calculus
derivative
calculus computations

Let f be twice differentiable on (0,infinity) and suppose that f(x) -> A as x -> infinity. (A is any real #) If f'' is bounded on  (0,infinity), prove that f'(x) -> 0 as x -> infinity.

I think I get this intuitively, but cannot prove it. Also, I don't really get why f'' should be bounded.

[quote="car202"]Also, I don't really get why f'' should be bounded.[/quote]
Let $ f(x)\equal{}\frac1x\sin(x^2).$ This is twice differentiable (and more), but $ f'(x)\equal{}2\cos(x^2)\minus{}\frac1{x^2}\sin(x^2)$ does not tend to zero (or any other limit) as $ x\to\infty.$ Of course, in this case, $ f''(x)\equal{}\minus{}4x\sin(x^2)\plus{}O(1)$ is unbounded for large $ x.$

That's why you need a condition on $ f''(x).$ And when you're trying to convince yourself intuitively through pictures, don't forget to visualize oscillating functions.

$ A$ might as well be zero.
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=176075

Ah, I thought I'd seen it. My example in this topic and mlok's example in that other topic aren't identical but are closely related.


Hello, I'm new here, I hope I'll not broke any rules of this forum with this thread. ;) I have a task from "Nonlinear Dynamics and Chaos" which looks like this:
[img]http://img430.imageshack.us/img430/1325/task9vr.jpg[/img]

Can anyone help me, since I don't know even where to start. :( Any suggestions would be highly appreciated.

P. S. Sorry for my English.
P. P. S. Maybe someone knows other forums who maybe are more related to my subject?

[quote="calc rulz"]I'm guessing that would go somewhere in our "College Playground" section. This area is for high school math.[/quote]
I guess that your guess is wrong. :) It's not so easy as it looks like. If it does look easy to you, then maybe you could help to solve it, please? :) Btw, this task is given in High school.

Steve10, I think you are misinterpreting his response.

He said the problem is too hard for this section and it should be moved to a more advanced section where people who know how to do this stuff would see it sooner.

I guess that you think "College Playground" means easier. College tends to mean what we call junior high in languages like French, Spanish, and Italian. And playground makes it seem to be easier.

College means university in English, and the type of math that this needs is math at a college level. Basically, the "College Playground" section has more advanced math than here.

PS: Problem means what you probably meant by task.


algebra
polynomial
articles
binomial theorem

The sum of the numerical coefficients in the expansion of the binomial $ (a\plus{}b)^8$ is:

$ \textbf{(A)}\ 32 \qquad
\textbf{(B)}\ 16 \qquad
\textbf{(C)}\ 64 \qquad
\textbf{(D)}\ 48 \qquad
\textbf{(E)}\  7$

[hide="Click for solution"]
Let $ a\equal{}b\equal{}1$ so the $ x$'s and $ y$'s have no effect.  Then $ (1\plus{}1)^6\equal{}2^6\equal{}64$, or $ \boxed{\textbf{(C)}}$.
[/hide]

[quote="AIME15"][hide="Click for solution"]
Let $ a \equal{} b \equal{} 1$ so the $ x$'s and $ y$'s have no effect.  Then $ (1 \plus{} 1)^6 \equal{} 2^6 \equal{} 64$, or $ \boxed{\textbf{(C)}}$.
[/hide][/quote]

ohh AIME15 this is very very elegant solution ,but what about the general case .How can we calculate the sum ?

[hide="Click for...general case?"]
By the Binomial Theorem, the expansion will look like:

$ \binom{8}{0}a^8 \plus{} \binom{8}{1}a^7b \plus{} \binom{8}{2}a^6b^2 \plus{} \ldots \plus{} \binom{8}{8}b^8$.

The sum of the coefficients is thus $ \binom{8}{0}\plus{}\binom{8}{1}\plus{}\binom{8}{2}\plus{}\ldots\plus{}\binom{8}{8}\equal{}2^8\equal{}64$, or $ \boxed{\textbf{(C)}}$.[/hide]

@enndb0x: There isn't really a "general case." Setting variables to 1 in a problem like this will (probably) always work.

This is a consequence of the binomial theorem (as AIME15 showed above), which states that $ (x\plus{}y)^n\equal{}\sum_{k\equal{}0}^{n}\binom{n}{k}x^{n\minus{}k}y^{k}$, where $ \binom{n}{k}\equal{}\frac{n!}{(n\minus{}k)!k!}$.

Shouldn't it be $ (a \plus{} b)^6$ in the problem? $ 2^8 \equal{} 256\not\equal{} 64$.

If that were true, then none of the answer choices would work because $ 256$ is not an answer choice.

[quote="AIME15"]$ 2^8 \equal{} 64$ [/quote]
yeah, the question should either be $ (a\plus{}b)^6$, or $ 256$ should be an answer choice.

This is basically using the formula for the sum of combinations/coefficients.
[hide="General case"]

$ \binom{n}{0}\plus{}\binom{n}{1}\plus{}\binom{n}{2}\plus{}\dots\plus{}\binom{n}{n\minus{}1}\plus{}\binom{n}{n}\equal{}2^n$

Proof (going backwards):

$ 2^n\equal{}(1\plus{}1)^n\equal{}\binom{n}{0}*1^n*1^0\plus{}\binom{n}{1}*1^{n\minus{}1}*1^1\plus{}\dots\plus{}\binom{n}{n}*1^0*1^n\equal{}\binom{n}{0}\plus{}\binom{n}{1}\plus{}\binom{n}{2}\plus{}\dots\plus{}\binom{n}{n\minus{}1}\plus{}\binom{n}{n}$

So we proved the general case:

The sum of the coefficients of $ (a\plus{}b)^n$ is $ 2^n$
[/hide]

So the answer is 256 in AIME15's problem, or the question is $ ...^6$.

[quote="PowerOfPi"]This is basically using the formula for the sum of combinations/coefficients. [/quote]
No, this is one way to [b]prove[/b] that formula.  You don't have to compute the coefficients of a polynomial to know that you can find the sum of its coefficients by substituting $ 1$ for every variable.

Fun way of solving: so it's basically $ (a\plus{}b)(a\plus{}b)(a\plus{}b)(a\plus{}b)(a\plus{}b)(a\plus{}b)(a\plus{}b)(a\plus{}b)$
So if you're expanding this, you can make it just a whole lot of terms with coefficient 1, and not group them. Then you just count number of terms, and it's easy. So number of terms is $ 2^8\equal{}256$ since there are 2 choices for each parentheses. :)

Its due to inflation.. you can read more about it in the news

Or maybe we are looking at base 42. Never forget bases! P.S. Remember to read AkshajK's article on converting to bases if you forgot! It is a must read!
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=721&t=566684

[quote="blasterboy"][quote="Disdain"][quote="1023megabytes"]Is it just me, or does the problem say 2^8 = 64? Last time I checked, 2^8 = 256.[/quote]

Yes, you're entirely correct. $2^8$ is equal to $256$ now. However, 5 years ago when this problem was posted, this was not the case. The original posters were completely correct in saying that $2^8 = 64$. Thanks for bringing the thread up, I hope everything makes sense now.[/quote]

Is this a troll post or something, because how does the value of $ 2^8 $ change from 2009 to 2014???[/quote]
No it's proof  by contradiction. In this case it's pointing out an error.

I just searched "Is $2^8=64?$ on google... and expected news...


geometry algebra polynomial trigonometry

Is it possible to construct an isosceles triangle inscribed in a given circle of radius 1 so that the center of the circle is in the interior of the triangle and the area of the triangle is 1? Explain.

Let $ ABC$ be the triangle with $ AB \equal{} AC$ and side lengths $ a, b, c$ (so $ b \equal{} c$). Let $ AD$ be the median/altitude with $ D$ on $ BC$ with length $ h$.

1. We know the area of the triangle is $ \frac {abc}{4R}$, so therefore $ abc \equal{} 4$. Since $ b \equal{} c$, $ ab^2 \equal{} 4$.
2. We know that the area is one half base times height. The area of our triangle is $ ah/2$, so $ ah \equal{} 2$.
3. $ a^2 \plus{} 4h^2 \equal{} 4b^2$ by the Pythagorean theorem.
4. Since the center of the circle is inside the triangle, $ h > 1$.

Combining the first three, we get $ a^4 \minus{} 16a \plus{} 16 \equal{} 0$. Using 2 and 4, we get that $ a < \frac {1}{2}$. However it is easy to prove that the polynomial has no roots in the interval $ (0,\frac {1}{2})$.

EDIT: Woops, sorry, I did screw that last step up.

MellowMelon, thanks for the explanation. I just want to clarify one thing.
How did you figure out that by using 2 and 4, we get that a is less than 1/2, and not that a is less than 2?

I'm guessing he made a mistake as $ a < 2$ and there is a solution.

Another way to see that there is a solution, is by drawing radii and defining an angle and equating areas.

You reach,
$ [ABC] \equal{} 1 \equal{} \frac{1}{2}\cdot 1\cdot 1\cdot \sin \theta \plus{} \frac{1}{2}\cdot 1\cdot 1\cdot \sin \theta \plus{} \frac{1}{2}\cdot 1\cdot 1\cdot \sin (360 \minus{} 2\theta), \; \theta \in [\frac{\pi}{2}, pi]$

$ \implies 1 \equal{} \sin \theta \minus{} \frac{1}{2} \sin (2\theta)$

Which has a solution. This doesn't show that it's constructible though.

Suppose that we have triangle $ ABC$ with $ AB\equal{}AC\equal{}a$, and $ \angle BAC\equal{}2\theta$ where $ \theta\in(0,90)$. Since the triangle is acute, we must have $ \theta\in (0,45)$. We are given that $ \frac{1}{2}\cdot a^2\sin 2\theta\equal{}1$ so\[ a^2\sin 2\theta\equal{}2\] Additionally, $ \frac{a}{2R}\equal{}\sin 90\minus{}\theta$, so \[ a\equal{}2\cos\theta\]
Therefore \[ \cos^2\theta\sin2\theta\equal{}\frac{1}{2}\]
This obviously has a solution at $ \theta\equal{}45$. There is another solution as well, I am just not sure how to show it yet...perhaps we can square and consider a polynomial...

Because x^4-16x+16=(x^3+2x^2+4x-8)(x-2),
can we prove that x^3+2x^2+4x-8 does not have any rational root, which means that it has no constructible root?


Let $ ABC$ is triangle, $ \alpha, \beta, \gamma > 0$ proof that:

$ P_{ABC} \equal{} \left(\frac {a \plus{} b \plus{} c}{2}\right)^2tg{\frac {\alpha}{2}}tg{\frac {\beta}{2}}tg{\frac {\gamma}{2}}$

Did you mean "$ P_{ABC}$" is area of triangle?

[b][color=blue][u]Proof:[/u][/color][/b]
$ tg{\frac {\alpha}{2}} = \sqrt {\dfrac{(p - b)(p - c)}{p(p - a)}}$
$ tg{\frac {\beta}{2}} = \sqrt {\dfrac{(p - a)(p - c)}{p(p - b)}}$
$ tg{\frac {\gamma}{2} = \sqrt {\dfrac{(p - a)(p - b)}{p(p - c)}}}$

So $ p^2 \cdot tg{\frac {\alpha}{2}}tg{\frac {\beta}{2}}tg{\frac {\gamma}{2} = p^2 \sqrt {\dfrac{(p - b)^2(p - c)^2(p - a)^2}{p^3(p - a)(p - b)(p - c)}} = \dfrac{p^2}{p^2}\sqrt {p(p - b)(p - c)(p - a)} = S_{ABC}}$, q.e.d.


Find all right -angled triangles that each have integral side lengths and the area is equal to the perimeter.

[hide="Solution"]
Since the triangle is a right triangle we know that(when $ c$ is the hypotenuse),
\[ a^2+b^2=c^2\]
Also we have that the perimeter equals the area,
\[ a+b+c=\dfrac{ab}{2}\]
From here we get,
\begin{eqnarray*}
4(a+b+c)&=&2ab\\
4(a+b+c)&=&2ab+a^2+b^2-c^2\\
4(a+b+c)&=&(a+b)^2-c^2\\
4(a+b+c)&=&(a+b+c)(a+b-c)\\
4&=&a+b-c
\end{eqnarray*}
We then have,
\begin{eqnarray*}
2(a+b)&=&(a+b-c)+(a+b+c)\\
2(a+b)&=&4+\dfrac{ab}{2}\\
4(a+b)&=&8+ab
\end{eqnarray*}
Which we can manipulate even further,
\[ 4(a+b)=8+ab\implies ab-4a-4b+8=0\implies ab-4a-4b+16=8\]
Which implies that,
\[ (a-4)(b-4)=8\]
Solving this equation in positive integers gives,
\begin{eqnarray*}
&a=5,b=12\implies c=13&\\
&a=6,b=8\implies c=10&
\end{eqnarray*}
And those are all the right triangles that fit the requirements(you can swap $ a$ and $ b$ for different but basically the same triangles)... there probably is a faster way to do this.
[/hide]

Following anthony's start
$ a\plus{}b\minus{}\frac{ab}2\equal{}c$
squaring
$ (a\plus{}b\minus{}\frac{ab}2)^2\equal{}c^2$
substitute
$ (a\plus{}b\minus{}\frac{ab}2)^2\equal{}a^2\plus{}b^2$
Simplify
$ 8\minus{}4b\minus{}4a\plus{}ab\equal{}0$
Factor
$ (a\minus{}4)(b\minus{}4)\equal{}8$
for which we get the same conclusion as him :P


How did ya'll do on the proof section? Scores should be on your teachers' emailboxes by Monday.

I got a 46, composite 246.

Wow, I actually made the top 30 with a composite of 102 + 3 * 29 = 189 (top 30 cut-off was 153). Good job on being nearly 100 points above the cut-off, solafidefarms.

[quote="solafidefarms"]Well, I have a suspicious feeling that the top score was 300 and that at least 2 people scored above 290, so I don't know how well I'll actually place...[/quote]
What makes you think that??  By "supsicious" do you mean "unsubstantiated"?  According to rumor, Phillip had something like the highest part I score in the state at 144.  I had a 132 and got a 50 on part II to get 282, which was the highest Lassiter score so post if you know that someone who got over 132 on part I and pretty close to perfect on part II, or maybe I'll just have to wait until 14 Apr. to find out how I placed. :P  Good job to everybody who made the top 30 cutoff, anyway.

By the way, Mrs. Poss said I should be getting an email as well and I haven't gotten one; is this the case with anybody else?

I did laughably terribly  :( I didn't even score 100+ on the written  :noo:

Got results today.

$134+3\cdot29=221$

Note to self: learn to write proofs and not forget about the competition.

I got 150, missed the cutoff by 3 points, rats. Dunno how I only got 20 points (60) on the written, I thought I nailed two and had an insignificant error on one, which should have been 25 minimum... oh well.

The awards ceremony was interesting, but I don't understand what that weird clip from that movie was all about. I got laughed at when accepting my award because I am bad at smiling for cameras. I am disappointed that I didn't get to find out how I placed (they announced all but the top 3 in alphabetical order); I would have liked to know if I beat any of the 13 former ARMLers who were in the top 30 (8 former ARMLers who are still in high school were not in the top 30 and most likely did not participate). I only have to wait two more weeks to find out if I make ARML.


pigeonhole principle
algorithm
number theory unsolved
number theory

Can anyone prove this?

A real number is rational if and only if it has a periodic decimal expansion.

I am having trouble with this proof:

=> Assume a number has a periodic decimal expansion:

$a_{0}.a_{1}a_{2}a_{3}\sim a_{n}a_{1}\sim a_{n}a_{1}\sim a_{n}\sim$ for $n \in N$. This number equals $a_{0}+\frac{a_{1}}{10}+\frac{a_{2}}{10^{2}}+\ldots+\frac{a_{n}}{10^{n}}+\frac{a_{1}}{10^{n+1}}+\ldots+\frac{a_{n}}{10^{2n}}+\dots$. But how can express this in compact form? If I can express this in compact form, then I can probably write this as a rational number $\frac{m}{n}$.

<= Assume the number is rational:
so we have $x = \frac{m}{n}$, where $n \ne 0$. We must somehow change this to the form we have above, but I have no idea how.

Can anyone help me?

THanks.

I guess this should probably be moved to another appropriate forum.

Your statement isn't quite right; after all, the decimal expansion $1.2340000000000\dots$ isn't periodic, but that's definitely a rational number. A correct statement:
A real number is rational if and only if it has a decimal expansion which is eventually periodic.

Ignoring that "eventually" detail, here's the key step:

$a_{0}+\frac{a_{1}}{10}+\frac{a_{2}}{10^{2}}+\ldots+\frac{a_{n}}{10^{n}}+\frac{a_{1}}{10^{n+1}}+\ldots+\frac{a_{n}}{10^{2n}}+\dots$
$=a_{0}+\frac{10^{n-1}a_{1}+10^{n-2}a_{2}+\cdots+a_{n}}{10^{n}}+\frac{10^{n-1}a_{1}+10^{n-2}a_{2}+\cdots+a_{n}}{10^{2n}}+\cdots$
Think geometric series.

I'll leave the rest to you, including how to handle "eventually periodic".

Thanks a lot, jmerry. I think I've figured out how to prove that any eventually repeating decimal can be expressed as a rational number. However, I still have to prove the reverse, namely, any rational number $\frac{m}{n}$ can be written as a repeating decimal (in this problem, we can assume repeating decimal - we don't have to prove the "eventually" repeating decimal thing). This part is giving me the most trouble.

I divide into 3 cases:

if $m < n$, we need to prove $\frac{m}{n}$ is a repeating decimal.

$m = n$: the case is trivial => $\frac{m}{m}= 1 = 1.0000 \sim$

$m>n$, we can always refer back to the case $m<n$ b/c $\frac{m}{n}$ will then be an improper fraction.

So really the problem comes down to proving the first case with $m<n$.
THanks.

[b]Problem:[/b] Prove that every fraction ${m\over n}, m,n\in\mathbb{N}$ can be expressed either as a finite decimal number or as a "properly" or "improperly" periodic decimal number. ("Improperly" meaning that the period begins after some number of leading digits, and "properly" that the period begins right after the decimal point.)

[u]Lemma 1[/u]: For every natural number $N>1$ such that $\gcd(N,10)=1$ there exists a natural $M$ such that $N\mid10^{M}-1$

[u]Proof[/u]: Observe $N$ numbers $10^{1},10^{2},\dots,10^{N}$. If two of them give the same remainder modulo $N$ - say $10^{k}$ and $10^{l}$ where $k>l$ - then $N\mid 10^{k}-10^{l}\implies N\mid 10^{l}(10^{k-l}-1)$, but as $N$ is coprime with $10$, this gives $N\mid 10^{k-l}-1\implies M=k-l$. If all of them give pairwise different remainders modulo $N$, then by the Pigeonhole Principle one of those remainders must be $1$. Hence there exists $M$ such that $N\mid 10^{M}-1$

[u]Lemma 2[/u]: An "improperly" periodic number $r=0.qpppp...$, where $q$ is a string of $Q$ digits and $p$ is a string of $P$ digits, can be represented as $r=\frac{q(10^{P}-1)+p}{10^{Q}(10^{P}-1)}$. (Note: the same reasoning applies for "properly" periodic numbers if we put $q=p$, which gives $r=\frac{p}{10^{P}-1}$, and the rational numbers with finite decimal expansion are obtained by putting $p=0, P=1$, which gives $r=\frac{q}{10^{Q}}$.)

[u]Proof[/u]: Observing the sum $r=\frac{q}{10^{Q}}+\frac{p}{10^{P+Q}}+\frac{p}{10^{2P+Q}}+\frac{p}{10^{3P+Q}}+\dots$, we see that it can be evaluated as a sum of the first summand and an infinite geometric series. Doing so yields the given expression.

[b]The proof of the main statement[/b]: Let $r={m\over n}$ be a non-integer fraction with $m,n\in\mathbb{N}$. WLOG we can asume $m<n$ since otherwise we can always extract the integer part.

Write $n$ as $2^{a}5^{b}s$, where $a,b$ can be zero and $s$ is coprime with $10$. Then there exists $N_{1}$ such that $2^{a}5^{b}N_{1}=10^{u}$ for some natural $u$, and if $s>1$, then by Lemma 1 there exists $N_{2}$ such that $sN_{2}=10^{v}-1$ for some natural $v$. Expand the fraction by $N_{1}N_{2}$. Then the denominator will be $10^{u}(10^{v}-1)$, and according to Lemma 2, that leads to an "improperly" or "properly" periodic decimal number. If $s=1$, then expand the fraction by $N_{1}$ to obtain the denominator $10^{u}$, which gives the finite decimal expansion.

Sorry for bringing up such an old topic, but I had to solve this problem yesterday while studying ACoPS. I think I have a different solution, although not so rigorous.

Consider the fraction $ \frac{a}{q}$ with $ a$ and $ q$ integers, that have the decimal expansion $ b.b_1b_2b_3...$ We use repeatedly the division algorithm:

$ a\equal{}bq\plus{}r$
$ 10r\equal{}b_1q\plus{}r_1$
$ 10r_1\equal{}b_2q\plus{}r_2$

and so on. Note that $ r<q$, so the maximum possible value for $ 10r$ is $ 10(q\minus{}1)<10q$, and so the numbers $ b_1$, $ b_2$, ... are all less than $ 10$. It is easy to see that if one of the $ r_i$s repeat we get a periodic expansion. But this will happen, because there are only $ q$ possible remainders and the operation is repeated [i]ad infinitum[/i].


AMC
USA(J)MO
USAMO
search
inequalities unsolved
inequalities

If x,y,z are positive reals prove that
Σ 1/(x^3)+(y^3)+xyz <=1/xyz

didn't someone post this yesterday? it's USAMO 97.

[quote="silouan"]I do not know . I want a solution[/quote]
I think you really need to search before you post. Double-posting is highly disgusting and unwelcomed.
For your problem, use $x^3+y^3 \ge xy(x+y)$


How many integers between 1 and 2009 inclusive can be expressed as a sum of five consecutive multiples of 7?

[quote="Jongy"]Let the multiples of 7 be 7(x-2), 7(x-1), 7x, 7(x+1), and 7(x+2).
Notice that the average of these five numbers is the middle one: 7x.
Thus, the sum of the five multiples is 5*7x or 35x.
Now we know that the possible answers are multiples of 35.
Do the floor of 2009/35 and get [b]57[/b][/quote]

Please note that x has to be greater than two so that the numbers are positive, so there are $\boxed{55}$ ways.

[hide="Alternate Solution"]
Let $n$ be a positive integer. Our 5 integers are $7n$, $7n+7$, $7n+14$, $7n+21$ and $7n+28$. Adding these up yields $35n+70=35(n+2)$. We can only have numbers divisible by $35$. We see that $35(n+2)=35\implies n=-1, 35(n+2)=70\implies n=0$. So $n=3$ is the smallest. We also know that $1995\div 35=57$. So $n=57$ is our largest.

We have $57-3+1=\boxed{55\text{ numbers}}$.[/hide]

How come n must be greater than 2? Someone please help me :ewpu:

You can get a sum of 35 if the numbers are -7, 0, 7, 14, 21 or a sum of 70 if the numbers are 0, 7, 14, 21, 28. However, both of these cases, while they do lead to multiples of 35, do not use positive numbers.


For natural $ a, b$ we have
$ a|b \plus{} 1$ and $ b|a^2 \minus{} 2$.
Prove that $ \frac {b \plus{}1}{2}$ is square of integer.

[quote="Kamyar1991"]b+1/2 must be 1![/quote]
Don't just post unproven wrong claims  :wink:


trigonometry
inequalities
vector
search
inequalities proposed

i Think that the expression doesn't have a max , and the sup is $ 1$, probably I'm wrong because I'm sleepy now  :sleeping:

We have that $ \sqrt{a_ka_{k\plus{}1}} \leq \frac{a_k \plus{} a_{k\plus{}1}}{2}$ for $ k \equal{} 1,2,3,...,10$ by AM-GM. Therefore the maximum is when we have equality, which is when $ a_k \equal{} a_{k\plus{}1}$. This means that the maximum of $ P$ happens when $ a_1\equal{}a_2\equal{}a_3\equal{}...\equal{}a_{10}\equal{} \frac{1}{10}$.

$ \therefore \max (P) \equal{} 9 \cdot \frac{1}{10} \equal{} \boxed{\frac{9}{10}}$

That's wrong. $ P\leq 1 \minus{} \frac {a_1 \plus{} a_{10}}{2}$. This  can be also made a confirmation by Cauchy-Schwarz.

That's not he maximum kunny. The maximum is atypical .

alex ,I think  kunny is right , with your estimation  you get only  $ P \leq 1 \minus{} \frac {a_1 \plus{} a_{10}}{2}$  anyway I will try to find a counterexemple later :wink:

Yes , kunny is right , but his way doesn't find the desired value . The problem was posted before , but it has no solution.

[hide="Hint"]The maximum value is $ \cos \frac{\pi}{11}$[/hide]

thank you for your hint , the ineq  is similar to fan's  to inequality , and I think the solution is related  to vectors and dot space in $ R^n$ but I won't try anyway  :D  because I don't know very well about that subject  , please see:
[url]http://www.mathlinks.ro/viewtopic.php?search_id=1592523491&t=19556[/url]

Sorry for my stupid solution.. I guess I'll lay off this forum lol


Who's going to the Awards ceremony?
The speaker isn't as famous as last year's (JOHN CONWAY, OMG!) but should still be very good.
I can't come due to a scheduling conflict. Is anyone here going that can help me pick up my t-shirt and anything else I get? Maybe you could give it to me at SCVMA field day or whatever the next math contest is.
Thanks in advance!

[quote="bowei"]I can't come due to a scheduling conflict. Is anyone here going that can help me pick up my t-shirt and anything else I get?[/quote]

Same here...

I really wasn't expecting anything, but whatever :) I got the email today :)

Oh, and bowei, I was the guy who sat behind you and to the left who kept dropping his pencil :)

Do they only email to those who place right??If it is,that means I failed! :(

@bowei: I emailed Mr. Zucker and he said he could send me the prizes through mail.  You should email him about it.

@ghjk: Yes. :(

Yeah, I'll be going. I don't think I won anything substantial, but I'm going anyways :)

When is it again?
Also, does honorable mention win anything interesting?

Last year honorable mention got a book, I think.  Maybe a t-shirt also.

I'm really looking forward to hearing Prof. Su again. I really like how he talks about applications of math in human behavior.

What topic is the lecture going to be on?

so I'll probably go as well

Is there anything lower than hm which still earns a prize?

Wait, does everyone get a t-shirt? (I didn't take BAMO)

Darn. I was afraid that they were going to be EXTRA EXTRA rigorous on the first few problems, but I still got owned.

[quote="serialk11r"]Sorry team for being lazy and not writing real solutions, and losing us the participation award [sob].  :([/quote]
Hmm...thanks?

So does anyone have a somewhat complete account of the top placements and schools? I know how Paly did, but that's it.

So Paly got participation?? asdkjla;dfosaypwuerwtncm
John Boyle better not have signed up as being from "Paly".

No, he didn't. We hosted a lot of people from other schools, but the 11 students who contributed to participation were all ours (including this random freshman girl who *does not do math*...lucky for us that someone just dragged her in at the last minute...)

Congrats to quatto for winning the beast award!

Honorable mention got a book and a shirt featuring problem 4  :|

hm? i wasn't there.
which beast award? the beast proof award, or just plain high-score-beast award?
who else got awards that we know? bookaholic? ubemaya?

[quote="bowei"]hm? i wasn't there.
which beast award? the beast proof award, or just plain high-score-beast award??[/quote]
Both.

[quote="serialk11r"]No, because he got 4 points deducted on number 4 for not writing "ABE is isoceles" :O[/quote]
4 points deducted??? :o. I wrote this part but didn't get answer. Hmm... i probably get 2 points at most :rotfl: . And I'm so suspicious that I'm one of the 3 people who got #7! Did anyone hear a strange Vietnamese name?? Evan O Dorney got perfect???

No, Evan got 34, 1 point off somewhere.
The 4 points was a rigor deduction, meaning you wouldn't have gotten anything for just that, since it was pretty much implied and therefore not a critical step.

hahaha quatto = beast :)

where can I find more results?

That's what I want to know! Come on, people, who else got awards?

Fine, I'll start. Paly was 1st in participation and 2nd in team score. Avi Arfin and I tied for 3rd in 11th/12th. Nathan Pinsker was 3rd in 9th/10th. That's as much as I've been made aware of.

[quote="serialk11r"]Albert tells me:
Quatto = 31, 4 points deducted on number 4.
Albert tied for 3rd, among many others.
Someone from Homestead won 11-12, Taylor Han got 2nd.
Tons of HM from Gunn, then we have me...
I heard Saratoga won overall.[/quote]

I think I actually got 2nd (that's what Quatto told me), but maybe he lied to raise my self confidence.

BTW, all of that info that you just said was from Quatto.

Nobody talks about BAMO 8 :mad:

I heard that I got honorable mention from dragon96, and then he said my score must have been higher than 21 points.  However, I [b]know[/b] i misread number 4, so i'm pretty sure that's impossible........

ololololol i got 2nd place in 11/12
by like 1 point more than lynnelle
=P?
but wtf i got ripped 4 points (2 from 4 2 from 5)
and man i spent way too long on #5 and i know i couldve gotten the whole of 6 and partially 7 with more time blargh?

Ahh...got my score.

21 exactly, 7-7-7-0 :D

My bs trig from misreading didn't even get me one point :)


Let A(9,3), B(9,10) and C(k,10) be points in the cartesian coordinate plane. what is the smallest value for k such that the distance from A to C is 25 units?

Why are we given the point B?

[hide]distance formula.
$\sqrt{(k-9)^{2}+49}=25$
$k^{2}-18k+130=625$
$(k+15)(k-33)=0$
so $k=-15,33$ and the smallest value is $\boxed{-15}$[/hide]


I have found archives of previous AIME tests, but I've never been able to find any AMC 10/12.  Can anyone help me out?

[url=http://www.math.ksu.edu/main/events/hscomp/samples/amc12/sample.htm]This[/url] is probably the best resource as of this time. However, AoPS is building a more comprehensive database of problems as I type :D Click Contests at the top of the page. (highlighted in white)

[url=http://www.unl.edu/amc/mathclub/04,1-problems.html]Here are some more recent ones[/url].


Um quick q...
Isn't the moderator of this forum jeffery?
His last post was like 3~4 months ago on a different forum.
No offense jeffery but I think another moderator is called for.
The purpose of a moderator is to moderate the forum and if the last time the mod posted was on April 17th, some one needs to step in.

Another question.
How come this forum is so empty and unused.
We should be proud to be washingtonians.
Home of Darryl Wu, the youngest MATHCOUNTS winner.

Last question.
Jeffery, are u the one who calls out the name of the MIC winners?
U might have said my name a couple times...

A forum with no activity doesn't need a moderator. There just isn't that much to talk about of local interest, and there's even less during the summer.

Maybe we need more people in Washington to join AoPS
Or maybe their just to shy to post in this forum...
It seems like the older "generation" used this forum, but the newer people (Class of 2012-2014) just don't use AoPS.

Maybe the 2008-2009 year will yield better results.

[quote="LastWarrior"]
It seems like the older "generation" used this forum, but the newer people (Class of 2012-2014) just don't use AoPS. [/quote]

There just doesn't seem to be a ton of exposure. I know that I never heard of AoPS until I went to MC Nationals in 7th grade.

Another explanation: people don't see anyone post...and then they don't post...and then nobody else bothers to post...(neverending cycle)

-Joy

P.S. Dojo, LastWarrior, who are you?

There isn't much to do here, although exposure has something to do with it.  The only thing really worth communicating is information about local competitions, and it's easier to do that using the e-mail list of coaches (or talking to Norris).

There are also two "Jeffreys" from Washington that post here, and neither of them is currently the moderator.

I was born in Washington but don't live there. I don't even remember anything. :(

I wasn't born in WA but do live here.  I remember everything! :)


Putnam function calculus derivative trigonometry logarithms integration

Functions $ f,g,h$ are differentiable on some open interval around $ 0$ and satisfy the equations and initial conditions
\begin{align*}f'&=2f^2gh+\frac1{gh},\ f(0)=1,\\
g'&=fg^2h+\frac4{fh},\ g(0)=1,\\
h'&=3fgh^2+\frac1{fg},\ h(0)=1.\end{align*}
Find an explicit formula for $ f(x),$ valid in some open interval around $ 0.$

First, clear the fractions.
\begin{align*}f'gh&=2f^2g^2h^2+1\\fg'h&=f^2g^2h^2+4\\fgh'&=3f^2g^2h^2+1\end{align*}
Add those three equations together, and note that the left hand side is the derivative of a triple product. Let $ y=fgh$ and we have
\[ y'=6y^2+6,\ y(0)=1\]
This is a separable equation. Solving it, we get $ \arctan(y)=6x+C,$ with $ C=\arctan 1.$ In a neighborhood of zero, we can solve that to get
\[ y=\tan(6x+\arctan 1)=\frac{1+\tan 6x}{1-\tan 6x}\]
Now, return to the first equation"
\[ f'=2f(fgh)+\frac{f}{fgh}=2fy+\frac fy=f\left(2\,\frac{1+\tan 6x}{1-\tan 6x}+\frac{1-\tan 6x}{1+\tan 6x}\right)\]
\[ \frac{f'}{f}=\frac{3+2\tan 6x+3\tan^26x}{1-\tan^26x}=\frac{3\cos^26x+2\cos 6x\sin6x+3\sin^26x}{\cos^26x-\sin^26x}\]
\[ \ln f=\int\frac{3+\sin12x}{\cos12x}\,dx=\int3\sec12x+\tan12x\,dx\]
\[ \ln f=\frac3{12}\ln(\sec12x+\tan12x)+\frac1{12}\ln(\sec 12x)+C.\]
Give $ f(0)=1,$ we obtain that $ C=0.$ Exponentiating gives us our solution:
\[ \boxed{f(x)=\left[\sec12x+\tan12x\right]^\frac14\left[\sec 12x\right]^{\frac1{12}}}\]

For this question, I approached it almost exactly the same way as the above solution, but I never learned how to solve differential equations, so on the proof, I just said "If I could solve this equation, then this is what I would do to get a solution..." How many points could I possibly get for that (because the process leading to the differential equation was exactly as outlined above..)?

For fans of calculus,this problem shouldn't take more than 15 minutes i think to write a completed proof.
Too easy for A2,i think.The idea of the triple product comes immediately in your mind.

Took me more than 15 minutes, because I kept making mistakes. Somewhere, I have some scratch paper with $ \frac1{1\minus{}6x}$ on it - because I forgot about the constant terms.

You might be right.That's why, in my opinion,i believe that most of the contestants had trouble with the computational part in this problem
rather than finding the separable differential equation.

ps This kinda answers your question [b]Nawahd Werdna[/b]. I don't think not fully showing an explicit formula for the function
can give you more than 1 or 2 points.Since the majority of students have found that seperable equation.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=19629&search_id=191077245]System of Differential Equation 1[/url]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=212215&search_id=898826459#212215]System of Differential Equation 2[/url]

I didn't know the identity

$ \tan(6x \plus{} \arctan 1) \equal{} \frac {1 \plus{} \tan 6x}{1 \minus{} \tan 6x}$

so I ended up sticking with $ \tan(6x \plus{} \arctan 1) \equal{} \tan(6x \plus{} \frac {\pi}{4})$, which brought me to a final answer of:

$ y(x) \equal{} 2^{\minus{}1/12} \left[\sin(6x \plus{} \frac {\pi}{4})\right]^{1/6}\left[\cos(6x \plus{} \frac {\pi}{4})\right]^{ \minus{} 1/3}$

Which I believe is equivalent to yours.

Roosta: the identity in question was the addition formula:
\[ \tan(a\plus{}b)\equal{}\frac{\tan a\plus{}\tan b}{1\minus{}\tan a\tan b}.\]

I ended up with the same thing as you Roosta, it agrees with Professor Kent's solution in a neighborhood around 0, so hopefully it counts.

Sure. If you can verify that it's the solution, it's right. There can't be a preferred form here - they'll have to take anything that is correct. You could potentially write this function is several other forms, as well.

How many points can you expect to get if your method of solving it is correct, but you made a small algebra mistake? I accidentally got $ \frac{f'}{f}\equal{}2y^2\plus{}\frac{1}{y}$  :(.

Roosta my answer was also in your form but the constant is $ \frac{1}{\sqrt[12]{2}}$

That's the form that Kiran Kedlaya [i]et al.[/i] used for the answer in the solutions he posted on the AMC website. To be specific, he said
\[ f(x) \equal{} 2^{\minus{}1/12}
\left(\frac{\sin(6x\plus{}\pi/4)}{\cos^2(6x\plus{}\pi/4)}\right)^{1/6}\]

Whoops, I misremembered the answer. I actually wrote it on the exam as:

$ f(x) \equal{} \left(\frac {\sin(6x \plus{} \pi/4)}{\sqrt {2}\cdot \cos^{2}(6x \plus{} \pi/4)}\right)^{1/6}$

but accidentally pulled the $ \sqrt{2}$ when I typed it up for the forum.

Why don't you make sure that your answer is at least correct or not by plugging $ x \equal{} 0$ into the result?

I got the answer little different
$ \frac{\tan{(x +\frac{\pi}{4})} \sec{(x +\frac{\pi}{4})}}{\sqrt{2}} $

Got the same solution as #2
[hide][img]https://i.imgur.com/0v6igcg.jpg[/img] [img]https://i.imgur.com/zcxJmyX.jpg[/img][/hide]


trigonometry
geometry
search
analytic geometry
area of a triangle

You see an Equilateral triangle painted on the floor. From where you are standing, one corner is $3m$ away, another corner is $4m$ away and the third is $5m$ away.

How big is the triangle?

if the sides can only be integers, Is the answer..... $6$?

[quote="violetcraze"]if the sides can only be integers[/quote]

Where does it say that?

This is a very famous problem ... it has appeared on the forum multiple times, I am sure.

[quote="Gyan"]Yes it has appeared here before ... Also .. answer depends on whether you are inside the triangle or not.
If you are inside the triangle, and want to check your answer, it is about $6.7664m$[/quote]

Can you post your solution? Nice problem actually. Thanks

[quote="Gyan"]Jose : Sure, One of the method  (There are some other cute ones too)
[hide]Let  the triangle be $A B C$ and the point $P$ is “inside” such that  $PA = 5,PB=4,PC=3$  Now  “rotate” the triangle  $APC$  by $60^o$   keeping C fixed  and  in such a way that  A is now touching B. Let the new position of P be called Q.  (Note Q is outside the triangle) (Now  $QB=5, QC=3$). Triangle QPC is equilateral ($CP=CQ=3$ and the angle PCQ =$60^o$. and Triangle BPQ is right angle triangle (Sides being  3-4-5) so angle  BPQ = $90^o$ and thus angle BPC= $90^o+60^o=150^o$  so now (see triangle BPC – apply  cos formula)
 $BC^2 = 3^2 + 4^2 - 2 \cdot 3 \cdot 4 \cdot \cos 150^o = 25+12 \sqrt 3$

(if you are outside the triangle, similar argument would yield [hide] $\sqrt{25 - 12 \sqrt 3 }$  for the side)[/hide]

[hide]In general , if the segents are $a,b,c$ you get something like $\sqrt{\frac{a^2+b^2+c^2} 2 - 2\sqrt 3 \cdot A}$ where $A$ is area of a triangle whose sides are $(a,b,c)$ (There are few really cute geometric way to see this - search this forum and you will find it) [/hide]
 [/hide][/quote]

excellent answer. I ended up having $4th$ degree equations when tried solving using coordinate geometry.


function
logarithms
algebra unsolved
algebra

[quote="tranthanhnam"]find all continous functions $f: (0;\infty )\to (0;\infty)$ which satisfy:
$f(\frac{1}{f(xy)})=f(x)f(y)$[/quote]

[quote="tranthanhnam"]determine all real functions $f(x)$ which satisfy
$f(x-f(y))=f(x+y^{1999})+f(f(y)+y^{1999})+1$[/quote]

Both are problems from KOMAL,aren't they? :D  :)
I have solved the first one.Try to bring the problem to the form
$g(x+y)=g(x)g(y)$
then by taking logarithims of both sides
you get: $log{g(x+y)}=log{g(x)}+log{g(y)}$
which is additional function.


geometry incenter ratio calculus trigonometry trapezoid parallelogram

Dear Mathlinkers,
let ABC a A-rectangular triangle, I the incenter, PQR the contact triangle, P'Q'R' the ex-contact triangle and A' the midpoint of BC.
Prove that QR', RQ' and A'I are concurrent.
Sincerely
Jean-Louis

The only think I can do : a figure ! :maybe:

[size=117][color=darkblue][b]Thank you, [u]Jayme[/u], for your very nice problem ![/b][/color][/size]

[quote="jayme"] [color=darkred]Let $ ABC$ be an $ A$ - right triangle with the incircle $ C(I,r)$ . Denote the midpoint $ M$ of  $ [BC]$ and $ \{Q,Q'\}\subset (AC)$ ,

$ \{R,R'\}\subset (AB)$ so that $ AQ \equal{} AR \equal{} CQ' \equal{} BR' \equal{} p \minus{} a$ . Prove that $ QR'$ , $ RQ'$ , $ MI$ are concurrently.[/color] [/quote]

[quote="jayme"] [color=darkred]I used an ugly method to prove it (algebraic method). Just wonder if there is a beautiful method (pure geometry).

Can you find [b]a proof without cross ratio, calculus, report, congruent triangle, trigonometric a.s.o.[/b][/color] [/quote]
[hide="Here is a pure idea."]
[color=darkblue]Try to show that the lines $ QR'$ , $ RQ'$ , $ MI$ are concurrently in a point $ S$ which belongs

to the $ A$ - altitude and $ AS \equal{} r$ . [b]Sorry[/b], I"ll present here my hidden proof because [b]it is ugly[/b]

(I used [b]cross ratio, calculus, report, congruent triangle, trigonometric a.s.o.[/b]).[/color][/hide]
[hide="Here is my ugly proof."]
[color=darkblue]Suppose w.l.o.g. $ b > c$ . Denote $ \left\|\begin{array}{c} D\in BC\ ,\ AD\perp BC \\
 \\
T\in BC\cap QR' \\
 \\
S\in AD\cap QR'\end{array}\right\|$ . Apply the [b]Menelaus' theorem[/b] to $ \overline {TR'Q}/\triangle ABC\ :$

$ \frac {TB}{TC}\cdot\frac {QC}{QA}\cdot\frac {R'A}{R'B} \equal{} 1$ $ \implies$ $ \frac {TB}{TC}\cdot\frac {p \minus{} c}{p \minus{} a}\cdot\frac {p \minus{} b}{p \minus{} a} \equal{} 1$ $ \implies$ $ \frac {TB}{TC} \equal{} \frac {(p \minus{} a)^2}{(p \minus{} b)(p \minus{} c)}$ . Since $ \tan^2\frac A2 \equal{} \frac {(p \minus{} b)(p \minus{} c)}{p(p \minus{} a)} \equal{} 1$

obtain $ \frac {TB}{TC} \equal{} \frac {p \minus{} a}{p}$ $ \implies$ $ \frac {TB}{p \minus{} a} \equal{} \frac {TC}{p} \equal{} \frac {a}{a} \equal{} 1$ $ \implies$ $ TB \equal{} p \minus{} a$ , i.e. the $ C$ - excircle touches the sideline $ BC$ in $ T$ .

Thus, $ TD \equal{} TB \plus{} BD \equal{} (p \minus{} a) \plus{} \frac {c^2}{a}$ $ \implies$ $ TD \equal{} \frac {(b \plus{} c)(p \minus{} b)}{a}$ . Apply the [b]Menelaus' theorem[/b] to $ \overline {TR'S}/\triangle ABD\ \ :$

$ \frac {TB}{TD}\cdot\frac {SD}{SA}\cdot\frac {R'A}{R'B} \equal{} 1$ , i.e.  $ \frac {p \minus{} a}{\frac {(b \plus{} c)(p \minus{} b)}{a}}\cdot \frac {SD}{SA}\cdot\frac {p \minus{} b}{p \minus{} a} \equal{} 1$ , i.e. $ \frac {SA}{a} \equal{} \frac {SD}{b \plus{} c} \equal{} \frac {h_a}{2p}$ . Therefore, $ SA \equal{} \frac {ah_a}{2p} \equal{} \frac {2pr}{2p}$ ,

i.e. $ SA \equal{} r$ what means $ S\in MI$ . Prove analogously that the intersection between $ RQ'$ , $ AD$ belongs to $ IM$ .

[b][u]Remark.[/u][/b] Now you can find easily a "pure" synthetical proof starting from the tangent point  $ T$ of the

$ C$ - excircle with the sideline $ BC$ and the point $ S\in AD$ for which $ AS \equal{} r$ , i.e. $ S\in MI\cap AD$ . [b]SUCCESS ![/b] [/color] [/hide]

A solution not in datails :
Let C-excircle touches AC at $ Q''$  we know that $ RQ\perp R'Q''$ and $ R'A\perp AC$ therefore $ R$ is orthocenter of $ R'QQ''$ hence $ Q''R\perp R'Q$ also we know that $ Q''R\parallel{}I_cB$ hence $ R'Q$ is parallel to angle bisector of $ B$ simialry RQ' is parallel to C-angle bisector .
Let QR' intersect A-altitude at S by angle chasing it's easy to prove that $ AS \equal{} AQ \equal{} r$ hence RQ',R'Q and A-altitude are concurrent so according to well-know lemma IM passes through $ S$

It's well known that AR=AQ=IR=IQ=BR'=CQ'=r. See that BIQR' and ICQ'R are paralellograms, so QR'¡¡BI and Q'R¡¡CI. Note D and E intersections of QR' and respectively Q'R with BC, apply Menelaus in triangle ABC with QR'D and RQ'E transversals, get BD=CE, so triangles BIC and DES (S-intersection of QR' and Q'R) are similar, with parallel sides, have the same median A'I.

[color=darkblue] [b]Jayme[/b], here is a similar nice and easy problem from my friend, the student [b]Andrei Ciupan[/b]. I hope you like it. [/color]

[quote="Andrei Ciupan"][color=darkred]Let $ \triangle ABC$ with the incircle $ w = C(I,r)$ and the $ A$ - excircle $ w_a = C(I_a,r_a)$ . Denote the tangent points $ F$ , $ F_a$

of  $ w$ , $ w_a$ respectively with the sideline $ AB$ . Prove that $ IF_a$ , $ I_aF$ , $ BC$ are concurrently if and only if $ B = 90^{\circ}$ .[/color][/quote]

[color=darkblue][b][u]Method 1.[/u][/b] Denote $ D\in II_a\cap BC$ , $ S_1\in BC\cap IF_a$ and $ S_2\in BC\cap I_aF$ . Apply the [b]Menelaus' theorem[/b] to :

$ \blacktriangleright\ \overline {IS_1F_a}/\triangle ABD\ : \& \frac {F_aB}{F_aA}\cdot \frac {IA}{ID}\cdot\frac {S_1D}{S_1B} = 1$ $ \implies \& \frac {p - c}{p}\cdot\frac {b + c}{a}\cdot\frac {S_1D}{S_1B} = 1$ $ \implies\ \boxed {\frac {S_1B}{S_1D} = \frac {(b + c)(p - c)}{ap}}$

$ \blacktriangleright\ \overline {I_aS_2F}/\triangle ABD\ : \& \frac {FB}{FA}\cdot \frac {I_aA}{I_aD}\cdot\frac {S_2D}{S_2B} = 1$ $ \implies \& \frac {p - b}{p - a}\cdot\frac {h_a + r_a}{r_a}\cdot\frac {S_2D}{S_2B} = 1$ $ \implies\ \boxed {\frac {S_2B}{S_2D} = \frac {(b + c)(p - b)}{a(p - a)}}$

because $ \frac {h_a+r_a}{r_a}=\frac {ah_a+ar_a}{ar_a}=$ $ \frac {2(p-a)r_a+ar_a}{ar_a}=\frac {2(p-a)+a}{a}=\frac {(b+c-a)+a}{a}=\frac {b+c}{a}$ .

Therefore $ IF_a\cap I_aF\cap BC\ne\emptyset$ $ \Longleftrightarrow$ $ \frac {S_1B}{S_1D} = \frac {S_2B}{S_2D}$ $ \Longleftrightarrow$ $ \frac {p - c}{p} = \frac {p - b}{p - a}$ $ \Longleftrightarrow$ $ b^2 = a^2 + c^2$ $ \Longleftrightarrow$ $ B = 90^{\circ}$ .

[b][u]Method 2.[/u][/b] Denote : $ S\in IF_a\cap I_aF$ ; $ D\in BC\cap II_a$ ; $ X\in AB$ , $ Y\in AD$ so that $ S\in XY$ and $ XY\perp AB$ ;

$ Z\in AB$ so that $ DZ\perp AB$ . Prove easily that $ BD = \frac {ac}{b + c}$ and observe that $ DZ = BD\cdot\sin B$ . In the trapezoid

$ IFF_aI_a$ is well-known the relation $ \frac {2}{XY} =$ $ \frac {1}{IF} + \frac {1}{I_aF_a}$ . Thus, $ \frac {2}{XY} = \frac 1r + \frac {1}{r_a}$ . But $ S = pr = (p - a)r_a$ .

Thus, $ \frac {2}{XY} =$ $ \frac pS + \frac {p - a}{S} =$ $ \frac {b + c}{S}$ $ \implies$ $ XY =$ $ \frac {2S}{b + c} =$ $ \frac {ac}{b + c}\cdot\sin B$ $ \implies$ $ XY = BD\sin B$ $ \implies$

$ XY = DZ$ . In conclusion, $ S\in BC$ $ \Longleftrightarrow$ $ XY\equiv BD$ $ \Longleftrightarrow$ $ BD = DZ$ $ \Longleftrightarrow$  $ B = 90^{\circ}$ .[/color]

Let be the points $ P\equiv QR'\cap Q'R$ and $ S\equiv I_{b}Q'\cap I_{c}R',$ where $ I_{b},\ I_{c},$ are the $ B,\ C$-excenters respectively, of the given rectangular triangle $ \bigtriangleup ABC.$

Because of $ B'Q \equal{} B'Q'$ and $ C'R \equal{} C'R',$ where $ B',\ C',$ are the midpoints of $ AB,\ AC$ respectively, we conclude that the points $ I,\ A',\ S,$ are collinear $ ($ easy to prove by [b][size=100]Thales theorem[/size][/b], from $ QI\parallel B'A'\parallel Q'S$ and $ RI\parallel C'A'\parallel R'S$ $ ).$

Because of now, $ IQ\parallel SQ'$ and $ IR\parallel SR',$ in the configuration of the parallelogram ( rectangular here ) $ AR'SQ',$ based on the [b][size=100]lemma[/size][/b] already has been proved [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=107997]here[/url] ( see the schema [b][size=100]t=107997(e)[/size][/b] in the post #8 ), we conclude that the points $ P,\ I,\ S,$ are collinear.

Hence, the line segments $ QR',\ Q'R,\ IA',$ are concurrent at one point and the proof is completed.

Kostas Vittas.

In any triangle lines through tangency points of B-excircle with AC and BC    and C-excircle with AB and BC

   intersect on A-altitude along with the line through midpoint of BC and incenter. This concurrency point is at a distance  r

   from A.    When A=90 the tangency points of incircle are in-line with the tangency points transversals of

   the excircles.  To identify these lines using incircle tangency points only confuses the issue.

    M.T.


Does anyone have any tips or advice on how to best weigh the cost vs. quality of education for colleges?

On a personal level, I have been accepted to the honors programs at both Ohio State and University of Michigan.  On the one hand, the cost of Ohio State and Michigan are roughly $ \$$21,000 and $ \$$47,000 per year respectively (I am in-state for Ohio).  On the other hand, Michigan's honors math sequence has a fantastic reputation - certainly a better reputation than Ohio State (although Ohio State's program appears [to me] to be pretty solid as well).  Is it really worth the extra $ \$$26,000 per year to attend the school with a better math program?

This is probably the most difficult question faced by students deciding on a college. I had to face the same decision when deciding on a college many years ago. The answer, though, depends on a lot of personal variables - like what is the nature of your family's personal finances? do you have other siblings who are currently in, or will be going to college in the near future? do you qualify for any financial aid? if so, how much of that money is in student loans? how sure are you that mathematics will be your future major (or that you will be successful in the honors sequence/program at Michigan)? what are your long term career plans? and how important is money to you in your future?

What makes it more difficult, is that the answers to some of these questions you may only be able to speculate on at this point. I can tell you what I would do -- but ultimately I think it's a personal decision that only you and your family can make by weighing all of the personal variables.

An interesting thing to do (if you can find the information) would be to find out what a sample of past (honors) math majors from each school have gone on to do in their careers.

At this point, you've only mentioned the reputation of the school and its math program.  What you should also do is look at the specifics of each program and school, and ask how they would impact your life.

Can you talk with current students about the quality of the teaching and availability of outside help (office hours, study groups, and so on)?  What is the academic advising like?  Can you gather syllabi for the courses of interest to you, and use those to evaluate which courses are more rigorous?  What do the degree requirements look like, and how much freedom will you have to pursue your own interests?  Do you write a thesis?  Can you do research, or find people to supervise you?

Does either program offer extra perks, like travel/research money, social events, early registration, or greater access to the people who make things work?

There are also non-academic considerations.  Is the neighborhood around the university, and the neighborhood where you'll be living, safe?  Can you get quick and reliable access to medical care and counseling services?  These may or may not have a strong impact on your quality of life.  Are the students there people you see yourself making friends with?

Would going to the less expensive school give you the financial resources to do something cool, like study abroad or working on a project that interests you?

Doing more research into these sorts of considerations would give you a better idea into what the extra $ \$$26,000 is actually buying you.  It could be a little, or it could be something really significant.


Pump L, working alone, can fill a swimming pool in 4 hours. Either one of pumps M or N working alone can fill the same pool in 2 hours. If all three pumps, working at these rates, are used at the same time to fill the pool, what fraction of the pool is filled by pump L?

[hide]
L, M, and N pump at a ratio of 1/4:1/2:1/2 = 1:2:2, so L pumps 1/5 of the water.
[/hide]

[quote="Brut3Forc3"]
L, M, and N pump at a ratio of 1/4:1/2:1/2 = 1:2:2, so L pumps 1/5 of the water.
[/quote]

Nice solution...Haven't seen proportions used that way.

In terms of rates, pump L pumps at rate Q. Pumps M and N pump at rate 2Q.
Total pumping rate 5Q  -->  1Q/5Q = 1/5...


[quote="stvs_f"]let $ a_i and b_i(n \geq i \geq 1) \in [0,1]$
prove:$ \exists i,j: \frac {2}{n \minus{} 1} \geq a_ib_j \minus{} a_jb_i \geq 0$
$ (a_i and b_i \in R)$[/quote]

$ i\equal{}j\equal{}1$

This is a beautiful problem. Consider $ A_i(a_i,b_i)$ as a point in the plane , by the condition all points lie in square ABCD which A(0,0),B(0,1),C(1,1),D(1,0) . We can arrange  $ A_i$ such that $ \angle{A_1OA_2}\leq \angle{A_1OA_3}...\leq \angle{A_1OA_n}$ . Sum area of all triangle $ A_iOA_{i\plus{}1}$ is not greater than area of the square ABCD ,because  there are exactly n-1 triangle so exist at least one triangle that its area is not greater than 1/n-1 . We known that area of this triangle is $ \frac{1}{2}|a_ib_{i\plus{}1}\minus{}a_{i\plus{}1}b_i$ , that gives us the solution .


real analysis
linear algebra
matrix
function
abstract algebra
algebra
domain

Hi,

I have just learnt the convolution theorem:

F(f*g)=F(f)*F(g).

So in order to do the fast convolution, I did like this example, in MATLAB:

f= [1 2; 3 4] (2-dimension matrix)

g= [5 6; 7 8]

Get

fourier_f= fft2(f)
fourier_g=fft2(g)

w= fourier_f .* fourier_g (point-wise product)

result= ifft2(w) (inverse fourier transform).

The result i got is: r= [70 68; 62 60]

But the correct answer is: [60 40; 52 32] (use c= conv2(a,b,'same'))

So what's wrong with this calculation, could anyone please help me? :(

I did the convolution by hand and got $\begin{pmatrix}70 & 68 \\ 62 & 60\end{pmatrix}$. Maybe your "correct" answer is based on another definition of convolution?

Hi mlok,

Can I know how did you come out with that result?

I use the MATLAB function: conv2(a,b,'same'), base on the definition here: [url]http://www.mathworks.com/access/helpdesk/help/techdoc/ref/index.html?/access/helpdesk/help/techdoc/ref/conv2.html&[/url]. And when I did by hand I also got [60 40; 52 32].

The big matrix big_matrix  is:

     5    16    12
    22    60    40
    21    52    32

big_matrix(1,1) is calculated as:

big_matrix(1,1)= f(0,0)*g(1,1) + f(0,1)*g(1,0) + f(1,0)*g(0,1) + f(1,1)*g(0,0)
= 1*8 + 2*7 + 3*6 + 4*5 = 60.

Thank you. :)

First of all, let's number the rows and columns of a matrix starting with zero (unlike Matlab), since this simplifies formulas a bit.

conv2 is "convolution without wrap-around", which in general increases the size of a matrix. The whole $3\times 3$ matrix that you got is the result of conv2-style convolution of two $2\times 2$ matrices. The conv2-convolution of $A$ with $B$ has entries $c_{pq}=\sum a_{ij}b_{kl}$, where the sum is taken over all indices such that $i+k=p$ and $j+l=q$. (Recall that the indices start with $0$.)

Fourier transform can be applied to functions defined on an [url=http://en.wikipedia.org/wiki/Abelian_group]abelian group[/url]. In order to apply F.T. to an $m\times n$ matrix $A$, one considers $A$ as a function defined on $\mathbb Z_{m}\oplus \mathbb Z_{n}$ that takes value $a_{ij}$ on an element $(i,j)\in \mathbb Z_{m}\oplus \mathbb Z_{n}$. The cyclic groups $\mathbb Z_{m}$ and $\mathbb Z_{n}$ are equipped with [url=http://en.wikipedia.org/wiki/Modular_arithmetic]modular arithmetics[/url]. This is why the convolution you got from fft2 is "convolution with wrap-around". It only makes sense when both $A$ and $B$ are of the same size, say $m\times n$. Its definition is $c_{pq}=\sum a_{ij}b_{kl}$, where the sum is taken over all indices such that $i+k\equiv p \mod m$ and $j+l\equiv q \mod n$. The result is again an $m\times n$ matrix.

hi,

I found that to factorize a signal, we may need to find frequencies from -pi to pi only (or some interval [k,k+2pi], k in Z).

So are the 2-radix, 4-radix,.. FFTs (transform time domain[t0..t0+N-1] to [w0..w0+N-1]) just for Convolution calculating? Is there any where people apply it?

Could anybody tell me how to transform fast from time domain t in (1..N-1) to an arbitrary frequency domain, or at least to frequency domain Dw=(1*delta_w,..,X*delta_w). Where 1*X is the resolution X is just some positive integer (rather small in compared to N (X/N=1/10))?

Thanks for reading :-)


It's only a little over fifteen minutes away, and the turn up has been dismal till now.
As and when you appear here, say 'I'm present', so that you are not to blame later for not breaking the record. :P

Estoy aqui.

I'm here.

Looks like we're not going to break it then  :( .

this is FAILING

ONLY 157 PEOPLE (including 53 guests...)

It was 138 about a minute ago, so it's growing.  Even if it's hopeless, we have to stick it out to the end!

Hey cant we open up as a user in one tab and then open another window and just enter the forum as a guest, or would that be bad?
 :stink:
P.S. This reqally is not going too well.
P.S.S.  Still here

Okay, these are the people here at exactly 6:33 pm Pacific Time:

[hide="People"]33286, 501316, 7h3.D3m0n.117, abacus50, ac-king, agentpenguin, alanchou, Albanian Eagle, alkjash, Altheman, Anaxerzia, anise_117, Arvind_sn, asdfguy, auroraak, b-flat, balto, bball43, bboyggirl1345, Begoner, binomial_4eva, Black, BOGTRO, brademonster, bubka, cbk, chess64, Code1011011, cognos599, D3m0n Shad0w, deliky6328, Double.Integral, Dr Sonnhard Graubner, ejr322, element139, Eli, emilgouliev, Erken, E^(pi*i)=-1, fcatrina, fedja, funmath, gangjute, gauss202, genehsu910, GoogleBot, Hamster1800, helagha, humDdum, hxvlnmt321, Ihatepie, i_like_pie, jack4585, janecw, JotThisDown, kaz74, kev015, KING$ C$MONEY, kookamunga, krsattack, LadyKn1ght, LHM, lifeisacircle, LittleBecca, lotsofmath, maru, Math Geek, mathgirl_fractals, mathluvr, mathpassion101, Max300, michaelwong256, mingcn007, minicon, minsoens, miyomiyo, msecco, MSNBot, MustItAlwaysBeMe, Nawahd Werdna, not_trig, olorin, Onewholovesfrogs4795, panjia123, Patterns_34, peanutman, permutations, police_cap, qwang, randomdragoon, redcomet46, redvsblueownz, rem, samath, shaggy75, shentang, slettjohnson, solafidefarms, sos440, Swimmergirl_1998, t0rajir0u, Teki-Teki, Temperal, The QuattoMaster 6000, THEMATHONE, tjhance, valeriummaximum, WNick, worthawholebean, xpmath, Yahoo! Slurp, ytrewq, ZhangPeijin[/hide]

WE CAN STILL BREAK THE RECORD!!!!!!!!!

In spite of the jump increase from 120 to 134, I doubt if there's any chance left, so bye people!
Nice taking part in this - let's hope we have better luck next time. :)

Nope, no where close to 593 or whatever it is.
Well, at least I'm on the list!  :P

You didn't break the record...
You didn't break the record...
You didn't break the record...
You didn't break the record...
You didn't break the record...
You didn't break the record...
*chanting continues*

I had to do homework, so I didn't really give up.  But it was fun.  Next time we'll be less ambitious and go for 200.

Haha we got $ \pi$ users on Pi Day (that's in radians, by the way). Well maybe we can beat 184 next time =).

Lol I'm interested as to how the 593 users broke it the previous record. I don't think it was timed...

Eh not to be cynical(well, I am a cynical person, but...) but I had 0% confidence that this would work. There simply weren't enough people who would actually care enough to remember.

I thought so too. Everyone I told wasn't even on yesterday.

And I forgot, I thought it was today, but I still was on, with my other comps.

Wow, maybe we should make a list of users BEFORE the thing happens, so we can know if it will work before trying it. Tedious, but it would pretty much assure us success (assuming we got enough people)

Wow. AoPS is awesome, but based on people online at a time, not really. For example, smashboards has about 1300 people on at any given time. Their record is 4000 something (actually, it was really recent. it was the day after brawl's US release).

Actually, we're depending on a large number of guests to boost the count up. We'll only have maybe 50 signed users (and maybe another 2-3 dozen of random ones) so the guests are a large factor.

How about the ~10k registered people?
Most of them are dead, or only come on AoPS for classes anyways...
Meh, you guys ever look into the user database?
There's so many people with 0 posts.

I don't think many people actually go on AoPS anymore. Its kind of like myspace. Everyone used to go to it and then now myspace is in the past. I wonder if that will ever happen AoPS... :|


vector
linear algebra
algebra unsolved
algebra

Let n be a positive integer, P be a set of n primes and M a set of at least n+1 natural numbers, each of which is divisible by no primes other than those belonging to P.  Prove that there is a nonempty subset of M, the produt of whose elements is a square integer.

Could someone explain the basic methods to solving a problem like this?

There are no "basic methods" which can be applied universally.

For instance, I first thought "Pigeon Hole" should give this, but actually, we can do this more easily using Linear Algebra:

Let the primes be $p_{1}, p_{2}, ..., p_{n}$.

For each natural number (among the n+1 numbers) $a_{k}$ associate a 0-1 vector $v_{k}= (v_{k1}, v_{k2}, ..., v_{kn})$. $v_{kj}$ is 0 iff the highest power of $p_{j}$ which divides $a_{k}$ is even.

We can assume none of these vectors is $(0,...,0)$ as, we are done otherwise.

Now we have n+1 non-zero vectors over the binary field (i.e working modulo 2).

Since there are n+1 n-dimensional vectors, they are linearly dependent.

i.e there is some $r > 1$ such that

$v_{i_{1}}+v_{i_{2}}+..+v_{i_{r}}= 0$

This shows that the product $\prod_{m=1}^{r}a_{i_{m}}$ is a perfect square.


A, B, C, and D are four different weights.  When they are placed on a balance scale, the following observations are made:
A and B together weigh the same as C and D together
A and C together weigh more than B and D together
C is lighter than D

Arranged in order from heaviest to lightest, the weights are:

A.   A, D, C, B
B.   D, A, C, B
C.   D, A, B, C
D.   A, B, D, C
E.    A, D, B,C

[hide="Solution"]We have that $ D$ is heavier than $ C$. Also, since swapping $ C$ and $ B$ makes the $ C$ side heavier, that implies $ C$ is heavier than $ B$. Now, if $ D$ is heavier than $ B$, and $ C$ is heavier than $ B$, then $ A$ must be heavier than $ D$ in order for the $ A$ and $ B$ together to weigh the same as $ C$ and $ D$. The correct answer: $ \boxed{A}$

[/hide]


linear algebra
matrix
induction
integration
calculus
real analysis
real analysis unsolved

Le $ K \subset   R^n$, $ K$ is bounded and $ vol(K)$ exists. $ A$ is $ nxn$ matrix (and you can assume that $ det A\not\equal{}0$). Prove that $ vol(KA)$ exists and
\[ vol(AK)\equal{}|det A|vol(K),\]

Is there a short solution for this problem? I believe that this should be a well-known theorem but I dont know its name.

maybe I will prove some particular case  or maybe I eill post full solution.
it looks more Algebraic problem then analysis.
your figure $ K$ we can denote as $ P$ and in basis $ a_{1},...,a_{n}$ in $ R_{n}$ we can rewrite it as
$ P(a_{1},a_{2},...,a_{n})$ and say that $ P(a_{1},a_{2},...,a_{n} \equal{} \{\alpha_{1}a_{1} \plus{} \alpha_{2}a_{2} \plus{} ... \plus{} \alpha_{n}a_{n}\}$ $ 0\geq \alpha_{i}<1$
let's donet $ orta_{n} \equal{} h$  in $ < a_{1},...,a_{n \minus{} 1} >$
and define $ vol$ as
$ Vol P(a_{1},...,a_{n}): \equal{} Vol P(a_{1},...,a_{n \minus{} 1})\parallel{}h\parallel{}$
$ Vol a_{1}: \equal{} \parallel{}h\parallel{}$
$ Vol 0: \equal{} 1$

observation 1:
$ (Vol P(a_{1},a_{2},...,a_{n})^{2} \equal{} |G(a_{1}...a_{n})|$
where $ G$ is  Gramian matrix.
we know that $ \parallel{}h\parallel{} \equal{} \parallel{}ort a_{n}\parallel{}$  in $ < a_{1},...,a_{n \minus{} 1}$ hence
easly seen that $ \parallel{}h\parallel{}^{2} \equal{} \frac {|G(a_{1}...a_{n})|}{|G(a_{1}...a_{n \minus{} 1})|}$
suppose $ e_{1},...,e_{n}$ is basis of $ R^{n}$ such as
$ < e_{i},e_{j}>$ $ \equal{} \delta_{i,j}$
and $ \parallel{}e_{i}\parallel{} \equal{} 1$
hence
by induction
$ Vol P(a_{1},...,a_{n})^{2}: \equal{} Vol P(a_{1},...,a_{n \minus{} 1})^{2}\parallel{}ort a_{n}\parallel{}^{2} \equal{} Vol P(a_{1},...,a_{n \minus{} 1})\frac {|G(a_{1}...a_{n})|}{|G(a_{1}...a_{n \minus{} 1})|} \equal{} |G(a_{1}...a_{n})|$ done

observation 2:
let $ (b_{1}...b_{n}) \equal{} (a_{1},...,a_{n})A$
then $ \equal{} Vol P(b_{1}...b_{n}): \equal{} Vol KA \equal{} |A|Vol (K)$
really
$ G(b_{1}...b_{n}) \equal{} A^{T}G(a_{1}...a_{n})A$
hence
$ |G(b_{1}...b_{n}) |\equal{} |A|^{2}|G(a_{1}...a_{n})| \equal{} |A|^{2}Vol (K) ^{2}$by observation 1.
hence
$ Vol AK \equal{} |A|Vol (K)$

Another idea is to consider the special case where $ A$ is an elementary matrix.

By the way, what is your definition of volume? If it is $ \int_K 1$, I suppose change of variables formula is an overkill, but you can still use partitions that define the integral, and consider what happens after $ A$ acts on it. (A is elementary matrix)

[quote="Soarer"]

By the way, what is your definition of volume? If it is $ \int_K 1$, [/quote]
No,
At first I define figure so called paralelepiped (I don't  know how it will be in English  :(  )
paralelepiped is :
$ P(a_{1},a_{2},...,a_{n}) \equal{} \{\alpha_{1}a_{1} \plus{} \alpha_{2}a_{2} \plus{} ... \plus{} \alpha_{n}a_{n}\}$ $ 0\leq \alpha_{i} < 1$
and vol define by induction  as :
$ Vol P(a_{1},...,a_{n}): \equal{} Vol P(a_{1},...,a_{n \minus{} 1})\parallel{}h\parallel{}$ (in this case $ h$ is $ \equal{} ort a_{n}$ in $ < a_{1},...,a_{n \minus{} 1} >$)
$ Vol a_{1}: \equal{} \parallel{}h\parallel{}$
$ Vol 0: \equal{} 1$
and after I prove your statement for  paralelepiped.
as I note I consider particular case for figure paralelepiped, but it's not hard for  generalization  for all $ K$ from $ R^{n}$
you need to divide your figure in small paralelepipeds.  :)


[quote="Klaas"]a,b are given natural numbers such that number
(((a^2 - 1)/(b+1) + (b^2 - 1)/(a+1)) is natural.
Prove that:
(a^2 - 1)/(b+1) and (b^2 - 1)/(a+1) are also natural numbers.

 :?:[/quote]

Let $ c\equal{}\gcd(a\plus{}1,b\plus{}1)$ and so $ a\plus{}1\equal{}uc$ and $ b\plus{}1\equal{}vc$ with $ \gcd(u,v)\equal{}1$

We have $ N\equal{}\frac{a^2\minus{}1}{b\plus{}1}\plus{}\frac{b^2\minus{}1}{a\plus{}1}$ $ \equal{}\frac{(uc\minus{}2)u}{v}\plus{}\frac{(vc\minus{}2)v}{u}$ $ \equal{}\frac{(uc\minus{}2)u^2\plus{}(vc\minus{}2)v^2}{uv}$

Since $ N$ is an integer, $ v|(uc\minus{}2)u^2\plus{}(vc\minus{}2)v^2$ and so $ v|(uc\minus{}2)u^2$ and so $ v|(uc\minus{}2)$ since $ \gcd(u,v)\equal{}1$ and so $ \frac{a^2\minus{}1}{b\plus{}1}\equal{}\frac{(uc\minus{}2)u}{v}$ is an integer.

Since $ N$ is an integer, $ u|(uc\minus{}2)u^2\plus{}(vc\minus{}2)v^2$ and so $ u|(vc\minus{}2)v^2$ and so $ u|(vc\minus{}2)$ since $ \gcd(u,v)\equal{}1$ and so $ \frac{b^2\minus{}1}{a\plus{}1}\equal{}\frac{(vc\minus{}2)v}{u}$ is an integer.

Q.E.D.


Let $ a_{i,j}$,    $ 1 \le i \le n , 1 \le j \le p$  a set of np reals

compare:

$ A\equal{} \min_{1 \le i \le n} \max_{1 \le j \le p} a_{i,j}$

$ B\equal{} \max_{1 \le j \le p} \min_{1 \le i \le n} a_{i,j}$

just for more clarification:

$ A\equal{} \min \{ \max_{1 \le j \le p} a_{1,j} , \max_{1 \le j \le p} a_{2,j} , ... , \max_{1 \le j \le p} a_{n,j}  \}$

$ B\equal{} \max \{\min_{1 \le i \le n} a_{i,1},\min_{1 \le i \le n} a_{i,2},...,\min_{1 \le i \le n} a_{i,p}   \}$


How do you do that problem. I used guess and check.....[hide]AND I STILL GOT IT WRONG[/hide]

From mathcounts.org

Eli throws five darts at a circular target, and each one lands within one of the four regions. The point-value of a dart landing in each region is indicated. What is the least score greater than five points that is not possible when the point-values of the five darts are added together?
[hide]I remember having tons of time left on this question so I just did it the old fashioned way, figure out a solution for every method.
1-5 Problem said greater than 5
6-10 Combination of 1s and 2s
11-18 You have 2, 2, 2, 2, 2 for 10. Add 2 to a 2 [to make 4] and subtract 1 from another 2 [to make 1], repeat when necessary.
19 You have 4, 4, 4, 4, 2, from 18. Add 2 to a 4 [to make 6] and subtract 1 from a 2 [to make 1].
20 Add the 1 back onto the 1 to make 2 and add 2 to another 6. Repeat these steps for 19 and 20 until you get up to 26.
27 Impossible as you need a 5 or a 3. You need at least 4 6s to use a 3 and 3 6s to use a 5 both of which don't exist.
[/hide]

Yea.....but you have to think and write, hard and a lot, respectively. And.....well I am thinking that there should be a faster way to get it. I remember Kyyuanmathcuonts had sum way to solve it....

I posted a quick solution [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=449079#p449079]here[/url].  Because it really belongs in this thread, I will copy the solution below.

[hide="Solution"]
If you choose the five darts to be 2, 4, or 6, then you can get every even score from 10 to 30. If you choose one dart to be 1 and the other four darts to be 2, 4, or 6, then you can get every odd score from 9 to 25. It's easy to get the low scores of 6, 7, and 8. Thus the first score we can't get is 27.[/hide]

Wow, that is a good way to solve it :D I guessed and checked, working my way up the numbers until I got to 27.  Another kid used guess and check, but somehow got 47 :? Ravi, can you post that again in a little more detail, like how you instantly know how to eliminate evens 8-30 with 2 4 6?  Thanks

Yes, and the same when you add the one.  Now I really understand your solution :D Thanks!

[quote="Ravi B"]When I first saw the problem, I also worked my way brute-force from 6 to 27.  Only later did I see the quicker solution.

Let's suppose we have five darts chosen from 2, 4, or 6.  The smallest score we can get is 10.  By changing any 2 or 4 to a 4 or 6, respectively, we get a score that is 2 higher.  We only get stuck when we have all 6's, which is a score of 30.  So we can get any even score from 10 to 30.  Did that make sense?[/quote]

I still don't understand about the part of changing the 2 or 4 to a 4 or 6. How can you be sure that you can get every even score from 10 to 30?

So the lowest score you can have is $2(5) = 10$. The
highest is $6(5) = 30$. If you change a 6 to a 4, the score
will lower by a 2. Thus it is possible to get 28, 26, 24, 22, 20.
By the same process, you can achieve every even score from 10 to 20.
Thus you can achieve every even score from 10 to 30.

...

Can someone please actually post the problem? "The point-value of a dart landing in each region is indicated" does not help me with understanding what the problem is at all.

Here, with the point values:
[quote="Ignite168"]From mathcounts.org

Eli throws five darts at a circular target, and each one lands within one of the four regions. (1 point, 2 points, 4 points, or 6 points) The point-value of a dart landing in each region is indicated. What is the least score greater than five points that is not possible when the point-values of the five darts are added together?[/quote]


Let $a,b,c,d>0$ and $a+b+c+d=4$.
Prove that
$\frac{ab}{4-d}+\frac{bc}{4-a}+\frac{cd}{4-b}+\frac{da}{4-c}\leq \frac{4}{3}$.

Sorry to say that it's also my inequality. I formulate it in homogenous form. Prove that for nonegative real numbers $a,b,c,d$
1) $\frac{ab}{a+b+c}+\frac{bc}{b+c+d}+\frac{cd}{c+d+a}+\frac{da}{d+a+b}\leq\frac{a+b+c+d}3.$
2) $\frac{ab}{a+b+2c}+\frac{bc}{b+c+2d}+\frac{cd}{c+d+2a}+\frac{da}{d+a+2b}\leq\frac{a+b+c+d}4.$

I used four variables SOS to solve 1). Could you post your ways to solve 1)?

[hide="my interesting solution to part 1) without SOS"] \[\frac{a+b+c+d}3\geq\frac{ab}{a+b+c}+\frac{bc}{b+c+d}+\frac{cd}{c+d+a}+\frac{da}{d+a+b}=\sum_{cyc}^{a,b,c,d}\frac{ab}{a+b+c}\]  \[\large =\sum_{cyc}^{a,b,c,d}\frac{1}{\frac{1}{a}+\frac{1}{b}+\frac{c}{ab}}\] The last sum is nice in terms of AM-HM style: \[\frac{a+b+\frac{ab}{c}}{3}\geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{c}{ab}}\implies \sum_{cyc}^{a,b,c,d}\frac{a+b+\frac{ab}{c}}{9}\geq \sum_{cyc}^{a,b,c,d}\frac{1}{\frac{1}{a}+\frac{1}{b}+\frac{c}{ab}}\] Now consider rearrangement, starting from the other end: \[\large \sum_{cyc}^{a,b,c,d}a\geq \sum_{cyc}^{a,b,c,d}\frac{ab}{c}\implies \sum_{cyc}^{a,b,c,d}3a\geq \sum_{cyc}^{a,b,c,d}\frac{ab}{c}+a+b\] And now we are done: \[\frac{\sum_{cyc}^{a,b,c,d}a}{3}\geq \frac{\sum_{cyc}^{a,b,c,d}\frac{ab}{c}+a+b}{9}\geq \sum_{cyc}^{a,b,c,d}\frac{1}{\frac{1}{a}+\frac{1}{b}+\frac{c}{ab}}\]  \[\implies \frac{a+b+c+d}3\geq\frac{ab}{a+b+c}+\frac{bc}{b+c+d}+\frac{cd}{c+d+a}+\frac{da}{d+a+b}\] $QED_{OREGON}$[/hide]

[quote="me@home"]
Now consider rearrangement, starting from the other end: \[\large \sum_{cyc}^{a,b,c,d}a\geq \sum_{cyc}^{a,b,c,d}\frac{ab}{c}\implies \sum_{cyc}^{a,b,c,d}3a\geq \sum_{cyc}^{a,b,c,d}\frac{ab}{c}+a+b\] [/quote]

I don't think so.. Try $c \rightarrow 0$.

[quote="me@home"]
Now consider rearrangement, starting from the other end: \[\large \sum_{cyc}^{a,b,c,d}a\geq \sum_{cyc}^{a,b,c,d}\frac{ab}{c}\] [/quote]
How to apply rearrangement to get it?
I don't think it is right :(

Wow I am so sorry :(
I just realized that the exact opposite is true:
$\large \sum_{cyc}^{a,b,c,d}a\leq \sum_{cyc}^{a,b,c,d}\frac{ab}{c}\implies \sum_{cyc}^{a,b,c,d}3a\leq \sum_{cyc}^{a,b,c,d}\frac{ab}{c}+a+b$
Where I am using the sequences $\{ ab, bc, cd, da\}$ and $\{ \frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{d}\}$
This inequality is a really strict boud... by using just the AM-HM inequality on the less than side of the given we get a result greater than the greater than side...
$\implies \frac{\sum_{cyc}^{a,b,c,d}\frac{ab}{c}+a+b}{9}\geq \frac{a+b+c+d}3\geq^{?}\frac{ab}{a+b+c}+\frac{bc}{b+c+d}+\frac{cd}{c+d+a}+\frac{da}{d+a+b}$

[quote="Parkdoosung"]Let $ a,b,c,d > 0$ and $ a \plus{} b \plus{} c \plus{} d \equal{} 4$.
Prove that
$ \frac {ab}{4 \minus{} d} \plus{} \frac {bc}{4 \minus{} a} \plus{} \frac {cd}{4 \minus{} b} \plus{} \frac {da}{4 \minus{} c}\leq \frac {4}{3}$.[/quote]
See here: http://can-hang2007.blogspot.com/2010/01/inequality-22-v-q-b-can.html


If $a,b,c$ are positive numbers prove that :
$\sum \frac{a(3a-b)}{c(a+b)} \leq \frac{a^3+b^3+c^3}{abc}$

Okay, I won't say it is easy this time ;) , but still it is far from being difficult:

We have $a\left(a+b\right)-b\left(3a-b\right)=\left(a-b\right)^2$; this is clearly $\geq 0$. Thus, $a\left(a+b\right)\geq b\left(3a-b\right)$. Consequently, $\frac{a}{b}\geq\frac{3a-b}{a+b}$, and $\frac{3a-b}{a+b}\leq\frac{a}{b}$. Multiply this with $\frac{a}{c}$ to get $\frac{a}{c}\cdot\frac{3a-b}{a+b}\leq\frac{a}{c}\cdot\frac{a}{b}$. This rewrites as

$\frac{a\left(3a-b\right)}{c\left(a+b\right)}\leq\frac{a^3}{abc}$.

Similarly, we can find

$\frac{b\left(3b-c\right)}{a\left(b+c\right)}\leq\frac{b^3}{abc}$;
$\frac{c\left(3c-a\right)}{b\left(c+a\right)}\leq\frac{c^3}{abc}$.

Summing up these three inequalities, we get

$\frac{a\left(3a-b\right)}{c\left(a+b\right)}+\frac{b\left(3b-c\right)}{a\left(b+c\right)}+\frac{c\left(3c-a\right)}{b\left(c+a\right)}\leq\frac{a^3}{abc}+\frac{b^3}{abc}+\frac{c^3}{abc}=\frac{a^3+b^3+c^3}{abc}$.

QED.

  Darij


Rules:
1.Everyone starts with 25 HP, unless I state otherwise. Anyone can join at any point.
2.Each situation is short, stupid, pointless and is more then likely not related to any other situation.
3.Once you're dead, you can start again, although it'll say how many times you've died.
4.Doing something that's one of my choices will more then likely result in instant death. In other words, go for it.

Event #1:

You're walking down the road when suddenly OMG! a Gundam AKA extremely big robot thing  that is very shiny (yes I said shiny) drops out of nowhere. What do you do?

A.Steal it
B.Tell the police
C.Shrug and keep walking

I can't only tell you anymore info. It makes it more intresting :)

blow it into oblivion w/ my portable nuke and make a quick escape in my oversized ATV

open up a tear into the universe and send it hurtling through a wormhole.

Scream with delight! Because now I am not alone!!!

wow so whats the point of this it seems very meaningless

Math92 starts with 30 HP for being the first poster. From now on, the first poster after me gets +5 HP.

A.The pilot kicks you out. You get back in and punch the pilot. -5 HP. Add 1 MS to your inventory.

B.The police attack the MS and get killed. A cookie falls from the sky. You take it, and run away. Add 1 cookie to your inventory.

C.The MS steps on you. -10 HP

D.The pilot is Randomus McGruntus. He gives you a donut. +1 donut to your inventory.

E: (Exploding nukes/wormholes/ATVs) INSTANT DEATH!, but add cookie to your inventory when you respawn.

Event #2:
A Haro comes along and says "Haro Genki". What do you do?

A.Punch the Haro
B.Hug the Haro
C.Reply with "Haro Genki".

Items:
MS- Random Grunt MS
Cookie- Yummy
Donut- Summons a Kirby

[hide]B hug him and feed him a cookie[/hide]

^Hatake Kakashi could beat up him up :p. Just to let you know you don't have a cookie. You chose A "steal it".

Update!: Items do not do anything unless I say so. So unless you are in a situation where I say items are allowed, DON'T USE YOUR ITEM. That is all.

I will reply the same... hopefully it doesn't mean "hey strange ugly being" :P

[hide]say haro genki maybe it means hello....[/hide]

Take out my portable translator and figure out what it's saying...if it's friendly take out my cookie and hug it...if it's unfriendly then use my magical powers to melt it.

say the same thing
unless its a swear word it shouldnt hurt :D


I came across this question in a local competition (Hwa Chong Challenge, Singapore) sample paper:

Calculate the value of cot 15° cot 25° cot 35° cot 85° .

The answer is [hide]1[/hide].  I know of a solution using the trigonometric identity
tan (a+b) = (tan a + tan b)/(1 - tan a * tan b), but it is extremely long (the full solution is attached).  Moreover this question is in the buzzer round so it is supposed to take less than 1.5 minutes, and the answer is so simple that there must be shorter way, but I have no idea what.....

Does anyone know of a faster method to solve this?

[quote="Waynerius"] [color=darkred]Prove that[/color] $ \boxed {\ \tan 15\cdot\tan 25\cdot\tan 35\cdot\tan 85\ \equal{} \ 1\ }$  .[/quote]
[color=darkblue][b][u]Proof.[/u][/b][/color]

$ \left\|\begin{array}{c} 4\cdot\sin 25\cdot\sin 35\cdot\sin 85 \equal{} 2\cdot\cos 5\cdot\left(\cos 10 \minus{} \frac 12\right) \equal{} \cos 5\cdot(2\cdot\cos 10 \minus{} 1) \equal{} \cos 15 \plus{} \cos 5 \minus{} \cos 5 \equal{} \cos 15 \\
 \\
4\cdot\cos 25\cdot\cos 35\cdot\cos 85 \equal{} 2\cdot\sin 5\cdot\left(\frac 12 \plus{} \cos 10\right) \equal{} \sin 5\cdot (1 \plus{} 2\cdot\cos 10) \equal{} \sin 5 \plus{} \sin 15 \minus{} \sin 5 \equal{} \sin 15\end{array}\right\|$

$ \tan25\cdot\tan35\cdot\tan85 \equal{} \frac {\cos 15}{\sin 15}$ $ \implies$ $ \tan 15\cdot\tan 25\cdot\tan 35\cdot\tan 85\ \equal{} \ 1$ .

[quote="Virgil Nicula"][color=darkred][b][u]Generalization.[/u][/b] Prove that $ \boxed {\ \tan 3x\cdot\tan (30 \minus{} x)\cdot\tan (30 \plus{} x)\cdot\tan (90 \minus{} x)\ \equal{} \ 1\ }\ (*)$ .[/color][/quote]
[b][u]Proof I.[/u][/b]

$ \tan 3x \equal{} \frac {3\tan x \minus{} \tan^3x}{1 \minus{} 3\tan^2x}$ $ \implies$ $ \boxed {\ \frac {\tan 3x}{\tan x} \equal{} \frac {3 \minus{} \tan^2x}{1 \minus{} 3\tan^2x}\ } \equal{} \frac {3 \minus{} \frac {1 \minus{} \cos 2x}{1 \plus{} \cos 2x}}{1 \minus{} \frac {3(1 \minus{} \cos 2x)}{1 \plus{} \cos 2x}} \equal{} \frac {1 \plus{} 2\cos 2x}{2\cos 2x \minus{} 1} \equal{}$

$ \frac {\cos 60 \plus{} \cos 2x}{\cos 2x \minus{} \cos 60} \equal{}$ $ \frac {\cos (30 \plus{} x)\cos (30 \minus{} x)}{\sin (30 \plus{} x)\sin (30 \minus{} x)} \equal{}$ $ \cot (30 \plus{} x)\cot (30 \minus{} x)$ $ \implies$

$ \frac {\tan 3x}{\tan x} \equal{} \cot (30 \plus{} x)\cot (30 \minus{} x)$ $ \implies$ $ \tan 3x\cdot\tan (30 \minus{} x)\cdot\tan (30 \plus{} x)\cdot\tan (90 \minus{} x)\ \equal{} \ 1$ .

[color=darkblue][b][u]Proof II.[/u][/b]

$ \blacktriangleright\ \ \ 4\cdot\sin (30 \minus{} x)\cdot\sin (30 \plus{} x)\cdot\sin (90 \minus{} x) \equal{}$ $ 2\cdot\cos x\cdot\left(\cos 2x \minus{} \frac 12\right) \equal{}$

 $ \cos x\cdot (2\cdot\cos 2x \minus{} 1) \equal{}$ $ \cos 3x \plus{} \cos x \minus{} \cos x \equal{} \cos 3x$ .

$ \blacktriangleright\ \ \ 4\cdot\cos (30 \minus{} x)\cdot\cos (30 \plus{} x)\cdot\cos (90 \minus{} x) \equal{}$ $ 2\cdot\sin x\cdot\left(\frac 12 \plus{} \cos 2x\right) \equal{}$

 $ \sin x\cdot (1 \plus{} 2\cdot\cos 2x) \equal{}$ $ \sin x \plus{} \sin 3x \minus{} \sin x \equal{} \sin 3x$ .

In conclusion, $ \tan (30 \minus{} x)\cdot\tan (30 \plus{} x)\cdot\tan (90 \minus{} x) \equal{} \frac {\cos 3x}{\sin 3x}$ $ \implies$

$ \tan 3x\cdot\tan (30 \minus{} x)\cdot\tan (30 \plus{} x)\cdot\tan (90 \minus{} x)\ \equal{} \ 1$ .

[b][u]Proof III.[/u][/b]

If $ \tan x \equal{} t$ , then $ (*)\ \Longleftrightarrow\ \frac {t(3 \minus{} t^2)}{1 \minus{} 3t^2}\cdot\frac {\frac {1}{\sqrt 3} \minus{} t}{1 \plus{} \frac {t}{\sqrt 3}}\cdot\frac {\frac {1}{\sqrt 3} \plus{} t}{1 \minus{} \frac {t}{\sqrt 3}} \equal{} t$

$ \Longleftrightarrow\ (3 \minus{} t^2)\left(\frac 13 \minus{} t^2\right) \equal{} (1 \minus{} 3t^2)\left(1 \minus{} \frac {t^2}{3}\right)$ , what is truly.[/color]

it can be interesting to note that the problem it's equivalent to show that in the figure $ \angle CDB \equal{} 5^\circ$

[img]http://img20.imageshack.us/img20/1768/immagineditrigoqy2.jpg[/img]

Very nice the your remark, [b]¬[ƒ(Gabriel)³²¹º]¼[/b] !

See [url=http://www.mathlinks.ro/viewtopic.php?t=199591][size=117][color=red][b]here[/b][/color][/size][/url] a geometrical interpretation of the my generalization.


That is equivalent to
$ \frac{6!}{2!(6\minus{}2)!}\equal{}\frac{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{2\cdot 1\cdot 4\cdot 3\cdot 2\cdot 1}\equal{}15$

There is a detailed explanation in my blog, link's in my signature. Find the entry titled, "Combinations". Hope this helps!

[quote="isabella2296"]Nick, I'm not quite sure exactly what you mean by that. Here's the link to the specific entry titled Combinations: [url=http://www.artofproblemsolving.com/Forum/weblog.php?ppage=3&sep=%2F&w=1172&start=36]here[/url].

Nick, do you not agree that it would be easier to simply give a link rather than copy and paste an entire entry in one post?[/quote]

And to think I just found out about this now. *sigh. Better late then never...  :|

[quote="PiLuvah"]I forgot how to simplify those combinations, can someone remind me?

example- 6C2[/quote]

We can write: $ C_n^k  \equal{} \frac{{n!}}{{k!(n \minus{} k)!}} \equal{} \frac{{A_n^k }}{{k!}} \equal{} \frac{{n(n \minus{} 1)(n \minus{} 2).....(n \minus{} k \minus{} 1)}}{{k!}}$  :)  :)

Or you could always refer to Pascal's triangle if your oh-so-lazy.  It's the second number after the 1 in the [b]7[/b]th row of Pascal's triangle. (first row is Row 0)


algebra
polynomial
superior algebra
superior algebra solved

Prove that $[\mathbb{Q}(\sqrt{3},\sqrt[4]{2}): \mathbb{Q}]=8$.

Step by step:
$[\mathbb{Q}[\sqrt 3]: \mathbb{Q}]=2$
$[\mathbb{Q}[\sqrt 3, \sqrt 2]: \mathbb{Q} [\sqrt 3]]=2$
$[\mathbb{Q}[\sqrt 3,\sqrt[4]{2}]: \mathbb{Q}[\sqrt 3, \sqrt 2]] = [\mathbb{Q}[\sqrt 3, \sqrt 2, \sqrt{\sqrt 2}]: \mathbb{Q} [\sqrt 3, \sqrt 2]]=2$

The proves for them work by showing that the new element is not a square in the desired field.

[quote="ZetaX"]
$[\mathbb{Q}[\sqrt 3,\sqrt[4]{2}]: \mathbb{Q}] = [\mathbb{Q}[\sqrt 3, \sqrt 2, \sqrt{\sqrt 2}]: \mathbb{Q} [\sqrt 3, \sqrt 2]]=2$.[/quote]
you are wrong!But it is small :D

A other problem:There exitsts a extension algebra of $\mathbb{Q}$ which is extension infinite of $\mathbb{Q}$?

Thank you!

[quote="ZetaX"]I edited it now.

For the other problem:
Let be $\mathbb{Q}_n : = \mathbb{Q}[\sqrt[2^n]{2}]$. Then $\mathbb{Q}_0 = \mathbb{Q}$ and $[\mathbb{Q}_n : \mathbb{Q}]=2^n$ (the minimal polynomial $x^{2^n}-2$ is irreducible after Eisensteins criterion). So the union $U : = \bigcup_{n=0}^\infty \mathbb{Q}_n$ has a degree higher than every integer, so $\infty$.[/quote]

You are true.But in :Field and Galois Theory(Morandi) has problem :Let $\mathbb{A}$ be the algebraic closuere of $\mathbb{Q}$ in $\mathbb{C}$.Prove that $[\mathbb{A}: \mathbb{Q}]=\infty$.

And i can't slove it  :( .Can you?Thank you!

The roots of the equations given in my previous post are complex (indeed all roots of any polynomials are complex by the fundamental theorem of Algebra which is indeed not needed here) since $\sqrt[2^n]{2}$ is already real.
Now we get $\mathbb{Q} \subset U \subset \mathbb{A} \subset \mathbb{C}$ and also $[\mathbb{A}: \mathbb{Q}]=[\mathbb{A}: U]\cdot [U: \mathbb{Q}]=[\mathbb{A}: U] \cdot \infty = \infty$.


trigonometry
calculus
calculus computations

Solve $ (D^2\minus{}3D\plus{}2)y\equal{}e^{4x}\sin x$.

It can be solved by method of undetermined coefficients.


geometry
3D geometry
inequalities
geometry proposed

Show that the volume $V$ and the lateral area $S$ of any right circular cone satisfy the inequality

$\begin{equation*}
\left( \frac{6V}{\pi }\right) ^{2}\leq \left( \frac{2S}{\pi
\sqrt{3}}\right) ^{3}.
\end{equation*}$

When does equality hold ?

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


Jen thought of a word to describe a MATHCOUNTS competitor. Jen said that none of the letter in quick were in the word she thought of, but exactly one of the letters in smart was in her word. Which of the following four words was Jen thinking of: wise, ready, mathy, or able?

This is Logic. Logic would certainly be a branch of Math.

The four choices are

Wise
Ready
Mathy
Able

It doesn't have the letters q,u,i,c,k.

Wise --> Has the letter "i", so it can't be this word.
Ready --> Does not have the letters q, u, i, c, k. So It could be.
Mathy --> Does not have the letters q, u, i, c, k. So it oculd be.
Able --> Does not have the letters q, u, i, c, k. So it could be.

The word must have ONE of the letters s, m, a, r, t.

Ready --> Has two letters, so it can't be.
Mathy --> Has "a", "t", so it can't be.
Able --> Has "a" only, so it is.

$ \boxed{\text{Able}}$.


Try to convert these to fractions:

1. 0.71717171...

2. 0.245245...

3. CHALLENGE: 0.99999...

These come up frequently in classroom math, and it is always good to know how to convert those... :D

[hide="Hint"]
Keep in mind where x, y, and z are digits,
$\frac{x}{9} = .xxxxxxxxxxxxxxxxxxxxxxxxxx...$
$\frac{xy}{99} = .xyxyxyxyxyxyxy...$
$\frac{xyz}{999} = .xyzxyzxyzxyzxyzxyz...$
etc....
[/hide]
See hint for all my solutions
[hide="1"]
$\frac{71}{99}$[/hide]
[hide="2"]
$\frac{245}{999}$[/hide]
[hide="3"]
$1$ There are many proofs of this, it's a fun challenge for people who might not be familiar yet to prove.[/hide]


function
algebra
polynomial
vector
calculus
integration
projective geometry

Hello,

I am finding this problem particularly interesting, yet hard.

Let us take an algebraically closed field $k$, and the projective space $\mathbb{P}^{n}$.  Let $T$ be a set of homogeneous polynomials over $k$ in $n+1$ variables $x_{0},\cdots,x_{n}$
Let $Y=Z(T)$ be the projective variety defined by $T$

We call a function $f$ regular on a open set of $Y$, if for every point there is an open set around it where the function is equal to $\frac{a}{b}$ where $a$ and $b$ are homogeneous polynomials of the same degree, with $b$ not disappearing in that point.

Show that a function can only be regular in all points of $Y$ if it is a constant element of $k$. :huh:

PS: what is the best place to post stuff like this? :ninja:  Is this "Advanced"?

If $T$ is just any set of homogeneous polynomials, $Z(T)$ is not necessarily a variety (i.e., irreducible), e.g., $n=1$, $T=\{x_{0}(x_{0}-x_{1})\}$. The usual argument should be something along the lines: show that the ring of global regular functions is a finite-dimensional $k$-vector space and use some hypotheses like irreducibility to show that it is an integral domain. owk

anyway, getting to the solution: consider a regular function $f\in k[Y]$, and consider $F=f\circ i: Z\to \mathbb{P}^{1}$, where $i: \mathbb{A}^{1}\to\mathbb{P}^{1}$ is the inclusion (i think $f\circ i$ has something to do with pullbacks, has it?).
$F$ is a morphism between projective sets, so its image is closed*, and since it's contained in $\mathbb{A}^{1}\hookrightarrow\mathbb{P}^{1}$, it's finite.
but since $Y$ was irreducible, so it is $f(Y)$, so it's one single point.

* this was the big theorem involved in the proof...


It's important--obviously at some point you are going to have to not be indebted to other countries, but whether the current situation is critical is another issue.

Hmm...  so far, 100% say it's important.

Obviously the deficit is important.  However, it seems that most people either ignore it or complain about it being so high.  Every time Congress sets new limits on how high the national debt can get, they break them a few months later.

i just voted no. can you feel deficit? is the deficit affecting you? i think not. the real issue is the war in iraq. itis unnessacary and killing thousands of soldiers in the name of democracy and freedom. i say let the iraqis figure it out. it's stupid invading another country especially when the UN was negociating with Iraq. and america's stupid and incompetent president persisted with the weapons of mass destruction and terror levels into tricking america to go to war. can't wait till Jan. 20, 2009. also, it is the war that is creating the deficit with its million dollar bombs and other high-tech mumbo-jumbo. so yea, the deficit isn't the problem.

So ... the deficit isn't important because the war is causing the deficit?  ;-)

Also, our wars are definitely not causing our deficit.  The deficit is caused by spending more than we're taking in in taxes, which is partly a result of wars (spending) but also of a lot of other kinds of spending and tax giveaways and so on.

Finally, in response to the question, "Can you feel the deficit?" the answer is, "No, because I don't know what a deficit feels like; but just wait until it stops being propped up by cheap foreign money.  [i]Then[/i], you'll feel the deficit."

well if you stop the war and budget the economy reasonably, the deficit will slowly go away. clinton did a good job with budgeting. and then you won't have to worry about feeling the deficit.

Of course it's important. Money is important in the world. How can we not address the issue of the deficit seriously when it is directly affecting us? The war is important as well -- that is, important to get rid of. We don't need war and we can rid ourselves of the deficit by leaving (though I guess it may be too late now). That's all.

Well, considering the trends in American unilateralism, they could probably just not pay ever, and get away with it!

That would be true, if America didn't require continued loans in order to maintain our economy and relatively high standard of living, among other things.


[b][color=red]
Moderator says:
Introduce yourself in this thread.  Include any information about yourself, and your recent achievements in mathematics in this thread [/color][/b]
I'm from Kentucky... And so is AndrewLiu
[b][color=red] Moderator says:
Changed title of thread, and description.  [/color][/b]

And PastaBoy... Nobody checks this forum, then again, so...

Okay, so now we have:
Mathlete9999
AndrewLiu
PastaBoy
Butler

Sorry, I wasn't able to get here for a while.... There are more people here than I imagined...

Anyone here live near Frankfort? Well, who attended the Capital MC?

Add me on.
Mathlete9999
AndrewLiu
PastaBoy
Butler
Dsquared

Hey, DSquared, do you live near Frankfort?

Close...  I've been there before,  It's like a 1-2 hour drive away.  Do u???

Yeah, I live about 1 hour away. Scott County, I live there..

Cool....  Have you ever been to the Kentucky derby...

No, it's rained most of the time. I don't like horses either... Not to smart animals... Too crowded there too... I saw the Versailles Magical Musical 4th. It was the last one they will ever have. They're gonna change it next year. :(

Ok....  I have been to Lexington b4 and my chess teams biggest rivals are from Lexington(They won in 2006 then we won from 2007-2010  :)

What county are you in? I'm in Scott County

Oh, cool.  What part of Kentucky is that?

Cool, what landmarks is it near?
Louisville is famous for Louisville slugger, and for KFC, and Kentucky Derby, and thunder over Louisville.

Add me.

1. Mathelete9999
2. Butler
3. PastaBoy
4. Dsquared
5. dude01
6. halowarfare17
7. fries4guys
8. AndrewLiu

[quote="Butler"]Wait, Highway 64...I don't know.... YOu know where TFCA is?[/quote]
Oh I know what you are talking about now XD

[quote="ProfessorJellyBean"][quote="Butler"]I am from Kentucky, so  we have:

Mathlete9999
AndrewLiu
PastaBoy
Butler[/quote]
ADD ME!!![/quote]
 also me :starwars:

Hello everyone,
I will moderate the Kentucky Forum, and I would like to welcome everyone to the Kentucky forum.
Name:  Justin Stevens
Grade: 7th
Age:  12 (turning 13 on April 30th/4 days!)
Recent achievements in Mathematics:  Currently taking Algebra 2, 96 on AMC 10B, 2nd place in 7th grade in Trinity math competition.
Current list of AoPSers from Kentucky:

1. Mathelete9999
2. Butler
3. PastaBoy
4. Binomial-Theorem
5. dude01
6. halowarfare17
7. fries4guys
8. AndrewLiu
9. pi_3-14159
10.  RelexationUtopia
11. halowarfare17

I have deleted all spam posts in this thread.  There was about 2 pages filled (24 total deletes) of spam posts (about moderating, spamming, etc.).

Woah...it's funny that on AoPS, some 15 yo's sound like their 10 and b-t sounds like he's an adult.

Add me! I thought id never find anyone from our good 'ol state on here haha! Anyone here compete in governers cup (ive never been able to spell that word)?

add me even though i think it's too late...
Governers cup for the win !!:)


$A$, $B$, $C$ and $D$ are on circle $O$ (in this order) such that chords $AC$ and $BD$ intersect perpendicularly at point $P$. Show that $4r^2=PA^2+PB^2+PC^2+PD^2$.

[hide] Let $\angle ACB=\alpha$ then $\angle CBD=90 - \alpha$. By sine theoreme $AB=2r \sin \alpha$ and $CD=2r \sin (90 - \alpha) = 2r \cos \alpha$ $(PA^2+PB^2)+(PC^2+PD^2)=AB^2+CD^2=4r^2(\sin ^2 \alpha+ \cos^2 \alpha) = 4r^2$[/hide]


AMC
AIME
rotation
geometry
geometric transformation

Here it is. What you've all been waiting for... the [url=http://wangsblog.com/jeffrey/2007mockAIME6.pdf]2007 Mock AIME 6[/url]!

[b]Rules[/b]

* 3 hours.
* No calculators.

[b]Format[/b]

* 15 questions.
* All answers will be integers in the range 000-999, inclusive.

[b]Scoring[/b]

* 1 point per correct answer.

[b]Submitting Answers[/b]

* Submit answers via PM to paladin8.
* Format as shown (X's are digits):

1. XXX
2. XXX
...
15. XXX

[b]Deadline[/b]

* 11:59 PM PST, Sunday, March 4 (subject to change).

[b]Grading[/b]

* I will PM your result back to you as soon as I grade it. Feel free to ask about specific problems (I may or may not end up compiling a solutions document).

[b]Credits[/b]

* Thanks to Altheman for proofreading the test.

Will we be disqualified for using more than 3 hours? I normally spend as long as I need until I get bored on mock AIMEs

@arjunlandes: Really? Finally, I wrote a decently difficult one :P.

@Ubemaya: You could send your answers after working for about 3 hours, then send whatever answers you want again later but have the first one be "official." Doesn't really matter, though.

When you say subject to change for the deadline, do you mean sooner or later more likely?

Most likely later.  Usually it depends on how many people submit by the original deadline for things like that...

Yeah. Definitely not sooner. If people don't have time, just let me know and I'll push it back probably a week (that gets very close to the real AIME, though).

Directed angles are usually taken to be counterclockwise, I think.

[quote="arjunlandes"]Maybe I'm missing something, but in #10, how do you know which way to rotate the point?  You can rotate it either clockwise or counterclockwise, can't you?[/quote]

Good call. It's supposed to be counterclockwise.

EDIT: I can't edit my original post anymore, so either just read here that it's counterclockwise or go to the PDF on my website, which is updated.

you pretty much consider 90 degree rotation to mean counter clockwise...that is the standard for polar notation and stuff

anyway...the test is very good...*gives a thumbs up to paladin and his test*

*bump* Just a reminder that this is still here :wink:.

[quote="bpms"]How difficult is this compared to a normal AIME?[/quote]

Seems to be on the slightly harder side; I can't say for sure, though. I'm pretty sure it falls within the general AIME difficulty range.

I find this one harder than a lot of the actual AIME papers, I guess it's a matter of opinions.

Apologies to anyone who has taken the test so far. #9's problem statement yielded an answer that was too big, so it has been edited. Don't download the PDF here; go to the LINK.

[quote="chess64"]Ugh yeah I spent like half an hour wondering what was wrong...

edit: Wait uh isn't it still the same? I used the one at http://wangsblog.com/jeffrey/2007mockAIME6.pdf ...[/quote]

No, it changed. Try refreshing it, maybe.

*bump* A couple more days until the deadline.

Mm another Mock AIME...thankfully I can barely finish it by Sunday =)

[quote="paladin8"][quote="Ubemaya"]For #2, can you choose ANY two diagonals to form an angle?[/quote]

Yes. You only count all the distinct angle measures formed, though.[/quote]

And the intersection must be inside the octagon?  Ie do ones on the border count?

No, only the intersection points strictly inside the octagon count.

[quote="DanK"]Is there an extension possibility then?  I think I could take it on Saturday, but probably not before then...[/quote]

I'm thinking no. I can still grade if you want, but I'm going to start a discussion thread shortly.

[quote="paladin8"][quote="Altheman"]I ono...you might extend...Also there is some value in "advertising"...I usually put it a link in my sig and when I post, people see the link. That and I bothered my friends like a bazillion times to take it...[/quote]

Yeah, maybe. I tried putting it in my signature, but I currently have external links which aren't allowed anymore so if I change it those links go away... :roll:. We'll see, though.[/quote]
I'll add it to my sig.
I'll be able to submit tommorow if we get the time.

all problems have been posted in separate threads in HSM,  I just created the related post collection

[url=https://artofproblemsolving.com/community/c3751277_2007_mock_aime_6]here[/url]

enjoy  / start solving

USA mocks [url=https://artofproblemsolving.com/community/c2439870_usa_mocks]here [/url] (also mentioned [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url])
more AIME mocks [url=https://artofproblemsolving.com/community/c2439872_aime_mocks]here[/url]
more User-created contests [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url] (contains all above)


analytic geometry
geometry
geometry proposed

Square $ ABCD$ and $ AEFG$,$ E,G$ lie on $ AB,AD$ respectively.If $ EF,CD$ intersect at $ H$,Prove that:$ AH,BF,CG$ concurrent at one point.

[quote="Johan Gunardi"]A(0,0), B(b,0), C(b,b), D(0,b)
E(a,0), F(a,a), G(0,a), H(a,b)

The three lines CG, BF, BF pass through $ (\frac {a^2b}{a^2 \minus{} ab \plus{} b^2},\frac {ab^2}{a^2 \minus{} ab \plus{} b^2})$..

we can also prove it be using menelaus twice.[/quote]

Nice ,but can you make it out in a more beautiful way.I 'm waiting for your beautiful solutions. :lol:

Actually the solution with coordinate geometry is shorter. :D  Here is a more beautiful solution.

[hide]Let $ FG$ and $ BC$ intersect at $ I$, $ AH$ and $ CG$ intersect at $ X$, $ BF$ and $ CG$ intersect at $ Y$. Also let $ a$ and $ b$ be the sides of $ ABCD$ and $ AEFG$ respectively. By Menelaus on triangle $ CDG$ and line $ HXA$, we get \[ \frac{DA}{AG}\frac{GX}{XC}\frac{CH}{HD}\equal{}1\] so \[ \frac{a}{b}\frac{GX}{XC}\frac{a\minus{}b}{b}\equal{}1\] which gives $ \frac{GX}{XC}\equal{}\frac{b^2}{a^2\minus{}ab}$. By Menelaus on triangle $ ICG$ and line $ YFB$, we get \[ \frac{CB}{BI}\frac{IF}{FG}\frac{GY}{YC}\] or \[ \frac{a}{b}\frac{a\minus{}b}{b}\frac{GY}{YC}\equal{}1\] so $ \frac{GY}{YC}\equal{}\frac{b^2}{a^2\minus{}ab}\equal{}\frac{GX}{XC}$, which implies that $ X\equal{}Y$, as desired.[/hide]

Thank you, but I mean that I need a solution without calculation (although I did it also by calculation :D )


algebra
polynomial
irrational number
number theory unsolved
number theory

Given any constant $ c$, prove that there exists an irrational number $ E$ and infinitely many rational numbers $ h/k$ such that

$ \left|E\minus{}\frac{h}{k}\right| < \frac{1}{k^c}$.

Take $ E \equal{} \sum_{n \equal{} 1}^\infty \frac {1}{10^{n!}}$, this works for all $ c$.

It's Liouville's constant. In fact, a number that works for every $ c$ is transcendental by Liouvilles approximation theorem, especially this one.
To get such a number, just try to create a number that is very close to some rationals but isn't rational itself; so taking a number of typ $ 1,100...0100...00100...$ with very fast increasing amounts of zeros should work (it suffices that the number of zeros between consecutive ones grows faster than any polynomial, $ n!$ is just one way out of uncountably many).