I have found this limit in a contest given in Italy (1998):

Let (a_n) n>=1 be a series of pozitive real numbers which for any n natural\{0} satisfies the relation: (a_(n+1)-a_n)^2=a_n and we also know that a_n has a limit. Compute the limit of a_n.

bye!

Well, probably everybody has noticed this, but l is either 0 or infinity. The question is which one... :D

It's strange I remember that I saw this problem last year in Romania, but I don't know where... anyway it doesn't matter :D :

Solution:

We consider x_n=a_(n+1)/a_n; so  (a_(n+1)-a_n)^2=a_n becomes : (x_n-1)^2=1/a_n. we have that  x_n>=|x_n-1|-1=1/sqrt(a_n)-1 --> infinity so we get that a_n--> infinity, which is indeed a contradiction. so we get that lim a_n=infinity, because we get that lim a_n does not belong to R, and a_n>0 so we get infinity!!!

cheers! :D

Yeah, I thought this looked familiar. Here's what I did. If an tends to 0 then a(n+1)-an also tends to 0, so there is an m such that |a(n+1)-an|<1 for any n>=m. Then an=(a(n+1)-an)^2<|a(n+1)-an| for any n>=m. If an>a(n+1) then we get an<an-a(n+1), so a(n+1)<0, false. Then a(n+1)>an, so an< a(n+1)-an, so 2*an< a(n+1) for any n>=m, but a sequence which satisfies this cannot tend to 0, so we got a contradiction. an tends to infinity.

I found this problem given at the local olypics in Bucharest last year!!

cheers!

There are two difference equation the solution of which is $a_{n}$...

a) $\Delta_{n}=a_{n+1}-a_{n}= \sqrt{a_{n}}$

b) $\Delta_{n}=a_{n+1}-a_{n}= -\sqrt{a_{n}}$

... and only in case b) a finite limit can exist and this limit is 0. The most simple case is given by $a_{0}=1$ that produces the sequence 1,0,0,...

Kind regards

$\chi$ $\sigma$


I think it's not so hard:

There are 10 player in a ping-pong game. Let us denote by x_1 the numbers of the victorys and by  y_1  the numbers of defeats of the first player. Denote by  x_2 the numbers of the victorys and by  y_2 the numbers of defeats of the second one and so on. Prove that
x_1^2+x_2^2+...+x_10^2=y_1^2+y_2^2+...+y_10^2.

Source: unknown. :(  :(

The way it stands now, it's impossible...

because, a simple counterexample shows that if x_1=3, x_2=3, x_3=3, etc. and y_1=1, y_2=1, y_3=2, y_4=2, y_5=3, y_6=3, etc.

Which is possible, because the total number of wins is the same as the total number of losses

then it isn't true.

I will try now with the additional assumption that each player plays the same number of games, i.e. x_1+y_1=x_2+y_2=x_3+y_3...


Express the following sum as a simple fraction in lowest terms.

\[ \frac{1}{1\times2} \plus{} \frac{1}{2\times3} \plus{} \frac{1}{3\times4} \plus{} \frac{1}{4\times5} \plus{} \frac{1}{5\times6}\]

Easy way: The sum telescopes.

Think about 1/(1*2) = 1/1-1/2, 1/(2*3) = 1/2-1/3, etc. Thus it's just 1/1-...-1/6=5/6

Hard way: add them up bruteforce-ly.

I still don't get it and can you please use Latex.

[quote="mcqueen"]Here is the long-ish way. $\frac {1}{2}+\frac {1}{6}=\frac {2}{3}$. $\frac {1}{12}+\frac {1}{20}=\frac {2}{15}$. Add those fractions [2(a+b)/ab] and add that to $1/30$ o get your answer.[/quote]
I know you can do that, but I want to know the shortcut way.

[quote="Mrdavid445"]Like what James said, that IS the easy way.[/quote]

Um... What, that is NOT the easy way....  (that = what mcqueen did I'm assuming [judging by the quotes])

[hide="The easy way"]
Take a few starting examples....

$\frac{1}{1\cdot 2} = \frac{1}{2}$

$\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} = \frac{2}{3}$

$\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} = \frac{3}{4}$

Notice anything?

In general:

$\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + ... + \frac{1}{n\cdot (n+1)} = \frac{n}{n+1}$ (Why?)

Therefore the sum of $\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}$ is $\boxed{\frac{5}{6}}$.

[/hide]

[quote="Mrdavid445"]Like what James said, that IS the easy way.[/quote]
I know but I don't understand it that well.

[quote="Mrdavid445"][quote="james4l"]Easy way: The sum telescopes.

Think about 1/(1*2) = 1/1-1/2, 1/(2*3) = 1/2-1/3, etc. Thus it's just 1/1-...-1/6=5/6

[b]Hard way: add them up bruteforce-ly.[/b][/quote]

So what you just did was the easy way.

Edit: Oh oops, I see. Mcqueen did the long way.

Edit #2: Ok, then what do you not understand about it Andrew[/quote]
After looking at the problem one more time I get it. Sorry for all the confusion.

I know you can telescope the sum or there a couple variations of this, but I used a simple way that you can do relatively quick. It will not take you $that$ much longer to do it my way than the short way.


Find the area of a heptagon with apothem length $ 15$ feet.

[hide="Solution"]
 [geogebra]9f9b7d01411609d394ba844611aa13dc5a38987e[/geogebra]

$ \tan\theta\equal{}\frac{15}{x}$.

Since one angle of a regular heptagon is $ \frac{5\pi}{7}, \theta\equal{}\frac{5\pi}{14}$.

$ \therefore \tan\frac{5\pi}{14}\equal{}\frac{15}{x} \implies x\approx 2.08$

We wanted $ 15\cdot2x\cdot\frac{1}{2}\cdot 7\equal{}105x$.

Hence, the area of the heptagon is $ 2.08\cdot105\equal{}218.4$.

Answer: 218.4[/hide]


algebra
function
domain
abstract algebra
superior algebra
superior algebra solved

hi everyone,

Let G a finite group anf f a morphism so that f(f(x))=x for every x in G.
How can we prove that f(x)=x^(-1) ???

I really think that you forgot to add something to the problem.

If it is the case that f(x) = x only for x = 1, then it follows that f(x) = x^(-1) and that G is abelian.

Consider the mapping x -> x^(-1)f(x).  This mapping is injective, as x^(-1)f(x) = y^(-1)f(y) if and only if f(xy^(-1)) = xy^(-1) if and only if xy^(-1) = 1 if and only if x = y.  This mapping is then necessarily surjective since the domain and codomain are finite and equal.  Thus given any z in G, there exists an x such that z = x^(-1)f(x).  Hence

f(z) = f(x^(-1)f(x)) = f(x^(-1))f(f(x)) = f(x)^(-1)x = (x^(-1)f(x))^(-1) = z^(-1)

To show that G is abelian, we have that

xy = f(x^(-1))f(y^(-1)) = f(x^(-1)y^(-1)) = f((yx)^(-1)) = yx

for any x,y in G.

I'm not sure yet about the case where f(x) = x for x's other than 1...

- Jeff

the problem's statement is unclear. The title of the thread is "automorphism", but alekk is speaking about a "morphism".

Further on, it is obvious that f(x)=x is the identity automorphism, and f(f(x))=x for every x in G. Obviously f(x) need not be x<sup>-1</sup> !

So, what is the real problem, alekk?

First of all, if f is a morphism and f(f(x))=x, then f is an automorphism because f must be a bijective morphism. I saw a similar problem in a book, but it had the extra condition f(x)=x => x=e. Like Valentin said, I don't think all of this is enough.

The prb I saw asked for this:

a) prove that g(x)=x^(-1)*f(x) is surjective (JayEll did it)
b) prove that G is abelian (JayEll did that too)
c) if |G|=n, then n is odd (this is easy if we consider the extra condition f(x)=x=>x=e)

This was actually given at an OLD final round of the Romanian NMO.

yeah .. i forgot to mention that f(x)=x => x=e ... sorry ..

Very funy !  It was moved in the solved problems .. Frankly , I dont see any post with correct solution ...

the problem was solved by JayEll. Look carefull at this post.


I received this notice by email. I just looked to the web and it seems interesting, but I haven't carefully looked at it. Let's see what does the people in the forum things about it! ;)

                [size=150] [color=blue][b]TurkZeka 2005/2 Weekly Puzzle Competition.[/b][/color]
 [/size]

TurkZeka 2005/2 Puzzle Competition wil start at 14 May 2005  22:00:00 GMT.

This year, each competition (total 4 times in a year-this one is second competition) is going to have 10 puzzles and these puzzle competitions are open to all country.

Competitions will be published in only Turkish, English and Japanese.

You will have exactly 7 days to submit the answers of the puzzles via Answer" form page, beginning from the puzzle's publishing date.

You should read information about answer format in puzzle text carefully before answering. Each correct answer is 100 points and additionally there are two types of  bonus points.

One of the bonus points are given to the first 3 competitors who give the correct answer (30, 20 and 10 point) and the other bonus points are given depending on answering day, beginning from current puzzle's publishing date (30, 25, 20, 15, 10 and 5 point).

Correct answers and solutions will be given when the next puzzle is published. Your score will be updated when you post your answer. You can see whether your answer is correct or not on the "Score Anlayze Table". To compete, you have to sign up to the site and log in.

Please tell friends who are interested in puzzles about these competitions and the site.

Our web address : http://www.turkzeka.com

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Aim : The site TurkZeka.Com was established with a purpose of helping puzzle fans improve their ability of imaging, thinking, finding winding key, heading conclusion, finding out the hidden properties etc. and having nice and useful time at the same time.

Nonograms :
Total 5 new nonograms are published every day.
Nonograms' sizes : 10x10 - 15x15 - 20x20 - 25x25 - 30x30
New types of puzzles will posted coming soon. (Diamond, Mines...)

Just Solve & Enjoy...
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That was fun, I hope the latter rounds are harder though.


i want to take the SAT2 Chem subj test, which book or books should i use to study?
i want to buy Princeton review's Cracking the SAT II Chemistry book but the online comments are negative, saying "this book's practice tests are nothing like the real SAT..."
any suggestions? thanks

I studied off of Baron's (I think)

The test isn't too hard

I was never that great at chemistry, but I did relatively okay when I took the test

i like the barrons books... they are rather hard and overprepare people for the exams  :D , you won't find something you didn't expect but sometimes the practices are much harder than the real test.

I took the SATII Chem this year.  I studied from the Princeton Review Book.  I got an 800 :)

Here's what I say:

1) This test is not very hard.  You'll do fine.
2) DEFINITELY do a review book.  You will definitely want to study.
3) The Princeton Review Book is great for preparing you for the test!  However, the practice tests are nothing like the real test.
4) Good luck!

which is the lastest ed. for barron? i only found 2002


7. Through the orthocenter $H$ of the triangle $ABC$, is driven two perpendicular lines, one of them meets side $BC$ at $X$, second one meets the side $AC$ at $Y$. The lines $AZ,BZ$ are parallels to the lines $HX,HY$ respectively. Prove that the points $X,Y,Z$ are collinear.$\clubsuit$

Denote by $S=BZ\cap HX$ and $T=CZ\cap HY$. We will show that: $\frac{HS}{SX}=\frac{YT}{TH}=k$. Then we are done beacuse we will know that $BS$ and $CT$ both pass trough  a point $D$ for which $\frac{YD}{XD}=k$, and hence $Z=S$.

Let's prove that equality. By angle chasing we find $HBS=TCY=x$ and $XBS=TCH=90-A-x$. From sine law we get that $\frac{HS}{SX}=\frac{BH \sin x}{BX\sin 90-A-x}$ and $\frac{YT}{TH}=\frac{CY \sin x}{CH\sin 90-A-x}$, to finish the problem we need to prove that $\frac{BH}{BX}=\frac{CY}{CH}$. But from sine law in the triangles $XBH$ and $YCH$ we have that this is equivalent with $\frac{\sin 90-A-x}{\sin 90+x}=\frac{\sin A+x }{\sin 90-x}$ where $x=\angle TCY$. The last equality is definitely true, because $\sin a=\sin 180-a$.  :lol:

Remark: The transformation $X\to Y$ is a projective transformation.


calculus integration algebra polynomial derivative AMC USA(J)MO

I'm quite a beginner at calculus,so I wondered if we could post problems strting from the most basic differetial and integral calculus and proceed along the way and learn new things.It won't only help me but even you.(and we all know,to serve god,serve mankind)

Since you wanted the most basic problems from differential and integral calculus, here are two easy ones:

1. Show that every cubic polynomial with real valued coefficients agrees with it's derivative at some point.

2. Find $\int\frac{x}{\sqrt{1-x^{4}}}\,dx$.

By the way, nice idea.  It would be pretty cool if we could keep one of these going :).

[hide="1"]
Take a polynomial of odd degree.  The difference between that polynomial and its derivative is also a polynomial of odd degree, which must have a real root.
[/hide]

[hide="2"]
Substitute $u = x^2$.  The problem becomes

$\frac{1}{2} \int \frac{du}{ \sqrt{1 - u^2} } = \frac{1}{2} \arcsin (x^2)$.
[/hide]

Something a little easier, maybe:

[size=150]Problem 3[/size]

Prove that a cubic polynomial is rotationally symmetric about its point of inflection.

I don't understand what rotationally symmetric means.

If you translate the function so that the inflection point is at the origin, it would be an odd function. In fact, that's probably the proof too.

That's basically the idea.

Since nobody else has posted another, here is a fairly easy one:

[size=150]Problem 4[/size]

Find $\sum_{n=1}^{\infty}\frac{n^{2}}{4^{n}}$.

1. $\int\cos^5(2x)\textsl{dx}$

2. $\int\(x^2}\ln{x}\textsl{dx}$

[quote="shadysaysurspammed"]Ummm..thats simpler?:D,could u tell the solution,maybe start from liitle more simpler like trhe formulas and substituiton thing then proceed,calculus is fun!!!

BTW-how was USAMO toirajiou?[/quote]

The solution doesn't actually need calculus unless you want to use calculus to define the inflection point.

[hide="Solution to my problem"]
Let $f(x) = x^3 + bx^2 + cx + d$ (we can WLOG set the first term to $1$)
$f'(x) = 3x^2 + 2bx + c$
$f''(x) = 6x + 2b = 0$

So the inflection is at $x = - \frac{b}{3}$.  Transform $f(x)$ as follows:

$f(x) = (x + \frac{b}{3})^3 + (c - \frac{b^2}{3})(x + \frac{b}{3}) + (d - \frac{4b^3}{27} - \frac{bc}{3} )$

From which we can clearly see the result.  QED.
[/hide]

And USAMO was bad.  :(

Lets have a simple one

Prove that

$\frac{d e^x}{dx}=e^x$ :D

[hide="Solution"]$\frac{d}{dx}e^{x}=\lim_{h\rightarrow 0}\frac{e^{x}\left(e^{h}-1\right)}{h}$

$e=\lim_{h\rightarrow 0}\left(1+h\right)^{\frac{1}{h}}\implies e^{h}=\lim_{h\rightarrow 0}\left(1+h\right)$

Plugging this in gives

$\frac{d}{dx}e^{x}=\lim_{h\rightarrow 0}\frac{e^{x}\left(1+h-1\right)}{h}=\lim_{h\rightarrow 0}e^{x}=e^{x}$[/hide]

I derived a non-rigorous proof of the quotient rule that didn't apply the chain rule once in 8th grade...

[hide="Here it is"]
Define $df = f(x + dx) + f(x)$ and $dg = g(x + dx) + g(x)$.

\begin{eqnarray*} \frac{d}{dx} \frac{f(x)}{g(x)} &=& \lim_{dx \to 0} \frac{ \frac{f(x) + df}{g(x) + dg} - \frac{f(x)}{g(x)} }{ dx} \\ &=& \lim_{dx \to 0} \frac{ ( f(x) + df) g(x) - f(x) ( g(x) + dg ) }{ g(x) g(x + dx) dx} \\ &=& \lim_{dx \to 0} \frac{ f(x) g(x) + g(x) df - f(x) g(x) - f(x) dg}{g(x) g(x+dx) dx} \\ &=& \lim_{dx \to 0} \frac{ g(x) \frac{df}{dx} - f(x) \frac{dg}{dx} }{ g(x) g(x+dx) } \\ &=& \boxed{ \frac{ g(x) f'(x) - f(x) g'(x) }{ g(x)^2} } \end{eqnarray*}[/hide]

And similar logic can be applied to the chain rule,

[hide="Like so"]
\begin{eqnarray*} \frac{d}{dx} f( g(x) ) &=& \lim_{dx \to 0} \frac{ f(g(x + dx)) - f(g(x))}{dx} \\ &=& \lim_{dx \to 0}\frac{f(g(x) + dg) - f(g(x))}{dx} \\ &=& \lim_{dx \to 0}\frac{f(g + dg) - f(g)}{dg} \cdot \frac{dg}{dx} \\ &=& f'(g) \cdot \lim_{dx \to 0} \frac{dg}{dx} \\ &=& \boxed{ f'(g(x)) g'(x) } \end{eqnarray*}[/hide]

That's exactly the sort of intuitive reasoning that the hyperreal number system lends rigor to. (If you look carefully, you'll see that you pretty much did what I was talking about in your chain rule proof). I think it is very interesting that it is so difficult to make these arguments rigorous -- I wish I understood the reason for that better.

On a vaguely related note, Conway once challenged my analysis class with the following question: why is there such a big difference between a function $f(x)$ being continuous on $(a,b)$ versus continuity on $[a,b]$? It is not hard to think of examples that make the difference obvious -- any function continuous on $[a,b]$ is integrable on the same interval, for example, but this is not necessarily the case for a function continuous only on $(a,b)$. But $a$ and $b$ are just two points! When you learn Lebesgue integration, you learn that you can pretty much ignore large (edit: large in comparison to two points, I mean. They still have to be fairly small [measure 0, but possibly still uncountable]) sets of points in many situations (like when integrating or taking derivatives). (That's right, you can take a derivative of a function that's not continuous. It's pretty sweet.) But in this particular situation, just two points make a huge difference. In fact, even just one of those points has the same importance. Why is this one point such a big deal?

Conway didn't have an answer for this; I think he just wanted us to think about it. I think it is a good "elementary" example of how analysis can be very subtle.

Here's a really easy exercise to go along with my babbling: come up with a function $f(x)$ that is continuous on $[a, b)$ but such that $\int_a^b f(x)dx$ is divergent.

Well, one answer is that $[a,b]$ is closed with $(a,b)$ is not. Due to Heine-Borel's theorem local properties of a function in $[a,b]$ can be made into global properties, while this is not true for $(a,b)$.

I mean, it's really not just about the two points. If we have a function that is defined on $[a,b]$ then we will have no problems considering it on $(a,b)$. But when we have a function that is not defined on the endpoints it might become very large close to the endpoints. I don't really see how this is a property of two points, it's really a property of the function.

[quote="Xevarion"]The quotient rule is really just an application of the product rule and the chain rule. The product rule is not difficult; I think someone posted about proving that the limit of a product is the product of the limits just a little while ago here and Professor Merryfield gave a hint describing how to approach it.

I actually don't remember the classical proof of the chain rule, but the trick of saying $g(x) = y$, so $f(g(x)) = f(y)$, and then $df/dy = f'(y)$, and then multiplying by $dy/dx$ is not a legitimate rigorous argument using the normal definitions.

You can prove the chain rule using that argument if you define differentials $dx$ appropriately using the hyperreal numbers. The differentials are infinitesimals, and it can be shown that $df/dy$ (for example) differs from the standard definition of the derivative only by an infinitesimal quantity, so you just look at the nearest real. Similarly, multiplying two "derivatives" together as above is legit because you will only create an error on the order of a real times an infinitesimal, which is still an infinitesimal.

I think this is really cool.

edit: since you only need to use a specific case of the chain rule, you don't necessarily need to prove it in general to prove the quotient rule. It would be cool to see a proof of it that doesn't require assuming the chain rule. I'd try to do it now but I have to do research for my history class...[/quote]

If you say that $f$ is differentiable at $x$ if and only if $f(x+h)=f(x)+hf'(x)+o(h)$, then the chain rule falls out pretty much right away.

[hide="Product & Quotient Rules"]
$\frac{d}{dx}f\left(x\right)g\left(x\right)=\lim_{h\rightarrow 0}\frac{f\left(x+h\right)g\left(x+h\right)-f\left(x\right)g\left(x\right)}{h}$

$=\lim_{h\rightarrow 0}\frac{f\left(x+h\right)g\left(x+h\right)-g\left(x+h\right)f\left(x\right)+g\left(x+h\right)f\left(x\right)-f\left(x\right)g\left(x\right)}{h}$

$=\lim_{h\rightarrow 0}\left(g\left(x+h\right)\frac{f\left(x+h\right)-f\left(x\right)}{h}+f\left(x\right)\frac{g\left(x+h\right)-g\left(x\right)}{h}\right)$

$=f'\left(x\right)g\left(x\right)+f\left(x\right)g'\left(x\right)$

To derive the quotient rule, consider $\varphi\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}$.  Therefore we have $f\left(x\right)=\varphi\left(x\right)g\left(x\right)$.  Using the product rule we have just derived, we get

\[ f'\left(x\right)=\varphi '\left(x\right)g\left(x\right)+\varphi\left(x\right)g'\left(x\right)  \]

\[ \varphi '\left(x\right)=\frac{f'\left(x\right)-\varphi\left(x\right)g'\left(x\right)}{g\left(x\right)}  \]

Now use the definition of $\varphi\left(x\right)$ and multiply the numerator and denominator by $g\left(x\right)$ to get

\[ \varphi '\left(x\right)=\frac{f'\left(x\right)g\left(x\right)-f\left(x\right)g'\left(x\right)}{g\left(x\right)^{2}}  \]
[/hide]

[hide="Chain Rule"]
I suppose this is basically the same as previous posts, but here's another version of it (the first one I learned).  However, there is a huge hole in this proof and it would take forever to write out a real proof.

$\frac{d}{dx}f\left(g\left(x\right)\right)=\lim_{h\rightarrow 0}\frac{f\left(g\left(x+h\right)\right)-f\left(g\left(x\right)\right)}{h}$

Now let $g\left(x+h\right)=g\left(x\right)+k$, and as $h\rightarrow 0$, $k\rightarrow 0$, so we have

$=\left(\lim_{k\rightarrow 0}\frac{f\left(g\left(x\right)+k\right)-f\left(g\left(x\right)\right)}{k}\right)\left(\lim_{h\rightarrow 0}\frac{k}{h}\right)$

$=f'\left(g\left(x\right)\right)g'\left(x\right)$

Because $k=g\left(x+h\right)-g\left(x\right)$.
[/hide]


How many colors are needed to color the regions in three dimensional space, where the regions are obtained by partitioning the space using planes ?

This type of  question is new to me so I hope you guys can help me.

I assume you want to use different colors for regions that have a "face" in common", otherwise one color is enough! It can be done with two colors only, by induction. For one plane only, color the two regions white and black. Assume you can do it for $ n$ planes.
One more plane will "cut" through some of the previous regions. Just reverse the colors for all the regions situated on one side of this new plane. Same proof works in the plane, partitioned by lines, so you may want to check it there, it is easier to visualize.

Or interpret as a graph, note every minimal cycle is of even length, so it's bipartite.


One ticket to a show costs $ \$20$ at full price. Susan buys 4 tickets using a coupon that gives her a $25\%$ discount. Pam buys 5 tickets using a coupon that gives her a $30\%$ discount. How many more dollars does Pam pay than Susan?

$ \textbf{(A)}\ 2 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 20$

They've already been posted, sorry :|

http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=44&year=2007

Please edit your post to remove the extra info, so it can go in the Resources section.  Thanks!

[quote="King_James23"]This problem is simple, the answer is [b](C.) 10[/b][/quote]Refrain from posting uninformative posts like most of your other posts elsewhere, the '23 guy' or 'rohit lebron'.
[hide]If $p=\$20$ is price of ticket, Susan spends $(1-25\%)(4p)=3p$ and Pam spends $(1-30\%)(5p)=3.5p$. So Pam pays $3.5p-3p=0.5p=\$10$ more than Susan. $\boxed{C}$[/hide]


$ x^3y dy \equal{} (x^4 \plus{} 3x^2y^2 \plus{} y^4)dx$

They're homogeneous right?  What do I do when they're homogeneous though?

:idea: Search for "homogeneous differential equations"
http://en.wikipedia.org/wiki/Homogeneous_differential_equation


Determine all pairs $(a, b)$ of positive integers such that $ab^2+b+7$ divides $a^2 b+a+b$.

[quote="Peter"]Determine all pairs $ (a, b)$ of positive integers such that $ ab^2 \plus{} b \plus{} 7$ divides $ a^2 b \plus{} a \plus{} b$.[/quote]
$ ab^2\plus{}b\plus{}7|a^2b\plus{}a\plus{}b$
$ \Rightarrow ab^2\plus{}b\plus{}7|a^2b^2\plus{}ab\plus{}b^2$
$ \Rightarrow ab^2\plus{}b\plus{}7|a(ab^2\plus{}b\plus{}7)\plus{}b^2\minus{}7a$
$ \Leftrightarrow ab^2\plus{}b\plus{}7|b^2\minus{}7a$
Because $ ab^2\plus{}b\plus{}7>b^2\minus{}7a \rightarrow b^2\equal{}7a$
Let $ b\equal{}7k,a\equal{}7k^2$
Replace to the condition :
$ 7^3k^4\plus{}7k\plus{}7|7^3k^5\plus{}7k^2\plus{}7k$
$ \Rightarrow 7^2k^4\plus{}k\plus{}1|k(7^2k^4\plus{}k\plus{}1)$
It is true for all $ k\in N$
So the solution is $ (a,b)\equal{}(7k^2,b\equal{}7k)$

[quote="TTsphn"][quote="Peter"]Determine all pairs $ (a, b)$ of positive integers such that $ ab^2 \plus{} b \plus{} 7$ divides $ a^2 b \plus{} a \plus{} b$.[/quote]
$ ab^2 \plus{} b \plus{} 7|a^2b \plus{} a \plus{} b$
$ \Rightarrow ab^2 \plus{} b \plus{} 7|a^2b^2 \plus{} ab \plus{} b^2$
$ \Rightarrow ab^2 \plus{} b \plus{} 7|a(ab^2 \plus{} b \plus{} 7) \plus{} b^2 \minus{} 7a$
$ \Leftrightarrow ab^2 \plus{} b \plus{} 7|b^2 \minus{} 7a$
Because $ ab^2 \plus{} b \plus{} 7 > b^2 \minus{} 7a \rightarrow b^2 \equal{} 7a$
Let $ b \equal{} 7k,a \equal{} 7k^2$
Replace to the condition :
$ 7^3k^4 \plus{} 7k \plus{} 7|7^3k^5 \plus{} 7k^2 \plus{} 7k$
$ \Rightarrow 7^2k^4 \plus{} k \plus{} 1|k(7^2k^4 \plus{} k \plus{} 1)$
It is true for all $ k\in N$
So the solution is $ (a,b) \equal{} (7k^2,b \equal{} 7k)$[/quote]

This solution is incomplete; when $ b^2 \minus{} 7a < 0$, two solutions $ (11,1)$ and $ (49,1)$ are yielded.

It is easy to prove that those two are the only ones satisfying the equation.


when i got this prob... i 4got 2 ask it on the 4um immediately  :blush:

it was the motorcycle prob (if any of u hav ever seen it)

a motorcycle is valued at 6,000. The first year, its value decreases by 20%. Each year after, the value decreases by 10%. What is the value of the motorcycle after 10 years? i know the ans is 3,149.28, but is ther an ezer way?

ya
there is something that i vaguely remember
something like: (1-__)^__
the __ is the number that you put in
forgot though
something like that
look for it though :)

Well you didn't give us your method but here is how I would do it:

After 1 year, it goes down 20%; simple enough.  Lets call this new price $ C$.
After that, its price is modeled by $ P=C(1+\Delta\text{Price})^{t}$.  In this case, $ \Delta\text{\prime}=-.1$ and $ t=9$ years is the time so our equation is
$ P=C(.9)^{9}$.

How would you go about solving this in the countdown or sprint round because the expression would simplify to $ 4800*\frac {9^9}{10^9}$
EDIT: @ Ernie do you mean 5 years, because I finished the computation and my expression did not equal your given answer but i discovered that 4800*(9/10)^4= 3149.28

I believe that I have reduced it as much as possible:
$ \frac{3^9*2^2}{5^2}$

thx 4 ur help!  :D

@mathemagician1729- no, it was 10 years

oh, duh! i know how 2 do that  :blush:

but i dont no how to find (9/10)^9 really quickly

can sum1 revive ernie?  :D


A custom made window is a single pane of glass consisting of a semicircular sheet of glass on top of the narrow end of a rectangular sheet of glass. The rectangle has dimensions of 4 feet by 3.5 feet. How many feet of weather-stripping will be needed to go around the glass?

[hide]The rectangular part is 4+4+3.5 = 11.5 . the semicircle is $\frac{1}{2}(3.5\pi)$ or about 5.5. You need about 17.5 feet of strip. [/hide]


Here's a problem from the 2004 National Target Round:

An edgy spider walks only along the edges from A to B of the dodecahedron formed by the net shown.  What is the number of edges in the shortest path that the spider could take?

(The picture is attached)

[hide]I think the shortest path is 3 edges. Remember that this is a web, and when it is folded into a dodecahedron, some of the edges join into one edge, and new paths are formed.[/hide]

i seem to remember the ans. was [hide] 3 , so i think you're right, i_like_pie. Here's the path that I found:
1)Go across the edge that is not connected to vertex A in the web, but is when you fold it.
2) Go to the vertex from there that is also not connected on the web, but is when folded. This is the second vertex counterclockwise from B.
3) Go to the vertex next to B
4) Go to B

That is 4 vertices, so there are 3 edges the spider needs to got through. [/hide]

For those with difficulty viewing a DOC file, here's the picture.

To help with the solution, I've colored the lines you move across a thick red.  The green lines mean the points are coincident when you fold the net.


It's up now!  Good luck on round 4 everybody!  :)

[quote="nsato"]You can thank us by writing great solutions.   :D[/quote]

4 solutions written...and already a pretty messy one.

i dont know why, but the last round seems to be easier than others.
i did 4 in the night the problems came out. it leaves me a full month to do one more math problem.

[quote="K81o7"][quote="zanttrang"]Okay, we're getting dangerously close to discussing unauthorized aspects of the round here.  Let's let our comments lie until we turn in the contests in March  ;)[/quote]

Would it then be better if this thread were locked until then?[/quote]
Probably, but I'm not a mod, so... *shrugs* one will do it eventually if they see the need.

Ok, yeah, I'm gonna lock it.

Please do not further discuss Round 4 until after it's over.  Thanks.


The median of the set of integers {5, 8, 12, 7, x} is the same as the mean. What is the sum of all the distinct possible values of x?

[hide="Answer"]$\frac{5+8+7+12+x}{5}=\frac{32+x}{5}$ is an integer. The median can never be $5$ or $12$, since there can only be at most one integer ($x$) that is less than or greater than $5$ or $12$, respectively. Therefore, the options are $7$ and $8$ as medians, since $x$ cannot be between the two, and therefore cannot have two numbers smaller and two larger in the set. If $7$ is both the median and the mean, then $x=3$, and if $8$ is both the median and the mean, then $x=8$, and since the numbers do not have to be distinct, this is a possible value. Therefore, the sum of the two values is $3+8=11$.[/hide]

[hide]
This is a good problem.  The hardest part is to notice that median can either be 7 or 8, depending on whether x is lower than 7 or greater than 8.  If it's 7, then x has to be 3, and if it's 8, then x has to be 8.  The sum is thus 3+8=11.[/hide]


Michael walks at the rate of $ 5$ feet per second on a long straight path. Trash pails are located every $ 200$ feet along the path. A garbage truck travels at $ 10$ feet per second in the same direction as Michael and stops for $ 30$ seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leaving the next pail. How many times will Michael and the truck meet?

$ \textbf{(A)}\ 4\qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 8$

Sorry for that, but I don't even know about the answer sheet, do you have that?

For solution, I just solve it by brute force. Suppose Michael start at zero, I tried and found they will meet at
1.1000
2.1200
3.1300
4.1400
5.1600
since Michael pass through every 200 feet using 40s, truck pass through 200 feet and a trash can using 50s, so finally Michael goes forwards truck and they never meet again. (since the truck stop at 1600 until Michael go to 1750, so Michael can escape the truck before the truck reach 1800). Obviously, they will never meet again :)

This is a somewhat fudgy way but it works.

Notice that the truck travels at a rate of 50 seconds per garbage stop and Michael travels at a rate of 40 seconds per garbage stop. This means that Michael will eventually reach a garbage stop just as the truck leaves it. That's intersection #1. Michael gets to the next garbage stop 20 seconds after the truck does and leaves 10 seconds before it does. That's intersection #2. The truck passes him halfway to the next stop (this is the only tricky part) for intersection #3. Michael arrives at the next stop 10 seconds after the truck does, meaning he leaves 20 seconds before. That's intersection #4. Finally, the truck arrives at the next stop right as Michael leaves, for #5.

[quote="Phelpedo"]Michael travels at a rate of 40 seconds per garbage stop[/quote]

Well, the reason I got this one wrong was because I found that he travels at a rate of 42 seconds per garbage stop.  So I wasted a lot of time on this problem while numbers like 40 and 50 are so much nicer.  And in the end, I got it wrong.  Oh well.

Here's a pretty clean solution:

Michael walks 200 feet in 40 seconds, and the garbage truck moves 200 feet in 20 seconds, and then waits for 30 seconds.  Let's find which pails they meet at first, and then we'll worry about the other meetings.  Suppose Michael starts at pail 0, and the truck starts at pail 1.  Then Michael passes pail $ n$ after $ 40n$ seconds, and the garbage truck arrives at pail $ n$ after $ 50(n \minus{} 1) \minus{} 30$ seconds and leaves pail $ n$ after $ 50(n \minus{} 1)$ seconds.  Therefore they meet at pail $ n$ iff $ 50(n \minus{} 1) \minus{} 30\le 40n\le 50(n \minus{} 1)$.  Dividing by 10, we get $ 5n \minus{} 8\le 4n\le 5n \minus{} 5$.  Subtracting $ 4n$, we get $ n \minus{} 8\le 0\le n \minus{} 5$.  Therefore they meet at pails 5, 6, 7, and 8.

Michael is at pails 5 through 8 at 200, 240, 280, 320 seconds respectively, and the truck is at pails 5 through 8 at 170-200, 220-250, 270-300, and 320-350 seconds respectively.  Therefore, Michael and the truck also meet between pails 6 and 7, since Michael leaves pail 6 first and arrives at pail 7 last.  Therefore, there are 5 total meetings.

The first time they meet is conveniently right when the truck starts leaving...

20 sec later: Michael is 100 behind.

30 sec later: 50 ahead. so 2nd meeting.

20 sec later: 50 behind. 3rd meeting because they must have met sometime along the way

30 sec later: 100 ahead. 4th meeting.

20 sec later: TIED: 5th meeting.

No more...

I forgot to include the meeting in the middle -_-.

dang...

ironically the first time i did, i got 5 (number 25 on 10) (but i did it the wrong way, like its not even totally close to what the problem was asking)...

when i did it the right way, I got 4 but I forgot to include the one in the middle...

maybe i shouldnt check my answers.

If you make a line with dots it helps.
You realize that for every time period for the garbage truck (time from leaving pail a to leaving pail b) that Michael gets 50 closer, or more ahead
so they meet exactly at one stop, he's 50 ahead at the next, and so on, leaving 5 stops
0, 50, 100, 150, 200 (Amount Michael is ahead per time period in during meeting periods)

you can also do it this way.

The distance between michael and the truck is $ 200$.

Michael GAINS $ 150$ feet every time the truck stops (which I'll signify by STOP).

Michael loses $ 100$ feet every time the truck goes (which I'll signify by GO).

200 + 100 = 300  GO
300 - 150 = 150  STOP
150+100= 250  GO
250-150 = 100  STOP...etc.
100+100 = 200
200-150=50
50+100=150
150- 150 = 0 <--MEET
0+100 = 100
100-150 = -50 <----MEET (in between because it crossed 0)
-50+100 = 50 <-----MEET(in between because it crossed 0)
50-150 = -100 <----MEET(in between because it crossed 0)
-100+100 = 0 <----MEET
0-150 = -150
-150+100=-50
-50-150 = -200 and etc.

wow.... of all the questions, this is the only one i actually solved....

I HATE GEOMETERY.

Think of it like a game: distance truck travels vs. distance kid travels. We start out with kid losing by 200, truck goes first. Truck and kid take turns: when the truck goes, it gains 100. When the kid moves, he gains 150. Also, nothing interesting happens until the kid first meets the truck:
0 -> -100 -> 50 -> -50 -> 100 -> 0

The score changes sign $ \boxed{5}$ times  :P

[quote="aidan"]Best way would be to draw a distance-time graph if you had graph paper. The scale doesn't matter so long as the ratios are correct. It gives 5 intersections, so that's the answer  :P[/quote]
Yeah... my school gave us graph paper and I didn't think of this. I ended up spending 10 minutes and putting down A finally. Yuck.

[quote="aidan"]Best way would be to draw a distance-time graph if you had graph paper. The scale doesn't matter so long as the ratios are correct. It gives 5 intersections, so that's the answer  :P[/quote]

Yep. I read the problem with 10 minutes left, spent maybe 20 seconds trying to come up with an algebraic solution, then just grabbed some graph paper and a ruler and blitzed it.  :) Easiest AMC #25 ever.

[quote="allanyangtl"]:maybe: What is the official answer? Why did I use diagram to slove it and got the answer they met 4 tmes?
first start at the point they first meet,sign it as No.1
then the boy use 40s and truck use 20s and wait 30s in the next dustbin,so they can meet too,sign it as No2
the boy reached the next dustbin at the 80 second,truck get there in the 70th second and wait for 30s,so they can meet,sign it as No3
the forth dustbin the boy get in the 120 second and the truck get in the 120 second ,they can meet,sign in as No4
the fifth dustbin the boy get in the 160 second and the truck get in the 170, they can't meet,
the sixth dustbin the boy get in the 200 second and the truck get in the 250.....
......
so they can only meet 4 times,isn't it?[/quote]

No, since between 40s and 80s the truck passed the boy(who left first but arrived at the next dustbin last). So they meet 5 times.

Wait this is a pretty non-rigorous solution but: [hide="Non-rigorous Solution"] Notice that the truck takes 50 seconds per garbage stop and Michael takes 40 seconds per garbage stop. Through some simple estimation, we realize that the first time they meet, they're both leaving the same garbage stop at the same time. So then, since I ran was running out of time, I just did the method that : $\frac{50}{50-40}=\boxed{5}$. In hindsight this worked only because they were leaving the stop at the same time.[/hide]

Also #25 on 2008 AMC 10B

[quote=archimedes1]Or you could realize that Mike's overall speed is faster than the truck's, so the number of intersection [i]must[/i] be odd.  Then you could guess with a 50-50 chance, save youreslf pain and move on.[/quote]

Or you could solve the problem, save yourself pain of losing an easy $6$ points (for AMC 12) and move on.

Can someone give me a nice solution to do this, I am confused looking at the others. How do you do these types of problems without considering the individual steps...?

Thanks

@below That is what I did when I was solving this, but I was just wondering if there is a 'synthetic' way to solve this without having to consider any of the individual steps, I feel whenever I solve a problem like this, it isn't a 'legitimate' solution.

I believe the easiest solution is simply to graph their positions


A) if x is the second quadrant angle for which cos(x) = -8/17
determine tan((3x)/2).

B) if sin(2x) = 21/25, and cos(x) > sin(x) then compute cos(x)-sin(x).

[hide="B"]
$ 1\minus{}sin2x\equal{}(cosx\minus{}sinx)^2 \\
4/25\equal{}(cosx\minus{}sinx)^2 \\
cosx\minus{}sinx\equal{} \pm 2
\ but\ cosx>sinx\\
cosx\minus{}sinx\equal{}2$ [/hide]

[hide="Solution to A"]Draw a right triangle in QII and use Pythagorean to find that the third side is 15. Therefore, $ \tan x \equal{} \frac {15}{ \minus{} 8}$, $ \cot x \equal{} \frac { \minus{} 8}{15}$, and $ \csc x \equal{} \frac {17}{15}$.

Use this identity:
$ \tan\left(\frac {3\theta}{2}\right) \equal{} \tan\left(x \plus{} \frac {x}{2}\right) \equal{} \frac {\tan x \plus{} \tan\left(\frac {x}{2}\right)}{1 \minus{} \tan x\tan\left(\frac {x}{2}\right)}$.

And this identity:
$ \tan\left(\frac {x}{2}\right) \equal{} \csc x \minus{} \cot x$

To get:
$ \frac {\frac { \minus{} 15}{8} \plus{} \frac {17}{15} \plus{} \frac {8}{15}}{1 \plus{} \frac {15}{8}\left(\frac {17}{15} \plus{} \frac {8}{15}\right)}$

Which simplifies down to $ \boxed{ \minus{} \frac {5}{99}}$[/hide]


[b]1.[/b] When I enter mathlinks.ro I see the following text:

[quote]Bad Request
Your browser sent a request that this server could not understand.
Size of a request header field exceeds server limit.

--------------------------------------------------------------------------------

Apache/1.3.26 Server at http://www.artofproblemsolving.com Port 80[/quote]

[b]2.[/b] Reloading doesn't help, hence I go to http://www.artofproblemsolving.com, then log in on the forum. Now, after having entered the forum, I click on "Advanced Section" in the left upper corner, but instead of getting in the Advanced Section, I arrive again at the AoPS main site.

Any ideas on these two bugs?

  Thanks...
  Darij

PS. Please delete this thread when the bugs are fixed.

[quote="darij grinberg"][b]1.[/b] When I enter mathlinks.ro I see the following text:[/quote]

I should actually have said it happens when I enter http://www.mathlinks.ro. Without www everything works well!

  Darij

[hide="test"]
[hide="see below"]
[hide="experiment"]
Now I am misusing this obsolete thread for testing some possibilities on the new board.

$a^2+b^2=c^2$ [/hide][/hide][/hide]

1.  http://www.mathlinks.ro works for me....

For 1., try deleting your cookies.  That same exact error happened to me, but when I deleted my cookies, everything worked fine.

But does anyone know why that happened?

Deleting the cookies seems not to be sufficient for me :(

Pierre.

Still having problems?  I reworked how mathlinks.ro accesses the Forum about 30 min after you posted.

Yeah!! It seems ok now :)  :)

Thanks again.

Pierre.


let $ N\equal{}111111.....11(1998 ones)$
find the 1000th digit after the decimal point of $ \sqrt{n}$

a much better approach would be to experiment with small values of n.then we observe a pattern in the digits after the decimal point

There is no "much better approach", the one given already is in some sense "general" (and thus better than a guess ;)).

But the main reason why I respond: "nice problem" is a terrible title. Try to describe some aspect of the problem in the titles instead of your feelings towards the problem (I don't think decimal-problems are nice, by the way). "1000th digit of sqrt(111.11)" doesn't sound that bad, does it¿

[quote]Message
ZetaX There is no "much better approach", the one given already is in some sense "general" (and thus better than a guess ).

But the main reason why I respond: "nice problem" is a terrible title. Try to describe some aspect of the problem in the titles instead of your feelings towards the problem (I don't think decimal-problems are nice, by the way). "1000th digit of sqrt(111.11)" doesn't sound that bad, does it¿
 [/quote]
how can you assume that if you dont like the problem others wont.
as for the title sorry :oops:

[quote="ZetaX"]
[b]I don't think[/b] decimal-problems are nice, by the way
[/quote]
Just like how you like the problem , he doesnt like decimal problems ,it is his view right ?. He never said that no one should like the problem ,he just said [b]he[/b] doesnt like them .  Even you said
" a much better approach " ,so even I can ask you "How can you assume that yours is a better approach in general" , :P :lol:
No offence my friend   :lol: ,sorry for interruption . As far as "stepehencheng's " approach ,it is fine ,if you think you have a better approacb please post it .
And again No offence.

[quote]how do you solve this aadil? even if you find the pattern, you need to prove that your pattern is correct. i'm curious how do you prove your pattern, if it's different from stephencheng's proof.[/quote]
find a pattern in the digits after the decimal place and induct on it


algebra
polynomial
function
calculus
derivative
integration
calculus computations

I haven't been able to solve this problem after 30 minutes. Can someone help me?

"Let P(x) = 7 - 3(x-4) + 5(x-4)^2 - 2(x-4)^3 + 6(x-4)^4 be the fourth-degree Taylor polynomial for the function f about 4. Assume f has derivatives of all orders for all real numbers."

1. Write the second-degree Taylor polynomial for f' about 4 and use it to approximate f'(4.3)

2. Write the fourth degree Taylor polynomial for g(x) = \displaystyle \int_4^x f(t) dt.

EDIT: Took out some useless '.

isn't that just termswise integration/derivation? :?

Peter VDD- its not just termwise integration/differentiation.

$P(x)$ is the Taylor Approximation of a function. the questions wants you to find a function $f$ that is approximated by $P(x)$

once you find $f$, then for #1 you do its derivative and then use the formula $P_{n}(x) = \sum^{\infty}_{k=0}\frac{f^{(k)}(a)}{k!}(x-a)!$ for two terms (since its asking for second order) then you will plug in 4.3 for $x$ in order to find the approximation.

you use the same formula for the second problem.

im still, however,  trying to figure out which function $f$ is approximated by the $P(x)$ taylor polynomial given by Tarento...lol

No, you're not supposed to find f.  You are given information about f in the form of its partial Taylor exapansion and you are supposed to use that to find the other Taylor expansions you are asked for -- you absolutely don't need f to do this.

I thought you could just derivative the P(x), but it didn't work when f(x) = ln x.  :(

did you ever figure out the solution to this problem? if you did, can you explain it? how can you approximated f'(4.3) without finding f(x)?

One of the purposes of taylor polynomials is to approximate values of functions that can't be written as a combination of elementary functions (i.e. antiderivative of e^(-(x^2))).  If they give you the third degree Taylor polynomial for f(x), just take the derivative to get the second degree Taylor polynomial for f'(x) and plug in 4.3 for x to get an approximation.


I just checked the AMC website and saw that the AIME's this year will be on MUCH earlier dates than last year. Does anyone know whether this is correct?

Yes, if by much you mean two days earlier than last year.

If you're referring to the dates posted [url=http://www.unl.edu/amc/whatswhen.html]here[/url], then yes.  (I don't know about the ones in 2005) The ones for 2004 are:

AIMEA - Tuesday, March 23, 2004
AIMEB - Tuesday, April 6, 2004

USAMO - Late April or Early May 2004

Oh! Haha! I was looking at the wrong column, at the dates for next year! :D :D :D

I freaked out because if the dates were much later, then my school would take the AIME A, and I wouldn't have spring break to work on past AIME's.

YAY!

Oh, are you sure you're not looking at the 2004-2005 dates?  Somebody forgot to change the box that says "2003-2004
Tentative Dates" to "2004-2005 Tentative Dates".

[quote="Fierytycoon"]Oh wait. I just looked again and BOTH columns were for 03-04. What's up with this?[/quote]

Like I said, someone forgot...


Let a,b be real numbers, a<b. Prove that, if a<=x<=y<=b then

            0<= 2xy - (a+b)(x+y) + a^2 +b^2 <= (b-a)^2.

Just see that if we increase all the variables by t (x'=x+t etc.), values of the terms will be the same. So we can WLOG assume that a=0.
Then we have to prove
$0\leq 2xy-bx-by+b^{2}\leq b^{2}$. We can write the left part like this:
$0\leq xy+(x-b)(y-b)$, since $0\leq x,y$ the first term is nonnegative and since x-b and y-b are both negative, the second term is nonnegative as well.
The right part can be written like this:
$y(x-b)+x(y-b)+b^{2}\leq b^{2}$, substracting $b^{2}$
$y(x-b)+x(y-b)\leq 0$
Because $x,y\leq b$ both the terms on LHS aren't positive.

Just a little hint for dina013: putting a $\$$ sign before and after math formulas will make your posts another bit better :)

thanks a lot Kondr :wink:  nice solution.


How about lets make an AoPS MC test. Maybe not even mathcounts format. i have no idea if this will work. Just post problems you like and we can organize them into something. if there is enough interest, then we can make another topic on for it.

Actually, many people from my school already made one, though mine is still in revision. If they want they can share with Aopsers.

Wait, is it everyone make one or one person make one and shares it?

This has already been done by some others, including #H34N1.

MAKING one..... no.

TAKING one...... yes.

I need all the practice I can get my hands on.

There already was one. If one was to be made, it should be around the right level.

I am currently making a mock MC test right now (it only uses Mathcounts subjects, but is pretty hard, maybe a slight bit above national level).

I just need to make up a few more sprint questions at the end and 1 more target question.

[quote="PI-Dimension"]include lots of tricks that need some ingeniuty, for example...

given that:

2x+y+z=12
x+2y+z=15
x+y+2z=21

find (x,y,z)

[hide="solution"]

adding the three equations gives:

4x+4y+4z=48
x+y+z=12

therefore, by subtracting that from the given equations, we have x=0, y=3, z=9

therefore the solution is (0,3,9)

[/hide]

Yeh... everyone knows that one..... but its a start![/quote]
Even without that insight, you could use matrices and still solve it.
$ 2^x \equal{} x^2$
EDIT: I fixed it.

is it me, or could the answer to that be anything? since a power to a power means it is multiplied and y*z always equals z*y.

[quote="Lawrence Wu"]
$ 2^x \equal{} x^2$
EDIT: I fixed it.[/quote]

I think that by looking at the graphs, there's only 2 solutions, x=2 or x=4, and then $ 2^x$ just grows too quickly

[quote="xpmath"]I will randomly guess 2.

A Mock MC would be useful, but I don't really want it yet.[/quote]

who told you to guess 2??? thats our school thing. no i am not having a spaz.

I think the answers are 2, 4, and whatever -0.766664696 expressed as fraction is, or so my calculator says.

$ x^2\equal{}2^x$

Graph $ x^2$.

Graph $ 2^x$

Obviously there must be some intersection point where $ x<0$.

Then the next intersection point is at $ 2$.  Then $ 2^x$ grows exponentially. Obviously, when $ x\equal{}3, 2^x<x^2$.  After that, $ 2^x>x^2$ for all $ x$.  Thus it will intersect once more, and blah blah.


Prove or disprove: there is an inner product on $ R^2$ such that the associated norm is given by ||(x,y)||=|x|+|y| for all (x,y) in $ R^2$

[quote="Kent Merryfield"]Does it satisfy the parallelogram law?[/quote]

The problem doesn't say it.

[quote="Kent Merryfield"]I know the problem doesn't say it. I was suggesting that you discover whether or not it satisfies this law.

For instance: let $ v \equal{} (1,0)$ and $ w \equal{} (0,1)$

$ \|v\| \equal{} 1,\ \|w\| \equal{} 1,\ \|v \plus{} w\| \equal{} 2,$ and $ \|v \minus{} w\| \equal{} 2.$

Do those numbers satisfy the parallelogram law?[/quote]

Ah, I understand what you mean now. Thank you very much!


geometry
trigonometry
vector
linear algebra

$C$ is a helix iff $k/\tau$ is constant. I was reading the proof and i don't have understand that line:
$<b,w> =-\frac{k}{\tau}\cos \theta$
then
$w=Tcos \theta-\frac{k}{\tau}\cos \theta \, b$
$T$ is the unitary tangent vector, $\tau$ is the torsion ,$k$ is the curvature and $b$ is the binormal vector
I remember that teacher said something like take inner product to both sides to isolate $w$. I wondered how this is possible but i really don't get it... Tell me if i'm wrong please. Thanks for any help.

oh.... i think now i got were he used the inner product...sorry for the above bad explanation :(
we had earlier found that $w=aT+cb$. I can take the inner product to both sides like this?
$<w, T> = <aT+cb, T>$?
Thanks


Given that $\{ a_{n}\}$ is an arithmetic sequence ( Common difference $d \neq 0$). The sequence $a_{k_{1}}$, $a_{k_{2}}$, $\cdots$, $a_{k_{n}}$  formed by some terms of $\{ a_{n}\}$ is geometrical. If $k_{1}=1$, $k_{2}=5$ and $k_{3}=17$, then find the value of $\sum_{i=1}^{n}k_{i}$.


I think for this problem we have to know the velocity at the beginning, otherwise we can just get an ugly equation for the distance.
Sorry for give you guys such an ugly question. I have another question:there are two charges whose mass are m1 and m2, and velocities v1, v2, and charges q1,q2 (Q1*Q2<0), and initial distance d. how will you decide if the charges will saparate at last or not?

i made a seperate thread so that u will get more views
 :)

Thank you. Is there something called relative energy?

no...atleast i haven't heard of it...well what's the context :?:

My physics teacher gave us a physics quantity called relative energy E=total kinetic energy + total potential energy - the kinetic energy of the mass center

If the velocities point towards each other, then they can never seperate. Otherwise use energy conservation.

P.S. $ \displaystyle\sum_i U_i+\sum_i T_i-\frac{1}{2}M_T v_{\text{com}}^2$ is the energy of a system of particles in their center of mass frame.


Let $\{x_k\}_{k\ge1}$ be a sequence of reals such that $x_1=1$ and $x_kx_{k+1}=k$ for $k\ge1$. Prove that:
\[ \sum_{k=1}^n \frac{1}{x_k}\ge 2\sqrt{n} -1. \]

well i have not finished yet but it can be shown that $\frac{1}{x_n}=\frac{1}{(n-1)!}$
and not im working on showing that the sum is greater than or equal to  $2\sqrt{x}-1$

EDIT: i dont think your inequality is true. try n=4 and u get $\frac{8}{3}{\ge}{3}$ which is obviously not true.

also if you do $\displaystyle \lim_{n\to\infty}\sum_{k=1}^n\frac{1}{x_k}\ge 2\sqrt{n}-1$
you get

$e\ge\infty$

[quote="maokid7"] EDIT: i dont think your inequality is true. try n=4 and u get $\frac{8}{3}{\ge}{3}$ which is obviously not true.
[/quote]

Umm...When I plug in n=4, I get $1+1+\frac{1}{2}+\frac{2}{3}=\frac{19}{6}{\ge}3$

[quote="maokid7"]well i have not finished yet but it can be shown that $\frac{1}{x_n}=\frac{1}{(n-1)!}$
and not im working on showing that the sum is greater than or equal to  $2\sqrt{x}-1$

EDIT: i dont think your inequality is true. try n=4 and u get $\frac{8}{3}{\ge}{3}$ which is obviously not true.

also if you do $\displaystyle lim_{n\to\infty}\sum_{k=1}^n\frac{1}{x_k}\ge 2\sqrt{n}-1$
you get

$e\ge\infty$[/quote]

Up to $n=6$, for $\frac{1}{x_n}$, I've gotten $1, 1, \frac{1}{2}, \frac{2}{3}, \frac{3}{8}, \frac{8}{15}$, which is definately not $\frac{1}{(n-1)!}$.
And as far as I can tell $\displaystyle\sum_{k=1}^n\frac{1}{x_k}$ diverges, so $\displaystyle\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{x_k}\ge 2\sqrt{n}-1$ is $\infty\ge\infty$, which... :D

[quote="maokid7"]did i do something wrong? here is what i did
$x_{K+1}=\frac{k}{x_k}$ so $\frac{1}{x_{k+1}}=\frac{x_k}{k}$
and since $x_1=1$ we have
 $\frac{1}{x_2}=\frac{1}{1}$
$\frac{1}{x_3}=\frac{1}{1*2}$
$\frac{1}{x_4}=\frac{\frac{1}{1*2}}{3}=\frac{1}{1*2*3}$ and so on...[/quote]

Here is the problem:
$\frac{1}{x_4}=\frac{x_3}{3}\not=\frac{\frac{1}{1*2}}{3}$

What is there to explain?  :huh:

Well, I guess one way that might work is
[hide]to prove by induction that $\frac{1}{x_k}+ \frac{1}{x_{k-1}} \geq 2\sqrt{k}-2\sqrt{k-2}$ for odd $k \geq 3$ and do something similar for even $k$...[/hide]

[quote="maokid7"]could you please explain why  :?[/quote]

$\frac{1}{x_4} = \frac{x_3}{3} \neq \frac{\frac{1}{x_3}}{3}$. You plugged in $\frac{1}{x_3}$ for $x_3$.

[quote="ashwinrk_jain"]there must be some mistake in the problem. please check it.[/quote]

Why? If it's not true, can you disprove it?

Since I haven't seen a solution, it's time to give a little hint. Hope someone can finish it off.

[hide="hint"]
express $\frac{1}{x_k}$ as a difference of two terms,...
[/hide]


We flip a fair coin 10 times.  What is the probability that we get heads in at least 8 of the 10 flips?

[hide="?"]Could somebody explain the technique Kouichi used, or explain why you can just add $\binom{10}{8}+\cdots+\binom{10}{10}$ and subtract from $2^{10}=1024$? Why isn't there a $\frac12$ factor because of heads/tails?[/hide]

[hide="@Aequilipse"]
Each of the terms is the number of ways to choose $r$ of the $10$ coins, times the probability that those $r$ are heads, times the probability that the other $10-r$ are tails. Of course, this is just the number of ways to choose $r$ coins, times the probability of getting any single result.

He summed from $r=8$ to $r=10$, because we want at least $8$ heads.

I don't know where you're seeing subtraction, and there is a $\frac{1}{2}$ factor, except it's $\frac{1}{2^{10}}$ since there are $10$ flips.
[/hide]


Let   $ a,b,c,t$ positive reals and $ t\le \frac {5}{7}$. Prove that

$ \frac {a}{\sqrt {a^2 \plus{} t(b \plus{} c)^2}} \plus{} \frac {b}{\sqrt {b^2 \plus{} t(c \plus{} a)^2}} \plus{} \frac {c}{\sqrt {c^2 \plus{} t(a \plus{} b)^2}}\le \frac {3}{\sqrt{4t \plus{} 1}}$.

[quote="hedeng123"]Let   $ a,b,c,t$ positive reals and $ t\le \frac {5}{7}$. Prove that

$ \frac {a}{\sqrt {a^2 \plus{} t(b \plus{} c)^2}} \plus{} \frac {b}{\sqrt {b^2 \plus{} t(c \plus{} a)^2}} \plus{} \frac {c}{\sqrt {c^2 \plus{} t(a \plus{} b)^2}}\le \frac {3}{\sqrt {4t \plus{} 1}}$.[/quote]

I think, just need to check 2 case:

(1), $ a \to 0$ ;

(2), $ b\equal{}c$ .

[quote="kuing"][quote="hedeng123"]Let   $ a,b,c,t$ positive reals and $ t\le \frac {5}{7}$. Prove that

$ \frac {a}{\sqrt {a^2 \plus{} t(b \plus{} c)^2}} \plus{} \frac {b}{\sqrt {b^2 \plus{} t(c \plus{} a)^2}} \plus{} \frac {c}{\sqrt {c^2 \plus{} t(a \plus{} b)^2}}\le \frac {3}{\sqrt {4t \plus{} 1}}$.[/quote]

I think, just need to check 2 case:

(1), $ a \to 0$ ;

(2), $ b \equal{} c$ .[/quote]

the inequality  holds

how to prove it

[quote="kuing"][quote="hedeng123"]Let   $ a,b,c,t$ positive reals and $ t\le \frac {5}{7}$. Prove that

$ \frac {a}{\sqrt {a^2 \plus{} t(b \plus{} c)^2}} \plus{} \frac {b}{\sqrt {b^2 \plus{} t(c \plus{} a)^2}} \plus{} \frac {c}{\sqrt {c^2 \plus{} t(a \plus{} b)^2}}\le \frac {3}{\sqrt {4t \plus{} 1}}$.[/quote]

I think, just need to check 2 case:

(1), $ a \to 0$ ;

(2), $ b \equal{} c$ .[/quote]and why?

without loss of generality, we can assume that $ a \plus{} b \plus{} c \equal{} 1$, and inequality becomes

$ f(a) \plus{} f(b) \plus{} f(c) \le \frac {3}{\sqrt {4t \plus{} 1}},$

where $ f(x) \equal{} \frac {x}{\sqrt {x^2 \plus{} t(1 \minus{} x)^2}}.$

but if I calculate correct, when $ 0<t<\frac{5}{7}$, $ f(x)$ will be concave in $ (0,1)$ or half convex and half concave in $ (0,1)$, so just need to check that 2 case...


calculus
integration
trigonometry
geometry
calculus computations

Evaluate  $ \int_0^{n\pi} \sqrt{1\minus{}\sin t}\ dt\ (n\equal{}1,\ 2,\ \cdots).$

We have
$ \sqrt{1-\sin t}$
$ =\sqrt{\cos^2\frac{t}{2}+\sin^2\frac{t}{2}-2\sin\frac{t}{2}\cos\frac{t}{2}}$
$ =\left|\cos\frac{t}{2}-\sin\frac{t}{2}\right|=\sqrt{2}\left|\cos\left(\frac{t}{2}+\frac{\pi}{4}\right)\right|$
which is periodic with period $ 2\pi$.
Accordingly, if $ n=\ \mathrm{even}=2m$, then there are $ m=n/2$ periods in the interval zero to $ n\pi$. Hence, the integral equals n/2 times the area in the positive "wave" of the curve $ \sqrt{2}\cos(t/2)$, which is
$ 2\int_0^\pi \sqrt{2}\cos(t/2)\ \mathrm dt =4\sqrt{2}$
Hence,
$ \int_0^{n\pi}\sqrt{1-\sin t}\ \mathrm dt=\dfrac{n}{2}\,4\sqrt{2}=2n\sqrt{2}$  if $ n$ is even.

On the other hand, if $ n=\ \mathrm{odd}$, then first of all we have $ \left(\dfrac{n-1}{2}\right)$ positive "waves" of  the curve $ \sqrt{2}\cos(t/2)$ plus twice the area under the curve $ \sqrt{2}\cos\left(\frac{t}{2}+\frac{\pi}{4}\right)$ from 0 to $ \pi/2$.
Hence, for $ n=\ \mathrm{odd}$,
$ \int_0^{n\pi}\sqrt{1-\sin t}\ \mathrm dt=2(n-1)\sqrt{2}+2\int_0^\frac{\pi}{2}\sqrt{2}\cos\left(\frac{t}{2}+\frac{\pi}{4}\right)\mathrm dt$
$ =2(n-1)\sqrt{2}+2(\sqrt{2}-1)$
$ =2(n\sqrt{2}-1)$

Hence,
$ \boxed{\int_0^{n\pi}\sqrt{1-\sin t}\ \mathrm dt =\left\{ \begin{array}{ll} 2n\sqrt{2}-2 & \mathrm{if}\ n=\mathrm{odd}\\[2ex] 2n\sqrt{2} & \mathrm{if}\ n=\mathrm{even}\end{array}\right.}$
P.S. Here integers mean positive integers.

You have almost solved the problem , but in case of $ n\equal{}odd$, your answer is incorrect.

Well yes I made a slight mistake in the algebra:
For n = odd, we would have
$ \int_0^{n\pi}\sqrt{1-\sin t}\ \mathrm dt=2(n-1)\sqrt{2}+2\int_0^\frac{\pi}{2}\sqrt{2}\cos\left(\frac{t}{2}+\frac{\pi}{4}\right)\mathrm dt$
$ =2(n-1)\sqrt{2}+\boxed{4}(\sqrt{2}-1)$
Hence,
$ \boxed{\int_0^{n\pi}\sqrt{1-\sin t}\ \mathrm dt =\left\{ \begin{array}{ll} 2(n+1)\sqrt{2}-4 & \mathrm{if}\ n=\mathrm{odd}\\[2ex] 2n\sqrt{2} & \mathrm{if}\ n=\mathrm{even}\end{array}\right.}$
P.S. I made the mistake of writing 2 in that boxed place.

Now the problem has been solved completely. :lol:


I have to walk between my house at (0,0), my library at (1,3) and my school at (5, 2). What’s the minimum distance I have to walk?

a  27       b   2.5       c  7.3        d  5.2     e  6.6

Use the distance formula and add the distances from home to the library, then from the library to school.

$ \sqrt{1^2\plus{}3^2}\plus{}\sqrt{(5\minus{}1)^2\plus{}(2\minus{}3)^2}$
$ \equal{}\sqrt{10}\plus{}\sqrt{17}$
$ \approx 7.3$

So the answer is $ \boxed{(C) 7.3}$.


$ S_1\equal{} 1\plus{}C_n^1cos \alpha\plus{}C_n^2cos2\alpha\plus{}...\plus{}C_n^ncos n\alpha\equal{}2^ncos^n\frac {\alpha}{2}cos(\frac{n\alpha}{2})$
$ S_2\equal{}C_n^1sin\alpha\plus{}C_n^2sin2\alpha\plus{}...\plus{}C_n^n sin n\alpha\equal{}2^ncos^n\frac{\alpha}{2}sin(\frac{n\alpha}{2})$
i got to this, using Moivre
$ S_1\plus{}iS_2\equal{}1\plus{}C_n^1(cos\alpha\plus{}isin \alpha)\plus{}C_n^2(cos\alpha\plus{}isin\alpha)^2\plus{}..\plus{}C_n^n(cos\alpha\plus{}isin\alpha)^n$, but i don't know how to finish...i must use this method..pls help
thanx

Does that last equality remind you of any theorems whose names start with B?  :)


Just wondering.........

If other, please post your choice.

what happened to GBSP???

even though the DS can play GBA games and more, so that was my choice

sorry I kinda thought Game Boy would cover GBA, GBASP, GBC, and GBP.  I had a blue GBSP but now I have a DS.

lol 1=2

DS>all of those other ones (except maybe DSi, but I already had a DS)

i have a DS and a DSi (i bought one and then bought the other because it looked kewl), and i still say the Touch is the best. you have a very wide selection of games, half of which are free. and i also heard that there's going to be a new version coming up thats going to have a camera (or is that the iPhone? i don't know...), so that it covers the capabilities of the DSi too. i think it can also access the internet through wifi, so that would cover almost all the capabilities of previous Nintendo systems. but Nintendo makes awesome games, so that good enough for me.

 :D  :D  :D  :D  :D  :D  :D


An exterior angle of a regular polygon is 1/35 th of the interior angle. How many
sides does the polygon have?

The exterior angle of a regular polygon is $ \frac{360}{n}$, where n is the number of sides. Thus

\[ 35\frac{360}{n}\equal{}180\minus{}\frac{360}{n}\]
\[ 36\frac{360}{n}\equal{}180\]
\[ \frac{1}{n}\equal{}\frac{180}{36\cdot360}\]
\[ \frac{1}{n}\equal{}\frac{1}{72}\]
\[ n\equal{}\boxed{72}\]


geometry
trapezoid
blogs
MATHCOUNTS
AMC
AMC 10
AMC 10 B

How did you do?

I got 28/14=42  :D  #25 was soooo tricky.  I got #4 on the target wrong; scratch paper revealed that I somehow listed one possibility twice -__-.  Oh well, I'm still happy with the score.

It was a hard test, definitely.  The target was normal maybe, but the last 10 sprint problems were pretty intense.

I got 16/12=28
I could have gotten the last 2 targets right, but for the second one, I didn't notice a 2 in the prime factorization, so I got the ages as 14,14,15,15, and 8?!?!

I messed up so badly on sprint. Easy mistakes

You're in 6th grade AND you're in a harder state than most.  I didn't make states in 6th grade (got 31 at chapters -__-).  You've got 2 years to improve!

Haha...last year, I was the one watching 8th graders complain about how they don't have another year and now, I'm getting semi-depressed at having only one more competition left.

All I've got left is the big one :(.

Agree with kyyuan =/

Test Notes:

Did anyone else cash in on #30 on the sprint being the same as like #19 on the AMC 10B, except with numbers twice as big?

I felt guilty so I started to do the problem, then 30 seconds later went... "oh what the heck am I doing" and just wrote the answer I remembered from the AMC 10B times 2 down on the answer blank... and it was right  :lol:

My favorite problem was probably the one with the quilt, where you had to trim off one from 27 and 24. Like 20 people around me all got 24 and I got worried for a second. I still don't get how they get 24!

#25 was tricky but it boiled down into a really easy problem if you took into consideration:

242 = 121*2. You know the denominator can't be even or else the end result won't be a common fraction. Thus it has to be divisible by 11.

Options:

649
847
946 (ruled out because it's even)
748 (ruled out because it's even)
341
143

and simple guess and check gives you $341$ as your answer.

I bombed the target. Only got a 12. Kept reading the trapezoid one wrong as "the other base" instead of "the legs" so I got that one wrong and I totally blanked on the tens/units digits thing. Was freaked because in our state they have the target round first and then the sprint, so by the time the written portion of the competition is actually done, they have the target checked so they post those rankings only. I freaked because I knew I bombed it and I was in 12th. :( Then I had a "good" sprint (at least by Wisconsin standards) with a 24. Was the only one in whole state to get #30 right, which is why I won the tiebreaker. The cutoff for nats was 32. :( I got a 36. Our team isn't the greatest, but I guess we'll just have to improve by the time the national competition comes around! BTW, I'm the only "veteran" from nats on that team. Top 10 written in WI were (official CD)

10th-Shan Sivanushanthan (29) (10th CD)
9th- Suhas Kodali (30, but had a perfect target, so lost tiebreakers) (9th CD)
8th- Abraham Shin (30) (5th CD)
7th- Laura Xu (31, but also had a perfect target, and lost tiebreakers) (8th CD)
6th- Sohil Shah (31) (7th CD)
5th- Can't remember. It was a hypenated name (32) (6th CD)
4th- Killian Kvalvik (32) (3rd CD)
3rd- Eric Wang (32) (4th CD)
2nd- Cutler Lewandowski (36) (2nd CD)
1st- Yours Truly (36) (1st CD)

oh yeah and by the way my grandma was there so I couldn't lose :P

Abraham Shin had a nice run, but was just stopped short.

OUCH

ack...he's a 7th grader. he's still got next year. In the last 3 years in Wisconsin, there's been someone with a great run who's been stopped just short.

2005- Peter Wear (mustafa) (8th-6th, but he could have beat the 4th place kid cuz he had a really lame afro):rotfl: check out my blog for the whole story
2006- Edward Hou (8th-5th)
2007- Abraham Shin (8th-5th)

I got a 23/14=37, 4th place in Massachusetts. And yeah, the last problem was very similar to the AMC 10B and took me 10 seconds to get.

lol me too. like i said, only one to get that one right. strange.

I did HORRENDOUS.

I got 19 on the sprint

and I got 2 on the target...if I had a 6 I would've gone to nats

the 6 that I made stupid mistakes on...

#2: forgot to take the sqrt
age one: mixed up the ages (omg!)
power of 2 one: multiplied the answer by 2 accidently
trapezoid: didn't add correctly
i remember getting the abcd thing right (was it DABC?)

omg

jorian

edit:
now that i see the test...

nvm, i didn't get the abcd thing right. i put it down backwards (oh my gawsh!) as dabc

the crispy rice thing really had me looking at the paper for like 30 seconds before I made sense out of it. i got 1.17 within seconds and thought it was too obvious

jorian

[quote="Fanatic"]
My favorite problem was probably the one with the quilt, where you had to trim off one from 27 and 24. Like 20 people around me all got 24 and I got worried for a second. I still don't get how they get 24!
[/quote]

Wasn't the quilt problem 22%? Since it is 4 squares out of a total 18=2/9=22.22222%?

I got a 18/12
So a 30
I got 10th
it was like a 5 way tie for 10th (and a scholarship) :o
but i broke all the tiebreakers :lol:
My mood swing from depressed to relieved killed me at countdown
even when i had an extra round to relax
The team got second
We got 8/10 right on the team round
I edited it from 7 to 8

Notes:
The team lost on tiebreakers
Hardest last 10 to 15 questions i've ever seen in a State MATHCOUNTS
I improved by 0 on my score from last year
I only improved by one place from last yeer
4th place got a score of either 32,33, or 34
My teammate got 5th
First place in our state got a 42 (i don't know the split) :omighty:
I was depressed after everyone on my team told me how they felt they did after the tests (they all said they missed like 5 or 6 on the sprint and two/three on target)
Then it turned out they were mostly liars  :furious:  :cursing:  :censored:
I almost felt like giving up in the last 15 minutes :surrender:
but i didn't
I was off lack of sleep (not blaming it either because it's entirely my fault for not having good sleeping habits)
I felt relieved when countdown came after me and my other teammate were the only ones to make countdown (and the only ones that felt depressed after the test)
I could've gotten number 20 and 30 right on sprint if I had
1) guessed and checked with the right number
2) remember niumber 30 was on AMC10
I think i missed number one... (i know. i'm sad)
Could've gotten number 7 on target right if I had not miscounted by one
Could've gotten number 6 on target if I hadn't spent so much time triple checking if number 5 was right
My mood is overall:
RELIEVED :sleep2:

wow jorian you really did do bad. 2 on target?!?!?!?!?

15/10 :blush: , wow i bombed the sprint.

yeah, and i got that stupid knight/knave thing wrong which ended my wonderful 16 in a row streak on sprint

jorian

I guess that makes sense... at least the second reason...

19/14=33

Not as good as you guys but I more then doubled the competition average and beat Alaska's cutoff by nine points.

actually fanatic, it's 3/4. you and mathcrazed should count too!

If I could divide by 5, I would have made it to nationals...(I'm talking about #8 on the target). I decided to do it out on paper instead of use my caculator, and when I wrongly divided by 5 I got a prime number, so my ages were like 16 & 8191 or something. Then I guessed 14.7. As you might know, the answer was 14.8.

Mathcrazed and I live 1.5 hours away from Mr. Fischer, so we shouldn't count.

[quote="jhredsox"]they don't mean that no one can know. they just want the general public not to know. it's not like the new york times is investigating about this!

jorian[/quote]

hypocrite.
remind me, who again made like 20 guesses as to their scores when Fanatic kept telling you to shut up?

ok, lol, good point (i still think i was within 3 for each score, perfect...i'm still challenging you for the 3)

ok, but didn't i stop?

now guessing and telling are two different things

ok, but seriously, INDIANA *JUST* DOESN'T WANT TO SAY THEIR SCORES

honor what they want

NO MORE QUESTIONS PLEASE... :)

jorian

for all those people who think their states is hard come to texas and we will show you what evil means!!! XD
anyways, 38-40, perfect target, = 6th place written (me)
cutoff was like 40 or 41 in texas

Do you guys have a copy of the test? Are they posted somewhere? Am I being dense??   :blush:

Also, I wish somewhere there'd just be a chart made of all state's cutoffs -- hey, anyone attempt to do that? If so, please pm me!!

Uh, the test was posted on http://www.mathcounts.org, if that's what you mean.

Also, I think that some people got their actual test scores (as in which problems they got wrong) back.

[quote="Karth"]I didn't make nats :(  :(  :mad:[/quote]

What? I thgouht in the woot thread you said you turned 16 on July 6, 2006?!

well he didn't make nats did he??? :rotfl:

I think that was actually Arvind_sn, Karth's brother, but they forgot to switch accounts; Arvind is a 7th grader and did not make nats.


For $\triangle ABC$ with medians $AD$, $BE$, and $CF$, let $m=AD+BE+CF$, and let $s=AB+BC+CA$. Prove that $\frac32s>m>\frac34 s$.

The given inequality can be written $\frac{3}{4}\sqrt{2}<\frac{\sqrt{b^2+c^2-a^2}+\sqrt{c^2+a^2-b^2}+\sqrt{a^2+b^2-c^2}}{a+b+c}<\frac{3}{2}\sqrt{2}.$

Can the left hand side be made even stronger? ($s>m$)

Cool. I solved this problem in the Olympiad Inequalities course, so I was wondering what the source was...(there aren't too many problems I can solve in that class ;))

[quote="rzsolt"]Ummm, this is basicaly this months olymon. question one, with a minor change. Its very double plus bad to do olymon publicly last time I checked.

( the actual question is bounded by 1, not 3/2 )

( this months olymon: http://www.cms.math.ca/Competitions/MOCP/2006/prob_apr.pdf )[/quote]

I would like to say that some problems are in many different sources. This problem was given as one of the homework assignments at MathCamp 2005, so it is definitely not new to they olymon. I would say that it is a coincidence that this problem was posted at the same time that a similar problem was given in the olymon.

[quote="rzsolt"]Ummm, this is basicaly this months olymon. question one, with a minor change. Its very double plus bad to do olymon publicly last time I checked.

( the actual question is bounded by 1, not 3/2 )

( this months olymon: http://www.cms.math.ca/Competitions/MOCP/2006/prob_apr.pdf )[/quote]Actually, I got it from Chellenging Problems in Geometry. I've never heard of that contest prior to reading this post.


[color=purple]Pa` kon va` kac' anh chi em ko' biet' DDTH la`m sao ko a? Em va`o thay' no' hien ra trang vo' van? em ko sao vao` dc buc the'[/color]

ah` co le ban nen thay duoi .net = .org
   minh thay the thi` vo binh thuong con ko thi` chiu :D

Sao em vào đuôi .org hôm trước được mà hôm nay lại không đuợc?

[color=purple] Vua` vao` dc http://www.diendantoanhoc.org/forum  neus ko ko' forum thi` chiu[/color]

Chủ đề này có thể tiếp tục ở đây http://www.mathlinks.ro/Forum/viewtopic.php?t=40582 Tôi sẽ khoá topic này lại.


vector
algebra
polynomial
analytic geometry
linear algebra
matrix
trigonometry

In each of the following, find the dimension of the subspace of P_3 spanned by the given vectors.

$ a. \ \ x, \ x\minus{}1, \ x^2\plus{}1 \\ b. \ \ x, \ x\minus{}1, \ x^2\plus{}1, \ x^2\minus{}1 \\  c. \ \ x^2, \ x^2\minus{}x\minus{}1, \ x\plus{}1 \\ d. \ \ 2x, \ x\minus{}2$

I know I must be thinking with too much complexity.  Are x, x-1, etc. vectors or elements of a vector.
Note:  this is exactly how they are listed in my textbook, without the T symbol for transverse which usually denotes that a horizontal list is in fact a vector.
Thank You

Do you know what $ P_3$ is? Apparently it is the space of polynomials of degree at most $ 3$. You see, polynomials are vectors too - they can be added and multiplied by a scalar. There is a natural basis of $ P_3$, namely $ 1,x,x^2,x^3$. When you expand a polynomial in this basis, the coordinates are just coefficients. This allows you to represent the polynomial $ x^2\plus{}1$ as $ (1,0,1,0)$. With T if you wish.

So that helps,  I guessed that the resulting number of rows that result when making those associations and reducing the resulting matrix is the dimension of the list of vectors.  The only thing missing from my textbook is a list of the standard bases for common linear algebra problems, and how to use them.  If you don't mind I would like to know the standard basis for $ C[\minus{}\pi,\pi]$ for another problem. vectors: $ 1, \ \cos {2x}, \ \cos^2{x}$
Thanks

Never mind "standard basis"  in this instance. Note that $ 1$ and $ \cos2x$ are pretty clearly independent. Now ask this: is it possible to write $ \cos^2x$ as a linear combination of $ 1$ and $ \cos2x?$


integration
limit
function
calculus
calculus computations

compute the radius of convergence of:

$ \sum_{n\ge 0} a_n z^n$
where $ z \in \mathbb{C}$
and $ a_n\equal{} \int_{0}^{1} (1\plus{}t^2)^n dt$

A well-known analysis fact (the justification of the notion of $ \infty$ norm):

For continuous $ f$ on $ [a,b],$ $ \lim_{n\to\infty}\left[\int_a^b|f(t)|^n\,dt\right]^{\frac1n}\equal{}\max |f(t)|.$

For this particular function, $ \lim_{n\to\infty}a_n^{1/n}\equal{}2,$ and hence by the root test, the radius of convergence of the power series is $ \frac12.$

ok, what's the name of this fact?
I've got a nice solution to the problem

[hide="The hint"]$ 2t \le 1 \plus{} t^2 \le 2$[/hide]

thanks

please I want the name of this theorem or a proof to it if someone knows it.

thanks


compute the following:

$ \frac{tan1}{cos2}\plus{}\frac{tan2}{cos4}\plus{}.....\plus{}\frac{tan2^n}{cos2^{n\plus{}1}}$

I posted this problem to pre-olympiad section but no reply from there(i think this one is bit hard) so i decided to post here. :)  :blush:

We have
$ \frac{\tan{a}}{\cos{2a}}\equal{}\frac{\tan{a}(1\plus{}{\tan{a}}^2)}{1\minus{}{\tan{a}}^2}\equal{}\frac{2\tan{a}\minus{}\tan{a}(1\minus{}{\tan{a}}^2)}{1\minus{}{\tan{a}}^2}$

$ \equal{}\frac{2\tan{a}}{1\minus{}{\tan{a}}^2}\minus{}\tan{a}\equal{}\tan{2a}\minus{}\tan{a}.$
So the sum in the problem becomes
$ \tan{2}\minus{}\tan{1}\plus{}\tan{4}\minus{}\tan{2}\plus{}\ldots\plus{}\tan{2^{n\plus{}1}}\minus{}\tan{2^n}\equal{}\tan{2^{n\plus{}1}}\minus{}\tan{1}$.


geometry
algebra
polynomial
abstract algebra
calculus
integration
topology

Let $F: \mathbb{A}^{n}\to\mathbb{A}^{n}$ an injective unramified morphism. Then it is birational.

I have already a proof, which I don't understand: since $F$ is unramified, $k(X)/k(F)$ is separable, so a generic fibre of $F$ has $[k(X): k(F)]$ elements (why?). Since $F$ is injective, $[k(X): k(F)]=1$.

Let $f: k[y_{1},\ldots,y_{n}]\to k[x_{1},\ldots,x_{n}]$ be the algebra homomorphism induced by $F$.

$F$ is a morphism of finite type because $f$ is of finite type. This means that $ImF$ is a constructible subset of $\mathbb A^{n}$. Let $W$ be the closure of $ImF$ in $\mathbb A^{n}$. $ImF$ and $W$ have the same dimension. Now look at $F: \mathbb A^{n}\to W$. It is injective, so by the Dimension of fibres theorem, $dim W=dim\mathbb A^{n}$, so $W=\mathbb A^{n}$. This proves that $F$ is dominant, so $f$ is injective.

Depending on your definition, you either get directly that $k(X)/k(F)$ is separable, or if you use a scheme theoretic definition, to the same end notice that because $f$ is injective, $F$ sends the generic point of $\mathbb A^{n}$ to itself.

The extension $k(X)/k(F)$ is also algebraic. Since $F$ is dominant, its image contains an open subset of $\mathbb A^{n}$. Call it $U$. Let $V=F^{-1}(U)$. $k(X)=k(V)$ and $k(F)=k(U)$. Since $U$ and $V$ are $n$-dimensional, $k(U)$ and $k(V)$ have transcendence degrees $n$ over $k$, so $k(X)/k(F)$ has transcendece degree $n-n=0$, hence it is algebraic.
I had a scheme theoretic proof for this too, but I am not sure that injectivity of $F$ in the classical algebraic geometry sense implies injectivity on the level of schemes.

$k(X)/k(F)$ is separable and algebraic, so it is a simple extension, i.e. $k(X)=k(F)(z)$ for some $z\in k(X)$ that we can take in $k[X]$.

Now every $x_{i}\in k[x_{1},\ldots,x_{n}]=k[X]$ is the root of a polynomial with coefficients in $k(F)=k(f(x_{1}),\ldots,f(x_{n}))$. Let $L$ be the product of the denominators of the leading coefficients of these $n$ polynomials. Take $U$ an open set on which $1/L$ is regular. Let $V=F^{-1}U$. Then the extension $k[U]\to k[V]$ is a finite extension i.e. $k[V]$ is finitely generated as a module over $k[U]$. This just means that $F$ restricted to $V$ and co-restricted to $U$ is a finite morphism.

Now we use this theorem, which is the step smullyan asked why at:
Let $\varphi: W\to V$ be a finite surjective morphism of irreducible varieties and assume that $V$ is normal. Define $\deg\varphi=[k(W): k(V)]$.
Then
(a) Every fibre has at most $\deg\varphi$ elements.
(b) The set of points on $V$ which fibre has $\deg\varphi$ elements, is open. If $k(W)/k(V)$ is separable, this open set is non-empty.

The proof for this is quite nice. You can find the result as Theorem 8.9/pag143 in Milne's Algebraic Geometry book that is distributed freely on his web page.

It is clear that $F$ injective now implies $[k(X): k(F)]=1$, so $F$ induces an isomorphism between the fields of rational functions. This means that $F$ is birational.

Thank you, amfulger.

In the paper 'The Jacobian conjecture..." by H. Bass, E. Connell and D. Wright (Bull. Amer. Math. Soc. 1982) you can find a sketch of the following:

THM: Let $k$ be a field of characteristic $0$ and let $F: \mathbb{A}_{k}^{n}\to\mathbb{A}_{k}^{n}$ an unramified morphism. Then, the following conditions are equivalent:
(a) $F$ is invertible, i.e. $k[X]=k[F]$;
(b) $F$ is birational, i.e. $k(X)=k(F)$;
(c) $F: k^{n}\to k^{n}$ is injective;
(d) The integral closure $\overline{k[F]}$ of $k[F]$ in $k[X]$   in unramified over $k[F]$;
(e) $k[X]$ is a finitely generated $k[F]$-module;
(e') $F$ is proper;
(f) $k[X]$ is a projective $k[F]$-module;
(g) $k(X)$ is Galois over $k(F)$.


calculus
integration
calculus computations

A strangely ignored post [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=108806]complex integrals calculations[/url] has recieved my singular attention, and so I promptly made this post.

See you there,

       -Ben

[quote="benorin"]Thanks for the team effort and the quick response to my plea, --Ben[/quote]

 Thanks for reminding  :)


Show that if $n \in N$ so that $2^n+1$ is a prime number, then n is a power of 2.

[hide]Notice for all odd n other than 1, 2^n+1 is a composite multiple of 3.  This is because multiples of 2 are either 1 or 2 mod 3.  It starts at 2 with n=1.  Then, at n=2, we have 1.  At n=3, we have 2.  Since we're adding 1, whenever n is odd, we have a number 0 mod 3.  3 is prime and 1 is a power of 2 so there's no violation.  Now, what happens when n is not a multiple of 2 but is even?  N is 2^a*various odd primes to various powers.  But remember, this is n which is put in another power of 2.  We set this as 2^(2^a*odd primes to powers).  Similarly 1 can be set as 1^(2^a*odd primes to powers).  But since the odd primes to powers part is being multiplied against the exponent, we can make it look like (2^(2^a))^odd primes to powers and (1^(2^a))^odd primes to powers.  This is basically x^(2e+1)+1=(x+1)(x^(2e)-x^(2e-1)+...+1), where x is 2^(2^a).  This shows that those numbers aren't prime.  Lastly, we show that when n is a power of 2, it is prime.  We have 2^(2^a)+1.  This is rewritten x^2+1, which can't be factored without imaginaries.  Same goes with x^4+1 or x^8+1 or...  [/hide]

Other solution :
[hide]
if $n=2p+1$ is odd then $2^n+1$ is divisible by $3$, it's just algebra $x^n+1=(x+1)(x^{n-1}-....)$[/hide]


geometry
3D geometry
sphere
tetrahedron
circumcircle
incenter
AMC

Three mutually tangent spheres of radius 1 rest on a horizontal plane.  A sphere of radius 2 rests on them.  What is the distance from the plane to the top of the larger sphere?

This problem has been bothering me for some time now.  I just don't even know how to begin to approach it.

Note that connecting the centers of the four spheres gives you a tetrahedron. We will call the tetrahedron ABCD, with base ABC an equilateral triangle of side-length 2 and AD = BD = CD = 3.

To find the altitude from vertex D, we let E = the point where it meets the base, and notice that E is the centroid of ABC (and also the circumcenter and incenter and orthocenter). So, we can make a right triangle AED with right angle at E. AE = 2/3(rad(3)/2)*s = 2/3(rad(3)/2)*2 = 2rad(3)/3. AD = 3. So, ED = AD - AE = 9  4/3 = 23/3 --> ED = rad(69)/3.

Then notice that distance from plane to base (triangle ABC) is 1, and distance from center of top sphere (point D) to top of top sphere is 2. Thus, the total distance from plane to top of top sphere is rad(69)/3 + 2 + 1 = 3 + rad(69)/3.


integration
trigonometry
limit
geometric series
algebra proposed
algebra

[quote="ehsan2004"]Can  you calculate $\frac 13-\frac 15+\frac 17-\frac 19+...$ ??? :?[/quote]
Yes :)  It is $1+\frac 13-\frac 15+\frac 17-\frac 19+...=ln(1+3)=ln(4)$

In my opinion, you're wrong , my solution is $ \frac{\pi}{4}$;
   I know that $ \frac{1}{2}-\frac{1}{4}+\frac{1}{6}-...... = (1/2)ln(2)$, but it doesn't  mean my sum is $ ln(4)$.
    And you can use some program to calculate this sum, i think it will give the resulf
$ \frac{\pi}{4}$.

this sum is exact this   $1+\frac 13-\frac 15+\frac 17-\frac 19+...=ln(1+3)=ln(4)$ because we know that
$ln(a+x)=lna+\frac xa-\frac{x^2}{2a^2}+\frac{x^3}{3a^3}-.............$
In our case x=1 and a=3 thats why I write $ln(1+3)$

Alternatively, one may consider
\[I_n=\int_0^{\frac {\pi}{4}}\tan^{2n} {x} dx.\] Integrating by parts  gives
\[I_n=\frac{1}{2n-1}-I_{n-1}\]
and this yields
\[I_n=(-1)^n\cdot \left( \frac{\pi}{4}-1+\frac13 -\frac15+\ldots +(-1)^n\frac {1}{2n-1}\right).\]
On the other hand, it is a simple task to prove that
\[\lim_{n\to \infty}{I_n}=0,\]
hence the result.

Yes, enescu, your proof  is similar to my proof.

[quote="darkmaster"]You can try to solve it only with high-school tools:
     Calculate the infinite sum:$ 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+.................$[/quote]

THis is Libniz series.

[quote="jhaussmann5"]By geometric series,
\[\frac{1}{1 + x^2} = 1 - x^2 + x^4 - x^6 + \dots.\]
Integrating, we get
\[\int \frac{dx}{1 + x^2} = \int (1 - x^2 + x^4 - x^6 + \dots) \: dx,\]
which gives
\[\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \dots.\]
Substituting $x = 1$ gives
\[\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots.\][/quote]

You need to limit $|x|<1$, that is to say, in Taylor expansion you should confirn the convergency of the term of residue.


THE MOST AMAZING MONTH OF THE YEAR!!! JUNE!!!! :D

[quote=AppleYum][quote=HumanCalculator9]March is cool![/quote]

Yes![/quote]

AGREED!


group theory
abstract algebra
superior algebra
superior algebra solved

Let H and N be normal subgroups of a group of G, with N as a subset of H. Define
Phi: G/N-->G/H by Phi(xN)=xH, for all cosets xN in G/N.
a) Show that Phi is a well-defined onto homomorphism
b) Show that (G/N)/(H/N) is isomorphic to G/H.

now to show that $\pi : G/ N \rightarrow G/ H$ defined by $\pi(xN)=xH$ is well defined let $xN=yN$, then this implies that $xy^{-}1\in N$ but $N\subset H$ so $xy^{-}1\in H$. in other words $xH=yH$ or $\pi(xN)=\pi(yN)$ , so this shows the first part. $\pi$ is clearly a onto homomorphism. the kernel of this homomorphism is $H/N$ ( i leave the above 2 as excersises for you, if you cant get them please ask) so use the first isomorphism theorem to get the result you wanted

thanks for your help. But I'm still having trouble with 2.

Do you know the homomorphism theorem¿ (or also called $x$-th isomorphism theorem for some $x \in \mathbb{Z}$)
Namely:
Let $f: G \to H$ be a homomorphism, then $G/\ker(f) \cong \mathrm{Im}(f)$.

Now try to use it or tell where you have problems.


function
calculus
derivative
integration
analytic geometry
graphing lines
slope

That has to be the wrong quantifier. You must have meant "show that $ |f''(x)|\ge 4$ for [b]some[/b] $ x\in[0,1].$"

[quote="liam_gallagher"]Demonstrate that if $ f$ it is a function twice diferentiable with $ f (0) \equal{} 0$, $ f (1) \equal{} 1$ and $ f ' (0) \equal{} f ' (1) \equal{} 0$, then $ \left | f " (x) \right | \geq 4$ for some $ x \in [0,1]$

EDIT: thx kent merryfield[/quote]

Only we can prove that for some $ x \in [0,1] : \; |f^{''}(x)| \geq 2 $ , am I right ? :maybe:

It looks as though equality is achieved only for the function $ f(x) \equal{} \begin{cases}2x^2 & (0\leqslant x\leqslant1/2) \\
\minus{} 2x^2 \plus{} 4x \minus{} 1 & (1/2 < x\leqslant1)\end{cases}$
(in other words, any other function satisfying these conditions must have $ |f''(x)| > 4$ for some $ x$).

[b]Edit.[/b] That function admittedly does not have a second derivative at the point x=1/2. But I think it can probably be used to prove the required result, using PeterP's variational method as in [url=http://www.mathlinks.ro/viewtopic.php?p=1256910#1256910]this thread[/url].

Write $ f' \equal{} g$.
The problem can be rephrased in terms of the function $ g$ as follows:
Show that if $ g$ is differentiable with $ g(0) \equal{} g(1) \equal{} 0$ and $ \int_0^1 g(x) \equal{} 1$, then $ |g'(x)|\ge 4$ for some $ x\in [0,1]$.

The tentfunction $ h: [0,1] \to \mathbb{R}$ with
$ h(x) \equal{} 4x$ for $ x\in [0,\frac12]$ and $ h(x) \equal{} 4 \minus{} 4x$ for $ x\in [\frac12,1]$ fullfills the requirements, except that the function is not differentiable in $ x \equal{} \frac12$.
A function $ g$  which fullfills all the requirements has to have points on the graph which are situated above the graph of $ h$, otherwise the integral is smaller than 1. But then the line through such a point and point $ (0,0)$ has a slope that is larger that 4 or the line through such a point and point $ (1,0)$ has a slope that is smaller that -4. This shows that there is a point $ x \in [0,1]$ with $ | g'(x)|>4$.

I posted essentially this problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123255]in the physics forum[/url] here, after it had appeared previously [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123244]in the analysis forum.[/url]


Apparently, in many categories in "Allowed Extensions and Sizes," there are quite a few the require a "0 byte maximum"...I'm confused. I was attempting to upload a .ppt file, which happened to be in one of those categories, and it came out as "General Error: attempt to upload empty file." What should I do?

So 0 bytes=can't upload? Then how come I remember dragon96 uploaded a ppt once in the FTW forum on something about starting an answer with a zero?

Wait, 0 bytes? I just made an empty powerpoint file and it took up 30 KB (MS 07). The "Error: loading empty file part seems strange." In fact, I just attached an impty document here fine:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1587649#1587649

So why is the uploading not working for me?

http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=270875

Is where I upload an actual ppt file.

Thinking again, could the "empty file" mean that my file is too large? It is a pretty large file...2.3 megabytes.

Likely.

Like, in the Games forum, I tried to c/p 100000! (which is a couple hundred kilobytes long), and it said, "Your message is too short".


geometry
combinatorics proposed
combinatorics
Polinomials

1st player has the winning strategy.

Considering the leftmost lower point of the triangle as (0,0) and the top of the triangle as (4,4), we observe that vertices (3,1),(4,2) and (5,1) are shared by 6 triangles, rest of the vertices except (0,0),(8,0) and (4,4) are shared by 3 triangles and (0,0),(8,0) and (4,4) are a part of exactly one triangle. The strategy will now be to acquire the points that are included by maximum number of triangles. (It is not difficult to see that p1 has already won the game).

Let p1 acquire (4,2) and (5,1) while p2 acquires (3,1) wlog( p1has already ahead by 2 triangles). From here p2 can go ahead in two ways.

1.Aggressively- p2 acquires (6,2) (in order not to let p1 acquire it and thereby get two triangles). In that case p1 always has a counterattack by acquiring one vertex among (2,2),(1,1),(2,0),(4,0) (preferably (2,0) or (1,1)) thereby incurring equal loss to p2. And so on.

2.Passively- P2 acquires one among (2,2),(1,1),(2,0),(4,0) (preferably (2,0) or (1,1)) getting 2 triangles. Clearly p1 always has a vertex so that by acquiring it he/she can get two triangles. There is going to be a situation when p2 exhausts the last vertex that can give two triangles (as p2 starts acquiring after 3rd move). So invariably p1 will have to get just one triangle on the next turn, but it does not matter to him/her as s/he is already two triangles ahead. Hence s/he will always be 1 or more triangles ahead.


USAMTS articles algebra polynomial LaTeX

[quote="banblink182"]If we used a computer program that we found on a website (written by someone else) to help us, how should we give credit to it? I assume a formal MLA Bibliography is not necessary, but should we at least give the URL, and the name of the site?

Also, what about programs like Microsoft Excel that we used to help us quickly and accurately calculate numbers and arrange data?[/quote]

Your submission for the USAMTS is not an article to be published, so you do not have to be formal.

The main consideration in giving credit is that we graders want see your work. If the program you found is faulty or you entered your data wrong, you could have erroneous results. If we blame the error on the program rather than on your reasoning, we assign you more points. So say something like, "I found the roots of the polynomial with Rossum's Universal Math-Robot, available at http://www.obscureliteraryreference.org/rossum." We are more concerned with which steps the program accomplished than with which program you used. Be warned that too heavy a reliance on computer programs rather than your own reasoning will prevent you from earning a commendation for your solution.

A second consideration applies if you submitted your solution in LaTeX in hopes that it would be selected as an official solution. Official solutions are selected largely to explain how to solve the problem to the participants who could not solve it. So you have to include enough of an explanation how to find and use that program that the readers can duplicate your work. However, chances are that if you rely on an outside program that might not be there next year, we won't select your problem as an official solution.

Erin Schram
USAMTS grader


Solve in the set of real numbers the equation :

    $5^x + 5^{x^2} = 4^x + 6^{x^2}$

 [u]Babis [/u]

Equivalent form:

$5^x-4^x=6^{x^2}-5^{x^2}$

we can apply mean value theorem to the function $f(t)=t^x$ and $g(t)=t^{x^2}$ with $x$ fixed to obtain:

$xt_1 ^{x-1}=x^2 t_2 ^{x^2-1}$ (where $t_1 \in (4,5)$ and $t_2 \in (5,6)$)

we get two obvious solutions: $x=0$ and $x=1$ so we can rewrite it now as:

$(\frac{t_1}{t_2 ^{x+1}})^{x-1}=x$

now:

for $x<0$ rhs is negative while lhs is positive
for $x \in (0,1)$  lhs is greater than 1 (due to values of $t_1, t_2$)
for $x>1$ lhs is much bigger than rhs (easy to check)

hence the only solutions are $x=1$ and $x=0$

Note: Funny that very similar problem appeared in Romanian MO in 1978.  :)


Dear Can:

Here is my solution (different from the solution you said to me :P )
Let $b/a=x, c/b=y, a/c=z$ then $xyz=1$ and the desired inequality reads
\[\frac1{\sqrt{1+7x+x^{2}}}+\frac1{\sqrt{1+7y+y^{2}}}+\frac1{\sqrt{1+7z+z^{2}}}\geq1. \]
Now we put $x=wv/u^{2}, y=uw/v^{2}, z=uv/w^{2}$, we wish to prove that
\[\sum_{\rm cyclic}\frac{u^{2}}{\sqrt{u^{4}+7u^{2}vw+v^{2}w^{2}}}\geq1. \]
By Cauchy Schwar inequality, it suffices to show that
\[(u^{2}+v^{2}+w^{2})^{2}\geq \sum_{\rm cyclic}\sqrt{u^{8}+7u^{6}vw+u^{4}v^{2}w^{2}}. \]
Using Cauchy again, we have
\begin{eqnarray*}\Bigl(\sum_{\rm cyclic}\sqrt{u^{8}+7u^{6}vw+u^{4}v^{2}w^{2}}\Bigl)^{2}&=&\left(\sum_{\rm cyclic}u \sqrt{u^{6}+7u^{4}vw+(uvw)^{2}}\right)^{2}\\ &\leq& (u^{2}+v^{2}+w^{2})(u^{6}+v^{6}+w^{6}+7uvw(u^{3}+v^{3}+w^{3})+3(uvw)^{2}). \end{eqnarray*}
It reduces to proving that
\[(u^{2}+v^{2}+w^{2})^{3}\geq u^{6}+v^{6}+w^{6}+7uvw(u+v+w)+3(uvw)^{2}, \]
which is simple enough.

Hmm, that last should be $(u^{2}+v^{2}+w^{2})^{3}\geq u^{6}+v^{6}+w^{6}+7uvw(u^{3}+v^{3}+w^{3})+3(uvw)^{2}$. But it is still wrong. What a pity! Can you post yours, toanhocmuonmau.

Let $b/a=x, c/b=y, a/c=z$ then $xyz=1$ and

$f(x,y,z)=\frac1{\sqrt{1+7x+x^{2}}}+\frac1{\sqrt{1+7y+y^{2}}}+\frac1{\sqrt{1+7z+z^{2}}}-1.$, So we have to prove that $f(x,y,z)\geq 0$
Because ${f(x,\sqrt{yz},\sqrt{yz}})\geq 0$ then it's enough to prove that:
$f(x,y,z)\geq f(x,\sqrt{yz},\sqrt{yz})$ $\leftrightarrow$ $\frac{2}{\sqrt{1+7\sqrt{yz}+yz}}\leq \frac{1}{\sqrt{1+7y+y^{2}}}+\frac{1}{\sqrt{1+7z+z^{2}}}$,but i can't check it right now

Appling holder inequality,we have:
$A^{2}B \geq (a+b+c)^{3}$
with $B=\sum a(a^{2}+7ab+b^{2})$
Then,we slove : $(a+b+c)^{3}\geq \sum a(a^{2}+7ab+b^{2})$
But the Inequality that true with AM-GM inequality

[quote="vo thanh van"]
Then,we slove : $(a+b+c)^{3}\geq \sum a(a^{2}+7ab+b^{2})$
But the Inequality that true with AM-GM inequality[/quote]
Are you sure? :wink:

[quote="manlio"]Toanhocmuonmau, can you please give a hint for a solution?

Do you use Cauchy-Scwarz or another method?

Thank you very much.[/quote]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=147675]see here my friend.[/url]:)


perfect cube
Diophantine equation
modular arithmetic
IMO Shortlist
number theory
IMO Longlist

Does there exist an integer such that its cube is equal to $3n^2 + 3n + 7,$ where $n$ is an integer.

[b]Please post your solutions.[/b] [url=http://www.mathlinks.ro/LaTeX/AoPS_L_About.php]Use $\LaTeX$ please![/url] Please omit jokes/smilies etc. Comments and generalizations are welcome. If you noticed that the comment has been discussed elsewhere, please provide a link. If you don't know the link(s) do NOT post and state that the problems has been discussed many times. If the provided solution is not complete, right or in $\LaTeX$-style I would be happy if you could (re-) post your/the solution in this thread again! At the end of your (re-)written solution post the links to those insufficient solutions as follows:

[hide="Remark"]
- first link, e.g. http://www.mathlinks.ro/Forum/viewtopic.php?p=346917#p346917
- second link
etc.
[/hide]

[b]Happy Problem-Solving![/b] :)

We have \[ 3n^2 + 3n + 7 \equiv 4\ \textrm{ or }\ 7\ (\textrm{mod}\ 9)  \] while perfect cubes are always congruent to 0, 1 or 8 (mod 9).

So, there are no solutions.

[b]Solution by[/b] [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=26]grobber:[/url]

Assume such a number $k$ exists.

We have $k^3-1=3(n^2+n+2)$, so $k\equiv 1\pmod 3$. Let $k=3a+1$. We then get $27a^3+27a^2+9a=3(n^2+n+2)\iff 9a^3+9a^2+3a=n^2+n+2$, but $n^2+n+2$ cannot be divisible by $3$. We have a contradiction, so no such $k$ exists.


Mr. Subby makes sub sandwiches. Each small sub has 2 slices of ham. Each medium sub has 5 slices of ham and each large sub has 7 pieces of ham. If each piece of ham costs $ \$$0.27 plus $ \$$0.10 per dollar tax, how much would the ham cost for one small, medium, and large sub?

[hide]the answer is $ \$$4.13.[/hide]

Just a note, it's not really a counting problem, it's really just arithmetic

No offense, just saying.

Well since the answer is obvious

$ (2 \plus{} 5 \plus{} 7).37 \equal{} \$5.18$

Somehow the answer she gave is wrong.

edit: I can't press a key :(

that would still be wrong, even number times an odd number is even.

[quote="hunter34"]Well since the answer is obvious

$ (2 \plus{} 5 \plus{} 7).37 \equal{} \$5.13$

Somehow the answer she gave is wrong.[/quote]

The question states that the tax is 10 cents [u]per dollar[/u].

Ok.

$ .27\cdot14\equal{}3.78$ Now there are 3 dollars so there are $ \$.30$ added giving $ \$4.08$

actually i did give the wrong answer. sorry. when you find the cost of all the ham without tax you get $ \$$3.78. so you add $ \$$0.378 getting a total of $ \$$4.158 and you round to the nearest cent so your answer should be $ \$$4.16. I was 3 cents off my first answer.