Given a line segment $AB=7$, $C$ is constructed on $AB$ so that $AC=5$. Two equilateral triangles are constructed on the same side of $AB$ with $AC$ and $BC$ as a side. Find the length of the segment connecting their two circumcenters.

$AC = 5$ and $BC = 2$. Let $H$ and $K$ be the foot of the perpendicular bisectors of the first and the second triangle respectively, and $C_{1}$, $C_{2}$ their circumcenters. So $C_{1}H = \frac{5}{6}\sqrt{3}$ and $C_{2}K = \frac{\sqrt{3}}{3}$. Then $C_{1}C_{2} = \sqrt{13}$.

if centeres  of triangle with base AC and CB are M and N respectively,then we can seeangle MCN =120.so using cosine rule on MCN we get MN=13^(1/2)

Let $BCD, CAE, ABF$ be equilateral triangles with $AB$ separating $F$ of $DE$. From [b]Torricelli[/b] to "triangle $ABC$ we get $AD=BE=CF$.
Find $CF$ by Stewart for $\Delta ABF$, see that $MN= \frac{AD}{\sqrt{3}}$ ($M, N$ as before).

Best regards,
sunken rock


MATHCOUNTS
AMC
AMC 12
AIME
AMC 12 A
AoPS Books

This year is my last and only shot at Nationals, and i dont want to waste my chance.  What do you guys use to prepare?  I worked through most of the old tests last year.

Hmm...well, once you've done problems so much that you're muttering about problems in your sleep, (and not the ones related to any state of mental health), you should go back to the concepts. Once you know how the problems make you apply them, it's a matter of being prepared and knowing the concepts behind it. Maybe once you get closer to the competition, you can go over the tests again, as well as periodically leading up to the that point.

I guess you can re-learn some concepts, reinforce your weak points, work through some AoPS books, stuff like that. AMC and stuff, also.

I got nothing else....:)

[quote="perfect628"]You're about the millionth one to ask, but I'll still tell you that you need to buy the old nats tests. They are not publically available.[/quote]
Oh..I am sorry. Should've known better..

How do you even get the past national tests? I surfed all over the Web and couldn't find any..

Is the only option buying them from MATHCOUNTS?

[quote="Ihatepie"]well the best way is old tests. Try to find all of the ones that you didn't do; and do them. At your first meeting, ask your state coach which years you are gonna do at your meetings and do the other ones. Try to do everything, including sprint target team and countdown and be sure to understand how to do every single problem.

Btw, this is my first (I first heard of it in October) and last year (I'm in 8th grade) and my only shot at nats.[/quote]

You could always get held back. You're allowed to participate in mathcounts while in middle school at most 3 times. =P

[quote="Ihatepie"][quote="DonkeyKong"]Do you guys have a link that provides the old nats problems? I couldn't find any besides 92 and 93. I would appreciate it if you post one.[/quote]

you found 92 and 93?
Where?[/quote]
You could find 1990s MATHCOUNTS tests from the Internet??? :huh:

Mathcrazed said something about being able to participate in mathcounts for three years in middle school. I know someone who has skipped one(or more) grades. they are currently in 8th grade, but they should be(by their age) in 7th,6th or lower(trying not to be too specific). that person has already done mathcounts for three years. So i was wondering if they could do mathcounts for more years based on their age and their special situation.

thanks for any information.... :D
by the way, how do you quote someone??

You get 3 years and none past 8th grade....period.  No special "I skipped a grade" rules.

If you did it when you were 7 but had skipped grades and were in 6th, that was your call.  Younger kids aren't supposed to be doing MATHCOUNTS.  Otherwise they could just let anyone under 8th grade do it andsome genius would move to a weaker state and go 7 straight years :lol:

To quote someone click on the word "quote" at the bottom right of the message you want to quote.  It automaticaly opens up a response.

No way.
Still, do past Nats tests, and understand how to do each problem.

The All Time Greatest Mathcounts Problems book also helps.

[quote="ZhangPeijin"]All old Nats tests, probably go through AMC 10 and 12 and be sure you can do the hard problems, (20-25).[/quote]
amc10/12 is almost completely useless in preparing for nats imo
100 greatest mathcounts problems, running through old cds, and practicing old states/nats tests is probably best...

The hard AMC questions are way beyond the level of Mathcounts. Some hard AMC12 questions are around the same level as 1-6 on the AIMe

1-6? Some of them are harder than that...1-6 are easy. Example: 2008 AMC12A #24 (or was it 23? the one with the tangent).

24, 25 amc12 in recent years can be compared to ~10+ on aime
1-7 aime aren't even close to 24/25 amc12 level; more like... 18-22ish?
back on track though, doing states and shooting for perfects also helps, and even if you have already done nats tests once, i see no problem with doing them again...


[hide="Dunno"]
...but it has 14 letters when spelled out :P
[/hide]

The "77" is correct after all, because "seventy-seven" has 12 letters.  EDIT: --> NOT TRUE for the answer because another number beats it

The answer should be 104, because "one hundred four" has the required 14 letters.

EDIT:  The number "73" is correct, because "seventy-three" has
12 letters and it occurs before "77".

[quote="box of rocks"]... because "seventy-seven" has 12 letters.[/quote]

As does 73.

I placed an EDIT in a post above.

I don't see how I miscounted the number of letters in the number
for "73".  At least twice I slowly used fingers from both hands to do it.  So it was NOT a rush job.

[quote="mustafa"]Yeah, but 73 is the right one because it is the smallest one with 12 letters.[/quote]

As an aside, this sequence is not "the first occurrence of a positive integer with $ n \geq 3$ letters (when following a particular convention)" -- rather, each term is the smallest positive integer larger than the previous term having the appropriate number of letters.  It doesn't actually matter in this case because there are no numbers smaller than 73 with 12 letters, but it does matter at some points in the sequence, e.g. the beginning is not "1, 4, 3, 11, ...."


did anyone get any scholarships for the math camp?

Yep, quite a bit here. They were very generous I think.

Or do you know when we should expect one?

I didn't get a scholarship (alumni usually don't) which was crucial to my attendance, so I can't go. :(

I did get the fundraising kit though... sigh.

I suppose that you got the fundraising kit along with your registration forms?  (I'm trying to decide whether I asked for a kit when applying, because i haven't gotten one yet).

How successful were you with fundraising last year?  From the website it sounds like you can get a lot of money... maybe that will help you ?

I got $400. Not too shabby, but part of it will have to come out of my pocket.

I just got accepted today. :)   Such a long wait on the waiting list!
I didn't get any scholarship though, since "their scholarship budget is fully committed at present".
I am really dependent on the fundraising now.  Can any of you tell me how much you generally get from fundraising from your previous experiences?

congrats on getting in off the waitlist, I still havent found out if I got in off the waitlist. I guess maybe I didn't since I havent found out yet, but I can't be sure.
fingers crossed.


How does one write belongs to,there exists?

thanks

PS I didnt want to start a new topic for this

Steve you can use the Wiki link notation [code][[URL]][/code] to link to stuff on the Wiki (actually it's the recommended way).

Unfortunately that only gives [[http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Symbols]] instead of [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Symbols]LaTeX:Symbols[/url].
The wiki way of using a text link doesn't work here [[http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Symbols|LaTeX:Symbols]]

I edited your message above. I don't understand your objection?  :maybe:
It's an internal link, so it goes like any internal link within the wiki
[[LaTeX:Symbols]]
[code][[LaTeX:Symbols]][/code]

[quote="stevem"]Ah right, I didn't realise that it's an internal link. Now I see why it's easier to use the double square brackets :)[/quote]Anything within the site is an internal url :) And since we use two different URLs (mathlinks and artofproblemsolving) we need to feed the user the correct one based on his current location. Using square brackets ensures that is done correctly.

[quote="Valentin Vornicu"]Anything within the site is an internal url [/quote]I assume by 'site' that you mean the Wiki? I presume it won't work within the forum, for example with [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]LaTeX FAQ[/url].

Yes, it's only for the Wiki namespaces. For the site just write the normal urls they will be processed.


Four  pupils  decided  to buy  some  mathematical  books  so  that
(a) everybody  buys exactly  three different books , and
(b) every  two  of the pupils  buy  exactly  one  book in  common.

What are the  greatest  and smallest  number  of  different  books  they  can  buy?

As each student buys 3 books there are bought 12 books.
We can form 6 pairs of students so the repeated books can be between 1 and 6 books.
If there's 1 repeated book then there are 4 copies (3 repeated copies), so we have 12-3=9 different books.
If there are 6 repeated books then there are 2 copies of each (1 repeated), so we have 12-1*6=6 different books
They can buy between 6 and 9 books.


linear algebra
matrix
linear algebra unsolved

Suppose that $A$ is a real matrix with nonnegative elements and having the sum of the elements of every line equal to 1. Prove that for any $n$ the largest element in every column of the matrix $I+A+A^2+...+A^n$ appears on the main diagonal.

Does "every line" mean the rows, the columns, or both?  (i.e. is it stochastic or double-stochastic matrix)

This has been reposted and solved meanwhile: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=349&t=346412


RMT = MR RM CC

[size=200]HINT NUMBER ONE[/size]

[img]http://img126.imageshack.us/img126/4769/rmtclue1py6.gif[/img]

[size=200]This has nothing to do with the random matrix theory[/size]

OH MAN

I FIGURED IT OUT

ITS A CAT

AND HE LIKES TECHNOLOGY

Hint 2:

There is absolutely no numbers involved. Not a single number will be involved in order to solve this problem. There are no calculations involved, there is "research."  Do the research and you may discover the answer to this problem. I say again, this is not an easy problem, it is extremely hard, if you feel you're wasting your time then quit now. I guarantee no one will solve it in the next few weeks as it takes A LOT of research to solve it. If someone does, I will be very surprised.

[quote="alkjash"]Does it have something to do with the  fact that I can only find "RMT = MR RM CC" on Diablo 2 forums? I guess the cat symbolizes playfulness, so it likes playing on technology (aka computers)?

What is it like rady, mephy, and ... t ... who the heck starts with t????

Ok, never mind...[/quote]

You can find it on Diablo 2 forums because there is some contest on some Diablo 2 forum where if you figure this out, you get Diablo 2 money, or something like that, and iceyjdp wants the cash :wink:

Oh.... I get it.
lol  :rotfl:  :rotfl:

good luck!

Btw, why would anyone want D2 money?
It's so easy to get, and it's not that useful.
You can't really find rares or anything with it, unless you gamble.


A 10-digit arrangement $ 0,1,2,3,4,5,6,7,8,9$ is called [i]beautiful[/i] if (i) when read left to right, $ 0,1,2,3,4$ form an increasing sequence, and $ 5,6,7,8,9$ form a decreasing sequence, and (ii) $ 0$ is not the leftmost digit. For example, $ 9807123654$ is a beautiful arrangement. Determine the number of beautiful arrangements.

[quote="Albanian Eagle"]he meant 126.[/quote]It's correct. How is your method?

[quote="Johan Gunardi"]It's wrong. :wink:  How do you get $ 146$?[/quote]
Oops I'm sorry,I've made a mistake in calculations
If we arrange numbers $ 0,1,2,3,4$,then other numbers will be arranged automatically.
After that I considered several cases and inductively determined the needed value.
For example if numbers are arranged like
$ .,0,1,....$ then there are exaclty $ (5 \plus{} 4 \plus{} 3 \plus{} 2 \plus{} 1) \plus{} (4 \plus{} 3 \plus{} 2 \plus{} 1) \plus{} (3 \plus{} 2 \plus{} 1) \plus{} (2 \plus{} 1) \plus{} 1$ ways to arrange $ 2,3,4$.

[hide]The leftmost digit cannot be $ 0,1,2,3,4$. So $ 0,1,2,3,4$ can only be placed on the other nine positions. For every increasing configuration of $ 0,1,2,3,4$, there is only one decreasing configuration of $ 5,6,7,8,9$. So the number of beautiful arrangements is equal to the number of ways to choose five positions out of nine possible positions, that is $ \dbinom{9}{5}\equal{}126$.[/hide]

The solution can be even simpler if we relize that the first digit must be 9, as 0 cannot be the leftmost digit and so the "team" 98765 must preceed. So we  have to determine the number of new beautiful arrangements that can start with 0. First of all we consider the must-arrangement 8765. The digits 01234 must satisfy a certain condition and that allows as to consider the distribution of 5 identical "apples" , whose order wiil be determined by an additional indepedent condition, into 5 bins formed by the first must-arrang. But we can cosider, instead of bins,  4 barriers. So the answer is Po(4,5).


geometry
complex analysis
geometry proposed

Let $ z_1 , z_2 , \cdots$ be a infinite countable set of distinct complex numbers.

If $ |z_j \minus{} z_k |$ is an integer for every $ j, k$ (the integer may depend on $ j$ and $ k$)

then prove that the $ \{ z_j \}$ lie on a single straight line.

Very nice of you to revive this very old, very nice, and very well-known result of Erdos. Why not then also ask to find an as large as wanted finite collection without collinearity?

Also, it is not "complex analysis", nor "geometry", but rather "combinatorics" in some setting.

Thanks you for information. I didn't know that it is one of Erdos's work.  :lol:


Solve these equation with $ x\in R$:

[u][b]equation 1[/b][/u]

$ \sqrt [4] {1 \minus{} x} \plus{} \sqrt [8] {x^2 \plus{} x \minus{} 1} \plus{} \sqrt [6] {1 \minus{} x^2} \equal{} 1$

[u][b]equation 2[/b][/u]

$ 2(x^2 \plus{} x \plus{} 1)^2 \minus{} 7(x \minus{} 1)^2 \equal{} 13(x^3 \minus{} 1)$

[u][b]equation 3[/b][/u]

$ \sqrt {2x^2 \plus{} 3x \minus{} 2} \plus{} \sqrt {2x^2 \plus{} 11x \minus{} 6} \equal{} 4 \plus{} 3\sqrt {x \plus{} 6} \plus{} 3\sqrt {x \plus{} 2}$

[u][b]equation 4[/b][/u]

$ \sqrt {x \plus{} 1} \plus{} \sqrt {3x \plus{} 4} \equal{} x^2 \plus{} 3x \plus{} 3$

[color=red][u][size=150][b]Note:[/b][/size][/u][/color] [b][color=darkblue]Please post full solution. Thank you very much.[/color][/b]

[quote="thanhnam2902"]Solve these equation with $ x\in R$:
[u][b]equation 2[/b][/u]
$ 2(x^2 \plus{} x \plus{} 1)^2 \minus{} 7(x \minus{} 1)^2 \equal{} 13(x^3 \minus{} 1)$
[/quote]

$ 2x^4\minus{}9x^3\minus{}x^2\plus{}18x\plus{}8\equal{}0$
$ (x\plus{}1)(x\minus{}2)(x\minus{}4)(2x\plus{}1)\equal{}0$
$ \boxed{x \in \{\minus{}1,\minus{}\frac{1}{2},2,4\}}$


limit
real analysis
calculus
calculus computations

The only possible real value for $L$ is $L=0.$ This can be seen simply by taking the limit in the recursive relationship to obtain $(L-L)^2=L.$

But, in fact, it is not possible for the sequence to tend to zero.

We have $a_{n+1}=a_n\pm\sqrt{a_n}.$ Since the "$\pm$" offers a choice, this does not completely determine the sequence. But note that if $0<a_n<1$, then $\sqrt{a_n}>a_n.$ Thus, once we have $a_n<1,$ we must choose $a_{n+1}=a_n+\sqrt{a_n}$ in order to maintain the sequence as positive.

Hence the sequence must increase, at least until it becomes greater than 1.

We have not ruled out the possibility that $\liminf a_n=0,$ but in that case we would also have to have $\limsup a_n\ge1.$

If we're willing to accept extended real numbers as limits, then it is possible for $\lim a_n=\infty.$

why do we need to exclude the posibility $\lim \inf a_n = 0$ ?

anyway, your solution is a lot shorter than the official one

We didn't exclude the possiblity of $\liminf a_n=0.$ I think that could happen, although I don't have the example to show it.

Let me try a different phrasing of the same idea:

In every case for which $a_n\le1$, there exists $m\ge1$ such that

$a_n<a_{n+1}<\cdots<a_{n+m}$ with $a_{n+m}>1.$

We know that the increasing sequence $a_n,a_{n+1},\dots$ must eventually exceed 1 because if it didn't, it would converge to a limit $L$ with $0<L\le1$ and we know that the sequence cannot have a nonzero real limit.

Hence we've shown that for any $N,\,\exists\,n>N$ such that $a_n\ge 1.$ (Informally, either the terms are always bigger than 1, or if they ever dip below 1, they must come back up later.) That is, $\limsup a_n\ge 1.$

Now it doesn't matter what the liminf is; it's already impossible for the limit to be zero.

I tried a little numerical simulation with this rule: if $a_n\le1,$ let $a_{n+1}=a_n+\sqrt{a_n}$ and if $a_n>1,$ let $a_{n+1}=a_n-\sqrt{a_n}.$ Using that rule and a variety of initial conditions in the interval $(0,2),$ it looked like we would always get $\liminf a_n=0$ and $\limsup a_n=2.$

That's not a proof, just a suggestion.

[quote="perfect_radio"]Let $(a_n)_{n \geq 1}$ be a sequence of positive reals s.t. \[ \left( a_{n+1}-a_n \right)^2 = a_n, \ \forall n \in \mathbb{N} \ . \ \]

It is known that $L=\lim_{n \to \infty} a_n$ exists. Compute $L$.[/quote]

It seems this problem is making lot of rounds.
I found this problem at
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=296 too.


Έστω $ A$ πραγματικός τετραγωνικός πίνακας τάξης $ 2n , n\in N^* : A^2\equal{}0$ και $ P(x)\equal{}det(A\minus{}xI) , x\in R$. Τότε δείξτε ότι $ P'(0)\equal{}0$

[quote="Demetres"]Δεν βλέπω γιατί ο Α χρειάζεται να έχει τάξη 2n. Νομίζω πως το θεώρημα δουλεύει αν η τάξη του Α είναι n όπου $ n \geq 2$. (Αν όχι τότε μάλλον κάποιο λάθος έχω στην απόδειξή μου.)[/quote]

Υποθέτω ότι θέλουμε να λυθεί η άσκηση χωρίς Cayley-Hamilton (που την κάνει τετριμμένη αν βλέπω καλά)? :huh:

Cheerio,

Durandal 1707

[quote="r_boris"]Φαντάζομαι πως δεν έχεις λάθος Η άρτια τάξη του πίνακα ίσως δίνει κομψότερες λύσεις σχολικές!!![/quote]
Να μια λύση που μου άρεσε
$ P(x)P(\minus{}x)\equal{}det(A^2\minus{}x^2.I)\equal{}x^{4n} (1)$
Το $ P$ ως συνεχής και μη έχοντας ρίζα στα διαστήματα του $ R^*$ έχει σταθερό πρόσημο για $ x>0,x<0$
Το πρόσημο αυτό είναι ίδιο επειδή κοντά στο $ \pm \infty$ έχει ομόσημες τιμές διότι είναι αρτίου βαθμού ίσου με $ 2n$
Άρα επειδή λόγω της $ (1) P(0)\equal{}0$ θα ισχύει το Θ Fermat $ \Rightarrow P'(0)\equal{}0$

Αν δεν θέλετε παραγώγους υπάρχουν και  άλλες εύκολες λύσεις αλλά η προηγούμενη μου φάνηκε πολύ όμορφη


I'm trying to compile a list of the top ten women mathematicians and I've got eight, but can't make a decision about the last two I should add.

In no particular order:

1. Hypatia of Alexandria
2. Winifred Edgerton
3. Emmy Noether
4. Ada Lovelace
5. Sofia Kovalevskaya
6. Sophie Germain
7. Maria Agnesi
8. Mary Somerville

Any suggestions would be appreciated!

Well, as any other list of THE BEST .... EVER,  this list is very subjective. It is just that there are a lot more of ten remarkable females, in so different situations on their lifes and their times, that it is hard to say who is more remarkable.


[quote="TripleM"]Find the 5th digit from the end of the number 5^(5^(5^(5^5))).[/quote]
the last digit is 5

the last two digits are 25

the last three digits cycle between 125-625

the last four digits cycle between 3125-5625-8125-0625

the last five digits cycle between 15625-78125-90625-53125-65625-28125-40625-03125

There is a pattern; the second half of the cycle will be the first half of the cycle repeated but with the first number altered by 5. Each half of  the (n+1)st cycle will have the last n digits cycling through the n-cycle.

Since 5^(5^5) corresponds to 1 mod 4, and 5^4:equiv:1 mod 8, 5^(5^(5^5)):equiv:5^1:equiv:5 mod 8. This corresponds to the last five digits being 03125, so the fifth to last digit should be 0.

um, mystic
where did u get those numbers from?

Yeah...when you don't have a calculator...

o
so u did do brute force
is there a more elegant method?

im thinking dirichlet, but i don't know how to actually use it

[quote="MysticTerminator"]it's really not that hard to do in my head, especially since I've already memorized the first few powers of 5...[/quote]that's amazing. I mean, you could do it but it's a lot of work...if you have a computer, than there must be a calculator in there somewhere :)

*sigh* there probably is a calculator, but I did it in my head. I just remembered that 625*5=3125 and 125*5=625 and 15*5=75 and basic stuff like that.

You can solve this using modulos and things.. but MysticTerminators answer is [hide] correct, though it would take some explaining in writing up a full proof  Can anyone find a nicer solution?[/hide]

so what, u would taket this thing mod100,000 and look for a pattern?

actually, this looks really familiar to one of the PROMYS application questions, except it was:

find the last three digits of 3^3^3^3

I think that one is easier than the one TripleM gave

Anyone care to try it?

(i actually solved this one)


induction
number theory unsolved
number theory

[quote="Chen241290"][quote="ice_age"]k is an odd positive integer.
if (2+sqrt(3))^k=m+n*sqrt(3), where m and n are positive integers,
show that m-1 is a perfect square.

anybody can help on this problem?
many thanks.[/quote]{}
$(2-\sqrt{3})^k=m-n\sqrt{3}$
$1=m^2-3n^2$
$3n^2=(m+1)(m-1)$
But gcd(m+1,m-1)= 1, since k is odd $\Rightarrow$ m is even and n is odd (it prove by induction)
$\Rightarrow m+1=3u^2, m-1=v^2$ or $m+1=u^2, m-1=3v^2$ ;gcd(u,v)=1
If $m+1=u^2, m-1=3v^2$
m is even, therefore u,v  are odds, therefore $u^2=8k+1, v^2=8j+1$
m=8k=8(3j)+4(contradiction)
Hence $m-1=v^2$[/quote]

thanks so much!


logarithms
calculus
calculus computations

Solve for $ t$. Assume $ a$ and $ P_{0}$ are positive.

$ P\equal{}P_{0}a^t$

Divide out by $ P_{0}$:
$ \frac{P}{P_{0}}\equal{}a^{t}$.
Take the logarithm base $ a$:
$ \log_{a}\frac{P}{P_{0}}\equal{}t$.

hello, dividing both sides of the equation by $ P_0$ and we get
$ \frac{P}{P_0}=a^t$, taking the logarithm of both sides
$ \ln(\frac{P}{P_0})=t\ln(a)$ we get $ t=\frac{\ln(\frac{P}{P_0})}{\ln(a)})$ with $ a\ne 1$.
Sonnhard.


A deli sells three kinds of sandwiches. Easyas3.14159... comes in and makes an order for 16 sandwiches (pi people get hungry). How many different possible orders are there, if the arrangement of the sandwiches doesn't matter?

no. *cough*balls *cough* and boxes *cough*

is it distinguishable boxes containing indisguishable balls, if it is that, than I think I was right, or not...

the balls are distinguishable.<-that's a mouthful!

the types of sandwiches are the boxes and the number of sandwiches in them are the balls...or am I doing the wrong problem as usual?...

[hide]I got 132...[/hide]
What's the deal with boxes and balls...? I don't really know what your talking about.  :P Explanation would be appreciated.  :)

how are the balls disguishable if the order in which your order them doesn't matter...
[hide][hide]*ehehheehee ponders*[/hide]
*ehehheehee feels dumb*[/hide]

Alrighty lets say you have n different kinds of things (boxes) and you want m of them (balls).This is how to use that for this problem:
[hide="solution"]There's n=3 different kinds of sandwiches(boxes), and you want m=16 of them(balls). The formula for that is m+n-1 choose n-1, or 18 choose 2, and you get...[/hide]

please excuse me if I mixed up the balls and boxes, I always do.

153 what I said

*ehehheehee is growing angry* not really

ehehhehee thinks: I put that as my first answer above, but I did not believe in myself

[quote]Click to reveal hidden content
153? not so sure [/quote]

[hide]
Yes, you use balls and boxes.
$\binom{16+3-1}{16}=\binom{18}{2}=153$[/hide]


[i]Solve [/i]

[color=red]1)[/color]  $ cos^{2}x \minus{} 2\sqrt3sinxcosx \minus{} sin^{2}x \equal{} \sqrt3$

[color=red]2)[/color] $ 2^{sin^{2}x} \plus{} 2^{cos^{2}x} \equal{} 2$

[color=red]3)[/color]$ sin^{x}a \plus{} cos^{x}a \equal{} 2$

[color=red]4)[/color]$ cosx \minus{} cos2x \plus{} cos3x \equal{} 1$

[color=red]5)[/color]$ ln(sinx) \equal{} x^2$

[color=red]6)[/color]$ sinx \plus{} cosx > \frac {\sqrt2}{2}$

[quote="giannis"][i]Solve [/i]

[color=red]1)[/color]  $ cos^{2}x \minus{} 2\sqrt3sinxcosx \minus{} sin^{2}x \equal{} \sqrt3$

[color=red]2)[/color] $ 2^{sin^{2}x} \plus{} 2^{cos^{2}x} \equal{} 1$

[color=red]3)[/color]$ sin^{x}a \plus{} cos^{x}a \equal{} 2$

[color=red]4)[/color]$ cosx \minus{} cos2x \plus{} cos3x \equal{} 1$

[color=red]5)[/color]$ ln(sinx) \equal{} x^2$

[color=red]6)[/color]$ sinx \plus{} cosx > \frac {\sqrt2}{2}$[/quote]

[hide="hint for 4"]$ cos2x \equal{} cos^{2} \minus{} sin^{2}$
$ cos3x \equal{} cos(2x \plus{} x)$[/hide]

and is #2 possible?
[hide]
$ 2^{sin^{2}x}\plus{}2^{cos^{2}x}\equal{}2^{0}$
$ sin^{2}x\plus{}cos^{2}x\equal{}0?$[/hide]

1. $ A\cos \theta \plus{}B\sin \theta \equal{}\sqrt{A^2\plus{}B^2}\cos (\theta \plus{}\alpha)$.

4. $ \cos A\plus{}\cos B\equal{}2\cos \frac{A\plus{}B}{2}\cos \frac{A\minus{}B}{2}$.

[hide="2 is most definitely possible"]
$ 2^{\sin^2x}\plus{}2^{1\minus{}\sin^2x}\equal{}1$
$ 2^{\sin^2x}\equal{}y$
$ y\plus{}\frac{1}{y}\equal{}1$
[/hide]

[hide="6"]
I'll assume you mean acute angles, because otherwise it's not always true:
If it's acute both sinx and cosx are nonnegative.  That means we can square both sides.
$ 1\plus{}\sin{x}\cos{x}>\frac{1}{2}$
$ \sin{x}\cos{x}>\minus{}\frac{1}{2}$
Obviously true.
[/hide]

[hide="6"]$ \sin x \plus{} \cos x \equal{} \sqrt{2} \sin \left(x \plus{} \frac{\pi}{4}\right)$. But for angles x such that $ 0 \leq x \leq \frac{\pi}{2}$ we have $ \sin \left(x \plus{} \frac{\pi}{4}\right) \geq \frac{1}{\sqrt{2}}$ and so $ \sin x \plus{} \cos x \geq 1 > \frac{\sqrt{2}}{2}$.[/hide]


Fie $a\in(0,4)$ şi $x, z, y>0$ astfel încât \[xyz+x^2+y^2+z^2=a.\]
Să se arate că \[x+y+z<3.\]

Marius te sfatuiesc sa inveti sa scrii in $\LaTeX$. Urmareste link-urile din semnatura mea pentru asta! ;)


geometry perimeter parallelogram rhombus

[color=darkred]Let $ ABCD$ be a convex quadrilateral. Denote  $ O\in AC\cap BD$ . Prove that if the

 perimeters of  $ AOB$ , $ BOC$ , $ COD$ , $ AOD$ are equally, then  $ ABCD$ is a rhombus.[/color]

WLOG assume that $ AO\geq CO$ and $ BO\geq DO$. Let B', C' be points symmetric to B,C with respect to O. We see that BCB'C' is parallelogram => ABCD also must be a parallelogram and the end of the proof is trivial.

[quote][color=darkred]Let $ ABCD$ be a convex quadrilateral . Denote  $ O\in AC\cap BD$ . Prove that if the

perimeters of $ AOB$ , $ BOC$ , $ COD$ , $ AOD$ are equally, then  $ ABCD$ is a rhombus.[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b] $ \left\|\begin{array}{c} AB \plus{} AO \equal{} CB \plus{} CO\ \ (1) \\
 \\
BC \plus{} BO \equal{} DC \plus{} DO\ \ (2) \\
 \\
CD \plus{} CO \equal{} AD \plus{} AO\ \ (3) \\
 \\
DA \plus{} DO \equal{} BA \plus{} BO\ \ (4)\end{array}\right\|$ . Adding the relations $ (1)$ , $ (3)$ obtain $ AB \plus{} CD \equal{} AD \plus{} CB$ ,

i.e.   $ ABCD$ is a tangential quadrilateral. Denote the tangent points  $ M\in (AB)$ , $ N\in (BC)$ , $ P\in (CD)$ ,

$ R\in (DA)$ of the incircle of $ ABCD$ with the its sides. It is well-known $ O\in MP\cap NR$ . Since $ BM \equal{} BN$ ,

the relation  $ (1)$ becomes $ AM \plus{} AO \equal{} CN \plus{} CO$ . But $ AM \equal{} AR$ and for $ \triangle AOR$ si $ \triangle CON$ we have

$ \widehat {AOR}\equiv \widehat {CON}$ and $ m\left(\widehat {ARO}\right) \plus{} m\left(\widehat {CNO}\right) \equal{} 180^{\circ}$ . Hence $ \frac {AO}{CO} \equal{} \frac {AR}{CN} \equal{} \frac {AM}{CN} \equal{} \frac {AM \plus{} AO}{CN \plus{} CO} \equal{} 1$ .

In conclusion, $ OA \equal{} OC$ . Prove analogously that $ OB \equal{} OD$ , i.e.  $ ABCD$ is a parallelogram.

But alone tangential parallelogram is a rhombus ! [/color]


La   OLIMPIADA  nacional mexicana  se  celebrará del  6 al  12  de  noviembre  en  campeche  yo  voi  a  participar  y  tengo  que sacar  oroooooo"!!!!   yo acá  les  pongo  luego  los problemas esperenlos!!!!! :D

AQUI  PUEDEN   encontrar  los problemas  me da  mucha  flojera  escribirlos  yo  saque  lo  siguiente:
761777=35 OROOOOOOOOO!!!!!

http://erdos.fciencias.unam.mx/omm/

 disfruten  y  espero  poder ver  a  algunos de  ustedes en  la  ibero  o  la imo   :D !!!!!

Bueno, .......... :lol:

este parece ser el primer evento desde la reorganizacion de la Comunidad.......
y donde participan miembros de esta.

echale ganas y a traer el oro!! ;)

PD:Procura difundir la comunidad en este evento entre los chicos que estaran alla!!
Y gracias por adelantado por colgar las preguntas de este evento!!

Carlos Bravo :)
Lima-PERU


If x,y,z are reals and x^2+y^2+z^2=2
prove that x+y+z=<xyz+2

[quote="Ulanbek_Kyzylorda KTL"]If $ x,y,z$ are reals and $ x^2\plus{}y^2\plus{}z^2\equal{}2$
prove that $ x\plus{}y\plus{}z\le xyz\plus{}2$[/quote]

This is Poland 1991 .

[hide="Here is my solution"]

$ WLOG  \ x\le y\le z \Rightarrow xy\le 1$

By Cauchy-Schwartz

$ (xyz \minus{} (x\plus{}y\plus{}z))^2\equal{} (z(xy\minus{}1)\minus{}x\minus{}y)^2 \le (z^2\plus{}(x\plus{}y)^2)((xy\minus{}1)^2\plus{}1)
\equal{} (2\plus{}2xy)(2\minus{}2xy\plus{}(xy)^2) \equal{} 4 \minus{} 2(xy)^2(1\minus{}xy) \le 4$
[/hide]

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32399

[quote="Ulanbek_Kyzylorda KTL"]If $ x,y,z$ are reals and $ x^2\plus{}y^2\plus{}z^2\equal{}2$
prove that $ x\plus{}y\plus{}z \le xyz\plus{}2$[/quote]
we can prove the stronger ineq:
$ x\plus{}y\plus{}z \le \frac{xyz}{2}\plus{}2$
if $ x;y;z$ be nonegative real numbers
 :)

[quote="Toan_VN_LC"][quote="Ulanbek_Kyzylorda KTL"]If $ x,y,z$ are reals and $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 2$
prove that $ x \plus{} y \plus{} z \le xyz \plus{} 2$[/quote]
we can prove the stronger ineq:
$ x \plus{} y \plus{} z \le \frac {xyz}{2} \plus{} 2$
if $ x;y;z$ be nonegative real numbers
 :)[/quote]
It isn't stronger, Toan  :wink: . The first ineq is trues for all real $ a, b, c$ and equality is more special.
With the second, pqr solve it easly. Equality occurs when $ a\equal{}b\equal{}1$;$ c\equal{}0$


Ok,
At 8:00 am you leave your car and hike into the woods. Six hours and five miles later you stop to make camp. The next morning, you leave at 7:00 am and hike back to the car. I claim there is a time T during the hike back at which your are at the same spot on the path as on the previous day's hike out at time T. How can I be so sure?

tips: I dont know that answer but I know that if it is more than 2 or 3 lines long your going to far in depth.

Suppose it happens  on the same day: - You started at 8AM  from the car towards the camp and your clone started at 7AM from the camp (in the reverse direction towards the car) (following the same timeline as your next day's trip - just a day earlier) .. would   at some time you meet your clone? ...QED.

Not if I can hike 5 miles in 58 minutes.

Well the original problem says (If it did not  say that explicitly, then it ought to say that :)) if you finish the walk in 58 minutes ... you stay at the car  for the next two minutes.. and  everyone knows that  interval of a hike is always closed  IOW -  that is spot of car is part of the hike path. :)


inequalities
real analysis
real analysis unsolved

Find a sequence $ s_{k}$ with the following properties:
1. It satisfies the inequality $ (n_{k}-n_{j})(2n_{j}+n_{k-1}-2n_{k}+2k-j)\leq n_{j}(n_{j}-n_{j-1}-2j)$ for any $ j<k$
2. $ n_{2}\geq 7$
3. $ n_{i}\geq n_{i-1}+2i$ for any natural number $ i$
4. $ e^{\sqrt{k}}\leq s_{k}\leq 8^{\sqrt{k}}$
5. $ s_{k}= O (a_{k})$, where $ a_{k}$ is any other sequence with the same properties ( i mean from 1 to 4)

Nick

Are $ s_{k}$ and $ n_{k}$ the same thing? It seems that $ n_{k}=Ce^{\sqrt{k}}$ with large $ C$ may work.

yes they are the same thing and i am sorry for this bad notation. now you are right! i can tell you that this $ s_{k}$ is somewhere between $ 1.35^\sqrt{k}$ and $ 4^\sqrt{k}$.
i didn't say it before because these days i made the progress.

but can u find the exact $ s_{k}$? it would be really helpful...oh and even if not this, can u find a lower bound for it (i mean better than the known i am giving you  which is roughly $ 1.35^{\sqrt{k}}$???

I have made some changes, because my last post was a mess! And also I have found some thing since then...I think the problem becomes easier now.
What do u think?


On the sides $AB,AD$ of the rhombus $ABCD$ are the points $E,F$ such that $AE=DF$. The lines $BC,DE$ intersect at $P$ and $CD,BF$ intersect at $Q$. Prove that:

(a) $\frac{PE}{PD} + \frac{QF}{QB} = 1$;

(b) $P,A,Q$ are collinear.

[i]Virginia Tica, Vasile Tica[/i]

(Note: The diagram is attached.)

Denote $s=AB$

(a) $\triangle PEB\sim\triangle PDC\implies {PE\over PD}={EB\over CD}$ and $\triangle QFD\sim\triangle QCB\implies {QF\over QB}={FD\over BC}.$ Therefore ${PE\over PD}+{QF\over QB}=\frac{BE+FD}{s}$

$AE=FD\implies BE+FD=BE+AE=s$, hence the conclusion.

(b) Denote $\alpha={PE\over PD}={PB\over PC}$ and $\beta={QF\over QB}={QD\over QC}.$ In (a) it has been proven $\alpha+\beta=1.$ Now

$\alpha={PB\over PC}={{PC-BC}\over PC}=1-{s\over PC}\implies PC={s\over{1-\alpha}}={s\over\beta}$

Similarly $QC={s\over\alpha}$

Now $\overrightarrow{CP}={1\over\beta}\overrightarrow{CB}$ and $\overrightarrow{CQ}={1\over\alpha}\overrightarrow{CD}.$ Also, by the rule of parallelogram, $\overrightarrow{CA}=\overrightarrow{CB}+\overrightarrow{CD}$

Next, $\overrightarrow{PA}=\overrightarrow{CA}-\overrightarrow{CP}=\overrightarrow{CB}+\overrightarrow{CD}-{1\over\beta}\overrightarrow{CB}=\frac{\beta-1}{\beta}\overrightarrow{CB}+\overrightarrow{CD}=-{\alpha\over\beta}\overrightarrow{CB}+\overrightarrow{CD}$ and
$\overrightarrow{QA}=\overrightarrow{CA}-\overrightarrow{CQ}=\overrightarrow{CB}+\overrightarrow{CD}-{1\over\alpha}\overrightarrow{CD}=\overrightarrow{CB}+\frac{\alpha-1}{\alpha}\overrightarrow{CD}=\overrightarrow{CB}-{\beta\over\alpha}\overrightarrow{CD}$

Now $\overrightarrow{PA}=-{\alpha\over\beta}\overrightarrow{CB}+\overrightarrow{CD}=-{\alpha\over\beta}\left(\overrightarrow{CB}-{\beta\over\alpha}\overrightarrow{CD}\right)=-{\alpha\over\beta}\overrightarrow{QA}$

Hence the vectors $\overrightarrow{PA}$ and $\overrightarrow{QA}$ are collinear, so the points $P, A, Q$ are too.

[b]A generalization (for the middle-scholars).[/b]

Let $E,\ F$ be two points on the sides $AB,AD$ of the parallelogram $ABCD$ respectively.

Denote $P\in BC\cap DE$ and $Q\in CD\cap BF$. Prove that :

the points $P,A,Q$ are collinear $\Longleftrightarrow$ $\frac{EA}{EB}=\frac{FD}{FA}\Longleftrightarrow \frac{PE}{PD} + \frac{QF}{QB} = 1$.

[quote="perfect_radio"][color=darkred]On the sides $AB,AD$ of the rhombus $ABCD$ are the points $E,F$ such that $AE=DF$. The lines $BC,DE$ intersect at $P$ and $CD,BF$ intersect at $Q$. Prove that:

(a) $\frac{PE}{PD}+\frac{QF}{QB}= 1$;

(b) $P,A,Q$ are collinear.[/color]

[color=indigo][i]Virginia Tica, Vasile Tica[/i][/color] [/quote]
[color=darkblue][b]Proof. [/b]
$1.$ By [i]Thales Theorem[/i], we have: $\frac{PE}{PD}+\frac{QF}{QB}=\frac{BE}{CD}+\frac{DF}{BC}=\frac{BE}{BA}+\frac{AE}{AB}=1$

$2.\quad\frac{DQ}{DC}=\frac{DF}{CB-DF}=\frac{AE}{AB-AE}=\frac{AE}{BE}\Longrightarrow \frac{AE}{QD}=\frac{BE}{CD}=\frac{PE}{PD}$
On the otherhand, $P,\,E,\,D$ are collinear, so $P,\,A,\,Q$  are also collinear. [/color]

1)$PE/PD=BE/CD$,($BE \parallel CD$),also $QF/QB=FD/BC$,add both equation,$PE/PD+QF/QB=BF/CD+FD/CD=AB/CD=1$.
2)$\angle PAB=\angle PQC,\angle QAD=\angle QPC$,but $\angle BAD=\angle BCD$ and $\angle BCD+\angle BPA+\angle DQA=180$,hence $\angle PAB+\angle BAD+\angle DAQ=180$,hence done.

$1)$ is clear and easy.
$2)$: $\Delta BPF\sim\Delta ADE\implies\frac{PB}{AD}=\frac{BF}{AE}$; $\Delta DQF\sim\Delta ABF\implies\frac{AB}{DQ}=\frac{AF}{FD}$, consequently $\frac{PB}{AD}=\frac{AB}{DQ}$ hence $\Delta ABP\sim\Delta QDA$, done.

Best regards,
sunken rock

For the second part, it suffices to show that $\frac{PB}{PC} =\frac{AB}{CQ}$. This is equivalent to $$\frac{PB}{PC} = \frac{BE}{CD} = \frac{AF}{AB} = \frac{AE}{QD} = \frac{AB}{CQ},$$ or $AE \cdot CD = QD \cdot EB$. Writing $QD = AB \cdot \frac{DF}{AF}$, $$\frac{AE}{QD} = \frac{AE \cdot AF}{AB \cdot DF} = \frac{AF}{AB} = \frac{BE}{CD},$$ as needed.


Let $ ABCDE$ be an arbitrary convex pentagon. Suppose that $ BD\cap CE \equal{} A'$, $ CE\cap DA \equal{} B'$, $ DA\cap EB \equal{} C'$, $ EB\cap AC \equal{} D'$ and $ AC\cap BD \equal{} E'$. Suppose also that $ eABD'\cap eAC'E \equal{} A''$, $ eBCE'\cap eBD'A \equal{} B''$, $ eCDA'\cap eCE'B \equal{} C''$, $ eDEB'\cap eDA'C \equal{} D''$, $ eEAC'\cap eEB'D \equal{} E''$. Prove that $ AA'', BB'', CC'', DD'', EE''$ are concurrent. (Here $ l_1\cap l_2 \equal{} P$ means that $ P$ is the intersection of lines $ l_1$ and $ l_2$. Also $ eA_1A_2A_3\cap eB_1B_2B_3 \equal{} Q$ means that $ Q$ is the intersection of the circumcircles of $ \Delta A_1A_2A_3$ and $ \Delta B_1B_2B_3$.)

[quote="Wizzy"][quote="lchserious"]$ eABD'\cap eAC'E \equal{} A''$[/quote]
What does that mean?[/quote]
I have explained that at the last sentence.  :)

That is the intersection of the circumcircles.


[hide]
Let $a,b,c,d$ be the sides and $x,y$ be the diagonals. By Ptolemy's, $ab+cd = xy$. But $x, y \le 2$ since the diameter is the longest chord. So $ab+cd \le 4$. If $a,b,c,d > \sqrt{2}$, then $ab+cd > 4$. Contradiction.[/hide]

[quote="JBL"]That's nice -- I would have done it by adding up arclengths.[/quote]

Ah, yes. That's a very simple solution as well.

[quote="chess64"]Show that for any quadrilateral inscribed in a circle of radius 1, the length of the shortest side is less than or equal to $\sqrt{2}$.[/quote]

[hide = I don't know if this is rigorous]
 but an inscribed quadrilateral must satisfy ptolemy's theorem. The maximum length of a diagonal in a circle of radius $1$ is $2$(the diameter; so the maximum product of diagonals is $4$. However, if the shortest side is greater than $\sqrt{2}$ then the sum of the product of the sides is necessarily greater than 4 as $\sqrt{2}\sqrt{2} + \sqrt{2}\sqrt{2} = 4$, which means ptolemy's theorem can't be fulfilled and is a contradiction with the problem statement.[/hide]

let the chords intercept arcs of a,b,c,d

$a+b+c+d=360$

$a\ge b\ge c\ge d\ge0$

the sum to get $4d\le a+b+c+d$, then $d\le 90$, from which we can say that the short chord is

$2*\sin\frac{d}{2}$, and clearly max is $\sqrt{2}$

[hide="Different from all the above"]Since the circle has radius $1$, for a side to have a sidelength of less than $\sqrt {2}$, the two endpoints of the side must be less than $90$ degrees apart.

Assume the the minimal sidelength of this quadrilateral is greater than $\sqrt {2}$. Then all the points are spaced more than $90$ degrees apart.

Label the points of the quadrilateral $A, B, C, D$ counterclockwise in this manner. Let $f(A)$ be the angle that $OA$ makes with the $x-axis$. WLOG, let $f(A) = 0^{\circ}$. Then we have:
\[f(B) - f(A) > 90^{\circ}\]
\[f(C) - f(B) > 90^{\circ}\]
\[f(D) - F(C) > 90^{\circ}\]
\[f(A) - f(D) > 90^{\circ} \]

Adding these equations, we get
\[0^{\circ} > 360^{\circ}\]

Since $360^{\circ}$ is the same as $0^{\circ}$, this is like saying

\[0^{\circ} > 0 ^{\circ}\]

which is clearly absurd.

Thus, the length of the shortest side is less than or equal to $\sqrt {2}$.
[/hide]

This is not noticably different from the idea suggested in the third post on this thread, except that it's sloppy in two ways.  First, 360 is not "the same" as 0, even when we add a degree symbol -- if it were, why couldn't I just rewrite the equation from which you draw a contradiction as $360^\circ > 0^\circ$?  Second, one of your three equations is wrong (it should be $f(A) - f(D) < -270$, which is not exactly helpful).  The way to fix these errors is to bring your argument more closely in line with what I suggested, lo those many years ago.

Reviving only because I actually do have a different solution than the ones already mentioned.

[hide="Law of cosines"]Let $O$ be the midpoint of the circle that circumscribes quadrilateral $ABCD$. WLOG let $AB$ be the shortest side and $\theta := \angle{AOB}$. The angles $\angle{AOB},\angle{BOC},\angle{COD},\angle{DOA}$ all add up to $360^{\circ}$ so there is one angle less than $90^{\circ}$.
By the law of cosines in $\triangle ABO$ we have \[ |AB|=\sqrt{2-2 \cos{\theta}}. \] Obviously $|AB|$ is minimal when $\theta$ is minimal and thus $\theta \leq 90^{\circ}$. Hence, $\cos{\theta} \geq 0$ and we're done.[/hide]

[quote=sanurose]Can we also proceed by proving that the largest value that the shortest side of the quadrilateral can take is less than $2^{1/2}$? I thought that, if I use this idea, then the shortest side can take the maximum value when the quadrilateral is a square, in which case all $4$ sides are equal to $2^{1/2}$. So the length of the shortest side should be $<=2^{1/2}$.
Anyone? Is this correct? Or am I missing something?[/quote]

Well, this is kind of like circular reasoning — The main point is actually [i]proving[/i] that any other quadrilateral would have a maximum of sqrt2, you cannot just claim that because equality holds at a square other quadrilaterals would have a shortest length less without proof.

[hide=bad sol]
Let the quadrilateral be $ABCD.$ Assume for the sake of contradiction that all sides are of length greater than 2. Now, let the center of the circle be $O.$ Therefore, $AB>\sqrt{2}\implies\angle{AOB}>90^{\circ},BC>\sqrt{2}\implies\angle{BOC}>90^{\circ},CD>\sqrt{2}\implies \angle{COD}>90^{\circ},$ and $DA>\sqrt{2}\implies \angle{DOA}>90^{\circ}.$ So, $\angle{AOB}+\angle{BOC}+\angle{COD}+\angle{DOC} >90^{\circ}+90^{\circ}+90^{\circ}+90^{\circ} = 360^{\circ}.$ But, clearly, $\angle{AOB}+\angle{BOC}+\angle{COD}+\angle{DOC} = 360^{\circ},$ so we are met at a contradiction. We conclude that at least one side is of length less than or equal to $\sqrt{2},$ or in other words, the length of the shortest side is less than or equal to $\sqrt{2},$ as desired.
[/hide]


probability
superior algebra
superior algebra unsolved

Prove or disprove the following:
For each $N \in \mathbb{N}$, there exists $n \geqslant m \geqslant N$, s.t. $\binom{n}{m}$ is squarefree.

Definition: A number is said to be squarefree if its prime decomposition contains no repeated factors.

Take $n$ squarefree and $m=n-1$¿ ;)
[or even more simple: $n=m$]

ZetaX,
You are right, I missed one important part of the question.

Prove or disprove the following:
For each $N \in \mathbb{N}$, there exists $n \geqslant m \geqslant N$, s.t. $\binom{n}{m}$ is squarefree and $\binom{n}{m}>n^{3}$.

Okay. Just use $m=n-4$; $\binom{n}{4}$ is squarefree with positive asymptotic probability $(1-\frac{4}{16})(1-\frac{8}{27})(1-\frac{4}{5^{2}})(1-\frac{4}{7^{2}})\cdots$.
In fact, $\binom{n}{n-k}=\binom{n}{k}$ is squarefree with positive asymptotic probability for each fixed $k$.


Esto  $ G$  mia simpagi Hausdorff topologiki omada. Na deiksete oti o xoros $ G/\sim $  ton sizigon klaseon (i.e congugacy classes) einai Hausdorff (in the quotient topology !)

Isxuei to idio apotelesma xoris tin ipothesi peri simpagias tou $ G$ ?

Stergios

Ιδέα που ίσως οδηγεί σε λύση αν κάποιος βρει το χρόνο να δει αν οι λεπτομέρειες δουλεύουν...

Ummm, αυτό που έχω στο μυαλό μου είναι το εξής: κατ'αρχήν, μπορούμε να υποθέσουμε ότι η $ G$ έχει τετριμμένο κέντρο (otherwise, descend to $ G/Z(G)$).

Tότε, με μεταφορά, αρκεί να βρεθεί περιοχή $ U$ του ταυτοτικού στοιχείου $ e\in G$ τόσο μικρή ώστε αν $ g,g'\in G$ να μην υπάρχουν $ u,u'\in U$ ώστε τα $ gu, g'u'$ να ανήκουν στην ίδια τάξη συζυγίας (btw, έτσι μεταφράζεται το conjugacy class?).

Tώρα όμως, εφόσον η $ G$ είναι συμπαγής, οι τροχιές $ O_g, O_{g'}$ των $ g,g'$ υπό τη δράση $ a.g \equal{} aga^{\minus{}1}$ θα είναι επίσης συμπαγείς. Έτσι, μπορείς να βρεις disjoint περιοχές τους και, αν το neigbourhood chasing μου είναι σωστό, τελικά καταλήγεις στη ζητούμενη $ U\ni e$.

Ωστόσο, το έχω κάνει πολύ βιαστικά μέσα σε ένα τέταρτο και είναι πολύ πιθανό να κάνω λάθος... :( :( Eίναι όντως στοιχειώδης η απόδειξη ή έχω χάσει κάτι σημαντικό στην πορεία και θέλει κάτι αρκετά πιο δυνατό?

Cheerio,

Durandal 1707

[quote="Anto"]I apodeiksi pou exo den xreiazete tipote dinato alla sigoura den einai tetrimeni.
Den mporo na katalabo ta bimata sou. Ean thes koitakse to me prosoxi kai ean pragmati douleuei parousiase tin ligo pio analitika.  :)

Stergios[/quote]

Ναι, δυστυχώς το χρόνο που έχω για όμορφα και διασκεδάστικα προβλήματα τον εξήντλησα εδώ και αρκετή ώρα... αν δε μου είχες ανοίξει τα μάτια ότι είχα ξεχάσει ολόκληρο Cantor στο προηγούμενο topic, δεν υπήρχε περίπτωση να ξαναμπώ στο forum σήμερα.

Οπότε, για να είμαστε λίγο πιο τίμιοι, το έκανα edit το post ώστε να χρησιμεύει σαν αρχική ιδέα για όποιον τυχόν ασχοληθεί... μακάρι να είχα το χρόνο να δώσω στα προβλήματά σου τη σημασία που τους χρειάζεται :(

Anyway...

Durandal 1707


It is desired to choose $n$ integers from the collection of $2n$ integers, namely,   $0,0,1,1,2,2,\ldots,n-1,n-1$ such that the average of these $n$ chosen integers is itself an integer and as minimum as possible. Show that this can be done for each positive integer $n$ and find this minimum value for each $n$.

When $n$ is even, the sum of the first $n$ numbers is $0+0+1+1+2+...+\frac{n-2}{2}+\frac{n-2}{2} = \frac{n-2}{2}\frac{n}{2}$.
When $4|(n-2)$, we are done. Otherwise the sum is of the form $\frac{un}{2}$ with odd $u$, so we lift the sum by $\frac{n}{2}$ by changing one summand $\frac{n-2}{2}$ into $n-1$ and are done again.
When $n$ is odd, the sum of the first $n$ numbers is $\frac{n-1}{2}\frac{n-1}{2}=\frac{n^2-2n+1}{4}$. By adding $\frac{n-1}{4}$ to that, we would get $\frac{n (n-1)}{2}$ which is divisible by $n$. But that only works when $4|(n-1)$, and otherwise, we simply add $\frac{3n-1}{4}$.


Let $ r_{1},r_{2},\ldots ,r_{n}$ be $ n$ real numbers each greater than zero. Prove that for any real number $ x>0$,

$ (x+r_{1})(x+r_{2})\cdots (x+r_{n})\geq (x+\sqrt[n]{r_{1}r_{2}\cdots r_{n}})^{n}$.

[hide="hint"]Expand and use AM-GM[/hide]

Posted here before...
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1373786427&t=106526[/url]

From art of problem solving vol.2 i think...


Experience, forte instrument, current projects, etc.?

I'm a 6th year, primarily marimbist/vibraphonist, and currently working on The Process of Invention by Dean and Footpath by Dave Samuels.

The community orchestra I belong to (I play viola, myself) this year featured its youngest member, a teen-aged percussion/double reed whiz, on Paul Creston's Concertino for Marimba and Orchestra. Terrific piece (crazy difficult for the orchestra), terrific young player.

I practice on vibraphone at school.  I have an adams 4.3 octave RVS artist series Marimba at home.
When I get older, I think that I would rather own a vibraphone, simply because it seems to be a far more versatile instrument (and a bit smaller than most marimbas).

I think Creston's Marimba Concertino is good, despite many people's claim that it is overused.
Another good Concerto is Ney Rusuaro's (which I got to play in the accompanying orchestra for soloist Mark Ford this past year).  A friend at UKentucky told me that a recently released Marimba Concerto by Eric Ewazen was pretty good too.

I am a 6th year percussionist, Marimbist, Drummer and my specialty  latin hand drums. I am playing some of Steve Reich pieces in the marimba. Also i've been playing some pices of Mithcell Peters. I also intersted in arab, african and in brazilian percussion.

You don't happen to know anything about Udu drums do you?  I've been looking for a good brand to purchase recently,  but I know no one with expertise in the area.

I don't have udu drums but I think I've seen some LP udu drums you should check http://www.lpmusic.com/Product_Showcase/Udu_Drums/lp_udu_drum_claytone.html

Also 6th year percussionist, I play marimba.
During marching season I don't have to march, I'm a pit player  :P

Hexiditidom:  interested in drum corps at all?

4th year drumset player, 2nd year percussionist (not very good at mallets though lol), and 1st year marching snare drummer.  I specialize in jazz drumset and so I'm not well versed in any keyboard pieces.  I will be trying out for the Indiana all state honor band as a percussionist though...

I'm a 5th year percussionist. I used to play marimba mainly but i play rudimental snare most now--looking into DCI if anybody has experience.

I've played percussion for the last nine and a half years. I mostly do concert snare drum, but I also have lots of experience in rudimental snare, timpani, and some experience in mallets (both jazz vibraphone and orchestral stuff). I would do DCI, but that stuff is really expensive. Maybe if I was going to be a musician I would be willing to spend that much money to do work all summer.

Lund,
This reply is about two months late, but I marched with The Cadets for '03 and '04.  If you want more info about DCI, You can message me or something.

I'm an 8th year percussionist.  I specialize in jazz vibraphone and I absolutely love improv - it's just about the only thing i still practice.


AMC
AIME
MATHCOUNTS
geometry
AMC 10
number theory

Hey guys, for those of you with the AOPS Volumes 1 and 2, how useful do you think they are with the AMC and AIME's? I've seen volume 2 from another person, and it seems to be pretty comprehensive. I just want to have some feedback from people who have used it to get results, since my goal is to do well on AIME next year. Thanks.

Vol. 2  is awesome!! I just got the books this week.

Volume 1 is really helpful for Mathcounts and the AMC's.  It is really good for AMC 10 geometry type questions.  Basically IMO it gets you ready to take the AMC's and gives you a good base for the AIME.  If you go through this book carefully you should make the AIME.  Volume 2 picks up where volume 1 left off and takes it a whole lot further.  This also helps you with the AMC 10/12 (especially the last 5 or so questions).  It will get you good enough to pass the AIME and it will probably even give you a very good background for the USAMO.

What grade level are the books at? (If you were in public school, in what grade would you learn most of the material?) I think Volume 1 is a little hard for 7th grade

I think it is hard to put a grade level on these books.  Most of the stuff in these books you don't learn at a normal school.  I would say that volume one spans from middle school up to 10 grade.  Volume 2 would probably be at 11 or 12 grade.

The left of the screen.  Here is a link [url]http://www.artofproblemsolving.com/Books/AoPS_B_About.php[/url]

[quote="white_horse_king88"]Lets put it this way. Before volume 2 - practice AIME score - 3. After practice AIME score - 8. It makes a huge difference.[/quote]

Before volume 1- AIME 6, after volume 1- AIME 10

I've always believed that volume 1 alone is enough for competitions up to the olympiads, although most people seem to disagree with me.

Actually, I never did get very far in volume 2.

I've always had difficulties with series and sequences, and Volume 1 really doesn't go in depth - same with number theory for the most part, and it almost never touches trig.


function algebra domain LaTeX calculus integration derivative

Let f, g and h be functions of x and let f = - dh/dx. f and g are defined for x>=0. Let h be defined so that h(0) isn't equal to 0.
Now let f^2 + 2Ch = g^2 for some constant C. Problem: If possible, eliminate dependancy on h in this equation. Now, if you don't take into accaount the domains, you get f^2 - 2C int f(x) dx = g^2. With respect to domains and limits, this becomes f^2 - 2C int f(x') dx' [from x'=0 to x'=x] = g^2. Now, int f(x') dx' [from x'=0 to x'=x] = h(x) - h(0) = h(x) => h(0) = 0, but it's not! So this equation only holds if h(0) = 0. Or..?
Can anyone help me eliminating dependancy on h here?

The thing is: h(x) = - int(0->x)f(x')dx' = h(x) - h(0), but since h(0) isn't zero, this is a contradiction! It's like saying x = x + C for some non-zero constant C. Does this mean that the equation, given the condition f = - dh/dx, only holds if h(0) is zero???  :?:

[hide="hint"]maybe if you consider using $\LaTeX$ someone will understand and help you[/hide]

But I haven't got Latex, nor do I hae the time to get/install it!
int(0->x)f(x')dx' is the integral of f(x')dx' from x'=0 to x'=x.

f^2 + 2Ch = g^2, f and g are defined for x greater than or equal to 0, and f = - dh/dx, f equals minus the derivative of h wrt x. Eliminate dependancy on h in the equation. That's not hard to understand, is it???

No downloading is necessary to use $\LaTeX$ on these forums.  Just put the code between dollar signs.

Check out the link labeled "LaTeX" on the left sidebar.  There is also a forum for discussing $\LaTeX$.

OK:
$f(x)^{2}+2Ch(x)=g(x)^{2}$,
$f(x)$ and $g(x)$ are defined for $x\geq 0$,
and $f(x)=-dh/dx$
Eliminate,if possible, dependancy on h in this equation.


function
search
number theory proposed
number theory

Please, someone can tell me where I can find problems olympic on the divisor function?

It depends on what you mean with divisor function, maybe $ \sigma_k(n)\equal{}\sum_{d \mid n}{d^k}$ with $ k\equal{}0$ or $ k\equal{}1$ or, in general with k in R?

Here there are a few examples:
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=401796479&t=246662
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=401796479&t=245271
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=401796479&t=225460
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=401796479&t=196334

Thanks for your help, jin thynj. :)
The problems are also about the function $ \tau (n)$.

[url=http://www.mathlinks.ro/Forum/index.php?f=484]Section J[/url] of PEN is all dedicated to the topic "Divisor function".
Have fun  :wink:

Thanks again for your help, jin thynj :)


FTW
MATHCOUNTS
AMC
AMC 8
factorial
trigonometry

I was just wondering. Piggybacking off calculator thread.

sometimes, im not a wiz like mewto, so i like my calc on the target problems.

Well on like problems that have radicals, I need to compute them to like the nearest tenth or on problems with pi and stuff, I use calculators. I try not to use calculators on normal problems.

[quote="Mewto55555"]Never. In countdown, you don't get a calculator. Practicing with one ruins the point.[/quote]

Yeah, I definitely agree. I don't ever use calcs on FTW. By the way, I am shocked that 5 people said they always use calcs!

WOW! CHEATERS!

I would go against calling people cheaters...

As jonathanchou711 said, if a problem requires you to round $ 75\sqrt{13}$ to the nearest tenth, it's probably a Target problem.

[quote="SnowHawk"]Yeah, I definitely agree. I don't ever use calcs on FTW. By the way, I am shocked that 5 people said they always use calcs!

WOW! CHEATERS![/quote]
Please don't call people cheaters who are not cheating. There is no rule against calculators in FTW.

[quote="Mewto55555"]Never. In countdown, you don't get a calculator. Practicing with one ruins the point.[/quote]

I agree with the fact that when doing countdown problems, you shouldn't use calculators.  However, on "For the Win", there are mixed targets, too, so I use calculators when necessary (like with radical rounding and stuff).

FTW includes problems from all rounds of mathcounts (excluding masters round). On the Target and Team rounds, there are some problems that are meant for calculator use. Many of these problems are obviously target or team. On these problems, there is nothing wrong with using a calculator.

Some problems will be from Sprint or CD. You can usually tell which those are. If you wish to use a calculator, you are free to, but it will not help you in the actual sprint where there are no calculators allowed.

To use a calculator or not is your decision, but using a calculator is not cheating.

the way i see it, i just try to practice for countdown rounds/math bowl so there's no point in using a calculator
plus sometimes it takes longer to input the numbers than to do it in y head
XD just my opinion if anyone disagrees sorry

Yes, I agree. Everyone has different views on calculators, but I think we should remember FTW is for 2 reasons: fun and practice. no, not to get a high rating.

I don't ever use calculators b/c they're kinda slower than my head. Anyway, I think people should establish before a game if they're going to use calcs. I think that not using calcs is definitely better if you want to get better at math. If you want to stay about the same level, then go ahead, use calcs. Sometimes, though, on target and team round questions, there is usually a faster way than using a calc.

No, on many target problems calculators make the problem much easier, and in some cases, are almost required.

I normally only use calculators on problems with square roots or $ \pi$ that require rounding, or on those annoying compound interest problems.

[quote="connaissance"]I don't ever use calculators b/c they're kinda slower than my head.[/quote]

ok

lets find 1337*42*pi*sqrt2 to the nearest tenth. you can use only ur head. i will use a calculator. lets see who wins :dry:

calculators are necessary sometimes, like problems with sin^-1(angle other than a multiple of 30 or 45)

i highly doubt u can use ur head for that

so, yea, calculators are necesary sometimes, but that doesnt change the fact that i wont use them

COMPLETELY agree with mz94. Sometimes there are problems that are calc problems, but I don't use them, because I feel it is morally wrong. I have no grudge against people who use calcs, though.

[quote="Mewto55555"]Never. In countdown, you don't get a calculator. Practicing with one ruins the point.[/quote]
Yes, but FTW includes AMC8 and MC Target and Sprint Round problems, often which you are ALLOWED to use a calculator.  Of course, not using a calculator improves number sense, and it's often faster for easy problems.  But for simplifying square roots, big calculations, or huge factorials, I use a calculator.

In AMC 8 and Sprint Round you're not allowed to use calculators :wink:

Actually on the AMC 08 before 2008, you were previously allowed to use a calculator.

I believe so.

What's with these one-line posts?
Anyway, I don't use a calculator for FTW, but then, I don't play it competitively. I usually just click give up if there's some Trig or Sqrt expression which you really need a calculator to evaluate.

I never do in a CD but in a game i only use it when it is nessesary

I don't use calculators. In the real competition, you don't use calculators. If the question specifically states to use a calculator and punch in these numbers, I will use it. In FTW, however, you probabily will never need to use a calculator.

*Gasp* I didn't know that there were that many cheaters! :lol:

[quote="r15s11z55y89w21"]There is nothing against using calcs on FTW. (Statement backed by levans)[/quote]

I know, I was joking :)

Of course, most of the questions are designed, so that you can do it quicker with mental math than with a calc(of course, there are exceptions to this.)

Obviously, calculators are sometimes needy for target questions.

I use calculators only when I absolutely have to...like on the Target problems and some AMC problems (before 2008). Mostly I just use mental math...but my mental math is an epic fail. I wonder what my rating would be like if I used calculators all the time! :lol: But I don't use calculators unless I have to.

Only when necessary, hehe :P
Yeah, just when it needs like a square root I don't know of the top of my head or something.


.A music academy has to send 10 members to 5 shows ,two in each show. two of the shows are in city and the 3 are outside .two of the 10 members prefer to work in city while 3 prefer to work outside city..the number of ways in which five shows can be held if preference are to be satisfied .
options are 5200, 5400,5472,5250

[quote="befuddlers"]ur question is unclear. u told there are 10 members and next line u post 2 of the 5 members prefer to perform in the city. :rotfl:[/quote]
sorry it is 10 members  not 5 members


Online Math Open
LaTeX
geometry
geometric transformation

[quote="vangelismak"]Παιδιά δεν ξέρω αν γνωρίζετε το παρακάτω πρόβλημα αλλά αν δεν το γνωρίζετε πραγματικά είναι πολύ ωραίο.

Το πρόβλημα είναι στο attachment...
Προσωπικά με κούρασε πολύ να το λύσω :(  μακάρι κάποιος από σας να βρεί καμία πιό εύκολη λύση γιατί πραγματικά μου "άλλαξε τα φώτα" !

( Με συνθετική γεωμετρία ... )[/quote]
Είναι ερώτημα στο σύστημα Vecten του τριγώνοου ΑΒΓ .Ο Κώστας Βήττας θα μπορεί να σου πει πολλά περισσότερα

Αν δεν κάνω λάθος, το ABD είναι ομοιοθεσία των ομοίων CBH, CAE.

Μανιέ είναι ομοιοθεσία με κέντρο πιό σημείο ? Δεν βλέπω κάτι καθαρά ...
Κοίτα Μανώλη ένας στο διεθνές την έλυσε με στροφή κατα π/2
αλλά να σας πώ ότι αυτή η άσκηση δώθηκε σε τμήμα Α Λυκείου που έχει διδαχτεί μέχρι και το 6ο κεφάλαιο ( παραλλη/γραμμα - εγγεγραμμένες - εγγράψιμα κ.λ.π )

[color=red][i]Βαγγέλη , δεν ξέρω αν είναι κάτι σαν αυτό που ζητάς , αλλά σου στέλνω μια λυσούλα, μάλλον καλούτσικη. Εσύ θα μου πεις:[/i][/color]

 Να φέρεις $ EK,DN, CP, HL \perp AB$ .Είναι :

  $ \triangle EKA = \triangle CPA , \triangle CPB = \triangle BHL$ , $ KA = CH = BL$ ,

  $ NK = NL$ , οπότε $ NA = NB$. Αρα το τρίγωνο [b]ΑDΒ[/b] είναι ισοσκελές.

 Επίσης είναι :

   $ DN = \frac {EK + HL}{2} = \frac {AP + BP}{2} = \frac {AB}{2}$ , οπότε το τρίγωνο ΑΜΒ είναι και ορθογώνιο.

 [color=red][b]΄Αλλη  σκέψη [/b][/color]

 Αν $ K$  είναι το συμμετρικό του $ A$  ως προς το $ D$  , τότε τα τρίγωνα $ ABC , BKH$ είναι ίσα διότι

    $ AC = AE = KH, CB = BH , \angle ACB = \angle KHB$ ,

 μια και   το  $ AEKH$ είναι παραλληλόγραμμο  και  $ KH // EA , EA \perp CA , HB \perp CB$

 Επομένως είναι  $ BK = BA$  και έτσι  $ BD \perp AK$.

 Επειδή  $ \angle KBH = \angle ABC$ και  $ \angle CBH = 90$ , είναι και  $ \angle KBA = 90$ ,

 οπότε  $ BD = \frac {AK}{2} = AD$.

 Νομίζω ότι κάτι κάναμε :) .

[color=red]edited - today[/color]

[u] Μπάμπης[/u]

Κύριε Μπάμπη δεν ξερω αν είναι κάτι που δεν βλέπω αλλά τί είναι τα σημεία
Κ, Ρ, L   στον πρώτο τρόπο  ?

Το σύστημα [b][size=100]Vecten[/size][/b] όπως το ξέρουμε, αφορά στα τετράγωνα που κατασκευάζονται επί των πλευρών δοσμένου τριγώνου και προς το εξωτερικό μέρος αυτού.

(Τα ορθογώνια ισοσκελή τρίγωνα $ \bigtriangleup ACE$ και $ \bigtriangleup BCH,$ είναι τα μισά των αντίστοιχων τετραγώνων με βάσεις τις πλευρές $ AC,$ $ BC,$ αντιστοίχως $ ).$

Αν αντί των τετραγώνων θεωρήσουμε όμοια ορθογώνια παραλληλόγραμμα (θυμίζουμε ότι τα τετράγωνα είναι μία ειδική περίπτωση όμοιων ορθογώνιων παραλληλογράμμων, με λόγο πλευρών ίσον με $ 1$ ), αποδεικνύεται ότι αληθεύουν επίσης πολλές από τις ιδιότητες που ισχύουν, όταν έχουμε τετράγωνα.

Αληθεύουν επίσης και κάποιες από τις ιδιότητες, όταν έχουμε τυχαία ορθογώνια παραλληλόγραμμα επί των πλευρών του δοσμένου τριγώνου. Οι αποδείξεις δυσκολεύουν περισσότερο, αλλά πιστεύω είναι πιο ενδιαφέρουσες.

Ενδιαφέρον επίσης είναι, ότι τα σχήματά μας επί των πλευρών του δοσμένου τριγώνου ( τετράγωνα ή όμοια, ή τυχαία ορθογώνια παραλληλόγραμμα ), μπορεί να είναι είτε προς το εξωτερικό μέρος αυτού, είτε προς το εσωτερικό του, είτε άλλα προς το εξωτερικό μέρος του και τα υπόλοιπα προς το εσωτερικό του.

$ \bullet$ Στην περίπτωσή μας τώρα, το πρόβλημα αληθεύει και όταν αντί τετραγώνων έχουμε όμοια ορθογώνια παραλληλόγραμμα, επί των πλευρών του $ \bigtriangleup ABC.$

Έχει τύχει να αναφερθώ παλιότερα στο παρακάτω πρόβλημα, χρησιμοποιώντας το σαν Λήμμα, για την απόδειξη του προβλήματος [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=125520]A problem with quadrilateral[/url] και με έκπληξή μου διαπίστωσα ότι δεν έχω στείλει τόσο καιρό, την απόδειξή του ως όφειλα.

Ας είναι, κάλιο αργά παρά ποτέ.

[b][size=100][color=DarkBlue]ΠΡΟΤΑΣΗ. – Επί των πλευρών $ AB,$ $ AC,$ δοσμένου τριγώνου $ \bigtriangleup ABC$ και προς το εξωτερικό μέρος αυτού, κατασκευάζουμε τα όμοια ορθογώνια παραλληλόγραμμα $ ABDE$ και $ ACFH,$ αντιστοίχως. Αποδείξτε ότι η μεσοκάθετη της πλευράς $ BC,$ διχοτομεί επίσης και το ευθύγραμμο τμήμα $ DF.$[/color][/size][/b]

[b][size=100]ΑΠΟΔΕΙΞΗ[/size][/b]. - Δια του μέσου $ M$ της πλευράς $ BC,$ φέρνουμε κάθετη ευθεία επί την ευθεία $ BC,$ η οποία τέμνει την ευθεία $ DF,$ στο σημείο έστω $ N.$

Για να είναι το σημείο $ N,$ μέσον του $ DF,$ αρκεί να αποδείξουμε ότι το $ M$ είναι μέσον του  ευθυγράμμου τμήματος $ D'F',$ όπου $ D,$ $ F,$ οι ορθές προβολές των $ D,$ $ F$ αντιστοίχως επί την ευθεία $ BC.$

Γράφουμε τους κύκλους $ (K),$ $ (L),$ τους περιγεγραμμένους περί τα $ ABDE,$ $ ACFH$ αντιστοίχως, οι οποίοι τέμνουν την ευθεία $ BC,$ στα σημεία $ E',$ $ F',$ ως ορθές προβολές των $ E,$ και $ H$ αντιστοίχως, επί την ευθεία $ BC$ $ ($ $ \angle EE'A \equal{} 90^{o} \equal{} \angle HH'C,$ από εγγράψιμα τετράπλευρα $ ABE'E,$ $ ACH'H$ $ ).$

Έστω $ K',$ $ A',$ $ L',$ οι ορθές προβολές των $ K,$ $ A,$ $ L$ αντιστοίχως, επί την ευθεία $ BC.$

Από τα εγγράψιμα τετράπλευρα $ ABE'E,$ $ ACH'H,$ συμπεραίνουμε ότι $ \angle AE'B \equal{} \angle AEB \equal{} \angle AHC \equal{} \angle AH'C$ $ ,(1)$ και άρα το τρίγωνο $ \bigtriangleup AE'H'$ είναι ισοσκελές με $ AE' \equal{} AH'$ και επομένως έχουμε $ A'E' \equal{} A'H'$ $ (2)$

Επίσης έχουμε τις εύκολες ισότητες $ K'B \equal{} K'E' \equal{} x$ $ ,(3)$ $ K'A' \equal{} K'D'$ $ \Longrightarrow$ $ D'E' \equal{} A'B \equal{} z$ $ ,(4)$ και $ L'C \equal{} L'H' \equal{} u$ $ ,(5)$ $ L'A' \equal{} L'F'$ $ \Longrightarrow$ $ H'F' \equal{} A'C \equal{} w$ $ ,(6)$

Και βέβαια έχουμε ότι $ MB \equal{} MC \equal{} \frac {\alpha}{2}$ $ ,(7)$

Από $ (2)$ $ \Longrightarrow$ $ t \plus{} \frac {\alpha}{2} \plus{} 2x \equal{} \frac {\alpha}{2} \minus{} t \plus{} 2u$ $ \Longrightarrow$ $ u \equal{} x \plus{} t$ $ ,(8)$ όπου $ A'M \equal{} t$

Έτσι, $ MD' \equal{} MF'$ $ \Longleftrightarrow$ $ \frac {\alpha}{2} \plus{} 2x \plus{} z \equal{} \frac {\alpha}{2} \plus{} 2u \plus{} \frac {\alpha}{2} \minus{} t$ $ \Longleftrightarrow$ $ u \equal{} x \plus{} t$ $ ($ γιατί $ D'E' \equal{} A'B$ $ \Longrightarrow$ $ z \equal{} \frac {\alpha}{2} \plus{} t$ $ ).$

Άρα η ευθεία $ MN,$ ως μεσοπαράλληλη του τραπεζίου $ DD'F'F,$ διχοτομεί το ευθύγραμμο τμήμα $ DF$ και η πρόταση έχει αποδειχθεί.

$ \bullet$ Όσον αφορά στη γωνία $ \angle BNC,$ του ισοσκελούς τριγώνου $ NBC,$ αποδεικνύεται ότι είναι γνωστή και ίση με $ \angle BNC \equal{} \angle AKB \equal{} \angle AKC \equal{} 2\angle \omega$ $ ,(9)$

Επομένως στην περίπτωση των τετραγώνων επι των πλευρών του $ \bigtriangleup ABC,$ προκύπτει ότι $ \angle BNC \equal{} 90^{o}$ και άρα το ισοσκελές τρίγωνο $ \bigtriangleup NBC,$ είναι και ορθογώνιο.

Για την απόδειξη της $ (9),$ αρκεί να αποδείξουμε την ομοιότητα των τριγώνων $ \bigtriangleup MBN$ και $ \bigtriangleup ABE.$

Αποδεικνύεται εύκολα ότι $ \frac {MB}{AB} \equal{} \frac {MN}{AE}$ αν λάβουμε υπόψη μας ότι $ D'E' \plus{} H'F' \equal{} BC$ $ ($ από $ D'E' \equal{} A'B$ και $ H'F' \equal{} A'C$ $ ),$

καθώς επίσης και ότι $ MN \equal{} \frac {DD' \plus{} EE'}{2}$ από το τραπέζιο $ DD'F'F.$

Επίσης είναι εύκολο να αποδειχθεί η ομοιότητα των ορθογωνίων τριγώνων $ \bigtriangleup D'DE$ και $ \bigtriangleup ABE.$

Δώστε μου παρακαλώ λίγο χρόνο γιατί είμαι πολύ κουρασμένος και θα βάλω αύριο περισσότερες λεπτομέρειες.

Κώστας Βήττας.

Αντπωσιάστηκα από την παραπάνω λύση. Θα παραθέσω άλλη μία:
Έστω M το μέσο της AB, CG το ύψος του τριγώνου ABC και K,L οι προβολές των σημείων E,H στο AB σντίσοιχα τότε:

τα τρίγωνα AEK και BHL είναι ίσα (αν δε βλέπετε πώς, ρωτήστε με)
άρα CG=AK
ομοίως CG=BL

άρα AK=BL άρα, επειδή είναι και MA=MB έχουμε MK=ML
άρα M=μέσο του KL

EKLH=τραπέζιο άρα (επειδή M,D μέσα μη παράλληλων πλευρών)
MD=διάμσος άρα
MD=παράλληλο στο EK
άρα MD=κάθετος στη AB

το τρίγωνο ABD έχει διάμεσο και ύψος ταυτόχρονα το MD άρα είναι ισοσκελές

ακόμα (επειδή MN=διάμεσος του τραπεζίου EKLH) είναι MN=(EK+HL)/2=(από την αρχική ισότητα τριγώνων) (AG+GB)/2=AB/2
άρα στοτρίγωνο ABD έχουμε τη διάμεσο ίση με το μισό της πλευρά άρα είναι ορθογώνιο q.e.d. :D

[size=150][color=red]Άσκηση:[/color][/size]
Αν Ζ το δεύτερο σημείο τομής του περιγεγραμμένου κύκλου του τριγώνου DAB με την ευθεία EH, να βρεθεί το MZ συναρτήσει των πλευρών και του ύψους του τριγώνου ABC.

Πολύ δύσκολο! :blush: Βέβαια μπορούμε να το υπολογίσουμε εξ' ολοκλήρου από τις πλευρές αν εφαρμόσουμε μετά τον τύπο του ήρωνα για το ύψος τριγώνου.  :lol:

[hide]Να φέρετε την κάθετη από το E στο HL και μετά δύναμη σημείου ως προς κύκλο.[/hide]

[quote="maniopas"]
[size=150][color=red]Άσκηση:[/color][/size]
Αν Ζ το δεύτερο σημείο τομής του περιγεγραμμένου κύκλου του τριγώνου DAB με την ευθεία EH, να βρεθεί το MZ συναρτήσει των πλευρών και του ύψους του τριγώνου ABC.

Πολύ δύσκολο! :blush: Βέβαια μπορούμε να το υπολογίσουμε εξ' ολοκλήρου από τις πλευρές αν εφαρμόσουμε μετά τον τύπο του ήρωνα για το ύψος τριγώνου.  :lol:

[hide]Να φέρετε την κάθετη από το E στο HL και μετά δύναμη σημείου ως προς κύκλο.[/hide][/quote]

Κατ' αρχήν θέλω να πω ότι ήταν πολύ ωραίες οι λύσεις όλων σας :)

maniopas, όσον αφορά την παραπάνω άσκηση μήπως ζητάς κάτι άλλο? Θέλω να πω το να βρούμε την $ MZ$ είναι πολύ εύκολο αφού είναι η ακτίνα του κύκλου περί το $ ADB$ η οποία είναι $ AB/2$ (αφού εκ του αποτελέσματος $ M$ είναι το κέντρο αυτού του κύκλου.

@vangelismak: νομίζω ότι η πρώτη λύση του κ. Στεργίου είναι βασικά αυτή του maniopas.

Ακόμα θέλω να κάνω και τις εξής παρατηρήσεις σε σχέση με αυτό το ωραίο πρόβλημα και το σχήμα γενικότερα που αφήνει πολλά περιθώρια:

1)Απ' το αποτέλεσμα του προβλήματος φαίνετια ότι το $ D$ ε[i]ίναι ανεξάρτητο της επιλογής του σημείου $ C$.
[/i]
2)Ισχύει ότι $ CZ\perp {EH}$, όπου $ Z$ αυτό που όρισε ο maniopas (δηλαδή σημείο τομής οτυ περιγεγραμμένου κύκλου του $ ADB$ με την $ EH$).

3)Αν τα $ P,R$ είναι σημεία τέτοια ώστε τα $ CAEP, CBHR$ να είναι τετράγωνα τότε ισχύει επίσης ότι $ EH, BP, AR$ συντρέχουν στο $ Z$ (μάλιστα $ BP\perp \equal{}AR$).

[quote="stergiu"]
Να φέρεις $ EK, DN, CP, HL \perp AB$ .Είναι :

$ \triangle EKA = \triangle CPA , \triangle CPB = \triangle BHL$ , $ KA = CP = BL$ ,

$ NK = NL$ , οπότε $ NA = NB$. Αρα το τρίγωνο $ \bigtriangleup DAB$ είναι ισοσκελές.

Επίσης είναι :

$ DN = \frac {EK + HL}{2} = \frac {AP + BP}{2} = \frac {AB}{2}$ , οπότε το τρίγωνο $ \bigtriangleup DAB$ είναι και ορθογώνιο.[/quote]

Βρήκα και μια άλλη λύση σε αυτό το ωραίο πρόβλημα :)

Εστω $ O_1, O_2$ τα κέντρα των υπάρχοντων τετραγώνων και $ O_3$ το κέντρο του (εξωτερικού) τετραγώνου με πλευρά $ AB$.

Τότε είναι αρκετά γνωστό ότι $ O_1O_2\perp \equal{}CO_3$. Για την απόδειξή του μπορούμε να χρησιμοποιήσουμε σύνθεση 2 spiral similarity, μία με κέντρο $ C$, γωνία 45 και λόγο $ \frac{1}{\sqrt{2}}$ και την άλλη με κέντρο $ A$ γωνία 45 και λόγο $ \sqrt{2}$.

Τώρα ξέρουμε ότι $ DH//\equal{}O_1O_2\perp \equal{}CO_3$. Αρα τα τρίγωνα $ CO_3B, BHD$ είναι ίσα αφού:
$ DH\equal{}CO_3$, $ BH\equal{}BC$, $ \angle{BHD}\equal{}\angle{O_3CB}$(αφού έχουν πλευρές κάθετες).

Αρα $ BD\equal{}BO_3$ kai $ \angle{BDH}\equal{}BO_3C$. Oμοια προκύπτει $ AD\equal{}AO_3$ και $ \angle{ADE}\equal{}AO_3C$.

Omos $ AO_3\equal{}BO_3$ kai $ \angle{AO_3B}\equal{}90$ κι άρα έπεται το ζητούμενο.

[quote="vittasko"][quote="stergiu"]
Αν $ K$  είναι το συμμετρικό του $ A$  ως προς το $ D$  , τότε τα τρίγωνα $ ABC , BKH$ είναι ίσα διότι

$ AC \equal{} AE \equal{} KH, CB \equal{} BH , \angle ACB \equal{} \angle KHB$ ,

μιας και   το  $ AEKH$ είναι παραλληλόγραμμο  και  $ KH // EA , EA \perp CA , HB \perp CB$

Επειδή  $ \angle KBH \equal{} \angle ABC$ και  $ \angle CBH \equal{} 90^{o}$ , είναι και  $ \angle KBA \equal{} 90^{o}$ ,

οπότε  $ BD \equal{} \frac {AK}{2} \equal{} AD$.[/quote][/quote]

 Κώστα ,
 σε ευχαριστώ για τα [color=red][b]ωραία σχήματα[/b][/color].Αν καμιά φορά φτιάξω καμιά γεωμετρία , θα σε παρακαλέσω να μου φτιάξεις όλα τα σχήματα ! Παρακαλώ λοιπόν το Θεό να σε έχει καλά , για έναν παραπάνω λόγο :) (ο πρώτος είναι γιατί είσαι φίλος και σπάνιος άνθρωπος!).

[u] Μπάμπης[/u]

Κύριοι, συγχωρέστε με που δεν σας προλαβαίνω.

[quote="vittasko"]$ \bullet$ Όσον αφορά στη γωνία $ \angle BNC,$ του ισοσκελούς τριγώνου $ NBC,$ αποδεικνύεται ότι είναι γνωστή και ίση με $ \angle BNC \equal{} \angle AKB \equal{} \angle AKC \equal{} 2\angle \omega.$

Επομένως στην περίπτωση των τετραγώνων επι των πλευρών του $ \bigtriangleup ABC,$ προκύπτει ότι $ \angle BNC \equal{} 90^{o}$ και άρα το ισοσκελές τρίγωνο $ \bigtriangleup NBC,$ είναι και ορθογώνιο.

Αποδεικνύεται εύκολα ότι $ \frac {MB}{AB} \equal{} \frac {MN}{AE}$ αν λάβουμε υπόψη μας ότι $ D'E' \plus{} H'F' \equal{} BC$ $ ($ από $ D'E' \equal{} A'B$ και $ H'F' \equal{} A'C$ $ ),$

καθώς επίσης και ότι $ MN \equal{} \frac {DD' \plus{} EE'}{2}$ από το τραπέζιο $ DD'F'F.$

Επίσης είναι εύκολο να αποδειχθεί η ομοιότητα των ορθογωνίων τριγώνων $ \bigtriangleup D'DE$ και $ \bigtriangleup ABE.$[/quote]

Για την απόδειξη ότι $ \angle BNM \equal{} \angle BEA \equal{} \angle \omega,$ αρκεί να αποδείξουμε ότι τα τρίγωνα $ \bigtriangleup MBN,$ $ \bigtriangleup ABE$ είναι όμοια.

Δηλαδή αρκεί να αποδειχθεί ότι $ \frac {MB}{MN} \equal{} \frac {AB}{AE}.$ (σχήμα [b][size=100]t=188591[/size][/b])

Από εγγράψιμα τετράπλευρα $ BEDE'$ και $ CHFH'$ έχουμε $ \angle DE'D' \equal{} \angle DEB \equal{} \angle FHC \equal{} \angle FH'F'$ $ ,(1)$

Από $ (1)$ και από ίσα ορθογώνια τρίγωνα $ \bigtriangleup DEB,$ $ \bigtriangleup ABE,$ συμπεραίνουμε ότι τα ορθογώνια τρίγωνα $ \bigtriangleup D'E'D,$ $ \bigtriangleup ABE$ και $ \bigtriangleup F'H'F,$ είναι όμοια.

Στην προέκταση της ευθείας $ DD'$ και προς το μέρος του $ D',$ ορίζουμε το σημείο $ X$ έτσι ώστε $ D'X \equal{} FF'$ και έστω $ \bigtriangleup XYD'$ το ορθογώνιο τρίγωνο, ίσο με το $ \bigtriangleup F'H'F.$

Από $ \angle XD'Y \equal{} \angle D'DE' \equal{} \angle \omega$ $ \Longrightarrow$ $ D'Y\parallel DE'$ $ ,(2)$

Εάν $ Z,$ είναι το σημείο τομής των ευθειών $ DE',$ $ XY,$ έχουμε $ D'E' \equal{} YZ \equal{} z$ $ ,(3)$

Έτσι από $ D'E'\parallel XYZ$ $ \Longrightarrow$ $ \frac {D'E'}{XZ} \equal{} \frac {DD'}{DX}$ $ \Longrightarrow$ $ \frac {D'E'}{w \plus{} z} \equal{} \frac {DD'}{DD' \plus{} D'X} \equal{} \frac {DD'}{DD' \plus{} FF'}$ $ \Longrightarrow$ $ \frac {D'E'}{2(MB)} \equal{} \frac {DD'}{2(MN)}$ $ \Longrightarrow$ $ \frac {MB}{MN} \equal{} \frac {D'E'}{DD'} \equal{} \frac {AB}{AE}$ $ ,(4)$

Από $ (4)$ συμπεραίνουμε ότι τα ορθογώνια τρίγωνα $ \bigtriangleup MBN$ και $ \bigtriangleup ABE$ είναι όμοια και άρα το ζητούμενο $ \angle BNM \equal{} \angle BEA \equal{} \angle \omega,$ έχει αποδειχθεί.

Κώστας Βήττας.

O Κύριος Μπάμπης έγραψε:   "Να φέρεις ΕΚ, DN, CP, HL $ \perp$ στην ΑΒ"

Τώρα τι να σας πώ. Οτι διάβασα "Να φέρεις ΕΚ, DN, CP, HL, ΑΒ" ??? :mad:
Ωρες ώρες πραγματικά δεν βλέπω τίποτα... :mad:  :mad:
Γι΄αυτό έκανα στο παραπάνω post αυτή την ηλίθια ερώτηση για το τί είναι τα σημεία Κ,Ρ,L  και όχι μόνο αυτό αλλα βιάστηκα να ρωτήσω χωρίς να το κοιτάξω μερικές φορές ακόμη μπας και ξεστραβωθώ.  Συγγνώμη ρε παιδιά ...

Κύριε Μπάμπη ευχαριστώ για τις πολύ ωραίες λύσεις. ( Προσωπικά μου άρεσε περισσότερο η πρώτη λόγω της απλότητας του σχήματος )

Κύριε Κώστα τι να πώ? Φοβερή δουλειά με τα σχήματα και πολύ ωραία αυτά που γράψατε για τις γενικές ιδιότητες στο σύστημα Vecten. Ευχαριστώ εκ μέρους όλων  :)
Επίσης θα ήθελα αν γίνεται να δώσετε κάποια πρόταση που να ζητάει να αποδείξουμε κάποια ιδιότητα όταν κατασκευάζουμε στο εξωτερικό των πλευρών τριγώνου τυχαία ορθογώνια , έτσι για εξάσκηση .

Παραθέτω τη λύση που έκανα στο αρχικό πρόβλημα που έκανα post. Τα γράμματα που χρησιμοποιώ είναι τα ίδια με αυτά που έβαλα στο attachment.
Ρίξτε μια ματιά μην έχω κάνει καμία "πατάτα" :)

Λοιπόν φέρνω τις  ΕC και CH και έτσω Κ και Ρ τα μέσα τους.

Από το τρίγωνο ECH προκύπτει ότι  DR=//KC.
Από το τρίγωνο CEH προκύπτει ότι  KD=//CP.
Άρα το CKDR είναι παραλληλογραμμο.

Στη συνέχεια δείχνουμε ότι τα τρίγωνα  KAD και DPB είναι ίσα.
Πράγματι έχουν
1. ΚD=PB,   2. AK = DP,  3. τις γωνίες  AKD kai DPB ίσες.

Άρα ΑD = DB.

Για την ορθή γωνία εργάστηκα ώς εξής.

Αν πούμε Μ το σημείο τομής των ΑD και CB ( Στο σχήμα μου το D, βγήκε στο εσωτερικό του τριγώνου BCP ) , αρκεί να δ.ο.
$ \angle$DMB +$ \angle$MBD = 90.

Χρησιμοποίησα ότι :

1.  $ \angle$MBD = 45 - $ \angle$DBP
2.  $ \angle$DBP=$ \angle$KDA,  $ \angle$DPB=$ \angle$AKD,..
     $ \angle$BDP=$ \angle$DAK  ( λόγω τις ισότητας των τριγώνων KAD και DPB )

3. $ \angle$AKD=$ \angle$ACB  και βγήκε το άθροισμα 90 μοίρες.

Αυτό που με προβληματίζει λίγο είναι ότι δεν ξέρω αν το D  θα βρίσκεται απαραίτητα στο εσωτερικό του τριγώνου BCP.  Αλλά φαντάζομαι  ότι η απόδειξη είναι ανάλογη και στις άλλες περιπτώσεις αν δεν συμβαίνει αυτό.

Ποιά είναι η γνώμη σας ?

[quote="vangelismak"]O Κύριος Μπάμπης έγραψε:   "Να φέρεις ΕΚ, DN, CP, HL $ \perp$ στην ΑΒ"

Τώρα τι να σας πώ. Οτι διάβασα "Να φέρεις ΕΚ, DN, CP, HL, ΑΒ" ??? :mad:
Ωρες ώρες πραγματικά δεν βλέπω τίποτα... :mad:  :mad:
Γι΄αυτό έκανα στο παραπάνω post αυτή την ηλίθια ερώτηση για το τί είναι τα σημεία Κ,Ρ,L  και όχι μόνο αυτό αλλα βιάστηκα να ρωτήσω χωρίς να το κοιτάξω μερικές φορές ακόμη μπας και ξεστραβωθώ.  Συγγνώμη ρε παιδιά ...

[/quote]

Μην αδικείς τον εαυτό σου !

 Είχα πράγματι παραλείψει με το Latex να βάλω την καθετότητα και  για αυτό  δίκαια δεν το έβλεπες. Ζητάω συγνώμη που σε παίδεψα.
 Σήμερα το πρωί που είδα το μήνυμά σου , το διόρθωσα.Γι αυτό όταν το  ξαναδιάβασες το είδες σωστό ! :) .
 Η παράλειψη ήταν δική μου , αλλά ... περνάνε και τα χρόνια και η όραση  στην οθόνη με απατά ! ( μακάρι βέβαια να είναι και το μόνο που με απατά , αλλά αφού γεννηθήκαμε άντρες πάντα διατρέχουμε αυτόν τον κίνδυνο :wink: ) !!!.
 [u]Μπάμπης[/u]

[quote="bilot4"]
maniopas, όσον αφορά την παραπάνω άσκηση μήπως ζητάς κάτι άλλο? Θέλω να πω το να βρούμε την $ MZ$ είναι πολύ εύκολο αφού είναι η ακτίνα του κύκλου περί το $ ADB$ η οποία είναι $ AB/2$ (αφού εκ του αποτελέσματος $ M$ είναι το κέντρο αυτού του κύκλου.
[/quote]

Ναι, μπέρδεψα τις ονομασίες στο πρόχειρο με τις ονομασίες της official λύσης μου. Εννοούσα, να βρεθεί το DZ συναρτήσει του ύψους και των πλευρών του ABC.

[quote="vangelismak"]....
Κύριε Μπάμπη ευχαριστώ για τις πολύ ωραίες λύσεις. ( Προσωπικά μου άρεσε περισσότερο η πρώτη λόγω της απλότητας του σχήματος )

Κύριε Κώστα τι να πώ? Φοβερή δουλειά με τα σχήματα και πολύ ωραία αυτά που γράψατε για τις γενικές ιδιότητες στο σύστημα Vecten. Ευχαριστώ εκ μέρους όλων  :)
Επίσης θα ήθελα αν γίνεται να δώσετε κάποια πρόταση που να ζητάει να αποδείξουμε κάποια ιδιότητα όταν κατασκευάζουμε στο εξωτερικό των πλευρών τριγώνου τυχαία ορθογώνια , έτσι για εξάσκηση .....[/quote]
Έβαλα μία πρόταση με τυχαία ορθογώνια παραλληλόγραμμα επί των πλευρών δοσμένου τριγώνου, για εξάσκηση όπως λες.

http://www.mathlinks.ro/Forum/viewtopic.php?t=188881

Βαγγέλη και λοιποί φίλοι του φόρουμ, θερμά σας παρακαλώ, κάντε και μια εξάσκηση στη χρήση του ενικού, ιδιαίτερα όταν αναφέρεστε σε μένα. Απλά, αφαιρείτε τη λέξη ''κύριος'' πριν από το όνομα και όλα γίνονται πιο οικεία και πιο όμορφα.

Πουθενά δεν αναφέρεται ότι ο ενικός δεν ταιριάζει με την ευγένεια το ήθος και το σεβασμό που πρέπει να διέπει τις σχέσεις όλων μας σ' αυτή την κοινότητα, που μας συνδέει με ισχυρούς φιλικούς δεσμούς.

Κώστας Βήττας.

[quote="vangelismak"]....
Λοιπόν φέρνω τις  ΕC και CH και έτσω Κ και Ρ τα μέσα τους.

Από το τρίγωνο ECH προκύπτει ότι  DR=//KC.
Από το τρίγωνο CEH προκύπτει ότι  KD=//CP.
Άρα το CKDR είναι παραλληλογραμμο.

Στη συνέχεια δείχνουμε ότι τα τρίγωνα  KAD και DPB είναι ίσα.
Πράγματι έχουν
1. ΚD=PB,   2. AK = DP,  3. τις γωνίες  AKD kai DPB ίσες.

Άρα ΑD = DB.

Για την ορθή γωνία εργάστηκα ώς εξής.

Αν πούμε Μ το σημείο τομής των ΑD και CB ( Στο σχήμα μου το D, βγήκε στο εσωτερικό του τριγώνου BCP ) , αρκεί να δ.ο.
$ \angle$DMB +$ \angle$MBD = 90.

Χρησιμοποίησα ότι :

1.  $ \angle$MBD = 45 - $ \angle$DBP
2.  $ \angle$DBP=$ \angle$KDA,  $ \angle$DPB=$ \angle$AKD,..
     $ \angle$BDP=$ \angle$DAK  ( λόγω τις ισότητας των τριγώνων KAD και DPB )

3. $ \angle$AKD=$ \angle$ACB  και βγήκε το άθροισμα 90 μοίρες.

Αυτό που με προβληματίζει λίγο είναι ότι δεν ξέρω αν το D  θα βρίσκεται απαραίτητα στο εσωτερικό του τριγώνου BCP.  Αλλά φαντάζομαι  ότι η απόδειξη είναι ανάλογη και στις άλλες περιπτώσεις αν δεν συμβαίνει αυτό.

Ποιά είναι η γνώμη σας ?[/quote]
Βαγγέλη, προβληματίστηκα λίγο με το σημείο $ M\equiv AD\cap BC$ και επειδή έθεσες το θέμα για το που μπορεί να βρίσκεται το σημείο $ D$ $ ($ στο εσωτερικό του τριγώνου $ \bigtriangleup BCP$ ή όχι $ ),$ προτείνω την ακόλουθη απόδειξη ότι $ \angle ADB \equal{} 90^{o},$ με βάση τις ισότητες των γωνιών που ανέφερες.

$ \angle ADB \equal{} 360^{o} \minus{} (\angle ADK \plus{} \angle KDP \plus{} \angle BDP)$ $ ,(1)$

$ \angle ADK \equal{} \angle DBP$ και άρα $ \angle ADK \plus{} \angle BDP \equal{} 180^{o} \minus{} \angle BPD \equal{} 180^{o} \minus{} \angle ACB$ $ ,(2)$

$ \angle KPD \equal{} \angle KCP \equal{} 90^{o} \plus{} \angle ACB$ $ ,(3)$

Από $ (1),$ $ (2),$ $ (3)$ $ \Longrightarrow$ $ \angle ADB \equal{} 90^{o}$ και το ζητούμενο έχει αποδειχθεί.

Κώστας Βήττας.

Κώστα  ( χωρίς το κύριε μπροστά, όπως μου είπες  :)   )

το τσέκαρα και είναι όλα Ο.Κ.  και είμαστε απαλλαγμένοι από το Μ.
Η διαίσθησή μου μου έλεγε ότι σίγουρα θα έβγαινε με αυτόν τον τρόπο η γωνια ADB = 90  και ευχαριστώ και για τη λύση και για το σχήμα. Τωρα μελετάω τα άλλα που έγραψες  :)  Ενδιαφέροντα πράγματα βγαίνουν από το σύστημα Vecten. Αλήθεια πόσες προτάσεις περίπου να υπάρχουν γύρω από αυτό;
Μπορούμε να γενικεύσουμε τις κατασκευές στο εξωτερικό των πλευρών του τυχαίου τριγώνου ΑΒC με κανονικά πολύγωνα ?  Για παράδειγμα ξέρω μια πρόταση που λέει ότι αν εξωτερικά των πλευρών τυχαίου τριγώνου ABC κατασκευάσουμε ισόπλευρα τρίγωνα, τότε το τρίγωνο που έχει κορυφές τα βαρύκεντρα των ισοπλέυρων τριγώνων είναι και αυτό ισόπλευρο ! ( Νομίζω ότι η πρόταση αυτή είναι γνωστή σαν το Θεώρημα του Ναπολέοντα )

Τέλος πάντων ,...πολλές ερωτήσεις έκανα   :)

Στο  συνημμένο στέλνω μια ακόμη λύση στην πράγματι ωραία αυτή άσκηση.


Hi all, I am currently pursuing under-graduation (I'm in the final year of school). I have some queries regarding Mathematical/Quantitative Finance, they are as follows:

[b]Query 1 [/b]

Can anyone suggest some universities/schools(if they are less reputed)in the U.S that offer an M.S in Quantitative/Mathematical Finance?

[b]Query 2[/b]

Assuming I managed to graduate in 2012,How are the job prospects for a person (particularly at Wall Street) who has acquired an M.S. in Quantitative Finance (say, from a less reputed school, not necessarily from Princeton, Berkeley etc)?

I am an international student;I have a pretty ordinary GPA, and my GRE score is 1400 (Quants-700; Verbal- 700;AWA- 4.0); I am planning of re-taking the GRE solely to bolster my Quant score (I feel it is too low for an M.S. in Quantitative Finance);My baccalaureate degree is in Computer Science, I have good grades in Math related courses at the under-graduate level.

In a nut-shell, I would like to know -- Is it feasible to pursue an M.S. in Quantitative/Mathematical Finance, are there any jobs that I may land up with (after graduation) at Wall Street or any other place?

Do provide some advice/replies/suggestions to my queries, Thanks in advance!

the timing is really bad. There are still some jobs available, but for fresh graduates to land a job in this market is near to impossible. There are plenty of experienced people on the street now. and the prospects are quite bad  :maybe:

do a degree in some real science instead (maths/physics).

Can anyone post a more effective/elaborate reply?

You could start by Googling job prospects in finance.

That doesn't mean you should be discouraged altogether from finance; the markets [i]will[/i] recover, and they [i]will[/i] be on the lookout for fresh minds again -- eventually, anyway.

Different, how is a degree in pure mathematics or physics going to help in a bad economy? Academic jobs are very hard to land right now.

[quote="TZF"]Different, how is a degree in pure mathematics or physics going to help in a bad economy? Academic jobs are very hard to land right now.[/quote]

having a degree in math or physics qualifies you for a lot of non-academic jobs as well!

[quote="TZF"]

Different, how is a degree in pure mathematics or physics going to help in a bad economy? Academic jobs are very hard to land right now.[/quote]

I've got a better question for you: how is a degree in financial engineering going to help in a bad economy?

to put it simply, if you have a degree in Finance you're pretty much screwed up in the times similar to those we are experiencing now.
If one would have a degree in maths/physic, at least there is an option to go back to academia, as well as options to do something else... and the other way around, if one has a degree in maths/physics it's not a problem (assuming one has brains, not just a diploma) to switch to finance.

I would like to ask people to provide useful advice/suggestion to my query before they comment (this is probable) upon another person's post.

I need your suggestions/advice, and I take them pretty seriously, so I ask people who reply to my thread not to digress.

Query 1: NYU, Columbia (2 programmes: math finance, and fin. engineering), Baruch College (CUNY) is probably the best option (cheapest and easier to get an admission);
UK schools: Imperial college, LSE, Oxford

Query 2: Improving you GRE score is a good idea.

... in our team of 11 quants no one has a degree in quant finance / fin. engineering, although a few people (including me) have a second degree in finance (on top of msc/phd in maths/physics).
I am not a clairvoyant, but, as I already said,  the prospects of getting a job within the next couple of years are bad. ...not to mention the chance of becoming unemployed while having a degree that is pretty much useless elsewhere...

Okay, I had perused the site of NYU (It seems it admits really smart people!).

Since there doesn't exist many (I mean very few) jobs in the Financial market, then in what course should I opt for an M.S. in? (I should be guaranteed that I might/will land up with a fairly good/average paying job) I already hold a baccalaureate degree in Computer Science, so does pursuing an M.S. in computer science (at an average/above average school) appear feasible or should I opt for a completely new field like Finance etc?

[b] I would like to tell that I (in a team of four) managed to get a prize in a Business plan competition which was held at the renowned Indian technological school IIT-Madras ( I managed to participate again in a global conference which was held over there), so does it have any kind of special "weightage" in my M.S. application ( along with GPA, GRE/TOEFL scores, LOR's, and SOP)? [/b]

Can anyone answer these two queries of mine?

Thanks!

Is anyone "alive" in this particular section of AOPS? There isn't a reply to my query.

nobody really cares b/c its not math-related really


linear algebra
matrix
vector
search
combinatorics unsolved
combinatorics

A magic square is a square matrix such that the sums of entries in all its rows and columns are equal. In a magic square nxn composed from the number

1,2,.....n^2.The center of any two squares are joined by vector going from smaller number to bigger one.Prove that the sum of all these vectors is zero.

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1964293625&t=22867

Pierre.


Define f(n)=(15+sqrt(220))^n+(15-sqrt(220))^n, and note that binomial theorem guarantees that f(n) is an integer for each n. Secondly, we may notice (by experimentation or binomial considerations) that f(n) is always divisible by 10, so it ends in a 0. So, f(19)+f(82) ends in a 0.

But how big is (15-sqrt(220))^n? Well, 15-sqrt(220)<1, so raising it to high powers makes it very very small. So, the desired value is just a bit under something that ends with a 0, so the value must end as 9.999---, and its units digit is 9.

(update) Mathematica says that the number ends with 49999.999999999999998---

a bit different method:
\sqrt{220} is very very close to 15. Thus 15+\sqrt{220} is very close to 30. Let it be 30-k
Since [tex]k^{19}+k^{82}[/tex] is veryveryvery small(less than 1 for sure), [tex](30-k)^{19}+(30-k)^{82}[/tex] ends in 9.


number theory
relatively prime
number theory open

Show that if $ [x]$ denotes the greatest integer $ \leq x$ then $ C(n,r) \minus{} [\frac{n}{r}]$ is a natural number divisible by $ r$ for all values of $ n$.Note $ r < n$.

ur statement is true if $ r$ is $ 1$ or a prime number. i think it's false for all other $ r$.

it's obvious if $ r \equal{} 1$

lets first deal with the option $ r$ is prime
than $ n$ is of a form $ ar \plus{} b$ where $ a,b$ are two natural numbers ($ b < r$)
$ C(n,r) \equal{} C(ar \plus{} b,r) \equal{} \frac {(ar \plus{} b)(ar \plus{} b \minus{} 1)(ar \plus{} b \minus{} 2)....(ar \plus{} 1)(ar)(ar \minus{} 1)...(ar \plus{} b \minus{} r \plus{} 1)}{r!} \equal{} \frac {a(ar \plus{} b)(ar \plus{} b \minus{} 1)...(ar \plus{} 1)(ar \minus{} 1)...(ar \plus{} b \minus{} r \plus{} 1)}{(r \minus{} 1)!}$
now calculating by modulo $ r$ and using wilsons teorem we have $ (ar \plus{} b)(ar \plus{} b \minus{} 1)...(ar \plus{} 1)(ar \minus{} 1)...(ar \plus{} b \minus{} r \plus{} 1) \equal{} (r \minus{} 1)! \mod r$ therefore $ \frac {a(ar \plus{} b)(ar \plus{} b \minus{} 1)...(ar \plus{} 1)(ar \minus{} 1)...(ar \plus{} b \minus{} r \plus{} 1)}{(r \minus{} 1)!} \equal{} a \mod r$
now obviouly $ [\frac {n}{r}] \equal{} [\frac {ar \plus{} b}{r}] \equal{} a$ so $ C(n,r) \equal{} [\frac {n}{r}] \mod r$ and $ r$ divides $ C(n,r) \minus{} [\frac {n}{r}]$

i'm also trying to prove that the statement is false for all composite $ r$ but some cases prove themselves anoying. i managed to prove it false for all composite nubers $ r$ for which there exists a prime number $ p$ such that $ p$ devides $ r$ but $ p^2$ does not. i will try some more

ok i did it. i've been doing it wrong way.

if $ r$ is composite it is of a form $ p^xk$ where $ p$ is some prime number dividing $ r$, $ x$ is a natural number and $ k$ is some natural number relatively prime to $ p$. ($ k\equal{}1$ and $ x\equal{}1$ don't happen at the same time or $ r$ is prime)

we will show that the statement is false if we take $ n\equal{}p^xk\plus{}p$
$ C(p^xk\plus{}p,p^xk)\equal{}C(p^xk\plus{}p,p)\equal{}\frac{(p^xk\plus{}p)(p^xk\plus{}p\minus{}1)(p^xk\plus{}p\minus{}2)...(p^xk\plus{}1)}{p!}\equal{}(p^{x\minus{}1}k\plus{}1)\frac{(p^xk\plus{}p\minus{}1)(p^xk\plus{}p\minus{}2)...(p^xk\plus{}1)}{(p\minus{}1)!}$
now note that for every $ i,1\leq i\leq p\minus{}1$ $ p^xk\plus{}i\equal{}i \mod p^x$ and relatively prime to $ p$ therefore:
$ (p^{x\minus{}1}k\plus{}1)\frac{(p^xk\plus{}p\minus{}1)(p^xk\plus{}p\minus{}2)...(p^xk\plus{}1)}{(p\minus{}1)!}\equal{}p^{x\minus{}1}k\plus{}1 \mod p^x$
since $ x>1$ or $ k>1$, $ (p^{x\minus{}1}k\plus{}1)$ can't equal $ 1$ by modulo $ p^x$
obviously $ [\frac{p^xk\plus{}p}{p^xk}]\equal{}1$ and therefore $ C(p^xk\plus{}p,p^xk)$ can't be equal to $ [\frac{p^xk\plus{}p}{p^xk}]$ by modulo $ p^x$ and consequently by modulo $ r\equal{}p^xk$

therefore the statement is false for all composite $ r$

The proofs seem Ok and yes I forgot to mention that $ r$ is a prime!


inequalities
geometry
geometric transformation
reflection
triangle inequality
geometry proposed

We have $ab + ac > b^2 - c^2$ (since that is equivalent to $a > b - c$, which is true by the triangle inequality)

Also $ab + (b^2 - c^2) > ac$ (since that is equivalent to $(a + b + c)(b - c) > 0$ which is true since $b > c$)

And finally $ac + (b^2 - c^2) > ab$ (since that is equivalent to $(b + c - a)(b - c) > 0$ which is true since $b > c$ and $b + c > a$ by the triangle inequality.

Hence these numbers can be the sides of a triangle.

As for the geometric interpretation, I don't really see one  :?

The geometrical interpretation is the interesting part of this problem and what solves and the first part.

Arne, should I give a hint about the geometric interpretation?

[hide="Hint"]Reflect the point C in the A-altitude...[/hide]

Nice problem, Levi!

  Darij

Haha, thanks Darij, I see now. Very nice.

I have to go now, I might type it up later.

We can apply any remarkable inequalities to the triangle $\{ab,ac,b^2-c^2 \}$ !!!
Try you.

[quote="Virgil Nicula"]Let ABC be a triangle (b>c). Prove that $ab,\ ac,\ b^2-c^2$ can be the lengths of the sides of a certain triangle. Give a geometrical interpretation.[/quote]

[u]A geometrical interpretation.[/u] I note the point $E\in AC,\ AE=AB=c$, i.e. $EC=b-c$, $DB=DE$ and the point $D\in (BC),\ \frac{DB}{DC}=\frac cb$, i.e. $DB=\frac{ac}{b+c}$, $DC=\frac{ab}{b+c}$. We observe that for the triangle $DCE$ we have $DE=\frac{ac}{b+c}$, $EC=b-c$, $CD=\frac{ab}{b+c}$, i.e. $\frac{DC}{ab}=\frac{CE}{b^2-c^2}=\frac{DE}{ac}=\frac{1}{b+c}$. Therefore, the triangle $DEC$ is similarly with the demanded triangle.


can anyone make up ten

base 10 logarithmic function equations

and/or ten inverse function equations

i need to graph them but i'll take care of that

ive just been really busy with other things and havent been able to do this

Anyoe who can do this for me i sadly cannot give you anything in return but it will be GREATLY APPRECIATED!!!!!!

Can you clarify?
Do you mean functions like $f(x)=\log x$?

It takes about 5 seconds to make up a function
That means it will take you about 2 minutes to do this assignment

you are not that busy

$f(x) = log( Fill in random stuff here )$

Don't ask us to do your homework for you. This is obviously a homework assignment.

Pointless? Hardly.  All answers, including this one, are food for the troll.  Let it end here.


let a;b;c in R; prove that:
$ \frac {\frac {a^3 \minus{} b^3}{(a \plus{} b)^3} \plus{} \frac {b^3 \minus{} c^3}{(b \plus{} c)^3} \plus{} \frac {c^3 \minus{} a^3}{(c \plus{} a)^3}}{\frac {a \minus{} b}{a \plus{} b} \plus{} \frac {b \minus{} c}{b \plus{} c} \plus{} \frac {c \minus{} a}{c \plus{} a}}\geq 0$

[quote="mitdac123"]$ \sum \frac {a^3}{(a \plus{} b)^3} \minus{} \frac {3abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \equal{} \frac {1}{2}( \sum \frac {a}{a \plus{} b} )( \sum \frac { (ab \minus{} c^2)^2}{(a \plus{} c)^2(b \plus{} c)^2} )$
:)[/quote]

I don't understand


Prove that for all $n\geq 1,$ there exist $a_1<a_2<...<a_n$ a psotive integer numbers such that for $1\leq i<j\leq n$ we have$(a_j-a_i)|(a_j+a_i).$

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=native&t=16469

Pierre.

Too fast for me, Pierre  :)
Here's my solution:
Let $(u_n)_{n>0}$ such that $(u_1;u_2)=(1;2)$ and $u_{n+1}=u_n+\prod_{k<n}(u_n-u_k)$
It's easy to prove that : $u_q-u_p|u_n-u_p$ for $n>q>p$
So we can choose $a_i=2^{u_{n+1}}-2^{u_{n+1-i}}$, $1 \leq i \leq n$
In fact, we have more generally : $a_j-a_i|a_i$ for $1 \leq i <j \leq n$


induction
combinatorics unsolved
combinatorics

Let $A_{1}, ..., A_{n}$ be different subsets of an $n$-element set $X$. Show that there exists $x\in X$ such that the sets
$A_{1}-\{x\}, A_{2}-\{x\}, ..., A_{n}-\{x\}$ are all different.


What is the length of the shortest string that contains all strings of size 9 with characters ABCDEFGHI (repetitions are allowed)? Example: "AABBA" contains all strings of size 2 (AA,AB,BA,BB) with characters AB.

uugh, there is a problem like this in my schools discrete math book, that problem is stated where it is what is the minimum string length where there are 9 digits, and it gives every 4 digit code, but they only solved it with 2 digits, and it is pretty difficult, there was some special name for these sequences, but i don't remember, and i don't have the book, also this is a graph theory problem...also, i was thinking that there would be some way to use induction to find a formula but i don't know

the book said that these are the brujin cycles

Well, in the example using only A and B, the string begins with one string of size 2 (AA).  Then, each letter that is added to the example creates one more string of size 2 (A[b]AB[/b]).  If the string using ABCDEFGHI can be created in the same way (i.e. no repeating combinations), then the length of the string would be
$9+(9!-1)=\boxed{9!+8}$
Still, we would have to prove that the string can be created without repeating any combinations.


function
complex analysis
complex analysis unsolved

please help me to determine the number of zeros of the following function in the disk D(0,2) by using Rouche's Th:

f(z)=(z^2)*(e^z)-z, z\in D(0,2)

z is a complex number!~

[quote="LkNsngth"]Let $ h(z) \equal{} z^2e^z$ and let $ g(z) \equal{} \minus{} z$ Then on the circle of radius 2 centered around 0, the absolute values of the two functions are as follows:
$ |h(z)| \equal{} |z^2\parallel{}e^z| \equal{} |z|^2|e^x\parallel{}e^{iy}| \equal{} 4e^x$. We want to show that $ |h(z)| > |g(z)|$ so consider the minimum of $ |h(z)|$, this happens when $ x \equal{} 0$ and we get $ |h(z)| \equal{} 4$. Now $ |g(z)| \equal{} 2$ Therefore, we can apply Rouche's theorem telling us that $ g \plus{} h$ and $ h$ have the same zeros in $ D_2(0)$. Now what are the zeros of $ h(z) \equal{} z^2e^z$? Well $ e^z$ is never equal zero, so the only zeros are when $ z^2 \equal{} 0$. Thus there is a zero of multiplicity 2 at $ z \equal{} 0$[/quote]

I am afraid you are wrong, if you take z=-2, then |h(z)|=4e^(-2), and |g(z)|=2, obviouly |h(z)|<|g(z)| at this point, and z=-2 is right on the circle of |z|=2.

Anyway, I managed to solve the question by my own, thank you all!