algorithm
number theory
prime numbers
number theory open

I apologize if this is an inappropriate place for this question.  Is there a way to accurately test for prime numbers that are so large they exceed the limit of the programming langauge?  I am running a PERL program to find primes and I beleive PERL has a 16 digit limit. Is there a way mathematically (or by other means) to work with numbers that are more than 16 digits?

You need a package (data types, operations) that can handle large integers.The default types use a fixed amount of memory storage, and thus have an upper limit. Very large integers can be represented with structures that use more memory as the numbers get larger.
The details of that are a computer science question- I'm sure there are standard packages, and I don't know the details of the structures used.
The simplest version of this I can think of would do calculations in base $N$, where $N$ is the size of a standard unsigned integer.

Once you can handle large integers, you can use standard algorithms to test for primality.

Create an array representing the digits of the large number, and code the division algorithm yourself.


inequalities AMC USA(J)MO USAMO induction geometry

I didn't use that highpowered of a result but now time to solve yours :)

Heres the solution to the problem

[hide]
For ThAzN1's problem, we first apply cauchy

 \[ \frac{1}{1+2x_1} + \frac{1}{1+2x_2} + \cdots + \frac{1}{1+2x_n} \geq \frac{(1 + 1 + ... + 1)^2}{(1 + 1 + .... + 1) + 2(x_1 + x_2 + ... + x_n)} \vskip .2in = \displaystyle\frac{n^2}{n + 2(x_1 + x_2 + ... + x_n)} \]

now by am-gm,

\[ \frac{x_1 + x_2 + ... + x_n}{n} \geq (x_1x_2x_3...x_n)^{\displaystyle\frac{1}{n}} = 1 \Rightarrow x_1 + x_2 + ... + x_n \geq n \]

Combining this with the result above, we get

\[ \frac{n^2}{n + 2(x_1 + x_2 + ... + x_n)} \geq \frac{n^2}{n + 2(n)} = \frac{n}{3} \]

and we are done.
[/hide]

For the next problem... how about

Let a,b,c be real numbers greater than or equal to zero. Prove that

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

[hide="solution"]First square both sides and simplify:

$a + b + c \geq \sqrt{a}\sqrt{b} + \sqrt{a}\sqrt{c}+ \sqrt{b}\sqrt{c}.$

This statement is true by applying the re-arrangement inequality to $\sqrt{a}, \sqrt{b}, \sqrt{c}$ and assuming WLOG that $a \geq b \geq c$.[/hide]

A new problem: Let $a,b,c > 0$ and $abc = 1$.  Prove that

$1 + \frac{3}{a + b + c} \geq \frac{6}{ab + ac + bc}$

[quote="DPopov"]Given $x+y+z=1$ and $x,y,z>0$ Prove
\[ \frac{x}{\sqrt{1-x}}+\frac{y}{\sqrt{1-y}}+\frac{z}{\sqrt{1-z}} \geq \sqrt{\frac{3}{2}}  \][/quote]
[hide="my solution"]Jensen's inequality also works for this, btw.  Since if we let

$f(x) = \frac{x}{\sqrt{1-x}} \Rightarrow f''(x) = \frac{1}{(1-x)^{3/2}} + \frac{3x}{4(1-x)^{5/2}} > 0$

for $x > 0$, we have by Jensen:

$\frac{x}{\sqrt{1-x}}+\frac{y}{\sqrt{1-y}}+\frac{z}{\sqrt{1-z}} \geq 3 f\left(\frac{x + y + z}{3}\right) = 3 f\left(\frac{1}{3}\right) = \frac{3}{\sqrt{6}} = \sqrt{\frac{3}{2}}$[/hide]

[quote="The Original Pi Guy"]Heres the solution to the problem

[hide]
For ThAzN1's problem, we first apply cauchy

 \[ \frac{1}{1+2x_1} + \frac{1}{1+2x_2} + \cdots + \frac{1}{1+2x_n} \geq \frac{(1 + 1 + ... + 1)^2}{(1 + 1 + .... + 1) + 2(x_1 + x_2 + ... + x_n)} \vskip .2in = \displaystyle\frac{n^2}{n + 2(x_1 + x_2 + ... + x_n)}  \]

now by am-gm,

\[ \frac{x_1 + x_2 + ... + x_n}{n} \geq (x_1x_2x_3...x_n)^{\displaystyle\frac{1}{n}} = 1 \Rightarrow x_1 + x_2 + ... + x_n \geq n  \]

Combining this with the result above, we get

\[ \frac{n^2}{n + 2(x_1 + x_2 + ... + x_n)} \geq \frac{n^2}{n + 2(n)} = \frac{n}{3}  \]

and we are done.
[/hide][/quote]

That's actually not correct. You applied AM-GM, but when you took the reciprocal you didn't flip the inequality. In fact

$\frac{n^2}{n+2(x_1+\cdots+x_n)} \le \frac{n}{3}$.

oh...  :blush:

um... evidently cauchy isn't strong enough...

but you're definitely right

 :wallbash:

My solution to the first problem basically reduced it to Nesbitt's from which you can procede in w/e fashion you would like.

The Original Pi Guy wrote:

Even quicker with Cauchy's.

[hide]

Take square root and you're done.

[/hide]

[quote="ThAzN1"]

Marathon? Ok, here's a prob:

Let $x_1, x_2, \dots , x_n$ be positive reals with product equal to $1$.  Prove that

\[ \frac{1}{1+2x_1} + \frac{1}{1+2x_2} + \cdots + \frac{1}{1+2x_n} \geq \frac n3  \][/quote]

Shouldn't $\sum x_i=n$? This way the result follows immediately from a direct application of Jensen's.

*Perhaps you can multiply both sides by the product in some fashion so that the inequality is homogenized.*

Wow...woops, yeah, AM-GM *ahem* Well in that case it again follows immediately from Jensen:

[hide="Solution Perhaps"]
Let $f(x)=\frac{1}{ax+b}$. With quick computation we see that $f''(x)>0$. It follows from Jensen's that
\[{ \sum_{i=1}^n f(x_i) \geq n f (\frac{x_1+...+x_n}{n}})  \]
By AM-GM $\sum x_i \geq n$. Hence ${\frac{x_1+...+x_n}{n}} \geq 1$. Using this fact we see that \[{ n f (\frac{x_1+...+x_n}{n}} ) \leq \frac{n}{a+b}  \] In this case $a=2, b=1$ so we have
\[{ \sum_{i=1}^n f(x_i) \geq n f (\frac{x_1+...+x_n}{n}} )  \]
\[ \sum_{i=1}^n f(x_i) \geq \frac{n}{a+b}  \]
\[ \sum_{i=1}^n f(x_i) \geq \frac{n}{3}  \]

as desired.

*Equality holds when $\sum x_i=n$*

This was a bit rushed so I might have some errors  ;)

[/hide]

[quote="DPopov"][hide]Wow...woops, yeah, AM-GM *ahem* Well in that case it again follows immediately from Jensen:

Let $f(x)=\frac{1}{ax+b}$. With quick computation we see that $f''(x)>0$. It follows from Jensen's that
\[{ \sum_{i=1}^n f(x_i) \geq n f (\frac{x_1+...+x_n}{n}})  \]
By AM-GM $\sum x_i \geq n$. Hence ${\frac{x_1+...+x_n}{n}} \geq 1$. Using this fact we see that \[{ n f (\frac{x_1+...+x_n}{n}} ) \leq \frac{n}{a+b}  \] In this case $a=2, b=1$ so we have
\[{ \sum_{i=1}^n f(x_i) \geq n f (\frac{x_1+...+x_n}{n}} )  \]
\[ \sum_{i=1}^n f(x_i) \geq \frac{n}{a+b}  \]
\[ \sum_{i=1}^n f(x_i) \geq \frac{n}{3}  \]

as desired.[/hide][/quote]
Nope - that doesn't work.  Your second-to-last step doesn't follow from your given inequalities (you have $\sum f(x_i) \geq n \ f(...) \leq n / 3$, which doesn't tell you anything).  Jensen isn't powerful enough for this inequality.

Yeah woops, I just realized but I think this is the correct approach? ;)

[quote="probability1.01"]Huh? They are the same problem![/quote]

#14 and #15?

Amazing solution on #12.

For #15 I did some work on it (note all products are cyclic)

[hide]

From the given condition we can substitute  and so on such that . Now we have

The only problem is that plugging in a couple values into the end equation showed me that it wasn't true, so I'm at a loss right now. The beginning is fine, I believe.

[/hide]

[quote="Elemennop"]Amazing solution on #12.

For #15 I did some work on it (note all products are cyclic)...

[/quote]

Your last inequality is true for acute triangle. I actually created #15 from that. But there is a simple algebraic solution for #15 without using any substitution.

[quote="Elemennop"]Shouldn't it be true for all triangles, though? Or maybe that's a stipulation of the substitution.[/quote]

If the triangle is not acute, your 2nd last step doesn't imply the last step because you were dividing a negative number. Anyway, since $x,y,z>0$, the triangle has to be acute with your substitution.

Aha, I see now.

Woot, I actually did an inequality correct.

edit> bah, I can't finish it up. I don't know how to prove that the final inequality is true for acute triangles.

Yeah $x,y,z>0$ basically implies acute-ness

#14 and #15 (they're equivalent) I'll just show #15 here:
[hide]Note that $x^2 = (\sqrt{x^2+1}-1)(\sqrt{x^2+1}+1)$ etc so the inequality becomes

$\prod (\sqrt{x^2+1}-1) \geq 1$ (which in fact is the secant inequality above) Note that

$\sqrt{yz}\sqrt{x^2+1} = \sqrt{x(x+y+z) + yz} = \sqrt{(x+y)(x+z)} \geq x+\sqrt{yz}$ (by Cauchy)

$\Rightarrow \sqrt{x^2+1} -1 \geq \frac{x}{\sqrt{yz}}$ and yeah...[/hide]

Anyone care elaborating how #14 and 15 are equivalent?

So basically just normalize $xy+yz+zx = 1$ in #14 then replace everything with their reciprocals and voila

Ok, I see it now.

Is there any specific guide to normalizing? Like for this one, we basically show that if (x,y,z) is a solution than (tx,ty,tz) is a solution as well, then we can normalize...but can you normalize any symmetric sum? ie. could we also have normalized x+y+z=1 or xyz=1 ??

I've moved this topic into the Olympiad Inequalities forum.

Please only start topics in the AMC forum that are [i]directly relevant[/i] to the AMC contests.  The AMC forum has way too many topics as is without having general mathematical discussions in there as well.

Besides, the AMC forum is going to be locked tomorrow for the Alternate AIME, but there's no reason that this topic needs to be locked as well!

[quote="jin thynj"]So no one new problem?[/quote]
For you, jin thynj :)
Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that
\[ \frac {a^3 \plus{} b^3 \plus{} c^3}{ab \plus{} ac \plus{} bc} \plus{} \frac {4abc}{a^2 \plus{} b^2 \plus{} c^2}\geq\frac {7}{9}(a \plus{} b \plus{} c)\]

I have done that we can assume $ a \plus{} b \plus{} c \equal{} 1$, and considering that:
$ \sum{a^3} \equal{} (\sum{a})(\sum{a^2} \minus{} \sum{ab}) \plus{} 3abc \equal{} (1 \minus{} 3\sum{ab}) \plus{} 3abc$,
the statement is equivalent to:

"let a,b,c>0 with sum 1 then $ \frac {1 \plus{} 3abc}{\sum{ab}} \plus{} \frac {4abc}{1 \minus{} 2\sum{ab}} \ge \frac {34}{9}$."

But nothing else (I have tried with fixing abc and derivate all respect to sum(ab) with regating to bound, but it would be not olympic and ugly..), any idea?

[quote="ThAzN1"]By Holder, we have
$\left(\sum_{\text{cyc}} \frac{1}{\sqrt{1-x}} \right)^2 \left( \sum_{\text{cyc}} (1-x) \right) \geq 3^3 \Longrightarrow \sum_{\text{cyc}} \frac{1}{\sqrt{1-x}} \geq \frac{3\sqrt{6}}{2}$
By Chebyshev,
$\sum_{\text{cyc}} \frac{x}{\sqrt{1-x}} \geq \frac 13 \sum_{\text{cyc}} x \sum_{\text{cyc}} \frac{1}{\sqrt{1-x}}$
Combining these yields the result  :)
Jensen probably works too...

Marathon? Ok, here's a prob:

Let $x_1, x_2, \dots , x_n$ be positive reals with product equal to $1$.  Prove that

\[ \frac{1}{1+2x_1} + \frac{1}{1+2x_2} + \cdots + \frac{1}{1+2x_n} \geq \frac n3  \][/quote]
Jensen works here


If  a * b is defined as “half the average of a and b”, determine the positive difference between: (3 * 5) * 6 and 3 * (5 * 6). Express answer as a fraction in simplest form or as an exact decimal.

if    a * b  =  ½ (a+ b)/2  then a * b = a + b /4

so  3 * 5 = 3 + 5 /4  = 2 and 2 * 6  = 2+6/4 =2

5 * 6  = 5+6/4  = 11/4      3 * 11/4   =   (3 + 11/4)/4  = 23/16

positive difference between  =  2 - 23/16  =   9/16  = 0.5625 = 0.57

I've checked, and I do not see any problem (though be warned, my arithmetic is awful), except that I don't think you were supposed to round the decimal in the last step (it says "exact decimal").

Incidentally, you might want to learn LaTeX if you're going to post any more math expressions. The main thing being that it looks a bit nicer, and sometimes clearer.

$ a * b \equal{} \frac{1}{2}\cdot\frac{a \plus{} b}{2}$


[quote="spinos"]Δίνεται ο πίνακας Α=$ \left( \begin{array}{c c c} 2 & 1 & 0 \\
3 & 0 & 0 \\
0 & 0 & 1 \end{array} \right)$

α) να βεθεί η κανονική μορφή Jordan
β) να βρεθεί ο αντιστρέψιμος πίνακας P τέτοιος ώστε $ P^ - ^1AP$ να
    είναι ίσος με την μορφή Jordan του Α

*Βασικά παιδιά δεν θυμάμαι πως βρίσκουμε τον αντιστρέψιμο πίνακα P, δηλαδή ποιά είναι η γενική μεθοδολογία που ακολουθούμε για να βρίσκουμε τα διανύσματα που τον δημιουργούν?

Παρακαλώ βάλτε τα στο κατάλληλο μέρος[/quote]
http://en.wikipedia.org/wiki/Jordan_normal_form#Example


logarithms
number theory
greatest common divisor

How many distinct four-tuples $ (a,b,c,d)$ of rational numbers are there with

$ a \log_{10} 2 \plus{} b \log_{10} 3 \plus{} c \log_{10} 5 \plus{} d \log_{10} 7 \equal{} 2005$?

$ \textbf{(A)}\ 0\qquad
\textbf{(B)}\ 1\qquad
\textbf{(C)}\ 17\qquad
\textbf{(D)}\ 2004\qquad
\textbf{(E)}\ \text{infinitely many}$

I'm not exactly sure as to how you got the answer... i only get the part until you solve the entire thing (right before the last step).

Um, I got the first couple of steps until you actually solved the equation. How do you know that there [hide]is only one solution[/hide] and not infinitely many?  :?

I'm assuming you understood at least up until $\log_{10}2^a3^b5^c7^d=2005$. From there, by one of the basic properties of logarithms, we have $10^{2005}=2^a3^b5^c7^d$. $10^{2005}=2^{2005}5^{2005}$, so we only have one solution: $(a,b,c,d)=(2005,0,2005,0)$. :)

No, I understand all the stuff about logarithms, but I don't understand why there can be only one solution.  :?

well, b, d have to equal 0 because 3, 7 are not factors of 10 to any power.  Now with a and c, try to find another pair of numbers which satisfies that...

if you think about it in your head, 2 * 5 = 10, right?

so to get 2005 10's you need 2005 2's and 2005 5's, and since 2 and 5 dont have any common factors, there is only one solution, which is 2^2005 * 5^2005

[hide]We can rewrite this as  $\log_{10}(a^2b^3c^5d^7)=2005$.

$a^2b^3c^5d^7=10^{2005}$. Since this is a power of 10, we want the 3 and the 7 to stay out of this, so b and d must be 0. If a and c are both 2005, this works.

tHus there is one solution set. [/hide]

Does "four-tuples $(a,b,c,d)$ of rational numbers" mean a, b, c, and d have to be integers?

No, rational means they must be able to be written in fraction form (i.e. no sqare roots or transcendentals).

uhhh 236factorial, I think you meant $ log_{10} {2^a3^b5^c7^d} \equal{} 2005$.  :D

[quote="Justong"]What would be the rigorous proof to show that b,d must be 0? That is, how can we show that there is no way for $3^{b}5^{d}$ to work in this problem for any [b]rational[/b] values of b and d?

In advance, sorry for reviving this thread.[/quote]

Just consider the GCD of the denominators of all the fractions, let it be $m$. Then, we have $2^a3^b5^c7^d=10^{2005m}$ for $a,b,c,d,m$ integers, and the result follows.


induction
combinatorics unsolved
combinatorics

Let $ n$ be a positive integer, and let $ A_1$, $ A_2$, ..., $ A_{n\plus{}1}$ be non-empty subsets of $ \left\{1,2,...,n\right\}$.
Prove that: we can choose two non-intersecting groups of $ A_i$'s such that the union of the sets in one group is equal to the union of the sets in another group.

Is it really hard?
well, let U be the set of all subsets of M={A1,...,An+1} not including M and empty set.
there are 2^(n+1) -2>n such subsets, and therefore there two subsets with the same union. remove the common elements and you will end up with two non-empty and disjoint groups of Ai with same union.

i don't think that's quite what he means by the question :?
anyway what u consider a "union" (sorry for my bad english) ?

I consider the union of two sets A and B two be the set of all elements that belong either to A or B or both.

[quote="ali"]remove the common elements and you will end up with two non-empty and disjoint groups of Ai with same union.[/quote]

No.

For example, Take A1={1,2},A2={1,2,3},A3={2,3}
Then we have A1 \cup A2=A2 \cup A3, but A1!=A3.

China West Mathematical Olympiad, sort of a district olympiad, before the actuall All Chinese Mathematical Olympiad.

I have a solution.
Call degree of an element the number of sets it belongs to.
Firstly we should analyse the case all elements of A have degree at least 2.
(1)I claim there are two groups of subsets with union the set A.
If a set has just one element, we can easy delete it and use induction( as there is another set containing it).
So suppose every set has at least two elements.
(2)I claim there is a set A[k] such that at least |A[k]|-1 elements have degree at least 3.
Suppose not. Then every subset A[i] correspond at least 2 elements with degree 2 contained in it.
Summing and dividing by 2( each element belongs to just 2 subsets) we get at least n+1 elemnts, contradiction. The claim is proven.
Now let this subset be A[n+1]. Take x belonging to A[1] with smallest degree.
Then there is another set containing x, let it be A[n]. If we delete x and A[n+1]
we respect the induction hypothesis so there are two groups with union A/{x}.
If we add x, we have three cases:
*neither of groups contains x. Then we add A[n],A[n+1] to them
**one of groups contains x. Then we add A[n+1] to the another
***Both groups contain x. Then we are done.
This claim is also proved.
Now pass to the general case. Let Z be the set of all elements with degree at least 2.
(3)I claim that at least |Z|+1 of A[i] are subsets of Z.
If not, then to at least n+1-|Z| subset we put in correspondence an element with degree 1 contained in it. Summing we get at least  n+1-|Z| elements with degree 1 and |Z| elements with degree 2 so at least n+1 elements contradiction.
The claim is proved.
Now we can apply (1) to set Z and make the conclusion.

That's a nice solution, Andrew! I also found it in a book. I don't really understand iura's solution, but statement (1) seems wrong to me. For example, you could take n=4 and A_1={1,3},A_2={1,2},A_3={2,3},A_4={4},A_5={4}. You can't divide these sets into 2 groups both having the union [4] ([4]:={1,2,3,4}). Maybe I missunderstood the whole thing in iura's solution...

This problem was again discussed in this forum.
"Set of subsets" . I think there my solution is written more explicitely.
Note that I supposed any 2 subsets are distinct and any subset has at least 2 elements( or else delete it an use induction). So your example doesn't work according to this criteria.

I'm sorry, iura, but I still don't really understand. As far as I understand, your conditions are: every element has degree >=2, #A_i>=2 and A_i!=A_j (I mean different by !=). Now you could take n=6 and A_1={1,2}, A_2={1,3}, A_3={2,3}, A_4={4,5}, A_5={4,6}, A_6={5,6}, A_7={4,5,6}. If I still don't understand, maybe I'll ask you at the IMO  :D ...

Yes, I finally must assume you are right.
But I have abother idea: suppose the set is s.t. the union of any k subsets has at least k elements( if not, use induction). Can't we prove now there are 2 groups of sets with union A. I mean we can delete one element as in my previous proofs and to hold the condition?
Please anwer me Dzeta, I see you are the most interested in this problem. Thank you.

That’s a very nice idea. I really think your claim is true, but I still think there’s a small gap in your induction argument. I think the problem is the case when there are no weak elements. It is possible that by removing an arbitrary element of A_1, not to keep that property of the sets. For example, take n=4, A_1=A_2=A_3={1,2,3,4} and A_4=A_5={1,2}. By removing 1 and A_1, you get the sets A_4’={2} and A_5’={2}, which have union {2}.


complex analysis
function
advanced fields
advanced fields unsolved

Consider a sequence of distinct numbers $a_{k}\in \mathbb{C}$ with $|a_{k}| < 1$. Let $e_{k}= \{1, a_{k}, a_{k}^{2}, a_{k}^{3}, ...\}$. Let $S$ be the closure of the span of the $e_{k}$. Is $S$ all of $l^{2}$?

Observe: $\exists f \in S^\bot$. If $f=\{c_{0}, c_{1}, c_{2}, ...\}$, then we have $F(x) = \sum_{n}c_{n}x^{n}$ and $F(a_{k}) = 0$ for all $k$. [b]IF[/b] the $a_{k}$ have an accumulation point in the open ball of radius 1, then $F$ is a holomorphic function on the ball with an accumulation point of zeros, so it must be 0. Then $f \equiv 0$, and $S = l^{2}$ as desired.

The question: do we still have the result if $a_{k}\to 1$ (and has no other accumulation point)?

The question is equivalent to the description of possible zeroes of a function in the Hardy class $H^{2}$ in the unit disk. The answer is well-known: the condition $\sum_{k}(1-|a_{k}|)<+\infty$ is necessary and sufficient for existence of a non-trivial function with zeroes $a_{k}$.

In other words, the $a_{k}$ have to go to 1 "fast enough." I can see why that might be the condition, but could you please give me a hint about how to prove it? (Or a reference; I wouldn't be surprised if it's already been discussed on the forum. I don't have Halmos' book, though.)

The proof with all details is a bit too long to post, so I'll just indicate the main ideas.
To construct such a function, just consider the infinite Blaschke product $B(z)=\prod_{k}\frac{|a_{k}|}{a_{k}}\frac{a_{k}-z}{1-\bar a_{k}z}$. The condition $\sum_{k}(1-|a_{k}|)<\infty$ ensures that it converges to some non-zero function bounded in the unit disk.
To show that a function $f\in H^{2}$ with zeroes $a_{k}$ satisfying $\sum_{k}(1-|a_{k}|)=\infty$ must be identically $0$ is a bit harder. Basically, you have to apply some version of maximum principle to the function $f(z)/B_{n}(z)$ where $B_{n}(z)$ are the finite Blaschke products corresponding to the first $n$ zeroes $a_{k}$. Note that $|B_{n}(z)|=1$ on the unit circumference and $B_{n}(z)\to 0$ as $n\to\infty$ for all $|z|<1$.
As to the references, just take any book on Hardy spaces. You'll find all the details there.

Thanks. The girl who asked the question understood that. I guess I'll have to wait till I take a course on complex analysis (I'd teach myself, but I already have too long a list of things I'd like to teach myself...).


I consider myself okay at math. I can easily get an A in school (i am currently taking Alg II).  I just really want to get GOOD at math.  I want to be able to do well in competitions at maybe one day take the USAMO.  Whats the best way to get good at math? I practice everyday, but there's just so much stuff to learn, and I dont know where to start.  Should I get a tutor to teach me or just buy a textbook and work through the problems myself, or should I just practie old tests from competitions?

You should probably start by getting AoPS I and II and possibly the introduction series books. Also, try to solve problems that are posted on the forum, post questions that you need help on, and look at past contests.

I've got both AoPS books.  I'm going to start the second one in the summer.

Practice tests are useful because they help you identify subjects you are not yet familiar with so that you know what to study. The alternative is using a textbook, presumably AoPS as you have mentioned, because you get a pretty comprehensive overview of high school mathematics.

wut grade are you taking alg 2? b/c if it's middle school, give yourself a pat on the back. not many ppl can do that!( whoops! I forgot that this is AOPS!)


All the squares of a drawn into squares board with $(n + 1) (n - 1)$ squares are painted with $3$ colours
such that, for each two different columns and each two different rows, the four squares
of their intersections are not painted with the same colour. Find the maximum value
of $n$.

nothing about the number of diferent colors?  ;)

[quote="M4RI0"]All the squares of a drawn into squares board with $(n + 1) (n - 1)$ squares are painted
such that, for each two different columns and each two different rows, the four squares
of their intersections are not painted with the same colour. Find the maximum value
of $n$.[/quote]
How many colours in your problem?


[quote="zkw"]Let $ x,y,z\ge0, x+y+z=2$, prove that $ \frac{1}{1+x^{2}}+\frac{1}{1+y^{2}}+\frac{1}{1+z^{2}}\ge2$[/quote]
I wish to show the Zhaobin's method. :wink:
Let $ x\leq y\leq z$ and $ f(x)=\frac{1}{1+x^{2}}.$ Hence, $ f''(x)=\frac{2(3x^{2}-1)}{(x^{2}+1)^{3}}.$
Id est, $ f$ is convex function on $ [\frac{1}{\sqrt3},2]$ and $ f$ is concave function on $ [0,\frac{1}{\sqrt3}].$
Case 1. $ \frac{1}{\sqrt3}\leq x.$
Hence, $ \sum_{cyc}f(x)\geq3f\left(\frac{x+y+z}{3}\right)=3f\left(\frac{2}{3}\right)>2.$
Case 2. $ x\leq\frac{1}{\sqrt3}\leq y.$
Hence, $ \sum_{cyc}f(x)\geq f(x)+2f\left(\frac{y+z}{2}\right)=f(x)+2f\left(\frac{2-x}{2}\right)=\frac{1}{1+x^{2}}+\frac{8}{x^{2}-4x+8}.$
Thus, it remain to prove that $ \frac{1}{1+x^{2}}+\frac{8}{x^{2}-4x+8}\geq2,$ which true for $ x\in[0,\frac{1}{\sqrt3}].$
Case 3. $ x\leq y\leq\frac{1}{\sqrt3}$ and $ x+y\geq\frac{1}{\sqrt3}.$
Hence, $ f(x)+f(y)\geq f\left(x+y-\frac{1}{\sqrt3}\right)+f\left(\frac{1}{\sqrt3}\right)=f\left(2-z-\frac{1}{\sqrt3}\right)+f\left(\frac{1}{\sqrt3}\right).$
Hence, it remains to prove that $ f\left(2-z-\frac{1}{\sqrt3}\right)+f\left(\frac{1}{\sqrt3}\right)+f(z)\geq2,$
 where $ z\in [2-\frac{2}{\sqrt3},2-\frac{1}{\sqrt3}],$ which easy to check.
Case 4.  $ x\leq y\leq\frac{1}{\sqrt3}$ and $ x+y\leq\frac{1}{\sqrt3}.$
Hence, $ f(x)+f(y)\geq f(0)+f(x+y)=1+f(2-z).$
Thus, it remains to prove that $ f(2-z)+f(z)\geq1,$ where $ z\in[2-\frac{1}{\sqrt3},2],$ which triviality.

:rotfl:
Are you sure? When does equality hold?

What is not clear in my proof?
Equality holds for (1,1,0) and permutations.

[quote="andyciup"]Well.. $ \frac{1}{1+x^{2}}=1-\frac{x^{2}}{1+x^{2}}\geq 1-\frac{x}{2}$  :)
Sum it up and you`re done :wink:[/quote]
This is easy and right...thank you!

See also here the Drij's post:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=82172


function
calculus
integration
trigonometry
Euler
calculus computations

For each $ f : [0, \pi] \to \mathbb{R}$ satisfying $ f(0) \equal{} f(\pi) \equal{} 0$, define the integral

$ I(f) \equal{} \int_{0}^{\pi} f'(t)^2 \plus{} 2\sin(t)f''(t) \,dt$

Find a function $ f$ that minimizes this integral.

$ f'(t)^2 \plus{}f(t)f''(t) \equal{} (f(t)f'(t))'$

$ \int_0^\pi (f(t)f'(t))'dt \equal{} f(\pi)f'(\pi)\minus{}f(0)f'(0)\equal{}0$

so letting $ f(t)\equal{}2\sin t$ seems logical?

From your computation it follows that letting $ f(t) \equal{} 2\sin t$ gives $ I(f) \equal{} 0$, but this could be achieved with $ f\equiv 0$ anyway.

Let $ y \equal{} f'$ and $ L(t,y,y') \equal{} y^2 \plus{} 2y' \sin t$, so that we're looking to minimize $ J(y) \equal{} \int_0^{\pi}L(t,y,y')dt$.  The relevant [url=http://en.wikipedia.org/wiki/Euler-Lagrange_equation]Euler-Lagrange equation[/url] is, formally, $ \frac {\partial L}{\partial y} \equal{} \frac {d}{dt}\frac {\partial L}{\partial y'} \Rightarrow y \equal{} \cos t$ which yields a stationary value for $ J$, and hence our solution is $ f \equal{} \sin t$ for which $ I(f) \equal{} \minus{} \frac {\pi}{2}$.

Hm, can you explain more thoroughly how you used the $ \frac {\partial L}{\partial y} \equal{} \frac {d}{dt}\frac {\partial L}{\partial y'} \Rightarrow y \equal{} \cos t$ part? I'm not too familiar with the Euler-Lagrange equations.


The common application for undergraduate applications has a brief section for academic honors and gives National Merit and Cum Laude Society as two examples. What are other examples of academic honors high school students should shoot for?

Pretty much anything that you can arguably describe as an academic honor.  That's the place to put your valedictorian status, AMC scores, scores on other academic competitions, and so forth.

Oh.

Would it be inappropriate to ask about cum laude society in this thread? I checked the website and got really confused about what it is and such...

its just a high honor organization for hi skool, similar to phi beta kappa in college. all you have to do to get nominated for it is to be in the top 20% of your class , which is pretty easy. Its basically a national honor society thats more selective and serious. At least thats what i heard.


This is equivalent to $ ((10000\cdot .3)\cdot .2)\cdot .1\equal{}(3000\cdot .2)\cdot .1\equal{}600\cdot .1\equal{}60$

Or you could just do:

$ (0.1 \times 0.2 \times 0.3) \times 10000 \equal{} \boxed{60}$


Does anyone know if it is possible to get a copy of our solutions from 2004-2005? Are they filed somewhere, or have they all been discarded? I would be highly grateful for any information on this (in private message or public post). Thanks.

I'm sorry for not being clear. I was referring to my own solutions, which I sent by fax and have since misplaced.


superior algebra
superior algebra unsolved

Prove that if $R$ is a finite ring, then every prime ideal of $R$ is a maximal ideal.

sorry, found it's been on the forum before, so disregard it.


geometry
parallelogram
trigonometry
similar triangles
geometry proposed

i)we easily get that, by sines law, $\frac{sin(MBA)}{sin(MBC)}= \frac{sin(MDA)}{sin(MDC)}$!

ii)$\angle{MBA}+\angle{MBC}= \angle{MDA}+\angle{MDC}$

with this, you can kill your problem.

Yeah, with trigo. we can kill all things easily  :mad:

[quote="April"][color=darkred]Let be given a parallelogram $ABCD.$ A point $M$ inside its so that $\angle MAB=\angle MCB$
Prove that: $\angle MBC=\angle MDC$ [/color][/quote]
[color=indigo][b]Proof. (See the figure)[/b] $(EF\parallel AB,\,GH\parallel BC)$
We have:
$\angle AGM = \angle CFM\Longrightarrow\triangle AGM\sim\triangle CFM\Longrightarrow\frac{AG}{CF}=\frac{MG}{MF}$ $\Longrightarrow\frac{DH}{MH}=\frac{BF}{MF}\quad (1)$
On th otherhand, we have: $\angle DHM = \angle MGB = \angle BFM\quad (2)$
From $(1)$ and $(2)$ we have: $\triangle DHM\sim\triangle BFM$
Hence: $\angle MDH = \angle MBF$ i.e. $\angle MDC = \angle MBC$  [/color]

I had the same proof (but you posted first!!  :ninja: )

Angry¿?

Why should I?

Send me a PM please, 'cause I don't know what you're talking about.

first solution
$AM$ meets $CD$ at $K$, $CM$ meets $AB$ at $N$
so triangle $ADK$ and $NBC$ are similar. Thus $KD/BC=AK/CN$
But we also have $CM/CN=KM/KA$ so triangle $DMK$ and $BMC$ are similar
Thus $\angle MDC=\angle MBC$ $q.e.d$

second solution
$AM$ meets $BC$ at $E$, $CM$ meets $AB$ at $N$
we have triangles $BNF$ and $DAC$ are similar
triangles $AMC$ and $NMF$ are similar
so $CM/MF=CA/NF=CD/FB$
but $\angle MCD=\angle MAD=\angle MFB$ so triangle $CMD$ and $FMB$ are similar
Thus $\angle MDC=\angle MBC$ $q.e.d$


linear algebra
matrix
superior algebra
superior algebra solved

Let A,B\in M_N(C) s.t. 5A-2B=AB.
a)Prove that (A-B)(A+B)=A^2-B^2.
b)Prove that for any n\geq 2 there exists an infinity of matrix A,B\in M_n(Z) s.t. 5A-2B=AB.

:D
i found a little generalisation of this one moubi...it sounds like this:
Let A,B \in M_p{C} satisfying the relation : aA+bB=AB, a,b \in C*. Prove that there exists an infinity of matrix A,B and a,b \in C with the property that aA+bB=AB.

for the first one the proof is like yours moubi... for the second one(quite easy:)) ).
if aA+bB=AB then
we have that (A-bI_2)(aI_2-B)=-abI_2=>A-bI_2 is invertible
we choose the matrix B=a(A-bI_2)^(-1).A which veryfies the hypothesis . so there exists an infinity of matrices with the property that aA+bB=AB.
:D:D

[quote="Moubinool"]
This problem should be in College Playground -> Superior Algebra[/quote]

done that! :D


MIT college blogs Support Princeton Stanford calculus

Thanks to all who came and stayed through the nearly 3 hours of tonight's MIT Math Jam.  We'll post a link to the chat transcript here when it goes online.

I'm happy to take more questions here or elaborate on the answers I provided.  Some were incomplete and at least one was incorrect (thanks to the student who called my attention to this): I recommended the NASA SHARP summer program, but two weeks ago NASA announced it is cutting funding to the program, retiring it.  I am disappointed to say the least.

I'm also happy to take questions on my MIT Admissions web log: [url]http://matt.mitblogs.com[/url]

Thanks again!

Thank you Matt for participating in the Math Jam. I will be telling a lot of friends about the transcript after the transcript is posted.

Your blog already mentions that you will turn to the issue of homeschooled students, and as a leader of a homeschooling support group I'm sure we'd like to hear all about it. I recall a remark by Marilee Jones, quoted in some place that can be found by Google, that some homeschooled applicants are "extraordinary" while others "are not even close." I think it would be interesting to explore how much pursuit of personal passions results in adequate preparation for MIT and what kind of preparation might be plainly insufficient.

In general, I admire your school's commitment, from the admissions director and all of the admissions staff, to open the windows on the process and make MIT admissions less of a "black box." Keep up the good work.

As promised, the link to this week's Math Jam transcript:

[url]http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=114[/url]

[quote="frankthetank"]Thanks.  I missed a large part of the Math Jam because I was at my MIT interview.[/quote]

How ironic is that?

Matt,
I just sent my early application, and i'm looking foward to go to MIT (with a lot of luck of course)
I have some questions:
I have some friends in other universities like Princeton and Stanford and they can take the classes they want since their freshmen year (like there are not many required classes). I've seen and heard that in MIT its quite different. I want to study math and I wanted to know if i could take 2 math classes every year, and if i can take any math class I want even in the freshmen year. I am really interested in combinatorics and I've seen some great courses MIT offers for undergradautes so I was wondering if I can take at least one of them the first year and the required math class (i gues it is calculus or something like that...though i am taking calculus this year...)
I think MIT has a research group in combinatorics for undergrads., can I be part of this group even in my freshmen year? if not, when?

thanks for answering them!

I'm not sure of the answers to your questions, manuel, but I'll note that few (very few) universities give undergraduates as many research opportunities as MIT does...

That's the main reason why MIT is my first choice!  ;)

[quote="manuel"]
I want to study math and I wanted to know if i could take 2 math classes every year, and if i can take any math class I want even in the freshmen year. [/quote]

The requirements for a math major are clearly spelled out on the math dept website. They differ somewhat depending on which type of math major you are going for, and you should go to the website for an "official" answer to your question. http://math.mit.edu/undergraduate/degree-options.html

A general math major (Option I) only has 3 required courses (single and multivar calc, plus diff eq). You can test out of any of those that you feel you have already mastered, and then you take at least 8 more "substantially different" math courses of your choice.

As long as you take your required 8 semesters of humanities, there doesn't seem to be anything keeping you from taking all math courses with the rest of your time. That would potentially be 3 math courses per semester as a freshman if you wanted, and no limit after freshman year. It also seems very easy to take advanced courses, including grad courses, without proving that you have the prereq.

Hi manuel,
rrusczyk and texas137 covered nicely for me while I was away from the forum.  The short answers to your questions are:
There's little preventing you from taking 2+ math classes each term at MIT, with the exception of the need for sleep and sanity.  There is no "per unit" cost at MIT so exceptional students may take as many classes as they wish, though I generally do not recommend more than 5 each term.  In the first year, there is a credit limit (~4.5 classes/term), but in theory that may include more than one math class, including advanced classes.
And generally, research (or 'UROP,' for the Undergraduate Research Opportunities Program) is open to all students, including first term freshmen.  I don't know whether specific groups will necessarily have openings, though most students can find a number of groups that meet their needs.
Good luck with your EA application!

And hi tokenadult,
Thanks for the comment.  I do want to write more about homeschooled applicants, though I feel I would write a better post about them after early action, when we'll all be back in the application-evaluating mindset.  How's that sound?

It's starting to get quite busy here in Cambridge.  It's an exciting time of year...

[quote="MattMcGannMIT"]I do want to write more about homeschooled applicants, though I feel I would write a better post about them after early action, when we'll all be back in the application-evaluating mindset. How's that sound?[/quote]

It's perfectly understandable that this is a busy time of year for you. When you post your collected thoughts about homeschooled applicants on your blog, I'll be sure to let lots of parents know.

Quick question: is it worth mentioning a 5 on last year's AIME, or in MIT's pool would that be detrimental?

Thanks,
sran

i think that would be a plus.  just because mit has many applicants who have scored almost perfectly on the aime and are usamo qualifiers, the majority of the applicant pool is not like that.  the aime is a very difficult exam.  just being able to qualifiy for it is impressive.  a 5 out of 15 on the aime is certainly helpful.

Here is the statistics from the last AIME (2005), which showed about 2200 contestants got 5 or more. About 680 of them among the 11th grade contestants (current seniors who are supposedly applying for colleges), got 5 or more.

[url=http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2005-aa/05aimestats.html]AIME stats[/url]

sran,
A 5 on the AIME is definitely worth mentioning.  It's a nice score, good for you.

Interestingly, I've already seen some mentions of AoPS on this year's applications...

Matt, I will be visiting MIT in NOV 19-24. I would like to talk to somebody of the math departmenet because i am comparing the math depts. of different universities. Do you think I should meet someone from the admission office too? Also, I am interested in playing tennis in college so it would be nice if I could meet the coach or someone from the team. So it would be nice if you could tell me what to do during my visit and who to contact for the math dept. info..

Thanks!

PS. I didn't mention AOPS in my app...  :(  if you read the app. of a guy named Manuel from PR interested in maths, remember he is a member of AoPS  :D

[quote="manuel"]Matt, I will be visiting MIT in NOV 19-24. I would like to talk to somebody of the math departmenet because i am comparing the math depts. of different universities. Do you think I should meet someone from the admission office too? Also, I am interested in playing tennis in college so it would be nice if I could meet the coach or someone from the team. So it would be nice if you could tell me what to do during my visit and who to contact for the math dept. info..[/quote]Although I believe that Matt reads this board fairly frequently, I would recommend that you contact him directly (at MIT) if you have a specific request such as this.

Hi everyone! I'm Maria, and am currently a frosh at MIT... I didnt' go on AoPS for a while cause things have been pretty busy, but the term is winding down and i shoudl have more free time till the end of January or so. I really like it here- the people are great and the professors are quite interested in the education of the studnets. i haven't gotten a UROP, but know some first term frosh who have... if you have any questions feel free to PM me (or post on this thread...)

to manuel, the core requirement is 8 terms of humanities, 2 terms of calc (single and multi variable), 2 terms of physics (maechanics and E&M), a term of chem and a term of bio. you can place out of calc I with BC, mechanics with physics C, bio with AP bio. you can also place out of the other sciences by taking a placement test, though you have to take all classes if you dont' place out. most people take them as soon as they get here, but you dont' have to...

thanks mariag!

two weeks ago i was hanging out in mit! i liked it a lot!!! def, my first choice..
i stayed in simmons hall and in burton third..
also, it was cool to visit wellesley too........... :D


mikham inja ba ejazeye hameye bozorgan basaate esteghra va lane kabootari ro pahn konam ta hame lezzat bebearim

$n$ pich o mohre az nazare zaheri shabih be ham hastand,dade shode ast.midanim ke har pich tanha be yek mohre mikhorad(ba an ham andaze ast) va hich do pichi ham andaze nistand.
amale "azmoon" yani bardashtane yek pich va yek mohre va emtehan kardane anha.ba in kar tashkhis midahim ke pich az mohre bozorgtar ast ya mohre az pich bozorg tar ast ya ham andaze hastand.mikhahim ba anjame tedaadi amale "azmoon" koochek tarin pich va koochek tarin mohre(ke mosallaman be ham mikhorand) ra peyda konim.tavajjoh konid ke nemitavan  do mohre ba do pich ra mostaghiman ba ham moghayese kard.
a)neshan dahid ke baraye $n=2$ masale ra dar badtarin halat mitavan ba do azmoon hal kard.
b)raveshi erae dahid ta betavan masale ra dar haalate kolli ba $2n-2$ azmoon hal kard.(injash esteghra dare!)
bedrood

tromino ine ke ye khane bechasbooni be ye tarafe domino (na dar emtedadesh) va shabih be L shavad albatte ba se khane.

dar yek mehmaani n nafar hozoor darand.saabet konid $2$ nafar vjood daran be tori ke betavan az miyan $n-2$ nafare baghi mande haddeaghal $[\frac n{2}]-1$ nafar joda kard ke harkodam yahar do in afraad ra beshnasand va hich kodam ra nashnasand.dar nazar dashte bashid ke ashnayi rabetei do tarafe ast.
($[x]\leq{x}<[x]+1$ yani joze sahihe $x$ ast)

hala ham ye soaale bi rabt be topic va dar mabhase "dirty mathematics":
ma yek noshaabe darim ke mikhahim oon ro be daste $10$ nafar beyne hamoon $10$ nafar taghsim konim.hame ham ba ham sare taghsim kardan ekhtelaaf darand (masalan taghsim bandie hich kas ro baraye khodeshoon ghabool nadaran.)hala shoma rah kari eraae konid ke betoonim nooshabe ro taghsim konim.albatte deghat konid ke eddeaye har kas bayad manteghi bashad va masalan kasi nemitoone bege ke man hameye nooshabe ro mikham.

inam jalebe albatte be topic kheili rabt nadare:
do nafar bazie zir ra ba zrafi ke shaamele n keshmesh ast anjaam midahand:
har bazikon dar nobate khod mitavanad $1,3$ ya $5$ keshmesh bar daarad.bazi koni ke aakharin keshmesh ra bardaarad baazande ast.aya bazikone avval esterategie bord darad?

heyfe ke jaee bahs esteghra beshe va in masale kheily ghashang az esteghra matrah nashe az tot
there is a large pile of cards. on each card one of numbers 1,2,3,...,n is written. it is know that thesum of all numbers of all the cards is equal to k.n! for some integer k. prove that it is possible to arrange cards into k stacks so that the sum of numbers written on teh cards in each stack is equal to n! .

agha ma mokhlese hameye combinatorics baazaye roozegarim.manam age ye hafte tarkibiyat nakhoonam degh mikonam.
ye soale dige:
farz konid $x_1$ va $x_2$ rishe haye moaadele ye $x^{2}+px-1=0$ ke $p$ adadi fard ast bashand.gharar dahid $y_n={x_1}^n+{x_2}^n$ .saabet konid ke $y_n$ va $y_{n+1}$ be ham avval hastand.

in ham badak nist:
$51$ hashareye riz daroone yek morabba be zale $1$ gharar darand.neshan dahid dar har lahze hadde aghal $3$ hashare vojood darand ke mitavan anha ra daroone dayerei be shoaae $\frac 1{7}$  ja dad.

saabet konid ke tamaame adad be shekle $12008$,$120308$,$1203308,...$ bar $19$ bakhsh pazir hastan.
va hamin tor saabet konid ke tamaame adaade be shekle $1007$,$10017$,$100117,...$ bar $53$ bakhsh pazir hastand.

say konid chand leme fibonacci ro ke dar ketaabe sterategi hast sabet karde va sepas oona ro ba ye jozi taghirat dar in donbale ham be vojood biyarin va ba esteghra saabet konid
${ g_n }$ ra in tori tarif mikonim:
$g_{n}+g_{n-1}+g_{n-2}=g_{n+1}, n\geq2$ va $g_0=g_1=0$ va $g_2=1$
shayad donbale jalebi bashe.

[quote="mehdi"]avvalin soaal:
haddeaghal chand nemoone az shekle "tromino" ra bayad dar morabbae $8*8$ khanei ja dad ta digar natavan hatta yeki az in shekl ha ra bedoone in ke rooye digaran gharar girad dar an mostaghar kard?(shoravi)[/quote]

age manzoorte "domino" e,
hamoon mostatil e 2*1 e. ;)

[quote="mohamadhosein"]na manzooresh triminoe ke age az yek morabae $2*2$yek khane ra hazf konim hasel mishavad. ;)[/quote]

Be tore kolli be har majmoye hamband az $n$ moraba $n$-mino ggoyand. Masalan be yek mostatile $3\times1$ ham trimino gooyand. ;)  ;)

in ham ye soale toooooop.

be ezaye har addae tabei manande $k$,yek $n$ vojod darad ke be ezaye har efraz az addade $1,2,\dots,n$ be $k$ daste ,
dastei daraye se ozve$x,y,z$ bashad toori ke $x+y=z$ :D

[quote="BaBaK Ghalebi"]yek dayere be shoae $1$ darim
shakhsi ba gam be toole $\alpha$ ke adadi gong ast dore an ghadam mizanad dar yek ghesmat az in dayere godali be toole $\epsilon>0$ vojood darad,neshan dahid $\epsilon$ har ghadr ham ke koochak bashad in shakhs dir ya zood dar godal mioftad :?:[/quote]
ebteda ghaziei ra esbat mikonim:
[b]ghazie:[/b]
farz konid $\alpha<0$ adadi gong bashad hamchenin $a,b \in [0,1]$,$a<b$
dar in soorat $\exists n \in \mathbb {N}$ ke $a<\{n\alpha\}<b$
[b]esbat:[/b]
dayerei be mohite $1$ dar nazar migirim va yek noghte an ra mabda ekhtiar karde va an ra taghsim bandi mikonim
noghate $A_1,A_2,A_3,...$ ke faseleye anha az mabda $\alpha,2\alpha,3\alpha,...$ hastand ra dar nazar migirim
hich dotai az in noghat yeki nistand
ziora agar masalan $A_2,A_s$ yeki shavand natije mishavad:
$r\alpha-[r\alpha]=s\alpha-[s\alpha],\alpha=\frac{[r\alpha]-[s\alpha]}{r-2}$
ke ba gong boodane $\alpha$ tanaghoz darad
pas baraye har bazeye har ghadr koochake $I$ khosoosan $I=(a,b)$ , do noghte $A_p,A_p_+_q$ vojood darand ke faseleye anha az yekdigar az $I$ kamtar ast
zira agar faseleye har donoghte az toole $I$ bishtar bashad tanha motanahi noghte rooye dayere mitavan gereft ke khalafe farze ma ast
az tarafi darim:
$(p+q)\alpha-p\alpha=(p+2q)\alpha-(p+q)\alpha=(p+3q)\alpha-(p+2q)\alpha=...$
dar natije faseleye noghate $A_p_+_q,A_p+{}_2{}_q$ az yekdigar az toole bazeye $I$ kamtar ast va be hamin tartib $A_p_+{}_2{}_q,A_p_+{}_3{}_q,...$
banabarin donbaleye $A_p,A_p_+_q,A_p_+{}_2{}_q,A_p_+{}_3{}_q,...$ daste kam yek noghte darone baze $I$ darad
esbate in gozare ham be vasile asl;e hojreha mohaia mishavad biaid dayere ra be bazehaye be toole $I$ taghsim konid be tori ke yeki az in bazeha baze morede nazar bashad hal agar gharar bashad noghtei dar in baze naioftad angah fasele $2$ noghte motavali bishtar az $I$ mishavad ke in ham tanaghoz darado hokm sabet mishavad
hal be soraghe halle masale miravim:
farz konid godale morede nazar az $2$ te noghteye $s$ rooye dayere emtedad dashte bashad ke $r-s=\epsilon$ tebghe ghazie esbat shode adade $n$ vojood darad ke $s<\{n\alpha\}<r$ yani $s+[n\alpha]<n\alpha<r+[n\alpha]$
banabarin shakhs pas az $n$ gam dar godal mioftad
 ;)  :D
agha sadra didi idasho
man ke khodam ba in kheili hal kardam

[quote="amir2"]sabet dar beine har $2n-1$ addade tabiei .daghighan $n$ adda vojod darad ke majomeshan bar $n$ bakhshpazir ast.

$\mathrm{Paul\ Erdos}$[/quote]
bebinam in hamoon ghazie chevally nist baraye $2n-1$ adade sahih??

[quote="BaBaK Ghalebi"]neshan dahid majmooei az noghat rooye dayere vojood darad be tori ke tahte $davaran$ be ghesmati az khod miravad :?:[/quote]
dayere be mohite yek ra dar nazar begirid yek noghte rooye an ra be onvane mabda entekhab konid
farz konid $\alpha$ adadi gong va mosbar bashad
noghate $\alpha,2\alpha,3\alpha,...$ ra rooye dayere dar nazar migirim in majmooe $(S)$ ba $davaran$ be ghesmati az khod tabdil mishavad zira agar in majmooe ra be andaze $m\alpha$ davaran dahim majmooe $S-\{0,\alpha,...,(m-1)\alpha\}$ be dast miaiad
 ;)  :D

[quote="BaBaK Ghalebi"]farz konid $n\geq 4$ noghte rooye yek dayere be fasele barabar az ham entekhab shodeand
neshan dahid agar yek zir majmooeye $k=[\sqrt{2n+\frac{1}{4}}+\frac{3}{2}]$ ozvi az in noghat dashte bashim angah in zir majmooe shamele $4$ rase yek $zoozanaghe$ mibashad :?:[/quote]
azla va  ghotrhaye yek $n$ zeli montazam $n$ jahat (emtedad) moshakhas mikonand (tavajoh dashte bashid ke aghtare movazi yek jahat ra moshakhas mikonand)
agar $k$ ras az roose $n$ zeli ra dar nazar begirim noghate entehai $\left(\begin{array}{c}{1cr}k\2\end{array}\right)$vatar ra moshakhas mikonand
ba tavajoh be asle hojreha natije mishavad agar bish az $n$ vatar dashte bashim haddeaghal do ta az anha movaziand
hal chon $\left(\begin{array}{c}{1cr}k\2\end{array}\right)>n$ hokme masale sabet mishavad
 ;)  :D

[quote="BaBaK Ghalebi"]khanehaye yek safhe moshabake $7*7$ (ke az $49$ khane tashkil shode ast) ra ba do rang rang kardeim
neshan dahid haddeaghal $21$ mostatil ba $4$ goosheye yek rang  va azlae movazie azlae safheye moshabak dar in safhe vojood darad :?:[/quote]
yek joft az khanehaye yek satr ra joft $khobb$ minamim har gah az yek rang bashand
farz konid dar yek satr $k$ khane sefid va $7-k$ khane siah bashad
dar in soorat dar in satr tedad jofthaye khoob barabar ast ba:
${{\frac{k(k-1)}{2}}+\frac{(7-k)(6-k)}{2}}=k^2-7k+21$
kamtarin meghdare ebarate fogh $9$ ast va be ezaye $k=2,k=4$ be dast miaiad
banabarin dar har satr haddeaghal $9$ jofte $khoob$ darim
pas dar kolle safhe haddeaghal $63$ jofte khoob darim
do jofte $khoob$ ra $aali$ minamim har gah dar sotoonhaye yeksan bashand
be vozooh do jofte $aali$ yek mostatile moredenazar masale ra moshakhas mikonand
tedad halathai ke baraye yek jofte $khoob$ az lahaze mogheiat sotooni vojood darad barabare $\left(\begin{array}{c}{1cr}7\2\end{array}\right)=21$ ast
banabarin bishtar az $42=2*21$ zoje $khoob$e gheire $aali$ nadarim pas haddeaghal $21$ jofte $aali$ vojood darad(lazem be zekr ast ke adade $21$ daghigh mibvashad)

[quote="BaBaK Ghalebi"]aadade tabii $a_1\leq a_2\leq...\leq a_n\leq 2n$ tori hastand ke $k.m.m$ har do ta az anha az $2n$ bozorgtar ast
neshan dahid $a_1>[\frac{2n}{3}]$ :?:[/quote]
[b]$Lemma:$$n+1$ adade az $\{1,2,...,2n\}$ entekhab shode ast.az miane aadade entekhab shode yeki bar digari bakhshpazir ast.
borhan:farz konid $n+1$ adade mafrooz $a_1,a_2,...,a_n_+_1$ bashand anha ra be shekle $a_i=2^{k_i}b_i$ minivisim ke $b_i$ adadi fard bashad dar in soorat $b_1,b_2,...,b_n_+_1$ $n+1$ adade far mibashand ke beine $1,2n$ gharar darand chon dar in baze $n$ adade fard bishtar nadarim pas tebghe asle hojreha do ta az $b_i$ ha manande $b_p,b_q$ barabarand banabarin az do adade $a_p,a_q$ yaki bar digari bakhshpazir ast[/b]
hal be soraghe masale miravim ba borhane kholf farz konid $a_1\leq [\frac{2n}{3}]$ dar in soorat $3a_1\leq 2n$ majmooe $\{2a_1,3a_1,a_2,...,a_n\}$ ke az $n+1$ adade sahihe koochektar ya mosavie $2n$ tashkil shode ast hichkodam bar digari bakhshpazir nist,in matlab ba $Lemma$ bala tanaghoz darad
pas be tanaghoz residim
pas $a_1>[\frac{2n}{3}$
 ;)  :D

[quote="BaBaK Ghalebi"]noghate safhe ra be vasileye do rang rang kardeim neshan dahid yek mostatil ba roose tak rang mitavan peida kard :?:[/quote]
$7$ noghte hamkhat dar nazar migirim
tebghe asle hojreha haddeaghal $4$ ta az in noghta az yek rang masalan ghermez mibashand in noghat ra $R_1,R_2,R_3,R_4$ minamim
in noghat ra rooye do khate movazie khate avalie $tasvir$ mikonim $tasvir$e in noghat rooye khate aval ra $S_1,...,S_4$ va rooye khate dovom $T_1,...,T_4$ minamim agar do noghte az $S_i$ ha ghermez bashand masale hal ast va agar do noghte az $T_i$ ha ghermez bashand baz ham masale hal ast
pas har kodam az do khate $T,S$ haddeaghal $3$ noghte abi darand
dar natije yek mostatil ba rooose abi be dast miaiad va baz ham masale hal ast
 ;)  :D

[quote="amir2"][quote="BaBaK Ghalebi"][quote="amir2"]sabet dar beine har $2n-1$ addade tabiei .daghighan $n$ adda vojod darad ke majomeshan bar $n$ bakhshpazir ast.

$\mathrm{Paul\ Erdos}$[/quote]
farz konid $n+1$ adade mafrooz $a_1,a_2,...,a_n_+_1$ bashand anha ra be shkle $a_i=2^{k_i}b_i$ minivisim ke $b_i$ adadi far bashad
dar in soorat $b-1,b_2,...,b_n_+_1$ $n+1$ adade fard mibashand ke beine $1,n$ gharar darand va chon dar in baze $[\frac{n}{2}]$ ta adade fard vojood darad pas tebghe asle hojreha haddeaghal do ta az $b_i$ha manande $b_r,b_s$ barabarand banabarin az do adade $a_r,a_s$ yeki bar digari bakhshpazir ast
doroste??
 ;)  :D[/quote]
 :o  :o

[quote]farz konid $n+1$ adade mafrooz $a_1,a_2,...,a_n_+_1$[/quote]
kodoom $n+1$ addade mafrooz.

az inke $2n-1$ adad darim che estefadei kardi?

[quote] dar in soorat $b-1,b_2,...,b_n_+_1$ $n+1$ adade fard mibashand ke beine $1,n$ gharar darand  [/quote]
Chera?  beine har $2n-1$ addad ,pas in addad mitoonand kheili bozorg bashand.[/quote]
oh
sharmande man masalaro eshtebahi gereftam :wallbash_red:  :stretcher:

plz!!!!!!!!!!!!!!!!!!!!!!!!
age momkene hale in soval ru baram bezarid
man montazer hal haye shoma hastam  ;)
 $14.$  The number $1$ or $-1$ is written in each cell of a $2000 * 2000$ array. The sum of all the numbers in the array is non-negative. Show that there are $1000$ rows and $1000$ columns such that the sum of the numbers at their intersections is at least $1000$???????  :?:

Rahnamiyi  baraye masaleye zibaye Erdos. Kafi ast barye aadade avval sabet konid. tamame $(x_{i_1}+\dots+x_{i_{p}})^{p-1}$ ha ra begirid va ba ham jam konid. Intori hokme ghavitari sabet mishe va sabet mishe teedae zirmajmeehaye $n$ ozvi az $\{1,\dots,n\}$ mesle $\{i_1,\dots,i_n\}$ ke $p|x_{i_1}+\dots+x_{i_{n}}$ be peimaneye $p$ barabar ba $1$ ast.


Find the smallest positive angle $\theta$ in degrees satisfying

\[\sin^2(2004\theta) +\cos^2(2005\theta)=1.\]

$\sin^2(2004\theta)=\sin^2(2005\theta)$
$\sin(2004\theta)=\sin(2005\theta)$ since $\sin\theta$ is positive for $0<\theta<180^\circ$.

$\sin(2004\theta)=\sin(2005\theta)\rightarrow180^\circ-2004\theta=2005\theta$ since we want the smallest value of $\theta$.
$\theta=\frac{180}{4009}^\circ$


calculus
geometry
geometric transformation
algorithm
linear algebra
superior algebra
superior algebra solved

I agree with Myth.I hope that someone will bring up a proof for this one  :)

I have a proof based on identity
(x+k)k=\sum {i=0..k-1} (-1)k+1-iC(k,i)(x+i)k + k!.

Using this identity we can reduce initial determinant to standart Wan der Warden one.

Enjoy!

Myth,can you show us the introductory part of your proof?Thank you.

What do you mean?

Unfortunately, I have no time to write detailed proof.

I will wait until you're not so busy  :)

Take this idea:
A=(a_ij)1 \leq i,j \leq n, where
a_ij=(a_i+a_j) n-1 . If a_i=i, we have our problem.

Lets compute detA.
Take Pi(X)=detAi(X), where A_i(X) is obtained by replacing a_i by X.
P_i has degree n-1 in X.

Notice that P_i(a_j) is 0 if j<>i (because the determinant has 2 equal lines) and det(A) if a_j=a_i. This means, by Bezout, that (X-a_1)*...*(X-a_(i-1))(X-a_(i+1))*...*(X-a_n)|P_i(X).
You continue from here, I'm lazy and I should study for a Differential Equations Exam.

quite by accident, i found that this problem is included in a book of linear algebra problem set.

Russians have some famous problem set books, the most famous one in China is a math analysis problem set (i don't know how to spell its author in English)

i have the translation..and i don't know its original name...

[quote="Valentin Vornicu"]and I should sleep for the same exam :D:D God have mercy on us, please! ...[/quote]
well :( he didn't have too much mercy. we passed but not with full-score, neither me, nor amfulger, nor pliviu. Fortunately we will get a second chance in autumn ...

[quote="Valentin Vornicu"]well :( he didn't have too much mercy. we passed but not with full-score, neither me, nor amfulger, nor pliviu. Fortunately we will get a second chance in autumn ...[/quote]

do all of you require full-score?

[quote="siuhochung"]chinese version?
May I knwo the name?
See if I can get it.[/quote]

they are old ones... i don't know if they are out of print now

the linear algebra problem set:
线性代数习题集 by И. B.普罗斯库列柯夫

the calculus problem set:
数学分析习题集 by Б. П. 吉米多维奇

i don't know how to translate the Russian names....�ion op_start() { windowop_start() { window

[quote="liyi"]do all of you require full-score?[/quote] we like to belive that we are capable of perfect scores at least in a usual math college exam :D unfortunately it wasn't the case. we under-prepared for the test and now we sufer the consequences.

[quote="PiDeltaPhi"]If you guys,dont achieve perfect scores,i bet that most of the other students failed the whole exam  :D .Siochung,give me that translation on MSN,thx.[/quote]

i dunno how to translate russian names too...

[quote="PiDeltaPhi"]If you guys,dont achieve perfect scores,i bet that most of the other students failed the whole exam  :D .Siochung,give me that translation on MSN,thx.[/quote]
yeah unfortunately many papers were graded 2/10. The passing note is 5/10. We got 8/10 :(

Were the exercises that hard?Or did the prof consider that your given solution was not flawless?

The exam was not all problems. It had a theoretic part also. That was a problem for us if we didn't study. That was not all. Me and Valentin got 8/10 and Pliviu got 9. 2 of our colleagues got 10 and another colleague 9, so there is room for improovement.
The problems were easy but I have managed to err in each of them. Valentin made the exercises correctly but he made less contact with the theoretic subjects.
Our results were bettered by our active presence in seminaries, so the real results are smaller. I don't know about the others, but I am ashamed :(. I have ignored the course over the entire semester and have comed, int he end, to realize that it is quite hard. There is a big difference between the courses and the seminaries.

To put it simple I can solve almost any ODE, but I couldn't prove correctly why I can apply the algorithms I use :D funny isn't it?

The author of the Calculus problem set is B. P. Demidovich (Б.П.Демидович).  I can't find the author of the linear algebra problem set on Google :?

Demidovich's analysis problem set is very famous among all the math dept students. Everyone knows it -- but few of them do the problems.  :D
In Western world, his book "5,000 problems of mathematical analysis" is famous, but this books seems to be designed for engineering courses, not math specialty. The book which is popular in China contains less than 5000 problems. :)

[quote="amfulger"]Take this idea:
A=(a_ij)1 \leq i,j \leq n, where
a_ij=(a_i+a_j) n-1 . If a_i=i, we have our problem.
Lets compute detA.
Take Pi(X)=detAi(X), where A_i(X) is obtained by replacing a_i by X.
P_i has degree n-1 in X.
Notice that P_i(a_j) is 0 if j<>i (because the determinant has 2 equal lines) and det(A) if a_j=a_i. This means, by Bezout, that (X-a_1)*...*(X-a_(i-1))(X-a_(i+1))*...*(X-a_n)|P_i(X).
[/quote]

I must correct the idea. a_ij=a_i+b_j. I must do this to have P_i(X) of degree n-1. I don't actually know how to finish the proof.
It must be smth like P_i(X)=c*(X-a_1)*...*(X-a_(i-1))(X-a_(i+1))*...*(X-a_n).

Was it ODE using Lie groups or something? Because it's pretty strange that you learn about ODE's "again"? So I assume it has to be some other way, like Lie groups.

The solution to the determinant:

[url]http://www.blacksun.de.tc/math/determinant.pdf[/url]

Enjoy!

Here is a texed web version of the determinant:

[url]http://www.sosmath.com/matrix/determ3/determ3.html[/url]


Jack is walking around in Luna Park, and notices an alleyway called 'Infinite Ice Cream'. Jack notes that the 'Infinite Ice-Cream' appears to possess an infinitely large number of people selling ice-cream. Upon walking outside any particular shop, Jack feels a huge compulsion to purchase an ice-cream. For every shop that Jack visits, he is 37% less likely to purchase an ice-cream then at the previous shop. After purchasing an ice-cream, Jack leaves Luna Park. What is the probability of Jack purchasing an ice-cream at the second shop he visits?

Thanks guys !

This depends on how much the probability is that he will buy an ice cream :icecream: at the first store and how the probability goes down (linear, exponential, etc.)

ohhh sorry i left out that accidently lol i have edited my post, it is 37%


I need a sanity check, please.

I'm testing this for uniform convergence (using Weierstrass M-Test):

$\sum_{n=1}^{\infty} \frac{n}{n+1} x^n$

It looks like it would converge for $0<x<1$.

Setting up for the M-test is where my confusion begins.

$\vert \frac{n}{n+1} x^n \vert \leq$ [color=blue]?[/color] $= M_n$ and then show that $\sum{M_n}$ converges.

It's the question mark ([color=blue]?[/color]) that's getting me. Any ideas?

Thanks,
Bailey

[quote="Bailey"]It looks like it would converge for $0<x<1$[/quote]
That's your problem; it doesn't converge uniformly on the whole interval, since the sum is unbounded at one end. Try it on something like $[0,1-\epsilon]$.


trigonometry
real analysis
real analysis theorems

$1> \epsilon > 0$
prove that there are infinitely many integers $n$ for wich $\cos n \geq 1 - \epsilon$

A well known theorem (  :D  ) says that every $x \in [0;1]$ is 'cluster point' of the series $\{an\}$, where a is irrational. Since $\pi$ is irrational, you find by this that $0$ is cluster point of the series $n \mod \pi$, which gives immediatly the result.

Yes the density of {$a_n$} follows from Weyl's criterion.

This is Weyl theorem. It works for $ cos(P(n))$ with $P$ having at least one coefficient incomensurable with $\pi$, I think.


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Let $ \mathbb{K}$ be a field, $ \Phi : \mathrm{GL}_n(\mathbb{K}) \to \mathbb{K}^*$ a morphism.
Suppose that for $ M\equal{}(m_{i,j})$, $ \Phi(M) \in \mathbb{K}(m_{1,1,}, \ldots, m_{n,n})$.
Find $ \Phi$.

P.S.: It might (should) be in this forum already, but I couldn't find it.

The transvections $ T_{ij}=I+E{ij}$, where, if $ i\not=j$, $ E_{ij}$ is the elementary matrix with the $ {i,j}$ element is $ 1$ and the other $ 0$, span $ SL_n(\mathbb{K})$.
$ T_{ij}T_{jk}T_{ij}^{-1}T_{jk}^{-1}=T_{ik}$ if $ i\not=k$. Then $ \phi(T_{ik})=1$ and $ \phi=1$ on $ SL_n(\mathbb{K})$.
Now $ GL_n(\mathbb{K})$ is the semi direct product of $ SL_n(\mathbb{K})$ and ${ \mathbb{K}}^*$ with the product $ (n,k)(n',k')=(n\bar{k}n'\bar{k}^{-1},kk')$ where $ \bar{k}=diag(k,1,\cdots,1)$. $ \phi((n^{-1},1)(n,k))=\phi(1,k)=\phi(n^{-1},1)\phi(n,k)=\phi(n,k)$ and $ \phi(n,k)=\phi(\bar{k})\in{\mathbb{K}(k)}$.
We assume that $ \mathbb{K}(k)$ is the field of fractions.
There exists a morphism $ \psi: \mathbb{K}^*\rightarrow{\mathbb{K}}^*$ s.t. $ \phi(\bar{k})=\psi(k)$. Here $ \psi$ is a rational function. Then there exists $ \alpha\in{\mathbb{Z}}$ s.t. $ \psi(k)=k^{\alpha}$.
Conclusion: $ \phi(A)=(det(A))^\alpha$.

Proof of $ \psi(k)\equal{}k^\alpha$.
Case 1: $ K$ is an infinite field:
There exists $ \alpha \in\mathbb{Z}$ s.t. $ \phi(k)\equal{}k^\alpha{P(k)/Q(k)}$ where $ P,Q$ are polynomials s.t. $ P(1)\equal{}Q(1)\equal{}1,P(0)\not\equal{}0,Q(0)\not\equal{}0$ and $ gcd(P,Q)\equal{}1$.
$ \phi(kk')\equal{}\phi(k)\phi(k')$ gives $ P(kk')Q(k)Q(k')\equal{}P(k)P(k')Q(kk')$ for all $ k,k'\in\mathbb{K}^*$ i.e. for an infinity of values of $ k,k'$. Then this relation is also satisfied when $ k\equal{}0$: $ P(0)Q(0)Q(k')\equal{}P(0)P(k')Q(0)$; thus $ P(k')\equal{}Q(k')$ for all $ k'\not\equal{}0$ and $ \psi(k)\equal{}k^\alpha$.
Case 2: $ K$ is a finite field:
Then the supplementary hypothesis upon $ \phi$ is useless. $ \mathbb{K}^*\equal{}<a>$ (cyclic group). Let $ \psi(a)\equal{}a^\alpha$; then $ \psi(a^u)\equal{}\psi(a)^u\equal{}(a^u)^\alpha$ and $ \psi(k)\equal{}k^\alpha$.


Hey I was working on some problems which I don't have the answers and I would like to check if I got them right (which I don't think so, they seem strange)

[hide="Problem 1"]In triangle $ ABC$ with $ AB\equal{}c,AC\equal{}b,BC\equal{}a$ , find the distance between the baricenter and side $ \overline{BC}$

Answer I found: $ d\equal{}\frac{\sqrt{4a^2c^2\minus{}b^2\plus{}a^2\plus{}c^2}}{4a}$[/hide]

[hide="Problem 2"]Find the relation between cevians $ x$ and $ y$ of a right triangle with hypothenuse equal to $ 9$ knowing that the feet of the cevians divide the hypothenuse in three equal parts.

I found $ 4y^2\minus{}x^2\equal{}135$[/hide]

[hide="Problem 3"]In a right isosceles triangle with leg $ a$ find the distance between incenter $ I$ and baricenter $ G$

I found $ IG\equal{}a\left(\frac{6\minus{}5\sqrt{2}}{3}\right)$[/hide]

[quote="triplebig"]Hey I was working on some problems which I don't have the answers and I would like to check if I got them right (which I don't think so, they seem strange)

[hide="Problem 1"]In triangle $ ABC$ with $ AB \equal{} c,AC \equal{} b,BC \equal{} a$ , find the distance between the baricenter and side $ \overline{BC}$

Answer I found: $ d \equal{} \frac {\sqrt {4a^2c^2 \minus{} b^2 \plus{} a^2 \plus{} c^2}}{4a}$[/hide]

[hide="Problem 2"]Find the relation between cevians $ x$ and $ y$ of a right triangle with hypothenuse equal to $ 9$ knowing that the feet of the cevians divide the hypothenuse in three equal parts.

I found $ 4y^2 \minus{} x^2 \equal{} 135$[/hide]

[hide="Problem 3"]In a right isosceles triangle with leg $ a$ find the distance between incenter $ I$ and baricenter $ G$

I found $ IG \equal{} a\left(\frac {6 \minus{} 5\sqrt {2}}{3}\right)$[/hide][/quote]

 i think that for the problem 1 coordinate geometry will do the trick.


integration
calculus
algebra
function
domain
probability and stats

I'm not sure what is going on in this class so I would appreciate it if someone would explain how to do this to me...

1. Let the random variables X and Y have the joint pmf
a) p(x,y) = 1/3, (x,y) = (0,0), (1,1), (2,2), zero elsewhere
b) p(x,y) = 1/3, (x,y) = (0,2), (1,1), (2,0), zero elsewhere
c) p(x,y) = 1/3, (x,y) = (0,0), (1,1), (2,0), zero elsewhere

2. Let f(x,y) = 2, 0<x<y<1, zero elsewhere, be the joint pdf of X and Y. Show that the conditional means are, respectively, (1+x)/2, 0<x<1, and y/2, 0<y<1. Also show that the correlation coefficient of X and Y is 0.5.

It is hard to comment on #1 because the question of the problem is missing.

As to #2, if $ f(x,y)$ is the joint density of $ X$ and $ Y$, then the conditional expectation of $ X$ under the condition $ Y\equal{}y$ is just $ \frac{\int_{\mathbb R} f(x,y)x\,dx}{\int_{\mathbb R} f(x,y)\,dx}$. I leave it to you to find the integral in question.

The correlation coefficient is $ \frac{E[(X\minus{}EX)(Y\minus{}EY)]}{\sqrt {E[(X\minus{}EX)^2]E[(Y\minus{}EY)^2]}}$. You can also find $ E[(X\minus{}EX)^2]$ as $ E[X^2]\minus{}(EX)^2$.

These are "cookbook" answers. As you see, all you need to do in each case is to perform a few integrations over the triangular domain where $ f>0$.

If you want to know why it all should be done this way, ask your teacher first. If you still have some trouble after that, post your questions here :).


trigonometry
algebra
polynomial
function
geometry
LaTeX
cyclic quadrilateral

Can someone explain to me the following quote and derviations of the "18n trig family", very slowly and in simple words? :wink:
[quote="Mildorf"]The quickest way I know to get into the "18n trig family" is to use Ptolemy on a regular pentagon. You come up with $\cos{108^{\circ}}= \frac{1-\sqrt{5}}{4}$. Then the complete set $\frac{\pm 1+\sqrt{5}}{4}, \sqrt{\frac{5\pm\sqrt{5}}{8}}$ of sines and cosines follows easily. If you don't memorize exact values, at least try to become familiar with some derivation.[/quote]

Also, what is the solution to the following related problem?
[quote] Using the roots of $x^{4}+x^{3}+x^{2}+x+1$ find $\sin 18^\circ$ and $\cos 18^\circ.$[/quote]

Maybe this will help: http://www.math.fau.edu/yiu/EuclideanGeometryNotes.pdf

Look at page $11$

Thank you! Do you have a solution for that polynomial problem?
Also, how do you show (ii) in the the third exercise?

In the exercises, I can't find the segment lengths, but I know how to continue from there to find the trig values, I think. Could someone explain this?

(page 11 of the packet Mario linked to)

How do I find all of the 18n values or each of the three main functions? Or are there only a few that can be found easily? Because Mildorf implies that there are more...

[color=darkblue]Let $ABC$ be an isosceles triangle with $AB=AC=1$ and $A=36^{\circ}\ .$ Denote the point $M\in (AC)$ so that $BM=BC=x\ .$ Thus,  $AM=BM=BC=x$ and $CM=1-x\ .$ From the similarity $\triangle ABC\sim\triangle BCM$ obtain $\frac{1}{x}=\frac{x}{1-x}\ ,$ i.e. $x^{2}+x-1=0$ $\Longrightarrow$ $x=\frac{-1+\sqrt 5}{2}\ .$ Denote the projection $N$ of the point $M$ to the line $AB$ (the middlepoint of the side $[AB]$). Then $\cos 36^{\circ}=\frac{AN}{AM}=$ $\frac{1}{2x}=\frac{1}{-1+\sqrt 5}=$ $\frac{1+\sqrt 5}{4}\ ,$ i.e. $\boxed{\ \cos 36^{\circ}=\frac{1+\sqrt 5}{4}\ }\ .$

[b]Remark 1.[/b] $\|\begin{array}{c}2\sin^{2}18^{\circ}=1-\cos 36^{\circ}\Longrightarrow\sin 18^{\circ}=\frac{-1+\sqrt 5}{4}\\\\ 2\cos^{2}18^{\cic}=1+\cos 36^{\circ}\Longrightarrow \cos 18^{\circ}=\frac{\sqrt{10+2\sqrt 5}}{4}}\end{array}$

[b]Remark 2.[/b] Otherwise, $x=36^{\circ}\ .$ $\Longrightarrow$ $\|\begin{array}{c}3x+2x=\pi\\\ \sin 3x=\sin 2x\end{array}$ $\Longrightarrow$ $\sin x(3-4\sin^{2}x)=2\sin x\cos x$ $\Longrightarrow$ $3-4\sin^{2}x=2\cos x$ $\Longrightarrow$ $4\cos^{2}x-2\cos x-1=0$ $\Longrightarrow$ $\cos x=\frac{1+\sqrt 5}{4}\ .$[/color]

Thank you! How do you find the cosine and sine values for 18, 54, 72, and 108 geometrically? I'm on a computer without a pdf reader, but doesn't the packet have a derivation of the 108 or 18 ones? If so, could a person expliciting type that up on the wiki so I could see it?

EDIT: I did not mean through the regular method (addition formula), I ment geometrically, though now I see the other derivation is nothing special and I rescind my question.

[quote="Virgil Nicula"][color=darkblue][b][u]A geometrical proof of the relation[/u][/b] $\boxed{\ \cos 2x=2\cos^{2}x-1\ }\ .$ Let $ABC$ be an $A$-isosceles triangle and denote $m(\widehat{BAC})=2x<90^{\circ}\ .$ Define the middlepoint $D$ of the side $[BC]$ and the projection $E$ of the point $B$ on the line $AC\ .$ Observe that $\cos 2x=\frac{AE}{AB}$ , $\cos x=\frac{AD}{AB}$ , the quadrilateral $ABDE$ is cyclically , $BD=DE$ and $BE=2\cdot DE\cdot \cos x\ .$ Apply the [u]Ptolemeu's theorem[/u] in the cyclic quadrilateral $ABDE\ : \ AB\cdot DE+BD\cdot AE=BE\cdot AD$ $\Longleftrightarrow$ $DE+BD\cdot\cos 2x=BE\cdot \cos x$ $\Longleftrightarrow$ $DE+DE\cdot \cos 2x=2\cdot DE\cdot\cos^{2}x$ $\Longleftrightarrow$ $2\cos^{2}x=1+\cos 2x\ .$

[b][u]A geometrical proof of the relation[/u][/b] $\boxed{\ \sin (x+y)=\sin x\cos y+\sin y\cos x\ }\ ,$ where $\{x,y\}\subset (0,90^{\circ})\ .$ Let $ABCD$ be a quadrilateral inscribed in the circle with the diameter $[AC]$ and $m(\widehat{BAC})=x$ , $m(\widehat{DAC})=y\ .$ Prove easily that $\|\begin{array}{c}AB=AC\cdot\cos x\ ,\ BC=AC\cdot\sin x\\\ AD=AC\cdot\cos y\ ,\ DC=AC\cdot\sin y\\\ BD=AC\cdot\sin (x+y)\end{array}\|\ .$ Apply the [u]Ptolemeu's theorem[/u] for this quadrilateral : $AB\cdot CD+AD\cdot BC=AC\cdot BD$ $\Longleftrightarrow$ $AC^{2}\cdot\cos x\cdot\sin y+AC^{2}\cdot \cos y\cdot\sin x=AC^{2}\cdot\sin (x+y)$ $\Longleftrightarrow$ $\sin (x+y)=\sin x\cdot\cos y+\sin y\cdot\cos x\ .$

[b]Remark 1.[/b] $18^{\circ}=\frac{1}{2}\cdot 36^{\circ}\ ;\ 54^{\circ}=18^{\circ}+36^{\circ}\ ;\ 72^{\circ}=2\cdot 36^{\circ}=18^{\circ}+54^{\circ}$ a.s.o.

[b]Remark 2.[/b] $\boxed{\ \|\begin{array}{c}x>0\ ,\ y>0\\\ x+y=180^{\circ}\end{array}\Longrightarrow\|\begin{array}{c}\sin x=\sin y\\\ \cos x=-\cos y\end{array}\ }$[/color][/quote]

I should have remembered that those can add...

Funny though, I remembered it for the sin 15 derivations in the packet to get sin 30, 45, 105, etc. Their method with the polygon is a good way to memorize those quickly for a contest if you know the addition formula.

Anyway, I find the demonstration in AoPS 2 of the addition formula to be much clearer and elementary, also the first relation can be easily derived from the second given sinx squared plus cos x squared equals one.

Let ABC be a general triangle. Turn it so BC is the base and draw a perpendicular to BC at H. Set AH=1 and HAB=a, HAC=b. Area ABC=X.

X=1/2*AC*AB*sinBAC=1/2*1/cosb*1/cosa*sinBAC.
X=Area ABH plus Area AHC=1/2*AH*BH+1/2*AH*CH=1/2(tan a + tan b)
Since sinBAC=sin(a+b), sin(a+b)/cosa*cosb=sina/cosa+sinb/cosb and multiply by cosacosb to get the desired relation. Quick and easy to memorize.

[quote="mna851"]Quick unrelated question:  How do you put something in a box as Virigil has done using LaTex?[/quote]

You can find out yourself by hovering your cursor over his LaTeX or quoting his text (you don't have to post it, just click quote and look at it, of course) :D

How do we get BC=AC*sin(x+y) is the cyclic quadrilateral?


Find a positive integer $n$ with five non-zero different digits, which satisfies to be equal to the sum of all the three-digit numbers that can be formed using the digits of $n$.

If the digits are all different, then the sum of all possible three-digit numbers is

$n = 10000a+1000b+100c+10d+e = 12 \cdot 111(a+b+c+d+e)$

Note that $\bmod 9$ we conclude that $9 | a+b+c+d+e$.  Let $a+b+c+d+e = 9d$; then

$n = 11988d$

When $d = 3$ we have $n = \boxed{ 35964}$, with digit sum $27 = 9d$.


AMC USA(J)MO USAMO AIME MATHCOUNTS Alcumus geometry

Hello!!!

I am your host, Dojo, in Survivor.

Sign up now and I will put you into 3 (or 4) different tribes.
Each tribe will be given a set of questions. I will grade the questions and/or proofs.
The tribe with the least score will have to vote off one unproductive player.

Your goal is to be the last person standing and to have your team win.

Rules:

1. You must not use any resources other than each other.
2. You may talk to other team mates though (not restricted to) PM, forum, FTW.
3. You must select only [b]one[/b] team member to submit your answers.
4. You must turn in your answers by the due date.
5. You may not help team members even if you have been voted off.

Remember, have fun!!!

I need 12 or 16 people, depending on turnout by a certain date.

Do we PM you to join, or just post on this topic?

Either way, 007math joins Survivor...AoPS Style!

JOIN JOIN JOIN!

Wait, you put us into tribes?  Use random.org then.

BTW, because this is the MC forum, I don't think there should be proofs :wink:

I'm in.

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

Join. It would be good if we could pick our own tribes.

@AIME_is_hard: No I don't want any "stacked teams".

@AIME15: It's okay if there are no proofs on MATHCOUNTS. This forum is Middle School Tournaments not MATHCOUNTS.  :wink:

[u][i][b]Sign ups!!![/b][/i][/u]
1. 007math
2. iphailedusamo
3. BOGTRO
4. isabella2296
5. AIME15
6. high_flyer10
7. AIME_is_hard

We need 5 more.

I have finished writing the tests, and need to type them out.

The test will consist of 25 normal answer response questions: No explanation will be needed.
There will be one proof question. This is worth 5 points.

In the case of a tie, the team with more of the challenge questions (last five) correct will win.

Dojo, it's a subforum of the MathCounts forum.  Look at some other tourneys; mods have given warnings about level.

All the problems should be solvable. If I can do them, you probably can too. :P

Just don't want to get this cancelled...

As long as all the problems can be solved using mc techniques we'll be fine.

All righty got all the questions all nicely typed up in LaTeX.

Just get more peeps!!!

i'll join

i passed u up in alcumus points btw dojo :P

[u][i][b]Sign ups!!![/b][/i][/u]
1. 007math
2. iphailedusamo
3. BOGTRO
4. isabella2296
5. AIME15
6. high_flyer10
7. AIME_is_hard
8. theprodigy

Not definite on how many more people, possibly 1 or 4, we will see by tomorrow.

[quote="BOGTRO"]Oh I forgot to say I won.

[b] BOGTRO vs. nikeballa96:

BOGTRO wins! 3 games to 2 [/b][/quote]

Hm...doesn't nikebella have to state this is true or false now?

W/E, but i don't htink any1 was watching

nike... where are you?  Thi needs to get clarified.

Okay get it done.

[b]nikeballa96[/b] please choose a challanger.

BTW: You will have to play against everyone else.

The FTW idea for Round 2 was a very bad idea.  I think you should just make a test...by the speed this is going, tis season will be dead...

As always, pm me the answers.
These are due Jan. 14th.

lol..did you make these or are they homework....(you probably made them,i think)

[quote="Dojo"]In genetics, a Punnett Square is used determine the probabilites of a certain trait in an organism.

There are two basic types of inheritance that can happen within punnett squares.
a) Mendelian
b) Incomplete Dominance

Mendelian Inheritance is where if there are two different traits, one dominates over the other.
Mendel's famous pea plant experiment showed just that.
For example, there were two purebred plants. One with green pods and one with yellow pods.
Those two plants were then crossbred.

(Y=yellow, y=green)
The yellow parent is represented by YY while the green is represented by yy.

[asy]draw((0,0)--(0,2));
draw((1,0)--(1,2));
draw((2,0)--(2,2));
draw((0,0)--(2,0));
draw((0,1)--(2,1));
draw((0,2)--(2,2));
label("$Y$",(0.5,2),N);
label("$Y$",(1.5,2),N);
label("$y$",(0,0.5),W);
label("$y$",(0,1.5),W);[/asy]

When the plants were mixed the resulting genotypes (or genetic types) were Yy, Yy, Yy, and Yy.

[asy]draw((0,0)--(0,2));
draw((1,0)--(1,2));
draw((2,0)--(2,2));
draw((0,0)--(2,0));
draw((0,1)--(2,1));
draw((0,2)--(2,2));
label("$Y$",(0.5,2),N);
label("$Y$",(1.5,2),N);
label("$y$",(0,0.5),W);
label("$y$",(0,1.5),W);
label("$Yy$",(0.5,0.5),N);
label("$Yy$",(0.5,1.5),N);
label("$Yy$",(1.5,0.5),N);
label("$Yy$",(1.5,1.5),N);[/asy]

All the plants now had the yellow and the green trait. However, the yellow trait dominated the green trait resulting in all yellow pods. (Which is why it is represented by a capital.)

b)
Later when scientists discovered Mendel's work, they tried to put it to work. Using flowers that were red and white, they expected either red or white flowers:

[asy]draw((0,0)--(0,2));
draw((1,0)--(1,2));
draw((2,0)--(2,2));
draw((0,0)--(2,0));
draw((0,1)--(2,1));
draw((0,2)--(2,2));
label("$R$",(0.5,2),N);
label("$R$",(1.5,2),N);
label("$W$",(0,0.5),W);
label("$W$",(0,1.5),W);[/asy]

However, the got:

[asy]draw((0,0)--(0,2));
draw((1,0)--(1,2));
draw((2,0)--(2,2));
draw((0,0)--(2,0));
draw((0,1)--(2,1));
draw((0,2)--(2,2));
label("$R$",(0.5,2),N);
label("$R$",(1.5,2),N);
label("$W$",(0,0.5),W);
label("$W$",(0,1.5),W);
label("$RW$",(0.5,0.5),N);
label("$RW$",(0.5,1.5),N);
label("$RW$",(1.5,0.5),N);
label("$RW$",(1.5,1.5),N);[/asy]

Which resulted in all pink flowers. (Which is why there were R and W.)

Now for the questions:

1. Assume that there are only two colors in pea plants, green and yellow.

a) If we have a purebred yellow plant (YY) crossbred with a mixed yellow plant (Yy), what are the resulting genotypes and phenotypes, with probabilities (phenotypes are the final result you can see, such as yellow)?

(List all possible types, even if they don't show up).

b) If we have a mixed yellow (Yy) and another mixed yellow plant, what are the resulting genotypes and phenotypes, with probabilities?

(List all possible types, even if they don't show up).

c) If you have two green pods, what is the probability that the offspring will be yellow? Green?

2.
a) If we have a red flower and a white flower that are crossbred, what are the resulting genotypes and phenotypes, with probabilities?

(List all possible types, even if they don't show up).

b) Using the offspring from above, we crossbred 2 of them. What are the resulting genotypes and phenotypes, with probabilities?

(List all possible types, even if they don't show up).

3.
Blood types are a little different then the above. There are three different traits (or alelles).
A, B and O. O is the only recessive (or non-dominant) trait and when A and B are put together, they result in incomplete dominance.

a) If you have a parent that is mixed A and a parent that is full A, what are the resulting genotypes and phenotypes, with probabilities?

(List all possible types, even if they don't show up).

b) If you have a parent that is mixed B and a parent that is AB, what are the resulting genotypes and phenotypes, with probabilities?

(List all possible types, even if they don't show up).

c) If you have a parent that is mixed B and a parent that is mixedl A, what are the resulting genotypes and phenotypes, with probabilities?

(List all possible types, even if they don't show up).

4)
Let there be a planet with flowers that have 3 parents. The traits carried by them are T which is T and short, which is t. If three parents are, respectively, TTT, TTT and TTt, what are the resulting genotypes and phenotypes, with probabilities?

Table:

TTT: Tall
TTt: Medium Tall
TtT: Medium Tall
tTT: Medium Tall
Ttt: Medium Short
ttT: Medium Short
ttt: Short

(List all possible types, even if they don't show up).[/quote]

For me, this is 6th grade HW...

We did this recently and I added a twist on to it.
(See question 4)


While waiting for the bishop to arrive at St Stephen's Anglican Church to lead a service, every person present greeted every other person with a hand shake. Arriving late, the bishop shook hands with only some of the people as he made his way in. Altogether 1933 handshakes had taken place. How many people shook hands with the bishop?

[hide="Solution"]If every person shook hands with every other person, then $ \binom{n}{2}$ handshakes took place. We are looking for a value of n which satisfies both:

$ \binom{n}{2} < 1933$

$ \binom{n}{2} \plus{} n > 1933$

The only value of n which suits is 62.

We now solve $ \binom{n}{2} \plus{} x \equal{} 1933$

$ \binom{62}{2} \plus{} x \equal{} 1933$

$ 1891 \plus{} x \equal{} 1933$

$ x \equal{} 42$

Thus 42 people shook hands with the bishop.[/hide]


For $a=b=c=1$  the ineq becomes  $\frac{3}{\sqrt{5}}\geq 1$ which is true.

I think the ineq is true. Also it is from Serbia 2004 MO

[quote="silouan"]For $a=b=c=1$  the ineq becomes  $\frac{3}{\sqrt{5}}\geq 1$ which is true.
[/quote]
OK. Sorry.

[quote="silouan"]If  $a,b,c>0$ and $abc=1$ prove that

$\frac{1}{\sqrt{b+\frac{1}{a}+\frac{1}{2}}}+\frac{1}{\sqrt{c+\frac{1}{b}+\frac{1}{2}}}+\frac{1}{\sqrt{a+\frac{1}{c}+\frac{1}{2}}}\geq\sqrt{2}$[/quote]

Isn;t this a classical one??

We  use $\sqrt{2x+1}<x+1$ etc.... :?:


"Three men play the old coin-matching game "odd man wins". As usual, in the case of a tie, they just toss again. If the stakes of the players consist of x,y, and z coins, what is the average (or expected) number of tosses that will occur before one of them goes broke?"

If all three flip the same (either heads or tails), nothing happens.  If two flip the same (heads or tails) and the third flips different (tails or heads), the third (the "odd man out") gains one coin from each of the others.


Errr, calculus? Pretty sure theres no other easy way...

[quote="lotrgreengrapes7926"][quote="cincodemayo5590"]Find, with proof, the volume of a 3-dimensional ellipse with foci at $(0,0,0)$ and $(x,y,z)$ and with major axis $a$.[/quote]
[hide] Is a "three-dimensional ellipse" an ellipse rotated about its major axis?

We have the well-known property that $b^{2}+c^{2}=a^{2}$, where $b$ is the semiminor axis, $c$ is half the distance between foci, and $a$ is the semimajor axis.
$b^{2}+(\frac{\sqrt{x^{2}+y^{2}+z^{2}}}{2})^{2}=a^{2}$
$b^{2}=a^{2}-\frac{x^{2}+y^{2}+z^{2}}{4})$
The volume of a prolate spheroid (3-D ellipse) is $\frac{4}{3}\pi ab^{2}=\frac{4}{3}\pi\frac{a}{2}(a^{2}-\frac{x^{2}+y^{2}+z^{2}}{4})=\boxed{\frac{2\pi a}{3}(a^{2}-\frac{x^{2}+y^{2}+z^{2}}{4})}$[/hide][/quote]

Nice  :D  Where'd you find that formula?

(btw i just posted the problem because it seemed interesting, i really did not know how to solve this at the time)

why have such nasty foci??? the volume would be just the same if you had the foci be $(0,0,c)$ and $(0,0,-c)$

[quote="cincodemayo5590"]Nice  :D  Where'd you find that formula?[/quote] Wikipedia and MathWorld. Duh :roll: . :D

http://en.wikipedia.org/wiki/Prolate_spheroid


[hide="Answer"]First, we find the possible combinations for the numbers, which is $10^4$. Next, we find the possible ways to have initials, which is $26^2$ (although I can't think of a persons' initials that are qq or xx

Then, we find the number of ways to insert the initials in the numbers, which is ${5 \choose 1}=5$

To get the answer, we multiply all this stuff together to get $10^4*26^2*5=33800000$[/hide]

If you're interested in a harder problem of this kind : http://www.mathlinks.ro/Forum/viewtopic.php?t=58792 ;)

Just as an addition we can see that there are only 50,000 possibilities for the password of one particular person.

Also, I do know someone whose initials are qq.

[quote="Altheman"]At my school, people's passwords consist of 6 characters, they passwords are made in the following fashion
1)the person's 2 initials are put in the password anywhere such that the letters are right next to each other
2)the other 4 spots consist of random numbers between 0 and 9
ex. 12ab354 would be a password, but 1a2b345 would not be a password
How many unique passwords are there?

i know it is easy, but somewhat interesting since it is a real world application[/quote]

[hide]there are 26^2 ways for the initials

you can treat the initials as a pair so there are 5 places to place the initials

the rest of the numbers can be 10 #'s so 26^2*5*10^4[/hide]


geometry
circumcircle
conics
projective geometry
geometry unsolved

In triangle ABC , I is incenter. A1,A2 are points on BC that:
BI is perpendicular to IA1 and CI is perpendicular to IA2.
Suppose A' is midpoint of arc BC of circumcircle of ABC.
A'A1 and AC intersect at A1'. A'A2 and AB intersect at A2'.
Similiarly construct B1',B2' and C1',C2'.
Prove line passing through A1' and A2', line passing through B1' and B2' and the line passing through C1' and C2' are concurrent.

[color=red][Moderator edit: corrected a little typo][/color]

Construct the circle tangent to BA and BC in C2 and A1 respectively and show tht it's also tangent to the circumcircle. Call the point of tangency B". In the same way we obtain A" and C". We have to show some other things as well:

(1) AA", BB" and CC" are concurrent (call the point X).
(2) A1'A2', B1'B2' and C1'C2' are concurrent in X.

I haven't proved (1) but I imagine it's not that hard.

Proof of (2):

First of all, show that A'A1 cuts the cicumcircle of ABC in A" and that A'A2 cuts it in C". Then use Pascal's theorem on the cyclic hexagon ABB"A'C"C and you'll get that lines BB", CC" and A1'A2' are concurrrent.

I know I didn't prove everything I used, but it would take too much. Maybe I'll post the rest of it in a while. Anyway, this is the main idea.

Very nice problem, Omid, and very good idea, Grobber. I can give a further step of the proof. Let (kb) be the circle tangent to the lines BA and BC in the points C2 and A1, respectively. Similarly, define the circles (kc) and (ka). Then, we must begin by proving that

(0) the circles (ka), (kb), (kc) internally touch the circumcircle of triangle ABC.

Once this is proven, there is a good way to derive (1). In fact, the point X is the external center of similitude of the circumcircle and the incircle of triangle ABC. This is not just a curious remark but it's a fact very useful in our proof. In fact, (1) is very easy to show by [i]defining[/i] the point X as the external center of similitude of the circumcircle and the incircle of triangle ABC, and showing that this X lies on the lines AA", BB" and CC". Why does X lie on these lines? Well, the point X is the external center of similitude of the circumcircle and the incircle. The point B is the external center of similitude of the incircle and the circle (kb) (since both circles touch the lines BA and BC). The point B" is the external center of similitude of the circle (kb) and the circumcircle (since B" is the point of tangency of the two circles). But the Monge theorem states that the three pairwise external centers of similitude of three circles are collinear. Hence, our points X, B and B" are collinear, i. e. the line BB" passes through X. Similarly, the lines CC" and AA" pass through X, too, i. e. the point X lies on the lines AA", BB", CC", proving the assertion (1).

Note that the circles (ka), (kb) and (kc) are called [i]mixtilinear incircles[/i] of triangle ABC, and the point X is the triangle center X(56) in Kimberling's ETC.

Okay, it remains to prove (0). I have seen a trigonometric proof in several places, but a synthetic proof would be much better to my opinion. Your turn, Grobber :-))

  Darij

Nice :D

Actually, when I posted my first message on this I had shown (0), but was too lazy to post the proof :). Here it is now:

I looked at the problem backwards: I took the circle (kb) and it's tangency point with the circumcircle, B" (as in my first message), and tried to show that B"A1 cuts the circumcircle at A', the midpoint of arc BC, which is equivalent to B"A1 being the nisector of BB"C. For a proof of this fact check this:

[url]http://www.mathlinks.ro/viewtopic.php?t=5131[/url]

Look at the third message, the one starting with "Here's a proof of (2)".

I hope it's correct.

I think it's clear that proving it like this (sort of "backwards") also proves (0).

i have noticed that
If i pick any point P inside the triangle.
And AP intersect the circumcircle at A' , also similarly we have B' C'
then the line perpendicular to AP at P cut AB AC at C1 B2
As in the problem before we have A1 A2 B1 B2 C1 C2
and A'A1 A'A2 intersect the other two sides at
A1' A2', samely the three lines A1'A2'  B1'B2'  C1'C2' are concurrent.

[color=red][Moderator edit: Little typo corrected][/color]

Very interesting! Maybe it suffices that the points A, B, C, A', B', C' lie on one conic and the points A1, A2, B1, B2, C1, C2 lie on another conic (in this case we would have a much more general projective result)... but maybe this is not enough. I don't know.

  Darij

I don't know what you want tell.You can write problem so easy that I can know it .Thank! :(

I wanted to say that chaogold's result is a special case of a more general theorem (or, better, conjecture):

Let P be a point in the plane of a triangle ABC; an arbitrary conic k through A, B, C meets the lines AP, BP, CP at the points A', B', C' (apart from A, B, C). Now, let one line through P cut AB and AC at C1 and B2, let another line through P cut BC and BA at A1 and C2, and let yet another line through P cut CA and CB at B1 and A2, such that the points A1, A2, B1, B2, C1, C2 lie on one conic k'.

Now let the lines A'A1 and A'A2 meet the sides AC and AB at A1' and A2', and similarly define B1' and B2', and C1' and C2'.

Then, the lines A1'A2', B1'B2', C1'C2' meet at one point.

In the case of chaogold's theorem, the conic k is the circumcircle of triangle ABC. In order to apply the above result, one must show that points A1, A2, B1, B2, C1, C2 lie on one conic k'. Because of Pascal's theorem, this is equivalent to the fact that three points B2C1 /\ A1A2, C2A1 /\ B1B2 and A2B1 /\ C1C2 are collinear, where the sign "/\" denotes the point of intersection of two lines. But these three points are just the intersections of the perpendiculars to AP, BP, CP through P with the lines BC, CA, AB, respectively, and it is pretty known that these intersections are collinear; hence, the points A1, A2, B1, B2, C1, C2 lie on one conic, and we can apply the above result.

Remains to prove the above general result.

Now everything clear?

  Darij

Glady I discover at present this nice problem (Iran selection test for IMO-2002)
and I compose a other problem with the apllication often the Menelaus's theorem.

I note $D\in AA'\cap BC,\ X\in BC\cap A'_1A'_2,\ Y\in CA\cap B'_1B'_2,\ Z\in AB\cap C'_1C'_2$. [b]

I will prove that [/b] $Z\in XY$. There is my solution.

$\frac {A'A}{A'D}=\left ( \frac {b+c}{a}\right) ^2,\ \frac {A_1B}{A_1C}=\frac {c(s-b)}{S},\ \frac {A_2B}{A_2C}=\frac {S}{b(s-c)}\Longrightarrow \frac {A_1D}{A_1C}=\frac {4R(s-b)-ac}{a(b+c)},$

$\frac {A_2D}{A_2B}$ $=\frac {4R(s-c)-ab}{a(b+c)}\Longrightarrow $

$\frac {A_1'A}{A_1'C}=\frac {b+c}{a^3}\cdot [4R(s-b)-ac],\ \frac {A'_2A}{A'_2B}=\frac {b+c}{a^3}\cdot [4R(s-c)-ab]\Longrightarrow \frac {XB}{XC}$ $=\frac {4R(s-b)-ac}{4R(s-c)-ab}.$

Similarly, $\frac {YC}{YA}=\frac {4R(s-c)-ab}{4R(s-a)-bc},\ \frac {ZA}{ZB}=\frac {4R(s-a)-bc}{4R(s-b)-ac}$. Therefore, $\frac {XB}{XC}\cdot \frac {YC}{YA}\cdot \frac {ZA}{ZB}=1$, i.e. $Z\in XY.$

[b]Remark[/b]. $4R(s-a)-bc\ .s.s.\ 4Ra(s-a)-abc\ .s.s.\ a(s-a)-S=a(s-a)-(s-a)r_a\ .s.s.\ a-r_a$,

where $x.s.s.y\Longleftrightarrow x=y=0\ or\ xy>0$. Thus, $\frac {XB}{XC}=\frac cb\cdot \frac {s-b}{s-c}\cdot \frac{b-r_b}{c-r_c},\ a.s.o.$

But I consider that the [b]Grobber's idea [/b]and the complete [b]Darij's solution [/b] of the initiated problem are the [u][b]BEST[/b][/u] !.


integration
geometry
3D geometry
sphere
calculus
trigonometry
function

Let F be $ F \equal{} ( x^2 z^2 ) i \plus{} (sin xyz) j \plus{} (e^x z) k.$$ Find \int\int \nabla \times F \cdot n dS$

where the region E is above the cone $ z^2 \equal{} x^2 \plus{} y^2$ and inside the sphere centered at (0,0,1) and with radius 1. (so it is $ x^2 \plus{} y^2 \plus{} (z\minus{}1)^2 \equal{} 1).$. I know that they intersect at z = 1, z = 0. So the boundary is at z = 1.

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So I know that I can apply Stokes Theorem here or evaluate the surface integral directly.

If I try Stokes Theorem, I apply it to the boundary $ x^2 \plus{} y^2 \plus{} (z\minus{}1)^2 \equal{} 1$, which can be parametrized by $ r(\theta) \equal{} cos(\theta) i \plus{} sin(\theta) j \plus{} k, r'(\theta) \equal{} \minus{}sin(\theta) i \plus{} cos(\theta) j \plus{} 0 k$

But when I try $ F(r(\theta)) \cdot r'(\theta)$, I get $ \int_0^{2 \pi} \minus{}sin(\theta) cos(\theta)^2 \plus{} cos(\theta) sin( cos(\theta) sin(\theta)).$. Which is pretty much impossible to evaluate, since I never heard of integrating a sine function within a sine function

as for trying to evaluate the surface integral directly... I get the cross product which is...

$ \nabla \times F \equal{} (\minus{}x y cos(xyz))i \plus{} (2x^2 z \minus{} e^x z)j \plus{} (y z cos(xyz) k).$

So I try to evaluate the surface integral through the northern hemisphere. Of $ x^2 \plus{} y^2 \plus{} (z\minus{}1)^2 \equal{} 1)$, I get...

$ \frac{\partial g}{\partial x} \equal{} 2x, \frac{\partial g}{\partial y} \equal{} 2y$

So...

$ P \frac{\partial g}{\partial x}\minus{} Q \frac{\partial g}{\partial y} \plus{} R$
$ \equal{} 2x xy cos(xyz) \minus{} 2y 2x^2 z \plus{} 2y e^y z \plus{} yz cos(xyz)$... which is like totally impossible to evaluate with a surface integral.

So what am I doing wrong?

$ \iint {\nabla  \times \overrightarrow F .ndS}$ mean? there is some misprints, isnt it?

You mean that $ \iint {\frac{\partial }
{{\partial x}}\left( {x^2 z^2 } \right)dydz \plus{} \frac{\partial }
{{\partial y}}\left( {\sin xyz} \right)dzdx \plus{} \frac{\partial }
{{\partial y}}\left( {e^x z} \right)dzdx}$?

No. The notation refers to the [url=http://en.wikipedia.org/wiki/Curl]curl[/url].


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number theory unsolved
number theory

$p^{q}+p^{r}= x^{2}$. wlog $q \leq r$. Then $p^{q}(1+p^{r-q}) = x^{2}$. Since the two factors are coprime and $p$ is a prime, then $q = 2$ and it rewrites as $p^{2}(1+p^{r-2}) = x^{2}$. So $(1+p^{r-2}) = y^{2}$ and $p^{r-2}= (y-1)(y+1)$. If $p$ is odd it's clear the only solutions are $(3, 2, 3)$ and $(3, 3, 2)$.
If $p = 2$ then $(2, 2, 5)$ and $(2, 5, 2)$ are the only solutions.

1)$r=q$ then we have $2p^{q}=x^{2}\longrightarrow x=2k \longrightarrow p^{q}=2k^{2}\longrightarrow p=2 \longrightarrow 2^{q}=2k^{2}\longrightarrow 2^{q-1}=k^{2}$ $k=2^{\frac{q-1}{2}}$ , in this case we have $(p,q,r)=(2,q,q)$ where $q$ is odd prime number .
2)WLOG$q>r$ then we have $p^{r}(p^{q-r}+1)=x^{2}$ but $(p^{r}, p^{q-r}+1)=1 \longrightarrow p^{r}=x^{2}$ and $p^{q-r}+1=y^{2}\longrightarrow r=2$ . we get $p^{q-2}+1=y^{2}\longrightarrow p^{q-2}=(y-1)(y+1)$, now if $p\neq2$ then $(y-1,y+1)=1 \longrightarrow y-1=1 , y+1=p^{q-2}\longrightarrow p=q=3$ which gives 2 solution $(p,q,r)=(3,3,2)$ or $(p,q,r)=(3,2,3)$ if $p=2$ then $y=2k+1 \longrightarrow 2^{q-2}=(2k)(2k+2)$. if$q>4$then we have $2^{q-4}=k(k+1)$ but one of $k$ or $k+1$ is odd $\longrightarrow k=1$ which gives $(p,q,r)=(2,5,2)$ or $(p,q,r)=(2,2,5)$
if $q=3$ then $4k(k+1)=2$ which gives contradiction :)

[quote="mayhem"]Let $p,q,r$ are primes number. Find all $(p,q,r)$ triples make $p^{q}+p^{r}$ is a square.[/quote]

you can use [url=http://www.mathlinks.ro/viewtopic.php?search_id=1855663038&t=145692]this[/url]. also, it can be showed using the same idea that if $p^{a}+p^{b}$ is a perfect square and $a\ge b$, then either 1.$p=2,a=b=2k+1$ or 2.$p=2,a=2k,b=2k-3$ or 3.$p=3,a=2k+1,b=2k$. so your problem has soluitons $(2,q,q),(2,2,5),(3,3,2)$ for an odd prime $q$.


I was just wondering... not that I'll do that, but the USAMTS website says "high enough". Not very specific  :wink:

It varies from year to year. This past year, it was 68/75.

[quote="Adunakhor"][quote="Ignite168"]What would your AMC score be if it becomes necessary during USAMO qualification?[/quote]

I think it's treated as a score of 100 on the AMC 12.[/quote]

Look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=105637]here[/url].

Although Richard Rusczyk doesn't know exactly what the scores become, they actually do base your index on your USAMTS score.


can u restart alcumus

like make a new account

thanks!!! :lol:  :rotfl:

to get a new account, you'd have to make a new account on AoPS, which is forbidden

anyway, you should use Alcumus to help you learn math
if you want to restart, the most likely reason i can think of is that you want to get higher on the scoreboard
Alcumus should not be used in a competitive way
just keep doing problems, learn from them, and get better at math

no its not because of the competition
its because all my problems are chapters ahead of what i am doing

no i dont want to restart because of the competition

but i want to start again because the problems were suited for the lessons

If you haven't already, post your request [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=497&t=429784]here[/url].

I don't think they reset stuff anymore because I made a request and my account still hasn't been resetted.

[quote="tmathman"]It can take a long time because Paul could get many requests a day, and yours could take a couple months.[/quote]

Unless Paul or whoever else manages this stuff decides that your account does not need to be reset.

[quote="paulhryu"]You are allowed to reset your student progress on Alcumus on a case-by-case basis. To avoid abuse of this feature, I have made it such that resetting your progress not only resets your subject and topic scores, it will also [b]erase your quests, secret honors, settings, and levels[/b]. Really, for all purposes, it will be like you never existed on Alcumus before.

[b]This process is completely irreversible. Please do not take this lightly.[/b]

If you would like your student progress reset, please [url=http://aops.com/Alcumus/Reset][b]click here[/b][/url] (don't worry, there's a confirm dialog) and then choose the "Yes, delete my progress!" button.

I'll only allow one reset per year. Zero exceptions.[/quote]

Actually, you could because if you click the "click here," it will show you to a page that you can restart.