geometry
trapezoid
symmetry
conics
geometric transformation
reflection
trigonometry

Let C (O, R) be a circle and PQ a chord of this
circle. Let M be the midpoint of PQ and A, C points on the circle C
such that the chord AC intersects PQ at M. The quadrilateral ABCD is
a trapezium inscribed in the circle C and AB, CD are parallel to PQ.
Prove that the intersection point X of the lines AD, BC does not
depend on the choice of A on the circle C.

$X$ is the pole of $MT$, where $M = AC \cap BD$ (easy to prove from symmetry) and $T=AB \cap CD$. But $AB \parallel CD$, so $T$ is a point at infinity that also lies on $PQ$. Therefore $X$ is the pole of $PQ$. Thus, $X$ is a fixed point :) (it is the intersection of the tangents to the circle at $P$ and $Q$).

Excellent solution but you can prove it without poles?

[quote="silouan"][quote="perfect_radio"]$X$ is the pole of $MT$, where $M = AC \cap BD$ (easy to prove from symmetry) [/quote]
I cannot prove this please prove it.[/quote]

If you are talking about $M = AC \cap BD$: the isosceles trapezium $ABCD$ is symmetric with respect to $OM$. Thus, $AC$ is symmetric to $BD$ wrt $OM$. Since $M \in AC$, and the symmetric of $M$ is $M$, $M \in BD$.

If you are talking about the other thing (with poles), Darij gave a link in another thread.

[quote="perfect_radio"]If you are talking about the other thing (with poles), Darij gave a link in another thread.[/quote]
Where?

Here http://www.cut-the-knot.org/cgi-bin/dcforum/ctk.cgi?az=read_count&om=390&forum=DCForumID6 ( Theorem 1 ).

Use that theorem twice and obtain this consequence: If $A,B,C,D$ are concyclic, $T=AB \cap CD$, $U=AC \cap BD$, $V=AD \cap BC$, then $T$ is the pole of $UV$. (question: $A,B,C,D$ can be on any conic, not just on a circle?)

As for a elementary soln, I can't think of any. Another projective solution would involve the fact that the (i hope projective) map $A \mapsto D$ is an involution, hence $AD$ passes through a fixed point, but I guess it's more complicated than the one with polars.

Of course, there is an elementary solution, although the problem trivially follows from poles and polars.

[color=blue][b]Problem.[/b] Let k be a circle with center O, let P and Q be two points on this circle, and let M be the midpoint of the segment PQ. Also, let A, B, C and D be four points on the circle k such that AB || CD || PQ and such that the line AC passes through the point M. Let G be the point of intersection of the lines AD and BC. Prove that the position of this point G depends only of the circle k and of the points P and Q, but not of the points A, B, C and D.[/color]

[i]Solution.[/i] The segments AB, CD and PQ are parallel to each other; thus, their perpendicular bisectors are also parallel to each other. On the other hand, these perpendicular bisectors all pass through the point O, since the segments AB, CD and PQ are chords in the circle k whose center is O (and the perpendicular bisector of a chord of a circle always passes through the center of the circle). Now, if three parallel lines pass through one and the same point, then they must coincide, i. e. they are one and the same line; hence, the perpendicular bisectors of the segments AB, CD and PQ are one and the same line. This line passes through the point O (as we just saw) and through the point M (since it is the perpendicular bisector of the segment PQ and thus passes through the midpoint M of this segment). Hence, the perpendicular bisectors of the segments AB, CD and PQ all coincide with the line OM. Hence, the reflection with respect to the line OM maps the points A, B, C, D, P and Q to the points B, A, D, C, Q and P, respectively. In particular, it follows that this reflection maps the line AC to the line BD, and thus the lines AC and BD intersect on the axis of reflection OM. In other words, the lines AC, BD and OM concur at one point; this point must be M (since it lies on the lines AC and OM). Thus, we see that the point M lies on the line BD.

On the other hand, the reflection with respect to the line OM maps the line AD to the line BC (since it maps A to B and D to C). Hence, the lines AD and BC intersect on the line OM. In other words, the point of intersection G of the lines AD and BC lies on the line OM.

Now, we will use directed angles modulo 180°. Since the line OM is the perpendicular bisector of the segment AB, we have $OM\perp AB$, so that < (OM; AB) = 90°. On the other hand, AB || CD yields < (AB; AD) = < (CD; AD). Thus,

  < OGA = < (OM; AD) = < (OM; AB) + < (AB; AD) = 90° + < (CD; AD) = 90° + < CDA.

On the other hand, since the point O is the center of the circle k which passes through the points A, C and D, the central angle theorem for directed angles modulo 180° yields < OAC = 90° - < CDA, and thus - < OAC = - 90° + < CDA = 90° + < CDA (since - 90° = 90°, as we are working with directed angles modulo 180°). Hence, < OGA = - < OAC. In other words, < OGA = - < OAM. Also, trivially < GOA = - < AOM. Hence, the triangles OGA and OAM are inversely similar, and thus OG : OA = OA : OM, so that $OG=\frac{OA^2}{OM}$. Since the points O and M and the length OA depend only of the circle k and of the points P and Q (in fact, the point O is the center of the circle k, the point M is the midpoint of the segment PQ, and the length OA is the radius of the circle k), this yields that the position of the point G on the line OM depends only of the circle k and of the points P and Q. And the problem is solved.

  darij

Thank you very very much . Your solution is extremely good

M is the polar of point X.then it doesnt mather where the point A is. :D

[quote="ashegh"]M is the polar of point X.then it doesnt mather where the point A is.
[/quote]
you are wrong I think.
I solved yesterday this problem with analitical geometry with vectors. Does anyone  want to see this solution?

Hm, I don't think many people would be interested to see a solution with analytical geometry (sine that is probably the ugliest possible way to do it) but you can post it anyway if you want :)

Yes I agree with you, but If anyone interest....to see this solution I will put it

[b]A solution without pole and polar-line.[/b] I note the intersections $X\in AD\cap BC,\ R\in AD\cap PQ,\ S\in BC\cap PQ$. From the relations $\frac {RA}{RD}=\frac {XA}{XD}=\frac {AB}{CD}$ results that the points $X,R$ are conjugate w.r.t. the points $A,D\ \ [\Longleftrightarrow $ (A,D,R,X) is a harmonical division$]$. Analogous the points $X,S$ are conjugate w.r.t. the points $B,C$.

[b]Lemma. [/b]Let $w=C(O,r)$ be a fixed circle, a fixed point A for which $OA>r$ and a mobile line $d,\ A\in d$, for which $\{M,N\}=d\cap w$. Then the geometrical locus of the point $L\in (MN)$ for which $\frac {LM}{LN}=\frac {AM}{AN}$ is the segment $[CD]$, where $C,D\in w,\ A\in CC\cap DD$.
[i]Remark. [/i]The line $CD$ is the [u][i]polar[/i][/u] of the point $A$ w.r.t. the circle $w$ and the point $A$ is the [u][i]pole[/i][/u] of the line $CD$ w.r.t the circle $w$.

[u]A solution.[/u] I note the midpoint $T$ of the segment $[AL]$. From the relation $TA^2=TM\cdot TN$ results that the point $T$ belongs to the radical axis of the circles $w$ and $a(tom)=C(A,0)$, a.s.o.

\[.\ \ A.S.O\ \ .\]

siluan i think im not wrong here is my proof:
its clear that O and M and X  are in a line .lets PQ cut AD  in point Y.
and let PQ cut XC in point Z.
we have a theorem in harmonic wich says:XAYD and XBZC are harmonic points.
let XM intersect circle (C) in points K and L.
again we have :(XKML ).i mean X and K and M and L are harmonic.
look at point M.we know its stable.and we know K and L are stable too.because KL is perpendicular to PQ.
then if three points of four points are stable,and the four points are harmonic,it means that X should be stable too.
if (XKML) we know thatOM.OX=OK.OK
and it means that PQ is the polar of point X. :D .

Sorry your solution is OK I think. I said before that it was wrong because you have post only your result

yes u are right dear siluan the mistake was from me :( .

siluan ,i want to see u solution ,because i think it will be so nice .then post it plz.

My solution for ashegh.
Let an right-angled system of co-ordinates such that the axis $y'y$ is perpendicular at the midpoint of the chord $PQ$.Now without lose of genarality we assume that the circle has radius $r=1$.
The trapezoid $ABCD$ is isosceles and because $AB//PQ//CD$ the sengement $OM$ with be perpendicular at the midpoint of $AB$ and $CD$. Also the trapezoid $ABCD$ has axis of symmetry the axis $y'y$. So the symmetric of $AC$ is $BD$, but $M\in AC$ and $M\in ME$ so $BD$ passes from $M$.
Now let $G$ the pole of $PQ$. $G\in y'y$ so $G(0,y_0)$ and let $A(k,l)$,$ B(-k,l)$,$D(m,n)$,$C(-m,n)$.
We also have that $(PQ):$ $0x+y_0y=1$ so $PQ:$ $y=\frac{1}{y_0}$ so $M(0,\frac{1}{y_o})$
We have $k^2+l^2=1$ and $m^2+n^2=1$, and $\vec{BM}=r\vec{MD}$ <=>
$(k,\frac{1}{y_0}-l)=r(m,n-\frac{1}{y_0})$ <=> $k=rm$ and $y_0=\frac{m+k}{lm+kn}$
we want to prove that $AD,BC$ passe from $G$.So we will find $AD$.
$AD:$ $y-l=\frac{l-n}{k-m}(x-k)$. this for $x=0$ and $y=y_0$ turrns to:
$y_0-l=-k\frac{l-n}{k-m}$ <=>  $(k-m)y_0-lk+lm=-kl+kn $ <=> $(k-m)y_0=kn-lm$ <=> $(k-m)\frac{m+k}{lm+kn}=kn-lm $ <=>....<=>$k^2m^2-m^2k^2=0$ which is true. So $G\in (AD)$.
Now from symmetry $G\in (BC)$ so $AD,BC$ passes from the same fixed  point $G$ which is the pole of $PQ$ and we are done.

thanks alot siluan.
i didnt know how to solve problem by using analytic gemotry.and here is a good lesson for me.


search
superior algebra
superior algebra theorems

Hello, everybody. I have collected over 200 problems on algebra, and published some little book on my homepage. Here is the [url=http://omidhatami.googlepages.com/book.pdf]link[/url]. I would greatly like to hear about yours opinions or any errors you find.

Nice book. Maybe you should add MathLinks as a reference at the end?  :D

Yes, you're right. I forgot to do this. I will do it right now. :)

[quote="kidmath"]It's great. But I think I will be better if it has solutions for all exercises, because readers are not always good at algebra, solutions will help them when they really meet difficulties in solving exercises.

But that's a great book. Thanks alots.[/quote]

If someone has difficulty with a problem, they could just post it on AoPS :)

[quote="calc rulz"]If someone has difficulty with a problem, they could just post it on AoPS :)[/quote]Or even quicker: search for it. Most likely someone has already posted it ;)

I have picked up some of the problems from Mathlinks, but not all of them. You can not find some of the problems on Mathlinks.

Does anybody know what [b]IMS[/b] stand for?  :oops:

Thanks a lot for replying, dear Sir. The new inquiry is: how did those IMS-labeled problems reach you? Is there some type of publication sponsored by the IMS where one can find them?  :D Really curious about it... Thanks in advance for taking the time to reply.

[quote="Valentin Vornicu"]Or even quicker: search for it. Most likely someone has already posted it ;)[/quote]

A one-liner that must definitely be included in the AoPS T-shirt...  :lol

[quote="Omid Hatami"]Problems of IMS are released each year in the in the Internet only in "Persian". I translated the problems in English. Problems of a few years is available on Mathlinks.[/quote]

It is always good to know...  :lol:

Downloaded your file already. [b]Great compilation[/b], dear Sir.

I have several suggestions for the Second Edition of the text. Is there an e-mail address where I can reach you?

All the Best Regards,
José

I could not open the file....Please post another link with  him!!!Thanks!!!

[color=blue][b]Why don't you try to download it from the following mirror [url=http://ndikandi.utm.mx/~lm2002070425/archivo.html]site[/url]?[/b][/color]  :D

[color=blue][b]Best of Luck,[/b][/color]

OK., here is another link:

http://math.sharif.edu/~omidhatami/pdfs/book.pdf

[i]P.S You can find my e-mail in my Mathlinks profile.[/i]

Wow really really nice and perfect collection of Algebra Problems... Thanks Omid :)


complex analysis
function
algebra
domain
trigonometry
integration
calculus

Assume $ G$ be a star-shaped domain in $ \mathbb{C}$,  $ f$ and $ g$ are holomorphic functions on $ G$ such that $ f^2\plus{}g^2\equal{}1$. Prove that there exists a holomorphic function $ \varphi(z)$ such that $ f(z)\equal{}\cos\varphi(z)$ and $ g(z)\equal{}\sin\varphi(z)$.

Note that $ f(z) \plus{} ig(z)$ is never zero, since $ (f \plus{} ig)(f \minus{} ig) \equal{} f^2 \plus{} g^2 \equal{} 1\ne 0.$

Let $ z_0$ be the point with respect to which $ G$ is star-shaped, and find $ w$ such that $ e^{iw} \equal{} f(z_0) \plus{} ig(z_0).$

Now, define $ \varphi(z) \equal{} w \minus{} i\int_{z_0}^z\frac {f'(\zeta) \plus{} ig'(\zeta)}{f(\zeta) \plus{} ig(\zeta)}\,d\zeta.$

The integral is well-defined since the region is star-shaped, and we have that $ \varphi' \equal{} \minus{} i\,\frac {f' \plus{} ig'}{f \plus{} ig}.$

We now claim that $ e^{i\varphi(z)} \equal{} f(z) \plus{} ig(z)$ for all $ z\in G.$

To this end, let $ h(z) \equal{} (f(z) \plus{} ig(z))e^{ \minus{} i\varphi(z)}.$

$ h' \equal{} (f' \plus{} ig')e^{ \minus{} i\varphi} \plus{} (f \plus{} ig)( \minus{} i\varphi')e^{ \minus{} i\varphi}$

$ \equal{} e^{ \minus{}i \varphi}\left[(f' \plus{} g') \plus{} (f \plus{} ig)( \minus{} i)( \minus{} i)\frac {f' \plus{} ig'}{f \plus{} ig}\right] \equal{} 0.$

Hnece, $ h(z)$ is constant, and $ h(z_0) \equal{} (f(z_0) \plus{} ig(z_0))e^{ \minus{} iw} \equal{} 1.$ So $ h(z) \equal{} 1$ on $ G.$

Now let's finish this off. Starting with $ e^{i\varphi} \equal{} f \plus{} ig,$ we have $ e^{ \minus{} i\varphi} \equal{} \frac1{f \plus{} ig} \equal{} \frac {f \minus{} ig}{f^2 \plus{} g^2} \equal{} f \minus{} ig.$

Hence $ \cos\varphi \equal{} \frac {e^{i\varphi} \plus{} e^{i\varphi}}{2} \equal{} \frac {(f \plus{} ig) \plus{} (f \minus{} ig)}{2} \equal{} f$

and $ \sin\varphi \equal{} \frac {e^{i\varphi} \minus{} e^{i\varphi}}{2i} \equal{} \frac {(f \plus{} ig) \minus{} (f \minus{} ig)}{2i} \equal{} g$

Dear Kent Merryfield,
I understand, very good solution, thank you so much  for your help.

Great solution by Kent Merryfield!
Here is my solution:

If one of $ f,g$ is constant then proposition is trivial. Assume that both of them are not constant.

$ 1^{st}$ case.   Assume that $ f^2(z)\neq 1$ for all $ z\in D$. Then $ \varphi(z) \equal{} \sqrt {1 \minus{} f^2(z)}$ is well-defined, analytic in $ D$. Then for any $ z_0\in D$ the function $ h(z): \equal{} \minus{} \displaystyle\int_{z_0}^{z}\frac {f'(\zeta) d\zeta}{\varphi(\zeta)}$ satisfies proposition of the problem.

$ 2^{nd}$ case. Assume that there exists $ z_0\in D$ such that $ f^2(z_0) \equal{} 1$. Then by the uniqueness theorem such kind of points are local finite in $ D$. W.L.O.G assume that $ f(z_0) \equal{} 1$. Then $ f(z) \minus{} 1 \equal{} (z \minus{} z_0)^kf_1(z)$, where $ f_1(z)$ is analytic with $ f_1(z_0)\neq 0$ and $ k$ is even. Then the function $ \frac {f'(\omega)}{\sqrt {1 \minus{} f^2(\omega)}}$ has only removable singularities and this function can be well-defined on $ D$.Then for any $ z_0$ with $ f^2(z_0)\neq 1$ the function $ h(z): \equal{} \minus{} \displaystyle\int_{z_0}^{z}\frac {f'(\omega) d\omega}{\sqrt {1 \minus{} f^2(\omega)}}$ satisfies proposition of the problem.  :wink:


geometry
circumcircle
conics
ellipse
geometry proposed

Dear everyone.

For a right-angled triangle ABC, let points D, E be on BC, AB respectively such that $ \frac {CD}{DB} \equal{} \frac {BE}{EA}$.

Prove the following.

(1) the circumcircle (of the triangle BDE) passes through a fixed point (call it F)
(2) $ \angle{EDF}$ is constant

And when D moves on BC, find the locus of the intersection point of AD and CE.

For the locus, please No analytic method  :blush:
Thanks. :)

The point F=[ac^2/(a^2+c^2),a^2c/(a^2+c^2)]

cos(D)=a/b

LG is an ellipse


linear algebra
matrix
superior algebra
superior algebra solved

Let K be a commutative field and Let Mn(K) be the ring of square matrices with elements in K and dimension n. Find all ring morphisms from Mn(K) to K.

There aren't many for n>1.

Let n>1.
Let f : Mn(K) -> K be a morphism.

If A is invertible and A' is its inverse, then 1=f(I)=f(AA')=f(A)f(A'), so f(A) is not 0 in K.

If A is nilpotent, then there exists m such as Am=0, so f(A)m=0, so f(A) is 0.
Because f is a morphism f(A)=0 for all A that are sums of nilpotent matrices.

The matrix
P=
0 1 0...0
0 0 1...0
............
0......0 1
1 0..... 0
is invertible and it is the sum of two nilpotent matrices
0 1 0...0
0 0 1...0
............
0..... 0 1
0 ....... 0
and
0.........0
0.........0
............
1 0......0

So f(T)=0. But T is invertible, so f(T) is invertible in K (a contradiction).

I think f(A)=0 for all A works just fine. It's a RING morphism, which means that we don't need f(I)=1.

It depends on your definition of a ring morphism. In most books, you will find the condition f(1)=1 stated in the definition.
The ring 0 is special and in some places it is not considered a field.

Well, it's not really that important, it's merely a detail. However, if a ring morphism generally has f(1)=1, then what about morphisms which don't have f((1)=1? How are those called? I think ring morphisms for which f(1)=1 are called unit ring morphisms (or something like that; in Romanian: "morfism unitar").

Anyway, thanks again for the solution. :)


function
inequalities
calculus
search
derivative
logarithms
real analysis

$ f(t)\equal{}\frac{1}{(t\plus{}1)^{\frac{1}{z}}}$, $ t\ge 0$, $ z\ge 1$.
let $ x_1,\dots,x_n>0$ reals und $ x_1\dots x_n\equal{}m^n$.
show that $ f(x_1)\plus{}\dots\plus{}f(x_n)\ge\min(1,nf(m))$

$ f^{\prime\prime}(t)=\frac{1}{z}(\frac{1}{z}+1)\frac{1}{(t+1)^{-\frac{1}{z}-2}}$

Hence for $ t \geq 0$ we have $ f^{\prime\prime}(t)>0$. Therefore $ f$ is convex function on $ \mathbb{R}^{+} \cup \{0\}$.

Therefore we apply Jensen`s inequality for convex functions:

\[ \frac{f(x_1)+\cdots+f(x_n)}{n}\geq f(\frac{x_1+\cdots+x_n}{n}) \geq f(\sqrt[n]{x_{1}\cdots x_{n}})=f(m)\]
where we used $ AG$ inequality.

Hence \[{ f(x_1)+\dots+f(x_n)}>nf(m).\]

@tmrfea:

$ f \left( \frac {x_1 \plus{} \dots \plus{} x_n}{n} \right) \le f( \sqrt [n]{x_1\dots x_n} )$

Jensen. Apply it to $ g(s)\equal{}f(e^s)\equal{}(e^s\plus{}1)^{\minus{}k}$, where $ 0< k\equal{}\frac1z\le 1$.

We have $ g''(s)\equal{}\minus{}k(\minus{}k\minus{}1)e^{2s}(e^s\plus{}1)^{\minus{}k\minus{}2}\minus{}ke^s(e^s\plus{}1)^{\minus{}k\minus{}1}$
$ \equal{}\left(\minus{}k(\minus{}k\minus{}1)e^{2s}\minus{}ke^s(e^s\plus{}1)\right)(e^s\plus{}1)^{k\minus{}2}\equal{} (k^2e^{2s}\minus{}ke^s)(e^s\plus{}1)^{k\minus{}2}$

This is positive where $ ke^s>1$ and negative where $ ke^s<1$. The extreme points with a possible minimum are where either all of the $ s_i$ are equal or where some $ s_i$ goes to $ \minus{}\infty$. The latter gives $ 1$ and the former $ nf(m)$.

It is more complicated than the simple convex case.

[quote="FelixD"]$ m,n$ are real numbers, integers...?[/quote]
$ n$ positive integer,
$ m$ positive real

[quote="jmerry"]

The extreme points with a possible minimum are where either all of the $ s_i$ are equal or where some $ s_i$ goes to $ \minus{} \infty$. The latter gives $ 1$ and the former $ nf(m)$.
[/quote]

may I ask you to write a more detailed explanation of this ?

I don't understand what do you mean by this ........ (obviously I know classical Jensen inequality)

Actually, just scrap the whole classical inequality mindset and work with the calculus approach to finding extrema. We search for extreme values of $ \sum g(s_i)$ under the constraint $ \sum s_i\equal{}c$. Using Lagrange multipliers, the only possible extreme points are those where the $ g'(s_i)$ are all equal and the boundaries in which all $ s_i\in \{\minus{}\infty,\infty\}$. From analysis of the second derivative, $ g'$ has a negative minimum somewhere on the real line, and increases monotonically to zero in both directions from there. If $ g'(s)\equal{}g'(t)$ for some unequal pair $ s<t$, these are the only two points with that value of the derivative, and any critical point featuring those two values has only those two values. Note that $ g''(s)<0<g''(t)$.
If each of those values appears at least twice, the critical point is a saddle point after adjusting for the constraint and can be ignored. If $ t$ appears exactly once and $ s$ more than once, the point is a maximum if $ |g''(s)|<\frac1{n\minus{}1}|g''(t)|$ and a saddle point otherwise. If $ s$ appears exactly once and $ t$ more than once, the point is a minimum if $ |g''(t)|<\frac1{n\minus{}1}|g''(s)|$ and a saddle point otherwise. If only $ s$ appears, it's a maximum. If only $ t$ appears, it's a minimum.

Based on this analysis, the possible minima are on the boundary (easily seen to have value 1), at the critical point $ s_1\equal{}s_2\equal{}\cdots\equal{}s_n\equal{}\ln m$ when $ m>1$, and at a critical point $ s_1\equal{}s<t\equal{}s_2\equal{}\cdots\equal{}s_n$.

That last possibility needs more analysis, which my intuitive look the first time missed. These critical points do exist in many cases, but they're not minimums. Note that $ |g'(\minus{}x)|\le|g'(x)|$ if $ x>0$ with equality when $ k\equal{}1$, so $ |s|\le|t|$ in any such pair. Now consider the $ n\equal{}2$ case; if the constraint is $ s_1\plus{}s_2\equal{}s\plus{}t$, we get a minimum at $ (\frac{s\plus{}t}{2},\frac{s\plus{}t}{2})$ (as long as $ k<1$), and there can't be a second minimum, since there is no critical point inbetween to be a maximum. This makes $ |g''(s)|>|g''(t)|$, and this pair never leads to an extra minimum for larger $ n$.


number theory
greatest common divisor
algebra
polynomial
relatively prime
Rational Root Theorem

Here's a challenge!

(a) Show that a/b + b/c + c/a = 4 has no solutions in positive integers a,b,c.

(b) In general, when does a/b + b/c+ c/a = N have solutions in positive integers a,b,c?

This problem seems particularly difficult. Here's one way that I did it for a special case where gcd (a,b) = 1. Maybe someone can extend it without using this condition.

--------------------------------------------------------------------------------------
If gcd (a,b,c) = h, then h can be divided out of each numerator and denominator, and give a smaller solution. So w/o loss of generality assume that gcd (a,b,c) = 1. Now make the further assumtion that gcd (a,b) = 1, although this does not necessarily follow from gcd (a,b,c) = 1.

Multiplying the original equation through by abc we have:
a^2c + b^2a + c^2b = nabc

Looking at this equation mod a, mod b, and mod c respectively, we get the following relations:

c^2b = 0 (mod a)
a^2c = 0 (mod b)
b^2a = 0 (mod c)

Now using the fact that gcd (a,b) = 1, we must have c^2 = 0 (mod a) and c = 0 (mod b). Combined with the third congruence this means that c must equal either ab or ab^2. Can you see why? (think about unique factorization).

Then there are two cases to consider.

[u]Case c = ab:[/u]
a/b + b/c + c/a = a/b + b/(ab) + (ab)/a
= a/b + 1/a + b
And hence a/b + 1/a must be an integer. Or in other words, (a^2 + b) = 0 (mod ab), where a and b have gcd = 1. This can only happen when a = 1 and b = 1, and hence c  = 1.

[u]Case c = ab^2:[/u]
a/b + b/c + c/a = a/b + b/(ab^2) + (ab^2)/a
= a/b + 1/(ab) + b^2
And hence a/b + 1/(ab) is an integer. That means (a^2b + b) = 0 (mod ab). Factoring out the b that gives (a^2 + 1) = 0 (mod a). And this is only possible when a = 1. That then means that b = 1 or 2, and that c = 1 or 4, respectively.

Therefore this gives the only solutions as (a,b,c) = (1.1.1) or (1,2,4), or multiples or cyclic permutations of these. And that would make only possible values of n as 3 and 5.

Hmm. I *think* I found an arguement that reduces all the cases to gcd (a,b) = 1.

Start out with the three congruences, c^2b = 0 (mod a), a^2c = 0 (mod b), and b^2a = 0 (mod c).

Then consider the unique factorizations of a, b, and c. Since c divides b^2a, all of it's prime factors must come from a or b. So assume c = (p_1^s_1*p_2^s_2...p_n^s_n)*(q_1^t_1*q_2^t_2...q_m^t_m), for primes p_i and q_i. Let the p_i's be the prime factors coming from a and the q_i's be the prime factors coming from b. Note that there cannot be any prime factors coming from both a and b, because gcd (a,b,c) = 1.

Call the product the p_i powers p and the product of the q_i powers q. Then c = pq, where p and q are relatively prime.

Similarly. a splits into powers of prime factors coming from b and c, and b splits into powers of prime factors coming from a and c. Call the product of the powers in common with c, r. Then a = pr and b = qr, again where p and r and q and r are relatively prime. How do I know the p's, q's and r's are the same here?

Draw a Venn Diagram for the prime factors of a, b, and c. There cannot be anything in the intersection of all three circles because gcd (a,b,c) = 1. There also can't be anything in the non-intersecting parts of the circles or the congruences won't work (i.e. a, b, and c won't split into powers of prime factors coming from the other two). So all the prime factors must come from the pairwise intersections of the circles. Call the product of all those prime power factors in the pairwise intersection of a and b, r. Those between b and c, q. And those between a and c, p. It is clear to see that p, q, and r are well defined, and all pairwise relatively prime.

So then a/b + b/c + c/a = (pr)/(qr) + (qr)/(pq) + (pq)/(pr)
= p/q + r/p + q/r. So any solution where gcd (a,b,c) = 1 reduces to a solution where gcd (p,q) = 1, and the arguement of my post above shows that n must be 3 or 5.

Yes that's essentially how I attacked the problem.  Now I wonder, what integers are presentible as x + y + z if xyz = 1?  where x,y,z are rationals.  Or for that matter, is there a least integer n such that for any integer k, there exist rational numbers a_1, a_2, ..., a_n such that a_1 * a_2 * ... * a_n = 1 and a_1 + a_2 + ... + a_n = k.

I haven't thought 'bout the above problem yet so consequently I wouldn't know how difficult it is.  Intuitively there seems to be enough possibilities for x,y,z to make all integers just with n = 3.  Might turn out to be a boring problem.

Nice job for your solution.

I'm not sure I understand well gauss202 solution but n=3 or 5 is not the only solution :

2/12 + 12/9 + 9/2 = 6
12/63 + 63/98 + 98/12 = 9
2/36 + 36/81 + 81/2 = 41

[quote="gauss202"]Looking at this equation mod a, mod b, and mod c respectively, we get the following relations:

c^2b = 0 (mod a)
a^2c = 0 (mod b)
b^2a = 0 (mod c)

Now using the fact that gcd (a,b) = 1, we must have c^2 = 0 (mod a) and c = 0 (mod b). [i]Combined with the third congruence this means that c must equal either ab or ab^2.[/i]
[/quote]

Doah! you're right. This last statement is not necessarily true. I think the other parts are right, that you can reduce the problem to one where gcd (a,b) = 1. But it's definitely not true that c has to equal either ab or ab^2.

Hmm. back to the drawing board.

[quote="ComplexZeta"]How did you find those, i/3?[/quote]

Brute force 10 lines computer program !

Here are some others (a,b,c,N) :

225, 81, 5, 19
350, 196, 5, 41
588, 9, 14, 66
28, 637, 338, 14
175, 882, 1620, 10
52, 3549, 882, 21
28, 1323, 1458, 53

I suspect there are many possible values for N.

Hmm. scratch that second post I made also. That's wrong too. i/3 can you post the code that you used to find these?

However my first post does work when a and b are distinct primes. So whatever the solutions are, at most one of a, b, c can be prime.

[quote="gauss202"]i/3 can you post the code that you used to find these?[/quote]

Here it is.
Surely you will be very disapointed, it's a brute force, blind and unoptimized, C++ program. Just a quick and dirty try to see how things were going on. It needs a few minutes on my PII to complete.

#include <iostream>
void main() {
   const long long int NB = 5000;
   for (long long int a = 1; a < NB; a++) {
      if ((a % 20) == 0) cout << "a:" << a << endl;
      for (long long int b = 1; b < NB; b++) {
         for (long long int c = 1; c < NB; c++) {
            const long long int v = a*a*c + b*b*a + c*c*b;
            const long long int m = a*b*c;
            if ((v % m) == 0) {
               const long long int N = v/m;
               if (N == 3) continue;
               if (N == 5) continue;

               cout << a << ' ' << b << ' ' << c << ' ' << N << endl;
            }
         }
      }
   }
}

The more I worked on this the more I suspected this was not an elementary problem. So I posted it over on the sci.math newsgroup and got some interesting responses.

--------------------------------------------------------------------------------------
Subject: Re: Problem: a/b + b/c + c/a = n
Author: Dan Hoey <haoyuep@aol.com>
Date Posted: Jul 13 2003 9:18:56:000AM

On 10 Jul 2003, Marcus wrote:

> I'm stuck on this problem that was posed on another
> message board.  The problem is to find which positive
> integers n can be the result of a/b + b/c + c/a, where
> a, b, and c are positive integers.

>Marcus

I don't know why this question hasn't shown up on Usenet or
Google, but that may be why you've gotten so few answers.  I
think the problem is well-known (though not to me).

Anyway, an approach is to assume gcd(a,b,c)=1, though a, b,
and c may not be pairwise relatively prime.  Then it turns
out that we can find integers t,u,v that are pairwise
relatively prime, and for which a=t^2 u, b=u^2 v, c=v^2 t.
The same (t,u,v) also yields a triple a=t^2 v, b=v^2 u,
c=u^2 t, which is different if the t,u,v are different
(which can only occur with (1,1,1) and (1,1,2)).

Anyway, n=(t^3+u^3+v^3)/(tuv).  My program tries
u=1,2,...,50 (though it could easily go further on a better
Lisp).  For each u it tries all 1 <= t <= u with
gcd(t,u)=1.  For each t and u it tries all divisors
v|t^3+u^3 with v >= u.  The result (sorted by n) is:

      n   t   u     v
      3   1   1     1
      5   1   1     2
      6   1   2     3
      9   2   3     7
     10   5   7    18
     13   9  13    38
     14   2   7    13
     17   5  18    37
     18  13  42    95
     19   1   5     9
     21   2  13    21
     26   9  38    91
     29  27  43   182
     30   2  21    31
     41   2  31    43
     41   1   5    14
     41   1   2     9
     51   9  13    77
     53   2   7    27
     54   2  43    57
     66   1   3    14
     83   5   9    61
    106   1  35    54
    149   1  14    45
    154   2  13    63
    161  11  38   259
    174   5   7    78
    178   2  27    97
    195   7  15   143
    250   2   9    67
    261   3   7    74
    269   1  14    61
    323   9  49   377
    326   5  14   151
    451  23  31   567
    478  13  23   378
    633   3  14   163
    978  33  43  1178
    978   2   7   117
   1011   7  11   279
   1410   3  22   305
   1658  19  26   905
   1718   9  19   542
   1769   2   5   133
   1875  15  19   731
   1893   3  43   494
   1914  31  45  1634
   2163   1  49   325
   2309   2  13   245
   2369   1   9   146
   2681   1  49   362
   4018  14  45  1591
   5246  13  31  1454
   5430  19  25  1606
   6638  26  37  2527
   7061   2  35   703
   8346  31  37  3094
  10995   3  23   871
  12105   7  18  1235
  13971   3  11   679
   14803   1   9   365
   15722   7  38  2045
   17859  37  45  5453
   18014   1  35   794
   20778   2  13   735
   21529   1  14   549
   24761   7  31  2318
   31130  14  43  4329
   33081   3  31  1754
   39290   7  45  3518
   59802   1  14   915
   82949   5  13  2322
   84178   9  19  3794
   92454    5  39  4246
  112901   5  34  4381
  118366   2  31  2709
  187001   2  13  2205
  191974   9  46  8915
  196026   7  26  5973
  203654   2  43  4185
  941491   1  45  6509
2000041   1  29 24390

Dan

Subject: Re: Problem: a/b + b/c + c/a = n
Author: Hugo Pfoertner <nothing@abouthugo.de>
Date Posted: Jul 13 2003 5:42:12:000PM

On 13 Jul 2003, Dan Hoey wrote:
>On 10 Jul 2003, Marcus wrote:
>
>> I'm stuck on this problem that was posed on another
>> message board.  The problem is to find which positive
>> integers n can be the result of a/b + b/c + c/a, where
>> a, b, and c are positive integers.
>
>>Marcus
>
>I don't know why this question hasn't shown up on Usenet or
>Google, but that may be why you've gotten so few answers.  I
>think the problem is well-known (though not to me).

The thread seems to be invisible on Usenet and Google.

>
>Anyway, an approach is to assume gcd(a,b,c)=1, though a, b,
>and c may not be pairwise relatively prime.  Then it turns
>out that we can find integers t,u,v that are pairwise
>relatively prime, and for which a=t^2 u, b=u^2 v, c=v^2 t.
>The same (t,u,v) also yields a triple a=t^2 v, b=v^2 u,
>c=u^2 t, which is different if the t,u,v are different
>(which can only occur with (1,1,1) and (1,1,2)).
>
>Anyway, n=(t^3+u^3+v^3)/(tuv).  My program tries
>u=1,2,...,50 (though it could easily go further on a better
>Lisp).  For each u it tries all 1 <= t <= u with
>gcd(t,u)=1.  For each t and u it tries all divisors
>v|t^3+u^3 with v >= u.  The result (sorted by n) is:
>
>      n   t   u     v
>      3   1   1     1
>      5   1   1     2
>      6   1   2     3
>      9   2   3     7
>     10   5   7    18
>     13   9  13    38
>     14   2   7    13
>     17   5  18    37
>     18  13  42    95
>     19   1   5     9
>     21   2  13    21
>     26   9  38    91
>     29  27  43   182
>     30   2  21    31

[...]

I had cross-posted the problem to
The SeqFan Mailing List
http://www.ccr.jussieu.fr/gmpib/seqfan/seqfan.html

There were various contributions (e.g. by Hans Havermann,
Jud McCranie, Dean Hickerson, Olivier Gerard) confirming
the terms given above.

There is already a corresponding sequence in the
On-Line Encyclopedia of Integer Sequences:
http://www.research.att.com/projects/OEIS?Anum=A072716

This sequence will probably get an update during the next
few days, with a conjectured continuation
...26,29,30,38,41,51,53,54,57,66,...

A yet unsolved problem seems to be how far up you have to
go with t,u,v to be sure that there are no further terms
below some given n. To give an example, Dean Hickerson found
{t,u,v,n}={133, 2502, 4607, 74}

>
>Dan

Hugo Pfoertner

Subject: Re: a/b+b/c+c/a=n
Author: Dave Rusin <rusin@vesuvius.math.niu.edu>
Date Posted: Jul 14 2003 3:02:40:000PM

In article <3F11C053.A83DB1A6@abouthugo.de>,
Hugo Pfoertner  <nothing@abouthugo.de> wrote:

>>On 10 Jul 2003, Marcus wrote:
>>> I'm stuck on this problem that was posed on another
>>> message board.  The problem is to find which positive
>>> integers n can be the result of a/b + b/c + c/a, where
>>> a, b, and c are positive integers.

>>Anyway, an approach is to assume gcd(a,b,c)=1, though a, b,
>>and c may not be pairwise relatively prime.  Then it turns
>>out that we can find integers t,u,v that are pairwise
>>relatively prime, and for which a=t^2 u, b=u^2 v, c=v^2 t.

>>Anyway, n=(t^3+u^3+v^3)/(tuv).

Perhaps I'm simply retreading ground covered in another
conversation about this problem but...

That equation describes (for each  n)  an elliptic curve.
You want a (positive) integer solution on the curve.
Equivalently, you can set, say, v=1  and look for rational solutions to
    x^3+y^3+1 = n xy
and then scale back to integers.

You can write the elliptic curve in canonical form if you like;
a transformation which works is
   x = -1/2*(X*n^2-2*Y)/n/(3*X-27*n+n^4)
   y = -1/2*(X*n^2+2*Y)/n/(3*X-27*n+n^4)
giving the form
   Y^2 = X^3 + 1/4*n*(n^3-108)*X^2 - 9*n^2*(n-3)*(n^2+3*n+9)*X
                 -n^3*(n^2+3*n+9)^2*(n-3)^2
We can then check for the existence of rational points. There is
always torsion of order 3 corresponding to solutions with
{t,u,v}={1,-1,0}.

The curve is singular when n=3.
It has rank 0 but additional torsion when n=5.
It has rank 0 (no nontrivial rational points) when n=1,2,4,7,8,11,12,...
It has rank 1 (infinitely many nontrivial solutions) when n=6,9,10,13,14, ...
These are in agreement with the table posted, which shows solutions:

>>      n   t   u     v
>>      3   1   1     1
>>      5   1   1     2
>>      6   1   2     3
>>      9   2   3     7
>>     10   5   7    18
>>     13   9  13    38
>>     14   2   7    13
>>     17   5  18    37

This skips over are the cases n=15 and n=16, which also have rank 1.
For example, for n=15, we have the generator  (X,Y)=(11160, 1674000),
which gives x = 1/3, y = -7/3, z = 1 (and then t=1,u=-7,v=3).
You wanted _positive_ t,u,v, which means we need to find a multiple
of the generator [*] which makes  x>0 and y>0. I believe there
are no such points. (All multiples of a point on the unbounded portion
of an elliptic curve in normal form will also be in the unbounded portion.)
The same thing happens for n=16,20,31,35,36,40,44,...

The case n=41 is the first with rank 2, which doesn't mean very much
in this setting except that there will be "more than the usual"
number of solutions (t,u,v). (The next rank-2 case is n=69.)

The behaviour for all the values of n < 67 mentioned in the previous post
follows the pattern of one of the preceding paragraphs, except that
I was unable to verify that there were no solutions when n=62 and n=64
(I suppose the ranks are one but my software found no generator for the
curves in the little time I allotted, so I couldn't check to see whether
the generator lay on the bounded part of the curve or not.)

dave

[*] Torsion doesn't help; adding one of the elements of order 3 just
permutes t,u,v cyclically. And taking negatives on the elliptic curve
corresponds to performing a permutation of odd sign on these three.

Ooh

PROOF . GENERAL case too!

[hide]

Let m = a/b

     n = b/c

     p =  c/a.

Then m+n+p = N

        mn + np + mp = N

        mnp = 1

then the polynomial in x:

P(x) = (x-m)(x-n)(x-p) = x^3 - Nx^2 + Nx - 1.

But m, n, p are rational. By the rational root theorem, m, n, and p are all +/- 1, since the polynomial is monic.

Therefore a=b=c yields that N = 3 works.

suppose not all three are equal. Then two must be equal, since our choices for a, b, c are +/- k for some k in the integers without 0. WLOG a=b=-c.

Then N = -1.

So N = 3, N = -1 are the only solutions

[/hide]

Hmm...seems very simple and elegant, but I don't think that's the answer.  First of all, tons of mathematicians have looked at this and are stumped, or at least don't completely know the answer - not that that's to say that you couldn't have arrived at a proof for it, but I don't think it's likely.

Besides, I don't think N can equal -1, since a, b, and c are all positive integers - perhaps unless you're using modular arithmetic.

Finally, there are plenty of examples of a,b,c that work out there that your proof doesn't arrive at.  Gauss202 himself found that n could be 5, and the people on sci.math have found countless others, which are currently being added to the Encyclopedia of Integer Sequences.

I think its the mn+np+pm bit which is not equal to N.

oh yeah. lets hear it for milid dyslexia!

either that or me not checking my work when something looked like it might work... heh.


geometry
MATHCOUNTS
area of a triangle
Heron\u0027s formula

Oh yeah, there's $\le$ signs, not $<$ signs. That makes it a lot easier: $A=\frac{\sqrt{15}}{4}$ (Heron's Formula)

[quote="236factorial"]What is the greatest area a triangle with sides a, b, and c could have if:

0 <= a <= 1 <= b <= 2 <= c <= 3[/quote]
[hide]
Assuming you mean integer values..
a=1
b=2
c=2
Apply Heron's Formula
$\sqrt{2.5(2.5-1)(2.5-2)(2.5-2)}$
$=\sqrt{2.5(1.5)(.5)(.5)}$
$=\sqrt{2.5(1.5)(.5)(.5)}$
$=\sqrt{.9375}$
$~0.9682$
[/hide]

why  :sqrt:6/4?
the problem never said about integer values,

for it to have the greatest area, the side would all have to be 3

so it is going to be an equalateral triangle.

$\frac{\sqrt{3}3^2}{4}=\frac{9\sqrt{3}}{4}$

I think ccy is right, I got .968 too.

You can't have all 3s, a has to be no more than one, and b has to be no more than 2.

I tried that once too (1,2,2), but I have no answer key

The maximum should occur at:

$-(a^4+b^4+c^4)+2a^2b^2+2a^2c^2+2b^2c^2$

(I derived that from Heron's Formula)

There has got to be an easier way.

well from what I know, the site gives you a month to do 8 or 9 problems like these.
not really mathcounts

It's from the 1998 ABACUS International Math Competition for 7th and 8th graders. You can view the rest of the problems [url=http://www.gcschool.org/pages/program/abacus/8_2_98.html]here.[/url]

They should have the solutions posted up sometime.

a=1, b=2, c=2 isn't a triangle. Triangle inequality...

Never mind. Don't know why I said that...

i've gotta question. a=1, b=2, c=just under 3 would have a smaller area than if c were 2. i think...why?? ?
?? doesn't make sense to me...

-rorri :?

Ok, put 2 as the base.  1 is the length of one side and x is the length of the other.  When x is just over one, the height is minimul because the angle between the side length 1 and side length 2 is small.  A x gets bigger, the height gets bigger and so does the angle between the two sides of 1 and 2. When that angle reaches 90 degrees, the height is at its maximum.  x, then would be  :sqrt: 5.  After that, when x gets bigger and the angle becomes obtuse, the height gets smaller.

Max area: 1, 2,  :sqrt: 5,  Area=1

Wouldn't C have to be infinitesimally smaller than 3, but still smaller?

No, because a 1 2 <3, triangle, where <3 is very clos to three is almost just a line segment.  There is a very obtuse angle, making the height minimul.  The maximum height is 1 when the base is 2.


When $x^{13}+1$ is divided by $x-1$, the remainder is
\[
\textrm{(A)} 1\qquad
\textrm{(B)} -1\qquad
\textrm{(C)} 0\qquad
\textrm{(D)} 2\qquad
\textrm{(E)} \textrm{none of these}
\]

[hide]\[x^{13}+1=(x^{13}-x^{12})+(x^{12}-x^{11})+(x^{11}-x^{10})...(x^3-x^2)+(x^2-x)+(x-1)+2\]

Each of the expressions in parentheses is divisible by $x-1$, so the remainder is $2$.  Therefore, the answer is $D$.[/hide]

[hide="alternative solution"]
using the remainder theorem, $(1)^{13}+1=2$
so the answer is D
[/hide]


I have a question for somebody who is familiar with the very last inequalities problem in AoPS Vol. 2 (chapter 14 i think). It was some IMO problem... anyway in the solution they determine the angles of PQR without explanation, and I don't see how one could prove that the specific quadrilaterals were cyclic... can someone explain (tomorrow I will look up the problem when I have time and post a proper question...)

This is the last problem, but it's from Bulgaria 1993, not the IMO, so I'm not sure if it's the one he mentioned, although most of the variables are indeed the same.

Let $M$ be an interior point of the triangle $ABC$ such that $\angle AMC=90^\circ$, $\angle AMB=150^\circ$, and $\angle BMC=120^\circ$. The circumcenters of the triangles $AMC$, $AMB$, and $BMC$ are $P$, $Q$, and $R$ respectively. Prove that the area of $\bigtriangleup PQR$ is greater than or equal to the area of $\bigtriangleup ABC$.

I have to second the OP. I can't quite undestand how angle RPQ is just said to be 90, angle PQR=angle BCR (?? why?))=30, and then obviously QRP=60 (assuming the previous two are true).

Could someone explain how to get these angles.

Come on...

$PQ \bot AM$ and $PR \bot CM$ (Do you see why?)

You can find all the other angles of $\Delta PQR$ in a similar way.

Thanks, I knew I was missing something incredibly obvious. :blush: Focused too much on PQR=BCR.


LaTeX
special factorizations
number theory
Diophantine equation

I told my best [url=http://www.artofproblemsolving.com/Forum/images/avatars/45961890041e4a6671adbd.jpg]friend[/url] ,  that he must learn Latex if he is going to take AoPS forms seriously. I told him, using  Latex  on these  forms is really not hard, “all you have to do is to type a “$\$$”  on both sides.” And ask him to practice, giving him two two digit numbers ($a$ and $b$) and asked him to try  typing out “a^2+b^2”.  which should look like $a^2+b^2$

($a$, and $b$ were two  two digit natural numbers).

Like a true lazy lab, mumbling something about ‘using keyboard with paws is hard’  he just put the two number together, making it a four digit number.

When asked what happened to  keying symbols like “^" and "2"  etc.. he said “Makes no difference”

He was right of course,  the four digit number was actually the value of the expression.
What is the four digit number. (which was form by putting a and b right next to each other)

(As always, please hide the solution)

(And no smart solutions like a=b=0 , (both a and b are >9  and  less than 100 )

Entertaining as always :D but a little hard to understand. Is the four digit number "a2b2" where a and b are digits?

[quote="boxedexe"]No...a and b are both 2 digit numbers, so it I believe if $(cd)_{10}$ is the digits of $a$ and $(ef)_{10}$ are the digits of $b$, then $a^2+b^2=(cdef)_{10}$.  Is this right?

Masoud Zargar[/quote]
Yes,  That is  correct. [hide]you have to solve $a||b = a^2+b^2$  where '$a||b$' means ,  concating the two (both 2- digit numbers) numbers $a$ and $b$  (IOW  = Hint: [hide]$100a+b$[/hide][/hide]

(w/o lookin' at your hint  :D )
[hide="Not sure..."]
We know that $a,b \in \mathbb{N}$ and $\in [10,99]$.

$a^2+b^2=100a+b$
$a^2-100a+b^2-b=0$
$(a^2-100a+250)-250+(b^2-b+\frac{1}{4})-\frac{1}{4}=0$

Completing the square, we get
$(a-50)^2+(b-\frac{1}{2})^2=\frac{1001}{4}$.
$b$ is maximized when $a=50 \implies b \leq \frac{1}{2}+\sqrt{\frac{1001}{4}} \implies b < 17$. Thus $10 \leq b < 17$.

However, when testing the values for b (10,11...16), none of them give integer values for $a$. So i'm not sure if there is a solution... does anyone see a flaw in my logic or work?[/hide]

[hide="Some progress"]
$a^2 + b^2 = 100a + b$
$a^2 - 100a + b^2 - b = 0$
$a(a-100) + b(b-1) = 0$
$-(a(100-a)) + b(b-1) = 0$
$b(b-1) = a(100-a)$

Since b is an integer, either b or b-1 must be even, so b(b-1) is even. Therefore a(100-a) is even, therefore a is even.
Hope that helps someone[/hide]

[quote="minsoens"][quote="cincodemayo5590"]$(a^2-100a+250)-250+(b^2-b+\frac{1}{4})-\frac{1}{4}=0$[/quote]
$50^2\not= 250$?
hmm[/quote]

riiiight...  :D

(i cant do basic math at 1 in the morning :P )

[hide="Then it would be..."]
$(a-50)^2 + (b-\frac{1}{2})^2 = \frac{10001}{4}$
$b \leq \sqrt{\frac{10001}{4}}+\frac{1}{2}$
So we have $10\leq b <51$ . That leaves 41 values of $b$ to be tested, looking for $2500-b^2+b$ to be a perfect square ( $(a-50)^2$ ).

Some quick testing yields [hide]$b=33$ and $a=12,a=88$ giving two numbers: $\boxed{1233 \text{ and } 8833}$.[/hide][/hide]

[quote="Gyan"]cincodemayo5590:  Also ...
[hide]Hint:  [hide]10001=73*137[/hide][/hide][/quote]

Gyan, i'm confused as to how knowing that $10001=73 \cdot 137$ would help solve this problem...

I'm wondering how you solvd the problem with the factorization of 10001. Would you like to explain your solution, or at least your approach? :)

From where Hexaditidom left off:

[hide]$b(b-1)=a(100-a)$, which we are solving as a Diophantine equation with the restriction $9<a,b<100$. We see that $(a,b)=(12,33)$ works, and since $100-12=88$, $(88,33)$ works as well. [/hide]

Sorry, not an elegant solution, but an answer nonetheless. Does the author claim an elegant solution exists?

[quote]Does the author claim an elegant solution exists? [/quote]

Yes,  I have given one hint (hidden)  and I do expect someone would  post a  full solution  here .. else I would post it here ... there is a simiar problem (same technique -but numbers are a little larger) as mentioned above..

If you post a solution please hide it so others can work on it too.

Okay: Here is one solution:

[hide]We use $(a^2+b^2)(c^2+d^2) = (ac \pm bd)^2 + (ad \mp bd)^2$
Thus if you want to write a  number (eg 10001) as sum of two squares, factor the number  and decompose each smaller number as sum of two squares
and then use the above trick to de compose it as sum of two squares.

For example $10001 = 73 \cdot 137 = (8^2+3^2)(11^2+4^2) = (88 \pm 12)^2 + (33\mp 32)^2$
Thus $1001$ can be written as $100^2+1^2$ or $76^2 + 65^2$

We need
$a^2+b^2 = 100 a +b$
or $a^2 -100a +b^2 - b = 0$
or $4a^2 - 4 \cdot 100 a + 100^2 + 4 b^2 -4 b + 1 = 100^2+1 = 73 \cdot 137$  (See my hint 10000=73*137)
or $(2a-100)^2+(2b-1)^2 = (8^2+3^2)(11^2+4^2)= 100^2+1^2 = 76^2+65^2$

So our souutions $(2a-100) = \pm 100$ and $(2b-1) = \pm 1$  gives trivial solution for $a = (100,0)$ and $b = (1,0)$
And  $(2a-100) = \pm 76$ gives  $a = 88 \ or \ 12$   and $2b -1 = \pm 65$ gives $b = 33 \ or -32$

The solution thus is $1233$ or $8833$[/hide]

[quote="shadysaysurspammed"][quote="Gyan"]cincodemayo5590:  Also ...
[hide]Hint:  [hide]10001=73*137[/hide][/hide][/quote]

How would that help?[/quote]

See above  ....


Let be given a triangle $ ABC$.$ M \in BC$.
$ r(AMB)=r(AMC)$. Prove that $ m_a \geq AM \geq l_a$
Note: $ r(XYZ)=$ radius of incircle of triangle $ XYZ$

Actually, I don't think we need any formulae. It suffices to show that if $AB<AC$, then $r(AMB)<r(AMC)$ when $M$ is the foot of the bisector from $A$, and the opposite inequality takes place when $M$ is the midpoint of $BC$. This is easy to do.

[quote="nttu"]I think this problem will be easy if we prove that : [tex] AM^2 = S. cotg\frac{A}{2} [/tex] .[/quote]
Yes, this 's Mr'Le problem. The lemma is $ AM=\sqrt{p(p-a)}$.


Prove that

Prove (pi^2)/6=1/1^2+1/2^2+1/3^2+...

is true.

Help me out!


Find all integral $x$ such that $x^2+2$ is prime.

i know that $x$ has to be odd but not all odd numbers will work.

$(5^2+2)$ is not prime

Post solution here  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80736]Same problem posted[/url]


Hi, i was just wondering if anybody could help me with this question..

The following null and alternative hypotheses are being considered
H0:µ≤2000, HA:µ>2000

It is required to decide on the sample size n so that the risk of making a type 1 error is at most 1%, and also so that if the value of µ is at least 2300, there is a 90% chance that the Null hypothesis would be rejected. Taking the population standard deviation to be 600, find the value of n.

I have no clue on how to start this question.. Do i get the value of the test statistic, z= (X-µ)/(s/√n )
with s=600 and µ=2300???


probability
function
calculus
integration
probability and stats

Sorry, still don't know how to do the math type stuff.

Y is distributed uniformly on [0,1]
f(x|Y=y)=1/y for 0<=x<=y

Find f(Y,X)
Find E(X)

Let's go back to the notation as it was on the exam:

$Y_{1}$ is distributed so that its (marginal) density is uniform on $[0,1].$ Suppose that the conditional distribution of $Y_{2}$ given $Y_{1}=y_{1}$ is uniform on $[0,y_{1}].$

That is, $f(y_{2}|y_{1})=\frac1{y_{1}}$ for $0\le y_{2}\le y_{1}.$

(a) Compute the conditional expectation $E(Y_{2}|Y_{1}=y_{1}).$

Answer: this is the mean of a distribution that is uniform on $[0,y_{1}].$ The mean of a uniform distribution is simply the midpoint of the interval, so $E(Y_{2}|Y_{1}=y_{1})=\frac{y_{1}}2.$

(b) Find the joint density function $f(y_{1},y_{2}).$

Answer: Multiply the appropriate marginal density by the appropriate conditional density.

$f(y_{1},y_{2})=f_{1}(y_{1})\cdot f(y_{2}|y_{1})=1\cdot\frac1{y_{1}}= \frac1{y_{1}}$
This formula is valid on the triangle $0\le y_{2}\le y_{1}\le 1.$

(c) Compute $E(Y_{1}),E(Y_{2}),\text{Var}(Y_{1}),\text{Var}(Y_{2}),$ and $\text{Cov}(Y_{1},Y_{2}).$

Answers:

$E(Y_{1})=\frac12.$ Just the mean of a uniform $[0,1].$

The simplest way to get $E(Y_{2})$ is by using the conditional expectations. Use what we did in part (a).

$E(Y_{2})=E(E(Y_{2}|Y_{1}))=E\left(\frac{Y_{1}}2\right)= \frac12E(Y_{1})=\frac14.$

You could also do the integral:

$E(Y_{2})=\int\int y_{2}\,dy_{1}\,dy_{2}=\int_{0}^{1}\int_{0}^{y_{1}}\frac{y_{2}}{y_{1}}\,dy_{2}\,dy_{1}$

$=\int_{0}^{1}\left.\frac{y_{2}^{2}}{2y_{1}}\right|_{0}^{y_{1}}\,dy_{1}= \int_{0}^{1}\frac{y_{1}}2\,dy_{1}=\frac14.$

I'll let you finish the other parts; all the other double integrals should have the same setup.


algebra
functional equation
real analysis
real analysis unsolved

Hi to all!

I would like to know if there is any paper or research about the functional equation $ f(f(x))\equal{}x$.
Or maybe someone could tell me here something about the solutions? For example: is it completely solved?
Thank you!

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=245906]Very recently posted[/url].  fedja gave the general solution; note that $ f$ must be a bijection in order for this to be true and then look at its fixed and non-fixed points.


The two spinners shown are spun once and each lands on one of the numbered sectors. What is the probability that the sum of the numbers in the two sectors is prime?
[asy]unitsize(30);
draw(unitcircle);
draw((0,0)--(0,-1));
draw((0,0)--(cos(pi/6),sin(pi/6)));
draw((0,0)--(-cos(pi/6),sin(pi/6)));
label("$1$",(0,.5));
label("$3$",((cos(pi/6))/2,(-sin(pi/6))/2));
label("$5$",(-(cos(pi/6))/2,(-sin(pi/6))/2));[/asy]
[asy]unitsize(30);
draw(unitcircle);
draw((0,0)--(0,-1));
draw((0,0)--(cos(pi/6),sin(pi/6)));
draw((0,0)--(-cos(pi/6),sin(pi/6)));
label("$2$",(0,.5));
label("$4$",((cos(pi/6))/2,(-sin(pi/6))/2));
label("$6$",(-(cos(pi/6))/2,(-sin(pi/6))/2));[/asy]

$ \textbf{(A)}\ \frac {1}{2} \qquad \textbf{(B)}\ \frac {2}{3} \qquad \textbf{(C)}\ \frac {3}{4} \qquad \textbf{(D)}\ \frac {7}{9} \qquad \textbf{(E)}\ \frac {5}{6}$

[hide] There a total of $ 3(3)\equal{}9$ combinations the spinners could land on. Of these. $ 7$ have prime sums ($ 1\plus{}2$, $ 1\plus{}4$, $ 1\plus{}6$, $ 3\plus{}2$, $ 3\plus{}4$, $ 5\plus{}2$, $ 5\plus{}6$). Hence, we have a probability of $ \boxed{\frac79}$, or D.[/hide]


To be continuous at $x = -1$, the value of $\frac{x^3+1}{x^2-1}$ is taken to be:
\[ \text{(A)}\ -2 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ \frac{3}{2} \qquad \text{(D)}\ \infty \qquad \text{(E)}\ -\frac{3}{2} \]

answer:
[hide]simply use L'hopitals Rule, and get:
$\lim_{x\rightarrow -1}\frac{3x^2}{2x}=\frac{-3}{2}$[/hide]


1) $ \sum_{cyclic} \frac{x\minus{}1}{2x\plus{}1}\le 0$
2) $ \sum_{cyclic} \frac{1}{x\plus{}y\plus{}1} \le 1$

I beleive it is sufficient to solve $ 1$, because $ 1\implies 2$.

EDIT: This claim was stated due to incorrect algebraic manipulation in an effort to relate the two inequalities.

Can you please show me how are they connected and how to solve them?

[hide="1"]
We have $ \frac {x \minus{} 1}{2x \plus{} 1} \plus{} \frac {y \minus{} 1}{2y \plus{} 1} \plus{} \frac {y \minus{} 1}{2y \plus{} 1} \equal{} \frac32\left(1 \minus{} \left(\frac1{2x \plus{} 1} \plus{} \frac1{2y \plus{} 1} \plus{} \frac1{2z \plus{} 1}\right)\right)$.

By AM-GM, $ x \plus{} y \plus{} z\geq3\sqrt [3]{xyz} \equal{} 3$.

From this, $ xyz \equal{} 1$, and $ x,y,z\in\mathbb R^ \plus{}$, we obtain $ \frac1{2x \plus{} 1} \plus{} \frac1{2y \plus{} 1} \plus{} \frac1{2z \plus{} 1} \equal{} \frac {4(xy \plus{} yz \plus{} zx) \plus{} 4(x \plus{} y \plus{} z) \plus{} 3}{8xyz \plus{} 4(xy \plus{} yz \plus{} zx) \plus{} 2(x \plus{} y \plus{} z) \plus{} 1}$

$ \equal{} \frac {4(xy \plus{} yz \plus{} zx) \plus{} 4(x \plus{} y \plus{} z) \plus{} 3}{4(xy \plus{} yz \plus{} zx) \plus{} 2(x \plus{} y \plus{} z) \plus{} 9}\geq\frac {4(xy \plus{} yz \plus{} zx) \plus{} 2(x \plus{} y \plus{} z) \plus{} 9}{4(xy \plus{} yz \plus{} zx) \plus{} 2(x \plus{} y \plus{} z) \plus{} 9}$

$ \equal{} 1$.

Plugging this into our first equation, we get $ \frac {x \minus{} 1}{2x \plus{} 1} \plus{} \frac {y \minus{} 1}{2y \plus{} 1} \plus{} \frac {y \minus{} 1}{2y \plus{} 1} \equal{} \frac32\left(1 \minus{} \left(\frac1{2x \plus{} 1} \plus{} \frac1{2y \plus{} 1} \plus{} \frac1{2z \plus{} 1}\right)\right)\leq0$, as desired.

Equality occurs when $ a \equal{} b \equal{} c \equal{} 1$.
[/hide]

Edit: Oh lol, I was reluctant to expand. :P

2. Set $ x\equal{}a^3, y\equal{}b^3, z\equal{}c^3$, we get $ abc\equal{}1$.
So, $ LHS \equal{} \sum \frac{1}{a^3\plus{}b^{3}\plus{}abc} \leq \sum \frac{1}{a^2b\plus{}ab^2\plus{}abc} \equal{} \sum \frac{c}{a\plus{}b\plus{}c} \equal{}1$

$ \sum_{cyclic} \frac {x \minus{} 1}{2x \plus{} 1}\le 0\Leftrightarrow \sum_{cyclic} \frac {1}{2x \plus{} 1}\ge 1$
Let $ x \equal{} \frac {a}{b},y \equal{} \frac {b}{c},z \equal{} \frac {c}{a}$
So $ \sum_{cyclic} \frac {1}{2x \plus{} 1}\ge 1\Leftrightarrow \frac {b^{2}}{b^{2} \plus{} 2ab} \plus{} \frac {c^{2}}{c^{2} \plus{} 2cb} \plus{} \frac {a^{2}}{a^{2} \plus{} 2ac}\ge 1$which is easily  done by Cauchy :)

[quote="dgreenb801"]Just expanding #1, we have
$ \sum (x \minus{} 1)(2y \plus{} 1)(2z \plus{} 1) \le 0 \implies x \plus{} y \plus{} z \ge 3$, true by AM-GM

For number 2, $ \sum \frac {1}{a \plus{} 1} \le 1 \implies a \plus{} b \plus{} c \plus{} 2 \le abc$
So we have to show
$ 2(x \plus{} y \plus{} z) \plus{} 2 \le (x \plus{} y)(y \plus{} z)(z \plus{} y) \implies$
$ 2(x \plus{} y \plus{} z) \equal{} 2(xyz)^{2/3}(x \plus{} y \plus{} z) \le x^2y \plus{} x^2z \plus{} y^2x \plus{} y^2z \plus{} z^2x \plus{} z^2y$, true by Muirhead.[/quote]

Remember, AM-GM only works for positive reals.

Yes, but the inequality is false if x,y,z can be negative, try x=y=-1 z=1


If 3 machines can produce 1000 widgets in 4 days, how many hours would it take 9 of these machines to produce 250 of these widgets?  Assume all machines are working continuously at the same rate.

[hide]Since we need our answer in hours, we convert 4 days into 96 hours (24 hours in a day).  Since the rate of the machines is the same, the time it takes the job to be done is directly proportional to the widgets and inversely proportional to the number of machines.  With some scaling, we get:

3 machines, 1000 widgets, 96 hours
9 machines, 1000 widgets, 32 hours
9 machines, 250 widgets, 8 hours

Therefore, the time it takes for 9 machines to make 250 widgets is $\boxed{8}$ hours.[/hide]


There are three consecutive integers x,y,z. What is the largest prime factor of 3x+3y+3z?
Thanks

Because they are 3 consecutive integers, they will be relatively prime to each other (unless the first and last are even, in which case it would be 2).  Then, when you add 3 of each of these numbers, write it as $3(n-1)+3n+3(n+1)=3n-3+3n+3n-3=9n$.  The largest number you know is a prime factor of $9n$ is 3.  There maybe others, depending on what $n$ is.

[hide]think about it, if since x,y,z are consecutice it they have 2 be relatively prime, ubnless x is an even number. then you are trying to find the greatest prime factor of 3(x+y+z). your greatest prime factor would have to be 3.[/hide]

Yikes man look at the problem more carefully. You factor it so you get 3(x+y+z) and their prime factor is 3. Also, since there are no other factors in the equation 3y+3x+3z you can take 3 from it.........


Hi, I guess this is a Theorem or some proposition wonder if anyone would help me understand more clearly, thx.

[b]If $A$ is nonsingular and has $LU$ decomposition, then $A_k$ is nonsingular for $k=1,....n$[/b]

If I want to prove it, what do I need to start with? Please Help, Thx again.

I think that $A_k$ is the matrix obtained from the first $k$ lines and first $k$ columns of $A$.

$A_k$ is the leading princepial sunmatrices when it preform the Gaussian elimination


In ABC triangle the bisector of <B intersect the line that pass fom the midpoint of AC and the midpoint of altitude AH
in ABC triangle in point M. N is the midpoint of bisector of <B.prove that the the bisector of <C is the bisector of
<MCN too.

Are you sure the problem is right? :huh:
According to my figure, it's not true. :maybe:

I am sorry that i answer to you late.
if you think this problem is not true please tell me why?

What is wrong with this problem dear akashnil?

The line through $ C$ which is between the lines $ CM$ and $ CN$ is the bisector of $ \angle C$. But clearly it is not the bisector of $ \angle MCN$, which is asked to prove dear behdad.

And the picture was generated by a software and with the exact constraints. These computer diagrams are very precise and celdom goes inaccurate, that too very slightly. :wink:

Now i am sure that the question is wrong,but i think if we draw BH altitude instead of AH the question will be true.
would you think about it?

[quote="mr.danh"]Let a triangle ABC. BH is the altitude, BD is the bisector. Denote M,N,Q be the midpoints of BH,AC,BD. BD meets MN at P. Then, $ \angle PCD \equal{} \angle QCB$ [/quote]
yes, the above problem is true.
Let $ \frac {AD}{AB} \equal{} \frac {CD}{BC} \equal{} \frac {AC}{AB \plus{} BC}\equal{}k$
BD paases through the midpoint K of arc AC of the circle (ABC). BH is parallel to KN, $ \frac {ND}{NH} \equal{} \frac {KD}{KB}$.
Apply Ptolemy's theorem to the cyclic quadrialteral ABCK, BK.AC=AK(AB+AC)  => KA/KB=AC/(AB+BC)=k
Apply Menalaus's theorem to triangle BHD with transversal (MPN),
$ \frac {PD}{PB} \equal{} \frac {ND}{NH} \equal{} \frac {KD}{KB} \equal{} \frac {KD}{KA}.\frac {KA}{KB} \equal{} \frac {CD}{BC}.\frac {KA}{KB} \equal{} k^2 \equal{} (\frac {CD}{CB})^2$.
Hence result.

Dear Behdad,

The question you posed is true only if ABC is a right angle triangle with Angle A as 90 degrees. :)


Use latex tags:
\begin{bmatrix}a \qquad b \\ c \qquad d\end{bmatrix}
to produce \[\begin{bmatrix}a \qquad b \\ c \qquad d\end{bmatrix}\]

And I'm not sure what your question is asking, do you just want to show that \[\begin{bmatrix}a \qquad b \\ c \qquad d\end{bmatrix}\times \begin{bmatrix}d \qquad-b \\-c \qquad a\end{bmatrix}\times \frac{1}{ad-bc}= I \]

yes
i have to show somehow
that the first equation's inverse is the second equation

by using the three elementary row operations
interchange two rows
multiply
add
*things that u have to do to each row
to get the inverse

this is how im confused

[quote="kryms3n"]hello
i am really confused how to do inverse
and i have to show how [color=red]this [/color]matrix is an inverse of A

A= [a b]
     [c d]
it is all one matrix ( 2x2)
(a and c are in the first column and bd are in the other)
i know that the answer is

the inverse of A = (1/ ad -bc) [d -b]
                                         [-c a]
(d and -c are in the first column and -b and a are in the second column)

ANY HELP?[/quote]

which matrix?  the first one or the second?

[quote="kryms3n"]hello
i am really confused how to do inverse
and i have to show how this matrix is an inverse of A

A= [a b]
     [c d]
it is all one matrix ( 2x2)
(a and c are in the first column and bd are in the other)
i know that the answer is

the inverse of A = (1/ ad -bc) [d -b]
                                         [-c a]
(d and -c are in the first column and -b and a are in the second column)

ANY HELP?[/quote]
maybe i will help maybe i will confuse
i can 't latex this but...
in a 2*2 matrix A(A=  [a b]
                             [c d]the inverse is
A^{-1}= [d/(ad-bc) -b(ad-bc]
             [-c/(ad-bc) a/(ad-bc)]
you can substitute ad-bc with the determinant of A but i am not sure if you know that

Hmm I'm not sure what you guys are doing... just verify what I have posted above and then it will be proven... :maybe:

Also just copy and or edit my code if you want to do the latex :P

I think he means find the inverse of A using "matrix identity"


Prove that the product of the two roots of x4 + x3 - 1 = 0 is a root of x6 + x4 + x3 - x2 - 1 = 0.

[quote]Do you mean the real roots?[/quote]
I don't think they have to be real.

You would have to mean the real roots surely -  otherwise you wouldn't know which two roots out of the 4 you are talking about.

No, it means any two roots.  Basically, show that the pairwise products of the roots of the first equation are the roots of the second.

Let a, b, c, and d be the 4 roots of x^4+x^3-1=0. This tells us

a+b+c+d = -1
ab+ac+ad+bc+bd+cd = 0
abc+abd+acd+bcd = 0
abcd = -1

Set q = a+b, r = c+d, s = ab, and t = cd. Then the system becomes

q+r = -1
s+t+qr = 0
sr+tq = 0
st=-1

Equation 1 and 4 tell us that

r = -1-q

and

t = -1/s.

Putting these values into equation 3 gives

q = -s^2/(1+s^2).

Now, we can sub these values of q, r, and t into equation 2, and after a bit of algebra we get

s^6+s^4+s^3-s^2-1 = 0,

as desired. (So  it actually doesn't matter which 2 roots we choose.)

(I left out the algebra work. When writing this solution on the USAMO, however, it's probably a good idea to show the steps.)

Show that it's true for any two roots.

Nevermind. JBL and zabelman beat me to it.


geometry
projective geometry
geometry solved

There's one (maybe not so hard) problem, but actually I can't see the solution.

Let [tex]ABC[/tex] be an acute triangle and [tex]P, Q, R[/tex] be the feet of altitudes on sides [tex]BC, CA, AB[/tex] in this order. Let [tex]O[/tex] be the orthocentre of this triangle and then let's define the points [tex]S, X, Y[/tex] as follows in [url=http://us.f3.yahoofs.com/users/4295791cz634298e0/mail/__sr_/35c9.jpg?ph9GYlCBNWcNttFg]the picture[/url]. Prove that points [tex]O, X, Y[/tex] are collinear.

I've tried to use the Menelaos' Theorem, but I wasn't successful. Thanks for a help.

Have a look at [url=http://www.cut-the-knot.org/pythagoras/Pappus.shtml]this[/url]. The problem folows immediately from Pappus' Theorem.

It's trivial if you understand the theorem....
BTW , I couldn't access your picture more than two times....... I drew it on a piece of paper in the  , so no problem anyways ... The next time you post , mind adding the picture as an attachment here ?  :)

I see, now it's very easy, but I haven't worked in the projective geometry much before... although I've heard about this theorem, I didn't notice it. :roll:


Let $ a,b,c$ be positive reals such that $ a\plus{}b\plus{}c\equal{}1$. Prove that

$ \sum\frac{3b^{5}\plus{}3a^{3}\plus{}3c^{5}\plus{}abc}{b^{3}\plus{}c^{3}}\geq7$

[quote="Ligouras"]Let $ a,b,c$ be positive reals such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that

$ \sum\frac {3b^{5} \plus{} 3a^{3} \plus{} 3c^{5} \plus{} abc}{b^{3} \plus{} c^{3}}\geq7$[/quote]
$ a\equal{}b\equal{}c\equal{}\frac{1}{3}$ gives $ LHS \equal{} 3 \cdot \frac{9 \cdot \left(\frac{1}{3}\right)^{5} \plus{} \left(\frac{1}{3}\right)^3}{\left(\frac{1}{3}\right)^3\plus{}\left(\frac{1}{3}\right)^3} \equal{} 3$. Should $ RHS \equal{} 3$?

[quote="Mathias_DK"][quote="Ligouras"]Let $ a,b,c$ be positive reals such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that

$ \sum\frac {3b^{5} \plus{} 3a^{3} \plus{} 3c^{5} \plus{} abc}{b^{3} \plus{} c^{3}}\geq7$[/quote]
$ a \equal{} b \equal{} c \equal{} \frac {1}{3}$ gives $ LHS \equal{} 3 \cdot \frac {9 \cdot \left(\frac {1}{3}\right)^{5} \plus{} \left(\frac {1}{3}\right)^3}{\left(\frac {1}{3}\right)^3 \plus{} \left(\frac {1}{3}\right)^3} \equal{} 3$. Should $ RHS \equal{} 3$?[/quote]

My Friend Mathias:

$ LHS \equal{} 3\cdot\frac{6\cdot\left(\frac{1}{3}\right)^{5}\plus{}4\left(\frac{1}{3}\right)^{3}}{\left(\frac{1}{3}\right)^{3}\plus{}\left(\frac{1}{3}\right)^{3}}\equal{} 7$$ \equal{}RHS$

[quote="Ligouras"][quote="Mathias_DK"][quote="Ligouras"]Let $ a,b,c$ be positive reals such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that

$ \sum\frac {3b^{5} \plus{} 3a^{3} \plus{} 3c^{5} \plus{} abc}{b^{3} \plus{} c^{3}}\geq7$[/quote]
$ a \equal{} b \equal{} c \equal{} \frac {1}{3}$ gives $ LHS \equal{} 3 \cdot \frac {9 \cdot \left(\frac {1}{3}\right)^{5} \plus{} \left(\frac {1}{3}\right)^3}{\left(\frac {1}{3}\right)^3 \plus{} \left(\frac {1}{3}\right)^3} \equal{} 3$. Should $ RHS \equal{} 3$?[/quote]

My Friend Mathias:

$ LHS \equal{} 3\cdot\frac {6\cdot\left(\frac {1}{3}\right)^{5} \plus{} 4\left(\frac {1}{3}\right)^{3}}{\left(\frac {1}{3}\right)^{3} \plus{} \left(\frac {1}{3}\right)^{3}} \equal{} 7$$ \equal{} RHS$[/quote]
Oh. I missed $ a^{\textcolor{red}{3}}$ :blush:


logarithms
algorithm
function
calculus
calculus computations

Hello, Stephanie! You do not need to post the same topic twice. It irritates people and If you want to change something, there is an [b]Edit[/b] button   :).

I can suggest three possible lines in the essay:

1) Using $\ln$ for computations like multiplication and especially raising numbers to non-integer powers. You can even write a short historical survey here about how $\ln$s were used before invention of computers. Why natural $\log$s and not $\log$s base 2, say? However paradoxal it may sound, natural $\log$s are easier to compute.
One way is to use the Taylor formula for computation of $\ln(1-x)$ for $x$ between $1/2$ and $1$ (all other numbers can be reduced to these by multiplication or division by an appropriate power of $2$). Now there exist even faster algorithms but they still use the fact that one deals with natural $\log$.

2) There are many formulas in mathematics where $\ln$ appears. The most famous one is, probably, the asymptotic formula for the number of primes less than $n$. You can fill several pages along this line finding various formulas, explaining how interesting was the problem they answer and going into various historical remarks. The main point is that if there were no $\ln$s at our disposal, we would not even be able to formulate the answers to some "simple" questions.

3) $\ln$ is the inverse function to the $e^x$ and many processes in nature are described by the exponential function. And if you use $e^x$ to find the population after time $t$, say, then you use $\ln$ to find the time after which the population doubles.

I hope someone will extend this list  :)

[quote="stephanie"]why are natural logs so important? i need to write an essay on this and have no idea what to write please help me!!!!![/quote]

Stephanie,

Usually there is no need to post twice in the same forum, in the same day the exact same question (actually, there is no need to post the same question twice ever!).

On the other hand, I hope you notice that the question is totally ambiguous. Important according to what criteria? May be you can start in [url=http://mathworld.wolfram.com/NaturalLogarithm.html]this website[/url]. If you have any other question (hopefully more specific questions) post it in this thread, and if possible, delete by yourself the other post (if you can't, ask a moderator to do it!)

Best,


Which of the following statements must be true?
i:   If x>0, then x^3 > x^2.
ii:  If x<0, then x^3 < x.
iii: If x<0, then x^3+x^2<0.

(A) i only.
(B) ii only.
(C) iii only.
(D) ii and iii only.
(E) None of the above.

I think the answer is [hide]E, because the first varriant is also true! For instance $let x=4, than 4^3>4^2 as 64>16$[/hide]

I don't think any of the choices have to be true. Just let x 1 or -1.

[hide]if you substitute 1 in for x in i, -1 in for x in ii, 1 in for x in i, -1 in for x in iii, none of them are true so the answer is E.[/hide]


The parallel sides of a trapezoid are 3 and 9. The non-parallel sides are 4 and 6. A line parallel to the bases divides the trapezoid into two trapezoids of equal perimeters. The ratio in which each of the non-parallels is divided is?

[hide]Split the leg with length 4 into two parts: $x$ and $4-x.$

The perimeter of the top trapezoid is $3+x+\frac{3x}{2}+\frac{9x+3(4-x)}{4},$ or $4x+6.$

The perimeter of the bottom trapezoid is $9+(4-x)+\frac{3(4-x)}{2}+\frac{9x+3(4-x)}{4},$ or $22-x.$

$4x+6=22-x\Longrightarrow x=\frac{16}{5}$

The ratio is $\frac{16}{5}: \frac{4}{5},$ or $\boxed{4: 1}.$[/hide]

[quote="i_like_pie"]The ratio is $\frac{16}{5}: \frac{24}{5},$ or $\boxed{2: 3}.$[/quote]

Actually, this should be ${16\over 5}: \left(4-{16\over 5}\right)=4: 1$

For some reason I thought $4\cdot5=40.$ :rotfl:


$ M$ is a point inside the square $ ABCD$, so that $ AM\equal{}\sqrt{2}, BM\equal{}1, DM\equal{}\sqrt{5}$. Compute the length of the side of the square.

Let $ \angle MAB\equal{}\theta$.
So, $ \cos \angle MAB \equal{} \frac{x^2\plus{}2\minus{}1}{2\sqrt{2}x}$, $ \cos \angle MAD \equal{} \frac{x^2\plus{}2\minus{}5}{2\sqrt{2}x}$.
And, $ \cos \angle MAB \equal{} \cos \theta$, $ \cos \angle MAD \equal{} \cos (90^\circ \minus{} \theta) \equal{} \sin \theta$.
Therefore, $ \cos^2 \theta \plus{} \sin^2 \theta \equal{} \left(\frac{x^2\plus{}1}{2\sqrt{2}x}\right)^2 \plus{} \left(\frac{x^2\minus{}3}{2\sqrt{2}x}\right)^2 \equal{} 1 \iff x\equal{}\pm1, \pm\sqrt{5}$.

Thus, the length of the side of the square is $ \boxed{\sqrt{5}}$.

draw $ {M'A }$⊥ $ {MA}$ with$ {M'A\equal{}MA}$ ;join$ {M'M , M'D   }$ we easily get :$ {triangle AM'D}$≌$ {triangle AMB}$; $ {\longrightarrow}$ $ {M'M\equal{}\sqrt{2}MA\equal{}2}$;$ { M'D\equal{}1}$;$ {MD\equal{}\sqrt{5}}$; With $ {MM'^2\plus{}M'D^2\equal{}MD^2}$;we get $ {\angle{MM'D}\equal{}90}$   $ { \longrightarrow}$     $ {\angle{AM'D}\equal{}\angle{AMB}\equal{}45\plus{}90\equal{}135}$;in $ {triangleAMB }$ we use law of cosine :$ {AB\equal{}\sqrt{AM^2\plus{}BM^2\minus{}2*AM*BM*cos135}\equal{}\sqrt{2\plus{}1\plus{}2*\sqrt{2}*1*\frac{\sqrt{2}}{2}}\equal{}\sqrt{5}}$.And we have done !  :)

Prove easily that (for example, using the [u][b]Nakagawa[/b]'s nice method[/u]) the following extension :

[quote]Let $ M$ be an interior point for the square $ ABCD$ so that $ AM \equal{} a$ , $ BM \equal{} b$ , $ DM \equal{} d$. Then

$ |b \minus{} d|\le a\sqrt 2\le b \plus{} d$ and in this case $ 2\cdot AB^2 \equal{} b^2 \plus{} d^2 \plus{} \sqrt {\left[(b \plus{} d)^2 \minus{} 2a^2\right]\cdot\left[2a^2 \minus{} (b \minus{} d)^2\right]}$ [/quote].


trans-1,2-divinylcyclohexan-1,2-diol + heat -----> ?

This problem is very interesting. Should I post an hint?

I am guessing something like a fused trans-decalin structure where if carbons 1, 2 are the two carbons on both rings and their neighbors on one of the rings are 3 and 4, then 1 3 and 2 4 are double bonds

Yes, I can. The molecular formula of the product is $ \ce{C10H14O}$. :spider:

Is the answer by any chance this? I think this is what mystic terminator roundabout meant :maybe:

Just pure intuition! I think it may involve a C-O-C or a C-C=C-C bridge in the molecule :maybe:

A reasonable mechanism is essential. :juggle:

doesnt an oxycope rearrangement occur initially?

A complete and reasonable answer please. :play_ball:

ok. initially you get cyclodecan-1,6-dione.
then u get somethng like this i think-
i've drawn it not bcoz im gr8 at drawing but bcoz my nomenclature sux even more. :oops:
and thats only the second time i've seen u using smileys. :)

That's correct! But how about a mechanistic explanation?


Is there a way to train yourself to stop making stupid mistakes in math? I do that all the time, and it's really annoying. I know some people who just DON'T EVER make stupid mistakes. How do you do it? Do you train yourself by playing games that have extreme punishments when you mess up? If so, what games are they?

Maybe try slowing down a little when doing problems. You'd lose time if it's on a contest, but it'd be quicker than going through it again to fix a small mistake.

I find that if I slow down, then I make even more stupid mistakes due to falling asleep  :(

Well, don't slow down so much that you'd fall asleep. Just enough so that you can think about what you're writing down and don't copy the wrong number or something. It works for me sometimes.

I don't think there is a magical formula for careless mistakes :( , but by practice/ doing a lot of problems / going over your careless mistakes / organizing your work should reduce the number of careless mistakes.

[quote="zeb"]Here's a game: Find 168^34 with only a pencil and paper, and every time you make a mistake kill yourself.[/quote]

But then I'd be dead over $34^2=1122$ times before I can really mess up on a test... :?  :o

What I would do before a contest is do some basic calculations like $123\times 45$ by hand to warm up. (I usually do this about 10-15 minutes before the contest begins.) This helps me concentrate in a contest, which prevents me from making careless mistakes.

i dont think theres any sure-fire way to prevent any and all mistakes, but you should definitely slow down a problem, especially towards the end. And always check  your work. If its an easy problem, make sure it makes sense (quickly) - like make sure u dont get $127x49=8$. those kind of things. And for the more difficult problems, i say CHECK IT.

Mr. Rusczyk wrote a great article: [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php]Stop making stupid mistakes[/url].

Write from $1$ to $1000$ (or less) on a piece of paper. And see where do you get the mistake.

Yeah, I'm really well organized in math - it definitely helped me a lot and I reduced my stupid mistakes by a huge number. But I still make a lot of them...

[Edit] this is in response to mel

Do lots and lots and lots and . . . lots . . . of problems. Straightforward practice is the best way I know of.
Also, try writing down as much as you can when you solve problems. In my experience, most stupid mistakes occur while doing seemingly simple calculations mentally and not taking the time to write them down.

[quote="ZennyK"]Most stupid mistakes occur while doing seemingly simple calculations mentally and not taking the time to write them down.[/quote]
Good point.


You and Bob are competing in a multiple choice trivia game. You are each given a buzzer.
[hide="Rules of the Game"]a. If you buzz in before Bob and answer correctly, you get ten points and Bob gets none. The same rule applies for Bob where he wins ten points if he answers correctly first.
b. If you answer incorrectly, you may not buzz in again until the next question and Bob has a chance to answer the question correctly after you. Assume he does not pick the incorrect answer choice you picked.
c. You and Bob only get one chance to buzz in per question.
d. All of the questions have the same number of answer choices
e. The person with the most points after ten rounds wins.
[/hide]
[b]1. Supposing you have no idea what the answer is to a problem, is it better to guess immediatly or wait and hope Bob messes up? [/b]
[b]2.Do you and Bob have equal chances of winning? [/b]
[b]3. If one wins the game by scoring over nine thousand points first, is the probability that you win one half? [/b]

[hide="Part 1"]Let $ n$ be the number of answer choices.

[b][color=darkred]Case 1:[/color][/b] I guess first.

In this case, the probability of guessing correctly is simply $ \frac {1}{n}$.

[b][color=darkred]Case 2:[/color][/b] I let Bob guess first.

The probability of Bob getting it wrong is $ \frac {n \minus{} 1}{n}$. The probability of me then getting it right is $ \frac {1}{n \minus{} 1}$.

Thus, if I let Bob guess first, the probability of winning that round is $ \frac {n \minus{} 1}{n}\cdot\frac {1}{n \minus{} 1} \equal{} \frac {1}{n}$.

As shown, it does not matter whether I guess first or I let Bob guess first. Note that the number of answer choices is an independent variable, and does not affect this probability.[/hide]

[hide="Part 2"]As demonstrated above, the probabilities of winning a round are equal for both players, no matter who goes first.

Since the probabilities don't change over time, our chances of winning the entire game are also equal.[/hide]

[hide="Part 3"]As demonstrated in the solutions to parts 1 and 2, the chances of winning each round are equal for Bob and me. Thus, the expected values of our scores will always be the same, and the probability of me winning is $ \frac {1}{2}$.[/hide]

Yeah. Awesome. I like the first result alot.

For the first problem even if the number of answer choices varies per problem, does it even matter if the number of answer choices vary per problem? The chance of winning when you guess or wait will always be the same, right?

Also, I'm not sure of the following is doable, but if there are a large number of students all with a buzzer playing this game together, is it beneficial to wait instead of going ahead and guessing?

That is, everyone sees the same question at the same time, there is a finite number of answer choices and the buzzer is very precise and won't accept simultaneous buzzes?

[hide="General Probability"]Let $ k$ be the number of people and $ n\ge k$ be the number of answer choices. Also, let the set $ S$ represent the people in the game, where $ S_1$ through $ S_k$ are people, in the order that they respond to the question. Finally, let $ p(S_x)$ be the probability that person $ S_x$ guesses correctly. Note that $ 1\le x\le k$.

$ S_1$ guesses first, and answers correctly with a probability of $ \frac {1}{n}$.

To determine the probability that some person $ S_x$ guesses correctly, we must first determine the probability of everybody before $ S_x$ guessing incorrectly. This can be expressed as a product of the complementary probabilities of everybody up to $ S_x$ answering correctly:
\[ \prod_{i \equal{} 1}^{x\minus{}1}1 \minus{} p(S_{i})
\]
Now, we must determine the probability of $ S_x$ answering their question correctly, ignoring all other factors. There are $ n \minus{} x \plus{} 1$ answer choices left, so the probability is $ \frac {1}{n \minus{} x \plus{} 1}$.

Finally, we multiply the two probabilities:
\[ p(S_x) \equal{} \frac {1}{n \minus{} x \plus{} 1}\prod_{i \equal{} 1}^{x\minus{}1}1 \minus{} p(S_{i})
\]
[/hide]

Wait... so, using your mathematical expression if you were in such a game with many people and you had no idea what the answer was to a question, would you go ahead and guess? Would you wait until there are a certain number of answer choices left?

Also, what if n<k?

[quote="aufha"]Also, what if n<k?[/quote]
I made $ n\ge k$ to make sure that everybody would get a chance to guess in the worst case scenario.

This is possible, but then my formula needs a bound at $ n < k$ because the probability will become negative once $ x > n$.

I also fixed a little error in my formula (it had $ k \minus{} 1$ as the limit of the product, whereas it's actually $ x \minus{} 1$).


How many square inches are in
the area of the regular hexagon
shown? Express your answer in
simplest radical form.

[asy]size(60); path outpath; for(int i = 0; i<6; ++i) outpath = outpath -- dir(60i); draw(outpath--cycle); draw("12''",dir(-60)--dir(120),NE,fontsize(10pt));[/asy]


$ a,b,c \in Z^ \plus{}$  and $ \frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a} \in Z^ \plus{}$ .Prove that:
$ abc$ is a perfect cube :)

[hide="solution"]
Let  $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \equal{} k$

The LHS is homogeneous so assume $ \gcd(a,b,c) \equal{} 1$

Assume there exists some prime $ p$  that divides  $ a$ and $ b$ , but not $ c$, $ (\because (a,b,c) \equal{} 1)$

Write $ a \equal{} xp^n, b \equal{} yp^m, c \equal{} z$  and  $ (x,p) \equal{} (y,p) \equal{} (z,p) \equal{} 1$

Therefore   $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \equal{} \frac {x^2p^{2n}z \plus{} z^2yp^m \plus{} y^2p^{2m}xp^n}{xyzp^{m \plus{} n}}$

Since the expression is an integer and $ p^{m \plus{} n}|y^2xp^{2m \plus{} n}$ we know that

$ p^{m \plus{} n}|x^2zp^{2n} \plus{} z^2yp^m$

If $ 2n > m$ then $ p^{n \plus{} m}|p^m(x^2zp^{2n \minus{} m} \plus{} z^2y)$, which is impossible as: $ p^{n \plus{} m}\not | p^m$ because $ p^{m \plus{} n} > p^m$ and $ p\not |x^2zp^{2n \minus{} m} \plus{} z^2y$ because $ (z,p) \equal{} (y,p) \equal{} 1$.

If $ m > 2n$ then $ p^{m \plus{} n}|p^{2n}(x^2z \plus{} y^2p^{m \minus{} 2n})$ which again is impossible.

Hence $ m \equal{} 2n$ and  $ p^{2n}|x^2z \plus{} z^2y$

$ \therefore m \plus{} n \equal{} 2n \plus{} n \equal{} 3n \equiv 0 \mod 3$

 So if a prime, $ p$ divides any number $ a,b,c$ then it divides the product $ abc$ a number of times that is divisible by three. hence $ abc$ is a perfect cube.
[/hide]


$ f$ is a function such that for every $ x$ $ \in$ $ R$ $ \minus{}$ {$ \minus{} 1,0,1$} :

$ f$ $ (\frac {x \minus{} 1}{x \plus{} 1})$ $ \plus{}$ $ f$ $ (\frac { \minus{} 1}{x})$ $ \plus{}$ $ f$ $ (\frac {1 \plus{} x}{1 \minus{} x})$ $ \equal{}$ $ x$

find $ f(x)$

Let $ x\equal{}u$, $ x\equal{}\frac{1\plus{}u}{1\minus{}u}$, $ x\equal{}\frac{\minus{}1}u$, and $ x\equal{}\frac{u\minus{}1}{u\plus{}1}$, respectively. You can then solve $ f(u)$


Compute the sum

$ \sum_{n\equal{}0}^{\infty} \frac{F_n}{n!}$

where the Fibonacci numbers are defined recursively by $ F_0 \equal{} 0$, $ F_1 \equal{} 1$, and $ F_n \equal{} F_{n\minus{}1} \plus{} F_{n\minus{}2}$ for $ n \geq 2$.

Well you can just use the fact that 1+x/1! + (x^2)/2! + .... = e^x, and then apply Binet's formula to get the sum.

What Kamil suggested:

\begin{eqnarray*} \sum_{n = 0}^{\infty} \frac {F_n}{n!} & = & \sum_{n = 0}^{\infty} \frac {\frac {1}{\sqrt {5}}(\phi^n - ( - \frac {1}{\phi})^n)}{n!} \\
& = & \frac {1}{\sqrt {5}}\left( \sum_{n = 0}^{\infty} \frac {\phi^n}{n!} - \sum_{n = 0}^{\infty} \frac {( - \frac {1}{\phi})^{n}}{n!}\right) \\
& = & \frac {1}{\sqrt {5}}\left(e^{\phi} - \frac {1}{\sqrt [\phi]{e}} \right) \\
& = & \frac {e^{\phi}\sqrt [\phi]{e} - 1}{\sqrt {5}\sqrt [\phi]{e}} \end{eqnarray*}

I think there's a way to do it by just manipulating the sum? Or maybe I'm confusing it with another problem where instead you have a $ 3^n$ in the denominator.

You are.  Instead of computing a value of the ordinary generating function $ F(x) \equal{} \sum_{n \equal{} 0}^{\infty} F_n x^n \equal{} \frac {x}{1 \minus{} x \minus{} x^2}$, you're computing a value of the exponential generating function $ G(x) \equal{} \sum_{n \equal{} 0}^{\infty} \frac {F_n}{n!} x^n \equal{} \frac {e^{\phi x} \minus{} e^{\varphi x} }{\phi \minus{} \varphi}$.

Conclusion:  the Fibonacci numbers are much easier to study with the ordinary generating function.


What is the remainder when the sum of the first 100 positive integers is divided by 9?

the sum of the first $ n$ integers is represented by the expression $ \frac {n(n \plus{} 1)}{2}$

we plug $ 100$ in for $ n$ to get $ \frac {100(101)}{2} \equal{} 50\times101 \equal{} 5050$

when we divide this by 9, we get a remainder of $ \boxed{1}$

Another way:
the first 99 numbers are divisible by nine.
Therefore, 100/9 = 11 r 1.
Our answer is 1


A straw is placed into a rectangular box that is 3 inches by 4 inches
by 8 inches. If the straw fits
exactly into the box diagonally from the bottom left front corner to
the top right back corner, how long is the straw, to the nearest tenth
of an inch?

[hide="Hint"]Use the extended Pythagorean Theorem.[/hide]

How 'bout let's let Interval work this one out himself, with the hint.

Can you set up the equation?

I can take it from there.

The dimensions are 3, 4, and 8.

Think about the rectangular face of the box that has sides a and b. What is the diagonal of that? $\sqrt{3^{2}+4^{2}}$. This is NOT the answer.

Now think of the edge with length 8. Draw a triangle with one leg along the plane diagonal, and one leg along that edge. Find the hypotenuse.


search algorithm algebra polynomial \/closed

what are those green question marks next to your name when u post something?  how do you get one of those pictures next to your name whne you post things?

those pictures are "avatars" that can be uploaded...click on "profile" and scroll to the bottom

the green question mark thingys are kinda of like "ranks"...you get additional ones after a certain number of posts...you should have another one added on at 50 posts...

the FAQ section should answer both these questions...

what happens when u get so many question marks? do they mean anything?  do you get a prize?

It means you have a lot of posts.  I believe this is explained in the FAQ, but I'm not positive.  Anyhow, as you post more, you will eventually get more and more question marks.  I think the largest number comes after 2,500 posts or so, something only two people have so far.

What about the thing under your name?  Mine says P versus NP.

That's the name of your "rank". Each one is a famous not-yet-proven theorem I believe.

It is one of 7 Millenium Prize problems given by The Clay Mathematics Institute. Basically, if anyone solves any one of these, they are rewarded $1 million. They are also the names of the "ranks" that AoPS has:

P versus NP

The Hodge Conjecture

The Poincar Conjecture

The Riemann Hypothesis

Yang-Mills Existence and Mass Gap

Navier-Stokes Existence and Smoothness

The Birch and Swinnerton-Dyer Conjecture

[quote]is there a website that explains these problems?
nobody knows how to solve them?
i could become a millionaire if i can solve all of these?[/quote]

Sure, the Clay Institute has a site with explanations... just google "Millenium Prize Problems". No one has solved one of the seven yet, an exception being Perelman proving the Poincare Conjecture. As for your last statement, yes, but, no offense intended, unlikely.

U can become a millionaire if u solve ANY of them...not "all"...

just look up any one of the problems or the CMI Millenium Prize on any search engine...

these problems have been tried by many many mathematicians and scientists...there's a reason they're giving a milllion dollars to someone who proves them...

I'm having trouble just understanding the problems and what it's asking u to prove...

maybe i'll drop out of school so i can work on the like that 17 year old kid...   http://www.indianexpress.com/full_story.php?content_id=42806

[quote]I'm having trouble just understanding the problems and what it's asking u to prove...[/quote]

Well, they arent problems from elementary math, so to get a good grip on what the problem statements are, (i imagine) you need a good base in the field itself.

I can explain P vs. NP to you.  It's really a computing problem -- the question is about the speed of certain algorithms.  P algorithms are those which run in polynomial time -- that is, as the size of the computation goes up, the amount of time the computation takes goes up like a polynomial.  So, for example, doubling the size of the computation to be done might always result in a computation that takes 8 times as long -- this would be an algorithm that runs in n3 time.  These are "fast" algorithms, relatively speaking.  Another class of algorithms is those that run in exponential time -- every time you double the length of the problem, the amount of time the computation takes gets squared.  These are "slow" algorithms.

NP (or nondeterministic polynomial, I think) problems aren't known to be polynomial or exponential time algorithms -- people aren't sure if they're fast or slow.  Also, they can be checked quickly.  For example, think of the problem of scheduling large numbers of college students for final exams in such a way that no student has to take two exams at once.  This is a very difficult thing to do with 100% effectiveness (although it is possible to quickly design a schedule that will work for [i]most[/i] students), but it is very easy to check if the schedule of exams does in fact cause conflicts.  The P vs. NP problem is whether these NP problems can be solved in polynomial time.  I believe that most people who work with this kind of thing are convinced the answer is "no," but no one has proven it.

SirErnest wrote:what if u prove that they r not provable? it kind of reminds me of americas most wanted. whos the other person with over 2500 posts?

Tare has well over 4500 posts. [hide]Most of which are...well...that's another topic...[/hide]

In any case, there are other prizes (or u may be given the full million) if you find a counterexample or something of the sort...however, u should note that many many mathematicians have tried over the years to solve these problems so finding a counterexample would be very difficult if even possible...

Finding a counterexample probably won't be possible if the conjecture you are working on is actually true. I learned what the Birch Swinnerson-Dyer Conjecture is today, and it's a really neat problem.

I think you get the $1,000,000 whether you prove that P = NP or disprove it.

On an interesting side note, the first time I saw this question, it was titled "When does P = NP?" and the very first thought in my head was "When N = 1 or P = 0, of course."

Anyhow, the nature of the problem is such that if you prove an "NP-complete" question to be P (or not to be P), then [i]all[/i] NP problems also are P or NP.

could all the members of AoPS work these problems and then split the prize?

Well, there are two different ways to interpret that.  It would be within the rules of the prize, yes.  It is almost certainly not actually possible, however. :)

[quote="JBL"]It is almost certainly not actually possible, however. :)[/quote]

haha...i'm sure there are plenty of teams of mathematicians who are far more experienced than us working on this...I doubt that we could prove any of these problems...

Understanding them would be a feat...

[quote="SirErnest"]but there are many bright mathematicians that congregate here on AoPS.  200 heads are better than one  :)[/quote]

we might need like 2000 or 20,000 of these "heads" to do this problem....

It's a quantity vs. quality thing. I don't think any of us have studied any one of these problems seriously, and I don't think a finite number of people who know nearly nothing about a really hard problem can solve it. Hopefully I'll learn a thing or two about the Birch Swinnerson-Dyer Conjecture in the next year.


calculus search trigonometry integration derivative probability induction

I was wondering what the general curriculum is for a advanced placement pre-calculus course in the U.S.? Any help would be greatly appreciated.

The is not such thing as an Advanced Placement Pre-Calculus class in the U.S.  :wink: However, if you search the High School Basics forum and the Intermediate forum, there are plenty of results as this question has been asked before.

My class this year was a review of algebra 2 (for about 2 weeks) and trig (ended with inverse trig functions).

Um...Some schools can still place you in Pre-Calc, even if you're not at that grade level.

well this post is now officially out dated cause i used the precalc book that we used for the trig class at the univ and completed precalc and i am now doing calc  :)

If your high school is nice (like mine), they'll let you skip right to AP Calc if you show them that you know your "stuff" ;)

Poincare, BC in 6th grade??  Are you crazy?  I have no idea... I am actually taking BC this year as a senior.  Anyway, the material would be single variable differential and integral calculus.  Limits, derivatives, chain rule, quotient and product rules, optimization problems, substitution rule, integration by parts, trig substitutions, etc.  I'm sure the College Board website has a much more detailed description, but you should definitely check with your school to see how they organize the curriculum - at my school, 2nd semester honors precalc is almost entirely differential calculus.

My Honors Precalc class:

1st semester:
Sequences, series, counting, premutations, combinations, probability
Induction
TRIG TRIG TRIG!  Law of sines, law of cosines... and almost every identity in our text (pythagorean, sum and difference, reciprocal, half angle, double angle, etc.)
solving trig equations
complex numbers, deMoivre's theorem, complex roots
Polar coordinates

2nd semester:
more polar stuff - polar functions
Limits / rational function asymptotes review
Differential Calculus - power rule, chain rule, quotient and product rules, implicit differentiation, related rates problems, optimization problems, derivatives of exponential and logarithmic functions, trig derivatives
Intro to integration - substitution rule

I'm sure many schools have different curriculums...

You can find a sort of curriculum for AP calc AB in the College Board website. If you can take the AP test (AB or BC) in 6th grade and pass it, more power to you. IMHO teachers and counselors tend to underestimate what a student can do.

[quote="Poincare"]slightly tangent, but i was wondering what the curriculum is in AP calc AB. Cause i was wondering whether i could take it or not in sixth grade(this year).[/quote]

You'll learn all the derivative stuff such as polynomials, trig, exponential, and all the rules such as product and chain. That stuff is pretty easy and will not take long. You'll do integration afterwards. There will be substitutions (such as trig substitutions) and integration by parts (although my BC book says this will only appear in the BC clac test). Then there will be some volume stuff with surfaces of revolutions and surface area. I don't know if you go into sequences and differential equations.


Does anyone have a solution for #3? I am very much interested.

Thanks in advance.  :)

No, i only solved 1 and 2. I'm also very much interested.  :D

Do you remember this thread: http://www.mathlinks.ro/Forum/viewtopic.php?t=20244 ? It contained a discussion of a more general inequality than problem 3 of this round. Unfortunately, all messages from #7 on were lost in the hacker attack. I have stored messages #16-#22 (I believe), and one of them contains a quote of a former message without the author being named:

[quote="someone, probably Vasc"]I think for $r>0$ and $a,b,c\geq 0$ such that $a+b+c=3$, the following inequality holds

$\left(ab\right)^r+\left(bc\right)^r+\left(ca\right)^r \leq \max\left(3,\left(\frac94\right)^r\right)$.[/quote]

For $a=x^3$, $b=y^3$, $c=z^3$ and $r=\frac{4}{3}$, this yields the inequality in question (problem #3 of the MathLinks contest round 6).

I don't really know what was done in the thread but I don't think that the inequality above was proven there.

  Darij

You must be mean first version of pr.3. Download latest version.

I changed that inequality two weeks ago, because a solution was presented somewhere on the forum, and another solution was given on spot :)

Harazi was kind enough to supply me the generalisation :D

I made an attempt at it, anyway. I think the problem can be simplified by showing that replacing two variables by their harmonic mean lowers the LHS:RHS ratio except under certain conditions, namely that all but one of the variables are the same.

However, it was a bit messy; in the end it seemed to work, but I couldn't really say why.