Let $ x,y,z>0$ and $ \frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\equal{}1$. Prove that

$ (x\plus{}y\minus{}z\minus{}1)(y\plus{}z\minus{}x\minus{}1)(z\plus{}x\minus{}y\minus{}1)\leq 8$.

[quote="jbmorgan"]Let $ x,y,z > 0$ and $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} 1$. Prove that

$ (x \plus{} y \minus{} z \minus{} 1)(y \plus{} z \minus{} x \minus{} 1)(z \plus{} x \minus{} y \minus{} 1)\leq 8$.[/quote]
Let:$ x\plus{}y\minus{}z\minus{}1\equal{}2c,y\plus{}z\minus{}x\minus{}1\equal{}2a,z\plus{}x\minus{}y\minus{}1\equal{}2b$ then
It's eqivalent to:
Let $ abc\equal{}1$
prove that:$ \frac{1}{1\plus{}a\plus{}b}\plus{}\frac{1}{1\plus{}c\plus{}b}\plus{}\frac{1}{1\plus{}c\plus{}b} \leq 1$

[quote="Hong Quy"][quote="jbmorgan"]Let $ x,y,z > 0$ and $ \frac {1}{x} + \frac {1}{y} + \frac {1}{z} = 1$. Prove that

$ (x + y - z - 1)(y + z - x - 1)(z + x - y - 1)\leq 8$.[/quote]
Let:$ x + y - z - 1 = 2c,y + z - x - 1 = 2a,z + x - y - 1 = 2b$ then
It's eqivalent to:
Let $ abc = 1$
prove that:$ \frac {1}{1 + a + b} + \frac {1}{1 + c + b} + \frac {1}{1 + c + b} \leq 1$[/quote]
It's equivalent to:
$ \sum_{cyc} (1+a+b)(1+b+c) \le \sym_{cyc} (1+a+b)(1+b+c)(1+c+a)$ $ \iff$
$ \sum_{cyc} (1+2b+c+a+ab+ac+bc+b^2) \le \sym_{cyc} 1+2(a+b+c)+3(ab+bc+ca) + (a^2+b^2+c^2) + (a^2b+a^2c+b^2a+b^2c+c^2a+c^2b) + 2abc$ $ \iff$
$ 2 + 2\sum_{cyc} a+  \le \sum_{cyc} (a^2b+a^2c) + 2abc$ $ \iff$
Since $ abc = 1$ then it's equivalent to:
$ 2\sum_{cyc}a^{5/3}b^{2/3}c^{2/3} \ge \sum_{cyc} a^2b+a^2c$.
Which is true by muirhead.

[quote="Hong Quy"][quote="jbmorgan"]Let $ x,y,z > 0$ and $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} 1$. Prove that

$ (x \plus{} y \minus{} z \minus{} 1)(y \plus{} z \minus{} x \minus{} 1)(z \plus{} x \minus{} y \minus{} 1)\leq 8$.[/quote]
Let:$ x \plus{} y \minus{} z \minus{} 1 \equal{} 2c,y \plus{} z \minus{} x \minus{} 1 \equal{} 2a,z \plus{} x \minus{} y \minus{} 1 \equal{} 2b$ then
It's eqivalent to:
Let $ abc \equal{} 1$
prove that:$ \frac {1}{1 \plus{} a \plus{} b} \plus{} \frac {1}{1 \plus{} c \plus{} b} \plus{} \frac {1}{1 \plus{} c \plus{} b} \leq 1$[/quote]
Why it's equivalent? $ a$ can be negative. :wink:
[quote="Mathias_DK"]
$ 2\sum_{cyc}a^{5/3}b^{2/3}c^{2/3} \ge \sum_{cyc} a^2b \plus{} a^2c$.
Which is true by muirhead.[/quote]
Are you sure? :wink:

[quote="arqady"]
[quote="Mathias_DK"]
$ 2\sum_{cyc}a^{5/3}b^{2/3}c^{2/3} \ge \sum_{cyc} a^2b \plus{} a^2c$.
Which is true by muirhead.[/quote]
Are you sure? :wink:[/quote]
I switched sign in the last one.. It should be $ 2\sum_{cyc}a^{5/3}b^{2/3}c^{2/3} \le \sum_{cyc} a^2b \plus{} a^2c$, which is what we want to prove.

[quote="jbmorgan"]Let $ x,y,z > 0$ and $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} 1.$ Prove that

$ (y \plus{} z \minus{} x \minus{} 1)(z \plus{} x \minus{} y \minus{} 1)(x \plus{} y \minus{} z \minus{} 1)\leq 8.$[/quote]$ 8 \minus{} \prod{(y \plus{} z \minus{} x \minus{} 1)} \equal{} \frac {8x^3y^3z^3}{(yz \plus{} zx \plus{} xy)^3} \minus{} \prod{\left(y \plus{} z \minus{} x \minus{} \frac {xyz}{yz \plus{} zx \plus{} xy}\right)}$

$ \equal{} \sum\frac {(y \minus{} z)^2\left[xyz(y \plus{} z \minus{} x)^2(yz \plus{} 2zx \plus{} 2xy \plus{} x^2) \plus{} y^3z^3(y \plus{} z)\right]}{(yz \plus{} zx \plus{} xy)^3}\geq0.$
[quote="arqady"]The following stronger inequality is true too.
Let $ x,$ $ y$ and $ z$ are positive numbers such that $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} 1$. Prove that
\[ \sum(z \plus{} x \minus{} y \minus{} 1)(x \plus{} y \minus{} z \minus{} 1)\leq 12.\]
[/quote]Why it's stronger?

Your inequality holds for [b]real[/b] numbers $ x,y,z$:

$ 12 \minus{} \sum{(z \plus{} x \minus{} y \minus{} 1)(x \plus{} y \minus{} z \minus{} 1)}$

$ \equal{} \frac {12x^2y^2z^2}{(yz \plus{} zx \plus{} xy)^2} \minus{} \sum{\left(z \plus{} x \minus{} y \minus{} \frac {xyz}{yz \plus{} zx \plus{} xy}\right)\left(x \plus{} y \minus{} z \minus{} \frac {xyz}{yz \plus{} zx \plus{} xy}\right)}$

$ \equal{} \frac {[\sum{x^4(y \plus{} z)} \minus{} \sum{x^3(y^2 \plus{} z^2)}]^2 \plus{} (y \minus{} z)^2(z \minus{} x)^2(x \minus{} y)^2\sum{y^2z^2}}{(\sum{yz})^2(\sum{x^4} \minus{} \sum{y^2z^2})}\geq0.$

I just wonder how we can get the expression like Jichen by hand because the factuality here is it takes time and energy to establish that such equality. So thrilling though!  :|

[quote="Hong Quy"][quote="jbmorgan"]Let $ x,y,z > 0$ and $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} 1$. Prove that

$ (x \plus{} y \minus{} z \minus{} 1)(y \plus{} z \minus{} x \minus{} 1)(z \plus{} x \minus{} y \minus{} 1)\leq 8$.[/quote]
Let:$ x \plus{} y \minus{} z \minus{} 1 \equal{} 2c,y \plus{} z \minus{} x \minus{} 1 \equal{} 2a,z \plus{} x \minus{} y \minus{} 1 \equal{} 2b$ then
It's eqivalent to:
Let $ abc \equal{} 1$
prove that:$ \frac {1}{1 \plus{} a \plus{} b} \plus{} \frac {1}{1 \plus{} c \plus{} b} \plus{} \frac {1}{1 \plus{} c \plus{} b} \leq 1$[/quote]

 Denote $ a \equal{} A^3,b \equal{} B^3,c \equal{} C^3$ so

 $ \sum \frac {1}{1 \plus{} a \plus{} b} \equal{} \sun\frac {1}{1 \plus{} A^3 \plus{} B^3}\le\sum\frac {1}{1 \plus{} AB(A \plus{} B)} \equal{} \sum\frac {C}{A \plus{} B \plus{} C} \equal{} 1$

[quote="jbmorgan"]Let $ x,y,z > 0$ and $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} 1$. Prove that

$ (x \plus{} y \minus{} z \minus{} 1)(y \plus{} z \minus{} x \minus{} 1)(z \plus{} x \minus{} y \minus{} 1)\leq 8$.[/quote]
let $ x \equal{} \frac {a \plus{} b \plus{} c}{a},y \equal{} \frac {a \plus{} b \plus{} c}{b},z \equal{} \frac {a \plus{} b \plus{} c}{c}$
$ \iff \prod_{cyc}{((a \plus{} c)ac \plus{} (a \plus{} b)ab \minus{} (b \plus{} c)bc)}\leq 8a^3b^3c^3$
$ \iff\sum_{cyc}(a^6(b \plus{} c) \plus{} 2a^5(b^2 \plus{} c^2 \plus{} bc) \plus{} a^5(b \plus{} c)^2 \minus{} a^3bc(ab \plus{} bc \plus{} ca) \minus{} a^3b^2c^2)(b \minus{} c)^2 \geq 0$
Let $ a\geq b \geq c$ ,So $ S_a\geq0,S_b\geq0$
$ S_b \plus{} S_c \equal{} a^2(3b^3 \plus{} 3c^3 \minus{} bc^4 \minus{} cb^4 \minus{} 2b^3c^2 \minus{} 2c^3b^2) \plus{} (b^6(a \plus{} c) \plus{} c^6(a \plus{} b) \plus{} b^5(3c^2 \plus{} 4ac) \plus{} c^5(3b^2 \plus{} 4ab) \minus{} b^4c^2a \minus{} c^4b^2a) \geq 0$
done :)

[quote="arqady"][quote="jbmorgan"]Let $ x,y,z > 0$ and $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} 1$. Prove that

$ (x \plus{} y \minus{} z \minus{} 1)(y \plus{} z \minus{} x \minus{} 1)(z \plus{} x \minus{} y \minus{} 1)\leq 8$.[/quote]
The following stronger inequality is true too.
Let $ x,$ $ y$ and $ z$ are positive numbers such that $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} 1$. Prove that
\[ \sum_{cyc}(x \plus{} y \minus{} z \minus{} 1)(x \plus{} z \minus{} y \minus{} 1)\leq 12\]
[/quote]
let $ x \equal{} \frac {a \plus{} b \plus{} c}{a},y \equal{} \frac {a \plus{} b \plus{} c}{b},z \equal{} \frac {a \plus{} b \plus{} c}{c}$
$ \sum_{cyc}(x \plus{} y \minus{} z \minus{} 1)(x \plus{} z \minus{} y \minus{} 1)\leq 12$
$ \iff \sum_{cyc}((a \plus{} c)ac \plus{} (a \plus{} b)ab \minus{} (b \plus{} c)bc)((a \plus{} b)ab \plus{} (b \plus{} c)bc \minus{} (c \plus{} a)ca)\leq 12a^2b^2c^2$
$ \iff (\sum_{sym}a^4b^2\minus{}\sum_{sym}a^4bc)\plus{}2(\sum_{cyc}a^3b^3\minus{}\sum_{cyc}a^3b^2c\plus{}\sum_{cyc}3a^2b^2c^2) \geq 0$
using Muirhead and Schur inequalities ,we know easily that the above is true.


abstract algebra
factorial
Ring Theory
superior algebra
superior algebra unsolved

Give example of noetherian closed ring which is not factorial.

What is a closed ring?

The ring $\mathbb Z[x, y]/ (x^2 = y^3)$ is an exaple of a Noetherian ring, which is not factorial.
The ring is Noetherian by Hilbert's basis theorem. It is not factorial, since the element $x^2 = y^3$ has two non-equivalent factorizations: $x\cdot x$ and $y \cdot y \cdot y$.


Βρειτε την ελαχιστη τιμη της

$ |sinx\plus{}cosx\plus{}tanx\plus{}cotx\plus{}secx\plus{}cscx|$

για πραγματικους αριθμους $ x$.

[hide]
Σωστά βγάζω $ 2\sqrt {2} \minus{} 1$ ?
[/hide]
Cheerio,

Durandal 1707

Σωστος!!! Αν μπορεις γραψε την λυση σου.

Η λύση μου είναι λίγο άσχημη και ντρέπεται να βγει γιατί είναι κάπως brute force μετά από την παρατήρηση κάποιων στοιχειωδών συμμετριών και την αντικατάσταση $ y\equal{}\sin x$... ήμουν απλώς περίεργος να δώ αν τελικά είχα βγάλει σωστό αποτέλεσμα ;)

Ας μην το σκοτώσουμε λοιπόν ακόμα το πρόβλημα :)

Cheerio,

Durandal 1707

$ |sinx+cosx+tanx+cotx+secx+cscx|=$ $ \left|sinx+cosx+\frac{1}{sinx\cdot cosx}+\frac{sinx+cosx}{sinx\cdot cosx}\right|$

Θετουμε $ y=\frac{3\pi}{4}+x$ οποτε

$ sinx+cosx=-\sqrt{2}cosy$

$ tanx+cotx=\frac{2}{cos^2 y+sin^2 y}$

${ secx+cscx=-\frac{2\sqrt{2}cosy}{cos^2 y+sin^2 y}}$

Μετα θετουμε $ a=\sqrt{2}cosy$ και ετσι θα εχουμε

$ |sinx+cosx+tanx+cotx+secx+cscx|=$ $ \left|a+\frac{2}{a+1}\right|$

Απο AM-GM εχουμε οτι $ a+1+\frac{2}{a+1}\geq 2\sqrt{2}$ για $ a+1>0$ και $ \leq -2\sqrt{2}$ για $ a+1<0$

Επομενως η ελαχιστη τιμη της ειναι η $ 2\sqrt{2}-1$

ps. Ισοτητα εχουμε για $ a=\sqrt{2}-1$

Πολύ ωραία άσκηση, αν και νομίζω δεν είναι τόσο "για φοιτητές"

Η άσκηση αυτή εχει τεθεί στο Putnam. Επίσης τα $ secx$ και $ cscx$ δεν διδάσκονται στο σχολείο. Αλλά θα συμφωνήσω οτι είναι ωραία άσκηση.  :)


What is the largest [tex]\lambda_{r}[/tex] for which we can be sure that for all non-negative reals [tex]a, b, c[/tex], we have

[tex]\displaymode{
2\left(a^{r+2} + b^{r+2} + c^{r+2}\right) + 2\lambda_{r}\left(a^rbc + b^rca + c^rab\right) \ge \left(1+\lambda_{r}\right)\left(a^{r+1}(b+c) + b^{r+1}(c+a) + c^{r+1}(a+b)\right)
}[/tex]

Schur's inequality gives us [tex]\lambda_{r} \ge 1[/tex] for all positive [tex]r[/tex].

Ok, so setting [tex]a = 0, b = c[/tex] trivially demonstrates that [tex]\lambda_{r} \le 1[/tex] so that [tex]\lambda_{r} = 1[/tex] for all [tex]r[/tex], and we are done. :D  But what about

[tex]\displaymode{
2\left(a^{r+2} + b^{r+2} + c^{r+2}\right) + \lambda_{r}\left(a^r(b^2 + c^2) + b^r(c^2 + a^2) + c^r(a^2 + b^2)\right) \ge (1 + \lambda_{r})\left(a^{r+1}(b+c)+b^{r+1}(c+a)+c^{r+1}(a+b)\right)
}[/tex]

again for [tex]a, b, c \in \mathbb{R}_{0}^{+}, r \in \mathbb{R}^{+}[/tex]?


If $ f: [0,1] \rightarrow [0,1]$ is continuous, prove that there is an $ x$ in $ [0,1]$ such that $ f(x) \equal{} x$.

The diagonal from $ (0,0)$ to $ (1,1)$ cuts every curve from $ (0,y_1)$ to $ (1,y_2)$  because $ y_1\geq 0$ and $ y_2\leq 1$. I know this is not what you are looking for because it is only intuitive.

Consider the function $ F(x) \equal{} f(x) \minus{} x$ $ F(0)\geq 0$ and $ F(1)\leq 0$ if either $ F(0) \equal{} 0$ or $ F(1) \equal{} 0$  then we are done if not, then $ F(0) > 0$ and $ F(1) < 0$ but because $ f$ and the identity are continuous, $ F$ is continuous. This means that at some point its graph will cut the $ x$-axis and $ F(x_i) \equal{} 0$ i.e. $ f(x_i) \equal{} x_i$

if $ f(0) \equal{} 0$ or $ f(1)\equal{}1$ then we are done, so assume not

$ g(x)\equal{}f(x)\minus{}x$ is continuous with $ g(0)\equal{}f(0) \in (0,1]$ and $ g(1)\equal{}f(1)\minus{}1 \in [\minus{}1,0)$

then it should be easy to see!


limit
integration
calculus
calculus computations

Evaluate,
\[ \lim_{n \rightarrow \infty} \int_{a}^{\infty} \frac {n}{1 \plus{} n^2x^2} dx
\]

[quote="beezees222"]Evaluate,
\[ \lim_{n \rightarrow \infty} \int_{a}^{\infty} \frac {n}{1 + n^2x^2} dx
\]
[/quote]

Here is what I think, I could be very wrong though. I am a newbie here, so sorry if I am really off  :blush: .
[hide] Let

$ L = \lim_{n\to\infty}n\int_a^{\infty}\frac {dx}{1 + (nx)^2}$

Now let $ \varphi = nx\Rightarrow\frac {\varphi}{n} = x$

So $ dx = \frac {d\varphi}{n}$

So we have then

$ \lim_{n\to\infty}n\int_{na}^{\infty}\frac {\frac {d\varphi}{n}}{1 + \varphi^2} = \lim_{n\to\infty}n\cdot\frac {1}{n}\int_{na}^{\infty}\frac {d\varphi}{1 + \varphi^2}$

$ = \lim_{n\to\infty}\int_{na}\frac {d\varphi}{1 + \varphi^2} = \lim_{n\to\infty}\arctan(\varphi)|_{na}^{\infty} = \lim_{n\to\infty}[\frac {\pi}{2} - \arctan(na)]$

So then we look at the three cases

If $ a > 0$ then $ L = \lim_{n\to\infty}[\frac {\pi}{2} - \arctan(na)] = \frac {\pi}{2} - \frac {\pi}{2} = 0$

If $ a = 0$ then $ L = \lim_{n\to\infty}\frac {\pi}{2} - \arctan(0n) = \frac {\pi}{2}$

Et finalement if $ a < 0$ we have $ L = \lim_{n\to\infty}[\frac {\pi}{2} - \arctan(na)] = \frac {\pi}{2} - \frac { - \pi}{2} = \pi$

So
${ L = \{\begin{array}{cc}0 & \mbox{ if }a > 0 \\
\frac {\pi}{2} & \mbox{ if }a = 0 \\
\pi & \mbox{ if }a < 0 \end{array}}\quad\blacksquare$
[/hide]

[quote="facis"]Very well done.[/quote]
Thank you very much!  :D


[quote="CEM_College_Champ"]How does MathChallengers relate to other math contests aimed at 8th and 9th grade students (like Mathcounts in the case of 8th graders)?[/quote]

It is Mathcount in BC.

[quote="CEM_College_Champ"]Is MathChallengers harder than its U.S. counterpart?

Or is it the other way around?[/quote]

I'm pretty sure its about the same


Putnam
function
algebra
polynomial
probability
geometry
combinatorial geometry

These questions all come from old (more than 40 years, for the most part) Putnam exams.

1. Let f(n) be the number of ordered pairs of positive integers (a, b) such that ab/(a + b) = n.  If d(x) = the number of divisors of a positive integer x, express f in terms of n and the function d.

2. The real polynomial p(x) is such that for any real polynomial q(x), we have p(q(x)) = q(p(x)). Find p(x).

3.  Let f(m, n) = 3m + n + (m + n)2.  Find the sum of all numbers of the form 2-f(m, n) where m and n are non-negative integers.

4. Two points are selected independently and at random from a segment length β. What is the probability that they are at least a distance α (< β) apart?

5. The sequence ai (i in the positive integers) of non-negative reals satisfies an+m ≤ anam for all m, n. Show that ai1/i is bounded (that is, that there exist two real numbers j and k such that for all i, j < ai1/i < k.)

6. 5 points lie in a plane, no 3 collinear. Show that 4 of the points form a convex quadrilateral.

Quote:1. Let f(n) be the number of ordered pairs of positive integers (a, b) such that ab/(a + b) = n. If d(x) = the number of divisors of a positive integer x, express f in terms of n and the function d.

[hide]

ab-na-nb=0

(a-n)(b-n)=n^2

So f(n) = d(n^2)[/hide]

Quote:2. The real polynomial p(x) is such that for any real polynomial q(x), we have p(q(x)) = q(p(x)). Find p(x).

[hide]Let q(x)=c. Then p(c)=c for all c, so p(x)=x.[/hide]

Quote:4. Two points are selected independently and at random from a segment length β. What is the probability that they are at least a distance α (< β) apart?

[hide]Let the endpoints of the segment be (0,0) and (β,0), and let the randomly chosen points be (p, 0) and (q, 0).

Since we are dealing with nondiscrete probability, we can try using a square to represent our possible outcomes. Let the width of the square be p ranging from 0 to β, and let the height of the square be q ranging from 0 to β. If p = 0, then q=0 to q=a doesn't work. If p = k, then q=k-a to q=k+a doesn't work. After trying to sketch this out for all p and q, we find that the region that includes order pairs(p,q) that work has an area of (β-a)^2  (the region is two right triangles at the corners of the square).[/hide]

JBL wrote:5. The sequence ai (i in the positive integers) of non-negative reals satisfies an+m ≤ anam for all m, n. Show that ai1/i is bounded (that is, that there exist two real numbers j and k such that for all i, j < ai1/i < k.)

[hide]a_1=q. We have a_k<=a_(k-1)*a_1, so a_2<=q^2, a_3<=q^3, etc... So a_k^(1/k)<=q.

[/hide]

Are these looking good Joel? Some of my solutions have been pretty short for Putnam problems... Still working on 3 and 6 though..

Good so far, except for number 5 on which you made a read-o.  The question isn't quite that trivial :).  (Actually, it's significantly simpler than what the actual Putnam question was -- the Putnam question asked you to show that the sequence a[size=50]i1/i[/size] converged, a trickier task.  Also, if you could enable HTML for that post, it would make the quote much more readable.

Mm...yeah, a couple of *read-o's*, but its a minor thing I think. Unless I'm missing something else.

Quote:3. Let f(m, n) = 3m + n + (m + n)^2. Find the sum of all numbers of the form 2^(-f(m, n)) where m and n are non-negative integers.

[hide]Ok, I claim that f(m, n) can generate all nonnegative even integers, and no odd numbers. If this is so, then the sum is simply 2^0 + 2^-2 + 2^-4 + ... = 1 + 1/4 + 1/16 + ... = 1/(1-1/4) = 4/3.

To prove this claim, we see that f(0,0)=0. If n>0, then we can replace m by m+1 and n by n-1 to yield f(m+1,n-1)=f(m,n)+2. If n=0, then f(0,m+1)=m^2+3m+2=f(m,0)+2. Thus we can generate all of the nonnegative even integers. A simple parity check shows that no odd numbers can be generated, so our proof is complete.[/hide]

One more to go.

Fierytycoon -- good for number 3, except for one problem:  [hide]you have to show that f can't generate any even number twice. [/hide] (Also, there is a typo when you wrote out f(m + 1, n - 1).)

Good fix for number 5, as well.  (Although, technically speaking, you didn't prove it was bounded from below, but I'm just nit-picking.)

[quote] you have to show that f can't generate any even number twice. [/quote]

Not quite. Read the problem statement again.

[quote]Let f(m, n) = 3m + n + (m + n)^2. Find the sum of all numbers of the form 2^(-f(m, n)) where m and n are non-negative integers. [/quote]

I only have to sum the expression for all numbers of the form <expr>. If a number can be generated with more than one ordered pair (m, n), then that doesn't matter.

zscool wrote:here's a hint for 6:
[hide]
if the convex hull contains 4 or 5 points, you are done, so what happens if it contains 3 points, draw yourself a few pictures and see what happens.
[/hide]

Mmm...too bad I don't know what convex hulls are :/

Okay, valid point for number 3.  Prove it anyways :).  (I think the original wording probably required you to sum it over all pairs (m, n), but the combination of my rewording Kalva's rewording managed to wipe that bit out.)

The convex hull of some set of points is the smallest convex region that contains them.  Generally, for a finite set of points, it will be a n-dimensional polyhedron with some of the points at its vertices.

[quote]Prove it anyways[/quote]

Well, I fixed my typo and added something that I forgot initially (the case when n=0). I think that my current solution proves the integers are increasing through the process that I described...I don't know how I would show it otherwise.


Show that for arbatrary positive numbers a1,a2,...an (n>=2), the following inequality holds:
(a1^3+)(a2^3+1)....(an^3+1)>=(a1^2a2+1)(a2^3a3+1)...(an^2a1+1)


abstract algebra
algebra
polynomial
superior algebra
superior algebra unsolved

Find the kernel and describe its cosets in Z[x] for an evaluation of the homomorphism...?
The homomorphism is
phi: Z[x] -> Z
phi(p(x))=p(1). Find kernel. Describe cosets in Z[x]...

--------MY NOTES on it----------------

Ive tried thinking of it as a polynomial with specific coefficients to help me outt... so, p(x)=a0+a1x+....+an x^n

Also for a(x) belonging to Z[x], considerb(x)=x-1 and so write... a(x)=(x-1)q(x)+r(x) ... and r(x) is clearly the remainder...
If the remainder goes down from deg(x) then 0<=deg(r(x))<1

then.... ker(phi) = H
a(x)+H

Kernel: I need to find when all the polynomials will vanish... so the remainder is 0... How much of a difference betweenthe poly not inthe kern and other not in the kern...

Cosets: I know that they are infinite... now working on simplest representation...

Can you help me solve this? Thanks!!!

We have $ \ker\phi\equal{}\{p(x)\in\mathbb{Z}[x] : p(1)\equal{}0\}$. I'm not sure how else to do this except to say that the factor $ x\minus{}1$ must be in each of the polynomials, so try to show that the ideal generated this factor equals the kernel: Show:  $ (x\minus{}1)\equal{}\ker\phi$. Show subsets both ways.


I have the following recurrence

F(n) = n/2 * F(n/2) + 2

with a boundary of

F(1) = 1

so....

F(1) = 1
F(2) = F(1) + 2
F(4) = 2( F(1) + 2 ) + 2
F(8) = 4( 2( F(1) + 2) + 2 ) + 2
F(16) = 8( 4( 2( F(1) + 2 ) + 2 ) + 2 ) + 2

etc...

I'm stumped on the simplification, tried many combinations.  I'm aware though that there will be a summation due to the addition of 2 multiple times.  Any help would be greatly appreciated.

I've also been having difficulty with the induction method, for example

ex1) T(2^k) >= 2^[(k−1)k/2] for all k >= 0
ex2) F(2^k) <= 2^(k^2) for all k >= 0

Can anyone help me with these examples in order for me to understand how to use induction on my solution further up in this post.  Thanks

Do not double post.  Continue your discussion here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=196462[/url]


Does anyone have a solution? I need it... thanks

[quote="hydro"]Find the greatest constant M such that $(a+b+c+d)^2 \geq M(ab+bc+cd)$ for $a,b,c,d \in \mathbb{R^+}$[/quote]

Shoudn't it be $(a+b+c+d)^2 \geq M(ab+bc+cd+da)$ instead of $(a+b+c+d)^2 \geq M(ab+bc+cd)$ ??

Anyway the answer for the first is $4$ and $\frac{16}{3}$ for the second..

The first was solved in the forum and the second is very ugly.

No, I didn't get wrong, the text was the one I wrote... I hope you have time to post your sol or say to me where I can find it!
thanks!

[quote="conceive"]
Shoudn't it be $(a+b+c+d)^2 \geq M(ab+bc+cd+da)$ instead of $(a+b+c+d)^2 \geq M(ab+bc+cd)$ ??

Anyway the answer ... $\frac{16}{3}$ for the second..
[/quote]
I think you are wrong: $d\rightarrow0$ and $a=b=c=1.$ ;)

[quote="hydro"]Find the greatest constant M such that $(a+b+c+d)^2 \geq M(ab+bc+cd)$ for $a,b,c,d \in \mathbb{R^+}$[/quote]
If $d=c\rightarrow0^{+}$ then $M<4.$
But $(a+b+c+d)^2 \geq4(ab+bc+cd+da)>4(ab+bc+cd).$


OK PEOPLE, WE WILL BE VERSING EACH OTHER AT 7:30 PM EST ON THURSDAY THIS WEEK.

Wat if i cant make it?
Hav u thot about that "anirudh"
that's wat i thot.


A rectangle with length x has area 3 and   x ^3 d = 20 , where d is the length of the diagonal of the rectangle. Find d .

(A)4 (B) 3.5 (C) 2.75 (D) 2.5 (E) 2.25


algebra
polynomial
geometry
3D geometry
linear algebra
matrix
linear algebra solved

$A\in M_{3}(\mathbb C)$

$x_n=det(A^n-I_{3})$

we suppose $x_1=1 ; \;x_2=x_3=0$

1/ Find the characteristic polynomial of $A$

2/ Does $A$ is diagonalisable ?

$A$ is diagonalizable.
$A$ is not real-valued.
There are two possiblities for the characteristic polynomial of $A$. One is

$P(x)=-x^3+\left(-\frac14+\frac{7\sqrt{3}}{12}i\right) +\frac32x + \left(\frac34-\frac{7\sqrt{3}}{12}i\right)$

The other is the conjugate of that.

Explanations: both a square root of 1 and a cube root of 1 are eigenvalues of $A$ but 1 is not an eigenvalue. Thus we assume that the roots of the characteristic polynomial are $-1, \omega,$ and $r,$ where $\omega=-\frac12+\frac{i\sqrt{3}}2$ or its conjugate. The condition that $P(1)=1$ is sufficient to determine that $r=\frac54 + \frac{i\sqrt{3}}{12}$ (or its conjugate.) Since the three roots of the characteristic equation are distinct, the matrix is diagonalizable.

Finding the characteristic polynomial is an easy exercise, isn't it? We get $P(1)=-1$ (I took $P(x)=\det(xI_3-A)$), then $P(-1)=0=P(\varepsilon)\cdot P(\bar{\varepsilon})$, where the $\varepsilon$'s are the roots of $x^2+x+1$. I think we get two distinct cases, depending on whether we take $\varepsilon$ or its conjugate to be an eigenvalue of $A$. In any case, we get a polynomial with distinct roots, so $A$ is diagonalizable.

Am I missing something here? :? Sorry if I am.

[Edit: I was writing when Kent posted that :)]

My edits were just to fix typos. Yes, I did exactly what grobber said to do, although it took some time to stamp out the arithmetical errors - the detailed calculations aren't pretty. Grobber and I chose different    sign conventions for $P(x),$ but that makes no difference in the result.

Here another solution less elegant, but it is an occasion to use any $A\in M_3(C)$ is trigonalisable.

 A is similar to

$\left[\begin{array}{ccc}a&*&*\\0&b&*\\0&0&c\end{array}\right]$

where $a,b,c$ are the eigenvalues, the conditions gives

$(a-1)(b-1)(c-1)=1$
$(a^2-1)(b^2-1)(c^2-1)=0$
$(a^3-1)(b^3-1)(c^3-1)=0$

solving you get $a,b,c \in \{-1;t ; t/6+4/3 \}$ where $t^2+t+1=0$


algebra
function
domain
calculus
integration
Ring Theory
number theory

Let $ R$ be an integral domain. Prove that if the following two conditions hold then $ R$ is a Principal Ideal Domain:
(i) any two nonzero elements $ a$ and $ b$ in $ R$ have a greatest common divisor which can be written in the form $ ra\plus{}sb$ for some $ r,s\in R$, and
(ii) if $ a_1,a_2,a_3,\cdots$ are nonzero elements of $ R$ such that $ a_{i\plus{}1}|a_i$ for all $ i$, then there is a positive integer $ N$ such that $ a_n$ is a unit times $ a_N$ for all $ n\geq N$.

for any ideal I, pick $ r_1\in I$. if there exists x, that $ r_1$ doesn't divide x, than $ r_2 \equal{} (x,r_1)\in I$. do this again for $ r_2$.
we have $ r_{i\plus{}1}|r_{i}$. so there is r, (r)=I.

[quote="mathiq"]for any ideal I, pick $ r_1\in I$. if there exists x, that $ r_1$ doesn't divide x, than $ r_2 \equal{} (x,r_1)\in I$. do this again for $ r_2$.
we have $ r_{i \plus{} 1}|r_{i}$. so there is r, (r)=I.[/quote]

I.e. if not a|b then there is a "smaller" R-linear combination, namely gcd(a,b),
i.e. R is a PID by the Dedekind-Hasse PID criterion  [1].  Here we take "smaller"
to mean proper divisor, which is well-founded by the ACCP hypothesis

This Bezout & ACCP => PID is a 1-dim version of GCD & ACCP => UFD.

--Bill Dubuque

[1] http://google.com/groups?selm=y8z3awla74e.fsf%40nestle.csail.mit.edu


calculus
integration
function
number theory proposed
number theory

\[A=3\sum_{m=1}^{n^2}(\frac12-\{\sqrt{m}\})\]
where $n$ is an positive integer. Find the largest $k$ such that $n^k$ divides $[A]$.

Just checking, { :sqrt: m} and [A] denote the fractional and integral part of  :sqrt: m and A, respectively, right?

My solution:
Let $\sigma=\sum_{k=1}^{n^2-1}$ then $A=2n^3-\frac n2-3\sigma$
Function $f(x)=\sqrt x$ is concave. Thus $\int_k^{k+1}f(x)dx>\frac{\sqrt k+\sqrt{k+1}}2$ ($*$) and $\int_{k-\frac12}^{k+\frac12}f(x)dx<\sqrt k$ ($**$).
From ($*$) we conclude that $\sigma<\int_0^{n^2-1}f(x)dx+\frac12\sqrt{n^2-1}=\frac23(n^2-1)^{\frac32}+\frac12\sqrt{n^2-1}<\frac23n^3-\frac12n$
the last ineq is equivalent to $n^6-\frac12n^4+\frac1{16}n^2-n^4+\frac12n^2-\frac1{16}<n^6-\frac32n^4+\frac9{16}n^2$.
From ($**$) we conclude that $\sigma>\int_{\frac12}^{n^2-\frac12}=\frac23(n^2-\frac12)^{\frac32}-\frac23\cdot\frac1{2^{\frac32}}>\frac23n^3-\frac12n-\frac13$
the last ineq is obtained from $-\frac1{2\sqrt2}>-\frac12$ and $(1-\frac1{2n^2})^{\frac32}>1-\frac3{4n^2}$ (Bernoully).
Thus $n<A<n+1$ and $[A]=n$.


calculus
geometry
integration
3D geometry
sphere
analytic geometry
function

What is the formula for the intersection of two cylinders, whose axes are perpendicular to each other? I have some ideas using calculus (multivariable).

if you want the formula for the volume it is 16r^3/3 for both cylinders oto have same radius if i remeber correctly if you want the surface area it is 16r^2
i'll take the intergral of the squares tommorow if nobody does it before me i need to sleep.

The best way to do it -- put a sphere down next to it and compare cross-sections of the two -- what do you see?  (The sphere should be of the same radius as the two cylinders, obviously.)

i was wondering if someone could demonstrate it using multivariable. i was thinking polar coordinates, but im not sure what the equation would be, or the limits on the integrals. thanks

Nope, don't use polar coordinates.  I would recommend looking at cross-sections in planes parallel to the axes of both cylinders and integrating in that direction.

i'm not in multivariable caclulus but i learned how to take integral of many simple shapes.
r is a constant and is the radius of both cylinders.
Int((2*(sqrt(r^2-h^2))^2,h,-r,r)=16*r^3/3
as you see it forms layers of squares
there are many more ways to take the integral by looking at the problem differently..

definitely a multivar calc  problem, find the volume of the intersection of 3 cylindars at right angles to each other. We just solved it with all of them radius 1. Try that, and then find a formula with radii a, b, c.

yeah, that's what i was thinking. i was just unsure about what the equation is of the cylinders is. would it be something like [hide]x^2 + y^2 or y^2 + z^2, etc. [/hide] ? I guess that's the part im confused about. From there, i think it's pretty easy to find the limits on [hide] r,  :theta: , z or x, y, z[/hide]. Thanks again for your contributions to this problem.

8 (2 - sqrt(2)) r^3 for three cylinders of the same radius
its crazy i can't even visualize the shape.

i don't know how you get it?

wow, i must say, that's an amazing graphic representation. before i do any more work on this though, i'd like to know whether or not my ideas for the function are correct. for instance, would (for integrating purposes) a cylinder with an x-axis of rotation be [hide]y^2 + z^2[/hide] ??? once again thanks for your help

i didn't create that one.
but you can create all sorts of cool stuff using MAPLE

[b]this gives you a cylinder in a sphere. [/b]

with(plottools):

r := 1: h := 1.2*r:

cyl := cylinder([0,0,-h/2], sqrt(r^2-h^2/4), h, shading=xyz, style=patchnogrid):

ball := sphere([0,0,0], r, color=green):

plots[display]([cyl,ball], scaling=constrained, style=wireframe, orientation=[45,74]);

[img]http://www.mapleapps.com/categories/graphics/gallery/graphics/L8-maxmin11.gif[/img]


algebra
polynomial
search
Rational Root Theorem
algebra proposed

Find all polynomials with rational coefficients that gets irrational value for all irrational argument
(My teacher W.Sobchuk problem)

Look at [url]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=467278524&t=279780[/url]

For good measure, here is a solution
[hide]Assume that there exists a polynomial $ P(x)$ with $ \deg P \equal{} n > 1$ satisfying the condition.

Amplify $ P(x)$ by some non-zero integer to get $ Q(x) \equal{} a_nx^n \plus{} a_{n\minus{}1}x^{n\minus{}1} \plus{} \cdots \plus{} a_1x \plus{} a_0$ with integer coefficients and positive leading coefficient. Clearly, for any large enough $ p$, the equation $ Q(x)\equal{}p$ has (exactly) a positive real root.

Let $ p$ be chosen such that $ q \equal{} p \minus{} a_0$ is a prime. Clearly there are infinitely many such primes.

Then, by the Rational Root Theorem, a positive rational root of $ Q(x)\equal{}p$ is of the form $ \frac{q}{d}$ or $ \frac{1}{d}$, where $ d$ is some factor of $ a_n$.

Now, the $ \frac {1} {d}$ are only finitely many, so they are roots for a finite only number of chosen $ p$'s. For the other form, it means $ a_nq^n \plus{} a_{n\minus{}1}dq^{n\minus{}1} \plus{} \cdots \plus{} a_1d^{n\minus{}1}q \minus{} d^nq \equal{} 0$, hence $ Q_d(q) \equal{} a_nq^{n\minus{}1} \plus{} a_{n\minus{}1}dq^{n\minus{}2} \plus{} \cdots \plus{} d^{n\minus{}1}(a_1 \minus{} d) \equal{} 0$.
But the $ Q_d(x)$ are a finite set of non-constant polynomials, and so
they only have a finite number of roots, hence there are only a finite number of chosen $ p$'s that qualify.
Therefore, for any large enough $ p$, the equation $ Q(x)\equal{}p$ has no rational roots, contradiction.[/hide]


[b]1.
[/b]
$ tg(\pi tgx) = ctg(\pi ctgx)$

$ tgx = ?$

[b]2.[/b]
$ \alpha, \beta, \gamma \in (0, \frac {\pi}{2})$
$ cos{\alpha} = tg{\beta}$
$ cos{\beta} = tg{\gamma}$
$ cos{\gamma} = tg{\alpha}$

$ sin{\alpha} = ?$

For the second one, just use the identity:

$ \tan^2 \phi \equal{} \frac{1}{\cos^2 \phi}\minus{}1$

After following transformations
\[ \tan{\left( \pi \tan{x} \right)} \equal{} \cot {\left( \pi \cot{x} \right)}
\]
\[ \sin{\left( \pi \cot{x} \right)} \sin{\left( \pi \tan{x} \right)} \equal{}
\cos{\left( \pi \cot{x} \right)} \cos{\left( \pi \tan{x} \right)}
\]
\[ \cos{\left( \pi \cot{x} \right)} \cos{\left( \pi \tan{x} \right)} \minus{}
\sin{\left( \pi \cot{x} \right)} \sin{\left( \pi \tan{x} \right)} \equal{} 0
\]
we get
\[ \cos{\left( \pi \cot{x} \plus{} \pi \tan{x} \right)} \equal{} 0
\textrm{.}
\]
It follows
\[ \tan{x} \plus{} \frac{1}{\tan{x}} \equal{} \frac{1}{2} \plus{} k
\textrm{, \ \ }
k \in \mathbb{Z}
\textrm{,}
\]
which becomes
\[ \tan^{2}{x} \minus{} \left( \frac{1}{2} \plus{} k \right) \tan{x} \plus{} 1 \equal{} 0
\textrm{,}
\]
so we finally have
\[ \tan{x} \equal{} \frac{1}{4} \left( 2k \plus{} 1 \pm \sqrt{4k^{2} \plus{} 4k \minus{} 15}\right)
\textrm{, \ \ }
k \in \mathbb{Z} \setminus \left\lbrace \minus{}2, \minus{}1, 0, 1 \right\rbrace
\textrm{.}
\]

Let's define positive numbers
\[ A : \equal{} \cos^{2}{\alpha} \textrm{, \ \ } B : \equal{} \cos^{2}{\beta} \textrm{, \ \ } C : \equal{} \cos^{2}{\gamma} \textrm{.}
\]
Using identity $ \displaystyle \tan^{2}{x} \equal{} \frac {1}{\cos^{2}{x}} \minus{} 1$ we can get
\[ A \equal{} \frac {1}{B} \minus{} 1 \textrm{, \ \ } B \equal{} \frac {1}{C} \minus{} 1 \textrm{, \ \ } C \equal{} \frac {1}{A} \minus{} 1 \textrm{.}
\]
From $ \displaystyle C \equal{} \frac {1 \minus{} A}{A}$ follows $ \displaystyle B \equal{} \frac {2A \minus{} 1}{1 \minus{} A}$, so
\[ A \equal{} \frac {1 \minus{} A}{2A \minus{} 1} \minus{} 1
\]

\[ A^{2} \plus{} A \minus{} 1 \equal{} 0
\]

\[ A \equal{} \frac {\sqrt {5} \minus{} 1}{2}
\]
Finally we get
\[ \sin{\alpha} \equal{} \sqrt {1 \minus{} A} \equal{} \sqrt {\frac {6 \minus{} 2\sqrt {5}}{4}} \equal{} \frac {\sqrt {5} \minus{} 1}{2} \textrm{.}
\]


inequalities
trigonometry
circumcircle
inequalities proposed

By Stewart's theorem, or whatever (I used trigonometry), you can get that
$l_{a}=\frac{2\sqrt{s(s-a)bc}}{b+c}$, where $s$ is a semiperimetar of
triangle. So $l_{a}l_{b}l_{c}=\frac{8\cdot s \cdot P \cdot abc}{(a+b)(b+c)(c+a)}$. Now all we have to prove is that \[8\cdot s \cdot P> \frac{9abc\sqrt{P}}{2R}.\] $\frac{abc}{2R}=2P$ so this inequality becomes $4s>9\sqrt{P}$ or $\frac{16}{81}s^{2}>P$. As we all know $\frac{s^{2}}{3\sqrt{3}}\geq P$ ($\frac{s}3=\frac{(s-a)+(s-b)+(s-c)}{3}\geq \sqrt[3]{(s-a)(s-b)(s-c)}$ etc.), but $\frac{16}{81}>\frac1{3\sqrt{3}}$. So $\frac{16}{81}s^{2}>\frac{1}{3\sqrt{3}}s^{2}\geq P$. Q.E.D.

[quote="bodan"]Suppose $a,b,c$ are sides of a triangle, $R$ the radius of its circumscribed circle, $l_{a},l_{b},l_{c}$ the bisectors of the angles $\alpha,\beta,\gamma$. Prove the inequality \[l_{a}l_{b}l_{c}>\frac{9\sqrt{P}}{2}\frac{a^{2}b^{2}c^{2}}{R(a+b)(b+c)(c+a)}.\] [/quote]
i think if you use $A,B,C$ for ALPHA and BETA and GAMMA it will be better. :blush:


inequalities quadratics conics parabola algebra

[quote="ElChapin"]The roots of the equation

                           $ ax^{2}-4bx+4c$,          $ a>0$

are on the segment $ [2,3]$.

Prove that

i) $ a \leq b \leq c < a+b$
[/quote]
[hide="Incomplete :("]

Let $ x_{1},x_{2}$ be two (not necessarily distinct) roots then:
$ 4\leq x_{1}+x_{2}\leq 6$
So,
$ 4\leq \frac{4b}{a}\leq 6\implies 4a\leq 4b \leq 6a \implies a\leq b\leq 1.5a$

Also,
$ 4\leq x_{1}x_{2}\leq 9$
So,
$ 4\leq \frac{4c}{a}\leq 9 \implies 4a\leq 4c\leq 9a \implies a\leq c\leq 2.5a$
[/hide]

Can you believe this is Category D(13-14 years)???

I really don't know what do they eat.

 :huh:

Lets go ahead and scale the coefficients [we can see that this changes nothing] so that a=1.

$ 2\le 2(b-\sqrt{b^{2}-c})$; $ 3\ge 2(b+\sqrt{b^{2}-c})$.

$ \sqrt{b^{2}-c}\le b-1$ [here we get that $ b\ge 1$]
$ b^{2}-c\le b^{2}-2b+1$
$ 2b\le 1+c$. [here we get $ c>0$]

$ 3\ge\frac{2c}{b-\sqrt{b^{2}-c}}$ [since $ c$ is positive, so is the denominator]
$ 3b-2c\ge 3\sqrt{b^{2}-c}$ [here we get $ 3b\ge 2c$].
$ 9b^{2}-12bc+4c^{2}\ge 9b^{2}-9c$
$ -12b+4c\ge-9$

I think I have reduced this to linear inequalities...I will get back to this tomorrow.

[quote="ElChapin"]The roots of the equation

                           $ ax^{2}-4bx+4c$,          $ a>0$

are on the segment $ [2,3]$.

Prove that

i) $ a \leq b \leq c < a+b$

ii)  $ \frac{a}{a+c}+\frac{b}{b+a}>\frac{c}{c+b}$

 :)[/quote]

i) $ ax^{2}-4bx+4c=0\ (a>0)\ \Longleftrightarrow x^{2}-4px+4q=0$ where $ \frac{b}{a}=p,\ \frac{c}{a}=q$.

The condition for which quadratic equation $ f(x): =x^{2}-4px+4q=0$ have roots on $ [2,\ 3]$

$ \Longleftrightarrow f(2)\geq 0$ and $ f(3)\geq 0$ and the axis: $ 2\leq 2p\leq 3$ and $ D/4\geq 0$

$ \Longleftrightarrow q\geq 2p-1$ and $ q\geq 3p-\frac{9}{4}$ and $ 1\leq p\leq \frac{3}{2}$ and $ q\leq p^{2}$.

On the other hand the given inequality in question is $ 1\leq p\leq q<p+1$.

Let $ P=\{(p,\ q)| q\geq 2p-1$ and $ q\geq 3p-\frac{9}{4}$ and $ 1\leq p\leq \frac{3}{2}$ and $ q\leq p^{2}\}$,
$ Q=\{(p,\ q)| 1\leq p\leq q<p+1\}$, sketching these region on $ p-q$ plane gives $ P\subseteqq Q$ which completes the proof.

Especially two lines $ q=2p-1$ and $ q=3p-\frac{9}{4}$ are the tanget lines respectively at the points $ (1,\ 1),\ \left(\frac{3}{2},\ \frac{9}{4}\right)$ which are on the parabola $ q=p^{2}$.

ii) We are to prove that $ \frac{1}{1+q}+\frac{p}{p+1}>\frac{q}{p+q}$ with $ 1\leq p\leq q<p+1$.

$ \frac{1}{q+1}+\frac{p}{p+1}>\frac{1}{p+2}+\frac{p}{p+1}$

$ \Longleftrightarrow\frac{1}{q+1}+\frac{p}{p+1}>\frac{1}{p+2}+1-\frac{1}{p+1}$

$ \Longleftrightarrow\frac{1}{q+1}+\frac{p}{p+1}>1-\frac{1}{(p+1)(p+2)}\ \cdots [1]$.

Now $ p(p+1)(p+2)-(2p+1)=p^{3}+3p^{2}-1>0\ \because p\geq 1$, by $ q<p+1$, we have $ 0<p+q<2p+1<p(p+1)(p+2)$

$ \therefore \frac{1}{(p+1)(p+2)}<\frac{p}{p+q}\ \cdots [2]$. Thus from $ [1],\ [2]$ we obtain $ 1-\frac{1}{(p+1)(p+2)}>1-\frac{p}{p+q}$, yielding

$ \frac{1}{q+1}+\frac{p}{p+1}>1-\frac{p}{p+q}=\frac{q}{p+q}$, which completes the proof.


Find all functions $ f: \mathbb{R} \rightarrow  \mathbb{R}$ that satisfy \[ f(m\plus{}nf(m))\equal{}f(m)\plus{}mf(n)\] for all $ m$ and $ n$.

[quote="james digol"]Find all functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ that satisfy
\[ f(m \plus{} nf(m)) \equal{} f(m) \plus{} mf(n)\]
for all $ m$ and $ n$.[/quote]

$ m,n \in  \mathbb{R} ?$ Usually we assume that $ m,n \in N$...

[quote="james digol"]Find all functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ that satisfy
\[ f(m \plus{} nf(m)) \equal{} f(m) \plus{} mf(n)\]
for all $ m$ and $ n$.[/quote]

[i](already posted, I think)[/i]

Let $ P(x,y)$ be the assertion $ f(x \plus{} yf(x)) \equal{} f(x) \plus{} xf(y)$

$ f(x) \equal{} 0$ $ \forall x$ is a solution and let us consider from now that $ \exists a$ such that $ f(a)\ne 0$

1) $ f(x) \equal{} 0$ $ \iff$ $ x \equal{} 0$
=======================
$ P(1,0)$ $ \implies$ $ f(0) \equal{} 0$
Let $ u$ such that $ f(u) \equal{} 0$ : $ P(u,a)$ $ \implies$ $ uf(a) \equal{} 0$ and so $ u \equal{} 0$
Q.E.D.

2) $ f( \minus{} 1) \equal{} \minus{} 1$ and $ f(1) \equal{} 1$
==========================
$ P( \minus{} 1, \minus{} 1)$ $ \implies$ $ f( \minus{} 1 \minus{} f( \minus{} 1)) \equal{} 0$ and so, according to 1) above : $ \minus{} 1 \minus{} f( \minus{} 1) \equal{} 0$ and so $ f( \minus{} 1) \equal{} \minus{} 1$

Let then $ u \equal{} 1 \minus{} f(1)$ : $ P(1, \minus{} 1)$ $ \implies$ $ f(u) \equal{} \minus{} u$
Then $ P(u,1)$ $ \implies$ $ f(u \plus{} f(u)) \equal{} f(u) \plus{} uf(1)$ $ \implies$ $ 0 \equal{} u(f(1) \minus{} 1) \equal{} \minus{} u^2$ and so $ u \equal{} 0$ and $ f(1) \equal{} 1$
Q.E.D.

3) $ f(px) \equal{} pf(x)$ $ \forall p\in\mathbb Q$, $ \forall x$
=====================================================

$ P(1,x)$ $ \implies$ $ f(x \plus{} 1) \equal{} f(x) \plus{} 1$ and so $ f(x \plus{} n) \equal{} f(x) \plus{} n$ and $ f(n) \equal{} n$ $ \forall n\in\mathbb Z$, $ \forall x$
$ P(n,x)$ $ \implies$ $ n \plus{} f(nx) \equal{} n \plus{} nf(x)$ and so $ f(nx) \equal{} nf(x)$ $ \forall n\in\mathbb Z$, $ \forall x$
Then, using $ f(\frac pqx) \equal{} f(p\frac xq) \equal{} pf(\frac xq)$ and $ f(x) \equal{} f(q\frac xq) \equal{} qf(\frac xq)$, we get the result.
Q.E.D.

4) $ f(x) \equal{} x$
===========
Let $ x\ne 0$ :
$ P(x,\frac {y \minus{} x}{f(x)})$ $ \implies$ $ f(y) \equal{} f(x) \plus{} xf(\frac {y \minus{} x}{f(x)})$
$ P(2x,\frac {y \minus{} x}{f(2x)})$ $ \implies$ $ f(x \plus{} y) \equal{} f(2x) \plus{} 2xf(\frac {y \minus{} x}{f(2x)})$ $ \equal{} 2f(x) \plus{} xf(\frac {y \minus{} x}{f(x)})$

And so, subtracting these lines : $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$, obviously still true for $ x \equal{} 0$

Then $ P(x,y)$ may be written $ f(yf(x)) \equal{} xf(y)$ and so $ f(f(x)) \equal{} x$ and so $ f(xy) \equal{} f(x)f(y)$

And we got the very classical system :
$ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$
$ f(xy) \equal{} f(x)f(y)$

Whose unique non constant solution is $ f(x) \equal{} x$ which, indeed, is a solution of original equation.

5) Synthesis of solutions :
=========================
$ f(x) \equal{} 0$ $ \forall x$
$ f(x) \equal{} x$ $ \forall x$

Job well done pco... :lol:
The solution that you have is way too simple and beautiful than mine..so I don't have to post it coz it's really long..i'm slow with latex..but i'll try posting it if anybody cares..Anyway, the problem was originally proposed by Wu Wei Chao (the lecturer) and I'm not sure if it's already posted somewhere here in ML..but sure there are a lot of similar problems like this in which the solutions are the zero and identity functions...

Very nice problem! I liked the difficulties presented in this problem.

I find that part (4) of pco's solution is quite magical... maybe the following can be motivated more easily?

[hide="Solution"]Let $(x,y)$ denote the assertion that $f(x+yf(x))=f(x)+xf(y)$

1) $f(a)=0 \Leftrightarrow a=0$   (apart from when $f(x)=0 \forall x$)

$(x,0): xf(0)=0$
Thus $f(0)=0$.

If $f(a)=0$ for some $a$,
$(x,a): f(x+af(x))=f(x)$
Thus either $f$ is constant (in which case we see that $f(x)=0 \forall x$ is indeed a solution, or $a=0$.

2) $f$ is injective.

Suppose $f(x)=f(x+k)$ (for some $x\neq 0$, since $x=0 \Rightarrow k=0$.
$(x,\frac{k}{f(x)}): f(x+k) = f(x) + \frac{k}{f(x)}x$
$\Rightarrow \frac{kx}{f(x)} = 0$
Since $x,f(x)\neq 0$, thus $k=0$ and hence $f$ is injective.

3) $f(1)=1$

$(1,y): f(1+yf(1))=f(1)+f(y)$
If $f(1)\neq 1$, then set $y=\frac{1}{1-f(1)}$ which is equivalent to $1+yf(1)=y$.
$\Rightarrow f(1)=0$, which is clearly false due to injectivity.

4) $f(x+1)=f(x)+1, f(2x)=2f(x)\forall x$

$(1,y): f(y+1)=f(y)+1$
$(1,1): f(2)=2$
$(2,y): f(2y+2) = 2+2f(y)$
$\Rightarrow 2+2f(y) = f(2y+2) = f(2y+1) + 1 = f(2y) + 2$
$\Rightarrow f(2y)=2f(y)$

5) $f(x)=x\forall x$

$(x,\frac{x}{f(x)}): f(2x)=f(x)+xf(\frac{x}{f(x)}$
$\Rightarrow f(\frac{x}{f(x)})=\frac{f(x)}{x}$

$(y,y): f(y+yf(y))= y+yf(y)$
If $yf(y)=1$, then $f(y)=y$.

Combining the two, we see that $f(x)^2 = x^2$.

If for some $b\neq 0$, $f(b)=-b$, then
$(b,b): b^2-b = 0$
$\Rightarrow b=0 \text{ or } 1$, either of which will contradict with $f(b)=-b$.

Thus either $f(x)=x\forall x$ hhor $f(x)=0\forall x$[/hide]

4) $ f(x) \equal{} x$
===========
Let $ x\ne 0$ :
$ P(x,\frac {y \minus{} x}{f(x)})$ $ \implies$ $ f(y) \equal{} f(x) \plus{} xf(\frac {y \minus{} x}{f(x)})$
$ P(2x,\frac {y \minus{} x}{f(2x)})$ $ \implies$ $ f(x \plus{} y) \equal{} f(2x) \plus{} 2xf(\frac {y \minus{} x}{f(2x)})$ $ \equal{} 2f(x) \plus{} xf(\frac {y \minus{} x}{f(x)})$
@Pco , how did you get up this magical substituion ?

Is this comparable to this year's IMO #5 in beauty and difficulty?

IMO #5 was neither beautiful nor difficult

Yet it only had 30 perfects  :P ! Would this problem (not the IMO one) be harder then?


Given positive integers $m$ and $n$ so there is a chessboard with $mn$ $1 \times 1$ grids. Colour the grids into red and blue (Grids that have a common side are not the same colour and the grid in the left corner at the bottom is red). Now the diagnol that goes from the left corner at the bottom to the top right corner is coloured into red and blue segments (Every segment has the same colour with the grid that contains it). Find the sum of the length of all the red segments.

Look at Balkan Math Olympiad, 2003, Problem 4.
There it is the same problem but in softer variant. It says
that |red|-|blue|= $\frac{\sqrt{m^2 +n^2}} {mn}$.
But R + B= $\sqrt{m^2 +n^2}$.
From here R= $\frac{mn+1}{2mn} \sqrt{m^2 +n^2}$


algebra
polynomial
calculus
derivative
superior algebra
superior algebra unsolved

in which finite field if we have polynomial with coefficient in this field its derivative has roots alternatively between the root of polynomial.

i meant that when there exists between any two roots of polynomial ,at least a root of its derivative

What I don't know is what you mean by "between". What is the order you're using? Am I missing something obvious? :?

That's exactly what [b]grobber[/b] said. ;)

And, logically, the "between" should be an "among", but, we have to wait for [b]sam-n[/b] to be sure.

excuse me i have explain it with intuition of real i specifically meant that in which field if any polynomial  splits to linear factor so does it derivative.


If the ratio of $ 2x \minus{} y$ to $ x \plus{} y$ is $ \frac23$, what is the ratio of $ x$ to $ y$?

$ \textbf{(A)}\ \frac15 \qquad \textbf{(B)}\ \frac45 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ \frac65 \qquad \textbf{(E)}\ \frac54$

$ \frac{2x\minus{}y}{x\plus{}y}\equal{}\frac{2}{3} \longrightarrow 6x\minus{}3y\equal{}2x\plus{}2y \longrightarrow 4x\equal{}5y \longrightarrow \frac{x}{y}\equal{}\boxed{\frac{5}{4}}$

$ \frac{2x\minus{}y}{x\plus{}y}\equal{}\frac{2}{3}$ cross multiply

$ 2x\plus{}2y\equal{}6x\minus{}3y$

$ 4x\equal{}5y$ divide 4y

$ \frac{x}{y}\equal{}\frac{5}{4}$

$ \fbox{E}$

If $ \frac{x}{y}\equal{}t$ then: $ \frac{2}{3}\equal{}\frac{2x\minus{}y}{x\plus{}y}\equal{}\frac{2\frac{x}{y}\minus{}1}{\frac{x}{y}\plus{}1}\equal{}\frac{2t\minus{}1}{t\plus{}1} \Rightarrow 6t\minus{}3\equal{}2t\plus{}2 \Rightarrow 4t\equal{}5 \Rightarrow t\equal{}\frac{5}{4} \Rightarrow \boxed E$


Three people, Alan, Beth and Cindy, pooled their money to purchase lottery tickets. Alan invested $\$25.00$, Beth invested $\$20.00$ and Cindy invested $\$35.00$. Their winnings totaled 6.4 million dollars. How many dollars was Beth's share of the winnings if the amount received was proportional to the amount invested?

Edit: I can't read.... Beth has 20 dollars. (sorry ckck)

[quote="Quevvy"]Three people, Alan, Beth and Cindy, pooled their money to purchase lottery tickets. Alan invested $\$25.00$, Beth invested $\$30.00$ and Cindy invested $\$35.00$. Their winnings totaled 6.4 million dollars. How many dollars was Beth's share of the winnings if the amount received was proportional to the amount invested?[/quote]
[hide]So Beth invested 1/3 of the money. So 1.3 of 6.4 million is $\$$2133333.34.[/hide]

[hide]So Beth invested 1/4 of the money.  1/4 x 6400000 = 1600000.[/hide]


Since MathLinks was re-arrenged, I starting seeing that my account started to decrese some post... as a result, I started to write in a paper the number of posts, to see what happened... but, I passed from having 1484 to have 1481!!! what's going on?

Post count is not important. Do not post to increase your post count only, as you will achieve the opposite effect ;)

I probably deleted some vacuous posts from the Round Table.


[quote="N.T.TUAN"]For what positive values of $ a$ is $ \log_{a}b<b\;\forall b\in (0,\infty)$?[/quote]
$ \frac{ln(b)}{ln(a)}<b.$
$ f(b)=\frac{ln(b)}{b}<ln(a)$. Maximum value of $ f(b)$ is $ f(e)=\frac{1}{e}$, so $ a>e^{\frac{1}{e}}.$


I'm attending UPenn next year and planning on majoring in pure math. Can anyone here speak to the math culture at Penn?

I'm just a rising sophomore, so I don't know how helpful I can be. The culture here is definitely not cutthroat or superstar based, like it would be at a Harvard/Princeton/MIT. There are a few professors who are prominent in their fields. The building is a shithole, though -- looks like it was built in the 60s and that it hasn't been redone since then. Many of the math majors that I met were very arrogant about their abilities (I remember one kid in my 202 class announce to everyone before class that he "was getting an A") and antisocial.

If you have any specific questions, though (since my answer here sucked), PM me or fire away here.

Thanks for responding. I guess I'm wondering if the math culture is at least positive; do people expect to rise high in their areas of study, are professors interested in helping guide aspiring mathematicians, do mathematicians feel like their area of study is valued by the school or is math somewhat of an afterthought to other areas of liberal arts?

Penn math is generally not a feeder to Harvard/Pton/MIT/Chicago/Berkeley, etc...although I suppose that many math majors do want to get PhDs due to the academic nature of the subject (I plan on going into i-banking after I finish). The math program in general seems to be an afterthought -- shitty building, few notable professors -- although the dean of the College is a math prof.

However, despite Penn's ignorance of math, the professors are all incredibly helpful and interested (especially Herman Gluck, who is also one of the more notable faculty).

I'd say that overall the culture is very nice, perhaps because math is such a downtrodden major at Penn. The people in the program really feel like they're a part of something.


function
limit
integration
real analysis
real analysis unsolved

A sequence $(f_{i})$ of measurable functions is said to converge towards the measurable function $f$ in sence of measure if every $\varepsilon>0$:
\[\lim_{i\to \infty}\mu(\{x\in X: |f_{i}(x)-f(x)|>\epsilon\})=0.\]
Prove that if $(f_{i})$ converges in $L^{p}$, it converges in sense of measure. Give a counterexample showing that the converse does not hold.

$\int_{X}|f_{i}(x)-f(x)|^{p}d\mu \ge \int_{\{x: |f_{i}(x)-f(x)|>\varepsilon\}}|f_{i}(x)-f(x)|^{p}d\mu \ge \int_{\{x: |f_{i}(x)-f(x)|>\varepsilon\}}\varepsilon^{p}d\mu = \varepsilon^{p}\mu(\{x: |f_{i}(x)-f(x)|>\varepsilon\})$

in other hand $f_{j}(x)=\sqrt[p]{j}, x\in [0,1/j]$ and $f_{j}(x)=0, x\nin [0,1/j]$ is a counterexample for the second part ($f(x)=0$)


Sorry$ S(n)$ is the sum of digits :oops:

Where did you find this problem? Because proving that the sum of digits of a number is larger than sth might be very hard. Didnt try to get the second part of this problem, tho.

[quote="mufc"]Let $ n \equal{} \left[\frac {5^{5^5}}{4^{4^4}}\right]$
Prove $ 2013 < S(n) < 2059$[/quote]

I think this result is false :

$ 5^5\log(5)\minus{}4^4\log(4)\equal{}2030,1539057701\cdots$. So $ n$ has $ 2031$ digits.

This means that the mean value for digits would be between $ \frac{2013}{2031}\equal{}0.99113$ and $ \frac{2059}{2031}\equal{}1.01378$ !!

More :
Since $ 10^{0.1539057701}\equal{}1.42529830971024\cdots$, the sum of the $ 15$ first digits is $ 57$
So, in order to have $ S(n) < 2059$, the $ 2016$ next digits must have a sum $ <2002$, which means that the mean value of digits less than 1 !!

And this seems very unlikely.

Sorry Sorry!! :blush: $ S(n)$ is the number of digits


[hide="hint"]Draw a picture  :D [/hide]

[hide="bigger hint"]What is the diameter of the circle if the square is inscribed in it?[/hide]

[quote="Interval"]An adjustable water sprinkler that sprays water in a circular pattern is placed at the center of a square field whose area is 1250 square feet.  What is the shortest radius setting that can be used if the field is to be completely enclosed in the circle?[/quote]
[hide]Since it's a square field, each side is $\sqrt{1250}=25\sqrt{2}$ Then, the radius is half of the diagonal you need to figue that out $\frac{25\sqrt{2}*\sqrt{2}}{2}=\frac{50}{2}=25$ so then the radius of the circle is 25 feet.[/hide]

[quote="Interval"]So, you're saying that 25(sqrt{2}) is the diagonal?

I would have to take 1/2 of that, right?[/quote]
No, the sides of the field is $25\sqrt{2}$ and then the diagonal is always $x\sqrt{2}$ so then that would be $25\sqrt{2}*\sqrt{2}=25*2=50$ so the diagonal is 50 so that's the diameter of the circle and then 25 is the radius.

I got it now.

Diagonal = 50 after solving for x.

So, x = diagonal = 50.

Then, 50 is DIVIDED by 1/2 to obtain 25 feet = shortest radius.

Good job!


AMC
USA(J)MO
USAMO
geometry
angle bisector
geometry proposed

http://www.artofproblemsolving.com/Wiki/index.php/1991_USAMO_Problems/Problem_1
I understand the solution but have one question:

It says from the angle bisector theorem we have BD = ac/(b+c), but I
obtained a different result.
Using the angle bisector theorem we should have BD = (ac - bc²/2)/b
= (a²c-bc²)/ab = bc/a; manipulate both sides of (a²c-bc²)/ab = bc/a
gives a²-bc = b².

My question is how to transform (a²c-bc²)/ab into ac/(b+c), and I
don't see how to obtain ac/(b+c) = BD using the angle bisector
theorem.

EDIT;

Thank you very much.  Is it possible to algebraically transform (a²c-bc²)/ab into ac/(b+c)?  They should equal since they are both equivalent to BD = bc/a.

we have a triangle ABC and D the feet of the angle bisector from A. Call $ \overline{BD}\equal{}x$ and $ \overline{DC}\equal{}y$. Then for the angle bisector theorem we have $ \frac{x}{y}\equal{}\frac{c}{b}$ and we have $ x\plus{}y\equal{}a$, so $ y \plus{} y \frac{c}{b} \equal{} a \ \Longleftrightarrow \ y \equal{} \frac{ab}{b\plus{}c}$ and $ x\equal{}\frac{ca}{b\plus{}c}$


Two tour guides are leading six tourists. The guides decide to split up. Each tourist must choose one of the guides, but with the stipulation that each guide must take at least one tourist. How many different groupings of guides and tourists are possible?

$ \textbf{(A)}\ 56 \qquad \textbf{(B)}\ 58 \qquad \textbf{(C)}\ 60 \qquad \textbf{(D)}\ 62 \qquad \textbf{(E)}\ 64$

[hide]
There are $2^{6}=64$ total ways to split up the tourists.  However, this includes the two arrangements inwhich all 6 go with one guide.  So the answer is $64-2=62$.  [b]D.[/b]
[/hide]

It's Pascal's Triangle.  Since there are two tour guides, we are finding the distinct ways that they can be split up, or the combinations.  The possibilities are simply the combinations in the 6th row of Pascal's triangle.  The sum of the nth row is 2^n.  However, two of the combinations (both of value 1) include the ways for no tourists to go with a tour guide, which is not possible.  So we subtract those from 64.

Um, more simply, there are two choices for each person as to what guide. So $2^{6}$

[hide="Easy Solution"]Let the tour guides be $a$ and $b.$ Out of the six tourists, we know that $1, 2, 3, 4,$ or $5$ must go with $a.$ Note that $0$ and $6$ aren't included, because this would imply that no tourists would go with $a$ or $b.$ Thus, the answer is simply: $\binom{6}{1}+\binom{6}{2}+\cdots \binom{6}{5}= 2^{6}-2 = 62.$ So the answer is $\boxed{62}$[/hide]

[hide="Solution"]  Each of the 6 tourists can choose between the two guides, for a total of $2^{6}=64$ arrangements.  However, if all the tourists choose the same guide, then one guide will have nobody to guide.  This can happen in two ways, so our answer is $64-2=62$.[/hide]

This problem screwed me up on the test, but i eventually got it.


logarithms
trigonometry
AMC
AIME
algebra
log identities

Given that $\log_{10} \sin x + \log_{10} \cos x = -1$ and that $\log_{10} (\sin x + \cos x) = \frac 12 ( \log_{10} n -1)$, find $n$.

[hide="Nice problem"]
Since $\log a+\log b = \log (ab)$, we can simplify the first equation to get \[ \log (\sin x\cos x)=-1 \Longleftrightarrow \sin x\cos x = \frac{1}{10}. \] Since $(\sin x+\cos x)^2=\sin^2x + \cos^2 x + 2\sin x \cos x=1+\frac{2}{10}$, we find that \[ \sin x + \cos x = \sqrt{\frac{12}{10}} \Longleftrightarrow \log (\sin x + \cos x) = \frac{1}{2} (\log 12 - \log 10), \] so $n=\boxed {12}$.
[/hide]

[hide="Answer"]By the first equation, we have $\sin x\cos x=\frac{1}{10}$. We have $(\sin x+\cos x)^2=\sin ^2x+\cos ^2x+2\sin x\cos x=1+\frac{2}{10}=\frac{12}{10}$, so $\log_{10}(\sin x+\cos x)=\log_{10}\left(\frac{12}{10}\right)^\frac{1}{2}=\frac{1}{2}\left(\log_{10}\frac{12}{10}\right)$. Simplifying, we have $\log_{10}12-\log_{10}10=\log_{10}n-\log_{10}10\Rightarrow n=\boxed{12}$.[/hide]

[hide="Solution"]
Start with the first equation. Using log identities, $\log_{10} \sin x \cos x = -1 \Rightarrow \sin x \cos x = \frac{1}{10}$.
Now look at the left side of the second equation. We see $\log_{10} (\sin x + \cos x)$. Since we know that $\sin x \cos x = \frac{1}{10}$, we can solve for $\sin x + \cos x$ by squaring the quantity.

$(\sin x + \cos x)^{2} = \sin^{2} x + \cos^{2}x + 2(\sin x \cos x)$

Substituting, we get $(\sin x + \cos x)^{2}= \frac{12}{10} \Rightarrow \sin x + \cos x = \left(\frac{12}{10}\right)^{\frac{1}{2}}$

Going back to the original second equation, we now can substitute the previous statement into the left side. So:
$\log_{10}{\left(\frac{12}{10}\right)^{\frac{1}{2}}} = \log_{10} (\sin x + \cos x) = \frac{1}{2} ( \log_{10} n -1)$

Since $\log_c{a^{b}}=b\log_c{a}$,  $\frac{1}{2}\left(\log{\frac{12}{10}\right)=\frac{1}{2}\left(\log_{10}n -1\right)}$

Cleaning things up, we get $\log{12} - 1 = \log{n} - 1 \Rightarrow \boxed{n=12}$ [/hide]
Sorry if this is a bit rough it's been a while since I've used TeX... also been a while since I've had to explain myself on an AIME problem so I guess my solution was miles longer than what I had thought out.

EDIT: I just figured out how to make big parentheses!

If you had just multiplied both sides by 2 to begin with, you wouldn't have to bother with the square roots because $2\log_{10}{(\sin{x}+\cos{x})}=\log_{10}{(\sin{x}+\cos{x})^2}$.

Yeah, but the square roots don't make this problem that complex. If you take out all the words from my solution, it comes out as like a four line solution.. or something like that.

yes, excellent problems..coincidentally, all the AIME problems that I've done in the last two days have been appearing on this forum...including this one.


algebra polynomial rotation trigonometry calculus geometry LaTeX

Hello,  I have recently become interested in complex numbers and their applications, so I woudl like to run a complex marothon.  You know how it goes, so I'll begin with a problem.

[b]1.[/b] A number of points are given on a unit circle so that the product of the distances from any point on the circle to the given points does not exceed 2.  Prove that the points are the vertices of a regular polygon.

Masoud Zargar

It appears that no one has answered for long.  Do you have any interesting problems?

[quote="boxedexe"]
[b]1.[/b] A number of points are given on a unit circle so that the product of the distances from any point on the circle to the given points does not exceed 2.  Prove that the points are the vertices of a regular polygon.

[/quote]

[hide="somewhat of a solution"] say we have our point at the origin where we take the product of the distances connecting to that point, consider the polynomial $((z+1)e^{i\theta})^n=1$, this has roots that are evenly spaced around a circle with a radius of 1, centered at (-1,0), and rotated $\theta$ radians clockwise from the normal roots of unity, if (0,0) is the point from where we take our product of lengths from, consider our constant term, it is $\prod z_i=a_0$, take absolute values, and distribute, $\prod |z_i|=a_0$ or our product of lengths is that, expanding, the product  is $\left|\frac{e^{i*n*\theta}-1}{e^{i*n*\theta}}\right|=|e^{in\theta}-1|$ which clearly has to be less than 2 since it is just the distance from the origin to any point on a circle at (1,0) with a radius of 1, (the product of 2 is attained when $\theta=\frac{\pi}{n}$

say that the above points (as a regular polygon) are rotated on $(x-1)^2+y^2=1$ so that the product distance from (0,0) was at a maximum (of 2 as shown above), say we move any one of those points ever so slightly so it is farther away, this makes the product above 2, thus the points cannot move, and must be in a regular polygon

any comments, perhaps not fully rigirous, but i think the principle i used in the second paragraph is rigirous, it is similar to the principle used to prove the am-gm in aops vol 2, show that nothing can occur but the optimal arrangment[/hide]

[quote="seamusoboyle"]Heres a pretty good one. A person starts at the orgin of the complex plane. He goes one unit along the positive x-axis, turns by an angle of $\theta$, walks half a unit, turns another $\theta$, walks a quarter of a unit..... and so on.

1 for what $\theta$ does he end up away from the real line as much as possible?

2 for what $\theta$ does he end up farthest from the origin?[/quote]

Wow, that seems interesting. But does it necessarily have to be in the complex plane?

Well, since its a complex numbers marathon!

But no, as some of you may know, in the complex plane the shortest distance between two points might not be a straight line. In this question, however, this is not an issue, it works out ok. He walking in the complex plane because it is a clue about how to solve this problem!

[quote="seamusoboyle"]Heres a pretty good one. A person starts at the orgin of the complex plane. He goes one unit along the positive x-axis, turns by an angle of $\theta$, walks half a unit, turns another $\theta$, walks a quarter of a unit..... and so on.

1 for what $\theta$ does he end up away from the real line as much as possible?

2 for what $\theta$ does he end up farthest from the origin?[/quote]

he moves:

$1*e^{i\theta*0}+\frac{1}{2}e^{i\theta*1}+...$
$=\sum_{k=0}^{\infty}\frac{e^{i\theta*k}}{2^k}=\frac{1}{1-\frac{e^{i\theta}}{2}}$
$=\frac{2-\cos\theta+i\sin\theta }{4-4\cos\theta+\cos^2\theta+\sin^2\theta}=$

1. obviously we want to maximize the imaginary part, $\frac{\sin\theta}{5-4\cos\theta}$ i am not sure how one could do this without calculus, but now i will leave this for someone to finish

2. the maximum distance from the origin is also the same, we want maximum absolute value, which is

$\frac{\sqrt{ 5-4\cos\theta}}{5-4\cos\theta}=\frac{1}{\sqrt{ 5-4\cos\theta}}$

where the maximum occurs when the bottom is at the smallest, cos at its largest, $\cos\theta=1 \implies \theta=\frac{\pi}{2}+\pi k \qquad \forall k\in\mathbb{Z}$

Kinda a bump, but I felt like doing the calc and got the angle to be $\arccos \frac15$.

Someone post a new problem...

$\displaystyle\frac{1}{5}$ is an unusual angle  :)  Do you mean $\cos^{-1} \frac{1}{5}$?

Here's a strange problem I've given in modified form to give it more of a complex number flavor, although I'm not sure whether the original problem could've been solved by pure geometry or not.

The $7^{th}$ roots of unity are plotted in the complex plane, along with $P = 1 + i$.  The distances (in the complex plane) from each of the roots to $P$ are given by $a_1, a_2, ... a_7$.  Find $(a_1 a_2 ... a_7)^2$.

[quote="t0rajir0u"]$\displaystyle\frac{1}{5}$ is an unusual angle  :)  Do you mean $\cos^{-1} \frac{1}{5}$?[/quote]
Yeah that's actually what I had I just did the latex wrong... thought \acos was a function...

And do you mean $|a_1a_2\cdots a_7|^2$?

[hide]
Let the roots be $a,b,\cdots,g$.  They are the solutions to $z^7 -1 = 0$.  Since they are the roots, we can say $z^7 - 1 = (z-a)(z-b) \cdots (z-g)$.  The distance $a_1 = |P-a|$ and $a_2 = |P-b|$, etc.  Then $a_1a_2 \cdots a_7 = |(P-a)(P-b)\cdots (P-g)| = |P^7 - 1|$.  Since $P= \sqrt 2 e^{\frac{\pi i}{4}}$, the expression is $|8\sqrt 2 (\frac{\sqrt 2 }{2} - i \frac{ \sqrt 2 }2) - 1| = |7 - 8i| = \sqrt{113}$.  Then squaring we get the solution, $\boxed{113}$
[/hide]


[quote="Omid Hatami"]Suppose that $\gamma: S^1\rightarrow \mathbb R^2$ is a smooth curve, which is contained in a circle of radius $r$. Prove that there exist a point which curvature at that point is at least $\frac 1r$[/quote]Look at the [i]smallest[/i] such circle. This circle must have at least one point in common with $\gamma(S^1)$. That's the point to look at.


The AAPT has asked us not to discuss the USAPhO until March 27.  This includes problems as well as comments like "It was easy/hard" or "I think I did well/badly."

Thank you!

Ah, thanks for the heads-up!  Ditto.

On an unrelated note, does anyone have official solutions/scores for the Prelim?

I've only seen my Semis listing...but yeah...it's nice to know how I did...

Can I say stuff like "my school is going to take it Monday, I think we'll have 2-3 finalists this year?"

???

well, we can talk now.

i thought the exam was much, much easier than last year's.  i think i made the team again unless i managed to misunderstand like every question or if everyone else did really well, too.

[quote="durt"][quote="PTynan89"]so easy, guys.[/quote]

Yeah, I agree that it was easier. Though that might be because I know more stuff. Anyway, I didn't ace it, so I won't get my hopes up.[/quote]

Way to misquote me...

But yeah, I think I got about twice the score I did last year (but still made some retarded mistakes on A1).

yeah, i made a dumb error on A1 (a), but i think i will get at least half credit.  i thought A2, A3, A4 were pretty easy.

B1 was alright.  one small mistake in (b), but that's all i see in retrospect.  i kind of ran out of time when i was doing part (d), but i think i did enough of it to get at least 30-35 points overall.  B2 should have been a 50/50, but i made a numerical error in the biot-savart calculation that made like most of the calculations in part 2 of that question either impossibly ugly to crank out or completely trivial.  if i'm lucky, i'll get a 25/50 overall on it.

even though i figure i scored around a 140 +/- 10 points (twice last year's score haha), i am worried because everyone else seems to be saying that they beasted the exam, so this might be one of those years where the cutoff is very high.  i think shurtz said that in one year the score for seniors/returning finalists was like a 156...

[quote="iostream.h"]
even though i figure i scored around a 140 +/- 10 points (twice last year's score haha), i am worried because everyone else seems to be saying that they beasted the exam, so this might be one of those years where the cutoff is very high.  i think shurtz said that in one year the score for seniors/returning finalists was like a 156...[/quote]

Sorry to bring back really old thread, but is there a seperate cutoff for seniors/returning finalists?

I think that the cut-off for seniors is higher than for juniors. Don't know if there is a higher cut-off for returning camp members, but it doesn't seem like there would be.


Anybody here taking the final exam?

I am but I still have maybe 18 chapters left to do in a little over a month  :( .

A final exam for school? Or the AP exam?

My class is on chapter..28. Protists! We did 50-55 over the summer, though--so that's what--only 22 chapters to go! Totally feasible. Plenty of time. (not)

there--------u can rely on guesses i suppose

[quote="7h3.D3m0n.117"][quote="Harmonikdiskorde"]A final exam for school? Or the AP exam?[/quote]

The College Board AP Exam  :) .

We still have to finish plant structures, and start on organ systems.[/quote]

Alas--I think I get a nice extra few days to study, since my Music Theory exam forces me to the late testing date, May 23. Which...is also the day of senior prom. Hmm.

Wow, I can't believe how time flies. I only have a week left of studying left before the exam; we just finished our biology book today and we're going to wrap up the last two labs sometime...later.

Any tips on what I should study on? I'm already concentrating on cell respiration/photosynthesis (wrapping that up now), mitosis/meiosis, and I plan on starting a quick review of organismal (especially the protostomes vs deuterostomes) later.

So how did everyone do? Keep in mind we're "not allowed to discuss the multiple choice nor the written response sections with other people." I guess you can just say what you think you will get, etc.

Apparently everything I concentrated on in studying didn't appear in the exam  :o .

The essays weren't bad so I'm expecting at least a 4, if not a 5.


Tom, Dick and Harry started out on a 100-mile journey.  Tom and Harry went by automobile at the rate of 25 mph, while Dick walked at the rate of 5 mph.  After a certain distance, Harry got off and walked on at 5 mph, while Tom went back for Dick and got him to the destination at the same time that Harry arrived.  The number of hours required for the trip was:
\[ \text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E) none of these} \]

I don't want to explain an incorrect answer, but after a three variable equation, i get D, or 8 hours.

well nor do I but I am getting $\text{(B)}$.
Correct?

Let's have the administrator clear this up.

The choices look like natural numbers. I doubt there are decimals.

[quote="shadysaysurspammed"]Is the answer coming perfectly(no decimal)?[/quote]
the answer is perfect

Im sorry but Im too lazy too latex it maybe some other time. But I can ask you how did you get 8?

[quote="shadysaysurspammed"]The part i think i'm gettin wrong is when tom goes back for dick...do u use relative speed or somethin in that?[/quote]
I thought that was obvious. use relative speed NILESH and then you might get the same answer as me.

Hint for my solution;

let x, y, and z be the time intervals, and three equation for each person's motion.

[quote="shadysaysurspammed"]I consulted eminem on this problem and i realised where i went wrong.one stupid thing i missed out on.eminem got it.the ans is 6.
riddler,did u consider that dick moves some distance forward when tom comes back from him?if u dont ,then ur gettin 8 :lol:[/quote]

I'm the one whose getting 8, I checking my equations.

i dunno what u ppl did but i got 8. i got tom and harry going 75 miles.

We have A, B, and C for short. A is Tom, B is Harry, C is Dick.

Tom and Harry traveled $d$ miles. Dick traveled $\frac{d}{5}$ miles. At this time, Harry is dropped off $100-d$ miles from the end.

Now, Dick and Tom are both moving, Tom going back, but both moving simultaneously. Between them, $\frac{4}{5}d$ of mileage. So, Tom travels $\frac{2}{3}d$ before he meets Dick since Dick is walking and Tom is on the Moped (Dick travels one-fifth of Tom). So, after that, he has the rest of $\frac{2}{3}d$ that he came back on the Moped and the remaining $100-d$ miles to go. You divide that huge sum by 25 to get the time traveled by Tom on the Moped (Since he is always on the Moped).

Harry has merely traversed $\frac{d}{25} + \frac{100-d}{5}$ in hours of time.

Tom has traveled the time of $\frac{d + \frac{2}{3}d + \frac{2}{3}d + (100 - d)}{25}$ in hours.

If you simplify, $\frac{16}{3}d = 400$.

D = 75.

Thus, 8 is the answer.

Same method as whitehorse,guess calculations went wrong.ans is 8 :blush:

I set up a long equation that would take forever to explain, but I got 8. If anyone is dying to see my tedious solution, tell me  ;)


inequalities
Euler
inequalities unsolved

[quote="L_Euler"]With $ a,b,c (a \ne b,b \ne c,c \ne a)$, prove that: $ \frac {{a^3 \minus{} b^3 }}{{(a \minus{} b)^3 }} \plus{} \frac {{b^3 \minus{} c^3 }}{{(b \minus{} c)^3 }} \plus{} \frac {{c^3 \minus{} a^3 }}{{(c \minus{} a)^3 }} \ge \frac {9}{4}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=37317


Given $ F$ is the Fermat point of $ \triangle ABC$, is it possible to derive a formula for the distance $ AF \plus{} BF \plus{} CF$?

what is a fermat point?  :blush:  :blush:

[quote="pythag011"]If I remember correctly, it is the point P such that m<APB = m<APC =m<BPC = 120 degrees.

Or was that Torricelli Point?[/quote]
The Torricelli Point is the same thing.  :D

hmmm. . . is this in any way related to Olymon #569? I think you can do this problem without finding an explicit distance.

The Fermat-Torricelli Point $ F$ is a relatively easy thing to look up for yourself... it coincides with the point $ P$ that minimizes the quantity $ AP\plus{}BP\plus{}CP$. $ AF\plus{}BF\plus{}CF$ happens to be equal to $ CC'$ where $ C'$ is the point such that $ ABC'$ is an exterior equilateral triangle of triangle $ ABC$. A proof can be found at
cut-the-knot
[url]http://www.cut-the-knot.org/Generalization/fermat_point.shtml[/url]

Or, when $ \max \{A,B,C \} < 120^o$, let $ k\equal{}\frac12 (a^2\plus{}b^2\plus{}c^2)\plus{}2\sqrt3 S_{\triangle}$, then

$ PA\equal{}\frac{b^2\plus{}c^2\minus{}2a^2\plus{}k^2}{3k}$


geometry
circumcircle
cyclic quadrilateral
geometry solved

We are given a cyclic quadrilateral $ABCD$. Let $E$ and $F$ be points of intersection of lines respectively $AB,CD$ and $AD,BC$. Let $E'$ be the inverse of $E$ w.r.t. circumcircle of $ABCD$. Prove, that $E'F\perp E'E$.

[quote="dondigo"]We are given a cyclic quadrilateral $ ABCD$. Let $ E$ and $ F$ be points of intersection of lines respectively $ AB,CD$ and $ AD,BC$. Let $ E'$ be the inverse of $ E$ w.r.t. circumcircle of $ ABCD$. Prove, that $ E'F\perp E'E$.[/quote]

First, I rewrite your problem, renaming A, B, C, D, E, F, E' into A, C, B, D, P, R, P', respectively:

[color=blue][b]Problem.[/b] Let A, B, C, D be four points on a circle k. Let $ P \equal{} AC\cap BD$ and $ R \equal{} BC\cap DA$. Let P' be the image of P under the inversion with respect to the circle k. Prove that $ P^{\prime}R\perp P^{\prime}P$.[/color]

[i]Solution.[/i] We adopt the terminology of Theorem 9 from my note [i][url=http://www.cip.ifi.lmu.de/~grinberg/Butterfly.zip]On cyclic quadrilaterals and the butterfly theorem[/url][/i]. According to Theorem 9 [b]f)[/b], the point M is the image of P under the inversion with respect to the circle k. Thus, M = P'. Hence, $ P^{\prime}R\perp P^{\prime}P$ is equivalent to $ MR\perp MP$. But this is clearly true since the point M is the orthogonal projection of the point P on the line QR, as Theorem 9 [b]g)[/b] states. QED.

  darij


Does anyone sing?

I used to play the piano and saxophone, but I am wondering if there are any singers out there- any genres. I sing a variety of genres- pop, broadway, soft rock, country, classical/ opera, etc.  :ninja:
And, btw, singing in the shower counts!

I don't REALLY sing... but I like to sing for fun. I usually just sing along with a cd or something. I sang in a chorus once at a music camp, but everyone who went had to... lol so that doesn't mean much. But anyway, yay singing! :)

i can sing, and i do sing
but that doesn't mean i'm really good at it

i'm ok...not that good, not that bad

I sing bass in a choir, and I'm the lead singer for a band.  I've been working hard on extending my upper range, and I'm starting to get somewhere.

I sing in the shower (heh.)
I was a top soloist in choir,
growing up I used to pwn karaoke night (nowadays my voice cracks :().
I prefer catchy melodies lacking high notes.
Singing those make me sound professional.
I'm working on my upper range,
if I sing in public (as in a gig) I do it in C major.

i sing a little, mostly hard rock/heavy metal.

I sing, mostly to the radio or my iPod, but I'm also starting to teach myself guitar and am singing while epic-failing at playing it...

i love singing!!!!

too bad i'm not any good at it...  :blush:

I sing classical Indian music, and for fun.

I try to sing a good Justin Bieber, but my voice keeps cracking :(

I also like to sing along to the radio :D

My singing is OK.  And yes, I hate to admit it but I sing in the shower.   :dry:

I good at singing, though I'm a boy... My voice is still great... (*cough* deeper voice *cough)

I really like to sing, and I think I have a pretty good voice. I used to be in a chorus but now I mostly try to sing whatever songs I play on the violin, which is kind of impossible...
I sing in the shower... even at camp :P

[color=#FF0000]I have an airy voice, but I'm not loud. I can't belt out things very well. But I'm pretty darn good at soprano.
I love singing pop songs in different keys[/color].

I sing, but just free sing. I sing whatever I want to.

Hmm. I guess I can do a good opera. High and low.  :D

Dude I like to rap, and I'm as good as Eminem.

I wish.


A single atom of Uranium-238 rests at the origin.  Each second, the particle has a 1/4 chance of moving one unit int he negative $ x$ direction and a 1/2 chance of moving in the positive $ x$ direction.  If the particle reaches $ (\minus{}3, 0)$, it ignites a fission that will consume the earth.  If it reaches $ (7,0)$, it is harmlessly diffused.  The probably that the particle will be diffused can be expressed as a common fraction $ \frac{m}{n}$.  Find the remainder when $ m\plus{}n$ is divided by $ 1000$.

1/4+1/2=3/4, so what happens when it neither moves left or right?

Then you wait till the next second.

[hide="Hint on this matter"]
You can disregard the probability that it doesn't move, since it will move eventually, and just say that it is twice as likely to move right as to move left, so the probability of moving in the + direction is 2/3, and the probability of moving in the - direction is 1/3.
[/hide]

Ok sorry this doesn't have to do with math...

but "diffuse" and "defuse" are different words, you see.

err any thoughts?
i dont quite understand the authors solution.

its a bit above me  :D

Could somebody post a solution?

EDIT: Never mind, I found the solution at
http://www.tjhsst.edu/~tmildorf/math/PracticeAIME4Solutions.pdf


geometry
circumcircle
LaTeX
cyclic quadrilateral
geometry unsolved

[quote]If a quadrilateral with sides od distinct lengths is inscribed into a circle of radius
$ R$, then there exist two more quadrilaterals not equal to it with the same lengths of sides
inscribed in the same circle. These quadrilaterals have not more than three distinct lengths
of diagonals: $ d_1$, $ d_2$ and $ d_3$. Prove that the area of the quadrilateral is equal to $ \dfrac{d_1d_2d_3}{4R}$[/quote]
 :(

First you need a lemma to prove the theorem :

If $ a,b,c$ are the side lengths of a triangle and $ R$ is the length of the circumradius of the triangle then the area $ A$ of the triangle is $ A \equal{} abc/4R$. (Can you prove this?)

Now for the three cyclic quadrilaterals which have diagnols equal in pairs as suggested by your statement, we will apply this above stated lemma to a pair of triangles.

Denote the length of the three diagnols by $ A,B,C$. and the lengths of the sides of one of the cyclic quadrilateral by $ a,b,c,d$. Let $ C$ be the diagnol in one of the three cyclic quadrilateral which separates the sides $ a,b$ and $ c,d$. (That is here $ a,c$ and $ b,d$ are opposite pairs of sides)Now the area of the triangle bounded by $ a,b,C$. This area $ D_1$ = $ abC/4R$ where $ R$ denotes the common circumradius of the three cyclic quadrilaterals. Similarly area of the triangle bounded by $ c,d,C$ is
$ D_2$ = $ cdC/4R$. Adding the area of the two triangles $ D_1 \plus{}D_2$ = $ C*(ab \plus{} cd)/4R$. Sorry for the latex trouble but here the whole expression is divided by $ 4R$. Now consider another cyclic quadrilateral with opposite pairs of sides $ a,b$ and $ c,d$ and diagnols $ A,B$. By Ptolemy's theorem to this cyclic quadrilateral we have $ ab \plus{} cd \equal{} AB$. Substituting this in the above expresssion we have $ D_1 \plus{} D_2 \equal{} CAB / 4R$


Here's the rules:

We have person A and person B. A posts a word, and hides the definition, like this:

[quote="A"]

antebellum

[hide="definition"]A time before a war, such as in 1905, before World War I

Example: After the war, the French found it hard to return to their past antebellum way of life.[/hide]

[/quote]

Then person B says

[quote="B"]

answer: [hide="look in here"](insert random definition here)[/hide]

did I get it right??

[/quote]

If B has made the correct definition (It doesn't have to be exact) a random person says so, and B does what A does, except B says a different word, and it's definition. You don't have to make an example.

But if B is wrong, play continues, with the first word still on, and a different person guesses, and he/she is right, he/she posts the next question.

ALSO: Ix-Nay on looking up the definition or peeking at the definition!

Here we go!

First word:  incriminate

[hide="definition:"]To find evidence that strongly points to a suspect[/hide]

To accuse or make it seem like someone has done something wrong.

dude what is this?
A TOPIC IN GAMES FORUM TO IMPROVE ENGLISH?
 :mad:

[quote="mustafa"]It's a learning game :rotfl:[/quote]
IN GAMES FORUM?

err. if it is a noun-realllllllly old person!
if descriprion thingy-err-adjective......reallllllllllly oolllldddd! YAY!

E. A. Poe's "favorite word" JK Funcia thinks it is and has no proof...PHANTASMAGORIA!

P.S. I love the Fall of the house of usher and the martian chronicles' "fall" too!

wont=habit thing can't expl. well but hope you get idea
mein= i dunno is it german?

umm you were basically right more of illusions though like going insane..i'm sure there is some website with a very accurate definition if it is'nt in your dictionary at home  :)

[quote="randomdragoon"]phantasms?

wont and mein are a couple of cool four-letter words that few people know the meanings of.[/quote]
hmm isnt it spelled mien?
it means facial expression or something like that.
(lol "He had a mean mien")

Whoops, it is spelled mien. Silly typo  :blush:
It means demeanor.

antediluvian means from before the Flood, or from a very long time ago; I don't think it means a very old person.

probity=honesty, trustworthyness.

ananym.

Phantasmagoria is like...lots of random, surreal scenes flashing by in quick succession. I used the word phantasmagoric in my essay on Hitchhiker's guide to the galaxy.

yes that is a wonderful definition i love it!

[quote="randomdragoon"]Incoherent [speech]! My English teacher a couple years ago used to use this word a lot.

Strengths has 9 letters but only one vowel; twelfth has only one vowel and no "s"! Queueing has five vowels in a row.

Quixotic[/quote]

Wootwoot, one of my Lesson 3 vocabulary words--idealistically romatic

Dirndl :harhar:

a full, wide skirt with a tight waistband.

modus vivendi :wink:

How about these?

wan
cur
ilk
fen

Collectively they have about the same number of letters as your average vocabulary word...


algebra
polynomial
linear algebra
linear algebra unsolved

Let $A$ be a nilpotent matrix.
What is the characteristic polynomial of $A$ ? Prove that $\mathrm{trace}(A^p)=0$ for all positive integers $p$.
Please justify because i can only find the answers when the field is the set of complex numbers.

you can use your proof in an algebraic closure of the field.. (characteristic polynomial, Trace, etc .. do not change with field extensions. It has already been discussed many times on the forum. Try to search..)

but do every field has an extension which is algebrically closed? if no how can we proove these properties?

you don't need to use algebraic closures...

extending bases of $ker(A) \subseteq ker(A^{2}) \subseteq ...$ shows that $A$ is similar to an upper triangular matrix. from this, the rest is obviuos. remark that $A^{p}$ is also nilpotent.


complex numbers
inequalities unsolved
inequalities

Let $z_1,z_2,...,z_n$ be complex numbers such that $|z_1|+|z_2|+...+|z_n|=1$. Prove that the sum $S$ of a subset of ${z_1,z_2,...,z_n}$ satisfies $|S|>=1/6$

I think it has been discussed several times on the forum that one can replace $\frac{1}{6}$ with $\frac{1}{\pi}$. Unfortunately, we cannot further improve this.  :D

I think we can improve this with $\frac{1}{4}$.Do you agree with me? :?

And I guess you are answering to the proposer, Cezar.  :D

[quote="probability1.01"]1/6 here seems to fit well with the solution by pigeonholing the quadrants; maybe if you divide into more parts and take the limit you can get the $\frac{1}{\pi}$...[/quote]

I don't want to seem an ignorant  :blush:  , but what are quadrants?

[quote="cezar lupu"]Why, you don't agree with me? :D  ;)[/quote]

don't you read harazi's improvement to $\frac{1}{\pi}$?  :D


modular arithmetic
quadratics
geometry
3D geometry
arithmetic sequence
number theory unsolved
number theory

The difference between consecutive squares gets arbitrary large (in fact $ (n\plus{}1)^2 \minus{} n^2 \equal{} 2n\plus{}1$), thus at some point it will be larger than the common difference of the arithmetic sequence.

[quote="winter"]i'm thinking about these questions:

1) are there infinitly many perfect squares in an arithmetic progression?
2) any conditions so it would be so?
3) if yes, how to prove it?

would appreciate any input or references you might provide :) and sorry if this is wrong place to post or the question is stupid :P[/quote]

You want to have $ p^2\equal{}a\plus{}bn$, and so $ a\equal{}p^2\pmod{b}$ and so two cases :

1) If $ a$ is a quadratic residue modulus $ b$, you have at least one perfect square in the sequence, and then infinitly many squares :
$ p_0^2\equal{}a\plus{}bn_0$ implies $ (p_0\plus{}kb)^2\equal{}a\plus{}b(n_0\plus{}2kp_0\plus{}k^2b)$

Exemple : $ a\equal{}10$ $ b\equal{}3$ : $ (3p\plus{}1)^2\equal{}10\plus{}3(3p^2\plus{}2p\minus{}3)$

2) If $ a$ is a not quadratic residue modulus $ b$, you have no perfect square in the sequence.

Exemple : $ a\equal{}11$ $ b\equal{}3$ : no perfect square at all.

[quote="winter"]i'm thinking about these questions:

1) are there infinitly many perfect squares in an arithmetic progression?
2) any conditions so it would be so?
3) if yes, how to prove it?

would appreciate any input or references you might provide :) and sorry if this is wrong place to post or the question is stupid :P[/quote]
I think your problem is:
Prove that if a arithmetic sequence contain infinite perfect square if an only if it contain an perfect square.
Consider an arithmetic $ x_{n\plus{}1}\equal{}ax_n\plus{}b$ .
The sequence contain an perfect square if and only if a is quadric reside mod b.
Other problem:
1)Prove that if an arithmetic sequence contain an perfect an a cube then it contain an term is the from]
$ a^6$ where $ a\in Z$

[quote="pbornsztein"]It is an old and famous result from Fermat that there do not exist four distinct squares in arithmetic progression.

Pierre.[/quote]

I think the question is not to have infinitly many squares in arithmetic progression but infinitly many squares in [u][b]an [/b][/u]arithmetic progression.

So, for example, in the arihmetic progression $ 10\plus{}3n$, we have infinitly many squares $ 16, 49, 121, ...$

Btw, could the guy who rated all this thread average or below give an explanation of his quotation ?

thank you all for your replies! :)
I dont really understand ZetaX's reply.. the squares need not be consecutif?

i fully understand pco's solution, thank you :)
TTsphn: that's an interesting way of looking at it, thinking about your question :)

pbornsztein: the squares are not arithmetic progression on their own, they appear in the terms of some arithemtic progression.

Concerning TTsphn's other problem -- look at the solution to
 AMM problem 11182 in the June-July 2007 issue of the
 American Mathematical Monthly.

For example, the problem is quite easy when the common
difference in the progression is prime.

Schmu

This other problem appeared on the  $ 1997$ ISL.

Pierre.

It wasn't very clear how the problem was meant to be...

For the additional one see http://www.mathlinks.ro/Forum/viewtopic.php?p=950156#950156 .


At a shotgun factory in Ouagadougou, Burkina Faso, the $N$ employees don't feel like working. Unfortunately, they ran out of luck since the new manager, Oleg Protasov, wants work to get done. He gave each of the workers a task and he computed how long each task lasts. The task of employee $i$ lasts for $T_i$ days.

To waste as much time as possible, the employees started to make pertinent comments: "I can't start working until some other employees of my choice finish their work... Off with Protasov's head." Thus, every employee $i$ has a list of employees on which he depends, i.e. those workers must finish their tasks before $i$ can start his own. Obviously, some of them do not depend on anybody, so they can start right away.

Protasov, in order to mantain his head on his shoulders, accepted the sindicate's proposal, but he also computed the minimum time required to complete the tasks. Furthermore, it was commonly agreed that every employee will start his/her work as late as possible, but so that the time required is the minimum time computed a sentence ago.

Input:
"protasov.in": on the first line, the number $N$; on the next $N$ lines, the $i$th line describes the task of the $i$th employee by $T_i \ \ D_i \ \ S_{i 1} \ \ S_{i 2} \ \ \ldots \ \ S_{i D_i}$, where $T_i$ is the time required to fulfill the task, $D_i$ is the number of employees on which he/she depends and $S_{i1},\ldots,S_{i D_i}$ are the respective employees.

Output:
"protasov.out": $N+1$ lines; on the first line is the minimum time required to complete the job; on the next $N$ lines are the maximum delay time of each employee.

Restrictions:
$0 < N , T_i < 101$.

Observations:
There are no circular dependencies.

Construct an oriented graph and use greedy to calculate the minimum total time - keep removing leaves and set the starting time for each removed vertex to the maximum of ending times of its dependencies.

After you calculate the ending time of the last vertex, take again the original graph but this time remove vertices from the top (i.e. the ones on who nobody depends first). Set the ending times as late as possible. And that's it!


Any ideas? I though on this problem some time ago. I suspect that technique similar to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=19496]Combinatorial Nullstellensatz[/url] may be applied.

[quote="manuel"]Suppose a graph G has 4n vertices with minimum degree at least 2n. Prove that G has n vertex-disjoint cycles of 4 vertices.[/quote]

Open problem (65) in http://math.ucsd.edu/~fan/ep.pdf , with reference to

[88] P. Erdös, Some recent combinatorial problems, Technical Report, University of Bielefeld, Nov. 1990.

I cannot exclude that this was recently proven, but until someone finds out more I'm moving this into Open Questions.

  darij


vector
invariant
geometry
3D geometry
geometric transformation
rotation
trigonometry

I just can't progress on the following questions. Could I get some help?  :(
   In $R^3$ consider the set $A$ of those vectors $x=(x_1,x_2,x_3)$ in an orthonormal basis such that $x_1\geq\sqrt{x_2^2+x_3^2}$. Let $N$ be the set of linear maps of $R^3$ such that $f(A)$ is a subset of $A$ and such that they cannot be written as a sum of two linearly independent linear maps with the same property (that $A$ is invariant under $f$). Prove or disprove:
   1) There are no maps in $N$ of rank 2;
   2) Find all maps in $N$ of rank 3.
   3) Is every linear map for which $A$ is invariant a finite sum of elements of $N$?
    I don't know why, I find these questions terribly difficult.

3) is clear: the set N is a convex cone in a 9-dimensional space, so every element of N is a sum of at most 9 extremal elements.

Here's the ideas I found until now. Sorry I didn't finish them up, I will do this when I return from Cyprus.

I will better note $A=\{(x,y,z)|z\geq \sqrt{x^2+y^2}\}$
We note that the rotations $R_t$ that map $(x,y,z)$ to $(x\cos t+y \sin t, -x \sin t+y \cos t,z)$ belong to $M$. They actually belong to $N$, and here's why:

Assume that $R_t=S+T$ where $S,T$ belong to $M$. We shall assume $t=0$ (other cases are just rotations of this case). So $(x,y,z)=(x_1,y_1,z_1)+(x_2,y_2,z_2)$ where $S(x,y,z)=(x_1,y_1,z_1), T(x,y,z)=(x_2,y_2,z_2)$
Then if $x^2+y^2=z^2$ then $(x_1+x_2)^2+(y_1+y_2)^2=(z_1+z_2)^2$. However $x_1^2+y_1^2\leq z_1^2, x_2^2+y_2^2\leq z_2^2, x_1x_2+y_1y_2\leq z_1z_2$ (CBS). So actually $(x_1+x_2)^2+(y_1+y_2)^2\leq(z_1+z_2)^2$. Therefore equalities hold everywhere which implies that $S,T$ are actually proportional to $R_1$ QED

Now consider $T \in N$. Let $T(x,y,z)=(a_1x+b_1y+c_1z,a_2x+b_2y+c_2z, a_3x+b_3y+c_3z)$. Denote by $v_x$ etc. the $x$-coordinate of $v$.

Let $\mathcal{B}$ be the border of $A$ i.e. $B=\{(x,y,z)|z=\sqrt{x^2+y^2}\}$. Let $B$ be the subset of $\mathcal{B}$ which contains the points whose $z$-coordinate is $1$, i.e. $B=(\sin t, \cos t,1)$. Denote $v(t)==(\sin t, \cos t,1)$

Firstly let's show that $Tv\neq 0$, where $v \in B$. Indeed, assume so. WLOG $v=(1,0,1)$ (otherwise rotate). Let $(a,b,c)$ with $a<c\neq 0$ and$T(a,b,c)=(m,n,p)$, Then for sufficiently small $k>0$, $v+k(a,b,c)$ will belong to $A$ hence $T(v+k(a,b,c))$ belongs to $A$ so $(m,n,p)=T(a,b,c)$ belongs to $A$ whenever $a<c$. If $b_3 \neq 0$ for $a<c$ we can choose $b$ that $p=0$ which implies that $m=n=0$. This implies that $T$ is of rank $1$, which is not the case here, because it is easy.

So we've shown that $Tv\neq 0, v\in B$ so $Tv_x>0$ hence for some $c$ $Tv_\geq c$(this is from the continuity of $T$).

It's easily seen that $\delta T=T_z^2-T_x^2-T_y^2$ is a quadratic in $x,y,z$ therefore we must deduce that it is positive and increasing in $z=\sqrt{x^2+y^2}$ (as a function of $z$)

We claim for some $v \in B$ we have $Tv \in \mathcal{B}$. Indeed. let $I=R_1$ be the identity transformation.  By the continuity of $T$ we can find $d>0$ s.t. $Tv_x\geq d+\sqrt{Tv_y^2+Tv_z^2}$  We know that $T-kI \notin M$ so for some $v_k$ $Tv_k-kv_k \notin A$. We may assume $v_k \in B$ so $v_k \in \mathcal{B}$. Now taking $k \rightarrow 0$, extract a convergent subsequence from $v_k$ and we've fount some $v(p)$ with $T(v(p)) \in \mathcal{B}$. We may assume $T v(p)=q$.

Now we can assume $p=q=\frac{\pi 2}$ otherwise work with $R_{\frac{\pi}2-q} \circ T \circ R_{\frac{\pi}2+p}$. So we have $T(1,1,0)=(1,0,1)$.

It's clear that $\delta T (v_t)$ can be written as a quadratic polynomial in $\sin t, \cos t$. Moreover it has the root $\frac{\pi 2}$. We can write $\delta T$ as $f(x) \pm \sqrt{1-x^2} g(x)$ where $x=\sin t$, and the sign depends on whether $\cos t=\sqrt{1-x^2}$ or $\cos t=-\sqrt{1-x^2}$. Aslo $deg f\leq 2, \deg g \leq 1$. As $\delta T\geq 0$ we conclude that $f^2(x)-(1-x^2)g^2(x)\geq 0$. However this polynomial is divisible by $1-x$ which means $1-x|f$ then $1-x|g$ .

If we set $f=(1-x)f_1, g=(1-x)\lambda$ then we have $\\delta T (v_t)=(1-x)(f_1(x)+\lambda \sqrt{1-x^2})$.

We claim that $f_1(x)+\lambda \sqrt{1-x^2}$ also has a root in $(0;1)$. Indeed, if not, then we can replace $T$ by $T-\epsilon T'$ where $T'$ maps $(x,y,z)$ to $(0,0,z-y)$.
If $\epsilon$ is small enough then we can deduce that the new transformation will cleraly satisfy the above polynomial requirements, contradicting the extremality of $T$.

So $f_1^2(x)-\lambda^2(1-x^2)$ has a root, which is double root since the polynomial is non-negative. This root can be $1$ only if $\lambda=0$. We shall now assume that the root is not $1$. (the case when the root is $1$ shall be investigated later).

This implies that there is one more $t$ such that $T(\sin t, \cos t,1)=a(\sin u, \cos u, 1)$.

Now let $T(0,0,1)=(m,n,p)$. Set $v=(1,0,1), v'=(\sin t, \cos t,1), v''=(0,0,1)$

We see that $\alpha v+ (1-\alpha )v'+k v' \in B$, where $k=1-\sqrt{1-2 \alpha (1-\alpha )(1-\sin t)}$.

So, the condition that $T(\alpha v+(1-\alpha)v'+kv') \in A$ can be wriiten polynomially after substitution as

$2a\alpha (1-\alpha )(\sin u-1)+k^2(m^+n62-p^2)+2k(\alpha(p-m)+(1-\alpha)a(p-m \sin u -n \cos u)) \leq 0$.

By substituting $2\alpha (1-\alpha)$ with $\frac{2k-k^2}{1-\sin t}$ we transform this condition into:

$k^2(m^2+n^2-p^2+\frac{a(1-\sin u)}{1-\sin t})+2k(\alpha(p-m)+(1-\alpha)a(p-m \sin u -n \cos u)+\frac{a \sin u}{1- \sin t})\leq 0$.

We claim that for $\alpha =0,1$ the coefficient at $k$ vanishes. Indeed, if it would be positive, by taking $\alpha %Error. "tends" is a bad command.
0, \alpha<0$ we would obtain $k$ for which the expression is positive, and anlogously for the other cases.

Thus the coeefficient vanishes at $\alpha=0,1$ and so at any $\alpha$ since int's linear in $\alpha$. Thus our polynmial has for $uk^2\leq 0$.

We deduce $u=0$, otherwise we could replace $m,n,p$ by $r m, rn,rp$ for $r<1$, constructing a "smaller" transformation which contradicts the extremality of $T$.

So actually this polynomial is zero.

This implies that $T$ maps $B$ into a subset of $\mathcal{B}$.

Now writing this condition as an explicit polynomial, it becomes $(a_1^2+a_2^2-a_3^2-b-1^2-b_2^2+b_3^2)x^2+(c_1^2+c_2^2-c_3^2+b_1^2+b_2^2-b_3^2)+(a_1c_1+a_2c_2-a_3c_3)x+(a_1b_1+a_2b_2-a_3b_3)\sqrt{1-x^2}x+(b_1c_1+b_2c_2+b_3c_3)\sqrt{1-x^2}=0$.

However we know that $c_1=1-a_1, c_2=-a_2, c_3=1-a_3$. Algebraic manipulations now lead to $b_1=b_3, a_1-a_3=b_2^2, (a_1-a_3)b_1+a_2b_2=0$ etc.

From this relations we can completely characterize the transformations.

The second case, when the second root is $1$, gives us $(a_1^2+a_2^2-a_3^2-b-1^2-b_2^2+b_3^2)x^2+(c_1^2+c_2^2-c_3^2+b_1^2+b_2^2-b_3^2)+(a_1c_1+a_2c_2-a_3c_3)x+(a_1b_1+a_2b_2-a_3b_3)\sqrt{1-x^2}x+(b_1c_1+b_2c_2+b_3c_3)\sqrt{1-x^2}=u(1-x)^2$.

and can be handled analogously.

Ok, I understood two facts: why nobody in that ens exam managed to solve more than 4 questions ( from 14) and why I could not solve these 2 ( the third one I finally managed to observe that it is more or less trivial). Thanks, Iura.


function
algebra
domain
logarithms
advanced fields
advanced fields unsolved

Find all $f:[0,+\infty) \rightarrow \mathbb{R}$ of $C^2$ ( twice differentiable with f'' continous) such that $F(x_1,x_2 \ldots x_n)= f( \sqrt{x_1^2+\ldots +x_n^2})$ verify Laplacian F = 0 or $\sum_{i=1}^{n} \frac{\partial ^2F}{\partial x_i^2} = 0$

Let $r$ be the function $r(x_1,...,x_n)=\sqrt{x^2_1 + ... + x^2_n}$
 $\frac{\partial F}{\partial x_i} = \frac{\partial r}{\partial x_i}f'(r) = \frac{x_i}{r} f'(r)\ ,\  \frac{\partial^2 F}{\partial^2 x_i}=(\frac{1}{r} - \frac{x_i^2}{r^3})f'(r) + \frac{x_i^2}{r^2}f''(r)$ , so :
$\sum_{i=1}^{n} \frac{\partial ^2F}{\partial x_i^2} = \frac{n-1}{r}f'(r)+f''(r) = 0 \Longrightarrow ln|f'(r)| = A -(n-1) ln(r)$ $ \Longrightarrow f(r)= \frac{B}{r^{n-2}}+C $
$f$ has to be continous in $0$ , so finally $f(r) = Const.$
 :cool: Laplace haven’t need the god hypothesis :cool:

I would be of more physical interest to leave the origin out of the domain of the function and let $0<r<\infty$ in the statement of the problem.

If we do that, then Diogene's solution shows that for $n>2$ the solution must be the "Newtonian potential": $\frac{B}{r^{n-2}}+C.$

Diogene did tacitly assume $n>2.$ If $n=2$ the solution is $B\ln r+C.$ and if $n=1$, the solution is $B|x|+C.$


What is the explcit relationship between sine and cosine?  (Meaning how can you replace $\sin x$ with $\cos y$, other than $\sin x = \pm \sqrt{1-\cos^{2}x}$).

um dude that is the relationship, unit circle, pythagorus?

If $x$ and $y$ are just arbitrary, there's not really anything you can say about the relationship between $\sin x$ and $\cos y$.

$\sin x=\cos\left(\frac{\pi}{2}-x\right)$ , $\sin^{2}x+\cos^{2}x=1$ and so much !
$\sin x=\cos y\Longleftrightarrow y\pm z\in 2k\pi\pm\frac{\pi}{2}\ ,$ where $k\in Z$ (plus with plus, minus with minus).