calculus integration function derivative algebra polynomial ratio

is there a general formula for integrating expression like
1/(1+x^n) ,the case n=2 is particularly easy.i tried n=3 and tried factoring it over complex numbers and now i am hopelessly lost
plz help.
thank

In french, there's a distinction between "integration" and "primitivation".  Integration means computing the value of a definite integral.  Primitivation means finding a function whose derivative is a given function.  For your problem, that function might not be "primitivable" (that is, there are no "elementary" function whose derivative is 1/(1+x^n) for n >= 3).  But for certain cases, it's possible to compute the integral.  For example, you can compute it over ]-\infty, \infty[ using the residue theorem, or over any closed curve in the complex plane.

I'll have a look on it.

Well, I didn't think long enough.

Actually, I don't see the difficulty.  It's just basic stuff.  Polynomials and ratio of polynomials (of ANY degree) are not a problem at all.
Let p(x) = 1 + x^n.  This is a polynomial with real coefficients.

If n is even, it has no real roots, but every complex root has a conjugate root.  So, you can factorize p(x) with a product of quadratic polynomials.
Exemple : n = 4.  x^4 + 1 has e^(i*pi/4), e^(-i*pi/4), e^(3*i*pi/4) and e^(-3*i*pi/4).  So, you can easily compute x^4+1 = (x^2 + sqrt(2)*x + 1)(x^2 - sqrt(2)*x + 1)
Easy, no?

If n is odd, it's exactly the same thing, except that -1 is also a root (so, you'll have a (x+1) factor)

Next, you just have to rewrite 1/(1+x^n) as a function of the form
(A_1x + B_1)/(first quadratic polynomial) + (A_2x + B_2)/(second quadratic polynomial) + ...

The computation of the A_i and B_i is really easy, needing just some basic algebra, but it's quite tedious for large n.  Maybe there's a fast method, but I don't care...  Logically, you'll get complex A_i and B_i.

Next, you have to integrate every terms.  I guess you can do it.  Then, you're done.

For your information, I found, in my tables of integrals, that

for n = 3 :
(1/6)*ln((1+x)^2/(1-x+x^2)) + (sqrt(3)/3)*arctg((2x-1)/sqrt(3))

for n = 4 :
(sqrt(2)/8)*ln ((x^2 + sqrt(2)x + 1)/(x^2 - sqrt(2)x + 1)) + (sqrt(2)/4)*arctg(sqrt(2)x/(1-x^2))

Primitive of 1/(1+x^n)

first take the partial fractions of 1/(1+x^n) in C(X)
second take the primitive

 F(x) = \int 1/(1+x^n) dx

[color=blue]
First case n  a positive even number

F(x) =  -2/n \sum _{ k=0}^{n/2 -1}P_k(x).cos( \pi (2k+1)/n) +

2/n \sum_{k=0}^{n/2 -1}Q_k(x)sin(( \pi (2k+1)/n))

[/color]

[color=blue]
Second case n a positive odd number

F(x) = 1/n.ln(1+x) -2/n \sum_{k=0}^{(n-3)/2}P_k(x).cos( \pi (2k+1)/n)

+ 2/n \sum _{k=0}^{(n-3)/2)}Q_k(x).sin( \pi (2k+1)/n)

[/color]

where

P_k(x) = 1/2.ln(x 2 -2x.cos( \pi (2k+1)/n) +1)

Q_k(x) = arctg((x.sin( \pi (2k+1)/n)/(1-x.cos( \pi(2k+1)/n))

Table of integrals,series,and products by
Gradshteyn page 75

 :D [color=red]o[/color][color=orange]u[/color][color=green]bi[/color] :D


If $ \alpha$ and $ \beta$ are the roots of the equation $ x^{2} \minus{} px \plus{} q \equal{} 0$,then  the value of

$ (\alpha \plus{} \beta)x \minus{} \frac {(\alpha^{2} \plus{} \beta^{2})x^{2}}{2} \plus{} \frac {(\alpha^{3} \plus{} \beta^{3})x^{3}}{3} \minus{} \frac {(\alpha^{4} \plus{} \beta^{4})x^{4}}{4}.......$

a)$ \log_{e}{(px^{2}\plus{}qx\plus{}1)}$
b)$ \log_{e}{(px^{2}\minus{}qx\plus{}1)}$
c)$ \log_{e}{(qx^{2}\minus{}px\plus{}1)}$
d)$ \log_{e}{(qx^{2}\plus{}px\plus{}1)}$

hello, we get
$ \sum_{i=1}^\infty\frac{(\alpha^i+\beta^i)x^i}{i}=-\ln(1-x(\alpha+\beta)+\alpha\beta x^2)=-\ln(1+px+qx^2)$
So the right answer is $ \text{d}$.
Sonnhard.


calculus
function
real analysis
real analysis unsolved

Suppose that $ f(x)$ is a twice continuously differentiable function on $ \mathrm{R}$, such that $ |f(x)|\le 1$ and $ f(0)^2\plus{}f'(0)^2\equal{}4$. Then there exists $ x_0$ such that $ f(x_0)\plus{}f''(x_0)\equal{}0$.

I noticed $ (f^2\plus{}(f')^2)'\equal{}2f'(f\plus{}f'')$. But I cannot see why $ f^2\plus{}(f')^2$ has a extremum and why on that point $ f'\neq 0$.

You've already shown that $ f'(x)(f(x)\plus{}f''(x))\equal{}0$ for all $ x.$

So at every point, either $ f'(x)\equal{}0$ or $ f(x)\plus{}f''(x)\equal{}0.$ The only way to even possibly have that the latter is never true is to have the former always true. But if $ f'(x)\equal{}0$ always, then $ f$ is constant. If $ f(x)\equal{}C$ is constant, then $ f(0)^2\plus{}f''(0)^2\equal{}C^2\plus{}0\equal{}4,$ so $ C\equal{}\pm 2.$ But that would violate $ |f(x)|\le 1.$

So we can't have $ f'(x)\equal{}0$ everywhere, so we must have $ f(x)\plus{}f''(x)\equal{}0$ somewhere.

My brain must be so rusted. why $ f'(f\plus{}f'')\equal{}0$ every where? I did not say $ f^2\plus{}(f')^2$ was a constant...

Sorry - I misread the hypotheses. So my proof doesn't work. Back to the drawing board.

Let $ E(t)\equal{}f(t)^2 \plus{} f'(t)^2$. Since $ f$ is bounded by $ 1$ and $ \minus{}1$, it must oscillate between maxima and minima. What do we know about the values of $ E$ at a critical point of $ f$? They're at most $ 1\plus{}0\equal{}1$, which is less than the value $ 4$ at zero. Sandwich zero between two critical points, and $ E$ must have a maximum in that interval. At that maximum, we know $ f'$ is nonzero and also that $ E'$ is zero, so $ f\plus{}f''$ is zero.

Okay, that's not quite true as stated. We could grow slowly instead, never actually reaching a minimum/maximum on one side. If we do, then $ f'\to 0$ and $ |f|\le 1$ in that direction, making the limiting value of $ E$ at most $ 1$. That's still smaller than $ 4$, and $ E$ must have a maximum somewhere.


Can anybody give the names of good olympiad books that I can buy? I mean, books related to olympiad topics such as Combinatorics, Algebra, Geometry and Number Theory. If it is possible, I like Titu Andrescu's books.

Subject has been discussed many times. If you know already what you like, go to Amazon.com and get some ;)
Topic locked.


Well, a^x + b^x cannot equal c^x where x > 2...

That should help quite a bit.

[quote="somedude"]4^x+25^x=100^x...
[hide]is it possible?  I mean take the xth root of both sides and you're left with 29=100 implying the answer is some imaginary number.  That is assuming x isnt zero, and I already tried that.  I don't know if I can work well enough with i to do this...[/hide][/quote]

[hide]$\sqrt[x]{4^x+25^x}\ne 4+25$.
For example, $\sqrt{3^2+4^2} = 5\hbox{, not } 7.$[/hide]

What is x's domain?  We can't use Fermat if it's R or some extension of R....

[quote]
[hide]$\sqrt[x]{4^x+25^x}\ne 4+25$.
For example, $\sqrt{3^2+4^2} = 5\hbox{, not } 7.$[/hide][/quote]
lol duh, I can't think straight

treething I don't get how that helps cause i already know 4+25 is not equal to 100.

well here's all I know how to do can anyone help maybe?[hide]25 to any whole power will give an odd number, likewise 4 will always yeild an even number.  odd + even is odd and 100 is even so this doesn't work.  that means x is NOT a whole number.  I though about doing logarithms... but that gets really complicated.  could anyone give me some insight?[/hide][/hide]

I think it should be in N, not in R. By the way, you can transform it into $(25^x-1)(4^x-1)=1$ and from it flows that ther's only one solution from R, but hard to find :)


function
vector
Functional Analysis
combinatorial geometry
real analysis
real analysis solved

In any infinitely dimensional Banach space there exists a set whose convex hull is not relatively compact.

What did you really want to say, harazi? :) As stated, the problem is trivial: we simply take our set to be unbounded. At first I thought you wanted the set to be compact, but in that case the result is no longer true:

Let $X$ be a compact topological space and let $C(X)$ be the space of real continuous functions on $X$. We give $C(X)$ (which will be our Banach space) the $\sup$ norm. By the Arzela-Ascoli Theorem, a subset of $C(X)$ is compact iff it's closed, bounded and equicontinuous. It's a simple exercise to check that if a set satisfies these three requirements, then so does the closure of its convex hull, so we're done.

Sorry if I'm missing something obvious.

Edit:

I see you've shown [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Banach&t=42275]here[/url] that every Banach space works, so my counterexample is actually the rule, not the exception :).

Sorry, I was in a hurry and I've written stupid things. The question is to prove the existence of a compact set whose convex hull is not closed.

Oh, Ok, thaks for clarifying it.

I think we can even take this set $\mathcal S$ of ours to be a sequence together with its limit. Clearly, such a set is compact (in any metric space). I won't get into all the details, but it should be easy enough to construct a sequence of vectors $(x_n)_n\to 0$ such that $\sum\frac 1{2^n}x_n$ converges absolutely to a limit $\ell$ s.t. the system $(*)=(\{x_n\}_n,\ell)$ is linearly independent. The sequence $\left(\frac{\sum_{i=1}^n\frac 1{2^i}x_i}{\sum_{i=1}^n\frac 1{2^i}}\right)_n$ converges to $\ell$ and it is contained in the convex hull of $\mathcal S=\{x_n\}_n\cup\{0\}$. However, $\ell$ does not belong to this convex hull, which consists only of finite linear combinations of $x_i$'s, because of the linear independence of $(*)$.

Sorry, but I think the construction of such a sequence if the most interesting part of the problem.

Choose $x_1$ with norm $\le 1$ arbitrarily. After that, take a ball $\mathcal B_1$ of radius $\le 1$ around $\frac 12x_1$ and choose $x_2$ with norm $\le\frac 12$, not collinear with $x_1$, s.t. $\frac 12x_1+\frac 1{2^2}x_2\in\mathcal B_1$. Then take a ball $\mathcal B_2\subset\mathcal B_1$ of radius $\le\frac 12$ around $\frac 12x_1+\frac 1{2^2}x_2$ which does not intersect the space spanned by $x_1$, and pick $x_3$ with norm $\le\frac 13$ s.t. $\frac 12x_1+\frac 1{2^2}x_2+\frac 1{2^3}x_3\in\mathcal B_2$, and so on. I think the next steps are clear: in the $n$'th step we choose a ball $\mathcal B_{n-1}\subset\mathcal B_{n-2}$ of radius $\le\frac 1{n-1}$ around $\sum_{i=1}^{n-1}\frac 1{2^i}x_i$ which does not intersect the space spanned by $x_1,\ldots,x_{n-2}$, and choose an $x_n$ with norm $\le\frac 1n$ s.t. $\sum_{i=1}^n\frac 1{2^i}x_i\in\mathcal B_{n-1}$.


[quote="Jhyon8149"]When I read this, I didn't know whether to believe it or not.....  :D

[url=http://news.yahoo.com/s/ap/burrito_lockdown]Buritto[/url][/quote]

lol....ppl r so dumb

[quote]The drama ended two hours later when the suspicious item was identified as a 30-inch burrito filled with steak, guacamole, lettuce, salsa and jalapenos and wrapped inside tin foil and a white T-shirt.[/quote]

Yeah well I thought he wrapped some hand grenades in there.

i wonder if they tested the burrito- chemically analyzed- u never know- someone might mix the burrito with gunpowder... im too paranoid

Yea i'm pretty sure its real since it also showed up in Newsweek.

OMG.... what a story.

btw, a bunch of funny stories by [url=http://www.cbsnews.com/stories/2005/05/04/national/main692982.shtml]CBS[/url]... some are gross.... yuck.

I don't remember the last time I laugh that much about something that pretends to be a real story! :rotfl:

Hmm, that gives me an interesting idea...

That's really funny.....and well , I don't believe it. :?


AMC
AMC 12
AMC 12 A
geometry
3D geometry
sphere
trigonometry

My coach didn't get a chance to give us the answers yet, so I was wondering when the AMC 10 A answers will be posted?  Does anyone have them?

Thanks,
lyra

AMC12A:[hide]
c
d
a
a
d
d
c
c
b
a
d
e
b
c
e
c
b
d
e
b
a
d
a
d
e[/hide]

There.

[quote="solafidefarms"][hide]AMC12A:
c
d
a[/hide]
...
There.[/quote]

THanks, anyone have the AMC 10A :) pretty please???

~lyra

where did you guys get all of your answers?  :maybe:

coaches are often given them
and who knows maybe those are their answers  :maybe:

Sadly, i don't meet my instructor until Monday.  :(
hmm. I would never be able to remember all of those answers.  :roll:

THANK YOU!!! No, I didn't write them down, my teacher took the test, but I remeber most of them.  DO you think the top 1% will be less than or greater than 120?

~lyra

I don't know. I thought it was easier this year, so I'm guessing it'll be the 120 cut. Number 25, I just did an educated guess and check lol,  :lol:

[quote]NOOOOO!
How do you do #25?![/quote]

I am assuming that is #25 amc 10A because I did not take the 12.

The sum of the digits of a number is equivalent to that number in (mod 9).  Let $n\equiv x$ (mod 9).  Since n, S(n), and S(S(n)) are all equivalent in mod 9, $3x\equiv 0$ (mod 9) or x is divisible by 3.

The second (and more important thing) to realize is that $n$ is bounded from below since the sum of the digits cannot exceed 1+9+9+9 and the sum of the sum of the digits cannot exceed 2+8 or 1+9.  And it is bounded from above since n cannot be greater than 2007.

So we can test every multiple of 3 from 1969 to 2004.

If anybody has a more direct way, post it, I would like to see it.

Someone tell me why number question 21 is C....

Number 21 (not in exact words)

There is a cube with a surface area of 24 cm2. Inside of the cube, there's an inscribed sphere and inside of that sphere, there's an inscribed cube. What is the surface area of the inner cube?

I don't know all the choices but C was 8 and E was 12. I chose 12. Can someone tell me why it is C?

Thanks

just because of 21,  i didnt pass....

i could have passed with one more minute! i was so close to getting number 20.

this was my first time (without prep) so i dont feel that bad...

Well you couls still take amc 10b if u want to. thats wat im gonna do cause on this one i got 115.5

I got in the same argument on 21 with a friend, but of course the trick is the rad 3.

I ended up with a 132 I guess, even worse than my second projected score.... after the test I was sure about a 150. Then, I found the typo on 15 (4rad 2 in one place instead of 6... and my answer was listed)
then, I realize I forgot to divide by 2 in 14.
Then I realize I should have spent more than a glance on number 16 (1/2 just couldn't be the answer, i guess)

oh well
same as last year

(at least I didn't mis a #3 this time... but might still be best in oregon)

For the person asking about 3-d geometry, I just draw 2-d shapes and visualize them 3-dimensionally. Works fairly well for me.

Are you sure 6 is D.

I remember it being pretty easy, and I got A...

i got 102 but no one in my school got over 132 from what i heard

yuck...

20 right
2 wrong
3 blank

>> 124.5

if only i hadn't gotten those two wrong...

is this still sufficient for usamo?

any remarks about the index? what do you think it will be compared to last year?

[quote="jasonwingofcarmelin"]is the amc 12b or 10b easier to pass for a 9th grader[/quote]
It depends on two things.
1) How good are you at solving problems?
2) How prone are you to careless mistakes?

well i was just wondering because i signed up for the amc 12 because everyone else said it was easier to pass.  I cant do a lot of problems (like probably 16 max on amc 12, but maybe less), but i ususally dont make any mistakes on the problems i do.


Princeton
college
factorial
search
number theory

Post practice here.  Spam will be strictly deleted:

Problem #1: Algebra

Let x be a real number such that $x^{3}+4x = 8$.  Determine the value of $x^{7}+64x^{2}$.

[quote="mysmartmouth"]Problem #1: Algebra

Let x be a real number such that $x^{3}+4x = 8$.  Determine the value of $x^{7}+64x^{2}$.[/quote]

That was tricky...

[hide="Weird solution"]
$x^{7}+64x^{2}=k$
$x^{7}+4x^{5}=8x^{4}$
$k=4x^{2}(-x^{3}+2x^{2}+16)$
$=4x^{2}(-(8-4x)+2x^{2}+16)$
$=4x^{2}(2x^{2}+4x+8)$
$=8x^{2}(x^{2}+2x+4)$
$=8x(x^{3}+2x^{2}+4x)$
$=8x((8-4x)+2x^{2}+4x)$
$=8x(2x^{2}+8)$
$=16x(x^{2}+4)$
$=16(x^{3}+4x)$
$=128$
[/hide]

There's probably a different way :D

[size=150]Problem 2[/size]

(Number Theory) Find all four numbers that are equal to the sum of the factorials of their digits.

Do you want to move this to the princeton forum maybe?

OK, fine.  I was trying to save the SC forum, but it appears to be dead no matter what.  I think Shobhit gave up math after USC (only other big contest left for SC seniors is ARML).

Uh, I guess somebody can post a different problem. This one appears to be too difficult :P

i have no clue how to approach this. but i found [hide]145[/hide] works.  is there some sort of pattern?

after wasting the last few hours of my life picking factorials to add, i found the 4:

[hide]1, 2, 145, 40585[/hide]

but im not too intelligent, so whoever wants to post the next one, go ahead.

The only number which is equal to the sum of subfactorial of its digits is $148349$

[quote="8parks11"]$1 = 1!$
$2 = 2!$
$40,585 = 4!+0!+5!+8!+5!$

    $ABC = A!+B!+C!$

u didnt need to find it since it's a notorious problem on the net..[/quote]

yea, i realized that only after i found the answer which took me about 4 hours to get...

[quote="joml88"]There isn't a pattern.  Two of the numbers are trivial to find.  The one you found is a little harder to find and the last one's even harder.  However, if you're systematic about it and use logic you can narrow the search down.

A better exercise would be to find a bound on how big such a number that is equal to the factorial of its digits can be.[/quote]
Anybody know what logic to use to narrow the search down?
Btw chess, I liked your solution for prob 1 :wink:
Well this question is out of my league, but you guys are smart so I'll post it for

[b]Problem 3[/b]
Let $n, n>1$, be a positive integer. Determine if there exist $2n$ distinct positive integers $a_{1}, a_{2}, ... ,a_{n}, b_{1}, b_{2}, ... , b_{n}$ such that

(a) $a_{1}+a_{2}+...+a_{n}=b_{1}+b_{2}+...+b_{n}$;
(b) $n-1>\sum_{i=1}^{n}\frac{a_{i}-b_{i}}{a_{i}+b_{i}}>n-1-\frac{1}{1998}$


inequalities
geometry
perimeter
circumcircle
geometric transformation
reflection
vector

The following problem was proposed in [i]Mathlinks Contest, 1-st Editions,  Second round[/i]

Prove that in any acute triangle with sides a,b,c circumscribed in a circle of radius R the following inequality holds

\[
\frac{{\sqrt 2 }}{4} < \frac{{Rp}}{{2aR + bc}} < \frac{1}{2}
\]

where p represents the semi-perimeter of the triangle.

Does anybody have a solution ?

Leon

I have an ugly solution
I post it tomorow

Let s denote the area of the triangle
$\Leftrightarrow   \frac{{Rp}}{{2aR + bc}} < \frac{1}{2}$
$\Leftrightarrow   R(a+b+c)<2aR+bc$
$\Leftrightarrow   (b+c-a)R<bc$
$\Leftrightarrow   (b+c-a)\frac{abc}{s}<4bc$
$\Leftrightarrow   16s^2>a^2(b+c-a)^2$
$\Leftrightarrow   (a+b+c)(b+c-a)(c+a-b)(a+b-c)>a^2(b+c-a)^2$
$\Leftrightarrow   2a^3-(b-c)^2a-(b-c)^2(b+c)\ge 0$
put $f(a)=LHS$where $a\ge \sqrt{b^2-c^2}$(assume that $b\ge c$)
$f'(a)=6a^2-(b-c)^2>0$
then
$f(a)>f(\sqrt{(b-c)(b+c)})=(b-c)\sqrt{(b-c)(b+c)}[2(b+c)-(b-c)-\sqrt{(b-c)(b+c)}>(b-c)\sqrt{(b-c)(b+c)}[b+c-\sqrt{(b-c)(b+c)}]>0$
(note that $b+c\ge b-c$)

Ah...! I have to go to school now
The other ineq. can be solve by similar way

That's a nice problem, from the good old times when MathLinks contest problems were still solvable. And I dare say that I have a nice solution ;) . But first I rewrite the problem, since I just cannot understand how one can designate such a nice value as the semiperimeter of a triangle with such an ugly letter as p.

[color=blue][b]Initial problem.[/b] Let ABC be an acute-angled triangle with sides a, b, c, semiperimeter $s=\frac{a+b+c}{2}$ and circumradius R. Prove that

$\frac{\sqrt2}{4}<\frac{Rs}{2aR+bc}<\frac12$.[/color]

The first what I will do with this inequality is sharpening the left bound:

$\frac25\leq\frac{Rs}{2aR+bc}<\frac12$.

This is stronger than $\frac{\sqrt2}{4}<\frac{Rs}{2aR+bc}<\frac12$, since $\frac25>\frac{\sqrt2}{4}$.

So we have to prove the inequality $\frac25\leq\frac{Rs}{2aR+bc}<\frac12$. Actually, we split it into two assertions:

[b]Assertion 1.[/b] We have $\frac25\leq\frac{Rs}{2aR+bc}$ for [b]any[/b] triangle ABC (not necessarily acute-angled!).

[b]Assertion 2.[/b] We have $\frac{Rs}{2aR+bc}<\frac12$ for any triangle ABC with acute angles B and C (the angle A need not necessarily be acute).

Once these two assertions will be proven, the problem will follow.

[i]Proof of Assertion 1.[/i] First note that, by Lemma 1 of post #2 at http://www.mathlinks.ro/Forum/viewtopic.php?t=19000 , we have $2Rh_c=ab$, where $h_c$ is the altitude of triangle ABC issuing from the vertex C. Similarly, if $h_a$ is the altitude of triangle ABC issuing from the vertex A, then $2Rh_a=bc$. Thus

$\frac{Rs}{2aR+bc}=\frac{Rs}{2aR+2Rh_a}=\frac{s}{2\left(a+h_a\right)}$.

Since $s=\frac{a+b+c}{2}$, this becomes

$\frac{Rs}{2aR+bc}=\frac{\left(\frac{a+b+c}{2}\right)}{2\left(a+h_a\right)}=\frac{a+b+c}{4\left(a+h_a\right)}$.

Now, let X be the orthogonal projection of the point A on the perpendicular bisector of the side BC of triangle ABC, and let U be the reflection of the point A in this perpendicular bisector. Then, of course, the point X is the midpoint of the segment AU. Thus, using vectors, $\overrightarrow{BX}=\frac{\overrightarrow{BA}+\overrightarrow{BU}}{2}$, so that $2\cdot\overrightarrow{BX}=\overrightarrow{BA}+\overrightarrow{BU}$. Thus, $2\cdot BX=\left|2\cdot\overrightarrow{BX}\right|=\left|\overrightarrow{BA}+\overrightarrow{BU}\right|$. But, by the triangle inequality, $\left|\overrightarrow{BA}+\overrightarrow{BU}\right|\leq\left|\overrightarrow{BA}\right|+\left|\overrightarrow{BU}\right|$. Thus, $2\cdot BX\leq\left|\overrightarrow{BA}\right|+\left|\overrightarrow{BU}\right|=BA+BU$.

Now, the point U is the reflection of the point A in the perpendicular bisector of the side BC of triangle ABC. Also, trivially, the point B is the reflection of the point C in the perpendicular bisector of the side BC of triangle ABC. Thus, since reflections preserve distances, we have BU = CA. Hence, the inequality $2\cdot BX\leq BA+BU$ becomes $2\cdot BX\leq BA+CA$. In other words, $2\cdot BX\leq c+b$. Equivalently, $b+c\geq 2\cdot BX$.

Now, let H be the foot of the altitude of triangle ABC issuing from the vertex A, and let M be the midpoint of the side BC. Then,

< AHM = 90° (since the line AH is an altitude of triangle ABC);
< HMX = 90° (since the point X lies on the perpendicular bisector of the segment BC);
< MXA = 90° (since the point X is the orthogonal projection of the point A on the perpendicular bisector of the segment BC).

Thus, the quadrilateral AHMX has three right angles, and is therefore a rectangle. Consequently, XM = AH. But since the segment AH is the altitude of triangle ABC issuing from the vertex A, we have $AH=h_a$, and thus $XM=h_a$. Also, since the point M is the midpoint of the segment BC, we have $BM=\frac{BC}{2}=\frac{a}{2}$. Thus, in the right-angled triangle BMX, the Pythagoras theorem yields

$BX^2=BM^2+XM^2=\left(\frac{a}{2}\right)^2+h_a^2=\frac{a^2}{4}+h_a^2=\frac{a^2+4h_a^2}{4}$,

so that $\left(2\cdot BX\right)^2=4\cdot BX^2=4\cdot\frac{a^2+4h_a^2}{4}=a^2+4h_a^2$, and $2\cdot BX=\sqrt{a^2+4h_a^2}$.

Since $\frac{Rs}{2aR+bc}=\frac{a+b+c}{4\left(a+h_a\right)}$, $b+c\geq 2\cdot BX$ and $2\cdot BX=\sqrt{a^2+4h_a^2}$, we have

$\frac{Rs}{2aR+bc}=\frac{a+b+c}{4\left(a+h_a\right)}\geq\frac{a+2\cdot BX}{4\left(a+h_a\right)}=\frac{a+\sqrt{a^2+4h_a^2}}{4\left(a+h_a\right)}$,

so that the inequality we have to prove,

$\frac25\leq\frac{Rs}{2aR+bc}$,

will immediately follow once we will show the inequality $\frac25\leq\frac{a+\sqrt{a^2+4h_a^2}}{4\left(a+h_a\right)}$. So it remains to show this inequality.

Well, let's set $x=\frac{a}{h_a}$. Then,

$\frac{a+\sqrt{a^2+4h_a^2}}{4\left(a+h_a\right)}=\frac{\frac{a}{h_a}+\frac{\sqrt{a^2+4h_a^2}}{h_a}}{4\left(\frac{a}{h_a}+1\right)}=\frac{\frac{a}{h_a}+\sqrt{\left(\frac{a}{h_a}\right)^2+4}}{4\left(\frac{a}{h_a}+1\right)}=\frac{x+\sqrt{x^2+4}}{4\left(x+1\right)}$.

And thus, the inequality in question, $\frac25\leq\frac{a+\sqrt{a^2+4h_a^2}}{4\left(a+h_a\right)}$, rewrites as $\frac25\leq\frac{x+\sqrt{x^2+4}}{4\left(x+1\right)}$. Upon multiplication with 20 (x + 1), this becomes $8\left(x+1\right)\leq 5\left(x+\sqrt{x^2+4}\right)$. After subtraction of 5x, this simplifies to $3x+8\leq 5\sqrt{x^2+4}$. Now square this and get $\left(3x+8\right)^2\leq 5^2\left(x^2+4\right)$. But this is true, since $5^2\left(x^2+4\right)-\left(3x+8\right)^2=4\left(2x-3\right)^2$, what is trivially $\geq 0$. Hence, the inequality $\frac25\leq\frac{a+\sqrt{a^2+4h_a^2}}{4\left(a+h_a\right)}$ is proven. In other words, the proof of Assertion 1 is complete.

Note that, in the inequality $\frac25\leq\frac{Rs}{2aR+bc}$, equality holds if and only if the triangle ABC is isosceles with base BC and base angle $B=C=\mathrm{arccot}\ \frac34$. This can be easily seen using the above argumentation.

[i]Proof of Assertion 2.[/i] Since the angles B and C of triangle ABC are acute, the foot H of the altitude of triangle ABC issuing from the vertex A lies inside the side BC. Clearly, $AH=h_a$. Now, by the triangle inequality in triangles ACH and ABH, we have AC < CH + AH and AB < BH + AH. Adding up these two inequalities, we get

  $AC+AB<\left(CH+AH\right)+\left(BH+AH\right)=\left(CH+BH\right)+2\cdot AH=BC+2\cdot AH$.

In other words, $b+c<a+2h_a$. Thus, $a+b+c<a+\left(a+2h_a\right)=2\left(a+h_a\right)$, and thus, using one of the formulas obtained in the proof of Assertion 1, we have

$\frac{Rs}{2aR+bc}=\frac{a+b+c}{4\left(a+h_a\right)}<\frac{2\left(a+h_a\right)}{4\left(a+h_a\right)}=\frac12$.

And Assertion 2 is proven.

  darij

[quote="Valentin Vornicu"][quote="harazi"]Oau! Darij, why are you saying that this is from the period when the Mathlinks problems were still solvable?  ;)[/quote]I think that he means that the level of difficulty of the MathLinks Contest has increased since then :P :D  :blush:[/quote]

Yes, Valentin, you understood me correctly.  ;)  I mean, to that time, there were no open problems proposed in the MLC yet (or have I forgotten something?), and most of the questions were comparable to the IMO level. In the last edition of the MLC, most of the problems were absolutely hopeless to me; I even could not solve the algebraic inequalities. I find this somehow depressing...

  darij

[quote="darij grinberg"]
This is stronger than $\frac{\sqrt2}{4}<\frac{Rs}{2aR+bc}<\frac12$, since $\frac25<\frac{\sqrt2}{4}$.
 darij[/quote]

The typo error $\frac25<\frac{\sqrt2}{4}$ should to be corrected in $\frac25>\frac{\sqrt2}{4}$.

Leon

[quote="Leon"]The typo error $\frac25<\frac{\sqrt2}{4}$ should to be corrected in $\frac25>\frac{\sqrt2}{4}$.[/quote]

You are right, thanks. Now I have corrected the typo.

  darij

Sorry, I do not understand why you introduced the $\frac{2}{5}$. Can someone explain this?

Because $\frac25$ is the actual minimum of the expression $\frac{Rs}{2aR+bc}$ ?

  darij

[quote="Darij Grinberg"][color=darkred][b]Prove that in any triangle $ ABC$ there is the inequality $ \frac{2}{5}\le\frac{{sR}}{{2aR+bc}}$.
If the angles $ B$ and $ C$ are acute, then $ \frac{sR}{2aR+bc}<\frac{1}{2}$. (standard notations).[/b][/color][/quote]

[quote="Virgil Nicula"][size=134][color=darkred][b][u]An easy extension[/u]. $ \boxed{\ \{\begin{array}{c}\triangle\ ABC\\ \\ m > 0\ ,\ n > 0\end{array}\|\implies\frac{1}{m+n}\le\frac{sR}{naR+bc\sqrt{mn}}\ }$ with equality iff $ b = c$ and $ \tan\frac{A}{2}=\frac{n-m}{2\sqrt{mn}}$.

If the angles $ B$, $ C$ are acute and $ n\le 4m$, then there are the strict inequality[/b] $ \boxed{\ \frac{sR}{naR+bc\sqrt{mn}}<\frac{1}{n}\ }$.[/color][/size][/quote]
[color=darkblue][size=134][b]Proof.[/b][/size]

$ 1\blacktriangleright$ Denote $ \boxed{\ x =\tan\frac{B}{2}=\frac{r}{s-b}\ }$ and $ \boxed{\ y =\tan\frac{C}{2}=\frac{r}{s-c}\ }$. Prove easily that $ x+y =\frac{4(s-a)R}{bc}$ and $ xy =\frac{s-a}{s}$.

From the inequality $ (x+y)(m+nxy)\ge 4xy\sqrt{mn}$ obtain : $ \frac{4(s-a)R}{bc}(m+n\cdot\frac{s-a}{s})\ge\frac{4(s-a)\sqrt{mn}}{s}$ $ \implies$

$ R[(m+n)s-na]\ge bc\sqrt{mn}$ $ \implies$ $ \frac{1}{m+n}\le\frac{sR}{naR+bc\sqrt{mn}}$ with equality iff $ x = y$ and $ nxy = m$, i.e.

$ x = y =\sqrt{\frac{m}{n}}\implies b = c$ and $ \tan\frac{B}{2}=\sqrt{\frac{m}{n}}$, i.e. $ \tan\frac{A}{2}=\frac{n-m}{2\sqrt{mn}}$. More precisely, $ \frac{a}{2(n-m)}=\frac{b}{m+n}=\frac{c}{m+n}=\frac{h_{a}}{2\sqrt{mn}}$.

$ 2\blacktriangleright$ Prove easily that in this case $ a > s-h_{a}> 0$ and using the well-known relation $ bc = 2Rh_{a}$ obtain :

$ \frac{sR}{naR+bc\sqrt{mn}}=\frac{s}{na+2h_{a}\sqrt{mn}}<$ $ \frac{s}{n(s-h_{a})+2h_{a}\sqrt{mn}}=$ ${ \frac{s}{ns+h_{a}(2\sqrt{mn}-n})}\le\frac{1}{n}$, i.e. $ \frac{sR}{naR+bc\sqrt{mn}}<\frac{1}{n}$.

[b]Remark.[/b] For $ m = 1$ and $ n = 4$ obtain the [b]Darij[/b]'s proposed problem.[/color]


$a^{4}+b^{2}c^{2}\geq 2\sqrt{a^{4}b^{2}c^{2}}=2a^{2}bc=2a$ by AM-GM. Sum this cyclically to get your answer.

[quote="N.T.TUAN"][quote="me@home"]$2\sqrt{a^{4}b^{2}c^{2}}=2a^{2}bc$[/quote]
But $a,b,c\in\mathbb{R}$ :P[/quote]
$(a^{2}-bc)^{2}\geq 0$ is true for all $a,b,c \in R$
so $a^{4}+b^{2}c^{2}\geq 2a^{2}bc$ is true for all $a,b,c \in R$ :P

[quote="N.T.TUAN"][quote="me@home"]$2\sqrt{a^{4}b^{2}c^{2}}=2a^{2}bc$[/quote]
But $a,b,c\in\mathbb{R}$ :P[/quote]
Okay so this proves in the case $a,b,c\geq 0$. Now suppose some of $a,b,c$ are negative; since the lhs is positive, we have:
\[a^{4}+b^{4}+c^{4}+a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}\ge \geq 2|abc(a+b+c)|\geq 2(a+b+c)\]
where we apply our inequality proven for positives to the numbers $|a|,|b|,|c|$ for the first and apply $|a|\geq a$ etc. for the second :)

The ones posted by Hung and Arqady seem similar to Popoviciu's but I can't exactly remember it...?
\[\sum a^{4}+33(abc)^{4/3}-12\sum a^{5/3}b^{5/3}c^{2/3}\geq 0 \]
Which is the same as
\[\sum a^{12}+33a^{4}b^{4}c^{4}-12\sum a^{5}b^{5}c^{2}\geq 0 \]
Which is \[[12,0,0]+33[4,4,4]\geq 12[5,5,2] \] and can be solved by Conceive's methods.
What's your factorization?


Solve in positive reals

$\sqrt[5]{x}=[\sqrt[5]{3x}]$

where $[a]$ denotes the greatest integer which does not exceed  $a$.

[b]Problem 2 from current Polish MO due October 10![/b]

Shame on you narenx...

Don't try it again, narenx.  :mad: Topic locked.

I think we should ask valentin to check IP's... I start thinking all these "different" people posting polish problems are maybe not all different. (rather on someone's malicious purpose)

Last time we had a guy with armenian flag posting the problem, but his email address ended in .pl :roll: that is not something you'd call accidently.


floor function
number theory
relatively prime
number theory solved

(by positive i mean zero or more)

consider the equation a*x+b*y=c with a,b,c positive and a and b coprime

it can easily be proven that if c>a*b-a-b , positive solutions x and y can be found for the equation

but for some reason i can't seem to prove this fact :

exactly half of the integers :1,2,3,4,......a*b-a-b has positive solutions

some experiment suggest that k has positive solutions iff (a*b-a-b)-k has none

but how to prove that problem? anyone?
plz help me?

I'm pretty sure I've seen this problem again in some russian olympiad (for anyone recalling that particular problem: it considered two classes of positive integers, the "good" and "bad" numbers, depending on whether they admitted of some particular representation). And as far as I can recollect, the "official" solution entailed a few geometric arguments (not horribly obvious, though!). Anyway, I'll post again if I find time to fish my old notes on these russian problems.

Cheers,
Thanos

Ta ta ta ta!
Here is the source TeX code of a write-up of mine a long time ago. Fredbel6, your problem is question b).

Cheers,
Thanos

4. (USSR 1965)

Let $p, q$ be two fixed relatively prime positive integers. We call an integer $n$ \emph{good} if it can be represented in the form
\[n=px+qy \mbox{\ , where $x, y$ are non-neqative integers}\]
Otherwise, n is called \emph{bad}. Find an integer $c$ such that, for any
\emph{good} integer $n$, $c-n$ is \emph{bad}. How many bad non-negative numbers are there in total?

\vspace{5 mm}
\textbf{Solution}

By our hypothesis, $p$ and $q$ are  relatively prime integers, so any integer $z$ can be expressed in the form $z=px+qy$. It is
known that any such pair $(x,y)$ is obtained by some some \emph{fundamental} pair $(x_0,y_0)$, according to the formula
$(x,y)=(x_0-qt, y_0-pt)$, where $t$ is a suitable integer. Obviously, there is a unique $t$, so that $0 \leq x \leq q-1$ (namely,
$t= \lfloor \frac{x_0}{q}\rfloor$).

We can map any integer $z$ to the lattice point $(x,y)$, chosen so that $0 \leq x \leq q-1$, and $z=px+qy$. Due to the uniqueness
of $t$, this map is a bijection and therefore $z$ will be good if and only if $y \geq 0$ (because, if $z$ is a good
number with $x \geq q$, then we may write $x=qt+r$, where $0 \leq r \leq q-1$ and so $z$ is mapped to one of the points mentioned,
namely to $(x-qt,y+pt)$).

Now, if number $z=px+qy$ is good, then number $z'=p(q-1-x)+q(-1-y)$ is bad and vice versa. Points $(x,y)$ and
$(q-1-x,-1-y)$ are symmetric with respect to point $(x_0,y_0)=(\frac{q-1}{2},\frac{-1}{2})$, while $z$ and $z'$ are symmetric
with respect to point $z_0=px_0+qy_0=\frac{pq-p-q}{2}$ (since $z+z'=2z_0=c$). Therefore, any good number $z$ can be
identified with the bad number $z'= (pq-p-q)-z$ and vice versa. The least good number is $0$, so the greatest bad
number is $pq-p-q$. Hence, there must be precisely $\frac{pq-p-q+1}{2}=\frac{(p-1)(q-1)}{2}$ bad numbers. QED

[color=red][[b]Moderator edit (Darij, 2017):[/b] Replaced "mean" by "bad", since the word "bad" was used for the same thing in the same sentence.][/color]

Or you can have a look at

http://www.kalva.demon.co.uk/soviet/ssoln/ssol6513.html

:)


calculus
integration
real analysis
real analysis unsolved

Show that
\[ \sum\limits_{n \equal{} 2}^{ \plus{} \infty } {\frac{1}
{{n^2 \left( {n \minus{} 1} \right)}}}  \equal{} \frac{{\pi ^2 }}
{6} \minus{} 1.
\]
 :lol:

$ \sum _{n \equal{} 2}^{ \plus{} \infty } \frac 1{n^2 (n \minus{} 1) } \equal{} \sum _{n \equal{} 2}^{ \plus{} \infty } (\frac 1{n\minus{}1}\minus{}\frac 1n \minus{}\frac 1{n^2})\equal{} 2\minus{}\frac{{\pi ^2 }}6$
 :cool:

[quote="kunny"]bookworm_vn, please edit your problem. Diogene' s answer is right.[/quote]

I totally do not solve the problem. This comes from an integral and then I used Maple v11. It seems that Maple cannot computer this integral exactly  :| .


Let $\bigtriangleup ABC$ be a triangle and $\omega$ be its incircle. Let's draw parallel lines to the sides of $\bigtriangleup ABC$, these lines are also tangents to $\omega$. (In this figure, there are four triangles, one is  $\bigtriangleup ABC$ ). Let $r$ be the radii of $\omega$ and $r_a, r_b, r_c$ be the inradii of small triangles. Prove that $r = r_a + r_b + r_c$.

Denote ${s=\frac{{a+b+c}}{2}}$ and $\Delta=[ABC]$. From similarity of $\triangle ABC$ and $\triangle AB'C'$ we have

$R_a  = \frac{{h_a  - 2r}}{{h_a }}r = \frac{{\frac{{2 \Delta}}{a} - \frac{{2 \Delta}}{s}}}{{\frac{{2 \Delta}}{a}}}r = \frac{{s - a}}{s}r$

In similar way we can find $R_b$ and $R_c$. Thus

$R_a  + R_b  + R_c  = \left( {\frac{{s - a}}{s} + \frac{{s - b}}{s} + \frac{{s - c}}{s}} \right)r = r$

and the problem is solved.

Leon


[quote="hien"]Let $a,b,c>0$ satify condition: $ac\geq 12$. Prove that:
$a+b+c+3\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\geq\frac{105}{8}$[/quote]

Try $(a,b,c)=(3,1,4)$


inequalities
geometry
geometric transformation
search
inequalities proposed

[quote="arqady"][quote="FOURRIER"]$ \boxed{(a^2 \plus{} b^2 \plus{} c^2)^2 \geq 3(a^{3}b \plus{} b^{3}c \plus{} c^{3}a)}$

It's equivalent to
$ 0\le \sum \frac {1}{2}(a^{2} \minus{} 2ab \plus{} bc \minus{} c^{2} \plus{} ca)^{2}$
but I want a simpler proof
 :maybe:[/quote]
I think this is the simplest proof.[/quote]
Here http://diendantoanhoc.net/forum/index.php?showtopic=35373

[quote="FOURRIER"]Thanks, dduclam,

can you translate me [u] Bùi Việt Anh[/u]'s post ?

Does he mean to apply am gm on $ A,B,C$?[/quote]

He using ABC method. But my English is very bad,sorry  :( . Can N.T.TUAN or pvthuan help me,please? Thanks a lot!

Here is the translation as required.

Let \[ A\equal{}\frac{a^4\plus{}2b^2c^2}{a^3b\plus{}b^3c\plus{}c^3a},\;\; B\equal{}\frac{b^4\plus{}2a^2c^2}{a^3b\plus{}b^3c\plus{}c^3a},\;\;C\equal{}\frac{c^3\plus{}2a^2b^2}{a^3b\plus{}b^3c\plus{}c^3a}.\]
We need to prove that $ A\plus{}B\plus{}C\geq3$.

It suffices to consider the case when two among $ A, B, C$ are equal. WLOG, suppose that $ A\equal{}B$. Then $ (a^2\minus{}b^2)(a^2\plus{}b^2\minus{}2c^2)\equal{}0$. That is, $ a\equal{}b$ or $ a^2\plus{}b^2\equal{}2c^2$. Now, the problem is reduced to one variable.

Thanks a lot pvthuan  :)
[quote="pvthuan"]
It suffices to consider the case when two among $ A, B, C$ are equal[/quote]

Do you mean to use this
http://www.mathlinks.ro/viewtopic.php?search_id=1432448505&t=165132
?

[quote="Erken"][quote="FOURRIER"]
[quote="pvthuan"]
It suffices to consider the case when two among $ A, B, C$ are equal[/quote]

I really wonder why  :maybe: , if you could explain it, it will be great  :) (or give me a link)[/quote]
Maybe because he used ABC theorem :wink:[/quote]

Yes I know he used it  :)

[quote="arqady"][quote="FOURRIER"]$ \boxed{(a^2 \plus{} b^2 \plus{} c^2)^2 \geq 3(a^{3}b \plus{} b^{3}c \plus{} c^{3}a)}$

It's equivalent to
$ 0\le \sum \frac {1}{2}(a^{2} \minus{} 2ab \plus{} bc \minus{} c^{2} \plus{} ca)^{2}$
but I want a simpler proof
 :maybe:[/quote]
I think this is the simplest proof.[/quote]
My solution :D :
$ \boxed{(a^2 \plus{} b^2 \plus{} c^2)^2 \geq 3(a^{3}b \plus{} b^{3}c \plus{} c^{3}a)}$

$ \Leftrightarrow \sum ((a^{2} \plus{} b^{2} \minus{} 2c^{2}) \minus{} 3c(a \minus{} b))^{2} \geq\ 0$


I have recently attended a seminar which went quite fast. The second table deals with $ x^2\equiv4\pmod{15}$ and finding roots. You construct this by following the arrows (somehow). I am not sure why numbers are circled. This has something to do with equivalent classes and counting the elements in that congruence class.

The first also deals with mods and roots in someway, but i am not sure what you are trying to actually find here.

These are hieroglyphics to me!  :(

I see the second table gives $ \phi(15)$, once the numbers not coprime to 15 are cancelled out.


trigonometry
trig identities
Law of Cosines
geometry
Law of Sines

[hide="Law of Cosines"] IF you're not familiar with it, see [url=http://www.artofproblemsolving.com/Wiki/index.php/Law_of_Cosines]The Wiki[/url]. Using it, we have $ 49=81+64-144\cos{\angle{BAC}}$. We simplify to find $ \cos{\angle BAC}=\frac{2}{3}$. I doubt most people know this without using a calculator, so you could merely use Arccos, or use a calculator fo find it is around 48.2 degrees.[/hide]
I hope I didn't mess up.

[quote="shaun ong"]Triangle ABC. AC= Hypotenuse  that is 9 cm , AB=8cm , BC=7cm. Angle BAC is x. What is x? Is it something to do about cos , tan and sin? It is not a right angle triangle by the way.[/quote]

If ABC is not a right triangle, AC is not a hypotenuse, even though it is the longest side.

There is a law of sines way. It is really messy, so I will not post it here.

Can you find the other angles?

After finding one angle, you can just use the Law of Sines, find another, and subtract both from 180.

LOL !!!!!DAMN i forgot that if its not a right angle tirangle there wont be a hypotenuse......damn I kept thinking that the longest side of a triangle is the hypotenuse....... so how to do it?

[quote="shaun ong"]so how to do it?[/quote]

[quote="xpmath"][hide="Law of Cosines"] IF you're not familiar with it, see [url=http://www.artofproblemsolving.com/Wiki/index.php/Law_of_Cosines]The Wiki[/url]. Using it, we have $ 49 = 81 + 64 - 144\cos{\angle{BAC}}$. We simplify to find $ \cos{\angle BAC} = \frac {2}{3}$. I doubt most people know this without using a calculator, so you could merely use Arccos, or use a calculator fo find it is around 48.2 degrees.[/hide]
I hope I didn't mess up.[/quote]


pigeonhole principle
calculus
irrational number
combinatorics proposed
combinatorics

Let $a\in \mathbb{R}$. Prove that for every $Q\in \mathbb{N}$ , there exist integers $p$ and $q$ such that $Q>q>0$ and $|qa-p|\le \frac{1}{Q}$

Seems trivially right from an analysis point of view, but there's a nice combinatorial proof to this.

assume the numbers $0,a-[a],2a-[2a],...,Qa-[Qa]$ these numbers are $Q+1$ numbers in the interval $[0,1)$ so the difference of two of them must be less than $\frac{1}{Q}$ (by pigenhole) so there exists $i,j$ s.t $i>j$ and $|ia-ja+[ja]-[ia]|<\frac{1}{Q}$
so $|(i-j)a+([ja]-[ia])|<\frac{1}{Q}$ now put $q=i-j$ and $p=[ja]-[ia]$ so $p,q \in \mathbb{Z}$,now we get that:
$|qa+p|<\frac{1}{Q}$...

Actually this is used to prove Kronecker's density theorem.
Btw, how exactly would you prove that with calculus? :roll:

I can't see a straightforward proof, but the construction of the real numbers makes the fact that we can get an estimate by rationals (as good as we please) to an irrational number seem quite trivial.

The thing is that sometimes calculus problems require elementary solutions. I think that this is one of the times in which you can not give a solid calculus argument. The reason why I asked is because I wanted to know if this problem can be dissmised by using some other density argument.


Is there any other way for an AoPSer to get to the PEN forum without having to "accidentally" click on Home after posting a message, and then clicking on the announcement in "Recent events" on the MathLinks Portal page?

Hmm i have been thinking about the same question for sometime now.

Currently, the other options are:

1: the mathlinks portal is [url]http://www.artofproblemsolving.com/Forum/[/url].  I'm not sure if this works in the AoPS skin.

2: The PEN forum is [url]http://www.artofproblemsolving.com/Forum/index.php?f=456&sid=1b0533a94e491d112f63a6010c15e85e[/url]

I agree, however, that the PEN forum should be a bit more accessible.  I first learned of it by noticing that it appeared in the drop-down list of forums in the search command.

It works on the AoPS skin as well. In fact, this makes it a bit awkward.

This is indeed a reason that so few people find PEN. It has been on the main page, but was quickly pushed back by the IMO news.

Maybe it should be re-announced, or the forum should be made publicly accesible through the forum structure.

Suggestions always welcome, I'll discuss them with Valentin when he's back from his break. :)


Show that:

$ \frac{2009}{2008}>\sqrt[2008]{2008}$

without using a calculator.

consider the function
$ f(x)\equal{}x^{\frac{1}{x}}$

[quote="exoeragonius"]Show that:

$ \frac {2009}{2008} > \sqrt [2008]{2008}$

without using a calculator.[/quote]
You are wrong.
$ \frac{2009}{2008}<\sqrt[2008]{2008}.$


advanced fields
advanced fields unsolved

Give an example of a linear mapping between two normed spaces with closed graph which is not continuous.

The idea is that work with the space which is not Banach, let $ D: C^1[0,1] \rightarrow C^0[0,1]$ be $ D(f) \equal{} f'$, so $ G \equal{} \{ (f,f') \mid \;f \in C^1 \} \subset C^1 \times C^0$ is closed because if $ (f_n,f'_n) \rightarrow (f,g) \in C^1 \times C^0$ then $ \|f_n \minus{} f \|_{\infty} \rightarrow 0$ and $ \|f'_n \minus{} g\|_{\infty} \rightarrow 0$ hence $ f' \equal{} g$ and $ (f,g)\in G$, $ D$ is not continuous, just remember that $ \|x^n\|_{\infty} \equal{} 1$, though $ \|D(x^n)\|_{\infty} \equal{} n$.


vector
trigonometry
trig identities
Law of Cosines

Is the dot product of 2 vectors  $ \vec u \equal{} (x_1 \ y_1)$ and $ \vec v \equal{} (x_2 \ y_2)$ given as $ x_1x_2 \plus{} y_1y_2$?

When I plug in those 2 vectors to the formula $ \vec u \cdot \vec v \equal{} \frac{\parallel{} \vec u\parallel{}^2 \plus{} \parallel{} \vec v\parallel{}^2 \minus{} \parallel{}\vec u \plus{} \vec v\parallel{}^2}{2}$, I get $ \minus{}x_1x_2 \minus{} y_1y_2$. I can't see what I am doing wrong

That's because the formula is actually $ \vec u\cdot\vec v \equal{} \frac {\left\|\vec u\right\|^2 \plus{} \left\|\vec v\right\|^2 \minus{} \left\|\vec u \minus{} \vec v\right\|^2}2$.

oh ok thanks.

I think there is some sort of discreptancy. In aops volume 2, the formula is what I put in my first post, but perhaps it depends on where the angle is defined (this is still a guess, and I still need to clear up a little confusion). Ok, so I am confused about what $ \theta$ refers to in the following excerpt of aops volume 2: \[ \parallel{}\vec v \plus{} \vec w\parallel{}^2 \equal{} \parallel{}\vec v\parallel{}^2 \plus{} \parallel{}\vec w\parallel{}^2 \minus{} 2\parallel{}\vec v\parallel{} \ \parallel{}\vec w\parallel{}\cos \theta\] "where $ \theta$ is the angle between $ \vec v$ and $ \vec w$.""

Is $ \theta$ the angle between the vectors when the vectors have the same starting point, or when the starting point of one vector is the the head of the other?

I think that either definition is OK, but you would have to switch the formulas (math's formula and my formula) for the 2 definitions.

If either definition really is fine, then which is more common?

Thanks math154 again.

[quote="hurdler"]Is $ \theta$ the angle between the vectors when the vectors have the same starting point[/quote]
Yes.  This is the definition consistent with the Law of Cosines and the definition that makes two vectors at an angle of $ 0$ degrees with respect to each other parallel and pointing in the same direction.


function
vector
real analysis
real analysis theorems

I need help with proving this statement:
Suppose f is a function from A to B and is a one-to-one correspondence. Show that the inverse of f is also a one to one correspondence and that the inverse of and inverse is equal to f.   :(  :?:

If $f: A\rightarrow B$ is a bijection, then for each $b\in B$, there exists $a\in A$ such that $f\left(a\right)=b$.  Furthermore, $f^{-1}\left(b\right)=a$ is distinct (since it is injective).  So we have $\textrm{Im}\left(f\right)=B$ and $f^{-1}\left(B\right)=A$.  Now just use the definition of a bijection to get the desired result.

I believe there is also a theorem that says $f: A\rightarrow B$ is one-to-one if and only if $\textrm{Ker}\left(f\right)=\left\{e_{1}\right\}$.

[quote="amcavoy"]I believe there is also a theorem that says $f: A\rightarrow B$ is one-to-one if and only if $\textrm{Ker}\left(f\right)=\left\{e_{1}\right\}$.[/quote]
That's for algebraic morphisms ([i]i.e.,[/i] group homomorphisms if we're talking about groups, linear transformations if we're talking about vector spaces, and so on). JRClegg said nothing about $f$ having any algebraic properties.


[quote="tranquoc"]1/Prove  if [b]a,b,x,y>0 ;x >= y;x >= a[/b];$ x^{2}$ +$ y^{2}$ >=  $ a^{2}$ + $ b^{2}$
  then        $ x^{3}$ + $ y^{3}$  >= $ a^{3}$ + $ b^{3}$

2/Let  a,b,c,x,y,z>0  Find  the minimal value of the expression
    P= $ \frac {( a^{3} \plus{} x^{3} \plus{} 1)( b^{3} \plus{} y^{3} \plus{} 1)( c^{3} \plus{} z^{3} \plus{} 1)}{(a \plus{} b \plus{} c)(x \plus{} y \plus{} z)}$

3/Prove if a,b,c>0 and abc=1  then     $ \sqrt {1 \plus{} 3 a^{2} } \plus{} \sqrt {1 \plus{} 3b^2} \plus{} \sqrt {1 \plus{} 3c^2}$  <= 2(a+b+c)

4/  Prove if a,b,c>0  and  a+b+c=1  then   $ \sqrt [3]{a \minus{} b \plus{} c^3} \plus{} \sqrt [3]{b \minus{} c \plus{} a^3} \plus{} \sqrt [3]{c \minus{} a \plus{} b^3}$  <= 1

5/ let m is a integer pos   Prove   $ 2001.m^{2000}$+$ (m \minus{} 1)^{2001}$   >= $ m^{2001}$

6/let  0<a<1   and x>0  CMR   $ \frac {1 \minus{} x^a}{1 \minus{} x}$ <=$ (1 \plus{} x)^{a \minus{} 1}$

7/Let a,b,c >0 such that a+b+c=3  Prove  $ \sqrt [3]{a} \plus{} \sqrt [3]{b} \plus{} \sqrt [3]{c}$  >= $ ab \plus{} bc \plus{} ca$.
8/ Let a,b,c>0  Prove :$ \sum\frac {a \plus{} b}{2a^2 \plus{} bc}$>=$ \frac {6}{a \plus{} b \plus{} c}$
9/ let a,b,c>0   such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2$  Prove
$ a^3 \plus{} b^3 \plus{} c^3 \minus{} abc$<=2$ \frac {2}|$[/quote]
i think you forgot to write a condition tranquoc in [b]1).[/b]

[quote="tranquoc"]aSFcAS<VHvcfDBdfbfbf[/quote]
I am sorry to tell you something, tranquoc.
First, which language you are using? "aSFcAS<VHvcfDBdfbfbf" I don't understand anything.
Second, I saw you posted many wrong problem in this forum, I think you should check out them before posting them, that is the "polite" in Mathematics, I think. If you are not sure about them, you should write the name of the topic as "a conjecture"
Third, You have posted the problems many times. You should check about that before posting!

See you.
Vo Quoc Ba Can

maybe, tranquoc don't know anything about English so he couldn't used English to exchanging idea with mathlinker :D The best solution is tranquoc should learn more about English. (I said like that but my English is also terriable. Enought to understand  althought having a lot of mistakes in grammar :P
I total agree with Can's opinion.  :wink:


limit
integration
trigonometry
calculus
function
real analysis
real analysis unsolved

Calculate $ \lim_{n \rightarrow \infty} a_n \int_{0}^{1}x^{2n}\sin\frac {\pi x}{2}dx$, where $ a_n \equal{} \sum_{k \equal{} 1}^{n}\sin \frac {\pi k}{2n}$.

You can divide it into two limits:

$ \lim_{n \to \infty} \frac {1}{n}\sum^n_{k \equal{} 1}\sin\frac {\pi k}{2 n} \cdot \lim_{n \to \infty} n \int^1_0 x^{2 n} \sin\frac {\pi x}{2} \,dx$

The first one is the Riemann integral sum, and it equals:

$ \lim_{n \to \infty} \frac {1}{n}\sum^n_{k \equal{} 1}\sin\frac {\pi k}{2 n} \equal{} \int^1_0 \sin\frac {\pi x}{2} \,dx \equal{} \minus{} \frac {2}{\pi} \cos\frac {\pi x}{2}\Big|^1_0 \equal{} \frac {2}{\pi}$

For the second one, expand the sine with Taylor's expansion. I will do it more generally, so you need not to worry what's the exact expansion. Just suppose that the function you are expanding can be expanded around some point $ a$. Then:

$ \lim_{n \to \infty} n \int^1_0 x^{k n} f(x) \,dx \equal{} \lim_{n \to \infty} n \int^1_0 x^{k n} \sum^\infty_{i \equal{} 0} \frac {(x \minus{} a)^i}{i!} \left(\frac {d^i f}{d x^i}\right)_{x \equal{} a} \,dx \equal{}$

$ \equal{} \sum^\infty_{i \equal{} 0} \frac {1}{i!} \left(\frac {d^i f}{d x^i}\right)_{x \equal{} a} \lim_{n \to \infty} n \int^1_0 x^{k n}(x \minus{} a)^i\,dx$

$ \lim_{n \to \infty} n \int^1_0 x^{k n}(x \minus{} a)^i\,dx \equal{} \sum^i_{j \equal{} i} \binom{i}{j} (\minus{}a)^{i \minus{} j} \lim_{n \to \infty} n \int^1_0 x^j x^{k n} \,dx \equal{}$

$ \equal{} \sum^i_{j \equal{} i} \binom{i}{j} (\minus{}a)^{i \minus{} j} \lim_{n \to \infty} \frac {n}{j \plus{} k n} \equal{} \frac {1}{k} \sum^i_{j \equal{} i} \binom{i}{j} (\minus{}a)^{i \minus{} j} \equal{} \frac {(1 \minus{} a)^i}{k}$

Plugging that back we get:

$ \sum^\infty_{i \equal{} 0} \frac {1}{i!} \left(\frac {d^i f}{d x^i}\right)_{x \equal{} a} \lim_{n \to \infty} n \int^1_0 x^{k n}(x \minus{} a)^i\,dx \equal{}$

$ \equal{} \frac {1}{k} \sum^\infty_{i \equal{} 0} \frac {(1 \minus{} a)^i}{i!} \left(\frac {d^i f}{d x^i}\right)_{x \equal{} a} \equal{} \frac {1}{k} f(1)$

So, to your limit, this is:

$ \lim_{n \to \infty} n \int^1_0 x^{2 n} \sin\frac {\pi x}{2} \,dx \equal{} \frac {1}{2}\sin\frac {\pi}{2} \equal{} \frac {1}{2}$

This makes the initial limit equal to:

$ \boxed{\lim_{n \to \infty} \sum^n_{k \equal{} 1}\sin\frac {\pi k}{2 n} \int^1_0 x^{2 n} \sin\frac {\pi x}{2} \,dx \equal{} \frac {1}{2}\cdot\frac {2}{\pi} \equal{} \frac {1}{\pi}}$

I must admit I don't fully understand your solution because I don't know much about Taylor's expansion  :blush: . This limit was given to a simulated admission test in my country and I really doubt that the official solution contained this method (in our country, we don't learn about Taylor's expansion in high-school). I wonder if someone knows a different solution. Thanks :)

You worked integrals but didn't mention Taylor expansion!? That seems odd. But, if you say so, it must be true.. I will try to come up with a solution not involving expansions.. Do you know Riemann integral sums? Like the one I used?!

How about the representation of sine and cosine functions using complex exponential function? Like this:

$ \cos x \equal{}\frac{e^{i x} \plus{} e^{\minus{} i x}}{2}$ and $ \sin x \equal{} \frac{e^{i x} \minus{} e ^{\minus{} i x}}{2 i}$

Hi :) !!!

Good solution [b] milin [/b] !!!

my solution:

$ a_n \int_0^1 x^{2n}sin\left( \frac {\pi x}{2} \right)dx \equal{} \frac {a_n}{n} \int_0^1 nx^{2n}sin\left( \frac {\pi x}{2} \right)dx$

by integration by part we find that:

$ \equal{} \frac {a_n}{n} \left( \frac {n}{2n \plus{} 1} \minus{} \frac {\pi n}{2(2n \plus{} 1)} \int_0^1 x^{2n \plus{} 1}cos \left( \frac {\pi x}{2} \right) dx \right)$

the suit of functions $ g_n(x) \equal{} \frac {\pi n x^{2n \plus{} 1}}{2(2n \plus{} 1)}cos(\frac {\pi x}{2})$

are continues in $ [0,1]$ than they are measurable and $ |g_n(x)| \leq \frac {\pi n x^{2n \plus{} 1}}{2(2n \plus{} 1)}$ .... and converging to $ g(x) \equal{} 0$
than theorem of Lebesgue Domainant give that $ \lim_{n \to \plus{} \infty} \int_0^1 g_n(x) dx \equal{} \int_0^1 \lim_{n \to \plus{} \infty} g_n(x)dx \equal{} 0$

than

 $ \lim_{n \to \plus{}\infty}a_n \int_0^1x^{2n}sin(\frac{\pi x}{2})dx\equal{}$
 $ \lim_{n \to \plus{} \infty}\frac {a_n}{n} \left( \frac {n}{2n \plus{} 1} \minus{} \frac {\pi n}{2(2n \plus{} 1)} \int_0^1 x^{2n \plus{} 1}cos \left( \frac {\pi x}{2} \right) dx \right) \equal{} \lim_{n \to \plus{} \infty} \frac {a_n}{2n \plus{} 1} \equal{} \frac {1}{2} \ \frac {2}{\pi} \equal{} \frac {1}{\pi}$

because:(

$ \lim_{n \to \plus{} \infty} \frac {a_n}{2n \plus{} 1} \equal{} \lim_{n \to \plus{} \infty} \frac {a_n}{n} \frac {n}{2n \plus{} 1}$

with $ \lim_{n \to \plus{} \infty} \frac {a_n}{n} \equal{} \int_0^1 sin \left( \frac {\pi x}{2} \right) \equal{} \frac {2}{\pi}$

and $ \lim_{n \to \plus{} \infty} \frac {n}{2n \plus{} 1} \equal{} \frac {1}{2}$ )

than finaly:

$ \boxed {\lim_{n \to \plus{} \infty} a_n \int_0^1 x^{2n}.sin \left( \frac {\pi x}{2} \right)dx \equal{} \pi^{ \minus{} 1}}$

and thanks :)


[quote="#H34N1"]How thick must a coin be to half exactly $ \frac 13$ probability to land on its side?[/quote]

[hide]
reduce the coin to a rectangle, 2d.  it should work the same.

it will exactly balance if when it lands, the vertex on the bottom is directly below the opposite vertex.
the angle between the edge of the coin and the surface must be 60 for there to be a 1/3 chance
so it must be $ \frac{\sqrt{3}}{3}$ times as thick as it is wide.

probably wrong, oh well.
[/hide]


geometry
geometric transformation
homothety
geometry proposed

Let AH be altitude of given triangle ABC. The incircle (I) touches three sides $ BC,CA,AB$ at $ D,E,F$, the excircle $ (I_a)$ touches $ BC,CA,AB$ at $ M,S,N$, respectively. Show that $ AH,ED,MN$ are concurrent, $ AH,DF,MS$ are concurrent.

Assume that $ MN, DE$  meet  $ AH$ at $ P,P',$ respectively. We shall prove $ P \equiv P'$

By Menelaus' theorem for $ \triangle AHB$ cut by the transversal $ \overline{NMP},$ we get

$ \frac {NB}{NA} \cdot \frac {AP}{PH} \cdot \frac {MH}{MB} \equal{} 1  \Longrightarrow \ \frac {AP}{PH} \equal{} \frac {NA}{MH} \equal{} \frac {p}{MH} \ \ (1)$

By Menelaus' theorem for $ \triangle AHC$ cut by the transversal $ \overline{DP'E},$ we get

$ \frac {DH}{DC} \cdot \frac {CE}{EA} \cdot \frac {AP'}{P'H} \equal{} 1  \Longrightarrow \ \frac {AP'}{P'H} \equal{} \frac {EA}{DH} \equal{} \frac {p \minus{} a}{DH} \ \ (2)$

Antipode $ D'$ of $ D$ in the incircle is collinear with $ A$ and $ M.$ Then from $ \triangle D'MD \sim  \triangle AMH$ we have:

$ \frac {MH \minus{} DH}{MH} \equal{} \frac {2r}{h_a} \Longrightarrow \  \frac {DH}{MH} \equal{} \frac {h_a \minus{} 2r}{h_a} \equal{} \frac {p \minus{} a}{p}$

Therefore, from $ (1)$ and $ (2),$ it follow that $\frac {AP}{PH} \equal{} \frac {AP'}{P'H}$ $\Longrightarrow$ $ P \equiv P'$

By similar reasoning,  we prove that $ AH,DF$ and $ MS$ concur. $\square$

Another idea is using the radical center :D

Dear Mathlinkers,
I know that the beloved Juan Carlos Salazar has worked on this situation
Salazar J. C., Mathlinks 08/01/2005.
an somebody give a more complete reference?
A prove without calculation will be nice.
Sincerely

Also, we can say that the points $ A,\ D',\ M$ are collinear, because of $ ID'\parallel I_{a}M$ and $ \frac{AI}{AI_{a}} \equal{} \frac{IE}{I_{a}S} \equal{} \frac{ID'}{I_{a}M},$ applying the [b][size=100]Thales theorem[/size][/b].

Kostas Vittas.

If $AB=AC$ , the result is obvious .
WLOG : $ AB < AC$
$FD$ meets $AH $ at $K$
we'll prove: $ KM ||IC$

Apply Thales' theorem ,we get : $\frac{HD}{HM}=\frac{ID}{I_aM}$(1)
Beside , we have $\angle KDH= \angle MBI_a(=90-\frac{\angle{B}{2}) \Rightarrow {HK}{HD}=\frac{I_aM}{MB}$(2)
$\Rightarrow {HK}{HM}= \frac{ID}{I_aM}.\frac{I_aM}{MB}=\frac{ID}{MB}=\frac{ID}{CD}$( $ID=BM$)
And since $\frac{HK}{HM}=\frac{ ID}{CD} \Rightarrow KM||IC$

From now on , we'll consider triangle $ KDM$ and see that:
$AH, DE,DM$  are its altitude hence they will concurrent .
QED

well, I solved the problem by lot of calculations but I think they are simple.So let's calculate..

Let $AX$ be the altitude and $AH,ED$ intersects at $X$.So now $AE=s-a,KD=bcosC-s+c$.Now in the triangle $ECD$ with triad $AX$ applying Menelaus we get $\frac {AE}{AC}.\frac {CK}{KD}.\frac {XD}{XE}=1$.Putting the values of $QE,CK,KD$(easy to find out) we get $\frac {XD}{XE}=\frac {bcosC-s+c}{(s-a)\cosh}$

Now let $ED$ and $MN$ intersect at $X'$ and $MN$ cuts $AC$ at $T$.
$\frac {AT}{TC}=\frac {s}{s-b}$ gives $AT=\frac {bs}{2s-b}$ and $ET=\frac {2s(s-c)-ab}{2s-b}$.

Now in $\triangle EDC$ with triad $MN$ we get by Menelaus $\frac {ET}{TC}.\frac {CM}{MD}.\frac {X'D}{X'E}=1$

Again putting the values of $ET,TC$ we get $\frac {X'D}{X'E}=\frac {b(b-c)}{a^{2}+b^{2}-c^{2}}$ and now its easy to show that $\frac {XD}{XE}=\frac {X'D}{X'E}$ and hence the result follows.

[b]My solution: [/b]Let $DF$ cuts $AI$ at $V$ then $V$ is the midpoint of $C$ on $AI$, hence $AHCV$ is concyclic. Now, let $AH$ cuts $MS$ at $T$, we will prove that $T$ lies on $DF$. Indeed, we have $L,V,M,N,C,S$ lie on the same circle then $VMCS$ is concyclic, this leads to $\angle{TSV}=\angle{VLM}=\angle{HAI}=\angle{HAV}$, so $ATVS$ is cyclic, i.e $\angle{AVT}=\angle{ASM}=\angle{ICV}=\angle{AVD}$. Therefore $V,D,T$ are collinear or we can say $T$ lies on $DF$ $\blacksquare$


for the second, only the owner has that problem. No one else does. I see it too on my blog in the Aops blogs page

[quote="batteredbutnotdefeated"]My blog isn't allowing me to change my background.... Also, how do you get your picture to show instead of your username in the box titled "blog owner"[/quote]

Can you provide details on what you want to change in your background? You have already changed it from the default.

I'm not sure what you mean by showing your picture instead of your user name.

I believe that he means that in the blog list, instead of seeing his username he wants to see his avatar, which, as EggyLv.999 pointed out, is only an issue for the owner.


geometry
rhombus
perpendicular bisector
Pythagorean Theorem

Two circles of radius 10 cm overlap such that each circle passes through the center of the other, as shown. How long, in cm, is the common chord (dotted segment) of the two circles? Express your answer in simplest radical form.

[asy]draw(Circle((0,0),10),linewidth(1));
draw(Circle((10,0),10),linewidth(1));
dot((0,0));
dot((10,0));
draw((5,8.66)--(5,-8.66),linetype("0 4")+linewidth(1));[/asy]

The 2 centers and the 2 end points of the chord are forming a rhombus with angles 60 and 120 (because we can divide it into 2 equilateral triangles). That means if side of the rhombus is a, then the length of the longer diagonal is $ a\sqrt3$, and as $ a\equal{}10cm$, $ d\equal{}10\sqrt3$

Here is a different way to arrive to the answer:

Notice that the common chord is a perpendicular bisector of the radius the two circles that has both endpoints at the centers of both circles.  Using the Pythagorean Theorem and Let $x$=length of common chord, we have:
$(\frac{10}{2})^2+(\frac{x}{2})^2=10^2$

$\frac{100+x^2}{4}=100$

$100+x^2=400$

$x^2=300$

$x=\boxed{10\sqrt{3}}$


geometry
circumcircle
incenter
Euler
geometry unsolved

Let $ ABC$ be triangle and $ D,E,F$ are points on its side $ BC,CA,AB$. Circumcircle of triangle $ AEF,BFD,CDE$ with center $ X,Y,Z$ meet at $ M$.(It is known that this three circle meet one point,called miquel point). Circumcircle of triangle $ DEF$ meet $ BC$ at $ S$ and $ T$ be symmetric of $ T$ wrt $ EF$. $ K$ is isogonal conjugate point of $ T$ wrt $ \triangle AEF$. Let $ P,Q$ be inverse of $ M$ wrt circumcircle of $ \triangle ABC,\triangle XYZ$. Show that $ P,Q,K$ are collinear.

remark. If triangle $ DEF$ is pedal triangle of $ M$, then $ K$ lie on $ MO$ where $ O$ is circumcenter of triangle $ ABC$.

How did you find this raletion?

a typo:
[quote="Leonhard Euler"]...and  $ T$ be symmetric of $ T$ wrt $ EF$. [/quote]
should be"  $ T$ is symmetric of $ S$ with respect to $ EF$."

and a picture:

[quote="Amir.S"]
[quote="Leonhard Euler"]...and  $ T$ be symmetric of $ T$ wrt $ EF$. [/quote]
should be"  $ T$ is symmetric of $ S$ with respect to $ EF$."[/quote]
Sorry for mistake
I found by generalizing bulgaria 1997 problem: Let $ ABCD$ be a convex quadrilateral such that $ \angle DAB \equal{} \angle ABC \equal{} \angle BCD$(this condition is same as $ \measuredangle DAB \equal{} \measuredangle ABC \equal{} \measuredangle BCD$ mod 180 where $ A,B,C$ are arbitrary points). Let $ H$ and $ O$ denote the orthocenter and circumcenter of the triangle $ ABC$. prove that $ H,O,D$ are collinear.(can be solved by denot $ M \equal{} AB\cap DC,N \equal{} BC\cap DA$ and apply http://www.mathlinks.ro/viewtopic.php?t=181573)
In my problem, take $ M \equal{} I$ where $ I$ is incenter of $ \triangle ABC$ and $ D,E,F$ as tangency points with incircle and side of triangle $ ABC$. Then it is easy to see $ K$ satisfy $ \measuredangle KFD \equal{} \measuredangle FDE \equal{} \measuredangle DEK mod 180$. Line $ PQ$ is $ OI$ where $ O$ is circumcenter of $ \triangle ABC$. Hence $ K$ lie on $ OI$. Using fact that $ OI$ pass orthocenter of $ DEF$(ie, Euler line of $ \triangle DEF$ pass $ O)$, this become above problem.

Dear [b][size=100]Leonard Euler[/size][/b],  [b][size=100]Amir.S[/size][/b] and all.

I am confused in this problem, because I don't understand how are defined the points $ P,$ $ Q$ (probably I must to draw again my schema).

Because of the point $ P,$ is the inverse of $ M,$ with respect to the circumcircle of $ \bigtriangleup ABC,$ does it means that $ (OM)\cdot (OP) \equal{} (OA)^{2},$ where $ O$ is the circumcenter of $ \bigtriangleup ABC?$

Thank you and regards, Kostas Vittas.

[quote="vittasko"]... Because of the point $ P,$ is the inverse of $ M,$ with respect to the circumcircle of $ \bigtriangleup ABC,$ does it means that $ (OM)\cdot (OP) \equal{} (OA)^{2},$ where $ O$ is the circumcenter of $ \bigtriangleup ABC?$
....[/quote]
yeah.that's its definition .


function
limit
irrational number
calculus
calculus computations

Let f(x) which:
f(x)=1, if xeQ
f(x)=0, if x does not belong in Q

You must prove that lim f(x) while x-->x0 does not exist for every x0eR

PS:Sorry for not using latex..
I got this in my cuntinuation of functions lesson but i cannot solve it

$ \lim_{x \to q} f(x)$ exists if and only if for every sequence $ x_n$ such that $ \lim_{n \to \infty} x_n \equal{} q$ the limit $ \lim_{n \to \infty} f(x_n)$ exists and is the [b]same[/b] for all such sequences.  So take $ x_n$ to be a sequence of rationals and then a sequence of irrationals.

yeah for rationals we can take x(n)=[n/n+sqrt(2)]x0?
But for irrationals what are we gonna take?

You seem to have pretty much answered your own question.  For rational $ x_0$ the sequence $ x_n \equal{} x_0 \plus{} \frac{1}{n}$ is a sequence of rationals that converges to $ x_0$ and the sequence $ x_n \equal{} x_0 \plus{} \frac{1}{n \plus{} \sqrt{2} }$ is a sequence of irrationals that converges to $ x_0$.

Of course, this is obscuring what's really going on:  given fixed $ x_k$ there exists a rational number and an irrational number in the ball of radius $ \frac{|x_0 \minus{} x_k|}{2}$ around $ x_0$.

Now i see the solution to my problem and my mistake..Thank you.
Do you know any books with examples and exercises?


Is it possible to get the LaTeX code for a pdf file?  For example, if someone else made a pdf but I want to add something to it, is it possible to convert the file back into LaTeX?


Don't ask me how, because I can't remember and I wouldn't be posting this but I'd been given a photocopy of a single chapter on 'Fourier series, Fourier transforms and Laplace transforms',
specifically chapter 13 page 167, it gives an introduction and moves on to Dirichlet's theorem and Fourier integrals, etc.
The text as far as I read was brilliant and since I have an avid interest in orthogonal systems I'm considering getting the full text.
One small problem, I don't know the name of the text, it's a slim chance but if anyone on this forum has seen it, please post a reply.
Thanks!

Have you tried using Google's book search feature to search specifically for some of the text you've read?

Thanks but I'd already checked there, no luck.


h is the harmonic mean of the equation given (that is the definition of the harmonic mean).  Here is the proof:
[hide] Since $\frac{1}{h}=\frac{1}{2}(\frac{1}{a}+\frac{1}{b})\Rightarrow \frac{1}{h}=\frac{1}{2}(\frac{a}{ab}+\frac{b}{ab})\Rightarrow \frac{1}{h}=\frac{a+b}{2ab}\Rightarrow h=\frac{2ab}{a+b}.$ Now say that h?a and h??b, where the ? can be replaced with >,<, or = and the ?? can be replaced with >,<, or =.  Substitute the values of h into the equation, we have that $\frac{2ab}{a+b}?a$ and $\frac{2ab}{a+b}??b.$  Dividing the first equation by a and the second by b, we get $\frac{2b}{a+b}?1$ and $\frac{2a}{a+b}??1.$  Multiplying both equations by a+b, we have $2b?a+b\Rightarrow b?a$ and $2a??a+b\Rightarrow a??b$.  Now, since a<b (which implies that b>a), we can substitute the ? with > and the ?? with <.  This means that h>a and h<b, so a<h<b.  [/hide]

[quote="Interval"]What exactly is meant by the term HARMONIC MEAN?

I think of the word harmonic in terms of music but not in terms of mathematics.

QUESTION:

For 0 < a < b, let h be defined by

(1/h) = (1/2)((1/a)+(1/b)).

Show that a < h < b. The number h is called the harmonic mean of a and b.

Also, is there another purpose for the harmonic mean other than just proving the above?[/quote]
the harmonic mean of two numbers is the reciprical of half the sum of the recipricals of the numbers
in other words, the harmonic mean of $a$ and $b$ is
$\frac{1}{\frac{\frac{1}{a}+\frac{1}{b}}{2}}$
simplified, it is
$\frac{2ab}{a+b}$

and as it stated in the problem,  h is defined by (1/h) = (1/2)((1/a)+(1/b)).
[hide="why a < h < b"]
$a=\frac{a^{2}+ab}{a+b}$
$b=\frac{b^{2}+ab}{a+b}$
$h=\frac{2ab}{a+b}=\frac{ab+ab}{a+b}$
if since $0<a<b$, then $a^{2}<ab<b^{2}$.
[/hide]

harmonic means are very useful.
suppose you have the problem:
Bob traveled from his home to Teapotland at the rate of 40mph.
He traveled back at the rate of 60mph. what was his average rate?

with this all you have to do is find the harmonic mean of the two speeds, which is 48, and youre done!! :)


Cauchy : $\frac{1+x+y}{9}\geq  \frac{xy}{x+y+xy}$.

[quote]pohoatza wrote : $\frac{1+x+y}{9}\geq \frac{xy}{xy+x+y}$[/quote]

How to find it ???


find all $ (x,y)\in N^{*}$ such that $ 2^{x} \equal{} 3^{y} \plus{} 5$

The only solutions are $ (x,y)\equal{}(3,1)$ and $ (5,3)$.
Non-existence of other solutions can be proved considering the given equation modulo 5440.
For general approach to this kind of problems, see http://www.mathlinks.ro/Forum/viewtopic.php?t=48431

[quote]modulo 5440. [/quote]

LOL ! Can you say me how can we have this idea ?  :maybe:

what is the order of 3 mod 5440=2^5*17?
It seems like it must be rather big

[b]Ramanujan47[/b], see the link above.

[b]me@home[/b], the order of 3 mod 5440 is 16.

How can that be? \[ 3^8\equal{}6561 \not \equal{} \minus{}1\]

[b]me@home[/b], because 5440 is not prime.

[quote="maxal"]For general approach to this kind of problems, see http://www.mathlinks.ro/Forum/viewtopic.php?t=48431[/quote]

Very very nice indeed.

Thanks for the link and great congrats for this idea!


no. of persons were engaged to do a piece of work which would have occupied them 24 hrs if they had commenced at the same time but instead of doing so, they commenced at equal intervals and then continued to work till the whole was finished , the payment being proportional to the work done by each . The first comer received 11 times as much payment as the last . Find the time occupied.

I have wasted around 10 days on this but Iam still unable to find the answer :(

The total amounts of work will be 24x hours of works, where x is the number of people.  If the first person received 11 times the pay of the last one, then that person worked for 11 times as long.  Let the work time of the last person be n.  Then the first and last person worked a total of 12 hours.  The second to last person and second person together also worked for 12n hours, because the timing was as regular intervals.  So then the total amount of work done was
12xn/2=6xn
Since the total amount of work remains constant, 6xn=24x, and then n=4.  So the last person worked for 4 hours and so the first worked for 44 hours, so the total time was 44 hours.


Let $ a,b,c\ge0$ such that $ ab \plus{} bc \plus{} ca \equal{} 1$ .Prove that

$ \frac {1 \plus{} bc}{2bc \plus{} a^2} \plus{} \frac {1 \plus{} ca}{2ca \plus{} b^2} \plus{} \frac {1 \plus{} ab}{2ab \plus{} c^2}\ge3$

Equality holds for $ (a,b,c)\sim (1,1,0)$  :)

And we have a hard problem

Let $ a,b,c\ge0$ such that $ ab \plus{} bc \plus{} ca \equal{} 1$.Find the minimum value of

$ P\equal{}\frac {1 \plus{} bc}{kbc \plus{} a^2} \plus{} \frac {1 \plus{} ca}{kca \plus{} b^2} \plus{} \frac {1 \plus{} ab}{kab \plus{} c^2}$

[quote="dduclam"]Let $ a,b,c\ge0$ such that $ ab \plus{} bc \plus{} ca \equal{} 1$ .Prove that

$ \frac {1 \plus{} bc}{2bc \plus{} a^2} \plus{} \frac {1 \plus{} ca}{2ca \plus{} b^2} \plus{} \frac {1 \plus{} ab}{2ab \plus{} c^2}\ge3$

Equality holds for $ (a,b,c)\sim (1,1,0)$  :)

And we have a hard problem

Let $ a,b,c\ge0$ such that $ ab \plus{} bc \plus{} ca \equal{} 1$.Find the minimum value of

$ P \equal{} \frac {1 \plus{} bc}{kbc \plus{} a^2} \plus{} \frac {1 \plus{} ca}{kca \plus{} b^2} \plus{} \frac {1 \plus{} ab}{kab \plus{} c^2}$[/quote]
How do you do? Mr dduclam. Your first inequality is equivalent to
\[ \sum\limits_{cyc}\frac {(a\minus{}b)^2(ab \minus{} 2bc \minus{} 2ac)}{(2ca \plus{} b^2)(2bc \plus{} a^2)}\geq 0
\]
i think it's not true, if i'm not wrong. Can you show me your proof?

[quote="Sunjee"][quote="dduclam"]Let $ a,b,c\ge0$ such that $ ab \plus{} bc \plus{} ca \equal{} 1$ .Prove that

$ \frac {1 \plus{} bc}{2bc \plus{} a^2} \plus{} \frac {1 \plus{} ca}{2ca \plus{} b^2} \plus{} \frac {1 \plus{} ab}{2ab \plus{} c^2}\ge3$

Equality holds for $ (a,b,c)\sim (1,1,0)$  :)

And we have a hard problem

Let $ a,b,c\ge0$ such that $ ab \plus{} bc \plus{} ca \equal{} 1$.Find the minimum value of

$ P \equal{} \frac {1 \plus{} bc}{kbc \plus{} a^2} \plus{} \frac {1 \plus{} ca}{kca \plus{} b^2} \plus{} \frac {1 \plus{} ab}{kab \plus{} c^2}$[/quote]
How do you do? Mr dduclam. Your first inequality is equivalent to
\[ \sum\limits_{cyc}\frac {(a \minus{} b)^2(ab \minus{} 2bc \minus{} 2ac)}{(2ca \plus{} b^2)(2bc \plus{} a^2)}\geq 0
\]
i think it's not true, if i'm not wrong. Can you show me your proof?[/quote]
I think you have mistake, Sunjee. Since the original inequality doesn't have equality when $ a\equal{}b\equal{}c$.

[quote="Sunjee"][quote="dduclam"]Let $ a,b,c\ge0$ such that $ ab \plus{} bc \plus{} ca \equal{} 1$ .Prove that

$ \frac {1 \plus{} bc}{2bc \plus{} a^2} \plus{} \frac {1 \plus{} ca}{2ca \plus{} b^2} \plus{} \frac {1 \plus{} ab}{2ab \plus{} c^2}\ge3$

Equality holds for $ (a,b,c)\sim (1,1,0)$  :)

And we have a hard problem

Let $ a,b,c\ge0$ such that $ ab \plus{} bc \plus{} ca \equal{} 1$.Find the minimum value of

$ P \equal{} \frac {1 \plus{} bc}{kbc \plus{} a^2} \plus{} \frac {1 \plus{} ca}{kca \plus{} b^2} \plus{} \frac {1 \plus{} ab}{kab \plus{} c^2}$[/quote]
How do you do? Mr dduclam. Your first inequality is equivalent to
\[ \sum\limits_{cyc}\frac {(a \minus{} b)^2(ab \minus{} 2bc \minus{} 2ac)}{(2ca \plus{} b^2)(2bc \plus{} a^2)}\geq 0
\]
i think it's not true, if i'm not wrong. Can you show me your proof?[/quote]

I don't think so,Sunjee  :(  My proof using SOS method. And the best (great) constant for the inequality

$ \sum\frac {1 \plus{} bc}{kbc \plus{} a^2}\ge \frac{12}{k\plus{}1}$

is $ k\equal{}2\minus{}\sqrt3$.

[quote="dduclam"]Let $ a,b,c\ge0$ such that $ ab \plus{} bc \plus{} ca \equal{} 1$ .Prove that

$ \frac {1 \plus{} bc}{2bc \plus{} a^2} \plus{} \frac {1 \plus{} ca}{2ca \plus{} b^2} \plus{} \frac {1 \plus{} ab}{2ab \plus{} c^2}\ge3$

Equality holds for $ (a,b,c)\sim (1,1,0)$  :)
[/quote]
The first problem, we can prove it as
setting $ x\equal{}ab,y\equal{}bc,z\equal{}ca$ then $ x\plus{}y\plus{}z\equal{}1$, the inequality becomes
\[ \sum \frac{x(x\plus{}1)}{2x^2\plus{}yz} \ge 3\]
\[ \Leftrightarrow \sum \frac{x(2x\plus{}y\plus{}z)}{2x^2\plus{}yz} \ge 3\]
\[ \Leftrightarrow 2(x\minus{}y)^2(y\minus{}z)^2(z\minus{}x)^2\plus{}3xyz(x\plus{}y\plus{}z)(xy\plus{}yz\plus{}zx) \ge 0\]
which is true. :)


I don't know for sure, but I thought the 12 this year was easy (seeing as how we normally have 4-5 people qualify for AIME, but this year, we had 10).  Was it true just for the first 15 or so problems?  In any case, I predict an impossible AIME or a mind-blowing USAMO index.

However there was an expansion which should mitigate the scores some. My understanding of the AMC 12s was that a lot of people qualified with 100-120 scores, but not too many people got beyond that, as the last few were tricky. I don't know too much though as I took the 10s.

Well if you haven't noticed it was only first 16 problems that were easier, the later problems are about the same.
Well there's no way the USAMO index is going to be even 220+ with the USAMO expansion.
And I don't think AMC is a good indication of the difficulty of AIME/USAMO.

In fact I think 2006 was like a practice for the 2007 AMC12's when the guessing penalty drops to 1.5 and I remember that they said they were going to make first half of the AMC easier in 2007 so the AIME qualifying score stays at 100 with the lower blank award .

Yeah ummm... I think the index is actually going to be a bit [i]lower[/i] than usual this year (even if the expansion didn't happen) because I felt the last 5 or so problems were a lot harder than usual. I expect a large number of 130s who will set the index to around 200.

The first 15 or so were really easy, which is probably why so many people from your school qualified for AIME. I expect that next year a lot fewer people will qualify.

I have to agree with Probability.  Again, the easy first 16 will mean more AIME qualifiers, but the difficult last nine (and especially 16-20) is going to mean fewer really high scores.

IF the AIME is like last year's AIME, and IF there was no expansion, given this year's AMC I'd put the index at about 225.

JB

[quote="probability1.01"]Yeah ummm... I think the index is actually going to be a bit [i]lower[/i] than usual this year (even if the expansion didn't happen) because I felt the last 5 or so problems were a lot harder than usual. I expect a large number of 130s who will set the index to around 200.
[/quote]

Hmm I thought #22 and 25 were pretty hard for AMC problems but the rest are okay.

Key words there were "than usual" :P. The AMC always has one or two tricky problems, but this year I thought that all the problems between 19 and 25 required more than 2 minutes to do and check, which was definitely not the case last year. Anyways my signature doesn't seem to be working...

Yeah, a bunch of scores between 100 and 115 doesn't mean that the USAMO cutoff will be higher.  Perhaps a few people that wouldn't have made it to AIME will now make USAMO, but this won't raise the cutoff.  This will probably mean that the AIME average will be a bit lower than it would have been, (2.5 vs 2.45 maybe) but it doesn't have as much of an effect near the top.


Hi all,

With the following code:
[code]\def\term#1#2{A^1_{{#1}:\vbox{\hrule height .1pt
\kern 1pt
\hbox{$\scriptstyle{#2}\kern 1pt$}}
\vrule width .1pt}}

\def\pure#1#2{A^{\hspace{20pt}1}_{{#1}:\vbox{\hrule height .1pt
\kern 1pt
\hbox{$\scriptstyle{#2}\kern 1pt$}}
\vrule width .1pt}}

\term{[1000]}{2}
\pure{[35]}{123456789}
[/code]

The output is shown in the image below.

The example is a little stupid as realistically I wouldn't need to use numbers that large, but I think it illustrates the point I'm trying to make.

In the first case, I would like the "1" to be centred over the "[1000]", and in the second case, I would like it to be centred over the "123456789"

Is there any way to make LaTeX automatically align it over the middle of the relevant bit of the subscript, or do I have to do it manually?

Thanks.

I am not sure that I correctly understand what you want to get. Anyway, here is what you can do.

$ A_{[1000: \overline{2}]}^{\phantom{[1000: }1} A_{[35: \overline{123456789}]}^{\phantom{35: [1234}1}$

Is that what you want?

Thanks for your reply Yury!

Oops. I'll try to clarify exactly what I'm trying to do.

In the first case, I want the "1" to appear [i]directly above [/i]the centre of the number [i]before [/i]the colon, whereas in the second case, I want the "1" to appear [i]directly above [/i]the centre of the number [i]after [/i]the colon.

Of course, I could do it as you did in the second case with \phantom or \hspace, but if possible I would like LaTeX to automatically figure out where the centre of the number is, and place the "1" above it. This is because I'm going to be using these symbols a lot, which is the reason I'm trying to get a function to do it automatically, rather than fiddling around with it every time I use that symbol.

The point is that with each of the functions, there are two arguments: the number before the colon, and the number after the colon. Because in general the numbers might have 1, 2, 3, 4, etc. digits, the centre of those numbers, and hence the alignment will vary depending on the numbers used, so I can't just use say \hspace{10pt} in general.

I wanted it to be more like this in the first case $ A_{[1000]: \overline{2}|}^{\phantom{[1}\,\,1}$
The second case is basically what I want, although I would (ideally) like the "1" to be a little more to the right like
$ A_{[35]: \overline{123456789}|}^{\phantom{35: [1234\,\,}1}$

So to sum up: is there some function in LaTeX that would be able to examine a number, and find the "centre" of that number? Or is there some other possible method to tackle this problem in general?

I hope this elucidates matters a little.

Edit: Here are another couple of examples of what I would like it to do if there were less digits:

$ A_{[3]: \overline{123456789}|}^{\,\,1}
\\
A_{[1236]: \overline{12}|}^{\phantom{1236: [1\,}1}
\\
A_{[123]: \overline{12}|}^{\phantom{[1}1}$

[quote="jshen"]So to sum up: is there some function in LaTeX that would be able to examine a number, and find the "centre" of that number? Or is there some other possible method to tackle this problem in general?[/quote]
\frac does that for you so you can use it. For example: instead of your $ A_{[3]: \overline{123456789}|}^{\,\,1}$ you could use $ A_{\genfrac{}{}{0pt}{2}{}{[3]: }\!\frac{1}{123456789}\kern \minus{}0.8pt\rule[\minus{}2.5pt]{.2pt}{0.9ex}}$ adjusting the numbers to suit your document.

Thanks stevem!

That should work for the second case.

In the first there isn't meant to be a bar below the "1" though. Is it possible to use \frac then somehow delete the viniculum?

Do you mean $ A_{\genfrac{}{}{0pt}{2}{1}{[1000]: }\!\frac{}{2}\kern \minus{}0.8pt\rule[\minus{}2.5pt]{.2pt}{0.9ex}}$ ?
[quote="jshen"]Is it possible to use \frac then somehow delete the viniculum?[/quote]Yes - look how it was done in the previous example with $ \genfrac{}{}{0pt}{2}{}{[3]: }$ see [url=http://tug.ctan.org/tex-archive/info/math/voss/mathmode/Mathmode.pdf]Math Mode[/url] page 68

Ah yes, that should be perfect!

Thanks for all of your help!


calculus
integration
inequalities
function
limit
real analysis
real analysis unsolved

Let $f: [0,\infty)\to\mathbb{R}$ a function such that \[ |f(x)-f(y)|\leq |x-y|,\forall x,y\geq 0.  \]
  Prove that \[ \int_{a}^{b} f(x)dx\leq bf(b)+\frac{b^{2}}{2}-af(a)-\frac{a^{2}}{2},\forall a, b\in[0,\infty), a<b.  \]
  If $f$ is derivable, find when equality occurs.

  [i] Alexandru Rosoiu [/i]

Since $\left|\lim_{y\rightarrow x} \frac{f(x)-f(y)}{x-y}\right|\leq 1\Longleftrightarrow |f'(x)|\leq 1.$

Thus use integral by parts we have $\int_a^b f(x)\ dx=\left[xf(x)\right]_a^b -\int_a^b xf'(x)\ dx$

$\leq[xf(x)]_a^b +\int_a^b x\ dx=bf(b)-af(a)+\frac{b^2-a^2}{2}.$

The equality holds when $f'(x)=-1.$

[b]Alternative Solution[/b]

Let $a$ fix as constant numbers, set $g(b)=bf(b)+\frac{b^2}{2}-af(a)-\frac{a^2}{2}-\int_a^b f(x)\ dx.$

Differentiate this with respect to $b$ for $b>a>0,$ we have $g'(b)=f(b)+bf'(b)+b-f(b)$
$=b\{f'(b)+1\}\geq 0\ \because |f'(b)|\leq 1,$ yielding $g(b)\geq g(a)=0.$

[quote="perfect_radio"]The hypothesis only tells us that $\displaystyle f$ is Lipschitz continuous.

It is differentiable only for "point b".[/quote]

f is not certanly derivable:)

Cezar Lupu, if you say it's that [b]easy[/b], could you give some sort of hint?

It results by integrating the inequality $f(x)\leq-x+a+b+f(b)+a \frac{f(b)-f(a)}{b-a}$ which is true because $f(x)-f(b)\leq b-x\leq b-x+a \left( 1+\frac{f(b)-f(a)}{b-a}\right)$ because $\frac{f(b)-f(a)}{b-a}\geq-1$.

[quote="zhaobin"]I think it can be proved by write $\int_{a}^{b}f(x)dx$ as Remmain Sum.[/quote]
Exactly :)

Let's use Abel's Identity, a.k.a. summation by parts:
\[\sum_{k=0}^{n-1}f_{k}\left( g_{k+1}-g_{k}\right) = f_{n}g_{n}-f_{0}g_{0}-\sum_{k=0}^{n-1}g_{k+1}\left( f_{k+1}-f_{k}\right) .\]
The main reason for doing that is that we want to hide under carpet the integration by parts done by kunny.
Letting $f_{k}= a+\frac{b-a}{n}\cdot k, \, g_{k}= f \left( a+\frac{b-a}{n}\cdot k \right)$, we get that
$\sum_{k=0}^{n-1}\left( a+\frac{b-a}{n}\cdot k \right) \left[ f \left( a+\frac{b-a}{n}\cdot (k+1) \right)-f \left( a+\frac{b-a}{n}\cdot k \right) \right] =$
$= b \cdot f(b)-a \cdot f(a)-\sum_{k=0}^{n-1}f \left( a+\frac{b-a}{n}\cdot (k+1) \right) \cdot \frac{b-a}{n}$.
Using the hypothesis, we get that $\left| LHS \right| \leq \frac{b-a}{n}\cdot \sum_{k=0}^{n-1}\left( a+\frac{b-a}{n}\cdot k \right) \to \frac{b^{2}-a^{2}}2$ as $n \to \infty$.
The $RHS$ is $b \cdot f(b)-a \cdot f(a)$ minus a Riemann sum that tends to $\int_{a}^{b}f(t) \, dt$ as $n \to \infty$.
The conclusion follows.


Hi,
Does anyone know if the Intermediate Number Theory Seminar will be offered next fall or when the book will be available.

Thanks very much

I don't know about next fall, but they have it this fall.  Starts oct. 21.  IDK about the book though.

We will likely have the Interm NT next summer, but not next fall.  We're not sure when we'll have a book for this material.


geometry trapezoid algebra polynomial calculus AMC USA(J)MO

Lets now discuss Round 2! What score did you think you guys got, as predictions?

You can also discuss what you did to approach solutions.

#1: Perfect
#2: Perfect. Eh, I thought I had to prove that the centers formed a trapezoid, but I guess not...
#3: ARGH. I had the solution by the casework way, but I thought that it wouldn't work, so I gave up after finding p(x)=x/2...stupid me. Hoping for 2.
#4: Perfect
#5: Nothing even close...

so 55250=17

My predictions:

#1: Perfect
#2: Perfect
#3: Epic fail. My method was sssooooo stupid, and I highly doubt I'd get it right...
#4: Sorta confident on this, though my method was, again, kinda dumb.
#5: Brute forcy here in this problem, but otherwise I seemed to discover a neat pattern.

So, 5/5/1/4/4=19...At least I should do better than I did on Round 1.

[quote="the future"]You know for problem 4, couldnt you just use Catalan's Conjecture?  (Instead of doing all that mod)[/quote]

Heh, that's what I did.

I cited Wolfram in the unlikely case that it's not well-known, but I doubt that that was necessary.

23/24
1. Full credit.
2. Full credit unless i made a stupid mistake.
3. I think i got this one at least part right. I proved that no polynomials of degree 2 or more could exist but did not explain why my linear solutions were the only solutions.
4. Full credit, but the proof was uglyish.
5. Full credit by mass summation and then application of the multiple angles formula.
Doe anyone who took the test last year think that this year's test is following a similar pattern? Round one had four relatively easier problems and then one hard problem(number 5 for this year and last year). On round two, there's always one problem that was really challenging which i missed and one problem which i brute forced(by calculus or summation).

1. Good
2. Good
3. Good
4. Good
5. I found a nice proof using directed areas, which allowed me to solve the problem really quickly.

1: good
2: good
3: almost good, I ran out of time, and hopefully my explanation is good enough for showing $ x\minus{}1$ and $ x/2$ work, but i didn't have time to show that $ x/4\plus{}1/2$ does not work. it was [i]almost [/i]a complete solution.
4: good, mostly the same as official, slightly longer
5: good, used identity $ \sum_{k\equal{}1}^{n\minus{}1} \sin(k\pi/n) \equal{} \cot(\pi/(2n))$

55455 = 24 probably.

Dang...after relooking the solutions i think i might have made a stupid mistake on #2... :mad:

Hmm... did anyone solve #5 using directed areas? I did that and it allowed me to have a solution a lot shorter than the official one... except I did a bunch of unnescessary complex number stuff instead of just one simple application of sinx = sin(180-x)...grr...

Nope. I didn't. I just used brute force like I almost always do :(

24-25
1. Perfect(Too easy)
2. Perfect(Too easy)
3. Perfect(Nice and easy once I found the procedure, but it took 6 pages. I tried to shrink it. There where probably some Theorems thaat I could have used that I don't know.)
4. Perfect(Catalan's conjecture(Mih$ \breve{a}$ilescu's theorem)  made it easy.)
5. I got an expression that had a 2006 term sum. And I outlined how to sum it, and gave the result. If I had summed it, I would have used about 10 extra pages (I think. 'cause I stayed up till 10:30 and got 1/3 of the way through (3.5 pages) 'till I realized it would take untill the deadline. I did like 3 hours :sleeping:  :fool:  for that 1/3!!!!! I'm slow :( at typing messsssssy stuff in LaTeX.)
Aaaaah. pianoforte, I didn't know that identity. Typical of me I guess.

Could I get full credit for it still?
I want gold :winner_first: soo bad. 10th grade :roll:
Oh well if I don't.

#5 can we written as a constant times the sum in http://www.artofproblemsolving.com/Forum/viewtopic.php?p=353049#353049

interestingly, this was the first USAMO problem I ever solved...

Forgot the $ \frac{1}{2}$ in the area triangle formula in #5.

All solutions are pretty similar to the official ones. My 2, 3, 4 are considerably longer than the official ones because of over-proving some sort-of obvious stuff (like the trapezoid)...I guess I can hope for a 25.

Personal rating: 3>4>5>2>1. In #3, it took me a while to realize that $ p$ can only be linear or constant. It also took me a while to realize that I can use limit to prove this (but isn't limit sorta calculus?). #4 is a bunch of casework and stuff. #5 is kinda routine if you know how to use complex number clearing the trig summation...

[quote="Math Geek"]Um, did anyone else use java to solve number 5 like I did? :P Cuz, like apart from the time it took to run the program, I thought my solution was easier to carry out (no thinking involved  :lol: ).

For number three, I didn't do anything with trapezoids.  I plotted the circles on the coordinate plane and solved using really basic application of the distance formula.[/quote]
Java would've taken too long for me to program, so I programmed with VB. I used the program to check my answer. I found the solution separately since I knew the program wasn't accurate enough to be the solution.

1) messed up casework...maybe a 3 or 4
2) perfect!
3) perfect!
4) perfect!
5) eh...maybe I'll get away with it. 3 or 4

I did worse this round, I'm pretty sure......................

[quote="Yongyi781"][...] Thus it will not suffice just to show that two numbers are "approximately equal." Especially since it will be off after 2013021 calculations![/quote]

Yes, you're right it can't show equality, but you'd be surprised that it isn't as far off as you might think after the 2000000+ operations: it's accurate to any decimal place that both Java and the TI-89 can display.

[quote="xpmath"][...] and used cyclic quadrilaterals for #5.[/quote]

Can you explain further what you did?

1- 5
2- 5
3- 2-3(forgot x - 1 goddang!!!)
4- Perfect used alot of mod and fermat euler
5- ran out of time- 0
so 15 + 3 = 18?

deproved a bit........

Well, I had almost no time left for #5, so I just hurriedly wrote some random claim down that I'm almost sure isn't even correct. Basically, I chose two points, found a point on one side of the line determined by those two points (that was also on the circle) and also chose the point directly "opposite." Then I tried summing them by using the ridiculous claim I mentioned before :P  :roll:

[quote="jvenezuela716"]
[quote="xpmath"][...] and used cyclic quadrilaterals for #5.[/quote]

Can you explain further what you did?[/quote]

I think any triangle can be extended past the center to the opposite side of the circle and form a cyclic quadrilateral. so each pair of triangles is actually equivalent to two RIGHT triangles, bam.

1. fail. i assumed that rotations/reflections are equal. hopefully i get credit since it was "thinkable" and correct in that way
2. perfect. coordinate bash FTW!
3. fail. did not regard when a_n were polynomials itselves and totally screwed up
4. perfect. bunch of mods and stuff to solve it
5. epic fail. just bashed in order to get some partial credit (or hope so)

lol maybe 19-22. i spent the last two days working on it. i think between those two days, i got a total of 5-6 hours of sleep (total).

23, 24 maybe?

1. good
2. good, but confidence brought down by Mr. Schram's post in the other topic
3. good, but made a typo one place.
4. good, used Catalan's conjecture
5. not as good, used a TI-83 Plus program to formulate a conjecture, then compare that conjecture against the result given by the formula

Combined w/ the 22 of round 1 to give a 45 or 46. Blah not much chance for a gold.

1: 5 (no problem)
2: 5 (I'm almost sure about this one)
3: 4 (I can sense already some lack of rigor, but correct)
4: 4 (use of Catalan's conjecture is questionable)
5: 2 (correct answer, but that's about it)

score: 20

This is actually the first time I think I got a 24-25. Considering last time I got an 18... : /

#1: Perfect
#2: Perfect
#3: Kind of Odd. Proved that p(x) had to be linear (which was a huge waste of time) and created bounds for a_n and p(a_n). Then I got to the casework.
#4: Perfect
#5: Summed up the area left over after the triangles were taken out of the circles and subtracted that from the total area of the circles.

Hopefully they won't take off for my unnecessary rambling in 3.

I didn't know USAMTS even existed until the day before we had to hand in round 2, so I had to work really quickly and didn't have much time. I'd say:
#1 perfect
#2 can't remember if i got the exact same answer as in the solution, but i did exactly what they did there, so maybe i made a stupid error. Maybe 4?
#3 counted all the cases, but i don't know why i got 3x/2 instead of x-1 in the second case. A 3?
#4 completely perfect. Did the exact same thing as the official solution
#5 really didn't have time. when i actually got to this one, i had like 1 hour until the deadline. i might have gotten a 1 maybe.

on #2 i simply assumed that the center of circle C1 was C1, the center of C2 was C2, the center of C was C, etc. Hope they don't take off points for that, since in the official solution, they named the centers of the circles explicitly.

Um...will they take off points for that?
sigh...the only other problem i actually solved RIGHT besides #1...i am such an idiot...

#1: Bah, wrong assumption (accounted for rotations) and managed to completely miss a case, so hopefully 2 if i'm really lucky...
#2: Good. Hoping 4,5
#3: No progress
#4: PHAIL
#5: No progress.

So like 6,7. fail. Last round was 13 so.......


This question is originally from AHSME but I have NO clue on this problem so I need someone to explain me by step by step.  :D Before giving the problem, thanks!

[b]A large cube is formed by stacking 27 unit cubes. A plane is perpendicular to one of the internal diagonals of the large cube and bisects that diagonal. The number of unit cubes that the plane intersect is:

A) 16
B) 17
C) 18
D) 19
E) 20
[/b]

take out a rubiks cube will help visualize much better than a drawing. Although im not sure about the answer, make sure you are looking at internal diagonal, not face diagonal.

Um.

That didn't help me a lot. Anymore?


[quote="metafor"]“Which planet is usually closest to the [dwarf] planet Pluto? For purposes of analysis, it may be assumed that the orbits are concentric circles in a plane, and that each planet has a different constant angular velocity. For a given position of any planet, each of the other planets has an equal probability of being at any point on its orbit.”

Complete solutions, please. :P[/quote]

Uh, could you repeat that in English please????

 :?: :?:

Basically asking, which planet is usually closest to Pluto? Assume that each of the planet's orbits are circles on a 2-d plane, and that each planet has a different speed. A planet also has an equal probability of being anywhere in it's orbit.

Could you give some info such as planet speed and distance from the sun?

Anyways, my guess would be

[hide]
Mercury since it can never be [b]that[/b] far away, while all of the others can be.[/hide]

The exact speed or distance of the planets is irrelevant; assume the planets' orbits are perfect circles that lie on the same plane, and have the same center (the Sun). Consider Pluto as stationary.

[hide="As for your guess..."]Can you explain, using probability theory, [i]why[/i] you think it is Mercury?[/hide]

hey dude u hav to lok into eccentricity as well so for a period of timr usually in angstorm parrallaxs pluto closes to uranus but for most of the time neptune is its partner

The orbits are assumed to be concentric circles, therefore the eccentricity of Pluto's orbit (as well as those of the other planets) is zero.

then thats not a possible scenario and the answer implifies me!
 :wink:

[hide]I don't think it's Mercury because it is very rare that out of the 7 other planets (besides Mercury) none of them are near Pluto. This is a very difficult question.[/hide]

Its definitely not Mercury (at least in general). If you put one planet arbitrarily close to the sun, and 100 planets arbitrarily close to Pluto, then the one by the sun is only closest (2/3)^100 of the time. Which means one of the others must be closer more often.

I'm intrigued to what the real answer is.

[quote="AIME15"]im guessing neptune?[/quote]

Well, no, it can't be neptune either, at least in general. Same case as above but only one planet near pluto; then the one near the sun is closer 2/3 of the time.

[quote="moogra"]Since it is concentric and there are other unrealistic constraints, can you give more information?[/quote]

Kepler's First Law of Planetary Motion:

A planet will sweep out the same area in the same amount of time as every other planet.

Its probably not Neptune because if the answer was obvious, metafor wouldn't post it. It must be a trick question, or else the author of the book wouldn't include it in his book.

Im gonna throw one out there and say its the (dwarf) planet Eris which lies just behind pluto (relative to the size of the solar system). and since Eris has a slower and wider orbit than Pluto and assuming that the planets closer to the sun have quicker orbit times then, in theory, Eris would stay closer for longer.

Assumptions - that all the 'wandering stars' (planets) start from the same alingment

[quote="Ihatepie"]Anyways, my guess would be

[hide]
Mercury since it can never be [b]that[/b] far away, while all of the others can be.[/hide][/quote]

Oh you're saying [hide="that"]Mercury could be on the same side of the sun as Pluto, and all the others are on the other side?[/hide]

Clarification: Only the 8 well-known planets (Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune) of the solar system are to be considered.

Could you explain your reasoning, Lawrence?

I'll try to post the author's solution soon; I've been having problems with uploading the image, so I might have to draw it.

[quote="Lawrence Wu"]the average distance from pluto is the same for every planet because the orbits are concentric circles and therefore have all their centers at the sun[/quote]

That's the AVERAGE distance. We're asking which planet is nearest Pluto for most of the time, which might not be the same planet as the one with the shortest average distance from Pluto.

[quote="stevenmeow"]i am guessing that a balance of having an orbit far away from pluto's and having few planets between pluto and it would result in the optimal planet (english: a planet close to the middle)[/quote]

Yeah, I think thats exactly the idea, though its impossible to know the real answer without the actual details of the radii. I'd guess it would be around venus or earth, as a rough guess.


function
geometry
geometric transformation
reflection

Graph y = |x^3 + x|.

MY WORK:

After graphing the above equation, the graph I created is very similar to the graph of f(x) = |x|.  It looks like the letter V.

However, the book tells me that the graph of y = |x^3 + x| actually looks like the graph of f(x) = x^2.  In other words, that it actually looks like a parabola.

How can this be?

How can I PROPERLY graph y = |x^3 + x| using paper and pencil and not a graphing calculator?

Maybe you just didn't pick enough points to base your graph upon? Try plugging in -2, -1, 0, 1, and 2 into the function and graphing those points.

Well it is $x^{3}$ so the try substituting more points so that it would look like the actual graph:

When the function encounters negative values for x it will output a negative value due to it being cubic. Because the function is absolute valued (what is this actually called?) these values now become positive leaving the graph in only the first two quadrants.

Just look at the graph of the function by itself and reflect the third quadrant in the x axis and you should have your graph.