Caelestor is working in the chemistry lab with a classmate of his, Millinova. Today's assignment requires analyzing the Pylumilt fungus, a potentially volatile species. After an hour of observation, Millinova discovers that the mass of the sample in kilograms can be measured in the Pylumilt function P(t), where t is in seconds:

$ 1250\displaystyle\sum_{k = 1}^\text{t} \frac {k}4 (\frac {3}8)^{k - 1}$.

Caelestor then notices that the small container will most likely shatter if the fungus continues to grow at this rate, so he tells Millinova to obtain a new container to store the specimen.

4. If the new container can hold a substance with a density of 2 kg/L (i.e. a 2 L container can hold up to 4 kg of stuff), what is the minimum volume that the container should be, so that the amorphous fungus will never be able to break its container?

Millinova then finds that adding a sample of Irdivos fungus to the Pylumilt fungus creates a certain reaction, in which at a moment t seconds, the ratio of Irdivos to Pylumilt is designated by the relation ${ a_{t + 1} = a_t + 3t^2}$.

5. Given $ a_1 = \frac {4}9$, $ a_{20}$ can be expressed as $ m/n$, where m and n are relatively prime. What are the last 3 digits of m?

6. With the 8 species of fungus in the lab, Caelestor wants to throw at least 2 of them into the pot and record their reactions. However, he knows that mixing the Beta and Omega samples together could be hazardous. Thus, how many possible safe combinations of reactants can Caelestor create (this includes the one mentioned in the previous problem)?

4) first, find a better representation for $ P(t)$.

$ P(t) \equal{} 1250\sum_{k \equal{} 1}^t \frac {k}{4} (3/8)^{k \minus{} 1}$
$ 3/8 P(t) \equal{} 1250\sum_{k \equal{} 1}^t \frac {k}{4}(3/8)^k$

but we also have
$ P(t) \equal{} 1250\sum_{k \equal{} 0}^{t \minus{} 1} \frac {k \plus{} 1}{4} (3/8)^k$
$ P(t) \minus{} 3/8P(t) \equal{} 1250\sum_{k \equal{} 0}^{t \minus{} 1} \frac {k \plus{} 1}{4}(3/8)^k \minus{} 1250\sum_{k \equal{} 1}^t \frac {k}{4}(3/8)^k$
$ 5/8 P(t) \equal{} 1250(1/4 \minus{} t/4 (3/8)^t \plus{} 1/4\sum_{k \equal{} 1}^{t \minus{} 1} (3/8)^k)$
$ 5/8 P(t) \equal{} 1250(1/4 \minus{} t/4(3/8)^t \plus{} 3/32\frac {1 \minus{} (3/8)^{t \minus{} 1}}{1 \minus{} 3/8})$
$ P(t) \equal{} 2000(1/4 \minus{} t/4(3/8)^t \plus{} 3/20 \minus{} 3/20(3/8)^{t \minus{} 1})$
$ P(t) \equal{} 2000(2/5 \minus{} (3t/32 \plus{} 3/20) (3/8)^{t \minus{} 1})$

now $ \lim_{t \to \infty} 2000(2/5 \minus{} (3t/32 \plus{} 3/20) (3/8)^{t \minus{} 1}) \equal{} 800$ (rewrite $ (3/8)^{t \minus{} 1}$ as $ 1/(8/3)^{t \minus{} 1}$ and then l'hopital's rule shows the second part limits to zero)

so the new container must be able to hold $ 800$ kg, so it must be $ 400$ Liters.

5) $ a_{t \plus{} 1} \equal{} a_t \plus{} 3t^2$ so $ a_{t \plus{} 1} \minus{} a_t \equal{} 3t^2$.  Now $ \binom{t}{2} \equal{} \frac {t(t \minus{} 1)}{2}$ and $ \binom{t}{1} \equal{} t$ so $ 3t^2 \equal{} 6\binom{t}{2} \plus{} 3\binom{t}{1}$.  So we have $ a_k \minus{} a_1 \equal{} \sum_{t \equal{} 1}^{k \minus{} 1}a_{t \plus{} 1} \minus{} a_t \equal{} \sum_{t \equal{} 1}^{k \minus{} 1} 6\binom{t}{2} \plus{} 3 \binom{t}{1} \equal{} 6\binom{k}{3} \plus{} 3\binom{k}{2}$ by hockeystick.

so $ a_k \equal{} k(k \minus{} 1)(k \minus{} 2) \plus{} 3/2k(k \minus{} 1) \plus{} 4/9$ so $ a_20 \equal{} 6840 \plus{} 570 \plus{} 4/9 \equal{} 7414/9$ so the last three digits of M are $ 414$

6) all combinations but 1 are safe, so there are $ \binom{8}{2} \minus{} 1 \equal{} 8*7/2 \minus{} 1 \equal{} 27$ safe combinations.

facis, please be more careful when adding an integer to a fraction. Also, remember that the last question allows for a combination of fungi, as long as the number is or exceeds 2.

hmmm... I guess I miscomputed in #5.  The ratio will be $ 66694/9$, so the answer is $ 694$.

As for #6, I think I just misread it.  If all combinations were safe, we would have $ 2^8\minus{}8\minus{}1\equal{}255$ possible combinations - for each of the $ 8$ fungus, we can either include or not include it, and subtract out the cases where only 1 or no fungus's are used.  But this is not the case.  For beta and omega, we can either include one of the fungus's or neither, but not both, so we have $ 3$ choices using just beta and omega.  Then we have $ 6$ remaining fungi which we can either throw in or not, for a total of $ 3*2^6 \minus{} 8 \minus{} 1$ - just as before, we subtract out combinations with only 1 fungus and the combination with no fungus's used.  This gives an overall total of $ 183$ safe combinations of 2 or more fungi.


[quote="jackdillon"]X=5 because 5-3=2  and 5-2=3.  3+2=5[/quote]

You forgot about $x = 0$.


logarithms
ratio
number theory proposed
number theory

According to Bertrand's postulate, there is always at least one prime between $n$ and $2n$ where $n \in \mathbb{Z}_+$. But how small can number $c$ be so that there are always prime between $n$ and $cn$ for all $n$ "sufficient large"? At least numbers between $n!+2$ and $n!+n$ are composites, but on the other hand
\[\frac{n!+n}{n!+2}\to 1\]
as $n\to \infty$.

It's true for all $c>1$, by the prime number theorem: $\pi(n)\sim\frac n{\log n}$. From here we can conclude that as $n\to\infty$, the ratio $\frac{\pi(cn)}{\pi(n)}$ approaches $c>1$, so it's certainly true that for large enough $n$ we have $\pi(cn)>\pi(n)$.

Indeed, from the prime number theorem we have that for an $\epsilon > 0$, $\pi(x+\epsilon x)-\pi(x)=\frac{x+\epsilon x}{\log(x)+\log(1+\epsilon) } - \frac{x}{\log(x)} + o( \frac{x}{\log(x)}) = \frac{\epsilon x}{\log(x)} + o( \frac{x}{\log(x)})$. This last expression is positive provided that x is big enought. So there is always a prime p such that $x<p<(1+\epsilon )x$ for all big x.


The H isn't moving anymore; now it's rotating :-)

you've got three identical bars and you put them together to form the letter H. Now you put the H-structure in horizontal position and then let it rotate, making one of the two parallel bars remain fixed on its original axis. Each bar's lenght is $l$. Find the angular speed $\omega$ when the H-structure is in vertical position.

I apologize for my horrible english; if you have doubts about the text, please ask me.

[quote="bus"]Is there still any el/mag field?[/quote]

No, the only field is gravity.

Then it should be easy, you only use energy. Is the right answer $\sqrt{\frac{12g}{5l}}$?

Uhm, i got something different... $\frac{3}{2} \sqrt \frac{g}{l}$ i used energy too

$3mg\frac{L}{2} = \frac{1}{2} (mL^2 + \frac{1}{3} mL^2) \omega^2$ I get Bacco's answer?


An unfair coin lands on heads with probability $ \frac35$, on tails with probability $ \frac15$, and on its edge with probability $ \frac15$.  If it comes up heads, I win 4 dollars.  If it comes up tails, I lose 1 dollar.  But if it lands on its edge, I lose 10 dollars.  What is the expected winnings from one flip?  Express your answer as a dollar value, rounded to the nearest cent.

We have $ \frac{3}{5}$ probability that it's heads, so $ \frac{12}{5}$ expected winnings from heads.  For tails, its' $ \frac{\minus{}1}{5}$, and for edge, its $ \minus{}2$.  Adding yields $ \frac{1}{5}\equal{}0.2$.

[quote=davidninjaking]What is the difference between the probability of landing heads and the expected winnings from heads?[/quote]

The probability times money


AMC
USA(J)MO
USAMO
AIME
AMC 10
AMC 10 A
geometry

I created this poll just to get an idea of how many potential points people were losing because of stupid mistakes.  To calculate the "Silly Mistake Index," add the amount of points you lost on the AMC to ten times the amount of points you lost on the AIME.  For example, if your USAMO index is 200, but could have been 240, then your Silly Mistake Index is 40.
Use your best judgment to determine what qualifies as a stupid error.

well 0 since nothing was stupid.

Although I solved couldve solved 3 more AIME problems had I tried them (i didnt look at 12-15, 12 and 14 were fairly understandable and easy to do) and #10 after looking at the solution...and number 25 on the AMC 10, BUT those problems arent qualified as stupid errors, its just I didnt get it during the test...even though the solution was trivial
(i consider stupid=careless only NOT if you look at the solution after the test and think "wow i couldve done that")

I'd have got about 46 to 56 points higher if I simply spent more time checking. I'm kinda too impatient and excited and handed in both the AMC10A and the AIME early. I'm always excited with math competitions, and the excitement often makes me want to finish early.

0  :)

I skipped a lot of easy problems though...  :(

141/141 --> no stupid mistakes on 12A
7/15 --> 8 stupid mistakes on AIME

Change in USAMO Index: 80

Here, I define a stupid mistake as an error in adding, subtracting, multiplying, or dividing a pair of closed-form integers.

[quote="The Zuton Force"]
Here, I define a stupid mistake as an error in adding, subtracting, multiplying, or dividing a pair of closed-form integers.[/quote]
I personally would also include misbubbling, forgetting that $ n$-digit numbers cannot begin with a zero, concluding that $ a_0, a_1, \dots , a_{999}$ consists of 999 terms, forgetting to divide by 2 when using triangle area formulas, etc.

Using the more strict definitions of "dumb mistake" posted above, my AMC 12A - 10.5 > AIME - 10.

12 + 30 = 42.

On the 12B, I got both questions 1 and 2 wrong and everything from 3-20 correct.  :huh:
On the AIME, I had two off-by-one errors (the one with counting squares and the one with the number of cars passing the camera); also, I took the altitude to the wrong side of the square on the second question (and obtained 15 instead of 25).

$ \fbox{0}$, actually.
I mean, technically, I [i]could've[/i] gotten #25 on the 10B and #10 on the AIME I in the time given, but it wasn't because I was making dumb mistakes as defined above; more like me kinda wasting time on futile strategies.

Three questions on the AMC 12A (the percentages one, the letterboxing one, the quadrilateral-in-the-complex-plane one), one question on the AIME (the triangular array).

$ 18\plus{}10\equal{}\boxed{28}$

No silly mistakes for me, though I was dumb on comprehending some problems.

A good solid $ \boxed{0}$.

You should have some negative numbers in the poll, just in case someone got points they didn't deserve (i.e. guessing)

Had I taken AIME I and done exactly as I did when I did it for practice, I would have a silly mistake index of 10. On #10, I mixed up EA and ED and so E was on the wrong side of the trapezoid. Then again, that would have been the difference between 280 and 290, so it wouldn't have mattered much.

20 on the AIME, 10+ or so on the AMC.

Although in retrospect...my mistakes on the AIME were pretty funny.
I took the AIME II, so for number 10 (problem dealing with the paths), I didn't calculate all the possible lengths systematically, so I forgot two was a possible segment length. So I did the problem right...but I did the wrong problem. For number 11, I miscalculated the last step (solving the quadratic), and I even knew what I did wrong, but by the time I realized it there were 5 min remaining and my answer was written on scratch work that I had so foolishly tossed into the trashcan behind me.... If I had remembered the actual answer (and not the a+bc junk) I could have gotten it in the time remaining.

Shame...I tossed my chances of making USAMO...quite literally.

Missed #1 on the AIME

I lost my chance, too...

0+0=0 because I didn't take the 2008 AMC

0+0=0 because I wasn't born for the 2008 AMC

[size=50]why da bump doe[/size]

I got 2 wrong cuz of a silly mistake, and I'm going to go to 6th grade. I took printed AMC.

[quote=Sw1Ft_A][quote=MONTGOMERY_BLAIR_STUDENT][quote=eduD_looC]bump bump[/quote]

May I inquire what the purpose of your post is?

Sent from Montgomery Blair HS[/quote]

lol[/quote]

I return my question to you...

Sent from Montgomery Blair HS

By the way, you don't need to type your name in at the end, but carry on doing so if you want to.

[quote=OlympusHero][quote=eduD_looC][quote=OlympusHero]Guys, what is going on here? Why is this bumped? Please act maturely, thanks.[/quote]

-_- Just wanted to see some more results.[/quote]

This is for [b]2008[/b]. If you want to see results from a newer test, please make a new thread.[/quote]

ok..... i asked

back to the point, the larger the sample size the better ;-;

the fact that this thread is older than me and still alive is XD


Coloring
graph theory
Extremal combinatorics
combinatorics
IMO
IMO 1990

Let $ n \geq 3$ and consider a set $ E$ of $ 2n \minus{} 1$ distinct points on a circle. Suppose that exactly $ k$ of these points are to be colored black.  Such a coloring is [b]good[/b] if there is at least one pair of black points such that the interior of one of the arcs between them contains exactly $ n$ points from $ E$.  Find the smallest value of $ k$ so that every such coloring of $ k$ points of $ E$ is good.

We form a graph on the $2n-1$ points by connecting two of these points iff one of the arcs they determine has exactly $n$ points in its interior. If $3\not|2n-1$, then this graph is a cycle, and the question becomes: what's the minimal $k$ s.t. whenever we color $k$ vertices of a cycle of length $2n-1$, there are two adjacent colored vertces? Obviously, the answer is $k=n$. On the other hand, when $3|2n-1$, the graph is a disjoint union of three cycles of length $\frac{2n+1}3$, and the answer will be, in this case, $3\cdot\left\lfloor\frac{2n-1}6\right\rfloor+1$.

Take a look at this closed formula, which seems to confirm every such k.

$k_{min}=\frac{2n-1- gcd(2n-1,n-2)+2}{2}$

I think I have a correct proof for the above formula, which is rather long but still uses the same elements of the very nice solution provided by grobber

(graph theory, that is to say!)

Cheers,
Nick

You can easily see that the closed formula is the same as that of the first post (by Grobber) once you note that gcd (2n-1,n-2) = 1 or 3. It is 3 iff n is 2 mod 3.

We are considering the graph on these $2n-1$ vertices and we will join $2$ vertices iff  there are exactly $n$ vertices between them (in any one of the 2 arcs).So in our graph every point has degree $2$. So it is the union of disjoint circle.
 Now if we mark the vertices $1,2,3,..2n-1$ then $1$ will be connected to $n+2$ $n+2$ will be connected to $2n+3$ i.e. $4$.So $1,4$ will in the same cycle.
If $2n-1$ is not divisible by 3 the graph consists of only one circle.So the maximum number of independent set is $(2n-1)-1/2=n-1$.So if we colour $n$ vertices then the graph will have at least 2 consecutive coloured vertices and $k=n$ in this case.
If $2n-1$ is divisible by 3 then the graph is union of 3 disjoint circles each of length $2n-1/3$.So the maximum number of independent set in each cycle is $f(n)=\frac{\frac{2n-1}{3} -1}{2}$.So total number of independent set is $3f(n)=n-2$ So in this case we need to colour $n-1$ vertex.


geometry
algorithm
graph theory
\/closed

What do you think of having a forum (subforum/category whatever it may be) for computer algebra systems? A place where people working with mathematica, maple, etc can come find advanced syntax help. I've seen many posts in the improper area about (mathematica) syntax, but these don't exactly fit into the CSI & Informatics category.

I'm more than aware that mathematica has a detailed documentation, but sometimes human assistance can be a whole new kind of help. For instance, I can't figure out how do increase the resolution on a fractal image in mathematica and the online info isn't really helping.

[quote="Dojo"]Is this what you are looking for?: [url]http://www.artofproblemsolving.com/Forum/index.php?f=331[/url] (I really don't know)[/quote]

Yes, correct opinion that you don't know!  :D I think the [url=http://www.mathlinks.ro/index.php?f=331]Computer Science and Informatics[/url] forum is supposed to discuss cs competitions at the school and university level and the problems those competitions ask which are mainly based on the creative applications of algorithms, data structures, graph theory etc.

What the original posters aims at is help with tools, i.e. with computer algebra systems, but this forum may also be used to discuss these things from other programming languages. I just do not know whether ML/AoPS is the correct place for that as there are mailing lists or newsgroups that already allow for that kind of discussion.


number theory
algebra
calculus
IMO Shortlist
quadratics

Consider the system \begin{align*}x + y  &= z + u,\\2xy & = zu.\end{align*} Find the greatest value of the real constant $m$ such that $m \leq x/y$ for any positive integer solution $(x,y,z,u)$ of the system, with $x \geq y$.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

I get $|x+yi|=|z-u\omega|$.
Can we solve the problem from this?

[quote="orl"]Consider the system $ x + y = z + u,$ $ 2xy \& = zu.$ Find the greatest value of the real constant $ m$ such that $ m \leq x/y$ for any positive integer solution $ (x,y,z,u)$ of the system, with $ x \geq y$.[/quote]
[hide="Solution"]
We notice that, by the Four-Numbers Theorem, there exist integers $ a$, $ b$, $ c$, and $ d$ so that $ ab = x$, $ cd = y$, $ z = 2bc$, and $ u = ad$ ($ u$ could equal $ 2ad$ and $ z$ could equal $ bc$, but the cases are analagous.)  Then, we have that $ ab + cd = 2bc + ad$, so $ b(a - c) - d(a - c) = bc$, which means that $ (b - d)(a - c) = bc$, and by the Four-Numbers Theorem, we have that there exist integers $ k$, $ l$, $ m$, and $ n$ so that $ b - d = kl$, $ a - c = mn$, $ b = lm$, and $ c = kn$.  This yields that $ x = ab = (ml)(n)(k + m)$ and $ y = cd = l(m - k)kn$, meaning that $ \frac {x}{y} = \frac {mnl(k + m)}{knl(m - k)}$, so $ \frac {x}{y} = \frac {m(k + m)}{k(m - k)}$.  Clearly, the numerator is larger, so $ x\ge y$.  Yet, $ y > 0$, so $ m > k$.  Let $ m = k + t$.  This gives that $ \frac {x}{y} = \frac {(k + t)(2k + t)}{tk} = 3 + \frac {2k^2 + t^2}{tk}\ge 3 + 2\sqrt {2}$ the last by AM-GM.  $ \frac {x}{y}\ne 3 + 2\sqrt {2}$ since $ \sqrt {2}$ is irrational, but it is dense, so $ \frac {x}{y}$ can get as close as possible to $ 3 + 2\sqrt {2}$.  This seems a little large... I hope I made no mistakes  :read: [/hide]

i have another solution... note that $ x,y$ are the solutions to the equation $ a^2 \minus{} a(z \plus{} u) \plus{} \frac {zu}{2} \equal{} 0$..

calculating its discriminant it follows that it equals $ \Delta \equal{} z^2 \plus{} u^2$, so $ (z,u) \equal{} (m^2 \minus{} n^2,2mn)$ or $ (2mn,m^2 \minus{} n^2)$ for som positive integers $ m,n$, $ m > n$...

in any of the cases $ x \equal{} \dfrac{z \plus{} u \plus{} \sqrt\Delta}{2} \equal{} m^2 \plus{} mn$ and $ n \equal{} \dfrac{x \plus{} u \minus{} \sqrt\Delta}{2} \equal{} mn \minus{} n^2$... it follows that $ q \equal{} \dfrac{x}{y} \equal{} \dfrac{\left(\frac {m}{n}\right)^2 \plus{} \left(\frac {m}{n}\right)}{\left(\frac {m}{n}\right) \minus{} 1}$...

note that $ q\geq (1 \plus{} \sqrt2)^2$  is equivalent to $ \left(\frac {m}{n} \minus{} (1 \plus{} \sqrt2)\right)^2\geq 0$ which is true... if $ m \equal{} (1 \plus{} \sqrt2)^2 \plus{} \epsilon$ for some $ \epsilon > 0$ we would have that $ \dfrac{\left(\dfrac{m}{n} \minus{} (1 \plus{} \sqrt2)\right)^2}{\frac {m}{n} \minus{} 1}\geq \epsilon$...

however we can take a sequence of rationals that converges to $ 1 \plus{} \sqrt2$ and thereferore the last inequality is not satisfied by every rational number $ \frac {m}{n}$ and we're done...
 :D

Does anyone find two numbers  $ m,n$ for which $ \frac{x}{y}<5.9$?

[b]Official Solution:[/b]

Squaring the first equation and then subtracting four times the second, we obtain $x^2-6xy + y^2 = (z-u)^2$, from which $(x/y)^2-6(x/y)+1=((z-u)/y)^2$. The quadratic $w^2-6w+1$ takes the value $0$ for $3\pm2\sqrt2$, and is positive for $w>3+2\sqrt2$. Because $x,y\geq1$ and the right side of $(*)$ is a square, the left side of $(*)$ is positive, and we must have $x/y>3+2\sqrt2$.

We now show that $x/y$ can be made as close to $3+2\sqrt2$ as we like, so the desired $m =3+2\sqrt2$. We prove this by showing that the term $((z-u)/y)^2$ in $(*)$ can be made as small as we like.

To this end, we first find a way to generate solutions of the system. If $p$ is a prime divisor of $z$ and $u$, then $p$ is a divisor of both $x$ and $y$. Thus we may assume, without loss of generality, that $z$ and $u$ are relatively prime. If we square both sides of the first equation, then subtract twice the second equation we have $(x-y)^2 = z^2 + u^2$. Thus $(z,u,x-y)$ is a primitive Pythagorean triple, and we may assume that $u$ is even. Hence there are relatively prime positive integers $a$ and $b$, one of them even and the other odd, such that $z = a^2-b^2; u = 2ab$; and $x-y = a^2 + b^2$. Combining these equations with $x + y = z + u$, we find that $x = a^2 + ab$ and $y = ab-b^2$.

Observe that $z-u = a^2-b^2 -2ab = (a-b)|^2 -2b^2$. When $z-u = 1$, we get the Pell equation $1 = (a-b)^2-2b^2$, which has solution $a-b = 3, b = 2$. By well known facts, this equation has infinitely many positive integer solutions $a-b$ and $b$, and both of these quantities can be made arbitrarily large. It follows that $y = ab-b^2$ can be made arbitrarily large. Hence the right side of $(*)$ can be made as small as we like, and the corresponding value of $x/y$ can be made as close to $3+2\sqrt2$ as we like.

[hide="Soln with andersonw"]
$x^2+2xy+y^2=z^2+2zu+u^2 \implies (x-y)^2=z^2+u^2$

As these are integers, it is a pythagorean triple. If it is nonprimitive, we can divide through by the GCD, so WLOG, it is a primitive triple.

Then, $z=2ab$ , $u=a^2-b^2$ , $x-y=a^2+b^2$

$x+y=x-y+2y=a^2+b^2+2y=2ab+a^2-b^2 \implies y=ab-b^2$

Similarly, $x=a^2-ab$

We are then seeking the minimum value of $\frac{x}{y}=\frac{a^2-ab}{ab-b^2}$

Using a method of one's choice (partial derivative bashing works nicely), we see that the minimum for real $a,b$ occurs when $a=(1+\sqrt2)b$ , giving a value of $3+2\sqrt{2}$ Applying an obvious density argument, $\frac{x}{y}$ can get arbitrarily close, so that's our value.

[/hide]

[quote="Mewto55555"]

Then, $z=2ab$ , $u=a^2-b^2$ , $x-y=a^2+b^2$

$x+y=x-y+2y=a^2+b^2+2y=2ab+a^2-b^2 \implies y=ab-b^2$

Similarly, $x=a^2-ab$

[/quote]

It is a simple solution, but there is a mistake, we get $x=a^2+ab$ for having $x+y=u+z=a^2+2ab-b^2$

Now $ \frac{x}{y}=\frac{a^{2}+ab}{ab-b^{2}} $ is homogenous and so
$b=1$ and $a^2+(3+2\sqrt{2} \ge 2(1+\sqrt{2})a$ of AM-GM.
Hence the solution was true, with those mistakes.

[quote="eraydin"]I dont understand how can we get arbitrarily close to  $ 3+2\sqrt2 $. Somebody can explain?[/quote]

Take $a=[(1+\sqrt{2})b]$ where $[]$ is to take the greatest smaller integer (entier)

For $b$ very high, it works very good, for $10^5$ we get
$x=82426406871,y=14142135624$ for ex. and that is a good thing.

[hide = Solution]
I claim that $m = 3+2\sqrt{2}$.

Let $x+y=z+u=p$ and $2xy=zu=2r$ for some positive integers $p,r$. $x$ and $y$ are the roots of the equation $t^2-pt+r=0$, and $z$ and $u$ are the roots of the equation $t^2-pt+2r=0$. For these roots to be integers, $p^2-4r$ and $p^2-8r$ must both be perfect squares. Let $p^2-4r=h^2$ and $p^2-8r=k^2$; then $\frac{x}{y} = \frac{p+h}{p-h} = 1+\frac{2h}{p-h}$.

Observe that the two equations above give $p^2+k^2=2h^2$, with $p>h>k$. From this we deduce that $p<h\sqrt{2}$ and so $1+\frac{2h}{p-h}>1+\frac{2h}{h\sqrt{2}-h} = 3+2\sqrt{2}$, so $\frac{x}{y}\ge 3+2\sqrt{2}$.

Now I show that $\frac{x}{y}$ can get arbitrarily close to $3+2\sqrt{2}$. Let $k=1$; now as long as $p,h$ are both odd we can find an $r$ that will satisfy our constraints. In particular, any $p,h$ satisfying $p^2-2h^2=-1$ will give us what we want. But notice that we can take the sequence $u_{k+1}+v_{k+1}\sqrt{2} = \left(u_k+v_k\sqrt{2}\right)(3+2\sqrt{2})$ with $u_0=v_0=1$ to generate the family of integral solutions to this Pell equation. This recursion simplifies to $$u_{k+1}=3u_k+4v_k,v_{k+1}=2u_k+3v_k$$ from which it follows that $u_k = \frac{v_{k+1}-3v_k}{2}$ and $v_{k+2}=6v_{k+1}-v_k$. Now $v_k$ takes the general form $c_1(3+2\sqrt{2})^k+c_2(3-2\sqrt{2})^k$ for some constants $c_1, c_2$. In particular, as $k$ grows large $v_k \approx c_1(3+2\sqrt{2})^k$ since $3-2\sqrt{2}<1$.

It will suffice to show that $\lim_{k \to \infty} \frac{2v_k}{u_k-v_k} = 2(\sqrt{2}+1)$. Plugging in the information above, this is
$$\lim_{k\to \infty}\frac{4v_k}{v_{k+1}-5v_k}$$
$$=4 \lim_{k\to \infty}\frac{1}{3+2\sqrt{2}-5}$$
$$=\frac{4}{2\sqrt{2}-2} = 2(\sqrt{2}+1)$$ as desired. Therefore $\frac{x}{y}$ is always greater than $3+2\sqrt{2}$ but can get arbitrarily close to it, so $m=3+2\sqrt{2}$.
[/hide]

[hide="solution"]
$xy=\frac{z}{2}\cdot u\implies (x, y, z, u)=(ab, cd, 2bc, da)$
$x+y=z+u\implies ab+cd=2bc+da\implies (a-c)(b-d)=bc$
$\implies (a-c, b-d, b, c) = (ij, kl, jk, li)\implies (a, b, c, d) = (i(j+l), jk, li, k(j-l))$
$\dfrac{x}{y}=\dfrac{ab}{cd}=\dfrac{ijk(j+l)}{ikl(j-l)}=\dfrac{j(j+l)}{l(j-l)}=3+\dfrac{j-l}{l} + \dfrac{2l}{j-l}\stackrel{\text{AM-GM}}{\ge} \boxed{3+2\sqrt{2}}$
Taking $j=\lfloor (\sqrt{2}+1)l\rfloor$ as $\ell\to\infty$ brings $\dfrac{x}{y}$ arbitrarily close to this constant, so it is the best possible. $\Box$
[/hide]

Please check my (bit different) solution :help:

[hide=Proof] Let the lower bound be $l$.

[b]Claim=0[/b]: $f(x,y) := 2(x-y)^2 -(x+y)^2$ is a perfect square greater than $0$.
[hide = Proof] WLOG assume $z \geq u$. So, $z-u = \sqrt{(z+y)^2-4zy} = \sqrt{(x+y)^2-8xy} = \sqrt{2(x-y)^2+(x+y)^2} \geq 0$. Now, if that term is not a perfect square, then atleast one of $z,u$ is noninteger, which contradicts their integer-ness, so $f(x,y) := 2(x-y)^2 -(x+y)^2$ is a perfect square greater than $0$ [/hide]

[b] Claim - 1 [/b] The recurrence equation is defined by $a_0 = 1, a_1 = 6, a_i = 6a_{i-1} - a_{i-2}$. Then  $f(a_i, a_{i-1}) = 1$ for all $i$, and $a_i+a_{i+1} \equiv 1 (mod 2)$.
[hide = Proof] The base case $i=1$ is trivial to check, and $f(x,y)$ is invariant under the transformation $x \rightarrow 6x-y$ and $y \rightarrow x$, so it holds for all $i$.   [/hide]

[b]Claim-2[/b] $l = 3+2\sqrt{2}$.
[hide=Proof] As shown before $f(x,y) \geq 0 \Rightarrow \sqrt{2}(x-y) \geq x+y \Rightarrow \frac{x}{y} = \frac{\sqrt{2}+1}{\sqrt{2}-1} = 3 + 2\sqrt{2} \Rightarrow l \geq 3 + 2\sqrt{2} $. Now, $(x,y,z,u) = (a_i,a_{i-1}, \frac{a_i + a_{i-1} + 1}{2},\frac{a_i + a_{i-1} - 1}{2} )  $ in the previous recurrence relation produces a working pair of triplets, as both $z+u = a_i + a_{i-1}$ and $z-u = f(a_i, a_{i-1}) = 1$ are integers and of the same parity.

As it's well known that for the recurrence relation given in Claim-1 $\lim_{i \to \infty} \frac{a_i}{a_{i-1}} = 3 + 2\sqrt{2}$, no tighter bound is possible. [/hide]

Here is a solution with the slope trick :)

In fact we will go further and characterize all solutions $(x,y)$; notice that if we scale $x,y,z,u$ all by the same constant, both equations still hold, so assume that we are working in $\mathbb Q^+$ and scale so that $x+y = z+u$, and $2xy=zu$; let $x+y = z+u = 1$, and $2xy=zu = 2p$; then, we see that \[
1-8p = a^2, 1-4p = b^2,
\]for squares $a,b$ by considering the discriminants of $w^2-w+p$ and $w^2-w+2p$, which are squares of rational numbers as the quadratics have rational roots; so, $8p\leq 1$; this implies that \[
y \leq \frac{1-\sqrt{1-4p}}{2} \leq \frac{1}{4}(2-\sqrt 2),
\]so \[
4+2\sqrt 2 \leq \frac{1}{y} = 1+\frac{x}{y},
\]so $\frac{x}{y} \geq \boxed{3+2\sqrt 2}$. Now, in order to show that $\frac xy$ can get arbitrarily close, it suffices to show that $a$ can get arbitrarily close to $0$ in the equation $2b^2-a^2 = 1$, where we take $(a,b)$ over $\mathbb Q^2$. As long as this holds, we can get $(1-a^2) = 2(1-b^2)$, so setting $8p = 1-a^2$ works, and since $a\to 0$, $p \to \frac 18$ and $y\to \frac 14(2-\sqrt 2)$, which will imply the desired.

Let $\mathcal C$ denote the locus of $\{(a,b)\colon 2b^2=1+a^2\}$, then $O = (1,1)\in \mathcal C$. Furthermore, for any other point $P$ on $\mathcal C$, notice that $P$ is rational if and only if $PO$ has rational slope! So, let this slope be $s$, and intersecting we have $1+a^2 = 2b^2$, $(1-b) = s(1-a)$; since $(1-a^2) = 2(1-b^2)$, this implies $(1+a) = 2s(1+b)$. But solving yields \[
a = \frac{2s^2-4s+1}{2s^2-1},
\]so taking $s$ to be an arbitrarily good approximation of $1+\frac{1}{\sqrt 2}$ is sufficient.

Here is a solution with the chain rule.

[hide = solution] We square both sides of $x+y=z+u$ to obtain that $x^2 + 2xy + y^2 = z^2 + 2zu + u^2$. Subtracting two times the second equation from both sides, we find that $$(x-y)^2 = z^2 + u^2.$$ By the properties of Pythagorean triples, there exist $k$, $m$, and $n$ with $m> n$, $m$ and $n$ coprime, and at least  one of $m$ and $n$ divisible by $2$ such that $k(2mn) = u$, $k(m^2 - n^2) = z$, and $k(m^2 + n^2) = |x-y|$. Because $x \ge y$, this gives that $k(m^2 + n^2) = x-y$. Thus, the information that $u+z = x+y$ is equivalent to $$k(2mn + m^2 - n^2) = x-y + 2y = k(m^2 + n^2) + 2y.$$ Simplifying, this gives $$k(mn - n^2) = y.$$ Using this and our information about $x-y$, we see that $x = k(m^2 + mn)$. We therefore have that $$\frac{x}{y} = \frac{m^2 + mn}{mn - n^2} = \frac{(m+n)m}{(m-n)n} = \frac{m+n}{m-n} \cdot \frac{m}{n} = $$ $$\left(1 + \frac{2n}{m-n}\right) \cdot \frac{m}{n} = \frac{m}{n} + \frac{2m}{m-n} = m \left(\frac{1}{n} + \frac{2}{m-n}\right).$$ Letting $\frac{n}{m} = p$ and $m$ be constant yields that $$\frac{x}{y} = m \left(\frac{1}{pm} + \frac{2}{m - pm}\right) = \frac{1}{p} + \frac{2}{1-p}.$$ Differentiating this to find the minimum as a function in $p$, we obtain that $$\left(\frac{1}{p} + \frac{2}{1-p}\right)' = \left(\frac{1}{p}\right)' + \left(\frac{2}{1-p}\right)' = $$ $$-\frac{1}{p^2} + 2 \cdot \left(\frac{1}{1-p}\right)'.$$ We apply the chain rule on $f(x) = 1/x$ and $g(x) = 1-x$ to obtain that $$\left(\frac{1}{p} + \frac{2}{1-p}\right)' = -\frac{1}{p^2} + 2 \cdot \left(\frac{1}{1-p}\right)' = $$ $$-\frac{1}{p^2} + 2 \cdot f(g(p))' = -\frac{1}{p^2} + 2 \cdot \left(-\frac{1}{(1-p)^2}\right) \cdot (-1) = $$ $$\frac{2}{(1-p)^2} -\frac{1}{p^2} = \frac{2p^2 - (1-p)^2}{p^2(1-p)^2} = \frac{p^2 + 2p - 1}{p^2(1-p)^2}.$$ For this to be $0$, we must therefore have that $0 = p^2 + 2p - 1 = (p+1)^2 - 2$, and thus $p = -1 \pm \sqrt{2}$. As $p$ is positive, we must have that $p = \sqrt{2} - 1$. We note that because every irrational number can be the limit of an infinite string of rational numbers, which we may choose such that they are all $\frac{n}{m}$ with $\mbox{gcd}(n,m) = 1$ and with at least one of $n$ and $m$ even, we can thus approach $\sqrt{2} - 1$ with a string of $p$'s, which must therefore approach $$\frac{1}{\sqrt{2} - 1} + \frac{2}{2 - \sqrt{2}} = \sqrt{2} + 1 + 2 + \sqrt{2} = 3 + 2\sqrt{2}.$$ By the definition of $p$ and the observation that for $p = \frac{1}{2}$ we have $\frac{x}{y} = 2 + 4 = 6 > 3 + 2\sqrt{2}$, $3+2\sqrt{2}$ must be our minimum $m$. $\fbox{}$ [/hide]

Right... this bound seems so random, but somehow it works- anyways, here's my solution-
First, notice that, as $x^2+y^2+2xy=z^2+u^2+2uz=>(x-y)^2=u^2+z^2$. As $\frac{(x-y)^2}2=\frac{z^2+u^2}2 =>x-y>\sqrt{2uz}$ (by RMS-GM) $=> x-y>2\sqrt{xy}=>x^2+y^2-6xy>0=>m^2-6m+1>0$ (here $m=\frac{x}{y}$)... hence $m>3+2\sqrt{2}$.

After I got here, I tried to find some nicer bound, because such an arbitrary value couldn't be the bound, right... well, not rather unexpectedly, I didn't get anywhere, and only when I tried to bash out the problem using the fact that $(u,z,x-y)$ is a Pythagorean triplet did I realize that somehow this bound must be the correct one.

After I realised that, it was just a matter of noticing that, as $x-y=m^2+n^2$ and $u,z=2mn,m^2-n^2$  (using Pythagorean triplet $(z,u,x-y)$ and $x+y=u+z$) $=> \frac{x}{y} =\frac{m(m+n)}{n(m-n)} =\frac{l^2+l}{l-1}$ where $l=\frac{m}{n}$- notice that we can take rational $\frac{m}{n}=l$ as close to $\sqrt{3+2 \sqrt{2}}=1+\sqrt{2}$ as we want from the right while still satisfying the conditions of the problem, we can bring $\frac{x}{y}$ as close to $3+2 \sqrt{2}$ as we want, and thus we're done.

[hide=Note] Really, for me, the hardest part of the problem was convincing myself as $3+2 \sqrt{2}$ was the correct bound... also, at least looking at the way most people solved it, doesn't this seem more of a real analysis problem than an NT one, or am I just deluded? [/hide]

[hide=Solution]
We will find the general solution to this system. Let
\[a=x+y=z+u \qquad \textrm{ and } \qquad b=2xy=zu\]
So \(x,y\) are roots to the quadratic equation
\[t^2 - at + b/2 = 0\]
and \(z,u\) are roots to the quadratic equation
\[t^2 - at + b = 0\]
Since the discriminants of both of these equations must be perfect squares, we get that
\[a^2 - 2b = k^2\]
\[a^2 - 4b = l^2\]
for some positive integers \(k,l\). Thus, we have
\[(2k)^2 = (a-l)^2 + (a+l)^2\]
Now we can make use of the general solution of the Pythagorean equation to find that
\[(a,b)=\left(\frac{p(m^2 + 2mn - n^2)}{2}, \frac{p^2mn(m+n)(m-n)}{2}\right)\]
for some positive integers \(p,m,n\), where \(\gcd(m,n)=1\). So we can check that
\[(x,y) = \left(\frac{p(m^2 + mn)}{2}, \frac{p(mn-n^2)}{2} \right)\]

This is the general solution, and hence we obtain
\[\frac{x}{y} = \frac{m^2 + mn}{mn - n^2} = 3 + (r-1) + \frac{2}{r-1}\]
where \(r=m/n\). By AM-GM this expression is at least \(3 + 2\sqrt{2}\). Since we can get \(r\) as close we want to \(1+\sqrt{2}\) (which is the equality case), we are done.
[/hide]

The answer is $\boxed{3+2\sqrt{2}}.$\\\\ $\emph{Proof of Bound: }$ Let $z=x-d$ and $u=y+d.$ Then, the second equation yields $$2xy=xy+dx-dy-d^2$$$$\implies (d+x)(d-y)=2d^2.$$ Write $x=da$ and $y=db$ for rational $a,b.$ We have $$(1+a)(1-b)=2\implies a=\frac{1+b}{1-b}.$$ Hence, $$\frac{x}{y}=\frac{a}{b}=\frac{1+b}{b(1-b)}.$$ Now it is easy to check that this expression has a minimum of $3+2\sqrt{2}$ on the interval $(0,1).$\\\\ $\emph{Proof of Maximality: }$ Fix a positive integer $k,$ and let $m=\lfloor(10^k(2+\sqrt{2})).$ Check that $$(d,x,y)=(10^km, m^2-10^km, m-2\cdot 10^{2k})$$ satisfies the equation.\\\\ Moreover, as $k$ becomes large, $\frac{x}{y}$ becomes arbitrarily close to $3+2\sqrt{2}.$ Thus, our bound is the best possible.

The answer is $3+2\sqrt{2}$. Note that $$xy = \frac{zu}{2} \le \frac{(\frac{z+u}{2})^2}{2} = \frac{(z+u)^2}{8} = \frac{(x+y)^2}{8}$$ thus $$(x+y)^2 \ge 8xy \Rightarrow x^2 - 6xy + y^2 \ge 0 \Rightarrow
\left(\frac{x}{y}\right)^2 - 6 \frac{x}{y} + 1 \ge 0.$$ Solving for $\frac{x}{y}$ (noting that $\frac{x}{y}\ge 1$) gives us the desired bound, proving sufficiency.

It suffices to prove necessity. WLOG let $u>z$ and scale down so that $z=1$. It suffices to prove that we can find rational triples $(x,y,u)$ so that the equation is satisfied and $u$ can get arbitrarily close to $1$. Note that $x$ and $y$ are roots of the quadratic $a^2 - (u+1)a + \frac{u}{2} = 0$. For $x$ and $y$ to be rational, we must have the discriminant $(u+1)^2-4\cdot \frac{u}{2} = u^2+1$ be the square of a rational number.
--------------
[color=#f00][b]Claim:[/b][/color] We can find Pythagorean triples $(a,b,c)$ where $c>a,b$ and $\frac{a}{b}$ is arbitrarily close to $1$.
[i]Proof.[/i] Consider only primitive triples because others won't help. Using Euclid's formula we have $a = m^2-n^2, b=2mn$ for some integers $m,n$. Therefore we need to prove we can find $m,n$ so that $\frac{m^2-n^2}{2mn} = \frac{1}{2} \left(\frac{m}{n} - \frac{n}{m}\right)$ is arbitrarily close to $1$, which is equivalent (by solving) that $\frac{m}{n}$ is arbitrarily close to $1+\sqrt{2}$. It is clear that an arbitrarily close rational approximation to $1+\sqrt{2}$ can be found. $\blacksquare$
--------------
For a Pythagorean triple, scale it down so that the smallest number is $1$ and the second leg length is $u$. Thus there are valid values of $u$ sufficiently close to $1$, finishing the problem.

We call $x+y=z+u$ as $(1)$ and $2xy=zu$ as $(2)$

$(1)^2-4(2)$ leads to $x^2-6xy+y^2=(z-u)^2, (\frac{x}{y})^2-6 \frac{x}{y}+1=(\frac{z-u}{y})^2\geq 0$ by solving the quadratics we can find $\frac{x}{y}\geq 3+2\sqrt{2}$, now we need to prove that $\frac{z-u}{y}$ can approach $0$

Then, we do similar thing, $(1)^2-2(2)^2$ tells that $(x-y)^2=z^2+u^2$ which is a pythagorean triple. Where there must be an number is divisible by $2$. Let $u=2ab, z=a^2-b^2, (x-y)=a^2+b^2$. As $x+y=z+u=a^2+2ab-b^2$, we just let $x=a^2+ab, y=ab-b^2$ Firstly, if $z=u, a^2-2ab-b^2=0$, divide $b^2$ for each term, $(\frac{a}{b})^2-2\frac{a}{b}-1=0$, there are only irrational roots which doesn't satisfy the question. Then we check if $y$ can approach infinity

We let $z-u=1, (a-b)^2-2b^2=1$ which is the general form of a pell equation. It's obvious that the elementary solution for the equation is $a-b=3, b=2$ so this equation can have infinity solutions so $ab-b^2$ can definitely approach infinity

So the maximum $\frac{x}{y}=3+2\sqrt{2}$

We have $x^2+2xy+z^2=z^2+2zu+u^2$ so $x^2-6xy+y^2=(z-u)^2\ge 0$ and so if $r=\tfrac{x}{y}$ then $r^2-6r+1\ge 0$ which means $r>3+2\sqrt 2.$ We claim that we can get the ratio to be arbitrarily close to $3+2\sqrt 2.$
\\$~$\\
We have $(x-y)^2=z^2+u^2$ so let $z=a^2-b^2,u=2ab,x-y=a^2+b^2.$ Note that $x+y=a^2+2ab-b^2$ so $x=a^2+ab,y=ab-b^2.$ Let $a/b=s$ where $s>1.$ Then, we can make $s$ arbitrarily close to $1+\sqrt{2}$ which makes $\frac{x}{y}=\frac{s^2+s}{s-1}$ arbitrarily close to $3+2\sqrt2$ which is what we wanted.

The answer is $3+2\sqrt2$.
Note that $(x-y)^2=u^2+z^2$. This means $x-y=k(a^2+b^2)$, $u=2kab$, and $z=k(a^2-b^2)$. We have that $x+y=u+z=k(a^2+2ab-b^2)$. This means $x=k(a^2+ab)$, and $y=k(ab-b^2)$. This means $a > b$. We have that $$ \frac{x}{y} = \frac{\left( \frac{a}{b} \right) ^2 +\frac{a}{b}}{\frac{a}{b}-1} = 3+\left( \frac{a}{b}-1 \right) + \frac{2}{\left( \frac{a}{b} -1 \right)} \ge 3+2\sqrt2,$$ so we have that $3+2\sqrt2$ works. Taking $\frac{a}{b}$ to be arbitrarily close to $\sqrt{2}+1$, and $k=1$, we get a working quadruple and $\frac{x}{y}$ arbitrarily close to $3+2 \sqrt2$, so $3+ 2 \sqrt2$ is the optimal bound and we are done.

Taking the discriminant of the quadratic with roots $u,z$ we easily obtain $\frac{x}{y}\ge \boxed{3+2\sqrt{2}}$ so it's enough to show that we can pick $x,y$ so that $\frac{x}{y}$ becomes arbitrarily close to this value. To show this, we show that we can pick arbitrarily large $x,y$ such that $x^2-6xy+y^2=1$. Note that $(3y+k,y)$ is always a solution where $k^2=8y^2+1$, and since there are infinitely many solutions to this we are done. $\blacksquare$

We have $x^2-6xy+y^2=(x+y)^2-8xy=(z+u)^2-4zu=(z-u)^2\ge 0$. Then if v=x/y we have $LHS/y^2=v^2-6v+1\ge 0$, or that v>3+2$\sqrt{2}$. Now it remains to find such a v.

Note that $(a^2+b^2)^2=(x-y)^2=(x+y)^2-4xy=(z+u)^2-2zu=z^2+u^2=(a^2-b^2)^2+(2ab)^2$, so $(x-y,z,u)=((a^2+b^2),(a^2-b^2),(2ab))$. We also have that $a^2+b^2+2y=x-y+2y=x+y=z+u=a^2-b^2+2ab$, so $(x,y)=(a^2+ab,ab-b^2)$. If a/b=w, we can make w infinitely close to $1+\sqrt{2}$ which would then satisfy $v=\frac{x}{y}=\frac{a^2+ab}{ab-b^2}=\frac{a^2/b^2+a/b}{a/b-1}=\lim_{w->1+\sqrt{2}}\frac{w^2+w}{w-1}=\boxed{2\sqrt{2}+3.} \blacksquare$

The answer is $3+2\sqrt{2}$.\\

Suppose that $z+u=4s$. Then, $$2xy=zu\leq 4s^2,$$ so $$xy\leq 2s^2.$$ Since $x+y=4s$, given that $x\geq y$, the minimum possible $x/y$ occurs when $x=s(2+\sqrt{2})$ and $y=s(2-\sqrt{2}),$ as we need to spread at least this far to get the product at most $2s^2$, so thus $m=3+2\sqrt{2}$ works since $x/y\geq 3+2\sqrt{2}.$\\

Now, consider $z=2s+2,u=2s-2$. Then, we need $$xy=2s^2-2,x+y=4s.$$ Using Vieta and then solving the quadratic, we get $$x,y=2s\pm \sqrt{2s^2+2},$$ and by Pell on $k^2-2s^2=2$, there exist arbitrarily large $s$ for which $x,y$ are both integers. Furthermore, $$\frac{x}{y}=\frac{2s+ \sqrt{2s^2+2}}{2s- \sqrt{2s^2+2}}=\frac{6s^2+4s\sqrt{2s^2}+2}{2s^2-2}.$$ As $s$ gets large, this approaches $3+2\sqrt{2}$, hence done.


July 4, 1903, was a thursday. On what day of the week was July 4, 1904?

1904 was not a leap year. Thus, there were $ 365$ days in that year.

Each week has $ 7$ days. \[ 365\equiv1\mod{7}\] This means we "add $ 1$" to Thursday.

Our answer is [b]Friday[/b].

How was 1904 not a leap year? It's $ \boxed{\text{Saturday}}$.

[quote="akhil0422"]1904 was not a leap year. Thus, there were $ 365$ days in that year.

Each week has $ 7$ days.
\[ 365\equiv1\mod{7}
\]
This means we "add $ 1$" to Thursday.

Our answer is [b]Friday[/b].[/quote]

Unless I'm being extrodinarily stupid, 1904 was a leap year.

akhil0422,

Divide 1904 by 4 --- you should get 476. This means that 1904 is a leap year and you add two days on to Thursday to get Saturday.

_______________
~ pi man :roll:

Any year that is a multiple of four is a leap year. Thus, the answer to the problem is $ \boxed{\text{Saturday}} $.


Um.. Are there any sample problems for Advanced Mathcounts? I want to know what I'm going to do in the class before I take it!

That course is designed for students preparing for State and National MATHCOUNTS.  If you're confident you will pass Chapter and have a chance at States, then it is the right class for you.


$x^{9}-2x^{7}+1>0$

Any way to attack this without going through factor theorem/long division till the end of time? I'll be honest, it's a homework question somebody asked me for help with, but if there is a smart solution I figured it would be appropriate HSB material.

Doubtful.  The only easy root to see is $x = 1$.  There are at least two more relevant ones that probably won't be nice.

Actually, I suspect this is one of those "just graph it and estimate the roots" that I hate classroom math books for doing.

[quote="max_tm"]if there is a smart solution I figured it would be appropriate HSB material.[/quote]Nice solution does not mean HSB material. If you are unsure of the level, post it in Intermediate Topics. If doubtful, please look at the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=32837]guidelines[/url].

As for your problem, $x^{9}-2x^{7}+1=x^{5}(x^{2}-1)^{2}-(x^{5}-1)$ making $x=1$ one of the roots. By Descartes Sign, there are at least 2 positive real roots and one negative real root...

What I meant by nice solution is that if it was simply approximation  or some brute force method then it would be useless to people browsing the forum (as the forum isn't a discussion area for how to solve random equations). I don't understand how if this problem were to have had a clean answer it wouldn't be HSB material? (Although now there is a pretty general consensus that there isn't anything nice about this problem  :P).


induction
number theory unsolved
number theory

[quote="mustafa-yafes"]Prove that there exist distinct pozitive integers $ n_1,n_2,...,n_{50}$ such that

$ n_1 \plus{} S(n_1) \equal{} n_2 \plus{} S(n_2) \equal{} ... \equal{} n_{50} \plus{} S(n_{50})$ where S(x) is the sum of all digits of x in decimal representation.[/quote]

Let $ P(m)$ be the proposition that there exist distinct pozitive integers $ n_1,n_2,...,n_{m}$ such that
$ n_1 \plus{} S(n_1) \equal{} n_2 \plus{} S(n_2) \equal{} ... \equal{} n_{m} \plus{} S(n_{m})$

For $ m\equal{}2$ , $ n_1\equal{}100$ , $ n_2\equal{}91$ satisfies the condition. So $ P(2)$ is true.

Assume that $ P(k)$ is true for any $ k \in \{2,3,4,... \}$. Let an example of the numbers be $ n_1,n_2,...,n_k$
Then $ (n_1',n_2',...,n_{k\plus{}1}')\equal{}(1000n_1\plus{}100 , 1000n_2\plus{}100 , ... , 1000n_k\plus{}100 , 1000n_1\plus{}91)$ satisfies the condition. So $ P(k\plus{}1)$ is also true.

By induction , $ P(m)$ is true for all $ m \in \{2,3,4,...\}$ . Put $ m\equal{}50$ and the result follows.

[quote="stephencheng"][quote="mustafa-yafes"]Prove that there exist distinct pozitive integers $ n_1,n_2,...,n_{50}$ such that

$ n_1 \plus{} S(n_1) \equal{} n_2 \plus{} S(n_2) \equal{} ... \equal{} n_{50} \plus{} S(n_{50})$ where S(x) is the sum of all digits of x in decimal representation.[/quote]

Let $ P(m)$ be the proposition that there exist distinct pozitive integers $ n_1,n_2,...,n_{m}$ such that
$ n_1 \plus{} S(n_1) \equal{} n_2 \plus{} S(n_2) \equal{} ... \equal{} n_{m} \plus{} S(n_{m})$

For $ m \equal{} 2$ , $ n_1 \equal{} 100$ , $ n_2 \equal{} 91$ satisfies the condition. So $ P(2)$ is true.

Assume that $ P(k)$ is true for any $ k \in \{2,3,4,... \}$. Let an example of the numbers be $ n_1,n_2,...,n_k$
Then $ (n_1',n_2',...,n_{k \plus{} 1}') \equal{} (1000n_1 \plus{} 100 , 1000n_2 \plus{} 100 , ... , 1000n_k \plus{} 100 , 1000n_1 \plus{} 91)$ satisfies the condition. So $ P(k \plus{} 1)$ is also true.

By induction , $ P(m)$ is true for all $ m \in \{2,3,4,...\}$ . Put $ m \equal{} 50$ and the result follows.[/quote]

Unfortunately, this is wrong : $ n_1\plus{}S(n_1)\equal{}n_2\plus{}S(n_2)$ does not imply $ 1000n_1\plus{}S(n_1)\equal{}1000n_2\plus{}S(n_2)$

[quote="mustafa-yafes"]Prove that there exist distinct pozitive integers $ n_1,n_2,...,n_{50}$ such that

$ n_1 \plus{} S(n_1) \equal{} n_2 \plus{} S(n_2) \equal{} ... \equal{} n_{50} \plus{} S(n_{50})$ where S(x) is the sum of all digits of x in decimal representation.[/quote]

Here is another induction.

Let the numbers $ A_p\equal{}10^{10^p\plus{}p\plus{}1}\minus{}10^p$ and $ B_p\equal{}10^{10^p\plus{}p\plus{}1}\plus{}8\cdot 10^p$

$ S(A_p)\equal{}9(10^p\plus{}1)$ and $ A_p\plus{}S(A_p)\equal{}10^{10^p\plus{}p\plus{}1}\plus{}8\cdot10^p\plus{}9$

$ S(B_p)\equal{}9$ and $ B_p\plus{}S(B_p)\equal{} 10^{10^p\plus{}p\plus{}1}\plus{}8\cdot 10^p \plus{}9$

So $ A_p\plus{}S(A_p)\equal{}B_p\plus{}S(B_p)$

And so, if we have $ k$ $ n_i$ such that $ n_i\plus{}S(n_i)$ are identical, so $ A_p\plus{}S(A_p)\plus{}n_i\plus{}S(n_i)$ and $ B_p\plus{}S(B_p)\plus{}n_j\plus{}S(n_j)$ are identical (for $ p$ great enough such that $ 10^p>\max(n_i)$

So $ \{A_p\plus{}n_i\}\cup\{B_p\plus{}n_j\}$ is a set of $ 2k$ ad hoc numbers

Hence the induction

thank you everyone who write answers this question...


geometry
inequalities
calculus
number theory
complex numbers
IMO
IMO 2007

Well, IMO 2004 was from 4/7 to 18/7, IMO 2005 was from 8/7 to 19/7, and IMO 2006 was from 6/7 to 18/7, so you can see more or less the date, but I don't know the exact day. The country hosting is Vietnam.

Yes, I heared it would be a bit later, so that's well possible. How do you know in fact, tim?

[quote="mattilgale"]does anyone know what is the address of the official web site of IMO 2007?[/quote]
Here is the address:
http://imo48.edu.net.vn/index.php
However most parts of it are under construction so sometimes some information quite non-related to IMO and Viet Nam pops up(ie info about Mexican Math Olympiad...)
EDIt: I just visited the webs-ite and they changed it again: now it is all in Vietnamese :( (before many parts of it were in English)

What's exactly covered by the 2007 IMO? I know about `[color=blue]geometry, number theory, algebra, and combinatorics[/color]` but i'd like a more detailed list.

There is no fixed list. I'd add inequalities to that, then you have it all, however it's officially part of algebra.

The is no official syllabus for IMO. However, it is generally accepted calculus questions are not allowed (though you are not banned from using calculus), all geometry questions should be able to be done using Euclidean geometry, you do not need complex numbers etc.

The best way to find out what is covered would be to try out recent questions (from both the actual papers and the shortlists). You can access them by clicking on the resources icon on this site.

[quote="tim1234133"]all geometry questions should be able to be done using Euclidean geometry, you do not need complex numbers etc.
[/quote]

That's a very valuable information, thank you very much

Thanks

and u think what country will be the first :?:

[quote="BogG"]unfortunately like in gymnastics certainly china... but romania'll be second :)[/quote]
Russia will be second for sure, if not first. :!:

Are perfect scorers coming? I wish 3 of them to come, :)))) concurention:)) It is great when you study and go for 42, not for some kind of medal...

Russia can't be second!!! US can't fail so much:)))

[quote="gelash"]Are perfect scorers coming? I wish 3 of them to come, :)))) concurention:)) It is great when you study and go for 42, not for some kind of medal...

[/quote]
I know two perfect scorers that coming;Iurie Boreico(42-IMO 2005,2006) and Peter Sholtze(42-IMO 2005).I don't think there will be any other perfect scorers,because in IMO 2006 there where only three perfects and two of them,bisides Iura,has graduated.
I wish you to be a concurent for them,in the IMO 2007,of cours if you are going to be there. :wink:

Thanks, you too:)) it can sound loud from me but why a silver medalist shouldn't be planning gold? good luck!

--
I am maximalist...

how come nobody mentions Korea.
they got 2nd place in medals.
Korea could go for 1st..

i guess China is unbeatable :p

[quote="Davron"]@all: well this is not the right place to discus WHICH TEAM IS GOING THE FIRST?

 I think that China will have a good result maybe the best! But Russia team, I have no idea about this new team, because you can easily the difference Russia is going down year by year hence maybe they will not be the first but I am sure Russia team is one of the best! Hmmm, maybe the main problem of the Russia results is in the problem three of the IMO 2006, in my opinion because of this problem they had a bad result, since problem three was gift for Chiness guys [well it is really INequality China type problem], this types of problems a least solved in the countries like Russia. Ahhh about IMO 2006 US team: people were saying that this year, actually they are saying that this members of the team they are not the best olympiad students of this year, they said that there are two more guys which are really good in solving MO problems, but they couldn't enter IMO because they had invitiation to conference or something like that.  between: KOREA, had a good results again the reason is problem three, since it was created by Mr.Lee  ...

So finally I want to say that WE CAN'T SAY SOMETHING SURE LIKE, THIS OR THAT TEAM WILL BE THE FIRST etc! BUT WE WILL SEE ... SEE YOU AND GOOD LUCK GUYS! THANKS!

DAVRON LF.[/quote]

the problem 3 of imo 2006 was not created by Hojoo Lee. it was even not proposed by Korea. and i am little upset: why are you thinking the good result of korea was due to the proposer of the number 3? in case there are some people thinking so, i just wanna say that there has been no contact between Hojoo Lee and the contestants.

Well guys I think you misunderstand each other.Probably Davron didn't ment that Korean guys knew the problem 3 beforhand(it would be too acute announcment,wouldn't it ? :) ) but the fact that Korean guys liked that problem is obviouse :lol: (I liked it too although I am not a Korean :P ). You see,In the national Olympiads of such a countries as China or Korea we can often see problems of that kind and I think there is nothing wrong.I am sure that Korea's results were good because Korea had one of the best teams in the IMO.I am sure that the teams of counteries like Russia or USA shouldn't have loved or unloved problems if they want to be a competitor for such a strong team as China's. The matter that Russians fall behaind China doesn't mean that Russians are weak;it is because Chinas are too strong :wink: .

You know,even in IMO 2000 when 3 problems(Q1,Q5 and Q6) were proposed by Russia,China beat them again.If I am not mistaken the in the last 12 years China was beaten only twice: in IMO 1996(mostly because of problem 5 proposed by Armenia :D ) and in IMO 2003(Bulgarians were first).In fact in IMO 1998 Chinas weren't partisipating in the IMO because of political problems,and,also, in IMO 1999 Chinas and Russians results were same.So,I think that $8parks11$ guesses correctly that China is(mostly :P )unbeatable :ninja: .

[quote="Davron"]Assalamu aleykum!

I didn't say that Mr.Lee did that... since I know the value for the professors like Mr.Lee. But I am sure if he was teaching to his students before the Olympiads, that  means the ideology he gives to his students is the ideology he has in his mind. So probably we can say that it helped them ... But you are saying that this problem is not created by him even by Korea. Yeah, sounds cool, since I was waiting a problem from him. Anyway maybe you will tell us who was the creator? [maybe Poland]. The main thing I want to say in my message it was: don't doubt yourself guys, we  all don't know, can't be sure about which team is going to be the 1st and which one is going to be the last... Thank you one more time!

Sincerely
Davron Latipov[/quote]

Aleykumu selam.

The third problem was proposed by Ireland... If I my memory still serves me... Korea (=Hojoo Lee :D ) proposed problem 1...

sorry for my rudeness. i think i misunderstood and became upset.

[quote="armanf"]Tiks are you sure Iurie Boreico is coming this year?! :o
[/quote]

I am not sure about that, but I am sure that he is now in his senior year at the school and if everything goes well(i.e. if he will be included in the Moldovian team :P  :P ) he will probably be in Vietnam in July :wink: .

If I am not mistaken another perfect-scorer is still at school this year. I am talking about Peter Scholze from Germany. He got perfect in IMO 2005 and 35point in IMO 2006.

[quote="Tiks"]
If I am not mistaken another perfect-scorer is still at school this year. I am talking about Peter Scholze from Germany. He got perfect in IMO 2005 and 35point in IMO 2006.[/quote]

Yes, Peter is his final grade this year, 13. grade in Germany. As long nothing unfortunate happens to either Peter or Iurie both are most likely to participate.

[quote="armanf"]Tiks are you sure Iurie Boreico is coming this year?! :o
problem 3 was proposed by Ireland and problem 1 was Mr.Lee's
but do you think problem 1 need any help davron  ? :wink:[/quote]

??

d.

Well
Be sure that Iurie is going to take a part in IMO 2007, too. Because, he is in the last class this year. Between: I remember he told me that he is going to be in the next IMO, during the IMO 2006... So we can say that 99%ly he will! Peter, also going be... But I think that this is not our problem who will participate and who will not, isn't it?

@armanf: birader salam! I didn't get your message?!

all the best!
d.

do you know who is dzafar jafarov , he is maximalist from azerbaidzhan  :P

I know Jafar Javarov. He was the perfect scorer ($42$) with Turea Lucian Gabriel (Romania) in Zhautykov Olympiad 2007, Kazakstan... Well, I was participating, too.. But I only got the first silver medal with the score of $26$ where I misread $1$ question !  :mad:
However, I will participate in IMO 2007, too.. And I've targeted a gold medal..  :)

Oh no, Iurie Boreico is gonna take his 3rd perfect score this year! And with two more medals (silver in 2003 and gold in 2004) will definitely beat Ciprian Manolescu's perfomance. :( :D


For $ a,b,c$ positive reals prove that $ a^2 \plus{} b^2\plus{}c^2 \plus{}d^2\equal{}4$ prove that

$ (\frac{a\plus{}b\plus{}c\plus{}d}{4})^3 \leq (1\plus{}abcd)(\frac{1}{a} \plus{}\frac{1}{b}\plus{}\frac{1}{c}\plus{}\frac{1}{d})$

OK, we know that:
$ 1\equal{}\sqrt{\frac{a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}d^{2}}{4}}\geq \frac{a\plus{}b\plus{}c\plus{}d}{4}$ and then
$ a\plus{}b\plus{}c\plus{}d\leq 4$
So $ LHS\leq 1$
We also knowfrom harmonical-arithmetical mean $ \sum \frac{1}{a}\geq \frac{4^{2}}{\sum a}\geq \frac{4^{2}}{4}\equal{}4$
So we acquire:
$ RHS\equal{} (1\plus{}abcd)(\frac{1}{a} \plus{}\frac{1}{b}\plus{}\frac{1}{c}\plus{}\frac{1}{d})\geq 4(1\plus{}abcd)$
So it suffices to prove:
$ \frac{1}{4}\leq 1\plus{}abcd$ which is obvious
I have a feeling that it went too easy, so where is my mistake? :wink:

[quote="polskimisiek"]OK, we know that:
$ 1 \equal{} \sqrt {\frac {a^{2} \plus{} b^{2} \plus{} c^{2} \plus{} d^{2}}{4}}\geq \frac {a \plus{} b \plus{} c \plus{} d}{4}$ and then
$ a \plus{} b \plus{} c \plus{} d\leq 4$
So $ LHS\leq 1$
We also knowfrom harmonical-arithmetical mean $ \sum \frac {1}{a}\geq \frac {4^{2}}{\sum a}\geq \frac {4^{2}}{4} \equal{} 4$
So we acquire:
$ RHS \equal{} (1 \plus{} abcd)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \plus{} \frac {1}{d})\geq 4(1 \plus{} abcd)$
So it suffices to prove:
$ \frac {1}{4}\leq 1 \plus{} abcd$ which is obvious
I have a feeling that it went too easy, so where is my mistake? :wink:[/quote]

For $ a,b,c$ positive reals and d is not.

[quote="polskimisiek"]
So it suffices to prove:
$ \frac {1}{4}\leq 1 \plus{} abcd$ which is obvious
I have a feeling that it went too easy, so where is my mistake? :wink:[/quote]
Here. :wink:

Aaah, yeah, I mechanically assumed that all $ a,b,c,d \in R_{\plus{}}$ :wink:
Next time I'm gonna read more carefully. So, what'd be the solution? :wink:

[quote="manlio"]For $ a,b,c$ positive reals prove that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 4$ prove that

$ (\frac {a \plus{} b \plus{} c \plus{} d}{4})^3 \leq (1 \plus{} abcd)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \plus{} \frac {1}{d})$[/quote]
[hide="hint"] $ 9\sum a \le 4abcd \plus{} 32$[/hide] :)


How would I do the division symbol that is taught in elementary school in latex (the dash with a dot over and below it)?

[quote="stevem"]By looking at the [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php]AoPS Symbols Guide[/url] and finding $\div$[/quote]
Oh thanks :oops:
BPMS


logarithms
trigonometry
calculus
function
complex numbers
imaginary numbers
complex analysis

How would you find something like $5^{i}$, $-3^{i}$, or $i^{i}$?

We know the following things:

$e^{\ln x}= x$
$e^{ix}= \cos x+i\sin x$

So we do the following:

$x^{i}= (e^{\ln x})^{i}= e^{i\ln x}= \cos{\ln x}+i\sin{\ln x}$

Umm.... someone correct me if I am wrong. Doesn't raising a number to the power of a complex or imaginary number involve some complex analysis (which is not really high school stuff).

[quote="anirudh"]read this:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=26438[/quote]
Why did you want us to read that? There's no calculus involved, unless you're refering to the derivation for Euler's Formula. Also, [quote="tetrahedr0n"]Euler's Theorem isn't strictly a calculus theorem, so don't worry about it! I think the point of this message is to avoid calculus problems and exercises - but using a theorem related to calculus isn't a big deal.[/quote]
Anyway, this exercise is in AoPS Vol. 2.

[quote="anirudh"][quote="cincodemayo5590"]Well, for $1 cis \theta$ (ie $1,i, \text{ etc}$) we have

$(1 \ cis (\theta+2k\pi))^{i}= 1^{i}\ cis [i(\theta+2k\pi)]$ with $k \in \mathbb{Z}$. Then, since $1^{m}=1$ for complex $m$ (I think...), we have $e^{i^{2}(\theta+2k\pi)}=\frac{1}{e^{\theta+2k\pi}}$.

So, for example, $i^{i}= [1 \ cis (\frac{\pi}{2}+2k\pi)]^{i}= \frac{1}{e^{\frac{\pi}{2}+2k\pi}}$.

Im not sure how to do anything larger than that, though.[/quote]

what about this?[/quote]
That is definitely not calculus (though it may look confusing enough to be). Finding $i^{i}$ is actually an exercise in Chapter 9 of AoPS 2. (I checked :D )

[quote="anirudh"][quote="cincodemayo5590"]Well, for $1 cis \theta$ (ie $1,i, \text{ etc}$) we have

$(1 \ cis (\theta+2k\pi))^{i}= 1^{i}\ cis [i(\theta+2k\pi)]$ with $k \in \mathbb{Z}$. Then, since $1^{m}=1$ for complex $m$ (I think...), we have $e^{i^{2}(\theta+2k\pi)}=\frac{1}{e^{\theta+2k\pi}}$.

So, for example, $i^{i}= [1 \ cis (\frac{\pi}{2}+2k\pi)]^{i}= \frac{1}{e^{\frac{\pi}{2}+2k\pi}}$.

Im not sure how to do anything larger than that, though.[/quote]

what about this?[/quote]

What i did was just DeMoivre's Theorem ($a \ cis \theta)^{n}= a^{n}\ cis (n\theta)$ and $e^{i \theta}= 1 \ cis \theta$... not calculus.

[quote="t0rajir0u"][quote="cincodemayo5590"]What i did was just DeMoivre's Theorem ($a \ cis \theta)^{n}= a^{n}\ cis (n\theta)$ and $e^{i \theta}= 1 \ cis \theta$... not calculus.[/quote]

This is extremely misleading.  While it's true that nowadays Euler's and DeMoivre's are frequently taught before calculus, especially in a problem-solving context (because it's very useful when dealing with complex numbers), Euler's Formula is only a definition until you see the motivation behind it, which you see through [b]calculus.[/b]

I concede that DeMoivre's Theorem actually can be proven in an elementary fashion.

But really, has anyone ever told you [b]why[/b] $e^{i \theta}= \cos \theta+i \sin \theta$?[/quote]

Not really.  :oops:  But still, using Euler's is fine for these forums, right?

the elementary proof for demoivre's theorem only holds for rational exponents, muchless complex numbers

$e^{i\theta}=\cos\theta+i\cdot\sin\theta$ is surely a calculus based theorem, it is proven using taylor series

try better to understand the no calculus rule of the forum, they are just trying to get rid of the calculus exercises, and really difficult problems that calculus covers, there is no problem with simple theorems like this one

be careful with: $(ab)^{i}\ne a^{i}\cdot b^{i}$, $x^{i}$ is not a function, also $x^{\frac{1}{m}}$ is not a function unless we restrict the output to real numbers, but there is no similar rule for $x^{i}$

sam saying 1^m=1 does not make sense, right in your post, you show the contrary, according to your result, taking $\theta=0$, then $1^{i}=e^{2k\pi}\, \forall \, k\in\mathbb{Z}$


Asymptote
function
algebra unsolved
algebra

i seriously have no idea on how to do these problems. i would really appreciate anyone who can do these, then i could learn from looking at these examples

1] use descates rule to find the # of positive and negative zeros of
f(x) = x^3 + 5x^2 + 8x + 6

2] find the vertical and horizontal asymptotes of
f(x) = [u]3x^3 - 1[/u]
[color=white]xxxxx[/color]         x^2 - x - 6

3] for the function f(x) = x^2 + 2x + 1, find
[f(x) - f(2)] / 4

4] find the inverse function of f(x) = [2x - 3] / f

thanks, any help is greatly appreciated

1)I think you mean the Decarte's rule of signs here. If so then for the positive roots you look the number of the inverses in f(x).
(the inversion is any sign change in the sequence of coefficints of f)
Since this number is zero there are no positive roots.For the negative you look the number of the inverses in f(-x).This number is 3,so unfortunately from here you can tell only
that the number of negative roots is 1 or 3.
THe number is actually 1 since:f(x)=(x+2)(x^2+2x+2).

2)The horizontal asympote of f exist if and only if lim f(x) is not infinity (the limit is when x tends to infinity) since this limit is infinity the horizontal asymptote does not exhist.
the vertical is such line x=c for which lim f(x)=infinity (this limit is when x  tends to c) such[b] c [/b]here are -2 and 3 so the asymptotes are [b]x=-2 [/b]
and [b]x=3[/b]
I,m sorrry but for the next i just don't understand you.


Στο σύνδεσμο :

 http://www.mathlinks.ro/viewtopic.php?t=223164

 έχω βάλει μια άσκηση. Έχουν ασχοληθεί πριν και άλλοι, κυρίως ο Κώστας Βήττας. Επειδή δεν προλαβαίνω να μελετήσω όλη την αλληλογραφία, θα ήθελα κάποιος από σας που θα διαβάσει  το θέμα καλά  και θα ξεκαθαρίσει το τοπίο, κάνοντας ίσως και παρεμβάσεις , να μας στείλλει την πιο σύντομη συνθετική λύση στην ασκηση που βάζω . Αξίζει τον κόπο.
 Την άσκηση(ένα μέρος της) πήρα από το βιβλίο  [color=red][b]Bulgarian Olymbiads 2003 - 2006 [/b][/color]που μου έφερε ο Σιλουανός από την ΙΜΟ στο Βιετ Νάμ(Εκδόσεις GIL).Τον ευχαριστώ και από αυτή τη θέση.

 Μπάμπης

Ας δούμε μία απλή λύση βασισμένη στο  [b][size=100]θεώρημα του Μενελάου[/size][/b], όπως την υποδεικνύει ο φίλος μας [b][size=100]Fang-jh[/size][/b] από τη μακρινή Κίνα, [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=164092]εδώ[/url].

$ \bullet$ Έστω $ AD$ το ύψος του τριγώνου $ \bigtriangleup ABC$ και τα σημεία $ R\equiv AD\cap C_{2}C_{3},\ R'\equiv AD\cap B_{1}B_{3}.$ Αρκεί να αποδέιξουμε ότι $ R'\equiv R.$

Στο τρίγωνο $ \bigtriangleup ADB,$ με διατέμνουσα την ευθεία $ C_{2}C_{3}R,$ σύμφωνα με το [b][size=100]θεώρημα του Μενελάου[/size][/b], έχουμε : $ \frac {RA}{RD}\cdot \frac {C_{2}D}{C_{2}B}\cdot \frac {C_{3}B}{C_{3}A} \equal{} 1$ $ ,(1)$

Ομοίως, στο τρίγωνο $ \bigtriangleup ADC,$ με διατέμνουσα την ευθεία $ B_{1}B_{3}R',$ έχουμε : $ \frac {R'A}{R'D}\cdot \frac {B_{1}D}{B_{1}C}\cdot \frac {B_{3}C}{B_{3}A} \equal{} 1$ $ ,(2)$

Από $ (1),$ $ (2)$ $ \Longrightarrow$ $ \frac {RA}{RD}\cdot \frac {C_{2}D}{C_{3}A} \equal{} \frac {R'A}{R'D}\cdot \frac {B_{1}D}{B_{3}A}$ $ ,(3)$

$ \bullet$ Από $ I_{b}B_{1}\parallel AD\parallel I_{c}C_{2}$ $ \Longrightarrow$ $ \frac {C_{2}D}{B_{1}D} \equal{} \frac {AI_{c}}{AI_{b}}$ $ ,(4)$

Από όμοια ορθογώνια τρίγωνα $ \bigtriangleup C_{3}AI_{c},\ \bigtriangleup B_{3}AI_{b}$ $ \Longrightarrow$ $ \frac {C_{3}A}{B_{3}A} \equal{} \frac {AI_{c}}{AI_{b}}$ $ ,(5)$

Από $ (4),$ $ (5)$ $ \Longrightarrow$ $ \frac {C_{2}D}{B_{1}D} \equal{} \frac {C_{3}A}{B_{3}A}$ $ \Longrightarrow$ $ \frac {C_{2}D}{C_{3}A} \equal{} \frac {B_{1}D}{B_{3}A}$ $ ,(6)$

Από $ (3),$ $ (6)$ $ \Longrightarrow$ $ \frac {RA}{RD} \equal{} \frac {R'A}{R'D}$ $ \Longrightarrow$ $ R'\equiv R$ και η πρόταση έχει αποδειχθεί.

Με παρόμοιο τρόπο, αποδεικνύεται ότι τα σημεία $ P\equiv C_{1}C_{2}\cap B_{1}B_{2}$ και $ Q\equiv A_{1}C_{2}\cap B_{1}A_{2},$ ανήκουν επίσης στην ευθεία του ύψους $ AD$ του τριγώνου $ \bigtriangleup ABC.$

Κώστας Βήττας.

Κώστα , ευχαριστώ πολύ !

 Αυτή τη λύση προσπάθησα λίγο να καταλάβω, αλλά δεν είχα σχήμα καλό και με δυσκόλεψαν οι λόγοι των παραλληλιών. Εξαιρετικό θέμα για την απλότητά του, αν και σχετικά δύσκολη λύση.

 Νάσαι καλά  - Μπάμπης

Συγχωρέστε με αν έχει ειπωθεί κάτι παρόμοιο με τα παρακάτω . Θα χρησιμοποιήσω του συμβολισμούς του κ.Μπάμπη
Έστω ότι $ (MM_1)\cap (LL_1)\equiv S$  . Έστω επίσης $ (ME)\cap (LL_1)\equiv N$ .
Επειδή τώρα έχουμε ότι $ I_cM//I_aL$ , $ MN//AI_a$ και $ LN//AI_c$ , οι τομές αυτών των τριών είναι συνευθειακά σημεία στο άπειρο άρα από το αντίστροφο του θεωρήματος του Πάππου , τα σημεία $ I_a,N,I_c$ είναι συνευθειακά και όμοια τα σημεία
$ I_a,Q,I_c$ είναι συνευθειακά όπου $ (LD)\cap (MM_1)\equiv Q$ και αφού επιπλέον το $ B$ ανήκει στη $ I_aI_c$ θα έχουμε ότι τα σημεία . $ I_a,Q,B,N,I_c$ είναι συνευθειακά . Εύκολα τώρα κάποιος μπορεί να δει ότι η $ AQ$ είναι η πολική του $ N$ ως προς τον $ (I_a)$ επομένως $ AQ\bot I_aI_c$ και όμοια $ CN \bot I_aI_c$ . Επιπλέον εύκολα βλέπει κανείς ότι $ MN\bot SL$ και
$ LQ\bot SM$ άρα $ SP\bot AC$ επομένως αρκεί να δείξουμε ότι η $ SP$ τέμνει την $ I_aI_c$ στο σημείο $ B$ ή ισοδύναμα αρκεί να δείξουμε ότι αν $ (I_aI_c)\cap (BC)\equiv B'$ τότε τα σημεία $ B,Q,B',N$ σχηματίζουν αρμονική τετράδα .
Πράγματι : έχουμε $ \frac {B'Q}{B'N} \equal{} \frac {AQ}{CN}$ (1) . Ὀμως τα τρίγωνα $ QBA$ και $ BCN$ είναι όμοια επομένως
$ \frac {AQ}{CN} \equal{} \frac {BQ}{BN}$ (2)  . Από (1),(2) το ζητούμενο .

Η απόδειξη αφιερώνεται στον κ.Βήττα

Μετά από πολύ καιρό απουσίας, βάζω και εγώ μια συνθετική λύση σε αυτή την πολύ ωραία άσκηση :)

Εστω $ I$ το έγκεντρο του $ ABC$ και $ P$ το μέσο της  $ BC$. Τώρα, τα τρίγωνα $ MSL$ και $ BIC$ είναι ομοιόθετα γιατί έχουν παράλληλες πλευρές (λόγω γωνιών). Ετσι εύκολα βλέπουμε ότι το $ P$ είναι το κέντρο ομοιοθεσίας τους αφού είναι μέσο του $ BC$ και ταυτόχρονα μέσο του $ LM$ (αφού $ MB \equal{} CL \equal{} \tau \minus{} a$).

Οπότε τώρα έχουμε $ \frac {PS}{PI} \equal{} \frac {PM}{PB} \equal{} \frac {\tau \minus{} a/2}{a/2} \equal{} \frac {b \plus{} c}{a}$. (από εδώ κατ’αρχάς έχουμε προσδιορίσει τη θέση του $ S$).

Στη συνέχεια ας θέτουμε $ Q, R$ τις προβολές των $ I, Q$ αντίστοιχα επί της $ BC$. (θεωρώ χωρίς βλάβη της γενικότητας ότι $ Q$ είναι ανάμεσα στα $ B, P$.) Ξέρουμε έτσι ότι $ PQ \equal{} PB \minus{} BQ \equal{} \frac {a}{2} \minus{} (\tau \minus{} b) \equal{} \frac {b \minus{} c}{2}$.
Tέλος από ομοιότητα των τριγώνων $ IQP$, $ SRP$ παίρνουμε: $ \frac {PR}{PQ} \equal{} \frac {PS}{PI}\iff PR \equal{} \frac {(b \plus{} c)(b \minus{} c)}{2a}\iff b^2 \minus{} c^2 \equal{} 2aPR$,
που σημαίνει, λόγω του δεύτερου θεωρήματος διαμέσων ότι το $ R$ είναι το ίχνος της καθέτου απ’το $ A$ στη $ BC$. Συνεπώς το ζητούμενο!
QED


Out of a pack of 36 cards (without jacks, queens, kings, aces, or jokers), 5 cards are chosen randomly. Out of these, a card is picked randomly, and the rest are returned to the deck. This process is repeated with 4, 3 ,and 2 cards. What's the probability that the sum of the numbers on these four cards is equal to or greater than 24?

[quote="ArcticMonkey"][quote="Inspired By Nature"]Out of a pack of 36 cards (without jacks, queens, kings, aces, or jokers), 5 cards are chosen randomly. Out of these, a card is picked randomly, and the rest are returned to the deck.[/quote]

My intuition usually fails me in combinatorics or probabilities, but isn't this process equal to just picking a card randomly from the deck?[/quote]
yes, just picking 4 random cards. :P

[hide]

it shouldn't matter whether you choose 5 cards or not.  its all equivalent to choosing 1 card 4 times without replacement.  There are 4 for every type of card.  Thus we can average the numbers as if they did not have 4.  So we get  6 as the average card number.  And since we choose 4 cards, $4 \times 6=24$ There is a .5 probability that the sum will be greater than or equal to 24.

[/hide]

[quote="JBL"]That doesn't work.  Suppose I roll 2 standard dice.  The "average" outcome is a 7.  The chance of getting a 7 or greater is 7/12, not 1/2.

It's also not at all clear that the question with replacement will give the same answer as the question without replacement.[/quote]
that's true. :)


$ p_1^2  ,p_2^2 ,...,p_n^2$ are quadrates of n numbers different from each others.
So prove that $ (1\minus{}\frac{1}{p_1^2 })(1\minus{}\frac{1}{p_2^2 })*...*(1\minus{}\frac{1}{p_n^2 })\ge\frac{1}{2}$

Are you sure your statement is true?

For instance, if there is $ i$ such that $ p_i\equal{}1$, then $ \prod_{k\equal{}1}^n \left(1\minus{}\frac1{p_i^2}\right)\equal{}0$.

Besides, for a big $ n$ we have $ \prod_{k\equal{}1}^n\left(1\minus{}\frac1{p_i^2}\right)\to 0$

Sorry my friend ifai. I don't think as u think. This statement is surely true. But of course I agree with your statement $ p_i\ne1$


linear algebra
matrix
vector
geometry
geometric transformation
reflection
linear algebra unsolved

Let D be an n*n matrix for which D^(k+1) is the zero matrix (k is a possitive integer),show that D^k+D^(k-1)+...+D^2+D+I(n) is invertible ,and find an expression for its inverse.

[quote="Semigroups"]If U is a unit vector in R3, and we have A=I-UU^T(means transpose of U) ,we can prove :A^2=A, how to use this result to prove that A is not invertible?

(I found det(A)=0 or 1, but how to show that det(A) is not 1?)[/quote]

In fact, $A$ annihilates a nonzero vector, namely $U$ (because $AU=\left(I-UU^{T}\right)U=U-UU^{T}U=U-U\cdot 1=U-U=0$, where we used that $U^{T}U=1$ what is because $U$ is a unit vector) and is therefore not invertible.

  Darij

$A=I-uu^{T}$ is a projection (an idempotent, with $A^{2}=A$) onto the orthogonal complement of the vector $u.$ It annihilates $u$ and fixes the plane perpendicular to $u.$

$B=1-2uu^{T}$ is a reflection. We can directly compute that $B^{2}=I.$ It is a reflection across the plane perpendicular to $u,$ as it fixes that plane but sends $u$ to $-u.$


After one-fifth of the girls left a school dance, the ration of girls to boys was $2: 3$. Then, 44 boys left and the ration of boys to girls was 2:5. How many students remained at the dance?

[quote="Ubemaya"]How can the ratio of girls to boys be less after a bunch of boys left the dance?[/quote]
The mathcounts ppl are trying to trick you, the terms are switched...

[quote="Ubemaya"]You can't switch terms in a ratio.

From the ratio 2:3 to the ratio 2:5, either the second term grew or the first term lost, which means that either more boys came or girl left.[/quote]
read the problem
first it says the ratio of [i]girls to boys [/i]is 2:3
after it changed,
then it says ratio of [i]boys to girls [/i]is 2:5

I found this problem to be rather simple if you understand algebraic solutions.

[hide="Solution"] 3:2 is the original ratio of boys to girls. After subtracting 44 from the boys side of the ratio, it becomes 2:5. Thus,

3x : 2x is the original ratio

2y : 5y is the new ratio

3x-44 : 5y is the new ratio (with boys in terms of x) after 44 boys left. Since the number of girls is unchanged, 2x = 5y.

because of this, we can now solve 3x-44 = 5y using the substitution method.

3x-44 = 5y
7.5y-44 = 2y
5.5y = 44
y = 8

Since the new ratio was 2y : 5y, the number of students remaining is 56 (2*8 + 5*8)[/hide]

[hide] Say that before the boys left, that there were 2x girls and 3x boys.  This imples that \[\frac{3x-44}{2x}=2/5\Rightarrow 15x-220=4x\Rightarrow 11x=220\Rightarrow x=20\]  This means that there were 60 boys and 40 girls before the boys left.  This totals to 100 people.  Now, since 44 of them are left, only $56$ remain.  [/hide]


algebra
polynomial
inequalities proposed
inequalities

The polynomial $ f(x)\equal{}ax^2\plus{}bx\plus{}c$ has real coefficients and satisfies $ \left|f(x)\right|\le 1$ for all $ x\in [0, 1]$. Find the maximal value of $ |a|\plus{}|b|\plus{}|c|$.

[quote="linh DHKHTN"]this problem is very easy . The maximal is 3 :D[/quote]
What about $ f(x)\equal{}8x^2\minus{}8x\plus{}1$ $ ?$ :wink:

The proof that the maximum is 17 (e.g. attained by arqady's example) can be found here:
www.artofproblemsolving.com/community/u212018h625626p3751529
www.artofproblemsolving.com/community/c6h348219

This problem is trivial by using lagrange interpolation formula

We claim the answer is $\boxed{17}$, attained at $(a, b, c) = (8, 8, 1)$, which can be verified to work. Now we claim this is the maximum -- in fact, we claim that all greater values fail if we consider $f(0), f(1)$ and $f\left(\frac 12\right)$. In particular,
\begin{align*}
-1 &\leq c \leq 1 \\
-1 &\leq a+b+c \leq 1 \\
-1 &\leq \frac 14a+\frac 12b+c \leq 1.
\end{align*}
We claim that the maximum of the desired expression, 17, can be achieved regardless of $c$. WLOG assume $c$ is positive by symmetry; consider the equation for $c=1-\varepsilon$, where the desired locus on the $xy$-plane is a parallelogram, and we can verify by that the largest solution occurs when $x+y = \varepsilon$ and $x+2y = -8+\varepsilon$. Here $y=-8$ and $x=-8-\varepsilon$, so $|a|+|b|+|c| = 8+(1-\varepsilon)+8+\varepsilon = 17$ still.


here is a cool little thing that i noticed...

let $F_n$ denote the $n$th number in the fibonacci sequence (where $F_0=0$ and $F_1=1$)

prove (or disprove) that

$\begin{tabular}{|ccc|} F_{n}&F_{n+3}&F_{n+6}\\ F_{n+1}&F_{n+4}&F_{n+7}\\ F_{n+2}&F_{n+5}&F_{n+8}\\ \end{tabular}=0$

For all whole numbers $n$

i am not 100% sure that it is true, but i have seen a pattern

Last row is the sum of the first two rows.


limit
Pascal\u0027s Triangle
algebra solved
algebra

We have the sequnces [tex]x_{n}[/tex] and [tex]y_{n} [/tex] which verify: [tex] (1+\sqrt{2})^{n}=x_{n}+y_{n}\sqrt{2}[/tex] forall n in N. Compute  [tex] \lim_{n\to \infty} \sum_{k=1}^{n} \arctan \frac{1}{6x_{k}y_{k}}  [/tex].

please help! :) thank you very much for your answer!

sorry it's suppose to be:

We have the sequnces [tex]x_{n}[/tex] and [tex]y_{n} [/tex] which verify: [tex] (1+\sqrt{2})^{n}=x_{n}+y_{n}\sqrt{2}[/tex] forall n in N. Compute  [tex] \lim_{n\to \infty} \sum_{k=1}^{n} \arctan \frac{1}{6x_{k}^{2}y_{k}^{2}}  [/tex].

Are we to assume [tex]x_n[/tex] and [tex]y_n[/tex] are integers?

They must be, right?  Well at least sensibly.

Not sure where to go with this, but here's a start:

[tex]x_1=1, y_1=1[/tex]

[tex]x_n=2y_{n-1}+x_{n-1}, y_n=y_{n-1}+x_{n-1}[/tex]

To see this, try doing out the first few terms. Hint: [hide]Try using the Pascal's Triangle[/hide]

Also, could anyone do anything with this information?

Also, [tex]\frac{x_n}{y_n}=[a_0, a_1 \cdots a_{n-1}][/tex]

Where [tex][a_0, a_1 \cdots a_{n-1}][/tex] denotes a continuous fraction representation of [tex]\sqrt{2}[/tex]

boy, am I stupid..

I forgot: a_n and b_n are rational numbers.

this problem doesn't love me or vice-versa! :)

This may help. We can write

[tex]\displaystyle x_n=\frac{1}{2}\left((1+\sqrt{2})^n+(1-\sqrt{2})^n\right)[/tex]

[tex]\displaystyle y_n=\frac{1}{2\sqrt{2}}\left((1+\sqrt{2})^n-(1-\sqrt{2})^n\right)[/tex]

What I've done so far is used arctan(a)+arctan(b)=arctan[(a+b)/(1-ab)] to try to find a pattern in the partial sums. I did find the pattern, and the answer I got is

spoiler [hide]arctan(3-2 :rt2: )[/hide]

I know that I could prove the pattern I found using the square root stuff I posted above, but that seems kinda ugly. I'll look for a different way. (BTW, I know my answer is right, I just need to find an easy way to prove it.)

Can you please post this solution, I would be very interested.

Thank you very much.


search blogs videos algebra function domain Support

Disclaimer: This thread is not meant for people like valeriummaximum and madness who do not the spelling of music nor do they what it means. It is for normal people who listen to music while doing problems or to relax and not people who do chemistry to relax. :P

With that disclaimer out.

I listen to loads of music, english, tamil and Hindi (though my hindi isn't so good. :P). I have 12GB collection but my favs are Linkin Park, Shakira, MLTR, Savage Garden, Scorpions, Flipsyde.

Tamil: ARR rulez (in all languages :D) and sometimes harris jeyaraj. I also like Yogi--B!

Hindi: I like sufi-ish music, dunno why. Don't listen to too much of Himesh though. ;)

listen to lots of Rahman...favorite -Uyire Uyire from Bombay...any song from aayutha ezhuthu ,kalayil Dhinamum from New,...,Any song from Alai Payuthey(especially Evano Oruvan)....lots of others
and not to forget Mannile(from may madham)

next comes Illayaraja-Mandram vantha...from Mouna Ragam,Sundari Kannal from Thalapathi,Poo Potahthu from Mumbai Express,Thani Thotti from... ,Then Pandi Seemayile from Nayagan
and some songs from Kadhaliku Mariyathey

but [b]My favorite-Loosu Penne From Vallavan[/b]

Harris Jeyaraj-Iruvizhi(minnale),Uyirin Uyirin(KK)

some misc...like theradi Vidhiyil....

From hindi prob like Swades ....

don't like english songs

you don know abt me shreyas a.k.a Cyber Shreyas

I know how to sing,play the violin,play the violin and to play the keyboard
My favorite songs are Kurai ondrum illai       by rajaji
                              katrilvarumgeethame by illairaja
i have many favorites....
i list goes on...
Shreyas -you listen only to english songs but i listen to both though English only a few like Bon jove,blue,etc

[quote="valeriummaximum"]
Shreyas -you listen only to english songs but i listen to both though English only a few like Bon jove,blue,etc[/quote]

Cant you read. I listen to all languages. I even some japanese songs (w/o vocals :D:RoFL)

i do listen to some bhajans ...my father plays them...his favorite alai Payuthey...hence by default mine also

my breath is music.......
ok i know lot of songs to sing, and to listen o course...
btw madness-you are always at war with your sister

[quote="madness"]and dont leave me out bcos i listen only to classical music except some old Hindi Classics like kuch kuch hota hai and dilwahle..... and btw i can speak a lot on music bcos my sister's crazy abt music  :D[/quote]

I remember you telling me listening to music is a waste of time but you are forced to cause of your sister.

im talking abt my own sister and not my cousin
Btw i am at war with my own sis also  :D  :D  :D

Ah music! Lol.

I am addicted to music... I listen to english, tamil, hindi mainly... But also like spanish, punjabi and bengali... Multilingual :D

I love Linkin Park, Shakira, Bon Jovi, Backstreet Boys, Blue, and of course Enrique, my personal favourite... In tamil and hindi the list is endless :D

And ranga listens to music, and plays the violin n keyboard? Wow... lol

I have been learning the keyboard for the past hmm.... 7.5 years too..

N I have to say this here, ranga's favourite song is yaakai thiri, from aayudha ezhuthu.... no qualms about it at all

Yeah, he loves it.. But then, there is also an inside joke about that... :D

and can someone Pm(private message...click on the letter kind of thing on lower left corner of my post) me the joke :rotfl: ...will pm him later

Ranga sang it for me and i recorded the whole thing on my mobile.. I had it as my ringtone for a few days till i got IN DINO from METRO.. in those days, ranga underwent maximum torture.. :D
Btw, I don listen to much of foreign stuff..
Hindi I like all from Jatin lalit to shankar, ehsaan loy to pritamm.. I love in dino, ya ali, tu hi meri shab hai, kank, kuch kuch hota hai among many many others.
Any language ARR rocks.. Simply rocks.. So after Jodha Akbar, Im waitin for Robot..
Tamil I like almost all Ilaiyaraja stuff.. And pardesi, I simply luvv Mandram Vandha too...
Harris Jeyaraj has been great in Majnu, Minnalae, PKMC, Bheema..
I like all great songs..
Ive fallen in love with GV prakash too.. And Joshua Shridhar.. Singers like SPB, Karthik, Chinmayee, Chitra, Haricharan also feature i nmy fav too...

i know that GV is addicted to music..............
he listens to music even if he is writing a test........................
often tortures us(me and rohit) by singing songs of himesh...

i am at my aunts house now and will hate to head towards iit m
godav sux da!!!!! :P  :P

wats godav?
dude !
and quite a time since i hav chatted with u
hope that now ull remain!

the moderator is in the US. now Hi u aren't across seas from me now! HI!

godav is the name of the hostel i live( :rotfl:  :rotfl:  :rotfl: )  at iit m
@razy4math when did i go to the us da(i) :mad:  :o  :o  :o  :o  :P  :P

[quote="rituraj007"][hide]
[hide]
[hide]

  :o
 :P
[size=200][color=darkred]CHUMMA JUJUBI[/color][/size]

 :rotfl:

[/hide]
[/hide]
[/hide][/quote]
[hide]
[hide]
[hide]

  :o
 :P
[size=200][color=darkred]thalai u rok da ritu :oops:  :oops: [/color][/size]

 :rotfl:

[/hide]
[/hide]
[/hide]
[/hide]

[quote="pardesi"][quote="rituraj007"][hide]
[hide]
[hide]

 [size=200][color=darkred]CHUMMA JUJUBI[/color][/size]

 :rotfl:

[/hide]
[/hide]
[/hide][/quote]
[hide]
[hide]
[hide]

  :o
 :P
[size=200][color=darkred]thalai u rok da ritu :oops:  :oops: [/color][/size]

 :rotfl:

[/hide]
[/hide]
[/hide]
[/hide][/quote]

the best way is to use Quote option :rotfl:  :D

Exactly.

Godav sucks? :P Ensoi!

And I am not a moderator for this forum. Pardesi is, and he is VERY much in India.

why in the world do u have the virgin islands flag then??? whatever HI MR.MODERATOR.
the hidden content thing..............both of u cool but a waste of my time cuz, i  don't know what u mean. clever way to put things though.

ROFL. He did that just for fun, a LONG time back :)

He is not, in fact, in the Virgin Islands.

okayyyyyy. i'm bored.....we should have brain tesers here, i love brain teasers. like

one fact here is true
two facts here are true
three facts her are true
Four facts here are true

how many facts are true? pretend here is in the invisible box the four statements are in.

yeah one fact, good job rt!!!!!!!!!!!!!! :winner_first:

now that was not actually a TOUGH BRAIN TEASER!

hey, it is one i memorized want a hard one.hold on give me some time buddy, ur going DOWN :D


Try using the multinomial theorem backwards.

what's there to solvre? there's non equals sign.

Read the topic he wants help factoring that.

[quote="tridecalogism_lover"]solve:

$ x^{2}-y^{2}+2yz-2zx-4x+2y+2z+3$[/quote]

I'm not sure if there is a set method. You just have to use your creativity!

[hide]That $ 2z$ looks suspicious. Let's factor it out:
$ x^{2}-y^{2}+2yz-2zx-4x+2y+2z+3$
$ =x^{2}-y^{2}-4x+2y+3+2z(y-x+1)$

Treat $ x^{2}-y^{2}-4x+2y+3$ as a polynomial in $ x$:
\begin{eqnarray*}x^{2}-y^{2}-4x+2y+3 &=& x^{2}-4x+(-y^{2}+2y+3)\\ &=& x^{2}-4x+(3-y)(y+1)\\ &=& (x-3+y)(x-y-1) \end{eqnarray*}

Okay, we got lucky. But the transition from the 2nd to the 3rd step should be clear. The sum of $ 3-y$ and $ y-1$ is $ 4$, so that indicates that they might be the factors. After this crucial step, the problem falls apart quite easily:

\begin{eqnarray*}x^{2}-y^{2}+2yz-2zx-4x+2y+2z+3&=&x^{2}-y^{2}-4x+2y+3+2z(y-x+1)\\ &=&(x-3+y)(x-y-1)+2z(y-x+1)\\ &=&\boxed{(x-3+y-2z)(x-y-1)}\end{eqnarray*}[/hide]

What let you know that:

[hide]
$ x^{2}-4x+(3-y)(y+1)$ was equal to $ (x-3+y)(x-y-1)$ (What hints did you have to be able to change the addition into multiplication)

erdogankerem, it's just a quadratic in $ x$!  Just use the normal method for factoring quadratics. :)


ten people post a phail photo
vote for each photo through a pm
after a week, votes will be counted
winner mods round 2

Make it clearer. Do you want it to say PHAIL on it? Or does it have to be a "fail" photo?
Well
I'll post both
Here's the PHAIL one:[img]http://www.motifake.com/image/demotivational-poster/small/0901/phail-sports-demotivational-poster-1231087368.jpg[/img]
[img]http://www.lolz.se/fun/fail-1.jpg[/img]
that's the other one

it doesn't have to say fail (yes, phail was a deliberate typo)
i once saw a fail pic; it has a man on a bridge contemplating suicide and in the distance he sees a nike swoosh and just do it in big block letters
i'll look 4 it and post it and please post ur own pic and make it appropriate; the link with the oral *** light etc. was very vulgar
u have been blocked from pix posting, Aristocrat
c'mon we need 10

hey, i wanna post a different one, cuz i didn't understand it!!!!!!!!!
here it is:
[img]http://i123.photobucket.com/albums/o308/MCRocks08/FAIL.jpg[/img]

sorry no can do
it is not neccesary to have fail in the pic to qualify as "fail "
pm your pic 2 me
i will approve it
and then post it again in this manner
Phail Photo #1
photo underneath
the week deadline has been reset
all pix have been reset
u must post the pic
not a link
all the rules are the same unless stated otherwise or contradictory 2 these

Why do we have to get your approval :mad:


How do you think the Milky Way will end?

The Milky Way will never end, matter does not end.  :P

If you are trying to hint on it the future collisions it will have with other neighboring galaxies, like M31 (Andromeda), that will not bring about an end to the galaxy either, the galaxies would just combine.  Though, that may then mean the end of Earth.

With that being said what will we call that galaxy?
Androway Galaxy?

mmm, well M32 is sceduled for collsion with the milky way in the FAR FAR FAR future if I remember correct. but from all the research that we have on galactic collisions, we have noted that they are more of a "cosmic dance" than a collision, there isnt much of a chance that "earthlike" planet or any star for that matter will get in the way of another as cosmic distances are too vast.

We do however know that "collisions" generally mean new star formation, as when the total mass of interstellar material increases, more mass accumilates and form new starts. a good example of this is the Anteannae.

after time the wounds will heal and the galaxy will return to its previous form, just bigger. (You can go check recent articles on M32, as it is now proved that it did have a collision with the smaller M31 and now it looks like a normal spiral)

Well, the Milky Way would not necessarily return to its" previous" form.  It will be more massive and may take on a different galactic shape other than a spiral.
M32 did collide with M31, though it is [i]M31[/i] that is the larger of the two and [i]M32[/i] is a satellite of of M31.
As for the the first part, the Milky Way is suppose to collide with [i]M31[/i] not M32.  The distances between stars is too large, as Vader said, for them to collide during a collison but, the heat and change of the distribuition of mass would not be ideal for Earth.

[quote="AstroPhys"] Though, that may then mean the end of Earth.[/quote]

Is'nt Earth "gone" long before that will happen? I mean our own star can get
quite unpleasant in the more near (?) future.

The collision with M31 is expected to occur in 3 billion years, while the sun is expected to be viable for over another 5 billion years.  Though I guess the sun could become unpleasant towards the Earth before then.

supposedly the sun is going to turn into a red giant, enveloping mercury and venus. at that point the earth will be fried

Yeah, that is exactly what is suppose to happen.  As the sun begins to use up it fuel of hydrogen and helium its core of carbon begins to expand and the sun's surface becomes hotter, changing to red.


what is the time difference between GMT and ET? i have looked on the inernet but cant find it and would really ike to take a few courses this year. thanks  :)

Right now, 4 hours.  But after daylight savings time ends, it will be 5 hours.

thanks.

looks like i cant take any classes then.  :(  i was really looking forward to them aswell.


calculus
integration
function
limit
Euler
number theory
totient function

f continuous function

Prove that

$\lim_{n\rightarrow +\infty}\frac{1}{n^2}\sum_{k=1}^{n}f(k/n)g(k)=\frac{6}{\pi^2}\int_{0}^{1}tf(t)dt$

g(k) is Euler indicator

[quote="Moubinool"]
g(k) is Euler indicator[/quote]
Should it rather be $\frac 1{n^2}$ in front of the sum? (if I understand it right, "Euler indicator" is the same as "totient function", i.e., the number of numbers not exceeding $k$ and relatively prime to $k$. It is comparable to $k$ for most $k$, so the sum should be normalized like $\sum_{k=1}^n f(k/n)k$, i.e., with the factor $1/n^2$ to get a meaningful limit...):?

[quote="fedja"]
Should it rather be $\frac 1{n^2}$ in front of the sum? [/quote]

Yes of course


Prove that:
\[ \sqrt{\frac54 a^2-a+1}+2\sqrt{1-\frac{a^2}{4}} \le 3  \]
for any value $a \in [0,2]$.

let's try
[hide]
$\sqrt{\frac{5}{4}a^2 -a +1} +2\sqrt{1 - \frac{a^2}{4}} \leq 3 \Rightarrow \frac{5}{4}a^2 -a +1 + 4 - a^2 \leq 9$
Now we have:

$5a^2 - 4a^2 - 4a \leq 16$

Finally $a^2 - a \leq 4$, which is obvius since a^2 will never be > 4
[/hide]

[quote="JBL"]Two problems: first, you're going in the wrong direction.  Second, $(x + y)^2 \neq x^2 + y^2$.[/quote]

???

That was an enlightening response.

1. You're attempting to start with the inequality-to-prove and prove something true.  That doesn't work: you need to start with true things and arrive at the inequality-to-prove.

2. In your very first step, you "squared both sides."  But you did not, in fact, square the left hand side.

3. I didn't notice this before, but the step from your next-to-last to your last line doesn't make any sense, either.

[hide]
$2\sqrt{1-\frac{a^2}{4}}=2\sqrt{ \left( 1-\frac{a}{2} \right) \left( 1+\frac{a}{2} \right) } \le 1+\frac{a}{2}+1-\frac{a}{2}=2$ by AM-GM

$\sqrt{\frac{5}{4}a^2-a+1}=\sqrt{\left (\frac{a}{2}-1 \right )^2+a^2} \le \sqrt{\left (\frac{a}{2}-1 \right )^2}=\left | \frac{a}{2}-1 \right |$,  cause square root is a monotonic function

so

$\sqrt{\frac{5}{4}a^2-a+1}+2\sqrt{1-\frac{a^2}{4}} \le \left | \frac{a}{2}-1 \right | +2 \le 3 \Longleftrightarrow -\frac{a}{2} \le 0$, since for $a \in [0,2], \left | \frac{a}{2}-1 \right |=1-\frac{a}{2}$[/hide]

[quote="hydro"]
$\sqrt{\frac{5}{4}a^2-a+1}=\sqrt{\left (\frac{a}{2}-1 \right )^2+a^2} \le \sqrt{\left (\frac{a}{2}-1 \right )^2}=\left | \frac{a}{2}-1 \right |$,  cause square root is a monotonic function
[quote]
$\sqrt{\left (\frac{a}{2}-1 \right )^2+a^2} \le \sqrt{\left (\frac{a}{2}-1 \right )^2}$ is not true

After expanding and simplifying, i got:
$81a^3-56a^2-272a+384\geq 0$

from cauchy ,

$\sqrt{5a^2-4a+4}+\sqrt{4-a^2}+\sqrt{4-a^2}\leq\sqrt{3(3a^2-4a+12)}\leq 6$

where the last inequality is equivalent to

$a(3a-4)\leq 0$ which is true for $0\leq a\leq \frac{4}{3}$


Sophia has 15.25 dollars worth of nickels and dimes in her piggy bank. If she has an odd number of dimes and if she has more dimes than nickels, what is the largest number of nickels that she can have?

[hide="Solution"]The most number of dimes possible is $ 151$, meaning the least number of nickels possible is $ 3$. For every $ 2$ dimes we lose, we must replace them with $ 4$ nickels. So, we want $ 151 \minus{} 2x \ge 3 \plus{} 4x$, so $ 148 \ge 6x$, so $ x \le 24$. $ 3 \plus{} 24 \times 4 \equal{} \boxed{99}$ nickels.

[/hide]

If she had the same number of each, this would give $ 101.66666..$, so she must have $ 103$ dimes and $ 99$ nickels.


How do you do this?
 Rocky and Bullwinkle are playing Risk. Rocky rolls one six-sided die, while Bullwinkle rolls two of them. What is the probability that Rocky's roll is as high as Bullwinkle's largest number?

Here's the way I see it. (I'm probably so off)

There is $ 1$ way for Bullwinkle to have a highest of $ 1$, $ 3$ ways for a highest of $ 2$, $ 5$ ways for a highest of $ 3$, and so on.

Case1.
If Bullwinkle gets a sum of $ 1$ ($ \frac {1}{36}$ chance), Rocky is sure to tie/beat him.

Case 2.
Bullwinkle's sum=$ 2$ ($ \frac {3}{36}$ chance), Rocky has a $ \frac {5}{6}$

And so on.

The answer is $ \frac {1}{36}\times\frac {6}{6} \plus{} \frac {3}{36}\times\frac {5}{6} \plus{} \frac {5}{36}\times\frac {4}{6}\cdots$.
or
$ \frac {6 \plus{} 15 \plus{} 20 \plus{} 21 \plus{} 18 \plus{} 11}{216} \equal{} \boxed{\frac {91}{216}}$

I have no idea if theres a general formula for this, sorry, I didn't understand your post mszew  :blush: .

EDIT: @cf249 oops! :oops: It's fixed now.

Textangle, your method will work, but there are a few errors.

First of all, $ \frac{3}{36}\cdot\frac{5}{6}$ is $ \frac{15}{216}$, not $ \frac{8}{216}$.

Also, there are 9 ways to have a highest roll of 5, and 11 ways to have a highest roll of 6 (the work you have done seems to indicate 8 and 9, respectively).

With these corrections, your answer is the same as mszew's.

[quote="mszew"]$ P(1,2) = \frac {W(1) + W(2) + \cdots + W(6)}{6^3} = \frac {1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2}{6^3} = \frac {91}{216}$
(on the numerator the winning cases with a $ 1$, $ 2$, ... , $ 6$ for example: with $ 4$ there are $ 4 \cdot 4$ cases
each dice any number from ${ \{1,2,3,4}\}$

Therefore for $ P(1,k) = \frac {1^k + 2^k + \cdots + 6^k}{6^{k + 1}}$[/quote]

More details for that solution:

Let $ P(R,B)$ be that probability with $ R$ dices for Rocky and $ B$ dices for Bullwinkle.

Calculate $ P(1,k)$
Denominator:
The total amount of results are $ 6^1$ for Rocky and $ 6^k$ for Bullwinkle, both independent so the denominator is $ 6^{1}6^{k} = 6^{k + 1}$

Numerator:
There are $ 6$ mutually exclusive cases depending on the result of Rocky's dice, so the numerator is $ W(1) + W(2) + \cdots + W(6)$
with $ W(i)$ the number of cases when Rocky gets $ i$ on his dice, in that case each of $ k$ Bullwinkle's dices has to be lower or equal to $ i$
so the amount of cases is $ i \cdot i \cdots i = i^k$ getting the following numerator $ 1^k + 2^k + \cdots + 6^k$

With the same logic but a little more complicated formula you can extend for any $ P(R,B)$

Oh thank you. Your method is much simpler  :) .