trigonometry
calculus
integration
real analysis
real analysis unsolved

[img]http://img201.imageshack.us/img201/8844/calculuswl6.jpg[/img]

This question was for extra challenge and I have not been able to solve it. Help or solution would be appreciated.[/b]

Define $ f_n(x)\equal{}\cos^n(x)\, \cos(nx)$ .

Because of $ f_n(\pi\minus{}x)\equal{}f_n(x)$ your integral $ I_n$ equals $ \frac12 \int_0^\pi f_n(x) dx$ .

$ \equal{}\frac12 \int_0^\pi \frac{(e^{ix}\plus{}e^{\minus{}ix})^n}{2^n} \, \cos(nx) \, dx\equal{}\frac{1}{2^{n\plus{}1}}\, \int_0^\pi \sum_{k\equal{}0}^n {n\choose k}\, e^{i (n\minus{}2k) x}\, \cos(nx) \, dx$

$ \equal{}\frac{1}{2^{n\plus{}1}} \sum_{k\equal{}0}^n {n\choose k} \left[ \underbrace{\int_0^\pi \cos\Big((n\minus{}2k)x\Big) \cos(nx) \, dx}_{\equal{}\frac{\pi}{2} \, \text{for} \, k\equal{}0,n \qquad \equal{}\, 0 \, \text{otherwise}}\plus{}i\underbrace{\int_0^\pi \sin\Big((n\minus{}2k)x\Big) \cos(nx)\, dx}_{\equal{}0}\right]$ .

Hence $ I_n\equal{}\frac{\pi}{2^{n\plus{}1}}$ .

[color=darkblue][b][u]Proof.[/u][/b] Denote $ \boxed {a_n\equiv\int_0^{\frac {\pi}{2}}\cos^nx\cos nx\ \mathrm{dx}}$ . Then

$ 2a_n \equal{} \int_0^{\frac {\pi}{2}}\left[\cos(n \plus{} 1)x \plus{} \cos(n \minus{} 1)x\right]\cos^{n \minus{} 1}x\ \mathrm{dx}$ . Thus,

$ 2a_n \equal{} a_{n \minus{} 1} \plus{} \int_0^{\frac {\pi}{2}}\cos^{n \minus{} 1}x(\cos x\cos nx \minus{} \sin x\sin nx)\ \mathrm{dx}$ $ \implies$

$ 2a_n \equal{} a_{n \minus{} 1} \plus{} a_n \minus{} A$ , where $ A \equal{} \int_0^{\frac {\pi}{2}}\sin nx\sin x\cos^{n \minus{} 1}x\ \mathrm {dx}$ .

Now I"ll apply the integration by parts to the definite integral $ A$ :

$ \boxed {\ \begin{array}{ccc} u(x) \equal{} \sin nx & \implies & u'(x) \equal{} n\cos nx \\
 \\
v'(x) \equal{} \cos^{n \minus{} 1}x\sin x & \implies & v(x) \equal{} \minus{} \frac 1n\cdot\cos^nx\end{array}\ }$ $ \implies$ $ A \equal{} a_n$ .

Therefore, $ a_0 \equal{} \frac {\pi}{2}$ and for any $ n\in \mathcal N^*$ , $ a_{n} \equal{} \frac 12\cdot a_{n \minus{} 1}$ .

In conclusion, $ \boxed {\ a_n \equal{} \frac {\pi}{2^{n \plus{} 1}}\ ,\ n\in \mathcal N\ }$ .[/color]

It's tedious, but the indefinite integral can be computed via the exponential substitution Dr. Doe used.


algebra
polynomial
Galois Theory
algebra unsolved

Solve this equation in R:

$ 16x^4\plus{}5\equal{}6\sqrt{4x^3\plus{}x}$

[quote="Dr Sonnhard Graubner"]hello, squaring both sides and factoring we get
$ \left( 2\,x \minus{} 1 \right) \left( 128\,{x}^{7} \plus{} 64\,{x}^{6} \plus{} 32\,{x}^{5} \plus{} 16 \,{x}^{4} \plus{} 88\,{x}^{3} \minus{} 28\,{x}^{2} \minus{} 14\,x \minus{} 25 \right) \equal{} 0$
from here we get the solutions
$ x_1 \equal{} 0.5$
$ x_2\approx0.6675550667172057665549844573345132913176851282843684092636002078783803133984116828289411631696168458$
Sonnhard.[/quote]

Dear Dr Sonnhard Graubner! Please don't post the solution by use soft, because every Body known that.

hello, how would you solve this equation?
Sonnhard.

Hi thanhnam2902, your post
"Please don't post the solution by use soft, because every Body known that"
is absolutely ridiculous for $ 5$ (or more ?) reasons:
1) The Sonnhard's solution $ \approx{0.66755}$ is correct and is a root of a polynomial of degree 7 whose Galois group is $ S_7$ that is this solution can't be calculed with radicals. Moreover there don't exist any solutions in the form $ cos(k\pi/n)$.
2) Nowadays using calculator is a very important part of mathematics. That gives ideas for solving problems.
3) Here it is important to know that there exists a trivial solution $ 0.5$ and a non trivial solution.
4) Very often the problems you have posted couldn't be solved without soft. Thus if you want to be in agreement with your post, don't send such problems.
5) Do not discourage the good wills; if not, only few people will answer your exercises


calculus
integration
trigonometry
calculus computations

Evaluate

\[ \int_\frac{\pi}{4}^\frac{\pi}{3} \frac{2\sin ^ 3 x}{\cos ^ 5 x} dx. \]

The problem is not difficult because $I(-sinx,-cosx)=I(sinx,cosx)$

[quote="zhaobin"][b]Solution[/b]
\[ I=\int_\frac{\pi}{4}^\frac{\pi}{3} \frac{2\sin ^ 3 x}{\cos ^ 5 x} dx =-\int_\frac{\pi}{4}^\frac{\pi}{3} \frac{2\sin ^ 2 x}{\cos ^ 5 x} d\cos x  \]
Let $t=\cos x$
we get:
\[ I=\int_{\frac{1}2}^{\frac{\sqrt 2}2}\frac{2-2t^2}{t^5}dt=\int_{\frac{1}2}^{\frac{\sqrt 2}2}\frac{2}{t^5}dt-\int_{\frac{1}2}^{\frac{\sqrt 2}2}\frac{2}{t^3}dt=6-2=4  \] :)[/quote]

zhaobin, your answer is correct. :)

[quote="du0_hell"]$I=\int \frac{2\sin ^ 3 x}{\cos ^ 5 x} dx= \int 2 tan^3 x sec^2 x dx = \frac{ tan^4 x} 2 + C$

I think u can do it that way too, just pop in the proper limit[/quote]

This solution is same as mine, that is to say, we should remark tangent. :lol:


calculus
integration
function
inequalities
real analysis
algebra
domain

Let $f: [0,1]\longrightarrow \mathbb{R}$ a [b]continuous by parts function[/b], and let $g$ be a continuous and convex function.
Proof the Jensen's inequality:
\[ g\left(\int_{0}^{1} f(t)dt\right)\leq \int_{0}^{1} f(g(t))dt  \]

The condition on $f$ can be weakened; what you need is that $\int_0^1f(t)\,dt$ exists; as a Lebesgue integral would be enough.

The key lemma: for all $x_0$ there exists $m$ such that the function $L(x)=m(x-x_0)+g(x_0)$ satisfies the condition

$L(x)\le g(x)$ for all $x.$

Note: if $g$ is differentiable at $x_0,\,m$ must be chosen to be $g'(x_0),$ which makes $L$ the tangent line. There are circumstances in which $m$ is not uniquely specified; all the lemma requires is that there exist at least one.

I like expressing Jensen's inequality this way:

Suppose $X$ is a random variable such that $E(|X|)<\infty.$ Then

$g(E(X))\le E(g(X)).$

Note that that includes the version above. A random variable is a (measurable) function whose domain is a probability sample space. The $f$ is a function whose domain is $[0,1].$ The set $[0,1],$ with its usual measure, is a probability sample space.

[hide="Proof."]Let $x_0=E(X).$ (Alternatively, let $x_0=\int_0^1f(t)\,dt.$)

Choose $L$ as in the lemma. Then

$E(g(X))\ge E(L(X))=E(mX+b)=mE(X)+b$

$=mx_0+b=L(x_0)=g(x_0)=g(E(X)).$

Alternatively (and it's the exact same argument):

$\int_0^1g(f(t))\,dt\ge\int_0^1L(f(t))\,dt=\int_0^1(mf(t)+b)\,dt$

$=m\int_0^1f(t)\,dt+b=mx_0+b=L(x_0)=g(x_0)=g\left(\int_0^1f(t)\,dt\right).$[/hide]

Thank you :)
But, how can I proof this lemma?

Tanks again

For each point $x_0,$, consider $m^-=\sup_{x< x_0}\frac{g(x_0)-g(x)}{x_0-x}$ and $m^+=\inf_{x> x_0}\frac{g(x)-g(x_0)}{x-x_0}.$

We need to show that $m^-$ and $m^+$ both exist and that $m^-\le m^+.$


LaTeX algebra function domain polynomial Asymptote

$ \text{Solve}$ $ f(x)\equal{}\frac{x^2}{1\minus{}x}$ $ \text{for x and y(which is also f(x))}$

PS- I figured out how to write actual text in latex with out it running together!!!
PSS- Its not all of the stop starting with the dollar signs either

\text{type your text here but don't forget ur dollar signs on either side}

Are you asking for $ f(x)$ to be expressed in terms of $ x$ and $ y$?

And please take a look at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=143044]this[/url].

oh crap, DANKE!!!

and i need the domain and rang of that function

Well, $ x \neq 1$, so that's obvious. Also, because the power of the polynomial is larger in the numerator than the denominator, then there is no horizontal asymptote. To find vertical asymptotes, we set our denominator to zero and solve for $ x$. There are no common factors between the numerator and denominator, so there are no holes in the function.

Therefore, the domain is $ ( \minus{} \infty, 1) \cup (1, \infty)$ and the range is $ y \in \mathbb R$

EDIT: The all-reals thing doesn't look as it should be - does anyone know what the code should be? Thanks.

EDIT2: Thanks, math154!

[code]\mathbb R[/code]

produces

$ \mathbb R$.


Let a,b,c>0 such that $ a\plus{}2b\plus{}3c\equal{}1$. Prove that:
$ (1\minus{}a)(1\minus{}b)^2(1\minus{}c)^3\geq 5^6ab^2c^3$


abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ a finite group , $ H\leg G$ such that $ [G: H] = 2$ (i.e. $ H \triangleleft G$)  ,

let $ x\in H$.

Prove that :

a) if $ C_G(x) \leq H$ , then $ |c\ell_H(x)| = \frac {1}{2}|c\ell_G(x)|$.

b) if $ C_G(x)$ is not contained in $ H$, then $ |c\ell_H(x)| = |c\ell_G(x)|$.

Note that :

$ c\ell_G(x)$: conjugacy class of $ x$ in $ G$.

$ c\ell_H(x)$: conjugacy class of $ x$ in $ H$.

I think we need to use First Isomorphism theorem in both $ a$ and $ b$, what do u think guys ?  :)

We do not need isomorphism theorems or $ H$ normal, (coset decomposition and corresponding) index/counting formulas sufficient.

a) $ |cl_H(x)|\equal{}|H: C_H(x)|\equal{}\frac{1}{2}|G: C_H(x)|\equal{}\frac{1}{2}|G: C_G(x)|\equal{}\frac{1}{2}|cl_G(x)|$

b) $ C_G(x)\not\subseteq H$ implies $ G\equal{}HC_G(x)$ and $ |G|\equal{}|HC_G(x)|\equal{}\frac{|H\parallel{}C_G(x)|}{|H\cap C_G(x)|}\equal{}\frac{|H\parallel{}C_G(x)|}{|C_H(x)|}$, i.e. $ [C_G(x): C_H(x)]\equal{}2$ and we get
$ |cl_H(x)|\equal{}|H: C_H(x)|\equal{}\frac{1}{2}|G: C_H(x)|\equal{}|G: C_G(x)|\equal{}|cl_G(x)|$

Thank you hjbrasch  :)

I have one note only : how did you "get" if $ C_G(x) \not\subseteq H$ , then $ G = HC__G(x)$ ??

Since $ H$ has index $ 2$ in $ G$, we have the right coset property $ \forall g\not\in H: \,H\cup Hg\equal{}G$. Since $ C_G(x)$ contains some element $ g$ not in $ H$, we must have $ G\equal{}H\cup Hg\equal{}H\cdot \{1,g\}\subseteq H\cdot C_G(x)\subseteq G$, i.e. $ H\cdot C_G(x)\equal{}G$.


Find the value of $ \lim_{n \to \infty} (a^{n} \plus{} b^{n})^{\frac{1}{n}}$ where $ a \ge b > 0$.

$ a\le\left(a^{n}\plus{}b^{n}\right)^{1/n}\le\left(2a^{n}\right)^{1/n}\equal{}2^{1/n}a$. Now apply the squeeze theorem.

$ (a^n\plus{}b^n)^{\frac{1}{n}}\equal{}\left(a^n\left(1\plus{}\left(\frac{b}{a}\right)^n\right)\right)^{\frac{1}{n}}\equal{}a\left(1\plus{}\left(\frac{b}{a}\right)^n\right)^{\frac{1}{n}}$.

Because $ 1<\left(1\plus{}\left(\frac{b}{a}\right)^n\right)^{\frac{1}{n}}<1\plus{}\left(\frac{b}{a}\right)^n$, $ \lim_{n\rightarrow\infty}\left(1\plus{}\left(\frac{b}{a}\right)^n\right)^{\frac{1}{n}}\equal{}1$ by squeeze theorem. ($ a\neq{b}$)

If $ a\equal{}b,\quad(a^n\plus{}b^n)^{\frac{1}{n}}\equal{}2^{\frac{1}{n}}\cdot{a}\rightarrow{a}$  when $ n\rightarrow{\infty}$.

So, the value we want to find is $ a$.


function
search
number theory unsolved
number theory

Let $ m$ and $ n$ are positive integer. Prove that
$ \phi(5^{m}\minus{}1)\equal{}5^{n}\minus{}1$  $ \implies$  $ (m,n)>1$

This has been solved on the forum before.Try to search :wink:


The sides of triangle $BAC$ are in the ratio $2: 3: 4$. $BD$ is the angle-bisector drawn to the shortest side $AC$, dividing it into segments $AD$ and $CD$. If the length of $AC$ is $10$, then the length of the longer segment of $AC$ is:

$\text{(A)} \ 3\frac12 \qquad \text{(B)} \ 5 \qquad \text{(C)} \ 5\frac57 \qquad \text{(D)} \ 6 \qquad \text{(E)} \ 7\frac12$

[hide]The angle bisector theorem yields $C$ as the answer.[/hide]

The angle bisector theorem states that if a line bisects an angle of a triangle, then it divides the opposite side into two segments that are proportional to the other two sides of the triangle.  Applied to this problem, the proportion would be AD/AB=DC/CB.  Plug in the values of the sides and x for the side AD and 10-x for the side DC.  This gives you the proportion x/15=10-x/20.  Solving the propotion gives you 40/7, or 5 5/7.  C.


[quote="shfdfzhjj"]a+b+c=3；a，b，c>0;
prove：sigma a/(ab+1)>=1.5[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21689


Let $ a,b,c>0$. Show that:

\[ 2\sqrt{\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2}}\plus{}\sqrt{\frac{a\plus{}b}{c}\plus{}\frac{b\plus{}c}{a}\plus{}\frac{c\plus{}a}{b}\minus{}2}\ge 4\]

[quote="Pain rinnegan"]Let $ a,b,c > 0$. Show that:
\[ 2\sqrt {\frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}} \plus{} \sqrt {\frac {a \plus{} b}{c} \plus{} \frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b} \minus{} 2}\ge 4\]
[/quote]
By AM-GM $ 2\sqrt {\frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}} \plus{} \sqrt {\frac {a \plus{} b}{c} \plus{} \frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b} \minus{} 2}\equal{}$
$ \equal{}2\left(\sqrt {\frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}} \plus{} \sqrt {\frac{1}{4}\left(\frac {a \plus{} b}{c} \plus{} \frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b} \minus{} 2\right)}\right)\geq$
$ \geq4\sqrt[4]{\frac {ab \plus{} bc \plus{} ca}{4(a^2 \plus{} b^2 \plus{} c^2)}\left(\frac {a \plus{} b}{c} \plus{} \frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b} \minus{} 2\right)}.$
Hence, it remains to prove that $ \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}\left(\frac {a \plus{} b}{c} \plus{} \frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b} \minus{} 2\right)\geq4,$ which is obvious by $ uvw.$

[quote="arqady"][quote="Pain rinnegan"]Let $ a,b,c > 0$. Show that:
\[ 2\sqrt {\frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}} \plus{} \sqrt {\frac {a \plus{} b}{c} \plus{} \frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b} \minus{} 2}\ge 4\]
[/quote]
By AM-GM $ 2\sqrt {\frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}} \plus{} \sqrt {\frac {a \plus{} b}{c} \plus{} \frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b} \minus{} 2} \equal{}$
$ \equal{} 2\left(\sqrt {\frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}} \plus{} \sqrt {\frac {1}{4}\left(\frac {a \plus{} b}{c} \plus{} \frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b} \minus{} 2\right)}\right)\geq$
$ \geq4\sqrt [4]{\frac {ab \plus{} bc \plus{} ca}{4(a^2 \plus{} b^2 \plus{} c^2)}\left(\frac {a \plus{} b}{c} \plus{} \frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b} \minus{} 2\right)}.$
Hence, it remains to prove that $ \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}\left(\frac {a \plus{} b}{c} \plus{} \frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b} \minus{} 2\right)\geq4,$ which is obvious by $ uvw.$[/quote]

$ \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}\left(\frac {a \plus{} b}{c} \plus{} \frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b} \minus{} 2\right)\geq4$

$ \iff [3,2,0] \geq [3,1,1]$

which is true by Muirhead... :)

or $ \iff \sum{a^3(b \minus{} c)^2} \ge 0$, clear true... :lol:

[quote="kuing"]
$ \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}\left(\frac {a \plus{} b}{c} \plus{} \frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b} \minus{} 2\right)\geq4$

$ \iff [3,2,0] \geq [3,1,1]$

which is true by Muirhead... :)

or $ \iff \sum{a^3(b \minus{} c)^2} \ge 0$, clear true... :lol:[/quote]
I did not try it. :D


trigonometry
ratio
complex numbers
geometry proposed
geometry

Construct the convex quadrilateral $ABCD$, given the lengths $AB,BC,CD,DA$ and the sum $\angle{A}+\angle{C}$

[color=brown][b]P.S.[/b]
This was the subject of an old (lost) topic. I searched but I can't find it. It is possibly deleted.
I found a solution which is based on complex numbers, but the construction can be done with ruler and compass.
If there will be no replies in 2 months I'll post my solution[/color]

Well, I wrote that I had a solution with complex numbers. And it was really complex, I mean, very complicated  :D
I found that there is an elegant solution, with no trig, no complex numbers at all

$ WLOG$ assume that $ A+C=\phi\leq 180^\circ$ (if not, then we rename the vertices $ A,B,C,D \to B,C,D,A$)

Suppose that we have constructed the quadrilateral. We will draw a triangle congruent to $ CBD$
On the ray $ AB$ take a point $ B'$ such that $ AB'= CB$
And take a point $ D'$ out of $ ABCD$ such that $ AD' = CD$ and $ B'D'= BD$.
Clearly the triangles $ CBD,A'B'D'$ are congruent.
And $ \angle DAD' = \phi$.
[img]9818[/img]

Now let $ E\in AD'$ such that $ BE \parallel B'D'$.

We have $ \frac{BE}{B'D'}= \frac{AB}{AB'}\Rightarrow  \frac{BE}{BD}= \frac{AB}{CB}$ = known

[size=150]Construction[/size]

Construct the angle $ \widehat{DAx}= \phi$

On the ray $ Ax$ take the point $ E$ such that $ \frac{AE}{CD}= \frac{AB}{CB}$

Draw the circle $ w$ which divides the segment $ ED$ in the ratio $ \frac{AB}{CB}$ (apollonius circle)

Draw a circle $ b$ with center $ A$ and radius equal to the given $ AB$.

Then the point $ B$ is an intersection point of $ w$ and $ b$, the one which is in the interior of the angle $ DAD'$

And $ C$ is constructed [b]s-s-s[/b] from the already constructed points $ B,D$ and the given lengths $ BC,CD$


geometry
AMC
AIME
AoPS Books
number theory
\/closed

I think I am past all of the introductory classes as of now.
I am ready for Intermediate Algebra, Intermediate Counting, and Intermediate Geometry.
Will Intermediate Geometry and Intermediate Counting be offered this summer?
I am pretty sure I will enroll in 2 classes this summer, so I was just wondering which ones will be offered.
Also, if Intermediate Geometry is not offered, then what chapters do you suggest working through in AoPS Volume 2 that will cover it? I'd really like to take Olympiad Geometry next year, that's why.
Thanks!
Also, what subject will those new AopS books come out in? Are they only beginner-level books, or will they go past AOPS Volume 2?

The classes listed on the "Enroll" page are all of the classes that we'll be offering.  (I know that it currently says "more classes will be listed soon", but in fact all of the Summer 2005 classes have been listed.)

The only intermediate subject class that we're offering this summer is Intermediate Algebra.  The AIME Problem Series class is also intermediate level.  We are also offering two different Intermediate level Independent Study curricula.

We actually don't have an Intermediate Geometry class at the present time.  I would suggest looking at the Olympiad Geometry pre-test (go to the Enroll page, scroll down to Olympiad Geometry, and click "Are You Ready?") to get some idea of what topics you'll want to know before enrolling in Olympiad Geometry -- and if you've mastered the pre-test, then you may be ready to take Olympiad Geometry this summer.

The new AoPS books will come out in all subjects at the intro level this year if all goes as planned (this will be four different texts in algebra, number theory, counting, and geometry).  Intermediate books will follow in the next year (portions of these will be beyond Volume 2).  Olympiad books sometime thereafter (nearly all beyond Volume 2).

I'll make sure to buy all the intermediate and olympiad books for my personal library, then! :lol:

Actually, we may offer an Intermediate Number Theory course late in the Summer.  If not, it will probably be ready for the Fall.


geometry
trapezoid
ratio
AMC
USA(J)MO
USAMO
cyclic quadrilateral

Let ABCD be a cyclic quadrilateral and let E and F be the feet of perpendiculars from the intersection of diagonals AC and BD to AB and CD, respectively. Prove that EF is perpendicular to the line through the midpoints of AD and BC.

http://www.unl.edu/amc/a-activities/a7-problems/questions/q-tse/-pdf/tse00.pdf

Let O be the intersection of the diagonals, and M and N be the midpoints of AD and BC respectively. Also, let P be the midpoint of EF.

If AB || CD, then ABCD is an isosceles trapezoid since it is cyclic. Then MN is parallel to AB since it is the midsegment, and EF is perpendicular to MN since it is the line of symmetry. Hence, the result.

If AB and CD are not parallel, let them intersect at X, and assume WLOG that X is nearer to B and C. Finally, extend FO and EO to intersect lines XA and XD at G and H respectively.

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

Lemma 1: If Q and S are on sides AB and DC of quadrilateral ABCD such that BQ/CS = BA/CD, then the midpoint of QS lies on line MN.

Proof: Let k be the common ratio BQ/BA = CS/CD. Then with an arbitrary origin, Q = B+k(A-B) and S = C+k(D-C). Then the midpoint of QS is

1/2(B+k(A-B)+C+k(D-C)) = 1/2(k(A+D)+(1-k)(B+C)),

which is a weighted average of N = 1/2(B+C) and M = 1/2(A+D). So, the midpoint of QS is on line MN as desired.

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

Back to the proof. From similar triangles BEO and CFO and triangles BOA and COD, BE/CF = BO/CO = BA/CD, and so E and F satisfy the conditions of the lemma, meaning P is on line MN. Likewise, from similar triangles BGO and CHO and triangles AOG and DOC, we have BG/CH = BO/CO = BA/CD, and so the midpoint of GH (call it Z) is also on MN.

Since <MEG = <MFG = 90, GEFH is a cyclic quadrilateral, and the center of the circumscribed circle must be the midpoint of GH. Therefore, ZE = ZF since these are both radii of this circle. Now, since ZE = ZF, PE = PF, and PZ equals itself, we have triangles ZEP and ZFP congruent. Therefore, <EPZ = <FPZ, and since they add to 180 they each must be 90. So EF is perpendicular to PZ = MN as desired.

(Oh ya... the triangles that I claimed similar are easily proven similar by angle chasing. I was too lazy to put it in now, but I'll make sure to do it on USAMO!)

Hm....... That wasnt too bad for a TST question.


Prove that for every positive integer $ n$, $ (n^2)!$ is divisible by $ (n!)$^(n+1)

[hide="One line"] The group of permutations of $ \{ 1, 2, ... n^2 \}$ has as a subgroup the [url=http://en.wikipedia.org/wiki/Wreath_product]wreath product[/url] $ S_n \wr S_n$.  The copies of $ S_n$ are the permutations of $ \{ kn \plus{} 1, ... (k \plus{} 1)n \}, k \equal{} 0, 1, 2, ... n \minus{} 1$ and $ S_n$ acts on these copies by... well, permutation.  This subgroup has order $ (n!)^{n \plus{} 1}$.

More generally, $ (ab)!$ is divisible by $ a!^b b!$ for the same reason.[/hide]


Prove: out of $ 79$ consecutive positive integers, one can find at least one whose sum of digits is divisible by $ 13$. Show that this isn't true for $ 78$ consecutive integers.


Just wonderin what anyone would think the Orals for A/2A would be for next year...
Isn't it usually the same as 3AA/4AA the previous year...
just wonderin...

well if it is what 3AA/4AA had this year then its relations and functions, but im not sure that it is.


I think the area is 64/45 and the perimeter is 736/585.

Tare wrote:Pascal's Pizza Parlor is changing the size (diameter) of its circular pizza from 12 inches to 16 inches and increasing the number of slices per pizza from 8 to 10 congruent pieces. What is the ratio of increase in area of each new slice? What is the ratio of increase in perimeter of the size of each new slice?

[hide]If the radius = r, then

Perimeter = 2r + 2pi*r/(#slices)

Area = pi*r^2/(#slices)

Old area = pi*36/8 = 9pi/2

New area = pi*64/10 = 32pi/5

Ratio of increase = (32/5)/(9/2) -1 = 64/45 - 1 =  19/45

Old perimeter = 2*6 + 2*6*pi/8 = 12 + 1.5pi

New perimeter = 2*8 + 2*8*pi/10 = 16 + 1.6pi

Ratio of increase = (16+1.6pi)/(12+1.5pi) - 1 = (approx.) 0.258

Ugh.  Not pretty.  Whoops.  [/hide]

The second one you're both right but the first I agree with topper's answer, check the statement "Old area = pi*36/8 = 9pi/4"


Pat leaves a small town A at 10:18am and walking at uniform speed arrives in town B at 1:30pm.  On the same day Chris leaves town B at 9:00am and walking along the same route at uniform speed arrives at A at 11:40am.  The road crosses a wide river and they both arrive at the bridge on their respective sides at exactly the same momne t.  Pat leaves the bridge one minute later than Chris.  When did they arrive at the bridge?

[hide]

Let's say that the distance from A to B is $1000$ units. Pat travels this in $192$ minutes and so he is travelling at $125 \over 24$ units per minute. Similarly, Chris is travelling at $25 \over 4$ units per minute. So we see that they are $512.5$ units apart at 10:18.

Let's say that the bridge is $b$ units long, and we know that Chris traversed it  $1$ minute quicker. Hence, ${{24b}\over{125}}-{{4b}\over{24}}= 1$. Therefore $b = 31.25$.

Finally, lets call $t$ the time (in minutes) elapsed since 10:18. We know the combined distance travelled is $s = 512.5-b = 481.25$ units, and their combined speed is $u ={{25}\over{4}}+{{125}\over{24}}={{275}\over{24}}$. As there is no acceleration, we know that $s=ut$, hence $t = 42$ minutes. Hence, they arrived at the bridge at $10: 18+42 = 11$ am.

[/hide]


$ a,b\in \mathbb{R}_\plus{}$, $ n\geq2, n\in\mathbb{N}$
prove that:
$ \sum_{i\equal{}1}^{n}{\frac{1}{a\plus{}ib}}<\frac{n}{\sqrt{(a\plus{}\frac{b}{2})(a\plus{}\frac{2n\plus{}1}{2}b)}}$


find $ m \in \mathbb{R}$ so this system
$ x^{2}+y^{2}=z$
$ x+y+z=m$
has a unique real solution

the same for this system :
$ x+y=m$
$ x^{2}+z^{2}-2y+2z=0$

Replacing $ y=m-x-z$ in the first equation we have
$ 2x^{2}+2(z-m)x+z^{2}-2mz+m^{2}-z=0$
$ \triangle_{x}=b^{2}=4ac=0\Rightarrow z^{2}-2(m+2)z+m^{2}=0$
$  \triangle_{z}=0\Rightarrow 8m+4=0\Rightarrow m=-\frac{1}{2}$

For the second system we can proceed in the same way.

what should we replace in the second one ?
$ x=m-y$ i guess


inequalities
ratio
geometry
geometry solved

In triangle ABC, the sides BC, CA, AB are equal to a, b, c. Medians AD, BE, CF are equal to p, q, s. Prove that
$ p(bc \minus{} a^2) \plus{} q(ac \minus{} b^2) \plus{} s(ab \minus{} c^2)\geq 0$

Only hint I know is that you should use Ptolemeus Theorem.

Indeed, with the hint, the problem becomes quite easy:

Consider our triangle ABC, and let D, E, F be the midpoints of its sides BC, CA, AB. Then, the segments AD, BE, CF are the medians of triangle ABC. We have to prove that

$AD\cdot\left(bc-a^2\right)+BE\cdot\left(ca-b^2\right)+CF\cdot\left(ab-c^2\right)\geq 0$.

Well, let the medians AD, BE, CF of triangle ABC intersect at its centroid G. Since the centroid of a triangle divides the medians in the ratio 2 : 1, we have AG : GD = BG : GE = CG : GF = 2. From CG : GF = 2, it follows that $CG=2\cdot GF$, so that $CF=CG+GF=2\cdot GF+GF=3\cdot GF$ and thus $GF=\frac13\cdot CF$. Similarly, $GE=\frac13\cdot BE$. From AG : GD = 2, it follows that $AG=2\cdot GD$, so that $AD=AG+GD=2\cdot GD+GD=3\cdot GD=\frac32\cdot\left(2\cdot GD\right)=\frac32\cdot AG$ and thus $AG=\frac23\cdot AD$. Also, since the point E is the midpoint of the side CA of triangle ABC, we have $AE=\frac{CA}{2}=\frac{b}{2}$. Similarly, $AF=\frac{c}{2}$. Finally, since the points E and F are the midpoints of the sides CA and AB of triangle ABC, we have $EF=\frac{BC}{2}=\frac{a}{2}$.

Now, the Ptolemy inequality, applied to the quadrilateral AEGF, yields $GE\cdot AF+GF\cdot AE\geq AG\cdot EF$. Using the above results, this becomes $\frac13\cdot BE\cdot\frac{c}{2}+\frac13\cdot CF\cdot\frac{b}{2}\geq\frac23\cdot AD\cdot\frac{a}{2}$. Multiplication with 6a yields

$BE\cdot ca+CF\cdot ab\geq 2\cdot AD\cdot a^2$.

Similarly,

$CF\cdot ab+AD\cdot bc\geq 2\cdot BE\cdot b^2$;
$AD\cdot bc+BE\cdot ca\geq 2\cdot CF\cdot c^2$.

Adding these three inequalities, we get $2\left(AD\cdot bc+BE\cdot ca+CF\cdot ab\right)\geq 2\left(AD\cdot a^2+BE\cdot b^2+CF\cdot c^2\right)$, so that $AD\cdot bc+BE\cdot ca+CF\cdot ab\geq AD\cdot a^2+BE\cdot b^2+CF\cdot c^2$. Hence,

$\left(AD\cdot bc-AD\cdot a^2\right)+\left(BE\cdot ca-BE\cdot b^2\right)+\left(CF\cdot ab-CF\cdot c^2\right)\geq 0$, i. e.
$AD\cdot\left(bc-a^2\right)+BE\cdot\left(ca-b^2\right)+CF\cdot\left(ab-c^2\right)\geq 0$,

and the problem is solved.

  darij

He he, this was our school geometry problem.. rather cool isn't it??


geometry
circumcircle
trigonometry
analytic geometry
trig identities
Law of Sines
Law of Cosines

$E$ is a point on the diagonal $BD$ of the $\square{ABCD}$. Show that the points $A$, $E$ and the circumcenters of $\triangle{ABE}$ and $\triangle{ADE}$ form a square.

We know that $AFEG$ is a kite (two pairs of congruent sides), with $F$ and $G$ as the two circumcenters.  We can show this because we have two pairs of radii in two circles.  But can you prove that the quadrilateral is a square?  In order for that to be the case, the two circumcircles would have to have the same radii.  I don't think that's always the case...

look deeper into the problem......
it is pretty easy actually. :D

[quote="K81o7"]In order for that to be the case, the two circumcircles would have to have the same radii.  I don't think that's always the case...[/quote]
Apply Law of sines to $\triangle ADE$ and $\triangle ABE$, and see what happens.

[quote="K81o7"]Oh yeah...as $E$ mvoes towards $B$, the diameter of each circle tends towards $AB$...
So then...
Oh...here we go...
We know that $ADE$ and $ABE$ are half of a right angle each.  $ACE$ intercepts the same arc as $AGE$, so $AGE$ is right.  $ABE$ intercepts the same arc as $AFE$, so that one is right.  From there, the conclusion follows, because $GAF$ is congruent to $GEF$...[/quote]
 there is no need for congruency K81o7, think why is $E$ located on the diagonal, why not anywhere else? remember it is a square so diagonal bisects the angle.... $45^0$.
[quote="frt"]Apply Law of sines to  and , and see what happens.[/quote]
no need for law of cosines aswell.

[quote="riddler"][quote="K81o7"]Oh yeah...as $E$ mvoes towards $B$, the diameter of each circle tends towards $AB$...
So then...
Oh...here we go...
We know that $ADE$ and $ABE$ are half of a right angle each.  $ACE$ intercepts the same arc as $AGE$, so $AGE$ is right.  $ABE$ intercepts the same arc as $AFE$, so that one is right.  From there, the conclusion follows, because $GAF$ is congruent to $GEF$...[/quote]
 there is no need for congruency K81o7, think why is $E$ located on the diagonal, why not anywhere else? remember it is a square so diagonal bisects the angle.... $45^0$.
[/quote]

What congruency?...That is actually how i got that those two angles are 45 degrees...
But I was just finishing off the proof with $GAF=GEF$, because that fact is necessary to know that it is a square...

[quote="riddler"][quote="frt"]Apply Law of sines to $\triangle ADE$  and $\triangle ABE$, and see what happens.[/quote]
no need for law of cosines aswell.[/quote]
Ok, yeah, now I found the easier way to prove.
Since $\angle ADE=45^\circ=\frac{1}{2}\angle AGE$, $\angle AGE=90^\circ$.
For the same reason, $\angle AFD=90^\circ$.
Therefore, $AFEG$ is a kite with two opposite angles $90^\circ$, which means $AFEG$ is a square.

The reason I thought Law of sines is useful is that you can easily prove that two circles have the same radii.
$\frac{AE}{\sin ADE}=\frac{AE}{\sin ABE}$, so $\triangle ADE$ and $\triangle ABE$ have the same circumradii.

[quote="frt"]

The reason I thought Law of sines is useful is that you can easily prove that two circles have the same radii.
$\frac{AE}{\sin ADE}=\frac{AE}{\sin ABE}$, so $\triangle ADE$ and $\triangle ABE$ have the same circumradii.[/quote]
That really doesn't help... it is kinda obvious that $AE=AE$ and $\angle{ADE}=\angle{ABE}=45$. I don't see how it helps

It wouldn't be too hard to use coordinates, just long and nasty  ;)

$DA=2R_1 \sin{\angle{DEA}}$ and $AB = 2R_2 \sin{\angle{AEB}}$

Since $DA=BA$ and $\sin{\angle{DEA}}= \sin{\angle{AEB}}$ Hence $R_1=R_2$.

So $AQ=QE=EP=PA$.

Construct $PH \perp AD$ and $QG \perp AB$. Hence: $AH=AQ$, $AP=AQ$, so according to $HL$, we have:
\[\triangle{PHA} \cong \triangle{QGA}\]
Hence: $\angle{PAH}=\angle{QAG}$
So: $\angle{PAQ}=\angle{PAH}+\angle{HAQ}=\angle{QAG}+\angle{HAQ}=\angle{HAG}=90^{o}$

Therefore $\text{PEQA}$ is a square.

Note: This is the easiest way to get the answer though involves trig. So anyone can post a purely geomtrical proof?

[quote="frt"][quote="shobber"]$DA=2R_1 \sin{\angle{DEA}}$ and $AB = 2R_2 \sin{\angle{AEB}}$

Since $DA=BA$ and $\sin{\angle{DEA}}= \sin{\angle{AEB}}$ Hence $R_1=R_2$.

So $AQ=QE=EP=PA$.
[/quote]
Yeah, that's what I meant by the purpose of Law of sines :)[/quote]
when you know that the quadrialteral is formed by $2$ $45^o$ right angled triangles then you know it is a square.Simple. ;)

[quote="shobber"]How? Can you show your steps?

BTW, it is no need to make such a huge sketch.[/quote]

 in the $1$st circumcircle we know that $\angle{DAL}=90^0$ because that angle at the top is $45^0$ why? because angle subtended thing...
similarily for the $2$nd triangle and if we know that a quadrilateral is formed by putting $2$ [$45^0$,$90^0$,$45^0$] triangles together on the hypotenuse then we know it is a square.

[quote="riddler"][quote="shobber"]Indeed. This is the easier than the trig way...also much more beautiful.[/quote]
and you wouldn't believe it was from an olympiad. :huuh:[/quote]
Which contest?

[quote="riddler"][quote="shobber"]Indeed. This is the easier than the trig way...also much more beautiful.[/quote]
and you wouldn't believe it was from an olympiad. :huuh:[/quote]

Is it ?  :huh:  .... Really cant believe it .... by the way , nice approach riddler !

[quote="K81o7"]Riddler, I don't quite understand the diagram...are you sure those are the circumcenters?  I believe that one circumcenter will be outside the square, and one inside...
But yeah...you just have to show that the two angles from the circumcenters are right angles, and the rest of the information is right there...but i'm guessing you'd need to put down your logic...from what i can see you and i followed the same logic...
What olympiad is this from?[/quote]
Im not sure about olympiad but it is from a contest..... $\text{1st CIS 1992}$
take a look [url]http://www.kalva.demon.co.uk/soviet/sov92.html[/url]
the $2$nd question...


Let D be an interior point in acute triangle ABC, and satisfy: \[DA \cdot DB \cdot AB+DB \cdot DC \cdot BC+DC \cdot DA \cdot CA= AB \cdot BC \cdot AC\] Find the geometrical position of $D$ and prove your conclution. (CMO 1998)

orthocentre
 using complex nos. one can show l.h.s. greater than equal to r.h.s and eq. iff. p is h.there is sol. using ptolemy .  :)


inequalities
number theory proposed
number theory

[quote="pco"][quote="HUY VAN"]Find all triples $ (x, y, z)$ of integers such that:
$ \frac{1}{x^{2}}\plus{}\frac{2}{y^{2}}\plus{}\frac{3}{z^{2}}\equal{}\frac{2}{3}$[/quote]

I'll consider obviously only positive values for $ x,y,z$.
We have immediatly that $ x > 1$, $ y > 1$, $ z > 2$ and $ \frac{1}{\min(x,yz)^{2}}\plus{}\frac{2}{\min(x,y,z)^{2}}\plus{}\frac{3}{\min(x,y,z)^{2}}\geq\frac{2}{3}$ and so $ \min(x,y,z)\leq 3$.
So we have five possibilities :
$ \min(x,y,z) \equal{} x \equal{} 2$, $ \min(x,y,z) \equal{} x \equal{} 3$, $ \min(x,y,z) \equal{} y \equal{} 2$, $ \min(x,y,z) \equal{} y \equal{} 3$ or $ \min(x,y,z) \equal{} z \equal{} 3$.
For each, we can then find an inequality for the min of the two others numbers :
$ \min(x,y,z) \equal{} x \equal{} 2$ $ \implies$ $ 2\leq\min(y,z)\leq 3$ and so either $ y \equal{} 2$, either $ y \equal{} 3$, either $ z \equal{} 2$, either $ z \equal{} 3$
$ \min(x,y,z) \equal{} x \equal{} 3$ $ \implies$ $ 3\leq\min(y,z)\leq 3$ and so either either $ y \equal{} 3$, either $ z \equal{} 3$
$ \min(x,y,z) \equal{} y \equal{} 2$ $ \implies$ $ 2\leq\min(x,z)\leq 4$ and so either $ x \equal{} 2$, either $ x \equal{} 3$, either $ x \equal{} 4$, either $ z \equal{} 2$, either $ z \equal{} 3$, either $ z \equal{} 4$
$ \min(x,y,z) \equal{} y \equal{} 3$ $ \implies$ $ 3\leq\min(x,z)\leq 3$ and so either $ x \equal{} 3$, either $ z \equal{} 3$
$ \min(x,y,z) \equal{} z \equal{} 3$ $ \implies$ $ 3\leq\min(x,y)\leq 3$ and so either $ x \equal{} 3$, either $ y \equal{} 3$

And so very few ($ 13$) possibilities which can be checked :
$ (2,2,z)$, $ (2,3,z)$, $ (2,y,2)$, $ (2,y,3)$, $ (3,3,z)$, $ (3,y,3)$, $ (3,2,z)$, $ (4,2,z)$, $ (x,2,2)$, $ (x,2,3)$, $ (x,2,4)$, $ (x,3,3)$ or $ (3,y,3)$

And so the only solution $ (3,3,3)$[/quote]
This is easy .A same problem
Find all $ (x,y)>0$
$ (x\minus{}1)(y\minus{}1)|2xy\minus{}1$


inequalities
induction
4-variable inequality

[quote="hungkhtn"]For the first one, notice that
\[\sum (a-b)^{2}=4\left(\sum a^{2}\right)-(a+b+c+d)^{2}\]
So the inequality is equivalent to
\[\left(\sum a\right)\left(\sum \frac{1}{a}\right)\ge \frac{48(a^{2}+b^{2}+c^{2}+d^{2})}{(a+b+c+d)^{2}}+4. \]
WLOG, assume that $a+b+c+d=4$, the inequality becomes
\[4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right) \ge 3(a^{2}+b^{2}+c^{2}+d^{2})+4. \]
This is well-known. We may solve by three way as "induction", SMV theorem, EV-theorem.[/quote]

I have an idea  :blush:
$4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right) \ge 3(a^{2}+b^{2}+c^{2}+d^{2})+4$
we can factor expression. Then above inequality become:
$\sum \frac{(b+c+d-3a)^{3}}{4a}\ge 0$ (3)
How to solve the third inequality?  :roll:

[quote="hungkhtn"]Hoang, you can see this in my Vietnames book. Every will be clear. It is a problem in the middle of chapter II. I don't remember exactly now. In this book, I solve the general by induction, while VASC uses LCF theorem to solve it in his book.

Alternatively, this is problem of 4 variable only, so you can use SMV theorem, too. By SMV, a solution can made up in one-minute. Try $a\ge b\ge c\ge d$ then mix $b$ and $d$.

Notice that the equality holds for $a=b=c=\frac{2}{3}, d=2$[/quote]

How can i buy your book?

Thanks for you concern, but It is not available in English, just Vietnamese only. The English version is only available before IMO 2007. If you want to know something about it, Please send a private message to me and please never post it here in another topic.


Prove that if $0<a_{1} \leq a_{2} \leq \ldots \leq a_{n}$ the we have:

$ \sum_{k=1}^{n} \frac{1}{a_{k}} \leq \frac{1}{a_{1}a_{n}}[n(a_{1}+a_{n})- \sum_{k=1}^{n} a_{k}]$

made by Berkolaiko L.K.

We prove this lemma : (xz)/y + y  \leq x+z ( 0<x \leq y \leq z ) (1)
We denote : a = x/y , b = z/y => 0<a \leq 1 \leq b
(1) <=> ab + 1 \leq a+b
<=> 0 \leq (a-1)(1-b) (obvious)
Now , we prove this problem :
n=3 :
1/a1+1/a2+1/a3 \leq 1/(a1*a3)[3(a1+a3)-a1-a2-a3]
<=>(a1*a3)/a2 + a2 \leq a1 + a3 (lemma)
we suppose the inequality is right with n=m-1  \geq 3
We prove that the inequality is also right with n=m
We note :
sum(k=1,m)(1/ak)-1/a2 \leq 1/(a1*am)[(m-1)(a1+am) - (sum(k=1,m)(ak) - a2)] ( using the inequality with m-1 number 0<a1 \leq a3 \leq a4 \leq ... \leq am )
(a1*am)/a2 + a2  \leq a1+am (lemma)
So:sum(k=1,m)(1/ak) \leq 1/(a1*am)[m(a1+am) - sum(k=1,m)(ak)]
=>q.e.d


We have this formula \[ \sigma = \frac{L}{\Delta S} . \]

$L$ is the mechanical work, measured in $N \cdot m$. $\Delta S$ the area. $\sigma$ the coefficient of superficial tension. (probably I messed the names up. I just translated them from Romanian to English word by word)

Now, in the case of a spherical drop of rain, the teacher said to use \[ \sigma = \frac{L}{2 \Delta S} . \] Why is that? She wouldn't give any explanations  :(

Actually, the '2' factor comes in the formulae for soap bubbles. this is because it has 2 surfaces. One surface surrounds the inner air... then there is a layer of the chemicals soap is made up of , then another surface seperating the outside water from this layer... hence the factor of '2'!!!

In case one considers a drop of a rain, what's the correct formula?

I think it should be definitely the first one.


geometry
inradius
circumcircle
incenter
geometry proposed

Let $\triangle  ABC$ be an acute triangle with inradius $r$ and circumradius $R.$ $ r_0$ is the inradius of its orthic triangle, $ \lambda_a, \lambda_b, \lambda_c$ are the radii of its three  [url=http://www.xtec.cat/~qcastell/ttw/ttweng/definicions/d_Jenkins_c.html]Jenkins circles[/url] and $ \phi $ is the radius of its [url=http://www.xtec.cat/~qcastell/ttw/ttweng/definicions/d_Conway_c.html]Conway circle[/url]. Show the following relation:
\[ \phi^2 \left (\frac {1}{\lambda_a} \plus{} \frac {1}{\lambda_b} \plus{} \frac {1}{\lambda_c} \right) \equal{} 8R \plus{} 4 \left (r \plus{} r_0 \plus{} \frac {r}{R^2} \right) \]

Let $ S$ and $ I$ be the Spieker point and incenter of $ \triangle ABC.$ $ \varrho$ denotes the radius  of the Apollonius circle. We know that

$ \varrho =\frac{p^2+r^2}{4r} \ \ (1) \ \ , \ \   \phi^2=p^2+r^2 \ \ (2)$

Inversion with center $ S$ and power $ k^2$ equal to the power of $ S$ with respect to the excircles takes the sidelines $ BC,CA,AB$ into the Jenkins circles $\mathcal{J}_a,\mathcal{J}_b,\mathcal{J}_c.$ Their  centers lie then on the perpendiculars $SX,SY,SZ$ to $ BC,CA,AB.$ Such inversion also takes the 9-point circle $(N)$ into the Apollonius circle $(U,\varrho).$

$\frac{2\varrho}{R}=\frac{2k^2}{Rr} \ \Longrightarrow \  \varrho =\frac{k^2}{r} \ \ (2.1)$

On the other hand, we have $ k^2=SX \cdot 2\lambda_a=SY \cdot 2\lambda_b=SZ \cdot 2\lambda_c$

But distance $ SX$ is half the distance from the incenter $ I$ to the parallel through $ A$ to $ BC.$ Similarly $ SY$ and $ SY,$ hence

$ k^2=(h_a-r) \cdot \lambda_a=(h_b-r) \cdot \lambda_b=(h_c-r) \cdot \lambda_c$

Combining this latter expression with $(2.1)$ yields

$ \varrho \left (\frac{1}{\lambda_a}+\frac{1}{\lambda_b}+\frac{1}{\lambda_c} \right )=\frac{h_a+h_b+h_c}{r}-3$

But according to [1] we have $h_a+h_b+h_c=2R+4r+r_0+ \frac{r}{R^2}$

$\Longrightarrow \ %Error. "vahrro" is a bad command.
\left (\frac{1}{\lambda_a}+\frac{1}{\lambda_b}+\frac{1}{\lambda_c}\right )=\frac{2R+4r+r/R^2}{r}-3$

Combining this latter relation  with $(2)$ gives:

$ \phi^2 \left (\frac{1}{\lambda_a}+\frac{1}{\lambda_b}+\frac{1}{\lambda_c} \right )=8R+4 \left (r+r_0+\frac{r}{R^2}\right )$

[1] [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=259185[/url]


probability
combinatorics proposed
combinatorics

A set of $ \frac{n(n\plus{}1)}{2}$ distinct numbers is arranged at random in a triangular array:
$ \: \: \: \: \: *$
$ \: \: \: \: * *$
$ \: \: \: * * *$
$ \: \: ...........$
$ \: * * ... * *$

If $ M_k$ denotes the largest number in the $ k$-th row from the top, find the probability that $ M_1<M_2<...<M_n$.

[quote="moldovan"]A set of $ \frac {n(n \plus{} 1)}{2}$ distinct numbers is arranged at random in a triangular array:
$ \: \: \: \: \: *$
$ \: \: \: \: * *$
$ \: \: \: * * *$
$ \: \: ...........$
$ \: * * ... * *$

If $ M_k$ denotes the largest number in the $ k$-th row from the top, find the probability that $ M_1 < M_2 < ... < M_n$.[/quote]

We choose the $ n$ th row first. To satisfies $ M_n>M_1,M_2,...,M_{n\minus{}1}$ we have the largest number among the set of $ \frac {n(n \plus{} 1)}{2}$ distinct numbers must be in the $ n$ th row. And the probability of which is $ \dfrac{n}{d\frac{n(n\plus{}1)}{2}}\equal{}\frac{2}{n\plus{}1}$.
Similarly the probability of $ M_{n\minus{}1}>M_1,M_2,...,M_{n\minus{}2}$ is $ \dfrac{2}{n}$
.....

As $ M_1 < M_2 < ... < M_n \Leftrightarrow \begin{cases} M_n>M_1,M_2,...,M_{n\minus{}1} \\ M_{n\minus{}1}>M_1,M_2,...,M_{n\minus{}2} \\ ... \\ M_3>M_2,M_1 \\ M_2>M_1 \end{cases}$,
we have the required probability $ \equal{}\frac{2^{n\minus{}1}}{(n\plus{}1)(n)(n\minus{}1)...(4)(3)}\equal{}\frac{2^n}{(n\plus{}1)!}$


trigonometry
inequalities
inequalities open

Let a,b,c it 3 positive real number such that : $ 2abc\plus{}ab\plus{}bc\plus{}ca\equal{}1$
Show that : $ \sqrt{a\plus{}b\plus{}c\plus{}3}\geq\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}$

[quote="Harry Potter"]Let a,b,c it 3 positive real number such that : $ 2abc\plus{}ab\plus{}bc\plus{}ca \equal{} 1$
Show that : $ \sqrt{a\plus{}b\plus{}c\plus{}3}\geq\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}$[/quote]

$ \sqrt{a\plus{}b\plus{}c\plus{}3}\geq\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}\Leftrightarrow 3\ge2\sum\sqrt{ab}$(*)

Let $ \sqrt{ab}\equal{}x,\sqrt{bc}\equal{}y,\sqrt{ca}\equal{}z,x,y,z>0$ , so we have $ x^{2}\plus{}y^{2}\plus{}z^{2}\plus{}2xyz\equal{}1$,(1)

(*)$ \Leftrightarrow x\plus{}y\plus{}z\le\frac{3}{2}$ .

Because $ x^{2}\plus{}y^{2}\plus{}z^{2}\plus{}2xyz\equal{}1$ so have a traingle $ \text{ABC}$ so that

$ x\equal{}\cos{A},y\equal{}\cos{B},z\equal{}\cos{C}$.

Then the inequality is equivalent to $ \cos{A}\plus{}\cos{B}\plus{}\cos{C}\le\frac{3}{2}$ :)


algebra
polynomial
number theory open
number theory

hi,
I am new to finite fields .
I have some doubts on finite fields. Can anyone answer the following
questions?

we wish to encode the message using  F32... Define the cumulative
degree of a polynomial to be the sum of degrees of its monomials.

a)How many irreducible polynomials of degree 5 over Z/2Z are there?
what are they?

b)
Let Y be a root of the irreducible polynomial of degree 5 of lowest
cumulative degree. Then { 1,Y,Y^2,Y^3,Y^4} forms a basis for F32 over
Z/2Z. Writing each element  of F32 as a 5 tuple with respect to this basis,
find all the powers of Y.

c) We encode message (numbers between 0 and 31) by raising it Y to that
power and then returning the five-bit binary representation we get when
writing this element as a 5-tuple with respect to {1,Y,Y^2,Y^3,Y^4}.If
19 is the result m then what number was initially sent?.

 can anyone help ?.

(a) The (monic) irreducible polynomials of degree 5 are the minimal polynomials of those elements of $GF(32)$ which are not elements of any subfield: in this case, not elements of $GF(2)$.  There are $30 = 2^5 - 2$ such elements and each polynomial has 5 of them as roots.  So there are 6 irreducible monic quintics over $GF(2)$.


MATHCOUNTS
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The competition is over and here are some of the results (as best I understood them):

[b]Neal Wu[/b] (LA 7th grade) won the Count Down Round.

[b]Sergei Bernstein[/b] (MA) was 1st on the written test.

Nathan Benjamin (IN) was 2nd on the written test.

Mark Zhang (TX) was 3rd on the written test and 2nd after the Count Down.

Patricia Li (CA) was 4th on the written test [and plans to tell her sister all about it].

David Benjamin (IN) was 5th on the written test.

AoPSers [b]Piggipi[/b] (CA) and [b]cutiepi[/b] (FL) were 4th and 7th on the test and were both semifinalists in the Count Down Round!  This is the first time two girls have finished so high in the same year (correct me if this is wrong, but I've seen most nationals results).

[b]Hexahedron[/b] (NY) was also in the top 12.

[b]Texas[/b] won the team competiton
Indiana 2nd (with 3 of the top 12 students)
California 3rd (piggipi, bob123, NoSoupForYou, uberl33tmage)

Congratulations to all the winners and all the AoPSers competing at nationals!

wow...AMAZING job piggipi and cutiepi...and ofcourse Neal Wu...I've never even heard of him!

[quote="Max300"]YAY!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
I had a feeling Texas would win. Congrats! However, I just got a call from somebody who got an email from Mark's mom. Mark placed 3rd in the individual and 2nd in the countdown[/quote]

I've made a couple of mistakes that I'll correct.

[quote="MCrawford"]The competition is over and here are some of the results (as best I understood them):
[b]Neal Wu[/b] (LA 7th grade) won the Count Down Round.
[/quote]

Wowzerz... a 7th grader.  I guess that's how people named "Neal" do it  :lol:

Does anyone know how Amy Tai or anyone else from Oklahoma did?

[quote="MCrawford"]The competition is over and here are some of the results (as best I understood them):

...

Patricia Li (CA) was 4th on the written test [and plans to tell her sister all about it].

...
[/quote]

Hahahahaha  I know, i've heard.  But not from her, not yet.  :P

yeah
i got a call from cutiepi (karlanna) like sometime during my english presentation today but my teacher never found out  :lol: im so amazed, i didnt expect her to place so high
oh well... congratulations to all u mathcounts winners

7th grader as mathcounts champion? my goodness

Yesss!!!!! Texan pride!!!!  We are  :first:

A 7th grader won countdown? :what?:

I expected you to do better, California. I'm disappointed. :mad:

[quote="JesusFreak197"]
I expected you to do better, California. I'm disappointed. :mad:[/quote]

Texas obviously had a very strong team and the Benjamin brothers helped power the Indiana team.  I'm sure the competition between the top teams was tough.

Congrats everyone! The California team especially. Now I can get back to hoping (wishfully thinking) I'll represent California next year...

The Benjamins are extremely impressive, I was surprised that Indiana didn't pull it out. (I live in Indiana now, so that's all I hear about these days)

Congratulations to yet another 7th grade champion, and this is the first time the new Countdown has generated a somewhat "random" winner.

It ought to be interesting to see Neal Wu and Nathan Benjamin compete against each other next year...

Oh, yeah.  But if you had more people like 24 or something, then it would be good

Well it still should be ladder.  Countdown just shows how fast you are not how mathematically skilled you are, thats why the better you are at math, the more of an advantage you should get

But neither does 30 problems in 40 minutes. . .

but Nathan got nervous, so it should be a ladder

lol, so the probability of Nathan winning is __ compared to Neals probability (_) which would be lower

Is it possible to get a copy of the national test, without buying the book; I know Mathcounts does not post it on their website, but was wondering if anyone else does? I would understand if there was an issue with legaity, though.

[quote="sirorange"]There's an issue with legality[/quote]
:rotfl::rotfl::rotfl: yeah, ther probably is

[quote="noneoftheabove"]i dunno somehow my dad got a copy off the internet i dont know where though[/quote]
 :o  :o really? where, where, where??!! :D

[quote="noneoftheabove"]i dunno somehow my dad got a copy off the internet i dont know where though[/quote]cool, i want one.  what website was it?

My brother has a binderful of different things, I know for sure there's at least 95-96's national masters, countdown, sprint, and target. Other years have various pages of this stuff (e.g. maybe target and sprint but no masters and countdown). I asked him where he got these and he said "i really don't know" since he hasn't competed in about 4 years, but he thinks he got it from a coach. He won nats in 2002, so do you think he'd get like a prize binderful of problems or something?

I guess I could scan them at distribute them if I have the time & people really want them...

Oh, and he also has a bunch of states/chapters from '90 on.

[quote="xxreddevilzxx"]how did Washington do?[/quote]
i got 19th and our team got 15th, but i dont think anybody else got in the top 50.
we'll do better next year. i was one of two 7th graders that will be returning  :)


how do you do this problem

a triangle and a trapezoid are the same in area. if the median of the trapezoid is 24 inches long, find the length of the triangle's base in inches.

please include an explanation :nhl:

[color=blue]Moderator kyyuanmathcount's says: Please include more descriptive titles, its hard to tell the difference between "Geometry" and "Geometry"[/color]

Do you mean equilateral triangle? otherwise its impossible because non equilateral triangles have more than one base

I came up with a general solution.

If $a$ and $b$ were the two bases of the trapezoid, then we can find the altitude of the trapezoid in terms of $a$ and $b$.

Since the median bisects the two bases, we can form a right triangle with the median as the hypotenuse and the height of the trapezoid as one of the legs.

The other leg would be equal to $\frac{1}{2}(a - b)$ if $a>b$.

The height of the trapezoid would be $\sqrt{576 - \frac{1}{4}(a - b)^2}$

Using the formula for the area of a trapezoid we would have an area of $\frac{1}{2}(a + b) \sqrt{576 - \frac{1}{4}(a - b)^2}$

We would have an equivalent triangle whose base would be the sum of the two bases of the trapezoid $(a + b)$ with an altitude equal to the height of the trapezoid, which is $\sqrt{576 - \frac{1}{4}(a - b)^2}$

We could also have an equivalent triangle whose base is equal to the height of the trapezoid and whose altitude is equal to the sum of the two bases of the trapezoid.


We know that if $ A$ is a positive semidefinite Hermitian  matrix, then there is a positive semidefinite Hermitian matrix $ P$ with $ A \equal{} P*P \equal{} P^2$.
Question: Must $ P$ have the same rank as $ A$? Is $ P$ unique?

Yes, $ A$ and $ P$ must have the same rank. More specifically, they must have the same null space.

Clearly, if $ Px \equal{} 0,$ then $ Ax \equal{} 0.$

Now assume $ Ax \equal{} 0.$ Then

$ 0 \equal{} x^*Ax \equal{} x^*P^2x \equal{} x^*P^*Px \equal{} \|Px\|^2.$

I used that $ P \equal{} P^*$ is Hermitian. That implies $ Px \equal{} 0.$ Hence, $ A$ and $ P$ have the same null space.

Now, suppose $ P^2 \equal{} A$ and both $ P$ is positive semidefinite Hermitian.

$ A$ commutes with $ P,$ so $ A$ and $ P$ are simultaneously diagonalizable. Hence, WLOG, we may assume that $ A$ and $ P$ are diagonal. In this case, each diagonal element of $ P$ is the unique nonnegative square root of the corresponding element of $ A.$ Hence, $ P$ is unique.


if weighted am-gm can be used with real weights, this should work.  sorry i still haven't learnt how to use latex so bear with the notation.

replace the 1 on the RHS with (a+b+c)^2, then rearranging, it remains to show that sqrt(abc) <= ab+bc+ca,
or (ab+bc+ca)^2 => abc.
now LHS= [(ab+bc+ca)/(a+b+c)]^2
=> [(a^b*b^c*c^a)(a^c*c^b*b^a)]^(1/(a+b+c))
=> a^(b+c) b^(c+a) c^(a+b)
= a^(1-a) b^(1-b) c^(1-c)
> abc

as x^(1-x)> x for 0<x<1.

[quote="silouan"]If a,b,c>0 and $a+b+c=1$ prove that $a^2+b^2+c^2+2\sqrt {abc}\leq 1$[/quote]

Replace 1 with $(a+b+c)^2$ as above.  Use algebra to convert inequality to

$\sqrt{abc} \le ab + bc + ca$
$\sqrt{abc(a+b+c)} \le ab + bc + ca$

Substitute $x = ab, y = bc, z = ca$ to get

$\sqrt{xz + xy + yz} \le x + y + z$

which is actually strict...


trigonometry
function
real analysis
real analysis unsolved

$0<a_n<1$ such that $\sum_{n\geq 0}\frac{1}{\frac{2}{\sin a_n}+\frac{1}{\tan a_n}}$ is divergent

Prove that $\sum a_n$ is divergent

let me rewrite:\[ \displaystyle \sum_{n\geq 0}\frac{1}{\frac{2}{\sin a_n}+\frac{1}{\tan a_n}} = \sum_{k = 0}^{+\infty}\frac{\sin a_k}{2+\cos a_k}\]
now, $2+\cos a_k \ge 1$, and $\sin a_k < a_k$.
so we have \[ \displaystyle \sum_{k = 0}^{N}\frac{\sin a_k}{2+\cos a_k} < \sum_{k=0}^{N} a_k \forall N\]
being the first series divergent, also the second one is, and that should complete the proof ;)

As $x\to 0$, $\frac{1}{\frac2{\sin x}+\frac1{\tan x}}=\frac x3+O(x^3)$. Dividing by $x$, we have an analytic function, which is obviously continuous and strictly positive on $[0,1]$. Therefore, for all $x\in [0,1]$, there exist positive constants $a$ and $b$ such that $ax\le\frac{1}{\frac2{\sin x}+\frac1{\tan x}}\le bx$.
The problem is trivial from here.

here is my solution.
assume $\sum a_n$ is convergence.
then we can prove $\sum_{n\geq 0}\frac{1}{\frac{2}{\sin a_n}+\frac{1}{\tan a_n}}=\sum_{n\geq 0}\frac{\sin a_n}{2+\cos a_n }$ is convergerce,too.
because  $\sum_{n\geq 0}\frac{\sin a_n}{2+\cos a_n }>0,and,\sum_{n\geq 0}\frac{\sin a_n}{2+\cos a_n } \leq \sum a_n$
but it seems so easy for your problem,is there something wrong?


[quote="VANCHANH123"]have :a,b,c >0 and a+b+c=3.
 $\to$$\sum\sqrt{a}\geq \sum a\sqrt{a}$[/quote]
Firstly, it should be moved to Inequalities.
Secondly, the sign of the ineqaulity should reversed (try $(a,b,c)=(2,1,0)$). The true ineqaulity is
$\sum\sqrt{a}\leqslant \sum a\sqrt{a}$.
[b]Proof[/b]
Let's mark the power mean of numbers $a,b,c$ of order $t$ as $\mu_{t}$. Then we have to  prove
$3\left(\mu_{\frac12}\right)^{\frac12}\leqslant 3\left(\mu_{\frac32}\right)^{\frac32}$
for $\mu_{1}= 1$. The inequality simplifies to
$\mu_{\frac12}\leqslant \left(\mu_{\frac32}\right)^{3}.$
But it is true since
$\mu_{\frac12}\leqslant \mu_{1}=1 = \left(\mu_{1}\right)^{3}\leqslant \left(\mu_{\frac32}\right)^{3}.$
It was a very weak inequality. :(

One nice solution :D
$2a\sqrt{a}+1 \ge 3a$ etc...
$a+1\ge 2\sqrt{a}$
It's very simple!


trigonometry
geometry
cyclic quadrilateral

If ABCD is a cyclic quadrilateral, then: \[\sin{\angle{ADC}}\cdot \sin{\angle{BAD}}=\sin{\angle{ABD}}\cdot \sin{\angle{BDC}}+\sin{\angle{ADB}}\cdot \sin{\angle{DBC}}\]

[quote="shyong"]smells ptolemy theorem to me  :)[/quote]
Yes:
$\frac{sin(ABD)sin(BDC)}{sin(ADC)sin(BAD)}=\frac{sin(ABD)}{sin(BAD)}*\frac{sin(BDC)}{sin(ADC)}=\frac{sin(ABD)}{sin(BAD)}*\frac{sin(BAC)}{sin(ABC)}$(because $ABCD$ is cyclic)$=\frac{AD}{BC}*\frac{BD}{AC}$(by Sine Law)$=\frac{AD*BD}{BC*AC}$
Also, $\frac{sin(ADB)sin(DBC)}{sin(ADC)sin(BAD)}=\frac{sin(ADB)}{sin(BAD)}*\frac{sin(DBC)}{sin(ADC)}=\frac{sin(ADB)}{sin(BAD)}*\frac{sin(CAD)}{sin(ADC)}$(because $ABCD$ is cyclic)$=\frac{AB}{BC}*\frac{CD}{AC}$(by Sine Law)$=\frac{AB*CD}{BC*AC}$
By Ptolemy's theorem, $1=\frac{AD*BD}{BC*AC}+\frac{AB*CD}{BC*AC}$
So: $1=\frac{sin(ABD)sin(BDC)}{sin(ADC)sin(BAD)}+\frac{sin(ADB)sin(DBC)}{sin(ADC)sin(BAD)}$
Multiplying both sides by $sin(ADC)sin(BAD)$, get the result.


A sequence of positive integers is constructed as follows: the first term is $ 1$, the following two terms are $ 2$, $ 4$, the following three terms are $ 5$, $ 7$, $ 9$, the following four terms are $ 10$, $ 12$, $ 14$, $ 16$, etc. Find the $ n$-th term of the sequence.

This is called the "Connell sequence;" it's [url=http://www.research.att.com/~njas/sequences/A001614]A001614[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].


There is a finite number of towns in a country. They are connected by one direction roads. It is known that, for any two towns, one of them can be reached from another one. Prove that there is a town such that all remaining towns can be reached from it.

Consider the town which can be reached from maximum number of other towns.$ (1)$
(If there are more than one such town we consider one of them)

Call this town $ T$
Suppose it can be reached from $ k$ towns.

Assume that there is a town which doesn't lead to $ T$.$ (2)$
Call this town $ X$.

It is given that either $ T$ can be reached from $ X$ or $ X$ can be reached from $ T$.
Because of assumption $ (2)$, the former can't be true.
So the later must be true.

So $ X$ can be reached from $ T$
But in that case $ X$ can be reached from all those $ k$ towns which lead to $ T$.
So $ X$ can be reached from at least $ k\plus{}1$ towns.
This contradicts the definition of $ T$ (in $ (1)$)
So assumption $ (2)$ must be false.

Hence there is no town which doesn't lead to $ T$.
Proved.

Sorry for reviving an old thread, but I think there's a small mistake in the previous post. The question says:
[quote]Prove that there is a town such that all remaining towns can be reached [b]from[/b] it.[/quote]
The same approach works anyway.

For any towns $ x, y$ denote $ x \to y$ if town $ y$ can be reached from $ x$, and denote by $ C_x \equal{} \{y \ | \ x \to y\}$ the set of all towns that can be reached from $ x$ ($ C_x$ may even be empty). Clearly $ y \in C_x$ implies $ C_y \subseteq C_x$. With a finite number of towns, there exists (at least) a maximal set $ C_x$ with respect to the order relation $ \subseteq$ among sets. Assume $ y \not \in C_x$, so $ x \not \to y$, therefore $ y \to x$, so $ x \in C_y$, hence $ C_x \subseteq C_y$. By the maximality of $ C_x$ follows $ C_y \equal{} C_x$, whence $ x \to y$, contradiction.

For an infinite number of towns this proof fails (even Zorn's lemma does not apply). A simple counterexample is given by towns $ x_z$ indexed by $ z \in \mathbb{Z}$, with one-way roads leading from $ x_z$ to $ x_{z\plus{}1}$, so $ x_s \to x_t$ if and only if $ s < t$. There is no town from which all others can be reached, neither one reached from all others.


[quote="mai quoc thang"]
\[ \ a,b,c > 0
\]
.Prove that:
\[ \frac {a^{2} \plus{} 2bc}{(b \plus{} c)^{2}} \plus{} \frac {b^{2} \plus{} 2ca}{(c \plus{} a)^{2}} \plus{} \frac {c^{2} \plus{} 2ab}{(a \plus{} b)^{2}}\geq \frac {9}{4} .
\]
[/quote]
It's equivalent to $ \sum_{sym}(2a^6\plus{}8a^5b\minus{}5a^4b^2\minus{}5a^3b^3\plus{}3a^4bc\minus{}3a^2b^2c\plus{}3a^2b^2c^2)\geq0,$
which obviously true.

[quote="arqady"][quote="mai quoc thang"]
\[ \ a,b,c > 0
\]
.Prove that:
\[ \frac {a^{2} \plus{} 2bc}{(b \plus{} c)^{2}} \plus{} \frac {b^{2} \plus{} 2ca}{(c \plus{} a)^{2}} \plus{} \frac {c^{2} \plus{} 2ab}{(a \plus{} b)^{2}}\geq \frac {9}{4} .
\]
[/quote]
It's equivalent to $ \sum_{sym}(2a^6 \plus{} 8a^5b \minus{} 5a^4b^2 \minus{} 5a^3b^3 \plus{} 3a^4bc \minus{} 3a^2b^2c \plus{} 3a^2b^2c^2)\geq0,$
which obviously true.[/quote]
I'm sorry but you are wrong:3a^2b^2c ??????????????

[quote="mai quoc thang"]
I'm sorry but you are wrong:3a^2b^2c ??????????????[/quote]
It should have been $ 3a^3b^2c.$ Thank you!

It's enough to prove that
(by IRAN 1996)
\[ \sum (\frac{a^2\plus{}2bc}{(b\plus{}c)^2}\minus{}\frac{ab\plus{}bc\plus{}ca}{(b\plus{}c)^2})\ge 0\]
or
\[ \sum (a\minus{}b)(a\minus{}c)(\frac{1}{(b\plus{}c)^2})\ge 0\]
wich is schure :wink:


algebra
polynomial
Vieta
floor function
modular arithmetic
linear algebra
matrix

Πάντα και παντού έχουν θέμα από τη θεωρία αριθμών.

  [color=red][b]Καλές εμπνεύσεις !!! [/b][/color]

[color=red][b]ΑΣΚΗΣΗ 1[/b][/color][color=red]( Το προτείνει ο Σιλουανός Μπραζιτίκος)[/color]

Έστω $ p\geq 5$ ένας πρώτος αριθμός .Αν $ 1\plus{}\frac{1}{2}\plus{}\frac{1}{3}\plus{}...\plus{}\frac{1}{p}\equal{}\frac{a}{b}$ τότε $ p^{4}|(ap\minus{}b)$

 ΛΥΣΗ[color=red] ( Αλ. Συγκελάκης)[/color]

[i]Πολύ δυνατό πρόβλημα Σιλουανέ. Πού το βρήκες?[/i]

Θα θεωρήσω γνωστό το παρακάτω θεώρημα:

[b][u][color=red]θεώρημα του Wolstenholme[/color][/u][/b]

Εαν $ p$ πρώτος $ \geq 5$, τότε ο αριθμητής του κλάσματος $ 1\plus{}\frac{1}{2}\plus{}\frac{1}{3}\plus{}...\plus{}\frac{1}{p\minus{}1}$ διαιρείται από το $ p^{2}$.

Επιστροφή λοιπόν στην άσκηση.

Εαν λοιπόν $ 1\plus{}\frac{1}{2}\plus{}\frac{1}{3}\plus{}...\plus{}\frac{1}{p\minus{}1}\equal{}\frac{k}{l}$, με $ (k,l) \equal{} 1$, τότε $ p^{2}|k$ άρα $ \boxed{k \equal{} p^{2}k_{1}\ \ (1)}$.

Συνεπώς η δοσμένη γίνεται $ \frac{k}{l}\plus{}\frac{1}{p}\equal{}\frac{a}{b}$, με $ (a,b) \equal{} 1$ (Εαν $ (a,b)\neq 1$, τότε κάνουμε τις απλοποιήσεις και το φέρνουμε σε τέτοια μορφή ώστε να είναι πρώτοι μεταξύ τους). Άρα $ \frac{k}{l}\equal{}\frac{ap\minus{}b}{bp}$ δηλαδή (λόγω της $ (1)$)
$ \boxed{p^{3}k_{1}b \equal{} (ap\minus{}b)l\ \ (2)}$ και λόγω της $ (p,l) \equal{} 1$ (διότι $ (k,l) \equal{} 1$) παίρνουμε $ p^{3}|ap\minus{}b$ άρα $ \boxed{ap\minus{}b \equal{} p^{3}r\ \ (3)}$ (άρα $ b \equal{} ap\minus{}p^{3}r)$.

Συνεπώς αντικαθιστώντας στην $ (2)$ έχουμε $ k_{1}b \equal{} rl$ ή λόγω της $ (3)$ : $ k_{1}(ap\minus{}p^{3}r) \equal{} rl$ δηλαδή $ p|rl$ και λόγω της $ (p,l) \equal{} 1$, παίρνουμε $ p|r$ δηλαδή $ r \equal{} pr_{1}$ που λόγω της $ (3)$, ολοκληρώνει την απόδειξη.

Σιλουανέ (ή κάποιος άλλος) έχεις κάποια απλούστερη απόδειξη στο μυαλό σου?

Αλέξανδρος

Να βρεθούν όλες οι ακέραιες λύσεις της εξίσωσης:

$ (x^{7}\minus{}1)/x\minus{}1\equal{}y^{5}\minus{}1$

Βάζω μια υπόδειξη σε hidden διότι μπορεί κάποιοι να θέλουν ακόμα να την προσπαθήσουν. Δεν είναι και τόσο δύσκολη πάντως για θεωρία αριθμών και ΙΜΟ (αναλογικά πάντοτε).
 :wink:
[hide]
Nα δείξουμε πρώτα ότι $ p\equal{}1 (mod7)$ ή $ p\equal{}7$ για έναν πρώτο διαιρέτη p του πρώτου μέλους.
[/hide]

[color=red][b]ΑΣΚΗΣΗ 2[/b][/color]

Αποδειξτε οτι η εξισωση $ x^{2}\plus{}y^{2}\equal{}nz^{2}$ εχει ακεραιες λύσεις , μονο οταν ο $ n$ παριστανεται ως αθροισμα $ 2$ τετραγωνων.

Βάζω τη λύση που μου ήρθε πρώτη στο μυαλό, αλλά επιφυλλάσσομαι ότι θα υπάρχει και απλούστερη.

Θα δουλέψουμε στο $ Z[i]$. Έχουμε $ (x\plus{}iy)(x\minus{}iy) \equal{} nz^{2}$, οπότε σύμφωνα με το Θεώρημα των 4 Αριθμών (στο $ Z[i]$ ισχύει το Θεμελιώδες Θεώρημα Αριθμητικής) θα υπάρχουν $ p,q,r,s\in Z[i]$ ώστε $ x\plus{}iy \equal{} pq, x\minus{}iy \equal{} rs, n \equal{} pr, z^{2}\equal{} qs$. Όμως $ n, z^{2}\in R$, οπότε $ r \equal{} v\overline{p}, s \equal{} u\overline{q}$ με $ u, v\in Q$. Τότε $ x\minus{}iy \equal{} uv\overline{pq}\equal{} uv(x\minus{}iy)$, άρα $ uv \equal{} 1$. Συνεπώς αν $ u \equal{}\frac{k}{l}$ με $ (k,l) \equal{} 1$ θα είναι $ r \equal{}\frac{l}{k}\overline{p}, s \equal{}\frac{k}{l}\overline{q}$, οπότε $ p \equal{} k(a\plus{}ib), q \equal{} l(c\plus{}id)$. Επομένως, αν $ x^{2}\plus{}y^{2}\equal{} nz^{2}$, θα είναι $ x \equal{} t(ac\minus{}bd), y \equal{} t(ad\plus{}bc), n \equal{} t(a^{2}\plus{}b^{2}), z^{2}\equal{} t(c^{2}\plus{}d^{2})$ όπου $ t \equal{} kl$.

Τώρα χρησιμοποιούμε επαγωγή. Υποθέτουμε ότι κάθε $ m < n$ που γράφεται στη μορφή $ m \equal{}\frac{x^{2}\plus{}y^{2}}{z^{2}}$ γράφεται ως άθροισμα 2 τετραγώνων και θα δείξουμε ότι το ίδιο ισχύει και για το $ n$. Η σχέση $ z^{2}\equal{} t(c^{2}\plus{}d^{2})$ συνεπάγεται ότι $ z \equal{} hef, t \equal{} he^{2}, c^{2}\plus{}d^{2}\equal{} hf^{2}$, οπότε $ h \equal{}\frac{c^{2}\plus{}d^{2}}{f^{2}}$ και αφού $ h \equal{}\frac{t}{e^{2}}\leq t < n$ σύμφωνα με την επαγωγική υπόθεση $ h \equal{} w^{2}\plus{}j^{2}$. Έπεται λοιπόν ότι $ n \equal{} t(a^{2}\plus{}b^{2}) \equal{} e^{2}(w^{2}\plus{}j^{2})(a^{2}\plus{}b^{2}) \equal{} (eaw\plus{}ebj)^{2}\plus{}(eaj\minus{}ebw)^{2}$. QED

Εστω οτι $ x^{2}\plus{}y^{2}\equal{}nz^{2}$.
LEMMA 1
Καθε πρωτος διαιρετης $ >4$ του $ x^{2}\plus{}y^{2}$ ειναι της μορφης $ 4k\plus{}1$
LEMMA 2.
Καθε πρωτος της μορφης $ 4k\plus{}1$ γραφεται και ως αθροισμα $ 2$ τετραγωνων.

Αν ο $ n$ δεν ειναι τετραγωνο ακεραιου τοτε εστω $ (x,y)\equal{}d$ τοτε υπαρχουν $ a,b$ ωστε $ x\equal{}da$ & $ y\equal{}db$ με $ (a,b)\equal{}1$ αρα $ d^{2}(a^{2}\plus{}b^{2})\equal{}nz^{2}$ οποτε $ d^{2}/nz^{2}$.Θα διακρινω $ 2$ περιπτωσεις:
$ 1$)Αν ο $ n$ δεν ειναι πολλαπλασιο του $ d^{2}$ τοτε προφανος αφου $ d^{2}/nz^{2}$ τοτε $ d/z$ με $ z\equal{}dc$ αρα η εξισωση μας αναγεται στην $ a^{2}\plus{}b^{2}\equal{}nc^{2}$ με $ (a,b)\equal{}1$
$ 2$)Αν ο $ n$ ειναι πολλαπλασιο του $ d^{2}$ εστω οτι $ n\equal{}kd^{2}$ τοτε $ d^{2}(a^{2}\plus{}b^{2})\equal{}k(dz)^{2}$ αρα $ a^{2}\plus{}b^{2}\equal{}kz^{2}$ με $ (a,b)\equal{}1$
Μελεταμε λοιπον την εξισωση $ x^{2}\plus{}y^{2}\equal{}nz^{2}$ με την επιπλεον συνθηκη οτι $ (x,y)\equal{}1$ χωρις βλαβη της γενικοτητας οπως απεδειξα παραπανω.
Εστω οτι $ n\equal{}m^{2}p_{1}p_{2}....p_{k}$ τοτε συμφωνα με το πρωτο λημμα ολοι οι πρωτοι p_i θα ειναι της μορφης $ 4k\plus{}1$ γιατι $ (x,y)\equal{}1$ αρα συμφωνα και με το  δευτερο λημμα αποδειχθεικε.

Ωραια η λυση σου savvas_michael :wink: .Αν καποιος αλλος εχει διαφορετικη αποδειξη της ασκησης παρακαλω ας την στηλει. :)

[color=red]ΑΣΚΗΣΗ 3[/color]
Η παρακάτω είναι άσκηση που μπήκε στην προετοιμασία της Ρουμανίας το 2007 .
Υποθέτουμε ότι $ a$, $ b$ είναι θετικοί ακέραιοι για τους οποίους ο αριθμός $ A \equal{}\frac{a\plus{}1}{b}\plus{}\frac{b}{a}$ είναι ένας ακέραιος .

Δείξτε ότι  $ A \equal{} 3$ .Όταν λυθεί θυμήστε μου να πω κάτι ........

Έστω $ (a, b)\equal{}z$ και $ a\equal{}zx_{1}, b\equal{}zy$. Τότε $ A\equal{}\frac{zx_{1}^{2}\plus{}x_{1}\plus{}zy^{2}}{zx_{1}y}$ οπότε $ z/x_{1}$, δηλαδή $ x_{1}\equal{}zx$. Άρα $ A\equal{}\frac{z^{2}x^{2}\plus{}x\plus{}y^{2}}{zxy}$ και συνεπώς $ x/y^{2}$ και επειδή $ (x, y)\equal{}1$ θα είναι $ x\equal{}1$ και $ A\equal{}\frac{y^{2}\plus{}z^{2}\plus{}1}{yz}$. Όμως είναι γνωστό ότι αν $ yz/y^{2}\plus{}z^{2}\plus{}1$ τότε  $ A\equal{}\frac{y^{2}\plus{}z^{2}\plus{}1}{yz}\equal{}3$.

Απόδειξη: Θεωρούμε το σύνολο των ζευγών λύσεων $ (y, z)$ της εξίσωσης $ A\equal{}\frac{y^{2}\plus{}z^{2}\plus{}1}{yz}$ και έστω $ (y, z)$ το ζεύγος εκείνο ώστε το $ y\plus{}z$ να είναι ελάχιστο. Λόγω συμμετρίας μπορούμε να υποθέσουμε ότι $ y\geq z$. Έχουμε ότι $ y^{2}\minus{}Azy\plus{}z^{2}\plus{}1\equal{}0$. Το τριώνυμο αυτό ως προς $ y$ θα έχει από τους τύπους του Vieta ακόμα μια λύση την $ y_{1}\equal{}Az\minus{}y\equal{}\frac{z^{2}\plus{}1}{y}$. Πρέπει όμως $ y_{1}\geq y$, οπότε $ z^{2}\plus{}1\geq y^{2}$. Τώρα αν $ y>z$ θα είναι $ z^{2}\plus{}1\geq y^{2}\geq (z\plus{}1)^{2}$ δηλαδή $ z\leq0$ που είναι άτοπο. Άρα $ y\equal{}z$, απ' όπου εύκολα παίρνουμε $ A\equal{}3$.

Εξαιρετική λύση  :)  .
Κοιτάξτε κάτι όμως . Στην προετοιμασία τους φέτος οι Ρουμάνοι διδάχτηκαν αυτή την άσκηση με descent (Vieta Jumping) .Στην ΙΜΟ πέφτει θέμα με descent .Θα μπορούσε κάποιος από αυτούς να μην το γράψει ?  :wink:
Τελος πάντων

Δειξτε οτι ο
  $ {{{3^{2}}^{n}}+1}$
 ειναι πολ του 2 αλλα οχι του 4.

manoliss μπορείς να εξηγήσεις λίγο την άσκηση;

δεν καταλαβαίνω...αφού για $ n\equal{}1$ η παράσταση ισούται με 8 :huh:

Η σωστή εκφώνηση (εαν είναι η άσκηση που ξέρω) είναι η εξής:

Να αποδειχθεί ότι για $ n\geq 1$ ο αριθμός $ 3^{2^{n}}\plus{}1$ είναι πολλαπλάσιο του $ 2$ αλλά όχι του $ 4$.

Μάλιστα υπάρχει και μία ενδιαφέρουσα εφαρμογή αυτής της πρότασης στην εξής άσκηση:

Να βρεθεί η μεγαλύτερη δύναμη του $ 2$ που διαιρεί τον αριθμό $ 3^{1024}\minus{}1$.

Αξίζει οι μικροί φίλοι που παρακολουθούν το forum, να ασχοληθούν με αυτή την άσκηση.

Αλέξανδρος

Φίλε Αλέξανδρε φαντάζομαι ότι και εσύ εννοείς να βρούμε τη μεγαλύτερη δύναμη του 2 που διαρεί το
$ 3^{1024}\plus{}1$  :oops: ως εφαρμογή της παραπάνω άσκησης η οποία είναι μια καλή εφαρμογή στη μέθοδο της επαγωγής :wink:

Ένας μεγάλος αριθμοθεωρίστας μέλος του φόρουμ μου έδωσε αυτή την πανέμορφη άσκηση προς λύση . Μου είπε ότι είχε μια λύση αρκετά μεγάλη αλλά βρήκα κάτι πολύ σύντομο  :D  . Here it goes
Αν $ x_1,...,x_n$ είναι ακέραιοι αριθμοί τότε να δείξετε ότι
ο αριθμός
$ A\equal{}\prod_{1\leq i < j\leq n}\frac {x_{j} \minus{} x_{i}}{j \minus{} i}$  είναι ακέραιος .

Πολύ ωραία άσκηση:

Υπάρχουν ακέραιοι $ a,b,c$ ανά δύο πρώτοι μεταξύ τους και όλοι $ >1$ που να ικανοποιούν:

$ b|2^a\plus{}1$
$ c|2^b\plus{}1$
$ a|2^c\plus{}1$

[quote="Protonios"]Πολύ ωραία άσκηση:
[/quote]

Συμφωνώ . Αλλά είναι ταυτόχρονα υπερβολικά δύσκολη . Με όσους πολύ καλούς αριθμοθεωρίστες έχω μιλήσει την είχαν αναφορά ως πολύ δύσκολη .Την έχουν λύσει μέχρι τέλους πολλοί λίγοι αν και η ιδέα που χρησιμοποιεί φωνάζει
(αποδείξτε το πρώτα για πρώτους είναι εύκολο)

Την εχεις προσπαθησει η ειδες την λυση?Μια ασκηση ειναι δυσκολη μεχρι να την λυσουμε.....μετα γινεται ευκολη.... :wink: Υπαρχουν και απλες λυσεις αυτης....

Την έχω προσπαθήσει ναι (πέρσυ) και έφτασα σε ικανοποιητικό σημείο μπορώ να πω (ήμουνα πολύ κοντά γιατί κατάλαβα την ιδέα που ήθελε . Αλλά σε αυτό το σημείο έχουν φτάσει και οι άλλοι ... Υπάρχει διαφορά του να ξέρεις να προσεγγίσεις την άσκηση από το να προσεγγίσεις την άσκηση και από το να η λύσεις  :wink:

Αν έχεις σύντομη λύση πόσταρε την θα έχει ενδιαφέρον ...

Θα λυσω το συστημα για a,b,c πρωτους οπως ειπε και ο κυριος Σιλουανος.Εστω οτι:
$ q/2^p \plus{} 1$,$ r/2^q \plus{} 1$,$ p/2^r \plus{} 1$
Xωρις β.τ.γ θεωρω την διατεταγμενη τριαδα $ p > \equal{} q > \equal{} r$
Επειδη $ 2^p \equal{} \minus{} 1modq$ και $ 2^{q \minus{} 1} \equal{} 1modq$ τοτε θα εχουμε οτι $ 2^u \equal{} \minus{} 1modq$ oπου u το υπολοιπο της διαιρεσης του $ p$ με το $ q \minus{} 1$ με $ 0 < \equal{} u < q \minus{} 1 < p$ αρα $ (u,p) \equal{} 1$ οποτε υπαρχουν ακεραιοι $ x,y$ ωστε $ ux \plus{} py \equal{} 1$ αρα $ 2^{ux \plus{} py} \equal{} {1}^x({\minus{} 1}^y)modq$ αρα $ 2 \equal{} { \minus{} 1}^ymodq$ αρα μοναδικη λυση ειναι η $ (p,q,r) \equal{} (3,3,3)$
Μπορει καποιος να λυση ακριβως την ασκηση του protonios?Δεν ειναι και παρα πολυ δυσκολη.....προσπαθηστε την.... :wink:

[quote="Αρχιμήδης 6"] ο κυριος Σιλουανος.
[/quote]

Ε όχι και κύριος  :P   Πάντως η τριάδα λύσεων είναι $ (p,q,r)\equal{}(2,5,3)$  και η λύση που είχα υπόψη χρησιμοποιεί ord

Αλλη μια με πρωτους αριθμους:

Να βρεθουν ολοι οι τετραψηφιοι πρωτοι που τελειωνουν σε 7.

[quote="silouan"]Ένας μεγάλος αριθμοθεωρίστας μέλος του φόρουμ μου έδωσε αυτή την πανέμορφη άσκηση προς λύση . Μου είπε ότι είχε μια λύση αρκετά μεγάλη αλλά βρήκα κάτι πολύ σύντομο  :D  . Here it goes
Αν $ x_1,...,x_n$ είναι ακέραιοι αριθμοί τότε να δείξετε ότι
ο αριθμός
$ A \equal{} \prod_{1\leq i < j\leq n}\frac {x_{j} \minus{} x_{i}}{j \minus{} i}$  είναι ακέραιος .[/quote]

Έψαχνα κάποιες ασκήσεις το μεσημέρι και έπεσα σε αυτή την άσκηση του Σιλουανού την οποία βρήκα πολύ ενδιαφέρουσα!

Καταρχήν εαν κάποιο $ x_i \equal{} x_j$ για $ i\neq j$ το αποτέλεσμα είναι φανερό. Οπότε μπορούμε να υποθέσουμε ότι $ x_1 < x_2 < \cdots < x_n$.

Λοιπόν ως γνωστόν το γινόμενο $ \prod_{1\leq i < j\leq n}\left(x_{j} \minus{} x_{i}\right)$ είναι η λεγόμενη ορίζουσα Vandermonde και ισούται με  $ \Delta(x_1,x_2,\ldots,x_n) \equal{} \begin{bmatrix} 1 & x_1 & x_1^2 & \dots & x_1^{n \minus{} 1} \\
1 & x_2 & x_2^2 & \dots & x_2^{n \minus{} 1} \\
1 & x_3 & x_3 & \dots & x_3^{n \minus{} 1} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & x_n & x_n^2 & \dots & x_n^{n \minus{} 1} \\
\end{bmatrix}$

Επίσης είναι φανερό ότι $ \prod_{1\leq i < j\leq n}\left(j \minus{} i\right) \equal{} \prod_{i \equal{} 1}^{n} (i \minus{} 1)!$ . Οπότε αρκεί να δείξουμε ότι

$ \prod_{i \equal{} 1}^{n} (i \minus{} 1)! | \Delta(x_1,x_2,\ldots,x_n)$.

Παρατηρούμε ότι επειδή $ \Delta(x_1,x_2,\ldots,x_n) \equal{} \Delta(x_1 \plus{} 1,x_2 \plus{} 1,\ldots,x_n \plus{} 1)$ μπορούμε να θεωρήσουμε όλα τα $ x_i\geq 0$.

Επίσης εαν τα $ f_i(x)$ είναι μονικά πολυώνυμα βαθμού $ i$, $ 1\leq i \leq n \minus{} 1$, τότε στοιχειώδεις πράξεις μέσα στην ορίζουσα μας δίνουν ότι $ \Delta(x_1,x_2,\ldots,x_n) \equal{} \begin{bmatrix} 1 & f_1(x_1) & f_2(x_1) & \dots & f_{n \minus{} 1}(x_1) \\
1 & f_1(x_2) & f_2(x_2) & \dots & f_{n \minus{} 1}(x_2) \\
1 & f_1(x_3) & f_2(x_3) & \dots & f_{n \minus{} 1}(x_3) \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & f_1(x_n) & f_2(x_n) & \dots & f_{n \minus{} 1}(x_n) \\
\end{bmatrix}$

Εαν τώρα πάρουμε $ f_i(k) \equal{} k(k \minus{} 1)\cdots (k \minus{} i \plus{} 1)$ τότε για τους μη αρνητικούς ακεραίους $ k$ ισχύει $ f_i(k) \equal{} i! \binom{k}{i}$.

άρα κάθε αριθμός στην $ i$ στήλη αυτής της ορίζουσας είναι διαιρετός από το $ (i \minus{} 1)!$ και συνεπώς όλη η ορίζουσα είναι διαιρετή από το $ \prod_{i \equal{} 1}^{n} (i \minus{} 1)!$.

Αλέξανδρος

Υ.Γ. Σιλουανέ έχεις κάποια συντομότερη λύση ή μοιάζει με την παραπάνω?

Ωραία Αλέξανδρε αν και νομίζω ότι υπάρχει μια Vandermonde με τιν οποία βγαίνει κατευθείαν χωρίς τίποτα άλλο μετα  :wink:
Λοιπόν η λύση μου είναι βασισμένη σε ένα διαφορετικό πνεύμα

Θα χρησιμοποιήσω επαγωγωγή στον $ k = \max{(x_{1},x_{2},...,x_{n})}$.
Οι περιπτώσεις όπου $ k\leq n$ είναι προφανείς . Έστω $ k = m + 1(m\geq n)$  και έστω ότι η υπόθεση ισχύει για όλα τα  $ 1,2,...,m$.

Έστω τωρα $ x_{n}=m+1$ .Αν υπάρχει δείκτης $ i < n$ ώστε $ x_{i} = m + 1$ τελειώσαμε .
Αν  $ x_{i}\leq m$ για κάθε  $ i = 1,2,...,n - 1$ τότε έστω $ f(t) = Q(x_{1},x_{2},...,x_{n - 1},t)$, όπου
 $ Q(x_{1},x_{2},...,x_{n}) = \prod_{1\leq i < j\leq n}\frac {x_{j} - x_{i}}{j - i}$.

Τότε  $ f(x)\in\mathbb{Z}$  για κάθε  $ x = 1,2,...,m$  και επιπλέον  $ \deg{f} = n - 1 < m$ άρα από το Lagrange interpolating Polynomial έχουμε $ f(m + 1)\in\mathbb{Z}$ so qed  :D

Να πω σε αυτό το σημείο ότι μπορούμε νδο
για οποιουσδήποτε ακεραίους  $ a_1,...,a_n$ ο αριθμός
 ${ \frac {LCM(a_1,...,a_n)}{a_1a_2...a_n}}\cdot\prod_{i < j}{(a_i - a_j)}$ διαιρείται από τον $ 1! 2!...(n - 2)!$

Αυτό μπορεί να δειχτεί με έναν πολύ γενικό τρόπο με πολυώνυμα αλλά και απευθείας με ττην παρακάτω ορίζουσα
\[ \left| \begin{array}{cccc} \frac {L}{a_1} & \frac {L}{a_2} & \cdots & \frac {L}{a_n} \\
1 & 1 & \cdots & 1 \\
a_1 - 1 \choose 1 & a_2 - 1 \choose 1 & \cdots & a_n - 1 \choose 1 \\
\vdots & & & \vdots \\
a_1 - 1 \choose n - 2 & a_2 - 1 \choose n - 2 & \cdots & a_n - 1 \choose n - 2 \\
\end{array} \right|
\]

Η αρχαία φυλή των δράκων μιλούσε την αληθινή! γλώσσα, είχε όμως ένα μειονέκτημα. Η γλώσσα αυτή περιείχε πολύ λίγες λέξεις. Ο Καλεσίν και το παιδί του ήταν δυστυχισμένοι γιατί διέθεταν γνήσια ποιητική φλέβα, αλλά δεν είχαν λέξεις για να γράψουν τα ποιήματά τους. Να λοιπόν τι σκαρφίστηκαν: Μάζεψαν μια μικρή ομάδα λέξεων και μια δεύτερη μεγαλύτερη ή ίση, τις ξεφλούδισαν και τις πρόσθεσαν κι έτσι έφτιαξαν μια τρίτη ομάδα. Στη συνέχεια  πήραν την δεύτερη ομάδα και την τρίτη, τις ξεφλούδισαν πάλι και τις πρόσθεσαν και έφτιαξαν την τέταρτη ομάδα. Συνέχισαν έτσι μέχρι την ένατη ομάδα, γιατί τα πόδια τους είχαν δάχτυλα με εννέα νύχια και οι δράκοι μετρούσαν μόνο με τα δάχτυλα. Το ποίημα λοιπόν αποτελούνταν από τις λέξεις της ένατης ομάδας. Να βρείτε τον ελάχιστο αριθμό λέξεων που περιέχει κάθε δρακοποίημα, ώστε οι δράκοι να μπορούν να απολαμβάνουν τουλάχιστον δύο ποιήματα.
Θεωρήστε όλες τις λέξεις διαφορετικές και όλες να χρησιμοποιούνται.

Σε πιο μαθηματική γλώσσα

Να βρείτε τον μικρότερο θετικό ακέραιο k, ώστε οι παρακάτω συνθήκες:
Α) $ (a_n)$ είναι μια αύξουσα ακολουθία θετικών ακεραίων
Β) $ a_n \equal{} a_{n\minus{}1} \plus{} a_{n\minus{}2} ,  n>2$
Γ) $ a_9\equal{}k$
να ικανοποιούνται από περισσότερες από μία ακολουθίες.

Επειδή έχει μείνει άλυτο πολύ καιρό (ΜΆΛΛΟΝ σας φόβισε ο δράκος ) ποστάρω την λύση στο attachment που ακολουθεί

[quote="silouan"][color=red]ΑΣΚΗΣΗ 3[/color]
Η παρακάτω είναι άσκηση που μπήκε στην προετοιμασία της Ρουμανίας το 2007 .
Υποθέτουμε ότι $ a$, $ b$ είναι θετικοί ακέραιοι για τους οποίους ο αριθμός $ A \equal{} \frac {a \plus{} 1}{b} \plus{} \frac {b}{a}$ είναι ένας ακέραιος .

Δείξτε ότι  $ A \equal{} 3$ .Όταν λυθεί θυμήστε μου να πω κάτι ........[/quote]

Αφού είναι ακέραιος τότε έχουμε ότι $ \frac {a \plus{} 1}{b}$ και $ \frac {b}{a}$ είναι ακέραιοι. Άρα $ a\plus{}1\equal{}bx$ και $ b\equal{}ay$.

Με λύση του συστήματος έχουμε:
$ a\plus{}1\equal{}axy$
$ a\minus{}axy\equal{}\minus{}1$
$ a(1\minus{}xy)\equal{}\minus{}1$
Αφου a θετικός όμως:
$ a\equal{}1$,    $ 1\minus{}xy\equal{}\minus{}1 <\equal{}> xy\equal{}2$
Και αφού $ b\equal{}y$, $ b\equal{}1$ ή $ b\equal{}2$.
Όμως η $ b\equal{}1$ απορρίπτεται γιατι τότε θα είχαμε: $ b\equal{}ay <\equal{}> y\equal{}2$ άρα $ x\equal{}1$ οπότε $ a\plus{}1\equal{}bx <\equal{}> a\plus{}1\equal{}b <\equal{}> b\equal{}2$ αδύνατο.

'Αρα αντικαθιστούμε στην Α τα a,b και βγαίνει ίση με 3.

[quote="userresu"]
Αφού είναι ακέραιος τότε έχουμε ότι $ \frac {a \plus{} 1}{b}$ και $ \frac {b}{a}$ είναι ακέραιοι.[/quote]

Δεν είναι απαραίτητο να ισχύει αυτό .... :wink:


Prove that it is possible to find some whole number multiple of $ \sqrt{2}$ as close as you want to a positive integer, but never equal to one.

Hi KrazyFK, I think I have it - not sure if it's rigorous enough.

It seems that your question can be expressed as

Given $ N \in \mathbb{N}, \exists\ m, n \in \mathbb{N}$ such that $ \mid m\sqrt {2} \minus{} n \mid \le \frac {1}{N}$.

Proof:

Firstly, we subdivide the unit interval $ [0, 1]$ into $ N$ equal subintervals $ [0, \frac {1}{N}], [\frac {1}{N}, \frac {2}{N}], ..., [\frac {N \minus{} 1}{N}, 1]$.

Then we consider the $ N \plus{} 1$ real numbers $ m\sqrt {2} \minus{} \lfloor m\sqrt {2} \rfloor, 0 \le m \le N$, and place these $ N \plus{} 1$ numbers into the $ N$ subintervals.

Then, by Dirichlet's pigeon-hole principle, two of the numbers must occupy the same subinterval. let's call these two numbers $ a\sqrt {2} \minus{} \lfloor a\sqrt {2} \rfloor$ and $ b\sqrt {2} \minus{} \lfloor b\sqrt {2} \rfloor$ where $ 0 \le a < b \le N$.

Let $ m \equal{} b \minus{} a$ and $ n \equal{} \lfloor b\sqrt {2} \rfloor \minus{} \lfloor a\sqrt {2} \rfloor$.

Then $ \mid m\sqrt {2} \minus{} n\mid \equal{} \mid (b\sqrt {2} \minus{} \lfloor b\sqrt {2} \rfloor) \minus{} (a\sqrt {2} \minus{} \lfloor a\sqrt {2} \rfloor) \mid \le \frac {1}{N}$.

QED

[$ b \minus{} a > 0 \implies m \ge 1$ and $ m \equal{} b \minus{} a \le b \le N$.]

[quote="AndrewTom"]$ \exists\ m, n \in \mathbb{N}$ such that $ \mid m\sqrt {2} \minus{} n \mid \le \frac {1}{N}$ for any $ N \in \mathbb{N}$.[/quote]  Minor note: the way you've written this is really confusing (though of course I know exactly what you mean) since actually the choice of $ N$ comes [i]before[/i] the choices of $ m, n$ (since they are dependent on it).  As written, you appear to be suggesting that there is a choice of $ m, n$ such that your inequality holds for all $ N$, which is not true.  Quantifier order matters a lot  :)

Thanks JBL. I see where it is confusing and perhaps misleading. I've edited it.

Yes AndrewTom, you have it, that's basically exactly what I was after.


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geometry
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geometry unsolved

By the m-n theorem,

 $2 \cot 135^\circ = cot B - cot C$

 solving,
 $cot B = \sqrt{3} - 2$
 which gives $B= 105^\circ$

 yielding, by angle sum prop.
 $\angle MAB = 30^\circ$

Okay, okay, it's like this:
 consider any triangle $ABC$ and mark a point $M$
 on the base $BC$ such that $M$ divides it into the ratio $m: n$
 i.e. $BM: MC$ = $m: n$
 now, let $\angle AMC = \theta$
 then, by $m-n \ theorem$, we have :
 $(m+n) \cot{\theta} = m\ cotB - n\ cotC$
 Is that sufficient?

[quote="Beat"]Let $M$ be the midpoint of the side $BC$ of the $\triangle ABC$. If $\angle ACB=30^{\circ}$ and $\angle AMB=45^{\circ}$, then find $\angle BAM$[/quote]

or alternatively

Draw the following to complete the figure. By point $C$ draw a parallel to $AM$. Let $X$ be the perpendicular to this line drawn from $B$. Also extend median $AM$ to point $Q$ such that $\angle AQB=30$. Now

1) $ACQB$ is cyclic, cause $\angle ACB = \angle AQB = 30$. Hence $\angle QBC = 15$
2) $\angle ACX = 15 \Rightarrow \angle BCX = 45$. Hence triangle $BXC$ is isosceles
3) Hence $\angle XBC = 45$
4) From 1) and 4) $\angle QBX = 60$.
5) Also, if $H$ is the intersection point of $BX$ and $AM$ then clearly $\angle BHQ = 90$ and $BH = BQ/2$. Also $BH = BX/2$ so $BQX$ equilateral.
6) Now $\angle QXC = 30$ and as $\angle QAC = 15$ it runs out that $A$ belongs to the circumference with center $X$ and radius $XB$. That is $X$ actually is the circumcenter of triangle $ABC$
7) Finally $\angle BAQ = 30$ as $\angle BXQ = 60$

The trig solution is probably easier, I just couldn't resist to search for a purely geometrical one.

Daniel


let $ A$ and  $ B$ be the universal covering spaces of $ X$ and $ Y$ respectively? my question  : is $ A\times B$ a universal covering space of $ X\times Y$? is $ A\bigvee B$ is the universal covering of $ X\bigvee Y$? many thanks

Part 1: Yes

First we must see that $ A\times B$ is simply connected, this is trivial because the fundamental group of the product is just the product of the fundamental group. Now, we need to see that it is a covering space:

Pick a point $ p\equal{}(x,y)$ in $ X\times Y$. Let $ U$ be a neighborhood of $ x$ such that the projection map acts as a homeomorphism on this and one of the sheets. Let $ V$ be a neighborhood of $ y$ such that the projection is a homeomorphism. Then $ U\times V$ is homeomorphic to a neighborhood of $ A\times B$ by construction. Moreover this construction holds for any point $ p$. Therefore we have that $ A\times B$ is simply connected and locally $ X\times Y$ is homeomorphic to the sheets that cover it.

Part 2: do you mean the one point union of $ A$ and $ B$?

[quote="LkNsngth"]
Part 2: do you mean the one point union of $ A$ and $ B$?[/quote]

yes, I am talking about the wedge sum

I think the answer to the second question is no. Take $ X \equal{} Y \equal{} S^1$, and $ A \equal{} B \equal{} \mathbb{R}$. The natural map $ A\bigvee B \rightarrow X \bigvee Y$ is not even a covering.


Let $ a,b,c$ be the side lengths of a triangle. Prove that:

$ (a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\plus{}(\minus{}a\plus{}b\plus{}c)(a\minus{}b\plus{}c)(a\plus{}b\minus{}c)\ge 10abc$

using RAVI's transformation we get
$ x^{3}\plus{}y^{3}\plus{}z^{3}\plus{}3xyz\geq xy(x\plus{}y)\plus{}yz(y\plus{}z)\plus{}zx(z\plus{}x)$
which is schur for $ n\equal{}1$

[quote="Pain rinnegan"]Let $ a,b,c$ be the side lengths of a triangle. Prove that:

$ (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \plus{} ( \minus{} a \plus{} b \plus{} c)(a \minus{} b \plus{} c)(a \plus{} b \minus{} c)\ge 10abc$[/quote]

If we denote by $ x\equal{}\minus{}a\plus{}b\plus{}c,y\equal{}a\minus{}b\plus{}c,z\equal{}a\plus{}b\minus{}c$ the above inequality transforms into $ (x\plus{}y)^3\plus{}(y\plus{}z)^3\plus{}(z\plus{}x)^3\plus{}12xyz\geq  0$, right?

hello, setting $ a \equal{} y \plus{} z,b \equal{} x \plus{} z,c \equal{} x \plus{} y$ in your inequality we get
$ x^3 \plus{} y^3 \plus{} z^3 \plus{} 3xyz\geq x^2y \plus{} xy^2 \plus{} y^2z \plus{} z^2y \plus{} x^2z \plus{} xz^2$
this is equivalent with
$ (x \minus{} y)^2(x \plus{} y \minus{} z) \plus{} z(x \minus{} z)(y \minus{} z)\geq 0$
which is true for $ x\geq y\geq z$
Sonnhard.

[quote="Pain rinnegan"]Let $ a,b,c$ be the side lengths of a triangle. Prove that:

$ (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \plus{} ( \minus{} a \plus{} b \plus{} c)(a \minus{} b \plus{} c)(a \plus{} b \minus{} c)\ge 10abc$[/quote]

It is well-known ?


Let
$ A \equal{} \left[\begin {array}{ccc} 3 & 0 & 0\\
                \minus{}5 & 2 & 0\\
                  4 & 0 & 1 \end {array} \right]$

How can I found all the matrices that satisfy the relationship $ X^{ 2}\equal{} A$

I would be grateful if someone show me a step-by-step solution.

Thanks in Advance

Find one matrix $ P$ such that

$ P^{\minus{}1}AP \equal{} \left[\begin {array}{ccc} 3 & 0 & 0\\
0 & 2 & 0\\
0 & 0 & 1 \end {array} \right]$.

there exist $ P \equal{} \left[\begin {array}{ccc}1 & 0 & 0 \\
\minus{} 5 & 1 & 0 \\
2 & 0 & 1 \end {array} \right]$then
$ P^{ \minus{} 1} \equal{} \left[\begin {array}{ccc} 1 & 0 & 0 \\
5 & 1 & 0 \\
\minus{} 2 & 0 & 1 \end {array} \right]$
$ P^{ \minus{} 1}AP \equal{} \left[\begin {array}{ccc} 3 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 1 \end {array} \right]$
So $ P^{ \minus{} 1}X^2P \equal{} P^{ \minus{} 1}AP \equal{} \left[\begin {array}{ccc} \sqrt {3} & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 1 \end {array} \right]$,
We have
$ X^2 \equal{} P \left[\begin {array}{ccc} 3 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 1 \end {array} \right]P^{ \minus{} 1} \equal{} P \left[\begin {array}{ccc} \sqrt {3} & 0 & 0 \\
0 & \sqrt {2} & 0 \\
0 & 0 & 1 \end {array} \right]P^{ \minus{} 1}P \left[\begin {array}{ccc} \sqrt {3} & 0 & 0 \\
0 & \sqrt {2} & 0 \\
0 & 0 & 1 \end {array} \right]P^{ \minus{} 1}$,So
$ X \equal{} P\left[\begin {array}{ccc} \sqrt {3} & 0 & 0 \\
0 & \sqrt {2} & 0 \\
0 & 0 & 1 \end {array} \right]P^{ \minus{} 1} \equal{} \left[\begin {array}{ccc}1 & 0 & 0 \\
\minus{} 5 & 1 & 0 \\
2 & 0 & 1 \end {array} \right]\left[\begin {array}{ccc} \sqrt {3} & 0 & 0 \\
0 & \sqrt {2} & 0 \\
0 & 0 & 1 \end {array} \right]\left[\begin {array}{ccc} 1 & 0 & 0 \\
5 & 1 & 0 \\
\minus{} 2 & 0 & 1 \end {array} \right]$
$ \equal{} \left[\begin {array}{ccc} \sqrt {3} & 0 & 0 \\
5(\sqrt {2} \minus{} \sqrt {3} )& \sqrt {2} & 0 \\
2(\sqrt {3} \minus{} 1) & 0 & 1 \end {array} \right]$

If $ A\in{GL_n}(\mathbb{C})$ and if $ A$ has $ n$ distinct eigenvalues $ \lambda_i$ then there exist $ 2^n$ square roots of $ A$.
Indeed $ X$ has also $ n$ distinct eigenvalues $ \mu_i$ such that $ \mu_i^2\equal{}\lambda_i$. $ XA\equal{}AX$ then $ X,A$ are simultaneously diagonalizable and there exists $ 2^n$ correspondences $ \lambda_i\rightarrow\mu_i$. We can conclude with Lagrange interpolation.
   Here, for example, with the correspondence $ (1,2,3)\rightarrow(1,\sqrt{2},\sqrt{3})$ we obtain $ X\equal{}(\sqrt{3}/2\minus{}\sqrt{2}\plus{}1/2)A^2\plus{}(\minus{}3\sqrt{3}/2\plus{}4\sqrt{2}\minus{}5/2)A\plus{}(\sqrt{3}\minus{}3\sqrt{2}\plus{}3)I$.


calculus
integration
geometry
trapezoid
calculus computations

Find the volume of the solid obtained by rotating the region bounded by the given curves using disk or washer method.

the problem:
 y = x, y = 0 , x = 2, x = 4 about x =1

My solution:
pi(outer radius)^2 - pi(inner radius)^2

pi(2 + y)^2 - pi(1 + y)^2

thus we have:

integral of [pi(2 + y)^2 - pi(1 + y)^2]dy from 2 to 3

I end up with a final answer of 8pi.  Does this answer seem correct or is there something I am doing wrong using the disk method?

Thank you for your input!

Doesn't seem right to me. If I understood yo task right, u're to execute the volume of the solid obtained by rotating of the trapezium (base of trapezium is: $x \in [2, 4]$ and the sides are $y \in [0, 2]$ and $y \in [0, 4]$) around the line $x = 1$, right? Let's take a cylinder $[2, 4] \times [0, 2]$ which is smaller than the trapezium and calculate the volume of its surface of revolution. It will be $2 \cdot \pi \cdot 3^2 - 2 \cdot \pi \cdot 1^2 = 16 \cdot \pi$

Well, now I'm not sure what u call disk or washer methods, but if i were to calculate this volume using HM methods I'd do smth like this (check values, i'm no good in exact calculations :)):
$\int^{2}_{0}(\pi \cdot 3^2 - \pi \cdot 1^2 ) dy + \int^{4}_{2}(\pi \cdot 3^2 - \pi \cdot (y - 1)^2)dy$. What's this? I'm calculating the areas of sections  made by lines y = c, where c takes values from 0 to 4 and then sum these areas up (=integrate).


[hide]
(c) $log_{2}9=\frac{ln(9)}{ln(2)}=\frac{\frac{ln(9)}{ln(10)}}{\frac{ln(2)}{ln(10)}}=\frac{2b}{a}$, using the same method for the second, we find: $log_{3}8=\frac{3a}{b}$ so our answer is: $\boxed{\frac{2b}{a}+\frac{3a}{b}}$
[/hide]
*edited

this is how I solved the fourth one, correct me if I'm wrong...

$log_{\sqrt{45}}144=\frac{log(144)}{log(\sqrt{45})}=\frac{4log{(12})}{log{(45)}}=\frac{8log{(2)}+4log{(3)}}{1+2log{(3)}-log{(2)}}=$
$\boxed{\frac{8a+4b}{1+2b-a}}$

Also demji. For the third equation, you got to add not subtract

whats wrong with my answer for (b), calculating it out it fails, but why?

yeah, 2 dumb mistakes with that, might be a slight problem.  I really haqd no clue for (4).  I don't understand the breakdown of $log(45)$ I did check it and it works.

oh ok, so simple, didn't even consider using a constant.  Thx

demji: Fix (4) please

236factorial: Finish part (3) please

[quote="10000th User"]236factorial: Finish part (3) please[/quote]

What do you mean by "finish"?  :?

Edit: See, it's I, I mean Mr. Stupid, asking "unnecessary" questions

[quote="236factorial"][quote="10000th User"]236factorial: Finish part (3) please[/quote]

What do you mean by "finish"?  :?[/quote] :ninja:  Check your answer to part (3) with the original problem posted so that you don't have to [u]unnecessarily[/u] post and ask what "finish" means...


trigonometry
algebra
partial fractions
calculus
calculus computations

Find the inverse of $\frac{3s-4}{s(s-4)}$.  I did this two ways and I'm just wondering if both solutions are correct (or one or neither  :lol: ).
Method 1:
Break the fraction into: $\frac{3}{s-4}-\frac{4}{s(s-4)}$, then a table gives the inverse:$3e^{4t}-2e^{2t}\sin2t$.
Method 2: Decompose into partial fractions:
$\frac{1}{s}+\frac{2}{s-4}$ Then a table gives the inverse: $2e^{4t}+1$.
Are they both correct?

The first one is wrong.  Actually,

$\mathcal{L}\left(-2e^{2t}\sin 2t \right) =-\frac{4}{s^{2}-4t+8}$

oops.  The way I was doing it I should've used $\sinh$ instead so it should be $3e^{4t}-2e^{2t}\sinh2t$.  And then the answers are equivalent.  Thanks t0r.


What is the positive three-digit integer that is four times the product of its digits?
2004-05 workout 9 # 5

Let the number be $ \overline{abc}$ so that $ 100a \plus{} 10b \plus{} c \equal{} 4abc\implies10b \plus{} c \equal{} 4a(bc \minus{} 25)$. Since both sides are divisible by $ 4$, some inspection tells us that $ b,c$ are even. Also, $ bc \minus{} 25\ge0\implies bc\ge25$. Now we just test, remembering that $ 0\le b,c\le9$ and $ 2|b,c$. By the second requirement, it is clear that $ bc$ must have at least two factors of two, so we ignore the others (so only check multiples of four, since otherwise the number can have at most one factor of two). $ 28 \equal{} 2^2\cdot7$, but the only way to make both even is $ 2$ and $ 14$, which isn't in our bounds. $ 32 \equal{} 2^5$, and the only one that has a chance to work here is $ 4$ and $ 8$. Substituting these in gives $ 10b \plus{} c \equal{} 4a(32 \minus{} 25) \equal{} 28a$; the possibilities for $ 10b \plus{} c$ are $ 48$ and $ 84$, and $ 10b \plus{} c \equal{} 84\implies b \equal{} 8\implies c \equal{} 4\implies a \equal{} 3\implies\boxed{384}$ does indeed work. Of course, to be rigorous, we have to realize that $ 10b\plus{}c\le88$, so $ 88\ge4a(bc\minus{}25),a\ge1\implies22\ge bc\minus{}25\implies bc\le47,4|bc\implies bc\le44$. I won't check the remaining cases $ bc\equal{}36,40,44$, but they don't work.

Alternatively, use a program! :P

i would logically use guess and check here.

Wow, i finally got it after a lot of guess and check, i got the answer to be 384, 3*8*4=96 and 96*4=384!!

EDIT: dang it, too late.. :wink:

[hide="faster, maybe"]
$ \overline{abc}$ must be divisible by 4, hence c is even and $ \overline{abc}$ is divisible by 8; additionally, by math154's logic, b is even as well. now suppose $ a$ is even. then $ \overline{bc}$ must be divisible by 8; hence $ \overline{bc} \equal{} 24,64,88\Rightarrow 4bc \equal{} 32,96,256$, remembering that none of the digits can be $ 0$. but in each of these cases, $ \overline{abc}$ must be divisible by $ 64$ (since a is even), from which we can deduce from some units digit figuring that the number is $ 64k$ for $ k \equal{} 1,2,4,7,11,12$, and we can easily verify none of these work; hence $ a$ is odd. this means $ \overline{bc}\equiv 4\pmod{8}$, from which we find $ \overline{bc} \equal{} 28,44,84\Rightarrow bc \equal{} 16,16,32$. We already found that $ 6(64) \equal{} 384$ earlier, and so we're done. I guess we could check the others in a similar fashion to be rigorous.
[/hide]


calculus integration geometry vector inequalities probability

Hey everyone. I'm in AP calc and I'm doing an essay on the development and theory behind the Riemann integral. I'm having a bit of trouble with my conclusion, though.. I was wondering if anyone could think of some important uses of integrals? I've got aerodynamics, and Einstein's theory of general relativity.. but I need a few more examples! At this point, I don't even care if they're Riemann integrals or not, just integrals.. any thoughts are greatly appreciated!

AIM -- jjflyhighx
e-mail: juli.jones@gmail.com

field equations, area, mechanics, volume, proving different things in algebra, einstein's field equations, wave equations, quantum mechanics and complex integrals, n-dimentional geometry, vectors, evaluating special sums, fourier series, electricity, chemistry, planetary motion (kepler's law), error calculations, estemation of teh sum of series, compression of spacetime (relativity), Fermat's principle, inequalities (i.e. Cauchy Schwarz and AM-GM)rocket science, thermodynamics, probabiltiy, newtonian mechanics, gravitational force, etc.

Masoud Zargar

most mechanics can be done with integrals...also a classic thing is that one of the things that spurred newton to work on calculus was that if you have a glass, and make a hole, how long does it take for the water to leave, and this is difficult because the pressure changes along with the water level, so it is not a constant acceleration, (nonconstant acceleration problems are a central thing in calculus)

this would be an interesting anecdote to include...but you could say that integrals are central to physics in general...good luck on your essay, i would cry if i had to write essays in math, english has enough essays as it is

also this should go in another forum...this is not really the place for calculus

Both of you-- thanks soooo much!

Altheman-- the related rates tidbit about Newton actually works PERFECTLY in another history essay i'm writing, so thanks for that. haha as for crying.. i was the one who picked the topic; i actually find this super-interesting, and thoroughly enjoyed writing the theory part. :)

boxedexe-- you are my hero as well :)