1.given tha d(n) is the highest odd factor n.eg dn(90)=15.
find.d(1)+...+d(2^99)
if you solve this problem explain.
2.prove that (n)^5 -(n) is divisible by  15.
my soluting involved using fermat's and the chinese rem's.
(n)^5=n(mod5).
maybe you may use induction :D [quote]all laws of motion are the same in all uniformly moving frames of refrence[/quote]

1. Isn't $ d(90)\equal{}45$?
2. It's easy by Little Fermat's theorem, but it can also be done like this:
$ n^5\minus{}n\equal{}n(n^4\minus{}1)\equal{}(n\minus{}1)n(n\plus{}1)(n^2\plus{}1)$ Now try $ n\equal{}5a\plus{}k$ for $ k\in \{0,1,2,3,4\}$.

[hide="Three Solutions for the 2nd Part"]
1st solution: We know that $ 5 \mid n^5 \minus{} n$ by Fermat's Little Theorem. Now $ n^5 \minus{} n \equal{} n(n^4 \minus{} 1) \equal{} n(n \minus{} 1)(n \plus{} 1)(n^2 \plus{} 1)$, which clearly contains a multiple of 3.
2nd solution: 5 and 3 are small mods, we can just try all numbers $ \mod 5$ and $ \mod 3$.
3rd solution: $ n^5 \minus{} n \equiv n^5 \plus{} 5n^4 \plus{} 10n^3 \plus{} 10n^2 \plus{} 5n \plus{} 1 \minus{} 1 \minus{} n \equiv (n \plus{} 1)^5 \minus{} (n \plus{} 1) \pmod{5}$.
Since this is equivalent to $ 0$ for $ n \equal{} 1$, it is equivalent to $ 0$ for all integers.

Also, we can actually say that $ 30 \mid n^5 \minus{} n$, and for odd $ n$, $ 240 \mid n^5 \minus{} n$ (by considering this in $ \mod 16$)[/hide]

i'll take #1

so i'm assuming that $ d(1) \plus{} \cdots \plus{} d(2^{99})$ means, like, $ d(1) \plus{} d(2) \plus{} d(3) \plus{} ... \plus{} d(2^{99})$. if we're only summing powers of 2, the problem is pretty trivial

so between $ 1$ and $ 2^{99}$, there are $ 2^{98}$ numbers $ k_0$ such that $ 2^0$ is the greatest power of two going into them. so we sum the first $ 2^{98}$ odd numbers to get the sum of all $ d(k)$. that's $ (2^{98})^2 \equal{} 2^{196}$

similarly, there are $ 2^{97}$ numbers $ k_1$ such that $ 2^1$ is the greatest power of two going into them. so $ d(k_1)\equal{}\frac{k_1}{2}$. we sum the first $ 2^{97}$ odd numbers and we get $ 2^{194}$.

we continue this trend and our final answer is $ 2^0 \plus{} 2^2 \plus{} 2^4 \plus{} \cdots \plus{} 2^{196}$.


vector
inequalities
geometry
geometry solved

Let ABCD is a a quadrilateral. M,N is the midpoint of AB and CD. P is the point of intersection of AC and BD. Prove that:
            area(MNP) \leq 8/27 area(ABCD).
It is maybe not too difficult but it is so strange when I use vector and find a fantasy result. By this result, I can't say that inequality is right . However, help me prove it.

[quote="Anh Cuong"]In fact,that inequality must be:
area(MNP)  \leq  4/27 area(ABCD).[/quote]
with x,y be unit vectors
can we have (x-y)/((1+x)(1+y)) \leq 8/27
can you show me?

If you do well, master, my inequality seem wrong. I am not sure whether my problem is right , you can prove or not. :(

This thread is a mess. Let's put things at the right places.

In the following, I will denote by $\left[P_1P_2...P_n\right]$ the directed area and by $\left|P_1P_2...P_n\right|$ the non-directed area of a polygon $P_1P_2...P_n$.

At first, for the inequalities we are discussing here, the quadrilateral ABCD must be assumed to be convex; in fact, if the quadrilateral ABCD can be concave, there is no real number k such that $\left|MNP\right| \leq k \cdot \left|ABCD\right|$ holds for every quadrilateral ABCD. But still, the inequality $\left|MNP\right| \leq \frac{4}{27} \cdot \left|ABCD\right|$ holds not for every convex quadrilateral ABCD. On the other hand, the inequality $\left|MNP\right| \leq \frac{8}{27} \cdot \left|ABCD\right|$ does hold for every convex quadrilateral ABCD, and actually it can be replaced by a stronger inequality: $\left|MNP\right| < \frac14 \cdot \left|ABCD\right|$. In the following, I will prove this stronger inequality. Note that the constant $\frac14$ in this inequality cannot be replaced by a better one; in fact, when the point B approaches the point C, the values |MNP| and $\frac14 \cdot \left|ABCD\right|$ get arbitrarily close to each other.

So it remains to prove the inequality $\left|MNP\right| < \frac14 \cdot \left|ABCD\right|$. First I will prove a lemma:

[b]Lemma 1.[/b] If ABCD is a quadrilateral, if M and N are the midpoints of the segments AB and CD, respectively, and if P is the point of intersection of the diagonals AC and BD, then the directed areas of triangles ADP, BCP and MNP satisfy the following equation:

$\left[MNP\right] = \frac14 \cdot \left(\left[ADP\right] + \left[BCP\right]\right)$.

[i]Proof.[/i] There are many different ways to prove Lemma 1. For instance, you can note that Lemma 1 follows from applying Theorem 3.14 of H. S. M. Coxeter, S. L. Greitzer, [i]Geometry Revisited[/i], Toronto - New York 1967, to the (self-intersecting) quadrilateral CBDA. But one can also easily prove Lemma 1 using the cross product of vectors:

It is well-known that the directed area $\left[P_1P_2P_3\right]$ of a triangle $P_1P_2P_3$ equals $\frac12 \cdot \overrightarrow{P_3P_1}\times\overrightarrow{P_3P_2}$. Thus,

$\left[ ADP\right] =\frac{1}{2}\cdot \overrightarrow{PA}\times\overrightarrow{PD}$;
$\left[ BCP\right] =\frac{1}{2}\cdot \overrightarrow{PB}\times\overrightarrow{PC}$;
$\left[ MNP\right] =\frac{1}{2}\cdot \overrightarrow{PM}\times\overrightarrow{PN}$.

On the other hand,

$\left( \overrightarrow{PA}+\overrightarrow{PB}\right) \times \left( \overrightarrow{PC}+\overrightarrow{PD}\right)$
$=\overrightarrow{PA}\times \overrightarrow{PC}+\overrightarrow{PA}\times \overrightarrow{PD}+\overrightarrow{PB}\times \overrightarrow{PC}+\overrightarrow{PB}\times \overrightarrow{PD}$.

Now remember that the cross product of two collinear vectors is 0. Since the vectors $\overrightarrow{PA}$ and $\overrightarrow{PC}$ are collinear, we thus have $\overrightarrow{PA}\times \overrightarrow{PC} = 0$. Similarly, $\overrightarrow{PB}\times \overrightarrow{PD} = 0$. Thus,

$\left( \overrightarrow{PA}+\overrightarrow{PB}\right) \times \left( \overrightarrow{PC}+\overrightarrow{PD}\right) =\overrightarrow{PA}\times\overrightarrow{PD}+\overrightarrow{PB}\times \overrightarrow{PC}$.

But since the points M and N are the midpoints of the segments AB and CD, respectively, we have $\overrightarrow{PM}=\frac{\overrightarrow{PA}+\overrightarrow{PB}}{2}$ and $\overrightarrow{PN}=\frac{\overrightarrow{PC}+\overrightarrow{PD}}{2}$. Thus,

$\left[ MNP\right] =\frac{1}{2}\cdot \overrightarrow{PM}\times\overrightarrow{PN}=\frac{1}{2}\cdot \frac{\overrightarrow{PA}+\overrightarrow{PB}}{2}\times \frac{\overrightarrow{PC}+\overrightarrow{PD}}{2}$
$=\frac{1}{8}\cdot \left( \overrightarrow{PA}+\overrightarrow{PB}\right)\times \left( \overrightarrow{PC}+\overrightarrow{PD}\right) =\frac{1}{8}\cdot \left( \overrightarrow{PA}\times \overrightarrow{PD}+\overrightarrow{PB}\times \overrightarrow{PC}\right)$
$=\frac{1}{4}\cdot \left( \frac{1}{2}\cdot \overrightarrow{PA}\times\overrightarrow{PD}+\frac{1}{2}\cdot \overrightarrow{PB}\times\overrightarrow{PC}\right) =\frac{1}{4}\cdot \left( \left[ ADP\right] +\left[BCP\right] \right)$,

what proves Lemma 1.

Now, the equation

$\left[MNP\right] = \frac14 \cdot \left(\left[ADP\right] + \left[BCP\right]\right)$,

when converted to non-directed areas, yields

$\left|MNP\right| = \frac14 \cdot \left|\left|ADP\right| - \left|BCP\right|\right|$

(in fact, we get the - sign since, because of the convexity of the quadrilateral ABCD, the triangles ADP and BCP have different orientations and thus their directed areas differ in sign). But clearly

$\left|\left|ADP\right| - \left|BCP\right|\right| < \max\left\{\left|ADP\right|;\;\left|BCP\right|\right\}$.

But since the quadrilateral ABCD is convex, we have |ADP| < |ABCD| and |BCP| < |ABCD|, thus

$\max\left\{\left|ADP\right|;\;\left|BCP\right|\right\} < \left|ABCD\right|$.

As a result, $\left|\left|ADP\right| - \left|BCP\right|\right| < \left|ABCD\right|$, and thus

$\left|MNP\right| = \frac14 \cdot \left|\left|ADP\right| - \left|BCP\right|\right| < \frac14 \cdot \left|ABCD\right|$,

proving our initial inequality.

  Darij


Look at the title for the first sentence

But you have to sign up with a special form, and I turn 13 september 30th, so would it just be easier to sign up when I turn 13 or sign up now?

It's not going to matter whether you sign up now, or you sign up on Oct. 14th, so I guess you should do whatever is easier.
However, there is nothing stopping you from working on the problems now.

If you haven't yet registered on the USAMTS web site, then the simplest thing is for you to wait and register after you've turned 13.  However, if you register while you're still under 13, then we'll need the parental permission form.


Hi joml88 and Everyone,

[quote="joml88"]Does anyone know where I can find some good math screensavers?  I tried google but I couldn't find anything good.[/quote]

There are many screensavers that use fractal art that are pretty neat.  Try the search term "fractal screen savers" using < [url]http://www.yahoo.com[/url] >.   I got several hits.

I personally run a program called ZetaGrid as my screensaver.  ZetaGrid is a grid computing project to calculate the zeroes of the Reimann zeta function.  The aim is to either find statistics about the zeroes that might help generate ideas about how to prove the Reimann hypothesis, or to actaully find a zero of the Reimann zeta function that does not lie on the line [tex]1/2+bi[/tex] in [tex]\mathbb{C}. [/tex]  The program can be run as a screensaver.  So while you're not using your compuer, it can be doing some fancy mathematics! :lol:  My computer has computed over 170 million zeroes of the Reimann zeta function to date.  The URL for the ZetaGrid project is < [url]http://www.zetagrid.net/zeta/index.html[/url] >.

Since I brought up the topic of grid computing, one can find a list of active distributed computing projects at < [url]http://www.aspenleaf.com/distributed/ap-math.html[/url] >.

All the best,

mathfreak

Hi Peter VDD and Everyone,

[quote="Peter VDD"]how does the screensaver look like? (be4 I install it)[/quote]

The screensaver for ZetaGrid is nothing fancy.  It's a box that changes position on one's computer screen every few minutes.  It gives the current time and also shows your computer's progress on its current work unit.  It also tells you how many zeroes your computer is calculating per second (?).

Yours truly,

mathfreak


Given $ a, b, c$ are real numbers. Prove that: $ (a \minus{} b)^4 \plus{} (b \minus{} c)^4 \plus{} (c \minus{} a)^4 \geq\ 8(a \minus{} b)^2(c \minus{} a)(c \minus{} b)$

We only did (c-a)(c-b)>0.And we also can assume $ a \minus{} b\ge0$
Then we identify $ a \minus{} b \equal{} x,c \minus{} b \equal{} y,c \minus{} a \equal{} z$.Now $ x,y,z\ge0$
And x+z=y We need to prove $ x^4 \plus{} (x \plus{} z)^4 \plus{} y^4 \equal{} x^4 \plus{} y^4 \plus{} z^4\ge8x^2yz \equal{} 8x^2(x \plus{} z)z$
Then we need to prove:$ x^4 \plus{} z^4 \plus{} 2xz^3\ge 2x^3z \plus{} x^2z^2$.
We assume x/z=p.Then we need to prove $ p^4 \minus{} 2p^3 \minus{} p^2 \plus{} 2p \plus{} 1\ge0$
And I think this one maybe need computer to calculate the min.I guess it's not a simple digit.

I think U have a mistakes in calculus. Notice that equality occurs when $ a \equal{} b \equal{} c.$ In your poof, I don't see this case. But indeed, it's very simple and I said that "It's a joke". My solution is very short :)
Besides, equality occurs when $ a \equal{} 0; b \equal{} 1; c \equal{} \frac{1\plus{}\sqrt[]{5}}{2}$...
Do U think can_hang's lenma works here ?

[quote="nguoivn"]I think U have a mistakes in calculus. Notice that equality occurs when $ a \equal{} b \equal{} c.$ In your poof, I don't see this case. But indeed, it's very simple and I said that "It's a joke". My solution is very short :)
Besides, equality occurs when $ a \equal{} 0; b \equal{} 1; c \equal{} \frac {1 \plus{} \sqrt []{5}}{2}$...
Do U think can_hang's lenma works here ?[/quote]
I'm not wrong.But I didn't find it.
$ \equal{} (p^2 \minus{} p \minus{} 1)^2\ge0$

Sorry, your proof is trues. And my solution is here (it's also the way to creat it)  :)
http://www.maths.vn/forums/showthread.php?t=9244&page=15


The fractional part of a number is the difference between the number and the largest integer not exceeding the number. For example, the fractional part of $ 2\frac{3}{4}$ is $ \frac{3}{4}$, and the fractional part of 4 is 0.

What is the ratio of the fractional part of the sum of $ 2\frac{2}{3}$, $ 3\frac{3}{4}$, and $ 4\frac{4}{5}$ to the sum of their fractional parts?

The sum of these three mixed numbers is $ 11\frac{13}{60}$. The sum of their fractional parts is $ \frac{2}{3} \plus{} \frac{3}{4} \plus{} \frac{4}{5} \equal{} 2\frac{13}{60}$. Thus, the ratio is $ \frac{133}{673}$.

EDIT: Stupidity fail. It's easier to leave everything as a common fraction, which lets you do the ratio more easily.

Actually, it asks for the ratio of the fractional part of the sum (which would be $ \frac{13}{60}$) to the sum of their fractional parts (which is $ 2\frac{13}{60}\equal{}\frac{133}{60}$).  So our final answer would be $ \frac{\frac{13}{60}}{\frac{133}{60}}\equal{}\boxed{\frac{13}{133}}$


We have 2001 pieces of cheese and we know the weight of every piece. Is it possible to chose one piece, to cut it in two parts (not necessarily equal) and then to arrange the pieces of cheese obtained in two sack of the same weight everyone with 1001 pieces?

Yes.
[hide]Proof:
First, label the initial 2001 pieces $m_1, m_2, m_3, ..., m_{2001}$ in increasing order from smallest to largest.  Then make two piles, $P_1$ and $P_2$.  Put $m_1$ into $P_1$ and note that $P_1-P_2=m_1$.  Then put $m_2$ into $P_2$ and the difference changes to $m_2-m_1$, which is less than $m_2$.  Then put m_3 into $P_1$, and the difference changes to $m_3-m_2+m+1<m_3$.  Repeat until you get 1001 items in $P_1$ and 1000 items in $P_2$ with $P_1-P_2=m_{2001}-m_{2000}+m_{1999}-...<m_{2001}$.  Now take $m_2001$ and cut it into 2 pieces, one of weight $\frac{P_1-P_2}{2}$.  Putting that piece in $P_2$ and the remaining part in $P_1$ makes the new P_1 = $P_1-\frac{P_1-P_2}{2}= \frac{P_1+P_2}{2}$; the new P_2 becomes $\frac{P_1+P_2}{2}$ as well.  It remains to be shown that the largets piece can be cut into a weight of $\frac{P_1-P_2}{2}$.  However, the first thing we did shows that $P_1-P_2<m_{2001}$, therefore it can be cut into that weight and the process can occur.[/hide]


As many of you probably know, the Shuttle was sent to the ISS (International Space Station) to add a new section.
This Monday or Tuesday, depending on your location, you will be able to see both go across the night sky.

Since I always quote SpaceWeather.com
[quote="SpaceWeather"]
Space Weather News for Aug. 18, 2007
http://spaceweather.com

DOUBLE SPACESHIP FLYBYS:  Space shuttle Endeavour is scheduled to undock from the International Space Station on Monday, August 20th.  If that happens as planned, sky watchers across North America may be able to witness something rare and beautiful: a double-spaceship transit across the night sky. US cities favored for flybys on Aug. 20th or 21st include Atlanta, Boston, Chicago, Denver, Detroit, Las Vegas, New York and Phoenix.  The space station and shuttle will appear as separate, bright points of light moving in tandem.  Flyby times depend on where you live. ...

BONUS: The International Space Station is under construction, and with each new addition the sprawling complex becomes easier to see from the ground.  To the naked eye, the space station now resembles a super-bright star gliding slowly across the sky.  Backyard telescopes reveal much more: solar panels, laboratories and living modules, a robotic arm and docked supply vessels. Photos and videos are featured on this weekend's edition of http://spaceweather.com.
[/quote]

For times of viewing check [url=http://www.heavens-above.com/]Heavens Above[/url], just enter your location and one option is for times for the next ten days of the ISS.  Look for the viewing time for the $ 20^{\text{th}}$ or the $ 21^{\text{st}}$ with the greatest visible magnitude, or the negative numbers.


Let $a,b,c,d$ real numbers.Suppose $[na]+[nb]=[nc]+[nd]$ for all $n \in Z^{+}$.Show that at least one of $(a+b),(a-c),(a-d)$ is an integer.
I have the solution but it's so long and the method is the same with the problem $[na]+[nb]=[nc]$ for all $n \in Z^{+}$.Prove that at least one of a,b,c is an integer.But I'm interested in this problem.Can anyone have the solution with this problem?  :?

Seems strange because, as far as I know, if $a,b,c,d$ are any positive irrational numbers such that $a+b = c+d$ then $[na]+[nb] = [nc]+[nd]$ for all nonnegative integers $n$.

Therefore, choosing $a=\sqrt 5 - \sqrt 2,b = \sqrt 2 , c = \sqrt 5 - \sqrt 3 ,d = \sqrt 3$ gives a counter-example... :?

Pierre.

You have a mistake .If $n=2$ then $[2*a]+[2*b]$ different to $[2*c]+[2*d]$( my problem is true for all $n \in Z^{+}$ :)

Ooops...sorry, I didn't see that the common value had to be an integer one... :blush:

Pierre.


[quote="Maverick"]This is very easy:

$\displaystyle \sum{\frac{a^3}{a^2-ab+b^2}}\geq \sum{2a-b}=1\geq 3(ab+bc+ca)$[/quote]

How does the first ineq come about?
Are you claiming that $\frac{a^3}{a^2-ab+b^2}\geq 2a-b$, but this does seem to be correct.

$\displaystyle a^3\geq(2a-b)(a^2-ab+b^2)\\
a^3 \geq 2a^3-2a^b+2ab^2-ba^2+ab^2-b^3\\
b^3-a^3\geq3ab(b-a)\\
a^2+b^2\geq 2ab\\$
which is obviously true. Am I missing something? :?

[quote="Valiowk"][quote="Maverick"]$b^3-a^3\geq3ab(b-a)\\
a^2+b^2\geq 2ab\\$
which is obviously true. Am I missing something? [/quote]

I suppose that you divided it by $b-a$ on both sides?  This is only true if $b \geq a$.[/quote]

Indeed, but as this ineq (the very first) is symmetric we can assume that $b \geq a \geq c$

To Maverick:
Before dividing you must make sure that $(b-a)$ isn't $0$
 ($a=b$).

So we have to look at the case $a=b$ seperately

Well, proving LHS>=1 won't get any plausible results, trust me, I have been there already... :(

I guess the key is what to do with the a^2-ab+b^2...(I think everyone's noticed that it's equal to (a^3+b^3)/(a+b), but it doesn't seem to yield anything... :? )

[quote="billzhao"]First of all, it's not symmetric (it's only cyclic).

Secondly, assume an order won't help Maverick's argument.[/quote]

You're right! I mixed again everything up  :(

Don't worry DanielKirchner about your errors and remember that our experience is the amount of our errors, but probably to have too much experience, in live, is not too much pleasant ;)

However the original inequality I found on my notes was like that

$ \frac{a^3(a+b)}{a^3+b^3}+ \frac{b^3(b+c)}{b^3+c^3}+  \frac{c^3(c+a)}{c^3+a^3} \geq 3\frac{(ab+bc+ca)}{a+b+c}$

and this is a well known old conjecture never solved.

Probably my semplification is not a good thingh to perform :(

Probably Cauchy can help us to solve it, but perhaps it is very hard.

[quote="manlio"]For $a,b,c$ positive real numbers such that $a+b+c=1$ prove the ineq

$\frac{a^3}{a^2-ab+b^2}+\frac{b^3}{b^2-bc+c^2}+\frac{c^3}{c^2-ca+a^2} \geq 3(ab+bc+ca) $[/quote]

What about $a=\frac16$, $b=\frac13$, $c=\frac12$ ?

  Darij

[quote="darij grinberg"]

What about $a=\frac16$, $b=\frac13$, $c=\frac12$ ?

  Darij[/quote]

Indeed, that's a counterexample for both inequalities (the simplification by manlio and the other one)

$LHS= \frac {8}{9}$
$RHS= \frac {11}{12}$

Its obvious that $RHS > LHS$

[quote="manlio"]$ \frac{a^3(a+b)}{a^3+b^3}+ \frac{b^3(b+c)}{b^3+c^3}+  \frac{c^3(c+a)}{c^3+a^3} \geq 3\frac{(ab+bc+ca)}{a+b+c}$[/quote]

As I've said, this one is not correct, but it seems that the minimum of

$\frac{\frac{a^3(a+b)}{a^3+b^3}+ \frac{b^3(b+c)}{b^3+c^3}+  \frac{c^3(c+a)}{c^3+a^3}}{\left(\frac{(ab+bc+ca)}{a+b+c}\right)}$

is somewhere around 2.902. Exact value?

  Darij


Sprint Round
27. What is the greatest integer value of n such that (n 2  + 2n + 5)/(n - 3) is an integer?

Target Round
8. The sum of a list of positive integers is 20. What is the maximum possible product of the integers in that list?

I need help finding an easy, effective method for solving these kinds of problems.

[b]PLEASE INCLUDE SOLUTIONS, NOT JUST ANSWERS!! I ALREADY HAVE THE ANSWERS![/b]

Solutions that involve guess & check are fine - as long as they are educated and effective under time pressure.

27
[hide]Do the division: $\frac {n^2 + 2n + 5}{n - 3} = n + 5 + \frac {20}{n-3}$. The largest number that the denominator can be and the fraction still remain an integer is 20, so the largest value of $n$ is $23$.[/hide]

[quote="tarquin"]8
[hide]Know that in order to achieve the maximum product you must have as many 3's as possible. The answer is then $3^6 \cdot 2 = 1458$.[/hide][/quote]

How would you discover that three is the best given a short amount of time?

Hey, who wants to prove that $3^a > b^c$ where $3a = bc$ and $b$ is a prime number ?

Yeah really, it's just something you have to know.
But there are some times where as many threes as possible is not the best.  That is when the remainder is 1 when the sum is divided by three.  There you should instead use 4 instead of 3x1

[hide]Do the long division.  We find that $\frac{n^{2}+2n+5}{n-3}=n+5+\frac{20}{n-3}$.  So whenever we plug an integer into that, we will receive an integer, as long as $20$ evenly divides $n-3$.  The largest integer $n$ for that to work is $n-3=20$, so $n=23$.[/hide]

[quote="nat mc"]Actually you don't have to do long division, you can do synthetic, it's much easier[/quote]Whats synthetic division???

So if you have n 2 +2n+5 and you divide it by n-3, then you put this

3 l 1 2 5
   l
3 is the number you are subtracting and 1,2,and 5 are coefficients
Bring the 1 down

3 l 1 2 5
   l
     1

Now multiply 1 by 3 and put it under the 2

3 l 1 2 5
   l    3
     1

Now do 2+3 and put it under the last 3

3 l 1 2 5
   l    3
     1 5

Now multiple 5 by 3 and put it under the other 5

3 l 1 2 5
   l    3 15
     1 5
Now do 5+15 and put it under the 15

3 l 1 2 5
   l    3 15
     1 5 20
1, 5, and 20 are the coefficients of the answer 1 degree less than the first equation, so it is n+5+20/(n-3)
Answer to question=23

The reason that 3 gives you the greatest answer is because its closest to e.

[quote="Phelpedo"]How would you discover that three is the best given a short amount of time?[/quote]
[hide]We have already been told the answer is $2\times 3\times 3\times 3\times 3\times 3\times 3$.  Let's try to forget that, and pretend we don't know anything about the answer.

Suppose one member of the final set is $5$.  We could replace the $5$ with $2$ and $3$.  (I'll write it as $5\rightarrow 2+3.$)  The sum would remain the same, but the product would be larger.  Thus, there cannot be a $5$ in the final set.  Similarly, there cannot be any number larger than $5$.

Suppose one member of the final set is $1$.  We could combine the $1$ with any other member of the set.  The sum would remain the same, but the product would be larger.  For example, if the set contained a $1$ and a $3$, you could combine these elements and replace them with a $4$.  $(1+3\rightarrow 4).$  The sum is the same, but the product is larger.

All we have left to consider are $2$, $3$, and $4$.  Suppose the final set contains three 2's.  You could replace them with $2+2+2\rightarrow3+3$.  The sum is the same, but the product is larger.  Therefore, the final set cannot contain 3 or more 2's.

Suppose the final set contains a $2$ and a $4$.  You could replace them with $2+4\rightarrow3+3$.  The sum is the same, but  product is larger.  Therefore, the final set cannot contain both 2's and 4's.

Suppose the final set contains two 4's.  You could replace them with $4+4\rightarrow3+3+2$.  Again, the sum is the same, but the product is larger.  Therefore, the final set cannot contain two or more 4's.

What we are left with is a set that can contain nothing but 2's, 3's, and 4's; a maximum of two 2's; a maximum of one 4 (but not both 2's and 4's).

Now, there is a unique solution to the problem, given the above observations.[/hide]

Does this mean that for any positive a, b, and c, $e^a > b^c$, where $ae = bc$


induction
floor function
ceiling function
inequalities
combinatorics unsolved
combinatorics

Suppose we have $n$ one's in a row .In each move we can delete two numbers like $a,b$ and replace them with $(a+b)/4$
.We continue these moves till just one number remains in the row .Prove that the remaining number is not smaller than $\frac{1}{n}$.

Let $m_{n}$ be the smallest possible last resulting number. It is clear that \[m_{n}= \min_{s+t=n}\frac{m_{s}+m_{t}}{4}.\] Let's prove by induction that $m_{n}\geq \frac{1}{n}$.
The base case $n=1$ is trivial.
Suppose that for all $k<n$ we have $m_{k}\geq \frac{1}{k}$. Then \[m_{n}\geq \min_{s+t=n}\frac{\frac{1}{s}+\frac{1}{t}}{4}= \min_{s+t=n}\frac{n}{4st}\geq \min_{s+t=n}\frac{n}{(s+t)^{2}}= \frac{1}{n}.\]

[quote="ehsan2004"]nice solution maxal , but this problem was solved by using this ineq easily $\frac{1}{a}+\frac{1}{b}\geq \frac{4}{a+b}$ .[/quote]

This solution is identical to maxal's solution, if you think about it.

It would be interesting to compute the $m_{n}$ and determine the exact
behavior.

Your conjectures are correct. Indeed, we have $ m_{2^{n}} \equal{} 2^{ \minus{} n}$ and between two powers of two, $ m_{n}$ is linear. This is easy to prove by induction, using the convexity of the function. Explicitly, we have $ m_{n} \equal{} \frac {3\cdot 2^{k} \minus{} n}{2^{2k \plus{} 1}}$ where $ 2^{k}\leq n\leq 2^{k \plus{} 1}$.

Another question(asked by darij) is whether we achieve that value by always picking the maximal pair $ (a,b)$ - the same question can be asked for arbitrary starting values. In our case considered here, using what i have just shown and spending a little thought on it, we see that the "greedy" strategy is indeed best possible.

[Moderator edit by Darij: It seems that my question has also been thought about by Valentin: http://www.mathlinks.ro/Forum/viewtopic.php?t=5367 ]

[quote="maxal"]It is interesting that all $m_{n}\cdot 2^{n}$ are integer, namely:
2, 2, 3, 4, 7, 12, 20, 32, 60, 112, 208, 384, 704, 1280, 2304, 4096, 7936, 15360, 29696, 57344 [/quote]

I think the exact formula for $m_{n}$ is as follows.  The process of combining pairs of numbers can be represented as the construction of a rooted binary tree.  (Start with $n$ leaves with the number $1$ at each leaf, then when joining $a,b$ add a new vertex above them labelled by $(a+b)/4$).  The root of the tree will be labelled by the final number in the process, which the sum of $1/4^{d}$ over all the leaves, where $d$ is distance to the root.

So we are asking what type of tree minimizes $\Sigma 1/4^{d}$.  Such a tree must be [i]balanced[/i], in the sense that for some integer $k$, all the leaves have distance $k$ or $k+1$ to the root.     I think this determines the structure of the tree uniquely (modulo switching the left and right subtrees at various vertices, which does not affect the depth), and the tree is indeed formed by splitting the leaves into groups of size $n/2$, with $(2^{n})m_{n}$ an integer as you mention.

I came up with a different solution.  If we keep track of which sums to which the various original numbers belong as we combine them, the final contribution of the $ith$ original number will be $\frac{1}{4^{a_{i}}}$, where $a_{i}$ is the number of times the $ith$ number gets "combined".  Note that, if we perform the same operations, except use $a,b$ to $\frac{a+b}{2}$, the final number will be 1. So

$\sum_{i=1}^{n}\frac{1}{2^{a_{i}}}= 1$

and by the AM-RMS inequality,

$\frac{\sum_{i=1}^{n}\frac{1}{4^{a_{i}}}}{n}\geq (\frac{\sum_{i=1}^{n}\frac{1}{2^{a_{i}}}}{n})^{2}= \frac{1}{n^{2}}$.

I claim that the minimum occurs when all the $a_{i}$ are either $j$ or $j+1$, where $n = 2^{j}+k, k < 2^{j}$.  There will be $2k$ values of $j+1$, and $2^{j}-k$ values of $j$, so the total is

$m_{n}= \frac{2k}{4^{j+1}}+\frac{2^{j}-k}{4^{j}}= \frac{2^{j+1}-k}{2^{2j+1}}$

We can prove this claim by induction on $n$.  Suppose the above claim is true for all $m < n$.  As maxal observed,

$m_{n}= \min_{s+t=n}\frac{m_{s}+m_{t}}{4}$,

and for $s = 2^{x}+y$, with $y < 2^{x}$ and $s < n-1$,

$m_{s+1}-m_{s}=-\frac{1}{2^{2x+1}}$.

So if $s < 2^{z}< t$ for some $z$, we can decrease the sum $\frac{m_{s}+m_{t}}{4}$ be incrementing $s$ and decrementing $t$.  So the minimum occurs with $s = 2^{x}+y, y < 2^{x}$ and $t = 2^{x}+z, z < 2^{x}$.  By induction, for both $s$ and $t$, the minimum is achieved when the respective $a_{i}$ are set to either $x$ or $x+1$;  when the final values for $s$ and $t$ are combined, this will increase the $a_{i}$ for $n$ by 1, so they have either $x+1$ or $x+2$ as values, which matches $n = 2^{x+1}+y+z, y+z < 2^{x+1}$.

As a check, observe that

$\frac{m_{s}+m_{t}}{4}= \frac{2^{x+1}-y}{2^{2x+3}}+\frac{2^{x+1}-z}{2^{2x+3}}= \frac{2^{x+2}-(y+z)}{2^{2x+3}}$


Sides AB, BC, CD, and DA, respectively, of convex quadrilateral ABCD are extended past B, C, D, and A to points B', C', D', and A'. Also, AB = BB' = 6, BC = CC' = 7, CD = DD' = 8, and DA = AA' = 9; and the area of ABCD is 10.

Find the area of A'B'C'D'.

[hide]Draw the diagonals AC, BD of quad ABCD. Then draw A'B. Since AB is the median of triangle A'BD, so area of A'AB is equal the area of ABC. Also area A'AB is 1/2 of area of A'AB' because A'B is a median.

So areaABD=1/2(areaA'AB'). Same reasoning will get us

areaBCD=1/2(areaCC'D')

areaADC=1/2(areaD'A'D)

areaABC=1/2(areaBB'C')

AreaA'B'C'D'=Area sum of triangles +AreaABCD

                   =Area 2(ABC)+2(ABD)+2(BCD)+2(ACD)+10

Area of ABC+ACD= quad ABCD=10

Area of BCD+ABD=quadABCD=10

so area of total =2(10)+2(10)+10=50      [/hide]


We haven't covered any chemistry, but I would like to take it in high school.  When should I take it, and what level should I take?  I know you don't have any idea how smart (or dumb!) I am, but can you give a good reccomendation for a slighly above-average AAP student?

Depends on what year of highschool you're in.  If you can, take Chem Honors then AP Chem.  However, if you're entering your senior year, just take AP Chem, it's a real easy AP.

At my school its usually pre-ap chem in 10th grade, and most sciencey people take AP chem in 11th grade.

[quote="towersfreak2006"]Here was my high school science schedule:

[u]9th[/u]
Pre-AP Biology
Pre-AP Chemistry

[u]10th[/u]
Pre-AP Physics
AP Biology
AP Chemistry

[u]11th[/u]
AP Physics

[u]12th[/u]
Anatomy & Physiology
University class???

And yes, for above average honors students, biology and chemistry were recommended for 9th and 10th grades, respectively. Anything after that is your own choice.[/quote]
This looks good.  Thanks!

[quote]However, if you're entering your senior year, just take AP Chem, it's a real easy AP.[/quote]
Easy? Who said anything about easy? I'm an academic advisor at a masters-level state university (CSULB), and I see the AP credit that people bring. Our chemistry department accepts a 4 or a 5 on the AP Chem as credit for the first semester of General Chemistry. (A 3 earns credit for a non-majors general education course.) AP Chemistry is an uncommon AP exam. It's much rarer than Calculus AB and somewhat rarer than Calculus BC. It also appears that 5's on the Chemistry AP exam are less common than 5's in either AP Calculus exam, at least among students I see.

Our Biology department has adopted a "chemistry first" position with its curriculum. The Introductory Biology course for majors has at least the first semester of General Chemistry as a prerequisite. I would suggest that for anyone with a proposed major in some biological science, AP Chemistry is just about the most important possible AP exam.

The Anatomy & Physiology and Environmental Biology AP exams do not necessarily lie on the main line of courses that prepare one for that Biological Science major. If towersfreak is up to it, the one university class in the last year that would push the furthest along the path for a proposed biology/molecular biology/biochemistry/chemisty major would be Organic Chemistry. Be warned that that could be very time-consuming, especially is a lab is involved.

But there's another problem with AP exams - all AP exams, including Calculus and Physics: Some medical school admissions committees don't like them and require that students take college courses in those subjects anyway.

[quote="shobber"]well, what do you mean by AP?? :blush:[/quote]

AP stands for advanced placement.  It's a college level course so it let's you get a few college credits in high school.  The scores range from (1-5), and credit depends on the college.

Kent, I was wondering, what's the rarest AP you know of, I here the Physics C E&M is pretty rare.

[quote]Kent, I was wondering, what's the rarest AP you know of, I here the Physics C E&M is pretty rare.[/quote]
Of course, there are some unusual ones in arts and languages. If we limit your question to math and science, yes, I'd say AP Physics C E&M is the least common.


Find the sum of the odd numbers between 90 and 110.

Yes, the solution that Kouichi used works. Also, the fact the sum of the first $n$ odd numbers is $n^2$ can be used. $90/2 = 45, 110/2 = 55.$ Thus $55^2-45^2 = (55-45)(55+45) = (10)(100) = \boxed{1000}.$ This is why we can just divide by 2: For each of 90 and 110, we have to subtract 1 to get our case (89 for 90 to get 1+3+...89, and 109 for 110 to get 1+3+..+109). Then, since the $n^\text{th}$ odd number is $2n-1,$ we add 1 then divide by 2.

[hide="Dumb Way"]Add up the odd numbers one by one.[/hide]
[hide="Regular way"]Pair up the odd numbers like $(91+109)+(93+107)+(95+105)+\cdots$[/hide]
[hide="Smart Way"]Because that the average of these numbers is $100$ and we have $10$ numbers we can simply just do $100\times10$ in order to find the sum.[/hide]


geometry
circumcircle
geometric transformation
parallelogram
perpendicular bisector
geometry solved

Let $ABC$ be an acute angled triangle with circumcenter $O$. Let $M$,$N$ be two points on the line $AC$ such that $MN = AC$ (as vectors). Let $D$ be the orthogonal projection of $M$ on the line $BC$, and $E$ be that of $N$ on the line $AB$. Let $L$ be the midpoint of $AN$.

(a) Prove that the orthocentre $H$ of triangle $ABC$ lies on the circumcircle (with center $O'$) of triangle $BED$.
(b) Prove that the midpoint of $OO'$ is the midpoint of $LB$.

with $O$ as the origin, let $\vec{T} = \vec{AM} = \vec{CN}$. Consider the point $\vec{X} = \vec{H} + \vec{T} = \vec{A} + \vec{B} + \vec{C} + \vec{T}$. Then  $\vec{XM} \cdot \vec{BC} = (\vec{B} + \vec{C}) \cdot (\vec{B} - \vec{C}) = 0$, so $XM \perp BC$; similarly $XN \perp AB$, so $X$ is the intersection of NE and AD. As such, the circle with diameter $BX$ contains $D, E$ as $\measuredangle{BEX} = \measuredangle{BDX} = \frac{\pi}{2}$. Also, $\measuredangle{BHX} = \frac{\pi}{2}$, as $\vec{X} - \vec{H} = \vec{T}$, and $\vec{T}$ is perpendicular to $\vec{BH}$. Thus, $H$ also lies on this circle that has already been shown to be the circumcircle of $BED$, as desired. The circumcenter of this circle is the midpoint of $BX$, so $2\vec{O'} = \vec{B} + \vec{X}$. Thus, $2\vec{O} + 2\vec{O'} = \vec{B} + \vec{X} = 2\vec{B} + \vec{A} + \vec{C} + \vec{T} = 2\vec{B} + \vec{A} + \vec{N} = 2\vec{B} + 2\vec{L} \Rightarrow \frac{\vec{O} + \vec{O'}}{2} = \frac{\vec{B} + \vec{L}}{2}$, as desired.

Nice little triangle geometry problem, reminding me of my old days when I solved only problems like this one...

Part [b](a)[/b] is [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=19092]problem 1 of the 3rd German TST 2004[/url]. In order not to mess up the notations, let me repeat the solution here:

Since $\overrightarrow{MN}=\overrightarrow{AC}$, there exists a translation mapping the points A and C to the points M and N, respectively. Let this translation map the orthocenter H of triangle ABC to a point R. Then, since translation maps lines to parallel lines, we have MR || AH and NR || CH. But since H is the orthocenter of triangle ABC, we have $AH\perp BC$ and $CH\perp AB$, so that $MR\perp BC$ and $NR\perp AB$; thus, the point R lies on the perpendiculars to the lines BC and AB through the points M and N, respectively, i. e. on the lines MD and NE. Thus, < BDR = 90° and < BER = 90°. Also, since our translation maps the point C to N and the point H to R, we have $\overrightarrow{HR}=\overrightarrow{CN}$, so that HR || CN, or, equivalently, HR || CA. On the other hand, again since H is the orthocenter of triangle ABC, we have $BH\perp CA$. Thus, $BH\perp HR$, so that < BHR = 90°. Together with < BDR = 90° and < BER = 90°, this yields that the points H, D and E lie on the circle with diameter BR. In other words, the circumcircle of triangle BED is the circle with diameter BR and passes through the point H. So problem [b](a)[/b] is solved, and we have also got the extra information that the center O' of the circumcircle of triangle BED is the midpoint of its diameter BR.

Let Y be the midpoint of the segment CA; then, $\overrightarrow{AY}=\frac{\overrightarrow{AC}}{2}$; on the other hand, since L is the midpoint of the segment AN, we have $\overrightarrow{AL}=\frac{\overrightarrow{AN}}{2}$; thus,

$\overrightarrow{YL}=\overrightarrow{AL}-\overrightarrow{AY}=\frac{\overrightarrow{AN}}{2}-\frac{\overrightarrow{AC}}{2}=\frac{\overrightarrow{AN}-\overrightarrow{AC}}{2}=\frac{\overrightarrow{CN}}{2}=\frac{\overrightarrow{HR}}{2}$.

On the other hand, since O is the circumcenter of triangle ABC and thus must lie on the perpendicular bisector of its side CA, we have $OY\perp CA$ (as Y is the midpoint of this side CA); together with $BH\perp CA$, this yields OY || BH. Also, from http://www.mathlinks.ro/Forum/viewtopic.php?t=24870 post #4 Theorem 3, we know that $BH=2\cdot OY$ (with directed segments), so that $OY=\frac{BH}{2}$. Thus, $\overrightarrow{OY}=\frac{\overrightarrow{BH}}{2}$. This and $\overrightarrow{YL}=\frac{\overrightarrow{HR}}{2}$ imply

$\overrightarrow{OL}=\overrightarrow{OY}+\overrightarrow{YL}=\frac{\overrightarrow{BH}}{2}+\frac{\overrightarrow{HR}}{2}=\frac{\overrightarrow{BH}+\overrightarrow{HR}}{2}=\frac{\overrightarrow{BR}}{2}$.

Since O' is the midpoint of the segment BR, we have $\frac{\overrightarrow{BR}}{2}=\overrightarrow{BO^{\prime}}$, and thus this becomes $\overrightarrow{OL}=\overrightarrow{BO^{\prime}}$. So the quadrilateral OLO'B is a parallelogram; thus, its diagonals OO' and LB bisect each other, i. e. the midpoint of OO' is the midpoint of LB. So we have solved problem [b](b)[/b].

But in fact, the configuration is too nice to stop here. First note that the point L is not only the midpoint of the segment AN, but also the midpoint of the segment CM. This is because MN = AC yields AM = AN - MN = AN - AC = CN, so that AL = LN (since L is the midpoint of the segment AN) leads to ML = AL - AM = LN - CN = LC, and thus L is the midpoint of CM.

Now, since < AEN = 90°, the point E lies on the circle with diameter AN; the center of this circle must be the midpoint L of the segment AN. Thus, LA = LE. Hence, the triangle ALE is isosceles with apex L, so that < LAE = < AEL, where we use directed angles modulo 180°. On the other hand, by the sum of angles in triangle ALE, we have < LAE + < AEL + < ELA = 0°, and thus

$\measuredangle ELA=-\left(\measuredangle LAE+\measuredangle AEL\right)=-\left(\measuredangle LAE+\measuredangle LAE\right)=-2\cdot\measuredangle LAE$
$=-2\cdot\measuredangle\left(CA;\;AB\right)$.

Similarly, $\measuredangle DLC=-2\cdot\measuredangle\left(CA;\;BC\right)$. Thus,

$\measuredangle DLE=\measuredangle DLC-\measuredangle ELA=\left(-2\cdot\measuredangle\left(CA;\;BC\right)\right)-\left(-2\cdot\measuredangle\left(CA;\;AB\right)\right)$
$=2\cdot\left(\measuredangle\left(CA;\;AB\right)-\measuredangle\left(CA;\;BC\right)\right)=2\cdot\measuredangle\left(BC;\;AB\right)=2\cdot\measuredangle DBE$.

On the other hand, since the points D and E (and, of course, B) lie on the circle with diameter BR, whose center is O', the central angle theorem yields $\measuredangle DO^{\prime}E=2\cdot\measuredangle DBE$. Thus, < DLE = < DO'E, so the points D, E, L and O' lie on one circle. This is actually a kind of generalized nine-point circle of triangle ABC (it becomes the nine-point circle if M = A and N = C), although it doesn't directly yield the properties of the nine-point circle of triangle ABC since there are too few points on it. Maybe someone finds more properties of this funny configuration.

Anyway, I have almost forgotten to say, the condition that triangle ABC is acute-angled is absolutely inessential.

  Darij

The Darij's and Mis.Term.'s solutions are instructively for understanding of the immediate application of the vectorial geometry without many difficulties and with clarity.
I come back at the our problem. I note $H'\in NE\cap MD,\overline {BB'}=\overline {AM}$. The triangle ABC, through slide on the line AC, becomes the triangle MB'N. Therefore, HH'||AC, i.e. $HH'\perp BH$ and H' is orthocentre of the triangle MB'N. The points D, E, H belong to the circle with the diameter  BH' a.s.o.


Let $A,B,M,N$ be four points on a circle $\mathcal C$ of center $O$. The tangents at $\mathcal C$ in $A$ and $B$ meet at $C$, while those in $M$ and $N$ meet at $P$. Let $C^\prime,P^\prime$ be the feet of the perpendiculars of $C,P$ on $MN,AB$.

Prove that \[ \frac{CC^\prime}{PP^\prime} = \frac{OC}{OP} .  \]

This has been probably posted before, but I couldn't find it using the "Search" function.

Here is a synthetic solution. Warning: I rename the points.

[color=blue][b]Problem.[/b] Let k be a circle with center O, and A, B, A', B' four points on this circle k. The tangents to the circle k at the points A and B meet each other at P, and the tangents to the circle k at the points A' and B' meet each other at P'. Let R and R' be the orthogonal projections of the points P and P' on the lines A'B' and AB, respectively. Show that $\frac{PR}{P^{\prime}R^{\prime}}=\frac{OP}{OP^{\prime}}$.[/color]

[i]Solution.[/i] The points A and B are the points where the two tangents from the point P to the circle k touch this circle. Thus, these points A and B are symmetric to each other with respect to the line OP (since O is the center of the circle k). Hence, $AB\perp OP$, and the midpoint Q' of the segment AB lies on the line OP. Similarly, $A^{\prime}B^{\prime}\perp OP^{\prime}$, and the midpoint Q of the segment A'B' lies on the line OP'.

Since the line PA is the tangent to the circle k (whose center is O) at the point A, we have $PA\perp OA$, and thus < PAO = 90°. On the other hand, $AB\perp OP$ yields < AQ'O = 90°. Thus, < PAO = < AQ'O. Also, trivially, < POA = < AOQ'. Thus, the triangles PAO and AQ'O are similar, so that $\frac{OP}{OA}=\frac{OA}{OQ^{\prime}}$. In other words, $OP\cdot OQ^{\prime}=OA^2$. If we denote by r the radius of the circle k, then OA = r, so this becomes $OP\cdot OQ^{\prime}=r^2$. Similarly, $OP^{\prime}\cdot OQ=r^2$. Thus, $OP\cdot OQ^{\prime}=OP^{\prime}\cdot OQ$, or, equivalently, $\frac{OP}{OQ}=\frac{OP^{\prime}}{OQ^{\prime}}$. Together with < POQ = < P'OQ', this shows that the triangles POQ and P'OQ' are similar. In other words, there exists a similitude transformation which maps the points P, O, Q to the points P', O, Q', respectively.

Thus, this similitude transformation must also map the perpendicular to the line OQ at the point Q to the perpendicular to the line OQ' at the point Q'. But the perpendicular to the line OQ at the point Q is the line A'B' (since $A^{\prime}B^{\prime}\perp OP^{\prime}$), and the perpendicular to the line OQ' at the point Q' is the line AB (since $AB\perp OP$). Hence, our similitude transformation maps the line A'B' to the line AB. Since it maps the point P to the point P', it therefore transforms the orthogonal projection R of the point P on the line A'B' into the orthogonal projection R' of the point P' on the line AB.

Now, as we have seen that our similitude transformation takes the points P, O, R to the points P', O, R', respectively, it follows that $\frac{PR}{P^{\prime}R^{\prime}}=\frac{OP}{OP^{\prime}}$, and the problem is solved.

  Darij


geometry
3D geometry
modular arithmetic
quadratics
number theory proposed
number theory

Show that if the difference of two consecutive cubes is a square, then it is the square of the sum of two successive squares.

It sounds hard but it isn't :)

I have a solution.

The problem asks us to prove that if $(x+1)^3 - x^3 = 3x^2 + 3x + 1 = y^2$, then $y$ is the sum of $2$ consecutive squares, i.e. $y = a^2 + (a+1)^2 = 2a^2 + 2a + 1$ for some $a$. This is equivalent to showing that $2y-1 = 4a^2 + 4a + 1 = (2a+1)^2$ is a perfect square.
Note that $3x^2 + 3x + 1 = y^2 \Leftrightarrow (2x+1)^2 = \frac{4y^2-1}{3} = \frac{(2y+1)(2y-1)}{3}$, and note that $\gcd (2y+1, 2y-1) = 1$, since both $(2y+1)$ and $(2y-1)$ are odd. We now claim that $y \equiv 1 \pmod{3}$, then $3 \nmid 2y-1 \Rightarrow 2y-1$ is a perfect square, and we are done.

To prove that $y \equiv 1 \pmod{3}$, we utilise the law of quadratic reciprocity. (I can't think of another approach to show that $y \equiv 1 \pmod{3}$... if anyone else knows of a simpler approach, please share!!)
Let $p$ be an arbitrary prime dividing $y$. Now obviously $3 \nmid y$, thus taking $\mod p$, we have $p \equiv 1, 5, 7$ or $11 \pmod{12}$. We claim that $p$ has to be $\equiv 1 \pmod{3}$. Now suppose not, then $p \equiv 5$ or $11 \pmod{12}$. We note that from $3(2x+1)^2 = 4y^2-1$, taking $\mod p$, we have $3(2x+1)^2 \equiv -1 \pmod{p} \Rightarrow (\frac{3}{p}) = (\frac{-1}{p})$.

Case 1: $p \equiv 5 \pmod{12} \Rightarrow p \equiv 1 \pmod{4} \Rightarrow (\frac{-1}{p}) = 1 \Rightarrow (\frac{3}{p}) = 1$. By the law of quadratic reciprocity, we have $(\frac{p}{3})(\frac{3}{p}) = (-1)^{\frac{(p-1)(3-1)}{4}} = 1 \Rightarrow (\frac{p}{3}) = 1 \Rightarrow p \equiv 1 \pmod{3}$ (Contradiction)

Case 2: $p \equiv 11 \pmod{12} \Rightarrow p \equiv 3 \pmod{4} \Rightarrow (\frac{-1}{p}) = -1 \Rightarrow (\frac{3}{p}) = -1$. By the law of quadratic reciprocity, we have $(\frac{p}{3})(\frac{3}{p}) = (-1)^{\frac{(p-1)(3-1)}{4}} = -1 \Rightarrow (\frac{p}{3}) = 1 \Rightarrow p \equiv 1 \pmod{3}$ (Contradiction)

Thus $\forall$ primes $p$ dividing $y$, we must have $p \equiv 1 \pmod{3}$ and consequently $y \equiv 1 \pmod{3}$ and we are done.

I think there is a shorter way to prove this result :)

$(2y - 1)(2y + 1) = 3(2x + 1)^2$ and $2y - 1$, $2y + 1$ are coprime.

Hence $2y - 1 = 3a^2$ and $2y + 1 = b^2$ or $2y - 1 = c^2$ and $2y + 1 = 3d^2$, where $a$, $b$, $c$, $d$ are integers.

The first case is impossible since then $b^2 = 2$ (mod 3).

Hence $2y - 1 = c^2$ is a perfect square... :)

Hahaha... How could I not see such an easy approach? Didn't know it could be so simple :):):)
Guess there's more than one way to skin a chicken and you can say I used a machine gun to skin this chicken.  :P (Sorry to the vegetarians, I mean it as a harmless joke)  :lol:


How did everyone do?

I got an 800 on Chem which was the only one I took.  That was the most surprising thing that has happened to me this year..by far.

I'm soph and haven't taken any subject this school year due to ARML ....

I have two questions regarding SAT II
 1. What is the benefit of taking SAT II?
 2. Is SAT II easier than AP generally?

Thanks,

SAT II tests are required for admission to some colleges, including the University of California campuses. They are helpful information to have in your file even for colleges that don't require them.

SAT II tests are tests of high school knowledge, unlike AP tests, but some of them are fairly hard. Not all high schools do a good job of teaching high school knowledge.

2150

790 Math
690 Verbal
670 Writing (11 Essay)

Hm, retake.

I still haven't gotten mine!  Apparently i'm part of the "small percentage" that are not available on the first score release date.  They said check back in a week, and it has been a week!  I'm a patient person, but I want to know  :(   Tomorrow.  It has to be tomorrow.


number theory
prime factorization
number theory unsolved

Let $m$ be a possitive integer and $p_1,p_2,...,p_n$ are all the prime that $\le m$ :
Proof the ineq : $\frac{p_1}{p_1-1}.\frac{p_2}{p_2-1}.....\frac{p_n}{p_n-1}>1+\frac{1}{2}+...+\frac{1}{m}$

Each positive integer not greater than $m$ has prime factorization which only uses primes not greater than $m$. Moreover, using the unicity of the prime factorization it is clear that for each $k \leq m$ the term $\frac 1 k$ appears exactly once in the expansion of $P = (1 + \frac 1 {p_1} + \frac 1 {p_1^2} + \cdots) (1 + \frac 1 {p_2} + \frac 1 {p_2^2} + \cdots) \cdots  (1 + \frac 1 {p_n} + \frac 1 {p_n^2} + \cdots).$
Moreover, it is also clear that some terms of that expansion are of the form $\frac 1 k$ where $k > m.$
It follows that $S > 1 + \frac 1 2 + \cdots + \frac 1 m.$
Since $ (1 + \frac 1 {p_i} + \frac 1 {p_i^2} + \cdots) = \frac {p_i} {p_i - 1}$ the result follows.

Pierre.


inequalities
calculus
derivative
function
inequalities proposed

If $ a$, $ b$, $ c$ are positive reals, prove that

$ \left(a^{2}\plus{}b^{2}\plus{}c^{2}\right)\left(\frac{a^{2}}{\left(a^{2}\plus{}bc\right)^{2}}\plus{}\frac{b^{2}}{\left(b^{2}\plus{}ca\right)^{2}}\plus{}\frac{c^{2}}{\left(c^{2}\plus{}ab\right)^{2}}\right)\geq\frac{9}{4}$.

I think it's weaker than Iran1996's .
 Put $ a^2=yz , b^2=xz , c^2=xy $
 The ineq is equivalent to $ A= (xy+yz+xz) . \sum(\frac{1}{(x+\sqrt{yz})^2}) \geq 9/4 $
 By AM-GM : $ \sqrt{yz} \leq (y+z)/2 $
 So $ A \geq (xy+yz+xz). \sum(\frac{4}{(2x+y+z)^2}) \geq 9/4 $ : It's just a different formation of I96

The last inequality is not true (see x=y=0). :P

vasc x,y,z cannot be 0  because a,b,c>0.

[quote="Vasc"]The inequality is very hard. I think the following is less hard (but also very hard):
$\sum \frac{1}{6a^2+bc} \ge \frac{9}{7(ab+bc+ca)}$.
Has somebody a solution ?[/quote]
$ \sum \frac{1}{(k-1)a^2+bc} \ge \frac{9}{k(ab+bc+ca)} $.
  Is it still true for k =6 ? I "proved" the ineq is wrong for k= 1,2,3,4,5
  Do you have a solution , Vasc ?

[quote="nthd"]I think with a,b,c >0 andc=min(a,b,c)
(ab+bc+ca)(2/(c+(ab)^(1/2))^2+1/(a+b)^2)>=9/4
Is it right?[/quote]
It is not right (see $c=0$).

Now, related to
$\displaystyle \sum \frac{1}{pa^2+bc} \ge \frac{9}{(p+1)(ab+bc+ca)}$,
I think that it is true for $p\ge 3+\sqrt{7}$, but I have no solution.

mda.........vasc  $c>0$. you can not take $c=0$  :?

he can take $c\rightarrow 0$.  But could anyone solve this very hard inequality ?

view   [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=55584&highlight=#[/url]

I found...

   $(a^2+b^2+c^2)[\frac{a^2}{(a^2+bc)^2}+\frac{b^2}{(b^2+ac)^2}+\frac{c^2}{(c^2+ab)^2}] \ge \frac{9}{4}$ <->
   $(a^2+b^2+c^2)[(\frac{a}{a^2+bc})^2+(\frac{b}{b^2+ac})^2+(\frac{c}{c^2+ab})^2] \ge \frac{9}{4}$ <->
   $(a^2+b^2+c^2)[(\frac{a^2}{a^3+abc})^2+(\frac{b^2}{b^3+abc})^2+(\frac{c^2}{c^3+abc})^2] \ge \frac{9}{4}$.

   Consider now $f(x)=(\frac{x^2}{x^3+P})^2$, where we can assume that $P=abc$ is a constant. The problem it´s demonstrate that the second derivative of this functions is always positive for all x positive real. But it´s possible and you must believe me...
   Now we make:

   $(\frac{a^2}{a^3+P})^2+(\frac{b^2}{b^3+P})^2+(\frac{c^2}{c^3+P})^2 \ge 3(\frac{S^2}{S^3+P})^2$ where we have $S=\frac{a+b+c}{3}$ and we must note that $S \ge \sqrt[3]{P}$ -> $S^3 \ge P$ and now we have $2S^3 \ge S^3+P$ -> $\frac{1}{S^3+P} \ge \frac{1}{2S^3}$ -> $\frac{S^2}{S^3+P} \ge \frac{1}{2S}$ -> $(\frac{S^2}{S^3+P})^2 \ge \frac{1}{4S^2}$ -> $(a^2+b^2+c^2)3(\frac{S^2}{S^3+P})^2 \ge 3(a^2+b^2+c^2)\frac{1}{4S^2}$ and so it´s proved because by Jensen´s Inequalitie again we have $(a^2+b^2+c^2) \ge 3S^2$

[quote="mffmartinelli"]Consider now $f(x)=(\frac{x^2}{x^3+P})^2$, where we can assume that $P=abc$ is a constant. The problem it´s demonstrate that the second derivative of this functions is always positive for all x positive real. But it´s possible and you must believe me...[/quote]

I don't believe you, because calc101.com does not believe you. I don't understand why so many people claim a function is convex.

Yes. You are write, I had a mistake when I calculate the second derivate of this function. I´m sorry. In some interval this function changes concave. But is a "small" interval...   :lol:


The positive integrer number $n$ has $1994$ digits. $14$ of its digits are $0$'s and the number of times that the other digits: $1, 2, 3, 4, 5, 6, 7, 8, 9$ appear are in proportion $1: 2: 3: 4: 5: 6: 7: 8: 9$, respectively. Prove that $n$ is not a perfect square.


Can anyone share a procedure or a reference to a procedure to prepare and observe sperm cells under a microscope?

What magnification would be appropriate?

Thanks.

Hi,

This site has a link to microscope procedures.

http://www.freewebs.com/ads-online/

and this site has information on magnification

http://ask.metafilter.com/mefi/20996

Hope that helps.


i think it doesn't hold for $n=3$ allready. try $1,1+i,1+2i$ and you get
$\frac{(1+(1+i))((1+i)+(1+2i))((1+2i)+1)}{1(1+i)(1+2i)}=$
$2\frac{(2+i)(2+3i)}{1+2i}=2\frac{1+8i}{1+2i}=\frac{2}{5}(1+8i)(1-2i)=\frac{2}{5}(17+6i)$
doesn't work for $n\geq 4$ because you put these three numbers for $z_{1},z_{2},z_{3}$ and $1$ for all the rest...
but it holds for $n=2$  :wink:

ok,i'm an idiot. :blush:
but if it holds for all $n$ then we would have that $\frac{(z_{n}+z_{1})(z_{n}+z_{n-1})}{z_{n}(z_{n-1}+z_{1})}$ must be real for those kind of $z$ or $\frac{(1+a)(1+b)}{a+b}$ is real where $a=\frac{z_{1}}{z_{n}}$ and $b=\frac{z_{n-1}}{z_{n}}$ and $|a|=|b|=1$ and then
$\frac{1+ab}{a+b}$ real
$(1+ab)(a^{*}+b^{*})$ real ($aa^{*}=1$)
$2(a^{*}+b^{*})$ real which is not true...
could you find my mistake?

Let $U = \prod_{i=1}^{n}\frac{z_{i}+z_{i+1}}{z_{i}}= \prod_{i=1}^{n}\left( 1+\frac{z_{i+1}}{z_{i}}\right)$.
Then, we have that $\overline U = \prod_{i=1}^{n}\left( 1+\frac{z_{i}}{z_{i+1}}\right)$, since $\frac{z_{i+1}}{z_{i}}$ has absolute value $1$, i.e. its conjugate is its inverse.
That product is nothing but $U$, so we're done.


[color=darkred][size=134][u]An easy extension of the [b]Eyeball's theorem.[/b][/u][/size] Let $ \{\begin{array}{c}w_{1}= C(O_{1},R_{1})\\ \\ w_{2}= C(O_{2},R_{2})\end{array}$ be two circles, $ O_{1}\not\equiv O_{2}$.

Denote the line $ d = O_{1}O_{2}$. Let $ M\not\in O_{1}O_{2}$ be a point so that $ \frac{MO_{1}}{MO_{2}}=\frac{R_{1}}{R_{2}}$.

Choose the points $ \{\begin{array}{c}X\in MO_{1}\cap w_{1}\\ \\ Y\in MO_{2}\cap w_{2}\end{array}$ for which denote $ \{\begin{array}{c}P\in d\ ,\ XP\perp d\\ \\ Q\in d\ ,\ YQ\perp d\end{array}$. Prove that $ XP = YQ$.

[b]Remark.[/b] You can find this theorem at the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=127056 (the fifth question - user [b]Ilthigore[/b])[/color]

[hide]Draw perpendicular from M to d and label it A.
$ \triangle{XO_{1}P}\sim\triangle{MAO_{1}}$ and $ \triangle{MAO_{2}}\sim\triangle{YQO_{2}}$ since MA is parallel to XP and YQ.
$ MO_{1}/MO_{2}\equal{}R_{1}/R_{2}\equal{}XO_{1}/YO_{2}$
By similarity and the above, MA/YQ=MA/XP so YQ=XP[/hide]


Let's start an Indiana Mathcounts marathon.

New Problem:

$ ab\equal{}3,bc\equal{}9,ac\equal{}12$ What is $ a$? If $ a,b,c>0$

2.

fine i won't be a jerk.

[hide](abc)^2 = 3 * 9 * 12 so abc = 18 so a = 18/9 = 2[/hide]

Let's leave this to middle school students cough cough

[hide="NP"]
If 1 diagonal of a rhombus is 4 meters greater than the other, and the perimeter of the rhombus is 40, what is the area of the rhombus? [/hide]


modular arithmetic
number theory unsolved
number theory

Find all primes $ p$ such that $ p\equal{}a^2\plus{}b^2\plus{}c^2$ and $ p|a^4\plus{}b^4\plus{}c^4$ for some $ a,b,c \in N$.

[quote="KDS"]Find all primes $ p$ such that $ p \equal{} a^2 \plus{} b^2 \plus{} c^2$ and $ p|a^4 \plus{} b^4 \plus{} c^4$ for some $ a,b,c \in N$.[/quote]

Let $ a\ge b\ge c\ge 1$ and so $ p\ge 3$

We get $ p|(a^2\plus{}b^2\plus{}c^2)^2\minus{}(a^4\plus{}b^4\plus{}c^4)$ and so $ p|2(a^2b^2\plus{}a^2c^2\plus{}b^2c^2)$ and so $ p|a^2b^2\plus{}a^2c^2\plus{}b^2c^2$ (since $ p$ is prime $ \ge 3$)

$ \implies$ $ p|a^2(a^2\plus{}b^2\plus{}c^2)\minus{}(a^2b^2\plus{}a^2c^2\plus{}b^2c^2)$

$ \implies$ $ p|a^4\minus{}b^2c^2\equal{}(a^2\minus{}bc)(a^2\plus{}bc)$. So :

If $ a^2\ne bc$ then $ 0<a^2\minus{}bc<a^2\plus{}b^2\plus{}c^2\equal{}p$ and so $ p|a^2\plus{}bc$, which is impossible since $ p\equal{}a^2\plus{}b^2\plus{}c^2>a^2\plus{}bc$
So $ a^2\equal{}bc$ and so $ a\equal{}b\equal{}c$ and then $ p\equal{}3a$ and the only solution is $ p\equal{}3$ (else $ p$ would not be prime).

Hence the unique solution $ (a,b,c)\equal{}(1,1,1)$ and $ p\equal{}3$ which, indeed, is a solution.

Equivalently, $ \minus{}c^4 \equiv a^4 \plus{} b^4 \equal{} (a^2 \plus{} b^2)^2 \minus{} 2a^2b^2 \equiv c^4 \minus{} 2a^2b^2 \pmod{p}$, hence $ p \mid 2(ab\minus{}c^2)(ab\plus{}c^2)$, etc.


college
search
real analysis
real analysis theorems

I'm sure you guys know the book "Berkeley Problems in Mathematics."

Do you guys know about more books like this one?

I just know there exists someting called the Berkely Maths circle. What is this book?

Go to Spriger's web , and try to search to find the book of your. And I want to know about this book on your opion , give me please. I had to find list of all book on this series , but It's vary on country to country, China Edition and western.

 :mad:  I like the problem in analytical form , not set form.

Kong

Thailand

Problems in Mathemaical analysis I,II,III,  W.J.Kaczor-M.T.Nowak
Polya-Szego, Problems in analysis, Springer


geometry
perimeter
geometric transformation
reflection
rotation
function
analytic geometry

Are you counting triangles as equivalent under reflection? For example is a triangle with sides (3, 4, 5) the same as a triangle with sides (4, 3, 5), with both read clockwise?

[color=cyan]I think the problem is better considered when we look at odd perimeters and even perimeters seperately.  For odds, the first few values (starting with 3) are 1, 1, 2, 3, 4, 5, 7, 8, . . .   and for evens (starting with 4), 0, 1, 2, 3, 4, 5, 7, . . .  which seems to me to have a definite posibility for a nice pattern, if you make your lists twice or thrice (a great word!) as long as mine are.  Although, also maybe not.[/color]

[color=red]

in fact, there does exist such a formula. however, it is not particularly nice.

here it is:

let a+b+c=n. then, let 8N = n :^2: -6n+8.

if the number of incongruent triangles is T, then

(if n even) T=N(n :^2: /48)
(if n odd) T=N((n :^2: +6n)/48)

to prove this, one would first count the N=number of (ordered) triples (a,b,c) such that a+b+c=n and a+b>c, b+c>a, a+c>b (which can be done very nicely by considering lattice points inside triangles in the coordinate plane), and then subtract off the number of (triples, or septuples) of (a,b,c).

[/color]


a woman invested 10,000 in two accounts. One of the accounts pays 8% interestand the other 15%. how much is invested in each account if the total interest equals 1,080?

I need some help coming up with the formula to solve.. I thought I had it figured out but now I am clueless... Any help would be great!! Thanks!

I think I know how to do this. Let's see...

Set the amount of interest money in each account  as variables...

Clues
[hide]If the amount of interest in one account is 0.15 x... What will be the amount of interest in the other account in terms of x?[/hide]

[hide]If there's 10 fruits consisting of apples and bananbas, and there's x apples, how many will be bananbas?...[/hide]

[hide]0.08(10,000-x)[/hide]

[hide]So the amount of interest in one account is [b]0.15x [/b]and the amount of interest in the other account is [b]0.08(10,000-x). [/b]Agree So Far?[/hide]

What does the problem say about the interests of the accounts

[hide]Total Interest = 1,080 [/hide]

Set it up as an equation

[hide]0.15x+0.08(10,000-x)=1080[/hide]

Got it?

Then solve for x AND ________

[hide] I got x=4000, so 10,000-x=6000[/hide]

[hide] so 4000 is invested in the account that pays 15% interest and the 6000 is invested in the account that pays 8%[/hide]
CHECK YOUR WORK

Any question?

sorry, I think you know where the dollars signs should be placed... so make sure to include them

I didn't read the previous post because it took too long.  ;) No offense of course.  ;)

Let x be the money invested in the account with 8% interest. Let y be for 15%.
$x+y=10000$ and $.8x+.15y=1080$

Now, just solve it as a linear equation to get that y=4000 and x=6000.

Done!  :)

Awesome!! Thank you both for your help!! :)

[hide="equations"]x+y=10000
.08x+.15y=1080.

Solve to get 4000 and 6000.

Oh well, someone else posted an answer so don't mind this too much. [/hide]


Let $P_1$ be a regular r-gon and $P_2$ be a regular $s-gon$ $(r\ge s\ge3)$ such that each interior angel of $P_1$ is $59/58$ as larger as each interior angle of $P_2.$  What is the largest possible value of $s$?

[hide]117? solution later, if its right.[/hide]


Find all of the real and complex solutions (if any exist) of the equation $x^{7}+7^{7}=(x+7)^{7}$.

Thanks

See [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=116936[/url] with $a=7$.

this problem is the same as

$\begin{eqnarray*}(x+1)^{7}-x^{7}-1 &=& x^{6}+3x^{5}+5x^{4}+5x^{3}+3x^{2}+x \\ &=&(x)(x+1)(x^{4}+2x^{3}+3x^{2}+2x+1)\\ &=&(x)(x+1)(x^{2}+x+1)^{2}$

where we get the second line because $0$,and $-1$ are clearly roots, then we get the third line from a common factorization

the solutions to the proposed problem are just the solutions to the above equation, just they are multiplied by $a$

[quote="mathgeniuse^ln(x)"]Wait, so it doesn't make a difference whether or not it is 7^7 instead of 1^7...[/quote]

$x^{7}+a^{7}=(x+a)^{7}$
$\left(\frac{x}{a}\right)^{7}+1=\left(\frac{x}{a}+1\right)^{7}$
$y^{7}+1=(y+1)^{7}$

so they are equivalent problems (if $a=0$, then any $x$ is a solution)

If you have $a$ instead of $1$, solutions for $1$ must be multiplied by $a$ at the end.


In a given regular polygon, the ratio of the number of degrees in an interior angle to the number of degrees in an exterior angle is 7:2. How many sides does the polygon have?

[hide]The number of degrees of each interior angle is $\frac{180(k-2)}{k}$. Each exterior angle is $\frac{360}{k}$. Given the ratio of 7:2, we have

$2(\frac{180(k-2)}{k})=7(\frac{360}{k})$

The k's cancel.

$2(180k-360)=7(360)$

$360(k-2)=360(7)$

$k=9$[/hide]

Thanks.  I knew that I needed to use 180(n-2)/n and 36/n but just did not know how to set up the proper equations.
I got it now.


geometry
circumcircle
power of a point
radical axis
geometry unsolved

Let B and C be arbitrary points on sides AP and PD respectively of an acute triangle APD. The diagonals of the quadrilateral ABCD meet at Q, and $ H_1,H_2$ are orthocenters of triangle APD and BPC respectively. Prove that if $ H_1H_2$ pass through point X which is the intersection of the circumcircles of triangles ABQ and CDQ other than Q, then it must also pass through Y which is the intersection of the circumcircles of triangles BCQ and ADQ other than Q.

$\angle APD \equiv \angle BPC$ is not right $\Longrightarrow$ $H_1 \not\equiv H_2$ are different from each other. $\mathcal U, \mathcal V$ are circles on diameters $AC, BD.$ Powers of $H_1$ to $\mathcal U, \mathcal V$ are both equal to half the power of $H_1$ to circumcircle $\odot(APD),$ while powers of $H_2$ to $\mathcal U, \mathcal V$ are both equal to half the power of $H_2$ to cirumcircle $\odot(BPC)$ $\Longrightarrow$ $H_1H_2$ is radical axis of $\mathcal U, \mathcal V.$ Inversion $\mathbb Q$ with center $Q$ takes quadrilateral $ABCD$ into quadrilateral $A'B'C'D',$ the opposite sides of which intersect at inversion images $X' \equiv A'B' \cap C'D'$ of $X$ and $Y' \equiv B'C' \cap D'A'$  of $Y.$ Circles $\mathcal U, \mathcal V$ on diameters $AC, BD$ go to circles $\mathcal U', \mathcal V'$ on diameters $A'C', B'D'.$ $\mathcal Z'$ is a circle on diameter $X'Y'.$ Powers of the orthocenter of $\triangle A'X'D'$ to $\mathcal U', \mathcal V', \mathcal Z'$ are all equal to half its power to the circumcircle $\odot A'X'D.'$ Similar statements hold for orthocenters of $\triangle B'X'C', \triangle A'Y'B', \triangle C'Y'D'$ $\Longrightarrow$ these circles are coaxal (and their collinear centers $U', V', Z'$ make Newton line of the quadrilateral $A'B'C'D'$). Let $\mathcal Z$ be inversion image of $\mathcal Z'$ under  $\mathbb Q.$ Then $\mathcal Z$ is a circle through $X, Y$ coaxal with $\mathcal U, \mathcal V$ and $H_1H_2$ is their common radical axis.

Assume $X \in \mathcal Z$ is on the radical axis $H_1H_2$ of the pencil $\{\mathcal U, \mathcal V, \mathcal Z\}$ $\Longrightarrow$

1. Either $\mathcal Z$ is a line, identical with the radical axis $H_1H_2$ of this pencil (regardless of whether the pencil is intersecting or not) $\Longrightarrow$ $Y \in \mathcal Z$ is also on $H_1H_2.$

2. Or the pencil is intersecting and $X \in \mathcal Z$ is identical with one intersection of this pencil. Feet of perpendiculars from $X$ to $AQC, BQD, ABP, PCD$ are collinear on a common Simson line $x$ of $\triangle ABQ, \triangle CDQ$ with the pole $X$ $\Longrightarrow$ $x$ is also common Simson line of $\triangle  APC, \triangle BPD$ with the pole $X$ $\Longrightarrow$ circles $\mathcal U, \mathcal V$ are circumcircles of the last two triangles  $\Longrightarrow$ $P$ is the other intersection of the circles $\mathcal U, \mathcal V$ on diameters $AC, BD$ (Clifford circle theorem) $\Longrightarrow$ $AP \perp PD$, which contradicts the problem condition that the $\triangle APD$ is acute.


How's everyone's summer AMC Preparation going? I haven't been doing too much lately, just a weekend problem solving seminar but I need to do more work to prep, that's the only way to get good IMO, solve problems... Gah, once summer assignments are in I'm gonna go straight up on math...

[b]Vaya con Dios![/b]

IW.OAV

I've been doing problems and reading out of the AoPS everyday, and once I am done with that, Volume 2 and once I am done with that ACoPS.  My goal for this year is to just make USAMO which I think is a very realistic goal plus I plan to participate in every single math oppurtunity there is (in the bay area).

[quote]Vaya con Dios![/quote]

I like that comment.  Hablaste espanol?

The Bay Area is an excellent place to live in terms of mathematical opportunities. There are at least nine Saturday competitions for you to go to.

Is there a list of Bay Area contests and math oppurtunities?

There's a list of competitions with sample problems on my website at [url]http://www.albanyconsort.com/simon[/url]. Then click on the LOCAL MATH COMPETITIONS link. There's a bunch of other fun stuff there. Enjoy.

Math contests by state:
http://www.mualphatheta.org/Links/MAO_Links_Contests_Regional.htm

Great resource for pple in and around SFBA:
http://www.albanyconsort.com/contests/contests.html

Harker (private school in Bay Area) goes to almost every math contest. See their site for some contests - http://www.harker.org/announcements/2001_2002/math0102.htm will help, I hope.

Yay. I'm on Hann's math contest reference list.

Haha - took me a long time to realize that was your site. I found it through google last year... Nice site, btw. Only problem I have with it is the home page...too colorful for me. =P

Yes, it is a bit colorful, but I wanted a nice math background. Any other suggestions?


I like this one :

Let $n$ be an integer.
Prove that there exists a finite graph with no triangle and with chromatic number greater than $n$.

Pierre.

From a k-chromatic triangle free graph G  lets construct a k+1 chromatic triangle free graph G'. Let [i]X[/i](G) be the chromatic number of G. Let V(G)= {v1,...,vn}. Beginning with G add vertices X={x1,...,xn} and one more vertex y. Add edges to make x_i adjacent to all of N(v_i), and let N(y)= X. Clearly X is an idependet set in G'. Hence the other vertices of any triangle containing x_i belong to V(G) and are neighbors of v_i. This would complete a triangle in G, which can't exist. Therefore G' is triangle free.
A proper k-coloring f of G extends to a proper k+1 coloring of G' by setting f(x_i)=f(v_i) and f(y)=k+1. Therefore [i]X[/i](G')<=[i]X[/i](G) + 1. So to obtain equality we show that [i]X[/i](G) < [i]X[/i](G'). So we consider any proper coloring of G' and obain from it a proper coloring of G using fewer colors.
Let g be a proper k-coloring of G'. By changing the names of colors we may assume that g(y)=k. This restricts g to {1,...,k-1} on X. In V(G), it may use all k colors. Let A be the set of vertices in G on which g uses color k; we change the colors used on A to obain a proper k-1 coloring of G.  For each v_1 in A we change the color of v_1 to g(x_i). Because all the vertices in A have color k under g, no two vertices of A are adjacent.  Thus we need only check edges of the form v_i,v' with v_i in A and v' in V(G) - A. If v' is adjacent to v_i then by construction also v' is adjacent to x_i, which yields that g(v') and g(x_i) are different. Since we change the color on v_i to g(x_i) our change does not violate the edge v_i,v'. So the modified coloring of V(G) is a proper k-1 coloring of G.

My solution is different :
Let $k=2^n$, let's consider the grid of the points of the plane with integer and positive coordinates. Let $\Delta_i$ be the line with equation $x=i$, where $i=1,...,k$.
Now consider the graph whose vertices are the grid points where $1 \leq a,b \leq k$. And join $(a,b)$ and $(c,d)$ by an edge iff $a+b=c$ or $c+d = a$, that is join $(a,b)$ to all the points of the line $\Delta_{a+b}$ and only to them.
Assume that $(a,b),(c,d),(e,f)$ form a triangle. Wlog, we may assume that $a \leq c \leq e$. In that case $a+b=c=e$ and $c+d=e$, which forces $d=0$, a contradiction. Thus, there is no triangle.

Now, let $a,b$ be two positive integers both not greater than $k$, and assume that $a<b$. It follows that the point $(a,b-a)$ is joined to $(b,y)$, for all $y$, so that each point of the line $\Delta_b$ must have a color distinct from the color of $(a,b-a).$ It follows that the set of colors determined by $\Delta_a$ is distinct from the set of colors of $\Delta_b$.  Thus, we must have at least $2^n$ pairwise distinct available sets of colors, and clearly none is the empty set. It follows that the number of available colors is greater than $n$, and we are done.

Pierre.


In an office at various times during the day, the boss gives the secretary a letter to type, each time putting the letter on top of the pile in the secretary's in box. When there is time, the secretary takes the top letter off the pile and types it. There are nine letters to be typed during the day, and the oss delivers them in the order 1,2,3,4,5,6,7,8,9.

While leaving for lunch, the secretary tells a colleague that letter 8 has already been typed, but says nothing else about the morning's typing. the colleague wonders which of the nine letters remain to be typed after lunch and in what order they will be typed. based upon the above info, how many such after-lunch typing orders are possible? (that there are no letters left to be typed is one of the possibilities.)

This one is a toughie

[hide="Hint"]
Suppose there are now (at lunch) $ k$ (ordered) elements on the secretary's pile, ignoring $ 9$. In how many places can we add $ 9$ to the pile?
[/hide]

Solution on wiki: [[1988 AIME Problems/Problem 15]]

I posted a solution from a book.

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=524430745&t=80060

 And for next time, if you're going to post a problem from contest, look it up on wiki and/or search the forum. Locked.


inequalities
geometry
perimeter
inequalities unsolved

Prove or disprove the inequality:

If $ a,b,c$ are the lengths of a triangle with $ s = \frac{1}{2}$ then
\[ a^{2}(1-a)+b^{2}(1-b)+c^{2}(1-c) \geq \frac{1}{8}+\frac{3}{2}abc, \]
where $ s$ is the semiperimeter.

[i][b]by Maicol D'Avolio[/b][/i]

[quote="Maicol D'avolio"]Prove or disprove the inequality:

If $ a,b,c$ are the lengths of a triangle with $ s = \frac{1}{2}$ then
\[ a^{2}(1-a)+b^{2}(1-b)+c^{2}(1-c) \geq \frac{1}{8}+\frac{3}{2}abc, \]
where $ s$ is the semiperimeter.[/quote]
Maybe $ s = \frac{1}{4}$ and $ a^{2}(1-a)+b^{2}(1-b)+c^{2}(1-c) \geq \frac{1}{16}+\frac{3}{2}abc$ $ ?$ :wink:
If so it's true for all non-negative $ a,$ $ b$ and $ c.$

[quote="Maicol D'avolio"]No. It's like I wrote it above![/quote]
Id est, perimeter equal $ 1$ $ ?$ Something not OK. :wink:

[quote="Maicol D'avolio"]Yes, $ a+b+c=1$.

[quote="arqady"]
 Something not OK. :wink:[/quote]

Is the inequality false? I could not find a counter-example.[/quote]
No. It's true, but very ugly. And equality does not hold.


trigonometry
inequalities unsolved
inequalities

Let $ ABC$ be a triangle . Prove that:

$ \sin^2\frac Ak\plus{}\sin^2\frac Bk\plus{}\sin^2\frac Ck\ge\frac 32\left(1\minus{}\cos\frac{2\pi}{3k}\right)\ ,\ k\in\mathbb{R},\ k\ge 4$ .

[quote="Mateescu Constantin"]Let $ ABC$ be a triangle . Prove that:

$ \sin^2\frac Ak \plus{} \sin^2\frac Bk \plus{} \sin^2\frac Ck\ge\frac 32\left(1 \minus{} \cos\frac {2\pi}{3k}\right)\ ,\ k\in\mathbb{R},\ k\ge 4$ .[/quote]

$ LHS \equal{} \frac{1 \minus{} \cos\frac {2A}k}{2}\plus{}\frac{1 \minus{} \cos\frac {2B}k}{2}\plus{}\frac{1 \minus{} \cos\frac {2C}k}{2}\equal{}\frac{3}{2}\minus{} \frac{1}{2}(\cos\frac {2A}k\plus{}\cos\frac {2B}k\plus{}\cos\frac {2C}k)$ $ \ge \frac{3}{2} \minus{} \frac{3}{2}\cos(\frac{2A\plus{}2B\plus{}2C}{3k})$


Given points $ P(\minus{}2,7)$ and $ Q(4,y)$ in a coordinate plane, for what value of $ y$ is the slope of the line through $ P$ and $ Q$ equal to to $ \frac{\minus{}3}{2}$?

slope = $ (y2\minus{}y1)/(x2\minus{}x1)$
slope=$ (7\minus{}y)/(4\minus{}[\minus{}2])$
slope =$ (7\minus{}y)/6\equal{}3/2$
If the above is true, then $ 7\minus{}y\equal{}9$. $ y\equal{}\minus{}2$.


abstract algebra
trigonometry
real analysis
real analysis unsolved

\[ z \equal{} \frac{{1 \plus{} e^{2\alpha i} }}
{{1 \plus{} e^{2\beta i} }}
\]

find real, imaginary (part)... and his module

thanks

$ \frac{1\plus{}e^{2\alpha i}}{1\plus{}e^{2\beta i}}\equal{}e^{(\alpha \minus{}\beta )i}\left( \frac{e^{\alpha i}\plus{}e^{\minus{}\alpha i}}{e^{\beta i}\plus{}e^{\minus{}\beta i}}\right) \equal{}\frac{\cos \alpha }{\cos \beta }\left( \cos (\alpha \minus{}\beta )\plus{}i\sin (\alpha \minus{}\beta )\right)$

So from here, you can have all that you wanted :ninja:

That's right.

$ 1 \plus{} e^{2\alpha i} \equal{} 1 \plus{} \cos 2\alpha \plus{} i\sin 2\alpha \equal{} 2\cos ^ 2 \alpha \plus{} i\cdot 2\sin \alpha \cos \alpha$
$ \equal{} 2\cos \alpha (\cos \alpha \plus{} i\sin \alpha) \equal{} 2(\cos \alpha) e^{i\alpha}$, Similarly $ 1 \plus{} e^{2\beta i} \equal{} 2(\cos \beta )e^{i\beta}.$

$ \therefore z \equal{} \frac {\cos \alpha}{\cos \beta}e^{i(\alpha \minus{} \beta)}.$


induction
modular arithmetic
inequalities
articles
pigeonhole principle
function
AMC

The terms of the sequence $ (a_i)$ defined by $ a_{n \plus{} 2} \equal{} \frac {a_n \plus{} 2009} {1 \plus{} a_{n \plus{} 1}}$ for $ n \ge 1$ are positive integers. Find the minimum possible value of $ a_1 \plus{} a_2$.

[hide="rigorous solution"]
We prove that $ a_1a_2 \le 2009.$ Assume that $ a_1a_2 > 2009.$ Then, $ a_3 < a_1$ is equivalent to $ \frac{a_1\plus{}2009}{a_2\plus{}1} < a_1,$ or multiplying out, $ a_1\plus{}2009 < a_1a_2 \plus{} a_1,$ which is true, so $ a_3 < a_1.$ However, this means that the sequence $ b_i$ with $ b_1 \equal{} a_2$ and $ b_2 \equal{} a_3$ will also work but have smaller sum of first two numbers.

Now, we prove that $ a_1a_2 \ge 2009.$ Assume that $ a_1a_2 < 2009.$ Then similarly as last time, we can prove that $ a_1 < a_3.$ Also, we can prove that $ a_2a_3 < 2009,$ or $ \frac{(a_1\plus{}2009)a_2}{a_2\plus{}1} < 2009,$ or $ a_1a_2 \plus{} 2009a_2 < 2009a_2\plus{}2009$ which is true, so $ a_2a_3 < 2009.$ Thus, induction gives us that $ a_ia_{i\plus{}1} < 2009$ and $ a_i < a_{i\plus{}2}.$ This is implies that the $ a_i$ will get arbitrarily large because they are integers, so $ a_ia_{i\plus{}1} < 2009$ cannot be satisfied.

Thus, $ a_1a_2 \equal{} 2009,$ and the minimum sum for this is $ a_1 \equal{} 41, a_2 \equal{} 49$ which works because the $ a_i$ just goes 41,49,41,49...[/hide]

Is this the easiest solution. It seems to be the hardest problem to actually prove on this AIME.

Simple proof by showing that $ a_n \equal{} a_{n \plus{} 2}$
[hide="Another Approach"]
Rearranging the given relationship, we have
\[ a_{n \plus{} 2} \plus{} a_{n \plus{} 2}a_{n \plus{} 1} \equal{} a_n \plus{} 2009 (E1)
\]
Changing $ a_{n \plus{} 2}$ to $ a_{n \plus{} 3}$, we now have
\[ a_{n \plus{} 3} \plus{} a_{n \plus{} 3}a_{n \plus{} 2} \equal{} a_{n \plus{} 1} \plus{} 2009 (E2)
\]
E2-E1 gives us,
\[ a_{n \plus{} 3} \minus{} a_{n \plus{} 1} \equal{} \frac {a_{n \plus{} 2} \minus{} a_n}{a_{n \plus{} 2} \plus{} 1}
\]
Assuming all terms are positive integers,
\[ a_{n \plus{} 2} \plus{} 1 > 1
\]
Let $ D_n$ be $ a_{n \plus{} 2} \minus{} a_n$, if $ D_1$ is not 0, then $ D_1 > D_2 > D_3....$ forms a contradiction since all Ds must always be positive integers. (Fermat's method of infinite descent)
Then $ D_1 \equal{} D_2 \equal{} D_3 \equal{} ... \equal{} D_n \equal{} 0$
This question hence becomes $ a_n \equal{} a_{n \plus{} 2}$[/hide]

Could you show the general idea?

Edited:very beautiful

Copied the typing regulation line by line... DONE

[hide]Let $ a_1 \equal{} a$ and $ a_2 \equal{} b$. Then compute $ a_3 \equal{} \frac {a \plus{} 2009}{b \plus{} 1}$ and $ a_4 \equal{} \frac {(b \plus{} 2009)(b \plus{} 1)}{a \plus{} b \plus{} 2010}$. $ a_4$ is an integer, so $ (b \plus{} 2009)(b \plus{} 1) \equal{} b^2 \plus{} 2010b \plus{} 2009\equiv 0\pmod{a \plus{} b \plus{} 2010}\Rightarrow b^2 \plus{} 2010b \plus{} 2009 \minus{} b(a \plus{} b \plus{} 2010) \equal{} 2009 \minus{} ab\equiv 0 \pmod{a \plus{} b \plus{} 2010}\Rightarrow ab \equiv 2009 \pmod{a \plus{} b \plus{} 2010}$.

If $ ab \equal{} 2009$, the minimum of $ a \plus{} b$, $ a$ and $ b$ are integers is $ 41$ and $ 49$, and we can verify that this yields $ a_1 \equal{} a_3 \equal{} a_5 \equal{} \ldots \equal{} 41$ and $ a_2 \equal{} a_4 \equal{} a_6 \equal{} \ldots \equal{} 49$.

If $ ab\neq 2009$, then $ ab\ge 2009 \plus{} (a \plus{} b \plus{} 2010) \equal{} a \plus{} b \plus{} 4019$. Then $ \sqrt {ab} > \sqrt {4019} > 63$. By AM-GM, $ a \plus{} b\ge 2\sqrt {ab} > 126$. This doesn't give a minimum, so the answer is when $ a\plus{}b \equal{} 41\plus{}49\equal{}\boxed{090}$.
[/hide]

this is what i did:

you can see that some sequence of integers has to repeat
if a_1 = a_2,
then a_i = sqrt(2009), not integer

test a_1 = a_3

you get a nice expression of a_1 * a_2 = 2009
a_1 + a_2 is minimum when they are as close to each other as possible
so you get min (a_1 + a_2) = 41 + 49 = 90

a_1 = a_4 gets too complicated..so i ignored

not a full solution, but still worked :D

[quote="gf4848"]:( ....
DPatrick has STOLEN my solution.[/quote]

Alas, the horrors of AoPS admins.  :o  :(

But really, you should be proud of having such an elegant solution.

This one seems rather different from the ones already posted.
[hide]
Let $ b_k: \equal{} a_k \plus{} 1$, to simplify notation. The text now reads
\[ b_{n \plus{} 1}b_{n \plus{} 2} \equal{} 2008 \plus{} b_n \plus{} b_{n \plus{} 1},
\]
from which we can infer
\[ b_{n \plus{} 1} \mid 2008 \plus{} b_n \plus{} b_{n \plus{} 1},
\]

\[ b_{n \plus{} 2} \mid 2008 \plus{} b_n \plus{} b_{n \plus{} 1}.
\]
Both are valid for all $ n$. Let us shift the indices down by one in the second formula, and subtract it from the first, to get $ b_{n \plus{} 1} \mid b_{n \plus{} 1} \minus{} b_{n \minus{} 1}$. This implies that $ b_{n \plus{} 1} \mid b_{n \minus{} 1}$, and thus from a certain point on we have $ b_n \equal{} b_{n \plus{} 2} \equal{} b_{n \plus{} 4} \equal{} \dots$, that is, the succession $ b_n$ is made with two alternating values, $ a,b,a,b,a,b,\dots$. Since the relation can be inverted to yield $ b_n$ in terms of $ b_{n \plus{} 2}$ and $ b_{n \plus{} 1}$, this implies that the succession is in that form starting from the $ b_1$. Now it is clear that the only two conditions needed are $ a \equal{} \frac {a \plus{} 2009}{b \plus{} 1}, b \equal{} \frac {b \plus{} 2009}{a \plus{} 1}$, from which we get $ ab \equal{} 2009$.
(look, ma, no inequalities! :))
[/hide]

Can you please explain the logic behind Fermat's method of infinite descent (used in [b]gf4848[/b]'s solution)? I read a wikipedia article on it with an example about the rationality of $\sqrt{2}$, but I still don't completely understand it. Thank you :)

What you call it doesn't really matter. I'd call it the Pigeonhole Principle.

[hide]Define the integer sequence $D_n = a_{n+2} - a_n$, and assume for the sake of contradiction that $D_1 \neq 0$.

We know the sequence $D$ satisfies the equality $ D_{n+1} = \frac{D_n}{a_{n+2}}$.

Taking the magnitude and using $a_{n+2} > 0$, we get: $0 < | D_{n+1}| < |D_{n}|$

$\blacktriangleright$ Transitively: $0 < |D_i | < |D_j|$ for all $i > j \; $.

We know $|D_i | \in \mathbb{N}$, so $|D_i| \in \{ 1 \ldots |D_1| -1 \}$ for all $i > 1$.

Consider that set of values as $|D_1| - 1$ "holes", with each of $+\infty$ sequence term $|D_i|$ as a "pigeons". By the Pigeonhole Principle, there must exist two distinct equal terms $|D_i | = |D_j|$ for some $i > j$, which contradicts the statement at $\blacktriangleright$.

It follows that our assumption of $|D_1| \neq 0$ must be false, and hence, $|D_1| = 0 \implies D_1 = 0 \implies \boxed{\forall i \in \mathbb{N}, \; : \; D_i = 0}$.[/hide]

[b]@ gf4848:[/b][i] It's not immediately clear that $D_i$ must be positive.[/i]
[b]Edit:[/b] You're right, I've fixed my post.

I think I understand it :). (I think $i > j$ though for fifth line? o.o unless I misunderstood your detailed response XD)

Thank you very much! :) :) I think I understand it much better now :)

Sure. By the way, a proof by "infinite descent" is basically a particular type of proof by contradiction, saying that you cannot have an infinite subset of a finite set. In that sense, it can always be expressed in terms of the Pigeonhole Principle. In the example here, we wanted an infinite set of unique $| D_i |$'s that had to be a subset of $ \{ 1, \; \ldots \| D_1 \| -1 \}$, which is impossible.

In the $\sqrt{2}$ irrationality case, the argument was that if $(p_0 , q_0)$ is a solution, then $(p_1, q_1) = \left( \frac{p_0}{2} , \; \frac{q_0}{2} \right)$ is another, strictly smaller solution. Thus, we can generate an infinite set of correct solutions no larger than $(p_0, q_0)$.

However, the set of solutions no larger than $(p_0, q_0)$ is clearly a subset of the set of ordered pairs no larger than $(p_0, q_0)$, and the latter is a finite set. Thus, we've arrived at a contradiction.

Note that I haven't defined "smaller" -- I could have used $(w, x) < (y, z) \iff w<y \wedge x < z$, or if $w+x < y+z$, or whatever. There are many choices in this case, though in other cases, finding the right definition of "smaller" might be nontrivial. In fact, we can just abandon the notion of "smaller" for a more general notion of some ordering function that allows us to assert that the infinitely many terms we're generating are in fact accumulating into an infinitely large set.

Here is another solution:

[hide]$a_3=\frac{a_1+2009}{a_2+1}$ and $a_4=\frac{(a_2+2009)(a_2+1)}{a_1+a_2+2010}$, so $a_3a_4=\frac{(a_1+2009)(a_2+2009)}{a_1+a_2+2010}=\frac{(a_1+2009)(a_2+2009)}{(a_1+2009)+(a_2+2009)-2008}$ is an integer. Let $x=a_1+2009$, $y=a_2+2009$, $k=a_3a_4$. Then $xy=kx+ky-2008k$, or $(x-k)(y-k)=k(k-2008)$. Substituting back gives $(a_1+2009-k)(a_2+2009-k)=k(k-2008)$, and thus $k \geq 2009$. If $k=2009$, then $a_1a_2=2009$, and the minimum value of $a_1+a_2$ is achieved when $a_1=41$ and $a_2=49$(the sequence will just shift back and forth between $41$ and $49$). If $k \geq 2010$, then the minimum possible value of $a_1+a_2$ is clearly greater than $90$(just use AM-GM). Thus the minimum possible value of $a_1+a_2$ is $90$.[/hide]

this idea is not new: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3037976&sid=01ba5e67f75f1ef104f92b531887561f#p3037976

[hide=Comment on gf4848 solution]Infinite descent is useful, but not necessary to finish. Once we have
$a_{n+3}-a{n+1}=\frac{a_{n+2}-a_n}{a_{n+2}+1}$ Clearly the LHS is an integer, thus the RHS is an integer. If the RHS is an integer then $a_{n+2}-a_n\ge{a_{n+2}+1}$ (Note if the numerator cannot be less than the numerator because the sequence would see turn to negative integers. Rearranging we get $-1\ge{a_n}$ which is clearly wrong. Thus the numerator must be $0$. Many others have commented on how to finish from here[/hide]

[hide=probably similar to others but wtvr]
Define $b_k=a_k+1$. This means that for all $k$, $b_k\geq 2$. Rearranging and manipulating the relation, we get \[b_{n+2}b_{n+1}-b_{n+1}-b_n=2008\]
Now consider the equations when $n=1$ and $n=2$, and subtract them. We get the relation \[b_3-b_1=b_3(b_4-b_2)\iff b_1=b_3(b_2-b_4+1)\] thus, $b_3|b_1$. Similar reasoning gives that for all $k$, $b_{k+2}|b_k$.

I will show that $b_k=b_{k+2}$ for all $k$: FTSOC, suppose that $b_2-b_4+1>1 \iff b_2>b_4$. Since $b_2=b_4(b_3-b_5+1)$, and $b_2>b_4$, we have $b_3-b_5+1>1 \iff b_3>b_5$. Continuing this reasoning gives that if for there exists some $k$ such that $b_k>b_{k+2}$, then the sequence of odd indices and even indices are strictly decreasing, and for large enough $k$ we can get a contradiction. Thus, $b_k=b_{k+2}$ for all $k$.

Answer extraction is easy from here: we now have \[b_1+b_2+2008=b_1b_2\iff (b_1-1)(b_2-1)=a_1a_2=2009\] so the minimum of $a_1+a_2$ is $41+49=\boxed{090}$.
[/hide]

Can I have a hint on how to start this problem? It seems rather difficult!  :read:

[quote=channing421]Try to show that the sequence has to have period of length 2.[/quote]

I'm just revisiting this now for some reason, but I found several terms and didn't find a period, I don't really know how to show $a_n = a_{n+2}$..

It doesn't hold for all sequences; you must use the condition that all terms in the sequence are positive integers (it is intuitively true that the sequence must be periodic; otherwise, it'd be hard to make all of the terms positive integers!).


Giati kaneis den edose prosoxi se aftin ?
Loipon, iparxei mia lisi se 2 seires  :wink:

Ναι Στέργιο ακριβώς  :D .Καλώς ήρθες πάλι  :lol:

Ena hint?

einai kleisth, isn't it?

 :maybe:

Ένα hint : Μετασχηματισμός   :D

Αν κι θα ήθελα την λύση σε δυο γραμμές του Στέργιου θα ήθελα να ρώτησω μήπως βγαίνει με απαγωγή σε άτοπο δηλαδη αν μπορουμε να θεωρήσουμε το Α μέλος >1 ? θα ήθελα να μας πεί ο Σιλουανός αν γίνεται έτσι νομίζω οτι έτσι μπορεί να βγεί

Λοιπόν μιας και πολύς καιρός πέρασε......
Θέτουμε $x=\frac{2a}{a^{2}+1}$  άρα $0\leq x\leq 1$ .
Όμοια και για τα άλλα .Έτσι η ανισότητα μετασχηματίζεται στην :
$\frac{x}{1+yz}+\frac{y}{1+zx}+\frac{z}{1+xy}\leq 2$

την οποια θέλω να τη λύσετε όπως δεν την έχουν τα βιβλία   :wink:
Τα βιβλία λένε το εξής :(τουλαχιστον αυτά που έχω εγώ.....)
Υποθ ΧΒΤΓ ότι  $0\leq x\leq y\leq z\leq 1$
Άρα
$\frac{x}{1+yz}+\frac{y}{1+zx}+\frac{z}{1+xy}\leq\frac{x+y+z}{1+xy}\leq 2$

που είναι μια ωραία λύση αλλά γιατί να μην ψάξουμε για κάτι το διαφορετικό...... :)

Νομίζω Σιλουανέ οτι εννοείς
χ= 2*α/α^2+1 κι 0<χ=<1
διοτι έτσι όπως το έγραψες ειναι σαν μια εξισωση με βάση το χ
δεν το κάνω κακοπροεραιτα αλλα για την αποφυγη παρεξηγησεων
ευχαριστώ Σιλουανέ αν θέλεις διορθωσέ το

εξυπνο σιλουανε.αμβροσιε αυτο εννοει προφανως.υπαρχει αλλη λυση ?γιατι διαφορετικα το θεμα δυσκολευει.αληθεια ποιο ηταν ποιο δυσκολο η ανισοτητα του σιλουαν ή το φετινο τεταρτο θεμα του μικρου αρχιμηδη?

προφανώς το πρώτο

Το διόρθωσα το τυπογραφικό..........
Στέργιο η λύση σου είναι διαφορετική???? :)

σιλουαν θες μια λυση χωρις να τους διαταξεις σε αυξ.σειρα? γιατι χ/zy+1<=x/xyz+1 to idio isxyei kai gia ta alla 2 klasmata αρα προσθετεις και τις 3 σχεσεις και λυνεται

Γι'αυτο που ρωτησε ο Κωστας δηλαδη πια ασκηση απο τις 2 ειναι πιο δυσκολη πιστευω οτι απαντηση δεν μπορουμε να δωσουμε!Ηλυση για την ασκηση του Σιλουαν με μετασχηματισμο ηταν πολυ καλη!Βεβαια για την φετινη ανισοτητα που επεσε στους μικρους μπορω να πω οτι ηταν πολυ καλη και η λυση της ΕΜΕ ακομα καλυτερη!Αν και υπαρχουν και πιο απλες λυσεις... (χωρις χρηση τριγονωμετρικων αριθμων)... :)

Φίλε Δημήτρη εγω την έχω λύσει χωρις τργωνομετρία αποδεικνυεται οτι 1/3>=χ*υ> -1 άρα 2/9>=Κ>=-2 η αποδειξη είναι ευκολή με τριώνυμο

Καλη σκεψη φιλε μου Αμβροσιε.Εγω εχω βρει την εξης απλη αποδειξη:θετω x+y=a kai x-y=b οποτε η προς αποδειξη ανισοτητα μετασχηματιζεται και αποδ.ευκολα. :)

αν λυνεται ετσι το 4ο θεμα του αρχιμηδη πιστευω και εγω οτι του σιλουαν ειναι δυσκολοτερο.γιατι θελει μεγαλη προσοχη στο μετασχηματισμο. :o

Ναι πιστεύω οτι έχεις δίκιο φίλε Κώστα καθότι αυτή ειναι τεράστια ανισότητα για να την μετασχηματίσεις εύκολα


In the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=3184&view=next#,
[quote="manlio"]
Let $d_1 \leq d_2 \leq d_3$ where $d_1,d_2,d_3$ are the distances of the circumcenter of a triangle to its [b][u]excentres[/u][/b]. Prove that:
\[ 1)\ R<d_1<2R \\ 2) R<d_2<R\sqrt{5} \\ 3) 2R<d_3<3R.  \]
[/quote]
What does [b][u]excentres[/u][/b] mean?
I tried to find in dictionary but failed.
Help!
Thanks!

It means the intersection of two external angle bisectors of a triangle (which is the center of a circle that is outside the triangle but tangent to all the 3 sides).

The American English spelling would be "excenter" (as in "center" rather than "centre"). That might or might not help the search.


N noghte too faza darim, be torike bare har mosalas yek zavie bozorgtar az 120 bashe.sabet konid mishe in noghteha ro tori shomare gozari kard ke baraye har i, j. k ke dashte bashim i<j<k angah zaviyeye AiAjAk bozorgtar az 120 beshe.

kasi nemikhad nazar ya ide ee bede, hadeaghal begid asoone ya sakhte. :|

yeki az soalaye doreye sovome(sale 1969) olympiade shoraviye va be nazare man kami ta ghesmati abry sakhte!

sigari,az etelaati ke dadi mamnoonam.
oon soala dashte bashin, ye soale ghashange dige mizaram
soal: tamame N ha ee ro peida konid ke ye gerafe kamele N rasi,N ta king dashte bashe(rasi az geraf ra king minamim ke az rasaye digar betavan be tor mostaghim ya hadeaksar ba yek vasete be an raft)

fekonam manzooret az graphe kamel, graphe kamele jahatdare!
ba esteghra rooye n sabet mikonim baraye har adade tabiyi be joz 1,2,4 chenin graphi vojood darad:

sabet mikonim age hokm baraye n dorost bashe baraye n+2 ham doroste.
mikhahim ba rOose a(1),...a(n+2) ye graph ba chenin khasiyati besazim.farzkonid a(1),...,a(n) n ras az ye graph bashand ke in khasiyato darand. jahate yalhayi ra ke rase a(n+1) ra be rOose a(1),...,a(n) vas mikonand ra az a(n+1) be a(1),...,a(n) begirid va jahate yalhayi ra ke rase a(n+2) ra be a(1),...,a(n+1) vasl mikonand ra az a(n+2) be a(1),...a(n+1) begirid in graphe n+2 rasi khasiyate masale ra dare. baraye 3,6 ham besadegi(makhsoosan baraye 6) mitavan chenin graphi mesal zad.

zemnan, man SIGARY am, na SIGARI. :mad:


algebra
polynomial
quadratics
calculus
derivative
real analysis
real analysis unsolved

Let $ a,b,c\in\mathbb R$ and $ a\not =0$ such that $ f(x)=ax^{2}+bx+c$ has real root. Prove that there exist $ d\in\mathbb R$ such that polynomial $ \frac{ax^{3}}{3}+\frac{bx^{2}}{2}+cx+d$ has all  roots are reals.

If the quadratic has two different real roots then the cubic has two points of zero derivative.  Between the horizontal tangents at these two points, any horizontal line cuts the cubic in three points, so we choose $ d$ such that the $ x$-axis is such a line and we'll have 3 real roots.

If the quadratic has a double root, the cubic has a point where the first and second derivatives are both zero.  Choosing $ d$ so that the $ x$-axis is tangent to the cubic at this point, we have a root of order 3 and so 3 real roots.


Hi,
i want example about homomorphism in BCK-algebra and dual ideal., please

That's an algebraic structure, part of abstract algebra. This is an inappropriate forum for it. A better place to post this is in [url]http://www.artofproblemsolving.com/Forum/index.php?f=61[/url].


I took it for the first time this year.  Wish I knew about it last year.  Anyways, my teacher told me results should be up by February 27th.  And I don't remember exactly what the question was concerning rate law...was it the one where they told you the constant was equal to .6 something and what was the half life?

I thought it was much easier this year than last year.

I agree. It was easier than last year. Unfortunately that will probably result in an ugly cutoff.

yes, i believe the results will come out on the 27th...as i heard from my teacher.

wow...
will the list of semi-finalists be posted online on the cee website?

Yeah, it looks like scores are in. How did you guys do?

wait, so how can we actually find out the results online?

Talk to your coach. Only way, they have a login/password to use.

I was a semi-finalist last year, but my school's bio teacher "forgot" to register this year. Oh well...  :(

I got a 31...hopefully the semifinal exam is nice too.

I just took the semifinal exam. It was really hard, much worse than last year.

As not all students have taken the Semifinal exam,  (some take it later because they have spring break this week) please avoid discussing particulars; it is important that the exam's integrity is maintained.

hrm... when is the mail-in deadline for the semi's, anyway?

Haha, I thought the semi-final exam was pretty hard.  Chance of moving on as top 20 is pretty much nil.

It was hard for me this year too. Unfortunately, I didn't study as much as last year so I'm not expecting anything great. I think the overall difficulty of the test was much greater than last year's.

I heard finalists have been notified.  :lol:  Good luck to all finalists!

I was [b][i]really[/i][/b] close to becomming a finalist this year... but i didn't make it.  :(
Congratulations to all finalists though!

I got a 102...I wonder how close that was to making finalist.

From what I've gathered from browsing CollegeConfidential, you needed a 103 to make finalist.

edit: Wait...just from looking at your sig, would you happen to be biomath2?

My teacher forgot to register in time for even the first test.  She told me that she was disappointed about that because I would have at least made semis.  Oh well...


salmon often jump waterfalls to reach their breeding grounds. One salmon starts 2m from a waterfall that is .55m tall and jumps at an angle of 32degrees. what must be the salmon's minimum speed to reach the waterfall?

i have no clue on how to find the initial velocity!! please help

[hide="hint"]
the initial speed should be such that at a certain time $ t$ the horizontal disp must be $ 2$ and the vertical height $ 0.55$ now frame two equations :) [/hide]

can u use the range equation (r=Vi^2sin(2X)/g)?

not exactly that but the projectile equation....do u know that?

um i dnt think so i m sryy its kinda like a self learning class so he doesnt really teach.

priya

have  a look at [url=http://en.wikipedia.org/wiki/Trajectory_of_a_projectile]this[/url]

have a look at the equation that is $ y$ in terms of $ x$ and initial speed
$ y \equal{}  x \tan \theta \minus{} \frac {gx^2}{2(v\cos\theta)^2}$
wehere $ v$ is the initial speed.
now we know $ y\equal{}0.55$ $ x\equal{}2$ and $ \theta\equal{}32$....using that get $ v$

dats a bit complicated can you like use regular equations to solve it?


Two strips of width $ n$ are cut off an 8-meter-by-10-meter rectangle, one off the width and off the length. How wide, in meters, should the strips be so that the area of the remaining rectangle is one-tenth the area of the original rectangle?

The original area is $ 8\times10\equal{}80$. We want it to be $ \frac{80}{10}\equal{}8$. We could just guess and check to get an answer of $ \boxed{6}$, or use a more advanced approach.
Let the width of the cut be $ w$. Now $ (8\minus{}w)(10\minus{}w)\equal{}8\implies w^2\minus{}18w\plus{}80\equal{}8\implies w^2\minus{}18w\plus{}72\equal{}0$ Factoring gives $ (w\minus{}6)(w\minus{}12)\equal{}0$, so it looks like there are $ 2$ solutions! But there aren't, think about why ;). The only solution is $ \boxed{6}$

$ 80 \minus{} n(10 \minus{} n) \minus{} n(8 \minus{} n) \minus{} n^2 \equal{} 8$
$ n^2 \minus{} 18x \plus{}72\equal{}0$

n is positive, so n=6.


find all pairs of positive integers $(n,k)$, which satisfies:
${n \choose k}=k^3+1$

[hide]The only solutions are $(n, k) = (n, 0), (2, 1), (9, 5)$ and $(14, 10)$.
For $k = 0$, it is clear any $n$ suffices.  For $k = 1$ we have ${n \choose 1} = 2$ so $n = 2$.  For $k = 2$ we have ${n \choose 2} = 7$ which is readily seen to have no solutions.
$(k^3 + 1) - {k + 3 \choose k} = \frac16(5k^3 - 6k^2 - 11k) = \frac 16 k(k + 1)(5k - 11)$, so $k^3 + 1 > {k+3 \choose k}$ for $k \geq 3$.  Thus, we must have $n > k + 3$ when $k \geq 3$.  Also note that
${k + 4 \choose k} - (k^3 + 1) = \frac1{24}(k+1)((k+4)(k+3)(k+2)-24(k^2-k+1)) = \frac1{24}k(k+1)(k-5)(k-10)$
This final expression is positive when $k > 10$, so in those cases we must have $n < k + 4$.  Combining this result with that of the previous line, there are no solutions for $k > 10$.  We also note from this factorization that $k = 5, 10$ give us solutions.  The only cases left to check are $5 < k < 10$, and in these it suffices to note that ${11\choose6} > 6^3+ 1, {12\choose 7} > 7^3 + 1, {13\choose 8} > 8^3 + 1$ and ${14\choose 9} > 9^3+1$, so there are no solutions in these cases, either.[/hide]


limit
logarithms
calculus
calculus computations

Prove that $ \lim_{x \to 0} x^{x} \equal{} 1$.

hello, rewriting $ x^x$ in the following form
$ x^x = e^{\frac {\ln(x)}{\frac {1}{x}}}$ and applying L'Hospitals rule we get
${ e^{\lim_{x \to 0 + }\frac {\frac {1}{x}}{{\frac { - 1}{x^2}}}} = e^{\lim_{x \to 0 + } - x}} = e^{0} = 1$
Sonnhard.

Thanks Sonnhard. (You've written $ x$ instead of $ e$ just after the $ \equal{}$ sign in your second line.)

I was hoping for something without l'Hospital's rule (spelling your way, as the circumflex, which I can't do, indicates that "s" was once there).

hello, thank you for your hint, i hope it is correct now.
Sonnhard.

Does this means proving the limit exist? I hesitated as I was not sure if the limit even exists or not.

Hi KenHungKK. I guess so. Do you have a proof (assuming that the limit exists)?

Hi Sonnhard. I can't see what you've edited but your typo is still there.

$ x^{x} \equal{} e^{\ln x^{x}} \equal{} e^{x \ln x} \equal{} e^{\frac{\ln x}{\frac{1}{x}}}$.

So, $ x^{x} \equal{} e^{\frac{\ln (x)}{\frac{1}{x}}}$; on your second line you've written $ x^{x} \equal{} x^{\frac{\ln (x)}{\frac{1}{x}}}$.

[u]Another Way:[/u] $ \lim_{x \to 0^ \plus{} } x^x \equal{} \lim_{x \to 0^ \plus{} } e^{x \ln x} \equal{} e^0 \equal{} 1$

hello, but here one must know that $ \lim_{x \to 0\plus{}}x\ln(x)\equal{}0$.
Sonnhard.

And how do we show that $ \lim_{x \to 0^{\plus{}}} x \ln x \equal{} 0$?

How about first showing that $ \lim_{x \to \infty} \frac{1}{x} \ln x \equal{} 0$?

Let $ x\equal{} e^{t}$. Then $ \frac{1}{x} \ln x \equal{} e^{\minus{}t}\cdot t \equal{} te^{\minus{}t} \to 0\  (t \to \infty)$

Now let $ x \equal{} \frac{1}{u}$. Then $ u \ln u^{\minus{}1} \equal{} \minus{}u\ ln u \to 0 \ (u \to 0^{\plus{}})$.

Therefore $ x \ln x \to 0 \ (x \to 0^{\plus{}})$.