Jeremy buys several item from a catalog. The bolded regions in the order form indicate information Jeremy must supply. Including shipping and handling (S&H), what is the total cost of his order?

\begin{tabular}{|l|l|l|l|l|}\hline
 Item $ \#$ & Item Description & Price per unit & Quantity & \textbf{Subtotals} \\
 V55-005 & Gumball Refill & $ \$4.95$ & 2 &  \\ \hline
 V69-064 & Gumball Machine & $ \$19.95$ & 1 & \\ \hline
\end{tabular}

\begin{tabular}{|l|l|}\hline
 Merchandise Total & S$ \&$H Costs \\
 $ \$0.01 \minus{} \$10.00$ & add $ \$4.95$ S$ \&$H  \\ \hline
 $ \$10.01 \minus{} \$25.00$ & add $ \$5.95$ S$ \&$H  \\ \hline
 $ \$25.01$ and above & add $ \$6.95$ S$ \&$H  \\ \hline
\end{tabular}
\\
\textbf{Merchandise Total:} \\
\textbf{S&H:} \\
\textbf{Order Total:}\\

This is a little hard question to read but the trick is to ignore everything that follows a backslash
on a line.
So James is ordered to buy 2 Gumball Refills (9.9)
and 1 Gumball Machine (19.95)
For a total of 29.85 dollars.
Then refering to the table below, we see that anything that costs more than 25 dollars has a S&H of 6.95 dollars.

In total then, we have a total cost of 29.85+6.95=36.80 dollars.

I hope i read this right... :maybe:


Prove that for every positive integer $n$ can be represented uniquely in the form \[n=b_{0}+b_{1}\cdot 3+b_{2}\cdot 3^{2}+...+b_{k-1}\cdot 3^{k-1}+3^{k},\] where \[b_{i}\in\{0,1,-1\}.\]

[hide="One line"] $(x^{-1}+1+x)(x^{-3}+1+x^{3})(x^{-9}+1+x^{9})... = ...+x^{-2}+x^{-1}+1+x^{1}+x^{2}+...$

The proof follows by taking the product finitely and multiplying by sufficiently large $x^{k}$.  Then the generating function becomes equivalent to the one for base three:

$(1+x+x^{2})(1+x^{3}+x^{6})(1+x^{9}+x^{18})... = 1+x+x^{2}+...$

Which is proven by multiplying both sides by $1-x$.  [/hide]

This is called balanced ternary, one of the more beautiful results handled by generating functions in my opinion :lol:


topology
geometry
3D geometry
sphere
symmetry
abstract algebra
group theory

I am teaching myself algebraic topology for this summer and this is an exercise 3.2.1 of Hatcher's Algebraic Topology.

Assuming as known the cup product structure on the torus $ S^1 \times S^1$, compute the cup product structure in $ H^*(M_g)$ for $ M_g$ the closed orientable surface of genus g by using the quotient map from $ M_g$ to a wedge sum of g tori. The picture for the quotient map is shown in the book (Hatcher's Algebraic Topology p 228). The part of book for the picture is also available online http://www.math.cornell.edu/~hatcher/AT/ATch3.2rev.pdf

My attempt,

Let X be $ M_g$ and A be the subspace of X which will be identified from a sphere with g disks removed to a point.
We see that X/A is homotopy equivalent to a wedge sum of g tori and A is homotopy equivalent to a wedge sum of g-1 circles.

Now, we have the following exact sequence for a pair (X, A).
$ 0 \rightarrow H^0(X) \rightarrow H^0(A) \rightarrow H^1(X, A) \rightarrow H^1(X) \rightarrow H^1(A) \rightarrow H^2(X,A) \rightarrow H^2(X) \rightarrow H^2(A) \rightarrow 0$.

Since  $ H^0(A) \equal{} 0$ and $ H^2(A) \equal{} 0$, we have
$ 0 \rightarrow H^1(X, A) \rightarrow H^1(X) \rightarrow H^1(A) \rightarrow H^2(X,A) \rightarrow H^2(X) \rightarrow  0$.

A $ H^k(S^1 \times S^1)$ for k=1,2 has a cup product structure with a basis $ \alpha$ for  $ H^1(S^1 \times S^1)$ and  $ \alpha \smile \beta$ for $ H^2(S^1 \times S^1)$ where generators $ \alpha, \beta \in H^1(S^1 \times S^1)$.

How do I proceed from here?

Any help will be appreciated.

It is clearly intended to use the fact that if $ q$ is the quotient map, then $ q^*$ is a homomorphism of graded cohomology rings. Therefore the cup product structure of $ \bigvee^g T$ tells you all you need to know about the cup product structure of $ M_g$, provided you know where $ q^*$ sends [i] elements [/i] of the cohomology ring. The solution I have seen does this directly by looking at a $ \Delta$-complex structure on $ M_g$; although that seems a little impure to me.

Try showing (this ends up being true) that $ q^*$ maps the obvious $ 2g$ generators of $ H^1(\bigvee^g T)$ onto the $ 2g$ generators of $ H^1(M_g)$, and that $ q^*$ maps each generator of $ H^2(\bigvee^g T)$ to the same generator of $ H^2(M_g)$. Then the cup product structure is determined by naturality of $ q^*$. I would like to see a nice proof why $ q^*$ maps elements in this way, though!

Hmm, if you're comfortable asserting $ H^2(M_g) \equal{} \mathbb{Z}$ (orientability + being a manifold), then since you get the long exact sequence

$ 0 \to H^1(X) \stackrel{q^*}{\to} H^1(M_g) \stackrel{i^*}{\to} H^1(A) \stackrel{\delta}{\to} H^2(X) \stackrel{q^*}{\to} H^2(M_g) \to 0.$

(where: $ X \equal{} \bigvee^g T$, $ A \equal{} \bigvee^{g\minus{}1} S^1$) and the tail end of the sequence is isomorphic to $ \mathbb{Z}^{g\minus{}1} \to \mathbb{Z}^g \to \mathbb{Z} \to 0$, a rank argument shows that $ \delta$ must be injective. Therefore $ i^*$ must be the zero map in dimension 1, and so $ q^*$ is an isomorphism in dimension 1. (This is key.) By symmetry each generator of $ H^2(X)$ maps to the same element of $ H^2(M_g)$ under $ q^*$; yet this map is surjective, so each must map to the generator of $ H^2(M_g)$.

From here, $ H^1(X) \equal{} \bigotimes_{i\equal{}1}^g \bigwedge [a_i,b_i]$ for some generators $ a_1,\dots,a_g,b_1,\dots,b_g$. Let $ H^2(X)$ be generated by $ c_1,\dots,c_g$, with one generator for each copy of $ T$; orient such that $ a_i \smile b_i \equal{} c_i$. Then the $ q^*(a_i) \equal{}: \alpha_i$ and $ q^*(b_i) \equal{}: \beta_i$ form generators for $ H^1(M_g)$, and $ q^*(c_i) \equal{} \gamma$ is a generator for $ H^2(M_g)$; we have $ \alpha_i \smile \beta_j \equal{} q^*(a_i \smile b_j) \equal{} q^*(c_i \delta_{ij}) \equal{} \gamma \delta_{ij}$, where here $ \delta$ is the Kronecker delta (not the coboundary map). Since for $ i \ne j$, $ a_i \smile a_j \equal{} b_i \smile b_j \equal{} 0$, $ \alpha_i \smile \alpha_j \equal{} \beta_i \smile \beta_j \equal{} 0$; this determines the ring structure on $ H^*(M_g)$.

[quote="Nukular"]Hmm, if you're comfortable asserting $ H^2(M_g) \equal{} \mathbb{Z}$ (orientability + being a manifold), then since you get the long exact sequence

$ 0 \to H^1(X) \stackrel{q^*}{\to} H^1(M_g) \stackrel{i^*}{\to} H^1(A) \stackrel{\delta}{\to} H^2(X) \stackrel{q^*}{\to} H^2(M_g) \to 0.$

(where: $ X \equal{} \bigvee^g T$, $ A \equal{} \bigvee^{g \minus{} 1} S^1$) and the tail end of the sequence is isomorphic to $ \mathbb{Z}^{g \minus{} 1} \to \mathbb{Z}^g \to \mathbb{Z} \to 0$, a rank argument shows that $ \delta$ must be injective. Therefore $ i^*$ must be the zero map in dimension 1, and so $ q^*$ is an isomorphism in dimension 1. (This is key.) By symmetry each generator of $ H^2(X)$ maps to the same element of $ H^2(M_g)$ under $ q^*$; yet this map is surjective, so each must map to the generator of $ H^2(M_g)$.

From here, $ H^1(X) \equal{} \bigotimes_{i \equal{} 1}^g \bigwedge [a_i,b_i]$ for some generators $ a_1,\dots,a_g,b_1,\dots,b_g$. Let $ H^2(X)$ be generated by $ c_1,\dots,c_g$, with one generator for each copy of $ T$; orient such that $ a_i \smile b_i \equal{} c_i$. Then the $ q^*(a_i) \equal{} : \alpha_i$ and $ q^*(b_i) \equal{} : \beta_i$ form generators for $ H^1(M_g)$, and $ q^*(c_i) \equal{} \gamma$ is a generator for $ H^2(M_g)$; we have $ \alpha_i \smile \beta_j \equal{} q^*(a_i \smile b_j) \equal{} q^*(c_i \delta_{ij}) \equal{} \gamma \delta_{ij}$, where here $ \delta$ is the Kronecker delta (not the coboundary map). Since for $ i \ne j$, $ a_i \smile a_j \equal{} b_i \smile b_j \equal{} 0$, $ \alpha_i \smile \alpha_j \equal{} \beta_i \smile \beta_j \equal{} 0$; this determines the ring structure on $ H^*(M_g)$.[/quote]

I am trying to follow your proof and I have some questions.

It seems like my previous exact sequene has an error. It should be $ H^0(A)\equal{}\mathbb{Z}$ rather than $ H^0(A)\equal{}0$. If that is the case, is it still true that both $ q^*$ below is an isomorphism?

$ H^1(X) \stackrel{q^*}{\to} H^1(M_g) \stackrel{i^*}{\to} H^1(A) \stackrel{\delta}{\to} H^2(X) \stackrel{q^*}{\to} H^2(M_g) \to 0.$

The second question is,
Is it true that the resulting ring structure on $ H^*(M_g)$ is $ \bigwedge[\alpha_1, \beta_1, ... , \alpha_g, \beta_g] \equal{}  \bigotimes_{i \equal{} 1}^g \bigwedge [\alpha_i, \beta_i]$ ? I guess $ H^1(X)$ alone is an abelian group that does not have any multiplicative structure.

The last question is,
You mentioned that $ H^2(M_g) \equal{} \mathbb{Z}$ (orientability + being a manifold). Is it the result of "Poincare duality theorem"?
In wiki, it says  $ H^k(M) \cong H_{n\minus{}k}(M)$. Since $ M_g$ is a two-dimensional manifold and $ M_g$ has a single connected component, $ H^2(M_g) \cong H_0(M_g) \cong \mathbb{Z}$. I have not studied "Poincare duality theorem" yet, so I am trying to use the theorem without a proof.

Thank you so much for your help. It helped me a lot.

[quote="algtop"]
I am trying to follow your proof and I have some questions.

It seems like my previous exact sequene has an error. It should be $ H^0(A) \equal{} \mathbb{Z}$ rather than $ H^0(A) \equal{} 0$. If that is the case, is it still true that both $ q^*$ below is an isomorphism?
[/quote]
Actually, that part of the exact sequence is fine, since the map $ H^0(M_g) \to H^0(A)$ is an isomorphism, so the map $ H^0(A) \to H^1(X)$ must be the zero map.

[quote="algtop"]
The second question is,
Is it true that the resulting ring structure on $ H^*(M_g)$ is $ \bigwedge[\alpha_1, \beta_1, ... , \alpha_g, \beta_g] \equal{} \bigotimes_{i \equal{} 1}^g \bigwedge [\alpha_i, \beta_i]$ ? I guess $ H^1(X)$ alone is an abelian group that does not have any multiplicative structure.
[/quote]

That's very close; but you also need the fact that all the cross-term products $ \alpha_i \smile \beta_j$ are zero, and also that you always have $ \alpha_i \smile \beta_i \equal{} \gamma$; it's always the same $ \gamma$. So... very close.

[quote="algtop"]
The last question is,
You mentioned that $ H^2(M_g) \equal{} \mathbb{Z}$ (orientability + being a manifold). Is it the result of "Poincare duality theorem"?
In wiki, it says  $ H^k(M) \cong H_{n \minus{} k}(M)$. Since $ M_g$ is a two-dimensional manifold and $ M_g$ has a single connected component, $ H^2(M_g) \cong H_0(M_g) \cong \mathbb{Z}$. I have not studied "Poincare duality theorem" yet, so I am trying to use the theorem without a proof.
[/quote]

Yes, it follows from Poincare duality; a bit heavy-handed but it's a really useful theorem.

[quote="algtop"]
Thank you so much for your help. It helped me a lot.[/quote]

You're very welcome! I'm trying to review all this stuff this summer myself, and it actually helps to answer questions like this :).


Hello Everyone,

I was just wondering if it was worth buying the AoPs books...I just bought the full set today (volumes I and II with solutions), and I was wondering if I did the right thing, because my ARML coach reccommended these books..what do you guys think? Is the way they present the math easy to understand, even if you don't have a lot of experience in higher-up math competitions?

Thanks

ABSOLUTELY they're worth it. Easy to follow, easy to understand, and occasionally humorous (which is more than you can say for the average math textbook) ...well at least the first one is. I don't have the second one  :(

The books are definitely worth the money.  The way the book teaches material is the best out of any math book that is geared toward competition math that I have found.  This book is truly one-of-a-kind.  Volume 1 is accessible to students who don't have any competition experience.  So just about anyone can understand it.  Volume 2 expands upon volume 1 and adds a few things.  Volume 1 is about Mathcounts to Amc 12 level whereas volume 2 is from about Amc 12 to the doorstep of USAMO level.

Ok..thanks guys..can't wait to get them in the mail then  :)

lol.. ur asking this question in a aops forum...
more biased than fox news

I don't have the first one;I started directly on the second. Bit hard, though. Still, it taught me a lot. It requires common sense.

I got the full set in the mail today, and after giving everything a quick glance, I have to say I am impressed with the material. I can't wait to get started on these books, and look forward to learning a lot :)

You say they're pretty good?

*is thinking about buying one*

[quote="gameworld7"]can u give us some things what they say in the books?[/quote]

Look on this page [url]http://www.artofproblemsolving.com/Books/AoPS_B_About.php[/url]

That's just a link to another part of this website but it has the table of contents and a few sample problems.

Yea, they have problems from all sorts of competitions.  Very cool.


For the three identical squares shown, points $ A$, $ B$
and $ C$ are vertices, and $ AB$ = $ 2\sqrt{5}$ cm. What is the
length of $ AC$, in centimeters? Express your answer
in simplest radical form.

[asy]draw((0,0)--(0,10)--(30,10)--(30,0)--cycle);
draw((10,0)--(10,10));
draw((20,0)--(20,10));
draw((0,10)--(20,0),linewidth(1));
label("$A$",(0,10),NW);
label("$B$",(20,0),S);
label("$C$",(30,0),S);[/asy]

The formula for hypontenuse of a triangle with legs of x and 2x is $ x\sqrt{5}$ THe formula for legs of x and 3x is $ x\sqrt{10}$ x in this case is 2, using the given info. THerefore AC is $ 2\sqrt{10}$


Show that there are no functions $f: N\to N$ for wich:
$f(3n+1)\leq f(3n+4)-1$ and $f(5n+2)\geq f(5n+7)-1$


Prove that if $a,b,c,d$ are positive real and $a+b+c+d=a^2+b^2+c^2+d^2$ then :

$a(a^2+1)+b(b^2+1)+c(c^2+1)+d(d^2+1) \le 8$

 :)
_____

Good Luck.

my computations, if not wrong, give the initial ineq is equivalent to $\sum (2-ab)(a-b)^2 \ge 0$ (6 stuff)
we know that $4 \ge a^2+b^2+c^2+d^2 \ge 2ab$ so $(2-ab) \ge 0$, same for the rest, so $\sum (2-ab)(a-b)^2 \ge 0$

Yeah... it's not difficult ay all. But my first problem is:

Find the maximum of:

$P=a^3+b^3+c^3+d^3$

(It's unsolved with me. I add $a+b+c+d$ to it possible solve  :( )...

Now try the following with $P=a^3+b^+c^3+d^3$. I hope it can be cracked.

By Equal Variable Theorem, it suffices to consider $a=b=c \leq d$ (for both inequalities).
If I am not wrong, then
$a+b+c+d+4(a^3+b^3+c^3+d^3) \leq 20$.

Oh.. Is there anything other than $n-1 EV$? It's not good to find an elementary solution. I hope someone will find that. (maybe prove that $3$ of $4$ numbers equal, (it's the key), but the solution doesn't use anything too high.

  Argady, where can you find this value?

[quote="hungkhtn"]
  Argady, where can you find this value?[/quote]
Through Lagrange. :blush:

And It's really hard to find an easier solution?... :(

Thank God! I've found an elementary solution (mixing Variable) to prove three number equal. Any one tries it again?

My first solution used it identify:

$a^3+a-2=(a-1)(a^2+a+2)=(a^2-a)(a+1+\frac{2}{a})$

Then Use Trebyseff....

It's easy but still vaild for the open question with $S=a^3+b^3+c^3+d^3$. Use lenma:

If $d\ge b \ge c \ge a$

And $a+b+c-3m=a^2+b^2+c^2-3m^2$($m\le 1$) .

Then we have $f(a,b,c,d) \le f(m,m,m,d)$

The remain is :

Find $max$ of $f(m,m,m,d)$ if $3m+d=3m^2+d^2$.!

It's not too hard.


How many three-digit numbers can be formed given that the sum of the three digits is equal to 5?

List out the partitions of $ 5$ that contain $ 3$ numbers:

${ \{5,0,0}\}$ - $ 1$ possible number

${ \{4,1,0}\}$  -  $ 4$ possible numbers

${ \{3,2,0}\}$- $ 4$ possible numbers

${ \{3,1,1}\}$ - $ 3$ possible numbers

${ \{2,2,1}\}$ - $ 3$ possible numbers

Therefore, the total number of three digit numbers whose digits sum to $ 5$ is $ 1+4+4+3+3=\boxed{15}$

you can also you sticks and stones for this, and since the first digit cant be 0, you add 2 instead of 3 for each of the numbers. 5+2=7 -1 for the overcount gives 6 slots and 2 dividers. 6 choose 2=15


[quote="Makaveli"]Find all pairs of distinct positive integers a, b such that b^2+a devides a^2+b and b^2+a is a power of a prime.
I came up with a solution according to which b^2+a has to be power of 3 and then showed there are no such numbers, but there is something fishy about it.Namely, ifb^2+a=p^n then p^n devides a^2+b-(b^2+a)=(a-b)(a+b-1) so a=b(modp)Let x=a=b(modp)0<=x<p p devides b^2+a=x^2+x=x(x+1)(modp)so x=0 or x=p-1 .For x=p we find a contradiction. p devides a+b-1=-3(modp) so p=3 and the rest is optained by the infinite decrease of powers method when you put a=3a1-1 and
 b=3b1-1 (I used = for is congruent to)I am not sure if the case a-b=1 produces something[/quote]

Solved this problem some time ago, there is one solution $(5, 2)$. I proved that it is a power of 3 and that it divides $a^3-b^3=(a-b)(a^2+ab+b^2)$ and proved that if $b^2+a=3^\alpha$ then 9 doesn't divide $a^2+ab+b^2$, and somehow (can't remember that part) prvoed that only solution is one mentioned before.. I don't have time now, I'll write down whole proof in few days..

this is what I think:

Given b^2 + a = p^ n and a^2+ b = p^n * z
Then (a-b)(a+b-1) = p^n (z-1)

First, you show that a and b are relatively prime.
Then if p=2, this leads to one solution namely (1,1), but this is not what we want.

So say p>2,

1) we show that a and b are of different parity. note z is odd
2) a - b= odd  (a+b -1 ) = even
3) a - b = p^c , a+b-1 = p^(n-1) * z
4) substitute a = b + p^ c into b^2+a = p^n

You get b(b+1) = p^c (p^(n-c) - 1)
Then one can reason that there are two cases.

two cases:
(1) b = p ^ c, c<=n and b+1 = p^(n-c) -1
this leads to p =3, c = 1, n=2  => b=3 and a=0, but  this is not what we want.
(2) b +1 = p ^ c, c<=n and b = p^(n-c) -1
this leads to c = n - c or n = 2c

b^2 + a = p^ n =(p^c)^2 = (a-b) ^2 =a^2 - 2ab + b^2
=> a = 2b+1
substitute this into (a-b)(a+b-1) = p^n (z-1)
then you get (1+b)(3b)=p^n (z-1), so p= 3.

so then, b= 3^c - 1.
substitute that in:
(1+b)(3b)=p^n (z-1)
=> 3^(c+1) (3^c - 1) = 3^(2c) (z-1)
this leads that c+1 = 2c or c=1

so b= 2, n=2, then a= 5.

So the only solution is (5,2).

I hope this is correct.


Find $ y$ if $ 3^{x\plus{}y} \equal{} 81$ and $ 32^{x\minus{}2y} \equal{} 2$.

My bad, this is supposed to be for middle school students and teachers only. Oops.

Well, this is part of middle school curriculum (exponents) so enjoy anyway!  :)

[hide]
$ 3^{x\plus{}y}\equal{}81\implies x\plus{}y\equal{}4$ and $ 32^{x\minus{}2y}\equal{}2\implies x\minus{}2y\equal{}\frac{1}{5}$. Subtract the two equation to get that $ 3y\equal{}\frac{19}{5}\implies y\equal{}\frac{19}{15}$
[/hide]

How would you solve this without logarithms though (for middle school curriculm)? Brute force?

Convert $ 3^{x \plus{} y} \equal{} 81$ to $ 3^{x \plus{} y} \equal{} 3^4$. Then notice that the bases are the same, so the powers must also be the same.

I don't think that counts as using logarithms, even though the rigorous method behind it is logs.

These problems won't come up often in middle school classroom math... :maybe:

[quote="i_like_pie"]Convert $ 3^{x \plus{} y} \equal{} 81$ to $ 3^{x \plus{} y} \equal{} 3^4$. Then notice that the bases are the same, so the powers must also be the same.[/quote]

That is the way I learned to solve that, but I'm wondering if there is another solution?


[hide="Original Post"]I attempted to make a simple rendering program for the mandelbrot set:

[code]import java.awt.event.MouseListener;
import java.awt.event.MouseEvent;
import java.applet.Applet;
import java.awt.Graphics;
import java.awt.Color;

public class Mandelbrot extends Applet
{
	private static boolean rendered = false;

	private static int incr = 1;

	private int RGB = 0;
	private int RG, B;
	private int iter = 0;
	private double x, y, a, b;
	private int pxlx = 0, pxly = 0;
	private int left = -2, right = 2, up = 2, down = -2;

	public void paint(Graphics g)
	{
		pxlx = 0;
		for(x = left; x < right; x += 0.01 /*(right - left) / 400.0*/)
		{
			pxlx++;

			pxly = 0;
			for(y = up; y > down; y -= 0.01 /*(down - up) / 400.0*/)
			{
				pxly++;
				iter = 0;

				a = 0;
				b = 0;

				while(Math.sqrt(a * a + b * b) <= 2 && iter <= 10000)
				{
					a = a * a - b * b + x;
					b = a * b * 2 + y;

					iter++;
				}

				RGB = (iter % 1023);

				B = 511 - Math.abs(511 - RGB);
				RG = 0;

				if(B > 255)
				{
					RG = B - 256;
					B = 255;
				}

				if(iter == 10001)
				{
					B = 255;
					RG = 255;
				}

				g.setColor(new Color(RG, RG, B));
				g.drawRect(pxlx, pxly, 1, 1);
			}
		}
	}
}[/code]

However, it does not work; the figure it produces is not even close to what the Mandelbrot set looks like; I have debugged it for hours, but I still could not find the problem. Would anyone mind venturing a guess as to why it is not rendering correctly?
[/hide]

EDIT> WOOT! The program now works as intended. Thanks for the help!

Everything is done in one class, it is fully zoomable (albeit only zooming-in is supported). There is only one color scheme (Blue) and the mandelbrot set is rendered in white.

I made some modifications to support zooming in and fixed the flickering problem by using a JLabel with a BufferedImage instead of rendering within paint(). Performance was also increased substantially. There are still problems with loss of precision at deep zooms due to the limited bitsize of (double), but that is a minimal issue.

[code]import javax.swing.*;
import java.awt.*;
import java.applet.*;
import java.awt.image.*;
import java.awt.event.*;

public class Mandelbrot extends Applet implements MouseListener
{
	private static final int SIZE = 512;

	private BufferedImage fractal = new BufferedImage(SIZE, SIZE, BufferedImage.TYPE_INT_RGB);

	private double precision = 40;
	private double preincr = 40;
	private int intpre;

	private boolean rendered = false;

	private int incr = 1;

	private int RGB = 0;
	private int RG, B;
	private int iter = 0;
	private double x, y, a, b, aa, incrx, incry;
	private int pxlx = 0, pxly = 0;
	private double left = -2, right = 2, up = 2, down = -2;
	private double dist = 2;
	private int mouse_x,mouse_y;

	private double ctrx, ctry;

	private JLabel container;

	//Code to zoom in by a factor of 2 every click
	public void mouseClicked(MouseEvent e)
	{
		mouse_x = e.getX();
		mouse_y = e.getY();

		ctrx = left + (2 * dist * (mouse_x / (double)SIZE));
		ctry = up - (2 * dist * (mouse_y / (double)SIZE));

		dist /= 2;

		System.out.println("Click registered! X=" + mouse_x + " Y=" + mouse_y); //DEBUG

		left = ctrx - dist;
		right = ctrx + dist;
		up = ctry + dist;
		down = ctry - dist;

		System.out.println("Left:" + left + " \nRight:" + right + " \nUp:" + up + " \nDown:" + down);

		precision += preincr;
		preincr *= 1.1;

		System.out.println("Drawing...");
		System.out.println("Current precision: " + precision);

		redraw(fractal);

		System.out.println("Done!");

		container.setIcon(new ImageIcon(fractal));
		container.repaint();

	}
	public void mouseEntered(MouseEvent e) {}
	public void mouseExited(MouseEvent e) {}
	public void mousePressed(MouseEvent e) {}
	public void mouseReleased(MouseEvent e) {}

	public void init()
	{
		redraw(fractal);

		container = new JLabel(new ImageIcon(fractal));

		add(container);

		setSize(SIZE, SIZE);

		addMouseListener(this);
	}

	//Renders the set and stores it to the BufferedImage
	public void redraw(BufferedImage z)
	{
		pxlx = -1;

		incrx = (right - left) / (double)SIZE;
		incry = (down - up) / (double)SIZE;

		for(x = left; x < right; x += incrx)
		{
			pxlx++;

			pxly = -1;
			for(y = up; y > down; y += incry)
			{
				pxly++;
				iter = 0;

				a = 0;
				b = 0;

				intpre = (int)precision;

				while(Math.sqrt(a * a + b * b) <= 2 && iter < intpre)
				{
					aa = a * a - b * b + x;
					b = a * b * 2 + y;
					a = aa;

					iter++;
				}

				RGB = (iter % 102);

				B = 51 - Math.abs(51 - RGB);
				RG = 0;

				if(B > 25)
				{
					RG = B - 26;
					B = 25;
				}

				RG *= 10;
				B *= 10;

				if(iter == intpre)
				{
					B = 255;
					RG = 255;
				}

				z.setRGB(pxlx, pxly, RG * 65792 + B);
			}
		}
	}
}[/code]

Oh man, this seriously puzzled me for good half hour. The problem is really quite simple:

double orig_a = a;
a = a * a - b * b + x;
b = orig_a * b * 2 + y;

Whoops, that was stupid >.>

Thank you very much! (I ran into new problems though...)


function
floor function
inequalities
inequalities proposed

For the first problem, I don't think it is correct. Besides the obvious typo, what if y >>> z so that $\frac{y+1}{z+1}>> 2$?

Ahh, indeed forgot the condition $x+y+z = 1$ for part one. :)

Let's say some stupid things.
i)Suppose without loss of generality $\{x,y,z\}=\{x',y',z'\}$ with  $x'\geq y'\geq z'$.
Then by rearrangement: $\left(\frac{x+1}{y+1}\right)^{n}+\left(\frac{y+1}{z+1}\right)^{n}+\left(\frac{z+1}{x+1}\right)^{n}\leq \left(\frac{x'+1}{z'+1}\right)^{n}+\left(\frac{y'+1}{y'+1}\right)^{n}+\left(\frac{z'+1}{x'+1}\right)^{n}=$
$=\alpha^{n}+\left(\frac{1}{\alpha}\right)^{n}+1$.
Where $\alpha=\frac{x'+1}{z'+1}>1$. Since the expression is an increasing funtion of $\alpha$ ($\alpha>1$) it will reach its maximum when $\alpha$ itself is maximum, and since the last is increasing with $x'$ and decrasing with $z'$ that will happen when $z'=0$ and $x'=1$, so:
$LHS\leq 2^{n}+2^{-n}+1.$

ii) If we put $LHS=f(x_{1},x_{2},\ldots x_{n})$ we get:
$\frac{\delta f(x_{1},x_{2}\ldots x_{k}\ldots x_{n})}{\delta x_{k}}=\frac{\delta[\frac{1}{x_{k+1}+1}-\frac{1}{(x_{k+1}+1)(x_{k}+1)}+\frac{x_{k-1}}{(x_{k-1}+1)(x_{k}+1)}]}{\delta x_{k}}=\frac{1}{(x_{k}+1)^{2}}\frac{(1-x_{k+1}x_{k-1})}{(x_{k-1}+1)(x_{k+1}+1)}$.
Since the sign of the last expression doesn`t depend on $x_{k}$, $\exists \;l_{1},l_{2}\ldots l_{n}\in \{ 0,\+\infty\}/\;\;S=sup \{f(x_{1},x_{2}\ldots x_{n})|x_{i}\in\mathbb{R}^{+}\forall i \}=lim_{x_{i}\to l_{i}\;\forall i}\;f(x_{1},x_{2}\ldots x_{n})$.
But   $L_{k}=lim_{x_{i}\to l_{i}}(\frac{x_{k}}{(x_{k}+1)(x_{k+1}+1)})\in \{0,1\}\;\;\forall k$  and  $L_{k}=1\Rightarrow( l_{k}=\infty \wedge l_{k+1}=0)\Rightarrow L_{k-1}=L_{k+1}=0$, so $S\leq \lfloor\frac{n}{2}\rfloor$.  By letting $l_{2i}=0$, $l_{2i+1}=\infty$ we see that actully we have equality.

PS I have edited the post according to Rzeszut observations.

notice that for the second part putting $\frac{1}{x_{i}+1}=a_{i}$
we get the famous problem:
$a_{i}\in [0,1]$ show that $a_{1}+a_{2}+\cdots+a_{n}-a_{1}a_{2}-a_{2}a_{3}-\cdots-a_{n}a_{1}\le \left[ \frac{n}{2}\right]$

[quote="Mimoide"]Suppose without loss of generality $x\geq y\geq z$.[/quote]
Correct me if I'm wrong, but I don't think that it's possible. The inequality isn't symmetric; it's only cyclic.

PS. You can use \left( and \right) to type powers of fractions in nicer form, compare $(\frac{x+1}{y+1})^{n}$ and $\left(\frac{x+1}{y+1}\right)^{n}$.

[quote="Mimoide"]Suppose without loss of generality $\{x,y,z\}=\{x',y',z'\}$ with  $x'\geq y'\geq z'$.
Then by rearrangement: $\left(\frac{x'+1}{y'+1}\right)^{n}+\left(\frac{y'+1}{z'+1}\right)^{n}+\left(\frac{z'+1}{x'+1}\right)^{n}\leq \left(\frac{x'+1}{z'+1}\right)^{n}+\left(\frac{y'+1}{y'+1}\right)^{n}+\left(\frac{z'+1}{x'+1}\right)^{n}$[/quote]
[quote="Mimoide"]Maybe now it's right, thank you.[/quote]
At first, you say that $x'=\max(x,y,z)$, $z'=\min(x,y,z)$ and $y'$ is the middle of numbers $x,y,z$. Then you can't say that $f(x,y,z)=f(x',y',z')$, where $f(a,b,c)=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$.
Let $a=(x+1)^{n}$ etc. Then in the first version of your post you wanted to prove $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\leqslant \frac{a}{c}+1+\frac{c}{a}$. It's equivalent to $(b-a)(b-c)\geqslant 0$. But it can be proved without rearrangement if you assume WLOG in the inequality from the beginning that $b=\max(a,b,c)$ (you can do it, becaus the inequality was cyclic), so the first version of your post needed only assumption $b\geqslant a,c$.
Then you put $\alpha= \frac{x+1}{z+1}$. Now we can't say anything about relation between $\alpha$ and $1$. But it also can be easily repaired if you set $\alpha=\max\left(\frac{x+1}{z+1},\frac{z+1}{x+1}\right)$. Then $\alpha\geqslant 1$ and the reasoning follows as stated in your post.
I think it will be right if you correct it.  :D

Ok, it's just a typing error during the correction; I didn't mean the variables with $'$ in the first expression.

[quote="dyd"][quote="Mimoide"]$\left(\frac{x+1}{y+1}\right)^{n}+\left(\frac{y+1}{z+1}\right)^{n}+\left(\frac{z+1}{x+1}\right)^{n}\leq \left(\frac{x'+1}{z'+1}\right)^{n}+\left(\frac{y+1}{y'+1}\right)^{n}+\left(\frac{z'+1}{x'+1}\right)^{n}=$
$=\alpha^{n}+\left(\frac{1}{\alpha}\right)^{n}$.
Where $\alpha=\frac{x'+1}{z'+1}>1$. [/quote]

???
$\alpha^{n}+\left(\frac{1}{\alpha}\right)^{n}=\left(\frac{x'+1}{z'+1}\right)^{n}+\left(\frac{z'+1}{x'+1}\right)^{n}$, excluding $\left(\frac{y+1}{y'+1}\right)^{n}$.[/quote]
It should be $\leqslant 1+\alpha^{n}+\alpha^{-n}$ a.s.o., but the reasonig, except of the obvious typo, is good.


Sorry, I edit it.

For real numbers a,b,c,

1) prove that $ (a \plus{} b)^6 \plus{} (b \plus{} c)^6 \plus{} (c \plus{} a)^6 \ge \frac {16}{31}( a^6 \plus{} b^6 \plus{} c^6)$

2)prove or disprove that $ (a \plus{} b)^6 \plus{} (b \plus{} c)^6 \plus{} (c \plus{} a)^6 \ge a^6 \plus{} b^6 \plus{} c^6$

(maybe disproving will need calculator!)


search
number theory unsolved
number theory

Prove that:
$ ( \minus{} 1)^\frac {p \minus{} 1}{2}$$ C_{p \minus{} 1}^{\frac {p \minus{} 1}{2}}$ $ \equiv$ $ 4^{p \minus{} 1}$   (mod $ p^3$)   where $ p$ is a prime number $ \geq$5. :ninja:

It's in PEN, search the congruences section.


There are more than 12 but fewer than 32 students in Al orithm's class. When grouped by twos, there was one student left. When grouped by threes, there were two students left. When grouped by fours, there were three students left. How many students are in the class?
Answer-23
How do u do this in terms of mod tho?

so $ x \equal{} 2k \plus{} 1 \equal{} 3k' \plus{} 2 \equal{} 4k'' \plus{} 3$

$ x \plus{} 1 \equal{} 2(k \plus{} 1) \equal{} 3(k' \plus{} 1) \equal{} 4(k'' \plus{} 1)$

so $ x \plus{} 1$ is multiple of 24.

so $ x \equal{} 24k''' \minus{} 1$

and $ 12 < x < 32$.

answer : 23

Notice that if you add $ 1$ more student to the class, you can group the students by $ 2$s, $ 3$s, and $ 4$s. If you can group by $ 4$s, then you can also group by $ 2$s. So, you want the least number between $ 12$ and $ 32$ that can by be grouped by $ 3$s and $ 4$s (namely multiple of $ 12$). The possibilities are $ 12$ and $ 24$. But then you take out one student since you added one in the beginning. Now, the possibilities are $ 11$ and $ 23$ students. You know that there are more than $ 11$ students, so the answer is $ 23$.

Look at The Chinese Remainder Theorem for a more formal, generalized explanation.

How about:

x is 1mod2
x is 2mod3
x is 3mod4

so

x is (-1)mod2, (-1)mod3, (-1)mod4...

this leads to x is (-1)mod12, x = 11,23,35,... x = [b]23[/b]

thats exactly same way as mine.  :D

EDIT: oops, oh yea

well anyways it has form of $ 12k\minus{}1$ and the smallest number that satisfies the condition is 23 so blahblahblah

Oh, btw, Fantasy, your solution has one minor hitch in it:

x+1 does not necessarily have to be a multiple of 24, it only has to be a multiple of 12.

But since the constraints applied effectively remove those possibilities, that gives the correct answer.

If instead Al's class has 11 students, that satisfies the grouping constraint, but fails the inequality constraint.

Just a tiny problem.


What score is usually required on the USAMO to qualify for MOSP?(particularly red and blue)

Red is usually around 9, Blue around 20.

wow, i can't even do one of them except a few of the exams where its mainly algebra or counting :D

2005: 9/21/28
2006: 9/19/25
2007: 9/17/25
2008: 10/20/28

The above are the Red/Blue/Black cutoffs for their respective years. There may be a mistake in one or two of them, but I'm quite sure the above numbers are right.

[quote="cognos599"]i'm pretty sure 2007 was 9/19/23[/quote]

Oh yeah, you're right about that black cutoff. However, 19 was the HM cutoff, not the blue cutoff, because I know some 2007 blue MOPpers who got 17.

Yeah, 2007 was 9/17/23 (although, like 5/12 of the people who originally qualified for black got demoted to blue).

Also, 2006 was originally announced as 9/19/25, but the blue MOP cutoff got lowered to 18 after some people decided not to come.

(I would also point out that 2007 was a [i]very[/i] hard year...)


ratio
combinatorics unsolved
combinatorics

An airport contains 25 terminals which are two on two connected by tunnels. There is exactly 50 main tunnels which can be traversed in the two directions, the others are with single direction. A group of four terminals is called [i]good[/i] if of each terminal of the four we can arrive to the 3 others by using only the tunnels connecting them. Find the maximum number of good groups.

it can be solved in the same way as this china 1999 problem:

there are 99 space stations.each pair of space stations is connected by a tunnel.there are 99 two-way main
tunnels and all the other tunnels are strictly one-way tunnels.a group of four space stations is called
connected if one can reach each station in the group from every other station in the group without using
any tunnels other than the 6 tunnels which connect them.determine the maximum number of connected groups. :D

[quote="mlm95"]it can be solved in the same way as this china 1999 problem:

 :D[/quote]
it's not exactly the same.in this problem the ratio of the mainroads and the stations is $2$,not $1$.


Let $ X_{b}$ and $ X_{c}$ be the points of tangency of the $ B$- and $ C$- excircles of $ \Delta ABC$ with $ CA$ and $ AB$, respectively. Let $ X_{b}I_{c}$ and $ X_{c}I_{b}$ meet at $ P$. Find $ \dfrac{[PBC]}{[ABC]}$.

I don't feel like calculating it, but this is really easy with barycentric coordinates...


February $ 1$, $ 2008$ is a Friday.  The last Friday of February $ 2008$ will be February $ x$, $ 2008$.  If $ 2008$ is a leap year, what is the value of $ x$?

\[ \begin{array}{|c|c|c|c|c|c|c|}\hline
\multicolumn{7}{|c|}{\text{February} \ 2008} \\ \hline
\color{red}\text{Sun}\color{black} & \text{Mon} & \text{Tue} & \text{Wed} & \text{Thu} & \text{Fri} & \color{blue}\text{Sat}\color{black} \\ \hline \hline
  &   &   &   &   & 1 & 2 \\ \hline
3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline
10 & 11 & 12 & 13 & 14 & 15 & 16 \\ \hline
17 & 18 & 19 & 20 & 21 & 22 & 23 \\ \hline
24 & 25 & 26 & 27 & 28 & 29 &  \\ \hline
\end{array}\]

Thus $ x\equal{}\boxed{29}$.

Faster way: If February 1st is a Friday, then the multiple of 7 that's closest to 29 would be 28th, so you might be tempted to say that the 28th would be a Friday. However, we must add one, so the last Friday would be the 29th.

We know the date is 1 mod 7. We also know that it is in the domain [1,29]. Clearly the only numbers that satisfy this are 1,8,15,22,29 of which $ \boxed{29}$ is the largest.


function
logarithms
limit
real analysis
real analysis unsolved

For each integer n>=2, set
\[\zeta (n) = \sum_{k = 1}^\infty{\frac{1}{{k^{n}}}}\]

Prove that, for -1 < x <= 1,

\[\log \Gamma (1+x) =-\gamma x+\sum_{k = 2}^\infty{\frac{{(-1)^{k}}}{k}\zeta (k)x^{k}}\]

$ \Gamma(1+x)=\lim_{n\to\infty}\frac{n! \, n^{x}}{(x+1)\cdots (x+n)}=\lim_{n\to\infty}\frac{e^{x \ln n-x H_{n}}\, e^\frac{x}{1}\cdots e^\frac{x}{n}}{\left(1+\frac{x}{1}\right)\cdots \left(1+\frac{x}{n}\right)}=e^{-\gamma x}\prod_{k=1}^\infty \frac{e^\frac{x}{k}}{1+\frac{x}{k}}$

$ \ln \Gamma(1+x)=-\gamma x+\sum_{k=1}^\infty \left[\frac{x}{k}-\ln\left(1+\frac{x}{k}\right)\right] =-\gamma x+\sum_{k=1}^\infty \sum_{n=2}^\infty \frac{1}{n}\left(-\frac{x}{k}\right)^{n}$

$ =-\gamma x+\sum_{n=2}^\infty \frac{(-1)^{n}}{n}\underbrace{\sum_{k=1}^\infty \frac{1}{k^{n}}}_{=\zeta(n)}x^{n}$


geometry
circumcircle
ratio
projective geometry
angle bisector
geometry proposed

ABC is an acute-angled triangle. The incircle (I) touches BC at K. The altitude AD has midpoint M. The line KM meets the incircle again at N. $ E\equal{}BN\cap (I), F\equal{}CN\cap (I)$. Show that $ EF\parallel BC$

[quote="mr.danh"]ABC is an acute-angled triangle. The incircle (I) touches BC at K. The altitude AD has midpoint M. The line KM meets the incircle again at N. $ E \equal{} BN\cap (I), F \equal{} CN\cap (I)$. Show that $ EF\parallel BC$[/quote]
[hide="Solution"]
By http://www.artofproblemsolving.com/Forum/viewtopic.php?p=118682#118682, we have that the circumcircle of $ \triangle BNC$ is tangent to the incircle at $ N$.  Now, inverting around $ N$ with arbitrary radius, say radius $ R$, maps the inverted image of the incircle, which is the circumcircle of $ \triangle NEF$, parallel to the inverted image of the circumcircle of $ \triangle NBC$.  Now, let $ B$ map to $ B'$, $ C$ map to $ C'$, $ E$ map to $ E'$, and $ F$ map to $ F'$.  The inverted image of the circumcircle of $ \triangle NBC$ is $ B'C'$ and the inverted image of the circumcircle of $ \triangle NEF$ is $ E'F'$.  Thus, $ B'C'\parallel E'F'$.  Also, $ NB\cdot NB'\equal{}R^2\equal{}NC'\cdot NC$, so $ B'BCC'$ is cyclic.  Similarly, $ E'EFF'$ is cyclic.  Thus, $ \angle NFE\equal{}\angle NE'F'\equal{}\angle NB'C'\equal{}\angle NCB$, so $ EF\parallel BC$. [/hide]

We can solve it easily.
See at here [url]http://www.mathlinks.ro/viewtopic.php?p=118682#118682[/url]
NK is the bisector of angle BNC, hence $ \frac{KB}{BN}\equal{}\frac{KC}{CN}\equal{}k$
$ \triangle BKE\sim\triangle BNK\Rightarrow\frac{BE}{BK}\equal{}\frac{BK}{BN}\equal{}k$
$ \triangle CKF\sim\triangle CNK\Rightarrow\frac{CF}{CK}\equal{}\frac{CK}{CN}\equal{}k$
Then $ \frac{BE}{BN}\equal{}\frac{CF}{CN}\equal{}k^2$
Hence result.  :wink:

Let $ P,\ Q,$ the tangent points of the incircle $ (I),$ with $ AB,\ AC$ respectively and we denote the point $ S\equiv BC\cap PQ.$

Let $ K'$ be, the antidiametric (= diametric opposite) point of $ K$ and we denote the point $ N\equiv (I)\cap SK'.$

Because of the concurrency of the line segments $ AK,\ BQ,\ CP,$ as well, we have that the points $ S,\ B,\ K,\ C,$ are in harmonic conjugation $ ($ from the complete quadrilateral $ AQG_{e}P,$ where $ G_{e}\equiv AK\cap BQ\cap CP,$ as the [b][size=100]Gergone point[/size][/b] of $ \bigtriangleup ABC$ $ )$ and so, the pencil $ N.SBKC,$ is in harmonic conjugation too.

Because of now, $ NK\perp NS,$ we have that the ray $ NK,$ bisects the angle $ \angle BNC$ and so, we conclude that $ EF\parallel BC,$ where $ E\equiv (I)\cap BN$ and $ F\equiv (I)\cap CN.$

$ \bullet$ Let be the point $ M\equiv AD\cap KN,$ where $ AD,$ is the altitude of $ \bigtriangleup ABC$ and it is enough to prove that $ MA \equal{} MD.$

We draw the line $ SX,$ through the point $ S$ and perpendicular to $ SB.$

Because of $ IK \equal{} IK'$ and $ KK'\parallel SX,$ we conclude that the pencil $ S.KINX,$ is in harmonic conjugation.

Because of now, $ KK'\perp SK$ and $ KA\perp SI$ $ ($ because of the line segment $ PQ,$ as the polar of $ A,$ with respect to $ (I),$ passes through the point $ S,$ we also have that the line segment $ KA,$ is the polar of $ S$ with respect to $ (I)$ $ )$ and $ KN\perp SN$ and $ KS\perp SX,$ we conclude that the pencil $ K.K'ANS,$ is in harmonic conjugation too.

$ ($ Because of the angles formed by the lomologous rays of the pencils $ S.KINX$ and $ K.K'ANS,$ are equal $ ).$

Hence, because of $ AD\parallel KK',$ we conclude that $ MA \equal{} MD$ and the proof is completed.

Kostas Vittas.

Dear Mathlinkers,
1. it is to day known that the circumcircle of BNC is tangent to the incircle of ABC at N
2. Applied the Reim's theorem to this two tangent circles and we are done
Sincerely
Jean-Louis

My solution is quite complicated.

Let $ X,Y$ be the points of tangent of $ (I)$ and $ AB,AC$ and $ T\equiv XY \cap BC$. It is well-known that $ (TKBC) \equal{} \minus{} 1$. Let $ S$ be the midpoint of $ TK$. Since $ AK$ is the polar of $ T$ with respect to $ (I)$, $ IT \bot AK$, thus $ \angle KAD \equal{} \angle KTI$, which yields $ \triangle KTI$ and $ \triangle DAK$ are similar. But $ IS, KM$ are respectively the median of $ \triangle KTI$ and $ \triangle DAK$ so $ \triangle ISK$ and $ \triangle KMD$ are also similar. By this, we have $ IS\bot NK$. However, $ SK$ is the tangent of $ (I)$ so either is $ SN$. By Newton's identity in a harmonic division we have $ SN^2 \equal{} SK^2 \equal{} \overline{SC}\cdot\overline{SD}$. This means $ SN$ is also the tangent of the circumcircle of $ \triangle NBC$ at $ N$. Therefore, the triangles $ \triangle SNC$ and $ \triangle SBN$ are similar, form the ratios, we deduce that $ \frac {SC}{SB} \equal{} \frac {NC^2}{NB^2}$. Finally, $ (TKBC) \equal{} \minus{} 1$ so by Mc Laurin's identity, take the geometrical length, we have $ BC\cdot BS \equal{} BK\cdot BT$ and $ CB \cdot CS \equal{} CK \cdot CT$, divides those two equations we deduce that $ \frac {SC}{SB} \equal{} \frac {KC}{KB} \cdot \frac{TC}{TB} \equal{} \frac {KC^2}{KB^2}$. Hence $ \frac {KC^2}{KB^2} \equal{} \frac {SC}{SB} \equal{} \frac {NC^2}{NB^2}$, which means that $ \frac {NC}{NB} \equal{} \frac {KC}{KB}$, id est $ NK$ is the angle bisector of $ \angle BNC$. So $ \angle KFE \equal{} \angle KNB \equal{} \angle KNC \equal{} \angle CKF$. By this, $ EF\parallel BC$. Q.E.D.

[quote="Mashimaru"]My solution is quite complicated.

Let $ X,Y$ be the points of tangent of $ (I)$ and $ AB,AC$ and $ T\equiv XY \cap BC$. It is well-known that $ (TKBC) \equal{} \minus{} 1$. Let $ S$ be the midpoint of $ TK$. Since $ AK$ is the polar of $ T$ with respect to $ (I)$, $ IT \bot AK$, thus $ \angle KAD \equal{} \angle KTI$, which yields $ \triangle KTI$ and $ \triangle DAK$ are similar. But $ IS, KM$ are respectively the median of $ \triangle KTI$ and $ \triangle DAK$ so $ \triangle ISK$ and $ \triangle KMD$ are also similar. By this, we have $ IS\bot NK$. However, $ SK$ is the tangent of $ (I)$ so either is $ SN$. By Newton's identity in a harmonic division we have $ SN^2 \equal{} SK^2 \equal{} \overline{SC}\cdot\overline{SD}$. This means $ SN$ is also the tangent of the circumcircle of $ \triangle NBC$ at $ N$. Therefore, the triangles $ \triangle SNC$ and $ \triangle SBN$ are similar, form the ratios, we deduce that $ \frac {SC}{SB} \equal{} \frac {NC^2}{NB^2}$. Finally, $ (TKBC) \equal{} \minus{} 1$ so by Mc Laurin's identity, take the geometrical length, we have $ BC\cdot BS \equal{} BK\cdot BT$ and $ CB \cdot CS \equal{} CK \cdot CT$, divides those two equations we deduce that $ \frac {SC}{SB} \equal{} \frac {KC}{KB} \cdot \frac {TC}{TB} \equal{} \frac {KC^2}{KB^2}$. Hence $ \frac {KC^2}{KB^2} \equal{} \frac {SC}{SB} \equal{} \frac {NC^2}{NB^2}$, which means that $ \frac {NC}{NB} \equal{} \frac {KC}{KB}$, id est $ NK$ is the angle bisector of $ \angle BNC$. So $ \angle KFE \equal{} \angle KNB \equal{} \angle KNC \equal{} \angle CKF$. By this, $ EF\parallel BC$. Q.E.D.[/quote]

From the beginning of the above proof, we have learned that $ (BNC)$ and $ (I)$ internally tangents at $ N$, further $ SN$ is also the common tangent of these $ 2$ circles. Therefore, we have $ \angle SNE\equiv$ $ \angle SNB$ $ \equal{} \angle NFE$ $ \equal{} \angle NCB$, which imples that $ EF\parallel BC$. $ \square$.


geometry
3D geometry
sphere
trigonometry
induction
geometry proposed

Given $ n$ points in space, it is known that any three of them are the vertices of a triangle with an angle $ >120^{\circ}$. Prove that these points can be labelled $ A_{1},A_{2},\dots,A_{n}$ such that each angle $ A_{i}A_{j}A_{k}$, with $ 1\leq i<j<k\leq n$ is greater than $ 120^{\circ}$.

[hide]On Kalva http://www.kalva.demon.co.uk/soviet/sov69.html the problem seems to be incorrectly stated.[/hide]

[edit]The original was deleted because it was horribly wrong.. hopefully this one isn't[\edit]

Consider the pair of points $ (X,Y)$ with the largest distance
Let $ S$ be the region defined by the intersection of the family of solid spheres $ T$ centred at $ O$ for all points $ O$ such that $ OX\equal{}OY$ and $ \angle XOY\equal{}120^{\circ}$ with radius $ OX$
Since for any other point $ P$, $ \angle XPY>120^{\circ}$, all the other points must lie within $ S$
Consider the pair of points $ (Z,W)$ with the second largest distance

If each of $ Z$,$ W$ is distinct from $ X$,$ Y$,
WLOG $ XZ>YZ$
Consider the sphere $ U$ in $ T$ with centre $ N$ in the same plane as $ XYZ$.
Let $ V$ be the intersection of $ ZW$ and the boundary of $ U$ with $ W$ between $ Z$ and $ V$
$ ZW < ZV \equal{} NX^2 \plus{} NZ^2 \minus{} 2(NX)(NZ) \cos \angle ZNW$ $ < NX^2 \plus{} NZ^2 \minus{} 2(NX)(NZ) \cos \angle ZNX \equal{} ZX$ $ (\Rightarrow\Leftarrow)$

Thus WLOG $ W \equal{} X$
If the $ (n \minus{} 1)$ points excluding $ Y$ (but including $ X$,$ Z$) can be ordered satisfying the conditions with $ X$ in front, append $ Y$ at the back
It is easy to see that this new ordering also satisfies the conditions because for any points $ (P,Q) \not\in \{X,Y\}$, $ \angle XPQ > 120^{\circ} \Rightarrow \angle YPQ < 120^{\circ} \Rightarrow \angle PQY > 120^{\circ}$
Therefore by induction we are done

let $ S$ be the set of our points...
no let $ A,B$ be two points among these points having the maximum distance...
and show that $ AX\not\equal{} AY$ for $ X,Y\in S$,then let $ A_1\equal{}A$ and $ A_2$ be the nearest point to $ A$ in  $ S\minus{}\{A_1\}$ (which is unique because of what we said before),...,$ A_{k\plus{}1}$ be the nearest point to $ A$ in $ S\minus{}\{A_1,...,A_k\}$...
so we will have $ A_n\equal{}B$
now show that this labelling has the conditions of the problem...


calculus
integration
inequalities
function
LaTeX
real analysis
real analysis unsolved

Let Ø:B1->R be a measurable function s.t. |Ø(x)|≦M, for every x in B1.
Here M is a positive constant and B1 is the unit open ball of R^n center at the origin.
Show that
[img]http://img8.imageshack.us/img8/3159/problemm.gif[/img]
Thanks for any help. :)

Roger that!
Thanks.

Start with $ (X-Y)(e^{X-Y}-e^{Y-X})\ge 0$, plug in $ X=\varphi(x)$, $ Y=\varphi(y)$, open the parentheses, integrate with respect to $ x$ and $ y$, and try to learn $ \text{\LaTeX}$ :).


Who here is doing Cody Bowl IV?  (I want to scope out the competition, heh.)

[u]Team Pythagorean[/u]
mathfanatic
ComplexZeta
confuted
applepuppy88

[u]ACTON AoPS BOWLERS[/u]
Rep123max
beta
JSRosen3
tetrahedr0n

Turns out that they already filled all the spots, so Im going to be in teh draft.

Wow, looks like the competition will be pretty tough.  (Did I say that?  I meant 'We'll kill you.  Twice over.')

wait...how many AoPSers are gonna be doing this?...

besides myself, the 8 listed for those two teams, and jelyman?

I saw a person with the name Magnara on the msg boards, so perhaps he is as well.

[quote="JBL"]I just registered for the draft, I think.  See you all on the playing field, or whatever :)[/quote]

The draft is still open?...Wow...

It looks like they were doing it one pick a day.  Anyhow, I don't know if Cody will put me in, since I only sent him the sign-me-up e-mail just now.


In [u]Mathlinks Contest, first editions, second round[/u]
(see http://www.mathlinks.ro/Forum/viewtopic.php?t=28185 )
appears the following (difficult and interesting) inequality: $\frac 25\le \frac{Rs}{2aR+bc}$.
This inequality is equivalently with the following symmetrical inequality:
$\max \{a+h_a,\ b+h_b,\ c+h_c\}\le \frac 58 (a+b+c).$ Here is the my solution
($2s\equiv a+b+c,\ S^2=s(s-a)(s-b)(s-c),\ bc=2Rh_a,\ldots$):

[b]Lemma.[/b] $a^2+4h^2_a\le (b+c)^2.$

[u]Proof.[/u] $a^2+4h^2_a\le (b+c)^2 \Longleftrightarrow h^2_a\le s(s-a)\Longleftrightarrow 4S^2\le a^2s(s-a)\Longleftrightarrow$
$4(s-b)(s-c)\le a^2\Longleftrightarrow$ $0\le (b-c)^2.$ We have equality iff b=c.

[b]The proof of the initial inequality.[/b] I used the [i]Tchebychev's[/i] inequality.

$36(b+c)^2\ge 36(a^2+4h^2_a)=[9\cdot (2a)^2+16\cdot (3h_a)^2]\ge \frac{1}{25}(9\cdot 2a+16\cdot 3h_a)^2\Longrightarrow$
$25(b+c)^2\ge (3a+8h_a)^2\Longleftrightarrow 5(b+c)\ge 3a+8h_a\Longleftrightarrow \overline {\underline {\left| 5s\ge 4(a+h_a)\right| }}\Longleftrightarrow$
$5sR\ge 4aR+2bc\Longleftrightarrow \frac 25\le \frac{Rs}{2aR+bc}.$

[u]Remark.[/u] $a+h_a\le \frac{5s}{4}$ a.s.o. $\Longrightarrow \max \{a+h_a,\ b+h_b,\ c+h_c\}\le \frac 58 (a+b+c).$

Your proof is mostly the same as my proof of Assertion 1 in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=187381#p187381]http://www.mathlinks.ro/Forum/viewtopic.php?t=28185 post #4[/url], except of the part where you derive the initial inequality from the crucial Lemma, which is more elegant in your proof.

  Darij


IMO Shortlist
number theory unsolved
number theory

Do there exist two distinct positive intergers $ x,y$ such that $ y^n\plus{}n|x^n\plus{}n$ for every positive interger $ n$?

No, there such integers do not exist. That's (slightly reworded) problem N6 from IMO shortlist 2005. See http://www.mathlinks.ro/Forum/viewtopic.php?p=466215 .


number theory
relatively prime
Recursive Sequences

Let $x_{1}$ and $x_{2}$ be relatively prime positive integers. For $n \ge 2$, define $x_{n+1}=x_{n}x_{n-1}+1$.[list=a][*] Prove that for every $i>1$, there exists $j>i$ such that ${x_{i}}^{i}$ divides ${x_{j}}^{j}$. [*] Is it true that $x_{1}$ must divide ${x_{j}}^{j}$ for some $j>1$? [/list]

[quote="Peter"]Let $ x_{1}$ and $ x_{2}$ be relatively prime positive integers. For $ n \ge 2$, define $ x_{n \plus{} 1} \equal{} x_{n}x_{n \minus{} 1} \plus{} 1$.[list=a][*] Prove that for every $ i > 1$, there exists $ j > i$ such that $ {x_{i}}^{i}$ divides $ {x_{j}}^{j}$. [*] Is it true that $ x_{1}$ must divide $ {x_{j}}^{j}$ for some $ j > 1$? [/list][/quote]
http://www.kalva.demon.co.uk/short/soln/sh94n6.html


function
calculus
derivative
real analysis
real analysis theorems

Consider the open interval $(a,b)$ :

let f be a function on that interval to $\mathbb{R}$

under what conditions does f have a primitive : under what conditions is there a differentiable function on the same interval with f as derivative?

It seems like such a basic question but I have never seen a solution.

I know by Darboux that it has to satisfy the intermediate value property.
I know continuous functions all have primitives, but not all derivatives are continuous.

So it will be a set of functions ( a space) between intermediate value and continuous, but which exactly?

They also have to be measurable.
I do not think such an easy characterization exists.


It is possible to sb to explain me how the vandermonde determinat works? with an example maybe?

Start by looking [url=http://en.wikipedia.org/wiki/Vandermonde_matrix]here[/url] and [url=http://www.mathematics-online.org/inhalt/erlaeuterung/erlaeuterung282/]here[/url].


Hello.  I'm Bill, an 8th Grader from NE.  I won my Chapter Competition (1st in Countdown too, and first in team   :lol:), and I'm preparing for the State Competition this weekend.

My team practiced by using the 2002 state competition, and we came across a problem that my coach and a math teacher at my school were stumped by.  Here it is (number 26 in Sprint):

"What is the remainder when 2^87 (+ 3) is divided by 7?"

I think that I may be coming across these problems at state, so could you help.  Thanks!

EDIT: Oops, I just saw that this was a repeat post.  I still don't quite get the explanation though, so could you help?  Here's another problem I have a question on (and stumped by coach).  Number 8 in 2002 Target:

"Debbie has an ample supply of the square ceramic tiles with the patterns shown below.  She intends to create a horizontal, rectangular wall decoration by cementing three of the squares to the wall in a row.  How many different designs can she create in this way that have at least one line of symmetry?

[url]http://img144.imageshack.us/my.php?image=mathcountsfb5.png[/url]"

Thanks!!!

:welcomeani: Welcome to AoPS!

[hide]Note that $2^{3}$ when divided by 7 leaves a remainder of 1. Thus, $(2^{3})^{2}9=2^{8}7$ also leaves a remainder of [b]1[/b].[/hide]

Thanks!  I also posted another problem, since I saw that this was already in discussion.

Are you sure? This is my answer. It uses mods, and i'm not sure they're appropriate for this forum. however, i have seen them used, so here goes:

[hide]$2^{87}+3 \equiv x \mod 7$

$2^{{3}^{29}}\equiv (x-3) \mod 7$

So, evaluating two to the third power, we have:

$8^{29}\equiv (x-3) \mod 7$

Now, we know that 8 mod 7 is 1 , so we have: (this is in conjunction with a theorem in AoPS volume II that states that [b]A^b mod c is congruent to (A-c)^b[/b] where the ^ means to the power of)

$1^{29}\equiv (x-3) \mod 7$

Since one to the anythingth power is always one,

$1 \equiv (x-3) \mod 7$

So, adding the three to both sides,

$4 \equiv x \mod 7$

Our final answer is four. (this is also the final answer in the answer key). [/hide]

Can someone explain in a way that doesn't use mods? That would probably be helpful too. :ninja:

 :welcome: AoPS is fun!

yes. all of you before forgot to add 3. [hide]You could just look for a pattern.
2^1/7=0 R2
2^2/7=0R4
2^3/7=1R1
2^4/7=2R2. So we see the repeating pattern of remainders. 2,4,1...We see that this repeats every three numbers. 87 is divisible by 3, so it will end up at the end of a complete cycle, at 1. Then add 3 and get 4! (not 4 factorial)[/hide]

[quote="zserf"][quote="easyas3.14159..."]:welcomeani: Welcome to AoPS!

[hide]Note that $2^{3}$ when divided by 7 leaves a remainder of 1. Thus, $(2^{3})^{29}=2^{87}$ also leaves a remainder of [b]1[/b].[/hide][/quote]

[hide]I [i]think[/i] that that logic is flawed say you were asking what $2^{6}$/7 remainder is. by your logic $2^{2}$ has a remainder of 4 therefore $(2^{2})^{3}$ has a remainder of 4, but $2^{3}$ has a remainder of 1 so $(2^{3})^{2}$ should have a remainder of one.[/hide][/quote]

[hide]No, what it actually should say is that since $2^{3}$ divided by 7 has a remainder of one, $(2^{3})^{29}$ divided by 7 has a remainder of $1^{29}$ , which is, obviously, one. Then in your argument $(2^{2})^{3}$ divided by seven would have a remainder of $4^{3}$ which is 64 which is congruent to one mod 7.[/hide]

For the second problem
[hide="hint"]
Remember that she has an unlimited supply of tiles. This means she can use a tile more than once.
[/hide]

[quote="zserf"][quote="easyas3.14159..."]:welcomeani: Welcome to AoPS!

[hide]Note that $2^{3}$ when divided by 7 leaves a remainder of 1. Thus, $(2^{3})^{29}=2^{87}$ also leaves a remainder of [b]1[/b].[/hide][/quote]

[hide]I [i]think[/i] that that logic is flawed say you were asking what $2^{6}$/7 remainder is. by your logic $2^{2}$ has a remainder of 4 therefore $(2^{2})^{3}$ has a remainder of 4, but $2^{3}$ has a remainder of 1 so $(2^{3})^{2}$ should have a remainder of one.[/hide][/quote]
$2^{2}$ has a remainder of 4, but it has to be cubed
4 cubed has a remainder of 1
however, $2^{3}$ has a remainder of 1. after the 1 is squared, it is still a 1.
so either way the remainder is 1 when $2^{(2)(3)}$ is divided by 7

Ok.  Thanks for the help.  I see how the powers problem can be figured out.  Also, abouit the adding three, I saw that I forgot to add + 3 to the problem the first time, so I just put it in parentheses.  Thanks!!!


integration
trigonometry
calculus
calculus computations

Compute:

$ \int_{0}^{1} \frac{x}{\sqrt{1\plus{}x^4}}dx$

put $ x^2\equal{}\tan\,\theta$ to get $ \frac{1}{2}\int_0^\frac{\pi}{4}\;\sec\,\theta\;\textbf d\theta$

Write $ x^4\equal{}(x^2)^2$ and make $ z\equal{}x^2,$ the resultant integral can be tackled via trigonometric, hyperbolic or another kind of substitution.

Hello, write your integral in the form
$ \int_{0}^{1}\frac{x}{\sqrt{1\plus{}(x^2)^2}}dx$ and set $ x^2\equal{}t$ and you will
get the folowing integral: $ \frac{1}{2}\int_{0}^{1}\frac{1}{\sqrt{1\plus{}t^2}}$ which is
$ \frac{1}{2}arsinh(t)\plus{}C$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]Hello, write your integral in the form
$ \int_{0}^{1}\frac {x}{\sqrt {1 \plus{} (x^2)^2}}dx$ and set $ x^2 \equal{} t$ and you will
get the folowing integral: $ \frac {1}{2}\int_{0}^{1}\frac {1}{\sqrt {1 \plus{} t^2}}$ which is
$ \frac {1}{2}arsinh(t) \plus{} C$.
Sonnhard.[/quote]

I did like this.


Hey guys. This is a tough combinatorics problem our teacher gave to us as an extra problem to do - I'm sure I've seen it somewhere, but I can't quite grasp how to solve it.

"5 pairs of cats are sitting on a fence. There are two red cats, two blue cats, two yellow cats, two green cats, and two purple cats. However, these cats do not like other cats of the same color, and will attack each other if place next to each other. How many ways can these ten cats be arranged on the fence (in a line) so that none of the cats is sitting next to a cat of the same color?"

I'm not really sure how to approach this problem. A little help, guys?

Hmm...interesting problem. I'm sure there's some clever bijection, but I can't find it.  :|

Isn't this a simple derangement problem? It is the next step after learning all about PIE (inclusion/exclusion)

um

recursively i think it works

suppose you have a string of n pairs of cats that satisfies the conditions (we'll call this good)

Then to make a good string with n+1 pairs, you put the first cat of the new color in any of the 2n+1 positions in between the other 2n cats; after that, you have 2n+2 positions for the second cat, but it can't be in the 2 positions next to the same color cat, so yo uhave 2n choices for the second cat to go into

so you have $a_{2}= 2$ and $a_{k}= (2k-1)(2k-2)a_{k-1}$ where $a_{i}$ is the number of arrangements with $i$ pairs.

A couple problems

Not sure if cats of the same color are distinct or indistinct. We can just assume their indistinct even though that makes less practical sense. In this case, the recursion should be ${2n-1 \choose 2}$ because the placing of the two cats is overcounted.

If it's distinct just start with $a_{n}= 8$ and don't change the recursion.

The other problem is that "red blue blue red," for example isn't counted in $a_{n}$ but is perfectly valid if we insert a yellow to get "red blue yellow blue red."

So instead we do a big PIE:

The first term is all possible arrangements. The second is found by grouping one pair (5C1 possible pairs) and arranging that pair and all other cats (9! ways divided by 2^4 for the indistinct pairs.) And the other terms follow similarly.

$\frac{10!}{2^{5}}-\frac{{5 \choose 1}9!}{2^{4}}+\frac{{5 \choose 2}8!}{2^{3}}-\frac{{5 \choose 3}7!}{2^{2}}+\frac{{5 \choose 4}6!}{2^{1}}-\frac{{5 \choose 5}5!}{2^{0}}$

This is indistinct cats. I think distinct is just multiplying through by 2^5.


So how good are you at math?

Id give myself a 3

**Ok sry bout that...messed up when creating the poll :oops: ***

I own everyone in my math class at math.

My class sucks at academics.  :no:

CLOSED because I messed up when creating the pole :oops:


number theory
greatest common divisor
number theory solved

Hello, I just created my own problem and I think it's easy. I just want to check if I got it right.

If $a,b,c,d,e,f$ are 6 distinct integers, let $S(a,b,c,d,e,f) = (a-b)(a-c)(a-d)....(e-f)$ be the product of pairwise difference.

Let $T$ be the set of all such $S$, as $a,b,c,d,e,f$ go over all possible integers. That is $T = \{S(a,b,c,d,e,f) | a,b,c,d,e,f$ are 6 distinct integers $\}$.
What is the gcd of all elements in $T$?
(Note that all element in $T$ can't be zero, and it can't be 1 either, so we know the gcd exists)

My answer is [hide]$2^63^35$[/hide]

I get $2^{8}3^{3}5$ for the GCD.

Anyway, the problem is to show that $\prod_{1\leq i<j\leq 6}\left(j-i\right)\mid\prod_{1\leq i<j\leq 6}\left(a_{j}-a_{i}\right)$ for any six integers $a_{1}$, $a_{2}$, $a_{3}$, $a_{4}$, $a_{5}$, $a_{6}$, because this will automatically yield that $\prod_{1\leq i<j\leq 6}\left(j-i\right)$ is a common divisor of all products of the form $\prod_{1\leq i<j\leq 6}\left(a_{j}-a_{i}\right)$, and it must be their greatest common divisor, since a greater common divisor would contradict the fact that, taking $a_{i}=i$ for every i, we have $\prod_{1\leq i<j\leq 6}\left(a_{j}-a_{i}\right)=\prod_{1\leq i<j\leq 6}\left(j-i\right)$.

As for the relation $\prod_{1\leq i<j\leq 6}\left(j-i\right)\mid\prod_{1\leq i<j\leq 6}\left(a_{j}-a_{i}\right)$, this generalizes to $\prod_{1\leq i<j\leq n}\left(j-i\right)\mid\prod_{1\leq i<j\leq n}\left(a_{j}-a_{i}\right)$ for any n integers $a_{1}$, $a_{2}$, ..., $a_{n}$, and this was proven in http://www.mathlinks.ro/Forum/viewtopic.php?t=4858 .

  Darij

nice.... thanks
My solution is using pigeon hole, showing that it must be divisible by 5, by 3 three times, and by 2 eight times. Then exhibit the case that 1 2 3 4 5 6 is in fact this number.


geometry
3D geometry
sphere
integration
calculus

[b]What is graviational presuure?[/b]

[hide]I want the definiton and the algebraic formula for it in the general case.[/hide]

Also calculate the gravitational pressure in a gas cloud condensing to form a star if at any moment it is a sphere of volume $ \Omega $ and Mass $ M$ (INPHO 2005 question)

[hide]I am not writing much know but once somebody gives a response I would like to discuss sth[/hide]

Nobody has any idea???

Well I read this definiton. But using this I'm not able to get the result for the above question!Any ideas???

Consider a (a not very realistic) star of uniform density and temperature. The graviational pressure  can be obtained from the gravitational force  using the relation

 $ P_g -= \frac{F_g}{4 \pi r^2} $

To calculate the net gravitational pressure we consider the interaction of thin shells of matter with mass interior to each shell and add up contribnutions from all shalls. Thus,

$P_g = - \int_0 ^{R} G \frac{ (4 \pi r^2 \varrho ) ( \frac{4}{3} \pi r^3 \varrho dr )}{4 \pi r^4 } $

still nothing!!!!! Please post at least your opinion on the formula I have given!!!

Sounds quite believable. It is just some pressure caused by the gravitational forces between the particles of the cloud or whatever, right? But what is the problem then? You just integrate your formula and you should get the gravitational pressure in the center of the star. To get the gravitational pressure at distance $d$ from the center, you change the lower bound from 0 to $d$.

The problem is that my approach (which I also found at a no. of sites as the definiton of gravitational pressure) doesn't give the answer in the original INPhO 2005 question!!!

So you're saying that you don't know the exact distribution of $\rho$, only the mass and volume of the star? Hm... interesting. And are you sure that this can be done?

I'm afraid that this problem requires that you first calculate the distribution of density all over the cloud and then integrate it using your gravitational pressure formula - otherwise it could be anything. Well, there's a new problem for you...

I believe we do assume a unifrom mass distribution!!!

In that case, what is the problem? Radius of the star is

$R=\sqrt[3]{\frac{3\varOmega}{4\pi}}$

and its density is

$\varrho=\frac{M}{\varOmega}$.

Gravitational pressure in the center of the cloud is

$P_g=\int_0^R\!\!\!G\frac{(\frac{4}{3}\pi r^3\varrho)(4\pi r^2\varrho)\,dr}{4\pi r^4}=\frac{4}{3}\pi\varrho^2G\int_0^R\!\!\!r\,dr=\frac{2}{3}\pi\varrho^2GR$

So what do we do now? Are you saying that this is not the correct answer?

Yes, as per the question in the INPhO paper, the answer is a lot different !!(I don't exactly remember but it had a 5 in the denominator)

Thats the whole problem!!! Can anybody plz help!!!

Here's the original paper.(complete). My problem is with the last question. Maybe this will make things clear and maybe I'm interpreting wrongly!!

OK, I've got it! :)
First thing, I would like to correct some errors from my previous post:

$P_g=-\int_0^R\!\!\!G\frac{(\frac{4}{3}\pi r^3\varrho)(4\pi r^2\varrho)\,dr}{4\pi r^4}=-\frac{4}{3}\pi\varrho^2G\int_0^R\!\!\!r\,dr=-\frac{2}{3}\pi\varrho^2GR^2$

Let's plug in $R$ and $\varOmega$:

$R=\sqrt[3]{\frac{3\varOmega}{4\pi}}$

$\varrho=\frac{M}{\varOmega}$

$P_g=-\frac{2}{3}\pi G\!\left(\frac{M}{\varOmega}\right)^{\!\!2}\!\!\!\left(\frac{3}{4\pi}\right)^{\!\!\frac{2}{3}}\!\!\!\varOmega^\frac{2}{3}=-\frac{2\pi}{3}\frac{GM^2}{\varOmega^\frac{4}{3}}\!\left(\frac{3}{4\pi}\right)\!\!\!\left(\frac{4\pi}{3}\right)^{\!\!\frac{1}{3}}=-\frac{1}{2}\!\left(\frac{4\pi}{3}\right)^{\!\!\frac{1}{3}}\!\frac{GM^2}{\varOmega^\frac{4}{3}}$

Now it looks almost like what we want to get, only there's 1/2 instead of 1/5. The problem is that this is the gravitational pressure in the center of the cloud. It seems that what we want to get is the average gravitational pressure throughout the whole cloud. There's probably a much easier way to find it but we can for example integrate the gravitational pressures in all points of the cloud and then divide this by the volume of the cloud. Let $P_g(b)$ be gravitational pressure in the distance $b$ from the center of the cloud:

$P_g(b)=-\int_b^R\!\!\!G\frac{(\frac{4}{3}\pi r^3\varrho)(4\pi r^2\varrho)\,dr}{4\pi r^4}=-\frac{2}{3}\pi\varrho^2G(R^2-b^2)$

The average gravitational pressure $P_a$ is:

$P_a=\int_0^R\!\!\frac{P_g(b)(4\pi b^2)\,db}{\frac{4}{3}\pi R^3}=\int_0^R\!\!\!P_g(b)\frac{3b^2\,db}{R^3}=-\frac{2}{3}\pi\varrho^2G\int_0^R\!\!\!(R^2-b^2)\frac{3b^2\,db}{R^3}=-\frac{2}{3}\pi\varrho^2G(R^2-\frac{3}{5}R^2)=-\frac{2}{3}\pi\varrho^2GR^2\frac{2}{5}$

There's the 2/5 coefficient we need! Hurrah!

But why do we take this 'average'???

What's wrong with my approach????

Frankly, I don't really know :) . But if you ask "What is the gravitational pressure inside of this cloud?", to me it makes much more sense to answer the average gravitational pressure than to answer only the value of gravitational pressure in some one single point inside of that cloud. (In your case, this one single point was the center of the star.)

But how did you find $ P_g $ at a general point. Isnt that needed and how did you find that???

Yes, it is needed, and it is

$P_g(b)=-\int_b^R\!\!\!G\frac{(\frac{4}{3}\pi r^3\varrho)(4\pi r^2\varrho)\,dr}{4\pi r^4}=-\frac{2}{3}\pi\varrho^2G(R^2-b^2)$

or in other words, you change the integration bounds from $0$ .. $R$ to $b$ .. $R$. ($b$ is the distance of your general point from the center of the cloud.) Now you may ask "why does it work like this?" Well, you shouldn't have problems understanding that if you understand how we got the first integral. Let's start at the surface of the cloud - if $b=R$ then the pressure is obviously zero. If we move $dr$ "deeper", to a point $r$ away from the center, the pressure will be by definition

$\frac{F_g}{4\pi r^2}=G\frac{(\varrho4\pi r^2dr)(\varrho\frac{4}{3}\pi r^3)}{r^2}\frac{1}{4\pi r^2}$.

If we move another $dr$ depper, the pressure in this $2dr$ depth will be the pressure that's in depth $dr$ plus some more

$G\frac{(\varrho4\pi r^2dr)(\varrho\frac{4}{3}\pi r^3)}{r^2}\frac{1}{4\pi r^2}$

etc. etc. until we get

$\int_R^b G\frac{(\varrho4\pi r^2dr)(\varrho\frac{4}{3}\pi r^3)}{r^2}\frac{1}{4\pi r^2}$

and that's it.

Basically, $ P_g $ is the pressure force(total) on a sphere of some rasius $ >0 <R $. But how did we find $F_g $ on that sphere due to external mass's forces and internal mass???

OK, I'll try to show a nicer explanation. We want to calculate the pressure in distance $b$ from the center of the cloud. Let's have a particle of mass $m$ in distance $r$ from the center of the cloud. What is the gravitational force acting on this particle? It's not hard to show that it is exactly

$F_p=G\frac{m(\frac{4}{3}\pi r^3\varrho)}{r^2}$

in the direction to the center of the cloud. Therefore, one "layer" of gas in distance $r$ from the center $dr$ thick creates force

$F_l=G\frac{(4\pi r^2dr\varrho)(\frac{4}{3}\pi r^3\varrho)}{r^2}$.

Particles further than $b$ press on the sphere of radius $b$ while particles closer to center than $b$ don't. Total force $F_g$ is:

$F_g=\int_b^RG\frac{(4\pi r^2dr\varrho)(\frac{4}{3}\pi r^3\varrho)}{r^2}$.


This question was in Mr. Frosts's wonderful MC practice book.
But I dont get it.
How many positive three digit integers have at least one seven in them?
Could someone explain this answer more indepthly? And present a quicker solution?
[hide="solution"]From 100-199, there are 19 numberas that have at least one 7 as one of their digits. (There are ten that have a seven in the tens place 70-79 and ten that have a units place 107, 117, ...197. That would be 20 numbers, except that 177 would be counted twice so 19.) There will also be 19 numbers in the 200s, 300s, 400s, 500s, 600s, 700s, 800s, 900s, while all hundre numbers in the 700s have a 7 in them. That is a total of 19*8+100=252 [/hide]

I really cant see a quicker solution than that, so I will try to explain it in more depth:

[hide="solution"]Three digit numbers range from 100-999, so first lets consider how many 7's there are from 100-199.
CASE 1: 7 in the units place from 100-199
we have 10 possibilities: 107,117,127,137.....197

CASE 2: 10 in the tens place from 100-199
We have 10 possibilities: 170,171,172,173......179

CASE 3: 7 in the hundreds place from 100-199
We have no possibilities

[b]BUT[/b] we counted one twice:177. We counted this twice since it has both a 7 in the ones and a 7 in the tens. Therefore there are only 10+10+0-1=19 numbers that have at least one 7 from 100-199.

The same goes for 200-299. By the same exact way as we used above, there are 19  numbers with at least one 7. the same is true for all such intervals:100-199,200-299,300-399....900-999 [b]EXCEPT[/b] for 700-799. Do you see why? All of the numbers from 700-799 have at least one 7; it is always the hundreds digit. Therefore, there are 100 numbers from 700-799 that have at least one 7.

Adding everything up, we have 19*8+100 (its not 19*9 because the 9 includes the 700-799, which we showed has 100 numbers, not 19, so instead its 19*8+100)[/hide]

Hope that helped :lol:

There is a faster way: Complementary Counting

You count the numbers that don't have any sevens in them and subtract that from the total numbers and get the answer.

To count the total numbers, we have 9 choices for the first, 10 for the second, and 10 for the last. Thus, a total of $ 9 \cdot 10 \cdot 10\equal{}900$ numbers.

Now, we count the numbers with no sevens. There are 8 choices for the first number, 9 for the second, and 9 for the last for a total of $ 8 \cdot 9 \cdot 9\equal{}648$.

Now, we subtract and get: $ 900\minus{}648\equal{} \boxed{252}$


Here is an interesting but really easy problem that I solved recently. Its level is about MC so I'm putting here.

We have a triangle P that its vertices are (0,0), (a,0), and (0,b) where a and b are both positive. If perimeter and area of the P is both 30, what is a+b?

Well...
[hide="solution"]
Since the perimeter is an integer, it's most likely a right triangle.

3-4-5? Nope.
Well, 7.5-10-12.5 has a perimeter of 30. But its area is 37.5.

5-12-13? Yep.
5 + 12 + 13 = 30
5 x 12 / 2 = 30
[/hide]

What do you mean by "it's most likely a right triangle"? It has to be a right triangle!  :D  ;)

Is there a more mathematical way of solving this instead of just guess and check

Well, to have a positive integer as x,y and perimeter = area, you really don't have a lot of choice.

And answer is just one of the ppt (primitive pythagorean triple).

There is a way mathematically... It takes A LOT longer though.

[hide="See What I mean"]
$\frac{1}{2}ab=30$
$ab=60$

$a+b+c=30$

$c=\sqrt{a^2+b^2}$

$a+b+\sqrt{a^2+b^2}=30$

$\sqrt{a^2+b^2}=30-a-b$

So on and so forth...

Total waste of time.
[/hide]

Well...

x + y + sqrt(x^2 + y^2) = 30
xy = 60

So yeah :P .


geometry
rectangle
combinatorics unsolved
combinatorics

A disc is placed on each field of an $ n\times n$ board. Each disc has a red and a blue face. Two players play the following game. They move alternately. The one who moves may choose among the following 4 options:

(i) to turn a blue disc over;
(ii) to turn a blue disc over, and at the same time to turn over all discs lying in the same row to the right of the selected disc;
(iii) to turn a blue disc over, and at the same time to turn over all discs lying in the same column below the selected disc;
(iv) to pick a blue disc out and turn every disc over in the rectangle whose upper left corner is the selected disc, lower right corner is the lower right corner of the whole board.

The player who cannot move, loses the game. Determine those initial positions which admit the first player to have a winning strategy.


i tried with  tang A/2  =  s/2(p(p-a))  but  i don't understand the  p(p-a) =segment

 :(  :(  :(


A triangular region is enclosed by the lines with equations
y  =  1/2 x +3  y = −2x + 6 and y = 1. What is the area of the triangular region? Express your answer as a decimal to the nearest hundredth.

Three straight lines that enclose a triangle are given, so we have to find three vertices that make up this triangle. This can be solved by setting any two equations  equal to each other.

$ \dfrac{1}{2}x \plus{} 3$ = $ \minus{}2x\plus{}6$

$ \dfrac{1}{2}x \plus{} 3$ = $ 1$

$ \minus{}2x\plus{}6$ = $ 1$

If we solve for $ x$ and plug the values back in, we get the vertices

$ (\dfrac{6}{5}, \dfrac{18}{5})$

($ \minus{}4$,$ 1$)

($ \dfrac{5}{2}$, $ 1$), respectively.

$ b$= $ \dfrac{5}{2}\minus{}(\minus{}4)$=$ \dfrac{13}{2}$

$ h$=$ \dfrac{18}{5}\minus{}1$= $ \dfrac{13}{5}$.

$ A$=$ \dfrac{169}{20}$ = $ 8.45$


Find the integral solutions of the equation $7(x+y)=3(x^2-xy+y^2)$

[quote="ychjae"][quote="ng_hei"]Well I heard some people using delta to confine the range of x,y
But it is not as beautiful as yours, Lee[/quote]
I think you are talking about Ken Wong,brother of mecrazywong.[/quote]
in fact i did it in this way as well.

[quote="Valentin Vornicu"][quote="Soarer"][quote="shobber"]it seems that we suddenly have a lot of Hong Kong mathmeticians here.[/quote] :D In fact before IMO, the forum is only known to most of hk team members, but after that, I think it becomes more famous among the younger group of the training team.[/quote]That is very good for them! :)[/quote]
Yes, this is very good for me! :D  :D  :D

this site is not so famous in China. I think it is because the language problem...

It seems whichever way you approach this problem, the result is going to come.
Even trying to find the discriminant in the original equation (use $x$ as variable), which is $-27y^2+126y+49$, and setting it $\geq 0$, we can limit $y\in \{0,1,2,3,4,5\}$.


geometry
perimeter
incenter
congruent triangles

Prove that if two triangles have the same area and the same perimeter, then their incircles are congruent.

Note: There are infinitely many non-congruent triangles with the same area and the same perimeter. (Prove it!)

[hide]

Consider a triangle $ \bigtriangleup ABC$ with sides $ a,c,b$ and incenter $ O$
The incircle is tanglet to side $ AB$, $ BC$ and $ AC$ and at points $ M,N,$ and $ P$ respectivly
$ r = OM = ON = OP$
The area of the triangle, $ A$, is equal to the sum of the area of triangles $ \bigtriangleup ABO, \bigtriangleup ACO$ and $ \bigtriangleup BCO$, so..

$ A = \frac{1}{2}a.r +\frac{1}{2}b.r +\frac{1}{2}c.r$

$ \dfrac{2A}{a+b+c} = r$

Therefore every trianlge with the same area and perimeter has an incircle with the same radius, and thus a congruent incircle.[/hide]

[hide="Shorter, cheap proof"]We know that $ A\equal{}rs \iff \frac{A}{s}\equal{}r$, where $ A$ is the area of the triangle, $ s$ is the semiperimeter, and $ r$ is the radius of the incircle. Thus, having the same area and perimeter preserves the radius of the incircle.[/hide]


inequalities
function
three variable inequality

Hey, this is problems.
Latvian 1987:
  Let a,b be real numbers, n be nature number. Prove that:
  1/(a + b) + 1/(a + 2b) + ... + 1/(a + nb) < n/(a(a + nb))1/2.
I've solution, and you?

Are a and b just real or are they positive real numbers?
I think you need them to be positive.

I think I have a solution for positive numbers(in fact I think you need them to be positive):

We use induction. It's easy to prove it for n=1 and n=2. Then let us assume it's true for n-1 so if we apply the same inequality to a+b and b, then we get 1/(a+2b)+1/(a+3b)+..+1/(a+nb)<(n-1)/((a+b)(a+nb))^1/2 so our sum is smaller than (n-1)/((a+b)(a+nb))^1/2 + 1/(a+b) which is smaller than n/(a(a+nb))^1/2. We can prove this by multiplying with a and naming b/a=x. We get a function of x with f(x)=0 and f'(x)>0 so f(x)>0 for any positive x. It takes some time to do the calculations but it's not hard at all.

[quote="hxtung"]Let $ a,b$ be positive real numbers, $ n$ be nature number. Prove that

$ \frac{1}{a\plus{}b}\plus{}\frac{1}{a\plus{}2b}\plus{}\cdots\plus{}\frac{1}{a\plus{}nb}<\frac{n}{\sqrt{a(a\plus{}nb)}}.$[/quote]$ x>0\Longrightarrow 1\plus{}x\minus{}\left(\frac{4\plus{}3x}{4\plus{}x}\right)^2\equal{}\frac{x^3}{(4\plus{}x)^2}>0\Longrightarrow \sqrt{1\plus{}x}>\frac{4\plus{}3x}{4\plus{}x}\Longrightarrow$

$ \frac{k}{\sqrt{1\plus{}kc}}\minus{}\frac{k\minus{}1}{\sqrt{1\plus{}(k\minus{}1)c}}$

$ \equal{}\cfrac{\cfrac{k^2}{1\plus{}kc} \minus{} \cfrac{(k\minus{}1)^2}{1\plus{}(k\minus{}1)c}} {\cfrac{k}{\sqrt{1\plus{}kc}}\plus{}\cfrac{k\minus{}1}{\sqrt{1\plus{}(k\minus{}1)c}}}>\cfrac{\cfrac{k^2}{1\plus{}kc} \minus{} \cfrac{(k\minus{}1)^2}{1\plus{}(k\minus{}1)c}} {\cfrac{k(4\plus{}kc)}{4\plus{}3kc}\plus{}\cfrac{(k\minus{}1)\left[4\plus{}(k\minus{}1)c\right]}{4\plus{}3(k\minus{}1)c}}$

$ \equal{}\frac{1}{1\plus{}kc}\plus{}\frac{c\left[8(2k\minus{}1)\plus{}2(k\minus{}1)(2k^2\plus{}7k\minus{}2)c\plus{}3k(k\plus{}1)(k\minus{}1)^2c^2\right]}{(1\plus{}kc)\left[1\plus{}(k\minus{}1)c\right]\left[16(2k\minus{}1)\plus{}4(8k^2\minus{}8k\plus{}1)c\plus{}3k(k\minus{}1)(2k\minus{}1)c^2\right]}$

$ >\frac{1}{1\plus{}kc}\Longrightarrow$

$ \sum_{k\equal{}1}^{n}{\frac{1}{1\plus{}kc}}<\sum_{k\equal{}1}^{n}\left[\frac{k}{\sqrt{1\plus{}kc}}\minus{}\frac{k\minus{}1}{\sqrt{1\plus{}(k\minus{}1)c}}\right]\equal{}\frac{n}{\sqrt{1\plus{}nc}},$

where $ c\equal{}\frac{b}{a}$.

By the way, the following stronger inequality

$ \sum_{k\equal{}1}^{n}{\frac{1}{1\plus{}kc}}<n\left(1\plus{}\cfrac{5}{6}nc\right)^{\minus{}\frac{3}{5}}$

holds for positive real number $ c$ and nature number $ n$.

Latvian Mathematical Olympiad 1997 , $1^{st}$ Selection Test , Problem 4
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=418740&hilit=inequality]here[/url]


inequalities
function
inequalities solved

The main luck in this problem is right interpretation.

Let ABC is triangle with sides AB=x+y, AC=x+z and BC=y+z.
S=S_ABC, Sx=1/2*x*x*sin A and so on.

Then inequality is rewriting as
\sqrt Sx/(\sqrt Sx+\sqrt S)+\sqrt Sy/(\sqrt Sy+\sqrt S)+\sqrt Sz/(\sqrt Sz+\sqrt S)\leq 1  <==>
2\leq 1/(1+\sqrt(Sx/S))+1/(1+\sqrt(Sz/S))+1/(1+\sqrt(Sz/S)).

Now we apply well-known inequality
3/(1+(abc)^(1/3))\leq 1/(1+x)+1/(1+y)+1/(1+z)
(Jensen's inequality for function 1/(1+ex))

New inequality has form
2\leq 3/(1+(Sx.Sy.Sz/S^3)^(1/6))

Now it is necessary to prove
Sx.Sy.Sz/S^3<1/2^6

Actually Sx.Sy.Sz/S^3=xyz/(x+y)^2(x+z)^2(y+z)^2\leq xyz/(2\sqrt(xy))^2(2\sqrt(xz))^2(2\sqrt(yz))^2=1/2^6

The End.


If n is any natural number not equal to 2, 5, or 13, prove that there will always be one pair (a,b) from the set {2, 5, 13, n}, a and b being distinct, such that ab-1 is not a perfect square.

I am having trouble finding the solution to this problem. If anyone finds a solution, please post it - but I would greatly appreciate it if you could explain how you found that solution as well. What tipped you off, so that you knew that was the procedure?

[hide]If $ n\equiv0,2\pmod4$ then $ 2n \minus{} 1\equiv3\pmod4,$ while if $ n\equiv3\pmod4$ then $ 5n \minus{} 1\equiv2\pmod4.$ These can’t be perfect squares.

Suppose $ n \equal{} 4k \plus{} 1$ for some $ k.$ Then $ 5n \minus{} 1 \equal{} 4(5k \plus{} 1)$ and $ 13n \minus{} 1 \equal{} 4(13k \plus{} 3).$ We will show that these cannot both be perfect squares. We have that if $ k\equiv0,3\pmod4,$ then $ 13k \plus{} 3\equiv3,2\pmod4,$ while if $ k\equiv1,2\pmod4,$ then $ 5k \plus{} 1\equiv2,3\pmod4;$ hence $ 5k \plus{} 1$ and $ 13k \plus{} 3$ cannot both be perfect squares.[/hide]

This was an old IMO problem, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366557]IMO 1986 #1[/url] to be exact.

A good way to prove that some number is not a square is to look at it mod 8: it can only be 0, 1, or 4.  Also this can be reduced to mod 4, when it can only be 0 or 1 mod 4.  This is the method ValentineA used, and is fairly common.


[hide="Hint for 2"]
Every parallelogramm is transformed into itself if you return it by 180° around the intersection of its diagonals.
[/hide]

[hide]
A parallelogram can always be divided into two congruent figure by a line past its symmetric center (the intersection of the two diagonals).

Therefore, for question 1, it's the line past both centers of the squares; for question 2, it's the line past both symmetric centers of the parallelograms.[/hide]

[quote="xxxyyyy"][u]$ Question 1$[/u]
Take a square and draw a smaller square inside it. The smaller
square can have any orientation, its sides need not be parallel to
the sides of the larger square. Now look at the area between the two
squares. Can you give a simple method to draw a few lines to divide
this area into two regions of equal areas?

[u]$ Question 2$[/u]
Draw two parallelograms on a sheet of paper. Give a simple method
to draw a line such that it divides each of the parallelograms into
two regions of equal areas.[/quote]
[hide="1"] The line that we seek is the line that connects the centers of the two squares.  Say that the big square is $ ABCD$, it's center is $ O$, and the small square is $ A'B'C'D'$ and it's center is $ O'$.  Now, say that $ OO'$ meets $ ABCD$ at $ F$ and $ E$ and say that it meets $ A'B'C'D'$ at $ E'$ and $ F'$.  Now, we can notice that $ [ABEF]\equal{}[CDEF]$ and $ [A'B'E'F']\equal{}[C'D'E'F]$.  We subtract the equations to find that $ OO'$ bisectst the area of $ ABCDD'C'B'A'$. [/hide]


inequalities
geometry
3D geometry
sphere
pigeonhole principle
perpendicular bisector
geometry proposed

Consider a sphere with radius 1.
a) Suppose we have 7 points on the sphere. Prove that we can always choose two of them such that their distance is less than$ \sqrt{2}$.
b) Suppose we have 5 points on the sphere.  Prove that we can always choose two of them such that their distance is less or equal $\sqrt{2}.$

I'll just solve the second one.

Consider one of the $5$ points. The perpendicular bisector plane of the diameter passing through this point divides the sphere in two hemispheres. The other $4$ points, $A,B,C,D$, must lie in the open hemisphere which does not contain the point we initially chose. If we assume that, for instance, all the angles $\angle DOA,\angle DOB,\angle DOC$ are $>\frac\pi 2$, then we find that $O$ (the center of the sphere) is on the same side of the plane $(ABC)$ as $D$. The same holds for all the other combinations of points, which means that $O$ is inside $ABCD$, but this is impossible, since $A,B,C,D$ are on a hemisphere.

Actually, grobber, can't we just divide the sphere into 4 equal quarter-spheres, (by drawing two perpendicular diameter-bisecting planes). Now by Pigeonhole, at least 1 quarter-sphere contains more than 1 point, and we are done... :)

You can have two points with distance $2$ between them in a "quarter-sphere", man :). Think about the endpoints of the diameter which is the intersection of your "two perpendicular diameter-bisecting planes".

Sorry, grobber, silly me, it should be 8 not 4... :blush:


For $ f : R\to R : f(20x\minus{}11y)\equal{}20f(x)\minus{}11f(y)\forall x\in R$
Prove that :  $ f(x)\equal{}ax,\forall x\in R$,  $ a\in R$

$ f(x\alpha\plus{}y\beta )\equal{}\alpha f(x)\plus{}\beta f(y)$

See http://en.wikipedia.org/wiki/Cauchy_functional_equation

The functional equation is a linear congruence since $ f(\alpha x\plus{}\beta y) \equal{}\alpha f(x)\plus{}\beta f(y).$. QED


Could someone please provide a hint for the following problem (no solutions, please)? I was unable to find this with the search. Thanks!

Debbie has an ample supply of the square ceramic tiles with the patterns shown below. She intends to create a horizontal, rectangular wall decoration by cementing three of the squares to the wall in a row. How many different designs can she create in this way that have at least one line of symmetry?

I solved this by just listing them out, but I was wondering if there was a faster and easier way.

Note that if you have the "X" or "+" on one end, you must have it on the other end.  Furthermore, note that you must have an "X" or a "+" in the middle.  Also, if you have a "/" on one end, the other end must be "\" and vice versa.

Thanks.

Is this right:

If the first tile is an X, we can have two tiles in the middle - X or + - and one tile at the end, so 2 possibilities. If the first tile is a +, it's virtually the same thing, giving us another two possibilities. If the first tile is a /, the second one must be a X or +, and the last must be \, so that's two possibilities. Finally, if the first tile is \, then the second is X or +, and the last is /, giving us 2 possibilities, so that's 8.

Wait, what did I do wrong? The answer key says it's 12.

Hint: If the first tile is X, the last tile doesn't have to be X. Why?

Oh right, it could be a +, because the line of symmetry could be horizontal too. So in addition to the 8 afore-mentioned possibilities, there's also two more because of X_+, plus another two because of +_X, giving a total of $ \boxed{12}$.

Thanks everyone.  :)


geometry
3D geometry
sphere
circumcircle
Iran

Call two circles in three-dimensional space pairwise tangent at a point $ P$ if they both pass through $ P$ and lines tangent to each circle at $ P$ coincide. Three circles not all lying in a plane are pairwise tangent at three distinct points. Prove that there exists a sphere which passes through the three circles.

We use a spatial inversion (just like in 2D) which has one of the 3 pts of tangency as pole. It turns the 2 circles which pass through the pole into lines, which are both tangent to the circle which is the image of the 3'rd circle, so the images of the 3 circles are coplanar (a circle and 2 lines tangent to it), which means that the points were originally on a sphere (which passes through the point chosen as pole).

It's a nice problem!

nice solution [b]Grobber [/b] but more elementary one is :

let $w_1,w_2,w_3$ be three circles ,pairwisely tangent to each others at $P_3,P_2,P_1,$ respectively.

[b]Lemma:[/b] let $X,Y$ be two distinct points on the circle $Q$ in $3-D.$ let $S$ be an arbitrary sphere passing through $X,Y.$ then there exist a set of points common in $Q,S.$

[b]proof:[/b] just let $O$ be the circle with diameter of $S$ which passes through $X,Y$;so lemma proved.

[b]case 1:[/b] not all of the circles tangent externaly,

so there exist two circles [b]WLOG [/b] called $w_1,w_2$ which the projection of them on the plane of $w_3,$ is completely into the circle $w_3.$ as we know not all of them are coplanar, so $P_1,P_2,P_3$ make a triangle. let $W$ be the circumcircle of this triangle.rotate $W$ from one of it's diameters to build a sphere>>>[b]Lemma[/b]

[b]case 2:[/b] all of the circles are externaly tangent to each others,
which means projection of no two circles into the plane of the third one,is completely in it. again because not all of the circles are coplanar, so $P_1,P_2,P_3$ make a triangle.again  let $W$ be the circumcircle of this triangle.rotate $W$ from one of it's diameters to build a sphere>>>[b]Lemma[/b].

[b]clarify of an assumption :[/b]
 [i]because not all of the circles are coplanar, so $P_1,P_2,P_3$ make a triangle.[/i] there's a case when the centers of $w_1,w_2,w_3$ are collinear,and so $P_1,P_2,P_3$ are collinear. but circles are not coplanar,so the assumption is correct.

grobber,can you tell me what spatial inversion is? BTW is inversion also applicable in three dimensional space?


induction
number theory solved
number theory

Find all nonempty sets $S$ of positive integers such that $\frac{i + j}{\gcd(i, j)}$ is in $S$ for all $i$, $j$ in $S$.

I was confused about that too :)
If we can choose i = j the question seems to be too simple.
But the problem statement does not say it is not allowed to choose i = j...

I agree : If i = j is allowed, then the problem is simple.
But in the other case, there are many solutions (for exemple, each set {a}, where a is a positive integer...) and subcases so that it makes it a little boring...

Pierre.

Ok, let's give a complete solution before moving it :

Let a be in S. Then (a + a)/a = 2 is in S.
Note that S = {2} satisfies the statement of the problem. Now assume that |S| > 1.

- If 1 is in S.
Then, for each x in S, we have (x+1)/1 = x+1 is in S. So that easily S = |N *  = {1,2,3,...}, which clearly is a solution.

- Now assume that 1 is not in S.
Let m be the least element in S-{2}.

If m = 2k is even : Then k > 1, and (m + 2)/2 = k+1 is in S.
But 2 < k+1 < 2k = m, which contradicts the minimality of m.
Thus m is odd.
It follows easily by induction that for each nonnegative integer k, the number m + 2k is in S. (1)
Thus m + 2m is in S. So that ( (m+2m) + m)/m = 4 is in S. The minimality of m, and since m is odd, leads to m = 3.
From (1), we deduce that each odd integer k > 1 is in S.
Moreover from (1), for each integer k \geq 0, we have m + 2km is in S. Thus, ((m + 2km) + m)/m = 2k+2 is in S, therfore each even positive integer is in S.
Then S = {2,3,4,5,...} = |N * - {1}, which clearly is a solution.

The solutions are {2}, |N *  and |N * - {1}.

Pierre.

of course on the actual contest it said all finite sets, which made the problem even easier.


The student locker numbers at Olympic High are numbered consecutively beginning with locker number $ 1$. The plastic digits used to number the lockers cost two cents apiece. Thus, it costs two cents to label locker number $ 9$ and four centers to label locker number $ 10$. If it costs $ \$137.94$ to label all the lockers, how many lockers are there at the school?

$ \textbf{(A)}\ 2001 \qquad
\textbf{(B)}\ 2010 \qquad
\textbf{(C)}\ 2100 \qquad
\textbf{(D)}\ 2726 \qquad
\textbf{(E)}\ 6897$

[hide="Click for solution"]
The locker labeling requires $ \frac{137.94}{0.02}\equal{}6897$ digits.  Lockers $ 1$ through $ 9$ require $ 9$ digits total, lockers $ 10$ through $ 99$ require $ 2 \times 90\equal{}180$ digits, and lockers $ 100$ through $ 999$ require $ 3 \times 900\equal{}2700$ digits.  Thus, the remaining lockers require $ 6897\minus{}2700\minus{}180\minus{}9\equal{}4008$ digits, so there must be $ \frac{4008}{4}\equal{}1002$ more lockers, because they each use $ 4$ digits.  Thus, there are $ 1002\plus{}999\equal{}2001$ student lockers, or answer choice $ \boxed{\textbf{(A)}}$.
[/hide]

[hide="Solution"]9 one digit lockers = 18 cents
90 two digit lockers = 360 cents
900 three digit lockers = 5400 cents
$ n$ four digit lockers = $ 8n$ cents.

$ 5778\plus{}8n\equal{}13794\implies n\equal{}1002$
$ 999\plus{}1002\equal{}\boxed{\textbf{(A)}\ 2001}$[/hide]


A 12-slice pizza was made with only pepperoni and mushroom toppings, and every slice has at least one topping. Only five slices have pepperoni, and exactly eight slices have mushrooms. How many slices have both pepperoni and mushrooms?

I was a bit confused by their explanation, wouldn't it make more sense to say that
8+5=13
13-12=1 slice of pizza

There are 12 total slices, and 6 of them have pepperoni and 10 have mushrooms. Let $n$ be the number of the slices that have both toppings, so there are $6-n$ with just pepperoni and $10-n$ with just mushrooms. Since every slice has at least one topping, and there are 12 slices, we must have $(6-n) + (10-n) + n = 12$, which gives $n=\boxed{4}$.

You're wrong; the question says that there are only 6 pepperoni slices and 8 mushroom slices

[hide=Solution]If we imagine $x$ as the area that is in both Venn diagrams, then we have $5-x$ slices that have pepperoni and $8-x$ slices that have mushrooms. This means we have $5-x+8-x+x=12$. Solving we get $x=\boxed1$.[/hide]


function calculus derivative inequalities geometry integration algebra

Given function $f:\mathbb{R}{\rightarrow}\mathbb{R}$ which has derivatives up to the $p$-th order $p{\geq}2$ is a natural number.Let $M_{k}=\sup_{\mathbb{R}}{|f^{k}(x)|}$,for $k=0,1,\ldots,p$.show that if $M_{0},M_{p}<+\infty$ then $M_{k}<+\infty$.Moreover $M_{k}{\leq}2^{\frac{k(p-k)}{2}}M_{0}^{1-\frac{k}{p}}M_{p}^{\frac{k}{p}}$, for $k=1,2,\ldots,p-1$.

Is this problem easy,hard or just boring? :?

It is enough to prove that $M_1\leqslant \sqrt 2 M_0^{\frac 12}M_2^{\frac 12}$. This can be done by a nice geometric argument, which I leave to someone else to find.

Dear [b]fedja[/b],I think you mean to prove the inequalities by induction.But then don't you have first to show that from the condition $M_{0},M_{p}<+\infty$,follows that $M_{k}<+\infty$ for $k=1,2,\ldots,p-1$.And it is very interesting to see the geometric argument you told about :) .

I think the fact that they are finite is not difficult ( but I was amazed by the result when I first saw it) and has been discussed on the site.

And what about the geometric proof [b]harazi[/b]?

Probably Fedja refers to this, but I'm not sure: assume that $ |f(x)|\leq A$ and $|f''(x)|\leq B$. Then for any $x$ such that $ f'(x)>0$ (for example, it is clearly wlog), the triangle of height $ f'(x)$ and corresponding edge $ \frac{2f'(x)}{B}$ is beneath the graph. It means that the area under the graph of $f'$ between $ x-\frac{f'(x)}{B}$ and $x+\frac{f'(x)}{B}$ is at least $\frac{f'^2(x)}{B}$ and from here we find $ 2AB\geq f'^2(x)$. Am I right, Fedja?

[quote="LevonNurbekian"]But then don't you have first to show that from the condition $M_{0},M_{p}<+\infty$,follows that $M_{k}<+\infty$ for $k=1,2,\ldots,p-1$?[/quote]
Not quite. We can always use the standard trick of convolving our function with some compactly supported mollifier $0\leqslant \varphi\in C^\infty$ with total integral $1$. Then we shall have all derivatives of $f*\varphi$ bounded and the bound for the $0$-th and the $p$-th derivatives will be the same as before. Now we can use the inequality between three consequtive derivatives to get the desired estimate for the derivatives of $f*\varphi$ (here, as Levon correctly noted, finiteness of all $M_k$ plays essential role) and then, choosing a sequence of mollifiers $\varphi_n$ supported by shorter and shorter intervals around $0$, pass to the limit as $n\to\infty$. It is easy to see that $(f*\varphi_n)^{(k)}=f^{(k)}*\varphi_n\to f^{(k)}$ for $k=0,\dots,p-1$.

[quote="harazi"] Am I right, Fedja?[/quote]
Yes, that's what I meant.

BTW, when $p$ is large, the factor $2^{\frac {k(p-k)}{2}}$ is very far from optimal. If I remember it right, one can get a factor like $C^k$ for some constant $C>0$ instead but it requires a completely different proof.

[quote="fedja"]
Not quite. We can always use the standard trick of convolving our function with some compactly supported mollifier $0\leqslant \varphi\in C^\infty$ with total integral $1$. Then we shall have all derivatives of $f*\varphi$ bounded and the bound for the $0$-th and the $p$-th derivatives will be the same as before. Now we can use the inequality between three consequtive derivatives to get the desired estimate for the derivatives of $f*\varphi$ (here, as Levon correctly noted, finiteness of all $M_k$ plays essential role) and then, choosing a sequence of mollifiers $\varphi_n$ supported by shorter and shorter intervals around $0$, pass to the limit as $n\to\infty$. It is easy to see that $(f*\varphi_n)^{(k)}=f^{(k)}*\varphi_n\to f^{(k)}$ for $k=0,\dots,p-1$.

[/quote]
Nice.For that part I found clean algebraic proof.Fix $h_{1},h_{2},\ldots,h_{p-1}$ different from zero and each other.Then for any $x$ we have
$f'(x)h_{i}+\frac{f''(x)}{2!}h_{i}^{2}+\cdots+\frac{f^{(p-1)}(x)}{(p-1)!}h_{i}^{p-1}=f(x+h_{i})  -f(x)-\frac{f^{(p)}(x+{\theta}_{i}h_{i})}{p!}h_{i}^{p}$,for $i=1,2,\ldots,p-1$.We get a linear equation respect to derivatives of $f$ with nonzero determinant,so we can just find $f^{(k)}$-s and as you see they will be bounded.


Give an example of a graph G satisfying 2 conditions:
1) G has no bridges.
2) $G^2$ is non-hamiltonian.
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$G^2$ is a square of G.

It's known that the graph definitely exists. In order to avoid misunderstanding I let you know beforehand: [u]If you look attentively, you'll see that there's [b]no[/b] contradiction to Fleischner theorem.[/u]
 Can anyone construct something??

What is $G^2$ exactly? I don't see what meaning the square of a graph has...

in general $G^k$ is defined to be the graph with vertex set $V(G^k)=V(G)$ and ,and for the edges $x,y\in V(G^k), x\leftrightarrow y \Leftrightarrow d_G(x,y)\le k$ where $d$ denote the distance on the graph $G$.
It is a theorem of Fleischner(proved sometime in the 90's) that if a graph is $2$-connected, then $G^2$ is hamiltonian.