err....what is the midpoint of a circle?  :?

I take it he means center

[hide]
The segments between the centers form an equilateral triangle with side length $1$.  To find the total area, we can find the area of the triangle, then find the area of the segment  in the circle outside of the triangle.  The area of an equilateral triangle is $\frac{s^2\sqrt 3}{4} = \frac{\sqrt 3}{4}$.  Since the angles in an equilateral triangle are $60$ degrees, the area of the whole sector is going to be $\frac \pi 6$, subtracting the triangle, we get the area of the curved segment to be $\frac{2\pi - 3\sqrt 3}{12}$.  Multiplying by $3$, because there are three of them, and then adding the triangle back in, our total area is $\frac{\pi - \sqrt 3}{2}$
[/hide]

just a note,
you can do this with calculus as well--(polar coordinates)

I got $\frac{\pi - \sqrt 3}{2}$ too

 :D

How about when there are four circles?  Obviously they cannot all intersect each other's centers, but say the centers are the vertices of a square side length 1.

Is the answer to 4 circles
[hide]
$1 - \sqrt{3} + \frac{\pi}{3}$
[/hide]
?

I didn't post my solution because I am too tired and I doubt this is right as I probably made an error in calculation due to my tiredness.  I will post it tomorrow morning.

[quote="Silverfalcon"]Three circles of radius 1, intersect such that each circle passes through the midpoints of the other two circles. Find the area that is intersection of three circles.

Express your answer in terms of $\pi$.[/quote]

Yes, More Geometry!  :omighty:  :jump:  :trampoline:

Okay, anyway, you know that when circles intersect, they form equalateral triangles. So, thus, you have 3 equalateral triangles in which each side is 1. Then you also know that since each equalateral triangle has 60 degrees in each side, then the area of the sector is 1/6 of the whole circle or pi/6. Therefore, you take pi/6-1^2(3^1/2)/4=pi/6-3^1/2/4=
(2pi-3(3^1/2))/12. Now you have to multiply this by 3 because there are 3 equalateral triangles if 3 circles intersect and we also have to add in the area of the area of the triangle, so you should get [b](pi-3^1/2)/2.[/b]

Here is my solution for 4 circles:
From the image,
$4X + 4A + N = 1$
The area of one of these quarter circles is $\frac{\pi}{4}$, so we have:
$2X + 3A + N = \frac{\pi}{4}$
We can then find the area of the intersecting regions of two of these circles.  (The region bounded by the circle which starts in the bottom left corner of the square, the one in the bottom right, and a side of the square)
We can draw in an equilateral triangle of side 1 whose area is then $\frac{\sqrt{3}}{4}$  , and we are then left with the 2 areas between the triangle and the circle:
$2{\frac{1}{6}(1^2\pi) - \frac{\sqrt{3}}{4}}$
$\frac{\pi}{3} - \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{4}$
$\frac{\pi}{3} - \frac{\sqrt{3}}{4}$
So we can write
$X + 2A + N = \frac{\pi}{3} - \frac{\sqrt{3}}{4}$
So, we have 3 equations, 3 unknowns.  Solving this system for N gives
$\boxed{1 - \sqrt{3} + \frac{\pi}{3}}$

Yeah I also got [b]1-3^1/2+pi/3[/b] Fun guys, fun, keep the geometry rollin  :wow:

Yes that is the correct answer.  At least that's what I got.

I did it a different way.  I wrote equations for the graphs on the cartesian plane, then picked two equations and solved them for the intersection points.  Then I used those intersection points to make a triangle to find out the amount of degrees in the "circular segments" outside the square that's in the inside.  So basically I solved it almost the exact same way as the 3 circles one.  It was just harder finding the angle measure with 4 circles.  Anyway that angle turned about to be 30 degrees, and then I found the area the same way as before, and there is a square in the middle, inside the four arc segments, and I just used found the length of a chord angled 30 degrees to find the side length.


What is the maximum number of regions in the plane created by three distinct lines, no two of which are parallel?

7. Draw a triangle and extend its sides. There are 3 regions out of a side, 3 regions out of a corner, and 1 region inside the sides.

As a general rule, n lines define tri(n)+1 regions.


probability
Alcumus
Counting Probability

2 diagonals of a regular heptagon (a 7-sided polygon) are chosen.  What is the probability that they intersect inside the heptagon?

[geogebra]ab4d0f4a37a6f02763e03eb51db80b3d46c54638[/geogebra]

For every vertex (there are 7 of them), 4 diagonals "work" while 3 don't. The required probability is $ \boxed{\frac{4}{7}}$.

there are a total of 7*4/2=14 diagonals, therefore there are 14C2= 91 ways to choose 2 diagonals. there are a total of 7C4= 35 ways to chose 2 diagonals that intersect.
therefore the probability is $ \frac {35}{91}\equal{} \frac {5}{13}$

so the solution given by alcumus says that any four points on the heptagon uniquely determine a pair of intersecting diagonal... but this isn't always true right!? i mean i could point out two pairs of points that form parallel lines... meaning they don't intersect!! how did they know exactly there is this one to one bijection relationship?!!

In any four points you choose, we get a quadrilateral. In that quadrilateral, the opposite points when connected intersect in one place. This is a unique intersection point for every quadrilateral, so we only need to choose 4 points to get a unique set of intersecting diagonals.

[hide=Solution]Probability is, for those know don't know, $\dfrac{\text{Outcomes we want}}{\text{Possible Outcomes}}$. Let us first find possible outcomes: There are $\dbinom72=\dfrac{7\cdot6}{2}=21$ ways to choose the two endpoints of a diagonal, but seven of these are edges of the heptagon, so there are $21-7=14$ diagonals. We want to choose 2 diagonals such that they intersect inside the regular heptagon. There are a $\dbinom{14}{2}=91$ ways to do this. How many of these intersect inside the polygon? To find this, we remember that for every convex quadrilateral, there is 1 pair of diagonals that intersects in it. There are $\dbinom74=35$. So, our probability is $\dfrac{35}{91}=\boxed{\dfrac5{13}}$[/hide]

Sorry for the bump, but I have a question -- does this trick work even if the heptagon is convex but not necessarily regular? I feel like it does because the four points still uniquely determine a convex quadrilateral. However, can we be guaranteed of 14 diagonals?

[quote=cw357]I got this problem wrong on the first try because I thought that $7C2 = 15$.  :wallbash_red:[/quote]

That is sad. Once I thought that 6choose2 was 20 when I was playing FTW.

Basically the same as this problem:

[quote=Alcumus]2 diagonals of a regular nonagon (a 9-sided polygon) are chosen. What is the probability that their intersection lies inside the nonagon?[/quote]

Anyway, I have a much easier [hide=sol]Choose any diagonal from the set of 7c2-7=14 diagonals of the heptagon. From here, we have 13 options for the second one, and we can manually count that 5 diagonals intersect with the chosen diagonal. By symmetry, the answer is 5/13.[/hide].

Very beautiful for an Alcumus problem :love:

I tried to do complementary counting, but didn't count it exactly right :| , I got 8/13. But nice problem!  :laugh:

For a general $n$-sided convex polygon, it is just $$\frac{\binom{n}{4}}{\binom{\frac{n(n+3)}{2}}{2}}.$$


geometry
geometric transformation
rotation
circumcircle
reflection
ratio
geometry proposed

Let $ABCD$ be a square with $AB=1$. Trough $C$ passes two lines which intersect  $BD$ in $M$ and $N$,  $AB$ and $AD$ in $P$ and $Q$.
The following statements are equivalent:
[color=blue]a)[/color] $m(\angle PCQ)=45^0$;
[color=blue]b)[/color] $P_{APQ}=2$;
[color=blue]c)[/color] $PQ=BP+QD$;
[color=blue]d)[/color] $MN^2=BM^2+ND^2$;
[color=blue]e)[/color] $PQ$ is tangent at the arc $BD$ of the circle by centre $C$ and radius $CB$;
[color=blue]f)[/color] $CT=1$, were $T$ is the point of tangency from e);
[color=blue]g)[/color] $S_{CMN}=\frac{1}{2} S_{CPQ}$.
[img]http://i19.photobucket.com/albums/b200/mariusdamian/secantala45.gif[/img]

Use a rotation $R: (C, 90^o)$. Then $B \to B'$. It is easy to see from here that $a \Longleftrightarrow b \Longleftrightarrow c$.

Let the circumcircle (Z) of the triangle $\triangle CPQ$ meet AB, AD again at Q', P'. Since CPP'Q is cyclic, $\angle AP'P = \angle PCQ = 45^\circ.$ Thus P' is a reflection of P in AC and similarly, Q' is a reflection of Q in AC, so that the circle (Z) is the centered on AC and AP' = AP, AQ' = AQ, DP' = BP, BQ' = DQ. Let (Z) meet AC, BC again at R, S. Since RC is a diameter of (Z), the angle $\angle CSR = 90^\circ$ is right and $RS \parallel AB,$ so that $AR = BS \sqrt 2.$ Power of B, A to (Z) are

$BP \cdot DQ = BP \cdot BQ' = BS \cdot BC$

$AP \cdot AQ = AP \cdot AQ' = AR \cdot AC = BS \sqrt 2 \cdot BC \sqrt 2 = 2 BS \cdot BC = 2 BP \cdot DQ$

Conversely, if $AP \cdot AQ = 2 BP \cdot DQ,$ then $\angle PCQ = 45^\circ.$ (The argument is reversible and the problem was posted before.) Denoting $p = AP, q = AQ, r = PQ, a = AB = BC = CD = DA = 1,$ we have

$pq = 2(a - p)(a - q),\ \ \ pq - 2a(p + q) + 2a^2 = 0$

$(2a - p - q)^2 = 4a^2 + p^2 + q^2 - 4a(p + q) + 2pq = p^2 + q^2 = r^2$

$AP + AQ + PQ = p + q + r = 2a$

Again, this is equivalent to $pq = AP \cdot AQ = 2 BP \cdot DQ = 2(a - p)(a - q).$ But then and only then the excircle of the triangle $\triangle APQ$ against the vertex A  touches AP at B and AQ at D, because $a = AB = AD = \frac{p + q + r}{2},$ hence this excircle is centered at C and has radius CB = CD = a, which is also the distance of the tangency point T of PQ with this excircle (C). This in turn is equivalent to

$PQ = r = 2a - p - q = (a - p) + (a - q) = BP + DQ$

From the basic relation $pq = 2(a - p)(a - q),$ we get $p = \frac{2a(a - q)}{2a - q},\ q = \frac{2a(a - p)}{2a - p}.$ BM, DN are bisector legths of the right angle triangles $\triangle BCP, \triangle DCQ$ with the legs BP - a - p, BC = a resp. DQ = a - q, DC = a, hence

$BM = \frac{BC \cdot BP \sqrt 2}{BC + BP} = \frac{a(a - p) \sqrt 2}{2a - p} = \frac{q}{\sqrt 2}$

$CN = \frac{DQ \cdot DC \sqrt 2}{DQ + DC} = \frac{a(a - q) \sqrt 2}{2a - q} = \frac{p}{\sqrt 2}$

$BM^2 + CN^2 = \frac{p^2 + q^2}{2} = \frac{r^2}{2} = \frac{PQ^2}{2}$

Consequently,

$NM = BD - (BM + DN) = a \sqrt 2 - \frac{p + q}{\sqrt 2} = \frac{2a - (p + q)}{\sqrt 2} = \frac{r}{\sqrt 2} = \frac{PQ}{\sqrt 2}$

Let the square diagonals meet at X. Since $NM \equiv BD \perp AC \equiv CX$, $CX = \frac{a}{\sqrt 2}$ is the C-altitude of the triangle $\triangle CNM,$ while CT = a is the C-altitude of the triangle $\triangle CPQ.$ Since $\frac{CX}{CT} = \frac{NM}{PQ} = \frac{1}{\sqrt 2}$ and the angle $\angle PCQ \equiv MCN = 45^\circ$ is common, the triangles $\triangle PCQ \sim \triangle NCM$ are similar with similarity coefficient $k = \frac{1}{\sqrt 2},$, so that their areas are in the ratio $k^2 = \frac 1 2.$ While numerous other ways can be used to prove implications between variuous statements (a), (b), ..., (g), the above approach maintains equivalence of all to the basic relation $pq = 2(a - p)(a - q).$


geometry
geometric transformation
reflection
trigonometry
projective geometry
trig identities
Law of Sines

Let a circle $O$ be inscribed in quadrilateral $ABCD$, with the tangent points on $AB,BC,CD,DA$ being $P,Q,R,S$ respectively. Let $l$ be the line perpendicular to $BD$ and passing through $O$ (Say it intersects $BD$ at $N$). Let $K$ be the intersection of $l$ and $AC$ and let $Q'$ be the reflection of $Q$ across $AC$.
(i) Prove that $\frac{AK}{KC}= \frac{AP}{CQ}$
(ii) Prove that $P,N,Q'$ are collinear.

Firse, we show that K is the pole of BC.
Let K' be the pole of BC. Then K' lies on the polar of B and on the polar of C, that is, K' is the intersection of PQ and RS. From Brianchon's theorem, AC,BD,PR,QS concour in Z But from Desaregues' theorem applied to the perspective triangles APS,CRQ, we have that PQ,RS,AC concour. Then K' lies on AC. And also the line joining K' to the circumcentre is perpendicular to BC => K' = K.

From some known theorem about projective geometry, we know that $(ACZK) =-1$, or $\frac{AK}{KC}= \frac{AZ}{ZC}$ It remains to show that $\frac{AZ}{ZC}= \frac{AP}{CQ}= \frac{AP}{CR}$ But from the law of sines:
$\frac{AP}{AZ}= \frac{\sin \angle AZP}{\sin \angle APZ}= \frac{\sin \angle CZR}{\sin \angle CRZ}= \frac{CR}{RZ}$ (note that $\angle BPZ = \angle CRZ$).

By the way, in my figure $P;N;Q'$ are not collinear. Maybe I misunderstood the problem.

I'm sorry; let $Q'$ be reflected across $NK$ rather.

Ok! It is sufficient to prove $\angle PNK = \angle Q'NK \Leftrightarrow \angle PNK =-\angle QNL \Leftrightarrow \angle PNK-90 = 90-\angle QNL \Leftrightarrow \angle PNB = \angle QNB$, iff BP is the angle bisector of $\angle PNQ$.
But this follows from Apollonius' theorem, the fact that (let's call $X = BN \cap PQ$) $(PQXK) =-1$ and the fact that $BN$ and $NK$ are perpendicular.

If you would like to see more details, ask please, because answering I'll understand better what I've written  :)


in proving "every even number is a sum of two primes" can i use the fact that "every odd number is a sum of three primes"  :maybe: ?

In 1937 Vinogradoff proved every sufficiently large odd integer is the sum of three odd primes, but it is not known if every odd intger is the sum of three primes. If every even ineger is the sum of two primes then its easy to see every odd integer is the sum of three odd primes.


hi,

I have an example of finding the forecast error for ARIMA (0,1,1):

$ (1\minus{}L)*X_{t}\equal{}(1\plus{}D*L)*z_{t}$

$ z_{t}\equal{}(1\plus{}D*L)^{\minus{}1}$*$ (1\minus{}L)*X_{t}$

$ z_{t}\equal{}(1\minus{}L)$*$ \sum_{i\equal{}0}^{\infty}(\minus{}D)^{i}*L^{i}*X_{t}$

$ z_{t}\equal{}X_{t} \minus{} (1\plus{}D)*\sum_{i\equal{}1}^{\infty}(\minus{}D)^{i\minus{}1}*L^{i}*X_{t}$

I now need to find an expression for $ z_{t}$ for SARIMA $ (0,1,1)*(1,1,0)_{5}$

$ (1\minus{}B*L^{5})*(1\minus{}L)*(1\minus{}L^{5})*X_{t}\equal{}(1\plus{}D*L)*z_{t}$

$ z_{t}\equal{}(1\plus{}D*L)^{\minus{}1}$*$ (1\minus{}B*L^{5})*(1\minus{}L)*(1\minus{}L^{5})*X_{t}$

$ z_{t}\equal{}(1\minus{}B*L^{5})*(1\minus{}L)*(1\minus{}L^{5})$*$ \sum_{i\equal{}0}^{\infty}(\minus{}D)^{i}*L^{i}*X_{t}$

that's how far i've got so far and i am a bit stuck.

any help would be appreciated.

just to give a bit more detail:

$ SARIMA$ is a seasonal autoregressive integrated moving average

$ L$ is a lag operator

http://en.wikipedia.org/wiki/Lag_operator

i need to bring all $ L's$ inside the series similar to the example with ARIMA.

any help would be appreciated. thanx.

managed to find the solution. help is no longer required.  :lol:


LaTeX
trigonometry
geometry
algebra
polynomial
area of a triangle
Heron\u0027s formula

Okay folks, discuss week 0 problems and solutions. Hold off on Vertical problems until after CX is complete.

*nudge Sammy to post sols*

So uh...here's #9, (although there are probably an easier ways), with $ \text{\LaTeX} $:

Let the length of $ AB $ be $ c $, the length of $ BC $ be $ a $, and the length of $ AC $ be $ b $.
Similarly, denote the lengths of $ OA,OB,OC $ by $ r,s,t $ respectively. Then what we are after is $ (OA)^2+(OB)^2+(OC)^2=r^2+s^2+t^2 $.

From Law of Cosines, \[ \begin{equation*} a^2=s^2+t^2-2(s)(t)(\cos{120^\circ})\\ b^2=r^2+t^2-2(r)(t)(\cos{120^\circ})\\ c^2=r^2+s^2-2(r)(s)(\cos{120^\circ}) \end{equation*} \] Noting that $ \cos{120^\circ}=-\frac{1}{2} $, the sum of the three equations shows that \[ a^2+b^2+c^2=2(r^2+s^2+t^2)+(r)(s)+(r)(t)+(s)(t). \] Since the sides of the triangle are the roots of $ x^3-16x^2+85x-149=0 $, \[ a+b+c=16, \] \[ ab+ac+bc=85, \] \[ abc=149, \] \[ a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)=86, \] \[ a^2b^2+a^2c^2+b^2c^2=(ab+ac+bc)^2-2abc(a+b+c)=2457, \] \[ a^4+b^4+c^4=(a^2+b^2+c^2)^2-2(a^2b^2+a^2c^2+b^2c^2)=2482 \] From the expansion Hero(n)'s formula with $ a,b,c $, the area of $ \bigtriangleup ABC $ is \[ \left(\frac{1}{4}\right)\sqrt{2(a^2b^2+a^2c^2+b^2c^2)-(a^4+b^4+c^4)}=2\sqrt{38} \] Hence, \[ r^2+s^2+t^2=43-\left(\frac{rs}{2}+\frac{rt}{2}+\frac{st}{2}\right). \]
$ \frac{(rs)(\sin{120^\circ})}{2} $ is the area of $ \bigtriangleup OAB $.
$ \frac{(rt)(\sin{120^\circ})}{2} $ is the area of $ \bigtriangleup OAC $.
$ \frac{(st)(\sin{120^\circ})}{2} $ is the area of $ \bigtriangleup OBC $.
Thus, the area of $ \bigtriangleup ABC $ is $ (\sin{120^\circ})\left(\frac{rs}{2}+\frac{rt}{2}+\frac{st}{2}\right) $. That is, \[ \left(\frac{\sqrt{3}}{2}\right)\left(\frac{rs}{2}+\frac{rt}{2}+\frac{st}{2}\right)=2\sqrt{38}, \] \[ \left(\frac{rs}{2}+\frac{rt}{2}+\frac{st}{2}\right)=\frac{4\sqrt{114}}{3}. \] Consequently, \[ r^2+s^2+t^2=\frac{129-4\sqrt{114}}{3}. \]
So $ 129-4+114+3=\boxed{242} $, our answer.

For anyone interested, problem 9:

Call the sides of the triangle a, b, c, with respect to their angles A, B, C.  From the polynomial, you can deduce that:

$ a + b + c = 16 $

First, find the area of the triangle using Heron's formula.

$ s = \frac{a + b + c}{2} = 8 $

$ A_t = \sqrt{s(s-a)(s-b)(s-c)} $

You can save yourself the trouble of expanding if you notice that since a, b, c are the roots of the original polynomial, the coefficients will be the same.

$ A_t = \sqrt{8(8^3 - 16(8)^2 + 85(8) - 149)} = \sqrt{152} $

Now, find the area of the smaller triangles, add them up, and set equal to the area of the large triangle.

$ A_1 + A_2 + A_3 = A_t $

$ A_1 = \frac{bh}{2} = \frac{OA \cdot OB \sin{A}}{2} = \frac{\sqrt{3}}{4} OAOB $

$ A_2 = \frac{\sqrt{3}}{4} OBOC $

$ A_3 = \frac{\sqrt{3}}{4} OCOA $

$ \frac{\sqrt{3}}{4}(OAOB + OBOC + OCOA) = \sqrt{152} \rightarrow OAOB + OBOC + OCOA = 4\sqrt{\frac{152}{3} }$

Lastly, use law of cosines to bring everything together:

$ \frac{OA^2 + OB^2 - a^2}{2OAOB} = \cos{A} = -\frac{1}{2} $

$ OA^2 + OB^2 - a^2 = -OAOB $

$ OB^2 + OC^2 - b^2 = -OBOC $

$ OC^2 + OA^2 - c^2 = -OCOA $

$ 2(OA^2 + OB^2 + OC^2) - (a^2 + b^2 + c^2) = - (OAOB + OBOC + OCOA) = -4 \sqrt{\frac{152}{3}} $

A quick way to find $ a^2 + b^2 + c^2 $ using Newton's Sums:

$ s_2 - 16s_1 + 2(85) = 0 \rightarrow s_2 = 86 $

Finally,

$ OA^2 + OB^2 + OC^2 = 43 - 2\sqrt{\frac{152}{3}} = \frac{129 - 2\sqrt{456}}{3} = \frac{129 - 4\sqrt{114}}{3} $

$ 129 + (-4) + 114 + 3 = 242 $

Unfortunately, when I was actually doing the problem, I forgot the last simplification of the radical, lol.

edit: haha, a little late.  basically same method as above.

Wait... what's question #9? PDF has expired on my computer.

Ugh, I was on exactly the right path on #9 and would have gotten it but just ran out of time with my computations!   :wallbash_red:

#9:

Suppose that the lengths of the sides of the triangle $ \bigtriangleup ABC $ are the roots of the cubic polynomial $ x^3-16x^2+85x-149 $, and $ O $ is a point inside $ \bigtriangleup ABC $ such that $ m\angle AOB=m\angle BOC=m\angle COA=120^\circ $. The sum $ OA^2+OB^2+OC^2 $ can be written in the form \[ \frac{a+b\sqrt{c}}{d}, \] where $ a,b,c, $ and $ d $ are integers, and $ c $ and $ d $ are positive, but as small as possible. Compute $ a+b+c+d $.

[quote]#7. Suppose that $ABCD$ is a quadrilateral, inscribed in a circle, such that $AB=5, BC=10, CD=11$, and $DA=14$. Find the integer closest to $AC^2$.[/quote]
Note that \[ AB^2+DA^2=5^2+14^2=25+196=221=100+121=10^2+11^2=BC^2+CD^2. \] Hence, \[ m\angle BAD=m\angle BCD=90^\circ. \] The length of $BD$, then is \[ \sqrt{AB^2+DA^2}=\sqrt{BC^2+CD^2}=\sqrt{221}. \] From Ptolemy's Theorem, \[ AB\cdot CD+AD\cdot BC=AC\cdot BD, \] \[ 5\cdot11+14\cdot10=\sqrt{221}\cdot AC, \] \[ AC^2=\dfrac{195^2}{221}\approx172.06. \] Desired answer is $\boxed{172}$.

Happy Nathan? Not a single one in sight.  :)

Shoot. I was really really excited about just being able to solve the problem since I usually freak out on anything that has to do with trig, but forgot to account for the fact that the 4 had a minus sign in front of it, and ended up with 250. :(

interestinly enough the given polynomial in question 9 has only one real root.  But that's not the focus of the solution

[quote="towersfreak2006"]
Note that \[ AB^2+DA^2=5^2+14^2=25+196=221=100+121=10^2+11^2=BC^2+CD^2. \] Hence, \[ m\angle BAD=m\angle BCD=90^\circ. \] The length of $BD$, then is \[ \sqrt{AB^2+DA^2}=\sqrt{BC^2+CD^2}=\sqrt{221}. \] From Ptolemy's Theorem, \[ AB\cdot CD+AD\cdot BC=AC\cdot BD, \] \[ 5\cdot11+14\cdot10=\sqrt{221}\cdot AC, \] \[ AC^2=\dfrac{195^2}{221}\approx172.06. \] Desired answer is $\boxed{172}$.[/quote]

Curse you. Ugh. I spent 20 minutes on Ptolemy and side ratios. *face-palm*

Why didn't I think of the Pythagorean Theorem. *smack self*

On a different note, could someone post their answers *cough sammy cough* so we can try to get a list of the correct ones?

Cody posted the answers: 390 169 155 084 322 476 172 820 242.

If you didn't see that the angles were $90^\circ$ in #7 you could have just used straight up Law of Cosines, used the fact that $\cos(180^\circ-\theta)=-\cos(\theta)$ and solved.

[quote="joml88"]Cody posted the answers: 390 169 155 084 322 476 172 820 242.[/quote]

Where was this?

And I see what you mean about the LoC. Thanks. :) Missed that shortcut too.

It's at the very end of the "Week 0 Individual Honor Roll" news post.

[quote]
#8. Give the smallest integer multiple of $10$ that is greater than the value of the expression \[ \sqrt\dfrac{1\cdot2+3\cdot4+5\cdot6+\cdots+2003\cdot2004}{2004}}. \][/quote]
Define $f(x)=\sum\limits_{n=1}^{x}(2n-1)\cdot2n$ for $x\in\mathbb{N}$. Then the value of the expression is \[ E=\sqrt{\dfrac{f(1002)}{2004}}. \] But we know that $\sum\limits_{n=1}^{x}n^2=\dfrac{x(x+1)(2x+1)}{6}$ and $\sum\limits_{n=1}^{x}n=\dfrac{x(x+1)}{2}$. Hence, \[ f(x)=\sum\limits_{n=1}^x(4n^2-2n)=4\left(\dfrac{n(n+1)(2n+1)}{6}\right)-2\left(\dfrac{n(n+1)}{2}\right). \] Thus, the value of the expression is \[ E=\sqrt{\dfrac{2(1002)(1003)(2005)}{3\cdot2004}-\dfrac{1002\cdot1003}{2004}, }\] \[ E=\sqrt{\dfrac{4007\cdot1003}{6}}=\sqrt{4007\left(167+\dfrac{1}{6}\right)}=\sqrt{669836}, \] Since $800^2=640000$, we start there and go upwards. \[ 810^2=656100, \]\[ 820^2=672400. \] Noting that $810^2<E^2<820^2$, we know the desired answer must be $\boxed{820}$.

Doh! I misread the question for 8 and wrote 810!
I also missed the right angles in 7 and waded through huge numbers, but at least I got that one.


[quote="keshav_57"]Ny1 here who has taken part in a previous INOI?[/quote]

Well @keshav , you can talk to bruteforce (NADEEM MOIDU) who represented India At the IOI 2008 , and won a Bronze Medal too..

Bye the way have you made it to the INOI this year ???????????

i did not appear for ZIO...
was a bit too late for it to register...

im appearing in ZCO tomorrow...
mayb by the time u will be reading this post... i might be competing
:)

n it was nadeem moidu and pradeep mathias who told me about ZCO
:)

[quote="bruteforce"]Those who are used to turbo c++ can have a look at [url=http://web.iiit.ac.in/~nadeem.moiduug08/gcc.htm]this[/url][/quote]

Thanks Nadeem , it was really helpful . I use  Turbo C++ 3.0 and this is definitely better than that.

Bye the way , have you joined IIIT for your Graduation????


Using Cauchy-Schwarz we have :

$ 9\left(a^{3} \plus{} 3b^{3} \plus{} 5c^{3}\right)\left(a \plus{} 3b \plus{} 5c\right)\ge9\left(a^{2} \plus{} 3b^{2} \plus{} 5c^{2}\right)^{2}\Leftrightarrow9\left(a^{3} \plus{} 3b^{3} \plus{} 5c^{2}\right)\ge\frac {9\left(a^{2} \plus{} 3b^{2} \plus{} 5c^{2}\right)^{2}}{a \plus{} 3b \plus{} 5c}$.

It is enough to prove that :

$ \frac {9\left(a^{2} \plus{} 3b^{2} \plus{} 5c^{2}\right)^{2}}{a \plus{} 3b \plus{} 5c}\ge\left(a^{2} \plus{} 3b^{2} \plus{} 5c^{2}\right)\left(a \plus{} 3b \plus{} 5c\right)\Leftrightarrow9\left(a^{2} \plus{} 3b^{2} \plus{} 5c^{2}\right)\ge\left(a \plus{} 3b \plus{} 5c\right)^{2}$.

This inequality is true using Cauchy :

$ 9\left(a^{2} \plus{} 3b^{2} \plus{} 5c^{2}\right) \equal{} \left(1 \plus{} 3 \plus{} 5\right)\left(a^{2} \plus{} 3b^{2} \plus{} 5c^{2}\right)\ge\left(a \plus{} 3b \plus{} 5c\right)^{2}.$

This is an inequality of M.Chirciu and O.Stroe in G.M.

[quote="Mathias_DK"][quote="peine"][quote="Mateescu Constantin"]Show that:
$ 9(a^{3} \plus{} 3b^{3} \plus{} 5c^{3})\geq(a^{2} \plus{} 3b^{2} \plus{} 5c^{2})(a \plus{} 3b \plus{} 5c)$, for all $ a,\ b,\ c\geq 0$.[/quote]
assume that $ a\geq b\geq c$ and your inequality is direct applicatiob of Chebychev.[/quote]
Why would you want to prove $ a \geqslant b \geqslant c$ seperatly?[/quote]
sorry my solution was wrong because I supposed that $ a\geqslant b \geqslant c$ and the inequality isn't symetric, this is another solution,
we have $ \sqrt[3]{\frac{a^3\plus{}3b^3\plus{}5c^3}{9}}\geqslant \frac{\sqrt{a^2\plus{}3b^2\plus{}5b^2}}{3}$
$ \Leftrightarrow \frac{a^3\plus{}3b^3\plus{}5c^3}{9}\geqslant (a^2\plus{}3b^2\plus{}5b^2)\left(\frac{\sqrt{a^2\plus{}3b^2\plus{}5b^2}}{27}\right)$
by Cauchy-Schwarz, $ \sqrt{a^2\plus{}3b^2\plus{}5b^2}\geqslant \frac{a\plus{}3b\plus{}5c}{3}$
then  $ 9(a^{3}\plus{}3b^{3}\plus{}5c^{3})\geqslant (a^{2}\plus{}3b^{2}\plus{}5c^{2})(a\plus{}3b\plus{}5c)$


AMC
USA(J)MO
USAMO
induction
number theory

37th Canadian Mathematical Olympiad Winners

FIRST PRIZE and the Sun Life Financial Cup
[b]Peng Shi[/b], Sir John A. MacDonald Collegiate Institute, Agincourt, Ontario.

SECOND PRIZE
[b]Richard Peng[/b], Vaughan Road Academy, Toronto, Ontario

THIRD PRIZE
[b]Yufei Zhao[/b], Don Mills Collegiate Institute, Don Mills, Ontario

HONOURABLE MENTIONS
[b]Boris Braverman[/b], Sir Winston Churchill High School, Calgary, Alberta
[b]Elyot Grant[/b], Cameron Heights Collegiate Institute, Kitchener, Ontario
[b]Zheng Guo[/b], Western Canada High School, Calgary, Alberta
[b]Oleg Ivrii[/b], Don Mills Collegiate Institute, Don Mills, Ontario
[b]Lin Fei[/b], Don Mills Collegiate Institute, Don Mills, Ontario
[b]Dong Uk (David) Rhee[/b], McNally School, Edmonton, Alberta
[b]Shaun White[/b], Vincent Massey Secondary School, Windsor, Ontario

I did worse than last year! Should have solved #5...
Yufei, where did you get this result?(CMS website doesn't seem to have the result).

We found out after getting phone calls from the school board press.

The news release can be found [url=http://www.newswire.ca/en/releases/archive/May2005/12/c8090.html]here[/url].

Alas!  I'm not up there.  Oh well.

Congrats everyone!
Oh - and to everyone who made IMO - Congrats again!  Of course, there's no official word until May 27ish, but I wouldn't be too scared to hazard a guess at who's going...

Btw, is this: "the winners of the 2004 Open and the 2005 CMO will be honoured at the CMS Awards Banquet" (from that linked website) actually true or did they just forget to delete it?

hey geoffrey, i attended seminars last year... and i thought my friend john would get honoured for his open score, but it never happened... he was there because of his euclid score.  so i don't think they'll do it this year... but i think they should, since the open is a significant math competition...

i hope u get to come to waterloo this year... i didn't get my invite yet... wow but i got an HM on cmo... should that be enough?

who are the members of the Canadian team for IMO?

shhh... it's a secret... anyone who made the imo team shouldn't be allowed to tell for a couple more weeks... telling now would be no fun lol

Forgot to congraturate all the winners.
Congraturation to all the winners of CMO 2005! Good job! :first:

By the way, is "congraturation" spelled correct?

[quote="lightrhee"]

By the way, is "congraturation" spelled correct?[/quote]

Close, I can never spell that word.

Congratulations!

[quote="Elyot"]Cool.  Having written NOTHING for #5, I didn't expect to do so well.  My #1-4 must have been really good.  Sweet.

Contrats to everyone else.  Interesting to see that, like APMO, CMO was totally penguinized.[/quote]
Penguinized? Interesting word!

Haha~the legendary penguins~

Congratulations to everyone! (Sarah, thanks for the spelling tip.)

By the way, where can I find the APMO solutions and results?

[quote="blahblahblah"][quote="mel"][quote="Elyot"]Contrats to everyone else.  Interesting to see that, like APMO, CMO was totally penguinized.[/quote]

Hmmm...what does "penguinized" mean? Totally black and white? Either good or bad? Doesn't have a "nice" statistical distribution?

Congrats to everyone, btw![/quote]

hint: look at the first/last names of the first/second places.[/quote]

That and one of the honorable mentions is the third penguin...so all three penguins are on there.

I think the name of "penguins" was made by Keigo last summer in national camp...though I'm not sure.

Do you guys study math in Olympiad training programs or something?

i don't quite understand what the question means. can you reword it?

[quote="YY"][/quote][quote="blahblahblah"][quote="mel"][quote="Elyot"]Contrats to everyone else.  Interesting to see that, like APMO, CMO was totally penguinized.[/quote]

Hmmm...what does "penguinized" mean? Totally black and white? Either good or bad? Doesn't have a "nice" statistical distribution?

Congrats to everyone, btw![/quote]

hint: look at the first/last names of the first/second places.[/quote][quote="YY"]

That and one of the honorable mentions is the third penguin...so all three penguins are on there.

I think the name of "penguins" was made by Keigo last summer in national camp...though I'm not sure.[/quote]
Who's the third penguin?  *scratches head*

As for selection, it's mostly APMO and CMO, with USAMO as more of a tiebreaker or a last resort than a major factor, I think - I may be wrong.  (Any 2005 IMOers want to correct me? :P )

[quote="YY"][/quote][quote="blahblahblah"][quote="mel"][quote="Elyot"]Contrats to everyone else.  Interesting to see that, like APMO, CMO was totally penguinized.[/quote]

Hmmm...what does "penguinized" mean? Totally black and white? Either good or bad? Doesn't have a "nice" statistical distribution?

Congrats to everyone, btw![/quote]

hint: look at the first/last names of the first/second places.[/quote][quote="YY"]

That and one of the honorable mentions is the third penguin...so all three penguins are on there.

I think the name of "penguins" was made by Keigo last summer in national camp...though I'm not sure.[/quote]

lol!  :rotfl:

[quote="Sunny"][quote="shobber"]well, can anyone tell me how does your selecting program work?[/quote]

It is basically based on the 3 contests APMO, CMO, and USAMO.  There may be other factors in there, but I don't know for sure.[/quote]
 In Canadian system, people decides who are going to be on IMO team, not like America where everything is strictly decided by numbers(scores on USAMO and so on). But the scores still plays important roll when it comes to choosing members of the team.
 APMO counts about 30%, CMO counts about 40%, and USAMO counts about 30%. They also look at MOCP(Mathematical Olympiads Correspondence Program) and winter camp.
 As I said, they do not use the percentages. They consider it about that much when choosing a team.

hehe... then i never had a chance of making the imo team... i never attended winter camp, and i haven't done mocp since... 2003 maybe?  i don't know how you imo people can dedicate so much of your time to practising problem solving... i love math so much, but i could never do that... it's just not fun for me... i'd rather just do number theory all day long :)

congrats to all the cmo winners, and good luck with all your future endeavours.

[quote="Elyot"]If you guys haven't figured out the penguin thing yet, just look at the names of the 1st and 2nd place winners of CMO ;)[/quote]
Somebody said, "That and one of the honorable mentions is the third penguin...so all three penguins are on there. "
Who's the third penguin?

i was wondering that too. help us out guys

Ah, well~
Boris was unofficially the third penguin, mainly because he always wanders around the other two during the camp. It was announced a few hours after the initial "penguins".

And I just remembered...someone else from the camp proved (by induction) that all humans are penguins...

Anyway...never thought the "penguins" would be such a hot topic here ^^

[quote="YY"]Ah, well~
Boris was unofficially the third penguin, mainly because he always wanders around the other two during the camp. It was announced a few hours after the initial "penguins".
[/quote]

Hahaha I can actually see that...lol..

[quote="lotus87"]hehe... then i never had a chance of making the imo team... i never attended winter camp[/quote]
You don't need to be in the winter camp to make the IMO team. Peng Shi wasn't in winter camp in 2004 but he made the team.

[quote="lotus87"]and i haven't done mocp since... 2003 maybe?[/quote]
You definitely should work on MOCP, if you want to make the IMO team. It is an necessary thing to do.


I guess I'll be the first one to post a problem in this section of this new forum :)

Let U be a finite set and let f,g:U->U be 2 bijections. A={x \in U | f(f(x))=g(g(x))} and B={x \in U | f(g(x))=g(f(x))}. Show that for x \in U f(x) \in A <=> g(x) \in A.

Well, the way you posted it, there seems to be no sense in introducing the set $B$, but comparing with http://www.kalva.demon.co.uk/short/sh96.html, I guess you just forgot the condition $U=A\cup B$.

So, here's my solution: We know that for any $x\in U$, either $f(f(x)) = g(g(x))$ or $f(g(x)) = g(f(x))$. Setting $y=f(x)$, this may be rewritten as $(g^{-1}\circ f)(x) = (g\circ f^{-1})(x)$ or $(g\circ f^{-1})(x) = (g^{-1}\circ f)^{-1}(x)$. This somehow says that $h=g\circ f^{-1}$ can choose whether it is equal to $k=g^{-1}\circ f$ or its inverse. However, it may not do so for every single element $x\in U$, as we will now show:

Let $x,k(x),k^2(x),...,k^{-1}(x),x$ be an orbit of $U$ under the action of $k$ (this works as $k$ is a permutation).
We will show

Lemma 1: Either $h(y)=k(y)$ for all $y$ in one orbit or $h(y)=k^{-1}(y)$ for all $y$ in one orbit.

Proof:

1. case: $h(x)=k(x)$ and $h(k(x))=k^2(x)$. Then $h(k^2(x))$ cannot be $k(x)$ since $h$ is a permutation, therefore it is $k^3(x)$. Continuing that way, we see that $h(y)=k(y)$ for all $y$ in that orbit.

2. case: $h(x)=k^{-1}(x)$ and $h(k(x))=x$. This case is completely analogous to case 1.

3. case: $h(x)=k(x)$ and $h(k(x))=x$ or $h(x)=k^{-1}(x)$ and $h(k(x))=k^2(x)$(in that case, we immediately have $h(k^{-1}(x))=x$ since there is some $z$ with $h(z)=x$ which is only possible for $z=k^{-1}(x)$ or $z=k(x)$). It follows in that case that $h(h(x))=x$. But this means that $(f^{-1}\circ g\circ f^{-1}\circ g\circ f^{-1})(x)=f^{-1}(x)$ and therefore $k(k(f^{-1}(x))) = f^{-1}(x)$.
It follows immediately from the condition on $h$ that for every $z$ with $k(k(z))=z$, we have $h(h(z))=z$ (we must have $h(z)=k(z)=k^{-1}(z)$ and $h(k(z))=z=k^2(z)$). This means that we have just shown that if $h(h(x))=x$, then $k(k(f^{-1}(x)))=f^{-1}(x)$ and $h(h(f^{-1}(x)))=f^{-1}(x)$. Since there is some $n$ with $f^{-n}(x)=x$, we can prove by induction that $k(k(x))=x$. But knowing this, we immediately see that the lemma is true.

Lemma 1 also shows that orbits of $k$ are the same as orbits of $h$.

Lemma 2: If $O$ is any orbit of $k$, then $f(O)=g(O)$ is again an orbit.

Proof: Let the orbit be $x,k(x),...$. Then $f(O)=\{f(x),f(k(x)),...\}$. But $f(k(x))=(f\circ g^{-1}\circ f)(x) = h^{-1}(f(x))$. Therefore $f(O)$ is an orbit of $h^{-1}$, which is the same as an orbit of $k$. The equation $f(O)=g(O)$ follows from $f(x)=g(k(x))$.

Lemma 3: If $f(f(x))=g(g(x))$, then this is also true for all $x^{\prime}\in O$, the orbit generated by $x$.

Proof: $g(g(x))=f(f(x))=g(k(f(x)))=g(k(h^{-1}(g(x))))$ is true iff $k(h^{-1}(z))=z$ for some $z\in g(O)$. Since $g(O)$ is an orbit of $k$, this means that the equation holds for all $z\in g(O)$ by Lemma 1. But this is just what we had to show.

Since $f(x)$ and $g(x)$ are in the same orbit, Lemma 3 solves the problem.


I've solved all of the 50 problems on the AHSME 50th anniversary sheet. Is there another way to do this last problem besides brute force?

The letters of the word A N N I V E R S A R Y  can be arranged 4,989,600 different ways.  When these 'words' are arranged in alphabetical order, what is the tenth letter of the 50th word.

$\text{(A) N}$
$\text{(B) R}$
$\text{(C) S}$
$\text{(D) V}$
$\text{(E) Y}$

Is the answer D? If it's right, I'll post my solution (not brute force)

[hide="my solution"]

First off, it's talking about distinct ways to arrange the letters.  You can tell by double checking the total number of ways for arrangement.  The first "word" would be: AAEINNRRSVY.  Taking just the first six letters, you have AAEINN, and five slots to fill with the remaining letters.  The number of ways to arrange the remaining letters is:

$\displaystyle{{5\choose2} \cdot {3\choose1} \cdot {2\choose1} \cdot {1\choose1} = \frac{5!}{2!} = 60}$

The first 60 "words" arranged in alphabetical order begin with AAEINN.  The first subset of this would begin with AAEINNR, with which there are this many words:

$4! = 24$

The next subset would begin with AAEINNS, and there are this many "words" in that set:

$\frac{4!}{2!} = 12$

So the total is up to 36 words.  The next subset begins with AAEINNV, with the same number of words in that set, making AAEINNYRRSV the 49th word.  The 50th is then AAEINNYRRVS, making the 10th letter V.

Answer: D
[/hide]

edit: made a mistake in calculations.  Bob was right.


counting
derangement
combinatorics proposed
combinatorics

if A is a sequence of 5 numbers 1,2,3,4,5 and B is its one of the possible derangments Ex:-3,1,2,5,4. how many possible derangments of A and derangments of B intersect?

Understand I correctly: You are looking for the number of pairs of derangements of $1,2,3,4,5$ being not derangements of each other itself¿

No i mean
if A=1,2,3,4,5  and B=3,1,2,5,4 is a derangment of A , the derangment 5,4 1,2,3 is one possible permutation which is in the set of derangments of A and also in the derangments of B. here I need to count to number of possible permutations which are in derangements of both A and B like 5,4,1,2,3?

The number of permutations $p$ of $\{1, 2, \ldots, n\}$ in which $p(i) \neq i$ and $p(i) \neq i+1 \mod n$ are given by [url=http://www.research.att.com/~njas/sequences/A000179]A000179[/url].  These give the answer to your problem, taking our first derangement as $\{2, 3, 4, 5, 1\}$.  The Encyclopaedia gives both a recursion and a summation form for the sequence.  In the case $n = 5$, there are 13 such derangements.


geometry
geometric transformation
reflection
trigonometry
similar triangles
geometry proposed

Let $ \Gamma$ be a circumference with center $ O$ tangent to sides $ AB$ and $ AC$ of triangle $ ABC$ in points $ E$ and $ F$, respectively. The line perpendicular to $ BC$ passing through $ O$ intersects $ EF$ in point $ D$. Let $ M$ be the midpoint of $ BC$. Prove that $ A$, $ D$, $ M$ are collinear.

I basically posted the same problem before. http://www.mathlinks.ro/viewtopic.php?t=287158
By considering the central similarity with center A that carries the incircle of ABC (I) ---> (O) the collinearity follows  :)

very nice prblem！ :lol: Here is my solution :blush: : Draw QP ∥BC; then we can easily get that :point :Q,E,D,O and point O,D,F,P are concyclic→∠OED=∠OQD ;∠OFD=∠OPD=∠OED;∠DEA=∠DOQ;∠DOP=∠DFA=∠DEA→∠QOD=∠DOP ,∠OQD=∠OPD→QD=PD;join AD meet BC at M';By similar triangles ; we can get :M'B=M'A;so M' is just the point M ;That's points M,D,A are collinear ...Q.E.D

[b]CCmath1[/b] your solution is very nice  :coolspeak:

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=904674#904674]An other solution[/url]  :D

Draw the parallel to $ BC$ tangent to $ \Gamma$, which cuts $ AB$ and $ AC$ at $ B'$ and $ C'$, respectively. ($ \Gamma$ is the incircle of $ AB'C'$). Let $ G$ be the point of tangency of $ B'C'$ and $ \Gamma$. Let $ P$ be the reflection of $ G$ over the midpoint $ M'$ of $ B'C'$. $ AP$ cuts $ \Gamma$ at $ P_1$ and $ P_2$(with $ P_1$ between $ A$ and $ P_2$). $ EF$ cuts $ AP$ and $ AG$ at $ P'$ and $ G'$ respectively. $ AG$ cuts $ \Gamma$ again at $ G_1$. Let $ AD$ cut $ B'C'$ at $ D'$.

It is known that $ P_1 \in GD$(consider the hometethy with center in $ A$ which sends the incircle to the excircle). Projecting from $ A$ gives $ (B',C';D',G) \equal{} (E,F;D,G')$ .Projecting from $ G$ gives $ (E,F;D,G') \equal{} (E,F;P_1,G_1)$. Projecting from $ A$ gives $ (E,F;P_1,G_1) \equal{} (E,F;P_2,G)$. Projecting from $ P_1$ gives $ (E,F;P_2,G) \equal{} (E,F;P',D)$. Finally, projection from $ A$ gives $ (E,F;P',D) \equal{} (B',C';P,D')$.

Then, we have $ (B',C';D',G) \equal{} (B',C';P,D')$. as $ G$ and $ P$ are simmetrical over $ M'$, we have $ \frac {C'G}{B'G} \equal{} \frac {B'P}{C'P}$.  Using both equalities we get $ \frac {B'D'}{C'D'} \equal{} \frac {C'D'}{B'D'}$. This can only happen if $ D' \equal{} M'$.

Then $ A$,$ D$ and $ M'$ are collinear, and the result follows.

This is an easy problem. I solved it using the same approach that CCMath1. The official solution is very messy, as it uses tons of trigonometry.

Was the official solution the one that calculates $ \angle CAD$ and $ \angle CAM$ trigonometrically and proves they are equal? After seeing that solution in an article I started looking for an alternate solution (which ended up being CCMath1's), since I thought the one with trigonometry was too ugly to be an official solution...


LaTeX
integration
algebra unsolved
algebra

Umm...
Continue in the other thread, writing on the board is better than this...
Oh, and use $\LaTeX$ please...

[quote="Rust"]$B=0.006,A=0.06$.[/quote]
Rust, can you tell my _any_ reason why you posted this¿
We are not the we-do-your-homework-for-you-forum!
Especially answering these kind of questions, and especially with answers only, is absolutely senseless!

Explain me please what's your intention in doing this. It helps nobody, it's simple showing off!
[b]STOP IT![/b]


A person takes a computer based test for getting a job. The test is indeed structured in 3 exams consisting in theoretical and applicative questions. The ratios between the theoretical questions and the applicative ones are 5/3 , 4/6 and 4/4 in those 3 exams, respectively.
The questions are chosen randomly, so a question can be repeated. If I chose five questions from  each of those 3 exams, what is the probability for the event A : From one exam we got 3 theoretical questions, and from the other two any combination of questions.

[quote="kenn4000"]
$ X_i \equal{}$ number of theoretical questions from test i

$ P(A) \equal{} P(X_1 \equal{} 3 or X_2 \equal{} 3 or X_3 \equal{} 3) \equal{} \sum_i P(X_i \equal{} 3)$
[/quote]

Are these events disjoint?
Shouldn't have been
$ P\left( {B_1  \equal{} 3 \cup B_2  \equal{} 3 \cup B_3  \equal{} 3} \right) \equal{} \sum\limits_i {P\left( {B_i  \equal{} 3} \right)}  \minus{} \sum\limits_{i < j} {P\left( {B_i  \equal{} 3 \cap B_j  \equal{} 3} \right)} \plus{} P\left( {B_1  \equal{} 3 \cap B_2  \equal{} 3 \cap B_3  \equal{} 3} \right)$
?


Let a,b,c be positive number such that a+b+c=1.
Find the minimum of
$ P\equal{}\frac{a}{1\plus{}b^2\plus{}c^2}\plus{}\frac{b}{1\plus{}c^2\plus{}a^2}\plus{}\frac{c}{1\plus{}a^2\plus{}b^2}$

[quote="8826"]Let a,b,c be positive number such that a+b+c=1.
Find the minimum of
$ P \equal{} \frac {a}{1 \plus{} b^2 \plus{} c^2} \plus{} \frac {b}{1 \plus{} c^2 \plus{} a^2} \plus{} \frac {c}{1 \plus{} a^2 \plus{} b^2}$[/quote]
This inequality by PHAM KIM HUNG :wink:
$ P_{min} \equal{} \frac {4}{5}$ when $ a \equal{} b \equal{} \frac {1}{2} ,c \equal{} 0$ and any permutations

no quykhtn-qa1,Ithink you mean $ P_{min}\equal{}\frac{4}{5}$, that's what I found me too, if you don't agree with me, write your solution please, this is my solution:
bu Cauchy-Schwarz,
$ S\equal{}P\left(a\plus{}b\plus{}c\plus{}\sum ab(a\plus{}b)\right)\geq (a\plus{}b\plus{}c)^2\equal{}1$
by Schur we can prove that, $ \sum ab(a\plus{}b)\leq a^3\plus{}b^3\plus{}c^3\plus{}3abc\leq a^3\plus{}b^3\plus{}c^3\plus{}6abc$
$ \Leftrightarrow 4\sum ab(a\plus{}b) \leq a^3\plus{}b^3\plus{}c^3\plus{}3\sum ab(a\plus{}b) \plus{}6abc \equal{} (a\plus{}b\plus{}c)^3\equal{}1$
$ \Leftrightarrow \sum ab(a\plus{}b)\leq \frac{1}{4}$
then $ P\left(1\plus{}\frac{1}{4}\right)\geq S \geq 1$
and finally, $ P_{min}\equal{} \frac{4}{5}$

[quote="peine"]no quykhtn-qa1,Ithink you mean $ P_{min} \equal{} \frac {4}{5}$, that's what I found me too, if you don't agree with me, write your solution please, this is my solution:
bu Cauchy-Schwarz,
$ S \equal{} P\left(a \plus{} b \plus{} c \plus{} \sum ab(a \plus{} b)\right)\geq (a \plus{} b \plus{} c)^2 \equal{} 1$
by Schur we can prove that, $ \sum ab(a \plus{} b)\leq a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc\leq a^3 \plus{} b^3 \plus{} c^3 \plus{} 6abc$
$ \Leftrightarrow 4\sum ab(a \plus{} b) \leq a^3 \plus{} b^3 \plus{} c^3 \plus{} 3\sum ab(a \plus{} b) \plus{} 6abc \equal{} (a \plus{} b \plus{} c)^3 \equal{} 1$
$ \Leftrightarrow \sum ab(a \plus{} b)\leq \frac {1}{4}$
then $ P\left(1 \plus{} \frac {1}{4}\right)\geq S \geq 1$
and finally, $ P_{min} \equal{} \frac {4}{5}$[/quote]
Yes,your solution is true.My solution is same you

okay, what about $ P_{max}$ can you find it?

Quykhtn and peine,You can see the prolem,"a,b,c is positive"  :) .I think P don't have min  :wink:

why?the expression have $ min$, if you don't agree with me, give me a value for $ a,b,c$ where we can find $ P<\frac{4}{5}$

[hide]and I think we can also find $ max$[/hide]

[b]secrets[/b] is right. Since a,b and c are positive none of the could be equal to 0. So we can't reach $ \frac {4}{5}$. P could approach infinitely many to $ \frac {4}{5}$ but it will never reach it.

[quote="secrets"]Quykhtn and peine,You can see the problem,"a,b,c is positive"  :) .I think P don't have min  :wink:[/quote]
[quote="peine"] $ \sum ab(a\plus{}b)\leq a^3\plus{}b^3\plus{}c^3\plus{}3abc\leq a^3\plus{}b^3\plus{}c^3\plus{}6abc$[/quote]
Secrets is right, as $ a, b, c$ are STRICTLY POSITIVE, peine's this step quoted above is not right as we should have $ a^3\plus{}b^3\plus{}c^3\plus{}3abc<a^3\plus{}b^3\plus{}c^3\plus{}6abc$.

In case of NONNEGATVE reals, both peine and Quykhtn are right.
@quy: in 2nd post, you said that equality occurs iff $ a\equal{}b\equal{}0, c\equal{}\frac{1}{2}$ and it's cyclic permutations: how come $ a\equal{}b\equal{}0$ if $ a, b>0$? :oops:

So this is the correct argument, $ a, b, c\ge 0$ will be the correct condition in the original problem.

I hope both of you understand what I mean. :blush:

@peine: positive does not mean non-negative.

[quote="peine"]hello potla! in the original problem, [b]8826 [/b] writes: "Let $ a,b,c$ be positive number" not :"Let $ a,b,c$ be strictly positive number". then $ a,b,c$ can be equals as $ 0$. that's why my solution and [b]quykhtn-qa1[/b]'s one are true.[/quote] 0 isn't positive number. That's what all of us want tell you.

mmm... I inderstand now, thank's for you all.


probability
group theory
abstract algebra
superior algebra
superior algebra unsolved

prove $Pr(<\sigma_1,\sigma_2> = A_n or S_n)\rightarrow 1$ as $n \rightarrow \infty.$

Can you explain some symbol that you used in your problem?

$Pr(X)$: probability of $X$.
$<a_1,a_2,...a_n>$: subgroup generated by $a_1 , a_2 , ... , a_n$.
$S_n$: group of permutations on $n$ elements.
$A_n$: alternating subgroup of $S_n$, the even permutations.
$1$: the $1$ in the field $\mathbb{R}$ of real numbers, the element fulfilling $1 \cdot x = x \cdot 1 = x$.
$=$: the symbol for identity.
$or$: $X$ $or$ $Y$ means that at least one of them is true.

this is a theorem of Dixon from the sixties.


superior algebra
superior algebra unsolved

Show that if a∈R (which is ring) is nilpotent, then a is quasi-regular. Further, if R has a unity, then 1-a is invertible.

Assuming that the definition is "$ a$ is [u]quasi-regular[/u] iff there is some $ b\neq 0$ with $ a \plus{} b \equal{} ab \equal{} ba$" then $ b \equal{} \minus{} \sum_{n\ge 1}a^n$ could be a good candidate. Is it non-zero? ... but I am not very sure if this is your definition.

For $ 1 \minus{} a\in R^*$ think of the geometric series.

$ b\neq 0$ if $ a\neq 0$ since $ b(1\minus{}a)\equal{}\minus{}a$ (using geometric series again).


trigonometry
ratio
geometric series
complex numbers

Given that $ w_n \equal{} 3^ {\minus{}n}cos2n\theta$ for $ n \equal{} 1,2,3...$ use De Moivre's Theorem to show that

$ 1 \plus{} w_1 \plus{} w_2 \plus{} w_3 \plus{} ... \plus{} w_{N \minus{} 1} \equal{} \frac {9 \minus{} 3cos2\theta \plus{} 3^{ \minus{} N \plus{} 1}cos2(N \minus{} 1)\theta \minus{} 3^{N \plus{} 2}cos2N\theta}{10 \minus{} 6cos2\theta}$

I am hopelessly lost on this question.
Help is greatly appreciated.
Thanks in advance.

$ \sum_{n\equal{}0}^{N\minus{}1}3^{\minus{}n}\cos 2n\theta\equal{}\text{Re}\sum_{n\equal{}0}^{N\minus{}1}3^{\minus{}n}e^{2n\theta i}$

Note that $ 3^{\minus{}n}e^{2n\theta i}\equal{}\left(\frac{e^{2\theta i}}{3}\right)^{n}$, making the sum geometric. Go from there.

I don't know how right this is or how good an approach this is; it's just the first thing I thought of, and I'm feeling way too tired at the moment to work through it :roll:

The first term of the geometric series is $ a = 1$, the common ratio is $ r = \frac {e^{2i\theta}}{3}$, so the sum to $ n$ terms is

${ \frac {1 - \frac {e^{2iN\theta}}{3^{N}}}{1 - \frac {e^{2i\theta}}{3}}}$

$ = \frac {1}{3^{N - 1}} \frac {(3^{N} - \cos 2N\theta - i \sin 2N\theta)}{(3 - \cos 2\theta - i \sin 2\theta)}$

$ = \frac {1}{3^{N}} \frac {(3^{N} - \cos 2N\theta - i\sin 2N\theta)}{(3 - \cos 2\theta - i\sin 2\theta)}\frac {(3 - \cos 2\theta + i \sin 2\theta)}{(3 - \cos 2\theta + i \sin 2\theta)}$.

We want the real part of this which is

$ \frac {1}{3^{N - 1}} \frac {3^{N + 1} - 3^{N} \cos 2\theta - 3 \cos 2N\theta + \cos 2N\theta \cos 2\theta + \sin 2N\theta \sin 2\theta}{9 - 6\cos 2\theta + \cos^{2}\theta + \sin^{2}\theta}$.

$ = \frac {9 - 3\cos 2\theta - 3^{ - N + 2}\cos 2N\theta + 3^{ - N + 1}\cos 2(N - 1)\theta}{10 - 6\cos 2\theta}$

Hope this helps.


Hey! Anyone here know books/websites/magazines where I can find a lot of information about Homo Neanderthalensis (Neanderthal Man)?

;) I'd say a magazine would be [u]Dig[/u],it is an archaeology magazine that I don't read alot though I see the cover all the time and it looks like it almost specializes in Homo neanderthalensis.
I'd try google,msn or some other search engine for websites.

Funcia

[i]Homo neanderthalensis[/i], actually were featured on Discovery's latest online article:

[quote]Neanderthal Bone Yields Complete Mitochondrial Genome
Jennifer Viegas, Discovery News
Aug. 7, 2008 -- DNA extracted from a 38,000-year-old Neanderthal bone has just enabled scientists to sequence the complete mitochondrial genome for the human-like species, according to a paper that will be published tomorrow in the journal Cell.

The remarkable feat, which has led to at least three major discoveries about the extinct stocky European individuals, represents a breakthrough for studies on the human family.

"This is the first complete mitochondrial genome sequence from an extinct hominid," lead author Richard Green explained to Discovery News.

Mitochondria, which an individual inherits from his or her mother, are cellular powerhouses that possess their own DNA and include 13 protein-coding genes. The researchers sequenced the Neanderthal mitochondria 35 times to ensure their findings were as accurate as possible.

After studying the newly completed genome, Green, a researcher at the Max-Planck Institute for Evolutionary Anthropology in Germany, and his team first concluded that the Neanderthal mitochondria falls outside the range of variation found in humans today, offering no evidence that interbreeding occurred between them and us.

The researchers are quick to add that such interbreeding could still have happened and that the Neanderthals' "exact relationship with modern humans remains a topic of debate."

Clearer is the fact that Neanderthals and humans split from a common ancestor around 660,000 years ago. The researchers based this initially upon prior research that determined humans and chimpanzees diverged from each other six to eight million years ago.

They calculated mtDNA sequence changes for both humans and Neanderthals since that time. These accumulated changes then "let us calculate how long ago was the most recent common ancestor of humans and Neanderthals," Green said.

He added, "This common ancestor likely looked something like Homo erectus." This extinct hominid is believed to have been super strong with a relatively large head and brain.

What most surprised the scientists was how little purification acted upon the Neanderthal's DNA, meaning that the elimination of slightly deleterious alleles, or variant gene forms, didn't occur very often within the population.

"One sensible explanation for this could be a very small effective population size," Green said, explaining that only a few thousand Neanderthals may have roamed Europe around 40,000 years ago, close to when they went extinct.

It's unclear if this was a general feature of the Neanderthal population, perhaps explained by the fact that they had to deal with repeated glaciations, or if some population bottleneck "happened late in the game," he said.

Perhaps the biggest modern human revelation to come out of the project is that there was an explosion of certain amino acid substitutions within the human genome after the Neanderthal/human split.

"What we can say is that there was a lot of change in a very short time within modern humans," Green said. "Further work will be necessary to say what the consequences of these changes were."

John Hawks, assistant professor of anthropology at the University of Wisconsin, Madison, told Discovery News that he has "been waiting a long time for this sequence to come out," adding that "all previously reported sequences of Neanderthal mtDNA" were fragmentary when compared to this one.

Geneticist David Reich at the Harvard Medical School also agrees that the newly sequenced genome "is exciting and important."

"The most striking thing about the paper is that it shows that the authors are able to get an extremely reliable DNA sequence out of a (38,000-year-old) Neanderthal fossil especially when they do a large amount of DNA sequencing," Reich told Discovery News, mentioning that it then "becomes obvious that the sequence the authors are obtaining is correct."

Green and his team are already at work on yet another Neanderthal genome project -- sequencing the complete Neanderthal nuclear genome -- that should be finished by the end of the year. It should answer, once and for all, whether or not modern humans and Neanderthals interbred to such a degree that the mixing would have resulted in a Neanderthal genetic contribution to the modern human gene pool.[/quote]
Source: [url]http://dsc.discovery.com/news/2008/08/07/neanderthal-dna.html[/url]

Other links for information on [i]homo neanderthalensis [/i]include:
[url]http://en.wikipedia.org/wiki/Neanderthal[/url]
[url]http://anthropology.si.edu/HumanOrigins/ha/neand.htm[/url]
[url]http://www.archaeologyinfo.com/homoneaderthalensis.htm[/url]

Google "homo neanderthalensis" to get more websites.

it wud b better if u ask some doubt/particular pt
otherwise its a wide topic.......


Suppose that $ a,b,c,d$ are real numbers such that -
\begin{eqnarray*}
a^2+b^2=1\\
c^2+d^2=1\\
ac+bd=0
\end{eqnarray*}

then show that -

\begin{eqnarray*}
a^2+c^2=1\\
b^2+d^2=1\\
ab+cd=0
\end{eqnarray*}

[hide="Straight Linear Algebra"]Define $ A \equal{} \binom{a \; \; \; c}{b \; \; \; d}$.

Note that the columns of $ A$ form an orthonormal basis for $ \mathbb{R}^2$.

Note that $ A^{ \minus{} 1}$ must also then be orthonormal, so $ A^{ \minus{} 1} \equal{} A^T \equal{} \binom{a \; \; \; b}{c \; \; \; d}$, and the columns of this matrix form another orthonormal basis for $ \mathbb{R}^2$.

[i]* - $ A^{\minus{}1} \equal{} A^T$ for all orthonormal matrices.
[/i][/hide]

i am sorry but is school expected to know that? if so i dont  :(
where do i learn such stuff.. ?

That stuff is part of linear algebra, which is a college-level topic (although easier to learn than subjects like calculus).

You can do trig substitutions as follows: $ a \equal{} \sin{x}, b \equal{} \cos{x}, c \equal{} \sin{y}, d \equal{} \cos{y}$ for $ 0 < x,y < \frac{\pi}{2}$ as a consequence of the first two equations.

$ \sin{x}\sin{y} \plus{} \cos{x}\cos{y} \equal{} 0\implies \cos(x \minus{} y) \equal{} 0\implies x \minus{} y \equal{} \frac{\pi}{2}\implies x \equal{} y \plus{} \frac{\pi}{2}$.

Using the relationship between $ x$ and $ y$, it becomes fairly easy to show those equations.


function
AMC
USA(J)MO
USAMO
algebra unsolved
algebra

Prove that for non negative reals $x,y$ we have:
$[5x]+[5y] \geq [3x+y]+[3y+x]+[x]+[y]$

It is USAMO 1975.
http://www.kalva.demon.co.uk/usa/usoln/usol751.html


graph theory
combinatorics unsolved
combinatorics

in a tournament each of 2n+1 teams plays with any other exactly once and there are no ties.we say that x,y,z is a cyclic triple if x beats y ,y beats z and z beats x.find the maximum number of cyclic triples.

Let $M$ be the maximum number of cyclic triples
Construct the complete graph with a vertex $i$ for team $i$ and a directed edge from vertex $i$ to vertex $j$ if team $i$ beat team $j$
Then $M$ is the maximum number of cycles of length $3$

Let vertex $i$ have in-degree $a_{i}$ and out-degree $2n-a_{i}$
Consider any two in-edges or two out-edges from vertex $i$
It eliminates exactly one possible cyclic triple
Furthermore each possible cyclic triple is eliminated exactly $2$ times
Thus exactly $\frac{1}{2}\sum_{i=1}^{2n+1}{\left(\binom{a_{i}}{2}+\binom{2n-a_{i}}{2}\right)}$ possible cyclic triples are eliminated
Since $\binom{x}{2}+\binom{2n-x}{2}= (x-n)^{2}+n^{2}-n \geq n^{2}-n$ with equality at $x=n$,
$M \leq \binom{2n+1}{3}-\frac{(2n+1)(n^{2}-n)}{2}= \frac{n(n+1)(2n+1)}{6}$

We have the following construction
Let there be a directed edge from vertex $i$ to vertex $(i+j)$ for all $i$ and all $1 \leq j \leq n$ where vertex $x+(2n+1)$ is defined as vertex $x$
Consider each directed edge $(i,i+j)$
There is a directed edge $(i+j,i+j+k)$ for any $1 \leq k \leq n$
There is a directed edge $(i+j+k,i)$ iff $j+k > n$ (note that $j+k < 2n+1$)
So there are exactly $\sum_{j=1}^{n}{(n+1-j)}= \frac{n(n+1)}{2}$ cyclic triples for each vertex
Also each cyclic triple is counted exactly $3$ times
Thus there are exactly $\frac{1}{3}(2n+1)\frac{n(n+1)}{2}= \frac{n(n+1)(2n+1)}{6}$

Therefore $M = \frac{n(n+1)(2n+1)}{6}$


[quote="xzlbq"]Let $ a,b,c$ be real positive numbers  and $ a \plus{} b \plus{} c \equal{} 1$, Prove  that

$ \frac {9}{2}(2 \minus{} a)(2 \minus{} b)(2 \minus{} c) \geq (3 \minus{} a)(3 \minus{} b)(3 \minus{} c) \plus{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)$

$ \geq \frac {104}{25}(2 \minus{} a)(2 \minus{} b)(2 \minus{} c)$

BQ[/quote]
$ (\frac {5}{6}\sqrt {6} \plus{} \frac {1}{6}\sqrt {2})\sqrt {(2 \minus{} a)(2 \minus{} b)(2 \minus{} c)}$

$ \geq \sqrt {(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)} \plus{} \sqrt {(3 \minus{} a)(3 \minus{} b)(3 \minus{} c)}$

$ \geq \frac {3}{2}\sqrt {2}\sqrt {(2 \minus{} a)(2 \minus{} b)(2 \minus{} c)}$
BQ


I want to ask you the detail solution of these problems:
$ 1)$ Let $ k,n$ be positive interger numbers. Denote $ A \equal{} \{a\in \mathbb {N}*|a\le n, k|a\}$. Find $ |A|$.
$ 2)$ Given $ n$ a positive interger real number, and $ k$ numbers $ a_1,a_2,...,a_k$ such that $ (a_i,a_j) \equal{} 1$ for all $ i$ different from $ j$, $ k$ is a postitive interger number. Denote $ A \equal{} \{a\in \mathbb {N}*|a\le n$,$ a$ cannot divided by any $ a_i\}$. Find $ |A|$.
$ 3)$ Let $ n$ be positive interger number. Let $ A$ be the set of numbers $ a$ such that $ a\le n$ and $ (a,n) \equal{} 1$. Find $ |A|$.


trigonometry
calculus
derivative
ratio
geometry
trig identities
Law of Sines

1. Find the maximum value of $ \frac{a\plus{}x}{\sqrt{b^2\plus{}x^2}}$ for real values of $ x$.

1. $ x \equal{} b\tan \theta \left( \minus{} \frac {\pi}{2} < \theta < \frac {\pi}{2}\right)$.

2. Cauchy Schwarz

3. Set $ \left(\frac {a \plus{} x}{\sqrt{b^2 \plus{} x^2}}\right)^2 \equal{} k$ to use Discriminant.

4. Differentiation

[hide="Hint"]Law of Sines[/hide]

[hide="Geometric Argument"]Consider the case of positive $ a$ and positive $ x$.

Let $ O \equal{} (0,0)$, $ B \equal{} (b,0)$, and $ A \equal{} (b,\minus{}a)$ be fixed, and let $ C \equal{} (b,x)$ vary with $ x$. We want to maximize the ratio $ \frac{AC}{OC}$ in $ \triangle OAC$.

Let $ \theta \equal{} \text{m} \angle OAC$ be fixed, and let $ \phi \equal{} \text{m} \angle AOC$ vary. We know $ \frac{AC}{OC} \equal{} \frac{\sin \phi}{\sin \theta}$, which is maximized when $ \sin \phi \equal{} 1 \implies \angle AOC \; \text{is right}$.

Setting $ x$ to the optimal value, we now have, by similar right triangles, $ \frac{AC}{OC} \equal{} \frac{OA}{OB} \equal{} \frac{\sqrt{a^2\plus{}b^2}}{b} \equal{} \boxed{\sqrt{1 \plus{} \left( \frac{a}{b} \right)^2}}$

[i]Left to do is to consider the case of negative $ a$ and negative $ x$, which are equivalent (why?), and finally, the case of both negative, which is actually very similar to the given situation.[/i][/hide]


calculus
integration
inequalities
function
invariant
real analysis
real analysis unsolved

Let $ P\subseteq \mathbb{R}^m$ be a non-empty compact convex set and $ f: P\rightarrow \mathbb{R}_{ \plus{} }$ be a concave function. Prove, that for every $ \xi\in \mathbb{R}^m$
\[ \int_{P}\langle \xi,x \rangle f(x)dx\leq \left[\frac {m \plus{} 1}{m \plus{} 2}\sup_{x\in P}{\langle\xi,x\rangle} \plus{} \frac {1}{m \plus{} 2}\inf_{x\in P}{\langle\xi,x\rangle}\right] \cdot\int_{P}f(x)dx.\]

We can assume $\xi$ is an unit vector. Since the inequality is invariant under the transformations $x\to x-a\,,\, x\to ax$, we also can assume WLOG $0\in P$, $P$ lies between two supporting hyperplanes: $\langle \xi, x \rangle =0$ and $\langle \xi, x \rangle=1$. Thus, it's enough to prove:
\[\int_{P}\langle\xi,x\rangle f(x)dx\leq \frac{m+1}{m+2}\sup_{x\in P}{\langle\xi,x\rangle}\cdot\int_{P}f(x)\,dx  \,\,\,\,\,\,\, \text{(1)}\]
For any $t, t\in [0,1]$, denote $S_t=\{x\in \mathbb{R}^m \mid x\in P, \langle \xi,x\rangle =t\}$
Let's set $\phi(t)=t^{1-m}\int_{S_t} f(x)\, dx\,,\, t\in(0,1]$. We claim that $\phi$ satisfies the following property:
\[ \phi(t_1 t)\ge t_1 \phi(t)\,,\forall t, t_1\,,  0< t,t_1 \le 1 \,\,\,\,\,\,\,\,\, \text{(2)} \]
Indeed, changing variable $u=t_1 x$, using $f$ is a concave and positive, and $ t_1 S_t \subset S_{t_1 t}$ (since $P$ is convex) implies $\text{(2)}$. Thus we have:
\[ \int_{S_t} f(x)\,dx = t^{m-1}\phi(t) \,\,\,\,\,\,\, \text{(3)} \]
Now, we can rewrite $\text{(1)}$ in the form:
\[\int_{0}^1 t \int_{S_t} f(x)\,dx \, dt \le \frac{m+1}{m+2} \int_{0}^1 \int_{S_t} f(x)\,dx \, dt \]
Plugging $\text{(3)}$ into the above inequality, it's enough to prove:
\[ \int_0^1 t^m \phi(t)\,dt \leq  \frac{m+1}{m+2} \int_0^1 t^{m-1} \phi(t) \,dt \,\,\,\,\,\,\, \text{(4)}\]
Let us fix $\alpha, 0<\alpha<1$. Using $\text{(2)}$, we consecutively obtain:
\[ \int_0^1 t^m \phi(t)\,dt \le \int_0^1 t^{m-\alpha} \phi(t^{\alpha+1})\,dt =\]
\[ =\int_0^1 u^{(m-\alpha)/(1+\alpha)}\phi(u)(1+\alpha)^{-1}u^{-\alpha/(1+\alpha)}\,du =\]
\[=(1+\alpha)^{-1} \int_0^1 u^{(m-2\alpha)/(1+\alpha)}\phi(u)\,du \]
We want $(m-2\alpha)/(1+\alpha)$ to be equal to $m-1$, so we set $\alpha=(m+1)^{-1}$. Thus, it yields $\text{(4)}$ and we are done.


Show that 2008 divides an integer K of the form 11...10...00.

Ok, thanks, that helps.  How do you know this, though?

[url=http://mathworld.wolfram.com/EulersTotientTheorem.html]Euler's Totient Theorem[/url].  Also consider the following much simpler argument:

Consider the sequence $ 1, 11, 111, ... \bmod 2008$.  Among the first $ 2009$ terms at least two must have the same remainder $ \bmod 2008$ by Pigeonhole.  Their difference is in the desired form and is divisible by $ 2008$.

[quote="t0rajir0u"]Consider the sequence $ 1, 11, 111, ... \bmod 2008$.  Among the first $ 2009$ terms at least two must have the same remainder $ \bmod 2008$ by Pigeonhole.  Their difference is in the desired form and is divisible by $ 2008$.[/quote]

 :clap2:  :first:


geometry algebra function domain Support

The following was taken from the Los Angeles Times:
[quote]
The Supreme Court gave local governments broad power Thursday to bulldoze homes and other private property to make way for business development, a ruling that could encourage more city-backed plans to replace small stores with big-box retailers.
The 5-4 ruling upheld a plan by officials in a coastal Connecticut town to condemn nine homes of longtime residents that would be replaced with an office complex and a marina.
The dispute between the homeowners and the city officials became a classic test of government power versus individual rights. It pitted a community's hopes for economic rebirth against an individual's right to keep one's home.
Economic development won.
The Supreme Court's opinion goes further than before in allowing the government to invoke its "eminent domain" and to seize private property from unwilling sellers.
The Constitution says government may take private property "for public use" if it pays the owners "just compensation." Originally, public use meant the land was used for roads, canals or military bases. In the 19th century, railroads were permitted to take private lands because they served the public.
In 1954, the court said officials could condemn homes and stores in "blighted" areas as part of a redevelopment plan.
In Thursday's decision, the court went a step further and said officials need not claim they are condemning blighted properties or clearing slums. Now, as long as officials hope to create jobs or raise tax collections, they can seize the homes of unwilling sellers, the court said. This "public purpose" is a "public use" of the land, the court said in Kelo v. New London.
"Promoting economic development is a traditional and long accepted function of government," Justice John Paul Stevens said.
Justices Anthony M. Kennedy, David Souter, Ruth Bader Ginsburg and Stephen Breyer joined him.
The dissenters said the court was ignoring the basic rights to private property written into the Constitution.
Justice Sandra Day O'Connor said all property was now potentially subject to seizure.
"Nothing is to prevent the state from replacing any Motel 6 with a Ritz-Carlton, any home with a shopping mall, or any farm with a factory," she said.
[/quote]

Comments, please. I would say more, but...  *ahem* :censored:

I'm kind of torn on this decision.  While I don't think the ruling contradicts the Constitution (and I applaud the justices for not taking a strict-constructionist view in this case), it sounds like this ruling could in theory be used to justify some pretty outrageous acts by local governments.  I doubt, however, that such acts will be very widespread; I have more faith than that in local governments.

I personally would have opted for some sort of test (like the "Lemon test" and others the Court has established in previous rulings) to determine whether or not a given seizure of private property is "for public use" to a great enough extent to make it legal.

Sounds like socialism to me. What I understand is that the governement can take away private lands in order to help the public for the greater good. Is the government that desperate to create jobs?

Basically, the government is taking PRIVATE lands and giving them to other PRIVATE owners. This is NOT public land use, folks. This land is owned by PRIVATE sanctions. BAD. Bad bad bad bad bad.

In response to Sly Si, you have trust in your local government? This is a rare thing. The local governments I am used to are extremely untrustworthy. Let me cite a couple of examples for you:
1. A local government decides to build houses along a river. Problem? Where they want to build these houses is upriver of a very valuable water supply. Their proposed building plans would greatly damage the river, which is bad not only from an environmental standpoint, but from an economic one as well, as the water supply would be even more polluted, meaning more money would have to go into filtering, etc. Reason? More houses=>more tax money for the city=>bigger paychecks for city officials.
2. A classic example of local government gone mad is in Africa. Many, many of the local governments in Africa are so corrupt that it makes the United States look pristine by comparison. For instance, in Africa (as well as in the U.S., in some areas) officials pay family members or friends to perform tasks for the city, such as building roads, regardless of price or quality of previous work. Now, in these U.S. cities, the workers will merely do a shoddy job at the task, well under the price paid. In Africa, nothing will happen. The roads will remain unpaved, but thousands of public dollars will be spent for [i]nothing[/i]. Realize that I say "Africa" very generally. I doubt that this is typical of the entire continent, although I am aware that it is very much the norm in many countries.

These are just two accounts that I can give you off of the top of my head. Still trust your local government? Want to rethink? I could probably give you more examples if you requested them, with time and research.

Well, I live in Connecticut so I've been hearing about this on the news a lot.

I think it's wrong for the government to take the land away from the owners even though they were unwilling to sell. They should have bought the land, instead of just putting a price tag on it and giving the owners the money. It's like saying anything in America can be bought by the government if they put a fair price on it, whether it's for sale or not.

Although, I see why they're doing this. New London's a pretty cruddy city.

Excuse me, but how exactly do you think all your nice highways (that let you Conneticuters work in New York City but live far away) got to be there?  Eminent domain has always been something that governments can do: it is a tool that allows the good of the majority to preside without being thwarted by single stubborn individuals.  Those who have their property confiscated are paid full market value for it (which, in the current real estate climate, is probably far more than they would be paid a couple of years from now).  The justices made a really good call in not saying that the right to private property is an absolute that overrules everything else.

Now, any system designed to overrule a minority interest is prone to abuse.  However, there are a good number of safeguards in place in this situation.  (Kennedy's concurring opinion covered this ground fairly well, I think.)  In bleumoose's first scenario, for example, state and perhaps federal environmental regulations would limit the potential for abuse.  State courts can still excercize oversight powers.  Number 2 has nothing whatsoever to do with eminent domain, so I don't really know what to say about it.

The worry I do share is in the giving of private lands into other private hands -- it troubles me, even with potential oversight.  The other alternatives, however, are far worse: forbidding the government to ever take an eminent domain action would make infrastructure creation nearly impossible.  There has probably never been a highway, subway system or sewer/sewage system that didn't require it.  Without it, NIMBY becomes an even more impenetrable force.  Not cool at all.

I don't think you understand "other alternatives" quite well enough:

I do not have a problem with eminent domain(I think we are going to be using this a lot, so I'm going to shorten it to ED, k?) when it is going for public works (in most cases). You, also, do not see a problem with this. Your examples are all of ED being used for TRUE public works: highways, and the like. The problem I have with Thursday's decision is the way they defined "public." They defined the use of land to benefit INDIVIDUALS as public. Granted, some jobs will be gained by this. SOME tax money will be raised. However, the majority of the money made from this endevor will go to a very very small group of very very rich people. So, instead of X acres being used to make Y amount of money for one family, X acres is being used to make 100Y amount of money for a different individual, plus .3Y amount of money for the community. I don't find this fair at all. The problem is that someone with 10Y amount of money [b]will be[/b] more influential in the community than the one that is making Y amount. Therefore, 10Y can... convince government officials that using ED to change the X acres into a strip mall is a good idea, and there is very little that Y can do about it. Money talks, guys. That's just all there is to it.

In response to bleumoose:
I completely agree with your second post.  I think it is completely wrong for the government to use ED to take private property for individual benefit.  However, I think that in most cases, local governments (at least in the U.S.) will not use this ruling to do anything like that.  I think they will use their discretion to determine what uses of land carry enough benefits to the general public to justify use of ED.  However, the problem arises when the local government has poor judgement; that's where state court oversight comes in.  And like I said, I would have preferred that the Supreme Court set some standards rather than just trusting local governments with a blank check.

As for my faith in government, I can't really justify it.  It's probably a product of my (liberal) political ideology; I tend to believe that government officials generally have good intentions.  It may be that I am rather naive in these matters.

bleumoose:
Okay, then, we both agree that it is rather creepy that the government could take my property and give it to someone else, that there is (substantial) potential for abuse, and that such a redistribution would be "unfair."

However, "unfair" is not equivalent to "bad."  I could suggest a significant number of "unfair" things that would be, for the public as a whole, quite good.  (Indeed, most of society agrees that a number of popular social programs are "good," while Republicans (evidently a majority, at least among certain groups like voters) frequently complain that the taxes required to fund them are "unfair" as they fall most heavily on the wealthy.)  My political beliefs make it difficult for me to think that it is "best" for the government to involve itself in such an ED taking/handover, but that doesn't necessarily mean that it will always (or even usually) be "bad" for government to orchestrate such a deal.

Frankly, if I were in government and I wanted to do someone a favor, an ED proceeding does not at all seem like the most likely approach I would take.  After all, that subjects me to a great deal of scrutiny (as the person or body orchestrating the ED).  Far easier to simply give people large contracts to do nothing, for example, or "suggest" that lobbying firms should hire all my friends in order to get a piece of the action.  There are far more simple and subtle methods of corruption than a proceeding likely to result in a significant (and expensive) court battle.

ok, this question may seem completely irrelevant, but does anyone here have strong irish heritage?

I don't know what exactly you're referring to, but I can pre-empt it: the abuse of a system does not necessarily indicate that the system itself is bankrupt.  Corporate income tax shelters do not mean we should scrap the corporate income tax;  insurance fraud does not mean we should scrap insurance.  Even the eggregious abuse of a system (say, prisons in Nazi Germany and Soviet Union) merely indicate that we should make sure our prisons are not subject to such abuse, via oversight and not setting up prisons in foreign countries run by the CIA.  It isn't a condemnation of the entire concept of a prison.  (I can't make the appropriate analogy for your comment because I don't know what you're talking about exactly.  I hardly think the treatment of a conquered nation is a good proxy for what elected governments do to/for their own citizens.)

Heh, you're forcing me to sound much more pro-ED than I really am.  I guess that's what I get for arguing too much :-)

actually, i was attempting to pull at someone's emotionals rather than their logicals, which has gotten me into trouble on the rare occasions i try. its rather complicated to explain if you don't already have a sense of what i am talking about, when i refer to irish blood. there's something about having a family that has had strong ties to the land for a long time that makes land- one's OWN land- so precious, that i don't think you suburbanites can understand. the idea of taking something like that from someone and making it into just another subdivision, another stripmall, is something... nuaseating.

but i digress. i think i should offer you a few more examples of ED being used inappropriately. i don't think many of you understand how often it happens.
1. this case, itself. do you realize that the people that lived there... some of them had lived there their ENTIRE LIVES? try and cope with that losing that much of your identity for a moment... a place you had gone home to for 87 years, gone.
2. similar case, different city. low-cost housing is condemned to build a super walmart. LOW-COST HOUSING. many of these people have NOWHERE to go. nowhere. they were turned out because their tax money could not compete with the revenue provided by a super walmart. is this fair? is this justice? is this RIGHT? turning people who make very little out of their own homes just so people that make a lot can make MORE?

you keep saying "what is best for the majority." THIS IS NOT WHAT IS BEST FOR THE MAJORITY. a new walmart is put up. the small businesses in the neighborhood suffer. the people whose houses were condemned suffer. the community, which now has to provide public housing for these newly homeless people, suffers. the quality of life in another small town, america, suffers, as they become generic and connected to mass U.S. sure, it may be better for the stockholders in their cozy little homes in their cozy little subdivisions, but as a weighted percentage, is doing a small portion of the population a GROSS injustice while doing another (rather small) portion of the population a service.... is that "best for the community"? i can harldy see how this could be a good thing.

So, first of all, I am a proud resident of New York City, so your complaints about suburbanites in an argument with me are unwarranted (and indeed, a bit insulting :) ).  And I have as strong a tie to my city, my home and my neighborhood as anyone I know.

Now, for your objections:
1) Yes, fine, it's a shame for those 9 people.  I am, however, generally willing to trade the memories of nine people (since, after all, they aren't losing anything economically, it is their physical memories) for real jobs and recreation for the people of New London.  Is that crass and uncaring of me?  Perhaps, but I think that it is the correct choice.  (There is an implicit assumption on my part here that the development will actually be productive, one that I cannot in fact prove.)
2) Once again: abuses do not necessarily invalidate an entire system.  Argument by anecdote is very emotionally appealing but proves nothing.  Indeed, it would be required to compare all your anecdotes to those in which a little well-applied ED led to the revival of a town or a successful municipal rebuilding.  Frankly, a ruling in the opposite direction would have been a victory primarily for those who already have.

Your last paragraph, in particular, is irrelevant: there is no Walmart going up in New London, there will be an office tower and a marina.  Now, we can argue whether municipalities make good decisions about what they do with their ED power -- I agree with you that in many cases, they do not, especially when they pass ownership into private hands.  But that wasn't what this case was about.  To rule the way you wanted, the SC would have had to have held one of two things: (1) that it is [i]never[/i] in the public benefit (I think public use is actually the relevant phrase, but whatever) to use ED to benefit a private individual, or (2) that it is always appropriate to bring cases about this to federal courts, which then have the discretion to decide whether or not a proposed project is for public use or benefit.  Frankly, I see neither of these as desirable outcomes, especially when each potential project could be evaluated by local authorities, state authorities and state courts.

Sorry if I insulted you by throwing out the word "suburbanite" but as a city-dweller, you still don't really understand the true value of land.

And I don't understand how arguing by example doesn't prove anything. You say you trust your government. I say your trust is misplaced and give you examples. You say "but if it is truly well-placed" and I say that will rarely happen because the government has proven itself untrustworthy. A few well-placed EDs do not justify 50 misused ones. You don't take a chocolate bar from the youngest kid on the playground, give him a penny for it, and then hand it off to the biggest person there. It's wrong. It's morally unjust. I don't see why your opinions on what is good for the land is better than what the owner's opinion on what is good for the land.

First, wrt "land," perhaps you just believe there is more value in dirt than there really is.  I see value in dirt, to the extent that it contains minerals that are necessary for life, supports plants and the general ecosystem, and provides homes, livelyhood or recreation to people and other life forms.  I am willing, under appropriate conditions, to switch among those various uses.  The sentimental value of land is very much not high on my list, for many reasons.  I don't particularly see why the sentimental feelings of 9 individuals should be able to override the potential for jobs and recreation for dozens or hundreds.

Listen: some people on Long Island became convinced that high-voltage electric lines and a variety of environmental pollutants were causing a "breast cancer bubble" in the area.  They found dozens of women exposed to pollutants and high voltage power over the long term.  A $30 million federal study was conducted: the result was no correlation at all, and a finding that the "bubble" wasn't statistically meaningful.  That's why anecdotal evidence is no good.  I have no way of knowing whether your anecdotes represent a good sampling of projects of this sort.  All it does is supports my (pre-existing) belief that this isn't always a good thing to do.  It does not support the proposition that this is [i]never[/i] a good thing to do, and that's the other position available here.

If I take a candy bar from a kid (little, big or otherwise, thank you), and give him the market value of his candy bar, I have not done him any economic harm.  Depending what I do with that candy bar, I may or may not do quite a bit of good.  If you want to challenge every single ED procedure of this sort: go for it.  You can organize politically, you can bring state court cases, you can send letters to your politicians (local or state), and you can try to vote them out.  You still have ways of fighting this; what you've lost is the ability to fight it simply because of the particular nature of it, without regards to what is actually going to be done with the land.

Well, let's see... how will the justices take... THIS?

http://news.yahoo.com/s/ap/20050629/ap_on_re_us/souter_property

HATE it. </sarcasm>

I think this is a very difficult decision to defend on the grounds 'the government knows what's the best use of the land.'  Already this decision is spawning all sorts of government infringement on individuals.  Witness [url=http://www.ij.org/private_property/connecticut/6_29_05pr.html]these[/url].  Simply put, seizing land from one private entity and handing it to another private entity for 'the good of the people' is tyrrany, even if the intentions are noble (which, of course, they almost always will not be - we out in San Diego can testify to the corruption of government all day long; those in CT (home of the Kelo case) last year booted their governor due to the type of corruption that Kelo will make rampant).

Moreover, there's not much the government does better than private industry.  This site exists in part due to this simple fact.

This is a dangerous decision, but you won't hear too many Republicans decry it too loudly, as their champion likely made a great deal of his personal wealth in part due to similar usage of eminent domain.

thanks richard, for being so eloquent when i find it hard to be.

[quote="JBL"]
If I take a candy bar from a kid (little, big or otherwise, thank you), and give him the market value of his candy bar, I have not done him any economic harm. [/quote]

That's not necessarily true.  In Norwood OH recently (it's an independent municipality contained in Cincinnati) the City Council used eminent domain to buy out some residences and small businesses in order to construct a shopping center.  (Irrelevant to this argument is the painful fact that the location chosen was across the street from another shopping center...)  Two of the businesses were located on a busy street and had successfully been operating for decades; one was health-related and the other was a tutoring center.  Giving the businesses the market value of their buildings does them economic harm because they lose their locations---and that's actually two harms because they'll have to move to less prominent locations so will take in fewer new customers, and they are likely to lose some of their current customers in the move.

Of course I think ED is a good thing in theory (highways, etc.) but I'm disturbed that it's legal for a municipality to give land to corporations.  Also recently in Cincinnati, the city paid to build a parking garage for Kroger corporation because they threatened to move out of the city if not.  What's now legal is for Kroger to demand that the city take private homes so it can build something-or-other.  JBL, you can say that we should be activists, vote people out, etc., but when after all the petitions and such are done, and everyone who runs for Council is on record as agreeing with this type of ED action that doesn't leave much action left to take.

At the same time, I don't hold this decision against the Supreme Court.  They interpret the law rather than making it, and given my (admittedly not deep) understanding of ED law, this decision was consistent.  My view is that in response to the decision, lawmakers should draft/pass legislation which refines and restricts current ED law.

I don't think it was consistent, actually. The 5th Amendment of the Constitution states that "...nor shall private property be taken for public use, without just compensation." Now, for the last two hundred and something years, "public use" has been interpreted as highways, public buildings, etc. In Kelo vs. New London, see the following:

[quote]Moreover, while the city is not planning to open the condemned land--at least not in its entirety--to use by the general public, this "Court long ago rejected any literal requirement that condemned property be put into use for the ... public." Id., at 244. Rather, it has embraced the broader and more natural interpretation of public use as "public purpose." [/quote]

This brings us back to the age old question of how the Constitution should be interpreted. Now, in my opinion, if the writers of the Bill of Rights had meant "public purpose" instead of "public use," they would have written "purpose" instead of "use." However, this was not the case. They put use, and I think they meant it. There has been some discussion on this topic as to whether or not the Supreme Court overstepped their bounds in this case; that is, some believe that this decision was an act of making the law rather than interpreting it. Is this the case? Would anyone like to comment? Another comment I have heard is that this decision is "the worst since the Dred Scott decision." This led me to wonder- is there any hope then, for redemption? Could this case, like the Dred Scott trial, eventually be overturned?

[quote="bleumoose"] Another comment I have heard is that this decision is "the worst since the Dred Scott decision." This led me to wonder- is there any hope then, for redemption? Could this case, like the Dred Scott trial, eventually be overturned?[/quote]

Notice that few, and perhaps no, politicians are grandstanding over this.  You'd think at a time when Republicans, particularly alleged small-government conservatives, are bashing liberal judges, we'd see some of them going bonkers over this.  But we see, by and large, silence from them.  Why?  Follow the money. . .  This is yet another example of how the 'small-government Republicans v. big-government Democrats' is patently false.  'Big-government party in power' v. 'obstructionist opposition' is more accurate.

As for an answer to your question, I think you'll see it become an issue as more local governments abuse their power.  And eventually some politicians will start grandstanding aggressively.  But not as aggressively as they do over lifestyle issues, which both parties have glommed onto as ways to rally the troops.


5 people each brought a gift to a birthday party. Then the boxes are given back to the 5 people randomly. What is the probability that
(a) 3 people will get back their own gifts
(b) every person will get another person's gift

This is a classical example using inclusion-exclusion.


Let $ K$ be midpoint of $ BD$. Then $ AKCD$ and $ ANCD$ have the same area, so that $ AC\parallel KN$. Similarly, then $ ABCK$ and $ MBC$ have the same area, so that $ AC\parallel MK$. Therefore $ AC\parallel KN\parallel MK$, so $ M,K,N$ are collinear which proves the segment $ MN$ bisects the diagonal $ BD$.

[quote="April"]$ ABCD$ is a convex quadrilateral for which $ AB$ is the longest side. Points $ M$ and $ N$ are located on sides $ AB$ and $ BC$ respectively, so that each of the segments $ AN$ and $ CM$ divides the quadrilateral into two parts of equal area. Prove that the segment $ MN$ bisects the diagonal $ BD$.[/quote]

$ Fact:$(This is even in the textbook of our country so I am not proving it)
To construct the point $ M$  we have to draw line M'N' parallel to $ AC$ and construct the midpoint of $ M'B$.(and use the same way $ N$)

So, It is clear that $ M$ and $ N$ are the midpoints of $ M'B$ and $ N'B$ respectively.So, $ MN \parallel  M'N'$ and $ D \in M'N'$. So,$ MN \cap BD\equal{}K$ is the midpoint of $ BD$

[quote="suleymansoyler"]Let $ K$ be midpoint of $ BD$. Then $ AKCD$ and $ ANCD$ have the same area, so that $ AC\parallel KN$. [/quote]

Could someone explain why $AKCD$ and $ANCD$ necessarily have the same area?

Consider the affine transformation which takes $A$ to $(1,0)$, $B$ to the origin, and $C$ to $(0,1)$. This necessarily preserves all the ratios we care about. Then, if $D$ is at $(2p,2q)$, $p,q>0$ the area is clearly $[ABD]+[CBD]=p+q$, whence $M$ is located at $(2p+2q,0)$ and $N$ at $(0,2p+2q)$.

The line $MN$ has equation $x+y=2p+2q$, so the midpoint of $BD$, $(p,q)$, trivially lies on it.

Draw a parallel line to $AC$ through $D$.
Let $E$ and $F$ be the intersections of $BC$ and $BA$ to the parallel line through the point $D$,now $N$ be the midpoint of $BE$ because $[AXD]=[CXE]$,($X=AE\cap CD$),so $[ABCD]=[ABE]$,hence $N$ be the midpoint of $BE$,similarly we get $M$ be the midpoint of $BF$,so $MN\parallel EF$ with $M,N$ are the midpoints of $BF,BE$,hence we get the result.

We now have that $A_{MADC}=\frac{A_{ABCD}}{2}=A_{NADC}$, or $A_{ANC}=A_{AMC} \implies MN \parallel AC$. Now, let`s define line $x$ such that $x \parallel AC, MN$ and let $x$ intersect $AB$ at $E$ and $BC$ at $F$. Now:
$A_{MEC}=A_{MAC}+A_{CAE}=A_{MAC}+A_{CAD}=A_{MADC}=A_{BMC} \implies BM=ME, BN=NF$. This means that $MN$ is the midline of triangle $BEF$, so it bisects $BD$ :)


limit
real analysis
real analysis theorems

From the definition of limit ($ \delta , \epsilon$), it seems like the limit exists if the approach from the left equals the approach from the right.
So, what is the definition of $ lim_{x \rightarrow 0^{\plus{}}}$ and $ lim_{x \rightarrow 0^{\minus{}}}$ ?
I understand, myself, that it means the approach form the right and the left, respectively? (Do I understand correctly?)
And then the limit exists if the approach form the right and the left equals. (again, Am I correct?)

And what about $ lim_{x \rightarrow \infty}$? What is the definition? Thank you.  :huh:

$ \lim_{x \to 0^{\plus{}}} f(x) \equal{} L$ if for every $ \epsilon > 0$ there exists $ \delta > 0$ such that $ 0 < x < \delta$ implies $ |f(x) \minus{} L| < \epsilon$.  In other words, we take $ f$ to be defined on $ (0, a)$ for some $ a$ and ignore negative values of $ x$.  Similarly for the left limit.

$ \lim_{x \to \infty} f(x) \equal{} L$ if for every $ \epsilon > 0$ there exists $ N$ such that $ x > N \implies |f(x) \minus{} L| < \epsilon$.

One can understand the last definition with the following definition:  a neighborhood of infinity is defined to be a set of the form $ \{ x : a > x \}$.  Then the definition says that for every $ \epsilon$ we can find a neighborhood of $ \infty$ such that $ x$ is at most $ \epsilon$ away from $ L$ everywhere in that neighborhood, which agrees with the usual definition of a limit.


algebra
function
domain
algebra proposed

Find all continuous funtions $f: (0,\infty) \rightarrow (0,\infty)$ such that:
                            $f(a)+f(b)+f(c)=f(a)f(b)f(c)$ whenever $ab+bc+ca=1$

If $\ f(x)=\frac{1}{x}$

$\rightarrow$ $\ f(a)+f(b)+f(c)=\frac{ac+ab+bc}{abc}$, and $\ f(a)f(b)f(c)=\frac{1}{abc}$

Then $\ f(a)+f(b)+f(c)=f(a)f(b)f(c)$, if $\ ac+ab+bc=1$

But I don't know how to prove uniqueness (in case it's unique that is).

Oh, it's not unique.

Proof: Dividing both sides by $f(a)*f(b)*f(c)$, we get:
$\frac 1{f(a)f(b)}+\frac 1{f(a)f(c)}+\frac 1{f(b)f(c)}= 1$.

So then, there are 2 possibilities: 1) $f(x)$ varies with $x$,
and 2) $f$ is a constant function.

Taking case 1, and substituting in $ab+ac+bc$ for $1$, there are clearly 3 terms on each side.  Also, there is only 1 term that does not depend on $a$ on each side (ditto for b and c).

Therefore, we can conclude that $\frac 1{f(x)f(y)}= xy$,
for $x, y$ $\epsilon$ {$a, b, c$}.

Thus, $xf(x) = \frac 1{yf(y)}$, which must be constant, since x and y can independently vary.

Therefore, $xf(x) = k = \frac 1{xf(x)}$,
and therefore, $f(x) = \frac{k}{x}$ and $f(x) = \frac 1{kx}$.

Then, $k = \frac{1}{k}$, therefore $k = 1$ or $k =-1$.

But, if $k =-1$, $f(x)$ is not in the range of $f$ for any $x$ in the domain of $f$.  Therefore, $k = 1$.

So, the first solution is $f(x) = \frac{1}{x}$.

Now, treating case 2, let $f(x) = t$, where $t$ is some constant.

Then, $t+t+t = t*t*t$, so $3t = t^{3}$.

Therefore, $t(t^{2}-3) = 0$, so $t = 0$ or $t = \sqrt 3$.

However, $t$ cannot be $0$, since 0 is not in the range of f.
Therefore, $t = \sqrt 3$.

So our second function is $f(x) = \sqrt 3$.


Prove that $GL_{n}(Z)$ has infinitely many classes of isomorphisms of its subgroups.

$\mbox{Gl}_{n}(\mathbb Z),\ n\ge 2$ contains an isomorphic copy of $\mbox{Sl}_{2}(\mathbb Z)$, which, in turn, contains a free group on two generators (this has been discussed before on the forum, I think; the newest topic is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=117095]this one[/url]). Since all free groups on at most countably many generators can be embedded in the free group on two generators, we're done.


Let $ n \equal{} 3k^2 \plus{} k \plus{} 1$ $ (k\in\mathbb N, k\ge 1)$. Find the min value of $ S(n)$ when k moves on $ N$\{0} where $ S(n)$ is the sum of digits of $ n$.

the minimum is $ m\equal{}3$ for $ k\equal{}8$, $ n\equal{}201$.

Notice that $ n\ge5$

$ m\equal{}1?$
$ n\equal{}3k^2\plus{}k\plus{}1\equal{}10^a$ for $ a\ge1$ however $ 3k^2\plus{}k\plus{}1\equal{}k(3k\plus{}1)\plus{}1$ always odd and $ 10^a$ is always even. Contradiction

$ m\equal{}2?$
$ n\equal{}3k^2\plus{}k\plus{}1\equal{}2 \times 10^a$ Similarly, Contradiction
$ n\equal{}3k^2\plus{}k\plus{}1\equal{}10^a\plus{}10^b$ for $ a>b\ge 1$ Similarly, Contradiction

$ n\equal{}3k^2\plus{}k\plus{}1\equal{}10^a\plus{}1$ for $ a\ge 1$
$ k(3k\plus{}1)\equal{}2^a \times 5^a$
notice that $ k$ and $ 3k\plus{}1$ have different parity then either:
$ 3k\plus{}1\equal{}2^a \times 5^x$ and $ k\equal{}5^{a\minus{}x}$
$ 2^a \times 5^x\equal{}3 \times 5^{a\minus{}x}\plus{}1$ taking mod $ 5$ Contradiction for $ x>0$
$ 2^a \equal{}3 \times 5^a\plus{}1$ the RHS is always bigger than LHS. Contradiction

or $ k\equal{}2^a \times 5^x$ and $ 3k\plus{}1\equal{}5^{a\minus{}x}$
$ 3\times 2^a \times 5^x\plus{}1\equal{}5^{a\minus{}x}$ taking mod $ 5$ contradiction for $ x>0$
$ 3\times 2^a \plus{}1\equal{}5^a$ no solution for $ a\equal{}1$ and for $ a\ge 2$ the RHS is always bigger than LHS. Contradiction

Your solution is also mine :P
Now, let try with a similar problem which was posted in VMO
Let $ n$ be a positive integer satisfying $ n$ is divisible by 2003. Find the min value of $ S(n)$ where $ S(n)$ is the sum of demical digits of $ n$ :roll:


$ ABCD$ is a convex quadrilateral.$ A'$ and $ C'$ are points on $ BD$ such that $ AA'$ and $ CC'$ are perpendicular to $ BD$.$ B'$ and $ D'$ are points on $ AC$ such that $ BB'$ and $ DD'$are perpendicular to $ AC$.Prove that $ BCD$ and $ BAD$ are similar to $ B'C'D'$ and $ B'A'P'$ respectively.

Very easy! Apply some properties of cyclic quadrilateral.


Consider a square, the side length of which equals 12. Rotate this square over 30 by a vertex. What's the area of the overlapping of the original square and the new square ?

Let ABCD be the initial square, which is rotated about vertex A. The new square is AMNP. The intersection between MN and CD is X. It's pretty obvious that the whole figure is symmetric with respect to line AX, so the area we're looking for (which is the area of AMXD) is twice the area of XDA, which is a right triangle with angle XAD=30 deg.

Our area is AD*DX=(12*12*3^1/2)/3=48*3^1/2. Is it right?

Of course.

The Dutch olympiad problems seemed surprisingly easy to me this year.
So it's strange that the 4 best scores were 34, 33, 31, 27/50.
(Each problem is wort ten points.)
Our Belgian olympiads are more difficult :)