very diffucult question  :(
[img]http://img217.imageshack.us/img217/2678/adszur9.png[/img]

Oh! Problem is nice. I will solution it.

:what?: i didnt solve...i think not nice problem  :(

Thank you for your help..Can you show me your solution way?
Thanks and best regards!

$ \bullet$ Let $ P$ is midpoint of $ EC$, let $ AB \equal{} EC \equal{} a$ and $ BE \equal{} b$ $ \Longrightarrow BC \equal{} a \plus{} b$.

$ \bullet$ In $ \triangle ABC$ we have $ AC^2\equal{}AB^2\plus{}BC^2\minus{}2AB.BC.cos120^0\equal{}a^2\plus{}(a\plus{}b)^2\plus{}a(a\plus{}b)$

$ \bullet$ We have $ DP // AE \Longrightarrow EF//DP \Longrightarrow \frac{BF}{BD}\equal{}\frac{EF}{DP} \Longleftrightarrow  \frac{x}{BD}\equal{}\frac{4}{DP}$.

$ \bullet$ $ BD^2\equal{} \frac{AB^2\plus{}BC^2}{2} \minus{} \frac {AC^2}{4} \equal{} \frac{a^2\plus{}(a\plus{}b)^2}{2} \minus{} \frac {a^2\plus{}(a\plus{}b)^2\plus{}a(a\plus{}b)}{4}\equal{}$ $ \frac {a^2\plus{}b^2\plus{}ab}{4}$
$ \Longrightarrow BD \equal{} \frac {1}{2} \sqrt {a^2\plus{}b^2\plus{}ab}$ , (1).

$ \bullet$ But $ DP \equal{} \frac {1}{2}AE \equal{}  \frac {1}{2} \sqrt {AB^2\plus{}BE^2\minus{}2AB.BE.cos120^0}\equal{}\frac {1}{2}\sqrt {a^2\plus{}b^2\plus{}ab}$ , (2).

$ \bullet$ Since (1) and (2) $ \Longrightarrow BD \equal{} DP$ $ \Longrightarrow x \equal{} 4$.

Ok!  :rotfl:

[hide="Geometrical solution"]
Extend BD to X such that DX=BD.
Join X to E. Clearly triangle CEX is equilateral.
So /_ XEB=120

Now, triangles ABE and BEX are congurent
Hence /_ XFE=/_AEB
=>FB=FE [/hide]

I am new to this forum, can u tell me how to add pictures and other symbols to posts.[/hide]

[quote="guru_bhai"][I am new to this forum, can u tell me how to add pictures and other symbols to posts.[/hide][/quote]

You can learn about [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX]LaTeX[/url] and [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote_%28Vector_Graphics_Language%29]Asymptote[/url] from the AoPS Wiki.

LaTeX can be enclosed in dollar signs, while Asymptote needs to be enclosed in [ a s m ] tags (no spaces).

[quote="fishythefish"]Why can't I see OP?[/quote]

Probably your browser. Use Firefox.

Dear Guru_bhai can you show me your solution by picture  ?
Thanks and best regards!i added by imageshack  :lol:

You can use sketpad to draw picture and post it.

Oh! I have other solution for problem. I will post it soon.

Really? :o
I wonder your another way solution...

[hide="Complete solution :"]

Construction :

Extend $ AD$ to $ H$ such that $ AD=DH$.
Join $ E$ and $ C$ to $ H$.

[asy]size(300);
    import olympiad;
    pair A,B,C,D,E,F,G,H;
    A=(0,0);
    B=10*dir(120);
    C=(15,0);
    G=(-8,0);
    D=midpoint(B--C);
    E=(5,0);
    F=intersectionpoint(B--E,A--D);
    draw(A--G,blue,MidArrow);
    draw(A--C--B--A,blue);
    draw(A--D,blue);
    draw(B--E,blue);
    fill(anglemark(B,A,G,40),green);
    dot(A,red);dot(B,red);dot(C,red);dot(D,red);dot(G,red);dot(E,red);dot(F,red);
    label("$G$",G,S);
    label("$A$",A,S);
    label("$B$",B,W);
    label("$C$",C,S);label("$E$",E,S);
    label("$D$",D,NE);label("$F$",F,N);
    label("$x$",A--F,NW);label("$4$",F--E,NE);
    label("$60^\circ$",anglemark(B,A,G),4*NW);
    add(pathticks(A--B,2,0.5,0,25,red));
    add(pathticks(E--C,2,0.5,0,25,red));

    H=2*D-A;
    draw(E--H--D,dashed+blue);
    draw(H--C,dashed+blue);
    dot(H,red);
 add(pathticks(E--H,2,.5,0,25,red));add(pathticks(H--C,2,0.5,0,25,red));

 label("$H$",H,NE);
fill(anglemark(E,H,C,40),green);fill(anglemark(H,C,E,40),green);fill(anglemark(C,E,H,40),green);
label("$60^\circ$",anglemark(E,H,C),4*S);label("$60^\circ$",anglemark(H,C,E),4*NW);label("$60^\circ$",anglemark(C,E,H),4*NE);[/asy]

Clearly,
$ \triangle ABD \cong \triangle HCD$
$ \Rightarrow \angle BAD = \angle CHD$
$ \Rightarrow AB \parallel CH$
$ \Rightarrow \angle HCE = 60^\circ$

Now in $ \triangle CEH, EC=HC \Rightarrow \triangle CEH$ is equilateral.
$ \Rightarrow \angle HEA = 120^\circ$

Now clearly $ \triangle ABE \cong \triangle AEH$
$ \Rightarrow \angle FAE = \angle FEA$

So, In $ \triangle FAE, FA=FE.$

Hence Proved[/hide]

[quote="guru_bhai"][asy]size(300);
import olympiad;
pair A,B,C,D,E,F,G,H;
A=(0,0);
B=10*dir(120);
C=(15,0);
G=(-8,0);
D=midpoint(B--C);
E=(5,0);
F=intersectionpoint(B--E,A--D);
draw(A--G,blue,MidArrow);
draw(A--C--B--A,blue);
draw(A--D,blue);
draw(B--E,blue);
fill(anglemark(B,A,G,40),green);
dot(A,red);dot(B,red);dot(C,red);dot(D,red);dot(G,red);dot(E,red);dot(F,red);
label("$G$",G,S);
label("$A$",A,S);
label("$B$",B,W);
label("$C$",C,S);label("$E$",E,S);
label("$D$",D,NE);label("$F$",F,N);
label("$x$",A--F,NW);label("$4$",F--E,NE);
label("$60^\circ$",anglemark(B,A,G),4*NW);
add(pathticks(A--B,2,0.5,0,25,red));
add(pathticks(E--C,2,0.5,0,25,red));
add(pathticks(D--B,2,.5,7,25,red));add(pathticks(D--C,2,0.5,7,25,red));[/asy]

[b] Hepl me to draw $ DH$ such that $ AD\equal{}DH$[/b]

[code][asy]
size(300);
import olympiad;
pair A,B,C,D,E,F,G,H;
A=(0,0);
B=10*dir(120);
C=(15,0);
G=(-8,0);
D=midpoint(B--C);
E=(5,0);
F=intersectionpoint(B--E,A--D);
draw(A--G,blue,MidArrow);
draw(A--C--B--A,blue);
draw(A--D,blue);
draw(B--E,blue);
fill(anglemark(B,A,G,40),green);
dot(A,red);dot(B,red);dot(C,red);dot(D,red);dot(G,red);dot(E,red);dot(F,red);
label("$G$",G,S);
label("$A$",A,S);
label("$B$",B,W);
label("$C$",C,S);label("$E$",E,S);
label("$D$",D,NE);label("$F$",F,N);
label("$x$",A--F,NW);label("$4$",F--E,NE);
label("$60^\circ$",anglemark(B,A,G),4*NW);
add(pathticks(A--B,2,0.5,0,25,red));
add(pathticks(E--C,2,0.5,0,25,red));
add(pathticks(D--B,2,.5,7,25,red));add(pathticks(D--C,2,0.5,7,25,red));
[/asy][/code][/quote]

add
[code]
H=2*D-A
[/code]

[asy]

pair A, B, C, D, EE;

EE=(0,0);
A=(-1,0);
B=(0,-1.5);
D=(0,.8);
C=(1,0);

draw(A--B--C--D--A--C);
draw(D--B);
label("$A$",A,W);
label("$B$",B,S);
label("$C$",C,E);
label("$D$",D,N);
label("$E$",EE,SE);

//pathticks(path, how many ticks, where on segment (0=start, 1=end), ?, size of ticks)
add(pathticks(A--B, 1, .5, 6, 2));
add(pathticks(B--C, 1, .5, 6, 2));

add(pathticks(A--D, 2, .5, 1, 2));
add(pathticks(C--D, 2, .5, 1, 2));

[/asy]


:maybe: i think it was on poland math olympiad.

Determine all possible functions $f: Q\to Q$
which satisfies

$f(x^{2}+y) =xf(x)+f(y)$

$x,y are rational numbers$ :


geometry
parallelogram
perpendicular bisector
geometry proposed

Let $ ABCD$ be a convex quadrilateral, such that

$ \angle CBD\equal{}2\cdot\angle ADB, \angle ABD\equal{}2\cdot\angle CDB$ and $ AB\equal{}CB$.

Prove that quadrilateral $ ABCD$ is a kite.

[color=darkblue]Let $ P \in AD$, $ Q \in CD$, such that $ BP$ and $ BQ$ bisects angles $ ABD$ and $ DBC$ respectively. Then $ \angle PBD=\angle CDB \Rightarrow PB \parallel CD$, and, similliarly $ AD %Error. "parralel" is a bad command.
BQ$. So $ DPBQ$ is a paralelogram and $ BD$ bisects $ PQ$. Also

$ \dfrac{DQ}{QC}=\dfrac{BD}{BC}=\dfrac{BD}{AB}=\dfrac{DP}{PA}$

So $ PQ\parallel AC$ and $ BD$ bisects $ AC$. Thus, $ BD$ is perpendicular bisector of $ AC$. The conclusion follows. [/color]

[hide="Solution in short notes"]
First, note $ E$ and $ F$ be the intersections of $ BD$ with the circle $ (B,AB)$, which contains $ C$, such that $ E$ is on segment $ BD$. Looking central and inscribed angles, we get easily $ AFCD$ is a parallelogram, so its diagonals are bisecting each other, and we get $ BD$ is a bisector of $ AC$.
Let $ \{S\}\equal{}AC\cap BD$. Then $ \triangle ASB\cong\triangle CSB$, so $ BD\bot AC$. [/hide]

$AB=b$ $BD=k$ accoring $ \triangle ABD$ and $ \triangle BCD$
$\frac{k}{sin(\alpha+2\lambda)}=\frac {b}{sin\alpha}$
$\frac{k}{sin(\lambda+2\alpha)}=\frac {b}{sin\lambda}$
divede by side-by side
$cos(\alpha+\lambda)-cos(\lambda+3\alpha)=cos(\alpha+\lambda)-cos(\alpha+3\lambda)$ then $\boxed{\alpha=\lambda}$
NOTE: $\angle{ADB}=\alpha$ $\angle{ABD}=2\lambda$ $\angle{BDC}=\lambda$ $\angle{DBC}=2\alpha$


complex analysis
function
complex analysis unsolved

Suppose that $ f$ in holomorphic inside and on $ \alpha(0;1)$ , with taylor expansion $ \sum_{n>\equal{}0}c_{n}z^{n}$. give that $ f$ has $ m$ zeros inside $ \alpha(0;1)$, prove that

\[ min\{|f(z)|: |z|\equal{}1\}<\equal{}|c_{0}|\plus{}........\plus{}|c_{m}|\] :maybe:

Assume the opposite and apply the Rouche theorem to the pair $ f(z)$, $ f(z)\minus{}\sum_{k\equal{}1}^m c_k z^k$. Note that the second function has a zero at the origin of multiplicity $ m\plus{}1$.


LaTeX
inequalities unsolved
inequalities

Let x, y, z be real numbers in the interval (0; 1). Prove that
\frac{1}{x(1-y)}+\frac{1}{y(1-Z)}+\frac{1}{Z(1-x)}\geq \frac{3}{xyz + (1-x)(1-y)(1-z)}

[quote="lnnvnu"]Let $ x, y, z$ be real numbers in the interval $ (0; 1).$ Prove that
$ \frac{1}{x(1\minus{}y)}\plus{}\frac{1}{y(1\minus{}z)}\plus{}\frac{1}{z(1\minus{}x)}\geq \frac{3}{xyz \plus{} (1\minus{}x)(1\minus{}y)(1\minus{}z)}$[/quote]

Use the dollar sign to write in LATEX dear friand.....

It's equivalent to
$ \displaystyle (xyz\plus{}(1\minus{}x)(1\minus{}y)(1\minus{}z))\sum_{cyc}\frac{1}{x(1\minus{}y)}\geq3$
$ \displaystyle\Leftrightarrow \sum_{cyc}\frac{yz}{1\minus{}y}\plus{}\sum_{cyc}\frac{(1\minus{}x)(1\minus{}z)}{x}\geq3$
$ \displaystyle\Leftrightarrow \sum_{cyc}\frac{(1\minus{}(1\minus{}y))z}{1\minus{}y}\plus{}\sum_{cyc}\frac{(1\minus{}x)(1\minus{}z)}{x}\geq3$
$ \displaystyle\Leftrightarrow \sum_{cyc}\left(\frac{z}{1\minus{}y}\minus{}z\right)\plus{}\sum_{cyc}\left(\frac{1\minus{}z}{x}\minus{}(1\minus{}z)\right)\geq3$
$ \displaystyle\Leftrightarrow \sum_{cyc}\frac{z}{1\minus{}y}\plus{}\sum_{cyc}\frac{1\minus{}z}{x}\geq6$ which is obviously true by AM-GM.


algebra
polynomial
function
complex analysis

Hi all,

I have a homework question I could use some help with. I don't want anyone to give the full answer so I won't type out the whole question but I will give the gist of it and maybe some kind soul out there can let me know what theorem or approach would be helpful in solving the problem.

Anywho, the question asks to find the # of zeros in the closed, right half complex plan (x >=0) of a degree 4 polynomial. I have only ever solved problems like these asking for # of zeros in a disk, and so I usually just apply Rouche's theorem. I can't think of a way to generalize my approach to this problem though. Any thoughts?

Thanks for your time.

Try the [url=http://en.wikipedia.org/wiki/Routh-Hurwitz_theorem]Routh-Hurwitz theorem[/url] - it's based on the same idea (the argument principle) as Rouche's theorem.

Another possibility is to create some continuous family $ P_t$, $ t\in[0,1]$ of polynomials such that none of them has any roots on the imaginary axis, $ P_1$ is your polynomial, and $ P_0$ is a polynomial for which the answer is obvious. For instance, if your polynomial is something like $ z^4\plus{}10z^3\plus{}5z\plus{}1$, you can take $ P_t(z)\equal{}z^4\plus{}1\plus{}t(10z^3\plus{}5z)$ (note that $ \Re P_t(z)>0$ on the imaginary axis for all $ t$). The numbers of roots of $ P_0$ and $ P_1$ in the right half-plane are the same. This doesn't work always, of course, but it is a useful trick nevertheless. Note that it is even better than Rouche's theorem: if you can represent your polynomial as $ Q\plus{}S$ with $ |Q|>|S|$ everywhere on the imaginary axis, you can always take $ P_t\equal{}Q\plus{}tS$ but in the example I gave above $ |z^4\plus{}1|$ does not dominate $ |10z^3\plus{}5z|$.

Thanks for the responses. Fedja, what theorem is that, do you know?

Essentially it is the same argument principle. (By the way, I forgot to mention one little but important detail: the degree of all polynomials $ P_t$ should be the same. In practice you can always do it by fixing the leading coefficient). The idea is that if you have a closed contour and a continuous family $ f_t$, $ t\in[0,1]$ of nice analyic functions none of which vanishes on the contour, then the increment of the argument over the contour should be continuous on one hand and can take only a discrete set of values on the other hand, so it must stay constant, which implies that the number of zeroes inside the contour also does not change. Of course, the right half plane isn't bounded by a closed contour, but you can take a long interval on the imaginary axis, complement it by a semicircle, let the radius of the semicircle be large enough, etc.


Find a formula f(x) where x is the number of posts, and f(x) is the number of your stars. I doubt if it is a linear function.

In what way is this related to the American Mathematics Competitions?

Well...

0-19: 0 stars
20-49: 1/2 star
50-99: 1 star
100-249: 2 stars
250-499: 5/2 stars
500-999: 3 stars
1000-2499: 4 stars
2500+: 5 stars

It's kind of weird, now that you look at it. Someone else want to come up with a formula? ;)

i don't think it is linear...
can't see any "mappings" from it...

[quote="dts"]The way I see it, it $f$ will either have to be piecewise defined, or use the floor function.   :wacko:[/quote]

The floor function is piecewise defined.

Formula? well, henryyangrui is bored of his prepration for USAMO.

What exactly does this have to do with AMC?  Was this an old USAMO problem? :P

You guys are also forgetting that administrators, regardless of the number of posts, get 6 stars automatically.


geometry
geometric transformation
rotation

Link is coloring a triforce, which consists of four equilateral triangles and is depicted below. He has three colors to use: gold, black, and green. So that it remains recognizable, he doesn't want to color any two triangles the same color if they share a side. How many different ways can he color the triforce?
[asy]size(50);
draw((0,0)--(2,0)--(1,1.732)--cycle);
draw((1,0)--(1.5,.866)--(.5,.866)--cycle);[/asy]

I didn't realize you were supposed to count rotations.

It never says that rotations should not be distinct.

the answer is 24. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=540&t=357525&p=1950491&hilit=Link+is+coloring+a+triforce#p1950491]here[/url]

It's always black in the middle and golden in the three others.

[hide]There are $3$ ways to color the middle triangle.  Since the other triangles cannot be the same color as the middle triangle, there are $2$ ways to choose the color for each of the $3$ triangles (excluding the middle one).  The total ways to color is $3\times 2^3=\boxed{24}$ ways.[/hide]


geometry
rhombus
geometric transformation
reflection
circumcircle
angle bisector
geometry unsolved

Hello

I don't know if this has already been posted, but this problem is good, and I need help.

Let ABC be a triangle with measure of angle ABC 80 degrees.   Let BD be the angle bisector of angle ABC with D on AC.  If AD=DB+BC, determine angle A, using only a geometric argument.

Thanks

Let $ A_0BC_0D$ be a rhombus with $ \angle A_0BC_0 \equal{} 80^\circ.$ Let a line $ b$ through $ D$ cut the rays $ BA_0, BC_0$ at $ A, C,$ respectively, forming a $ \triangle ABC$ with bisector $ BD$ of the angle $ \angle ABC \equal{} 80^\circ.$ The angle $ \angle ADB$ must be in $ (40^\circ, 140^\circ),$ otherwise $ b$ does not intersect both rays. We demand $ DA \equal{} DB \plus{} BC.$ Let $ P$ be foot of the perpendicular from $ D$ to $ BA_0$ and $ A_1 \in BA_0$ reflection of $ B$ in $ DP$ so that $ DA_1 \equal{} DB.$ Because of our demand $ DA > DB \equal{} DA_1,$ it must be $ BA > BA_1$ and this further limits the angle $ \angle ADB > \angle A_1DB \equal{} 100^\circ,$ or to $ (100^\circ, 140^\circ).$ Since $ \angle BAD < \angle BA_1D < BPD \equal{} 90^\circ,$ $ DA$ is strictly increasing and $ BC$ strictly decreasing with increasing $ BA.$ Thus $ DA \equal{} DB \plus{} BC$ can hold only for one position of $ A$ and a unique angle $ \angle ADB.$

Let $ \triangle ADE$ be isosceles triangle with $ DA \equal{} DE$ and $ \angle ADE \equal{} 120^\circ,$ let $ M$ be the midpoint of $ AE,$ $ DM \perp AE,$ and let $ N$ be a point on the segment $ DM,$ such that $ \angle NAE \equal{} \angle NEA \equal{} 10^\circ.$ Let $ AN$ cut $ DE$ at $ B.$ Since

$ \angle BNE \equal{} \angle NAE \plus{} \angle NEA \equal{} 20^\circ \equal{}$
$ \equal{} 30^\circ \minus{} 10^\circ \equal{} \angle DEA \minus{} \angle NEA \equal{} \angle DEN \equal{} \angle BEN,$

the $ \triangle BEN$ is isosceles with $ BE \equal{} BN.$ Let $ C$ be reflection of $ N$ in $ DE,$ so that $ BD$ bisects the angle $ \angle ABC,$ $ B$ is circumcenter of the isosceles $ \triangle ENC,$ and $ DA \equal{} DE \equal{} DB \plus{} BE \equal{} DB \plus{} BC.$ Due to $ \angle ADE \equal{} 120^\circ,$ the line $ DA$ is a reflection of the line $ DN$ in the line $ DE,$ therefore $ C \in DA$ and $ \angle ABC \equal{} 2 \angle NBD \equal{} 4 \angle NED \equal{} 80^\circ.$ As for the remaining angle(s) of the $ \triangle ABC,$ $ \angle CAB \equal{} \angle DAN \equal{} 20^\circ.$


videos
Alcumus
pigeonhole principle
rate problems
Support

What happens when you run out of Karma?
(no, i have not run out, i'm just curious.)

You need to go back and rate solutions and such to build karma back up. Otherwise you won't have access to more problems. So far, no one has ever run out of karma.

I have once  :D
because I never rated problems :D

Oh wow!  I guess I should have said, "as far as I know, no one has run out of karma." :)  Did you stop getting the problems?

If you miss all your questions, even on your second chance and first, and rate the solution, you can still run out of karma.  You'll lose 5 points for every problem.  Then when you run out, you can't rate any solutions, you can't watch a video and rate that, or anything else.  So it is possible to run out of karma (if you are not good at Alcumus).

You get more karma for rating solutions then getting problems.

[quote]
Answering a problem	Spend 5
Rate problem/solution	Receive 5
Submit bug report	Spend 10
Bug report determined valid	Receive 25
Watch video	Spend 5
Rate video	Receive 5
Comment on video	Receive 5

[/quote]

SO if you answer a question twice, you lose 10, but gain 5 for rating, so a loss of 5.

wait, [b]lets say [/b]i ran out of karma, and have 0 karma. if i rate a problem, i must first answer a question, and that takes karma, which i dont have any.

if i watch a video, and rate it, i spend karma for wattching the video.

how do i earn the karma then??

So...

you can gain up to 15 karma per problem (10 because spending 5)
since you can rate 3 different questions?

or would you just have to rate all the Questions to get your 5 karma back?

wait, how do u earn 10 (or 15?) karma per problem?

i dont see it right now......

and wat do u mean by "since you can rate 3 different questions?

or would you just have to rate all the Questions to get your 5 karma back?"?

I think in your situation you have to click on the subject (the line with the bar next to it) so you can go back and rate problems and gain karma to get more problems.

oh......

im gonna earn xtra karma that way!

For each problem you can rate three things: The difficulty of problem, the quality of solution, and something else.

The "something else" is how much you learned from the solution.

Unfortunatley you won't be able to do anything
did you actually run out of karma!!!  :D

If you ever run out of karma, you can go to the report and rate unrated problems (click the category and then the magnifying glass).
That way, you won't run out of karma (I think).

I have run out of Karma before!! I used to be on the leaderboard in Karma I think... But then I ran out. I had to go back to my report, and then select a problem and then rate it... Thanks for the little graph there 007math!   :lol:

What is karma? is 100 a lot? how do you rate problems?

Karma is the currency of Alcumus. You need to spend a little bit of karma to answer a question or watch an Alcumus video.

100 karma is very little. If you only have 100 karma, that means you have really been spending a lot of karma and not earning any back. If you continue with 100 karma, you could only afford 20 more questions or Alcumus video views (since they cost 5 karma each).

To rate problems, go to the bottom of the solution screen; there, there are three questions you can choose a rating on. For each one of those you answer, you receive 5 karma. Thus, if you answer all three questions for each solution, you receive 15 karma. That means that if you rate all three questions for each solution and answer each question only once, you will end up with a net profit of 10 karma. This is the main way karma is gained. For more information, consult the instructions.

Oh ok yea i found out. So you have to rate the questions when you see the answer.

ok cool. try [url]http://www.artofproblemsolving.com/Alcumus/Instructions.php#karmapoints[/url]

Haha, I only have 700
I think I got negative karma once...somehow

I don't like Karma. It's a good idea, but I don't like it because I usually forget to rate....

Please understand the great quality of Alcumus in that it is free; the administrators of AoPS created Alcumus as a service brought to you free of charge; thus, don't you think you should at least do them a favor by rating the questions so as to help them improve the site?

once you start rating problems, and keep it up, you will start rating problems naturally.

Yeah, is the difficulty of the questions affected by how you rate them?

Why bother using Karma as "currency" in AOPS?  Most people will probably just do problems regardless of right or wrong, and just rate the problems....people will try to get Karma instead of trying to get the questions right.  Why do you have Karma anyways?  You really need to depend on a type of "currency" to get kids to study???  The point of Alcumus is to do questions and learn....while some people thin it is a goal to get the most Karma!!

 :ninja:  :ninja:  :ninja:  :ninja:  :ninja:  :ninja:  :ninja:  :ninja:  :ninja:  :ninja:

BTW, what do you use Karma on anyway?  do you spend it on like a certificate or something?  Because if you just use it to watch videos, or something, people won't want to watch videos anymore....they want to get RICH!

 ______________________________________________________________________________________________

The Math Kid

Let's see...I'll take it apart piece by piece for you.

[quote="faush101"]The point of Alcumus is to do questions and learn....while some people thin it is a goal to get the most Karma!![/quote]

Those that think that way aren't going to be very successful, and most AoPSers would know that learning is the point.

[quote="faush101"]people will try to get Karma instead of trying to get the questions right[/quote]

Absolutely not true. Ask most any experienced AoPSer that frequently does Alcumus. It is a greater pleasure to get questions right than to get more Karma.

[quote="faush101"]Why do you have Karma anyways?  [/quote]

Note that Alcumus is free. The makers of Alcumus are just asking for a little in return from you in the form of feedback on the quality of the questions. That way, they know which questions are preferred by users, and will make more of those, which makes users happy, and also attracts more users, so that users would regard AoPS as a better learning tool (which it really is) and buy more AoPS material. It's a win-win situation.

[quote="faush101"]people won't want to watch videos anymore....they want to get RICH! [/quote]

Apparently, you do not know the "people" of AoPS. Please refrain from assumptions when you do not know what is actually going on. Not everybody is like you, and just because you think a certain way doesn't mean everyone else thinks along the same lines.

click the laughing dude!
[hide=" :rotfl: "]then you cannot answer the questions, to get Kamara, you haz to rate the answers![/hide]

i have run out of Kamara, but that is cuz i didnt know about kamara, LOLZ!

We have a temporary maintenance right now, but we're able to watch videos, so do we still spend karma when we watch them?

I Dont think so because when you watch it youre not allowed to rate the video so you would not get any karma for that.

You can still post comments though, but I don't think you get awarded Karma for that.  And think of this as your free opportunity to watch as many videos as you like without having to worry about karma at all! :)

Also, how come there are no videos on the Pigeonhole Principle in Alcumus?

[quote="maybach"]Also, how come there are no videos on the Pigeonhole Principle in Alcumus?[/quote]

Same reason with having no Pigeonhole Principle Alcumus problem, they haven't made any yet.  :)

[quote="nsun48"]I think they might be making it right now[/quote]

Probably not, because

[quote="DPatrick"]We have nothing to add to what is already stated.[/quote]

[quote="rrusczyk"]We're thinking 1-2 more weeks. We will still have all your work from before, but will be using a different model to determine where you are. That will be the main initial change, with several others coming later in the fall (adding algebra, more counting problems in [b]existing topics[/b], and giving students more control over what topics they see).[/quote]

So we don't get to have additional topics.

Oh... :huh: Then maybe it's a glitch. I'm sure it will come back. I also had a time when there were Hockey Stick Identity videos available and soon disappeared, but after a while it came back.

[quote="maybach"]But when Alcumus was actually up, under my report it said
                     Answered Correctly         Out of
Basic Pigeonhole     NA                           NA

So was there once Pigeonhole, and then they removed it or what?[/quote]

No, they inputted the category before they made the questions, and they still haven't made the questions yet.

[quote="scfliu"]i have a lot of karma!!!!! :starwars:  :o[/quote]

Please do not revive very old topics, especially for irrelavent spam like this.


geometry
circumcircle
trigonometry
incenter
angle bisector
geometry proposed

Nice problem!

Denote $\angle{\frac{A}{2}}=t$, and $\angle{OKC}=x$, therefore after a simple angle chasing we obtain:
$\angle{OCK}=t-30, \angle{CDK}=\angle{CKD}=30+t$
and also $\angle{OKD}=30+t+x, \angle{ODK}=2t-60$.

Now:
1) From the sin law in $\triangle{OCK}$ we obtain: $\frac{OC}{\sin{x}}=\frac{KC}{\sin{(x+t-30)}}$.

2) From the sin law in $\triangle{OCD}$ we obtain: $\frac{OC}{\sin{(90-t)}}=\frac{CD}{\sin{2t}}$.

By dividing them and from $CD=CK$, we obtain:

$\sin{2t}\cdot \sin{x}= \cos{t}\cdot \sin{(x+t-30)}$

$\iff \sin{2t}\cdot \sin{x}= \cos{t}\cdot \sin{(t+[x-30])}$

$\iff \sin{2t}\cdot \sin{x}= \cos{t}\sin{t}\cos{(x-30)}+\cos^{2}{t}\sin{(x-30)}$

$\iff \sin{2t}\cdot (\sin{x}-\frac{1}{2}\cdot \cos{(x-30)})=\cos^{2}{t}\cdot \sin{(x-30)}$

$\iff \sin{2t}\cdot (\sin{x}-\sin{30}\cdot \cos{(x-30)}) =\cos^{2}{t}\cdot \sin{(x-30)}$

$\iff 2\sin{t}\cdot (\frac{\sin{x}-\sin{(60-x)}}{2}) = \cos{t}\cdot \sin{(x-30)}$

$\iff 2\sin{t}\cdot \sin{(x-30)}\cdot \cos{30}= \cos{t}\cdot \sin{(x-30)}$.

From here we have 2 cases:

1) $\sin{(x-30)}=0$, therefore $x=30^\circ \iff \angle{OKC}=30^\circ$.

2) $\tan{t}=\frac{1}{\sqrt{3}}\iff A=60^\circ$, therefore $ABC$ is equilateral, and then it remains us to check again, but this time the problem is trivial.

Therefore the problem is solved.

[quote="コリネアル"][color=darkred]Given $\triangle ABC$, where $\measuredangle\ B=60^\circ$. Angle bisector of $\measuredangle\ BAC$ meets $\overline{BC}$ in point $D$. Point $K$ lies on $\overline{AC}$ such that $\measuredangle\ ADK=30^\circ$.
Find $\measuredangle\ OKC$ where $O$ is the circumcenter of $\triangle ADC$[/color][/quote]
[color=darkblue][b][u]Proof.[/u][/b] Suppose that $A\ne 60$. $A=2k\ (k\ne 30)\ , m(\widehat{OKC})=x\Longrightarrow$ $\left\{\begin{array}{c}m(\widehat{CDK})=30+k\ ,\ m(\widehat{OCK})=30-k\\\ C=120-2k\ ,\ m(\widehat{ADC})=60+k\\\ b=2\cdot OC\cdot\sin (60+k)\end{array}\right\|$ $\Longrightarrow$ $\frac{KA}{KC}=$ $\frac{DA}{DC}\cdot\frac{\sin\widehat{KDA}}{\sin \widehat{KDC}}=$ $\frac{\sin (120-2k)}{\sin k}\cdot \frac{\sin 30}{\sin (30+k)}=$ ${\frac{\sin (60+2k)}{2\sin k\sin (30+k)}}=$ ${\frac{\cos (30+k)}{\sin k}}$ $\Longrightarrow$ $\frac{KA}{\sin (60-k)}=\frac{KC}{\sin k}=\frac{b}{\cos (30-k)}$ $\Longrightarrow$ $\boxed{KC=\frac{b\sin k}{\cos (30-k)}}$. Therefore, $\frac{\sin (x+30-k)}{\sin x}=$ $\frac{\sin \widehat{COK}}{\sin \widehat{OKC}}=$ $\frac{KC}{OC}=$ $\frac{KC}{b}\cdot 2\sin (60+k)=$ $\frac{2\sin k\cos (30-k)}{\cos (30-k)}=2\sin k$ $\Longrightarrow$ $\sin (x+30-k)=2\sin k\sin x$ $\implies$ $\sin (x+30-k)=\cos (k-x)-\cos (k+x)$ $\implies$ $\cos (k+x)=\cos (x-k)-\cos (k-x+60)$ $\implies$ $\cos (k+x)=2\sin 30\sin (k-x+30)$ $\implies$ $\cos (k+x)=\cos (60+x-k)$ $\implies$ $\emptyset$. In conclusion, the triangle $ABC$ is equilateral !?[/color]

Yeah, it's a really nice problem.

Does anybody has a synthetic proof?

Indeed, it is a very nice problem, but pretty computational, i hope to see a synthetic solution, but i doubt, that it will be a very short and nice one :).

@Virgil Nicula: i think you misstyped something somewhere at the beggining, but your method, similar to mine, is very straightforward.

[quote="pohoatza"]... $\angle{OCK}=t-30$ ...[/quote]
Pohoatza, it is wrong. Correctly, $m(\widehat{OCK})=30-t$ !!!!!

[quote="pohoatza"]... $\angle{OCK}=t-30$ ...[/quote]
Pohoatza, it is wrong. In the my opinion, $m(\widehat{OCK})=30-t$ is correctly !!!!!

Let M be the midpoint of CA, R' circumradius of $\triangle ADC$ and usual notation for $\triangle ABC.$ $\angle ADC = \frac{\angle A}{2}+60^\circ,$

$|OM| = \left|R' \cos \left(\frac{A}{2}+60^\circ\right)\right| = \left|\frac{b}{2 \tan \left(\frac{A}{2}+60^\circ\right)}\right| = \frac{b}{2}\left|\frac{1-\sqrt{3}\tan \frac{A}{2}}{\sqrt{3}+\tan \frac{A}{2}}\right|$

Substituting

$\tan \frac{A}{2}= \frac{r}{s-a}= \frac{\triangle}{s(s-a)}= \frac{ca \sqrt{3}}{(b+c)^{2}-a^{2}}$

$|OM| = \frac{b}{2 \sqrt{3}}\cdot \left|\frac{(b+c)^{2}-a^{2}-3ca}{(b+c)^{2}-a^{2}+ca}\right| = \frac{b}{2 \sqrt{3}}\left|1-\frac{4ca}{(b+c)^{2}-a^{2}+ca}\right|$

Substituting $ca = c^{2}+a^{2}-b^{2}$ in the denominator yields

$|OM| = \frac{b}{2 \sqrt{3}}\left|1-\frac{2a}{b+c}\right|$

$\angle CKD = \frac{A}{2}+30^\circ = \angle CDK,$ $\triangle CKD$ is isosceles with $CK = CD = \frac{ab}{b+c},$

$|KM| = \left|\frac{b}{2}-\frac{ab}{b+c}\right| = \frac{b}{2}\left|1-\frac{2a}{b+c}\right|$

$\tan{\widehat{OKM}}= \frac{|OM|}{|KM|}= \frac{1}{\sqrt{3}},\ \ \ \angle{OKM}= 30^\circ.$

If c > b, then b > a, b + c > 2a, $\frac{b}{2}-\frac{ab}{b+c}> 0,$ and C, K, M, A follow on CA in this order. In addition, $\angle A < 60^\circ,$ $\angle ADC < 90^\circ$ and O is inside $\triangle ADC.$ Then $\angle OKC = 180^\circ-\angle OKM = 150^\circ.$ On the other hand, if c < b, then b < a, ...,  and C, M, K, A follow on CA in this order. In addition, $\angle A > 60^\circ,$ $\angle ADC > 90^\circ$ and O is outside $\triangle ADC.$ Then $\angle OKC = \angle OKM = 30^\circ.$ If c = b, then b = a, ..., $O \equiv M \equiv K$ are identical and $\angle OKC$ undefined.

My solution:
Let $I$ be the incenter of $ABC$ and $P$ be the point outside triangle $ABC$ s.t. $ACP$ is equilateral. We have $AKD \sim AIB$ so $AKI \sim ADI$ thus $\angle AIK=60$. $\angle AIC=120$ so points $A$, $I$, $C$, $P$ lie on a circle so $I$, $K$, $P$ are colinear. By angle chasing $IP$ is perpendicular to $OC$ and $CK$ is perpendicular to $OP$. Thus $K$ is orthocenter of $COP$ so $\angle OKC=180-\angle OPC=150$.


geometry
geometric transformation
rotation
3D geometry
calculus
sphere
reflection

[quote="unimpossible"]Find the area of the solid created by rotating Triangle ABC with A(0.0) B(9,0) C(9,40) first about the line x=9 and then the y-axis.[/quote]

uhh i know how to do from just x=9.

I dont what a rotation of a cone looks like (calculus?)

[quote="uldivad9"]Correct me if I'm wrong, but rotating the cone around the x-axis would produce two cones stacked on top of each other, I think.

Edit: Yes, it would, since the sphere formed by rotating the base is completely enveloped by the shape formed by rotating the rest of the cone.[/quote]

No it woundt..thats if you directly reflect it across the x-axis

A sphere would work because everything is equidistant from a point, but thats the not the case with a sphere.

Try rotating it only 90 degrees around hte x-axis.  Where you stopped, you would get another cone 90 degrees away from it, so if it were rotated 360 degrees, you would get like inifnity many cones.

if you rotate the figure about the line x=9, you get two cones that meet at the vertex.


I seem to have made a mistake in my Online Classes enrollment...I thought I was signed up for the unevaluated enrollment, but I recently received an e-mail asking for my Midterm challenges...does this mean that even if you do not have the evaluated enrollment, you should still send in your Midterm challenge? Or did I make a mistake when I signed up for the courses?

Our records show that you do indeed have an evaluated enrollment in both the Counting and the Algebra classes.


Si a y b son enteros positivos que cumplen:
$(\sqrt{b+a^{3}}+29)a^{3}=b$
Determinar todas las soluciones $(a,b)$ de la ecuacion.

[quote="DarthTenebrus"]Este problema vino en el examen selectivo IBERO 1998 en PERU como problema # 4 .[/quote]
si, postea , mi pregunta es que si tiene o no solucion, ya q ami me salio de un paso :blush:  y para comprovar que esta bien o no. gracias, espero soluciones. :D

hay 4 pares ordenados que resuelven dicha ecuacion

bueno ami mi me salio que la unica solucion es b=35, k=6, a=1, (a menos que me hayan faltado reemplazar otros valores admisibles) ;en el transcurso de la semana pondre la solucion  :D

La primera vez que lo resolvi me salieron las 4 soluciones pero lo que no hice muy bien debido a mi inexperiencia de aquel entonces fue probar que no habian mas soluciones osea no
sabia resolver muy bien las ecuaciones diofanticas

Espero ver las 4 soluciones  .

Si a y b son enteros positivos que cumplen:
$(\sqrt{a^{3}+b}+29)a^{3}=b$
Determinar  las soluciones $(a;b)$  de la ecuacion.
$Solucion$
Como a y b son enteros positivos -> $b=ka^{3}$(k entero positivo) de aqui reemplazandolo en la ecuacion:$\sqrt{a^{3}(k+1)}=k-29$ ....($k>29$), luego elevando al cuadrado a ambos miembros:$a^{3}(k+1)=k^{2}-2.29+29^{2}$->$0=k^{2}-(a^{3}+2.29)+(29^{2}-a^{3})$ de donde: $k=\frac{a^{3}+58 ±\sqrt{a^{6}+120a^{3}}}{2}$ Ahora $k>29$ ->$\frac{a^{3}+58 ±\sqrt{a^{6}+120a^{3}}}{2}>29$ ....$a^{3}±\sqrt{a^{6}+120a^{3}}>0$ $...(f)$; como $a^{3}<\sqrt{a^{6}+120a^{3}}$ entonces $(f)$ tan solo se cumple cuando el signo es positivo de donde: $k=\frac{a^{3}+58+\sqrt{a^{6}+120a^{3}}}{2}\[: 147b3ff598....(i)$
Como a y k son enteros ->$a^{6}+120a^{3}=n^{2}$...$a^{4}\frac{(a^{3}+120)}{a}=n^{2}$->$\frac{a^{3}+120}{a}=m^{2}$
de donde: $120=a(m-a)(m+a)$, ahora sea $(m+a)=X$, $(m-a)=Y$ (X e Y enteros positivos)->$\frac{X-Y}{2}=a$
entonces: $120=\frac{X-Y}{2}XY\]: 147b3ff598...(ii)$; como $(X-Y)<X$ y siendo $(X-Y)$ y $X$ divisores de 120 ->$(X-Y)<\sqrt{120}$->$\frac{X-Y}{2}<6$; ademas $\frac{X-Y}{2}|120$-> $\frac{X-Y}{2}=a$ puede tomar los valor: 1, 2, 3, 4 y 5
En $(ii)$::: con a=1, se tiene que X=12 y Y=10 (admite el valor de 1); Con a=2, X=10 y Y=6(admite 2); con a=3, X=10 y Y=4(admite 3); con a=4, no hay soluciones de X e Y en enteros(no admite); con a=5, X=12 y y=2
Ahora reemplazando los valores de a en $(i)$:
$a=1$->$k=35$ de ahi $b=35$
$a=2$->$k=49$ de ahi $b=49.8$
$a=3$->$k=74$ de ahi $b=74.27$
$a=5$->$k=179$ de ahi $b=125.179$
Entonces solo hay 4 soluciones.
Creo que ha sido algo extenso :P
$Américo Arone$

[quote="aev5peru"]Si a y b son enteros positivos que cumplen:
$(\sqrt{a^{3}+b}+29)a^{3}=b$
te falto para $a=4$. aca la explicacion:D
dando $b=k.a^{3}$, y $k=t^{2}-1$, donde $t$ pertenece a los enteros.
reemplasando.
$(\sqrt{a^{3}(t^{2})}+29)a^{3}=a^{3}.(t^{2}-1)$
$\Rightarrow , a=4, 8t+29=t^{2}-1$
$t^{2}-8t-30=0$, su discriminante, es igual a $64+120=184$, el cual no es un cuadrado perfecto, entonces, $t$, no pertenece a los enteros.
por lo tanto, para $a=4$, no hay solucion.[/quote]
Cuando a=4 se tiene:
$120=4XY$    $X-Y=2.4$ y $XY=30$, no hay solucion.
Américo Arone :wink:

[quote="DarthTenebrus"]Veo que tiene un serio problema de teoria de numeros :
porque para empezar no se puede aceptar que k es entero y ustedes lo afirman como si fuese obvio . Espero que sepan remendar ese error .[/quote]
hola DarthTenebrus ...me interesa mucho que postees los errores cometidos en la solucino, xfavor y tambien como poder corregirlo para soluciones posteriores...
Gracias  :roll:


geometry
parallelogram
geometry proposed

#Let, $ ABCD$ be a cyclic quadrilateral. diangonal intersect at $ O$. Prove that centroids of $ \triangle AOB$, $ \triangle BOC$, $ \triangle COD$, $ \triangle AOD$ forms a parallelogram.

#A quadrilateral $ ABCD$. the centroids of $ \triangle ABD$, $ \triangle ABC$, $ \triangle BCD$, $ \triangle ACD$ forms a quadrilateral $ A'B'C'D'$. prove: $ [A'B'C'D'] \equal{} \frac {[ABCD]}{9}$
ext. for a cyclic quadrialteral $ A'B'C'D'$ is cyclic.

I am going to assume that you are taking about ''Centroids''

to the first part.

It is not necessary that ABCD has to be cyclic. See that this quadrilateral is centrally similar to the parallelogram of the midpoint of the sides of ABCD.Center of similarity is the diagonal intersection O and the coefficent is 2/3.


Sa se determine functia $f : \mathbb{N}\rightarrow \mathbb{N}$ astfel incat:

$f(1)=1$
$f(3)=3$
$f(2n)=f(n)$
$f(4n+1)=2f(2n+1)-f(n)$
$f(4n+3)=3f(2n+1)-2f(n)$

scriind $n$ si $f(n)$ in baza 2 pt cateva valori ale lui $n$, se observa ca $f(n)$ inverseaza cifrele lui $n$: de exemplu, $f(110_{2}) = 011_{2}$. demonstram prin inductie. pt fiecare nr binar $x$, notam cu $x'$ nr-ul obtinut prin inversarea cifrelor lui $x$. pt. numere binare de o singura cifra sau de doua cifre, ipoteza se verifica usor. fie acuma $n$ un numar binar de cel putin 3 cifre, si sa presupunem ca pt orice $k$ de mai putine cifre ca $n$, $f(k) = k'$. daca $n = x0$ (un sir binar $x$ urmat de 0), atunci $f(n) = f(x0) = f(2\cdot x) = f(x) = x' = n'$. daca $n = x01$, atunci $f(n)=f(x01)=f(4\cdot x+1)=2f(x1)-f(x)=$$2\cdot 1x'-x'=1x'+1x'-x' = 10x'=n'$. si daca $n = x11$, atunci $f(n) = f(x11) = f(4\cdot x+3) = 3f(2\cdot x+1)-2f(x) = 3f(x1)-2f(x) =$ $3\cdot 1x'-2\cdot x' = 1x'+1x'+1x'-x'-x' = 1x'+2\cdot 10...0$ (unde sunt atatea 0'uri cat sunt cifre in $x'$) $= 1x'+10...0$ (unde sunt atatea 0'uri cat sunt cifre in $x'$, plus 1) $= 11x' = n'$


calculus
inequalities
analytic geometry
graphing lines
slope
continued fraction
real analysis

[size=125]I don't understand this part in my advanced calculus textbook:

For S = {m/n | 0 <= m/n and m² < 2n²}
it is almost obvious that the smallest upper bound for this set has to be \sqrt {2} and hence outside [b]Q[/b], for if m/n is an element of S, then
m/n < 2(m+n)/(m+2n) and {2(m+n)/(m+2n)}² < 2...

My problem is: how do we derive the inequalities m/n < 2(m+n)/(m+2n) and {2(m+n)/(m+2n)}² < 2?  :huh: [/size]

[b]Lemma:[/b]  Let $ a, b, c, d$ be positive integers.  If $ \frac {a}{b} < \frac {c}{d}$, then $ \frac {a}{b} < \frac {a \plus{} c}{b \plus{} d} < \frac {c}{d}$.

Proof:  consider the slopes of the lines from the origin to the lattice points $ (b, a), (d, c), (b \plus{} d, a \plus{} c)$.

Then $ \frac {2m}{2n} < \frac {2m \plus{} 2n}{2n \plus{} m} < \frac {2n}{m}$.  I do not have a similarly elegant argument for the other inequality (of this type), but I believe both inequalities are in some sense a consequence of the [url=http://en.wikipedia.org/wiki/Continued_fraction#Periodic_continued_fractions]continued fraction[/url] expansion of $ \sqrt {2}$.


Determine a formula for

$ 1^{2} \minus{} 2^{2} \plus{} 3^{2} \minus{} 4^{2} \plus{} ... \plus{} ( \minus{} 1)^{n \plus{} 1} n^{2}$

for every possible integer $ n$.

[hide="answer"]
$ (\frac{n^{2} \plus{} n}{2}) (\minus{}1)^{n\plus{}1}$
[/hide]


trigonometry
geometry
similar triangles
geometry proposed

Let $ABC$ be a  triangle with  $b\neq c$. Points $D$ is the midpoint of $BC$ and let $E$ be the foot of angle $A$ bisector. In the exterior of the triangle we construct the similar triangles $AMB$ and $ANC$ . Prove:
a) $MN\bot AD \Longleftrightarrow MA \bot AB$
b) $MN\bot AE \Longleftrightarrow M,A,N$ are colinear.

[quote="felixmann"][color=darkred]Let $ABC$ be a  triangle with  $b\neq c$. Denote the middlepoint $D$ of the side $[BC]$. In the exterior of the triangle $ABC$ construct the points $M$ and $N$ so that $\widehat{BAM}\equiv\widehat{CAN}$ and $\frac{AM}{c}=\frac{AN}{b}$. Prove that $MN\bot AD \Longleftrightarrow AM \bot AB$.[/color][/quote]
[color=darkblue][u][b]Proof.[/b][/u] Denote $\{\begin{array}{c}m(\widehat{BAM})=m(\widehat{CAN})=x\\\ m(\widehat{DAB})=u\ ,\ m(\widehat{DAC})=v\end{array}$. Observe that

$DB=DC\Longleftrightarrow$ $\frac{\sin u}{\sin v}=\frac{b}{c}$ $\Longleftrightarrow$ $\frac{AM}{AN}=\frac{\sin v}{\sin u}$ $\Longleftrightarrow$ $\frac{\sin\widehat{ANM}}{\sin\widehat{AMN}}=\frac{\sin v}{\sin u}$.

Thus, $MN\perp AD$ $\Longleftrightarrow$ $\{\begin{array}{c}m(\widehat{AMN})=90^{\circ}-(x+u)\\\ m(\widehat{ANM})=90^{\circ}-(x+v)\end{array}\|$ $\Longleftrightarrow$

$\frac{\cos (x+v)}{\cos (x+u)}=\frac{\sin v}{\sin u}$ $\Longleftrightarrow$ $\sin u\cdot\cos (x+v)=\sin v\cdot\cos (x+u)$ $\Longleftrightarrow$

$\sin (x+v-u)=\sin (x+u-v)\ \ (u\ne v)$. But $b\ne c$ $\Longleftrightarrow$ $u\ne v$.

Therefore,  $(x+v-u)+(x+u-v)=\pi$ $\Longleftrightarrow$ $x=\frac{\pi}{2}$, i.e. $AM\perp AB$.[/color]


A bowl contains only red marbles, blue marbles, and green marbles. The probablilty of selecting a red marble from the bowl is $\frac{3}{13}$. The probablilty of selecting a blue marble is $\frac{2}{5}$. There are fewer than 100 marbles in the bowl. What is the probability of selecting, at random and without replacement, a green marble and then a red marble from the bowl on the first two selections?
Edit: sorry.  :blush: I thought I was in the mathcounts forum. Can a mod move this? Or it this okay where it is?

[hide]There must be 65 marbles, because 65 is the only positive multiple of lcm(13,5) less than 100. $\frac{3}{13}$ red marbles becomes $\frac{15}{65}$, and $\frac{2}{5}$ blue marbles becomes $\frac{26}{65}$. 24 marbles remain, so there are $\frac{24 \cdot 15}{65 \cdot 64}$, which simplifies to our answer, $\boxed{\frac{9}{104}}$[/hide]

[quote="diophantient"][hide]
Obviously there are an integer number of marbles. The number of marbles must be a multiple of 13. It also must be a multiple of 5.
So the number of marbles must be a multiple of 65.
Since our answer is under 100, $\boxed{65}$[/hide][/quote]
The question wasn't for how many marbles there were. It asked for the probablity of pulling a green and then a red without replacement.


For example, does 3 to the power of -2 = -9 or positive 9?

I've heard an explanation on negative exponents that I'm not sure is true. If it's an even negative exponent, then you'll get a positive answer, and if it's an odd negative exponent, you'll get a negative answer. Is that true?

Neither.
$ 3^{\minus{}2} \equal{} \frac{1}{3^2} \equal{} \frac{1}{9}$, with nothing to do with positive/negative.
Basically, when you have a negative exponent, it's the reciprocal of the positive exponent of the same number.

[quote]it's the reciprocal of the positive exponent of the same number.[/quote]
So it's the reciprocal of a positive 2? What do you mean by reciprocal?

3 to the power of negative 2 turns into 1/9th?

[quote="leroyjenkens"]3 to the power of negative 2 turns into 1/9th?[/quote]

Yes because if $ 3^2 \cdot 3^{\minus{}2} \equal{} 3^{2 \plus{} (\minus{}2)} \equal{} 1$ (by the rules of exponents), then $ 3^{\minus{}2}$ must be equal to $ 1/9$.

[quote="gauss202"][quote="leroyjenkens"]3 to the power of negative 2 turns into 1/9th?[/quote]

Yes because if $ 3^2 \cdot 3^{ \minus{} 2} \equal{} 3^{2 \plus{} ( \minus{} 2)} \equal{} 1$ (by the rules of exponents), then $ 3^{ \minus{} 2}$ must be equal to $ 1/9$.[/quote]
Ok thanks. One more question. Is -5^2= -25 or 25? I'm getting two different answers from different places. Is a negative base always negative regardless of the exponent?

By writing $ (\minus{}5)^2$, you're taking the number -5, and squaring it.

By simply writing $ \minus{}5^2$, you are implying that you are taking the negative of the square of 25, so the answer would be $ \minus{}25$.

Moved to Classroom Math.

The reciprocal means the number that, when multiplied by the original, equals 1. The easiest way to obtain the reciprocal is putting it as the denominator, and having one as the numerator.

e.g:

The reciprocal of $ a$ equals $ \frac{1}{a}$

Also, memorize the formulas given above; this will help avoid further confusion.

[quote="7h3.D3m0n.117"]By writing $ ( \minus{} 5)^2$, you're taking the number -5, and squaring it.

By simply writing $ \minus{} 5^2$, you are implying that you are taking the negative of the square of 25, so the answer would be $ \minus{} 25$.

Moved to Classroom Math.[/quote]
So if it's written as -5^2 or whatever the exponent is, it's going to be a negative number as the answer. But if it's written as (-5)^2 or any even number will give you a positive answer and any odd number will give you a negative answer?
[quote]The reciprocal means the number that, when multiplied by the original, equals 1. [/quote]
Multiplied by the original what?
[quote]The easiest way to obtain the reciprocal is putting it as the denominator, and having one as the numerator. [/quote]
And I do that with every number that has a negative exponent?

Oh and I have one more exponent question. Say for example I have the equation (5 + 5)^2. I'm supposed to add the 5's together to make (10)^2, instead of making it (5^2 + 5^2), right? Doesn't that contradict the order of operation?

Sorry: The reciprocal of a number is what you multiply that number by to get 1. It's easier to think of it as $ \frac{1}{a}$

The order of operations is not contradicted because it goes:

Parentheses

Exponents

Multiplication/Division

Addition/Subtraction

Exponents do not distribute.

[quote="fishythefish"]Sorry: The reciprocal of a number is what you multiply that number by to get 1. It's easier to think of it as $ \frac {1}{a}$[/quote]
edit: Oh nevermind, I'm dumb. I edited cause I figured this out in my sleep. lol. I woke up thinking "oh crap I hope he didn't respond yet"

[quote]The order of operations is not contradicted because it goes:

Parentheses

Exponents

Multiplication/Division

Addition/Subtraction

Exponents do not distribute.[/quote]
Oh yeah. I got mixed up because you can simplify the expressions.
Thanks.

Actually I have another question. This is in regards to the parentheses around the negative base.

For example -2^2(3). The answer to that will be -12. But if I had put it in parentheses, like this (-2)^2(3). It would be a positive 12 instead of a negative. So putting parentheses can change the answer, so putting parentheses needs to be done with caution. But that's not what I was taught. I wasn't taught that the parentheses around a single number made a difference like that.
So when do I know when I can put the parentheses around the number or not?

I'll show you what I'm referring to.
On this page http://www.purplemath.com/modules/evaluate.htm the top question is a^2b  a= -2, b= 3
He puts the answer like this (-2)^2(3). If I had done it, I wouldn't have put the 2 in parentheses and I would've got the wrong answer. How do I know when and when not to use parentheses in a situation like that?
Thanks.

[quote="jackdillon"]You put parentheses around numbers that are operated apon, so if you are substituting numbers into an equation, you should always use them.[/quote]
Thanks for the response.

So I should always put parentheses around numbers that I substitute into an equation? Because as you can see from my example, the parentheses make the difference between having the right answer and the wrong answer.
What I don't understand, really, is why I need to put parentheses around the first number.

parentheses are extremely important. parentheses are arguably as important as the order of operations. here are a few examples of what can happen if you dont pay attnetion to your parentheses...

5-(3-2)=5-1=4
(5-3)-2=2-2=0;4 doesnt equal 0.

also, about the (5+5)^2 problem, theres this cool formula about a general (a+b)^2.

$ (a\plus{}b)^2\equal{}a^2\plus{}2ab\plus{}b^2$

so (3+4)^2=7^2=49obviously. but using the formula....let a=3 and b=4

(3+4)^2=(3)^2+2(3)(4)+(4)^2=9+24+16=49.
but in general, unless you've had some exposure to algebra, try to follor order of operations instead of using shortcuts like this lol.
you could use this trick to evaluate 1001^2. 1001^2=(1000+1)^2=1000^2+2(1000)(1)+1^2=1,000,000+2,000+1=1002001 NEATO!!

Basically, the reason you put parentheses around the -2 was because of the confusion of whether the "-" applied to the  whole quantity, or just the "2". If it was part of an expression, without the parentheses it would have been subtraction.

BTW: (-2)^2(3)=(-2)^6, not (-2)(6).

(-2)^2(3)=(-2)^6=64  :lol:

A note on the original question:

The motivation for defining negative exponents this way is so that a familiar law of exponents holds:  in particular, we want $ a^{x \plus{} y} \equal{} a^x a^y$ even if one (or both) of $ x, y$ are negative.  In order for this to hold, it is necessary that

$ a^{x \minus{} x} \equal{} a^x a^{\minus{}x} \equal{} a^0 \equal{} 1 \implies$
$ a^{\minus{}x} \equal{} \frac{1}{a^x}$.

Thanks for the responses. I think Mathking misunderstood my problem. My problem is I don't know when I'm supposed to put parentheses around negative numbers. Like if I'm given the problem -2^2+8 for example. The original problem was A^2+B, but I filled in the blanks where the variables were. If I fill it in and put (-2)^2 for the beginning part instead of -2^2 for the beginning part, it changes the answer. (-2)^2 gives you a different answer than -2^2.

[quote]#16Basically, the reason you put parentheses around the -2 was because of the confusion of whether the "-" applied to the whole quantity, or just the "2". If it was part of an expression, without the parentheses it would have been subtraction.
[/quote]
Why would the minus sign in -2^2(3) apply to the whole thing and not just the 2 in front?
[quote]BTW: (-2)^2(3)=(-2)^6, not (-2)(6).

(-2)^2(3)=(-2)^6=64[/quote]
The 3 is a big 3, it's not part of the exponent.

[quote="fishythefish"]If you want the 3 to be a "big 3", then consider starting to use LaTeX. Go to the guide in the Wiki for help.

I'll leave the explaining about the negative sign to someone else. I'm having a little trouble making it this basic.[/quote]
I can't see why. I'm breaking it down by asking simple questions about a step I don't understand in the process. Each step has a reason for why it is how it is and if I don't understand that step, how else can I begin to understand it unless I ask?

[quote="leroyjenkens"]My problem is I don't know when I'm supposed to put parentheses around negative numbers. [/quote]

When in doubt, [b]always.[/b]

[quote="fishythefish"]Exponents do not distribute.[/quote]
Here's question #1 on a the multiple choice section of a high school contest that I'm responsible for:

$ (a^{-2}+b^{-2})^{-1}$ equals to

\[ \mathbf{(A)} \ a^2b^2 \quad \&
\mathbf{(B)} \ a^2+b^2  \quad \&
\mathbf{(C)} \  \frac{1}{a^2+b^2} \quad \&
\mathbf{(D)} \ \frac{a^2+b^2}{a^2b^2} \quad \&
\mathbf{(E)} \ \frac{a^2b^2}{a^2+b^2} \]
Let's just say that this wasn't even the easiest question on the test - there were people who got it wrong.

[quote="Kent Merryfield"][quote="fishythefish"]Exponents do not distribute.[/quote]
Here's question #1 on a the multiple choice section of a high school contest that I'm responsible for:

$ (a^{ - 2} + b^{ - 2})^{ - 1}$ equals to
\[ \mathbf{(A)} \ a^2b^2 \quad \& \mathbf{(B)} \ a^2 + b^2 \quad \& \mathbf{(C)} \ \frac {1}{a^2 + b^2} \quad \& \mathbf{(D)} \ \frac {a^2 + b^2}{a^2b^2} \quad \& \mathbf{(E)} \ \frac {a^2b^2}{a^2 + b^2}
\]
Let's just say that this wasn't even the easiest question on the test - there were people who got it wrong.[/quote]
It's not C?
[quote]When in doubt, always.[/quote]
So always put the parentheses around the negative numbers. I don't need to do that for non-negatives though, right?

[quote="long14893"](E) is an right answer .It seems to be very easy ...[/quote]
I didn't say that it was supposed to be hard, but people did miss it. Yes, (E) is the correct answer. The most commonly given incorrect answer was (B).

In case some people are still having trouble with it, I'll break it down.

[hide="Clickness"]

You have $ (a^{\minus{}2} \plus{} b^{\minus{}2})^{\minus{}1}$.

First, according to the Order of Operations, we do the operation inside of the parentheses.

At first, you may think to simplify the problem as $ \frac{1}{a^2\plus{}b^2}$.

However, this is incorrect, as you add the terms together and they cannot be grouped like that. You do them individually and then use a common denominator to group them together.

[u]Therefore, the process then goes like this:[/u]
First, you simplify each exponent.
$ \frac{1}{a^2} \plus{} \frac{1}{b^2}$

The common denominator is fairly easy to find. You simply multiply both of the denominators to get $ a^2b^2$

Multiply to get the common denominators on both. It now looks like this:
$ \frac{b^2}{a^2b^2} \plus{} \frac{a^2}{a^2b^2}$

Now we add:
$ \frac{a^2\plus{}b^2}{a^2b^2}$

We're not done yet. Remember that $ \minus{}1$ exponent? Now we flip what we currently have for the final answer of: $ \boxed {\frac{a^2b^2}{a^2\plus{}b^2}}$ or $ \boxed{(E)}$ [/hide]

Fairly simply once you know what you're doing. Just remember the basics. Small mistakes could easily get you an answer of $ \boxed{(B)}$ or something else. It never hurts to double-check your work. :) Hope that helped.

@leroy: Just remember that subtracting is the same as adding a negative. Now see if you can figure out why this would cause confusion if there were no parentheses.

Take this as an example:

$ 3\minus{}(\minus{}2)^2$

Now, work that out. The answer is -1, right? That is because the parentheses say "treat this as a whole quantity". In this case, the parentheses are telling you to square the -2, sign and all. That's what parentheses are for. They tell you to perform all operations on the quantity as a whole.

Another way of expressing this would be:

$ X\equal{}\minus{}2   \rightarrow$
$ 3\minus{}X^2$

Compare this to:

$ 3\minus{}\minus{}2^2$

It's the same thing without the parentheses.
Because the parentheses are gone, this is not interpreted as "3 minus -2 squared", it's interpreted as 3 minus the negative of $ 2^2$.  Now work out the new answer. It's completely different. (Remember, subtracting a negative means adding the positive. e.g. 2--2=2+2=4)
Basically, due to the order of operations, anytime you need to perform any operations on parentheses, you need to perform that operation on the quantity inside.

Thanks for the response, fishy.
My problem is that with your example, the -2 is right next to the minus sign on the right side (which means the two minus signs are right next to each other), so I was told always to put parentheses around negative numbers to keep negative signs seperate from other signs. Like in your example of 3 - (-2). But for instance if the 2 comes first in the equation, ie -2 - 3, is it still necessary to put the parentheses?
Which brings me to my next point. If that 2 has an exponent, parentheses around that -2 would cause the answer to be different than if it didn't have parentheses.

For a short question and answer, should I just put parenthses around all negative numbers? Because I'm wondering if there would ever be an instance in which the negative number isn't supposed to be in parentheses.

Hope all this makes sense. Thanks.

Well, generally, when you have 3--2. (3 minus a negative two), you don't necessarily [i]need[/i] parentheses. However, some schools tell kids to put parentheses around the negative 2, just for clarity. You don't need them. That eliminates one problem right there. You see, when you put the negative two in front, it eliminates all this confusion. :)

If you really still need the parentheses, just think them or something else in stead. Learn to diifferentiate. Since you don't need them, it would be an added burden of constantly adding them. Break that bad habit. :lol:

for your case, 3 - -2 does not have to be changed to 3 - (-2), it just makes it easier to visualize that way. If it was the negative of 2^2, just do

3 - (-2^2)

Yeah, a lot of people get confused with the negtive signs, but just remember, parenthesis around a number means that whole term squared or cubed, etc..
i.e. (-2)^2 means negative 2 squared as opposed to -2^2 which means the negative of two squared aka -4

[quote]Well, generally, when you have 3--2. (3 minus a negative two), you don't necessarily need parentheses. However, some schools tell kids to put parentheses around the negative 2, just for clarity. You don't need them. That eliminates one problem right there. You see, when you put the negative two in front, it eliminates all this confusion.[/quote]
This is what I was confused about because...

(-2)^2= 4
But
-2^2= -4
The parentheses are the difference between the right and wrong answer. See how one results in negative 4 and the other in positive 4. So my original question was just asking how do I know when to put the parentheses and when not to? If the answer is always, then I'm cool with that.
[quote]i.e. (-2)^2 means negative 2 squared as opposed to -2^2 which means the negative of two squared aka -4[/quote]
Exactly.
So if I'm substituting variables, like this A = 2
And the question is -A^a, when I substitute the 2 for the A, I need to put parentheses around that negative 2, right?
Or what if the question wants me to get a negative 4? Which means I'm not supposed to put parentheses? That's where I'm confused. How do I know which answer the question is looking for?

Review PEMDAS if you're not sure.

[url]http://en.wikipedia.org/wiki/Order_of_operations#Examples_from_arithmetic[/url]

[quote]5. Evaluate negation on the same level as subtraction, starting from the left[/quote]

The question has been answered. Be confident in the answer!

[quote="leroyjenkens"]
[quote]i.e. (-2)^2 means negative 2 squared as opposed to -2^2 which means the negative of two squared aka -4[/quote]
Exactly.
So if I'm substituting variables, like this A = 2
And the question is -A^a, when I substitute the 2 for the A, I need to put parentheses around that negative 2, right?
Or what if the question wants me to get a negative 4? Which means I'm not supposed to put parentheses? That's where I'm confused. How do I know which answer the question is looking for?[/quote]

basically assume as if the whole number is squared when you insert

so with your question. You insert 2 into the equation.

So [b]only[/b] the two is squared, since that is what A is.

so -2^2=-4

Try some others out.

-b^2 b=3
-b^2 b=-2
-b^2 b=-5
-b^2 b=4

[quote]basically assume as if the whole number is squared when you insert

so with your question. You insert 2 into the equation.

So only the two is squared, since that is what A is.

so -2^2=-4

Try some others out.

-b^2 b=3
-b^2 b=-2
-b^2 b=-5
-b^2 b=4[/quote]
Oh, duh. Thanks.

So had it been A= -2, the answer would be 4 instead of -4.

For the ones you gave me
-b^2 b=3 The answer is -9 because the negative stays there while you insert the positive 3 and square it, so the answer becomes negative.

-b^2 b=-2 The answer I believe is -4. I insert the -2, square it, that becomes a positive 4, then the negative sign that was there before makes it become a negative 4.

-b^2 b=-5 Same story as above, but the answer is -25.

-b^2 b=4 As I did in the first one, the 4 is positive which I insert, but the negative stays there and makes the answer become a negative number, which is -16.

So are my answers right and is my reasoning correct?

Thanks.

[quote="Ihatepie"]yep, it's all right.

just a bit more to make sure.

-b^3 b=7
b^2 b=5
-b^3 b=-6
(-b)^2 b=4
-b^2 b=2
(-b)^3 b=-5
(-b)^3 b=3
b^3 b=6
(-b)^2 b=-6[/quote]

Alright let's see here.

1) -343
2) 25
3) 216
4) 16
5) -4
6) 125
7) -27
8) 216
9) 36  <- Not so sure about this one. I don't know if I square the -6 and then add the - to it after I get my answer of positive 36, or does the -6 automatically become a positive 6 once I put it in the quotations with that other minus sign?

[quote="leroyjenkens"]I don't know if I square the -6 and then add the - to it after I get my answer of positive 36, or does the -6 automatically become a positive 6 once I put it in the quotations with that other minus sign?[/quote]

Where are the parentheses?

well you are right with 1-8.

I'll let you figure out if you are right for number 9 yourself. (read the post above.)

Remember the order of operations.

First I do what's in the parentheses, which is make the negative of a negative a positive and then I do the exponents?

[quote="Ihatepie"]yep, so you change -6 to 6 by taking it's negative, and you square it.[/quote]

Cool, thanks for all your help.


The students in Mrs. Reed's English class are reading the same
760-page novel. Three friends, Alice, Bob and Chandra, are in the
class. Alice reads a page in 20 seconds, Bob reads a page in 45
seconds and Chandra reads a  page in 30 seconds. \\ \\
Chandra and Bob, who each have a copy of the book, decide that they
can save time by `team reading' the novel. In this scheme, Chandra
will read from page 1 to a certain page and Bob will read from the
next page through page 760, finishing the book. When they are
through they will tell each other about the part they read. What is
the last page that Chandra should read so that she and Bob spend the
same amount of time reading the novel?

Since the problem only mentions Bob and Chandra, we can kick Alice out of the reading club for now. Their reading ratios are 2:3, so Chandra will read 3/2 times as many pages as Bob, so Chandra should read $ \frac{3}{5}\cdot 760\equal{}456$ pages. Pages 456 happens to be the 456th page in the book, so Chandra will stop on page 456. Now, we can let Alice back in.

Quit spamming. This was already answered. Thank you.

[quote] Chandra will read 3/2 times as many pages as Bob[/quote]

how do you know that

Their reading ratios are $ 2$ to $ 3$, so she reads $ \frac{3}{2}$ as much as he does.


function
number theory proposed
number theory

Let $n$ be a positive integer which ends with exactly $k$ zeros but $n>10^k$ holds.
For an integer $n$, only $k=k(n)\geq 2$ is known. At least how many ways (as function of $k=k(n)$ ) can $n$ be written as the difference of two perfect squares?


trigonometry
algebra
binomial theorem
combinatorics unsolved
combinatorics

Compute the sum: $ S_n\equal{}\displaystyle\sum_{k\equal{}0}^{n}q^k \binom {n}{k} \cos (a\plus{}kx),$ where $ a,q,x$ are real numbers.

I would approach this as follows.

$ S_n \equal{} cos(a) \sum_{k \equal{} 0}^{n} q^k {n\choose k} cos(kx) \minus{} sin(a) \sum_{k \equal{} 0}^{n} q^k {n\choose k} sin(kx)$

Further, one can evaluate using binomial theorem with $ sin (a) \equal{} \frac {e^{ika} \minus{} e^{ \minus{} kia}}{2i}$ and $ cos(a) \equal{} \frac {e^{ika} \plus{} e^{ \minus{} kia}}{2}$

Do we have a combinatorial interpretation ?


i am thinking about getting one of the AoPS books (ex, vol 1), which one do your reccomend?

and whats the difference between the 11$one and the 30$ one?, (it says solutions text)  does that mean that one has solutions and one doesnt?

The solution manual and the text are two different books.  The solution manual contains full solutions (not just answers) to all the questions in the text.  Almost everyone who buys the text for their own individual work buys the solutions as well.


Hi
I should have 3-4 way of solving this problem.
I appriciate everyonr who help me
thank you.

(sorry for bad english)

What is the problem?
Anyway, it is known that any ordinary (ruler and compass) construction can be performed with compass only (Mascheroni's theorem).

I know.
but i'm looking for diffrent ways of solving the problem not the mascheroni's way.
i'm really need help!!!thank you :)
thanks.

[quote="shinytick"]I know.
but i'm looking for diffrent ways of solving the problem [/quote]

Any problem ? Or do you have some specific problem that you're referring to ?

khoobi  :!:  :!:  :?:  :!:  :?:
ina migan masalehat koo :?:  :?:
what and where is your problem :?:  :?:  :?:  :?:  :?:  :mad:

the problem is just this:
How can we carry out the constructions with a compass only:
this means:
1-finding the intersection of two lines
2-foinding the intersection of a line and circle

and i know the macheroni's way for solving the problem.i;m looking for new and diffrent ways of solving this problem.

kheyli mamnoon man motavajjeh shodam ina chi migan ina nefimahman man chi migam.

:rotfl:  :rotfl:  :rotfl: Sorry but let me just Speak persain:
kheli bahal bod.Shoma masalaro matrah nakardi chon fekr kardid az mozoye masale mifahman :?:  :?:
Be nazaret ye zare ajib nist ke begi :

[quote]Hi
I should have 3-4 way of solving this problem.
I appriciate everyonr who help me
thank you.
[/quote] bad nagi masale chiye ? ;)  ;)  ;)
vali rast migi , moshkele in hast , khob nemifahman. :D

PS. aga in alamataro ( :rotfl: ) jedi nagir , shokhiye  :)  :) your sincerely


We choose 5 points $A_1, A_2, \ldots, A_5$ on a circumference of radius 1 and centre $O.$ $P$ is a point inside the circle. Denote $Q_i$ as the intersection of $A_iA_{i+2}$ and $A_{i+1}P$, where $A_7 = A_2$ and $A_6 = A_1.$ Let $OQ_i = d_i, i = 1,2, \ldots, 5.$ Find the product $\prod^5_{i=1} A_iQ_i$ in terms of $d_i.$

We have $ \frac {\overline{A_iQ_i}}{\overline{Q_iA_{i + 2}}} = \frac {\bigtriangleup A_iPA_{i + 1}}{\bigtriangleup A_{i + 2}PA_{i + 1}},i = 1,2,3,4,5$

Mutiply all these five equations and we can get $ \prod_{i = 1}^{5}\frac {\overline{A_iQ_i}}{\overline{Q_iA_{i + 2}}} = 1$ ..........(i)

However, suppose $ \overrightarrow{A_iO}$ meets the circle at $ {A}'_i$, then the circle power of $ Q_i = \overline{A_{i + 1}Q_i}\cdot\overline{Q_i{A}'_{i + 1}} = (1 + d_{i + 1})(1 - d_{i + 1}) = \overline{A_iQ_i}\cdot\overline{Q_iA_{i + 2}}$

That is ${ \overline{A_iQ_i}\cdot\overline{Q_iA_{i + 2}} = (1 - d_i^2}),i = 1,2,3,4,5$
Mutiply all these five equations and we can get ${ \prod_{i = 1}^{5}\overline{A_iQ_i}\cdot\overline{Q_iA_{i + 2}} = (1 - d_i^2})$ .........(ii)

$ \sqrt {(i)\times (ii)}\Rightarrow \prod_{i = 1}^{5}{\overline{A_iQ_i}} = \prod_{i = 1}^{5}\sqrt{(1 - d_i^2)}$

QED#


Can someone help me come up with the right formula to calculate the number of subscriptions that we'd be getting paid for on a given month k given the following assumptions:

    Here are my assumptions:

    1. Assume 250 users sign up to use a subscription-based service during
    the 1st month and pay for their subscriptions.
    2. Monthly subscription is automatically renewed unless the user cancels.
    3. Assume that 20% of them (50) will drop out, i.e. their membership will not be renewed during the second month, so only 200 previous subscribers stay on.

    4. Assume that 12% more subscribers start using our service during the
    second month as compared to the first month– which means that we will
    have 280 (250 * 1.12) new subscriptions during that second month.
    5. Add to that the 200 users that stayed on from the previous month (see
    #3) and get 480.
    6. The same pattern continues repeating itself, where the 20% of the previous month's new subscribers drop out during a given month and 12% more sign up each month than have signed up the previous month.

    What is the formula to calculate the number of subscriptions that we will be getting paid on a given month k?

[hide="Solution"]

Each month, there will be 80% of the subscribers from the last month; there will also be 112% new subscribers, so the number of subscribers is 192% of the old number of subscribtions.  Then the number of subscribers after $k$ months is $250\times (1.92)^{k}$.[/hide]

That formula can't be the right one because it seems to grow too fast.

No, I think it works.
You can check the numbers for the first few months if you want.

If you plug in 10, the answer is 170,197.1548
It's highly unlikely that a business can go from 250 subscriptions to 170,197 in the span of 10 months.

Perhaps it's your assumptions that are wrong, then.  Why did you multiply by $1.12$, and then add on the users from the previous month again?

It intuitively seems to make sense to me that if there's automatic subscription renewal, then 80% of the previous month's subscribers will stay on during the present month.
The only question then is at what rate the amount of new subscribers grow.  It seems logical that a business (through advertising, etc.) can attract 12% more people than it attracted the previous month.
What do you think?

The problem is customer base. If I own a store and I have 12% subscriber growth then I get (say) 100 customers this month, 112 the next, 125 the third month, etc. This means perhaps 20% customer loss rate and 32% customer gain rate. However, in this example, we have a 20% loss and 112% gain rate. In other words, If I have X subscribers this month I will gain more than X new people over the month.

New problem: using the store data from this post, at what month will less then 1% of the current customers be among the originals?

[b]Bill The Maniac:[/b]  I believe the 112% growth refers to the number of [i]new[/i] subscribers, not the number of [i]total[/i] subscribers.

The assumptions notwithstanding, therefore, samath's approach is quite off, but the whole statement is quite confusing.

[quote="mathman25"]the 20% of the previous month's [i]new[/i] subscribers drop out during a given month and [i]12% more[/i] sign up each month than have signed up the previous month.[/quote]

So, let me see if I get this straight.  In a given month, some number of new subscribers join, which is 12% more than the number of new subscribers that joined last month (so $1.12$ times as many new subscribers).  But then 20% of the subscribers from [i]only the previous month[/i] (and [i]not[/i] any of the other months) drop out?

The way it's worded here, if a subscriber does not drop out within a month, then they are assumed to stay forever.  Was that your intention?

My solution matches exactly with the problem :P

We all know that. However, the information given in the problem is a little odd, giving a 92% growth rate. My growth rate converges to 12% in my alternate interpretation.

The 112% growth definitely refers to the number of new subscribers, not the number of total subscribers.

I'm trying to say the following:  the amount of people paying for month k subscriptions are from two categories:
1) 80% of the people who paid last month (k-1)
2) The new subscribers.  The amount of [i]new[/i] subscribers in month k are determined by the amount of [i]new[/i] subscribers in month (k-1) * 1.12

So if month 1 has 100 subscribers.  Then 20 of these drop out by month 2.  Also, 112 [i]new [/i]subscribers sign up in month 2.  So the total paying subscribers in month 2 is 112+80=192.  The 80% of 192 stay on for month 3 = 153.6 (I'm not rounding).  The amount of [i]new[/i] subscribers in month 3 is 112 * 1.12 = 125.44.  The total paying subscribers in month 3 is 125.44+153.6=279.04. etc.

In that case, it's my formula that should work, not 1.92^i.

[quote="mathman25"]1) 80% of the people who paid last month (k-1)[/quote]

Okay, that's what I thought.  And that isn't what you said before.

[hide="Solution"] We have the recursion

$s_{k+1}= 0.8s_{k}+250(1.12)^{k}$

Where $s_{1}= 250$.  Clearly $s_{k}$ is in the form $A \cdot 0.8^{k}+B \cdot 1.12^{k}$.  Substituing and simplfying,

$1.12B = 0.8B+250 \Leftrightarrow$
$B = 781.25$

Then $s_{1}= 0.8A+1.12B = 250 \Leftrightarrow$
$A =-781.25$

So we have

$s_{k}= \boxed{-781.25 \cdot 0.8^{k}+781.25 \cdot 1.12^{k}}$[/hide]
The number of subscribers in month $10$ is then $\approx 2343$, much more reasonable.

I must have posted it somewhere else. *looks for it* *can't find it*

T_i=N_0(1.12^(i-1))(1-K^i)/(1-K) where

T_i=ith total number of customers
N_0=250
K=.8/1.12


geometry
integration
real analysis
real analysis unsolved

Find the area of the minor segment cut off from $ r \equal{} a(1 \plus{} cos \theta)$ by the half line $ \theta \equal{}\pi/2$.

I did it like,
  $ \frac {1}{2}\int r^2 d \theta \equal{} \frac {a^2}{2}\{\frac {3}{2} \theta \plus{} 2 sin \theta \plus{} \frac {sin 2 \theta}{4}\}$

But I can't get the limits of this, means $ \theta$ where to where.

By "minor segment" I'm assuming they mean the bump that occurs from $ \frac{\pi}{2}$ to $ \pi$. (Try sketching it, that is the only thing that makes sense to me).


I need an advice. I need a program to make any kind of animations ( i saw some on Art of problem solving).
I'm writing an scientific article and i need it.

Indeed, all of our [url=http://www.artofproblemsolving.com/Gallery/AoPS_G_Gallery.php]animations[/url] are written using [url=http://www.macromedia.com/software/flash/flashpro/?promoid=BINT]Macromedia Flash[/url].


I sent my solutions to USAMTS and got it postmarked at 8:00 AM Oct. 13, 2009

I see a check on the "My USAMTS" page, so my solutions were received.

However, I am a little worried. No problems have been graded yet.

Just wondering, usually, how long does the grading take?

nobody has been graded yet. scores aren't released until NOV.

The scores should be available in the second week of November.

Are they all uploaded at the same time? Or do you post the scores as you grade them?

They will all be posted at the same time.  Not just scores, but commentary as well.

[quote="mathwizarddude"]I sent in my solution by mail before the day it is due, but I haven't even received the check mark yet... is there something wrong?[/quote]Please email with usamts@usamts.org with the details, and include your USAMTS ID#.

To our knowledge, all of the round 1 solutions that we have received have been logged in, but we did get a small handful with no ID# on them.

Hmm, I'm pretty sure I have my ID number on every page of the solution... so could there be another reason for that? My USAMTS ID number is 11825, so have the staff received the solution with this ID number? Thanks.

We have received it.  Although it was mailed on time, we only received it this weekend.

Update: due to the unexpectedly large number of participants this year (up over 20% from last year), the grading is taking longer than expected.  We are aiming for the week of Nov 16-20 for the Round 1 scores and comments to be available.

Please keep in mind that one of the unique features of the USAMTS is that you get written comments to your solutions, not just numeric scores, and this takes time.


You are blindfolded, and there are $6$ cubes in front of you. Of these $6$ cubes, $3$ are red, $2$ are blue, and $1$ is yellow. If you pick four cubes at random, what is $m^n$ if $\frac{m}{n}$ is the probability that you will obtain $2$ red cubes and $2$ blue cubes?

[hide="my solution"]$\frac{m}{n} = \frac{5*3*2*2!}{30} = \frac{1}{6}$
$m^n = 1$[/hide]

[hide="Err..."]
There are $\frac{4!}{2!2!} = 6$ different ways to pick $2$ red and $2$ blue, and the odds of any one of these configurations is $\frac{ 3 \times 2 \times 2 \times 1}{6 \times 5 \times 4 \times 3} = \frac{1}{30}$, so the probability is $6 \left( \frac{1}{30} \right) = \frac{1}{5}$, so $m^n = 1^5 = \boxed{1}$...?

This seems like a very bad problem because $m^n = 1$ whenever $m = 1$ so it is very possible to get the wrong value of $n$ and still get the problem right.

Also, it seems too easy for Intermediate.
[/hide]

Edit:  Your solution seems wrong.  It seems like the $5$ term is for picking either red or blue, but as there aren't an equal number of reds and blues this way of counting will result in an incorrect answer later on.

Edit:  Furthermore, $\frac{5 \times 3 \times 2 \times 2!}{30} = 2$...  :huh: however, if you used $\frac{5 \times 3 \times 2 \times 2!}{6 \times 5 \times 4 \times 3}$ then you would get $\frac{1}{6}$, which still disagrees with what we have gotten...

[hide="hmm"]$\frac{m}{n}$ is also the probability that, in picking two cubes at random, you get 1 red 1 yellow.  So label the cubes as $R_1$, $R_2$, $R_3$, $B_1$, $B_2$, $Y$.  Then there are $6*5=30$ ways to pick two cubes, order mattering, and the number of ways to pick the yellow and 1 red is 6: $(Y, R_1), (Y, R_2), (Y, R_3), (R_1, Y), (R_2, Y), (R_3, Y)$.  Thus, \[ \frac{m}{n}=\frac{6}{30}=\frac{1}{5}\implies m^n=\boxed{1} \][/hide]
Our final answers were the same, but something went wrong.  How did you get your answer?

Doesn't $\frac{4!}{2!2!}$ treat R1R2B1B2 and R2R1B2B1 as the same permutation? Because the sample space treats it as two seperate permutations.

This is the part I am confused on.

EDIT: Nevermind, I get can visualize this now.


HMMT
algebra
polynomial
analytic geometry
function
calculus
derivative

The polynomial $3x^5 - 250x^3 + 735x$ is interesting because it has the maximum possible relative extrema and points of inflection at integer lattice points out of any quintic. Find the the sum of the x coordinates of said points.

[hide="this is what confuses me"]The relative extrema will occur when $15x^4 - 750x^2 + 735 = 0$, which is at $x = 1, 7, -1, -7$. The points of inflection will have x-coordinates that satisfy $60x^3 - 1500x = 0$ which are $x = 0, 5, -5$. Yet the answer is $75$.[/hide]

Tha looks like it might be a sum of squares. Are you sure you have the right statement?

Since the function $f=3X^5-250X^3+735X$ is odd, it's derivative is even(simetric about x-axis), and it's second derivative is odd again(simetric about origin). So the x-coordinates of their roots is $0$ even without knowing their value. Probably the problem is wrong.


Two strips of width 1 overlap at an angle of $\alpha$ as shown. The area of the overlap (shown shaded) is

[asy]
pair a = (0,0),b= (6,0),c=(0,1),d=(6,1);
transform t = rotate(-45,(3,.5));
pair e = t*a,f=t*b,g=t*c,h=t*d;
pair i = intersectionpoint(a--b,e--f),j=intersectionpoint(a--b,g--h),k=intersectionpoint(c--d,e--f),l=intersectionpoint(c--d,g--h);
draw(a--b^^c--d^^e--f^^g--h);
filldraw(i--j--l--k--cycle,blue);
label("$\alpha$",i+(-.5,.2));
//commented out labeling because it doesn't look right.
//path lbl1 = (a+(.5,.2))--(c+(.5,-.2));
//draw(lbl1);
//label("$1$",lbl1);[/asy]

$\text{(A)} \ \sin \alpha \qquad \text{(B)} \ \frac{1}{\sin \alpha} \qquad \text{(C)} \ \frac{1}{1 - \cos \alpha} \qquad \text{(D)} \ \frac{1}{\sin^2 \alpha} \qquad \text{(E)} \ \frac{1}{(1 - \cos \alpha)^2}$

Please hide your answer. :dry:

Anyway, it would be nice if you gave some explanation.

[hide="Answer"]I think that the figure formed is a rhombus... so, $\sin \alpha=\frac{h}{x}$, where $x$ is the side length of the rhombus. However, we know that $h=1$, and $A=xh=x$. From the previous equation, we have $x=\frac{h}{\sin \alpha}=\frac{1}{\sin \alpha}\Rightarrow \boxed{B}$.[/hide]


function
linear algebra
linear algebra unsolved

Hi,
I have an interesting problem on my hands and as it is of general interest - I suppose - I hoped you guys could give me some pointers.

Here is a particular example that would help me explain it. Let's say that in a company you decide to test and assess your employees, by giving them each a mark based on single skills. The skills are structured in a tree, and each node in the tree can get a grade, like in the following image:

http://www.interact-studio.com/alex/skill-tree.jpg

(sorry for the poor originality while choosing the example :) )

Now, when it comes to grading the employees, I use the following grades (this is a must for me):
0 - unknown
1 - poor, the lowest grade
2 - basics
3 - advanced
4 - expert, the best grade

Assume that after grading two employees, Tom and Jerry, their skills and grade look like this (please note that 0 means that they have not been graded for that skill). Just look at the red numbers, I'll explain next what the blue ones are

http://www.interact-studio.com/alex/tom-jerry.jpg

Thanks for your patience, as now comes the true problem. I hope I made myself clear until now. As you may see, Tom (the first guy) has 0 (unknown) at Windows XP, as no-one has graded him on that particular skill. However, by using the children nodes of Windows XP (SP1 and SP2) for which he has a 3 and a 2, we could approximate that for Windows XP he has a skill of 2.5 (the blue number :) ). By generalising, for each unknown node we could use some sort of an average (I am now using the Arithmetic mean) based on its children's grades.

We can go all the way up to the root (please look at the image again), and after some simple computation, we can see that Tom gets an average of 2,16, while Jerry gets an average of 2,33. To some this may seem right, but this is not the case. Just compare the known grades from Tom to those of Jerry, which indicate that Tom is better than Jerry and really this is the case. It's just that Tom didn't get graded on the other skills Jerry has been.

So, my question is what kind of mean do you suggest I would use ? (as the arithmetic one doesn't seem to do the job). Please suggest anything which comes to mind, as I am open to interesting ideas.

Thanks for your patience and I hope I will wake up some grey cells :),
Alex

Well I think there is some problem in your rating system... let me try to explain:

Suppose you have a function that does the mean of a set of numbers, let's call it $f$. For example, the arithmetic mean is
$f(a_{1}, \ldots, a_{n}) = \frac{a_{1}+\cdots+a_{n}}{n}$
and the geometric mean is
$f(a_{1}, \ldots, a_{n}) = \sqrt[n]{a_{1}a_{2}\cdots a_{n}}$.
But generally, you want a mean to satisfy the condition
$\text{min}(a_{1}, \ldots, a_{n}) \le f(a_{1}, \ldots, a_{n}) \le \text{max}(a_{1}, \ldots, a_{n})$

According to your tree, Jerry has a 3 for Windows XP even though he has a 2 for both of the children of that node. But no mean will give $f(2, 2) = 3$. Actually any mean should give $f(2, 2) = 2$. So if your rating system allows for a parent node to be larger than all of its children (or smaller than all of its children), then you need to account for that in your estimation of unknown skills.

It's hard to tell what kind of function you want -- it should be based on your rating system. Think about what kind of factors go into the ratings for parent nodes. Do you give a bonus to Windows XP because Jerry is familiar with both service packs? Then you might want to include the number of non-1 children of a node in your estimation function for that node.

I hope this helps.

PS. Another thing I just noticed is that when you do Tom's mean for the Windows node, you counted 0 as 0. Shouldn't you count it as 1? Why should "unknown" be worse than "unfamiliar with the software"? Then, Tom gets a 2.5, and is still better than Jerry even though Jerry somehow has a 3 for Windows XP even though his skill with each SP is 2. And if Tom turns out to know Windows 2000 as well, you got a great deal for hiring him.


Diamonds increase in value by 10\% each year. If you inherit your grandmother's diamond pendant which is worth $ \$36,000.00$ in 1994, what will its value be in 1998?

Hello
new prize =initial prize$ {\frac{11}{10}}^4
              \equal{}$52707.6
thank u.


Another way(a bit shorter)[But unfortunately it is wrong :( ]:
Using AM-GM,we get that $x\leq \frac{1+x^{2}}{2}$ so \[\frac{x^{2}}{x^{2}+x+1}\geq\frac{x^{2}}{x^{2}+1}\cdot\frac{2}{3}\] and the other two cyclic results.Now we simply set $x^{2}=\frac{a}{b}$ etc and we get that the ineq is equivalent to the famous \[\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\geq\frac{3}{2}\] The end :)

[quote="andyciup"] \[\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\geq\frac{3}{2}\] [/quote]
I'm sorry but it is wrong :( (maybe you mistoke it with Nesbitt). Set $a=1$, $b=2$, $c=4$. Then you get $\mathrm{LHS}=\frac{22}{15}<\frac32$.

OOh..sorry bout that..it seemed so nice,I didn`t check it :blush:


The question isn't specific enough, so I think it is safe to assume that the number of terms in each case doubles each time, and uses the next set of terms. Thus, the answer is $ \fbox{B}$, which has $ 8\times2\equal{}16$ terms in the first group and $ 16\times2\equal{}32$ terms in the second group.


1.) Show by modulo division that $ a^3 \equiv a (mod 6)$.
2.) Show by modulo division that $ a^2 \equiv 1 (mod 8)$.

[hide]For the first, $ a^3 \equiv a \pmod{6} \iff a^3 \minus{} a \equiv 0 \pmod{6} \iff a(a\plus{}1)(a\minus{}1) \equiv 0 \pmod {6}$

Clearly, at least one of the three is $ \equiv 0 \pmod{2}$, and one of the three is $ \equiv 0 \pmod{3}$.

For the second, I assume you mean, prove that $ a \equiv 1 \pmod{2} \implies a^2 \equiv 1 \pmod{8}$:

$ a: \equal{}2b\plus{}1 \implies a^2 \equal{} 4b^2\plus{}4b\plus{}1 \equal{} 4b(b\plus{}1)\plus{}1$

Since either $ 2|b$ or $ 2|b\plus{}1$, $ 8| 4b(b\plus{}1)$, and $ a^2 \equiv 0\plus{}1 \equiv 1 \pmod{8}$[/hide]


Let $ x,y,z>0$ such that $ (2\minus{}x)(2\minus{}y)(2\minus{}z)\equal{}xyz$. Prove that:

$ x\plus{}y\plus{}z\le 3$

[quote="Pain rinnegan"]Let $ x,y,z > 0$ such that $ (2 \minus{} x)(2 \minus{} y)(2 \minus{} z) \equal{} xyz$. Prove that:

$ x \plus{} y \plus{} z\le 3$[/quote]
$ x\equal{}y\rightarrow2,$ $ z\rightarrow0.$ :wink:

[quote="arqady"][quote="Pain rinnegan"]Let $ x,y,z > 0$ such that $ (2 \minus{} x)(2 \minus{} y)(2 \minus{} z) \equal{} xyz$. Prove that:

$ x \plus{} y \plus{} z\le 3$[/quote]
$ x \equal{} y\rightarrow2,$ $ z\rightarrow0.$ :wink:[/quote]

Yes, sorry i was careless. I meant the following one:

Let $ x,y,z > 0$ such that $ (2 \minus{} x^2)(2 \minus{} y^2)(2 \minus{} z^2) \equal{} x^2y^2z^2$. Prove that:

$ x \plus{} y \plus{} z\le 3$

This one should be true.


$ f: R\longrightarrow R$                    $ f(xf(y))\plus{}f(yf(x))\equal{}2xy$           find all functions. :!:

I don't think this problem is easy at all... :maybe:
[hide]first of all prove that $ f(1) \equal{} 1\textrm{ or} \minus{} 1$,then prove that $ f$ is injective,then prove that $ f(\frac 1x) \equal{} \frac 1{f(x)}$ from which it follows that $ f(x) \equal{} x$ or $ f(x) \equal{} \minus{} x$ for every $ x\in\mathbb{R}$...[/hide]

yes you're right. maybe isnt  easy . sorry!  ..  ..  ..  ..  :(


induction
combinatorics proposed
combinatorics

Given n different positive numbers a_1, a_2, ... , a_n.
We construct all the possible sums (from 1 to n terms).
Prove that among those sums there are at least n(n+1)/2 different ones.

This one is really easy. Just use induction on $n$.
The result clearly holds for $n=1$.
Now assume it holds for any set of $n$ positive real numbers.
Lert's give a set of $n+1$ positive real numbers. Wlog, we may assume that $a_1 < \cdots < a_{n+1}$.
Then, let $s_i$, where $i=1, \cdots, \frac {n(n+1)} 2$ be the distinct sums from the set $\{a_1, \cdots , a_n \}$ (by the induction hypothesis).
. Then, the sums $a_1, a_2, \cdots, a_{n+1}, a_{n+1}+s_1, a_{n+1} + s_2, \cdots, a_{n+1} + s_{\frac {n(n+1)} 2}$ are pairwise distinct and we are done.

Pierre;


well you know that
$x^5+x+1=(x^2+ax+1)(x^3+bx^2+cx+1)$ expand this out and solve the system of equations and you get that
$a=1$, $b=0$, and $c=-1$.

im not sure what you get if you try, but go ahead and give it a shot.
$x^5+x+1 = (x^2+ax-1)(x^3+bx^2+cx-1)$

you see that $x^2+x+1$ is a factor of the LHS by plugging in 3rd roots of unity.  the motivation to do that is that the powers of each term on the LHS are all distinct mod 3.

[quote="kimby_102"]Source: 360 PMC
Prove that for any integer $n$ the number
$5^{5^{n+1}}+5^{5^n}+1$ is not prime.

Ok..I let $5^{5^n}=x$ and got
$x^5+x+1$..and tried to factor it out, but couldn't...
And on the solution it just showed that
$x^5+x+1=(x^2+x+1)(x^3-x^2+1)$. Can anyone plz show me how to factor like that? Thanks,[/quote]
but what about $x=1$ it doesn't gives natural $n$ indeed but... :?


Your inequalities can give me some inspiration. I like the first, and from this one we can write down more. The second is one of them. Vasc, have you got any generalizations for such?


For a,b,c,d real positive numbers prove that

(a^4+b^4+c^4+d^4)/(abcd)  + (abcd)^(1/4)/(a+b+c+d) \geq 1

Did I miss something or the same argument as in m14 gives a lower bound equals to 2, which is better than desired...

Pierre.

indeed, it's done in the same way!

cheers! :D :D


search
email
inequalities unsolved
inequalities

It's really IMO short list 2000. I guess it's quite well-known.

Anyone can post the solution.I don't know this problem

[quote="James Potter"]Prove that $\forall$ $x_{1} , x_{2} ,....., x_{n} \in R$ ,we have:
$\frac{x_{1}}{1+x_{1}^2} +\frac{x_{2}}{1+x_{1}^2+x_{2}^2}+.....+ \frac{x_{n}}{1+x_{1}^2+x_{2}^2+....+x_{n}^2} \leq \sqrt{n}$[/quote]

[b]1.[/b] Stop spamming.
[b]2.[/b] Don't post from two accounts.
[b]3.[/b] Use the "search" button. A search for *\frac{x_n}{1+x_1^2*, for instance, will give you this topic: http://www.mathlinks.ro/Forum/viewtopic.php?t=48859 . And there you can find some links to solutions of the problem.

  Darij

[b]1)[/b]I cannot find any solution from your three links
[b]2)[/b]Buihaiha_2000 is not me.

[b]1.[/b] The second link leads to a solution. (The first one leads to a thread where I hope a solution will be posted one day.)

[b]2.[/b] This is a bit strange, as your email addresses differ by 1 letter.

  Darij


Find the remainder of the division $ \displaystyle\frac {69!( 60! \plus{} 1 )}{71}$

Greetings  :D

[hide]$ 71$ is prime, so $ 70! \equiv 70$ modulo $ 71$. $ 69! \equiv 1$ modulo $ 71$ as a result, so we need only consider $ (60!\plus{}1)$ modulo $ 71$.

$ 70! \equiv (\minus{}1)(\minus{}2)(\minus{}3)(\minus{}4)(\minus{}5)(\minus{}6)(\minus{}7)(\minus{}8)(\minus{}9)(\minus{}10)60!$ modulo $ 71$, so if we can compute $ (\minus{}1)(\minus{}2)(\minus{}3)(\minus{}4)(\minus{}5)(\minus{}6)(\minus{}7)(\minus{}8)(\minus{}9)(\minus{}10)$ modulo $ 71$, since $ 70! \equiv \minus{}1$ modulo $ 71$, we can compute $ 60!$ modulo $ 71$.

$ (\minus{}1)(\minus{}2)(\minus{}3)(\minus{}4)(\minus{}5)(\minus{}6)(\minus{}7)(\minus{}8)(\minus{}9)(\minus{}10) \equiv (1)(2)(3)(4)(5)(6)(7)(8)(9)(10) \equiv 61$ modulo $ 71$. Therefore, $ 60! \equiv 64$ modulo $ 71$. Therefore, $ 60! \plus{} 1 \equiv 65$ modulo $ 71$, and the desired answer is $ 65$.[/hide]


geometry
geometric transformation
rotation
analytic geometry
combinatorics unsolved
combinatorics

Given a n*n grid filled with the numbers from 1 to n^2, in the obvious order with the exception of 1 and 2 that are swapped, determine whether a sequence of cyclic rotations of rows or columns can result in a fully ordered grid.

More precisely, "obvious order" means that the cell at (x, y), with coordinates in the [0, n) interval , contains y * n + x + 1 and "cyclic rotation of a row" means that for a given y, for each x (x, y) is moved to (x + k mod n, y) while cyclic rotations of a column are similarly defined.

If $n$ is odd, I think the permutations generated by cyclic rotations of rows/columns are even, so we will never be able to do that. However, when $n$ is even, it might be possible in the general case, I'm not sure. Think about $n=2$: it's clearly possible to interchange $1$ and $2$.


Hi, I'm a new member.  I'm curious, how does one get stars and/or the name of a cool mathematical formula under their member name.

That all depends on the number of posts you have.


Let $ x_1,x_2,...,x_n$ be real numbers such that $ x_1>1,x_2>2,...,x_n>n$. Find the minimum value of

$ \frac{(x_1\plus{}x_2\plus{}...\plus{}x_n)^2}{\sqrt{x_1^2\minus{}1^2}\plus{}\sqrt{x_2^2\minus{}2^2}\plus{}...\plus{}\sqrt{x_n^2\minus{}n^2}}$

Here $ \sum$ denotes $ \sum_{i\equal{}1}^n$. Let $ y_i>0$ such that $ y_i^2\equal{}x_i^2\minus{}i^2$, $ i\equal{}1,2,\dots,n$. Then we are looking for the minimum value of
\[ Y\equal{}\frac{\left(\sum\sqrt{y_i^2\plus{}i^2}\right)^2}{y_1\plus{}y_2\plus{}\dots\plus{}y_n}\]
Using $ \sqrt{y_i^2\plus{}i^2}\ge\frac{1}{\sqrt 2}(y_i\plus{}i)$ we deduce that
\[ Y\ge\frac{\left(\sum y_i\plus{}\sum i\right)^2}{2\sum y_i}\ge 2\sum i\equal{}n(n\plus{}1)\]
where the last inequality follows from $ (a\plus{}b)^2\ge 4ab$. Equality holds iff $ y_i\equal{}i$ for all $ i$. That is, when $ x_i\equal{}i\sqrt 2$, $ i\equal{}1,2,\dots,n$.


calculus
integration
trigonometry
logarithms
limit
complex numbers
real analysis

OK, here goes....... show :D

$ \int_{0}^{\infty}\;\;\frac{4\cdot\tan^{\minus{}1}\,(4\,x)}{e^{2\pi\,x}\minus{}1}\;\;\textbf dx\;\;\equal{}\;\;\boxed{\ln\left(\Gamma^{2}\left(\frac{1}{4}\right)\right)\minus{}2\ln 2\minus{}\ln\pi\plus{}\frac{1}{2}}$

For all complex numbers $ z\neq 0$ with $ \text{Re}(z)\ge 0$ define

$ \text{Ei}(z) \equal{} \gamma \plus{} \ln(z) \plus{} \sum_{k \equal{} 1}^\infty \frac {z^k}{k\, k!} \quad , \quad \text{Ci}(z) \equal{} \gamma \plus{} \ln(z) \plus{} \sum_{k \equal{} 1}^\infty ( \minus{} 1)^k$$ \frac {z^{2k}}{2k\, (2k)!} \quad , \quad \text{Si}(z) \equal{} \sum_{k \equal{} 0}^\infty ( \minus{} 1)^k \frac {z^{2k \plus{} 1}}{(2k \plus{} 1) \, (2k \plus{} 1)!}$ .

Because of $ \ln(iz) \equal{} \ln(i) \plus{} \ln(z) \equal{} i\frac {\pi}{2} \plus{} \ln(z) \qquad \boxed{\text{Ei}(iz) \equal{} \text{Ci}(z) \plus{} i\left(\text{Si}(z) \plus{} \frac {\pi}{2}\right)}$

For $ a,s > 0$ define $ f(a,s) \equal{} \int_0^\infty \frac {e^{ \minus{} sx}}{x \minus{} ia} \, dx$ and substitute $ x \equal{} \minus{} \frac {1}{s} t$

$ \minus{} \int_{ \minus{} \infty}^0 \frac {e^t}{t \plus{} ias} \, dt \equal{} \minus{} \int_{ \minus{} \infty \plus{} ias}^{ias} \frac {e^{t \minus{} ias}}{t} \,dt \equal{} \minus{} e^{ \minus{} ias}\, \Big(\text{Ei}(ias) \minus{} i\pi\Big) \equal{}$ $ \minus{} \Big(\cos(as) \minus{} i\, \sin(as)\Big)\, \Big(\text{Ci}(as) \plus{} i\left(\text{Si}(as) \minus{} \frac {\pi}{2}\right)\Big)$

Thus $ \int_0^\infty \frac {a\, e^{ \minus{} sx}}{x^2 \plus{} a^2} \, dx \equal{} \frac {f(a,s) \minus{} \overline{f(a,s)}}{2i} \equal{} \text{Im}\big(f(a,s)\big) \equal{}$ $ \sin(as)\, \text{Ci}(as) \minus{} \cos(as)\,\Big(\text{Si}(as) \minus{} \frac {\pi}{2}\Big)$

whereat $ \int_0^\infty \frac {a\, e^{ \minus{} sx}}{x^2 \plus{} a^2} \, dx$ is the same as $ \int_0^\infty s \, e^{ \minus{} sx} \, \arctan\left(\frac {x}{a}\right)\, dx$ after integration by parts.

Therefore $ \boxed{\int_0^\infty e^{ \minus{} sx} \, \arctan\left(\frac {x}{a}\right)\, dx \equal{} \frac {1}{s} \left[\sin(as)\, \text{Ci}(as) \minus{} \cos(as) \left(\text{Si}(as) \minus{} \frac {\pi}{2}\right)\right] }$

For $ n\in \Bbb{N} \quad I_n: \equal{} \int_0^\infty \frac {\arctan\left(\frac {x}{n}\right)}{e^{2\pi x} \minus{} 1} \, dx \equal{} \sum_{k \equal{} 1}^\infty \int_0^\infty e^{ \minus{} 2\pi k x} \, \arctan\left(\frac {x}{n}\right) \, dx \equal{} \sum_{k \equal{} 1}^\infty \frac {1}{2\pi k} \left[0 \minus{} 1\cdot \left(\text{Si}(2\pi kn) \minus{} \frac {\pi}{2} \right)\right]$

$ \equal{} \sum_{k \equal{} 1}^\infty \frac {1}{2\pi k} \int_{2\pi kn}^\infty \sin(t) \, \frac {dt}{t}$ . Replace $ t$ by $ kx \quad I_n \equal{} \frac {1}{2\pi} \sum_{k \equal{} 1}^\infty \int_{2\pi n}^\infty \frac {\sin(kx)}{k} \, \frac {dx}{x}$ and split $ \int_{2\pi n}^\infty$ up to $ \sum_{j \equal{} n}^\infty \int_{2\pi j}^{2\pi (j \plus{} 1)}$

to get  $ I_n \equal{} \frac {1}{2\pi} \sum_{j \equal{} n}^\infty \sum_{k \equal{} 1}^\infty \int_{2\pi j}^{2\pi (j \plus{} 1)} \frac {\sin(kx)}{k} \, \frac {dx}{x}$ then use the formula $ \sum_{k \equal{} 1}^\infty \frac {\sin(kx)}{k} \equal{} \frac {(2j \plus{} 1)\pi \minus{} x}{2}$ for $ 2\pi j < x < 2\pi (j \plus{} 1)$ .

$ 2 I_n \equal{} \frac {1}{\pi} \sum_{j \equal{} n}^\infty \int_{2\pi j}^{2\pi (j \plus{} 1)} \frac {(2j \plus{} 1)\pi \minus{} x}{2} \, \frac {dx}{x} \equal{} \lim_{m\to\infty} \sum_{j \equal{} n}^{m \minus{} 1} \left[\left(j \plus{} \frac12\right)\, \ln\left(\frac {j \plus{} 1}{j}\right) \minus{} 1\right]$ , whereas

$ \sum_{j \equal{} n}^{m \minus{} 1} j \, \ln\left(\frac {j \plus{} 1}{j}\right) \equal{} \ln \prod_{j \equal{} n}^{m \minus{} 1} \left(\frac {j \plus{} 1}{j}\right)^j \equal{} \ln\left(\frac {m^m}{m!}\right) \minus{} \ln\left(\frac {n^n}{n!}\right) \;\, , \;\, \sum_{j \equal{} n}^{m \minus{} 1} \ln\left(\frac {j \plus{} 1}{j}\right)$ $ \equal{} \ln(m) \minus{} \ln(n)$ and $ \sum_{j \equal{} n}^{m \minus{} 1} 1 \equal{} m \minus{} n$ .

So $ 2I_n \equal{} \lim_{m\to\infty} \left(\ln\left(\frac {m^m}{m!}\right) \plus{} \frac12 \ln(m) \minus{} m\right) \minus{} \left(\ln\left(\frac {n^n}{n!}\right) \plus{} \frac12 \ln(n) \minus{} n\right)$ , whereas $ \lim_{m\to\infty} (...) \equal{} \minus{} \ln \sqrt {2\pi}$ after Stirling's formula.

$ \Rightarrow \frac12 \ln(2\pi) \plus{} 2I_n \equal{} \ln(n!) \minus{} n\ln(n) \minus{} \frac12 \ln(n) \plus{} n \; \Rightarrow \;$ $ \ln(\Gamma(n)\cdot n) \minus{} \frac12 \ln(n) \equal{} n\ln(n) \minus{} n \plus{} \frac12 \ln(2\pi) \plus{} 2I_n$ ,

which shows at least [url=http://mathworld.wolfram.com/BinetsLogGammaFormulas.html]Binet's Log Gamma Formula[/url] $ \ln\Gamma(z) \equal{} z\, \ln(z) \minus{} \frac12 \ln(z) \minus{} z \plus{} \frac12 \ln(2\pi) \plus{} 2I_z$ for $ z \equal{} \frac {1}{n}$.

Should it be true for all $ z$ with $ \text{Re}(z) > 0$ then write $ 4I_z \equal{} \ln\Big(\Gamma^2(z)\Big) \minus{} 2z\ln(z) \plus{} \ln(z) \plus{} 2z \minus{} \ln(2\pi)$ and set $ z \equal{} \frac14$ .