analytic geometry
parameterization
calculus
derivative
vector
trigonometry
function

Suppose we have a curve parametrized like
$x = x[t] , y = y[t] , z = z[t]$

And we want to make surface that is at same distance from this line. in anothers words, we want to make a cilinder with radius $r$ on this curve.

how can we get the coordiantes of this cillinder?!

If we have $r=$ we get the same curve.

i thougth this. Make a circle with radius $r$ normal to a segment $ds$
but i cannot continue... someone have some idea?

please, help me

thanks

You do exactly what you said: one parameter is $t$, and the second parameter, for a fixed $t$, traces a circle of radius $r$ in the normal plane to the curve.  (This is easiest to do if you have a parametrization of the curve so that the derivative is always a unit vector; then the second derivative is normal to the curve, and you can get the other normal by crossing the first and second derivatives.)

HI JBL  thanks for the post

the question is. How can I trace that circle?

I thougth another way.

we have a vector $ds$ in the same direction of the curve.
We can find all vectors that are perpendicular to it.
$A.ds =0$
And the modulus of this vector is $|A|=r$.
With this, the vecto $A$ is definied by only one parameter. just like you said.

I'm trying to solve this only in my mind, to find the easiest solution.

Your solution is one of i thougth, but is more harder than these new one. I think...

sorry If I'm wrong. hum...

thanks

In order to trace the circle easily, you need a coordinate system in the plane of the circle: given perpendicular unit vectors $u$ and $v$ in that plane, the circle is just given by the equation $r\cos \theta u+r\sin \theta v$, where $\theta$ is your new parameter.  The stuff about normalizing the curve, finding a second derivative, etc., is just a convenient way to find $u$ and $v$.

Hm...  Thanks JBL

I did not thougth in this equation.

Solution

$x,y,z$ Are functions of only $t$. We define

${s=\{x,y,z\}}$

${\text{Ta}=\frac{\frac{ds}{dt}}{\sqrt{\frac{ds}{dt}.\frac{ds}{dt}}}}$

${\text{No}=\frac{\frac{d\text{Ta}}{dt}}{\sqrt{\frac{d\text{Ta}}{dt}.\frac{d\text{Ta}}{dt}}}}$

${\text{Pa}=Ta\times \text{No}}$

So, The cilindrical Curve is given by

${C=s+( r \cos (\theta ) \text{No}+r \sin (\theta ) \text{Pa})}$

To Simplify the calculations it's easy to see that:

${v=\frac{ds}{dt}}$ ,  ${a=\frac{d^{2}s}{dt^{2}}}$

${\text{Ta}=\frac{v}{\sqrt{v.v}}}$

${k=\frac{a v.v-v v.a}{(v.v)^{3/2}}}$  ,  ${\text{No}=\frac{k}{\sqrt{k.k}}}$

${j=v\times a}$ ,  ${\text{Pa}=\frac{j}{\sqrt{j.j}}}$

The final solution

${s+r \left(\frac{ (a v.v-v v.a)}{\|a v.v-v v.a\|}\cos (\theta )+\frac{v\times a\text{ }}{\|v\times a\|}\sin (\theta )\right)}$

NOTE: If you prefer you can use the identity ${(v\times a)\times v=a (v.v)-v (v.a)}$
http://mathworld.wolfram.com/VectorTripleProduct.html

Adios


Give x,y,z>0 and 2xyz \geq 2x+4y+7z.
Find the smallest value of A=x+y+z.
I have a sol for it but it has a mistake so that i want another beautifull sol.I think you can. Good luck! :)  .

Yes, we can.

Optimal triple (x,y,z) is (3,2.5,2). Let us prove it.

We have 2xyz \geq 2x+4y+7z = 30*(1/5*x/3+1/3*y/2.5+7/15*z/2) \geq C*x1/5y1/3z7/15 where C is a constant.
It follows that x4/5y2/3z8/15 \geq C'.

In the other hand, A = x+y+z = 7.5*(2/5*x/3+1/3*y/2.5+4/15*z/2) \geq C''*x2/5y1/3z4/15.

Thus A2 \geq C''*(C')2 = C'''.

Final step: for triple (3,2.5,2) all inequalities become equalities (!).

you will have A(Min)=15/2.
I think i have too.
After a long time :maybe: , i had a good solution for this problem but it too long. I need a short time to make it shorter.

Myth (or anybody else who solved this problem), how do you come up with a solution like that? Is this a common technique? I'm just baffled by how one is supposed to invent such a proof.

The main part of solution is to invent right supposition about optimal triple.


i'm very interested in taking an introductory modern algebra course at a local communitty college. i have taken cal 1, cal 2, first half of cal 3, and i've looked into about the first 1/3 of linear algebra.

i've heard from some people that linear algebra is required, but the university that is offering the course usually doesn't usually have really rigorous courses (out of about 100 math classes this fall, 40 or so are college algebra). i've had some experience with proofs and i mainly am concerned with gaining familiarity with the modern algebra.

do you think it would be a bad idea to try to take this course at this type of university (i would never try this at a place like stanford, mit, etc. without the prerequisites)

oh, almost forgot. i'll be taking lin. alg. concurrently

Definitely not a problem.  Lose no more sleep over this issue.


geometry
vector
topology
advanced fields
advanced fields theorems

Can any one tell some literature where I can get mapping class groups treated on algebraic curves.i have found approaches like teichmuller theory and stuff but am not getting it.Please help me because I want to write my BSC thesis on it.

Are you asking for a purely algebro-geometric treatment of the mapping class group of an algebraic curve?  I'm sure that this is possible, but my humble guess is that it would involve some very fancy math.

Regarding your other post about moduli spaces: the question is moduli spaces of what?  Moduli spaces are used in many areas of mathematics; it's not really a subject of its own.  Based on this post, perhaps you're interested in moduli spaces of complex structures on surfaces?

Well you are right.Yes I am looking for algebro-geometric treatment of MCG of an algebraic curve.Also I am interested in moduli spaces of complex structures on surfaces.Is some fairly expository or introductory literature possible.

Actually, I mentioned this to a friend, and he said that you can't really define the mapping class group in pure algebraic geometry.  In particular, you cannot even define the fundamental group.  The best you can do is the pro-finite completion of the fundamental group.  So, unless you want to deal with the pro-finite completion of the mapping class group, you should stick with the topological definition.

Representations of the fundamental group correspond to vector bundles and these can be studied algebraically.  There is more recent "cutting edge" stuff such as geometric Langlands correspondence and some of the algebraized Teichmueller theory recently developed and related to cluster algebras.  This is not at the level of a BSC thesis if you don't first digest the classical analytic theory.


Fie A, B doua comutative inele. A[x] izomorf cu B[x] implica A izomorf cu B ?

English version :

Let us consider A, B two commutative rings. A[x] izomorph with B[x] implies that A izomorph with B ?

Mama frate, pai tu postezi probleme dintr-astea sadice. Poate ca Gigel ar trebui sa dea mai mult de 2 puncte in plus la examen. Ce crezi? :)  :?

eh eu's mai pervers din fire ;) cat despre examen si gigel nush ce sa zic, probabil ca da punctele alea, dar de preferat e sa-l intrebi pe el :)


abstract algebra
topology
group theory
superior algebra
superior algebra unsolved

hello all,

i cannot solve the following, maybe it is harder that it seems at the first glance or maybe i just got messed up & stuck a little with it :) appreciate your help a lot:)

show that if each element in a group G is a square of some element, then G is abelian

Is that correct?
Consider a group of odd order that is not abelian (that exists, there is one of order 21 actually)

now take a random element x, it will have an order g, such that x^g=e with e the identity
however g must divide 21 and thus is odd

thus : x^(2*k-1)=e
or x^(2*k)=x
or x=(x^k)^2

or am i confused now?

That looks fine to me. Could you give some more information, or a link, about that non-Abelian group having 21 elements?

On the other hand, it is obvious that if $x^2 = e,\ \forall x \in G$, then $G$ must be Abelian. ...

No wonder you couldn't prove it: it's not true :).

Take any non-abelian finite group $G$ with an odd number of elements $n$. Assume that for some $x,y\in G$ we have $x^2=y^2$. Let $p,q$ be integers for which $2p+nq=1$. Then, from $x^2=y^2$ and $x^n=y^n=1$, we get $x=x^{2p+nq}=y^{2p+nq}=y$, so the map $x\mapsto x^2$ is injective. Since the group is finite, it's also surjective, so every element is the square of some other element. However, the group was chosen to be non-abelian, so that's that :).

Edit:

Those posts weren't there when I started writing :).

That's actually a very wide interesting theory, which is also explained at mathreference : http://www.mathreference.com/grp,sdp.html

You have to construct something called a semidirect product
You start with a group G and X, and a homomorphism sigma from X to Aut ( G) (note the abstractness, sigma goes from the group X to the group of automorphisms of G)

let us agree that x^(a*b) denotes (x^b)^a (some authors differ on the convention )

now you define this operation law on G * X
(g1,x1)*(g2,x2)=(g1+g2^sigma(x1),x1*x2)

one easily checks all axioms of a group are fulfilled
now this new group is abelian if and only if G and X are abelian , and sigma is the trivial map to Aut(G) (one checks that easily)

well, take for G the group C7 (so modulo seven, additive)
and for X the group C3

map zero in C3 to the trivial automorphism of G
map 1 in C3  to the 'the automorphism in G that multiplies with two"
map 2 in C3  to the 'the automorphism in G that multiplies with four"

as usual , some numbers are special in group theory, in this case 21, and it is always interesting to wonder why
well there is this theorem :

if the order is p, then it is cyclic
if the order is p^2, then it is Cp*Cp or C(p^2)
if it is p*q, with p and q primes and q bigger than p , then there are two cases :
p does not divide q-1  ---> only cyclic group ( this is the case for fifteen)
p does divide q-1--> cyclic group and one more : a nonabelian semidirect product

If you have more questions, or want more details, post it here or do a PM

thanks a lot for all the replies!
i've just realized x -> x^2 must be a homeomorphism in my problem besides being surjective :) which really makes a difference :))

The problem¿ (let $f: x \mapsto x^2$ be a homomorphism, show that the group is abelian)


inequalities
trigonometry
inequalities unsolved

Let $a,b\in R^+$ and $x\in R$.Prove the inequality

$\sqrt [\sin^2x]{a^2}+\sqrt [cos^2x]{b^2}\geq \frac {2a^2b^2}{2-\sin ^2{2x}}$.

Rewrite in [tex]\displaystyle (\sin^4(x)+\cos^4(x))(\sqrt [\sin^2x]{a^2}+\sqrt [cos^2x]{b^2})\geq a^2b^2[/tex]
Then apply Cauchy Shwartz and then Young:
[tex]\displaystyle (\sin^4x+\cos^4x)(\sqrt [\sin^2x]{a^2}+\sqrt [cos^2x]{b^2})\geq[/tex]
[tex]\displaystyle (\sin^2x\sqrt [\sin^2x]{a}+\cos^2x\sqrt [cos^2x]{b})^2 \geq[/tex]
[tex]\displaystyle (ab)^2[/tex]

That's it!


Problem 3
Let $ H$ be the orthocenter of the triangle $ ABC$ which is inscribed in a circle with center $ K$ and radius$ R\equal{}1$ ($ \angle B,\angle C\leq 90^{o}$ is given). If $ S$ is the intersection point of the lines that are defined from the segments $ HK$ and $ BC$ and holds that $ SK\cdot (SK\minus{}SH)\equal{}1$, evaluate the area of the quotation $ ABHCA$. .

see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=190777]here[/url]  :wink:


function
logarithms
number theory unsolved
number theory

Question

The Mangoldt function $ \Lambda$ is defined by:

$ \Lambda (n) = \log{p}$, if $ n = p^{k}$, where $ p$ is a prime and $ k \geq 1$ or $ \Lambda (n) = 0$ otherwise.

Prove that    $ \, \, \,\Lambda (n) = \sum_{d|n}\mu \left(\frac{n}{d}\right) \log{d}=-\sum_{d|n}\mu (d) \log{d}$

It shouldn't matter as long as you take a consistent base.

It is much easier to think about Dirichlet products:
$ f*g(n)=\sum_{d\mid n}f(d)g(n/d)$.
The arithmetic functions under this operation $ *$ form a group.  So, it is equivalent to prove that $ u*\Lambda=\log N$ where $ u(n)=1$ for $ n$ and $ N(n)=n$ for all $ n$.  This should be easier.

Note:  I am using $ u$ (not $ \mu$) so moreover $ u*\mu=I$ where $ I(n)=1$ if $ n=1$ and 0 elsewhere; that is, $ I$ is the identity function in this group and also $ u$ and $ \mu$ are inverses.

It is in fact $ \ln{x}$, it is assumed that if a base is not specified then we have a natural logarithm. But as t0rajir0u correctly points out, it doesn't actually matter what base we take, just as long as it is consistent.

By the way, the mobius inversion formula is very useful...


Suppose A + B + C = 3 where A,B,and C belong to the set of integers,

and  -10 < A,B,C < 10.

A is a perfect cube, B is a perfect square, and C is an integer
(not necessarily a perfect power of any number).

In how many ways can 3 be partitioned into triples (A,B,C)?

Hints/info as to what the perfect cubes &
the perfect squares for this problem are,
if you're wanting to know:
[hide]perfect cubes: -8,-1,0,1,8
perfect squares:  0,1,4,9[/hide]

"Jim, he's dead.", that is the answer, unless there are overlooked cases.


Prove that a positive integer can be written as the sum of two perfect squares if and only if it can be written as the sum of the squares of two rational numbers.

This problem is posted before ,the main idea of proof is showing that a number $ n$can represent as sum of two squares if and only if $ v_p(n)$ is even for all prime p of the form 4k+3

The main idea is http://www.mathlinks.ro/viewtopic.php?t=146761 .


geometry
trapezoid
analytic geometry
circumcircle
graphing lines
slope
quadratics

i have to solve geometry problem  :( :

(picture is attatched)

in the half circumference (sfere) you draw in trapezium with two sides even (b=d) and side a is diameter of circumference (2R). than in this trapezium you have to draw in another circumference with radius r, that thuches all sides of trapezium - in draw circumference.

i have to  express r with R - r(R).

i aready tried solve this problem. i think that there is one possbility of costruct this, but i dont find it.

(sorry for my bad english, i didnt find the appropriate words.) :blush:

tjasa

$r=\frac{\sqrt{abcd}}{s}$ where $s=\frac{a+b+c+d}{2}$ and if you know $a,b,c,d$ it's done.

Assume that the problem has been solved and that $ABCD$ is the desired bicentric trapezoid with the parallel bases $AB \parallel CD$, $AB > CD$. Since all cyclic trapezoids are isosceles, $BC = DA$. Put the coordinate origin at the trapezoid circumcenter $O$ (the midpoint of the base $AB$) and the positive x-axis along the ray $OB$. The points $A, B$ have the coordinates $A = (-R, 0),\ B = (+R, 0)$. The lines $DA, BC$ passing through the points $A, B$ have equations

$DA:\ \ y = m(x + R)$

$BC:\ \ y = -m(x - R)$

where $+m > 0, -m < 0$ are the unknown slopes of the lines $DA, BC$, and the equation of the trapezoid circumcircle $(O)$ is

$(O):\ \ x^2 + y^2 = R^2$

The intersections $D, B$ of the lines $DA, BC$ are determined by solving the set of 2 equations for the circumcircle and the corresponding line. Substituting the equation of the line $DA$ into the circumcircle equation,

$m^2(x + R)^2 + x^2 = R^2$

$x^2(m^2 + 1) + 2m^2Rx + R^2(m^2 - 1) = 0$

The roots of this quadratic equation are the x-coordinates of the points $D, A$ (we need only $D$):

$x_{D,A} = \frac{-2m^2R \pm \sqrt{4m^4R^2 - 4R^2(m^2 + 1)(m^2 - 1)}}{2(m^2 + 1)} =$

$= \frac{-m^2R \pm R \sqrt{m^4 - (m^4 - 1)}}{m^2 + 1} = R\ \frac{-m^2 \pm 1}{m^2 + 1}$

The root $x_A = -R$ with the minus sign corresponds to the point $A$ and the other root

$x_D = -R\ \frac{m^2 - 1}{m^2 + 1}$

with the plus sign corresponds to the point $D$. We can find the x-coordinate of the point $C$ in the same way, but from the symmetry of the isosceles trapezoid $ABCD$, it is clear that

$x_C = -x_D = +R\ \frac{m^2 - 1}{m^2 + 1}$

$y_D = y_C = m(x_D + R) = m \left(R - R\ \frac{m^2 - 1}{m^2 + 1}\right) = \frac{2mR}{m^2 + 1}$

Since the trapezoid $ABCD$ is tangential (has an incircle), the sums of the opposite sides are equal:

$AB + CD = BC + DA = 2BC$

$AB = 2R,\ \ CD = x_C - x_D = 2R\ \frac{m^2 - 1}{m^2 + 1}$

$BC = \sqrt{(x_C - x_B)^2 + (y_C - y_B)^2} =  \sqrt{\left(R\ \frac{m^2 - 1}{m^2 + 1} - R\right)^2 + \left(\frac{2mR}{m^2 + 1}\right)^2} =$

$= R\ \frac{\sqrt{\left(m^2 - 1 - (m^2 + 1)\right)^2 + 4m^2}}{m^2 + 1} = 2R\ \frac{\sqrt{m^2 + 1}}{m^2 + 1} =  \frac{2R}{\sqrt{m^2 + 1}}$

As a result, we get a quadratic equation for square of the slope $m$:

$AB + CD = 2R + 2R\ \frac{m^2 - 1}{m^2 + 1} = \frac{4R}{\sqrt{m^2 + 1}} = 2BC$

$1 + \frac{m^2 - 1}{m^2 + 1} = \frac{2}{\sqrt{m^2 + 1}},\ \ \frac{2m^2}{m^2 + 1} = \frac{2}{\sqrt{m^2 + 1}},\ \ \frac{m^2}{\sqrt{m^2 + 1}} = 1$

$m^4 - m^2 - 1 = 0,\ \ m^2 = \frac{1 \pm \sqrt 5}{2}$

Since $m^2 > 0$, the negative root is not acceptable and

$m = \pm \sqrt{\frac{1 + \sqrt 5}{2}} = \pm \sqrt G$

where $G$ is the golden ratio. The negative slope corresponds to the same bicentric trapezoid turned upside down, hence we can ignore it. The radius of the inscribed circle $(I)$ is obviously equal to

$r = \frac{y_C}{2} = \frac{mR}{m^2 + 1} = R \frac{\sqrt G}{G + 1} = R\ \frac{\sqrt G}{G^2} = \frac{R}{\sqrt {G^3}} = \frac{R}{\sqrt{2G + 1}} = \frac{R}{\sqrt{2 + \sqrt 5}}$

The construction of the bicentric trapezoid $ABCD$ begins like the construction of a regular pentagon (which is full of golden ratios). Construct a circle $(O)$ centered at a point $O$ and with diameter $AB = 2R$. Let the normal to the line $AB$ at the point $O$ intersect the circle $(O)$ at a point $N$. Find the midpoint $M$ of the segment $OB$. By Pythagorean theorem, the segment $MN = \frac{R \sqrt 5}{2}$. Construct a circle $(M)$ centered at the point $M$ and with radius $MN$. This circle intersects the line $AB$ at points $P, Q$ left and right from the point $M$. The segment $PN$ is the side of a regular pentagon with the circumcircle $(O)$, but we do not need it. The segments

$OA = R,\ \ OQ = OM + MQ = OM + MN = \frac{R}{2} + \frac{R \sqrt 5}{2} = R\ \frac{1 + \sqrt 5}{2}$

are in the ratio $\frac{OQ}{OA} = \frac{1 + \sqrt 5}{2} = G$. To construct the square root of the golden ratio, find the midpoint $K$ of the segment $AQ$ and construct a circle $(K)$ centered at the point $K$ and with radius $KA = KQ$. This circle intersects the normal $ON$ at a point $X$, such that $OX = R \sqrt G = R \sqrt{\frac{1 + \sqrt 5}{2}}$. The lines $AX, BX$ intersect the circle $(O)$ at the points $D, C$ different form $A, B$.

Nikdy nepriznavej, ze tvuj problem je domaci ukol. O tohle forum se staraji vetsinou ucitele a ti to nemaji radi.

Ahoj, Yetti.

Thank you so much, Yetti. :)   It was more difficult homework that I thought.

goska

PS: I' m not from Slovakia, so I don't understand the sentence that you wrote at the bottom.  :?

There is a slightly simpler solution, based on Slizzel's idea in post #2.

[color=blue][b]Problem.[/b] Let ABCD be an isosceles trapezoid, with AB || CD and BC = AD. As every isosceles trapezoid, this trapezoid ABCD must have a circumcircle; assume that the segment AB is a diameter of this circumcircle. Let R be the radius of this circumcircle. Also, assume that the trapezoid ABCD has an incircle, and let r be the radius of this incircle. Prove that $\frac{R}{r}=\sqrt{2+\sqrt5}$.[/color]

[i]Solution.[/i] Let a = AB and c = CD be the lengths of the bases of the isosceles trapezoid ABCD, and let b = BC = AD be the length of its equal legs.

Also, by symmetry, the diagonals AC and BD of the isosceles trapezoid ABCD are equal; denote e = AC = BD.

Since the trapezoid ABCD has a circumcircle, Ptolemy's theorem yields $AB\cdot CD+BC\cdot AD=AC\cdot BD$. In other words, $a\cdot c+b\cdot b=e\cdot e$, so that $ac+b^2=e^2$. On the other hand, since the segment AB is a diameter of the circumcircle of the trapezoid ABCD, while the point C lies on this circumcircle, we have < ACB = 90°; thus, the triangle ACB is right-angled, and Pythagoras yields $AC^2=AB^2-BC^2$. In other words, $e^2=a^2-b^2$. Comparing this with $ac+b^2=e^2$, we get $ac+b^2=a^2-b^2$. Equivalently, $2b^2=a^2-ac$. But the quadrilateral ABCD has an incircle; hence, AB + CD = BC + AD, so that a + c = b + b. So we have a + c = 2b. Thus, $2b^2=\frac12\left(2b\right)^2=\frac12\left(a+c\right)^2$. Thus, the equation $2b^2=a^2-ac$ becomes $\frac12\left(a+c\right)^2=a^2-ac$. Considering this as a quadratic equation for a and solving it, we get $a=\left(2+\sqrt5\right)c$ (the other possible solution $a=\left(2-\sqrt5\right)c$ doesn't come into question, since a must be positive). Hence, $\frac{a}{c}=2+\sqrt5$.

Since the area of a quadrilateral which has an incircle equals the semiperimeter $\cdot$ the inradius, the area F of the quadrilateral ABCD must equal F = sr, where $s=\frac{AB+BC+CD+AD}{2}$ is the semiperimeter of the quadrilateral ABCD. On the other hand, by the Brahmagupta formula (see http://www.mathlinks.ro/Forum/viewtopic.php?t=15973 post #1), since the quadrilateral ABCD is cyclic, its area F is $F=\sqrt{\left(s-AB\right)\left(s-BC\right)\left(s-CD\right)\left(s-AD\right)}$. Thus, $sr=\sqrt{\left(s-AB\right)\left(s-BC\right)\left(s-CD\right)\left(s-AD\right)}$. Now, since a + c = 2b, we have $s=\frac{AB+BC+CD+AD}{2}=\frac{a+b+c+b}{2}=\frac{\left(a+c\right)+2b}{2}=\frac{2b+2b}{2}=2b$, so that s - AB = 2b - a = c (since a + c = 2b), s - BC = 2b - b = b, s - CD = 2b - c = a (again since a + c = 2b), and s - AD = 2b - b = b. Thus, $sr=\sqrt{\left(s-AB\right)\left(s-BC\right)\left(s-CD\right)\left(s-AD\right)}$ becomes $2br=\sqrt{cbab}$. In other words, $2br=b\sqrt{ac}$. Upon division by b, this becomes $2r=\sqrt{ac}$.

On the other hand, since the segment AB is a diameter of the circumcircle of the trapezoid ABCD, its length a = AB equals twice the radius R of this circumcircle; so we have a = 2R. Thus,

$\frac{R}{r}=\frac{2R}{2r}=\frac{a}{\sqrt{ac}}=\sqrt{\frac{a}{c}}=\sqrt{2+\sqrt5}$.

And the problem is solved.

  Darij


Prove that every polynomial with complex coefficients is a factor of a polynomial with real coefficients.

[hide]
The idea is simple, but I lack the ability to express it nicely, so this'll have to do.

As the polynomial (let's call it $q(x)$) has complex coefficients, it must have at least one complex root (as a polynomial with all real roots has all real coefficients). For each complex root $a+ib$, multiply the polynomial by $(x-a+bi)$. So, the polynomial becomes:

$p(x) = (x-a_{1}-ib_{1})(x-a_{1}+ib_{1})...(x-a_{n}-ib_{n})(x-a_{n}+ib_{n})s(x)$
where $s(x)$ is a polynomial with all real roots (and so real coefficients), and where $a_{1}+ib_{1},...,a_{n}+ib_{n}$ are the complex roots of $q(x)$.
This becomes:
$p(x) = (x^{2}-2a_{1}x+a_{1}^{2}+b_{1}^{2})...(x^{2}-2a_{n}x+a_{n}^{2}+b_{n}^{2})s(x)$
But this is just the product of several polynomials with real coefficients. As the product of two polynomials with real coefficients is another polynomial with real coefficients, $p(x)$ has all real coefficients. Also, $q(x)$ is a factor of $p(x)$ (as all roots of $q(x)$ are roots of $p(x)$ by construction), so we're done.
[/hide]

[quote="gauss202"][quote="rjt"]
As the polynomial (let's call it $q(x)$) has complex coefficients, it must have at least one complex root (as a polynomial with all real roots has all real coefficients).[/quote]

This is not true. For example, $q(x) = ix^{2}-4i$ has complex coefficients, but two real roots.

For the rest of your solution, you're on the right track though.[/quote]

Ah, indeed.

Would I be correct in saying instead (at the start of my argument):
As $q(x)$ has complex coefficients, then either:
a) $q(x)=is(x)$, in which case, either $iq(x)$ has all real coefficients (and we're done), or $iq(x)$ has at least one complex coefficient (and we then we can proceed as above).
b) $q(x)$ has at least one complex root, proceed as above.

How about this alternative argument?

Suppose our polynomial is $p(x) = \alpha_{0}x^{n}+\alpha_{1}x^{n-1}+\ldots+\alpha_{n}$ where the sequence of $\alpha$ can have complex elements. Write $\alpha_{0}= a+bi$ for real $a$ and $b$. If we multiply this polynomial by $(\frac{a-bi}{a^{2}+b^{2}}x+1)$, then we get a monic polynomial of degree n+1. By FTA this polynomial must be the product of $n+1$ factors of the form $(x+\beta)$ where $\beta$ is a real or complex number. If the polynomial has all real coefficients, then we're already done. If not, then it has a complex coefficient and one of those roots must be complex. Then we proceed with the proof above.

Both arguments overcomplicate for my tastes...

[hide="The intended solution"] Every polynomial $p(x)$ with complex coefficients can be written in the form $p(x) = a(x)+i b(x)$ where $a(x), b(x)$ are polynomials with real coefficients.  Then the polynomial

$a(x)^{2}+b(x)^{2}$

Has $p(x)$ as a factor, and has all real coefficients. [/hide]

suppose we have p(x), with a root as a, then consider the polynomial,q(x), with roots conj(a), then p(x)q(x)=prod (x-a)(x-conj(a)), but we know conj(a)+a is real and conj(a)*a is real, so p(x)q(x) is a polynomial with real coefficients

rjt, your corrected version still misses the case $q(x) = (1+i)x^{2}-(4+4i)$, I believe.  (MellowMellon has one fix; another is simply to divide out by the leading coefficient, which won't change the roots.)

Altheman, your solution fails, too, with the same counter-example -- you've neglected the leading coefficient.  Similarly, there is no "the polynomial with roots ...," there are (infinitely) many such polynomials.

good point; then, suppose that p(x) has roots r_i for i=1,2,...,n, then take q(x) with roots s_i=conj(r_i) for i=1,2,...,n, the lead coefficient of q(x) is the conjugate of the lead coefficient of p(x), it follows that p(x)q(x)=a polynomial with real coefficients by considerint the product of the factors and lead coefficient


[quote="e.lopes"]Nice!  :smile:

the idea is prove that $\sqrt{a_{n+2}} = 4\sqrt{a_{n+1}} - \sqrt{a_n}$

and this is easy!  :lol:[/quote]
Maybe it should be ${a_{n+2}} = 4{a_{n+1}} - {a_n}$ :wink:

And maybe you should all post solutions instead of ideas. :P

hawk tiger: i don't no where i find this \sqrt, thanks for your correction :grin:  :oops:

peter: sorry peter, i will post complete solutions now  :razz:

[quote="Peter"]The sequence $\{a_{n}\}_{n \ge 1}$ is defined by \[a_{1}=1, \; a_{n+1}=2a_{n}+\sqrt{3a_{n}^{2}+1}.\] Show that $a_{n}$ is an integer for every $n$.[/quote]

Well, as nobady has written a complete solution I will post  bellow a solution for a generalization of the problem.

[color=blue]Problem:The sequence $\{a_{n}\}_{n \ge 1}$ is defined by $a_{1}=1, \; a_{n+1}=ma_{n}+\sqrt{(m^{2}-1)a_{n}^{2}+1}$,where $m\in{N}$. Show that $a_{n}$ is an integer for every $n$. [/color]

[color=red]Solution:[/color] Let us rewrite the condition in the following way and square both sides: $a_{n+1}-ma_{n}=\sqrt{(m^{2}-1)a_{n}^{2}+1}$, so $a_{n+1}^{2}+m^{2}a_{n}^{2}-2ma_{n}a_{n+1}=(m^{2}-1)a_{n}^{2}+1$,thus $a_{n+1}^{2}+a_{n}^{2}=2ma_{n+1}a_{n}+1$ and $a_{n}^{2}+a_{n-1}^{2}=2ma_{n}a_{n-1}+1$. Now subtracting the last two equations we get that $a_{n+1}^{2}-a_{n-1}^{2}=2ma_{n}(a_{n+1}-a_{n-1})$, therefore

\[a_{n+1}=2ma_{n}-a_{n-1}\]

And we can easily see that $a_{1}=1$ and $a_{2}=2m$, consequently all the members of the sequence are integers, as the problem calemed.

[quote=Peter]The sequence $\{a_{n}\}_{n \ge 1}$ is defined by \[a_{1}=1, \; a_{n+1}=2a_{n}+\sqrt{3a_{n}^{2}+1}.\] Show that $a_{n}$ is an integer for every $n$.[/quote]
$$\bf\color{red}My \ solution$$$\color{green}\textbf{Claim.}$ \begin{align*}
&4a_{n+1}-a_n=a_{n+2}
\end{align*}for all $n\in \mathbb{N}.$
$\color{green}\textbf{Proof}.$ Before starting the proof of the claim, we need to prove some facts.
------------------------------------
$\color{blue}\textbf{Fact 1}.$ $a_n$ is positive for all $n\in \mathbb{N}.$
$\color{blue}\textbf{Proof of fact 1}.$ We will use induction. For $n=1,$ $a_1=1>0.$ Let's assume that $a_n$ is positive. Then $a_{n+1}=2a_n+\sqrt{3a^2_n+1}>0.$ So, the induction completed.
----------------------------------------------
$\color{red}\textbf{Fact 2}.$ The sequence is increasing.
$\color{red}\textbf{Proof of fact 2}.$ Since $a_n$ is positive for every $n\in \mathbb{N},$ we have
\[a_{n+1}=2a_n+\sqrt{3a^2_n+1}=a_n+(a_n+\sqrt{3a^2_n+1})> a_n,\]as desired.
-------------------------------------------
Now, we can turn into the proof of the claim. Starting from the given condition
\begin{align*}
&a_{n+1}=2a_n+\sqrt{3a^2_n+1}\ (\bigstar)
\end{align*}$\iff$
\[a_{n+1}-2a_n=\sqrt{3a^2_n+1}\]$\iff$
\[a^2_{n+1}+4a^2_n-4a_{n+1}a_n=3a^2_n+1\]$\iff$
\[a^2_{n+1}+a^2_n-4a_{n+1}a_n=1.\]Since $(\bigstar)$ holds for all $n\in \mathbb{N},$
\[a^2_{n+1}+a^2_n-4a_{n+1}a_n=1\]holds for all $n\in \mathbb{N},$ too.
Then \[a^2_{n+2}+a^2_{n+1}-4a_{n+2}a_{n+1}=1\]is also true. $\Longrightarrow$
\[a^2_{n+1}+a^2_n-4a_{n+1}a_n=a^2_{n+2}+a^2_{n+1}-4a_{n+2}a_{n+1}\]$\Longrightarrow$
\[a^2_n-4a_{n+1}a_n=a^2_{n+2}-4a_{n+2}a_{n+1}\]$\Longrightarrow$
\begin{align*}
&(a_{n+2}-a_n)(a_{n+2}+a_n)=4a_{n+1}(a_{n+2}-a_n).\ (\clubsuit)
\end{align*}In the $\color{red}\textbf{fact 2},$ we proved that the sequence is increasing. So,
\[a_{n+2}-a_n>a_{n+1}-a_n>0,\]which means that we are able to cancel $a_{n+2}-a_n$ in $(\clubsuit).$ Canceling $a_{n+2}-a_n,$ we obtain
\[a_{n+2}+a_n=4a_{n+1}\]$\Longrightarrow$
\[4a_{n+1}-a_n=a_{n+2},\]so the claim has been proved.
Setting $n=1$ in
\[a_{n+1}=2a_n+\sqrt{3a^2_n+1}\]and using $a_1=1,$
we get that $a_2=4.$
Then, the first $2$ terms of the sequence are integers, and there is a linear relationship between $3$ consecutive terms, so all the terms are integers, as desired. $\square$

Similar problem:
Define the sequences $(a_n)_{n\geq1} $ by $a_1=1$ and $a_{n+1}=2a_n+\sqrt{3a_n^2-2}$ for all integers $n\geq1.$ Prove that $a_n$ is an integer for all $n.$
$\bf{Solution:}$
We have $(a_{n+1}-2a_n)^2=3a_n^2-2,$ $n\geq1.$
Then $a_{n+1}^2+a_n^2-4a_{n+1}a_n=-2$
And $a_{n+2}^2+a_{n+1}^2-4a_{n+2}a_{n+1}=-2,$ $n\geq 1.$
Subtracting these relations yields
$a_{n+2}^2-a_n^2-4a_{n+1}(a_{n+2}-a_n)=0$
And $(a_{n+2}-a_n)(a_{n+2}+a_n-4a_{n+1})=0,$ $n\geq 1.$
Since the sequence $(a_n)_{n\geq1}$ is increasing,it follows that $a_{n+2}=4a_{n+1}-a_n,$ $n\geq1.$
Taking into account that $a_1=1,a_2=3$ and inducting on $n$ we reach the conclusion.


[quote="tranthanhnam"]$prove $$that$
$a+d+bc+3\sqrt{(1+a^2+b^2)(1+c^2+d^2)} \geq 2(1+c+d)+2(ac+ad+bd)$[/quote]

what's conditions for a,b c,d ???????

He didn't mention any conditions so maybe there are none.

The inequality looks pretty odd though...

[quote="Arne"]He didn't mention any conditions so maybe there are none.[/quote] There must be, since it's false for $a=b=c=1,d=2$. :?


advanced fields
advanced fields unsolved

Does any body knows if given a prime $p$ such that $p\equiv1\mbox{ mod }4$ then it is possible to express $p=a^2+b^2$ where $a$ and $b$ are both integer numbers??

Best,

yes it is possible, i will give you a clue: prove, p is not irreducible in $Z+Zi$, using the fact that -1 is a square modulo p

Yeah, man, it's really well-known and it's been discussed here many times. The last topic dealing with this is, I believe, [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=26548]this one[/url].


inequalities
vector
inequalities proposed

If $ a, b, c$ are positive numbers, then

$ \sqrt{a \plus{} \frac {1}{b}}\plus{}\sqrt{b \plus{} \frac {1}{c}} \plus{} \sqrt{c \plus{} \frac {1}{a}}\leq\frac{3}{2}\sqrt{\left(a \plus{} \frac {1}{b}\right)\left(b \plus{} \frac {1}{c}\right)\left(c \plus{} \frac {1}{a}\right)},$

with equality if and only if $ a\equal{}b\equal{}c\equal{}1.$

By am-gm and Cauchy-Schwarz we get
$ \frac{\sum{\sqrt{a\plus{}\frac{1}{b}}}}{\prod{\sqrt{a\plus{}\frac{1}{b}}}}\equal{} \sum{\frac{1}{\sqrt{a\plus{}\frac{1}{b}}\cdot \sqrt{b\plus{}\frac{1}{c}}}}\equal{} \sum{\frac{1}{\sqrt{ab\plus{}1}\cdot \sqrt{\frac{1}{bc}\plus{}1}}}\leq$
$ \leq\sqrt{\sum{\frac{1}{ab\plus{}1}}}\cdot \sqrt{\sum{\frac{1}{\frac{1}{bc}\plus{}1}}}\equal{} \sqrt{\sum{\frac{1}{ab\plus{}1}}}\cdot \sqrt{\sum{\frac{bc}{bc\plus{}1}}}\equal{}$
$ \equal{}\sqrt{\left(\sum{\frac{1}{ab\plus{}1}}\right)\left(3\minus{}\sum{\frac{1}{ab\plus{}1}}\right)}\leq \frac{3}{2}$
It remains to examine the case of equality. Suppose that both inequalities we use are equalities. From the first one, the vectors $ \mathbf{v}\equal{}\left(\frac{1}{\sqrt{ab\plus{}1}}, \frac{1}{\sqrt{bc\plus{}1}}, \frac{1}{\sqrt{ca\plus{}1}}\right)$ and $ \mathbf{w}\equal{} \left(\frac{1}{\sqrt{\frac{1}{bc}\plus{}1}}, \frac{1}{\sqrt{\frac{1}{ca}\plus{}1}}, \frac{1}{\sqrt{\frac{1}{ab}\plus{}1}}\right)$ are linearly dependent. Their entries are positive, so the exists a positive number $ k$ such that $ \mathbf{w}\equal{}k\mathbf{v}$. Let $ \mathbf{e}\equal{}\left(1,1,1\right)$ and $ L\left(x,y,z\right)\equal{}\left(y,z,x\right)$. Then $ L^{3}\equal{}I$, $ L\left(\mathbf{e}\right)\equal{}\mathbf{e}$ and $ \mathbf{w}\equal{}\mathbf{e}\minus{}L\left(\mathbf{v}\right)$. Thus $ L\left(\mathbf{v}\right)\equal{} \mathbf{e}\minus{} k\mathbf{v}$, $ L^{2}\left(\mathbf{v}\right)\equal{} \mathbf{e}\minus{} k\left(\mathbf{e}\minus{} k\mathbf{v}\right)\equal{} \left(1\minus{}k\right)\mathbf{e}\plus{} k^{2}\mathbf{v}$ and $ \mathbf{v}\equal{} L^{3}\left(\mathbf{v}\right)\equal{} \left(1\minus{}k\minus{}k^{2}\right)\mathbf{e}\plus{} k^{3}\mathbf{v}$. Therefore $ \left(1\minus{}k\minus{}k^{2}\right)\mathbf{e}\plus{} \left(k^{3}\minus{}1\right)\mathbf{v}\equal{}\mathbf{0}$. The vectors $ \mathbf{v}$ and $ \mathbf{e}$ can not be linearly independent, because it would imply $ 1\minus{}k\minus{}k^{2}\equal{}0\equal{}k^{3}\minus{}1$. Thus they are linearly dependent. It follows that all the entries of $ \mathbf{v}$ are equal and consequently $ a\equal{}b\equal{}c$. From the second inequality being an equality we get $ \frac{3}{a^{2}\plus{}1}\equal{}\frac{3}{2}$, meaning that $ a\equal{}b\equal{}c\equal{}1$.


integration
calculus
analytic geometry
graphing lines
slope
calculus computations

Solve:
$ y''\equal{}a\cdot y \sqrt{1\plus{}y'^2}$

$ a$ is constant, independent variable is $ x$

[quote="yeahnigga"]Solve:
$ y'' \equal{} a\cdot y \sqrt {1 \plus{} y'^2}$

$ a$ is constant, independent variable is $ x$[/quote]

$ y'' \equal{} a\cdot y \sqrt {1 \plus{} y'^2}$

$ \Rightarrow a \;\;y\; \equal{} \;\dfrac{y''}{\sqrt {\;1 \plus{} y'\,^2\;}}$

multiply by $ 2\,y'$ to get

$ 2a\;y\,y'\; \equal{} \;\dfrac{2y'\,y''}{\sqrt {\;1 \plus{} y'\,^2\;}}$

$ \Rightarrow a \;\;d(y^2)\; \equal{} \;2\;\;d\left(\sqrt {\;1 \plus{} y'\,^2\;}\right)$

integrating both sides gives

$ a\;\;y^2\; \equal{} \;2\;\;\sqrt {\;1 \plus{} y'\,^2\;}\;\; \plus{} \;\;C_1$

$ \Rightarrow \;\int\;\dfrac{dy}{\sqrt {\;(a\;y^2 \minus{} C_1)^2 \minus{} 4\;}}\; \equal{} \;\pm\dfrac{1}{2}\;\int\;dx\; \equal{} \;\pm\dfrac{1}{2}\;(x \plus{} C_2)$

$ \pm\dfrac{a}{2}\;(x \plus{} C_2)\; \equal{} \;\int\;\dfrac{dy}{\sqrt {\;(y^2 \minus{} \dfrac{C_1}{a})^2\; \minus{} \;\left(\dfrac{2}{a}\right)^2\;}}$

looks like the RHS will yield an elliptic integral of the $ 1^{st}$ kind...  :maybe: do you know what they are ?

Another possibility is the qualitative approach.  Write it as the system:

$ y' \equal{} u$

$ u' \equal{} ay\sqrt {1 \plus{} u^2}$

Now, plot the phase diagram for this system (the slope field in the y-u plane is close enough).  You'll find it has two states depending on the value of a.   If you linearize this system and calculate the eigenvalues, the origin is either a saddle or center depending on $ a$ and giving rise to either periodic motion or motion which tends then moves away from the origin.


geometry
incenter
angle bisector
geometry solved

Let ABC be a triangle and let (T) be a circle passing through A and C. Sides AB and BC meet (T) again at D and E respectively. Let (U) be the incircle of the circular triangle EBD (which touches two sides BD ,BE and the arc DE) and let S be its center. Suppose that (U) touches the arc DE at M. Prove that the angle bisector of angle AMC passes through the incenter of triangle ABC

http://www.mathlinks.ro/Forum/viewtopic.php?t=6086

  darij


Prove the following inequality$:$ $(x+y+z)(xy+yz+zx) \geq 9xyz+(y-x)(z-x)^{2}$, where $0<x \leq y \leq z$.

Let [tex]f(x,y,z)=(x+y+z)(xy+yz+zx)-9xyz[/tex]. Then [tex]f(x+a,y+a,z+a)-f(x,y,z)=2a(x^2+y^2+z^2-xy-yz-zx)[/tex] is minimal when [tex]a[/tex] is minimal. Hense we may assume [tex]x=0[/tex]. Then it's obvious.

I think [tex](y-x)(z-x)^2[/tex] is too weak,since:
[tex](x+y+z)(xy+xz+yz)-9xyz=z(x-y)^2+x(y-z)^2+y(z-x)^2[/tex] ;)


Consider a triangle $ABC$.Let  $D,E,F\in (ABC)$  such that :
([b][u]Notes:[/u][/b]$(XYZ)$ is the circumcircle of a triangle $XYZ$)
$\frac{AF}{FB}=\frac{CE}{EA}=\frac{BD}{DC}=k>0$
Construct the triangle $A_1B_1C_1$such that:$A_1B_1=CF;B_1C_1=AD;C_1A_1=BE$
Let $D_1,E_1,F_1\in (A_1B_1C_1)$ such that :
$\frac{A_1F_1}{F_1B_1}=\frac{C_1E_1}{E_1A_1}=\frac{B_1D_1}{D_1C_1}$
Construct the triangle $A_2B_2C_2$such that :$A_2B_2=C_1F_1;B_2C_2=A_1D_1;C_2A_2=B_1E_1$
Prove that :a) $\exists$ the triangle $A_1B_1C_1$and$A_2B_2C_2$
                 b)$ABC$and $A_2B_2C_2$ are the similar triangles.

Please help me ! this problem is very difficult. :(


geometry
incenter
ratio
Euler
angle bisector
area of a triangle

In triangle $ABC$ with incenter at $I,$ excenters at $I_a, I_b, I_c$ prove that \[\dfrac{AI}{AI_a} + \dfrac{BI}{BI_b} + \dfrac{CI}{CI_c} = 1.\]

Except for the incircle, there are three more circles tangent to the lines on which the sides of the triangle lie. The centers of these three circles are called excentres, and that's because, unlike the incenter, they lie outside the triangle.

Hint: Prove that
\[\frac{AI}{AI_a} = \frac{r}{r_a},\]
where $r_a$ is the radius of the excircle opposite vertex $A$.

I think there is a simpler solution:

Since the internal angle bisector of an angle is perpendicular to the corresponding external angle bisector, the line $AII_a$ is perpendicular to the line $AI_bI_c$. Hence, the segment AI is an altitude in triangle $II_bI_c$, and the segment $AI_a$ is an altitude in triangle $I_aI_bI_c$. Now, if we denote the area of a triangle XYZ by [XYZ], then, since the area of a triangle equals $\frac12 \cdot$ side $\cdot$ corresponding altitude, we have

$\left[ II_{b}I_{c}\right] =\frac{1}{2}\cdot I_{b}I_{c}\cdot AI$ and
$\left[ I_{a}I_{b}I_{c}\right] =\frac{1}{2}\cdot I_{b}I_{c}\cdot AI_{a}$.

Thus,

$\frac{AI}{AI_{a}}=\frac{\left[ II_{b}I_{c}\right] }{\left[ I_{a}I_{b}I_{c}\right] }$.

Similarly,

$\frac{BI}{BI_{b}}=\frac{\left[ II_{c}I_{a}\right] }{\left[ I_{a}I_{b}I_{c}\right] }$

and

$\frac{CI}{CI_{c}}=\frac{\left[ II_{a}I_{b}\right] }{\left[ I_{a}I_{b}I_{c}\right] }$.

Hence,

$\frac{AI}{AI_{a}}+\frac{BI}{BI_{b}}+\frac{CI}{CI_{c}}=\frac{\left[ II_{b}I_{c}\right] }{\left[ I_{a}I_{b}I_{c}\right] }+\frac{\left[ II_{c}I_{a}\right] }{\left[ I_{a}I_{b}I_{c}\right] }+\frac{\left[ II_{a}I_{b}\right] }{\left[ I_{a}I_{b}I_{c}\right] }$
$=\frac{\left[ II_{b}I_{c}\right] +\left[ II_{c}I_{a}\right] +\left[ II_{a}I_{b}\right] }{\left[ I_{a}I_{b}I_{c}\right] }=\frac{\left[ I_{a}I_{b}I_{c}\right] }{\left[ I_{a}I_{b}I_{c}\right] }=1$.

  Darij

darij -- your solution is the same as the one proposed by the hint (and so not actually easier).  Those area ratios are exactly the same as the radii ratios.

Not exactly - the idea is similar, but it's not entirely equivalent. The solution proposed by the hint (at least, in the form I understand it) goes as follows: If we draw the perpendiculars from the points I and $I_a$ to the line CA, then Thales immediately shows us $\frac{AI}{AI_{a}}=\frac{r}{r_{a}}$; similarly, $\frac{BI}{BI_{b}}=\frac{r}{r_{b}}$ and $\frac{CI}{CI_{c}}=\frac{r}{r_{c}}$. Hence, in order to prove $\frac{AI}{AI_{a}}+\frac{BI}{BI_{b}}+\frac{CI}{CI_{c}}=1$, it remains to establish $\frac{r}{r_{a}}+\frac{r}{r_{b}}+\frac{r}{r_{c}}=1$, or, equivalently, $\frac{1}{r_{a}}+\frac{1}{r_{b}}+\frac{1}{r_{c}}=\frac{1}{r}$. And this is readily done using the formulas $S=sr=\left( s-a\right) r_{a}=\left( s-b\right) r_{b}=\left( s-c\right) r_{c}$, where S is the area of triangle ABC, and $s=\frac{1}{2}\left( a+b+c\right) $ is the semiperimeter. You see, this solution uses Thales (my solution didn't), and it uses the area of triangle ABC (my solution used those of triangles $II_bI_c$, $II_cI_a$, $II_aI_b$ and $I_aI_bI_c$ instead). The ideas are pretty alike but there is no complete equivalence.

Note that the problem can be generalized:

If ABC is a triangle, and P is a point in its interior, and if the lines AP, BP, CP intersect the sides BC, CA, AB of the triangle ABC at the points X, Y, Z, respectively, then

$\frac{PX}{AX}+\frac{PY}{BY}+\frac{PZ}{CZ}=1$.

This is called the [i]Gergonne-Euler theorem[/i] (and, indeed, it was discovered by Leonhard Euler, in contrast to the Euler circle which has not much to do with him).

To prove this theorem, you can apply a slight generalization of the solution I gave in my previous message. The initial problem is a particular case of this if you take the triangle $I_aI_bI_c$ as the base triangle and I as the point in its interior.

  Darij


Find the smallest number such that when it is divided successively by the number 45,454,4545,45454, it leaves the remainder 4,45,454,4545 respectively. In other words, if the number is x,

x / 45 has remainder 4.

x / 454 has remainder 45.

x / 4545 has remainder 454, and

x / 45454 has remainder 4545.

Good luck!

Answer: [hide]35641667749[/hide]

Hmm...Maybe I should've posted this in the intermediate problems board...

[quote="Rep123max"]I am guessing just doing the Chinese Remainder Theorem alot (which we learned in the MATHCOUNTS/AMC Number Theory Course), but there is probably a slicker way because the numbers have that pattern.[/quote]

Chinese Remainder theorem? Someone please help?


If between the elements of a triangle we have the relation $a+b=tg{\frac{C}{2}}(a*tgA+b*tgB)$, than the triangle is isosceles.


calculus
integration
vector
geometry
trigonometry
calculus computations

If $ D$ is a portion of the plane $ 4x\plus{}y\plus{}z\equal{}1$ which lies in the first octant oriented upward
How do you find the flux of the vector field $ F\equal{}4{i}\plus{}{j}\plus{}4{k}$ across surface $ D$

Since vector field $ F$ is a set of constant vectors, the flux you want to find is just the product of $ \text{norm}(\bold F)$ and the area of $ D$. Three vertexes of $ D$ is $ (0,0,1), (0,1,0)$ and $ (\frac{1}{4},0,0)$, so by three-dimensional Phytagorian theorem, the area of $ D$ is $ \sqrt{(1/2\cdot 1\cdot 1)^2\plus{}(1/2\cdot 1/4\cdot 1)^2\plus{}(1/2\cdot 1\cdot 1/4)^2}\equal{}\frac{3}{8}\sqrt{2}$. So, the answer is $ \frac{3}{8}\sqrt{2}\cdot\text{norm}(\bold F)\equal{}\frac{3}{8}\sqrt{2}\cdot\sqrt{4^2\plus{}1^2\plus{}4^2}\equal{}\frac{3}{8}\sqrt{66}$.

Hello thank you but the final answer is incorrect.
The area is $ D\equal{}\displaystyle\int^1_0 \int^\frac {1\minus{}y}{4}_0 \sqrt{1\plus{}(\minus{}4)^2\plus{}(\minus{}1)^2}\,dxdy \equal{} \frac {\sqrt{18}}{8}$ with $ \vec{F} \cdot \hat{x}\equal{} \frac {21}{\sqrt{18}}$ therefore the flux is $ \frac {21}{\sqrt{18}}\frac {\sqrt{18}}{8}\equal{} \frac {21}{8}$

Sorry, I made a mistake. The flux you want to find is $ \int_{D}\vec{F}\cdot d\vec{A}\equal{}|\vec{F}|\cdot\cos\theta\cdot\int_{D}d\vec{A}\equal{}|\vec{F}| |D|\cos\theta$ where $ |D|$ is the area of $ D$, and $ \theta$ is the angle between $ \vec{F}$ and the normal vector of $ D$. I didn't product $ \cos\theta$ to $ \frac{3}{8}\sqrt{66}$.

Since $ \cos\theta\equal{}\frac{(4,1,1)\cdot (4,1,4)}{\sqrt{4^2\plus{}1^2\plus{}1^2}\sqrt{4^2\plus{}1^2\plus{}4^2}}\equal{}\frac{21}{\sqrt{33}\sqrt{18}}$, the right answer is $ \frac{3}{8}\sqrt{66}\cdot\frac{21}{\sqrt{33}\sqrt{18}}\equal{}\frac{21}{8}$.


summer program
PROMYS
AwesomeMath
AMC
AIME

has anyone ever been to the National Young Leaders Conference in washington DC, or is planning to go this summer? i just received a package in the mail and i'm wondering whether to apply or not.

also, has anyone attented [i]Lead[/i]America? would you suggest it as a good program?

I'm not so sure about the National Young Leaders Conference ... a bunch of people got packets, including some that I completely think do not have any hope of becoming leaders or role models in society. The way the mailed the packages out, it seems selective, but in reality it's not.

Personally (and if I wasn't going to be a senior next year), I would go to PROMYS, AwesomeMath, Math Camp, or some other related place that you actually heard about before you got informational brochures. If you hang around here a lot the places I named are probably going to be around the ones you'll hear about.

Yeah, I got both of those packets in the mail.  They seemed like a ripoff for the money.  My impression was it was something rich kids can do to fluff their college application.

randomdragoon: what's wrong with going to a math camp if you're going to be a senior?

i personally think leadamerica is a waste of time, since it's basically just the same as a normal college class.

but NYLC sounds interesting... any other opinions?

Oh darn. I enrolled in NYLC...

 :cursing:  :censored:  :wallbash_red:

If NYLC is the one with the brochure that lists a bunch of former US presidents as honorary members/chairmen (forgot title the brochure used), then yeah... I haven't heard good things about it.

Well I guess I'll cancel my registration if that's the case. At least that way I won't waste 10 days (and miss an AIME Problem Series class  :P ).


ARML
geometry
3D geometry
calculus
integration
AMC
AIME

Okay... so this is how I originally did the problem, except I messed it up at the end and COULD NOT figure out my trivial mistake until today when I decided to generalize the problem. I'm providing the generalization here.

[hide="We have a triangle with consecutive sides..."]
... of length $k-1$, $k$, and $k+1$. Now apply Heron's to get
\[\sqrt{s(s-a)(s-b)(s-c)}= \sqrt{\frac{3k}{2}\left(\frac{k}{2}+1\right)\left(\frac{k}{2}\right)\left(\frac{k}{2}-1\right)}= \frac{k}{4}\sqrt{3(k+2)(k-2)}= \frac{k}{4}\sqrt{3k^{2}-12}\]
.
We are looking for the first $k>14$ where $k$ is an integer and $3k^{2}-12$ is a perfect square.
\[3k^{2}-12 = n^{2}\]

\[n=(3m)^{2}\]

\[k^{2}-4 = 3m^{2}\]

\[k^{2}-3m^{2}= 4 \]

\[\frac{1}{4}k^{2}-\frac{3}{4}m^{2}= 1 \]

\[\left(\frac{k}{2}\right)^{2}-\left(\frac{m\sqrt{3}}{2}\right)^{2}= 1 \]

\[\left(\frac{k}{2}-\frac{m\sqrt{3}}{2}\right)\left(\frac{k}{2}+\frac{m\sqrt{3}}{2}\right) = 1 \]
Hunting for trivial solutions, we find that $(k, m) = (4, 2)$ works.
\[(2-\sqrt{3})(2+\sqrt{3}) = 1 \]
Now for a sneaky trick that I learned (and would appreciate if someone explained to me how it worked and if there are other possible ways to solve Pell's Equations): Take this equation and take it to any power! Because we have an integer plus/minus a square root, we end up with another integer plus/minus a square root, and furthermore since these two terms are conjugates the new terms will be conjugates, and satisfy the equation and become the next solution. So for example, the corresponding $k$ for this solution is $4$, which we have already been told is the smallest solution! Squaring, we get
\[(7-4\sqrt{3})(7+4\sqrt{3}) = 1 \]
The corresponding $k$ is $14$, which we have been told is the next smallest solution. Continuing in this fashion, when we cube it we get
\[(26-15\sqrt{3})(26+15\sqrt{3}) \]
The corresponding $k$ is $52$, and we have our answer.

Anecdote:
Obviously, I didn't think of solving for general solutions, I chose $4k-1, 4k, 4k+1$ to be my sides. It amounted to the exact same thing because it gave solutions for real $k$, not just integral, but you had to multiply by 4 so the actual length of the side was integral. When figuring out the smallest solution, I squared it to get the next smallest, and then squared that and couldn't figure out why it wasn't a multiply of 4 :rotfl: . It wasn't until the last few seconds that I thought of cubing it, and I did so, getting $k = 26$. I hastily wrote down 26, not reading the question (it was wrong anyway). I kept thinking and thinking about it and wondered why my k was wrong because when multiplied by $4$ you get $103, 104, 105$ as the side lengths, which is wrong. It wasn't till today that I realized my equation was $(2k-m\sqrt{3})(2k+m\sqrt{3}) = 1$, so I had to divide k by 2...[/hide]

dude wtf
that is pretty easy
how come i didnt get it during the test
bah must be my lack of number theory practice
i never do number theory
speaking of which
i will do some number theory after homework
its like
if you know number theory(by that, i mean you know all basic stuff, and had decent practice), then boom you're done
too bad i dont know number theory and dont have much practice
but
pell equations?
dude those pop out EVERYWHERE
awesome stuff
i used it once on a mock aime GEOMETRY problem i think : |(well geo-nt kinda thing i dont really remember)

that's the main property of pell equations, hunt for minimal solution then take to any power and that's a solution too
in fact, ALL solutions are found that way

That's a nifty trick.  I just set $k=4a$ and hunted for values of $a$ that worked.  Didn't take too long, especially once I eliminated multiples of 3 from consideration (because $4a^{2}-1$ must be divisible by 3).

Who is Sly Si?
What score did you get two days ago?

DUDE WHEN DID POST-RATING BECOME AVAILABLE?

[quote="Ubemaya"]Who is Sly Si?
What score did you get two days ago?
[/quote]
He's Silas...one of the assistants?
[quote]
DUDE WHEN DID POST-RATING BECOME AVAILABLE?[/quote]
um today

I'm Silas, one of the Stanford kids that comes to help out on Sundays.  I wasn't there this week because one of my friends was doing his senior recital, though.  I'll be going to ARML in June with you guys, as will Nathan, the other Stanford assistant.

Oh, and my score is a 1 as of now, as #8 is the only problem I've tried off of that test.

Can someone tell me how to get rid of the ginormous spaces in between the centered TeX in my solution?

And what do you people think of it? Good/bad? Better/worse than the guess and check one?

can someone post the question? just wanna see it.

I think this is how it goes:

The smallest triangle with an integer area and consecutive integer sides has sides 3, 4 and 5. The next such triangle has sides 13, 14 and 15. Find the three sides of the next such triangle (minimal area), given that its "middle" side is a multiple of 4.

Btw, this was solved on [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=139263&start=20]page 2 of the ARML Problem Marathon[/url] (ARML Forum).


Assume in a college, the average number of girl friends of a male student is $ m$, and the average number of boy friends of a female student is $ n$. Some students may have no boy friend or girl friend. Then, is it necessarily true that the proportion of male and female in the college are respectively $ \frac{n}{m\plus{}n}$ and $ \frac{m}{m\plus{}n}$?

Well, there's the ambiguity of translating English into mathematics. If we take it to mean the averages over all, including those with no opposite-sex partner, and assume a closed system with no relationships with outsiders counting, yes. Relationships are generally considered symmetric (although I wouldn't assume a bipartite graph), so it's the same numerator in the averages, and the denominators must be scaled to fit.
Oh yeah- you also have to assume that "male" and "female" partition the set of all people. This is not a safe assumption in the real world.


geometry
perimeter
rectangle
geometry proposed

There is given convex polygon with S=2007 and P=223. Prove that there always exist circle with radius equal to 9 such that every point of it is internal for the polygon.

[quote="SAPOSTO"]P is perimeter and S is surface.[/quote]what is a surface? :blush:

I am wondering why there is no solution for this problem yet. Try it.

well known problem.
im sure it was posted before but i cant find it so i write a solution:
suppose $ x_{i}$ are sides of polygon.now  for each side make rectangle $ x_{i} \times \frac {S}{P}$ with side parallel to
the sides of polygon for each $ i$.put these rectangle in polygon such that every side is a side of one recangle.put the area of rectangles $ S_{i}$ we have $ \sum S_{i} \leq S$ so there exist a point in polygan that is in no rectangle.now put this point center of circle
with radius $ \frac {S}{P}$.


Αν $ a>0$ και n φυσικος, να αποδειξετε οτι

$ \frac{a^n\minus{}1}{a\minus{}1}>\equal{}na^\frac{n\minus{}1}{2}$

Γιατι ειναι $ LHS\equal{}a^{n\minus{}1}\plus{}a^{n\minus{}2}\plus{}...\plus{}a\plus{}1$ ?

Βγαίνει με παραγώγους εύκολα θεωρώντας την $ f(x)\equal{}x^n\minus{}1\minus{}(x\minus{}1)nx^{n\minus{}1/2}$ διακρίνοντας δυο περιπτώσεις $ x>1,0<x<1$ και παραγωγίζοντας δυο φορές

Υπάρχει και σε εφαρμογή του σχολικού της Β΄ Λυκείιου, Άλγεβρα, σ.71 Παράδειγμα 2ο:.
Για τη γενική μορφή:

Σχήμα Horner με διαιρετέο $ a^n \minus{} b^n$ και διαιρέτη το $ a \minus{} b$ δίνει:

$ 1 \ \ \ \ 0 \ \ 0 \ \ \ \ \ldots \ \ 0 \ \ \ \ \ \ \minus{} b^n \ \ \boxed{b}$
$ \minus{} \ \ \ \ b \ \ b^2 \ \ \ldots \ \ b^{n \minus{} 1} \ \ \ \ b^n$
$ 1 \ \ \ \ b \ \ b^2 \ \ \ldots \ \ b^{n \minus{} 1} \ \ \ \ 0$

Επομένως το υπόλοιπο της διαίρεσης $ \left(a^n \minus{} b^n\right) : \left( a \minus{} b \right)$ είναι μηδέν, ενώ το πηλίκο είναι το πολυώνυμο
$ \pi \left(a\right) \equal{} a^{n \minus{} 1} \plus{} ba^{n \minus{} 2} \plus{} b^2a^{n \minus{} 3} \plus{} \cdots \plus{} b^{n \minus{} 1}$

Τέλος, από την ταυτότητα της διαίρεσης προκύπτει ότι: $ a^n \minus{} b^n \equal{} (a \minus{} b)\pi(a) \plus{} 0$ ή
$ a^n \minus{} b^n \equal{} (a \minus{} b)(a^{n \minus{} 1} \plus{} ba^{n \minus{} 2} \plus{} b^2a^{n \minus{} 3} \plus{} \cdots \plus{} b^{n \minus{} 1} )$

Τώρα για $ b \equal{} 1$ προκύπτει το ζητούμενο.

η και πιο απλα
$ A(A^{n\minus{}1}\plus{}A^{n\minus{}2}B...\plus{}B^{n\minus{}1})\equal{}A^{n}\plus{}A^{n\minus{}1}B\plus{}...\plus{}AB^{n\minus{}1}$
$ B(A^{n\minus{}1}\plus{}A^{n\minus{}2}B...AB^{n\minus{}2}\plus{}B^{n\minus{}1})\equal{}BA^{n\minus{}1}\plus{}B^2A^{n\minus{}2}...\plus{}AB^{n\minus{}1}\plus{}B^n$
afairesi kai to zitumeno. :)

η ταυτοτητα παντως αυτη ειναι ιδιαιτερα χρησιμη ακομα και στην θεωρια αριθμων . για παραδειγμα παραθετω μια πολυ ωραια ασκηση στην οποια φαινεται πραγματικα η αξια της(χωρις βεβαια να αποκλειω το ενδεχομενο αλλων λυσεων).
ΑΣΚΗΣΗ
εστω ν ενας φυσικος που εχει ολα τα ψηφια του ισα με 1.Αν ο ν ειναι πρωτος ν.δ.ο και το πληθος των ψηφιων του ειναι πρωτος αριθμος.
δικο σας.

[color=red][b][size=100]ΑΣΚΗΣΗ[/size][/b][/color]

Το γράφω λίγο σε $ \text{\LaTeX}$ αν και πιστεύω πρέπει να το βάλεις σε άλλο topic.

$ \nu = \underbrace{111111111\ldots 11}_{\mu}$
Νδο αν $ \nu$ πρώτος, τότε και $ \mu$ πρώτος


Let $ y\plus{}z\equal{}2$, $ x \geq 1$, $ y \leq 0$, $ z \geq 2$ and $ t \geq 2$, prove that

$ \frac{x^{2(t\minus{}1)}}{x^{t\minus{}1}\plus{}x^2y} \geq \frac {z}{x^2\plus{}1}$.

I make mistake before. This should be:

Let $ x,t \geq 1$, prove that

$ x^{2t}(x^2\plus{}1) \plus{}x^2 \geq x^{t}(x\plus{}1) \plus{}x^4$

From the above inequality, can make another inequality:

Let $ 0<y \leq 1, p \geq 1, 0<q \leq 1$. Prove that

$ \frac{1\plus{}D_2}{1\plus{}D_1} \geq y^{p\plus{}q}$ ,

where $ D_n  \equal{} y^{nq} \minus{} y^{n(p\minus{}q)}$ for $ n \in N$.


MATHCOUNTS
probability
geometry
rectangle
ratio
inequalities
3D geometry

These questions were posted by MCrawford.
I'm importing this from Penn.Forum.
I'm going to try to solve them,too..

1.Darts are thrown randomly at a board showing three concentric circles of radii 2, 3, and 4.  If a dart hits inside one of the circles, what is the probability, expressed as a common fraction, that it is in the interior of the circle of radius 3 but not in the interior of the circle of radius 2?

2.
Three tennis balls are tightly sealed in a right cylindrical can, so that the balls cannot move inside the can.  If the radius of each ball is 4 cm, find the number of cubic centimeters in the volume of the space within the cylinder, not taken up by the three tennis balls.  Express your answer in terms if $ \pi$.

3.
A custom made window is a single pane of glass consisting of a semicircular sheet of glass on top of the narrow end of a rectangular sheet of glass.  The rectangle has dimensions of 4 feet by 3.5 feet.  How many feet of weather-stripping will be needed to go around the glass?  Express your answer to the nearest integer.

4.
The numbers from 1 through 8 are separated into two sets, A and B.  The numbers in A are multiplied together to get $ a$.  The numbers in B are multiplied together to get $ b$.  The larger of the two numbers $ a$ and $ b$ is written down.  What is the smallest possible number that can be written down using this procedure?

5.
Eric's Florist Shop sells roses for 20 dollars per dozen, daises for 5 dollars per dozen, and chrysanthemums for 10 dollars per dozen.  At these rates how much would an arrangement of 9 roses, 6 daises, and 18 chrysanthemums cost in dollars?

6.
Given the equation $ \displaystyle \frac {1}{m} + \frac {1}{n} = \frac {2}{15}$, where $ m$ and $ n$ are positive integers, find the value of $ m + n$ such that $ m > n$ and $ m$ is not a multiple of $ n$.

7.
Use the pattern below to express $ 100^2$ in the form $ a^2 + b^2 - c^2$.  Then find $ a + b + c$.
\begin{eqnarray*} 12^2 = 8^2 + 9^2 - 1^2 \\
14^2 = 10^2 + 10^2 - 2^2 \\
16^2 = 12^2 + 11^2 - 3^2 \\
18^2 = 14^2 + 12^2 - 4^2 \end{eqnarray*}

8.
The ratio of the length of a rectangle to its width is the same as that of the diagonal to the length.  If the width is 2, how many units are in the length of the diagonal?

9.
Suppose a number $ x$ satisfies $ x^2 - 4x - 1 = 0$.  How much greater is $ x$ than its reciprocal?

10.
What is the difference between the largest and the smallest values of $ x$ satisfying the following system of inequalities?
Express your answer as a mixed number.
\begin{eqnarray*} 2x - 6 \le 6 \\
4x + 3 \ge 13 \end{eqnarray*}

First 5 answers.
Tell me if I got them wrong.

[hide="#1"]Total Area:$ 16\pi$.
The area we want:$ 9\pi - 4\pi = 5\pi$

So the answer is $ \boxed{\frac {5}{16}}$[/hide]

[hide="#2"]The area of cylinder: $ 16\pi\times24cm = 384\pi cm^2$

The formula for area of a sphere:$ \frac {4}{3}r^3\pi$
The areas of 3 tennis balls:$ \frac {4}{3}r^3\pi\times3 = 256\pi$

The area we want:$ 384\pi - 256\pi = \boxed{128\pi}$[/hide]

[hide="#3"]The perimeter of a rectangle sheet of glass:$ 2(3.5) + 2(4) = 15$
The perimeter of a semicircle:$ \frac {4\cot\pi(3.14)}{2} = 6.28$

The perimeter(measure of weathing tape) we want: $ {15 + 6} = \boxed{21}$[/hide]

[hide="#4"]The smallest I could find was: $ 1\cdot8\cdot6\cdot5 = \boxed{240}$
EDIT: xpmath's answer : $ \boxed{210}=5\cdot6\cdot7\cdot1$
[/hide]

[hide="#5"]Given$ \longrightarrow$What we want

$ R$=Roses, $ D$=Daises, $ C$=Chrysanthemums.

$  20 = 12R\longrightarrow \frac {20}{12}\cdot9 = \boxed{15} $
$  5 = 12D\longrightarrow \frac {5}{12}\cdot6 = \boxed{2.5} $
$ 10 = 12C\longrightarrow \frac {10}{12}\cdot18 = \boxed{15} $

We add the boxed numbers to get $ \ {\$32.5}$[/hide]

[hide="#4"]The total product is $ 8!\equal{}40320$.So we keep checking til we get 2 pairs that work.210=5x6x7x1.[/hide]

[quote="xpmath"][hide="#4"]The total product is $ 8! \equal{} 40320$.So we keep checking til we get 2 pairs that work.210=5x6x7x1.[/hide][/quote]
Oh.. Thanks... I'll fix mine..

Finally!!
I'm seeing somethings!! ( solutions and stuff... :D )
Even though they're easy, i'll just post 'em in case.... :)

[hide="#9"] it's just algebra, i guess, and then since $ \ {x^2 - 4x = 0}$, I knew that $ x$ had to equal $ 4$(guess and check :lol: ). So if $ \ {x = 4}$, the reciprocal result -$ 8$, in which the difference is $ \boxed{ - 8}$. [/hide]

[hide="#10"] It's asking for the difference between the greatest and the least value of $ x$, so the least value should be $ 2\frac {1}{2}$, since $ 4x\ge 10$, and the greatest value of $ x$ has to be  $ 2\frac {1}{2}$..( if you do the calculation right..)  And to get the greatest value, you would get $ 6$ for $ x$ , cause $ \2x - 6\le 6$ =$ \2x\le 12$ and $ 12$ is the greatest you can get.. so, if you just subtract $ 6$ and $ 2\frac {1}{2}$, you should get  $ \boxed{3\frac {1}{2}}$[/hide]

joga, good job on #10.

On 9, where did you get x^2 - 4x = 0? It should equal 1, no?
But there a few ways to do this. Find the roots through quardratic formula (remember that from yesterday?) and recipricol them and subtract. The easier way is the flip the coeffiecients so that the resulting equation has roots that are the recipricols (does that make sense?)

If that doesn't make sense... say, the roots are 2 and 3. The equation should be (x-2)(x-3) = 0
The recipricols are 1/2 and 1/3. The equation is (2x-1)(3x-1) = 0
Multiply those two out.

1. x^2 - 5x + 6 = 0
2. 6x^2 - 5x +1 = 0

Do you see something? When you flip the coefficients of the first equation, you get the second equation.

I am not sure on #9... I'll have to recheck my work..

EDIT:New solution
[hide="#9"]Let $ a$ be what we are trying to get.
$ 2\pm\sqrt{5}\minus{}\frac{1}{2\pm\sqrt{5}}\equal{}a$

Get rid of 2.

$ 4\pm 2\sqrt{5}\minus{}\frac{1}{2\sqrt{5}}\equal{}2a$

Get rid of $ 2\sqrt{5}$.

$ 8\sqrt{5}\pm20\minus{}1\equal{}4\sqrt{5}a$

Get rid of $ 4\sqrt{5}$.

$ 2\pm19\equal{}a\longrightarrow \boxed{a\equal{}2\pm19}$
[/hide]

[quote="Chinaboy"]joga, good job on #10.

On 9, where did you get x^2 - 4x = 0? It should equal 1, no?
But there a few ways to do this. Find the roots through quardratic formula (remember that from yesterday?) and recipricol them and subtract. The easier way is the flip the coeffiecients so that the resulting equation has roots that are the recipricols (does that make sense?)

If that doesn't make sense... say, the roots are 2 and 3. The equation should be (x-2)(x-3) = 0
The recipricols are 1/2 and 1/3. The equation is (2x-1)(3x-1) = 0
Multiply those two out.1. x^2 - 5x + 6 = 0
2. 6x^2 - 5x +1 = 0

Do you see something? When you flip the coefficients of the first equation, you get the second equation.[/quote]

Oh! sorry... I made a mistake.. :ewpu:
Anyways, I think I figured it out

[hide="#9"]since it's $ x^2 \minus{} 4x \minus{} 1 \equal{} 0$, I used the quadratic formula, which

states $ x \equal{} \frac { \minus{} b\pm\sqrt {b^2 \minus{} 4ac}}{2ac}$.  So I just plugged numbers in, which gave me $ \frac { \minus{} 4 \plus{} \sqrt {20}}{ \minus{} 2}$ (I used

$ \plus{}$ instead of $ \minus{}$). Then I simplified that to $ 2 \plus{} \sqrt {20}$, and then

I simplified again $ \sqrt {20}$ to $ 2\sqrt {5}$, and added $ 2$ from

$ 2 \plus{} \sqrt {20}$. So, I guess if you do the subtraction and calculation right, you'd get $ \boxed{\frac {\sqrt {5}}{4}}$ as the answer... :lol: Am I right?? :w00t: [/hide]


You have a square that has a side length of $ 1$ centimeter. Each minute, $ 2$ centimeters are added to the side length. After $ 10$ minutes, what will be the difference between the area of the last square and the area of the initial square in square centimeters (do not include the units in your answer)?

After 10 minutes, the side length of the square would be $ 1\plus{}2(10)\equal{}21$ centimeters. Its area is $ 21^2\equal{}441$ square centimeters.

Difference between area of last square and that of the initial square would be $ 441\minus{}1\equal{}\boxed{440}$.


topology
Functional Analysis
advanced fields
advanced fields unsolved

Prove that a Banach space is reflexive iff its closed unit ball is compact in the weak topology.

The easy half (if you qoute a famous result):

It is known that the closed unit ball of any dual Banach space is compact in the weak* topology.

A reflexive Banach space is a dual space, and the norm as a dual of its dual is the same as the standard norm (that's what reflexive means, after all.). The weak topology is the same as the weak* topology, so the closed unit ball is compact in the weak topology.

As I said, that's the easy direction.

the other part follows from http://www.mathlinks.ro/Forum/viewtopic.php?t=84091


Hello,

I can't figure out the problem below. Any help would be greatly appreciated. Thank you!

Let R be a relation on the set N of natural numbers defined by

R = {(a,b) (element symbol) N x N ׀ a divides b in N }

Is R a partial order of N? Explain.

Is N with the divisibility relation given above a totally ordered set? Explain.

If you know what the words "partial order" and "total order" mean, you should be able to write down a list of axioms.  If you know what $N$, $\in$, and $a\mid b$ mean, you should for each axiom be able to verify that it holds or give a counter-example.

Okay, thank you I have been working on the problem and understand it (I believe) except the antisymmetric part. Here is what I have, please tell me if I am right so far:

Reflexive: so, a <= a always holds.

For example, 2 <= 2
3 <= 3
...etc.

Antisymmetric: so, if a <= b and b <= a holds, then a = b

[b]Don't see this yet? What values to choose to show? Can a and b be the same value? [/b]

Transitive: so, if a <= b and b <= c hold, then a <= c also holds.

For Example, since 2 <= 3, and
3 <= 6, then
2 <= 6

** I need to know the antisymmetric part before I can tell if it holds and therefore is a poset.

To be totally ordered every pair of elements a,b in the set is comparable (a <=b or b <= a). The set of natural numbers is not totally ordered since for example, 3 and 5 are not comparble, neither 3 <= 5 nor 5 <= 3 holds.

Thank you!

[quote="JBL"]You can disprove a "for all" statement with a counter-example, so you have correctly demonstrated that this ordering is not total.  However, you can't prove a "for all" statement with examples: you have not shown that this ordering is reflexive or transitive [i]for all[/i] choices of the relevant variables.  This is also why you are finding the antisymmetric condition confusing.[/quote]

Then how would I go about that.  I really am not sure.

Let's do reflexivity.  Our order relation is "divides," so the general statement of the reflexivity axiom ("For all $x$, $x \leq x$") becomes in our case, "For every $x \in \bf N$, $x$ divides $x$."  Then you either have to prove that that's true or show that it's false.  And then you have to do the same thing for the two other axioms.

If you're still lost, you should probably go talk to your teacher about it.


Is it possible to select 102 17-element subsets of a 102-element set,such that the intersection of any two of the subset has at most 3 elements?

Do you ever bother thinking a bit before posting? I'm just curious...

We can chose subsets with 3 common terms. Exactly (3 fixed terms) we cyoose 14 terms from 102-3, and $ \binom{99}{14}>102$.


limit
algebra
functional equation
real analysis
real analysis unsolved

Let $ (x_n)_{n \geq 1}$ be a sequence such that $ x_1 \equal{} a \in \mathbb{R}$ and $ x_{n \plus{} 1} \equal{} x_n^2 \plus{} x_n \plus{} 1$ for all $ n \geq 1$. Evaluate the limit:
\[ \lim_{n \to \infty} \frac { (x_{n\plus{}1} \plus{} 1)^2 }{(x_1^2 \plus{} 2)(x_2^2 \plus{} 2) \dots ( x_n^2 \plus{} 2)}.
\]

A few numerical experiments with a spreadsheet suggest that this converges (and converges rapidly) to $ 2\plus{}(1\plus{}a)^2.$

If we denote the partial product associated with $ x_1 \equal{} a$ by $ S_n(a)$, then $ S_{n \plus{} 1}(a) \equal{} \frac {S_n(a^2 \plus{} a \plus{} 1)}{a^2 \plus{} 2}$. If the limit is $ S(a)$, then S satisfies the functional equation $ (a^2 \plus{} 2)S(a) \equal{} S(a^2 \plus{} a \plus{} 1)$.
Kent's limit satisfies the equation. Are there any other solutions? Also, intuitively, $ S$ should be continuous, so we should look for continuous solutions.

Edit - Putting $ g(a)\equal{}\frac{S(a)}{2\plus{}(1\plus{}a)^2}$ yields $ g(a)\equal{}g(1\plus{}a\plus{}a^2)$. Showing that the only continuous solutions are constant ones and calculating $ S(a)$ at one point should be enough.

Set $ f(x)\equal{}x^2\plus{}x\plus{}1$ and note, that $ [2\plus{}(1\plus{}f(x))^2]\equal{}(x^2\plus{}2)[(2\plus{}(1\plus{}x)^2]$ and $ x_{n\plus{}1}\equal{}f(x_n)$ hold. Using the $ S_n$-notation and the result $ S_{n\plus{}1}(a)\equal{}\frac{S_n(f(a))}{a^2\plus{}2}$ of the previous post we derive the following recursive formula:

$ \frac{S_{n\plus{}1}(a)}{2\plus{}(1\plus{}a)^2}\equal{}\frac{S_n(f(a))}{(a^2\plus{}2)[2\plus{}(1\plus{}a)^2]}\equal{}\frac{S_n(f(a))}{2\plus{}(1\plus{}f(a))^2}$

It translates after $ n$ steps into:

$ \frac{S_{n\plus{}1}(a)}{2\plus{}(1\plus{}a)^2}\equal{}\frac{S_1(x_n(a))}{2\plus{}(1\plus{}x_n(a))^2}\equal{}\frac{[f(x_n(a))\plus{}1]^2}{(x_n(a)^2\plus{}2)[2\plus{}(1\plus{}x_n(a))^2]}\equal{}\frac{[f(x_n(a))\plus{}1]^2}{2\plus{}(1\plus{}f(x_n(a)))^2}$.

Since obviously $ x_{n\plus{}1}(a)\equal{}f(x_n(a))$ tends to infinity, the product $ S_n$ converges to $ 2\plus{}(1\plus{}a)^2$.


function
superior algebra
superior algebra unsolved

Let $ G$ be a group of size $ n$ and let $ \varphi_{k}: G\times G\rightarrow G$ be a function (in this case, binary operation) on $ G$.  Are the following statements true?:

1. There are $ n^{n^{2}}$ of these functions ($ k\equal{}n^{n^{2}}$).

2. There are $ n^{n\left(n\minus{}1\right)}$ commutative functions.

3. There are $ n^{\left(n\minus{}1\right)^{2}\plus{}1}$ functions with the identity element.

4. There are $ n^{\frac{n\left(n\minus{}1\right)}{2}\plus{}1}$ commutative functions with the identity element.

Also, is there any way to say how many associative functions exist for such a set?

How many functions are there from $ \phi: X\mapsto Y$? I think it is $ |Y|^{|X|}$.
So if $ |G| \equal{} n$ then $ G\times G \equal{} n^{2}$ and now use the above results ...

Here are my computations:

1. It is right there are $ n^{n^{2}}$ functions $ \phi : G\times G\to G$.

2. There are $ n^{\frac{n(n\plus{}1)}{2}}$ commutative of such functions.

3. There are $ n^{2}(n\minus{}1)^{n^{2}\minus{}1}$ functions where the identity element appears.

4. There are $ \frac{n(n\plus{}1)}{2}(n\minus{}1)^{\frac{n(n\plus{}1)}{2}\minus{}1}$ commutative functions with the identity element.

Sorry, I have a mistake in clausules 3 and 4 (I've computed the number of functions in which the identity appears exactly once).


modular arithmetic
number theory
relatively prime

McDonalds sells Chicken McNuggets in boxes of 19, boxes of 17, and boxes of 13. What is the highest number of chicken nuggets that you can't obtain?

I won't answer this question until round 1 of Amazing Race is over.

[quote]On my playschool dart board, there are two sections with values 12 and 19 points. What is the highest unattainable sxore with an unlimited number of darts? [/quote]

This may help you...

EDIT: I mean that i will answer this question tomorrow after the Amazing Race Round 1 is foficially over.

[quote="anirudh"]I won't answer this question until round 1 of Amazing Race is over.

[quote]On my playschool dart board, there are two sections with values 12 and 19 points. What is the highest unattainable sxore with an unlimited number of darts? [/quote]

This may help you...[/quote]

It's already over though.

[quote="anirudh"]I won't answer this question until round 1 of Amazing Race is over.

[quote]On my playschool dart board, there are two sections with values 12 and 19 points. What is the highest unattainable sxore with an unlimited number of darts? [/quote]

This may help you...[/quote]

What are you talking about?

Was another case of the "hackers" ?

[quote="anirudh"]I won't answer this question until round 1 of Amazing Race is over.

[quote]On my playschool dart board, there are two sections with values 12 and 19 points. What is the highest unattainable sxore with an unlimited number of darts? [/quote][/quote]

well for only 2 i know it is 12*19-12-19 but what about 3

There is no formula, you have to integrate a way of using mods to help you in this problem.

I think I know a long way to do this, but is there a quick way (under 4 min.) to this?

[u]More Conjectures:[/u]

Well, we know that for rel. prime $x,y$ the formula is $(x-1)(y-1)-1$, which can also be written as $ab-\sum^\text{cyc.}a$.

So maybe it's one of these:
$\text{ Guess 1: }(a-1)(b-1)(c-1)-1$
or
$\text{ Guess 2: }abc-\sum^\text{cyc.}(ab)-\sum^\text{cyc.}a = abc-ab-bc-ac-a-b-c$

Either gives a value lower than $4150$. The first gives $3455$, and the second gives $3359$.

(Anyone wanna eat alot of nuggets??)

I'd love to see a way to do this every time it comes up, but in the mean time I think this discussion is headed in the wrong direction.

If using only the 13s and 17s you can make anything bigger than 13(17)-13-17 = 191, than why are we looking at 4000+ as being the largest we can't make.  We KNOW we can make everything bigger than 191 using just 13 & 17.  We should be looking at whether we can make 191 and less if we incorporate 19s.
191 = 19(2) + 17(9)
190 = 19(10)
189 = 19(1) + 17(10)
188 = 19(1) + 13(13)
.
.
.
.
What is the biggest you can't make?  I'd say start with the 2 smallest numbers and then work your way back!  Just a thought.

By the way 4150 = 17(4) + 13(314)

[quote="frost13"]
If using only the 13s and 17s you can make anything bigger than 13(17)-13-17 = 191, than why are we looking at 4000+ as being the largest we can't make?
[/quote]

Oh man, now I feel $\text{REALLY}^{10}$ stupid.

don't feel stupid, I've been working on this for 40 more minutes, and just now think I have the answer......

[quote="i_like_pie"][quote="mysmartmouth"]Let's use the Chicken McNugget Theorem.  12 AM, so more details tomorrow:

19, 17, and 13.

So we do mn-m-n on 19 and 17, and get 287.  This is relatively prime to 13, so we do it again, and get 287*13 - 13 - 287.

However, I think that is wrong, because we didn't take into account the fact that 13 could be used to get lower than 287 in the first place.  So let's try something else:

Conjecture:

Expand the Chicken McNugget Theorem to lmn - l - m -n:

$13\cdot19\cdot17-13-19-17 = \boxed{4150}$

I have no idea, does that work?  Can someone make 4150 with the given numbers?

I don't think it will work, but we can see.[/quote]
$4150=13\cdot5+19\cdot215$

However, your conjecture is incorrect. If all numbers after and including $4150$ can be made with $13,\ 17,$ and $19,$ then $4149$ can't be. But it can be: $4149=13\cdot1+17\cdot3+19\cdot215$[/quote]
frost13 killed that theory already ;)

[quote="jhredsox"]guys, i think u can only add chicken nuggets, not subtract them
-jorian[/quote]
No one is subtracting nuggets...

jorian is probably referring to how people are subtracting numbers, i.e. pq-p-q.

I don't think that there exists a nice simple formula for the 3-nugget case, but I could be wrong.

jorian,

When we do mn - m -n, we are not actually counting (or subtracting) nuggets.  This is just a formula for finding a number that we cannot attain by adding nuggets.

ya, i saw that right after i posted
ty guys
-jorian

Guys all I can say is that you are wasting your time looking for a formula.

So, I have a suggestion...let us try to look for a way of doing this without formulae.

[hide]Well, we know that 13*17-13-17=191, cannot be achieved with 13s and 17s, but

Let us consider mod13.

We agree that

0(mod 13)=13x
1(mod 13)=53+13x
2(mod 13)=93+13x
3(mod 13)=55+13x
4(mod 13)=17+13x
5(mod 13)=57+13x
6(mod 13)=19+13x
7(mod 13)=72+13x
8(mod 13)=34+13x
9(mod 13)=87+13x
10(mod 13)=36+13x
11(mod 13)=76+13x
12(mod 13)=38+13x

So, we see the one that has the largest integer which is 2(mod 13)=93+13x

Then we just subtract 13 from this to get the answer of 80.

80 we see cannot be done...

$80$[/hide]

[quote="bpms"][quote="i_like_pie"][quote="kyyuanmathcount"]I believe the CMNT (typing the name is getting annoying) is derived by PIE, doesn't it look like it?

Thus, I conjecture that the triple formula is $lmn-mn-lm-ln+n+m+l$.[/quote]
Testing: 1, 3, 5

$15-15-3-5+5+3+1=1$

I don't think $lmn-mn-lm-ln+n+m+l$ works. :lol:[/quote]
1 throws everything off. Let's try 2,3, 5.
30-6-10-15+5+3+2=9. No it doesn't work.[/quote]

Hm.. yeah.  I agree with Neil, though I'm sure it is possible to find a formula.

Um...maybe if we found a proof for the 2 mcnuggets, we could prove it for 3, but I still doubt it.

Well, it's easy to prove the 2-mcnugget case. Just consider the multiples of m mod n

m
2m
3m
...
(n-1)m

And note that because m and n are relatively prime, this forms a complete residue class mod n. Thus, we can form all numbers equal to $m \pmod{n}$ after $m$, all numbers $2m \pmod{n}$ after $2m$, etc...continuing in this way, the last residue class that we can form is $(n-1)m \pmod{n}$, so the last one should be $nm-m-n$.

Well, yeah, but any proof for the 3 case...I hope not...because it probably would be lengthy.

[quote="mathgeniuse^ln(x)"]

Well, we know that 13*17-13-17=191, cannot be achieved with 13s and 17s, but

Let us consider mod13.

We agree that

0(mod 13)=13x
1(mod 13)=53+13x
2(mod 13)=93+13x
3(mod 13)=55+13x
4(mod 13)=17+13x
5(mod 13)=57+13x
6(mod 13)=19+13x
7(mod 13)=72+13x
8(mod 13)=34+13x
9(mod 13)=87+13x
10(mod 13)=36+13x
11(mod 13)=76+13x
12(mod 13)=38+13x

So, we see the one that has the largest integer which is 2(mod 13)=93+13x

Then we just subtract 13 from this to get the answer of 80.

80 we see cannot be done...

$80$[/quote]

mathgeniuse, I agree looking for a formula is not practical as it probably would have been found by now by an eager AoPSer.  That doesn't mean there is no formula(unless it has been proven there is none), just not the most practical method for this problem.

I too got 80 as my answer, but with more work.  can you explain though your "We agree that"  0(mod13) = 13x, 1(mod13)= 53 + 13x, 2(mod 13)=93+13x.  I love modular arithmetic, but while I agree your statements are true, where are you getting the 53, 93 parts of these.  They seem arbitrary to me.  yes,  2(mod 13)=93+13x  but 2(mod13) =15 +13x as well, why 93 and not 15.  If you can explain that one I'd be excited about a simpler way to solve this one!!!

Thanks,

Frost

I programmed it and got 99. But perhaps the code is wrong.
[code]
#include <stdio.h>

int main(int argc, char *argv[]) {
        char done = 0;
        char obtainable = 0;
        int num, i, j, k;
        for (num = 191; num > 1 && !done; num--) {
                for (i = 0; i < 15; i++) {
                        for (j = 0; j < 12; j++) {
                                for (k = 1; k < 11; k++) {
                                        if (13 * i + 17 * j + 19 * k == num) {
                                                obtainable = 1;
                                        }
                                }
                        }
                }
                if (!obtainable) {
                        printf("%d is the highest number of nuggets not obtainable\n", num);
                        done = 1;
                }
                obtainable = 0;
        }
        return 0;
}
[/code]
Here's the output:
[code]
andrew@roflbirds:~/code/problems/3nuggets$ gcc -o three_nuggets three_nuggets.c
andrew@roflbirds:~/code/problems/3nuggets$ ./three_nuggets
99 is the highest number of nuggets not obtainable
[/code]
And in Java, if anyone is more used to that.
[code]
public class ThreeNuggets {
        public static void main(String[] args) {
                boolean done = false;
                for (int num = 191; num > 1 && !done; num--) {
                        boolean obtainable = false;
                        for (int i = 0; i < 15; i++) {
                                for (int j = 0; j < 12; j++) {
                                        for (int k = 1; k < 11; k++) {
                                                if (13 * i + 17 * j + 19 * k == num) {
                                                        obtainable = true;
                                                }
                                        }
                                }
                        }
                        if (!obtainable) {
                                done = true;
                                System.out.println(num + " is the highest number of nuggets not obtainable.");
                        }
                }
        }
}
[/code]
The Java program also yields 99.
[code]
andrew@roflbirds:~/code/problems/3nuggets$ javac ThreeNuggets.java
andrew@roflbirds:~/code/problems/3nuggets$ java ThreeNuggets
99 is the highest number of nuggets not obtainable.
[/code]

Please don't revive old topics. Thank you.


Αν  $ x,y,z > 0$ και  $ x \plus{} y \plus{} z \equal{} 1$, να αποδειχθεί ότι : $ \frac {x^2 \plus{} 3xy}{x \plus{} y} \plus{} \frac {y^2 \plus{} 3yz}{y \plus{} z} \plus{} \frac {z^2 \plus{} 3zx}{z \plus{} x}\le 2.$

[b] Μπάμπης[/b]

(Έχουμε και λύση !)

η αφου ισχυει, $ \frac{ x^2\plus{}3xy}{x\plus{}y}\equal{} 3x \minus{} \frac{2x^2}{x\plus{}y}$
εχουμε νδο ισοδυναμα
$ \sum \frac{x^2}{x\plus{}y} \geq \frac{1}{2}$, προφανες απο andreescu

Με παρομοιο τροπο

  $ \frac{x^2\plus{}3xy}{x\plus{}y}\equal{}\frac{x^2\plus{}xy\plus{}2xy}{x\plus{}y}\equal{}x\plus{}\frac{2xy}{x\plus{}y}$.

  Αρκει  $ x\plus{}y\plus{}z\plus{}\frac{2xy}{x\plus{}y}\plus{}\frac{2yz}{y\plus{}z}\plus{}\frac{2zx}{z\plus{}x}\leq2$

  Αυτο προκυπτει απο την $ \frac{2xy}{x\plus{}y}\leq\frac{x\plus{}y}{2}$

Βαζω και εγω μια χειροτερη λυση....
$ \sum \frac {x^2 \plus{} 3xy}{x \plus{} y} \equal{} \sum\frac {(x \plus{} y)^2 \plus{} xy \minus{} y^2}{x \plus{} y} \equal{} 2(x \plus{} y \plus{} z) \plus{} \sum\frac {y(x \minus{} y)}{x \plus{} y}$
Αρα αρκει νδο $ \sum\frac {y(x \minus{} y)}{x \plus{} y}\leq 0$
Εαν πολλ με $ \minus{} 1$ εχουμε $ \sum\frac {y(y \minus{} x)}{x \plus{} y} \geq 0$
Θεωρουμε την $ f(x) \equal{} \frac {1}{x}$
$ f''(x)\geq 0$
Aρα απο Jensen $ \frac {\sum y(y \minus{} x)f(x \plus{} y)}{x^2 \plus{} y^2 \plus{} z^2 \minus{} xy \minus{} yz \minus{} zx}\geq f(\frac {\sum y(y \minus{} x)(y \plus{} x)}{x^2 \plus{} y^2 \plus{} z^2 \minus{} xy \minus{} yz \minus{} zx})$
Ara to $ LHS \geq {(x^2 \plus{} y^2 \plus{} z^2 \minus{} xy \minus{} yz \minus{} zx)^2}\frac {1}{\sum y(y \minus{} x)(y \plus{} x)}$
Αρα αρκει το $ \frac {1}{\sum y(y \minus{} x)(y \plus{} x)}\geq 0$
Αλλα $ \sum y(y \minus{} x)(y \plus{} x) \equal{} x^3 \plus{} y^3 \plus{} z^3 \minus{} x^2y \minus{} y^2z \minus{} z^2x\geq 0$......
Please check it ,γιατι μπορει να εχω κανει καμια βλακεια :)

[quote="silouan"]Σχετικά απλό  :wink:
$ \frac {x^2 \plus{} 3xy}{x \plus{} y}\leq\frac{3x\plus{}y}{2}$ .Εργαζόμενοι όμοια και προσθέτοντας έχουμε το ζητούμενο.[/quote]
$ \frac {x^2 \plus{} 3xy}{x \plus{} y}=x+\frac{2xy}{x+y}\leq x+\frac{2}{x+y}(\frac{x+y}{2})^2=\frac{3x\plus{}y}{2},$

$ \sum \frac {x^2 \plus{} 3xy}{x \plus{} y}\leq  \sum \frac{3x\plus{}y}{2}=2.$


Given are two sequences:

(i) $ x_{n}$ defined by  $ x_{1}\equal{}1996$ and $ x_{n\plus{}1}\equal{}\frac{1}{4\minus{}3x_{n}}$ and

(ii) $ y_{n}\equal{}1\plus{}\frac{1}{(2!)^{2}}\plus{}\frac{1}{(3!)^{2}}\plus{}...\plus{}\frac{1}{(n!)^{2}}$

Find $ \ lim x_{n}$ and $ lim y_{n}$ (when n tends to infinity).

Correct me if I'm wrong, ii is the series:

$ \sum_{n \equal{} 1}^{\infty}\frac {1}{(k!)^2}$

which is

$ I_0 (2)$ where I denotes Bessel I function (see http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html ,  last equation on the page).

(EDIT: actually, it's $ I_0 (2)\minus{}1$, because the series begin with $ n\equal{}1$, not $ n\equal{}0$)

[quote="delegat"]

(i) $ x_{n}$ defined by  $ x_{1} \equal{} 1996$ and $ x_{n \plus{} 1} \equal{} \frac {1}{4 \minus{} 3x_{n}}$

Find $ \ lim x_{n}$(when n tends to infinity).[/quote]

Let $ V_n\equal{}\frac{3U_n\minus{}1}{U_n\minus{}1}$

Then $ V_{n\plus{}1}\equal{}\frac{1}{3}V_n$

Then $ U_n\equal{}1\plus{}\frac{2}{\frac{1}{3^{n\minus{}1}}.V_1\minus{}3}$

Then $ Lim U_n\equal{}\frac{1}{3}$


search
geometry
analytic geometry
geometry solved

Circle $C$ (with center $O$) does not have common point with line $l$. Draw $OP$ perpendicular to $l$, $P \in l$. Let $Q$ be a point on $l$ ($Q$ is different from $P$), $QA$ and $QB$ are tangent to circle $C$, and intersect the circle at $A$ and $B$ respectively. $AB$ intersects $OP$ at $K$. $PM$, $PN$ are perpendicular to $QB$, $QA$, respectively, $M \in QB$, $N \in QA$. Prove that segment $KP$ is bisected by line $MN$.

After making a sketch, I have a strong feeling that this problem can be solved by using analytic geometry. But I am having troubles finding the equations of the tangants......

Grobber gave a solution long time ago . Try to search with the keyword : " circle_and_tangent "

[color=red][Moderator edit: In fact, see http://www.mathlinks.ro/Forum/viewtopic.php?t=5198 .][/color]

Let QA and QB intersect OP at X and Y respectively. Then the problem is equivalent to the fact P,X,K,Y are harmonic conjugate(why ;) ), which is easy to justify.

dear shober i love solving problem by using analytic geometry.but i think :
solving  th e problems with out using analytic geometry is more powerfull and nice.
dont u think so?
 :D

Yes, I love to solve a problem with pure geometry. But the perpendicular lines in this problem really reminded of the coordinate system.

draw PX perpendicular to AB ( X is a point on AB)
We know that O, A, P, Q, B are cyclic.
Then apply Simson's theorem from point P to the triangle ABQ.
K, N, M are on the same line.
angle QOP = angle QAP = angle NXP = angle KPX (because OQ are parallel to XP)
So KP is bisected by line NM

[quote="ashegh"]dear zaoli ithink u question is wrang :!: .
be cause u know by moving Q on line L the locus of point X (inter section ofAQ andKP)is only a point .
and it is stable.but by by changing Qthe locus of Kwont be stable .and it change on the line OP.
then how can line QA bisect KP :?:  :!:  :?:  :!:[/quote]

Why on earth is X stable?

And yes, K is stable. This is something most solvers showed in their solutions as an auxiliary fact. Please read the solutions posted by others before posting such comments, since they could clear up your doubts.

And, by the way, it's MN and not QA that bisects KP.

  Darij

ya ya i finally found where was the mistake.
but itsreallyanicequestion .isnt it?
because as u know the locus of point X is fixed(X is the inter section ofKP and MN).
and again the lucas of point K,will be stable.
because K is the pole of line L.and for each point like Q on L we have:
the polar of Q will go through the pole of L wich is K.and it means k is stable :D .

[b][color=#00f]Lemma:[/color][/b] Let $\Delta ABC$ be an isosceles triangle with $AB=AC$, $H$ as its orthocenter and $D$ as $A-$antipode. Take a point $F$ on $\odot(ABC)$. Let $FD\cap BC=G$. Then the simson line of $F$ is parallel to $GH$
[b][color=#f00]Proof:[/color][/b] Assume, $F$ on minor arc $AB$. Let $N$ be foot from $F$ to $AB$ and $DF$ intersects $F-$simson line at $O$. $$\angle FON=\angle 180^{\circ} -\angle FAN-\angle FAC=180^{\circ}-2\angle FAD=\angle FGH \qquad \blacksquare$$
-----------------------------
Using the fact, of orthocentric bisection by simson line, the problem is proved


AHHH!! This is driving me nuts.. can anybody simplify:  :sqrt: (75-24 :sqrt: 6)? I've tried for like hours and hours and I can't figure it out.. Is there even one?  I'm a little wary of my own brain.

Mr.Ocax

Merry Christmas!  :lol:

nothing except that you can pull the 3 out in the radical @@


For the set of numbers $ \{a, b, c, d, e\}$, the following five
inequalities are true: $ c < d, a < b, d < a$, and $ e < d$. Which
number in the set is the median?

We want to try and arrange these numbers in order to find the median.

Using the inequalities, we find that the order of the numbers is(left is less, right is more):

$ c, e, d, a, b$

There are other possibilities, such as:

$ e, c, d, a, b$

but in all of them, the middle number is $ \boxed{d}$.


In triangle $ ABC$, $ m_a, m_b, m_c$ are the length of the median lines, and $ r_a, r_b, r_c$ are the radius of the escribed circles.

Prove that

$ \frac{r_ar_b}{m_am_b}\plus{}\frac{r_br_c}{m_bm_c}\plus{}\frac{r_cr_a}{m_cm_a}\ge3$.

[quote="Stephen"]In triangle $ ABC$, $ m_a, m_b, m_c$ are the length of the median lines, and $ r_a, r_b, r_c$ are the radius of the escribed circles.

Prove that

$ \frac {r_ar_b}{m_am_b} \plus{} \frac {r_br_c}{m_bm_c} \plus{} \frac {r_cr_a}{m_cm_a}\ge3$.[/quote]
This inequality by Jichen :)

I have a proof that used $ m_a\equal{}\frac{1}{2}\sqrt{2b^2\plus{}2c^2\minus{}a^2}, r_a\equal{}\sqrt{\frac{s(s\minus{}b)(s\minus{}c)}{s\minus{}a}}$.

So it's ugly. Any other ideas?

[quote="Stephen"]In triangle $ ABC$, $ m_a, m_b, m_c$ are the length of the median lines, and $ r_a, r_b, r_c$ are the radius of the escribed circles.

Prove that

$ \frac {r_ar_b}{m_am_b} \plus{} \frac {r_br_c}{m_bm_c} \plus{} \frac {r_cr_a}{m_cm_a}\ge3$.[/quote]

By $ Ptolemey$ : $ m_am_b \le \frac{c^2}{2}\plus{}\frac{ab}{4}$ , $ m_bm_c \le \frac{a^2}{2}\plus{}\frac{bc}{4}$ , $ m_cm_a \le \frac{b^2}{2}\plus{}\frac{ca}{4}$ ;

$ r_ar_b\equal{} p(p\minus{}c)$  ,  $ r_br_c \equal{} p(p\minus{}a)$  ,  $ r_cr_a\equal{} p(p\minus{}b)$ ;

So : $ LHS \ge 4p ( \frac{p\minus{}a}{2a^2\plus{}bc}\plus{}\frac{p\minus{}b}{2b^2\plus{}ca}\plus{}\frac{p\minus{}c}{2c^2\plus{}ab})$ $ \ge \frac{4p^3}{2p(a^2\plus{}b^2\plus{}c^2)\minus{}2(a^3\plus{}b^3\plus{}c^3)\minus{}3abc\plus{}p(ab\plus{}bc\plus{}ca)} \ge 3$

$ \Leftrightarrow 28p^3 \ge 27p(ab\plus{}bc\plus{}ca) \minus{} 27abc$ , ($ p\equal{}\frac{1}{2}(a\plus{}b\plus{}c)$ , $ ab\plus{}bc\plus{}ca\equal{} r^2\plus{}p^2\plus{}4rR$ , $ abc\equal{}4prR$)

$ \Leftrightarrow p^2 \ge \frac{27}{2}Rr \ge 27r^2$ -which is very well known ;

[quote="Stephen"][quote="Sergic Primazon"][quote="Stephen"]In triangle $ ABC$, $ m_a, m_b, m_c$ are the length of the median lines, and $ r_a, r_b, r_c$ are the radius of the escribed circles.

Prove that

$ \frac {r_ar_b}{m_am_b} \plus{} \frac {r_br_c}{m_bm_c} \plus{} \frac {r_cr_a}{m_cm_a}\ge3$.[/quote]

$ LHS \ge 4p ( \frac {p \minus{} a}{2a^2 \plus{} bc} \plus{} \frac {p \minus{} b}{2b^2 \plus{} ca} \plus{} \frac {p \minus{} c}{2c^2 \plus{} ab})$ $ \ge \frac {4p^3}{2p(a^2 \plus{} b^2 \plus{} c^2) \minus{} 2(a^3 \plus{} b^3 \plus{} c^3) \minus{} 3abc \plus{} p(ab \plus{} bc \plus{} ca)}$ [/quote]

Why this step works?[/quote]

$ 4p ( \frac {p \minus{} a}{2a^2 \plus{} bc} \plus{} \frac {p \minus{} b}{2b^2 \plus{} ca} \plus{} \frac {p \minus{} c}{2c^2 \plus{} ab}) \ge$ $ 4p(p\minus{}a\plus{}p\minus{}b\plus{}p\minus{}c)^2 \frac{1}{(p\minus{}a)(2a^2\plus{}bc)\plus{}(p\minus{}b)(2b^2\plus{}ca)\plus{}(p\minus{}c)(2c^2\plus{}ab)}$


Physics is more concered with fundamental theories rather than applications. However Engineering Physics makes use of physical theories (for example quantum mechanics or nanotechnology) in real-world applications like in electronic circuits. Also if you really like mathematics, physics (especially theoretical physics e.g. superstring theory) is much closer to mathematics than engineering physics.

[quote="martha"]
And.... which one has an easier life in school? :P

Thanks![/quote]
That would probably be engineering physics. though it still depends at what level you're working...

Engineering students require usually 6 or 7 required courses per semester, while someone is a science program such as physics will coast along with 4 required courses and 1 elective. Engineers learn as much or more math as any physicist, and combine it with classical mechanics and real world design applications, engineering is a profession career. Physics is not.

Engineering is a much more difficult and time consuming area of study.

i'm an engineering student at waterloo, i say life of math/science programs is soooo much easier than engineering. but don't go for the easiness? do it because u want to! to get to the top is not easy, regardless of any program! (although workload for math or science is significantly less than that of engineering, try competing with the high calibre students for the top spots! ie. ex-IMOers, IPHOers, etc... even if not, ur dealing with top students in various competitions! and trying to top them is not easy!  :rotfl: ) we got a lot of those on this forum.) unless u just want to pass, thats another story.  :P hope this helps!

[quote="martha"]Applying for university soon. Just heard UBC isn't that good at math. Decided to go for physics instead.

There are 2 different programs called "Physics" and "Engineering Physics". What are the differences of them? I am much confused.

And I heard they are equally good at UBC....

And.... which one has an easier life in school? :P

Thanks![/quote]

UBC is probably the third best university in Canada for math.  They have a very good math department.