[quote="Peter VDD"]hey where is the text mode style? seems gone since the merge :?

will it ever come back or is it gone forever?[/quote]I do not plan to re-install that style for now. It's too much a pain in the [beep] :D ... maybe later, when all the other problems remaining are solved, we will think about it ...  :maybe:

Problems? site? Oh yes! :)

hey anyway for the voting system: if I gotta send it which version do I send
- minimal size, voting goes very fast, view goes normal (=no sort of cache)
- small size, voting goes fast, view goes lightning speed (=some sort of cache)

I think the CPU will be glad if you take the second one but i'll first ask before i start working on it :D


Only the alums should have theirs right now or get them soon.

[quote="Demon"]1. On the letter I recieved with all the forms it says that the registration has to be submitted by May 5th. However, the online registration requires travel itinerary, I don't plan to know that quite yet. What should we do?

2. Should the checks for the American Airlines waiver and Camp down payment be separate or joint?

Thanks[/quote]

With these sorts of technical questions, you should probably email us at camp@mathcamp.org.


[quote="1=2"]Let $F_{n}$ be the nth term of the fibbonacci sequence(let me know if I spelled that wrong.).

Define $F_{1}=F_{2}=1$.

Also define $F_{n-1}=F_{n+1}-F_{n}$

For all integers n.

Assume that x is negative.

Compute $F_{x}$ in terms of $F_{-x}$[/quote]
If x is even then they're opposites, if x is odd then they're the same.


There is a triangle ABC and its circumcircle O . A circle that is tangent to AB , AC and circle O meets the circle O on the point A_1.
We define B_1 and C_1 the same way. Show that the lines AA_1, BB_1, CC_1 are concurrent.

Dear Mathlinkers,
see for example
http://perso.orange.fr/jl.ayme   vol. 4  A new mixtilinear incircle adventure p. 20-23
Sincerely
Jean-Louis

You also have to say that the circle tangent to $ AB, AC$ is tangent to the circumcircle $ \mathcal O$ of the $ \triangle ABC$ internally or externally in the angle $ \angle CAB.$

$ \mathcal I$ is the triangle incircle and $ \mathcal K_A$ is a circle tanget to $ AB, AC$ and internally to $ \mathcal O$ at $ A_1.$ $ A$ is the external similarity center of $ \mathcal I, \mathcal K_A$ and $ A_1$ is the external similarty of $ \mathcal O, \mathcal K_A$ $ \Longrightarrow$ the external similarity center $ X$ of $ \mathcal I, \mathcal O$ is on the line $ AA_1.$ Likewise, $ X \in BB_1, X \in CC_1.$


geometry
circumcircle
incenter
analytic geometry
Euler
geometry solved

The incircle of triangle $ABC$ touches its sides $AB, BC, CA$ in points $C_1, A_1, B_1$ respectively. The point $B_2$ is symmetric to $B_1$ with respect to line $A_1C_1$, lines $BB_2$ and $AC$ meet in point $B_3$. points $A_3$ and $C_3$ may be defined analogously. Prove that points $A_3, B_3$ and $C_3$ lie on a line, which passes through the circumcentre of a triangle $ABC$.
[b]
proposed by L. Emelyanov[/b]

I always love the triangle geometry problems of the 239MO.

For this one, the line through the points $A_3$, $B_3$, $C_3$ is the famous OI line of triangle ABC (this is the line joining the circumcenter with the incenter of triangle ABC). See [url=http://forumgeom.fau.edu/FG2004volume4/FG200411index.html]Lev Emelyanov, [i]On the Intercepts of the OI-Line[/i], Forum Geometricorum, 4 (2004) p. 81-84[/url] for a trigonometric solution, [url=http://forumgeom.fau.edu/FG2003volume3/FG200312index.html]Lev Emelyanov and Tatiana Emelyanova, [i]A note on the Schiffler point[/i], Forum Geometricorum, 3 (2003) p. 113-116[/url] for a solution using barycentric coordinates, and [url=http://groups.yahoo.com/group/Hyacinthos/messages/7256?expand=1&viscount=1]Hyacinthos message #7256[/url] for a synthetic proof. Must the problems of the 239MO always be new problems?  :D

[b]PS.[/b] The links to Forum Geometricorum above sometimes don't work - the FAU servers are unpredictable...

  Darij

I know this is now redundant, but here's the sketch of a proof that $A_3$ lies on $OI$ (without some computations, which are pretty mild).

Let's turn the problem around alltogether, and look at it from the point of view of the intouch triangle. It's well-known that the orthocenter of the intouch triangle lies on $OI$, so we have the following problem:

Given a triangle $ABC$, let $UVT$ be its tangential triangle. If $A'$ is the symmetric of $A$ wrt $BC$, show that $VT$ and $UA'$ intersect on the Euler line of $ABC$. [$U,V,T$ correspond to $A,B,C$ respectively]

Let $M=UO\cap VT$, and $D=AH\cap BC$. After we observe that $A'H\|UO$, all we need to show is that $\frac{A'H}{HA}=\frac{UO}{OM}$. This is easily reduced to $\frac{UB}{VT}=\frac{DH}{HA}$, and from here it's rather easy.


http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=43014

Someone's abusing the system. Can someone IP ban them? The person also posted in my blog (http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=203382).

Thanks a lot

There are also a few instances of stuff like this. Someone made an account "mathcrazed," with the c accented but very subtlely, to mime the user mathcrazed. Today, I noticed a user "CatalystOfNostalgia" posted twice in a blog (xpmath's) thread that the real CatalystOfNostalgia (me) had started. It consisted of quite rude comments that even I wouldn't make. I don't know if it's an actual account though, because I clicked on the name in both comments. One didn't do anything, the other redirected me to the same page. But I definitely did not make those comments, as in my one real comment, it goes to my profile.

I'm not pointing the finger at anyone, I suspected the OP but he said it wasn't him, and seeing that someone impersonated him too, I believe him.

Well it was clearly not as subtle or clever, but there also exists a perfect629...They posted in my shout box at one point.

I must confess that, while I didn't do the things described as above, I created a clone of my own account, which is not exactly conducive to solving the "abuse" problem. I think those who created the initial clones did not do so to outright insult or ridicule or slander as above, but instead to just make harmless fun. I think it was all in good fun at first, and I am sorry to see people actually abusing it by saying mean things. :(

[quote="perfect628"][url]http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=203388[/url]
It's getting downright pathetic and disrespectful in this blog...[/quote]
The lasts two posts, by "rrusczyk" and "Valentin Vornicu," are NOT real, and are probably posts by that same person. I think the admins should take care of this immediately, before it gets nasty.

To everybody - I suggest not pointing fingers at anybody until this gets sorted out, as this can cause lasting effects, even after everything is dealt with.

This totally ruins the point of AoPS...........

Come on people, what's the point?

Impersonating is not good.

Stop discrediting people.

Seriously.

There was also an InhibitorOfNostalgia at one point in order to mock me, but that was more of a friendly joke and it was indeed someone from my school.

Solution?

Don't let guests post comments under pseudonyms. That'll take out half of the abuse, guarenteed.

I don't know why everyone is attacking me. IP check me, I don't care. Its not me!!!\

EDIT: In fact, I'd really like it if an admin could actually run an IP check so that I can be publically indemnified

[url]http://www.mathlinks.ro/profile.php?mode=viewprofile&u=43053[/url].
Familiar?  :dry:

[quote="i_like_pie"][quote="#H34N1"]I don't know why everyone is attacking me. IP check me, I don't care. Its not me!!![/quote]
Who's "everyone"? Only one person in this thread ever mentioned it being you, and they rejected the thought later.

Also, I'm not saying it's you (in fact, I'm pretty sure it's not you), but for future reference, saying "It's not me!" and begging to be identified as innocent sometimes looks suspicious. :wink:[/quote]

I'm talking about Temperal

[url]http://www.mathlinks.ro/profile.php?mode=viewprofile&u=43052[/url].
They have too much time wasted.

wow that dude even created a gmail for that s/n...

freakay..

YO I AM REALLY SORRY FOR WHAT I DID CAN SOMEONE UNBAN ME


parameterization
algebra
polynomial
inequalities unsolved
inequalities

\[ \ x\in R ;n\in N^{*}.\] Prove  that:
\[ \sum\limits_{k\equal{}1}^{n} C^{k}_{n}(k\minus{}nx)^{2}x^{k}(1\minus{}x)^{n\minus{}k} \leq \frac{n}{4}\]

This is not quite true ($ n\equal{}1$, $ x$ big). You probably want to let the sum
start at  $ k\equal{}0$. Then the LHS is equal to $ nx(1\minus{}x)$, but $ x(1\minus{}x) \le \frac 1 4$.

To see the simplification, note that the LHS is the variance for a Binomial random
variable with parameters $ n$, $ x$, which is known to be $ nx(1\minus{}x)$.
(Of course this only works for  $ 0 \le x \le 1$,  but two polynomials are equal
if they are equal on $ [0,1]$.)


[color=darkblue]Let $ABC$ be a fixed triangle and let $w$ be a variable circle which passes through the points $B$ and $C$. Denote : $M\in BB\cap CC$ ; the second intersections $X$ , $Y$ of the circle $w$ with the lines $AB$ , $AC$ respectively ; $N\in CX\cap BY$. Prove that the line $MN$ passes through a fixed point.[/color]

[quote="Virgil Nicula"][color=darkblue]Denote : $M\in BB\cap CC$ [/color][/quote]
Please explain :o

$BB$ means the tangent at $B$ to the circle. See the figure below.

[img]http://i77.imagethrust.com/i/356658/virgilnicula.png[/img]

The fixed point should be $A$.


What are some other good contests besides Mathcounts, USAMTS, and AMC. Are they free? How do you get into them?

ARML Power, local math leagues, NEML (for new england anyways, different places have different names). Tons of competitions.

You should look through this forum to see what contests people here have talked about.


inequalities
calculus
calculus computations

If  x,y,z are non-negative real numbers and $ x\plus{}y\plus{}z\equal{}1$ and $ a,b,c$ are positive real numbers prove the inequality:
$ a.x\plus{}b.y\plus{}c.z\plus{}2\sqrt{(x.y\plus{}y.z\plus{}z.x).(a.b\plus{}b.c\plus{}c.a)}\leq a\plus{}b\plus{}c$

It is a simple corollary of the Cauchy-Schwartz Inequality.

$ (ax + by + cz + \sqrt {(ab + bc + ca)(xy + yz + zx)} + \sqrt {(ab + bc + ca)(xy + yz + zx))^2}$
$ \leq (a^2 + b^2 + c^2 + (ab + bc + ca) + (ab + bc + ca)) (x^2 + y^2 + z^2 + 2xy + 2yz + 2zx) = (a + b + c)^2$


function
induction
algebra unsolved
algebra

Well here are the steps to solution:
a) if f(x) is constant then f(x)=0
b) assume f(x) is not constant
c) show there is a p such that f(p)=0
d) if f(p)=0 then p=0
e) for x>=0 f(x)=x by induction
f) for x<=0 f(x)=x by induction
g) so f(x)=x for x an integer
h) show f(1/k)=1/k
i) show f(p/q)=p/q
j) show f(x) is increasing sequence.
h) So by sandwich theorem f(x)=x for all the real x.

And we are done.

Though it seems that the soln is very long but actually this is not.  :D

substitue: (x,y)=(x,1/x) and then (x,y)=(1/x,x)

u will get two equations. play with it!

if u can't tell me!

Mahbub

[quote="ali"]could you elaborate this:
"c) show there is a p such that f(p)=0 "
that would be the critical point of the solution.
-a[/quote]

$x=y=-1 \Longrightarrow f(-1+f(1))=0$ :D
  :cool:

hmm u rearranged the process a bit to get the advantage of f(1)=1. Not bad! ;)

[quote="mahbub"]Diogene, Sorry that is wrong. As the proof of f(1)=1 comes from the base case f(0)=0 so you can't use that to prove f(1)=1.[/quote]

Man, I don't understand your objection ! I have to prove that there is $p$ such that $f(p)=0 $.
My proof  uses only  $f(x+f(xy))=f(x)+xf(y)$  :
$x=y=-1 \Longrightarrow f(-1+f(1))=f(-1)-f(-1)=0$
 :cool:  Do you follow me ? :cool:

how do u show it's increasing over the reals?

Isn’t it too easy? Say x>y then f(x)>f(y) [for x and y being rational]. Hence it is monotonic on rationals! Now do the last step that sandwich theorem and say for any x, f(x)=x.

You need not to show the monotonicity for real!
 :D

sorry, i don't understand.
don't you need either continuity or monotonicity over the reals? so what if it's monotone over the rationals? we don't have continuity, so that doesn't tell us anything about the irrationals...

if u can prove that f() is increasing for rational then by sandwich theorem you can say that for real number.

For more information about this method you can consult with the mathematical excalibur of volume 8 and i don't remember which issue.

Ofcourse not.!!
sandwhich theorem is irrelevant here.
Take this function:
f(x)=x for rationals.
f(x)= y for all irrational numbers.

Here is my original proof. I was only stuck at proving f(x)=0 for some x.
but thatnx to the very simple idea of diogone it is now solved.

Here it goes:

f(x)=0 iff x=0.
so f(0)=0.
x is arbitrary and:
Now take x+f(xy)=0.(*)
Then f(x)+xf(y)=0.
So  x+f(x)f(y)=0
and so f(x)f(y)=f(x)/f(y) so f(y)=+/- 1.
Hence f(x)=+/- x
if f(x)=-x.
It is clear that f(1)=1. (why?)
also f(-1)=-1.
f(1+f(y))=1+f(y)
f(x+1) = f(x) +x f(1/x)
so for x in R(f): f(1/x)=1/x.but if f(x)=x in (*)  then y=-1 and so f(-x)=-x
so R(f)=R and so f(x)=x for all x.

maybe this will finish up the long discussion.
hahaha :D  :D

you should read the sandwich theorem at the source that i stated so that you can have a clear idea on sandwich theorem. f(x)=x for all the irrationals as they are between two rationals and by proving the fact of increasing we can say f(x)=x.

mahbub you seem to have some fundamental problems in mathematics...
 :?

yeah, you might find that u need monotonicity over the reals in order to apply the sandwich theorem.
unfortunately, monotonicity over the rationals doesn't tell us very much about the irrationals.

OOPS! SORRY!!!  :oops_sign:

I have not noticed that!  :fool:

[quote="ehsan2004"]Find all function $ f:R\Longrightarrow\ R$ such that for any\[x,y \in\ R\] we have
$f(x+f(xy))=f(x+f(x)f(y))=f(x)+xf(y)$[/quote]


We now have a private forum just for MOEMS Teachers.  Students, please do not request to join this forum!  It is for teachers only to discuss teaching issues.

In order to join the forum, log in, then follow these steps:

1) Join the MOEMS Teachers Usergroup by [url=http://www.artofproblemsolving.com/Forum/groupcp.php?g=9579]clicking here[/url], and clicking 'Join Group'.

2) Send a private message to user RichKal-MOEMS by [url=http://www.artofproblemsolving.com/Forum/privmsg.php?mode=post]clicking here[/url].  To complete that form, enter the following:

Username: RichKal-MOEMS
Subject: MOEMS Teachers
Message Body: Please give the name of the school at which you teach.

This private message will be sent to Richard Kalman, MOEMS Executive Director and is the moderator of the MOEMS Teachers Forum.

After completing the form, click Submit.

3) RichKal-MOEMS will periodically review the requests to join the MOEMS Teachers forum and add those he determines to be MOEMS Teachers.  Again, students should *not* try to join this group.

If you have any questions, please post here or send a message to me or to RichKal-MOEMS.

Oh, wow, I am excited to have a MOEMS forum here on AoPS!   We did MOEMS Elementary for the first time this year (as well as MATHCOUNTS).  Next year I plan to do both levels of MOEMS.  It is great to have a forum with appropriate questions and a relevant Teachers' forum as well :D

[quote="anirudh"]Are students allowed to post here?[/quote]

Students can post in the MOEMS forum, but the MOEMS Teachers forum is for teachers of MOEMS material.

Oh, so what you mean is taht you don't want us to join the moems teachers usergroup.

Correct.  The reason for the forum is to have a private forum distinct from public users, meaning students.

Am i too late to join the group?  I click to join and it says your group membership pending?

What next?

I have a group of students in Michigan who are eager math students.  I would like to cultivate their interest in math and problem solving which aren't always available in regular school curriculums.

It's never too late to join the group.  It just takes time for the forum moderator (or one of the admins such as me) to approve your request.  I will go ahead and add you to the usergroup now.  Enjoy!


Lets define f(A) as a one digit number that we can get if we keep multiplying all the digits.

For example, f(24373)=f(1344)=f(48)=f(32)=f(6)=6

Find all number A that f(A)=1.

Thanks.

If f(A) = 1,we should start working backwards. Take A when it reaches the step as a 2 - digit number. The only number A as a 2 - digit integer that when its digits are multiplied equals 1 is 11. For when its a 3 - digit number, its digits have to multiply to 11. However, this is impossible since the only factors of 11 are 1 and 11, but 11 is not a single digit number. Therefore, $ A \equal{} 11.$

we will prove that a number in type 111..111 have a divisor which is greater than 10.
let B(t) is set of t 's prime divisors
If exist a number t = 111..111 (n digits and n>1) such that B(t) is subset of {3,5,7} we have
B($ 10^n \minus{}1$)is subset of {3,5,7} ( because $ 10^n \minus{}1$=9.111...111).
let p is smallest prime divisors of n
if p = 2 we have $ 10^n \minus{}1$ divide by 11.
if p = 3 we have $ 10^n \minus{}1$ divide by 37.
if p \geq 5 we have $ 10^n \minus{}1$ divide by $ 10^p \minus{}1$ which divide by q ( q is prime and q-1 divde by p)
and q>2.p+1>10.
So that we can't creat 111...111 from keeping multiplying all digits .
so the solution is A = 1111...111

Edit: Nevermind, misread the question ;)

[quote="math man"]if p \geq 5 we have $ 10^n \minus{} 1$ divide by $ 10^p \minus{} 1$ which divide by q ( q is prime and q-1 divde by p)
and q>2.p+1>10.[/quote]

Could you elaborate this please?

[quote="greentreeroad"][quote="math man"]if p \geq 5 we have $ 10^n \minus{} 1$ divide by $ 10^p \minus{} 1$ which divide by q ( q is prime and q-1 divide by p)
and q $ \geq$ 2.p+1>10.[/quote]

Could you elaborate this please?[/quote]

because $ 10^{q \minus{} 1} \minus{} 1$ divide by q
             $ 10^{p} \minus{} 1$ divide by q
10 - 1 don't divide by q ( because p $ \geq$ 2)
so p divide q-1
so q-1 $ \geq$2.p   .And .....

Solution of modularmarc101 is obviously wrong.

Solution of math man is wrong also because we cannot always find a prime $ q$ so that $ q\minus{}1|p$, for example , when $ p\equal{}5$, we get $ q$ must be $ 2$, which is not what we want.

[b]My solution:[/b]

Obviously, we move one step backward we get $ 111....1$ (n digits, n \ge 1)

So what we need to prove now is we can always find a prime $ p \neq 3,7$ that divides $ 11...1$ for every $ n \ge 2$

When $ n\equal{}2$, this is easy to check. Now for other $ n$ , assume that we can find a $ 11...1$ so that it is not divisible by primes other than $ 3,7$, then

[b]Case 1[/b]:$ 3|11....1$($ n$ digits) $ \Leftrightarrow 3|n \Leftrightarrow 111|11.....1 \Leftrightarrow 37|11....1$, contradiction.
[b]Case 2[/b]:$ 7|11.....1$($ n$ digits) $ \Leftrightarrow 6|n \Leftrightarrow 111111|11.....1 \Leftrightarrow 37|11....1$, contradiction.
[b]Case 3[/b]:$ 11.....1$($ n$ digits) are not divisible by $ 3,7$ , then it must be divisible by a larger prime, contradiction again, done.

So $ 11.....1$($ n$ digits, where n is any natural number) are the only possible values of $ A$

I don't know why all of you said my solution is wrong

@stephencheng: p=5 and q=2 ?
are you mad ?
my solution is very clearly.
It 's use order of a prime(smallest k such that a^k-1 divide by p)

[quote="greentreeroad"]How do you be sure that there exists such a q that p|q-1?[/quote]

bacause of Fermat 's little theorem
p is smallest such that 10^p -1 divide by q


Hamvatanah....salam.
komak mikham bare ye math problem.

1.) Show that the equation  1 + x + (x^2)/2 + (x^3)/6 = 0 has exactly one root. Then find all the roots of the equation.

2.) Find all roots of the equation: 5x^7 - 21x^6 - 51x^5 + 141x^4 + 264x^3 + 342x^2 -560x + 96 = 0
    bache ah man mikham solutioneh een soala ba dast bashe na ba program.

Merci!!

I just give you a hint, and the solution is left to yourself.
The first polynomial is of an odd degree and hence has at least one real root.(do you know the reason?)

One can find the (rational) roots of such polynomials with finite experiments, using Gauss' lemma.
You may search the web if needed.

But i have to use methods like bisection or newton-raphson method and am having trouble applying them on the problem.

Baba dige komak am kon.
Khahesh mikonam.


Let $ p,q$ be two primes and $ k$ a natural number. Show that $ \sigma(p^{k})\equal{}q$ has [i]at most[/i] 1 solution $ (p_{1},k_{1})$.

Let $ k$ be a natural number. Show that if $ p$ and $ \sigma(p^{k}) \equal{} q$ are [i]primes[/i], then so is $ k \plus{} 1$.

Proof: Let $ k \plus{} 1 \equal{} rn; r,n \in N$. Then $ q \equal{} \sigma(p^{k}) \equal{} \sigma_{r}(p^{n \minus{} 1})\sigma_{1}(p^{r \minus{} 1})$ is non-trivially factored for $ r > 1$.

Let $ p$, $ q$ be primes. Is the following reasoning valid?
$ \sigma(p^{q\minus{}1}) \equal{} 19$ has no solution because there is no element of the "congruence class modulo 19" whose "order" divides $ 18\equal{}\varphi(19)$.

Note that $ p^{q}\minus{}1\equal{}19(p\minus{}1)$. Since we must have $ p^{q\minus{}1}<19$, $ p<19^{1/(q\minus{}1)}$.

$ 2$ is a "primitive root modulo $ 13$" (and "modulo $ 19$" too). Thus $ 2^{12}\equiv1$ (modulo $ 13$). $ 12$ has 2 distinct prime divisors and can only be factored (according to our purpose) as $ 4\times3$ and $ 6\times2$. A solution for $ \sigma(p^{q \minus{} 1}) \equal{} 13$ must satisfy:
$ (2^{4})^{3}\equiv3^{3}\equiv1$ (modulo $ 13$); and
$ (2^{6})^{2}\equiv12^{2}\equiv1$ (modulo $ 13$).
The former is a solution, because $ 3(\equal{} p \equal{} q)$ is a prime.
The later is not, because $ 12$ is not a prime.

Is the solution of this problem equivalent to those of Problem 10 ("Problem Set 7.7"; p.450-451) in the book cited on the title of this posting? Namely the number of solutions of the congruence $ x^{n}\equiv 1$ (mod $ p$) where $ n|(p\minus{}1)$ is equal to the number of [b]characters[/b] modulo $ p$ whose [b]orders[/b] are equal to $ n$. Note that $ n$ must be a prime in the problem we are trying to solve.


AMC
AMC 10
induction
modular arithmetic
number theory

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

$ \textbf{(A)}\ 0 \qquad
\textbf{(B)}\ 2 \qquad
\textbf{(C)}\ 4 \qquad
\textbf{(D)}\ 6 \qquad
\textbf{(E)}\ 8$

[hide]First, let's find the units digit of $ k$.  $ 2008^2$ ends in a $ 4$, because $ 8^2 \equal{} 64$.  Now we need to find the units digit of $ 2^{2008}.$  We can easily show that $ 2^n$ ends in a $ 6$ is $ n$ is divisible by $ 4$.  Thus, $ 2^{2008}$ ends in a $ 6$.  So $ k$ ends in $ 0$.

Therefore, $ k^2$ also ends in $ 0$.  Since $ k$ is divisible by $ 4$, $ 2^k$ ends in $ 6$.  So our final answer is $ 0\plus{}6\equal{}6$.[/hide]

[quote="lokesh"]Was this really number 24 on the AMC 10A? If so, then this was way to easy.[/quote]

Yes, but AMC10s are generally much easier than 12.
[hide="solution even though there is one"]
$ 2008^2$ ends in $ 4$.
$ 2^{1}$ ends in $ 2$
$ 2^{2}$ ends in $ 4$
$ 2^{3}$ ends in $ 8$
$ 2^{4}$ ends in $ 6$
$ 2^{5}$ ends in $ 2$
...and so on
Since it repeats every $ 4$, $ 2^{2008}$ ends in $ 6$.
So $ k$ ends in $ 4\plus{}6\equal{}0$
$ k^{2}$ ends in $ 0$ because anything that ends in $ 0$ squared ends in $ 0$.
Also $ 2008$ divides $ 4$, so $ 2^{k}$ is divisible by $ 4$. That makes the last digit $ 6$.
$ 0\plus{}6\equal{}6$[/hide]

[quote="moogra"][quote="lokesh"]Was this really number 24 on the AMC 10A? If so, then this was way to easy.[/quote]

Yes, but AMC10s are generally much easier than 12.
[hide="solution even though there is one"]
$ 2008^2$ ends in $ 4$.
$ 2^{1}$ ends in $ 2$
$ 2^{2}$ ends in $ 4$
$ 2^{3}$ ends in $ 8$
$ 2^{4}$ ends in $ 6$
$ 2^{5}$ ends in $ 2$
...and so on
Since it repeats every $ 4$, $ 2^{2008}$ ends in $ 6$.
So $ k$ ends in $ 4 \plus{} 6 \equal{} 0$
$ k^{2}$ ends in $ 0$ because anything that ends in $ 0$ squared ends in $ 0$.
Also $ 2008$ divides $ 4$, so $ 2^{k}$ is divisible by $ 4$. That makes the last digit $ 6$.
$ 0 \plus{} 6 \equal{} 6$[/hide][/quote]\
Much harder? Nah.
But yes, this was really easy for a number 24...

regardless of whether or not AMC 10s are easier than AMC 12s, number 24s should require at least a little bit of cleverness/ingenuity. However, this is not the case with this problem. :/

yes and it was also a number 15 on the 12...and i got this one right...in like 1 minute...

which makes me regret taking the 12 since i was eligible to take the 10...since i didnt think it would be that big of a gap...

actually most AMC 10 number 24s dont require that much cleverness/ingenuity

[quote="lokesh"]Was this really number 24 on the AMC 10A? If so, then this was way to easy.[/quote]

just because the problem is at the end, doesn't mean that it has to be hard.

I know a few certain ppl who got this wrong beceause they tried to find the units digit of $ k$, instead of whatever it was they wanted... :rotfl:

I actually got it right :)

i got it right however how do we know k is divisible by 4[/i]

[quote="johnnash"]i got it right however how do we know k is divisible by 4[/quote]

$ 2008^2\equal{}4^2\cdot 502^2$ and $ 2^2008\equal{}4^1004$

[quote="moogra"][quote="lokesh"]Was this really number 24 on the AMC 10A? If so, then this was way to easy.[/quote]

Yes, but AMC10s are generally much easier than 12.
[hide="solution even though there is one"]
$ 2008^2$ ends in $ 4$.
$ 2^{1}$ ends in $ 2$
$ 2^{2}$ ends in $ 4$
$ 2^{3}$ ends in $ 8$
$ 2^{4}$ ends in $ 6$
$ 2^{5}$ ends in $ 2$
...and so on
Since it repeats every $ 4$, $ 2^{2008}$ ends in $ 6$.
So $ k$ ends in $ 4 \plus{} 6 \equal{} 0$
$ k^{2}$ ends in $ 0$ because anything that ends in $ 0$ squared ends in $ 0$.
Also $ 2008$ divides $ 4$, so $ 2^{k}$ is divisible by $ 4$. That makes the last digit $ 6$.
$ 0 \plus{} 6 \equal{} 6$[/hide][/quote]
How do we PROVE that pattern?

Alternatively, you could use induction.

Induction usually works for proving patterns.

OK could someone please give me a demonstration of the proof?
I need to learn a lot on proofs...

Well, you could examine the first few until it repeats, and then since you keep multiplying by the same number, the pattern repeats.

[hide="Proof by modular arithmetic"]$ 2^n$ is always even, so $ 2^{n\plus{}1}\equiv 2k\pmod {10}$ for some integer $ k$.
Therefore, $ 2^n\equiv k\pmod 5$. Now, $ (2,5)\equal{}1$, so $ 2^{\varphi(5)}\equiv 1\pmod 5$. $ \varphi(5)\equal{}4$, so $ 2^n\equiv 2^m\pmod 5$, where $ m\equiv n\pmod 4$.
The residue class modulo 4 is $ \{0,1,2,3\}$, so $ m\in\{0,1,2,3\}$. Therefore, $ 2^m\equiv 1, 2, 4, 8\equiv 1, 2, 4, 3\pmod 5$.

Therefore, $ 2^{n\plus{}1}\equiv 2, 4, 8, 6\pmod {10}$.[/hide]

Here's a link to another solution I created if anybody is interested:  [url=https://docs.google.com/document/d/1Ytx_-2jPonDjJqgdhP4cr3zx9O4i4XKqrHI1EpCQeo8/edit?pli=1]link[/url].

$k\equiv 9+1\equiv 0\pmod 5$
$k\equiv 0+0\equiv 0\pmod 4$
since k is a multiple of 20, k^2 has units digit 0 and 2^k has units digit 6
our answer is 6


1) $75 is invested at 7.5% compounded continuously. When will the amount in the bank be quadrupled? Answer correct to the nearest hundredth of a year.

2) How many negative roots do you expect for the equation
8x^6-3x^4+2x^2-4=0.

3)Solve for x if 1+i is one of the roots of x^3-5x^2+8x-6=0.

4) If lnA=x, lnB=y, and lnC=w, then ln(B^4C^2/A^(1/2))=?.

How do you solve them all?

for the first problem,

M = P( 1 + i )n

(by the way, n is supposed to be an exponent)

anyways, try the compound interest formula.....or go to this site:

[hide][url]http://www.mathworld.com[/url][/hide]

hope this works![/url]

On number 3, note that if $1+i$ is a root, its conjugate $1-i$ is a root.  Also, by Vieta, the sum of the roots is 5.  So the last root must be $\boxed{3}$.


Give an elementary proof that if the length of two angle bisectors in a triangle are equal, then the triangle is isosceles.

This is Steiner-Lehmus theorem. See  [url=http://www.mathlinks.ro/viewtopic.php?t=266981]here[/url], [url=http://www.mathlinks.ro/viewtopic.php?t=114607]here[/url], [url=http://www.mathlinks.ro/viewtopic.php?t=91314]here[/url] or in Wikipedia.


[quote="Drunken_Math"]Prove that the products of any two perfect squares must be another perfect square.

In algebraic terms: Prove that $x^2*y^2=z^2$, where $x$ and $y$ can be anything, but $z$ is dependant on $x$ and $y$.

[hide="Hint"]
Find the prime factorization of the two numbers and rearrange.
[/hide][/quote]

[hide]
$xy$ is an integer is $x$ and $y$ are integers.
$(xy)^2)=x^2\cdot y^2$ and must be an integer since xy is an integer.
[/hide]

Well, $x$ and $y$ are integers, and $x^2y^2=(xy)^2=z^2$.  Since $x$ and $y$ are integers, $xy$  is a integer. As a result, $(xy)^2$ is a perfect square, which means that according to $xy=z\rightarrow (xy)^2=z^2$, $z^2$ is also a perfect square.

Masoud Zargar

Proofs belong in the High School forum...It's a great question though  :D


The square ABCD is to be decomposed into $n$ non-overlapping triangles all of whose angles are acute.find the smallest n for which such a decomposition is possible and construct one such decomposition for that n.

it's an old puzzle.
http://www.ics.uci.edu/~eppstein/junkyard/acute-square/postings.html

is there a more elegant solution? surely there has to be a better way than cosine rule and stuff like that


Let: $ abc\equal{}1$and $ a,b,c>0$. Prove that :
$ \frac{a^2b^2}{a\plus{}b\plus{}2}\plus{}\frac{b^2c^2}{b\plus{}c\plus{}2} \plus{}\frac{c^2a^2}{c\plus{}a\plus{}2} \geq\frac{3}{4}$

Yes it is easy :)

[hide]
From cauchy-schwrarz we have that:
$ LHS \ge \frac{(ab\plus{}bc\plus{}ca)^2}{2(a\plus{}b\plus{}c)\plus{}6}$.
So we only need to prove that:
$ \frac{(ab\plus{}bc\plus{}ca)^2}{(a\plus{}b\plus{}c)\plus{}3} \ge \frac{3}{2} \Longleftrightarrow 2\sum a^2b^2\plus{}4abc(a\plus{}b\plus{}c) \ge 3(a\plus{}b\plus{}c)\plus{}9 \Longleftrightarrow  2\sum a^2b^2\plus{}(a\plus{}b\plus{}c) \ge 9$
which true by am-gm.[/hide]

$ \sum \frac{a^{2}b^{2}}{a\plus{}b\plus{}2} \equal{} \frac{ab}{ac\plus{}bc\plus{}\frac{2}{ab}}$

Let $ ab\equal{}x, bc\equal{}y, ca\equal{}z, xyz\equal{}1$

$ L.H.S \equal{} \sum \frac{x^2}{xy\plus{}xz \plus{} 2}$

By Cauchy

$ 2(xy\plus{}yz\plus{}zx \plus{}3) LHS \ge (x\plus{}y\plus{}z)^2$

So we only have to prove

$ \frac{(x\plus{}y\plus{}z)^2}{2(xy\plus{}yz\plus{}zx\plus{}3)} \ge \frac{3}{4} \Leftrightarrow 2(x\plus{}y\plus{}z)^2 \ge 9 \plus{} 3(xy\plus{}yz\plus{}zx)$

but $ (x\plus{}y\plus{}z)^2 \ge 3(xy\plus{}yz\plus{}zx)$ and from $ xyz\equal{}1$ we get $ (x\plus{}y\plus{}z)^2 \ge 9$. Done

this is true by the am gm ineq. directly

[quote]Aadil, you are simply spamming around in the preolympiad section. This has happened many times.
If you have a solution, please post. I dont think the inequality is so direct using AM GM. So unless you post a proof, you are indeed spamming here.
You have been doing this for several topics.[/quote]
sorry.you need to apply am gm 2-3 times to arrive at $ \frac{3}{4}$

[quote]iYes fine. But I would like to see your complete solution. Any solution for inequalities like these with AM GM alone is very elegant. So please tell us the complete solution time permitting.[/quote]
it is too long.i'll show you on friday in class

[quote="cnyd"]$ \frac {(ab \plus{} ac \plus{} bc)^{2}}{2.(a \plus{} b \plus{} c \plus{} 3)}\geq\frac {3}{4}$

$ 2(ab \plus{} ac \plus{} bc)^{2}\geq 3(a \plus{} b \plus{} c \plus{} 3)$

$ 2.(a^{2}b^{2} \plus{} b^{2}c^{2} \plus{} a^{2}.c^{2} \plus{} 2(a \plus{} b \plus{} c))\geq3(a \plus{} b \plus{} c \plus{} 3)$

$ 2(a^{2}b^{2} \plus{} b^{2}c^{2} \plus{} a^{2}c^{2}) \plus{} a \plus{} b \plus{} c\geq9$ (AM-GM)[/quote]

Look at the previous posts before you post a new one!!! :wink:


Two circles have a common internal tangent of length 7 and a common external tangent of 11. What is the product of the radii of the two circles.

I can't seem to be making any headway on this, will someone help me?
Thanks.

All I got was that the sum of the angles between the two radii on each of the two circles (the radii connect the centers and the point of tangencies) is $ 180$.

Try the HSB forum.

I am pretty certain that the answer is [hide="this"]18[/hide], but I have not come up with anything better than the following cheap solution:

[hide]Take these two simple cases:

1:  congruent circles, in which case the internal and extrenal tangents bisect each other, making two right triangles with leg 3.5 and hypotenuse 5.5 ... making the radii $ sqrt{18}$.
2:  One circle of radius 2 and the other 9, positioned such that the internal and external tangents are perpendicular to one other ... the internal tangent is then 9-2=7 and external is 9+2=11.

I will work on a more rigorous explanation ... but at least that is an (admittedly cheap) answer:  18.[/hide]

[hide="Hint"] Let $ X$ be the distance between the centers and $ r1$ and $ r2$ be the radius of the circles. [/hide]

[hide="Solution"] Using the hint, for the external tangent, draw it and draw the perpendiculars from the radius. Then do what the picture says. Draw $ X$.

[geogebra]6b1daba4354f46d752444f75f746ac8b063c9e16[/geogebra]

You will get a right triangle with legs $ 11$, $ r2 \minus{} r1$  and hypotenuse $ X$. So $ X^2$ = $ 121$ + $ (r2 \minus{} r1)^2$.

Then for the internal tangent, I can't explain it so look at the picture.  [geogebra]cb2fba41219fd8e186c0de9767eda04ddecd9176[/geogebra]

It is a rectangle and the width is 7. The length is $ r2 \plus{} r1$ and it's diagonal is $ X$.
So you get $ X^2$ = $ (r1 \plus{} r2)^2$ + $ 49$. Solve those two equations and you will get $ 4*r1*r2 \equal{} 72$ so $ r1*r2$ = $ 18$

[/hide]

Rate my post, so my 1.2 post rating goes up.

Not nearly as pretty as the one above ... but makes the solution fairly simple:
If I knew how to hide this image I would, it pretty much gives it away ... mods, hide the image for me if possible...

[hide][img]http://i33.tinypic.com/29202t3.png[/img]

DB = 7+x and AB = 11-x so x = 2, and 2/r = R/9, making Rr = 18.[/hide]

[color=red][size=67]Edited. -mod[/size][/color]


There is a hole in the ground.
There is a ping pong ball at the bottom of the hole.
Get the ping pong ball.

(I'm not listing the things that you can't do because there are too many.  Once you start asking I'll tell you.  Keep in mind: this is a hidden assumptions problem meaning the answer isn't the first thing that pops into your head.)

Posted by MrMcGeek

once you start giving me solutions that use something which you can't do, i'll tell you that you can't do that

Posted by MrMcGeek

:roll: um stick your hand in and get the ball?

you can't reach, i forgot to mention that
also you can't dig, the ground is cement
you don't have any tools (like a jackhammer)

etc.

Posted by MrMcGeek

why do you need to dig? in your question it already states that there is a hole and a golf ball at the bottom.
if you can't reach can someone else reach for you?

drop a molotov cocktail in there and run

so during the explosion the golf ball is suppose to get blown out?

Haha.  How about gather a very large amount of mass above the hole, so that the force of gravity pulls the golf ball out.

the reason u can't reach into the whole is because it is too deep and you can't blow it up

Posted by MrMcGeek

I mean, like, if you just bring a black hole nearby, that would solve your problem really quick.

Fill the hole with mercury and stop spamming.

Or better yet, let's move this back to my house and have more Fun and Games si? :)

that was Neal

How about this...take a rope, tie it to something nearby*, and climb down the rope, grab the ping-pong ball and climb back out...

*If nothing is nearby, have a person hold the rope**

**if there is no one nearby, steal the pick-axe and wait for the policeman...

If the hole was thet big then a vacuum cleaner would've worked (because then a vacuum won't form inside the hole) but oh well, lets keep the golf ball rolling! (and if that could be tied into the Saga somehow that'd be great.)

Okay, well if the hole isn't big, you can dig a tunnel to the bottom of the hole from another spot in the ground and then blow REALLY hard and then the ball could come flying out and hit someone on the head.

This was my contribution to the saga of the golf/ping-pong ball.

Thank you raccoon, this could be the primary weapon for the rebels since they don't got no money :lol:

Yeah... Fill it with water... and since the ping- pong ball is lighter than the water, it should float up to the surface.  and then you can reach in and get it out.
You can also use a vaccum cleaner, perhaps?
But if the hole is too deep, it won't work.
I guess to go to ll of this trouble to get a measly ping- pong ball is just to much.  Go to the store, mcgeek, and get another for like a few dollars, I guess.

Addendum: The ping-pong ball is less dense than water, the mass does not necessarily determine whether it floats or not because the volume might differ (M/V=D)

Also if it's only big enough to fit a ping-pong ball it's going to be hard to suck it out unless you have a vacuum cleaner with a smaller nozzle than the diameter of the ball.
...I need to come up with a cool problem like this...

never really mentioned the ping pong ball had to be a ping pong ball when we got through with it right?

Well, then there are about a dozen more ways - plus if your really bent on solving this one a different way dig a new hole and put another ping pong ball in that hole. They come in cheap packs of six anyhow.

I just realized ping pong balls are hollow and extremely not dense...no wonder we had to stick with a ping pong ball and not a gold ball...

Blow into the hole
Lubricate it then hit it over and over again until it bounces enough to be caught
stab it with a spear and bring it back up
stab it with a hypodermic needle and put air into it so the air pressure gets high enough for it to pop up (might be better to lubricate it first)
do the above except suck in the air inside so it will be small enough for it to be pulled by hand or something else
lower a small suction cup, attach it, then pull it back up
Cause extreme wind in the opening of the whole thus lowering the pressure near the entrance of the hole and let the ball float upwards
Ask ******* to push it out the hole (though now he can't) :twisted:

If it was a golf ball then get the Star Wars Kid :)

use a straw to suck it out of the hole
stick it with clay and pull it out
if it's cold, lick it
fill the hole with apple juice for variety

What is this doing in Getting Started? Somebody move this to Fun and Games, please.

We said that years ago and yet it's still not moved so I'm guessing the only thing we can do is refrain from posting in this thread anymore and let it die in peace...


Here's a problem that I came up with but you may have seen it or something similar before.  Either way, I don't think it is too difficult.  If it is let me know and I'll move it:

Given triangle ABC with orthocenter P and circumradius R prove

APxBPxCP :le: R :^3:

Let $AH_1, BH_2, CH_3$ be altitudes.
\[
AP = \frac{AH_2}{\cos H_1AC} = \frac{AB \cos A}{\cos H_1AC} = \frac{2R\sin C\cos A}{\sin A_1CB} = 2R\cos A
\]
thus
\[ AP\cdot BP\cdot CP = 8R^3\cos A\cos B\cos C \]
we want to prove that
\[
\cos A\cos B\cos C\leq \frac{1}{8}
\]
Without loss of generality, suppose $C \leq \pi/2$, thus
\[
\cos A\cos B\cos C = \frac{1}{2}\cos C(\cos(A+B)+\cos(A-B))\leq \frac{1}{2}\cos C(-\cos C+1)\leq \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8}
\]


What is the remainder when $ 5^{137}$ is divided by 8?

hello, we have $ 5^2\equiv1 \mod 8$, so we get $ (5^2)^{68}\equiv1^{68}\equiv1 \mod 8$ and from here we get
$ 5^{137}\equiv 5\mod 8$.
Sonnhard.

Or, we use the well-known divisibility trick that we only have to look at the last 3 places of a number to determine its divisibility by 8. Because 137 is odd, 5^137 will end in 125, and as 8x15=120, 125=5^137=5mod8

the last 3 digits are 125 because 137 is odd so 125/8 has remainder 5

and does GameBot get all the post count for these :maybe:


1.	Αν $f: R \to R$  παραγωγίσιμη συνάρτηση , να δείξετε ότι : υπάρχει τριάδα αριθμών a , b , c με a<b<c έτσι ώστε να ισχύουν ταυτόχρονα:$f(a)-f(b)=(a-b)f'(k) , f(a)-f(c)=(a-c)f'(m)$ με k<m αν f κυρτή αντιστρόφως αν  f  κοίλη και $k \neq m$αν f  τίποτε από τα δυο (Δείξτε ότι : Αν στην Cf  δεν υπάρχουν τρία συνευθειακά σημεία, τότε η f είναι κυρτή ή κοίλη )

[quote="r_boris"]1.
[color=red][size=150]ΑΣΚΗΣΗ 1[/size][/color]
Αν $f: R \to R$  παραγωγίσιμη συνάρτηση , να δείξετε ότι : υπάρχει τριάδα αριθμών a , b , c με a<b<c έτσι ώστε να ισχύουν ταυτόχρονα:$f(a)-f(b)=(a-b)f'(k) , f(a)-f(c)=(a-c)f'(m)$ με k<m αν f κυρτή αντιστρόφως αν  f  κοίλη και $k \neq m$αν f  τίποτε από τα δυο (Δείξτε ότι : Αν στην Cf  δεν υπάρχουν τρία συνευθειακά σημεία, τότε η f είναι κυρτή ή κοίλη )[/quote]


How many ten digit telephone numbers, using any one of the ten digits as many times as possible, are there such that the leading digit is not a 0 or a 1, and the seqence of 911 (exact order) does not appear anywhere in the phone number

For example, 189-875-2345 wouldnt work since a 1 appears as the first digit, also 981-909-1187 woundnt work also since a "911" appears in its phone number.

[hide=Answer]Without 0 or 1 as the first digit, there is 8 X 10^9 ways.  Now, we have to subtract the number of ways that include 911.  If it is the first 3 digits, then the # of ways are 1 X 10^7.  If it is the next three digits, there are 8 X 10^6 ways.  Similarly, for the next 6 spots where the 9 can go, there are 8 X 10^6 ways per.  Now of course, our subtraction is overcounting because when it ends the whole number consists of more than 1 911, more than 1 of your subtractions is counting this.  I have to go, so I'll show the answer later...

dont forget that you overcounted the ones with more than on set of "911"

actually you can make it  7 digits theres not really much point to ten digits than its more calculation,

so 7 digits with the same rules is fine.  You can do 10 if you wnat but thats boatloads of calculation..


linear algebra
matrix
linear algebra unsolved

Prove that the $n\times n$ matrix $A_n=(a_{ij})$ which $a_{ij}=\frac1{i+j-1}$ is invertible and all elements of $A_n^{-1}$ are integers.


inequalities
function
calculus
derivative
inequalities unsolved

Prove that $ (x\plus{}4)^{\frac{1}{n}}\plus{}(2x)^{\frac{1}{n}}\geq (2x\plus{}2)^{\frac{1}{n}}\plus{}(x\plus{}2)^{\frac{1}{n}}$ for all real $ x\geq 2$ and n is natural $ \geq 2$

uhm , Use Karamata inequality .

We have the function $ f(x) \equal{} x^{\frac{1}{n}}$ is a concave function for all $ x\in\[2,\plus{}\infty)$.

-$ x\le 4$

And $ (2x\plus{}2,x\plus{}2)\succ(x\plus{}4,2x)$ so

$ (x\plus{}4)^{\frac{1}{n}}\plus{}(2x)^{\frac{1}{n}}\ge (x\plus{}2)^{\frac{1}{n}}\plus{}(2n\plus{}2)^{\frac{1}{n}}$..

-$ x\ge 4$
$ (2x\plus{}2,x\plus{}2)\succ(2x,x\plus{}4)$.Simirlary . :)

Consider function $ f(y) \equal{} (y\plus{}2)^{\frac{1}{n}}\minus{}y^{\frac{1}{n}}$.  Its first derivative is $ (y\plus{}2)^{\frac{1\minus{}n}{n}}\minus{}y^{\frac{1\minus{}n}{n}}$, clearly negative since $ n > 1$.  Or, since $ 2x\geq x\plus{}2$ when $ x\geq2$, then $ f(2x)\leq f(x\plus{}2)$, with equality iff $ x \equal{} 2$, and we are done.  Note that the inequality still holds if we substitute $ n$ by any real larger than or equal to 1.  (Trivially $ n \equal{} 1$ leads to the equality holding for any $ x$).


algebra
polynomial
superior algebra
superior algebra unsolved

x^3+x+1

what are the roots of this polynomial?

You could use the [url=http://mathworld.wolfram.com/CubicFormula.html]cubic formula[/url] if you really wanted to.  They're not nice and won't help you solve any problems related to the extension they generate, if that's what you're asking.

Let $ t\equal{}\sqrt[3]{108\plus{}12\sqrt{93}}$. The roots of $ x^3\plus{}x\plus{}1\equal{}0$ are $ \frac{2}{t}\minus{}\frac{t}{6}$, $ \frac{t}{12}\minus{}\frac{1}{t}\pm \frac{\sqrt{3}}{2}\left (\frac{t}{6}\plus{}\frac{2}{t}\right )i$

Let x=$ (\sqrt [3] {n}\minus{}(1/(3\sqrt [3] {n})))$. Everything will cancel nicely to a quadratic.


Well, my team used to simply do a "who wants problem 5?" strategy(not very formal) and the newer strategy we adopted was to make groups, like if members were a,b,c, and d, a and c would work together on the evens, checking eachother, while b and d would work on odds. etc.

so my question is, How did(does) your team do team round?

When I was in MATHCOUNTS we would simply solve the problems we knew how to solve. It may not be the best strategy, but it prevents people from wasting time thinking about one problem and not getting anywhere.

lets say the team members, in rank of intelligence, are a, b, c, d.

my team did:

a-1,3,5
b-2,4,6
c-7,9
d-8,10

My team strategy last year was: two people do the evens, and the two others do the odds (i.e. one person starts from the beginning, the other from the end). We never ended up following it, though.

I think it would be a good idea to place the second strongest and weakest together to do the first five and the others to work on the last 5. After fifteen minutes they should simply switch places. By work together I mean two people on the same problem collaborating and sharing ideas.

For all tedious problems (like casework, especially), at least two people should be working on them at the same time; otherwise, I think it's fine to not collaborate and just check later.

Person A: Probs 1,2,3,4,5 (6,7,8,9,10) (done in that order)

Person B: Probs 1,3,5,7,9 (odds)

Person C: Probs 2,4,6,8,10 (evens)

Person D: Probs 10,9,8,7,6, (5,4,3,2,1) (done in that order)

Yeah, that's what I meant. But the only problem with this method is that the last few problems are too dependent upon one person, which can really help the team if that person is feeling well, but can also do some damage otherwise. :wink:

Just let people choose problems that they know how to do (make sure they all get done though).

last year we had one person start from the beginning and one start at the end, another start from problem 5 and work up (to problem 1) and lastly another start from problem 6 and work down (to 10)

it actually works

we also did an "i'll do this" for one of the levels
didn't work as well as the first method

In my team:

Most intelligent person = #10 - #5 (that order)
2nd most intelligent person = #5 - #10
3rd most intelligent person = #5 - #1
... = #1 - #5

After we were done, most intelligent to least intelligent worked in this order: checked over the problems uncompleted, then problems unsure of, hard to easy and just for the fun of it.

My opinion:

Last year we did what akhil0422's team did. I personally believe that oversimplifies mathematical ability (some people are better at algebra, others are better at probability, etc.).

Now, my team uses a topic-based approach

In my team:

Call the members of my team a, b, c, and me (where I am me).

C&P, NT, anything discrete math related-me
Algebra-a
Geometry-b
Checking our work, logic, arithmetic-c
Anything anyone else can't solve-me

We then trade problems that we either can't solve or need fresh insight into.  We keep a sheet where we communicate who solves each problem and who checks each problem.

I wonder what Bobby Shen's TX team does?

We have two students do the evens and two people do the odds.  One person works from 2 top 10 the other works 10 to 2.  the same with the odds.  If someone gets a problem that they know they cannot do they will trade with someone.  The method that you use depends on the abilities of the four team members.  You have to do what ever works for your team.

Us:
Me: 10, 9, 8, 7, 6, 5, 4, 3, 2, 1
2nd person: 10, 9, 8,7,6
3rd: 1,2,3,4,5
4th: 4,5,6,7,8

And we get a 10 on all practices...

I wonder why?

[color=#F8F8F8]No seriously, that's basically what we do, no joke.[/color]

People: a, b, c, d.

A - 1, 3, 5, 7, 9, 10
B - 2, 4, 6, 8, 10
C - 1, 2, 5, 8, 10
D - 2, 3, 4, 9, 10

This way the work can be checked multiple times.

My team:

2nd and 3rd best people do 1-5

other two do 6-10

if any time left, switch roles.

[quote="isabella2296"]People: a, b, c, d.

A - 1, 3, 5, [b]7[/b], 9, 10
B - 2, 4, [b]6[/b], 8, 10
C - 1, 2, 5, 8, 10
D - 2, 3, 4, 9, 10

This way the work can be checked multiple times.[/quote]

#6 and #7 are only checked once.

Adding on to that: I think you can make a chart to better organize the round.

On the left side there is the Problem # and on the top there is each person's name.

:\ For our team, this helped.

the problem with following an organized method is:
1. it takes time to draw a chart and do everything else. time is money.
2. each person might happen to do all the problems they are not good at.
3. in actual competition, it is difficult to perform an organized method as practiced.
my suggestion is just to work on the math and play to your strengths. your team will end up solving all the problems multiple times anyway, as is the goal of the aforementioned approaches. trust me.

My mathcounts team had people from several different classes so we'd divide the problems up by the class skills it took to solve them. Hope that helps ;)

Interesting strategies. I don't think it is a great idea to divide problems strictly by number, because people have strengths and weaknesses in different topics. Sometimes partner working gets less done for us. We tend to do a choosing-a-problem-you-like method. We all work alone (except asking for a formula or something) at the beginning, and then help check the problems together when all the problems are done. On Chapter, we get everyone to check all the problems.

Chapter- finish all and start checking in 4 min
State- 7 min
Nats- never finish all of them, just keep working!!!!


How do you simplify \[ \sqrt[3]{2 \plus{} \sqrt 5} \plus{} \sqrt[3]{2 \minus{} \sqrt5}?\]

I know the answer but cannot solve it.
[hide="answer"] 1 [/hide]

I would appreciate any help. Thanks.

let it be x, then cube both sides. Solve x then you get it.
 :P

A similar way (essentially the same) is to let $ x \equal{} \sqrt [3]{2 \plus{} \sqrt5}$ and $ y \equal{} \sqrt [3]{2 \minus{} \sqrt5}$ such that

$ x^3 \plus{} y^3 \equal{} 4$ and $ xy \equal{} \sqrt [3]{ \minus{} 1} \equal{} \minus{} 1$.

We want to find $ x \plus{} y$, so note that $ (x \plus{} y)^3 \equal{} x^3 \plus{} y^3 \plus{} 3xy(x \plus{} y) \equal{} 4 \minus{} 3(x \plus{} y)\implies(x \plus{} y)^3 \plus{} 3(x \plus{} y) \minus{} 4 \equal{} 0$

$ \implies((x \plus{} y) \minus{} 1)((x \plus{} y)^2 \plus{} (x \plus{} y) \plus{} 4) \equal{} 0$, and since $ x \plus{} y\in\mathbb R$, we must have

$ x \plus{} y \equal{} \boxed1$.

Observe that $ 2 \plus{} \sqrt{5} \equal{} \left( \frac{1 \plus{} \sqrt{5}}{2} \right)^3$.

That's a pretty cheap solution because you sort of have to be looking for it, but once you make the right assumptions this will pop out.  Certainly there are more methodical ways to do this; write $ 2 \plus{} \sqrt{5} \equal{} 1 \plus{} 2 \phi$ and learn this nice identity:

$ \phi^n \equal{} F_n \phi \plus{} F_{n\minus{}1}$.

This is still pretty cheap because it's quite specific to this scenario, so in general I would go with math154's method.


geometry
incenter
circumcircle
inradius
trigonometry
parallelogram
angle bisector

Let $ ABC$ be a triangle, and $ P,Q,R$ the points where its inscribed circle touches $ BC,CA,AB$ respectively. Suppose the orthocenter $ H$ lies on the segment $ PQ$.

(a) Prove that $ RH \perp PQ$.
(b) If $ I$ is the incenter, $ O$ the circumcenter of $ ABC$, and $ N$ is the point where the excircle relative to $ AB$ touches $ AB$, prove that $ I,O,N$ are collinear.

I've found some nice things but I think I'm still far from a solution... could you help me please? I'd prefer hints rather than a complete solution, if it's possible...  :)

For question (a), note that:

Denote by $ T$ the intersection of $ PQ$ and $ AB$. It is easy to verify that the four points $ (B, S, A, T)$ establish a harmonic division. Also note that

$ \angle AHQ \equal{} \angle BHP$

Hence, it should be the case that $ HR$, $ HT$ are both angle bisector of $ \angle (AH, BH)$. And hence, they are perpendicular to each other. It means $ HR \perp PQ$. (QED).

Let $ R_0$ be circumradius, $ r$ inradius, $ r_c$ C-exradius.

$ \cos A \plus{} \cos B \plus{} \cos C \equal{} \frac{r}{R_0} \plus{} 1$
$ \cos A \plus{} \cos B \minus{} \cos C \equal{} \frac{r_c}{R_0} \minus{} 1$

$ \Longrightarrow \cos C \equal{} 1 \plus{} \frac{r \minus{} r_c}{2 R_0}$ for any triangle. For the given triangle, $ CIRH$ is parallelogram, $ r \equal{} IR \equal{} CH \equal{} 2 R_0 \cos C \Longrightarrow r_c \equal{} 2 R_0.$ If $ CI$ cuts $ (O)$ at $ M,$ then $ MO \parallel I_cN$ is midline of the $ \triangle II_cN$ and $ O$ the midpoint of $ IN.$

To [b]mathangel[/b]: thanks, using your idea I found a solution using only a similarity and the bisector theorem.

To [b]yetti[/b]: I think I did not completely understand that... could you explain where do those identities come from? Or much better, is there a non-trigonometric proof of the fact that $ 2R\equal{}R_c$?

Thank you in advance

Let [b]M[/b] is the intersection of [b]HI[/b] and [b]AB[/b].Prove that [b]M[/b]is the middle of [b]AB[/b].

for part (b) we know that [b]M[/b] is the middle of [b]NR[/b].On the other hand, it is easy to prove that OM=r/2. so it is clear [b]O[/b]is midpoint of [b]IN[/b].


[quote="Arne"]Is a G.P. a geometric progression?[/quote]
yes.


Hi, So I found this really interesting question..

I was wondering if anybody can help me prove it...

The question is:

$\sqrt[2+\sqrt{5}]{3}+\sqrt[2-\sqrt{5}]{3}= 1$

I tried squaring both sides..and that does not work..

Are you sure you have it right? Unless I am mistaken, the first term is greater than one, and the second is positive so that would be impossible

i agree w/ the previous post.
did u get this by plugging it into a calculator? because if you did then there is going to be rounding errors.

Sorry you guys...I tried to correct it but I had problems...

And my latex code didnt come up properly...

The Question was

cubed-root(2 + sqrt(5)) + cubed-root(2 - sqrt(5)) = 1

I was told that we should use:

a^3 + b^3 + C^3 = 3abc  if a + b + c = 0.

so...

(2 + sqrt(5))^(1/3) + (2 - sqrt(5))^(1/3) -1 = 0

and from here...im lost  :(

Question in LaTeX
$\sqrt[3]{2+\sqrt 5}+\sqrt[3]{2-\sqrt 5}=1$

I used (a+b)^3= a^3+3(ba^2)+3(ab^2)+b^3, then multiply everything out, and letting a+b=x, you get

4-3x=x^3

x^3+3x-4=0
(x-1)(x^2+x+4)=0

only real root is x=1

[quote="player"][quote="Elemennop"]Either cube the equation (most of it cancels out), or let $a=\sqrt[3]{2-\sqrt{5}}$, $b=\sqrt[3]{2+\sqrt{5}}$ and $c=-1$ in your identity, so you cube all those and let it equal three times the product.[/quote]

I like your ideas...but it dosnt really solve the problem

So I want to show that

$\sqrt[3]{2-\sqrt{5}}+\sqrt[3]{2+\sqrt{5}}= 1$   (1)

$\sqrt[3]{2-\sqrt{5}}+\sqrt[3]{2+\sqrt{5}}-1 = 0$

I also have the identity that...

$if a+b+c = 0, then ... a^{3}+b^{3}+c^{3}= 3abc$

So if I let $a=\sqrt[3]{2-\sqrt{5}}$, $b=\sqrt[3]{2+\sqrt{5}}$ and $c=-1$ ... that dosn't tell me anything at all...everything the left hand side is exactly equal to the right hand side...

and cubing the entire equation (1) is very messy...

Any other ideas?  :idea:[/quote]

Notice that, with the substitutions suggested,
$a^{3}+b^{3}+c^{3}=3abc$
$(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

It shouldn't be hard to show the second expression here can't be 0, and so the first should be.

Let $a=^{3}\sqrt{2+\sqrt{5}}$ and $b=^{3}\sqrt{2-\sqrt{5}}$

Let $c=-a-b$

Then a+b+c=0, so

$a^{3}+b^{3}+c^{3}=3abc$

The value of $c$ pops out immediately

This has the advantage of actually determining $c$ if we do not know in advance what $c$ should be.


So here's the deal.

I post a rule, and the next person posts a rule. The next person posts a rule, etc.

You must abide by the rules when posting, and one rule may not contradict another rule. Ready? We are playing the game

[i][b][u]NOW![/u][/b][/i]

Rule 1: You must post all of the past rules.

P.S. I lost the game.

Rule 1: You must post all of the past rules.
[b]Rule 2: You must always make your rules bold[/b]
[b]Rule 3: You may not use a compound word when you post.[/b]

Also, you can post comments, to make the game interesting.

Rule 1: You must post all of the past rules.
[b]Rule 2: You must always make your rules bold[/b]
[b]Rule 3: You may not use a compound word when you post.[/b]
[b]Rule 4: You must use exactly one "a" (word, not letter) in the rule.[/b]
[b]Rule 5: You must use at least one mathematical formula in your post.[/b]

$ 1\plus{}2\plus{}3\plus{}\cdots \plus{}n\equal{}\dfrac{n(n\plus{}1)}{2}$

Thief...

Rule 1: You must post all of the past rules.
[b]Rule 2: You must always make your rules bold
Rule 3: You may not use a compound word when you post.
Rule 4: You must use exactly one "a" (word, not letter) in the rule.
Rule 5: You must use at least one mathematical formula in your post.
Rule 6: All rules must reference a previous poster from this thread, like 1=2.[/b]

$ \binom{n}{0}\plus{}\binom{n}{1}...\plus{}\binom{n}{n}\equal{}2^n$.

P.S. 1=2 broke Rule 4...

Rule 1: You must post all of the past rules.
[b]Rule 2: You must always make your rules bold
Rule 3: You may not use a compound word when you post.
Rule 4: You must use exactly one "a" (word, not letter) in the rule.
Rule 5: You must use at least one mathematical formula in your post.
Rule 6: All rules must reference a previous poster from this thread, like 1=2.
Rule 7: All rules must be seven words long.
[/b]

$ e^{\pi i}\equal{}\minus{}1$

1=2, what happens if we lose??

You broke [i]two[/i] rules, well done.

Rule 1: You must post all of the past rules.
[b]Rule 2: You must always make your rules bold
Rule 3: You may not use a compound word when you post.
Rule 4: You must use exactly one "a" (word, not letter) in the rule.
Rule 5: You must use at least one mathematical formula in your post.
Rule 6: All rules must reference a previous poster from this thread, like 1=2.
Rule 7: All rules must be seven words long.
Rule 8: We can't post a picture of shaggy75.[/b]

$ 1\minus{}1\equal{}2\minus{}2$

Rule 1: You must post all of the past rules.
[b]Rule 2: You must always make your rules bold
Rule 3: You may not use a compound word when you post.
Rule 4: You must use exactly one "a" (word, not letter) in the rule.
Rule 5: You must use at least one mathematical formula in your post.
Rule 6: All rules must reference a previous poster from this thread, like 1=2.
Rule 7: All rules must be seven words long.
Rule 8: We can't post a picture of shaggy75.
Rule 9: All rules must break a previous rule.[/b]

$ 9|123456789$

Rule 1: You must post all of the past rules.
[b]Rule 2: You must always make your rules bold
Rule 3: You may not use a compound word when you post.
Rule 4: You must use exactly one "a" (word, not letter) in the rule.[/b]
[color=red][i]Rule 5: You must use at least one mathematical formula in your post.[/i][/color]
[b]Rule 6: All rules must reference a previous poster from this thread, like 1=2.
Rule 7: All rules must be seven words long.
Rule 8: We can't post a picture of shaggy75.
Rule 9: All rules must break a previous rule.
Rule 10: Reformatting a previous rule is necessary to post.[/b]

$ \lim_{x\rightarrow0} \frac {\sin x}{x}\equal{}1$

[quote="pythag011"]Rule 1: You must post all of the past rules.
[b]Rule 2: You must always make your rules bold
Rule 3: You may not use a compound word when you post.
Rule 4: You must use exactly one "a" (word, not letter) in the rule.
Rule 5: You must use at least one mathematical formula in your post.
Rule 6: All rules must reference a previous poster from this thread, like 1=2.
Rule 7: All rules must be seven words long.
[/b]

$ e^{\pi i} \equal{} \minus{} 1$

1=2, what happens if we lose??[/quote]

we start over.

rule 7 does not incorporate rule 6, so we lose.

A lot of the posts made have made us lose.

So we start over.

Rule 1: A rule must not break another rule.

Rule 1: A rule must not break another rule.
Rule 2: you must post all previous rules

Rule 1: A rule must not break another rule.
Rule 2: You must post all previous rules.
Rule 3: You must have at least one smiley in your post.  :)
Rule 4: A rule must have a word start with the letter 'w'. :wink:

[hide="MustItAlwaysBeMe thinks 142857 broke a rule."]Rule 1: A rule must not break another rule.
Rule 2: You must post all previous rules.
Rule 3: You must have at least one smiley in your post.  :)
Rule 4: A rule must have a word start with the letter 'w'. ;)
Rule 5: Rules must be hidden with <poster> thinks <previous poster> broke a rule. :rotfl: [/hide]

Also, smileys are better known as emoticons and this game was copied from a task on the mole: AoPS deception.

[quote="MustItAlwaysBeMe"]12|_|L3 1: 411 12|_|L35 4F7312 7|-|15 0N3 |V||_|57 13 \/\/121773|\| 1|\| 1337. (All rules after this one must b written in leet.)
$ \pi$|_|L3 2: 411 3\/3|\| |\||_||\/||33$ \pi$3D $ \pi$|_|L35 |V||_|57 13 \/\/$ \pi$1773|\| 1|\|  1337 |_|51|\|G 4T L3457 0|\|3 L473>< L3TT3$ \pi$. (All even numbered rules must be written in leet using at least one latex letter.)
12|_|L3 3: 411 0|)|) 13773123|) 12|_|135 |V||_|57 133 \/\/121773|\| |_|51|\| 4|\| 3|\|G115|-| 1377312. (All odd lettered rules must be written [size=59]"[/size]usin[size=59]"[/size] an English letter.)  (I am going to guess you meant odd [i]numbered[/i] rules)
[color=green]$ \color{green}\pi$|_|l3 4:  $ \color{green}\pi$|_|L35 |\/||_|5T 133 ||\| |)1|=|=3$ \color{green}\pi$3|\|T (0L0$ \color{green}\pi$5.[/color][/quote]
That contradicts rules 1, 2, 3, so we need to start over.
Rule 1: There are no rules.

Zhangpeji's rule is impossible, so...

Rule 1: .sdrawkcab nettirw eb tsum selur llA

Rule 1: .sdrawkcab nettirw eb tsum selur llA
Rule 2: backwards. written be must word each in letters All

[heheh no contradiction here]

Rule 16 has bad punctuation, but we will keep going.

Rule 1: .sdrawkcab nettirw eb tsum selur llA
Rule 2: backwards. written be must word each in letters All
Rule 4: sequence. geometric a form will numbers rule on now From
Rule 8: on. now from 3 mod 2 is that rule each in theorem mathematical a say must You Theorem! Pythagorean By
Rule 16: on now from rule every in once least at $ \text{\LaTeX}$ use must You.
Rule 32: "theorem". word the contains that rule every in "theorem" misspell must you $ Theorum$, Stewart's in Like

[quote="ZhangPeijin"]
That contradicts rules 1, 2, 3, so we need to start over.[/quote]Actually, it didn't contradict any rules; you just made that up! >:(

[hide="$ .$"]Rule 1: .sdrawkcab nettirw eb tsum selur llA
Rule 2: backwards. written be must word each in letters All
Rule 4: sequence. geometric a form will numbers rule on now From
Rule 8: on. now from 3 mod 2 is that rule each in theorem mathematical a say must You Theorem! Pythagorean By
Rule 16: on now from rule every in once least at $ \text{\LaTeX}$ use must You.
Rule 32: "theorem". word the contains that rule every in "theorem" misspell must you $ Theorum$, Stewart's in Like
Rule 64: 4. of $ multiple$ a be should etc., emoticons, symbols, numbers, words, English including words, of number the '4', a has number rule the If
Rule 128: dot. $ \text{\LaTeX}$ a in rules the hide must You  Theoruses. Pythagorean the prove can't people Some[/hide]
^Click to view hidden content :) ^

\/ Click to view hidden content \/

[hide="$ .$"]Rule 1: .sdrawkcab nettirw eb tsum selur llA
Rule 2: backwards. written be must word each in letters All
Rule 4: sequence. geometric a form will numbers rule on now From
Rule 8: on. now from 3 mod 2 is that rule each in theorem mathematical a say must You Theorem! Pythagorean By
Rule 16: on now from rule every in once least at $ \text{\LaTeX}$ use must You.
Rule 32: "theorem". word the contains that rule every in "theorem" misspell must you $ Theorum$, Stewart's in Like
Rule 64: 4. of $ multiple$ a be should etc., emoticons, symbols, numbers, words, English including words, of number the '4', a has number rule the If
Rule 128: dot. $ \text{\LaTeX}$ a in rules the hide must You  Theoruses. Pythagorean the prove can't people Some
Rule 256: Used $ be$ must! punctuation and capitalization: incorrect.
[/hide]

^Click to view hidden content ^

Huh?

This was an awesome game. It's sad to see it locked.


I need to know how many presents are received throughout the song "Twelve Days of Christmas" and if an algebraic formula to solve this problem could be provided, thanks!

the number of gifts is obviously $ 1\plus{}2\plus{}3\plus{}\cdots \plus{}12$.

use the formula $ 1 \plus{} 2 \plus{} \cdots \plus{} n \equal{} \frac {n(n \plus{} 1)}{2}$

[quote="ak5528"]actually, the gifts on every preceding day repeat on the next day and the day after that and so on. so, if n were 12, which is the last day, it would be:

n(n+1)/2 + n-1(n)/2 + n-2(n-1) ... n-11(n-10)= 364[/quote]

$ t_{n}\equal{}\frac{n(n\plus{}1)}{2}\Rightarrow$

$ \sum_{1}^{12}t_{n}\equal{}\sum_{1}^{12}\frac{n(n\plus{}1)}{2}\equal{}\frac{12(12\plus{}1)(2*12\plus{}1)}{12}\plus{}\frac{12*(12\plus{}1)}{4}\equal{}364$

We are just computing the sum of the first $ 12$ triangular numbers.


geometry
perpendicular bisector
geometry unsolved

Supose that A and B are two fixed points on the circle (O) . Find the locus of M such that the line segment AM intersects the circle again at C , distinct from A , and $AM=MC+CB$

so easy,

 let $D$ be point on AC such $CD=CB$ (C is between M and D).locus of $D$ is a cirlcle and $M$ is midpoint of $AD$ so the locus of $M$ will be a circle too.

[quote="tranthanhnam"]Supose that A and B are two fixed points on the circle (O) . Find the locus of M such that the [color=red]line segment [/color]AM intersects the circle again at C , distinct from A , and $AM=MC+CB$[/quote]

I will check the case that [color=red]the line segment [/color]AM intersects the circle at C

So $M$ is out of the circle (and C is between A and M).
$AM=MC+CB$
$AM-MC=CB$
$AC=CB$

So, $C$ is a point on the circle, and also belongs to the perpendicular bisector of $AB$.
Let $PP'$ the perpendicular diameter to $AB$.

$C$ would be some of $P,P'$.

If $Ax,Ay$ are rays such that $P \in Ax$ and $P' \in Ay$ then the locus for $M$ consist of two rays $Px,P'y$ (including points $P,P'$)

Let's see amir2's solution

So, we have that [u]the ray[/u] $AM$ intersects the circle at $C$ and $M$ between $A$ and $C$
Let $PP'$ the diameter, perpendicular to $AB$.

also let $\phi=\widehat{APB} \Rightarrow  \widehat{AP'B} = \pi - \phi$

if the point $C$ is closer to $A$ than $B$ then the chord $CB$ is longest than $AC$, so the midpoint $M$ of $AD$ is out of the circle, so we'll not have a solution.

So $C$ moves on the semicircle with diameter $PP'$ which contains the point $B$
We'll have two cases:
One when $C$ is on the arc $PB$ and the other when $C$ is on the arc $P'B$

For every position of $C$ on one of those arcs, we have a constant measure for $\angle ACB$.

If $C$ lies on the arc $PB$, then $\widehat {ACB} = \phi$

The triangle $CBD$ is isosceles $(CB=CD)$ and $\widehat {ACB} = 2 \widehat{CDB} \Rightarrow$
$2 \widehat{CDB} = \phi$
$\widehat{ADB} = \widehat{CDB} = \frac{\phi}{2}$

Hence, the point $D$ lies on the arc of a circle from wich the segment $AB$ is visible under an angle $\frac{\phi}{2}$

and since $\widehat{APB}=\phi$ we get that the arc is centered at $P$

(not the opposite arc; from the triangle $ABD$ we know that $D$ is in the same half plane (w.r.t $AB$) with $C$ and $P$)

similarly, if $C$ is on the arc $P'B$, then the locus of $D$ is the arc $AB$ centered on $P'$.

at any case, it's easy to find the locus of $M$ since it is the midpoint of $AD$


[quote="007math"][quote="doggyspin"]Find all positive integers less than 200 that have exactly 6 factors.[/quote]

For a number to have exactly $ 6$ factors, its prime factorization must be in the form $ p^5$, $ p^2q$.

[i]Case 1: form of $ p^5$:[/i]  There are only $ 2$ number that satisfy this which are under $ 200$, $ 1$ and $ 2$.

[i]Case 2: form of $ p^2q$:[/i]  Since $ p \neq q$, let's try $ p \equal{} 2$ and $ q \equal{} 3$.  We get $ 2^23 \equal{} 12$, which works.  Here are the rest that work with $ p \equal{} 2$: (including this one)

$ 2^2\{3, 5, 7, .... 47\}$ for a total of 14.

Now let $ p \equal{} 3$, we get $ 3^2\{2, 5, 7, 11, 13, 17, 19\}$ for a total of 7.

Let $ p \equal{} 5$ and we get $ 5^2\{2, 3, 7\}$ for a total of 3.

Let $ p \equal{} 7$ and we get $ 7^2\{2, 3\}$ for  total of 2.

Add them up to get the answer of $ 14 \plus{} 7 \plus{} 3 \plus{} 2 \plus{} 2 \equal{} 28$[/quote]

This reasoning is the result of the formula for the number of divisors/factors a number has; that is, for any natural number $ n$ with prime factorization:
\[ n \equal{} p_1^{e_1} \cdot p_2^{e_2} \cdots p_m^{e_m}\]
has $ t(n)$ positive divisors:
\[ t(n) \equal{} (e_1 \plus{} 1)(e_2 \plus{} 1) \cdots (e_m \plus{} 1)\]

try looking for the reason behind this; it's more important than the formula.

[quote="enndb0x"]what do you mean "the reason behind this "  ?[/quote]

why it works

[quote="ThinkFlow"][quote="007math"]
[i]Case 1: form of $ p^5$:[/i]  There are only $ 2$ number that satisfy this which are under $ 200$, $ 1$ and $ 2$.
[/quote]
How do $ 1$ and $ 2$ each have exactly six factors? :maybe:[/quote]

They don't.  Put $ 1$ and $ 2$ into $ p$ in $ p^5$.  Then they will have six factors.


OK, so I'm using the fancyhdr package for headers, but how does one have a header that has the section name one is currently on on the left, and some text on the right. Ideally I want my headers to look like:

(Section Name Here)                             (My Name Here)
--------------------------------------------------------

Hi,

have a look at the [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/fancyhdr/fancyhdr.pdf]fancyhdr documentation[/url], the necessary commands are described accompanied by examples.
If you have problems using those commands or don't understand something in the documentation feel free to ask.

Stefan

You can use
[code]\pagestyle{fancy}
\renewcommand{\sectionmark}[1]{\markright{#1}}
\lhead{\rightmark \quad My Name}[/code]
You'll find lots more about this in the [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/fancyhdr/fancyhdr.pdf]fancyhdr documentation[/url].


calculus
derivative
trigonometry
complex analysis
complex analysis unsolved

The following problem appears in my complex analysis text and I was wondering how you solve it in this context.
Find the 10th derivative of [e^(0.5x)]*sin((sqrt(3)/2)x) (obviously without taking the derivative ten times).

Thanks for any input.

What's that symbol you used after the equal sign?

Isn't e^[(x/2)(1 + isqrt(3))] = e^(x/2)[cos((sqrt(3)/2)x) + isin((sqrt(3)/2)x) ?


this is a problem created last year at the final round of the swiss mathematical olympiad. at the smo contestants write the test in two rooms which share the same toilet. so three of the participants (Mathlinks-Usernames: EliasWeber, fcsequeda, ndb) came up with the following problem:

An [b]SMO-Contest[/b] consisting of [i]n[/i] problems is written in [i]2 rooms[/i]. [i]In each room[/i] there are [i]n contestants[/i] solving exactly one problem per person on their own such that exactly every problem gets solved once in each room.
Now the following happens:
In both rooms a contestant is allowed to go to the toilet. At the toilet these two have [b]a meeting[/b] with a short talk in which they share all the solutions they know. How many [b]meetings[/b] are necessary at least such that every contestant has a [i]perfect score[/i] in the end?

This probelm is funny. I like it. a probelm with a good sense of humour :lol:

I think some clarification is needed, concerning information available to contestants. There are at least three possible variants:
1) Each contestant knows who solved each problem (in both rooms)
2) Information about solved problems and results of meetings can be shared in each room, but not between rooms
3) No information sharing is possible, except through meetings; but contestants have common strategy

The following strategy seems to be optimal (at least for most n): first, one contestant must learn solution to all problems. This is done with n-1 meetings in first case (contestant from room 1 who solved problem 1 meets with contestants from room 2 who solved problems 2 to n). In second and third cases this is done with n meetings, as contestants who solved the same problem can meet. After there is one person who knows all the solutions (actually, one in each room), he tells solutions to other contestants, which is done with 2(n-1) meetings in each case.

This strategy requires 3n-3 meetings in first case, and 3n-2 meetings in other cases.

If we consider only first case, then this strategy can be improved at least for n=2 and n=4. For n=2, only two meetings are required (people from different rooms who solved different problems). For n=4, we can split contestants in two groups: solving problems 1 or 2; solving problems 3 or 4. After 4 meetings, each group knows solutions to all their problems, and then each member of first group meets with one member from second group, so everyone knows all problems after 8 meetings (as opposed to 3*4-3=9 if using general strategy).

An interesting generalization of this problem will be to limit maximum number of meetings per contestant :)

We can assume that the contestants know who solved which problems and which person of the other room goes to the toilet, and we can assume that they have a common strategy if you want so.
Because we are [b]in search of the least number of meetings needed[/b]. And I think one can construct a solution $3*n-4$ for all $n$. But I search for a proof to show that $3*n-4$ is the minimal number respectively to lead to a contradiction with a smaller number.

Hi Andreas tomorrow you'll be on mathlinks for 1 year  :coolspeak:

See you at the barbecue..
 :spam:   :rotfl:


By AM-GM, we have \[ \frac{1 + 1 + 1 + 1 + 1 + \frac{9}{2}a^6 + \frac{9}{2}a^3 + \frac{9}{2}b^6 + \frac{9}{2}b^3}{9} \geq \sqrt[9]{\left(\frac{9}{2}\right)^4} ab \] but alas, $\sqrt[9]{\left(\frac{9}{2}\right)^4} = 1.9512668\cdots < 2$ so this doesn't work :(

Where did you see the inequality, and are you sure it is correct? :) It's an odd one.

Yeah I'm absolutely sure it's correct ( the proof takes $3$ elementary lines ). Also you have the following numerical approximation:

$\forall a,b \geq 0, \frac{5}{9} + \frac{(a^{6}+b^{6}+a^{3}+b^{3})}{2} > 2ab + 0.029$, and
$\exists a,b \geq 0, \frac{5}{9} + \frac{(a^{6}+b^{6}+a^{3}+b^{3})}{2} < 2ab + 0.030$

Hope my computation 's right !
 By AM-GM : $a^6+b^6 \geq 2a^3b^3$
                   $a^3+b^3 \geq 2\sqrt{a^3b^3}$
  Put $t = \sqrt{ab}$
 We need to prove : $f(t) = t^6+t^3-2t^2+\frac{5}{9} \geq 0$
 $f'(t) = t(6t^4+3t-4) =t(t-c)g(t)$ , $g(t) \geq 0$
   I checked with my calculator : $f(c) >0$ -> Q.E.D
  If my computation is right , 3 problems of julien can be solved similarly

nttu: no calculator involved, everything can be done the elementary way (using only standard inequalities, with absolutely no calculus).

$\frac{5}{9}+\frac{a^{6}+b^{6}+a^{3}+b^{3}}{2} > 2ab$
Yeas. It is a very strict inequality after all. I can do it without using caculus. So... I will use AM-GM .
Your inequality can be write in other way:
$a,b\geq 0$
Prove $10+9(a^6+b^6+a^3+b^3)-36ab \geq 0$
--> $(9a^6+9a^3-18a^2+5)+(9b^6+9b^3-18b^2+5)+18(a-b)^2 \geq 0$
So we have only proved that
$f(a)= a^6+a^3+\frac59-2a^2 \geq 0$
I use AM-GM here.
$\frac{a^3}{2}+\frac{a^3}{2}+a^6+ \frac{5}{27}+ \frac{5}{27}+ \frac{5}{27} \geq 6 \sqrt[6]{(\frac{5}{27})^3\frac14 a^{12}}$
Because $6 \sqrt[6]{(\frac{5}{27})^3\frac14 }=2.0493 >2$
Done!
Well probably I addicted Mathlinks already. I can not survive if I don't check ML for a week

[quote="Gibbenergy"]
Well probably I addicted Mathlinks already. I can not survive if I don't check ML for a week[/quote]

Well, that's excellent, keep going I'd say :D I check it a few times a day ;)

I had:

$\frac{a^{3}}{3}+2\frac{b^{\frac{3}{2}}}{3} \geq ab$ (Young inequality), then applying AM-GM (the equality case being ruled out easily):
$\frac{1}{2}(a^{6}+b^{3})+\frac{5}{9}>ab$. Now since the inequality is symmetric we have the result.


If and $ \frac{a}{b}\equal{}12$ , $ \frac{b}{c}\equal{}20$ then what is the value of $ \frac{a}{c}$?

Another way that I used:

I replaced ${c}$ with 1 to make the problem simpler, and then replaced ${b}$ with $20$, because $\frac{x}{1} = {x}$. Then, since ${b}$ is $20$, then ${a}$ must be $12\cdot{b}$, where ${b}$ is $20$. Therefore, ${a}$ has to be $240$. Using these replacements for the variables, we can then do $\frac{a}{c}$, where ${a} = 240$ and ${c} = 1$, resulting in $\fbox{240}$.

[quote=circubed]Thank you I was wondering..[/quote]

Are you talking to me? If so, no problem.


geometry
parallelogram
analytic geometry
ratio
similar triangles
geometry proposed

Given triangle $ABC$ with arbitrary point $P$ let $B',B''$ and $C',C''$ be midpoint of $CA,PB$ and $AB,PC$ resp, let $CC'\cap BC''=A_{1},PC'\cap AC''=A_{1}'$ similar $BB'\cap CB''=A_{2},PB'\cap AB''=A_{2}'$, prove that three lines connecting $A_{1},A_{1}'$ and $A_{2},A_{2}'$ and midpoints of $PA$ and $BC$ are parallel !

Let the opposite sides of the quadrilateral ABPC meet at $K \equiv BP \cap CA,\ L \equiv AB \cap PC.$ Project the line KL to infinity, the quadrilateral ABPC going into a parallelogram (or even a square)$A^{*}B^{*}P^{*}C^{*}$ and midpoints C', C'' of the opposite sides AB, PC go into some arbitrary points $C'^{*},\ C''^{*}$ on the the opposite sides $A^{*}B^{*}\parallel P^{*}C^{*}.$ Let parallels to $A^{*}B^{*}\parallel P^{*}C^{*}$ through $A_{1}^{*},\ A_{1}'^{*}$ meet $C^{*}A^{*}$ at $X_{1},\ X_{1}',$ i.e. $A_{1}^{*}X_{1}\parallel A^{*}B^{*}\parallel A_{1}'^{*}X_{1}'.$ Either by similar triangles or by coordinate geometry, it is easy to show that $A_{1}^{*}X_{1}= \frac{A^{*}C'^{*}\cdot C^{*}C''^{*}}{A^{*}C'^{*}+C^{*}C''^{*}-A^{*}B^{*}}= A_{1}'^{*}X_{1}'$ and consequently, $A_{1}^{*}X_{1}X_{1}'A_{1}'^{*}$ is a parallelogram, $A_{1}^{*}A_{1}'^{*}\parallel X_{1}X_{1}' \equiv C^{*}A^{*}\parallel B^{*}P^{*}.$ Thus the line $A_{1}^{*}A_{1}'^{*}$ passes through the intersection $K^{*}\equiv C^{*}A^{*}\cap B^{*}P^{*}$ at infinity, which means that the original points $A_{1},\ A_{1}',\ K$ are collinear.

Let $A',\ D,\ Q,\ M,\ N_{1},\ N_{1}'$ be midpoints of the segments $BC,\ C'C'',\ PA,\ KL,\ A_{1}L,\ A_{1}'L.$  $MN_{1}N_{1}' \parallel KA_{1}A_{1}'$ is the common midline of the $\triangle LKA_{1},\ \triangle LKA_{1}'.$ QA'M is Newton line of the complete quadrilateral ABPC with diagonals AP, BC, KL. $DA'N_{1}$ is Newton line of the complete quadrilateral C'BC''C with diagonals $C'C'',\ BC,\ A_{1}L.$ $DQN_{1}'$ is Newton line of the complete quadrilateral C'PC''A with diagonals $C'C'',\ PA,\ A_{1}'L.$ In addition, $D \in QA'M$ is centroid of the quadrilateral ABPC lying on its Newton line, which makes all 6 midpoints collinear, hence $A'Q \equiv MN_{1}N_{1}' \parallel KA_{1}A_{1}'.$ The proof of $A'Q \parallel A_{2}A_{2}'$ is similar.

Thanks for your solution yetti, it is a hard and good solutioni, It takes me a lot of times to understand and check it, actually here we don't need $B', B''$ are midpoint of $CA,PB$ we can change them by weaker condition $\frac{B'C}{B'A}=\frac{B''P}{B''C}$ and similarly for $C',C''$
In my opinion about plane geometry, I don't like definition of quadrilateral and polygon,triangle is special shape in plane, I always consider quadrilateral as a triangle $ABC$ and arbitrary point $P$, similarly for polygon. Actually  this problem is extension of a problem in convex quadrilateral.
Given quadrilateral $ABCD$, $M,N$ is midpoint of $AC,BD,AN\cap BM=I,CN\cap DM=K$ prove that $IK$ is parallel with the line connecting mid point of $AB,CD$ and it is hard to recognize two problems are the same!

We can have another approach, based on the below Lemma.

[b][size=109][color=blue]LEMMA.[/color][/size][/b] [color=blue]– Two segments $AB,$ $CD$ are given, in an arbitrary configuration with respect to each other and their edges. Let $E,$ $F$ be, two fixed points on the segment lines $AB,$ $CD$ respectively $($ inwardly or outwardly to the corresponded segment $),$ such that $\frac{EA}{EB}= \frac{FD}{FC}.$ Prove that the collinear points $K\equiv AF\cap DE,$ $L\equiv AC\cap BD$ and $M\equiv BF\cap CE$ $($ as the Pappus’s theorem states $),$ lie on a line parallel to the line connecting the midpoints of the segments $AD,$ $BC.$[/color]

$\bullet$  In the configuration of your problem now, if $K,$ is the intersection point of $AC,$ $PB,$ then we have $($ by Pappus’s theorem $),$ that the points $A_{1}\equiv BC''\cap CC',$ $K\equiv AC\cap PB$ and $A'_{1}\equiv AC''\cap PC',$ are collinear and based on the above Lemma, we conclude that these points lie on a line parallel to the segment line $MN,$ where $M,$ $N,$ are the midpoints of the segments $AD,$ $BC,$ respectively.

We consider now the segments $PB$ and $AC,$ of which the points $B'',$ $B',$ are as their midpoints respectively. So, by Pappus’s theorem, we have again that the points $L\equiv AB\cap PC,$ $A_{2}\equiv BB'\cap CB''$ and $A'_{2}\equiv AB''\cap PB',$ are collinear and
based on the above Lemma, we conclude that these points lie on a line parallel also to the segment line $MN.$

That is we have that $A_{1}A'_{1}\parallel MN\parallel A_{2}A'_{2}$ and the proof is completed.

Kostas Vittas.

PS. I will post here the proof I have in mind, of the above Lemma.

[quote="vittasko"]
[b][size=109][color=blue]LEMMA.[/color][/size][/b] [color=blue]– Two segments $AB,$ $CD$ are given, in an arbitrary configuration with respect to each other and their edges. Let $E,$ $F$ be, two fixed points on the segment lines $AB,$ $CD$ respectively $($ inwardly or outwardly to the corresponded segment $),$ such that $\frac{EA}{EB}= \frac{FD}{FC}.$ Prove that the collinear points $K\equiv AF\cap DE,$ $L\equiv AC\cap BD$ and $M\equiv BF\cap CE$ $($ as the Pappus’s theorem states $),$ lie on a line parallel to the line connecting the midpoints of the segments $AD,$ $BC.$[/color][/quote]

[b][size=109]PROOF OF THE LEMMA.[/size][/b] – Through the points $A,$ $D,$ we draw two lines parallel to $CD,$ $AB$ respectively and let $N$ be, their intersection point.

We denote as $H,$ the intersection point of the segment lines $AB,$ $CD$ and let $P$ be, the midpoint of the segment $HL.$

From the complete quadrilateral $HALDBC,$ by Gauss’s theorem, we have that the points $P,$ $Q,$ $R,$ are collinear, where $Q,$ $R,$ are the midpoints of the segments $AD,$ $BC,$ respectively.

Because of the points $H,$ $Q,$ $N,$ are collinear $($ from the parallelogram $HAND$ $),$ from the triangle $\bigtriangleup HLN,$ we conclude that $PQR\parallel LN$ and it is enough to prove that the point $N,$ lies on the line $KLM.$

Because of $CD\parallel AN$ $\Longrightarrow$ $(A.DFCN) = (D,F,C)$ $,(1)$

We symbolize as $(A.DFCN),$ the double ratio of the pencil $A.DFCN$ and as $(D,F,C),$ the simple ratio $\frac{CD}{CF}.$

Similarly because of $AB\parallel DN$ $\Longrightarrow$ $(D.AEBN) = (A,E,B)$ $,(2)$

But $(D,F,C) = (A,E,B)$ $,(3)$

From $(1),$ $(2),$ $(3)$ $\Longrightarrow$ $(A.DFCN) = (D.AEBN)$ $,(4)$

From $(4)$ and because of the pencils $A.DFCN,$ $D.AEBN,$ have the segment line $AD,$ as their common ray, we conclude that the points $K\equiv AF\cap DE,$ $L\equiv AC\cap DB$ and $N\equiv AN\cap DN,$ are collinear, as the intersection points of the other homologous rays.

So, we have that the lines $KLN$ and $KN,$ are coincided $($ they have two common points $).$

Hence, we conclude that $KLM\parallel QR$ and the proof is completed.

$\bullet$  This proof is dedicated to [b]Nikos Rapanos[/b].

Kostas Vittas.

PS.  When the points $E,$ $F,$ on the segment lines $AB,$ $CD$ respectively, lie the one inwardly to the corresponded segment and the other outwardly to the other segment, then the line $KLM,$ is parallel to the line connecting the midpoints of the segments $AB,$ $CD.$


In acute-angled triangle $ABC$ side $AC$ has minimal length. Points $D\in [AB]$, $E\in [BC]$, so that $AC=CE=AD$. Prove that $OI$ is equal to the radius of the circumscribed circle of $BDE$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14655]http://www.mathlinks.ro/Forum/viewtopic.php?t=14655[/url]
Sasha, I'm quite astonished you posted this problem! Well...I cannot tell you why now :(
Myth, I think you still remember that :)  but can it really be a coincidence :?


Let a sequence $ a_1 \equal{} 2009,a_{n \plus{} 1} \equal{} \frac {a_n}{1 \plus{} a_n^2}$. Find $ \lim\limits_{n\rightarrow \infty} na_n^2$

It's easy to conclude that:$ \lim\limits_{n\rightarrow\infty} a_n \equal{} 0$
We have:
$ \lim\limits_{n\rightarrow\infty} (\frac {1}{a_{n \plus{} 1}^2} \minus{} \frac {1}{a_n^2}) \equal{} \lim\limits_{n\rightarrow\infty} \frac {(a_n^2 \plus{} 1)^2 \minus{} 1}{a_n^2} \equal{} \lim\limits_{n\rightarrow\infty} a_n^2 \plus{} 2 \equal{} 2$
Hence,by Stolz Theorem:
$ \lim\limits_{n\rightarrow\infty} \frac {1}{na_n^2} \equal{} 2$
So $ \lim\limits_{n\rightarrow\infty} na_n^2 \equal{} \frac {1}{2}$


The vertices of a right triangle $ ABC$ inscribed in a circle divide the circumference into three arcs. The right angle is at $ A,$ so that the opposite arc $ BC$ is a semicircle while arc $ AB$ and arc $ AC$ are supplementary. To each of the three arcs, we draw a tangent such that its point of tangency is the midpoint of that portion of the tangent intercepted by the extended lines $ AB$ and $ AC.$ More precisely, the point $ D$ on arc $ BC$ is the midpoint of the segment joining the points $ D'$ and $ D''$ where the tangent at $ D$ intersects the extended lines $ AB$ and $ AC.$ Similarly for $ E$ on arc $ AC$ and $ F$ on arc $ AB.$
Prove that triangle $ DEF$ is equilateral.

please post your solution here http://www.mathlinks.ro/Forum/viewtopic.php?search_id=35943252&t=130652  :wink:


linear algebra
matrix
combinatorics proposed
combinatorics

Suppose that $ T\subseteq\{(1,2),(2,1),(2,3),(3,2)\}$ and suppose that we have a directed triple system $ DT(v,\lambda)$ constructed using a set $ X,\ |X|\equal{}v$, $ Y\equal{}\{\left< a,b,c\right>|a,b,c\in X\}$, where $ \left<a,b,c\right>\subseteq X^2\ \text{T\minus{}contains}\ \{(a,b),(b,c)\}$ if $ T\equal{}\{(1,2),(2,3)\}$ and so on, and each $ (a,b)\in X^2$ is $ \text{T contained}$ in exacty $ \lambda$ such $ \left<a,b,c\right>$.  What are some algebraic conditions that $ v$ must satify if $ DT(v,\lambda)$ exists?

I cnstructed a $ v(v \minus{} 1)$ by $ |Y|$ 0,1-matrix and places a 1 in $ (i,j)$ if the i-th pair in $ X^2$ is $ \text{T \minus{} contained}$ in the j-th triple in $ Y$, and 0 otherwise.  So each row has $ \lambda$ 1s, and so the entire matrix has $ \lambda v(v \minus{} 1)$.  Similarly, each column contains $ |T|$ 1s, and so the matrix has $ |T\parallel{}Y|$ 1s.  Therefore, we have the identity $ \lambda v(v \minus{} 1) \equal{} |Y\parallel{}T|$.  Now, this is too easy of a deduction to have much value.  Are there other identities that the system must satisfy?


[color=darkblue]Slove this equation: $ \sqrt{4x^2\plus{}5x\plus{}1}\minus{}2\sqrt{x^2\minus{}x\plus{}1}\equal{}9x\minus{}3$[/color]

[quote="thanhnam2902"][color=darkblue]Slove this equation: $ \sqrt {4x^2 \plus{} 5x \plus{} 1} \minus{} 2\sqrt {x^2 \minus{} x \plus{} 1} \equal{} 9x \minus{} 3$[/color][/quote]
Obsrerve that $ (4x^2 \plus{} 5x \plus{} 1) \minus{} (4x^2 \minus{}4x \plus{} 4) \equal{} 9x \minus{} 3$
Calling $ 4x^2 \plus{} 5x \plus{} 1$ as $ a^2$ and $ 4x^2 \minus{}4x \plus{} 4$ as $ b^2$, we have $ a\minus{}b\equal{}a^2\minus{}b^2$


Prove that the equation
x^x + y^y = z^z has no positive integral solutions.

I think I got it.

If there are positive integer solutions to x^x+y^y = z^z, then both x and y are less than z. Thus, the maximum value of x or y is z-1.

Thus, for there to be solutions to the original equation, there must be solutions to (z-1)^(z-1)+(z-1)^(z-1) >= z^z.
Thus, 2*(z-1)^(z-1) >= z^z --> 2*(z)^(z-1) >= (z)^z --> z <= 2.

Because x > 0 and y > 0, z > 1. Thus z = 2 or nothing at all.

Case 1: z = 2
x and y are less than z, so x = y = 1. But, 1^1+1^1 < 2^2. Thus, no solution.

So, in conclusion, z^z is always greater than x^x+y^y if x and y are less than z (in positive integers). Thus, there are no positive integer solutions to x^x+y^y = z^z.


trigonometry
calculus
calculus computations

Let $ x_n$ be a sequence of real numbers such that $ lim_{x\rightarrow\infty} (x_n \minus{} x_{n \minus{} 1}) \equal{} k$. Prove that $ lim (x_n)/n \equal{} k$.

If $ lim (x_n)/n \equal{} k$, then is it true that $ lim (x_n \minus{} x_{n \minus{} 1}) \equal{} k$?

For the first question just apply [url=http://en.wikipedia.org/wiki/Stolz-Ces%C3%A0ro_theorem]Stolz Cesaro[/url]'s theorem.
As for the second question try $ x_n\equal{}\sin \frac{n\pi}{3}$ and you will notice that it's false.

Using a theorem to prove itself is bad form.

Actual proofs have been posted here before.

[quote="jmerry"]Using a theorem to prove itself is bad form.
[/quote]

Haven't we got here just a particular case of the theorem ? What are you trying to say ?

Until I started posting here, I had never even heard of the theorem. It is not taught in the USA. This form is also the one I saw first and most often; it's general enough to prove any other form easily, so it might as well be the standard form of the theorem.


Tony has two candles. Both are 30cm tall, but one is thicker than the other. He lights them at the same time. Due to the difference in thickness, the height if the thicker candles decreases by just 3/4 of the amount the hieght of the thinner candles loses in each minute. If after 30 minutes, the height of the thinner candle os 2/3 of the thicker, what is the height of the thicker candle 15 minutes after it has been lighted?

If the thinner one loses $ x$ cm a minute, the thicker one loses $ \frac{3}{4}x$ cm a minute.

So in $ 30$ minutes, the height of the thicker one is $ 30 \minus{} \frac{90x}{4}$ cm and the height of the thinner one is $ 30 \minus{} 30x$.

Therefore $ 30 \minus{} 30x \equal{} \frac{2}{3} ( 30 \minus{} \frac {90x}{4})$

$ x\equal{}\frac{2}{3}$

Therefore after $ 15$ minutes the height of the thinner one is $ 30 \minus{} (15 \times \frac{2}{3}) \equal{} 30 \minus{} 10 \equal{} 20$ cm.

Sorry again orangefronted: I tend to rush and not read a question carefully. You asked for the height of the thicker candle but I gave the height of the thinner one.

The thicker one loses height at a rate of $ \frac{3}{4}x \equal{} \frac{1}{2}$

Therefore its height after $ 15$ minutes would be $ 30 \minus{} 15\times \frac{1}{2} \equal{} 30 \minus{} 7.5 \equal{} 22.5$cm, as reuired.

I hope this is OK now.


calculus
integration
algebra
polynomial
trigonometry
calculus computations

Hi, here is the question
integrate $ \frac{x^{3}}{x^{5}-1}$
Is there anyway to do this without using computer based calculators?

It can be done by partial fractions.  It's rather ugly, though.

But how? using complex numbers? because $ x^{5}-1$ can be only factor into $ (x-1)(x^{4}+x^{3}+x^{2}+x+1)$

Let $ w = e^{2\pi i/5}$, then

\begin{eqnarray*}x^{5}-1 & = & (x-1)(x-w)(x-\bar{w})(x-w^{2})(x-\bar{w}^{2})\\ & = & (x-1)(x-2\cos(2\pi/5)+1)(x-2\cos(4\pi/5)+1) \end{eqnarray*}

(In fact, every polynomial in $ \mathbb{R}[x]$ can be factored into the product of polynomials in $ \mathbb{R}[x]$ of degree 1 or 2.)

Also,

$ \frac{1}{x^{5}-1}\; = \; \frac{1}{5}\left( \frac{1}{x-1}+\frac{2\cos(2\pi/5)x-2}{x^{2}-2\cos(2\pi/5)x+1}+\frac{2\cos(4\pi/5)x-2}{x^{2}-2\cos(4\pi/5)x+1}\right)$

Just to make the statement complete, we should mention that $ \cos\left(\frac{2\pi}5\right)=\frac{\sqrt{5}-1}4.$

Also, $ \cos\left(\frac{4\pi}5\right)=\frac{-\sqrt{5}-1}4.$


geometry
3D geometry
combinatorics unsolved
combinatorics

The image below shows cubic grid in space being composed from unit segments of edge length 2. An ant moves along these segments from $A_0$ to $K_2$. How many of such shortest ways are there for the ant, if it

a.) moves along arbitrary segments,
b.) only along the edges on the cube's surface ?

I don't know if this is correct, but it seems so.

The shortest paths are those with just three different directions: right (r), front (f) and up (u). The path is $6$ segments long ($2$ from each direction), so all the cases are covered by the positions of these directions in the sequence of steps ($6$ steps). This means tht we find the answer to (a) to be $C_6^2\cdot C_4^2\cdot$, because there are $C_6^2$ ways in which we can place the two r segments, and after we choose the positions for those there are $4$ positions left in which we can place the two f segments, and after we place those the positions of the two u segments is fixed.

For (b) we just have to count those paths which don't pass through $E_1$ (the center of the cube). We apply the same technique as above to count the paths from $A_0$ to $E_1$. We count the paths from $E_1$ to $K_2$ afterwards, we multiply the two numbers and we get the number of paths which do pass through $E_1$. We subtract this number from whatever we found in (a) and we get the desired number for (b).


trigonometry
quadratics
inequalities
geometry proposed
geometry

Prove that $ \minus{}1<6\cos A\plus{}3\cos B\plus{}2\cos C<7\forall\Delta ABC$.

[hide="hint"] 2($ xy\text{cos}A\plus{}yz\text{cos}B\plus{}zx\text{cos}C)\leq x^{2}\plus{}y^{2}\plus{}z^{2}$[/hide]

[quote="TTsphn"][hide="hint"] $ xy\text{cos}A\plus{}yz\text{cos}B\plus{}zx\text{cos}C\leq x^{2}\plus{}y^{2}\plus{}z^{2}$[/hide][/quote]
Why? Are you sure?

$ \cos{A}.xy\plus{}\cos{B}.yz\plus{}\cos{C}.zx\leq 1.xy\plus{}1.yz\plus{}1.zx \equal{} xy\plus{}yz\plus{}zx\leq x^{2}\plus{}y^{2}\plus{}z^{2}$
But it cannot be used for ours problem.
$ A\plus{}B\plus{}C \equal{}\pi$ is not used. Trig formulaes may be used then probably and quadratic function...

$ e_{1}\parallel AB$
$ e_{2}\parallel BC$
$ e_{3}\parallel AC$
$ (xe_{1}\plus{}ye_{2}\plus{}ze_{3})^{2}\geq 0$
$ 2(\sum xy\text{cos}A)\leq x^{2}\plus{}y^{2}\plus{}z^{2}$

Ok, then $ 6\cos A\plus{}3\cos B\plus{}2\cos C\equal{}2.3\cos A\plus{}1.3\cos B\plus{}1.2\cos C\leq\frac{1^{2}\plus{}2^{2}\plus{}3^{2}}{2}\equal{}7$.

But with 1st inequality?  :maybe:

A quite easy method is use Cosin law
$ \text{cos}A\equal{}\frac{b^{2}\plus{}c^{2}\minus{}a^{2}}{2bc}$
Then use Ravi
$ a\equal{}x\plus{}y$
$ b\equal{}y\plus{}z$
$ c\equal{}x\plus{}z$


Determine all pairs of prime numbers $(p, q)$, with $2 \leq p, q < 100$, such that $p+6, p+10, q+4, q+10$ and $p+q+1$ are all prime numbers.

This wasnt a very nice question  :(
Everyone came out of the exam with different answers.

The key was to reuce the casework by writing the primes in the form $6m-1$ or $6n+1$

And nothing the neither of them can be 2, and only q can be 3.
Also p can only be in the form $6m-1$ and q can only be $6n+1$
Then its casework!

(Alternatively do what i actually did.... i think you can guess)


Homer began peeling a pile of 44 potatoes at the rate of 3
potatoes per minute.  Four minutes later Christen joined him
and peeled at the rate of 5 potatoes per minute.  When they
finished, how many potatoes had Christen peeled?

$ 3x\plus{}5(x\minus{}4)\equal{}44 \iff 8x\minus{}20\equal{}44 \iff 8x\equal{}64 \iff x\equal{}8 \ \text{minutes}$.
Thus, $ 5(8\minus{}4)\equal{}5\times4\equal{}\boxed{20}$.


geometry
3D geometry
tetrahedron
algebra
polynomial
rectangle
geometric transformation

VietnamMO92
Day1
1.A tetrahedron$ABCD$ is such that
                       $\angle ACD+\angle BCD=\angle BAC+\angle CAD+\angle DAB=\angle ABC+\angle CBD+\angle DAB=180^0$
Given that \angleACB=\alpha and $AC+CB=k$,compute the total area of the surfaceof the tetrahedron $ABCD$
2.For a positive number$n$ ,$f(n)$ is the number of divisors of n which are congruent to 1 or -1 mod 10,and $g(n)$ is the number of divisors which are congruent to 3 or -3 mod 10.Provethat $f(n)\geq  g(n)$
3.Three real sequences $(a_n),(b_n),(c_n)$ are constructed as follows
(i)$a_0=a,b_0=b,c_0=c$ where $a,b,c$ are given real number.
(ii)$a_{k+1}=a_k+\frac{2}{b_k+c_k}, b_{k+1}=b_k+\frac{2}{c_k+a_k}, c_{k+1}=c_k+\frac{2}{a_k+b_k},$ for all k
Prove that $a_n$ tends to infinity as $n$ tends to infinity.
Day2
1.Let $n_1<n_2<...<n_k$ be positive integers .Prove that all real roots of the polynomial
$P(x)=1+x^{n_1}+x^{n_2}+...+x^{n_k}$ are greater than $\frac{1-\sqrt5}{2}$.

\[\text{Day 1}\]

1. http://www.mathlinks.ro/Forum/viewtopic.php?p=758756#758756
2. http://www.mathlinks.ro/Forum/viewtopic.php?p=758753#758753
3. http://www.mathlinks.ro/Forum/viewtopic.php?t=134073

\[\text{Day 2}\]

4. http://www.mathlinks.ro/Forum/viewtopic.php?p=758769#758769
Let $9<n_{1}<n_{2}<...<n_{s}<1992$ be positive integers and $P(x)=1+x^{2}+x^{9}+x^{n_{1}}+...+x^{n_{s}}+x^{1992}$. Prove that if $x_{0}$ is real root of $P(x)$ then $x_{0}\leq\frac{1-\sqrt{5}}{2}$.

5. http://www.mathlinks.ro/Forum/viewtopic.php?p=758762#758762
Let $H$ be a rectangle with angle between two diagonal $\leq 45^{0}$. Rotation $H$ around the its center with angle $0^{0}\leq x\leq 360^{0}$ we have rectangle $H_{x}$. Find $x$ such that $[H\cap H_{x}]$ minimum, where $[S]$ is area of $S$.

6. http://www.mathlinks.ro/Forum/viewtopic.php?p=758767#758767
Label the squares of a $1991 \times 1992$ rectangle $(m, n)$ with $1 \leq m \leq 1991$ and $1 \leq n \leq 1992$. We wish to color all the squares red. The first move is to color red the squares $(m, n), (m+1, n+1), (m+2, n+1)$for some $m < 1990, n < 1992$. Subsequent moves are to color any three (uncolored) squares in the same row, or to color any three (uncolored) squares in the same column. Can we color all the squares in this way?


Find the largest number of the infinite sequence
$ 1,2^{\frac{1}{2}},3^{\frac{1}{3}},.......,n^{\frac{1}{n}},..........\infty$
And prove your answer.


AMC
USA(J)MO
USAMO
inequalities
algebra solved
algebra

for non-negative xi, xj show ∑ min(xixj, yiyj) ≤ ∑ min(xiyj, xjyi)

To my mind, this is the most difficult inequality ever proposed in a contest. It is simply impossible. I've tried two weeks to prove it and I failed.

you can find the solution at

http://www.kalva.demon.co.uk/usa/usoln/usol006.html

The comment at the end of the solution on kalva says it all about this problem ! :D :D :D

[b]No one made any headway with this in the USAMO exam. As an olympiad problem, it is exceptionally (and unreasonably) difficult.[/b]

cheers! :D :D

i always wondered how did it get on the usamo .... :?


[color=darkblue]Prove the following:

(a)
$\sin \frac{\pi}{2n} \cdot \sin \frac{2\pi}{2n} \cdots \sin \frac{(n-1)\pi}{2n} = \frac{\sqrt{n}}{2^{n-1}}$

(b)
$\sin \frac{\pi}{2n+1} \cdot \sin \frac{2\pi}{2n+1} \cdots \sin \frac{n\pi}{2n+1} = \frac{\sqrt{2n+1}}{2^{n}}$

[/color]

These two are basically the same and well-known.
I apologize if they were posted before  ;)

[quote="Nené"]First: Use the factorization of the polynomial $x^{2n}-1.$

Second:  write $\frac{x^{2n}-1}{x^2-1}$ in two different ways and then put $x = 1$.

Third: Use the trigonometric identity $ 2-2\cos\alpha=4\sin^2\frac{\alpha}2$ and all is OK.[/quote]

you're right :)

i don't think that other methods will work

A few interesting ideas are

[url=http://www.artofproblemsolving.com/Resources/Papers/TrigIdent.pdf]here.[/url]

I didn't understand it.Could you explain it?
 :(

Mildorf posted some interesting trig problems at:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12775