geometry
analytic geometry
graphing lines
slope
inequalities
vector
triangle inequality

show that the shortest way between 2 points is the straight line that pass between them. you need to refer not only to broken straight lines, but also to curved lines.

[hide]
i would use the fact that in euclidian geometry the shortest way is the way which has the most even change in slope.  Since the most even way to do this is through a straight line the straight line must be the shortest way.  Other curves have slopes which change, sometimes greater and sometimes less, thus it isn't the most efficient.
[/hide]

[quote="pkerichang"][hide]
first we have the triangle inequality, put it in vector form to get $|a| + |b| \ge |a+b|$, with equality only when $a$ and $b$ are collinear.  So a random line (can be curved), and break it up into bunch of small vectors connected tip-to-tip.  Then put them in groups of 2 (consecutive), then we can draw a shorter line by taking each groups of 2 and "connect the missing side of the triangle", a.k.a. replace each two consector vectors by their sum, therefore reducing the total amount of vectors by factor of 2.  Continue this process we'll end up with only 1 vector, which is the one connecting the starting point and the ending point.
[/hide][/quote]
correct!  :)

This question is absolutely ridiculous unless you go down to the very basic definitions of geometry. For broken lines this is the very meaning of the triangle inequality. For curves you need a solid definition of length. And for there to really be definitions that make sense we need continuous curves and the concept of supremum of sets... and we quickly leave the vicinity of of high school level.


calculus
integration
function
limit
floor function
real analysis
real analysis solved

Let
\[
I(\alpha,\beta)=\displaystyle\int_0^1\sqrt[\alpha]{1-x^{\beta}}\,dx,\qquad \Gamma(n)=\int_0^{+\infty}e^{-x}x^{n-1}\,dx
\]
Prove
I)
\[
I(\alpha,\beta)=I(\beta,\alpha)=\displaystyle\frac{\Gamma(\frac 1 {\alpha})\Gamma(\frac 1 {\beta})}{(\alpha+\beta)\Gamma(\frac 1 {\alpha}+\frac 1{\beta})}
\]
II)
\[
I(\alpha,\beta)=\displaystyle \lim_{n \to +\infty}\displaystyle\frac {\sqrt[\alpha]{n-1^{\beta}}+\sqrt[\alpha]{n-2^{\beta}}+\sqrt[\alpha]{n-3^{\beta}}+\cdots}
{n^{\frac 1 {\alpha}+\frac 1{\beta}}}
\]

Preliminary note (which everyone should already know): The gamma function satisfies the recursion $\Gamma(z+1)=z\Gamma(z).$ Prove that by integration by parts.

Next step: make the substitution $u=x^{\beta}$ to get

$I(\alpha,\beta)=\frac1{\beta}\int_0^1u^{\frac1{\beta}-1}(1-u)^{ \frac1{\alpha}}\,du.$

Now, define $B(a,b)=\int_0^1x^{a-1}(1-x)^{b-1}\,dx.$

We have that $I(\alpha,\beta)= \frac1{\beta}B\left(\frac1{\beta},\frac1{\alpha}+1\right).$

Now let's calculate $B(a,b)$:

$\Gamma(a)\Gamma(b)=\int_0^{\infty}x^{a-1}e^{-x}\,dx \int_0^{\infty}y^{b-1}e^{-y}\,dy$

$=\int_0^{\infty}\int_0^{\infty}x^{a-1}y^{b-1}e^{-(x+y)}\,dx\,dy.$

Make the change of variables $x=vu,\,y=(1-v)u$ so that $x+y=u.$ The Jacobian of this change of variables is $\frac{\partial(x,y)}{\partial(u,v)} = u$ so that $dx\, dy = u\,du\,dv.$ The integral becomes

$\int_0^{\infty}\int_0^1v^{a-1}u^{a-1}(1-v)^{b-1}u^{b-1}e^{-u}u\, dv\,du$

$=\int_0^{\infty}u^{a+b-1}e^{-u}\,du\int_0^1v^{a-1}(1-v)^{b-1}\,dv = \Gamma(a+b)B(a,b).$

Hence, $\Gamma(a)\Gamma(b)=\Gamma(a+b)B(a,b)$ or $B(a,b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}.$

Finally, $I(\alpha,\beta)=\frac{\frac1{\beta}\Gamma\left(\frac1{\beta} \right)\Gamma\left(\frac1{\alpha}+1\right)}{\Gamma\left(\frac1{\alpha} + \frac1{\beta}+1\right)} =$ $ \frac{\Gamma\left(\frac1{\beta} + 1\right)\Gamma\left(\frac1{\alpha}+1\right)}{\Gamma\left(\frac1{\alpha} + \frac1{\beta}+1\right)} $

$=\frac{\frac1{\alpha}\cdot\frac1{\beta}}{\frac1{\alpha}+\frac1{\beta}} \cdot \frac{\Gamma\left(\frac1{\alpha}\right)\Gamma\left(\frac1{\beta} \right)}{\Gamma\left(\frac1{\alpha}+\frac1{\beta}\right)} = \frac{\Gamma\left(\frac1{\alpha}\right)\Gamma\left(\frac1{\beta} \right)}{(\alpha+\beta)\, \Gamma\left(\frac1{\alpha}+\frac1{\beta}\right)}$

As for part (II): the sum must be a finite sum before we take the limit. We are taking the limit as $n\to\infty$ of

$n^{-\frac1{\alpha}-\frac1{\beta}}\sum_{k=1}^{\lfloor n^{1/\beta}\rfloor}\sqrt[\alpha]{n-k^ {\beta}} = n^{-\frac1{\beta}}\sum_{k=1}^{\lfloor n^{1/\beta}\rfloor}\sqrt[\alpha]{1-n^{-1} k^{\beta}} = $

 $=n^{-\frac1{\beta}}\sum_{k=1}^{\lfloor n^{1/\beta}\rfloor}\sqrt[\alpha]{1-\left(n^ {-\frac1{\beta}} k\right)^{\beta}}$

If we use the points $\{n^ {-\frac1{\beta}} k\}$ for $1\le k \le{\lfloor n^{1/\beta}\rfloor}$ as the points of a partition of $[0,1]$, then this is a Riemann sum for the integral that defines $I(\alpha,\beta).$

Fine Kent! A side remark on simmetry of mathematical object $I(\alpha,\beta)$.
Given the curve $y^{\alpha}+x^{\beta}=1$ we have:
\[
I(\alpha,\beta)=\displaystyle\int_0^1 y\,dx=\displaystyle\int_0^1\sqrt[\alpha]{1-x^{\beta}}\,dx=\displaystyle\int_0^1x\,dy=\displaystyle\int_0^1\sqrt[\beta]{1-y^{\alpha}}\,dy=I(\beta,\alpha)
\]


Samantha won the election for class president over Samuel by a margin of 32 votes. Given that they received a total of 332 votes, how many votes did Samuel receive?

We know that the amount of votes samantha recieved, $ x$, is equal to the amount of votes samuel recieved, $ y$, +32.

We have two equations:
$ x\plus{}y \equal{} 332$
$ x\equal{} y\plus{}32$

Solving for these two equations, we get that $ y \equal{} \text{ the number of votes samuel recieved } \equal{} \boxed{150}$.

Or we note that Samantha + Samuel=332. SUbtracting 32 we get 300, dividing by 2 gets us 150. The other number is 182. Obviously samuel got the 150,.


group theory
abstract algebra
inequalities
superior algebra
superior algebra unsolved

Let G be a group and p a prime such that p^n divides G but p^(n+1) does not divide G.  Fix an integer 1<=k<=n. Prove that any subgroup A in G of order p^k is contained in a Sylow p-subgroup of G.

Hint: Use the reasoning to show that if G is a Sylow p-subgroup of G then there must be some x in G such that A is contained in xBx^-1.

It's pretty hard to write an adaptation of a proof when you haven't told us the proof.  :roll:

Lemma: If $ P\in Syl_{p}(G)$ and A a p-subgroup of G, then $ N_{G}(P) \cap A \subseteq P$.

Proof: Since p does not divide $ [G: P]$, it cannot divide part of it so p does not divide $ [N_{G}(P): P]$.*
Set $ H \equal{} N_{G}(P) \cap A$. Then $ [H: P \cap A]$ must be a power of p since $ H \leq A$. If $ H$ is not a subset of $ P$, then $ P$ is properly contained in $ PH$.
Hence $ [PH: P] \equal{} \frac {|PH|}{|P|} \equal{} \frac {|H|}{|H\cap P|}$, which is a power of p and so p does divide $ [PH: P]$, which contradicts *. So $ H \equal{} N_{G}(P) \cap A \subseteq P$.

I'm a little worried about one point here that may confuse the less experienced readers.  This proof depends in a slightly subtle way on the observation that $ PH$ is a group.  Anyone trying to understand this proof should make sure that they can prove this fact.

Okay, that $ PH$ is a group follows from the following assuming the reader is familiar with it:
Proposition: If $ H \leq G$ & $ K \leq G$, then  $ HK\leq G \iff HK \equal{} KH$.

Set $ P \equal{} K, G \equal{} N_{G}(P)$. Then $ P$ is normal in $ N_{G}(P)$ and
$ H\leq N_{G}(P)$.

Choose $ x \in P, h \in H$. Then $ hxh^{ \minus{} 1} \in P$ & $ h^{ \minus{} 1}xh \in P$ so we have:
$ hx \equal{} (hxh^{ \minus{} 1})h \in PH \implies HP \subseteq PH$.
$ xh \equal{} h(h^{ \minus{} 1}xh) \in HP \implies PH \subseteq HP$.
So $ HP \equal{} PH$ and that $ PH$ is a group follows from the proposition.

[quote="rawkerrxx"]Hint: Use the reasoning to show that if G is a Sylow p-subgroup of G then there must be some x in G such that A is contained in xBx^-1.[/quote]

I think you meant:
[quote]Hint: Use the reasoning to show that if [b]B [/b]is a Sylow p-subgroup of G then there must be some x in G such that A is contained in xBx^-1.
[/quote]

Anyway, let's break $ G$ down into double cosets $ AxB$. The order of such a coset is \[ \frac{(|A|)(|B|)}{|A \cap x B x^{\minus{}1}|} \equal{} \frac{(p^k)(p^n)}{|A \cap x B x^{\minus{}1}|}\] Then $ p^{n\plus{}1} \nmid |AxB|$ iff $ A \cap x B x^{\minus{}1} \equal{} A$ or $ A \le x B x^{\minus{}1}$. If $ A$ is never contained in $ x B x^{\minus{}1}$, then all double cosets have order divisible by $ p^{n\plus{}1}$, but the sum of their orders is $ |G|$, which is not divisible by $ p^{n\plus{}1}$. This gives a contradiction, meaning that for some $ x$, we have $ A \le x B x^{\minus{}1}$.

[quote="calc rulz"]
Then $ p^{n \plus{} 1} \nmid |AxB|$ iff $ A \cap x B x^{ \minus{} 1} \equal{} A$ or $ A \le x B x^{ \minus{} 1}$. If $ A$ is never contained in $ x B x^{ \minus{} 1}$, then all double cosets have order divisible by $ p^{n \plus{} 1}$, but the sum of their orders is $ |G|$, which is not divisible by $ p^{n \plus{} 1}$. This gives a contradiction, meaning that for some $ x$, we have $ A \le x B x^{ \minus{} 1}$.[/quote]

How did you get that all the double cosets have order divisible by p^(n+1)?

A is never contained in $ xBx^{\minus{}1} \implies |A \cap xBx^{\minus{}1}| \equal{} 1$ from which it follows that $ |AxB| \equal{} p^{k}p^{n} \equal{} p^{k \plus{} n}$ for all $ 1 \leq k \leq n$.

p^(n-1) *always* divides p^(k+n)?  How do we know this?

If $ k \ge 1$, then $ p^{n\plus{}1} \mid p^{n\plus{}k}$. Simple inequalities  :wink:


Let $ A_m = \{x \in R: (\exists a \in R) x = \frac{a^2+a+m}{a + 1}\}$

Prove that $\bigcap _{m\in R} A_m = \emptyset$

Here, R is the set of all real numbers.

Assume that there an $x\in R$ which belongs to all $A_m$. Then for every m there exists an $a\in R$ with $x=\frac{a^2+a+m}{a+1}$ which is equivalent to
$a^2+a(1-x)+m-x=0$. Now the discriminant of this quadratic equation must be non-negative for the fixed x and all m. But the discriminant is: $(1-x)^2-4m+4x=(1+x)^2-4m$. And it is negative for $m=(1+x)^2+1$. It is the contradiction we wanted to achieve.

Misha

I think that we have to say that for any real x, there exists m such that $x\notin A_m$ but don't know how to go about finding that $A_m$ for each given x.

Consider $m=(x+1)^2+1$ and read my reply again ;)

Misha


[quote="jensen"]find the value of $\frac{1+tan15}{1-tan15}$[/quote]

[hide]Rewrite as $\frac{\tan{45}+\tan{15}}{1-\tan{45}\tan{15}}= \tan{60}= \sqrt{3}$[/hide]

[quote="jensen"]find the value of $\frac{1+tan15}{1-tan15}$[/quote]
my first solution it's like your's (1234567890 )
[hide="other solution"]
$\frac{1+tan15}{1-tan15}$
=$\frac{sin15+cos15}{cos15-sin15}$
=$\frac{\frac{\sqrt{2}}{2}(sin15+cos15)}{\frac{\sqrt{2}}{2}(cos15-sin15)}$
=$\frac{sin(15+45)}{cos(15+45)}$
$=tan60$
$=\sqrt{3}$ [/hide]

We draw the trigonometric circle. The points $ A,B,C$ on the line $ x=1$ corresponds to the angles $ 45^{o},15^{o},-45^{o}$ respectively.
Then $ B(1,tan 15^{o})$ and we want to find the ratio $ \frac{1+\tan 15^{o}}{1-\tan 15^{o}}= \frac{BC}{BA}$

[asy]size(150); pair A=(1,1), B=(1,2-sqrt(3)), C=(1,-1), O=(0,0); pair X1=(1.1,0), X2=(-1.1,0); pair A1=(1,1.2), C1 = (1,-1.2); draw(X1--X2); draw(A1--C1); draw(O--A); draw(O--B); draw(O--C); draw(circle(O,1)); dot(O^^A^^B^^C); label("O",O,NE); label("A",A,NE); label("B",B,NE); label("C",C,SE);[/asy]

Sines law in $ \triangle OBC: \frac{BC}{\sin 60^{o}}= \frac{OB}{\sin 45^{o}}$

Sines law in $ \triangle OBA: \frac{BA}{\sin 30^{o}}= \frac{OB}{\sin 45^{o}}$

Then $ \frac{BC}{\sin 60^{o}}= \frac{BA}{\sin 30^{o}}\Rightarrow  \frac{BC}{BA}= \frac{\sin 60^{o}}{\sin 30^{o}}= \sqrt 3$


function
limit
real analysis
real analysis unsolved

Find all differentiable function at $ x\equal{}0$ that statisfies $ f(2x)\equal{}2f(x)$ for all $ x \in R$. Thanks in advance.

Let $ x \sim y$ if $ \frac{x}{y}$ is a power of $ 2$ (positive or negative) and partition the reals into equivalence classes.  Then $ \frac{f(x)}{x}$ is constant on all of these equivalence classes.  If $ f$ is differentiable at zero, then $ \lim_{x \to 0} \frac{f(x)}{x}$ must be defined and the same regardless of the way in which $ 0$ is approached, and we can approach $ 0$ along any of the equivalence classes.  It follows that they all share the same value of $ \frac{f(x)}{x}$, so $ f(x) \equal{} ax \forall x$.


On the left of the site, there is a proof without words.

Can anyone explain the one with the Pythagorean Theorem?
I don't seem to get it that clearly.

I think it's showing that the area of a square with side $ a$ plus the area of a square with side $ b$ is the area of a square with side $ c$, where $ a,b,c$ are the legs and hypotenuse of a right triangle.

It doesn't really look like a rigorous proof but rather a demonstration of how to interpret the Pythagorean theorem geometrically.

I think that may have been the original way that Pythagoras proved it.  My recollection is that the Greeks placed a lot of importance on equal-areas in proofs, and I think the proof in question does it just that way.


function
quadratics
algebra proposed
algebra

Let $k\in \mathbb{R}$ and $f: \mathbb{R}\rightarrow\mathbb{R}$ an injective function such that $f(xy)=f(x)f(y)+k$, for every real numbers $x,y$. Prove that $f$ is an odd function.

$f(1) = f(1)^2 + k$
So $f(1)$ can be either of the solutions to the quadratic $x^2 - x + k$.

$f(x) = f(x) f(1) + k \implies (1 - f(1)) f(x) = k$, so we can write $f(x) = \frac{k}{1 - f(1)} = \frac{f(1) - f(1)^2}{1 - f(1)} = f(1)$, which is constant, which is clearly an even function, and so it seems the problem is invalid unless $f(1) = 1$.

In that case, $k = 0$ and the function need only satisfy $f(xy) = f(x) f(y)$, which still does not imply $f(x)$ is an odd function - take $f(x) = 1$, for example.

t0rajir0u: Read the problem again: the function has to be injective.

But to finish it directly here: we already know $f(xy)=f(x)f(y)$ (t0rajir0u's prove works for that). So take $x=y=-1$ to get $1=f(1)=f(-1)^2$, but $f(-1)=1$ is impossible since that would contradict injectivity. So $f(-1)=-1$ and using the given equation we get $f(-x)=f(-1)f(x)=-f(x)$, thus $f$ is odd.

The fact that $k=0$ could be shown easier, considering $x,y,z$ distinct real numbers and computing $f(xyz)$ in all ways, and because $f$ injective it would result that $k=0$. From now on, my solution was exacly as yours.


As a system increases in volume, it absorbs 52.5 J of energy in the form of heat from the surroundings. The piston is working against a pressure of 0.500 atm. The final volume of the system is 58.0 L. What was the initial volume of the system if the internal energy of the system decreased by 103.1 J?

I'm not sure what I did wrong:

I used $ w \equal{} P\Delta V$, which can be expressed as $ \Delta E \minus{} q \equal{} P(V_F \minus{} V_I)$

Plug in: $ \minus{}103.1\text{ J} \minus{} 52.5 \text{ J} \equal{} 0.500 \text{atm} (58.0 \text{ L} \minus{} V_I)$

Convert joules to L atm using $ 101.3 \text{ J} \equal{} 1 \text{ L atm}$

$ \minus{}1.536 \text{ L atm} \equal{} 0.500 \text{atm} (58.0 \text{ L} \minus{} V_I)$

Then solve:

$ V_I \equal{} \boxed{61.1 \text{ L}}$

Why is this incorrect?

Thanks in advance!

I found my own mistake: I used the wrong equation. w = -P delta V.  :blush:


Martha purchases 240 feet of fence for enclosing her rectangular garden.  Her plan called for the width to be 40 feet.  However, her neighbor Mrs. Euler convinced her to make the width 60 feet.  Using the same 240 feet of fence, how many square feet larger is the rectangular garden Mrs. Euler suggested compared to the one Martha had planned?

A.0 ft2	B.240 ft2	 C.400 ft2   D.600 ft2   E. 2800 ft2

The original land has a width of 40, so a height of $ 120 \minus{} 40 \equal{} 80$. That means the area would be 3200. Ms. Euler's suggestion means the height is $ 120 \minus{} 60 \equal{} 60$, so the area is 3600. Thus the difference is $ 400$, or C.


inequalities
function
geometry
circumcircle
inequalities proposed

Let a, b, c be arbitrary positive reals such that abc=1. Show that
$
a^2+b^2+c^2+3\geq 2(ab+bc+ca)
$
I've already had a nice proof(In fact, I get this problem by the idea of the proof). But I am wondering whether there is a proof without that idea.
Should I give my proof first? :maybe:

Hate to spoil the fun, but this is just Schurs.

For aesthetic reasons, let's set $a= x^3$, $b=y^3$, $c=z^3$. Then ineq becomes
\[
x^6 + y^6 + z^6 + 3x^2y^2z^2 \geq 2(x^3y^3 + y^3z^3 + z^3x^3)
\]
which comes from
\[
x^6 + y^6 + z^6 + 3x^2y^2z^2 \geq \sum_\mathrm{sym}x^4y^2
\]
and you can finish it up with Muirhead (or if AM-GM if you with)

Just some possible approaches: Schur in the form bilzhao used, Schur in the form $ 2(ab+bc+ca)-(a^2+b^2+c^2)\leq\frac{9abc }{a+b+c}$ and then AM-GM, mixing variables, geometrical approach...

From the generalized form of Popoviciu's Inequality for convex function $f=e^x$, we get the following more general result:

If $a_1, a_2, ...,a_n$ are positive numbers such that $a_1a_2...a_n=1$, then
$a_1^{n-1}+...+a_n^{n-1}+n(n-2)\geq (n-1)(\frac{1}{a_1}+...+\frac{1}{a_n})$.

Vasc, could you please tell me how the generalized form of Popoviciu looks like? I am very interested in this kind of inequalities...

Harazi, you are right, there is a nice geometric interpretation:

We rename a, b, c as $a^2$, $b^2$, $c^2$, with positive a, b, c; then, the condition rewrites as $a^2b^2c^2=1$, what is, of course, equivalent to abc = 1. Now, it remains to prove:

[i]Let a, b, c be arbitrary positive reals such that abc = 1. Then, the inequality

  $a^4+b^4+c^4+3 \geq 2\left(b^2c^2+c^2a^2+a^2b^2\right)$

holds.[/i]

In order to prove this one, we rewrite it as

  $3 \geq 2\left(b^2c^2+c^2a^2+a^2b^2\right) - \left(a^4+b^4+c^4\right)$.

But

  $2\left(b^2c^2+c^2a^2+a^2b^2\right) - \left(a^4+b^4+c^4\right) \geq \left(a+b+c\right)\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)$;

thus, we have to prove

  $3 \geq \left(a+b+c\right)\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)$.

Now, if the right hand side of this inequality is $\leq 0$, then it is trivial. So we only need to consider the case when the right hand side of this inequality is > 0. Then, since a + b + c > 0 par definitionem, we have (b + c - a) (c + a - b) (a + b - c) > 0. This easily yields that there exists a triangle ABC with sides a, b, c. By the Heron formula, the area S of this triangle satisfies $16S^2 = \left(a+b+c\right)\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)$. And thus, it remains to show that

  $3 \geq 16S^2$.

Equivalently, $27 \geq 16^3S^6$. But since abc = 1, we also have $\left(abc\right)^4=1$, and thus we can rewrite this as $27\left(abc\right)^4 \geq 16^3S^6$. Now, if R is the circumradius of triangle ABC, we have $S=\frac{abc}{4R}$, so that abc = 4RS. Thus, the inequality in question is equivalent to $27\left(4RS\right)^4 \geq 16^3S^6$. After cancellations, this becomes $27R^4 \geq 16S^2$, or, after taking the square root,

  $3\sqrt3 R^2 \geq 4S$.

But this inequality is nothing but the algebraic incarnation of the well-known fact that among all triangles inscribed in a given circle, the equilateral triangle has the greatest area.

[b]EDIT:[/b] See also http://www.mathlinks.ro/Forum/viewtopic.php?t=18430 for a proof of this inequality (in a homogenized version).

  Darij

Thanks! This will be another nice exercise to prove...  :)

  Darij

And what an exercise! Here is another useful exercice: deduce from this inequality a short solution for Mathlinks Contest problem 3, third round, third edition.

The one about the friendly integers?!

  Darij

Yes, Darij, that one!  ;)  No, round 6 ,sorry.  :blush:

[quote="Vasc"]If $f$ is a convex function on I and $a_1,...,a_n\in I$, then

$f(a_1)+...+f(a_n)+n(n-2)f(a)\geq (n-1)[f(b_1)+...+f(b_n)]$,

where $a=(a_1+...+a_n)/n$ and $b_i=(na-a_i)/(n-1)$ for all i.[/quote]

It seems that this can come very useful one day! I'll put it in Theorem and Formula section for future references.


geometry
trigonometry
ratio
similar triangles
trig identities
Law of Sines
AMC 10

I seriously need help on geometry in questions #20-#25. When there is a geometry question there, I never get it right. I consistently get numbers #1-#19 correct. Any ideas on how to improve?

Really the only way you can learn is by practicing the problems in the database.
Other than that, know Stewarts's theorem, similar triangles, cyclic quadrilaterals, law of sines and cosines, when to drop a perpendicular, when to extend lines, double and triple angle formulas, angle sum and difference formulas, when to make a point such that a certain triangle is isoceles/equilateral, when to draw a parallel line, to use $ 1/2 \cdot ab \cdot sin C$ for area, area ratios, occasionaly Ptolemy Ceva and Menelaus. Get comfortable using these, don't revert to old methods which take longer and are harder to see. It comes with experience. Do easier problems that you can do first, then gradually increase the difficulty.

There are a lot of things you can do.

1) Highly recommended: take the introduction to gemoetry class
2) read the introduction to geometry book
3) read the solutions to those problems you're missing: if you don't know a technique, learn it. if you didn't see the "crux" move, do more 20-25 geometry problems.
4) learn trig. most 20-25 amc 10 geo problems can be easily bashed by law of sines, cosines, or other trig identities.
5) read Silverfalcon's intermediate geometry guide. its on [url]hdmatholympiad.com[/url].

I hope that helps

for the amc 10, i would imagine that only very basic formulas would be needed... know the different ways to show areas though. and the best advice anyone is gonna be able to give you is to continue to do problems.

Knowing many advanced formulas and trig isn't really necessary for the AMC 10, the methods used to solve the problems really just need to be learned through practice.

but knowing trig can seriously trivialize many amc 10 problems. however, nothing beats learning the basics and applying them to countless problems.

Don't learn to bash with trig. It won't serve you well when you try harder problems.

But it can help for the AMC 10, which was the original question.


Αν για την ακολουθια ακεραιων $ b_n$ με $ b_n$ διαφορο του $ \minus{} 1$  gia καθε $ n$ kai $ b_1\neq 1$ και για καθε $ n$ στο $ N^*$ ισχυει
$ b_n(b_n^2 \minus{} b_{n \plus{} 1}) \equal{} b_{n \plus{} 1} \minus{} 1$  νδο για κάθε $ k,l$ με $ k > l$ στο $ N^*$ ,  ο $ \frac {b_k}{b_l}$  δεν ειναι ακεραιος

η δοθεισα σχεση ειναι ισοδυναμη με $ b_{n \plus{} 1} \equal{}b_n(b_n \minus{}1)\plus{}1(1)$που ειναι η ταυτοτητα της ευκλειδιας διαιρεσης για το b_n-1
αρκει λοιπον να αποδειξουμε οτι η διαιρεση του b_k kai b_l (η και γενικα με $ b_n,0<n<k$)αφηνει παντα υπολοιπο και στην προκειμενη περιπτωση ισο με 1.εστω οτι $ l\equal{}k\minus{}w$,opou w fysikos ,mikroteros tou k
τοτε με βαση την (1) παιρνουμε:
$ b_k\equal{}b_{k\minus{}1}(b_{k\minus{}1} \minus{}1)\plus{}1\equal{}(b_{k\minus{}2}(b_{k\minus{}2}\minus{}1)\plus{}1))(b_{k\minus{}2}(b_{k\minus{}2}\minus{}1)) \plus{} 1\equal{}$
$ \equal{}b_{k\minus{}2}[(b_{k\minus{}2}\minus{}1)(b_{k\minus{}2}(b_{k\minus{}2}\minus{}1)\plus{}1)] \plus{}1\equal{}....\equal{}b_{k\minus{}w}p\plus{}1$
οπου p ακεραιος και πηλικο της ταυτοτητας και $ p\equal{}(b_{k\minus{}w}(b_{k\minus{}w}\minus{}1) \plus{}1)(....)$
το αποτελεσμα προκυπτει αφου σε καθε προηγουμενη σχεση η παρενθεση της παρακατω μορφης που βρισκεται στην σχεση πηλικου*διαρετη ειναι ισο με(παντα βασισμενοι στην (1))
$ (......)(b_{k\minus{}x} \minus{}1)\equal{}(......)(b_{k\minus{}(x\plus{}1)}(b_{k\minus{}(x\plus{}1)}\minus{}1) \plus{}1\minus{}1)\equal{}$
$ \equal{}(b_{k\minus{}(x\plus{}1)}(b_{k\minus{}(x\plus{}1)}\minus{}1)(......)\equal{}b_{k\minus{}(x\plus{}1)}p_1$  :wink:

Σωστος .Εγω εχω εναν τροπο τηλεσκοπικο επισης συντομο.


limit
function
inequalities
geometry
geometric transformation
reflection
calculus

$f(x)$ is defined on $[a,b]$ and $f(a) < f(b)$.
Let $g(x) = \lim\limits_{h\to 0} \frac{f(x+h)-f(x-h)}{h}$. If $g(x)$ exists for all $x\in(a,b)$, show that there exists $\xi\in(a,b)$ such that $g(\xi) \geq 0$.

This is a strange choice for a wording. Do we have to show that $g$ cannot always be $<0$? (what you wrote translates to "show that $g(x)<0,\ \forall x\in(a,b)\Rightarrow$ a contradiction")

I think, you meant $f(0)=0$ and $f(1)=1$.

And where is your easy proof, Killer?
Should I again do whole work myself?

Moreover, you still didn't write Hadamard-Landau inequality proof. And is it again work for me?

[quote="Killer"]Isn't there any continuity to be specified for the function $f$ ?.Take $f(0)=0, f(1)=1, f(x)=1-x.$We get that $ g(x)=-2$.
For f continous it can be easily proved.[/quote]

yes, it is continuous.

The limit which is considered is(with little deflection) [b]Swartz[/b] or symmetric derivative of first degree.It is
$f'_{s}(x)=\lim_{h{\rightarrow}0}{\frac{f(x+\frac{h}{2})-f(x-\frac{h}{2})}{h}}.$
There are few properties of [b]Swartz[/b] derivative which are very similiar to famous theorems of differential calculus.I'll give a hint and a some time ago I will post the proof of more general theorem which is very similiar to the [b]Lagrange[/b] theorem.It is
---------------------------------------------------------------------------------------------------
Given function $f:[a,b]{\rightarrow}\mathbb{R}$,s.t
[i]a)[/i]$f$ is continious on $[a,b]$.
[i]b)[/i]It has [b]Swartz[/b] derivative everywhere in $(a,b)$.
Then there exist two points $\zeta_{1}$ and $\zeta_{2}$ from interval $(a,b)$,s.t.
$f'_{s}(\zeta_{1})\leq{\frac{f(b)-f(a)}{b-a}}\leq{f'_{s}(\zeta_{2})}$.
---------------------------------------------------------------------------------------------------
Now hint for the original problem.First of all we'll try to prove that for every $[c,d]\subset[a,b]$ s.t. $f(c)<f(d)$ then there exists a point $\zeta$ s.t. $f'_{s}(\zeta)\geq{0}$.For proving it we will assume that $f'_{s}(x)<{0}$
for every $x$ from $[c,d]$.Now the main idea:prove that in such case it can't be that $f(x)<f(b)$ $\forall x\in{[a,b)}$.

Indeed, initial problem is not difficult, but I want [b]Killer[/b] write its solution  ;)

[quote="LevonNurbekian"]
Given function $f:[a,b]{\rightarrow}\mathbb{R}$,s.t
[i]a)[/i]$f$ is continious on $[a,b]$.
[i]b)[/i]It has [b]Swartz[/b] derivative everywhere in $(a,b)$.
Then there exist two points $\zeta_{1}$ and $\zeta_{2}$ from interval $(a,b)$,s.t.
$f'_{s}(\zeta_{1})\leq{\frac{f(b)-f(a)}{b-a}}\leq{f'_{s}(\zeta_{2})}$.
[/quote]
Can you write a proof?

Actually, I have encountered some problems proving that. Maybe I don't see some obvious thing?

Yes,[b]Myth[/b] there is sreie of problems which bring us to the proof of the theorem posted by me.The first problem of that serie is the problem posted by [b]liyi[/b].So I'll prove it first.We can find $[c,d]\subset(a,b)$,s.t. $f(c)<f
(d)$ and $f(x)>f(d)$ $\forall x\in(d,b]$.Assume that $f'_{s}(x)<0$ $\forall x\in[c,d]$.Now consider the set $X=\{x\in[c,d]|f(x)\geq{f(d)}\}$.It is clear that the set $X$ is bounded so there is finite $\inf{X}=m$.First let us prove that $m<d$(it is clear that $m>c+\delta,\delta>0$).If $m=d$ then $f(x)<f(d)$ for all $x\in[c,d)$.Now $f'_{s}(d)<0$   that means that there exists $\delta>0 $s.t. as $|h|<min(\delta,d-c)$ then $f(d-h)>f(d+h)$.So $f(d-h)>f(d)(h>0)$ that contradicts to $m=d$.So $c<m<d$.It is not difficult to show that $f(m)=f(d)$,so $f(m)>f(c)$.So $f(x)<f(m)$ for all $x\in[c,m)$.It can't be right if $f'_{s}(x)<0$(it can be proved as above).So there exists $\zeta\in(a,b)$ s.t. $f'_{s}(\zeta)\geq{0}$.
Now the second problem of the serie is the theorem posted by me with an extra condition $f(a)=f(b)$ i.e. there exist two points $\zeta_{1},\zeta_{1}\in(a,b)$
s.t. $f'_{s}(\zeta_{2})\geq{0}\geq{f'_{s}(\zeta_{2})}$.Since $f$ is continious
then it admits its maximum and minimum:$m$ and $M$.Since $f(a)=f(b)$ there exist two possible cases
a)$f$ is constant.Then the statement is trivially true.
b)$f$ admits at least one of its extremums in the interval $(a,b)$.For example $f(x_{0})=M,x_{0}\in(a,b)$.Then using the first problem there exist points $\zeta_{1}\in(a,x_{0})$ and $\zeta_{2}\in(x_{0},b)$ s.t. $f'_{s}(\zeta_{2})\geq{0}\geq{f'_{s}(\zeta_{2})}$.The rest is clear.
Don't blame me [b]Myth[/b] that I wrote everything in such details. ;)

Hi! Thank you for your proof, but...
[quote="LevonNurbekian"]So $c<m<d$.It is not difficult to show that $f(m)=f(d)$,so $f(m)>f(c)$.So $f(x)<f(m)$ for all $x\in[c,m)$.It can't be right if $f'_{s}(x)<0$(it can be proved as above).[/quote]
I think, here is error. Am I right? :?

Hi [b]Myth[/b].Can you tell me what actually is error?

Your last statement: why if $f(x)<f(m)$ for all $x\in[c,m)$ then it can't be $f'_s(x)<0$ for all $x$.

Oh,I see [b]Myth[/b].Thank you for your remark.
Now look at this one please.
Take a $\delta>0$ s.t. $f(x)>f(a)$ for all $x\in[b-\delta,b]$.Then choose $d$ the point where $f$ admits its minimum on the segment $[b-\delta,b]$.Then $f(d)>f(a)$ and
the set $M=\{x\in[a,d)|f(x)\geq{f(d)}\}$ is nonempty.Now consider $m=\inf{M}$.The problem is that $m$ can be an isolated point for $M$.Remove all isolated points of $M$.The set got after removing denote by $M'$.The worst case is that the set $M'=\{d\}$.We consider that case after.Now consider the case when $M'$ contains not only $d$.Denote $m'=\inf{M'}$.Again it is not difficult to show that $f(m')=f(d)$.Since $f'_{s}(m')<0$(by our assumption) then there exist a $\zeta>0$ s.t. $f(m'-h)>f(m'+h)$ for all $0<h<\zeta$.As $m'\in{M'}$ and $m'$ is a condensation point then there exists a point $t=m'+h'\in{M'}$,s.t. $0<h'<h$(we can also assume that the interval $(m-h,m+h)$ does not contain any of removed isolated points) so $f(m'-h')>f(t)\geq{f(d)}$ but it is a contradiction to the $m'=\inf{M'}$ because $m-h'$ can't be one of the removed isolated points and $m'-h'<m'$.What about the case $M'=\{d\}$.Consider $d'$ very close to $d$($d'<d$ and $d'$ closer to $d$ than at least one of the isolated points of $M$) s.t. $f(a)<f(d')<f(d)$.Then the set $N=\{x\in[a,d')|f(x)\geq{f(d')}\}$.Now the set $N$ is non empty and contains condensation points different from $d'$ because in isolated points of $M$ $f$ admits value $f(d)>f(d')$.Now we can end the consideration as above.How is this one [b]Myth[/b]?
Sorry that I didn't react on your post quickly.I had a credit from topology.
By the way it is a very abstract but at the time very nice discipline.

I think it is Ok. Though I can make some improvements.  :)

I will be glad to see your improvements,because in my solution there are
many different consideration of cases and for reading my solution one needs
to be very patient(it applies at least me).Waiting for your post.  :)

I think, I willn't write any remarks, since your proof is correct.

Thank you, Levon!

[quote="LevonNurbekian"]I will be glad to see your improvements,because in my solution there are
many different consideration of cases and for reading my solution one needs
to be very patient(it applies at least me)[/quote]
Indeed, Levon's solution is correct but quite hard to understand. Here is a slight simplification: Let $c$ be any number strictly between $f(a)$ and $f(b)$. Let $\xi=\inf\{x\in[a,b]\, : \,f\geqslant c\text{ on }[x,b]\}$. By continuity, we have $\xi\in(a,b)$ and $f\geqslant c$ on $[\xi,b]$. If $g(\xi)=f_s'(\xi)<0$, we must have $f(\xi-h)\geqslant f(\xi+h)\geqslant c$ for all sufficiently small $h>0$. But this contradicts minimality of $\xi$.


What is the 123,999th digit after the decimal in the decimal expansion of $ \frac{123}{999}$?

The digits repeat every 3 digits. We quickly see that 123999 is divisible by 3, so our desired answer is $ \boxed{3}$.


search
superior algebra
superior algebra solved

Let $p_1,...,p_n$ be different primes. Prove that $\sqrt(p_1),...,\sqrt(p_n)$ are independent over $\mathbb{Q}$

pbornzstein has already posted a proof for this one. Try to search ...

The more general version was proved by vess in http://www.mathlinks.ro/Forum/viewtopic.php?highlight=irrational+sum&t=27089


logarithms
real analysis
real analysis unsolved

Prove that the sequence $\ds a_{n}=\frac{3}{2}\cdot\frac{5}{4}\cdot\frac{9}{8}\cdot\ldots\cdot\frac{2^{n}+1}{2^{n}}$ converges to some number $a\in(\frac{3}{2}\sqrt[4]{e},\frac{3}{2}\sqrt{e}).$


Prove that $ 3^{4n\plus{}2}\plus{}2^{3n\plus{}6}$ is divisible by $ 73$ for, of course, sufficiently large $ n$.

[quote]You simplified it incorrectly. Check it again.[/quote]
 :(   I expected something like that.

$ 3^{4n \plus{} 2} \plus{} 2^{3n \plus{} 6} \equal{} 9^{2n \plus{} 1} \plus{} 8^{n \plus{} 2} \equal{} 9(81^n) \plus{} 64(8^n)$.

Taking this mod 73, we get that

$ 9(81^n) \plus{} 64(8^n) \equiv 9(8^n) \plus{} 64(8^n)$
$ \equiv 9(8^n) \plus{} (\minus{}9)(8^n)$
$ \equiv 0 \mbox{ mod } 73$.

By the way, "suffficiently large" could have been replaced with integer.

[quote="Zhero"]
By the way, "suffficiently large" could have been replaced with integer.[/quote]
Just to nit-pick, natural number (positive integer).

[hide="Hint"]Cast it as a sequence with a straightforward recurrence. This is an effective technique for many, many problems.[/hide]

[hide="Solution"]The sequence $ a_n \equal{} 9 \cdot 81^n \plus{} 64 \cdot 8^n$ satisfies the recurrence $ a_{n \plus{} 1} \equal{} (81 \plus{} 8) a_n \plus{} (81 \cdot 8) a_{n \minus{} 1}$, thus $ 73 | a_{n \minus{} 1}, \; a_n \implies 73 | a_{n \plus{} 1}$.

It suffices to establish any two consecutive base cases, preferably lower. Here, $ n \equal{} 0,1$ work just fine. If you know how to justify it, you can use $ \minus{} 1,0$ for lighter computation.[/hide]


Are you sure that the left side is not a cyclic sum???

Misha

If the left side is really cyclic then we have:

$\sum_{cycl}\frac{x^3+1}{y^3}\geq\sum\frac{x^3+1}{x^3}=3+\sum\frac{1}{x^3}\geq3+\frac{3}{xyz}\geq6=\frac{6(x+y+z)^2}{9}\geq\frac{6\cdot3(xy+yz+zx)}{9}=2(xy+yz+zx)$.

Misha

Indeed easy.
No [b]Misha123[/b] it is not necessary left side to be cyclic.
Just note that
$(\frac{x}{z})^{3}+(\frac{z}{x})^{3}+1\geq3$(from Am-Gm) and
$\frac{1}{x^{3}}+\frac{1}{y^{3}}+\frac{1}{z^{3}}\geq\frac{1}{9}\*{({\frac{1}{x}+\frac{1}{y}+\frac{1}{z}})}^{3}$,
but $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq{\frac{9}{x+y+z}}=3$.
On the other hand $2(xy+yz+zx)\leq\frac{2}{3}\*(x+y+z)^{2}=6$.So we are done.


linear algebra
matrix
probability
invariant
expected value
probability and stats

A white rat is put into the maze at time 0. At each instant n = 1, 2, . . ., it moves to a
new compartment, or stays in the same compartment, with probabilities 1/3 and 2/3 respectively. If
it moves, and there are k ways to leave the compartment, it chooses one of these at random. If the rat
starts out in compartment 8, find the expected time at which it returns to compartment 8.

I didn't include the maze's picture, but here is the transition matrix I composed based on the compartments:
*zer is 0 so that they line up evenly
2/3 1/3 zer zer zer zer zer zer zer
1/6 2/3 1/6 zer zer zer zer zer zer
zer 1/6 2/3 zer zer 1/6 zer zer zer
zer zer zer 2/3 zer zer 1/3 zer zer
zer zer zer zer 2/3 zer zer 1/3 zer
zer zer 1/3 zer zer 2/3 zer zer zer
zer zer zer 1/3 zer zer 2/3 zer zer
zer zer zer zer 1/9 zer 1/9 2/3 1/9
zer zer zer zer zer zer zer 1/3 2/3

How do I find the expected value above based on this?

How do you find the probability that we're in a specific spot at time $ n$? Take the $ n$th power of the matrix. In that spot for the first time? Now we have to adjust the transitions so that that spot is a black hole; zero out the corresponding (row?/column?). Since it's also our starting point, we should start the whole procedure after one step without the zeroed (row?/column?).
That's how to calculate the probability we reach this compartment at time $ n$. Given these probabilities, can you find the expected time?

No, I'm not really sure what to do. I appreciate the explanation though. Can't I do it by counting the number of non-zero probabilities in each row?

The sizes of the probabilities matter. "Nonzero" is inadequate information.

I believe I need to compute the invariant distribution.

If I restrict the chain to the closed class {7,8,9}, then this restricted chain is ergodic and has transition matrix
zer zer zer 1/3 zer zer 2/3 zer zer
zer zer zer zer 1/9 zer 1/9 2/3 1/9
zer zer zer zer zer zer zer 1/3 2/3
Then, I think the unique invariant distribution is (1/4, 1/2, 1/4), which implies that the mean recurrence time of state 8 is 2. Is this correct?


geometry
AMC
AIME
MATHCOUNTS
AwesomeMath
summer program

Wat should i do??
i dont no...
Awesome Math sounds really fun and i have like 5 friends that got in too
but
Geometry... i will skip a grade in school and that is kind of important for college
so any information on student life and any input from Alumni would be wanted!!
THX
ILIKEPI

OMG. go to amsp! ;]
seriously you'll have an amazing time. and learn so much more stuff. ><
geometry [school geometry] can be pretty dull [at least at my school it was] . and if you really wanted. learning geometry by yourself isnt too hard..
but amsp really gives you the opportunity to excel at something else besides school math. it really focuses on prepping you to do better on national tests like amc, aime, usamo. etc. or mathcounts if you're still in middle school . and you get to meet so many cool people that are talented at math. the math you learn at amsp is so much better & more fun than regular school math. [my opinion]
i know college is important. but truthfully. skipping geometry isnt going to make that big of a difference. high school /college apps have so much more to it then just saying im x years ahead in math. b/c other ppl have that as well. amsp gives you the chance to say ' hey i did ___ well at amc/aime/usamo/mathcounts' which is a lot more prestigious. if thats what you're going for.
but i feel like the experience of amsp is just so much more worth while. ;] my two bits.

life at amsp is pretty sweet. there's not many rules, everything's pretty sweet. you get to hang out a lot. socialize etc. you should check out other topics under the awesomemath section cuz student life is mentioned a lot :D

If you have the choice between a math camp and skipping a course in school, you should definitely take the math camp.

The only bad thing is you have to sit through a year of Geometry class, which isn't that fun.

AWESOMEMATH looks good for college as well!

yea.. I sat through a year of geo (this year).. very boring class of the day.. I almost find it bad when I dont do the homework but still get 100+ on the tests...
Now that I think about it, the only reason to skip Geometry is to take another course (non-math) and only have 3 years of math instead of 4.. since the only really important ones are Alg II,  and AP calc BC pt1 and pt 2. Then after that, I dont really see what else you'd go for.

If you discover that you are just 1 "slot" away from being able to "complete" a certain elective then you should go skip geo (and dont go to amsp :() (like you want to reach ap laws and politics or something, but you need 3 years of learning beforehand, and you dont have enough "room" for the course).

AMSP would be a better choice. I'm just saying the only reason you would want to skip geometry is if you have another course you wish to take after your schools math courses - getting an A+(or A) in geometry honors is considered very hard by like most people. If you go to AMSP, you can take a geometry course or something, and all of it would be easy anyway. If you don't take geometry, I assume you are smart enough to just breeze through that class anyway.

well, in my school district , after ap calc bc. u can move onto college math.... but school districts differ greatly.

but i still HIGHLY recommend amsp over geo.

well in my school after calc bc we have multi. and then i'm going to do stats. lol... epic fail =D... Then I guess I have no math for senior year o.o? Maybe "Math Studies" lol =D. should make a class just for me... Ohh I know. I'll take discrete.

To the OP: You can easily skip out of most courses by talking to the teacher who oversees that subject. Just take the final, pwn it, and bam, skipped out.

AMP>school

Make an agreement with your school to take AOPS Geometry as your class and take part in AMSP.


Provided a convex equilateral pentagon. On every side of the pentagon We construct equilateral triangles which run through the interior of the pentagon. Prove that at least one of the triangles does not protrude the pentagon's boundary.

[b]Fact 1.[/b] Pentagon $ABCDE$ contains at most 1 angle $<60^\circ$.
Suppose contrary. It is clear that two angles $<60^\circ$ can't be two neighbour vertices. Consider Figure 1: $B,E<60^\circ$. Then $AC,AD<1$. Construct two points $C'$ and $D'$ s.t. $AC,AD<AC'=AD'<1$. Since $\angle BAC,\angle EAD>60^\circ$ we conclude $\angle CAD<60^\circ$, so $C'D'<1$. So $CD$ lies inside triangle $AC'D'$, whose sides $<1$, so $CD<1$. Contradiction.

[b]Fact 2.[/b] Pentagon $ABCDE$ contains at most 2 angles $>120^\circ$.
Suppose contary, then we can find two neighbour angles $>120^\circ$. Consider Figure 2: $A,E>120^\circ$. It immediately follows that projection $BD$ to line $AE$ is greater than 2, since $B'D'>1+\cos 60^\circ+\cos 60^\circ$. Contradiction, since $B'D'\leq BD\leq 2$.

Suppose pentagon $ABCDE$ doesn't contain angles, which is less than $60^\circ$. Then, due to [b]Fact 2[/b], we can find two neighbour angles $\leq 120^\circ$. Consider Figure 2. Suppose $A$ and $E$ such angles. Than $BB',DD'>\sin 60^\circ$. It follows that all points of lines $BD$ are remote from $AE$ by at least $\sin 60^\circ$. It means, that regular triangle constructed on side $AE$ lies inside pentagon.

Suppose pentagon $ABCDE$ contains angle $<60^\circ$. Consider Figure 3: $A<60^\circ$. We will show that $C,D\leq 120^circ$. Indeed, suppose contary, i.e. $C>120^\circ$. Than from triangle $EDC$ we obtain $EC>\sqrt{3}$, on the other hand, it is necessarily $\angle EBC<120^\circ$, so from triangle $EBC$ we obtain $EC<\sqrt 3$ (we use $EB<1$).
From [b]Fact 1[/b] we conclude $C,D\geq 60^\circ$. So we can state that regular triangle constructed on $CD$ lies inside pentagon.


If A, B and C are positive integers such that AsqrtB/C = 9/2sqrt3 is the value of A + B + C given that A and C have no common prime factors, and B has no perfect-square factors other than 1?

Well, to me, the second fraction looks a bit suspicious.  So to try something, let's divide the numerator and denominator by $ \sqrt3$.  Then, you get the fraction $ \frac{3\sqrt3}{2}$, which satisfys the above needs.  Thus, the answer is $ 3\plus{}3\plus{}2$, or $ \boxed{8}$.

I call it smart guess and check. :wink:

Wait, I don't see the guess and check in this problem.

Yeah, it's pretty much the most obvious step you can possibly do: rationalize the denominator, and there's the answer...


function
geometry
geometric transformation
reflection
complex analysis
complex analysis unsolved

Find all entire functions $ f: \mathbb{C} \rightarrow \mathbb{C}$ which satisfy $ f(\partial D)\subseteq \partial D$
where $ D$ is an arbitrary square on the plane and  $ \partial D$ its boundary

[quote="Kalle"]Well, what does this mean? If it's just for one fixed $ D$ it seems like the solutions are complicated. If it's for all $ D$ then $ f$ increases linearly and not too much happens[/quote]
It's just for one fixed square

Do you have a solution for this problem?

It is actually not that hard. Just use the Schwarz reflection principle a few times ;)


Prove or disprove:
If p is a prime, then so is $2^{p}-1$.

[b]Salvage:[/b]  If $2^{p}-1$ is a prime, then so is $p$.

[hide="Hint"] The contrapositive:  if $a$ is composite, then so is $2^{a}-1$. [/hide]

[hide="Solution for t0rajir0u"]Suppose $p$ is not prime. Then $p = ab$ for two integers $a, b > 1$. $2^{p}-1 = 2^{p-1}+2^{p-2}+\ldots+2+1$
$= (2^{a-1}+2^{a-2}+\ldots+2+1)(2^{a(b-1)}+2^{a(b-2)}+\ldots+2^{a}+1)$. Since both $a$ and $b$ are greater than 1, both factors are positive integers greater than 1.[/hide]

I think the hypothesis is wrong some ways...
MellowMellon: You just prove that if p is a prime then 2^p - 1 is a prime, not if 2^p - 1 is a prime then p is a prime

No, Mellow proved that if p is NOT a prime, then 2^p-1 is NOT a prime, which is correct.

Sorry about that.
But I'm sure there's a mistake in his solution, because 2^11 - 1 is not a prime.

I think your missing what mellowmelon is proving.  He is proving that is $p$ is NOT prime, then neither is $2^{p}-1$.  Therefore the only way that $2^{p}-1$ can ever be prime, is if $p$ is prime.  Obviously this doesn't mean $2^{p}-1$ is prime when $p$ is, but $p$ being prime is a necessary condition.  There is no known formula or equation that predicts primes or has only prime outputs, so any question like this is really asking to disprove the statement.

It's like saying all people on AOPS are nerds (which is true) and then concluding that all people not on AOPS are not nerds (which is false).

[quote="miyomiyo"]So basically he proved that if $p$ is prime, then the only integer solutions $x$ to $\sqrt[x]{p+1}= x$ are prime?[/quote]

... What?  I thought my original wording was clear enough:

$2^{p}-1\text{ is prime}\Rightarrow \text{p is prime}$

Now:

If $2^{n}+1$ is prime, then $n$ is a power of $2$.

[hide="Hint1"]$x^{mn}=(x^{m})^{n}$[/hide]
[hide="Hint2"]For odd $n$ we have $x^{n}+1=(x+1)(x^{n-1}-x^{n-2}+... )$[/hide]

The only thing is remaining: find all such primes.  :lol:

Well let $p=2^{2^{n}}+1$ be any Fermats prime. Prove:

1) if $n\ge 1$ then $3$ is primitive root $\pmod{p}$.
2) if $n\ge 2$ then $5$ is primitive root $\pmod{p}$.
3) if $n\ge 2$ then $10$ is primitive root $\pmod{p}$.

Sorry for my english: I mean $a$ is primitive root $\pmod{p}$ if $a$ generates multiplicative group $Z_{p}^{*}$


search
combinatorics unsolved
combinatorics

49 students solve a set of 3 problems. Each problem is marked from 0 to 7. Show that there are two students A and B such that A scores at least as many as B for each problem.

Note: see also IMO2001_3

Nice problem. Any ideas ?

[Mod edit: this problem has also been posted at http://www.artofproblemsolving.com/Forum/viewtopic.php?p=361321 but no discussion was made.]

Is "A scores at least as many as B for each problem" in other words " A majorize B"?

I gave it to my supervisor one year ago, and as far as I remember he solved the problem, but I didn't ask him for details  :(

[quote="grobber"]I kind of remember this being posted again and being solved on the forum, but I have no idea by whom, or how the topic was named.. :([/quote]
Really? I didn't see it.

At least search for "49 3 0 7" or "49 7" gives nothing  :(

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16461[/url]

Haven't a clue about the apparent double post, before anyone asks. I admit, it is slightly unwieldy.

[quote="Fiachra"]Haven't a clue about the apparent double post, before anyone asks. I admit, it is slightly unwieldy.[/quote]
I am not sure that I understand this.

I guess you can remove double post in the mentioned thread.

I have no idea why "search" didn't show me it ?????

The solution, or the two lines you quoted?

If the latter, what I mean is:
a) My post on that topic seems to have been copied; I don't know why this is.
b) My solution is somewhat long and complicated.

I were not able to translate what you said (quoted text).


I don't think this one is as easy as the other ones, but I may be wrong.

Note that, if you suppose that the b_i's are nonnegative, then the inequality may be generalized to k-th power instead of just squares.

A hint : Use Abel's transformation...

Pierre.

I have this solution.
After summation by a_k-a_k+1 we have

 (a_k-a_k+1) \sum (i=1..k)a_i=(a_k-a_k+1) \sum (i=1..k)b_i

for k=1..n  Summing both sides for k=1..n we get

 \sum (i=1..n)(a_i) 2  \leq  \sum (i=1..n)a_ib_i

By Cauchy we obtain the inequality

 ( \sum (a_i) 2 ) 2  \leq ( \sum a_ib_i) 2  \leq ( \sum (a_i) 2 )( \sum (b_i) 2 ) 2


f(x)=1,xeQ
f(x)=0,x is not eQ
 it's easy to understand that this function does not have limit as x tends to 0,but how we can prove it with mathematical formulas.


AMC
AIME
floor function
calculus
search
analytic geometry
graphing lines

Find the smallest positive integer n for which there are no integer solutions to $\displaystyle \left \lfloor \frac{2002}{x} \right \rfloor = n$

This is a rare instance when even a basic understanding of calculus makes life just a bit easier on the AIME :roll: Of course, calculus is not necessary, just helpful.

[hide]Let's try a few values of $n$ first.

$\frac{2002}{1} = 2002 \Rightarrow x = 2002$
$\frac{2002}{x} = 1 R 0$

$\frac{2002}{2} = 1001 \Rightarrow x = 1001$
$\frac{2002}{x} = 2 R 0$

$\frac{2002}{3} = 667 R 1 \Rightarrow x = 667$
$\frac{2002}{x} = 3 R 1$

$\frac{2002}{4} = 500 R 2 \Rightarrow x = 500$
$\frac{2002}{x} = 4 R 2$

Obviously, the remainder will be the same in both divisions, unless it is greater than the quotient in the first division. So, we wish to find an $n$ such that the remainder when 2002 is divided by $n$ is greater than the quotient. Clearly, the remainder must be less than the divisor, so the divisor must be greater than the quotient. We can thus start our search from $\sqrt{2002}$, which is between 44 and 45, so we start with 45.

$\frac{2002}{45} = 44 R 22$

$\frac{2002}{46} = 43 R 24$

$\frac{2002}{47} = 42 R 28$

$\frac{2002}{48} = 41 R 34$

$\frac{2002}{49} = 40 R 42$

So, $n = 49$ is our answer.[/hide]

I don't see how calculus makes this easier than it is.

We know immediately that:

$\frac{2002}{n} - \frac{2002}{n+1} < 1$ and it follows that
$n(n+1) > 2002$, so $n \ge 45$.

Trying 45, 46, 47, 48, 49, 50, we see that we get 44, 43, 42, 41, 40, 40. Though it's not necessary to try all of those. Only enough to see that least $n$ is $n = 49$.

Calculus...?

I remember this problem really well since it was on my first AIME in 8th grade and helped push me into USAMO. And believe me, I didn't know any calculus back then. Basically I realized the answer was around $\sqrt{2002}$, since $\displaystyle \frac{2002}{x}$ starts to get big pretty fast as $x$ gets small.

Plug and chug ... psh on the fancy stuff. Those were the good days.

[quote="behemoth571"]I even told you at school :P...

Let $y = \frac{2002}{x} = 2002x^{-1}$

We are looking for values of x where the slope of y is less than -1, such that it is possible for the curve to "skip" an integer.

$\frac{dy}{dx} = -2002x^{-2} < -1$
$2002>x^2$

So x < 45. Trying decreasing values for x, we try $\left \lfloor \frac{2002}{41} \right \rfloor$ and $\left \lfloor \frac{2002}{40} \right \rfloor$ and see that they respectively equal 48 and 50. So, $n=49$.[/quote]

 :roll:

Just playing with your head. But still, after Calculus, it's still plug and chug, same as if you just saw that originally as the inequality.

Calculus can be used, but it is simple enough to find the right area to test on this problem that it just takes longer.

[quote="behemoth571"]We are looking for values of x where the slope of y is less than -1[/quote]
Why do we want the slope of y to be less than -1?

We want the slope to be less than -1, since this is the only way that it can decrease by more than one when x increases by 1.


algebra
polynomial
superior algebra
superior algebra theorems

It is well known that 2nd order recursive series R_n +b*R_(n-1) +c*R_(n-2) can be usually solved by us of the auxiliary equation x^2 +bx + c = 0 having roots r_1 and r_2 using the formula R_n = A*(r_1)^n + B*(r_2)^n, but I find this is not always the case when b^2= 4c.  How do you solve those recursive series when the auxiliary equation and formula does not work?

example

R_(0) = 0   R_(1) = m   R_(n) = 2R_(n-1) - R_(n-2) - 2

This can be put into the form

R'_(n) - 2*R'_(n-1) +R'_(n-1) = 0 by adding -1/2 to each term of R_(n)

But then we have R'_n = A*1^n + B*1^n  = A+B which is useless.

Yes if it has $ x_1\equal{}x_2$ then
$ x_{n}\equal{}(A\plus{}Bn)x_1^n\plus{}f(n)$

Is this the right place for this post? I think it should be moved to pre-olympiad forum or maybe algebra forum.
Anyway, think about how that "equation" is derived: we propose that it works and then show it works and that it is the only one that works based on how many "points" (or variables) it takes to define the form uniquely. When roots collapse in on each other, you still have the same basic equation but since number of variables to play with is decreased, we have to account for this somewhere. The solution here is to just consider polynomials which cancel out, etc. The logic behind it is really easy, I suggest you look into the proof of the general recursive series auxiliary equation "idea" (no, I won't call it a theorem)

[quote="ramsey2879"]
$ R_0 \equal{} 0,\ R_1 \equal{} m,\  R_n \equal{} 2R_{n \minus{} 1} \minus{} R_{n \minus{} 2} \minus{} 2$[/quote]

$ R_n \equal{} 2R_{n \minus{} 1} \minus{} R_{n \minus{} 2} \minus{} 2\Longleftrightarrow R_{n} \minus{} R_{n \minus{} 1} \minus{} (R_{n \minus{} 1} \minus{} R_{n \minus{} 2}) \equal{} 2$

$ \therefore R_n \minus{} R_{n \minus{} 1} \equal{} (R_1 \minus{} R_0) \plus{} 2(n \minus{} 1) \equal{} m \plus{} 2(n \minus{} 1)\ (n\geq 1).$

$ \therefore \boxed{R_n \equal{} R_1 \plus{} \sum_{k \equal{} 1}^{n \minus{} 1} (R_{k \plus{} 1} \minus{} R_k)\ (n\geq 2)}$

$ \equal{} m \plus{} \sum_{k \equal{} 1}^{n \minus{} 1} (m \plus{} 2k)$

$ \equal{} m \plus{} m(n \minus{} 1) \plus{} 2\cdot \frac {(n \minus{} 1)n}{2}$

$ \equal{} n(n \plus{} m \minus{} 1)\ (n\geq 0).$


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Find a,b,c are three positive reals satisfying two conditions follow:
$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {9}{c} \equal{} 3$

$ a \plus{} b \plus{} c \leq 12$

thanks a lot

I think it must is $ \frac{1}{a}\plus{}\frac{4}{b}\plus{}\frac{9}{c}\equal{}3$

hello, i have found $ a\equal{}2,b\equal{}4,c\equal{}6$ , with these numbers are both conditions fulfilled,
$ \frac{1}{2}\plus{}\frac{4}{4}\plus{}\frac{9}{6}\equal{}3$ and $ 2\plus{}4\plus{}6\le 12$.
Sonnhard.

Let $a,b,c >0 $ and $ \frac {1}{a} + \frac {1}{b} + \frac {9}{c} = 3.$ Prove that$$ a + b + c \geq \frac {25}{3}$$
Equality holds when $a=b=\frac{5}{3},c=5.$

Let $a,b,c >0 $ and $ \frac {1}{a} + \frac {4}{b} + \frac {9}{c} = 3.$ Prove that$$ a + b + c \geq 12$$
Equality holds when $a=2,b=4,c=6.$

[quote name="hungnsl" url="/community/p1414463"]
Find a,b,c are three positive reals satisfying two conditions follow:
$ \frac {1}{a} \plus{} \frac {4}{b} \plus{} \frac {9}{c} \equal{} 3$

$ a \plus{} b \plus{} c \leq 12$

thanks a lot
[/quote]
By Cauchy-Schwarz inequality, we have
$$3=\frac{1^2}{a}+\frac{2^2}{b}+\frac{3^2}{c} \geqslant \frac{6^2}{a+b+c} \Rightarrow 12\leqslant a+b+c \le 12$$
 Hence $\boxed{a=2, b=4,c=6}$


inequalities trigonometry LaTeX symmetry conics parabola calculus

If $ a,b$ are positive real numbers such that $ a\plus{}b\equal{}1$ then the minimum value of $ ab\plus{}\frac{1}{ab}$ is
a)$ 2$
b)$ 4$
c)$ 6$
d)none of these

[hide]According to the source option  $ a$ is not the correct answer[/hide]

Of course it's not (a): what's the equality condition for AM-GM?  (Note that in our case we have $ ab < 1 < \frac {1}{ab}$.)  In fact, it's easy to see (e.g. by AM-GM and some manipulation) that $ \frac {1}{ab} \geq 4$ all by itself!  (And this is a big hint for the answer, since we're given quite a limited set of answer choices.)

Answer will be $ \frac{17}{4}$ so correct option $ d$

Through AM-GM inequality we see:
a+b/2>=sqrtab.
so 1>=2sqrtab
1>=4ab
1/4>=ab
1/ab is minimized when ab is maximized so the answer is 1/4+4=17/4 which concludes that the answer is D.

I could get a convincing (that is to say, no hand-waving) answer only by appealing to calculus. If you let $ a \equal{} \sin^2 \theta$ and $ b \equal{} \cos^2 \theta$, then

$ f(\theta) \equal{} \frac {\sin^2 2 \theta}{4} \plus{} 4 \csc^2 \theta$ is minimised when $ \theta \equal{} \frac {\pi}{4}$

i.e. $ a \equal{} b \equal{} \frac {1}{2}$

So the min is $ \frac {17}{4}$

[quote="varunrocks"]1/ab is minimized when ab is maximized so the answer is 1/4+4=17/4[/quote]  This does not follow.  Also, please learn some [[LaTeX]] if you want people to be able to decipher your posts.

To get that the answer is (d) is easy, as I noted: we have that $ \frac{1}{ab} \geq 4$ and $ ab > 0$, so $ ab \plus{} \frac{1}{ab} > 4$, but we can achieve values less than 6.

I agree with hsbhatt that actually proving that $ \frac{17}{4}$ is the minimum seems trickier.  It reminds me of someone who wandered around the inequalities section for a while (before being banned) who would just answer each question by saying, "by symmetry, it's trivial that ...."   :rotfl:

why isn't 1/ab minimized when ab is maximized?

It is, but that doesn't imply that the answer is $ \frac{17}{4}$.  (The minimum of the sum of two quantities need not occur at a minimum value for either quantity.)

Well... this might be a bit of an unorthodox solution but here it goes... We want to find the minimum value of $ ab\plus{}\frac{1}{ab}$. We know (this is trivial to prove) $ f(x)\equal{}x\plus{}\frac{1}{x}$ is monotonically increasing as $ x$ drifts from $ 1$. Then, since $ ab<1$, we want to find the maximum possible value of $ ab$. Let $ b\equal{}1\minus{}a$. Then we want to find the maximum of the parabola $ f(a)\equal{}a(1\minus{}a)\equal{}\minus{}a^{2}\plus{}a$. This is when $ a\equal{}\frac{1}{2}$ and, consequently, $ b\equal{}\frac{1}{2}$. Then it follows that the maximum value of $ ab$ is $ \frac{1}{4}$ and the minimum value of $ ab\plus{}\frac{1}{ab}$ is $ \frac{17}{4}$. Answer is d.

A non-calculus approach that occurred to me later :

$ f\left(\frac {1}{2}, \frac {1}{2} \right) \minus{} f(a,b) \equal{} f\left(\frac {a \plus{} b}{2}, \frac {a \plus{} b}{2} \right) \minus{} f(a,b)$

$ \equal{} \frac {(a \plus{} b)^2}{4} \minus{} ab \plus{} \frac {1}{\frac {(a \plus{} b)^2}{4}} \minus{} \frac {1}{ab} \equal{} \frac {(a \minus{} b)^2}{4} \left(1 \minus{} \frac {4}{ab} \right) \le 0 \ (\because \frac{1}{ab} \ge 4)$


ratio
algebra
system of equations
number theory open
number theory

Let $ x \plus{} y \equal{} 1$, $ x^{3} \plus{} y^{3} \equal{} 4$. Let also $ S(n) \equal{} x^{n} \plus{} y^{n}$. Is it true that if $ n$ is not divisible by $ 3$ $ S(n)$ is prime?
I noticed that more than year ago but I don't know if it is true.

$ S(n)$ must be Lucas numbers .

So see http://www.research.att.com/~njas/sequences/A001606.

Solve the system of equations: $ x\plus{}y\equal{}1$, and $ x^3\plus{}y^3\equal{}4$ and you get: $ x\equal{}\phi$, $ y\equal{}\frac{\minus{}1}{\phi}$. So, $ x^n\plus{}y^n$ doesn't need to be an prime or integer perhapse.

Show example where $ S(n)$ is not integer. Then I will agree with you.

[quote="SAPOSTO"]If 0, 2, 4, 5, 7, 8, 11, 13, 16, 17, 19, 31, 37, 41, 47, 53, 61, 71, 79, 113, 313, 353, 503, 613, 617, 863, 1097, 1361, 4787, 4793, 5851, 7741, 8467, 10691, 12251, 13963, 14449, 19469, 35449, 36779, 44507, 51169, 56003, 81671, 89849, 94823, 140057, 148091, 159521, 183089, 193201, 202667, 344293, 387433, 443609, 532277, 574219 are the first Lucas numbers $ S(n)$ can't be Lucas numbers because for example $ S(17) \equal{} 3571$ which as I see from the numbers above isn't a Lucas number.[/quote]

There has been a misunderstanding: this sequence isn't Lucas sequence: it's the sequence of indices of Lucas primes (meaning L(0), L(2), L(4)... are prime). Your sequence is definitely Lucas sequence: solving the system gives $ x\equal{}\frac{1}{2} \left(1\minus{}\sqrt{5}\right),y\equal{}\frac{1}{2} \left(1\plus{}\sqrt{5}\right)$, and it's well known that if you resolve the Lucas recurrence, the solution is

$ \left(\frac{1}{2} \left(1\plus{}\sqrt{5}\right)\right){}^n\plus{}\left(\frac{1}{2} \left(1\minus{}\sqrt{5}\right)\right){}^n$

Thank you for your explanation [b]hsiljak[/b]. I don't know the Lucas sequence.

I didn't knew it also. By the way my $ \phi$ is [b]golden ratio[/b]($ 1.618...$) which is equal to $ \frac{\sqrt{5}\plus{}1}{2}$.


calculus
derivative
function
calculus computations

1, Let $ f$ is a continuous function in $ [a,b]$ and $ f$ have derivation on $ (a,b)$ such that : $ f(a) \equal{} f(b)\equal{}0$ . Show that with all real number ( $ \alpha$ ) we always have $ x\in (a,b)$ such that : $ \alpha.f(x) \plus{} f'(x) \equal{} 0$

2, Let $ f$ is a continuous function in $ [a,b]$ with $ a > 0$ and $ f$ have derivation on $ (a,b)$ . Show that we have $ x_{1} \in (a,b)$ such that : $ \frac {bf(a) \minus{} af(b)}{b \minus{} a} \equal{} f(x_{1}) \minus{} x_{1}f'(x_{1})$

[quote="Harry Potter"]1, Let $ f$ is a continuous function in $ [a,b]$ and $ f$ have derivation on $ (a,b)$ such that : $ f(a) \equal{} f(b)$ . Show that with all real number ( $ \alpha$ ) we always have $ x\in (a,b)$ such that : $ \alpha.f(x) \plus{} f'(x) \equal{} 0$[/quote]

Wrong : choose $ f(x)\equal{}1$ $ \forall x\in[a,b]$ and obviously, for all $ \alpha\ne 0$, we cant find $ x\in (a,b)$ such that : $ \alpha.f(x) \plus{} f'(x) \equal{} 0$

[quote="pco"][quote="Harry Potter"]1, Let $ f$ is a continuous function in $ [a,b]$ and $ f$ have derivation on $ (a,b)$ such that : $ f(a) \equal{} f(b)$ . Show that with all real number ( $ \alpha$ ) we always have $ x\in (a,b)$ such that : $ \alpha.f(x) \plus{} f'(x) \equal{} 0$[/quote]

Wrong : choose $ f(x) \equal{} 1$ $ \forall x\in[a,b]$ and obviously, for all $ \alpha\ne 0$, we cant find $ x\in (a,b)$ such that : $ \alpha.f(x) \plus{} f'(x) \equal{} 0$[/quote]

i am sorry , i've edited the post , we must add the condition : $ f(a)\equal{}f(b)\equal{}0$ .  :o

[quote="Harry Potter"]1, Let $ f$ is a continuous function in $ [a,b]$ and $ f$ have derivation on $ (a,b)$ such that : $ f(a) \equal{} f(b) \equal{} 0$ . Show that with all real number ( $ \alpha$ ) we always have $ x\in (a,b)$ such that : $ \alpha.f(x) \plus{} f'(x) \equal{} 0$[/quote]

Consider $ g(x) \equal{} f(x)e^{\alpha x}$ : $ g(x)$ is $ C_1$ and $ g(a) \equal{} g(b)$ so $ \exists x_1\in(a,b)$ such that $ g'(x_1) \equal{} 0$ $ \iff$ $ (f'(x_1) \plus{} \alpha f(x_1))e^{\alpha x_1} \equal{} 0$

Hence the result.


Ok, this the the correct version, I am sorry for the mistakes (it was my first post  :D )

Prove for every real numbers a,b and c:

a/(b+c) + b/(c+a) + c/(a+b)  \geq  3/2

And prove for every real numbers a,b,c and d:

a/(b+c) + b/(c+d) + c/(d+a) + d/(a+b)  \geq  2

It's better to edit your previous post, instead of making a new post.

for the first one... it is true only for positive numbers (try a=-1,b=2,c=-1)
1)denote with x=a+b,y=b+c,z=c+a
now the inequality becomes (-x+y+z)/x+(x-y+z)/y+(x+y-z)/z>=3 <=>
x/y+y/x+x/z+z/x+y/z+z/x>=6 which is true by AM>=GM.
2)with C.B.S.
(\sum a/(b+c)) (\sum a(b+c)) >=(\sum a)^2
results \sum a/(b+c) >= (\sum a^2 +2\sum ab)/(2\sum ab)=1+1/2\sum a^2/\sum ab>=1+1/2=3/2.

[quote="dEUS"]Ok, this the the correct version, I am sorry for the mistakes (it was my first post  :D )

Prove for every real numbers a,b and c:

a/(b+c) + b/(c+a) + c/(a+b)  \geq  3/2 [/quote]

Hey, does anybody know what was the name of the mathematician who proved it first? Actually, I know it`s sth like Nesbit/Nesbitte, but I couldn`t chech it anywhere. I need his name, don`t ask why.

What`s the idea of this topic? To put the most used inequalities at one place or sth else?

Sorry about bad English.

[quote="Valentin Vornicu"][quote="idioteque"] I need his name, don`t ask why.

What`s the idea of this topic? To put the most used inequalities at one place or sth else?

Sorry about bad English.[/quote] Your English seems prety good :) :P I know we are not supposed to ask you why do you need his name, but I just can't help myself :P:D so why do you need his name? :)[/quote]

I`m sending a solution of one inequality to one magazine, and it mustn`t be over one page (totally stupid - isn`t it?). So, I have to know his name, it`s in the last line.

Thanks, Lagrangia.

Maybe you should put here Cauchy-Schwartz-Buniakowski, Bernoulli, inequality between the middles (I don`t know how is it called in English)...

Ok, this is the Nesbitt inequality and has a nice and easy solution by cauchy as follows:
Sum 3 to both sides and you get:\[\begin{array}{{c}{cc}(a+b+c)\left(\frac1{b+c}+\frac1{c+a}+\frac1{a+b}\right)\geq\frac92}\\Rightarrow((b+c)+(c+a)+(a+b))\left(\frac1{b+c}+\frac1{c+a}+\frac1{a+b}\right)\geq9\end{array}\]
And the last is true by Trusty Cauchy :D

[quote="Valentin Vornicu"]inequalities between middles ... hmmz ... maybe you mean the mean inequalities?
arithmetic mean $ \frac { a_1+a_2+\cdots + a_n } n $, geometric mean $ \sqrt[n]{a_1a_2\cdots a_n}$, harmonic mean, etc. ? :) (also known as AM-GM inequalities).[/quote]

Yes, I mean that, but all in one.

I would write it if I knew how...

Looks like I`m the only beginner here.  :(

you mean you want this:

H- harmonic mean $=\frac{n}{\frac{1}{a_{1}}+\frac{1}{a_{2}} + \cdots +\frac{1}{a_{n}}}$
G- geometric mean $=\sqrt[n]{a_{1}a_{2} \cdots a_{n}}$
A - arithmetic mean $=\frac{a_{1}+a_{2}+ \cdots+a_{n}}{n}$
Q- quadratic mean $=\sqrt{ \frac{a_{1}^{2}+a_{2}^{2}+ \cdots + a_{n}^{2}}{n}}$

The relation is well known:
$min(a_{1},a_{2},\cdots,a_{n}) \leq H \leq G \leq A \leq Q \leq max(a_{1},a_{2},\cdots,a_{n})$.

[quote="idioteque"] Looks like I`m the only beginner here.  :([/quote] :) don't worry about it. Peter VDD was also a beginner a few months ago, but he quickly learned much stuff. I'll do too :)

And probably you are not the only beginner, but you may be the only one posting :) other are either lazy, or too shy :)

[quote="Lagrangia"]you mean you want this:

h<=g<=a<=q

[/quote]

$M(r)=\sqrt[r]{\frac{w_{1}a_{1}^{r}+w_{2}a_{2}^{r}+ \cdots+w_{n}a_{n}^{r}}{w_{1}+w_{2}+ \cdots +w_{n}}}$

$w_{n}$ is a constant number for every n, n is a natural number.

For $r=0$, $M(0)$ is defined as:

$M(0)=\sqrt[n]{a_{1}a_{2} \cdots a_{n}}$ (geometric mean)

$a_{n}>0$ for every n.

If $p>q$, then $M(p)>M(q)$.

I think I finally wrote this as it should be.

How is this called in English?


geometry
parallelogram
geometric transformation
rotation
analytic geometry
geometry unsolved

Let $ \triangle ABC$, construct point $ D$ as picture with $ AB \perp AD$ and $ AB\equal{}AD$, construct point $ E$ as picture with $ AC \perp AE$ and $ AC\equal{}AE$, let $ M$ is midpoint of $ DE$. Prove that $ AM \perp BC$.

Nice problem!
We need to show $ \angle MAE \equal{} \angle ACB$
But $ \frac {sin ACB}{AB} \equal{} \frac {sin ABC}{AC}$
Also, $ [AME] \equal{} 1/2 \cdot AM \cdot AC \cdot sin MAE$
Also, $ [AME] \equal{} [AMD]$
Also, $ [AMD] \equal{} 1/2 \cdot AM \cdot AB \cdot sin DAM$
Combining them, we find that if the problem is true on the right side, then it is true on the left side, in other words
$ \angle MAE \equal{} \angle ACN \implies \angle MAD \equal{} \angle ABC$
But then if you follow along the proof, you will find $ \angle MAE > \angle ACN \implies \angle MAD > \angle ABC$ and $ \angle MAE < \angle ACN \implies \angle MAD < \angle ABC$, a contradiction as $ \angle DAE$ and $ \angle BAC$ must add up to 180.
[geogebra]7749fbc2ed4b1e9c4dd8b291445352840a0fdb67[/geogebra]

Let $ P$ be a point satisfy that $ PBAC$ is parallelogram , $ AP$ meet $ BC$ at $ T$ ;$ O$ be midpoint of $ EC$
$ R_{(90^o ,O)}: A \rightarrow E , P \rightarrow D,C \rightarrow A$ .
Therefore ,$ AM \perp CT$

Construct Q such as AEQD be a paralellogram; obviously, DQ is perpendicular to AC, M is the middle of AQ. A 90 deg anticlockwise rotation around A brings C to E and B to B', the simmetrycal of D w.r.t. A, consequently, AQ and BC are equal and perpendicular.

Best regrards,
Sunken Rock

[hide="Coordinates"]Let $ A \equal{} (0,0)$, $ B \equal{} (x_B,y)$, and $ C \equal{} (x_C,y)$. $ D$ is a $ \minus{} \frac {\pi}{2}$ rotation of $ B$ about the origin, and similarly $ E$ is a $ \frac {pi}{2}$ rotation of $ C$ about the origin. Thus, $ D \equal{} (y, \minus{} x_B)$ and $ E \equal{} ( \minus{} y,x_C)$. Their midpoint is on the $ y$-axis, so line $ AM$ is equivalent to the $ y$-axis, which is perpendicular to $ BC$.[/hide]

Wow, your proofs are much better than mine!


AMC
AIME
algebra
polynomial
geometric sequence

Has there been a thread about this one yet? If so, please point me to the link. If not...

Let P(x) = 24x^24 + Sum[j = 1...23] (24 - j)(x^(24-j) + x(24 + j)). Let z_1, z_2, ..., z_r be the distinct zeros of P(x), and let (z_k)^2 = a_k + i(b_k) for k = 1, 2, ..., r, where a_k and b_k are real numbers. Let Sum[k = 1...r] |b_k| = m + n*Sqrt(p), where m, n, p are integers. Find m + n + p.

This problem looked really intimidating...

A hint about a method of attack, nothing deep.  But probably helpful if you look at that problem and simply move on to the next one . . .

[hide]Try writing out what the polynomial actually looks like, then maybe play around with smaller versions of the same thing, see if you can get a feel . . .[/hide]

Hmm, well, I tried expanding some similar polynomials of smaller degrees, and the results were actually quite nice. I don't know how I would go about finding the sum of the squares of the roots though. Newton's Sums would work, but since they are recursive, and since the polynomial is one of 47th degrees, that method would not be feasible...

[color=cyan]First of all, the sum of the squares of the roots won't help you, because you need absolute values.  So that will be problematic, unless you can prove all the roots have the property that their squares have positive imaginary parts, which would certainly be harder than actually solving the problem.  If it was even true.  Which we don't know.  So your best bet is to look for patterns.  And this problem is full of them.  So try things.  Look for specific roots, for instance.  Or look for patterns in the roots of smaller, similar-looking polynomials.  Or factorizations.  Or anything.  Just keep playing around.[/color]

Yea, looking at smaller versions definitely works.

If its hard, don't worry. THere's a reason why is #15.

OK I'm just trying this quickly, still got a lot of new posts to read

Heres a few things I did:

[hide]

First of all write it out in full without condensed sums. Now you can cancel out the root of 0 as that won't affect the answer.. now the rest looks like you may be able to use geometric sequences.. say (1+x+x^2 + ...) + (x+x^2+x^3+...) etc. That should get the coefficients right I think.. now if you add them up with geometric sequence stuff it may help. May not, no idea, I'm just trying things.. I'm guessing you may just be able to find the roots generally and then they'll add up nicely or something..

[/hide]

[color=cyan]What you suggested should work.  There are quicker ways to get the same result, but that'll work fine.[/color]

Yeah it does work..

[hide]

A little bit of algebra and you get that being equal to (x(x^24 - 1)^2)/(x-1)^2. So we want x(x^24-1) = 0, but x is not equal to 1. Then the roots are 0 and the 24th roots of 1.. except for 1. So when we square these I think we get the 12th roots of 1 all repeated twice.. (and the 1s won't count in the sum of imaginary parts).. and I think I'll leave it there

[/hide]


Let be $ x,y,z\ge 0$ such that $ x\plus{}y\plus{}z\equal{}1$ . Show that :

$ \frac{x\plus{}1}{x\plus{}yz}\plus{}\frac{y\plus{}1}{y\plus{}zx}\plus{}\frac{z\plus{}1}{z\plus{}xy}\ge 9$

[quote="alex2008"]Let be $ x,y,z\ge 0$ such that $ x \plus{} y \plus{} z \equal{} 1$ . Show that :

$ \frac {x \plus{} 1}{x \plus{} yz} \plus{} \frac {y \plus{} 1}{y \plus{} zx} \plus{} \frac {z \plus{} 1}{z \plus{} xy}\ge 9$[/quote]

Indeed it is:

$ \frac {x \plus{} 1}{x \plus{} yz} \plus{} \frac {y \plus{} 1}{y \plus{} zx} \plus{} \frac {z \plus{} 1}{z \plus{} xy}\equal{} \frac{(x\plus{}y)\plus{}(x\plus{}z)}{(x\plus{}y)(x\plus{}z)}\plus{}\frac{(x\plus{}y)\plus{}(y\plus{}z)}{(x\plus{}y)(y\plus{}z)}\plus{}\frac{(x\plus{}z)\plus{}(y\plus{}z)}{(x\plus{}z)(y\plus{}z)}\equal{}2(\frac{1}{x\plus{}y}\plus{}\frac{1}{x\plus{}z}\plus{}\frac{1}{y\plus{}z}) \ge 2(\frac{9}{2(x\plus{}y\plus{}z)})\equal{}9$

it's easy but nice,
we have $ S\equal{}\sum\frac{x\plus{}1}{x\plus{}yz}\equal{} \sum\frac{x\plus{}1}{1\minus{}y\minus{}z\plus{}yz}\equal{}\sum\frac{x\plus{}1}{(1\minus{}x)(1\minus{}y)}$
we have $ (1\minus{}x)(1\minus{}y) \leq \left(\frac{2\minus{}(x\plus{}y)}{2}\right)^2 \equal{} \left(\frac{x\plus{}1}{2}\right)^2$
then $ S\geq \sum \frac{x\plus{}1}{\left(\frac{x\plus{}1}{2}\right)^2}\equal{}\sum\frac{4}{x\plus{}1}$
By Cauchy-Schwarz, $ \left(\sum\frac{4}{x\plus{}1}\right)\left(\frac{x\plus{}y\plus{}z\plus{}3}{4}\right)\geq9$
then $ S\geq9$


Let $ p$ be an odd prime number and $ a_1,a_2,...,a_p$ and $ b_1,b_2,...,b_p$ two arbitrary permutations of the numbers $ 1,2,...,p$ . Show that the least positive residues modulo $ p$ of the numbers $ a_1b_1, a_2b_2,...,a_pb_p$ never form a permutation of the numbers $ 1,2,...,p$.

Sorry, but this makes no sense...  :maybe:

The two $ p$'s must be at the same place (otherwise we already get two $ p$'s in $ a_1b_1, ..., a_pb_p$). W.l.o.g. let $ a_p \equal{} 0 \equal{} b_p$.
If now $ a_1b_1, ..., a_{p \minus{} 1}b_{p \minus{} 1}$ would be a permutation of $ 1,2,...,p \minus{} 1$ $ \mod p$, then
\[ \minus{} 1 \equiv (p \minus{} 1)! \equiv \prod_{i \equal{} 1}^{p \minus{} 1} a_i b_i \equal{} \prod_{i \equal{} 1}^{p \minus{} 1} a_i \cdot \prod_{i \equal{} 1}^{p \minus{} 1} b_i \equiv (p \minus{} 1)!^2 \equiv 1 \mod p,
\]
being a contradiction.


Given three distinct unit circles, each of which is tangent to other two, find the radii of the circles which are tangent to all three circles.

[hide="Solution"]

The problem said "circles" but I only can find only one.

Let three circles have center $A,B,C$. Since three circles are tangent and same size, we have $\triangle ABC$ is an equilateral. Moreover, let $D$ be the centroid. $D$ will also be the radius of the circle which will be tangent to all three circles. To find the radius of circle with center $D$, we note that radius equals radius of smaller circle plus $AD, BD$ or $CD$ (They're all same in length so it doesn't matter. We will use $AD$. To find $AD$, we draw point $E$ in $AB$ so that $DE \perp AB$. $AE = 1$ and $AD$ bisects angle $A$ so $\angle DAE = 30^\circ$. Thus, we have 30-60-90 triangle and $AD = \frac{2 \sqrt 3}{3}$. The radius of large circle is therefore:

$1 + \frac{2 \sqrt 3}{3} = \frac{2 \sqrt 3 +3}{3}$. [/hide]

I wonder if this is right..  :huh:

I think you took your radius by AD + 1. What happens if you take AD - 1?

Oh I see.

There is one really tiny circle that is tangent to all three circles.. Length is basically AD-1 like you said.  :D

I guess so there are only two circles then. Isn't it??

Yeah, there are only two circles. Nice solution, as well.

[quote="qweretyq"][quote="Silverfalcon"]$D$ will also be the radius of the circle which will be tangent to all three circles.  [/quote]

A point can't be a radius  ;)

Was this really an Olympiad problem?
Wow that was easy[/quote]

I meant "center" instead of "radius." It was just a typo.  ;)  And yes, it was an Olympiad problem.

I'm guessing it was #1 on the CMO since they have a 1 day test with 5 questions.

This is more like first 5 on AIME or final 10 on AMC-12.

CMO is typically easier than USAMO (no offense to them canadians)

also how do you rigorously prove that there are only two circles? you can't just stare at the diagram and be like well that's sort of obvious

like how do you know there isn't one which externally tangent to two and internally tangent to the third or something

[quote="Silverfalcon"]Given three distinct unit circles, each of which is tangent to other two, find the radii of the circles which are tangent to all three circles.
[/quote]

Would you just use Pythagorean to find the distance between two centers of the small circles and add 2, and divide the whole thing by 2?

This was from the 70's, what did you expect? Olympiad style problem solving has changed a lot just in the last 10 years. For example, consider #2 on IMO 1995, a problem that completely owned the US team (only one person got any points at all), but is now known as the "easiest" inequality of all time. (It's not actually the easiest, but since you can solve it using approximately 10000 methods, give or take 9985....)

Or how about IMO 1988 #6, considered quite difficult back then, and given a nice couple of pages of discussion in Engel's Problem Solving Strategies, but a problem that would be at worst a #2 now (and is noticeably easier than a similar #2 from IMO 2003).

As more Olympiad style problems are written, new techniques become more widely known and practiced, and hard problems become easier. #3 on this year's IMO (2005) simply involved a direct rearrangement of the problem, and then  an easy Cauchy to substitute in the denominator, in which case the rest is obvious. On paper, this makes the problem only slightly more complicated than #5 on the 1997 USAMO, in which a direct substitution in the denominator yielded a solution. However, it was still a relatively different concept and stumped many contestants.

Indeed, the problems are getting harder, even if the IMO's don't always get harder from year to year (but they do very noticeably from decade to decade).

descartes's circle theorem comes in handy here.

it tells you that there are only 2 circles.

and then it tells you how to find their radii.


[quote="NeverOddOrEven"]if i just took the amc 10A, is it possible for me to take the amc 10B somewhere?[/quote]
Today's the last day to register.
I suggest you email your coach quickly.
Or you can take it at a nearby location.

I already signed up with the teacher that runs the test, but I was wondering how would I take it in another location?

If, for example, my school gotten snowed out (which it did the day before and after the test) how would I take it in a different place?

AMC website has a list of colleges that administer AMCs on date B.
When I checked, you do not even have to be in their state.

is there a deadline for registering at colleges?

I think depends on the colleges,whether they have all their spots filled.
They order certain number of bundles and that's it. So does not hurt to call them or e-mail them. They respond pretty fast in my experience.


Hello everybody. I suppose some (if not many) of you have player the third Fallout game that came out not a long time ago and therefor are familiar with the problem. If not, this is the problem I'm referring to: a player is supposed to "crack the password" on a computer by choosing one of many given words. Any time a word is chosen, the player is given the information of how many letters in the chosen words are in the right place in reference to the unknown password, but not giving the positions of the letters. Now, what would be a good strategy for obtaining the password? Naturally, the password itself is within the set of given words, the words are of the same length and the number of tries is fixed (but this can be ignored since one can always exit the "computer terminal" and try again with a new set of words).

Now, this seems to be a problem of entropy, but I am unable to see the connection. I tried to use the C4.5 algorithm, but it searches the highest entropy, and it seems the problem here is connected to the lowest entropy. So, any thoughts?


Let $ a,b,c,d$ be positive reals s.t$ abcd\equal{}1$.Prove that   $ \dfrac{1}{1\plus{}a^2}\plus{}\dfrac{1}{1\plus{}b^2}\plus{}\dfrac{1}{1\plus{}c^2}\plus{}\dfrac{1}{1\plus{}d^2}\geq2$

[quote="Inequalities Master"]Let $ a,b,c,d$ be positive reals s.t$ abcd \equal{} 1$.Prove that   $ \dfrac{1}{1 \plus{} a^2} \plus{} \dfrac{1}{1 \plus{} b^2} \plus{} \dfrac{1}{1 \plus{} c^2} \plus{} \dfrac{1}{1 \plus{} d^2}\geq2$[/quote]
Try $ a\equal{}b\equal{}c\equal{}2$ and $ d\equal{}\frac{1}{8}.$ :wink:

[quote="Inequalities Master"]Let $ a,b,c,d$ be positive reals s.t$ abcd \equal{} 1$.Prove that   $ \dfrac{1}{1 \plus{} a^2} \plus{} \dfrac{1}{1 \plus{} b^2} \plus{} \dfrac{1}{1 \plus{} c^2} \plus{} \dfrac{1}{1 \plus{} d^2}\geq2$[/quote]
  problem   transfer no exactly.
 $ \frac{1}{(1\plus{}a)^2}  \geq 1$
 it's old  and   solve  by  ineq
$ \frac{1}{(1\plus{}x)^2}\plus{}\frac{1}{(1\plus{}y)^2}  \geq \frac{2}{1\plus{}xy}$


Euler
number theory
number theory unsolved

The question is:

Find the least residue of a mod m
a = 43 and m = 13

I typed it wrong.. it should read

The question is:

For each pair a and m below, find the least residue of a modulo m
a = 43^43 and m = 13

We use Euler's Theorem, with $\phi(13) = 12$, to get that $43^{12} \equiv 1$ (mod 13). From this, we know that $43^{36} \equiv 1$ (mod 13). Then we need to find $43^{7}$ which in mod 13 is the same as $4^7$ which is 4 in mod 13.

I still need some more clarification on this problem please. thank you

What is Euler's Theorem? How did you get 43^23 = 1 (mod 13)? I am not doubting your answer, just asking for clarification.

Eulers theorem:
Let be $a,n$ two coprime integers, $n$ positive, and define $\phi (n)$ as the number of positive integers smaller or equal to $n$ that are coprime to $n$ (formal: $\phi (n) := \left| \{ x | 1\leq x\leq n, gcd(x,n)=1 \} \right|$)
Then you have:
$a^{\phi (n)} \equiv 1 \mod n$

ZetaX please explain your signature. I don't get it. Sorry.

You haven't to worry about getting it, it's only an identity from analytic number theory that proves Jacobi's formula for the number of representations of an integer as sum of four squares
(number of representations of $n$ is $\mathrm{A}_4 (n) = 8 \cdot \sum_{ \substack{ d \in \mathbb{N} \\ 4 \nmid d \mid n } } d $ )


Give $ a,b,c$ be nonnegative real numbers such that $ ab\plus{}bc\plus{}ca\plus{}6abc\equal{}9$.
Prove that $ a\plus{}b\plus{}c\plus{}3abc\ge 6$

[quote="akai"]Give $ a,b,c$ be nonnegative real numbers such that $ ab \plus{} bc \plus{} ca \plus{} 6abc \equal{} 9$.
Prove that $ a \plus{} b \plus{} c \plus{} 3abc\ge 6$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=117363


inequalities
number theory
Diophantine equation
number theory unsolved

Find all the natural numbers $x,y,z,n$ such that  $x^3+y^3+z^3=nx^2y^2z^2$

[quote=Solar Plexsus]The only solution of the Diophantine equation

$(1) \;\; x^3 + y^3 + z^3 = nx^2y^2z^2$

where $x$, $y$, $z$ and $n$ are natural numbers, is $x=y=z=1, \, n=3$.

[u]Proof:[/u] Since (1) is symmetric in the variables $x$, $y$ and $z$, we may without lost of generality assume that $x \geq y \geq z$. In the case where $x=1$ the solution of (1) is obviously $x=y=z=1, \, n=3$.
Next suppose $x \geq 2$. Then by (1)

$x^3 + y^3 + z^3 \geq x^2y^2z^2,$

which implies that

$(2) \;\; \frac{y^2z^2}{x} - 1 \; \leq \; \frac{y^3 + z^3}{x^3} \; \leq \; \frac{y^3 + y^3}{x^3} \; = 2 \; (\frac{y}{x})^3 \; \leq \; 2$

because $x \geq y \geq z$. Hence $\frac{y^2z^2}{x} \leq 3,$ i.e.

$(3) \;\; x \geq \frac{y^2z^2}{3}$.

Moreover from (1) we observe that $x^2 | \: y^3 + z^3$, thus

$(4) \;\; x^2 \leq y^3 + z^3 \leq 2y^3$.

Combining the inequalities (3) and (4), the result is the following series of implications:

$(\frac{y^2z^2}{3})^2 \; \leq \; x^2 \; \leq \; 2y^3$

$\frac{y^4z^4}{9} \; \leq \; 2y^3$

$yz^4 \; \leq \; 18$

$z^5 \; \leq \; 18$

$z < 2$.

Therefore $z = 1$, which inserted in (2) gives

$\frac{y^2}{x} - 1 \; \leq \; \frac{y^3 + 1}{x^3} \; \leq \; (\frac{y}{x})^3 + \frac{1}{x^3} \; \leq \; 1 + \frac{1}{x^3},$

i.e.

$\; y^2 \; \leq \; x + \frac{1}{x^2} \; \leq \; x + \frac{1}{4} \; < \; x + 1,$

so

$(5) \;\; x \geq y^2.$

On the other hand,

$(6) \;\; x^2 \leq y^3 + z^3 = y^3 + 1$

by (4). Now $y^3 \leq x^{3/2}$ according to (5), which combined with (6) implies that

$x^2 \leq x^{3/2} + 1$.

Hence $x < 2$, a contradiction which completes the proof.[/quote]

But what about x,y,z=3,2,1, n=1?


number theory
relatively prime
number theory unsolved

Does there exist a natural number with the sum of digits of its $ kth$ power being
equal to $ k$, if a) $ k \equal{} 2004$; b) $ k \equal{} 2006?$

a) $ n^2004$ is congruent to $ 6 mod 9$

n obviously has to be relatively prime to 9, since otherwise $ n^2004$ is congruent to 0 mod 9, but then since phi(9) is 6, $ n^6$ is congruent to 1 mod 9, so $ n^2004$ is congruent to 1 mod 9, so no solution.

b) $ n^2006$ is congruent to 8 mod 9

again, n is relatively prime, blah, blah, blah, $ n^2$ is congruent to 8 mod 9. But the quad. res. mod 9 are 1,4,0,7, contradiction!


A circle is tangent to two sides of a square and its diagonal. Given that the length of a side of the square is four units, how many units are in the radius of the circle? Express your answer in simplest form.
[asy]draw(unitsquare);
draw((0,0)--(1,1));
draw((.293,1)..(.5,.5)..(0,.707)..cycle);
draw((.293,.707)--(.5,.914));
label("$r$",(.3965,.8105),NW);
label("4",(1,.5),E);[/asy]

$ (\sqrt {2} + 1)r = 2\sqrt2$

$ r = \frac {2\sqrt {2}}{\sqrt {2} + 1} = \boxed{4 - 2\sqrt {2}}$

[color=blue] Added $ \text{\LaTeX}$. In the future, please use LaTeX for the sake of clarity. It's very simple to learn, just look at other people's codes...if you want I can private message you the tutorials I sent ernie. :) -Izzy[/color]


advanced fields
advanced fields unsolved

Prove that

$ f(a,b,c)=\begin{cases}
1 & \text{ if } A(a,b)=c\\
0 & \text{ otherwise }
\end{cases}$

is primitively recursive (where $ A$ denotes the Ackermann function).

Sorry for bumping, but I'd still like to see a solution, if someone knows it.


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Let $ B\equal{}{(b_{ij})}_1 \leq i,j \leq n \in M_n(R)$, and $ b_{ij}\equal{}p_ip_j ,  1 \leq i,j \leq n , p_1,...,p_n \in R$. Prove that $ A\equal{}B\plus{}I_n$ is an invertible matrix, and find its invertible.

Let $ p\equal{}[p_1,p_2,\dots,p_n]^T.$ We have defined $ B\equal{}pp^T.$ Then

$ B^2\equal{}pp^Tpp^T\equal{}p(p^Tp)p^T\equal{}(p^Tp)pp^T\equal{}\|p\|^2\,B.$

If $ I\plus{}B$ is invertible, its inverse should be a polynomial in $ B.$ Let's try out the possibility that its inverse is a linear polynomial in $ B:$

$ (I\plus{}B)(I\plus{}t B)\equal{}I\plus{}B\plus{}tB\plus{}t\|p\|^2B.$

For that to be $ I$ we should have $ 1\plus{}t\plus{}t\|p\|^2\equal{}1,$ which happens when $ t\equal{}\frac{\minus{}1}{1\plus{}\|p\|^2}.$ Since we can always do that, we find that $ I\plus{}B$ is invertible and

$ (I\plus{}B)^{\minus{}1}\equal{}I\minus{}\frac1{1\plus{}\|p\|^2}\,B.$

---
Another demonstration of invertibility: We've shown that $ B^2\equal{}\|p\|^2B.$ $ B$ is also singular (in fact, rank 1), so $ B$ satisfies $ x(x\minus{}\|p\|^2)$ as its minimal polynomial. The only eigenvalues of $ B$ are $ 0$ and $ \|p\|^2.$ In particular, $ \minus{}1$ is not an eigenvalue, so $ I\plus{}B$ must be invertible.


geometry
circumcircle
incenter
power of a point
radical axis
geometry proposed

Given the circle (O) and BC is a chord of (O). A point D moves on the segment BC such that BD<DC. OD cuts the bigger arc BC of (O) at A. E,F are the circumcenter and incenter of triangle OBD. Prove that the perpendicular line from A to EF always passes through a fixed point when C moves.

I think this problem is not bad. Why do you post your comments, dear Mathlinkkers ?

[quote="vittasko"]I think you mean the point $ D$ as mobile one, instead of $ C.$

If yes, then the line through the point $ A$ and perpendicular to $ EF,$ ia parallel to $ BD$ and so ( easy to prove ), always passes through the point $ B',$ as the antidiametric point of $ B$ $ ($ $ B'\equiv (O)\cap BO$ $ ).$

kostas Vittas.[/quote]
First, yes, "when D moves" instead of "when C moves"
Second, BD is [b]not[/b] a chord of (O), why is EF is perp to BD. i wrote, BC is the fixed chord of (O) and D moves on segment BC such that BD<BC.
I think you made a mistake, is that true?

Sorry dear [b][size=100]mr.danh[/size][/b], for such my blindness.

I thought the point $ D,$ as mobile on the arc $ BC.$ :blush:

Kostas Vittas.

Condition $ BD < BC$ is irrelevant. Sides of the $ \triangle OBD$ are $ o \equal{} BD, b \equal{} DO, d \equal{} OB.$ $ d$ is also radius of the given circle $ (O).$ Perpendicular to the internal bisector $ BF$ of the angle $ \angle OBD$  through $ O$ cuts $ BD$ at $ P.$ $ Z \in (O)$ is diametrically opposite to $ A.$ Then $ \overline {PD} \equal{} \overline {BD} \minus{} \overline {BP} \equal{} o \minus{} d$ and $ \overline {ZD} \equal{} \overline{ZO} \minus{} \overline {DO} \equal{} d \minus{} b.$ External bisectors of the angles $ \angle OBD, \angle DOB$ cut the lines $ BD, DO$ at $ X, Y.$ $ \overline{DX} \equal{} \frac {ob}{d \minus{} b},$ $ \overline{DY} \equal{} \frac {ob}{o \minus{} d},$ $ \frac {\overline{DP}}{\overline{DZ}} \equal{} \frac {\overline{PD}}{\overline{ZD}} \equal{} \frac {o \minus{} d}{d \minus{} b} \equal{} \frac {\overline{DX}}{\overline{DY}},$ $ PZ \parallel XY.$ Let $ Q$ be a point on the ray opposite to $ (OP,$ such that $ \overline{OQ} \equal{} \minus{} \overline{OP}.$ Then the $ \triangle OAQ \cong \triangle OZP$ are congruent by SAS, $ AQ \parallel ZP \parallel XY.$ Since $ O, P$ are fixed, $ Q$ is also fixed. The fact that $ XY \perp EF$ (where $ E$ is the circumcenter and $ F$ the incenter of the $ \triangle OBD$) has been proved multiple times, $ XY$ is radical axis of the circumcircle $ (G)$ of the excentral triangle of the $ \triangle OBD$ and its 9-point circle $ (E),$ their center line $ GE$ passing through the orthocenter $ F$ of the excentral triangle.


Prove that $ \forall x,y,z \geq 0$ we have:
$ \sum(x-y)^2*(x+y-z) \geq 0$.


does anyone think AOPS should have a class on how to solve rubiks cube?

Well I might be wrong, but I think it would be hard to teach people that have trouble visualizing things how to solve a Rubiks cube without actually demonstrating various moves to them in person.  However, many people do have the ability to learn from instructions rather than demonstrations. Even when someone taught me how to solve a Rubik's cube in person, I had trouble completing the second level. :( It would, however, be interesting to learn more about the math behind a Rubik's cube.

This should be in the Questions/Suggestions forum..

Google is your friend. I'm not going to pay 150 for something off the internet.