calculus
MIT
college
percent
Harvard
linear algebra
real analysis

How rare is this for people applying to top tech schools? By the time I am a senior next year, I will have completed multivariable calculus, linear algebra, ordinary differential eqations, partial differential equations, and intro to real analysis by the end of my senior year (at my local university). Does this give me a head up over other applicants when applying to MIT or Cal Tech.

           On a side note, my twin sister, who doesn't like math as much as I, self-studied Calculus BC AP this year. She is obsessed with taking AP's, and is taking Stats AP and Art History AP to pad her resume, when she could be taking multivariable calc at our local university. What should she take?

My opinion is that Cal Tech is a bit different from most schools.  It is far more highly specialized than other top schools.  They are likely to be more interested in students who have shown a deep focus in math, computer science, and other sciences.

MIT is probably very similar.  Most of the people I knew who went to school there were those who were concentrating very hardcore in one or more of the same kinds of areas.

It is important to be well-rounded no matter where you go to school.  Unfortunately APs are some of the only reasonable tests of high level abilities in a large number of subjects.  Having 8 or 10 APs probably helps more with an Ivy League school than a Cal Tech or an MIT, though the technical schools would certainly like to see the APs in math/sci/tech areas.

Overall I think the APs are a bit over-rated, though I'm not sure there is anything better yet (other than doing things like taking national Olympiads in math/sci/computers).

One thing you might do is try creating dialogue with students who go to those schools.  This will do a number of things for you.  First it's a communication of your interest.  This is always important when trying to find an academic (or any other kind of) home in life.  Second, it creates information flow.  You'll find out more about what those programs are really like.  You'll have a better and better idea of what you want out of each program.  You'll also find out more and more about what it takes to get into those programs.

I have seen a figure, I think in one of Michele Hernandez's books, that roughly 40 percent of admitted applicants to Ivy League universities have sufficient college credits already earned to have sophomore standing if one simply counted their credits. So having college credit before applying is fairly common at the top schools. Here in Minnesota, hundreds of young people participate in a [url=http://www.mheso.state.mn.us/mPg.cfm?pageID=797]state program[/url] allowing high school juniors and seniors to [url=http://www.pseo.org/]go to college as part of their high school[/url] program.

Harvard tells applicants that previous college credits can be used to fulfill course prerequisites, but can't be used as credits toward graduation with a Harvard degree. Evidently the top schools face this question all the time. As Mr. Crawford said, one of the best things you can do if want to know more about a particular college is contact students at that college (which will become increasingly easy to do through this forum in the next few years).

Hope this helps!

[quote="samus"]I will have completed multivariable calculus, linear algebra, ordinary differential eqations, partial differential equations, and intro to real analysis by the end of my senior year (at my local university). [/quote]

How did you choose those courses?  I too have taken courses at a university during high school, but I opted to pursue analysis courses instead of differential equaltions because I thought analysis courses would give me a head start in math placement in college (and I found the course material interesting).

I spoke with the department head at My local university, and he mapped these courses out for me.

It's not uncommon for hardcore math/science students to have taken some college classes before applying to college. It certainly helps, because Universities like seeing students who reach out to challenge themselves with demading courses. But the quality of these classes can vary greatly depending on where you take them, and who teaches them.

A lot of the students that I know who took Linear Algebra or Analysis (at their high school or a local college) before going to college full-time, found that they needed to retake it at a more rigorous level when they got to college. But it still helped that they had some exposure to it beforehand.

But there are also a lot of very good math and science students who don't take college courses in high school - and they do just fine also. It's all about taking advantage of the opportunities that you have available to you. Some students do not live near a college or University, so the AP program provides the best outlet for them.

Basically though, I think that too many students right now are judging themselves by what others do. If you see an interesting and challenging course that you want to take at your high school or a local university, then take it. As long as you are learning and thirsting for knowledge, you will be successful in life.


[url=http://www.freedomunderground.org/memoryhole/pentagon.php]link[/url]
about Pentagon on 9/11. interesting point of view.

It was a well made video. Interesting Idea.

I've seen this before...I think it was made up as propaganda for certain political views.

I hate this kind of stuff. As a liberal, I think it makes us look awfully bad.


If I apply to Harvard under early decision and get rejected, is it possible for me to apply again under regular decision? (By the regular decision deadline, perhaps I will have more awards such as the siemens westinghouse competition, national science olympiad, etc.)

Harvard doesn't have "early decision," but it has what is called single-choice early action. If you are rejected, you are rejected. If you apply in the early round, the single most likely result is that your application will be deferred for further consideration in the regular round.


If x, y, z are pozitive numbers, than
 \sqrt(x + y + z)(1/x + 1/y + 1/z) \geq 1 + \sqrt1 + \sqrt(x2 + y2 + z2)(1/x2 + 1/y2 + 1/z2).

This one is really lovely. It is of course equivalent to the following one:
  (x+y+z)(1/x+1/y+1/z)>=2sqrt((x+y+z)(1/x+1/y+1/z))+sqrt((x^2+y^2+z^2)(1/x^2+1/y^2+1/z^2)) and is a direct consequence of the fact that sqrt((a+b)(c+d))>=sqrt(ac)+sqrt(bd) (just put a=x^2+y^2+z^2, b=2(xy+yz+zx), c=1/x^2+1/y^2+1/z^2, d=2/xy+2/yz+2/zx).


conics
parabola
geometry
function
analytic geometry

The lower two vertices of a square lie on the x-axis and the upper two vertices lie on the parabola $y=15-x^{2}$. What is the area of the square?

[hide] Say the two lower verticies are A (a,0) and B (b,0), where b>a.  Say the two upper verticies are C(b,c) and D(a,c).  Firstly, since AB=AD, we have that b-a=c.  Plugging these values in, we get that C is (b,b-a) and D is (a, b-a).  Now, C and D are on the parabola with function $y=15-c^{2}$.  Plug these values in:
Point C: $b-a=15-a^{2}$
Point D: $b-a=15-b^{2}$.
Set the two equal to each other; $15-a^{2}=15-b^{2}\Rightarrow a^{2}=b^{2}\Rightarrow a=\pm(b)$.  However, $b\ne(a)\Rightarrow a=-b$.  Take point C now and plug in a=-b.  This means that $2b=15-b^{2}\Rightarrow b^{2}+2b-15=0\Rightarrow b=-5, b=3$.  We earlier defined that a>b so if b was -5, then a>b.  Therefore, b=3.  This means that the verticies are A(-3,0), B(3,0), C(3,6), and D(-3,6).  Therefore, each side of the square is 6 and hence the area of it is $36$. [/hide]

[hide]You may easily see from a simple construction of y = 15 - x^2 and a simple square that follow the problems guide lines that the coordinate values for x must be double that of y.  Also from the construction you would note that y values are less than 15 and in essence x values are less than 15/2.  Guess and check works are this to just find a x value that has a doubled y value.  that is x = 3, y = 6;  From that the square follows, x = -3, y = 6; x = 3, y = 0; and x = -3, y = 0.  This is a sqaure with length 6, hence area = 36
<p>
the bad thing is however, this method won't work efficiently always. :)[/hide]

[hide="Hmm"] I think it is pretty clear that the vertices must take the form $(-a, 0), (a, 0), (-a, 15-a^{2}), (a, 15-a^{2})$.  Hence we need to solve $2a = 15-a^{2}\implies (a-3)(a+5) = 0 \implies a = 3$.  The side length is $2a = 6$, so the area is $6^{2}= \boxed{ 36 }$. [/hide]


geometry
3D geometry
floor function
logarithms
AMC

I read the answers on the Harold Reiter's Page but some of the questions weren't that helpful. I got 133.5 this time (second time I took this) which is a lot better than last time, where I got only 102 (yay!). Some questions appear on mock amc, I think but still, these are some that I need help from this forum:

14. A charity sells 140 benefit tickets for a total of 2001 dollars. Some tickets sell for full price (a whole dollar amount), and the rest sell for half price. How much money is raised by the full-price tickets?

[hide="Steps"]
Let p = the original price
$xp+(140-x)p*1/2 = 2001$
$2xp+140p-xp = 4002$
$xp+140p = 4002$
$p(x+140) = 4002$
From here, do I have to use a guess-and-trial? I prefer not to use it because it's not that much accurate on AMC.[/hide]

21. A right circular cylinder with its diameter equal to its height is inscribed in a right circular cone. The cone has diameter 10 and altitude 12, and the axes of the cylinder and cone coincide. Find the radius of the cylinder.

[hide="WHAT?"]I read the answer and well, I'm weak at this 3D geometry and I don't know how they have similar triangle in there. If I know how, that should help me but I don't understand how they got there.[/hide]

25. How many positive integers not exceeding 2001 are multiples of 3 or 4 but not 5?

[hide="easyway?"]I wonder if there's easy way to do this. My answer was 767 and I put 768 which was close to it (it was logical, in my opinion) and I got wrong. And when I saw the answer sheet and it was really complicate. I know there are good chances of AMC appearing in this way and I need to memorize them but I just wondered if there were any EASIER way.[/hide]

Thanks for many good helps! :lol:

[quote="Silverfalcon"]these are some that I need help from this forum:

14. A charity sells 140 benefit tickets for a total of 2001 dollars. Some tickets sell for full price (a whole dollar amount), and the rest sell for half price. How much money is raised by the full-price tickets?[/quote]
[hide="14.  Guess and check is not required"]
Let p = the original price
Let x = Number of full-price tickets
$xp+(140-x)p*1/2 = 2001$
$2xp+140p-xp = 4002$
$xp+140p = 4002$
$p(x+140) = 4002$
$p=\frac{4002}{x+140}$

Now, since we know $0 < x < 140$, we can get bounds on $p$:
If $x=0$, then $p=\frac{4002}{0+140}$
So, $p < 28.59$

If $x=140$, then $p=\frac{4002}{140+140}$
So, $p > 14.29$

$15 \le p \le 28$

But $p$ must also be a factor of $4002=2\times 3\times 23\times 29$

The only factor of 4002 in the allowable range is 23.  So $p=23,~x=34,~px=782$.[/hide]

For 25.... question is confusingly stated... (as you probably know) the question really is asking how many positive integers less than or equal to 2001 are divisible by 3, or 4, but not 5.
 I'm not going to solve the question here, but the methdo is relatively easy
[hide]1. Start by dividing 2001 by 3 then 4
2. Subtract the number of double coutned number
3. See how many of these numbers are divisible by five and subtract[/hide]

[quote="Silverfalcon"]25. How many positive integers not exceeding 2001 are multiples of 3 or 4 but not 5?[/quote]

Well, inclusion-exclusion.

Take $\lfloor {2001/3} \rfloor + \lfloor {2001/4} \rfloor - \lfloor {2001/12} \rfloor - \lfloor{2001/15} \rfloor - \lfloor {2001/20} \rfloor $. That's your answer, I believe.

Ok. Thanks, guys for helping me out here.

Hmm... Number 21 is still bothering me. I don't understand where $\frac {2r}{5}$ came from. I got 2r for height since height=diameter and if we set r = radius of cylinder, 2r = diameter = height. But how did you find the upper cone's height?

That's where I'm confusing.

Help! ;)

Ohohohohohoho

Yeah, I got it

Thanks for word "similar." That helped me a lot!

What is the value of expression:

${{{{\frac {1}{\log_{2} {100!}}} + \frac {1}{\log_{3} {100!}}} + \frac {1}{\log_{4} {100!}}} + .. + \frac {1}{\log_{100} {100!}}}$


logarithms
function
real analysis
LaTeX
irrational number
real analysis unsolved

A perfect set is one which is equal to the set of its limit points. Is there a nonempty perfect set in $ \mathbb{R}$ (defined under the usual metric) which contains no rational number?

[hide="My solution"]Yes. The previous exercise involves showing the subset of $ [0,1]$ consisting of decimal expansions with only $ 4$s and $ 7$s is perfect. If you add something like Liouville's constant to every element, perfectness is preserved since this is an isometry, but all of the elements become irrational since it's impossible for any decimal to become repeating.[/hide]

Posting this here because I suspect there is a simpler construction, and because I thought the problem was interesting anyway.

If you did the previous exercise, it seems like a valid solution.  Otherwise, I would prove that previous exercise.  You may also want to prove that sums of irrational numbers are irrational, sums of rationals and irrationals are irrational...

There are other, more natural constructions (at least if you have seen cantor sets before).  see [url=http://planetmath.org/encyclopedia/ANonemptyPerfectSubsetOfMathbbRThatContainsNoRationalNumber.html]here[/url] for more details.

[quote="JRav"]If you did the previous exercise, it seems like a valid solution.  Otherwise, I would prove that previous exercise.[/quote]

without wanting to word this too strongly, why do you feel the need to post replies to so many threads where you have nothing useful to contribute?  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=248946]this thread[/url] is another 'good example'.  he's not asking for someone to verify his proof, he's asking if there are other constructions.

[quote]You may also want to prove that sums of irrational numbers are irrational[/quote]

on the other hand, this isn't even true.

[quote]Quote:
You may also want to prove that sums of irrational numbers are irrational

on the other hand, this isn't even true.
[/quote]
This is true...

Yes, I caught that mistake a while back... :)

[quote="JBL"]$ \minus{} \sqrt {2}$ and $ \sqrt {2}$ are both irrational, but their sum is rational.   :roll: For any rational number $ q$ and any irrational number $ i$, there is an irrational number $ x \equal{} q \minus{} i$ such that $ x \plus{} i \equal{} q$ is rational.[/quote]

In fact, there are infinitely many such rationals (any would do). Suppose $ a$ is irrational. Then for any rational number $ b$, $ b\minus{}a$ is irrational and $ a \plus{} (b\minus{}a) \equal{} b$ is a rational number, sum of two irrationals.

That's exactly what the second sentence of the post you quoted says.    :huh:  :wink:

Αnd a question: exist [i]irrational[/i] numbers [i]a,b[/i] such as the number $ a^b$ be rational?
Good night from Greece...

$ a^{\log_a 2} \equal{} 2$ for every $ a$.

Asume a fuction D(x) as valued 1 at rationals in [0,1] and 0 at irrationals in [0,1].Then D(x) is a finite lebesgue measureable fuction.According to Lusin Theorom:we can find a compact subset C of [0,1]  such that  D(x)  is continue on C, let E denote the intersection of C and irrationals,naturely D(x) is continue on E. Now denonting F be the set of E with its isolated point deleted,and we also have that D(x) is also continue on F.It is easy to see that F is the desired set:Since F does not centain rationals and isolated point,if  F is a compact set,then F is a perfect set with no rationals.In fact, choose a infinitely squesence An in F and sign its limit point as A.If A is rational ,recalling  that D(x) is continue on C,we concluded a contradiction;then A is irrational ,recalling C is compact and the construction of F, we concluded that the limit point A belongs to F.As is desired.

I had in my mind the $ \sqrt {2}^sqrt{2}^sqrt{2}$ :P

p.s. if someone could replace it using latex i'll thank him... :blush:

See [[LaTeX]] and http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 -- you can edit your own posts for up to a day or two.

i will be appreciated if someone looks my example and tell me it true or false...


function
integration
real analysis
real analysis unsolved

$f \in L^{\infty}([0;1])$ a function $f: \mathbb{R}\to \mathbb{C}$.
Is $S=\{ \int_{E}f(x) dx : E \textrm{ mesurable subset of [0;1]}\}$ a convex subset of $\mathbb{C}$

First consider a special case: $f\ge 0$. In this case $f$ defines an absolutely continuous measure $fdx$, and the convexity of $S$ is just a restatement of the [url=http://en.wikipedia.org/wiki/Atom_(measure_theory)]intermediate value theorem for measures[/url]

Now to the general case: given $\epsilon>0$, partition $[0,1]$ into measurable subsets $A_{i}$ such that the oscillation of $f$ on $A_{i}$ (i.e. the diameter of $f(A_{i})$) is at most $\epsilon$. Pick $a_{i}\in A_{i}$. For any measurable set $E\subset[0,1]$ we have \[\left|\int_{E}f(x)dx-\sum_{i}f(a_{i})|E\cap A_{i}|\right|\le \epsilon\] The set $\left\{\sum_{i}f(a_{i})|E\cap A_{i}|\colon E\subset[0,1]\right\}$ is convex by the same "Int.value theorem". Since $\epsilon>0$ is arbitrary and $S$ is closed, it follows that $S$ is convex.

I just tried editing your post to see if I could get the url to work, without success. Apparently the system doesn't like parentheses - but in this case, the parenthesis are a vital part of the url.


integration
limit
calculus
calculus computations

Help me to check the solution right or wrong?

\[ \left[ \int\limits_{0}^{\plus{}\infty }{\frac{{{e}^{\minus{}ax}}.dx}{x}} \right]_{a}^{'}\equal{}\int\limits_{0}^{\plus{}\infty }{\minus{}{{e}^{\minus{}ax}}.dx}\equal{}a.\left[ {{e}^{\minus{}ax}} \right]_{0}^{\plus{}\infty }\equal{}{{\left( \minus{}1 \right)}^{2}}a\,\,\,\,\left( a>0 \right)\]

\[ \Rightarrow \int\limits_{0}^{\plus{}\infty }{\frac{{{e}^{\minus{}ax}}.dx}{x}}\equal{}\int{a.da}\equal{}\frac{{{a}^{2}}}{2}\]

\[ \left[ \int\limits_{0}^{\plus{}\infty }{\frac{{{e}^{\minus{}ax}}.dx}{{{x}^{2}}}} \right]_{a}^{''}\equal{}\int\limits_{0}^{\plus{}\infty }{{{\left( \minus{}1 \right)}^{2}}.{{e}^{\minus{}ax}}.dx}\equal{}{{\left( \minus{}1 \right)}^{3}}a\Rightarrow \int\limits_{0}^{\plus{}\infty }{\frac{{{e}^{\minus{}ax}}.dx}{{{x}^{2}}}}\equal{}{{\left( \minus{}1 \right)}^{3}}\int{\frac{{{a}^{2}}}{2}}\equal{}{{\left( \minus{}1 \right)}^{3}}.\frac{{{a}^{3}}}{3!}\]

\[ \Rightarrow \int\limits_{0}^{\plus{}\infty }{\frac{{{e}^{\minus{}ax}}.dx}{{{x}^{n}}}}\equal{}{{\left( \minus{}1 \right)}^{n\plus{}1}}.\frac{{{a}^{n\plus{}1}}}{\left( n\plus{}1 \right)!},\,\,\,\,\underset{a\to 1}{\mathop{\lim }}\,\int\limits_{0}^{\plus{}\infty }{\frac{{{e}^{\minus{}ax}}.dx}{x}}\equal{}\int\limits_{0}^{\plus{}\infty }{\frac{{{e}^{\minus{}x}}.dx}{x}}\equal{}\frac{1}{2}\]

\[ But\,\,\int\limits_{0}^{\plus{}\infty }{\frac{{{e}^{\minus{}ax}}.dx}{x}}\equal{}\int\limits_{0}^{\plus{}\infty }{\frac{{{e}^{\minus{}ax}}.d\left( ax \right)}{ax}}\equal{}\frac{1}{2}\,\,?\]

$ \int_0^{\infty}\frac {e^{ \minus{} ax}}{x}\,dx \equal{} \infty.$

As $ x\to0,$ the integrand behaves like $ \frac1x,$ and $ \int_0^1\frac1x\,dx \equal{} \infty.$

Hence the given integral diverges by limit comparison.


integration
real analysis
real analysis unsolved

Let $\{r_n\}$ be a sequence in $(0,1)$ with the property that if $r_n=r_m$ then $n=m$
Define $f: [0,1] \rightarrow R$ by $f(x)=$ $\sum_{\{n| r_n \leq x\}} 2^{-n}$

1. Show that $f$ is integrable and determine $\int_{0}^{1} f$

2. Show that $f$ is discontinuous at every point of the set $\{ r_n | n\in N*\}$ and continuous at every other point

$f$ is increasing so Rieman integrable.
$f=\sum f_n$ where $f_n(x)=0$ if $x \in [0;r_n[$ and $f_n(x)=2^{-n}$ if $x \in [r_n;1[$.
Hence $\int_{0}^{1}f(x)dx=\sum \frac{1-r_n}{2^n}<2$.

The sequence $S_n=\sum_{i=1}^n f_k$ converges uniformly to $f$ and for every $n \geq 1$, $S_n$ is continuous
at the point $x$ if  $x \in ]0;1[-\{r_1,r_2,..,r_k,..\}$.


Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=1$. Prove that:
$6(a^{3}+b^{3}+c^{3}+d^{3})\geq (a^{2}+b^{2}+c^{2}+d^{2})+\frac{1}{8}$

Fitim Dika   :wink:

Use Cauchy-swatch ,It 's easy to have solution :D


Time was when I was weapon and warrior; Now the young hero hoods me with gold,
and twisted silver. At times men kiss me. At times I speak and summon to battle
Loyal companions. At times a courser, Bears me o'er marchland. At times a ship
Bears me o'er the billows, brightly adorned. At times a fair maiden fills me with breath;
At times hard and headless I lie on the board, Bereft of beauty. At times I hang
Winsome on wall, richly embellished, Where revelers drink. At times a warrior
Bears me on a horse, a battle adornment, And I swallow, bright-shining, the breath from his bosom.
At times with my strains I summon the heroes, Proudly to wine. At times I win back
Spoil from the spoiler, with sounding voice, Put foemen to flight. Now ask what I'm called.


geometry
geometric transformation
reflection
angle bisector
geometry unsolved

Let $ (O_1)$ and $ (O_2)$ be two cirlces.$ (O_1)$ cuts $ (O_2)$ at $ A$ and $ D$
Suppose that $ BC$ is a chord through $ D$($ B \in (O_1);C \in (O_2)$)
$ BA$ cuts $ (O_2)$ at $ E$;$ CA$ cuts $ (O_1)$ at $ F$
$ M;N$ is two points in $ BF;CE$,respectively such that $ \frac {MB}{MF} \equal{} \frac {NC}{NE}$
$ CM$ cuts $ BN$ at $ K$
Prove that $ \angle KAC \equal{} \angle DAB$

[hide="Solution"]
We consider the points $ X$ and $ Y$ on $ AD$ so that $ \angle BCM \equal{} \angle ACX$ and $ \angle ABY \equal{} \angle NBC$.  Now, since $ \angle MBC \equal{} \angle XAC$, we have that $ \triangle XAC\sim \triangle MBC$, which means that since $ \triangle ACD\sim \triangle BCF$, we have that $ \triangle DXC\sim \triangle FMC$.  We may thus say that $ \frac {AX}{XD} \equal{} \frac {MB}{FM}$.  Similarly, $ \frac {AY}{YD} \equal{} \frac {NC}{EN} \equal{} \frac {MB}{FM} \equal{} \frac {AX}{XD}$, so $ X \equal{} Y$.  Thus, $ X$ is the intersection between the reflection of $ BN$ over the bisector of $ \angle ABC$ and $ CM$ over the angle bisector of $ \angle ACB$.  It follows that $ K$ and $ X$ are isogonal conjugates of $ \triangle ABC$, so $ \angle DAB \equal{} \angle KAC$. [/hide]

My solution:

Let $ T $ be a point on $ AD $ such that $ \frac{AT}{TD}=\frac{BM}{MF}=\frac{CN}{NE} $ .

Since $ \triangle CAD \cup T \sim \triangle CBF \cup M $ ,
so $ CT, CM $ are isogonal conjugate of $ \angle ACB $ .
Similarly we can prove $ BT, BN $ are isogonal conjugate of $ \angle CBA $ ,
so $ T $ is the isogonal conjugate of $ K $ WRT $ \triangle ABC \Longrightarrow AT \equiv AD, AK $ are isogonal conjugate of $ \angle BAC $ .

Q.E.D


inequalities
rearrangement inequality
inequalities proposed

[quote="Marius Mainea"]If a,b,c>0 ,then
\[ \frac {a^5}{a^3 \plus{} b^3} \plus{} \frac {b^5}{b^3 \plus{} c^3} \plus{} \frac {c^5}{c^3 \plus{} a^3}\ge \frac {a^2 \plus{} b^2 \plus{} c^2}{2}
\]
[/quote]
The inequality is equivalent to:

$ \frac {a^2(a^3 \minus{} b^3)}{a^3 \plus{} b^3} \plus{} \frac {b^2(b^3 \minus{} c^3)}{b^3 \plus{} c^3} \plus{} \frac {c^2(c^3 \minus{} a^3)}{c^3 \plus{} a^3} \ge 0$

since $ \frac {a^2 \plus{} ab \plus{} b^2}{a^2 \minus{} ab \plus{} b^2} \ge 1,$

it suffices to prove that:

$ \sum \frac {a^2(a \minus{} b)}{a \plus{} b} \ge 0,$

which is equivalent to:

$ \frac {a(a^2 \plus{} b^2)}{a \plus{} b} \plus{} \frac {b(b^2 \plus{} c^2)}{b \plus{} c} \plus{} \frac {c(a^2 \plus{} c^2)}{a \plus{} c} \ge ab \plus{} ac \plus{} bc$

but we know that:

$ a^2 \plus{} b^2 \ge \frac {1}{2}(a \plus{} b)^2,$

similarly for others then it suffices to prove that:

$ a^2 \plus{} b^2 \plus{} c^2 \ge ab \plus{} ac \plus{} bc$ which is true.

 :)

[quote="rachid"][quote="Marius Mainea"]If a,b,c>0 ,then
\[ \frac {a^5}{a^3 \plus{} b^3} \plus{} \frac {b^5}{b^3 \plus{} c^3} \plus{} \frac {c^5}{c^3 \plus{} a^3}\ge \frac {a^2 \plus{} b^2 \plus{} c^2}{2}
\]
[/quote]

since $ \frac {a^2 \plus{} ab \plus{} b^2}{a^2 \minus{} ab \plus{} b^2} \ge 1,$

it suffices to prove that:

$ \sum \frac {a^2(a \minus{} b)}{a \plus{} b} \ge 0,$

[/quote]

Why?

It's equivalent to:
$ a^2\plus{}b^2\plus{}c^2 \geq\ \sum\ \frac{2a^2b^3}{a^3\plus{}b^3}$ $ \Leftrightarrow a^2\plus{}b^2\plus{}c^2 \geq\ \sum\ \frac{2ab^2}{a\plus{}b}$
But:$ \sum\ \frac{2ab^2}{a\plus{}b}  \leq\ \frac{a^2\plus{}b^2\plus{}c^2\plus{}ab\plus{}bc\plus{}ca}{2} \leq\ LHS$
$ \Rightarrow  q.e.d$

 :)

[quote="Marius Mainea"]If a,b,c>0 ,then
\[ \frac {a^5}{a^3 + b^3} + \frac {b^5}{b^3 + c^3} + \frac {c^5}{c^3 + a^3}\ge \frac {a^2 + b^2 + c^2}{2}
\]
[/quote]
$ \sum_{cyc}\frac{a^5}{a^3+b^3}\geq\frac {a^2 + b^2 + c^2}{2}\Leftrightarrow\Leftrightarrow\sum_{cyc}\left(\frac {a^5  }{a^3+b^3}-\frac{a^2+b^2}{4}\right)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac{(a-b)(3a^4+3a^3b+2a^2b^2+ab^3+b^4)}{a^3+b^3}-\frac{5(a^2-b^2)}{2}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{(a-b)^2(a^3+2a^2b+6ab^2+3b^3)}{a^3+b^3}\geq0.$

[quote="rachid"]The inequality is equivalent to:

$ \frac {a^2(a^3 \minus{} b^3)}{a^3 \plus{} b^3} \plus{} \frac {b^2(b^3 \minus{} c^3)}{b^3 \plus{} c^3} \plus{} \frac {c^2(c^3 \minus{} a^3)}{c^3 \plus{} a^3} \ge 0$[/quote]
Please detail this step.... otherwise your proof seems OK.

[quote="Marius Mainea"]If $ a,b,c>0,$ then

\[ \frac {a^5}{a^3 \plus{} b^3} \plus{} \frac {b^5}{b^3 \plus{} c^3} \plus{} \frac {c^5}{c^3 \plus{} a^3}\ge \frac {a^2 \plus{} b^2 \plus{} c^2}{2}.
\][/quote]Using the arithmetic-mean — geometric-mean inequality,

$ \sum_{cyc}\frac {a^5}{a^3 \plus{} b^3}\plus{}\sum_{cyc}\frac{9a(a^3\plus{}b^3)}{4(a\plus{}b\plus{}c)^2}\geq\frac{3}{a\plus{}b\plus{}c}\sum_{cyc}{a^3},$

it remains to prove quartic inequality :

$ 12(a\plus{}b\plus{}c)\sum_{cyc}a^3\geq(a\plus{}b\plus{}c)^2\sum_{cyc}a^2\plus{}9\sum_{cyc}a\left(a^3\plus{}b^3\right).$

This comes down to the following identity :

$ 12(a\plus{}b\plus{}c)\sum_{cyc}a^3\minus{}(a\plus{}b\plus{}c)^2\sum_{cyc}a^2\minus{}9\sum_{cyc}a\left(a^3\plus{}b^3\right)$

$ \equal{}\sum_{sym}a^4\plus{}8\sum_{cyc}a^3b\minus{}\sum_{cyc}ab^3\minus{}4\sum_{sym}b^2c^2\minus{}4\sum_{sym}a^2bc$

$ \equal{}\frac{1}{10}\sum_{cyc}(2a\minus{}b\minus{}c)^2\left(a^2\plus{}18ab\plus{}6b^2\right)\geq0.$

WHY? $ a^2\plus{}b^2\plus{}c^2 \geq\ \sum\ \frac{2a^2b^3}{a^3\plus{}b^3}$ is equivalent to $ \Leftrightarrow a^2\plus{}b^2\plus{}c^2 \geq\ \sum\ \frac{2ab^2}{a\plus{}b}$

[quote="matex (L)(L)(L)"]WHY? $ a^2 \plus{} b^2 \plus{} c^2 \geq\ \sum\ \frac {2a^2b^3}{a^3 \plus{} b^3}$ is equivalent to $ \Leftrightarrow a^2 \plus{} b^2 \plus{} c^2 \geq\ \sum\ \frac {2ab^2}{a \plus{} b}$[/quote]

I think that it follows from

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

which is

$ (a\plus{}b)(a\minus{}b)^2 \geq 0$

[quote="nguoivn"]It's equivalent to

$ a^2 \plus{} b^2 \plus{} c^2 \geq\ \sum\ \frac {2a^2b^3}{a^3 \plus{} b^3}$  :)[/quote]$ \sum_{cyc}\frac {a^5}{a^3 \plus{} b^3} \minus{} \frac {a^2 \plus{} b^2 \plus{} c^2}{2} \equal{} \sum_{cyc}\left(a^2 \minus{} \frac {a^2b^3}{a^3 \plus{} b^3}\right) \minus{} \frac {a^2 \plus{} b^2 \plus{} c^2}{2}$

$ \equal{} \frac {a^2 \plus{} b^2 \plus{} c^2}{2} \minus{} \sum_{cyc}\frac {a^2b^3}{a^3 \plus{} b^3}\geq \frac {a^2 \plus{} b^2 \plus{} c^2}{2} \minus{} \sum_{cyc}\frac {a^2b^3}{2\sqrt {a^3b^3}}$ (arithmetic-mean — geometric-mean inequality)

$ \equal{} \frac {1}{2}\left(\sum_{cyc}\sqrt {aa^3} \minus{} \sum_{cyc}\sqrt {ab^3}\right)\geq0.$ ([url=http://en.wikipedia.org/wiki/Rearrangement_inequality]rearrangement inequality[/url])


[quote="mathmaster46"]If (1/2) ^ x = x, solve for x.
Could someone please tell me how to evaluate this without using a calculator?[/quote]

$\left(\frac{1}{2}\right)^{x}=x$

well then x is all real numbers =)

What are you talking about?

$x$ can't be evaluated without a calculator.

My teacher used a calculator to do it and he said it was about $0.64$. He also said there was a way to simplify it without at least graphing and finding the intersection.

[quote="mathmaster46"][quote="7h3.D3m0n.117"]wait so are you saying find all values of x so that the equation is true?[/quote]
He said there's only one answer.[/quote]

well there might only be one solution of x that makes it true =)

[quote="7h3.D3m0n.117"]wait so are you saying find all values of x so that the equation is true?[/quote]

What else would he mean?

[quote="mathmaster46"]He also said there was a way to simplify it without at least graphing and finding the intersection.[/quote]

Well, I doubt your teacher means $\frac{ W (\ln 2) }{ \ln 2}$.

http://mathworld.wolfram.com/LambertW-Function.html


geometry
3D geometry
algorithm
geometric transformation
reflection

[quote="ch1n353ch3s54a1l"][quote="nat mc"]i'm a beginner at cubing.  I've looked at all the "beginners solutions" and stuff, but I can never get the first layer correct. I do the cross and put in the corners, but around the sides of the first layer, there arent three in a row of the other colors.  Help me, please![/quote]
When the cube is solved, each piece is distinct. Find the correct pieces and line the sides of the first layer pieces with the corresponding center while twisting each piece correctly so that the top face is also solved...yeah. Hope this isn't too confusing..[/quote]
hmmm... I don't really understand.  Do you mean there is only one spot where each can go, so sometimes even the cross is messed up. How do you know how to get it right the first time? I still cant get it right at all...

EDIT: I just solved my first first layer!

EDIT2: Yay! Got the F2L completed!

EDIT3: I have the F2L and the face opposite the top has all the edges and corners in, but the outsides of the LL aren't right. What do I do?

[quote="nat mc"]
EDIT3: I have the F2L and the face opposite the top has all the edges and corners in, but the outsides of the LL aren't right. What do I do?[/quote]
You've oriented all the pieces, now you have to permute all the pieces. You can either use the beginner algorithms and have to orient them again, or look up the fridrich method algorithms (There are only 13 I think....minus reflections and inverses).


As the title implies I am looking for a mathematics drawing program, and I have to stress it should be a mathematics drawing program. It should be a program where you can easily draw all kind of stuff, like circels, triangles and many other things, it should be a program where you can draw sketches that can help you in answering the problem. It should be suitable for Windows and freeware. I really hope to get an answer, and I would be very gratefull, if you can help me.

Thanks!

Come on help me!  :ninja:  :ninja:  :ninja:

If you want it for $\LaTeX$ document, I advice you to use [url=http://perso.orange.fr/Fradin.Patrick/TeXgraph/fichiers/TeXgraph1-8.zip]TeXgraph[/url]. Is the better! :)

try smartdraw, it has a 30 days trial unfortunatly but it is very usefull for me


induction
function
combinatorics proposed
combinatorics

Let $S$ be a set {$1^k, 2^k, \cdots, n^k$} and $2^{k+1} | n$.

Prove that we can divide set $S$ into two subsets such that sum of elements in each set is same.

[quote="Chang Woo-JIn"]and $2^{n+1} | n$.[/quote]

I think this is a wrong hypothesis, because this implies that $n=0$.  ;)

Pierre.

we partition the set $I_{n,k+1}=\{n+1,n+2,\ldots,n+2^{k+1}\}$ for a general $n$ first.
we induct on $k$.for $k=0$ we partition as $\{(n+1)^0\}\cup\{(n+2)^0\}$.
we inductively partition the set $I=A\cup B$ so that $\sum_{a\in A} a^r=\sum_{b\in B} b^r$ for all $r\leq k$.suppose we have done that inductively for the set $I_{n,k}=\{n+1,n+2,\ldots,n+2^{k}\}$. we now write $I_{n,k+1}= I_{n,k}\cup(2^k+I_{n,k})$. bu induction we have a partition $I_{n,k}=A_1\cup B_1$. this induces a 'similar' partition on  the set $I_{n,k+1}\setminus I_{n,k}=A_2\cup B_2$.
 we claim that the partition $A_1\cup B_2, A_2\cup B_1$ will do the job for $I_{n,k+1}$.
i.e. we claim $\sum_{a\in A_1\cup B_2} a^r=\sum_{b\in A_2\cup B_1} b^r$.Now firstly when we say similar partition all we mean is that $A_2=2^k+A_1,B_2=2^k+B_1$ where $x+A=\{x+a|a\in A\}$. now by induction $\forall r<k+1 \sum_{a\in A_1} a^r =\sum_{b\in B_1} b^r$.
now \[\sum_{a\in A_1\cup B_2} a^r -\sum_{a \in B_1\cup A_2} b^r=\sum_{a\in A_1} a^r +\sum_{a\in A_2} (2^k+a)^r-\sum_{b\in A_1} (2^k+b)^r-\sum_{b\in A_2} b^r\].
$\sum_{a\in A_2} (2^k+a)^r=\sum_{m}\sum_{a\in A_2} \binom{r}{m} 2^{k(r-m)}a^m$.plugging this back we see
that the sum we have in the earlier line is simply
 $\sum_{m\neq r}\{ \sum_{a\in A_2} \binom{r}{m} 2^{k(r-m)}a^m-\sum_{b\in A_2} \binom{r}{m} 2^{k(r-m)}b^m\}$.but by induction since $m<r\leq k+1$ we have that $\sum_{a\in A_1} a^m=\sum_{b\in A_2} b^m$ and so the above difference is zero, so the induction is complete.
now if in general $2^{k+1}\vert n$ we can cut $\{1,2,...,n\}$ into subsets of size $2^{k+1}$ and by the above each such subset can be partitioned into 2 sets with equal $k^{th}$ power sums.putting these together finishes the proof.

Nice Proof!  :D . Using generating funtion can be one of the proofs.  :D  :D


calculus
integration
trigonometry
LaTeX
logarithms
function
calculus computations

Find the closed form of the integral in terms of a and b.

$ \int_{0}^{\pi} \ln(a \plus{} b \cdot \cos \ x)dx$

$ \text{\LaTeX}$ notes: use \ln,\cos,etc. for function names; it looks better. Also, \cdot (centered) is best for multiplication.

Are there any assumptions on the size of $ a$ and $ b$? For the integral to make sense as written, we need $ a>0$ and $ |b|\le a$.

[quote="jmerry"]For the integral to make sense as written, we need $ a > 0$ and $ |b|\le a$.[/quote]
How do you see these conditions?

That's the condition for which anti logarithm should be defined.

[hide="answer"]
$ \int_{0}^{\pi} \log(1 \plus{} \alpha \cos \ x)dx \equal{} \pi \log \left(\frac {1}{2}\left(1 \plus{} \sqrt {1 \minus{} \alpha^2}\right)\right)$ if $ 0\leq \alpha\leq 1$.  Let $ \alpha \equal{} \frac {b}{a}$ and add $ \log a$ to get the desired integral.

One way to derive this is to first integrate $ \int_0^\pi{\log(1 \plus{} \cos x)dx}$ using the dilogarithm.  After that use inverse trig functions to get the $ \alpha$.  It gets pretty ugly.
[/hide]

It is an example of Jensen's Theorem application: \[ I\equal{}\int_{0}^{\pi} \log(1 \plus{} \alpha \cos \ x)dx\equal{}\frac{1}{2}\int_{0}^{2\pi} \log(1 \plus{} \alpha \cos \ x)dx\equal{}\frac{1}{2}\int_0^{2\pi}\ln|f(e^{i\phi})|d\phi,\] where $ f(z)\equal{}1\plus{}\alpha\frac{z\plus{}z^{\minus{}1}}{2},$ so \[ I\equal{}\frac12\int_0^{2\pi}\ln \left|1\plus{}\alpha\frac{e^{i\phi}\plus{}e^{\minus{}i\phi}}{2}\right|d\phi\equal{}\frac{1}{2}\int_0^{2\pi}(\ln \left|\alpha e^{2i\phi}\plus{}2e^{i\phi}\plus{}\alpha\right|\minus{}\ln2)d\phi.\]
Let $ g(z)\equal{}\alpha z^2\plus{}2z\plus{}\alpha$. We are to compute $ \int_0^{2\pi}\ln|g(e^{i\phi})|d\phi$ now. Function $ g(z)$ has two zeros at $ z_1\equal{}\frac{\minus{}1\plus{}\sqrt{1\minus{}\alpha^2}}{\alpha}\in(\minus{}1,1)$ and $ z_2\equal{}\frac{\minus{}1\minus{}\sqrt{1\minus{}\alpha^2}}{\alpha}\not\in[\minus{}1,1]$. Hence, due to Jensen's Theorem, we obtain \[ \int_0^{2\pi}\ln|g(e^{i\phi})|d\phi\equal{}2\pi\ln\frac{|g(0)|}{|z_1|}\equal{}2\pi\ln\frac{\alpha^2}{1\minus{}\sqrt{1\minus{}\alpha^2}}\equal{}2\pi \ln(1\plus{}\sqrt{1\minus{}\alpha^2}).\]
Finally we get the same answer as JoeBlow pointed.


inequalities
trigonometry
integration
function
calculus
real analysis
real analysis solved

prove that for any $n \in N$, and for every $k_1,k_2,...,k_n \in N$ following inequality:
\[\min {\sum_{i=1}^{n} {\cos k_i x}} < - \frac{1}{2}\]
where $x \in [0,2\pi]$.

[quote="alexilic"]Can someone prove this?[/quote]
Yes, this is a classical problem  :) . Here is a small [hide="hint"]Denote the sum by $f(x)$. All you need to know is that $f$ is continuous, $f(x)\leqslant n$ for all $x$, $\int_0^{2\pi}f(x)\,dx=0$ and $\int_0^{2\pi}f(x)^2\,dx\geqslant \pi n$[/hide]
As usual, I'll post a complete solution if nobody solves it in a week.

Here goes:

Assume that a continuous function $f$ is bounded by $-\frac12$ from below and by $n$ from above, and $\int_0^{2\pi}f(x)\,dx=0$. Our aim is to show that then $\int_0^{2\pi}f(x)^2\,dx<\pi n$. Let $E=\{f<0\}$ and let $m$ be the measure of the set $E$. Then, since $f$ is continuous, we have $\int_E |f(x)|^2\,dx<\frac m4$, $\int_{[0,2\pi]\setminus E}|f(x)|^2\,dx<(2\pi-m)n^2$ and, therefore, $\int_0^{2\pi}|f(x)|^2\,dx<\frac m4 + (2\pi-m)n^2\leqslant \pi n^2$ if $m\geqslant 2\pi\frac{2n}{2n+1}$. So, assume that $m<2\pi\frac{2n}{2n+1}$. Then $\int_{[0,2\pi]\setminus E}|f(x)|^2\,dx\leqslant n\int_{[0,2\pi]\setminus E}|f(x)|\,dx=n\int_{E}|f(x)|\,dx<n\frac m2$ (recall that the integrals of the positive and the negative parts of $f$ are equal), so $\int_{[0,2\pi]}|f(x)|^2\,dx<\frac m4+n\frac m2<\pi n$ again.


Find the smallest integer $n$ > 1 such that $\frac{1^2 + 2^2 + 3^2 + ... + n^2}{n}$ is a square.


probability
logarithms
expected value
number theory
relatively prime
probability and stats

Hey
Imagine I have 3 options to choose from. I have only one coin ( I can toss this any number of times). Devise a scheme to choose any option with equal probability.

I hope I have been clear.

Regards

Hint:  any strategy that uses a bounded (that is, finite) number of coin tosses can only simulate probabilities with a power of $ 2$ as a denominator, so anything that works needs to involve an unbounded number of coin tosses.  (The strategy should still terminate with probability $ 1$ and the [i]expected[/i] number of coin tosses needed for it to end should still be finite.)

Flip the coin twice, in order:

HH: option 1
HT: option 2
TH: option 3
TT: start over, and flip the coin twice more.

As t0rajir0u indicated, the possible number of flips is unbounded, but the number of pairs of flips required to reach a decision is a geometrically distributed random variable with $ p \equal{} \frac34$ and hence expected value $ \frac 43$ (which would be $ \frac83$ individual coin flips.)

There's a good thread on the xkcd forums including a [url=http://forums.xkcd.com/viewtopic.php?f=3&t=32920&p=1287583&hilit=coin+flips#p1287583]proof that this is optimal[/url].

Suppose we wanted to create an event of probability $ \frac{p}{q}$ where $ p$ and $ q$ are relatively prime integers. We would find the smallest $ k$ such that $ 2^k\ge q.$ We flip a coin $ k$ times. Divide the $ 2^k$ possible outcomes into three sets: set $ A$ of cardinality $ p,$ set $ B$ of cardinality $ q\minus{}p,$ and set $ C$ of cardinality $ 2^k\minus{}q.$

If one of the outcomes in set $ A$ occurs, we say "yes"; if one of the outcomes in set $ B$ occurs, we say "no", and if one of the outcomes in set $ C$ occurs, we start over.

The expected number of flips needed to decide this is $ \frac{k2^k}{q}.$ Since $ q>2^{k\minus{}1},$ this is less than $ 2k,$ which is at worst a constant times $ \log q.$


The graph of the polynomial \[P(x) \equal{} x^5 \plus{} ax^4 \plus{} bx^3 \plus{} cx^2 \plus{} dx \plus{} e\] has five distinct $ x$-intercepts, one of which is at $ (0,0)$. Which of the following coefficients cannot be zero?

$ \textbf{(A)}\ a \qquad \textbf{(B)}\ b \qquad \textbf{(C)}\ c \qquad \textbf{(D)}\ d \qquad \textbf{(E)}\ e$

[hide]
$ d$ is the sum of the products of all sets of $ 4$ of the roots.  All but one of these includes $ 0$ and is $ 0$.  The other must be nonzero since the roots are distinct.  So $ d\neq 0$ and the answer is $ D$.
[/hide]

P(0)=e=0, then we can reduce the polynom by x(x=0 is a root of the polynom). As a result, f(x)=x^4+ax^3+bx^2+cx+d has 4 distinct roots and no root is equl to 0. So, d!=0;

$ P(x) \equal{} (x \minus{} r)(x \minus{} s)(x \minus{} t)(x \minus{} u)(x \minus{} v)$
$ \therefore d \equal{} rstu \plus{} rstv \plus{} rsuv \plus{} stuv$

Suppose that $ r \equal{} 0, \; s,t,u,v \neq 0$; then, $ d \equal{} stuv \neq 0$

The answer is $D$, due to Vietas. If the roots are $lmnop$, one root must be zero, so WLOG $l=0$. Vietas says, $d=lmno+lmnp+lmpo+lpmo+mnpo$. The values of these are $0,0,0,0,mnpo$. (Note: $E$ is definitely zero because if you plug in $0$, you get $E=0$). Since none of $mnpo$ is zero, it must be a non-zero number. Therefore $D$ cannot be zero because $D=mnpo$ and that does not equal $0$.

The given condition implies e=0. Now if d=0 then 0 is a root of P with multiplicity 2, so P has at most 4 distinct roots. Thus d cannot be zero.

[quote=Bob314159265]If a=0, then the polynomial can't have 5 roots, so isn't $A$ another possible answer?[/quote]

a is the coefficient of the x^4 term, not the x^5


Has anyone who applied through late admission heard back yet? When should I found in if I got in or not?

So Nervous

The [url=http://www.awesomemath.org/application.shtml]website[/url] says June 3rd.

Thank you for the information. I will have to wait...

lol
dont be nervous.

how many questions did you answer? and are you sure you answered them correctly? o.0

I submitted solutions for all and I hope they are right :blush: But some of them like 9 don't really show the motivation for what I did. So I have no clue.

Is there still regular admission going on or are people just put on a waitlist at this point?


A square poster is replaced by a rectangular poster that is 2 inches wider and 2 inches shorter.  What is the positive difference in the number of square inches in the areas of the posters?

[asy]size(101); draw((0,0)--(0,-8)--(8,-8)--(8,0)--cycle); draw((0,-6)--(10,-6)--(10,0)--(8,0),linetype("3 3"));
 label("2''",(0,-7),W); label("2''",(9,0),N);[/asy]

The way I would do this problem is first let the side of the square be a nice easy number, such as 10.

This means the rectangle will have dimensions $ (10 \plus{} 2) \equal{} 12$ by $ (10 \minus{} 2) \equal{} 8$. The area of the new rectangle is $ 12 \times 8 \equal{} 96$ and the original square is $ 100$, so the answer is $ 100 \minus{} 96 \equal{} \boxed{4}$.

Personally, I find this way simpler, but you could also let the side of the square be $ s$ and proceed from there.

If the side of the square is n, the area of the square is n^2. Then the area of the rectangle is (n+2)(n-2)=n^2-4, which is apparently 4 less than n^2.


[color=darkred]Let $ABC$ be a right-angled triangle ($AB\perp AC$) and let $M,N$ be two points of the line $BC$ so that $B\in (MC)$, $C\in (NB)$ and $BM=CN=x$. Denote the circumcenters $O_{1},O_{2}$ of the triangles $ABM$, $ACN$. Prove that the area of the triangle $O_{1}AO_{2}$ is equal to $\frac{a^{2}x(x+a)}{8bc}$.[/color]

[color=darkblue][b][u]A particular case.[/u][/b] If $x=p-a$, then $x(x+a)=p(p-a)=\frac{bc}{2}$ and the points $M,N$ are the tangent points with the line $BC$ of the $B$- exincircle, the $C$- exincircle respectively of the triangle $ABC$ and the area of the triangle $O_{1}AO_{2}$ is equal to $\left(\frac{a}{4}\right)^{2}$ ([b]constant[/b]) for any point $A$ of the circle with the fixed diameter $[BC]$.[/color]

Let O, K, L be midpoints of BC, BM, CN and D foot of the A-altitude.

$|\triangle AO_{1}O_{2}| = |O_{1}KLO_{2}|-(|O_{1}KDA|+|O_{2}LDA|) =$

$= \frac{O_{1}K+O_{2}L}{2}\cdot KL-\left(\frac{O_{1}K+AD}{2}\cdot KD+\frac{O_{2}L+AD}{2}\cdot LD \right) =$

$= \frac{O_{1}K}{2}\cdot (KL-KD)+\frac{O_{2}L}{2}\cdot (KL-LD)-\frac{AD}{2}\cdot (KD+LD) =$

$= \frac{O_{1}K}{2}\cdot LD+\frac{O_{2}L}{2}\cdot KD-\frac{AD}{2}\cdot KL$

$KL = x+a,$ $AD = \frac{bc}{a},$ $KD = KB+BD = \frac{x}{2}+\frac{c^{2}}{a},$ $LD = LC+CD = \frac{x}{2}+\frac{b^{2}}{a},$

$O_{1}K = OK \cdot \frac{c}{b}= \frac{x+a}{2}\cdot \frac{c}{b},$ $O_{2}L = OL \cdot \frac{b}{c}= \frac{x+a}{2}\cdot \frac{b}{c}.$

$|\triangle AO_{1}O_{2}| = \frac{x+a}{4}\cdot \left[\frac{c}{b}\cdot \left(\frac{x}{2}+\frac{b^{2}}{a}\right)+\frac{b}{c}\cdot \left(\frac{x}{2}+\frac{c^{2}}{a}\right)-\frac{2bc}{a}\right] =$

$=\frac{x(x+a)}{8}\cdot \left(\frac{c}{b}+\frac{b}{c}\right) = \frac{a^{2}x(x+a)}{8bc}$


to find the exact area of a circle without using pi is impossible but how close can you get? .000000001% ?

the egyptian octagon something is one way to go within about 1 %

because i dont know latex i will attempt to describe it:
make a 3 by 3 unit square with gridlines.
draw a circle tangent to all its INSIDE sides.
draw and octagon that's area is 7 units it should have 4 sides touching the middle thirds of the 4 sides of the square. the area of the octagon should be within 1 % of the area of the actual circle.

BUT is there a better and more accurate way?

Well, an inscribed regular polygon is a good estimate. As the number of sides goes up, the closer your estimate will be.
That's why it's not a good idea to do this with a triangle (it doesn't seem to work does it?) but your octagon is a much closer estimate.

Techincally, isn't a circle a polygon with infinate sides?

yes, technically it is
however, we just assume that the line is curved


2. A function $f$ is defined for all positive integers and satisfies $f(1)=1996$, and $f(1)+f(2)+...+f(n)=n^2f(n)$ for all $n>1$.
Calculate the exact value of $f(1996)$.

Two mistakes.  The big one, from a theoretical point of view, is that you generalized from two examples.  The smaller one is that your examples are wrong.  In particular, $2 \cdot 3 \neq 8$.

Calculate $f(n)$ for some small values of $n$. Guess a formula and prove it by induction.


factorial
function
calculus
complex analysis
number theory

Factorial is not defined for negative integers. There is an extension of factorial to the complex plane excluding negative integers. See the [url=http://en.wikipedia.org/wiki/Gamma_function]gamma function[/url].

As Hamster1800 wrote, you've picked the single set of values (the negative integers) on which it is not possible to naturally extend the factorial function.  In fact, this is not too surprising: the recurrence $ (n\plus{}1)! \equal{} (n\plus{}1)\cdot n!$ can't possibly be extended to $ n \equal{}\minus{}1$, since to solve it for $ (\minus{}1)!$ would require division by 0.  Once you've learned calculus (and even better, a little complex analysis or analytic number theory), you'll be in a much better position to understand what the gamma function is and how it works.

The factorial is only defined everywhere except the non-positive integers like JBL explained. And you are not able to extend it to these bad points analytically.  :(

[quote="binomial_4eva"]what is (-1)! equal to ?
in general what is (-n)! equal to?

please dont say this doesnt have an answer,[/quote]
[quote="[url=http://en.wikipedia.org/wiki/Gamma_function]Wikipedia[/url]"]The Gamma function is in fact defined for all complex numbers z except for the nonpositive integers (z = 0, − 1, − 2,...).[/quote]


A man is at one corner of a square pool. He can run at a constant rate $r$ and swim at a constant rate $s$. If he wants to move to the opposite corner of the pool, what path should he take?

Shouldn't these be somewhere else?  Using the fact that the shortest distance between two points is a straight line, we need only consider paths from

$(0,0)\to (t-a,0)\to (t,b)\to (t,t)$ where $t$ is the length of the square, and where $0<a,b<t$

We see that the time required is $T=(2t-(a+b))/r+\sqrt{a^2+b^2}/s$

The time required to run the entire way is $2t/r$; the time required to swim all the way is $\sqrt{2}t/s$

Say that $\sqrt{2}t/s \geq (2t-(a+b))/r+\sqrt{a^2+b^2}/s$.  I claim that this is true if $r/s\geq \sqrt {2}$.

Indeed, we need only show that $\sqrt{2}(\sqrt{2}t-\sqrt{a^2+b^2})\geq 2t-(a+b)$, which reduces to $2(a^2+b^2)\geq (a+b)^2$, which is trivial from AM-GM.

So if $r/s\geq \sqrt{2}$, it's best to run the whole way.  Now, say that

$2t/r \geq (2t-(a+b))/r+\sqrt{a^2+b^2}/s$.  I claim that this is true if $r/s\geq \sqrt{2}$.  This is done by more of the same.  The result follows


Let $A$,  $B$,  $C$,  $D$ be (not necessarily square) real matrices such that $A^{T}=BCD$, $B^{T}=CDA$, $C^{T}=DAB$, $D^{T}=ABC$. For $S=ABCD$ prove that  a) $S^{3}=S,$ b) $S^{2}=S.$
(V. Brayman)

See all the problems from [color=orange]Kyiv Mechmat Competition, 2004[/color] [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=99484]here[/url]

[quote="rogue"][quote="zhaobin"]a)$S=ABCD=AA^{T}$
$S^{3}=D^{T}C^{T}B^{T}A^{T}=(BCD)^{T}A^{T}=AA^{T}$
B)$S^{2}=ABCDABCD=A(CDA)^{T}CDABCD=AA^{T}D^{T}C^{T}CDABCD=ABCDABCDABCD=S^{3}=S$[/quote]
a) is solved! :lol:   b) Why $AA^{T}D^{T}C^{T}CDABCD=ABCDABCDABCD$? I think it may be wrong   :huh:[/quote]
 :blush: sorry it is wrong

Hint[quote]Let $A$,  $B$,  $C$,  $D$ be (not necessarily square) [color=orange][b]real[/b][/color] matrices such that $\ldots$[/quote]
The statement is wrong in gereral even for $1\times1$ [color=orange][b]complex[/b][/color] matrices! :wink:

As zhaobin noticed S is symetric real and positive
it is diagonalisable via an orthogonal matrix.
With $S^{3}=S$ and the positivity of S we deduce the eingenvalue of S are
0 or 1

$S=Pdiag(s_{1},...,s_{n})P^{t}$ where $s_{k}=0$ or $s_{k}=1$
and P is an orthogonal matrix.

we have
$S^{2}=S$

$S$ is a projector


Can we prove that there exists a translate of a Cantor set which contains no rational number (preferably using some sort of measure-theoretic argument)?

Let $C$ be the Cantor middle-third set. Given a rational $q$, we have $q\in t+C\iff t\in q-C$. This means that the set consisting of those real $t$ for which $q\in t+C$ is precisely $q-C$. The set of reals $u$ such that $\mathbb Q\cap(u+C)\ne\emptyset$ is thus $\bigcup_{q\in\mathbb Q}(q-C)$, and this is a countable union of sets of measure zero, so it can't contain every real number (or, if you prefer a categorical argument, it's a set of category one, so it cannot cover the whole real line :)).


induction
trigonometry
limit
real analysis
real analysis unsolved

can you show $ \textbf{LHS}\equiv\textbf{RHS}$ ?  :maybe:

$ 1\minus{}5\left(\frac{1}{2}\right)^{3}\plus{}9\left(\frac{1\cdot 3}{2\cdot 4}\right)^{3}\minus{}13\left(\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)^{3}\plus{}\;\;\;\ldots$
$ \equal{}\;\;\;\frac{\sqrt2}{2}\cdot\frac{\sqrt{2\plus{}\sqrt2}}{2}\cdot\frac{\sqrt{2\plus{}\sqrt{2\plus{}\sqrt2}}}{2}\cdot\frac{\sqrt{2\plus{}\sqrt{2\plus{}\sqrt{2\plus{}\sqrt2}}}}{2}\cdot\;\;\;\ldots$

$ \;\;\;\boxed{\left(\equal{}\frac{2}{\pi}\right)}$

for the second indentity:
by induction its easy to prove that $ \underbrace{\sqrt{2\plus{}\sqrt{2\plus{}\sqrt{2}\plus{}\cdots\plus{}\sqrt{2}}}}_{k}\equal{} 2\cos\frac{\pi}{2^{k\plus{}1}}$

now $ \lim_{n\rightarrow\infty}\cos\frac{\pi}{4}\cdot\cos\frac{\pi}{8}\cdot\ldots\cdot\cos\frac{\pi}{2^{n}}$

$ \equal{}\lim_{n\rightarrow\infty}\cos\frac{\pi}{4}\cdot\cos\frac{\pi}{8}\cdot\ldots\cdot\cos\frac{\pi}{2^{n}}\cdot\sin\frac{\pi}{2^{n}}\cdot\frac{1}{\sin\frac{\pi}{2^{n}}}$

$ \equal{}\lim_{n\rightarrow\infty}\frac{\sin\frac{\pi}{2}}{2^{n\minus{}1}\sin\frac{\pi}{2^{n}}}$

$ \equal{}\lim_{n\rightarrow\infty}\frac{\frac{\pi}{2^{n}}}{\sin\frac{\pi}{2^{n}}}\times\frac{2}{\pi}\equal{}\frac{2}{\pi}$

yes sinajackson,my soln is the same.but how do you do the first part?


floor function
ceiling function
induction
inequalities
logarithms
number theory unsolved
number theory

Let $X$ be a non empty subset of $\mathbb{N} = \{1,2,\ldots \}$. Suppose that for all $x \in X$, $4x \in X$ and $\lfloor \sqrt{x} \rfloor \in X$. Prove that $X=\mathbb{N}$.

Main idea: (I'll hide it if you don't want to see it) :
[hide]
$1 \in X$ and $\forall k \in \mathbb{N}^*$, $4^k \in X$.
And for $x,k,n \in \mathbb{N}^*$,    $\quad \frac{\textrm{ln}(1+\frac{1}{x})}{\textrm{ln}(2)} \geq \frac{1}{2^{n-1}}\Longrightarrow
x^{2^n}\leq 4^k < {(x+1)}^{2^n}$.
Take $n$ sufficiently large and after applying step 2 on $4^k$     $n$ times we get $x \in X$.
[/hide]

[quote="Igor"]Define $[u]$ as the integer part of any real $u$.
Let $X$ be a non empty subset of $\mathbb{N}^*$.
Suppose that for all $x \in X$, $4x \in X$ and $[\sqrt{x}] \in X $.
Prove that $X=\mathbb{N}^*$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=242638]Link to the test[/url][/quote]

what is $X=\mathbb{N}^*$ ?

Surely it is not the group of unit natural numbers i.e. the trivial group $\{1\}$.... :P

Why do you put * next to $\mathbb{N}$:?:

Obviously for any integer [tex]k[/tex] we have [tex]4^k \in X[/tex]. Thus [tex]\lfloor \sqrt{4^k} \rfloor = 2 k \in X[/tex], so for any integer [tex]m[/tex], we also have [tex]4 m \in X[/tex]. Thus [tex]\lfloor \sqrt{4 m} \rfloor = \lfloor 2 \sqrt{m} \rfloor \in X[/tex], for any integer [tex]m.[/tex] Let's take [tex]m = \lceil \frac{a^2}{4} \rceil[/tex] for a given integer [tex]a \geqslant 2[/tex] then [tex]\lfloor 2 \sqrt{m} \rfloor = a[/tex], the conclusion follows... :)

Maybe I don't understand something, but $\lfloor \sqrt{4^k} \rfloor = 2^k$ not $2 k$. So you can say that only powers of 2 are in $X$, and nothing more. After that Igor's solution is the only way I think.

Yes, of course... sorry :oops:
(I really don't know why I tought that sqrt(4^k)=2*k  :P )

My solution is mainly using first principle of mathematical induction

Part 1:
[hide="to prove that 1 is in X"]
this is basically a rigorous representation of common sense
let $ f\left( x \right) = \left[ {{x^{1/2}}} \right]$

$ \& f\left( x \right) \le {x^{1/2}} \\
\& f \circ f\left( x \right) \le \left( {f{{\left( x \right)}^{1/2}}} \right) \le {\left( {{x^{1/2}}} \right)^{1/2}} \le {x^{1/{2^2}}} \\
\& \implies {f^n}\left( x \right) \le {x^{1/{2^n}}}$

Also we  know that
$ {\text{If }}x \in X \implies {f^n}\left( x \right) \in X{\text{ }}\forall {\text{ }}n \in {\Bbb N}$

for any particular $ x \in X$
let $ g(n) = f^n(x)$

$ \& 1 \le g\left( n \right) \le {x^{1/{2^n}}} \\
\& {\text{as }}n \to \infty ,{x^{1/{2^n}}} \to 1 \\
\& \implies {\text{ For some }}n \in {\Bbb N}{\text{ we have }}{{\text{x}}^{1/{2^n}}} < 2 \implies g\left( n \right) = {f^n}\left( x \right) = 1$
$ %Error. "since" is a bad command.
g(n) \in X \ \forall \ \n \in \Bbb N \implies \boxed{1 \in X}$
[/hide]

Part 2.
[hide="to prove that if x in X then (x+1) is also in X"]
If $ x \in X \implies 4^nx \in X \ \forall n\in \Bbb N$

Now our task is to show that there exists a number of form $ 4^nx$ for every $ x\in X$ such that after repeatedly rooting and finding greatest integer, we get $ x + 1$.

thus we must prove that there exists a number $ 4^nx$ such that $ (x + 1)^{2^n} \le 4^kx < (x + 2)^{2^n}$ where $ n ,k \in N$
[hide="Why so?"]
$ %Error. "eqalign" is a bad command.
{ \& {(x + 1)^{{2^n}}} \le {4^k}x < {(x + 2)^{{2^n}}} \\
\& \implies {(x + 1)^{{2^{n - 1}}}} \le \sqrt {{4^k}x} < {(x + 2)^{{2^{n - 1}}}} \\
}$
${ \& \implies {(x + 1)^{{2^{n - 1}}}} \le \left[ {\sqrt {{4^k}x} } \right] < {(x + 2)^{{2^{n - 1}}}} - - - - - \left[ {\because {{(x + 1)}^{{2^{n - 1}}}}{\text{ and }}{{(x + 2)}^{{2^{n - 1}}}} \in {\Bbb N}} \right] \\
\& \dots \\
\& \dots \\
\& \dots \\
\& \implies (x + 1) \le \left[ {\sqrt {\left[ {\sqrt {\dots \left[ {\sqrt {{4^k}x} } \right]} } \right]} } \right] < x + 2 \\
\& \implies \left[ {\sqrt {\left[ {\sqrt {\dots \left[ {\sqrt {{4^k}x} } \right]} } \right]} } \right] = x + 1 \in X }$

[/hide]

the inequality can be rewritten as-
\[ \& {4^{{2^n}{{\log }_4}\left( {x + 1} \right)}} \le {4^{k + {{\log }_4}\left( x \right)}} < {4^{{2^n}{{\log }_4}\left( {x + 2} \right)}} \\
\& \Leftrightarrow {4^{{2^n}{{\log }_4}\left( {x + 1} \right) - {{\log }_4}\left( x \right)}} \le {4^k} < {4^{{2^n}{{\log }_4}\left( {x + 2} \right) - {{\log }_4}\left( x \right)}}\]
consider the case of $ k = \left[ {{2^n}{{\log }_4}\left( {x + 1} \right) - {{\log }_4}\left( x \right)} \right] + 1 = \left[ {{2^n}{{\log }_4}\left( {x + 1} \right) - {{\log }_4}\left( x \right) + 1} \right]$

we know that $ k < {2^n}{{\log }_4}\left( {x + 2} \right) - {{\log }_4}\left( x \right)$
so,
\[ \left[ {{2^n}{{\log }_4}\left( {x + 1} \right) - {{\log }_4}\left( x \right) + 1} \right] < {2^n}{\log _4}\left( {x + 2} \right) - {\log _4}\left( x \right) \\
\Leftarrow {2^n}{\log _4}\left( {x + 1} \right) - {\log _4}\left( x \right) + 1 < {2^n}{\log _4}\left( {x + 2} \right) - {\log _4}\left( x \right)\]

\[ \& \Leftrightarrow 4{\left( {x + 1} \right)^{{2^n}}} < {\left( {x + 2} \right)^{{2^n}}} \\
\& \Leftrightarrow 4 < {\left( {1 + \frac {1} {{x + 1}}} \right)^{{2^n}}}\]
which is true if we take arbitrarily high values of $ n$
hence, for every value of x, we have such a number $ 4^kx$
hence $ \because x \in X \implies 4^kx \in X \implies (x + 1) \in X$

hence $ x\in X \forall x \in \Bbb N$
[/hide]


What is the sum of all values of $ z$ such that $ z^2\equal{}12z\minus{}7$?

We can change
\[ z^2 \equal{} 12z \minus{} 7\]
into
\[ \color{red}(1)\color{black}z^2 \color{green} \minus{} 12\color{black}z \plus{} 7 \equal{} 0\]
By Vieta's, the sum of the values of $ z$ is $ \minus{}\left( \frac {\color{green} \minus{} 12}{\color{red}1}\color{black}\right) \equal{} \boxed{12}$


LHS is equal to $\frac{\ln(a-1)+\ln(a+1)}{\ln a}$
RHS is equal to $\frac{\ln a}{\ln(a+1)}+\frac{\ln a}{\ln(a-1)}= \ln a\cdot\left(\frac{\ln(a-1)+\ln(a+1)}{\ln(a-1)\ln(a+1)}\right)$
So we want to prove that $\ln^{2}a > \ln(a-1)\ln(a+1)$, i.e. $\log_{a-1}a > \log_{a}(a+1)$. To prove this, I refer to [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=583891#583891]this post[/url].

The last inequality can also be proven using the fact that $\ln{\ln{x}}$ is concave. Jensen's says that

$\frac{1}{2}\ln{\ln{(a-1)}}+\frac{1}{2}\ln{\ln{(a+1)}}< \ln{\ln{a}}$,

as desired.

[b]Remark.[/b] $\boxed{\ 1<x<y\ ,\ z>0\ }$ $\implies$ $1<\frac{y+z}{x+z}<\frac{y}{x}$ $\implies$ $\log_{x}\frac{y}{x}>\log_{x}\frac{y+z}{x+z}>\log_{x+z}\frac{y+z}{x+z}$ $\implies$

$\log_{x}y-1>\log_{x+z}(y+z)-1$ $\implies$  $\boxed{\ \log_{x}>\log_{x+z}(y+z)\ }\ .$

[b]Exercise[/b] (from [u]Dan Seclaman, Craiova[/u]) :

$1.\blacktriangleright\boxed{\ 0<a<b\implies \ln\left(1+\frac{a}{b}\right)\cdot\ln\left(1+\frac{b}{a}\right)<\ln^{2}2\ }\ .$

$2.\blacktriangleright\boxed{\ a>0\ ,\ b>0\implies (1+a^{2})^{b}\cdot(1+b^{2})^{a}\le (1+ab)^{a+b}\ }\ .$


For which natural $ n$ numbers the $ 2007 \plus{} 4^n$ is a perfect square of an integer? :)
Though it's easy very few contestants have solved this problem!! :(

$ \equal{}2007\equal{}x^2\minus{}2^{2n}\equal{}(x\minus{}2^n)(x\plus{}2^n)$
i don't know if 2007 is prim or now,
but if 2007 is prime then $ x\equal{}2^{n}\plus{}1$ and $ 2^{n\plus{}1}\plus{}1\equal{}2007$ impossible

3 mod 4
 :P
[quote="Γιώργος"]
Though it's easy very few contestants have solved this problem!! :([/quote]
btw
as far as i know most of the contestants solved this problem....

[hide]Is it just as easy as taking each side of $ 2007 \plus{} 4^n \equal{} m^2$ mod 4?[/hide]

2^2n + 2007 = (2^n+k)^2 => k(2^(n+1)+k) = 2007
k can take positive integer value divisors of 2007 less than 2007^0.5
=> k=1,3,9.
for these k value, 2^(n+1) =2006, 666, 214.
None of these are powers of 2.
So, the given expression has no solution.


algebra
polynomial
function
calculus
integration
complex analysis
complex analysis unsolved

1. Let $ f$ be an entire function and suppose there is a constant $ M$, and $ R>0$, and an integer $ n\geq 1$ such that $ |f(z)|\leq M|z|^n$ for $ |z|>R$. how that $ f$ is a polynomial of degree $ \leq n$.

think about Cauchy integral formula, then take the limit

Take a look at my post here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=306732

[quote="$ math$"]1. Let $ f$ be an entire function and suppose there is a constant $ M$, and $ R > 0$, and an integer $ n\geq 1$ such that $ |f(z)|\leq M|z|^n$ for $ |z| > R$. how that $ f$ is a polynomial of degree $ \leq n$.[/quote]

Suppose $ f(z) \equal{} \sum_{n\equal{}0}^\infty a_nz^n$. Write $ \int_0^{2\pi} |f(re^{it})|^2 dt$ in terms of the $ a_n$'s and think about $ r\rightarrow \infty$.


$ a,b,c,d,e > 0, abcde \equal{} 1$. prove the following inequality
$ 81(a^3 \plus{} 2) (b^3 \plus{} 2) (c^3 \plus{} 2) (d^3 \plus{} 2) (e^3 \plus{} 2) \geq \left( 2 \sum ab \plus{} 7 \right)^3$
$ \left( \sum ab \equal{} ab \plus{} ac \plus{} ad \plus{} a e \plus{} bc \plus{} bd \plus{} be \plus{} cd \plus{} ce \plus{} de \right)$

I found it accidentally. :lol:

why ?  i  think it's hard and beatifiul.


I'm just wondering, what's the AoPS-Mathlinks forum's policy about double posting?

[quote="nr1337"]Watch this:

EDIT:  See?  No double post!  Just click the "edit" button![/quote]

Yeah, I know about editting and deleting posts, but I'm just wondering because I see a lot of people doing it. I guess it's not much of a matter here, because everyone's not as concerned about their post counts.

Well it may not really be worse but at one hand it's kinda unfair that double posters get a higher post count than others, but on the other hand it's often handy that one can verify that your other post is unedited. Like when it comes to "you said this" or "you said that" I really don't want a mod to have edited my post...

The one problem with the post count system here is that it emphasizes quantity over quality, big-time. It is possible to implement a collaborative post-QUALITY rating system,  but perhaps not with phpBB technology. There is, um, a certain notorious example of a person responding in precisely the wrong way to the incentives of the current system here at AoPS.

I'm curious - how would your quality-rating system work?

And it's not like post count really matters, anyway.

[quote="Wumbate"]I'm curious - how would your quality-rating system work?[/quote]

Well, it's not "my" system, and indeed I actually hardly ever hang out on boards with quality ratings (which are rather rare, because they are hard to implement). The basic idea usually is that a trusted initial group of members are given votes to vote on which posts they like. On some boards, it's always the same initial group of voters, and on others, people can join the voting group by being voted on as a people who post good posts. One example of a board with rated posts is [url=http://slashdot.org/]Slashdot[/url], which may be the most visited board with that kind of technology.

I agree with everybody. Double posting isn't such a bad thing except when it is in the form of SPAM. In that case, it should be deleted whether the poster double posted or not.

plus another reason why it is necessary is if you post half a sol, and later you find the rest of the sol

then you want that people see there's a new post

if not no one will ever come read your sol so it's useless even te post it ;)

[quote="tokenadult"][quote="RC-7th"]Besides, aren't moderators there to delete extraneous posts?[/quote]

That's something I do fairly aggressively on the two forums I moderate or co-moderate, but some of the forums are rather lightly patrolled by all of their moderators. To me, a double post isn't as bad as a just plain time-wasting post, which fortunately I don't see a lot of in the forums I moderate.[/quote]

Why do I somehow think that you are refering to the Games and Fun Factory forum?  If you are that forum is out a bit because 2 out of the 4 mods are gone elsewhere and with only 2 mods trying to keep up with it it gets a little overlooked sometimes.

[quote="Peter VDD"][quote="yif man12"][quote="Peter VDD"]if not no one will ever come read your sol so it's useless even te post it ;)[/quote]

I don't know. I think a solution is very useful even if nobody ever reads it. You figured out a problem and that's not wasted time.[/quote]

1. After you figure it out it you still have to type it, that's loss of time
2. Sometimes you just solve a problem to help the person asking it, so in that case i'm also right

But I can understand your point of view as well ;)[/quote]

Same here. :)


function
superior algebra
superior algebra unsolved

1. Let R be any ring, and define a function D:R[X] →R[X]  by setting D[∑a_nX^n]=∑na_nX^(n-1)].

a) check that if f(X) = ∑a_nX^n and g(X) = [∑b_nX^n, then D[f+g]= D[f]+D[g].

b) check that D[a_ib_jX^(i+j)]=D[a_iX^i]b_jX^j + a_iX^iD[b_jX^j].

c)Now using the additivity in a), show that D[f(X)g(X)]= D[f(X)]g(X)+f(X)D[g(X)].

2. Using the results from 1, show inductively that D[f(X)^n] = nf(X)^(n-1)D[f(X)] for all n ≥ 1.

3. Again using 1 and 2, Let R be any ring. Suppose that f(X) ∈R[X] has a repeated root, in the sense that f(X) = g(X)(X-r)^2 for some r ∈R and g(X)∈R(X). Prove that D[f(x)] has value 0 at r.

i have shown true for all of question 1, and now i am stuck on questions 2 and 3


:blush: cómo borro esto. Cometí un error

¿[color=blue]No hay una opción, cerca de "Quote", que lleve el nombre "Delete"?[/color]

dice delete/edit pero dentro de esta no la encuentro. No es lo mismo un topic que un post.

[color=blue]Pide, entonces, a los moderadores que eliminen este tópico.
Ellos tienen la potestad de eliminar todo lo que consideren necesario.[/color]

okey .. pues entonces lo pido ... Sr. Carlos Bravo, sería tan amable?
Gracias


A rhombus is formed by two radii and two chords of a circle whose radius is 16 feet. What is the area of the rhombus in square feet?

[quote="pianoforte"][hide] about 221.7.25 ft^2[/hide][/quote]

Where's the solution? Show your work. (I knwo you're new, so....)

Good Job!  :D
I hope you find you time spent on the AoPS forum is fun and worthwhile!


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

I am stuck on a few questions. If anyone would like to help, that would be awesome.
I means identity matrix
1. let A be an n x n matrix with all entries 1's. Show that det((k-m)I+mA)=(k+(n-1)m)(k-m)^(n-1)
2. let A=CBC^-1. Show that det(A-mI)=det(B-mI)
3. Let A be an n x n matrix with an eigenvector u and eigenvalue m. Assume M is invertible and has u as its first column. What is the first column of B. Show that B has an eigenvalue m as well.
Thanks

[quote="jmerry"]1. This is nicer if you write $M=mJ+lI$ for the matrix we're taking the determinant of; you're finding the characteristic polynomial of $J$, slightly generalized. Do some row reduction: subtract the first row from each other, so that only the first row, the first column, and the diagonal have nonzero elements. Then expand by the first column or by the first row.
(Your statement has a mistake)

2. $A-mI$ and $B-mI$ are similar.

3. The statement is incomplete. The problem cannot be parsed.[/quote]

Thanks for your reply.
For
1. isn't it extremely tedious to do row reduction after the second pivot?
2. Is it true that det(A)=det(B)? and how do you deduce that A-mI and B-mI are similar?
3. The question is exactly as it is worded

I'm not going to give you everything- think about how these work. The row reduction in #1 is not at all tedious; I did it in my head. You're not trying to complete the process, or even zero out the first column.


If a and b be  any two roots of x^4 + 2x +1 = 0.
Prove that  a+b  is a root of the equation x^6 - 4*(x^2) – 4 = 0


[hide]Jamie = 15 min/lap
Holly = 40 min/lap
Liz = 5 min/lap

$\text{lcm}(5,15,40)=120$

$\boxed{2\text{ hours}}$[/hide]

[quote="mr. math"]Jamie, Holly, and Liz are running around a track. Jamie runs 12 laps per 3 hours, Holly runs 6 laps per 4 hours, and Liz runs 24 laps per 2 hours. If they all start together, how many hours will it take them to all return to the begginning at the same time?[/quote]
[hide]
Jamie takes 1/4 hour to run a lap, holly takes 2/3 hour to run a lap, and liz takes 1/12 hour to run a lap.
after 24/12 hour, which is 2 hours, they all completed an integer amt of laps.
it is 24/12 because 24/12 is the LCM of 3/12, 8/12, and 1/12 (ignoring the denominator)

the answer is $2$ hours.
[/hide]


Let $p$ denote a prime number of the form $4k+3$ and $x,y$ be positive integers. Prove that $x^2-xy+\frac{p+1}{4}y^2$ does not have any divisor of the form $kp-1$.

We need to assume that $x,y$ are coprime or something similar.
When both have some other common divisors, we just forget them and thus assume $x,y$ coprime.
Let $q \neq p$ be a odd prime divisor of $x^2-xy+\frac{p+1}{4}y^2=\left(x-\frac{y}{2}\right)^2+\frac{p}{4}y^2$, so by coprimality especially $q\nmid y$.
Then $-p \equiv \left(\frac{2x-y}{y}\right)^2 \mod q$.
Thus $1=\left( \frac{-p}{q} \right) = \left( \frac{q}{p} \right)$, thus $q$ is a square $\mod p$.
Similar we can treat $q=2$ and get the same.
Since squares are closed under multiplication, we cannot reach $-1 \mod p$ (by multiplication of the prime divisors) which is a nonsquare.


Is a little bit of memorization good? I mean, I just can't remember all the formulas and theorems covered in intro to geo.

[quote="pythag011"]Well, if use them a lot, you will automatically remember them...[/quote]Quoted for truth.

You have to do the practice problems. If you do them you will remember everything [b]that comes up frequently[/b].

My friend: "How did you do #14 on the AMC 12? You had to use shoelace to find the area of the triangle, right? because I forgot shoelace"

Me: "Um, base times height divided by two?"

[quote="Poincare"]but, i don't have enough time to do all those problems![/quote]

Didn't you say you were in 5th grade or something?  That's a lot more time than almost anyone else here.

When it comes to something like doing problems to drill formulas or whatever into your brain, it doesn't matter how many of them you solve. If you spend a lot of time (hours) really struggling with one problem just outside of your reach, then regardless of whether you solve it or not, you will know all the related concepts upside down and backwards. (said from personal experience)

I don't know what kind of memorization you're referring to. Mnemonic-type stuff is almost always total garbage unless it's a bashing thing like Stewart's; yay dad + man = bomb + sink. For harder problems, you need to have a [i]lot[/i] of intuition behind anything you learn, or you will never see when to use it in an actual problem. Memorization is then a moot point.


search
combinatorics unsolved
combinatorics

[quote="polers"]Hi all,

Can somebody help me on this?

If let's say we are interested to find out the number of simple paths from (0,0,0,0) to (1,1,1,1) in a tesseract, how should we go about doing it?

Hope that you guys can help me on this... thanks =)[/quote]

It is always better to define terms like tesseract. Everybody doesn't care to search in wikipedia ( http://en.wikipedia.org/wiki/Tesseract ) like me.

And define where the tesseract is? (In this case it is obvious that its vertices are $ \{(a,b,c,d)|a,b,c,d\in \{0,1\}\}$
And what are simple paths?

One more thing: This should go to a combinatorics forum.

Does simple paths mean going parallel to the axes and turning only at lattice points? (A lattice point is a point $ (a,b,c,d)$ with $ a,b,c,d$ integers)
In that case exactly one of $ a,b,c,d$ in $ (a,b,c,d)$ changes by one. Since the paths are confined within the tesseract, one of $ a,b,c,d$ will change from $ 0$ to $ 1$(not the other way around, because in that case the path returns to a previous position). For the first move there are $ 4$ choices, Then $ 3$, then $ 2$ and then $ 1$, Thus total $ \boxed{4!}$ paths.

That would be true if all moves were in the same direction (from 0 to 1). Actually, the answer is 6432 and the only way I could get it was to write a program and let the computer do the actual computation. Needless to say, this brute force approach is rather useless in higher dimensions :(.

Oops :oops: . Sorry. I thought simple paths meant that.


Equilateral triangles $ ABQ$, $ BCR$, $ CDS$, $ DAP$ are erected outside of the convex quadrilateral $ ABCD$. Let $ X$, $ Y$, $ Z$, $ W$ be the midpoints of the segments $ PQ$, $ QR$, $ RS$, $ SP$, respectively. Determine the maximum value of
\[ \frac {XZ\plus{}YW}{AC \plus{} BD}.
\]

This is Problem four in CGMO 2008 proposed by XiongBing  :wink:
[b]lemma:[color=blue]if $ {\triangle QBA }$and $ {\triangle PAD}$are Equilateral triangles denote $ {E,H,A,X}$ are the midpoints of segment $ {BA,AD,EH,QP}$ Then we have :$ {XT\equal{}\frac{\sqrt{3}}{4}DB}$;proof: denote the midpoint of $ {QA,PA}$ are $ {A',A''}$ We can easily get :$ {\triangle A'XE\cong \triangle A''HX \cong \triangle AHE}$ $ { \Longrightarrow}$$ {\triangle HXE}$ is an equilateral triangle ,the result easily follows.[/color]
Then we begin to solve this problem:[color=brown]
denote $ {E,H,F,G,T,U}$ are the midpoints of segments :$ {AB,AD,BC,DC,EH,FG}$;
Easy to get :$ {XZ\le XT\plus{}TU\plus{}ZU}$; with these midpoints we can get :$ {TU\equal{}EF\equal{}\frac{1}{2}AC}$;and by our lemma :$ {TX\equal{}ZU\equal{}\frac{\sqrt{3}}{4}BD}$;So we get :$ {XZ\le\frac{\sqrt{3}\plus{}1}{2}AC}$;
with the same way ,we similarly get that :$ {ZW\le\frac{\sqrt{3}\plus{}1}{2}BD}$
FINALLY:we get the inequality:$ {\frac {XZ\plus{}WY}{AC\plus{}BD}\le\frac{\sqrt{3}\plus{}1}{2}}$;
So the maximum value of $ {\frac {XZ\plus{}WY}{AC\plus{}BD}}$ is $ {\frac{\sqrt{3}\plus{}1}{2}}$; the equality reach when $ {ABCD}$ is a square (just like the image shows) .
[/color]
[/b]


number theory
prime numbers
number theory proposed

EDIT: [hide="Easy"]
This is equivalent to saying $ p_{i}\equiv\pm 1\mod{p_{i\plus{}1}}$.  So just take the smallest $ p_{i}$: this is clearly impossible unless that $ p_{i}$ is $ 2$.
Then, WLOG, $ p_{1}\equal{} 2$.  Then $ p_{2}\equal{} 3$, $ p_{3}\equal{} 2$, $ p_{4}\equal{} 3$, ...
This is impossible as $ 2007$ is odd.

[/hide]


Let $ c$ be the constant coefficient of $ P(x)$. We have $ c^2\equal{}c$.

(i) $ c\equal{}0$
$ P(0)\equal{}0$ $ \implies$ $ P(2)\equal{}0$ $ \implies$ $ P(6)\equal{}0$ $ \implies$ $ P(38)\equal{}0$ $ \implies\ldots$. Since $ x^2\plus{}2>x,\forall x\in\mathbb{R}$, then $ 0,2,6,38,\ldots$ is an increasing sequence of real numbers which are roots of $ P(x)$. Then $ P(x)$ has infinitely many roots, which implies that $ P(x)$ is constant. $ \boxed{P(x)\equal{}0,\forall x\in\mathbb{R}}$.

(ii) $ c\equal{}1$
Let $ P(x)\equal{}xP_1(x)\plus{}1$. Let $ c_k$ be the constant coefficient of $ P_k(x)$ for all $ k\in\mathbb{N}$.
$ (xP_1(x)\plus{}1)((x\plus{}1)P_1(x\plus{}1)\plus{}1)\equal{}(x^2\plus{}2)P(x^2\plus{}2)\plus{}1$
$ c_1\equal{}2c_1\implies c_1\equal{}0$
Let $ P_1(x)\equal{}xP_2(x)$
$ (x^2P_2(x)\plus{}1)((x\plus{}1)^2P_1(x\plus{}1)\plus{}1)\equal{}(x^2\plus{}2)^2P(x^2\plus{}2)\plus{}1$
$ c_2\equal{}4c_2\implies c_2\equal{}0$
Let $ P_2(x)\equal{}xP_3(x)$.
Continuing this process, it must continue indefinitely. So $ P_1(x)$ must be constant, i.e., $ P_1(x)\equal{}0$. Thus $ \boxed{P(x)\equal{}1,\forall x\in\mathbb{R}}$.

your solution is similiar to mine but shorter and beter.

[quote=jgnr]Let $ c$ be the constant coefficient of $ P(x)$. We have $ c^2\equal{}c$.[/quote]
How is that true? And what about the solution $P(x)=x^2-x+2$?
In fact, the only solutions are $P(x)=(x^2-x+2)^n$ for $n=0,1,2,\dotsc$ and $P \equiv 0$. See [url=http://artofproblemsolving.com/community/c6h392562p2239953]here[/url].


Let ABC be a triangle and a,b,c  length sides of it.If a(1-2cosA)+b(1-2cosB)+c(1-2cosC)=0  then show that ABC be an equilateral triangle.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=48627 .

  darij


limit
trigonometry
calculus
calculus computations

Find limit:
\[ \lim_{x\to 0}\frac{\sqrt{1\plus{}tanx}\minus{}{}\sqrt{1\plus{}sinx}}{x^{3}}\]

$ O(x^{3})$ ... what does this means?
I can not understand what you wrote  :(
I have edited my post ... sorry  :) Rust, can you explanin again?

Here is the solution for Japanese high school students, because they don't learn Landau sign.

$ \lim_{x\rightarrow 0}\frac{\sqrt{1\plus{}\tan x}\minus{}\sqrt{1\plus{}\sin x}}{x^{3}}$

$ \equal{}\lim_{x\rightarrow 0}\frac{\tan x\minus{}\sin x}{x^{3}(\sqrt{1\plus{}\tan x}\plus{}\sqrt{1\plus{}\sin x})}$

$ \equal{}\lim_{x\rightarrow 0}\frac{\tan x}{x}\cdot\frac{1\minus{}\cos x}{x^{2}}\cdot\frac{1}{\sqrt{1\plus{}\tan x}\plus{}\sqrt{1\plus{}\sin x}}$

$ \equal{}1\cdot\frac{1}{2}\cdot\frac{1}{1\plus{}1}\equal{}\boxed{\frac{1}{4}}$


AMC
AMC 12
AMC 12 A
AMC 10
AMC 10 A
USA(J)MO
USAMO

So who is the freshman that made the 120 on the A? Someone from Northview perhaps?

This is a wild guess, but maybe Robert Grosse, formerly of Chamblee Middle?

I'm thinking that's a good guess. Where is he now? Northview?

I don't know. I saw him at Rockdale, and his team didn't have anyone I recognized, but then they'd all be young people anyway who I probably wouldn't know if I had seen them before.

The other possibility was that it was one of the Sculleys. I don't know how they did, but they're good.

EDIT: Wasn't young Mr. Sculley, who got 130.5 on the AMC10A and 4 on the AIME.

Does anyone know who the 144 on the AMC10A was?

Yea, I think he was from Northview. Benjamin Hu, if it was, got it. He might have made USAMO this year, since the cutoff's going to be really low.

[quote="Benjamin Hu"]No, I only got 99 on AMC12 A. What is the cutoff anyway?[/quote]
If you mean USAMO cutoffs, those haven't been determined yet.  If you mean the AIME cutoffs, this is what they were this year :
AMC 12A :  97.5
AMC 12B :  100
AMC 10A :  117
AMC 10B :  115.5

[quote="2007 USAMO teacher manual"]
Selection for the USAMO will be made according to the following rules:
1. The goal is to select about 500 of the top scorers from this year’s AIME and AMC 12A, AMC 12B, AMC 10A and AMC 10B contests to participate in the USAMO.
2. Selection will be based on the USAMO index which is defined as 10 times the student’s AIME score plus the student’s score on the AMC 12 or the AMC 10.
3. The first selection will be the approximately 330 highest USAMO indices of students taking the AMC 12A or AMC 12B contest.
4. The lowest AIME score among those 330 first selected will determine a floor value. The second selection of approximately 160 USAMO participants will be among students in the 10th grade and below who received an AIME score at least as high as the floor value. If there are more than 160 young students with a score above the floor value, then approximately 160 students will be selected from this group by using the USAMO index.
5. The student with the highest USAMO index from each state, territory, or U.S. possession not already represented in the selection of the first and second groups will be invited to take the USAMO.
6. To adjust for variations in contest difficulty, the number of students selected from A & B contests will be proportional to the number of students who took the A & B Contests.
7. In advising young students (in grade 10 or below) who desire to be selected for the USAMO whether to take the AMC 12 contest or the AMC 10 contest, please be aware of the following facts:
a. In 2006, among 427 students invited to take the USAMO, 218 were in 10th grade and below. Those students had scored 8 or greater on the AIME.
b. Among those 218 students, 106 had their AIME qualifying high score based on the AMC 12 and 112 had their AIME qualifying high score based on the AMC 10.
c. In 2006, among 24,445 students invited to take the AIME, 3,994 were in grades 10 and below. Of those, 2,496 qualified for the AIME from the AMC 12 and 1,498 qualified from the AMC 10.
[/quote]

[quote="Xantos C. Guin"][quote="Karth"][quote="Benjamin Hu"]No, I only got 99 on AMC12 A. What is the cutoff anyway?[/quote]

WHOA

dude, that's quite a turkey

you were pushing it...[/quote]

You call that pushin it? At my school we had two 97.5's on the 12A and one 117 on the 10A. They made up 75% on the AIME qualifiers.[/quote]

okay, you win

but duke still sucks


inequalities
triangle inequality
topology
real analysis
real analysis unsolved

The short version is that if you use define the distance as $ \|f\minus{}g\|_p,$ then the triangle inequality fails, and if you define the distance as $ \|f\minus{}g\|_p^p,$ then that gives us a metric space, but you can't properly factor scalars out of the norm - it doesn't have the scaling property for norms. The unit ball, which is to say $ \{f: \|f\|_p<1\},$ is not convex.

The entertaining question: do there exist any nonzero continuous linear functionals on $ L^p$ for $ p<1?$

(The answer to this does depend somewhat on the properties of the underlying measure space.)


algebra
polynomial
abstract algebra
superior algebra
superior algebra unsolved

[quote="akech"]If F is a finite field with k elements, and p(x) is a polynomial of positive degree n over F, find a formula for the number of elements in the ring F(x)/p(x).[/quote]

I think it is $p^{k}$ if p(x) is irreducible over F.
But the general case sounds more interesting.

Should it not be expressed in terms of n and k, where n and k are as defined above?

I think sylow_theory meant $k^{n}$ (since this is the case with $p$ irreducible and the given notation). Also, I think the problem may be missing the premise that $p(x)$ is irreducible? Do we have a gaurantee that $F(x)/p(x)$ is a field otherwise? The factor ideal needs to be maximal I believe.

F(x)/p(x) is a commutative ring with unity. The problem does not require F(x)/p(x) to be a field.

Ah, I should read the question before trying to answer it!   :o

it does not play any role if $p$ is irreducible or not. the monomials $x^{0}, .. , x^{n-1}$ form a basis of $F[x]/p$ as a $F$-module, so we have $k^{n}$ elements. we may even replace $F$ by a ring and $p$ by a polynomial, whose leading coefficient is a unit.


Support
geometry
geometric transformation
reflection

When you think of misunderstood Economics issues, what do you feel are the explanations that would clear up misunderstandings?

Personally, I feel that most bad Economics is a matter of misunderstanding, misapplying, or being otherwise ignorant of mathematics.  This includes the inexperience of most people with creating realistic models (even among learned academics).

And the largest misunderstanding amongst people who create poor models is that incentives drive everything economically.  People can think what they want to think, support one type of economic policy over other, believe in any particular religion, and play any political game, but ultimately people follow their own personally constructed or perceived set of incentives.  Even laws are only part of the set of incentives for a market participant.  Perhaps this is why market forces are often referred to as "the invisible hand" -- we cannot quite devine the incentives of the masses -- or even influence them without dramatic means.

What are your thoughts?

I think that one of the reason some economists get different results from others is that there opinions on life are different, not neccessarily that they don't know economics. Some economists might think that the rich deserve their wealth but others think they don't. Maybe that is the source of their mistakes. I am not sure if I am responding to the right question.

I agree that incentives are not the ONLY force driving economic activity.  However, economics, being one of the most indistinct academic pursuits, has to rely on simplifications of the real world in order to provide for effective analysis.  Therefore, models based on knowledge of incentives, since they are undoubtedly the [i]primary[/i] motivational force in a market economy, are often the most effective.  But I do agree that the most effective economic models are those that take other variables into account and that classical economic laws are becoming increasingly inaccurate and outdated due to bias and change within our economic system.

I recommend Ludwig von Mises [u]The Free Market and Its Enemies: Pseudo-Science, Socialism, and Inflation[/u], I think you can get it at http://www.fee.org. It is a series of lectures that von Mises gave to high-school students, its only 80 pages small type, and he spends one lecture on almost this exact issue. What I say now is probably a pretty direct reflection of that book, because he really explains everything very clearly:

There certainly are many people who practice bad economics. I would argue that economics, being a science, should seek only to say the result of a particular action, not whether the action is right or wrong. Effectively, economics is the study of human action, not the study of whether its better for rich or poor to have money, which is obviously an unobjective statement. One profound example of bad economics is Marx, who's bad theories led to one of the most destructive governments in history. Marx argued that the foundation of existence was strictly material; and he extrapolated from this to argue that history was determined by economics. He thought history was completely determined, so to speak, that economically states would progress from feudalism to captialism to socialism, the greatest state. Marx's main error is that history is not solely economical, and the foundation of all economics and history itself is the fact that individuals can make choices, free of any economical reason, and not predestined by an already existing future. Not accounting for man's individuality is bad economics.


function
real analysis
real analysis solved

Let $f:R->R$ a continuous function.Show that if $f$ admits a local maximum and local minimum then there exist $(a,b) \in R^2$ ,distinct numbers,such that $f( \frac {a+b}{2})=\frac{f(a)+f(b)}{2}$.

It follows from a few basic facts:

Since $f$ is continuous, if the conclusion doesn't hold, then $\frac{f(x)+f(y)}2-f(\frac{x+y}2)$ must have a constant sign. Let's assume it's always $>0$. In this case, using, again, the continuity of $f$, we can show that $f$ must be convex, but not constant on any interval, and it's easy to show now that such a function cannot have a local maximum (we do pretty much the same thing if $f$ is concave).

The problem is already solved by [b]grobber[/b],so I suggest a little bit other solution,but the basic ideas are same.Let $x_{1},x_{2}$ are the points where function f approaches its local minimum and local maximum,respectively.Then there exist ${\delta}_{1}>0$ and ${\delta}_{2}>0$,such that $f(x_{1})\leq{f(x)},\forall x\in{[x_{1}-{\delta}_{1},x_{1}+{\delta}_{1}]}$,and   $f(x_{2})\geq{f(x)},\forall x\in{[x_{2}-{\delta}_{2},x_{2}+{\delta}_{2}]}$.So $f(x_{1})\leq\frac{f(x_{1}-{\delta}_{1})+f(x_{1}+{\delta}_{1})}{2}$ and  $f(x_{2})\geq\frac{f(x_{2}-{\delta}_{2})+f(x_{2}+{\delta}_{2})}{2}$.So we found four points $a,b,c,d$ such that $f(\frac{a+b}{2})\leq\frac{f(a)+f(b)}{2}$,and $f(\frac{c+d}{2})\geq\frac{f(c)+f(d)}{2}$.Now it is easy to prove that as $f$ is continious there exist(as [b]grobber[/b] noticed) $u,v\in{R}$ such that $f(\frac{u+v}{2})=\frac{f(u)+f(v)}{2}.$


geometry
incenter
inequalities
trigonometry
geometry proposed

This follows from Toricelli's problem for the case $max(A, B, C) \leq 120^0$ since the let $F$ be the Toricelli or Fermat point, then $FA + FB + FC = RHS. r$ and $LHS = \frac{IA+IB+IC}{r}$, here $I$ is the incenter of $ABC$.

 For the case $max(A, B, C) \geq 120^0$, we have $b+c \geq \sqrt{ \frac{a^2+b^2+c^2 + 4\sqrt{3}S}{2}}$.  Hence $IA + IB +IC \geq b + c \geq RHS.r$.

I have a different proof. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=509693#p509693]here[/url].

I will prove the following lemma:

  Let $ABC$ be an acute-angled triangle and $M$ be a point in triangle's plane. The following inequality is valid:

                       \[ MA+MB+MC\geq\sqrt{\frac{a^{2}+b^{2}+c^{2}+4S\sqrt{3}}{2}}.  \]

  [i] Solution. [/i]

      It is well-known that, if $M$ is a point in triangle's plane then the sum $MA+MB+MC$ is minimum when $M$ is Toricelli-Fermat's point, i.e \[ MA+MB\geq AA_{1}.  \]
      In triangle $AA_{1}B$, $\angle ABA_{1}=B+\frac{\pi}{3}$, $AC=c, BA_{1}=BC=a,$ so we have

       \[ AA_{1}=c^{2}+a^{2}-2ca\cos\left(B+\frac{\pi}{3}\right)=c^{2}+a^{2}-2ca\cos\left(\frac{1}{2}\cos B-\frac{\sqrt{3}}{2}\sin B\right)=  \]
       \[ =c^{2}+a^{2}-ac\cdot\frac{a^{2}-b^{2}+c^{2}}{2ca}+\sqrt{3}ca\sin B=\frac{a^{2}+b^{2}+c^{2}}{2}+2S\sqrt{3}=\frac{a^{2}+b^{2}+c^{2}+4S\sqrt{3}}{2}.  \]


[i]Prove that if the sequence $ \{ a_n \}_ { n \geq 1 }$ is defined by :

   $ a_n \ \equal{} \ \left[n \sqrt{3} \right]  \forall n \in N$

then  it occupies infinite many elements which are the powers of $ 2$ .[/i]

Let $ y_k\equal{}\{\frac{2^k}{\sqrt 3}\}\equal{}\frac{2^k}{\sqrt 3}\minus{}[\frac{2^k}{\sqrt 3}],n_k\equal{}[\frac{2^k}{\sqrt 3}]\plus{}1.$
Then $ n_k\sqrt 3 \equal{}2^k\plus{}(1\minus{}y_k)\sqrt 3$ and $ [n_k\sqrt 3]\equal{}2^k$ if $ 1\minus{}\frac{1}{\sqrt 3}\equal{}\alpha<y_k<1,\alpha <\frac 12.$
If $ \frac{1}{2^{l\plus{}1}}<y_k<\frac{1}{2^l}$, then $ n_{k\plus{}l}$ work. For any $ k>K$ exist $ l$, suth that $ n_{k\plus{}l}$ work.


$x,y,z$ are positive and $x+y+z=1$ prove that:
$0\le xy+yz+zx-2xyz\le\frac{7}{27}$

there is another inequality for $united kingdom 99$ and it's
let $a,b,c$ be positive real and $a+b+c=1$ prove that
$7(ab+bc+ac)\le2+9abc$
we can simply solve this ineq by usiny that ineq :D


geometry
incenter
geometric transformation
rotation
ratio
angle bisector
geometry unsolved

let $ CH$ be the height of $ \triangle ABC$ where $ H$ is on $ AB$ and $ \angle C$ is a right angle and $ \odot O_1$ and $ \odot O_2$ are the incircles of $ \triangle CAH$ and $ \triangle CBH$, respectively. let $ P$ (on $ CA$) and $ Q$ (on $ CB$) be two points such that $ PQ$ is tangent to  $ \odot O_1$ and $ \odot O_2$ . prove that $ P$, $ Q$, $ A$, $ B$ are cyclic.

sorry i was wrong...

but now it's right

now it's a easy problem...

sorry

I think you missed one condition: triangle ABC is right at C.

Prove that the angle bisector of <ACB is perpendicular to O_1O_2 and you are done.
To achieve this goal, you may considerate M and N the intersections of CO_1, CO_2 with AB and prove MNO_2O_1 cyclic.
I did not solve it actually.

Best regards,
sunken rock

i think the condition c=90 is missed ,  P,Q,B,A are not always on one cyclic

A $ 90^\circ$  clockwise rotation around H, followed by a dilatation of ratio BH/CH maps the triangles AHC and CHB, hence $ AO_1$ and $ CO_2$ are perpendicular, and so are $ BO_2$ and $ CO_1$, therefore I, the incenter of $ \triangle ABC$ is the orthocenter of $ \triangle CO_1O_2$, hence $ CI \perp O_1O_2$. But these lines are the angle bisectors of $ \angle ACB$ and $ \angle (AB,PQ)$, so PQBA is cyclic.

Best regards
sunken rock

It's so easy considering $ \angle(CA,O_1O_2) \equal{} \angle(O_1O_2,AB)$.


trigonometry
geometry
projective geometry
geometry unsolved

Let X be a circle centered at O, and let ABP be a line segment such that A and B lie on X and P is a point outside X. Let C be a point on X such that PC is tangent to X and let D be the point on X such that CD is the diameter of X and intersects AB inside X. Suppose the lines DB and OP intersect at E. Prove that AC is perpendicular to CE.

Let $C'$ be, the second tangent point of $(X),$ from the line through $P,$ the other than $C.$

It is easy to show that the quadrilateral $PEBC',$ is cyclic, because of $\angle C'BD = \angle C'CD = \angle CPO = \angle C'PO.$

So, we have $\angle PC'E = \angle PBE$ $,(1)$ $\Longrightarrow$ $\angle PBE = \angle PCE$ $,(2)$ $($ because of $\angle PC'E = \angle PCE$ $)$

and because of $\angle PBE = \angle ABD = \angle ACD$ $,(3)$ we have that $\angle PCE = \angle ACD$ $,(4)$

From $(4)$ $\Longrightarrow$ $\angle ACE = \angle PCD = 90^{o}$ $\Longrightarrow$ $AC\perp CE$ and the proof is completed.

Kostas Vittas.

Good solution, vittasko
here is an another solution
Let K be the midpoint of AB. We have KO and AB are perpendicular so O,K,C,P are cyclic
So angles CKP=COP and then AKC=DOE . This means triangle AKC and DOE are similar
And so that triangle ABC and DCE are similar
Thus angles ECD=CBA=CDA. so EC//AD. this means EC and AC are perpendicular q.e.d

[color=darkred][b][u] An equivalent enunciation.[/u][/b] Let $ABC$ be an acute triangle inscribed in the circle $w=C(O,R)$ with the diameter $[CD]$.
Denote the intersections $P\in AB\cap CC$ and $E\in DB\cap OP$. Prove that $CE\perp CA$.[/color]

[color=darkblue]$1\blacktriangleright$ [b][u] A synthetical proof (Vittasko).[/u][/b] Denote the second tangent $PC'$ from the point $P$ to the circle $w$, where $C'\in w$, the other than $C$.
Thus, $\widehat{EPC'}\equiv$ $\widehat{OPC'}\equiv$ $\widehat{OPC}\equiv$ $\widehat{OCC'}\equiv$ $\widehat{DCC'}\equiv{DBC'}$ $\Longrightarrow$ $\widehat{EPC'}\equiv\widehat{DBC'}$ $\Longrightarrow$ the quadrilateral $PEBC'$ is cyclically.
Therefore, $\widehat{ECP}\equiv$ $\widehat{EC'P}\equiv$ $\widehat{EBP}\equiv$ $\widehat{ABD}\equiv$ $\widehat{ACD}$ $\Longrightarrow$ $\widehat{ECP}\equiv\widehat{ACD}$ $\Longrightarrow$ $\widehat{DCP}\equiv\widehat{ACE}$, i.e. $CE\perp CA$ because $CP\perp CD$.

$2\blacktriangleright$ [b][u]A metrical proof (Virgil Nicula).[/u][/b] Denote the intersection $F\in CD\cap AB$ and $m(\widehat{ECP})=x$. Thus, $\frac{EP}{EO}=\frac{CP}{CO}\cdot\frac{\sin\widehat{ECP}}{\sin\widehat{ECO}}\Longrightarrow \boxed{\frac{EP}{EO}=\frac{CP}{R}\cdot\tan x}\ \ \ (1)$.
From the relations $\{\begin{array}{c}\frac{FB}{FC}=\frac{\sin\widehat{FCB}}{\sin\widehat{FBC}}=\frac{\cos A}{\sin B}\\\\ \frac{DF}{DO}=\frac{BF}{BO}\cdot\frac{\sin\widehat{DBF}}{\sin\widehat{DBO}}=\frac{BF}{R}\cdot\frac{\cos B}{\sin A}\\\\ \frac{BP}{BF}=\frac{CP}{CF}\cdot\frac{\sin\widehat{BCP}}{\sin\widehat{BCF}}=\frac{CP}{CF}\cdot\frac{\sin A}{\cos A}\end{array}$ and the Menelaus's theorem apllied to the transversal $\overline{DBE}$ in $\triangle POF$ obtain :

$\frac{DF}{DO}\cdot\frac{EO}{EP}\cdot\frac{BP}{BF}=1$ $\Longleftrightarrow$ $\frac{EP}{EO}=$ $\frac{DF}{DO}\cdot\frac{BP}{BF}=$ $\frac{BF}{R}\cdot\frac{CP}{CF}\cdot\frac{\cos B}{\cos A}=$ $\frac{CP}{R}\cdot\frac{\cos A}{\sin B}\cdot\frac{\cos B}{\cos A}=$ $\frac{CP}{R}\cdot \cot B$ $\Longrightarrow$ $\boxed{\frac{EP}{EO}=\frac{CP}{R}\cdot\cot B}\ \ \ (2)$.

From the relations $(1)$ and $(2)$ obtain $\widehat{ECP}\equiv \widehat{ACD}$ $\Longleftrightarrow$ $\widehat{DCP}\equiv\widehat{ACE}$, i.e. $CE\perp CA$.

$3\blacktriangleright$ [b][u]A "lightning" proof.[/u][/b] Denote the second intersection $F$ of the line $CE$ with the circle $w$.

Apply the Pascal's theorem to the degenerated cyclic hexagon $BAFCCD$ :

$\{\begin{array}{c}P\in BA\cap CC\\\ S\in AF\cap CD\\\ E\in FC\cap DB\end{array}\|$ $\Longrightarrow$ $S\in PE\cap AF\cap CD$ $\Longrightarrow$ $S\in PE\cap CD$ $\Longrightarrow$ $S\equiv O$, i.e. $CE\perp CA$.

[b]P.S.[/b] I"ll dedicate this nice and very short proof to the Vittasko for his geometry passion !

[b][u]A good-looking generalization.[/u][/b] Let $ACBD$ be a convex quadrilateral inscribed in the circle $w=C(O)$ so that the center $O$ is a inside point for the given quadrilateral. Denote the intersections $P\in CC\cap AB$, $E\in BD\cap PO$, $I\in CD\cap PO$. Prove that $|90^{\circ}-m(\widehat{ACE})|=m(\widehat{OAI})$.[/color]

Thank you dear  Virgil for a beautiful solution with magic ( and lovely to me ) applying of the [b][size=109]Pascal’s theorem[/size][/b] (in the inscribed non-convex degenerated hexagon $BAFCCD$ !! ).

This is exactly what I like very much in geometry. When we have found a proof of a problem, another unexpected one, it appears.

Best regards, Kostas Vittas.