calculus
derivative
trigonometry
function
limit
real analysis
real analysis theorems

In all rate of change problems involving $\frac{d\theta}{dt}$. The unit is always rad/s. Why rad/s and not grad/s or any other unit?

The definition of a derivative shows that the units of $\frac{dy}{dx}$ are $\frac{\text{unit of }y}{\text{unit of }x}$. So the question is equivalent to: why do we measure angles in radians and not degrees or something else? One possible answer: because $(\sin x)'=\cos x$ looks better than $(\sin x)'=\frac{\pi}{180}\cos x$.

[quote="mlok"]...because $(\sin x)'=\cos x$...[/quote]

As I told my calculus class today-
this is [i]the[/i] reason we use radians at all, for anything.

Here's a homework assignment I used to give in a differential equations course. This is from 15 years ago or so, before everyone had graphing calculators - I'd probably phrase it differently now:

[quote]	Let’s consider the shock absorber system of a car. Oversimplifying (to do anything else would land us in Chapter 7, and we’re not ready for that yet!), we consider the car to be a mass on a spring, capable of vertical motion. Assume that the car has a mass of $1000 \text{ kg}$ (a “metric ton”), and that the “spring” (in truth, the sum of four springs) has a strength of $8.1\times10^{4}\ \frac{\text{N}}{\text{m}},$ and that the fluid-filled cylinder(s) in which the shock absorber moves constitutes a “dashpot” that causes a resistive force that opposes motion.. We have three choices for the viscosity of the fluid that we fill the cylinders with, giving us three possible values for the coefficient of resistance:
	i)	$c = 1.2 \times 10^{4}\frac{\text{N}\cdot\text{sec}}{\text{m}}$
	ii)	$c = 1.8 \times 10^{4}\frac{\text{N}\cdot\text{sec}}{\text{m}}$
	iii)	$c = 2.4 \times 10^{4}\frac{\text{N}\cdot\text{sec}}{\text{m}}$
Assume that at time zero, the car is in its equilibrium position and has an upward velocity of $2\frac{\text{m}}{\text{sec}}.$
a)	Solve for the position as a function of time in each case.
b)	Compute enough points in each case to be able to draw a careful graph (use graph paper or computer-assisted graphing - I want accuracy!) showing these positions for the first second. One graph with three curves on it should look the best.
c)	In which case does the car reach the greatest height? How high is that? In which case does the position return to equilibrium the “fastest”?
[/quote]
If you're too lazy to actually work the homework exercise, (i) is underdamped, (ii) is critically damped, and (iii) is overdamped. This means that the three solutions have very different algebraic forms - the overdamped case has only exponentials, the critically damped case is $t$ times an exponential, and the underdamped case contains a trigonometric function, a sine. The point I was trying to get across with this exercise is that, despite the three different algebraic forms, the solutions aren't all that different over that first second (and much beyond that second, all three have decayed to such small values that the motion may not be noticeable.) But I had a persistent difficulty with the answers students handed in for this assignment. I would keep getting graphs in which the answer to (iii) is a nice little hump that goes up and comes back down, the answer to (ii) is similar, but rises a little higher and gets down to near zero a little quicker - but the answer to (i) would look totally different, smashed flat against the horizontal axis. In fact (i), the underdamped case, is supposed to rise a little higher than the other two, and to get back down to zero fast enough that you can see it go negative.

What went wrong? Why did the graph of (i) turn out so wrong? Because the students in question had the mode on their calculators set to degrees rather than radians. If a trigonometric function (in this case, a sine) appears as the result of a calculus problem - or a differential equations problem, which counts as the same thing - then you'd better be computing in radians!

But I still don't get the answer. I asked all math teachers I know but none gave me a [color=blue]clear[/color] or a [color=blue]convincing[/color] solution. Suppose the following example:

A person is standing at point A, as shown in the figure, is watching a boat approximating towards him at a speed of 20 feet/s. If the person is 250 feet over the sea, find $\frac{d\theta}{dt}$ when the boat is at 250 feet from the person.

solution:

$tan(\theta)=\frac{x}{250}$

$\therefore sec^{2}(\theta). \frac{d\theta}{dt}=\frac{1}{250}.\frac{dx}{dt}$

$\Longleftrightarrow \frac{d\theta}{dt}=\frac{-1}{25}=-0.04$ rad/s :?:

The formula $(\tan\theta)'=\sec^{2}\theta$ is valid only when $\theta$ is measured in radians. The proof of this formula is based on the fundamental limit $\lim_{\theta\to 0}\frac{\sin \theta}{\theta}=1$. This limit being $1$ is essentially the definition of a radian.

Find a calculus textbook. There is a derivation of the derivatives of $\sin$ and $\cos$, using the fundamental limits $\lim_{x\to 0}\frac{\sin x}{x}=1$ and $\lim_{x\to 0}\frac{1-\cos x}{x}=0$. Those limits require a geometric derivation.

Yes, I agree. They do both need to be calculated, of course.

[quote="mlok"]The first limit seems more fundamental, since the second one can be derived from it: $\frac{1-\cos x}{x}=\frac{\sin^{2}x}{x(1+\cos x)}\sim \frac{\sin x}{1+\cos x}\to 0$.[/quote]

But what does this have to do with Radian??

What they have to do with radians is that the results aren't true if you treat $\theta$ as an angle in degrees.  That is, the result $\lim_{\theta \to 0}\frac{\sin\theta}{\theta}= 1$ only holds if $\sin\theta$ means the sine of $\theta$ radians.


Let $ k$ be a positive integer. Cities of a certain country are connected by air routes (each connecting two cities) served by $ 2k\plus{}1$ companies, where the first company serves one route, the second serves two, etc. According to a law in this country, a company cannot serve more than one route from the same city. Some day, the companies agreed to redistribute the routes between themselves so that each company serves the same number of routes. Prove that this can be done without violating the law.


[quote="Wickedestjr"]Why did you make this thread? This is spam.[/quote]
I put it on this forum by mistake. It was not supposed to be here.

I'm waiting for an explaination!
(And if it's at any way intended to get locked, no it won't)


Find the number of ways to spell "BIGELOW" below (starting from the top, you may only move down and to the left or down and to the right [b][i]one[/i][/b]; for example, from the first "E" in line 4, you can [b]only[/b] move to the first "L" in line 5):

Yeah, many of these block walk questions can be done with Pascal's.

Speaking of Pascal, find the number of paths to write Pascal such that it is like this

..................P
................A A
...............S S S
................C C
...............A A A
................L L

this is a little trickier.


I was looking over my old trig book and I found these questions easy trig problems for anyone to try out.  :D

1. An angle of depression found by the top of lighthouse that is 75 feet tall and ship in the water is 31 degrees. If the rope was attached directly from ship to the top of lighthouse, what is the length of this rope?

2. When an airplane leaves the runaway, its angle of climb is 18 degrees and its speed is 275 feet per second. Find the plane's altitude after two minutes.

3. While traveling across the flat land, you notice a mountain directly in front of you. The angle of elevation to the peak is 3.5 degrees. After you drive 13 miles closer to the mountain, the angle of elevation is 9 degrees. Approximate the height of the mountain.

I'll add more later~ :D  :D

But enjoy these till then! :P

Just wondering, but are these problems solved by using Sinus and Cosinus??

[quote="Silverfalcon"]Do you mean "sine" and "cosine"? :?

They're really simple right triangle problems. Just set them in drawing and use basic trig.

 :lol:[/quote]

Oh ok.

And yeah i guess thats what i meant.
Im using the danish names and am not too familiar in the english terms.

Sorry for interupting the thread.

Could you explain what a trig is.

You don't have to though, i can wait till someone comes and solves the problem.

trig is short for trigonometry which literally means the study of triangles.

Since no one else tried, I'll give my answer.

[hide="1!"]

Let x be hypotenuse of the triangle and then 75 is the opposite side of the angle 31.

$\frac {75}{x} = \sin {31}\circ$

Then it comes out that x = 146.[/hide]

[hide="2!"]

$\frac {x}{3300} = \sin {18}\circ$

And so $x = 10198$ rounded.[/hide]

[hide="3!"]

Let $d$ be the distance between two points of observation. And let $x$ be the height of a mountain.

$\tan {9}\circ = \frac {x}{d}$ and $\tan {3.5}\circ = \frac {x}{13+d}$

Set both equal to $d$ and we'll get:

$d = \frac {x}{\tan {9}\circ}$ and $d = \frac {x}{\tan {3.5}\circ} - 13$

$\frac {x}{\tan {9}\circ} = \frac {x}{\tan {3.5}\circ} - 13$ and solving it, we'll get:

$x = 1.3$ approximately.[/hide]

I hope it's right :P


This one is wrong. I guess you mean

http://www.mathlinks.ro/Forum/viewtopic.php?t=59834 .

  darij


inequalities
LaTeX
inequalities unsolved

If $ x,y,z>0$  then prove:
$ 2^9(x\plus{}y\plus{}z)^7 \ge 3^6 (x\plus{}y)(y\plus{}z)(z\plus{}x) (x^2\plus{}y^2\plus{}z^2\plus{}xy\plus{}yz\plus{}zx)$.

[quote="soruz"]If $ x,y,z > 0$  then prove:
$ 2^9(x \plus{} y \plus{} z)^7 \ge 3^6 (x \plus{} y)(y \plus{} z)(z \plus{} x) (x^2 \plus{} y^2 \plus{} z^2 \plus{} xy \plus{} yz \plus{} zx)$.[/quote]

it is wrong  $ x\equal{}y\equal{}z\equal{}\frac{1}{2^{3}}$

[quote="arqady"][quote="soruz"]If $ x,y,z > 0$  then prove:
$ 2^9(x \plus{} y \plus{} z)^7 \ge 3^6 (x \plus{} y)(y \plus{} z)(z \plus{} x) (x^2 \plus{} y^2 \plus{} z^2 \plus{} xy \plus{} yz \plus{} zx)$.[/quote]
Maybe $ 2^4(x \plus{} y \plus{} z)^5\geq3^4(x \plus{} y)(x \plus{} z)(y \plus{} z)(x^2 \plus{} y^2 \plus{} z^2 \plus{} xy \plus{} xz \plus{} yz)$ $ ?$
It's true. :wink:
But $ 2^5(x \plus{} y \plus{} z)^7 \ge 3^5 (x \plus{} y)(y \plus{} z)(z \plus{} x) (x^2 \plus{} y^2 \plus{} z^2 \plus{} xy \plus{} yz \plus{} zx)^2$ is still wrong.[/quote]
I think using the following method will work, Arquady!
Since the homogenous of this inequality, we have: x+y+z=1
Then we need to prove:2^4>=3^4(xy+yz+xz-xyz)(1-xy-yz-xz)
Then I think we come up with solution(consider that (xy+yz+xz)^2=x^2y^2+y^2z^2+z^2x^2+2xyz and xyz=<1/27,xy+yz+xz=<1/3)
Luckily, the inequality occurs when x=y=z :P
PS: Can you help me fix my LATEX in my post, Arquady?I still don't have time to learn anything about LATEX! And would you mind doing my foloowing beautiful inequality which is fixed in LATEX: :oops:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=983783#983783


quadratics
number theory proposed
number theory

Prove that the eq. $x^2y^2=z^2(z^2-x^2-y^2)$ has no solution in positive integers.

let $a=x^2+y^2$ and $b=2xy$ we have that $a^2-b^2=x^4-2x^2y^2+y^4$ a square. But looking our equation as a quadratic in $z^2$ we get that its discriminant is $a^2+b^2$ so we get $a^4-b^4$ is a square and this is well known to be imposible, (unless trivial cases, namely $x=y=0$ or $x=0$)


geometry
3D geometry
prism
Pythagorean Theorem

What is the total surface area of this right prism with triangular bases?

[asy]import three;
triple A,B,C,D,E,F,G,H,I;
currentprojection=orthographic(1/2,-1,1/2);

A=(0,0,0);
B=(3,0,0);
C=(0,0,4);

D=(0,16,0);
E=(3,16,0);
F=(0,16,4);

draw(A--B--C--cycle3);
draw(A--D, dashed);
draw(B--E);
draw(C--F);
draw(D--E,dashed);
draw(E--F);
draw(F--D,dashed);

G=.5*(A+C);
H=.5*(A+B);
I=.5*(C+F);

label("4",G,S);
label("3",H,W);
label("16",I,N);[/asy]

First, let's find the area of the two triangles on the end. This is simply $ \frac {4 \times 3}{2} \equal{} 6$, and since there are two triangles, we multiply by 2 to get $ 12$.

Now find the area of the slanty rectangle. By the Pythagorean theorem, the width is 5, so it's $ 16 \times 5 \equal{} 90$.

Finally, let's find the area of the last rectangle. This is simply $ 16 \times 4 \equal{} 64$.

Our final answer is $ 12 \plus{} 90 \plus{} 64 \equal{} \boxed{166}$.

EDIT: Oops.

The combined area of the triangular bases is $ \frac {4\times3}{2}\times2 \equal{} 12$. The area   of the lateral face is $ \sqrt {3^2 \plus{} 4^2}\times16 \equal{} 5\times16 \equal{} 80$. Then, the area of the underside is $ 3\times16 \equal{} 48$. Finally, the area of the backside is $ 4\times16 \equal{} 64$. Adding these up gives $ 12 \plus{} 80 \plus{} 48 \plus{} 64 \equal{} \boxed{204}$]

EDIT: semi-beaten :D

[quote="isabella2296"]First, let's find the area of the two triangles on the end. This is simply $ \frac {4 \times 3}{2} \equal{} 6$, and since there are two triangles, we multiply by 2 to get $ 12$.

Now find the area of the slanty rectangle. By the Pythagorean theorem, the width is 5, so it's $ 16 \times 5 \equal{} 90$.

Finally, let's find the area of the last rectangle. This is simply $ 16 \times 4 \equal{} 64$.

Our final answer is $ 12 \plus{} 90 \plus{} 64 \equal{} \boxed{166}$.[/quote]

you forgot a side?


probability
expected value
probability and stats

This is a well known result in Markov chain theory that in a symmetric random walk every state is null recurrent. And null recurrence is a class property.

What you said would be fine if the result were also true on $ \mathbb Z^{100}$. But it is fairly simple to show that it isn't (just estimate the probability that you hit the origin at the $ n$-th step and see that the series converges ;)

To elaborate, this only holds in $ \mathbb{Z}$ and $ \mathbb{Z}^2$ :)

[quote="manuel"]show that for a random walk in the plane it is plausible that the origin is visited infinitely often. Also, show that it is plausible for every point to be visited infinitely often.[/quote]

this random walk needs to be simple (the transition probabilities need to be identical)

start with the one-dimensional case, and compute $ p_{2n}$, the probability that you are back at the origin after $ 2n$ steps.  by stirling's formula this is about $ Cn^{\minus{}1/2}$, which diverges if you sum it.  this implies that you hit the origin infinitely often (why?)

now write the two-dimensional random walk as a product of one-dimensional random walks on the diagonal lattice (that is, at each step, move up/down with equal probability and then left/right with equal probability).  for this random walk $ p_{2n}$ is $ Cn^{\minus{}1}$, which still diverges if you sum it.  so you hit the origin infinitely often again.

in dimensions $ d\geq 3$ and higher $ p_{2n}$ is about $ Cn^{\minus{}d}$ which converges - this is (essentially) why the corresponding random walks are transient.

finally, if you hit the origin infinitely often, then you should hit every point infinitely often, since the probability of hitting a given point before returning to the origin is nonnegative

[quote="blahblahblah"] which diverges if you sum it.  this implies that you hit the origin infinitely often (why?)
[/quote]
Why, indeed? :? It seems like you are hiding some general assumptions that need to be checked. I mean, suppose we have an infinite graph $ G$ and some random walk on it. Suppose also that $ p_k$ is the probability that we hit the origin at the $ k$-th step and suppose that $ \sum_k p_k\equal{}\plus{}\infty$. I doubt that it always follows that the origin is hit infinitely often with probability $ 1$ then though I do not see an immediate counterexample.

you are right, i really showed that the expected number of returns is infinite, not that the random walk hits the origin infinitely often with probability 1

e: on the other hand, this [i]does[/i] imply recurrence, since if the random walk returns to $ 0$ with probability $ p$, then there is a probability $ p^{k - 1}(1 - p)$ of it being at $ 0$ a total of $ k$ times, and $ \infty = \sum kp^{k - 1}(1 - p) = (1 - p)^{ - 1}\implies p = 1$.

e2: i checked my probability text, and it gave the following

$ \textbf{Theorem.}$  For any random walk, the following are equivalent

\begin{eqnarray}
P(\tau_1<\infty) &=& 1 \\
P(S_m=0 \ \text{i.o.}) &=& 1 \\
\sum^\infty_{m=0} P(S_m=0) &=& \infty
\end{eqnarray}

with essentially the same proof as what i wrote above.  am i missing something? :(

[quote="fedja"]No, it was I who was missing something. Nice theorem! :)[/quote]

i get very nervous when i find myself disagreeing with you :blush:


algebra
polynomial
number theory
number theory theorems

Is there exit polynomial $ P(x)$ with $ \deg P\equal{}k\geq 2$ such that there exit infinite prime number $ q$ such that equation $ P(x)\equal{}q$ have integer root .

[quote="math10"]Is there exit polynomial $ P(x)$ with $ \deg P \equal{} k\geq 2$ such that there exit infinite prime number $ q$ such that equation $ P(x) \equal{} q$ have integer root .[/quote]

This is an open question in general.  As an example, for $ k\equal{}2$, the polynomial $ n^2\plus{}1$ is not known to take prime values infinitely often.  The best known is that $ n^2\plus{}1$ is a product of at most 2-primes infinitely often.  This is due to H. Iwaniec proved using sieve methods.

-evo

[quote="evoluciona"][quote="math10"]Is there exit polynomial $ P(x)$ with $ \deg P \equal{} k\geq 2$ such that there exit infinite prime number $ q$ such that equation $ P(x) \equal{} q$ have integer root .[/quote]

This is an open question in general.  As an example, for $ k \equal{} 2$, the polynomial $ n^2 \plus{} 1$ is not known to take prime values infinitely often.[/quote]
I know open problem with $ P(x)\equal{}x^2\plus{}1$ but problem requires that "Is there exit polynomial".
In year 1971,Mitijasevic had prove that have polynomial $ P(x_1,...,x_n)$ s.t set value positive integer of $ P(x_1,..,x_n)$ equal set prime number.I think that also there exit polynomail $ P(x)$ has property problem.
 [quote="evoluciona"][ The best known is that $ n^2 \plus{} 1$ is a product of at most 2-primes infinitely often.  This is due to H. Iwaniec proved using sieve methods.

-evo[/quote]
a beautiful result.Do You know a solution nice for this result?

[quote="math10"]
I know open problem with $ P(x) \equal{} x^2 \plus{} 1$ but problem requires that "Is there exit polynomial".
In year 1971,Mitijasevic had prove that have polynomial $ P(x_1,...,x_n)$ s.t set value positive integer of $ P(x_1,..,x_n)$ equal set prime number.I think that also there exit polynomail $ P(x)$ has property problem. [/quote]

Ah I see.  I was looking at
http://primes.utm.edu/glossary/xpage/MatijasevicPoly.html
So the set of positive values of $ P(x_1, ..., x_n)$ where $ x_i\in\mathbf{Z}$ is contained in the primes.  Apparently the smallest degree known is degree 5 with 42 variables proved in 1976.

[quote="math10"]
 [quote="evoluciona"][ The best known is that $ n^2 \plus{} 1$ is a product of at most 2-primes infinitely often.  This is due to H. Iwaniec proved using sieve methods.

-evo[/quote]
a beautiful result.Do You know a solution nice for this result?[/quote]

Iwaniec's solution to this problem is unfortunately not simple nor short and I think it's the simplest known.  It depends on some heavy duty machinery in sieve methods.  The first page is here
http://www.springerlink.com/content/q577674g12011317/

I would highly recommend

Unsolved problems in number theory By Richard K. Guy

for information about elementary problems - it probably has this one.

You can find it here also:
http://books.google.com/books?id=1AP2CEGxTkgC&dq=richard+guy+unsolved+problems+in+number+theory

Good luck,
-evo


To conserve the contents of a 16oz bottle of tonic, a castaway adopts the following procedure. On the first day, he drinks 1 oz from the bottle and refills it with water. On the 2nd day, he drinks 2oz from the bottle and refills it with water. On the 3rd day, he drinks 3oz and refills it with water. This continues until no tonic is left. how much water (not tonic) did he drink?

[hide]The first day he puts 1oz of water into the bottle, the second day 2oz, and so on until the 15th day when he puts in 15 oz of water. So 15(15+1)/2=120, 120oz of water.[/hide]

But the question asks how much does he [i]drink[/i]

Yes, I know. Since he drinks everything in the bottle in the very end (16th day; he drinks 16 oz, which is the whole bottle), you only have to think about how much [b]he puts in[/b] to the bottle.

but on the first day, he puts in 1 oz, but on the second day, he doesn't drink 1 oz.

:?: First day: Drinks 1 oz, puts in 1oz water
Second day: Drinks 2 oz, puts in 2 oz of water
Third day: Drinks 3 oz, puts in 3 oz of water
...
Sixteenth day: Drinks 16 oz, stops.

Since there's no liquid going out of the bottle, the only thing he drinks is the water he puts in and the liquid that was already in the bottle in the beginning (16 oz of tonic water).

Tare -- you said in your last post something about the 16 ounces tonic -- where was that tonic reflected in your calculations?  All you've made is one minor error, I believe.

It's not reflected since the problem specifically asks for the amount of water he drank.

...I still don't get where I went wrong...

Here's a different question:  [hide]how much total stuff did this guy drink?[/hide]

Tare, you have my complete apologies -- you are absolutely right.  The answer is 120.  I was making a mistake that I thought was exactly the mistake you were making.  Again, very sorry.


Let $ a_1\equal{}a_2\equal{}1$ and $ a_{n}\equal{}\frac{a_{n\minus{}1}^{2}\plus{}2}{a_{n\minus{}2}}\; \forall n\geq 3$. Prove that all $ a_n$ are integers.

I'll prove another recursive formula   :)

[quote="Yuriy Solovyov"]Let $ a_1 \equal{} a_2 \equal{} 1$ and $ a_{n} \equal{} \frac {a_{n \minus{} 1}^{2} \plus{} 2}{a_{n \minus{} 2}}\; \forall n\geq 3$. Prove that all $ a_n$ are integers.[/quote]

We will prove that $ \boxed{\boxed{a_n \equal{} 4a_{n \minus{} 1} \minus{} a_{n \minus{} 2}}} \ (1)$

It is $ a_3 \equal{} \frac {1^2 \plus{} 2}{1} \equal{} 3$, so for $ n \equal{} 3$ it holds

Se assume that $ (1)$ holds for any natural number $ \leq n$. We'll prove that it holds for $ n \plus{} 1$

So we are to prove that $ a_{n \plus{} 1} \equal{} 4a_n \minus{} a_{n \minus{} 1}$

We have: $ a_{n \plus{} 1} \equal{} \frac {a_n^{2} \plus{} 2}{a_{n \minus{} 1}} \equal{} \frac {(4a_{n \minus{} 1} \minus{} a_{n \minus{} 2})^{2} \plus{} 2}{a_{n \minus{} 1}}$

$ \equal{} \frac {16a_{n \minus{} 1}^2 \plus{} a_{n \minus{} 2}^2 \minus{} 8a_{n \minus{} 1}a_{n \minus{} 2} \plus{} 2}{a_{n \minus{} 1}}$

$ \equal{} 16a_{n \minus{} 1} \minus{} 8a_{n \minus{} 2} \plus{} \frac {a_{n \minus{} 2}^2 \plus{} 2}{a_{n \minus{} 1}}$

$ \equal{} 16a_{n \minus{} 1} \minus{} 8a_{n \minus{} 2} \plus{} \frac {a_{n \minus{} 2}^2 \plus{} 2}{a_{n \minus{} 3}} \cdot \frac {a_{n \minus{} 3}}{a_{n \minus{} 1}}$

$ \equal{} 16a_{n \minus{} 1} \minus{} 8a_{n \minus{} 2} \plus{} a_{n \minus{} 1} \cdot \frac {a_{n \minus{} 3}}{a_{n \minus{} 1}}$

$ \equal{} 16a_{n \minus{} 1} \minus{} 8a_{n \minus{} 2} \plus{} a_{n \minus{} 3}$

$ \equal{} 16a_{n \minus{} 1} \minus{} 8a_{n \minus{} 2} \plus{} (4a_{n \minus{} 2} \minus{} a_{n \minus{} 1})$

$ \equal{} 15a_{n \minus{} 1} \minus{} 4a_{n \minus{} 2}$

$ \equal{} 15a_{n \minus{} 1} \minus{} 4(4a_{n \minus{} 1} \minus{} a_n)$

$ \equal{} 4a_n \minus{} a_{n \minus{} 1}$


A triathlete competes in a triathlon in which the swimming, biking, and running segments are all of the same length. The triathlete swims at a rate of $ 3$ kilometers per hour, bikes at a rate of $ 20$ kilometers per hour, and runs at a rate of $ 10$ kilometers per hour. Which of the following is closest to the triathlete's average speed, in kilometers per hour, for the entire race?

$ \textbf{(A)}\ 3 \qquad
\textbf{(B)}\ 4 \qquad
\textbf{(C)}\ 5 \qquad
\textbf{(D)}\ 6 \qquad
\textbf{(E)}\ 7$

[hide="Solution"]Let each segment be of length $ d$.
Then, the total distance traveled is $ 3d$. The total time it takes is:
$ \frac{d}{3}\plus{}\frac{d}{20}\plus{}\frac{d}{10}\equal{}\frac{29d}{60}$
Since the average speed is distance over time, the answer is:
$ \frac{3d}{\frac{29d}{60}}\equal{}\frac{180}{29}$, which is closest to $ 6 \Rightarrow \boxed{D}$.[/hide]

Or you can use Harmonic mean: $ \frac{3abc}{ab\plus{}ac\plus{}bc}$
And you get $ \frac{3*3*20*10}{3*20\plus{}3*10*\plus{}20*10}\equal{}\frac{1800}{290}\equal{}6 \frac{6}{29}$
So answer is C

I'm sorry, but could you explain to me how to use the harmonic mean?

The original Harmonic mean eaquation is:

$ \frac{n}{\frac{1}{a_{1}}\plus{}\frac{1}{a_{2}}\plus{}...\plus{}\frac{1}{a_{n}}}$
But with little rearengments we can get $ \frac{n*a_{1}*a_{2}*...*a_{n}}{a_{1}*a_{2}\plus{}a_{2}a_{3}*...*a_{n}a_{1}}$


limit
inequalities
function
calculus
calculus computations

Hello everyone,

Could someone please comment if my proof below suffices? In my textbook, when $ \lim_{x \rightarrow 0} \frac {sin}{x} \equal{} 1$ was proved using Squeeze Theorem, the inequalities determined were solved explicitly for x. However, my proof below doesn't involve that.

Thank you for your help!

---
[img]http://img34.imageshack.us/img34/2015/ps43q3she25q46dirichlet.jpg[/img]

Perfect. More generally, for any real function $ f$, bounded in a neighborhood of zero, by a similar proof $ \lim_{x \to 0} xf(x) \equal{} 0$.

Thank you for your response, mavropnevma.

I just want to clarify one question of mine: For squeeze theorem to be applied, the inequality does not have to be strict? Is this because if the inequality is weak, then the limit of the function squeezed (I'll call it $ f$) between the functions ( $ g$ and $ h$) can be the same as the limits of $ g$ and $ h$?

For the proof that $ \lim_{x \rightarrow 0} \frac {sin}{x} \equal{} 1$, the inequalities are strict so I want to understand that strict inequalities may not be compulsory for the Squeeze Theorem to be used.

Of course not. The inequalities may be given (or computed) strict, or not - it makes no difference. The only thing to be careful about is that a strict inequality may become an equality when passing to the limit. The simplest example is $ \frac {1} {n} > 0$, but $ \lim_{n \to \infty} \frac {1} {n} \equal{} 0$.


geometry
inequalities
AMC
USA(J)MO
USAMO
graph theory

I've been thinking about this for a while, and I'm curious to hear everyone's proof (or just an outline). I really don't know what makes a graph theory problem like this "rigorous." I've been under the impression for a long time that for graph theory, you have to show that everything works, not just that it passes certain tests. You have to show that a graph [b]does[/b] exist, not that a graph [b]can[/b] exist. However, after reviewing this problem, I'm not sure this is the case. It would be nice if a higher authority in the Bay Area, or someone who has experience with this type of stuff, to post.

So to start off, this is my solution:
We can set a lower bound by doing some algebra. I said there were $ n$ teams outside CA, so there were $ \binom{n\plus{}251}{2}$ games, which is greater than or equal to $ 127(n\plus{}250) \plus{} 126$. From this, we get the bound $ n >\equal{} 4$, so we're done. However, I didn't prove that Alcatraz *can* win exactly 126 games, I just proved that this is the minimum.

[quote="Ubemaya"]I've been thinking about this for a while, and I'm curious to hear everyone's proof (or just an outline). I really don't know what makes a graph theory problem like this "rigorous." I've been under the impression for a long time that for graph theory, you have to show that everything works, not just that it passes certain tests. You have to show that a graph [b]does[/b] exist, not that a graph [b]can[/b] exist. However, after reviewing this problem, I'm not sure this is the case. It would be nice if a higher authority in the Bay Area, or someone who has experience with this type of stuff, to post.

So to start off, this is my solution:
We can set a lower bound by doing some algebra. I said there were $ n$ teams outside CA, so there were $ \binom{n \plus{} 251}{2}$ games, which is greater than or equal to $ 127(n \plus{} 250) \plus{} 126$. From this, we get the bound $ n > \equal{} 4$, so we're done. However, I didn't prove that Alcatraz *can* win exactly 126 games, I just proved that this is the minimum.[/quote]

How many total teams did you get? (my top-of-head solution said 256 was good)

[quote="Ubemaya"]The answer is actually 255.[/quote]

Naw dangit. :(
I know why though : I forgot to count that they play each other!  That allows for a few extra losses against the CA teams.

Ubemaya, I've had your exact feelings before about graph theory.  With those problems I usually end up making things way harder than they actually are, and being unsatisfied by the real solution.

How did you come up with your inequality?  Could someone please provide a walkthrough  :(

My proof that 126 was the minimum was different from Ubemaya's, but I'm sure it still works.

As for 126 being possible, imagine this scenario:
Alcatraz: 126 wins, 128 losses
Some random team: 128 wins, 126 losses
253 other teams: 127 wins, 127 losses

archimedes1, I'm so glad that I'm not the only lunatic here :D. Guess I'm not completely crazy.

As for my inequality: Let $ n$ be the number of teams outside of CA. Assume that Alcatraz won exactly 126 games. Then each other team, and there are $ n \plus{} 250$ of them, won at least 127. However, the total number of games they won is $ \binom{n\plus{}251}{2}$, because in each game, there's one winner. Thus $ 127(n \plus{} 250) \plus{} 126 \leq \binom{n\plus{}251}{2}$, and when you plug in 4, you get $ 127\cdot 255 \minus{}1 \leq 127\cdot 255$, which just barely fits the inequality. So $ 4 \plus{} 251$ is the lower bound.

i_like_pie, your solution is [b]exactly[/b] the type of thing I'm talking about. You proved that 126 was the minimum for Alcatraz to win. However, unless I'm completely misunderstanding your solution, you did not prove that it is possible to have a graph with 255 vertices with degrees 126-127-127-127-...-127-128. You just assumed that it was possible. Furthermore, you did not prove that from Alcatraz having 126 games and the rest of CA having 125/124, that it is even possible, by adding four vertices, to get to the 126-127-...-127-128 graph.

So far, I think these three things must be proven (I might be completely mistaken, but this is what my intuition tells me):
1) It is possible having a graph with 251 vertices with one vertex of degree 124, one of degree 126, and 249 of degree 125
2) It is possible having a graph with 255 vertices with one vertex of degree 126, one of degree 128, and 253 of degree 127
3) It is possible getting from a graph of the former to a graph of the latter, without changing any previously drawn edges.
4) That 255 is the minimum number for the problem. (Everyone proved this one).

From what I've gathered, most people either assumed 1 and proved 2/3/4, or assumed 1, proved 4, and assumed 2 and 3. I could be completely wrong about this also, because I only skimmed most solutions.

Well right now, I still really don't know what the rigorous standards for a graph theory problem like this is. I'm taking a look at some of the WOOT graph theory problems (which I didn't even pay attention to :oops: ) to try to get a feel, but most of their problems are of a different flavor.

All your solutions could very possibly be completely rigorous, but I can't really tell :( .

[quote="junggi"][quote="serialk11r"][quote="i_like_pie"]Well, if you look at it that way, then my proof isn't really a proof at all, but rather a basic outline of a proof.

:wallbash_red:[/quote]
With Ubey's standards for rigor, a lot of people have illegitimate proofs, but hey, that's good USAMO preparation ;)[/quote]
his standards are usually the typical standards for proofs dont be cynical[/quote]
Yes...so in fact everyone at Gunn had incorrect proofs right?

[quote="Ubemaya"]Well based on what [b]torajirou[/b] said, it looks like everyone still had pretty rigorous proofs, probably more than rigorous enough for a 7 on the BAMO.[/quote]
In which thread?

[quote="Ubemaya"]Well based on what torajirou said, it looks like everyone still had pretty rigorous proofs, probably more than rigorous enough for a 7 on the BAMO.[/quote]
torajirou said it had to be proved that N=255 is possible, and N cannot be <255. The thing is, people didn't prove N<255 rigorously by your previous arguments and complaints, which I completely agreed with. To get the inequality that ended up being N>=255, you had to have that alcatraz had 126 wins. However, not many people actually proved that 126 was the unique number that minimized N (its not THAT hard to prove, but still, you can't assume it). I think it was necessary to prove that that was true, and stephen ge questioned brian zhang about his assumption of 126 wins in CA for alcatraz too, so I'm pretty sure you need to include a proof of that for it to be rigorous.

[quote="serialk11r"][/quote][quote="junggi"][quote="serialk11r"][quote="i_like_pie"]Well, if you look at it that way, then my proof isn't really a proof at all, but rather a basic outline of a proof.

:wallbash_red:[/quote]
With Ubey's standards for rigor, a lot of people have illegitimate proofs, but hey, that's good USAMO preparation ;)[/quote]
his standards are usually the typical standards for proofs dont be cynical[/quote][quote="serialk11r"]
Yes...so in fact everyone at Gunn had incorrect proofs right?[/quote]
iirc brian had a pretty rigorous proof
he didnt use alcatraz had 126 wins or whatever, he said somethin like wins>=126 then did some mumbo jumbo to end up at an inequality where only x>=255 satisfied it
that rigorously proves x<255 does not work
he also gave an example for 255 showing that 255 works

other than brian, yes everyone at gunn had incorrect proofs (all of them are probably way off, with mountains of nonrigor)

Wait, what would be valid for showing 255 works?

Ubey confundled.

[quote="junggi"][quote="Ubemaya"]I heard from Andrew that Brian's proof assumed the 126-125-125-125-...-125-124 case though, and he didn't give an example for n = 255. I'm confused.[/quote]
eh maybe i remembered wrong whatever

or maybe he misheard brian saying he cited that as an example for the 255 case not assumed it
that would be valid for showing 255 works[/quote]
Now I think I remember brian using the >=126 thing, but not the example..., but then I remember stephen saying something about assuming things...whatever. Boylie and pinsker won this one, that's for sure.


function
integration
real analysis
calculus
derivative
Functional Analysis
real analysis unsolved

H is a hilbert space, $ L^2(0,T;H)$ is defined as the function space contains all $ u: [0,T]\minus{}> H$ such that $ \int_0^T \|u(t)\| dt< \infty$

If $ \int_a^b (v,u(t)) dt <\equal{} C_1 \|v\| |b\minus{}a|$ for all $ v \in H$, how to show that:

$ ess\sup_{0<\equal{}t<\equal{}T} \|u(t)\| <C$ for some constant $ C$?

Shouldn't the condition for $ L^2$ be that $ \int_0^T \|u(t)\|^2 dt< \infty?$

If the set $ E\equal{}\{\|u\|>C\}$ has positive measure, then let $ t_0$ be its Lebesgue point and let $ (a,b)\equal{}(t_0\minus{}\epsilon,t_0\plus{}\epsilon)$. Try $ v\equal{}u/\|u\|$ on $ E\cap (a,b)$ and $ 0$ otherwise.

please use

[code]\mathop {\text{ess sup}}\limits_{0 \leqslant t \leqslant T} \| {u\left( t \right)} \|[/code]

to demonstrate
\[ \mathop {\text{ess sup}}\limits_{0 \leqslant t \leqslant T} \| {u\left( t \right)} \|
\]

[quote="Kent Merryfield"]Shouldn't the condition for $ L^2$ be that $ \int_0^T \|u(t)\|^2 dt < \infty?$[/quote]

No, $ L^p(0,T;X)$ consits of all measurable functions $ u : [0,T] \to X$ with
\[ \left\| u \right\|_{L^p \left( {0,T;X} \right)} \equal{} \left\{\begin{gathered} \left( {\int_0^T {\left\| {u\left( t \right)} \right\|_X^p dt} } \right)^{\frac {1} {p}} ,1 \leqslant p < \infty , \hfill \\
\mathop {\text{ess sup}}\limits_{0 \leqslant t \leqslant T} \left\| {u\left( t \right)} \right\|_X , \qquad p \equal{} \plus{} \infty . \hfill \\
\end{gathered} \right.
\]

thanks for Kent Merryfield and bookworm_vn to point out my typo: it should be
$ \int_0^T \|u(t)\|^2 dt< \infty$ and thanks for bookworm_vn for giving "ess sup" gramma.

Now for mlok's hint: I guess you mean to use Leb differentiation Thm. However, notice that $ v$ is a fixed function, indenpendent of $ t$.  So let $ v\equal{}u/\|u\|$ doesn't make sense...
Here we may assume that $ H \equal{} L^2 (U)$ for $ U \in R^n$ if this helps to construct $ v$.

[quote="mlok"]How about $ v \equal{} {u(t_0)}$ then?[/quote]

I think this works. :lol:

just let $ t_0$ be any leb point in $ [0,T]$
using the given condition: $ \frac{1}{2\epsilon} \int_{t_0\minus{}\epsilon}^{t_0\plus{}\epsilon} (u(t_0),u(t))dt
<\equal{} \frac{1}{2\epsilon} C_1 \|u(t_0)\| 2\epsilon\equal{}C_1 \|u(t_0)\|$

Then using Leb diff Thm, letting $ \epsilon\minus{}> 0$, we have $ \|u(t_0)\| ^2 <\equal{} C_1 \|u(t_0)\|$
i.e. $ \|u(t_0)\| <\equal{} C_1$.

Since $ t_0$ can be any leb point in $ [0,T]$, QED.


number theory
inequalities proposed
inequalities

[quote="Mahan73"]i hope that the question be a new one.......

$ Let\ {{\rm a}}_{{\rm 1}},{{\rm a}}_{{\rm 2}},\cdots ,{{\rm a}}_{{\rm n}}{\rm \ }{\rm \ be\ integers\ such\ that\ for\ any\ int}{\rm eger\ k\ \ we\ have: \ \ \ }{{\rm a}}_{{\rm 1}}{{\rm a}}_{{\rm 2}}\cdots {{\rm a}}_{{\rm n}}{\rm |}{\rm \ }\left({\rm \ }{{\rm a}}_{{\rm 1}}{\rm \plus{} k}\right)\left({{\rm a}}_{{\rm 2}}{\rm \plus{} k}\right)\cdots \left({{\rm a}}_{{\rm n}}{\rm \plus{} k}\right){\rm \ }$Prove that at least on of   $ a_i\ \left(1\le i\le n\right)\ is\ equal\ to\ 1\ or \minus{} 1$[/quote]
I think it must be moved into the Pre-Olympiad or Number Theory forum......
Same for http://www.artofproblemsolving.com/Forum/viewtopic.php?t=308496

[quote="Mahan73"]yes it must!
but i don't know how should i do this
would you help, please?[/quote]
Only a moderator can do it .... :rotfl:


function
real analysis
real analysis unsolved

Hi can any1 pls help with this question

Let f and g be real functions that are continuous on [a,b] and differentiable on(a,b).
Show that there exists c $\in$ (a,b) such that

$f'(c)(b-c)+g'(c)(c-a) = (f(c)-f(a)) + (g(b)-g(c)).$

HINT :Use Rolle's Theroem and the function F to defined for x $\in$ [a,b] by

$F(x)=(f(x)-f(a))(x-b)-(g(x)-g(b))(x-a).$

Thanks in advance

I may be missing something... but what do you need help with?

I do not know how to approach and solve the above problem :(

Applying Rolle`s theorem for the function $F$ then there exists $c\in(a,b)$ such that $F'(c)=0$ just derivate $F$


prove
$ 39pqr\equal{}4q^2r^2\plus{}27p^2$
where
$ p\equal{}abc,
q\equal{}a\plus{}b\plus{}c,
r\equal{}ab\plus{}bc\plus{}ca$

oh sorry there will be $ >$sign in place of $ \equal{}$ and $ a,b,c$ are positive real.
no one interested???? :(

[quote="ishfaq420haque"]prove
$ 39pqr \equal{} 4q^2r^2 \plus{} 27p^2$
where
$ p \equal{} abc, q \equal{} a \plus{} b \plus{} c, r \equal{} ab \plus{} bc \plus{} ca$[/quote]
Are you mean that we need to prove that $ 39pqr \leq4q^2r^2 \plus{} 27p^2$ $ ?$ :wink: If so it's equivalent to
$ \sum_{sym}(4a^4b^2 \plus{} 4a^3b^3 \plus{} 4a^4bc \minus{} 7a^3b^2c \minus{} 5a^2b^2c^2)\geq0,$ which true by Muirhead.


the teacher whispers positive integer a to anna, b to brett, and c to chris. the students don't know one another's number, but they do know that the sum of the numbers is 14. anna says: "i know that brett and chris have different numbers." then brett says : "i already knew that all three of our numbers were different." finally, chris announces, "now i know all three of our numbers."

what is the product abc?

i can figure out the anna has an odd integer, brett has an odd integer greater than or equal to 7, but then i dont know how to proceed...

any advice/solutions would be appreciated. :D

[hide]Sum = 14
By what Anna knows, she has an odd integer.
By what Brett knows, he has an odd integer larger than or equal to 7.
By what Chris knows, his number must mean that there is only one possible case for Anna and Brett if what they said was true. Since Anna and Brett both have odds, Chris must have an even. Brett has a number larger than 7, so Chris's must be less than 7. If Chris has 6, then Brett [i]must[/i] have 7 and Anna [i]must[/i] have 1, so this must be the case. Thus abc = 42.[/hide]

[hide]
Note that Chris says "[i]Now[/i] I now all three of our numbers"
[/hide]

thanks for both solutions! :D

i like uldivads better though.. :P

[quote="stevenmeow"]ok firecricket, if your going to do that plz include a solution, and besides, there is no way 3 positive numbers adding up to 14 multiply to 8[/quote]
cough*cough*

haha owned!


limit
integration
function
Support
real analysis
real analysis unsolved

Show that if $ f\in L^2(R)$ then $ T_z f(x) \equal{} f(x \minus{} z)\in L^2(R)$ for all $ z \in R$ and that $ \lim_{|z| \minus{} > 0} \int |T_z f \minus{} f|^2 \equal{} 0$

can you prove it for continuous functions with compact support ?

if u is a continuous function with compact support then u is in L^2?

Easily. Suppose $ u$ is continuous and supported on $ [a,b].$ Since it's continuous on a compact set, it is bounded - say, $ |u(x)|\le M$ everywhere. Then trivially, $ \|u\|_2^2\equal{}\int_{\mathbb{R}}|u(x)|^2\,dx\le M^2(b\minus{}a)<\infty.$

But that's not really the point of alekk's post. His point was that such nice functions are [b]dense[/b] in $ L^2(\mathbb{R}).$


(1-3sinx)/2=?
answer is 2  :blush:
Thanks and best regards!

if it is to find x
then clearly sinx=-1.finish it up

but if they dont give us answer ..how can we will find? :blush:

[quote="roserose18"](1-3sinx)/2=?
answer is 2[/quote]

I assumed the problem to be this, mainly because the answer comes out cleanly:

$ \frac {1 \minus{} 3\sin x}{2} \equal{} 2$

[hide="Solution."]
$ \frac {1 \minus{} 3\sin x}{2} \equal{} 2$
$ 1 \minus{} 3\sin x \equal{} 4$
$ \minus{} 3\sin x \equal{} 3$
$ \sin x \equal{} \minus{} 1$
$ \boxed{x\in \left\{\frac {3\pi}{2} \plus{} 2\pi k: k\in \mathbb{Z}\right\}}$
[/hide]

[quote="Anikaa"]ooops! sin is never lower than 0 and it is never more than 1 !

so sinx can't be -1![/quote]

The range of $ \sin$ is $ [\minus{}1,1]$. For example, $ \sin \frac{3\pi}{2}\equal{}\minus{}1$.


Find the numbers between 1 and 1000 having maximum number of divisors?

I can find one with 32 factors... is that the highest?

[hide="probably incomplete solution"]
If possible, we want to have the product of the first primes because each new prime will multiply the number of factors by 2 whereas adding an already existing prime will not increase it by as much.. However, that only leads us to $2\times3\times5\times7 = 210$. By adding duplicate prime factors, we can increase the total number of factors. The best prime factor to add now is a 2 because it will allow us to add more as well.
$2^2 \times 3\times 5\times 7 = 420$
This is still rather low, but adding a factor of 3 or higher increases it too much. Therefore, we can only add another 2 giving us the ideal number:
$2^3\times 3\times 5\times 7 = 840$ with 32 integer factors.
[/hide]
What needs to be filled in?


What is the sum of the reciprocals of 1, 2, 3, 4 and 5? Express your answer as a common fraction.

hello, we have $ 1\plus{}\frac{1}{2}\plus{}\frac{1}{3}\plus{}\frac{1}{4}\plus{}\frac{1}{5}\equal{}\frac{137}{60}$.
Sonnhard.

Is there a faster way to add this besides common denominator?

Well, just set common demoniator as 120. thus its 120+60+40+30+24 all over 120. and you get 137/60

Or find the LCM (60) and then you get $\frac{\frac{1}{1} \cdot 60 +\dfrac{1}{2} \cdot {60} + \frac{1}{3} \cdot 60 + \frac{1}{4} \cdot 60 + \frac{1}{5} \cdot 60}{60}$

That simplifies to $\frac{60+30+20+15+12}{60}$ And that goes to $\rightarrow$ $\boxed{\frac{137}{60}}$


Square $S_1$ is $1\times 1.$  For $i\ge 1,$ the lengths of the sides of square $S_{i+1}$ are half the lengths of the sides of square $S_i,$ two adjacent sides of square $S_i$ are perpendicular bisectors of two adjacent sides of square $S_{i+1},$ and the other two sides of square $S_{i+1},$ are the perpendicular bisectors of two adjacent sides of square $S_{i+2}.$  The total area enclosed by at least one of $S_1, S_2, S_3, S_4, S_5$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers.  Find $m-n.$

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=445[/img]

[hide="Answer"]We simply have the difference of two sums, but most of the terms will cancel. We are left with $1+\frac{1}{4}-\frac{1}{1024}=\frac{1279}{1024}\Rightarrow 1279-1024=\boxed{255}$.[/hide]

[hide]The area of the first square is simply $1$, and the area of the second is $\frac 14$. Their overlapping area is $\frac{1}{4^2}$. Notice that the sum of the areas of squares $3$, $4$, and $5$ (without overcounting the overlap) is $\frac{1}{4^5}$ more than the overlap between squares 1 and 2.

Therefore, the total area is $1+\frac14+\frac{1}{4^5}=\frac{1279}{1024}$ and our answer is $1279-1024=\boxed{255}$. [/hide]


inequalities
inequalities proposed
algebra

[quote="mathmanman"]Obviously it is a typo, and the problem is actually with positive [i]reals[/i].[/quote]

Yes indeed, I have edited the problem.

[quote="mathmanman"]Obviously it is a typo, and the problem is actually with positive [i]reals[/i].

[b]Cezar[/b], we can also use the substitution $a = \frac 1x$, $b = \frac 1y$ and $c = \frac 1z$. :)[/quote]

I don't quite think so. :?

And, may I know why you "don't quite think so"?  :roll:

Actually, using this substitution gives that $\frac 1{xyz} = 1$.
And then, if we put all on the same denominator, in the end, all we're left to prove is the exact same thing as in your first solution, that is :
$yz + y + xz + z + xy + x \geq 6$, but, it can be rewritten (from the condition) :
$x + \frac 1x + y + \frac 1y + z + \frac 1z \geq 6$, which is true. ($x + \frac 1x \geq 2$, a.s.o.)

But it is almost the same with my first solution. ;)

Of course it is, but I just wanted to show you that we could use the substitution I said.. :)

Oh, and, by the way, the equality case is : $a=b=c=1$.

supposing:$a\leq{b}\leq{c}$
According to the inequality of réordonnement we have:
$\frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)}\geq\frac{a}{(b+1)(c+1)}+\frac{b}{(c+1)(a+1)}+\frac{c}{(a+1)(b+1)}=W$
and:$\frac{1}{(b+1)(c+1)}\leq{\frac{1}{(c+1)(a+1)}}\leq{\frac{1}{(a+1)(b+1)}}$
According to the inequality of shebyshev:
$W\geq{\frac{1}{3}\((a+b+c)\((\frac{3+a+b+c}{(1+a)(1+b)(1+c)})}$
supposing : $\frac{a+b+c}{3}=\alpha$
according to the inequalities between averages :
$3\alpha=a+b+c\geq3{(abc)}^{1/3}=3$
and, $(1+a)(1+b)(1+c)\leq{[\frac{(1+a)+(1+b)+(1+c)}{3}]}^{3}={(1+\alpha)}^{3}$
so: $W\geq\alpha\frac{6}{{(1+\alpha)}^{3}}$
it is enough to prove  that: $\frac{6\alpha}{{(1+\alpha)}^{3}}\geq\frac{3}{4}$
we know that $\alpha\geq{1}$and$f(\alpha)=\frac{6\alpha}{{(1+\alpha)}^{3}}$ is strictly increasing on $R^{+}$
we have also $f(\alpha)\geq{f(1)}=\frac{3}{4}\longleftrightarrow{W}\geq{\frac{3}{4}}$ :)

let $a=m/n, b=n/p, c=p/m$... then, the inequality is equivalent to

$\sum_{cyc}\frac{mp}{(m+n)(p+n)}\geq \frac{3}{4}$

$\Leftrightarrow 4\sum_{cyc}mp(m+p)\geq (3\sum_{cyc}m^{2}p+mp^{2})+6mnp$

$\Leftrightarrow \sum_{cyc}mp(m+p)\geq 6mnp$

which is evidently true... equality holds iff $m=n=p$, i.e iff $a=b=c=1$.

edit: i'm sorry... i didn't see that cezar lupu had proposed the same solution...

I think i have a straightout Muirhead solution that is not bad at all.

We start with $ \dfrac{a}{(a\plus{}1)(b\plus{}1)}\plus{}\dfrac{b}{(b\plus{}1)(c\plus{}1)}\plus{} \dfrac{c}{(c\plus{}1)(a\plus{}1)} \geq \dfrac{3}{4}$. Multiply the common denominator $ 4(a\plus{}1)(b\plus{}1)(c\plus{}1)$. This yields $ 4(ab\plus{}bc\plus{}ca) \plus{} 4(a\plus{}b\plus{}c) \ge 3(a\plus{}b\plus{}c) \plus{} 3(ab\plus{}bc\plus{}ca) \plus{} 3abc \plus{} 3$. This becomes $ \frac{1}{2}T[1,1,0] \plus{} \frac{1}{2}T[1,0,0] \ge \frac{1}{2}T[1,1,1]\plus{}\frac{1}{2}T[0,0,0]$. Since $ abc \equal{} 1$, we have this equivalent to $ \frac{1}{2}T[1,1,0] \plus{} \frac{1}{2}T[1,0,0] \ge \frac{1}{2}T[\frac{2}{3},\frac{2}{3},\frac{2}{3}]\plus{}\frac{1}{2}T[\frac{1}{3},\frac{1}{3},\frac{1}{3}]$ and the result follows by Muirhead. The equality occurs if and only if $ a \equal{} b \equal{} c \equal{} 1$.

For: $ a\equal{}\frac{y}{x}, b\equal{}\frac{z}{y}, c\equal{}\frac{x}{z},$ we get to:
$ LHS\equal{}\frac{y^2}{(x\plus{}y)(y\plus{}z)}\plus{}\frac{z^2}{(y\plus{}z)(z\plus{}x)}\plus{}\frac{x^2}{(z\plus{}x)(x\plus{}y)}\stackrel{Titu's\  lemma}{\geq}$

$ \geq\frac{(x\plus{}y\plus{}z)^2}{(x\plus{}y)(y\plus{}z)\plus{}(y\plus{}z)(z\plus{}x)\plus{}(z\plus{}x)(x\plus{}y)}\equal{}$

$ \equal{}\frac{x^2\plus{}y^2\plus{}z^2\plus{}2(xy\plus{}yz\plus{}zx)}{x^2\plus{}y^2\plus{}z^2\plus{}3(xy\plus{}yz\plus{}zx)}\stackrel{(*)}{\geq}\frac{3}{4}\equal{}RHS$

We took into account that $ (*)$ is equivalent to:
$ 4(x^2\plus{}y^2\plus{}z^2)\plus{}8(xy\plus{}yz\plus{}zx)\geq3(x^2\plus{}y^2\plus{}z^2)\plus{}9(xy\plus{}yz\plus{}zx)\ \Leftrightarrow$

$ \ \Leftrightarrow\ x^2\plus{}y^2\plus{}z^2\geq xy\plus{}yz\plus{}zx.$

Obviously, we have equality for: $ x\equal{}y\equal{}z\ \Leftarrow\ \ a\equal{}b\equal{}c\equal{}1$

For Vasc's inequality I found that $ p\equal{}2$. Is it right, Vasile?

Marin Bancoș - ROMANIA

[quote="marin.bancos"]
For Vasc's inequality I found that $ p \equal{} 2$. Is it right, Vasile?
Marin Bancoș - ROMANIA[/quote]
Yes, it is right.

[quote="Vasc"]An interesting generalization is the following:
Find the largest positive real number $ p$ such that
\[ \frac {a}{(a \plus{} 1)(b \plus{} p)} \plus{} \frac {b}{(b \plus{} 1)(c \plus{} p)} \plus{} \frac {c}{(c \plus{} 1)(a \plus{} p)}\geq \frac {3}{2(1 \plus{} p)},\]
for any positive real numbers $ a,b,c$ such that $ abc \equal{} 1$.[/quote]
See my proof here, Mr Vasile: http://canhang2007.wordpress.com/2009/11/08/inequality-54-v-cirtoaje/

Similarly, we can show that

$ \sum \frac a{(a\plus{}p)(b\plus{}q)} \ge \frac 3{(1\plus{}p)(1\plus{}q)}$

for any $ q\ge \frac 1{2}$ and $ p\ge \frac {2q\minus{}1}{q\plus{}1}$.

[quote="Cezar Lupu"]I remember that I posted this inequality long long time ago. I have two solutions for this nice inequality.
Anyway, here there are:

[i] Solution 1. [/i]

The given inequality is equivalent with
\[ \sum a(c + 1)\geq\frac {3}{4} (1 + a)(1 + b)(1 + c)\Leftrightarrow\]

\[ \sum a + \sum ab\geq \frac {3}{4}(1 + a)(1 + b)(1 + c)\Leftrightarrow\]

\[ 4\sum a + 4\sum ab\geq 3 + 3\sum a + 3\sum ab + 3abc\Leftrightarrow\]

\[ a + b + c + ab + bc + ca\geq 6,\]
a trivial application of AM-GM. :)

[i] Solution 2. [/i]

We let $ a = \frac {x}{y}, b = \frac {y}{z},c = \frac {z}{x}$. The inequality becomes
\[ \sum\frac {\frac {x}{y}}{\left(\frac {x}{y} + 1\right)\left(\frac {y}{z} + 1\right)}\geq\frac {3}{4}\Leftrightarrow\]

\[ \sum\frac {zx}{(x + y)(y + z)}\geq\frac {3}{4}.\]
This inequality can be proved in many ways. For example you can use Muirhead ( stupid method and I hate it :| ) or AM-GM. ;)[/quote]

how to solve it using am-gm?

[quote="Martin N."][quote="tkrass"][quote="Cezar Lupu"]
\[ \sum\frac {zx}{(x + y)(y + z)}\geq\frac {3}{4}.\]
This inequality can be proved in many ways. For example you can use Muirhead ( stupid method and I hate it :| ) or AM-GM. ;)[/quote]

how to solve it using am-gm?[/quote]
\[ \sum\frac {zx}{(x + y)(y + z)}\geqslant\frac {3}{4}\Longleftrightarrow \sum\frac {zx}{\frac {x + y}{2}\cdot\frac {y + z}{2}}\geqslant 3\]
With $ \frac {x + y}{2}\geqslant\sqrt {xy}$, we have to prove
\[{ \sum\frac {zx}{y\sqrt {zx}} = \sum\frac {\sqrt {zx}}{y}}\geqslant 3\mbox{,}\]
but this is obvious as, by AM-GM again,
\[ \frac {\sum\frac {\sqrt {zx}}{y}}{3}\geqslant\sqrt [3]{\prod\frac {\sqrt {zx}}{y}} = 1\]
:wink:[/quote]

Martin N. your proof is wrong.

[quote="Vasc"]An interesting generalization is the following:
Find the largest positive real number $p$ such that
\[ \frac{a}{(a+1)(b+p)}+\frac{b}{(b+1)(c+p)}+\frac{c}{(c+1)(a+p)}\geq \frac{3}{2(1+p)}, \]
for any positive real numbers $a,b,c$ such that $abc=1$.[/quote]

How could we find the solutions or your generalisations? (both aren't solved yet with proof)

$\frac{a}{(a+1)(b+1)}+\frac{b}{(c+1)(b+1)}+\frac{c}{(a+1)(c+1)}\ge3/4$
I found something interesting.
If we assume the statement to be true then I find that
$a(c+1)+b(a+1)+c(b+1)\ge3/4(abc+ab+ac+bc+a+b+c+1)$
$ab+ac+bc+a+b+c\ge3abc+3$
$a+\frac{1}{a}+b+\frac{1}{b}+c+\frac{1}{c}\ge6$
WE know by AM-GM
$x+\frac{1}{x}\ge2$
NOW we are done.

Take $a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$. Inequality becomes:

$\sum_{cyc}\frac{xz}{(x+y)(y+z)}=\frac{\sum_{cyc}xz(x+z)}{(x+y)(y+z)(z+x)}\geq\frac{3}{4}$.

$4(\sum_{symmetric}xy^2)\geq 3(2xyz+\sum_{symmetric}xy^2)$

$\sum_{symmetric}xy^2\geq 6xyz$, witch is true because of AM-GM.

[hide="Solution"]
First off, we expand the LHS and put it under a common denominator.

We get $\frac{a(c+1)+b(a+1)+c(b+1)}{(a+1)(b+1)(c+1)}\ge \frac34$
$\implies \frac{ab+ac+bc+a+b+c}{abc+ab+ac+bc+a+b+c+1}\ge \frac34$
$\iff ab+ac+bc+a+b+c\ge 3(abc+1)=6$.

Which is true by AM-GM.

Equality holds in AM-GM when $ab=ac=bc=a=b=c$.  $abc=1$ is given, so $a\neq 0$ and $ab=a$ gives $b=1$, and similarly $(a,b,c)=(1,1,1)$ is where equality holds.  [/hide]

[quote="kunny"]$a+1\geq 2\sqrt{a}$ etc, we have $(a+1)(b+1)(c+1)\geq 8abc$, yielding $(a+1)(b+1)(c+1)\geq 8.$

$\therefore \frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)}$

$=1-\frac{2}{(a+1)(b+1)(c+1)}\geq 1-2\cdot\frac 18=\frac 34.$[/quote]
beautiful.

Let $a,b$ and $c$ be positive real numbers such that $abc=1$. Prove that \[\frac{a}{(b+1)(c+1)}+\frac{b}{(c+1)(a+1)}+\frac{c}{(a+1)(b+1)} \geq \frac{3}{4}. \]
Have a similar simple proof?

[quote="Cezar Lupu"]I remember that I posted this inequality long long time ago. I have two solutions for this nice inequality.
Anyway, here there are:

[i] Solution 1. [/i]

The given inequality is equivalent with \[ \sum a(c+1)\geq\frac{3}{4} (1+a)(1+b)(1+c)\Leftrightarrow  \]
\[ \sum a+\sum ab\geq \frac{3}{4}(1+a)(1+b)(1+c)\Leftrightarrow  \]
\[ 4\sum a+4\sum ab\geq 3+3\sum a+3\sum ab+3abc\Leftrightarrow  \]
\[ a+b+c+ab+bc+ca\geq 6,  \]
a trivial application of AM-GM. :)

[i] Solution 2. [/i]

We let $a=\frac{x}{y}, b=\frac{y}{z},c=\frac{z}{x}$. The inequality becomes

\[ \sum\frac{\frac{x}{y}}{\left(\frac{x}{y}+1\right)\left(\frac{y}{z}+1\right)}\geq\frac{3}{4}\Leftrightarrow  \]
\[ \sum\frac{zx}{(x+y)(y+z)}\geq\frac{3}{4}.  \]
This inequality can be proved in many ways. For example you can use Muirhead ( stupid method and I hate it :| ) or AM-GM. ;)[/quote]
Your first solution is very beautiful and very natural. Congratulations!

Let $a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$.Then, Clearing the denominator, we get

$6S(4,2,0)+6S(4,1,1)+6S(3,3,0) \geq 12S(3,2,1)+6S(2,2,2)$

Obviously this holds by Muirhead inequality :)

[quote=Igor]Let $a,b,c$ be three positive real numbers such that $abc=1$. Show that:

\[ \displaystyle \frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+ \frac{c}{(c+1)(a+1)} \geq \frac{3}{4}.  \]

When is there equality?[/quote]
$$\iff$$
$$ \frac{1}{ab(b+1)(c+1)}+\frac{1}{bc(c+1)(a+1)}+\frac{1}{ca(a+1)(b+1)}\geq\frac{3}{4}.$$
Where $a,b,c$ be three positive real numbers such that $abc=1$.
[url=https://artofproblemsolving.com/community/c6h1217275p6065841]2005 Czech
[/url][url=https://artofproblemsolving.com/community/c6h1812768p12082679]here[/url]

[quote=mihaig][quote=sqing][quote=Igor]Let $a,b,c$ be three positive real numbers such that $abc=1$. Show that:

\[ \displaystyle \frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+ \frac{c}{(c+1)(a+1)} \geq \frac{3}{4}.  \]

When is there equality?[/quote]
$$\iff$$
$$ \frac{1}{ab(b+1)(c+1)}+\frac{1}{bc(c+1)(a+1)}+\frac{1}{ca(a+1)(b+1)}\geq\frac{3}{4}.$$
Where $a,b,c$ be three positive real numbers such that $abc=1$.
[/quote]
It's obvious that the problems are not equivalent.
Greece implies Czech,but the other implication is not true.[/quote]
Let $a,b,c$ be three positive real numbers such that $abc=1$. Show that:

\[ \displaystyle \frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+ \frac{c}{(c+1)(a+1)} \geq \frac{3}{4}.  \]
$$\iff$$
 $$\frac{a}{(a+1)(c+1)}+\frac{c}{(c+1)(b+1)}+ \frac{b}{(b+1)(a+1)} \geq \frac{3}{4}$$
$$\iff$$
$$ \frac{1}{ab(b+1)(c+1)}+\frac{1}{bc(c+1)(a+1)}+\frac{1}{ca(a+1)(b+1)}\geq\frac{3}{4}.$$
Where $a,b,c$ be three positive real numbers such that $abc=1$.

Let $a,b,c$ be three positive real numbers such that $abc=1$. Show that:
\[ \displaystyle \frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+ \frac{c}{(c+1)(a+1)} \geq \frac{3}{4}.  \]
$$\iff$$
$$\frac{y^2}{(x+y)(y+z)}+\frac{z^2}{(y+z)(z+x)}+\frac{x^2}{(z+x)(x+y)}{\geq} \frac{3}{4}.  $$
Where $x,y,z>0.$
[url=http://www.artofproblemsolving.com/community/c275h247275p1357373]Moscow[/url]
[hide=h][quote=mihaig]Really?
Are you sure that is the meaning of "equivalent"?
Didn't realize that abc=1 changes all in my inequality?
Because it's non-homogenuous,so it's not allowed to replace by 1 how you wish and to claim the equivalency.
For example,why didn't you replace all 1's from Czech by abc,then in my inequality again.
Rookie mistakes and affirmations.[/quote][/hide]

[quote=WolfusA]Indeed mihaig, you're right as the second inequality should be replaced with $\frac{xz}{(x+y)(y+z)}+\frac{xy}{(y+z)(z+x)}+\frac{yz}{(z+x)(x+y)}{\geq} \frac{3}{4}. $[/quote]

Thank you. I wish I wasn't right...

Let $a=x/y,b=y/z,$ and $c=z/x.$ Since $(2,1,0)\succ(1,1,1),$ Muirhead yields
\begin{align*}\sum_{\text{sym}}x^2y\ge6xyz&\implies4\sum_{\text{sym}}x^2y\ge3\left(2xyz+\sum_{\text{sym}}x^2y\right)\\&\implies
4\sum_{\text{cyc}}xz(x+z)\ge3(x+y)(y+z)(z+x)\\&\implies\sum_{\text{cyc}}\frac{xz(x+z)}{(x+y)(y+z)(z+x)}\ge\frac{3}{4}.\end{align*} Noting $$\frac{a}{(a+1)(b+1)}=\frac{\frac{x}{y}}{\frac{x+y}{y}\cdot\frac{y+z}{z}}=\frac{xz}{(x+y)(y+z)}$$ finishes. $\square$


Can anyone give me a list of easy Olympiad Questions from any country's olympiad?

http://www.artofproblemsolving.com/Forum/resources.php

:D

(Just browse through randomly and find some easy ones.)

You could also try Kalva and do the first problem on USAMOs, or if those are too hard do Canada's olympiads.

.......should start after ARML!

Not that I'm particularly experienced with olympiad math (I just wanted to make that joke)  I would look at easy USAMO questions first, and some APMO questions too. (Asian Pacific or Atlantic Pacific, I can't remember)

AoPS have good resources, as does Kalva, or you can even buy the books from AoPS.

For the easiest olympiad problems,see the indian RMO's,specially 1990-2000,they're simple and even INMO(Indian national...).Both of themn are simpler than AIMEs.


Prove that for any $a,b,c\in (1,2)$ the following inequality holds
$\frac{b\sqrt{a}}{4b\sqrt{c}-c\sqrt{a}}+\frac{c\sqrt{b}}{4c\sqrt{a}-a\sqrt{b}}+\frac{a\sqrt{c}}{4a\sqrt{b}-b\sqrt{c}}\geq1$
Is there any one who can give me a nice solution?

Hey, that's not that hard. Anyway, it is easy to see that $\frac{b\sqrt{a}}{4b\sqrt{c}-c\sqrt{a}}\geq\frac{a}{a+b+c}$, so the conclusion follows. :D

I know, Cezar, everything is easy when you have the official solution in front of your eyes ( I know, because this problem was created by Dzeta). At least for this kind of problems we see solutions from you...

From where do I have this solution? :?  Actually, did Dzeta send this problem to magazine? :?
 I "copied" the solution from there.Who knows? :?  I really don't care anymore. From now on you can say everything. :(  Anyway, can you tell me where did this problem appear?

Sorry for being rude, but such attitude makes me mad. And yes, from what I saw till now, I don't think you would have ever found this trick. I hope I'm wrong.

Ok, if this is your opinion about me......what can I say? :? You still avoid my question?
You can send me a pm with the source to avoid spamming. :)

[quote="harazi"]And by the way, here's a natural solution (excuse me for being so dumb, I could not find that obvious lemma):
  Write ${\frac{b\sqrt{a}}{4b\sqrt c-c\sqrt a}}=\frac{ (\sqrt a)^2}{4\sqrt{ac}-\frac{ ac}{b}}$ and apply Cauchy-Schwarz. After that it becomes obvious.[/quote]
Very nice solution...
But can you tell me how did you guess it ? ;)

[quote="silouan"][quote="harazi"]  Write ${\frac{b\sqrt{a}}{4b\sqrt c-c\sqrt a}}=\frac{ (\sqrt a)^2}{4\sqrt{ac}-\frac{ ac}{b}}$ and apply Cauchy-Schwarz. After that it becomes obvious.[/quote]

Can anyone compplete the solution because I don't understand why itis obvious. Thanks. Ceasar could you post the proof of your leema[/quote]
$(\frac{ (\sqrt a)^2}{4\sqrt{ac}-\frac{ ac}{b}}+\frac{ (\sqrt b)^2}{4\sqrt{ba}-\frac{ ba}{c}}+\frac{ (\sqrt c)^2}{4\sqrt{cb}-\frac{ cb}{a}})(4\sqrt{ac}-\frac{ ac}{b}+4\sqrt{ba}-\frac{ ba}{c}+4\sqrt{cb}-\frac{ cb}{a})\geq(\sqrt{a}+\sqrt{b}+\sqrt{c})^2$
And now we sould prove:
$(\sqrt{a}+\sqrt{b}+\sqrt{c})^2 \geq (4\sqrt{ac}-\frac{ ac}{b}+4\sqrt{ba}-\frac{ ba}{c}+4\sqrt{cb}-\frac{ cb}{a})$
Or:
$a+b+c+\frac{ ac}{b}+\frac{ ba}{c}+\frac{ cb}{a}\geq 2\sqrt{ac}+2\sqrt{cb}+2\sqrt{ba}$
Now it is easy to prove ;)


trigonometry
limit
algebra unsolved
algebra

The problem is:
Find $\sum_{n=1}^\infty 3^{n}\cdot \left( \sin\left( \frac{\pi}{3^{n}}\right) \right)^{3}$.

Please always post them directly on the forum, not by attachments!

could you give me a link that i can learn how can i write mathematical terms?

See http://www.mathlinks.ro/Forum/viewtopic.php?t=5261  :)


calculus
derivative
algebra
system of equations
advanced fields
advanced fields theorems

The equations: $ 2xyz_x \plus{} z_y \equal{} z$ and $ 2xyz_x \plus{} z_y \minus{} z\equal{} 0$
look remarkably similiar to me.

The reasoning is unclear how to determine the proper set up the system of equations:

$ \frac{dx}{2xy} \equal{}\frac{dy}{1} \equal{}\frac{dz}{z}$
or
$ \frac{dx}{2xy} \equal{}\frac{dy}{1} \equal{}\frac{dz}{\minus{}z}$

I don't understand how to determine when to include the minus sign in the 3rd equality?

An equation of form:

$ Pu_x \plus{}Qu_y \plus{} Ru_z \equal{} 0$

is set up like $ \frac{dx}{P} \equal{}\frac{dy}{Q} \equal{}\frac{dz}{R}$

Sometimes, I work through the problem to a solution and then sub the solution into the original PDE only to find I need a minus sign.
As much as I enjoy working problems 4 or 5 times, I would rather understand when to include it from the start?

Forget the PDE and think of a plane with equation $ Ax\plus{}By\plus{}Cz\equal{}0$. A normal line to this plane has equation $ \frac{x}{A}\equal{}\frac{y}{B}\equal{}\frac{z}{C}$. If you had $ Ax\plus{}By\equal{}Cz$, the normal would be $ \frac{x}{A}\equal{}\frac{y}{B}\equal{}\frac{z}{\minus{}C}$.

Excellent...  Thank you.

$ \vec V \equal{} <2xy,1,z>$

$ \nabla z \equal{} <z_x, z_y, \minus{}1>$

I think I understand...


SFFT
videos
articles
AMC
AIME
special factorizations

Simon's Favorite Factoring Trick:

Factor $xy+x+y=a$, where a is some constant.
$x(y+1)+y=a$
$x(y+1)+y+1=a+1$
$(x+1)(y+1)=a+1$

...or something like that.

[quote="i_like_pie"][quote="ajai"]What is Simon's Factoring Trick? What kind of problems use it?[/quote]
http://www.artofproblemsolving.com/Wiki/index.php/Simon%27s_Favorite_Factoring_Trick[/quote]

To see SFFT in action, check out the video at the bottom of that article link. It was very good from my point of view and helped me understand it better.

A question that would use this is like

xy+2x+2y=38
xz+2x+2z=31
yx+2y+2z=26

What is (x,y,z)
I remember an old AIME problem (like #5 from the 1980s, early 1990s period) that used it.


The equation of the chord of contact is given by

$\frac{y-btan\theta}{x-asec\theta}$ =${b\cos\frac{\theta-\gamma}{2}}$ $\div$ ${{a\sin\frac{\gamma-\theta}{2}}}$

what would be the next step on proving the chord of contact eqn?

[hide="Shooting blindfolded."]I'm not familiar with this formula, or even an objective function you may be trying to reach, but if you're trying to simplify I noticed that

$cos(\theta) = cos(-\theta)$
So, let $\gamma-\theta = \beta$, if there is a difference in the angles so long as they are given in their positive form, $\gamma-\theta =-(\theta-\gamma)$.
Applying Cosine's property, $cos(\gamma-\theta) = cos(\theta-\gamma)$

I simplified to $\frac{b\cot(\gamma-\theta)}{a}$
[/hide]Apply double angle formula if you wish? I don't know if that will help much, I forgot most of trig already.  :P


vector
topology
real analysis
real analysis solved

Let $E$ be a normed vectorial space and let $L$ be the familly of liniar independent vectors $(x_1,...,x_m)$.Show that $L$ is an open set of $E^m$.First prove it for a finite dimensional vector space.

Map the $m$-uple $(x_1,\ldots,x_m)$ to $\inf\left\|\sum a_ix_i\right\|$, where the infimum is taken after all the $m$-uples $(a_1,\ldots,a_m)$ of reals (I assumed it to be a real vector space) with $\max|a_i|=1$. It's easy to show that this is a continuous map from $E^m$ to $[0,\infty)$, and that $L$ is precisely the preimage of $(0,\infty)$, which means that $L$ must be open.

I hope it's correct.

P.S.

Did I interpret that $E^M$ correctly by assuming you meant $E^m$?

Yes.It's seems ok.Another approach is to show that its complementary in $E^m$is a closed set.Take a sequence $(x_{n1},..,x_{nm})$ of dependent vectors which tends to $x=(x_1,..,x_m)$.Then there exists a sequence $(\lambda_{n1},..,\lambda_{nm})$ such that $x_{n1}*\lambda_{n1}+...+x_{nm}\lambda_{nm}=0,\forall n$.We can normalize it such that ..$|(\lambda_{n1},..,\lambda_{nm})|=1$ ....and now we can extract a subsequence which converges to $(\lambda_{1},..,\lambda_{m})$ .It's easy to see that $x_{1}*\lambda_{1}+...+x_{m}\lambda_{m}=0$.So $x\in C(L)$ showing that L is open.


[quote="tpulak"]Here is the problem: " To pass a 30 question test , Johnny needs to answer at least 60% of the questions correctly. When Johhny received his grade test back, he saw that he neeed to have answered correctly exactly 2 more questions to have passed the test. How many questions did he answer correctly?"[/quote]

[hide]
Uh, 60% of 30=18, so he answered 16
[/hide]

USE PROPORTIONS

X/30 is equal to 60/100

then cross multiply

100*X and 30*60

100X=1800

X=18

18-2=[u]16[/u]

[quote="tpulak"]Here is the problem: " To pass a 30 question test , Johnny needs to answer at least 60% of the questions correctly. When Johhny received his grade test back, he saw that he neeed to have answered correctly exactly 2 more questions to have passed the test. How many questions did he answer correctly?"[/quote]
[hide]Johnny needs to answer at least 18 questions right to get a 60% and pass the test. Therefore, he answered 16 questions correctly.[/hide]


How many different numbers can be expressed as the sum of exactly three different numbers from the set {1,  2,  3,  9,  10} ?

A.6	B.7	C.8	D.9	E.10

[hide=" click to reveal hidden content"]

I think this is $ 8$, since $ 13$ occurs twice: $ 1\plus{}2\plus{}10\equal{}1\plus{}3\plus{}9\equal{}13$ and $ 14$ occurs twice: $ 1\plus{}3\plus{}10\equal{}2\plus{}3\plus{}9\equal{}14$. So we can make $ 6, 12, 13, 14, 15, 20, 21, 22$.

[/hide]


USAMTS
LaTeX
geometry
geometric transformation
reflection

Is it cheating to get someone (whos not doing USAMTS) to LaTeX up your solutions for you?

I think this issue has been addressed before on a previous [url=http://www.artofproblemsolving.com/Forum/topic-12736-20.html]thread[/url], and I'm pretty sure the answer was no.

We have changed our policy on this issue - all work you submit must be your own work.  I will edit that old post to reflect this.

Please note that you do [b]not[/b] have to use LaTeX.  You can use a different program, create a PDF file, and send that, or you can hand-write your solutions.  If you fax handwritten solutions, please use a [b]dark ink pen [/b]to write your solutions - pencil will not come through clearly.


pigeonhole principle
geometry
rectangle
integration
parameterization
rotation
search

A piece of string of length $ 1$ lies in the plain.Prove that there is a rectangle containing the string that has area
not exceeding $ \frac {1}{4}$

For our purpose it is convenient to think of $ \gamma$ lying in the complex plane. Let the equation of the string be $ \gamma(t)$, $ t\in [0,1]$. We have that $ \gamma$ is absolutely continuous (because it's rectifiable), so $ \gamma(t) \equal{} \int_0^t \zeta(s) ds$. By changing the parameter to the natural one $ t\to \int_0^t |\zeta(s)| ds$ we can assume that the equation of the curve is natural, i.e. $ |\zeta(t)| \equal{} 1$ a.s.

Now, rotating our curve by angle $ 2\pi a$ we want to find the minimal straight rectangle it fits in. The equation of rotated curve is $ \gamma_a(t) \equal{} e^{i2\pi a}\gamma(t) \equal{} : x_a(t) \plus{} i y_a(t)$. This should certainly fit into rectangle with width $ W_a \equal{} \max_{[0,1]} x_a \minus{} \min_{[0,1]} x_a$ and height $ H_a \equal{} \max_{[0,1]} y_a \minus{} \min_{[0,1]} y_a$.

It is obvious that $ W_a\le R_a$ and $ H_a\le I_a$, where
\[ R_a \equal{} \int_0^1 |x'_a(t)| dt, I_a \equal{} \int_0^1 |y'_a(t)| dt\]
So it is enough to prove that there exists $ a$ such that $ R(a)I(a)\le 1/4$.

Further I've made a mistake showing that $ \int_0^1 R_a I_a da\le 1/4$ :) But I think it's true.

What I can prove in fact with a slight modification of former argument that a [i]closed[/i] curve can be covered by a rectangle of area $ 1/\pi^2$.

Thanks for your solution,zhoraster,though it's hard for me to understand it.

Can anyone prove it with more elementary ways?

Unfortunately, I didn't prove it, [b]skywalkerJ.L.[/b]

You cannot beat $ \frac {1} {4}$ as a bound, since when the string is made by two straight halves of length $ \frac{1} {2}$ each, at a right angle, the minimal area of an enclosing rectangle is $ \frac {1} {4}$.

I am very close of a really elementary and nice solution - unfortunately a pesky sub-case resists proof (yet!).

The conjecture $ \int_0^1 R(a) I(a) \leq \frac{1}{4}$ is, unfortunately, not true.  If the string is a line parallel to the real ($ x$) axis, then $ R(a)\equal{}|cos(2 \pi a)|$, $ I(a)\equal{}|sin(2 \pi a)|$, and
\[ \int_0^1 R(a) I(a)\equal{}\frac{1}{2} \int_0^1 |sin(4 \pi a)|\equal{}\frac{1}{\pi}.\]

Thanks, [b]kevinatcausa[/b], I already know it's not true. Stuck, as [b]mavropnevma[/b] with a subcase (but my partial solution is not very elementary).

In fact, I'm able to prove it when the diameter of the curve is (roughly) between $ 0,57$ and $ 0,87$.

I once posted a hint [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1693114740&t=99039]here[/url]. See if you can follow it ;).

Does anyone here actually care that you can't use calculus in the Olympiad forums?


geometry
3D geometry
prism
sphere
topology
geometric transformation
reflection

I posted this here a few days ago, and I just found it in the Fun Factory.  :( Guess JBL deemed the topic not serious enough because of my added comments.

But I want serious answers.  :)

So, again, what is mathematics?

Haha, haven't you been here long enough to know? :D

Well... I'm in the same boat as you. So I can only describe it, I'm afraid.

Let's say you have a rectangular prism, with all its perfectly perpendicular faces. Let's say you said, out of nowhere, that its "volume" was the product of the lengths of its sides.

Ok, so now we have volume.
But from that, it follows that the volume of a cube is s^3, and, even without any derivations or proofs, if we filled a bunch of spheres with water and then measured it, the volume would be about 0.52 ($ \frac {\pi}{6}$) times the distance across the sphere... And of course, this always holds because all spheres are similar— in short, a rule or formula for the volume of a sphere.

And we keep building on that, and it's hard to tell whether or not there's ever an end to it.

Think about the foundations of algebra.
Simple identities like $ a \plus{} b \equal{} b \plus{} a$. There's really no sense in saying why those things are true, because it's so obvious. That it was only logical, for 2 apples and 3 apples are the same as 3 apples and 2 apples. These foundations are there, and math is just exploring them. Why numbers? Didn't numbers, the pure foundations, stem from the fact that there can be more than one of an object and didn't variables exist because there are more than 1 types of objects in this world? In this way, God created the integers.

So, what is math? The natural relations between numbers. The explorations and creative process of changing and applying these relations for new ones. No two math problems are the same.

And typing this let me see what math really is (haha, I never knew either).

I consider mathematics to be an extension of logic. Unlike almost everything else, mathematics is inherently logical. Numbers always mean the same thing and statements are always true, false, or impossible to determine.

The experimental sciences are marred by the imperfection of human observation while the humanities use language that can be interpreted in many different ways. Mathematics does suffer somewhat from the ambiguity of language (which many students take advantage of when they learn to write proofs).

Thus, mathematics is not perfectly logical because we have no way of defining basic concepts and basic rules of logic without using language. However, nothing that humans can comprehend is purely logical, and math is as close as we can get.

Mathematics started out as a tool of the sciences and as a type of philosophy. In the past two centuries, however, mathematics has separated itself from the sciences,  (mostly) from ambiguity. Mathematics has become abstract and rigorous, and in this process it has been united with logic. By now, I believe that mathematics and logic are two different views of the same thing.

Modern mathematics now depends entirely on proof by logical principles. The meaning of the word "mathematics" has evolved over the years and it will continue to do so. That is part of the ambiguity of language. Mathematics itself will have to remain an imprecise term without a rigorous definition.

The only logical question to ask, then, is, "What is logic?"  :lol:

Inevitably, at least when a pathetic individual named leoxnlin with very little intellectual capability attempts to create one, definitions seem to include "thought", "judgment", or "reasoning". Unfortunately this are intertwined hopelessly with brains, which are a part of these atrocious, abominable creatures called humans. A dubious way to squirm out of this is to say, "Logic is that which consists of the principles of inference." Hmmm....

Others may strongly disagree, but sometimes I feel the spatial and visual mathematics, such as geometry or topology, are a little removed from the plane of logic. They require humans to visualize... GAH I want to justify geometry intuitively for myself so badly, but I can't. Help me out here... Maybe I'm too formalist...

I consider mathematics to be the foundation of the universe  :D
The universe is where it is right now because of math. Math leads to physics, which leads to chemistry (because of the way atoms form), which leads into all the other categories of science.
In addition, the Big Bang was determined by very precise equations.
Finally, math, and its derivations (science) are the only universal languages.

[quote="evilhamster"]I consider mathematics to be the foundation of the universe
The universe is where it is right now because of math. Math leads to physics, which leads to chemistry, which leads into all the other categories of science...Finally, math, and its derivations (science) are the only universal languages.[/quote]

http://www.xkcd.com/435/  :D

[quote="leoxnlin"]That xkcd should have the mathematician infinitely further to the right, in my opinion.  :D

 :o evilhamster, you are yet another remarking that math fits the world so perfectly... which is nice, but totally misses the point (again in my opinion)...[/quote]
I don't think so...
My definition is that math is the foundation of the universe. Is that a little more clear? The rest of my post was documenting my reasons.
And yes, I did read the xkcd comic (before I posted).

[quote="evilhamster"]My definition is that math is the foundation of the universe.[/quote]  What does this mean?  Do you expect that when the [url=http://lhc.web.cern.ch/lhc/]LHC[/url] collides two proton beams and scientists look at the scattering that results, instead of the [url=http://www.exploratorium.edu/origins/cern/ideas/higgs.html]Higgs boson[/url] they'll find the Chinese Remainder Theorem?

Not the Chinese remainder theorem, of course, but still some structure that is purely mathematical and cannot be described properly in any other terms (I mean, if you do something with two unitary representations and afterwards try to decompose the resulting representation into the sum of irreducible ones, you do not get Chinese remainder theorem either but certain particle transformations are described exactly like that, if my memory does not betray me).

The question whether and up to what extent the physical Universe is a manifestation of mathematical structures and the mathematical structures are reflections of the physical Universe is quite subtle. So far, it seems that mathematics is reacher in the sense that there is no physical process that does not allow a mathematical description in principle and there are mathematical theorems that do not have direct manifestations in the Universe but I may be wrong here on both accounts. At any rate, so far mathematics as a whole remains our best understandable model of the Universe as a whole (modern theoretical physics is really just "experimental mathematics" with somewhat different standards of rigor and all other sciences get more and more "mathematized").

The question about the exact relation between "model" and "reality" always remains but don't forget that no model is assumed to capture everything at once. It will suffice to ensure that every particular effect can be captured. So, in a sense, there is no clear borderline between the process and the full set of its models (including the ones not invented (discovered?) yet). In principle, we can hold the view that a physical process is identical to the set of its descriptions, provided that this set is capable of answering every meaningful question one might bother to ask. That we do not possess such a set now (or even won't possess it at any particular moment in the future) doesn't mean that this set does not exist.

But if we hold this view and the above set of models is there, the Universe becomes just a part of mathematics. The only way to show that it is not the case is to discover a process and a question about this process that can be posed experimentally such that no mathematical model will be capable of answering it but even then, how are you going to demonstrate the impossibility of such an answer? Of course, a long chain of disappointing attempts may be an indication but so far we have exactly the opposite, namely, a long chain of spectacular successes. The same Higgs boson, if it is finally discovered, will merely indicate that the standard model captures correctly more effects than it was initially designed to explain. Most likely, eventually we'll come to the limitations of the model but it will merely mean that the next approximation to the truth is a bit different from what we thought, i.e., the underlying mathematics is not exactly what we think it is, not that it is not there at all.

As to the original question, I think that an idea to give everything a short precise definition is flawed from the very beginning (for too many reasons to list here). So, I don't know what mathematics really is (or what water really is). I fancy I can recognize it under normal circumstances but (taking the same water example), if you show a snowflake to a native of the tropical zone, there will be a decent chance he will not associate it with what he means by "water". You've, probably, seen enough of mathematics to get the same recognition skill to some extent. Beyond that, any attempt to answer the question will be rather futile (though there are many thick books written on the subject). It all started with the idea that we may have the knowledge not only of $ A$ and $ B$ but also of $ A\Rightarrow B$ and the latter may preceed the knowledge of $ B$. The rest is history. :)

Yes, the statement I quoted is not meaningless if it is attached to the belief that a physical object is the same as the collection of abstract descriptions of it, but without this premise it sure looks ridiculous.

(I personally am very happy with the notion that abstract things exist, but I'm not so happy with the notion that the existence of abstract things is the same kind of existence as the existence of the physical universe in the way that your third paragraph suggests -- but I don't have any interesting philosophical commentary to add, or any particularly compelling argument to support this preference beyond the assertion that it's self-evident :).)

As I said in some other post, I think that $ 5$ in "$ 5$ eggs" exists in a stronger sense than "eggs" but, of course, everything boils down to what exactly one means by "existence". This seems to be one of the least agreed upon words in the language. :) I more or less equate existence to the ability to influence the things that I assume existing with the base assumption that I do exist myself both as a thinking entity and as an electrochemical machine (I'm not sure which one of them I refer to more often when I say "I"). This may be a bit simplistic but I haven't seen a much more satisfactory attempt to define the term. Anyway, this is clearly a sideline in this thread (though, as I said, I'm afraid that there can be no meaningful "mainstream" :( )

I think math is a type of science that attempts to explain abstract ideas of our minds that are motivated both by the physical world and previously studied abstract ideas. I would somewhat disagree that math is much more pure than the other sciences (look at geometry, for instance: no one knew Euclid's axioms were insufficient to describe what he wanted to describe rigorously until people like Hilbert decided to make matters more clear), and I would strongly disagree that math research is driven by a desire to prove theorems (many times understanding a new concept or coming up with a new concept is the hard part, and then proving it is much easier; I am not an expert on this, but I think that statements such as the Riemann hypothesis are in the minority and even the proof to that might be trivial with a deeper understanding of some theory that's not been stated yet). Mathematicians do many experiments too: their lab is paper and pencil (and possibly a computer), and "numerical" examples of concepts studied are many times just as important if not more important than the proofs.

[quote]Mathematics is the academic discipline, and its supporting body of knowledge, that involves the study of such concepts as quantity, structure, space and change. The mathematician Benjamin Peirce called it "the science that draws necessary conclusions". Other practitioners of mathematics maintain that mathematics is the science of pattern, and that mathematicians seek out patterns whether found in numbers, space, science, computers, imaginary abstractions, or elsewhere. Mathematicians explore such concepts, aiming to formulate new conjectures and establish their truth by rigorous deduction from appropriately chosen axioms and definitions.[5]

Through the use of abstraction and logical reasoning, mathematics evolved from counting, calculation, measurement, and the systematic study of the shapes and motions of physical objects. Knowledge and use of basic mathematics have always been an inherent and integral part of individual and group life. Refinements of the basic ideas are visible in mathematical texts originating in the ancient Egyptian, Mesopotamian, Indian, Chinese, Greek and Islamic worlds. Rigorous arguments first appeared in Greek mathematics, most notably in Euclid's Elements. The development continued in fitful bursts until the Renaissance period of the 16th century, when mathematical innovations interacted with new scientific discoveries, leading to an acceleration in research that continues to the present day.

Today, mathematics is used throughout the world as an essential tool in many fields, including natural science, engineering, medicine, and the social sciences such as economics and psychology. Applied mathematics, the branch of mathematics concerned with application of mathematical knowledge to other fields, inspires and makes use of new mathematical discoveries and sometimes leads to the development of entirely new disciplines. Mathematicians also engage in pure mathematics, or mathematics for its own sake, without having any application in mind, although practical applications for what began as pure mathematics are often discovered later.[/quote]

[quote="JBL"]What does this mean?  Do you expect that when the [url=http://lhc.web.cern.ch/lhc/]LHC[/url] collides two proton beams and scientists look at the scattering that results, instead of the [url=http://www.exploratorium.edu/origins/cern/ideas/higgs.html]Higgs boson[/url] they'll find the Chinese Remainder Theorem?[/quote]
No, they won't. But if you read how the LHC works, you'll find that they use quite a bit of math to determine what is left behind and what goes.

[quote="cf249"][quote="leoxnlin"]Inevitably, at least when a pathetic individual named leoxnlin with very little intellectual capability attempts to create one, definitions seem to include "thought", "judgment", or "reasoning". Unfortunately this are intertwined hopelessly with brains, which are a part of these atrocious, abominable creatures called humans. [/quote]
Are you saying that logic, by definition, should be entirely devoid of human thought?  If so, why do you feel this way?[/quote]

That's Platonism for you...

Does the thought of something so fundamental as logic should be sullied and limited by such things as [i]humans[/i] not make you shudder? (*steps down from stage and takes off disguise*)

And what are you without your disguise? An alien life-form bent on conquering the world?

Humans are the only sentient beings we have knowledge of (unless leoxnlin [i]is[/i] an alien life-form), therefore logic must be conducted by humans.
Though this is not necessarily always true, as octopi, dolphins, apes, and dogs have demonstrated basic deductive reasoning in laboratory experiments.

The real question is, what is quantity? It cannot be defined without using itself in the definition (try to do it). This puzzling question is at the heart of mathematics. We cannot define it, but we can describe it, and that's where the axioms come into play. Then we use logic, which also cannot be defined, but only described. Math is a combination of the two.

Well, that's the general problem with words: the definition of [b]every[/b] basic word is circular. I once heard that if you try to construct the "definition graph" using the standard English dictionary, you get something like a fully connected directed graph of about 300 words in the base which are defined through themselves and everything else is built upon them (I haven't tried to verify the exact number but I am inclined to believe the general conclusion). So, we can really define nothing using any language but we can describe quite a lot and language constructs are currently our only way to pass information.

The question is in how good agreement our constructs and formal ways to operate them correspond to what we consider the objects and the relations in what we call "reality". Mathematics turns out so efficient at that that it becomes not unreasonable to assume that it in an exact one-to-one correspondence with the "language of reality" and some people are even inclined to stretch it a bit further and assume that the reality is just a manifestation of purely mathematical structures.

This is akin to a religious belief, of course, but the point is that here we have a religion with well-established, immediate and verifiable predictive value. The only problem with this religion that prevents it from overtaking the world is that you can pass the rewards and punishments to other people too easily, so you do not need to know anything about electricity to chat over the phone, you do not need to understand particle physics to be killed by a nuclear explosion and you do not need to know organic chemistry to take headache pills. So, this religion treats the whole mankind as a single entity and doesn't offer any other individual relationship with its God than the process of invention/discovery or any other afterlife than the changes in the ways of humanity that your discoveries might bring, which is not what a normal person is looking for. As the result, the mankind as a whole is currently a strong adept of this religion but almost every individual human being rejects it despite all the prophets are right here right now and all the miracles are real, repeatable, and performed on a daily basis.

This brings us to a funny sideline of the main discussion. I suspect that neither a person who doesn't know mathematics well, nor the one who does can tell what it really is: the first one will not even understand what the question is really about, the second one will tend to say things that just do not make sense from the normal standpoint if he tries to answer honestly because for him the question is almost equivalent to "what is God?", and we've had a few attempts to discuss that on this forum with quite dissatisfactory results. So, who is really qualified to answer then?

Of course, when I'm talking about "mathematics", I rather mean the natural sciences as a whole. An attempt to make a clear distinction between the two will result in a long and rather futile attempt to draw the borders between various sciences. If we learned anything from the 20th century at all, it is that those borders merely fail to exist. I wish the 21st century to see the spectacular fall of the Berlin wall between "physics" and "lyrics", which is still there, but this requires us to survive the 21st century in the first place, which seems sort of problematic given the current human mentality :(.


What is the length of the tangent segment from the origin to the circle that passes through the points (3,4), (6,8), (5,13).

I tried doing this and I ended up solving the equation for the circle-but its all complicated and involves huge fractions and stuff-so I dont think that was the best way to go

[hide="hint"]The secant passing through (3,4) and (6,8) also happens to pass through the origin.[/hide]
[hide="solution"]Using PPT, $T^{2}=(5)(10)$.
$T=5\sqrt{2}$[/hide]


geometry
perimeter
inequalities
geometry solved

The perimeter of a convex 1419-gon is 2005, prove that there are three points among these 1419 vertices, such that the area of the triangle formed by this three points is less than 1.

Consider three consecutive vertices. Let $a$ and $b$ be corresponding sides of 1419-gons. If $ab<2$ then we are done (since area of the sected triangle is at most $ab/2$). If $ab>2$ then $a+b\geq 2\sqrt{2}$. If this inequality holds for arbitrary triple of consecutive vertices then perimetr of 1419-gons is at least $1419\sqrt 2>2005$.


So, what state do you live in, and what did 4th place in your state competition score? I'm in Massachusetts, and 4th place here got a 39. I had a 35 (7th place) but would have gotten a second 39 if I hadn't made a stupid problem-lengthening mistake in target which cost me two points. Oh well, what's over is over, and there is less potential for stupid mistakes in later competitions, especially the more important ones. (I mean, I can't imagine dying of stupid mistakes in the USAMO) So, what about everyone else? (note the poll)

I live in CA, I got 2nd with a 40 and 4th place got a 38.

The top four Texas scores were 46, 40, 38, and 37.

I don't know. didn't notice.  For the chapter, second got over 45 and less than 46.

I believe it was 43, 37, 36, and 35. Could be wrong, though.

[quote="NoSoupForYou"]CA: 43,40,39,37[/quote]

It's 43, 40, 39, 38. I checked ealier... Sorry about misinforming you.

Actually there were 2 31's in GA, and x=33.

Which state had 45-46 cutoff... Whoever it is, I think they're lying.

[quote="YSJeong"][quote="biffanddoc"]Not sure but I think it was <28[/quote]

wow that 28 jumps out from everybody else's.  What state?[/quote]
Delaware only 97 people in the whole state :P  :)

Mathmom, NH is pretty close to you in score and location :)   30, 29, 26, 26, 26 were top 5.  Hopefully our efforts to work together on the team round will pay off at Nationals though!

Wow, Frost, that last mathlete must be pretty bummed about missing that closely. 5th in VT was 22, so the tie-break rules didn't have to come in to decide the Nationals team.  Countdown is not official in VT; is it in NH?

Is there some general rule of thumb about which states do official countdowns (like, the larger states do?)

The kid who finished 5th had a brother who took 4th a couple years ago beating out his best friend who was teid and finished 5th, so I guess the universe was just balancing things out for that family :) 6th and 7th place were both 25, so one question could have moved them all the way to 3rd.

We don't use the Countdown thankfully.  My kid who took first lost in first round of unofficial(even though he won the countdown last year as a 7th grader who was 7th)  So I guess for him anyway he still would have earned one tate championship.  We do Countdown the way it was done the very first year I think.  All 12 kids sit 7 or 8 rows up in an auditorium.  The first 7 kids to get a question right move down to the next row.  The 5 of those 7 advance a row with one question.  Then they start advancing with 2 correct, and all but one kid advances so there is a definite ranking for 1st through 5th.  It does take 3 right to win the final.

If countdown is official, I think they have to use the ladder format, in which case no one can lose more than one place, so he still would have been ok :)

The format NH uses sound kind of cool though.  VT (and our chapter) did an unofficial bracket format.

I know that I'm glad UT does CountDown--I got 5th indiv, then got to 2nd. :)

grr. I'm not.  If it didn't count I would still be second...well that's not too good either.  Stupid tie-breaker round....

What I want to know is which state(s) had a National cutoff of 45/46.

Hm...I thought CA would have that cut-off but it didn't...maybe Illinois?  They're pretty strong.

That might've only been for last year...who knows.

I doubt that IL's cutoff was 41, and how would you know from CA?

[quote="Xantos C. Guin"][quote="Secret Asian"]gar...I am.  But seriously...how did that .25=2^-x what is 2^x? get us both wrong?[/quote]

[hide]2^x is the reciprocal of 2^-x.

ANS = 4[/hide][/quote]

2^-x=1/2^x=1/4, 2^x=4, x=2...


Aight... question again

vol 2... p 85-86. I dont understand this whole thing about finding each coefficient without going through all the cubing and squaring and such. I do understand how to get the constant. I also understand the way to get to the x term... subtracting the constant and dividing by x... but thats where my problem lies (because this is in general the same way to solve all the other coefficients, the solution to this problem that i have would be necessary and sufficient (hehe) for me. )

So we write f(x) = (x+3)(q(x)) + f(-3). And then that leads to
f(x-3) = xq(x-3) + f(-3).

Then it goes to (f(x-3) - f(-3))/x = q(x-3). After this, i dont really understand any of the remaining paragraph, especially as it starts with setting x = to 0, which would make it undefined.  But even besides that, it doesnt make sense.

Thanx in advance to all who help.

I only have time to tackle the division by zero issue; hopefully someone else can pick up the rest (or maybe this will clarify enough that you can continue on your own).

The expression f(x-3) - f(-3) is a polynomial with no constant term (since the constant term of f(x-3) is f(-3), since putting in x=0 will give you the constant).  Therefore, when we divide f(x-3) - f(-3) by x, we will get a polynomial (in the text, this polynomial is q_1(x-3))and we won't have a division by 0 problem when we let x=0.  Does that make any sense?

As a simpler example, find the polynomial g(x) such that

x^2-2x+7 = xg(x) +7.

We find g(x) by subtracting 7 and dividing by x.  This is essentially what we're doing here in the text.


symmetry
geometry
conics
parabola
3D geometry
sphere

I'm kinda having trouble on understanding the loci and symmetry from my geometry book.  Can anyone help?

A locus is a set of all points that satisfy a cerain condition.  For example, the locus of all points that are r away from a point p is the circle with center p and radius r, since a circle is the set of all points equidistant from a point.

Is it such an important thing because I don't see why it's so important and having its own chapter in geometry book.  Still, I'm confued.

The concept of a locus is important because a large number of geometric objects (circles, elipses, parabolas, spheres, etc., etc.) are defined as loci of various things.

yeah, all conic sections are based on loci