[quote="Phelpedo"]Prove that there are infinitely primes of the form
(a) $6n+1$
(b) $6n+5$ (hint: ATTC that there are a finite number. Square the product and subtract 1)
(c) $4n+1$[/quote]

a) 7 is prime and so are 13 and 19
b) 11 is prime and so are 17 and 23
c) 5 is prime and so are 13 and 17

trivial

i think i used to know how to solve these types of problems correctly but i was too busy listening to my ipod during josh's lectures about primes and mods =/

Treething is just another wannabe spammer at this point.

What I got so far

b) If there were a finite number, you could multiply them together.  If the result were 1mod 6, you could add 4, and if 5mod6, you could add 6.  The resulting number would be 5mod6, and would be divisible by none of the 5mod 6 primes.  But then it would be a product of 1mod 6 primes, and such a product is always 1mod 6.  Contradiction.

[quote]I don't see how this constitutes a proof that there are infinitely many such primes.[/quote]
Sorry to go on a tangent, but an old  story is:

[hide]The question was asked to different students.

(The question,  of course,  was  a little stronger  theorem, that is  to prove all numbers of the form $6n+1$ are prime.)

Anyway here is a sample of the answers.

Math student:

Well 7 is prime, 13 is  prime, hence by induction all such numbers are prime.

Physics student:

Well 7 is prime, 13, is prime, 19 is prime , 25  (oh well .. experimental error) but then 31 is prime, 37 is prime ... Yeah all are primes..

Engineering student:

7 is prime, 13 is prime, 19 is prime, 25 is prime ....:)  :) [/hide]

[hide](Anyway,  if you are serious, the problem has appeared here a few times . if getting stuck ...  use search function creatively  to get some help) [/hide]

Yes, these have been asked several times before. Just a note on your hint for (b), another (perhaps simpler?) way:

[hide]
As you said, assume to the contrary (let's call the list of primes of this form $p_{1},p_{2},...,p_{n}$), and look at $6p_{1}p_{2}...p_{n}+5$ which must have at least one prime divisor of the form $6k+5$.
[/hide]

Thanks for the solution, but I'm looking for a way described by my hint.

Yes  that will work for the case b).

For  a) and c) one  way to think about it is:

[hide="Hint"]For a: . All (prime)  factors  of  $n^{2}+1$  which are greater than $2$  are $4k+1$ types.

[hide="because"]if $q > 2$ is a factor, we have, $n^{2}\equiv-1 \bmod q \implies n^{4}\equiv+1 \bmod q$
but we do know that  (due to F_Littele_T)

$n^{q-1}\equiv 1 \bmod q$
which implies (q-1) must be a  multiple of 4. QED
[/hide]
[hide="And"]And you can find something like $(2p_{1}p_{2}p_{3}\dots)^{2}+1$ whose all factors have to be  of the form $k+1$ [/hide]
[/hide]

[hide=" For c "]For c: . All (prime) factors of $n^{3}+1$ which are greater than $3$  are of the form $6k+1$[/hide]

(Similar logic as above )

[quote="Phelpedo"][quote="t0rajir0u"][quote="Treething"]a) 7 is prime and so are 13 and 19
b) 11 is prime and so are 17 and 23
c) 5 is prime and so are 13 and 17

trivial
[/quote]

 :huh:

I don't see how this constitutes a proof that there are [b]infinitely[/b] many such primes.[/quote]

This is the usual treethin method of proof[/quote]

Ah.  The MathCounts method.  I see  :P

[quote="K81o7"]But Gyan, for the 6k+1 case, suppose $n \equiv-1\mod{q}$?[/quote]
Yeah ... there is a hole in the argument given in the hint. It can be remedied.

[hide](How about, considering $(n^{3}+1)/(n+1)$ .. Can you make  that work?
 . even simpler could be:
with the  $P = (6 p_{1}p_{2}p_{3}\dots )$   use $P^{2}+P+1 = (P^{3}-1)/(P-1)$[/hide]

[quote="t0rajir0u"][quote="Treething"]a) 7 is prime and so are 13 and 19
b) 11 is prime and so are 17 and 23
c) 5 is prime and so are 13 and 17

trivial
[/quote]

 :huh:

I don't see how this constitutes a proof that there are [b]infinitely[/b] many such primes.[/quote]

For $a^{2}\equiv 1 \mod n$, there is an elementary (but really long) proof that there are infinitely many primes $\equiv a \mod n$ iff there is at least one such prime.  :wink:  :P


A ball is thrown straight up at an initial velocity of 54 feet per second. The height of the ball t seconds after it is thrown is given by the formula h(t) = 54t - 12t^2. How many seconds after the ball is thrown will it return to the ground?

(1) 9.2            (3) 4.5
(2) 6               (4) 4

Solve the equation!
[hide]$54t-12t^{2}=0$
$9t^{2}-2t=0$
$t\cdot (9-2t)=0$
$9-2t=0$
$t=4.5$
[/hide]
My best Rubik's cube time is 2:11:00. Just over two minutes![/hide]


algebra
polynomial
number theory open
number theory

suppose that $ P$ is a polynomial,and $ P\in\mathbb{Z}[X]$ and $ P$ don't have rattional root.
we say that $ P$ is good,if for every $ n\in\mathbb{N}$
there exist $ m\in\mathbb{Z}$ such that $ n|P(m)$.
and we say that $ P$ is weakly good if for every prime number $ n$,there exist $ m\in\mathbb{Z}$ such that $ n|P(m)$.
1)suppose that $ P$ is good,what is the smallest possible value for $ \deg P$?
2)suppose that $ P$ is weakly good,what is the smallest possible value for $ \deg P$?
3)suppose that $ P$ is good,is it possible that $ P$ be irreducible over $ \mathbb{Z}[X]$?if yes,what is the smallest possible value for $ \deg P$?
3)suppose that $ P$ is weakly good,is it possible that $ P$ be irreducible over $ \mathbb{Z}[X]$?if yes,what is the smallest possible value for $ \deg P$?

All this has already been discussed on the forum in many variants. The question is far from being open, but unfortunately proofs are very nontrivial for some of your questions (unless you consider Chebotarev theorem and class field theory easy...).

thanks dear [b]harazi[/b]... :)
can you answer  these questions,please?only answer,without proof...

1) 5
2) 5
3) doesn't exist.
4) doesn't exist.

Too late. In any case, Zetax is surely much more likely to answer such a question than me.  :)

Now I had some time, here are the links:
1),2): http://www.mathlinks.ro/viewtopic.php?t=110231
3),4): http://www.mathlinks.ro/viewtopic.php?t=90638

Related ones:
http://www.mathlinks.ro/viewtopic.php?t=34935
http://www.mathlinks.ro/viewtopic.php?t=64273


logarithms
factorial
number theory unsolved
number theory

$x=2005!$
Can you prove that there are more than 500 0's in $x$?

The number of trailing zeroes is $401 + 80 + 16 + 3 = 500$.  Then the problem is to show that there is a non-trailing zero in the decimal expansion.

This is not simple, but because $2005!$ has a lot of digits, it is very unlikely that there are no non-trailing $0$s in it  :D

t0rajir0u, the number of trailing zeros I know, I need to prove somehow that there is at least one zero that except
this 500 zeroes.

Hi t0rajir0u, if you care, this question comes after : http://www.mathlinks.ro/Forum/viewtopic.php?p=392799

I said there was indeed more than 500 0's, but, I'm unable to prove it rigorously  :oops:

Note that by taking logarithm and stupid calculation we get that the first digits from the left are $1069...$, thus there is another $0$.

you mean logarithm of $2005!$ ?!
Thanks ;)

ZetaX, I'm sorry , but I didn't get the answer like you,
Can you please explain to me how exactly have you calculated this ?
 :?

[quote="shyong"]something like calculating $\sum^{2005}_{k=1}\log k$ ? then take its fractional part and apply the antilog thing  :?[/quote]
Yes, this is what I have done , but I didn't get the right answer.

Yes, thats what I meant.
You get something like $5752.0290547543$ for that sum, take $10^{0.0290547543}$ and you will get $1.06918967034136$, thus the result.

i had a mistake in my calculation, now it's OK! :)

Ok, cool :)
And now, a question from me : is it possible to know the number of 0's in 2005! ? ;)

What's the CAS? :)

Moreover, I've coded a proggy that gives me the factorial of a number less than 2147, that's why I could tell there were more than 500 0's, but, I want to know if there is a rigorous way to find the number of 0's ;)

CAS = Computer Algebra System (like e.g. Maple with which I calculated that monster)

Ok, thanks :)
I'll have to get my hands on Maple someday!


email
LaTeX
function
geometry
incenter
cyclic quadrilateral
angle bisector

Hi guys,

here's the topic of the "[b]Oliforum contest[/b]", the competition for training that i'm arranging from some weeks    :)  .
i'll try to sumarize all details about it, be carefully.

[b]1-[/b] The contest is made of 2 rounds.

[b]2-[/b] Each round is made of some problems (from 3 to 5..), and i will give points 0 to 7: 6 for all correct solutions, 7 for [i]only[/i] elegants and correct solutions (similar to ML contest).

[b]3-[/b] It is tought for high-school student, and also for undergraduate or higher student, but you cannot use non-elementary-theory, otherwise the solution is invalid.

[b]4-[/b] All begins from italian-oliforum and the classification is unique.

[b]5-[/b] The first round starts on 17:00 of 28 september and it ends on 17:00 of 30 seprember, so you have 48 hours :) .
        [b]NB.[/b] All times (date and hours) are according to [i]ROME MERIDIAN (+2GTM), [/i]please attention!

[b]6-[/b] Enrollment is not necessary, it sufficies to send me solution.

[b]7-[/b] Time of arriving solution is not important.

[b]8-[/b] [i]Way of sending solutions[/i].
-          You need to send solutions to both email address: pao_leo88@hotmail.it and leonettipaolo@gmail.com in a [i]unique pdf file[/i] (so, with attachment)  of a right size;
-          The pdf file must be written with [i]Latex [/i] (or a way that I can understand);
-          You [i]rename the pdf file [/i]as your nickname (the same of yours in ML);
-          You need to write nothing in the email;
-          Try to be [i]clear[/i] in solutions!

{If [i](and only if)[/i] you are not able to create a pdf in latex model, you can send me a email in .doc format with solutions, as latex or in a way that i (and all correctors) can understand, or also with private message to me}.

Later, together with texts, I'll make a list below this topic where i'll write all names of who sent me solutions in right way.

For all doubts and questions, ask me in private messages :wink:

Thanks for the attentions and good luck

[i]Paolo Leonetti[/i]

[b]Problem 1[/b]
Let $ \sigma(\cdot): \mathbb{N}_0 \to \mathbb{N}_0$ be the function from every positive integer $ n$ to the sum of divisors $ \sum_{d \mid n}{d}$ (i.e. $ \sigma(6) \equal{} 6 \plus{} 3 \plus{} 2 \plus{} 1$ and $ \sigma(8) \equal{} 8 \plus{} 4 \plus{} 2 \plus{} 1$).
Find all primes $ p$ such that $ p \mid \sigma(p \minus{} 1)$.
[i](Salvatore Tringali)[/i]

[b]Problem 2[/b]
Define $ \phi$ the positive real root of $ x^2 \minus{} x \minus{} 1$ and let $ a,b,c,d$ be positive real numbers such that $ (a \plus{} 2b)^2 \equal{} 4c^2 \plus{} 1$.
Show that $ \displaystyle 2d^2 \plus{} a^2\left(\phi \minus{} \frac {1}{2}\right) \plus{} b^2\left(\frac {1}{\phi \minus{} 1} \plus{} 2\right) \plus{} 2 \ge 4(c \minus{} d) \plus{} 2\sqrt {d^2 \plus{} 2d}$ and find all cases of equality.
[i](A.Naskov)[/i]

[b]Problem 3[/b]
Let a cyclic quadrilateral $ ABCD$,  $ AC \cap BD \equal{} E$ and let a circle $ \Gamma$ internally tangent to the arch $ BC$ (that not contain $ D$) in $ T$ and tangent to $ BE$ and $ CE$. Call $ R$ the point where the angle bisector of $ \angle ABC$ meet the angle bisector of $ \angle BCD$ and $ S$ the incenter of $ BCE$. Prove that $ R$, $ S$ and $ T$ are collinear.
[i](Gabriel Giorgieri)[/i]

[b]Problem 4[/b]
Let $ m$ a positive integer and $ p$ a prime number, both fixed. Define $ S$ the set of all $ m$-uple of positive integers $ \vec{v} \equal{} (v_1,v_2,\ldots,v_m)$ such that $ 1 \le v_i \le p$ for all $ 1 \le i \le m$. Define also the function $ f(\cdot): \mathbb{N}^m \to \mathbb{N}$, that associates every $ m$-upla of non negative integers $ (a_1,a_2,\ldots,a_m)$ to the integer $ \displaystyle f(a_1,a_2,\ldots,a_m) \equal{} \sum_{\vec{v} \in S} \left(\prod_{1 \le i \le m}{v_i^{a_i}} \right)$.
Find all $ m$-uple of non negative integers $ (a_1,a_2,\ldots,a_m)$ such that $ p \mid f(a_1,a_2,\ldots,a_m)$.
[i](Pierfrancesco Carlucci)[/i]

[b]Problem 5[/b]
Let $ X: \equal{} \{x_1,x_2,\ldots,x_{29}\}$ be a set of $ 29$ boys: they play with each other in a tournament of Pro Evolution Soccer 2009, in respect of the following rules:
[list]i) every boy play one and only one time against each other boy (so we can assume that every match has the form $ (x_i \text{ Vs } x_j)$ for some $ i \neq j$);
ii) if the match $ (x_i \text{ Vs } x_j)$, with $ i \neq j$, ends with the win of the boy $ x_i$, then $ x_i$ gains $ 1$ point, and $ x_j$ doesn’t gain any point;
iii) if the match $ (x_i \text{ Vs } x_j)$, with $ i \neq j$, ends with the parity of the two boys, then $ \frac {1}{2}$ point is assigned to both boys.
[/list]
(We assume for simplicity that in the imaginary match $ (x_i \text{ Vs } x_i)$ the boy $ x_i$ doesn’t gain any point).
Show that for some positive integer $ k \le 29$ there exist a set of boys  $ \{x_{t_1},x_{t_2},\ldots,x_{t_k}\} \subseteq X$ such that, for all choice of the positive integer $ i \le 29$, the boy $ x_i$ gains always a integer number of points in the total of the matches $ \{(x_i \text{ Vs } x_{t_1}),(x_i \text{ Vs } x_{t_2}),\ldots, (x_i \text{ Vs } x_{t_k})\}$.
[i](Paolo Leonetti)[/i]

[i]Good luck![/i]   :wink:
For all doubts ask me in mp.

Name of users who send me solutions (if you are not here contact me): Ahwingsecretagent, NickNafplio, Zhero, Bugi, azjps, Palina.41, mavropnevma, geda, mod_2, TBPL, dario2994, exodd, Giuseppe_R, kn, Maioc92, String.
Here there are links where you can post your solution:
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=303518]Problem1[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=1642315#1642315]Problem2[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=1642318#1642318]Problem3[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=1642319#1642319]Problem4[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=1642312#1642312]Problem5[/url].

Hi Bugi, I hope to publish result up to some days, together with official solutions.. as I hope to get the second round on next monday at same hours  :wink:

Best regards
Paolo

Thanks for remembering me  :lol:  Here are the results for the first round:
mavropnevma- 7/7/0/7/7
TBPL- 7/7/0/3/7
exodd- 7/7/0/5/0
azjps- 7/5/0/7/0
Zhero- 7/3/0/7/0
Maioc92- 7/7/0/0/0
kn- 7/0/0/7/0
NickNapflio- 7/7/0/0/0
dario2994- 2/4/0/7/0
mod_2- 7/1/0/5/0
Bugi- 6/6/0/0/0
String- 7/0/0/0/0
Palina.41- 7/0/0/0/0
geda- 7/0/0/0/0
Ahwingsecretagent- 7/0/0/0/0
Giuseppe_R-1/0/0/0/0

And my personal congrats to mavropnevma-pdf-solutions!
Second round will be taken next week (I'll open a new topic for that one  :wink: )
Thanks again for partecipating!

Total sum on no. 3: 0 points!
12 points :) enough, although I over-estimated that no. 4...

[quote="Bugi"]Total sum on no. 3: 0 points![/quote]
Yes, I know.. although on ML that problem received a lot of solutions, I think that for the second round at least geometry problem will be simpler..

[quote="Bugi"].. although I over-estimated that no. 4...[/quote]
Sure  :P

Cheers Bugi and all other!
Ps. For all doubt about points contact me with pm

Ps1. I cannot edit the previous message, however this is a edit about Nicknapflio points: 7/7/0/2/0.

Will there be a second round?
If so, when will it start? :)

It'll start this Friday  :wink:

Look [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=1653289#1653289]here[/url]  :lol:


function
search
real analysis
real analysis unsolved

I need to solve this problem, I would appreciate any help.

Let F: Rn -> Rn be a continue function such that:

|| F(x) – F(y) ||   ≥   a|| x – y ||    for all  x,y ϵ Rn   and some a  > 0.

Show that F is a bijection and has a continue inverse function.

THANK YOU.

$ F(x)\equal{}F(y)\Rightarrow \parallel{}F(x)\minus{}F(y)\parallel{}\equal{}0\geq a\parallel{}x\minus{}y\parallel{}\Rightarrow \parallel{}x\minus{}y\parallel{}\equal{}0\Rightarrow x\equal{}y$, hence $ F$ is injection.

$ F$ has inverse function $ F^{\minus{}1}: F(\mathbb R^n)\rightarrow\mathbb R^n$, which is continuous becouse it is Lipschitz with constant $ a$.

[b]Ognjen09[/b],  :roll:  you've prooved the easiest part. Can you proove that this is surjection?
PS the problem is not so simple as someone can think.

it has already been proved on the forum, by Grobber if I remember it right. Use the search button ...


How many different ways for constructing a sequence of 16 positive integers that has no increasing or decreasing subsequence of 5 items
thx for your nice solution  :)

A sophisticated solution that may mean nothing to you: if $ w \in S_{16}$ has no five-term increasing or decreasing subsequence, then $ w \overset{\textrm{RSK}}{\longrightarrow} (P, Q)$ where $ \textrm{sh}(P) \equal{} \textrm{sh}(Q) \equal{} (4, 4, 4, 4) \equal{}: \lambda$ (since this is the unique partition of 16 with all parts of size at most 4 and at most 4 parts), and conversely.  Thus, the answer is $ (f^\lambda)^2$, and by the hook length formula we have $ f^\lambda \equal{} \frac{16!}{1 \cdot 2^2 \cdot 3^3 \cdot 4^4 \cdot 5^3 \cdot 6^2 \cdot 7}$.

The associated sequence in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url] is [url=http://www.research.att.com/~njas/sequences/A079402]A079402[/url].


[quote="Nbach"]Let $ a,b,c$ be three ral numbers such that $ abc =1$ Prove that:
      $ \frac{1}-{a^{2}+ab}+\frac{1}-{b^{2}+bc}+\frac{1}-{c^{2}+ca}\geq \frac{3}-{2}$[/quote]

is this

\[ \frac{1}{a^{2}+ab}+\frac{1}{b^{2}+bc}+\frac{1}{c^{2}+ca}\geq \frac{3}{2}\]

yes,I'm so sorry.You are true. How can we pass this problem? It has a gernerlize with 4 numbers.Please show me if you know it's solution.

[quote="Nbach"]Let $ a,b,c$ be three ral numbers such that $ abc =1$ Prove that:
      $ \frac{1}{a^{2}+ab}+\frac{1}{b^{2}+bc}+\frac{1}{c^{2}+ca}\geq \frac{3}{2}$[/quote]
It's wrong. Try $ a=-6,$ $ b=\frac{1}{3}$ and $ c=-\frac{1}{2}.$ :wink:
If you are mean that $ a,$ $ b$ and $ c$ are positive numbers then see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=81588


A point is selected inside an equilateral triangle. From this point perpendiculars are dropped to each side. Show that the sum of the lengths of these perpendiculars is equal to the altitude length.

[hide]
Denote side length as $a$.  Draw segments from the point to each vertice to form three triangles within the equilateral triangle.  The three perpendiculars are $p_1, p_2, p_3$.  $h$ is the altitude.

$\frac{a}{2} p_1 + \frac{a}{2} p_2 + \frac{a}{2} p_3 = \frac{a}{2} h$

$\frac{a}{2} (p_1 + p_2 + p_3) = \frac{a}{2} h$

$p_1 + p_2 + p_3 = h$

The sum of the areas of the three triangles is equal to the area of the entire equilateral triangle, making the sum of the perpendiculars equal to the altitude.[/hide]

[quote="MCrawford"]This is another problem that does not belong in Getting Started.

Remember that Getting Started is meant for problems that are on the approximate level of the middle 15 problems on an AMC exam -- problems that would be good for students who are currently getting into problem solving.[/quote]
I remember this problem was solved in Middle School section.
Any way, this is definitely a middle school problem in my opinion.

shobber,

Do you really think the typical middle school student is writing geometry proofs of this kind?

Even most of the top middle school students are not yet.

We have clear forum guidelines and we ask all users to follow them.  Sorry if you disagree with them, but it's much more useful to students to have problems grouped by difficulty and even if you disagree with the agegroup, please follow the grouping by difficulty.


Let $R$ be the midpoint of the hypotenuse ($AB$) of triangle $ABC$. Prove that the line joining the incenters of triangles $ACM$ and $BCM$ divide the area of the triangle $ABC$ evenly.

See the attached diagram for notation. The incenters are $P,Q$.

$r_{a}=\frac{{1\over 2}\cdot(ab/2)}{(a+{c\over 2}+{c\over 2})/2}={ab\over 2(a+c)}, r_{b}={ab\over 2(b+c)}$

$\triangle PP_{1}M\sim\triangle QQ_{2}M\implies \frac{r_{b}}{x}=\frac{{a\over 2}}{{b\over 2}-r_{a}+x}$

${a\over 2}x={ab\over 2(b+c)}\left(x+{b\over 2}-{ab\over 2(a+c)}\right)$

${a\over 2}x={ab\over 2(b+c)}\left(x+{bc\over 2(a+c)}\right)$

$x\left({a\over 2}-{ab\over 2(b+c)}\right)=\frac{ab^{2}c}{4(a+c)(b+c)}$

$x\frac{ac}{2(b+c)}=\frac{ab^{2}c}{4(a+c)(b+c)}$

$x=\frac{b^{2}}{2(a+c)}$

Then $CM={b\over 2}+x=\frac{b(a+b+c)}{2(a+c)}$

By symmetry, $CN=\frac{a(a+b+c)}{2(b+c)}$

Then

\begin{eqnarray*}[CMN] &=&{1\over 2}CM\cdot CN\\ &=&{ab\over 2}\cdot\frac{(a+b+c)^{2}}{4(a+c)(b+c)}\\ &=& [ABC]\cdot\frac{a^{2}+b^{2}+c^{2}+2ab+2ac+2bc}{4(ab+ac+bc+c^{2})}\\ &=& [ABC]\cdot\frac{2a^{2}+2b^{2}+2ab+2ac+2bc}{4(a^{2}+b^{2}+ab+ac+bc)}\\ &=&{1\over 2}[ABC]\end{eqnarray*}

[color=red]I liked very much the your proposed problem, [b][u]Theawesomest23 ![/u][/b]. Whence have it you ? Thank you a lot for it.[/color]

[quote="theawesomest23"][size=117][color=darkred]Let $M$ be the middlepoint of the hypotenuse ($BC$) of the right triangle $ABC$.
Prove that the line joining the incenters of the triangles $ABM$ and $ACM$ divides the area of the triangle $ABC$ evenly.[/color][/size][/quote]

[color=darkblue][b]Remark.[/b] Prove easily that in a right triangle $ABC\ ,\ AB\perp AC$ there are the identities : $\|\begin{array}{c}(a+b+c)^{2}=2(a+b)(a+c)\ .\\\\ (b+c-a)^{2}=2(a-b)(a-c)\ .\end{array}$

[b]Lemma[/b] ([u]one's own[/u]). In the rectangle $ABCD$ denote the points : $\|\begin{array}{c}M\in CD\ ,\ \widehat{MAC}\equiv\widehat{MAD}\\\\ N\in BC\ ,\ \widehat{NAB}\equiv\widehat{NAC}\\\\ X\in AB\cap MN\ ,\ Y\in AD\cap MN\end{array}$ $\Longrightarrow$ $[XAY]=[ABCD]\ .$
[hide="Proof of the lemma."]
[b]Proof 1.[/b] Denote : $\|\begin{array}{c}AB=b\ , AD=c\ , AC=a\\\\ b^{2}+c^{2}=a^{2}\end{array}$ $\Longrightarrow$ $\|\begin{array}{c}\frac{MD}{MC}=\frac{AD}{AC}\Longrightarrow \frac{MD}{c}=\frac{MC}{a}=\frac{b}{a+c}\Longrightarrow \boxed{\ MC=\frac{ab}{a+c}\ }\\\\ \frac{NB}{NC}=\frac{AB}{AC}\Longrightarrow \frac{NB}{b}=\frac{NC}{a}=\frac{c}{a+b}\Longrightarrow \boxed{\ NC=\frac{ac}{a+b}\ }\end{array}\ .$ Therefore,

$\|\begin{array}{c}\frac{DY}{NC}=\frac{MD}{MC}\Longrightarrow DY=\frac{ac}{a+b}\cdot\frac{c}{a}=\frac{c^{2}}{a+b}\Longrightarrow AY=AD+DY=c+\frac{c^{2}}{a+b}\Longrightarrow \boxed{\ AY=\frac{c(a+b+c)}{a+b}\ }\\\\ \frac{BX}{MC}=\frac{NB}{NC}\Longrightarrow BX=\frac{ab}{a+c}\cdot \frac{b}{a}=\frac{b^{2}}{a+c}\Longrightarrow AX=AB+BX=b+\frac{b^{2}}{a+c}\Longrightarrow \boxed{\ AX=\frac{b(a+b+c)}{a+c}\ }\end{array}$ $\Longrightarrow$ $AX\cdot AY=\frac{bc(a+b+c)^{2}}{(a+b)(a+c)}\ .$

Using the first identity from the above remark obtain $AX\cdot AY=2bc\ ,$ i.e. $[XAY]=[ABCD]\ .$

[b]Remark.[/b] $[XAY]=[ABCD]$ $\Longleftrightarrow$ $[DMY]+[BNX]=[MCN]$ $\Longleftrightarrow$ $DM\cdot DY+BN\cdot BX=CM\cdot CN$ $\Longleftrightarrow$

$\frac{bc}{a+c}\cdot\frac{c^{2}}{a+b}+\frac{bc}{a+b}\cdot \frac{b^{2}}{a+c}=\frac{ab}{a+c}\cdot\frac{ac}{a+b}$ $\Longleftrightarrow$ $bc^{3}+b^{3}c=a^{2}bc$ $\Longleftrightarrow$ $b^{2}+c^{2}=a^{2}\ ,$ what is truly.

[b]Proof 2[/b] (with areas). [/hide]
\[\blacktriangleleft\maltese\blacktriangleright \]
[b]Proof of the proposed problem.[/b] Denote the incircles $w_{1}=C(I_{1},r_{1})$ , $w_{2}=C(I_{2},r_{2})$ of the triangles $ABM$ , $ACM$ respectively , the touch-points $P\in AB\cap w_{1}$ , $N\in AC\cap w_{2}$ and the intersections $X\in AB\cap I_{1}I_{2}$ , $Y\in AC\cap I_{1}I_{2}\ .$  Apply the above lemma to the rectangle $APMN$ ( the ray $[AI_{1}$ is the bisector of the angle $\widehat{PAM}$ and the ray $[AI_{2}$ is the bisector of the angle $\widehat{NAM}$ ) : $AX\cdot AY=2\cdot AP\cdot AN$ $\Longrightarrow$ $AX\cdot AY=\frac{AB\cdot AC}{2}\ ,$ i.e. $[XAY]=\frac{1}{2}\cdot [ABC]\ .$[/color]


USAMTS
AoPSwiki
search
articles
MATHCOUNTS
AMC 8
\/closed

Just thinking...maybe people who get 100s on USAMTS (or golds) should get AoPS community awards.

I'm not sure what you mean by community awards, but I had a similar idea.  Like we establish a page for every person (perhaps on AoPSWiki page: [b]User:Username/Community Awards[/b]) where they can list all the Mock AMCs and stuff they have won.  I just thought it would be a cool incentive-type thing...

It really reminds me of an online resume in an odd way.  :)

mysmartmouth, I like that idea. Can we write pages about ourselves and our accomplishments? ( I probably won't since I haven't done anything in life )

Yea sure, I don't see any reason why anyone would object.  Standardizing would be the hard part.  Just type this in the AoPSWiki search, and then edit when it gives you the option:

User:mysmartmouth/Community Awards

Where mysmartmouth is replaced by your username

I think he's talking about [url=http://www.artofproblemsolving.com/Wiki/index.php/AoPS_Community_Awards]this[/url] but your idea sounds cool too.

You can easily create such a page for yourself on AoPS Wiki if you wish (in fact you seem to have done so already, mysmartmouth :P ), but there isn't a need to force others to do likewise.  I wouldn't feel comfortable keeping a public bulletin board of my activities.

If you were indeed talking about the community awards page on the wiki, you should have posted on the discussion page for the community awards article, or simply added a USAMTS section to the article without antecedent.  I happen to think that would be a bit redundant, since almost everybody in USAMTS has an AoPS account.  But that shouldn't stop you from adding it.

I think that there would be way way too many people who get golds from aops to list them all though. You want the awards listed to be specific methinks.

Oh, on the AoPSWiki Community Awards thingy, I don't think they would like it if we add USAMTS.  They took down AMC 8 and MathCounts because the lists would get too large, and I'm thinking USAMTS would be an even larger list...

If the list gets too long for a certain category, make a new page and link to it.  I don't see any problem with that.


Mine are simply,what you call "fighting,adventure,and racing" games. Here are some examples of them:
NAMCO
GUMBY VS THE ASTROBOTS
Ravensburger
Mario.
I bought all these games online from Planetonline at great discount. What are your favorite games.

Brother Bear  :D
though i am admittedly yet to play it

Real-time strategy. Age of Empires, Command and Conquer, etc.

Unless someone can convince me that turn-based strategy is better.

Something that's interesting

Like chess. I found this website with many varients of chess.

[quote="mathnerd314"]Unless someone can convince me that turn-based strategy is better.[/quote]

um chess? but thats probably it. turn-based strategy is only soo limited.

Well, chess, of course. But I mean stuff like Civilization. (I played it once and found it too complicated...)

I loved Civ once, I even played Galactic Civ. But the Galactic civ was a little too buggy.

Before my windows hard drive crashed, I played Civ II, test of time  a lot. I even tried the mission, but that was too hard, even on cheat mode...


geometry
circumcircle
trigonometry
geometry proposed

Let $ \triangle ABC$ with $ G$-centroid and $ O$-circumcenter.
The midperpendiculars of $ GA,GB,GC$ cut each other at $ A_1,B_1,C_1$.Prove that $ O$ is the centroid of $ \triangle A_1B_1C_1$.

Sorry if it is already posted.

we can easily see that $ A_1$ is the circumcenter of $ BGC$ and so $ OA_1$ passes through the midpoint of $ BC$.
so we have $ \angle GBC \equal{} \angle OA_1C_1$ and $ \angle OA_1B_1 \equal{} \angle GCB$.so obviously the triangle constructed with the medians of $ ABC$ is similar to $ \Delta A_1B_1C_1$ but we have
$ \frac {\sin \angle OA_1C_1}{\sin \angle OA_1B_1} \equal{} \frac {CG}{BG} \equal{} \frac {CC'}{BB'} \equal{} \frac {A_1B_1}{A_1C_1}$
$ A',B',C'$ are the midpoints of the sides of triangle $ ABC$.
so we r done!


hi:

Let a(n) be a sequence defined: a(0)=2, a(n+1)=2*a(n)^2 - 1
prove if a prime p divides a(n) then 2^(n+3) divides p^2 - 1.

This was solved before (actually, I like to think it's solved, but I'm not sure). However, I forgot the name of the topic. It was proposed for some TST and a forum member posted the problem. Sorry I can't give you more info..

I found it:

[url]http://www.mathlinks.ro/viewtopic.php?t=5174[/url]

I don't know if you consider it finished, but there are some helpful ideas over there.


Find [tex] f:R \to R [/tex] that satisfying the equation [tex] f(x)=ax+b [/tex] has roots [tex] \Leftrightarrow a^2+4b \geq 0 [/tex] :)

Consider family of linear functions $l_a(x)=2a(x-a)+a^2=2ax-a^2$, $a\in\mathbb{R}$ (these are tangents to $x^2$). We see, that $f(x)=l_a(x)-\delta$ has no roots for all $\delta>0$, so $f(x)\geq l_a(x)$ for all $a$, but our family envelops $y(x)=x^2$, so $f(x)\geq x^2$. Moreover, if $f(a)>a^2$ for some $a$ then equation $f(x)=l_a(x)$ can't have roots. Therefore $f(x)=x^2$.

I have thought that you will give chance for another people. :D
Here is another idea to prove that [tex] f(x_0) \geq {x_0}^2 \forall x_0 \in R [/tex].
We consider the equation [tex] f(x)=2{x_0} \cdot x-2{x_0}^2+f(x_0) [/tex] have [tex] x_0 [/tex] as a roots. Hence [tex] (4x_0)^2-8(x_0)^2+4f(x_0) \geq 0 \Rightarrow f(x_0)\geq{x_0}^2 [/tex]  :)

[quote="Anh Cuong"]I have thought that you will give chance for another people. :D [/quote]
Don't wait gifts from me next time.
 :lol:  :lol:  :lol:  :lol:  :lol:


number theory
greatest common divisor
induction
algorithm

[b]Prove than any two consecutive Fibonacci numbers $ F_{k}$ and $ F_{k+1}$ are relatively prime[/b]

enjoy!

[hide="hint"]Assume the opposite.  What do you know about $ F_{k-1}$ and $ F_{k+2}$?[/hide]
[hide="Solution"]Assume the opposite.  Let $ (F_{k}, F_{k+1})=d$ for d is a positive integer greater than 1.  Then, we have $ F_{k+1}=d\cdot x$ and $ F_{k}=d\cdot y$.  Now, we have:

$ F_{k-1}=F_{k+1}-F_{k}=d(x-y)$
Similarly, $ F_{k+2}=d(x+y)$

Thus, all Fibonacci numbers are divisible by d.  But the first Fib. number is 1, so $ d=1$ is a contradiction.

*Note: If k was positive, we would not need to look at $ F_{k+2}$, but this was never specified in the original problem.[/hide]

very good and rigorous solutions both of you!
but i have a question...how do u know when to use contradiction in a given proof?...like how did u know to use contradiction for this one?

[quote="LideR_M"]In general case we have: $ \textrm{gcd}\Big(f_{m},f_{n}\Big)=f_{gcd(m,n)}$
[hide="solution"]
http://www.math.hmc.edu/funfacts/ffiles/20004.5.shtml[/hide]

But in your case:
[hide="solution"]
Assume that there are some two consecutive Finnobacci numbers say $ f_{m},f_{m+1}$ and they have a common disivor $ d$, where $ d>1$. Becouse: $ f_{m+1}=f_{m}+f_{m-1}$, so $ f_{m-1}$ must have common disivor $ d$ too, so by induction all Finnobacci numbers lowers than $ f_{m}$ must also be divisible by $ d$... but it's false, becouse $ f_{1}=1$...[/hide][/quote]

You should note that to prove your general theorem, you use the lemma that the statement is true for $ m=n+1$. That website takes that as a lemma and does not prove it. The induction method is the best.

@krsattack: A slightly more clean way of proving the induction step is the following:

$ (F_{k},F_{k+1})=(F_{k}+F_{k+1},F_{k+1})=(F_{k+2},F_{k+1})$.

p.s. you were at AMSP right?

[quote="Altheman"]You should note that to prove your general theorem, you use the lemma that the statement is true for $ m=n+1$. That website takes that as a lemma and does not prove it.[/quote]
Ohh yes, I didn't observe that.
[quote="Altheman"]The induction method is the best.[/quote]
I agree, imho the most elementary proof is the best ;) but not always it's induction, for example to prove sth like $ \sum_{k=1}^{n}k^{2}=\frac{n(n+1)(2n+1)}{6}$ I would prefer to 'show it', not just confirm with induction.
[quote="mihail911"]how do u know when to use contradiction in a given proof?...like how did u know to use contradiction for this one?[/quote]
I don't know, you must see that :P
experience  :lol:

[quote="mihail911"][b]Prove than any two consecutive Fibonacci numbers $ F_{k}$ and $ F_{k+1}$ are relatively prime[/b]
[/quote]
[hide="solution"]
it is very easy if we use Eucleadian algorithm .
$ (F_{k+1},F_{k})=(F_{k+1}-F_{k},F_{k})=\cdots = (F_{2},F_{1})=1$
so they are relatively prime.
[/hide]


i posted a question here:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1138514#1138514]here[/url]

but they said it might belong here too

The "burning" sensation you feel is due to a compound called capsaicin.  Spicier foods and peppers have more capsaicin.  Water is not very effective in neutralizing the capsaicin and actually serves only to spread it out even more.  That is why water is not a good idea.  I'm not 100% sure why milk is so effective, but all dairy products work well.  In fact, I have also read the lemons and other acidic fruits help relieve the sensation.

[quote="JRav"]I'm not 100% sure why milk is so effective, but all dairy products work well.  In fact, I have also read the lemons and other acidic fruits help relieve the sensation.[/quote]
I'm definitely no bio or chem wiz, but I thought milk is effective because the fat somehow absorbs it. Although of course, that doesn't explain the lemons and stuff (actually never knew about that). Then again (on a practical level), extreme sour feels kinda similar to spiciness, so I don't know if lemons would be too helpful.

The fat neutralizing it is definitely a possibility.  The lemon thing I just read and was equally surprised.


We have triangle ABC. We draw the bisectors: BB',CC' which intersect themselves in point I. We also draw the bisector ID of angle BIC. Prove that the triangle B'C'D is equilateral if and only if
A=60.

Thank you very much!

PART I:

:ang: A = 60 => B'C'D equilateral:

spoiler:

[hide]

:ang: A = 60 => :ang: B + :ang: C = 120.

:ang: AB'B = :ang: B / 2 + :ang: C.

:ang: CC'A = :ang: C/2 + :ang: B.

:ang CC'A + :ang: AB'B = 3/2(:ang: B + :ang: C) = 180 => AB'IC' is cyclic.

=> :ang: CIB = 120 deg => :ang: BIP = 60 = :ang: PIC.

=>:ang: B'IC = 60 deg.

But then IPC  :cong: IB'C = > PB' :perp: IC = > :ang: IPB = :ang: IB'P = 30 deg.

Similarly, :ang: IC'P = :ang:IPC' = 30 deg => :ang: C'PB' = 60 deg.

Furthermore, :ang: C'B'I = :ang: B'C'I = :ang: IAB' = 30 deg since AB'IC' is cyclic. Therefore, :ang: C'B'P = :ang: B'C'P = 60 deg, and we have that C'PB' is equilateral.

[/hide]


I can't get this problem right. It seems easy but I keep getting the wrong answer.

$ln(1+6x)$

Write a partial sum for the power series which represents this function consisting of the first 5 nonzero terms. Also indicate the radius of convergence.

Power series I got:
$6x-((6x)^{2})/2+((6x)^{3})/3-((6x)^{4})/4$

The radius of convergence i'm getting is $1/6$ but the system tells me it's wrong. Did I miss something here?

Thanks for help!

Well, that's the first four nonzero terms and you should add $\frac{(6x)^{5}}5.$ But that does look correct and the radius of convergence is $\frac16.$


$N$ is an integer not divisible by $81$ which can be expressed as $a^{2}+b^{2}+c^{2}$ where $a,b,c$ are integers divisible by $3$. Prove that $N$ can be expressed in the form $x^{2}+y^{2}+z^{2}$ where $x,y,z$ are integers not divisible by $3$


Show that if $ n$ is a positive integer ad $ x_1,x_2,..,x_n$ are nonegative real that sum to $ 1$,then:
  $ \sum_{j\equal{}1}^{n}{\sqrt{x_j}}\sum_{k\equal{}1}^{n}\frac{1}{1\plus{}\sqrt{1\plus{}2x_k}}\leq \frac{n^2}{\sqrt{n}\plus{}\sqrt{n\plus{}1}}$

this is a problem difficult, hope everyone made comments
I can not apply the methods to solve problems TST for this problem.


Lawyers?  Trust me I'm a filthy rich spoiled teen so money ain't a problem. :D

Well, is it harvard?

Trust me, this is a really short question.

In the United States, law school is a postgraduate program, so there is no particular reason to choose colleges based on what the law school affiliated with the college is like--you still have to apply to law school separately as you finish your undergraduate degree program. There are, however, some law firms that prefer to hire only new lawyers whose UNDERGRADUATE degrees came from Ivy League colleges.


Suppose there is a polynomial function in my mind, a polynomial function of degree $ n$ such that each coefficient is a POSITIVE INTEGER .

Give me any integer value and I will tell you the corresponding value of the polynomial function.

$ Question 1$
How many such questions do you need to ask me to determine the polynomial expression completely?

$ Question 2$
Will your answer be any different if you can now ask for the value of the function for any real or complex number?

is the value of $ n$ given to us?

if so, then the answer to the 1st question is $ n$. Though i doubt the coefficients being +ve has any bearing on the answer  :)

[b]I'm bored:[/b] Suppose xxxyyyy thinks of a polynomial of degree n, and harsh_include guesses the polynomial with certainty after asking the values of the polynomial at k*n input numbers (any kind). Given only this information, find the set of possible values of k achievable in such a situation.

[hide]
Part 1:
Ask what p(1) is, then ask for what p(k) is, where k is an integer and k > p(1) . Now find the base k representation of this number, the digits are the coefficients of the polynomial. yay.  We need to see what p(1) is so that k is greater than all the coeficients and they can be base k digits. So 2 questions total.

Part 2:
Hmm... as for the second question, where we can ask for any real or complex value I have a feeling we may be able to do it in one question. Here's the idea: take a transcendental number, lets say pi.  Now I claim that p(pi)=q(pi) where p and q have positive integer coefficients  implies that p(x) = q(x).  Proof: suppose the opposite, then r(x)= p(x)-q(x) has pi as a root, but pi is transcendental so we have a contradiction. So I ask for the value at pi and there is only one such polynomial.[/hide]

Clever  :)

I was thinking along the lines of complex conjugates, but I think your solution is correct.


A $ \emph{rep\minus{}date}$ occurs when the number formed by the number of the month and the number of the day is the same as the last two digits of the year. For example, 9/8/98 is a rep date since $ 98\equal{}98$. In which year of the 21st century will the first rep-date occur?

Obviously it cannot have a tens or units digit of zero.  Thus, the answer is $ \boxed{2011}$.


trigonometry
geometry
trig identities
Law of Sines
Law of Cosines

To find side $ c$, first find $ \angle A$ using the Law of Sines, then find $ \angle C$ using the fact that the angles add up to 180, and then use Law of Cosines to find side $ c$.

Although I haven't done the problem, I would assume there are two answers because SSA isn't a valid congruence method when the angle is acute. Also, how do you find what angle a is when you know sine A? Use a calculator?

The questions tells. But has two answers ?

It tells what A is when Sine A=$ \frac{3\sqrt{21}}{4}$? If it does, could you either include that in your original post or say we can use calculators?

And yes, I'm 90% sure it has two answers.

I don't know, it's not my question, and I find is question error, because it want only side C.

How can it have two answers? The angle is fixed using the law of sines, and ASA is a valid congruence method. Angle A will always be $ sin^{ \minus{} 1}(\frac {3sin(60)}{\sqrt {7}})$.

Read this: http://en.wikipedia.org/wiki/Law_of_sines#The_ambiguous_case

Anyways, could someone provide a full solution?

[quote="luiseduardo"]Find the value of side C.

[img]http://img12.imageshack.us/img12/8527/tesa.jpg[/img][/quote]
good question luiseduardo
draw it geometricaly and u will find the two values of C  :wink:
u can draw acircle with diameter scr 7  :ewpu:

Yes, almonkez is correct: there are two values of $ c$: $ 1$ and $ 2$.

[quote="luiseduardo"]Here my calculate:

$ (\sqrt {7})^{2} \equal{} c^{2} \plus{} 3^{2} \minus{} 2.c.3.\frac {1}{2}$

$ 7 \equal{} c^{2} \plus{} 9 \minus{} 3c$

$ c^{2} \minus{} 3c \plus{} 2 \equal{} 0$

$ \Delta \equal{} 9 \minus{} 8 \equal{} 1$

$ \frac { \plus{} 3\pm\sqrt {1}}{2}$

$ c' \equal{} 2$

$ c'' \equal{} 1$[/quote]

You are right. The problem can be solved just by using cosine law.

[quote="10000th User"]There are two answers to this problem, and that's because there are two values of $ \cos\angle A$, which are supplementary angles (it's nice to know that 'ambiguous case' pointed out by Ihatepie).

Using Law of Sines, $ \sin A \equal{} \frac {3}{\sqrt7}\sin B \equal{} \frac {3\sqrt3}{2\sqrt7}$. Note that $ \sin(180^\circ \minus{} \theta) \equal{} \sin\theta$ but $ \cos(180^\circ \minus{} \theta) \equal{} \minus{} \cos\theta$.

[b]Triangle 1[/b]: $ A \equal{} \sin^{ \minus{} 1}\left(\frac {3\sqrt3}{2\sqrt7}\right)$.

Using Law of Cosines, $ c^2 \equal{} 3^2 \plus{} (\sqrt7)^2 \minus{} 2(\sqrt7)(3)\cos(120^\circ \minus{} A)$,

$ c^2 \equal{} 16 \minus{} 6\sqrt7(\cos120^\circ\cos A \plus{} \sin120^\circ\sin A)$

$ c^2 \equal{} 16 \minus{} 6\sqrt7\left( \minus{} \frac12\cdot\frac1{2\sqrt7} \plus{} \frac {\sqrt3}2\cdot\frac {3\sqrt3}{2\sqrt7}\right) \equal{} 4$

$ c \equal{} 2$

[b]Triangle 2[/b]: $ A \equal{} 180^\circ \minus{} \sin^{ \minus{} 1}\left(\frac {3\sqrt3}{2\sqrt7}\right)$

Using Law of Cosines, $ c^2 \equal{} 3^2 \plus{} (\sqrt7)^2 \minus{} 2(\sqrt7)(3)\cos(120^\circ \minus{} A)$,

$ c^2 \equal{} 16 \minus{} 6\sqrt7(\cos120^\circ\cos A \plus{} \sin120^\circ\sin A)$

$ c^2 \equal{} 16 \minus{} 6\sqrt7\left( \minus{} \frac12\cdot \minus{} \frac1{2\sqrt7} \plus{} \frac {\sqrt3}2\cdot\frac {3\sqrt3}{2\sqrt7}\right) \equal{} 1$

$ c \equal{} 1$

Compare between triangle 1 and triangle 2.

Hope this 'demystified' all of your questions (but it's great to see you all discussing and learning from each other) :)[/quote]

good method10000th User
but as i said u can make it geometrically without computing angle A :wink:
greeting

[quote="10000th User"][quote="almonkez"]good method10000th User
but as i said u can make it geometrically without computing angle A :wink:
greeting[/quote]The point of my solution is to show why there are two solutions, not competing for short solutions. It's lengthy but it has some flavor to see what the angles are and fundamentally compare where it is that is different.[/quote]
ok " 10000th User "
my point of view :
there are two solution because the line of the angle 60 in vertic B will intersict with the circle which center is C and its diameter equal scr 7
greeting to your thought
mr mohamed
almonkez

[quote="10000th User"]There is also a 'Euclidean' geometric solution without trigonometry (Law of Cosines is trig). ;)[/quote]
[color=blue][b]do u mean this solution my dear :wink: [/b][/color][url=http://www.servimg.com/image_preview.php?i=317&u=12899977][img]http://i89.servimg.com/u/f89/12/89/99/77/almonk11.gif[/img][/url]
 :coolspeak:

Cool almokez ;)
Nice answer for this simple problem.  :D
Thanks to everybody here.

[quote="luiseduardo"]Mr. Mohamed, I'd like your method, but I don't know how to understand, I'm new in trigonometric and geometry too. Could you know how to explain me more detalhed and simple how to understand or do your method ?

Thanks.[/quote]

ok my dear luiseduardo
at first we will draw AD prendicular to BC
now traingle ADB ( 30 . 60 . 90 ) traingle the side opposite to 30 equale the have of hypothnes ..... BD= 1/2 AB = c/2
now  Bc= 3 , BD= c/2 then DC = 3 - c/2  from phithagorth theorem on traingle ADB,ADC u will reach easy :wink:
look again to the solution
 mr mohamed
almonkez

Ah ok ;)
Good mr. mohamed. Very good, know I understand, thanks for the wonderful explanation.


function
trigonometry
geometry
geometric transformation
reflection
calculus
integration

Is it true that if a real function $f$ is differentiable everywhere and $f'(x)$ is differentiable in some point, then $ f$ is $C^{1}$ class on some neighbourhood of this point?

Take the function $g$ s.t. for every $n\in\mathbb N^*$ its graph on the first half of the interval $\left(\frac 1{n+1},\frac 1n\right)$ is the graph of  on $\frac 1{n^2}\cdot \sin\frac 1x$ on $(0,1)$, properly scaled and translated (so that $0$ corresponds to $\frac 1{n+1}$ and $1$ corresponds to $\frac{\frac 1{n+1}+\frac 1n}2$), it's graph on the second half of that interval is the reflection of the graph on the first half of the interval in the perpendicular bisector of the segment, and it's value is $0$ in all points of the form $\frac 1n$. Of course, we also take $g$ to be $0$ in $0$.

Now take $f(x)$ to be the integral from $0$ to $x$ of $g$. It's easy to see that $f'=g$ everywhere, the derivative of $f'$ in $0$ is $0$, but $g=f'$ has a discontinuity point in every $\frac 1n$.

Grobber, how do you calculate the derivative of $f'$ in 0?


A cube showing the odd numbers 1, 3, 5, 7, 9
and 1 1 one number on each of its six faces is rolled three times, and the numbers on the top faces are added. What is the probability that the sum is even?

Any three odd numbers sum to an odd number, so the answer is 0. :rotfl:


trigonometry
algebra
binomial theorem
calculus
calculus computations

Hello everyone!

I found these series expansions of $ \sin kx$ and $ \cos kx$ and I really don't know where they come from  :( :

\[ \sin kx = k\sin x-k\frac{k^{2}-1^{2}}{3!}\sin^{3}x+\ldots+(-1)^{n}k\frac{\U(k^{2}-1^{2}\I)\U(k^{2}-3^{2}\I)\ldots\U(k^{2}-\U(2n-1\I)^{2}\I)^{2}}{\U(2n+1\I)!}\sin^{2n+1}x-\ldots\]

with $ k$ odd.

And

\[ \cos kx = 1-\frac{k^{2}}{2}\sin^{2}x+\frac{k^{2}\left(k^{2}-2^{2}\right)}{2\cdot 3\cdot 4}\sin^{4}x-\ldots+(-1)^{n}\frac{k^{2}\left(k^{2}-2^{2}\right)\ldots\left(k^{2}-(2n-2)^{2}\right)}{2\cdot 3\ldots 2n}\sin^{2n}x\pm\ldots\]

with $ k$ even.

The book I'm reading only states that these expansions come from the elements of trigonometry... but I don't find them anywhere...

Hope someone can bring some light to the subject  :)

Thank you

For example, for k = 3:

$ \sin 3x \equal{}\sin(x\plus{}2x) \equal{}\sin x\cos 2x\plus{}\cos x\sin 2x \equal{}\sin x(1\minus{}2\sin^{2}x)\plus{}$
$ \cos x (2\sin x\cos x) \equal{}\sin x\minus{}2\sin^{3}x\plus{}2\sin x(1\minus{}\sin^{2}x)$
$ \sin 3x \equal{} 3\sin x\minus{}4\sin^{3}x$.

For bigger k, the spirit is the same. Start by writing $ \sin(kx) \equal{}\sin[x\plus{}(k\minus{}1)x]$ and use the formula for the sine of a sum (note that (k-1) is an even number, with makes easier to expand its cosine.

You can use de Moiver's theorem.

$ (\cos x\plus{}i\sin x)^{k}\equal{}\cos (kx)\plus{}i\sin (kx)$
Now expand LHS by binomial theorem ... da ta!


floor function
search
number theory unsolved
number theory

[b]hey listen up:

I'm looking forward to prove that the number $ x \equal{} 17183$ is a prime number.

I've proved it using the following :
we said that n is prime if $ \forall n > 1,$ ,then there is no prime number p , such that  $ p \leqslant \sqrt n$ and $ p|n$ ,

I just looking to another proof. :)

thanks

[/b]

i don't think there is a simple proof that would lead you to a quick result

[b]
Hello Jure the frEEEk

Maybe someone know another method to prove it.
[/b]

If you really want something that badly you can have this, an equation producing the n-th prime number:
$ p_n\equal{}\sum_{k\equal{}2}^{2^n}k\left\lfloor\left(1\plus{}\left|n\minus{}\sum_{j\equal{}2}^k\left\lfloor\left(\sum_{i\equal{}2}^j\left\lfloor1\minus{}\frac ji\plus{}\left\lfloor\frac ji\right\rfloor\right\rfloor\right)^{\minus{}1}\right\rfloor\right|\right)^{\minus{}1}\right\rfloor \cdot\left\lfloor\left(\sum_{i\equal{}2}^k\left\lfloor1\minus{}\frac ki\plus{}\left\lfloor\frac ki\right\rfloor\right\rfloor\right)^{\minus{}1}\right\rfloor$

[b]Waw

It's so badly equation. :o

I'll wait until someone knows a good method.

Thanks polskimisiek  :lol:

[/b]

Well, I think that for your number the best method is simply the simplest, the one you showed. I only posted that equation to show you that there is no need for trying to search for complicated methods while we have a simple one :wink:

[b]Thanks  t0rajir0u:

I saw the website and I think the Fast deterministic tests are very useful here.

I'll try using them.

 :D [/b]


I donot have a complete solution. :blush:  But i reduced the problem to prove that the diagnols of the quadrilateral are perpendicular. :o Can anyone help me please?

[quote="ks1458"]ABCD is a quadrilateral with AB = CD. The diagonals intersect at X. Circles are drawn on diameters BC and AD. X lies outside both circles. Show that the four tangents from X to the circles have equal length.[/quote]
Consider a triangle we $ BC$ and $ AD$ are perpendicular.
Then let $ \triangle BCX$ and $ \triangle ADX$ be similar. If $ \triangle ADX$ is larger than $ \triangle BCX$, then the tangents from $ X$ to the circle on $ AD$ is longer than the tangent from $ X$ to $ BC$. Which is clearly the opposite of what's asked.

Do I misread the problem?

[quote="feliz"]How can the diagonals be perpendicular if $ X$ lies outside the circles? And how can $ AD$ be perpendicular to $ BC$ if $ AB \equal{} CD$?

But I also think the problem is wrong...[/quote]

Assuming that the question was correct this is what we had to prove :lol:


quadratics
algebra
number theory unsolved
number theory

Determine all pair of integers (a,b) such that (a^2)/(2*a^2*b-b^3 +1) is a positive integer.

djuro your problem is a bit different to doccocaubai's. check the square exponents. but i don't know if this is a mistake or not. anyway here's the solution.

if $ a \equal{} 0$ or $ b \equal{} 0$ we obviouslly get a solution so assume that's not the case.
$ \frac {a^2}{2a^2b \minus{} b^3 \plus{} 1} \equal{} k$ is equivalent to $ a^2 \equal{} \frac {k(b^3 \minus{} 1)}{2kb \minus{} 1}$ for a non-zero integer $ k$
because $ gcd(2kb \minus{} 1,k) \equal{} 1$ we need $ x \equal{} \frac {b^3 \minus{} 1}{2kb \minus{} 1}$ to be an integer.

here goes the calculations:D
$ b^3 \minus{} 1 \equal{} (b^2 \plus{} b \plus{} 1)(b \minus{} 1) \equal{} x(2kb \minus{} 1) \equal{} x((2k \minus{} 1)b \plus{} b \minus{} 1)$
$ (b^2 \plus{} b \plus{} 1 \minus{} x)(b \minus{} 1) \equal{} xb(2k \minus{} 1)$ which means, because the left side of the equation has to be devisable by $ b$, that $ x$ is of the form $ x \equal{} yx \plus{} 1$ for an integer $ y$ (if $ b \equal{} 1$ obviously no solutions). putting it into the expression

$ (b^2 \plus{} b(1 \minus{} y))(b \minus{} 1) \equal{} b(yb \plus{} 1)(2k \minus{} 1)$
$ (b \plus{} 1 \minus{} y)(b \minus{} 1) \equal{} (yb \plus{} 1)(2k \minus{} 1)$

which yealds the quadratic equation $ b^2 \minus{} 2kyb \minus{} 2k \equal{} 0$. analysing it we get the equation $ k^2y^2 \plus{} 2k \minus{} z^2 \equal{} 0$ for an integer $ z$
analysing this one to we get that $ 1 \plus{} y^2z^2$ is a perfect square which is clearely imposibble unless $ y \equal{} 0$ or $ z \equal{} 0$

if $ y \equal{} 0$ than $ b^3 \minus{} 1 \equal{} 2kb \minus{} 1$ which leads to $ 2a^2 \equal{} b^2$ which is clearely imposible
if $ z \equal{} 0$ than $ k \equal{} \minus{} 1$ and $ a^2 \equal{} \frac {b^3 \minus{} 1}{2b \plus{} 1}$. which again gives no new solutions

the only solutions are when either of $ a$ and $ b$ is $ 0$


Three congruent circles of radius $1$ intersect at a common point. A larger circle is circumscribed about them, tangent to each of them at one point. Consider the midpoint of one of the diameters of the smaller circles; call it $P$. What is the length of the arc along the bigger circle containing all the points that are within 2 units of $P$?

[hide][url=http://wangsblog.com/jeffrey/]See blog post[/url]

On a side note, these links to my blog actually pertain to the question at hand, in case the intentions of my posting these links was ever put into question...[/hide]

Anyone? I thought it was a pretty good problem... Intermediate problems die to fast if you don't revive them...

[hide="another hint"]
The big circle has radius $2$. Draw line segments of radius $2$ from $P$ to points $X$ and $Y$ the big circle. Then draw the radii of the big circle to $X$ and $Y$. You can calculate the angle intersected by the arc.[/hide]

OK if no one wants to do it, I'll post my solution...

Consider the point $X$ mentioned in my previous post. Form triangle $OXP$ where $O$ is the center of the big circle. We have $OX = XP = 2$ and $OP = 1$, so the triangle is isosceles. Then draw the perpendicular from $X$ to $OP$, to make a right triangle, from which we find $\sin{\angle XOP} = \frac{1}{4} \Rightarrow \angle OXP = \arcsin{\frac{1}{4}}$, which we will leave in that form.

So $\angle XOY = 2\angle XOP = 2\arcsin{\frac{1}{4}}$ so arc $XY$ has length $r \theta = 4\arcsin{\frac{1}{4}}$.


LaTeX
geometry
geometric transformation
reflection
radical axis
geometry proposed

let (C) be a circle and L is a line wich is tangent to circle in point N.
NM is the diameterof circle (C).A is a stable point on NM.draw a circle wich its center (O),is online L and goes through A. suppose that C and D are the intersectionsof this circle with line L.and MC and MD intersect circle (C)in pointsP and Q.
prove that the chord PQ goes through a stable point.

It isn't clearly ! My dear Darij, where are you ?

what do u mean levi ?
wich part of it u dont understand? :(

Please, learn and you a little LaTeX ! I tired reading and thus I don't understood.
See http://www.mathlinks.ro/Forum/viewtopic.php?t=52043

First of all, "stable" should be replaced by "fixed". I think the problem is understandable now :)

Let A' be the reflection of A in L. Then A' lies on (O). Suppose we have another circle (O') centered on L and passes through A, then it also passes through A'. Now consider the radical axes among the circles (C),(O),(O'), which are lines PQ,P'Q',MN. Thus the radical center is the fixed point we are looking for.


Let $x,y,z$ are the non -negative numbers and $xyz-x-y-z=2$. Find the minimum of $x^2+y^2+z^2$. :)

Of course, it is extremely easy. Direct approach with AM-GM gives $x+y+z\geq\frac{3}{2}$. You can also use the fact that there are positive numbers $a,b,c$ such that $x=\frac{b+c}{a}, y=\frac{c+a}{b}, z=\frac{a+b}{c}$. ;)

thank Mr.harazi but can you prove it with yensen inequality :huh:   :blush:

I don't know what you want. Why should I prove it with Jensen when it solves itself with AM-GM?


integration
trigonometry
calculus
logarithms
calculus computations

[quote="borislav_mirchev"]
3.$ \int \ \frac {cosx \plus{} xsinx}{(x \plus{} cosx)^{2}}\,dx$[/quote]

I observe that $ (x \plus{} \cos x)' \equal{} 1 \minus{} \sin x$ and little bit of trial and error
So:
$ \int \frac{ \cos x \plus{} x\sin x}{(x \plus{} \cos x)^{2}}\,dx$
$ \equal{} \int \ \frac{ \cos x  \plus{} x\sin x \plus{} \cos x\sin x \minus{} \cos x\sin x}{(x \plus{} \cos x)^{2}}\,dx$
$ \equal{} \int \ \frac{ \cos x (1 \minus{} \sin x) \plus{} (x \plus{} \cos x)\sin x }{(x \plus{} \cos x)^{2}}\,dx$
$ \equal{} \int \ d\left( \frac{\cos x}{x \plus{} \cos x}\right)$
$ \equal{} \frac{ \cos x}{x \plus{} \cos x} \plus{} C$

For the second integral I suppose Virgil Nicula may say something.
http://www.mathlinks.ro/Forum/viewtopic.php?t=168335 - here he mentioned Abel's substitution...

Excellent solutions but only on the third integral. What do you think about the first and the second integrals?

i think the first one is a typo :maybe:

I think the question has to be

$ \int \ x\sqrt{\frac{2sin(x^{2}\plus{}1)\minus{}sin2(x^{2}\plus{}1)}{2sin(x^{2}\plus{}1)\plus{}sin2(x^{2}\plus{}1)}}\,dx$

rather than what u have written

If that is right, i can give the soln :)

For the second question,

$ \int \ \frac {1}{(x^{2} \plus{} a^{2})\sqrt {x^{2} \plus{} b^{2}}}\,dx$

Put $ x \equal{} \frac {1}{t}$ Hence $ dx \equal{} \frac { \minus{} 1}{t^2}$

The given integral transforms to

$ \int \ \frac { \minus{} t}{(1 \plus{} a^2t^2)\sqrt {1 \plus{} b^2t^2}} \,dt$

Now put $ 1 \plus{} b^2t^2 \equal{} z^2$

The integral transforms to $ \int \ \frac {1}{b^2[ (1 \minus{} \frac {a^2}{b^2}) \plus{} \frac {a^2z^2}{b^2}]} \,dz$

which can be easily integrated by the substitution $ \frac {az}{\sqrt {b^2 \minus{} a^2}} \equal{} \tan \theta$

It is an excellent solution to this tricky integral. I'm wondering how did you get this solution.
What do you think is there simpler solution to this integral?

About the first integral .... I know it is not wrong. The source is:
http://www.sosmath.com/CBB/viewtopic.php?t=36202&highlight=int+frac+sin
Why do you think this integral is wrong? Integral you mentioned is also interesting.
Please, post your solution or only the steps.

Why i thought that the question should be

$ \int \ x\sqrt{\frac{2sin(x^{2}\plus{}1)\minus{}sin2(x^{2}\plus{}1)}{2sin(x^{2}\plus{}1)\plus{}sin2(x^{2}\plus{}1)}}\,dx$

was that ur question couldnt be simplified but this one neatly simplifies.

Here i use the identity $ \sin 2A \equal{}2 \sin A.\cos A$

$ \int \ x\sqrt{\frac{2sin(x^{2}\plus{}1)\minus{}sin2(x^{2}\plus{}1)}{2sin(x^{2}\plus{}1)\plus{}sin2(x^{2}\plus{}1)}}\,dx$

$ \equal{} \int \ x\sqrt{\frac{2sin(x^{2}\plus{}1)\minus{}2sin(x^{2}\plus{}1)\cos (x^2\plus{}1)}{2sin(x^{2}\plus{}1)\plus{}2sin(x^{2}\plus{}1)\cos(x^2\plus{}1)}}\,dx$

$ \equal{} \int \ x\sqrt{\frac{2sin(x^{2}\plus{}1)[1\minus{}\cos (x^2\plus{}1)]}{2sin(x^{2}\plus{}1)[1\plus{}\cos(x^2\plus{}1)]}}\,dx$

$ \equal{} \int \ x\sqrt{\frac{1\minus{}\cos (x^2\plus{}1)}{1\plus{}\cos(x^2\plus{}1)}}\,dx$

Now we can use the identities $ \frac{1\minus{}\cos A}{2} \equal{}\sin^2 \frac{A}{2}$
and  $ \frac{1\plus{}\cos A}{2} \equal{}\cos^2 \frac{A}{2}$

The given integral becomes

$ \int \ x\sqrt{\frac{2\sin^2(\frac{x^2\plus{}1}{2})}{2\cos^2(\frac{x^2\plus{}1}{2})}}\,dx$

$ \equal{} \int \ x.\tan (\frac{x^2\plus{}1}{2}) \,dx$

Now put $ \frac{x^2\plus{}1}{2} \equal{} \theta$

Hence $ xdx \equal{} d \theta$

The integral now becomes

$ \int \ \tan \theta \, d\theta$

$ \equal{}\log \mid \sec \theta \mid$

$ \equal{}\log \mid \sec (\frac{x^2\plus{}1}{2}) \mid$

[quote="borislav_mirchev"]It is an excellent solution to this tricky integral. I'm wondering how did you get this solution.
What do you think is there simpler solution to this integral? [/quote]

I dont think you can have an easier soln. :)
The condition $ \mid b \mid \ge \mid a \mid$ is essential bcoz in the last step  i have used the substitution $ \frac {az}{\sqrt {b^2 \minus{} a^2}} \equal{} \tan \theta$
i should have made sure that the quantity under the radical $ b^2\minus{}a^2$ is positive and the given condition $ \mid b \mid \ge \mid a \mid$ makes it positive always. :)

[quote="borislav_mirchev"]
About the first integral .... I know it is not wrong. The source is:
http://www.sosmath.com/CBB/viewtopic.php?t=36202&highlight=int+frac+sin
Why do you think this integral is wrong? Integral you mentioned is also interesting.
Please, post your solution or only the steps.[/quote]

Maybe there is a typo there too :wink:

:) who knows there may be really a typo. What about using sin(A+B)?
Thank you very much for the time spent with these integrals.


limit
real analysis
real analysis solved

If $x_n = \{\sum_{k=1}^{n} \frac{1}{\sqrt{k}}\}$, then for each x in [0, 1] exists $k_n$ such that $x_{k_n}$ converges to x. ({} means fractionary part)

Let $x_n=\sum_{k=1}^na_k$, where $a_k$ is a sequence with  these properties:

$a_k>0,\,\lim_{k\to\infty}a_k=0,\,\sum_{k=1}^{\infty}a_k=\infty.$

These are the only properties of $a_k=\frac1{\sqrt{k}}$ that we are using.

We want to show that $\forall\,x\in(0,1)$ and $\forall\,\epsilon>0,\,\exists\,n$ such that $|\{x_n\}-x|<\epsilon.$

To avoid some notational awkwardness, make $\epsilon <\min(x,1-x).$

Chose $N$ such that for $n>N,\,a_k<\epsilon.$ Let $m$ be an integer such that $m>x_N.$ Now let $n$ be the largest integer for which $x_n\le m+x.$ We know there is such an $n$ because $x_n\to\infty$ by the divergence of the series. Then $|x_n-(m+x)|<|x_n-x_{n+1}|<\epsilon.$

Note that $|x_n-(m+x)|=|(x_n-m)-x|=|\{x_n\}-x|$ and we are done.


Find all positive integers $n$ for which $-5^4 + 5^5 + 5^n$ is a perfect square. Do the same for $2^4+2^7+2^n$.

For the first case, $n = 0, 1, 2, 3, 4$ gives no solutions (calculate to verify), so $n > 4$.  This gives us $5^4(5 - 1 + 5^{n-4})$ is square, which implies $5^{n-4} = k^2 - 4 = (k+2)(k-2)$, so we need two powers of five that differ by four; the only two are $1, 5$, which gives us $n = \boxed{5}$ as our only solution.

For the second case, $n = 0, 1, 2, 3, 4$ gives no solutions (calculate to verify), so $n > 4$.  This gives us $16(9 + 2^{n-4})$ is square, which implies $2^{n-4} = k^2 - 9 = (k+3)(k-3)$ for some integer $k$.  This tells us that we need two powers of two that differ by six; the only two are $2, 8$, which gives us $n = \boxed{8}$ as our only solution.


probability
limit
logarithms
search
probability and stats

Given a Markov chain $ \{X_n\}_{n = 1}^{\infty}$,s.t.
\[ p_{kj} = \frac {1}{k},k + 1\le j\le 2k\]
.
Denote $ p_{n} = P(\exists m\in %Error. "mathbox" is a bad command.
{N},\textbf{s.t.} X_m = n)$. Show that
\[ \lim_{n\rightarrow\infty} np_n = \frac {1}{2\ln 2 - 1}\]
. :)

you could prove that $ p(n) = \sum_{[\frac{n}{2}] \leq k \leq n-1} \frac{p(k)}{k}$, where $ [x]$ is the smallest integer that is greater or equal to $ x$. Nevertheless, it does not seem so easy to conclude from that point.

Of course, seeing this problem, the first thing that one would like to use is the renewal theorem (we talked about a discrete version of that result [url=http://www.mathlinks.ro/viewtopic.php?search_id=580256576&t=183969]here[/url]): consider $ Y_n=\ln(X_n)$ and notice that this is ALMOST a Markov chain on $ \mathbb{R}$ such that $ Y_n = \sum_1^k U_k$ where $ (U_k)_k$ is an iid sequence of random variables distributed on $ [0;\ln(2)]$ as $ P[U \leq x] = \exp(x)-1$. The renewal theorem tells you that $ \lim_{x \to \infty} E[ \text{number of } Y_k \in [x;x+\delta]] = \frac{\delta}{E[U]} = \frac{\delta}{2 \cdot \ln(2) - 1}$. The conclusion then would follow since \[ P(n) = E[ \text{number of } X_k \in [n;n+1[ \, ] \approx E[\text{number of } Y_k \in [\ln(n); \ln(n+1)[ \,]\] and $ \delta = \ln(n+1)-\ln(n) \approx \frac{1}{n}$.

There are a lot of approximations, but it should not be so hard to make this rigorous. I was just curious if you had the right (in case this is an assignment) to use the renewal theorem (which is a non-trivial result). Btw, I am curious if one can easily conclude onely using the fact that $ p(n) = \sum_{[\frac{n}{2}] \leq k \leq n-1} \frac{p(k)}{k}$ ?


im sure we have to use the quadratic formula of t = -b +/- sqr(b squared - 4ac) / 2a

after expanding f(t)

This is SO close to being the Cauchy-Schwarz inequality.

Consider that a hint: look up some proof of Cauchy-Schwarz and make the appropriate modifications to turn it into this case.

[quote="probable"]im sure we have to use the quadratic formula of t = -b +/- sqr(b squared - 4ac) / 2a
[/quote]

No.  But you do need to use a fact about the discriminant.  If $ at^2\plus{}bt\plus{}c\geq 0$ for all real values of $ t$, what does this imply about the discriminant $ \Delta \equal{} b^2\minus{}4ac$?


geometry
circumcircle
geometric transformation
homothety
ratio
Euler
geometry unsolved

Another solution, easier to think of:

See that $ ADBC$ is an isosceles trapezium. Let $ DQ$ be perpendicular to $ BC$, with $ Q$ lying on $ BC$. Let $ M$ be the midpoint of $ BC$. Now $ PM \equal{} MQ \equal{} \frac 1 2 AD$. Let $ F$ be the intersection of $ AM$ and $ PD$. The triangles $ PMF$ and $ DAF$ are similar with the ratio $ \frac 1 2$ (since $ PM \equal{} \frac 1 2 AD$). Therefore, $ AF : FM \equal{} 2 : 1$, which means that $ F \equal{} G$ since the centroid divides the median in the ratio $ 2 : 1$. Now since $ P, F, D$ are collinear by assumption, we have the desired result.

[quote="Moonmathpi496"][quote="Grands"]In a triangle $ ABC$ are given the centroid $ G$ (cross point of the medians), $ AP$ is the height from $ A$ to $ BC$ ($ P$ lies on $ BC$). Point $ D$ lies on the circumcircle of $ ABC$ and $ AD\parallel{}BC$. Prove that $ P$, $ G$ and $ D$ lie on a line.

Any ideas how to solve it?[/quote]
[hide="Big Hint"]Think of nine point circle and homothety.
[/hide]

[hide="Solution"]A homothety with center $ G$ and ration $ \minus{} 2$ sends the nine point circle to circumcircle $ \omega$. This homothety would send P so some point $ D'$ on the circle $ \omega$ (As $ P$ is one the nine point circle).
But $ AD' \parallel BC$. So, $ D' \equiv D$
[/hide][/quote]

Thanks for the quick reply, [b]Moonmathpi496[/b]. I had also started with the same homothety, but have some questions.

A homothety with center $ G$ and ration $ \minus{} 2$ sends the nine point circle to circumcircle $ \omega$.
I suppose this could be easily proven with the Euler line?

But $ AD' \parallel BC$.
This was where I actually had doubts in my solution. How could I prove this? Thanks in advance for the help.

[quote="Grands"]In a triangle $ ABC$ are given the centroid $ G$ (cross point of the medians), $ AP$ is the height from $ A$ to $ BC$ ($ P$ lies on $ BC$). Point $ D$ lies on the circumcircle of $ ABC$ and $ AD\parallel{}BC$. Prove that $ P$, $ G$ and $ D$ lie on a line.

Any ideas how to solve it?[/quote]
see also here for reference application of this nice property: http://www.mathlinks.ro/viewtopic.php?p=1474886#1474886

[quote="Grands"]
A homothety with center $ G$ and ration $ \minus{} 2$ sends the nine point circle to circumcircle $ \omega$.
I suppose this could be easily proven with the Euler line?

But $ AD' \parallel BC$.
This was where I actually had doubts in my solution. How could I prove this? Thanks in advance for the help.[/quote]
Let $ A'$ be the midpoint of $ BC$. Also $ \frac{AG}{GA'}\equal{}2$.
Now homothety maps respective sides to parallel sides. So, that homothety maps $ A' \to A$ and $ P \to D'$. So, $ AB \equiv PA' \parallel AD'$ and $ P$ lies on the nine point circle implies that $ D'$ lies on $ \omega$.
I hope that it is more clear now. :)

Dear Mathlinkers,
1. Let H, O be the orthocenter, the center of the circumcircle of ABC
and A' the second meetpoint of AH with the circumcircle.
2. P is the midpoint of A'H
3. O is the midpoint of A'D
3. 2GO = GH
4. G is the centroid of HA'D   and we are done
Sincerely
Jean-Louis


group theory
abstract algebra
superior algebra
superior algebra unsolved

[quote="JackSchmidt"][quote="Dragonflare82"]Let S4 be the symmetric group of permutations over 4 objects.
Let A4 be the alternating group of S4 (subgroup of even permutation)

How could one show that every subgroup of order 12 in S4 is isomorphic to A4?[/quote]

Well S4 only has 24 elements.  Just write down a complete list of its subgroups.  In fact it is not a bad idea to carefully draw the subgroup lattice and frame it for a few years by your desk.

A general method: Every subgroup of index 2 contains the derived subgroup, but the derived subgroup of S4 is A4.[/quote]

I am not sure what a derived subgroup is. Can you further explain?


(Titu98) Let f : N -> N be such that f(n + 1) > f(f(n)) for all n E N. Prove that f(n) = n for all n.

It is from IMO 1977.it can be solved by induction.

[quote="danilorj"]ok. I've seen the solution at kalva. Anyway, I understood almost everything except this part :
f(n) > f(n-1) > ... > f(1), so f(n) ≥ n - 1 + f(1) ≥ n. [color=red]Why f(n)>=n-1 + f(1)>=n ?[/color] Where did this come from?[/quote]

Because these are integers and $ f(a)>f(b)$ means $ f(a)\geq f(b)+1$

So $ f(2)\geq f(1)+1$
So $ f(3)\geq f(2)+1\geq f(1)+2$
So $ f(4)\geq f(1)+3$

So $ f(n)\geq f(1)+n-1$

It came from Bulgaria,it was 6-th problem.

I was trying to prove like this: Suppose $ f(n)>n$ we have  $ f(n)\geq n+1$ then $ f(f(n))\geq f(n+1)$ what is a contradiction. Now put $ f(n)<n$ then $ f(n)\leq n-1$ implies that $ f(f(n))\leq n-1$ what is also a contradiction. Then $ f(n)=n$. Is there something wrong? In fact, I didn't understand why do I have to prove that f is an increasing function.
Any explanation would be grateful.

[quote="danilorj"]I was trying to prove like this: Suppose $ f(n)>n$ we have  $ f(n)\geq n+1$ then $ f(f(n))\geq f(n+1)$ what is a contradiction.[/quote]
this step is wrong without the case that $ f$ is increasing,because if $ f$ is not increasing then you cant say that $ f(n)\geq n+1$ implies $ f(f(n))\geq f(n+1)$...

and also for your last post:
$ f: \mathbb{N}\to \mathbb{N}$ so $ f(n) \in \mathbb{N}$ so $ f(n)\geq 1$ for all $ n \in \mathbb{N}$...
so $ f(1)\geq 1$
now add $ n-1$ to both sides of this inequality to get that:
$ f(1)+n-1\geq n$...

Solution : let $ d$ be the least element of the range of $ f$ .By the well ordering principle afore mentioned , such that an element $ d$ exists and it is unique . Let $ m\in\mathbb{N}$ be such that $ d=f(m)$ .If $ m>1$ ,then we see that $ d=f(m)>f(f(m-1))$. Thus we gat a new element $ f(m-1)$ whose $ f$- value is smaller than $ d$ . But this contracdicts the choise of $ d$ as the least element of $ {f(n): n\in\mathbb{N}}$ . We conclude that $ m=1$ .

Now consider the set $ {f(n): n\in\mathbb{N},n\ge2}$ .We can infer , as in the previous paragraph , that this the set has the least element and this least value in $ f(2)$.Moreover $ f(1)<f(2)$by the choise of $ f(1)$ . In fact $ f(1)=f(2)$ forces $ f(1)>f(f(1))$ contracdicting the choise of $ f(1)$ . Then can be continued to get :

$ f(1)<f(2)<...<f(n)<....$(1)
Note that $ f(1)\ge1$ .This bound along with (1) shows that $ f(k)\gek$ for all natural numbers $ k$ .Supose $ f(k)>k$ for some $ k$ .Then $ f(k)\ge k+1$ .Using (1) we have $ f(k+1)\le f(f(k))<f(k+1)$ contracdition .We conclude that $ f(k)=k$ for all natural numbers $ k$ .

[quote="danilorj"]ok. Now another point. In the solution at kalva. it is said that f(m-1) belongs to { n+1, n+2, n+3, .. }. Later on it is said that  f(m) is the smaller element of { f(n+1), f(n+2), f(n+3), ... }.  Why the set of elements are different ?
and why he takes m such $ f(m)\geq m+1$?[/quote]

he said that:
$ f(m-1) \in \{n+1,n+2,n+3,...\}$
$ \Rightarrow f(f(m-1)) \in A=\{f(n+1),f(n+2),f(n+3),...\}$...
now by the problem`s hypothesis we have that $ f(m)\geq f(f(m-1))$ so $ f(m)$ isnt the smallest value in the set $ A$...
now according to all these steps he claims that $ f$ is increasing...
so $ x>y$ implies that $ f(x)>f(y)$
now note that if for some $ a$ we had $ f(a) \not = a$ then we would have either $ f(a)>a$ or $ f(a)<a$...
but note that we had $ f(1)<f(2)<f(3)<...$ but in the case that $ f(a)<a$ the numbers $ f(1),...,f(a)$ are in the set $ \{1,2,...,a-1\}$ so two of them must be equal which is contradicting with the fact that $ f(1)<f(2)<...$
so we must have $ f(a)>a \Rightarrow f(a)\geq a+1$...


[b]AΣΚΗΣΗ 1-ΝΙΚΟΣ ΡΑΠΑΝΟΣ[/b]
Δίνεται κύκλος $C$ και σημείο $P$ του επιπέδου του(εξωτερικό).Θεωρούμε δύο μεταβλητές ημιευθείες με αρχή το σημείο $P$ που τέμνουν τον κύκλο στα σημεία $A,A'$ και $B,B'$.Αν οι περιγεγραμμένοι κύκλοι των τριγώνων $PAB,PA'B'$ τέμνονται σε σημείο $R$, να βρείτε το γεωμετρικό τόπο του $R$.

Καλησπερίζω τον φίλο μου Νίκο κι χρόνια πολλά

Βρήκα μία λύση .Έστω $O$ το κέντρο του $C$ και $R$ η ακτίνα του .
Θεωρούμε αντιστροφή $H$ κέντρου $P$ και ακτίνας $r^{2}=PA\cdot PA'=PO^{2}-R^{2}=ct$

άρα $r^{2}+R^{2}=PO^{2}$ . Άρα ο κύκλος αντιστροφής έχει ακτίνα
$r=OF$ όπου $F$ είναι το σημείο επαφής της εφαπτομένης από το $P$ στον $C$ .Θα έχουμε $H(R)=R'$ όπου το R' είναι το σημείο τομής  του περιγεγραμμένου κύλου του τριγώνου $B'BR$
(γιατί? :wink: ) .Τώρα από το θεώρημα Miquel έχουμε ότι οι ευθείες $PR,\ AB,\ A'B'$ συντρέχουν το οποίο μεταφράζεται όμως στο ότι οι ριζικοί άξονες των κύκλων συντρέχουν (εννοώ τους κύκλους $C$ και τους πειγεγραμμένους των $PAB$ και $PA'B'$  .To οποίο όμως σημαίνει ότι και ο ριζικός άξονας των ΣΤΑΘΕΡΩΝ κύκλων $C$ και του κύκλου αντιστροφής παιρνάει από το $R'$ .Άρα το $R'$ κνείται σε ευθεία και συνεπώς το $R$ κινείται σε κύκλο . :)

Νίκο πες μου τη γνώμη σου .

Άρα χθες μπλόφαρες ε??  :D  .(που μου έλεγες για λύση χωρίς μετασχηματισμό  :P )  .

Περιμένω να βάλλεις την άσκηση που δεν μπορούσες να λύσεις (ξέρεις ποια . :lol: )

Δίνεται ευθεία $l$ και εκτός αυτής δύο σταθερά σημεία $A,B$.Θεωρούμε δύο μεταβλητές ημιευθείες με αρχή το σημείο $A$ που είναι κάθετες μεταξύ τους και τέμνουν την ευθεία $l$ στα σημεία $C,D$.Να αποδείξετε ότι ο περιγεγραμμένος κύκλος του τριγώνου $BCD$ διέρχεται από δύο σταθερά σημεία.

[hide="ΑΣΚΗΣΗ 2 - ΛΥΣΗ"]Έστω $w$ ο περιγεγραμμένος κύκλος του $\triangle ACD$.

Έστω $K$ η προβολή του $A$ πάνω στην ευθεία $l$. Τότε θα είναι  $CK \cdot DK = AK^{2}$. Επειδή το $\triangle ACD$ είναι ορθογώνιο, το $K$ βρίσκεται στο εσωτερικό του $CD$. Άρα και στο εσωτερικό του κύκλου $w$

Η δύναμη του (σταθερού) σημείου $K$ ως προς τον $w$ είναι σταθερή και ίση με $-AK^{2}$. Και ο $w$ διέρχεται από σταθερό σημείο $B$. Επομένως θα τέμνει την ημιευθεία $BK$ (ξανά) σε σταθερό σημείο $B'$,  τέτοιο ώστε $BK \cdot KB' = AK^{2}$.
Τα σταθερά σημεία λοιπόν είναι τα $B,B'$

Λουκάς[/hide]

akrivos....
ps.lew na valo liges akomi oraies askiseis geometrias san ki autes giati pisteuo oti einai didaktikes kai antiprosopeutikes.


[i]Find all solutions in positive real numbers a, b, c, d to the following system:[/i]

[i]a[/i] + [i]b[/i] + [i]c[/i] + [i]d[/i] = 12,
[i]abcd [/i]= 27 + [i]ab[/i] + [i]ac[/i] + [i]ad[/i] + [i]bc[/i] + [i]bd[/i] + [i]cd[/i].

Use Newton's inequality, which holds for all real numbers a,b,c,d.

(d_1)^2>=d_0*d_2=d_2

((a+b+c+d)/4)^2>=(ab+ac+ad+bc+bd+ad)/6
144/16>=(ab+ac+ad+bc+bd+ad)/6
ab+ac+ad+bc+bd+ad>=54

So abcd = 27+ab+ac+ad+bc+bd+ad >= 81.

Yet a+b+c+d=12, we have 3>=(abcd)^(1/4) by AM-GM, abcd <= 81. So abcd=81, but equality occurs iff a=b=c=d=3.

BMO problems are usually much easier than problems from other national olympiads, for some reason...

[quote="fairyofwind"]
144/16>=(ab+ac+ad+bc+bd+ad)/6
ab+ac+ad+bc+bd+ad>=54
[/quote]

This is in err? Your inequality sign switched.

My solution:

Let k = ab + ac + ad + bc + bd + cd. By conditions of the question, k > 0.

By AM-GM,

(a + b + c + d) / 4 :ge: 4:sqrt:(abcd)
12 / 4 :ge: 4:sqrt:(abcd)
3 :ge: 4:sqrt:(abcd)
81 :ge: abcd
81 :ge: 27 + k
54 :ge: k.

However,

k / 6 :ge:  :sqrt:(abcd)
k / 6 :ge:  :sqrt:(27 + k)
k2 / 36 :ge: 27 + k
k2 - 36k - 972 :ge: 0
(k - 54)(k + 18) :ge: 0

So either k :ge: 54 or k :le: -18. However, k > 0, so k :ge: 54. By 54 :ge: k, k = 54. Thus by condition of equality of AM-GM, ab = ac = ad = bc = bd = cd, so a = b = c = d = 3.

my solution (essentially the same as osiris, but i dont use the extra variable)

by AM-GM, ab + ac + ad + bc + bd + cd >= 6 :sqrt:(abcd)

adding this to the first equation, we have

abcd - 6 :sqrt:(abcd) - 27 >= 0
( :sqrt:(abcd) + 3)(:sqrt:(abcd) - 9) >=0

so (abcd)^1/4 >= 3.

now 12 = a+b+c+d >= 4 * (abcd)^1/4

so 3 <= (abcd)^1/4 <=3, AM-GM implies a=b=c=d=3.


let x^5 + 5x^4 - 6x +4 = 0       Determine forall X

hello, this equation can only solved by a numerical way.
Sonnhard.


complex analysis
complex analysis unsolved

Discuss the convergence of series
$\sum_{n=1}^{\infty}\frac{1}{n^{p}}$   if:
(i) $Re(p)>1$
(ii) $Re(p)\leq 1$.

(i) converges absolutely, because $|n^{p}|=n^{\mathrm{Re}\,p}$.
(ii) does not converge absolutely (see above). But is convergent if $\mathrm{Re}\,p>0$ and $\mathrm{Im}\,p\ne 0$, by [url=http://en.wikipedia.org/wiki/Dirichlet%27s_test]Dirichlet's test[/url] with $a_{n}=1/n^{\mathrm{Re}\,p}$ and $b_{n}=1/n^{i\mathrm{Im}\,p}$.

[b]mlok[/b], I think you are wrong. The series diverges for $Re(p)\leqslant1$. This can be shown if one uses Abel's transform

Thanks for correction. I mixed up $n^{a}$ and $a^{n}$ again.

For global analytic continuations of this series, see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=56857]this post[/url].


In triangle ABC, points D, E, F are marked on sides BC, CA, AB, respectively, so that BD/DC=CE/EA=AF/FB=2.
Show that the triangle formed by the lines AD, BE, CF has area 1/7 that of the triangle ABC.

Isn't this very easy though? If $T_1,T_2,T_3$ are the "carpets" (:D) $ADC,BEA,CFB$ respectively, then the area we are looking for is $S_{ABC}-(\sum S_{T_i}-\sum_{\hbox{cyl}}S_{T_i\cap T_j})$. I'm sure you can easily compute the values present in this formula: for all $i\ne j$ we have $S_{T_i\cap T_j}=\frac 1{21}\cdot S_{ABC}$, while for all $i$, $S_{T_i}=\frac 13\cdot S_{ABC}$.


how to distinguish between permutations and combinations........

I think this would go in Mathcounts forum, but anyways, permutations is the number of ways you can assembe something if order matters. Combinations is when order doesn't matter.

An example of a permutation problem:

How many ways can five books be arranged on a shelf?

Something like ABCDE would be different from CDABE, because order does matter. That would mean it's a permutation problem with an answer of 5! = 120.

Comparatively, a combination problem:

How many ways can a 3-person committee be chosen out of five?

If the people were A, B, C, D, and E, a committee such as ABC is the same thing as BAC, because order doesn't matter. That's why it's a combination problem, with an answer of $ \binom{5}{3} \equal{} 10$.

[url=http://www.artofproblemsolving.com/Alcumus/Landing.php]Alcumus[/url] can be very helpful, not only in combinations and permutations but more advanced counting and probability strategies. There are video lessons by Richard Rusczyk himself, as well as many problems with full solutions.


geometry
analytic geometry
graphing lines
slope
trigonometry
ratio
ARML

Compute the area of the region enclosed by the x-axis, the y-axis, and the angle bisector of the larger angles formed by the lines y=x+5 and y=3x-1

Let $O$ be the coordinate origin, $A, B$ be the intersections of the line $k:\ \ y = x + 5$ with the x- and y-axes, $C, D$ the intersections of the line $l:\ \ y = 3x - 1$ with the x -and y-axes, $X$ the intersection of these 2 lines and $P, Q$ the intersection of the external angle bisector of the angle $\measuredangle AXC \equiv \measuredangle BXD$ with the x- and y-axes. The points $A, B, C, D$ have coordinates:

$A = (0, -5),\ \ B = (0, 5),\ \ C = (1/3, 0),\ \ D = (0, -1)$

Coordinates of the intersection $X$ are obtained by solving the set of 2 equation of the 2 lines: $X = (3, 8)$. The slopes of the lines $k, l$ are $\tan \alpha = 1, \tan \beta = 3$, the slope of the internal angle bisector of the angle $\measuredangle AXC \equiv \measuredangle BXD$ is $\tan{\frac{\alpha + \beta}{2}}$ and the slope of the external angle bisector of this angle is $-\frac{1}{\tan{\frac{\alpha + \beta}{2}}}$, because the internal and external angle bisectors are perpendicular to each other and the product of their slopes is equal to -1.

$\tan{\frac \alpha 2} = \frac{\sin{\frac \alpha 2}}{\cos{\frac \alpha 2}} = \sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}} = \frac{1 - \cos \alpha}{\sin{\alpha}} = \frac{\sqrt{1 + \tan^2{\alpha}} - 1}{\tan \alpha} = \sqrt 2 - 1$

$\tan{\frac \beta 2} = \frac{\sqrt{1 + \tan^2{\beta}} - 1}{\tan \beta} = \frac{\sqrt{10} - 1}{3}$

$\tan{\frac{\alpha + \beta}{2}} = \frac{\tan{\frac \alpha 2} + \tan{\frac \beta 2}}{1 - \tan{\frac \alpha 2} \tan{\frac \beta 2}} = \frac{3(\sqrt 2 - 1) + \sqrt{10} - 1}{3 - (\sqrt 2 - 1)(\sqrt{10} - 1)} =$

$= \frac{-4 + \sqrt 2(3 + \sqrt 5)}{2(1 - \sqrt 5) + \sqrt 2(1 + \sqrt 5)} = \frac{-4 + 2\sqrt 2G^2}{-\frac 4 G + 2\sqrt 2 G} = G = \frac{1 + \sqrt 5}{2}$

where $G$ is the golden ratio, $G^2 = \frac{3 + \sqrt 5}{2} = 1 + G$ and $\frac{-1}{G} = \frac{1 - \sqrt 5}{2} = 1 - G$. Thus the slope of the external angle bisector $PQ$ passing through the point $X = (3, 8)$ is

$-\frac{1}{\tan{\frac{\alpha + \beta}{2}}} = -\frac{1}{G} = 1 - G = \frac{1 - \sqrt 5}{2}$

and the equation of this angle bisector is

$y - 8 = -\frac 1 G (x - 3) = (1 - G)(x - 3) = \frac{1 - \sqrt 5}{2}\ (x - 3)$

The intersections $P, Q$ of this angle bisector with the x- and y-axes then have coordinates

$P = (3 + 8G, 0), \ \ Q = (0, 5 + 3G)$

and the area of the right angle triangle $\triangle POQ$ is

$|\triangle POQ| = \frac 1 2\ OP \cdot OQ = \frac 1 2\ (3 + 8G)(5 + 3G) = \frac 1 2\ (15 + 49G + 24G^2) =$

$= \frac{39 + 73G}{2} = \frac{151 + 73 \sqrt 5}{4} \doteq 78.56$

Another way to calculate the segments $OP, OQ$ is to get the sides $AC, CX, AX$ of the triangle $\triangle ACX$ and the sides $BD, DX, BX$ of the triangle $\triangle BDX$ and then use the fact that the external angle bisector $PQ$ cuts the sides $AC, BD$ of these 2 triangles externally in the ratio of the adjacent sides:

$AC = \frac{16}{3},\ \ CX = \frac{8 \sqrt{10}}{3},\ \ AX = 8 \sqrt 2$

$BD = 6,\ \ DX = 3 \sqrt{10},\ \ BX = 3 \sqrt 2$

$OP = AP - OA = \frac{AC \cdot AX}{AX - CX} - OA =$

$= \frac{\frac{16}{3} \cdot 8 \sqrt 2}{8 \sqrt 2 - \frac{8 \sqrt{10}}{3}} - 5 = \frac{16}{3 - \sqrt 5} - 5 = 7 + 4 \sqrt 5$

$OQ = DQ - OD = \frac{BD \cdot DX}{DX - BX} - OD =$

$= \frac{6 \cdot 3 \sqrt{10}}{3 \sqrt{10} - 3 \sqrt 2} - 1 = \frac{6 \sqrt 5}{\sqrt 5 - 1} - 1 = \frac{13 + 3 \sqrt 5}{2}$

$|\triangle POQ| = \frac 1 2\ OP \cdot OQ = \frac 1 4\ (7 + 4 \sqrt 5)(13 + 3 \sqrt 5) = \frac{151 + 73 \sqrt 5}{4}$

which is the same result as before.

This is from the Chicago ARML homework due next week... I'm not a moderator, but please stop posting these.

Let $(X,Y)$ be the point on the angle bisector in question, $(X,Y)$ satisfies the equation

$\frac{|X-Y+5|}{\sqrt{1^2+(-1)^2}}=\frac{|3X-Y-1|}{\sqrt{3^2+(-1)^2}}\Longleftrightarrow \sqrt{5} |X-Y+5|=|3X-Y-1|$.

By drawing the graph, we have the equation of the angle bisector in question is ${(3-\sqrt{5})x+(\sqrt{5}-1)y-(5\sqrt{5}}+1)=0$,

therefore the desired area is $\frac{1}{2}\cdot \frac{1+5\sqrt{5}}{3-\sqrt{5}}\cdot \frac{5\sqrt{5}+1}{\sqrt{5}-1}=\frac{151+73\sqrt{5}}{4}$, which is deduced by yetti.

my friend was asking me these questions, but I have no ideas how to do these, so I thought I can find some help here. sorry about that, ThAzN1


[quote="not_trig"]Can 1 be written in that form?[/quote]

$p=c^{2}+1=a^{2}+b^{2}$ obviously has solutions for prime $p=4k+1$

[quote="not_trig"]Can 1 be written in that form?[/quote]
Yes, $6^{2}+17^{2}-18^{2}=1$.

[quote="PenguinIntegral"]Proof all integers are in the form $a^{2}+b^{2}-c^{2}$ for a,b,c positive integers and $a<b<c$.[/quote]

Do you think induction might work? Since induction can also work backward, if we prove all integers in one direction we can prove the other half as well. Just a thought.

Ya know, this looks awfully like the Law of Cosines...

[hide="Solution"] Negative integers are trivial.  $b^{2}-c^{2}$ can be any integer $\equiv 0, 1, 3 \bmod 4$ so $a = 0$ or $1$.

For positive integers, let $c = b+1$ so that the expression is $a^{2}-2b-1$.  For an integer $n$ pick a large square $a^{2}$ of opposite parity and let the difference be $2b+1$, which ensures that $a < b$. [/hide]
Edit:  For $1$ I have the slightly smaller solution $1 = 4^{2}+7^{2}-8^{2}$ by this method.

Edit 2:  For the sake of completeness, $0$ can be expressed in this form by any Pythagorean triple.


Let $b$ a symmetric bilinear form which is nondegenerate in the $n$-th dimensional space $E$ over some algebrically closed field $K$. What is the condition that the space must satisfy (but also sufficient) in order that any self-adjoint endomorphism with respect to $b$ is diagonalisable.

harazi, it seems quite difficult.
Do you have simpler than what is presented in
[i]The Principal Axis Theorem Over Arbitrary Fields [/i], by
David Mornhinweg, Daniel B. Shapiro, K. G. Valente
American Mathematical Monthly, Vol. 100, No. 8 (Oct., 1993) , pp. 749-754?

It is extremely difficult, yet I saw it given at oral ens. Unfortunately, I have no solution for it.


The range of any set of data is the largest number subtracted by the smallest number. As this chart shows, the largest number is $ 70$ and the smallest is $ 10$, the range is $ 70\minus{}10\equal{}\fbox{60}$.


Let $ tan\alpha_1,tan\alpha_2,tan\alpha_3$ be $ 3$ roots of equation $ x^3 \plus{} ax^2 \plus{} bx \plus{} c \equal{} 0$ and $ tan\beta_1,tan\beta_2,tan\beta_3$ be $ 3$ roots of equation $ x^3 \plus{} cx^2 \plus{} bx \plus{} a \equal{} 0$.
Prove that $ \prod_{i \equal{} 1}^3tan(\alpha_i \plus{} \beta_i) \equal{} \sum_{i \equal{} 1}^3tan(\alpha_i \plus{} \beta_i)$

(post deleted)

Sorry I'm seeing things wrong. :blush:

[quote="nayel"]What should we prove? The left and right sides of your equation are already equal.[/quote]
Are u sure about this statement, nayel? THe LHS is the sum and RHS is the product. It doesn't look obvious though! :wink:

Someone helps me with this problem,plz! :?:

Note that if $ x + y + z = k\pi$, then it is known that
\begin{align*} \tan x \tan y \tan z & = \tan x \tan y \tan (k\pi - x - y) = - \tan x \tan y \cdot \frac {\tan x + \tan y}{1 - \tan x \tan y} \\
& = (\tan x + \tan y) \left[1 - \frac {1}{1 - \tan x \tan y}\right] = \tan x + \tan y - \frac {\tan x \tan y}{1 - \tan x \tan y} \\
& = \tan x + \tan y + \tan z \end{align*} By Vieta’s formulas, \begin{align*} - a & = \tan \alpha_1 + \tan \alpha_2 + \tan \alpha_3 = \tan \beta_1 \tan \beta_2 \tan \beta_3 \\
-c & = \tan \alpha_1 \tan \alpha_2 \tan \alpha_3 = \tan \beta_1 + \tan \beta_2 + \tan \beta_3 \\
b & = \tan \alpha_1 \tan \alpha_2 + \tan \alpha_2 \tan \alpha_3 + \tan \alpha_3 \tan \alpha_1 = \tan \beta_1 \tan \beta_2 + \tan \beta_2 \tan \beta_3 + \tan \beta_3 \tan \beta_1 \end{align*} Thus, by the sum of three angles formula, \[ \tan(\alpha_1 + \alpha_2 + \alpha_3) = \frac {1 - b}{c - a} = - \frac {1 - b}{a - c} = - \tan(\beta_1 + \tan \beta_2 + \tan \beta_3)
\] And the sum of two angles formula yields that $ \alpha_1 + \alpha_2 + \alpha_3 + \beta_1 + \beta_2 + \beta_3 = k\pi$. The desired result follows from the first paragraph, using the numbers $ \alpha_1 + \beta_1, \alpha_2 + \beta_2, \alpha_3 + \beta_3$.


calculus
integration
trigonometry
real analysis
real analysis unsolved

Calculate:

1. $ \int \frac {1\minus{}x}{e^{x}\plus{}x^{2}e^{\minus{}x}}\ dx$

2. $ \int \frac {x^{2}}{(xsecx\plus{}cosx)^{2}}\ dx$

3. $ \int \frac {sinx\minus{}sin^{3}x}{1\minus{}sin^{3}x}\ dx$

4. $ \int \frac {\sqrt{a^{2}\plus{}b^{2}cos^{2}x}}{cosx}\ dx$

5. $ \int \frac {\sqrt{a^{2}\plus{}b^{2}cosx}}{cosx}\ dx$

[hide="Number 3"]$ \int \frac {\sin x \minus{} \sin^3x}{1 \minus{} \sin^3x}\,dx \equal{} \int \frac {\sin x(1 \minus{} \sin x)(1 \plus{} \sin x)}{(1 \minus{} \sin x)(1 \plus{} \sin x \plus{} \sin^2x)}\,dx$

$ \equal{} \int \frac {1 \plus{} \sin x \plus{} \sin^2x \minus{} 1}{1 \plus{} \sin x \plus{} \sin^2x}\,dx \equal{} x \minus{} \underbrace{\int \frac {1}{1 \plus{} \sin x \plus{} \sin^2x}\,dx}_{ \equal{} \,P'}$

To find P' start by making $ x \equal{} \frac {\pi}{2} \minus{} t$, and so

$ P' \equal{} \minus{} \int \frac {1}{1 \plus{} \cos t \plus{} \cos^2t}\,dt \equal{} \minus{} \int \frac {1}{2 \cos^2\frac {t}{2} \plus{} \cos^2t}\,dt$.

Now make t = 2u,

$ P' \equal{} \minus{} 2 \int \frac {1}{2 \cos^2 u \plus{} \cos^22u}\,du \equal{} \minus{} 2 \int \frac {1}{4 \cos^4u \minus{} 2 \cos^2u \plus{} 1}\,du$

Now make $ \tan u \equal{} z$,

$ P' \equal{} \minus{}2 \int \frac{1}{\frac{4}{(1\plus{}z^2)^2} \minus{} \frac{2}{1\plus{}z^2} \plus{} 1} \cdot \frac{1}{1\plus{}z^2}\,dz \equal{} \minus{}2 \int \frac{z^2\plus{}1}{z^4\plus{}3}\,dz$

and you can continue from here.[/hide]

Excellent job as usual, Carcul!
I think this time integrals are more difficult or boring. btw ... I think the first one is an excellent excercise.

Write the first integral as $ \int \frac{(1-x)e^{-x}}{1+ (x/e{}^x){}^2}$, and note that $ xe^{-x}'$ is the numerator. so the answer is $ arctan xe^{-x}$

:lol: very good. When I grow up I'll become smart like you!


<asy>

draw((0,0)--(3,7));
dot((0,0)); dot((3,7));

</asy>

can someone tell me what i am doing wrong?

replace < and > by [ and ] respectively ?
add a command like [b]size(10cm);[/b] or [b]unitsize(1cm);[/b] ?

Using cse5's "composite function" capabilities:
[asy]size(100);
D(D((0,0))--D((3,7)));[/asy]
Notice that the points are dotted before the line is drawn if cse5 is used as above; the conventional approach draws the points on top:
[asy]size(100);
D((0,0)--(3,7));
D((0,0));
D((3,7));[/asy]


calculus
integration
trigonometry
function
calculus computations

I am not very sure whether I am putting the problem in the correct section.
Heres the problem which I cant solve
$dy/dx$=$x/y$+$tan(y/x)$

Plz help me in this.

Begin with the substitution, y = vx, solve the equation in v and x, and back-substitute.

Its not working that way bcoz of the tan function.

Where did you find this problem?  It's clearly homogenous, but I keep getting an integral I can't evaluate, $\int \frac{u \ du}{-u^2+u\tan u +1}$.

It was given to me by a frnd of mine.I dont know from where he got the problem.Well Jimmy i am also stuck at the same step.I cannot evaluate that integral.

Well, I wouldn't waste any time trying (Mathematica couldn't evaluate it either).  You could try a power series...

2 [b]the game[/b] : do you need to solve this equation for some application or just solve it?As for me, it's exist an connection between the function and variable in form this integral and this is important  ;)


MATHCOUNTS
probability
geometry
analytic geometry

I was just wondering if anyone had some examples or topics that are most used in MATHCOUNTS competitions, preferably state.

counting and probability are probably the most used
Also know algebra and GEOMETRY
number theory's usually not that useful

probability is big (especially in countdown). Algebra is also a lifesaver.

Oh. WOW that really sucks.....I am really bad at PROB. QUESTION: Who here actually thinks there good at probablity?

probability/counting is my best subject so yeah that would be me.

I'd say it's probably probability, no pun intended.

In all the state Mathcounts tests I have taken, this is the order of occurence of types of problems.  (Besides simple problem solving and simple operations)

1. Counting/Probability
2. Geometry (Also learn fast ways to find areas of figures on grids)
3. Algebra
4. Last Digit problems
5. Order of Operations!!!

Hope this helps!

Yeah, it does.  I'm studying for state and really want to go to nats this year.  Good luck to everyone else here as well.  Is anyone from Virginia on AoPS?

Hmmm... lots of it depends on your goals.

In chapter, I would say that geometry is far behind algebra and counting/prob., while in nationals, it can be just as important.

My worst areas are coordinate geometry, probability (my counting is somewhat decent, it depends on the problem), problems with venn diagrams, and problems with rotating thingies, like gears and planets being aligned.

Yeah, counting/probability shows up alot.

I'm really bad at problems that involve common sense  :blush:  My worst are rate/time problems. I HATE them. I'm ok at geometry and I'm ok at counting/probability. I'm not really good at anything.

I'm pretty good at guessing though  (including guessing and checking) :P  At the chapter I guessed in 3 sprint and 2 target questions and I got them all right :D

rate/time problems are really easy if you go about it the right way. make a table with rate, time and distance across the top, and rows for your two things, (or more), then fill in all the spaces with algebraic expressions and remember that rate*time equals distance. There should be enough information in the table to solve the problem.
Rate/time/distance problems are tricky, but they don't have to be. do a lot of them and get comfortable with the idea.

Usually if you put time into the problem, you'll come up with the right answer.   The AoPS book helped me with the counting problems, and everything else I would say I'm decent in.


How do I write the polar equation for a hyperbola with eccentricity 3/2 and directrix x = -2?

Thanks!


geometry
inradius
area of a triangle
Heron\u0027s formula
geometry proposed

Hi all!

What's the maximum value of the quotient of the areas of a triangle and its incircle? When do we get this value?

Hanno

The area of a triangle can be written as $\frac{r^2}2\left(\cot\frac A2+\cot\frac B2+\cot\frac C2\right)$, where $r$ is the inradius and $A,B,C$ are the angles. Since $\cot$ is convex on $(0,\frac\pi2)$, this area for fixed $r$ is minimized when the triangle is equilateral, which then gives the maximum ratio.

HI JMerry!

Here's my solution:

Let [tex]a,b,c[/tex] the sides of an arbitrary triangle. Then, by Heron's Formula, we have [tex]\Delta = \sqrt{s(s-a)(s-b)(s-c)}[/tex], where [tex]s[/tex] is the half-perimeter. Further, it's well-known, that [tex]\Delta = rs\Leftrightarrow r=\frac{\Delta}{s}[/tex]. Putting these things together, the quotient we are looking for becomes
[tex]q(a,b,c)=\frac{\pi r^2}{\Delta}=\frac{\pi A}{s^2}=\pi\sqrt{\frac{(s-a)(s-b)(s-c)}{s^3}}[/tex]. Since [tex]q(a,b,c)[/tex] is homogenous, we may assume that [tex]a+b+c=2[/tex]. Furthermore, [tex]\sqrt{x},\sqrt[3]{x}[/tex] are monotonically increasing, whence we the maximum value of [tex]q[/tex] is reached iff [tex]q'(a,b,c)=\sqrt[3]{\frac{(s-a)(s-b)(s-c)}{s^3}}=\sqrt{(1-a)(1-b)(1-c)}[/tex] takes its maximum value. Now, AM-GM yields [tex]q'(a,b,c)\leq\frac{3-(a+b+c)}{3}=\frac{1}{3}[/tex], where equality holds iff [tex]a=b=c[/tex], that is, iff [tex]a=b=c=\frac{1}{3}[/tex]. Calculating [tex]q\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)[/tex] we get the maximum value [tex]\frac{\pi}{\sqrt{27}}[/tex].

Hanno


[quote="111222333444555"]Let $ABC$ be a triangle, $O$ the circuncenter of $ABC$, $P$ the midpoint of $AO$, and $Q$ the midpoint of $BC$. If $\angle{CBA}= 4\angle{OPQ}$, and $\angle{ACB}= 6\angle{OPQ}$, find $\angle{OPQ}$.[/quote]

Let $O$ goes to $O'$ by the simmetry in BC.
Then $PQ$ is the midline of triangle $AOO' \Rightarrow$
$2\angle OAO'=2\angle OPQ=\angle ACB-\angle ABC = \angle OCA- \angle OBA= \angle OAC -\angle OAB\Leftrightarrow$
$\angle OAC-\angle OAO'=\angle OAB+\angle OAO'$
As $\angle ACB > \angle ABC \Rightarrow AB>AC \Rightarrow$
$\angle CAO'= \angle OAC-\angle OAO'=\angle OAB+\angle OAO'= \angle BAO' \Rightarrow$
$AO'$ is a bissector of the angle $BAC \Rightarrow$
$O'$ lies on the circumcircle of $ABC \Rightarrow \angle BAC=60^\circ \Rightarrow$
$10\angle OPQ=\angle ABC+ \angle ACB=120^\circ \Rightarrow$
$\angle OPQ=12^\circ$

V:)


Hi guys. I've been searching the internet today and found this site. Well, I am in 10th grade and doing a report on 'An Algabraic Approach to the 2nd dimension'. I've been using Google as my reference but can't seem to find anything good. Can someone please give me a site(s) on the 2nd dimension? Or even some information on it would be appreciated also. Thanks for reading this and hopefully I can get some usefull information for my report!  :)

I'm probably missing something, but your topic seems to be a fancy way of saying 2d 'Analytic Geometry'.

funny thing about dimensions

we know that 3D is like a real life object
A cube is a 3D
if you look it by the side you'll only see a plane
this is 2D
if you look at the plane by the side again, you see a line
this is 1D
if you look at the line by the side again, you see a dot
this is 0D

he he here's the thing
3D cube has 8 vertices
2D plane has 4 vertices
1D line has 2 vertices
0D line has 1 vertice of a dot itself

$2^3 , 2^2 , 2^1 , 2^0$


analytic geometry
geometry
geometric transformation
reflection
articles

In the lattice 2-D grid of integer coordinates, a lattice point (x,y) is colored iff 0≤y≤x.

From a colored point (x,y), one may next move to (x+1,y) or to (x, y+1), provided the destination point is colored.

How many valid paths are there from (0,0) to (n,n), in terms of n?

[hide="Answer"]
$ \frac{1}{n\plus{}1} \binom{2n}{n}$
[/hide]

Are these by any chance the Catalan numbers?

EDIT: yes they are

[hide="one proof"]
The total number of paths from $ (0,0)$ to $ (n,n)$ without the restriction of the point being colored is $ \binom{2n}{n}$

Now out of these points, we want to subtract the "bad paths" that cross over the line $ x\equal{}y$

Every single "bad path" crosses the line $ x\equal{}y$ at least one. One step after the first crossing, it will be at some point on $ y\equal{}x\plus{}1$
If you reflect all the subsequent points about $ y\equal{}x\plus{}\frac{1}{2}$ you will finally have a unique path to $ (n\minus{}1,n\plus{}1)$

This means that the total number of "bad paths" is $ \binom{2n}{n\plus{}1}$

That means the total number of good paths that stay on the colored lattice points is

$ \binom{2n}{n} \minus{} \binom{2n}{n\plus{}1}$

With some nice rearrangements, that comes out to $ \frac{1}{n\plus{}1}\binom{2n}{n}$
[/hide]

[quote="chickendude"]Are these by any chance the Catalan numbers?

EDIT: yes they are

[hide="one proof"]

If you reflect all the subsequent points about $ y \equal{} x \plus{} \frac {1}{2}$ you will finally have a unique path to $ (n \minus{} 1,n \plus{} 1)$

This means that the total number of "bad paths" is $ \binom{2n}{n \plus{} 1}$
[/hide][/quote]

Can someone further explain this?

To my surprise, I actually found a solution to this on Wikipedia under the article entitled "Catalan Numbers"

[url=http://en.wikipedia.org/wiki/Catalan_numbers#Second_proof]One Demonstration[/url]

[url=http://en.wikipedia.org/wiki/Catalan_numbers#Third_proof]Another Demonstration[/url]

It's one definition of the Catalan numbers
The series shows up everywhere in combinatorics

Basically, any time you have to count a situation in which you cannot have more of one thing than another, you will end up with catalan numbers.

For example:
You are standing at the edge of a cliff. You take 2n steps and end up back where you started. Each step is either forward or backward. How many different ways can you take those n steps?
The answer is once again $ C_n \equal{} \frac {\binom{2n}{n}}{n \plus{} 1}$

In the problem that the zuton force gave, moving one in the x direction is like taking a step backward. Moving in the y direction is like taking a step forward. Crossing over y = x is like falling off the cliff. In the end, they are essentially the same problem.

Richard Stanley's [url=http://www-math.mit.edu/~rstan/ec/]webpage[/url] for his text [u]Enumerative Combinatorics[/u] contains a link to a [url=http://www-math.mit.edu/~rstan/ec/catadd.pdf]pdf file[/url] with (currently) 161 combinatorial interpretations of the Catalan numbers (I believe in addition to a great number that actually appear in Chapter 6 of that text).


Hi;

Let $ABCD$ given a square with sidelength 1. A quadrilateral is inscribed in $ABCD$, it has the lengths $a,b,c,d$.
Prove that the following inequality holds:
\[2 \leq a^{2}+b^{2}+c^{2}+d^{2}\leq 4\]

[quote="SBHD"]Hi;

Let $ABCD$ given a square with sidelength 1. A quadrilateral is inscribed in $ABCD$, it has the lengths $a,b,c,d$.
Prove that the following inequality holds:
\[2 \leq a^{2}+b^{2}+c^{2}+d^{2}\leq 4 \]
[/quote]

Shouldn't it be $2\leq a^{2}+b^{2}+c^{2}+d^{2}<4$?

I also thought so, but my teacher told me that the inscribed quadrilateral could be the same as $ABCD$, and I thought she'd be right.


Four horizontal lines and four vertical lines are drawn in a plane. In how many ways can four lines be chosen such that a rectangular region is enclosed?

YOu pick 2 vertical and to horizontal lines so $ 6\times{6}\equal{}36$

I did it a different way.

So if you draw the lines out you should get what looks like a tic tac toe board with all of the lines extended.

There are 9 closed spaces and you multiply that by 4 because it says how many ways can 4 lines bee chosen to enclose a space.

$  9(4)= \boxed{\boxed36}

sorry for dat

$ 9(4)= \boxed{\boxed36}  $

I think you meant this:

$9\times4=\boxed{\boxed{36}}$

Also, edit your posts.