geometry
perimeter
induction
absolute value
combinatorics proposed
combinatorics

Suppose we have a simple polygon (that is it does not intersect itself, but not necessarily convex).
Show that this polygon has a diameter which is completely inside the polygon and the two arcs it creates on the polygon perimeter (the two arcs have 2 vertices in common) both have at least one third of the vertices of the polygon.

I'll sketch it very briefly, because I found it pretty difficult to write down.

The idea would be to triangulate the polygon, and choose the chord we're looking for among the sides of this triangulation. We pick one such side which divides the perimeter into two arcs with minimal difference in absolute value between the numbers of vertices, and prove, based on the fact that a triangle has at most three other triangles as neighbours, that as long as this difference is $<\frac n3$ ($n$ is the number of vertices of the polygon), one of the other two sides of one of the triangles which share the chosen side will be a better choice, contradicting the optimality assumption we made.

After triangulation, induction works just fine. Any triangulation has a triangle with 2 sides on the sides of the original polygon. Take this triangle out of the figure and let $AB$ be the diameter of the remaining $n-1$-gon which divides it in the way we need it to. Let $X$ be the vertice that we took out and suppose $AB$ doesn't work for the $n-gon$. This can only mean that $A...X...B$ has more than $2/3\cdot (n+1)+2$ vertices. There is a triangle $ABY$ in the triangulation with $Y\in A...X...B$. One of $A...Y$ and $Y...B$ has at least $[2/3\cdot (n+1)+3]/2$ vertices. Suppose it is $A...Y$. Then $AY$ is a diameter that works. It is clear that $A...Y$ has at least $1/3 \cdot (n+1)$ vertices. It is also clear that $A...B...Y$ has at least $\frac n3$ (induction) +1 vertices so at least $1/3 (n+1)$.

I found it quite not easy (AND uneasy...) to actually find a triangulation for a simple polygon. If anyone can illuminate me with an immediate solution I would be thankfull.

[quote="Hawk Tiger"]Can I  ask a quetion that what " the arcs it creats " means? and does the "diameter" means the "segment attaches the two points between which the distant is the largest".[/quote]
Arcs means the two parts created on the perimeter of the polygon.
And no, diameter is only the segment between two non-adjacent points.

My solution is same as ikap's but I think that grobber's better.


Let a and b be integers and p>3 a prime. Find all pairs (a,b) of integers satisfying:

a 2 +3ab+2p(a+b)+p 2 =0

First of all, we know that the part $a^2 + 3ab = a(a+3b)$ is some multiple of p.  Therefore, either a is a multiple of p or a+3b is a multiple of p.

Then see that if $a \not\equiv b \not\equiv 0 \mod p$, then only a+3b can be a multiple of p, and

$a^2 + 3ab + 2p(a+b) + p^2 \equiv a^2 + 3ab - 4pb + p^2$
$\equiv a(a+3b) -4pb \equiv -4pb \equiv 0 \mod p^2 \implies b \equiv 0 \mod p$

But this is a clear contradiction since $b \not\equiv 0 \mod p$.  Therefore, it must be that a or b is equivalent to 0 mod p.  We know for sure $a \equiv 0 \mod p$, because if only $b \equiv 0 \mod p$, then our first requirement does not hold.

Let a = kp.  We then have

$k^2p^2 + 3kpb + 2p(kp + b) + p^2 = 0$

$k^2p^2 + (3k+2)pb + 2kp^2 + p^2 = 0$

$(k+1)^2p^2 = -(3k+2)pb$

$(k+1)^2p = -(3k+2)b$

Note that now either $b \equiv 0 \mod p$ or $3k+2 \equiv 0 \mod p$

Case $b \equiv 0 \mod p$.  Let b = np.

-- $(k+1)^2p = -(3k+2)np$
-- $(k+1)^2 = -n(3k+2)$
-- $k^2 + 2k + 1 = -3nk - 2n$
-- $k^2 + (2+3n)k + 1 + 2n = 0$
-- $k = \frac{-2-3n \pm \sqrt{9n^2 + 4n}}{2}$

Since $(3n)^2 < 9n^2 + 4n < (3n+1)^2$ for all positive n, and $(3n)^2 > 9n^2 + 4n > (3n+1)^2$, no integral value for k exists for non-zero n.  The only possibility is n = 0 $\implies$ b = 0 $\implies$ k = -1 $\implies$ a = -p for this case.

Case $3k+2 \equiv 0 \mod p$.  Let 3k+2 = np.

-- $(k+1)^2p = -npb$
-- $(k+1)^2 = -nb$

-- $\big(\frac{np+1}{3}\big)^2 = -nb$

-- $(np+1)^2 = -9nb$
-- $n^2p^2 + 2np + 1 = -9nb$
-- $n^2p^2 + (2p+9b)p + 1$
-- $n = \frac{-2p-9b \pm \sqrt{36pb + 81b^2}}{2p^2}$

$81b^2 < 36pb + 81b^2 < (9b+2p)^2$ for positive b and $81b^2 > 36pb + 81b^2 > (9b+2p)^2$ for negative b.  Thus, the only possible integral value of $\sqrt{36pb + 81b^2}$ that is equivalent to 9b mod p is 9b + p.  Plugging that in for n, we don't get an integer, so this does not work.  b = 0 doesn't work either in this case.

Therefore, the only solution is b = 0, a = -p from the first case.  Horrible solution :).

[quote="probability1.01"]It was asking for all pairs (a, b)... no?  So I guess a and b are in terms of p?  I don't really know...[/quote]

Let me give you an example.
For instance, (-5,4,7) is a solution according to my solution, while according to yours, it isn't.


[quote="shfdfzhjj"]Is it old?[/quote]
My solution :)

[quote="shfdfzhjj"]I think my proof is better :lol:[/quote]
Yes, I know this solution :) It comes from Phan Thanh Nam, an admin of the website: http://www.diendantoanhoc.net/ but I like my solution more since I did it independently (It does not depend on anyone) :)


I know I sound a bit childish, but this is my first time and I'm curious on how I did.

If you mean when Round 4 results are going to be given out, it will be during the week of April 11th, according to DPatrick in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32351]this[/url] thread.

At this point, all of the papers have been graded.  We are simply waiting for the graded papers to return to us from the graders.  Once we get them (probably today or tomorrow), it will take 1-2 days to scan all of the papers into the computer and process the scores.

So, the grades should be available sometime later this week.


number theory
least common multiple
quadratics

Could anyone help me witht these problems, I have a test tomorrow. Also, in number 3) the 5 is seperate from the fraction. Also in number 4) the 3 is seperate from the fraction. Thanks.

1) [size=150][b]3/x - 2/x+1 = 4/x[/b][/size]

2) [size=150][b]2x/x+3     -      5 = 1/x+3[/b][/size]

3) [b][size=150]4/x - 1/x+2 = 2/x[/size][/b]

4) [size=150][b]5x/x-1      -       3 = 2x+5/x²-1[/b][/size]

5) [size=150][b]2x/x-2 - 4x-1/3x+2 = 17x+4/3x²-4x-4[/b][/size]

6) [size=150][b] x     -      24/x = 5[/b][/size]

[hide]just rationalize the denominators by multiplying by the LCM of the denominators. For example, the first one

$\frac{3}{x}-\frac{2}{x}+1=\frac{4}{x}$
$x(\frac{3}{x}-\frac{2}{x}+1)=x(\frac{4}{x})$
$3-2+x=4$
$x=3$[/hide]

Also,

[quote]Also, in number 3) the 5[/quote]

what 5?

Sorry, I meant number 2. Thanks for the help though.

So I am guessing that the LCD for:

2) LCD: 3x

3) LCD: x

4) LCD: -x

5) LCD: ?

6) LCD: x

Again, I am not really sure how to find the LCD.

wait.
i think i read the question wrong. for number 1, when you say 2/x+1,
do you mean $\frac{2}{x}+1$ or $\frac{2}{x+1}$?
if you mean the second one, put *parentheses* to make it clearer. do that for all of them.

I think I got the rest: For 3) -4; and for 5) -1 and 3.

Also for 2) -16/3 - I have no clue if this is right, and number 4) I did not get.

[hide="4"]
$\frac{5x}{x-1}-3=\frac{2x+5}{x^2-1} \Longrightarrow \frac{5x-3(x-1)}{x-1}=\frac{2x+5}{(x+1)(x-1)} \Longrightarrow 2x+3=\frac{2x+5}{x+1} \Longrightarrow (2x+3)(x+1)=2x+5 \Longrightarrow 2x^2+5x+3=2x+5 \Longrightarrow 2x^2+3x-2=0 \Longrightarrow (2x-1)(x+2)=0$
So the answers are $\boxed{\frac{1}{2}}$ and $\boxed{-2}$ :D
[/hide]
And I also got $-\frac{16}{3}$ for number 2.  :)

It's fairly simple. If there's an x in the denominator, multiply the entire equation by that denominator. With these, nothing should get bigger than a quadratic, and you know how to solve those. Dont be scared.


The price of an object is decreased two times with percent $p$ and then is increased three times with percent $p$. Find $p$ s.t. the price of the object at the end is the highest possible.

Maybe I never learned that but is P sepose to mean something?

I think you mean "percent" instead of "procent".

yes, you're right

i really don't know why i used "procent" (rom), when i knew how it was called in english :D

Is P sepose to be the ammount of percent?

Find the maximum of [tex]X(1-p)^2(1+p)^3[/tex].

Where [tex]X[/tex] is the initial price and [tex]0\le p< 1[/tex].

indeed, this is what the problem is asking for  :D

I forgot about this problem, i must get working on it again...

it's the local maximum of $p^5+p^4-2p^3-2p^2+p+1=0$. i don't know any ways to solve this by hand so i used a calculator and got
[hide="answer"]a max when $p=\frac{1}{5}$ which is $\frac{3456}{3125}X = 1.10592X$[/hide]

(hope i did it right)

Indeed, diffing that the roots are -1,1, and your answer

Actually, as p approaches 1, it could actually get bigger than 1/5. That could just be a local maximum, with a bigger one at 1.

yes, it's $\frac{1}{5}$

the (quite brilliant) solution given in the book from where i got it used am-gm only, no polynomials involved

You have this expression $(1-p)(1-p)(1+p)(1+p)(1+p)$. Try to transform it s.t. you obtain a constant sum (in order to apply am-gm nicely)

In more "words":
\begin{eqnarray*}\frac{1}{9}(3-3p)(3-3p) \cdot \frac18 (2+2p)(2+2p)(2+2p) &\leq& \frac1{72}\cdot \left( \frac{(3-3p)+(3-3p)+(2+2p)+(2+2p)+(2+2p)}5 \right)^{5}\\ \ &=& \frac1{72}\cdot \left( \frac{12}5 \right)^{5}, \end{eqnarray*}
with equality for $3-3p=2+2p$, i.e. $p = \frac15$.


function
quadratics
algebra
functional equation
algebra proposed

Let $ f: \mathbb{R}\rightarrow \mathbb{R}$ be a function such that
\[ f(2x)\leq 3f(x) \plus{} f( \minus{} x)\]
and
\[ f(x \plus{} y) \minus{} f(x) \minus{} f(y)\geq \phi(x,y)\]
with $ \phi: f: \mathbb{R}^2\rightarrow \mathbb{R}$ such that
\[ \phi(x, \minus{} y)\geq \minus{} \phi(x,y).\]
Prove that $ f$ can be written as $ a(x) \plus{} q(x)$ where A is an additive function and $ q(x)$ satisfies the quadratic functional equation
\[ q(x \plus{} y) \plus{} q(x \minus{} y) \equal{} 2q(x) \plus{} 2q(y).\]

The trick here was to write $ f$ as the sum of an odd and an even function and notice that those functions satisfy the required conditions.


[quote="xxxxx"]solve for x: 4/(x+1) + 6/(x-3)= (x squared + 14x -11)/(x+1)(x-3)
sorry i dunno LaTex[/quote]
No problem, I'll convert for you. :)
solve for $x$: $\frac{4}{(x+1)}+\frac{6}{(x-3)}= \frac{x^{2}+14x-11}{(x+1)(x-3)}$.

Try adding the two fractions on the left.

[quote="i_like_pie"][quote="Ihatepie"]put this in the high school forum. 2 hard. I'm in double advanced 7th grade(=advanced 8th grade=regular ninth grade=high school=2 hard for this forum)and can't figure this out.[/quote]
If you're in a ninth grade class, you should know how to solve this. All you need to know is how to factor and how to add fractions.[/quote]

we just started didn't get 2 factors yet doing algebra which is 9th grade math.

[quote="Ihatepie"]put this in the high school forum. 2 hard. I'm in double advanced 7th grade(=advanced 8th grade=regular ninth grade=high school=2 hard for this forum)and can't figure this out.[/quote]
I think it is more like one of the easier Mathcounts problems.


geometry
3D geometry
sphere
topology
projective geometry
advanced fields
advanced fields solved

How to show that $\pi_{1}(\mathbb{R}\mathbb{P}^{2})\cong \mathbb{Z}_{2}$? Is it known something more general about $\pi_{n}(\mathbb{R}\mathbb{P}^{m})$ for positive integers $n,m$?

The first one folows from covering space theory: if $p: \widetilde X\to X$ is a covering map, then the homomorphism $p_{*}: \pi_{1}(\widetilde X)\to\pi_{1}(X)$ induced by $p$ is a monomorphism, and the index of $p_{*}(\pi_{1}(\widetilde X))$ in $\pi_{1}(X)$ is equal to the number of sheets of the covering. In particular, we can take $X=\mathbb P^{n}$ and $\widetilde X=\mathbb S^{n}$ (for $n\ge 2$). This is a two-sheeted covering, and since the fundamental group of the sphere of dimension at least $2$ is trivial, it follows that the fundamental group of the projective space of dimension $\ge 2$ has two elements, and of course, the only such group is $\mathbb Z/2$.

A covering map induces isomorphisms on homotopy groups $\pi_{n},\ n\ge 2$, so those groups, for the projective spaces, coincide with the corresponding groups for spheres. It is notoriously difficult to compute higher homotopy groups; not even all the homotopy groups of $\mathbb S^{2}$ are known.

and why doesn't this work to compute $ \Pi_1 (\mathbb{CP}^n)$ ??

it can't work because the projection map $ S^{2n\plus{}1}\to \mathbb{CP}^n$ is not a covering map.

actually, it's a fibration with fiber $ S^1$, so the homotopy long exact sequence reads:
$ \dots\to\pi_{m\plus{}1}(\mathbb{CP}^n)\to\pi_m(S^1)\to\pi_m(S^{2n\plus{}1})\to\pi_m(\mathbb{CP}^n)\to\pi_{m\minus{}1}(S^1)\to\dots$
this yields in particular $ \pi_1(\mathbb{CP}^n)\equal{}0$, and $ \pi_2(\mathbb{CP}^n)\equal{}\mathbb{Z}$ (the latter could have been deduced from the former, together with hurewicz' theorem and the computation of $ H_*(\mathbb{CP}^n)$, which is standard).
for $ m\ge 3$, all we can say is that $ \pi_m(\mathbb{CP}^n) \equal{} \pi_m(S^{2n\plus{}1})$, and in particular for $ 3\le m\le 2n$ they're trivial, for $ m\equal{}2n\plus{}1$ it's $ \mathbb{Z}$ (for bigger $ m$, see grobber's remark).

and how about $ S \rightarrow \mathbb{CP}^n$ where S is the unit sphere in $ \mathbb C^{n\plus{}1}$ ?? still not a covering map?

that was the map i was talking about. the unity sphere in $ \mathbb{C}^{n \plus{} 1}$ is just $ S^{2n \plus{} 1}$. i'm sorry i've not been that clear about it.

anyway, there are (at least) three reasons explaining why this can't be a covering map: the first is [i]a priori[/i], while the others come from some computations:
1. the fiber of that projection is not discrete (in particular, any point in the projective space can't have an uniformly covered neighbourhood).
2. the projective spaces are simply connected, so their only covering are trivial.
3. if it was a covering map, it would induce an isomorphism between higher homotopy groups, but $ \pi_2$'s of spheres (except for $ S^2$, but we're not concerned with it) are trivial, while (assuming my previous post) those of projective spaces are isomorphic to $ \mathbb{Z}$.

is $ S^1$ the fiber of $ p: S^n (complex) \rightarrow \mathbb{CP}^n$ ??

it's homeomorphic to $ S^1$, yes, the homeomorphism being given by $ e^{i\theta}\mapsto e^{i\theta}(z_1,\dots,z_n)$, where $ (z_1,\dots,z_n)\in S^{2n\plus{}1}$ (and of course you're considering the fiber over its projection).


limit
trigonometry
real analysis
real analysis unsolved

One of the key step in the proof of $ \mathop {\overline {\lim } }\limits_{n \to \infty } \sin \left( {nx} \right) \equal{} 1$ for almost every $ x \in [0, 2\pi]$ is to show that the following set
\[ A \equal{} \left\{ {x \in \left( {0,2\pi } \right): \frac {x} {\pi } \notin \mathbb{Q}} \right\}\]is a measurable set of measure $ 2\pi$. How can we claim this?

Well, given $ x\in (0,2\pi)$, $ x/\pi\in\mathbb{Q}\iff x\equal{}\alpha\pi$ for some $ \alpha\in (0,2\pi)\cap\mathbb{Q}$.  So $ A$ is just $ (0,2\pi)$ (which is measurable) minus the countable set $ (0,2\pi)\cap\mathbb{Q}$.  Deleting a countable number of points can't change measurability nor the measure of the set.


Here is the problem. Prove $ (a^2\plus{}b^2)(a^3\plus{}b^3)(a^6\plus{}b^6) < 4(a^{11}\plus{}b^{11})$
It feels like chevyshev to me, but I dont know how to start.

It is only Chevychev like you say:

$ (a^2\plus{}b^2)(a^3\plus{}b^3)(a^6\plus{}b^6)\le 2(a^5\plus{}b^5)(a^6\plus{}b^6)\le 4(a^{11}\plus{}b^{11})$
And it is all. You have to study some books of inequalities.

:blush:
thanks! yeah I am reading a book right now..


Let $ S$ denote the area of a given triangle $ ABC$, and denote $ BC \equal{} a$, $ CA \equal{} b$, $ AB \equal{} c$. Prove the inequality
$ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2\ge 16S^2 \plus{} \frac {1}{2}[a^2(b \minus{} c)^2 \plus{} b^2(c \minus{} a)^2 \plus{} c^2(a \minus{} b)^2]$
When does equality hold?

Little algebraic transformation yields the equivalence of this inequality and Schur's Inequality.


function
real analysis
real analysis solved

Do there exist a bounded function $f$:$R\rightarrow R$ such that $f(1)>0$
$\forall x,y \in R$, $f^2(x+y)\geq f^2(x)+2f(xy)+f^2(y)$

I do not think so. If not, then $f^2(x+1/x)\geq f^2(x)+2a$ for all $x$, where $a=f(1)>0$. Then for all $|x|\geq 2$ we have $ f^2(x)\geq 4a$, then for all $x$ such that $ x=t+1/t$ for a certain $|t|\geq 2$ (we call this set $A_1$)  we have $ f^2(x)\geq 6a$. Then for all $x$ such that there exists $t\in A_1$ with $x=t+1/t$ we  have $ f^2(x)\geq 8a$ and so on. It is easy to see that actually the sets $A_i$ are not vide for any $i$.


WELL ARE THE ADMINS. LISTENING (ALTHOUGH YOU WON'T HEAR ME !) ? I WOULD LIKE TO POINT OUT THAT IN A CERTAIN FORUM, WE DO NOT HAVE TO WRITE THE LaTex . WE ONLY NEED TO SELECT THE PORTION TO BE PUT FORTH IN LaTex AND CLICK ON A BUTTON WHICH AUTOMATICALLY INSERTS THE DESIRED CODE. THEN WE NEEDN'T TYPE THE WHOLE LaTex COMMAND. IS THERE SOMETHING LIKE THIS HERE ?

You need to watch where you're posting rants. This is the POLLS forum, which is kind of like asking opinions and the like. If you have a LaTeX problem, go to the appropriate forum.

Methinks someone is just too lazy to learn LaTeX codes?


In the overlapping triangles $ \triangle{ABC}$ and $ \triangle{ABE}$ sharing common side $ AB$, $ \angle{EAB}$ and $ \angle{ABC}$ are right angles, $ AB \equal{} 4$, $ BC \equal{} 6$, $ AE \equal{} 8$, and $ \overline{AC}$ and $ \overline{BE}$ intersect at $ D$. What is the difference between the areas of $ \triangle{ADE}$ and $ \triangle{BDC}$?

[asy]
defaultpen(linewidth(0.8)+fontsize(10));size(200);
unitsize(5mm) ;
pair A=(0,0), B=(4,0), C=(4,6), D=(0,8), H=intersectionpoint(C--A, D--B);
draw(A--B--C--cycle) ;
draw(A--B--D--cycle) ;
label("E",(0,8), N) ;
label("8",(0,4),W) ;
label("A",A,S) ;
label("B",B,SE) ;
label("C",C,NE) ;
label("6",(4,3),E) ;
label("4",(2,0),S) ;
label("D",H,2*dir(85)) ;[/asy]

$ \textbf{(A)}\  2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\  5 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9$

The title was meant to deceive.  :rotfl:  :rotfl:

[hide="Solution"]B.

The answer is the area of ABE decreased by the area of BDC.[/hide]

[hide]The overlapping area doesn't matter; just subtract: you get 1/2(8*4) = 16, and 12 for the two areas of the two triangles; then subtract and you get 4...B[/hide]

[quote="4everwise"]The title was meant to deceive.  :rotfl:  :rotfl: [/quote]

:mad: Not anymore :D I disagree with whatever 4everwise does :P (Just kidding :))


geometry
circumcircle
trapezoid
geometric transformation
reflection
perpendicular bisector
geometry unsolved

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14402[/img]

 :(

The Nagel point $ M$ and the external similarity center $ X_{56}$ of the circumcircle $ (O)$ and incircle $ (I)$ are isogonal conjugates. This is well known but anyway: Let $ X, Y, Z$ be midpoints of the circumcircle arcs $ BC, CA, AB$ opposite to $ A, B, C,$ respectively. The circumcircle tangents at $ X, Y, Z$ are parallel to the triangle sidelines, they meet at points $ A', B', C',$ such that the circumcircle $ (O)$ is the incircle of the $ \triangle A'B'C' \sim \triangle ABC$ and their similarity center is $ X_{56}.$  In addition, let $ X'$ be midpoint of the circumcircle arc $ BC$ diametrically opposite to $ X.$ $ A'X'$ is the A'-Nagel line of the $ \triangle A'B'C',$ connecting the vertex $ A'$ with the tangency point of its A'-excircle with $ B'C',$ and the line $ A'A$ goes through $ X_{56}.$ $ AX' \perp AI$ is the external bisector of the $ \angle A$ and the tangency point line $ YZ$ is perpendicular to the internal bisector $ A'O$ of the angle $ \angle A'.$ From $ AI \parallel A'O,$ we have $ AX' \parallel YZ,$ $ AX'ZY$ is a cyclic trapezoid, isosceles, and $ A'$ lies on the perpendicular bisector of its base $ YZ.$ Therefore, $ \angle ZA'X' \equal{} \angle YA'A,$ etc., which proves the claim.

$ AX_{56}$ cuts the circumcircle $ (O)$ again at the tangency point $ U$ of the A-mixtilinear incircle with $ (O)$ (by Monge-d'Alembert theorem). Let a parallel through $ A$ to $ BC$ cut the circumcircle $ (O)$ again at $ A^*.$ Let $ D$ be the tangency point of the incircle $ (I)$ with $ BC.$ Since $ A^*D$ is a reflection of the A-Nagel line in the perpendicular bisector of $ BC,$ it cuts the circumcircle $ (O)$ again at $ U.$ The A-altitude $ AH$ cuts the circumcircle again at the reflection $ K$ of $ H$ in $ BC.$ $ A^*K$ is a diameter of $ (O)$ because $ AA^* \perp AK.$ Then the angle $ \angle DUK \equal{} \angle A^*UK$ is also right and $ DUKH_a$ is cyclic with a circumcircle $ (P)$ on diameter $ DK.$ Let tangents to $ (I)$ from $ K \in (O)$ cut $ BC$ again at $ E, F.$ According to the proof of the problem [url]http://www.mathlinks.ro/viewtopic.php?t=186117[/url], the mixtilinear A-incircle tangency point $ U$ lies on the circumcircle $ (Q)$ of the $ \triangle KEF.$ Thus $ U$ is the 2nd intersection of the circles $ (P), (Q)$ other than $ K.$ The circumcenter $ O$ is on the perpendicular bisector of $ KU.$

[b]Construction[/b] follows from this. $ h_a \equiv HH_a$ is the A-altitude line and perpendicular to $ h_a$ at $ H_a$ is the $ a \equal{} BC$ sideline ($ B, C$ unknown). $ D$ is the tangency point of the incircle $ (I)$ with the sideline $ a.$ Reflect $ H$ in $ a$ into $ K$ and this point is on the triangle circumcircle $ (O)$. Draw the circle $ (P)$ with diameter $ DK.$ Construct tangents from $ K$ to $ (I)$ intersecting $ a$ at $ E, F.$ Draw circumcircle $ (Q)$ of the $ \triangle KEF.$ $ U$ is the intersection of $ (P), (Q)$ other than $ K.$ Let $ ID \perp a$ cut $ (I)$ again at $ J.$ Draw parallel to $ DU$ through $ J$ and reflect it in $ JID.$ This is the A-Nagel line of the desired $ \triangle ABC,$ intersecting the A-altitude line $ h_a$ at $ A$ and the sideline $ a$ at the tangency point $ D_a$ of the A-excircle. To finish, either construct tangents to $ (I)$ from $ A$ intersecting $ a$ at $ B, C$ or find the circumcenter $ O$ as intersection of the perpendicular bisectors of $ DD_a, KU$ and draw the circle $ (O)$ with radius $ OA$ intersecting $ a$ at $ B, C.$

$ JD$ is a diameter of the incircle $ (I)$. Points $ J, I, D$ are on one line. I call this line $ JID$. A parallel to $ DU$ goes through $ J$. Reflection of this parallel in the line $ JID$ is the A-Nagel line $ AD_a,$ yielding the vertex $ A.$

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14674[/img]

 :thumbup:

i have nice solution . lemma if we have trangle ABC incircle I and line AK where K is on BC. if 'wright' tangent line to I intersect  BC at P and this line is parallel to AK then C-Mixtlinear circle is tangent CB at point P (easy to prove). if point H' is point on line HH_A and H_AH' = HH_A esy to prove that H' lies on circle around ABC, let's construct tagent line to I and parallel to AH_A and nearest to B if P is intersection of this line and BC then construct circle tangent BC at P and tangent to AH_A..

... than it is C-Mixtlineary circle like this we can construct B-Mixtlineary circle and then construct Apolony circle for point H' and two Mixt... circles so we construct circle around ABC intersection of this circle and line perpendicular to AH' (and if H' on this line) are points B and C and tangent lines throu B and C to I intersects at A.


A convex octagon is inscribed in a circle and has four consecutive sides of length 3 and four consecutive sides of length 2. Find its area in the form $r+s\sqrt{t}$.

[quote="asdff"]I dont know, bragumuphtas (spelling)?[/quote]
It only works for quadrilaterals; trying to use it here would require a lot of diagonal lengths and would be very nasty.
[hide="hint"]
It is a fact that you can rearrange the sides of a cyclic polygon without changing circumradius or area. Then consider the arrangement 3,2,3,2,3,2,3,2. It has lots of symmetry...
[/hide]

Im going to randomly guess
[size=200][color=green]29.97[/color][/size]
when rounded to the nearest hundredth.

Please tell me Im right................................

So anyone got an answer..btw 29.97 was rounded answer but I dont know about $r+s\sqrt{t}$, pluggint hat in for r,s,t should be that..I think?

[hide="A hint to follow scorpius"]After you have rearranged the sides in the way he suggested, notice that you have lots of parallel sides.  [hide="More hint"]Extend a few of them.[/hide][/hide]


someone is gotten passionate to the game theory? There is someone that can help me?  :blush:

i am reading a game theory textbook right now.  I am learning rectangular games :oops:


An acute angled triangle ABC is such that AC and BC are of different
lengths, and is inscribed in a circle "w". Let N be the midpoint of the arc
AC (and B does not lie on this arc). Let M be the midpoint of the arc BC
(and A does not lie on this arc). Let D be the point on the arc MN such
that DC parallel to NM. Let K be an arbitrary point on the arc AB (and C does not lie on this arc). Let O,O_1 and O_2 be the incentres of the triangles
ABC, ACK, CBK respectively. Suppose that L is the intersection of
the line DO and the circle "w" such that L /= D. Prove that the points
K,O_1;O_2,L are concyclic.

As we can see
angle O1KO2 =1/2 angle AKB
only need to prove angle O1LO2= 1/2 angle AKB
Its easy to see D is the midpoint of arcACB
then angle DLB=1/2 angle AKB
first we can see ALO is similar to OLB
then AL/OL=AO/BO=NO/MO
by useing NA=NO1=NO,MO=MO2
and also ANO1 , OMO2 are similar isosceles triangles
we have NO/MO=AO1/OO2
and there is angle O1AL=angle O2OL
so AO1L is similar to OO2L
then abgle O2O1L=angle OLB
then complete the proof.

 :) --------
this probelm is interesting.

Check this out:

[url]http://www.mathlinks.ro/viewtopic.php?t=555[/url]

chaogold's arguments are, of course, nicer (shorter and more elegant), but I think that from the proof which can be found at the link above we can derive that all the triangles LO1O2 obtained when K moves freely on the arc are similar (so more than just the fact that <O1LO2=(\pi-A)/2 for all K). We don't need that much to solve the problem, though :D...

[quote="StRyKeR"][quote="chaogold"]
and there is angle O1AL=angle O2OL[/quote]

Sorry, but how do you derive the above statement?

And what was the motivation for all your steps? There are quite many, and I am wondering just how you could come up with all of them.  :D[/quote]
well, just compute the angles.
O1AL=1/2(CAK)-LAK
=1/2arc CB -arc KL=arc BM-arc KL
O2OL=MOL-MOO2 =MOL-(90-1/2(OMO2))
=arc ML+arc DB-90=arc ML-arc KB
As for the motivation for the steps.I just noticed it looks like OO2L is similar to AO1L and by this can prove the problem.So all the steps are going around about the ratio and angle of the two triangles.


To me, I see four hills/mountains/cliffs, with valley in a cross shape in the middle.  In the center, there seems to be a depression. Note how the pattern repeats through the valley.

I'm not really scared to bump my album up.  This is a dead, hidden forum.

Does that make any sense?


MIT
college
Princeton
Harvard
articles
University of Chicago

what sports are you guys thinking of playing in college?

math... chess... :lol:  piano if it's considered one....

not really.

usually playing a sport in college is [b]very[/b] hard (well maybe not at a d2 or d3 school)

You can always play intramural.  Playing on the varsity team requires that you are good enough to make the team.  The requirements for doing this depend on where you're going to school.  It's easier to make the team at a DIII school since DIII schools don't give out athletic scholarships and therefore attract less of the top athletes.  However, you still usually have to be reasonably good at the sport to make the team even at a DIII.

Depends where I end up going.

If it's MIT or Caltech, I'll probably swim.

But a lot of the other schools I'm applying to are DI, and I'm not that good...

I will play basketball if possible. If I don't get recruited, I'll walk on. I am just going to try and have a productive year. I will throw down a lot of dunks this year.

[quote="quantum leap"]Cross country if I can.
So if you want to do a sport and fill out a recruiting form, can the school just not let you join the team, even if you get in the school, or do they accept anyone who's willing to come and practice?  My best 5K was just 18:11 this past season, but i think I can improve a lot in track.[/quote]

To be on a varsity sport in college, you have to make the team.  This depends on where you go.  Some schools might not even have cuts while others do.  You should look into this.  I can tell you that MIT's cutoff is that you have to run a sub 18 5K after the first week of practice and then a sub 30 8K after the second week of practice.  Coach is actually a little leniant on these times.  If you run a 30:15 or so and he thinks you'll contribute positively to the team, he'll let you stay.

Where are you thinking about by the way?  Are you a senior this year?

no matter wish sport you want to play, good luck.

[quote="joml88"]Where are you thinking about by the way?  Are you a senior this year?[/quote]

I am a senior this year and will apply to University of Chicago, MIT, Princeton, and Harvard, besides one or two others.  For the DI schools, should I complete the recruitment forms despite my (relatively) slow times?

it's definetly worth a shot. Just make sure you step your game up.

If any one ends up going to Caltech, then I don't think it would be too hard to get onto any of the teams there (unless it's for something like fencing)  :rotfl:

Just joking by the way, though I'm reminded of a particular Sports Illustrated article about Caltech basketball  :roll:

football for brown ;), aka Ivy league football champs (if i get in that is)

[quote="aznphatso"]Matt McGann joked about the MIT shooting team beating the Army at the MIT information session. Yay for MIT![/quote]

That Matt McGann, what a kidder!  [url=http://www-tech.mit.edu/V125/N56/56pistol.html]MIT beating the Army[/url]!  ROFL  Next thing you know they'll be ousting the Navy for the [url=http://web.mit.edu/pistol/www/news.html#mar22]National Title[/url].  :rotfl:  :rotfl:  :rotfl:

[quote="goldendomer"]Making a college team is really hard. Especially for a division one school.[/quote]

We know that, but it certainly isn't going to stop us.


linear algebra
matrix
vector
trigonometry
induction
calculus
derivative

Find the determinant of the nxn matrix $ A \equal{} [a_{ij}]$, where $ a_{ij} \equal{} \alpha$, if $ |i\minus{}j| \equal{} 1$ and $ a_{ij} \equal{} 0$ otherwise.

I believe that the vectors $ u_j$ whose entries are $ \sin\left(\frac{\pi jk}{n\plus{}1}\right)$ are the eigenvectors of this matrix. Given the eigenvectors, we should be able to find the eigenvalues, and given the eigenvalues, we can find the determinant.

Hmm... The only permutation $ \pi(i)$ that satisfies $ |i \minus{} \pi(i)| \equal{} 1$ is the one sending $ 2k$ to $ 2k \minus{} 1$ and $ 2k \minus{} 1$ to $ 2k$ (by induction, for example, and it exists only for matrix of even size). So, by Leibniz Formula, the determinant is just $ \alpha^n( \minus{} 1)^\pi$. We can explicitly count the number of inversions: it is $ \frac {n}{2}$, when $ A$ is a $ n$x$ n$ matrix.

for n = 2 determinant = -$ \alpha^2$
for n  =  3 determinant  =  0
for n > 3
take first line and use Lagrange only the second element a[1,2]  = $ \alpha$ counts all rest are zero and the determinant for this  +/- $ \alpha^n$ need to find the exact patern

Following up on what I wrote.

With those eigenvectors, the eigenvalues turn out to be $ 2\alpha\cos\left(\frac{\pi j}{n\plus{}1}\right).$

If $ n$ is odd, we can set $ j\equal{}\frac{n\plus{}1}2$ to see that $ 0$ is an eigenvalue and hence the determinant is zero.

For $ n\equal{}2k,$ we get that the determinant is

$ (2\alpha^{2k})\prod_{j\equal{}1}^n\cos\left(\frac{\pi j}{n\plus{}1}\right)\equal{}(\minus{}1)^k\left((2\alpha)^k\prod_{j\equal{}1}^{\frac n2}\cos\left(\frac{\pi j}{n\plus{}1}\right)\right)^2$

Calculation of this for a few values of $ k$ gives the determinant as $ (\minus{}1)^k\alpha^{2k}.$

And there must be a nicer way to get that. (Note that andreica  and bambaman mostly have this.)

Kent, how did you guess the eigenvectors?  :o

How did I "guess"? Familiarity. I know this matrix.

This is a famous matrix from a Fourier analysis/differential equations perspective, with maybe a little numerical analysis thrown in. Consider the following eigenfunction problem (itself a portion of many different PDE problems):

Find $ f(x)$ on $ [0,L]$ such that $ f''(x)\equal{}\lambda f(x)$ and $ f(0)\equal{}f(L)\equal{}0.$

The eigenfunctions for this turn out to be $ f_n(x)\equal{}\sin\left(\frac{n \pi x}L\right)$ for $ n\equal{}1,2,3,\dots.$

Now try discretizing the problem. Partition $ [0,L]$ as $ \left\{0,\frac{L}{n\plus{}1},\frac{2L}{n\plus{}1},\dots,\frac{(n\minus{}1)L}{n\plus{}1},L\right\}.$ Let $ u_k$ be the value of $ f(x)$ at $ x\equal{}\frac{kL}{n\plus{}1},$ noting that $ u_0\equal{}u_{n\plus{}1}\equal{}0$ because of our boundary values.

Use the second difference to approximate the second derivative - that is,

$ f''(x_k)\approxeq \frac{f(x_{k\plus{}1})\minus{}2f(x_k)\plus{}f(x_{k\minus{}1})}{\left(\frac{1}{n\plus{}1}\right)^2}.$

Hence looking for the eigenfunctions of the differential operator can be approximated by looking for the eigenvectors of this linear operator. Leaving off the $ k\equal{}0$ and $ k\equal{}n\plus{}1$ terms since they're always zero, we get that the matrix of that linear operator is a scalar multiple of
\[ \begin{bmatrix}\minus{}2&1&&&&\\1&\minus{}2&1&&&\\ &1&\minus{}2&1&&\\ &&\ddots&\ddots&\ddots &\\ &&&1&\minus{}2&1\\ &&&&1&\minus{}2\end{bmatrix}\]
Add a multiple of the identity to that matrix to get the matrix Carcul posted. Adding a multiple of the identity doesn't change the eigenvectors, only the eigenvalues.

Since the discrete problem isn't exactly the same as the differential equation, the eigenvectors don't have to be exactly the same as the eigenfunctions for the continuous problem - but when we try those sines, they do work out. (It helps to note that $ \sin(x\plus{}y)\plus{}\sin(x\minus{}y)\equal{}2\sin x\cos y.$)

This isn't exactly a [i]circulant[/i] matrix, but it a close enough relative. Circulant matrices are diagonalized by the discrete Fourier transform; this one incorporates boundary values and is diagonalized by a discrete sine transform.

One note: what I did is [i]not[/i] the efficient way to answer Carcul's problem. If all you want is the derivative, just expand by minors of the top row, then expand the resulting $ (n\minus{}1)\times (n\minus{}1)$ matrix by minors of the first column. You wind up with $ (\minus{}1)$ times the determinant of the $ (n\minus{}2)\times (n\minus{}2)$ version of the same problem. (I think I just tacitly assumed $ \alpha \equal{}1$ but that's a pretty harmless assumption.)

But combine the two methods and we get a nontrivial expression for a product of cosines:
\[ \boxed{\prod_{k\equal{}1}^m\cos\left(\frac{\pi k}{2m\plus{}1}\right)\equal{}2^{\minus{}m}}\]

We could actually turn it into a circulant by adding a row and column- to fit the original, the extra entry in the eigenvector must be zero.

That last identity can also be deduced from the existence of the [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=246555#comments]Chebyshev polynomials[/url].

[hide="Answer"]As was already derived above, $ det(A) \equal{} \cos\left(n \frac{\pi}{2}\right) \cdot \alpha^n$, where n is the order of the matrix.[/hide]


search
inequalities proposed
inequalities

Let $ a,b,c$ be positive real numbers such that $ a^3\plus{}b^3\equal{}c^3$.Prove that:
$ a^2\plus{}b^2\minus{}c^2>6(c\minus{}a)(c\minus{}b)$

Use the search function. [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=252171[/url]


[quote="indybar"]Let $a,b,c>0, ab+bc+ca=3$

Prove that $a^4+bc+1\geq 3a^2bc$[/quote]
$a^4+bc+1\geq 3a^2bc\Leftrightarrow a^4+\frac{bc(ab+ac+bc)}{3}+\frac{(ab+ac+bc)^2}{9}\geq3a^2bc\Leftrightarrow$
$\Leftrightarrow9a^4+5b^2ac+5c^2ab+a^2b^2+a^2c^2+4b^2c^2\geq25a^2bc.$ This is true by Cauchy. :)


So I don't know that much about matrices but am wondering if it is possible to have two $n$x $n$ matrices $A$ and $B$ such that $AB=I_{n}$ but $BA\neq I_{n}$?

It isn't possible- but finite-dimensionality is essential.

If $AB=I$, then $A$ and $B$ are both nonsingular. Since $A$ is nonsingular, there is some $B'$ with $B'A=I$. Now $B'=B'I=B'AB=IB=B$, and $B$ is the inverse of $A$.

It's the second sentence that uses the finite dimension; in particular, the row space is all of $\mathbb{R}^{n}$ if and only if the column space is all of $\mathbb{R}^{n}$.

There's also what can happen with non-square matrices. Suppose $m<n.$ One can find examples of matrices $A,B,$ with $A$ an $m\times n$ matrix and $B$ an $n\times m$ matrix such that $AB=I_{m}.$ But we must certainly have that $BA\ne I_{n}.$

Is this valid for any ring of matrices? I mean if there exists a ring  where $ab=1$ doesn't imply that $ba=1$ isn't there a ring of matrices where with this property?

Would you say that if V is an m-dimensional space and W is an n-dimensional space with linear transformations T:V->W and U:W->V,  then T has a left inverse if and only if T is non singular and U has a right inverse if and only if U is onto?

spix-If the entries of the matrices don't live in a field, you can transfer ring properties like that. It's not particularly interesting. On the other hand, you can't embed any ring that looks like that into $M_{n}(F)$, with $F$ a field.

§outh§tar- I wouldn't use the word "non-singular" for a nonsquare matrix. The condition for $T$ to have a right inverse $S$ ($ST=I$) is that $T$ is injective (one-to-one).
Constructing these left or right inverses (and thus proving the claim) is a nice exercise. You'll need to use the theorem that a linearly independent set in a finite-dimensional vector space can be extended to a basis.

sorry but i dont see how extending the set to a basis would help you create a left (or right) inverse.


geometry
rhombus
3D geometry
sphere
symmetry
similar triangles
Pythagorean Theorem

Three circles, A, B and C, are externally tangent to eachother and internally tanget to one large circle, D. Circle D has radius 2 while circle A has a radius of 1. Circle A also intersects the center of Circle D. If B and C are congruent, what is the radius of circle B?

It helps if there's a picture, but I'm not too good with computers and such. Anyways, I got a quartic for this, but I'm sure that's not the best solution. I looked at the official solutions and found that they assumed that the line through the center of circle D and the center of circle B passes through the point of tangency between circles D and B. I'm confused as to why we may assume this.

I'm not sure how to explain, but it seems pretty obvious to me that that's the case...

As for proving the problem, I'm not sure how to do it. I've never been too good with this kind of thing.

Let E be where circle B and circle D are tangent. Let k be the line that goes through E and is tangent to circle B and circle D. By definition of a tangent, BE is perpendicular to k and DE is also perpendicular to k. By the angles, it can be proven pretty easily that B, D, and E are collinear.

Uh, if k passes through E, how is it still tangent to B?

Because $E$ is on the edge of circle $B$.

Is the answer $\frac{\sqrt{3}}{3}$?

What I did:

I constructed a rhombus which contained the radii of circle $D$ and made a chord which was on the same line as the radius of Circle $A$ and the diagonals bisected each other, thus there were four $30,60,90$ triangles and then I formed a larger right triangle which included segments of one of the other circles of which we're trying to find the radius, and then I used similar triangles and made another rhombus in which the chord, which I found the length to be $1$ and then found the radius of the circle which was: $\frac{\sqrt{3}}{3}$.

Oh, I see. My bad, I was looking at the wrong circles. The answer is
[hide]8/9[/hide]

The answer is indeed $\frac {8}{9}$. This problem can also be extended to 3D, where a sphere of radius 1 is internally tangent to a sphere of radius 2 and there are three other spheres (congruent to each other), externally tangent to each other and the sphere of radius 1, and internally tangent to the sphere of radius 2. The task is the find the radius of each of the 3 spheres.

[quote="Iversonfan2005"]Um...can someone please make a detailed solution for this please? :([/quote]
My solution:

For convenience, let $A$, $B$, $C$, and $D$ be the center of each circle $A$, $B$, $C$, and $D$, respectively. Also, let $P$ be the point where the circles $B$ and $C$ are tangent.

Let $r$ be the radius of $B$, which we want to find.
Now, let's focus on a right triangle $ABP$.
(Note: $D$ is on $AP$ by symmetry)
We see that $AB=1+r$, $BP=r$.
By Pythagorean Theorem, $AP=\sqrt{(1+r)^2-r^2}=\sqrt{2r+1}$.
So $PD=AP-AD=\sqrt{2r+1}-1$.

Also, since $DE=2$ (for $E$, I'm referring to JesusFreak197's post) and $BE=r$, $BD=2-r$.

Then, we apply Pythagorean Theorem to $\triangle BDP$:
$r^2+(\sqrt{2r+1}-1)^2=(2-r)^2$
$\Longleftrightarrow r^2+2r+1-2\sqrt{2r+1}+1=r^2-4r+4$
$\Longleftrightarrow 6r-2=2\sqrt{2r+1}$
$\Longrightarrow 9r^2-6r-1=2r+1$
$\Longleftrightarrow 9r^2-8r=0$
$\therefore r=\boxed{\frac{8}{9}}$


Sorry, I guess I'm missing something.
I usually see capital sigma with a definitive range variable.
I must be missing something in your notation.

The ineq is equivalent to :
 $\sum (\frac{2a^3+abc}{b^2+bc+c^2}-a) \geq 0$
$\iff \sum (b^2-c^2)(\frac{b}{a^2+ac+c^2}-\frac{c}{a^2+ab+b^2}) \geq 0$
 Suppose $a \geq b \geq c$ , we change the ineq into :
 $(a^2-b^2)(\frac{2a}{b^2+c^2+bc}-\frac{b}{a^2+c^2+ac}-\frac{c}{b^2+a^2+ba}) + (b^2-c^2)(\frac{a}{b^2+c^2+bc}+\frac{b}{a^2+c^2+ac}-\frac{2c}{a^2+b^2+ba}) \geq 0$ , which is obviously true .

Hey!
   How can you transform the messy of mine to this? Is it a technique? For me,I have to prove a stronger one:
$\sum \frac{2a^3+abc}{b^2+c^2+bc} \geq 3 \sum \frac{a^2}{b^2+c^2+bc}$ then from here we can easily obtain the result by AM-GM and Nesbit.

   But is there anyway that we can use AM-GM or BCS or Holder,... to kill this one? Since it is a very weak inequality :lol:

In fact, we have
$\sum \frac{2a^3+abc}{b^2+bc+c^2} \geq a+b+c$
For it's well-known that
$\sum\frac{1}{a^2+ab+b^2} \ge \frac{9}{(a+b+c)^2}$
and it is easy to prove (e.g. using Schur) that
$2(a^3+b^3+c^3)+3abc \ge \frac{1}{3}(a+b+c)^3$
The result then follows by applying Chebyshev.

[quote="Gibbenergy"]Let $a,b,c$ are positive number such that $a+b+c=3$ Prove
$\sum \frac{2a^3+abc}{b^2+c^2+bc} \geq 3$
:D.[/quote]

Hey [b]ductrung[/b]
I think [b]Gibbenergy's[/b] inequality is not strict. With the same conditions as before , in fact, we have a more difficult one
\[ \sum_{cyc} \frac{a^3+2abc}{b^2+c^2+bc} \geq 3 \]

[quote="urun"]Let $a,b,c$ are positive number such that $a+b+c=3$ Prove

\[ \sum_{cyc} \frac{a^3+2abc}{b^2+c^2+bc} \geq 3 \][/quote]
It's obviously true by $uvw$. (this inequality is a linear inequality of $w^3$ :wink: ) .
Another way:
Let $a\geq b\geq c$.
Hence, $\sum_{cyc}\frac{a^3+2abc}{b^2+c^2+bc} -3=\sum_{cyc}\frac{a(a-b)(a-c)}{b^2+bc+c^2}\geq$
$\geq(a-b)\left(\frac{a(a-c)}{b^2+bc+c^2}-\frac{b(b-c)}{a^2+ac+c^2}\right)\geq0$.


calculus
integration
trigonometry
calculus computations

Solve this:

$ \int \frac{sinxcosx}{e^{3x}}$

$ \int \frac{sinxcosx}{e^{2x}}$

$ \int sin^3xcos^2x$

Thanks

hello
 I believe u forgot to write $ dx$
[hide="1,2"]
by using integration by parts we have this general result :
for $ int{e^{ax}(sinbx)dx=\frac{e^{ax}}{a^2+b^2}(asin(bx)-bcos(bx))}$
after writing $ sinxcosx=\frac{sin2x}{2}$

for ur first question $ a=-3$ and $ b=2$
        second question $ a=-2$ and$ b=2$

[/hide]
[hide="3"]
Just substitute $ cosx=t$
the answer -$ \frac{cos^3x(3cos2x-7)}{30}+C$

[/hide]
good to find u again in AoPS after such a long time.
Thank u.

[quote="Fitim"]
$ \int sin^3xcos^2x$
[/quote]
Write $ \cos^2 x$ as $ (1 \minus{} sin^2x)$ multiply into the bracket, and then use the reduction formula forn $ I_n \equal{} \int \sin^n x dx$:

$ I_n \equal{} \frac { \minus{} 1}{n} \sin^{n \minus{} 1}x \cos x \plus{} \frac {n \minus{} 1}{n} I_{n \minus{} 2}$

$ \int \frac {\sin x \cos x}{e^{px}}\ dx \equal{} \frac {1}{2} \int e^{ \minus{} px}\sin 2x\ dx$

$ \equal{} \minus{} \frac {e^{ \minus{} px}}{2(p^2 \plus{} 4)}(p\sin 2x \plus{} 2\cos 2x) \plus{} C$.

$ \int \sin ^ 3 x\cos ^ 2 x\ dx\equal{}\int (\cos ^ 4 x \minus{} \cos ^ 2 x)(\cos x)' \ dx$

$ \equal{} \frac {1}{5}\cos ^ 5 x \minus{} \frac {1}{3}\cos ^ 3 x \plus{} C$.


In a side of a triangle there's point M (M is arbitrary). Construct two lines which passes through M and separate the triangle in three regions with equal areas.

[color=brown]I suppose you mean that the point $M$ is [u]on a side[/u] of $ABC$
If $M$ is interior point then we have four regions
on the other hand, $M$ is not outside of $\triangle ABC$ because you said "M is in the triangle"
Also, I guess that the two lines should pass through $M$.[/color]

With these conditions, here is a solution:

Assume that $M$ is on the side $BC$.

$MP,MQ$ are the lines we are searching for
$P,Q$ are points on the sides of $\triangle ABC$
(if we can't have the points $P,Q$ in different sides of the triangle we will use a third point $R$, as we can see below)

We will see weither we can found a point $P$ on the side $AB$ such that $(ABC)=3(PBM)$

$3 = \frac{(ABC)}{(PBM)} = \frac{AB \cdot BC \cdot sinA}{PB \cdot BM \cdot sinA} = \frac{AB \cdot BC}{PB \cdot BM} \Leftrightarrow$

$\frac{PB}{AB} = \frac{1}{3} \cdot \frac{BC}{BM} \ \ (1)$

$P$ belongs to the side $AB$,so $AB\geq PB \Leftrightarrow$
$BM\geq \frac{1}{3}BC$

(if $BM$ is exactly a third of $BC$, then $P$ is the point $A$)

We can easily construct  the point $P$ given from $(1)$
The parallel line from $C$ to $MA$ intersects the line $BA$ at a point $Z$, and then we trisect the segment $BZ$ to find $BP$.

Similarly, there is a point $Q$ on the side $AC$ such that $(ABC)=3(QBM)$ iff $MC\geq \frac{1}{3}BC \Leftrightarrow BM\leq \frac{2}{3}BC$

So if $\frac{1}{3}BC\leq BM\leq \frac{2}{3}BC$ there is a solution with $P \in [AB]$ and $Q \in [AC]$

Now let's see the other cases

Say for example that $BM>\frac{2}{3}BC \Leftrightarrow \frac{BC}{BM}<\frac{3}{2} \ \ (2)$

Then we can construct the point $P$ but we cannot construct $Q$.

Then we must find a point $R$ on he side $AB$ such that $(RPM)=(BPM)=\frac{1}{3} (ABC)$

It's enough to find a point such that $RP=BP$, so $R$ is symmetric of $B$ w.r.t. $P$

We can be sure that $R$ is a point on $AB$. Truly:

$(1) \Rightarrow 2\frac{PB}{AB} = \frac{2}{3} \cdot \frac{BC}{BM} \Rightarrow$

$2\frac{PB}{AB} <1$, from inequality $(2)$

So the point $R$ belongs to the side $AB$ and we are done


geometry
angle bisector
geometry proposed

ABC is acute-angled and is not isosceles. The bisector of the acute angle between the altitudes from A and C meets AB at P and BC at Q. The angle bisector of B meets the line joining HN at R, where H is the orthocenter and N is the midpoint of AC. Show that BPRQ is cyclic.

This is a problem from the Russian Olympiad 2000. Has been discussed in this forum some weeks ago


How many multiples of 7 are between 100 and 500?
It is an easy question and i got the answer but is there a formula or something?

In any interval, get the lowest multiple of 7 and get the highest.  Divide them by 7, subtract the resulting numbers, and add 1.

For example how many 7's between 10 and 50?

Lowest is 14, Highest is 49.

14/7 = 2
49/7 = 7.

7-2+1=6.

How many 13's between 30 and 100?

lowest is 39, highest is 91.

39/13 = 3
91/13 = 7

7-3+1=5.

THanks Barnacle for showing me the formula.


inequalities solved
inequalities
3-variable inequality
Symmetric inequality
cauchy schwarz

If $a,b,c >0$ then prove that \[ \sum_{cyc}  \frac{a^2}{b+c} \geq \frac12(a+b+c)\]

We have by Cauchy Schwarz
\[\left(\sum_{cyc} \frac{a^2}{b + c}\right)\left(\sum_{cyc} (b + c)\right) \geq (a+b+c)^2\]
Cancelling the factor $2(a + b + c)$ on both sides we have the result.


probability
parameterization
linear algebra
matrix
limit
function
probability and stats

car parking spaces are labelled $ n\equal{}1,2,3, \ldots$, where the label indicates the distance to walk to a shop, and an arriving car parks in the lowest numbered free space. Cars arrive as a Poisson process of rate $ \nu$, and parking times are exponentially distributed with mean $ \mu$ (and are independant of each other and of the arrival process). For what fraction of time the parking space labelled $ i$ is occupied ?

If $ \mu<\frac1{\nu},$ then the number of cars parked at any time tends toward a steady state distribution, and I would guess that everything else about the model tends toward a steady state. If $ \mu>\frac1{\nu},$ then the number of parked cars, and hence the highest-numbered space occupied, grows linearly in time and there is no steady state - which doesn't automatically preclude the occupation time of each space tending toward some fixed ratio.

(I haven't worked anything out yet - just thinking aloud.)

I got a nasty expression for the fraction of time that the $ n$-th parking space is occupied:
\[ \pi(n) \equal{} \frac {\rho^n}{(n \minus{} 1)!(1 \plus{} \rho \plus{} \frac {\rho^2}{2!} \plus{} \ldots \plus{} \frac {\rho^n}{n!})}
\]
where $ \rho\equal{}\mu \nu < 1$, but there is a part of the 'proof' where I am still a little bit unsure. Anyone to confirm/infirm that ?

I got a different expression, but then I ran some numbers and the results didn't pass the smell test - I think my argument is so wrong that it's not worth posting.

As for your expression - I'm skeptical.

Here's my issue with it. Consider the parking spaces as unordered and unnumbered - just count how many cars are parked at any one time. This is a continuous-time Markov process, and it shouldn't be that hard to obtain an expression for the mean number of parked cars - and that number should misbehave as $ \rho\to1^\minus{}.$ For your expression, we can take the sum over all $ n$ of $ \pi(n),$ and it remains small even for $ \rho>1.$

Getting into that a little more:

For the model with the unordered parking spaces, there is a time-dependent process that tends towards a steady state as $ t\to\infty.$ In that steady state, the probability that $ k$ spaces are occupied is $ (1\minus{}\rho)\rho^k$ and the average number of spaces occupied is $ \frac{\rho}{1\minus{}\rho}$

For that reason, I think that $ \sum_{n\equal{}1}^{\infty}\pi(n)$ should be $ \frac{\rho}{1\minus{}\rho}.$

As for $ \pi(n):$ If we let $ s(n)$ be the probability that all $ n$ of the spaces $ 1,2,\dots,n$ are occupied simultaneously, then I believe we should have $ \pi(n)\equal{}\frac{\rho}{\rho\plus{}\frac1{s(n\minus{}1)}}.$ From that, it appears that we should have $ \pi(1)\equal{}\frac{\rho}{1\plus{}\rho}.$ If then $ s(1)\equal{}\pi(1),$ we would have that $ \pi(2)\equal{}\frac{\rho}{\rho\plus{}\frac1{\pi(1)}}\equal{}\frac{\rho^2}{1\plus{}\rho\plus{}\rho^2}.$

From there, I'm stumped, because I don't know what $ s(2)$ should be. Assuming that $ s(2)\equal{}\pi(1)\pi(2)$ leads to values of $ \pi(3)$ that seem to be too small - hence, I think there's an assumption of independence somewhere that doesn't hold. In fact, I'm worried that $ s(n)$ may not be a constant but rather something that depends on the state of higher-numbered spaces.

[quote="Kent Merryfield"]Getting into that a little more:

For the model with the unordered parking spaces, there is a time-dependent process that tends towards a steady state as $ t\to\infty.$ In that steady state, the probability that $ k$ spaces are occupied is $ (1 \minus{} \rho)\rho^k$ and the average number of spaces occupied is $ \frac {\rho}{1 \minus{} \rho}$

[/quote]
I think that in this case ( unordered parking spaces ), the stationary distribution is slightly different. Cars arrive with rate $ \nu$, and stay an exponentially distributed time of mean $ \mu$. But, when there are alredy $ n$ cars in the system, then the next event (a car arrives, or leaves) appears after an exponentially distributed time of parameter $ \frac {n}{\mu} \plus{} \nu$, and the number of cars increases with probability $ \frac {\nu}{\frac {n}{\mu} \plus{} \nu}$ (this is a $ M/M/ \infty$ queue). This gives a stationary distribution $ \pi(n) \equal{} \exp( \minus{} \rho) \frac {\frac {\rho^n}{n!}}{n!}$ for the fraction of time that $ n$ cars are in the system.

As regards the fraction of time that the $ n$ first parking spaces, I believe that is is
\[ \frac {\frac{\rho^n}{n}}{1 \plus{} \rho \plus{} \ldots \plus{} \frac {\rho^n}{n!}}
\]
which can be obtained by working out the stationary distribution of the continuous time Markov process with $ n \plus{} 1$ states labelled $ 0,1, \ldots, n$ which represent the number of parking spaces occupied among the first $ n$ parking spaces. The rate from $ k$ to $ k \plus{} 1$ is $ \nu$, and the rate from $ k$ to $ k \minus{} 1$ is $ \frac {k}{\mu}$. Detailed balance equations gives, I think, the above stationary distribution. This is where I got most part of the expression I wrote.

No, Kent. The cars arrive at a fixed rate and depart at the rate [i]proportional to the number of parked cars[/i], so the expectation of the number of parked cars is always finite!

Alekk, your answer certainly has some correct features in it. I guess you considered 2n states according to the number of cars parked in the first $ n$ lots and to whether the $ n$-th lot is occupied or not. I have no time now to check if it is entirely correct but it should be [i]something[/i] like that, so go ahead and post your argument. Even if there is a gap, we'll most likely be able to fix it :)

Edit: Oops, Alekk already posted when I was typing this :roll:

so the probability that $ k$ parking spaces among the first $ n$ are occupied is
\[ p(k,n) \equal{} \frac{\frac{\rho^k}{k!}}{1\plus{}\rho\plus{} \ldots \plus{} \frac{\rho^n}{n!}}\]

Nevertheless, once the first $ n$ parking spaces are occupied, this lasts an exponentially distributed time of parameters $ \frac{n}{\mu}$ (because there are $ n$ cars), and after that, the number of prking spaces occupied among the first $ n$ ones decreases to $ n\minus{}1$. Nevertheless, the parking spaces labelled $ n$ is occupied an exponentially distributed time of parameter $ \frac{1}{\mu}$. So (and here should be the flaw!), I guess that the fraction of time we were looking for is $ n \pi(n,n)$

@Fedja: at the beginning I considered the Markov chain you mentioned, but I have not been able to compute the stationary distribution. Can you ?

[quote="fedja"]No, Kent. The cars arrive at a fixed rate and depart at the rate [i]proportional to the number of parked cars[/i], so the expectation of the number of parked cars is always finite![/quote]
Oops! OK, you guys carry on without me. Never mind.

One thing that bothers me is that the matrix is clearly analytic in $ \mu$ and $ \nu$, so one would expect the same analyticity for the probabilities involved. But, if $ \mu\to\infty$, your expression tends to $ n$, not to $ 1$, which is ridiculous. As to the steady state, if we denote by $ p_k$ the probability that the $ n-th$ space is occupied and there are $ k$ cars in the previous spaces, we shall have the system
\[ \begin{aligned}
p_0(\mu^{-1}+\nu)&=p_1\mu^{-1}
\\
p_k[(k+1)\mu^{-1}+\nu]&=p_{k-1}\nu+p_{k+1}(k+1)\mu^{-1},\qquad 1\le k\le n-2
\\
p_{n-1}n\mu^{-1}&=\nu\left[\frac{\rho^{n-1}/(n-1)!}{\sum_{k=0}^{n-1}\rho^k/k!}-p_{n-1}\right]
\end{aligned}\]
The last equation is derived by looking at $ \pi(n-1,n-1)$.

The "typical" equation can be rewritten as $ p_{k+1}-p_k=\frac\rho{k+1}(p_k-p_{k-1})$. Combining this with $ p_1-p_0=\rho p_0$ (the first equation), we get $ p_k-p_{k-1}=\frac{\rho^{k}}{k!}p_0$ and
\[ p_k=\sum_{m=0}^{k}\frac{\rho^m}{m!}p_0\]
Now we can easily find $ p_0$, after which it'll remain to find the sum $ \sum_{k=0}^{n-1} p_k=p_0\sum_{k=0}^{n-1}(n-k)\frac{\rho^k}{k!}$.

So, the final answer I get is
\[ \frac{\rho^{n}/(n-1)!}{(n+\rho)\sigma_{n-1}^2}[n\sigma_{n-1}-\rho\sigma_{n-2}]\]
where $ \sigma_k=\sum_{m=0}^{k}\frac{\rho^m}{m!}$.

It survives the sanity check when $ \mu\to\infty$ (you get $ 1$ indeed) and trivial check $ n=1$ where you should get $ \rho/(1+\rho)$, so, I guess, it is correct :).

Thanks a lot Fedja! I realized yesterday that what I wrote was a total nonsense. I think that I got another answer that passes the sanity checks ( $ n = 1,2$ and $ \lim_{\rho \to \infty}$), but it seems a slightly different as yours, though I have not checked carefully if it is really that different.

So if $ p{k,n} = P[A_{k,n}]$ denotes the probability that exactly $ k$ parking spaces are occupied over the $ n$ first ones, one easily get:
\[ p(k,n) = \frac {\rho^k}{k!} \frac {1}{\sigma_n}
\]
where $ \sigma_k = \sum_{m = 0}^k \frac {\rho^m}{m!}$. Notice that $ \{A_{k,n}\}_k$ are disjoint and that $ A_{k,n} \subset A_{k,n - 1} \cup A_{k - 1,n - 1}$, and $ A_{k,n - 1} \subset A_{k,n} \cup A_{k + 1,n - 1}$. From here, if $ \pi_n(k)$ is the probability that the car park labeled $ n$ is occupied and that exactly $ k$ spaces are occupied among the first $ n$ ones, one gets:
\[ \pi_n(1) = p(0,n - 1) - p(0,n)
\]

\[ \pi_n(2) = [p(0,n - 1) + p(1,n - 1)] - [p(0,n) + p(1,n)]
\]

\[ \ldots
\]

\[ \pi_n(n) = [p(0,n - 1) + \ldots + p(n - 1,n - 1)] - [p(0,n) + \ldots + p(n - 1,n)]
\]
wich leads to the final answer, for the fraction of time the space labeled $ n$ is occupied:
\[ F_n = \sum_{k = 0}^{n - 1} (n - k) [p(k,n - 1) - p(k,n)] = (n \sigma_{n - 1} - \rho \sigma_{n - 2})(\frac {1}{\sigma_{n - 1}} - \frac {1}{\sigma_{n - 1}})
\]

[quote="alekk"]$ \equal{} (n \sigma_{n \minus{} 1} \minus{} \rho \sigma_{n \minus{} 2})(\frac {1}{\sigma_{n \minus{} 1}} \minus{} \frac {1}{\sigma_{n \minus{} 1}})$
[/quote]
:?  :what?: Obviously, there is a misprint here...

:D  I meant:
\[ \equal{} (n \sigma_{n \minus{} 1} \minus{} \rho \sigma_{n \minus{} 2})(\frac {1}{\sigma_{n \minus{} 1}} \minus{} \frac {1}{\sigma_{n}})\]

Arrrgh, my last equation in the system is slightly wrong: $ \nu p_{n\minus{}2}$ is missing from the right hand side :blush:. If you make this correction and recalculate, the answers agree but, of course, your derivation is much shorter and cleaner :).

Returning to the scene here. I was just thinking about the problem with the unordered spaces, and I realize now that the steady-state distribution there is that the number of occupied spaces is a Poisson distribution with mean $ \rho.$

Hence, for the original problem, we should have $ \sum_{n=1}^{\infty}\pi(n)=\rho.$

So in alekk's notation, $ p(k,n)$ is the probability that $ k$ out of the first $ n$ spaces are occupied. Then $ p(k,\infty)=\lim_{n\to\infty}p(k,n)$ should be the probability that $ k$ spaces are occupied, period, and we should have $ p(k,\infty)=\frac{e^{-\rho}\rho^k}{k!}$. Yes, alekk's formula is consistent with that.

Now, is it true that $ \sum_{n=1}^{\infty}\pi(n)=\rho?$

Great question, Kent! It is entirely non-transparent from either my or alekk's derivation, but the answer is actually equal to $ \rho[p(n\minus{}1,n\minus{}1)\minus{}p(n,n)]$ in the notation of the last alekk's post  :o Thus, there should be some reason for that (the expression in brackets is just the probability of the situation when $ n\minus{}1$ first places are filled but the $ n$-th isn't, i.e., when we can switch from the unfilled to the filled state, and the factor $ \rho$ is the intensity of arrivals times the expected length of stay; such a simple and natural formula cannot be a pure coincidence  :maybe:).

another proof found by a friend:
let $ 1_j(t)$ the indicator function of the event: parking space labelled $ n$ is busy at time $ t$. So, one is trying to find
\[ E[1_n(t)] \equal{} E[1_1(t) \plus{} 1_2(t) \plus{} \ldots \plus{} 1_n(t)] \minus{} E[1_1(t) \plus{} 1_2(t) \plus{} \ldots \plus{} 1_{n\minus{}1}(t)]\]
but \[ E[1_1(t) \plus{} 1_2(t) \plus{} \ldots \plus{} 1_m(t)]\] is just the average number of parking space busy among the first $ m$ ones, which is esily computable since one knows  that
\[ p(k,n) \equal{} \frac{\rho^k}{k!} \frac{1}{\sigma_n}\]
this leads to
\[ E[1_1(t) \plus{} 1_2(t) \plus{} \ldots \plus{} 1_m(t)] \equal{} \rho \frac{\sigma_{m\minus{}1}}{\sigma_{m}}\]
This is why, the desired fraction of time can also be written:
\[ F_n \equal{} \rho( \frac{\sigma_{n\minus{}1}}{\sigma_{n}} \minus{} \frac{\sigma_{n\minus{}2}}{\sigma_{n\minus{}1}})\]
ps: after some algebra one can see that this is effectively the same result as the one we got in the previous posts


Did you mis-type it? Would you not have used:

$f(-3x^3+1)$

Errrm, im having a little trouble filling in the gaps in that, do you mind expanding a bit more?

[quote="mateivld"]See that $-x^2+3x+1=x \leftrightarrow x^2-2x-1=0$ has two real solutions and $\epsilon$ be one of them.We have $f^2(\epsilon)-f(\epsilon)+2=0$ wich has no real solution,thus the problem is solved. :)[/quote]

The relation holds for all _[i]integers[/i]_ x. $\epsilon$ is not integer here.

Still cant get those gaps....

This property described dosnt hold becase $\epsilon$ isnt an integer

[quote="jin"]There doesn't exist such function $f: R\to R$ such that for all a in R,there exists x such that  f(x)=a,and for all integers x, $f(-x^2+3x+1)=(f(x))^2+2$.[/quote]

Write $g(x) = -x^2 + 3x + 1$.  If we can find an integer $n$ such that $g(g(n)) = n$ then we could write $f(n) = ( f(n)^2 + 2 )^2 + 2$, where the RHS is clearly larger than the LHS so no solutions.

Finding such an integer...

$g(0) = 1, g(1) = 3, g(3) = 1$

And we have it.  $f(1) = ( f(1)^2 + 2)^2 + 2$ for some real $f(1) = x$.  Now, the RHS is $\ge 2$, so $x \ge 2$, so the RHS is $\ge (2^2 + 2)^2 + 2$, so $x \ge ...$

This equation has no solutions - $f(1)$ doesn't exist - hence, no such function can exist.

(I admit I am a bit wary of this solution, as it did not use the first condition ...)

It's a easy problem,but at first I have some trouble to solve it.Thank you!


MOD PLEASE LOCK THIS FORUM/DELETE ALL POSTS WITHIN
This is a question to get into Stanford summer camp SUMaC, the application for which can be found here:
http://math.stanford.edu/sumac/Apply.html
And to JJ89, you should obey the rules on the application when it says "Do all work on your own and cite all external sources, all else is considered plagiarism."

Yes, I know, I intended to cite this source.

The rules state that I may recieve help, but I must acknowledge that I recieved it.

I will remove the problem right now.

I apologize if I offended you.

Receiving help from someone would be like a friend or math teacher that you know, not forums... Outside source would be like mathworld.com or wikipedia or a math book in your library, not forums... Forums are for discussions. We can discuss the problems, but that means all the other applicants who also use this forum also receive the same type of help, which means that you get lower chances of getting in the camp... If you cannot solve a problem and cannot receive other help, then acknowledge : I did the best I could and here are my solutions, I did not receive help... Avoid trying to show you can do superb because once you get in and if you are not a student of that quality, you cant follow the courses (not your level) and you cant learn the best that the camp can teach you. Show them how much you know is the true talent.

Ah, sorry if I overreacted a little. I was a tad offended because I mailed in my application just two days ago and didn't manage to solve this problem as well. I got just about as far as you (proving that it was divisible by 8). But yeah, good luck with the application and maybe I'll see you there! (If we both get in.)

I think 10000th User says it exactly right about outside resources.  I didn’t think about it that way.  Hope to see you at SUMaC.


Each card has either a circle or a star on one side and either a triangle or a square on the other side. In order to verify the statement ``every card with a star on it also has a triangle on it,'' which numbered card(s) must be turned over?
[asy]import graph;
size(300);
void drawRectangleAt(string l,pair A){
 draw(A--(A.x + 2,A.y)--(A.x + 2, A.y + 3)--(A.x,A.y + 3)--cycle);
 label(l,A--(A.x + 2,A.y),S);
}
drawRectangleAt("1",(0,0));
drawRectangleAt("2",(3,0));
drawRectangleAt("3",(6,0));
drawRectangleAt("4",(9,0));
draw(Circle((1,1.5),0.75));
draw(shift(3.25,0.8)*scale(1.5,1.5)*unitsquare);

pair A = (7,1.5);
pair[] pentPoints;
for(int i = 0; i < 5; ++i){
 pentPoints[i] = shift(A)*dir(18 + 72*i);
}
pentPoints.cyclic(true);
for(int i = 0; i < 5; ++i){
 draw(pentPoints[i]--pentPoints[i + 2]);
}
pair B = (10,1.5);

pair B = (10,1.5);
pair[] triPoints;
for(int i = 0; i < 3; ++i){
 triPoints[i] = shift(B)*(1*dir(120*i - 30));
}
draw(triPoints[0]--triPoints[1]--triPoints[2]--triPoints[0]);[/asy]

This is equivalent to the statement "If a card has a star on it, it also has a triangle on it."

The inverse and converse of this statement do not affect whether this statement is true or not.  However, the statement "If a card does not have a triangle on it, it does not have a star on it" does.

Clearly we need to check card $ 3$.  Flipping over card $ 1$ is equivalent to checking the inverse, and flipping card $ 4$ is equivalent to checking the converse.  We need to flip over card $ 2$ because if there's a star on the back, then the statement isn't true.  So we need to check $ \boxed{2\text{ and }3}$


Suppose that 50! is written in the form $ N$ $ \times 10^x$, where $ N$ is an
integer. What is the largest possible value of $ x$?

[quote="FantasyLover"]$ 10^x \equal{} 2^x*5^x$ so you need to find how many 5's are in 50!.

multiples of 5 = 5*1,5*2,...,5*10 (1)
multiples of 25 = 25*1,25*2 (2)
so (10-2)*1+2*2=12.

answer:12[/quote]

Can you explain more about your solution. After finding multiples of 5 and 25, you have (10-2)*1+2*2=12. How did you get it?

Well, in order to "make" a multiple of $ 10$, you need to multiply a $ 2$ and $ 5$ together. Clearly there are less $ 5$s, so we shall count them. Since $ 5(10)\equal{}50$, we know that we have $ 10$ factors of $ 5$. However, each multiple of $ 25$ contributes one additional $ 5$ (since $ 5^2\equal{}25$). There are two multiples of $ 25$, and $ 10\plus{}2\equal{}\boxed{12}$.


will anyone please explain the chinese remainder theorem?

hello, see here
http://www.math.hawaii.edu/~lee/courses/Chinese.pdf
Sonnhard.

I want to just know where this theorem is useful in solving problems like an example.

For $ m,\ n$ with $ (m,\ n)\equal{}1$, prove that $ \phi (mn)\equal{}\phi (m) \phi (n)$.

hello, here you will find examples
http://neworder.box.sk/files/CRT.pdf
Sonnhard.

Well tahnks a lot. it was really helpful.


Let $k$ be an integer with $k\geq2$ and let  :
$1<n_1<n_2<....<n_k$ be integer numbers. Positive integers $a,b$ are such that
$(1-\frac{1}{n_1})(1-\frac{1}{n_2})*...*(1-\frac{1}{n_k})\leq\frac{a}{b}\leq(1-\frac{1}{n_1})*...*(1-\frac{1}{n_{k-1}})$. Pove that $n_1*n_2*...*n_k\leq(4a)^{2^k-1}$


Source: http://www.q-e-d.be

In $\Delta ABC$, draw the angle trisectors of $\angle A$ and $\angle B$.
Two of those trisectors intersect in the midpoint of the circumscribed circle.
Prove that the other two trisectors intersect in the orthocenter of the triangle.

Just use the fact that the orthocentre and the circumcentre are (uniquely) isogonally conjugate, and you're done.

If you want a proof, <BAH = 90-<CBA = 90 - 1/2 (<COA)=<OAC
using assorted circle theorems.

[b]Very nice observation-problem ![/b] In this case : $\{\begin{array}{c}\frac{A}{6}=\frac{B}{3}=\frac{C}{5}=\frac{\pi}{14}\ \ \vee\ \ \frac{A}{6}=\frac{B}{5}=\frac{C}{3}=\frac{\pi}{14}\ ;\\\\ m(\widehat{ACH})=m(\widehat{BCO})=\frac{\pi}{14}\ \ \vee\ \ \frac{2\pi}{7}\ \ ,\ \ m(\widehat{OCH})=\frac{3\pi}{14}\end{array}\ ,$

i.e. the rays $[CH$ and $[CO$ arn't the trisectors of the  angle $\widehat{ACB}\ .$


Is there a function of two real variables $f$ which takes as values integers and such that $f(x,y)=f(y,z)$ implies $x=y=z$?

if such a function existed all real numbers would be equal. :P  :D

[quote="PTynan89"]Yes.

Consider the function $f: \mathbb{Z}\to \mathbb{Q}$, defined by $f(x, y) = 2^{x}3^{y}$.

In fact, any function $f(x, y) = p^{x}q^{y}$ would satisfy this, for $p, q \in \mathbb{Z}$, as long as neither $p$ or $q$ is a divisor of the other.

Unless I misunderstood the question...[/quote]
Must be $f\colon\mathbb{R}\times\mathbb{R}\to\mathbb{Z}$

It seems to me obviously impossible, because $f(x,x)=f(x,z) \iff x=z \, \forall \, x,z \in \mathbb{R}$, so by letting $g(x)=f(a,x)$ for some $a$, $g(x)=g(y) \iff x=y$ we see that reals are countably infinite $\Rightarrow\Leftarrow$. Am I missing something?

[quote="harazi"]:what?:[/quote]

Hmm... I am sorry, what did I do wrong?

[quote="TomciO"]
\[f(x,x)=f(x,z) \iff x=z \, \forall \, x,z \in \mathbb{R}\]
is obviously not true.[/quote]

That's the point! And that's why there is no such function. Or am I very very wrong?

Sorry, I haven't read the problem carefully, but anyway, I don't get the last step - I can't see how you reach contradiction from the fact that $g(x)$ is an injective?

[quote="harazi"]Weiquan, your function $g$ is obviously not injective, simply because your $f(x,x)=f(x,z)$ equivalent to $x=z$ gives no information about the injectivity of $g$.[/quote]

But doesn't $g(x)=g(y) \iff x=y$ mean that $g(x) \neq g(y) \forall x \neq y$?

[quote="harazi"]Man, you took an $a$ and you defined $g(x)=f(a,x)$, right? Now, you claim that this is injective. But the fact that $g(x)=g(y)$ says that $f(a,x)=f(a,y)$, which surely does not imply $x=y$.[/quote]

Oops... :blush: I am really sorry. I really wasn't thinking properly.

I think this is a soluton, but maybe it's just something stupid.Harazi please have a look...

I think such a function exists. Define $S$ as a set whose elements are subsets of $\mathbb{N}$ and no element of $S$ is a subset of another element of $S$. My "solution" is based on the fact that $S$ is $\aleph_{1}$.
So we can write $S=\{S_{x}|x\in \mathbb{R}\}$ .
Put $f(x,x)=0$ $\forall x\in \mathbb{R}$
Now fix $y\in \mathbb{R}$
Denote $M_{y}=\{f(x,y)\; | \forall x\in \mathbb{R}, x\doteq y\}$
and $K_{y}=\{f(y,x)\; | \forall x\in \mathbb{R}, x\not= y\}$
If we put $M_{x}=S_{x}$ and $K_{x}=\mathbb{N}-S_{x}$
Let $f(x,y)$ be any element from $K_{x}\cap M_{y}$ (this is never empty  :wink: ) then this function has the required property. ($K_{x}\cap M_{x}$ is always empty  :P )

for example I guess one could take $S$ like:
an element of $S$ conains one prime one number which is the ptoduct of two primes etc.
Harazi ?!
 :huh:

Sorry, I read it, thought it was ok and forgot to reply. I think it is ok.

then (the cliche) What is the idea of the solution you have? :)

I took an enumeration of $\mathbb{Q}$ in a sequence $x_{n}$ and defined $f(x,x)=x$ for all $x$ and $f(x,y)=2n$ if $x<y$ and $n$ is the first number for which $x_{i}$ meets $(x,y)$, while $f(x,y)=2n+1$ if $x>y$ and $n$ is similarly defined.

I think you mean $f(x,x)=0$ :D
and btw very nice solution.
and very interesting question !


calculus
derivative
function
real analysis
real analysis theorems

I'm doing a bivariate calculus problem from a textbook. I can reach all the answers, but am unsure about something -- what does an evaluated second partial derivative mean in a marginal analysis context?

Here's the question. (I won't show all the work in reaching the answers, as they are not problematic or controversial.)

Total annual profit f(x,y) is determined as

f(x,y) = 10x^3 + 20y^2 - 10xy,

when x = ad spending = 10 and
y = acres farmed = 5.

I do understand:
that f(10,5) = 10,000 = total annual profit;
that f[subscript]x = 30x^2 - 10y; and
that f[subscript]x (10,5) = 2950 = the approximate additional annual profit to be gained from increasing ad spending by one unit (i.e. from 10 to 11).

What I DON'T understand is something the author has us calculate, but which we never end up using in the analysis, and which she never comments on. Specifically, she asks for

f[subscript]xx (i.e the second partial) = 60x and
f[subscript]xx (10,5) = 600.

I can reach these answers, but I don't understand, what does 600 actually mean here?

I'd be grateful for any advice

That tells you the concavity of the function if you take a slice of it in the $x$ direction. It is just like the second derivative with respect to $x$ of that function when $y$ is fixed. But from an economic perspective, it is better to think of it from the perspective of the rate of change of your marginal profit. In this example, it is increasing with $x$, so adding to $x$ not only makes you more profitable, but your profit increases faster as you spend more on advertising.

The problem with my explanation is that the interpretation of that value depends on the question. I suggest that you just try to think about the mathematical relationship between the various derivatives, and see if you can describe that mathematical relationship in real-world terms based on the problem. The mathematical relationship will always be the same. You can rely on that. :P


I think it would be cool to have like a (taken from readers digest) "Quotable Quotes" forum or subforum somewhere.

Just a suggestion. :)

...But what does that accomplish?  :huh:

It would be kinda like G&FF except reserved for quotes. People could find some witty things to say to their friends.

Why a forum? Just make a thread in GFF ...


[quote="Nick Rapanos"][quote="Litlle 1000t"]On a plane given triangle $ABC$ and point $O$ .The lines which perpendicular with $OA, OB, OC$ intersect with $BC, AC, AB$ at ${A}_{1},{B}_{1},{C}_{1},$ respectively. Prove that ${A}_{1},{B}_{!},{C}_{1}$ are collinear[/quote]

These lines are any perpendiculars??[/quote]
I guess he meant that those lines are pass through $O$. If those perpendicular lines are taken arbitrary, the problem is wrong.

I thought he means perpendicular bisectors; then it's [b](a)[/b] of http://www.mathlinks.ro/Forum/viewtopic.php?t=1395 (if you're wondering what my posts does there: I have just found it in a different topic and merged it).

Maybe he also meant perpendiculars through O. Then, however, O could be any random point in the plane and we would get http://www.mathlinks.ro/Forum/viewtopic.php?t=51844 .

Finally, he may mean the generalization discussed in [url=http://forumgeom.fau.edu/FG2005volume5/FG200502index.html]K[b]h[/b]oa Lu Nguyen, [i]A Synthetic Proof of Goormaghtigh's Generalization of Musselman's Theorem[/i], Forum Geometricorum, 5 (2005) pp. 17-20.[/url]

[b]EDIT:[/b] Oops, O [b]is[/b] an arbitrary point in the plane!  :oops: I thought it were the circumcenter. For an arbitrary point O in the plane, the "perpendicular bisectors" assertion does [i]not[/i] hold, so I have nothing left but to assume that he meant perpendiculars at O.

I've moved the proof by PDatK40SP (which is by far not ugly) to http://www.mathlinks.ro/Forum/viewtopic.php?t=51844 .

  Darij

oh I am sorry it was perpendiculars through O.


algebra polynomial vector function calculus derivative topology

Let $n \ge 1$ be an integer, and write $M_{n}$ for the vector space of real n x n matrices. Suppose that $A_{0}\in M_{n}$, and $\lambda_{0}$ is a real root of multiplicity 1 of the characteristic polynomial of $A_{0}$. Fix a non-zero eigenvector $\mathord\over x}= (x_{1},x_{2},...,x_{n})$ of $A_{0}$, with one component, say $x_{j_{0}}$, equal to 1.

Prove that there exists an open set $U \subset M_{n}$ with $A_{0}\in U$, and $C^\infty$ mappings $\lambda : U \to R$, $\mathord\over y}: U \to R^{n}$ such that, for all $A \in U$, $A\mathord\over y}(A) = \lambda (A)\mathord\over y}(A)$ and $y_{j_{0}}(A) = 1$, and, at $A_{0}\in U$, $\mathord\over y }(A_{0}) = \mathord\over x$, $\lambda (A_{0}) = \lambda_{0}$.

Prove the analogous statement for complex matrices.

It's better to use $\vec x$ or $\mathbf{x}$ for vectors.

The [url=http://en.wikipedia.org/wiki/Implicit_function]Implicit Function Theorem[/url] is needed here. Think of $A\in M_{n}$, $\lambda\in\mathbb R$, and $\vec y\in \mathbb R^{n}$ as variables constrained by the vector equation $A\vec y-\lambda\vec y=0$. Add another equation to this system: $y_{j_{0}}=1$. You want to solve this system of $(n+1)$ equations for $\lambda$ and $\vec y$ in terms of $A$ (locally). The implicit function theorem says that this is possible (with the required smoothness) provided that the derivative of the map $(A,\lambda,\vec y)\mapsto (A\vec y-\lambda\vec y,y_{j_{0}})$ with respect to $(\vec y,\lambda)$ has rank $n+1$. Which it does.

Thanks for that great insight, I will definitely try and work through those steps.


geometry
geometric transformation
rotation
calculus
derivative
ratio

[color=darkblue][b]The problem of the Ness (coffee) can.[/b]

Ascertain the rotation cylinder for which the its volume is constant and the its surface (overall) area is minimum ([b][u]without calculus of derivatives[/u][/b]).

Precisely, ascertain the ratio between the radius and the altitude of the cylinder so that the its surface area is minimum when the its volume is constant.[/color]

[hide="Hint."][color=darkred]For a fixed positive number $k$ ascertain $\min_{x>0 , y>0 , x^{2}y=k}\left\{\ x\cdot (x+y)\ \right\}\ .$[/color][/hide]

r, h, etc, wlog volume=kpi

r^2h=k, maximize this: 2pi(r^2+rh)=2pi(r^2+k/r)

2r^2+k/r+k/r\ge 3(2k^2)^(1/3) (am-gm)

result follows

[quote="Virgil Nicula"][quote="Altheman"]r, h, etc, wlog volume=kpi

r^2h=k, [b][u]maximize[/u][/b] this: 2pi(r^2+rh)=2pi(r^2+k/r)

2r^2+k/r+k/r\ge 3(2k^2)^(1/3) (am-gm)

result follows[/quote] Not maximize, but minimize ! Read with the more attention the enunciation. For maximize exists only one solution : ... eat "pizza" or "sticks" ![/quote]

i mistyped...that does give the min


geometry
angle bisector
geometry proposed
U sviti matematyky

Since $ IM \equal{} IN$ and $ IP \equal{} IQ,$ it follows that the right $ \triangle IMP$ and $ \triangle INQ$ are congruent $ \Longrightarrow$ $\overline{MP} \equal{} \overline{NQ}.$ Let $ T \equiv PQ \cap MN.$ By Menelaus' theorem for $ \triangle APQ$ cut by the transversal $MTN,$ we have:

$ \frac {\overline{NQ}}{\overline{NA}} \cdot \frac {\overline{AM}}{\overline{MP}} \cdot \frac {\overline{PT}}{\overline{TQ}} \equal{} 1$

Since $ \overline{AM} \equal{} \overline{NA}$ and $ \overline{MP} \equal{} \overline{NQ}$ $ \Longrightarrow$ $ \overline{PT} \equal{} \overline{TQ}$ $ \Longrightarrow$ $ MN$ bisects $ PQ$

Assume that the second circle isnt the one that passes through $ A,M,I,N$

We have $ IQ\equal{}IP$. Let the circle with center at $ I$ intersect $ AC$ at $ R$(different from $ P$)
Let $ MN$ and $ PQ$ intersect at $ S$. Since $ QR$ and $ MN$ are both perpendicular to $ IA$, they are parallel.
In triangle $ PQR$, since $ PN\equal{}NR$ and $ RQ//NS$, we're done.

[img]http://img31.imageshack.us/img31/3303/imagegz.png[/img]

[quote="rogue"]Let $ \omega_1$ be the incircle of a triangle $ ABC.$ The circle $ \omega_1$ has center $ I$ and touches the sides $ AB$ and $ AC$ at points $ M$ and $ N.$ A circle $ \omega_2$ passes through points $ A$ and $ I$ and intersects the sides $ AB$ and $ AC$ at points $ Q$ and $ P$ respectively. Prove that the line segment $ MN$ passes through the midpoint of line segment $ PQ.$[/quote]

Let $ S'$ be the projection of $ I$ onto $ PQ$. Then by the [i]Simpson line [/i] theorem, we obtain that $ M,$ $ N,$ $ S'$ are collinear, which implies that $ S'\equiv S$ $ \Longrightarrow$ $ IS\perp PQ$. Also note that $ AI$ is the internal bisector of $ \angle PAQ$, thus $ IP\equal{}IQ$. As the consquence, $ S$ is the midpoint of $ PQ$. Our proof is completed then. $ \square$

Since $AI$ is an angle bisector, we get $PI=QI$, hence $\triangle QMI \equiv \triangle PNI$, hence $MQ=NP$. With $AN=AM$, applying the theorem of Menelaos to the $\triangle PAQ$ with the transversal $MN$, we get that $MN$ passes through the midpoint of $PQ$.

Best regards,
sunken rock


algebra
polynomial
number theory
number theory unsolved

Let $p$ be an odd prime number. Let $S$ be the sum of all primitive roots modulo $p$. Show that if $p-1$ isn't squarefree (i. e., if there exist integers $k$ and $m$ with $k>1$ and $p-1=k^2m$), then $S \equiv 0 \mod p$.

If not, then what is $S$ congruent to $\mod p$ ?

[color=red][[b]Moderator edit:[/b] This post refers to an older version of post #1. It is now obsolete.][/color]

[quote="sam-n"]show that If p-1 isn't empty of perfect square( I mean that we can show p-1 =k 2m)[/quote]

Use "square free" for "empty of perfect square", or "not square free" to deny it.

Let q be a prime s.t. q2 | p-1. Now let x be a residue (mod p) which has order q (this means that xq=1(p) and q is the smallest with this property). Now assume a is a primitive root but xa isn't. Then there exists t|p-1 s.t. xa has order t (t<p-1). From here we get (xa)t=1(p)=>(xa)qt=1(p)=>aqt=1(t), but ap-1=1(p) so a(qt,p-1)=1(p). a is a primitive root of p, so (qt,p-1)=p-1, so p-1 | qt. From the fact that t<p-1 it's easy to get qt=p-1, so t=(p-1)/q, which is divisible with q, so (x-1)t=1(p) and from the fact that (xa)t=1(p) we get at=1(p), which is false because a is a primitive root.

This shows that a=primitive root=>xa also primitive root, so all of the numbers a, xa, x2a, ..., xq-1a are primitive roots of p, but it's easy to check that their sum is 0(p), so the set of primitive roots of p can be partitioned in such sets of q elements each, each set having the sum of its elements 0(p), so the entire sum is 0(p).

I think that in all cases S=u(p-1)(p), where u is Mobius' function. This means that S=0(p) when p-1 isn't square-free (I've just showed that), S=1(p) when p-1 is square-free and has an even number of prime divisors and S=-1(p) when p-1 is square-free and has an odd number of prime divisors. I don't have the proof for the last 2 statements, but it loks that way (I checked some numbers on my computer). I might, of course, be wrong.

[color=red][[b]Moderator edit:[/b] I don't understand why "it's easy to get qt=p-1", but fortunately, all that is used later is the assertion that $t$ is divisible by $q$, and this assertion is obvious anyway (since $q^2 \mid p-1 \mid qt$). -- Darij][/color]

We can use cyclotomic polynomials. The primitive roots modulo p are the roots of the polinomial phip-1(x) in the field Zp(the p-1'th cyclotomic polynomial). I think that from here we can derive that their sum is u(p-1) (as I have already mentioned, u is Mobius' function). In order to do this we must show that the second coefficient of phin is -u(n). I think this shouldn't be too hard, right?

This would solve everything in just one step.

excuse me grobber ,I did'n understood what does cyclotomic polynomials&
phip-1(x)  mean?

The n'th cyclotomic polynomial, phin(x)= \pi{t is a primitive root of 1 of order n}(x-t), or, in other words,  \pi{(k,n)=1}(x-(cos(2k\pi)+isin(2k\pi))).

I found this problem in Hojoo Lee's book on number theory (a colection of problems):

Show that the sum of all the primitive roots mod p is congruent to u(p-1) (mod p).

It doesn't say that p is a prime, but I assume that's the case. This means that what I was saying a few posts ago is true (" I think that in all cases S=u(p-1)(mod p)"). The reference in Hojoo Lee's file for this problem is Km, problems sheet 1-3. Km means Keith Matthews, and you can find something here:

[url]http://www.numbertheory.org/courses/MP313/index.html[/url]

It might be helpful.

In order to show that S=u(p-1) (mod p) it's enough to use the fact that the primitive roots of p are the roots in Zp of the polynomial phip-1(x)=0 (phip-1 is, as before, the p-1'th cyclotomic polynomial), and then use the fact that phin(x)= \pi{d|n}(xd-1)u(n/d). Try it, this part isn't hard. What isn't exactly clear to me is why the primitive roots are the roots of that polynomial in Zp. This is true for the primitive roots of 1 of order p-1, but what does it have to do with the divisibility with p?

It's a bit unclear to me, but I still used it :D.


I don't know when the local chapter competitions are being held in Alabama, but are any of you competing?

for us it's Feburary 26... but i"m not sure it's the same for every district. We finish taking all the test last week, but i"m not sure whether or not I made it or not. THere's 6 eighth-graders, and they'll probably all make it, so there's room for 2 seventh-graders.... hopefully I'll make it :D  :?

Yes, there is no MATHCOUNTS in high school.  The AMC is the major problem solving contest in high school, but that's taken in-school.

[quote="LuCy4EvA"]yay!! I made district... hehe, this is my first year in Mathcount  :D[/quote]

Fun!  Good luck at the competition.


modular arithmetic
number theory proposed
number theory

Find all positive integers $ x$ and $ y$ such that $ 5^{x}-3^{y}= 16$.

First use mod 3 , we get $ 5^{x}\equiv 1{mod 3}\Longrightarrow x=2x'$
and then use mod 4 , we get $ 3^{x}\equiv 1{mod 4}\Longrightarrow y=2y'$
now rewrite as $ (5^{x'}-3^{y'})(5^{x'}+3^{y'})=16$ and we get $ 5^{x'}-3^{y'}=2$ , $ 5^{x'}+3^{y'}=8$ which gives :
$ x'=y'=1$ and we get $ x=y=2$

More generally :
the equation $ 3^{x}+4^{y}=5^{z}$ has only 2 solutions $ (0,1,1)$ and $ (2,2,2)$ for nonnegative integers $ x,y,z$ .

For the equation
$ 3^x\plus{}4^y\equal{}5^z$  $ (x,y,z \in \mathbb Z^\plus{})$

$ 3^x\equiv1 \pmod4 \implies x\equal{}2x_1$
$ 5^z\equiv1 \pmod3 \implies z\equal{}2z_1$

$ 4^y\equal{}5^z\minus{}3^x\equal{}5^{2z_1}\minus{}3^{2x_1}\equal{}(5^{z_1}\minus{}3^{x_1})(5^{z_1}\plus{}3^{x_1})$

$ \left\|\begin{aligned} 5^{z_1}\plus{}3^{x_1}\equal{}2^s \\
5^{z_1}\minus{}3^{x_1}\equal{}2^t\end{aligned}\right\}$  $ s>t$ and $ s\plus{}t\equal{}2y$

$ 2.5^{z_1}\equal{}2^s\plus{}2^t \implies 5^{z_1}\equal{}2^{s\minus{}1}\plus{}2^{t\minus{}1} \implies$

$ 5^{z_1}\equal{}2^{t\minus{}1}(2^{s\minus{}t}\plus{}1)$
$ 3^{x_1}\equal{}2^{t\minus{}1}(2^{s\minus{}t}\minus{}1)$

$ 5^{z_1}$ and $ 3^{x_1}$  are both odd numbers so $ \boxed{t\equal{}1}$

let $ s\minus{}t\equal{}k$  Then $ 3^{x_1}\equal{}2^k\minus{}1 \implies 2^k \equiv 1 \pmod3 \implies k\equal{}2k_1$

$ 3^{x_1}\equal{}\underbrace{(2^{k_1}\minus{}1)}_{3^\beta}\underbrace{(2^{k_1}\plus{}1)}_{3^\alpha}$

$ 3^\alpha\minus{}3^\beta\equal{}2$

This can only happen for $ \alpha\equal{}1 , \beta\equal{}0$ $ \implies x\equal{}y\equal{}z\equal{}2$


[quote="math10"]Prove that in any triangle ABC the following ineq holds:
  $ \cos(A) \plus{} \cos(B) \plus{} \cos(C) \geq \frac {1}{4}(3 \plus{} \cos(A \minus{} B) \plus{} \cos(B \minus{} C) \plus{} \cos(C \minus{} A))$[/quote]

See here : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1484942#1484942

See here for two more solutions:

http://www.mathlinks.ro/Forum/viewtopic.php?t=127520

Also I can post another trigonometric solution of this inequality and also a stronger version if you are interested.


Using Laplasian tranformations, solve the system

$ \left\{{\begin{array}{*{20}c}
{\dot{x}\plus{}x\plus{}y\equal{}f(t)}\vspace{0.2cm}   \\
{\dot{y}\minus{}3x\plus{}5y\equal{}0}\hfill\\
\end{array}}\right\}$

with initial values $ x\equal{}y\equal{}0$ at $ t\equal{}0$ and $ f(t)\equal{}\left\{{\begin{array}{*{20}c}
{{0}\hspace{0.1cm},\hspace{0.6cm}t<1}\hfill\vspace{0.2cm}   \\
{{e^{\minus{}t}}\hspace{0.1cm},\hspace{0.3cm}t\geq1}\hfill\\
\end{array}}\right.$

Thanks in advance

First, define:

$ X(s) \equal{} \mathcal{L}\{x(t)\}$

$ Y(s) \equal{} \mathcal{L}\{y(t)\}$

Now, you have:

$ \mathcal{L}\{\dot{x}(t)\} \equal{} s X(s)$

$ \mathcal{L}\{\dot{y}(t)\} \equal{} s Y(s)$

Also:

$ \mathcal{L}\{f(t)\} \equal{} \frac{1}{e^{s\plus{}1}} \frac{1}{s\plus{}1} \equal{} F(s)$

Plugging this in, you get:

$ \left\{{\begin{array}{*{20}c}
{s X(s) \plus{} X(s) \plus{} Y(s) \equal{} F(s)}\vspace{0.2cm} \\
{s Y(s)\minus{}3X(s)\plus{}5Y(s)\equal{}0}\hfill\\
\end{array}}\right\}$

This should be easy to solve..

[hide]

$ Y(s)(s\plus{}5)\minus{}3X(s)\equal{}0$

$ Y(s)\equal{}\frac{3}{(s\plus{}5)}X(s)$

$ s X(s) \plus{} X(s) \plus{} \frac{3}{(s\plus{}5)}X(s) \equal{} F(s)$

$ X(s)\left(s  \plus{} 1 \plus{} \frac{3}{(s\plus{}5)}\right) \equal{} F(s)$

$ X(s) \equal{} \frac{(s\plus{}4)(s\plus{}2)}{(s\plus{}5)}F(s) \equal{} \left(\frac{3}{2}\frac{1}{s\plus{}2} \minus{} \frac{1}{2}\frac{1}{s\plus{}4}\right)\frac{1}{s\plus{}1}e^{\minus{}(s\plus{}1)}$

$ X(s) \equal{} \frac{3}{2}\left(\frac{1}{s \plus{} 1} \minus{} \frac{1}{s \plus{} 2}\right)e^{\minus{}(s\plus{}1)}\minus{}\frac{1}{6}\left(\frac{1}{s\plus{}1}\minus{}\frac{1}{s\plus{}4}\right)e^{\minus{}(s\plus{}1)}$

$ X(s) \equal{} \frac{4}{3} \frac{1}{s\plus{}1} e^{\minus{}(s\plus{}1)}\plus{}\frac{1}{6}\frac{1}{s\plus{}4}e^{\minus{}(s\plus{}1)} \minus{} \frac{3}{2}\frac{1}{s \plus{} 2}e^{\minus{}(s\plus{}1)}$

Now, in general:

$ \mathcal{L}^{\minus{}1}\left\{\frac{1}{s\plus{}a} e^{\minus{}(s\plus{}1)}\right\} \equal{} \left\{{\begin{array}{*{20}c}
{{0}\hspace{0.1cm},\hspace{0.6cm}t<1}\hfill\vspace{0.2cm} \\
{{e^{\minus{}1\minus{}a(t\minus{}1)}}\hspace{0.1cm},\hspace{0.3cm}t\geq1}\hfill\\
\end{array}}\right.$

So, for $ t<$ you have $ x(t) \equal{} 0$, but for $ t\leq1$ you have:

$ x(t) \equal{} \frac{4}{3}e^{\minus{}t}\plus{}\frac{1}{6}e^{3\minus{}4t}\minus{}\frac{3}{2}e^{1\minus{}2t}$

Now, you can calculate $ y(t)$ from:

$ y(t) \equal{} f(t) \minus{} \dot{x} \minus{} x$

We have:

$ \dot{x}(t) \equal{} \minus{}\frac{4}{3}e^{\minus{}t}\minus{}\frac{2}{3}e^{3\minus{}4t}\plus{}3 e^{1\minus{}2t}$

So:

$ y(t) \equal{} e^{\minus{}t}\minus{}\left(\frac{4}{3}e^{\minus{}t}\plus{}\frac{1}{6}e^{3\minus{}4t}\minus{}\frac{3}{2}e^{1\minus{}2t}\minus{}\frac{4}{3}e^{\minus{}t}\minus{}\frac{2}{3}e^{3\minus{}4t}\plus{}3 e^{1\minus{}2t}\right)$

$ y(t) \equal{} e^{\minus{}t}\plus{}\frac{1}{2}e^{3\minus{}4t}\minus{}\frac{3}{2}e^{1\minus{}2t}$
[/hide]


calculus
integration
calculus computations

lim n-->infinity of:

$ \int^1_0 \ (n \plus{} 1)x^n \cdot \left(1 \plus{} x \plus{} x^{(x \plus{} 1)} \plus{} x^{{(x \plus{} 1)}^{(x \plus{} 2)}} \plus{} x^{(x \plus{} 1)^{(x \plus{} 2)^{(x \plus{} 3)}}}\right)$

With respect to x. Thanks!

[color=green][Fixed. See code. - moderator][/color]

I think the answer is infinity. And no need to calculate the integral, just estimate it.

Oops! I meant to integrate from 0 to 1, not 0 to infinity.


function
advanced fields
advanced fields unsolved

1.Find 2 different analytic structures on $ R$ whose underlying smooth structures are the same

2. Show that $ GL(n,C)$ is connected.

For #1: a simple stretch does it, if you're not conformal. Let $ 1$ and $ i$ be $ (1,0)$ and $ (0,1)$ or $ (1,0)$ and $ (0,2)$ respectively.

I didnot get your pt. I am supposed to find the structures for R. I guess I have to use the Cauchy's Function

A typical analytic function in one case: $ f(x,y) \equal{} x \plus{} iy$. In the second: $ f(x,y) \equal{} 2x \plus{} iy$. The transition function between them is not analytic, so they belong to different complex structures.
Complex conjugation is another cheap way. The key is to make the transition function between the two smooth but not analytic.

How can you take complex functions. I am supposed to give analytic charts on the [b]reals.[/b]


parameterization
algebra proposed
algebra

Given two equations $ 4^x\minus{}4^{\minus{}x}\equal{}2cosax (1)$,$ 4^x\plus{}4^{\minus{}x}\equal{}2cosax\plus{}4 (2)$, where the parameter a is choosen such that (1) has 2008 roots. Find the number of roots of (2).

Answer: 4016 roots
Can you explain about your solution?

Let $ 2b\equal{} ln 4\equal{}2ln 2$, then (1) equavalent to $ sh(2bx)\equal{}cos(ax)$ and (2) equavalent to
$ sh^2(bx)\equal{}cos^2\frac{ax}{2}$.
Obviosly if $ x$ is solution (1), then $ \minus{}x$ is not solution (1). Therefore all solutions (2) are 4016 solutions
$ \pm 2x_i$, were $ x_i, i\equal{}1,2,...,2008$ solutions (1).

Dear Rust, thanks for your interest and solution.
[hide="Another solution"]We can get some transformations:
$ 4^x \plus{} 4^{ \minus{} x} \equal{} 2cosax \plus{} 4$
$ \Rightarrow 4^x \minus{} 2 \plus{} 4^{ \minus{} x} \equal{} 2(1 \plus{} cosax)$
$ \Rightarrow(2^x \minus{} 2^{ \minus{} x})^2 \equal{} 4cosax \plus{} 4$
$ \Rightarrow 4^{x/2} \minus{} 4^{ \minus{} x/2} \equal{} 2cos\frac {ax}{2}$   (3) or $ 4^{\minus{}x/2} \minus{} 4^{x/2} \equal{} 2cos\frac {ax}{2}$    (4)
Since (1) has 2008 roots, each of equations (3),(4) has 2008 roots. (3) and (4) have no common root, so (2) has 4016 roots.[/hide]


What is the smallest four-digit whole number divisible by 9 which has two even and two odd digits?

We let the first digit be 1 and the second be 0.  If the digit sum is 9, then the last two digits sum to 8, meaning they're both even or both odd, which is impossible because of the restriction.

If the digit sum is 18, then we get the numbers 1089 and 1098, the smaller of which is $ \boxed{1089}$


Is this fact called Lagrange's Identity?

For any point $ M$ in the plane of triangle $ ABC$ and $ G$, the centroid of $ ABC$,
\[ MA^2\plus{}MB^2\plus{}MC^2\equal{}3MG^2\plus{}GA^2\plus{}GB^2\plus{}GC^2\]

furthermore, does anyone have a proof of this fact?

Yes; use vectors. Via translations, we can WLOG let $ M$ be the origin. Let $ A \equal{} a$, $ B \equal{} b$, $ C \equal{} c$; then $ G \equal{} \frac {a \plus{} b \plus{} c}{3}$.

Our equation then becomes

$ |a|^2 \plus{} |b|^2 \plus{} |c|^2 \equal{} 3|\frac {a \plus{} b \plus{} c}{3}|^2 \plus{} |\frac { \minus{} 2a \plus{} b \plus{} c}{3}|^2 \plus{} |\frac {a \minus{} 2b \plus{} c}{3}|^2 \plus{} |\frac {a \plus{} b \minus{} 2c}{3}|^2.$

After multiplication by 9, this becomes equilvanet to

$ 9|a|^2 \plus{} 9|b|^2 \plus{} 9|c|^2 \equal{} 3|a \plus{} b \plus{} c|^2 \plus{} | \minus{} 2a \plus{} b \plus{} c|^2 \plus{} |a \minus{} 2b \plus{} c|^2 \plus{} |a \plus{} b \minus{} 2c|^2.$

Observing now that $ |u \plus{} v \plus{} w|^2 \equal{} |u|^2 \plus{} |v|^2 \plus{} |w|^2 \plus{} 2 u \cdot v \plus{} 2 u \cdot w \plus{} 2 v \cdot w$. Using this identity on the RHS and grouping terms, we see that the RHS is actually the LHS, so we are done.

I'm not very sure about this, but I think Lagrange's identity is more generally used to refer to [url=http://en.wikipedia.org/wiki/Lagrange%27s_identity]this[/url]. I have heard it be used to refer to this identity (the one in this topic, that is) before, though.


[quote="Ánh xạ"]Let $ a,b,c$ be positive real numbers such that $ abc \equal{} 1$ Prove that
$ \sum\sqrt {\frac {a^2 \plus{} ab}{1 \plus{} ab}}\geq 3$[/quote]
Let $ a \equal{} \frac {x}{y},$ $ b \equal{} \frac {z}{x}$ and $ c \equal{} \frac {y}{z},$ where $ x,$ $ y$ and $ z$ are positive numbers.
Then $ \sum_{cyc}\sqrt {\frac {a^2 \plus{} ab}{1 \plus{} ab}} \equal{} \sum_{cyc}\sqrt {\frac {x^2 \plus{} yz}{y(y \plus{} z)}}\geq3\sqrt [6]{\frac {(x^2 \plus{} yz)(y^2 \plus{} xz)(z^2 \plus{} xy)}{xyz(x \plus{} y)(x \plus{} z)(y \plus{} z)}}$ by AM-GM.
But $ (x^2 \plus{} yz)(y^2 \plus{} xz)(z^2 \plus{} xy)\geq xyz(x \plus{} y)(x \plus{} z)(y \plus{} z)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x^3y^3 \plus{} x^4yz \minus{} x^3y^2z \minus{} x^3z^2y)\geq0,$ which obviously true.

Like it,
$ \sum_{cyc}\sqrt {\frac {a \plus{} b}{a \plus{} 1}}\geq 3$ for $ a,b,c > 0$ and $ abc \equal{} 1$ (Mathlinks Contests)