Dear Soarer, VASC, Arqady, and other Mathlinkers,

Following inequality is the most beautiful inequality I created in January 2007. It is hard I think but very very nice.

$\bigstar$ Problem (Hungkhtn's, SI inequalities).
Let $a,b,c$ be non-negative real numbers such that $a+b+c=3$. Prove that
\[\frac{a}{b^{2}+c}+\frac{b}{c^{2}+a}+\frac{c}{a^{2}+b}\ge \frac{3}{2}.\]

I never hear about it, Thuan. But my inequality is different. It is not only stronger but also more beautiful.  :)

Dear hungkhtn is the following true ?

$\frac{a}{b^{2}+c}+\frac{b}{c^{2}+a}+\frac{c}{a^{2}+b}\geq \frac{a}{b^{2}+1}+\frac{b}{c^{2}+1}+\frac{c}{a^{2}+1}$

Unfortunately, it is not true, silouan. Try $a=b=1.5,c=0$.

OK... It may be unsolved for weeks. You can prove next the following inequality, created with the same inspiration, but more easier

$\bigstar$
Let $a,b,c$ be non-negative real numbers such that $a+b+c=3$. Prove that
\[\frac{a^{2}+b}{a+b}+\frac{b^{2}+c}{b+c}+\frac{c^{2}+a}{c+a}\ge 3.\]

Certainly, my solution is still true for $p\ge 1$. Because If $p>1$, the inequality is slighter than $p=1$. However, your way is very ingenious, VASC.

[quote="Vasc"][quote="hungkhtn"]Let $a,b,c$ be non-negative real numbers such that $a+b+c=3$. Prove that
\[\frac{a}{b^{2}+c}+\frac{b}{c^{2}+a}+\frac{c}{a^{2}+b}\ge \frac{3}{2}. \]
[/quote]
We can prove by Cauchy-Schwarz Inequality that for $p\ge 1$ we have
\[\frac{a}{b^{2}+pc}+\frac{b}{c^{2}+pa}+\frac{c}{a^{2}+pb}\ge \frac{3}{1+p}. \]

[/quote]

Sorry, but I can't see the CS..  :(

Great! VASC! I also use your famous inequality
\[(a^{2}+b^{2}+c^{2})^{2}\ge 3(a^{3}b+b^{3}c+c^{3}a)\]
in my second proof (and also similar to toanhocmuonmau's proof). You are really excellent.

[quote="Vasc"]My solution for $p\ge 1$ is based on the inequality
\[\frac{a}{b^{2}+pc}+\frac{b}{c^{2}+pa}+\frac{c}{a^{2}+pb}\ge \frac{(a^{1.5}+b^{1.5}+c^{1.5})^{2}}{\sum{a^{2}(b^{2}+pc)}}=\frac{(a+b+c)(a^{1.5}+b^{1.5}+c^{1.5})^{2}}{3\sum{a^{2}b^{2}+p(a+b+c)\sum{ab^{2}}}}\]
and then on
\[(\sum{a^{2}})^{2}\ge 3\sum{ab^{3}}. \]
[/quote]

As for your previous question but deleted. I have two ways, one is directly expanding and use your inequality. One is EMV method. I will post my solution after 3 months.

Sorry, but I cannot see how to complete Vasc solution

[quote="Vasc"]My solution for $p\ge 1$ is based on the inequality
\[\frac{a}{b^{2}+pc}+\frac{b}{c^{2}+pa}+\frac{c}{a^{2}+pb}\ge \frac{(a^{1.5}+b^{1.5}+c^{1.5})^{2}}{\sum{a^{2}(b^{2}+pc)}}=\frac{(a+b+c)(a^{1.5}+b^{1.5}+c^{1.5})^{2}}{3\sum{a^{2}b^{2}+p(a+b+c)\sum{ab^{2}}}}\]
and then on
\[(\sum{a^{2}})^{2}\ge 3\sum{ab^{3}}. \]
[/quote]

Can anyone help me? Thank you very much. :blush:

My problem is how to solve

$(\sum{x^{3}})^{2}(\sum{x^{2}})\ge 9\sum{x^{4}y^{4}}$

and

$(\sum{x^{3}})^{2}\ge 9\sum{x^{2}y^{4}}$

using Vasc hint. I have tried without success. Sorry.

Thank you very much Arqady. I will study your expansion carefully. :)

$\sum_{cyc}\left(a^{4}+2a^{3}b+2a^{3}c+4a^{\frac{5}{2}}b^{\frac{3}{2}}+4a^{\frac{5}{2}}c^{\frac{3}{2}}+4a^{\frac{3}{2}}b^{\frac{3}{2}}c-14a^{2}b^{2}-3a^{2}bc\right)\geq0,$

How do you prove this one, Arqady. I have tried without success for two days using AM-GM and Shur. :(  Have you a special trick?

Thank you very much for your answer. :)

OK, I will try. :wink:

Thank you very much.

I used
$4\sum a^{\frac 3{2}}b^{\frac 3{2}}(a+b) \ge 8\sum a^{2}b^{2}$,
Schur's Inequality
$\sum a^{4}+abc\sum a \ge \sum ab(a^{2}+b^{2})$
and, finally,
$\sum (\frac{x-y}{z})^{2}(x+y-z) \ge 0$.

:oops:
I really need some help here please....I have to write the whole solution to this inequality ... and I'm kind of stuck here...can someone help me by posting the solution....or by telling me where I can find it....
please  :(
I can see this topic is old....but....HELP!

[quote="hungkhtn"]Great! VASC! I also use your famous inequality
\[(a^{2}+b^{2}+c^{2})^{2}\ge 3(a^{3}b+b^{3}c+c^{3}a) \]
in my second proof (and also similar to toanhocmuonmau's proof). You are really excellent.[/quote]

Can you give me the proof?
thx..

[quote="Vasc"][quote="hungkhtn"]Let $a,b,c$ be non-negative real numbers such that $a+b+c=3$. Prove that
\[\frac{a}{b^{2}+c}+\frac{b}{c^{2}+a}+\frac{c}{a^{2}+b}\ge \frac{3}{2}. \]
[/quote]
We can prove by Cauchy-Schwarz Inequality that for $p\ge 1$ we have
\[\frac{a}{b^{2}+pc}+\frac{b}{c^{2}+pa}+\frac{c}{a^{2}+pb}\ge \frac{3}{1+p}. \]
Actually, the inequality holds for any positive $p$. For $p=2$ we get a nice and easy inequality
\[\frac{a}{b^{2}+2c}+\frac{b}{c^{2}+2a}+\frac{c}{a^{2}+2b}\ge 1. \]
My solution for $p\ge 1$ is based on the inequality
\[\frac{a}{b^{2}+pc}+\frac{b}{c^{2}+pa}+\frac{c}{a^{2}+pb}\ge \frac{(a^{1.5}+b^{1.5}+c^{1.5})^{2}}{\sum{a^{2}(b^{2}+pc)}}. \][/quote]
For any positive $ p $ ,we can prove by use the Cauchy-Schwarz inequality:
\[ \left(\sum \frac{a}{b^2+pc}\right)\left(\sum a(b^2+pc)(2a+2c+b)^2 \right) \ge (2\sum a^2+3 \sum ab)^2. \]
It suffices to show that:
\[ (2\sum a^2+3 \sum ab)^2(1+p) \ge 3 \sum a(b^2+pc)(2a+2c+b)^2. \]
We can obtain this inequality by add two inequalities:
\[ \left(2 \sum a^2+3 \sum ab\right)^2 \ge 3 \sum ac(2a+2c+b)^2. \]
\[ \left(2 \sum a^2+3 \sum ab \right)^2 \left(\sum a \right) \ge  9 \sum ab^2(2a+2c+b)^2. \]
Both two inequalities can prove by $ p,q,r $ method.

For the inequality:
\[ \frac{a}{b^2+c}+\frac{b}{c^2+a}+\frac{c}{a^2+b} \ge \frac{3}{2} .\]
We can prove by use Cauchy-Schwarz inequality and the following inequality of pxchg1200:
\[ a\sqrt{a+b}+b\sqrt{b+c}+c\sqrt{c+a} \ge  3\sqrt{2}.\]
This is nice proof for this nice inequality  :lol:

[quote="quykhtn-qa1"][quote="Vasc"][quote="hungkhtn"]
For the inequality:
\[ \frac{a}{b^2+c}+\frac{b}{c^2+a}+\frac{c}{a^2+b} \ge \frac{3}{2} .\]
We can prove by use Cauchy-Schwarz inequality and the following inequality of pxchg1200:
\[ a\sqrt{a+b}+b\sqrt{b+c}+c\sqrt{c+a} \ge  3\sqrt{2}.\]
This is nice proof for this nice inequality  :lol:[/quote][/quote][/quote]
It isn't the best result. You can see my old results and the proof in here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=300129&hilit=nhocnhoc124
I think Problem nhocnhoc124 is the strongest and nice result with this type :).

[quote="hungkhtn"]$\bigstar$ Problem (Hungkhtn's, SI inequalities).   Let $a,b,c$ be non-negative real numbers such that $a+b+c=3$. Prove that
\[\frac{a}{b^{2}+c}+\frac{b}{c^{2}+a}+\frac{c}{a^{2}+b}\ge \frac{3}{2}.\][/quote]
[quote="hungkhtn"]$\bigstar$  Let $a,b,c$ be non-negative real numbers such that $a+b+c=3$. Prove that
\[\frac{a^{2}+b}{a+b}+\frac{b^{2}+c}{b+c}+\frac{c^{2}+a}{c+a}\ge 3.\][/quote]
 Let $a,b,c$ be non-negative real numbers such that $a+b+c=3$. Prove that
\[\frac{a+c}{b^{2}+c}+\frac{b+a}{c^{2}+a}+\frac{c+b}{a^{2}+b}\ge 3,\]\[\frac{a^{2}+b}{a+c^2}+\frac{b^{2}+c}{b+a^2}+\frac{c^{2}+a}{c+b^2}\ge 3.\]

Dear hungkhtn can you tell us your proof? :)
And is $ \sum_{cyc}^{ }ab^{2}+\sum_{cyc}^{ }ab \leq 6 $ ? :D

Can anybody tell me if this is true? [hide="I posted this to solve this problem :P "][url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=621506&p=3715704#p3715704[/url][/hide]

[quote=hungkhtn][hide=*]Dear Soarer, VASC, Arqady, and other Mathlinkers,

Following inequality is the most beautiful inequality I created in January 2007. It is hard I think but very very nice.

$\bigstar$ Problem (Hungkhtn's, SI inequalities).[/hide]
Let $a,b,c$ be non-negative real numbers such that $a+b+c=3$. Prove that
\[\frac{a}{b^{2}+c}+\frac{b}{c^{2}+a}+\frac{c}{a^{2}+b}\ge \frac{3}{2}.\][/quote]
Let $a,b,c$ be non-negative real numbers such that $a+b+c=3$. Prove that
\[\frac{a}{a+b^{2}}+\frac{b}{b+c^{2}}+\frac{c}{c+a^{2}}\geq 1.\]
[url=http://www.artofproblemsolving.com/community/c6h1200061p5895510]Ukrainian 2016[/url]

Let $a,b,c$ be non-negative real numbers such that $a+b+c=3$. Prove that
$$\frac{a}{a^2+8b}+\frac{b}{b^2+8c}+\frac{c}{c^2+8a}\geq \frac{1}{3}.$$
Hungkhtn,tell me if it is true.

[quote=sqing]
Let $a,b,c$ be non-negative real numbers such that $a+b+c=3$. Prove that
\[\frac{a}{b^{2}+c}+\frac{b}{c^{2}+a}+\frac{c}{a^{2}+b}\ge \frac{3}{2}.\][/quote]
Crux,Problems 3421:
Let $a,b,c$ be positive real numbers such that $ abc\leq1.$ Prove that$$\frac{a}{b^2+b}+\frac{b}{c^2+c}+\frac{c}{a^2+a}\geq \frac{3}{2}$$

[quote=luofangxiang]Let $a,b,c,k$ be positive real numbers such that $ abc\leq1.$ Prove that$$\frac{a}{b^2+kb}+\frac{b}{c^2+kc}+\frac{c}{a^2+ka}\geq \frac{3}{1+k}$$[/quote]
Very very easy.

[quote=JANMATH111]Let $a$, $b$ and $c$ be positive rea numbers such that $a+b+c=3$. Prove that
$$\frac{a^2}{a+b^2}+\frac{b^2}{b+c^2}+\frac{c^2}{c+a^2} \geq \frac{3}{2}.$$[/quote]
https://artofproblemsolving.com/community/c6h401290p2234611
https://artofproblemsolving.com/community/c6h1554403p9471045

[quote=sqing] Let $a,b,c$ be non-negative real numbers such that $a+b+c=3$. Prove that
\[\frac{a+c}{b^{2}+c}+\frac{b+a}{c^{2}+a}+\frac{c+b}{a^{2}+b}\ge 3,\][/quote]
Let $a,b,c$ be positive real numbers such that  $a+b+c=3$ or $abc=1 .$  Prove that
$$\frac{a+b}{a^2+b+c}+ \frac{b+c}{b^2+c+a}+ \frac{c+a}{c^2+a+b}\leq 2$$

[quote=hungkhtn][hide=*]Dear Soarer, VASC, Arqady, and other Mathlinkers,

Following inequality is the most beautiful inequality I created in January 2007. It is hard I think but very very nice.

$\bigstar$ Problem (Hungkhtn's, SI inequalities).[/hide]
Let $a,b,c$ be non-negative real numbers such that $a+b+c=3$. Prove that
\[\frac{a}{b^{2}+c}+\frac{b}{c^{2}+a}+\frac{c}{a^{2}+b}\ge \frac{3}{2}.\][/quote]
[url=https://artofproblemsolving.com/community/c6h2057481p14647802]MTRP 2019[/url]

[quote=pvthuan]But there was the following in 2006 :maybe:

Let $x,y,z$ be positive numbers such that $x+y+z=1$, prove that
\[\frac x{y^{2}+z}+\frac y{z^{2}+x}+\frac z{x^{2}+y}\geq\frac94.\][/quote]
[url=https://artofproblemsolving.com/community/c6h2523191p21416375]Serbia  2006 [/url]

Please someone give me a clear proof of this. Please :(


modular arithmetic
floor function
ceiling function
inequalities
number theory proposed
number theory

Let $ n \ge 2009$ be an integer and define the set:
\[ S \equal{} \{2^x|7 \le x \le n, x \in \mathbb{N}\}.
\]
Let $ A$ be a subset of $ S$ and the sum of last three digits of each element of $ A$ is $ 8$. Let $ n(X)$ be the number of elements of $ X$. Prove that
\[ \frac {28}{2009} < \frac {n(A)}{n(S)} < \frac {82}{2009}.
\]

The last three digits of $ 2^x$ must form a number divisible by 8 and its last digit is nonzero, implying that these three digits can take only values from the following set $ M=\{ 008, 152, 224, 512\}$.

Consider the congruence $ 2^x\equiv m\pmod{1000}$ that is equivalent to $ 2^{x-3}\equiv m/8\pmod{125}$ where $ m\in M$ and thus $ m/8\in \{ 1, 19, 28, 64 \}$. For each of them, the congruence has one solution modulo 125. Therefore, the total number of solutions modulo 125 is 4, implying that
\[ 4\lfloor \frac{n-6}{125} \rfloor \leq n(A) \leq 4\lceil \frac{n-6}{125} \rceil.\]
Since $ n(S)=n-6$, we have
\[ \frac{4}{125} - \frac{4}{n-6} \rfloor \leq \frac{n(A)}{n(S)} \leq \frac{4}{125} + \frac{4}{n-6}.\]
Together with $ n\geq 2009$ the latter inequality gives the required bounds for $ \frac{n(A)}{n(S)}.$


Chào cả nhà!
Mình đang chuẩn bị tham gia 1 cuộc thi Toán ở Mỹ, AMC và được biết đến quyển The Art of Problem Solving. Xem giới thiệu thì thấy cũng khá hay. Mình tìm cả tuần nay mà ko có, cả ở thư viện thành phố và thư viện toàn Bang cũng ko. Có 1 anh bạn đang làm nghiên cứu sinh ngành Toán ở UCLA tìm giúp, thế mà ở UCLA cũng ko có. Hỏi thầy cô ở trường thì chẳng ai biết.
Bạn nào có quyển đó thì share giúp nhé!

Cuốn này đang được bán mà... làm sao có ebook free dc...
Nếu bạn muốn mua thì lên trang http://www.artofproblemsolving.com hình như giá là 14 dollar
Trang đó còn có rất nhiều thông tin bổ ích, cậu nên liếc wa


Four bugs are situated at each vertex of a unit square. Suddenly, each bug begins to chase its counterclockwise neighbor. If the bugs travel at 1 unit per minute, how long will it take for the four bugs to crash into one another?

For those who haven't seen this problem before, try it! (And try to generalize for a regular $n-\text{gon}$, with a bug at each vertex.)

For those who have, please see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=106886]this thread[/url].


linear algebra
matrix
superior algebra
superior algebra solved

Let A={aI_2+bK; a,b\in R} where
K=
(0 1)
(-1 0)
Solve the system
X+Y=I_2
X^3 + Y^3 =7I_2 with X,Y \in A
--------------------------------------------------------
Any pair X,Y in A commute

(1) X+Y=I
(2) (X+Y)(X^2-XY+Y^2)7I

(1)  X+Y=I
(3)  X^2-XY+Y^2 = 7I

the square of (1) gives (4) X^2+2XY+Y^2=I
substract (4) by (3)

(1) X+Y=I
(2) XY=-2I

We look for
X=
(a b)
(-b a)
Y=
(c d)
(-d c)

(1) and (4) gives

ac-bd=-2
bc+ad=0
a+c=1
b+d=0

wich it is not difficult to find a,b,c,d

Hey!

Try using the following to prove it:
f:C-->A, given by:
f(a+bi)=
a b
-b a for any z=a+bi in C
we have that:
1)f((a+bi)(c+di))=f(a+bi)f(c+di)
2)f is a bijection
3)f((a+bi)+(c+di))=f(a+bi)+f(c+di) for any a,b,c,d reals.

cheers!!!

[quote="Moubinool"]
ac-bd=-2
bc+ad=0
a+c=1
b+d=0
[/quote]
I solved the system I found
X=2I and Y=-I
Is there  any solution ?
I see f^(-1) is a way to introduce complex number C, I 'll think how to use
it for the problem.

well there is another solution, as you can see the sistem is symetric so we also have that:
X=-I_2 and Y=2I_2

cheers!


inequalities
number theory unsolved
number theory

Let $ m$ and $ n$ be two distinct positive integers. Prove that there exists a real number $ x$ such that $ \frac {1}{3}\le\{xn\}\le\frac {2}{3}$ and $ \frac {1}{3}\le\{xm\}\le\frac {2}{3}$. Here, for any real number $ y$, $ \{y\}$ denotes the fractional part of $ y$. For example $ \{3.1415\} \equal{} 0.1415$.

Suppose the largest power of 3 that divides n is a, and that divides m is b, and wlog a>b.

Then, any x of the form y/3^(a+1) with y an integer will satisfy the first constaint. It remains to find a suitable y so that the second is satisfied.
However, this y must exist because as we stepwise increase y, the quantity ym/3^(a+1) takes on a number both below and above the range (mod 1): at y=1 it is guaranteed less than 1/3 and at y=3^(a-b) it is guaranteed greater than 2/3

We work in the first quadrant of the standard $ uv\minus{}plane$ studying the

$ R\equal{}\{(u,v)\equal{}(xm,xn): x>0\}$.

Observe that the problem is equivalent to showing that the ray $ R$ contacts at least one of the small $ \frac{1}{3}$ by $ \frac{1}{3}$ squares in the centre of the large standard 1 by 1 lattice squares(considered as $ 3  X  3$ checkerboard).[hide="proof"]For if (xm,xn) lies inside one of the small squares then ({xm},{xn}) lies in the small square {(u,v):$ \frac{1}{3}\le u,v \le \frac{2}{3}$} closest to the origin,as desired.[/hide]

To establish this contact,we assume, wlog, that 0<n<m, so that $ R$ is given by v=$ \frac{n}{m}$u , u>0.

Consider the sequence of rays $ v\equal{}\frac{1}{2} u,v\equal{}\frac{1}{5} u,v\equal{}\frac{1}{8} u,v\equal{}\frac{1}{11} u,\cdots$, u>0. these rays are determined by the lower right corners of the 1st,2nd,3rd,4th,... small central squares.

If,now it is apparent that our ray $ R$ lies between or on the ray v=u and v=u/2 ,or v=u/2 and v=u/5, or ..., and hence R will contact the first or the 2nd or... small square as required. :)

1. Let $ gcd(m,n) \equal{} 1$.
Without loss of generality, we can suppose that $ m$ is not divisible by 3.
Let $ a_1 \equal{} m$, $ a_2 \equal{} 4m$, ..., $ a_{n} \equal{} (3n \minus{} 2)m$. There are $ n$ numbers, and it's easy to prove that if $ a_i \equal{} a_j (mod{3n})$, then $ i \equal{} j$.
So, all of $ n$ rests are different by divisibility on $ 3n$.
And we have that $ a_i \minus{} a_j$ is divisible by 3, it means that all rests of divisibility by $ 3n$ in form $ 3b_i \plus{} C$, where $ C \equal{} 1$ or $ C \equal{} 2$.
So, the set of rests is $ \{2, 5, 8..., (3n \minus{} 1)\}$, or $ \{1, 4, 7,...,(3n \minus{} 2)\}$.
There is exist $ a_k \equal{} (3k \minus{} 2)m$ such that inequality $ n\leq a_k [mod(3n)]\leq 2n$ holds.
Taking $ x \equal{} \frac {a_k}{3mn} \equal{} \frac {3k \minus{} 2}{3n}$, we will have: $ \{xn\} \equal{} 1/3$, and $ \{xm\} \equal{} \frac {a_k}{3n}$,  $ \frac {n}{3n}\leq \{xm\}\leq \frac {2n}{3n}$, $ 1/3\leq \{xm\}\leq 2/3$.
2. If $ m \equal{} dm_1$, $ n \equal{} dn_1$ ($ gcd(m_1,n_1)\equal{}1$), we will find $ x_1$ for pair $ (m_1,n_1)$, and taking $ x \equal{} x_1/d$.


10% of product of a factory are broken down. someone wants to buy 200 goods if at max 5% of products are broken down. find the chance of  purchase?
1)0       2)0.0056   3)0.0068   4)0.0086

c(10,200)* 0.1^ 10 * 0.9 ^ 190 +..+c(1,200) * 0.1^1 * 0.9^199=~0.0086

this is wrong way but I dont know the reason.please solve this problem for me, it is really important .
Best Regards
Sheila

By my understanding, you mean "purchase" as in "buy at least one." 10% > 5%; is not the answer 0? If not, please clarify the question :?


I dunno much about the A flat sclaes and stuff, but I do know that C flat = B and F flat = E. So on the A and E string, first postition, C flat is one, and F flat is open.

All right, I got my scale book  :D

Okay...I looked in my scale book, but couldn't find the C-flat or F-flat major scales/arpeggios.

I have the Carl Flesh Scale System.

Sorry, but I can't help you. :(

[quote="xenon_nightmare152"]I have to play the A flat major arpeggio in 3 octaves on the violin, and as many of you know, every single note in this particular arpeggio is flat.
What is a C flat and what is an F flat?! Please reply in terms of fingering
i.e. low one, high three[/quote]

h_s_potter is right about the fingering. The A-flat major arpeggio is not flat on every note (A-flat, C, E-flat, A flat), although that's the case with G-flat major arpeggio.

Well, it might be written as that since its all flats.  It's kind of like how A# = Bb, but you use Bb in some scales, and A# in others.

[quote="xenon_nightmare152"]thanks h_s_potter and 236!

I didn't know that C flat was B or that F flat was E, and I was playing them natural. :blush:  In my scale book, it is written as an all flat scale.
Now I have to start practicing the Emajor arpeggio in 3 octaves   :dry:[/quote]

Hey, I'm practicing that too!!!


I haven't seen anyone else from Iowa posting in ages.  Where'd you all go?

HI!!!!!!!!!!!!!  lol...
Are you from Iowa?  Supposedly nobody got a 126 on the 10B here...

*walks over to console skyhog*

How many Iowans are registered in AoPS? And why are there so many Indianans??

I think there's a pretty decent group (maybe 10 or so) Iowans, but some graduated high school, and some just never post anymore.  Are [i]you[/i] from Iowa?  *crosses fingers*

No, sadly I'm a Georgian. However, my mom and dad used to live in Iowa (if that counts).

Hmm Ill be in Iowa in either late May/early June for ARML (havent looked up exact date)

lucky.
no middle schoolers allowed on the only ARML team in Virginia.
-_-

why?
even if you make USAMO?
hmm that sucks...


function
integration
real analysis
real analysis solved

P3

Determine all functions f:[0, 4]->[0, \infty ] integrable and concave s.t.

\int 31 f(x) dx=3f(2)/2.


analytic geometry
geometry
geometric transformation
dilation
parameterization

is anyone here good at geometer's sketchpad? i'm just learning. how do you create a segment of a given length or an angle of a given measure?

for example, say i want to draw AB=2 (exactly).

Um... I don't know exactly, but I think there is a coordinate system and you can set one point to an arbitrary location, and place the rest accordingly. For example, if you wanted to construct a triangle of 3-4-5 exactly, you could place A at 0,0; B at 3,0; and C at 0,4. You can usually calculate the coordinates.

Or if you don't want to use coordinate system, you can draw an arbitrary segment, measure it and then dilate it to construct the segment of the desired length.

Alternatively, you can set up a length parameter: Go to Measure / Calculate, then choose Values / New Parameter from the drop-down menu, then enter a label and value for the length of a segment (make sure to tick "cm") or for the angle (make sure to tick "degrees"). Then keep clicking OK till you return to the sketch. Now construct an arbitrary point, and with that point selected, select the parameter you just set up and choose Construct / Circle by Point and Radius. After that just choose any point on the circle, connect it to the center, hide the circle and presto - you're done. IMPORTANT: If you delete your parameter at any point, the segment will be deleted too, as it's Child - hence you must hide it in the final sketch.


a and b are integers such that 2^n.a + b is a square for all non-negative integers n. Show that a = 0.

THis was already solved. Anyway, since the solution is short I will write it again. Suppose a>0 and take x_n= \sqrt (a*2^n+b). Then x_(n+1)-2x_n tends to 0 and so from an n_0 it is 0. It follows immediately that b=0 and so a=0.

Sorry, but why does that limit tend to 0?


The whole numbers are arranged in an array as shown.  If the pattern continues going down one column and up the next, what is the number at the bottom of the 19$ ^{\text{th}}$ column?

\begin{tabular}{|c|c|c|c|c|c|c}
\hline
 $ 1$ & $ 8$ & $ 9$ & $ 16$ & ... & & \\\hline
 $ 2$ & $ 7$ & $ 10$ & $ 15$ & & & \\\hline
 $ 3$ & $ 6$ & $ 11$ & $ 14$ & & & \\\hline
 $ 4$ & $ 5$ & $ 12$ & $ 13$ & & & \\\hline
\end{tabular}

[quote="GameBot"]The whole numbers are arranged in an array as shown.  If the pattern continues going down one column and up the next, what is the number at the bottom of the 19$ ^{\text{th}}$ column?

$ \begin{tabular}{|c|c|c|c|c|c|c} \hline 1 & 8 & 9 & 16 & ... & & \\
\hline 2 & 7 & 10 & 15 & & & \\
\hline 3 & 6 & 11 & 14 & & & \\
\hline 4 & 5 & 12 & 13 & & & \\
\hline \end{tabular}$[/quote]

$ \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c} \hline 1 & 8 & 9 & 16 & & & & & & & & & & & & & & & \\
\hline 2 & 7 & 10 & 15 & & & & & & & & & & & & & & & \\
\hline 3 & 6 & 11 & 14 & & & & & & & & & & & & & & & \\
\hline 4 & 5 & 12 & 13 & 20 & & 28 & & 36 & & 44 & & 52 & & 60 & & 68 & & 76 & \\
\hline \end{tabular}$

bottom of the $ 1st$ column : $ 4 \equal{} 4\times1$
bottom of the $ 3rd$ column : $ 12 \equal{} 4\times3$
bottom of the $ 5th$ column : $ 20 \equal{} 4\times5$
bottom of the $ 7th$ column : $ 28 \equal{} 4\times7$

Thus, bottom of the $ 19th$ column : $ 4\times19 \equal{} \boxed{76}$


for $a;b;c\geq 0$
prove that :
                          $\sum \frac{1}{a^{2}+b^{2}}\geq \frac{5}{2(ab+bc+ca)}$

A moderaditor : spam
NO ONE CAN ????????????? :D

I can indeed, but why should I reveal to you my secret, lesser mortal?

i think it is wrong......

try a=1, b=3/4, c=0

 :P


AMC
AIME
number theory
greatest common divisor
AMC 12
prime factorization
AMC 10

The definition of divisor is: For integers $a$ and $b$, we say that $a$ is a divisor of $b$ if there exists an integer $k$ such that $ak=b$.

But I have seen many different definitions of divisor and factor on the AMC and AIME, so this term has become somewhat ambiguous.

Just to clear up this ambiguity, what would be the correct answer to this hypothetical question on the AMC 12?

What is the set of all divisors of 6?
(A) {-6,-3,-2,-1,1,2,3,6}
(B) {1,2,3,6}
(C) {1,2,3}
(D) {2,3}
(E) a monkey in a green shirt

I think the answer is probably E, hmm... :? :rotfl:

(A) but they would usually say positive integer divisors, which would be (B)

Really? In every AMC/AIME/USAMO/MOP test I have encountered, "divisors" is meant to be {1, 2, 3, 6} in the example. It may or may not have been explicit that they are positive.

I think the context should clue you in as to whether they mean positive integer divisors or integer divisors.  But $C, D$ are definitely wrong if they only say "divisors" without qualifying "proper divisors."

[quote="Philip_Leszczynski"]The definition of divisor is: For integers $a$ and $b$, we say that $a$ is a divisor of $b$ if there exists an integer $k$ such that $ak=b$.
[/quote]

Yeah this is the formal definition I think, therefore negative divisors are allowed.

Well say negative divisors are not allowed, then that means -7 does not have divisors, which creates a big problem (then GCD is restricted to the natural numbers only and we have to revise the entire number theory system, etc).

I'm not sure.  I would say B.  However, when they say "proper divisors," I'm fairly sure that they mean C (see, for example, 2005 AIME 1, No. 3).

[quote="Philip_Leszczynski"]Obviously the "correct" answer is (A) because it is consistent with the formal definition, but we want to know what is the correct answer ON THE AMC. (It would be really nice if an admin answered this and ended it quickly and painlessly.)[/quote]

lol definitely A because AMC uses formal definition I assume.

No, all AMC questions I have seen refer only to positive divisors; essentially, the number of divisors of a number $N=p_1^ap_2^b...$ is $(a+1)(b+1)...$. If you say proper divisors, it's obviously C, by definition. I think that if the AMC ever wanted to include negative divisors, they would specify.

A thorough study of past AMC12 and AIME questions might reveal a bit...

Positive divisors:

1) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367812#p367812
2) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=281801#p281801
3) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=385887#p385887
4) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=394493#p394493

Proper divisors:

1) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=394399#p394399
2) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=365485#p365485

Divisors:

1) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=370839#p370839
2) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=144466#p144466 (irrelevant case, I think)

We note that in Divisors Case 1, in which the divisors are specified to be neither positive nor proper, they are assumed to be positive. Solve the problem out (it is impossible if you consider negative divisors). So strangely enough, the AMC seem to have deviated from the "true" definition of divisors or they misprinted it, but it doesn't really matter I guess since it is in fact impossible using the "true" definition.

It is, however, clear that they will USUALLY specify positive in the problem statement (or proper, if that is the case). I would also say that if they mean for you to consider negative divisors, they will explicitly say so.

Mathematicians seem to be divided themselves on how to define the set of divisors of an integer.

I looked in several reference places and found the following:  On MathWorld.com, Weissstein appears to define divisors as postive divisors, although he acknowledges that negative divisors are possible.  He quotes from Sloane's Encyclopedia of Integer Sequences a table that suggests that Sloane also considers only positive divisors.  The theorem about the number of divisors (exressed in terms of the powers in the prime factorization) is true only if positive divisors are considered.   I grabbed a number theory book off my shelf and found the following quote "we shall usually limit ourselves to a consideration of posiive divisors."

On the other hand, in Herstein's Abstract Algebra book, the definition of "prime number" seems to admit both positive and negative divisors.  Another abstract algebra book does the same.

If I can generalize from this, it seems that mathematicians primarily interested in number theory consider divisors to be positive divisors, and mathematicians interested in abstract algebra consider both positive and negative divisors.  There are probably some mathematical reasons behind this dichotomy.

I will refer the issue to the AMC 10 and AMC 12 chairs, and ask for a definitive definition, and based on what they say, write into our style book a rule about what to say about divisors whenever the term appears.

Steve Dunbar
AMC Director

Did you write anything into the style book yet?


linear algebra
matrix
algebra
polynomial

Let $ A$ be a matrix with order $ n$. Suppose that exist a numer $ m >n$ and $ A^m$ =$ 0$.Prove that $ A^n$ =$ 0$.

every eigenvalue of $ A$ is 0, so $ A$'s characteristic polynomial is $ \lambda^n$. the result follows by Cayley-Hamilton.


Let $ a_{k}$ be the first digit of $ 2^{k}$, and let $ b_{k}$ be the first digit of $ 5^{k}$. Consider the sequences $ a_{k}$ and $ b_{k}$. Prove that for any block of consecutive terms in $ a_{k}$ there is a corresponding block of terms in $ b_{k}$ in reverse order.

[hide="hint"]$ b_{k}$ is the sequence of first digits of $ 2^{-k}$.  And for any $ \epsilon>0$, we can find $ a, b$ such that $ 10^{a}(1-\epsilon)<2^{b}<10^{a}(1-\epsilon)$, because the log base 10 of 2 is irrational.[/hide]


linear algebra
matrix
number theory
greatest common divisor
superior algebra
superior algebra unsolved

Hi

Let R=Z[sqrt(d)] where d is not 1 and is not divisible by the square of a prime. Then the norm is N(a +b(sqrt(d)))=I a^2 - db^2 I
Verify the following:
1) N(x)=0 <=> x=0
2) N(xy)=N(x)N(y)
3) x is a unit iff N(x)=1
4) If N(x) is prime then x is irreducible.

I did the first two ones but for 1) "=>" I got stuck at d=a^2/b^2 and can't conclude
For 3) I can't get "=>"
4) blank

Thanks for any help

Instead of writing elements as $ a \plus{} b\sqrt {d},$ try writing them as $ 2\times 2$ matrices:

$ x \equal{} \begin{pmatrix}a & bd \\
b & a\end{pmatrix}.$

Then $ N(x) \equal{} \det(x).$

For (1), we have that $ N(x) \equal{} 0$ implies $ a^2 \equal{} db^2.$ What if $ a$ and $ b$ are nonzero? (Either one being zero would make the other zero.) Factor out the square of the gcd of $ a$ and $ b$. We have left that $ u^2 \equal{} dv^2$ where $ u$ and $ v$ are relatively prime. Let $ p$ be a prime that divides $ u$ but not $ v.$ We would then have that $ p^2$ divides $ d,$ and that's prohibited by your stated hypotheses.

Note that (2) is immediate: this is a standard property of determinants.

As for (3), let's try inverting the matrix. We get

$ x^{ \minus{} 1} \equal{} \frac1{N(x)}\begin{pmatrix}a & \minus{} bd \\
\minus{} b & a\end{pmatrix}.$

This is only an integer matrix if $ N(x)\mid a$ and $ N(x)\mid b.$ Suppose $ p$ is a prime such that $ p^k$ is the largest power of $ p$ that divides $ N(x).$ Then if $ N(x)\mid a$ and $ N(x)\mid b,$ we would have that $ p^k\mid a$ and $ p^k\mid b.$ But this then gets us that $ p^{2k}\mid N(x).$ This contradicts that $ p^k$ was the largest such power unless $ k \equal{} 0$ Since we can say the same for any prime, there are no primes that divide $ N(x)$ and $ |N(x)| \equal{} 1.$

(4) follows from (2) and (3).  Let $ |N(x)| \equal{} p.$ Suppose $ x \equal{} uv.$ By (2), $ N(x) \equal{} N(u)N(v).$ But the only factors of a prime are units and itself. Hence, either $ |N(u)| \equal{} 1$ or $ |N(v)| \equal{} 1.$ By (3), that means that either $ u$ or $ v$ is a unit. Hence, $ x$ cannot be factored in any nontrivial way, so it is prime.

The Gaussian integers usually refer to the case when $ d\equal{}\minus{}1$. In other words $ G\equal{}\mathbb{Z}[i]$.

Thank you
I had not thought about using matrices. it actually makes more sense now.


geometry
trigonometry
quadratics
function
algebra
quadratic formula
trig identities

$ \triangle ABC$, median AD, $ \angle ABC$=$ \angle DAC$=$ 2\angle ACB$. Find $ \angle ACB$.
Edit:fixed

[quote="dgreenb801"]$ \angle ABC$=...=$ 2\angle ABC$.[/quote]

Is there a typo?

Sorry, I fixed it, it's ACB instead, and ACB is what you want to find.

Edit: I mis-read the problem. Corrected, I have the equation $ \sqrt {2}\sin 2\theta \equal{} \sin 3\theta$. Which gives me $ \theta \equal{} 15$?

My solution:
Applying the tangent formula we have: tg(ABD)=[AB^2+BD^2-AD^2]/4S(ABD)
                                                     tg(DAC)=[AD^2+AC^2-CD^2]/4S(ADC)
Since angle ABD=angleDAC and S(ABD)=S(ADC), we have: [AB^2-AC^2]/2+BC^2/4=AD^2(1)
Applying the median formula we have: AD^2=[AB^2+AC^2]/2 -BC^2/4(2)
From (1) and (2) we get the equation: AC/BC=1/square root(2)
Since BC=2CD we easily infer that: AC/CD=square root(2)
It's equivalent to: sin(CDA)/sin(DAC)=square root(2), which means: sin(2ACB)/sin(180-3ACB)=sin(3ACB)/sin(2ACB)=square root(2)
Now let x=cos ACB we will have the quadratic equation:2x.square root(2)=(4x^2-1)
Solving this equation, we will get two solutions, which is approximately rounded as: 0.965 and -0.258
Using inverse function to find out ACB, which is an acute angle, we will easily infer that -0.35 is eliminated(since cos(acute angle) is always positive). Therefore, angle ACB=cos^(-1)(0.965), which is nearly equal to 15 degrees(15.2036...)(Done!)

Edited: Oopps, my calculation is wrong! I edited it after seeing azip's solution. Anyway, both of us use the same idea

[asy]size(200);
pointpen = black; pathpen= black;
real theta = 15;
pair B=(0,0),C=(2,0),Aa=B+5*expi(2*pi*theta/180),Ab=C+5*expi(pi-pi*theta/180);
pair A=IP(B--Aa,C--Ab), D=(B+C)/2;
D(MP("A",A,N)--MP("B",B)--MP("C",C)--cycle);
D(A--MP("D",D)); MP("\theta",C+(-.34,0),NW);MP("2\theta",B+(.24,0),NE);MP("2\theta",A+(.2,-.09),SE);
D(anglemark(D,A,C));D(anglemark(C,B,A));D(anglemark(A,C,B));[/asy]
$ \triangle ADC \sim \triangle BAC \Longleftrightarrow \frac {DC}{AC} \equal{} \frac {AC}{BC} \Longrightarrow AC \equal{} \sqrt {2} DC$. By [[Law of Sines]] on $ \triangle ADC$ we have $ \frac {AC}{DC} \equal{} \frac {\sin (180 \minus{} 3\theta)}{\sin 2\theta} \equal{} \frac {\sin 3\theta}{\sin 2\theta} \equal{} \sqrt {2}$. So, expanding with double, triple angle formulas, we have
\[ 2\sqrt {2} \sin \theta \cos \theta \equal{} \sin\theta( \minus{} 4\sin ^2 \theta \plus{} 3) \Longrightarrow 4\cos^2\theta \minus{} 2\sqrt {2} \cos \theta \minus{} 1 \equal{} 0.
\]By quadratic formula, we have $ \cos \theta \equal{} \frac {2\sqrt {2} \pm \sqrt {8 \plus{} 4 \cdot 1 \cdot 4}}{8} \equal{} \frac {\sqrt {6} \pm \sqrt {2}}{4}$, so $ \theta \equal{} \boxed{15^{\circ}}$ (because the other roots are too large so don't work).

So now we have three different answers ..

[quote="azjps"][asy]size(200);
pointpen = black; pathpen= black;
real theta = 15;
pair B=(0,0),C=(2,0),Aa=B+5*expi(2*pi*theta/180),Ab=C+5*expi(pi-pi*theta/180);
pair A=IP(B--Aa,C--Ab), D=(B+C)/2;
D(MP("A",A,N)--MP("B",B)--MP("C",C)--cycle);
D(A--MP("D",D)); MP("\theta",C+(-.34,0),NW);MP("2\theta",B+(.24,0),NE);MP("2\theta",A+(.2,-.09),SE);
D(anglemark(D,A,C));D(anglemark(C,B,A));D(anglemark(A,C,B));[/asy]
$ \triangle ADC \sim \triangle BAC \Longleftrightarrow \frac {DC}{AC} \equal{} \frac {AC}{BC} \Longrightarrow AC \equal{} \sqrt {2} DC$. By [[Law of Sines]] on $ \triangle ADC$ we have $ \frac {AC}{DC} \equal{} \frac {\sin (180 \minus{} 3\theta)}{\sin 2\theta} \equal{} \frac {\sin 3\theta}{\sin 2\theta} \equal{} \sqrt {2}$. So, expanding with double, triple angle formulas, we have
\[ 2\sqrt {2} \sin \theta \cos \theta \equal{} \sin\theta( \minus{} 4\sin ^2 \theta \plus{} 3) \Longrightarrow 4\cos^2\theta \minus{} 2\sqrt {2} \cos \theta \minus{} 1 \equal{} 0.
\]
By quadratic formula, we have $ \cos \theta \equal{} \frac {2\sqrt {2} \pm \sqrt {8 \plus{} 4 \cdot 1 \cdot 4}}{8} \equal{} \frac {\sqrt {6} \pm \sqrt {2}}{4}$, so $ \theta \equal{} \boxed{15^{\circ}}$ (because the other roots are too large so don't work).

So now we have three different answers ..[/quote]
I obtain the same answer.  Here is my solution
[hide="Solution"]
Construct points $ E$ on $ AC$ and $ F$ on $ BC$ such that $ \angle ABE \equal{} \angle EBC \equal{} \frac {\angle ABC}{2}$ and $ \angle DAF \equal{} \angle FAC \equal{} \frac {\angle DAC}{2}$.  Now, let $ \angle ABC \equal{} \angle DAC \equal{} 2\alpha$ and $ \angle ACB \equal{} \alpha$.  It follows that $ \angle ABE \equal{} \alpha \equal{} \angle EBC \equal{} \angle ACF \equal{} \angle EAF \equal{} \angle DAF$.  Furthermore, $ \angle BFA \equal{} \angle FAC \plus{} \angle ACB \equal{} 2\alpha$, so $ \angle AFB \equal{} \angle ABC$ and $ \angle DAF \equal{} \angle ACB$, so $ \triangle DAF\sim \triangle ACB$.  Also, $ \angle AFB \equal{} 2\alpha \equal{} \angle ABC$, so $ AF \equal{} AB$ and $ \angle FAC \equal{} \angle FCA \equal{} \alpha$, so $ FC \equal{} FA$.  It follows that $ FD \equal{} CD \minus{} CF \equal{} \frac {BC \minus{} 2AB}{2}$.  Thus, from the similarity,
\[ \frac {AF}{BC} \equal{} \frac {DF}{AB}\implies \frac {AB}{BC} \equal{} \frac {\frac {BC \minus{} 2AB}{2}}{AB}
\]\[ BC^2 \minus{} 2\cdot AB\cdot BC \minus{} 2AB^2 \equal{} 0
\]\[ BC \equal{} AB\cdot (1 \plus{} \sqrt {3})
\]
(ignore the $ 1 \minus{} \sqrt {3}$ because that would make either $ BC$ or $ AB$ negative, which is impossible).  Furthermore, $ \angle ABC \equal{} \angle DAC$ and $ \angle ACB \equal{} \angle ACD$, so
\[ \triangle ABC\sim \triangle DAC\implies \frac {CA}{DC} \equal{} \frac {BC}{AC}\implies 2AC^2 \equal{} BC^2\implies AC\sqrt {2} \equal{} BC
\]\[ AC\sqrt {2} \equal{} AB\cdot (1 \plus{} \sqrt {3})
\]\[ 2\cos \alpha \equal{} \frac {\sin 2\alpha}{\sin \alpha} \equal{} \frac {AC}{AB} \equal{} \frac {\sqrt {2} \plus{} \sqrt {6}}{2}
\]
This gives that $ \cos \alpha \equal{} \frac {\sqrt {2} \plus{} \sqrt {6}}{4}\implies \angle ACB \equal{} \alpha \equal{} 15$.  [/hide]

EDIT: Here is another solution:
[hide="Solution"]
Let $ \angle ABC \equal{} 2\alpha \equal{} \angle DAC$, which would imply that $ \angle ACB \equal{} \alpha$.  Now, let $ E$ be on $ AC$ such that $ \angle ABE \equal{} \angle EBC \equal{} \alpha$, $ F$ be on $ BC$ such that $ \angle DAF \equal{} \angle FAC \equal{} \alpha$, and $ M$ be the midpoint of $ AF$.  Now, notice that $ \angle DAF \equal{} \angle ACB$ and $ \angle AFB \equal{} \angle FAC \plus{} \angle ACF \equal{} 2\alpha \equal{} \angle ABC$, so $ \triangle ADF\sim \triangle CAB$.  Since $ M$ is the midpoint of $ AF$ and $ D$ is that of $ BC$, we have that $ \triangle ADM\sim \triangle CAD\implies \angle ADM \equal{} \angle DAC \equal{} 2\alpha$.  Furthermore, $ \angle EAF \equal{} \alpha\equal{}\angle EBF$, so $ ABFE$ is cyclic.  This means that $ \angle EAF\equal{}\alpha\equal{}\angle ABE\equal{} \angle EFA$, so $ \triangle AEF$ is isoceles and we see that $ EM\perp AF$.  Also, $ \angle EBC \equal{} \alpha \equal{} \angle ECB$, so $ \triangle ECB$ is isoceles and we see that $ ED\perp BC$.  Thus, $ \angle EMF \equal{} \angle EDF \equal{} 90$, so $ EMDF$ is cyclic.  Thus, $ \angle EDM \equal{} \angle EFA \equal{} \alpha$.  Since $ \angle EDC \equal{} 90$ and $ \angle ADB \equal{} \angle DAC \plus{} \angle ACD \equal{} 3\alpha$, we have that $ \angle ADM \equal{} 180 \minus{} \angle ADB \minus{} \angle MDE \minus{} \angle EDF \equal{} 90 \minus{} 4\alpha$.  This means that $ 2\alpha \equal{} \angle ADM \equal{} 90 \minus{} 4\alpha$, so $ \alpha \equal{} 15$, which means that $ \angle ACB \equal{} 15$. [/hide]

[hide="Another solution (similar to previous)"]

WLOG, let BD=DC=1. As azjps showed, by similarity, $ AC \equal{} \sqrt {2}$. Let $ \angle ACD$=x. Let E on DC be the bisector of $ \angle DAC$. Let AD=n. By similarity, AE=$ n\sqrt {2}$ and DE=$ n^2$ (since DC$ \cdot$n=AD). Since $ \triangle AEC$ is isoceles, EC=$ n\sqrt {2}$. So we have DC=1=$ n^2 \plus{} n\sqrt {2}$. So n=$ (\sqrt {6} \minus{} \sqrt {2})/2$. By law of sines on $ \triangle ADC$, (sinx)/n=(sin2x)/1 from which it follows cosx=1/2n=$ (\sqrt {6} \plus{} \sqrt {2})/4$, x=15.[/hide]

I still think there's a non-trig solution shorter than the QuattoMaster's


The Little Twelve Basketball Conference has two divisions, with six teams
in each division. Each team plays each of the other teams in its own division
twice and every team in the other division once. How many conference games
are scheduled?

2 teams: Team A and Team B

games in own division

(5+4+3+2+1)*2*2(2 divisions)=60

games in other division

6*6=36

60+36=96


Not very good at probability... I think this would be how one solves it:
[hide]Given any pair of die, the chances of the sum being 7 is  $ \frac 16 $
Since there are three possible pairs, the probability that none of those pairs will add up to 7 is $ \frac 56 * \frac 56 * \frac 56 = \frac {125}{216} $[/hide]

Obfuscate, the problem with your reasoning is that those probabilities are not independent of each other.

A good approach for this problem is [hide]to consider the dice being rolled one at a time.  How many options do you have for the second and the third?[/hide]

Thanks JBL =) I didn't realize that

[hide="final answer"]For the second die, there is $ \frac 16 $ chance for die 1 and 2 to add up to 7.
For the third die, there is $ \frac 16 * \frac 16 + \frac 56 * \frac 26 = \frac {11}{36} $ since there is a $ \frac 16 $ chance the two previous die are equivalent
Adding $ \frac 16 $ and $ \frac 56 * \frac {11}{36} $ we obtain $ \frac {91}{216} $[/hide]

Did I get it right this time?  :?

Ooh, I wanna try :D

Ok, so for the first roll, it can be anything. Whatever that one dice is, it will have exactly 1 complementary value that will add up to 7 ($7-a$, for any roll $a$, obviously). It can be anything, so it is $\displaystyle \frac{\binom{6} {1}}{\binom{6} {1}} = 1.$ The next dice can be anything EXCEPT that number that is complementary to the first roll. There are 5 such possible rolls. So there are $\displaystyle \frac{\binom{5}{1}}{\binom{6}{1}} = \frac{5}{6}.$ Since there are now two numbers that are NOT complementary, there are now two cases for the third roll. If the first and second roll are different, there are 2 possible complementary numbers for the third roll, which we are trying to prevent. If the first two rolls were the same, there is only ONE complementary number that we are trying to prevent. So we can split this into two cases.
$\displaystyle CASE 1: (\frac{1}{1})(\frac{5}{6})(\frac{4}{6}) = \frac{20}{36},$

                   $\displaystyle CASE 2: (\frac{1}{1})(\frac{1}{6})(\frac{5}{6}) = \frac{5}{36}$

These are mutually exclusive events, so we add:
$\displaystyle \frac{20}{36} + \frac{5}{36} = \frac{25}{36}$

This is probably wrong, but...   :blush: It's my answer.

Ooh! I think I know why. One of my lines:
$\displaystyle \displaystyle CASE 1: (\frac{1}{1})(\frac{5}{6})(\frac{4}{6}) = \frac{20}{36}$
The second fraction is wrong. It should be $\displaystyle \frac{4}{6}$, since we can't have either the complementary roll OR the same number (that's case 2). So, making neccesary changes:
$\displaystyle \displaystyle CASE 1: (\frac{1}{1})(\frac{4}{6})(\frac{4}{6}) = \frac{16}{36}$

$\displaystyle \displaystyle CASE 2: (\frac{1}{1})(\frac{1}{6})(\frac{5}{6}) = \frac{5}{36}$
And, therefore adding,

$\displaystyle \displaystyle \frac{16}{36} + \frac{5}{36} = \frac{21}{36}$

So, the new answer is $\displaystyle \frac{21}{36} = \frac{7}{12}$
How's that? $\displaystyle \frac{7}{12}$... final answer...  :blush:  :blush:  :blush:

there you go. your answer is indeed correct!!!

[quote="amirhtlusa"]this is an easy question from NW math championship:
If three $6$-sided dice are rolled what is the probability that no two of the dice add up to seven?[/quote]

[hide="Solution"]
The probability that the first two don't add up to seven is 5/6.

Two cases - where the first two are equal, and when they're not. They're equal 1/6th of the time, and they're unequal 2/3rds of the time. The other 1/6th, they add up to 7.

If they are equal, then the third die just has to be anything but one number. If they aren't, the third die has to be anything but either of two numbers.

1/6(5/6) + 2/3(2/3) = 5/36 + 4/9 = 5/36 + 16/36 = 21/36 = [b]7/12[/b].
[/hide]


analytic geometry
induction
probability and stats

I have a question similar to #10 here http://books.google.ca/books?id=DyifaCLXxkIC&pg=PA166&lpg=PA166&dq=%22symmetric%22%2B%22walk%22%2B%22integers%22%2B%22distance%22&source=bl&ots=JuZcUNFCok&sig=phRLG5LmSUVYnglNWW0hjX6o6Sg&hl=en&ei=KfSqSfGeJYiSngft4Nj4Dw&sa=X&oi=book_result&resnum=5&ct=result#PPA166,M1

I have a 2D symmetric random walk starting from zero. Again, D_n is the distance from the origin after n steps. I need to show that E((D_n)^2) like #10 is equal to n.

So I know that euclidean distance = sqrt((x-0)^2 + (y-0)^2), so let X_n and Y_n be the coordinates of the particle.
It seems that E(X_n)=E(Y_n)=n/2, but how do I show this?
For n = 1, E(X_1)=(.25)(0^2 + 0^2 + 1^2 + -1^2) = .5 which holds
I can do this similarly for E(X_2), E(X_3) and I get the right answers (i.e 1 and 1.5 respectively), but I'm missing the pattern. Can someone show me how to do this? Does induction work?

Variances of independent distributions add. Each step is independent of position before that step.

if you want to do it by induction
\[ E[X_n^2]\equal{}E[E[X_n^2|X_{n\minus{}1}]]\equal{}E[E[X_{n\minus{}1}^2 \plus{} B_n^2 \plus{} 2B_nX_{n\minus{}1}|X_{n\minus{}1}]]\]
where $ B_n$ is independant of $ X_{n\minus{}1}$ and $ P(B_n\equal{}\minus{}1)\equal{}P(B_n\equal{}\minus{}1)\equal{}\frac12$. You should be able to conclude, isn't it ?


algorithm
combinatorics unsolved
combinatorics

Find the maximum natural number $ N$ such that any arrangement of $ 1,2,..,400$ in  $ 20\times 20$ chessboard
there exist two numbers in the same row or column that differ at least $ N$

I'm very interested in this problem. Could someone post a complete solution?

Let's conjecture that $ N<200$. Set $ 1$ in some cell, then the maximal element of the corresponding row is $ \leq 200$. But we have to put $ 400$ somewhere, therefore, $ N\geq  200$. Now, this is an idea of arranging the elements.

1   11   21   ...   191
2   12   22   ...   192
...
10   20   30   ...  200
201 211 221  ...  391
...
210 220 ... .....   400

The differences equal $ 209$ (columns) and $ 190$ (rows). Now let's find algorithm that reduces differences in columns, therefore lessening $ N$. I shall continue this tommorrow, I'm going to bed right now.

Estimation: $ 200\leq N\leq 209$.

We can get the difference $ \frac{1}{2}(n^{2}\plus{}n\minus{}2)$, for $ n$ x $ n$ table with $ n$ even.
(For $ n \equal{} 20$ the result is $ 209$).
The algorithm to prove this is the same as above. But still I don't know is it the best bound.

It's $ 209$ (for $ n$x$ n$ table, $ n$ even $ \frac{n^{2}\plus{}n\minus{}2}{2}$). OK we have the examples like you said. Let's prove that we can't have all differences in the columns and rows $ \le 208$. Take the first $ 91$ numbers (in the general case $ \left(\frac{n}{2}\right)^{2}\minus{}\frac{n}{2}\plus{}1$). We put them on the table using $ x$ rows and $ y$ columns. Then all the numbers in these rows and columns are $ \le91\plus{}208 \equal{} 299$. We have in these rows and colums $ f(x,y) \equal{} x\cdot20\plus{}y\cdot20\minus{}x\cdot y$ numbers with $ x\cdot y\ge 91$, $ x,y$ naturals between $ 1$ and $ 20$. It's easy to prove that $ f(x,y)$ is minimal with $ x \equal{} y \equal{} 10$, so $ f(x,y)\ge 300$. That means we have at least $ 300$  distinct positive integers $ \le 299$ false. For the general case it is just the same.


Can anyone spin a pencil? I've been trying to learn.

Sorry, this should be in the fun factory forum. How can I move it?

there is a thread about it... like ten lines below:

http://www.mathlinks.ro/viewtopic.php?t=251405

What are you talking about?  This [b]IS[/b] the Fun Factory forum.  Anyway...to spin a pencil, just push it w/ your middle finger so it goes around your thumb and back to where it started off.  Well, that's only one way to spin it...depends what your definition of "spin" is.


[size=200][b]The Rules(read them)[/b][/size]
[list][*]Votes must be in [b]bold[/b].  If you do not bold your vote, it will not be counted.
[*]Once a player has reached lynching state, his pleas are useless and any attempts to unvote will be unheeded. You may [b]vote: no lynch[/b] – majority votes of this kind are necessary to end the day without a death.
[*]The game is not to be discussed outside of the thread unless your role specifically states that you may do so.
[*]Once your death scene has been posted, you’re dead.  Stop typing. The only thing you can type is a "bah" post.
[*]Don’t edit/delete previously submitted posts
[*]Don’t quote any PMs from me. Neither true, nor false. Seriously.
[*]If you have a night choice to make, it is due by the posted deadline.  I will not wait for you – if you do not submit a choice to me, tough!  No choice will be made.
[*]Treat this game as a commitment.  Be considerate – don’t leave us hanging.
[*]Remember, it’s a game.  It’s supposed to played.  It’s supposed to be [i]fun[/i].  Do your part.[/list]

I shall add a few of my own: If you are going to be inactive for a time period please state that you are. If you will be inactive for more than 72 hours, then I shall either mod-kill you or find a replacement. But probably mod-kill.
Everyone has 48 hours to confirm that they have received there role in this topic.

Players:
1. isabella2296
2. crazypianist1116
3. pythag011
4. mz94
5. Bugi
6. abcak
7. james4l
8. Mewto55555
9. echinodermata
10. eggylv999
11. zapi2007
12. yaofan
13. nikeballa96
14. ernie
15. Dojo
16. Mathfreak123

Day starts now.

/confirm.

V/LA until Washington D.C. ceremonies finishes and I arrive at MOP and I get internet access.

Please don't replace me? If you want though, you can get a temporary replacement, but please don't get a permanent one...

I confirm. I'm guessing this game doesn't have a theme.

CONFIRM.

1=2, it takes you 2 week to come up with a no-theme mafia game? :P  :P

 :rotfl:

Players that still need to confirm:
mz94
james4l
Mewto55555
echinodermata
eggylv999
zapi2007
yaofan
nikeballa96
ernie
Dojo

Hello folks.

Question: V/LA = ? I know it has to do with inactivity, but what is it exactly?

i'm here

[quote="james4l"]Question: V/LA = ? I know it has to do with inactivity, but what is it exactly?[/quote]

i was actually wondering that too

[quote="Bugi"]Who looked at the setup :P [/quote]
Peter.
He said 3 was balanced towards town and 4 was balanced towards mafia. Apparently 4 wasn't that balanced towards mafia.

maybe we could have tried having 4 mafia and 1 extra town or something. or two

There is only one factor you need to weigh when taking into account who has an advantage:

Which team is Mewto55555 on?

I think if she gets killed, you skip a day and have consecutive nights, which is good for the mafia.

To anyone in the mafia:
Why did you NK the people that you did?

random.
random.
random.

just kidding

should we post our quicktopic?

Night 1. Killed crazypianist cuz we didn't like him  :P .
Night 2. Killed pythag cuz we didn't like him  :P .
Night 3. Killed Mewto cuz he was quite obviously cop. (Prevailing opinion was that the game couldn't [i]possibly[/i] be balanced with a doc...)
Night 4. Missed the deadline; random.org chose james from our shortlist.
Night 5. Only hope was to hit doc and a confirmed townie in one night, thus sent in Dojo.

I don't object to revealing the quicktopic... but first,
1. zapi please edit my name out of quicktopic post #2. (or i will use my quicktopic admin powers and delete the whole post for you)
2. wait for a few days and see if eggy objects (if he does not respond the assumption shall be "yes").

You know what I found really really funny?
[quote="james4l (post 303)"]I am actually going to partial-counter Mewto's bet. Given the abnormally long night it seems that eggylvl was the mainstay of the lynching squad and the other mafia are inactive.[/quote]
 :rotfl:  :rotfl:  :rotfl:
No, seriously eggy, you need to talk. And you too zapi.

[quote="echinodermata"]
Night 2. Killed pythag cuz we didn't like him  :P ..[/quote]

What... zapi knows me irl, and i think you do as well...?

You are so mean :(

nah, i'm ONLY in your fan club  :P

i'll work on deleting that post

EDIT:  can't make it budge

echino go ahead and nuke it

Good game.

Sorry I didn't play at the end.  I hate summer school.

lastpost (this is used to preserve order of locked topic)


trigonometry
geometry
angle bisector
geometry proposed

In any triangle $ ABC$, let $ AD$ be the interior angle bisector of angle $ A$ then prove that $ AD\equal{}\frac{2.BA.AC}{BA\plus{}AC}\cos\frac{A}{2}$.

Change to the lengthside: $ d_a\equal{}\frac{2\sqrt{bcp(p\minus{}a)}}{b\plus{}c}$ then apply [b]Stewart[/b]'s theorem.

Denote $ AD\equal{}d_a$ and $ E\in AB$ for which $ CE\parallel AD$ . Therefore,

$ [BDE]\equal{}[ABC]$ $ \Longleftrightarrow$ $ BE\cdot AD\cdot\sin\widehat {BAD}\equal{}AB\cdot AC\cdot\sin\widehat {BAC}$ $ \Longleftrightarrow$

$ (b\plus{}c)d_a\sin\frac A2\equal{}bc\sin A$ $ \Longleftrightarrow$ $ (b\plus{}c)d_a\equal{}2bc\cos\frac A2$ $ \Longleftrightarrow$ $ \boxed {\ d_a\equal{}\frac {2bc}{b\plus{}c}\cdot\cos\frac A2\ }$ .


parameterization
geometry
3D geometry
number theory
least common multiple
algorithm
Euclidean algorithm

Find the number of positive integer solutions for the equation

$ x^{2} \plus{} y^{3} \plus{} z^{5} \equal{} t^{7}$

[hide="Solution"]
One solution is $ 10^2 \plus{} 3^3 \plus{} 1^5 \equal{} 2^7$. We can multiply this equation by $ k^{210}$ to get a new solution $ (10k^{105})^2 \plus{} (3k^{70})^3 \plus{} (k^{42})^5 \equal{} (2k^{30})^7$, and there is an infinite number of solutions.
[/hide]

[hide="Three parameters"] Given positive integers $ a, b, c$, let

$ x \equal{} a(a^2 \plus{} b^3 \plus{} c^5)^{45}$
$ y \equal{} b(a^2 \plus{} b^3 \plus{} c^5)^{30}$
$ z \equal{} c(a^2 \plus{} b^3 \plus{} c^5)^{18}$
$ t \equal{} (a^2 \plus{} b^3 \plus{} c^5)^{13}$.

This doesn't cover math154's solutions, although it does give a one-parameter family by setting $ a \equal{} b \equal{} c$.   [/hide]

[hide="Here is my solution"]

Let us consider this identity $ a \plus{} a \plus{}a \equal{} 3a$

Thus $ 3^{k} \plus{} 3^{k} \plus{} 3^{k} \equal{} 3^{k\plus{}1}$

Thus now we can set

$ x^{2} \equal{} 3^{k}$
$ y^{3} \equal{} 3^{k}$
$ z^{5} \equal{} 3^{k}$
 and $ t^{7} \equal{} 3^{k\plus{}1}$

It is clear that $ 3^{k}$ is a square, cube as well as a power of five.

That means that k is a multiple of 2,3 as well as 5.

Also k should be one less than a multiple of 7. {Can you think why?}

So we are looking for a number which is a multiple of 2,3,5 and is 1 less than a multiple of 7.

The least number with this property I can think of is 90.

So, $ (3^{45})^{2} \plus{} (3^{30})^{3} \plus{} (3^{18})^{5} \equal{} (3^{13})^{7}$

And multiplying this equation by $ k^{210}$ gives us the desired result.

[/hide]

Is my solution correct?

yes i think harekrishna's solution is correct

It is correct.
I would do like math154, just I couldn't find one solution with small numbers ( :rotfl: ).
In general, if you find one solution, multiply each of the numbers by $ k^n$, for any $ k$ and $ n$ being the $ LCM$ of all exponents and you have a family of solutions.

[quote="t0rajir0u"][quote="harekrishna"]I am not too sure I understand t0rajir0u's solution. I would be much obliged if he could elucidate.[/quote]
Have you done the algebra?  It's basically a generalization of harekrishna's solution.  I don't know what else there is to elucidate.[/quote]

You meant math154?
And, judging by harekrishna's problems, he doesn't probably know that method...

[hide="Discussion"] With definitions as in my previous post,

$ x^2 \plus{} y^3 \plus{} z^5 \equal{} (a^2 \plus{} b^3 \plus{} c^5)(a^2 \plus{} b^3 \plus{} c^5)^{90}$
$ t^7 \equal{} (a^2 \plus{} b^2 \plus{} c^5)^{91}$.

That's it.  Setting $ a \equal{} b \equal{} c \equal{} 1$ gives harekrishna's solution. [/hide]

Maybe he wanted (and also I want, I don't know many methods for solving Diophantines) to explain how you got those numbers and exponents?

The basic idea is that it's very hard to avoid common factors, so instead we try to use common factors to our advantage and get them to factor out nicely.  I first saw this particular construction used by Rust, but scorpius119 is fond of similar ideas.  The exponents are there for the same reason they are there in harekrishna's solution.  One way to think about it is via the guess or "ansatz"

$ x \equal{} ak^p$
$ y \equal{} bk^q$
$ z \equal{} ck^r$
$ t \equal{} k^s$

for some choice of $ a, b, c, k, p, q, r, s$ such that the powers of $ k$ cancel out nicely.

Another way to think about such Diophantine equations is in contrast to Fermat's Last Theorem:  the reason it's relatively easy to generate solutions is that the exponents are different enough that you can do some kind of Euclidean algorithm on them.

[quote="t0rajir0u"]One way to think about it is via the guess or "ansatz"[/quote]
Sorry to be deviating from the topic but
what is "ansatz"?


Is it possible for a perfect square to end in 11? Prove your answer. :D

[hide="Solution"]
If a number ends in $ 11$ this implies that the number is congruent to $ 3\pmod{4}$.

We know that in order for a number to be a perfect square it must be either $ 0$ or $ 1\pmod{4}$, and since $ 3\not\equal{} 0,1\implies$ a number that ends in $ 11$ cannot be a perfect square.
[/hide]

[quote="calculatorwiz"]no it is not possible for a perfect ssquare to end in 11

if you take a pattern of 11^1ends in 11

11^2=>21

11^3=>31

.......11^9=>91

11^10=>01

11^11=>11

but the question is asking perfect squares.

the power of 11  is an odd numberso it is not a perfect square.[/quote]

How does this prove anything at all concerning the question?

I am sorry @ngoctu77 but I don't understand how that has anything to do with the question.

It proves that any perfect square cannot end in a 11 because 11^x. x has to equal an even number to equall 11 at the end but only an odd number value ends in a 11 but that is not an perfect square

and there are no other palindromic squares,cubes,x^4 ... besides 11

and also all squares  have a even number as there tens digit

is that enough proof for you pascal 12

tonypr's solution is very clean and nice.

however, I still don't understand why you choose the number 11 as the root. This doesn't prove cases for other numbers as the root.


symmetry
combinatorics unsolved
combinatorics

[quote="aidan"]The relation $ S$ contains the relation $ R$ if $ aSb$ whenever $ aRb$. Let $ R$ be the relation on $ \mathbb{Z}$, "$ aRb \textup{ if } b \equal{} a \plus{} 3$". How many equivalence relations on $ \mathbb{Z}$ contain $ R$?[/quote]

none, $ R$ isn't an equivalence relation, it violates reflexivity. so, any relation containing it will violate reflexivity.

[quote="JBL"]You probably want to rethink that.  ([i]Any[/i] relation can be made reflexive by just adding the relations $ aRa$ to it for every $ a$ in the set you care about.)[/quote]

yeah, you're right. i guess i was typing before thinking.

the way to start would probably be to find the smallest equivalence relation containing $ R$. close it under equivalence relations so to speak. i.e. reflexivity and symmetry mean you must have the lines $ (x,x \plus{} 3),(x,x\minus{}3)$ and $ (x,x)$ in $ S\subset \mathbb{Z}\times \mathbb{Z}$. transitivity is the harder part.

[i]edit: fixed another stupid mistake[/i]

after a second look, i think it might just be $ (x,x\plus{}3k)$ for $ k \in \mathbb{Z}$.

I thought the only possible equivalence relation $ aSb$ was $ a \equiv b \, (\textup{mod 3})$. If that is true, how do you prove it?

Also, is this the same as $ 2a \equiv 2b \, (\textup{mod 6})$? Otherwise there might be an infinite number of valid equivalence relations containing $ R$ - essentially I want to ask what it means to say that two equivalence relations are the same.

The relations $ R$ and $ S$ are the same if $ aRb \Longleftrightarrow aSb$, so the two in your preceding post are certainly the same.  (Equivalence relations are just a structure of pairs: they don't care how you refer to the pairs at all.)

Note that "everything is equivalent to everything else" is an equivalence relation that contains all other relations on a given set.  shallow is heading in the right direction now.  The equivalence relation aidan gives in the last post is the minimal equivalence relation with the given equivalences, and it has three equivalence classes (things congruent to 0, things congruent to 1, and things congruent to 2).  Where do we go from here?

So we can have a equivalence relation "$ aSb$ if everything is equivalent to everything else"?

As for the 3 congruence equivalence classes - $ a$ and $ b$ must always be in one of them only, right?

Huh?  For any set $ U$, there exists a stupid relation $ S$ on $ U$ given by $ aSb$ for all $ a, b \in U$.  This relation is an equivalence relation.  It also contains all other (binary) relations on the set $ U$.

The equivalence classes of every equivalence relation form a partition of the underlying set.  Two elements of the set can be in the same part (class) for some relation and in different parts (classes) for some other relation.

Probably it would help if you sat down and thought about what this question is asking when we replace the underlying set $ \bf Z$ with the set $ \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.  (Actually, for an even simpler case you might try replacing $ \bf Z$ with $ \{1, 2, 3\}$.)

[quote="JBL"]Huh?  For any set $ U$, there exists a stupid relation $ S$ on $ U$ given by $ aSb$ for all $ a, b \in U$.  This relation is an equivalence relation.  It also contains all other (binary) relations on the set $ U$.[/quote]

I see.. Misunderstood earlier.

[quote="JBL"]The equivalence classes of every equivalence relation form a partition of the underlying set.  Two elements of the set can be in the same part (class) for some relation and in different parts (classes) for some other relation.

Probably it would help if you sat down and thought about what this question is asking when we replace the underlying set $ \bf Z$ with the set $ \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.  (Actually, for an even simpler case you might try replacing $ \bf Z$ with $ \{1, 2, 3\}$.)[/quote]

But if we use the set $ \{1, 2, 3\}$ then if $ a \equal{} 3$, $ b \equal{} 6$ is not in the set. Basically we are trying to find partitions other than partitioning into classes modulo 3?

I think there are $ 10$ such equivalence relations. The minimal, that was already given and then also the relations which we get, when we add the elements $ (1,2), (1,0), (2,1), (2,0), (0,1), (0,2)$ to the minimal relation.

I think there are 5 o.0 Let the equivalence classes be $ A, B, C$ where $ A$ are numbers divisible by 3, $ B$ are numbers $ 1 \textup{ mod }3$, and $ C$ are numbers $ 2 \textup{ mod }3$. Then we can have the following equivalence relations: $ (A \cup B, C)$; $ (A \cup C, B)$; $ (A, B \cup C)$; $ (A \cup B \cup C)$; $ (A, B, C)$.


Prove that [tex]n^2+1>(\sum_{i=1}^{n}t_i).(\sum_{i=1}^{n}\frac{1}{t_i})[/tex], for all [tex]n\geq3[/tex], and [tex]t_i>0[/tex], and [tex]t_i>t_k+t_j[/tex] .

see here :
http://www.mathlinks.ro/Forum/viewtopic.php?t=14091


function
integration
logarithms
calculus
real analysis
real analysis unsolved

Following is not the standard Marcinkiewicz's theorem: Assuming $ b\minus{}a < 1$, show that the function
\[ \int_a^b {\frac{{dy}}
{{\left| {x \minus{} y} \right|\log \frac{1}
{{\delta \left( y \right)}}}}}
\]is finite a.e. in $ F$ where $ F$ is a closed subset of $ (a,b)$ and $ \delta(y)$ is the distance from $ y$ to $ F$.

Any hints?  :|

Integrate with respect to $ x$ over $ F$ and swap the integrals ;)

Thanks fedja but how to estimate $ \log$ function? We actually have to show that
\[ \int\limits_{\left( {a,b} \right)\backslash F}{\left( {\int\limits_F {\frac{{dx}}
{{\left| {x \minus{} y} \right|\log \frac{1}
{{\delta \left( y \right)}}}}} } \right)dy} < \infty.
\]

$ \int_F\frac{dx}{|x\minus{}y|}\le 2 \log\frac 1{\delta(y)}$, so the $ \log$ factor just cancels out.

Thanks.
\[ \int_F {\frac{{dx}}
{{\left| {x \minus{} y} \right|}}}  \equal{} \int_F {\frac{{d\left( {x \minus{} y} \right)}}
{{\left| {x \minus{} y} \right|}}}  \leq 2\int_{\delta \left( y \right)}^1 {\frac{{dt}}
{t}}  \equal{} 2\log t\left| {\begin{array}{*{20}l}
   1  \\
\\
   {\delta \left( y \right)}  \\

 \end{array} } \right. \equal{} 2\log \frac{1}
{{\delta \left( y \right)}}.
\]Now the topic can be closed  :) .


Inequality
algebra
n-variable inequality
IMO
IMO 1971

Let \[ E_n=(a_1-a_2)(a_1-a_3)\ldots(a_1-a_n)+(a_2-a_1)(a_2-a_3)\ldots(a_2-a_n)+\ldots+(a_n-a_1)(a_n-a_2)\ldots(a_n-a_{n-1}). \] Let $S_n$ be the proposition that $E_n\ge0$ for all real $a_i$. Prove that $S_n$ is true for $n=3$ and $5$, but for no other $n>2$.

Take $a_1 < 0$, and the remaining $a_i = 0$. Then $E_n = a_1(n-1) < 0$ for $n$ even,
so the proposition is false for even $n$.

Suppose $n \ge 7$ and odd. Take any $c > a > b$, and let $a_1 = a$, $a_2 = a_3 = a_4= b$,
and $a_5 = a_6 = ... = a_n = c$.
Then $E_n = (a - b)^3 (a - c)^{n-4} < 0$.
So the proposition is false for odd $n \ge 7$.

Assume $a_1 \ge a_2 \ge a_3$.
Then in $E_3$ the sum of the first two terms is non-negative, because $a_1 - a_3 \ge a_2 - a3$.
The last term is also non-negative.
Hence $E_3 \ge 0$, and the proposition is true for $n = 3$.

It remains to prove $S_5$.
Suppose $a_1 \ge a_2 \ge a_3 \ge a_4 \ge a_5$.
Then the sum of the first two terms in $E_5$ is
$(a_1 - a_2){(a_1 - a_3)(a_1 - a_4)(a_1 - a_5) - (a_2 - a_3)(a_2 - a_4)(a_2 - a_5)} \ge 0$.
The third term is non-negative (the first two factors are non-positive and the last two non-negative).
The sum of the last two terms is:
$(a_4 - a_5){(a_1 - a_5)(a_2 - a_5)(a_3 - a_5) - (a_1 - a_4)(a_2 - a_4)(a_3 - a_4)} \ge 0$.
Hence $E_5 \ge 0$.

Here is an intellectually honest solution to this problem.

Let $p(x) = (x - a_1)...(x - a_n)$ then $E_n = p'(a_1) + ... + p'(a_n) $.

If $p$ has repeated roots $a_i=a_j$ then $p'(a_i) = p'(a_j) = 0$. Note also that if $a$ is root of $p$ with multiplicity $m$ then it will stay on the same side of $x$ if $m$ is even but will cross it if $m$ is odd. Further if $a$ is a singular root then $p'(a) < 0$ if the polynomial passes from the positive side of $x$ to the negative side and $p'(a) > 0$ when it passes from negative to positive. This now informs thinking about the different cases. (For clarity let's assume $a_1 \leq a_2 \leq ... \leq a_n$).

If $p$ is even then the polynomial starts in the upper plane and can pass to the lower plane next to a singular root (where $p'$ is negative), and then go back to the upper plane via the remaining (odd number of) repeated roots.

Then if $p$ is odd with degree larger than $7$, then it starts in the lower plane can pass to the upper plane via a $3$ times repeated root, pass back to the lower plane via a singular root and then back to the positive plane via an odd number of repeated roots.

The remaining two cases can then be verified via size considerations.


In Korea the school year begins in March, and ends in Mid-February.

So in 2010, a Korean student in the Class of 2012 will be able to take the AMC 10, as he/she is in 10th grade.

However, if he/she qualifies for the AIME, (and maybe USAMO/USAJMO), by the time he/she takes those tests, he/she will be in 11th grade. So in this case, will this student be eligible for USAJMO?

형 안녕하세요  :P

Blue MOP 된거 축하드려요 ^^

다시 대원외고로 돌아가세요...? ㅇㅅㅇ


calculus
function
LaTeX
calculus computations

Find which of the following numbers is greater ${15^{16} {or} {16^{15}}}$

I solved it very easily by using calculus. Do you have any other way of solving the problem i.e, without using calculus.

Take the $(15\cdot 16)$th root of both sides, so that you need only calculate where $x^{\frac{1}{x}}$ is increasing and decreasing...

[quote="FMako"]How do you solve it using "calculus"?

I tried a few things, but I didn't calculate anything with them.

1)logarithmic expansion
2)binomal expansion
3)approximation of $n^{1/n}$[/quote]

I solved it in the same way as is suggested by blahblahblah.

take $f(x) = x^{\frac 1{x}}$ Then check where the $f(x)$ is increasing or decreasing. You would find that $f(x)$ is decreasing for $x>e$ .
So, $15^{\frac 1{15}} > 16^{\frac 1{16}}$
i.e, $(15^{\frac 1{15}})^{15.16} > (16^{\frac 1{16}})^{15.16}$. So you have the result.

Try to solve using other method.
Can you send me the codes of writing all the mathematical functions in $LATEX$.

Bye. :)

[quote="FMako"]The $n^{1/n}$ method does seem to be the best approach.
What do you mean by codes of latex?
Check the "latex" link on the sidebar.[/quote]
I meant that the different codes like frac for fraction etc in $LATEX$. CAN you send me the link to it.

Bye :)

how would u solve with calculus. unless the $x^{\frac{1}{X}}$ way was what u meant.


inequalities
geometry
incenter
circumcircle
inradius
search
inequalities unsolved

LetI be the incentre of $\triangle ABC$, and let $R$ and $r$ be its circumradius and inradius, respectively. Prove that
$6r\le AI+BI+CI\le \sqrt{12(R^{2}-Rr+r^{2})}$

The left side isn't hard it is true for all $M$ in plane, try to search it in ML
For the right side
we have $(bc+ca+ab)abc=(bc+ca+ab)(\sum aIA^{2})\ge abc(IA+IB+IC)^{2}\Rightarrow (IA+IB+IC)^{2}\le ab+bc+ca\le 9R^{2}$
Thus we need to prove
$9R^{2}\le 12(R^{2}-Rr+r^{2})\Leftrightarrow R^{2}-4Rr+4r^{2}\ge 0$
which is true .


How do you use sigma notation in display style?

The display style section in the aops latex guide has the same code for the display and nondisplay style ways to use sigma notation...http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Math

In general, if something about latex isn't on aops's wiki guide, where do you look?

[[Latex:Math]]

$ \frac{\sum_{i\equal{}sth}^{\pi} f}{t}$ - to show how it looks non-displaystyled

$ \displaystyle \sum_{i\equal{}sth}^{\pi} f$ with displaystyle

It's the same command; just if displaystyled sum is pretty, otherwise not pretty.

There's an error in the Wiki which will need correcting*. There are a number of ways to use displaystyle but things differ between documents and the forum.
1. Defaults
a) In a document you always get display mode when you use \ [...\ ], the equation, align and similar environments. You get text mode when using $ \$$...$ \$$
b) In the forum you get displaystyle if you use $ \$$...$ \$$ or \ [...\ ]
c) As solafidefarms noted, things change if sums are inside fractions

2. Overriding defaults
a) In a document you can force display mode for the whole equation using \displaystyle
b) In the forum you can force text mode by using \textstyle eg ${ \textstyle \lim_{\delta x \to 0}\sum_{i = 1}^n f(x)\delta x} = \int_a^b f(x)\;dx$ instead of ${ \lim_{\delta x \to 0}\sum_{i = 1}^n f(x)\delta x} = \int_a^b f(x)\;dx$
Note how \textstyle affects the whole equation
c) you can change the default behaviour of just one part of an equation by using \limits or \nolimits in a document or the forum. For example, changing the way lim behaves
${ \textstyle \lim\limits_{\delta x \to 0}\sum_{i = 1}^n f(x)\delta x} = \int_a^b f(x)\;dx$ and ${ \lim\nolimits_{\delta x \to 0}\sum_{i = 1}^n f(x)\delta x} = \int_a^b f(x)\;dx$
You can do the same for the sum and the integral.

* I have now corrected the Wiki code.

[quote="solafidefarms"][[Latex:Math]][/quote]
Link should be [[LaTeX:Math]]. Maybe a redirect should be created?


limit
real analysis
real analysis unsolved

For $a_{1},a_{2},\cdots,a_{n},\cdots \in \mathbb R$ and $\sum_{k=1}^{\infty}ka_{k}$ is convergent.Prove that for every $n \ge 1$,$\sum_{k=n+1}^{\infty}(k-n)a_{k}$ is convergent,too.And define $t_{n}=\sum_{k=n+1}^{\infty}(k-n)a_{n}$,then $\lim_{n\to \infty}t_{n}=0$

We are given that the remainder $r_{n}: =\sum_{k=n+1}^{\infty}ka_{k}$ tends to $0$ as $n\to\infty$. Let's try to express $t_{n}$ in terms of $r_{n}$, that is, find $\alpha_{m}$ such that $t_{n}=\sum_{m=n}^{\infty}\alpha_{m}r_{m}$. In order to have \[\sum_{k=n+1}^{\infty}(k-n)a_{k}=\sum_{m=n}^{\infty}\alpha_{m}\sum_{k=n+1}^{\infty}ka_{k}=\sum_{k=n+1}^{\infty}ka_{k}\sum_{m=n}^{k-1}\alpha_{m}\] we must make sure that $\frac{k-n}{k}=\sum_{m=n}^{k-1}\alpha_{m}$, which forces $\alpha_{k}=\frac{k+1-n}{k+1}-\frac{k-n}{k}=\frac{n}{k(k+1)}$. Therefore, $t_{n}=\sum_{m=n}^{\infty}\frac{n}{m(m+1)}r_{m}$. This gives a nice estimate for $t_{n}$: $|t_{n}|\le \sup_{m\ge n}|r_{m}| \sum_{m=n}^{\infty}\frac{n}{m(m+1)}=\sup_{m\ge n}|r_{m}|$.

Complaints about rearrangement of terms in a conditionally convergent series can be sent to the management. :P

[quote="mlok"]$t_{n}=\sum_{m=n}^{\infty}\frac{n}{m(m+1)}r_{m}$[/quote]
Thank you,But I think we should need a strick proof for the formula above.Maybe we should need to prove the difinition of $t_{n}$ is valid first,but this can be proved by $\sum a_{n}=\sum \frac{1}{n}na_{n}$ which is convergent by abel's criterium.


Someone can give me a hand in this equation

$ \ x^2\plus{}\sqrt{x}\equal{}18$

Let $ y\equal{}\sqrt{x},$ giving the quartic equation $ y^4\plus{}y\minus{}18\equal{}0.$

Try looking for integer roots of that (rational root theorem). There is one - and it's the only solution for $ y\ge 0,$ hence the only real root for the original equation.

hello, writing your equation in the form $ \sqrt{x}\equal{}18\minus{}x$ and squaring both sides
and putting all terms to the right hand-side and factoring we get
$ (x\minus{}4)(x^3\plus{}4x^2\minus{}20x\minus{}81)\equal{}0$. From here we will get all possible solutions, but we must make a test if they fulfill our original equation.
Sonnhard.


Do you know how it determines that a subject is mastered or passed? One topic it says I mastered
(Mastery of Counting Fundamentals Level 4) I only did two problems, and I got them both wrong.

alcumus determines whether or not you have passed a subject based on your percentage and other people's percentage

if you have a high percentage while other people have a low percentage in one subject, it'll show as passed
if you have a low percentage while other people have a high percentage, it will show as not passed

as for your specific example, not many people have made it to Mastery of COunting FUndamentals Level 4 (i don't think anyway), which is why it might say mastered


limit
number theory
relatively prime
number theory unsolved

Let $ a$ and $ b$ be relatively prime positive integers. Let $ J$ be the set of all primes of the form $ an\plus{}b$ with $ n$ a nonnegative integer and $ L$ the set of all primes.

Let $ J_n \equal{} \displaystyle\sum_{p\le n\; p\in J}{\dfrac{1}{p}}$ and $ L_n \equal{} \displaystyle\sum_{p\le n\; p\in L}{\dfrac{1}{p}}$ for each positive integer n.

Is it true that $ \displaystyle\lim_{n\rightarrow\infty}{\dfrac{J_n}{L_n}} \equal{} \dfrac{1}{\phi(a)}$ ??

:)

Thank you. By the way, do you know where I can look up for the proof?

:)

For example: J. Neukirch, Algebraic Number Theory, chapter 7.
But something like http://www.math.umass.edu/~isoprou/pdf/primes.pdf is probably a better source (as it's more direct).


Let a_n= sqrt(n+1)− sqrt(n). Prove that a_n >= 0.

Um, wouldn't it work just to say that $ \sqrt{n+1}>\sqrt{n}$? Although taking the conjugate is a bit cooler...


search
graph theory
combinatorics proposed
combinatorics

Let $ K_6$ be the complete graph with 6 vertices.
Prove that for any edge-coloring with 2 colors (we color the edges of $ K_6$ [u]without[/u] restrictions) in $ K_6$, we can always find a monochromatic 4-cycle (a cycle of length 4).

So basically, you have 6 vertices and you connect each pair, then color the edges with 2 colors, and you want to prove that there is a quadrilateral of one color?
The teacher in math club brought this up 2 months ago or something, it seems to be pretty well known.

Yes, that is the problem.
I am searching for an "elegant" method to prove it avoiding a "blind" enumeration (if, then).

Any suggestions?

Ok so let's say the colors are R and G. It is know that there exists a monocolored triangle, say ABC. Assume ABC is R. Now each of the other three vertices, say X,Y,Z when connected to A,B,C give two G edges and one R edge (otherwise we hav a 4-cycle). Moreover, no two of X,Y,Z are connected with the same side of ABC by two G edges (otherwise we hav a 4-cycle). Now if any side of XYZ is R we get a R 4-cycle. Otherwise all are green and one can spot a G 4-cycle. A figure would do.

[quote="Buddy Rich"]That is an elementar "blind" search strategy. It is a pretty elementar proof. Indeed, I'm searching for a "theoretical" solution...
Vertex degree, Ramsey number...
At the moment I have no good ideas.  :(

Does anyone know papers or notes about it?
Thanks.[/quote]

I tried some counting but that didn't work out nice. Is just my opinion, but I don't think this kind of problems can have a solution without a if/then part, since there are many ways to assign colors. I also don't see how ramsey numbers would help you arrive at a solution since they involve cliques and empty graphs (not cycles)...


analytic geometry
graphing lines
slope
algorithm

A mountain range is a polygonal curve from $(a,0)$ to $(b,0)$ in the upper half plane. Hikers $A$ and $B$ begin at $(a,0)$ and at $(b,0)$ respectively. Prove that A and B can meet by traveling on the mountain range in such a way that at all times their hights above the horizontal axis are the same.

(A polygonal curve is a curve in the plane made of stright line segments)

Art and craft of problem solving?
Beautiful question!

I don't understand the question. Just let them walk in such a way...

You have indeed misunderstood the question. Exapmle, draw the most complicated 'curve' you can. Now try to find a route subject to the conditions.

Maybe I'm not getting it either. It seems simple - too simple. Call the hikers A and B. Have them start climbing so that they proceed always at the same altitude. If both reach a peak at the same time, they continue down at the same rate. If one of them, say A reaches a peak first, when he goes down, B will go down the slope he just climbed until A reaches a trough and starts going back up again, when B starts going up as well. Using this algorithm, one of the two hikers will always cross a peak, and neither will be able to go back over a peak they've already scaled. So the peaks between them will eventually go to zero and they will meet.

This algorithm wont work for any mountian range, i can think of a few, but prove whatever you are trying to prove rigoursally!

I think tetrahedr0n's algorithm would work. Since the curve is continuous, each hiker would always have an option of moving backwards or forwards, and the other could move correspondingly. There are many cases to consider, so it would be difficult to describe the process for each of them individually. For example, A reaches a peak, and B goes backwards as A advances, but B hits a trough before A does. I guess the hard part is showing that there can't be a situation where they're both going back and forth perpetually.
For some reason, this reminds me of Rolle's Theorem, but I'm not sure if that could be applied.

I think I am thinking about the same problem/concept from Art and Craft as Seamusboyle.

[hide="hint"]
Think about the Monk traveling the Mountain problem (called something like that) from Art and Craft ----- this problem presents same algorithm as that problem. Consider making a graph of some sort, like that monk problem, and this problem will get easier.[/hide]

Can anyone show me a curve that could not work using the algorithm I detailed above?

[quote="tetrahedr0n"]If one of them, say A reaches a peak first, when he goes down, B will go down the slope he just climbed until A reaches a trough and starts going back up again, when B starts going up as well. [/quote]

And what if B reaches a truogh before A does?


integration
FTW
calculus
calculus computations

A: A ball of radius r is cut into two pieces by a horizontal plane a units above the center of the ball. Determine the volume of the upper piece by using the shell method.

B:A hole is drilled through the center of a ball of radius r, leaving a solid with a hollow cylindrical core of height h. show that the volume of this solid is independent of the radius of the ball.

does anyone know how to solve this problem? plz help. thx for helping.

[hide="A"]top of the ball is defined on the xy plane by $ y\equal{}\sqrt{r^2\minus{}x^2}$

$ 2\pi \int_0^a rhdx \equal{}2\pi \int_0^a x (\sqrt{r^2\minus{}x^2}\minus{}a) dx$

$ \equal{}2\pi \int_0^a x\sqrt{r^2\minus{}x^2}dx \minus{} 2\pi a\int_0^axdx$

$ \equal{}\pi \int_{r^2\minus{}a^2}^r z^{\frac{1}{2}}dz \minus{} \pi a^3\equal{} \frac{2\pi(r^\frac{3}{2}\minus{}(r^2\minus{}a^2)^\frac{3}{2})}{3}\minus{}\pi a^3$
[/hide]

[hide="B"]

$ 2\pi \int_0^h (\sqrt{r^2\minus{}y^2})^2 \minus{} (\sqrt{r^2\minus{}h^2})^2 dy\equal{} 2\pi \int_0^h h^2\minus{}y^2dy\equal{}\frac{4\pi h^3}{3}$
no r in the answer ftw

[/hide]


[quote="Nplusnequals"]Well, I recently encountered triangular numbers, eg. the number of points you can arrange into an equilateral triangle. The first triangular numbers are 1, 3, 6, 10,  15, ....
They can be found with the formula n(n+1)/2.
Then threre are square numbers, the number of points you can arrange into a square, their formula is obviosly n^2.
Then there are pentagonal numbers, the number of points you can arrange into a regular pentagon. The first such numbers are 1, 5, 12, 22, .... Their formula, significantly less obvious, is n(3n-1)/2.
I set out to find a formula that could find the nth x-gonal number....and I found one! This quite interesting formula is

[n(s-2)(n-1) + 2n]/2

where s is the number of sides of the x-gon.
For example, using the formula, we can fin the 1000th 100-gonal number, which is

48,952,000.[/quote]
show how did u found it!
the formula exist a long time I believe...
http://en.wikipedia.org/wiki/Polygonal_number


A circle with radius $ \sqrt {2}$ was divided in eight equal parts. Find the colored area.

[hide="Hint 1"]

Try to see a regular octagon.[/hide]

[hide="Hint 2"]

For alternate solution.

Diagonals of the rectangle are equal to diameter and the angle between these diagonals is ...[/hide]

$ \left( \sqrt{\left(\sqrt{2}\right)^2 \plus{} \left(\sqrt{2}\right)^2 \minus{} 2 \cdot \left(\sqrt{2}\right) \cdot \left(\sqrt{2}\right) \cdot \cos \dfrac{\pi}{4}} \right)\left( \sqrt{\left(\sqrt{2}\right)^2 \plus{} \left(\sqrt{2}\right)^2 \minus{} 2 \cdot \left(\sqrt{2}\right) \cdot \left(\sqrt{2}\right) \cdot \cos \dfrac{3\pi}{4}} \right)$
$ \equal{}\left( \sqrt{4\minus{}2\sqrt{2}} \right)\left( \sqrt{4\plus{}2\sqrt{2}} \right)$
$ \equal{}2\sqrt{2}$

[b]different solution[/b]
$ 2\left(\dfrac{1}{2} \cdot \sqrt{2} \cdot \sqrt{2} \cdot \sin \dfrac{\pi}{4} \plus{}\dfrac{1}{2} \cdot \sqrt{2} \cdot \sqrt{2} \cdot \sin \dfrac{3\pi}{4}\right)$
$ 2\sqrt{2}$

[hide="outline of solution"]The dots divide the circumference into 8 equal parts therefore the opposite verticies of the quadrilateral form the diameter since they are directly opposite each other. Therefore the quad is a rectangle since each angle subtends to a semicircle. the smaller leg is the base of an isoscles w/ vertex angle 45 and side lengths $ \sqrt{2}$ so that can be found. And the larger side is the leg of the right triangle w/ hypotenuse $ 2\sqrt{2}$ and other leg as the shorter side of the rectangle. [/hide]


How many different positive four-digit integers contain each of the digits 0, 1, 2, and 3 exactly once?

We can't put 0 in the first digit, so we find the arrangement of the other digits and multiply by the number of places that 0 can be.
$ 3! \equal{} 6$
0 can be in 3 places:
$ \_\_\_0$
$ \_\_0\_$
$ \_0\_\_$
So $ 6* 3 \equal{} \boxed{18}$

Alternatively, we can also find the number of 4 digit "numbers" and subtract the ones that don't work (complementary counting).
$ 4! \equal{} 24$
$ 0\_\_\_$ does not work.
There are 6 possibilities that don't work.
So the number of possibilities is $ \boxed{18}$.

Cookie, please do not multi-post. Thanks.

1. post count goes up more
2. if you multi post its harder to scroll up and down
3. please don't revive 6-year old threads with almost-same solutions
thanks! :)


geometry
rectangle
combinatorics unsolved
combinatorics

How many ways can a m*n grid can be partitioned into two distinct subregions? Any kind of cuts are possible if the cuts are made through the edges of the smaller 1 * 1 squares....

I am not sure how can I find the answer. Clearly I am missing something. Even worse, I can't be sure if it has a closed form. In that case I need a recurrence solution but I am not sure how can I progress. A friend told me about the problem, who was solving some similar problems. Sorry if I post it to a wrong place but at least plz post a link where I can find some help. Thanks. :)

Are cuts along the outer edges also permissible? This would allow for one of the resulting two regions to be disconnected, which seems undesirable, but isn't excluded by your phrasing.

Actually, what I wanted to say that the 2 distinct subregions will be of positive area. Any cut obeying this would be acceptable.

[quote="JBL"]What requirements precisely are you making on the subregions?  If one region is a square in the middle somewhere and the other region is everything else, is this okay?  If one region is a single strip and the other region is two (disconnected) rectangles, is this okay?[/quote]

Well.... how can I state the problem more clearly?  :?:  if one region is a square in the middle somewhere, that is ok. Actually the subregions must be two and distinct....
 If one region is a single strip and the other region is two (disconnected) rectangles, it actually is of 3 distinct subregions.
If you can place a zigzag cut along the grid so that the grid is divided in only two regions, that will be acceptable.


combinatorics
Coloring
graph theory
Ramsey Theory
Extremal combinatorics
IMO Longlist
IMO Shortlist

Given $5$ points in a plane, no three of them being collinear. Each two of these $5$ points are joined with a segment, and every of these segments is painted either red or blue; assume that there is no triangle whose sides are segments of equal color.

[b]a.)[/b] Show that:

[i](1)[/i] Among the four segments originating at any of the $5$ points, two are red and two are blue.

[i](2)[/i] The red segments form a closed way passing through all $5$ given points. (Similarly for the blue segments.)

[b]b.)[/b] Give a plan how to paint the segments either red or blue in order to have the condition (no triangle with equally colored sides) satisfied.

a)1) Assume that there are at least three red segments with endpoint $A_1$, the others endpoints being say (at least ) $A_2,A_3,A_4$.
Then, if one of the segments $A_2A_3,A_3A_4,A_4A_2$ is red, we have a red triangle, with $A_1$. Otherwise, $A_2A_3A_4$ form a blue triangle. In any case, we have a contradiction.
Thus, there are at most two red segment from any point.
Similarly, we have at most two blue segment from any point. Since there are exactly 4 segments from any point, we are done.

2) Now, consider the graph whose vertices are the five points and the edges are the red segments (that is, we delete the blue ones). From a), we know that each vertex has degree 2, which is well-known to be an even number. Thus, the graph has an eulerian closed path (which is a closed path which uses each edge only once).

b) Am I missing something or it is more trivial than the preceeding questions?

Pierre.


analytic geometry
graphing lines
slope
calculus
calculus computations

Hi Guys,
   Here's my prob:   Consider the graph below. If f(x) is the red sketch and g(x) is the blue sketch, let
u(x) = f(g(x)), v(x) = g(f(x)) and w(x) = g(g(x)). Then find:

a. u ‘(1) b. v ‘(1) c. w ‘(1)

Ok, I'm stumped on this one, can you guys point me in the right direction?

Here's what I have found so far:
Use algebra to find the line equations: On the interval from 0 to 2, f(x) = 6 - 3x and g(x) = 2x.  Does that sound right?
Then f'(x) = -3 and g'(x) = 2 (on those intervals).

Now where do I go from here?  How would I plug into the chain rule here?  Sorry, just really confused. Thanks in advance.
Chris

[img]http://img206.imageshack.us/img206/81/image1135xw.gif[/img]

You are given

$u(x) = f(g(x))$

Using chain rule, this means that

$u'(x) = f'(g(x))(g'(x))$

Use this and plug in that x = 1

$g(1) = 3$

So $u'(1) = f'(3) \cdot g'(1)$

To find those two values, evaluate the slope at the respective points of the f graph and g graph.

For part B, you have $v'(x) = g'(f(x))(f'(x))$ at $x=1$

$f(1) = 2$

So you need to find $g'(2)$ and $f'(1)$

However, the graph of $g(x)$ is not differentiable at $x=2$ since the graph comes to a point at that value. Hence, $v'(x)$ does not exist.

For the last one, it is another application of the chain rule.
$w(x) = g(g(x))$
so
$w'(x) = g'(g(x))(g'(x))$ at $x=1$

$g(1) = 3$, so what you need to find out is $g'(1)$ and $g'(3)$, which can be determined by taking the slope of the graph at those values.

$w'(1) = g'(1) \cdot g'(3)$

I hope that helped. If I didn't explain it clearly, I will try again

[b]Edit: I treated f(x) as the red sketch, and g(x) as the blue one by accident, if this confuses you...I will re post a new solution....

ankur,

   that was most helpful! thanks very much! my only question is this... looking at the graph shouldn't f(1) = 3 since it's the red graph, and g(1)=2 since it's the blue graph?  do you have these backwards or am I having a brain cramp? :)  thanks again.  Oops, I just saw your edit, so you did do it backwards... Would you mind reposting a new solution?  It is slightly confusing.  THanks again!
Chris

Sorry for the confusion....

$u'(x) = f'(g(x))(g'(x))$

We want to find $u'(1)$

We first find $g(1) = 2$

Now we know $u'(1) = f'(2)(g'(1))$

$f'(2)$ is the slope of f at x=2. This value can be determined by looking at the graph
However, we see that f is not differentiable at $x=2$, since the graph comes to a point at $x=2$. This means that $u'(1)$ has no solution.

The next part

$v'(x) = g'(f(x))(f'(x))$ at $x=1$

First we find $f(1) = 3$

so $v'(x) = g'(3)(f'(1))$

$g'(3)$ is = to the slope of g at $x=3$. This value is $-\frac13$.

$f'(1)$ is = to the slope of f at $x=1$. This value is $-3$.

Using $v'(x) = g'(3)(f'(1))$
We can calculuate $v'(x) = -\frac13 \cdot -3$ = 1

$w(x) = g(g(x))$
so
$w'(x) = g'(g(x))(g'(x))$ at $x=1$

First we find $g(1) = 2$

So $w'(1) = g'(2)(g'(1))$

Once again, $g'(2)$ does not exist because the graph of g has a point at the value of $x=2$. Like the first question, this, too, does not exist...

Both graphs of f and g are continuous at x=2, but not differentiable.

It is good to know that though continuity does not always imply differentiability, differentiability does imply continuity.

I hope that helped.
Sorry about the confusion the first time....

_ankur


linear algebra
superior algebra
superior algebra unsolved

If $ \mathcal{U}$ is a finite dimensional $ K$-algebra and $ x\in \text{Aut}(\mathcal{U})$ (or respectively $ x\in \text{Der}(\mathcal{U})$) then $ x_s$ , the semisimple part of $ x$, is also part of $ \text{Aut}(\mathcal{U})$ (or respectively $ \text{Der}(\mathcal{U})$

Only the case of derivation is worth of proving..

If you like, you can check Humphrey's GTM 9, I think it is in Section 4, the subsection of Jordan-Chevalley decomposition.

The idea is using the primary decompostion in linear algebra of $ \mathcal{U}$: $ \mathcal{U}\equal{}\bigoplus_{t\in K}\mathcal{U}_t$, where $ \mathcal{U}_t\equal{}\cap_r \ker (x_s\minus{}t.1)^r.$
Then the algebra structure of $ \mathcal{U}$ makes $ \mathcal{U}_t\mathcal{U}_s\subset \mathcal{U}_{t\plus{}s}$ holds. Now $ x_s$, when restricted to each subspace, is a scalar. From this it is easy to deduce that $ x_s$ is in $ Der(\mathcal{U})$.


Evaluate $ 3x^y \plus{} 4y^x$ when $ x\equal{}2$ and $ y\equal{}3$.

well, you just substitute the value for x for x and the value of y for y.

The answer is 60.


geometry
geometric transformation
reflection

The 'decimal reflection' of a positive integer is the number obtained by reflecting about the decimal point. For example, the decimal reflection of 240 is 0.042. Find the sum of the decimal reflections of the integers 10 through 20.

[quote="vallon22"]well, you could just flip the numbers and add
10 becomes .01
11 becomes .11
12 becomes .21
etc.

if we had listed them all out, we'll have an expression $ .01\plus{}.11\plus{}.21\plus{}.31\plus{}.41\plus{}.51\plus{}.61\plus{}.71\plus{}.81\plus{}.91\plus{}.02$

we add that up and get $ 4.62$[/quote]

You could also see that in the tenths place of each decimal, you basically just have to add 1-9. You can use the formula for that: (9)(10)/2=45.  Seeing as you had added all the numbers in the tenths place, the four would carry to the ones place.  Now you have 4.5.
You then see that in the hundredths place, you simply add 10 "1"s and 1 2 to get 12. Add .12 to 4.5 and you end up with the answer - [b]4.62[/b].


Let $ a,b,c$ be nonnegative real numbers. Determine the greatest and least values of
\[ P\equal{}\frac{a\plus{}b\plus{}c\minus{}3\sqrt[3]{abc}}{\max\left\{ (\sqrt{a}\minus{}\sqrt{b})^2,(\sqrt{b}\minus{}\sqrt{c})^2,(\sqrt{c}\minus{}\sqrt{a})^2\right\}}\]
:)

Thank you very much  :)  I will try  :wink:

[quote="Allnames"]The result is $ \frac {2}{3}$[/quote]
Sorry, but wrong! Try again!

For $ n$ numbers, the greatest value is $ n\minus{}1$.

[quote="Honey_S"][quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers. Determine the greatest and least values of
\[ P \equal{} \frac {a \plus{} b \plus{} c \minus{} 3\sqrt [3]{abc}}{\max\left\{ (\sqrt {a} \minus{} \sqrt {b})^2,(\sqrt {b} \minus{} \sqrt {c})^2,(\sqrt {c} \minus{} \sqrt {a})^2\right\}}
\]
:)[/quote]

The least value is $ 1$ and the inequality $ \frac {{a \plus{} b \plus{} c \minus{} 3\sqrt {abc} }}{{\max \left\{ {\left( {\sqrt a \minus{} \sqrt b } \right)^2 ,\left( {\sqrt b \minus{} \sqrt c } \right)^2 ,\left( {\sqrt c \minus{} \sqrt a } \right)^2 } \right\}}} \ge 1$ is obviously true by AM-GM.[/quote]
Yes, that's it, honey. :)
[quote="Vasc"]For $ n$ numbers, the greatest value is $ n \minus{} 1$.[/quote]
Right! :)

[quote="Vasc"]For $ n$ numbers, the greatest value is $ n \minus{} 1$.[/quote]
Yes.  :)


Should we post alternate solutions in the forum or send them as a "report error with problem"?

Will they even change the solutions? I mean, I have alternate solutions for many problems, but they were just easier for me. However, the difference will probably be unnoticeable by smarter people than me (which is most of the people in AoPS).

I saw that for some problems, more than one solution was given.

you can try pming one of the admins with your alternate solution
be sure to include the problem with it though, otherwise, it'll be a little hard on the admins to figure out what you're talking about


AoPS and mathlinks routinely cop out sometime around 12:00AM ~ 2:00 AM for periods of around 30 minutes. Is this server maintenance? (I can visit the rest of the internet just fine even when I can't access AoPS)

FF 3.x
Windows XP Service Pack Something
Both boards

EDIT: I forgot to say that one time I left the page load going, and after a while I received a error message to the effect of "message_die called multiple times".

He's a Southern Californian, so Pacific time.


Consider an equilateral triangle $ ABC$ with the length of the sides equal with $ 10$. Consider $ D,E,M$ on $ (AB),(BC),(CA)$ such that $ AD\equal{}BE\equal{}CM\equal{}3$. If $ AE \cap BM\equal{}\{ Q \},BM \cap CD\equal{}\{ R \}$, and $ CD \cap AE\equal{} \{ P \}$, then the triangle $ PQR$ is equilateral.

It's easy to see that the triangles ABE, BCM and CAD are congruent (s.a.s.).
Hence <DAE = <ACD and <DAP + <ADP = <ACD + <ADC =  180 degs - <DAC = 120 degs, consequently <PAD = 60 degs.
Analogously for the other angles of the triangle PQR.

Best regards,
sunken rock


Help me solve this problem.
Along the border between states A and B there are N(N<=10000) defence outposts. For every outpost k, the interval [Ak,Bk] which is guarded by it is known. Because of financial reasons, the president of country A decided that some of the outposts should be abandoned. In fact, all the redundant outposts will be abandoned. An outpost i is redundant if there exists some outpost j such that Aj<Ai and Bi<Bj. Your task is to find the number of redundant outposts.

Just sort the intervals in order of their first coordinate. Then an interval is redundant if its second coordinate is smaller than one to the left of it. So just go through the intervals in order, keeping track of the maximum right coordinate as you go, counting how many times you get a coordinate less than the current max.


inequalities
geometry
circumcircle
inequalities proposed

[quote="Virgil Nicula"]Let $ ABC$ be a triangle with the circumradius $ R$. Prove that $ 6R^{2}\cdot\sum a^{2}\ge\sum a^{4}\plus{}\sum b^{2}c^{2}$.[/quote]
Let $ a^{2}\equal{}x,$ $ b^{2}\equal{}y$ and $ c^{2}\equal{}z.$ Then $ 6R^{2}\cdot\sum_{cyc}a^{2}\ge\sum_{cyc}a^{4}\plus{}\sum_{cyc}b^{2}c^{2}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x^{4}\minus{}x^{3}y\minus{}x^{3}z\plus{}x^{2}yz)\geq0,$ which Schur.


http://www.artofproblemsolving.com/Forum/viewtopic.php?t=288068

Please search.

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I've been taking multivariable calculus online starting this semester and it's for the most part self-paced, so I've been almost entirely slacking off my work. Before an exam deadline, I'll just pick up the textbook and read it for a while. This is only my junior year, so I definitely need to fix my work ethic, but it's difficult because I have no free periods in common with the few other students that are in "accelerated" math. Have any of you had this problem?

I don't understand the title of this thread.

Anyway, it's not a problem if all of the following hold:
1. You're truly understanding and grasping the material as well as you would have had you worked more proactively
2. You are able to identify situations in which you can't get by like this, and are able to employ a better work ethic in those situations as appropriate.
3. You're doing something else worthwhile (however you define that) with the downtime/slacktime in your MV Calc learning schedule.

If any of those don't hold, you should say which one(s), so that we can offer relevant and useful suggestions.

The title is from an article on this website arguing against accelerated math if all you learn in each math class is the base curriculum and not the underlying concepts, which is what I had been doing until I "discovered" competition math problem solving recently. I guess that I just wanted to know if anyone else out there was going through the same thing I was, to relate I suppose. I can never talk to anyone I know about math; in fact, that's the absolute last thing my friends would want to talk about. As for my class I'm understanding the concepts, but if they don't get stored properly in my long term memory, I'll retake the class in college. My college might prefer that anyways.

If you understand the underlying concepts, that's great. However, storing them in long term memory is just a matter of doing practice problems over and over again, even if they seem easy.