function
linear algebra
matrix
topology
vector
analytic geometry
real analysis

According to the definition , a Riemannian metric is a family of smooth function g(ij) with the follwing properties: (i) the matrix (g) is positive definite; (ii) g(ij) are tensors, so they are subject to the transformation rule of tensors.

  But in some physical textbooks, it seems that as long as a matrix is positive definite, it can serve as a Riemannian metric. Besides, some books say we can choose any physical meaningful metric to compute geodesics on a manifold.

I am a little confused. Please help me.

Let $M$ be the manifold that is to be equipped with a metric. The purpose of coefficients $g_{ij}$ is to give us a symmetric positive bilinear form $\langle v,w\rangle$, where the vectors $v,w$ belong to the tangent space $T_{p}M$ for some $p\in M$. This is done by expanding $v$ and $w$ in some coordinate basis, and then computing $\langle v,w\rangle=\sum_{i,j}g_{ij}v^{i}w^{j}$. However, the coordinates of $v$ and $w$ depend on the choice of a basis. Therefore, $g_{ij}$ must be transformed in a certain way with the change of basis, to keep $\langle v,w\rangle$ the same. This is the  requirement (ii).

Now, in physics one may way to consider a non-Euclidean Riemannian metric on the Euclidean space $\mathbb R^{n}$. This simplifies things, because (a) the tangent plane to $\mathbb R^{n}$ is $\mathbb R^{n}$ itself; (b) $\mathbb R^{n}$ has a standard basis, in which all computations can be done. As long as your basis is fixed, there is no need to worry about the way in which $g_{ij}$ depend on a basis.


[quote="Nexmus"]As you may have noticed before, subsets appear alot in mathcounts problems.
Lets start practicing for them by starting a thread of problems.
I'll start it off:

Problem 1

What is the sum of all the two-element subsets of {1,2,3,4,5,6}?[/quote]

Why don't you just post in the Mathcounts Problem Solving thread?

[hide="Anyways,"]Each element appears in six subsets. Hence, our sum is
$ 6(1+2+3+4+5+6) = 6\cdot \frac{6\cdot 7}{2}= 6\cdot 21 = \boxed{126}$[/hide]

[quote="Temperal"]Problem?

Anyways...
how many different 3-element subsets are there of the set {3,4,5,6,3,4}?[/quote]

A set does not include doubled elements, do you mean just $ \{ 3,4,5,6 \}$?

Oops, that last one was supposed to be 34.

Well, firstly, the problems thread includes all kinds of problems. This includes only one type of problem.
Also, vishalarul, your close, but check your casework again.

[quote="indianamath"]er each element appears in 32 subsets.[/quote]

Aren't there only $ \binom{6}{2}= 15$ subsets total?

oh oops i didin't see 2 element.

there would be 5 subsets since you already used up one of the numbers.
each appears 5 times.

[quote="Nexmus"]vishalarul was the closest, but I'll give you the answer:

We use each number $ 5$ times, giving us $ 5\cdot(1+2+3+4+5+6) = 105$

Answer: $ 105$[/quote]

oops misread question....i thought sum of product...
I thoguht question was

"Find the sum of the products of the two elemnts of all 2-element subsets of (1,2,3,4,5,6)"

Does anyone want to post the next subset problem? :stink:

"Subsets" is a very confined topic, and speaking of "subset problems" is very vague. Usually, problems like these don't take any "new" concepts, concepts that aren't applicable to other problems. Furthermore, these specific types of problems aren't that common (maybe once or twice every year between all four tests). I think if you want to discuss these problems, I daresay there aren't that many directions you can go in, so you should just use the marathon or just use sepreate threads.


Let ABC be a triangle with circumcircle (O).
Let (W) be a circle centered at G (centroid) and internally tangent to (O).
(W) cuts two point P and Q on side AB. let M be the midpoint of PQ.
Let L be any arbitary point on arc(AB).
Prove that if LG passes through M then simson line wrt. L passes through M.

Isn't that obvious? Since $ M$ is the middle of the chord $ PQ$ then $ GM\perp PQ$


Please explain your answer to the following problems before Monday Febuary 13th, when the answers will be posted.  For inequalities please describe the graph you would make.

Okay, here are the problems, they are relativly easy, so if you want more challenging ones, say so, or post your own problems...here we go...

Reduce:
a)~~~ (5v - 20v^3)/(10v^4 + 5v^3 -10v(1+2v))
b)~~~((2y +1)^4 (y^3 +4) (6y^2) - (y^3 +4)^2 (2y + 1)^3 (8y))/(2y+1)^8

Solve:
a) ((3/(x+1)) + (5/(x+3)))>0
b) (2x^5 - 32x) <=0

<= is less than or equal to

Have fun! ;)  :lol:  :D  :)

Can you please explain more fully how you got your answer, also can you simplify it more?

~lyra


hadeaghal mohrehaee ra ke mishe too jadvale m dar n gozasht ta hadeaghal  ye mostatili ijad konan che ghadre :?:

manzoram hadeaghal mohrehast ke har tori to jadval bezarim hadeaghal ye mostatil ijad kone

agar $k$ ta khoonaro mohre bezarim va $k=nq+r$ dar in soorat :
$r.\dbinom{q+1}{2}+\dbinom{q}{2}.(n-r)>\dbinom{m}{2}$
hala khodet hadeaghale $k$ ro bedast biar. :wink:

yeki be man bege m,n ba ham che farghi daran ke javabha motegharen nist? :huh:

farz kon $m$ sotoon bashe bad baghiash halle dige.

ta yadam narafte ino begam K=n+2m-3 hadeaksar tedad mohrehaee ke mostatil ijad nemikone :)


Many thanks in advance!!!!!!!!!!
In a ABC triangle the side BC=2AB and angle B=2C ,prove that the ABC triangle is a right triangle.
I feel very embarased i can't solve this problem!!!!!
Greetings from Athens

maybe it would help to draw a figure and use trig
[hide]= pretty small hint]sin,cos,tan[/hide]

Hello,
[hide="Solution"]Let $ AM$ be the median ($ M\in BC$) and $ BD$ bisect  $ \widehat{ABC}$
$ \Rightarrow \Delta ABM$ and $ \Delta BDC$ are the isoceles triangles.
It's easy to prove that $ BD \perp AM, DM \perp BC$
$ Rightarrow \widehat{BAM}\equal{}\widehat{BMA}\equal{}\widehat{BDM}$
$ \Rightarrow ABMD$ is a inscribed quadrilateral.
$ \Rightarrow \widehat{A}\equal{}90^o$[/hide]

ΒD bisect and AM median,triangle BDC isosceles,DM cathetus to CD and median,
triangles BDM=ADB so angle A=90,
thanks a lot


[quote="Vasc"]If $ x$ and $ y$ are positive numbers, then

$ \frac 1{x^{2}}+\frac 2{y(1+y)}>\frac{16}{1+8xy}$.[/quote]
Let $ x=5, y=2$ then $ LHS-RHS=-1.0812.. <0$  :D

Nice solution, NguyenDungTN.
My solution is the same. But with
$ (1+8xy)(x-y)^{2}+8xy(x-1)^{2}+8xy^{2}+x^{2}+y>6xy$.:P


[quote="GameBot"]The energy given off by a spherical star, such as our sun, is equal to $ \sigma A T^4$, where $ \sigma$ is a constant, $ A$ is the surface area of the star, and $ T$ is its temperature. Spherical star JZ114 has ten times the radius and half the temperature of our sun. What is the ratio of the energy JZ114 gives off to the energy our sun gives off? Express your answer as a common fraction.[/quote]

If the sun gives off energy=$ 1$, JZ114 gives off 100*1/16 energy, or 6.25.  Thus the answer is 25/4.

Sorry, I made a mistake because $ surface\equal{}r^2\pi$


geometry
geometric transformation
homothety
trigonometry
conics
parabola
radical axis

Circle  $w_{0}$ exist interior of circle $w$.($w$ and $w_{0}$ haven't common center) Each $w_{j}$ (${j}=\bar{1.4}$) circles tangents $w_{0}$ and $w$.Prove that intersection points of exterior common tangents of $w_{i}$ and $w_{j}$( $1\le{i,j}\le 4$)are concurrent

The circles $\omega_{1},\ ...,\ \omega_{4}$ have interior tangency with the outside circle $\omega.$ They also have to have the same type of tangency with the inside circle $\omega_{0}$ (all exterior or all interior), otherwise the proposition does not hold. Assume that this is the case. All 6 circles can be considered directed and remain tangent. For example, if $\omega_{1},\ ...,\ \omega_{4}$ have exterior tangency with $\omega_{0},$ we can consider $\omega_{0}$ directed clockwise and $\omega,\ \omega_{1},\ ...,\ \omega_{4}$ counter-clockwise. The common external tangents of $\omega_{i},\ \omega_{j}$ pairs for $i,\ j = 1,\ ...,\ 4$ are also directed and they meet at their pairwise homothety centers (a pair of directed circles has only one homothety center). We have to show that these homothety centers are collinear. Since all circles and lines are directed, we can solve the dual problem instead: The pairwise radical axes of $\omega_{i},\ \omega_{j}$ are concurrent (that's because homothety center of 2 directed circles is a dual of their radical axis).

Any triple $\omega_{i},\ \omega_{j},\ \omega_{k}$ for $i,\ j,\ k = 1,\ ...,\ 4$ of circles with non-collinear centers has a unique circle perpendicular to all 3, centered at their radical center $R_{ijk}$ and with radius equal to the tangent length from $R_{ijk}$ to $\omega_{i},\ \omega_{j},\ \omega_{k}.$ The inverted circles $\omega_{i}'$ for $i = 1,\ ...,\ 4$ have a common unique circle $\rho'$ perpendicular to all of them, which is concentric with $\omega',\ \omega_{0}',$ because $\omega_{i}'$ are congruent and the common center O' of $\omega',\ \omega_{0}',$ is the common radical center of all triples $\omega_{i}',\ \omega_{j}',\ \omega_{k}'.$ Let $\rho$ be the circle, which inverts into $\rho'.$ This circle is perpendicular to all 4 original circles $\omega_{i}$ for $i = 1,\ ...,\ 4,$ but each triple has a unique circle with this property. Thus the center R of $\rho$ is the common radical center of all 4 triples $\omega_{i},\ \omega_{j},\ \omega_{k}$ for $i,\ j,\ k = 1,\ ...,\ 4.$

Beautiful problem!

Let $P_{i,j}$ be the intersection of the exterior common tangents of $\omega_{i}$ and $\omega_{j}$.  We claim that all the $P_{i,j}$ lie on the radical axis of $\omega$ and $\omega_{0}$.

Let $O_{i}$ be the centre of $\omega_{i}$ (and $O$ the centre of $\omega$), and let $\omega_{i}$ be tangent to $\omega_{0}$ and $\omega$ at $S_{i}$ and $T_{i}$, respectively.

It can be shown that $P_{1,2}$ lies on $S_{1}S_{2}$ and $T_{1}T_{2}$ by applying Menelaus's theorem to triangles $O_{0}O_{1}O_{2}$ and $OO_{1}O_{2}$, respectively (akin to [url=http://mathworld.wolfram.com/MongesCircleTheorem.html]Monge's theorem[/url]).

Furthermore, points $S_{1}$, $T_{1}$, $S_{2}$, and $T_{2}$ are concyclic.  (See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=103319]Tangency points of four circles form a cyclic quadrilateral[/url].)  Hence, $P_{1,2}$ has the same power with respect to $\omega$ and $\omega_{0}$, so $P_{1,2}$ lies on the radical axis of $\omega$ and $\omega_{0}$.  By the same argument, all the points $P_{i,j}$ lie on this radical axis.

I've never seen the word "dual" used this way.  Can you explain more?

[quote="yetti"]Since all circles and lines are directed, we can solve the dual problem instead: The pairwise radical axes of $\omega_{i},\ \omega_{j}$ are concurrent (that's because homothety center of 2 directed circles is a dual of their radical axis).[/quote]

A directed line is tangent to a directed circle, if their directions are the same at the tangency point. Similarly for directed tangent circles. Directed lines remind properties of points closer than undirected lines. For example, a unique circle passes throuch 3 points, but 4 different circles touch 3 lines. On the other hand, a unique directed circle touches 3 directed lines. 2 circles intersect at 2, 1 or 0 points, but they can have up to 4 common (undirected) tangents. On the other hand, 2 directed circles have 2, 1, or 0 common directed tangents. If a line p intersects a circle at 2 points A, B, then the 2 directed angles between the line and the circle at the intersection points are equal in size and opposite in sign. Similarly, if 2 directed tangents a, b to a directed circle exist from a point P and if A, B are the tangency points, then the directed segments PA, PB are equal in size and opposite in sign. This suggests duality between directed angle and directed line segment.  A circle can be defined not only as a set of points, but as a set of tangent lines and directed circle as a set of directed tangents. If a directed line p is tangent to 2 directed circles at A, B, it is the common line of these 2 circles, and their tangent distance is the segment length AB. Considering circle as a set of points, let P be a common point of 2 circles (one of their intersections). Let a, b be tangents of these 2 circles at P. The pencil of lines between a, b and passing through P corresponds to the set of points between A, B on p and the directed segment AB to the $\measuredangle aPb.$

Let a line k through a point P meet a circle at A, B. Power $PA \cdot PB$ of the point P to this circle does not depend on the line k through P. Similarly, let K be a point on a line p and let a, b be directed tangents from K to a directed circle. Power $\tan \frac{\widehat{pKa}}{2}\cdot \tan \frac{\widehat{pKb}}{2}$ of the line p to this circle does not depend on the point K on p. (I used power of a directed line to a directed circle in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=85336]Geometry problem revisted (constant product with parabola)[/url], [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=115723]A constant product[/url], ...) Radical axis of 2 circles is a line, every point of which has the same power to these 2 circles. Homothety center of 2 directed circles is a point, such that every directed line through this point has the same power to these 2 directed circles. Looking at this differently, the segment lengths of tangents to 2 circles from any point on their radical axis are equal. Similarly, any directed line intersecting 2 directed circles through their homothety center forms equal directed angles with these circles. 2 circles have a unique radical axis, but 2 homothety centers. 3 circles have a unique radical center, but 4 homothety axes. On the other hand, 2 directed circles have a unique homothety center and 3 directed circles have a unique homothety axis.

This duality goes deeper. For example, circular transformations, which carry points to points and circles (or lines) to circles (or lines), such as ordinary inversion, preserve angles between curves, they are conformal mappings. On the other hand, circular transformation, which carry directed lines to directed lines and directed circles (or points) into directed circles (or points), preserve tangent distances between curves, they are equilong mappings.
_______________________________________________________

Since [b]nsato[/b] proved that the common homothety axis of the (directed) circles $\omega_{1},\ ...,\ \omega_{4}$ is the radical axis of $\omega,\ \omega_{0},$ by the duality principle, it immediately follows that the common radical center of $\omega_{1},\ ...,\ \omega_{4}$ is the homothety center of the directed circles $\omega,\ \omega_{0}.$ Since these 2 circles have opposite directions, their homothety center is the internal homothery center of the undirected circles $\omega,\ \omega_{0}.$


calculus
integration
limit
logarithms
calculus computations

[img]http://i158.photobucket.com/albums/t120/toughtop16/integral.jpg[/img]

Can someone help? Thanks

Use this:

$\int_{0}^{1}\int_{0}^{1}\left(\frac{1}{x+y}+\frac{1}{y+1}\right)dxdy=\lim_{\epsilon\to 0^{+}}\int_{\epsilon}^{1}\int_{\epsilon}^{1}\left(\frac{1}{x+y}+\frac{1}{y+1}\right)dxdy$

You should get $3\ln 2$ when its all done.


limit
logarithms
calculus
calculus computations

Condition of Equivalence for Variable Substitutions in the Limit Calculations?

Did not elaborate on the textbook for Variable Substitutions of limit.

What are the conditions that make it effective?

example:
$ \underset{x\rightarrow\infty}{\lim}\frac {x}{\left( \ln x\right) ^{n}}$ $ \equal{}$$ \underset{y\rightarrow\infty}{\lim}\frac {e^{y}}{y^{n}}$ $ \equal{} \infty$

If $ \lim_{s\to a}g(s)\equal{}M$ and $ \lim_{t\to M}f(t)\equal{}L$, $ \lim_{x\to a}f(g(x))\equal{}L$. If you try to build a converse, convergence is not guaranteed.

These $ a,M,L$ are, however, allowed to be in the extended real numbers.

[quote="valeriummaximum"]i think any valid substitution that leads u to the answer is allowed...[/quote]

Circular, non?  :maybe:

[quote="jmerry"]If $ \lim_{s\to a}g(s) \equal{} M$ and $ \lim_{t\to M}f(t) \equal{} L$, $ \lim_{x\to a}f(g(x)) \equal{} L$. [/quote]
Ah, but even this may be not true. Consider $ g\equal{}0$ and $ f(x)\equal{}\begin{cases}0,& x\ne 0\\ 1,& x\equal{}0\end{cases}$.

Right. That condition for $ f$ really needs to be continuity- at which point this is a standard fact in every calculus textbook.
If you don't actually hit $ L$, as happens when $ L\equal{}\infty$, my original statement works.

[quote="jmerry"]If $ \lim_{s\to a}g(s) \equal{} M$ and $ \lim_{t\to M}f(t) \equal{} L$, $ \lim_{x\to a}f(g(x)) \equal{} L$. If you try to build a converse, convergence is not guaranteed.

These $ a,M,L$ are, however, allowed to be in the extended real numbers.[/quote]

Would build a converse for continuous $ f$ and $ g$, when $ M \equal{} \infty$ ?

[quote="jmerry"]Can you state precisely what you want to prove? If you can, then look for a proof or a counterexample.[/quote]

how about this ? can it be prove?

Known: $ f$ is continuous  and  $ \underset{x\rightarrow\infty}{\lim}g\left( x\right) \equal{} \infty$ ,then: $ \underset{x\rightarrow\infty}{\lim}\left(  f\circ g\right)  \left(  x\right)
\equal{}\underset{y\rightarrow\underset{x\rightarrow\infty}{\lim}g\left(  x\right)
}{\lim}f\left(  y\right)$

KNOWN : $ f$ is continuous at $ l$  and $ \underset{x\rightarrow\infty}{\lim}g( x) = l) ) \rightarrow( \underset{x\rightarrow\infty}{\lim}( f\circ g) ( x) = f( l) = \underset{y\rightarrow\underset{x\rightarrow\infty}{\lim}g( x) }{\lim}f( y))$
PROOF:
( proved ) KNOWN : $ f$ is continuous at $ l$  and $ \underset{x\rightarrow p^{ + }}{\lim}g( x) = l) ) \rightarrow( \underset{x\rightarrow p^{ + }}{\lim}( f\circ g) ( x) = f( l) )$
LET : $ h( \frac {1}{x}) = g( x)$

$ f$ is continuous at $ l$  and $ l = \underset{x\rightarrow\infty}{\lim}g( x) = \underset {x\rightarrow0^{ + }}{\lim}g( \frac {1}{x}) = \underset {x\rightarrow0^{ + }}{\lim}h( x) )$

$ \Rightarrow f( l) = \underset{x\rightarrow0^{ + }}{\lim }( f\circ h) ( x) = \underset{x\rightarrow\infty }{\lim}( f\circ h) ( \frac {1}{x}) = \underset {x\rightarrow\infty}{\lim}f( h( \frac {1}{x}) ) =$ $ \underset{x\rightarrow\infty}{\lim}f( g( x) )$

$ \Rightarrow\underset{x\rightarrow\infty}{\lim}f( g( x) ) = f( l)$

[quote="Kent Merryfield"]That statement cannot be precisely what you mean. What should $ \lim_{x\to\infty}g(x)$ be?[/quote]

i want that it wolud be $ \infty$ . can it be prove? please teach me, thank you a lot.


AMC
AIME
quadratics
probability
number theory
least common multiple
floor function

[hide="Keys"]
1. 013
2. 022
3. 001
4. 011
5. 280
6. 625
7. 192
8. 005
9. 287
10. 782
11. 011
12. 029
13. 029
14. 242
15. 072
[/hide]

I did get some answers from people but I never responded to any of them because there were some mistakes and I found one when I was making solution packet.

Solution packet, unfortunately, is unavailable to the web. I'm not even sure I'll have it up here.

But you're free to discuss it. :)

After you graded your own, post your score here!

btw should this one be Mock AIME 5 since we already had a Mock AIME 4?

Thnx.

I counted wrong on how many mock AIMEs we had. :D

I HAD some bad mistakes. i need explanation for 14 though. it was terribly ugly. i didnt know what to do. but except that, i should have gotten 11 or 13. few stupid mistakes on counting.

11/15 yay what does that translate into, for a hard AIME? about 7?

I didn't try the Mock AIME but I'll post a couple solutions:

[hide="Solution 8"]

We have:

\[x+\sqrt{y}=\sqrt{22+\sqrt{384}}.\]

So we think to square both sides:

\[x^2+y+2x\sqrt{y}=22+\sqrt{384}.\]

Matching up rational and irrational parts we get the system:

\begin{eqnarray*}
x^2+y &=& 22\\
2x\sqrt{y} &=& \sqrt{384}
\end{eqnarray*}

We can square the second and divide by 4 to get:  $x^2y=96$.  Now to make thinks a little simpler to see define a variable:  $w=x^2$.   This makes our system look like:

\begin{eqnarray*}
w+y &=& 22\\
wy   &=& 96
\end{eqnarray*}

You could solve this by substitution or you can say that $a$ and $y$ are roots of a quadratic.  So, let's just make the quadratic:

\[a^2-22a+96=0.\]

This factors nicely and the roots are 16 and 6.  Thus we say that $y=6$ and $w=x^2=16\Rightarrow x=4.$  So our answer is $4+\sqrt{6}$.  You double check this answer and I believe it works out.  So our answer is $3+2=\fbox{5}$.[/hide]

[hide="Solution 13"]

The sum of the reciprocals is just ${\frac{a_{n-1}}{a_n}}$.  So just the reciprocal of this: $\frac{10}{19}$ and thus $10+19=\fbox{29}$

Notice how it didn't matter whether you asked for the sum or the reciprocal of the sum...but that shouldn't be a surprise.

This was really, really easy for #13 on any AIME...no offense  :P [/hide]

hey, is the aime (the questions) still posted somewhere? i looked in the thread and it seems like it was edited out. :(

http://artofproblemsolving.com/Forum/viewtopic.php?t=28055

They're still there.

i think on 13, since it said real roots, it could be kinda hard
but from that sign rule, i guess there arent any complex

:blush: Oops. It should've been harder.

Oh well. It was pretty nice mock AIME anyway. :D

O, i didn't see the 'real' part.  My answer still agrees with silverfalcon's but my justification would be wrong then.

i have some solutions to share

[hide="number 4"]since there are 16 balls, and we have to put them in 8 boxes so that none of them are empty, it is basically like this:

we have 7 "dividers" which divides the balls into 8 "parts"(or in this case, boxes). since we cannot have a "part" without any balls, there are only 15 places where these 7 dividers could go into. thus, the number of ways can be expressed as 15C7. 15C7 = 6435, 6435/22=292R11.

so it is 11. [/hide]

[hide="number 6"] to find the probability of outcomes having at least one 2, we could find that probability, $\frac{p}{q}$, by subtracting the probability of NOT having ANY 2's from 1.

SO: find the probability of NOT HAVING ANY 2's.

well, since we can have any of the 5 remaining numbers occur in any of the four times,  the total number of outcome that has NO 2's is: $5^4$. divide that by the total outcome, $6^4$, we have $\frac{625}{1296}$. Subtract one from it, and
$\frac{p}{q} = \frac{671}{1296}$

then the absolute value of $p - q$ is $1296 - 671 = 625$[/hide]

[hide="number 10"]
since $r + \frac{1}{s}$ does not exceed one, we assume that r has to be a fraction, and that $\frac{1}{s}$ has a denominator of 46. since 391 is divisible by 23, but not 46, i found that t is an improper fraction which would cancel out the 2 if it is flipped (1/t). From there, it was easy to find r being $\frac{17}{23}$ and t being $\frac{23}{6}$. Then rs = 34, 12t = 46, $lcm\left(34, 46 \right) = 782$[/hide]

number 10 was kinda guess-and-check, but i had the right idea about what the numbers mgiht be. and I was right.

thats was my complaint, why isnt it 15C8 and the answer is 15C8 when divided by 15+8 = 23

why does 7 has to be    b?

a can be 15 and b can be 8

Really easy ones:

[hide="6"]
The probability of getting at least one two is equal to 1 minus the probability of getting no twos.

Probability of not getting a 2=5/6, for doing that four times: $(5/6)^4$
$1-(5/6)^4=\frac{6^4-5^4}{6^4}$
$6^4-5^4-6^4=-5^4$, the absolute value of which is $5^4=625$
[/hide]

[hide="12"]
This problem was super easy, and I doubt anyone needs a solution...

It was pretty simple to just find 101x101x101x101+56 and add up the digits, but another way to do it was to note that it's like Pascal's powers of 11 triangle, with zeroes between the ones:
$101^2=10201 (11^2=121)$
$101^3=1030301 (11^3=1331)$
$101^4=104060401 (11^4=14641)$

Then $256+104060401=104060657$
$1+4+6+6+5+7=29$
[/hide]

#12

2^4 + 2+5+6 = 29

how does #15 work?

isnt it
$\lfloor log_2 53 \rfloor + \lfloor log_2 54 \rfloor + \lfloor log_2 55 \rfloor + \lfloor log_2 56\rfloor + \lfloor log_2 57 \rfloor
+ \lfloor log_2 58\rfloor + \lfloor log_2 59 \rfloor + \lfloor log_2 60 \rfloor + \lfloor log_2 61 \rfloor + \lfloor log_2 62 \rfloor
+ \lfloor log_2 63 \rfloor + \lfloor log_2 64 \rfloor  $
= 5+5+5+5+5+5+5+5+5+5+5+6
= 55 + 6
= 61 ?

why is the solution  72  = 6*12 ?

On the other hand,
#4,  the 2nd solution   15C8 mod23  = 18 is legal
   along with 15C7 mod22 = 11

[quote="isaacchao"]
how does #15 work?

isnt it
$\lfloor log_2 53 \rfloor + \lfloor log_2 54 \rfloor + \lfloor log_2 55 \rfloor + \lfloor log_2 56\rfloor + \lfloor log_2 57 \rfloor
+ \lfloor log_2 58\rfloor + \lfloor log_2 59 \rfloor + \lfloor log_2 60 \rfloor + \lfloor log_2 61 \rfloor + \lfloor log_2 62 \rfloor
+ \lfloor log_2 63 \rfloor + \lfloor log_2 64 \rfloor  $
= 5+5+5+5+5+5+5+5+5+5+5+6
= 55 + 6
= 61 ?

why is the solution  72  = 6*12 ?

On the other hand,
#4,  the 2nd solution   15C8 mod23  = 18 is legal
   along with 15C7 mod22 = 11[/quote]

Yeah, I agree with both of those points... but I don't remember my answers, so I could be wrong.

If those brackets are a ceiling function, then it's just $6+6+6+6+6+6+6+6+6+6+6+6=72$

If they're a floor function, then it's $5+5+5+5+5+5+5+5+5+5+5+6=61$

And if they are merely grouping symbols, then it's $log_2 (64*63*62*61*60*59*58*57*56*55*54*53) \approx 70.4141$

But you're not allowed to use a calculator.

what the heck is a ceiling function? is that the opposite of floor function?

ceiling( TT) = 4
floor(TT) = 3

so you are correct

But how were we supposed to know that?  It was a super-easy problem if we did; but I didn't look at that thread of errors/ambiguities until after I had finished the test.

$\lfloor x \rfloor$ is a floor function
$\lceil x \rceil$ is a ceiling function

Say $x=2.4$

Then $\lfloor x \rfloor = 2$
and     $\lceil x \rceil = 3$

why is it that when i clicked on rules and then the test, much of the text did not show up? :?
(and i do have latex)

[quote="xxreddevilzxx"]but Silverfalcon did not make it clear on his test because there are brackets, not the symbols that represent either floor value or the ceiling value.[/quote]

Exactly. I said it was super-easy [i]if we knew that those were supposed to represent a ceiling function[/i]

Can anyone gives #7 a solution?

I used mass point geometry but I'm pretty sure there's other ways to solve it. :D

I just don't see it.


The following result, is true or not:
For all $ k>1$ is the set $ A\equal{}(1, 3, ..., 4k\minus{}1)$ then, there are at least $ k\plus{}1$ numbers primes.

Thanks.

Excuse me, ZetaX, but then what is the minor constant $ N > k$ .
Because for the first values of $ k \equal{} 2, 3, 4, 5, ...$ is true.

Big means really big  :wink:
Well, you want that $ \frac{4k \minus{} 1}{\log(4k \minus{} 1)} < k \plus{} 1$ by the prime number theorem (and $ k$ a little bit bigger than that as this is only an asymptotic result).


There's more than one answer: anything of the form $ k(x\minus{}49)(x\minus{}a\minus{}bi)(x\minus{}a\plus{}bi)(x\minus{}c\minus{}di)(x\minus{}c\plus{}di)$ works.

[quote="binomial_4eva"]
Find $ L_1,L_2,L_3,L_4,L_5,L_b$ [b]in terms of $ a,b,c,d$[/b][/quote]

Emphasis mine.

I don't understand this problem; it requires nothing more than tedious polynomial multiplication.

EDIT: My apologies.

[quote]Math154 is right...
so if we assume $ L_1 \minus{} 1 \equal{} k$
then:
$ L_2 \equal{} 1 \minus{} k(49 \plus{} a \plus{} c)$
$ L_3 \equal{} 1 \plus{} k(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} 2ac 98(a \plus{} c))$
$ L_4 \equal{} 1 \minus{} k(49(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2) \plus{} 2c(a^2 \plus{} b^2) \plus{} 2a(c^2 \plus{} d^2) \plus{} 196ac)$
$ L_5 \equal{} 1 \plus{} k(98(ca^2 \plus{} cb^2 \plus{} ac^2 \plus{} ad^2) \plus{} (ac)^2 \plus{} (ad)^2 \plus{} (bc)^2 \plus{} (bd)^2)$
$ L_6 \equal{} 1 \minus{} 49k((ac)^2 \plus{} (ad)^2 \plus{} (bc)^2 \plus{} (bd)^2)$[/quote]

thank you ManuelS.thats the answer i was looking for.

[quote="math154"]There's more than one answer: anything of the form $ k(x \minus{} 49)(x \minus{} a \minus{} bi)(x \minus{} a \plus{} bi)(x \minus{} c \minus{} di)(x \minus{} c \plus{} di)$ works.[/quote]
you mean there are more than one answer for each $ L_i$
you mean $ L_2 \equal{} 1 \minus{} k(49 \plus{} a \plus{} c)$ and $ L_2 \equal{} something else?$

Guys, what I mean is that the answer depends upon the value of $ L_1\minus{}1$.

Only given the roots of $ P(x)$, we can not determine anything else. We can find $ L_2,L_3,L_4,L_5,L_6$ in terms of $ a,b,c,d,L_1$, but that's not what the question is asking.

[quote="ocha"][quote="Temperal"] it requires nothing more than tedious polynomial multiplication.
.[/quote]

I suppose, but fortunately if you know Vieta's formula its very easy  :)

[url]http://planetmath.org/encyclopedia/VietasFormula.html[/url][/quote]

Just as a hint; it's rather patronizing to assume someone doesn't know Vieta's Formulas, but thanks nonetheless for pointing that out.


superior algebra
superior algebra unsolved

Suppose $G$ is a group generated by a finite subset $X$
which is composed of all the elements of some fixed order $r$.
If $|X|=n$, can we say that $|\textrm{Aut}\,G| = n!$ ???

That's wrong for as good as any group, e.g. cyclic or symmetric ones.


geometry
3D geometry
geometric transformation
reflection
Support
set theory

[quote="_anusuya"]which of these two are more in quantity?rational numbers or irrational numbers?why?[/quote]

Are they same? (infinity :D )

it is true that there will be infinitely many of both type but in a sense ;take a sample space consisting of rational numbers only
then every square root of a non square rational no. is irrational and likkewise are its cube root 4 th roots upto nth roots and n is tending to infinity .plus dividing any irrational by rational gives another irrational.Irrational should win :?:

Confusing (that was just my guess before I go to sleep :| )...

Well, I asked someone, and here's what he says:

[b]"Integers, natural numbers, and rational numbers, these are all countable as sets; however, irrational numbers and real numbers are not really countable. You should go to college and learn set theory first to really understand it."[/b]

So it is definitely not "a very easy problem" nor a MOEMS level.

And.. you're right!

However, these are true:

1) # of irrational numbers = # of real numbers

2) # of integers = # of rational numbers.

Edit - You should have used Google first  :rotfl: ;see [url]http://mathforum.org/library/drmath/view/51913.html[/url]

I will prove that even set <0;1> -interval, is uncountable. Suppose that it is countable. In that case we can make bijective reflection of set of natural numbs on the interval. Let it be this:
1 -> 0,a11a12a13a14...
2 -> 0,a21a22a23a24...
3 -> 0,a31a32a33a34...

But we can construct number 0,b1b2b3b4b5... where bi=1 if and only if aii=/=1 and bi=2 if and only if aii=1.  This is contradiction.

This is definitely not an easy question.
Do you, [b]anusuya[/b] have an answer to it?

ps. so many new members

i believe also that both of the terms are equal overall because they both has an infinite amount of numbers...
however, let's see how many rational and irrational numbers support the following
they are more than 0
they are less than 1

well, obviously we know that there are an infinite amount of numbers between two finite points, so there are definitely more rational numbers...

at first, so then, you see that thew farther you go, the more irrational numbers you can have, so the answer to the problem, i believe, is undefined, because they are the same and they are infinite.

I was just taking a look around this forum (I'm in high school, I shouldn't be here) and this thread has a lot of incorrect statements in it.  If you want a very clear, gentle explanation of this, try reading this Math Jam transcript:

http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=182

ack...

They have the same quantity, even thouh irrational numbers seem more in quantity.


A new massive black hole has been discovered.

http://www.sciencedaily.com/releases/2007/10/071030112102.htm

[quote]the extra electrons of elements such as carbon and oxygen "feel" the outward pressure of light and are thus more susceptible to being swept away in stellar winds[/quote]
How?
Acceleration due to radiation pressure is more for hydrogen (if that's what they mean).
We know that Radiation force $ F_R\equal{}\frac{IA}{c}$ where $ A$ is the cross sectional area.
Then ratio of acceleration $ \frac{a_C}{a_H}\equal{}\frac{r_C^2}{r_H^2}\frac{m_H}{m_C}\approx 0.65$

If you want to question the article, I think you should contact the webmaster at Science Daily.

I am asking it here as it is extremely likely that I am missing some simple point etc, not because I think the authors are wrong. Also, I think this is a nice problem in itself. Could someone here check the calculation?


factorial
ratio
function
algebra
binomial theorem
real analysis
real analysis unsolved

Can anyone tell me if the following sums converge or diverge?  Is it possible to evaluate them if they converge?  If so, what are they?

$(1)$ $\sum_{n=0}^{\infty}\frac{(2n)!}{(n!)^{2}}$

$(2)$ $\sum_{n=0}^{\infty}\frac{(n!)^{2}}{(2n)!}$

$(3)$ $\sum_{n=0}^{\infty}\frac{(-1)^{n}(2n)!}{(n!)^{2}}$

$(4)$ $\sum_{n=0}^{\infty}\frac{(-1)^{n}(n!)^{2}}{(2n)!}$

Thanks!  I guess I'm just curious about the behavior of factorials.

With all the factorials, just use the ratio test. And remember that $\frac{[2(n+1)]!}{(2n)!}=(2n+2)(2n+1).$

Related question: determine the interval of convergence for the power series

$\sum_{n=0}^{\infty}\frac{(2n)!}{(n!)^{2}}\,x^{n}.$

An interesting fact: $\binom{2n}{n}= \frac{(2n)!}{(n!)^{2}}\sim \frac{4^{n}}{\sqrt{n \pi}}$

[quote="Kent Merryfield"]Related question: determine the interval of convergence for the power series

$\sum_{n=0}^{\infty}\frac{(2n)!}{(n!)^{2}}\,x^{n}.$[/quote]

Thanks!

Using the ratio test, I got that the sum converges for $-\frac{1}{4}< x < \frac{1}{4}$, but I can't tell if it converges at the endpoints.  Can anyone help me?

I wonder, can the interval of convergence be calculated for:

$\sum_{n=0}^{\infty}\frac{(kn)!}{(n!)^{k}}\,x^{n}$

where $k > 0$ is an integer...

or even for:

$\sum_{n=0}^{\infty}\frac{(n!)^{k}}{(kn)!}\,x^{n}$

To deal with the endpoints, you need the asymptotic estimate that sos440 gave, or something equivalent to that. There are several ways to derive what sos440 claimed; one interesting path is to consider it a consequence of the Central Limit Theorem as applied to binomial distribution with $p=q=\frac12.$

You can also calculate $\sum_{n=0}^{\infty}\binom{2n}{n}x^{n}$ exactly inside its interval of convergence; it's an algebraic function.

How can one calculate $\sum_{n=0}^{\infty}\binom{2n}{n}x^{n}$?  I don't know what is meant by "it's an algebraic function."

Via the extended binomial theorem; see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=143605

For algebraic function, see
http://mathworld.wolfram.com/AlgebraicFunction.html

I also calculated a close relative [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=143327]here[/url].


1. My sock drawer has the following socks by color: 3 green, 5 white, and 7 black. I hurriedly pack for a trip one morning, randomly grabbing 5 socks from the drawer. What is the probability that I will have at least one pair of black socks?

2. Bill and George are among 7 boys seated randomly in a row. What is the probability that exactly two boys are seated between Bill and George?

1.) So what is the probability that out of the 5 socks you grab, you don't grab only 1 or no black socks.
Probability of grabbing one black sock only: 7C1*8C4/15C5
Probability of grabbing no black socks: 8C5/15C5

So probability of grabbing at least two black socks: 1-(7C1*8C4 + 8C5)/15C5
... i don't have a calculator, so you're going to have to plug that in.  oh, and "xCy" is "x choose y"

2.) draw seven lines for seven seats:
_ _ _ _ _ _ _
so let's pretend that bill or george has to sit in the first seat, and the other has to sit in the 4th seat (i.e. with 2 boys between them)

2 _ _ 1 _ _ _ --> 2 because either bill or george can sit there, and 1 because that's the left over
2*5*4*1*3*2*1 --> you fill in the other five boys into the seats, 5,4,3,2,1.

So 240 ways the boys can sit if they have to be in the 1st and 4th seats.

However, they can be in 1&4, 2&5, 3&6, or 4&7.  Thus, there are 240*4=960 ways total they can sit with exactly two boys between them.

probability = 960/7! --> there are 7! ways to rearrange the boys without restrictions

once again, i don't have a calculator, so i can't give you a numeric answer.

I really have no idea what I'm doing, so be warned.

[hide="Number 1"]So you want B B _ _ _

There are $ \frac{(3\plus{}5\plus{}5)!}{3!5!5!}\equal{}72072$ ways to fill the other socks.

A total of $ \frac{(3\plus{}5\plus{}7)!}{3!5!7!}\equal{}360360$ ways.

So $ \frac{72072}{360360}\equal{}\boxed{\frac{1}{5}}$.[/hide]

[hide="Number 2"]
We have B _ _ G. There are $ 5\cdot 4\equal{}20$ ways to put other people in there. We could also have G _ _ B, so $ 2\cdot 20\equal{}40$ ways to create that 4-person unit.
Now we can "slide" this "unit" around within the row of 7 seats. There are 4 places the first person could go. Wherever we put it, there are 3 people left over.
So $ 40\cdot 4\cdot 3!\equal{}960$ ways.
Total of $ 7!$ ways, so the probability is $ \frac{960}{7!}\equal{}\boxed{\frac{4}{21}}$.
[/hide]


1. For $ a,b,c > 0$ and $ a \plus{} b \plus{} c \equal{} 1$.
Prove that
\[ \sqrt {{{2b} \over {3 \plus{} a}}} \plus{} \sqrt {{{2c} \over {3 \plus{} b}}} \plus{} \sqrt {{{2a} \over {3 \plus{} c}}} \le \sqrt {{1 \over {1 \plus{} 2\sqrt {ab} }} \plus{} {1 \over {1 \plus{} 2\sqrt {bc} }} \plus{} {1 \over {1 \plus{} 2\sqrt {ca} }}}
\]
2.For $ a,b,c > 0$.
Prove that
\[ {{1 \plus{} a \plus{} b} \over {a \plus{} ab}} \plus{} {{1 \plus{} b \plus{} c} \over {b \plus{} bc}} \plus{} {{1 \plus{} c \plus{} a} \over {c \plus{} ca}} \ge {6 \over {\sqrt {1 \plus{} abc} }}
\]
3.Denote $ a,b,c > 0$ and $ abc \equal{} 1$.
Prove that
\[ \sum\limits_{\scriptstyle cyclic \hfill \atop \scriptstyle a,b,c \hfill} ^{} {{a \over {\left( {\sqrt {ab} \plus{} \sqrt a \plus{} \sqrt b \plus{} 1} \right)^2 }}} \ge {3 \over {16}}
\]
4.For $ a,b,c > 0$.
Prove that
\[ {a \over {\sqrt {a^2 \plus{} 2bc} }} \plus{} {b \over {\sqrt {b^2 \plus{} 2ca} }} \plus{} {c \over {\sqrt {c^2 \plus{} 2ab} }} \le {{a \plus{} b \plus{} c} \over {\sqrt {ab \plus{} bc \plus{} ca} }}
\]

[quote="HelloWorld"]1. For $ a,b,c > 0$ and $ a \plus{} b \plus{} c \equal{} 1$.
Prove that
4.For $ a,b,c > 0$.
Prove that
\[ {a \over {\sqrt {a^2 \plus{} bc} }} \plus{} {b \over {\sqrt {b^2 \plus{} ca} }} \plus{} {c \over {\sqrt {c^2 \plus{} ab} }} \le {{a \plus{} b \plus{} c} \over {\sqrt {ab \plus{} bc \plus{} ca} }}
\]
[/quote]
It's weaker than my old resutt:
$ \sum\ \frac{a^2}{a^2\plus{}bc} \leq\ \frac{(a\plus{}b\plus{}c)^2}{2(ab\plus{}bc\plus{}ca)}$
 :)

1.By cauchy ; \[ LHS \le \sqrt{\sum_{cyc}\frac{2}{3\plus{}a}}\]
It's remain to prove that \[ \sum_{cyc}\frac{2}{3\plus{}a} \le \sum_{cyc}\frac{1}{1\plus{}2\sqrt{ab}}\]
from am-gm ; \[ \sum_{cyc}\frac{1}{1\plus{}2\sqrt{ab}} \geq \sum_{cyc}\frac{1}{1\plus{}a\plus{}b}\]
by cauchy ; $ \frac{1}{1\plus{}a\plus{}b}\plus{}\frac{1}{1\plus{}b\plus{}c} \geq \frac{4}{3\plus{}b}$
thus, \[ \sum_{cyc}\frac{1}{1\plus{}2\sqrt{ab}} \geq \sum_{cyc}\frac{1}{1\plus{}a\plus{}b} \geq \sum_{cyc}\frac{2}{3\plus{}a}\]
2. It's wrong see : http://www.mathlinks.ro/viewtopic.php?t=173131
3.Let $ a\equal{}x^2,b\equal{}y^2 and c\equal{}z^2$ so $ xyz \equal{} 1$
\[ \sum\limits_{\scriptstyle cyclic \hfill \atop
\scriptstyle a,b,c \hfill} ^{} {{a \over {\left( {\sqrt {ab} \plus{} \sqrt a \plus{} \sqrt b \plus{} 1} \right)^2 }}} \equal{} \sum_{cyc}\frac{x^2}{(x\plus{}1)^2(y\plus{}1)^2}\]
by Cauchy It's satisfy to prove that $ \sum_{cyc}\frac{x}{(x\plus{}1)(y\plus{}1)} \geq \frac{3}{4}$ which is easy.
4.see here : http://www.mathlinks.ro/viewtopic.php?t=173359

[quote="HelloWorld"]
4.For $ a,b,c > 0$.
Prove that
\[ {a \over {\sqrt {a^2 \plus{} bc} }} \plus{} {b \over {\sqrt {b^2 \plus{} ca} }} \plus{} {c \over {\sqrt {c^2 \plus{} ab} }} \le {{a \plus{} b \plus{} c} \over {\sqrt {ab \plus{} bc \plus{} ca} }}
\]
[/quote]
What if $ a\equal{}b\equal{}c$ $ ?$ :wink:

[quote="arqady"][quote="HelloWorld"]
4.For $ a,b,c > 0$.
Prove that
\[ {a \over {\sqrt {a^2 \plus{} bc} }} \plus{} {b \over {\sqrt {b^2 \plus{} ca} }} \plus{} {c \over {\sqrt {c^2 \plus{} ab} }} \le {{a \plus{} b \plus{} c} \over {\sqrt {ab \plus{} bc \plus{} ca} }}
\]
[/quote]
What if $ a \equal{} b \equal{} c$ $ ?$ :wink:[/quote]
Sorry, It is \[ {a \over {\sqrt {a^2 \plus{} 2bc} }} \plus{} {b \over {\sqrt {b^2 \plus{} 2ca} }} \plus{} {c \over {\sqrt {c^2 \plus{} 2ab} }} \le {{a \plus{} b \plus{} c} \over {\sqrt {ab \plus{} bc \plus{} ca} }}
\]

[quote="HelloWorld"]
Sorry, It is
\[ {a \over {\sqrt {a^2 \plus{} 2bc} }} \plus{} {b \over {\sqrt {b^2 \plus{} 2ca} }} \plus{} {c \over {\sqrt {c^2 \plus{} 2ab} }} \le {{a \plus{} b \plus{} c} \over {\sqrt {ab \plus{} bc \plus{} ca} }}
\]
[/quote]
It's also weaker than my old result:
$ \sum\ \frac{a}{a^2\plus{}2bc} \leq\ \frac{a\plus{}b\plus{}c}{ab\plus{}bc\plus{}ca}$
 :)

[quote="nguoivn"]
It's weaker than my old resutt:
$ \sum\ \frac {a^2}{a^2 \plus{} bc} \leq\ \frac {(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca)}$
 :)[/quote]
[quote="nguoivn"]
It's also weaker than my old result:
$ \sum\ \frac {a}{a^2 \plus{} 2bc} \leq\ \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca}$
 :)[/quote]
It kills by expanding.


What is going to be the cutoff for AMC 10B? I am wondering if it was easier or harder than AMC 10A. BECAUSE I GOT A 119.5. i'm so dang pissed off because i missed a problem due to careless mistake that was really easy!!!!!!!!!!!!!

i think it's much harder to get a perfect score on this amc 10 compare to the other past amc 10s...
i got #11 , 18, and 19 wrong!!!! all because of stupid mistakes...
i eventually tore the test apart....

*by the way, i think the cut off score is probably going to be around 118...

i voted 119+ because last time it was 110 and i think this round was easier than the A round..but about the level of last year's AMC10s..and the first one of those was 119.  Kudos for anyone who can find logic in what I just said..but my vote is 20% of the total and that sounds pretty good!

rep123max how do you get to do a AMC adn know your score and not be official

I voted 120... just because i did A WHOLE LOT BETTER on B...

personally, i thought of the B this way in comparison to the A:

the easy questions... were even easier -- up to 10 i pretty much didn't have to spend brain power to solve -- 11-15 were just counting and ez stuff -- 16-19 were not hard either

the harder questions... were even harder -- #20 was just brutal, terrible, i hate it -- #21 was another counting problem -- #22 was pretty easy, pythagorean theorem and knowledge of right triangles -- #23 wasn't hard i was stupid and counted wrong -- #24 was basically if you saw all those equal angles and wasn't too hard (though i missed it) -- #25 once again was super cool :D, a really good problem and brilliant applications of various simple techniques, all you needed to know was area of a sector/circle and area of equilateral triangles

overall, the 1% mark will be higher but i think the average score will be around the same place. i voted 115+

btw when do they release the cutoff range like how many weeks from the test date? i am so stressed.

Judging from the A test, it should be about 2 weeks...

i thought that on the outside of the amc 10 pamphlet it said 120 or 1% was the cutoff. im just happy cuz i got a perfect...kinda. i got a 132.5. 20 right 0 wrong and 5 blank and thats more than 120. #25 was easy

[quote="Leeyuhang"]sigh =( i just hope that the cutoff is 115+.
i forgot to change my andwer for #4. would've gotten 123 but now its 117 >.<.[/quote]

Wow, me too.  Did you choose 24 as well?  I'm still mad at myself at that careless mistake.  I took this as a freshman and expected to get 122, but then I got #4 wrong, ending up with 116.  Sigh.

none of u guys screwed up as bad as i did. i got a 112. answered 21, and got 4 of them wrong... the 4 easiest problems... #2, #4, #5, and #11...  :cry:

WHAT????? You should've come to my school to check out the worst score ever on AMC 10 in my region: 44 :shock: !!!!! (well, this is refering to the A contest)


Prove that if a,b,c>0 then sum ab/(a+b)^2<=4abc/((a+b)(b+c)(c+a))+1/4.

Multiplying two sides with  4(a+b)^2.(b+c)^2.(c+a)^2, expanding and collect terms wa have this inequality is equivalent to (a-b)^2.(b-c)^2.(c-a)^2 >= 0. It is interesting that by our proof, this inequality holds for all a, b, c (of course we must have (a+b)(b+c)(c+a)<>0).

Namdung

My solution was the following:
  (a/(b+a)+b/(b+c)+c/(c+a))*(b/(a+b)+c/(b+c)+a/(c+a))<=9/4 since the sum of the factors is 3. Expanding and grouping similar terms you will find the inequality.

Your solution is really genial and ingenious for its semplicity.
Another mathematical gem. Congratulation.


For some reason, in my blog, I can only post one youtube video at a time. I am using IE on a Windows XP and the AoPS Skin. When I tried to post two videos, only the second video showed up, and I double-checked both urls.

I've had this problem (and posted about it) as well, so it's a known bug. Probably it's not terribly high priority, although I also would like to be able to make posts including multiple youtube videos.


Let a,b,,c,d be any real number.prove that $ \sum {{\sqrt{\frac{x}{y\plus{}z}\plus{}\frac{1}{\sqrt 2}}}^2}\geq {6}$

[quote="xiaohui"]Let a,b,,c,d be any real number.prove that $ \sum {{\sqrt {\frac {x}{y \plus{} z} \plus{} \frac {1}{\sqrt 2}}}^2}\geq {6}$[/quote]
Can you explain your problem

[quote="Allnames"][quote="xiaohui"]Let a,b,,c,d be any real number.prove that $ \sum {{\sqrt {\frac {x}{y + z} + \frac {1}{\sqrt 2}}}^2}\geq {6}$[/quote]
Can you explain your problem[/quote]
I think it is:Let x,y,z be any positive real number.prove that $ \(\{(\sum {{\sqrt {\frac {x}{y + z} + \frac {1}{\sqrt 2}}}})}^2\geq {6}$


inequalities
trigonometry
geometry
inradius
perimeter
triangle inequality
inequalities unsolved

Prove that
\[ \sin(2\alpha) + \sin(2\beta) + \sin(2\gamma)\leq \sin\alpha + \sin\beta + \sin\gamma\]
if ${ \alpha + \beta + \gamma = \pi}$

[hide="Geometric Proof for alpha,beta,gamma>0"]Inscribe a triangle $ \Delta ABC$ in a circle with center $ O$ and radius $ R$ such that $ 2\alpha \equal{} \angle BOC$, $ 2\beta \equal{} \angle COA$, and $ 2\gamma \equal{} \angle AOB$.

(Note that $ 2\alpha \plus{} 2\beta \plus{} 2\gamma \equal{} 2\pi$).

Let $ A'$, $ B'$, and $ C'$ be the feet of the perpendiculars from $ O$ to $ BC$, $ CA$, $ AB$ respectively.

Then, $ \alpha \equal{} \angle BOA' \equal{} \angle COA'$, $ \beta \equal{} \angle COB' \equal{} \angle AOB'$, and $ \gamma \equal{} \angle AOC' \equal{} \angle BOC'$.

Also, let $ K$, $ s$, and $ r$ be the area, semiperimeter, and inradius of $ \Delta ABC$.

Notice that: $ K \equal{} [\Delta BOC] \plus{} [\Delta COA] \plus{} [\Delta AOB] \equal{} \dfrac{1}{2}R^2\sin(2\alpha) \plus{} \dfrac{1}{2}R^2\sin(2\beta) \plus{} \dfrac{1}{2}R^2\sin(2\gamma)$.

Thus: $ \sin(2\alpha) \plus{} \sin(2\beta) \plus{} \sin(2\gamma) \equal{} \dfrac{2K}{R^2}$.

Also, the perimeter of $ \Delta ABC$ is: $ 2s \equal{} BC \plus{} CA \plus{} AB \equal{} 2(A'C \plus{} B'A \plus{} C'B) \equal{} 2(R\sin(\alpha) \plus{} R\sin(\beta) \plus{} R\sin(\gamma))$.

Thus: $ \sin(\alpha) \plus{} \sin(\beta) \plus{} \sin(\gamma) \equal{} \dfrac{s}{R}$.

So, the inequality is equivalent to:
$ \dfrac{2K}{R^2} \le \dfrac{s}{R} \Longleftrightarrow \dfrac{2rs}{R^2} \le \dfrac{s}{R} \Longleftrightarrow 2r \le R$.
which is true by [url=http://mathworld.wolfram.com/EulersInequality.html]Euler's Triangle Inequality[/url].[/hide]

[quote="Xantos C. Guin"]$ 2\alpha \equal{} \angle BOC$, $ 2\beta \equal{} \angle COA$, and $ 2\gamma \equal{} \angle AOB$
[/quote]

Aren't you implying here that $ 2\alpha,2\beta,2\gamma\geq \frac{\pi}{2}$?

Are $ \alpha ,\ \beta ,\ \gamma \geq 0?$. If so, the inequality can be proved easily.

but here arent they supposed to be  angles   of  a triangle?

yes, they are the angles of a triangle and yes they are all positive

O.K.

Use the equality $ \sin A\plus{}\sin B\equal{}2\sin \frac{A\plus{}B}{2}\cos \frac{A\minus{}B}{2}$,

$ \sin 2\alpha \plus{} \sin 2\beta \equal{} 2\sin (\alpha \plus{} \beta)\cos (\alpha \minus{} \beta) \equal{} 2\sin (\pi \minus{} \gamma)\cos (\alpha \minus{} \beta)\leq 2\sin \gamma$, similarly we have

$ \sin 2\beta \plus{}\sin 2\gamma\leq 2\sin \alpha ,\  \sin 2\gamma \plus{}\sin 2\alpha\leq 2\sin \beta$. Adding these inequalities and divides by 2 gives the proof.

$ sin2A\plus{}sin2B\plus{}sin2C\equal{}4sinA.sinB.sinC$ and $ 4sinA.sinB.sinC \leq \frac{4{(sinA\plus{}sinB\plus{}sinC)}^3}{27} \leq sinA\plus{}sinB\plus{}sinC$  :wink:

[quote="huang"][quote="Xantos C. Guin"]$ 2\alpha \equal{} \angle BOC$, $ 2\beta \equal{} \angle COA$, and $ 2\gamma \equal{} \angle AOB$
[/quote]

Aren't you implying here that $ 2\alpha,2\beta,2\gamma\geq \frac {\pi}{2}$?[/quote]

I don't see any reason why that proof doesn't work when one or two of $ 2\alpha,2\beta,2\gamma$ is less than $ \dfrac{\pi}{2} \equal{} 90^\circ$.
(If all three are less than $ \dfrac{\pi}{2}$, then $ \alpha\plus{}\beta\plus{}\gamma \neq \pi$.)

However, if $ \alpha > \dfrac{\pi}{2}$, then:
[hide]The area is:
$ K \equal{} \minus{}[\Delta BOC]\plus{}[\Delta COA]\plus{}[\Delta AOB]$

$ K \equal{} \minus{}\frac{1}{2}R^{2}\sin(2\pi\minus{}2\alpha)\plus{}\frac{1}{2}R^{2}\sin(2\beta)\plus{}\frac{1}{2}R^{2}\sin(2\gamma)$

$ K \equal{} \frac{1}{2}R^{2}\sin(2\alpha)\plus{}\frac{1}{2}R^{2}\sin(2\beta)\plus{}\frac{1}{2}R^{2}\sin(2\gamma)$

And the perimeter is:

$ 2s \equal{} 2(A'C\plus{}B'A\plus{}C'B)$

$ 2s \equal{} 2(R\sin(\pi\minus{}\alpha)\plus{}R\sin(\beta)\plus{}R\sin(\gamma))$

$ 2s \equal{} 2(R\sin(\alpha)\plus{}R\sin(\beta)\plus{}R\sin(\gamma))$

Which are both exactly the same as before. The same result applies if $ \beta \ge \dfrac{\pi}{2}$ or $ \gamma \ge \dfrac{\pi}{2}$.

[hide="Diagram for alpha,beta,gamma<pi/2"][asy]draw(Circle((0,0),20));
pair A,B,C,D,E,F,O;
O = (0,0);
A = 20*dir(90);
B = 20*dir(-135);
C = 20*dir(-60);
D = 0.5*(B+C);
E = 0.5*(C+A);
F = 0.5*(A+B);
dot(O); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F);
label("O",O,S);
label("A",A,N);
label("B",B,SW);
label("C",C,SE);
label("A' ",D,S);
label("B' ",E,NE);
label("C' ",F,NW);
draw(A--B--C--A);
draw(O--A); draw(O--B); draw(O--C); draw(O--D); draw(O--E); draw(O--F);
label("$\alpha$",anglemark(B,O,D),4*dir(-116));
label("$\alpha$",anglemark(C,O,D),4*dir(-79));
label("$\beta$",anglemark(C,O,E),dir(-22));
label("$\beta$",anglemark(A,O,E),dir(52));
label("$\gamma$",anglemark(A,O,F),1.5*dir(124));
label("$\gamma$",anglemark(B,O,F),2*dir(-165));[/asy][/hide]

[hide="Diagram for alpha>pi/2"][asy]draw(Circle((0,0),20));
pair A,B,C,D,E,F,O;
O = (0,0);
A = 20*dir(90);
B = 20*dir(150);
C = 20*dir(10);
D = 0.5*(B+C);
E = 0.5*(C+A);
F = 0.5*(A+B);
dot(O); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F);
label("O",O,S);
label("A",A,N);
label("B",B,NW);
label("C",C,NE);
label("A' ",D,S);
label("B' ",E,NE);
label("C' ",F,NW);
draw(A--B--C--A);
draw(O--20*dir(260));
draw(O--A); draw(O--B); draw(O--C); draw(O--D); draw(O--E); draw(O--F);
label("$\alpha$",anglemark(B,O,D),2.5*dir(205));
label("$\alpha$",anglemark(C,O,D),2.5*dir(-45));
label("$\beta$",anglemark(C,O,E),2.5*dir(30));
label("$\beta$",anglemark(A,O,E),2.5*dir(70));
label("$\gamma$",anglemark(A,O,F),4*dir(105));
label("$\gamma$",anglemark(B,O,F),4*dir(135));[/asy][/hide]

[/hide]


let $ a,b,c,k\in Z$, the system of equation

$ abc\minus{}(a\plus{}b\plus{}c)\equal{}k$
$ ab\plus{}bc\plus{}ca\minus{}1\equal{}k$

does the number of solutions (a,b,c,k) is finite? If it's true finite, find out all the solutions.

If it's infinite, please prove it. :maybe:

[quote="tobeno_1"]let $ a,b,c,k\in Z$, the system of equation

$ abc \minus{} (a \plus{} b \plus{} c) \equal{} k$
$ ab \plus{} bc \plus{} ca \minus{} 1 \equal{} k$

does the number of solutions (a,b,c,k) is finite? If it's true finite, find out all the solutions.[/quote]
Why do you introduce $ k$?
You could have written
[quote]let $ a,b,c\in Z$, the equation

$ abc \minus{} (a \plus{} b \plus{} c) \equal{} ab \plus{} bc \plus{} ca \minus{} 1$

does the number of solutions $ (a,b,c)$ is finite? If it's true finite, find out all the solutions.[/quote]
It factors into
$ (a\minus{}1)(b\minus{}1)(c\minus{}1)\equal{}2(a\plus{}b\plus{}c\minus{}1)$
Replacing $ a\minus{}1\to a',b\minus{}1 \to b',c\minus{}1 \to c'$
$ a'b'c'\equal{}2(a'\plus{}b'\plus{}c'\plus{}2)$
$ (a',b',c')\equal{}(0,k,\minus{}k\minus{}2)$ are some of the solutions. Thus number of triples $ (a,b,c)$ is infinite.

If we exclude these, the number of solutions is finite.
[hide="To find them"]We can make $ a',b',c'$ positive by considering the equations
$ a'b'c'\equal{}2(a'\pm(b'\plus{}c')\plus{}2)$ or $ a'b'c'\equal{}2(a'\plus{}b'\plus{}\pm(c'\minus{}2))$ and permutations in $ (a,b,c)$
each for different permutations of signs.
In all cases $ a'b'c'\geq 2(a'\plus{}b'\plus{}c'\plus{}2)$
WLOG $ a'\geq b'\geq c'$
For $ (a',b',c')\equal{}(3,3,3)$, $ LHS>RHS$.Also if we add $ k$ to $ a$, $ LHS$ is increased by at least $ 9k$ while $ RHS$ by $ 2k$. So $ c'\leq 2$
Now we need to procede like this and consider all cases to find some finite number of solutions like $ (a,b,c)\equal{}(7,3,3)$[/hide]


http://higgs.phys.washington.edu/phy121b/HW/light%20speed%20changes.pdf

Above is what I need to do. So essentially, I have to show that in the new reference frame, that the new magnetic force will counter the old electric force. So then I have to use the equation of attraction due to a wire, and then some lorentz contraction. buit then i think i'm lost...Thanks!

[b]Problem:[/b] In reference frame S, a wire carrying current I and charge +q are at rest. The electrons in the wire move with the drift velocity $v_d$. In reference frame S assume the wire is electrically neutral -so there is no electrical force on the charge ($\overrightarrow F_E = 0$). Further, since the charge is at rest, there is no magnetic force on the charge, $\overrightarrow F_B = q \overrightarrow v \times \overrightarrow B = 0$. Now, consider reference frame, where electrons are at rest. There is still current and now the charge is moving - meaning $\overrightarrow F_B$ is no longer zero. Show that the electrical force in S' exactly counters the magnetic force, so there is still no net force on the charge.

By symmetry, the electric field of an infinite wire charged with uniform linear charge density $\lambda = \rho \cdot S_w$ ($S_w$ is the wire cross-section area and $\rho$ charge density) is everywhere perpendicular to cylindrical surface with the wire as its rotational axis. Its magnitude depends only on the distance r from the wire. Imagine a cylinder surface S of radius r and height x, the wire being its rotational axis. By Gauss' law,

$\int_S \overrightarrow E \cdot \text d\vec S = 2 \pi r x E(r) = \frac{\lambda x}{\epsilon_0}$

$E(r) = \frac{\lambda}{2 \pi \epsilon_0 r}$

By symmetry, the magnetic field of an infinite wire carrying current $I$ is everywhere tangent to a circle centered on the wire and in a plane perpendicular to the wire. Its magnitude depends only on the distance r from the wire.  Imagine a circular loop L of radius r centered on the wire in a plane perpendicular to the wire. By Ampere's law,

$\int_L \overrightarrow B \cdot \text d\vec l = 2 \pi r B(r) = \mu_0 I$

$B(r) = \frac{\mu_0 I}{2 \pi r}$

In the reference system S, put the x-axis along the wire in the direction of the electron drift velocity and the xy-plane through the positive charge +q at rest. By the problem condition, the wire is electrically neutral, hence, the linear densities of the positive and negative charges are equal in magnuitude, so that the total linear charge density is zero: $\lambda = \lambda_+ - \lambda_- = 0$. By convention, the current $I = \lambda_- v_d$ flows in the direction opposite to the electron drift velocity. Since the negative charge is moving with the drift velocity $v_d$ in S, the negative charge density $\lambda_- = \frac{\triangle q_-}{\triangle x}$ in S has greater magnitude than the negative charge density $\lambda'_- = \frac{\triangle q_-}{\triangle x'}$ in S' because of Lorentz contraction of $\triangle x = \triangle x' \sqrt{1 - \beta^2}$, so that $\lambda_- = \gamma \lambda'_-$ (where $\gamma = \frac{1}{\sqrt{1 - \beta^2}},\ \beta = \frac{v_d}{c}$). Likewise, since the positive charge is moving with the drift velocity $-v_d$ in S', the positive charge density $\lambda'_+ = \frac{\triangle q_+}{\triangle x'}$ in S' has greater magnitude than the positive charge density $\lambda_+ = \frac{\triangle q_+}{\triangle x}$ in S because of Lorentz contraction of $\triangle x' = \triangle x \sqrt{1 - \beta^2}$, so that $\lambda'_+ = \gamma \lambda_+$.  The total charge density in S' is

$\lambda' = \lambda'_+ - \lambda'_- = \gamma \lambda_+ - \frac{\lambda_-}{\gamma} = \gamma \lambda_- \left(1 - \frac{1}{\gamma^2}\right) =$

$= \gamma \beta^2 \lambda_- = \frac{\gamma \beta^2 I}{v_d} = \frac{\gamma \beta I}{c} > 0$

Thus in S', the wire is positively charged, in addition to carrying the current

$I' = \lambda'_+ v_d = \gamma \lambda_+ v_d = \gamma \lambda_- v_d = \gamma I$

The electric field of the linear charge density $\lambda' = \frac{\gamma \beta I}{c}$ in S' at the positive charge +q has only a positive y-component

$E'_y = \frac{\lambda'}{2 \pi \epsilon_0 y} = \frac{\gamma \beta I}{2 \pi \epsilon_0 c y}$

and so does the electrical force $\overrightarrow F'_E$ on the positive charge +q in S':

$F'_{Ey} = qE'_y = \frac{\gamma \beta I q}{2 \pi \epsilon_0 c y}$

i.e., this force is repulsive from the wire. The velocity of the moving positive charge +q in S' has only a negative x-component $v'_x = -v_d$. The magnetic field of the current $I' = \gamma I$ in S' at the positive charge +q has only negative z-component (by the right-hand rule):

$B'_z = -\frac{\mu_0I'}{2 \pi y} = -\frac{\gamma I}{2 \pi \epsilon_0 c^2 y}$

where $c = \frac{1}{\sqrt{\epsilon_0 \mu_0}}$ was used.

Thus the magnetic force $\overrightarrow F'_B = q \overrightarrow v' \times \overrightarrow B'$ on the positive charge +q in S' has only a negative y-component

$F'_{By} = -qv'_xB'_z = -\frac{\gamma I q v_d}{2 \pi \epsilon_0 c^2 y} = -\frac{\gamma \beta I q}{2 \pi \epsilon_0 c y} = -F'_{Ey}$

i.e., this force is attractive to the wire and it exactly cancels the repulsive electrical force. Note that no approximation was necessary (such as $\sqrt{1 - \beta^2} \approx 1 - \frac{\beta^2}{2}$ for small $\beta$ suggested in your link).


Suppose we call $S$ as the set containing all fibbonanci number as its elements .

$S=\{1,2,3,5,8,13,21,34,55,\cdots F_n\}$ where $F_n$ is the $n^{th}$ term of Fibbonanci sequence start with $F_1=1$ , $F_2=2$ .

Now , for any positive integer $k\leq F_n$ , if $k$ can be represent of sum of two distinct elements in $S$ , we call the number  "[i]lucky[/i]" . Prove that every [i]lucky[/i] number is unique i.e it cannot be formed by more than one type of sum of two distinct elements in $S$ .

[quote="seamusoboyle"]I think you can prove that every natural number is the sum of some amount fibbonachi numbers, and that this is unique... which is cool[/quote]

How ?  :P

[quote="ZetaX"][quote="seamusoboyle"]I think you can prove that every natural number is the sum of some amount fibbonachi numbers, and that this is unique... which is cool[/quote]
Correctly it should be: Every positive integer can unique be represented as a sum of distinct Fibonacci numbers, such that no two consecutive Fibonacci numbers are used.[/quote]

sum of distinct Fibbonanci Numbers means can be more than 2 ? By the way , if no two consecutive Fibbonanci numbers are used , then how should you form positive integer which is also a fibbonanci numbers i.e how you represent the number $8$ ?

[quote="ZetaX"]$8$ can be represented as a sum of one $8$ and nothing more ;)
And yes, there can be more than two.[/quote]

Oh , I thought that each positive integer can be written as sum of two or more distinct Fibonanci number . Guess I misunderstand it  :blush:

But is there a prove for that ?

Its trivial. Take some number k. Subtract the largest Fibonacci number that is less than k. Continue subtracting the largest possible one from the remainder until you get to 0 (which you will since 1 is a Fibonacci number). $F_{n+1} < 2*F_n$, so you never have two of the same. $F_{n+2} = F_{n+1} + F_{n}$, so you can never have two consecutive ones either (you've have been able to subtract $F_{n+2}$ instead).

To prove uniquness, assume it works for all numbers up to k. Then, the largest fibbanochi number less than k [b]must[/b] be used (Why?) So taking that away, were left with a number, which by induction, also has a unique sum.

Prime numbers are the basis of multiplication,
Fibonnachi numbers are the basis of addition.

Kinda co-incidental?


hello, let us consider the system a) By substituting $ y \equal{} \frac {1 \minus{} 2x^2}{x}$ for $ x \not \equal{} 0$ in the second equation and factorize this we get:$ (2x^2 \minus{} 4x \plus{} 5)\cdot(2x^2 \plus{} 2x \minus{} 1) \equal{} 0$. From here we have our solutions:
$ x_1 \equal{} \minus{} \frac {1}{2} \plus{} \frac {1}{2}\sqrt {3}, y_1 \equal{} 2 \newline$
$ x_2 \equal{} \minus{} \frac {1}{2} \minus{} \frac {1}{2}\sqrt {3}, y_1 \equal{} 2 \newline$
$ x_3 \equal{} 1 \plus{} \frac {1}{2}\sqrt {6}i, y_2 \equal{} \minus{} \frac {8}{5} \minus{} \frac {6}{5}\sqrt {6}i \newline$
$ x_4 \equal{} 1 \minus{} \frac {1}{2}\sqrt {6}i, y_2 \equal{} \minus{} \frac {8}{5} \plus{} \frac {6}{5}\sqrt {6}i \newline$
Sonnhard.


[quote="Astronum"]Show that for all integers $n$ the number $n^{3}-n$ always has 3 as a factor. (Consider the three cases: $n = 3k$, $n = 3k+1$, $n = 3k+2$)

 :lol:[/quote]
[hide] $n^{3}-n=n(n^{2}-1)=(n-1)n(n+1)$.
If $n\equiv0\pmod3$, $(n-1)n(n+1)\equiv-1\cdot0\cdot1=0\pmod3$.
If $n\equiv1\pmod3$, $(n-1)n(n+1)\equiv0\cdot1\cdot2=0\pmod3$.
If $n\equiv2\pmod3$, $(n-1)n(n+1)\equiv1\cdot2\cdot3=6\equiv0\pmod3$.
Therefore, $n^{3}-n$ always is divisible by 3. [/hide]


ABC is a triangle; D is a point on AC uch that AB=CD=1. If angleABD is 90 degrees and angle DBC is 30 degrees, what is AD?

[quote="pda"]ABC is a triangle; D is a point on AC uch that AB=CD=1. If angle ABD is 90 degrees and angle DBC is 30 degrees, what is AD?[/quote]

I am quoting a standard result here:
In a Triangle ABC, if D is any point on BC such that BD:DC=m:n and angles BAD, DAC and CDA are given by $ \alpha ,\,\,\beta ,\,\,\theta$ respectively, then $ (m+n)\cot \theta = m\cot \alpha-n\cot \beta$

Let angle BDA=$ \theta$
Plugging the values suitably for the given data, we get $ \cot \theta =\frac{\sqrt{3}}{x+1}$
So, from the triangle BDA, $ 1+\frac{3}{(x+1)^{2}}=x^{2}$
Solving for x, we have $ x=\sqrt[3]{2}$


conics
parabola
geometry
analytic geometry
probability
parallelogram
trigonometry

Just announcing that I'll post the COP Round 1B soon. So, those who didn't qualify on the A, you still have chance!! :)

When I compile problems (hopefully today.. I don't know..), I'll post a post to remind you guys. Thanks.

Sorry for the delay on the problem. I need two more but I haven't got time to find the "right" problems.  :)

I'll post very soon.

Here is the COP 5th Round 1B Problems.

Please note that these are due next Saturday which is 9/24/05. COPSD will be available soon as possible as well.

[b]Problems[/b]

1. If $a,b,c$ are positive real numbers, find the minimum value of:

\[\frac{(\sqrt{ab})^3+(\sqrt{bc})^3+(\sqrt{ac})^3}{abc}\]

2. A parabola with equation $y = x^2$ is drawn on a real plane. Line $l$ is constructed such that it passes through (2,2) and also touches the parabola at its vertex. Line $m$ with x-intercept (12,0) is drawn so that it is perpendicular to the line $l$ and passes through the parabola at $P$. If a triangle is constructed using (2,2), $P$, and vertex of parabola, find the area of this triangle.

3. In a game called "COP," your goal is to choose 3 points that will form a triangle. There are 16 lattice points and each x-coordinate and y-coordinate satisfy $0 \leq x \leq 3, 0 \leq y \leq 3$. If the probability that you will win is $\frac{p}{q}$ where $p$ and $q$ are relatively prime integers, find $q-p$.

4. In parallelogram $ABCD$, $AB || CD$ and $AD || BC$. Diagonal $BD$ and segment $AE$, where $E$ is a point on $BC$, intersect at $G$, inside of parallelogram $ABCD$. When $AE$ and $CD$ are extended, they meet at $F$. If $AG = 8$ and $GE = 6$, find $m+n$ given that $EF = \frac{m}{n}$.

5. Find $m+n$ in following systems of equations:

\[1+\sum^{89}_{i=1} \frac{\sin i}{\cos i} = \sum^{m}_{j=1} [\sin (j) \cos (j)]^n\]

Good luck! You only need to send me answers.

[quote="Silverfalcon"]...choose 3 points that will form a triangle...[/quote]

...randomly choose 3 distinct points that will form a triangle...

right? :)

Yes.

Randomly because otherwise, you have 100% of winning.  :D

Um, Silverfalcon, I think COP should be moved to Intermediate. ;) These don't look like getting started problems anymore. :noo:

In 4, $\frac{m}{n}$ - $m$ and $n$ are relatively prime?

How many do you need to get to qualify? :?

You need to get all 5 right. :(

I agree, the COPSD is about Getting Started level.

He said that you only need the answers.  If you had to write out proofs/thorough solutions, it would be intermediate, i think...
But yes, perhaps you're right...seems a bit hard for getting started.  But then again, it is a contest...

[quote="K81o7"]I don't understand #2.  How can the line pass through (2,2) AND be tangent to the parabola at its vertex?[/quote]

Why can't this be possible?

Say this line is $l$. Line $l$ basically passes through (0,0) (vertex) and (2,2). Note that (2,2) is NOT in the parabola.

Remember, $y = x^2$ for $x = 2$ gives $y = 4$.  :D

[quote="solafidefarms"]I agree. It isn't really getting started anymore; that's what COPSD is now.....[/quote]

It is definitely harder and maybe Intermediate. But again, this is a contest for high schoolers in all countries (there is no restriction but most are from U.S... I think that's just because since AoPS is U.S. math forum) and I tend to make it challenging.

Since many people commented, I'll give hints to the problems. Numbers do not necessarily correspond to the problem number!

1. Perpendicular slopes multiply to -1; you can find the two lines (not necessarily linear)' intersection by writing them as $y =$ and setting them equal.

2. The trivial inequality states that $x^2 \geq 0$. From there, we get another famous inequality called AM-GM which states:

\[\frac{a_1+a_2+a_3+ \cdots a_n}{n} \geq \sqrt[n] {a_1 a_2 a_3 \cdots a_n}\]

3. For any angle $\theta$, we have $\sin^2 \theta + \cos^2 \theta = 1$.

4. You can prove triangles are similar through AA and use of parallel lines.

5. Triangle Inequailty states that for triangle $ABC$ with $AC > AB,BC$, we have:

\[AB+BC > AC\]

So if $AB + BC \leq AC$, we CANNOT form a triangle.

[quote="dakyru"][quote="Silverfalcon"][quote="K81o7"]I don't understand #2.  How can the line pass through (2,2) AND be tangent to the parabola at its vertex?[/quote]

Why can't this be possible?

Say this line is $l$. Line $l$ basically passes through (0,0) (vertex) and (2,2). Note that (2,2) is NOT in the parabola.

Remember, $y = x^2$ for $x = 2$ gives $y = 4$.  :D
[/quote]

But you said tangent.  The line $y = x$ is not tangent to $y = x^2$ at (0,0).[/quote]

How about intersect then?

I thought term "tangent" means both line $l$ and $y = x^2$ contain (0,0).

Notice that $y=x^2$ and $y=x$ intersect at (0,0) and (1,1).

A line is tangent only to a circle if they meant once and only once.
[quote="Mathworld"]
The word "tangent" also has an important related meaning as a line or plane which touches a given curve or solid at a single point. [/quote]

[b]Single[/b] point.

Yeah that's what my question was.  So then the line $l$ is NOT tangent to the parabola?  Otherwise it would not be tangent at the vertex, because that would require a slope of 0, creating a contradiction.

Also note that in this case, the perpendicular to $l$ with an x-intercept of (12,0) intersects the parabola in two points.  Which do we use?  The greater one?

Shoot, I forgot (1,1).  :mad:

And as for the intersection, $P$ is the right side intersection.

[quote="shobber"][quote="Silverfalcon"]I changed #5 bit. I remove the second equation. Other things are same.[/quote]
So the $1~~89$ are in degrees now?[/quote]

Yes. It's in degrees.

Just for $\LaTeX$ tips, your code [1~~89] doesn't show up with ~ in it.

Please keep in mind that all answers must be turned to ME by PM today (11:59:59).

Thank you and good luck!

I should have just done A :roll: these ones are harder and i'm too lazy to do them :(

I'm going to extend the dealine for COP Round 1B problems for two reasons.

First, I only received answer from one people. I know this is hard contest now but I was very sad to see that only one person "tried." I'm going to wait another week (so the new deadline is 10/3/05) to hopefully see more people trying.

Second, I'm very busy now and this contest is actually making the situation worse. But, I like doing this contest and to make myself little bit more "not-so-busy," I decided to extend the deadline.

Thanks, Silvefalcon.

Answers are due tomorrow..

I haven't received ANY.  :?

Hopefully more support will come tomorrow..

I'll post the solutions soon.

By the way, someone unstickied About COP (Contest of Opportunity) thread. I don't know who did it but don't do it. That's my favorite thread and I want to be stickied as long as contest goes on (and it is going on).


geometry
trigonometry
algebra
difference of squares
special factorizations

In a triangle with sides of lengths $a,b,$ and $c,$ $(a+b+c)(a+b-c) = 3ab.$ The measure of the angle opposite the side length $c$ is

$\displaystyle \text{(A)} \ 15^\circ \qquad \text{(B)} \ 30^\circ \qquad \text{(C)} \ 45^\circ \qquad \text{(D)} \ 60^\circ \qquad \text{(E)} \ 150^\circ$

[hide="solution"]
Answer=(D) $60^\circ$
First, expand the given equ. Let $A=a+b$. Then our given equ. becomes $(A+c)(A-c)=A^2-c^2=a^2-ab+b^2-c^2=0$.
Thus, $c^2=a^2+b^2-ab$. This hints the law of cosine;
$c^2=a^2+b^2-2ab\cos C=a^2+b^2-ab$.
Simplifying, we get: $\cos x = \frac{1}{2}$.
Thus, $x=60^\circ$.  :)

[/hide]

[hide]From difference of squares, $(a+b+c)(a+b-c)=(a+b)^2-c^2=a^2+b^2-c^2+2ab=3ab$.
Rearranging, we get $c^2=a^2+b^2-ab$, which equals $a^2+b^2-2ab\cos{C}$ from Law of Cosines.
Thus, $\cos{C}=1/2 \implies C=60^\circ$. $\boxed{D}$.[/hide]


trigonometry
inequalities
inequalities unsolved

[quote="thanhak5"]For the traingle ABC
 Prove that
       $ 2m_c.c \le m_b.a \plus{} m_a.b$[/quote]
Let $ ABCD$ quadrangle with sides ( $ a,b,c,d$ ) and diagonals ($ p,q$) then true : $ pq \le ac\plus{}bd$;
[url=http://mathworld.wolfram.com/BretschneidersFormula.html]Bretshneider's formula[/url] :)

Sorry . I don't find solution for the problem . Can you post your solution for me

He means another Bretschneider's formula, this:
If $ a,b,c,d$ -sides of quadrilateral $ ABCD$, $ p,q$ - its diagonals, and $ \alpha\equal{}\angle ABC\plus{}\angle CDA$, then :
\[ p^2 q^2\equal{}a^2 c^2\plus{}b^2 d^2 \minus{}2abcd\cos{\alpha}\]
from which is followed (Ptolemy's?)inequality $ |ac\minus{}bd|\le pq\le ac\plus{}bd$

Simple proof of the Ptolemy's inequality:
[hide]
Let us consider quadrilateral $ ABCD$, and take such point $ C'$, that triangle $ AC'B$ is similar to $ ACD$ and have the same orientation. Then if $ AB\equal{}a,BC\equal{}b,CD\equal{}c,DA\equal{}d,AC\equal{}p,BD\equal{}q$, we have that $ AC'\equal{}p\frac{a}{d}, BC'\equal{}c\frac{a}{d}$. Note that triangle $ C'AC$ is similar to $ BAD$, so $ C'C\equal{}q\frac{p}{d}$. Then:
\[ C'C\le C'B\plus{}BC \Rightarrow q\frac{p}{d}\le c\frac{a}{d}\plus{}b \Rightarrow pq \le ac\plus{}bd\]
One more step and we'll prove the Bretschneider's formula :) [/hide]

It is Ptolemy . Can you use it to slove my problem

[hide]Consider quadrilateral $ CA'MB'$[/hide]

$ p^2 q^2\equal{}a^2 c^2\plus{}b^2 d^2 \minus{}2abcd\cos{\alpha}$
 $ m_a$


For some reason, most of the topics from the forums will load completely, but then they will just keep reloading indefinitely, which is really annoying since I have really slow internet access.  Is there any way to stop this?  I use a mac and safari.

I can't reproduce this myself, so I may be wrong here.

Safari does tend to try to reload pages that failed the first time. If it's doing that for pages that loaded properly, there's a bug somewhere.
I've always used Firefox for AOPS, and it seems to handle poor connections reasonably well. That might get around your problem.

It is reloading properly loaded pages, but I just reinstalled the whole system a few days ago, so I don't know if there could be a bug. :huh:  It's possible I guess, but I'll try Firefox.


Ring Theory
superior algebra
superior algebra unsolved

$ R$ is a commutative ring with $ R \neq 0$. Let $ p \in Spec(R)$. Show that $ R_p$ is a local ring with maximal ideal $ pR_p$.

Anything helpful is appreciated.

What does $ p \in Spec(R)$ mean? I don't know the Spec part :(

[url=http://www.mathlinks.ro/viewtopic.php?t=308813]Same problem?[/url]

$ Spec(R)$ means the space of prime ideals of $ R$ and yes it is the same problem.


Mathematicians have yet to find an exact calculation for pi.  But if you take a circle and measure the circumference with a piece of string, and divide that by the diameter, you will get an approximation of pi.

Another cool estimate is demonstrated by Buffon's needle...[url=http://www.mste.uiuc.edu/reese/buffon/bufjava.html]LINK[/url].

[quote="mysmartmouth"]Mathematicians have yet to find an exact calculation for pi. [/quote]

It would help if it wasn't irrational...

however, they have found certain series which are proven to converge on pi


vector
calculus
function
analytic geometry
parameterization
calculus computations

sorry this is part of my calculus homework, but I can't seem to find anything in my previous lessons..

[b]Make a sketch of the function defined by the vector $ v(t) \equal{} \langle3t \minus{} 1,t^2\rangle$ for $ t\in[0,10]$. Be sure to indicate the coordinates of the beginning and ending points as well as the direction of the graph.[/b]

I'm being homeschooled, so I kinda took a break from school while preparing for the AMC/AIME :P

in those two weeks I managed to forget what $ v(t) \equal{} \langle3t \minus{} 1,t^2\rangle$ was supposed to indicate. I have a feeling it's the $ \langle i,j \rangle$ thing, but how would I sketch a graph out of it?

Thanks in advance!

$ v(t)$ is velocity, so you would find the dislocation : $ r(t)\equal{}(x,\ y)$.

Those parameters represent, respectively, the X and Y values of the vector as it progresses with T.
You could look at the graph of $ 3t \minus{} 1$ as being the X values the vector takes on (from $ 0 \le t \le 10$) This means that your X-vector values range from -1 to 29. The Y values use the graph $ t^2$ and so range from 0 to 100. Try looking at the values for t = 1, 2, 3, 4 etc.. and interpolate from there. Make sure you get the direction correct as well :)

ahh.. I think I may remember now.. parametric graphs?

Thanks!! :lol:

Yes, parametric graphs.  Make your axes, and pick values of $ t$.  You then plot your points wherever the function tells you to.

mhmm.

$ \vec{r}(t)\equal{}x(t)\vec{\imath}\plus{}y(t)\vec{\jmath}\plus{}z(t)\vec{k}$

$ z(t)$ isn't always applicable though. A good refresher might be to recall that the length of the velocity vector for some value of $ t$ is given by

$ v(t)\equal{}x^{\prime}(t)\vec{\imath}\plus{}y^{\prime}(t)\vec{\jmath}\Longrightarrow \|\vec{v_T}\|\equal{}\sqrt{[x^{\prime}(t)]^2\plus{}[y^{\prime}(t)]^2}$

And so on.  :)


linear algebra
matrix
algebra
polynomial

Given matrix $ A\equal{}\left( \begin{array}{cc}\minus{}1&0\\k&\minus{}2\end{array}\right)$,where $ k$ is any integer.Show that $ A^2\plus{}3A\plus{}2I\equal{}0$ and use the result to find the two matrices $ B$ and $ C$ such that $ A\equal{}B^3\plus{}C^3$.

For the first part:
The charateristic polynomial is given by $ p(\lambda)\equal{}\det(A\minus{}\lambda I)\equal{}\det\left( \begin{array}{cc}\minus{}(1\plus{}\lambda)&0\\k&\minus{}(2\plus{}\lambda)\end{array}\right)$
$ \equal{}(1\plus{}\lambda)(2\plus{}\lambda)\equal{}\lambda^2\plus{}3\lambda\plus{}2$
Cayley-Hamilton now stated that A satisfies $ p(A)\equal{}0$ (with zero the identity matrix), so:
$ A^2\plus{}3A\plus{}2I\equal{}0$

$ A^{2} \plus{} 3A \plus{} 2I \equal{} 0$

implies

$ (A^{2} \plus{} 3A \plus{} 2I)(A \minus{} I) \equal{} 0.$

Hence,

$ 0 \equal{} (A^{2} \plus{} 3A \plus{} 2I)(A \minus{} I) \equal{} (A \plus{} 2I)(A \plus{} I)(A \minus{} I) \equal{} (A \plus{} 2)(A^{2} \minus{} I) \equal{} A^{3} \minus{} A \plus{} 2A^{2} \minus{} 2I.$

It follows that

$ A \equal{} A^{3} \plus{} 2(A^{2} \minus{} I) \equal{} A^{3} \plus{} 2( \minus{} 3A \minus{} 2I \minus{} I) \equal{} A^{3} \minus{} 6(A \plus{} I)$

and

$ 7A \equal{} A^{3} \minus{} 6I \equal{} A^{3} \plus{} ( \minus{} 6^{\frac {1}{3}}I)^{3}$.

The 2nd part of the result is an immediate consequence of the latter equality.


quadratics
calculus
integration
algebra
algebra proposed

Find the quadratic equation with integral coefficients whose roots are the fifth powers of the roots of $ 3x^2\minus{}5x\plus{}7\equal{}0$.


algebra polynomial calculus derivative induction inequalities Gauss

First thought: [hide]Multiply by the polynomial (1 - 4z + 6z^2 - 4z^3 + z^4).[/hide]

I got to the same place that Joel did, but I did it by doing (differentiate, then multiply by x) three times on the identity

1+x+...+x^n = (1-x^(n+1))/(1-x).

Another thought: [hide]apply transformation (1+z)/(1-z), which transforms unit circle {z: |z|<1} into left half-plane {z:Re z<0}. Now we can apply Routh-Hurwitz or other criterion for stable polynomial.[/hide]

Hunches in spoiler:  [hide]Cotan must be part-Engineer.  My hunch is that stability (Control Systems Theory) is a good prospect for solving this problem.  I didn't have luck with the Strelitz Theorem / Hurwitz Polynomials.  The polynomials got hairy as the degree increased, and I was unable to generalize.  Even though I didn't have any luck, don't rule these out.

It's been over 20 years, but the Inverse Z Transform is another transform that maps inside/outside the complex unit circle to the left/right half-planes.  I'm not sure if the transform is applicable in this case, because it is ordinarily used to transform from discrete (as in Digital Signal Processing) to analog.  If the transform is applicable, I wouldn't be surprised if there was a simple and elegant solution.[/hide]

[quote="probability1.01"]I'm no advanced problem solver, but would all the roots have to be on the unit circle, since if we divide by z, we get a polynomial whose roots' product is 1 or -1 (right?).  So if we had a magnitude below 1, we would have to have its inverse (greater than 1), no?  Don't want to work on this all based on a false idea :-).[/quote]

The product of the roots would be [tex]\displaystyle\frac{\pm 1}{n^3}[/tex]. (pesky leading coefficient!)

Very good point!

I hope you guys don't mind my unhelpful comments.

I played around with determining a formula for the polynomial based on the derivitive of a geometric series trick, and it got really messy, so I have a feeling that isn't the way to go. I have been playing around with another method, but it isn't rigorous at all...

Write z = r * (cos(x) + i*sin(x)).
Then z^n = r^n * (cos(nx) + i*sin(nx))

Now, we want the real and imaginary parts of the polynomial to disappear. Seeing as cos(nx) can be written as a polynomial in cos(x) and six(nx) can be written as a polynomial in sin(x), we have another polynomial equation we want to set to zero. At this point, I played around with these polynomials at r=1, and it seems that unless r<1 (implying all of the solutions are inside the unit circle) we won't have a solution (cos(x) is bounded by +/- 1).

Any ideas?

A solution, by someone who has had complex analysis. The solution is understandable at a calculus level.

Theorem: [hide] If F(z) is a polynomial, all of the zeros of F'(z) lie in the convex hull of the zeros of F.

Proof: Let F(z) = a*(z - r_1)(z - r_2)...(z - r_n). For z /= r_i,

F'(z) = F(z)*(1/(z - r_1) + 1/(z - r_2) + ... + 1/(z - r_n)). Let y_i = z-r_i.

If z is outside the convex hull of r_1, r_2, ... , r_n, all of y_1, y_2, ... , y_n lie in an open half plane Re(y/B) > 0. Taking reciprocals, Re(B/y) > 0 for all y in that half plane, so Re(B*(1/(z - r_1) + 1/(z - r_2) + ... + 1/(z - r_n))) > 0 and F'(z) /= 0[/hide]

Problem : [hide]Let n be a positive integer.

Let P_0(z) = 1 + z + z^2 ... + z^n and

P_k(z) = z + 2^k*z^2 + ... + n^k*z^n for k >= 1.

For all k >= 0, z*d/dz(P_k(z)) = P_k+1(z).

The zeros of P_0(z) = (z^n+1 - 1)/(z - 1) all lie on the unit circle, so by the theorem above the zeros of its derivative lie inside the unit circle. The additional zero at 0 is also inside the unit circle, so all zeros of P_1(z) are inside the unit circle. Using induction on k and repeating the same reasoning, all zeros of P_k(z) are inside the unit circle.

In particular, the zeros of P_3(z) = z + 8z^2 + ... + n^3*z^n are all inside the unit circle.[/hide]

What is the statement of Rouche's theorem? I may have stated and proved it in my solution, without using the name.

If F(z) is a polynomial, all of the zeros of F'(z) lie in the convex hull of the zeros of F.

If f, g are analytic in G (any open set) and the boundary is a closed rectifiable curve homotopoic to 0 (i.e. can be continuously deformed to a point) (i.e. any closed smooth curve that can be continuously deformed to a point) Say the curve is [tex]$\gamma$[/tex]. (in this case, it's the unit circle) Also the curve cannot pass through the zeros of f, g.

Also, if |f - g| < f on [tex]$\gamma$[/tex]. In our case, for |z| =1.
Note that the inequality must be strict.

Then f, g have the same number of zeros in G. (in our case, the unit disc)

All right, I still don't have a name for the theorem I used.

Here's what you can do with the Rouche's theorem idea:

Let P(z) = a_n*z^n + ... + a_1*z + a_0.
Suppose each a_k is real, a_n > 0, and a_n   :ge:  a_n-1  :ge: ...  :ge:  a_0  :ge:  0.
Then all of the zeros of P lie in the closed unit disc.

Proof: Let Q(z) = P(z)*(z-1) = a_n*z^n - b_(n-1)*z^(n-1) - ... - b_1*z - a_0.
(For 1  :le:  k  :le:  n-1, b_k = a_(k+1) - a_k).
Then, for |z| > 1, |a_n*z^n|  =  |z|*|a_n*z^(n-1)| = |z|*|b_(n-1)*z^(n-1) + b_(n-2)*z^(n-1) +...+ b_1*z + a_0|  :ge:  |z|*(|b_(n-1)*z^(n-1)| + ... + |b_1*z| + |a_0|)  :ge: |z|*|b_(n-1)*z^(n-1) + ... + b_1*z + a_0| = |z|*|a_n*z^n - Q(z)| > |a_n*z^n - Q(z)|
Putting the beginning and ending on one line, |a_n*z^n| > |a_n*z^n - Q(z)|, so Q(z) /= 0 and P(z) /= 0 if |z| > 1.
You don't actually need Rouche's theorem, since the leading term of the polynomial dominates for all large enough z. In fact, this proof uses nothing more advanced than the Triangle Inequality.

[quote="jmerry"]All right, I still don't have a name for the theorem I used.[/quote]

Once you stated the theorem, it sounded familiar.  I hunted down the name...

"The classical Gauss-Lucas Theorem states that all the critical points (zeros of the derivative) of a nonconstant polynomial p lie in the convex hull of the zeros of p."

It appears to me that you've applied the theorem correctly to prove this challenge.  Nice!  [However, rrusczyk has a knack for finding holes in proofs of his challenge problems.  :? ]

[quote="jmerry"]Here's what you can do with the Rouche's theorem idea:

Let P(z) = a_n*z^n + ... + a_1*z + a_0.
Suppose each a_k is real, a_n > 0, and a_n   :ge:  a_n-1  :ge: ...  :ge:  a_0  :ge:  0.
Then all of the zeros of P lie in the closed unit disc.
[/quote]

This is a [b]very nice theorem[/b]. But I have trouble to follow your proof. So I take the liberty (and trouble) to rewrite a proof here in [tex]\LaTeX[/tex] so that more readers can understand the theorem and the proof.

[b]Theorem.[/b] Let [tex]P(z) = a_nz^n + ... + a_1z + a_0[/tex]. Suppose each [tex]a_k[/tex] is real, [tex]a_n > 0[/tex], and [tex]a_n   \ge  a_{n-1}\ge\cdots\ge a_0  \ge  0[/tex]. Then all of the zeros of [tex]P(z)[/tex] lie in the closed unit disc.

[b]Proof:[/b] Let [tex]Q(z) = P(z)(1-z )=(a_nz^n + \cdots+ a_1z + a_0)[/tex] [tex] - (a_nz^{n+1} + \cdots+ a_1z^2 + a_0z)[/tex] [tex]=a_0 + (a_1-a_0)z + \cdots +(a_n-a_{n-1})z^n - a_nz^{n+1}[/tex].

If [tex]|z|>1[/tex] is a zero of [tex]P(z)[/tex], then [tex]a_0 + (a_1-a_0)z + \cdots +(a_n-a_{n-1})z^n = a_nz^{n+1}[/tex].  But

[tex]\displaystyle |a_0 + (a_1-a_0)z + \cdots +(a_n-a_{n-1})z^n|[/tex]

[tex]\displaystyle \le a_0 + (a_1-a_0)|z| + \cdots +(a_n-a_{n-1})|z|^n[/tex]

[tex]\displaystyle \le \left(a_0 + (a_1-a_0) + \cdots +(a_n-a_{n-1})\right)|z|^n[/tex]

[tex]=a_n|z|^n < |a_nz^{n+1}|[/tex]. A contradiction.

Q.E.D.

[quote="koopa"]Did anyone try to first do what zabelman did [i.e. (differentiate, then multiply by x) three times on the identity :1+x+...+x^n = (1-x^(n+1))/(1-x).]

and then use Rouche's Theorem?[/quote]

[u]Refering to my above post[/u]:

In fact, we can apply Rouche's theorem here right after we defined [tex]Q(z)[/tex]. Using notations in [b]koopa[/b]'s post above. Let [tex]\gamma[/tex] be a circle of radius [tex]r>1[/tex] and let [tex]f(z) = - a_nz^{n+1}[/tex], then the above inequalities imply that

[tex]|Q(z)-f(z)|<|f(z)|[/tex] on [tex]\gamma[/tex].

Therefore all the zeros of [tex]Q(z)[/tex] lie inside the circle [tex]\gamma[/tex] by Rouche's theorem, since [tex]f(z)[/tex] has all zeros inside [tex]\gamma[/tex] and [tex]\deg{Q}=\deg{f}=n+1[/tex].

Since the radius [tex]r>1[/tex] of the circle [tex]\gamma[/tex] is [b]arbitrarily chosen[/b], the zeros of [tex]Q(z)[/tex] must lie in the closed unit disc.

This reminds me something I did long long time ago ..., probably exactly the same thing.

[quote="zabelman"]Hm, this is the second AoPS challenge that used convex hulls in its solution. How interesting![/quote]

Another interesting thing is that we multiplied [tex]P(z)[/tex] by a mysterious factor [tex](1-z)[/tex] to define [tex]Q(z)[/tex] in the above post. This has something to do with the [b]Abel[/b] Summation Formula. Family tradition? :-)

[quote="fuzzylogic"]
[b]Theorem.[/b] Let $ P(z) \equal{} a_nz^n \plus{} ... \plus{} a_1z \plus{} a_0$. Suppose each $ a_k$ is real, $ a_n > 0$, and $ a_n \ge a_{n \minus{} 1}\ge\cdots\ge a_0 \ge 0$. Then all of the zeros of $ P(z)$ lie in the closed unit disc.
[/quote]

if I would replace the inequalities by $ >$, is the following theorem true:

[b]Theorem.[/b] Let $ P(z) \equal{} a_nz^n \plus{} ... \plus{} a_1z \plus{} a_0$. Suppose $ a_n > a_{n \minus{} 1} > \cdots > a_0 > 0$.
Then all of the zeros of $ P(z)$ lie in the OPEN unit disc.

??
It seems to be true, I don't have a counterexample, but I can't prove it.


How do you do those problems when you are asked if a sports team can score 3 or 7 points at a timee then what is the greatest score that can be attaines. TY.

This is called the Chicken McNugget Theorem. Basically it says that given 2 numbers $ x$ and $ y$, the highest number that cannot be obtained is $ xy \minus{} x \minus{} y$. The proof is pretty complicated, so I wont go there.

Also, xpmath made a pretty cool algorithm you can use with more than 2 numbers, so once I find it I will edit this post.

EDIT: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=287418]xpmaths beastly generalization[/url]...scroll down a bit


calculus
integration
floor function
logarithms
limit
function
calculus computations

i asked this as a homework problem (integral calculus):  :D

show

$\sum_{k=1}^\infty\;\frac{(-1)^{k}\;\lfloor \;\log_{2}k\;\rfloor}{k}\;=\;\boxed{\gamma}$

${\sum_{k=2}^{\infty }\left(\frac{(-1)^{k}}{k}\left\lfloor \log_{2}(k)\right\rfloor \right)=\sum_{k=1}^{\infty }\left(\sum_{i=2^{k}}^{2^{k+1}-1}\left(k \frac{(-1)^{i}}{i}\right)\right)}$

it's easy to show that, and this sum converge to the Euler–Mascheroni constant.

but i dont know how to calculate it using integral calculus

Look at this:

[url]http://www.fmat.cl/index.php?showtopic=8096[/url]

(is in spanish)

Bye :wink:

(Long post, be patient)

I'm not particularly good with theorems on series' convergence (i.e. haven't learned any), so I hope someone else will prove that the series in question does converge to some $L \in \mathbb{R}$.

Write $\sum_{k=1}^\infty \frac{(-1)^{k}\lfloor \log_{2}k \rfloor}{k}$ as $\lim_{N}\sum_{k=1}^{N}\frac{(-1)^{k}\lfloor \log_{2}k \rfloor}{k}$.

Since this converges, we might as well prove that one of its "subsequences" converges (treating it as a sequence), namely:

$\sum_{k=1}^{2^{M+1}-1}\frac{(-1)^{k}\lfloor \log_{2}k \rfloor}{k}$

Grouping terms of equal $\lfloor \log_{2}k \rfloor$ together in the sum, and naming the series $L_{M}$, we get:

$L_{M}= \sum_{k=1}^{M}k \left( \sum_{n=0}^{2^{k}-1}\frac{(-1)^{n}}{2^{k}+n}\right)$

Set $A_{k}= \sum_{n=0}^{2^{k}-1}\frac{(-1)^{n}}{2^{k}+n}$ and $B_{k}= \sum_{n=0}^{2^{k}-1}\frac{1}{2^{k}+n}$ for $k \geq 0$

Notice that $A_{k}+B_{k}= 2 \sum_{n \equiv 0 (\mbox{mod}2)}^{2^{k}-2}\frac{1}{2^{k}+n}= 2 \sum_{m=0}^{2^{k-1}-1}\frac{1}{2\cdot 2^{k-1}+2\cdot m}$

That is, $A_{k}+B_{k}= \sum_{m=0}^{2^{k-1}-1}\frac{1}{2^{k-1}+m}= B_{k-1}$ so $A_{k}= B_{k-1}-B_{k}$

Getting back to $L_{M}= \sum_{k=1}^{M}k A_{k}= \sum_{k=1}^{M}k(B_{k-1}-B_{k}) = \sum_{k=0}^{M-1}B_{k}-M B_{M}= \sum_{k=1}^{2^{M}-1}\frac{1}{k}-M B_{M}$
Instead of showing that $L_{M}\rightarrow \gamma$, we will show that $(\gamma-L)_{M}\equiv (\gamma_{M}-L_{M}) \rightarrow 0$ for a $\gamma_{M}\rightarrow \gamma$, defined as: $\gamma_{M}= \sum_{k=1}^{2^{M}-1}\frac{1}{k}-\ln(2^{M}-1)$

$(\gamma-L)_{M}= MB_{M}-\ln(2^{M}-1) = M \sum_{n=2^{M}}^{2^{M+1}-1}\frac{1}{n}-\ln(2^{M}-1)$

Note that: $\int_{2^{M}}^{2^{M+1}}\frac{dx}{x}< B_{M}< \int_{2^{M}-1}^{2^{M+1}-1}\frac{dx}{x}$

So: $M \ln 2-\ln(2^{M}-1) < MB_{M}-\ln(2^{M}-1) < M \ln \frac{2^{M+1}-1}{2^{M}-1}-\ln(2^{M}-1)$

$\Leftrightarrow \ln \frac{2^{M}}{2^{M}-1}< (\gamma-L)_{M}< \ln \frac{(2^{M+1}-1)^{M}}{(2^{M}-1)^{M+1}}$

Now, both the leftmost and the rightmost expression tend to $0$ as $M \rightarrow+\infty$. This is easy to show for the leftmost expression, but for the rightmost it's way longer and I'm running out of time here (hint: consider the function $f(x) = \frac{\ln(2^{x}-1)}{x}$). If you want I can post my solution to that as well.

Thus, by the squeeze theorem, $(\gamma-L)_{M}\rightarrow 0 \Rightarrow L_{M}\rightarrow \gamma$.

The [i]rightmost expression[/i] isn't long, either. We write:
\begin{eqnarray*}\ln \frac{\left( 2^{M+1}-1 \right)^{M}}{\left( 2^{M}-1 \right)^{M+1}}&=& \ln \frac{\left( 2^{M+1}-1 \right)^{M}}{\left( 2^{M+1}\right)^{M}}+\ln \frac{\left( 2^{M}\right)^{M+1}}{\left( 2^{M}-1 \right)^{M+1}}\\ \ &=& M \Big( \ln \left( 2^{M+1}-1 \right)-\ln \left( 2^{M+1}\right) \Big)+(M+1) \Big( \ln \left( 2^{M}\right)-\ln \left( 2^{M}-1 \right) \Big) \\ \ &=&-M \cdot \frac1{\alpha_{M}}+(M+1) \cdot \frac1{\beta_{M}},\end{eqnarray*}
where $\alpha_{M}\in \left( 2^{M+1}-1, 2^{M+1}\right), \, \beta_{M}\in \left( 2^{M}-1, 2^{M}\right)$. Both terms clearly go to $0$.

Oh, nice; my solution is not much longer than that, though. I was thinking: short = one-liner.


NH3 +HCl ---> NH4Cl (s)
Is it right? how can u write net ionic equation 4 that?

how about this:
NH4Cl(aq) + NaOH(aq) → NH3(aq) + NaCl(aq) + H2O(l)

$\text{NH}_3$ does not dissociate into nitride and hydrogen ions in water; it is a molecule, and so remains intact.  The correct equation is

$\text{NH}_3+\text{H}^++\text{Cl}^-\longrightarrow\text{NH}_4\text{Cl}_{(s)}$

For the second reaction, ammonium chloride is soluble, as is sodium chloride, and sodium hydroxide is a strong base:

$\text{NH}_4^++\text{Cl}^-+\text{Na}^++\text{OH}^-\longrightarrow\text{NH}_3+\text{Na}^++\text{Cl}^-+\text{H}_2\text{O}$

$\text{NH}_4^++\text{OH}^-\longrightarrow\text{NH}_3+\text{H}_2\text{O}$


For the nth pentagonal number, you add 3n-2 to the previous pentagonal number (which can easily be proved by adding n dots on each of the right and left most sides, and the amount needed left is n-2). In this case, n is 5, so you need to add 3(5)-2=13 to 22. The 5th pentagonal number is 35.

how did you get 3n-2..... or , how did you derive it
 :) :) :lol:

[hide]You could quickly solve this problem. [geogebra]9350eaec0f9ca97357cf935f7cd50b940bc09777[/geogebra] The answer is $ 35 $[/hide]


Let a sequence {u_n} be defined by u_1=5, and the relation
u_(n+1) - u_n=3+4(n-1), n=1,2,3,...

If u_n is expressed as a polynomial in n, the algebraic sum of its coefficients is....?

Solution (in spoiler):

[hide]Whoa there, we don't need all this u_(n+1) - u_n=3+4(n-1), n=1,2,3,...  stuff do we? Because the sum of the coefficients of any polynomial p(x) is simply p(1), the sum here is u_1=5 in this case.[/hide]

SHHHH!

I was trying to see if people would get tricked!!!!


function
algebra
domain
polynomial
real analysis
calculus
calculus computations

I am weak in understanding the concept of continous and uniform continous function..particularly in identifying ,whether the function is continous or uniform continous  ..please help me how i get over this ..give any links or books ..or any thing which help me..

 I also made mistake in choosing epislon and delta...

help me...

A function is [url=http://en.wikipedia.org/wiki/Continuous_function]continuous[/url] if for each $ \varepsilon>0$, there exists a $ \delta>0$ such that whenever $ \left|x\minus{}a\right|<\delta$, then $ \left|f(x)\minus{}f(a)\right|<\varepsilon$.  That is, for each point your domain, you can find a $ \delta$, which may depend on $ \varepsilon$ and $ x$.  All polynomial functions are continuous on $ \mathbb{R}$.

A function is [url=http://en.wikipedia.org/wiki/Uniformly_continuous_function]uniformly continuous[/url] if you can find a single $ \delta$ that "works" for every $ x$ in your domain, so it doesn't depend on $ x$.  Polynomial functions are [b]not[/b] uniformly continuous on $ \mathbb{R}$ since we have to adjust our $ \delta$ depending on where we are.  However, on a compact set, any continuous function is uniformly continuous.


calculus
derivative
function
limit
calculus computations

Okay - here is part II of trying to learn derivatives. Mainly, some of the concepts. For all purposes, let's assume that $f$ and $g$ are differentiable functions defined for all real numbers.

Based on this, is it possible that $f > 0$ everywhere $f' > 0$, and $f'' < 0$ everywhere?

Also, if $f'' < 0$ everywhere and $f' <0$ everywhere, then $\lim_{x \to \infty}f(x) =-\infty$. Is this possible or not?

Finally, my last question is this. If $f'(x) > 0$ for all $x$ and $f(x) > 0$ for all $x$, then $\lim_{x \to \infty}= \infty$. Is this possible or not?

Sorry for so many questions, it's just that I'm having a hard time understanding some of the concepts behind limits and derivatives.

Okay - here is part II of trying to learn derivatives. Mainly, some of the concepts. For all purposes, let's assume that $f$ and $g$ are differentiable functions defined for all real numbers.

[quote]Based on this, is it possible that $f > 0$ everywhere $f' > 0$, and $f'' < 0$ everywhere?[/quote]

I'd say that is NOT POSSIBLE

1. an increasing function with a lower bound has a limit at $-\infty$

2. a function with a limit at $\-infty$ has a derivative with a zero limit at $\infty$

If you need help for any of these things, ask me.

Now combining this : we'd have a strictly decreasing derivative, with a zero limit at $\infty$, but it's always above zero---> contradiction

[quote]Also, if $f'' < 0$ everywhere and $f' <0$ everywhere, then $\lim_{x \to \infty}f(x) =-\infty$. Is this possible or not?[/quote]

I think that is true.  The function is strictly decreasing.  If it doesn't have a limit $-\infty$ at $\infty$, it would have a limit in $\mathbb{R}$
As I said above : this would mean that $f'$ has a zero limit, but it is decreasing and yet under zero---> contradiction

[quote]Finally, my last question is this. If $f'(x) > 0$ for all $x$ and $f(x) > 0$ for all $x$, then $\lim_{x \to \infty}= \infty$. Is this possible or not?[/quote]

No that is not true.  Consider the inverse of the tangent function :arctan

$arctan(x)+\frac{\pi}{2}$ has a strictly positive derivative and all its values are strictly positive, yet its limit at $\infty$ is not that big at all!  It's $\pi$.
[quote]Sorry for so many questions, it's just that I'm having a hard time understanding some of the concepts behind limits and derivatives.[/quote]
Your questions are nice. :P   It's weird I never really thought about these things.

Hmm, I think I understand why you said it's not possible for the first one. However, do you think you could provide an example of an increasing function with a lower bound that has a limit at $-\infty$? It's probably obvious, but I'm not sure :blush:

I understood the other two solutions fairly well. Here are two others that you might be able to help me with, equally conceptual as the others (assuming that f and g are differentiable functions defined for all reals)

1.) $f$ can satisfy: $f'' > 0$ everywhere, $f' >0$ everywhere, and $f > 0$ everywhere. Is this possible or not?

2.) $f$ and $g$ can satisfy: $f'(x) > g'(x)$ for all $x$ and $f(x) > g(x)$ for all $x$. Is this possible or not?

Thanks so much for all your help!

[quote="fanaticsm"]Hmm, I think I understand why you said it's not possible for the first one. However, do you think you could provide an example of an increasing function with a lower bound that has a limit at $-\infty$? It's probably obvious, but I'm not sure :blush: [/quote]

Yes, and they are not that weird actually : $arctan(x)$ does the trick again  :wink:

[quote]1.) $f$ can satisfy: $f'' > 0$ everywhere, $f' >0$ everywhere, and $f > 0$ everywhere. Is this possible or not?[/quote]

Definitely, take $f=e^{x}$, then $f=f'=f''=f'''\ldots$

[quote]2.) $f$ and $g$ can satisfy: $f'(x) > g'(x)$ for all $x$ and $f(x) > g(x)$ for all $x$. Is this possible or not?[/quote]
Definitely, take $f=2 e^{x}$ and $g=e^{x}$ :oops:

[quote="fredbel6"][quote]Based on this, is it possible that $f > 0$ everywhere $f' > 0$, and $f'' < 0$ everywhere?[/quote]
I'd say that is NOT POSSIBLE
1. an increasing function with a lower bound has a limit at $-\infty$
2. a function with a limit at $-\infty$ has a derivative with a zero limit at $-\infty$
If you need help for any of these things, ask me.
Now combining this : we'd have a strictly decreasing derivative, with a zero limit at $\infty$, but it's always above zero---> contradiction.[/quote]
You need to be a little more careful. (2) is not generally true. See [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=505651&highlight=derivative+sequence#505651]this post[/url].
There is it proved that if $\lim f^\prime (x)$ exists, then it's equal to $0$.
But, as you noticed, from the hypothesis we get that the limit exists and is $>0$, contradiction.

perfect_radio - can you clarify yourself? Are you saying that the original statement is true or not true?

He's saying that my statement "IT IS NOT POSSIBLE" is correct, but my proof is incorrect (too short).
And I admit it :oops:  : he's right. :)

Okay then - can you explain the correct proof please? Sorry, I'm not sure what to do/fix...