function
topology
real analysis
real analysis unsolved

Show that an upper semicontinuous function on a compact space assumes a maximum value.
We say that $ f$ is upper semicontinuous if $ f$ is continuous when $ \mathbb{R}%Error. "union" is a bad command.
\{-\infty\}$ is given the topology $ T$ on $ \mathbb{R}$ by declaring the empty set, $ \mathbb{R}$, and the unbounded open intervals $ (-\infty, a) (a\in \mathbb{R})$ to be the open sets.

I did not know this definition of usc. :blush:

Consider the sets $ U_a\equal{}\{x\in X\colon f(x)<f(a)\plus{}1\}$ for all $ a\in X$.


find integer a,b,n \geq 2 so that
             (a^3+b^3)^n=4(ab)^ :)  :(  :?  :D 2004

Hi,Thanhtrung . ( a :^3: +b :^3:)^n=4(ab)^2004   (1)
this problem is almost indentical to problem 3 from the 1995 Vietnamese selection test for imo.That year,power of (ab)is 1995(this problem, power of (ab)is 2004).First,we have: a :^3: +b :^3:   :ge:    2(ab) ^3/2deduce (a :^3: +b :^3: )^n :ge: 2^n(ab)^3n/2 :ge: 4(ab)^3n/2(because n :ge: 2)(2)
from (1),(2)we have:3n :le: 4008;n<2004
put gcd(a,b)=d,a=da',b=db' where gcd(a',b')=1
to replace a,b with da',db' thus (1) becomes:
   d^3n(a' :^3: +b' :^3: )^n =4(d)^4008(a'b')^2004
or: (a' :^3: +b' :^3: )^n=4(d)^(4008-3n)(a'b')^2004
........a=b=d(it is easy)
we have:2^(n-2) =d^(4008-3n) deduce d=2^k with k :ge: 1
hence n=1336- 1334/(3k+1) deduce 3k+1=46 ;58;667
we have a,b,n=....
ok?


I'd like to know more about techniques for solving this problem:

In Aye Nye Township, 81% of the people were  using the traditional telephone system for long distance dialing, and 19% were using one of the new systems.  One year later, 20% had switched from the traditional system to one of the new systems, and 5% switched from one of the new systems to the traditional system.  If this pattern continues indefinitely, find, rounded to the nearest percent the percent of people in Aye Nye township who will, in the long run, be using one of the new systems.

In my solution, I decided to assume there were 100 people living here and let u(n) and v(n) denote the number of people using the traditional system and a new system, respectively. after n years, where u(0) = 81, v(0)=19, u(n) = .8u(n-1) + .05v(n-1), and v(n)= .2u(n-1) +.95v(n-1) where n is an integer >=0.

I used the seq mode on my TI-83 to arrive at the correct answer of 80%.  This seemed to be an effective and efficient way to solve the problem, but I am still wondering if and how this problem might be solved easily without a calculator.

Well you could say that the limit is the point where they aren't changing any more.  Therefore, 20% of u(n) is the same as 5% of v(n) as n goes to infinity.  This should be easy to solve by algebra: you get 80% and 20%.

Great observation --the problem seems so easy this way!  I'll appreciate it  even more after I stop kicking myself   :blush:


calculus
integration
inequalities
geometry
rectangle
algebra
binomial theorem

Does anyone have clever ways to evaluate these?.

$ \int_{0}^{3}\sqrt [4]{x^{4} \plus{} 1}dx$

and

$ \int_{1}^{3}\sqrt [4]{x^{4} \minus{} 1}dx$

I tried using the Beta function.  Perhaps a series?.

hello, for the second one mathematica gives
$ \int_{0}^{3}\sqrt [4]{x^4 \minus{} 1}\,dx \equal{} \minus{} \frac {4i\sqrt {\pi}\Gamma[5/4]}{\Gamma[ \minus{} 1/4]} \minus{} \frac {80}{27}5^{1/4}\left( \minus{} 1 \plus{} \text{Hypergeometric2F1}[1/2,1,1/4,81]\right)$.
Sonnhard.

Yes, I just ran them through Maple.  I thought maybe there was some cool way to do them by hand, perhaps via Gamma or Beta.  Apparently, it's too cumbersome.  Thanks for your input. I always like to see some clever way to do tough integrals.

Here is what Maple 10 gives:

$ \int_{0}^{3}\sqrt [4]{x^{4} \plus{} 1}dx \equal{} \frac {72\Gamma(\frac {3}{4})\text{hypergeo}\left([\frac { \minus{} 1}{2},\frac { \minus{} 1}{4}],[\frac {1}{2}],\frac { \minus{} 1}{81}\right) \plus{} \frac {2\sqrt {2}{\pi}^{\frac {3}{2}}}{{\Gamma}(\frac {3}{4})}}{16\Gamma(\frac {3}{4})}$

$ \int_{1}^{3}\sqrt [4]{x^{4} \minus{} 1}dx \equal{} 3.858396027$. It did not give me a nice solution as above.

The two of these added together gives a solution very close to 9.

5.1416468411+3.858396027=9.0000428681

They came from a problem I saw that wanted to prove that the sum of these two integrals lies between 9 and 9.0001.

Prove that:

$ 9\leq \int_{0}^{3}\sqrt [4]{x^{4} \plus{} 1}dx \plus{} \int_{1}^{3}\sqrt [4]{x^{4} \minus{} 1}dx\leq 9.0001$

I realize to solve we do not actually have to evaluate the integrals, but I thought it would be cool to do so if possible.

No it wouldn't be cool. Why should problems have closed forms? This is about estimation.

What can you get out of the binomial series? You do have to do things differently for $ x>1$ and for $ x<1.$

For $ x>1,$ we have $ (x^4\plus{}1)^{1/4}\plus{}(x^4\minus{}1)^{1/4}\equal{}x\left[(1\plus{}x^{\minus{}4})^{1/4}\plus{}(1\minus{}x^{\minus{}4})^{1/4}\right].$

$ (1\plus{}x^{\minus{}4})^{1/4}\equal{}1\plus{}\frac14x^{\minus{}4}\minus{}\frac{3}{4^2\cdot 4!}x^{\minus{}8}\plus{}\frac{3\cdot 7}{4^3\cdot 3!}x^{\minus{}12}\minus{}\frac{3\cdot 7\cdot 11}{4^4\cdot 4!}x^{\minus{}16}\plus{}\cdots$

$ (1\minus{}x^{\minus{}4})^{1/4}\equal{}1\minus{}\frac14x^{\minus{}4}\minus{}\frac{3}{4^2\cdot 4!}x^{\minus{}8}\minus{}\frac{3\cdot 7}{4^3\cdot 3!}x^{\minus{}12}\minus{}\frac{3\cdot 7\cdot 11}{4^4\cdot 4!}x^{\minus{}16}\plus{}\cdots$

Add those together and multiply by $ x$ to get

$ (x^4\plus{}1)^{1/4}\plus{}(x^4\minus{}1)^{1/4}\equal{}2\left[x\minus{}\frac{3}{32}x^{\minus{}7}\minus{}\frac{3\cdot 7\cdot 11}{4^4\cdot 4!}x^{\minus{}15}\minus{}\frac{3\cdot 7\cdot 11\cdot 15\cdot 19}{4^6\cdot 6!}x^{\minus{}23}\minus{}\cdots\right]$

Integrate that to get that

$ \int_1^3(x^4\plus{}1)^{1/4}\plus{}(x^4\minus{}1)^{1/4}\,dx\equal{}$

$ \equal{}2\left[4\minus{}\frac3{32}\cdot\frac{1\minus{}3^{\minus{}6}}{6}\minus{}\frac{3\cdot7\cdot 11}{4^4\cdot 4!}\cdot\frac{1\minus{}3^{\minus{}14}}{14}\minus{}\frac{3\cdot 7\cdot 11\cdot 15\cdot 19}{4^6\cdot 6!}\cdot\frac{1\minus{}3^{\minus{}22}}{22}\minus{}\cdot\right]$

In the same vein, we get that

$ \int_0^1(1\plus{}x^4)^{1/4}\,dx\equal{}\int_0^11\plus{}\frac14x^4\minus{}\frac{3}{4^2\cdot 2!}x^8\plus{}\frac{3\cdot 7}{4^3\cdot 3!}x^{12}\plus{}\cdots\,dx$

$ \equal{}1\plus{}\frac1{20}\minus{}\frac{3}{4^2\cdot 2\cdot 9}\plus{}\frac{3\cdot 7}{4^3\cdot 3!\cdot 13}\minus{}\frac{3\cdot 7\cdot 11}{4^4\cdot 4!\cdot 17}\plus{}\cdots$

Now add the two series together. We do get $ 9$ plus some pretty small stuff, so we do see why the answer should be near $ 9.$ But it's not really good enough. The "leading" non-integer terms of each series, combined, give us $ \frac1{20}\minus{}\frac{1}{32}\equal{}\frac{3}{160}.$ That's too big, by orders of magnitude - we want the error from $ 9$ to be less than $ 10^{\minus{}4},$ which is asking a lot. So we need several of these non-integer terms to mostly cancel each other out.

So maybe the binomial series isn't the slickest estimating approach - there must be something else to do.

I have come to the conclusion that this is not a nice problem and does not have a nice answer.

We have an integral in two pieces; from $ 0$ to $ 1$ and then from $ 1$ to $ 3.$ The integral from $ 1$ to $ 3$ is less than the nearest integer value $ (8)$ by a smallish but not minuscule amount.

(One other inequality for seeing that: for $ x>1,$ we have
\[ 2\sqrt[8]{x^8\minus{}1}<\sqrt[4]{x^4\minus{}1}\plus{}\sqrt[4]{x^4\plus{}1}<2x\ )\]
Then the integral from $ 0$ to $ 1$ exceed the nearest integer value $ (1)$ by a smallish but not minuscule amount.

But somehow these two smallish amounts cancel each other out nearly perfectly; the sum is a truly minuscule distance from the nearest integer.

Normally one would show something like this by showing how the amounts in question cancel out (mostly) in some algebraic fashion - but how? I don't see it. Why is that upper limit of integration $ 3?$ The most obvious things you'd do to the interval $ (0,1)$ by way of substitution don't map it onto $ (1,3).$

I'm now thinking that the near-perfect cancellation of the deviation from an integer is just a coincidence, and to show it, you might as well use your favorite means of numerical integration and leave it at that.

Thanks much. That is a good suggestion :omighty: .

$ \int_{0}^{3}\sqrt [4]{x^{4} \plus{} 1}dx$

Let $ t \equal{} x \plus{} \sqrt [4]{x^4 \plus{} 1}\Longrightarrow (t \minus{} x)^4 \equal{} x^4 \plus{} 1\Longleftrightarrow \{(t \minus{} x)^2 \plus{} x^2\}\{(t \minus{} x)^2 \minus{} x^2\} \equal{} 1$.

Thus we can set :

$ (t \minus{} x)^2 \plus{} x^2 \equal{} u$

$ (t \minus{} x)^2 \minus{} x^2 \equal{} \frac {1}{u}$, where $ u > 0$.

Solving the system equation gives :

$ (t \minus{} x)^2 \equal{} \frac {1}{2}\left(u \plus{} \frac {1}{u}\right)$

$ x^2 \equal{} \frac {1}{2}\left(u \minus{} \frac {1}{u}\right)$

From the interval of the integration $ 0\leq x\leq 3$,  we have $ x \equal{} \frac {1}{\sqrt {2}}\sqrt {u \minus{} \frac {1}{u}}$

$ \therefore \frac {dx}{du} \equal{} \frac {1}{2\sqrt {2}}\cdot \frac {u^2 \plus{} 1}{u^2}\sqrt {\frac {u}{u^2 \minus{} 1}}$

$ x : 0\longrightarrow 3$
$ t : 1\longrightarrow 3 \plus{} \sqrt [4]{82}$
$ u : 1\longrightarrow 9 \plus{} \sqrt {82}$

$ \therefore \int_0^3 \sqrt [4]{x^4 \plus{} 1}\ dx$

$ \equal{} \int_1^{9 \plus{} \sqrt {82}} \frac {1}{\sqrt {2}}\sqrt {\frac {u^2 \plus{} 1}{u}}\cdot \frac {1}{2\sqrt {2}}\cdot \frac {u^2 \plus{} 1}{u^2}\sqrt {\frac {u}{u^2 \minus{} 1}}\ du$

$ \equal{}\frac{1}{4}\int_1^{9 \plus{} \sqrt {82}} \frac {u^2 \plus{} 1}{u^2}\sqrt {\frac {u^2 \plus{} 1}{u^2 \minus{} 1}}\ du$

$ \equal{} \frac {1}{4}\int_1^{9 \plus{} \sqrt {82}} \left(1 \plus{} \frac {1}{u^2}\right)\sqrt {1 \plus{} \frac {2}{u^2 \minus{} 1}}\ du$

We can treat the first integral as the area in the first quadrant of $ y^4 \le x^4 \plus{} 1$ with $ x \le 3$.  We can treat the second integral as the area in the first quadrant of $ y^4 \ge x^4 \plus{} 1$ with $ y \le 3$.  If you put those two regions together, you get at least the rectangle $ [0,3] \times [0, 3]$.  So the area is at least 9.  There's a little part of those two regions that sticks out of that rectangle; that excess is contained in the rectangle $ [\sqrt [4]{80}, 3] \times [3, \sqrt [4]{82}]$.  Because $ \sqrt[4]{80} > 2.99$ and $ \sqrt[4]{82} < 3.01$, that rectangle has area less than $ 0.0001$, so we are done.

I hereby retract my pessimistic evaluation in post #7. Ravi has the answer, and I missed that argument. It [b]is[/b] a nice problem.

What happens if we replace 3 by an arbitrary $ p > 1?$

Let $ A \equal{} \int_1^p\sqrt [4]{x^4 \minus{} 1} \plus{} \int_0^p\sqrt [4]{x^4 \plus{} 1}\,dx$

Ravi's argument applies. This can be interpreted as the area of $ [0,p] \plus{} [0,p]$ plus a slight excess. The excess is a nearly triangular piece with vertices $ \left(\sqrt [4]{p^4 \minus{} 1},p\right),(p,p),$ and $ \left(p,\sqrt [4]{p^4 \plus{} 1}\right);$ certainly that excess is contained in the rectangle with those as three of its corners.

We have $ A > p^2$ and in particular $ A \minus{} p^2\approx\frac12\left(p \minus{} \sqrt [4]{p^4 \minus{} 1}\right)\left(\sqrt [4]{p^4 \plus{} 1}\minus{}p\right).$

By the binomial theorem, we have $ A \equal{} p^2 \plus{} \frac1{32 p^6} \plus{} o(p^{ \minus{} 6})$ as $ p\to\infty.$

In the case, we could use Young's inequality.

At first original problem, Young's inequality occured to me, but $ \sqrt [4]{x^4 \minus{} 1}\ (0\leq x\leq 1)$ was obstacle.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=21574&search_id=1278839442]Take a look this[/url]

Thank you all. That is very informative.

[quote="Kent Merryfield"]
We have $ A > p^2$ and in particular $ A \minus{} p^2\approx\frac12\left(p \minus{} \sqrt [4]{p^4 \minus{} 1}\right)\left(\sqrt [4]{p^4 \plus{} 1}\right).$
By the binomial theorem, we have $ A \equal{} p^2 \plus{} \frac1{32 p^6} \plus{} o(p^{ \minus{} 6})$ as $ p\to\infty.$[/quote]
And I think here we have:
$ A \minus{} {{p}^{2}}\approx \frac {1}{2}\sqrt {{{p}^{2}} \minus{} \sqrt [2]{{{p}^{4}} \minus{} 1}}.\sqrt {\sqrt [2]{{{p}^{4}} \plus{} 1} \minus{} {{p}^{2}}}$

It would help if I would write the expression correctly in the first place. The typo is now fixed.


Hi. I want to start preparing for next year AMC and AIME after this year's disasterous score. It's no fun getting the flu on the test date  :( . Anyway, I am wondering if Vol1 or Vol2 is more appropriate for AMC 12 and AIME? Vol1 is "supposedly" for Mathcounts and novice high school students but will it still help though? How much will one's score typically improve with both?

Thanks for any reply.

Anh Nguyen

Mastering Volume 1 is a good idea if you intend on doing well on the AMC12 and AIME. Many of the topics in Volume 2 are important as well, but first make sure you understand everything in Volume 1 because it will probably help you more than trying to go on to more advanced concepts (which you can do later, as the year progresses, I suppose).

I only bought 2, but knowing that it had stuff that we had to learn in 1, i went back to it. It actually deacreased my AMC scores, but increased my AIME scores...

I just have Volume 2, although I didn't really use it for practicing for the AIME. I don't think I really practiced for the AIME or AMC except for taking a few classes here on AoPS and doing 5-6 old tests.

What about AMC10? Should I get both, or just vol 1?

You probably won't need volume 2 for AMC 10.  But AoPS is good for general, long term problem solving.  If you want specifically AMC 10 stuff, it's still a good idea just to do the tests (they have a number of books containing old tests or similar problems and solutions, as well as various statistics).


function
limit
real analysis
real analysis unsolved

Prove that the set of points at which a sequence of measurable real functions converges is a measurable set.  :blush:

Denote $ f_n: (X,\mathcal{F})\rightarrow (Y,\mathcal{G}), n\in\mathbb{N}, A\equal{}\{x\in X: \lim_n f_n(x) \textrm{ exists }\}\subseteq X,$. Remember that $ \liminf_n f_n\equal{}\limsup_n f_n\equal{}\lim_n f_n$ on $ A$. Prove that $ \{\sup_n f_n\leq y\}$ is measurable, for $ y\in Y$, from that will follow that $ \inf_n f_n$ is measurable, and remember the definition of $ \liminf_n$, that is $ \limsup_n$.

gn :   C -------> R  , C = { x| {fn(x)} is  convergent   }  , gn(x) = |fn(x) –f(x) | ,  f(x) is the pointwise  limit of {fn}
gn is measurable .I guess (but not sure)  that  C=  union(m=1 to infinity )(intersection(n=N(m)+1 to infinity) (gn^ (-1) ((-1/m,1/m))   which N is defined as a function of m by the limit .
(-1/m,1/m)  is open  so gn^ (-1) ((-1/m,1/m))   is measurable  then C is also measurable .


Hello,

I have recently created a wiki page on wikispaces.com. I was wondering if there is any way to allow my page to be LaTex enabled.

Thanks! :)

See [url=http://blog.wikispaces.com/2007/05/math-in-wikispaces.html]Math in Wikispaces[/url] and [url=http://jaysscratchpad.wikispaces.com/Entering+equations+into+a+wiki+page]Entering equations into a wiki page[/url].


calculus
integration
real analysis
real analysis unsolved

Big Numbers Law again. The answer is $\frac23$ (actually one can say that the integrand is about $2/3$ most of the time and, since it is always between $0$ and $1$, the result follows).


Define $ A_k \equal{} \displaystyle\sum_{i \equal{} 1}^{k}\frac {i^2 \plus{} i}{i^4 \plus{} 2i^3 \minus{} 7i^2 \minus{} 8i \minus{} 2}$ and $ B_k \equal{} \displaystyle\sum_{i \equal{} 1}^{k}\frac {2i \plus{} 1}{i^4 \plus{} 2i^3 \minus{} 7i^2 \minus{} 8i \minus{} 2}.$ Also, let $ C_k \equal{} \frac {3A_k}{3B_k \plus{} 1}$ unless B_k = -1/3. If B_k = -1/3, then C_k = 0. Find the limit as k approaches infinity of C_k.

Are limit problems allowed in pre-olympiad?

As long as the solution isn't steeped in epsilon-delta or convergence issues, they're okay. There's a pretty elementary way to discuss sequence limits after all.

[hide="Sketch"]Don't feel like going through the computations, but if $ P(x) \equal{} x^4 \plus{} 2x^3 \minus{} 7x^2 \minus{} 8x \minus{} 2$, $ P(x\minus{}1/2)$ is even - has only $ x^4, x^2, 1$ coefficients - which means partial fractions become doable.[/hide]

[quote="MellowMelon"]As long as the solution isn't steeped in epsilon-delta or convergence issues, they're okay. There's a pretty elementary way to discuss sequence limits after all.

[hide="Sketch"]Don't feel like going through the computations, but if $ P(x) \equal{} x^4 \plus{} 2x^3 \minus{} 7x^2 \minus{} 8x \minus{} 2$, $ P(x \minus{} 1/2)$ is even - has only $ x^4, x^2, 1$ coefficients - which means partial fractions become doable.[/hide][/quote]

I'm not sure that partial fractions work, because apparently A_k approaches 0 and B_k approaches -1/3 when k approaches infinity... is there a way to deal with limits of quotients of series that both approach 0?

What does that have to do with whether or not partial fractions work?  Partial fractions will let you get a closed form for $ A_k$ and $ B_k$, which will eventually give you what you want.  In particular, it's useful to note that $ i^4 \plus{} 2i^3 \minus{} 7i^2 \minus{} 8i \minus{} 2 \equal{} (i^2 \plus{} 4i \plus{} 2)(i^2 \minus{} 2i \minus{} 1)$ and $ (i \plus{} 3)^2 \minus{} 2(i \plus{} 3) \minus{} 1 \equal{} i^2 \plus{} 4i \plus{} 2$, and then you don't need any halves involved.

This is not a question about general theory, it's a question about very special tricks.  In other words, it looks very much like a(n ugly) question designed around a particular solution method.

Isn't there a discrete version of L'hopital's rule that says if

a_n/b_n is of an indeterminant form (ex. a_n, b_n go to 0), then the sequence (a_n-a_{n+1})/(b_n-b_{n+1)) has the same limit as a_n/b_n

If this is true, then the problem follows immediately after we verify that A_n and 3B_n+1 go to zero.

hello, i have found
$ A_k\equal{}\frac{k(k\plus{}1)(k\plus{}2)(3k^2\plus{}6k\minus{}10)}{3(k^2\minus{}2)(k^2\plus{}2k\minus{}1)(k^2\plus{}4k\plus{}2)}$
$ B_k\equal{}\frac{k(k\plus{}2)(k^4\plus{}4k^3\plus{}2k^2\minus{}4k\minus{}6)}{3(k^2\minus{}2)(k^2\plus{}2k\minus{}1)(k^2\plus{}4k\plus{}2)}$
Sonnhard.

[quote="Altheman"]after we verify that A_n and 3B_n+1 go to zero.[/quote]  But to verify this is nontrivial, and essentially requires one to do all the hard work of the problem already.

hello, inserting my formuals and simplifying i have got
$ C_k\equal{}\frac{3A_k}{3B_k\plus{}1}\equal{}\frac{k(k\plus{}1)(k\plus{}2)(3k^2\plus{}6k\minus{}10)}{\minus{}4\minus{}12k\plus{}6k^2\plus{}12k^3\plus{}3k^4}$, the limit
$ \lim_{k \to\infty}C_k$ doesn't exist.
Sonhard.


calculus
integration
calculus computations

[quote="bookworm_vn"]Calculate
\[ \iiint\limits_D {xyz\left( {1 \minus{} x \minus{} y \minus{} z} \right) dx dy dz}
\]
where $ D$ is $ x \geq 0$, $ y \geq 0$, $ z \geq 0$ and $ x \plus{} y \plus{} z \leq 1$.[/quote]

$ I\equal{}\dfrac{\Gamma^3(2)}{\Gamma(6)}\;\int_0^1\;w^5\cdot(1\minus{}w)\;dw\equal{}\frac{1}{\Gamma(6)}\;\dfrac{\Gamma(6)\;\Gamma(2)}{\Gamma(8)}\equal{}\frac{1}{5040}$

$ \frac{1}{7!}$. You should post the full solution for everyone who cannot solve this easy integral.


calculus algebra polynomial function domain limit

[quote="page 79"]
$ \lim _{x \to \infty} \frac {f(x)}{g(x)}$...But if both $ \lim _{x \to \infty} g(x) \equal{} 0$ and $ \lim _{x \to \infty} f(x) \equal{} 0$, interesting things can happen. the top could go to zero [i]much faster[/i] than the bottom, resulting in a limit of $ 0$ [b](for example, try $ f(x) \equal{} x^2,g(x) \equal{} x$)[/b].
[/quote]

I have difficulty understanding the example (in bolded). Why would $ f(x) \equal{} x^2$ reach zero faster than $ g(x) \equal{} x$? I was under the impression that as you approach larger [integer] numbers, $ x^2 > x$.

EDIT: Please do [b]not[/b] move to the calculus forum - I'm only asking a conceptual understanding question.

Higher-order polynomials converge (or diverge) faster than lower-order ones.  So $ x^{2}$ converges (and diverges) faster than $ x$ does.  Logarithmic functions are the slowest-growing functions while exponential functions grow the fastest.

This is the range mapping to the domain in one sense. Consider x as a function of $ y$. This way, $ f(y)\equal{}\sqrt{y}$ and $ g(y)\equal{}y$. Then their rates of change as $ y$ tends towards zero can be found by $ f^{\prime}(y)\equal{}\frac{1}{2\sqrt{y}}$ and $ g^{\prime}(y)\equal{}1$. \[ \lim_{y\searrow 0} \frac{f^{\prime}(y)}{g^{\prime}(y)}\equal{}\lim_{y\searrow 0} \frac{1}{2\sqrt{y}}\equal{}\infty\] $ x\equal{}f(y)$ is approaching zero infinitely faster than $ x\equal{}g(y)$. This made sense at first but now it's starting to confuse me...

 :)

The book seems to have a typo.  The authors probably meant $ f(x) \equal{} \frac{1}{x^2}$ and $ g(x) \equal{} \frac{1}{x}$.

I agree with Ravi

$ \displaystyle \lim _{x \rightarrow \infty} \frac{f(x)}{g(x)} \equal{} \lim _{x \rightarrow \infty} \frac{x^{2}}{x} \neq 0$

Maybe they mistakenly switched the functions...

It sounds like $ \lim_{x \to 0}$ is what is really meant.

If they mean $ f(x)\equal{}x^{2}$ and $ g(x)\equal{}x$, then
$ \lim_{x\rightarrow 0}\frac{x^{2}}{x}\equal{}0$
but interestingly, the quotient is not defined at 0.


Hello!

Could anybody give me the idea
how to change label style in bibliography list without
changing the bibliography style? Is it possible?

Exactly I need the following changes:

instead
[1] Author1 Title1 ...
[2] Author2 Title2 ...
[3] Author3 Title3 ...
i need
1. Author1 Title1 ...
2. Author2 Title2 ...
3. Author3 Title3 ...

Thanks in Advance!

I have found the definition of the bib label:

\newcommand{\@biblabel}[1]{[#1]\hfill}

I suppose that redefinition of this comand will get what I want.
e.g.
\begin{document}
...
\renewcommand{\@biblabel}[1]{#1.\hfill}

but when I trying to do so the LaTex produces the error
"You can't use'macro parameter character #' in horizontal mode"

What's wrong?

Thank you.

But it isn't the reason why it does not work.
When I inserting the slash before # document becomes to be compiled
but result is wrong.

Typing
\renewcommand{\@biblabel}[1]{\#1.\hfill}
I get the empty page with the characters:
ibiblabel[1]#1.

http://www.tex.ac.uk/cgi-bin/texfaq2html?label=formbiblabel seems to be what you need


geometry
perimeter
inequalities
triangle inequality

if $ a,b,c$ are the sides of a triangle with perimeter 3 find the minimum value of:
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 3abc/4$

[hide]By Square Root Mean vs. AM, we have

$ \sqrt{\frac{a^2 \plus{} b^2 \plus{} c^2}{3}} \geq \frac{a\plus{}b\plus{}c}{3}$

$ a^2 \plus{} b^2 \plus{} c^2 \geq 3$

And then by AM - GM,

$ \frac{a \plus{} b \plus{} c}{3} \geq \sqrt[3]{abc}$

$ \frac{3}{4} \geq \frac{3abc}{4}$

But this will give you the min of $ a^2 \plus{} b^2 \plus{} c^2$ but the max of $ \frac{3abc}{4}$..oh no I've gone cross-eyed..[/hide]

I guess that it cannot have a minimum value  $ A \equal{} a^{2} \plus{} b^{2} \plus{} c^{2} \plus{} \frac {3abc}{4}$ $ >3$ , but A cannot be equal to 3  :wink:

hello, i think the searched minimum is $ 3.75$ and will be attained for $ a\equal{}b\equal{}c\equal{}1$.
Sonnhard.

Having equality would give us the [b]minimum[/b] for $ a^2 \plus{} b^2 \plus{} c^2$, but it would also give us the [b]maximum[/b] of $ \frac{3abc}{4}$.

Actually, the problem is to find the minimum value of $ a^2 \plus{} b^2 \plus{} c^2 \plus{} \frac {4abc}{3}$ rather than $ a^2 \plus{} b^2 \plus{} c^2 \plus{} \frac {3abc}{4}$  though it does not matter about whether it is $ \frac {3}{4}$ and $ \frac {4}{3}$.

Here goes my solution:

First I state a lemma {a previous inequality that I got on own- posted by geniusbliss sometime ago }: LEMMA:
If $ a,b,c$ are positive reals, $ a \plus{} b \plus{} c \equal{} 1$, Prove that :
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 3abc \ge \frac {4}{9}$.

Proof is by rewriting the inequality in this form:  $ (\sum a^2 )(\sum a ) \plus{} 3abc \ge \frac {4}{9}(\sum a ) ^3$
 Then you would finally end up with AM GM : $ \sum a^3 \ge 3abc$ and Schur Inequality for degree $ 1$ where : $ \sum_{cyc} a^3 \plus{} 3abc \ge \sum_{sym} a^2b$
--------------------End Lemma 1-------------------------

Now here $ a \plus{} b \plus{} c \equal{} 3$ whihc implies $ \sum \frac {a}{3} \equal{} 1$
Thus applying LEMMA 1 to $ \frac {a}{3}, \frac {b}{3},\frac {c}{3}$, we get that : $ \sum a^2 \plus{} abc \ge 4$

Multiplying both sides by $ \frac {4}{3}$, we have :

$ K \equal{} \frac {4}{3} (\sum a^2 ) \plus{} \frac {4}{3}abc \ge \frac {16}{3}$ Our original expression is $ K \minus{} \frac {1}{3} (\sum a^2 )$. Now till this point, we have not even used the condition that $ a,b,c$ are sides of a triangle. I use the very well known triangle inequality $ (a \plus{} b \plus{} c)^2 \ge 2(a^2 \plus{} b^2 \plus{} c^2)$ where $ a,b,c$ are sides of a triangle. This is simply obtained by writing $ ( \sum a )^2 \equal{} (\sum a^2 ) \plus{} \sum a(b \plus{} c) \ge \sum a^2 \plus{} \sum a^2 \equal{} 2 \sum a^2$ So this implies that $ (\sum a^2 ) \le \frac {(\sum a )^2}{2}$ This implies that $ \frac {1}{3} (\sum a^2 ) \le \frac {1}{6}{(\sum a )^2}$ Which implies $ \minus{} \frac {1}{3} (\sum a^2 ) \ge \minus{} \frac {1}{6}(\sum a )^2 \equal{} \minus{} \frac {3}{2}$ So therefore , our initial expression $ \equal{} K \minus{} \frac {1}{3} (\sum a^2 ) \ge \frac {16}{3} \minus{} \frac {3}{2} \equal{} \frac {23}{6}$ Thus, for $ a,b,c$ forming the sides of a triangle with perimeter 3, $ a^2 \plus{} b^2 \plus{} c^2 \plus{} \frac {4abc}{3} \ge \frac {23}{6}$ According to Aadil's post, the minimum value of $ a^2 \plus{} b^2 \plus{} c^2 \plus{} \frac {3abc}{4}$ is $ \frac {15}{4}$It is infact simpler because once you apply LEMMA 1, you end up with $ 3 \plus{} \frac {\sum a^2}{4}$ and directly use power mean inequality for getting $ \sum a^2 \ge \frac {(\sum a )^2}{3}$.

Equality in Aadil's inequality holds for $ a\equal{}b\equal{}c\equal{}1$ as pointed out by Sonnhard though he used MAPLE.
Equality in the $ \frac{4}{3}$ case holds for a degenerate triangle.
Thank you.

I actually want to see some other solution for this nice problem. Mine i think is somewhat complicated.

OH my god!! Maple is a mathematical software like mathematica, etc.

You can find other solutions at http://www.mathlinks.ro/viewtopic.php?t=161553


[color=blue]
$ A$  tosses $ \ 2$  fair coins  $ \ B$  tosses $ \ 3$  fair coins . The game is won by the person who throws greater numbers of heads. In case of a tie the game is continued under identical rules until someone wins the game. Find the probability of $ \ A$  winning the game.
[/color]

i get that in a given round, there is a 3/16 chance of A winning and a 5/16 chance of a tie.  So then the probability of A winning is 3/16*(1/(1-5/16)) = 3/7


logarithms
inequalities
inequalities open

Let $x\geq a\geq b\geq y\geq 0$ , and

$x^p+y^p=a^p+b^p$ (where $p=0$ is treated as $xy=ab$) and let

$f(x,y,k)=x^k+y^k$ (where $k=0$ is treated as $xy$) , then PROVE or DISPROVE that when

(i)$k\geq 0,k>p$ or $k<0,k<p$

$\implies$ $f(x,y,k)\geq f(a,b,k)$

(ii)$k\geq 0,k<p$ or $k<0,k>p$

$\implies$ $f(x,y,k)\leq f(a,b,k)$

I try out many cases and they all work well , so I think they are true ,but I dont have proof yet  :)

We can write the condition as following: Suppose $x\geq a\geq b\geq y$ and $x+y=a+b.$ Prove that \[x^{p}+y^{p}\leq a^{p}+b^{p},\;p\in[0,1],\;\;\;\;\;x^{p}+y^{p}\geq a^{p}+b^{p},\;p\in\mathbb{R}/[0,1].\] But this is obvious because the first condition is equivalent to $(x,y)\succ(a,b).$ Now if we have the case $xy=ab,$ $(\log x,\log y)\succ(\log a,\log b)$ and then by Karamata for $e^{x},$ the inequality $x^{p}+y^{p}\geq a^{p}+b^{p}$ is true for positive $p$ and false for negative $p.$


calculus
derivative
function
calculus computations

Let f(x, y) = (x^2)*arctan(y/x) - (y^2)*arctan(x/y) if xy does not equal to 0. And let it be 0 if xy = 0.

a.) Is the function f continuous at (0, 0)?

b.) Compute d f(x, y)/ dx and d f(x, y)/dy, if they exist.

c.) Compute d(df/dx)/dy at (0, 0) and d(df/dy)/dx at (0, 0) if they exist.

Note: d is the partial derivative.

$ |\arctan \theta | \le \frac{\pi}{2}$

so $ |f(x,y)|\le \frac{\pi}{2} |x^2\plus{}y^2| \to 0$ as $ (x,y) \to 0$

$ f_x\equal{}2x\arctan \frac{y}{x} \plus{}x^2 \frac{\minus{}y}{x^2\plus{}y^2}\minus{}y^2\frac{y}{y^2\plus{}x^2}\equal{}2x\arctan \frac{y}{x} \minus{}y$
$ f_y\equal{}x\minus{}2y\arctan \frac{x}{y}$

$ f_{xy}\equal{}2x\frac{x}{x^2\plus{}y^2}\minus{}1\equal{}\frac{x^2\minus{}y^2}{x^2\plus{}y^2}$
$ f_{yx}\equal{}1\minus{}2y\frac{y}{x^2\plus{}y^2}\equal{}\frac{x^2\minus{}y^2}{x^2\plus{}y^2}$

as $ y\equal{}mx \to 0$ for some fixed real m,
$ f_{xy} \to \frac{1\minus{}m^2}{1\plus{}m^2}$ therefore this function is not continuous at (0,0) so the mixed partial doesnt truely exist there


linear algebra
matrix
linear algebra unsolved

Let $ A,B \in M_n(\mathbb{C})$ and $ AB^2\minus{}B^2A\equal{}B$. Prove that $ B^n\equal{}0$.

Already discussed on the forum. For a thorough explanation you can refer to those posts. It suffices to show that $ \forall k\in \mathbb N\ tr(B^k) \equal{} 0$.

[quote="gordon_c"]But there shouldn't be  $ AB \equal{} BA$ to get $ tr (B^k) \equal{} 0 ?$[/quote]
No. Note that $ B$ commutes with every power of $ AB^2\minus{}B^2A$, and it suffices here.

Here we have $[B^2, [B^2, A]] = O$, since $[B^2, A] = B^2A-AB^2 = -B$. Therefore, $[B^2, A] = -B$ is a nilpotent matrix, and so also is B, and Cayley-Hamilton theorem implies $B^n = O$.


Out of nine different pairs of shoes, in how many ways could I choose a right shoe and a left
shoe, which should not form a pair?

[hide="Solution"]First, chose one of the shoes. You have $ 9$ choices. Then, chose the other, for which you have $ 8$ choices. $ 9\cdot8=72$.[/hide]

[quote="i_like_pie"][quote="buzzer11"][hide]I say that we use complementary counting.

How many ways can we pick two shoes out of the 18? $ \binom{18}{2}$.

There are $ 9$ ways to pick 2 correct shoes.

Thus we get $ \binom{18}{2}-9 = \frac{18 \cdot 17}{2}-9 = \boxed{137}$.
[/hide][/quote]
You didn't account for right and left, but the idea of complementary counting is nice.[/quote]

haha oops i misunderstood the question. i thought it was saying that you would be able to pick any two shoes...


Prove that if $ a,b,c$ are positive real numbers, then:

$ \frac{9}{a\plus{}b\plus{}c} \le 2 \left( \frac{1}{a\plus{}b}\plus{}\frac{1}{b\plus{}c}\plus{}\frac{1}{c\plus{}a} \right) \le \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}.$

[quote="moldovan"]Prove that if $ a,b,c$ are positive real numbers, then:

$ \frac {9}{a \plus{} b \plus{} c} \le 2 \left( \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \right) \le \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}.$[/quote]

the left inequality $ \Leftrightarrow 9 \le [(a\plus{}b)\plus{}(b\plus{}c)\plus{}(c\plus{}a)]\left( \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \right)$ which is true by Cauchy-Schwarz.

$ 4 \equal{}(1\plus{}1)^2 \le (a\plus{}b)(\frac{1}{a}\plus{}\frac{1}{b}) \Leftrightarrow 2(\frac{1}{a\plus{}b}) \le \frac{1}{2}(\frac{1}{a}\plus{}\frac{1}{b})$

Similarly $ 2(\frac{1}{b\plus{}c}) \le \frac{1}{2}(\frac{1}{b}\plus{}\frac{1}{c})$ and $ 2(\frac{1}{c\plus{}a}) \le \frac{1}{2}(\frac{1}{c}\plus{}\frac{1}{a})$

Add up these $ 3$ inequalities we get the right inequality.

hello, the left hande-side of your inequality is equivalent to
$ \frac{9}{2}\le \frac{a\plus{}b\plus{}c}{a\plus{}b}\plus{}\frac{a\plus{}c}{b\plus{}c}\plus{}\frac{a\plus{}b\plus{}c}{c\plus{}a}$
$ \frac{3}{2}\le\frac{c}{a\plus{}b}\plus{}\frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}$
setting
$ a\plus{}b\equal{}x$
$ b\plus{}c\equal{}y$
$ c\plus{}a\equal{}z$
then we get
$ 6\le\frac{x}{y}\plus{}\frac{y}{x}\plus{}\frac{x}{z}\plus{}\frac{z}{x}\plus{}\frac{y}{z}\plus{}\frac{z}{y}$
this is true because of $ \frac{x}{y}\plus{}\frac{y}{x}\geq2$ for all $ x,y>0$ etc.
Sonnhard.

hello, to prove the right hand-side of our inequality we can use the inequality $ \frac {2}{x \plus{} y}\le \frac {1}{2}\left(\frac {1}{x} \plus{} \frac {1}{y}\right)$.
Sonnhard.

[quote="moldovan"]Prove that if $ a,b,c$ are positive real numbers, then:

$ \frac {9}{a \plus{} b \plus{} c} \le 2 \left( \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \right) \le \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}.$[/quote]
Use Am-Gm inequality:
$ 2(a \plus{} b \plus{} c) \equal{} (a \plus{} b) \plus{} (b \plus{} c) \plus{} (c \plus{} a) \ge 3.\sqrt [3]{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$
$ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \ge \frac {3}{\sqrt [3]{(a \plus{} b)(b \plus{} c)(c \plus{} a)}}$
So $ 2(a \plus{} b \plus{} c).(\frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a}) \ge 9$
$ \Leftrightarrow 2(\frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a}) \ge \frac {9}{a \plus{} b \plus{} c}$
==========================================
Use AM-GM:$ \frac {1}{a} \plus{} \frac {1}{b} \ge \frac {2}{\sqrt {ab}}$ and $ a \plus{} b \ge 2\sqrt {ab}$
$ \Rightarrow (\frac {1}{a} \plus{} \frac {1}{b})(a \plus{} b) \ge 4 \Rightarrow \frac {1}{a} \plus{} \frac {1}{b} \ge \frac {4}{a \plus{} b}$
So $ 2(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}) \ge 4 (\frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a})$

Who can tell me  how to  solve the 2009  IRAN   TST  DAY   ONE  QUESTION 3 (an inequality)?


Atkins

At TJ, there is a rule, though:
You are not allowed to even participate in bio olympiad, USACO, USAMO, or any other olympiad (yeah "olympiad" so science bowl would be ok but USA junior math olympiad wouldn't) if you participate in chem olympiad.

I was not allowed to take the USAMO last year despite qualifying.

Also, you must qualify for the 2 slots in our region and I know at least three people who were very, very good last year excluding the camp qualifier who will probably participate again this year and are taking orgo.

If you just want to know for AP chem, I would actually use the AP chem textbook or Cliff's (Mr. Kauffman recommended Cliff's) and Atkins has some stuff not useful for AP chem.

wow...I never knew that.

Why is there a rule like that???

The USAChO website just says.

A student who participates in the Physics Olympiad study camp, the Mathematics Olympiad study camp, or the Biology Olympiad study camp is not eligible to participate in the Chemistry Olympiad study camp during the same summer.

The rule was made years ago by a teacher at TJ who ran chemistry olympiad for TJ students. I had to sign a letter saying that I would not participate in other olympiads. This rule is just for our school.

This is my guess why the rule was made at TJ: we can only send 2 people to take the chemistry olympiad test.

After the list of USAMO qualifiers was emailed, which included my name, Mr. Kauffman stopped me in the hall and asked me to choose. I know that you do bio olympiad and obviously math competitions so I just wanted to warn you.


Good job everyone!

However, I am now posting a new rule:
You cannot answer more than one question.

Now, here comes the next question!
[b]MATHCOUNTS 2008 Chapter Sprint #2[/b]
Each of the five numbers 1, 5, 9, 13 and 17 is placed in one of the five squares so that the sum of the three numbers in the horizontal row equals the sum of the three numbers in the vertical column. What is the largest possible sum of the three numbers in the horizontal row?
I am sorry; I cannot paste the picture.
However, it is exactly like this:
$ \plus{}$
1 square on the top, bottom, left, right, and middle.

Isn't this basically a marathon?

And [i]if[/i] you are going to do this, shouldn't it be done in one topic instead of 3-5?

BTW: The answer is 31.

[quote="akhil0422"]It's not exactly a marathon because I'm going to post only about 6-7 problems at most.
I would've stopped if it the problems have proven to be too difficult.
But reflecting on it, it would've cluttered up the forum. So I'll stop posting these from now on.

Thank you for telling me this; I might've gotten in trouble with a moderator or and administrator![/quote]

Maybe a handful on one topic...but 99% of the people who post in this forum are from the MathCounts forum. :P


number theory
relatively prime
number theory unsolved

Find all prime numbers $ p,q$ less than 2005 and such that $ q|p^2 \plus{} 4$, $ p|q^2 \plus{} 4$.

[quote="The QuattoMaster 6000"][hide="Solution"]
If $ p$ or $ q$ is $ 2$, then we may let $ p \equal{} 2$, so $ q|8$, so $ q \equal{} 2$.  This means that $ (p, q) \equal{} 2$.  Otherwise, let $ p$ and $ q$ be odd primes.  We have that $ q|(p^2 \plus{} 4)$, so $ p$ and $ q$ may not be equal.  Therefore, $ q|(p^2 \plus{} 4)$ and $ p|(q^2 \plus{} 4)$ if and only if $ \frac {p^2q^2 \plus{} 4p^2 \plus{} 4q^2 \plus{} 16}{pq}$ is an integer.  This is an integer if and only if $ \frac {4p^2 \plus{} 4q^2 \plus{} 16}{pq}$ is an integer.  Since $ p$ and $ q$ are relatively prime with $ 4$, we see that this is an integer if and only if $ \frac {p^2 \plus{} q^2 \plus{} 4}{pq}$ is an integer.  Let this value be $ k$.

Then, we have that $ p^2 \minus{} p(kq) \plus{} (q^2 \plus{} 4) \equal{} 0$.  In general, let us consider the equation $ a^2 \minus{} a(bk) \plus{} (b^2 \plus{} 4) \equal{} 0$.  Let $ (a, b)$ be the the solution to this equation in positive integers with the lowest possible $ a \plus{} b$, and without loss of generality, say that $ a\le b$.  Now, if $ (a, b)$ is a solution, then $ (\frac {b^2 \plus{} 4}{a}, b)$ is a solution, so $ (b, \frac {b^2 \plus{} 4}{a})$ must be a solution.  Thus, $ b \plus{} \frac {b^2 \plus{} 4}{a} \ge a \plus{} b$, so $ a^2 \minus{} b^2 \le 4\implies (a \plus{} b)(a \minus{} b) \le 4$.  Now, this means that $ a \equal{} b$ or $ (a, b)$ is $ (1, 2)$.  If $ a \equal{} b$, then we have that $ (k \minus{} 2)a^2 \equal{} 4$, so $ (a, b)$ is $ (1,1)$ or $ (2,2)$.  We find $ k$ for each of these cases.

If $ (a, b) \equal{} (1, 2)$, then $ k$ is not an integer.  If $ (a, b) \equal{} (1,1)$, then $ k \equal{} 6$.  If $ (a, b) \equal{} (2, 2)$, then $ k \equal{} 3$.  If $ k \equal{} 3$, then we have that $ \frac {p^2 \plus{} q^2 \plus{} 4}{pq}$ is odd, which is impossible since the numerator is even and the denominator is odd.  It follows that $ k \equal{} 6$.  Thus, $ p^2 \minus{} 6pq \plus{} (q^2 \plus{} 4) \equal{} 0$.  The discriminant of this is $ 16(2q^2 \minus{} 1)$, so $ 2q^2 \minus{} 1$ is a perfect square.  We now consider the Pell's Equation $ 2m^2 \minus{} n^2 \equal{} 1$.   We can compute the primes $ q$'s that work for the the Pell's Equation under $ 2005$ (they are solutions to the recurrence $ a_{n \plus{} 1} \equal{} 6a_n \minus{} a_{n \minus{} 1}$ with $ a_1 \equal{} 1$ and $ a_2 \equal{} 5$), which are $ 5$ and $ 29$.  Plugging this in, we get that $ p^2 \minus{} 30p \plus{} 29 \equal{} 0$, so $ p \equal{} 29$ or $ p \equal{} 1$ (but $ p \equal{} 1$ is impossible since $ 1$ is not prime, so $ p \equal{} 29$).  If $ q \equal{} 29$, then $ p^2 \minus{} 174p \plus{} 845 \equal{} 0$, so $ (p \minus{} 169)(p \minus{} 5) \equal{} 0$, so $ p \equal{} 5$ or $ p \equal{} 169$ (but $ p \equal{} 169$ is impossible as $ 169$ is composite).  Thus, the only solutions are $ (p, q) \equal{} (5, 29), (29, 5), (2,2)$.   [/hide][/quote]

But how can that be? We say that $ a\le b$. So, $ (a \plus{} b)(a \minus{} b) \le 4$ is obvious.

[quote="pluricomplex"]@Stephen: He means $ a\geq b$. It seems to be a typo.[/quote]

Then how can $ (a, b)\equal{}(1, 2)$?

http://www.mathlinks.ro/viewtopic.php?p=1353597#1353597

essentially the same problem, but a bit more complicated.

[quote="Stephen"]
Then how can $ (a, b) \equal{} (1, 2)$?[/quote]
Sorry, I meant $ (a, b) \equal{} (2,1)$ and $ a\ge b$.

guys please help me i want to understand this part but i can't.PLEASE HELP ME!!!!!! :( :(:( :( :(
" We now consider the Pell's Equation $ 2m^2 - n^2 = 1$. We can compute the primes $ q$'s that work for the the Pell's Equation under $ 2005$ (they are solutions to the recurrence $ a_{n + 1} = 6a_n - a_{n - 1}$ with $ a_1 = 1$ and $ a_2 = 5$), which are $ 5$ and $ 29$. "

[quote=fattypiggy123]See [url="http://www.imomath.com/index.php?options=615"] this [/url] for a detailed explanation on Pell's equation.[/quote]

Thanks very much

[quote=Anar24]i have one question as you found that k equals to 6,can't you apply Vieta jumping after it? I mean that can't you solve the equation $ p^2 - 6pq + (q^2 + 4) = 0$ by Vieta jumping and also when solving that by Vieta,you will find that there is no need to have p,q are less than 2005[/quote]

Actually you can. Just need to revers the step of vieta jumping I mean first consider all pairs that with one jump they reach the root solution $(1,1)$ then those who reach $(2,2)$ in one jump and so on. One can observe that every pair reaches to one of these root answers after some jumps so continuing this processes will give us all solutions. But obviously this is possible only if we have some bound on the solution pairs.


Two people are in a best-of-11 match.  Person A has a 5/7 chance of winning, and person B has a 2/7 chance of winning.  Given that person A already won 1 game, what is the chance that person A will win the match?

A somewhat related question is this:
A person starts out walking on a 2-D grid, each time flipping two coins to decide which way to go (left, right, up, or down with equal probability).  What is the chance that the person gets to the line x=-3 before he/she gets to the line y=5?

I think I got it, but I did it differently, and I don't know if we got the same thing.  Here's my reasoning (on the best-of-11 one):
It is essentially a problem of playing 10 games (since A already won one), and seeing the probability that A wins 5.  Therefore, that would be
$(5/7)^5*(10C5)$
The $(5/7)^5$ is the 5 games won and the 10C5 is 10!/(5!5!) since I don't know the code for combinations.

im still working on a nice clean solution but i dont think wombate's solution is accurate:

there is actually a very small chance that 10 more games will be played, since it is possible for either A or B to win before that.

more specifically, i doubt that (10C5) is correct because some combinations, such as winning the last 5 of ten games, is much less likely due to the fact that player B must win 5 games in a row beforehand...

or maybe all these factors cancel out and you DO have the right answer, in which case i apologize in advance.

If they quit after winning, then I think my solution is incorrect, but the problem didn't specify that. If that is so, this problem is very difficult - you half to account for when or when not to multiply by 2/7, the probability that B loses. One scenario: ABBAAAAA = (5/7)^5 * (2/7)^2, but AAAAAA, which is still 6 A wins, is only (5/7)^5, which makes the problem very tricky.

I realized later when thinking about simpler cases that my solution is incorrect.  I think you do have to divide it into cases.  I was hoping that wouldn't be the case.

I don't think it matters if they quit after winning, since if you just consider 11 games, only one person CAN win.
By the way, white horse knight, your last post's formula didn't show up for me. It said "[unparseable or potentially dangerous latex formula]".

yeah i think that's essentially what i have, xept i drew a huge probability tree so that branches can end at diff places when one person wins before finishing 10 games... and if i multiply and add the fractions i'll probably get something like that...

now, would anyone like to evaluate that? cuz the whole point of a problem is to get a useful answer.

The probability that he loses is a little easier to find cuz the 2/7 guy has to win at least six games.  I am going to assume that they keep playing to get a final score even after someone wins.  If you add the independent  probabilities that the 2/7 guy wins 10, 9, 8, 7, and 6 games I think you can figure out the probability that he wins.  for 10, (2/7)^10 for 9, (2/7)^9*(5/7)*10C9 and so on. Unequal probabilities suck.


So if you're like me, you already looked at the solutions pamphlet and calculated your score.

What was your score? :D

Note: This is only for the AMC 10; the AMC 12 results are in another thread.

Note #2: Gahh. I accidentally deleted the poll. I put a new one up.

OK, I uber phailed the poll.... therefore, no poll.

Edit: Uber poll phailure!!!!!!

See the other thread I made--it has both AMC 10/12 A discussion.

As for scores, 129. :(


If you have a negative number, what exponents can you raise it to and still get a real number?

(I mean non-integer exponents, and especially fractional/decimal exponents.)

Thanks.

[hide="answer"]A negative number raised to a power [u]can be[/u] a real number if and only if the exponent is a rational number that can be written with an odd denominator.

For example: $ ( \minus{} 8)^{2/3}$ can equal 4, [i]or[/i] it can equal $ \minus{} 2 \pm 2\sqrt {3}i$.

An interesting exercise: try to sketch a graph of $ y \equal{} x^x$ for positive and negative values of $ x$.[/hide]

How did you get the $ \minus{}2\pm2\sqrt{3}i$?

Also, when you have $ (\minus{}8)^{2/3}$, do you take the cube root or square it first?

What if you had $ (\minus{}4)^{2/4}$?
If you squared it first, you would get $ (16)^{1/4} \equal{} 2$. But if you took the fourth root first, you wouldn't get a real number.
Or do you have to reduce it to $ (\minus{}4)^{1/2}$ first?

[quote="tornado.adv4"]How did you get the $ \minus{} 2\pm2\sqrt {3}i$?

Also, when you have $ ( \minus{} 8)^{2/3}$, do you take the cube root or square it first?

What if you had $ ( \minus{} 4)^{2/4}$?
If you squared it first, you would get $ (16)^{1/4} \equal{} 2$. But if you took the fourth root first, you wouldn't get a real number.
Or do you have to reduce it to $ ( \minus{} 4)^{1/2}$ first?[/quote]
$ ( \minus{} 4)^\frac {2}{4} \equal{} ( \minus{} 4)^\frac {1}{2} \equal{} 2i$ :wink:

And

$ ( \minus{} 8)^\frac {2}{3} \equal{} 4( \minus{} 1)^\frac {2}{3}$ and use roots of unity to get $ \minus{} 2  \pm 2\sqrt {3}i$

[quote="tornado.adv4"]

Also, when you have $ ( \minus{} 8)^{2/3}$, do you take the cube root or square it first?

What if you had $ ( \minus{} 4)^{2/4}$?

If you squared it first, you would get $ (16)^{1/4} \equal{} 2$. But if you took the fourth root first, you wouldn't get a real number.
Or do you have to reduce it to $ ( \minus{} 4)^{1/2}$ first?[/quote]

=======================================================

In response to the real values part and the questions in the quote box
above...

When you have $ (\minus{}8)^{2/3}$, you take the cube root first, and then
you square that result:

$ [({\minus{}8})^{1/3}]^2 \equal{} (\minus{}2)^2 \equal{} 4$.

When you have $ (\minus{}4)^{2/4}$, you can rewrite as $ (\minus{}4)^{1/2}$.

In either form, you take the root first, and then raise it to the numerator
of the exponent:

The second form would be $ (\sqrt{\minus{}4})^1 \equal{} 2i$.

The discrepancies in answers are all due to the fact that root extraction has multiple answers in the complex space.

Consider $ 1\equal{}\sqrt{(\minus{}1)(\minus{}1)}\ne\sqrt{\minus{}1}\sqrt{\minus{}1}\equal{}\minus{}1$. Contradiction? Strange? This simply shows an identity may hold for positive real numbers but it may not hold for negative numbers.

So, an attempt to evaluate $ (\minus{}4)^{\frac24}$ is not possible in the real space and it yields 2 complex roots. Why 2 but not 4? You can simplify the exponents but the order of operations when dealing with negative base does not 'commute'.
$ (\minus{}4)^\frac24\equal{}(\minus{}4)^\frac12\equal{}2\cdot\text{cis}\left(\frac{\pi\plus{}2\pi k}2\right)_{k\equal{}0,1}\equal{}\pm2i$


[b]How high can the highest mountain on earth be?[/b]

[hide]Dont shreak!! YOU ARE NOT GIVEN ANY OTHER INFO . You will have to use a lot of approximations[/hide]

can you please give us some hints?  :spider:

[hide]Thermodynamics and latent heat[/hide]

[hide]And Pressure !!!
Sill nOthing !!?![/hide]

[size=150][b]What about the highest mountain on MARS ??!!!??[/b][/size]

Well, my qualitative solution is easy. If the mountain is too high, its base cannot support it and it collapses. Cannot support means that there is too much pressure on the bottom and it starts to disintegrate. The mountains however tend to be quite wide compared to the height, so the problem is melting of the rock under pressure.

I don't have access to specific latent heat constants od rock and the melting point dependencies on pressure etc., so I won't give an answer better than a guess ... how about 25 km?

kubus, I just know for sure now that you have the book or have done all its questions before.

Isn't that true!!!  :roll:  :thumbup:


Prove that $\sum_{n \in \mathbb{Z}} (-1)^n x^{n^2} = \prod_{n \geq 1} \frac{1-x^n}{1+x^n}$ for $|x|<1$.

It's an immediate consequence of the [url=http://en.wikipedia.org/wiki/Jacobi_triple_product]Jacobi triple product identity[/url]. Just take $y=i$.


geometry
rectangle
ratio
trigonometry
projective geometry
geometry proposed

A tirangle $\bigtriangleup ABC$ is given and let $BCAcAb,$ $ACBcBa,$ $ABCbCa$ be, three similar rectangles outwardly to it, erected on $BC,$ $AC,$ $AB,$ respectively. We denote the points $A'\equiv BBc\cap CCb,$ $A''\equiv BBa\cap CCa$ and  $B'\equiv AAc\cap CCa,$ $B''\equiv AAb\cap CCb$ and  $C'\equiv AAb \cap BBa,$ $C''\equiv AAc\cap BBc.$ Prove that the segment lines $A'A'',$ $B'B'',$ $C'C'',$ are concurrent at one point.

Kostas vittas.

Nice problem, vittasko!

My first proof is based on some projective remarks:

One knows that affine transformations preserve collinearity of points, parallelness of lines, concurrence of lines, the property of a point to be a centroid of others, ratios of segments and areas, and convexity.

Thus, we can perform an affine transformation that maps the initial triangle in an equilateral one.

(In fact, any non-collinear three points can be mapped to any other three non-collinear points by an unique affine transformation)

Thus, the problem becomes more like trivial now.

The second proof I had in mind was based on the following [i]lemma[/i]:

[i]Let $ABC$ be a triangle and the lines $d_{1}$, $d_{2}$, $d_{3}$, such that $\{M,N\}\in d_{1}\cap \{AB,AC\}$, $\{P,Q\}\in d_{2}\cap \{BC,BA\}$, $\{R,S\}\in d_{3}\cap \{CA,CB\}$. Denote with $m$, $n$, $p$, $q$, $r$, $s$ the ratios $\frac{MB}{MA}$, $\frac{NC}{NA}$, $\frac{PC}{PB}$, $\frac{QA}{QB}$, $\frac{RA}{RC}$ respective $\frac{SB}{SC}$, where the segments are oriented. Thus, the lines $d_{1}$, $d_{2}$ and $d_{3}$ are concurrent, if and only if, $1+mpr+nqs = mq+ps+rn$.
[/i]

Now the proof for this "atomic bomb" is very computational as it follows:

From the ratio characterizations, we have that:

$1\blacktriangleright$ $(1-m)\overrightarrow{M}= \overrightarrow{B}-m\overrightarrow{A}$

$2\blacktriangleright$ $(1-n)\overrightarrow{N}= \overrightarrow{C}-n\overrightarrow{A}$

$3\blacktriangleright$ $(1-p)\overrightarrow{P}=\overrightarrow{C}-p\overrightarrow{B}$

$4\blacktriangleright$ $(1-q)\overrightarrow{Q}=\overrightarrow{A}-q\overrightarrow{B}$

$5\blacktriangleright$ $(1-r)\overrightarrow{R}=\overrightarrow{A}-r\overrightarrow{C}$

$6\blacktriangleright$ $(1-s)\overrightarrow{S}=\overrightarrow{B}-s\overrightarrow{C}$.

Denote $V \in AB \cap d_{3}$, $Y \in d_{1}\cap d_{3}$, $Z \in d_{1}\cap d_{2}$. Thus, we must prove that $Y \equiv Z$, i.e $\frac{\overrightarrow{YM}}{\overrightarrow{YN}}=\frac{\overrightarrow{ZM}}{\overrightarrow{ZN}}$.

From Menelaus, $V \in RS \iff \frac{VB}{VA}\cdot \frac{RA}{RC}\cdot \frac{SC}{SB}=1 \Rightarrow \frac{\overrightarrow{VB}}{\overrightarrow{VA}}=\frac{s}{r}$. $(1)$

Now because $\overrightarrow{R}=\frac{\overrightarrow{A}-r\overrightarrow{C}}{1-r}$ and $\overrightarrow{N}= \frac{\overrightarrow{C}-r\overrightarrow{A}}{1-n}$, we obtain that $\frac{\overrightarrow{RA}}{\overrightarrow{RN}}=\frac{r(1-n)}{1-nr}$. $(2)$

From $(1)$ we have that $\overrightarrow{V}=\frac{s\overrightarrow{A}-r\overrightarrow{B}}{s-r}$, and because $\overrightarrow{M}=\frac{\overrightarrow{B}-m\overrightarrow{A}}{1-m}$ $(3)$, we have that $\frac{\overrightarrow{VM}}{\overrightarrow{VA}}=\frac{s-mr}{r(1-m)}$.

Now similar, from Menelaus,

$V \in YR \iff \frac{YM}{YN}= \frac{RA}{RN}\cdot \frac{VA}{VM}$, and from $(2)$, $(3)$, we obtain that $\frac{\overrightarrow{YM}}{\overrightarrow{YN}}=\frac{(1-n)(s-mr)}{(1-m)(1-nr)}$. $(4)$.

In a similar way, obtain $\frac{\overrightarrow{ZM}}{\overrightarrow{ZN}}=\frac{(1-n)(1-mq)}{(1-m)(p-nq)}$. $(5)$.

Therefore, by $(4)$ and $(5)$, $d_{1}$, $d_{2}$, $d_{3}$ are concurrent, i.e
 $\frac{\overrightarrow{YM}}{\overrightarrow{YN}}=\frac{\overrightarrow{ZM}}{\overrightarrow{ZN}}\iff \frac{(1-n)(s-mr)}{(1-m)(1-nr)}=\frac{(1-n)(1-mq)}{(1-m)(p-nq)}$

$\iff (1-mq)(1-nr)=(p-nq)(s-mr) \iff 1+mpr+nqs=mq+ps+rn$, and the lemma is proved.

Returning to the problem, denote for example $T_{a_{1}}\in A'A'' \cap C_{a}C_{b}$.

By the sin law,
$\frac{A'C}{A'C_{b}}=\frac{BC}{BC_{b}}\cdot \frac{\sin{B_{c}BC}}{\sin{B_{c}BC_{b}}}$, and similar for the other ratios of the same type.

Now by Menelaus theorem in $\triangle{CC_{a}C_{b}}$ for $A'-A''-T_{a_{1}}$, we have that

$\frac{T_{a_{1}}C_{b}}{T_{a_{1}}C_{a}}= \cos{x}\cdot \frac{\sin{B_{c}BC_{b}}}{\sin{B_{c}BC}}\cdot \frac{\sin{B_{a}BC}}{\sin{B_{a}BC_{a}}}$.

So, by replacing the line $A'A''$ with $d_{1}$ (from the lemma), we have that $m=\frac{T_{a_{1}}C_{b}}{T_{a_{1}}C_{a}}$.

Thus, by optaining similar the $n,p,q,r,s$, and replacing them into the lemma condition and by simple angle manipulations, the problem is solved.

[i]Remark[/i] : The [i]lemma[/i] presented above could also solve the beautiful Serbian problem, from [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=863788#863788[/url]

Thank you dear [b][size=100]pohoatza[/size][/b], for your interest and quickly answer, with a solution based on an interesting and new to me Lemma.

Because of I am not a specialist like you, the solution I have in mind is a synthetic way based on a Lemma easy to prove by [b][size=100]Desarques’s theorem.[/size][/b]

So, we begin from the result that the segment lines $AA',$ $BB',$ $CC'$ are concurrent at one point, the orthocenter of $\bigtriangleup ABC$ $($ because of, based on [b][size=100]jacobi’s theorem,[/size][/b] it is easy to prove that they are its altitudes.  Lately, [b][size=100]Virgil Nicula[/size][/b] posted a relative problem at http://www.mathlinks.ro/Forum/viewtopic.php?t=154845  $).$

 After that, we can solve the problem as a direct application of a Lemma which I like to say as [b][size=100] ''The second theorem of concurrent segments''[/size][/b], based on which we can also prove that the segment lines $AA'',$ $BB'',$ $CC''$ are concurrent at one point, so be it, $Q$ and also that the segment line $PQ,$ passes through the orthocenter of $\bigtriangleup ABC.$

I will sent you as soon is possible, the complete solution I have in mind, before its posting in the forum ( after two months ), as a minimum dept of honour.

Thank you again, Kostas Vittas.

PS. Have you in mind, a synthetic proof of the Lemma you used?

Many thanks for your appreciations, dear Vittasko. I was almost sure you had a crushing synthetic proof with Desargues for this very nice problem.

Actually, I don't think the relation, in the lemma I used in proof II, could be obtained without those Menelaus theorems, the construction beeing to "arbitrarily", but this is only my opinion and I really haven't look for anything synthetic by now.

Thank you again.

As you said, dear [b]Kostas Vittas[/b], we have $ AA'$, $ BB'$, $ CC'$ are concurrent at the orthocentre $ H$ of triangle $ ABC$. (I'm sure that the result $ AA'\perp BC$ was posted in Mathlinks Forums, but I'm sorry, I don't remember the link)

Now, because of the lines $ A'H$, $ B'C''$ and $ B''C'$ are concurrent at the point $ A$, by [b]Desarques' theorem[/b], we have that three point $ S\equal{} C'B'\cap B''C''$, $ B\equal{}B'H\cap C''A'$, $ C\equal{}HC'\cap A'B''$ are collinear.
From this collinearity, and notice that $ S\equal{}B'C'\cap B''C''$, $ B\equal{}C'A''\cap C''A'$, $ C\equal{}A''B'\cap A'B''$, again by [b]Desarques' theorem[/b], we conclude that $ A'A''$, $ B'B''$ and $ C'C''$ are concurrent.

Thank you dear [b]Kostas[/b] for this very nice and hard problem.
Best regards.

Here is the topic, what I mentioned http://www.mathlinks.ro/Forum/viewtopic.php?t=39101.

Dear [b][size=100]Pohoatza, April[/size][/b] and all my friends.

I would like to present the solution I have in mind of this problem, based on the useful, I think, [b][size=100]theorem 2 of concurrent segments[/size][/b].

[quote="vittasko"][b][size=100][color=DarkBlue]A tirangle $ \bigtriangleup ABC$ is given and let $ BCAcAb,$ $ ACBcBa,$ $ ABCbCa$ be, three similar rectangles outwardly to it, erected on $ BC,$ $ AC,$ $ AB,$ respectively. We denote the points $ A'\equiv BBc\cap CCb,$ $ A''\equiv BBa\cap CCa$ and  $ B'\equiv AAc\cap CCa,$ $ B''\equiv AAb\cap CCb$ and  $ C'\equiv AAb \cap BBa,$ $ C''\equiv AAc\cap BBc.$ Prove that the line segments $ A'A'',$ $ B'B'',$ $ C'C'',$ are concurrent at one point.[/color][/size][/b][/quote]
[b][size=100]PROOF.[/size][/b] – It is easy to show that the line segments $ BAb,$ $ BCb,$ are isogonal conjugates with respect to the angle $ \angle B.$ Also the line segments $ CAc,$ $ CBc,$ are isogonal conjugates wrt $ \angle C$ and of course the line segments $ ABc,$ $ ABb,$ are isogonal conjugates wrt $ \angle A.$

We denote as $ A_{\infty}$ the intersection point of the parallel line segments $ BAb,$ $ CAc,$ at infinity and so, because of the [b][size=100]Jacobi’s theorem,[/size][/b] we conclude the concurrency of the line segments $ AA_{\infty},$ $ BBc,$ $ CCb.$

That is we have that the $ AA'\perp BC$ and similarly $ BB'\perp AC$ and $ CC'\perp AB.$

So, we have the concurrency of $ AA',$ $ BB',$ $ CC',$ as the altitudes of the given triangle $ \bigtriangleup ABC.$

Hence, based on the below [b][size=100]theorem,[/size][/b] we conclude that the line segments $ A'A'',$ $ B'B'',$ $ C'C'',$ are concurrent at one point and the proof is completed.

[b][size=100]REMARK.[/size][/b] - Based on the below [b][size=100]theorem,[/size][/b] we also conclude that the line segments $ AA'',$ $ BB'',$ $ CC'',$ are concurrent at one point so be it $ Q$ and also that the line segment $ PQ,$ passes through the orthocenter $ H$ of $ \bigtriangleup ABC$ $ ($ the concurrency point of $ AA',$ $ BB',$ $ CC',$ $ ).$

[b][size=100][color=DarkBlue]THEOREM 2 OF CONCURRENT SEGMENTS.

Let $ AA',$ $ BB',$ $ CC'$ be, three concurrent segments at a point $ P.$ We denote  as $ A'',$ the intersection point of $ BC',$ $ B'C$ and as $ B'',$ the intersection point of $ AC',$ $ A'C$ and as $ C'',$ the intersection point of $ AB',$ $ A'B.$ Prove that :

a)  The line segments $ AA'',$ $ BB'',$ $ CC'',$ are concurrent at a point, so be it, $ M.$
b)  The line segments $ A'A'',$ $ B'B'',$ $ C'C'',$ are concurrent at a point, so be it, $ N.$
c)   The line connecting the points $ M,$ $ N,$ passes through the point $ P.$[/color][/size][/b]

We say three segments as concurrent, when their line segments are concurrent. Three parallel segments are concurrent at infinity.

[b][size=100]1.  PROOF.[/size][/b] - $ ($ In my drawing, [b][size=100]schema t=155218(1)[/size][/b], the points $ A,$ $ B,$ $ C,$ have been considered as vertices of a triangle $ \bigtriangleup ABC$ and the point $ P$ inwardly to it. $ BC \equal{} 16.0,$ $ AC \equal{} 14.0,$ $ AB \equal{} 11.5,$ $ PA \equal{} 6.5,$ $ PB \equal{} 7.2$ and the points $ A',$ $ B',$ $ C',$ on the extensions of $ AP,$ $ BP,$ $ CP$ respectively, such that $ PA' \equal{} 1.5,$ $ PB' \equal{} 3.8,$ $ PC' \equal{} 2.6$ $ ).$

Because of the segment lines $ A'P,$ $ B'C'',$ $ B''C',$ are concurrent at the point $ A,$ by [b][size=100]Desarques theorem,[/size][/b] we have that the triangles $ \bigtriangleup A'B''C'',$ $ \bigtriangleup PC'B',$ are perspective. Hence the points $ S\equiv B'C'\cap B''C'',$ $ B,$ $ C,$ are collinear.

Let $ M$ be, the intersection point of $ BB'',$ $ CC''.$ Because of the points $ S,$ $ B,$ $ C$ are collinear, we may also conclude that the triangles $ MB''C'',$ $ A''C'B',$ are perspective. Hence the line segments $ B''C',$ $ C''B',$ $ MA'',$ are concurrent and so, we have that the line segment $ MA'',$ passes through the point $ A.$  That is, the line segments $ AA'',$ $ BB'',$ $ CC'',$ are concurrent at the point $ M$ and the proof of proposition [b][size=100](a),[/size][/b] is completed.

Because of the points $ S,$ $ B,$ $ C,$ are collinear, we may also conclude that the triangles $ \bigtriangleup A'B''C'',$ $ \bigtriangleup A''B'C',$ are perspective. Hence the line segments $ A'A'',$ $ B'B'',$ $ C'C'',$ are concurrent at one point, so be it $ N$ and the proof of proposition [b][size=100](b),[/b][/size] is completed.

Because of the points $ S,$ $ B,$ $ C,$ are collinear, we may also conclude that the triangles $ \bigtriangleup MB''C'',$ $ \bigtriangleup PB'C',$ are perspective. So, the line segments $ B'B'',$ $ C'C'',$ $ MP,$ are concurrent at point $ N.$ Hence, the line connecting the points $ M,$ $ N,$ passes through the point $ P$ and the proof of proposition [b][size=100](c),[/size][/b] is completed.

$ \bullet$ We use the same words for the proof, in every configuration of the given concurrent segments and their edges, in the plane or in 3D space $ ($ see two alternative configurations in [b][size=100]schemas t=155218(2)[/size][/b] and [b][size=100]t=155218(3)[/size][/b] $ ).$

kostas vittas.

PS. I will post here later, some applications of this useful theorem, with notes for the proofs.

Dear [b]Kostas Vittas[/b],

I noticed that your problem is still true when we erect outside triangle $ ABC$ three similar isosceles trapeziums. Indeed, based on your theorem above (It also was mentioned as [url=http://www.cip.ifi.lmu.de/~grinberg/LamoenCross.zip]The Lamoen Theorem on the Cross-Triangle[/url] by [b]Darij Grinerg[/b], in his website), we only have to prove that lines $ AA_1$, $ BB_1$, $ CC_1$ are concurrent (the name of points, please see the attachment). Applied [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=154396][b]Jacobi theorem[/b][/url] to triangle $ ABC$ with erected triangles $ BCA'$, $ CAE'$, $ ABF$, we deduce $ AA'$, $ BE'$, $ CF$ are concurrent at a point, the point $ A_1$. Similarly, $ B_1\in BB'$ and $ C_1\in CC'$. Now, again applying [b]Jacobi theorem[/b] to triangle $ ABC$ and erected triangles $ BCA'$, $ CAB'$, $ ABC'$, we obtain $ AA'$, $ BB'$, $ CC'$ are concurrent, i.e. $ AA_1$, $ BB_1$, $ CC_1$ share a common point. This completes our solution.

Best regards, April.


$a_{n}$ is a sequence such that $4\cdot a_{n}=a_{2n}$ and $a_{2n}=2\cdot a_{2n-1}+\frac{1}{4}$ for all $n\in\mathbb{N}$. Find the sum $S=a_{1}+a_{2}+...+a_{31}$

[quote="nebula42"][hide]Well, can't we just plug in $n = 1$ into the two original definitions of the recursive sequence to obtain
\[a_{2}= 4a_{1}\]

\[a_{2}= 2a_{1}+\frac{1}{4}\]
And thus $a_{1}= \frac{1}{8}$?
(Or am I missing something important and making a very bad mistake?  :maybe:  :P )[/hide][/quote]

Yeah, you're right.

[quote="nebula42"][hide]Well, can't we just plug in $n = 1$ into the two original definitions of the recursive sequence to obtain
\[a_{2}= 4a_{1}\]

\[a_{2}= 2a_{1}+\frac{1}{4}\]
And thus $a_{1}= \frac{1}{8}$?
(Or am I missing something important and making a very bad mistake?  :maybe:  :P )[/hide][/quote]

:)

[quote="t0rajir0u"]Help!

[hide="Not cool :("] First off, let's establish clearly that

$a_{2n}= 4 \cdot a_{n}\implies a_{2^{k}}= 4^{k}\cdot a_{1}$

Well, notice that $31$ is one less than a power of two.  We hypothesize that that might be important, since the problem strongly suggests powers of two.  To that end:

$S_{k}= a_{1}+a_{2}+...+a_{2^{k}-1}$

Where $S_{1}= a_{1}$.  We want to find a crafty way of getting $S_{k+1}$ from this expression.  The recursions we are given suggest the following:

\begin{eqnarray*}4S_{k}&=& a_{2}+a_{4}+...+a_{2^{k+1}-2}\\ &=& (2a_{1}+\frac{1}{4})+(2a_{3}+\frac{1}{4})+...+(2a_{2^{k+1}-3}+\frac{1}{4}) \\ &=& 2(a_{1}+a_{3}+...+a_{2^{k+1}-3})+\frac{2^{k}-1}{4}\\ \end{eqnarray*}
$a_{1}+a_{3}+...+a_{2^{k+1}-3}= 2S_{k}-\frac{2^{k}-1}{8}$
$a_{2^{k+1}-1}= \frac{a_{2^{k+1}}-\frac{1}{4}}{2}= \frac{4^{k+1}a_{1}-\frac{1}{4}}{2}$

And now for the grand finale!

\begin{eqnarray*}S_{k+1}&=& a_{1}+a_{2}+...+a_{2^{k+1}-1}\\ &=& \left( 2S_{k}-\frac{2^{k}-1}{8}\right)+4S_{k}+a_{2^{k+1}-1}\\ &=& 6S_{k}-\frac{2^{k}-1}{8}+2 \cdot 4^{k}a_{1}-\frac{1}{8}\\ &=& 6S_{k}-\frac{1}{8}2^{k}+2a_{1}\cdot 4^{k}\\ \end{eqnarray*}

This is an ugly recursion, but it's workable.  We want to find constants $C, D$ such that

$\left( S_{k+1}-C \cdot 2^{k+1}-D \cdot a_{1}\cdot 4^{k+1}\right) = 6 \left( S_{k}-C \cdot 2^{k}-D \cdot a_{1}\cdot 4^{k}\right)$

Expanding, our equation for $C$ is

$-4C =-\frac{1}{8}\Leftrightarrow C = 32$

And our equation for $D$ is

$-2D = 2 \Leftrightarrow D =-1$

This is pretty tedious, but let $P_{k}= S_{k}-32 \cdot 2^{k}+a_{1}\cdot 4^{k}$.  Then

$P_{k+1}= 6P_{k}\implies P_{k+1}= 6^{k}P_{1}$

Where $P_{1}= S_{1}-64+4a_{1}= 5a_{1}-64$.  Hence

$S_{k+1}= 6^{k}(5a_{1}-64)+32 \cdot 2^{k+1}-a_{1}\cdot 4^{k+1}$

[i]Yikes![/i]  I seem to have found myself in a situation where $S_{5}$ clearly has an $a_{1}$ term in it.  :(
 [/hide][/quote]

the recursion is easier if you have $S_{n}=\sum_{1}^{2^{n}}a_{k}$

[quote="ch1n353ch3s54a1l"][quote="PenguinIntegral"]I'm not getting how T0r gets that first relation (a_2n=4*a_n). Yes,  I'm stupid.[/quote]

It's given in the problem statement...[/quote]
Sorry, the n's confuse my little brain, but that you for the clarification.

[quote="nebula42"][hide]Well, can't we just plug in $n = 1$ into the two original definitions of the recursive sequence to obtain
\[a_{2}= 4a_{1}\]

\[a_{2}= 2a_{1}+\frac{1}{4}\]
And thus $a_{1}= \frac{1}{8}$?
(Or am I missing something important and making a very bad mistake?  :maybe:  :P )[/hide][/quote]

Oh.

[quote="Altheman"]the recursion is easier if you have $S_{n}=\sum_{1}^{2^{n}}a_{k}$[/quote]

[i]Oh.[/i]

 :oops:

Can someone post a complete solution? I still don't get how that first relation is given in the statment, is it (2n)-1 or 2(n-1)?

Small words and slow explanations, please :blush:

I swear, my browser is selectively rendering latex to make me ask idiot questions involving the problem statment. Who can I contact to fix this?

Note the evidence linked to below.
[url]http://img162.imageshack.us/my.php?image=evidencesv8.png[/url]


topology
advanced fields
advanced fields unsolved

Quick question: so the circle with a point omitted is homeomorphic to the reals. But with the "arc length" metric on the circle, are the two isometric? My gut says no.

They are not isometric: one metric space is complete and another is not.

An aside: a uniform homeomorphism is a map $f$ such that $f$ and $f^{-1}$ are both uniformly continuous. Such maps preserve completeness of spaces, therefore $\mathbb T\setminus\{a\}$  and $\mathbb R$ are not uniformly homeomorphic.

en, also an unbounded space cannot be isometric to a bounded one.  Just assume the converse and take an isometry you can see immediately.


A rectangular garden 30 feet long and 20 feet wide is surrounded by a paved walk with with $w,$ creating a larger rectangle. The area of the walk equals the are of the garden. How wide is the walk ?

I think this is starter level... since we actually got exactly this question in primary school (5th grade)

(20+?)(30+?)=2*20*30
?*?+50*?=600
?=10

Peter

Having questions that are too [i]easy[/i] is rarely a serious problem here.  If you want a harder problem, solve in the case of an equilateral triangle.  Or, if that is too easy, make it any triangle.


geometry
geometric transformation
rotation
linear algebra
matrix
superior algebra
superior algebra solved

compute the limite of An when n-->+oo
An=(Bn)^n where

Bn=
[ 1   k/n]
[-k/n  1]

i don't know if I understood the problem very well but here is what I think that we can do
we know that
[cosA sin A]^n       [cos(nA) sin(nA)]
[-sin A cos A]      = [-sin(nA) cos(nA)]
and we also know that for x->0: sinx->x;
now I think that we can write lim(n->+inf)An=

                  [cos (1/n)    sin(k/n)]^n
lim(n->inf)  [-sin(k/n)    cos(1/n)]
=[cos 1  sin(k)] = [0  sin(k)]
  [-sin(k) cos1 ]    [-sin(k) 0]

Fratila Dragos

Hello Fratila Dragos,

Welcome to this forum :) , i think you wanted to write cos(k/n) instead cos1 just error typing, and you got the good idea   :D to use rotation matrix.

Bn =
[ 1  (k/n)]
[(-k/n) 1]
= \sqrt(1 +k/n).R(n.arctan(k/n))

where R(x)=matrix of rotations in dimension two=
[ cosx sinx]
[-sinx  cosx]

(Bn)^n=
(1+k/n)^(n/2).R(n.arctan(k/n))

lim_{n->+oo} (1+k/n)^(n/2)=exp((n/2).ln(1+k/n))=1

lim_{n->+oo} n.arctan(k/n)=k

so lim_{n->+oo}(Bn)^n = R(k)

cheers
Moubinool

Which grade are you ?

I'm in the 11th grade
But i didn't know that I used rotation matrix  :?
Fratila Dragos

Have a look for rotation matrix
http://mathworld.wolfram.com/RotationMatrix.html
chers
oubi

uuu... this is a nice site
i didn't know nothing about it
do you know other sites like this?
or some sites from were i can download some mathbooks
thank you
Fratila Dragos

try the math-links section (top-right on your screen).
I sugest planetmath.org


calculus
derivative
analytic geometry
function
calculus computations

"Using IVT and RT work out an estimate for the smallest distance between the point (0,0) and the curve
y = e^x "

Can anyone help me with this question? Thanks.

Note. IVT is "Intermediate Value Theorem" and RT is "Rolle's Theorem"

Not sure if this is any help (or whether or not I've implicitly used the Intermediate Value Theorem or Rolle's Theorem); perhaps it will stir someone else into action:

$ (x, e^{x})$ is a point on $ y\equal{}e^{x}$ and its distance from the origin is $ \sqrt{x^{2}\plus{}e^{2x}}$ which is minimised when $ x^{2}\plus{}e^{2x}$ is. Differentiating and equating to zero, this ocurs when $ e^{2x}\equal{}\minus{}x$. Drawing a rough sketch and trying some numbers on a calculator, it seems that the minimum is achieved at about $ x\equal{}\minus{}0.425$, with a corresponding distance of about $ 0.78$.

First, thank you for you interest "AndrewTom". I found this equation, like you ->  $ e^{2x}\equal{}\minus{}x$ But how can we solve this without using a calculator? Or like the question says, how will we use IVT and RT to solve this? Please help me about this. It will probably be my exam question tomorrow. :)

I believe IVT is the Indermediate Value Theorem, but what is RT? :?

I've already deduced that RT must be Rolle's theorem in my previous post. However, I can't see how to use it and the IVT to solve the problem. Sorry. I'll think about it some more. But, hopefully before then, someone else will post the answer you want.

I have a feeling that $ e^{x} \equal{} \minus{} x$ cannot be solved analytically and that we must therefore resort to numerical methods.

Please help me, aren't there anyone solved it?  :(

I'm guessing Kent Merryfield, t0rajri0u, JBL, Farenhajt, kunny, Sonnhard et al have seen this. Perhaps they can help.

We can help with mathematics but, generally speaking, not with mind-reading. The problem just doesn't make much sense to me as posed, so you should ask your professor what he/she really meant here. There is no room for Rolle's theorem and very limited number of ways to use the IVT. Besides, the word "estimate" may mean anything (the precision is not mentioned, etc.). This is one of those rare cases when the blame for student's inability to understand the question should be put on the teacher and not on the student. So, just make sure that you know IVT,RT and FT (Fermat theorem; not the big one but just the one that says that the derivative at the point of an extremum must be zero if it exists) with proofs and go to the exam. If you get something like that on the test, raise your hand and ask for clarification. If the teacher refuses to clarify, write something like:

If $ |x| > 1$, then the distance from the origin to $ (x,e^x)$ is at least $ 1$ (the difference in $ x$-coordinate. Otherwise it is at least $ e^{ \minus{} 1}$ (the minimal $ y$-coordinate on $ [ \minus{} 1,1]$. Thus, it is never less than $ e^{ \minus{} 1}$. On the other hand, at $ x \equal{} \minus{} 1$ it is $ \sqrt {1 \plus{} e^{ \minus{} 2}}$. Thus, $ e^{ \minus{} 1} < D < \sqrt {1 \plus{} e^{ \minus{} 2}}$.

Continue like

This is an estimate. It doesn't use either IVT or RT but here are the statements of IVT and RT: ...

Finish like:

To find the exact distance, we can ... (just do what you did). The resulting equation cannot be solved algebraicly but the root is on $ [ \minus{} 1,0]$ (from IVT), it is unique (the function $ x \plus{} e^{ 2x}$ is strictly monotone on $ [ \minus{} 1,0]$ and we can find it with arbitrary precision using either bisection or Newton's method (describe both)

That's the best advice I can give here now. Maybe Kent can come up with a better idea of what it all might mean but my mind-reading abilities are not enough :(.

Thanks, fedja. I hope this clears things up for erencan144. It's certainly cleared it up for me. I didn't see how Rolle's Theorem could help or, indeed, how the Intermediate Value Theorem could be of much help either and presumed this was due to my lack of understanding.


Two cards are dealt at random from a standard deck of 52 cards (13 hearts, 13 clubs, 13 spades, and 13 diamonds).  What is the probability that the first card is a 6 and the second card is a Queen?

These probabilities, respectively, are $ \frac{4}{52}\equal{}\frac{1}{13}$ and $ \frac{4}{51}$.  Multiplying gives us $ \boxed{\frac{4}{663}}$.

[hide=More lengthy solution] We see that there are $4$ cards that are $6$.
That makes:
$$4/52=1/13$$
There are $4$ Queens, so we have $4/51$. Now we multiply these probabilities together:
$$1/13*4/51=\boxed{4/663 }$$


parameterization trigonometry algebra polynomial Vieta

[hide]Should we find a general triple with a parameter or should we just give a solution that we have found, because there are infinitely many solutions, I believe.[/hide]

You should find solution.
[hide]Hint x,y have one solution but z has many solutions.[/hide]

I think there are infinite solutions, because any x,y that satisfy xy=1 and 1<x+y<3 will work, since -1<cos z<1.

Are we supposed to solve in positive integers?

Since $xy+1=2$, we get:
$x+y+\cos z = xy+1$ or $x+y-xy-1=-\cos z$.
This is factored into:
$(1-x)(y-1)=-\cos z$.
So when $x=y=2$, $z=0^\circ,360^\circ$.
when $x=y=1$, $z=90^\circ, 270^\circ$.
Is this correct?

[quote="kimby_102"]Are we supposed to solve in positive integers?
[/quote]

That would sure help.  :) Most likely that's what he meant.

[hide]$xy=1$ and  $x+y=2-cosz$ using Vieta formulas $x,y$ are the roots of $u^2+(cosz -2)u+1=0$.The discriminant is $D=(cosz-2)^2-4=cosz(cosz-4)$ but from $D \ge 0$ we obtain $cosz=0$.
So $z=90^\circ+k*180^\circ$, and $x=y=1$[/hide]


a,b >0 x,y,z>0 x+y+w=1
ν.δ.ο.

(a+b/x)(a+b/y)(a+b/z) >= (a+3b)αυτο στην τρίτη .

Μανιέ βάλε την άσκηση σε Latex plz

Αν $ a, b > 0$ και $ x, y, z > 0$ και $ x \plus{} y \plus{} z \equal{} 1$
Να δειχτεί ότι:
$ (a \plus{} \frac{b}{x})(a \plus{} \frac{b}{y})(a \plus{} \frac{b}{z}) >\equal{} (a \plus{} 3b)^{3}$

Θα δώσω τη λύση που βρήκα:
Από AM - GM έχουμε: $ \frac {1}{3} \equal{} \frac {x \plus{} y \plus{} z}{3} > \equal{} (xyz)^{\frac {1}{3}} < \equal{} > 3 < \equal{} \frac {1}{(xyz)^{\frac {1}{3}}} (I)$

Από την ανισότητα των βαρών έχουμε:
$ (a \plus{} \frac {b}{x})^{\frac {1}{3}}(a \plus{} \frac {b}{y})^{\frac {1}{3}}(a \plus{} \frac {b}{z})^{\frac {1}{3}} > \equal{} a^{\frac {1}{3}}a^{\frac {1}{3}}a^{\frac {1}{3}} \plus{} (\frac {b}{x})^{\frac {1}{3}}(\frac {b}{y})^{\frac {1}{3}}(\frac {b}{z})^{\frac {1}{3}}$
$ \equal{} a \plus{} \frac {b}{(xyz)^{\frac {1}{3}}} > \equal{}$ (από $ (I)$) $ a \plus{} 3b$
$ < \equal{} > (a \plus{} \frac {b}{x})(a \plus{} \frac {b}{y})(a \plus{} \frac {b}{z}) > \equal{} (a \plus{} 3b)^{3}$

Πολύ ωραία άσκηση.

Αριθμητικος μεσος-Γεωμετρικος μεσος.


calculus
derivative
logarithms
algebra unsolved
algebra

$ x, y$ are natural numbers such that $ x\plus{}y\equal{}100$. Find maximum value of $ x^y$.

@Rust:You meant you set a table to see the change of the value of x^y and finally you see x=24,y=76 gets the best maximum?I think we can make it by using calculus methods(as deriative and differential). By the way, can you explain ur solution more clearly?

Let $ f(x)\equal{}x^{100\minus{}x}$

 $ f'(x)\equal{}x^{100\minus{}x}\left(\frac{100\minus{}x}{x}\minus{}\ln\;x\right)$

$ f'(x)\equal{}0\Leftrightarrow \left[\begin{array}{ll}x\equal{}0 \\ \ln\;x\equal{}\frac{100\minus{}x}{x}\end{array}\right.$

Let $ g(x)\equal{}\frac{100\minus{}x}{x},h(x)\equal{}\ln x,t(x)\equal{}g(x)\minus{}h(x)$, $ g'(x)\equal{}\frac{\minus{}100}{x^2}\le 0$ then $ g(x)\searrow$ and $ h(x)\equal{}\ln x\nearrow$

So:

$ 0>t(27)> t(28)>...> t(100)$
and
$ t(1)> t(2)> ...> t(26)$

Hence :

$ f(x)$ get the minimum value in $ x_0\Leftrightarrow x_0\in[26,27]$

....  :wink:


Isn't this being posted for many times.....
$abcd=1$
then $\sum \frac{1}{(1+a)^2} \ge 1$
The trick is that $\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2} \ge \frac{1}{1+ab}$ is true. :lol:

This problem's proposed for CMO 04  :)  I think Vasc created it .


[quote="maradona"]can we fine a bijection between IR and IR².(in general between IR and IR^n)[/quote]

Yes, of course : these set are bijective sets.

Here is one example of such a bijection :

Let $ f(x,y)$ from $ (0,1)\times(0,1)\rightarrow(0,1)$ :
Build the binary representation (right of dot) of $ f(x,y)$ by taking the two "normal" binary representations of $ x$ and $ y$ (those not containing infinite trailings 1) and copying the binary representation of $ x$ up to and including the first 0, then copying the representation of $ y$ up to and including the first 0, then copying the binary representation of $ x$ up to and including the second 0, and so on :
$ f(\frac{5}{8},\frac{3}{8})\equal{}f(B0.101,B0.011)\equal{}B0.1001011$
$ f(\frac{1}{2},\frac{1}{3})\equal{}f(B0.1,B0.01010101...)\equal{}B0.1000100100100100...$
$ f(B0.0010111110...,B0.001110110...)\equal{}B0.0000101110111110110...$

It's rather easy to see that $ f(x,y)$ is a bijection from from $ (0,1)\times(0,1)\rightarrow(0,1)$.

Now, since it exists well known bijections from $ (0,1)\rightarrow\mathbb{R}$ (for example $ g(x)\equal{}\cot(\pi x)$), it is easy to build a bijection from $ \mathbb{R}^2\rightarrow\mathbb{R}$ :

$ z(x,y)\equal{}g(f(g^{\minus{}1}(x),g^{\minus{}1}(y))$

Note :
1) When building bijections from binary representations, always be careful not producing infinite trailing ones (for example, a common mistake is to suggest to take one digit of x, then one digit of y, then one digit of x, then one digit of y, and so on. This is a mistake since then $ B0.00010101010101...$ would correspond to $ x\equal{}0$ and $ y\equal{}B0.01111111111...$)

2) Building a bijection from $ \mathbb{R}^n\rightarrow\mathbb{R}$  when you have a bijection from $ \mathbb{R}^2\rightarrow\mathbb{R}$ is obvious, I think.   :)


The lengths indicated on the rectangle shown are in centimeters. What is the number of square centimeters in the area of the shaded region?

[asy]path rectangle = (0,0)--(0,8)--(14,8)--(14,0)--cycle;
path quadrilateral = (7,4)--(5,8)--(14,8)--(14,6)--cycle;
path twolines[];
twolines = ((0,2)--(14,6))^^((9,0)--(5,8));
draw(rectangle);
draw(twolines);
filldraw(quadrilateral);
label("5",(2.5,8),N);
label("5",(11.5,0),S);
label("2",(0,1),W);
label("2",(14,7),E);
label("6",(0,5),W);
label("6",(14,3),E);
label("9",(4.5,0),S);
label("9",(9.5,8),N);[/asy]

i believe the shaded region is 1/4 of the total
the total area is 14 times 8, so 1/4 of that would be $ \frac{14\times8}{4} \equal{} 14\times2\equal{}28$

the shaded area is $ 28 \textit{cm}^2$

it's difficult to use words to explain why the shaded area is 1/4, so if someone can post why that is, that would be appreciated

That's what I got(I was the one who asked the question), but the answer was 25, apparently.

this question has not been answered correctly yet
i don't know how to solve this quickly
anyone want to show?
the answer is supposed to be 25.

Edit: nevermind

I solved it, with the help of ilovepink ;)
[asy]path rectangle = (0,0)--(0,8)--(14,8)--(14,0)--cycle;
path quadrilateral = (7,4)--(5,8)--(14,8)--(14,6)--cycle;
path twolines[];
twolines = ((0,2)--(14,6))^^((9,0)--(5,8));
path bisectors[];
bisectors = ((7,0)--(7,8));
draw(bisectors);
draw(rectangle);
draw(twolines);
filldraw(quadrilateral);
label("5",(2.5,8),N);
label("5",(11.5,0),S);
label("2",(0,1),W);
label("2",(14,7),E);
label("6",(0,5),W);
label("6",(14,3),E);
label("9",(9.5,8),N);
label("4",(7,2),W);
label("2",(8,0),N);
label("7",(3.5,0),N);[/asy]
since the bottom left one is easier to see, i write the numbers there :)
so area of left trapezoid is:
$ \frac12*7*(4\plus{}2)\equal{}21$
the right little triangle area is:
$ \frac12*2*4\equal{}4$
$ 21\plus{}4\equal{}25$ :)

Pardon me AceofDiamonds, but how do you know that the height/base of the trapezoid is 4cm? i can't seem to get that... :(

[asy]path rectangle = (0,0)--(0,8)--(14,8)--(14,0)--cycle;
path quadrilateral = (7,4)--(5,8)--(14,8)--(14,6)--cycle;
path twolines[];
twolines = ((0,2)--(14,6))^^((9,0)--(5,8));
path straightlines[];
straightlines = ((0,4)--(14,4))^^((7,0)--(7,8));
draw(rectangle);
draw(twolines);
draw(straightlines);
filldraw(quadrilateral);
label("5",(2.5,8),N);
label("5",(11.5,0),S);
label("2",(0,1),W);
label("2",(14,7),E);
label("4",(0,6),W);
label("4",(14,2),E);
label("4",(7,2),W);
label("2",(0,3),W);
label("2",(14,5),E);
label("9",(4.5,0),S);
label("9",(9.5,8),N);[/asy]

Because both sides have opposite but equally divided segments, you can conclude that the intersection point is also the center of the rectangle. Therefore, the distance from the point of intersection to the side is equal to half of the parallel side, hence, 4.


function
inequalities
vector
quadratics
special factorizations

Let $a_{1},\ldots,a_{n}$ be real numbers, and define the function $f$ by

$f(x)=(x-a_{1})^{2}+\ldots+(x-a_{n})^{2}$

Prove that $f(x) \geq f(\frac{a_{1}+\ldots+a_{n}}{n})$ for all $x$.

[b]Solution.[/b] $f(x)$ is clearly a convex function since $n>0$.  Notice that
\[f(x)=\sum_{i=1}^{n}(x-a_{i})^{2}=nx^{2}-2(a_{1}+\hdots+a_{n})x+(a_{1}^{2}+\hdots+a_{n}^{2}). \]
By completing the square for $ax^{2}+bx+c=0$, we find that the minimum occurs at $x=-\frac{b}{2a}$.  As a result, our minimum occurs at $x=\frac{a_{1}+\hdots+a_{n}}{n}$.

And the conclusion follows.$\Box$

[b]Remark.[/b]  In order to find extremal points and proving convexity algebraically, one may [i]also[/i] use derivatives.

oh ok. that was simpler than i thought it was going to be  :)

i have a few questions though. why is the convexity needed?

also, you can prove that the function is convex by noting that $|x-a_{i}|^{2}$ is convex for all $i$, and then you can prove that the sum of two convex functions is also convex straight from the definition (the inequality definition), correct?

i'm looking at the generalization of this problems, which considers vectors instead of real numbers $a_{i}$. if we know that $f(x,y)=f_{1}(x)+f_{2}(y)$ for all $x$ and $y$, how do we prove that min$f$=min$f_{1}$+min$f_{2}$?

You need to prove convexity in order to determine whether the extremal point gives you a maximum or minimum of the function.  Clearly, you can have extremal points that only yield maximums, e.g. in a quadratic function with a negative leading coefficient.

it's a quadratic with a positive coefficient of x^2


LaTeX percent search function trigonometry AoPSwiki calculus \/closed

I just noticed that many people behave very ugly here. They don't post their posts in the write sections, they don't use LaTeX, they write intentionally wrong spelled names of theorems they use.

On the other hand it is true, that it is better to post something (and sometimes some interesting problem) not in LaTeX then not to post it.

But still not writing math problems in LaTeX is at least very uncomfortable. Also posting in wrong sections. And also wrong spelled theorems.

Some examples:

[quote="Alisher"]hi  gduruchhcdghcdgucguucfuuhufhduuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu[/quote]

[quote="kunny"]Use Cauchy-Schwalz's inequality, since $a,b,c,t>0$

$\frac{a^{2}}{ta+b}+\frac{b^{2}}{tb+c}+\frac{c^{2}}{tc+a}\geq \frac{(a+b+c)^{2}}{(ta+b)+(tb+c)+(tc+a)}=\frac{(a+b+c)^{2}}{(t+1)(a+b+c)}$[/quote]

[quote="myc"]How do you make a PDF file in crimson editor?  In detailed please.

Thanks
myc :in section The Round-Table:[/quote]

[quote="Davron"]I love Cauchy Swarz ---[/quote]

[quote="akhrif"]Muirhead is stroooong!![/quote]

[quote="Hawk Tiger"]I like Jenson.[/quote]

[quote="rem"]I love the Murihead inequality. :)
I also like Cauchy-Schwartz and Jensen a lot. :)[/quote]

I think there should be more strict rules here about this.

HUH, YEAH...UM, I'M PISSED. LET'S START A RAID OR SOMETHING...

wait..am i posting this in the wrong forum...err..idk

I agree with shobhit. Chill a little man. 99 percent of the people on here do not do stuff like that. Do not waste our time by posting complaints like that. If it bothers you so much, why don't you just private message those you have a problems with. I mean come on!

:welcomeani:  Welcome to the Art of Problem Solving. :)

Well, I am not sure if my first post was in LaTeX, but wrong spelled theorems are here Very often.

@DanielHStennis08: Let me get some rest. I was doing math about 15 hours today.. Then I take care about our little children - AoPS.  :)

you see, the reality is that on aops, there are so many new members constantly that many times they don't know all the rules of aops. Therefore, some irrelevant topics and posts come up many many times. But the Mods and Admins usually take care of it in time. I think the forums have been enforced and cleaned up fine as they are right now. Although, the truth is that with such a large population of AoPS, the rules will not always be followed, but they certainly will be enforced.  :)

[quote="ZetaX"]The most anyoing users are like that one I think: http://www.mathlinks.ro/Forum/viewtopic.php?t=90891[/quote]

And this one too:
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=darij+keyboard&t=91014

What can be done about it?

ok, I see how you are getting upset. But, like shobhit said, all the administrators can do is enforce...

what just happened here? aops soap opera? :?

[quote="amcavoy"][quote="conceive"]They don't post their posts in the write sections[/quote]
Before you flame others for their posts, you may want to preview your own.  :)[/quote]

I was always posting in right sections.

i agree with you amcavoy...this thread is not necessary at all, but you have to feel sorry for conceive. He was 7th in Japan and did not make the IMO team. Nevertheless, it was not necessary at all to have this thread.

[quote="amcavoy"]conceive, I was making a reference to your grammar and spelling.

Is this thread necessary?[/quote]

This topic is not so irrelevant. He is right about the wrong spelled names theorems. How many times did I try to find some theorem, whose name I have read on AoPS (wrong spelled ) and then try to find it in google with no results...

He is right about this. Anyway the right spelling of nolmal text is not important.

[quote="conceive"]I will not spend here any minute more. I wish you your dreams became true..

I will not be in the IMO 2006 Slovenia.[/quote]

Come on man, IMO is not the only thing in life! actually, it is not even the best one.............................................

Sorry to tell you my point of view, but what you're doing is for people "afraid" of loosing.....

conceive, I apologize for my previous post.  I just thought of the times when others have said "this has been posted xx times, search for it" and I still have no luck.

As José mentioned, not making the IMO isn't something to beat yourself up over.  After all, only 6 can make it (and to be a candidate is an accomplishment itself!)

[quote="amcavoy"]conceive, I apologize for my previous post.  I just thought of the times when others have said "this has been posted xx times, search for it" and I still have no luck.[/quote]

Well, I also answer sometimes with something like "use the search function"; the reason is that some people really don't even try, often already searching with some formula from the problem or in other cases some related/occuring terms/words work. Those that don't find it by first attempt should at least try one or two times more (then it's ok not to find it I think, so noone has to try till the end of his life ;) ).

Guys, come down. Everyone has different opinions so let's respect them. By the way, if you can, please do not use comment in this type of tone.

[quote]what just happened here? aops soap opera?[/quote]

I'm not talking about a specific person. There were some comments in this type.

Now, as for the original intention, I think it's natural for you to feel that way because there are some people who spell not only math theorems wrong but name of the users. But again, most of the times, it's no big deal. No one is perfect, especially when it comes to typing. Speaking of kunny, I've seen him posting for quite long time (I think I even PMed him once maybe). He's a great guy who is pretty intelligent. But just like many other people in this forum, not all people here are Americans. They may have different use of calling names (note: I used word "may"). Also, I think in some nations of Europe, they have different names of $\sec$ or something. I've seen people asking what $\sec$ is couple times. Maybe it's just those but anyway, let's not press so much charges on them.

As for Zetax's last comment, I heart with you on that one. Most people in here do not even use search button. People know it's there but they don't use it. For instance, I think I've seen about 6 different threads with problem that includes $\tan x$ of 1989 AHSME in this forum recently. Moreover, there are even some cases where the problem is already put into the Resources section and they [i]still[/i] post the problems, saying "how do I do this?"

Well, you can't always expect everything to be in way you want them, right? That's why moderators and senior members are here to help you.  :)

To conceive:

 :welcomeani: Hello!

There is a significant degree to which it is necessary for AoPS members to self-moderate.  Usually this means diplomatically encouraging users to behave well toward each other and in the forums.  Patience and a thick skin go a long way.  Hopefully better behavior is established over time through leadership -- which mostly means starting with the oldest and most senior members of the community.

Wrong spelling of names is something pretty natural if you know these names from an e. g. Cyrillic book. I think this is by far not the greatest problem of ML/AoPS.

Apart from some massive spammers (we are just discussing them in the moderators section), a problem are new users with little computer experience who, even if given the link to the search function, need half an hour for understand how to use it. They don't understand what LaTeX is about (and that they don't have to install it in order to use it here), what the edit button is, how the subfores are organized (though they are indeed organized rather chaotically in my opinion), and that bmp is not number 1 graphics interchange format on the internet, particularly if the graphic is an MS Word screenshot with text and formulas only. What is missing is a kind of "newbie tutorial" to this forum. The LaTeX tutorial was already a great idea (well... it made a n00b like me learn the basics of LaTeX), now I mean basic things like "how do I post" and what the different sections are about. The Wikipedia has such tutorials, we don't.

  darij

[quote="darij grinberg"]The LaTeX tutorial was already a great idea (well... it made a n00b like me learn the basics of LaTeX), now I mean basic things like "how do I post" and what the different sections are about. The Wikipedia has such tutorials, we don't.
[/quote]

The AoPSWiki will be opening to the public relatively soon.  Those types of tutorials would be perfect for it.

Regarding searching for similar posts before creating a new post, it is of course a good idea to search first.  Nonetheless, there are an enormous amount of people on this site.  Chances are that one person will end up posting a duplicate thread.  This cannot be prevented completely (you can tell users in a tutorial not to do it but it will still happen, albeit more likely less often).  The best thing is to just have patience with new/inexperienced users.

[quote="joml88"]
The AoPSWiki will be opening to the public relatively soon.  Those types of tutorials would be perfect for it.
[/quote]

Sounds like an excellent use of the AoPSWiki!

[quote="darij grinberg"]The Wikipedia has such tutorials, we don't.

  darij[/quote]

Well, it is a good idea. At least tutorials like

"how to do a post"
"how can I edit my message"
"reply with quote"

will be good. Because it is normal, that 10 old people uses this Forum.

[quote="conceive"]Well, but what can I do? To became a poster here for the rest of my life?[/quote]

Just continue posting, maybe not as much as before, but just for love to math, e.g. solving problems.

I don't think that leaving MathLinks will solve the problem.... you were 7th on an TST!!! I'd never go to one probably. So, if you like solving problems, continue coming and posting, but maybe log in less time than before that TST where you finish 7th... that's all

[quote="fedja"][quote="conceive"]
Well, but what can I do? To became a poster here for the rest of my life?
[/quote]
I agree with Jose. Every math lover is always welcome to read and post on this forum. Being here has nothing to do with making it to the IMO. After all, the main idea of problem solving is not that of a sport where your ultimate goal is to become one of the 6 best in your age group in your particular country. The main ideas are professional growth, making friends and just sheer fun. I know that it is hard when some particular dream does not come true, but you cannot dwell on it forever. You will have a lot of other chances to succeed in life and in mathematics.[/quote]
I agree with fedja. This is certainly an aspect of mathematician's life: to learn math and to live it. Unable to make it into the team doesn't mean the end of your math career. If that's what you think, you do not possess that spirit, that dream of becoming a mathematician. A mathematician NEVER gives up. Well, a lot of people don't come here anymore after they graduate from high school but that's another issue. But, if you give up this forum with the only reason that you did not make it to the team, you are wrong. Not only are you not supporting your teammates, you have become incompetent, which is not an attitude of success in math competitions

Aside from the sadness of conceive's leaving, there is another growing trend of behavior in AoPS in which I think is annoying. For example:
[quote="José"]What's exactly calculus??? [/quote]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=536066#p536066]http://www.mathlinks.ro/Forum/viewtopic.php?p=536066#p536066[/url]
I'm sorry José but yes I find that annoying. First, if you want to know more about calculus, perhaps go take a course? Or perhaps go search information on the Net? Your curiosity should not be wasting other people's time and ask them to post what exactly calculus is. Second, I strongly believe that's spam. The question is irrelevant to the topic and the question can be asked somewhere else. Third, and this message is global, AoPS definitely needs suggestions for every user (including myself) on how to use the forums responsibly. Perhaps this is an innate instinct that whenever we go online, we don't feel the need to be responsible on what we say and what we do because it is a virtual world. This is so not true. This is a very bad habit. I think we all need a lesson on how to self-moderate, as suggested by MCrawford.

Except for the sports reference, I agree as well.  (Why bash sports? LOTS of people play sports without any intention of becoming one of the best in their country.)

Yes, that post of Jose that you quoted is extremely annoying.  And it's certainly not the first time he's made a post of that sort.

[quote]Third, and this message is global, AoPS definitely needs suggestions for every user (including myself) on how to use the forums responsibly. Perhaps this is an innate instinct that whenever we go online, we don't feel the need to be responsible on what we say and what we do because it is a virtual world. This is so not true. This is a very bad habit. I think we all need a lesson on how to self-moderate, as suggested by MCrawford.[/quote]

This is one of the first things we, the moderators, will be working on when the AoPSWiki comes out.

its annoying, but what can you do about it?

I'm always guilty. Everybody, here in my country we speak spanish, not english. Then, when I entered on this forum, I saw a lot of times the word "calculus" and I looked it up in Google, but I could not find a definition of "calulus" as "geometry", it only said "calculating the number of sth, e.g. the area of a rectangle".... it was not wnough for me.... so I asked, nothing more........ please stop criticising everything I put

I think people should try harder to make their posts look proper... including stuff like spelling, grammar, double posting, titles, etc. I can understand that people from other countries might not be as fluent in english, but I really don't understand typos on a forum much... I mean, whatever you type appears right in front of you, how hard is it to change? And people need to use the

[b]PREVIEW[/b] and [b]EDIT[/b]

buttons.

On another note, stuff like "I didn't make IMO, I'm going to leave this forum forever" [paraphrased] is neither necessary nor fun to listen to. Especially when it is said multiple times. Everyone has their fair share of failures; they don't need to hear about other people's. This is reminiscent of the thread about people complaining about high scores.

I don't think questions stuff like asking what Calculus is or posting a problem that has already been posted before (by someone else, a decently long time ago) is nearly as bad as the stuff I mentioned above. At least in these situations people are trying to learn. Homework help doesn't qualify here; that's annoying.

@ José (this is also going to be added to a MathLinks posting FAQ for the wiki):

Go to http://en.wikipedia.org . Search for "Calculus". You get an article in English. If this is too hard to translate for you: on the left bar, there is a list of languages (scroll down to "In other languages"), and one of them is "Español". Click on "Español" to get to the respective Spanish article. It is not quite good, but it should give you an idea on what calculus is, at least to the extent that it is not geometry...

Or go to the "Calculus" forum in College Playground and see what the topics there look like...

Or use MathLinks search (less recommended).

Sometimes, searching is more than writing one word in a google bar.

  darij

hey, I am new to aops but I thought that there are actual calculus forums on the site as well. Maybe those would help answer some questions you have.