At time t=0 a drag racer starts from rest at the origin and moves along a straight line with velocity given by v=5t^2, where v is in m/s and t in s. What is the expression for the displacement of the car from t=0 to time t?

What did I go wrong with the following solution:
s(t)=1/2.a.t^2
Since v=v_0+at=at we have:a=5t
So the displacement from t=0 to time t is: 5t^3/2-0=5t^3/2

Hello
$ v \equal{} 5t^2$
$ s \equal{} \int{5t^2dt}$
after solving
$ s \equal{} \frac {5t^3}{3} \plus{}c$
where $ c\equal{}0$ as at t=0 s=0
Thank u

[quote="kabi"]Hello
$ v \equal{} 5t^2$
$ s \equal{} \int{5t^2dt}$
after solving
$ s \equal{} \frac {5t^3}{3} \plus{} c$
where $ c \equal{} 0$ as at t=0 s=0
Thank u[/quote]
You got the right answer. Your solution is also correct.However,can you point out my mistake in my solution above?

[quote="ghjk"]
What did I go wrong with the following solution:
s(t)=1/2.a.t^2
Since v=v_0+at=at we have:a=5t
So the displacement from t=0 to time t is: 5t^3/2-0=5t^3/2[/quote]
Hello
$ a\equal{}10t$
as acceleration is depending on $ t$ we can't use of equation of motion as they are for uniform motion.
Thank u.

By Uniform Motion , He Means Uniformly Accelerated Motion i.e When Acceleration is constant (even if  a=0)

Sorry Kabi , but i mentioned it so that she might not get confused.
Hope you don't Mind!!

Hello
Ya by uniform motion I mean uniformly accelerated motion only.
Thanks Aman for correcting me but kindly edit he to she in ur post.
Thank u


Do you mean you want to convert a sound file of some other format (i.e. WAV) to MP3? If so, all you need is a conversion program. You can probably find a lot of good, free ones at [url]http://sourceforge.org[/url]. If you're looking to record a file directly as an MP3, there are also programs for that (though they're rather unneccesary, to be honest).

Hope that helps! :)

I personally use dBpowerAMP Music Converter, it is free (or at least it was two years ago when I download it).

To rename a file just change the name, not the content! The wav format and the mp3 are very different!

What's the original file extension?

Renaming a file doesn't change it's format...just the name. And sometimes, you can't make something into an mp3...

If it is a .rm or .ra file, you might need Streambox Ripper...


A lamppost in a park can illuminate a circle with radius 40 feet.  The park is a rectangular region with side lengths $200\sqrt{2}$ and $120\sqrt{2}$.  What is the minimum number of lampposts needed in the park so that the entire region is lighted?

[hide]A corner needs to be covered.  So put a circle so that the corner is on the circle.  Now draw two radii so that they endpoints are at the long side of the rectangle.  this makes a 45-45-90 triangle.  so, the part of the edge of the 200sqrt2 side that is covered is $40\sqrt{2}$.  So, we need 5 of those circles for the 200 side.  Now, draw the same triangle touching the other side of the rect.  The hypotenuse, the part of the edge that is covered, is again $40\sqrt{2}$.  so we need 3.
Total:  3 x 5 = $15$.[/hide]

[hide] um, when I tried it, I got the length covered by one to be 20 sqrt of 2, i think, by the same method,  so I got 15*4 = 60 posts, instead  :maybe: i probably messed up somewhere [/hide]

Maybe you thought the diameter was 40 insted of the radius.

i found my mistake, it wasn't the diameter one, but just as stupid  :blush:  sorry.


Let$f(x)=-x+\sqrt{(x+a)(x+b)}$with $a,b>0,a\neq b$
Prove that $\forall s\in (0,1)\exists$only one positive real number$\alpha$such that
     $f(\alpha)=(\frac{a^s+b^s}{2})^{\frac{1}{s}}$
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=76271]View all problem from Vietnam NMO 2006[/url]

We have $f'(x)>0$ and $lim_{x \to 0}f(x)=\sqrt{ab}$,$lim_{x \to \infty}f(x)=\frac{a+b}{2}$.
Moreover,$\sqrt{ab}$ $\leq$ $(\frac{a^{s}+b^{s}}{2})^{\frac{1}{s}}$ $\leq$ $\frac{a+b}{2}$.Thus we have q.e.d.


The three medians of a triangle divide it into six smaller triangles.
If the inradii of four of them are equal,Prove that the original triangle is equilateral.

Let $ AA'$, $ BB'$, $ CC'$ the medians of the triangle and $ G$ the centroid.Note that $ [BGA']\equal{}S_1$, $ [GA'C]\equal{}S_1$, $ [AGB']\equal{}[GB'C]\equal{}S_2$, $ [AC'G]\equal{}S_3$
$ r_{BGA'}\equal{}\frac{S_1}{p_1}\equal{}\frac{S_1}{p_2}\equal{}\frac{S_2}{p_3}\equal{}\frac{S_3}{p_4}$ where $ p_1,p_2,p_3,p_4$ are the semiperimeters of the triangles $ BGA',GA'C,GB'C,AGB',AGC'$ therefore $ BG\equal{}GC$, $ GC'\equal{}GB'$ and so on so forth $ AG\equal{}GB\equal{}GB$ $ \implies$ $ G$ is the circumcenter


integration
trigonometry
calculus
calculus computations

[quote="Salvador"]Solve $ y' \equal{} sin (x \minus{} y)$[/quote]
use $ x\minus{}y\equal{}z$ :wink:

Once you have z = x - y and work the problem out a little, your next substitution could be $ w \equal{} tan(z/2)$ or you could use this hint: $ \int \frac{1}{1\minus{} sinz}dz \equal{} \frac{2}{cot(\frac{x}{2}) \minus{} 1}$.

[quote="akech"]you could use this hint: $ \int \frac {1}{1 \minus{} sinz}dz \equal{} \frac {2}{cot(\frac {x}{2}) \minus{} 1}$.[/quote]
Or just write

$ \int {\frac{1}
{{1 \minus{} \sin z}}\,dz}  \equal{} \int {\frac{{1 \plus{} \sin z}}
{{\cos ^2 z}}\,dz}$

Then it's done.


2. A point x is called an accumulation point of set S if $\forall r > 0$, $I(x,r) -{}x$ contains points in $S$.

Will you please explain what does it mean in a more understandable way?

Example:

{$S ={\frac{1}{n}| n \in Z^{+}}$}

Answer: 0 is an accumulation point of S since no matter how small r is, (0,r) –{0}is still in S.

My idea of an accumulation point is actually a limit.  Can you differentiate the two?

2. A point x is said to be an isolated point of S if x is in S and there exists r such that I(x,r) – {x} contains no point in S.

I don’t understand, please simplify the definition.

[quote="guile"]My idea of an accumulation point is actually a limit.  Can you differentiate the two?[/quote]

I don't think I like that definition of accumulation point (because it says that the sequence 1, 1, 1, 1, 1, ... has no accumulation points), but consider the sequence $(-1)^{n}(1-\frac{1}n)$.  This sequence has no limit, but it has two accumulation points.

An accumulation point is a limit of some subsequence of the sequence, but that doesn't mean it's a limit of the entire sequence.

Another, more complicated example: since the rational numbers are countable, they can be listed as a sequence.  That sequence certainly has no limit (because it includes arbitrarily large positive and negative terms, for instance) but accumulates everywhere.

[quote="JBL"]What do [i]you[/i] think the definition of isolated point means?[/quote]

Well according to the definition (Wikipedia) a point x of a set $S$ is called an isolated point, if there exists a neighborhood of $x$ not containing other points of $S$. Of course this is very obvious, and conceptually, everyone understands it. But how are you going to know if set $S$ has an isolated point.

For example the set, $1,\frac{1}{2},\frac{1}{3},\ldots$? has an isolated point $\frac{1}{n}$ Can you explain why? And why is $1$ not an isolated point since it also has the form $,\frac{1}{n}$, i.e. $,\frac{1}{1}$ and besides, it is just one point on the number line and away from other values of form $\frac{1}{n}$? Im confused.

[quote="guile"]And why is $1$ not an isolated point...[/quote]

Of the set $\{\frac1n\, |\, n\in{\bf Z}_{>0}\}$?  Of course it is, with the proof being a neighborhood of radius $\frac{1}{3}$ (or any other positive radius up to $\frac12$).  In fact, every point of that set is isolated -- you can figure out for yourself what a good radius is to show this.  (Hint: what's the nearest element of the set to $\frac1n$?)  Now, in the set $\{\frac1n\, |\, n\in{\bf Z}_{>0}\}\bigcup\{0\}$, the point 0 is not isolated because every neighborhood of 0 contains some other element of the set (0 is an accumulation point).

One other thing I should have said earlier: another difference between accumulation points and limit points is that a set (not a sequence) can have accumulation points, but there is no notion of limit points for a set.

[quote]But how are you going to know if set S has an isolated point. [/quote]
You're going to do one of two things: either find a neighborhood with no other points of the set, or prove that every neighborhood contains another point.


[hide]

If we construct a circle centered at $ P$ that goes through $ A$ and $ B$, we find $ C$ is on circle $ P$, as $ 2\angle{ACB}\equal{}\angle{APB}$. Hence, by Power of a Point, $ (AD)(CD)\equal{}(3\minus{}2)(3\plus{}2)\equal{}5\Longrightarrow\textbf{(A)}$

[/hide]

[hide="A different proof"]Drop a perpendicular from $P$ to $AB$, bisecting $\angle P$ into two angles equal to $\angle C$. Let the intersection of this perpendicular with $AC$ be $E$. Since $\angle EPB =\angle ECB$, points $E$, $P$, $C$, and $B$ are concyclic. By the Power of a Point applied to the chords $PB$ and $EC$, $ED*DC = PD*DB = 2$. By the angle bisector theorem applied to $\triangle APD$,$\frac{AE}{ED} = \frac{AP}{PD} = \frac{3}{2}$ so $\frac{AD}{ED} = \frac{5}{2}$ and $AD*CD = \frac 52 ED*CD = \frac 52 * 2 = 5$.[/hide]

This is more complicated than lifeisacircle's proof, but I feel it is more natural.

Two of my students suggested the following:

[hide="Nonrigorous method"]Guess that $\angle D = 90^\circ$. By considering the angles of $\triangle APB, \angle PBA = 90^\circ - \angle C$ so $\angle A = \angle C$. Then $\triangle ABC$ is isosceles with altitude $BD$ and $AD=DC$. By the Pythagorean Theorem, $AD^2+2^2=3^2$ so $AD = \sqrt{5}, AD*DC=5$.[/hide]

I don't see a simple justification for their first step.

@above If $\angle D = 90^{\circ}$, then triangle $ABC$ must be isosceles through simple angle chasing. I am pretty sure that isn't always true...so maybe yours students got lucky.

@monkey8: It's a use of the weak competition axiom, that if some information is not determined, then it doesn't affect the answer. So, you can specify it to something convenient.

[quote=TachyonPulse]Triangle $ ABC$ and point $ P$ in the same plane are given. Point $ P$ is equidistant from $ A$ and $ B$, angle $ APB$ is twice angle $ ACB$, and $ \overline{AC}$ intersects $ \overline{BP}$ at point $ D$. If $ PB \equal{} 3$ and $ PD \equal{} 2$, then $ AD\cdot CD \equal{}$

$ \textbf{(A)}\ 5\qquad \textbf{(B)}\ 6\qquad \textbf{(C)}\ 7\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 9$
[asy]defaultpen(linewidth(.8pt));
dotfactor=4;

pair A = origin;
pair B = (2,0);
pair C = (3,1);
pair P = (1,2.25);
pair D = intersectionpoint(P--B,C--A);

dot(A);dot(B);dot(C);dot(P);dot(D);
label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,N);label("$D$",D,NE + N);label("$P$",P,N);

draw(A--B--P--cycle);
draw(A--C--B--cycle);[/asy][/quote]

Wow sorry for the bump but I really liked this problem...

[hide = Solution]

Notice that there is a degree of freedom in this diagram - we are only given $PA=PB$ and $\angle APB = 2\angle ACB$. Specifically, the angle condition can hint at us to put $\triangle ABC$ on its circumcircle with $P$ as its circumcenter. We can see that this configuration is unique by seeing that the angle condition is only satisfied when we do so.

Extend $DP$ until it hits the circle at a point $X \ne B$. Since $XP$ is a radius of the circle, $XP=3$. Also, we are given that $PD=2$, $PB=1$ and $XP+PB=XB=3+2=5$.

By Power of a Point applied on point $D$, we get $AD \cdot DC = XP \cdot PB = \boxed{5}$.

[/hide]


1) Two blocks of mass $ m$ and $ M$ and a spring (spring constant $ k$) are arranged on a horizontal frictionless surface (assume $ m<M$ and $ m$ is placed on top of $ M$; a generic block attached to spring attached to wall picture, with another block on top).  The coefficient of static friction between the blocks is $ \mu$.  Find the maximum possible amplitude of the simple harmonic motion if no slippage is to occur between blocks.

2) A solid cylinder (mass $ M$) is attached to a horizontal massless spring so that it can roll without slipping along a horizontal surface.  The force constant of the spring is $ k$.  If the system is released from rest at a position in which the string is stretched by $ x$ meters, find the (a) translational kinetic energy and (b) rotational kinetic energy of the cylinder as it passes through the equilibrium position.  (c) Show that under these conditions the center of mass of the cylinder executes SHM with a period $ T\equal{}2\pi \sqrt{\frac{3M}{2k}}$.

thats my solution to the first problem:

both mass M and m move with the same angular frequence:
$ \omega\equal{}\sqrt {\frac{k}{m\plus{}M}}$
For mass m we have that $ \sum Fmax\equal{}\minus{}DAmax$ where Amax is the amplitude we want.
We subsequently have that $ Tmax\equal{}DAmax$ where Tmax is tha maximum friction force between
m and M. $ Tmax \equal{} \mu*N$ where μ is the coefficient of static friction between the masses

Moreover, $ T \equal{} m\omega^{2}Amax$ and so by solving this equation we conclude that the maximum amplitude
is $ Amax \equal{} \mu*g\sqrt {\frac{m\plus{}M}{k}}$

tell me if my sollution is wrong :blush:

solution to second problem

.5kx(initial)^2 is initial energy and is conserved.
therefore

constant= total ke+total pe
 ke = .5mv^2+ .5IW^2
i.e ke=.75 mv^2

pe=.5kx^2

by differentiating the energy relation wrt time we get 2kx=3m*(dv/dt)

therefore t= as wanted
.

QED

I actually don't need that solution anymore, it's been solved, :)


i'm not sure if this is the right place to put this... but..

i am 100% sure i will qualify for aime. i am confident i can make usamo but i would like advice about prep for the aime. on Usamo also.

How should i prepare for the Aime and Usamo? My goal is to make red MOP this year...

i read somewhere that a 10 on usamo is likely to qualify for red MOP-is that accurate?

thanks

[url=http://www.artofproblemsolving.com/Wiki/index.php/How_should_I_prepare%3F]AoPSWiki topic - How should I prepare?[/url] This topic is relatively new (it has 95 views as of now) but it should help answer some of your questions. If the wiki page does not satisfy you, there are links to numerous aops topics that have been posted by others seeking help.
10 on the USAMO is probably enough to qualify you for red MOP.


$ x,y,z >0$ and $ x\plus{}y\plus{}z\equal{}1$

fnd the minimum value of : $ \frac{x^{2}}{x\plus{}yz}\plus{} \frac{y^{2}}{y\plus{}zx}\plus{} \frac{z^{2}}{z\plus{}xy}$

 :)

[hide]$ \frac{x^{2}}{x(x\plus{}y\plus{}z)\plus{}yz}\plus{}\frac{y^{2}}{y(x\plus{}y\plus{}z)\plus{}zx}\plus{}\frac{z^{2}}{z(x\plus{}y\plus{}z)\plus{}xy}\ge\frac{(x\plus{}y\plus{}z)^2}{x^2\plus{}y^2\plus{}z^2\plus{}3xy\plus{}3xz\plus{}3yz}$, by Cauchy-Schwarz.  Letting $ a\equal{}x^2\plus{}y^2\plus{}z^2$, $ b\equal{}xy\plus{}xz\plus{}yz$, we want to minimize $ \frac{a\plus{}2b}{a\plus{}3b}\equal{}1\minus{}\frac{b}{a\plus{}3b}\equal{}1\minus{}\frac{1}{\frac{a}{b}\plus{}3}$, which is minimized when $ \frac{a}{b}$ is, which occurs when a=b.  Thus, the given sum is at least $ \frac{3}{4}$, and that value is attained when x=y=z=1/3.[/hide]

[hide]
By Cauchy Schwartz inequality ,and since $ x\plus{}y\plus{}z\equal{}1$

\[ S\equal{}\frac{x^2}{x\plus{}yz} \plus{}\frac{y^2}{y\plus{}zx} \plus{}\frac{z^2}{z\plus{}xy} \geq \frac{1}{1 \plus{}xy \plus{}yz \plus{}zx}\]

The expresion $ \frac{1}{1 \plus{}xy \plus{}yz \plus{}zx}$  is minimal when $ xy \plus{}yz \plus{}zx$ is maximal ,

$ (x\plus{}y\plus{}z)^2 \geq 3 (xy\plus{}yz\plus{}zx) \Rightarrow xy \plus{}yz \plus{}zx \leq \frac{1}{3}$

Then $ S_{min} \equal{}\frac{1}{1\plus{}\frac{1}{3} } \equal{}\frac{3}{4}$

[/hide]

enndb0x . I did like your solution  :wink:


I would definitely choose cold. too bad i live in FL.

[quote="evilhamster"]Well I seem to function better in cold weather than in hot weather
Though right now my house is 63 degrees...brrr...[/quote]

WHAT? You think 63 is cold? It is 60 in my house, and not too cold.

No its not. Especially in our music classroom, the AC is always on in early fall, so it is always 55-60 degrees.

I'm wearing a t-shirt and shorts. What do you expect?

At school I get sweaty in 50 degrees. There's a difference.

EDIT: On the subject of temperature, click [url=http://jokesfunny.wordpress.com/2008/11/07/new-england-temperature-conversion-chart/]here[/url].

Depends what I'm doing. If I'm playing asport, well, how about in between? If I'm moving a piano for no reason, definitely cold. Overall, I'm a colder person than I'm hot.

I'm surprised at 50-50 split.


algebra
polynomial
inequalities
geometry
algebra unsolved

some time ago, harazi posed the problem whether a polynomial which is a square at all powers of 2 is a square itself which i solved. one might ask if we may replace the powers of 2 by an arbritary infinite set of integers. however, this trivially fails for degree 1 and for degree 2, the pell-fermat equation is a well-known counterexample.
however, the result that it is true for irreducible polynomials of degree >= 3 is due to Siegel(at least i hope so...i know it is true for degree 3 and that he wrote some article with title "about integer solutions to $y^2=ax^m+...+b$" or something similar):

let $P$ be a polynomial with integer coefficients of degree $\geq 3$ which is irreducible over $\mathbb{Z}$. then the set of all $x\in \mathbb{Z}$ such that $P(x)$ is a square is finite.

Peter

This is a very interesting problem. Harazi can you share the solution with us?

No, I can't share it. But Peter has already shared with us his magnificient solution.

:?  I do not remember having seen a solution about the power of two case...
And I do not find it wih the 'search' icon.

Pierre.

If I am not wrong, in fact Peter posted the solution for the power 2case in the inequalities section. Sincerely I do not remeber very well where you can find it. Hope that Peter does.

Yup, it is in the open questions area, inequalities, a post by Peter.

[quote=pbornsztein]Ok, it is here  :)
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=polynomial+square&t=15102

I have to admit that the title is not a hint for a search... ;)

Pierre.[/quote]

I can't use this link, may be other ?


If I have a commutative diagram and I need to put a comma after that then should I put the comma in the beginning of the next line or after the diagram? Or is the correct way to choose the words such that I don't have to put comma after the diagram?

The last option is absolutely the best. The first option makes sense from the viewpoint of formal grammar but it will, most likely, look rather ugly in print. The second one is completely unacceptable. I've seen children books using words mixed with pictures in sentences. The comma after a really small picture looks reasonably OK but when the picture becomes large, it is perceived by the reader as a dirty spot on the picture rather than as a punctuation mark.


conics
parabola
geometry
geometry unsolved

Prove that a plane can't be covered by a finite number of parabolas.

This problem was,of course, posted before, but there are very few solutions.Besides that,one of them is wrong.
Does anyone have a nice solution? :(
I hope to see a few more now. :D
[b]P.S.[/b] Don't lock my topic, please, because all topics of this con not be bumped up before, as they are locked.
Please help. :D

The line at infinity is the common tangent of all parabolas. A union of a finite number of parabolas (their convex areas) contains only a finite number of the points at infinity, therefore, it cannot cover the entire plane.

Also discussed at: [url]http://www.mathlinks.ro/viewtopic.php?t=140462[/url].


analytic geometry
geometry
trigonometry
email

I want to improve my coordinate geometry and trigonometry skills, can someone suggest me a book with extensive theory/examples and neat problems(not hard but tricky). Ebooks would be better.

OK add some plane euclidean geometry topics also.

Anyone? Olympiaders you must have ideas. Intermediate geometry using elementary theorems.

S.L.Loney's book Plane Trignometry part I is best for trignometry. also, you can try part II which deals with complex numbers.


It was posted recently ;) : http://www.mathlinks.ro/Forum/viewtopic.php?t=82345 .

[quote="José"][quote="luimichael"]Find the minimum value of positive integer n>1 such that
$\sqrt{\frac{1^2+2^2+...+n^2}{n}}$ is an integer.[/quote]

$\sqrt{\frac{1^2+2^2+...+n^2}{n}}=\sqrt{\frac{\frac{1}{6}*n*(n+1)*(2n+1)}{n}}=\sqrt{\frac{1}{6}*(n+1)*(2n+1)}$.

Then, or $n+1$ or $2n+1$ must be a multiple of $6$.

Suppose that $\frac{2n+1}{6}$ is an integrer. Then, $2n+1=6k$, for some $k,n$ integrers. Then, $n=3k-\frac{1}{2}$.... Contradiction show us that $\frac{n+1}{6}$ is an integrer...... Well, I stop here, that's all my work. can anyone finish the problem???.[/quote]

since (2n+1)(n+1)=6a^2, and (2n+1) is odd, then (n+1) must be even and n must be odd. Substitute m+1 for n, m being even, and multiply out the left hand side,
2n^2+3n+1=6a^2
2(m+1)^2+3(m+1)+1=6a^2
2m^2+7m+6=6a^2
(1/3)m^2+(7/6)m+1=a^2
So not only must m be even, it must be a multiple of 3.  Substituting 6p for m, where p is any integer:
(1/3)(6p)^2+(7/6)(6p)+1=a^2
12p^2+7p+1=a^2                         *
Then I don't know if this helps...
12p^2+7p=a^2-1
p(12p+7)=(a+1)(a-1)                     *

The goal would be to use one of the * equations to find the lowest p such that the * equation was true, and then 6p+1 would be the answer...


In  ABC, E is any point on AC. Through vertex A draw AD   ||   BE
         meeting CB produced in D and draw  CF  || BE through C meeting AB
         produced in F  prove that:
                                   1/AD+1/CF=1/BE
PLZZZZZ GIVE A DIAGRAM

Here is a diagram and a solution (I think)


give $P(x)$ and $Q(x)$ be two polynomials with real coefficient satisfying both of which have at least on real root and
                            $P(1+x+(Q(x))^{2})=Q(1+x+(P(x))^{2})$ for all $x \in R$
Prove that:$P(x)=Q(x)$ $\forall x \in R$

[quote="doicu"]hallo, have the two polynominals the same root ? if
this is so, then i can poste my solution.  :wink:[/quote] I don't
think so,doicu. I have a solution. Lemma: $r(x)=(P(x))^{2}-(Q(x))^{2}$
have a real root. proof of the lemma:suppose the real root of
$P(x),Q(x)$ are $\alpha,\beta$ respectively then
$r(\alpha)\leq 0,r(\beta)\geq 0$ so there is a real root of $r(x)$.
Lemma2. $r(x)=0$ proof:if $r(x)\not=0$then let $\gamma$the greatest rootof $r(x)$
then$p(1+\gamma+(Q(\gamma))^{2})=Q(1+\gamma+(P(\gamma))^{2})$
then$1+\gamma+(Q(\gamma))^{2}$ is still a root of $r(x)$
but
$1+\gamma+(Q(\gamma))^{2}>\gamma$ a contradiction.
Now ,we have poove that $P(x)^{2}=Q(x)^{2}$
so it's easy to know that $P(x)=Q(x)$(because $P(x)\not=-Q(x)$)


A convex quadrilateral $ABCD$ is inscribed in a semicircle with diameter $AB$. The diagonals $AC,BD$ intersect at $S$, and $T$ is the projection of $S$ on $AB$. Show that $ST$ bisects angle $CTD$.

A very easy problem.
Observe that STAD and STBC are cyclic. The conclusion is follows.
 :lol:


interesting!!!
nothing's obvious or I am missing sth!!!

[quote="a_vakilian"]Prove that
$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1999}-\frac{1}{2000}=\frac{1}{1001}+\frac{1}{1002}+...+\frac{1}{2000}$
where the signs are alternating on the left side of the equation but are alike on the right side.[/quote]

[hide]Firstly, the fractions with odd denominators $>1000$ cancel out on both sides, so you are left with:

$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{1996}-\frac{1}{1998}-\frac{1}{2000}=\frac{1}{1002}+\frac{1}{1004}+...+\frac{1}{1998}+\frac{1}{2000}$

Moving the fractions on the left side with denominators $>1001$ to the right side...

$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{1000}=\frac{1}{501}+\frac{1}{502}+...+\frac{1}{999}+\frac{1}{1000}$

Again, odd denominators cancel out, except this time $>500$

$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{998}-\frac{1}{1000}=\frac{1}{502}+\frac{1}{504}+...+\frac{1}{998}+\frac{1}{1000}$

Moving odd denominators to the right side $>501$

$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{499}-\frac{1}{500}=\frac{1}{251}+\frac{1}{252}+...+\frac{1}{499}+\frac{1}{500}$

Proceeding with this... you are left with:

$1-\frac{1}{2}=\frac{1}{2}$

$\frac{1}{2}=\frac{1}{2}$

QED (that was a fun problem)[/hide]

[quote="a_vakilian"]Actually Drunken_Math was good. But this is an example of a problem where it is easier to prove a general resul than special case :lol:  :lol:[/quote]

Is it just me, or when you prove something in a really long and tiring proof, you then figure out that there's a much easier and less pain staking way to prove it...  :lol:

[quote="Drunken_Math"]
Is it just me, or when you prove something in a really long and tiring proof, you then figure out that there's a much easier and less pain staking way to prove it...  :lol:[/quote]
OK. but did you get the easier one ;)

I think Drunken_Math's first proof was pretty good  :?

but I think Drunken_Math's proof is good but it is obvious one, And it is the first proof that everyone will work on, but if you see it in general case it can easily proved by induction an also it so smart. ;)


Let $x,y,z$ be positive reals such that $x+y+z=1$, prove or disprove that
\[ \frac8{(1-x)(1-y)(1-z)}\leq\frac1{x^2}+\frac1{y^2}+\frac1{z^2}. \]

Since $9(x+y)(y+z)(z+x)\ge 8(xy+yz+zx)(x+y+z)$,

LHS $\le \frac{9}{(xy+yz+zx)(x+y+z)} = \frac{9}{xy+yz+zx}$

RHS $\ge \frac1{xy}+\frac1{yz}+\frac1{zx}\ge \frac{9}{xy+yz+zx}$

That's right. This implies that the inequality is not sharp.  :lol:


How could i find the sum of all positive integers not exceeding 500 which are not divisible by 13?

[quote="Brut3Forc3"][hide="Too lazy to give actual numbers"]Find the sum of the numbers from 1-500, then the sum of the multiples of 13, and subtract.[/hide][/quote]
ah of course...thanks =]


Suppose $P$ is an interior point of a triangle $ABC$ such that the ratios \[ \frac{d(A,BC)}{d(P,BC)} , \frac{d(B,CA)}{d(P,CA)} , \frac{d(C,AB)}{d(P,AB)} \] are all equal. Find the common value of these ratios. $d(X,YZ)$ represents the perpendicular distance fro $X$ to the line $YZ$.

[hide]It's a well known fact that the centroid cuts the medians into ratios of 1:2. Drop perpendiculars from the vertices to their opposites sides and from the centroid to each side. Each pair of parallel lines are parts of similar triangles, in the ratio of 3:1. So P is the centroid and the ratio is 3:1.[/hide]

Bictor, you have not shown that the centroid is the only point satisfying the given condition. To show this, consider the ratios of areas rather than lengths. We see that BPC, APC, APB all have equal areas, and then the result follows. In fact, the result even follows without noting that the point in question is indeed the centroid

We have
\[ \frac{(1/2) a d(A,BC)}{(1/2) a d(P,BC)} = \frac{(1/2) bd(B,CA)}{(1/2) bd(P,CA)} = \frac{(1/2) cd(C,AB)}{(1/2) cd(P,AB)}= \frac{3\Delta}{\Delta}=3, \]
where the last equality follows from summing the numerators and denominators.


What is the sum of the odd integers from 11 through 39,
inclusive?

There are a few ways to do this. I can show two. One of them is just to list the numbers out so $ 11\plus{}13\plus{}15\plus{}17\plus{}19\plus{}21\plus{}23\plus{}25\plus{}27\plus{}29\plus{}31\plus{}33\plus{}35\plus{}37\plus{}39\equal{} \boxed{375}$
Another way is to use the formula: The sum of the first $ X$ number of odd numbers= $ X^2$. You know that there are $ 39/2$ round up to $ 20$ odd numbers. So you do $ 20^2\equal{}400$. After you do that, you need to subtract the first $ 5$ odd numbers that are not $ 11$ through $ 39$. So $ 5^2\equal{}25$ After you do all that, just do $ 400\minus{}25\equal{}\boxed{375}$

We know that the number of odds squared is the sum. Since there are 20^2=400 Now we get rid of the ones we don't want there are 5 of them and their sum is 25. The answer is 400-25=375.

These problems can be solved much quicker and without that kind of gussy and checky way (not that its bad, because personally i think the logic is great). When presented with the first and last term of a series you can apply the formula:
$  s=\frac{n}{2} (first + last term)  $

Where n= number of terms:

$  \frac{39-11}{2} +1=15  $

$  \frac{15}{2} (39+11)=375  $

Unfortunately, the previous methods are not guess and check and when I solved this, 1st way was much faster.


ratio
combinatorics unsolved
combinatorics

A balance has a left pan, a right pan, and a pointer that moves along a graduated ruler. Like many other grocer balances, this one works as follows: An object of weight $ L$ is placed in the left pan and another of weight $ R$ in the right pan, the pointer stops at the number $ R \minus{} L$ on the graduated ruler. There are $ n, (n \geq 2)$ bags of coins, each containing $ \frac{n(n\minus{}1)}{2} \plus{} 1$ coins. All coins look the same (shape, color, and so on).  $ n\minus{}1$ bags contain real coins, all with the same weight. The other bag (we don’t know which one it is) contains false coins.  All false coins have the same weight, and this weight is different from the weight of the real coins. A legal weighing consists of placing a certain number of coins in one of the pans, putting a certain number of coins in the other pan, and reading the number given by the pointer in the graduated ruler. With just two legal weighings it is possible to identify the bag containing false coins. Find a way to do this and explain it.

I don't think the task is possible for $n=2$.  You have no way to tell which weight correspond to the real coins.  So, I'm giving a solution for $n \geq 3$.

Let $g$ be the weight of a genuine coin, and $c$ that of a counterfeit one.  For each $i$, $i=1,2,\ldots,n$, $w_i$ signifies the weight of each coin in the $i$-th bag. Suppose the $k$-th bag is the one containing only counterfeit coins.

[size=150][b]CASE 1: $n > 3$.[/b][/size]

[b]1st Weighing[/b]
[hide]
Left: For $1 \leq i \leq n-2$, use $i-1$ coins from $i$-th bag; for $i=n-1$, use $n-1$ coins from $(n-1)$-st bag
Right: $0+1+2+\ldots+(n-3)+(n-2) = 1+\frac{(n-1)(n-2)}{2}$ coins from $n$-th bag
The resulting weight measurement is $W_1:=\sum_{i=1}^{n-2} (i-1)\left(w_i-w_n\right) + (n-1)\left(w_{n-1}-w_n\right)$.
[/hide]

[b]2nd Weighing[/b]
[hide]
Left: For $1 \leq i \leq n-2$, use $i$ coins from $i$-th bag; for $i=n-1$, use $n$ coins from $(n-1)$-st bag
Right: $1+2+3+\ldots+(n-2)+(n-1)=1+\frac{n(n-1)}{2}$ coins from $n$-th bag.
Thus, we have gotten $W_2:=\sum_{i=1}^{n-2} i\left(w_i-w_n\right)+n\left(w_{n-1}-w_n\right)$.
[/hide]

If $1 \leq k \leq n-2$, then $W_1 = (k-1)(c-g)$ and $W_2 = k(c-g)$.  Therefore, $\frac{W_1}{W_2} = 1-\frac{1}{k}$.

If $k=n-1$, then $W_1= (n-1)(c-g)$ and $W_2=n(c-g)$.  Consequently, $\frac{W_1}{W_2} = 1-\frac{1}{n}$.

If $k=n$, then $W_1=\left(1+\frac{(n-1)(n-2)}{2}\right)(g-c)$ and $W_2 = \left(1+\frac{n(n-1)}{2}\right)(g-c)$.  This means
\[\frac{W_1}{W_2} = 1 - \frac{2(n-1)}{n^2-n+2}\,.\]

You can easily show that the ratio $\frac{W_1}{W_2}$ is always different for a different value of $k$.

[size=150][b]CASE 2: $n = 3$.[/b][/size]

[b]1st Weighing[/b]
[hide]
Left: For $i=1$ and $i=2$, use $i-1$ coins from $i$-th bag.
Right: $0+1=1$ coin from the third bag
The resulting weight measurement is $W_1:=\left(w_2-w_3\right)$.
[/hide]

[b]2nd Weighing[/b]
[hide]
Left: For $i=1$ and $i=2$, use $i$ coins from $i$-th bag.
Right: $1+2=3$ coins from the third bag
Thus, we have gotten $W_2:=\left(w_1-w_3\right)+2\left(w_{2}-w_3\right)$.
[/hide]

If $k=1$, then $W_1 = 0$ and $W_2 = (c-g)$.  Therefore, $\frac{W_1}{W_2} = 0$.

If $k=2$, then $W_1= (c-g)$ and $W_2=2(c-g)$.  Consequently, $\frac{W_1}{W_2} = \frac{1}{2}$.

If $k=3$, then $W_1=(g-c)$ and $W_2 = 3(g-c)$.  This means $\frac{W_1}{W_2} = \frac{1}{3}$.

for $n=2$ since there is one bag of real coin and one bag of false coin; unless we know whether the real coins are heavier or lighter ; we can't say anything.

so, consider $n\geq 3$

Let the real coin has weight $x$ and false coin has weight $x+d $ where $d\neq 0$ may be positive or negative.

$n=2k+1$

order the bags as $B_{1},B_{2},\cdots,B_{n}$

take $1+2+\cdots +(n-1)=\frac{n(n-1)}{2} $ balls from bag $B_{1}$ and put them on the left pan.

take $1,2,3,\cdots,(n-1)$ balls from bag $B_{2},B_{3},\cdots,B_{n}$ respectively and put them on the right pan.

take the reading $R_{1}-L_{1}$

Now,empty the right pan and take $(n-1),(n-2),(n-3),\cdots,1$ balls from bag $B_{2},B_{3},\cdots,B_{n}$ respectively and put them on the right pan.

take the reading $R_{2}-L_{2}$

$\text {conclusion}:$

if $R_{1}-L_{1}=R_{2}-L_{2}$ then $B_{1}$ contains false coins(also reversely).

if not say the $B_{i}$ contains the fake coins where $i\geq 2$

then $R_{1}-L_{1}=(i-1) d=(n-i+1)d=R_{2}-L_{2}$ so, $(i-1) =(n-i+1)$

or, $n=2(i-1)$ contradiction since n is odd.

and in this case  if $R_{1}-L_{1}>0$ then the fake coins are  lighter and  if $R_{1}-L_{1}<0$ then the fake coins are  heavier.

Now, if $R_{1}-L_{1}\neq R_{2}-L_{2}$ then some $B_{i}$ for $i\geq 2$ contains the fake coins and also reversely.

Now, $R_{1}-L_{1}=(i-1) d$ and$R_{2}-L_{2}=(n-i+1)d$

so, $i=\frac{n(R_{1}-L_{1})}{(R_{2}-L_{2})+(R_{1}-L_{1})}+1$

also, we can say if $R_{1}-L_{1}>0$ then the fake coins are  heavier

and  if $R_{1}-L_{1}<0$ then the fake coins are  lighter.

$n=2k+2$ where $k\geq 1$

again order the bags as $B_{1},B_{2},\cdots,B_{n}$

take $1+2+\cdots +(n-1)=\frac{n(n-1)}{2} $ balls from bag $B_{1}$ and put them on the left pan.

take $1,2,3,\cdots,(n-1)$ balls from bag $B_{2},B_{3},\cdots,B_{n}$ respectively and put them on the right pan.

take the reading $R_{1}-L_{1}$

Now,empty the right pan and take $2,3,4,\cdots,(n-1),1$ balls from bag $B_{2},B_{3},\cdots,B_{n}$ respectively and put them on the right pan.

take the reading $R_{2}-L_{2}$

$\text {conclusion}:$

if $R_{1}-L_{1}=R_{2}-L_{2}$ then $B_{1}$ contains false coins(also reversely).

if not say the $B_{i}$ contains the fake coins where $(n-1)\geq i\geq 2$

then $R_{1}-L_{1}=(i-1) d=id=R_{2}-L_{2}$ contradiction.

or,say $B_{n} $ contains the fake coins. then $(n-1)d=d$ contradiction.

so, $B_{1}$ contains the fake coins.

and in this case if $R_{1}-L_{1}>0$ then the fake coins are  lighter and  if $R_{1}-L_{1}<0$ then the fake coins are  heavier.

Now, if $R_{1}-L_{1}\neq R_{2}-L_{2}$ then some $B_{i}$ for $i\geq 2$ contains the fake coins and also reversely.

if $ R_{1}-L_{1}>0$ then the fake coins are heavier and if $R_{1}-L_{1}> R_{2}-L_{2}>0$ then the fake coin comes from bag $B_{n}$

else it comes from $B_{i}$ where $2\leq i\leq (n-1)$ and  $i=\frac{(R_{1}-L_{1})}{(R_{2}-L_{2})-(R_{1}-L_{1})}+1$

if $ R_{1}-L_{1}<0$ then the fake coins are lighter and if $R_{1}-L_{1}< R_{2}-L_{2}<0$ then the fake coin comes from bag $B_{n}$

else it comes from $B_{i}$ where $2\leq i\leq (n-1)$ and  $i=\frac{(R_{1}-L_{1})}{(R_{2}-L_{2})-(R_{1}-L_{1})}+1$

Let the real coin weight x and counterfit one y, and number bags 1,2,3,....Counterfit coins are in bag L. Divide in two cases: n even and n odd. If n is even put 1 coin  from every bag from 1 to n/2 to right and 1 coin from bags from n/2+1 to n.Now R1-R2=x+_y.Next on left put 1 from first bag,i from ith up to n, and on right put n+1-i for 1=1,2,..,n. There we get (x-y)(n-2L+1) from these two its trivial to find L.If n is odd then put  1 coin  from every bag from 1 to n-1/2 to right and 1 coin from bags from n+1/2 to n-1.If they are equal L=nth bag, else we get x-y and finish similarly as in n even case.


Hi, I'm not sure if this is what this board is for but I couldn't find anywhere else so..

I have to make my decision between these two schools by May 1st, yet with each passing day I get more and more confused.  If anyone here was/is an undergrad at either of these schools, could you please PM/AIM/email me as soon as possible?  I just visited both campuses and talked to math advisors but I'm still very unsure and I think getting a student's perspective would help a lot.

Thanks.

I did not attend either university; only visited UIUC a few times. Both have very fine reputation. I would go with in-state tuition, but that's just me.

[quote="tokenadult"]I hope some current students will jump in. Since your location shows up as Illinois, my tendency would be to advise moving out of state for college, which would mean Berkeley for you.[/quote]

Assuming you have money or get lots of money.

I loved the weather at Berkeley, but that's as much as I know.

Berkeley is a much nicer location than Urbana-Champaign, from the point of view of this city-dweller.  As universities for undergraduates, they probably aren't that different, and they both have very good very large math departments.  Thus ends my two cents.

But doesn't Berkeley have a clear reputation edge favoring its math department?

It does. But is this edge relevant to undergraduates?

In general, berkeley is more famous than UI UC, any money factor?


floor function
number theory unsolved
number theory

Let $ p$ be a prime greater than $ 3$ and $ k\equal{}\lfloor{\frac{2p}{3}}\rfloor$.

Prove that $ p^2$ divides

$ \binom{p}{1}\plus{}\binom{p}{2}\plus{}...\plus{}\binom{p}{k}$

[quote="outback"]Let $ p$ be a prime greater than $ 3$ and $ k \equal{} \lfloor{\frac {2p}{3}}\rfloor$.

Prove that $ p^2$ divides

$ \binom{p}{1} \plus{} \binom{p}{2} \plus{} ... \plus{} \binom{p}{k}$[/quote]
It is also a problem of Komal magazine (A.271.2001).or MYM (in my country) 365
The hint is use Wilson theorem

Link: http://www.nxbgd.com.vn/toanhoctuoitre/?p=7&id=8&ReportID=424&ph=33
This is T9/365  :wink:

Could you please post it here in English? I can't read it.  :)


function
inequalities
Injective
domain
triangle inequality
complex analysis

Let $f(z)=\sum_{n=0}^{\infty}a_nz^n$ be an analytic function in the open unit disk $D$. Assume that $\sum_{n=2}^{\infty}n|a_n| \leq |a_1|$ with $a_1 \neq 0$. Prove that the function is injective

[quote="eugene"]Assume that $\sum_{n=0}^{\infty}n|a_n| \leq |a_1|$ with $a_1 \neq 0$.[/quote]

There has to be a typo here somewhere. The sum on the left is $|a_1|$ plus some other non-negative terms, such that if this inequality holds then $a_n=0,\ \forall n\ge 2$.

P.S.

I could have checked what the problem really looked like myself if you had posted the year when it was used in the Berkeley exam :). Could you please do that from now on, if it's not too much trouble?

Yes, of course it is a type,sorry,i've edited it.That's no problem for me to post the year of the Berkeley exams,i'll do it with great pleasure

Thanks for editing it :).

Given some $|z_0|<1$, we look at the function $g(z)=\frac{f(z)-f(z_0)}{z-z_0}=a_1+a_2(z+z_0)+a_3(z^2+zz_0+z_0^2)+\ldots$. We can regard it as an analytic function with the same domain as $f$. The hypothesis tells us that $|g(z)-a_1|<|a_1|$ for all $z$ with $|z|<1$ (we simply use the triangle inequality, the inequality given in the hypothesis, and the fact that $|z|,|z_0|<1$).

By Rouche's Theorem applied to some disk with radius $\to 1$ (we apply it to a sequence of disks, actually, centered at the origin and with the sequence of radii increasing towards $1$), $g$ has the same number of roots as the constant function $a_1\ne 0$, i.e. $g$ has no zeros inside its domain, which means that $z_0$ is the only point where $f$ takes the value $f(z_0)$. Since $z_0$ was arbitrary, $f$ is injective.

Wow, difficult solution, grobber. But it is direct consequence of triangle inequality, we don't need such difficult theorems. ;)

[quote="removablesingularity"]Hi Harazi , what do you mean by a direct consequence of triangle inequality? :)[/quote]
I thinks this is what harazi means:
$f(z)-f(w)=(z-w)(a_1+a_2(z+w)+a_3(z^2+zw+w^2)+....)$
but $|z+w|\le|z|+|w|<1+1=2$ and $|z^2+zw+w^2|\le|z|^2+|z||w|+|w|^2<1+1+1=3$ and...
therefore:$|a_1+a_2(z+w)+a_3(z_2+zw+w^2)+...|\ge |a_1|-(2|a_2|+3|a_3|+...)\ge0$ and equality happens iff $a_i=0$ for all $i\ge1$ but$a_1\neq0$ so equality dosn't happen.
so if $f(z)-f(w)=0$ then $z-w=0$.and $f$ is injection.


heres another one that i havent understood...

An equilateral triangle is placed in its starting position with its vertices labeled as shown (1). On each move, it can either be reflected over a vertical axis through the top vertex (ie switching the placement of its 2 vertices) (2), or it can be rotated clockwise 120 degrees around its center (3). How many different sequences of 5 moves will return the triangle to its original, starting position?

(1) cant draw a pic, so try to imagine this for me... an equilateral tirangle laying on one of its sides, with C at the top, B at the bottom right, and A as the last vertex

(2) an equilateral tirangle laying on one of its sides, with C at the top, A at the bottom right, and B as the last vertex

(3) an equilateral tirangle laying on one of its sides, with A at the top, C at the bottom right, and B as the last vertex

[hide="answer :idea:"] [size=200]5 [/size][/hide]


[quote="shencaili"]Given $a,b,c>0$.Prove that ${\sum\frac{a^{2}}{a^{2}+ab+b^{2}}\geq1}$[/quote]
${\sum\frac{a^{2}}{a^{2}+ab+b^{2}}\geq1}\Leftrightarrow\sum_{cyc}(a^{2}b-b^{2}c)^{2}\geq0.$ :D :)

It's hard to check,but it's very nice.Thank you.

See here for a solution with Cauchy in Engel form  :wink:
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=xyz&t=110854

[quote="Nameless"] Let $x=\frac{a^{2}}{bc}$
[/quote]
It's nice, silouan! :lol:

Yeah it's nice . I conjecture that every three fraction inequality with equality case a=b=c could solved in a way by Cauchy in Engel form  :P  :D


topology
real analysis
real analysis unsolved

Let X,Y be 2 topological spaces.If there exists a homeomorphism from X to a subspace of Y & there exist another homeomorhism from Y to a subspace of X,then are X & Y homeomorphic??

The answer is no. For a counterexample, take a segment and a union of two disjoint segments in $\mathbb{R}$.

:blush: Sorry didnt think about it at all.the answer is quite trivial.


A particular convex pentagon has two congruent, acute angles. The measure of each of the other interior angles is equal to the sum of the measures of the two acute angles. What is the common measure of the large angles, in degrees?

A pentagon's angles' measures have a sum of $ 180(5\minus{}2)\equal{}504$. Let the two smallest angles have measure $ x$. Then, each of the three larger angles has measure $ 2x$. Thus, the total sum of the angles is $ x\plus{}x\plus{}2x\plus{}2x\plus{}2x\equal{}8x\equal{}504$. Thus, a small angle has measure $ x\equal{}63$ and a large angle has measure $ \boxed{126}$. These angle measures do fit all of the restrictions.

[quote="PowerOfPi"]A pentagon's angles' measures have a sum of $ 180(5 \minus{} 2) \equal{} 504$. Let the two smallest angles have measure $ x$. Then, each of the three larger angles has measure $ 2x$. Thus, the total sum of the angles is $ x \plus{} x \plus{} 2x \plus{} 2x \plus{} 2x \equal{} 8x \equal{} 504$. Thus, a small angle has measure $ x \equal{} 63$ and a large angle has measure $ \boxed{126}$. These angle measures do fit all of the restrictions.[/quote]
540 not 504

Oops, typo. Then we get $ 8x\equal{}540$, so $ 2x\equal{}\boxed{135}$.

[quote="PowerOfPi"]Oops, typo. Then we get $ 8x \equal{} 540$, so $ 2x \equal{} \boxed{135}$.[/quote]

'Think I get it...   :huh:


All right, it's what all of you have been waiting for, Mafia game 6! So, here's the stuff.

Number of players: 15
Math92
Paladin8
Gr8sk8r
dangvh
probability1.01
GoBraves
Secret Asian
Darkknight
noneoftheabove
Chigr
Gandalf
Tunio
Sheepwarrior
Ambierona
baseballfreak2

4 mafia, 1 inspector, 3 masons, 7 citizens

For more detail on roles, look at the stuff posted in the mafia sign-up topic.

Mafia and inspector, PM me your stuff during the night phase. I'll need the information at the most within 36 hours of the beginning of the night phase.

Also, no last words will be tolerated at all. If you say last words, I will try to have them deleted and I might pursue further punishment (like not being eligible for the next game, that'll be up to other people though.) Along with this rule will be the rule of no dead talking. If you have something to tell me, PM me.

I'll trust that you guys will not try to communicate with each other regarding the game outside of the forum. No PMs either, unless you're mafia and it's during the night phase and you're just discussing who to kill. Obviously, there's no way to find out, but it would be pretty darn pathetic if you tried to cheat at online mafia.

Roles will be PMed shortly.

All right, let's start! Day 1 begins.... now!

All PMs sent.

Oh, and try to be able to visit the forums to play at least once a day. It get's boring if some people repeatedly don't post. I'll implement a 72 hr policy for participation. If you're not active for 72 hours, your character will be terminated and we'll find a replacement. If you find yourself not wanting to post as part of your strategy, be sure to drop me an IM (sn: caocoprez) or a PM just to tell me you're active.

I'd like the game to not begin until I have a verification from everyone, so just post on this topic or give me a PM, either one's fine.

LOL I love the (new?) :bruce:

Yeah... so I'm ready too :P.  Do we have night first or a lynching?

You guys will be lynching someone first. Still need responses from gr8sk8er, dangvh, Secret Asian, noneoftheabove, Gandalf and baseballfreak2

I would have joined this game, but I'm leaving on June 12, which is very close. :(
Enjoy the game!

Your unique way of playing the game will be missed dearly, lightrhee.  :)

Hey this worked out perfectly. I got killed just before my plane flight! (whichis in 4 hours time). Kudos to mafia :ninja: !

[quote="DarkKnight"]Hmm.... The only one who said something about sheepwarrior and didn't vote Aye was Secret Asian. I think he's the last mafia but we should see what he has to say.[/quote]

Er..I don't check this board to often so that's why I say things but sometimes don't reply to them.  Besides I needed a reason why to vote.

[color=red]Aye[/color] he didn't vote for probability or sheepwarrior

btw i gave u a reason secret asian

and its better to kill quiet ppl becuz even if they are citzen they didnt contribute to the citzens(Sly Si's strategy)

lol uhh [color=red]Aye [/color]for secret asian. That makes 2 right?

Sheepwarrior:
Aye-2
Nay-0

Ugh, this is going slowly. Am I gonna have to initiate some discussion and tension myself, like in online mafia?

[color=red]Aye[/color] for Secret Asian.

since it takes 4 people for majority and since secret asian wont vote nay is there any way that u can make secret asian dead if no one votes for the next 24 hours

Sorry, unless another person votes, it's going to be stalled.

Still to vote: Secret Asian, Press Alt f-4, Gr8sk8r, Gandalf

I would replace someone, but it's already late in the game. I think this should be decided by people who have been playing throughout the game (though they might not be active...  :mad: )

Well  just in the sake of letting everyone know I'm still here. [color=red][b]Aye![/b][/color] He's dead!

FINALLY after 3 days of wait some activity now we turn to yif man 12

After long weeks of deliberation with many citizens falling asleep and then awake again, the townspeople finally decided to lynch Secret Asian, who is presumably sleeping right now. So, they administered a dose of sleeping-drugs on him which killed him. Very uncool death.

Secret Asian was...

Mafia.

And thus ends the (long) game. Congrats townspeople, you caught all the mafia. Be happy for yourselves. Thanks for all the people who actually participated and were active. Sorry about all the inactive people who appartently didn't care.

woo hoo then next person i would have lynched was gandalf cuz he didnt vote for probability and i thought it was highly unlikely math92 would give up his job if he were mafia so that ruled out press alt f4

i was mason

me-mason
darknight-mason
ambierona-mason
probability 1.01-mafia
chigr-mafia
sheepwarrior-mafia
secret asian-mafia
Paladin8-inspector

everyone else-citzen


So first off I will say, this [i]is[/i] a homework problem but I've been really thinking about it and I'm just not sure.

We had a lab (this is hs chemistry 1) to study heat of hydration, which involved dissolving copper sulfate, hydrate and anhydrous. Dissolving the hydrate, the temperature of the water changed from 21.8 deg C to 20.8 deg C. Dissolving the anhydrous, the temperature changed from 21.5 deg C to 26.3 deg C. First, we have to calculate the heat of solution of the hydrate. Would it be negative since the temperature went down?

Also, would the dissolving of the hydrate be exothermic or endothermic? I was under the impression that if delta H is negative (which I am assuming it is, but you may correct me), the reaction is exothermic, but I have been looking around and it appears that it is endothermic. Can you explain? Thanks.

The change in the temperature of the water is indicative of the heat absorbed/released by the [b]water[/b].  Thus, an equal amount of heat (but opposite in sign since it "flows" in the "opposite direction") is released/absorbed by the reaction.

For the situation you describe:

$ q_{\text{sys}} \equal{} \minus{} q_{\text{surr}}$

In your case, the surroundings is the water and the system is the copper sulfate.

Thus, if the reaction is endothermic, it absorbs heat from its surroundings (the temperature of the surroundings decreases); if the reaction is exothermic, it releases heat into its surroundings (the temperature of the surroundings increases).  You can tell whether the dissolving of the hydrate is exothermic or endothermic by looking at the temperature change in the water.


Let $ p(x)$ be a polynomial with rational coefficients. Prove that there is an integer $ k$ such that
$ p(x \plus{} k)\minus{} p(x)$ is a polynomial with integer coe±cients.

just pick $ k$ to be the least common multiple of the denominators of the coefficients of $ p(x)$


Let $a,b,c>0$.Prove that:

         $ab(a+b)+bc(b+c)+ca(c+a)\geq\frac{2}{3}\cdot(a+b+c)(ab+bc+ca)$ ;)

amgm on $\sum a^2 b \ge 6abc$
I think some of the inequalities you posted recently are of the same type.  :huh:

As Harazi said the other day, what is the point of posting such easy inequalities?

ok,Harazi and you are right.From now on I won't post such inequalities. :oops:


Find all functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying
\[ f(f(x \plus{} y)) \equal{} f(x \plus{} y) \plus{} f(x)f(y) \minus{} xy\]
for all real numbers $x$ and $y$.

[quote="Raja Oktovin"]Find all functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying
\[ f(f(x \plus{} y)) \equal{} f(x \plus{} y) \plus{} f(x)f(y) \minus{} xy\]
for all real numbers $ x$ dan $ y$.[/quote]
Let $ P(x,y)$ be the assertion $ f(f(x\plus{}y))\equal{}f(x\plus{}y)\plus{}f(x)f(y)\minus{}xy$
Let $ f(0)\equal{}a$

It's elementary first (just plug in the original equation) to see that the only solution of the type $ f(x)\equal{}bx\plus{}a$ is $ f(x)\equal{}x$

Subtracting $ P(x,y)$ from $ P(x\plus{}y,0)$, we get the new assertion $ Q(x,y)$ : $ af(x\plus{}y)\equal{}f(x)f(y)\minus{}xy$

Then $ Q(a,\minus{}a)$ $ \implies$ $ f(a)f(\minus{}a)\equal{}0$ and so $ \exists u\in\{f(\minus{}a),f(a)\}$ such that $ f(u)\equal{}0$

Then $ Q(x\minus{}u,u)$ $ \implies$ $ af(x)\equal{}\minus{}u(x\minus{}u)$

If $ a\equal{}0$, $ Q(x,y)$ implies $ f(x)f(y)\equal{}xy$ and so either $ f(x)\equal{}x$ $ \forall x$, which indeed is a solution, either $ f(x)\equal{}\minus{}x$ $ \forall x$, which is not a solution.

If $ a\ne 0$, this implies $ f(x)\equal{}\minus{}\frac uax \plus{}\frac{u^2}a\equal{}bx\plus{}a$ for some $ b$ and so (see beginning lines), $ f(x)\equal{}x$, which is impossible since $ f(0)\ne 0$ in this subcase.

Hence the unique solution $ \boxed{f(x)\equal{}x}$ $ \forall x$

I have another solution with much more calculation the the one of Mr.[b]pco[/b].

Let $ P(x,y)$ be the assertion $ f(f(x\plus{}y)) \equal{} f(x\plus{}y) \plus{} f(x)f(y) \minus{} xy$ and let $ a: \equal{}f(0)$.

We have $ P(x,0) \Rightarrow f(f(x)) \equal{} (a\plus{}1)f(x),\forall x\in\mathbb{R}$. Thus, let $ x\equal{}0$ we have $ f(a) \equal{} a(a\plus{}1)$. Moreover, the assertion $ P(x,y)$ becomes: $ af(x\plus{}y) \equal{} f(x)f(y) \minus{} xy,\forall x,y,\in\mathbb{R}$.

Suppose that $ a\neq 0$. We have:

$ P(a,a) \Rightarrow af(2a) \equal{} a^2(a\plus{}1)^2 \minus{} a^2$, yielding $ f(2a) \equal{} a^2(a\plus{}2)$.

$ P(2a,a) \Rightarrow af(3a) \equal{} a(a\plus{}1)a^2(a\plus{}2) \minus{} 2a^2$, yielding $ f(3a) \equal{} a^2(a\plus{}1)(a\plus{}2) \minus{} 2a \equal{} a(a^3 \plus{} 3a^2 \plus{} 2a \minus{}2)$.

$ P(3a,a) \Rightarrow af(4a) \equal{} a(a\plus{}1)a(a^3 \plus{} 3a^2 \plus{} 2a \minus{}2) \minus{} 3a^2$

$ P(2a,2a) \Rightarrow af(4a) \equal{} a^4(a\plus{}2)^2 \minus{} 4a^2$.

This means that $ a(a\plus{}1)a(a^3 \plus{} 3a^2 \plus{} 2a \minus{}2) \minus{} 3a^2 \equal{} a^4(a\plus{}2)^2 \minus{} 4a^2\Leftrightarrow a^2(a\minus{}3)(a\plus{}3) \equal{} 0$

But we have supposed that $ a\neq 0$ then $ a\equal{}3$ or $ a\equal{}\minus{}3$.

1. If $ a\equal{}3$ then $ P(x,y)$ becomes $ 3f(x\plus{}y) \equal{} f(x)f(y) \minus{} xy$ and hence $ f(3) \equal{} 3\cdot (3\plus{}1) \equal{} 12$.

$ P(x,1) \Rightarrow 3f(x\plus{}1) \equal{} f(1)f(x) \minus{} x$.

$ P(x\plus{}1,1) \Rightarrow 3f(x\plus{}2) \equal{} f(1)\frac{f(1)f(x) \minus{} x}{3} \minus{} (x\plus{}1)$

$ P(x\plus{}2,1) \Rightarrow 3f(x\plus{}3) \equal{} f(1)\frac{f(1)\frac{f(1)f(x) \minus{} x}{3} \minus{} (x\plus{}1)}{3} \minus{} (x\plus{}2)$.

$ P(x,3) \Rightarrow 3f(x\plus{}3) \equal{} f(x)f(3) \minus{} 3x \equal{} 12f(x) \minus{} 3x$.

Therefore, $ 12f(x) \minus{} 3x \equal{} f(1)\frac{f(1)\frac{f(1)f(x) \minus{} x}{3} \minus{} (x\plus{}1)}{3} \minus{} (x\plus{}2),\forall x\in\mathbb{R}$, contradiction.

2. If $ a\equal{} \minus{}3$, the same argument yields another contradiction.

Thus $ a\equal{}0$ and we have $ f(x)f(y) \equal{} xy,\forall x,y\in\mathbb{R}$. Put $ y\equal{}1$ then $ f(x) \equal{} kx,\forall x\in\mathbb{R}$, where $ k\equal{}\frac{1}{f(1)}$. Now the assertion $ P(x,y)$ yields that $ k\equal{}1$.

Therefore, the only solution is $ f(x) \equal{} x$.

Hello!!!!!!
[hide=" my much easier solution which i am doubtful of"]
Let $ y\equal{}0$
$ f(f(x))\equal{}f(x)\plus{}f(0)f(x)$
$ f(f(x))\equal{}(f(0)\plus{}1)f(x)$
thus its easy to see
$ f(x)\equal{}(f(0)\plus{}1)x$

Now to find f(0)
Let $ x\equal{}0$ as well
$ f(0)\equal{}(f(0)\plus{}1)*0$
$ f(0)\equal{}0$ and $ f(x)\equal{}x$

Is that valid?
[/hide]
Roobiks
[/hide]

[quote="RobRoobiks"]Hello!!!!!!
Let $ y \equal{} 0$
$ f(f(x)) \equal{} f(x) \plus{} f(0)f(x)$
$ f(f(x)) \equal{} (f(0) \plus{} 1)f(x)$
thus its easy to see
$ f(x) \equal{} (f(0) \plus{} 1)x$
[/quote]

Yes, easy to see, but wrong. You got $ f(x) \equal{} (f(0) \plus{} 1)x$  $ \forall x\in f(\mathbb R)$ and not $ \forall x\in\mathbb R$ (you did not show that $ f(x)$ is a surjection).

[quote="Mashimaru"]Therefore, $ 12f(x) \minus{} 3x \equal{} f(1)\frac {f(1)\frac {f(1)f(x) \minus{} x}{3} \minus{} (x \plus{} 1)}{3} \minus{} (x \plus{} 2),\forall x\in\mathbb{R}$, contradiction.[/quote]
How do you reach the contradiction? For example if $ f(1) \equal{} 0$ then it is equivalent to $ f(x) \equal{} \frac {x \minus{} 1}{6}$. If $ f(1) \neq \sqrt [3]{108}$ then it is equivalent to $ f(x) \equal{} ax \plus{} b$ for some $ a,b \in \mathbb{R}$. If $ f(1) \equal{} \sqrt [3]{108}$ then you can reach the contradiction at once. So you can only use it to see that $ f$ is a linear function. (Sorry if this was obvious)

$ f(f(x \plus{} y)) \equal{} f(x \plus{} y) \plus{} f(x)f(y) \minus{} xy$     (1)
     Denote $ f(0) \equal{} c$.
     Take $ y \equal{} 0$ to get    $ f(f(x)) \equal{} f(x)(1 \plus{} c)$.    (2)
     Set $ y \equal{} \minus{} x$ in (1):    $ f(f(0)) \minus{} f(0) \equal{} x^{2} \minus{} f(x)f( \minus{} x)$, but  by (2),  $ f(f(0)) \minus{} f(0) \equal{} c^{2}$.
Hence, $ f( \minus{} x)f(x) \equal{} c^{2} \minus{} x^{2}$   (3).   Take $ x \equal{} c$  in (3) to conclude there is a real number $ u$ satisfying $ f(u) \equal{} 0$. Take $ x \equal{} u$ in (2):   $ f(0) \equal{} f(f(u)) \equal{} f(u)(1 \plus{} c) \equal{} 0$. It means $ f(f(x)) \equal{} f(x)$, as $ c \equal{} 0$, but than from (1),  $ f(x)f(y) \equal{} xy$  for all $ x,y$.
     Easy to verify $ f(x) \equal{} 0$, for all $ x$ is not a solution. Take $ t$ such that $ f(t) \equal{} w$ and $ w$ is nonzero.
     $ wf(x) \equal{} xt$, which can be rewritten as $ f(x) \equal{} kx$, where $ k$ is a fixed real number. Plugging it in (1),
     $ (x \plus{} y)(k^{2} \minus{} k) \equal{} xy(k^{2} \minus{} 1)$, for all $ x,y$. If $ x \equal{} \minus{} y$ and they are both nonzero, the last equation would give $ k^{2} \minus{} 1 \equal{} 0$. Assume $ k \equal{} \minus{} 1$;  Than, for all pairs $ (x,y)$, $ 2(x \plus{} y) \equal{} 0$, which is clearly false. It means that $ k \equal{} 1$ and $ f(x) \equal{} x$, which obviously satisfies the given conditions.

Let $P(x,y):=f(f(x+y))=f(x+y)+f(x)f(y)-xy$

$P(0,0)$ yields $f(f(0))=f(0)+f(0)^2$
$P(-f(0),f(0))$ yields $f(f(0))=f(0)+f(-f(0))f(f(0))+f(0)^2\overset{\text{ from }P(0,0)}{\Longrightarrow}f(0)+f(0)^2=f(0)+f(0)^2+f(-f(0))f(f(0))\Longrightarrow f(f(0))f(-f(0))=0$
Thus $f(f(0))=0\text{ or }f(-f(0))\text{ or both of them are equal to }0$

[b][color=#f00]Case 1:[/color][/b] $f(-f(0))=0$
$P(0,-f(0))$ yields $f(f(-f(0))=f(-f(0))+f(0)f(-f(0))\Longrightarrow f(0)=0$

[b][color=#274E13]Case 2:[/color][/b] $f(f(0))=0$
Plugging this in $P(0,0)$ we obtain $f(0)+f(0)^2=0\Longrightarrow f(0)(f(0)+1)=0$ thus either $f(0)=0\text{ or }f(0)=-1$
[i][b]Subcase:[/b][/i] $f(0)=-1$
This implies $f(-1)=0$
$P(x,0)$ yields $f(f(x))=f(x)+f(x)f(0)\Longrightarrow f(f(x))=0$ plugging this in $P(x,y)$ yields $f(x+y)+f(x)f(y)=xy:=Q(x,y)$
$Q(x+1,-1)$ yields $f(x)+f(x+1)f(-1)=-(x+1)\Longrightarrow f(x)=-(x+1)$ however plugging this into $P(x,y)$ yields a contradiction.

Therefore since $f(0)=0$ in both of these cases we can conclude that $f(0)=0$

$P(x,0)$ yields $f(f(x))=f(x)$ plugging this in $P(x,y)$ yields $f(x+y)=f(x+y)+f(x)f(y)-xy\Longrightarrow f(x)f(y)=xy:=T(x,y)$
$T(x,1)$ yields $f(x)f(1)=x\Longrightarrow f(x)=\frac{x}{c}\text{ where }c=f(1)$
Pugging this in $P(x,y)$ yields $\frac{x+y}{c^2}=\frac{x+y}{c}+\frac{xy}{c^2}-xy\Longrightarrow(c-1)(x+y)=(c-1)xy\Longrightarrow(c-1)(x+y-xy)=0$ thus either $c=1\text{ or }x+y=xy$
However the latter clearly yields a contradiction. Therefore $c=1$ which implies that $f(x)=x$

In conclusion $\boxed{f(x)=x, \forall x\in\mathbb{R}}$ $\blacksquare$.


inequalities
calculus
integration
inequalities unsolved

Hi I have an inequality here, that I could not prove algebraically (not inductively).  I was wondering if someone here can please help me with this.  Thanks!

Question:  Prove that for any positive integers $ n$ and $ k$ that
$ \displaystyle\left(\frac{n\plus{}1}{n}\right)\left[ (k\plus{}1)^{\frac{n}{n\plus{}1}}\minus{}k^{\frac{n}{n\plus{}1}}\right] <\frac{1}{k^{\frac{1}{n\plus{}1}}}<\left(\frac{n\plus{}1}{n}\right)\left[ k^{\frac{n}{n\plus{}1}}\minus{}(k\minus{}1)^{\frac{n}{n\plus{}1}}\right]$.

Hi .zou thanks for replying, and yes, I understand that this can be shown by calculus.  However, I was wondering if anyone happens to know of a solution that uses simply algebraic techniques such as factoring the $ \displaystyle (k\plus{}1)^{\frac{n}{n\plus{}1}}\minus{}k^{\frac{n}{n\plus{}1}}$ and $ \displaystyle k^{\frac{n}{n\plus{}1}}\minus{}(k\minus{}1)^{\frac{n}{n\plus{}1}}$ terms.

Oh... I may have misunderstood your request. Sorry!

Although I'm unable to crack this, but how do you prove it using calculus?  I'm very interested.

P.S. Where did you find this interesting inequality (and it is not basic at all!)?


[u]cls XI[/u]

[b]1.[/b]
Fie $ a_{1},a_{2},...,a_{n}$ numere naturale nenule ordonate $ a_{1}\leq a_{2}\leq ...\leq a_{n}$
a)Sa se determine numarul functiilor $ f: \{1,2,..,n\}\rightarrow \mathbb N^*$ cu proprietatea

$ (P):$  $ f(k)\leq a_{k} ,$ $ \forall k \in \{1,2,...,n\}$

b) Care este numarul functiilor injective cu proprietatea $ (P)$ ?

[b]2.[/b]
Fie $ (a_{n})_{n}$ un sir de numerele naturale , $ a_{n}\geq 2$ , $ \forall n \in \mathbb N$
Sa se arate ca orice numar natural $ N$ poate fi scris in mod unic sub forma

$ N\equal{}c_{0}\plus{}c_{1}\cdot a_{1}\plus{}c_{2}\cdot a_{1}a_{2}\plus{}...\plus{}c_{n}\cdot a_{1}a_{2}\cdot..\cdot a_{n}$

unde $ c_{0},c_{1},...,c_{n}$ sunt numere naturale cu proprietatea :

$ 0\leq c_{i}<a_{i\plus{}1}$ , $ i\equal{}\overline{0,n\minus{}1}$ , $ c_{n}\neq 0$

[b]3.[/b]
Fie $ a,b$ numere naturale nenule si $ d\neq 1$ , c.m.m.d.c al lor.
a)Sa se arate ca exista o partitie a lui $ \mathbb N^*$ in doua submultimi $ A$ si $ B$ ($ A$ reunit cu $ B$= $ \mathbb N^*$ si $ A$ intersectat cu $ B$ = multimea vida) cu proprietatea :

$ (P):$  $ x \in A \Rightarrow x\plus{}a \in A$  si $ y\in B \Rightarrow y\plus{}b \in B$

b)Sa se arate ca daca $ (A,B)$ este o partitie a lui $ \mathbb N^*$ cu proprietatea $ (P)$ atunci pentru orice $ z\in \mathbb N^*$ numerele $ z$ si $ z\plus{}d$ sunt ambele in $ A$ sau ambele in $ B$.

[b]4.[/b]
Pe suprafata unui poligon de arie $ 13$ se aseaza $ 10$ poligoane de arie $ 6$.Sa se arate ca exista patru poligoane ce se suprapun dupa o arie mai mare ca $ \frac{1}{70}$

Timp de lucru : 4 ore .

ash vrea si eu sa shtiu..unde..pot vedea...rezultatele la al. papiu ilarian..??...:-?...ajutatima plz..:D..:*


For those of you on the NYC math team, I guess you have heard that the ARML practice session is scheduled for Saturday, May 21, from 10 am to 3 pm, at CCNY.

I'll be there, giving a talk on "State Diagrams and Probability".  We also have a special guest coming.  He'll be giving a motivational speech at the beginning, followed by a talk on "Homothecy".  Hint: If you bring your AoPS books, then I'm sure he would be happy to autograph them.

Post here if you will be attending...

well, considering that the email said "Attendance at the practice session is [b]required.[/b]"  :)

I met mr.rusczyk at mathcamp.
not only he's super smart, hes very cool person.
we played basketball together.

Id be happy if he remembers me(i doubt it tho)


function
inequalities proposed
inequalities

with $ a,b,c\in[1;2]$ Prove that
   $ \frac{5}{a\plus{}b\plus{}c\plus{}2} \leq \frac{1}{x\plus{}2}\plus{}\frac{1}{y\plus{}2}\plus{}\frac{1}{z\plus{}2} \leq \frac{6}{x\plus{}y\plus{}z\plus{}2}$

[quote="2424"]with $ a,b,c\in[1;2]$ Prove that
   $ \frac {5}{a \plus{} b \plus{} c \plus{} 2} \leq \frac {1}{a \plus{} 2} \plus{} \frac {1}{b \plus{} 2} \plus{} \frac {1}{c \plus{} 2}$[/quote]

 $ f: [1,2]\rightarrow\mathbb{R},f(x)\equal{}\frac{1}{x\plus{}2}$ is convex function

[quote="Marius Mainea"][quote="2424"]with $ a,b,c\in[1;2]$ Prove that
   $ \frac {5}{a \plus{} b \plus{} c \plus{} 2} \leq \frac {1}{a \plus{} 2} \plus{} \frac {1}{b \plus{} 2} \plus{} \frac {1}{c \plus{} 2}$[/quote]

 $ f: [1,2]\rightarrow\mathbb{R},f(x) \equal{} \frac {1}{x \plus{} 2}$ is convex function[/quote]
Can you post full your sulotion?
my sulotion is very short and easy!

$ f(a)\plus{}f(b)\plus{}f(c)\ge 3f(\frac{a\plus{}b\plus{}c}{3})\ge 5f(a\plus{}b\plus{}c)$


I have never taken an AoPS class and I plan to. I was wondering how well the Algebra 2 class is. I took Algebra 2 last summer but my skills in that subject are quite shabby. I know a lot of Algebra 2 is covered in Precalculus. Would it be better to take the Algebra 2 class, or the Precalculus class.

Thanks :D

This would go better in the Classes Info forum.

A description of Intermediate Algebra: [url]http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=4[/url]


Assume ABCD is a rectangle.Two lines PQ and XY are drawn parallel to AB and CD.Both the diagonals AC and BD are joined.Find the no. of right triangles in the figure so obtained. :?:

stop answering vague questions........
give me the diagram also

This question depends on whether or not the rectangle is a square.

Isn't it kind of odd how
Sharma and MathWizard type topics in the same way
(like thisssssssssssssss)
and how they are first to post in each other's threads.
All threads they make have very trivial problems (the ones I put up are even better)
Also they both use at least 1 emoticon in each post, either in the front or in the end.
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=192495

anubhav
plz tell me how 2 put raise 2 power

it seems that you are very observant ...
yeah , sharma knows me .........& we are classmates ...........
but anyway 'thissssssssss' is a slip of letters like the slip of words


To place the first paving stone in a path, Alex starts at the crate
of stones, walks two feet, places the stone, and returns to the
crate. For each subsequent stone, Alex walks three feet farther
each way. Alex will place the first 100 stones in a path. After
returning to the crate from placing the $ 100^\text{th}$ stone, what is the total distance Alex walked, in feet?

We see that the the distances form an arithmetic sequence of 2, 5, 8,... 299.  However, since he walks both ways, we multiply this by 2.  Therefore, the answer is:

$ 2(2\plus{}5\plus{}8\plus{}...299)$

$ 2(2\plus{}299)(\frac{1}{2})(100)$

$ 301(100)$

$ \boxed{30100}$


let $ a_1,a_2,...,a_n > 0$

prove that :

$ (\sum_{i\equal{}1}^{n}a_i)(\sum_{i\equal{}1}^{n}\frac{1}{a_i})\leq n^2\plus{}[\frac{n^2}{4}](\sqrt{\frac{M}{m}}\minus{}\sqrt{\frac{m}{M}})^2$

with $ M\equal{}max(a_i)$ and $ m\equal{}min(a_i)$

$ x\equal{}[x]\plus{}{x}$

:)


calculus
integration
derivative
function
induction
calculus computations

The integral is obviously zero, since it's the derivative of a function which is zero at both endpoints.

If you want the integral of the square, that's also routine; integrate by parts $ n$ times.

[quote="jmerry"]The integral is obviously zero, since it's the derivative of a function which is zero at both endpoints.
[/quote]
Why? Example $ n=1$!


Show that the product of four consecutive positive integers is always one less than a perfect square.

[hide="deriviation"]Consider $n(n+1)(n+2)(n+3)$. Expand the first and last terms to get $(n^{2}+3n)(n^{2}+3n+2)=((n^{2}+3n+1)-1)((n^{2}+3n+1)+1)=(n^{2}+3n+1)^{2}-1$, which is one less than a perfect square.[/hide]


inequalities
geometry
incenter
inradius
trigonometry
symmetry
trig identities

Let $ABCD$ be a quadrilateral circumscribed about a circle, whose interior and exterior angles are at least 60 degrees. Prove that
\[
\frac{1}{3}|AB^3 - AD^3| \le |BC^3 - CD^3| \le 3|AB^3 - AD^3|.
\]
When does equality hold?

I have never known, and i ahve asked thousands of times but i never remember, what is a circumscribed cuadrilateral? I know it is A or B but i dont know wich.....
 :blush:

Pascual

well it's the exact opposite of cyclic quadrilaterals: it's a quadrialetaral with a circle inside, which touches all the sides. Obviously the sum of the opposite sides is the same (because they are composed of 4 segments each being a tangent to the circle from one of the 4 vertices).

hey, can anyone post an elegant solution for this one, I foun a solution but it is horrible! and also full of details i have to improve, so ANY other solution would be welcome......

um ok. this solution may in fact be the "horrible solution" that you found but anyway, here goes:
let AB=a, BC=b, CD=c, AD=d. note it is sufficient to prove one side of the inequality, since the other follows directly by symmetry. So we want to prove: 1/3 (a^3-d^3) leq (b^3-c^3). note we can WLOG a>d, b>c, since a+c=b+d, a-d=b-c and thus a>d, b<c is contradiction. a-d=b-c, so that inequality is equivalant to (a^2+d^2+ad) <= 3(b^2+c^2+bc). since b^2+c^2+bc=a^2-2ad+d^2+3bc, that ienquality is equivalant to : 7ad <= 2a^2+2d^2+9bc. since 2a^2+2d^2 >= 4ad, it would now be sufficient to prove 3ad <= 9bc, i.e. 3bc >= ad (which is in fact true). My way of proving this is a ruthless bash:
let the angle bisectors meet at the incentre I of ABCD. Let <IAB=<IAD=p, <ABI=<CBI=q, <BCI=<DCI=r, <CDI=<ADI=s, inradius = n. Note that 30 < p,q,r,s < 60, so 1/sqrt 3 < tan p, etc < sqrt 3. since a = n (tan p + tan q), b = n(tan q + tan r), etc; proving 3bc >= ad is equivalant to proving that  (tan r + tan s)(tan q + tan r)/(tan p + tan q)(tan p + tan s) <= 3. now, we know that r+s+p+q=180, so using the horrible tan sum formula, we can deduce that the sum of the numbers tan p, tan q, tan r and tan s is equal to the sum of their product in threes. after using *lots* of algebra, we can simplify that expression to something much nicer : (tan r)^2+1 / (tan p)^2 + 1. this obviously has a maximum of 3, attained when r = 60, p = 30. so we are done. the condition for equality occurs when: firstly, 2a^2+2d^2 = 4ad, and 3bc = ad, i.e. a = d, r = 60 and p = 30 (which is a very wierd condition for equality).

I have this solution:

Note first that it suffices to show only one of the inequalities, since we can obtain one from another interchanging vertices $A$ and $C$.

Since the quadrilateral is circumscribed, $AB - AD = BC - CD$. Thus the inequality is equivalent to
\[\frac13(AB^2 + AB\cdot AD + AD^2) \leq BC^2 + BC\cdot CD + CD^2\]

The condition concerning angles on the quadrilateral is equivalent to its internal angles being between $60^\circ$ and $120^\circ$. Hence, by the law of cosines,
\[ BC^2 + BC\cdot CD + CD^2 = BC^2 + CD^2 - 2\cdot BC\cdot CD\cdot \cos 120^\circ \]
\[ \geq BC^2 + CD^2 - 2\cdot BC\cdot CD\cdot \cos C = BD^2\]

Thus we just have to show that $BD^2 \geq \frac13(AB^2 + AB\cdot AD + AD^2)$. But $BD^2 = AB^2 + AD^2 - 2\cdot AB\cdot AD\cos A$, hence, since
\[ AB^2 + AD^2 - 2\cdot AB\cdot AD\cos A\geq \frac13(AB^2 + AB\cdot AD + AD^2) \]
\[ \iff 2AB^2 + 2AD^2 \geq AB\cdot AD(1 + 6\cos A)\]
and $\cos A \leq \frac12$,
\[ AB(1 + 6\cos A) \leq 4AB \leq 2AB^2 + 2AD^2,\]
because $AB^2 + AD^2 \leq 2AB\cdot AD$ and the result follows.

[quote="cyshine"]Since the quadrilateral is circumscribed, $ AB \minus{} AD \equal{} BC \minus{} CD$.[/quote]Why?

I think I misunderstand "circumscribed quadrilateral", can someone make a dwawing, in particular showing what "exterior angles" are here?

This is pretty much the same as cyshine's but with more explanation.
We set $ a \equal{} BC$, $ b \equal{} CD$, $ c \equal{} DA$, and $ d \equal{} AB$.

By a well-known property of tangential quadrilaterals, the sum of the two pairs of opposite sides are equal; hence $ a \plus{} c \equal{} b \plus{} d \Rightarrow a \minus{} b \equal{} d \minus{} c \Rightarrow |a \minus{} b| \equal{} |d \minus{} c|$ Now we factor the desired expression into $ \frac {|d \minus{} c|(c^2 \plus{} d^2 \plus{} cd)}{3} \le|a \minus{} b|(a^2 \plus{} b^2 \plus{} ab)\le 3|d \minus{} c|(c^2 \plus{} d^2 \plus{} cd)$. Temporarily discarding the case where $ a \equal{} b$ and $ c \equal{} d$, we can divide through by the $ |a \minus{} b| \equal{} |d \minus{} c|$ to get the simplified expression $ (c^2 \plus{} d^2 \plus{} cd)/3\le a^2 \plus{} b^2 \plus{} ab\le 3(c^2 \plus{} d^2 \plus{} cd)$.

Now, draw diagonal $ BD$. By the law of cosines, $ c^2 \plus{} d^2 \plus{} 2cd\cos A \equal{} BD$. Since each of the interior and exterior angles of the quadrilateral is at least 60 degrees, we have that $ A\in [60^{\circ},120^{\circ}]$. Cosine is monotonously decreasing on this interval, so by setting $ A$ at the extreme values, we see that $ c^2 \plus{} d^2 \minus{} cd\le BD \le c^2 \plus{} d^2 \plus{} cd$. Applying the law of cosines analogously to $ a$ and $ b$, we see that $ a^2 \plus{} b^2 \minus{} ab\le BD \le a^2 \plus{} b^2 \plus{} ab$; we hence have $ c^2 \plus{} d^2 \minus{} cd\le BD \le a^2 \plus{} b^2 \plus{} ab$ and $ a^2 \plus{} b^2 \minus{} ab\le BD \le c^2 \plus{} d^2 \plus{} cd$.

We wrap up first by considering the second inequality. Because $ c^2 \plus{} d^2 \minus{} cd\le BD \le a^2 \plus{} b^2 \plus{} ab$, $ \text{RHS}\ge 3(a^2 \plus{} b^2 \minus{} ab)$. This latter expression is of course greater than or equal to $ a^2 \plus{} b^2 \plus{} ab$ because the inequality can be rearranged to $ 2(a \minus{} b)^2\ge 0$, which is always true. Multiply the first inequality by $ 3$ and we see that it is simply the second inequality with the variables swapped; hence by symmetry it is true as well.

Equality occurs when $ a \equal{} b$ and $ c \equal{} d$, or when $ ABCD$ is a kite.

Since quadrilateral $ABCD$ is circumscribed about a circle we have $AB+CD=BC+AD \Rightarrow |AB-AD|=|BC-CD|$.Let us consider the case when $AB \neq AD$ and $BC \neq CD$.Then the inequality is equivalent to
$\frac{1}{3}(AB^2+AD^2+AB*AD) \le BC^2+CD^2+BC*CD \le 3(AB^2+AD^2+AB*AD)$

Before giving the proof I first provide a prerequisite lemma:

[u][b]Lemma[/b][/u]:$\frac{BC*CD}{AB*AD}=\frac{1-cosA}{1-cosC}$

[b]Proof[/b]:
$BD^2=AB^2+AD^2-2AB*ADcosA=BC^2+CD^2-2BC*CDcosC \Rightarrow AB*AD(1-cosA)=BC*CD(1-cosC) \Rightarrow \frac{BC*CD}{AB*AD}=\frac{1-cosA}{1-cosC}$

We first show the left part of the inequality,which is equivalent to

$((AB-AD)^2+3AB*AD) \le 3((BC-CD)^2+3BC*CD)$

$\Leftrightarrow 2(AB-AD)^2+9BC*CD-3AB*AD \ge 0$

Now $9BC*CD-3AB*AD \ge 0$

$\Leftrightarrow 3BC*CD-AB*AD \ge 0$

$\Leftrightarrow 3\frac{BC*CD}{AB*AD}-1 \ge 0$

$\Leftrightarrow 3\frac{1-cosA}{1-cosC}-1 \ge 0$

$\Leftrightarrow  3cosA-cosC \le 2$

Now $60^{\circ} \le \angle{A},\angle{C} \le 120^{\circ} \Rightarrow cosA \le \frac{1}{2},cosC \ge \frac{-1}{2}$ from which the above inequality follows.Thus with this the left part of the inequality follows.

Similar simplications of the rightpart suffices to prove that $3AB*AD \ge BC*CD$ for which the proof is similar to the left part.
In the case when $AB=AD$ we get $BC=DC$ for which the equality holds(note that otherwise there is strict inequality)

Silly problem...

Clearly it suffices to show the left inequality.  Since $AB+CD = BC+AD \implies |AB-AD| = |BC-CD|$, it suffices to prove \[ \frac13(AB^2 + AB \cdot AD + AD^2) \le BC^2 + BC \cdot CD + CD^2. \] This follows by noting that \begin{align*} BC^2 + BC \cdot CD + CD^2 &\ge BC^2 + CD^2 - 2(BC)(CD)\cos(\angle BCD) \\ &= BD^2 \\ &= AB^2 + AD^2 - 2(AB)(AD)\cos(\angle BAD) \\ &\ge AB^2 + AD^2 - AB \cdot AD \\ &\ge \tfrac13(AB^2 + AD^2 + AB \cdot AD) \end{align*} the last line following by AM-GM.

WLOG the incircle of $ABCD$ has radius $1$.

Let $\alpha = A/2$, etc. It suffices to prove the LHS, and then the RHS will follow by symmetry of applying the LHS result with $ABCD$ changed to $BCDA$. After some manipulations we find that the inequality is equivalent to

$(\cot \alpha + \cot \beta)^2 + (\cot \alpha + \cot \delta)^2  + (\cot \alpha + \cot \beta)(\cot \alpha + \cot \delta) \le 3((\cot \gamma + \cot \beta)^2 + (\cot \gamma + \cot \delta)^2 + (\cot \gamma + \cot \beta)(\cot \gamma + \cot \delta))$. Translate everything back to $AB, BC, CD,DA$ it suffices to prove that
$$AB^2 + AD^2 + AB\cdot AD\le 3(BC^2 + CD^2 + BC\cdot CD).$$

Indeed, observe that $3(BC^2 + BD^2+BC \cdot CD) \ge 3BD^2 \ge 3(AB^2 + AD^2 - AB\cdot AD) \ge \frac{3(AB^2 + AD^2)}{2} \ge  AB^2 + AD^2 + AB\cdot AD$.

Edit: whoops shouldn't have substituted

WLOG, we only need to prove one of the inequalities. Note that $AB-AD=BC-CD$ by Pitot's. Note that $AB^2+AB\cdot AD+AD^2\ge AD^2$ because $\angle BAD\le 120^\circ$ and $BC^2+CD^2-BC\cdot CD\le AD^2$ because $\angle BCD\ge 60^\circ.$ Thus, we have $AB^2+AB\cdot AD+AD^2\ge BC^2-BC\cdot CD+CD^2.$ It remains to prove $3(BC^2-BC\cdot CD+CD^2)\ge BC^2+BC\cdot CD+CD^2$ which is true.

Nice geo ineq!\\

Clearly, by symmetry, we only need to show one side. Let's show the left side.\\

By Pitot, $|BC-CD|=|AB-AD|$, so we only need to show that $$3(BC^2+BC\cdot CD+CD^2)\geq BA^2+BA\cdot AD+AD^2.$$

Note that $$3(BC^2+BC\cdot CD+CD^2)= 3(BC^2+CD^2-2\cdot BC\cdot CD\cos{120})\geq 3BD^2.$$ We also have $$3BD^2\geq 3(BA^2+AD^2-2BA\cdot AD\cos(60))=3BA^2+3AD^2-3BA\cdot AD$$ We then wish to show that $$3BA^2+3AD^2-3BA\cdot AD\geq BA^2+BA\cdot AD+AD^2,$$ but this just rearranges to $$BA^2+AD^2\geq 2BA\cdot AD,$$ which is clearly true, so we are done.\\

Remarks: When I first tried this problem a few months ago, I failed. What I did back then was trying to fix the inradius as 1, letting the incircle tangency points to $AB,BC,CD,DA$ be $W,X,Y,Z$ respectively, and let $AW=AZ=a,BW=BX=b$, and so on. The angle condition becomes $1/\sqrt{3}\leq a,b,c,d\leq \sqrt{3}$, and then I tried to expand everything and prove the resulting inequality. However, the issue with this approach is that $a,b,c,d$ are not actually independent, and the resulting inequality is not true for all $1/\sqrt{3}\leq a,b,c,d\leq \sqrt{3}$, which forces you to use the very cumbersome condition $$\arctan{a}+\arctan{b}+\arctan{c}+\arctan{d}=180,$$ which is hard to incorporate into the inequality.

By symmetry, we only need to show the first inequality. We may also assume WLOG that $AB>AD$ and $BC>CD$. Then simply notice the inequality $$\frac 13(AB^2+AB \cdot AD + AD^2) \leq AB^2 - AB \cdot AD + AD^2 \leq BD^2 \leq BC^2+BC \cdot CD+CD^2$$ by Law of Cosines and AM-GM, as needed.


linear algebra
matrix
linear algebra unsolved

Let $ M$ and $ N$ be complex matrices such that $ N$ and the transpose of $ M$ have the same dimension.

Let $ A$ be $ \begin{pmatrix}0 & N \\
M & 0\end{pmatrix}$

(When I say multiplicity, I mean geometric multiplicity)
Prove that the following are equivalent :

1. $ \lambda\neq 0$ is an eigenvalue of $ A$ of multiplicity $ f$
2. $ \minus{} \lambda\neq 0$ is an eigenvalue of $ A$ of multiplicity $ f$
3. $ \lambda^2\neq 0$ is an eigenvalue of $ M N$ of multiplicity $ f$
4. $ \lambda^2\neq 0$ is an eigenvalue of $ N M$ of multiplicity $ f$

Proving that 1. and 2. are equivalent is fairly easy, and proving that 3. and 4. are isn't that hard either.

But how to connect 1/2 and 3/4? What I find especially hard is showing that the multiplicities are the same. :huh:

Thanks,
fredbel6

Let us write an eigenvector of $ A$ to be $ [v w]^T$, which matches the dimension of block matrix. When you multiply out, being an eigenvector means $ Nw \equal{} \lambda v$ and $ Mv \equal{} \lambda w$, which means $ MNw \equal{} \lambda^2 w$. Consider the projection map $ [v w]^T$ to $ w$. This can be shown to be a linear isomorphism of eigenspaces. (eigenspace of $ \lambda$ of $ A$ and eigenspace of $ \lambda^2$ of $ MN$)


induction
graph theory
combinatorics proposed
combinatorics

a) Find the number of maximal matchings in $ K_n$.

b) Find the number of maximal matchings in $ K_{m,n}$.

c) Find the number of maximal matching in [url=http://mathworld.wolfram.com/PetersenGraph.html]Petersen graph[/url].

For the maximal matchings: http://en.wikipedia.org/wiki/Maximal_matching

$ K_n$ is the complete graph on $ n$ vertices and $ K_{m,n}$ is the complete bipartite graph with $ m$ and $ n$ vertices in its parts.

a) Let's denote this number as $ f(n)$. We see that $ f(2)\equal{}1,f(3)\equal{}f(4)\equal{}3,f(5)\equal{}f(6)\equal{}15$. So we want to prove that $ f(2n\minus{}1)\equal{}f(2n)\equal{}(2n\minus{}1)!!$ (product of all odd numbers up to $ 2n\minus{}1$).
This follows from the fact that in $ K_{2n\minus{}1}$ a maximal matching is going to leave exactly one vertex out (as in that vertex is not incident with any edge in the matching). We have $ 2n\minus{}1$ choices of choosing that vertex and then have to find a maximal matching of $ K_{2n\minus{}2}$. This can be done in $ f(2n\minus{}2)$ ways and therefore $ f(2n\minus{}1)\equal{}(2n\minus{}1)f(2n\minus{}2)$. Similarly a matching of $ K_{2n}$ is going to have exactly $ n$ edges so it is a partition of the vertices in disjoint pairs. Vertex $ A$ will be in an edge of the matching and that edge can be chosen in $ 2n\minus{}1$ ways. What's left is a $ K_{2n\minus{}2}$ which has $ f(2n\minus{}2)$ maximal matchings. We conclude that $ f(2n)\equal{}(2n\minus{}1)f(2n\minus{}2)$ and proceed by induction.

b) We call this $ f(m,n)$. WLOG $ m\le n$. Every vertex on the part with $ m$ vertices has to be contained in one of the edges of every maximal matching. fix a vertex. There is $ n$ edges incident there and if we choose one the remaining graph has $ f(m\minus{}1,n\minus{}1)$ maximal matchings. By induction it follows that $ f(m,n)\equal{}\frac{\max(m,n)!}{(\max(m,n)\minus{}\min(m,n))!}$.

c) I can only think of a brute force approach. If I'm not mistaken there is $ 9$ maximal matchings of size $ 3$, $ 36$ ones of size $ 4$ and $ 6$ ones of size $ 5$, giving a total of $ 51$ maximal matchings.

are you sure there are 36 maximal matchings in case if the maximal matching is of size 4.

nope :roll: . I counted again and got 24 this time :P
What's your approach?
Can we conclude anything about the maximal matchings of the voltage graph of a graph we already know?


give an examples for continous,uniformly continous,equicontinous and also the function which is continous not uniform similarly uniform continous  not equi continous

equicontinuous is a property of a family of functions (usually a sequence)

the other two are applicable to a single function f

$ f(x)\equal{}x^2$ is continuous but not uniformly continuous.


this was proposed at the 8th grade in bucharest olympiad:

suppose a>b>c>d are positive integers such that
a+b+c+d=2004 and  a2 - b2+c2 - d2 = 2004.

a) find the minimal possible value of a;

b) how many possibilities are for b,c,d when a is minimal?

It's cute  :)

. 4d < 2004 < 4a so a > 501 and d < 501.

. 2nd equation can be rewritten (a+b)(a-b-1) + (c+d)(c-d-1)=0 so a=b+1 and c=d+1

So a+c = 1003, and smallest a is 503.

Unique solution is 503,502,500,499.

Hope is's correct.

another way is to use the fact that 2004 = a2 - b2 + c2 - d 2 = (a-b)(a+b) + (c-d)(c+d) \geq 1(a+b) + 1(c+d) = a+b+c+d = 2004 so a-b=1, c-d =1 and from here on the same thing.  :D


function
limit
integration
calculus
calculus computations

$ \lim_{n \to \infty} \frac{\Gamma (n\plus{}1,n)}{n!} \equal{} ?$

Maybe I am ignorant but isnt the answer zero?

$ \lim_{n\to\infty}\frac{\Gamma(n\plus{}1,n)}{n!}\equal{}\lim_{n\to\infty}\frac{\int_n^{\infty}t^ne^{\minus{}t}dt}{\int_0^{\infty}t^{n}e^{\minus{}t}dt}$

The top is going to zero and the bottom to infinity...so isn't the answer zero?

Even if the limit is 0 (I don't know the answer), such a ,,proof'' isn't proof indeed...

[quote="@petko"] such a ,,proof'' isn't proof indeed...[/quote]
I'm sorry, I don't understand what that means. Do you mean that even if the answer is zero that I have not formalized it? I dont't know how much more I can do to formalize it if my analysis is indeed correct. Just noting that

$ \int_a^{a}f(x)dx\equal{}0$

[quote="Mathstud28"]Do you mean that even if the answer is zero that I have not formalized it?[/quote]

Yes. Well... there are many limit examples which may seem obvious at first glance but aren't indeed. Hence your solution is not helpful for me, sorry.

I believe the answer is zero. It is quite possible I am wrong, and if so I am sure someone on here will point it out. But as for the formalization of this I would say this

$ \lim_{n\to\infty}\frac {\Gamma\left(n \plus{} 1,n\right)}{n!}$

$ \equal{} \lim_{n\to\infty}\frac {\int_n^{\infty}t^ne^{ \minus{} t}dt}{n!}$

$ \equal{} \lim_{n\to\infty}\frac {\int_0^{\infty}t^ne^{ \minus{} t}dt \minus{} \int_0^{n}t^ne{ \minus{} t}}{n!} \equal{} \lim_{n\to\infty}\frac {n! \minus{} n!}{n!} \equal{} 0$

The only place I am not positive of is we know that

$ \lim_{\varphi\to\infty}\int_0^{\varphi}t^ne^{ \minus{} t}dt \equal{} n!$

But does $ \lim_{n\to\infty}\int_0^{n}t^ne^{ \minus{} t}dt \equal{} \lim_{n\to\infty}n!$...it seems shaky at the best. Im sure someone can clarify. Sorry if I have made an analytical mistake.

With the $ \chi^2$ distribution you can show $ \frac{1}{n!}\int_{n\plus{}a\sqrt{n}}^{n\plus{}b\sqrt{n}}\frac{x^n}{e^x}\, dx\to \frac{1}{\sqrt{2\pi}} \int_a^b e^{\minus{}\frac{x^2}{2}}\, dx$ for $ n\to\infty$  .

In particular for $ a\equal{}0$ and $ b\equal{}\infty$ we get $ \frac{1}{n!}\int_n^\infty\frac{x^n}{e^x} \, dx\to\frac12$ for $ n\to\infty$ .

Great! :) Answer 1/2 goes well with my numerical predictions. Thanks.


LaTeX AMC AIME ARML algebra polynomial geometry

Discuss Mock AMC H here.

Here is my answer key.  I will accept challenges if you have any.

Mock AMC Laurendi II Answer Key

1. B
2. B
3. D
4. E
5. D
6. C
7. B
8. D
9. E
10. B
11.A
12. C
13. C
14. E
15. D
16. C
17. E
18. D
19. D
20. B
21. B
22. D
23. C
24. D
25. A

1st Place-  [color=#DAA520]JGeneson (118)[/color]
2nd Place-  [color=#808080]towersfreak2006 (113.5)[/color]
3rd Place(T)- [color=olive]pilot[/color]
3rd Place(T)- [color=olive]alan[/color]

Average score: 88.9
Number of Participants: 25

I have an Excel document attached with everything EXCEPT names.

For the solutions I am going to try to put together a solution packet.  But since I don't have enough time to do the whole thing I am asking that we everyone could pitch in a couple solutions.  If you one to submit a solution, type it up as if you were going to put it in a $\LaTeX$ document and PM it to me.  I will look at the solution and if it is good I will use it.

Thanks

Solutions are still needed for problems:

\[\begin{tabular}{|c|c|c|c|c|} \hline
&  & &  &  \\  \hline
&  &  &  &10\\ \hline
11 & 12 &  & 14 & \\ \hline
16 & 17 & 18 & 19 & \\ \hline
21 & 22 &  &  &  \\ \hline
\end{tabular}\]

How do you do the problem with the quadrilateral inscribed in a circle. The quad had side lengths of $\sqrt{2002}, \sqrt{2006}, \sqrt{2005}, \sqrt{2009}$. At least I think it was. And they wanted the diameter, which is expressed as $\sqrt{d}$,  think. Do you use Ptolemy? How would you apply it here if so?

Is the answer to #17
[hide]240131, so choice (e)?[/hide]
I got that after time was up and want to know if that's right... seems like a crazy answer to me.

Can someone give me a link to the pdf or the questions please? Thanks.

I still have some things that I need to wait a little while on, so it will be a little while before I post results.

I will put the test back on the original page.

[quote="1234567890"]How do you do 3?[/quote]

Did you notice he later edited it, so that the first equation is now $2^x\times2^y=1024$?  If not, do you want to try it again?

If you are still having trouble, here is the answer:
[hide]We can change the given equations into the following system:
\[x+y=10\]
\[xy=4\]
Rather than just solve these, we can square the first and subtract the second to get
\[x^2+y^2=92\]
\[\sqrt{x^2+y^2}=2\sqrt{23}\]
Since the hypotenuse (which was what we wanted) is equal to $\sqrt{x^2+y^2}$, the answer must be $2\sqrt{23}$.[/hide]

For the solutions I am going to try to put together a solution packet.  But since I don't have enough time to do the whole thing I am asking that we everyone could pitch in a couple solutions.  If you one to submit a solution, type it up as if you were going to put it in a $\LaTeX$ document and PM it to me.  I will look at the solution and if it is good I will use it.

Thanks

Solutions are still needed for problems[edit: the version that will be updated is at the top of this page]:

\[\begin{tabular}{|c|c|c|c|c|} \hline
 &  & &  &  \\  \hline
6 &  & 8 &  &10\\ \hline
11 & 12 &  & 14 &15 \\ \hline
16 & 17 & 18 & 19 & 20\\ \hline
21 & 22 & 23 & 24 & 25 \\ \hline
\end{tabular}\]

[quote="joml88"]For the solutions I am going to try to put together a solution packet.  But since I don't have enough time to do the whole thing I am asking that we everyone could pitch in a couple solutions.  If you one to submit a solution, type it up as if you were going to put it in a $\LaTeX$ document and PM it to me.  I will look at the solution and if it is good I will use it.

Thanks[/quote]

Which ones do you need solutions for?

Ah, I don't think that's fair. I got the problems and just looked at the PDF while I did it, not noticing the change to number 3. It appeared as 2^y * 2^y at first... so...  :no:

Heh... I think people who did question 3 like it was should get credit...

Can you name all the geometric progressions in problem 22?

Edit: Actually I think 22 is poorly defined. I mean, (1, 10, 100) is not (10, 1, 100), and they are all in A geometric progression - nothing stated about order in the problem. And then add to 111. So, I think E is the right answer.

#20:

Note that by Fermat's Theorem, if p not equal to 5, then

p^4=1(mod 5), so 13p^4-888=0(mod 5).

Note that 13p^4-888=5 have no solution

therefore p=5 is the only possible way that 13p^4-888 is prime.

When p=5, we see that 13(625)-888=7237 is prime, so there's only one possble solution. So B

[quote="joml88"][quote="h_s_potter2002"]Just to let you know, my score of 74.5 wasn't included.  That's also why there are only 24 scores on the file.[/quote]

Are you sure?  I gave everyone full credit for #3 so scores may be higher than I previously told you.[/quote]

Yeah, it's on there.  I forgot about everyone getting full credit on 3  :blush: .  But I still think you're missing someone's score - there are 24 scores, but 25 people took the test.

23 is in fact C.  Oberserve that $AE = AD + DE = \sqrt{5}$. (Why give us such a strange D by the way?) Let $OC$ and $BC$ be $x$ and $y$ respectively, so that $AC = 1-x$. Take $F$ on $\overline{OA}$ such that $\overline{EF}$ is perpendicular to $\overline{AF}$. By similar triangles, we have $OF = x\sqrt{2}$ and $EF = y\sqrt{2}$. By Pythagoras,

$x^2 + y^2 = 1$
$2y^2 + (x\sqrt{2}+1)^2 = 2x^2 + 2y^2 + 2x\sqrt{2} + 1 = 5$

giving $x = \frac{1}{\sqrt{2}}$ and $y = \frac{1}{\sqrt{2}}$. It is easy to see that $AB^2 = (1-x)^2 + y^2 = 2 - \sqrt{2}$, which leads to C.

On 24 I get D as well. Call $p_n$ the largest product of positive integers that can be obtained where the sum of the integers is $n$. By inspection, $p_1 = 1, p_2 = 2, p_3 = 3, p_4 = 4$. Now, it is pointless to take a 1, but any integer $m > 3$ can be replaced by $m-2$ and $2$, giving a product as least as large. Thus, $p_n$ is composed of 2's and 3's.  Now we may replace 3 2's with 2 3's. Thus, we should have 0, 1, or 2 2's in the maximal product. In fact, we get 2 2's and 667 3's, which leads to $2 + 2\cdot 2 + 3\cdot 3 + 4\cdot 667 = 2683$.

25 was an old ARML problem if I remember correctly. Note that $N_b = 1 + 2 + \cdots + b^2 = \frac{b^2\cdot (b^2+1)}{2}$. If $b = 2k > 2$, then we have $N_b = \frac{1}{2}0\frac{1}{2}00_b = k0k0_b$; it is clear that the sum of the squares of the digits is $2k^2$, which gives $k = 2, 3, \dots, 16$; 15 solutions. Otherwise $b = 2k+1$. Note that $\frac{1}{2} = \frac{k}{2k} = \frac{k}{b-1} = k \cdot \left(\frac{1}{b} + \frac{1}{b^2} + \frac{1}{b^3} + \cdots \right) = 0.kkkkkk\dots_b$ and, consequentially, $1 = .(2k)(2k)(2k)\dots_b$. Thus, we have $N_b = \frac{1}{2}0\frac{1}{2}00_b = kkkk.kkkkk\dots_b + kk.kkkkk\dots_b = k(k+1)00_b$. The sum of the squares of the digits is less than 512 for $k = 1, 2, \dots, 15$. A total of 30 solutions, giving an answer of A.

What others need solutions?

an explanation of why #19 is not (2+rad3)^2 (E) would be appreciated :) .

I hope no one minds, but if you post a solution here I might use it for the solution packet.  Of course, I will try to fix any little typos etc. you may have.

Thanks

I think if you guys don't mind, we should try to post the solutions to the problems.

I did mock AMC and it's really hard to answer all those questions by yourself with explanation.

2. How many even numbers less than are divisible by 3?

Another word, how many number less than are divisible by 6?

2004/6 = 334 however, 2004 shouldn't be included so it's 333. :)

Let $ABC$ be an equilateral triangle of side length 2x. Draw mutually tangent circles $C_A, C_B, C_C$, each of radius $x$, centered at $A, B$, and $C$ respectively. Let $O$ denote the center of $ABC$. Let $\omega$ and $\Omega$ be the two circles tangent to each of $C_A, C_B$, and $C_C$, with $\Omega$ containing the other circles. The question asks for $\frac{R_\Omega}{R_\omega}$.

Because $ABC$ is equilateral, $\frac{AO}{AB} = \frac{1}{\sqrt{3}}$ which gives $AO = \frac{2x}{\sqrt{3}}$. But it is clear that $R_\Omega = x + AO = x\frac{2+\sqrt{3}}{\sqrt{3}}$. Furthermore, $R_\omega = AO - x = x\frac{2-\sqrt{3}}{\sqrt{3}}$. Thus, $\frac{R_\Omega}{R_\omega} = \frac{2+\sqrt{3}}{2-\sqrt{3}} = \left(2 + \sqrt{3}\right)^2 = 7 + 4\sqrt{3}$, which is simply not a choice in the version I see at http://www.artofproblemsolving.com/Forum/topic-23163.html .

As for your question, I think "E - Cannot be determined from the given information" is distinct from "F - None of the above."

i agree that "none of the above" is different from "cannot be determined" which is why i omitted that one...

[quote="h_s_potter2002"][quote="furious"]yeah! i got 15th place on this with the magnificant score of 89. go me!  :)[/quote]

An 89 is better than an 80.5.[/quote]

Which, in turn, is also better than a 77.5 :)

So what's the cutoff score? ;)

I hope I made it.

hey mildorf, i didn't see no choice F (none of the above) for #19 ;) . But, no problem. This test was quite good overall. I particularly liked the polynomial problem where you get 2001(x-1)(x-2)(x-3)(x-4)(x-5)+2x-1.

How come I figure them out now...

#12
Let the trapezoid be ABCD, with AD as the base.  Sin A + sin D=1/3.
To minimize the legs, they will be equal length (let's call them x), so the base angles will be equal.
Sin A + Sin A=1/3.
2 sin A=1/3
sin A=1/6
X is the hypotenuse of the right triangle with height opposite A=20.
Sin A=20/x
X=20/sin A
X=20*6
X=120

There are two legs of length 120, so 120+120=240.  (C)

18)

first look at the sequence 012345.  6 digits.  Using the binomial theorem, we observe that 6 combinations can represent a one digit term.  15 can represent a two digit.  20 can represent a three digit.  15 combos can represent a four digit.  6 combos can represent a 5 digit.  1 combination can represent a 6 digit.  63 combinations in all.
We also observe the fact that.  1>1/4+1/16+.....  So in other words, 4^6>4^5+4^4+4^3+4^2+4^1+4^0.  4^6 is the 64th term in the sequence.  4^6+4^0 is 65th.  4^6+4^1 is 66th.  4^6+4^0+4^1 is 67th.  4^6+4^2 is 68th.  4^6+4^2+4^0 is 69th.  4^6+4^2+4^1 is the 70th term in the sequence.  so its 4116.  Nice problem!

Is there a formal solution packet anywhere?


What is the greatest positive integer $ n$ such that $ n^2\plus{}2009n$ is a square of an integer?

Whatever, I'll post my solution here (it's also in the other thread).

[hide]Line of thought: To prove a number isn't a perfect square, we want to bound it between two consecutive perfect squares. The obvious choice here is $ (n \plus{} 1004)^2 < n^2 \plus{} 2009n < (n \plus{} 1005)^2$, or $ n^2 \plus{} 2008n \plus{} 1004^2 < n^2 \plus{} 2009n < n^2 \plus{} 2010n \plus{} 1005^2$. The right-hand inequality is always true (for positive n), and the left-hand is true for $ n^2 \plus{} 2008n \plus{} 1004^2 < n^2 \plus{} 2009n \implies n > 1004^2$. So, if $ n > 1004^2$, the value will not be a perfect square. Note that if $ n^2 \plus{} 2009n \equal{} (n \plus{} 1004)^2$, which is a perfect square, then $ n \equal{} \boxed{1004^2}$, so this has to be the answer.[/hide]

This is quite nice! My approach was to complete the square but that is sort of overly general. You use the "find the largest" well.

I would write my proof by saying: we claim $ n\equal{}1004^2$ is the largest.
i) check that works
ii) then prove that if $ n>1004^2$, then that is not a square

The reason that I would write the proof this way is because the way you used the inequality was a bit more like scratch work. The inequality you stated $ (n\plus{}1004)^2<n^2\plus{}2009n<(n\plus{}1005)^2$ isn't always true. Also what you showed technically was that if that inequality was true, then $ n>1004^2$, whereas you wanted to show the converse (which follows because your steps are reversible but again it would be better to do it the other way).


Given three real numbers $p$,$q$, and $r$ where $0<p,q,r<1$. Show that
$pq+qr+rp-2pqr<1$

For positive reals $u,v,t$ it holds

$uvt+uv+ut+vt>0$

Adding $u+v+t+1+2$ to the both sides and simplifying we get

$(u+1)(v+1)(t+1)+2>u+v+t+3$

Substitute $a=u+1, b=v+1, c=t+1$ where $a,b,c>1$:

$abc+2>a-1+b-1+c-1+3\iff abc+2>a+b+c\iff a+b+c-2<abc$

Substitute $p={1\over a},q={1\over b}, r={1\over c}$ where $p,q,r<1$ and the result follows.