal triangle and the inequality $a'^{2}+b'^{2}+c'^{2}\geq 4\sqrt{3}S'$.


calculus
integration
logarithms
real analysis
real analysis unsolved

$ \int_{0}^{1}\text{arccot}\left( 1-x+x^{2}\right) \, dx = x \, \cdot \, \text{arccot}\left( 1-x+x^{2}\right) \Big|_{0}^{1}+\int_{0}^{1}x \cdot \frac{2x-1}{1+\left( 1-x+x^{2}\right)^{2}}\, dx$
$ = \frac{\pi}4+\int_{0}^{1}x \cdot \left( \frac1{x^{2}-2x+2}-\frac1{x^{2}+1}\right) \, dx$
$ = \frac{\pi}4+\left[ \frac12 \ln \left( x^{2}-2x+2 \right)-\frac12 \ln \left( x^{2}+1 \right)+\arctan \left( x-1 \right) \right]_{0}^{1}$
$ = \frac{\pi}2-\ln 2$.

The other one can be solved in the same way.

By the way, isn't this better fitted in the [i]Computations[/i] section?


If you write the six digits $ 1$,$ 2$,$ 3$,$ 4$,$ 5$, and $ 6$ in all possible orders you will find that there are $ 720$ ways to do this. If the numbers are written in order of least to greatest, the first number on the list is $ 123456$ and the last number is $ 654321$. What number will be the $ 417$th number on the list?

#1 to #120 on list have a hundred-thousands digit of 1

#121 to #240 on list have a hundred-thousands digit of 2

#241 to #360 on list have a hundred-thousands digit of 3

#361 to #480 on list have a hundred-thousands digit of 4

So we know that the number begins with a $ 4$

Using same logic,

#361 to #384 begin with $ 41$

#365 to #408 begin with $ 42$

#409 to #432 begin with $ 43$

Our number begins with $ 43$

Continuing this, we find that the $ 417$th number is $ \boxed{432516}$

120 begin with "1", another 120 begin with "2", another 120 begin with "3", and another 120 begin with "4". 417 is less than 120*4, so the first digit of 417 is 4.

Of the numbers beginning with 4, 24 have a second digit of 1, another 24 have a second digit of 2, and another 24 have a second digit of 3. We find that the 417th number has a second digit of 3.

Of the numbers beginning with 43, 6 have a third digit of 1, and another 6 have a third digit of 2. This means the 417th number has a third digit of 2.

Of the numbers beginning with 432, 2 have a fourth digit of 1, and another 2 have a fourth digit of 5.

Our number is then $ \boxed{432516}$. Hopefully I didn't fail.

EDIT: tinytim submitted that while I was writing this.


This is for anyone is interested in [url=http://www.mathkangaroo.com/2010page/kangur/index.htm]Math Kangaroo[/url]. There is center in the [url=http://www.mathkangaroo.com/MathKangaroo/LocationDetails.jsp?locationID=55]Twin Cities[/url] as well as a new one in [url=http://www.mathkangaroo.com/MathKangaroo/LocationDetails.jsp?locationID=91]Grand Forks, ND[/url] for anyone who lives in the western part of the state.

This year, it will be on Thursday, March 18, 2010. Registration must be done by January 20, 2010.

Math Kangaroo is a math contest for grades 1 through 12. It is a 75-minute multiple choice test with 24 or 30 questions, depending on grade level. Prizes are given to the top scorers.

Hey Kev, you from minnesota yourself?  It says antarctica...

Whose kev?

And no I am not from Minnesota, I am from Antarctica.

No, but Minnesota is within walking distance from my house. (and @abcak I've met you once before)

I have met both powerofpi and abcak.  Seems interesting (math kangaroo)

This is creepy.

UNFORTUNATELY (for you), you guys don't know wizeng :P

I think.

I really want to meet Abcak.
I haven't seen him before.

Hah.

So is this only people in Minnesota and North Dakota.

I really want to meet PowerOfPi; I haven't seen him before :wink:

[quote="ksun48"]I really want to meet PowerOfPi, AwesomeToad, and abcak, I haven't seen any of them before...[/quote]

LOLWUT WHO ARE YOU???

http://m.youtube.com/#/watch?v=jDGnfT_OaCM

The final summary of this topic

[quote="PowerOfPi"]*Bump... Similar thing this year, but the registration deadline is January 10, 2011 and the competition itself is on March 17, 2011.[/quote]

Deadline has passed, but maybe registration will still be possible...

This year's registration deadline is the 16th of January. Hurry up and register if you haven't already.


There are two samples of gas in a thermally insulated cylinder. One of them is Argon at a temperature of 300 K, and the other is Nitrogen at a temperature of 400 K. There is a thermally insulated wall between them. The two gases have the same pressure and the same volume. Then the wall between the gases is taken away, and the two gases mix.

a) Does the energy of the system change?

b) What will be the common temperature?

c) How does the pressure change?

d) Does the entropy of the system change?

a) Considering the cylinder as an isolated system and both gases as ideal, and as the gases don't react with each other, then $\Delta U = 0$.

b) about $355$K
d)The entropy of the system will grow.

[quote="Thaakisfox"]b) about $355$K
[/quote]

Can you show the work?

[quote="Thaakisfox"]You mean how to calculate the temperature?[/quote]

Yeah, I don't see how to do it.

its just conservation of energy:
$\frac32 pV+\frac{5}{2}pV = \frac{3}{2}\frac{pV}{T_{1}}T+\frac{5}{2}\frac{pV}{T_{2}}T$
form here the common temperature:
$T=\frac{8T_{1}T_{2}}{3T_{2}+5T_{1}}$
$T_{1}$ is the begining temp. of Argon, $T_{2}$ is the begining temp of nytrogen, $p$ and $V$ are the begining volume, and pressure.

[quote="cancer"][quote="Thaakisfox"] $\frac{3}{2}\frac{pV}{T_{1}}T$[/quote]

Why is this the new energy?[/quote]

Can someone explain? Why divide by $T_{1}$ and multiply by $T$?

according to the gas law:

$pV=n_{1}RT_{1}$ so from here $n_{1}R=\frac{pV}{T_{1}}$
so if we want to find the energy of the Argon after mixing: (since $n_{1}$ is constant, non of the argon leaves the system)
$E_{1}= \frac32 n_{1}R T = \frac32 \frac{pV}{T_{1}}T$


Princeton
college
FTW
function
logarithms
MATHCOUNTS
irrational number

Currently i am planning a huge celebration of pi day at my school. It will probably involve me saying ~100 digits of pi over the announcements. I have committed only 83 digits to memory, and plan on memorizing 17 more by March.

What about you? All pi fans are encouraged to respond.

Princeton has a pi recitation contest on Pi Day. I had just under 100 digits memorized then, but now I am not confident of more than 20 or so. The person who won had nearly 200 I think.

You should have put a 100+ option or something :P.

I had this one on the shelf, because I knew this day would come :

[img]http://users.telenet.be/frederic_bel6/Fun/refusepi.jpg[/img]

3.1415926

What is the point of memorizing pi? Is it just how nerds show their masculinity?

lets see:
3.141592653589793238462643383279502884 and umm... umm..
yeah.

If i had any reason to memorize, i could probably get to 150

Memorizing pi is a great way to attract girls.

Person: Hey baby, your curves are like a circle with a radius of pi. Pi is approximately 3.1415...(353784973489)

or perhaps, more direct

Person: Hey baby, wanna get with someone who knows the first 3489305834890 digits of pi?

LOL, you got that totally from the MI forum.

Hey baby, wanna go out w/  a guy who got a 34 on the MMPC?

I kinda tried that...not like...directly.

Don't try it in your life, it's not pleasant.

Anyways...I don't feel like memorizing it, cuase it has no poitn:

3.141592653589793238462643383279?

(20:52:33) Atheist Temple: Hey baby, wanna get with someone who knows the first 3489305834890 digits of pi?
(20:52:34) <name removed> <AUTO-REPLY> : math homework

darn i thought it would work better

Mostly I only remember $3.14$

No point, lah :D

Hey! I'm a girl here, and memorization of pi is a popular nerdy pastime. :mad:

correction: kstan013, there must be a picture of me in the song in order for it to be classified as nerdy -_-.   :D

[quote="funcia"] [color=blue]kyyuanmathcount: clip, i'd rather not clog up thread, just look posts above[/color][/quote]

"obvious" is my word, I think you are offending him, captain obvious should not have to come to rescue too many times in one thread.

Also, [quote="I"]I don't care[/quote]

LOL

OBVIOUS SHOULDN'T BE PUT HERE SO I SAY OBLIVIOUS THEN.
JK I was just upset after a relative died.

[quote="kstan013"][quote="kyyuanmathcount"]correction: kstan013, there must be a picture of me in the song in order for it to be classified as nerdy -_-.   :D   [/quote]Awww... don't be so hard on yourself :)[/quote]

 :D it's for the good of the rest of the world... and the asians in it :P.

$\pi = 4(\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+-\cdots )$

yeah Asians are so pretty and smart
when i told my Chinese friend i wanted to be Asian she
practically threw up on me

No need to brownnose now is there?

Asians include indian people.  When we want to refer to chinese, japanese, korean, malaysian etc., we say oriental :).

The only reason I know anything much of pi is because my math teacher in 4th grade had a poster on her wall that had like 100 digits of pi on it. 3.14159265358979323846.

i simply memorized the first 10 digits
3.14159265
i think it's 10...i never counted it
Edit-OMG ITS ONLY 9 digits! actually without the 3 it's 8...i think i might've forgot 1 or 2 digits

$\pi$ is DA BOMB

I've memorized about 180 digits, but it has been a long time since I've recited them and I can't say them as well as I used to be able to.

this is the key that makes us all wind and the beat comes up on it

[quote="modeler"][hide="I know a guy who can memorize pi better than anyone of you."]
My Computer  :D
[/hide][/quote]
you silly  :roll: , computers don't have genders

computers have genders.... just like they have generations... vacuum tubes, transistors, newly artificial intelligence.

so why not gender?

I'm sure they do.

Oh yeah, I know $\pi$

It's approximately $\frac{22}{7}$!

well or $3.14159$

My computer is a male. I call him Mac  :D


[quote="LuCy4EvA"][hide]9+63/3
9+21
30[/hide][/quote]

How did that minus between the 5 and the 7 change to a plus? I got
-12.

oops, wasn't really paying attention... sry!!!

so it's um...

[hide]
9-21 = -12[/hide]

thanx for finding my mistake, that's why i need more practice with arithmetic! :blush:

[hide="hinty hint"]ALWAYS!!! do muliplication and divisions frist , and do addition and subtraction next[/hide]

 :D

I hate arithmeatic problems, so I won't solve it...

The trick of solving this problems is to know the order of operation. PEMDAS.
1. parenthesis
2. exponent
3.Multiplication and Division
4. Addition and Subtraction

[quote="mathgeek2006"][quote="GoBraves"][quote="mathgeek2006"]A stupid acronym for PEMDAS:

Please
Excuse
My
Dear
Aunt
Sally

....[/quote]

Whats stupid about that??[/quote]
I learned it in school, for one.  Also, it's really kinda weird...[/quote]

I learn tons of stuff in school..........i learn all english, and social studies in school...and I dont think its that weird

[quote="GoBraves"][/quote][quote="mathgeek2006"][quote="GoBraves"][quote="mathgeek2006"]A stupid acronym for PEMDAS:

Please
Excuse
My
Dear
Aunt
Sally

....[/quote]

Whats stupid about that??[/quote]
I learned it in school, for one.  Also, it's really kinda weird...[/quote][quote="GoBraves"]

I learn tons of stuff in school..........i learn all english, and social studies in school...and I dont think its that weird[/quote]
This is something I learned in 4th grade, though.  Now that I'm in 8th, I think that it's stupid.

[hide]4+5-7*9/3

[b]-12. There's your answer.[/b][/hide]

[hide]Very Easy just do Bedmass
          9x7=63
         63 Divided 3 = 21
         9-21= -12 = The Answer   [/hide]

[quote="math92"]all these quotes are getting confusing

can someone tell me how to do them?[/quote]

to quote someone's post, there is a "quote" button on the bottom right of the post

but if u remembr the "pemdas" and forget the left to right rule than there is no point of remembering the pemdas rule because u r gonna get it wrong! RITE?

yeah, they should make a new acronym: DPEDMASLTR which means do PEDMAS left to right. [/hide]


Primer nivel, problema 1:

En un año que tiene 53 sábados, ¿qué día de la semana es el 12 de mayo? Dar todas las posibilidades.

Supongo que el problema se resuelve así:

Hay dos casos:

-Año normal   (1)

-Año bisiesto  (2)

En (1) tenemos que el año tiene 365 días. Dado que $365=52\cdot 7+1$, entonces al haber 52 semanas es obvio que hay 52 sábados, el problema pide un año con 53 sábados, por lo tanto ese $+1$ que aparece debe ser sábado. Como el año está compuesto por 52 semanas enteras y un sólo día de otra semana, entonces se concluye que la semana nuestra empieza con el día sábado. El día 365 (31 de diciembre) es sábado según lo dicho más arriba. El día 1 (1 de enero) es sábado también, ya que es congruente a 1 módulo 7, al igual que 365. El 12 de mayo es el día número 132 del año, como $132\equiv 6 (mod 7)$, se concluye que es jueves, ya que si el sábado es congruente a 1, entonces el jueves es congruente a 6, porque la secuencia es sábado, domingo, lunes, ..., viernes. Como verificación tomen el año 2011, es decir este año, y verán que se cumple todo (fue un problema futurista este).

En (2) tenemos que el año tiene 366 días, es decir $366=52\cdot 7+2$, el sábado número 53 puede ser $+1$ o el $+2$. Si es el $+1$, entonces se tiene que el día 365 es sábado, es decir el 30 de diciembre, entonces el 1 de enero también es sábado (día 1). El 12 de mayo en esta oportunidad es el día 133 del año, como $133\equiv 0(mod 7)$, entonces el 12 de mayo caerá viernes. Si el sábado cae $+2$, entonces el 31 de diciembre es sábado, y se tiene que para ser sábado debe ser congruente a 2 módulo 7. Como 133 es congruente a 0, entonces debe caer dos días antes que el sábado, es decir el jueves.
Para verificar pueden ver el calendario del año 1988

Conclusión: el 12 de mayo puede caer jueves o viernes en un año con 53 sábados.


FTW
AMC
AIME
geometry
geometric transformation
reflection
analytic geometry

Join today!  And if you so desire, you may refer to the rules [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=250456]here...[/url]  :D

I say everything I said in the other thread :D

okay I sign up I guess. I'm really only interested in the chapter one though since my goal is to make state this year.

i am on fanfans team, but not bogtros.

....

HA!

I don't care. Sign me up on any team. I just wanna focus on chapter. If you don't want to be brought down don;t put me on your team. I seriously am bad.(despite my 1721 FTW rating :P )

[u]No Team:[/u]

AIME15
BOGTRO
Discrete_Math
FantasyLover
isabella2296
jjx1
nikeballa96
pythag011

dangit I posted right after aops was down. i said nike want to be a team. that is if its okay if i bring you down :blush:

[quote="FantasyLover"][quote="mz94"]i am on fanfans team
[/quote]

lol ernie[/quote]

only you mean lol mz

[quote="ernie"][quote="FantasyLover"][quote="mz94"]i am on fanfans team
[/quote]lol ernie[/quote]only you mean lol mz[/quote]

Lol, I did mean you because you said:

[quote="ernie"]No Team:
FantasyLover
[/quote]

[quote="nikeballa96"]i join...anyone wanna join me?[/quote]

ill join ur team,,,if it is okay with you.  PAY NO ATTENTION TO MY FTW RATING, IT DOES NOT REFLECT MY ABILITIES

I MEAN IT

I thought the captains were supposed to submit the team rounds?

my assumption is that ernie wants to make this as close to the real competition as possible, which means any submissions for the team round not from the captain probably won't be graded at all

Nope.

Only team captain sends answers.  Anyone who doesn't abide to that rule will disqualify their whole team. :P

I VOTE FOR "SUPERMATHBOY"
as teh team captain

[quote="gauss1181"]Um, guys, suma_milli hasn't voted for team captain yet. (for team  :D )[/quote]

Oh sorry it was because I didn't really have an opinion..  So no vote from me.  I guess AIME_is_hard is captain then.

I hope this means the tournament can finally start now...

Check the other thread, it already has :wink:

[quote="jjx1"]Are these problems from past mc chapter? did you make them up yourself. They seem awfully hard for chapter :maybe:[/quote]
I guess he made them up.

Remember...these tests were made by fellow AoPSers...so they'd definitely be harder than normal.

of course, that doesn't bar you from taking the test altogether

you just won't be able to participate in the competition aspect

Hmmm... since I'm really fair, I'll let you participate as an individual.


Hehehe :).

$f(999)=\frac 1{999}, f(1000)=999$, so the image of $f$ includes $500\in [\frac 1{999},999]$. Since $f(x)=\frac 1x$ for all $x$ included in the image of $f$, we get $f(500)=\frac 1{500}$.


algebra
polynomial
Rational Root Theorem

$ P(x)\equal{}ax^{3}\plus{}bx^{2}\plus{}cx\plus{}d$
a,b,c,d are integers.
ad is odd,bc is even.
Prove that not all roots are rational.

Over $ \mathbb{F}_2$ the only polynomial of degree $ 3$ with nonzero constant term and linear factors is $ (x \plus{} 1)^3$.

I think t0rajir0u is talking about fields and groups.

I don't know anything in fields and groups.

But,this was how  BJV solved it.

[hide="solution"]Assume that all roots are rational.
Let them be $ x_1,x_2,x_3$
Now,
$ ax_1^3 \plus{} bx_1^2 \plus{} cx_1 \plus{}d\equal{}0$
Multiplying the above equation by $ a^2$,
$ {(ax_1)}^3 \plus{} b{(ax_1)}^2 \plus{} ca(ax_1) \plus{} da^2 \equal{}0$

Therefore the equation,
$ y^3 \plus{} by^2 \plus{} cay \plus{} da^2 \equal{}0$ -------I
has roots $ y_1,y_2,y_3$,
$ y_1\equal{}ax_1$
$ y_2\equal{}ax_2$
$ y_3\equal{}ax_3$

I ia a monic polynomial with integer coefficients.
Therefore, $ y_1,y_2,y_3$ are integers.

Now,
 $ y_1 \plus{} y_2 \plus{} y_3\equal{} \minus{}b$  -----II

 $ \sum{y_1y_2} \equal{} ca$    -----III

$ \prod{y_1} \equal{} \minus{}a^2d$     ----IV

Now,ad is odd.Therefore $ a^2d$ is odd.
So,from IV
 $ y_1,y_2,y_3$ are each odd.

Then from II b is odd.
From III,c is odd.
THerefore bc is odd.
But this is a contradiction,as bc is even.
Therefore our assumption that all roots are rational is wrong.

Thus proved.

[/hide]

That's essentially the same solution as mine.  $ \mathbb{F}_2$ is the mathematical structure you get when you ignore everything about the integers except their parity; that is, it has two elements $ \{ \text{Even}, \text{Odd} \}$ satisfying the usual rules for adding and multiplying even and odd numbers.  Over $ \mathbb{F}_2$, the problem condition reduces to

$ P(x) \equal{} x^3 \plus{} x^2 \plus{} 1 \text{ or } x^3 \plus{} x \plus{} 1$

where $ 1 \equal{} \text{Odd}, 0 \equal{} \text{Even}$.  On the other hand, by the rational root theorem (which holds in this setting!) the only possible rational root is $ 1$, so

$ P(x) \equal{} (x \plus{} 1)^3 \equal{} x^3 \plus{} x^2 \plus{} x \plus{} 1$;

contradiction.


[u][b]The author of this posting is : pvthuan[/b][/u]
____________________________________________________________________

If $a, b, c \geq 0$, then

\[\sqrt{a^4+b^4+c^4}+\sqrt{abc\left(a+b+c\right)} \geq 2\sqrt{a^2b^2+b^2c^2+c^2a^2}\].

This was solved by Pham Van Thuan, Hanoi University of Science, and here is my conjecture:

If $0<a\leq b\leq c$, then

\[\sqrt{a^4+b^4+c^4}+\sqrt{abc\left(a+b+c\right)} \geq 2\sqrt{a^3b+b^3c+c^3a}\].

Manlio: shouldn't it be $4$ instead of $2$ on the right side ? :?

[quote="manlio"]Before trying to solve the first very nice inequality I have solved
$a^4+b^4+c^4 +abc(a+b+c) \geq 2(a^2b^2+b^2c^2+c^2a^2)$
using Shur.
Now I am asking if this one can be correlated to the original inequality.[/quote]
I think original inequality is more stronger.

I find it very hard. I don't want to use brute force method but a nice method like that used to solve [url=http://www.mathlinks.ro/viewtopic.php?t=363]the famous KMO summer 2001 ineq[/url] which is very similar to this one.

Do you remember it?

It is true this week I am extremely idiot but if we take c=0 we find a good inequality?

Upss, sorry if my remark was absolutely late, but I saw this inequality now> :?

The inequality and the conjecture are not true for $b=c\rightarrow\infty$ .

Can anyone ,please, tell me if this ineq

$(a^4+b^4+c^4)(a+b+c)abc \geq (a^2b^2+b^2c^2+c^2a^2)^2$

is true?

Thanks a lot.

:blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:

Don't worry. We all make small mistakes from time to time.


Hi,  i am having trouble to solve this inequality when i pass 2 to other side of the equation. i know how to solve if multiple the sides.

  [(3-x)^(1/2)] /x  <= 2

solve doing it please:

   [(3-x)^(1/2)] /x   -2   <= 0

2. Equation

find X.         [9(7x-3)^2]/(2x+1) = [125(14x^2 +x -3)^1/2]/3

Thanx! :)

hello, writing your inequality in $ \text{\LaTeX}$ we have
$ \frac{\sqrt{3-x}}{x}\le 2$, because of the range of definition of the square root we have $ -\infty<x\le 3$ and $ x\ne0$.
Case $ 1$ multiplying by $ x$ for $ x>0$ we get $ \sqrt{3-x}\le 2x$ squaring this we have
$ 0 \le 4x^2+x-3$ and this is fulfilled for all $ x$ with $ \frac{3}{4}\le x\le3$
Case $ 2$ multiplying by $ x$ with $ x<0$ we get $ \sqrt{3-x}\geq 2x$, this is true because we have $ -\infty<x<0$.
Our solution is $ -\infty<x<0$ or $ \frac{3}{4}\le x\le 3$.
Sonnhard.

hello, to the second problem, do you mean
$ \frac{9(7x\minus{}3)^2}{2x\plus{}1}\equal{}\frac{125\sqrt{14x^2\plus{}x\minus{}3}}{3}$
Is this ok?
Sonnhard.

Squaring both sides and rearranging the terms,
$ 729(7x\minus{}3)^{4}\equal{}5^6(7x\minus{}3){(2x\plus{}1)^3}$. Notice that $ x\equal{}3/7$ is one solution. And now
$ 3^6(7x\minus{}3)^{3}\equal{}5^6{(2x\plus{}1)^3}$, so doing the factorizations
$ (63x\minus{}27\minus{}50x\minus{}25)[81(7x\minus{}3)^{2}\plus{}225(7x\minus{}3)(2x\plus{}1)\plus{}625(2x\plus{}1)^{2}]\equal{}0$
$ x\equal{}4$ is other solution. The complex solutions can be obtained by some simple calculations.

thanks!!

but the fisrt question, i ask to solve without  do it.....

how can i solve in this way :

[(3-x)^(1/2)] /x -2 <= 0

[with the number "2" in the left side]

Writing the equation in this manner ($ \frac {\sqrt {3 \minus{} x} \minus{} 2x}{x}\le 0$), you have to divide the problem in this cases "$ x\le 0$ and $ \sqrt {3 \minus{} x} \minus{} 2x\ge 0$" and "$ x\ge 0$ and $ \sqrt {3 \minus{} x} \minus{} 2x\le 0$". You have to put the existence's conditions of the inequality. Both cases were solved in the second post.

HELP ME WITH THIS  DOUBT!

In this other question

 [2x(x-2)^(1/2) - 4(x-2)^(1/2)] / (x-4)(x-2)^(1/2) <= 1

I just passed 1 to other side, made simplifications, i got:

   [x(x-2)^(1/2)] / (x-4)(x-2)^(1/2)

so because x>2(existence condition)

answer: (2,4)

and i cant see the difference between this question and the first
In this case, i didnt split in 2 cases as you. and got the correct answer.

please, clarify this for me.

Well, you can solve this sort of inequation by other methods, since it is does not involve any kind of fallacy. This shape of inequation permitted a more simple form that leads to a easier solution. I don't know what you mean clearly, but I hope your doubt was answered.

hello, do you mean $ \frac{2x\sqrt{x\minus{}2}\minus{}4\sqrt{x\minus{}2}}{(x\minus{}4)\sqrt{x\minus{}2}}\le1$?
Sonnhard.


[quote="MCrawford"]For what value of $x$ does
\[ \frac{0.\overline{09}}{x} = 11^{-1}? \][/quote]

[hide]$\frac{\frac{1}{11}}{x}=\frac{1}{11}$

$x=1$[/hide]

[quote="jli"][quote="A+MATH"][hide]11^-1 is 1/11. .0999999 is 1/11 so 1.[/hide][/quote]

[hide]A+MATH...

0.99999999 is not 1/11.

you misread the question, its .0909090909...[/hide][/quote]

He read the question right, he just didn't write it down right.


algebra
polynomial
calculus
integration
number theory
relatively prime

Find the least positive integer n satisfying the equation n3+2n=b, where b is an odd perfect square.

n will have to be odd
b² will have to be 0 mod n
a²=b and thus a is odd

so a= \sqrt(n³+2n)

a is an integer, that means n has common divisors with n²+2
it's not possible that n divides n²+2, so at least one of these must contain a perfect square in its factors

that cannot be n²+2 so it must be n
so n can be written as c.k² (both being odd, k being the largest possible)

a = \sqrt(c³.k^6+2c.k²)
so k divides a
a/k=d is integer

d = \sqrt (c³.k^4 + 2c)
as k was largest possible, c cannot be a square, thus c divides c².k^4+2 so 2 divides c

so either c=1
then k^4+2 is a perfect square, which is not possible

so c must be 2
but it had to be odd

so this is the third time I tried this problem, I wrote is as long and as clear as possible, always the same result that there are none, but you're asking for the lowest one? damn. :?

[quote="Peter VDD"]a is an integer, that means n has common divisors with n2+2
it's not possible that n divides n2+2, so at least one of these must contain a perfect square in its factors
that cannot be n2+2 so it must be n
so n can be written as c.k2 (both being odd, k being the largest possible)
[/quote]

Well Peter, in fact n does not have common divisors with n2 + 2. Because if d is a common divisor of n and n2 + 2, then d divides also (n2 + 2) - n x n = 2, so d|2. But d must be odd (as n is odd) so n and n2 + 2 are coprime. So now you know that the product of two relatively prime integers (i.e. n and n2 + 2) is a perfect square. What does that mean? :) Try to go from there.

I don't know

I thought exactly that way the first time, but scratched it away because I concluded from it there were no solutions

I know it looks very long and stupid like that, my previous solutions were much shorter and more powerful, but I kept getting no results, so I just threw them away. And I didn't feel like retyping the whole stuff.

Are you sure about the problem statement, 3X.lich? :?

Isn't it n3 + 2n2 = b, where b is an odd perfect square?

did you look the source up in your private collection, Arne? ;)

I didn't find it till now but I'm sure I saw it before :D
Well, my collection is becoming huge so some things are difficult to find :D

well it'd surprise me if you'd have Sergipanas OM 2nd round lv2/q6 (Brazil)

unless it's one of the famous things I never heard of again :D

No, I think I saw it on some British IMO mentoring scheme a long time ago, but I don't find it now.

well, we'll hear from lich where he found it ;)

Yeah maybe I posted this in the wrong forum, should be at the unsolved problems section... Anyways, that's how I was communicated. I believe it's a typo from that site's part because I found out last night that the equation cannot be satisfied by any positive integer (no solution!!). I thought the problem was interesting but here is the link:

http://www.obm.org.br/regionais/se99.htm

And again, I apologize for not double checking the problem first before posting it. Sorry.

oh well no problem

I thought I became stupid, solving it 3 times, and getting the same wrong solution 3 times :P

[quote="Arne"]Isn't it n3 + 2n2 = b, where b is an odd perfect square?[/quote]
Then answer is n=7

that would be too easy then

but I just got this nice idea: :D

i think it's n³ + 2^n = b with b a perfect odd square

now let's try to solve thàt :P

I'm quite sure it is n3 + 2n2 = b, but anyway, I'll think of your proposal Peter :D

The number must be large if there is one, I'm believing more that it has no solution  :(  But then, I need to prove that there are no solutions.
n3+2n=b
I'll check back later, gotta run to class.

Well I think it's pretty obvious the number needs to be large :D

It's not a fill-in excercice you'll get after trying 7 times...

Let's just stop trying ... the equation n3+2n=m2 have no integral solutions (rank = 0). I had to consult some literatures on elliptic curves before getting this answer. Seriously, we should forget about this version and think that they had n3+2n2=b as the problem.

Not anonymous, I posted the above reply.

[quote="Peter VDD"]strange excercice then

well it's beginner's level :D[/quote]
Hehe yeah I agree Peter. Thx for the suggestion though, at least I got to learn something new out of this  ;)


Let $ A$ and $ B$ be two fixed points in the circle $ (O,R).$ Determine the position of point $ M$ $ (M \in (O))$ such that the sum $ MA\plus{}MB$ attains the maximum value.

It is easy to see that $ M$ must be the midpoint of the bigger arc $ AB$ of the circle $ (O).$ Take $ B'$ on the ray $ AM$ such that $ MB \equal{} MB'.$ The triangle $ \triangle MBB'$ is isosceles with base $ BB'$ $ \Longrightarrow$ $ \angle MB'B \equal{} \frac {_1}{^2}\angle AMB,$ therefore $ B'$ moves on the circumference $(N)$ centered at the midpoint $N$ of the arc $ AB$ and radius $ NA \equal{} NB.$

The length of the chord $AB',$ $ AB' \equal{} MA \plus{} MB$ obviously attains its maximum when it is identical to the diameter of $ (N)$ passing through $ A.$ Hence, it follows that $ M \equiv N.$

Sorry, Why do you know B' moves on (N;NA)

Dear [b]Mr.luis[/b], the problem first stated that $ A,B$ are two points [b]in[/b] circle $ (O)$, not [b]on[/b] circle $ (O)$


calculus
integration
calculus computations

I tried do integration with  x, and the result came out as a number which is 81/8, while i havenot enter any interval.  I expect it would come out as x^2/2 ... What mode should i change in my calculator ? Thanks

Not the right forum (not even sure what it would go in), but you probably have a variable stored into x, making the calculator treat it as a number rather than a variable. Look in MEM-LINK and delete your x variable, and you'll be okay.


limit
function
real analysis
real analysis unsolved

let $A_{n}=\{n,n+1,n+2,...\}$
why is
$\cap A_{n}= \phi$ ?
i mean even at infinity $A_{n}$ and $A_{n+1}$ still have something
common.
or better asked
how much is $\lim_{n \to \infty}\left( |A_{n}\cap A_{n+1}| \right)$ ?

[quote]i mean even at infinity$A_{n}$ and $A_{n+1}$ still have something[/quote]

we arent concerned about just $A_{n}$ and $A_{n+1}$, when we talk abt intersection we want it to be in [b]ALL[/b] the $A_{n}$.

 suppose $x\in\cap A_{n}$ ,$x\geq n$ for all $n$. From here conclude that the intersection is empty

no i dont understand.
for a finite intersection $x\in I=\cap_{i=1}^{k}A_{i}$ it is sufficient
that $x\in A_{k}$ so that $x\in I$.

for an infinite intersection $I'=\cap A_{i}$ what happens ?
i think $|I'|>0$.
i am a little confused,i mean ,how can $|I'|=0$ ?
even at infinity there still remain natural numbers in that
intersection.

kouboy do you have a proof ?

i thought of something
consider $B_{n}=\{1,2,...,n-1\}$
first of all $\mathbb{N}-A_{n}= B_{n}$
now as $n\to \infty$ ,  $B_{n}\to \mathbb{N}$
so if $B_{n}$ will at infinity equal $\mathbb{N}$
so that means that by the definition of $B_{n}$ that
at infinity $A_{n}=\phi$ because $\mathbb{N}-\mathbb{N}= \phi$
but if one of the $A_{i}$ in the intersection is $\phi$ then the
whole intersection is $\phi$
any suggestions are welcome

I think your confusion arises from treating sets as real-valued functions; you seem to approach the question about $\bigcap_{n=1}^{\infty}A_{n}$ as some sort of a question about $\lim_{x\to\infty}f(x)$. But these are different games. In particular, your statement $B_{n}\to\mathbb N$ is meaningless until you define what convergence ($\to$) means for sets.

its rather intuitive notation
but $B_{n}\to \mathbb{N}$ obviously means that
as $n\to \infty$ ,$B_{n}$ is more and more closer to beein $\mathbb{N}$.

i tried to define it like so:
let us define a function $c(A,B)=1-\frac{1}{|A-B|}$ and call it similarity
of two sets(it can be easily seen that as two sets A,B have less difference
one from another the similarity between them is greater).
now we can make an analogous defintion as for the convergent
sequence( http://mathworld.wolfram.com/ConvergentSequence.html )
only in place of the modulus we use the $c$ function defined earlier
as follows so,we have $B_{n}\to \mathbb{N}$ if
for any $\epsilon > 0$ there exists $N$ so that for $n>N$ we have $1-c(B_{n},\mathbb{N})<\epsilon$

but then i realised that we are dealing with ordinals and i dont know
how much sense it would make to have a difference of ordinals...
i guess my conclusion is that i cant define it properly,ill have to think
more about it
but maybe it is already defined

Measuring similarity of sets is an interesting (and possibly difficult) problem. But I'd like to point out that the original question is not nearly as difficult. Intersections and unions are conceptually simpler than limits. You can see it from comparing their formal definitions:
\[x\in\bigcap_{n=1}^{\infty}A_{n}\Leftrightarrow \forall n\ x\in A_{n}\]
versus
\[L=\lim_{x\to a}f(x) \Leftrightarrow \forall \epsilon>0\ \exists \delta>0\ \forall x\in (a-\delta,a+\delta)\setminus\{a\}\ |f(x)-L|<\epsilon\]
Interpreting intersections as limits actually makes a simple concept harder to grasp.

so , mlok , how do you suggest i modify the "attempt at a proof" above
so that it does not imply limits of sets,and only use intersections ?

Suppose that $x$ is an element of $\bigcap_{n=1}^{\infty}A_{n}$. By definition of intersection, this means that $x\in A_{n}$ for all $n=1,2,3,\dots$. The definition of $A_{n}$ implies that $x$ is an integer such that $x\ge n$ for any $n=1,2,3,\dots$. Such $x$ does not exist; therefore, the set $\bigcap_{n=1}^{\infty}A_{n}$ has no elements.

so it is empty,because there is no $x>\infty$...
at least thats what i figgure


If $ |\minus{}2a\plus{}1|<13$, what is the sum of the distinct possible integer values of $ a$?

We have the two inequalities:

$ \minus{}2a\plus{}1<13$

$ \minus{}2a\plus{}1>\minus{}13$

Solving and combining, $ \minus{}6<a<7$. So, $ \minus{}5\plus{}\minus{}4\plus{}...\plus{}5\plus{}6\equal{}\boxed{6}$, as the -5 to 5 cancel out

Consider two cases.

1. $ \minus{}2a \plus{} 1 > 13$

$ a > \minus{}6$

2. $ \minus{}2a \plus{} 1 < \minus{}13$

$ a < 7$

$ \minus{}5 \plus{} \minus{}4 \plus{} \minus{}3 \plus{} \minus{}2 \plus{} \minus{}1 \plus{} 0 \plus{} 1 \plus{} 2 \plus{} 3 \plus{} 4 \plus{} 5 \plus{} 6 \equal{} \boxed{6}$

EDIT: Beaten to it.


Describe a 2-Sylow subgroup of $ S_8$. How many are there?

Consider the set X = { {1,2,3,4,5,6,7,8}, {1,2,3,4}, { 5,6,7,8 }, {1,2}, {3,4}, {5,6}, {7,8}, {1}, {2}, {3}, {4}, {5}, {6}, {7}, {8} }.  If you imagine its elements (which are 8-, 4-, 2-, and 1-element sets) as nodes, and draw edges between minimal containments, then you get a binary tree.

The group S8 acts on sets like this, where g in S8 takes X to { { z^g : z in Y } : Y in X }.

A sylow 2-subgroup of S8 is the stabilizer of X.  It is a self-normalizing subgroup; there are 315 conjugates.

You can see this by checking that there are (8 choose 4)/2 ways of choosing the 4-element sets, and for each one there are (4 choose 2)/2 * (4 choose 2)/2 ways of choosing the 2-element sets, and of course just one way to choose the 8-element and the 1-element sets.

Similarly, looking at stabilizers of levels of the tree gives a nice normal series for the Sylow and allows you to see its structure and order.


We consider the triangle ABC such that AB=c, BC=a and AC=b.Prove that
 $\frac{a+b+c}{2}$ $\sqrt 3$>$\ l_a +\ l_b+\ l_c$, where the notation are that used in an acute triangle.

See
http://www.mathlinks.ro/Forum/viewtopic.php?t=14431
http://www.mathlinks.ro/Forum/viewtopic.php?t=805
The > has to be replaced by $\geq$, and the triangle needs not be acute.

  darij

Thank you, Darij, but I think it was hard to guess that anywhay...

Which book is written by Valentin Vornicu?I'm eager to have a look.

It's a book from GIL.If you want it, you can contact Mircea Lascu on this forum by PM.It's a very good book.Normally, it is written by V.V.

[quote="Hawk Tiger"]Thank you very much,stancioiu.But what is the meaning of GIL?And I wonder that what the book's name is.[/quote]
http://www.gil.ro

$\frac{a}{\sqrt [b+c]}+\frac{b}{\sqrt (a+c)} +\frac{c}{\sqrt (a+b)}\geq \sqrt \frac{3}{2}$


Using U.S currency, in how many different ways can you pay 25 cents? In how many different ways can you pay 50 cents? (The way depends on how many coins of each kind you use) You must answer both parts correctly!

i got an idea.  Number of ways to make 25cents squared is 50 cents arrangements

qdnp
1000
0210
0205
0130
0125
01110
01015
005
004*
003*
002*
001*
00025

13 ways to make 25
so making 50 is like making 25 twice
169 for 50 except if you use half dollars 170.


vector
group theory
superior algebra
superior algebra theorems

I would appreciate some help with the decomposition of $ Sym^3 V$ representation of the symmetric group $ S_3$, where $ V$ is its standard 2-dim representation.

I think I got it right for $ Sym^2 V$. If $ \alpha$ and $ \beta$ are the basis vectors of $ V$, then $ \alpha \otimes \alpha$, $ \beta \otimes \beta$, and $ \alpha \otimes \beta \plus{} \beta \otimes \alpha$ are basis vectors for $ Sym^2 V$. Then, if $ \alpha$ and $ \beta$ are eigenvectors of 3-cycles in $ V$, $ \alpha \otimes \alpha$ and $ \beta \otimes \beta$ are eigenvectors of 3-cycles in $ Sym^2 V$, and $ \alpha \otimes \beta \plus{} \beta \otimes \alpha$ spans the trivial representation, so $ Sym^2 V \cong V \oplus U$, where $ U$ is the trivial representation.

Now, how to work it out in $ Sym^3 V$?

Thanks in advance!

Please disregard, already got help on another forum:
https://nrich.maths.org/discus/messages/7601/147966.html?1261852868


geometry
rectangle
vector
geometric transformation
geometry proposed

Perform a spiral similarity $L$ with center $B$, angle $\angle ABD$ and factor $\frac{BD}{BA}$

Clearly, $L(A)=D, L(K)=C$. For this reason, we deduce that $L(M)=P$.

So $\Delta MBP$ and $\Delta ABD$ are similar. So $\angle PMB = 90^{\circ}$.

Nice solution.  :)

BTW, my solution uses vectors .

Thanks silouan  :)

Can I see your solution please?

[quote="silouan"]Let a rectangle $ABCD$ and let a point $K$ on $AC$ such that $BK$ is perpendicular to $AC$ .Also let $M$ the midpoint of $AK$ and $P$ the midpoint of $CD$ . Find the angle
$\widehat{PMB}$[/quote]

Right triangles $ABK , BCD$ are similar, since $\angle BAK = \angle BDC = x$.But $BM, BR$ are medians , so triangles $BAM, BPD$ are similar. Hence $\angle ABM = \angle DBP = y$.But then
 $\angle BMC = \angle BPC = x+y$, which means that $BMPC$ is incribed.Thus $\angle PMB = 180-\angle PCB = 90$!

[u] Babis[/u]

Mr Stergiu this is also my synthetic proof(in fact it same as Jan's one but you don't mention the spiral similarity) .
Here is the solution with vectors
Let $\vec{BA}=\vec{a}$ , $\vec{CB}=\vec{b}$ and $\vec{BK}=\vec{c}$
We have $\vec{BM}=\frac{\vec{a}+\vec{c}}{2}$
and $\vec{PM}=\vec{PC}+\vec{CB}+\vec{BM}=\vec{b}+\frac{\vec{c}}{2}$
So $\vec{PM}\cdot\vec{BM}=\frac{1}{4}(\vec{a}+\vec{c})(2\vec{b}+\vec{c})=\frac{1}{4}(2\vec{a}\cdot \vec{b}+2\vec{c}\cdot \vec{b}+\vec{a}\cdot \vec{c}+(\vec{c})^{2})=0$

I think it is not ungly at all .What do you think ?

I think it is a very nice proof dear silouan.  :)


Fiind data functia continua$f: [0,1]\rightarrow R$, sa se calculeze limita sirului $(a_{n}) ,a_{n}=\int_{0}^{1}x^{n}f(x)dx$

Merită pomenită în acest context şi problema clasică:
Fiind dată funcţia continuă$f: [0,1]\rightarrow R$, să se calculeze limita şirului $(b_{n}) , b_{n}=n\int_{0}^{1}x^{n}f(x)dx$

[quote="radac_mail"]$b_{n}$ nu are limita pentru orice functie continua $f\: ,\: f: [0,1]\rightarrow R$ .
Exemplu:$f(x)=1,si \: f(x)=-1$[/quote]
Cred ca glumesti.
Eu cred ca $b_{n}\to f(1)$ dar nu am reusit sa demonstrez decat cu "Dominant Convergence Theorem" ( care nu sunt sigur ca e foarte usor de demonstrat).

Cred că s-a mai discutat... Oricum, presupunem "WLOG" că $f(1) = 0$ (altfel considerăm funcţia $f(x)-f(1)$).

In continuare, folosesc metoda lui spix:

Fie $\epsilon > 0$ şi $t \in (0,1)$ astfel încât $\left| f(x) \right| < \epsilon, \, \forall x \in [t,1]$. De asemenea, $M = \sup_{x \in [0,1]}\left| f(x) \right| <+\infty$.
Avem $\left| b_{n}\right| \leq \left| \int_{0}^{t}n x^{n}f(x) \, dx \right|+\left| \int_{t}^{1}n x^{n}f(x) \, dx \right| \leq M t^{n+1}+\left( 1-t^{n+1}\right) \cdot \epsilon$.
Luând $n \to \infty$, obţinem $\limsup \left| b_{n}\right| \leq \epsilon$.
Cum $\epsilon$ a fost ales arbitrar, deducem că $\lim b_{n}= 0$.