Congratulations to all those who have made Nationals!  This is an extreme honor, and no matter if you get first or last place, you should be extremely proud of your accomplishment. You are literally some of the best middle school math students in the nation! Good luck!

We have been allowing a lot of National MATHCOUNTS level problems appear in the forums to allow those participation to practice these more challenging problems. Starting Friday, please revert back to School and Chapter level problems so that a new generation of problem solvers can begin to develop their skills in MATHCOUNTS.

Good luck to everyone at nats! See you there in Denver in 18 hours!

Reminder:  Several questions posted on this Forum since Nationals are too hard.  Administrator levans asked that you guys dial back the level of problems in this Forum after Nationals.

We certainly don't want to chill discussion of MATHCOUNTS problems.  So, problems taken from actual MATHCOUNTS competitions are OK.  Discussions and questions about this year's National Competition are still OK.

But, problems you guys invent (that you may think are at National Level) have tended to be too hard.  Please take them to the appropriate High School Forum.

If you are a rising 9th grader, you would do a service to yourself by spending more time over on the High School Forums, and less time on the MATHCOUNTS Forum.    (Check out Intermediate Topics or Pre-Olympiad).


inequalities
trigonometry
inequalities unsolved

download this and solve the problems.  If you have any paper from:Dr.Titu Andreescu or Bogdan Enescu .about Algebra(only inequalities)Please send me.thanks

1)Rearrangement
2)Product rearrangement
3)Consider the triangle MNP with MP=a MN=b ,and altitude MR=c
Our inequality is equvalent to $1\geq \sin{NMP}$

if a,b,c is aside of the triangle you are right but if it is not you are not right?  use the triangle ineq with sharp substituton.

I don't need a,b,c to be the sides of a triangle. Read my post again I just use the fact that a>c b>c which is obvous otherwise the square roots would not be real. :wink:


Is the http://www.kalva.demon.co.uk can't be opened now?

I've heard that some AoPSers in China had the same trouble, but I don't know why it happens...


there are three different graphs.
$ f(y) \equal{} x^2$
$ f(y) \equal{} x^3 \minus{} 4$
$ f(x) \equal{} 4$
they intersect at 3 points.
what are these points?
:juggle:

$ f(y)$ is expressed in terms of $ x$...

[quote="asterday1"]there are three different graphs.
$ f(y) \equal{} x^2$[/quote]
I assume you mean $ y \equal{} x^2$.

[quote="asterday1"]
$ f(y) \equal{} x^3 \minus{} 4$[/quote]
I assume you mean $ y \equal{} x^3 \minus{} 4$.

[quote="asterday1"]
$ f(x) \equal{} 4$
they intersect at 3 points.
what are these points?
:juggle:[/quote]

By the way, those graphs don't all intersect together at three points. $ f(x) \equal{} 4$ intersects the $ y \equal{} x^2$ twice and $ y \equal{} x^3 \minus{} 4$ once. Those points are:

$ (\minus{}2, 4), (2, 4), (\sqrt[3]{4}, 4)$

asterday1, Alan is saying that $ f(y)$ should be expressed in terms of $ y$, not $ x$. However, I think you meant $ y$ instead of $ f(y)$.

i got:
(2,4)1st and 2nd
(4,16) and (4,64) 3rd with 1st and second, respectively.
its f(x)=4 because it was the only way i could show x=4
y=undifined dosent cut it

Um...no. $ f(x)\not\equal{}x$. It's another way of saying $ y$, at least in the Cartesian plane.

[quote="ajain"]I don't get this topic.  There's a poll on smileys and the posts have no relavance with the poll!  I'm just commenting.[/quote]

I'm assuming it's just a random thing...

... another random thing: Asterday1, $ \pi$ does not equal $ 3.141592653589793238462643383279$, it approximately equals $ 3.141592653589793238462643383279$. So it would be more accurate to put $ \pi \approx 3.141592653589793238462643383279$. :D

And, by, the way, $ \pi \approx 3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821$ also in my head :D

OMG.
I think we have a savant on our team.
Seriously...
Memorizing $ 2^{40}$ and so many digits of $ \pi$...
I can do $ 3.14159265358979$...

well... then again... someone at rj grey (school in acton)
memorized like 300 digits of pi or something like that....
* I can do:pi is about: 3   (yes, that's on the top of my head :P)
:ddr: :ddr: -my favorite smiley!!!

Yea, good point.
That person was on my bus.
Still....

Your poll doesn't make sense. If you're asking a graphing question, then why do you ask people's fav smilies?

[quote="Ihatepie"]he is just doing something completely random. Are 3d graphs in mathcounts?[/quote]

So far, I haven't seen any 3d graphs in MATHCOUNTS problems...

[b]Disclaimer:[/b] Just because I haven't seen 3d graphs in MATHCOUNTS problems doesn't mean there aren't 3d graphs in MATHCOUNTS problems. I am just stating that I haven't seen any 3d graphs in the problems I've seen so far.

[quote="Ihatepie"]he is just doing something completely random. Are 3d graphs in mathcounts?[/quote]

The most I've seen is is the diagonal of a 3d object, generally a cube.

The hardest 3d one I can think of is 1991 Nats Sprint 23. It's in all time Greatest MC problems.


number theory
greatest common divisor
modular arithmetic

$ \alpha\equiv \beta \pmod{m}$ then $ (\alpha,m)\equal{}(\beta,m)$. I suspect you use the division theorem, combined with maybe something else.

From $ \alpha\equal{}\beta\plus{}km\ (k\in\mathbb{Z})$, it is clear that any integer divides $ \alpha$ and $ m$ if and only if it divides $ \beta$ and $ m$.

hm. why are the gcd's $ (\alpha,m)$ and $ (\beta,m)$ the same?

Definition of gcd: The gcd of two integers $ a,b$ (nonzero) is the positive integer $ d$ such that
(i) $ d\mid a$ and $ d\mid b$
(ii) if $ c$ is any integer such that $ c\mid a$ and $ c\mid b$, then $ c\mid d$.

In the given problem, let $ d \equal{} \gcd{(\alpha,m)}$. Then $ d\mid \alpha$ and $ d\mid m\ \Rightarrow\ d\mid \alpha \minus{} km \equal{} \beta$. So we have (i): $ d\mid \beta$ and $ d\mid m$. If c is an integer such that $ c\mid \beta$ and $ c\mid m$, then $ c\mid \beta \plus{} km \equal{} \alpha$. So we have $ c\mid \alpha$ and $ c\mid m\ \Rightarrow\ c\mid\gcd{(\alpha,m)} \equal{} d$ – which is (ii). Hence $ d \equal{} \gcd{(\beta,m)}$.

How did you think of that? It is very clever  :oops:

Another way: use the Euclidean algorithm.  It says that gcd(a,b) = gcd(b,a%b), where a%b means the smallest nonnegative integer k such that $ k \equiv a \pmod{b}$.  Then gcd(a,m) = gcd(a,a%m) and gcd(b,m) = gcd(b,b%m).  Since b%m = a%m, both gcds are equal.


function
functional equation
algebra
IMO Shortlist
IMO Longlist

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

[quote="orl"]The function $\varphi(x,y,z)$ defined for all triples $(x,y,z)$ of real numbers, is such that there are two functions $f$ and $g$ defined for all pairs of real numbers, such that
\[\varphi(x,y,z) = f(x+y,z) = g(x,y+z)\]
 for all real numbers $x,y$ and $z.$ Show that there is a function $h$ of one real variable, such that
\[\varphi(x,y,z) = h(x+y+z)\]
 for all real numbers $x,y$ and $z.$[/quote]

Trivial:

$\varphi\left(0,x+y+z,0\right)=f\left(0+\left(x+y+z\right),0\right)=f\left(x+\left(y+z\right),0\right)=\varphi\left(x,y+z,0\right)$
$=g\left(x,\left(y+z\right)+0\right)=g\left(x,y+z\right)=\varphi\left(x,y,z\right)$.

Thus, defining a function $h\left(u\right)=\varphi\left(0,u,0\right)$, we have $h\left(x+y+z\right)=\varphi\left(x,y,z\right)$, and the problem is solved.

  darij


modular arithmetic
calculus
integration
algebra
polynomial
function
quadratics

Find all $ (x,y,z) \in \mathbb{Z}^3$ such that $ x^3 \minus{} 5x \equal{} 1728^{y}\cdot 1733^z \minus{} 17$.
[i](Paolo Leonetti)[/i]

[hide]Rearranging, we have $ x^3 \minus{} 5x \plus{} 17 \equal{} 1728^y 1733^z$. Since 1733 and 1728 are relatively prime, we must have $ y \geq 0$ and $ z \geq 0$, or else the right-hand side will not be an inteegr. Since $ x^3 \minus{} 5x \plus{} 17$ is always odd, we must have $ y \leq 0$, so $ y \equal{} 0$ (otherwise, the right-hand side would be even.) Thus, we seek all integers $ x$ and nonnegative $ z$ such that $ x^3 \minus{} 5x \plus{} 17 \equal{} 1733^z$.

Suppose $ z > 0$. Then we have that there exists some solution $ x$ to $ x^3 \minus{} 5x \plus{} 17 \equiv 0 \pmod 1733$. But no such solution exists, so $ z \equal{} 0$. However, the rational root theorem implies that this has no integral solutions, so we get that there are $ \boxed{\mbox{no solutions}}$.

On the actual contest, I did not provide a legitimate proof (meaning I printed all 1733 possible outputs of $ x^3 \minus{} 5x \plus{} 17$ into a file and copy/pasted it into my document) that the polynomial has no zeroes mod 1733. So that's a hole in the proof, I suppose. I have an inkling of a feeling that it follows from the fact that the term under the square root [url=http://en.wikipedia.org/wiki/Cubic_function#Roots_of_a_cubic_function]here[/url], that is, 197181, is not a quadratic residue mod 1733. In any case, if I'm not mistaken, rigorizing this argument will involve working in field extensions, which I'm not very familiar with, and which certainly does not qualify as elementary (as the rules asked for an elementary solution.)

I tried taking the equation $ x^3 \minus{} 5x \plus{} 17 \equal{} 1733^z$ random mods as well, and that didn't turn out so well. There's probably a better solution along that route. [/hide]

[hide]$ \mod{7}$ gives that $ z \equiv 0 \pmod{3}$, and then some bounding upon $ x > 1733^z$, \[ x^3 \minus{} (1733^k)^3 \equal{} (x \minus{} 1733^k)(x^2 \plus{} 1733^kx \plus{} 1733^{2k}) \equal{} 5x \minus{} 17\] wins. Though I computed all residues $ \mod{1733}$ also, for good measure [and lack of time  :( ].  [/hide]

Answer to a question in [color=red]Zhero's post[/color]. Imposing the perfect square condition on the discriminant of the depressed cubic $ x^2\minus{}5x\plus{}17$, as given by Cardano's formula, does not work. There are primes $ p$ for which that value is not a square residue, and still the cubic does have a root modulo $ p$, so it's not a necessary criterium (it is of course sufficient, since it readily yields a root).

[quote="Zhero"][hide]FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. [/hide]
Darn. Oh well. That is pretty nice.[/quote]
LOL thanks  :P
My question is, since this question is raised: there exist some method to see if a cubic polynomial is irriducible in $ F_p$? It seems not a trivial one:is it what you are referring mavropnevma? I am interested in, it can be a very powerful tool (infact it has proved that $ x^3\minus{}5x\plus{}17\equal{}1733y$ is impossible in $ \mathbb{Z}^2$  :o )

for the case $ z\equal{}3k$ we can put the expression between two squares,namely,
$ x^3$ and $ (x\minus{}1)^3$.

and we r done!

$ 2 \nmid f(x): \equal{}x^3\minus{}5x\plus{}17 \in \mathbb{Z} \implies \min\{y,z\} \in \mathbb{N}\text{ and } y\equal{}0$. If $ z\equal{}0$ there are no solutions, so $ f(x) \ge 1733 \implies x>10$ Now $ f(x) \in \{0,1,3,5,6\}$ in $ \mathbb{F}_7$ so we have the only case $ 3 \mid z$. So $ f(x)$ must be a cube, but $ (x\minus{}1)^3<f(x)<x^3$ for all $ x>10$.

[i]Remark:[/i] 7 is [i]not[/i] a magic conguence since 2.3+1=7, and $ x^3\equal{}\left(\frac{x}{7}\right)$ in $ \mathbb{F}_7$. Now $ 1733^z\equal{}4^z \in \{1,2,4\}$. 7 does not divide f(-1)f(0)f(1) and if f(x)=2 then $ x^3\minus{}5x\plus{}17\equal{}3 \pm 1 \implies$ $ \left(\left(\frac{x}{7}\right)\pm 1\right)\plus{}2x\equal{}0$: both addends are odd, and the first one has absolute value 2, so |x|=1, then is a contradiction.

[quote="azjps"]
\[ x^3 \minus{} (1733^k)^3 \equal{} (x \minus{} 1733^k)(x^2 \plus{} 1733^kx \plus{} 1733^{2k}) \equal{} 5x \minus{} 17\]
[/quote]

It's seem to be easy to check there is NO solutions $ x$ that $ x < 4$

If $ x > 3$, It's obvious that $ (x \minus{} 1733^k)(x^2 \plus{} 1733^kx \plus{} 1733^{2k}) \equal{} 5x \minus{} 17 > 0 \rightarrow x > 1733^k$ so $ 5x \minus{} 17 > x^2 \plus{} 1733x \plus{} 1733$ (Clearly no solutions)


geometry
geometric transformation
reflection
analytic geometry
vector
linear algebra
matrix

Is there an easy way to find the coordinates of the image of the point (x,y), after being reflected across the line y=2x?

I'm not sure if there is a definite formula, like for reflection across $ y\equal{}x$, but you could always find the distance from the point to the line, and since the intersection is the midpoint, find the other point by the midpoint formula.

There is a nice application of vectors to this problem, which yields the same result as finding equations of lines and calculating an intersection.

[hide="Vectors"][b]Problem statement in terms of vectors:[/b]

[i]Let's say we have $ \mathbf{p} \equal{} (x,y)$ (to be reflected), and $ \mathbf{a} \equal{} (n, 2n)$ (over which $ \mathbf{p}$ is reflected). In the case in which $ \mathbf{a}$ is perpendicular to $ \mathbf{p \minus{} a}$, we want $ \mathbf{p^*}: \; \mathbf{p \plus{} p^*} \equal{} 2 \mathbf{a}$.[/i]

[b]Procedure:[/b]

Since these are perpendicular and $ n \neq 0$ we have:
$ \mathbf{a} \cdot \mathbf{(p \minus{} a) } \equal{} 0 \implies n(x \minus{} n) \plus{} 2n(y \minus{} 2y) \equal{} 0$ $ \implies (x\minus{}n) \plus{} 2(y\minus{}2n) \equal{} 0 \implies x \plus{} 2y \equal{} 5n$

Hence, $ \mathbf{a} \equal{} \left( \frac {x \plus{} 2y}{5}, \; \frac {2x \plus{} 4y}{5} \right) \implies \mathbf{p^*} \equal{} 2 \mathbf{a} \minus{} \mathbf{p} \equal{} \left( \frac { \minus{} 3x \plus{} 4y}{5}, \; \frac {4x \plus{} 3y}{5} \right)$[/hide]

An even nicer and more general interpretation:

[hide="Matrices"] The reflection across $ y \equal{} x$ is represented by multiplication by the matrix

$ \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)$

We can think of this as the composition of a reflection across $ y \equal{} 0$ and a counterclockwise rotation of $ 90^{\circ}$; in other words,

$ \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right) \equal{} \left( \begin{array}{cc} 1 & 0 \\ 0 & \minus{}1 \end{array} \right) \cdot \left( \begin{array}{cc} 0 & 1 \\ \minus{}1 & 0 \end{array} \right)$

Similarly, a reflection across $ y \equal{} 2x$ is the composition of a reflection across $ y \equal{} 0$ and a counterclockwise rotation of $ 2 \theta$, where $ \tan \theta \equal{} 2$.  Recall that a counterclockwise rotation by $ \phi$ is given by the matrix

$ \left( \begin{array}{cc} \cos \phi & \sin \phi \\ \minus{}\sin \phi & \cos \phi \end{array} \right)$

We can easily calculate that $ \cos 2 \theta \equal{} \minus{} \frac{3}{5}, \sin 2 \theta \equal{} \frac{4}{5}$.  We conclude that our transformation matrix is given by

$ \left( \begin{array}{cc} \minus{} \frac{3}{5} & \frac{4}{5} \\ \frac{4}{5} & \frac{3}{5} \end{array} \right) \equal{} \left( \begin{array}{cc} 1 & 0 \\ 0 & \minus{}1 \end{array} \right) \cdot \left( \begin{array}{cc} \minus{} \frac{3}{5} & \frac{4}{5} \\ \minus{} \frac{4}{5} & \minus{} \frac{3}{5} \end{array} \right)$

which is the same result as The Zuton Force's.

Note:  Stretching and translation are both also representible as matrices.  [/hide]

Continuing t0rajir0's explanation, The coordinates of the image of the point $ (x,y)$, after being reflected across the line $ y \equal{} mx.$

Let $ A$ be the matrix about this, $ A \left( \begin{array}{c} 1 \\
m \end{array} \right) \equal{} \left( \begin{array}{c} 1 \\
m \end{array} \right),\ A \left( \begin{array}{c} m \\
\minus{} 1 \end{array} \right) \equal{} \minus{} \left( \begin{array}{c} m \\
\minus{} 1 \end{array} \right)$ we have $ A \left( \begin{array}{cc} 1 & m \\
m & \minus{} 1 \end{array} \right) \equal{} \left( \begin{array}{cc} 1 & \minus{} m \\
m & 1 \end{array} \right),$ solving $ A \equal{} \left( \begin{array}{cc} \frac {1 \minus{} m^2}{1 \plus{} m^2} & \frac {2m}{1 \plus{} m^2} \\
\frac {2m}{1 \plus{} m^2} & \minus{} \frac {1\minus{}m^2}{1 \plus{} m^2} \end{array} \right)$.

For $ m \equal{} \tan \frac {\theta}{2},$ we have $ \left( \begin{array}{cc} \cos \theta & \sin \theta \\
\sin \theta & \minus{} \cos \theta \end{array} \right).$

Indeed, kunny, I'd also considered that route.  In fact, we can find any transformation $ A$ given its eigenvectors and eigenvalues this way, yes?  :)


A college student receives an interest free loan of 8250 dollars from a relative.  The student will repay 125 dollars per month until the loan is paid off.

[b][color=blue]a)Express the amount P (in dollars) remaining to be paid in terms of time t (in months).[/color] [/b]

[b][color=violet]b)After how many months will the student owe 5000 dollars?[/color][/b]

[hide="hint"]How much money does the student pay off in 1 month?  In 2?  How about in $ t$ months?  If you can do this part, then you should be able to figure out how much the student still needs to pay (need to pay = money owned - amount paid).[/hide]

Also, as some general advise, always try to write what you know in terms of algebraic expressions.  Can you do this?

hmmm b-flat that is quite a helpful hint

(don't click unless you really need this)[hide="solution"]

a)$ p\equal{}8250\minus{}125t$

b)Solving:

$ 5000\equal{} 8250\minus{}125t$

$ 125t\equal{}3250$

$ t\equal{}26$ (months)

There. Now you've looked at the answers. I hope you at least tried first... :P
[/hide]


Two circles of radius r are each tangent to the hypotenuse and a leg of a right triangle, and tangent to each other, as shown below. If the sides of the triangle are 3, 4, and 5, compute r.

I got [hide]r = 5/7[/hide] but I'm not sure. Can anyone verify this?

How did you get that answer? (in spoiler of course)

And others should work on it before they look at your answer.

[hide]Call the center of upper circle point O, other circle point P. O is equidistant from AC and AB, so O lies on Bisector of Angle A. Call the point line AO intersect with BC A', A'C/(3-A'C)=4/5. A'C=4/3. Call the point of tangency of Circle O and line AB point M, and AM is 4/4/3 times the r, or 3r. Same thing for circle P. Point of intersection of line BP and AC B'. CB'/(4-CB')=3/5, CB'=3/2.  Point of tangency of line AB and Circle P point N. BN=3/3/2=2r. OPNM is a rectangle, so MN=2r. 3r+2r+2r=5. r=5/7[/hide]

Very nicely done! There's also another way to do it using [hide]Trigonometry.[/hide]

Here's another way to do it...Say our triangle is ABC, with B the right angle, AB = 3 and BC = 4.

[hide] Draw the radii to the hypotenuse. Call those points A' and B', and we know A'B' = 2r. We also know that AA' = r cot ((arctan 4/3)/2) and BB' = r cot ((arctan3/4)/2). Thus we get AA' + A'B' + BB' = 5, or 7r = 5. So r = 5/7. [/hide]

Tada.[/hide]

How do you know that: [hide]AA' + A'B' + BB' = 7r?[/hide]


Hmm, I'm sure half the people on here could do at least 3/4 of these.
In any case, I'm too lazy to do all but #1:
$1)\text{ }i,-1,-i, 1$

[quote="bndct01"]Sup! y'all  i have problem to test you first.... i am a new kid here
so I'll just get to the point.....answer the following and i may actually say and admire that your the best one in Problem solving here......

1.)Find the fourth roots of 1

2.)Find the number whose fourth roots are cis 0 degrees, cis 90 degrees, cis180 degrees, and cis 270 degrees

3.)Find the cube roots of 68.4321

4.)Show that for any complex number Z and any non-zero complex number W, [z/w] = [z/w]

5.)the cube roots of i

6.)the fourth roots of 16

7.)the cube roots of -i

8.)the fourth roots of -16

that's all folks.............have fun....thxz for reading...if there's anyone that dared....[/quote]

Number 1 and 5-8 are really easy, but I don't want to do them right now.  I don't understand what the problem is in number 4. 3 I could get with a calculator easily.

Presumably he made a typo and wants us to prove that the magnitude of $\frac{z}{w}$ is equal to the magnitude of $z$ divided by the magnitude of $w$.

[quote="bndct01"]Sup! y'all  i have problem to test you first.... i am a new kid here
so I'll just get to the point.....answer the following and i may actually say and admire that your the best one in Problem solving here......

1.)Find the fourth roots of 1

2.)Find the number whose fourth roots are cis 0 degrees, cis 90 degrees, cis180 degrees, and cis 270 degrees

3.)Find the cube roots of 68.4321

4.)Show that for any complex number Z and any non-zero complex number W, [z/w] = [z/w]

5.)the cube roots of i

6.)the fourth roots of 16

7.)the cube roots of -i

8.)the fourth roots of -16

that's all folks.............have fun....thxz for reading...if there's anyone that dared....[/quote]
[hide="1"]i, -i, 1, -1[/hide]
[hide="6"]2, -2, i, -i[/hide]
[hide="8"]Uhh, there is none. square root of i???[/hide]
I don't feel like doing the rest. These are too easy.

Err...the square root of $i$ is $\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$.  :wink:

To the original poster: I think you've vastly underestimated this forum's abilities.  :P These questions hardly qualify one as 'the best in Problem Solving'.

[quote="t0rajir0u"]Looks an awful lot like homework to me.  And I would hope that teachers would assign homework that their students could solve.[/quote]I agree. And we got too many good-hearted people here who likes to do someone's homework  :roll: :whistling: :rotfl:


inequalities
geometry
circumcircle
Euler
trigonometry
triangle inequality
geometry unsolved

Let $ABC$ be a triangle, $G$ it`s centroid, $H$ it`s orthocenter, and $M$ the midpoint of the arc $\widehat{AC}$ (not containing $B$). It is known that $MG=R$, where $R$ is the radius of the circumcircle. Prove that $BG\geq BH$.

[i]Proposed by F. Bakharev[/i]

Denote by $D$ midpoint of $AC$. We have that $MO \mid \mid BH$. From yhis is $\angle MOG= \angle GHB$. It is also $\angle BGH= \angle DGO \leq MCO = \angle MOG$. Hence $BH\leq BG$. Equality holds if $\angle DGM=0$, but then points $B,H,G$ lies on same line. SO there must be $H = G \Longleftrightarrow AB=AC=BC$.
q.e.d.

Ok, now realized my solution has some flaws in it.

what is tagiyev's  pisi lemma?can you send me this lemma?

another proof: Call O the circumcenter, K the projection of B on OH and $ \theta$ the angle between the euler line and AC. So we have $ \sin \theta \equal{} \frac{\overline{GO}}{2r}$ then $ \overline{KH} \equal{} \overline{BH} \cdot \sin \theta \equal{} 2R \cos \beta \sin \theta \equal{} \overline{GO}\cos \beta \le  \overline{GO} \equal{} \frac{\overline{GH}}{2}$. Then $ \overline{KH} \ge \overline{KG} \Longleftrightarrow \angle GHB \ge \angle HGB \Longleftrightarrow \overline{BG} \ge \overline{BH}$

Let $ K$ be the midpoint of segment $ AC$ and $ O$ be the circumcenter of triangle $ ABC$.$ BGH$ similar to $ KGO$ and $ BG\geq BH$ iff $ GK\geq OK$. $ MG\equal{}MO\equal{}R$ and by the triangle inequality $ GK\plus{}MK\geq MG$ , $ GK\plus{}R\minus{}OK\geq R$ and $ GK\geq OK$ . The proof is completed.


Alcumus
number theory
algebra
binomial theorem

I wanted to learn about these problems, and I know there's the AoPS Number Theory and Counting&Probability textbooks, but I was wondering if there was any good free source to explain it to me.  Also, is the binomial theorem covered in Alcumus at all?

Yeah. there are a lot of problems in alcumus that cover the binomial theorem.

I used to be terrible at applying the Binomial Theorem, but after reading the solutions and [i]seriously[/i] attempting the problems, I'd say I'm pretty good at it.


In any triangle  ABC  show that
$\Sigma\frac{m_a}{b+c}\geq\frac{3\sqrt3\cdot r}{2R} $

[quote="nickolas"]In any triangle  ABC  show that
[tex] \Sigma\frac{m_a}{b+c}\geq\frac{3\sqrt3\cdot r}{2R}  [/tex][/quote]

Since $\displaystyle m_a \geq \frac{b^2+c^2}{4R}$, etc.,

$\displaystyle \sum_{cyc}\frac{m_a}{b+c}\geq \frac{1}{4R}\sum_{cyc}\frac{b^2+c^2}{b+c} \geq \frac{1}{4R} \sum_{cyc}\frac{b+c}{2} = \frac{1}{4R} (a+b+c)$.

So it suffices to prove: $\displaystyle a+b+c \geq {6\sqrt3 r}$ ------ (*)

Multiplying $(a+b+c)$ on both sides of (*), (*) is equivalent to:

$\displaystyle (a+b+c)^2 \geq {12\sqrt3 S}$, but this follows from

$\displaystyle a^2+b^2+c^2 \geq ab+bc+ca \geq {4\sqrt3 S}$.

For the last inequality, see  http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15934


superior algebra
superior algebra theorems

Does the set {$ 1, \minus{} 1,i, \minus{} i$} , (where $ i \equal{} \sqrt{ \minus{} 1}$), form a group under the operation
"[b]x[/b]"(the usual multiplication) ?

Why, it's $ \mathcal{U}_4 \equal{} < i >$, the cyclic group of order $ 4$.


A total of 50 juniors and seniors were given a mathematics test. The 35 juniors attained an average score of 80 while the 15 seniors attained an average of 70. What was the average score for all 50 students who took the test?

[hide]The total score for juniors was $35 \cdot 80 = 2800$, while the total score for seniors was $15 \cdot 70 = 1050$. The total score for all of them is $3850$. Since there are $50$ of them, we see that the average score is $\frac{3850}{50}= 77$

Note that the answer is not $\frac{80+70}{2}= 75$![/hide]

[hide="solution"]The total score for the juniors is 35*80=2800, and the total for the seniors is 15*70=1050. The overall total is 3850. We divide by 50 to get 77.[/hide]


May be this is easy,but i have to ask.
+question 1:
If  A=>B=True ,das mean Negative B=>Negative A=True,too.
So am i rong if i use negative B=>A=False to prove that A=>B=True?
+question 2:
i had 2<=x<=4 so do it also means -1<=x<=8 too?
Thx for reading. ;)

For your first question:

You're asking if $\sim{B}\not\Rightarrow{A}$ implies that $A\Rightarrow{B}$. Can you explain why this would be true? It doesn't tell you much that $B$ does not imply $A$.

Second question:

You have $2\le{x}\le4$, and you're asking if that also means $-1\le{x}\le8$. You can even say $-1<x<8$, because $-1<2\le{x}\le4<8$.

[quote="AntonioMainenti"]For your first question:

You're asking if $\sim{B}\not\Rightarrow{A}$ implies that $A\Rightarrow{B}$. Can you explain why this would be true? It doesn't tell you much that $B$ does not imply $A$.
[/quote]

I think what hes saying is that $\sim{B}\Rightarrow\sim{A}$ is the same as $A\Rightarrow B$, which is correct.


algebra
function
domain
analytic geometry
graphing lines
slope
calculus

What is the main difference between continuous and uniform continuity? I have those definitions from my first analysis textbook, but an example would be appreciated.

Think of the term uniform as being the same across the domain of your function. When you pick a $ \delta$, it can usually be dependent upon the point at which you are checking for continuity, and $ \varepsilon$. In the case of uniform, you have to be able to pick a $ \delta$ dependent only upon $ \varepsilon$.

E.g. $ f(x)\equal{}x$. Let $ \varepsilon>0$. Then if $ \delta\equal{}\epsilon$ (no dependence upon the point), then for any $ x_0\in\mathbb{R}$, we can say that if $ |x\minus{}x_0|<\delta$ we have that $ |f(x)\minus{}f(x_0)|\equal{}|x\minus{}x_0|<\delta \equal{} \varepsilon$.

If this were $ f(x)\equal{}x^2$ then the choice of $ \delta$ would have to be dependent upon where you pick $ x_0$. This in general is due to the slope becoming increasingly large as you get away from $ x\equal{}0$.

Probably not the best explanation ever. Hopefully someone else will have some good insight.


What will happen to your blood stream and tissue if you drink lots of cold beer in like 10 min.?

So you drink lots of cold liquid. That lowers your core temperature a might. Alcohol is a vasodilator which also allows blood to flow to the extremities. Alcohol is metabolized in the liver.

Nerve cells are not killed off by alcohol as CW would have it. What does happen is that the nerve cells do not function normally. It is unclear if the loss in functionality is correctable.


inequalities
real analysis
complex analysis

$ 0 \leq x \leq \frac{\pi}2$ and $ 0 \leq y < \plus{} \infty$. prove that:
$ y(arctg y \minus{} x) \geq ln(cosx \sqrt{1\plus{}y^2})$.
when does the equality hold? :)

[hide]
Because $ 0\leq y\leq \infty$, then for all $ y$ there exist $ 0 \leq L \leq \frac {\pi}{2}$ such that $ y = tan(L)$. Rewrite the inequality in the form :
$ tan(L)(L - x) \geq ln(cos(x).sec(L))$
$ tan(L)(L - x) \geq ln(\frac {cos(x)}{cos(L)}) = ln(cos(x)) - ln(cos(L)$
Look that
$ tan(L) \geq \frac {ln(cos(x)) - ln(cos(L))}{L - x}$
$ - tan(L) \leq \frac {ln(cos(x)) - ln(cos(L))}{x - L}$
Without loss of generality supposed $ x < L$, by mean value theorem, there exist $ x \leq A \leq L$ such that :
$ \frac {ln(cos(x)) - ln(cos(L))}{x - L} = - tan(A)$
$ - tan(L) \leq - tan(A)$, $ tan(A) \leq tan(L)$. It is true because $ x < L$ by our assumption (Note that $ f(x) = tan(x)$ is monotonic increasing in the interval $ [0,\frac {\pi}{2})$.
Then we can reverse the step to obtain the statement in the problem above
[/hide]
Sorry if I am wrong. (But, I think that it is better if the problem is in Real Analysis section)


These problems are very much like the ones we did today:

1. I have 4 red socks, 5 white socks, and 5 green socks. If I choose 3 socks, what is the probability that I will have 3 different-colored socks?

2. There are 13 swimmers and 12 divers in my swim class. If I pick 2, what is the probability that I'll have one swimmer and one diver?

3. I have 4 green pens, 4 black pens, 4 blue pens, and 4 red pens. If I choose 4, what is the probability that I'll have 4 different-colored pens?

4. I have 4 red socks, 5 white socks, and 5 green socks. If I choose 3 socks, what is the probability that I will have at least 2 socks that are the same color?

Okay, the next problems are a variety of probability problems. Think before doing them. They may not be like the ones you did above.

5. If I pick 3 cards from a deck, what is the probability that they are all face cards?

6. If I have 4 girls and 4 boys on my Mathcounts team, what is the probability that both captain and vice-captain are boys?

7. If I randomly chose a point in an isosceles triangle ABC (AB=AC) and the probability that it is closet to the base (BC) is 1/4. What is the probability that the point is close to AB?

8. If I have 4 girls and 4 boys in my mathcounts class, what is the probability that my four-person team has at least 1 girl? What about at least 2 boys?

1.
[hide]25/104[/hide]

2.
[hide]13/25[/hide]

3.
[hide]64/65[/hide]

4.
[hide]173[/hide]

i think that's all i'll do
check #4 plz

ah.. my account time is running out... :(

[hide="#1"]$ \frac{25}{91}$[/hide]

[hide="#2"]$ \frac{13}{25}$[/hide]

[hide="#3"]$ \frac{64}{455}$ This one seems little off..[/hide]

I'll do some more at home.

[quote="spiders_hate_chestnut"]donkeykong, can you explain how you got the answer for number 1?[/quote]

Sure.

We want 3 socks from each type.
So let's say that we draw red socks first and it can be any of 4 reds.
Next, let's say that we draw white socks and it can be any of 5 whites.
Finally, we draw the green socks and it can be any of 5 greens.

So $ 4\cdot5\cdot5$ are the possible ways to pick 3 of different types.

But there are total of 14 socks and you are choosing 3 from them to achieve 3 different socks.

So the final equation would be $ \frac{4\cdot5\cdot5}{\binom{14}{3}}$

Solve.

I am not as sure as the first three... :wink:
And I didn't get #7..

[hide="#4"]$ \frac{66}{91}$ [/hide]
[hide="#5"]$ \frac{12\cdot11\cdot10}{\binom{52}{3}}\equal{}\boxed{\frac{11}{1105}}$[/hide]
[hide="#6"]$ \frac{4\cdot3}{\binom{8}{2}}\equal{}\boxed{\frac{3}{7}}$[/hide]
[hide="#8"]At least one girl:$ \frac{23}{35}$

At least two boys:$ \frac{36}{7}$   This one seems really off...[/hide]

[hide="my number 7"]
Hard to explain, but I got $ \frac{3}{8}$.
[/hide]

Yup, symmetry gives the answer to 7.

DK, I'll get back to your answers in a minute. I have to work them out haha (I don't create an answer list when I think up problems. Maybe I should next time.)

DK, you're #6 is incorrect because for the numerator, you had order matter and in the denominator, the order didn't matter.  At least, according to what I got, which is $ \frac{3}{14}$.

Donk, good on #4.

5. Your step shows you got the wrong concept, but your final answer is correct.

6. Alanchou is right.

8. Sorry, remember these problems aren't necessarily the same as #1-4. And the key words are "at least."

first part of #8 can be solved with complementary counting


What is the range of the parametric function

$x(t)=t^{2}-1$
$y(t)=t^{2}+1$
$z(t)=1-t$

What he means by range is what the y-value can be or can't be.  :rotfl:

It depends on the domain.
If we have a curve $f: \mathbb{R}\to\mathbb{R}^{3}$, defined for all $t\in\mathbb{R}$:
[hide]Since, $t^{2}\geq 0$, the range should be
$[-1,\infty)\times [1,\infty)\times (-\infty,\infty)$[/hide]

What you mean?
Avidently this curve is parabola, which lies in the plane y-x=2 and coudn't cover Cartesian product.

[hide="for x"]Since $t^{2}\geq0$, $t^{2}-1\geq-1$[/hide]

[hide="for y"]We know again that $t^{2}\geq0$, so therefore,  $t^{2}+1\geq1$[/hide]

[hide="for z"]t can be any real number, so the range for z is $\mathbb{R}$[/hide]

If we define the range of a function as "the set of all the output values of the function", then digger is right and I was wrong.
The most important thing in my post still remains: we can't know the range of a function if we don't know its domain (in this case, the possible values of $t$).

Well, in this case, the range of the function would be the portion of the parabola
$y-x=2$,
$y+x=2(z-1)^{2}$,
inside the box
$[-1,99]\times [1,101]\times [-9,1]$.


Bo, Coe, Flo, Jo, and Moe have different amount of money. Neither Jo
nor Bo has as much money as Flo. Both Bo and Coe have more than Moe.
Jo has more than Moe, but less than Bo. Who has the least amount of
money?

In order to solve this problem, you need to use the process of elimination. It is not Flo because she has more than Jo and Bo do. Bo and Coe have more than Moe, so it is neither of them.It also says that Jo has more money than Moe so it is not Jo. The only person left is Moe.

To make this a littler clearer, we can list them.

The first fact we are given is that Flo has more money than both Jo and Bo. So something like:

Flo
Jo
Bo

Next we are given that Bo and Coe have more than Moe.

Flo
Jo
Bo
Coe
Moe

We know that Jo has more money than Moe, but less than Bo.

Flo
Bo
Jo
Coe
Moe

We may not know the [i]exact[/i] order, but we do know that $ \boxed{\text{Moe}}$ has the least.


[quote="Svejk"]Here's a nice system,which I received it  from a friend.I don't know it's source.

Solve in $ \mathbb R$ the system:
$ x+y+z=6$
$ x^{2}+y^{2}+z^{2}=14$
$ x^{3}+y^{3}+z^{3}=36$[/quote]

[hide="Solution"]
1)Square equation #1 to get:
$ x^{2}+y^{2}+z^{2}+2(xy+xz+yz) = 36$
$ 14+2(xy+xz+yz) = 36$
$ xy+xz+yz=9$

2)Use the identity:
$ x^{3}+y^{3}+z^{3}-3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}-xy-xz-yz)$
$ 36-3xyz = 6(14-9)$
$ xyz = 2$

So we have,
$ x+y+z=6$
$ xy+yz+xz=9$
$ xyz=2$

The solutions can be obtained by Viete's theorem to polynomial:
$ f(t) = t^{3}+6t-9t+2$
Which is
$ -2,1,1$
And there premutations.
[/hide]

Another approach:
[hide]
Using Newton Sums, let $ S_{i}= x^{i}+y^{i}+z^{i}$ and $ a_{i}$ be the $ i^{th}$ coefficient of our monic polynomial containing roots $ (x, y, z)$

We know that:
$ S_{1}+a_{2}= 0$
$ a_{2}=-6$

$ S_{2}-6S_{1}+2a_{3}= 0$
$ a_{3}= 11$

$ S_{3}-6S_{2}+11S_{1}+3a_{4}= 0$
$ a_{4}=-6$

So $ P(x) = x^{3}-6x^{2}+11x-6$. Using the RRT, we can factor the polynomial into $ P(x) = (x-1)(x-2)(x-3)$ so the solutions are the permutations of $ (1, 2, 3)$[/hide]

[quote="K81o7"]Well sylow_theory, the answers are permutations of $ (3, 2, 1)$...

That's a nice solution, but you just made a calculation mistake in the first step...it should be $ xy+yz+zx=11$...
So $ xyz=6$.
So the poly is $ t^{3}-6t^{2}+11t-6=(t-1)(t-2)(t-3)$.
I'm just being picky...[/quote]

Thanks for the correction but that doesnt really bother me. I was in a rush (like usual) and I didnt care for careless errors because its the methods thats important not the actual answer.*

*)Unless you building bridges.  :rotfl:


function
trigonometry
real analysis
real analysis unsolved

According to intermediate value theorem if function f be continuous on interval I , a, b belongs to I and f(a) doesn't equal to f(b),then f takes any value between f(a) and f(b) in a point between a and b.

      Is it possible that f  takes each value in an infinite number of  points ?

     If it is the case ,provide an example ,unless prove the impossibility .

well i dont know how to prove the impossibility, but i sure believe it

but if we consider this family of functions:

$ f_n(x)\equal{} \cos (nx)$ on $ [0, 2\pi]$

it attains each value between (-1,1) at 2n different points...

but unfortunately, these wont converge as n goes to infinity ...


conics
parabola
symmetry
analytic geometry
graphing lines
slope
calculus

[quote="mdk"]the parabola y = x^2 is translated vertically to a parabola y = x^2 + k in such a manner that the translated
parabola has exactly two points of intersection with the parabola y = −2x^2 and that at each point of intersection the parabolas meet at a right angle, that is, the tangent lines are perpendicular. Find all values of k for which this happens.

Now suppose that the parabola y = x^2 is translated to a parabola y = (x − h)^2 + k.
Show that the only way the translated parabola can meet y = −2x^2 in right angles is if
h = 0, that is, the two parabolas share the same axis.[/quote]

I'm not sure if my method is correct...

[hide="My attempt at 1"]Note that $ k<0$ for them to intersect.
Solve for $ x$:
$ x=\pm\sqrt{\frac{-k}{3}}$.

By symmetry, if one pair of tangent lines is perpendicular, the other pair is perpendicular, too.
At the positive point, the slopes are
$ -4\cdot \sqrt{\frac{-k}{3}}$
$ 2\cdot \sqrt{\frac{-k}{3}}$

Since they are perpendicular, the product of the slopes is $ -1$, so $ \boxed{k=-3/8}$.[/hide]

I guess an analogous method works for 2.


Search cannot locate formula strings like n^2+1. It used to work before but now it is not possible. Also, there is a search limit? It says I need to wait a bit after 3 attempted searches because I was 'overusing' the Search? That's off and waste time or maybe Search uses too much server resource...

i have talked about this at http://www.mathlinks.ro/Forum/viewtopic.php?t=97725 .

as for latex strings, it is messed up now indeed, but i can repeat my proposal:

[quote="darij grinberg"]what about a special environment allowing to search for strings of arbitrary symbols? for instance, by searching for [string]\sin a + \sin b[/string], i get exactly those topics that contain the string "\sin a + \sin b"? this would be very useful for searching latex code...[/quote]

this would not only fix the current problem but would allow searching for latex easier than it was before the new server.

as for the search limit, valentin says it indeed overloads the server... but maybe the new search engine bots will make it possible to use e. g. google for searching on ML (formerly, only parts of the discussions were stored in google cache).

anyway, these problems are already on the to-do list of the admins, let's give them time.

  darij


Let $a,b,c$ be positive real numbers such that $a^{2}+b^{2}+c^{2}=1$. Prove that the following inequality holds
\[\sum\frac{a^{3}}{b^{2}-bc+c^{2}}\geq \sqrt{2}\]

[quote="toanhocmuonmau"]Let $ a,b,c$ be positive real numbers such that $ a^{2} \plus{} b^{2} \plus{} c^{2} \equal{} 1$. Prove that the following inequality holds
\[ \sum\frac {a^{3}}{b^{2} \minus{} bc \plus{} c^{2}}\geq \sqrt {2}\]
[/quote]
Because by CS we obtain:
$ \sum_{cyc}\frac {a^{3}}{b^{2} \minus{} bc \plus{} c^{2}}\geq\frac{1}{\sum(a^2b\plus{}a^2c\minus{}abc)}$
and $ (a^2\plus{}b^2\plus{}c^2)^3\geq2\left(\sum_{cyc}(a^2b\plus{}a^2c\minus{}abc)\right)^2,$ which is obvious after expanding.

[quote="toanhocmuonmau"]Let $ a,b,c$ be positive real numbers such that $ a^{2} \plus{} b^{2} \plus{} c^{2} \equal{} 1$. Prove that the following inequality holds
\[ \sum\frac {a^{3}}{b^{2} \minus{} bc \plus{} c^{2}}\geq \sqrt {2}\]
[/quote]
By cauchy-swarchz we have:
$ LHS \ge \frac{(a^2\plus{}b^2\plus{}c^2)^2}{\sum ab(a\plus{}b)\minus{}3abc}\equal{}\frac{1}{\sum ab(a\plus{}b)\minus{}3abc}$
We need prove that:
$ \frac{1}{\sum ab(a\plus{}b)\minus{}3abc} \ge \sqrt{2}$(*)
Let:$ p\equal{}a\plus{}b\plus{}c;q\equal{}ab\plus{}bc\plus{}ca;r\equal{}abc$
So:$ 2q\equal{}p^2\minus{}1$
$ (*)<\equal{}>2 \ge [p(p^2\minus{}1)\minus{}12r]^2$
It's easy by Shur inequality :)


Diophantine equation
number theory
equation
algebra
IMO Shortlist

Provided the equation $xyz = p^n(x + y + z)$ where $p \geq 3$ is a prime and $n \in \mathbb{N}$. Prove that the equation has at least $3n + 3$ different solutions $(x,y,z)$ with natural numbers $x,y,z$ and $x < y < z$. Prove the same for $p > 3$ being an odd integer.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1995405#p1995405]Posted here[/url].

And a nice solution by pco:

[quote="pco"]It's rather easy to check that $(x,y,z)=(\frac{p^nu+p^a\frac{p^a+1}u}t,p^{n-a}u,\frac{p^a+t}u)$ is a solution whenever these are positive integers.

For $n>1$
=========
 let us look at the following solutions :

A first group of S1 of $4n-4$ solutions :
Setting $t=1$, $a\in[1,n-1]$ and $u\in\{1,2,\frac{p^a+1}2,p^a+1\}$ gives solutions $(x,y,z)=(p^nu+p^a\frac{p^a+1}u,p^{n-a}u,\frac{p^a+1}u)$
In order to show that all these solutions are distinct, consider $(x,y,z)$ such a solution :
Two of the three integers are divisible by $p$ and the third is not. The littlest of the two element divisible by $p$ need to be $p^{n-a}u$. Since $u$ is not divisible by $p$, $a$ is uniquely determined as $n-v_p(\text{ this number})$.
Then $u$ is uniquely determined.

A second group of $n-1$ solutions :
Setting $t=p$, $a\in[1,n-1]$ and $u=1$ gives solutions $(x,y,z)=(p^{n-1}+p^{a-1}(p^a+1),p^{n-a},p^a+p)$
In order to show that all these solutions are distinct, consider $(x,y,z)$ such a solution :
Only one of the three numbers is a perfect power of $p$, and this uniquely gives $a$.
In order to show that all these solutions are distinct from S1 :
If $a>1$, the three numbers are divisible by $p$ and so cant be in S1.
If $a=1$, the solution is $(p^{n-1}+p+1,p^{n-1},2p)$ and both element divible by $p$ are $<p^n$ while in S1, the greatest is $>p^n$

Four other solutions :
S3 : $t=1,u=1,a=0$ $\implies$ solution $(x,y,z)=(p^n+2,p^n,2)$
S4 : $t=1,u=2,a=0$ $\implies$ solution $(x,y,z)=(2p^n+1,2p^n,1)$
S5 : $t=1,u=1,a=n$ $\implies$ solution $(x,y,z)=(p^{2n}+2p^n,p^n,p^n+1)$
S6 : $t=1,u=2,a=n$ $\implies$ solution $(x,y,z)=(2p^{n}+\frac{p^n(p^n+1)}2,2,\frac{p^n+1}2)$

It's immediate to see that these 4 solutions are distinct
And they are also distinct from S1 and S2 since they contain exactly one element divisible by p while group S1 always contains exactly 2 and group S2 contains 3 or 2 (in case of $a=1$)

So we got $5n-1$ solutions and since $5n-1\ge 3n+3$ $\forall n>1$, we got the result (in fact a stronger result)

For $n=1$,
=========
We need to find 6 solutions. Choose for example :
$a=0$ and $t=1$ and $u=1$ $\implies$ solution $(x,y,z)=(p+2,p,2)$

$a=0$ and $t=1$ and $u=2$ $\implies$ solution $(x,y,z)=(2p+1,2p,1)$

$a=0$ and $t=2$ and $u=1$ $\implies$ solution $(x,y,z)=(\frac{p+3}2,p,3)$

$a=0$ and $t=2$ and $u=3$ $\implies$ solution $(x,y,z)=(\frac{3p+1}2,3p,1)$

$a=1$ and $t=1$ and $u=1$ $\implies$ solution $(x,y,z)=(p^2+2p,1,p+1)$

$a=1$ and $t=1$ and $u=2$ $\implies$ solution $(x,y,z)=(2p+\frac{p(p+1)}2,2,\frac{p+1}2)$

All these 6 solutions are clearly distinct (order their element).[/quote]


I've been wondering for a week thinking of this problem..
Let's say U have a bag contain all real numbers. If that so what is the probability that you've picked up randomly real no. 2. I find that the probability of that reaches almost to zero. So does it means I do not have any chance of getting no. 2 from the bag I picked randomly?

The probability of picking any discrete number from a bag of real numbers is zero. This is because your bag is uncountably large.

It's like if you flip a coin continuously. The chance of getting any individual infinite sequence of heads and tails in 0.

But back to your problem, although the probability of picking any discrete number from the bag is 0, you can for example if your bag was a cotinuous random variable generator, get a probability of picking a number between a range of numbers.

Don't worry there's nothing wrong with saying if you pick any random real number, the probability it is number 2 is 0. The situation can never exist because you can never truly pick any random real number.


there is an arithmetic progression: 2,5,8,11............... Prove that it does not contain a perfect square.


algebra
polynomial
advanced fields
advanced fields unsolved

Let $ F$ be a nonconstant polynomial in $ k[X_1, . . . ,X_n]$ ,  $ k$ algebraically closed.
Show that $ \mathbb{A}^n(k) \backslash V (F)$ is infinite

For a nonconstant polynomial in one variable over $ k$, clearly there are infinitely many numbers that are not roots of the polynomial. (since there are only finitely many roots)

In general, we write $ F$ as a polynomial in $ x_n$, with $ x_1,\cdots,x_{n\minus{}1}$ as coefficients. WLOG, assume that $ F$ is not degree 0 in $ x_n$, for otherwise we can choose another variable to work with. Pick $ x_1 \equal{} a_1,\cdots,x_{n\minus{}1} \equal{} a_{n\minus{}1}$ such that the leading coefficient is nonzero. (This is possible because otherwise the leading coefficient is nilpotent in $ k[X_1,\cdots,X_n]$, meaning that it is 0) Then we get a polynomial $ F(a_1,\cdots,a_{n\minus{}1},x_n)$ in $ x_n$, then just pick $ x_n$ that are not roots of this polynomial.


function
integration
logarithms
limit
calculus
calculus computations

If function $ f: \mathbb{R}\rightarrow\mathbb{R},$ satisfies the condition $ f(x)\geq\frac{1}{x},$ for all $ x>0,$ prove that $ f$ has no primitive.

The so-called key lemma is:

[color=blue]Let $ F$ be a differentiable function on $ [a,b]$. Suppose that $ F^\prime (x) \geq h(x), \, \forall x \in [a,b]$, where $ h$ is a Riemann-integrable function on $ [a,b].$ In these conditions we have that $ F(b)-F(a) \geq \int_{a}^{b}h(x) \, dx.$[/color]

Now let $ F$ be a primitive of $ f$. By the above lemma, we have that $ F(1)-F(x) \geq \int_{x}^{1}\frac1{t}\, dt =-\ln x,$ for all $ x > 0$.
Thus, $ F(x) \leq F(1)+\ln x$. Let's see what happens as $ x \to 0^{+}: \ \lim_{x \to 0^{+}}F(x) =-\infty.$
That's a contradiction, since that limit must be the finite number $ F(0).$

I'll leave the pleasure of providing a nice proof to that lemma to somebody else.

Your function $ f$ does not have the Darboux property.

The case when $ f(0)<0$ is trivial. I will consider only the case when $ f(0)\geq 0$. Let $ n$ be the least positive integer such that $ f(0)<n$.

Then consider the interval $ [0,\frac{1}{n}]$. we have that $ f(0)<n\leq f(\frac{1}{n})$. Pich $ \lambda=\frac{n+f(0)}{2}$ located between $ f(0)$ and $ f(n)$.

We know that there is $ c_{\lambda}\in (0,\frac{1}{n})$ such that

$ f(c_{\lambda})=\frac{n+f(0)}{2}\geq \frac{1}{c_{\lambda}}$,

and hence

$ \frac{1}{n}\geq c_{\lambda}\geq \frac{2}{n+f(0)}$.

It follows that $ f(0)\geq n$ which is a contradiction.

Of course that if $ f(0)=k$ for some integer $ k$ then you apply the same technique on the interval $ [0,1/(k+1)]$ with $ \lambda=k+\frac{1}{2}$.


Hi,

I was wondering whether this proof which I have come up with is actually valid. All the steps seem to make sense, but I get the distinct impression that something is invalid or just weak

The question is, find all integer solutions to $ a^3\plus{}b^3\equal{}9$.

My 'weak' answer

$ (a\plus{}b)(a^2\minus{}ab\plus{}b^2)\equal{}3^2$

Hence either $ a\plus{}b\equal{}3,a^2\minus{}ab\plus{}b^2\equal{}3$ or $ a\plus{}b\equal{}1,a^2\minus{}ab\plus{}b^2\equal{}9$ or $ a\plus{}b\equal{}9,a^2\minus{}ab\plus{}b^2\equal{}1$

Case 1: $ a\plus{}b\equal{}3$

Therefore $ (a\plus{}b)^2\minus{}3ab\equal{}3$
Hence $ 9\minus{}3ab\equal{}3$
Therefore $ ab\equal{}2$
Hence $ (a,b)\equal{}(2,1) or (1,2) (we note that both negative would yield invalid solutions)$

Case 2: $ a\plus{}b\equal{}1$

Therefore $ (a\plus{}b)^2\minus{}3ab\equal{}9$
Hence $ 1\minus{}3ab\equal{}9$
Therefore $ 8\equal{}\minus{}3ab$
No integer solutions

Case 2: $ a\plus{}b\equal{}9$

Therefore $ (a\plus{}b)^2\minus{}3ab\equal{}1$
Hence $ 81\minus{}3ab\equal{}1$
Therefore $ 80\equal{}3ab$
No integer solutions

Hence the only integer solutions to $ a^3\plus{}b^3\equal{}9$ are $ (1,2)$ and $ (2,1)$

BTW, there is another part to the question, but I have proved that assuming this result.

Thanks

Another prove:
i) If $ a,b\geq 0$ then $ 0\leq a,b\leq 2$ then easy to see that $ a\equal{}1,b\equal{}2$ or $ a\equal{}2,b\equal{}1$.
ii) If exit a number $ <0$ and a number $ >0$ we have to find integer solution of $ a^3\minus{}b^3\equal{}9$, with $ a\geq 3,b>0$.
We have $ LHS\geq a^3\minus{}(a\minus{}1)^3\equal{}3a^2\minus{}3a\plus{}1\geq 3.3^2\minus{}3.3\plus{}1>9$ then not satisfy!
Then we have root $ (1;2),(2;1)$


Find the largest odd natural number that is a factor of $ 6!$

$ 6!\equal{}6\cdot5\cdot4\cdot3\cdot2\cdot1$, and in order for a number to be odd, it must have no even factors. Any factor of any number must consist only of the factors of the original number. Thus, the largest odd natural factor here would be $ 5\cdot3\cdot1\equal{}\boxed{15}$.

If we prime factorize $ 6!$, we have $ 2^4\cdot 3^2\cdot 5$.  The largest odd factor would be $ 3^2\cdot 5\equal{}\boxed{45}$.


The Art and Craft of Problem Solving, 2nd Edition
By: Paul Zeitz

[b]1. The problem statement, all variables and given/known data[/b]

1.3.4
In the xy-plane, what is the length of the shortest path from $ (0,0)$ to $ (12, 16)$ that does not go inside the circle $ {{(x - 6)}^{2}} + {{(y - 8)}^{2}} = {25}$

[b]2. Relevant equations[/b]

Standard form of the Equation of a Circle.

$ {{(x - h)}^{2}} + {{(y - k)}^{2}} = {{r}^{2}}$

Standard form of the distance between two points $ ({x}_{1}, {y}_{1})$ and $ ({x}_{2}, {y}_{2})$ in the xy-plane.

$ {d} = {\sqrt {{{{({x}_{2}} - {{x}_{1}})}^{2}} + {{({{y}_{2}} - {{y}_{1}})}^{2}}}}$

[b]3. The attempt at a solution[/b]

$ {P}_{m} \equiv$ midpoint
$ {d} \equiv$ the distance between $ ({x}_{1}, {y}_{1})$ and $ ({x}_{2}, {y}_{2})$
$ d_{r} \equiv$ diameter
$ {{D}_{min}} \equiv$ the length of the shortest path between $ ({x}_{1}, {y}_{1})$ and $ ({x}_{2}, {y}_{2})$ that does not go inside the circle, $ {{(x - 6)}^{2}} + {{(y - 8)}^{2}} = {25}$.
$ {C} \equiv$ circumference of the circle.

$ ({x}_{1}, {y}_{1}) = (0, 0)$
$ ({x}_{2}, {y}_{2}) = (12, 16)$
$ (h, k) = (6, 8)$
$ {r} = 5$

Let, $ Ax + By = C$ be the equation of the line between $ ({x}_{1}, {y}_{1})$ and $ ({x}_{2}, {y}_{2})$.

Note, that $ h$ and $ k$ in this problem are half of $ {x}_{2}$ and $ {y}_{2}$, additionally since $ ({x}_{1}, {y}_{1}) = (0, 0)$ the point $ (h, k)$ is the $ {P}_{m}$ of the line $ Ax + By = C$. Furthermore, this means that the length between $ ({x}_{1}, {y}_{1})$ and $ ({x}_{2}, {y}_{2})$ that goes inside the circle is the diameter. Lastly, this implies that the shortest length to avoid the inside of the circle is half of the circumference of the circle.

So, this problem is then simplified to finding the below,

$ {D}_{min} = {{d} - {{d}_{r}} + {\frac {C}{2}}}$

Which is,

$ {{D}_{min}} = {({\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}})}-{(2r)}+{(\frac{(2{\pi}{r})}{2})}$

$ {{D}_{min}} = {{\sqrt {{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}}}+{r({\pi}-2)}$

$ {{D}_{min}} = {10+5{\pi}}$

Is the above answer right?

[b]In addition, how would one solve a generalized form of this problem,[/b]

In the xy-plane, what is the length of the shortest path from $ ({x}_{1}, {y}_{1})$ to $ ({x}_{2}, {y}_{2})$ that does not go inside the circle $ {{(x - h)}^{2}} + {{(y - k)}^{2}} = {{r}^{2}}$

I tried figuring out the above but did not know how to approach the case if the circle was not "in the way" so to speak of the shortest length between the two points. How is it that one can apply constraints to this problem to tell whether the circle: would be or would not be in the way?

Thanks,

-APSStudent

Your solution is wrong. In particular, this: [quote="APSStudent"]Lastly, this implies that the shortest length to avoid the inside of the circle is half of the circumference of the circle.[/quote] is wrong. Your proposed map makes a sharp turn when it reaches the circle, and cutting off the corner is faster.

This belongs somewhere in the high school forums.


Let $a$ and $b$ be two given numbers and $x$ be any real number. Determine the maximum value of\[F=\frac{(x^{2n}-a^{2n})(b^{2n}-x^{2n})}{(x^{2n}+a^{2n})(b^{2n}+x^{2n})}\]

looks romanian?

Basically, you want to minimize $\frac{(x-a)(x-b)}{(x+a)(x+b)}$.  (notice that I flipped it.), where x,a,b are nonnegative.

This is $1 - 2(a+b) \frac{x}{(x+a)(x+b)}$, so we want to maximize $\frac{x}{(x+a)(x+b)}$.  Clearly this isnt negative.

So we want to minimize $\frac{(x+a)(x+b)}{x} = x + a + b + \frac{ab}{x}$.  So we want to minimize $\frac{ab}{x} + x$.

AMGM says that this is atleast $2\sqrt{ab}$, equality when $x = \sqrt{ab}$.

sry in a rush cant explain full but that should be way more than enough.


Assign the values 2, 3, -2, -3, to u, v, x, and y, such that the expression uv - xy has a minimun and maximum value (the v and y are raised powers)

Checking all cases (they are 24), we find the maximum value is $ u^v\minus{}x^y\equal{}2^2\minus{}(\minus{}3)^3\equal{}31$ and the minimum value is $ (\minus{}3)^3\minus{}2^2\equal{}\minus{}31$.


search
USAMTS
Harvard
college
MIT
articles
Science Olympiad

i'm living in VA and i'm trying to search for mathematical contests.

contests in VA : only MAth League

other contest: AMC 12, USAMTS, Harvard- MIT Math League (but ineed to go to MA to do this)

That was all i could find. I 'm a Junior now. Does anyone know any other contests a person living in VA could do individually. (no group contest)?????????????????

and is there any science contests like physics and stuff i could do. if there is how do you enter it?

I'm not exactly sure in VA, but if you live in a big city, there has got to be science olympiad and science fair  :) competitions. If they are not at your school then one of the schools in your community has got to have these competitions. Also, if you can get 3 other people to just sit with you, do the Mandelbrot competition!

There are many contests you can participate in as an individual if your school doesn't have many interested students, some of which are written by my company.  Our individual contests are the Fall Startup Event (which already happened this year) and the Ciphering Time Trials on December 8th.  All you need to do is find a coach-type person to register for them and then administer them.  Discussion of our contests takes place in its own forum.  You'd certainly be at a huge disadvantage in team contests, but so long as you're okay with getting a lower award (or not getting one at all) then team contests have a lot of great math on them and you might not want to disregard them.

[quote="yarakvar"]i'm living in VA and i'm trying to search for mathematical contests.

contests in VA : only MAth League
[/quote]

Not true: the September 2005 issue of [i]Math Horizons[/i] has an article on the Virginia Tech Regional Math Contest.  It says "Although we don't encourage high school students to take part, they are eligible to take the exam if a participating university is willing to proctor the exam... Many of the high school students who have taken part have performed excellently.  In fact only twice has a perfect score been obtained in the contest... and these were both achieved by high school students."

VTRMC's website is http://www.math.vt.edu/events/competitions/Vtregional/
There's a sample problem and solution at http://www.maa.org/mathhorizons/supplement/VTRMC_Problem.htm
and of course, I would be remiss if I didn't mention the [i]Math Horizons[/i] website, which is at http://www.maa.org/mathhorizons


function
trigonometry
logarithms
calculus
calculus computations

let f(x)=-x^3 +1

Which is the f^-1?

Which are the intersect points between f and f^-1

[hide="Hint"]You can obtain the all intersections from the equation $ f(x)=x$.[/hide]

f(0)=1 then f^-1(1)=0

f(1)=0 then f^-1(0)=1

then f(0)=f^-1(0) =1 then the point (0,1) isn't an intersect point ?

Also f(1)=0 =f^-1(1) then the point (1,0) isn't another  intersect point ?

[quote="Virgil Nicula"]You can obtain the all intersections from the equation $ f(x)=x$.[/quote]
That is incorrect: $ f(x)=f^{-1}(x)$ does not imply that $ f(x)=x$.

In this case $ f^{-1}(x)=(1-x)^{1/3}$. The equation $ 1-x^{3}=(1-x)^{1/3}$ can be written as $ (1-x^{3})^{3}=1-x$. Both sides are zero when $ x=1$. Dividing them by $ 1-x$, we get $ (1-2x^{3}+x^{6})(1+x+x^{2})=1$. There is another obvious root, namely $ x=0$. We can multiply the left side out, subtract $ 1$ and factor out $ x$ -- but the result is ugly. :maybe:

There appear to be two more roots: one in the interval $ (0,1)$ (the solution of $ x^{3}+x=1$), and the other is greater than $ 1$.

[color=darkblue]You are right generally, [b]Mlok[/b] ! For example, $ f(x) =\frac{1}{x}\ ,\ x > 0$. But in the our case I am right ! [b]Can say you why ?[/b]

For the our function $ f(x) = x^{3}+1$ we have $ f(x) = f^{-1}(x)$ $ \Longleftrightarrow$ $ f(x) = x$ which has only one zero $ x_{0}\in (-2,-1)$.[/color]

[color=darkred][b]Another exercises.[/b]

$ 1\blacktriangleright$ Consider the function $ f:\mathbb R\rightarrow\mathbb R$, where $ f(x) = x+\sin x$.
Prove that this function is bijective and solve the equation $ f(x) = f^{-1}(x)$.
[b]Answer[/b] : $ x\in\pi\mathbb Z$.

$ 2\ \blacksquare$ Ascertain $ a > 1$ for which the equation $ a^{x}=\log_{a}x$ has only one zero.
[b]Answer[/b] : $ a = e^{\frac{1}{e}}$ and $ x_{0}= e$.

$ 3\blacktriangleright$ Consider the [b]Keppler's function[/b] $ F:\mathbb R\rightarrow\mathbb R$, $ F(x) = x-\lambda\sin x$, where $ \lambda\in (0,1)$.
Prove that this function is bijective and solve the equation $ F(x) = F^{-1}(x)$.
[b]Answer[/b] : $ x\in\pi\mathbb Z$.[/color]

Our function is $ f(x)\equal{}1\minus{}x^{3}$.

O.K. ! The function $ f(x)\equal{}1\minus{}x^{3}$ is bijective decresing and you are right. I read wrong. Sorry !


rotation
geometry
geometric transformation
reflection
rectangle
geometry proposed

Consider two polygons $ P$ and $ Q$. We want to cut $ P$ into some smaller polygons and put them together in such a way to obtain $ Q$. We can translate the pieces but we can not rotate them or reflect them. We call $ P,Q$ equivalent if and only if we can obtain $ Q$ from $ P$(which is obviously an equivalence relation).
[img]http://i3.tinypic.com/4lrb43k.png[/img]
a) Let $ P,Q$ be two rectangles with the same area(their sides are not necessarily parallel). Prove that $ P$ and $ Q$ are equivalent.
b) Prove that if two triangles are not translation of each other, they are not equivalent.
c) Find a necessary and sufficient condition for polygons $ P,Q$ to be equivalent.

This is a classical problem in a field of plane geometry known as transformation theory. But without rotation perhaps is a little harder. Here is my solution to (a). (I think the lemma 2 divides in two cases and is incomplete).\\ \\

\bf{Lemma 1:} \normalfont A rectangle is equivalent to a square with same area.\\
Proof:  Let $ a=AB, b=BC$ the sides of the rectangle $ ABCD$, and consider three different cases.\\
Case I: ($ a=b$).  In this case the rectangle is a square and the result is immediately.\\

Case II: ($ b<a\leq 4b$). Let $ s=\sqrt{ab}$ the side of the square $ AB'C'D'$ with equal area, where $ B'$ lies on $ AB$ and $ D'$ lies on $ AD$. The line $ BD'$ cut the sides $ CD$ and $ B'C'$ in $ K$ and $ L$, respectively. Clearly the triangles $ KBB'$ is similar to $ D'LD$ and $ BCL$ is similar to $ KC'D'$. From the similar $ KB'B\sim D'AB$ we have\\
\centerline{$ {\frac{B'K}{AD'}=\frac{B'B}{AB}\Rightarrow HK=\frac{AD'\cdot AB}{AB}=\frac{s(a-s)}{a}= s-b}$}\\ \\
The hypotesis, $ a\leq 4b$ implies $ s^{2}=ab\leq 4b^{2}$ so $ s-b\leq b$. Therefore $ K$ lies inside the segment $ B'C'$ and we have\\
\centerline{$ KBB'\simeq D'LD$ and $ BCL\simeq KC'D'$}.\\ \\

Case III: ($ 4b<a$). Divide consecutively to half the base $ AB$ and duplicate the altitude $ BC$ of the rectangle until obtain the cases I or II. \\ \\

\bf{Lemma 2:} \normalfont Two squares with same area are equivalent.\\
Proof:  Let $ ABCD$ and $ AB'C'D'$ two squares with equal area, where the second is a rotation by an arbitrary angle $ 0<\phi<90$ from the first.\\
Line $ B'C'$ cuts $ CD$ on $ E$, $ F$ is the foot of the perpendicular to $ CD$ from $ C'$, $ ED$ cuts $ C'D'$ on $ G$ and the parallel on $ G$ to $ DA$ cuts $ D'A$ on $ H$. $ I$ is the point inside $ AD$ so $ AI=C'F$ and the parallel on $ I$ to $ AB$ cuts $ D'A$ on $ J$. $ K$ is the point inside $ AB'$ so $ AK=C'G$ and the parallel on $ K$ to $ B'C'$ cuts $ AB$ on $ L$. Parallel on $ B'$ to $ BC$ cuts $ KL$ and $ CD$ on $ M$ and $ N$, respectively.
It is easy to prove that $ EC'G\simeq LKA$, $ D'HG\simeq KMB'$ and $ AIJ\simeq B'NE$, so we can translate these triangles without rotation. One can takes points $ X$, $ Y$ and $ Z$ from segments $ IJ$, $ DG$ and $ BC$ to construct a pentagon $ GHJXY$ congruent to $ B'MLBZ$, with $ B'Z$ parallel to $ AB$.\\
Finally the residues $ XIDY$ and $ B'ZCN$ are two rectangles with the same area an parallel sides, so by the lemma 1 these are equivalent.\\ \\

\bf{Proof of the problem:} \normalfont Since cutting for figures is an equivalence relation, from the lemma 1 and lemma 2 we obtain the equivalence of any two rectangles with the same area.


inequalities
algebra
binomial theorem
inequalities solved

The inequalities
(a + 1)(b + 1) \geq (ab + 1)
(b + 1)(c + 1) \geq (bc + 1)
(c + 1)(a + 1) \geq (ca + 1)
can be verified by simple computation. Multiplying gives the result.

Myth, are you sure it's the right generalisation ?

[quote="Myth"]Arne!
Can you prove generalization:

(1+a1n)...(1+ann) \geq (1+a1...an)n[/quote]

Since I don't see an obvious link ... the n = 3 case is a bit different ... What do you think ?

I mean a generalization of
(1+a^3)2(1+b^3) \geq (1+a2b)^3

Oh, like this.
Expand ;
Then apply AM - GM to each term ;
Then you can recognise the binomial theorem ;
Factorize and you're done.

The fact that (a + 1)(b + 1) \geq (ab + 1) can also be proved using Holder.
But I don't know whether it's possible to prove the generalisation with Holder.
(I don't use Holder often ...)

So we have shown that for all a(i) > 0, 1 \leq i \leq n, and r > 0 we have

\pi(r + a(i)) \geq (r + g)^n

where g is the geometric mean of the n number a(i), 1 \leq i \leq n.

we can consider that c>=b>=a and lets prove that:

(a^3+1)(b^3+1)>=(a^2b+1)(ab^2+1) which with simple computation gives that: (a+b)(a-b)^2>=0

So we get that:

(a^3+1)(b^3+1)(c^3+1)>=(a^2b+1)(ab^2+1)(c^3+1).

we only have to prove that:
(b^2a+1)(c^3+1)>=(b^2c+1)(c^2a+1) which after some easy computation gives: (c+b)(c-b)(c-a)>=0 which is true!

ineq done! :D :D

cheers!


Are there any simple expressions of sum(1 to inf)( 1/(k^2 + 1) ) and sum(1 to inf)( 1/(k^2 + 2) ) as sums of elementary functions?

I want to express sum(1 to inf)( 1/(k^2 + 1)(k^2 + 2) ) into a sum of elementary functions.

Use this formula:

$\sum_{k=1}^{\infty}\frac{\cos kx}{k^{2}+\alpha^{2}}=\frac{\pi}{2\alpha}\frac{\cosh\alpha(\pi-x)}{\sinh \alpha \pi}-\frac{1}{2\alpha^{2}}$, valid for $0\leq x\leq 2\pi$, $\alpha\neq 0$.

Let $x=0$ and $\alpha =1$ and $\alpha=\sqrt{2}$ and you can calculate your sums.

Thx!
Is there any nice way of proving that formula?

Is it possible to maybe split the denominators into (k-i)*(k+i) and (k-sqrt(2)*i)*(k + sqrt(2)*i) and use the famous formula for pi*cot(pi*x) = 1/x + sum(k = 1 to inf)( 1/(x + k) + 1/(x - k) )? So we get an expressions with cot(pi*i) and cot(pi*sqrt(2)*i) which give us the hyperbolic expressions?


topology
real analysis
real analysis unsolved

Let X be a metric space, K a compact subset of X, F a closed subset of X, and such that K and F are disjoint. Then there exists \delta > 0 such that d(x,y) > \delta for all x in K and y in F. Okay, I know how to prove this, but can we weaken the hypothesis so that K and F are both assumed only closed? I have yet to find a counterexample or a proof.  :roll:

In $\mathbb{R}^2$, let $A$ be the $x$-axis. That is, let $A=\{(x,0): x\in\mathbb{R}\}.$ Let $B=\left\{\left(x,\frac1x\right): x\in(0,\infty)\right\}.$

$A$ and $B$ are both closed, but $d(A,B)=0.$

Note how important it is to the counterexample that neither $A$ nor $B$ is compact.


Euler
number theory
IMC
college contests

What number theory knowledge is required for IMC problem solving??
I mean, does it need to be in IMO level????
Thank you for your interest..........

I think only basic number theory is required: euler, wilson, chinese remainder etc.. That and stuff from abstract algebra..


here is the figure(way to go Mr. Vornicu :)):

 [geogebra]673043667a014d98ed9ef3de5a31bc9280e5f4d0[/geogebra]
EF is parallel to AC. AF is parallel to DE. Find CD/BD. The area of EDC is twenty five times that of FBD.

[hide="The actual question (hidden because contains a hint)"]So, its pretty clear that EDC is similar to FBD. but, then introduction to geometry solution manual somehow says that the corresponding sides of EDC are $ \sqrt{25}\equal{}5$ five times the sides of FBD. I don't really understand that. Can someone explain?[/hide]

Totally out of the subject: i upgraded to Firefox 3 :P

Ratio of corresponding lengths of similar (two-dimensional) figures is the square root of the ratio of there areas.

If I understand what you're asking, which I'm not completely sure I am.

is there a proof for you first statement? And, as far as i know, intro to geometry doesn't mention that yet... (i'm on page 121)

[hide="Proof"]

Let's do triangles first. Let the ratio of new sides to the old sides be d. New semiperimeter = old semipereimeter * d.

New semiperimeter-new side length of a = d(old semiperimeter-old side length) Because everythihng is multiplied by d.

Using Heron's, we get the area increases by d^2.

Let's say we have an other figure. If it is a polygon, split it into a lot of trianlges. If it is not a polygon, try to fit a lot of triangles in there so that the uncovered area is as small as possible. Then, use what we just got on each of the triangles to get the area increases by a ratio of d^^2 when all the sides increases by a ratio of d.[/hide]

i've not done heron's or any of that stuff yet (not even Pythagorean theorem)  :blush:
so how does the book do it?
I was thinking that you could arrange the 25 FBD triangles into a sqaure than you could have it such that the side lengths of the square would be $ (\sqrt{25})(\sqrt{[FBD]})\equal{}5\sqrt{[FBD]}$. Then, maybe you could work from that...

thanks,
Poincare(Dhaivat)

[quote="pythag011"]Hmm..

Or you can look at it like this: if you make it bigger d times, the altitude will increases its length by a ratio of d. So will the base. So the area wiill increase by d^2[/quote]

Yeah, that's the way I first learned it. Well, the first way I proved it I guess, since I just had someone tell me it at first.

well now that's exactly the same as the square solution!!!!!!!!
thank you everyone!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
ok, mods, i will slow down on the !'s but plz lemme keep them in this post!

EDIT: What do you mean by a ratio of d?


[quote="tonsoffun18204"]So in my math class we are given this project to work on and it has to deal with how many squares are on a chessbaord. We searched the internet and found a few different sites with answers but the question we have is over the equation given and if their are any other ways of finding the answer. The equation we were given is N(N+1)(2N+1)/6 and we know the answer is 204 squares. Any other ideas or techniques to find this answer[/quote]
[hide]
Set $\frac{n(n+1)(2n+1)}{6}=204$. Rearranging and expanding, $2n^3+3n^2+n-1224=0$. Factoring, $n=8$. [/hide]

[hide]Basically, you organize casework by counting the number of squares of each size:

Since a chessboard is $8 \times 8$,

There are $8^2 \ 1 \times 1$ squares
There are $7^2 \ 2 \times 2$ squares
There are $6^2 \ 3 \times 3$ squares...

So the answer is $8^2+7^2+6^2+...=204$.

The formula $\frac{n(n+1)(2n+1)}{6}$ is the formula for the sum of the first $n$ squares  ;) [/hide]

i understand the formula but what does the 6 mean in the equation


hmmm, well, you may first want to learn how to read music, and understand which notes correspond to which keys, etc.  (unless you plan to learn by ear, i can't do that, so I can't help u there)

after, I suggest you buy some beginner piano books and just start playing?

w/o a teacher, it would be hard to learn many things, so I still think that's the best way

Look up in wiki or something what little dot goes to which key (ha ha thats in simple terms).
After you do that take a look into the Alfred series, I used that. (Speaking of Alfred that does have what matches with what).
Once you have Alfred down, try the Suzki method. (I didn't use that but I use it for cello). Also in addition to Suzki, try Hanon for technique. :)

I started out playing kinda late.  As a result, most of the beginner books they use for 5 year olds are with huge pictures and etc, and you don't get much content out of the books themselves, for the price you pay until you buy the entire set.

What I did first of all, was go through a bit of a large Adult Beginners book, which taught me notes and even got me playing small pieces with two hands.  Once I was at the point where I could read the notes, what I did then was use the Piano Adventures books, geared still for kids, but older kids, so you go a lot faster and it's more fun.  Once you've done that...keep going =)

Yeah, same here. I only learned a little piano (just for experience, not for the sake of playing it). My little brother actually does play piano. After he got out of Piano Adventures, he did some simple songs and repertoire. After that, he worked on some "little pieces" (Fur Elise, etc.). After he went through that, he worked on other big pieces- Chopin, Mozart, Beethoven (he's working on the Moonlight Sonata now), etc.

[quote]That's probably out of the question for me. I don't think I'd be willing to spend a lot of money on private lessons. [/quote]

 Aren't there some websites that have lessons online? Just an idea...

Online Lessons!
http://www.mahalo.com/How_to_play_piano_for_beginners_(and_save_$ 500_in_lessons)
Copy paste it. I can't do URL because it has a$ in it.

Basically, just play some random notes and realize that the low keys are on the left, and high keys are on the right. You have to learn to read music before you can play music otherwise it's just pure memorization.

try creating simple tunes on the piano like happy birthday. hum it to yourself and then find the right notes. afterwords go online and fined the sheet music for that tunes and you will see how the note you found matched with the note on the music. also, watch sound of music and try to find where the notes she sings are on the piano. then you can  start to play simple songs.

[quote="Lulze"]I have a piano, and I really can't play it at all, although I would like to learn to. Could you guys give me some advice on the best way to get started? I realize that developing a decent level of skill takes a lot of devotion, but right now I'm just having trouble with getting a feel for which keys are which and finding good "beginner" pieces to practice to get better.[/quote]

Do you play any other instruments?


What is the minimum number of United States coins Samantha
needs (pennies, nickels, dimes, quarters, half-dollars) to ensure she is
capable of making change for any amount of
money from one cent to 99 cents?

I think the trick is to get as many of the lower coins and make sure it doesn't exceed the next value coin.

Samantha will need 4 pennies to make sure for the number of cents mod 5, she can pay it back.
She will also need 1 nickel to make sure for the number of cents mod 10, she can pay it back.
Samantha will also need 2 dime to make sure for the number of cents mod 25, she can pay it back.
She will need 1 quarter to make sure for the number of cents mod 50, she can pay it back.
At last, Samatha will need 1 half dollar to make sure for the number of cents mod 100, she can pay it back.

Thus, she will need at least 4+1+2+1+1=9 coins to make sure she has change for anything from 1-99 cents, but only if she takes the right coins.

EDIT: Added the half dollars (thanks!)

Yeah, well there's a glitch in this problem, so the answer on FTW is 9.

Don't ask me why.

Dragon's answer is correct.

[quote="dragon96"]I think the trick is to get as many of the lower coins and make sure it doesn't exceed the next value coin.

Samantha will need 4 pennies to make sure for the number of cents mod 5, she can pay it back.
She will also need 1 nickel to make sure for the number of cents mod 10, she can pay it back.
Samantha will also need 2 dime to make sure for the number of cents mod 25, she can pay it back.
At last, she will need 3 quarters to make sure for the number of cents mod 99, she can pay it back.

Thus, she will need at least 1+2+3+4=10 coins to make sure she has change for anything from 1-99 cents, but only if she takes the right coins.[/quote]

no its 9;

[quote](pennies, nickels, dimes, quarters, [b]half-dollars[/b])[/quote]

so we can replace 2 quarters with 1 half dollar to get 9 coins

What happens if the answer is $\boxed{\text{Censored}}$, and someone uses a curse and uses it as the answer? Will FTW! count it right?

just curious.


[quote="jhredsox"][hide]well u know there's nine times base 10 amount of certain digit numbers...to show this[code]
digit           one     two    three    four    five
amount      9         90        900    9000   90000[/code]

so w/ 3000, we know its in the four digit numbers...

so the digits before 1000 is 2889...and we need 112 more...so we divide that by 3 and get 37 and 1/3...this makes the 37th four digit number, which is 1036...does that help?[/hide]
-jorian[/quote]

You didn't answer the question.

i know, so what is your opinion on exactly how to do this problemo?

-jorian

[quote="Quevvy"]We were doing a digit problem today in math it was this:
If the natural numbers (1, 2, 3, ...) are written consecutively, what is the 3001st digit?
Is there any easy way to do this problem, or any digit problem?
 :trampoline:[/quote]
[hide]I have seen this one enough times to know that 1-999 is 2889 numbers (this can be calculated in under 15 seconds mentally anyways) so there are 112 left. This is the last digit of the 28th number after 999 which is 1027 so the answer is 7.[/hide]
Hope that helps :)

[quote="Quevvy"]We were doing a digit problem today in math it was this:
If the natural numbers (1, 2, 3, ...) are written consecutively, what is the 3001st digit?
Is there any easy way to do this problem, or any digit problem?
 :trampoline:[/quote]

[hide]The number of digits from 1-9 is 9.
The number of digits from 10-99 is 90*2=180.
The number of digits from 100-999 is 900*3=2700.

This leaves 3001-9-180-2700=112 digits.

Divide by 4 to get 28 so it is the last digit of 1027 which is 7[/hide]

[hide]9 digits: 1-9
180 digits:10-99
2700 digits:100-999

2889 digits from 1-999. Now we count backwards. Sort of.

3001-2889 = 112. Divide by 4 (since that is the # of digits in the 1000s) to get 28. 1027, so 7.[/hide]

What would you do it we wanted the $3003^{\textrm{rd}}$ digit? You can't divide by $4$ since you would get $28.5$.

This is what I would do:

$3001 \rightarrow 7$
$3002 \rightarrow 1$
$3003 \rightarrow 0$

So if faced with a problem where you can't evenly divide, you should do this.


B and C are on a circle and A is outside the circle. AB intersects the circle at D and AC intersects the circle at E. Must it be true that the angle bisector of BAC bisects both arcs DE and BC (where we take the minor arc in both cases)?

since the lines AB and AC intersect the circle each in two points, i believe we are working with two secants, in which case angle BAC=1/2(DE-BC). the angle bisector would divide BAC in such a way, that each piece would have to equal 1/4(DE-BC). so the parts that the arcs are divided into do not have to be equal, just the difference of the new arcs must be equal. (i hope you can understand what i mean because i do not know how to do angles on LaTeX. Anyone know how? also, it is hard because i can't do pictures. Anyone know how to do that also?) :rotfl:


Here 's a question I heard from the local university, and I am puzzled with it...
There are 100 prisoners in solitary cells. There's a central living room with one light bulb; this bulb is initially off. No prisoner can see the light bulb from his or her own cell. Everyday, the warden picks a prisoner equally at random, and that prisoner visits the living room. While there, the prisoner can toggle the bulb if he or she wishes. Also, the prisoner has the option of asserting that all 100 prisoners have been to the living room by now. If this assertion is false, all 100 prisoners are shot. However, if it is indeed true, all prisoners are set free.

The prisoners are allowed to get together one night in the courtyard, to discuss a plan. What plan should they agree on, so that eventually, someone will make a correct assertion?

I saw this in orkut.
I will give you the solution soon.

In the meeting, one person was chosen to monitor the bulb.Now the plan is whenever someone visit the room for the first time and if-
1)the bulb is off, toggle the switch.
2)the bulb is on, leave it as it is.
Only the monitor(person chosen initially) toggle the switch from on to off and count the number of time he do so. When his count reaches 99(he automatically count himself), he told the warden that all prisoners have visited the cell.

But the catch is this may take 27,4 years.


What is the sum of the coefficients in the expansion of $ (2x^2\plus{}x/3\plus{}2/3)^3$?

[hide]I'm not sure if this method works, but I think removing all the $ x$'s and computing the resulting expression should give the answer:

$ \left(2 \plus{} \frac {1}{3} \plus{} \frac {2}{3}\right)^3 \equal{} 3^3 \equal{} \boxed{27}$[/hide]

Finding the sum of the coefficients of a polynomial $ P(x)$ is the same as evaluating $ P(1)$. So your method is valid.

[quote="13375P34K43V312"][quote="DonkeyKong"]I don't know if this is right..but here's my answer.

I multiplied 3 to get rid of fraction and went from there, which got me $ \boxed{685}$[/quote]

You cannot just multiply by 3 at will, this changes the coefficients.[/quote]
Oh... :blush:  Thanks..

Well, if not, what should you do?

MellowMelon said what to do.
You just plug in $ x\equal{}1$.

[quote="DonkeyKong"]I don't know if this is right..but here's my answer.

I multiplied 3 to get rid of fraction and went from there, which got me $ \boxed{685}$[/quote]
As 1337 said, multiplying by something changes the coefficients.

Here is another (albeit slower) solution:

$ \left(2x^2 \plus{} \frac {x}{3} \plus{} \frac {2}{3}\right)^3 \equal{} \frac {27}{27}\cdot\left(2x^2 \plus{} \frac {x}{3} \plus{} \frac {2}{3}\right)^3 \equal{} \frac {\left(6x^2 \plus{} x \plus{} 2\right)^3}{27}$

If $ f(x) \equal{} \frac {\left(6x^2 \plus{} x \plus{} 2\right)^3}{27}$, then $ f(1) \equal{} \frac {\left(6 \plus{} 1 \plus{} 2\right)^3}{27} \equal{} \boxed{27}$.

Note: Remember to take $ \sqrt[3]{27}$ when bringing it into $ \left(2x^2 \plus{} \frac {x}{3} \plus{} \frac {2}{3}\right)^3$.


Probably I have sent it here months ago , but my search did not give me answer.

Here is the link for this nice problem:
http://www.mathlinks.ro/viewtopic.php?p=841255#p841255


number theory
greatest common divisor
number theory unsolved

Let a,b,c,d be positive real number s.t $ad-bc=k>0$ and gcd(a;c)=gcd(b;d)=1.prove that there are k pair $(x;y) \in [0;1)$ s.t $ax+by$ and $cx+dy$ are integers

It isn't clear what you mean by the gcd of a pair of positive real numbers.

I think he means that a,b,c,d,k are integers.  I assume that they are all positive. So now I prove that there are only k pairs of (x,y) such that

ax + by = n
cx + dy = t

where n and t are integers. By multiplying the first equation by d and the second by b and then substracting the second from the first we get:
adx - bcx = nd - bt => kx = nd - bt. From here we see that x can only take on k different values, namely 0, 1/k, 2/k, ..., (k-1)/k, since the RHS must be an integer. Now we have y = (t - cx)/d = (n - ax)/b => kx + bt = dn. We also see that cx <= t < cx + d, since y is in [0,1). So t, which is a positive integer, can take on d different values. So considering t modulus d we have that t can be 0,1,2,...,d-1 (mod d). Now consider the equation kx + bt = dn. For a fixed x this is equal to e + bt = dn, where e is an integer 0 <= e < k. Considering this modulus d we have bt = -e (mod d). Since (b,d) = 1 there is a unique t that solves that equation => we only get one t that can solve that equation for a fixed x => we get one y for one x. Since there are k different x, and for each x we have one y, we get that there are exactly k pairs of (x,y) that satisfy the problem.


Solve for x: $ (A^4 \minus{} 2A^2B^2 \plus{} B^4) \equal{} (A \minus{} B)^{2x}(A\plus{}B)^{\minus{}2}$

I think I can factor the left-hand side into: $ (A^2 \minus{} B^2)^{2(x\minus{}1)}$, but I'm not sure what is the best way to go from there.

[hide="Hint"]
Difference of Squares
[/hide]

...and then the A-B and A+B expressions combine/cancel, and I can solve for x.  Thanks :lol: I'm always forgetting these trivial ways to factor.


trigonometry
projective geometry
geometry
3D geometry
tetrahedron
Euler
vector

Given two non-degenerate triangles ABC and A'B'C', the perpendiculars to the lines BC, CA, AB through A', B', C' concur if and only if the perpendiculars to B'C', C'A', A'B' through A, B, C concur. Two triangles ABC and A'B'C' satisfying these relations are called [b]orthologic[/b].

Well, the above theorem is quite easy to prove (I know three different ways: similtude, Steiner aka Carnot theorem on concurrent perpendiculars, Ceva theorem in the sine form).

Okay, this was the warm-up exercise. Now, for two orthologic triangles ABC and A'B'C', let P be the point where the perpendiculars to BC, CA, AB through A', B', C' meet, and let P' be the point where the perpendiculars to B'C', C'A', A'B' through A, B, C meet. These two points P and P' are called the [b]orthologic centers[/b] of triangles ABC and A'B'C'.

Now, regard the case when triangles ABC and A'B'C' are not only orthologic, but also perspective: The lines AA', BB', CC' concur at a point Q, and (after the Desargues theorem) the intersections of the lines B'C', C'A', A'B' with the lines BC, CA, AB, respectively, lie on one line l.

The [b]Sondat theorem[/b] states:

The points P, P', Q lie on one line perpendicular to l.

Of course, it is enough to show that PQ is perpendicular to l; then, the similar fact for P'Q follows by analogy and the theorem is proven.

Also, with the help of orthology, the theorem to be proven can be reduced to the following fact: If the lines C'A' and CA meet at B", and the lines A'B' and AB meet at C", then the lines AP' and A'P and the perpendiculars from B" to BB' and from C" to CC' meet at one point. Of course, again, it is enough here to prove three of these four lines concurrent, and the fourth one will follow by analogy.

However, with all these proof ideas, I have not found a synthetic proof of the above. (Maybe, even all of my proof ideas are dead end observations.) Does anybody know or can anybody find a good synthetic proof?

An analytical proof for the Sondat theorem can be found in

|  Victor Thébault, [i]Perspective and orthologic
|  triangles and tetrahedrons[/i], American Mathematical
|  Monthly 1952 pp. 24-28.

The proof by Thébault uses the Menelaos and Stewart theorems and is not what I would call elegant. He justifies his proof by its generalization potential (in fact, it goes analogously for tetrahedrons); however, I am mainly interested in a synthetic proof for the planimetric case.

For the Sondat theorem, Thébault refers to

|  P. Sondat, L'intermédiaire des mathématiciens, 1894,
|  p. 10 [question 38, solved by Sollerstinsky, p. 94].

However, I don't have access to this reference and don't know what the proof given there was like.

Ah, let me finally remark that the Sondat theorem was rediscovered and proved in a similar way to the above Thébault paper by Mitrea and Mitrea in

|  D. Mitrea, M. Mitrea, [i]A Generalization of a Theorem
|  of Euler[/i], American Mathematical Monthly January
|  1994.

The Sondat theorem generalizes not only the "Theorem of Euler" (meaning the Euler line of a triangle), but also several triangle collinearities.

Finally, let me remark that there was another theorem of Sondat on perspective triangles with mutually perpendicular sidelines, but this is another story...

  Darij Grinberg

I am reading the proof of the Sondat theorem by Mitrea and Mitrea and I don't understand why in page 57 it says

Clearly any two of the vectors  $\vec a$, $\vec b$, $\vec c$ are linearly independient

Is this true?

If not, how can we terminate the proof in this case?

(The referred article is in The American Mathematical Monthly, vol 101, no. 1, (January 1994), 55-58.

Thanks

[quote="pacoga"]I am reading the proof of the Sondat theorem by Mitrea and Mitrea and I don't understand why in page 57 it says

Clearly any two of the vectors  $\vec a$, $\vec b$, $\vec c$ are linearly independient

Is this true?[/quote]

This is not true, but actually it is not necessary.

Mitrea and Mitrea pretend to require any two of the vectors $\vec a$, $\vec b$, $\vec c$ to be linearly independent, but in fact they use only the linear independence of the vectors $\vec a$ and $\vec b$ and the linear independence of the vectors $\vec a$ and $\vec c$. The linear independence of the vectors $\vec b$ and $\vec c$ is not really used, what becomes clear if you replace all the $\Longleftrightarrow$ signs on page 58 by $\Longleftarrow$ signs (in fact, only the $\Longleftarrow$ direction is necessary for the proof).

Now, you can always reduce the problem to the case when the vectors $\vec a$ and $\vec b$ are linearly independent and the vectors $\vec a$ and $\vec c$ are also linearly independent: If the point O is one of the vertices A, B, C of triangle ABC, then the Sondat theorem is trivial (or doesn't make sense - depends on how exactly you interpret the theorem in this case), so you can assume that the point O does not coincide with any of the vertices A, B, C of triangle ABC; thus, the point O lies on at most one of the three lines BC, CA, AB; so we can WLOG assume that the point O lies neither on the line CA nor on the line AB. Thus, the vectors $\overrightarrow{OA}=\vec a$ and $\overrightarrow{OC}=\vec c$ are linearly independent, and the vectors $\overrightarrow{OA}=\vec a$ and $\overrightarrow{OB}=\vec b$ are also linearly independent. And then the proof works.

  Darij

A beautiful theorem, I didn't know. 2 or 3 months ago, my friend charlydif tell me a nice problem about orthologics triangles: Let a circunference, a triangle and his "polar triangle". Prove that the triangles are perspective. I think this problems is, with a "orthologic vocabulary": if ABC and A'B'C' are orthologics, and P=P' (the orthologics centers are the same), then ABC and A'B'C' are (necessary) perspective. My solution of this uses Menelao's theorem and similar triangles.
My apologies for my pathetic and non-existent english :blush:

Well, I don't know a syntethic proof of Sondat theorem for orthological triangles, but there's a fourth way to prove your warm-up exercise, a vector one :

[i]The perpendiculars from A',B',C' to the sides BC, CA, AB concurr if and only if $\vec{XA'}\cdot \vec{BC}+\vec{XB'}\cdot \vec{AC} + \vec{XC'}\cdot \vec{AB}=0$ for every point $X$ in the plane.[/i]
(from now on I'll omit the arrows over the vectors...)
First of all, the expression is indipendent from $X$: let $Y$ be another point, then
\[(XA'-YA')\cdot BC + (XB'-YB')\cdot AC+(XC'-YC')\cdot AB=XY\cdot(AB+BC+CA)=XY\cdot 0=0\]
Now, if the perpendiculars concurr, then let $P$ be the point of concurrence; obviously $PA'\cdot BC+PB'\cdot AC + PC'\cdot AB=0$, so the sum is zero for every point.
Conversely, if the sum is zero for all $X$, let $T$ be the intersection of the perpendiculars from A', B' to BC, AC; then
\[0=TA'\cdot BC+TB'\cdot AC + TC'\cdot AB=TC'\cdot AB\]
so TC' is perpendicular to AB and the assertion 1) is proved.

Now, observe that $0=A'A'\cdot BC+A'B'\cdot AC + A'C'\cdot AB=A'B'\cdot AC + A'C'\cdot AB=B'C'\cdot AA+A'B'\cdot AC + A'C'\cdot AB=B'C'\cdot XA+A'B'\cdot XC+A'C'\cdot XB$ so the perpendiculars to the sides of ABC concurr if and only if the perpendiculars to the sides of A'B'C' concurr.

This caracterization of orthology (does this word exists?) gives also a proof of Sondet, but, obviously, not a synthetical one.

For warm-up exercise.....
ABC and A'B'C', P is the intersection of perpendiculars from A', B', C' to BC, CA, AB and P' is the intesection of the perpendiculars from A and C to B'C' and A'B'. We want prove: PB and A'C' are perpendiculars.

Well, let A''B''C'' the translation of ABC, in the direction of P'P ... it's clear that the perpendiculars from A', B' and C' to BC, CA and AB are 'concurrentes' in P. Well, the idea is to prove that A''B''C'' and ABC are orthologics (with the same center), 'es decir': prove that PB' is perpendicular to C'A'...
Let X, Y, Z the proyections from P to B''C'',C''A'',A''B''.
ZYB'A' and YB'C'X are cyclics (it's simple). Then, P is in the radical axis of this circles. Then, PA'.PZ=PC'.PX --> A'ZXC' is cyclic, too, and the warm-up exercise is prove.

Could anybody give me the complete Theory about orthological triangles, I need this too much. Thanks

[b]Problem:-[/b]
Given two triangles $ ABC $ and $ A'B'C' $ so that they are orthologic and perspective. Then prove that the two centers of orthology and the perspector are collinear.

[b]Proof:-[/b]

Consider a point $ P $, such that pedal triangle(call it $ P_aP_bP_c $) of $ P $ wrt $ ABC $ is homothetic to $ A'B'C' $. The perspector of $ ABC $ and $ A'B'C' $ be $ Q $. Through $ P_a,P_b,P_c $ draw parallels to $ AQ,BQ,CQ $. Suppose, they concurr at $ Q' $. If $ P* $ is the isogonal conjugate of $ P $ wrt $ ABC $, then we will prove that $ Q'P\parallel P*Q $.Suppose, $ H $ and $ H' $ are the orthocenters of $ ABC $ and $ A'B'C' $. Using the notation of Property 1 [url=http://www.artofproblemsolving.com/blog/68568]here[/url], for the triangles $ ABC $ and $ P_aP_bP_c $ we have $ \Gamma(ABC) $ is the conic passes through $ H,P* $. And $ \Gamma(P_aP_bP_c) $ passes through $ H',P $. Since both of them are rectangular hyperbola and they have parallel asymptotes, so they are homothetic. Now draw parallels to $ P_aP_b,P_aP_c $ through $ B,C $. Suppose, they meet at $ A_2 $. Then note that $ P* $ is the orthocenter of $ A_2BC $. So $ A_2P*\perp BC $. Suppose, the homothety that maps $ \Gamma(ABC) $ to $ \Gamma(P_aP_bP_c) $ maps $ A,P*,Q,A_2 $ to $ A_1,P_1,Q_1,A_2' $. Note that, tangent at $ P_a $ on $ \Gamma(P_aP_bP_c) $ is parallel to $ A_1A_2' $. So applying Pascal's theorem on $ P_aP*A_2'A_1PP_a $ we get that $ P_aP*\parallel PA_1 $. Now applying Pascal's theorem on $ PA_1QP*P_aQ' $ we get that $ PQ'\parallel P*Q $. So the lemma is proved.

Here is my proof of Sondat's Theorem.

[b]Theorem[/b] Let triangles $ABC$ and $A_1B_1C_1$. Suppose that
[list]
[*] Perpendiculars from $A, B, C$ to $B_1C_1, C_1A_1, A_1B_1$ are concurrent at $P$
[*] Perpendiculars from $A_1, B_1, C_1$ to $BC, CA, AB$ are concurrent at $P_1$
[*] Lines $AA_1, BB_1, CC_1$ are concurrent at $Q$.
[/list]
Finally, let $X=BC\cap B_1C_1, Y=AC\cap A_1C_1$ and $Z=AB\cap A_1B_1$. Then $P, P_1, Q$ are colinear and perpendicular to $\overline{XYZ}$.

[b]Proof :[/b] We begin with two lemmas.
------
[b]Lemma 1 :[/b] Given two (not homothetic) triangles $ \triangle ABC $ and $ \triangle DEF $. Let $ P $ be a point such that the line passing through $ D, $ $ E, $ $ F $ and parallel to $ AP, $ $ BP, $ $ CP $, respectively are concurrent. Then $ P $ lies on a circumconic of $ \triangle ABC $ or the line at infinity.

[b]Proof of Lemma 1 :[/b] By suitable translation, we can let $D=A$. Let $K = BP\cap AE$ and $L =CP\cap AF$ and the concurrency point be $Q$. Then $\Delta EFQ$ and $\Delta KLP$ are homothetic at center $A$, which implies $KL\parallel EF$.

Let $K, L$ move along $AE, AF$ such that $KL\parallel BC$ and fix three such points $(K_1, K_2, K_3), (L_1, L_2, L_3)$. Let $P_i$ be the resulting $P$ from $K_i, L_i$. Notice that
$$B(P_1, P_2; P_3, P) = (K_1, K_2 ; K_3, K) = (L_1, L_2 ; L_3, L) = C(P_1, P_2; P_3, P)$$
so $B, C, P_1, P_2, P_3, P$ lies on a conic which implies $P$ lies on a fixed conic. This conic clearly pass through $A$ so we are done.
-----
[b]Lemma 2 : [/b] Let $ABC$ be a triangle. Let $P, Q$ be two points on a rectangular circum-hyperbola $\mathcal{H}$. Let the perpendicular from $P$ to $AQ$ meet $BC$ at $A_1$ and define $B_1, C_1$ similarly. Then $A_1B_1C_1$ is lies on a line perpendicular to $PQ$.

[b]Proof of Lemma 2 :[/b] As $\mathcal{H}$ is rectangular, the orthocenters $H_B$ and $H_C$ of $\triangle BPQ$ and $\triangle CPQ$ are the second intersections of $\mathcal{H}$ with $PB_1$ and $PC_1,$ respectively.

By Pascal theorem for $PH_CCABH_B,$ the intersections $C_1 \equiv PH_C \cap AB,$ $B_1 \equiv PH_B \cap AC$ and $CH_C \cap BH_B$ (point at infinity of $\perp PQ$) are collinear, which implies $B_1C_1\perp PQ$.  Hence, we conclude that $A_1,B_1,C_1$ lie on a perpendicular to $PQ.$
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[b]Back to the main theorem[/b]

By Lemma 1, we see that points $P, Q$ and orthocenter $H$ of $\Delta ABC$ lies on a circumconic of $\Delta ABC$. So this conic must be rectangular hyperbola.

Let the line perpendicular from $Q$ to $BP, CP$ intersects $AC, AB$ at $Y_1, Z_1$ respectively. Notice that $\Delta DYZ$ and $\Delta QY_1Z_1$ are homothetic at center $A$. But by Lemma 2, $Y_1Z_1\perp PQ$ so $YZ\perp PQ$, which is enough to conclude the main theorem.

Dear Mathlinkers,
you can see also my proof dating from 2007 on

http://jl.ayme.pagesperso-orange.fr/Docs/Le%20theoreme%20de%20Sondat.pdf

Sincerely
Jean-Louis

Also see the proof of Theorem 1.2 [url=https://ijgeometry.com/wp-content/uploads/2015/10/1.pdf]here.[/url]


let x,y,z such that: x^2+y^2+z^2=12
prove that:
           xy/z+yz/x+zx/y$ \ge$ 3+(x+y+z)/2

this problem similar a problem of ddlam  to  quykhtn-qa1[/b][/i]