A set of seven integers has a median of 73, a mode of 79, and a mean of 75. What is the least possible difference between the maximum and minimum values in this set?

[hide="Answer"]The mode is $79$, so there must be at least two of those. The median is $73$, so there are three numbers larger and three smaller; we'll say there are 3 $79$'s, so we have $79\cdot 3+73=310\Rightarrow 75\cdot 7=525$, so the other three numbers will sum to $215$, which gives an average of $71\frac{2}{3}$. Therefore, we can have ${71,72,72,73,79,79,79}$, and the difference will be $\boxed{8}$.[/hide]

[b]EDIT:[/b] Fixed.

[quote="JesusFreak197"]Oh, whoops, they sum to 215, not 225. That definitely changes the answer.[/quote]

you mean 525, right

No, I was referring to the three terms less than 73.


calculus
integration
trigonometry
calculus computations

The sum is equivalent to $4\sum_{n=1}^{\infty} \left(\frac{-3}{n+4}+\frac{2}{n+3}+\frac{1}{n+2}\right)$, but I'm in a hurry and can't evaluate it right now.

From what Jimmy has done
[hide]$13/3$[/hide]

Yes, your answer is correct, dheardie. :)


function
induction
algebra unsolved
algebra

Find all functions $ f: \mathbb{Z} \rightarrow \mathbb{Z}$ for which we have $ f(0) \equal{} 1$ and $ f(f(n)) \equal{} f(f(n \plus{} 2) \plus{} 2) \equal{} n$ for every natural number $ n$

[hide="Solution"]
Plug in $ f(n \plus{} 2) \minus{} 2$ for $ n$.  This gives that $ f(n \plus{} 2) \minus{} 2 \equal{} f(f(f(n \plus{} 2) \minus{} 2 \plus{} 2) \plus{} 2) \equal{} f(n \plus{} 2 \plus{} 2) \equal{} f(n \plus{} 4)$.  Plugging in $ n \minus{} 2$ for $ n$, we  have that $ f(n) \equal{} f(n \plus{} 2) \plus{} 2$.  Then, $ f(n) \minus{} 2 \equal{} f(n \plus{} 2)$.  We show that, by induction, $ f(n) \equal{} 1 \minus{} n$ for nonnegative integers $ n$.  The base cases are proven as $ f(0) \equal{} 1 \equal{} 1 \minus{} 0$ and $ f(1) \equal{} f(f(0)) \equal{} 0 \equal{} 1 \minus{} 1$.  Then, $ f(n \plus{} 2) \equal{} f(n) \minus{} 2 \equal{} 1 \minus{} n \minus{} 2 \equal{} \minus{} 1 \minus{} n \equal{} 1 \minus{} (n \plus{} 2)$ (if $ n$ is odd and even are seperate cases here, but both are proven in the base case).  Now, we prove the assertion for negative $ n$ by noticing that $ 1 \minus{} n$ is positive and if $ f(x) \equal{} y$, then $ f(y) \equal{} f(f(x)) \equal{} x$, so we have that $ f(n) \equal{} 1 \minus{} n$.  This works since $ f(f(x)) \equal{} 1 \minus{} (1 \minus{} x) \equal{} x$ and $ f(f(n \plus{} 2) \plus{} 2) \equal{} f( \minus{} n \minus{} 1 \plus{} 2) \equal{} f(1 \minus{} n) \equal{} f(f(n)) \equal{} n$, so $ f(x) \equal{} 1 \minus{} x$ is the only solution. [/hide]

EDIT: I apologize to everyone; I neglected to read "every natural $ n$."  The above solution will only hold for integers $ n$.

Keep in mind that we only have relations for $ n>0$... you did not pay attention in the beginning.
For instance, all you proved is that $ f(n)\equal{}f(n\plus{}2)\plus{}2$ for $ n\geq 3$

Unless I'm mistaken, functions like $ f(x)\equal{}(1\minus{}x)(x\geq2) \plus{} g(x)(x<2)$ will work, as long as $ f(g(1))\equal{}1$

Dear [b] The QuattoMaster 6000 [/b]

Your solution does not work. It could be proven that $ f(n) \leq 0 \forall n \in \mathbb{N}$. It means that $ f(n\plus{}2)\minus{}2 \notin \mathbb{N}$. And hence, at the very beginning, you cannot plug it in the position of $ n$.

Regards.

I think that there are infinitely many solutions. May [b] me@home [/b] elaborates your solution.


geometry
3D geometry
function
algebra
domain

The strength of a rectangular wooden beam is proportional to the product of the width and the cube if its depth.  If the beam is to be cut from a log in the shape of a cylinder of radius 3 feet, express the strength S of the beam as a function of the width x.  Also, what is the domain of S?

[hide="Hints:"]
Draw a diagram.
Look at right triangles.
The proportionality just means that strength is $S(w,d)=k \cdot w \cdot d^{3}$, where $k$ is a constant. [/hide]

Your reply did not fully display.

ALso, your hint is not too clear nor helpful.

Thank you all, especially the last reply.


trigonometry
trig identities
Law of Cosines
Pythagorean Theorem
geometry

An arch is built in the form of an arc of a circle and is subtended by a chord 30 ft long. If a
cord 17 ft long subtends half that arc, what is the radius of the circle?

Here is my solution:
[hide]Using law of cosines, we get
$30^{2}=2r^{2}-2r^{2}\cos{2\theta}$ and
$17^{2}=2r^{2}-2r^{2}\cos{\theta}$
Divide those to get
$\frac{900}{289}= \frac{1-\cos{2\theta}}{1-\cos{\theta}}$
Use double angle identity
$\frac{900}{289}= \frac{1-(2\cos^{2}{\theta}-1)}{1-\cos{\theta}}$
Simplify to
$\frac{900}{289}= 2(1+\cos\theta)$
$\cos\theta = \frac{161}{289}$
Substituting $\cos\theta$ into the second equation, we get $r = 18\frac1{16}$
[/hide]

And here is what the answer is supposed to be:
[hide]$18\frac34$[/hide]

What am I doing wrong?

A bit simpler solution, without Cosine Law:

[hide="Solution"]Let $O$ be the center of the circle which the arc belongs to, $A,B$ its endpoints, $M$ the midpoint of the arc and $N$ midpoint of the chord $AB$. Then $AN=15, AM=17$ which by Pythagorean Theorem gives $MN=8$. Therefore $OA=r, ON=r-8, AN=15$, hence again by Pythagorean we get

$r^{2}=15^{2}+(r-8)^{2}$

$r^{2}=225+r^{2}-16r+64$

$16r=289$

$r=18{1\over 16}$[/hide]

which means that there's an error in your book.

[quote="Hoborobo"]I just realized that the answer was [hide]$18 ft \frac34 in$ which is $18\frac1{16}ft$ -.-[/hide][/quote]

Yeah, that is what I almost thought, but then I realized that it is 18 feet 3/4 inches.

This is from the MMPC isn't it.

Michigan Mathematics Prize Competition.


inequalities
trigonometry
inequalities proposed

$\blacktriangleright 1^{\circ}.\ \ \boxed {\ \cos A\cos B\cos C\le \frac{r(R-r)}{2R^2}\ }\le \frac 18\ .$
$\blacktriangleright 2^{\circ}.\ \ \boxed {\ \cos A\cos B+\cos B\cos C+\cos C\cos A\le \frac{3r}{2R}\ }\le \frac 13\left(1+\frac rR\right)^2\ .$

1)We can show that
$\cos{A}\cos{B}\cos{C}=\frac{p^2-(2R+r)^2}{4R^2}$
Then we use inequality $p^2\leq 4R^2+4Rr+3r^2$ :D


A, B,C,D are 4 points on the circle (O). The intersection of AC and BD is D, the intersection of AD and BC is E, the intersection of AB and CD is F. Prove that O is the othorcenter of triangle DEF

(Let's find a splendid drill for this problem by using "Butterfly")

UHM
WLOG suppose that ABCD be convex quadrateral
let P be intesection of AC and BD
Let K be intersect of (ABP)and (CDP) you can prove A,D,K,O are on the circle and B,C,K,O be too
Applying power of point with a circle we have   F.K.O are on the line and EP perpendicular to OF at K
So we have result

[quote="disorder"](Let's find a splendid drill for this problem by using "Butterfly")[/quote]

In my note [i][url=http://www.cip.ifi.lmu.de/~grinberg/Butterfly.zip]On cyclic quadrilaterals and the butterfly theorem[/url][/i], I did it the other way round: proving the butterfly theorem using your problem. For your direction, we would have to use a converse of Theorem 5 of the above note. Anyway, this should work but I prefer direct proofs of your problem.

  darij


geometry
3D geometry
sphere
search
integration
calculus
calculus computations

I hope these are not too easy (and sorry I could not find these on the search menu):

Find $ \int_{B(0,1)}1\mbox{d}x$ and $ \int_{S^{n\minus{}1}}1\mbox{d}\sigma$
where $ B(0,1)\equal{}\{x\in\mathbb{R}^{n}: |x|\equal{}\sqrt{x_{1}^{2}\plus{}\cdots\plus{}x_{n}^{2}}<1\}$, $ S^{n\minus{}1}\equal{}\{x\in\mathbb{R}^{n}: |x|\equal{}1\}$, and $ \sigma$ denotes the volume over the sphere.
Please explain and not just write the answers.  Thanks!

I found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=26191]this[/url] with a search; "measure" and "ball" were the words that worked.

It outlines the methods, but doesn't finish the actual computations.
See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=63476]here[/url].


There are 3 points given on a plain so that they are not on the same

straight line. How many straight lines are there on this plain that

are equidistant from all three of these points?

by the way, both my parents were born near seoul, too. my key board doesn't have korean so anyong!

[quote="bigboy82892"]within that plane, there are three lines because if a line is not between two points, it is impossible for it to be equidistant from all three of the points.[/quote]

Not correct. There are at least three such lines: perpendicular bisectors of the altitudes of the triangle determined by the given points.

It's not difficult to prove that there are no more such lines, but I'll leave that to the reader.

To prove that those are the only such lines, couldn't we just say that any such line must be parallel to a line created by two of the points, and that line must also be directly in between the third point and that other line (hence perpendicular bisector of the altitude)? If it wasn't perpendicular to any of the lines created by adjoining the points, it would have to be closer to some point than another depending on which way it slopes.

Is there a more rigorous way of going about this?

You're almost there. Let $F$ be the foot of the altitude $AF$ in triangle $ABC$. Since line $p$ is equidistant from $B$ and $C$, it must be parallel to the line $BC$, hence perpendicular to $AF$. Also, this means that $F$ is at the same distance from $p$ (because all points of $BC$ are at the same distance from $p$). And now you have that $F$ and $A$ are equally distant from a line perpendicular to $AF$ and the conclusion is straightforward. Also, by the property of uniqueness of a segment's perpendicular bisector, you're guaranteed that there are exactly three such lines.


Two fair dice are rolled. If the total is at least five, what is the probability that the total is seven?

There are 36-1-2-3=30 ways to make at least 5, and 6 to make 7 so 6/30=1/5


trigonometry
logarithms
calculus
calculus computations

Answer following question.

[1] $\frac{d}{dx} x^3\sqrt{1+x^2}$

[2] $\frac{d}{dx} \sin ax\cdot\cos ax$

[3] $\frac{d}{dx} xe^{2x}$

[4] $\frac{d}{dx} \ln (\ln x)$

[5] Let $f(x)=\cos x +1,g(x)=\frac{a}{bx^2+cx+1}$.Find the constant numbers $a,b$ and $c$ such that

$f(0)=g(0),f'(0)=g'(0),f''(0)=g''(0)$.


For how many integers $N$ between $1$ and $1990$ is the improper fraction $\frac{N^2+7}{N+4}$ not in lowest terms?

$\text{(A)} \ 0 \qquad \text{(B)} \ 86 \qquad \text{(C)} \ 90 \qquad \text{(D)} \ 104 \qquad \text{(E)} \ 105$

[hide]We want to get rid of the variable $n$. to find the remainder when $n^2+7$ is divided by $n+4$, we find $f(-4)$ (synthetic division shortcut).

The remainder is $21$. Therefore, our answer is $\lfloor \frac{1990}{21}\rfloor=86$.[/hide]

First of all the remainder when $\ N+4$ is divided into $\ N^2+7$ is $\ 23$ and not $\ 21$.  Also, your last step is a little bit ambiguous and also $\ \lfloor \frac{1990}{21}\rfloor=94$ and not $\ 86$.  Heres how I did it:

[hide]$\ \frac{N^2 + 7}{N+4} = N - 4 + \frac{23}{N+4}$.  Now since $\ 23$ is prime we don't have to worry about $\ N+4$ being a multiple of any of its factors except itself.  So this fraction is not in lowest terms when $\ N + 4$ is a multiple of $\ 23$.  This happens for $\ \lfloor \frac{1994}{23}\rfloor=86$ ways.[/hide]


We given several squares with the total area 1. Prove that You can arrange them in a square of area 2 such that there are no intersections.

This is a very nice problem. I will post the solution tomorrow if nobody else posts a solution.

[quote="rem"]Sorry, I comletely forgot about this topic. I think that there is a nicer solution but I could not come up with one(I think the nice solution is similar to this one: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=61586[/url])
Anyways, here is my solution( i left out proof of some half-obvious facts, so you guys can work on them on your own).
First, you prove that the two largest squares can be placed in such a way, so that leagest square has its corner in top left corner of the square with are $2$ and the second largest square will be placed in the bottom right corner of that square. Oe can prove with basic inequalities that these squares will not overlap.
Then we will increase the size of both these squares so that they touch at one vertice. Then it's easy to prove that two more of the second leargest new squares can be placed in the square with area $2$ so that the four squares do not overlap. Finally you use induction to prove that each of those four squares can be tiled with no overlapping with a distinct subset of the original squares(whose sum is area 1) so that there is still space left and all the original squares will be used. We upse induction by assuming that the problem is true for the number of squares being $k\le n$, where $n+1$ is the number of squares we are dealing with in the problem.
All those points can be proved with basic inequalities. :)
So, QED[/quote]
Nice proof. I think i'll have to read it again though. :lol:

The proof is ok, not that nice. thanks though :)

Do you know the official solution?(If i proved all the "obvious points/lemmas whatever u call them, then my solution would be at least two pages long)

no, i don't have the official solution, i'll look though.


Given positive reals a,b,c with $ a+b+c=3$,
Prove that $ \sqrt{3-ab}+\sqrt{3-bc}+\sqrt{3-ac}\ge 3\sqrt{2}$

[quote="nealth"]Given positive integers a,b,c with $ a+b+c=3$,
Prove that $ \sqrt{3-ab}+\sqrt{3-bc}+\sqrt{3-ac}\ge 3\sqrt{2}$[/quote]
Check $ a=b=c=1.$ :D
Maybe $ a,b,c$ are real numbers?
If so it's equivalent to my inequality:
Given positive reals $ a,b,c$ with $ a+b+c=1$,
Prove that $ \sqrt{1-3ab}+\sqrt{1-3bc}+\sqrt{1-3ac}\ge \sqrt6.$ :wink:
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=156002
I have found very nice proof for this inequality, but I'll discuss it only after 19.07.2007. :P

[quote="arqady"][quote="nealth"]Given positive integers a,b,c with $ a+b+c=3$,
Prove that $ \sqrt{3-ab}+\sqrt{3-bc}+\sqrt{3-ac}\ge 3\sqrt{2}$[/quote]
Check $ a=b=c=1.$ :D [/quote]

$ a=b=c=1$ gives equality : $ 3\sqrt{2}=3\sqrt{2}$  :wink:

You didn't get him. He's meant that we only need to check $ a=b=c=1$ since $ a, b, c$ are positive integers and $ a+b+c=3$ :D.

[quote="arqady"][quote="nealth"]Given positive integers $ a,b,c$ with $ a + b + c = 3$. Prove that
$ \sqrt {3 - bc} + \sqrt {3 - ca} + \sqrt {3 - ab}\ge 3\sqrt {2}$.[/quote]
Maybe $ a,b,c$ are real numbers?

If so it's equivalent to my inequality:

Given positive reals $ x,y,z$ with $ x + y + z = 1$. Prove that
$ \sqrt {1 - 3yz} + \sqrt {1 - 3zx} + \sqrt {1 - 3xy}\ge \sqrt6$.
I have found very nice proof for this inequality.[/quote]
$ \sum{\sqrt {(x + y + z)^2 - 3yz}\geq \sum{\frac {\sqrt {3}[6x^2 + 19x(y + z) + 7y^2 + 2yz + 7z^2]}{10\sqrt {2}(x + y + z)}} = \sqrt {6}(x + y + z)}$.
[hide="Proof."]$ 200(x + y + z)^2[(x + y + z)^2 - 3yz] - 3[6x^2 + 19x(y + z) + 7y^2 + 2yz + 7z^2]^2$

$ = (y + z - 2x)^2[23x^2 + 52x(y + z) + 2(y + z)^2]$

$ + 3(y - z)^2[9(y + z - x)^2 + 5x^2 + 4x(y + z) + 4(2y^2 + 13yz + 2z^2)]\geq 0$.[/hide]

Let a,b,c>0 such that $ a^2\plus{}b^2\plus{}c^2\equal{}1$ prove that
$ \sum\sqrt[3]{1\minus{}3ab}   \geq  \sqrt[3]{9}$

[quote="gagarin"]Let a,b,c>0 such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$ prove that
$ \sum\sqrt [3]{1 \minus{} 3ab} \geq \sqrt [3]{9}$[/quote]
Try $ a\equal{}b\equal{}c\equal{}\frac{1}{\sqrt3}.$ :wink:

I think this problem is
$ \sum\sqrt [3]{1 \minus{} ab} \geq \sqrt [3]{18}$


I have a bunch of files that are stored as .asx files that I want to download to a C.D., but I don't know if there is any way to do that.  I tried using Windows Media Player, and MusicMatch Jukebox, but they won't let you burn them to a C.D.  So does anyone know how to do that?  Btw, this is meant for listening with a C.D. player, not for storing data.

I've never heard of .asx -- are you sure that most CD players will be able to read such files?

I just found out what asx was, and realized that you can't use it to burn to a CD becuase it's sotred on another computer's server.  Sorry  :blush:

For the record, .asx is Windows Media's streaming audio format, so you can't save it to your hard drive or to any removable media such as CDs, for that matter.

However, there do exist programs that allow you to record the audio being streamed, so that you can save them on your hard drive.  One such program is Streambox VCR.  I think there are legal problems with using these, though.


quadratics
combinatorics proposed
combinatorics

For which n do there exist subsets of {1,2,3...n} A(1), A(2), ... A(n) s.t.
(1) : n is not in A(n)
(2) : m is in A(n) *iff* n is not in A(m)
(3) : any two sets from A(1) ... A(n) have a non-empty intersection.

Maybe I don't get something ?????? :(  :?

[quote="iura"]I think n>=7.
For n=7 j belongs to A_i iff i-j is NOTa quadratic residue mod 7.
If we have constructed example for n we add A_n+1={1..n} and get one for n=1.
Finally for n<=6 one checks manually (maybe it's possible for n<=6 I'm too lazy to check)[/quote]
yes you are correct iura, and we cant do it for n<=6. My soln was a bit different:
suppose some set had less then three elements. If A(k) had one element, say p, qhen A(k) has an empty intersection with A(p). If A(k) has 2 elements, say p and q, then A(p) has an empty intersection with A(q), because for all t, A(t) contains either p or q --> t isn't in A(p) or t isn't in A(q), so t cannot simultaneoulsy be in A(p) and A(q). So each set has at least 3 elements. Now we formulate an equivalant problem: consider a n*n table, and place 0 in the cell (i,j) if i isn't in A(j) and 1 if i is in A(j). From our conditions we know (i) the table has a diagonal of '0's, (ii) it is inversely symmetric about that diagonal (iii) any 2 rows have a common '1'
(iv) any row has at least 3 '1's. But the total number of '1's is n(n-1)/2, so
n(n-1)/2 >= 3*n, so n>=7. For n=7 we can find a set by trial and error. Further, if such a set is doable for n=k then by adding a row of '1's and a column of '0's  :D it is doable for n=k+1. so thus it is doable iff n>=7.


geometry
analytic geometry
graphing lines
slope
ratio
similar triangles
AMC

Consider the $ 12$-sided polygon $ ABCDEFGHIJKL$, as shown. Each of its sides has length $ 4$, and each two consecutive sides form a right angle. Suppose that $ \overline{AG}$ and $ \overline{CH}$ meet at $ M$. What is the area of quadrilateral $ ABCM$?
[asy]unitsize(13mm);
defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=4;

pair A=(1,3), B=(2,3), C=(2,2), D=(3,2), Ep=(3,1), F=(2,1), G=(2,0), H=(1,0), I=(1,1), J=(0,1), K=(0,2), L=(1,2);
pair M=intersectionpoints(A--G,H--C)[0];

draw(A--B--C--D--Ep--F--G--H--I--J--K--L--cycle);
draw(A--G);
draw(H--C);
dot(M);

label("$A$",A,NW);
label("$B$",B,NE);
label("$C$",C,NE);
label("$D$",D,NE);
label("$E$",Ep,SE);
label("$F$",F,SE);
label("$G$",G,SE);
label("$H$",H,SW);
label("$I$",I,SW);
label("$J$",J,SW);
label("$K$",K,NW);
label("$L$",L,NW);

label("$M$",M,W);[/asy]$ \textbf{(A)}\ \frac {44}{3}\qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ \frac {88}{5}\qquad \textbf{(D)}\ 20 \qquad \textbf{(E)}\ \frac {62}{3}$

[quote="Xaero"][hide]$\triangle{AMH}\sim\triangle{GMC}\Longrightarrow\frac{CG}{AH}=\frac{2}{3}$.
Call the altitude to CG be $h_{1}$ and the altitude to $AH$ be $h_{2}$. Then $\frac{h_{1}}{h_{2}}=\frac{2}{3}$ or $h_{1}=\frac{2}{3}h_{2}$. Then, $h_{1}+h_{2}=\frac{5}{3}h_{2}=4\Longrightarrow h_{2}=\frac{12}{5}$ and $h_{1}=\frac{8}{5}$. Then the area of $ABGH$ is $48$, $[AMH]=\frac{72}{5}$, $[CMG]=\frac{32}{5}$, and $[HMG]=\frac{48}{5}$, Then, $[ABCM]=\frac{88}{5}=\boxed{C}$.[/hide][/quote]
[hide="Small revision"]You don't have to do HMG and CMG separatelly. It's easier to do them together; you only have to find one of the heights.[/hide]

[quote="Karth"]Or you could plot everything onto a coordinate plane and use shoelace.[/quote]
That's what I did. At the end you don't even need to use shoelace once you get the coordinates, you can subtract out areas and whatnot.

Are you sure?
...

You do have to account for that odd point somehow, and subtracting areas doesn't do it...

On the test I forgot about variables and ended up putting some stuff in a coordinate system and solving the two lines equations (simplified) to find the height of the point. It was much shorter than I expected (like less than 30 sec) which is amazing for a brute force method.

Actually, to account for that odd point, you'd have to draw the two lines which pass through it and solve for it. And if you're really lazy, you can just use shoelace. It guarantees an answer with a not-so-bad calculation method.

[hide="How I did it when taking the test."]Make H the origin, find the equations of the 2 lines, get their intersection and shoelace.[/hide]

Hey guys r u indians-- so am i

I also did it in corrdinate plane

Just find the slope of the two lines and use pick's therom if you drew the fiqure right

Here's how I did it.. since some of the other methods are seemingly vague.

[hide]I thought it was distinct; if you look closely there are a couple of triangles, especially conspicuous is the triangles which share the triangle MHG. So I used ( self formed triangles ) triangle AGH,triangle CHG, triangle AMH, triangle CFG, and the "given" MHG.

The altitude of the triangles I used are not given but can be obtained due to prior knowledge ( sides are length 4).

Altitude given to AMH will be x and altitude to CMG will be x-4.
Using the data given AGH has area of 24 and CHG has are of 16.

If you look closely you will notice that if you subtract the area of  AMH from AGH or CMG from CHG you will yield the area of MHG.

Anyway, I have to eat dinner in a couple, going to rush now.

24-6x=16-16+4x
24=10x
12/5=x
72/5 = area of AMH, 32/5 = area of CMG
Combine : 104/5
Given : Area of AHG and area of CHG, combine : 40
MHG accounted for twice in yielding.
Convert : 200/5 - 104/5 = 96/5
Divide by 2 since 2x MHG was accounted for.
Add : 104/5 + 48/5 = 152/5
Subtract : 240/5 - 152/5 = 88/5

Sorry if it's a bit vague towards the end, I'm rushing quite a bit. :x[/hide]

[quote="King_James23"]can anyone post the 2007 AMC 10 B test          i need it             :|[/quote]
No one who has the test is willing to type it up, and no one who has it has a scanner to send it to people who are willing to type it up, as far as I know.

[quote="me@home"][quote="white_horse_king88"]Fast and efficient: With similar triangles and area ratios: you can do this in your head.

[hide]$[ACH] = 24$ (Half the rectangle...)

$CM/HM = CG/AH = 2/3$

$\Rightarrow [ACM]/[AMH] = 2/3$, so $[ACM] = 48/5$

$[ABC] = 8$

$8+48/5 = 88/5$, our answer.[/hide][/quote]
How do you figure $CM/HM = CG/AH = 2/3$?
I don't see any quick and elegant ways without detours...[/quote]

Similar triangles

Wait, when I use coords, It is not working. Did you guys get that M was (12/5,24/5)?

Yes, if you set H as the origin, M is (12/5, 24/5)

Hmmm, for some reason, the shoelace theorem is not working for me...

make sure when you set up the coordinates, it goes in order clockwise


Let $ (x_{n})$ be a bounded sequence of real numbers, how do we show that if $ E$ is the collection of all limits of converging subsequences of $ (x_{n})$, then lim inf $ (x_{n})$ = min E and similarly with lim sup $ (x_{n}) \equal{} max E$?

A standard analysis technique: To show that $ A\equal{}B$, show that $ A\le B$ and $ A\ge B$. Try that with the $ \liminf$/$ \min$; the other side follows by mirroring everything.

so by definition of infimum, the greatest lower bound k such that every element in $ x_{n}$ is greater than or equal to k, can I just descriptively say that by definition of E, every element in E has to be greater than k and there are no elements in E less than k?

[quote="jmerry"]A standard analysis technique: To show that $ A \equal{} B$, show that $ A\le B$ and $ A\ge B$. [/quote]
Of course the second line of that idea is that most of the time, the way to show that $ A\le B$ is actually to show that $ A<B\plus{}\epsilon$ for any $ \epsilon >0.$


vector
geometry
geometric transformation
reflection
LaTeX

What is the vector eqn for the law of reflection ? With explanation

i will delete ur other posts...
the equation being
$ \vec u_{n} *(\vec k_{1} \minus{} \vec k_{2}) \equal{} 0$
where $ \vec u_{n}$ is the unit vector perpendicular to point of incidence
$ \vec k_{i}$
are the unit vectors along propagation direction of incidnet and reflected rays

This is Correct. Good one my boy
But i would like to have an equation like R(unit vector)=............

I would like an equation for r(unit vector) involving i(unit vector) and n(unit vector).
I donot want the cross product eqn

To make myself clear I WANT A dot product eqn and NOT A CROSS PRODUCT eqn :wink:

I Just found the answer
[hide] $ \vec r = \vec u - 2(\vec u.\vec n)\vec n$[/hide]
Can Anybody explain it

[color=blue]MOD EDIT:i $ \text{\LaTeX}$ ified it [/color]

i thnk that's ob enuf just use the fact that the light ray's horizontal component is same while verticcal componet is increased by twice the value from negative to positive
taht is if it were -x it became x


calculus
integration
function
trigonometry
calculus computations

Define $ f(x,\ y)$ be a function of $ {\mathbb{R}}^{2}$ and $ A$ be a set as follows.

$ f(x,\ y)\equal{}\left\{\begin{array}{ll}2xy\cos\left(\frac{y^{2}}{x}\right)\ x\neq 0,\ y\in\mathbb{R}&\quad\\ 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x\equal{}0,\ y\in\mathbb{R}&\quad\end{array}\right.$

$ A\equal{}\{(x,\ y)\in{\mathbb{R}}^{2}\ |\ 0\leq y\leq\pi ,\ y\leq x\leq\pi\}$.

Find $ \int\int_{A}f(x,\ y)\ dxdy$.


Solve the equation $ 6\cdot 9^x\equal{}5\cdot 25^y \plus{}1$ for positive integer values of $ x,y$.

If we put $ p\equal{}\frac{3^{x}\plus{}1}{2}, s\equal{}\frac{5^{y}\plus{}1}{2}$, we get: $ 6(p\minus{}1)p\equal{}5(s\minus{}1)s$. Does it help?

I can't see how your identity helps since we don't know the factorization of the numbers $ p,p\minus{}1,s,s\minus{}1$.


let the roots m,n,k of t^3 + at + b be rational.what can you say about the roots of mt^2 + nt + K ?

Since m+n+k = 0, the quadratic is $ mt^2\plus{}nt\minus{}(m\plus{}n)$

Obviously t=1 is a root of the equation which makes the other root also rational

edit: elaborated the solution


Are we allowed to discuss the questions from nationals this year?

They never said we couldn't, so I guess we could...

Yes, you may discuss 2008 Nationals problems.  (Discussion of Sprint #28 and Sprint #30 have been split into their own threads.)

is there a place on the internet where we can find them?

national problems can never be found on the internet.


geometry
3D geometry
AMC
AMC 10
Support
articles
AMC 8

since today's after November 25 I guess we can discuss about the test.

that was the problem with all the sibling right? i think i put either a or e.
i missed the last one.:(
i forgot to divide by 2.

actually.... i have no idea but im guessin around 21-23... oh well since it wuz mi first time takin the test i guess i did ok...

I got a 23...I missed the cubes one (grrrrr... stupid mistake), and the cat/mouse one (grrrrrrrrrrr...another stupid mistake)

I got 21.
Missed clock thing (forgot hour hand moved)
Mouse/cat thing (hate those)
And I don't remember the rest.

P.S. Childern one was easy the kid with the same hear colors aree brothers/sisters.

I dunno what i got... but i know i missed the one with the graph showing the friends... asking how many she wouldn't invite to her birthday... urg.

i got a 23 also missed # 14 with the two + two=four
i forgot that 1 was already used.

I got 24. :( Missed the birthday problem. I forgot to not include friends and friends of friends. Stupid! Stupid! Stupid! Stupid!

Yah! That was what I did... ugh...but i don't know my score yet... :(

Well our teacher gave us the answer keys today. If you want I could PM the answers to you.

its okay... i lost what i answered anyways... thanks though... that question was extremely tricky!!! UGHUGHUGH!

22...everyone says thats good for a 7th grader... but im not sure...

It is good. I got a 22 last year. :D However last year sum got 24 and 25 (3 7th graders in all)

amc 8 really isnt a test of how much you know (well for us that is), but more of how many silly mistakes can you make! if the price is right!

i no i could have done better... f*****g stupid mistakes!!! thats how i always lose points! d*m* it!

[quote="topper"]Does anyone know how the next level for kids our age works?  If we get a certain score do we get to take the AMC10?[/quote]

ANYONE below the cut-off age can take the AMC 10 on his own dime. There isn't any younger age limit for any of the AMC tests, and there isn't any requirement to have tested previously for the AMC 8, the AMC 10, or the AMC 12. But if you want your school to pay for your registration, you may have to deal with school rules. (If you want your parents to pay for your registration, you may have to deal with their rules  :)  .)

As a homeschooling dad, I sign up my son for tests when I think they have educational or challenge value for him. Then, after I've paid for a minimum registration (ten test-takers), I try to round up other local participants to make it more interesting. It's always disappointing to me that I can't get more people interested in AMC tests. One homeschooling mom I spoke to this noon says that she describes the AMC tests as like the initial tests that qualify people for the National Spellling Bee. That seems to be an analogy that is understandable to people who have never heard of math contests before. My eleven-year-old son will take the AMC 10 this season (25 February 2004) with her fifteen-year-old son, who aced the AMC 10 last year. The fifteen-year-old will be taking the AMC 12 on 10 February 2004, and the AMC 10 once more on 25 February 2004, and the mom is doing as I do, trying to round up other local people who might be interested. Anyone who wants to get to Minnesota on the 10 February 2004 AMC 12 date or the 25 February 2004 AMC 10 date is welcome to "crash" our administration--I doubt we will use up all ten slots for either test.

If you are too far away to travel to Minnesota, try to form your own registration group. You only have to follow your school's rules if you use your school's money. I have had good luck asking local public schools in my district to provide space and proctoring for the AMC 8 for the last two years. I ask for their help as a parent and taxpayer in the school district, and offer to the school the chance for their students to take the test on my dime with the extra slots I have left over in our administration after I contact my homeschooling support group for interested participants. Last year, none of the public school kids joined the administration. This year, most of our group were actually students from the school where we got our proctoring--I'm as curious as can be about how everyone scored, which I guess I'll know after the official score reports come out.

Don't give up too soon. Try to take any test you can afford, whenever you think you are ready for it.

Oh, to answer your question, yes, if you score well on the AMC 8, you will be INVITED to take the AMC 10, but you don't have to get the invitation to register for the AMC 10. You can take that test (and the AMC 12) as long as you are under the upper age limit for each test.

Hope this helps!

heehee...I'm the only gurl in my state to get a 23...I feel special...I'm the third highest gurl in IN...

ugh... i got a 23 also... but i don't know what other question i missed besides the birthday girl one... There's 4 people ahead of me... that means time to work hard and to blow them out in MC

[quote="notfunny"]. . . stupid mistakes!!! thats how i always lose points! . . . [/quote]

Has everyone seen what this site's founder has to say about [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php]avoiding stupid mistakes[/url]? The advice in his article makes good sense to me.

http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php

Hope this helps! Better luck next time.

yeah well how does that help when you can't even see the notches :-x grr

I missed the cube one (stupid corners) and the cat/mouse one...I hate those...

i keep on forgetting to divide by half for triangles :(
not a big improvement from last year though. :(

What is the cutoff for AMC10 invitation? If I got invited, when should I know? Is it on the AMC site?

Yes, you can take the AMC 10 no matter how you did on the AMC 8.  But if you do well on the AMC 8, then you get "invited" to take the AMC 10.

you do?
 hmm, maybe my math teacher hassn't given it to me,

Or have they not been sent out or somethinG?

ml2, I have an idea that my math teacher didn't want to bother arranging for me to do the next test, so nothing was said.  When I read on the AMC website that I could, it kind of bugged me.  Even though I'm know I won't do well enough on the AMC 10 to do much beyond it, I think it would be really good practice.

I'd be curious if you ask your teacher if they sent any more info along with the scores about those that placed high.

It is fair to think about the cost to the school.  I just wish if the info was sent that it could have been shared.


MY avatar isn't loading.?
It comes like debug..and when I try to upload from my comp it says no avvie .....selected..?

Well, it's most probably either it's not an accepted extension (.png, .gif, and .jpg are accepted, I recommend .png) or it's not an accepted size. There may be other possible causes of the error, but these are the two most common.

Can you post a screenshot or a quote of the error?


geometry
ratio
projective geometry
geometry proposed

Hello!

Let $ ABC$ be a triangle, $ A'\in\left(BC\right)$, $ A_{1},A_{2}\in AA'$, $ M\in\left(BA'\right)$ and $ N\in\left(A'C\right)$.
If $ \left\{B_{1}\right\} \equal{} MA_{1}\cap AB$, $ \left\{B_{2}\right\} \equal{} MA_{2}\cap AB$, $ \left\{C_{1}\right\} \equal{} NA_{1}\cap AC$ and
$ \left\{C_{2}\right\} \equal{} NA_{2}\cap AC$ then prove that the lines $ B_{1}C_{1}$, $ B_{2}C_{2}$, $ BC$ are concurrent. :)

$ (AB,B_1B_2)\equal{}M(AB,B_1B_2)\equal{}(AA',A_1A_2)\equal{}N(AA',A_1A_2)\equal{}(AC,C_1C_2)$
$ \Rightarrow BC,B_1C_1,B_2C_2$are concurrent.

Sorry but I don't know how to write projective geometry solution and my bad English..

By Desargues' Theorem ([url]http://mathworld.wolfram.com/DesarguesTheorem.html[/url]):

Let $ P : \equal{} B_1B_2\cap{}C_1C_2$. It's sufficient to prove that $ P$ lies on the line $ BC$ ...

It's easy to see that triangles $ \triangle{}A_1B_1C_1, \triangle{}A_2B_2C_2, \triangle{}A_3B_3C_3$ are all perspective from the point A, ie $ \{A, A_1, A_2\}, \{A, B_1, B_2\}, \{A, C_1, C_2\}$ are collinear. The theorem then implies that $ A_1B_1\cap{}A_2B_2 \equal{} M$, $ A_1C_1\cap{}A_2C_2 \equal{} N$ and $ B_1B_2\cap{}C_1C_2 \equal{} P$ are collinear. As $ MN \equiv BC$ we have that $ P$ lies on the line $ BC$ just as we wanted.

The notation of points was inviting to consider triangles of the form $ \triangle{}A_iB_iC_i$. The problem would be a little more trickier (and maybe less elegant) if the names were picked more randomly.

[quote="Mashimaru"]We will use the cross ratio:

$ (ABB_1B_2) \equal{} M(ABB_1B_2) \equal{} (AA_1A'A_2) \equal{} N(ACC_1C_2) \equal{} (ACC_1C_2)$

Thus, $ BC,B_1C_1,B_2C_2$ concurs and we are done.

P/S: A very nice problem.[/quote]

Can you please explain your notation (give a link, short explanation, book?). I've never seen such a solution before...


Using a $ 6\times6$ square array of dots as vertices of a concave polygon, what is the greatest number of sides the polygon can have?

Each side has $ 2$ points on the array as endpoints. Each endpoint has $ 2$ segments leading out. Thus, there should simply be $ 6\times6\equal{}\fbox{36}$ maximum segments(, I think...).

According to FTW, your answer is correct, but I don't completely understand.  Can you explain in more detail?


LaTeX
linear algebra
matrix
parameterization

Solve the following system of linear equations:
$ 2x_{1}\minus{}2x_{2}\minus{}3x_{3}\ \ \ \ \ \ \ \ \equal{}\minus{}2\\
3x_{1}\minus{}3x_{2}\minus{}2x_{3}\plus{}5x_{4}\equal{}7\\
x_{1}\minus{}\ \ x_{2}\minus{}2x_{3}\minus{}\ x_{4}\equal{}\minus{}3$

I arrive at a solution in which I solve for $ x_{3}$ and $ x_{4}$ (4 and 0 I think) and then $ x_{1}\equal{}a_{2}\plus{}5$ which seems correct.  Then in the back of the book, it says the answer is $ r(1,1,0,0)\plus{}s(\minus{}3,0,\minus{}2,1)\plus{}(5,0,4,0): r,s\in R$.  Please help.

Show us what you get for a row echelon form of the augmented matrix. That way, we can see what you've done.

[hide="LaTeX hint"]If you don't already know the syntax, try this:

\begin{bmatrix}1&2&3&4&5\\ 5&4&3&2&1\\ 0&1&2&37&2\end{bmatrix}

becomes

$ \begin{bmatrix}1&2&3&4&5\\ 5&4&3&2&1\\ 0&1&2&37&2\end{bmatrix}$

Or, if you'd prefer, using pmatrix instead of bmatrix:

$ \begin{pmatrix}1&2&3&4&5\\ 5&4&3&2&1\\ 0&1&2&37&2\end{pmatrix}.$[/hide]

Well after subtracting rows, I got down to
$ \begin{pmatrix}1&\minus{}1&\minus{}2&\minus{}1&\minus{}3\\
0&0&1&1&4\\
0&0&4&8&16
\end{pmatrix}$
Then I solved for $ x_{4}$ and got 0, and then that showed that $ x_{3}\equal{}4$.  But obviously, we can't get both $ x_{1}$ and $ x_{2}$ because there aren't enough equations.  But I can't figure out what the answer given means, or how to get there.

That's not in row echelon form yet. We need to do one more step: subtract $ 4$ times the second row from the third row.

$ \begin{pmatrix}1&\minus{}1&\minus{}2&\minus{}1&\minus{}3\\  0&0&1&1&4\\  0&0&0&4&0  \end{pmatrix}$

We have pivot elements in the columns for $ x_1,x_3,$ and $ x_4,$ but not in the column for $ x_2.$ We would get a one-parameter family of solutions.

However, that back-of-the-book answer you quoted in your first post has a two-parameter family of solutions, so we have a disagreement here. And I see the problem. Your intermediate step, as reported above, should be

$ \begin{pmatrix}1&\minus{}1&\minus{}2&\minus{}1&\minus{}3\\  0&0&1&2&4\\  0&0&4&8&16  \end{pmatrix}$

Now, one more step gets us to

$ \begin{pmatrix}1&\minus{}1&\minus{}2&\minus{}1&\minus{}3\\  0&0&1&2&4\\  0&0&0&0&0  \end{pmatrix}$

Now, this is in row echelon form. There are pivots in the first and third columns, but the second and fourth columns are free. So, let $ x_2\equal{}s$ and $ x_4\equal{}t.$ From the second row, $ x_3\plus{}2t\equal{}4$ so $ x_3\equal{}4\minus{}2t.$ The first row now reads

$ x_1\minus{}s\minus{}2(4\minus{}2t)\minus{}t\equal{}\minus{}3.$
$ x_1\equal{}5\plus{}s\minus{}3t.$

Consolidating,

$ \begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}\equal{}
\begin{bmatrix}5\plus{}s\minus{}3t\\s\\4\minus{}2t\\t\end{bmatrix}\equal{}
\begin{bmatrix}5\\0\\4\\9\end{bmatrix}\plus{}
s\begin{bmatrix}1\\1\\0\\0\end{bmatrix}\plus{}
t\begin{bmatrix}\minus{}3\\0\\\minus{}2\\1\end{bmatrix}.$

Ah, thank you very much.  I'm not familiar with some of those terms (pivot elements, row echelon) so I didn't really know what you meant before.  But thank you for the help.


:?: What will we see on the screen in double-slit experiment if I use two radiation have wavelength $\lambda _1$ and $\lambda _2$. I know there will be two systems of interference fringe, call I and II. But what if one light region of I is coincident one dark region of II ?
And then how many light regions do we see on the screen ?

Is this a very stupid question ?

Well, suppose we have a red and a yellow wavelength. The first will give a red interference pattern, the second a yellow one. If a maximum of the red occurs on the same spot as a minimum of the yellow one, we'll see a red fringe. If a maximum of the yellow occurs on the same spot as a minimum of the red one, we'll see a yellow fringe. In between, we'll see orange fringes, varying from more red to more yellow, depending of the local strengt of the red/yellow pattern.

Hmm, someone told me that when a maximum of the red occurs on the same spot as a minimum of the yellow one, they vanish each other ?

No, that can only happen if both waves have the same wavelength.

Unless you mean that the red light goes trough one slit and the yellow light through the other one  :?:


analytic geometry
geometry
3D geometry
sphere
Putnam
college contests

A circle moves so that it is continually in contact with all three coordinate planes of an ordinary rectangular system. Find the locus of the center of the circle.

the locus is just one eighth of a sphere with  its center at the origin and of radius R*(2)^(1/2).

I found it easily because the distance between center of the coin and the origin  never changes no matter how you move the coin.


AMC
AMC 10
AMC 12
AoPSwiki
search
function
geometry

I was wondering if any of u guys have it, i just want your review for it. Also, is this book appropriate for someone in 8th grade preparing to take AMC10 in 9th?

i have it. i guess its amc10 level, despite the title. it's definitely not olympiad level, so nothing to worry about

W8 what do u mean by "guess." Is the materials easier than AMC10 or harder?

it starts at amc10 and gets harder, but not much. it ends at amc12 level
also, its not purely a problems book, it teaches you concepts

It's not worth $ \$$50. The sections are too short to include enough details about each topic.
Get AoPS Vol. 1 and 2 instead (if you don't have it already).

If you want to work on AMC problems, try AoPSWiki and search for AMC problems.

[quote="Smartguy"]It's not worth $ \$$50. [/quote]

I echo this sentiment, it's not a bad book, for for your money's worth, AoPS books are better

oh, ok thx. So the AoPS books are good for AMC10 stuff too?

Yeah Volume 1 will teach you everything you need to know.  If your shaky in a subject you can get the intro book for that subject too.

I have the book and I am a 12th grader. I can say that it was definitely helpful for the amc10, while some of its chapters ( Complex numbers and functions) also helped me on the amc12.

Can I get the AoPS books from like barnes and nobles or some other book stores?

you can order online on this site- it's probably cheaper to buy directly from aops. click on bookstore and follow the links.

I don't know about your area, but I'm pretty sure AoPS isn't available at my Barnes and Nobles.


[quote="liyi"]Let $\pi$ denote a permutation of $\{1,\dots,n\}$.
Find \[\min_\pi \sum_{k=1}^{n}|2k-\pi_{k}|.\] I think the answer is $n(n+1)/2$, but I can't find a proof.[/quote]
$\sum_{k=1}^{n}|2k-\pi_{k}|.\geq |\sum_{k=1}^{n}(2k-\pi_{k})|.=n(n+1)/2$


integration
logarithms
real analysis
real analysis unsolved

Ascertain  $ \int\frac {1 \minus{} \ln x}{x^2 \plus{}\ln^2 x}\ \mathrm {dx}$ , where $ x\in R^*_ \plus{}\ .$

Use the substitution:

$ \frac{\ln{x}}{x}\equal{}t$

This gives you:

$ I\equal{}\int\frac{dt}{1\plus{}t^2}\equal{}\arctan{t}\equal{}\arctan{\frac{\ln{x}}{x}}$


$A, B \in M_n(C), AB = BA, c \in C$ is eigenvalue for A + B. Then exists $a, b \in C$ eigenvalues for A, B such that c = a + b.

You make the two matrices upper triangular in the same basis (it's possible, since they commute), and the conclusion is now obvious.


Suelyn counts up from 1 to 9, and then immediately counts down
again to 1, and then back up to 9, and so on, alternately counting
up and down
\[ (1, 2, 3,4,5,6,7,8,9,8,7,6,5,4,3,2,1,2,3,4,\ldots ).\] What is
the $ 1000^{\text{th}}$ integer in her list?

We see that the sequence \[ (1, 2, 3,4,5,6,7,8,9,8,7,6,5,4,3,2\ldots ).\] repeats every $ 16$ terms. So the remainder is $ 8$. Thus the number we want is $ 8$? I think it might be $ 7$ or $ 6$ though.....  :maybe:

1-9 and back to 2 is 16 numbers

starting at 1 begins the next set

1000 integers means there are 62 of these sets with a remainder of 8 numbers

you would then count up to 8 and that's your answer

:ddr: :rotfl:
[hide=Solution]
It repeats every 17
1000 is 14 mod 17
14th one is 4
So 4
:cursing:

[hide=Solution] We can treat this list as a sequence with a repetitive pattern. We see the sequence repeats itself every 16 elements (from 1 to 9 then back to 2). Because 1000 divided by 16 is 62 with a remainder of 8, to get 1000 terms in this list, we repeat the block 62 times, and then go 8 more elements. This means that the $1000^{\text{th}}$ integer is the same as the $8^{\text{th}}$ integer, which is $\boxed{8}$.[/hide]


Elrond is organizing a council in Rivendell. He has invited two men, two dwarves, and two elves, and plans to seat them along with himself in a circle with seven seats. He knows that dwarves and elves do not get along, so he plans to seat them so that no dwarf is sitting next to an elf. The circle has a head chair. If Elrond (himself an elf) must sit at the head of the circle, in how many ways can he seat the other six guests?

im guessing 24 or 36, im horrible at this type of stuff, algebra and geometry is much better for me

Actually 24 is correct. How'd you get 24?

Make a circle with 7 spots and designate one spot at the top to be the head elf. Now, there is a line of 6 people that can be opened up to look like this:

_ _ _ _ _ _

The two end spots cannot have a dwarf, because these will curve back up to sit next to the elf. Thus, there are 4 spots in the middle to be occupied by 2 dwarfs. However, there are some restrictions, because there can be a situation where the elves cannot be seated.

Can you work from here?

[quote=Bapun2012rta][hide]Since the men must separate dwarves from elves, going around clockwise, the seating arrangement must look like EEEMDDM, where E denotes an elf, M denotes a man, and D denotes a dwarf, and the first elf and last man are seated next to each other. There are $3!\cdot2!\cdot2!=6\cdot2\cdot2=24$ ways to arrange the elves, men and dwarves within such an arrangement. Elrond must be in the head chair, and can then take each arrangement and seat the other people accordingly by placing everyone in the circle clockwise in the order written. Since each of these lists produces a unique arrangement in a circle, there are $\boxed{24}$ ways to seat the six guests.[/hide][/quote]

Why the down vote? This problem had no solution, so he gave one!


AMC
AIME
geometry
USA(J)MO
USAMO
AMC 10
AMC 12

Okay. well I am an 8th grader.

I would like to know what you were do if you had a choice of AMC 10 or 12 in 8th grade.

What are the pros and cons of each?

I know you need about a 120 in amc 10 to make AIME and 100 in amc 12. but is amc 12 more difficult if you are just in geometry?

Since my next math tournament ( vestavia) is coming up, it will be over soon, and I can stop thinking about it. Right now , I just think of top 5 top 5 top 5 yeah..

I want to make it as far as I can this year with endless studying since I have free time each day, but I also want to be successful in my high school years...

Personally, I would take the AMC 12. Like you said, all you need is a 100 to make it to AIME. AMC 10 and 12 are almost like the same difficulty except the last 7 problems, which the AMC 12 is harder. Lucky me, I get to take both! :D

yeah..

well..

I can take both next year.

I just want to make AIME this year and do well on it?

maybe...

anyway i think if I don't make careless mistakes AMC 10 is better I suppose... after asking people

[quote="13375P34K43V312"]Just take both.[/quote]

Well what if you can't?

[quote="goldendomer"][quote="now a ranger"]i think i will just ask my teacher to take the amc 10a and 10b i suppose[/quote]

You can't take the AMC 10 twice in the same year. :huh:[/quote]

Yes you can.

You can take one A test and one B test.

I would recommend taking the 10a and the 12b, if possible, unless you don't qualify using the 10a.

[quote="SplashD"][quote="bpms"][quote="SplashD"]I think what's he's trying to say is that what if your school only offers one testing day and you're in eighth grade, which one should you take?[/quote]
Well couldn't you go to a high school to take the other?[/quote]

Are you allowed to take the A test at one school and the B test at another?

furthermore, say you qualified for the AIME but your school only offers the first AIME date and you're sick that day, are you allowed to take the alternate AIME at another school?[/quote]
I have no clue, but I would think the first would be allowed. As for the second, I got no idea.

[quote="jmerry"][quote="now a ranger"]is amc 12 more difficult if you are just in geometry?[/quote]
Yes. There are a significant number of problems on the AMC-12 which are based on algebra that you haven't had yet. Once you have that course (next year?), you'll be better off with the AMC-12. For now, the AMC-10 is a better idea if you're only taking one.[/quote]

Oh I see.

Thanks

I will just take the AMC 10, and 12b also if I am able to take two.

Well, it depends on how much was taught in your algebra classes already, and how much you've learned by doing stuff on your own too.

I think the best way to check is just by looking at some of the past AMC 12 and AMC 10 tests.

Is it easier to qualify for the USAMO with the AMC 10 or AMC 12?(as an eighth grader)

yeah that's the question I was contemplating about  :)

[quote="h_s_potter2002"]I would say to take whichever one you feel more comfortable in qualifying for the AIME for.

The AMC says though that they recommend that people that have a choice between the AMC 10 and AMC 12, and want to qualify for the USAMO, take the AMC 12.[/quote]

I thought that didn't apply to middle schoolers? and they can qualify for USAMO regardless of AMC score if they get an 8 or higher on the AIME?

It [i]doesn't[/i] apply to middle schoolers (or any one in 10th grade or below).

The AMC just [i]recommends[/i] that those who want to qualify for the USAMO take the AMC 12.

You know what I think: go and try some AMC 12's on this website and see what you think.

You may not have learned some of the more advanced concepts of Algebra 2 and beyond, although most of the AMC 12 problems don't require specific knowledge in those areas. I think the key is whether you're able to solve the somewhat harder problems on the 12, so just go and see what it's like  :wink:

[quote="Boy Soprano II"][quote="SplashD"]and they can qualify for USAMO regardless of AMC score if they get an 8 or higher on the AIME?[/quote]
The AIME floor score isn't necessarily 8.  We could have a super-easy AIME in which everybody gets 8s and the floor is maybe 12.  Or we could have a super-hard AIME in which everybody who gets a question right qualifies for the USMO.  But that doesn't happen; since the existence of the floor happened, it's always been between 7 and 9, inclusive, if I remember aright, and 8 is about normal.  But yes, each year, $\exists n \in \mathbb{Z}\cap [0,15] | \{ x | x \mbox{ is a person in grade 10 or lower}\}\cap \{ x | x \mbox{'s AIME score}\ge n \}\subseteq \{ \mbox{USMO qualifiers}\}.$[/quote]

What is the USMO? The Usless Math Olympiad (I guess it would be pretty altruistic...)? If you meant USAMO, then yes, n = 15 would generally work.

[quote="Ubemaya"]You can't qualify for the USAMO in AMC10 or AMC12.
I think.
I'm pretty sure it's only for AIME.

For making it to AIME, I'm not sure which one is better. Last year AMC12 was probably better because you only needed to answer 14 right and leave the rest blank.[/quote]

And the AIME qualifies for USAMO.
Also, usually the AMC12 score is added to the AIME score (x10) to get the index to qualify for the USAMO.  This apparently didn't happen last year though in the end (because people complained that the AMC10 had the same qualifying index I think.)

If you are a 9th grader and you take the AMC 12, can you still make USAMO with the floor score, or is that just for people taking the AMC 10?

When people say that you have a higher chance of making the AIME from AMC 12 because you have to answer less questions, you forget that the AMC 10 questions are easier than the AMC 12 questions. You only have to answer 3 or 4 extra questions on the AMC 10, and they are easier. The reason to take the AMC 12 in 9th or 10th grade is more for a challenge, I would think.

[quote="SplashD"][quote="Boy Soprano II"][quote="SplashD"]and they can qualify for USAMO regardless of AMC score if they get an 8 or higher on the AIME?[/quote]
The AIME floor score isn't necessarily 8.  We could have a super-easy AIME in which everybody gets 8s and the floor is maybe 12.  Or we could have a super-hard AIME in which everybody who gets a question right qualifies for the USMO.  But that doesn't happen; since the existence of the floor happened, it's always been between 7 and 9, inclusive, if I remember aright, and 8 is about normal.  But yes, each year, $\exists n \in \mathbb{Z}\cap [0,15] | \{ x | x \mbox{ is a person in grade 10 or lower}\}\cap \{ x | x \mbox{'s AIME score}\ge n \}\subseteq \{ \mbox{USMO qualifiers}\}.$[/quote]

I just want this to be super clear.

Say the floor was 8.

Person A: AMC 10: 150 AIME: 7
Person B: AMC 10: 120 AIME: 8

So in this case, person B would qualify and person A wouldn't even though person A has a higher index?[/quote]
Well it depends on what the index is.
My question is that last year it said that out of the people who made the floor score of 8, those with 212 or above qualified for USAMO. So if someone is a 10th grader and gets a 101 on the 12 and an 11 on the AIME would they qualify?

At first, it was that you had to get at least a 212 USAMO index, and a floor score of 8 to qualify (10th graders and below).  But they changed it afterwards, to make it so that any 10th grader or with a AIME score of at least 8 made it.

So if someone got a score of 101 on the AMC 12, and a 11 on the AIME they wouldn't have made it at first but later would have.

At least, that's what I think would have happened.  The AMC Director said somewhere though that something like this didn't occur last year, so I dunno how they would have handled it.

$\mbox{U.S. Math Olympiad}= \mbox{USMO}\equiv \mbox{USAMO}\pmod{A}$

$\mbox{USMO}\neq \mbox{USEMO}= \mbox{Useless}$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=81430]Some pages[/url] [url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2006-ua/06usamo-qualstate.shtml]regarding last year's[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=82163]qualification details[/url]

I was wondering, will you get your AMC A score before the AMC B test day?

[quote="calc rulz"]To eliminate confusion, this is what happened last year:

At first they selected people who took the AMC 12 by the index as usual, and then selected 10th graders and below by the floor score, which was 8. However, this selected too many people, so they made a requirement of 212 or higher index (which is defined using either AMC 12 or AMC 10) for 10th graders or below to qualify by the floor score.

The problem here was that this system favored people who took the AMC 10 greatly; those who took the AMC 10 and made the floor score got much higher indecies simply because they took the easier test, while those who took the AMC 12 tended to have lower scores and thus were less likely to qualify. After realizing that the AMC 10 takers were unfairly favored, the CAMC decided to qualify all 10th graders or below who got at least the floor score on the AIME, adding an extra 57 students.

Example:
Student (10th or under) takes 12: 122 AMC 12, 8 AIME - 202 index, they don't qualify
Student takes 10: They should easily get 132+ if they take AMC 10, which qualifies them with their 8 on the AIME.

Now, some people may argue that the AMC 12 takers were favored, but this is not so. People argued that the AMC 12 takers had the advantage of being able to qualify by the index cutoff as well as by the floor score, as had been a benefit in previous years. However, this only benefits a person in the rare case that they would have set the floor themself had they taken the AMC 12, meaning that overall AMC 10 takers were much favored. Another arguement was that it's easier for those who take the AMC 12 to qualify for the AIME, which is of course necessary if one wants to qualify for USAMO - but this arguement is completely silly as someone who is looking to qualify for the USAMO most likely has no concern about whether they'll qualify for the AIME.

Example:

Say the first cutoff index is 217, but although it's possible for someone to make this with a 7 on the AIME, nobody does, so the floor is an 8.

Now let's say we have a 10th grader who can get a perfect score on either the AMC 10 or the AMC 12.

If he takes the AMC 10: He gets 150 on the AMC 10 and 7 on the AIME, but does not qualify because his AIME score is below the floor, which is 8.

If he takes the AMC 12: He gets 150 on the AMC 12 and 7 on the AIME, which gives him a qualifying index of 220 (and here the floor becomes a 7 because he set the floor).[/quote]

 I would think the best way to avoid all of this is to get a score of 9+ at least to avoid all this confusion and have to hope that they let you qualify. On the AIME, that is.

Yeah, they should change the amc 10 and 12 ways.

Please correct me if I am wrong here.

An 8th grader (Geometry) needs 120 on AMC 10 - 19 correct and the rest blank
100 on AMC 12 - 14 correct and the rest blank.

About 12 problems on AMC 10 and 12 overlap, so one needs to get 2 more harder problems on AMC 12 as opposed to 7 more easier problems on AMC 10.

There are only few strictly Algebra II problems on AMC 12 (like logs and stuff).

So I cannot tell which one gives better chances to a Geometry student, probably more or less the same.

But doing them both (one on date A and another on date B) probably doubles the chances of qualifying, especially if a lot of mistakes are careless. Right?

Anyway, a student can take a date B test in a college or University even if that college is not in his own state.

So AMC 10 in your own school on date A, 12 in a college on date B.

Any possible advantages of taking AMC 12 first and AMC 10 second?


function
trigonometry
inequalities unsolved
inequalities

[quote="chien than"]Let $x$ is real numbers satisfying $0 \leq x \leq 1$.Prove that:
$x(9\sqrt{1+x^{2}}+13\sqrt{1-x^{2}}) \leq 16$[/quote]

Just find the maximum value of the continous function,
${f(x)=x(9\sqrt{1+x^{2}}}+12\sqrt{1-x^{2}})$
On the interval.

does anybody have a pretty solution for this problem?

it seems like a trig substitution or something like that should help...


Hi,

I was thinking about taking Intermediate Trig/Complex Numbers upcoming fall but I realized that I need to get higher AMC-12 score to ensure my qualification for AIME (just in case). I've never taken problem series other than AIME seminar last year (which is AIME Problem Series B this year I believe) and I'm not sure what AMC-12 Problem Series does. Is this like a class with bunch of AMC-12 level problems and maybe a mock AMC-12 at the end? Will it help raising your AMC-12 score (say 110 to around 125-130)? If it can help me raise my score, I think it's a lot better to take AMC-12 Problem Series than Trig/Complex Numbers but I just want to know what AMC-12 Problem Series does.

Thanks a lot! :)

The AIME Problem Series B is not the same as the AIME Problem Seminar from last year.  It includes much of the class text (which was a total of 5 hours of class), but much of it is entirely new and is 18 hours of total class.  Essentially, it's the same as the AIME Problem Series A, but with different problems.  What you read in the classes descriptions was a mistake we left behind when editing that page last week.  Thanks for bringing it to our attention.

I think the AMC 12 or AIME Problem Series B would both be good problem series for you.  If you want to be absolutely certain you make the AIME, the AMC 12 is probably best.

I would recommend that you wait until you know you can get to the AIME and can easily pass the Intermediate Algebra Posttest before enrolling in the Intermediate Trig/Complex Numbers class as it is the hardest of our Intermediate classes.

The AMC 12 does not include a mock test.  It includes 12 90-minute classes which cover fairly thoroughly the areas of problem solving and mathematics covered on the AMC 12.  There is a message board where scores of practice problems will be discussed, and (as with all our classes nowadays) includes access to some open chat times in the virtual classroom.

Oh okay.

[quote]The AMC 12 does not include a mock test.[/quote]

 :( Oh man.

I think I'll choose to take AMC-12 Problem Series and AIME since they will help me more than taking Intermediate Trig/Complex Numbers (like you said, this is not an easy class.. a hard class actually).

Thank you for fast response!


calculus
ratio
LaTeX
arithmetic sequence
arithmetic series

I thought this up because I was bored.
If a1, a2, a3, ... form an increasing arithmetic sequence, prove that the sum of the recriprocals of a1, a2, a3, ... is divergent.

[i]Title edited by moderator and promoted to more appropriate level.[/i]

This is probably not within the scope of getting started.  At least they need to know what divergent means.

Sorry, I haven't been here in a while.  Last time I was here I think i recall fierytycoon discussing a problem with some calculus proof he had.

Divergent means tends to infinity (goes on forever), as opposed to a sequence like 1 + 1/2 + 1/4 + 1/8 + ... which tends to two.

OK now that this has been moved to Intermediate, I will give a hint

[hide]1+1/2+1/3+... is divergent[/hide]

I was going to give that hint, but I thought it would be too much of a giveaway then. :D

[hide]if 1+1/2+1/3+... is divirgent then that basicly equals the problem correct?  So you would know that it is divirgent if 1+1/2+1/3+... is.

I think I messed this up, but if someone would point me in the right direction...[/hide]

Please show every step, because otherwise we won't know how you did something (like you can't just arbitrarily say that 7^7 :equiv: 1 (mod 6) to a beginner).  It looks like you're on the right track, but you made a few typos, so it's a little hard to understand.

[hide] 1+1/2+1/3+... is divirgent we know this because... oh drat now I'm stuck. [/hide]

First assume that my hint is correct.  Don't bother proving it yet.  Can you do it then?  If you can, please show all your steps so we don't have to make all the jumps ourselves when trying to understand what you're doing.

I know the following proof is bad since it treats infinite as a number, but

[hide]

Assume that 1+1/2+1/3+1/4... is divergent, so

1+1/2+1/3+1/4...=  :inf: .

Note that if there is another arithmetic series that starts with x and has common ratio x, then we can find the sum of the recprocals by dividing both sides of the equation above by x.

1/x + 1/(2x) + 1/(3x)... =  :inf: /x.

Since no real numbers can be multiplied to get :inf:, the right hand side of the equation stays the same.

1/x + 1/(2x) + 1/(3x)...= :inf:,

so it diverges.[/hide]

I'll need to improve this somehow...

Also note a very minor flaw.  What if we had 1, 3, 5, 7...?

Not sure if this is really a good correction of that flaw, but I have to go so I'm just typing it quickly:

[hide]

We already proved that 1/2+1/4+1/6... is divergent.  Since 1/1>1/2, 1/3>1/4... 1/(xth odd number in the odd number series} > 1/(xth even number in the even number series),  1+1/3+1/5... is slightly greater than 1/2+1/4+1/6... which diverges, so 1+1/3+1/5+1/7... diverges.

[/hide]

OK.  nr1337 basically has it.  I think this is what he's trying to say:

[hide]Let an infinite arithmetic sequence go a, a+d, a+2d, a+3d, ....

Assume WLOG |a| :le: |d|, or add a finite number of terms (not affecting divergence) a-d, a-2d, ... in the beginning and substitute a-nd = a' such that this is true.

Taking the reciprocals, we have:

1/a + 1/(a+d) + 1/(a+2d) + ...

We compare this to 1/d + 1/2d + 1/3d + 1/4d + ...

by matching 1/(a+kd) =? 1/((k+1)d).  Multiplying through, we see that this equates to

kd + d =? a + kd, simplifying to d =? a

We already know by our assumption that d > a.

Therefore, LHS > RHS, and so we need only to show that 1/d + 1/2d + 1/3d + 1/4d + ... is divergent.

Since this is equal to a divergent sequence divided by a finite value d, the result is still infinite.[/hide]

Someone else can explain better.

Yep, that's what I'm trying to say.  How would I prove that 1+1/2+1/3+1/4+1/5... is divergent.  Here is what I've done so far:

[hide]

We can do this by grouping terms together:

1>=1/2

1/2 >= 1/2

1/3 + 1/4 >= 1/2.

In general, 1/x + 1/(x+1) ... + 1/(2x-2) > 1/2.

If I can prove this, then I can turn the original series into 1/2+1/2+1/2... The problem is, I can't prove that for all positive integers x, 1/x + 1/(x+1) ... + 1/(2x-2) > 1/2.

[/hide]

nr1337 wrote:[hide]

In general, 1/x + 1/(x+1) ... + 1/(2x-2) > 1/2.

[/hide]

[hide]There are x-1 terms.  Each term is greater than ____.  Also, I think AM-HM works, but there's no point in doing that if there's an easier way.[/hide]

probability1.01 wrote:[hide]There are x-1 terms.  Each term is greater than ____.  Also, I think AM-HM works, but there's no point in doing that if there's an easier way.[/hide]

Each term is greater than [hide]0[/hide].  I never thought of AM-HM.  That seems easy enough to me. I'm going to try it.

I figured it out.  I'm LaTeX-ing my proof and I am going to post it online.
...
I finished.  Here is a complete proof: home.comcast.net/~carlosrd/ArithmeticSequence.pdf .


function
complex analysis
algebra
polynomial
complex numbers

Hi all, I am just studying for my final exam Wednesday and am stuck on a few problems. Hopefully you guys can help me out!

1) Does there exist a function f(z) holomorphic on C \ {0} st. |f(z)| is greater then or equal to 1 / sqrt(|z|) for all z in C \ {0}?

I'll probably have more questions later so I'll edit this as I go along. Thanks for any advice you can offer.

$ g \equal{} 1/f$ is entire and $ |g(z)| \leq \sqrt {|z|}$. Far out, we have $ \sqrt {|z|} \leq |z|$, so it follows by Cauchy estimates or maybe Liouville that $ g$ is a first degree polynomial, which is not possible.

Ah yes thanks very much. Here's another question along the same vein.

Let f denote a holomorphic function for |z| < R such that |f(z)| <= M. Suppose f(z_0) = w_0 with |z_0| < R
Show that  | M*(f(z)-w_0) / (M^2 - w_0'*f(z))| <= |R*(z-z_0) / (R^2 - z_0'*z)|

where w_0' and z_0' denotes the complex conjugate

It's schwarz lemma applied to $ S \circ f \circ T$. Where $ S$ and $ T$ are two appropriate Möbius transforms.


we have three numbers a1,a2,a3 >0 and 0<a1*a2*a3<=1 we want to proove that a1/a2+a2/a3+a3/a1 >= a1+a2+a3

[color=blue][b]Problem.[/b] Let a, b, c be three positive reals such that $abc\leq 1$. Prove the inequality $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq a+b+c$.[/color]

[i]Solution.[/i] By the AM-GM inequality, we have $\frac{a}{b}+\frac{a}{b}+\frac{b}{c}\geq 3\sqrt[3]{\frac{a}{b}\cdot\frac{a}{b}\cdot\frac{b}{c}}=3\sqrt[3]{\frac{a^2}{bc}}$. Now, since $abc\leq 1$, we have $a^2\geq a^2\cdot abc$, and thus

$\frac{a}{b}+\frac{a}{b}+\frac{b}{c}\geq 3\sqrt[3]{\frac{a^2}{bc}}\geq 3\sqrt[3]{\frac{a^2\cdot abc}{bc}}=3a$.

Similarly,

$\frac{b}{c}+\frac{b}{c}+\frac{c}{a}\geq 3b$;
$\frac{c}{a}+\frac{c}{a}+\frac{a}{b}\geq 3c$.

Adding these three inequalities, we get

$\left(\frac{a}{b}+\frac{a}{b}+\frac{b}{c}\right)+\left(\frac{b}{c}+\frac{b}{c}+\frac{c}{a}\right)+\left(\frac{c}{a}+\frac{c}{a}+\frac{a}{b}\right)\geq 3a+3b+3c$
$\Longleftrightarrow\ \ \ \ \ 3\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\geq 3\left(a+b+c\right)$
$\Longleftrightarrow\ \ \ \ \ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq a+b+c$,

and the problem is solved.

  darij

I remember that  this problem was given to me by manuel on yahoo messanger. Anyway, here is my solution:
By Cauchy-Schwarz, if $ab+bc+ca\leq a+b+c$, we have:

                                  \[ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}=\frac{a^{2}c+b^{2}a+cb^{2}}{abc}\geq{(a+b+c)^{2}}{abc(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}\geq a+b+c.  \]

 If $ab+bc+ca\geq a+b+c$ we will do it like above. ;)


AMC
AIME
geometry
rectangle
analytic geometry
HMMT
quadratics

Marathons are generally successes, and since the last AMC Marathon died, I made a new one.

[b]Rules:[/b]

1. Answer the previous question [i]with a full solution[/i]. You don't have to put every trivial step, but make sure you don't just write the answer.

2. If you use a problem from an actual contest(note that it doesn't have to be an actual AMC problem, just something around the same level), state the contest and year.

3. After answering a problem, post one of your own! If you can't think of one, take one from another contest(see #2) and post that, or better yet, modify it a little bit.

4. The difficulty level can range from AMC 8 to USAMO(meaning you can post proofs.)

5. Please post what problem number it is.

[size=150]Problem 1[/size]

An arithmetic progression $ a_1, a_2, a_3, a_4, \dots, a_{1296}$ with common difference one has a sum of $ 893592$. Find the product of the first and last term.

$ a_{n}\equal{}a_{1}\plus{}(n\minus{}1)d$ for a generic arithmetic series.  In our case, this is $ a_{1296}\equal{}a_{1}\plus{}1295$.  The sum of the first $ n$ terms of an arithmetic series is
$ S_{n}\equal{}\frac{n(a_{1}\plus{}a_{n})}{2}$.
Set this to 893592:
$ 1787184\equal{}1296(a_{1}\plus{}a_{1296})\equal{}1296(2a_{1}\plus{}1295)\Rightarrow 1379\equal{}2a_{1}\plus{}1295\Rightarrow a_{1}\equal{}42$, $ a_{1296}\equal{}1337$.
Then the product is $ 56154$.

For my problem, see [url=http://math.mit.edu/~rstan/myputnam.pdf]12*[/url]

Before we go much further, I'd like to say that the reason the other one died is because the problems were TOO HARD.  If we keep the problems at a respectable level (like no harder than hard AMC 12 - EASY AIME) this will be a success, but if it's any harder it'll die just like the last one.

If you draw the line parallel to the side of length $ b$ thru the center you see it cuts the rectangle into two congruent pieces. So the height of the triangle is $ 1\minus{}\frac{h}{2}$. Set that equal to $ 2h$ to get $ h\equal{}2/5$.

Please post a new problem after solving!

[size=150]Problem 3[/size]

Find the area of the larger region(i.e., larger section of circle), given that the radius of the circle is 5 inches and the angle measure of the cut section is $ 60^\circ$.

 [geogebra]e65edff4a030505ff486604bb9e684f539ff8004[/geogebra]

[quote="AIME15"]You took that from the AIME Problem Series B Class.

An engineer said he could finish a highway section in $ 3$ days with his present supply of a certain type of machine.  However, with $ 3$ more of these machines the job could be done in $ 2$ days.  If the machines all work at the same rate, how many days would it take to do the job with one machine?[/quote]
Where $ x$ is the number of original machines and $ y$ is their rate:
$ 3\dfrac{x}{y} \equal{} 1$
$ 2\dfrac{x \plus{} 3}{y} \equal{} 1$
Thus:
$ 3x \equal{} y$ and $ 2x \plus{} 6 \equal{} y$
so
$ x \equal{} 6$ and $ y \equal{} 18$
When $ x \equal{} 1$
$ \dfrac{1t}{18} \equal{} 1$
$ t \equal{} 18$

Square $ ABCD$ has sides of length $ 4$ and $ M$ is the midpoint of $ \overline {CD}$. A circle with radius $ 2$ and center $ M$ intersects a circle with radius $ 4$ and center $ A$ at points $ P$ and $ D$. What is the distance from $ P$ to $ \overline {AD}$?

[img]http://www.artofproblemsolving.com/Wiki/images/6/61/5d50417537c6cddfb70810403c62787b889cdcb1.png[/img]
AMC 12 2003

Two perpendicular chords intersect in a circle.  The segments of one chord are $ 3$ and $ 4$; the segments of the other are $ 6$ and $ 2$.  What is the diameter of the circle?  Express your answer in simplest radical form.

Source: a 1957 test, won't say exactly which test :wink:

[quote="crazypianist1116"][quote="AIME15"]You took that from the AIME Problem Series B Class.

An engineer said he could finish a highway section in $ 3$ days with his present supply of a certain type of machine.  However, with $ 3$ more of these machines the job could be done in $ 2$ days.  If the machines all work at the same rate, how many days would it take to do the job with one machine?[/quote]
Where $ x$ is the number of original machines and $ y$ is their rate:
$ 3\dfrac{x}{y} \equal{} 1$
$ 2\dfrac{x \plus{} 3}{y} \equal{} 1$
Thus:
$ 3x \equal{} y$ and $ 2x \plus{} 6 \equal{} y$
so
$ x \equal{} 6$ and $ y \equal{} 18$
When $ x \equal{} 1$
$ \dfrac{1t}{18} \equal{} 1$
$ t \equal{} 18$[/quote]
Yea obvi it was from the class.
let the number of current machines be m and their rate by r. Then because d = rt
1 = 3 * r * m , and 1 = 2 * r * (m + 1). Dividing we have 3m = 2(m+3) or m = 6. Then 18r = 1 and r = 1/18. So one machine does 1/18 of the job a day and it would take 18 days.
Crazy's problem : Let A be the origin. Easy coordinate bash yields $ 3\sqrt{23}/4$ I may have messed up the calculations.

New problem: Suppose I have a bag with 8 slips of paper in it. Six of these have  a 1 and the other 2 have a 3 on them.
a) What is the expected value of the number shown when I draw a single slip of paper
b)How many threes do i have to add to make the expected value equal to 2? great than 2.5?

Expected value is $ \frac{6(1)\plus{}2(3)}{8}\equal{}\frac{3}{2}$.

We've had a bit of confusion, so this will be:

[size=150]Problem 6[/size]

Positive real numbers $ x, y$ satisfy the equations $ x^2 \plus{} y^2 \equal{} 1$ and $ x^4 \plus{} y^4 \equal{} \frac{17}{18}$. Find $ xy$.

HMMT, 2008

how about a soln to your problem. Issabella2296 posted this one a little bit ago anyways  :wink:

Eh, sorry, I'm not sure what you mean by that.

For the last posted problem -

Squaring the first equation substituting in $ \frac{17}{18}$ for $ x^4 \plus{} y^4$ yields $ 2x^2y^2 \equal{} \frac{1}{18}$

$ xy \equal{} \frac{1}{6}$ since $ x$ and $ y$ are positive.

The problem involving $ 2008{}^{2007}{}^{2006{^{.....}^{1}}}}}}$ has already been solved, back on page 4.

my bad  :blush:  :P

[hide="Solution?"]So by pigeonhole? there has to be at least $ 13 \choose 3$ triangles in the color with the most points, $ \equal{} \frac {13\cdot12\cdot11}{6} > 130$
[/hide]

I knew I should have bolded [b]scalene[/b] (the ones not isosceles or equilateral).

[quote="xpmath"][hide="solution?"]Fix one side of the octahedron. Then there are 7! ways to color the others, but there are three possible rotations, so $ \frac {5040}{3} \equal{} 1680$.[/hide]

Let $ S$ be the set of ordered triples $ (x,y,z)$ of real numbers for which
\[ \log_{10} (x \plus{} y) \equal{} z\text{ and }\log_{10} (x^2 \plus{} y^2) \equal{} z \plus{} 1.
\]
There are real numbers $ a$ and $ b$ such that for all ordered triples $ (x,y,z)$ in $ S$ we have $ x^3 \plus{} y^3 \equal{} a \cdot 10^{3z} \plus{} b \cdot 10^{2z}$. What is the value of $ a \plus{} b$?[/quote]

Why do I not get Simo's answer for this problem?

I think theres a mistake in Simo's solution, as he did x^3+y^3=(x+y)^3+3xy(x+y), instead of =(x+y)^3-3xy(x+y)

Hm, yeah, apparently that's not the right answer. Try again.

[quote="mathemonster"]$ 25\pi * 5/6 \equal{} 125\pi/6$ Seriously?
Ok a> b> c> d
$ a \plus{} b \plus{} c \plus{} d \equal{} 2048,$ and $ a^2 \minus{} b^2 \plus{} c^2 \minus{} d^2 \equal{} 2048$ What is the minimum value of a?[/quote]
Are $ a,b,c,d$ supposed to be real numbers, or what?

$ a^2 \minus{} b^2 \plus{} c^2 \minus{} d^2 \equal{} 2048 \equal{} (a\plus{}b)(a\minus{}b) \plus{} (c\plus{}d)(c\minus{}d)$
$ (a\plus{}b)(a\minus{}b\minus{}1) \plus{} (c\plus{}d)(c\minus{}d\minus{}1) \equal{} 0$

Obviously we need $ a \equal{} b\plus{}1$ and $ c \equal{} d\plus{}1$. If we restrict ourselves to integer $ a,b,c,d$ (seems the likely choice, especially given $ a>b>c>d$) then we have that $ a\equal{}b\plus{}1\equal{}c\plus{}2\equal{}d\plus{}3$ so $ 4a\minus{}6\equal{}2048$ and $ a\equal{}513.5$ but this is not an integer. We therefore must move to $ a\equal{}b\plus{}1\equal{}c\plus{}3\equal{}d\plus{}4$ to achieve an integer value, which gives $ 4a\minus{}8\equal{}2048$ or $ a\equal{}514$.

Working on fixing simo's solution, I'll post that when done.

[hide="Stuck..."]
By the Pigeonhole Principle, there must always be at least 13 points of one color.
So we need to prove that there is at least 130 scalene triangles among any 13 points.[/hide]

Let's try to keep this moving. Bugi, can you please provide a hint?

[quote="zserf"][quote="Bugi"]In order to this not getting locked:

There are 50 points in a plane, colored in 4 different colors. Prove that there are at least 130 scalene triangles with their vertices in one color[/quote]

You can't prove this, because it isn't always true. Put all 50 points in a line, it's clear that you can't have 130 triangles of any type, much less 130 scalene triangles with restrictions.

Next problem:

The surface area of 3 sides of a rectangular prism are in ratio 3:4:6 and the entire prism has a surface area of 208. What is the area of the prism?[/quote]

I believe that he meant "scalene possibly-degenerate triangles."

Oh sorry, I meant non-collinear points.
[hide="Hint to mine"]
First we have 50 points, 4 colors. That means we have at least ___ points in one color
Second, how many isosceles triangles can we make from 1 line and the vertexes in one color?
Third, how many triangles can we total make with ___ points in one color?
Fourth, subtract total and isosceles
[/hide]

[hide="Solution to zserfs"]
We easily find that surface areas of sides are 48,64,96.
ab=48,ac=64,bc=96, so $ V\equal{}abc\equal{}376\sqrt2$[/hide]

[hide="Solution"] Let the surface areas be $ 3x$, $ 4x$, and $ 6x$ respectively. Since there are three pairs of congruent sides, we get $ 2(3x\plus{}4x\plus{}6x)\equal{}208\Rightarrow13x\equal{}104\Rightarrow x\equal{}8$ So the surface areas are 24, 32 and 48 respectively. Multiplying these together and taking the square root, we get $ 192$.[/hide]

I don't know which ones have been posted, but I like this one :)

A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of lengths $ 3$, $ 4$, and $ 5$. What is the area of the triangle?


Including the line [b]import geometry_dev;[/b] results in an error and omitting this line results in geometry_dev routines being not recognized. As far as I understand, the package is supposed to be possible to use on the site. Can somebody look into the problem?

Hello Valentin, this still does not work.

Without geometry_dev:
[asy]import olympiad; size(100); defaultpen(linewidth(0.8));
draw(unitcircle);[/asy]

With geometry_dev:
[asy]import olympiad; import geometry_dev; size(100); defaultpen(linewidth(0.8));
draw(unitcircle);[/asy]

Thankss!


LaTeX
inequalities unsolved
inequalities

[quote="stancioiu sorin"]Can we prove that ((a+b)/(b+c))(a/(b+c+2a))=<1/2??? :blush:[/quote]

Use LaTeX!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
For some LaTeX help please visit [url=http://www.mathlinks.ro/LaTeX/AoPS_L_About.php]this page[/url].

Probably you meant this one:

$\frac{a+b}{b+c}\frac{a}{2a+b+c}\le\frac{1}{2}$

and for what numbers do you think it can be valid? Please post only reasonable posts.  :ninja:

Note that I think that if some AoPS user post more than 10 post without LaTeX and continue to do so after warning, such a user should be DELETED. Nice day.

[quote="stancioiu sorin"]
$\frac{a+b}{b+c}\frac{a}{2a+b+c}\le 1$[/quote]
First, there's no any condition for $a,b,c$.
Then, just expanding two sides and guess... if $a,b,c>0$ it's wrong.

[quote="stancioiu sorin"]No, it's 1/2 no 1.I think you are right.It's false.[/quote]

Did you mean $\frac{1}{2}$? :)

Dear [b]stancioiu sorin[/b], try to learn some LaTeX. Please.

[quote="stancioiu sorin"]$\frac{a+b}{b+c}$ $\frac{a}{b+c+2a}$<$\frac{1}{2}$  ?[/quote]

Dear friend stan. sorin. There is a [url=http://www.mathlinks.ro/Forum/index.php?f=224]test forum[/url], where you can try writing in LaTeX.

Be a successful man!  :lol:

conceive

Dear conceive,
  I've already learnt LaTex. :D


Find all $f: \mathbb R \to \mathbb R$

(1) there are only finitly many $s\in \mathbb R: f(s) = 0$
(2) $\forall x,y \in \mathbb R : f(x^4+y) = x^3f(x)+f(f(y))$

Taking $x = 0$, we get $f(f(y)) = f(y)$. Taking $x = 1,y = 0$ we get $f(0) = 0$. Taking $y = 0$, we get $f(x^4) = x^3f(x)$. Hence, if $x$ is a zero of $f$, then $x^4$ is a zero as well. Since there are only finitely many zeroes, we must have $f(x) = 0 \Rightarrow x \in \{0,1\}$. But we can't have $f(1) = 0$ either since then $f(2) = f(1) + f(1) = 0$, $f(3) = f(1) + f(f(2)) = 0$, ... contradiction. So if $f(x) = 0$, then $x = 0$. Now take $t = f(x^4) - x^4$ (and keep $x$ fixed). Then $f(x^4) = f(f(x^4)) = f(x^4 + t)= x^3f(x) + f(t) = f(x^4) + f(t)$ and hence $f(t) = 0,t = 0,f(x^4) = x^4$. So for all positive $x$, we have $f(x) = x$. It's easy to see that we must have $f(x) = x$ for all negative $x$ as well, and we're done.


hi i have a problem

Let $A$ and $B$ be $n\times n$ matrices with $A>0$ and $B>0$ prove that $(A+B)^{-1}<A^{-1}$

sri hari

assume both are real symmetric matrices. Sorry that I did not give this before.

srihari

(All matrices in this answer have real entries.)

Lemma: Suppose $U$ is symmetric and $V$ is invertible.

Then $U>0$ iff $V^TUV>0.$

Proof: just consider the quadratic form $x^TUx.$ Since we can always write $x=Vy,$ this is $y^T(V^TUV)y.$ If one is always positive, then so is the other.

We want to show that $A^{-1}-(A+B)^{-1}>0.$ Consider:

$(A+B)^T(A^{-1}-(A+B)^{-1})(A+B)$

$=(A+B)(A^{-1}-(A+B)^{-1})(A+B)$

$=(I+BA^{-1}-I)(A+B)$

$=BA^{-1}(A+B)=B+BA^{-1}B.$

But $B>0$ and since $A>0,$ we also have $A^{-1}>0$ and thus$BA^{-1}B>0.$

The sum of two positive matrices is positive.

Thank you very much Mr Kent for the proof but how do you prove the Lemma?

I am a novice in this field so please bear with me with these questions!

thank you

sri hari

Oh Sory for the reply I got the answer!

Sri Hari

[quote="bhupala"]can we prove the same result using eigen value decomposition method?[/quote]
I spent a few minutes looking in that direction, but didn't see a way to make it work. The problem is that the eigenvectors of $A$ are different than the eigenvectors of $B,$ which in turn are different than the eigenvectors of $A+B.$

There's one hidden place where the eigenvalues were used: the statement that if $A>0,$ then $A^{-1}>0.$

one more quwstion please!

If $\lambda_{A_i}$ and $\lambda_{B_i}$ are the eigen values of $A$ and $B$ respectively which are positive definite, can we say that eigen values of $A+B$ are $\lambda_{A_i}+\lambda_{B_i}$? If not what are the conditions on $A$ and $B$ to hold the above result?

thanx in advance

sri hari.

That'll only hold if they have the same eigenvectors. That happens if they commute, I think.


MATHCOUNTS
probability
geometry
3D geometry
ratio

This is a test that I wrote a while back. It is formatted to be almost like an official MATHCOUNTS sprint test. I borrowed some problems, but I wrote most of them. It is very unlikely that you will be familiar with some of the questions that I borrowed because I borrowed them from a person that only sees Texas kids. Anyway, this test would probably be most useful for people going to nationals. Feel free to dispute the answer key.
Oh yeah, don't be alarmed if your answer comes out slightly unfriendly. Some of the answers come out weird ;P

I have confirmed two mistakes on the key.
[hide]#5 is 5. #18 is .5.[/hide]

One more
[hide]#20 $\frac {15}{68}$[/hide]

Mistake on my correction of #5.
[hide]#5 is actually 4! FINAL.[/hide]
Thanks to Naga for pointing out some of these mistakes.

How did you format that; I like how you made it look like a MC test.  Did you use Acrobat, or some free program? Please let me know!

PS:  I like it so much that I want to know where to get test 1!  Where can I download it?

I used Microsoft Word, and um nothing else.

Since you asked, here it is. It's about... 10 times easier. Also, don't rely on the key because I made it in like 2 seconds, so feel free to dispute. I used Word to format it using a much harder method, so it took longer to make. You might be familiar with many of the problems because I borrowed a lot of them from various Mathcounts workbooks.

But how did you get that cool looking border?  And converting to PDF?  Well, I could do that part with Primo...

I used Primo to convert it to pdf. The cool looking border is just a page border. The way I lined up the questions in test 2 was that I created a table with three columns, for the question number, the question, and the answer space. This made it a lot easier to format correctly.

Or could you just post the Word document you used?  Before conversion to PDF?  I would like to use the cover page (with a few alterations) if you don't mind.

It must have taken a lot of work to get that Word document perfectly matching the MC PDF's!!!  How long did it take?

10 minutes
btw I'm curious as to some scores that people are making.

I will let you know as soon as I complete it.  The second one looks fun...

Taking account tarquin's corrections, I have different answers from the answer sheet for the following problems:

[hide]4. 31/12 -- Edit: Oops, (1+1/2+1/4+1/8+1/16[b]+1/32[/b])*(1+1/3)
10. 4/11
11. 2/3
12. 3*pi*sqrt(135)
13. 114
25. 3/1012
26. 484*pi/3[/hide]

I did this in like 30 minutes, and my mathcounts days were over last year, though.  I'll recheck those and see if I made a mistake.

All checked.  I eliminated some.  The rest I still believe are correct.  If they are, then I got a 27, but considering some probably aren't, my score would be lower.  I got a 24 on last year's sprint, so this is indeed around national level.

I will tell you how I arrived at each answer.
[hide]4. $96 = 2^5 * 3$. The sum of the reciprocals of the factors is equal to the sum of the factors divided by $96$. The sum of the factors is $(2^6 - 1)(4) = 252$. $\frac {252}{96} = \frac {21}{8}$.

10. The only way for the sum to be even is if they both draw even balls, or they both draw odd balls. Probability that Jack and Jill draw even balls is $\frac 12 * \frac 49 = \frac 29$. Probability that they both draw odd balls is equal to the probability that they both draw even balls because there's the same number of even balls as odd balls. Therefore, the answer is $\frac 29 * 2 = \frac 49$.

11. Put one green ball in one of the bags. The probability of picking a green ball is then $\frac 12 * 1 + \frac 12 * \frac 12 = \frac 34$.

12. The circumference of the cone is $\frac {72 \pi}{12} =  6 \pi$. Therefore, the radius of the cone is $\frac {6 \pi}{2 \pi} = 3$. The slant height of the cone is $36$. We can make a right triangle to find the height which is $\sqrt {36^2 - 3^2} = \sqrt {1287}$. Since $V = \frac {\pi r^2 h}{3}$, plugging in the values we get $3 \pi \sqrt {1287} = 9 \pi \sqrt {143}$.

13. For every single-digit hour, there are $6$ palindromes (e.x. $1:01, 1:11... , 1:51$). There are $9$ single-digit hours. For every double-digit hour, there is only $1$ palindrome, making the answer $9*6 + 3 = 57$.

25. Think of it this way: Children are randomly walking to tables in order to pick groups out, and each table is one group. It doesn't matter where the first person goes. The next two people have to go to the particular table that the first person went to. The probability that that table receives those two people is $\frac 2{23} * \frac 1{22} = \frac {1}{253}$.

26. The volume of a frustrum with a circular base is $\frac 13 \pi h (r_1^2 + r_2^2 + r_1 r_2)$. Plugging in the values, we get $\frac 43 \pi (64 + 81 + 48) = \frac {592 \pi}{3}$. The cone is $36 \pi$. Subtracting, we get $\frac {484 \pi}{3}$. This is the only one I agree with you on  ;) .[/hide]

OK for the clock one, didn't it say 24 hours??  For the green and red balls one, I thought there were 11 green and 20 red, so the remaining box has 10 green and 20 red?  Other ones yeah I agree now, sucky arithmetic skills :).

Hmm yes you are right on both of those. Good job haha.
So
[hide] 11 is actually $\frac 23$.
13 is actually 114.[/hide]

THANX RJ !!!!!!!!!!!
It finally works even though I dont know why it didn't when you tried to give it to me earlier.

tarquin, before posting the answers, you might want to actually check your answers twice(Instead of other people correcting you so many times.  After that, how high is your self-esteem?) ;)

How do you find the sum of the factors of a number?

Did you get all of the names from the Flour Bluff math team?
Where's your name?

Don't worry RJ, don't lose your self-esteem, your the only person who took the time to actually make a test.

I got a 25, assuming that the answer key is right (taking into account the answers posted afterwords!)

[quote="Magixmaniak"]How do you find the sum of the factors of a number? [/quote]

To avoid messy notation, I'll give you an example.  Let's say it's 96.  96 = 2^5*3.  Then we want to find a way to add all combinations of powers of 2 from 0 to 5 and powers of 3 from 0 to 1.  The way to do this is multiply:

(2^0 + 2^1 + 2^2 + 2^3 + 2^4 + 2^5)(3^0 + 3) = (1+2+4+8+16+32)(1+3)

And you can express geometric sums in closed form of course.  If you want the sum of the reciprocals you can just divide by 96 at the end or multiply  (1+1/2+1/4+1/8+1/16+1/32)(1+1/3), since I can add powers of 1/2 a little bit faster than powers of 2.

[quote="mysmartmouth"]Someone should make a new answer key....

Thanks for the .doc file, Tarquin!!![/quote]

Yeah, it's getting to be that there are too many mistakes.

I made the key in 10 minutes... so I don't really lose any self-esteem. Anyway the answers are here so just check...

21
[hide]Using mass point geometry, you have a weight of 6 on C and therefore a weight of 9 on A. If A is 9 then D is 13.5 making a total of 22.5 on E. Since C has a weight of 6, then the ratio of EF to FC is $\frac {6}{22.5} = \frac {4}{15}$.[/hide]

The quickest way to add powers of $\frac 12$ is almost exactly the same as adding powers of 2.  :P

[quote="Magixmaniak"]Does anyone else know what mass point geometry is?[/quote]

kinda....sorta....no


If the waffle cone supporting the spherical pint of ice cream is 4 inches tall and has a radius of 2 inches, what will be the volume in inches cubed of the waffle cone?

A) 50.2 B) 16.8 C) 5.3 D) 25.1 E) 18.8

Using the fact that the volume of a cone is $ \frac{\pi r^{2} h}{3}$, we have

$ \frac{\pi(2^2)(4)}{3}\equal{}\frac{16\pi}{3}\approx16.8$.

So the answer is B.


Prove that there are infinitely many natural numbers a with the property that, for any positive integer n, n^4+a is composite.

IMPORTANT EDIT: LET'S MAKE n>1 RATHER THAN ALL POSITIVE INTEGERS.

Supposition:[hide]"There are infinite number of composite numbers" and "There are infinite number of natural numbers"

Then "There are infinite number of composite numbers after x whereas x is a positive integer"

Therefore substituting n:^4: into x gives us "There are infinite number of composite numbers after n:^4: whereas n:^4: is a positive integer" which is true since the problem specifies that n is a positive integer and (positive integer):^4:=positive integer

"There are infinite number of composite numbers after n:^4: whereas n:^4: is a positive integer" could be worded as "There are infinite number of composite number which are greater than n:^4:+y"

Substituting a for y gives us "There are infinite number of composite number which are greater than n:^4:+a"

So the statement "a can be any natural number in the above statement" holds.

Substituting "infinite number of" for "any" and combining it with  "There are infinite number of composite number which are greater than n:^4:+a"gives us "There are infinite number of composite number which are greater than n:^4:+a whereas there are infinite number of a"[/hide]

There are also infinitely many prime numbers. For example, suppose a=1. Then if n=2, we get 2^4+1=17, which is prime. If a=2, then if n=3, we get 3^4+2=83, which is prime. Note that we're looking for it to be true for ALL n, not just infinitely many. On second thought, let's change that to n>1. That eliminates some need for funny cases where it's only prime when n=1. I'll edit the original problem.

I think it can be proved using an algebraic identity involving powers of four (if you don't know this algebraic identity it will probably be hard to solve this problem !)

The story of Sophie Germain (who found this identity) is well worth a visit to my opinion :

http://www.cs.appstate.edu/~sjg/womeninmath/SophieGermain.html

Hm. The way I know how to solve it doesn't involve powers of four.

When you multiply that factorization out, you don't have any ks. Perhaps just a minor fix can solve the problem though.

Sorry.

maybe...[hide]

if a = 4k^4

the it would be n^4 + 4(nk)^2 + 4k^4 - 4(nk)^2 so it would be

(n^2 +2k^2)^2 - (2nk)^2

(n^2 + 2k^2 - 2nk)(n^2 +2k^2 + 2nk) and that's composite

[/hide]

As an added extension to this problem,

prove that n^4+4^n will always be composite, except when n=1, for positive integers n.

HMm, I believe there is an easier way to prove the case if  n is odd, let's see...

[hide]Let n^4+4^n=x. Then, n^4+2^(2n)=x. Now, complete the square on the left, to get (n^2+2^n)^2=x+n^2*2^(n+1). Now, since n is odd, n+1 is even, which means 2^(n+1) is a perfect square, which means that x is the difference of two squares. Hence, the only possibility for which x is prime is if one of the factors is 1, and that is when n=1. [/hide]

Keep in mind. Completing the square is very, very helpful when it comes to number theory, especially to proof problems such as these.

The question I posted over on the Advanced Forum is MUCH like this one, but with a twist which makes it a bit harder..... if you guys did this problem, try to answer here:

http://www.artofproblemsolving.com/wforum/viewtopic.php?t=629

I'm still waiting for an attempt :(


When I try to download some PDF files like http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45

there, a blank page appears and at the bottom of the window where the loading information is it just says "done"

The download on the first year worked but then the rest didnt.

[quote="ShoeFactory"]VV and the guys are in slovenia apparently and he hasn't had time to fix stuff on the site.

Yeah this bug is confirmed, go ahead and do the 83 84 85 AIME's cuz 4everwise made nice PDF's for those (assuming those arent gone as well).[/quote]
[quote]
Not Found

The requested URL /Forum/download.php was not found on this server.[/quote]

Not sure if its my computer or something... :maybe:

[quote="ch1n353ch3s54a1l"]
Not sure if its my computer or something... :maybe:[/quote]

I have the same problem... I can`t download any pdf file. Are there other people that have the same problem?

PD: My computer uses Windows XP, IE 6.0, and the board style is Mathlinks

$Tipe$

The problem isn't on your side. I guess we just have to wait for VV to fix it.

Same problem here..

Is there are any progress on solving this?

Valentin hasn't logged in for some time. I'm guessing he's upset about AoPSWiki making the Contest Archive obsolete. (Who wouldn't be?) He spent a ton of time on that project...

Personally, I find the contest archive to be much more convenient.

I only archived about 15 sets of AMC and AIME problems, and already feel like my efforts have gone to waste. :(

To answer your question, doubt there's been much progress. All of the admins in San Diego are busy writing the Intermediate books and finishing up the Introduction series.

The blog system hasn't been fixed yet either. I haven't been updating much recently, since I'm fairly picky about aesthetics. :( :stretcher:

when i try to download pdf file and click on link like  "Download pdf version", i return to main page. Bcoz url is :" http://www.mathlinks.ro/Forum/"
How can I solve this problem? :maybe:

It just reloads the page for me. I am on:

Windows XP Prof. SP 2
Mozilla Firefox 2.0 Beta 2

It doesn't work on IE either.


The class forums eventually dissappear. (reducing our post number and causing us to save everything the tediuous way). Is there a way to mass download the forum site or have AOPS offer a zip file download that has all the threads?

The post counts aren't reduced - we move the posts into an archive.

As for compiling everything into a file, we'll do some thinking as to how that might be achieved.  It's not at all obvious how to do that.


inequalities
integration
vector
topology
function
triangle inequality
real analysis

$ d(x,y) \equal{}\|x\minus{}y\|_{L^p}$. Is d a metric for $ p\in (0,1)$?

The only question is the triangle inequality; for a norm, that's equivalent to asking whether $ \|x\|\plus{}\|y\|\le \|x\plus{}y\|$.

For convenience, I'll look at a discrete version $ \|x\|_p \equal{}\left(\sum_n (x(n))^p\right)^{\frac1p}$. Consider $ x(n)\equal{}\begin{cases}1&n\equal{}1\\0&\text{otherwise}\end{cases}$ and $ y(n)\equal{}\begin{cases}1&n\equal{}2\\0&\text{otherwise}\end{cases}$. We have $ \|x\|_p \equal{} \|y\|_p \equal{} 1^{\frac1p} \equal{} 1$ and $ \|x\plus{}y\|_p \equal{} 2^{\frac1p}$. If $ 0<p<1$, $ 2^{\frac1p}>1\plus{}1$, and the triangle inequality is violated. This is not a metric.

We can convert it into a metric; $ d(x,y)\equal{}\|x\minus{}y\|_p^p\equal{}\int |x\minus{}y|^p$ does satisfy the triangle inequality for $ 0<p<1$. Unfortunately, that loses the scaling properties of the norm. While $ L^p$ makes sense as a topological vector space and that topology is even metric, it is not a normed vector space.

Oh, thank you for your clear instructions, jmerry :lol:

I have another question in this regard.

Throughout  $ 0 < p < 1$.
Consider $ L^p$ with the norm $ \|f\|_{p}^{p} \equal{} \int |f|^{p}d\mu$. Then $ L^p$ is a Banach space.
Let $ E$ be a closed subset of nonnegative functions in $ L^p$ not containing $ 0$ and define $ E' \equal{} \left\{\frac {f^p}{\|f\|_{p}^{p}}: f\in E\right\}$.
Is $ E'$ also closed?
Thanks.

[quote="limsup"]I have another question in this regard.

Throughout  $ 0 < p < 1$.
Consider $ L^p$ with the norm $ \|f\|_{p}^{p} \equal{} \int |f|^{p}d\mu$. Then $ L^p$ is a Banach space.[/quote]

No, $ n(f) \equal{} \|f\|_{p}^{p}$ is not a norm.  The homogeneous property fails: for $ f \in L^p$ and $ a \in \mathbb{R}$ it does not follow that $ n(af) \equal{} |a|\;n(f)$.  At least $ L^p$ is a complete metric topological vector space.  It is not locally convex.  It is not a Banach space.

[quote]
Let $ E$ be a closed subset of nonnegative functions in $ L^p$ not containing $ 0$ and define $ E' \equal{} \left\{\frac {f^p}{\|f\|_{p}^{p}}: f\in E\right\}$.
Is $ E'$ also closed?
Thanks.[/quote]

Yeh sorry...  you are right. $ L^{p}$ is not a normed space. But $ d(f,g) \equal{} \|f \minus{} g\|_{p}^{p} \equal{} \int |f \minus{} g|^{p}$ defines a metric on $ L^{p}$.

With this can you answer my question?
[quote]Let $ E$ be a closed subset of nonnegative functions in $ L^p$ not containing $ 0$ and define $ E' \equal{} \left\{\frac {f^p}{\|f\|_{p}^{p}}: f\in E\right\}$.
Is $ E'$ also closed?
Thanks.[/quote]