find all functions from $ R_{\plus{}}$ to $ R_{\plus{}}$, that satisfy
    $ f(x)f(y(x))\equal{}f(x\plus{}y)$.
enjoy.

[quote="arkhammedos"]find all functions from $ R_{\plus{}}$ to $ R_{\plus{}}$, that satisfy
    $ f(x)f(y(x)) \equal{} f(x\plus{}y)$.
enjoy.[/quote]
$ y \equal{} 1$ gives $ f(x)^{2}\equal{} f(x\plus{}1)$
$ x \equal{} 1$ gives $ f(1)f(y) \equal{} f(y\plus{}1)$
then $ f(1)f(x) \equal{} f(x\plus{}1) \equal{} f(x)^{2}$
so $ \forall x\in\mathbb{R}:\ f(x) \equal{} 0\ or\ f(x) \equal{} a\ (where\ f(1) \equal{} a)$
if $ a \equal{} 0$ then $ \boxed{f(x) \equal{} 0}$
if $ a\neq 0$ and $ \exists b:\ f(b) \equal{} 0$ then $ a \equal{} f(1) \equal{} f(1.(b\minus{}1))f(b) \equal{} 0$ (imppossible)
so if $ a\neq 0$ $ f(x) \equal{} a$ $ \forall x\in\mathbb{R}$ and we find easly that $ \boxed{f(x) \equal{} 1}$

i think you want say $ \boxed{f(x)f(yf(x)) \equal{} f(x\plus{}y)}$(posted before)

:blush:  oops sory my question  was $ f(x)f(yf(x))\equal{}f(x\plus{}y)$.
thanks aviateurpilot :D

[quote="aviateurpilot"]$ f(x) \equal{} x,\frac{1}{1\plus{}x}$[/quote]

f(x)=x doesn't work for all x and y, and f(x)=(1/(1+x)) implies that y=0.

[hide="solution"]We plug in y=0:
$ f(x)f(0)\equal{}f(x)$

So either f(0)=1 or f(x)=0. If f(0)=1, then we can plug in x=-y:

$ f(x)f(\minus{}xf(x))\equal{}1$

$ f(x)\equal{}\frac{1}{f(\minus{}xf(x))}$

$ f(x)\equal{}\frac{1}{\frac{\minus{}x}{\frac{\minus{}x}{\cdots}}}$

Let's call

$ z\equal{}\frac{\minus{}x}{\frac{\minus{}x}{\cdots}}$

$ \frac{1}{z}\equal{}\frac{1}{\frac{\minus{}x}{z}}$

$ z\equal{}\frac{\minus{}x}{z}$

$ z^{2}\equal{}\minus{}x$

That implies x is negative.

$ z\equal{}\sqrt{\minus{}x}$

$ f(x)\equal{}\frac{1}{\sqrt{\minus{}x}}$

$ f(x)\equal{}\frac{\sqrt{\minus{}x}}{x}$

x=-n, where n is positive.

$ f(\minus{}n)\equal{}\minus{}\frac{\sqrt{n}}{n}$

But, when we plug it in, n has to be negative, which is obviously false.

Also, f(1) works.

The functions that work are
$ f(x)\equal{}0$
$ f(x)\equal{}1$[/hide]

i am sur that there exist some non-constant solution.
but no one would help :mad:

[quote="1=2"]
$ f(x) \equal{}\frac{1}{f(\minus{}xf(x))}$

$ f(x) \equal{}\frac{1}{\frac{\minus{}x}{\frac{\minus{}x}{\cdots}}}$
[/quote]

Care to explain this step more explicitly?  Also, do keep in mind we're only considering [b]positive[/b] reals in the stated problem.

Well then: my stab at a solution. (as aviateurpilot correctly stated, $ \frac{1}{1\plus{}x}$ is a possibility.)

[hide]I assume $ x, y\in\mathbb{R}^{\plus{}}$
If $ f(x)>1$, then let $ y\equal{}\frac{x}{f(x)\minus{}1}$.  This yields $ f(x)f(x\plus{}y)\equal{}f(x\plus{}y)\implies f(x)\equal{}1$.  This is clearly a contradiction.  Thus, $ f(x)\le 1$.

This tells us that $ f(x)\ge f(x\plus{}y)$, and thus, $ f(x)$ is a nonincreasing function.

More to come...[/hide]

The same equation :

Find all functions $ f:\mathbb{R^{\plus{}}}\to\mathbb{R^{\plus{}}}$ so that

$ f(y)f(xf(y))\equal{}f(x\plus{}y)$

[hide="solution"]$ f(x)\equal{}1,f(x)\equal{}\frac{a}{a\plus{}(1\minus{}a)x}$[/hide] :)

:rotfl:  u make the substitution $ x\equal{}y$
but schow me the solution please  :(

[quote="arkhammedos"]:rotfl:  u make the substitution $ x = y$
but schow me the solution please  :([/quote]

No, my friend, my problem different from arkhammedos'problem .

My problem'solution  :)

Suppose there exist $ y > 0$ so that $ f(y) > 1$.
Put $ x =\frac{y}{f(y)-1}\Rightarrow f(y)f(xf(y)) = f(y)f(x+y)$ so $ f(y) = 1$ contracdition .
Then $ f(x)\le 1,\forall x\in\mathbb{R^{+}}$

If there exist $ a\in\mathbb{R^{+}}$ so that $ f(a) = 1$

By inductive easy to show that $ f(na) = 1,\forall n\in\mathbb{N}$ then $ f(x) =\tex{constant}= f(a) = 1$.

Now we consider the last case $ f(x) < 1,\forall x\in\mathbb{R^{+}}$.
So $ f(x)$ is a monotonic function.Let $ f(1) = a,a > 0$.
Put $ y = 1$ , $ f(ax)a = f(x+1) = f(ax+1+x(1-a)) = f(ax)(f((1+x-ax)f(ax))$
Then $ f((1+x-ax)(f(ax)) = a = f(1)$
So $ f(ax) =\frac{1}{1+x-ax}$
Hence $ f(x) =\frac{a}{a+(1-a)x},\forall x\in\mathbb{R^{+}},0 < a < 1$

arkhammedos'problem can proved by the same way . :wink:


Not rly sure if this is hard enough to be in the intermediate, so dont jump on me if this makes me look ignorant...

In the book, it syays if we take f(x)/(x-c) for any c, c is greater than or equal to zero, and f(x) has a positive leading coefficient, and the resulting coefficients including the remainder, are positive, then there aren't any roots greater than c.

Why?

I can see why if c=0, it must be true, since no value of x>0 would yield f(x) = 0, but im not sure about other values of c...

also, if f(x) had a negative leading coefficient, how would i have to manipulate this technique?

What are the values of a polynomial with positive coefficients at positive values of $ x$?

As for your second question, multiply by $ \minus{}1$.

oh, i see.. i was thinking it might not be true if it was something like $ x^n$, where n < 0, but then that wouldn't be a polynomial...

Multiply the function by -1, right, since flipping a function over the x-axis wouldn't change the roots?


Let $ c_1,......,c_n, b_1,......,b_n (n\geq2)$ be positive reals. Prove that the equation $\sum_{i=1}^n c_i\sqrt{x_i-b_i}=\frac{1}{2}\sum_{i=1}^n x_1$
has an unique solution if and only if $\sum_{i=1}^n (c_1)^2=\sum_{i=1}^n b_i$

Write the equation in the equivalent form
\[\sum_{i=1}^n {\left(\sqrt{x_i-b_i}-c_i\right)^2}=\sum_{i=1}^n{c_i^2}-\sum_{i=1}^n{b_i}.
\]
The conclusion is obvious.

[quote="silouan"][quote="enescu"]Write the equation in the equivalent form
\[\sum_{i=1}^n {\left(\sqrt{x_i-b_i}-c_i\right)^2}=\sum_{i=1}^n{c_i^2}-\sum_{i=1}^n{b_i}.
\]
.[/quote]
what are the steps you follow to find equivalent the result you found?Express it analitical please.[/quote]

Hm?! How hard can it be to multiply by 2, subtract the $b_i$'s and add the $c_i^2$'s?


integration
trigonometry
calculus
algebra
partial fractions
calculus computations

We make the substitution $x = \tan(t/2)$. We have that $\cos t = \frac{1-x^2}{1+x^2}$ and $dt = \frac{2dx}{1+x^2}$. This turns the integral into
\[ \int_0^\infty \frac{2(1-x^2)dx}{(1+x^2)(x^2(u+1)^2 + (u-1)^2)}  \]
Converting to partial fractions we get
\[ \frac{1}{u}\int_0^\infty \frac{u^2+1}{(u+1)^2x^2 + (u-1)^2} - \frac{1}{x^2+1}dx  \]
This is
\[ \frac{1}{u}\left[\frac{u^2+1}{(u+1)^2}\cdot\frac{u+1}{1-u}\arctan\left(x\frac{u+1}{1-u}\right) - \arctan x\right]_0^\infty = \frac{\pi}{2u}\left(\frac{u^2+1}{1-u^2} - 1 \right)  \]

Simplifying, we get the very beautiful result:
\[ \int_0^\pi \frac{\cos (t) dt}{1-2u \cos t + u^2} = \frac{u\pi}{1-u^2}  \]

I know that it doesn't simplify the calculus in this situation but, can I use this method?:

$J(u)=\int_0^{\pi}\frac{\cos t}{1-2u\cos t+u^2}dt$.

Since $J(u)$ is continuous we can integrate $J(u)$ with respect to $u$ from $-1$ to $x$. Then swich the integrals and compute the integral "$du$" first and then derivate with respect to $x$?

we can use complex numbers as in the next post

I think my answer is correct. At least mine is right for the special case $u=1/2$.

I understand where i have error and edited. Thanks


inequalities
Euler
inequalities proposed

Let  AD , BE , CF  cevians of a  triangle  ABC  meeting  at point  P. Prove that PD+PE+PF<max{a,b,c}
Thanks!

That's funny...

I assume that the point P is supposed to lie inside the triangle ABC. [Else, the problem is wrong.]

Call m = max(a, b, c) the greatest side of the triangle ABC, and then we have to prove that PD + PE + PF < m.

Well, if H is the orthogonal projection of the point A on the side BC, then the segments BH and CH cover the side BC (in fact, if the angles B and C are acute, then the segments BH and CH just form together the side BC, and if one of the angles B and C is obtuse, then the segments BH and CH cover more than the side BC). Hence, since the point D is an interior point of the side BC, it must lie inside one of the segments BH and CH. If the point D lies inside the segment BH, then AD < AB, but $AB=c \leq m$ (since m is the greatest side of triangle ABC), and thus AD < m. If the point D lies inside the segment CH, then AD < AC, and we similarly get AD < m. Thus, we have AD < m in every case. Similarly, BE < m and CF < m.

Now, the Gergonne-Euler theorem states that $\frac{PD}{AD}+\frac{PE}{BE}+\frac{PF}{CF}=1$. Thus,

$PD+PE+PF=\frac{PD}{AD}\cdot AD+\frac{PE}{BE}\cdot BE+\frac{PF}{CF}\cdot CF$
$<\frac{PD}{AD}\cdot m+\frac{PE}{BE}\cdot m+\frac{PF}{CF}\cdot m=\left( \frac{PD}{AD}+\frac{PE}{BE}+\frac{PF}{CF}\right) \cdot m=m$.

And this ends the solution.

This method also proves yields a somewhat stronger inequality: PD + PE + PF $\leq$ n, where n is the greatest among the segments AD, BE, CF. In fact, it is clear that $AD\leq n$, $BE\leq n$ and $CF\leq n$, so that

$PD+PE+PF=\frac{PD}{AD}\cdot AD+\frac{PE}{BE}\cdot BE+\frac{PF}{CF}\cdot CF$
$\leq \frac{PD}{AD}\cdot n+\frac{PE}{BE}\cdot n+\frac{PF}{CF}\cdot n=\left( \frac{PD}{AD}+\frac{PE}{BE}+\frac{PF}{CF}\right) \cdot n=n$.

  Darij


inequalities
vector
linear algebra
matrix
linear algebra unsolved

$ A,B \in M_{n} (K)$.Prove that:

$ r(A\plus{}B) \leq r(A)\plus{}r(B) \leq r(A\plus{}B) \plus{}n$.

The range of $ A\plus{}B$ is contained in the vector sum of the ranges of $ A$ and $ B$. This gives the first inequality. The second one is false: let $ A$ be a matrix of full rank and $ B\equal{}\minus{}A$.

The second one can be a little tricky because you may misinterpret A + B to mean sum of vector spaces

maybe it should be $ r(A) \plus{} r(B)\leq r(AB) \plus{} n$?  anyways this is well known; the magic words are 'sylvester's inequality' and it's been discussed a couple of times in the forums at least.

yes ^^,blahblahblah is right.I took this inequality from a problem of vietnam undergraduate olympiad.The original is that:
given $ A \in M_n(K)$ such that $ A^2\equal{}I_n$.Prove that : $ rank(A\plus{}I_n)\plus{}rank(I_n\minus{}A)\equal{}n$


invariant
geometry
geometric transformation
rotation
3D geometry
sphere
symmetry

Find all Möbius transforms (fractional linear transformations) that leave the circle $|z| = R$ invariant.

All of these transforamtions are of the form $R\frac{z-\alpha R}{R-\bar{\alpha}z}$.

How do you prove it? I've thought about the problem a lot, but I can't do it.

Look at [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=78814]this topic[/url].

[quote="Omid Hatami"]All of these transformations are of the form $R\frac{z-\alpha R}{R-\bar{\alpha}z}$.[/quote]
Make that $e^{i\theta}R\frac{z-\alpha R}{R-\bar{\alpha}z}.$ Rotations are allowed.

I don't know much about complex analysis; we've just been studying the extended compact plane/riemann sphere and discussing möbius transforms before. The excercise comes after a section on defining symmetry between points wrt circles in the extended complex plane using Möbius transforms. Anyone know what to do? :(

Does your textbook define the cross ratio of four complex numbers?

I realize your answer probably intended to make me think about using it somehow, but I have already tried using the cross ratio without success.

I finally solved the problem on my own by first studying transforms leaving the circle invariant with two diametrically opposite fixpoints, arriving at Omit Hamadis formula; then showing that all other transforms that work are rotations of this one.


inequalities
algebra proposed
algebra
lreland

[quote="moldovan"]Prove that if $ x,y$ are nonnegative real numbers with $ x \plus{} y \equal{} 2$, then: $ x^2 y^2 (x^2 \plus{} y^2) \le 2$.[/quote]

$ x\plus{}y\equal{}2$ and $ x,y\geq 0$ $ \implies$ $ xy\equal{}2x\minus{}x^2\leq 1$

$ x^2y^2(x^2\plus{}y^2)\minus{}2\equal{}x^2y^2((x\plus{}y)^2\minus{}2xy)\minus{}2$ $ \equal{}4x^2y^2\minus{}2x^3y^3\minus{}2$ $ \equal{}2(1\minus{}xy)(xy\minus{}\frac{1\plus{}\sqrt 5}{2})(xy\plus{}\frac{\sqrt 5\minus{}1}{2})$ $ \leq 0$ since $ 0\leq xy\leq 1$
Q.E.D.

[hide]We can also normalize: The inequality becomes
\begin{align*}
\frac{(x + y)^6}{32} &\ge x^2 y^2 (x^2 + y^2) \\
\Leftrightarrow \quad (x - y)^2 (x^4 + 8x^3 y - 2x^2 y^2 + 8xy^3 + y^4) &\ge 0 \\
\Leftrightarrow \quad (x - y)^2 [(x^2 - y^2)^2 + 8x^3 y + 8xy^3] &\ge 0.
\end{align*}[/hide]

Let $xy = a$.
    Then by AM -GM, $a = x(2 - x) \leq 1$, so with
    $$
    x^2 + y^2 = (x + y)^2 - 2xy = 4 - 2a,
    $$
    so it suffices to prove that
    $$
    a^2(4 - 2a) \leq 1,
    $$
    with $0 \leq a \leq 1$.

    We have equality at $a = 1$, and $a^2(4 - 2a)$ is increasing over $0 \leq a \leq 1$, as $a^3 > a^2$ over this interval, so the inequality holds.

Prove that if $ x,y,z$ are nonnegative real numbers with $ x+y+z=2$, then: $$ (x^2+ y^2) (y^2+ z^2)(z^2+ x^2)(x^2+ y^2 +z^2) \le 4$$
[hide=z=0][url=https://artofproblemsolving.com/community/c6h542249p3128662]Prove that[/url] if $ x,y$ are nonnegative real numbers with $ x\plus{}y\equal{}2$, then: $$ x y (x^2\plus{}y^2) \le 2$$[/hide]
[quote=moldovan]Prove that if $ x,y$ are nonnegative real numbers with $ x\plus{}y\equal{}2$, then: $$ x^2 y^2 (x^2\plus{}y^2) \le 2$$[/quote]
[url=https://artofproblemsolving.com/community/c1068820h2028088p14282030]2001 Estonia[/url]
$$0\leq xy\leq 1$$
$$ x^2 y^2 (x^2+y^2)=2x^2y^2(2-xy) \leq 2xy(2-xy) \leq  2\left(\frac{xy+2-xy}{2}\right)^2=2$$
Equality holds when $x=y=1.$

[quote=moldovan]Prove that if $ x,y$ are nonnegative real numbers with $ x\plus{}y\equal{}2$, then: $ x^2 y^2 (x^2\plus{}y^2) \le 2$.[/quote]

$$ x=2\cos^2{x}, \quad y=2\sin^2{x}\Rightarrow x^2y^2(x^2+y^2)=16\sin^4{x}\cos^4{x}(4\sin^4{x}+4\cos^4{x})=$$
$$ =4\sin^4{2x}(1-\frac{1}{2}\sin^2{2x})\leq2 \Longleftrightarrow 2\sin^4{2x}-\sin^6{2x}\leq 1\Longleftrightarrow \sin^6{2x}-2\sin^4{2x}+1\geq 0\Longleftrightarrow$$
$$\Longleftrightarrow  (1-\sin^2{x})(\sin^4{x}-\sin^2{x}-1)\leq 0$$
the last inequality is easily proved....


$ \triangle ABC$ $ \implies$ $ 8p\le\boxed{\sum\frac{(b\plus{}c)^{2}}{a}\le\frac{4Rp}{r}}$, where $ 2p\equal{}a\plus{}b\plus{}c$, $ R$- circumradius and $ r$- inradius.

[hide="Hope I didn't make any mistakes"]We consider
$ P\equal{}\sum_{cyc}ab(a\plus{}b)^{2}\equal{}...\equal{}(a\plus{}b\plus{}c)^{2}(ab\plus{}bc\plus{}ca)\minus{}5abc(a\plus{}b\plus{}c)$
From here, we have to prove $ p^{2}\plus{}r^{2}\minus{}6Rr\leq 4R^{2}$
We alreday have $ p^{2}\leq 4R^{2}\plus{}4Rr\plus{}3r^{2}$
To wrap it up, we need $ 4R^{2}\plus{}4Rr\plus{}3r^{2}\leq 4R^{2}\plus{}6Rr\minus{}r^{2}$
Or $ 4r^{2}\leq 2Rr$ or $ 2r\leq R$. Done  :D [/hide]

My way is sooo long. What's yours, Mr.Virgil Nicula?

By Cauchy-Shwarz, we get:
                              $ (a\plus{}b)^{2}/c\plus{}(b\plus{}c)^{2}/a\plus{}(c\plus{}a)^{2}/b\geq 16p^{2}/2p\equal{}8p$, Q.E.D.

[quote="bibobeo"][color=darkred]My way is sooo long. What's yours, Mr.Virgil Nicula ?[/color][/quote]
[color=darkblue][b]Gladly, Biboleo ![/b] Here is the my proof. You"ll see it is very easy.

I"ll use the well-known relations $ l_{a}\equal{}\frac{2}{b\plus{}c}\cdot\sqrt{bcp(p\minus{}a)}$ and $ abc \equal{} 4Rpr$, $ h_{a}\equal{}\frac{2pr}{a}$, where $ 2p \equal{} a\plus{}b\plus{}c$ and

$ l_{a}$ is the length of the $ A$- bisector in $ \triangle ABC$. Therefore, $ l_{a}\ge h_{a}\Longleftrightarrow$ $ \frac{4bcp(p\minus{}a)}{(b\plus{}c)^{2}}\ge\frac{4p^{2}r^{2}}{a^{2}}$ $ \Longleftrightarrow$ $ \frac{(b\plus{}c)^{2}}{a}\le\frac{abc(p\minus{}a)}{pr^{2}}\equal{}\frac{4R(p\minus{}a)}{r}$.

In conclusion, , $ \sum\frac{(b\plus{}c)^{2}}{a}\le\frac{4R}{r}\cdot\sum (p\minus{}a) \equal{}\frac{4Rp}{r}$. With another words, this inequality is equivalently to the commonplace inequality $ l_{a}\ge h_{a}$ ! [/color]

Is it natural?  :blush:
I mean How can we think of $ h_{a}\leq l_{a}$ when we first see it?
Is it your inequality or you create it?
I also have another way to solve this (also long  :D ), applying the SOS Method (maybe shorter than the first one)

[color=darkblue]This ineq. is me. Indeed, the ineq. $ l_{a}\ge h_{a}$ and the my proof are naturallly ! You can use the "natural" ineq. $ \gamma_{a}\ge h_{a}$,
where $ \gamma_{a}$ is the length of the $ A$- [b]Gergogne[/b]'s line, i.e. $ \gamma_{a}\equal{} AD$, where $ D$ is the point where the incircle meets the side $ BC$.
Maybe you"ll obtain a more strong inequality than upper. Try and successfully !
Thus, you "ll can find the left inequality from http://www.mathlinks.ro/Forum/viewtopic.php?t=165410 [/color]


Daniel types 20 words per minute and Paul types 80 words per minute. They each type one half of an article. In words per minute, what is the rate at which the article was typed?

[hide]Let's say the article is $2x$ words, so Daniel and Paul each type $x$ words.  Daniel types for $\frac{x}{20}$ minutes and Paul types for $\frac{x}{80}$ minutes.  If you add them up, it is $\frac{x}{16}$ minutes.  So, the number of words per minute is $\frac{2x}{\frac{x}{16}}$ which is $32$. [/hide]


What's the difference between $\text{\TeX}$ and $\text{\LaTeX}$?

[quote="Dacicus"][url=http://acs.ucsd.edu/info/tex.php]This page[/url] gives a basic explanation.[/quote]

Aaah.  Thank you.  Maybe I should consider AMSTeX ;)


a,b are positive interger.gcd(a,b)=1,a^n+b^n divides to a^m+b^m.Prove n divides to m

daihiep285 wrote:
[quote] $ a,b$ are positive integers that $ gcd(a,b) \equal{} 1$ and $ a^n \plus{} b^n$ divides to $ a^m \plus{} b^m$.prove that $ n$ divides to $ m$.[/quote]

It's better!!!

First of all, let $ n\geq 2m$.
$ a^n \plus{} b^n \equal{} (a^{n \minus{} m} \plus{} b^{n \minus{} m})(a^m \plus{} b^m) \minus{} a^mb^m(a^{n \minus{} 2m} \plus{} b^{n \minus{} 2m})$.
$ gcd(a^mb^m,a^m \plus{} b^m) \equal{} 1$.
So, $ a^m \plus{} b^m$ divides $ a^{n \minus{} 2m} \plus{} b^{n \minus{} 2m}$.
It's easy to see that $ a^m \plus{} b^m$ divides $ a^{n \minus{} 4m} \plus{} b^{n \minus{} 4m}$,
$ a^m \plus{} b^m$ divides $ a^{n \minus{} 6m} \plus{} b^{n \minus{} 6m}$...

So, if $ n \equal{} km \plus{} l$, where $ 0\leq l < m$:
$ a^m \plus{} b^m$ divides $ a^l \plus{} b^l$ (if $ k$ is even) or $ a^m \plus{} b^m$ divides $ a^{m \plus{} l} \plus{} b^{m \plus{} l}$ (if $ k$ is odd).
First case is obvious.
The second case:
$ a^{m \plus{} l} \plus{} b^{m \plus{} l} \equal{} (a^l \plus{} b^l)(a^m \plus{} b^m) \minus{} a^lb^l(a^{m \minus{} l} \plus{} b^{m \minus{} l})$.
So, $ gcd(a^lb^l,a^m \plus{} b^m) \equal{} 1$, and $ a^m \plus{} b^m$ divides $ a^{m \minus{} l} \plus{} b^{m \minus{} l}$, which obviously false for $ l\neq 0$.

[quote] $ a,b$ are positive integers that $ gcd(a,b) \equal{} 1$ and $ a^n \plus{} b^n$ divides to $ a^m \plus{} b^m$. Prove that $ n$ divides to $ m$.[/quote]
We can prove that if there is $ (m,n)$ such that $ (a^n \plus{} b^n)\mid (a^m \plus{} b^m)$, then $ (m,n) \equal{} ((2k \plus{} 1)n,n)$ $ \forall$ $ n,k$ $ \in$ $ \mathbb {N}$. Indeed, we divide the problem into two cases.

[i]First case.[/i] If $ n\le m < 2n$. Suppose there is a $ (m,n)$ satisfied the problem. Then, we have,
\[ (a^n \plus{} b^n).(a^{m \minus{} n} \plus{} b^{m \minus{} n}) \minus{} (a^m \plus{} b^m) \equal{} (ab)^{n \minus{} m}.(a^{2n \minus{} m} \plus{} b^{2n \minus{} m})\]
Which leads to the fact that $ (a^n \plus{} b^n)\mid (ab)^{m \minus{} n}(a^{2n \minus{} m} \plus{} b^{2n \minus{} m})$. But since $ \gcd (a,b) \equal{} 1$, thus $ \gcd ((ab)^{m \minus{} n},a^n \plus{} b^n) \equal{} 1$. Then we obtain $ (a^n \plus{} b^n)\mid (a^{2n \minus{} m} \plus{} b^{2n \minus{} m})$, yielding that $ m\le n$ $ \Longrightarrow$ $ m \equal{} n$.

[i]Second case.[/i] If $ m > 2n$. Suppose there is $ (m,n)$ satisfied the problem. Then let $ m \equal{} 2kn \plus{} d$, we will prove that $ d \equal{} n$. Indeed, we will deduce on $ k$. For $ k \equal{} 0$, it is obviously true due to te [i]first case[/i]. Now suppose that for some $ k \minus{} 1$, the statement is still true, we will prove that it is also true for $ k$. We have,
\[ (a^n \plus{} b^n)(a^{(2k \minus{} 1)n \plus{} d} \plus{} b^{(2k \minus{} 1)n \plus{} d}) \minus{} (ab)^n.(a^{2(k \minus{} 1)n \plus{} d} \plus{} b^{2(k \minus{} 1)n \plus{} d}) \equal{} a^{2kn \plus{} d} \plus{} b^{2kn \plus{} d} \equal{} a^m \plus{} b^m\]
Which implies that $ (a^n \plus{} b^n)\mid (a^m \plus{} b^m)$, which is equivalent to $ (a^n \plus{} b^n)\mid (ab)^n.(a^{2(k \minus{} 1)n \plus{} d} \plus{} b^{2(k \minus{} 1)n \plus{} d})$, also note that $ \gcd ((ab)^n,a^n \plus{} b^n) \equal{} 1$ then according to the hypothesis of the deduction, this is true iff $ d \equal{} n$. Then our result is lead as followed.


geometry
analytic geometry
incenter
conics
hyperbola
trigonometry
parabola

(a) Let $ H$ be the orthocentre of acute triangle $ ABC$.  Let $ P$ be the point where the incircle of triangle $ HBC$ is tangent to side $ BC$, and define points $ Q$ and $ R$ similarly.  Show that $ AP$, $ BQ$, and $ CR$ are concurrent.

[hide="Hint/Comment"]The barycentric coordinates of the point of concurrency are $ \cot(45^\circ-A/2) : \cot(45^\circ-B/2) : \cot(45^\circ-C/2)$.  This point does not appear to be in the [url=http://faculty.evansville.edu/ck6/encyclopedia/ETC.html]ETC[/url].[/hide]

(b)* What other points can the orthocentre be replaced by?

In the paper "Triangle centers associated with the Malfatti circles" ([url]http://forumgeom.fau.edu/FG2003volume3/FG200308index.html[/url]), it is shown that the incentre has this property, and the point of concurrency is $ X_{1143}$.

The circumcentre trivially has this property, and the point of concurrency is the centroid.

I believe the locus of all points $ P$ satisfying the problems property is a part of an hyperbola (at last my sketchpad suggests so) and I don't think we can find other famous points that lie on it besides $ H,O$ and $ I$.

[quote="nsato"][color=darkred]Let $ H$ be the orthocentre of an acute $ \triangle\ ABC$.  Let $ P$ be the point where the incircle of $ \triangle\ HBC$ is tangent to the line  $ BC$.
Define similarly the points $ Q$ and $ R$. Show that the lines $ AP$, $ BQ$ and $ CR$ are concurrently.[/color][/quote]
[color=darkblue][b]Proof.[/b] Prove easily that $ \frac{PB}{PC}=\frac{BC+HB-HC}{BC+HC-HB}=$ $ \frac{\sin A+\cos B-\cos C}{\sin A+\cos C-\cos B}=$ $ \frac{\tan \left(\frac{\pi}{4}-\frac{B}{2}\right)}{\tan\left(\frac{\pi}{4}-\frac{C}{2}\right)}$ a.s.o. Therefore,
$ \cot\left(\frac{\pi}{4}-\frac{A}{2}\right)\cdot\overrightarrow{PA}+\cot\left(\frac{\pi}{4}-\frac{B}{2}\right)\cdot\overrightarrow{PB}+\cot\left(\frac{\pi}{4}-\frac{C}{2}\right)\cdot\overrightarrow{PC}=0$. Observe that $ \cot\left(\frac{\pi}{4}-\frac{A}{2}\right)=\frac{1+\sin A}{\cos A}$ a.s.o.[/color]

[color=darkred][b]Remark.[/b] Let $ M$ be an interior point of $ \triangle\ ABC$. Let $ P$ be the point where the incircle of $ \triangle\ BMC$ is tangent to $ BC$.
Define similarly the points $ Q\in (CA)$ and $ R\in (AB)$. Denote $ MA=x$, $ MB=y$, $ MC=z$. Prove that
$ AP\cap BQ\cap CR\ne\emptyset$ $ \Longleftrightarrow$ $ \frac{PB}{PC}\cdot\frac{QC}{QA}\cdot\frac{RA}{RB}=1$ $ \Longleftrightarrow$ $ \frac{a+y-z}{a+z-y}\cdot\frac{b+z-x}{b+x-z}\cdot\frac{c+x-y}{c+y-x}=1$ $ \Longleftrightarrow$
$ (a+y-z)(b+z-x)(c+x-y)=(a+z-y)(b+x-z)(c+y-x)$. Therefore,
$ \boxed{AP\cap BQ\cap CR\ne\emptyset\Longleftrightarrow (y-z)(z-x)(x-y)+a(b-c)x+b(c-a)y+c(a-b)z=0}$.[/color]

The locus is definitely is not a parabola, and probably not a cubic either; I'm thinking it's a fairly high-degree curve.  At any rate, it's  a nice property of the orthocentre.


limit
LaTeX
calculus
calculus computations

(i) Sketch the graph of $ y\equal{}\coth x$ and give the equations of any asymptotes.

(ii) $ f(x) \equal{} x\tanh x \minus{} 2$. Use the Newton-Raphson method, with a first approximation $ x_1 \equal{} 2$, to find the next three approximations $ x_2$, $ x_3$ and $ x_4$ to a root of $ f(x)\equal{}0$. Give the answers correct to $ 4$ decimal places.

(iii) If $ f(x) \equal{} 0$, show that $ \coth x \equal{} \frac{1}{2} x$. Hence write down the roots of $ f(x)\equal{}0$, correct to $ 4$ decimal places.

hello, for (i), we have $ \coth(x)\equal{}\frac{e^x\plus{}e^{\minus{}x}}{e^x\minus{}e^{\minus{}x}}\equal{}\frac{e^{2x}\plus{}1}{e^{2x}\minus{}1}$ hence we get $ \lim_{x\to\infty}\coth(x)\equal{}1$ and $ \lim_{x\to\minus{}\infty}\frac{e^{2x}\plus{}1}{e^{2x}\minus{}1}\equal{}\minus{}1$.
Sonnhard.

And the rest of the question?

Und der Rest der Frage?

For part (ii), I have (probably made some mistakes):

I can't find the $ \text{\LaTeX}$ command for sech and so have used $ \frac {1}{\cosh}$.

$ f'(x) = \frac {x}{\cosh ^{2}x} + \tanh x$

Substituting into the Newton-Raphson formula $ x_{n + 1} = x_{n} - \frac {f(x_{n})}{f'(x_{n})}$,

$ x_{2} = 2.065089059$

$ x_{3} = 2.065338136$

$ x_{4} = 2.065338139$,

giving

$ x = 2.0653$ to $ 4$ decimal places.

For part (iii), the equation is equivalent to that in part (ii) and since the graph of coth x is symmetric about the origin, I guess the roots are

$ \pm2.0653$ to $ 4$ decimal places.

Is this correct?

Am I correct? Are there other ways of doing part (iii)?

Could someone check this. I want to be sure it's correct.

I still don't know whether or not my answer is correct.

Are there other ways, besides Newton-Raphson, of answering this question?

hello, i have got $ x \approx 2.065338139$
Sonnhard.


10 at least...idk though i dont remember my answers

There's already a thread with a poll on this...lock?

Please refer to this thread:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=195692

[quote="buzzer11"]How about for 9th graders and below?[/quote]Probably a 7 or 8...I've been hearing 144/7 and 145.5/7....but if it's 7, expect a high index. Like 200.

looking at this poll makes me feel dumb. I made some mistakes on 2-3 problems, and i got an $ 8$.

That poll is immensely skewed.
That being said, I got a 9. Problems 1-9, no dumb mistakes, failed to do 10 and 12 in the last 30 minutes.


function
analytic geometry
graphing lines
slope
calculus
derivative

[hide]$\frac{f(\sqrt3)-f(\sqrt2)}{\sqrt3-\sqrt2} = \frac{-2\sqrt3+2\sqrt2}{\sqrt3-\sqrt2}$

Multiply top and bottom by $\sqrt3+\sqrt2$:

$\frac{-6+\sqrt{24}-\sqrt{24}+4}{1}=-2$[/hide]

[hide]you mixed up the signs in the third part of your answer. try again.[/hide]

to avoid confusion, i was referring to drunken math's answer. 236factorial and i posted at the same time. by the way, 236factorial is correct. :)

I fixed it.

Geez, i'm making a bunch of careless answers today... sorry.  :blush:

[quote="dakyru"]I'm not sure if anyone noticed, but it's simply the slope of the line: $\frac{dy}{dx} = -2$[/quote]

Yeah, take the deriative of -2x-3 to get -2. Let's not get there.

Well, I wasn't referring to the calculus aspect, I just used the notation $\frac{dy}{dx}$ to mean change in y over change in x.  Which is what the expression was, meaning it's the slope of the line, -2.  No need to actually evaluate the expression.


inequalities
geometry
circumcircle
trigonometry
geometry unsolved

For three points $ X,Y,Z$ let $ R_{XYZ}$ be the circumcircle radius of the triangle $ XYZ.$ If $ ABC$ is a triangle with incircle centre $ I$ then we have:

\[ \frac{1}{R_{ABI}} \plus{} \frac{1}{R_{BCI}} \plus{} \frac{1}{R_{CAI}} \leq \frac{1}{\bar{AI}} \plus{} \frac{1}{\bar{BI}} \plus{} \frac{1}{\bar{CI}}.\]

we have that $ AI\equal{}\dfrac{r}{\sin \frac{A}{2}}$ and $ R_{BCI}\equal{}2R\sin\frac{A}{2}$, then, the inequality is equivalent to

$ \dfrac{r}{2R}\sum\dfrac{1}{\sin \frac{A}{2}}\leq \sum \sin \frac{A}{2}$

recall that $ \dfrac{r}{R}\equal{}4\prod \sin\frac{A}{2}$... this implies that we have to prove that

$ 2\sum \sin\frac{B}{2}\sin\frac{C}{2} \leq \sum \sin\frac{A}{2}$

this one follows from the well-known inequalities $ (x\plus{}y\plus{}z)^2\geq3(xy\plus{}yz\plus{}zx)$ and $ \sum \sin\frac{A}{2}\leq \dfrac{3}{2}$, and we're done  :D

It's actually an older problem of mine (?) that this is in fact true for any point $P$ inside triangle $ABC$ (not necessarily $P=I$).


Write numbers (not all zero) in the square on a 5x5 grid so that the number in every square would be the sum of the numbers in those squares that share a side with it.

that problem looks nasty, because every single time I get stuck on the last few grids

OH i know!

[hide]

1 1 0 1 1
0 0 1 0 0
0 0 1 0 0
1 0 0 0 1
1 0 0 0 1

[/hide]

don't kill me for having so many zeroes. and teh ones could be any digit.

Sorry, but that answer is wrong...even though I don't know the answer, I know that 1+0 does not = 0
1+1 does not equal 0
1+1+1 does not equal zero.

Try negatives...the conditions didn't say you couldn't.

oh yea...i was just thinking about the ones....oops

so try this:
[hide]
1 1 0 -1 -1
0 0 0 0 0
-1 -1 0 1 1
0 0 0 0 0
1 1 0 -1 -1

can someone tell me how to do tables in latex?

let's see if this works:
$\displaystyle \begin{tabular}{|c|c|c|c|c|}
\hline 1   & 1     & 0 & -1 & -1 \\
\hline 0 & 0 & 0 & 0 & 0 \\
\hline -1 & -1 & 0 & 1 & 1\\
\hline 0 & 0 & 0 & 0 & 0 \\
\hline 1 & 1 & 0 & -1 & -1\\
\end{tabular}\\$

[/hide]
yay! it works :jump: now why doesn't the bottom line show up?

Your solution looks correct, and I don't think the bottom line matters too much.  :lol:


inequalities
number theory unsolved
number theory

find all positive integers $(x,n)$ such that $x^n + 2^n +1$ is a divisor of $x^{n+1}+2^{n+1}+1$

[quote="cefer"]Its from  romanian math olimpiad 1998
It is easy and nice
Represent $x^{n+1}+2^{n+1}+1=x(x^n+2^n+1)-x2^n-x+2^{n+1}+1$
Then $x2^n+x-2^{n+1}-1$ is divisible by $x^{n}+2^{n}+1$
Using inequalities we can get answers $(x,n)=(4,1),(11,1)$. :)[/quote]

Ok!  please can you explain your solution.

Davron

i have no time to explain now but i will explain later if you want?

Here is my solution Davron
If $x=1$ then $2^{n+1}+2$ is divisible by $2^n+2$ (no such $n$)
If $x\geq2$
If $n=1$ then $x^2+5$ is divisible by $x+3$
From here we have $(x,n)=(4,1),(11,1)$
If $n\geq2$ represent $x^{n+1}+2^{n+1}+1=x(x^n+2^n+1)-(x2^n+x-2^{n+1}-1)$
Then $x2^n+x-2^{n+1}-1\geq 0$ is divisible by $x^n+2^n+1$
So $2^n+1\geq x^{n-1}+\frac{3.2^{n}+2}{x}$
If $3\geq x\geq 2$ then $2^n+1\geq x^{n-1}+\frac{3.2^{n}+2}{x}\geq1+\frac{3.2^{n}+2}{3}\ge2^n+1$ (equality cannot occur)(contradiction)
If $x=4$ then $2^n+1\geq 4^{n-1}+\frac{3.2^{n}+2}{4}\rightarrow\ 2^{n+2}+4\geq2^{2n}+3.2^{n}+2\rightarrow\ 2^n+2\geq2^{2n}$ (contradiction)
If $x\geq5$ then $2^n+1\geq x^{n-1}+\frac{3.2^{n}+2}{x}\geq5^{n-1}\geq 4^{n-1}+1\geq2^n+1$(equality cannot occur)(contradiction)
So only solutions are $(x,n)=(4,1),(11,1)$
   Cefer

Thank you my brother Cefer for your soltion i enjoyed that  :)
And sorry that i took your time

Thankx

Davron


Find the greatest $k\in N$ such that among any $k$ consecutive integers, there is at least one which is not equal to the $4th$ power of any integers.

Are you sure about the question? It's kind of weird.

I think it should find least $k \in N$, and should be $k_{min} = 3$.
Right?

Yes, indeed, although it's completely trivial... ? Strange, where did you get the question from?

Maybe he meanse "sum of different 4th powers"? And that we would somehow have to prove that no 4?5? consequent numbers can be the sum of different 4th powers? However that already goes far off the original one :?


Let $ a, b, c, r,$ and $ s$ be real numbers. Show that if $ r$ is a root of $ ax^2\plus{}bx\plus{}c \equal{} 0$ and s is a root of $ \minus{}ax^2\plus{}bx\plus{}c \equal{} 0,$  then \[ \frac{a}{2} x^2 \plus{} bx \plus{} c \equal{} 0\] has a root between $ r$ and $ s.$

[quote="orl"]Let $ a, b, c, r,$ and $ s$ be real numbers. Show that if $ r$ is a root of $ ax^2 \plus{} bx \plus{} c \equal{} 0$ and s is a root of $ \minus{} ax^2 \plus{} bx \plus{} c \equal{} 0,$  then
\[ \frac {a}{2} x^2 \plus{} bx \plus{} c \equal{} 0
\]
has a root between $ r$ and $ s.$[/quote]
Its rather easy $ f(x) \equal{} \frac {a}{2} x^2 \plus{} bx \plus{} c$ so $ f(r) \equal{} \minus{} \frac {a}{2}r^2$ and $ f(s) \equal{} \frac {3a}{2}s^2$ so $ f(s)f(r) \equal{} \minus{} \frac {3a^2}{4}r^2s^2\le 0$ so there are root of f(x) between $ r$ and $ s$


function
algebra
domain
induction
geometry
perpendicular bisector
strong induction

Consider a regular $n$-gon with $n$ odd. Given two adjacent vertices $A_{1}$ and $A_{2},$ define the sequence $(A_{k})$ of vertices of the $n$-gon as follows: For $k\ge 3,\, A_{k}$ is the vertex lying on the perpendicular bisector of $A_{k-2}A_{k-1}.$ Find all $n$ for which each vertex of the $n$-gon occurs in this sequence.

This is a half of the solution... we exclude the case n is not a power of 3.

Instead of working on the vertices of a n-gon, we work in $\mathbb{Z}_{n}$. The sequence is defined as:
$a_{0}=0$
$a_{1}= 1$
$a_{n+2}= 2^{-1}a_{n+1}+2^{-1}a_{n}$

This is a standard recurrence... it characteristic equation is $x^{2}-2^{-1}x-2^{-1}$ with roots:
$2^{-1}(2^{-1}\pm (2^{-2}+2^{2}\cdot 2^{-1})^{\frac{1}{2}}) = 2^{-2}\pm 2^{-1}(2^{-2}(1+8))^{\frac{1}{2}}= 2^{-2}\pm 3\cdot 2^{-2}$
ok, the roots are $1,-2^{-1}$.

So, the solution is of the form $a_{n}= k(-2^{-1})^{n}+j$.
First, we suppose $3 \nmid n$.
Checking for n=0,1 we get $k=-2\cdot3^{-1}, j = 2 \cdot 3^{-1}$.
Finally we get:
$a_{n}=-2\cdot 3^{-1}\cdot(-2^{-1})^{n}+2\cdot 3^{-1}$.
A function of $\mathbb{Z}_{n}$ is surjective iff $af(x)+b$, with $a \neq 0$ is surjective.
Then we have only to check that $(-2)^{-n}$ is surjective, but this obviously false because it is never 0.
Finally, the number of vertices we get is the order of -2 modulo n. (if n is not a multiple of 3).

If n is a multiple of 3, we prove that the same sequence is not surjective in $\mathbb{Z}_{k}$, where k is the greatest divisor of n that is not a multiple of 3. To prove this, just use the method above. Since the sequence is not surjective in $\mathbb{Z}_{k}$, it is not also in $\mathbb{Z}_{n}=\mathbb{Z}_{k}\times \mathbb{Z}_{3^{i}}$.

The case n is a power of 3 remains open... I have checked that with n=3,9 it is surjective.

Let $f$ be a function whose domain is $\left\{A_1,A_2,\cdots,A_k\right\}$ and such that $f(A_1)=0$, $f(A_2)=1$ and $f(A_k)$ is the arithmetic mean of $f(A_{k-1})$ and $f(A_{k-2})$, for $k\ge3$.
Then it's easy to prove, using strong induction, that $f(A_{k+2})=\frac{2^{k+1}+(-1)^k}{3\cdot2^k}$, for $k\ge-1$.
Suppose $n$ is a power of $3$. Let $p\ne3$ be a prime divisor of $n$. Let $m$ be an integer such that $3m\equiv_p2$ and suppose there is a vertex $A_{k+2}$ such that $f(A_{k+2})\equiv_nm$.
Therefore we have: $f(A_{k+2})\equiv_pm\Leftrightarrow2^{k+1}+(-1)^k\equiv_p3m\cdot2^k\Leftrightarrow(-1)^k\equiv_p2^k(3m-2)\Leftrightarrow(-1)^k\equiv_p0$, which is impossible.
Therefore $n$ must be a power of $3$.
Lemma: $\mbox{ord}_{3^k}(-2)=3^{k-1}$, where $k$ is a positive integer.
Proof:
For $k=1$ it's trivial. Consider $k\ge2$. Since $3^1\parallel(-2)-1$ and $3^{k-1}\parallel3^{k-1}$, then, lifting the exponent, we have $3^k\parallel(-2)^{3^{k-1}}-1$, therefore $\mbox{ord}_{3^k}(-2)\mid3^{k-1}$. Suppose $\mbox{ord}_{3^k}(-2)\ne3^{k-1}$. Then $\mbox{ord}_{3^k}(-2)\mid3^{k-2}$, therefore $3^k\mid(-2)^{3^{k-2}}$. Since $(-2)^{3^{k-1}}-1=((-2)^{3^{k-2}}-1)((-2)^{2\cdot3^{k-2}}+(-2)^{3^{k-2}}+1)$, $3^k\mid(-2)^{3^{k-2}}-1$ and $3\mid(-2)^{2\cdot3^{k-2}}+(-2)^{3^{k-2}}+1$, then $3^{k+1}\mid(-2)^{3^{k-1}}-1$, a contradiction.
Now let's prove that, for every integer $m$, there is a vertex $A_{k+2}$ such that $f(A_{k+2})\equiv_nm$.
Therefore we have: $f(A_{k+2})\equiv_nm\Leftrightarrow2^{k+1}+(-1)^k\equiv_{3n}3m\cdot2^k\Leftrightarrow(-2)^k(3m-2)\equiv_{3n}1$.
Note that $3m-2$ covers all the remainders modulo $3n$ that are congruent to $1$ modulo $3$. From the Lemma, $(-2)^k$ covers all the remainders modulo $3n$ that are congruent to $1$ modulo $3$, from where the result follows.
Concluding, each vertex of the $n$-agon occurs on the sequence if and only if $n$ is a power of $3$.


For each natural numbers $ n\ge 2$, we define the following transformation

$ n\to \frac {n}{2}$ if $ n$ is even.
$ n \to 3n \plus{} 1$ if $ n$ is odd.

Do we know, as expressed in form of $ n$, how many transformation will be done before $ n$ become $ 1$?

I think this problem is quite famous in CS? right?

Hung,
See here //http://en.wikipedia.org/wiki/Collatz_conjecture  :D

Thanks, Tuan. That is very useful for me.

A simple Computer program gives it some insight

Please Enter a natural number : 13
13 is odd, so I multiply it by 3 and add 1: 40
40 is even, so I take half: 20
20 is even, so I take half: 10
10 is even, so I take half: 5
5 is odd, so I multiply it by 3 and add 1: 16
16 is even, so I take half: 8
8 is even, so I take half: 4
4 is even, so I take half: 2
2 is even, so I take half: 1
The process took 9 round to reach 1.

Please Enter a natural number : 27
27 is odd, so I multiply it by 3 and add 1: 82
82 is even, so I take half: 41
41 is odd, so I multiply it by 3 and add 1: 124
124 is even, so I take half: 62
62 is even, so I take half: 31
31 is odd, so I multiply it by 3 and add 1: 94
94 is even, so I take half: 47
47 is odd, so I multiply it by 3 and add 1: 142
142 is even, so I take half: 71
71 is odd, so I multiply it by 3 and add 1: 214
214 is even, so I take half: 107
107 is odd, so I multiply it by 3 and add 1: 322
322 is even, so I take half: 161
161 is odd, so I multiply it by 3 and add 1: 484
484 is even, so I take half: 242
242 is even, so I take half: 121
121 is odd, so I multiply it by 3 and add 1: 364
364 is even, so I take half: 182
182 is even, so I take half: 91
91 is odd, so I multiply it by 3 and add 1: 274
274 is even, so I take half: 137
137 is odd, so I multiply it by 3 and add 1: 412
412 is even, so I take half: 206
206 is even, so I take half: 103
103 is odd, so I multiply it by 3 and add 1: 310
310 is even, so I take half: 155
155 is odd, so I multiply it by 3 and add 1: 466
466 is even, so I take half: 233
233 is odd, so I multiply it by 3 and add 1: 700
700 is even, so I take half: 350
350 is even, so I take half: 175
175 is odd, so I multiply it by 3 and add 1: 526
526 is even, so I take half: 263
263 is odd, so I multiply it by 3 and add 1: 790
790 is even, so I take half: 395
395 is odd, so I multiply it by 3 and add 1: 1186
1186 is even, so I take half: 593
593 is odd, so I multiply it by 3 and add 1: 1780
1780 is even, so I take half: 890
890 is even, so I take half: 445
445 is odd, so I multiply it by 3 and add 1: 1336
1336 is even, so I take half: 668
668 is even, so I take half: 334
334 is even, so I take half: 167
167 is odd, so I multiply it by 3 and add 1: 502
502 is even, so I take half: 251
251 is odd, so I multiply it by 3 and add 1: 754
754 is even, so I take half: 377
377 is odd, so I multiply it by 3 and add 1: 1132
1132 is even, so I take half: 566
566 is even, so I take half: 283
283 is odd, so I multiply it by 3 and add 1: 850
850 is even, so I take half: 425
425 is odd, so I multiply it by 3 and add 1: 1276
1276 is even, so I take half: 638
638 is even, so I take half: 319
319 is odd, so I multiply it by 3 and add 1: 958
958 is even, so I take half: 479
479 is odd, so I multiply it by 3 and add 1: 1438
1438 is even, so I take half: 719
719 is odd, so I multiply it by 3 and add 1: 2158
2158 is even, so I take half: 1079
1079 is odd, so I multiply it by 3 and add 1: 3238
3238 is even, so I take half: 1619
1619 is odd, so I multiply it by 3 and add 1: 4858
4858 is even, so I take half: 2429
2429 is odd, so I multiply it by 3 and add 1: 7288
7288 is even, so I take half: 3644
3644 is even, so I take half: 1822
1822 is even, so I take half: 911
911 is odd, so I multiply it by 3 and add 1: 2734
2734 is even, so I take half: 1367
1367 is odd, so I multiply it by 3 and add 1: 4102
4102 is even, so I take half: 2051
2051 is odd, so I multiply it by 3 and add 1: 6154
6154 is even, so I take half: 3077
3077 is odd, so I multiply it by 3 and add 1: 9232
9232 is even, so I take half: 4616
4616 is even, so I take half: 2308
2308 is even, so I take half: 1154
1154 is even, so I take half: 577
577 is odd, so I multiply it by 3 and add 1: 1732
1732 is even, so I take half: 866
866 is even, so I take half: 433
433 is odd, so I multiply it by 3 and add 1: 1300
1300 is even, so I take half: 650
650 is even, so I take half: 325
325 is odd, so I multiply it by 3 and add 1: 976
976 is even, so I take half: 488
488 is even, so I take half: 244
244 is even, so I take half: 122
122 is even, so I take half: 61
61 is odd, so I multiply it by 3 and add 1: 184
184 is even, so I take half: 92
92 is even, so I take half: 46
46 is even, so I take half: 23
23 is odd, so I multiply it by 3 and add 1: 70
70 is even, so I take half: 35
35 is odd, so I multiply it by 3 and add 1: 106
106 is even, so I take half: 53
53 is odd, so I multiply it by 3 and add 1: 160
160 is even, so I take half: 80
80 is even, so I take half: 40
40 is even, so I take half: 20
20 is even, so I take half: 10
10 is even, so I take half: 5
5 is odd, so I multiply it by 3 and add 1: 16
16 is even, so I take half: 8
8 is even, so I take half: 4
4 is even, so I take half: 2
2 is even, so I take half: 1
The process took 111 round to reach 1.

Sorry, I dont want to spam. Just I am very impressed.


[hide="solution"]when i looked at this problem i was like  :huh:
then i realized 0 was in between those numbers.
any thing  times 0 is [b]0[/b][/hide]

same here i was a little unsure how to solve it until i realized 0 was in the middle. so yes my answer is 0.

in the beginning, I thought the result was
$-(30!)^2$, a monster of a number!!!
But that zero cured it all.

If 0 is one of the numbers...

then it's obvious..

0...  :)


Given $ a, b, c > 0$. Find the best constant for the following ineq trues:
$ \frac {a^3}{b} \plus{} \frac {b^3}{c} \plus{} \frac {c^3}{a} \geq\ (k \plus{} 1)(a^2 \plus{} b^2 \plus{} c^2) \minus{} k(ab \plus{} bc \plus{} ca)$

A similar and well-known problem of can_hang:
$ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \geq\ \frac {6(a^2 \plus{} b^2 \plus{} c^2) \minus{} 3(ab \plus{} bc \plus{} ca)}{a \plus{} b \plus{} c}$
I have proved the second by Am-Gm (can_hang also checked it) :)

[quote="nguoivn"]Given $ a, b, c > 0$. Find the best constant for the following ineq trues:
$ \frac {a^3}{b} \plus{} \frac {b^3}{c} \plus{} \frac {c^3}{a} \geq\ (k \plus{} 1)(a^2 \plus{} b^2 \plus{} c^2) \minus{} k(ab \plus{} bc \plus{} ca)$

[/quote]
I have obtained $ k_{max}\equal{}\min_{[1,\frac{4}{3})}\frac{9x^2\minus{}18x\plus{}6\plus{}\sqrt{12(x\minus{}1)}}{3x\minus{}4}\equal{}2.58141218...$


Anyone know how to prove this?

Distance of point $(x_1,y_1,z_1)$ from $Ax+By+Cz + D = 0$

$= \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$

Let the point P be (a,b,c).
    The equation of PM will be
           (x-a)/A = (y-b)/B = ...... = r
    Therefore the co ordinates of any pt. on this line is
            (a+Ar, b+Br, c+Cr) which gives
             r = -(Aa+Bb+Cc) / (A^2+B^2+C^2)
    Therefore
             PM = sq. root[ (a+Ar-a)^2 + .......]

                   =sq. root[ A^2 + B^2 + C^2]*|r|

     Substitute for r and simplify.

[hide="Solution"]
Call point $(x_1, y_1, z_1)$ the point $P$.  Take another point, $Q(x_2, y_2, z_2)$ in the desired plane, and construct the vector $\vec{V} = <x_1 - x_2, y_1 - y_2, z_1 - z_2>$ from $Q$ to $P$.

Next, consider the normal vector $\vec{N} = <A, B, C>$ of the plane.  The expression $\frac{|\vec{V} \cdot \vec{N}|}{|\vec{N}|}$ will give the length of the projection of $V$ onto $N$ - precisely the desired distance.

$\vec{V} \cdot \vec{N} = A(x_1 - x_2) + B(y_1 - y_2) + C(z_1 - z_2) = Ax_1 + By_1 + Cz_1 + D$ (remember that $Q$ is on the plane, so it satisfies the plane equation)
$|\vec{N}| = \sqrt{A^2 + B^2 + C^2}$ (Pythagorean theorem)

And the result follows.
[/hide]

heh, looks like I should learn some vectors.


superior algebra
superior algebra unsolved

Show that any prime ideal in a ring R contains minimal prime ideal . R is noncommutative.

This is a standard application of zorns lemma. The intersection of a chain of prime ideals is prime itself.


[quote="manuel"]Prove that in a room with (m-1)n+1 persons there are m mutual strangers or there is a person who knows n persons.[/quote]
As the administrators said, please stop writing 'easy' in the title of your topics. thanks

err i think im misunderstanding the problem

say there is a room with $(3-1)(4)+1 = 9$ people

couldn't there then be a room of $5$ mutual strangers, and then one person who knows $3$ other people?


I have one square inscribed in another such that the 4 vertices of the smaller square are on the edges of the larger and I have drawn 1 x 1 grids in the square. I want Asymptote to tell me how many intersections there are of both the 1 x 1 grids. I have already drawn the image. Is there a command that bullets the intersections? Thanks a lot.

Hello,

you are searching the command [b]intersectionpoints(path1,path2).length[/b].
Here an example:
[asy]size(10cm);
path p1=randompath(20), p2=randompath(20);

draw(p1^^p2);
pair[] inter=intersectionpoints(p1,p2);
dot(inter,5bp+red);
label("There are "+string(inter.length)+" intersection points");[/asy]

@pytheagle: I don't reply to private mail for coding; my answer can also help somebody else...
Here is a not so bad solution for your problem:
[asy]size(10cm);
path[] grid(int Nx, int Ny=Nx)
{
  path[] P;
  for(int i=0; i <= Nx; ++i) P.push((i,0)--(i,Ny));
  for(int j=0; j <= Ny; ++j) P.push((0,j)--(Nx,j));
  return P;
}

path[] G1=grid(17);
draw(G1);

path[] G2=shift(12E)*rotate(aCos(5/13))*grid(13);
draw(G2);

pair[] inter=intersectionpoints(G1,G2);
dot(inter,red);
label("There are "+string(inter.length)+" intersection points",8.5E,align=S);[/asy]

Is this OK for you ?


May I ask somebodies to give me some tips about solving these problems? I know these may be easy for the others but I really don't know how to solve them. Thanks a lot.

1. Let $v_{1}$ and $v_{2}$ be vectors in $R^{n}$. Prove the following set equalities by showing that each of the spans is contained in the other. a. sp($v_{1}$, $v_{2}$) = sp($v_{1}$, 2$v_{1}$+$v_{2}$) b. sp($v_{1}$, $v_{2}$) = sp($v_{1}$+$v_{2}$, $v_{1}$-$v_{2}$)

2. Let $W_{1}$, and $W_{2}$ be two subspaces of $R^{n}$. Prove that their intersection $W_{1}\cap W_{2}$ is also a subspace.

3. let A and C be matrices such that the product AC is defined.
a. Prove that rank(AC) $\leq$ rank(A).
b. Is it true that rank(AC) $\leq$ rank(C)? Explain.

4. Let T: $R^{n}\rightarrow R^{m}$ be a linear transformation and let U be a subspace of $R^{m}$. Prove that the inverse image $T^{-1}$[U] is a subspace of $R^{n}.$

[quote="RICHMATH"]thanks a lot.

3. let A and C be matrices such that the product AC is defined.
a. Prove that rank(AC) $\leg$ rank(A).
b. Is it true that rank(AC) $\leg$ rank(C)? Explain.[/quote]

w u mean?
write the formula readable.

[quote="RICHMATH"]thanks a lot.

3. let A and C be matrices such that the product AC is defined.
a. Prove that rank(AC) $\leq$ rank(A).
b. Is it true that rank(AC) $\leq$ rank(C)? Explain.

[color=green]["Less than or equal to" is spelled \leq or just plain \le, not \leg. That's a Q, not a G. - Moderator.][/color][/quote]

I am sorry about that............


Not really that hard...
Find $ \displaystyle\sum_{i=2}^{100} {\frac 1{log_i{100!}}}$.

[quote="Brut3Forc3"]Not really that hard...
Find $ \displaystyle\sum_{i = 2}^{100} {\frac 1{log_i{100!}}}$.[/quote]

[hide="answer"]
1

just change the terms to the form of logi/log100! and sum them up
[/hide]


inequalities
function
search
quadratics
inequalities unsolved

I have an inequalitie wich I think is easy, but I don't know to solve it. :blush: We take the numbers x-[0;1],y-[1;2] and z-[2;3].Prove that 3/4=<x^2+y^2+z^2-xy-yz-zx=<7.Can anyone help me?

[quote="spanferkel"]We can write equivalently $\frac32\le(y-x)^2+(z-y)^2+(z-x)^2\le14.$
So the expression in the middle is just a function of the differences. From there you can see that for its minimum, we must have $x$ biggest and $z$ smallest possible, i.e. we are looking for the minimum of $(y-1)^2+(2-y)^2$.
Likewise the maximum is attained if $x$ is smallest and $z$ biggest possible => search the maximum of $y^2+(3-y)^2$. :)[/quote]
   Yes.I understood.Thank you!

Or you can try it with the sign of the quadratic function.


Source: AIME 1990 #12
12. A regular 12-gon is inscribed in a circle of radius $12$. The sum of the lengths of all sides and diagonals of the 12-gon can be written in the form $a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}$, where $a$, $b$, $c$, and $d$ are positive integers. Find $a+b+c+d$.

It's $\sqrt{2}$ not $\sqrt{12}$
[hide]First, get the sides of the 12-gon by using $\sin 15=\frac{\sqrt6-\sqrt2}{4}$ from the difference formula.
The side length is $6\sqrt6-6\sqrt2$ and multiply it by 12, $72(\sqrt6-\sqrt2)$.
The next shortest diagonal is 12 since it is part of a hexagon. Each hexagon contains 6 sides and there are 2 different hexagons, $12\times6\times2=144$
The next shortest forms a squre with side length $12\sqrt2$. Each square has 4 sides and there are 3 different squares. $12\sqrt2\times4\times3=144\sqrt2$
The next shortest forms a triangle with side length $12\sqrt3$. Each triangle has 3 sides and there are 4 different triangles. $12\sqrt3\times3\times4=144\sqrt3$
We can get the next shortest by $\sin 75=\frac{\sqrt6+\sqrt2}{4}$ and the length is $6(\sqrt6+\sqrt2)$, there are 12 of them. $6(\sqrt6+\sqrt2)\times12=72(\sqrt6+\sqrt2)$
The longest which is the diagonal of the circle has length 12 and there are 12 of them. $12\times12=144$
Adding everything, $72(\sqrt6-\sqrt2)+144+144\sqrt2+144\sqrt3+72(\sqrt6+\sqrt2)+144$
$=288+144\sqrt2+144\sqrt3+144\sqrt6$ the answer is $144\times5=720$[/hide]

[hide]There are 6 diagonals that are diameters of the circle. The sum of their lengths is $6\times24$.

There are 12 diagonals that connect vertices that are seperated by two vertices in between. The sum of their lengths is $12\times(12\sqrt{2})$, since they form 45-45-90 triangles with the diameter.

There are 12 diagonals that connect vertices that are seperated by 3 vertices in between, and 12 diagonals that connect vertices that are seperated by 1 vertex in between. The sum of their lengths is $12\times(12+12\sqrt{3})$, since they form 30-60-90 triangles with the diameter.

There are 12 diagonals that connect vertices that are seperated by 4 vertices in between, and the 12 sides of the 12-gon. These form 15-75-90 triangles. We find that $\sin 75^\circ=\frac{\sqrt{6}+\sqrt{2}}{4}$, and $sin 15^\circ=\frac{\sqrt{6}-\sqrt{2}}{4}$, so the sum of their lengths is $12\times(2*6\sqrt{3})$, ($\sqrt{2}$'s cancel).

So our total is $12(12+12+12+12+12)=720$.[/hide]

Oh, sorry for the typo...it's edited now.

So the answer is 720? Then kalva has a wrong ans. for this pbm.

I've got 720 too and I read the kalva which says 594...

Thanks guys,


geometry
parallelogram
area of a triangle
Heron\u0027s formula

Show that the area of a triangle with medians m_1, m_2, and m_3 is

(4/3)Sqrt(s_m(s_m-m_1)(s_m-m_2)(s_m-m_3))

where s_m = (1/2)(m_1+m_2+m+3).

I would be fascinated to see a geometric proof of this !

[hide]Waiting for it, here is an ugly algebraic one :

4 *m_a = 2b+2c-a and the same by permutation.

Believe it or not

((4/3)Sqrt(s_m(s_m-m_1)(s_m-m_2)(s_m-m_3)))^2 with my notations simplifies (with efforts !) to

1/16*(-a^4-b^4-c^4 + 2ab + 2bc + 2ca) which is Heron's formula !!!!![/hide]

[hide]First by Heron's formula, the given expression is equal to 4/3 times the area of a triangle with side lengths m_1, m_2, and m_3. First, we will construct such a triangle.

Draw triangle ABC with medians AA', BB', and CC'. Complete parallelogram B'BA'D. We have A'D = BB' = m_2. Also, since B'D and A'C are parallel and equal (they both measure a/2), B'A'CD is a parallelogram, so that CD = A'B' = c/2 = AC'. Since CD || A'B' || AC', we have CD and C'A equal and parallel, so CDAC' is a parallelogram. Therefore, AD = CC' = m_3. Finally, since AA' = m_1, AA'D is a triangle with side lengths m_1, m_2, and m_3.

Now we can find it's area. First, we have [A'B'D]=[A'B'C]=K/4, where K is the area of triangle ABC. Also, [A'B'A]=[A'B'C']=K/4. Finally, [AB'D]=[B'CD]=[B'CA']= K/4. Thus, [AA'D]=[AB'A']+[A'B'D]+[DB'A]=3K/4. Therefore, K = 4/3[AA'D] as desired.[/hide]


[quote="sharkman"]Ulura or Ayers Rock is a sacred place for Aborigines of the western desert of Australia.
Chun-Wei uses a surveying device to measure the angle of elevation to the top of the rock to be 11.5 degrees.
He walks half a mile closer and measures the angle of elevation to be 23.9 degrees.
How high is Ayers Rock in feet?

This is my set:

y = height of Ayers Rock (in feet)

x = the total distance in feet from the point of the first measurement to the point directly under the highest point of Uluru.

tan11.5 degrees = y/x

tan23.9 degrees = y/x - 2640 feet

This is where I get stuck.  Can someone perform the algebra part and find the answer?
Also, is there an easier method for answering similar questions?[/quote]

[hide="Hint"]
Assuming the equations you wrote were right, then
\[ y \equal{} x \tan (11.5) \equal{} (x \minus{} 2640) \tan (23.9)
\]
Now collect like terms and solve for $ x$ by calculator (that is indeed the easiest way to do these sort of problems).
[/hide]

$ \tan(11.5)\equal{}\frac{y}{x}$ and $ \tan(23.9)\equal{}\frac{y}{x\minus{}2640}$
Now you have 2 equations and 2 unknowns.  Solve for $ y$ in terms of $ x$: $ y\equal{}x\tan(11.5)$.  Plug this into the second equation:
$ \tan(23.9)\equal{}\frac{x\tan(11.5)}{x\minus{}2640}$.  Now, use a calculator to evaluate the tangents (3 or 4 digits) so you can easily solve for $ x$.  Then just plug that into one of the equations and you get $ y$.


Goodmorning.

I think I'm messing something up here:

Given $\gamma(t) = e^{i\: t}$   with  $0\leq t \leq \frac{\pi}{2}$ compute:

$\int_{\gamma}z\ sin(z) dz$

Solving using integration by parts, I get some weird answer

$sin(i)-i\ cos(i) - sin(1) + cos(1)$

Can someone please gently push me in the right direction.

Thanks,
Bailey


I've seen a proof which goes something along the lines of..
You know that $e^x=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+...$.  From there, you do the following...
$e^{ix}=1+ix-\frac{x^2}{2}-\frac{ix^3}{3!}+\frac{x^4}{4!}+...$
Which you can then group into...
$(1-\frac{x^2}{2}+\frac{x^4}{4}-...)+i(x-\frac{x^3}{3!}+\frac{x^5}{5!}-...)=\cos{x}+i\sin{x}$
Now let $x=\pi$.  Then you get $-1$. :wow:

Stating that without the series:

$ e^{i \theta}=cos(\theta)+i sin(\theta)$

Therefore,

$ e^{i \pi}=-1 $


[quote="otaku"]Anyone know how i can join math circles in Toronto?[/quote]
Send E-mail to Dr. Larry Rice
Then he will reply to you around September or August
about when and where we have first meeting
Or you can just ask me at the beginning of September.

I'm not sure about when it will start this year
but I'm pretty much sure that it will be at UTS.

You can also come to Waterloo Math Circles! :rotfl:
The 1-hour drive is so worth it! ;)  :P

btw, I am glad you decided to start training, Greg! Good for you!

thanks kyle

and alex,
if you haven't noticed gas prices are skyrocketing these days...
and i'm asian... :rotfl:


Consider the numbers $ a_1,a_2,a_3,...,a_n>0, n \ge 2$. Prove the inequality:

\[ (1\plus{}a_1)(1\plus{}a_2)...(1\plus{}a_n) \left( \frac{1}{a_1}\plus{}\frac{1}{a_2}\plus{}...\plus{}\frac{1}{a_n} \right) \ge \frac{n^{n\plus{}1}}{(n\minus{}1)^{n\minus{}1}}.\]

[quote="moldovan"]Consider the numbers $ a_1,a_2,a_3,...,a_n > 0, n \ge 2$. Prove the inequality:
\[ (1 \plus{} a_1)(1 \plus{} a_2)...(1 \plus{} a_n) \left( \frac {1}{a_1} \plus{} \frac {1}{a_2} \plus{} ... \plus{} \frac {1}{a_n} \right) \ge \frac {n^{n \plus{} 1}}{(n \minus{} 1)^{n \minus{} 1}}.\]
[/quote]

by Holder Ineq

$ (1 \plus{} a_1)(1 \plus{} a_2)...(1 \plus{} a_n) \left( \frac {1}{a_1} \plus{} \frac {1}{a_2} \plus{} ... \plus{} \frac {1}{a_n} \right)$

$ \equal{}\left(a_1\plus{}\frac{1}{n\minus{}1}\plus{}...\plus{}\frac{1}{n\minus{}1}\right)\left(\frac{1}{n\minus{}1}\plus{}a_2\plus{}...\plus{}\frac{1}{n\minus{}1}\right)...\left(\frac{1}{n\minus{}1}\plus{}...\plus{}\frac{1}{n\minus{}1}\plus{}a_n\right)\left( \frac {1}{a_1} \plus{} \frac {1}{a_2} \plus{} ... \plus{} \frac {1}{a_n} \right)$

$ \ge \left(\sqrt[n\plus{}1]{\frac{1}{(n\minus{}1)^{n\minus{}1}}}\plus{}\sqrt[n\plus{}1]{\frac{1}{(n\minus{}1)^{n\minus{}1}}}\plus{}...\plus{}\sqrt[n\plus{}1]{\frac{1}{(n\minus{}1)^{n\minus{}1}}}\right)^{n\plus{}1}$

$ \equal{} \frac {n^{n \plus{} 1}}{(n \minus{} 1)^{n \minus{} 1}}.$


inequalities
trigonometry
inequalities proposed

[quote="pvthuan"]Prove that
\[\frac {a}{\sqrt{a^2+b^2}}+\frac b{\sqrt{9a^2+b^2}}+\frac{2ab}{\sqrt{a^2+b^2}.\sqrt{9a^2+b^2}}\leq \frac32.\]
I spent two days on this, but I failed, can you help?[/quote]

[hide]
The inequality is obviously true if $a,b$ are negative . When $a,b$ are positive

Use the substitution $\cos A=\frac {a}{\sqrt{a^2+b^2}}$ ; $\cos B=\frac b{\sqrt{9a^2+b^2}}$ , and $\cos C=\frac{2ab}{\sqrt{a^2+b^2}.\sqrt{9a^2+b^2}}$

where $A,B,C$ are angle of triangle $ABC$ . This substitution can be check by the identity $\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1$  for any triangle .

So it is left to prove $\cos A+\cos B+\cos C\leq \frac{3}{2}$ which can be done by jensen .  :)
[/hide]

Let $x=\frac{3a}{b}, y=\frac{b}{a}$
We get xy=3 and we need to prove that $\frac{1}{\sqrt{y^2+1}}+\frac{1}{\sqrt{x^2+1}}+\frac{2}{\sqrt{x^2+1}\sqrt{y^2+1}}\le\frac{3}{2}$
and we need $3\sqrt{x^2+y^2+10}\ge2(\sqrt{x^2+1}+\sqrt{y^2+1}+2)$
compare the square of 2 sides, we get it.
(It's such tough work so my solution is not so good)

I think I have a nice solution.

Notice first that
\[{a\over \sqrt{a^{2}+b^{2}}}+{b\over\sqrt{9a^{2}+b^{2}}}+{2ab\over \sqrt{a^{2}+b^{2}}\sqrt{9a^{2}+b^{2}}}\leq{|a|\over \sqrt{a^{2}+b^{2}}}+{|b|\over\sqrt{9a^{2}+b^{2}}}+{2|a\parallel b|\over \sqrt{a^{2}+b^{2}}\sqrt{9a^{2}+b^{2}}}\]
so we can suppose $a$ and $b$ are nonnegative.

Since ${a\over \sqrt{a^{2}+b^{2}}}$ and ${b\over\sqrt{9a^{2}+b^{2}}}$ are less or equal than $1$ and greater or equal than $0$, there are reals $0 \leq \alpha,\beta\leq \pi/2$ such that $\sin\alpha ={a\over \sqrt{a^{2}+b^{2}}}$ and $\sin\beta ={b\over\sqrt{9a^{2}+b^{2}}}$. Additionally, $\cos\alpha ={b\over \sqrt{a^{2}+b^{2}}}$ and $\cos\beta ={3a\over\sqrt{9a^{2}+b^{2}}}$, so
\[\cos\alpha\cdot\cos\beta = 3\sin\alpha\cdot\sin\beta\iff 2\sin\alpha\cdot\sin\beta = \cos\alpha\cdot\cos\beta-\sin\alpha\cdot\sin\beta = \cos(\alpha+\beta)\]

The inequality is equivalent to
\[\sin\alpha+\sin\beta+2\sin\alpha\cdot\sin\beta \leq{3\over 2}\iff 2\sin\left(\alpha+\beta\over 2\right)\cos\left(\alpha-\beta\over 2\right)+\cos(\alpha+\beta) \leq{3\over2}\]

But $0\leq{\alpha+\beta\over 2}\leq{\pi\over 2}$, so $\sin\left(\alpha+\beta\over 2\right) \geq 0$. Thus
\[2\sin\left(\alpha+\beta\over 2\right)\cos\left(\alpha-\beta\over 2\right)+\cos(\alpha+\beta) \leq 2\sin\left(\alpha+\beta\over 2\right)+\cos(\alpha+\beta)\]

Let $x = \sin\left(\alpha+\beta\over 2\right)$. So $\cos(\alpha+\beta) = 1-2x^{2}$ and we have to prove that
\[2x+1-2x^{2}\leq{3\over 2}\iff (2x-1)^{2}\geq 0\]
and we are done.

Sorry to bring such an old topic back, but my students showed a much improved solution.

Again, suppose $ a,b \geq 0$. Using the same ideas from the previous post, there are $ \alpha, \beta\in [0,\pi/2]$ such that $ \cos\alpha \equal{} \frac a{\sqrt{a^2\plus{}b^2}}$ and $ \cos\beta \equal{} \frac b{\sqrt{9a^2\plus{}b^2}}$. Then the inequality is equivalent to
\[ \cos\alpha \plus{} \cos\beta \plus{} 2\cos\alpha\cos\beta \leq \frac 32\]

Notice that, very much like the previous post, $ 3\cos\alpha\cos\beta \equal{} \sin\alpha\sin\beta \iff 2\cos\alpha\cos\beta \equal{} \minus{}\cos(\alpha\plus{}\beta) \equal{} \cos(\pi\minus{}\alpha\minus{}\beta)$. Let $ \gamma \equal{} \pi \minus{}\alpha \minus{} \beta$. Then $ \alpha, \beta, \gamma$ are the angles of a triangle and the inequality is equivalent to
\[ \cos\alpha \plus{} \cos\beta \plus{} \cos\gamma \leq\frac 32\]

This is a well-known inequality. Indeed, since $ \cos\alpha \plus{} \cos\beta \plus{} \cos\gamma \equal{} 1 \plus{}r/R$,
\[ 1 \plus{} \frac rR \leq \frac 32\iff R \geq 2r\]
which is true.


Prove that

$ h_a^2\plus{}h_b^2\plus{}h_c^2\leq11r^2\plus{}4R^2$


[url]http://politsk.blogspot.com/2008/09/sarah_13.html[/url]

[url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=232011]My name is Chop.[/url] :) What's yours?

Ammo. Ammo Canal Palin.
My Dad's name is Falter Locust Palin. My Mother's name is Comma Liberty Palin. My brother's name is Fleck Rookie Palin. My Dog's name is Gripper Carom Palin.

Lassie's name is Blaster Commando Palin.

[hide="long lists of names"]
a => Skunk Grunt Palin
b => Wrangler Tractor Palin
c => Churn Scorpion Palin
[b]Sarah Palin => Flack Gobbler Palin
LOL[/b]
[b]
John McCain => Steam Fangs Palin [/b]
LOL2

Barack Obama => Tarp Lazer Palin
Joe Biden =>Beans Harpoon Palin
Yeah, that's right. Democrats own with lasers and harpoons!

ax^2+bx+c=0 =>Krinkle Bearcat Palin
Our little quadratic guy.

x^2+bx+c=0 => Chase Rooster Palin
Has become monic.

EPIC FAIL => Luger Otter Palin
. => Can Lightning Palin
.. => Goalie Sanka Palin
... => Pistol Tanker Palin
.... => Steak Leather Palin
[/hide]

! :arrow: Wesson Scalper Palin
!! :arrow: Churn Scorpion Palin
!!! :arrow: Speck Backfire Palin
!!!! :ran out of arrows: Mullet Troll Palin

?-->Strike Chipper Palin
??-->Still Hardrock Palin
???-->Package Wichita Palin

(blank)-->Knife Pile Palin
 (a space)-->Mangle Blue Palin
  (two spaces)-->Shank Piston Palin

What happens if you plug names back in?
[hide="this happens"]
Your name-->Mustache Warthog Palin
Mustache Warthog Palin-->Shove Maggot Palin
Shove Maggot Palin-->Knife Pile Palin
Knife Pile Palin-->Sack Panther Palin
Sack Panther Palin-->Creation Schwarzkopf Palin
Creation Schwarzkopf Palin-->Mangle Blue Palin
Mangle Blue Palin-->Sport Grunt Palin
Sport Grunt Palin-->Speck Backfire Palin
Speck Backfire Palin-->Loin Falcon Palin
Loin Falcon Palin-->Pie Gallon Palin
Pie Gallon Palin-->Skein Chug Palin
Skein Chug Palin-->Nam Guadalupe Palin
Nam Guadalupe Palin-->Commando Coalfire Palin
Commando Coalfire Palin-->Froth Moonshine Palin
Froth Moonshine Palin-->Recoil Zoo Palin
Recoil Zoo Palin-->Flag Cobra Palin
Flag Cobra Palin-->Molten Contra Palin
Molten Contra Palin-->Beans Harpoon Palin
Beans Harpoon Palin-->Bigger Channel Palin
Bigger Channel Palin-->Strike Chipper Palin
Strike Chipper Palin-->Seam Marauder Palin
Seam Marauder Palin-->Drown Wing Palin
Drown Wing Palin-->Log Justice Palin
Log Justice Palin-->Mullet Troll Palin
Mullet Troll Palin-->Goalie Sanka Palin
Goalie Sanka Palin-->Shank Piston Palin
Shank Piston Palin-->Stag Tonnage Palin
Stag Tonnage Palin-->Bigger Channel Palin
And repeats[/hide]

I am Hose Hotrod Palin
[quote="ak909090"]o.0

Jesus Christ => Open Aircraft Palin...

God => Charcoal Sniper Palin

Buddha => Barrel McRaven Palin

Martin Luther King, Jr. => Speck Backfire Palin

Martin Luther => Creation Schwarzkopf Palin (whoa..)

Pope John Paul => Staff Wrench Palin

Pope Benedict => Copper Catfish Palin

King George III => Luger Otter Palin

George Washington => Fork Decoy Palin

Abraham Lincoln => Still Hardrock Palin

George W. Bush => Jeep Pike Palin

George H. W. Bush => Vise Peeper Palin

Adolf Hitler => Block Lionel Palin

and last but not least,

Richard Rusczyk => Rock Crane Palin[/quote]
You forgot the following:
John McCain =>Steam Fangs Palin
Barack Obama=>Tarp Lazer Palin
Joe Biden => Beans Harpoon Palin
And, last but not least,
Sarah Palin => Flack Gobbler Palin

You know, we should start a game where you put in your resulting name into the "Enter Your Name" field and do it over and over.
After five generations my name becomes: Loin Falcon Palin

I wonder what the formula is. My name produces [i]"Stag Tunnel Palin"[/i]

my name is......Hump Gizzards Palin  :rotfl:  :rotfl:  :rotfl:

Terrorist=>Rifle Panzer Palin  :rotfl:  :rotfl:  :rotfl:

I got Hump Gizzards Palin after cycling my name a few times...


geometry
angle bisector
geometry proposed

Let $(O) , (O_')$ be two circles . $(O)$ intersects $(O')$ at $A , B$ . Two circles $(O_{1}) , (O_{2})$  internally contact with the arc $AB$ of $(O)$ and $(O')$ and $(O_{1})$ externally contacts with $(O_{2})$ at $T$ . Prove that : $T$ belongs to a fixed circle

I am not really sure whether I understood your problem correctly since it is not too well formulated. Anyway, here is my interpretation of the problem - please tell me whether it is right:

[color=blue][b]Problem.[/b] Let k and k' be two circles intersecting each other at two points A and B. We draw two circles $k_{1}$ and $k_{2}$ such that

- each of the circles $k_{1}$ and $k_{2}$ touches the arcs AB of both circles k and k' internally;
- the two circles $k_{1}$ and $k_{2}$ touch each other at a point T.

Prove that, for all such pairs of circles $k_{1}$ and $k_{2}$, the point T belongs to a fixed circle.[/color]

[i]Solution.[/i] Let i be an inversion with center A. For every geometrical object x, we will denote by i(x) the image of this object x under the inversion i. Clearly, i(i(x)) = x for every object x.

Now, since the circles k and k' pass through the center A of our inversion i, their images i(k) and i(k') under the inversion i are lines; hence, the images of the arcs AB of these two circles are two rays. Denote these rays by r and r'. Since the circles $k_{1}$ and $k_{2}$ touch both of these arcs AB, their images $i\left(k_{1}\right)$ and $i\left(k_{2}\right)$ under our inversion i are two circles touching both of these rays r and r'. But if a circle touches two rays, its center must lie on the angle bisector of the angle between these two rays; hence, the centers of the circles $i\left(k_{1}\right)$ and $i\left(k_{2}\right)$ both lie on the angle bisector of the angle between the rays r and r'. Now, since the circles $k_{1}$ and $k_{2}$ touch each other at the point T, their images $i\left(k_{1}\right)$ and $i\left(k_{2}\right)$ under our inversion i touch each other at the image i(T) of the point T. Since the centers of the circles $i\left(k_{1}\right)$ and $i\left(k_{2}\right)$ both lie on the angle bisector of the angle between the rays r and r', their point of tangency i(T) will also lie on this angle bisector. Thus, the point T = i(i(T)) will lie on the image of this angle bisector under the inversion i; this image must be a circle, and, in fact, it is a fixed circle (since the inversion i is fixed and the line i(T) is fixed, being the angle bisector of the angle between the fixed rays r and r'). So the point T lies on a fixed circle, and the problem is solved.

  darij


The link on my blog to my blog's RSS feed is supposed to point to [url]http://www.artofproblemsolving.com/Forum/feed.php?w=1065[/url].

However, it instead points to [url]http://www.artofproblemsolving.com/Forum/feed.php[/url].

When I manually add the "?w=1065", it gives me the correct RSS feed, except it says that "This XML file does not appear to have any style information associated with it. The document tree is shown below." It appears to update correctly, but my Facebook blog RSS reader hasn't been able to read my entries since Oct. 3 or so. It also doesn't look as "nice" as /Forum/feed.php.

Could you look into this sometime?

Yes I will look into them tomorrow. Both bugs confirmed, thanks for reporting them.


A six digit number (base 10) is squarish if it satisfies the following conditions.

(i) none of its digits is zero
(ii) it is a perfect square and
(iii) the first two digits, the middle two digits and the last two digits of the number are all perfect squares when considered as two digit numbers.

How many squarish numbers are there?

Hmm... weird question...

Total square - last square gives that the middle number is even, thus either 16, 36 or 64

The other squares must be 16,25,36,49,64 or 81. So at most 108 options left.

if the number is abcdef is a square is ab0000+cd00+de is the sum of 3 squares... parity leaves us 54 options...

Hmm really I don't see the right way. Cuz I guess there's is a better way than just reducing the amount and then check them all. :)

I remember solving this as a problem in a selection test for JBMO in 2001, but I analyzed lots of cases (I used the algorithm for extracting square roots from a number). If I remember correctly, the only ones are $166464$ and $646416$. It's not very likely that I'll try to solve it again, though :).


geometry
geometric transformation
rotation
Support
calculus
integration
search

A simplified model of a bicycle of mass $ M$ has two tires that each comes into contact with the ground at a point. The wheelbase of this bicycle (the distance between the points of contact with the ground) is $ w$, and the center of mass of the bicycle is located midway between the tires and a height $ h$ above the ground. The bicycle is moving to the right, but slowing down at a constant rate. The acceleration has a magnitude $ a$. Air resistance may be ignored.

Assume that the coefficient of sliding friction between each tire and the ground is $ \mu$, and that both tires are skidding: sliding without rotating. Express your answers in terms of $ w$, $ h$, $ M$, and $ g$.

28. What is the maximum value of $ \mu$ so that both tires remain in contact with the ground?

29. What is the maximum value of $ a$ so that both tires remain in contact with the ground?

There are two issues that are confusing me:

1) I know that the normal force on the back tire must go to 0 in order for it to lift off the ground. But I can't identify any force that would cause that to happen. Also, friction acts perpendicular to the direction of rotation, so I don't understand how that would cause any torque.

2) The way the question is worded implies that the maximum values of $ \mu$ and $ a$ do not depend on each other. If that is true, then suppose for a certain $ \mu$ and $ a$, the tires barely remain on the ground. Then according to #28, if I increase $ \mu$, then the tires should come off. But #29 is unaffected, so the tires should remain on the ground. How do I resolve this apparent paradox?

I get $ \frac{a}{\mu} \equal{} g$.

This absurd answer might be because I didn't understand [b]"...moving to the right..."[/b]. Does that mean the two wheels are in a line skidding sideways with the bike making $ \frac{\pi}{2}$ with the direction of motion?

My above answer is in accordance with this assumption.

Could you draw a quick diagram to support your question?

if you're taking moment wrt to point C, what is $ mg\mu h$ ?

Wikipedian1337, I have the same message for you as I had for integration - consider doing a search of the forums, since these problems have already been discussed in the Introductory Physics forum.

Wait, so when the bicycle is traveling and the tires do not leave the ground, does this mean the torque is always greater than 0 and rotating the bike counterclockwise?

[quote="integration"]It's the fictitious force (F=-ma) due to the fact I chose a noninertial frame.  Since ma=mgu, then F=mgu, and Torque=mguh (taken from C)[/quote]

I understand what you did from a mathematics point of view, but can you specify the lever arm and how exactly $ h$ plays into the setup? Did you use $ h$ as your lever arm? I think the lever arm should be a line joining the bottom of the front wheel (your point C) to the center of mass.

[quote="integration"]I think I solved that problem some time back.  Don't look if you want to solve yourself.  I don't know how to hide my solution...

Tip1: Use the center of mass as the point where gravity is applied.
Tip2: choose your center of rotation as the tip of the wheel
Tip3: Set the torque equation, and linear motion equation.
By the way, you are dealing with a non inertial frame (I used the bycicle CM as my frame for linear motion, torque.. doesn't really matter if it's non-inertial), so you should add the Fictitious for ma in the direction of motion, applied to the CM, that's the force that lifts up the back wheel and makes the bycicle topple over.

I hope my solution is correct...[/quote]

Can you draw in the point of application of your torques and the lever arm? I have trouble seeing how $ mg\frac{w}{2}$ makes physical sense, since there is no lever arm joining point C with some imaginary point $ \frac{w}{2}$ to its left on the ground. Thanks.

Here's a solution that may be less confusing than the one provided above:

By representing the bicycle as a diagonal rod (from the center of mass to the bottom of the front wheel), we see that $ F_{friction}$ acts to the left on the bottom edge, while $ mg$ acts downwards on the top edge.

To prevent rotation, $ F_{friction}sin\theta \equal{} mgcos\theta$, so $ \mu \equal{} \frac{w}{2h}$

Basically the reason you need to create this pseudo force is because the frame of the bicycle is not an inertial frame. In order to be able to treat the frame as inertial and do the problem, we need to pretend there is a force acting on everything.

The classic example of this is gravity. Because Earth's surface itself is not an intertial frame, we create a force mg (that happens to magically know the mass of the object its acting on) that acts on everything so we can pretend that the Earth is an intertial frame.


This is no joke - the problem comes from the St Petersburg Mathematical Olympiad (1998, if I remember correctly). That's a very hard competition and this problem isn't trivial either...

Well it's not trivial Arne...

But one should be able to find the solution if one tries hard enough right?

There's no "Evil trick" here... it's natural...

[quote="Detective"]Are you joking?

$y^3 \geq 4y, z^3 \geq 4z, x^3 \geq 4x$

So we need to prove

$(x+4y)(y+4z)(z+4x) \geq 125xyz$

the last one is obvious from AM-GM since the LHS is cyclic.[/quote]
yes you are right.and here is a proof for it:
$x+4y \geq 5 \sqrt[5]{y^4x}$ and the others like this and if we times them we will get \[ (x+4y)(y+4z)(z+4x) \geq 125 \sqrt[5]{x^5y^5z^5} \] and we R done.

Well Arne it's not from St. Petersburg 1998...

i've checked titu andreescu's math olympiads from around the world 1998 and it's not there...

I am doubtful that it is from St. Petersburg since it's so easy...

 :)

[quote="Mathx"]For $x,y,z\geq 2$Prove that:

  $(y^3+x)(z^3+y)(x^3+z)\geq 125xyz$[/quote]

A stronger inequality was given at http://www.mathlinks.ro/Forum/viewtopic.php?t=32430 :

$8\left(y^3+x\right)\left(z^3+y\right)\left(x^3+z\right)\geq 125\left(y+z\right)\left(z+x\right)\left(x+y\right)$

(stronger because of the known inequality $\left(y+z\right)\left(z+x\right)\left(x+y\right)\geq 8xyz$).

  darij

I'm pretty sure it IS from St Petersburg.


function
induction
real analysis
real analysis unsolved

Suppose the series $ \sum{a_n x^n}$ and $ \sum{b_n x^n}$ converge in the segment S= (-R,R). Let E be the set of all $ x\in{S}$ at which
(20)  $ \sum^{\infty}_{n \equal{} 0} {a_n x^n} \equal{} \sum^{\infty}_{n \equal{} 0}{b_n x^n}$

If E has a limit point in S, then $ a_n \equal{} b_n$ for n =0,1,2,... Hence (20) holds for all $ x\in S$

Pf: Put $ c_n \equal{} a_n \minus{} b_n$ and
     f(x) = $ \sum^{\infty}_{n \equal{} 0} {c_n x^n}$
     Then f(x) = 0 on E.
     Let A be the set of all limit points of E in S, and let B consist of all other points of S. It is clear from the definition of "limit point" that B is open. Suppose we can prove that A is open. Then A and B are disjoint open sets. Hence they are separated . Since S = A $ \cup$ B and S is connected, one of A and B must be empty. By hypothesis A is not empty. so A=E=S.
Thus $ c_n \equal{} 0$ for n= 0,1,2...

end of pf.

does any1 no why "Thus $ c_n \equal{} 0$ for n= 0,1,2..."

Thx !  :)

Once you get to "$ f(x)\equal{}0$ for all $ x\in(\minus{}R,R)$", just differentiate the identity $ f(x)\equal{}\sum_{n\ge 0}c_n x^n$ at $ 0$ several times. A big chunk of proof (showing that $ A$ is open) is missing in your outline, of course, but I assume that you just had no trouble with that part and didn't include it for that reason.

By the way, abbreviations like "any1" are rather frowned upon here (for many reasons), so, please, use standard English spelling for words "anyone", "know", "thanks", and others :).

The function is analytic at the limit point, which is contained in the disk of convergence. I would use induction and the Peano form of Taylor's formula.


logarithms trigonometry ratio limit calculus algebra polynomial

Determine whether the following series converge absolutely, converge conditionally, or diverge:

1. $\sum_{n=1}^{\infty}\frac{\cos(3n+5)}{\sqrt{4n^3+3n^2+5(-1)^n}}$

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

3. $\sum_{n=2}^{\infty}\frac{n^8}{(\ln n)^{n/2}}$

4. $\sum_{n=1}^{\infty}(-1)^n\left(1-\cos\left(\frac1n\right)\right)$

I won't try to copy kunny and do these every day - just once in a while, and if anyone else wanted to add to the collection, that would be OK with me.

1 is it ok that this is going to give you an imaginary value for your first term?

2. converges absolutley by the ratio test..

[quote="skylarth"]1 is it ok that this is going to give you an imaginary value for your first term? [/quote]
I didn't mean for that to be an issue. I just changed the problem so it isn't.

actually.. i suppose it was alright..  i mean.. it would just converge to an complex number.... hm. never seen one of those.. :)

4. converges by alternating series test..

$\cos(\frac{1}{n}) \geq \cos(\frac{1}{n+1})$ for all $n > 0$ because:

$\cos(x)$ is decreasing and $0\leq \cos(x) \leq 1)$ on $(0,1)$

$\frac{1}{n}$ is decreasing on $(0, \infty)$, as well as: $ 0<\frac{1}{n} \leq 1$ on $(0, \infty)$.

 therefore $1-\cos(\frac{1}{n})$ is decreasing for all n in $(1, \infty)$

also, $\lim_{n\rightarrow \infty} (1-\cos(\frac{1}{n})) = 0$

Skylarth: on #4, is that absolutely convergent or conditionally convergent?

oops.. i forgot about that..

i showed it is condionally convergent..

i have to think about the absolute part..

EDIT:

it is absolutley convergent by limit comparison test to $\frac{1}{n^2}$

1) is absolutly convergent by comparision to $\frac{1}{n^{3/2}}$

#1, 2, and 4 have now been correctly answered.

On comment on your teminology, skylarth: the alternating series test does not prove conditional convergence - it proves convergence. It differs from every other test commonly taught in a first-year calculus course in that it is not a comparison tests. All of the other usual tests are comparison tests, and a successful comparison test always shows absolute convergence. But to show conditional convergence, you would have do show that the series does not converge absolutely. (Presumably by some comparison test.)

We can take some ugly estimates for $3$.  We have $\ln n\geq 2$ for sufficiently large $n$, and $2^{n/2}\geq n^{10}$ for large $n$ (since exponentials kill polynomials)

Hence for large $n$ the series is dominated by $1/n^2$, which leads to.....

Out of curiosity, skylarth, how did you get the comparison to $1/n^2$ for (4)?  It's of course easy if we use $\cos\left(\frac 1n\right)=1-\frac {1}{n^2}+O(n^{-4})$ or something similar, but did you get it in another way?

Just as a note on the alternating series test, the difference between it and the other tests is mainly that the other convergence tests follow either from some twist on the cauchy criterion and/or a comparison to (convergent) geometric series.  However, the proof for alternating series is derived mainly by an appeal to the Abel Summation Formula and the results can't really be adapted.

So far, we've seen use of the Comparison Test and/or Limit Comparison Test for #1 and #4 and the Ratio Test for #2. Blahblahblah invoked another direct (one-sided) comparison in his argument for #3. However, there are still two of the usual tests taught in an ordinary first-year calculus course (or AP Calculus BC) that haven't been invoked yet - and one of them may be useful in #3.

lucky guess...  :D

lol.. i guess looking at the taylor makes it easier.. didnt think about that one..

EDIT: number 3 is absolutley convergent by root test.

3 was a tricky one. :)

Yes, I was thinking of the root test. Thank you, skylarth and other; this topic is now finished.


[quote="piano1321"]If a prism has 12 vertices, how many edges does it have?

p.s. i think it has something to do with the Euler's Formula.[/quote]
[hide]
I got $20$. I just drew it in my head, so it might not be correct.
What's Euler's Formula? I searched it and I don't get any of the notation used...[/hide]

[hide]
Prism with 12 vertices is a hexagonal prism.
How many edges? Each hexagon has 6, and there are 6 edges connecting the hexagons.
18.[/hide]

Euler's formula for prisms is F+V=E+2, where F is number of faces, V is number of vertices, and E is number of edges. Just look for it in that crazy long formulas topic.

[hide]
You don't need Eulers formula for this.
12 vertexed prism is a hexagonal prism.

6 edges on each of the bases.
6 on the lateral faces.

6+6+6=18[/hide]


Hello, the Mecanical Energy within a  gravity field can be calculated using

E = 1/2 m v^2 - G m M/r

Where m is the mass of the current planet and M the mass of the other planet.

What if there are more planets around ? Are they taken in consideration?

Thanks.

well other planets are often assumed to be distant from the body you are calculating from.  Often gravitational potential energy will be reasonably close to the planet.  Once three bodies are being taken into account the the equations become very complex, nearly impossible so perturbation theory is often used where power series are used to attain a certain degree of accuracy.


Prove for $a, b, c > 0$

$\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} \geq a+b +c$

[hide]From Cauchy's Inequality, $\left( \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \right) (b+c+a)\geq (a+b+c)^2$.

Then just divide by $a+b+c$ to get the desired inequality.[/hide]

well opposite ordering so use rearrangement or chebyshev.

$\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} \geq a+b +c$

Multiply everything by $abc$

So we obtain:

$a^3c+b^3a+c^3b \geq abc(a+b+c)$

Which is true by rearrangement.


trigonometry
function
algebra unsolved
algebra

Solve in natural the equation  $ \arcsin x\plus{}\arcsin 2x\plus{}...\plus{}\arcsin 2009x\equal{}2010x$ :?:

_______________________________________
Azerbaijan Land of Fire
$ \Rightarrow\bigstar\Leftarrow$
 :rotfl:

[quote="Pirkuliyev Rovsen"]Solve in natural the equation  $ \arcsin x \plus{} \arcsin 2x \plus{} ... \plus{} \arcsin 2009x \equal{} 2010x$ :?: [/quote]

1) If $ 0\notin\mathbb N$ : $ x\ge 1$ and so $ 2x>1$ and so $ \arcsin(2x)$ is not defined and so no solution.

2) If $ 0\in\mathbb N$ : If $ x\ge 1$ then $ 2x>1$ and so $ \arcsin(2x)$ is not defined and so no solution for $ x\ge 1$. So a unique solution $ x\equal{}0$

Let's try for real number $ x$. I think the condition is meaningless.

The function $ y\equal{}\sum_{k\equal{}1}^{2009}\arcsin{kx}$ is defined in the cell $ [0,\frac{1}{2009}]$. In this interval, $ y$ strictly increases and convex, because $ y'\equal{}\sum_{k\equal{}1}^{2009}\frac{k}{\sqrt{1\minus{}k^2x^2}}>0$, $ y''\equal{}\sum_{k\equal{}1}^{2009}\frac{2k^3x}{1\minus{}k^2x^2}>0$ for all $ x\in [0,\frac{1}{2009}]$. Zeros of the given equation is the intersecting point of $ y$ and the line $ 2010x$ geometrically. Since $ y$ strictly increases, convex and the gradeient of tangent line of $ y$ at $ (0,0)$ is $ y'|_{x\equal{}0^\plus{}}\equal{}\sum_{k\equal{}1}^{2009}k\equal{}\frac{2010\cdot 2009}{2}>2010$, $ y$ and the line $ 2010x$ never intersects in the interval $ [0,\frac{1}{2009}]$ except $ (0,0)$. So, $ x\equal{}0$ is the unique solution of the given equation.

[quote="J.Y.Choi"] I think the condition is meaningless. [/quote]

I think you're right, but I hate problems which are bad copy of real problems. It means the poster either invented the problem, either is unable to correctly copy ten words ... .
And I avoid to solve double problems : "here is a wrong statement. Please, transform it in a correct statement and solve the correct statement you created ... "


Three spheres of radius 10 are placed on a hemispherical dish. The tops of the spheres are exactly level with the top of the dish. What is the radius of the dish?

[hide="not sure"]Since the surface of dish and the top of the spheres are at level, the spheres will all be tangent to the top of the dish at only one point. Now since spheres are congruent and placed in a hemispherical dish, they will roll upto a point where they are tanget to each other equally distant from the center of the hemisphere. Now if you join points of tangency, that forms an equilateral triangle with side length 20 (the center of the equilateral triangle is also the center of the hemisphere). Now draw a segment from the center of the equilateral triangle to a point of tangency( lenght is $ \frac{20\sqrt{3}}{3}$)  and draw a segment joining that sphere's center with the center of the larger hemisphere. To form a triangle join the point of tangency to the center of the smaller sphere. Since this is a right triangle the length of the hypotenuse is $ \sqrt{\frac{700}{3}}$ (by applying pythag) and to find the whole radius just add the lenght of the original sphere 10 to get a total of $ \boxed{\frac{10\sqrt{21}}{3}+10}$ [/hide]


A rectangular floor measures a by b feet where a and b are positive integers with b > a. An artist paints a rectangle on the floor with the sides of the rectangle parallel to the sides of the floor. The unpainted part of the floor forms a border of width 1 foot around the painted rectangle nad occupies half of the area of the entire floor. How many possibilities are there for the ordered pair (a,b)?

We must have $ 2(a\minus{}2)(b\minus{}2)\equal{}2ab\minus{}4a\minus{}4b\plus{}8\equal{}ab\implies ab\minus{}4a\minus{}4b\plus{}8\equal{}0$

$ \implies(a\minus{}4)(b\minus{}4)\minus{}16\plus{}8\equal{}0\implies(a\minus{}4)(b\minus{}4)\equal{}8$

$ \equal{}1\cdot8\equal{}2\cdot4\implies(a,b)\equal{}\{(5,12),(6,8)\}\implies\boxed2$.


quadratics
algebra
number theory
relatively prime
number theory unsolved

are there any $ a,b,c\in\mathbb{Q}$ s.t. $ abc\equal{}a\plus{}b\plus{}c\equal{}1$ ?

[hide="a thought"] There exist such $ a,b,c$ if and only if there exists a $ q\in \mathbb{Q}\backslash \{ 0 \}$ such that $ \sqrt {q^4 \minus{} 2q^3 \plus{} q^2 \minus{} 4q}\in \mathbb{Q}$.[/hide]

if $ a,b,c$ are all positive than by AM-GM $ \frac {1}{27} \equal{} (\frac {a \plus{} b \plus{} c}{3})^3\geq abc$ clearely has no solutions. so in order to have solutions exactly two must be negative. so we get equivalent
$ xyz \equal{} 1 \equal{} z \minus{} y \minus{} x$ for positive rational $ x,y,z$ from where we conclude $ (1 \plus{} x \plus{} y)xy \equal{} 1$, a quadratic equation which discriminant we can analyse, leading to $ x^4 \plus{} 2x^3 \plus{} x^2 \plus{} 4x$ must be a perfect rational square. now since $ x$ is a positive rational we can assume it is of  form $ \frac {p}{q}$ where $ p,q$ are two relatively prime positive integers. the upper equation is therefore equivalent to
$ p^4 \plus{} 2p^3q \plus{} p^2q^2 \plus{} 4pq^3 \equal{} k^2$ where $ k$ is also a positive integer
note that this is equivalent to $ k^2 \minus{} (p^2 \plus{} pq)^2 \equal{} 4pq^3$ further equivalent to $ (k \minus{} p^2 \minus{} pq)(k^2 \plus{} pq \plus{} p^2) \equal{} 4pq^3$ or in other words $ p$ has to divide $ (k \minus{} p^2 \minus{} pq)$ or $ (k^2 \plus{} pq \plus{} p^2)$ but $ p$ devides $ (k \minus{} p^2 \minus{} pq)$ if and only if it devides $ (k^2 \plus{} pq \plus{} p^2)$, meaning $ p$ devides $ k$ and consequently $ p^2$ devides $ k^2$
therefore $ p^2$ devides $ 4pq^3$ or $ p$ devides $ 4q^3$ and since $ gcd(p,q) \equal{} 1$ we may assume $ p$ devides $ 4$ or $ p \equal{} 1,2,4$

if $ p \equal{} 1$ than $ 1 \plus{} 2q \plus{} q^2 \plus{} 4q^3 \equal{} k^2$ leading to $ k \equal{} rq\pm1$ for a positive integer $ r$ which by putting the former into equation and analysing discriminant yields $ (r^2 \minus{} 1)^2\pm32(r\pm1) \equal{} l^2$ for a non-negative integer $ l$. equivalent $ 32(r\pm1) \equal{} (k \plus{} r^2 \minus{} 1)(k \minus{} r^2 \plus{} 1)$ which is obiviously impossible for $ a > 33$ since $ k \plus{} a^2 \minus{} 1 > a^2 \minus{} 1 > 32(a \plus{} 1)$ for $ a > 33$

if $ p \equal{} 2$ or $ p \equal{} 4$ u can evaluate it in a very simmilar way. i didnt bother to try to calculate solutions if there are any because it takes quite a lot of time to calculate the $ a,b,c$ from this $ r$ but i think it shows the number of "different" solutions is finite and all of them can be calculated in a concpeptually easy way unfortunately taking some time and computations.

i hope i didn't make some stupid mistake in second line or sth simmilar:D


calculus
integration
real analysis
real analysis unsolved

Suppose independent random variables X, Y and Z each has a uniform distribution U(0,1). Prove that (XY)^Z also has a unifrom distribution U(0,1).

PS. I've tried to show that an integral in the space where (XY)^Z<a is a, but I have no clue how to evaluate the integral.

It is much easier to count the moments: all you need to check is that $E (XY)^{kZ}=\iiint_{[0,1]^{3}}(xy)^{kz}\,dx\,dy\,dz=\frac 1{k+1}$ for all $k\ge 0$, which is not hard to do if you integrate with respect to $z$ last. I agree that it is a slightly roundabout way to solve the problem but it seems to be the shortest one.


list
arithmetic sequence
mean
median
mode

When the mean, median, and mode of the list
\[ 10, 2, 5, 2, 4, 2, x\]are arranged in increasing order, they form a non-constant arithmetic progression. What is the sum of all possible real values of $ x$?

$ \textbf{(A)}\ 3 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 17 \qquad \textbf{(E)}\ 20$

[hide="Solution"]
The mode is clearly $ 2$, regardless of $ x$.

The median will either be an integer or an integer plus 0.5. Either way, the mean must be an integer, or an integer plus .25.

Since the mean is $ \frac{25\plus{}x}{7}$, it must be integral, and $ x \equal{} 3\plus{}7n$.

If $ n\equal{}0 \implies \underline{x\equal{}3}$, then it becomes the median, and the mean is $ 4$. Thus, we have the arithmetic progression $ 2,3,4$

If $ n \geq 1$, then the median becomes $ 4$. Since we have an arithmetic progression with $ 2,4,?$, either the mean is $ 6$ or the mean is $ 0$. For the mean to be $ 6$, we must have $ \underline{x\equal{}17} \implies n \equal{} 2$. For the mean to be $ 0$, we must have $ x\equal{}\minus{}25 \implies n \equal{} \minus{}4$, which contradicts $ n>1$, so nothing in this case.

If $ n<0$, then the median becomes $ 2$, equal to the mode, which violates an assumption of the problem.

Thus, $ x\equal{}3$ or $ x\equal{}17$, and the sum is $ \boxed{E. \; 20}$
[/hide]

I don't really understand what "non-constant" means. Also, why must the mean be integral?

A constant arithmetic progression would be one with a common difference of 0 such as 1, 1, 1, 1. The mean must be integral because there's a 7 in the denominator of the mean, and the mode and median are integers.

But just because the mode and median are integers, why does the mean have to be integral?

[quote="TZF"]The median will either be an integer or an integer plus 0.5. Either way, the mean must be an integer, or an integer plus .25.[/quote]
I think you missed a case here.  What if $2 \le x \le 4$, so that $x$ is the median?  Then the median (i.e. $x$) need not even be rational.

Fortunately, this case can be resolved just by solving the two equations \[ x = \frac{2 + \frac{x+25}{7}}{2} \] and \[ \frac{x+25}{7} = \frac{2+x}{2} \] since both the mean and median exceed $2$ in this case.  These gives $x=3$ and $x=7.2$, the latter of which falls outside the range.

The mean is $\frac{10+2+5+2+4+2+x}{7} = \frac{25+x}{7}$, the median is either $2,4$ or $x$ (dependent on $x$), and the mode is $2$.
We apply casework based on the median:

CASE 1: Median 2
In this case, the arithmetic progression will be constant because the mode and median take the same value (2); this is not allowed.

CASE 2: Median 4
From here, we know that since the mode is $2$, the mean must either be $6$, $3$, or $0$. If the mean is $6$, $\frac{25+x}{7}=6 \implies$ $x=17$. This does indeed work. If the mean is $3$, then $\frac{25+x}{7}=3 \implies$ $x=-4$. This does not work; similarly, if the mean is $0$, $\frac{25+x}{7}=0 \implies$ $x=-25$. This does not work either.

Given that so far, our sum is $17$ (and will not get any smaller) and the only choice we missed out is when the median is $x$, we know that either our answer is $17$ or $20$. If our answer is $20$, then the only case we can possibly miss is $x=3$. Testing out $x=3$ proves that it does indeed work. Hence our answer is $3+17 = \boxed{20}$.


MATHCOUNTS
number theory
least common multiple
function

what is the least positive integer that meets [b][i][u]all[/u][/i][/b] of the following conditions?

a)when divided by 7, the remainder is 5
b)when divided by 5, the remainder is 3
c)when divided by 3, the remainder is 1

this was a national countdown round problem.

mr. crawford told me to post problems that i enjoy. and for some reason, i really love mods. i have no idea why

[hide]12,8, and 4?[/hide]

I don't know anything about mods, so I'm just guessing...

This a great problem for teaching clever problem solving.  The rest is in spoiler text:

[hide]You could solve this with mods.  The method for solving is the Chinese Remainder Theorem which I cover in the Intro Number Theory course that we're offering in a few weeks.  You could also say that the number you seek is 2 less than a common multiple of 3, 5, and 7.  The LCM of 3, 5, and 7 is 105.  So the smallest possible value is 105 - 2 = 103.[/hide]

im sorry, but no.

mods are just remainder function sphinks. like 7 :equiv: 1 (mod 6)
so its like 7/6=1 remainder 1. so the remainder is after the the lined thingie. ill put a solution in 10 minutes if nobody posts it

ah. i didnt see mr. crawford post it there. but heres the way i did it

[hide]x :equiv: 5 (mod 7)

x :equiv: 3 (mod 5)

x :equiv: 1 (mod 3)

you can add 2 to each side to get:

x+2 :equiv: 0 (mod 7)

x+2 :equiv: 0 (mod 5)

x+2 :equiv: 0 (mod 3)

they are equivalent to 0 because adding 3 to 1 (mod 3) would be 3 (mod 3), which is zero. so the least positive integer that can fit that is x+2=3 x 5 x 7, so x+2=105, so x=103. voila![/hide]

yeah. sorry for the confusion, ill edit that in a sec

[quote="Fierytycoon"]Tare, all of properties a, b, and c have to be for a single number, not three different ones.[/quote]
I realized that...
I'm considering taking Intro Number Theory, but $75...

[quote="Tare"]
I'm considering taking Intro Number Theory, but $75...[/quote]

You can always take the audited version.  You still get all the same material, but the price is more affordable.

Also, wait til you see what it costs to take that class at a college ;)

*uber sigh* The only thing preventing me from going to an American College... I don't feel like carrying the due date for my loan along with my diploma :)

[quote="Syntax Error"]there are many scholarship opportunities. and anyways, the mod stuff is in aops v1[/quote]
*sigh* I know.
I'm begging for AoPS book, but my mom's too lazy to get up, write just one check, and go to the post office...I have this feeling that summer will be over by the time I get my AoPS book, and I'll be too busy to read it...

indeed. but if you evar want to do well in amc/aime/usamo etc. you should get both books

[quote="Fierytycoon"][quote]but my mom's too lazy to get up, write just one check, and go to the post office...[/quote]

You can order online too.[/quote]
Thanks for the info.

though i havent even finished aops v1 yet. but im getting there!

I'm probably too late to post this, but anyways.

[hide]

It's -2 mod 7, 5, and 3. Therefore it's something two less than a multiple of 5, 7, and 3.

3*5*7 - 2 = 103

[/hide]


geometry
3D geometry
analytic geometry
graphing lines
slope
calculus
derivative

hey everyone. i'm stuck on this problem that has to do with rate of change, and i could use a little help.

1)  a cube whose edge is x is contracting. when its surface area is changing at a rate which is equal to 6 times the rate of change of its edge, then what is the length of its edge?

any help ASAP would be much appreciated. thanks!

sh0ebox

Surface area=6x^2

Lenght=x

Surface area rate =derivative(Surface area)=12x

rate x=derivative(lenght)=1

12x=6->x=1/2

sweet thanks... how do i find the minimum value of the slope of a curve of an equation such as:

what is the minimum value of the slope of the curve y = x^5 + x^3 - 2x?

sh0ebox

second derivative=20*x^3+6*x=0 <=>x=0 V 20*x^2+6=0 <=> x=0 V x=Squareroot[-3/10]=complex number.

third derivative=60*x^2+6>0 for x=0. We have a minimum for x=0.

Derivative=5*x^4+3*x^2-2=-2 for x=0

Minimum of slope =-2


Same rules, good luck all!!

Starting at 25 again..

[b]25[/b]

[b]13[/b]

I told Blazer that I made a new thread.. I'm not sure if he saw my message or not, though.

[b]1[/b]

one more post and girls will win!!!! :D

Yay!! Good job angel!!

I'm gonna wait a while before making another one, seeing that boys are all afk right now... :wink:

no, it depends how much star/stars u have.

srry blazer, i was in the library.


While I can do some AIME problems, ace Collegeboard math exams without a sweat, and self-study BC Calculus in barely any time at all, I can't do many of the simplest of proof-based textbook problems, to my immense frustration. I also thought that I could pursue no more mathematics. It was also a major frustration for me during SIMUW, where I was definitely able to understand non-proof-based material more than proof-based material. I am inspired of the example of another SIMUW student, who got 6 on his AIME and was also lost when it came to proofs. Proofs were so mysterious to me that out of all possible outcomes, I couldn't wait for the classes to end, whenever they covered proofs! It can be explained, perhaps, by my lack of training in proofs. But when I think of other people who have acquired understanding of proofs prematurely, I have to ask, where in the heck did they get their skills?

The question is, where and when do most of you AoPSers and talented math students acquire knowledge of proof methods? And is it a huge struggle up to the point in which it finally gets into your subconsciousness, as D. Solow explained? Most math competitions up to the level of AIME do not use the proofs. Nor are they taught in math classes, beyond the unmathematical two column proof of geometry.

I'm 16 years old now. D. Solow explains that proofs are best learned by 8th grade (one has to wonder how people get that knowledge). I do not know if it is too late. But I am nonetheless inspired by one particular person I knew, who went from not considering college to becoming a college mathematics superstar. It may take an entire read of D. Solow's entire book.

Definitely outside of the regular high school curriculum. You'd be lucky if the teacher even mentions proofs, let alone demonstrate one or even attempt to teach the methods.
I learned from summer programs. I can't say I'm proficient, though. Practice and experience in problem solving help a lot. I don't know if you can learn from reading a book.

OMG $17^2$ posts :interrobang: