inequalities
analytic geometry
geometry
geometric transformation
reflection

Q. Find the smallest value of the expression
$ sqrt(x^{2}+1)+sqrt[(y-x)^{2}+4]+sqrt[(z-y)^{2}+1]+sqrt[(10-z)^{2}+9]$
where x,y and z are arbitrary real numbers.

Find the smallest value of the expression

$ \sqrt{x^{2}+1}+\sqrt{(y-x)^{2}+4}+\sqrt{(z-y)^{2}+1}+\sqrt{(10-z)^{2}+9}$

where $ x$,$ y$ and $ z$ are arbitrary real numbers.

You get $ \sqrt{expression}$ by typing \sqrt{expression}

Ideally, we want four things:
 $ x^{2}=-1$, $ (y-x)^{2}=-4$, $ (z-y)^{2}=-1$, $ (10-z)^{2}=-9$

However, we can't get these, because the squares are always non negative.
Therefore, we want the following:
$ x^{2}= 0$, $ (y-x)^{2}= 0$, $ (z-y)^{2}= 0$, $ (10-z)^{2}= 0$

If this be attained, then we've found that the expression's minimum.
The roots are always positive, and they are going to be (at best)
$ \sqrt{1}+\sqrt{4}+\sqrt{1}+\sqrt{9}$ or $ 1+2+1+3 = 7$

That's if we can realize the bounds above.
We clearly cannot, however, because that would mean that z and y would have to be 10, but that means x is ten, and that x^2 = 100, not zero as we wanted.
So we pick values for x, y, and z that minimize the expression, and are less than ten.
$ 0\le x,y,z < 10$

and that's all I have time to work out.

Dr Sonnhard Graubner...can you give a proof?

[hide="Hint:"]
Think of it geometrically. Where can we place the last point to minimize the sum of distances? [/hide]

[hide="Note:"]
The points $ (0,0), \ \left(\frac{10}{7},1 \right) , \ \left(\frac{30}{7},3 \right), \ \left(\frac{40}{7}, 4 \right), \ \left(10,7\right)$ give a straight line segment with the total distance equal to $ \sqrt{149}$, as [b]Dr. Sonnhard Graubner[/b] suggests. But I'm not sure if this is the minimum.[/hide]

how could it not be the minimum? Any deviation from a stright line is going to increase the distance.

I had the same proof too.
[hide]The expression is the total distance of this path:
$ A(0,0)\to B(x,1)\to C(y,3)\to D(z,4)\to E(10,7)$
The shortest path between $ A$ and $ E$ is following a straight line (with distance $ \sqrt{149}$), and that will be indeed the shortest total path $ A\to B\to C\to D\to E$, because we can make sure that $ B$, $ C$, $ D$ are on the line too.[/hide]
Edit: How about this?
[hide]With an inductive extension of the triangular inequality: $ AB+BC+CD+DE\geq AE$[/hide]

[hide="Well, the problem is that"]
the last point is not necessarily $ (10,7)$, as each point can vary from the last in y-coordinate in either the positive or negative direction. So, unless we can prove that this is the absolute minimum, we'd have to figure out a way of efficiently churning through the casework of reflecting the different final points over each horizontal line we'd need to hit in order to end up, say, at $ (10,1)$ instead of at $ (10,7)$.

And example would have the path as:  $ (0,0) \rightarrow (x,1) \rightarrow (y,-1) \rightarrow (z,-2) \rightarrow (10,1)$.[/hide]

why would it matter if a line is reflected or not? distance is unaffected by reflection
all the reflections have the same distance

[quote="wonki_noh"]Ok,, I'm a bit lost here....

cincodemayo5590, could you please explain how you came up with the five points??

I don't understand how this expression can b interpreted geometrically.....

thanks![/quote]

[hide="Because"]
Imagine a cartesian plane, the expression $ \sqrt{x^{2}+1^{2}}$, by pythagoras theorem, is the distance from the origin to the point (x,1). $ \sqrt{(y-x)^{2}+2^{2}}$ can be thought of as the distance from (x,1) to (y,3) etc.
[/hide]

[hide="Solution"]
Geometrically, it's going to trace the lines through points starting from (0,0) and ending at (10,7). The shortest distance between two points is a straight line, so the minimum is $ \sqrt{10^{2}+7^{2}}= \sqrt{149}$
[/hide]


geometry
3D geometry
pyramid
vector
Pythagorean Theorem
area of a triangle
Heron\u0027s formula

On a $4 \times 4 \times 3$ rectangular parallelepiped, vertices $A$, $B$, and $C$ are adjacent to vertex $D$. The perpendicular distance from $D$ to the plane containing
$A$, $B$, and $C$ is closest to

$\text{(A)}\ 1.6 \qquad \text{(B)}\ 1.9 \qquad \text{(C)}\ 2.1 \qquad \text{(D)}\ 2.7 \qquad \text{(E)}\ 2.9$

[quote="4everwise"][hide="Solution"]First, let us examine pyramid $ABCD$. Notice that the perpendicular segment from $D$ is simply the altitude of the pyramid to base $\triangle ABC$. Because the volume of a pyramid is $\text{Volume}=\frac 13 (\text{base})(\text{altitude})$, we can find the length of the perpendicular segment if we find the two other variables.

We can use altitude $AD$ to $\triangle BCD$ to find the volume. $V=\frac 13 (8)(3)=8$. Now, we need to find the area of $\triangle ABC$. Applying the Pythagorean Theorem, $AC=AB=5$, and $BC=4\sqrt 2$. A quick application of Heron's Theorem gives $\text{area}\triangle ABC=2\sqrt{34}$.

Let the length of the perpendicular segment be $x$. We now have $8=\frac 13 (2\sqrt{34})(x)$ and solving, $x=\frac{12}{\sqrt{34}}\approx 2.1$[/hide][/quote]

Just because I don't enjoy using Heron's Formula I wanted to poitn out that the area of ABC could be found easily by Pythagorean Theorem because it is an isosceles triangle.  The median bisecting the base of the isosceles triangle would be the height of the triangle, with base 4 squareroot of 2, and would equal sqrt ( 25 - 8 ) = sqrt (17).  The area would be (1/2)(4 sqrt 2)(sqrt 17)= 2 sqrt (34)

this is also a good problem to test your ability in analytic geometry

[hide="easy vector solution"]

This parallelepied can be formed by the vectors (4,0,0), (0,4,0), and (0,0,3). This makes the points corresponding to these vectors three adjacent vertices, with (0,0,0) a vertex next to each of them.

Using the intercepts of this plane, we easily find that its equation is $3x+3y+4z=12$.

Note that a vector that starts at (0,0,0), ends at the plane, and is perpendicular to the plane will have magnitude equal to the distance from that point to that plane. We shall find the distance by finding this vector.

Based on this equation, (3,3,4) must be a vector that is perpendicular to the plane (well-known vector fact). In general, any vector perpendicular to the plane is of the form (3t,3t,4t), where t is a real number. Thus, we solve $3(3t)+3(3t)+4(4t)=12=9t+9t+16t=34t$, or $t=\frac{6}{17}$. Our vector is then $(\frac{18}{17},\frac{18}{17},\frac{24}{17})$, and its magnitude is $\sqrt{\frac{18}{17}^{2}+\frac{18}{17}^{2}+\frac{24}{17}^{2}}\approx 2.05$. This is cloesest to 2.1, so $\boxed{(C)}$.[/hide]

[hide="Another Solution"]Set $C$ to $(0,0,0)$ in the xyz-plane, and set $D=(0,0,3)$, $B=(4,0,3)$ and $C=(0,4,3)$. Then let the equation of the plane be $ax+by+cz+d=0$ (the plane containing $A,B,C$).

Then substituting in $A,B,C$ we get $d=0$
$4a+3c=0$ and $4b+3c=0$. An  equation of the plane is therefore $3x+3y-4z=0$.

Using the point-to-plane distance formula, the absolute value of the distance is

$\frac{3\cdot0+3\cdot0-4\cdot3}{\sqrt{3^2+3^2+4^2}}=-\frac{12}{\sqrt{34}}$.

And this is closest to $-2.1$, so [b](C) 2.1[/b] is the answer.[/hide]


I need help with this question, it's pretty tough:

Let n be a 5-digit number, and let q and r be the quotient and remainder, respectively, when n is divided by 100.  For how many values of n is q + r divisible by 11?

Also, if anybody could offer any hints instead of full solutions, that'd be great.  I'm pretty sure the solution involves Mods, but I'm not good at all with Mods so I'm trying to find a different way to solve it.  Thanks!

Big hint:

[hide]100(q)+r=n. If q+r is divisible by 11, also 99q is always divisible by 11.

Then q+r+99q=100q+r=n will be divisible by 11. So we want to find how many 5-digit numbers are divisible by 11[/hide]

Oh thanks Beta.  I have one question though.  When you added the 99r (I think you meant 99q), did you add it because it was a multiple of 11 and thus would not change whether q + r was divisible by 11 or not?
What I mean is, let's say q + r = 3.  3 is not divisble by 11, so if we add 11 or something that is a multiple of 11 to the equation, it'll still go back to being a matter of whether q + r  is divisible by 11?

That was pretty smart what you did right there.  I wonder if there are any other ways to solve it.

Beta never adds 99r or 99q anywhere. He does this:

Obviously,

[tex]n = 100q + r = 99q + q + r. [/tex]

Of course, 99 q = 11*9q, so its divisible by 11. And we want q+r to be divisible by 11. So then 100q+r = n will be divisible by 11.

Hope this helps!

Which is thus mods without naming the word mods :lol:

You see, mods are really simple though, once you got them.

i did this another way because i didnt notice the nice relationship:

[color=white]if [i]q[/i] is not divisible by 11, then for each [i]q[/i] there are 9 solutions for [i]r[/i], one from each of the ranges 0-10, 11-21, 22-32, etc.
if [i]q[/i] is divisible by 11, then for each [i]q[/i] there are 10 solutions for [i]r[/i]: 0,11,etc.

there are 900 possibilities for q (from 100 to 999), 900/11=~81 of which are divisible by 11

therefore the answer is 81*10+(900-81)*9 = [b]8181[/b]
...so this is also the # of 5-digit numbers divisible by 11[/color]


Find all the $ (a,b)\in \mathbb{N}\times \mathbb{N}$ such that $ a(a\plus{}1)\equal{}4b(b\plus{}1)$ .


Consider $ f: \mathbb{N}\to\mathbb{N}$ satisfying next three conditions:

$ (1)f(xy)\equal{}f(x)\plus{}f(y)\minus{}1$ for all $ x,y\in\mathbb{N}$.
$ (2)$There are only finite integers $ x$ such that $ f(x)\equal{}1$.
$ (3)f(30)\equal{}4$

Find $ f(14400)$.

Well........

[hide="Solution"]

I noticed $ 14400 \equal{} 30^{2} \times 16$, so if I can find what $ f(16)$ is then I should be fine. If I have a number $ x \in\mathbb{N}$ such that $ f(x)\equal{}1$ then it reasons that $ f(x^{2})\equal{}1$ as well and so on giving a list of numbers, $ x,x^{2},x^{4},......$ all of which give $ 1$ when substituted into the function. Since the criterion stated that $ f(x)\equal{}1$ can only have finite number of solutions then $ x\equal{}1$ is the only possible value (and it indeed gives $ 1$).

$ f(2)\plus{}f(15) \equal{} 5$ and since neither $ f(2)$ or $ f(15)$ can equal $ 1$ so $ f(15) \equal{} 2,3$ since if $ f(15)\equal{}2$ then we have the following $ f(3)\plus{}f(5) \equal{} 3$ and one of $ f(3)$ or $ f(5)$ must be $ 1$, a contradiction. Thus we have $ f(15)\equal{}3$ which means $ f(2)\equal{}2$.

So a bit of computing gives $ f(16)\equal{}5$ thus $ f(14400) \equal{} 11$. [/hide]

I hope I didn't make any computational errors.

[quote="BanishedTraitor"]Well........

[hide="Solution"]

I noticed $ 14400 \equal{} 30^{2} \times 16$, so if I can find what $ f(16)$ is then I should be fine. If I have a number $ x \in\mathbb{N}$ such that $ f(x) \equal{} 1$ then it reasons that $ f(x^{2}) \equal{} 1$ as well and so on giving a list of numbers, $ x,x^{2},x^{4},......$ all of which give $ 1$ when substituted into the function. Since the criterion stated that $ f(x) \equal{} 1$ can only have finite number of solutions then $ x \equal{} 1$ is the only possible value (and it indeed gives $ 1$).

$ f(2) \plus{} f(15) \equal{} 5$ and since neither $ f(2)$ or $ f(15)$ can equal $ 1$ so $ f(15) \equal{} 2,3$ since if $ f(15) \equal{} 2$ then we have the following $ f(3) \plus{} f(5) \equal{} 3$ and one of $ f(3)$ or $ f(5)$ must be $ 1$, a contradiction. Thus we have $ f(15) \equal{} 3$ which means $ f(2) \equal{} 2$.

So a bit of computing gives $ f(16) \equal{} 5$ thus $ f(14400) \equal{} 11$. [/hide]

I hope I didn't make any computational errors.[/quote]


hi everyone,

can somebody help me with the following question? I'm not so sure how to do them.

For Q1-3, indicate whether it is in NP.

1.) Given a graph G(V,E) and an integer k. Does G have at most k different Hamiltonian Cycles?
2.) Given a graph G(V,E) and an integer k. Does G have at least k different Hamiltonian Cycles?
3.) Given a list L of integers of length 2k and an integer x. Are exactly k items of L greater than or equal to x?

I believe 2. and 3. are both in NP while 1. is not.

Q4. Are there any other single propositional connectives that are complete other than nand and nor?

I can't find any, but how should I go about proving it?

Thanks for all your help!

[quote="mathbeauty"]
1.) Given a graph G(V,E) and an integer k. Does G have at most k different Hamiltonian Cycles?
2.) Given a graph G(V,E) and an integer k. Does G have at least k different Hamiltonian Cycles?
3.) Given a list L of integers of length 2k and an integer x. Are exactly k items of L greater than or equal to x?

I believe 2. and 3. are both in NP while 1. is not.
[/quote]
You are right: 2 and 3 are in $NP$. 1 is not in $NP$ assuming $NP\neq coNP$.
Hints:
I. Find a witness for (2). Construct an m-reduction from the Hamiltonian Cycle problem to (2). Conclude that (2) is $NP$-complete.

II. Therefore, (1) is $coNP$ complete. Conclude that (1) is not in $NP$ assuming $NP\neq coNP$.

III. Show that (3) is in $P$.

[quote]
Q4. Are there any other single propositional connectives that are complete other than nand and nor?
[/quote]
Sorry, I do not understand what you are asking.

[quote="mathbeauty"]
I can't find any, but how should I go about proving it?

Thanks for all your help![/quote]
XOR and XNOR are both complete. That would be why you are having problems proving that NOR and NAND are the only complete logic gates.

I don't think that either one of your connectives is complete [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=23035]WatsonLadd[/url] - how would you construct for example a negation?

NAND and NOR are indeed the only complete connectives there are. To prove this, you should realize that any complete connective $\diamond$ must necessarily yield $0\diamond0=1$ and $1\diamond1=0$. This has a very simple reason, think about it. Now you have only four possibilities - two of them are obviously not going to work (do you see why?) and the other two are NAND and NOR.


2001 coins, each value 1, 2 or 3 are arranged in a row. Between any two coins of value 1 there is at least one coin, between any two of value 2 there are at least two coins, and between any two of value 3 there are at least three coins. What is the largest number of value 3 coins that could be in the row?

in any $4$ adjacent coins we have at most one $3$ therefor we have $500$ boxes containing $4$ coins and some other coin so we have at most $501$ 3 coins:
now we give an example:
$3,1213,1213,1213,...,1213$
in this example we have $500$ boxes containing elements $1213$
so the maximum val8ue is $501$


limit
trigonometry
analytic geometry
graphing lines
slope
function
trig identities

[b]Calculate:[/b] $\lim_{x\to\frac{\pi}{2}}\left(\frac{1}{\cos x}-\tan x\right)$

Trig identities can be nice. Write it over a common denominator as $\frac{1-\sin x}{\cos x}.$ Now what happens if you "rationalize the numerator" by multiplying by $\frac{1+\sin x}{1+\sin x}\ ?$

Thank you very much. And anorther:
[b]Calculate:[/b] $\lim_{x\to 0}\frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^{3}}$

[quote="April"]Thank you very much. And anorther:[/quote]

$L = \lim_{x\to 0}\frac{\sqrt{1+tanx}-\sqrt{1+sinx}}{x^{3}}$

1st aproach - power series:

$\sqrt{1+tanx}= 1+\frac{1}{2}tanx = 1+\frac{1}{2}\left(x+\frac{x^{3}}{3}\right) = 1+\frac{x}{2}+\frac{x^{3}}{6}$

$\sqrt{1+sinx}= 1+\frac{1}{2}sinx = 1+\frac{1}{2}\left(x-\frac{x^{3}}{6}\right) = 1+\frac{x}{2}-\frac{x^{3}}{12}$

And so:

$L = \lim_{x\to 0}\frac{1+\frac{x}{2}+\frac{x^{3}}{6}-1-\frac{x}{2}+\frac{x^{3}}{12}}{x^{3}}= \lim_{x\to 0}\frac{\frac{x^{3}}{4}}{x^{3}}= \frac{1}{4}$

2nd aproach - algebric manipulation: let's multiply and divide $L$ by $(\sqrt{1+tanx}+\sqrt{1+sinx})$ to get

$L = \lim_{x\to 0}\frac{1+tanx-1-sinx}{x^{3}(\sqrt{1+tanx}+\sqrt{1+sinx})}= \lim_{x\to 0}\frac{sinx(1-cosx)}{x^{3}cosx(\sqrt{1+tanx}+\sqrt{1+sinx})}= \lim_{x\to 0}\frac{sinx}{x}\frac{1}{cosx}\frac{1-cosx}{x^{2}}\frac{1}{\sqrt{1+tanx}+\sqrt{1+sinx}}= 1\cdot1\cdot\frac{1}{2}\cdot\frac{1}{2}= \frac{1}{4}$

[quote="Kent Merryfield"]Trig identities can be nice. Write it over a common denominator as $\frac{1-\sin x}{\cos x}.$ [/quote]

$\frac{1-\sin x}{\cos x}=-\frac{\sin x-1}{\cos x-0}.$

You can imagine this equation as the slope of passing through two points the fixed point $A(0,\ 1)$ and the moving point $P(\cos x,\ \sin x)$ on the unit circle. This limit is seems to related to cycloid.

theres an easy way, instead of doing all that work jus plug in a value very close the limit and u'll get the answer. Like instead of putting 0 u put 0.00000000000000000000000000000000000000001. :o

Bad idea, =s. Just what do you mean by "plug in," anyway? I have to assume that you're talking about a calculator or other floating-point arithmetic device. What does that tell you about, say, $\lim_{x\to0}\frac{1-\cos x}{\sin^{2}x}\ ?$ That's not a difficult problem, and there are numerous ways of showing that the limit is $\frac12.$ But what happens if you put $x=10^{-20}$ into that function on a calculator? The calculator thinks that the quotient is zero.

Here's the exercise: Consider $\lim_{x\to0^{+}}\frac{x^{x}-1}{x}.$ On the basis of numerical exploration ("number-plugging") can you determine that limit? Can you tell even whether it is finite or infinite?

I doubt you can do it without a calculator and I equally doubt your calculator will give you anything close to the right answer if you attempt it with the exact number you mentioned (or rather with $\pi/2$ plus or minus it) ;) Besides, how do you know that 40 zeroes before the 1 are enough; maybe you need 1000 of them to get reasonably close to the true limit. Also, you'll never get the exact value this way, so how will you know that 0.24999...999 is in fact 0.25 and not  0.24999...9997532758...? Of course, the more digits you see in this case, the more plausible it becomes that your guess is correct but you'll never been able to prove it by just plugging numbers in. And imagine that the right answer were $17/191$. I doubt you would be able even to guess that number from seeing the first 20 decimal digits of it (which is much more than a typical calculator allows you to see). The best you would be able to do is to pick a right answer on a multiple choice test, but this problem is not about picking numbers on a multiple choice test...

[quote="Kent Merryfield"]$\lim_{x\to0^{+}}\frac{x^{x}-1}{x}.$[/quote]

Is the limit, by any way, minus infinity?

Yes, $-\infty.$

[quote="fedja"]... or rather with $\pi/2$ plus or minus it ...[/quote]

On a typical calculator, $\frac{\pi}2\pm 10^{-30}$ is exactly the same as $\frac{\pi}2.$ The tangent of that number is probably "error" or "overflow" but you'd have to push your own calculator buttons to be sure of that. The floating-point numbers, endowed with operations as defined by the keys on the calculator, are not a field.


let $a,b,c \geq -1$  and $a+b+c=\sqrt[3]{4}-1.$
 find the minimum value of $P = a^3+b^3+c^3$

By the Power Mean Inequality (using the cubic and arithmetic means), $\sqrt[3]{\frac{P}{3}} \geq 3(\sqrt[3]{4} - 1)$.  Rearranging gives $P \geq 81(\sqrt[3]{4}-1)^3$.

Is there a stronger bound?


algebra
polynomial
complex analysis
integration
inequalities

Let $ f(z)=a_0+a_1z+...+a_nz^n$ be a non-constant complex polinomial . Prove that there exists $|z|\leq 1$ such that $|f(z)|\geq |a_0|+\frac{|a_1|}{n}$

How do you think: is College Playground more suitable place for this problem?

This is one of the problems used for 2003 Chinese IMO team Training....

I also thought that it can be done using integrals and analysis, but I just could not find such a solution. Can anyone post a solution?

I have just realized that it is trivial! :blush:

If $a_1=0$ then, obviously, we are done :)
Suppose $a_1\neq 0$, also we may assume $a_0\in \mathbb R$ and $a_0\geq 0$. Consider equation \[(a_0+|a_1/n|)-(a_0+a_1z+...+a_nz^n)=0.\]
It is sufficient to prove that this equation has a root in $|z|\leq 1$. Indeed, it is equaivalent to \[na_nz^n+...+na_2z^2+na_1z-|a_1|=0.\]
Dividing to $|a_1|$ and substituting $z=1/x$ we obtain $x^n+b_1x^{n-1}+...+b_n=0$, where $|b_1|=n$. This equation has $n$ zeros, and their sum is $b_1$. It follows that there is the root $x_0$ s.t. $|x_0|\geq 1$, then $z_0=1/x_0$ is the required root of the initial equation.

Thanks a lot, Myth. It is always the same with this kind of problems: either you have the idea and then it's trivial, else you search like a blindman.

Actually, I think that the lower bound $ |a_{0}| + \frac{|a_{1}|}{n} $ can be
replaced by $ |a_{0}| + \gamma |a_{1}|$, for some universal constant
$ \gamma $, independent of $ n $.
Here are my arguments:
\\
Suppose we have a complex polynomial $ f(z) = a_{0} + a_{1}z + \ldots + a_{n}z^{n} $.
I would like to find constant $ \gamma_{n} $ such that there always exists $ |z| \leqslant 1 $
such that $ |f(z)| \geqslant |a_{0}| + |a_{1}|\gamma_{n} $. Let us find inductively those
$ \gamma_{n} $. For $ n = 1 $ it is clear that we can take $ \gamma_{1} = 1 $.
Now suppose we know $ \gamma_{n-1} $. Consider a polynomial
$ f(z) = a_{0} + a_{1}z + \ldots + a_{n}z^{n} $. Take a number
$ \xi = e^{\frac{2\pi i}{2n}} $. Define a polynomial $ g(z) = \frac{1}{2} (f(z) + f(\xi z)) $.
By this we have killed the leading coefficient of $ f(z) $, so $ g(z) $ is a polynomial of
degree less or equal to $ n-1 $, and $ g(z) = a_{0} + \frac{1 + \xi}{2} a_{1} z +
\frac{1 + \xi^{2}}{2} a_{2} z^{2} + \ldots + \frac{1 + \xi^{n-1}}{2} a_{n-1} z^{n-1} $.
Applying the already known inequality to $ g(z) $, we get that there exists  $ |z| \leqslant 1 $
, such that $ |g(z)| \geqslant |a_{0}| + |\frac{1 + \xi}{2}||a_{1}|\gamma_{n-1} $, so
if we will define $ \gamma_{n} = |\frac{1 + \xi}{2}|\gamma_{n-1} $ then
$ max(|f(z)|,|f(\xi z)|) \geqslant \frac{1}{2} |f(z) + f(\xi z)| = |g(z)| \geqslant
|a_{0}| + |\frac{1 + \xi}{2}||a_{1}|\gamma_{n-1} = |a_{0}| + |a_{1}|\gamma_{n} $.
So we see that we can take $ \gamma_{n} = |\frac{1 + \xi}{2}| \gamma_{n-1} $.
Now, $ |\frac{1 + \xi}{2}| =  |\frac{1 + e^{\frac{2\pi i}{2n}}}{2}| =
= |\frac{e^{\frac{\pi i}{2n}} + e^{- \frac{\pi i}{2n}}}{2}| = cos (\frac{\pi}{2n}) $.
So we see that we can take $ \gamma_{n} = cos(\frac{\pi}{4}) cos(\frac{\pi}{6}) \ldots
cos(\frac{\pi}{2n}) $, $ n \geqslant 2 $.
We see that the sequence $ \gamma_{n} $ is a decreasing
sequence of positive numbers, and denote by $ \gamma $ it's limit. Let us show that
$ \gamma > 0 $. We know that for small $ x > 0 $, we have $ cos(x) > 1 - x^{2}/2 $, so
it is enough to show that the infinite product
$ \prod_{n=2}^{\infty} (1 - \frac{1}{2} (\frac{\pi}{2n})^{2} ) =
\prod_{n=2}^{\infty} (1 - \frac{\pi^{2}}{8} (\frac{1}{n^{2}})) $
is non-zero, which is equivalent
to the fact that $ \frac{\pi^{2}}{8} \sum_{n=2}^{\infty} (\frac{1}{n^{2}}) < \infty $,
which is true.

We can extend your great idea in order to obtain $|f(z)|\geq |a_0|+\gamma_1|a_1|+\gamma_2|a_2|+...$. It is really nice!
Strange, I never see similar results. Maybe the specialists will say us something?
What is equivalence for $\gamma_k$?

I see someone have already placed it to Solved Section. I think, we should extract it back for further discussion ;)

I don't think one can ask about "equivalence for $\gamma_k$" here because if one tries to maximize one $\gamma_k$, he has to make other gammas smaller. Each particular $\gamma_k$ can be made as large as a fixed constant (the inequality $\max |f|\geq |a_0|+\gamma |a_1|$ immediately implies the same inequality with any $a_k$ in place of $a_1$ just because we may consider the function $g(z)=\frac1k\sum_{0\leq m< k}f\left(e^{2\pi i\frac mk}z\right)=a_0+a_k z^k+a_{2k}z^{2k}+\dots$). Leva1980's result can be proved in many ways, but his own proof is the most elementary one and really nice.
BTW, the sharp estimate independent of $n$ is $\max|f|\geq \sqrt{|a_0|^2+\frac{|a_1|^2}{4}}+\frac 12|a_1|$

Fedja, could you please some other nice proofs for the beautiful result Leva proved? And for the sharp estimate? Thank you.

Here is the proof of the sharp estimate. As to other proofs, try to find them yourself! I do not want to spoil the fun... ;)
[hide="click here for the proof"]
WLOG, $a_0=1$. Let $M=\max|f|$. Then $f$ analytically maps the unit disk $\mathbb D$ into the disk $M\mathbb D$ and $f(0)=1$. Now comes a mighty theorem: among all such mappings, the conformal mapping has the greatest derivative in absolute value (this is just the Schwarz lemma in disguise). The corresponding conformal mapping is $z\to \frac{Mz+1}{1+\frac z M }$. Its derivative at $0$ is $M-\frac1M$, so we get $M-\frac 1M\geq |a_1|$. The rest is clear.
[/hide]

[quote="Myth"]Actually, I meant that Leva's approach allows to construct some particular sequence $\{\gamma_k\}$, and question was what is the equivalence for this sequence.[/quote]
I apologize for misunderstanding, Myth. Can you then be more specific about how exactly you find $\gamma_2$ and so on in your generalization of Leva's approach?

I thought we can do it in the same way: define $\gamma_2^{(1)}=1$ and then recurrence as for $\gamma_1$. Does it work?

[quote="fedja"]Not quite: this way you get some $\gamma_2$ but you change your $\gamma_1$
(may lose it completely...)! The question is how you get the entire series...[/quote]
Sorry. Now I see my flaw: really I need to correctly choose initial datas $\gamma_1$ and $\gamma_2$ in order they satisfy $|f(z)|\geq |a_0|+\gamma_1|a_1|+\gamma_2|a_2|$ for some $z$ for all quadratic polynomials.

We will work with quadratic polynomials.
Consider $f'(z)=a_1+2a_2z$. We can find $z$ s.t. $|z|=1/2$ and $|f'(z)|=|a_1|+|a_2|$, but $f'(z)=\frac{1}{2\pi i}\int_C\frac{f(\xi)}{(\xi-z)^2}d\xi$, where $C$ is a circle with center $z$ and radius $1/2$. Therefore, there is $z'\in C$ s.t. $|f(z')|\geq \frac{|a_1|+|a_2|}{2}$.

It follows that $|f(z)|\geq \max({|a_0|+|a_1|,\frac{|a_1|+|a_2|}{2}})$ for some $z$. And we can choose any intermediate value to obtain $\gamma_1$ and $\gamma_2$.
But anyway, Fedja totally right: we should search for compromise each time. In this way we can construct mentioned sequence $\gamma_n$.

[quote="fedja"]Yes, that's what I meant. Sorry for not making it clear the first time. Actually, I just realized that one can completely describe all admissible non-negative $\gamma_k$ in the inequality $\max|f|\geq |a_0|+\gamma_1|a_1|+\gamma_2|a_2|+\dots$. The answer is $\sum_{k\geq 1}\gamma_k\leq \frac12$.[/quote]
It is very interesting!

[quote="Myth"]Can you prove your assertion, Fedor?[/quote]
Actually it is easy: if $\sum_k\gamma_k\leq \frac12$, we have $\max|f| \geq |a_0|+\frac12\max|a_k|\geq |a_0|+\sum_{k\geq 1}\gamma_k |a_k|$. On the other hand, if we take the conformal mapping from my post #14 for $f$ with $M\to 1+$, we shall have $\max|f|=\sqrt{1+\frac{|a_1|^2}4}+\frac12|a_1|=1+\frac12|a_1|+o(|a_1|)$ and also we shall have $|a_k|=|a_1|+o(|a_1|)$ for all $k\geq 1$. From here it is easy to derive that if $\sum_k\gamma_k>\frac12$, the inequality $\max|f| \geq |a_0|+\sum_{k\geq 1}\gamma_k |a_k|$  will be violated for all $M$ sufficiently close to $1$.


with $xyz \geq 1$
i think we can prove $\frac{x^a}{x^a+y+z}+\frac{y^a}{x+y^a+z}+\frac{z^a}{x+y+z^a} \geq 1$
but i wonder $\frac{x}{x^a+y+z}+\frac{y}{x+y^a+z}+\frac{z}{x+y+z^a} \leq 1$
for $a>1$

Indeed, both ineqaulities are true.
See http://www.mathlinks.ro/Forum/viewtopic.php?p=284460#p284460.
Moreover, I conjectured there that these inequalities are also valid for the better condition $x+y+z \ge 3$.

thanks.
but i can't find the solution for $ 1 \ge \sum\frac{x_1}{x_1^q+x_2+...+x_n}$
can you help me? Vase
thanks again


number theory
greatest common divisor
number theory unsolved

Prove that equation \[2x^{2}-1=y^{15}\] has no solutions in natural numbers with $x>1$.

For all x $5\not |2x^{2}-1$, therefore $5\not |y$.

Oooooops...

I probably wrote it wrong, will repair it soon

Bye

When $p=\top$ and $q=\bot$ then $q \Rightarrow p=\top$ so in a strange way this statement that I wrote is true, but never mind, I edited it, so try now...

a much more general form of this eqution is [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=28590]here[/url]
and as I promised I'll post the solution soon.

I suppose with the same kind of reasoning we could prove if there exists solution then, $\exists k,m \in \mathbb{N}/ \frac{y+1}{2}=15k^{2}, \;\frac{y^{3}+1}{2}=5m^{2}$
But, uhh, I think the best for me know should be go and check Albanian eagle's link. :roll:
Thanks.  :lol:

Um I'm not sure about this but just look mod 9:
All 15th powers are -1,0,1, right? So x=1,0 mod 9
but we can use infinite reduction on x=1 mod 9 case
and I think we can also on x=0 mod 9 case
thats the idea, i have to go

I wonder how this is supposed to work, since it would probably also work for $2x^{2}-1=y^{3}$, which has a solution $(x,y)=(78,23)$.

Good point, you probly can't get anywhere with the idea i posted...


FYI we are working on the rotational motion chapter...

- To drive a typical car at 40 mph on a level road for one hour requires about 3.2x10^7 J of energy. Suppose one tried to store this much energy in a spinning flywheel which was then coupled to the wheels of the car. A large flywheel cannot be spun too fast or it will fracture. If one used a flywheel of radius 0.60m and mass 400 Kg, what frequency of revolution would be required to store 2X10^7 J? Incidentally, 2500 rpm is about the maximum feasible rate of revolution with present materials technology for such a flywheel.

If someone could help me out, that would be great =)

Starting from simpler principles:

An object of mass $ m$ and velocity $ v$ has kinetic energy $ \frac12mv^2.$

Now suppose this mass is $ r$ units from the center of a circle and is in uniform circular motion. Then from the arclength formula $ s \equal{} r\theta,$ we can differentiate to get $ v \equal{} \dot{s} \equal{} r\dot{\theta} \equal{} r\omega$ where we're using $ \omega$ for the angular velocity in radians per second. Now the kinetic energy of our particle is $ \frac12\,mr^2\omega^2.$

Now decompose that cylindrical flywheel into shells (as in the "shell method" for computing volumes.) The depth of the cylinder is $ h$ and it is made of a material of density $ \sigma.$ That makes the mass element $ dm \equal{} 2\pi \sigma hr\,dr.$ Integrate that to get the mass, using $ R$ as the radius of the cylinder.

$ M \equal{} \int_0^R2\pi \sigma hr\,dr \equal{} \pi\sigma h R^2.$ But then $ \sigma h \equal{} \frac {M}{\pi R^2}.$ Put that into what we had before and we get

$ dm \equal{} \frac {2M}{R^2}\,r\,dr.$

Now calculate the kinetic energy of our spinning flywheel:

$ K \equal{} \int\frac12\,r^2\omega^2\,dm \equal{} \int_0^R\frac {2M}{2R^2}\,r^3\omega^2\,dr \equal{} \frac {MR^2\omega^2}{4}.$

In the problem, we know $ M,R,$ and $ K$ but do not know $ \omega.$ Solve the equation I just derived for $ \omega:$

$ \omega^2 \equal{} \frac {4K}{MR^2},$ so $ \omega \equal{} \sqrt {\frac {4K}{MR^2}}.$

Putting in the quantities we have, we get $ \omega\approx 9.4\times 10^2\,\frac {\text{radian}}{\text{sec}}.$

[color=green][Added in edit: the above calculations use $ K\equal{}3.2\times 10^7\text{ J}.$ For the correction to $ K\equal{}2\times 10^7\text{ J},$ see several posts below.[/color]

We convert this to rpm: multiply by $ 60$ to convert seconds to minutes, divide by $ 2\pi$ to convert radians to revolutions. We get an angular velocity of about $ 9\times 10^3\text{ rpm}.$ This is 3 or 4 times the stated practical limit for flywheel speed. Note also how big and heavy this flywheel is - two feet or so across, nearly 1000 pounds. That's a pretty severe addition to the weight of a passenger car.

In practice, a flywheel could be used as part of a regenerative braking scheme: when you need to stop, store the energy rather then dissipating it as heat, then recover it to accelerate back from a stop. But it's not that practical to store the energy for a long trip.

One other factor, also in your physics rotational motion chapter: So we're driving down the road with this honking big flywheel/gyroscope in the back of the car. What's going to happen when we try to steer around a corner? (The answer may depend on the orientation of the flywheel.) Also: will the act of acceleration or decelerating the flywheel (in other words, energy transfer between the flywheel or the drive train) put any torques on the car that might affect steering?

everything i get except for two thing: a) why is K=(1/4)m(r^2)(w^2) in stead of (1/2)m(r^2)(w^2)? all of my physics books say 1/2 but ive seen 1/4 used before. and b) the question wants an answer in hertz?? (thats what mainly confused me about it) the answer choices are a- 119 Hz b-688 Hz c- 455 Hz d- 66.7 Hz. and btw thanks a ton for helping me with this problem, especially in such great detail.

oh and in reguards to your other post, I dont think so. the problem doesnt include any other information and we have never dealt with a problem that had anything like that in it, so i think not.

I was trying to work things from scratch, and I won't swear I got everything right - but so far I haven't found a mistake.

As for the $ \frac14:$ The "rotational inertia" of the cylinder (rotated about its axis) is $ \frac12MR^2.$ So that's a $ \frac12$ there. But there's another $ \frac12$ in the equation for kinetic energy:
\[ K\equal{}\frac12I\omega^2\equal{}\frac12\left(\frac12MR^2\right)\omega^2\equal{}
\frac14MR^2\omega^2\]
I gave a result in radians/sec. To convert that to Hz (which is, in this context, revolutions per second), divide by $ 2\pi.$

Ah - I now see one mistake. Above, I was using $ 3.2\times 10^7\text{ J},$ which was in the preamble to the problem - but instead, the question asks about $ 2\times 10^7\text{ J}.$ Putting that number in, I get an angular velocity of about $ 750$ radians per second. This is about $ 7000\text{ rpm},$ still well outside the feasible range. Convert to Hz and I get $ 119\text{ Hz},$ which does match one of your answers.

By the way, there is a physics forum on this site, and I think I'll move this thread there. I think the original question is now closed, but I am still interested in comments on whether a car with a big, fast-spinning flywheel would have any noticeable steering problems.

hmmm good question, and thanks for all of ur help! i think it works out right now =) thanks.

Thank you Dr.Kent for moving it here ...i had seen this interesting problem the Calculus forum :)
well of course you are right the $ K \equal{} \frac {I \omega^{2}}{2}$
Coming to the point of steering problems while turning around at corners....there are two things taking here
1)the car can slip(that is it may skid on the surface)
2)the car may topple that is overturn

it depends on the car construction which will happen first if at all anything has to happen

for slipping doe not happen(Assuming the rod is not banked) if the friction by the ground is sufficient to turn the car in the particular turn say circular of radii $ R$(measured from car's centre of mass)
then we have the friction $ f_{max} \geq \frac {mv^{2}}{R}$ taking into consideration the motion of the centre of mass
but $ f_{max} \equal{} 2(N_{1} \plus{} N_{2})\mu$ where $ m$ is it's mass $ \mu$ is the coefficient of friction.Taking the normal force to be $ N_{1}$ on inner tyres and $ N_{2}$ on outer ones
Aslo note that the car shud not topple now in this condition depend on torque due to various forces must be balanced.
Also observe that the friction is greater at the outer wheel as compared to inner since it executes a a turn of greater radius with the same angular speed.
This means that the outer wheel gets a stronger normal force so the car has chance of overturning due to this torque
Let the width of the car be $ 2l$ and it's height from the ground of centre of mass be $ d$
now lets calculate the torque about the centre of mass
we have the torque's due to $ N_{2}$ on one direction and rest all forces on the other so
$ N_{2}l \equal{} N_{1}l \plus{} (f_{1} \plus{} f_{2})d \Rightarrow \frac {N_{2} \minus{} N_{1}}{N_{2} \plus{} N_{1}} \leq \frac {\mu d}{l}$ but as the car is stable on ground $ N_{2} \plus{} N_{1} \equal{} \frac {mg}{2}$
so we must have  $ N_{2} \minus{} N_{1} \leq \frac {mg \mu d}{2l}$
actually we can now find $ N_{1}$ in terms of $ N_{2}$ and put a bound on the normal forces...if we are given the maximum stress the 'springs' of the car can sustain this problem will get even interesting :)


geometry
circumcircle
geometric transformation
reflection
geometry proposed

Let $ ABC$ be an acute triangle, and $ AB$＜$ AC$, Let $ D$ be the foot of altitude of triangle $ ABC$ from vertex $ A$, A point $ P$ is on the segment $ AD$. Points $ E,F$ are the projections of point $ P$ on its sides $ CA,AB$,respectively.Denote by $ O_{1},O_{2}$ the circumcenters of triangle $ BDF,CDE$,respectively.Prove that $ O_{1},O_{2},E,F$ are concyclic if and only if $ P$ is the orthocenter of triangle $ ABC$.

Dilating by a factor of two around P, we something angle-chaseable...

We have the quadrilaterals $ BDPF$, $ CDPE$ are cyclic, so $ O_1$, $ O_2$ are the middle points of the segments $ PB$, $ PC$, respectively.

$ \bullet$ If $ P$ is the orthocentre of triangle $ ABC$, then $ B$, $ O_1$, $ P$, $ E$ are collinear and $ C$, $ O_2$, $ P$, $ F$ are also collinear. Hence we have $ \angle FO_2O_1 = \angle PCB = \angle PEF$, i.e. the quadrilateral $ O_1O_2EF$ is cyclic.

$ \bullet$ If the quadrilateral $ O_1O_2EF$ is cyclic.
We have
\begin{eqnarray*} 180^\circ = \angle O_1O_2E + \angle EFO_1 & = & \angle PO_2O_1 + \angle PO_2E + \angle EFP + \angle PFO_1 \\
& = & \angle PCB + 2\angle PCE + \angle PAE + 90^\circ - \angle PBF \\
& = & \angle ACB + \angle PCE + 90^\circ - \angle PCB + 90^\circ - \angle PBF \\
& = & 180^\circ + \angle PCE - \angle PBF\end{eqnarray*}
Hence, we conclude that $ \angle PCE = \angle PBF$.
Denote by $ K$ the reflection point of $ B$ across the segment $ AD$, then we $ \angle PBF = \angle PKA = \angle PCE$, so the quadrilateral $ APKC$ is cyclic, which implies that $ \angle PBC = \angle PKB = \angle PAC = 90^\circ - \angle ACB$, i.e. $ \angle PBC + \angle ACB = 90^\circ$, i.e. $ PB\perp AC$. Hence $ B$, $ P$ and $ E$ are collinear, i.e. $ BP$ is $ B$ - altitude of triangle $ ABC$. On the other hand, $ AP$ is also the $ A$ - altitude of triangle $ ABC$. Hence $ P$ is the orthocentre of triangle $ ABC$.


Do there exist 1,000,000 consecutive positive integers s.t. each one has at least one repeated prime factor?

Yes!

Just ... don't ask me to prove it.

[hide]Chinese Remainder Theorem![/hide]

I'll also decline to do the dirty work too.  Especially if you want an example.

Does "repeated prime factor" mean that there is a prime divisor of the number with exponent at least 2?

Well, the Chinese Remainder Theorem solution is pretty short and quick:

Let's take pi to be distinct primes, with i from 1 to 1000000. Then there exists x s.t. p12|x, x=-1 (mod p22), x=-2 (mod p32), .., x=-999999 (mod p10000002).


How do you recommend I find the value of $ c$ in the figure? The line tangent to the graph of $ y\equal{}2^x$ at $ (0,1)$ intersects the $ x$-axis.

Use, $ (y\minus{}y_{0})\equal{} (y_{0})^{\prime}(x\minus{}x_{0})$, where $ (x_{0},y_{0})\equal{}(0,1)$ and $ (y_{0})^{\prime}\equal{}\frac{dy}{dx}(x_{0}).$

You want to find the point $ (c,0)$ lying on the above tangent. So just put $ y\equal{}0, x\equal{}c$ into the equation and calculate $ c$.


If the sum of the second and fourth terms of an arithmetic sequence is $ 6$, what is the third term?

Let the second term be $ x$.  Let the common difference be $ a$.  We can express the sum of the second and fourth terms of this arithmetic sequence as

$ (x) \plus{} (x \plus{} 2a) \equal{} 6$

So, simplifying this, we get

$ 2x \plus{} 2a \equal{} 6$
$ x \plus{} a \equal{} 3$

$ x \plus{} a$ is the third term (the second term plus the common difference).  So the third term must be $ \boxed 3$.

Or perhaps you know that 2 plus 4 equals 6. So why don't u do 1, 2, 3, and 4? Therefore, the third term should be 3. As in arthimetic sequence, the numbers are consecutive.

[quote=SoccerPro12345]As in arthimetic sequence, the numbers are consecutive.[/quote]

Um you don't need consecutive numbers for an arithmetic sequence

For example, 1, 4, 7, 10, ... is also one


Here is a problem I made up, in order to try to understand electrostatics of conductors better.  Suppose two large square parallel conducting plates, are in space.  One has a charge of Q, and another has a charge of -q.  They are put close together, from a separation of L, to a separation of l (l < L).  (abs(Q) different of abs(q))

1.  Describe both the inital charge distribution of both plates, and, the electric field between and on the outer sides of both plates, once equilibrium is reached.

2. After waiting, you "inject" an extra amount of charge m to the plate with inital charge -q.  What happens to the charge distribution? and to the electric field?
Consider both cases, where m is positive, and where m is negative.

It would be really nice if you guys could provide explanations for your answers, as I am still trying to understand electrostatics.

PS: My idea of solving this is setting E=0 inside the conductor.  Considering charges can only be on the surface, I set up surface charge densities so that the E field contribution of both plates end up cancelling each other within each plates. My difficulty lies in the fact that how do I know if my distribution is unique?  And also, how do I find the possible polarization of the plates due to the other plate?

Thanks so much.

How urgent is your need? Why the urgency? Is it kind of homework or an spontaneous exercise?

I'm asking in order to know how much help would you like to have. I guess you don't want an complete resolution... or either the answers...

First of all, I think that putting a minus sign before an algebraic quantity, like q, doesn't mean its negative ( if q<0, -q>0); just a remark, since seems it's irrelevant in that case (ops, I just noticed at the and of item 2  " m >0 and m<0"; maybe you should specify the signs of q, and Q too...)

For the uniqueness, I'm not sure I got your intention, but when you model out the system of equation for the electrostatic conditions you can see that the solution is unique.

 If polarization means sign of charges, it'll become automatic , with the mentioned proceeding.

No, it's not for a class, and not a technical urgency, but more because I am stuck on several problems that I do for olympiads/and fun, and those problems need strong concepts on charge polarization.

Oh, and assume all values given are positive (q, Q...).  So -q is a negative number.

Did you find the charge distribution? Your idea is OK

Sorry, I don't know how to latex, so I'll just scan my work.  Hopefully, you can read my handwriting.  :(

Anyways, either my problem is really, really simple, or I made unforgivable simplifications :D
Also, is it OK, if I start posting problems I have difficulty with this way, then post my solution to check if my reasoning is right, so you guys can help me?

I have many, many problems I am stuck on.

Thanks so much.

The charge distribution must not depend on the distance of separation.  Here, I [i]think[/i] I assumed that Q> q, which is not necessarily true.  If Q<q, then sigma a and d will be equal, but negative in magnitude (that's the only way I could put it so that the two outer sides cancel each other)

Not so bad, but; In (1) and (3) you're considering the contribution to the field inside conductors (contribution to zero, he he he ) separated from the "inner surfaces" and the outer ones. Why? Superposition takes in account all the charges present.

Some of your conclusion are wright but from wrong, or not exact, assumptions.

Maybe it would be better to remake it "fixing" the concept with the above remarks in mind, don't you think so?

In (1), I just drew a cylinder, with both disk-shaped face in the "meat" of the conductor.  Thus, the E-field at those disk faces is zero, thus the E-flux is zero thorough those faces.  Then I have to find the flux through the lateral surface of the cylinder, which obviously is zero because the field lines are parallel to the sides of the cylinder.   Thus the total flux through my gaussian surface is zero, and since I only enclosed the "inner" faces in that gaussian surface, it follows that sigma b= sigma c.  I don't get where this is wrong?    :blush:

In (3), since sigma b= sigma c, it acts as a parallel plate capacitor. thus the E field contribution by those 2 surfaces is 0 inside the "meat of the conductor, which is on the outside of the "capacitor" shaped distribution of charges.  Then, the remaining charges have to cancel each other's field, otherwise the E-field inside the conductor wouldn't be zero...

I actually saw a mistake in my solution, but it seems slightly unrelated to what you pointed out?  Maybe I'm not understanding well what you mean.  THe mistake is assume sigma a and sigma d= 0.  They are equal, but not necessarily 0. Here is an updated solution.

THe last part, where I say sigma a= sigma c, I mean equal in magnitude but opposite in sign.

I thought you where using superposition of effects, but instead you're using Gauss Theorem, so let's put it like that;

(i) in (1) the total charge enclosed by the gaussian is zero so $ \; \sigma_b \equal{} \minus{}\sigma_c  \;$ i.e., you can't take it only in magnitude; it's important that they do have opposite signs. You have still to relate those values with the given's Q and -q...

(ii) For the "external"  (in a capacitor they are  sometimes called spurious) charges they do not have to be zero, besides this would violate the charge conservation (it's assumed the plates are isolated all the time); it's enough if their effect cancels out in the space in between .

(iii) in your first scheme, in the last part how you said, I didn't find $ \; \sigma_a \equal{} \sigma_c \;$ but doesn't matter; now you can try ending  the solution, no?

Here's my updated version.  I didn't plug back in the numbers in part 2, considering it's the same thing with modified value for q.

Some comments: it seems separation l does not matter, as long as they are very large squares compared to the separation. Secondly, I get the E field on the outside using gauss law, but, I can't get it using simple planar charge distribution formulas. What has gone wrong?

THanks a bunch.

Thanks!

Wow. 3 takes for such a simple problems.  It really makes me wonder if I'm cut it for physics  :oops:

Doesn't matter the "2 takes". Your fundamentals and starting ideas where quite all right. We fixed just some details (important, but details).


Find the sum:
$\sum_{k=1}^{n}k\cdot2^{n-k}$
Is there some method that might be useful in evaluating sums?


my book "proved" it simply by stating
[r(cos phi + i sin phi)]^2 = r^2 (cos 2phi + i sin 2phi)
I understand the r part, but not the phi becoming 2phi part.  could anyone please prove why thats true?

[hide="Hint"]Try expanding $(\cos\phi+i\sin\phi)^2$, taking into account that $i^2=-1$, and then apply double-angle trigonometric formulas.[/hide]

Farenhajt, in fact it is more useful to prove De Moivre's in terms of Euler's in order to [i]find[/i] the $\sin n\theta, \cos n\theta$ formulas  :D

[hide="Like so"]
$e^{ix} = \cos x + i \sin x$ (this is 'provable' through Taylor series, but more than anything else it is simply an extension of $e^{x}$ to complex inputs)

$e^{2ix} = \cos 2x + i \sin 2x = (\cos x + i \sin x)^2$ is now almost trivial, see?

Expanding the RHS gives a demonstration of the formulas for $\sin 2x, \cos 2x$, and similarly:

$e^{inx} = \cos nx + i \sin nx = (\cos x + i \sin x)^n$ (De Moivre's)

And expanding the RHS gives the formulas for $\cos nx, \sin nx$ by equating real and imaginary parts.
[/hide]

Very nice and true (and also quite familiar to me :)), but it's very likely they don't mention Euler at all in the textbook. I believe our solutions should correspond to the level of the question - it's very unlikely that a question like this would be presented by someone who's acquainted with Euler's formula :)


geometry
circumcircle
Triangle
homothety
IMO Shortlist

Let $O$ be the circumcenter of an acute-angled triangle $ABC$ with ${\angle B<\angle C}$. The line $AO$ meets the side $BC$ at $D$. The circumcenters of the triangles $ABD$ and $ACD$ are $E$ and $F$, respectively. Extend the sides $BA$ and $CA$ beyond $A$, and choose on the respective extensions points $G$ and $H$ such that ${AG=AC}$ and ${AH=AB}$. Prove that the quadrilateral $EFGH$ is a rectangle if and only if ${\angle ACB-\angle ABC=60^{\circ }}$.

[i]Proposed by Hojoo Lee, Korea[/i]

First of all, it's easy to see that $AD\perp EF,GH$. The triangles $ACF,ABE$ are isosceles, and have the same angles, so they are similar. This means that $AFE$ is similar to $ACB$, and thus to $AGH$.

Since $AFE,AGH$ share the same altitude  starting from $A$, it means that $AFE$ is obtained from $AGH$ by reflecting it through a line $\perp AD$, and then performing a homothety centered at $A$ of ratio $\frac{AF}{AG}$, so $EFGH$ will be a rectangle iff $AF=GH$, i.e. iff $AF=AC\iff AFC$ is equilateral. With the given conditions, this is equivalent to $\angle ADC=30^{\circ}$, and the conclusion follows easily.

This problem was problem 2 in the 3rd German TST 2005. Hence, I have a conjecture on where it is from, although some people repeatedly dispute it ;) . Anyway, here is the way the problem was posed on our TST:

[color=blue][b]Problem.[/b] Let ABC be an acute-angled triangle with A < B, and let U be the circumcenter of the triangle ABC. The lines CU and AB intersect at a point D. Let E and F be the circumcenters of triangles ACD and BCD. Choose points K and L on the rays AC and BC such that AK = BL = a + b. Prove that the quadrilateral EFKL is a rectangle if and only if B - A = 60°.[/color]

And here is the [i]solution[/i] I gave on the exam (just copied from my writeup, not simplified, hence it will be far more complicated than necessary):

We will use non-directed angles. See the accompanying sketch for the arrangement of the points.

Since U is the circumcenter of triangle ABC, the central angle theorem yields < BUC = 2 < BAC = 2A. On the other hand, BU = CU, again because U is the circumcenter of triangle ABC. Hence, the triangle BUC is isosceles, and its base angle is therefore

$\measuredangle UCB=\frac{180^{\circ}-\measuredangle BUC}{2}=\frac{180^{\circ}-2A}{2}=90^{\circ}-A$.

Analogously, < UCA = 90° - B. Similarly, for the circumcenters E and F of triangles ACD and BCD we can find

< ECA = < CDA - 90°;
< EAC = < CDA - 90°;
< FBC = 90° - < CDB;
< FCB = 90° - < CDB.

Hereby, in the proofs of the equations < ECA = < CDA - 90° and < EAC = < CDA - 90°, we have to apply the central angle theorem in the form

< CEA = 2 $\cdot$ acute chordal angle of the chord CA in the circumcircle of triangle ACD
= 2 $\cdot$ (180° - < CDA) = 360° - 2 < CDA,

since the angle < CDA is obtuse, and the point E lies outside of the triangle ACD.

Also, since E and F are the circumcenters of triangles ACD and BCD, we have CE = AE and BF = CF, so that the triangles CEA and BFC are isosceles.

Now, by the sum of angles in triangle CDA, we have

< CDA = 180° - < DCA - < CAD = 180° - < UCA - A = 180° - (90° - B) - A = 90° + (B - A).

Consequently,

< CDB = 180° - < CDA = 180° - (90° + (B - A)) = 90° - (B - A).

Thus,

< ECA = < CDA - 90° = (90° + (B - A)) - 90° = B - A;
< EAC = < CDA - 90° = (90° + (B - A)) - 90° = B - A;
< FBC = 90° - < CDB = 90° - (90° - (B - A)) = B - A;
< FCB = 90° - < CDB = 90° - (90° - (B - A)) = B - A.

Hence, in particular, < ECA = < FCB and < EAC = < FBC. Thus, the triangles ECA and FCB are similar. This yields CE : CA = CF : CB. On the other hand, the equality < ECA = < FCB yields < ECF = < ECB - < FCB = < ECB - < ECA = < ACB. This, combined with CE : CA = CF : CB, shows that the triangles CEF and CAB are similar, so that EF : AB = CE : CA.

On the other hand, the equation AK = a + b implies CK = AK - CA = (a + b) - b = a = CB, and similarly CL = CA. This, together with < KCL = < BCA, shows that the triangles CLK and CAB are congruent. Consequently, KL = AB. Hence, the equation EF : AB = CE : CA becomes EF : KL = CE : CA.

The problem asks us to prove that the quadrilateral EFKL is a rectangle if and only if B - A = 60°. Hence, in order to solve the problem, it is enough to show the following two assertions:

[b]Assertion 1.[/b] If B - A = 60°, then the quadrilateral EFKL is a rectangle.
[b]Assertion 2.[/b] If the quadrilateral EFKL is a rectangle, then B - A = 60°.

[i]Proof of Assertion 2.[/i] If the quadrilateral EFKL is a rectangle, then EF = KL. Since EF : KL = CE : CA, this yields CE = CA. Hence, the triangle CEA, which is already isosceles with CE = AE, must be equilateral. Hence, < ECA = 60°. Since < ECA = B - A, we therefore obtain B - A = 60°. This proves Assertion 2.

[i]Proof of Assertion 1.[/i] If B - A = 60°, then, since < ECA = B - A, we have < ECA = 60°. Hence, the triangle CEA, which is already isosceles with CE = AE, must be equilateral. Thus, CE = CA. Since EF : KL = CE : CA, this yields EF = KL.

On the other hand, we know that the triangles CEF and CAB are similar, and that the triangles CLK and CAB are congruent. Thus, the triangles CEF and CLK are similar. Moreover, these two triangles must be congruent, since EF = KL. Hence, CE = CL and CF = CK. Thus, the triangles ECL and FCK are isosceles. The base angle of the isosceles triangle ECL equals

$\measuredangle CEL=\frac{180^{\circ}-\measuredangle ECL}{2}=\frac{\measuredangle BCE}{2}=\frac{\measuredangle BCA+\measuredangle ECA}{2}$
$=\frac{C+\left(B-A\right)}{2}$         (since < BCA = C and < ECA = B - A)
$=\frac{\left(B+C\right)-A}{2}=\frac{\left(180^{\circ}-A\right)-A}{2}$        (since B + C = 180° - A by the sum of the angles in triangle ABC)
$=\frac{180^{\circ}-2A}{2}=90^{\circ}-A$.

Also, since the triangles CEF and CAB are similar, we have < CEF = < CAB. Thus,

< FEL = < CEF + < CEL = < CAB + (90° - A) = A + (90° - A) = 90°.

Similarly, < EFK = 90°.

Since the triangles CEF and CLK are similar, we have < CEF = < CLK, and since the triangle ECL is isosceles with CE = CL, we have < CEL = < CLE. Thus,

< FEL = < CEF + < CEL = < CLK + < CLE = < ELK.

Consequently, < FEL = 90° implies < ELK = 90°.

Similarly, < FKL = 90°.

Altogether, we have obtained < FEL = 90°, < EFK = 90°, < ELK = 90° and < FKL = 90°. Thus, the quadrilateral EFKL has four right angles, so that it must be a rectangle. This proves Assertion 1.

Now, as both Assertions 1 and 2 are proven, the problem is solved.

  Darij

Incidentally, this was also Taiwan 2nd TST 2005 final exam problem 5.

[hide="Trigonometry"]Easy to get $EF=\frac{RcosCsinA}{sinCcos(B-C)}$

So $EF=HG=BC$
$\Rightarrow \frac{RcosCsinA}{sinCcos(B-C)}=2RsinA$
$\Rightarrow 2sinAcosC[1-2cos(B-C)]=0$.

The result follows.....Converse is easy...[/hide]

Bye...

Sayantan....

In case you did not notice that $\triangle AFC$ was equilateral - see [b]Grobber's[/b], as I did, then from $\triangle ABO\sim\triangle ADF$ get $\frac{DF}{BO}=\frac{AD}{AB}$, and, with $DF=AC, BO-AO$ getting $AO\cdot AD=AB\cdot AC$.
If the internal angle bisector of $\angle BAC$ intersects $BC$ and the circumcircle of $\Delta ABC$ (second time) at $M,N$ respectively, then we know that $AB\cdot AC=AM\cdot AN$, or $DOMN$ is cyclic; from $\angle DMN=\angle DON$ we get the same $\hat C-\hat B=60^\circ$.

Best regards,
sunken rock

Another beautiful problem from Hojo Lee.
[hide=Solution]Let $\measuredangle BAE=\alpha $,$\measuredangle EAD=\beta$.
Using angle-chasing we get $\measuredangle AHF=\measuredangle FAD=\alpha +\beta $ and $\measuredangle DAF=\measuredangle FHA=\alpha +\beta $ $\implies$ $AHF$ is similar to $FAD$.
Then $\frac{AF}{FD}=\frac{AH}{AF}=\frac {AC}{AF}=1$ as $F$ is circumcenter of $\Delta ADC$.So $AC=AF$ which means $\Delta AFC$ is equaliteral. So $\alpha =60^{\circ}$.The rest is easy to check.
For second case we get $\Delta AFC$ is equaliteral. So $AF=AH$.Then we get $\measuredangle AHF=\alpha+\beta$ $\implies$ $\measuredangle GHF=90^{\circ}$.Similarly $\measuredangle HGE=90^{\circ}$.Since $AE=AG$ and $AH=AF$ we easily get other angles are also $90^{\circ}$[/hide]

[hide=Comment]Nice problem!!! Generally this year's Shortlist problems are very beautiful. The only NT that I like is N3.G8 is beautiful but easy. G1 is just drawing picture correctly. G2 is Pascal. G4 took me 3 days. But not difficult. A5 is.too easy for A5.Also A1,A2,A4 are beautiful.

[hide=Solution]
[quote=ISL 2004 G3]Let $O$ be the circumcenter of an acute-angled triangle $ABC$ with ${\angle B<\angle C}$. The line $AO$ meets the side $BC$ at $D$. The circumcenters of the triangles $ABD$ and $ACD$ are $E$ and $F$, respectively. Extend the sides $BA$ and $CA$ beyond $A$, and choose on the respective extensions points $G$ and $H$ such that ${AG=AC}$ and ${AH=AB}$. Prove that the quadrilateral $EFGH$ is a rectangle if and only if ${\angle ACB-\angle ABC=60^{\circ }}$.
[/quote]
[b][color=#000]Solution:[/color][/b] If $EFGH$ is rectangle $\implies$ $\angle HGF=90^{\circ}=\angle HGB+\angle AGF=C+\angle AGF$ ,Since, $BCGH$ is a Isosceles Trapezium. But, $\angle BAF$ $=$ $2(90^{\circ}-C)$ $\implies$ $AF$ $=$ $AG$ $=$ $AC$ $=$ $FC$ $\implies$ $AE$ $=$ $AB$ $=$ $AH$ $=$ $BE$. Hence,
$$\angle ADC=B+90^{\circ}-C=30^{\circ}\implies C-B =60^{\circ}$$
If $C-B=60^{\circ}$ $\implies$ $\angle ADC=30^{\circ}$ $\implies$ $\Delta AFC$ is Equilateral and similarly, $\Delta ABE$ is Equilateral. Hence, $A$ is the center of $\odot (FCG)$ & $\odot (BEH)$. Also, $BCGH$ is an Isosceles Trapezium. $$\angle BAF=\angle BAO+\angle OAF =2(90^{\circ}-C) \implies \angle HGF=\angle HGB+\angle BGF =90^{\circ}$$
Hence, $HE||FG$ and since, $\Delta AEH$ & $\Delta AFG$ are Isosceles $\implies$ $EFGH$ is Rectangle. $\qquad \blacksquare$
[/hide]

Solution from Twitch solves ISL stream:

We start with a few observations which are always true regardless of the condition. [list] 	[*]Quadrilateral $HGCB$ is always an isosceles trapezoid, 	and in particular $BC = GH$.
[*]By angle chasing $\overline{AO} \perp \overline{GH}$ always holds. 	(One clean way to see this 	is to note that $\overline{GH}$ and $\overline{BC}$ 	are antiparallel through $\angle A$.) 	This implies $\overline{EF} \parallel \overline{GH}$.
[*]By Salmon theorem, 	we always have 	\[ \triangle AEF \overset{+}{\sim} \triangle ABC. \] [/list]
[asy] size(6cm); pair B = dir(200); pair C = dir(-20); pair A = dir(30); pair O = origin; pair D = extension(A, O, B, C); pair E = circumcenter(A, B, D); pair F = circumcenter(A, C, D);
filldraw(A--B--C--cycle, invisible, blue); draw(A--D, blue); filldraw(A--E--F--cycle, invisible, red); pair G = A+(A-B)*abs(C-A)/abs(B-A); pair H = A+(A-C)*abs(B-A)/abs(C-A);
filldraw(A--C--G--cycle, invisible, blue); filldraw(A--B--H--cycle, invisible, blue); filldraw(A--G--H--cycle, invisible, blue);
dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$A$", A, dir(330)); dot("$O$", O, dir(270)); dot("$D$", D, dir(270)); dot("$E$", E, dir(E)); dot("$F$", F, dir(270)); dot("$G$", G, dir(G)); dot("$H$", H, dir(H));
/* TSQ Source:
!size(10cm); B = dir 200 C = dir -20 A = dir 30 R330 O = origin R270 D = extension A O B C R270 E = circumcenter A B D F = circumcenter A C D R270
A--B--C--cycle 0.1 lightblue / blue A--D blue A--E--F--cycle 0.1 lightred / red G = A+(A-B)*abs(C-A)/abs(B-A) H = A+(A-C)*abs(B-A)/abs(C-A)
A--C--G--cycle 0.1 lightcyan / blue A--B--H--cycle 0.1 lightcyan / blue A--G--H--cycle 0.1 lightblue / blue
*/ [/asy]
We begin now with:
[b][color=red]Claim:[/color][/b]  	We have $\triangle AEF \cong \triangle ABC$ 	if and only if $\angle C - \angle B = 60^{\circ}$.
[i]Proof.[/i]  	The congruence just means $FA = AC$. 	Since $FA = FC$ always, 	triangle $AFC$ is equilateral if and only if 	$\angle AFC = 60^{\circ} \iff \angle ADC = 30^{\circ}$. 	As $\angle ADC = (90^{\circ} - \angle C) + \angle B$, and 	the result follows. $\blacksquare$

[b][color=red]Claim:[/color][/b]  	We have $EFGH$ is a parallelogram 	if and only if $\triangle AEF \cong \triangle ABC$.
[i]Proof.[/i]  	Since we already know $\overline{EF} \parallel \overline{GH}$, 	the parallelogram condition is equivalent to $EF = GH$, 	but as $GH = BC$ we get the earlier congruence. $\blacksquare$
It remains only to show that if $EFGH$ is a parallelogram then it is also a rectangle. In the situation of the claims, note that $EG = FH$ by symmetry through the oppositely congruent triangles $\triangle AEF$ and $\triangle AHG$ as needed.

[quote=weaving2]@v_Enhance what is the name of the twitch channel?[/quote]

[url]https://twitch.tv/vEnhance[/url] which normally runs Friday 8pm ET. You can see a schedule at [url]https://web.evanchen.cc/videos.html[/url].

[quote]By Salmon theorem, we always have\[ \triangle AEF \overset{+}{\sim} \triangle ABC. \][/quote]

Why is this the case? Can you provide any link to further material on the subject?

[hide=Solution]
[asy]
unitsize(5cm);
pair A, B, C, D, E, F, G, H;
A=(-sqrt(3)/2,1/2);
B=(-1,0);
C=(1,0);
D=(0,0);
E=(-1/2,1-sqrt(3)/2);
F=(1/2,1+sqrt(3)/2);
G=(1/2-sqrt(3)/2,3/2+sqrt(3)/2);
H=(-1/2-sqrt(3)/2,3/2-sqrt(3)/2);
draw(D--A--B--C--H--A--G--H--E--F--G--A--E--(1/2,1-sqrt(3)/2)--F);
dot(E);
dot(F);
label("$A$", A, NW);
label("$C$", B, SW);
label("$B$", C, SE);
label("$D$", D, S);
label("$F$", E, S);
label("$E$", F, dir(0));
label("$H$", G, N);
label("$G$", H, W);
label("$X$", (1/2,1-sqrt(3)/2), SW);
[/asy]

Since $\angle HGA=\angle ACB$ and $\angle GAD=180^{\circ}-\angle DAB=90^{\circ}+\angle ACB$, this means that $GH\perp AD$. Since $EF$ is the perpendicular bisector of $AD$, this means that $EF\parallel GH$. Let $X$ be the point satisfying $EX\perp BC$ and $FX\parallel BC$. Then, $FX=\frac12BC$, and $EF=GH$ if and only if $EFGH$ is a parallelogram, so $\angle EFX=60^{\circ}$ if and only if $EFGH$ is a parallelogram. This is equivalent to $\angle ADC=30^{\circ}$. Since $\angle BAD=90^{\circ}-\angle ACB$, this means that $\angle ADB=30^{\circ}$ if and only if $\angle ACB-\angle ABC=60^{\circ}$.

Now, we claim that $\angle ACB-\angle ABC=60^{\circ}$ implies $EFGH$ is a rectangle. It suffices to show $FG\perp FE$. Since $\angle ADC=30^{\circ}$, this means that $\angle AFC=60^{\circ}$, so $AG=AC=CF=FA$. Therefore, $AGF$ is isosceles. Therefore, we have
\begin{align*}
\angle GFE&=\angle GFA+\angle AFE\\
&=90^{\circ}-\frac12\angle GAF+90^{\circ}-\angle FAD\\
&=180^{\circ}-\frac12\angle GAF-\frac12\angle FAB\\
&=180^{\circ}-90^{\circ}\\
&=90^{\circ}.\end{align*}
Therefore, $EFGH$ is a rectangle if $\angle ACB-\angle ABC=60^{\circ}$, so $EFGH$ is a rectangle if and only if $\angle ACB-\angle ABC=60^{\circ}$.

$EFGH$ is a rectangle if and only if $EFGH$ is a rectangle.

Assume GHEF is rectangle we will prove ∠C - ∠B = 60.
EF = GH = BC and we have AEF and ABC are similar so ABC and AEF are congruent so AEB and AFC are regular triangles.
∠AFC = 60 so ∠ADC = 30.
∠ADC = ∠B + ∠OAB = ∠B + 90 - ∠C so ∠C - ∠B = 90 - 30 = 60 as wanted.

Assume ∠C - ∠B = 60 we will prove GHEF is rectangle.
∠ADC = ∠B + 90 - ∠C = 30 so ∠AFC = 60 and FA = FC so AFC is regular triangle. we have AEF and ABC are similar and AC = AF so AEF and ABC are
congruent so EF = BC = GH. ∠BAF = 180 - 2C so ∠FGA = 90 - ∠C. ∠HGF = ∠HGA + ∠FGA = ∠C + 90 - ∠C = 90. same way ∠GHE = 90 so HE || GF so HEFG is rectangle.

Inefficient angle chasing solution... will improve later if I have time

Let $\angle BAC=\alpha$ and so on. Notice $\overline{AO}\perp\overline{EF}$ since $\overline{AO}$ is the radical axis of $(ABD)$ and $(ACD).$ Also, $\overline{AO}\perp\overline{GH}$ as $$\measuredangle(\overline{AO},\overline{GH})=\measuredangle(\overline{AO},\overline{AG})+\measuredangle AGH=\measuredangle BCA+\measuredangle OAB=90.$$ Finally, $$\angle CDA=\measuredangle BCA+\measuredangle OCA=\measuredangle BCA+90-\measuredangle ABC=90-(\measuredangle ACB-\measuredangle CBA).$$

[b][color=blue]If Direction:[/color][/b] $\gamma-\beta=60$ implies $EFGH$ is a rectangle.
[i]Proof.[/i] Notice $\triangle AEB$ is equilateral as $$\angle BEA=2\angle CDA=2(90-60)=60.$$ Hence, $$\angle EAO=60+90-\angle ACB=\angle ACB-\angle CBA+90-\angle ACB=90-\angle CBA=\angle OAC$$ so $\angle AEH=\tfrac{1}{2}\angle EAC=\angle OAC$ and $\overline{EH}\parallel\overline{AO}.$ Similarly, $\overline{FG}\parallel\overline{AO}.$ $\blacksquare$

[b][color=blue]Only If Direction:[/color][/b] $EFGH$ is a rectangle implies $\gamma-\beta=60.$
[i]Proof.[/i] We know $\triangle AEF\cong\triangle DEF$ so $$\angle AFE=\tfrac{1}{2}\angle AFD=\angle ACB.$$ Similarly, $\angle FEA=\angle CBA$ so $\triangle AEF\sim\triangle ABC.$ Then, $\triangle AEF\cong\triangle AHG$ so $\triangle AEB$ is equilateral. Thus, $$90-(\measuredangle ACB-\measuredangle CBA)=\angle CDA=\tfrac{1}{2}\cdot 60=30.$$ $\blacksquare$ $\square$

This took me way too long

It is easy to see that $BCGH$ is a isosceles trapezoid. Let $X$ be $OA\cap GH.$ We have \[\angle GAX=\angle OAB=90^\circ-\angle ACB=90^\circ-\angle AGX.\] Thus, $AX\perp GH.$ We also have $EF\perp AD$ so $EF\parallel GH.$ It is clear why F is inside and E is outside so now, $\angle CFD=2\angle CAD=\angle COD$ so $CFOD$ is cyclic. Similarly, $BEOD$ is cyclic. We have \[\angle FOE+\angle FAE=360^\circ-\angle FOD-\angle DOE+\angle FDE\]
\[=\angle FCD+\angle EBD+180^\circ-\angle FDC-\angle EDB=180^\circ\] which implies that $AFOE$ is cyclic. We have $\angle ACB-\angle ABC= 90^\circ-\angle OAB-(90^\circ-\angle OAC)=\angle OAC-\angle OAB.$ What's important to see is that \[\angle OAE=\angle OFE=\angle FOD-90^\circ=90^\circ-\angle FCD=\angle OAC\] and similarly, $\angle OAF=\angle OAB.$ Thus, $\angle OAC-\angle OAB=\angle OAC-\angle OAF=\angle CAF.$ Therefore, we have \[\angle ACB=\angle ABC\iff \triangle ACF \text{ is equilateral}\]
We know that $\triangle ABC\sim \triangle AEF$ and $\triangle ABC\cong \triangle AHG$ so $\triangle ACF$ is equilateral $\iff$ $\triangle AFE\cong \triangle AHG.$ Since $EF\parallel GH$, $\triangle AFE\cong\triangle AHG\iff EFGH$ rectangle, as desired.

[quote=Numbertheorydog][quote=Mahdi_Mashayekhi]we have AEF and ABC are similar
[/quote] how do you get that they are similar?[/quote]

angle chasing.

First we prove the necessity: It's clear that $$30=C-60+90-C=ABC+OAB=ADC\implies AFC=60\implies AG=AC=FA.$$ Now note the cyclicislscelestrapezoid since $$HAB=GAC,BHA=HBA=90-HAB/2=90-GAC/2=AGC=ACG,$$ and that $AFD=2C\implies DAF=90-C=BAO$. Now, $$GFE=GFA+EFA=90-DAF+AGF=90-1/2BAF+90-GAF=90,$$$HGF=HGA+AGF=C+90-C=90,$ where the last step follows from knowing that $$AGF=(180-GAF)/2=FAB/2=FAD=90-C.$$ Finally, we note that $AEF\cong ABC$, which follows immediately from $$FA=FC, EAF=EAB+BAF=60+BAF=FAC+BAF=BAC,AFE=90-AFG=90-AGF=C.$$ It is now evident that EF=BC=HG ($BAC\cong HAG$), whence the already known right trapezoid (EFG and HGF are 90 degrees) has EF=HG which makes it a rectangle! $\blacksquare$

    As for sufficiency, if EFGH is a rectangle, remark that BC=HG=EF, and $$AFE=AFD/2=ACB,AEF=AED/2=ABC\stackrel{SAS}{\implies}ABC\cong AEF\implies AF=AC,$$ whence AFC is equilateral, and $$60=AFC=2ADC=2(B+BAO)=2(B+90-C)=180+2B-2C\implies C-B=60.$$ $\blacksquare$

I'm really happy about this solution because it only took half an hour and it was straightforward, a very nice problem, and I did it on my own, with some nice observations! Obviously my necessity was overkill but I can't be bothered to shorten it lol

how do you come up with problems like these
[hide = solution]
First, observe that the conditions imply $\angle ADC$ is acute. First $BCGH$ is an isosceles trapezoid and therefore cyclic. Hence the $A$ altitude in $AGH$ passes through the circumcentre of $ABC$, that is, $AD \perp GH$. Next, remark that by a simple angle chase $DF \parallel AB$ and thus if $EFGH$ were a rectangle, we would have $FG \parallel AH$ and then $AGFD$ is a paralellogram. In particular, $DF = AG = AC$. But $DF = AC$ implies that $AC$ subtends an acute angle of $60^{\circ}$ in $\odot(ADC)$. This implies $\angle ADC = 30^{\circ}$. But $90^{\circ}-\angle ADC$ is just $\angle ACB - \angle ABC$ (by a simple angle chase), proving one side of the story.

Now let us assume that $\angle ADC = 30^{\circ}$. We obtain that $DF = AC$ and $BE = AC$ in triangles $ADB$ and $ADC$. Now because $DF = AG$ and $DF \parallel AG$ it follows that $DFGA$ is a parallelogram, implying $FG \perp GH$ as $AD \perp GH$. Similarily, one obtains $EG \perp GH$. Finally, since $AD$ is the common chord of circles $\odot(ABD)$ and $\odot(ACD)$ we have $EF \perp AD \perp GH$ and so $EF \parallel GH$. Combined with the angle condition from earlier, we deduce that $EFGH$ is a rectangle, proving the sufficiency.
[/hide]

Solved with [b]GrantStar[/b] :omighty: :omighty:

First, let $P$ be the intersection of the altitude from $A$ to $BC$ with the circumcircle of $ABC.$ Notice that since $EF$ and $OF$ are the perpendicular bisectors of $AD$ and $AC,$ we have $\measuredangle AEF=\frac{1}{2} \measuredangle AED=\measuredangle ABC=\frac{1}{2} \measuredangle AOC=\measuredangle AOF,$ so $AEOF$ is cyclic and $AO \perp EF.$ Also, by symmetry, we have $\measuredangle AFE=\measuredangle ACB,$ so $\triangle AEF$ and $\triangle ABC$ are similar and similarly oriented. Thus, there exists a spiral similarity at $A$ sending $\triangle AEF$ to $\triangle ABC.$ Notice that this spiral similarity also sends $O$ to $P,$ so we have $\frac{EF}{BC}=\frac{AO}{AP}.$

Next, Reim on lines $AB,AC$ and quadrilaterals $AABC$ and $BCGH$ gives us that $GH$ is parallel to the tangent to the circumcircle of $ABC$ at $A,$ so $GH \perp AD$ and $GH \parallel EF.$ Also, we have $\angle EAD=90-\frac{1}{2} \angle AED=90-\angle ABC=90-\frac{1}{2} \angle AOC=\angle OAC,$ so $AD$ bisects $\angle EAC,$ so the tangent to the circumcircle of $ABC$ at $A$ bisects $\angle EAH,$ and thus also bisects $\angle GAF$ by symmetry. Therefore, $EFGH$ being a rectangle is equivalent to $EF=GH.$

However, we know $GH=BC,$ and $BC=EF$ if and only if $AO=AP.$ Since $OA=OP,$ this holds if and only if $AOP$ is equilateral, or equivalently $\angle OAP=60.$ Now, let $A'$ be the point such that $AA'BC$ is an isosceles trapezoid with $AA' \parallel BC,$ and let $AO$ intersect the circumcircle of $ABC$ again at $Q.$ Then we have $\angle OAP=\angle QAP=\angle A'CA=\angle ACB-\angle BCA'=\angle ACB-\angle ABC,$ so we are done.

Obviously $EF$ is the perpendicular bisector of $AD$ and $OE\perp AB$, $OF\perp AC$.

For convenience let $L$ be the foot of $A$ to $BC$. Now
\[\measuredangle DAE=\measuredangle LAB\]
\[\measuredangle DAF=\measuredangle LAC\]
so $\triangle AEF\sim \triangle ABC$.

Now we prove that $EF\parallel GH$. Consider the acute angles formed by each of the lines with $BC$. It suffices to show:
\[90^{\circ}-\angle ADC=180^{\circ}-(\angle A+2\angle B)\]
\[90^{\circ}-(\angle B+90^{\circ}-\angle C)=180^{\circ}-(\angle A+2\angle B)\]
which is true.

Now since $\triangle AEF\sim \triangle ABC$, $\triangle ABC\cong \triangle AHG$ we must have $AF=AG=AC$ in order to have $EF=GH$.

So $\angle AFC=60^{\circ}$, $\angle ADC=30^{\circ}$, so
\[\angle ADC=\angle B+90^{\circ}-\angle C=30^{\circ}\implies \angle C-\angle B=60^{\circ}\]
and we are done.

Since $\triangle GAB \cong \triangle CAH$, we have $GH \parallel BC$ due to $\angle ABG = \angle AHC$. If $EFGH$ is a rectangle, then $GH =  EF$. Then consider the homothety $\mathcal{H}$ sending $(ABD) \to (ACD)$. Then $\mathcal{H}(E) = F$, and $\mathcal{H}(B) = C$, so $\triangle{AEF} \cong \triangle{ABC}$. Then notice that if $\triangle{AEF} \cong \triangle{ABC}$, then $AF = AC$ which implies that $\triangle AFC$ is equilateral. From here, we find $\angle FAC = \angle FCA = 60^{\circ} \implies \angle ADC = 30^{\circ}$. We can angle chase to find $\angle B = 90^{\circ} = \angle C + 30^{\circ}$, so we are done.


Let $f(f(x))=3x$ for a function '$f$' where one inputs a natural number and a natural number is "outputted" ($f: N \longrightarrow N$) for every $x$ in the set of natural numbers and f(x+1)>f(x). Find $f(153)$.


prove that there exists a  set $X \subset Z$ such that : $\forall n \in Z$,     $\exists ! (a,b)$ with properties:
1. $a,b \in X$
2. $a+2b=n$

I think it is an not-so-old problem from USAMO. You may find a solution in french here (problem 5) :
http://www.animath.fr/tutorat/dossier_02033sol.html

Pierre.

I don't know French, can someone translate?

Here is a solution in English; I believe it was 1996 B3 in a USAMO.  Here is the link:

[url]http://www.kalva.demon.co.uk/usa/usoln/usol966.html[/url]


I only noticed that they have been awarded a while ago. Here they are though:

$ \text{Physics: Albert Fert, Peter Grunberg}$
$ \text{Chemistry: Gerhard Ertl}$
$ \text{Medicine: Mario R. Capecchi, Sir Martin J. Evans, Oliver Smithies}$
$ \text{Literature: Doris Lessing}$
$ \text{Peace: Intergovernmental Panel on Climate Change (IPCC), Albert Arnold (Al) Gore Jr.}$

Can't find the Economics prize though. What do you think of the achievements this year?

I think that the medical developments on embryonic stem cells can really benefit future generations and it seems to be the big thing in medicine at the moment.

The Chemistry prize was awarded for works on chemical reactions on solid surfaces and in the announcement it was illustrated with the Haber-Bosch process. It seem to lack real applications because it merely explored the actual mechanisms behind surface catalysis and doesn't contribute to economical developments of ammonia or other chemicals.

Physics was on the development of GMR (Gian Magnetoresistance) which I am unfamiliar with so can say little about but the way they talked about using the concept in computers sound interesting.

I haven't read any of Doris Lessing's works but the summary gives much information. It says that:

'Indeed, perhaps her most famous novel, The Golden Notebook, a dissection of a woman's psyche which is torn between emotional, social and creative demands, has been very influential on feminism.'

Sounds like a very interesting lady.

They should promote these kind of things more instead of showing "AUSTRALIAN" of the year or Miss universe  :|

They certainly should. Also, I found the laureate for the Economics prize:

$ \text{Economics: Leonid Hurwicz, Eric S. Maskin, Roger B. Myerson}$

They got it "for having laid the foundations of mechanism design theory" which I have no idea about.

A few years ago, I visited the Nobel Museum in Stockholm, and I was surprised to learn that there is so much controversy surrounding the Nobel Prize (did you know that there are several people who refused to accept the Nobel Prize? :maybe: )à

Yeah, it's on the official website as well. Some are forced to refuse the prize by their country and some guy refused it because he refuses all prizes!

Here's an article titled, "Why you should care about the 2007 Economic Nobel": http://prospect.org/cs/articles?article=why_you_should_care_about_the_2007_economic_nobel

I strongly disagree with your opinions on the Chemistry nobel, BanishedTraitor.  I think surface chemistry is far more significant to modern industry than most people realize (catalysts are applied in industrial processes that account for 25-30% of the GNP of the US, for example, [F. Zaera, J. Phys. Chem. B, 106, 4043 (2002)]).  Despite this significance, catalysis and surface science is a relatively unexplored area of chemistry, and advances in its fundamental study have and will continue to be accompanied by improvements in catalysis' industrial application.  Ertl's work led to major advances in the search for a method of direct epoxidation of propylene by molecular oxygen, for example, which has huge implications for industrial efficiency.

Also, I thought the Nobel panel could have done a much better job with Literature.  Doris Lessing is certainly a prolific and talented writer, but her work is far less skillful or innovative than, say, Philip Roth or Yves Bonnefoy's.  (I was gunning for Roth, partially because he's gonna die soon, and probably sooner than lessing for that matter).  I didn't think it was completely unfounded, but I thought there were far more deserving candidates.  I do, however, appreciate and agree a little with Lessing's aesthetic approach to feminism.  I think it's a little funny that awarding her the prize was justified by her influences on that movement, despite her cynicism towards its goals and her desire not to be labeled as a feminist author.

Many of my scientifically inclined friends have balked at Al Gore's award, but I don't think I'm really well-informed enough to have a defensible opinion on him yet.  I'd say the same for the medicine and physics, though they both sound pretty cool.

[quote="miraculouspostmaster"]Many of my scientifically inclined friends have balked at Al Gore's award, but I don't think I'm really well-informed enough to have a defensible opinion on him yet. [/quote]
I sat in on a lecture of an environmental geosciences course at my university, and the professor spent a bit of time talking about Gore. I did not read the statements by the Nobel committee; I'll just say what I heard from this professor:

(1) Gore's famous movie includes a lot of scientific statements that are either completely false, or known with far less certainty than he implies. This applies more broadly to most people who present data about climate change for political purposes; but that's something we need to discuss here. Anyway, the point is that MPM's friends are probably right in that Gore uses scientifically inaccurate information to advance his goals.

(2) Gore's objective of raising awareness about the issue of global climate change is not misguided (regardless of point 1). The worst one can do to refute political arguments stating that action needs to be taken on global warming issues is to observe that we still don't really understand our climate that well and a lot of the projections and models are still pretty inaccurate. However, "we don't know for sure" does not give us license to ignore the potential for a climate crisis! Evidence suggests that it is not infeasible; therefore we should take some action. It is important to recognize the contribution of people like Gore who spearhead the cause of getting people to recognize the risk of major climate change, even if the methods these people use are not perfect.

I would view the Nobel Prize as a recognition of this effort, rather than recognition of a successful compilation of the best current information on global climate change (and human influence thereon). After all, the Peace prize is not given primarily for scientific contributions.

[quote="Xevarion"](1) Gore's famous movie includes a lot of scientific statements that are either completely false, or known with far less certainty than he implies.[/quote]

So, I've heard this a fair amount, but I don't buy it.  In particular, Gore was very, very careful in the movie not to say things that were false and fundamentally succeeded, with perhaps a couple of minor exceptions.  Good recent discussions of this (in light of a recent British court ruling & the award of the Nobel) are available at [url=http://www.realclimate.org/index.php/archives/2007/10/convenient-untruths/]Realclimate.org[/url] (my one-stop shop for global warming-related needs) and [url=http://scienceblogs.com/deltoid/2007/10/update_on_the_nine_alleged_err.php]Tim Lambert's Deltoid blog[/url].

Unfortunately, I did not take notes during the lecture, so I can't present specific points. However, my impression from the article you linked to (which explained alleged "errors" in Gore's movie) is that the objections are not to explicit scientific inaccuracies but to misleading implications (which is what politicians are trained to do anyway, right?).

For example, Gore certainly does not go into detailed analysis of the uncertainties of various predictions and correlations. It's not his fault, really -- it would take too much time, it would be boring, and it would be too technical for most viewers. But it often gives the impression that we (as in the scientific community) are more certain about what he says than we actually are. What's important is that [i]it is reasonable to expect these things to happen/to be the case[/i], and therefore we must take action. Not that some specific event will take place in a certain number of years, or that some specific piece of evidence actually is caused primarily by human intervention. The people who claim that Gore is wrong based on these latter things are the ones who are making a serious mistake; my point is that Gore's imprecision is minor and should not be considered reason to object to the Nobel.

By the way, the professor I mentioned works on atmosphere/ocean dynamics. I think his contentions with Gore's movie were mostly about the certainty of the parts about the relationship between the climate and the ocean and the atmosphere. Here's his [url=http://www.aos.princeton.edu/faculty/philander.htm]page[/url], although his little discussion about climate and the ocean is more of an overview of his interests than anything specific about what's going on now, so I don't know if the information there contributes much to this discussion.


[quote="stevenmeow"]I sometimes get redirected to the front page (the 3 panels).
I never get a second chancec if I need one which ultimately leads to more missed problems for me.
Perhaps my mouse likes to malfunction, but this happens several tiems as opposed to a few.[/quote]

This happens to me as well. It seems like that's what happens when you click "Submit" more than once.

I've never heard of this. If you figure out exactly what triggers it, let me know and we can take a look into it.

I dunno... It happens to me too, and I don't think I'm doing anything unusual.

Yeah, it happened to me too on a lot of occasions, mainly because I get impatient at my slow comp and I clicked the submit button twice.  I think that this is the problem that needs looked at.


Prove that if \[N = \frac{a^2 + b^2}{ab - 1}\] is an integer for positive integers $a$ and $b$, then $N = 5$.

I have found it http://www.mathlinks.ro/Forum/viewtopic.php?t=22810


Three relatively prime integers are the sides of a right triangle.  If the smallest leg has length 28, find the sum of all posible values of the hypotenuse, or prove that the sum is undefined.

Hint: [hide] This was on the 1990 MAO
(you thought I'd give you a better hint than that, didn't you?!?)[/hide]

All Pythagorean tripples $(a, b, c)$ are generated by 2-integer sets $(m, n)$, such that $m < n$. Pythagorean tripples that are relatively prime are generated by integer pairs $(m, n)$ that are relatively prime or $m = 1$ and $n$ is even:

$a = n^2 - m^2$
$b = 2nm$
$c = n^2 + m^2$

Since $(a, b, c)$ are relatively prime, $b = 28 = 2nm$. Thus we have 2 possibilities for $(m, n)$:

[b][color=blue](1)[/color][/b] $m = 1, n = 14$

$a = 196 - 1 = 195$
$b = 28$
$c = 196 + 1 = 197$

$a^2 + b^2 = 38025 + 784 = 38809 = c^2$

[b][color=blue](2)[/color][/b] $m = 2, n = 7$

$a = 49 - 4 = 45$
$b = 28$
$c = 49 + 4 = 53$

$a^2 + b^2 = 2025 + 784 = 2809 = c^2$

Both the set $(195, 28, 197)$ and the set $(45, 28, 53)$ of Pyhthagorean tripples satisfy the problem conditions and no more relatively prime tripples with one leg 28 exist. The answer is therefore $197 + 53 = 250$.


[quote="litbon"]pro 1: let a;b;c in R and $ a;b;c\geq0$ prove that:
$ \frac {2a^{b \plus{} c} \minus{} 1}{(b \plus{} c)^2} \plus{} \frac {2b^{a \plus{} c} \minus{} 1}{(a \plus{} c)^2} \plus{} \frac {2c^{b \plus{} a} \minus{} 1}{(b \plus{} a)^2}\geq \frac {(a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2}{ab \plus{} bc \plus{} ca}$

pro 2: let a;b;c;r in R and $ a;b;c;r\geq0$; ab+bc+ca=3 prove that:
$ \sum \frac {a^r}{(b \plus{} c)^2}\geq \frac {3}{4}$

pro 3: let a;b;c in R and $ a;b;c\geq0$ prove that:
$ \frac {a^{b \plus{} c}}{(b \plus{} c)^2} \plus{} \frac {b^{a \plus{} c}}{(a \plus{} c)^2} \plus{} \frac {c^{b \plus{} a}}{(b \plus{} a)^2}\geq \frac {3}{4}$[/quote]

you can see more here(write by Vietnamese):
[url]http://ineqdinhthi.wordpress.com/2008/07/07/cac-bai-toan-ch%c6%b0a-co-l%e1%bb%9di-gi%e1%ba%a3i/#comments[/url]


Prove that :
1)$ [sin{\frac{A}{2}}\plus{}sin{\frac{B}{2}}\plus{}sin{\frac{C}{2}}][cot{\frac{A}{2}}\plus{}cot{\frac{B}{2}}cot{\frac{C}{2}}]{\geq}{\frac{9{\sqrt{3}}}{2}}$
2)$ (cotA\plus{}cotB\plus{}cotC)(sin^2A\plus{}sin^2B\plus{}sin^2C) \geq {\frac{9{\sqrt{3}}}{4}}$

we can let tan($ \frac{A}{2}$)=x;
                tan($ \frac{B}{2}$)=y
                tan($ \frac{C}{2}$)=z
 and xy+yz+zx=1;
I think our inequality will be more simple!!


geometry
search
combinatorics unsolved
combinatorics

A tramp has a coat of area 1 with five patches; each patch has area >= 1/2. Prove that two patches exist with common area >= 1/5

[quote="JBL"]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1396904985&t=187645
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1701477842&t=154180
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=24486#24486

(I foudn these by searching for the word "patches.")[/quote]

I found it . Thanks for ur sincere answer.


Let $x,y,z \in R$ such that $x+y+z=1$. Let $k$ be natural number, $1\leq k\leq5$.
For each $k$,Find the maximum value of :
$P_k=\frac{x}{(x^2+1)^k}+\frac{y}{(y^2+1)^k}+\frac{z}{(z^2+1)^k}$.

Albanian Eagle - Your solution is wrong.
Nobody has any idea?

I think Albanian Eagle - solution is correct. :lol:


calculus
integration
logarithms
calculus computations

Evaluate the following integrals.

(1) Meiji University

$\int_{\frac{1}{e}}^{e}\frac{(\log x)^{2}}{x}dx.$

(2) Tokyo University of Science

$\int_{0}^{1}\frac{7x^{3}+23x^{2}+21x+15}{(x^{2}+1)(x+1)^{2}}dx.$

i know how to use partial fractions but could you please show how you got the coefficients of the numerators.


Does anybody can give me something about harmonical poligons, I'd like to get a little but complete explanation about the conditions of harmonical poligons.

 thank you  :D

i have writen some thing about harmonic quadrilateral...

it may halp u :  http://www.mathlinks.ro/Forum/viewtopic.php?t=57006&postorder=asc


MATHCOUNTS
quadratics
geometry
probability
function
algebra
Pythagorean Theorem
\/closed

Is there a pre-test for the MATHCOUNTS problem series?  Or just a list of topics with which a student is expected to be familiar? My son is thinking of taking it in the fall, and we'd like to be sure he is ready to get most he can out of it.

Thanks

We don't have a pre-test for the MATHCOUNTS Problem Series mostly because the participants seem to be aware of the difficulty level and subject matter of the MATHCOUNTS curriculum.

The Problem Series focuses on problems in the mid to difficult range relative to the problems seen on MATHCOUNTS contests, so it's best for students who have some experience -- understand basic equation manipulations, solve a quadratic equation, and know the basic tools of geometry (areas of most regular shapes, pythagorean theorem, etc.).

Hmm, even though we know the what's included in MATHCOUNTS, it isn't necessarily obvious to me which pieces would be expected as background, and which would be things you would include in the course.  For example, I wouldn't have guessed that solving quadratics would be something you would expect.  Would looking at the pre-tests for the Introductory courses give us a good idea of what might be assumed?

On the one hand, I can sort of figure out that all of the "basic skills" needed will be assumed, so that you can concentrate on the more "interesting" problem types, techniquies, etc.  But after a year of coaching MATHCOUNTS, I still don't have a good list of all the "basic skills" required. ;-)

Okay, I see.  Perhaps a course outline will give a little clearer picture:

Lesson I: Number Sense/Math Sense
Lesson II: Number Theory
Lesson III: Algebra
Lesson IV: Algebra
Lesson V: Sequences and Series
Lesson VI: Counting
Lesson VII: Probability
Lesson VIII: Geometry
Lesson IX: Geometry
Lesson X: Word Problems
Lesson XI: Logic and Imagination
Lesson XII: Count Down Round

While this doesn't get down into a detailed version of what is covered, it at least shows our organization of the topics we see covered in MATHCOUNTS materials.  Number sense and Math Sense are just names we give to using our familiarity with numbers and problem solving to solve problems in the simplest possible way.

Having some experience with word problems is also really helpful.  From my experience, one of the biggest leaps from students in the bottom halves of state pools to the top halves is the ability to make analogies from words to math.

Logic and imagination involves simple logic and also making the nicest model (analogies again) for solving some of the hardest MATHCOUNTS problems.  That day is one that many of the kids find most fun (until the Count Down Round that it).

The other topics are a little more self explanatory and using the MATHCOUNTS materials as a guide is probably easier in those cases.  The Number Theory, Geometry, and Counting/Probability are subsets of the material covered in our Intro classes (though covered more deeply in the subject classes), so the pre/post tests from those classes might also help.  The only problem with that is that MATHCOUNTS level material probably makes up only half of those classes.  The rest is a step higher (AMC exam level mostly).

Thanks, that's helpful.  I see that you have two classes on "Algebra" but I am guessing that you assume a fair familiarity with manipulating algebraic expressions, right?  If you get a chance to expand on what is expected in that area, I think that is the area where my son would be most likely to need to get up to speed before taking a class like this.  The topics look great. :)

Algebra: Factorizations of quadratics, setting up equations from work problems, proportions (direct, inverse, and combinations of those), quadratic formula, differences of squares/cubes, basic functions and operations (plugging numbers in, but it helps to know what an inverse function is), graphing simple equations (and those involving absolute values), systems of linear equations.

I don't recall offhand how complete this list is, but it's representative of the mechanics of the material.

A student doesn't need to be a complete master at most of these, but the basics should all be covered.  We work from there on the problem solving aspects and iron out flaws in reasoning.

Thanks.  Scanning through AOPS volume I, it looks like what you are looking for as a "prerequisite" to this course is basic familiarity with almost all of the topics in there.   Does that sound about right?

Familiarity, but certainly not 'mastery'.


function
limit
real analysis
real analysis unsolved

Let $f(0)>0$ and $f(1)<0$. Prove that $f(x)=0$ for some $x$ in $(0,1)$ under the assumption that there exists a continuous function $g$ such that $f+g$ is nondecreasing.

Thanks

I'm not sure this is correct but I think it can be done like this:

$f+g$ is monotonous, so it has lateral limits in every point, but then $f$ has lateral limits in every point(because $g$ is continuous) and \[f(x_{0})=(f+g)(x_{0})-g(x_{0})\geq \lim_{x \nearrow x_{0}}(f+g)(x)-\lim_{x \nearrow x_{0}}g(x)=\lim_{x \nearrow x_{0}}f(x)\] and doing similary for the other side we get $\lim_{x \nearrow x_{0}}f(x)\leq f(x_{0})\leq \lim_{x \searrow x_{0}}f(x)$.

If we consider the $M=\{x | f(x)\geq 0\}$ and $a=\sup M$ then $\lim_{x \searrow a}f(x)\leq 0$.
If $a\in M$ then  $0\leq f(a)\leq \lim_{x \searrow a}f(x)\leq 0$, thus $f(a)=0$.
If $a\notin M$ then there exists $x_{n}\to a$ such that $f(x_{n})\geq 0$ so $0\leq \lim_{x \nearrow a}\leq f(a)$ which is a contradiction because $f(a)<0$ so only the previous case is valid, thus $f(x)=0$ for some $x$.

But if this is correct,  $f$ must have the intermediate value property, and hence only points of discontinuity of the second kind, but $f$ has lateral limits in every point , so all the points of discontinuity are of the first kind. So $f$ would be continuous. That's why I'm not sure about the proof.

Got no time to read the previous proof, so I post mine as it is and I go sleep, lol.

Since $f+g$ is nondecreasing, it admits a left and a right limit at each point (except at $0$ and at $1$, where $f+g$ has a right (resp. left) limit; if $x=0$, then we set $f(x-)=f(0)$; likewise, we define $f(x+)=f(1)$), and $f$ has the same property by continuity of $g$.

If $f$ is discontinuous at point $x$, then necessarily $f(x+)>f(x-)$ (*). The reason for this is that otherwise we would have $f(x+) \leq f(x-)$ and $f(x+)+g(x+)=f(x+)+g(x) \leq f(x-)+g(x)=f(x-)+g(x-)$. This amounts to saying that $(f+g)(x+) \leq (f+g)(x-)$, which is absurd (the equality case is easily discarded by monotonicity of $f+g$).
Note also that we must have $(f+g)(x+) \geq (f+g)(x) \geq (f+g)(x-)$, which implies that $f(x+) \geq f(x) \geq f(x-)$ (**).

Let $z=inf(E=\{x;\ f(x) \leq 0\})$. If $f(z)=0$, the problem is solved.
If $f(z)<0$, then necessarily $f(z-) \geq 0$ (by definition of the $inf$), which is absurd by (**).
Thus, $f(z)>0$. But then $f(x+) < 0$ (because $z$ is an accumulation point of $E$), while $f(x-) \geq 0$. This is absurd by (*).


summer program
Mathcamp
PROMYS
Ross Mathematics Program
MIT
college
HCSSiM

Where is everybody going for summer programs this year?

> Im considering Hamshire and Canada - USA mathcamp, anyone else going to either of those?

I'm not going to either, but I'll plug Mathcamp: I went to Mathcamp twice, in 2000 (Vancouver) and 2001 (Maine), and I have to say that both years were incredibly enriching experiences for me. Some people may dislike it because of the cost or that you won't get any high school credit for the stuff you learn there, but I believe the latter is unimportant: the extremely wide variety of classes and such aren't about learning the standard math curriculum but about exploring the cool stuff in math.

Of course, I'll also add the required "Mathcamp isn't all about math - it's about freedom" and whatnot. But talk to other people who have gone to those programs, and choose what's best for you.

(I also have a friend from town who went once because of me, another time because of his parents, and maybe now because he wants to himself.)

I'm not going anywhere this year, but I heard  PROMYS was a good choice for high schoolers. It's basically advanced number theory and some calculus.

I'll be attending the Canada-USA Mathcamp this summer.  Has anyone else been there, or does anyone else plan to go?

Is anyone going to Ross?? I went before my freshman year and I highly recommend it...also...its better than PROMYS...lol

I go to AAST. I went to PROMYS for 2 years.  I went to big MOP last year. I'm going to MIT next year. So I can answer any questions about any of the programs/schools I have attended.  Since this is a summer program post, I'll talk more about PROMYS and MOP.  Basically, MOP is the best math program out there in terms of quality of students, level of material and prestige.  Unfortunately, it is very very selective and only invitational to the high scorers of the USAMO.
So, the majority of the people on the post will probably have to decide on PROMYS, ROSS, Hampshire, SUMAC, and MathCamp.  I have had friends and people I know go to every camp except for Hampshire so I will not talk about Hampshire.  SUMAC and MathCamp are good.  ROSS is better. PROMYS is best.
Any questions? e-mail me

Please.  Let's not get into petty camp rivalries. All these camps are quite good.  I personally favor MOP, but people who aren't as much into the olympiad/competition thing may well prefer other programs.  Some people like the flexibility of MathCamp, or the number theory emphasis of PROMYS and Ross, or the other distinctive qualities of other programs.  So instead of going on about "my camp is better than your camp", let's focus on what makes each camp unique and special.

--Alison

[quote="Davy"]SUMAC and MathCamp are good.  ROSS is better. PROMYS is best.
Any questions? e-mail me[/quote]

I really don't think the differences between the camps are that drastic.  A better comparison would be say, Harvard and a small liberal arts college.  Harvard is certainly more prestigious, has a greater selection, and all that.  But there are still people out there who choose the small liberal arts college, because it's better suited to them than Harvard.  But anyway.  Let's get back on topic.  Why is PROMYS the best?

Alison

[quote="Davy"]alison, say you get a full ride to harvard and a Community College.  Where would you go?  You would go to Harvard since it is a better school.  Using the same logic, we can conclude that the better student should go to a better program to help him learn and enrich his mind.  So, I am just telling everyone out there which programs are better. xP[/quote]

I great program that is similar to promys and ross is the swt honors summer math camp in texas. It is a number theory camp as well but with mathematica, research groups, and fun trips incoporated. Furthermore, differing from Ross ( and maybe promys...not really sure) it actually has girls.

[quote="poop"]Furthermore, differing from Ross ( and maybe promys...not really sure) it actually has girls.[/quote]

Are you saying I'm not a girl?  And my roommates weren't girls?  And Anna and Grace and Kathy weren't girls?  :P

But you're right.  The gender ratio was a bit... unhappy our year.  Still, it could have been much worse.  Don't avoid Ross just because of that.

PS - Your username is terrible.  It's no wonder you didn't see much of the girls at Ross.  Okay.  Done teasing.  :wink:

I went to SUMaC and MOP, and they were both very good. I can highly recommend either of these. I'm sure the others are good too, but I haven't gone to them, so I shouldn't state any opinions based on guesses.

After Brice and Grace became more feminine...(only for one night...)the ratios were pretty even, that is...counting JC's

My sister is taking biology there.  (And I took Intro to Lab Sci at the Carlisle location 4 years ag).  Although it hasn't been discussed in the thread about because it is not a specialized math camp, it is a strong all-around summer program.

Best wishes for the summer!
--Alison

[quote="andy17null"]I'm going to Johns Hopkins CTY at Skidmore college, Saratoga Springs this year. I'm taking a course on Probability and Game theory- anybody else involved with CTY?[/quote]

first off, PROMYS boy to girl ratio is like 60/40 so it is not that bad, plus there are lot of other programs that are going around at boston university so lack of girls is not a problem....
and alison, you have not been to the programs while i have, and i think i should have a better opinion than you..so I WIN =) eve knows i WIN cause i always win

I'm planning on going to PROMYS this summer.  It's going to be my first time going to a math camp.  Could someone tell me what they most liked about attending a math program?

More so than actual math...i loved being around ppl like me...smart...yet human...It was great at building communities etc...but some camps are better at this than others...

I'm glad that you say that PROMYS is the best.

I'm going there this summer.

Actually, I am somewhat surprised they even accepted me.

I found those application questions VERY challenging.

Though, I must admit I learned quite a bit.

Anyhow, what's it like at PROMYS?  :roll:

I'll be at Mathcamp - they accepted me early. I haven't heard much about the other programs, does anyone have info?

I just got accepted to Hampshire, is anyone else going there?

I couldn't go to PROMYS cuz im a little kid. Next year I'll see though.