analytic geometry
trigonometry
number theory
relatively prime
geometry
angle bisector
Pythagorean Theorem

Four lighthouses are located at points $ A$, $ B$, $ C$, and $ D$. The lighthouse at $ A$ is $ 5$ kilometers from the lighthouse at $ B$, the lighthouse at $ B$ is $ 12$ kilometers from the lighthouse at $ C$, and the lighthouse at $ A$ is $ 13$ kilometers from the lighthouse at $ C$. To an observer at $ A$, the angle determined by the lights at $ B$ and $ D$ and the angle determined by the lights at $ C$ and $ D$ are equal. To an observer at $ C$, the angle determined by the lights at $ A$ and $ B$ and the angle determined by the lights at $ D$ and $ B$ are equal. The number of kilometers from $ A$ to $ D$ is given by $ \displaystyle\frac{p\sqrt{r}}{q}$, where $ p$, $ q$, and $ r$ are relatively prime positive integers, and $ r$ is not divisible by the square of any prime. Find $ p\plus{}q\plus{}r$,

This was too easy for a problem 10.

Let $ A$ be the origin, $ B$ have coordinates $ (5,0)$, and $ C$ have coordinates $ (5,12)$. Let $ \theta$ measure the angle $ \angle BAC$. $ D$ lies on the angle bisector of this angle, so we have
\[ \tan \frac{\theta}{2} \equal{} \sqrt{\frac{1\minus{}\cos \theta}{1\plus{}\cos \theta}}\equal{}\frac{2}{3}.\]
Then the line from $ A$ through $ D$ is $ y\equal{}\frac{2}{3}x$.

Easily, we can write the equation for the line through $ C$ and $ D$ to be
\[ y\equal{}\minus{}\frac{5}{12}x\plus{}24.\]

Solving these two equations simultaneously, we can find the coordinates of $ D$ and thus the distance from $ A$ to $ D$ can be determined using the Pythagorean Theorem:
\[ AD\equal{}\frac{60\sqrt{13}}{23} \implies p\plus{}q\plus{}r\equal{}\boxed{096}.\]

meh >.<
Well, if you had a time issue like me then you would skip anything that looked as if it required a lot of calculation and maybe come back later.... Except i never came back to it....

I had it right... then when I square rooted to get the final answer, I forgot to square root the denominator  :mad:

[hide="Really .. one-line?"]
Let $ CD$ and $ AB$ meet at $ A'$; then $ \overline{AD}$ is an angle bisector of isosceles $ \triangle AA'C$. Then by the Angle Bisector Theorem, $ A'D \equal{} \frac {10}{10 \plus{} 13} \cdot 13 \equal{} \frac {130}{23}$, and $ \cos \angle DA'A \equal{} \frac {5}{13}$. By the Law of Cosines on $ \triangle AA'D$, we have
\[ AD^2 \equal{} 10^2 \plus{} \left(\frac {130}{23}\right)^2 \minus{} 2 \cdot 10 \cdot \frac {130}{23} \cdot \frac {5}{13} \equal{} \frac {2^4 \cdot 3^2 \cdot 5^2 \cdot 13}{23^2} \Longrightarrow AD \equal{} \frac {60\sqrt {13}}{23}
\] and the answer is $ 60 \plus{} 13 \plus{} 23 \equal{} \boxed{096}$.

[asy]size(120); pathpen = linewidth(0.7); pointpen = black; pen f = fontsize(10); pair B=(0,0), A=(5,0), C=(0,13), E=(-5,0), O = incenter(E,C,A), D=IP(A -- A+3*(O-A),E--C); D(A--B--C--cycle); D(A--D--C); D(D--E--B, linetype("4 4")); MP("5", (A+B)/2, f); MP("13", (A+C)/2, NE,f); MP("A",D(A),f); MP("B",D(B),f); MP("C",D(C),N,f); MP("A'",D(E),f); MP("D",D(D),NW,f); D(rightanglemark(C,B,A,20)); D(anglemark(D,A,E,35));D(anglemark(C,A,D,30));[/asy]
[/hide]

i coordinate bashed and got it right lol
angle bisector to find the equation of line AD. lineCD is pretty much given.

I will refer to azjps' diagram. We have, $ A'B \equal{} 5$ and $ CA' \equal{} 13$, by the angle bisector theorem we have $ CD \equal{} \frac {169}{23}$ and $ DA' \equal{} \frac {130}{23}$. Now Stewart's finds AB (a little messy but not too bad) is $ \frac {60 \sqrt {13}}{23}$

[hide="Solution"]
Refer to azjps's diagram, and let the intersection of $ \overline{AD}$ and $ \overline{BC}$ be $ E$. By the Angle Bisector Theorem, $ BE\equal{}\frac{10}3$, and by the Pythagorean Theorem, $ AE\equal{}\frac53\sqrt{3^2\plus{}2^2}\equal{}\frac53\sqrt{13}$. Again by the Angle Bisector Theorem, $ CD\equal{}\frac{13}{23}\cdot13$, so finally by the Angle Bisector Theorem $ \frac{AC}{AE}\equal{}\frac{CD}{DE}\equal{}\frac{AC\plus{}CD}{AD}\implies\frac{13}{\frac53\sqrt{13}}\equal{}\frac{13\plus{}\frac{13}{23}\cdot13}{AD}$
$ \implies AD\equal{}\left(1\plus{}\frac{13}{23}\right)\cdot\frac53\sqrt{13}\equal{}\frac{36}{23}\cdot\frac53\sqrt{13}\equal{}\frac{60}{23}\sqrt{13}\implies60\plus{}13\plus{}23\equal{}\boxed{96}$.
[/hide]

:wallbash_red:

I solved this question in less than 10 minutes (way too easy for #10) using the angle bisector theorem / mass points
but then at the end I somehow wrote 23*3 = 39............................................................... :rotfl:

smart, eh?

Here is a slightly different trig approach using the law of sines.
[asy]pathpen = linewidth(0.7);
pen f = fontsize(10);
size(144);
pair B=(0,0);
pair A=(5,0);
pair C=(0,12);
pair E=(-5,0);
pair abA = A+unit(C-A)+unit(B-A);
pair D   = extension(A,abA,C,E);
D(A--C);
D(C--B);
D(A--B);
D(C--D);
D(A--D);
D(D--E,dashed);
D(B--E,dashed);
D(rightanglemark(C,B,A,20),red);
MP("A",D(A),plain.SE,f);
MP("B",D(B),plain.S,f);
MP("C",D(C),plain.N,f);
MP("D",D(D),plain.NW,f);
MP("E",D(E),plain.SW,f);
MP("5",(A+B)/2,plain.S,f);
MP("12",(B+C)/2,plain.E,f);
MP("13",(A+C)/2,plain.NE,f);
MP("\alpha",A,4*(unit(D-A)+unit(C-A))/2,f);
MP("\alpha",A,4*(unit(B-A)+unit(D-A))/2,f);
MP("2\alpha",E,4*(unit(C-E)+unit(B-E))/2,f);
MP("\beta",C,6*(unit(A-C)+unit(B-C))/2,f);
MP("\beta",C,6*(unit(E-C)+unit(B-C))/2,f);[/asy]

So using the law of sines on $ \bigtriangleup ADE$,

\begin{align*}
AD  &= AE \cdot \dfrac{\sin(2\alpha)}{\sin(180^\circ-3\alpha)}\\
    &= AE \cdot \dfrac{\sin(2\alpha)}{\sin(\alpha+2\alpha)}\\
    &= 10 \cdot \dfrac{\sin(2\alpha)}{\sin(\alpha)\cos(2\alpha)+\cos(\alpha)\sin(2\alpha)}\\
    &= 10 \cdot \dfrac{\sin(2\alpha)}{\sqrt{\dfrac{1-\cos(2\alpha)}{2}}\cdot\cos(2\alpha)+\sqrt{\dfrac{1+\cos(2\alpha)}{2}}\cdot\sin(2\alpha)}\\
    &= 10 \cdot \dfrac{\dfrac{12}{13}}{\sqrt{\dfrac{8}{26}}\cdot\dfrac{5}{13}+\sqrt{\dfrac{18}{26}}\cdot\dfrac{12}{13}}\\
    &= 10 \cdot \dfrac{12\sqrt{13}}{10+36} = \dfrac{60\sqrt{13}}{23}
\end{align*}

$ \therefore$ the answer is $ 60+13+23=\boxed{086}$.

[quote=m1sterzer0]

$ \therefore$ the answer is [b]$ 60+13+23=\boxed{086}$. [/b][/quote]

ten out of ten

The hardest part of this problem is drawing the diagram.

We can extend $AB$ and $CD$ to intersect at a point $O$. Also notice by ASA, $\triangle ABC \cong \triangle OBC$.

By the angle bisector theorem, $OD = \frac{130}{23}$ and $CD = \frac{169}{23}$.

Now we can apply Stewarts theorem and figure out that $AD = \frac{60\sqrt{13}}{23}$

So our answer is $60 + 13 + 23 = \boxed{096}$

Wow so many angle bisectors and I still didn't get this problem.

The key idea here is to draw the image of $\triangle ABC$ when it is reflected over $BC$, yielding $D$ as the intersection of the angle bisector of $\angle A$ and the hypothenuse of the second triangle. Setting $B$ as the origin and coordbashing would be the optimal solution :P

When you forget the formula for the length of angle bisector from sidelengths :oops:
-----
Applying the Angle Bisector theorem twice, we can see that we are dealing with a $10-13-13$ triangle. Let's call the third point $E$ (then, this triangle is $\triangle ABC$). Our goal is to find the length of the angle bisector of $A$.

Apply Barycentrics wrt $\triangle ACE$. Since $I=(13:10:13)$ is the incenter, the equation of $AI$ is

$$\left|\begin{matrix}
1 & 0 & 0 \\
13 & 10 & 13 \\
x & y & z
\end{matrix}\right|=0\rightarrow 10z=13y.$$ We have this intersecting with $CE$, so the foot of this bisector (call it $P$) is $\left(0, \frac{10}{23}, \frac{13}{23}\right)$. Now, we get vector $\overrightarrow{AP}=\left(1, -\frac{10}{23}, -\frac{13}{23}\right)$, so upon applying the distance formula ([url=https://www.wolframalpha.com/input?i=sqrt%28-169%28-10%2F23%29%28-13%2F23%29-100%28-13%2F23%29-169%28-10%2F23%29%29]which is reasonably bashy[/url]), we get $\boxed{\frac{60\sqrt{13}}{23}}$.

[b]Solution:[/b] It is easy to see that $\triangle{ABC}$ is right. Notice that $\overline{AD}$ is just the angle bisector of $\angle{BAC}$. Also, we have that $\overline{CB}$ is the angle bisector of $\angle{ACX}$, where $X$ is the reflection of $A$ over $B$. This gives us that $AX = 10$. We already know that $AC = 13$. Then using angle bisector theorem, we have that $CD = \frac{13}{23} \cdot 13, DX = \frac{10}{23} \cdot 13$. Therefore, letting $AD = d$, we see that $13d^2 + \frac{130}{529} \cdot 13^3 = 130 \cdot 13 \Rightarrow d^2 + \frac{130}{529} \cdot 169 = 130 \Rightarrow d^2 = 360 \cdot \frac{130}{529} \Rightarrow d = \frac{60\sqrt{13}}{23} \Rightarrow \boxed{096}$.

[hide = 2 solutions:]
Diagram looks like this: [asy] size(120); pathpen = linewidth(0.7); pointpen = black; pen f = fontsize(10); pair B=(0,0), A=(5,0), C=(0,13), E=(-5,0), O = incenter(E,C,A), D=IP(A -- A+3*(O-A),E--C); D(A--B--C--cycle); D(A--D--C); D(D--E--B, linetype("4 4")); MP("5", (A+B)/2, f); MP("13", (A+C)/2, NE,f); MP("A",D(A),f); MP("B",D(B),f); MP("C",D(C),N,f); MP("A'",D(E),f); MP("D",D(D),NW,f); D(rightanglemark(C,B,A,20)); D(anglemark(D,A,E,35));D(anglemark(C,A,D,30)); [/asy]

[b]Solution 1: Angle Bisector Theorem + Law of Cosines[/b]
Let $CD$ and $AB$ meet at $A'$; then $\overline{AD}$ is an angle bisector of isosceles $\triangle AA'C$. Then by the Angle Bisector Theorem, $A'D=\frac {10}{10+13} \cdot 13=\frac {130}{23}$, and $\cos \angle DA'A=\frac {5}{13}$. By the Law of Cosines on $\triangle AA'D$, we have\[AD^2=10^2+\left(\frac {130}{23}\right)^2-2 \cdot 10 \cdot \frac {130}{23} \cdot \frac {5}{13}=\frac {2^4 \cdot 3^2 \cdot 5^2 \cdot 13}{23^2} \Longrightarrow AD=\frac {60\sqrt {13}}{23}\]and the answer is $60+13+23=\boxed{096}$.

[b]Solution 2: Law of SInes:[/b]

\begin{align*} AD  &= AE \cdot \dfrac{\sin(2\alpha)}{\sin(180^\circ-3\alpha)}\\     &= AE \cdot \dfrac{\sin(2\alpha)}{\sin(\alpha+2\alpha)}\\     &= 10 \cdot \dfrac{\sin(2\alpha)}{\sin(\alpha)\cos(2\alpha)+\cos(\alpha)\sin(2\alpha)}\\     &= 10 \cdot \dfrac{\sin(2\alpha)}{\sqrt{\dfrac{1-\cos(2\alpha)}{2}}\cdot\cos(2\alpha)+\sqrt{\dfrac{1+\cos(2\alpha)}{2}}\cdot\sin(2\alpha)}\\     &= 10 \cdot \dfrac{\dfrac{12}{13}}{\sqrt{\dfrac{8}{26}}\cdot\dfrac{5}{13}+\sqrt{\dfrac{18}{26}}\cdot\dfrac{12}{13}}\\     &= 10 \cdot \dfrac{12\sqrt{13}}{10+36} = \dfrac{60\sqrt{13}}{23} \end{align*}
$\therefore$ the answer is $60+13+23=\boxed{096}$.

[/hide]


AMC
AIME
geometry
trapezoid
number theory
relatively prime
similar triangles

In quadrilateral $ ABCD$, $ \angle{BAD}\cong\angle{ADC}$ and $ \angle{ABD}\cong\angle{BCD}$,$ AB\equal{}8$,$ BD\equal{}10$, and$ BC\equal{}6$. The length $ CD$ may be written in the form$ \frac{m}{n}$, where $ m$ and $ n$are relatively prime positive integers. Find $ m\plus{}n$.

Here's my solution, not sure if it's as nice as yours though.
[hide]
Extend $ AB$ and $ CD$ to meet at point $ E$. Clearly, $ EAD$ is isosceles, because angle EAD is equal to angle EDA.
Now, because $ \angle{ABD}\cong \angle{BCD}$, we have $ \angle{EBD}\equal{}180\minus{}\angle{ABD}\equal{}180\minus{}\angle{BCD}\equal{}\angle{ECB}$, so triangle ECB is similar to triangle EBD.
Let $ EB\equal{}x$. Then, $ ED\equal{}8\plus{}x$ by the isosceles triangle.
By the similar triangles, we have $ \frac{BC}{DB}\equal{}\frac{EB}{ED}$, or $ \frac{6}{10}\equal{}\frac{x}{8\plus{}x}\implies 48\plus{}6x\equal{}10x\implies x\equal{}12$. So, $ EB\equal{}12$ and $ ED\equal{}20$.
Now, using similar triangles again, we have $ \frac{BC}{DB}\equal{}\frac{EC}{EB}$, or $ \frac{6}{10}\equal{}\frac{EC}{12}$, so $ EC\equal{}\frac{36}{5}$.
Finally, we have $ CD\equal{}ED\minus{}EC\equal{}20\minus{}\frac{36}{5}\equal{}\frac{64}{5}$.
And 64+5=69.
[/hide]

[quote="JITTT"]Should I say it now or wait?[/quote]

I don't see why not, especially after andersonw has already given their solution.

Look at the picture

[geogebra]9ae4e61eeff0131b9f11183a4995d3f5835963c0[/geogebra]

Draw line parallel to $ AD$ and through $ B$. Then you will get an isosceles trapezoid, so $ DE \equal{} 8$ . From parallel line angle stuff, $ \angle{ADE}\cong\angle{BEC}$ and since $ \angle{ADE}\cong\angle{BAD}$ , $ \angle{BAD}\cong\angle{BEC}$ . They give you $ \angle{ABD}\cong\angle{BCD}$ , so you now have similar triangle and $ \frac{8}{10} \equal{} \frac{x}{6}$ and $ x \equal{} \frac{24}{5}$. $ CD \equal{} 8 \plus{} x$, so $ CD \equal{} \frac{64}{5}$ and $ 64 \plus{} 5$ = [b]$ 69$[/b]


real analysis
topology
calculus
integration
function
algorithm
real analysis unsolved

Suppose that $ K$ is a convex compact set in $ \mathbb{R}^{n}$ and the center of mass of $ K$ is the origin. Also $ \lambda_{K}$ is the Lebesgue measure restricted to $ K$. Let $ \Pi$ be a subspace of $ \mathbb{R}^{n}$ of dimension $ m < n$, and let $ P$ be the orthogonal projection of $ \mathbb{R}^{n}$ onto $ \Pi$.

Given the set $ P(K)$ and the push-forward measure $ \lambda_{P}$, what can we reconstruct of $ K$? What are the extremal points of the set of possible $ K$ (in the Hausdorff metric)?

How much information do we need in order to reconstruct $ K$ completely? That is, if we do it for all hyperplanes in some set $ \{\Pi_{j}\}$ and the span of the $ \Pi_{j}$ is $ \mathbb{R}^{n}$, can we always reconstruct $ K$? If not, what more do we need?

I just realized I might not have been perfectly clear about what the push-forward measure is supposed to be. It's the push-forward of $ \lambda_{K}$, which means that the measure of a subset $ E$ of $ P(K)$ is equal to the $ n$-dimensional Lebesgue measure of the slice of $ K$ whose projection onto $ \Pi$ is $ E$. So the information we have from the measure is how thick $ K$ is (in the subspace $ \Pi^{\bot}$) at each point of $ P(K).$

That plus the fact that $ K$ is convex does seem to put a fair amount of restriction on possibilities for $ K;$ the question is how much freedom is there?

About the other part of the question (how much of this sort of info do we need): clearly by the result jmerry referred to, if we just project onto every line we should be able to reconstruct the set. But that method should work for every set, not just compact convex ones with uniform density, so it seems like a huge overkill.

Now that I understand it- there's a reasonable numerical algorithm to approximate it using finitely many slices, and this algorithm works without the convex/uniform density assumption.
This doesn't address the question of exactness with minimal information.

[quote="pluricomplex"][quote="jmerry"]A related problem: given the integrals of some compactly supported "density" function $ \phi$ along lines, can you find $ \phi$?
The answer to that one is yes, and it's the mathematical basis for CT scanning.

We've got less information here- just a yes/no option.[/quote]

jmerry, could you explain more details your problem (I'm not good at English), it's interested me.[/quote]

[url=http://en.wikipedia.org/wiki/Filtered_back_projection]he's talking about the Radon transform[/url]


Prove that for $ x_1;x_2;...;x_n$ being positive real numbers, we' ve got

$ \sum_{k\equal{}1}^{n} x_k \plus{} \sqrt{\sum_{k\equal{}1}^{n} (x_k)^2} \leq \frac{n\plus{}\sqrt{n}}{n^2} (\sum_{k\equal{}1}^{n} \frac{1}{x_k})(\sum_{k\equal{}1}^{n} (x_k)^2)$

\begin{align*}
I: &\left(\sum x_i\right)\left(\sum \frac{1}{x_i\right)\geq n^2  &\textnormal{(AM-GM)}\\
II: &\sqrt{n\cdot \sum x_i^2}\geq  \sum x_i} &\textnormal{(CS)}\\
&\textnormal{multplication of I and II: }&\\
&\left(\sum \frac{1}{x_i}\right)\left(\sqrt{\sum x_i^2}\right)\geq n\sqrt{n}\\
&(\sqrt{n}+1)\left(\sum \frac{1}{x_i}\right)\left(\sum x_i^2\right)\geq n\sqrt{n}\left(\sqrt{n\cdot \sum x_i^2}+\sqrt{\sum x_i^2}\right)&(\textnormal{multplicated by} (\sqrt{n}+1)\sqrt{\sum x_i^2})\\
&(\sqrt{n}+1)\left(\sum \frac{1}{x_i}\right)\left(\sum x_i^2\right)\geq n\sqrt{n}\left(\sum x_i+\sqrt{\sum x_i^2}\right)&\textnormal{(II)}
\end{align*}
Divison completes the proof

quite ugly :wink:


I'd like some assistance with this problem:

A nine digit number is formed using each of the nine non-zero digits exactly once.
How many possibilities exist for the number if it is to be divisible by 11?

For some reason I'm having difficulty sorting this one out (using the divisibility rule for 11).

Thanks.

Well, a number with base ten representation abcd. . . is divisible by 11 if and only if a-b+c-d+... is divisible by 11.  Therefore, if the nine digit number is abcdefghi then a-b+c-d+e-f+g-h+i=n*11, and a+b+c+...+i=1+2+3+...+9=45 so the above gives 45-2*(b+d+f+h)=n*11, giving b+d+f+h=6 or 17 or 28 as the only possibilities which can occur.  Now then, I don't feel like sorting through the rest  of it right now, but the answer is the number of ways to get 6 or to get 17 or to get 28 as an ordered sum of four distinct integers from one to nine.

[quote="hiitsme"]I think that is untrue. $11|22$ but $2+2=4$ so by your method, 4 is divisible by 11?[/quote]

[hide]
He never said to add the digits: you take the first digit - second + third - fourth...
So 2-2 = 0, which is divisible by 11.
[/hide]


Suppose that A is finite and f is 1 - 1 from A onto B. Prove that B is finite.

Suppose $ B$ is infinite, then you have $ cardA < cardB$. But, using the Cantor-Bernstein theorem you have $ cardA \equal{} cardB$, which is contradiction.

$ A$ is finite iff there is a bijection mapping $ A$ onto $ \{ 1,\cdots, n\}$ for some $ n\in\mathbb{N}$.

Then you're done! Just let the bijection $ g$, then we have a bijection $ g\circ f^{\minus{}1}$ mapping $ B$ onto $ \{ 1,\cdots, n\}$. This finishes the proof.

When you say that "f is 1 - 1 from A onto B", do you mean that f is also surjective?


trigonometry
calculus
calculus computations

\[ \begin{gathered}
  f\left( x \right) \equal{} \sin \left( {\frac{x}
{{1 \plus{} x}}} \right) \hfill \\
  f'\left( x \right)? \hfill \\
\end{gathered}\]

I got this
\[ f'\left( x \right) \equal{} \frac{{1 \plus{} x \plus{} 1}}
{{\left( {1 \plus{} x} \right)^2 }} \cdot \cos \left( {\frac{x}
{{x \plus{} 1}}} \right)\]

but maple say another thing...
Halp again pls

Note: $ \frac {x}{1 \plus{} x} \equal{} 1 \minus{} \frac1{1 \plus{} x}.$

From that, it's easier to see that $ \frac {d}{dx}\left(\frac {x}{1 \plus{} x}\right) \equal{} \frac {1}{(1 \plus{} x)^2}.$

I think you were using the quotient rule but I'm not sure I can replicate your mistake.

aaaaaaa
I put "+" in definition of \[ \left[ {\frac{{f\left( x \right)}}
{{g\left( x \right)}}} \right]^'\]

I got it, thanks xd


geometry
incenter
circumcircle
angle bisector
geometry unsolved

$ I$ is the incenter oF $ \triangle ABC$.the incircle is tangent to $ AC$,$ AB$ at $ E$,$ F$ respectively.the circle that it s diameter is $ BC$, intersect $ EF$ at $ K$,$ L$.prove that the circumcircle of $ \triangle IKL$ is tangent to incircle Of $ \triangle ABC$ if and only if $ AB \plus{} AC \equal{} 3BC$

Let $ D \equal{} (I) \cap BC$.
as we know the line passing through the midpoints of $ BC,AB$ and the line $ EF$ and the angle bisector of $ \angle C$ are concurrent at $ K$ on $ (BC)$.
similarly for the point $ L$.
so the point of intersection of $ CK$ and $ BL$ is in fact $ I$.
it's obvious that  $ (IKL)$ is tangent to $ (I)$ iff $ 2R_{\Delta IKL} \equal{} r$ where $ r$ is the inradius.
let the point of intersection of $ CL,BK$ be $ S$ (which is in fact on $ (IKL)$ since $ SKIL$ is cyclic).
(in fact if they are tangent ,it's obvious that $ S$ is the point of tangency)
so we get that $ I$ is the orthocenter of $ \Delta SBC$ which implies $ SI \perp BC$.
and also $ S$ is the orthocenter of $ \Delta IBC$.
now see that $ SD \perp BC$ and also it's well-known that $ SD \equal{} r_a$ where $ r_a$ is the radius of the excircle $ (I_a)$.
the rest is obvious:
$ SD \equal{} r \plus{} SI \equal{} r_a \equal{} \frac {S}{p \minus{} a} \equal{} \frac {2S}{p} \equal{} 2r \Longleftrightarrow 3BC \equal{} AB \plus{} AC$


Dear Friends,
                  I should have posted this into "Suggestion"
category.Yesterday,i did post a message to the said category.It emphasizes the importance of using "Mathematical
Value Point System" in our present-day "Ranking/Rating System."
Today,added ingridient to my 'Proposal" is the total "Categori-
zing"of all "WOULD-BE-MEMBERS" in the present Art Of Problem
Solving Math Forum.(Probably,my proposal can also be adopted
by other "MATH FORUM WEBSITE.") Example,why not put into
category the following would-be-member:

         1)Professor/Professional Level.
         2)Mathematician
         3)Mathematics Enthusiasts
         4)Lay Person(Passing-by-Curiuosity.)
         5)Researcher
         6)Students

                 These are just suggestion/proposal.The main ideas
concerning my suggestion is to:

          1)To Make Math Forum Website More Competitive.
          2)To Make Math Forum A really Serious Educational
Web Forum.
          3)To Make A Future "Statistical Mathematical Gaugeing
System" related to Its Members.

                   Please,members,let me know "objectively"your
opinion,ideas,and impression related to my suggestion.There
are two,all-in-all.(Yesterday,i did the first part.Today is the
second part.)

                                                   Sincerly,
                                                  Mikhail S.

You can fill in your occupation in your profile.


geometry
geometric transformation
algebra
polynomial
linear algebra
matrix
superior algebra

Fie A \in M_2(R) astel ca detA= -1.Aratati ca det(A+I) \geq 4.
Pentru ce matrici din M_2(R) are loc egalitatea ?
Stefan Tudosie, Lista scurta ONM 2003
------------------------------------------------------------------------
Let A \in M_2(R) such that detA= -1.Proove that det(A+I) \geq 4
For wich matricies in M_2(R) we have equality ?
Hope translation is correct
----------------------------------------------------------------------
If you have a different solution post it, thanks.

(1) det(A+I) + det(A-I) =2(det(A) + det I)= 2((detA)+1)= 4
Cayley-Hamilton theorem A - (trA)A +detA.I=0
(2) A - I = (trA)A so det(A-I) = (trA)detA = - (trA)

(1) & (2) => det(A+I) = 4+ (trA)  \geq 4
Equality if and only if trA = 0

This problem can also be solved like so:

P(x)=det(A-xI_2)=x^2-tr(A)x-1.

so det(A^2+I_2)=P(i)*P(-i)=4+tr^2(A)>=4

cheers! :D :D

the translation is correct moubi. soon you will be able to understand romanian :D:D

use the char. polynomial for A. we must prove that f(i)f(-i) \geq 4 knowing that f(x)= x2 -tx-1, where t is the sum of elements on the main diagonal of A.

but f(i)f(-i)=(-2-ti)(-2+ti)=t^2+4 \geq 4 obviously :) equality occurs when t=0.

I have a very funny feeling that this problem was already discussed on this forum! Check the unsolved problems! If not, my mystake! :D

cheers! :D have fun with this problem

PS the translation is good. and the condition that Tr(A)=0 means that A^2=I_2 doesn't it? cheers!

[quote="Lagrangia"] the condition that Tr(A)=0 means that A^2=I_2 doesn't it? cheers![/quote]

A^2 =I_2, A is a matrix of a symetry in the plane R^2


calculus
integration
LaTeX
logarithms
real analysis
real analysis unsolved

S_n=1+1/2+1/3+1/4+...+1/n
prove
a)S_n is not integral for n>=2
b)there exists k-natural in [2,1000] such that {S_k}<=1/251, where {x}=x-[x].

What exactly must be proved for point a)?
For point b) one could prove the more general statement: $a_{n}\to \infty$, $a_{n+1}-a_{n}\to 0$ then $\{a_{n}\}$ is dense in $[0,1]$.

A) is not a whole number
B)does density provides that k is in the given interval

do u mean that S_n->infinity because is this true?
and also does this help proving 1+1/2^3+1/3^3+1/4^3+...<3/2

Of course it doesn't, I appologise. I think we should use that $S_{n}-\ln(n)\to c$.

well it is well known that the row's limit is infinite. You can prove this by using the geometrical progression 1/ 2+1/4+1/8+..... <1+1/2+1/3+.....+1/n. This progression has the limit infinite so...  :lol:

in don't see why S_n-lnn-c..ln(1+x)=x-x^2/x+x^3/3+...
a)2^k<n<2^(k+1), k>=1. 2^k/lcd(denominators) and 2^(k+1) doesn't divide lcd(denominators).1 is whole so look at 1/2+1/3+...+1/n.amopng the summands there is 1/2^k and so if we take lcd then in the numerators all but one summands will be divisible by 2.so it's not whole.that's right?
also 1/2+1/4+1/8+..is not infinity(i think)

The divergence of that series and the fact that it is never integral have both appeared many times on the forum.  Do a search.  Also, please learn to use LaTeX -- see http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php .  (None of the stuff about downloading is necessary to use it on the forum -- if you know the code, just put it between dollar signs.  If you don't know the code, mouse over someone else's code or use the quote button to see how they made it.  You can practice in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=224]Test forum[/url] and ask questions in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=123]LaTeX forum[/url].)

For part (b), you actually don't need to go to 1000 ($k = 83$ works nicely).  However, if you don't want to verify that by hand, you can note that it is sufficient to prove that $S_{1000}-S_{250}> 1$ (or even just that there exists an integer between $S_{250}$ and $S_{1000}$).  In fact, it is generally true that $S_{4n}-S_{n}> 1$ (with the limit of the difference being $\log 4$), and proving this fact is easy:
\[S_{4}-S_{1}= \frac{1}{2}+\frac13+\frac14 > 1 \]
and
\begin{eqnarray*}(S_{4n+4}-S_{n+1})-(S_{4n}-S_{n}) &=& \frac{1}{4n+4}+\frac{1}{4n+3}+\frac{1}{4n+2}+\frac{1}{4n}-\frac{1}{n+1}\\ &>& \frac{4}{4n+4}-\frac{1}{n+1}= 0,\end{eqnarray*}
so $D(n) = S_{4n}-S_{n}$ is increasing and $D(1) > 1$, and we're done.


geometry
vector
topology
function
linear algebra
analytic geometry

[quote="deusuk"]Hi everyone
First off - im sorry if this is posted in the wrong forum or i've missed any rules but i'm coming to the end of my tether with this one!

Basically im teaching myself linear algebra and basic geometry, and the problem i'm trying to understand is the simple definition of a plane.

The text im reading states that the definition *requires* a normal vector, a point on the plane, and a distance from the origin.
Now - when i picture this i can see that in fact all you need is a vector (normal) and a distance. I dont see the *need* for the arbitrary point on the line??

Can someone explain this for me?
My maths isn't to hot so simple explainations please  :blush:
Adam[/quote]It's time to learn the difference between [b]Euclidean plane[/b] and [b]Cartesian plane[/b].

An [i]Euclidean plane[/i] is simply a flat surface (note that this is not a definition as you would know that Euclid made it clear that there are [i]undefined terms[/i]. In topology, Euclidean plane is simply Euclidean space $ \mathbb{R}^2$ but that has another interesting definition.

A [i]Cartesian plane[/i] is a set of points for which it is normal to a given vector passing through a given point and containing this given point. There are other equivalent definitions like the intersection of vectors and so on.

The need for the given point determines uniquely the plane. When you talk about distance, it requires two points. One is the origin, the other point is the given point already.

For more info, read [url=http://en.wikipedia.org/wiki/Plane_(geometry)]here[/url]

[quote="deusuk"]Ok thanks!
But the problem i have is that the definition i have states that the distance is a real not two points?
So this boils down to one vector for the normal, a real for the distance then this extra point that i cant define?!?
 :(  thanks for your patience[/quote]There is always confusion when all the things you learn are not made specific and clear. What do you mean by 'real'? What is your definition of 'distance'?

The definition of distance between two points, in the Cartesian sense, is the length of the segment from one point to another (the endpoints are the [u]two[/u] points I am referring to). The definition of distance from a point to a line (plane) is the shortest length of the segment perpendicular from the given point to the line (plane). What 'extra point' are you talking about? There is a normal vector, a distance function and a given point on the plane for the normal vector to pass through. That's all you need.

Thanks again for your reply!
By 'real' i mean a real number, 1, 2, 4.56 etc...
And my understanding of distance is simply the length of a line drawn from the origin to the closest part of the plane.
I understand that this imaginary line would HAVE to be perpendicular to the plane else it wouldn't be the shortest? is that correct?
If it is correct then this line could also be the normal?
Therefore i concluded that with a single vector positioned at the origin pointing towards the normal of the plane, and it's length being equal to the distance then this is all you need?!?
Where am i going wrong!  :oops:

[quote="deusuk"]Thanks again for your reply!
By 'real' i mean a real number, 1, 2, 4.56 etc...[/quote]
That's right, it's reminiscent of the distance function $ d: S\times S\rightarrow\mathbb{R}^{\ge0}$.
[quote="deusuk"]And my understanding of distance is simply the length of a line drawn from the origin to the closest part of the plane.
I understand that this imaginary line would HAVE to be perpendicular to the plane else it wouldn't be the shortest? is that correct?
If it is correct then this line could also be the normal?[/quote]
Well, that's what the normal vector is for... to be perpendicular to the plane. Though, you'll need to use equivalents of the Fifth Postulate to prove that the shortest segment perpendicular to the plane is unique. It doesn't hold on non-Euclidean geometries but that's not part of our subject.
[quote="deusuk"]Therefore i concluded that with a single vector positioned at the origin pointing towards the normal of the plane, and it's length being equal to the distance then this is all you need?!?
Where am i going wrong!  :oops:[/quote]
'[i]Normal[/i]' simply means '[i]perpendicular[/i]'; and when you say 'single vector positioned at the origin pointing towards the normal of the plane', that's simply the '[i]normal vector[/i]'. Is that all you need? Yes. Whare are you going wrong? IDK, you tell me.

If you trace up from the origin a certain distance going along the vector, you'll eventually need to stop and that stopping point is your second point (apart from the origin). So that's two points, a vector, a given length. Then, there is exactly one plane normal to this vector at the stopping point. How to define this plane? This plane is the set of lines that are normal to this vector. Every line [b]on[/b] this plane is normal to the given vector (if not convinced, draw some lines on this plane). I hope this illustration of using sets to define a plane would appeal to you :)

[quote="deusuk"]This has gone a long way to helping me understand this problem.
One thing i have learned is that there are two types of plane. The one i think i am trying to define (which has a distance from the origin) is called an affine plane. Otherwise its orientated at the origin.
Is this correct?[/quote]I'm sorry if this has gone a long way to helping you. This is my last attempt: a plane can be defined by a normal vector orientated towards the origin at a given distance and this plane is unique because you have a determinate length position and angle position. To answer your original question: do you need the additional point? No, because you have the equation of the normal vector and the specified length.

Extension: affine plane, in terms of linear algebra, is a two dimensional vector space.


Find the maximum value of the real number m satisfying $ x^2$≥$ m[x]${x} for all x(x is a real number)

It's more convenient to denote $ [x]\equal{}n ,\ \{x\}\equal{}t$ where $ n$ is integer and $ 0 \leq t < 1$ . The inequality becomes $ (n\plus{}t)^2 \geq mnt\ \Longleftrightarrow \ (n\minus{}t)^2\geq (m\minus{}4)nt$
Let's make $ n\equal{}1,\ t\rightarrow 1$, so we need  $ 0 \geq m\minus{}4 \Longleftrightarrow m\leq 4$.  Manifestly  $ m\equal{}4$ is the required value .
 :cool:


let A, B be positive integers and
A(p) is the residue of A by modp
If  A(p)>=B(p),for every prime number p ,then A=B.


function
real analysis
real analysis unsolved

Here's a small piece: http://planetmath.org/encyclopedia/AlmostContinuousFunction.html

[quote="phani krishna"]What are almost continuous functions? Why are they called weak continuous functions?[/quote] thank you it is really useful


Let $ S\in \mathbb{R}^n$. Prove that if $ S$ is a convex set $ \implies \overline{S}$ is a convex set

You may want to take a look at the following [url=http://www.mathlinks.ro/viewtopic.php?p=1448597#1448597]post[/url] now.

Regards,

Quoting from the link JRav posted:

"These examples suggest that the closure of a convex set is convex. To prove this fact, consider two points $ x$ and $ y$ in the closure of a convex set $ S$. There exist sequences $ u_{(k)}$ and $ v_{(k)}$ from $ S$ converging to $ x$ and $ y$, respectively. The convex combination $ w_{(k)} \equal{} \alpha u_{(k)} \plus{} (1 \minus{} \alpha)v_{(k)}$ is in $ S$ as well and converges to $ z \equal{} \alpha x \plus{} (1 \minus{} \alpha) y$ in the closure of $ S$."

What more do you want? :)

We show that $ \overline{S}$ is convex by taking two points $ x$ and $ y$ from that set. Then for any $ \alpha \in [0,1]$, we wish to prove that $ \alpha x \plus{} (1\minus{}\alpha) y \in \overline{S}$. Just apply the above and you're done.


If f(x) = |(x^2-12)(x^2+4)|, how many numbers in the interval -2 <= x <= 3 satisfy the conclusion of the Mean Value Theorem?

A. None
B. One
C. Two
D. Three
E. Four

anyone can solve this for me quickly? much appreciated

For your problem, a number $ c$ that satisfies the Mean Value Theorem is such that:

$ f'(c) \equal{} \frac{f(3)\minus{}f(\minus{}2)}{3 \plus{} 2}$

which is

$ f'(c) \equal{} \minus{}1$

For $ \minus{}2 \leq x \leq 3$, $ (x^2\minus{}12)(x^2\plus{}4) < 0$ and $ f(x) \equal{} \minus{}(x^2\minus{}12)(x^2\plus{}4)$.

$ f'(c) \equal{} \minus{}1$ thus turns into:

$ \minus{}4c(c^2\minus{}4) \equal{} \minus{}1$

the number of solutions to which can be found using whatever method you like.


Post the funniest things that have ever happened to you.

Since I can't think of anything right now, I'll post something later.

Something that just happened a couple days ago...
me: "omg this band is like metallica! "
ccy: "are they metal?"

Hey, Alex, Why is air a lot like sex?

mrchen484: Uhh. Because it's no big deal unless you're not getting any....

I was talking to my friend in class and was like "if you had a friend name Jack and Jack was stuck on a horse, would you help Jack off a horse" My teacher walked by and was like, Hey I have a son name Jack..

Me [malapropism]: "Biology is the most promiscuous topic."

---===---
And some funny in-class observations:
---===---

Mr. Roos: What is the significance of the date 11/18?
Student: Uhhh...
Mr. Roos: It's the last time you turned in a homework assignment.

[During a programming test]
Student: What's a function???
Mr. Roos: It's what we've been doing for the last two weeks, where you copied all of the assignments from someone else.

My funniest moments are when I'm in a Bible Quizzing tournament, I get asked a question, and as soon as I sit down, I remember the answer.

So I might get asked: "What did Paul say after he wrote, 'To the only wise God be glory forever through Jesus Christ'?"

"Oh goodness...finished." *sits down*

And the answer turns out to be "Amen". Then I bang myself in the head because we had that question last round.

I was talking to some friends while walking and I walked into a lamp post.


hello guys  :)

I'm a new girl in this forum , hope I'm not breaking any rules  :D

I've a question and need some help from you
Well, I'm studying MANAGEMENT and gonna study operation research

PLS can you tell me where I can find tasks with answers like linear algebbra tasks and global optimization

especially such tasks

http://www.mathlinks.ro/Forum/viewtopic.php?t=225396&search_id=1318664881

Thank u very much hope I'll get useful info

You are probably interested in the method of [url=http://en.wikipedia.org/wiki/Linear_programming]Linear Programming[/url]

She certainly is, but, most likely, not at such level of abstraction. Also, as far as I understand, the question was about self-tests and similar things, not about general mathematical theory. Unfortunately, my attempts to search google for those brought up nothing of interest except some reviews of pricey textbooks :(.

[b]TZF[/b]

yes, I'm interested in linear programming , thanks for the link

[b]fedja[/b]

Thank you for the reply  :)

I think I'll go to the library

I've loved getting to know this forum    :)


vector
linear algebra
linear algebra unsolved

V is a finite dimensional space and T: V-> V is a linear operator over V. If a,b are eigenvectors and a,b are elements of V with corresponding eigenvalues p and q different from each other, then prove that x=3a-b is not an eigenv. of T.

Hi
Suppose $x$ is the eigen vector for $T$ with eigenvalue $\alpha$, then

$T(3a-b)=3pa-qb$
$3pa-qb=\alpha (3a-b)$

so we then have $3\alpha=3p$ and $-\alpha=-q$ or $p=q$ which is a cotradiction.


geometry
algebra
functional equation
complex numbers
\/closed

Soon mathlinks will be capable of celebrating its first birthday. Congratulation to Valentin Vornicu who created the forum and thus established a platform to participate in this math olympiad community at mathlinks here. Many thanks also to all the others who made it possible to get a vivid and active forum. Keep up the excellent work.

Don't forget that on February, the third, you all have to hand in your presents !!!  :D

yey! although the board shows not as birthday 27th of January 2004, it is actually 3 of February 2003 :D

thanks orl, and all of you for making this site one of the best math olympiad sites on the world ...

Yeah great idea to do mathlinks Valentin, this the new style
of doing mathematics on the XXI century.
Happy birthday mathlinks, I will not forget the 3th February  :D

cool!!! so mathlinks is gonna complete one whole year.congratulations to the valentin vornicu for working up the idea of this  wonderful website(though sumhow i got cut off from 5 months) this site has managed to induce into me a love (strong passion) for maths (that 's a difficult thing to do when you are a "hardcore lover" of physics :D  :D )
neway hope the website keeps the momentum going
best regards
vasu

I am glad that you are back galois, mathlinks kinna missed you :D

Happy Birthday MathLinks :D

//brings birthday cake in
//brings prezzies in :)

I thought that it might be nice to share some of the memorable stuff that we'll remember as part of the first year.  For me it would be harazi's "Simetry is sometimes tricky" and later "Simetry is always tricky" (I always suspect there's another thread to go with roughly the same title! :D ) and Fedor's "Polish functional equation" which has been stickied for ages, with 141 views and still waiting for a solution... :D.

Happpyyy Birthday :P yey - you guys will have to repeat that in 11 days :D:D:D:D I don't belive it ... a year of MathLinks forum (the Mathlinks site existed long before - in fact I plan to make links to old mathlinks sites, like a historical museum :D:D).

add 3 with 11 you'll get 14 -> 14 of February that is .. and we all know what that day means :D

Happy birthday Mathlinks! Thank you Valentin for your good work. And I wish you many roses from beautifull girls on Valentin's Day.

We hope that Mathlinks will growth from day to day.

Namdung

[quote="Valentin Vornicu"]....in fact I plan to make links to old mathlinks sites, like a historical museum :D:D).....[/quote]
Sound great math museum, happy birthday µathlinks.

i was searching in vain for a mathematician who shared the same birthday as "MATHLINKS" and finally i have come up with sumthing .its "gaston julia",the person who discovered all those nice properties abt fractals and after whom julia sets are named.hey valentin,here's a suggestion
HOW ABT INCORPORATING A BEAUTIFUL FRACTAL IN MATHLINKS LOGO (IF THAT CAN BE DONE SUMHOW,ANYWAYS DO GIVE IT A THOUGHT :D )

and as for the old website ,i've been there .infact i have one of the documents posted on that website "using complex numbers in geometry"
(a short sweet collection 5 problems,i guess) and it went defunct just before christmas

there are more stages in the development of mathlinks.
stragely when I searched something with [url=http://www.google.com]google[/url] today I saw the modified theme with mathematics equations and symbols and fractals :D what a nice coincidence :D

PS galios, the mathlinks logo is already a sort of a fractal :) . but I have decided to let this be the final logo, so I shall keep that thought :)

Happy Birthday MathLinks!
I've been around for 3 weeks and I think it is a great community with great members.This forum will be a great resource for people preparing for math contests.

(One more thingie:why can people which are not registered,post here?I think they should have read,but not write permissions for the board)

What can I say more,[img]http://www.tuningfreaks.sprintweb.de/sonstiges/ok.gif[/img]

the guests can only post in some regions of the forum. they can also read only the unsolved problems sections and the non-math topics.

this facility has been offered for let the guest see what Mathlinks.ro is about :)

yet few people know of this site's existance, and many who know, have no idea about what is Mathlinks mainly about :)

hopefully that will change in the future ... I only wish that this site existed 3 years ago when I qualified for the IMO :( ..


limit
function
calculus
algebra
functional equation
calculus computations

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

$ \lim_{x\to 0}\frac{f(x\plus{}x_{0})\minus{}f(x_{0}\minus{}x)}{x}\equal{}0,\forall x_{0}$  :)

[quote="quangpbc"]Find all $ f:\mathbb{R}\to\mathbb{R}$ so that :

$ \lim_{x\to 0}\frac{f(x\plus{}x_{0})\minus{}f(x_{0}\minus{}x)}{x}\equal{} 0,\forall x_{0}$  :)[/quote]

At least we have all functions such that $ f(x) \equal{} a$ for all real $ x$ except for $ x\in A$ where $ A$ is any subset of $ \mathbb{R}$ having no accumulation point and $ f(x)$ being any value for $ x\in A$

We also have some cases with A having accumulation points :
$ f(x)\equal{}x^{2}$ $ \forall x\in A\equal{}\{\plus{}/\minus{}2^{\minus{}n}$, n any positive integer$ \}$
$ f(x)\equal{}a$ everywhere else

[quote="pco"]
At least we have all functions such that $ f(x) \equal{} a$ for all real $ x$ except for $ x\in A$ where $ A$ is any subset of $ \mathbb{R}$ non dense in $ \mathbb{R}$  and $ f(x)$ being any value for $ x\in A$[/quote]

I don't understand your solution ,  :( . Can you post a complete solution , please  :)

[quote="quangpbc"]
I don't understand your solution ,  :( . Can you post a complete solution , please  :)[/quote]

Ok,
Let $ A$ be a subset or $ \mathbb{R}$ without any accumulation point.
Let $ f(x) \equal{} a$ $ \forall x\notin A$

1) if $ x_{0}\in A$, and since $ x_{0}$ is not an accumulation point of $ A$, it exists a real $ h > 0$ such that $ I \equal{} (x_{0}\minus{}h,x_{0}\plus{}h)\cap A \equal{}\{x_{0}\}$
So $ \forall x\in (\minus{}h,\plus{}h)$, $ f(x_{0}\minus{}x) \equal{} f(x_{0}\plus{}x)$ and so $ \frac{f(x_{0}\plus{}x)\minus{}f(x_{0}\minus{}x)}{x}\equal{} 0$ and $ \lim_{x\to 0}\frac{f(x_{0}\plus{}x)\minus{}f(x_{0}\minus{}x)}{x}\equal{} 0$ Q.E.D

2) if if $ x_{0}\notin A$, and since it does not exist any accumulation point in $ A$, it exists a real $ h > 0$ such that $ I \equal{} (x_{0}\minus{}h,x_{0}\plus{}h)\cap A \equal{}$ emptyset
So $ \forall x\in (\minus{}h,\plus{}h)$, $ f(x_{0}\minus{}x) \equal{} f(x_{0}\plus{}x)$ and so $ \frac{f(x_{0}\plus{}x)\minus{}f(x_{0}\minus{}x)}{x}\equal{} 0$ and $ \lim_{x\to 0}\frac{f(x_{0}\plus{}x)\minus{}f(x_{0}\minus{}x)}{x}\equal{} 0$ Q.E.D

[quote="pco"]$ f(x) \equal{} x^{2}$ $ \forall x\in A \equal{}\{\plus{}/\minus{}2^{\minus{}n}$, n any positive integer$ \}$
$ f(x) \equal{} a$ everywhere else[/quote]

Uhm, i think , $ f(x)\equal{}a\equal{}\text{constant}$ is only function such that condition  :D

[quote="quangpbc"][quote="pco"]$ f(x) = x^{2}$ $ \forall x\in A =\{+/-2^{-n}$, n any positive integer$ \}$
$ f(x) = a$ everywhere else[/quote]

Uhm, i think , $ f(x) = a =\text{constant}$ is only function such that condition  :D[/quote]

My example is OK :
Let $ A=\{2^{n},-2^{n},\forall n\in\mathbb{N}\}$
Let $ f(x)=x^{2}$ $ \forall x\in A$ and $ f(x)=1$ $ \forall x\notin A$

1) If $ x_{0}\in A$, it exists a real $ h > 0$ (for example $ \frac{|x_{0}|}{2}$) such that $ I = (x_{0}-h,x_{0}+h)\cap A =\{x_{0}\}$
So $ \forall x\neq 0\in (-h,+h)$, $ f(x_{0}-x) = f(x_{0}+x)=1$ and so $ \frac{f(x_{0}+x)-f(x_{0}-x)}{x}= 0$ and $ \lim_{x\to 0}\frac{f(x_{0}+x)-f(x_{0}-x)}{x}= 0$ Q.E.D

2) If $ x_{0}\neq 0$ and $ x_{0}\notin A$, it exists a real $ h > 0$ such that $ I = (x_{0}-h,x_{0}+h)\cap A =$ emptyset
So $ \forall x\in (-h,+h)$, $ f(x_{0}-x) = f(x_{0}+x)=1$ and so $ \frac{f(x_{0}+x)-f(x_{0}-x)}{x}= 0$ and $ \lim_{x\to 0}\frac{f(x_{0}+x)-f(x_{0}-x)}{x}= 0$ Q.E.D

3) If $ x_{0}=0$:
If $ |x|=2^{-n}$ for some positive integer $ n$, $ f(x_{0}-x) = f(x_{0}+x)=2^{-2n}$
If $ |x|\neq2^{-n}$ for any positive integer $ n$, $ f(x_{0}-x) = f(x_{0}+x)=1$
So $ f(x_{0}-x) = f(x_{0}+x)$ $ \forall x$ and so $ \frac{f(x_{0}+x)-f(x_{0}-x)}{x}= 0$ and $ \lim_{x\to 0}\frac{f(x_{0}+x)-f(x_{0}-x)}{x}= 0$ Q.E.D

It does not seem to me that this problem is "calculus" or "calculus computation"
It is a kind of functional equation.
 :?:

It is only an example, what is your proof for it, pco  :roll:  .I will to think about your example  :D

[quote="pco"]It does not seem to me that this problem is "calculus" or "calculus computation"
It is a kind of functional equation.
 [/quote]

It is a functional equation use limit, pco. :wink:

[quote="quangpbc"]It is only an example, what is your proof for it, pco [/quote]

I don't understand your question : my proof for this example is given in my post; What do you want more ?

[quote="quangpbc"]:D .I want to see your proof: if $ f$ so that the condition then $ f$ is the only function you have posted above[/quote]

You should read more carefully my messages : I never said this was the unique solutions.

I said that we have [u]at least [/u]:

1) functions constant on $ \mathbb{R}$ less a subset A of $ \mathbb{R}$ without accumulation point.
2) some functions constant on $ \mathbb{R}$ less a subset A of $ \mathbb{R}$ with accumulation point (example : my function with $ x^{2}$ for $ 2^{\minus{}n}$)

Notice that, in 1) above, if you take $ A$=empty set, then you have trivial solution $ f(x)\equal{}a$ $ \forall x$ which is one family of solutions.
Notice that there exist a lot of other solutions in the class 2) above.


analytic geometry
combinatorics unsolved
combinatorics

Several points are given in the plane. Suppose that for any three of them, there exists an orthogonal coordinate system (determined by the two axes and the unit length) in which these three points have integer coordinates. Prove that there exists an orthogonal coordinate system in which all the given points have integer coordinates.

Posted many times before, even still unsolved :
http://www.mathlinks.ro/viewtopic.php?t=201137

Pierre.


inequalities
LaTeX
geometry
circumcircle
incenter
inequalities proposed

Let ta，tb，tc be the length of the bisectors of triangleABC.Let ra,rb,rc be the length of the radiuses of the escribed circle of triangleABC.Prove that ra/ta+rb/tb+rc/tc>=3

[quote="yixinshen"]Let $ ta,tb,tc$ be the length of the bisectors of triangle $ ABC$.Let $ r_a,r_b,r_c$ be the length of the radiuses of the escribed circle of triangle $ ABC$.Prove that $ \frac{r_a}{t_a}\plus{}\frac{r_b}{t_b}\plus{}\frac{r_c}{t_c}\ge 3$[/quote]
Your problem in $ LaTEX$
I think it is not new

[quote="Allnames"][quote="yixinshen"]Let $ ta,tb,tc$ be the length of the bisectors of triangle $ ABC$.Let $ r_a,r_b,r_c$ be the length of the radiuses of the escribed circle of triangle $ ABC$.Prove that $ \frac {r_a}{t_a} \plus{} \frac {r_b}{t_b} \plus{} \frac {r_c}{t_c}\ge 3$[/quote]
Your problem in $ LaTEX$
I think it is not new[/quote]
$ t_a\le \sqrt{s(s\minus{}a)}$ , $ t_b\le \sqrt{s(s\minus{}b)}$ , $ t_c\le \sqrt{s(s\minus{}c)}$ ; => $ t_at_bt_c \le sS$;
 $ r_ar_br_c \equal{} Ss$;
So : LHS $ \ge 3$ ( by am-gm ) :)

[quote="Arsenaler"][color=red] We have a nicer problem :
  $ \frac {2R}{r} \minus{} 1\ge \frac {r_a}{t_a} \plus{} \frac {r_b}{t_b} \plus{} \frac {r_c}{t_c}\ge\sqrt {\frac {4R}{r} \plus{} 1}$
[/color][/quote]
1.$ t_a\plus{}r_a \le \frac{s}{cos(\alpha/2)}\equal{}> 1\plus{} \frac{r_a}{t_a} \le \frac{s}{t_acos(\alpha/2)}\equal{} \frac{s(b\plus{}c)}{cos^2(\alpha/2)2bc}\equal{} \frac{b\plus{}c}{2(s\minus{}a)}\equal{}>$ $ \frac{r_a}{t_a}\le \frac{a}{2(s\minus{}a)}$
so we have : $ \frac {r_a}{t_a} \plus{} \frac {r_b}{t_b} \plus{} \frac {r_c}{t_c}\le \frac{1}{2}(\frac{a}{s\minus{}a}\plus{}\frac{b}{s\minus{}b}\plus{}\frac{c}{s\minus{}c})\equal{} 2\frac{R}{r}\minus{}1$;

2. well known : (a) $ \frac{h_a}{t_a}\ge \sqrt{\frac{2r}{R}}      ,    ( b)\frac{2}{h_a}\equal{} \frac{1}{r_b}\plus{}\frac{1}{r_c}$;
so we have : $ \frac {r_a}{t_a} \plus{} \frac {r_b}{t_b} \plus{} \frac {r_c}{t_c} \ge \sqrt{\frac{r}{2R}}(\frac{r_a}{r_b}\plus{}\frac{r_c}{r_b}\plus{}\frac{r_c}{r_a}\plus{}\frac{r_b}{r_a}\plus{}\frac{r_b}{r_c}\plus{}\frac{r_a}{r_c})\equal{}$ $ \sqrt{\frac{r}{2R}}\frac{(r_a\plus{}r_c)r_ar_c\plus{}(r_c\plus{}r_b)r_cr_b\plus{}(r_b\plus{}r_a)r_br_a}{r_ar_br_c}\equal{}$
$ \equal{}\sqrt{\frac{r}{2R}}\frac{(r_a\plus{}r_b\plus{}r_c)(r_ar_b\plus{}r_br_c\plus{}r_cr_a)\minus{}3r_ar_br_c}{r_ar_br_c}\equal{} \sqrt{\frac{r}{2R}}(\frac{4R}{r}\minus{}2)\equal{}$ $ 2\sqrt{2}\sqrt{\frac{R}{r}}\minus{}\sqrt{\frac{2r}{R}} \ge \sqrt{\frac{4R}{r}\plus{}1}$; :)

[quote="Allnames"][quote="yixinshen"]Let $ w_{a},w_{b},w_{c}$ be the angle-bisectors of  $ \triangle ABC$. Let $ r_a,r_b,r_c$ be the radii of excircles of $ \triangle ABC$. Prove that

$ \frac {r_a}{w_a} \plus{} \frac {r_b}{w_b} \plus{} \frac {r_c}{w_c}\ge 3.$[/quote][color=red]We have a nicer problem :

$ \frac {2R}{r} \minus{} 1\ge \frac {r_a}{w_a} \plus{} \frac {r_b}{w_b} \plus{} \frac {r_c}{w_c}\ge\sqrt {\frac {4R}{r} \plus{} 1},$

where $ R$ is the radius of circumcircle and $ r$ is the radius of incircle.[/color][/quote]$ \frac {2R}{r} \minus{} 1 \equal{} \sum{\frac {r_a}{h_a}}\geq\sum{\frac {r_a}{w_a}}\geq\sum{\frac {R}{IA}}\geq\sum{\frac {3R}{2w_a}}\geq\sqrt {3}\sum{\frac {R}{a}}\geq\sqrt {\frac {4R}{r} \plus{} 1}$

$ \geq\frac {9\sqrt {3}R}{2s}\geq\sum{\frac {w_a}{h_a}}\geq\frac {3}{2}\sum{\frac {a^2}{w_{b}^{2} \plus{} w_{c}^{2}}}\geq\sum{\frac {2w_a}{3IA}}\geq\frac {4}{3}\sum{\frac {a \plus{} b \plus{} c}{2a \plus{} b \plus{} c}}\geq 3,$

where $ a,b,c$ are the sides $ BC,CA,AB$, $ h_a,h_b,h_c$ are the altitudes, $ I$ is the incentre and $ IA,IB,IC$ are the distances from $ I$ to the vertices of $ \triangle ABC$.


[hide]$2x+2y=42$
$xy=108$

from the first equation, $x=21-y$
substituting, $(21-y)(y)=108$
$21y-y^2=108$
$-y^2+21y-108=0$
$y^2-21y+108=0$.
$(y-12)(y-9)$
y can either be 9 or 12, x can be be 12 or 9.

The shorter side must be 9.
[/hide]

[hide]the shorter side is 9 feet in length. we know that 2x+2y=42, x and y being the lenghts of the sides. we also know that x=108/y. putting that into the first equation and multiplying by y gives us 2y^2-42y+216, leaving 12 and 9 as the roots to that equation. in this case, the side lengths are 12 and 9, meaning that 9 is the shorter side.[/hide]

[quote="aznness"][hide]9[/hide][/quote]

Would you like to show your work  ;)

[quote="anirudh"]9ft[/quote]

[quote="236factorial"][quote="anirudh"][quote="hwenterprise"]A rectangular room has a perimeter of 42 ft. and an area of 108 sq. ft. What is the length, in feet, of the shorter side?[/quote]

9 ft[/quote]

Thanks captain obvious  ;)
work can make your answer more meaning ful.  :roll:[/quote]

[quote="anirudh"]
it just popped in my mind, well here. 108=9x12=3x3x3x4. (9+12)x2=42. therefore , anser is 108 [/quote]

You did it by guess and check? That can be quicker sometimes.  :)

well, not really. i just factored(the way), but the ans. just popped in my midn when i saw it.


The smallest sum one could get by adding three different numbers from the set $ \{7,25,\minus{}1,12,\minus{}3 \}$ is

\[ \textbf{(A)}\ \minus{}3 \qquad
\textbf{(B)}\ \minus{}1 \qquad
\textbf{(C)}\ 3 \qquad
\textbf{(D)}\ 5 \qquad
\textbf{(E)}\ 21
\]

Add the three smallest numbers, and you get C.

[hide="Click for solution"]
The three smallest numbers in the set are $ \minus{}3$, $ \minus{}1$, and $ 7$.  Their sum is $ \minus{}3\plus{}(\minus{}1)\plus{}7\equal{}3$, which corresponds to answer choice $ \boxed{\textbf{(C)}}$.
[/hide]

Alternatively, the largest sum is obtained by adding the 3 largest elements, which would be 25+12+7 = 44.


geometry
rectangle
trigonometry
parallelogram
perpendicular bisector
geometry unsolved

Outside for the acute-angled $\triangle ABC$ are given the rectangles $ABB_{2}A_{1}$, $BCC_{2}B_{1}$ and $CAA_{2}C_{1}$ such that $AA_{1}: BB_{1}: CC_{1}=\frac{1}{\cos\alpha}: \frac{1}{\cos\beta}: \frac{1}{\cos\gamma}.$
Prove that the $S_{A_{1}A_{2}},S_{B_{1}B_{2}}$ and $S_{C_{1}C_{2}}$ go through a common point.

The condition $AA_{1}: BB_{1}: CC_{1}=\frac{1}{\cos\alpha}: \frac{1}{\cos\beta}: \frac{1}{\cos\gamma}$ is unnecessary, the perpendicular bisectors $S_{A_{1}A_{2}},\ S_{B_{1}B_{2}},\ S_{C_{1}C_{2}}$ of $A_{1}A_{2},\ B_{1}B_{2},\ C_{1}C_{2}$ are concurrent for arbitrary rectangles $ABB_{2}A_{1},\ BCC_{2}B_{1},\ CAA_{2}C_{1}.$ See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=22917]Germany 1996 for squares erected on sides of triangle[/url], the last reply.

This problem is coming from the past. We draw two lines through $B_{1},$ $C_{2}$ and perpendicular to $B_{1}B_{2},$ $C_{1}C_{2}$ respectively and we denote as $A',$ their intersection point. Similarly we define the points $B',$ $C'$ and it is enough to prove that the quadrilaterals $B'C'A_{1}A_{2},$ $C'A'B_{1}B_{2},$ $A'B'C_{1}C_{2},$ are rectangles.

$\bullet$ Through vertices $B,$ $C,$ of $\bigtriangleup ABC,$ we draw two lines perpendicular to $B_{1}B_{2},$ $C_{1}C_{2}$ respectively and we denote as $N,$ their intersection point. Also we denote as $A',$ the intersection point of the lines through $B_{1},$ $C_{2}$ and perpendicular to $B_{1}B_{2},$ $C_{1}C_{2},$ respectively.

We define the points $B',$ $C',$ such that the quadrilaterals $ANB'A_{2},$ $ANC'A_{1},$ to be parallelograms.Now, in this configuration we have the follows results:

The triangles $BNC$ and $B_{1}A'C_{2},$ are congruent because of $NB\parallel A'B_{1},$ $NC\parallel A'C_{2}$ and ${BC\parallel = B_{1}B_{2}}.$ Hence, the quadrilaterals $NA'B_{1}B,$ $NA'C_{2}C,$ are parallelograms and so, we have that $NA'\perp B_{1}C_{2}$ $\Longrightarrow$ $NA'\perp BC.$

From the parallelograms $ANB'A_{2},$ $ANC'A_{1},$ we conclude that $NB'\perp AC$ and $NC'\perp AB.$ Also we conclude that the triangles $\bigtriangleup B'NC'$ and $\bigtriangleup A_{1}AA_{2}$ are congruent, because of $NB'\parallel = AA_{2},$ $NC'\parallel = AA_{1}$ and $\angle B'NC' = \angle A_{2}AA_{1}.$ Hence, we have that $B'C'\parallel = A_{1}A_{2}.$

From $NA'\parallel = CC_{2},$ $NB'\parallel = AA_{2}\parallel = CC_{1}$ and $\angle A'NB' = \angle C_{2}CC_{1},$ we conclude that the triangles $\bigtriangleup A'NB'$ and $\bigtriangleup C_{2}CC_{1},$ are congruent and so, we have that $A'B'\parallel = C_{1}C_{2}.$ Hence, the quadrilateral $A'B'C_{1}C_{2},$ is a rectangle $($ because of $A'C_{2}\perp C_{1}C_{2}$ $).$ Similarly the quadrilateral $C'A'B_{1}B_{2},$ is also a rectangle.

Now, it is already has been proved that the lines through vertices $A',$ $B',$ $C',$ of $\bigtriangleup A'B'C'$ and perpendiculars to the sidelines $BC,$ $CA,$ $AB,$ of $\bigtriangleup ABC$ respectively, are concurrent at the point $N.$

So, the triangles $\bigtriangleup A'B'C',$ $\bigtriangleup ABC,$ are orthologic and hence we conclude that the lines through vertices vertices $A,$ $B,$ $C,$ of $\bigtriangleup ABC$ and perpendicular to the sidelines $B'C',$ $C'A',$ $A'B',$ of $\bigtriangleup A'B'C'$ respectively, are concurrent at one point, which is again the point  $N,$ as the intersection point of the constructed lines through $B,$ $C$ and perpendicular to $B_{1}B_{2}\parallel = A'C',$ $C_{1}C_{2}\parallel = A'B',$ respectively.

We conclude now, that $AN\perp B'C'$ $\Longrightarrow$ $AN\perp A_{1}A_{2}$ and so, the quadrilateral $B'C'A_{1}A_{2},$ is a rectangle, because of $B'A_{2}\parallel = AN\parallel = C'A_{1}.$

$\bullet$ Consequently, the midperpendiculars of the segments $A_{1}A_{2},$ $B_{1}B_{2},$ $C_{1}C_{2},$ as also the midperpendiculars of the sidelines $B'C',$ $C'A',$ $A'B',$ of $\bigtriangleup A'B'C',$ are concurrent at one point and the proof is completed.

Kostas Vittas.

Because of the orthologic triangles $\bigtriangleup ABC,$ $\bigtriangleup A'B'C',$ have their orthologic centers, with respect to each other, as coincided at one point, here the point $N,$ we conclude that these triangles are also perspective (as well known). Hence, the lines $AA',$ $BB',$ $CC',$ are concurrent at one point.

An interesting remark now, is that we can use the same words in order to prove the concurrency of the midperpendiculars as the problem states and also the one of the lines $AA',$ $BB',$ $CC',$ in every configuration of the erected rectangles $ABB_{2}A_{1},$ $BCC_{2}B_{1},$ $CAA_{2}C_{1},$ on sidelines $AB,$ $BC,$ $CA$ respectively, of a given triangle $\bigtriangleup ABC$ $($ All of them outwardly to it, or inwardly, or some of them outwardly and the other inwardly $).$

Kostas Vittas.

Denote by $ A'$, $ B'$, $ C'$ the intersections of the perpendicular bisectors of the segments $ AA_{1}$, $ BB_{1}$, $ CC_{1}$.

We have two triangles $ ABC$ and $ A'B'C'$ are perpective, it follows that $ AA'$, $ BB'$, $ CC'$ are concurrent. And we also see that $ A'$, $ B'$, $ C'$ lie on the perpendicular bisectors of $ A_{1}A_{2}$, $ B_{1}B_{2}$ and $ C_{1}C_{2}$, respectively.

We'll prove that the perpendicalr bisector of the segment $ A_{1}A_{2}$ and $ AA'$ are isogonal conjugate with respect to $ \angle B'A'C'$.
Ideed, denote $ H\equal{}C'A'\cap A_{1}A_{2}$, we have $ \angle A'A_{1}A_{2}\equal{}\angle A'A_{2}A_{1}\equal{}\angle A'AH$, it follows that the quadrilateral $ A'HAA_{1}$ is cyclic. Therefore $ \angle HA'A\equal{}\angle HA_{1}A\equal{}\angle B'A'K$, where $ K$ is the midpoint of $ A_{1}A_{2}$.
Similarly, $ BB'$ and the perpendicular bisector of $ B_{1}B_{2}$ are isogonal conjugate, $ CC'$ and the perpendicular bisector of $ C_{1}C_{2}$ are isogonal conjugate.

Since $ AA'$, $ BB'$ and $ CC'$ are concurrent, we conclude that the perpendicular bisectors of $ A_{1}A_{2}$, $ B_{1}B_{2}$ and $ C_{1}C_{2}$ are concurrent.

[quote="April"]Denote by $ A'$, $ B'$, $ C'$ the intersections of the perpendicular bisectors of the segments $ AA_{1}$, $ BB_{1}$, $ CC_{1}$.

We have two triangles $ ABC$ and $ A'B'C'$ are perpective, it follows that $ AA'$, $ BB'$, $ CC'$ are concurrent. And we also see that $ A'$, $ B'$, $ C'$ lie on the perpendicular bisectors of $ A_{1}A_{2}$, $ B_{1}B_{2}$ and $ C_{1}C_{2}$, respectively.

We'll prove that the perpendicalr bisector of the segment $ A_{1}A_{2}$ and $ AA'$ are isogonal conjugate with respect to $ \angle B'A'C'$.
Ideed, denote $ H \equal{} C'A'\cap A_{1}A_{2}$, we have $ \angle A'A_{1}A_{2}\equal{}\angle A'A_{2}A_{1}\equal{}\angle A'AH$, it follows that the quadrilateral $ A'HAA_{1}$ is cyclic. Therefore $ \angle HA'A \equal{}\angle HA_{1}A \equal{}\angle B'A'K$, where $ K$ is the midpoint of $ A_{1}A_{2}$.
Similarly, $ BB'$ and the perpendicular bisector of $ B_{1}B_{2}$ are isogonal conjugate, $ CC'$ and the perpendicular bisector of $ C_{1}C_{2}$ are isogonal conjugate.

Since $ AA'$, $ BB'$ and $ CC'$ are concurrent, we conclude that the perpendicular bisectors of $ A_{1}A_{2}$, $ B_{1}B_{2}$ and $ C_{1}C_{2}$ are concurrent.[/quote]

Very nice! Thank you dear [b][size=100]April[/size][/b] for a clever solution new to me. [b][size=100]"The simple is the most important."[/size][/b]

Kostas Vittas.


[quote="Magnara"]Prove that in an acute triangle $ABC$, \[\frac{a^{2}}{\sqrt{b^{2}+c^{2}}}\cos{\frac{A}{2}}+\frac{b^{2}}{\sqrt{a^{2}+c^{2}}}\cos{\frac{B}{2}}+\frac{c^{2}}{\sqrt{a^{2}+b^{2}}}\cos{\frac{C}{2}}\geq{\frac{9\sqrt{2}r}{2}},\] with $r$ as the inradius of $ABC$.[/quote]
$\sqrt{b^{2}+c^{2}}=2R\sqrt{\sin^{2}B+\sin^{2}C}=2R\sqrt{1-\frac{\cos2B+\cos2C}{2}=}$
$=2R\sqrt{1-\cos(B+C)\cos(B-C)}=2R\sqrt{1+\cos A\cos(B-C)}\leq$
$\leq2R\sqrt{1+\cos A}=2\sqrt2R\cos\frac{A}{2}$ since, $\Delta ABC$ is acute triangle.
Hence, $\sum_{cyc}\frac{a^{2}\cos\frac{A}{2}}{\sqrt{b^{2}+c^{2}}}\geq\sum_{cyc}\frac{4R^{2}\sin^{2}A\cos\frac{A}{2}}{2\sqrt2R\cos\frac{A}{2}}=\sum_{cyc}\sqrt2R\sin^{2}A.$
Remain to prove that $\sum_{cyc}\sin^{2}A\geq\frac{9r}{2R}\Leftrightarrow\sum_{cyc}\frac{4S^{2}}{b^{2}c^{2}}\geq\frac{9\cdot\frac{S}{p}}{2\cdot\frac{abc}{4S}}\Leftrightarrow$
$\Leftrightarrow\frac{1}{a^{2}b^{2}}+\frac{1}{a^{2}c^{2}}+\frac{1}{b^{2}c^{2}}\geq\frac{9}{abc(a+b+c)},$ which obviously true. :)

[quote=Magnara]Prove that in an acute triangle $ABC$,
\[\frac{a^2}{\sqrt{b^2+c^2}}\cos{\frac{A}{2}}+\frac{b^2}{\sqrt{a^2+c^2}}\cos{\frac{B}{2}}+\frac{c^2}{\sqrt{a^2+b^2}}\cos{\frac{C}{2}}\geq{\frac{9\sqrt{2}r}{2}},\]
with $r$ as the inradius of $ABC$.[/quote]

It's true for all triangle.

A stronger inequality that really uses the acute-angled triangle condition is the following:
\[\frac{a^2}{\sqrt{b^2+c^2}}\cos{\frac{A}{2}}+\frac{b^2}{\sqrt{a^2+c^2}}\cos{\frac{B}{2}}+\frac{c^2}{\sqrt{a^2+b^2}}\cos{\frac{C}{2}}\geq{\frac{(R+r)^2\sqrt{2}}{R}},\]

And even stronger than the last one:
\[\frac{a^2}{\sqrt{b^2+c^2}}\cos{\frac{A}{2}}+\frac{b^2}{\sqrt{a^2+c^2}}\cos{\frac{B}{2}}+\frac{c^2}{\sqrt{a^2+b^2}}\cos{\frac{C}{2}}\geq{\frac{\sqrt2(s^2-4Rr-r^2)}{2R}},\]

It holds in an acute-angled triangle! :-)


Let f be a function defined for all nonzero real numbers.
Suppose that f(1/x) + 2f(x) = x for all nonzero real x. Determine the function f .

[hide="hint"]try substituting something for x.
[hide="giveaway"]try substituting 1/x for x.[/hide]
[/hide]

[hide]$f(1/x)+2f(x)=x$.
Subtitute 1/x for x:
$f(x)+2(f(1/x)=1/x$ and solve for f(x) in terms of x:
Double the first equation and subtract the second giving $3f(x)=2x-1/x\implies f(x)=\boxed{\frac{2x^{2}-1}{3x}}$.[/hide]


function
LaTeX
limit
algebra
domain
induction
real analysis

Suppose f is real valued function which is not bounded over the interval [a,b]....a<b.

Prove that there exists a sequence {$ x_{n}$} in [a,b] which converges to xε[a,b] and for which sequence |f($ x_{n}$)|>n,for all nεN

[quote="JBL"][[LaTeX]] is enabled on this forum -- it will make your posts vastly more legible if you use it more (i.e., enclose entire expressions or equations, use commands like \in for $ \in$ instead of ad-hoc stuff, etc.).  See that link and http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690

For your question: first find a sequence $ (x_n)$ with the desired property (but it may not converge).  Then ....[/quote]

[b]How do we find that sequence[/b],what theorems ,axioms or whatever will help us ,find or form that sequence??

Thanks for the La Tex references,these links are always helpfully.

Suppose you cannot find some $ x_1$ such that $ |f(x_1)| > 1$.  Then what does this say about $ f$?

(A much broader comment: you're working here with a very small amount of information.  All you have is a function with three conditions on it.  Since you probably only have theorems about real functions, in fact you have only two useful conditions on $ f$.  So, you better use each of them somewhere!  And there are only so many ways to make use of the fact that, say, $ f$ is not bounded.)

O.K

Because f is not bounded over [a,b] we can prove: For all n belonging to natural Nos ,there exists an ,x belonging to [a,b] and such that |f(x)|>n.

For n= 1 ,we pick $ x_{1}$,such that |f($ x_{1}$)|>1.

For n=2 ,we pick $ x_{2}>x_{1}$,such that |f($ x_{2}$)|>2.

Suppose we have found:

$ x_{1}<x_{2}<x_{3}.......................<x_{n}$ all in [a,b] and such that |f($ x_{n}$)|>n .

We select $ x_{n\plus{}1}>x_{n}$ and such that |f($ x_{n\plus{}1}$)|>n+1 .We can thus proceed inductively to construct the sequence {$ x_{n}$},which is increasing and bounded from above by ,a.Hence there exists a,y in [a,b] and such,$ \lim_{n\rightarrow\infty}{x_{n}}\equal{}y$.

So we have proved the existence of a sequence {$ x_{n}$} in [a,b] such that |f($ x_{n}$|>n and $ \lim_{n\rightarrow\infty}{x_{n}}\equal{}y$.

[b]Correct[/b] ??

Frankly speaking ,i was expecting some criticism for the above inductive method ,particularly from JBL,SINCE he was the one to force me to form the said sequence ,in a way so to speak

.[b]Personally i do not trust this method which i picked from an analysis book[/b].

HOWEVER if some body could base that method on an axiom ,or theorem i will be very glad to know.

The alternative method using the axiom of Choice seams more rigid.

I THANK anybody who would care to comment on the matter

You are correct that the method does not work.  Consider, for example, the function $ f: [0, 1] \to {\bf R}$ given by $ f(0) \equal{} 0$, $ f(x) \equal{} \frac {1}{x}$ for $ x \neq 0$.  It should be clear that no matter what value of $ x_1$ you pick, you won't be able to choose the entire sequence.  (You'll get stuck in at most $ \frac {1}{x_1}$ moves.)

Instead, go back to what I said in the first post: choose the sequence first, without regard to convergence; then show (using that sequence) how to find a convergent sequence with the same property.  (A suggestion: the fact that the domain of $ f$ is a closed interval is probably important, since there are counter-examples for, say, open intervals.  Like the function $ g: (0, 1) \to {\bf R}$ defined by $ g(x) \equal{} \frac {1}{x}$.)

[b]Are you suggesting the following?[/b]:

Using the formula ,for all natural Nos ,n there exists an x belonging to the interval [a,b] and such that,|f(x)|>n,

we can form the following formula inductively:

For ,n=1 ,we pick an $ x_{1}$ belonging to [a,b] and such that |f($ x_{1})|$>1.

After k steps we have formed the sequence :$ x_{1},x_{2},.............x_{k}$ and such that |f($ x_{k}$)|>k

WE proceed then inductively ,for k+1 to pick $ x_{k\plus{}1}$ belonging to [a,b] and such that ,|f($ x_{k\plus{}1}$)|>k+1 .

SO we have formed the sequence,{$ x_{n}$} belonging to [a,b] and such that |f($ x_{n}$)|>n.

Now, that sequence is bounded and hence according to the Bolzano Weiestrass theorem has an accumulation point ,v ,in [a,b]

And thus we proceed now to form its following convergent subsequence ,inductively:

We pick   $ n_{1}$ such that |$ x_{n_{1}}$-v|<1/1.We then pick $ n_{2}>n_{1}$ such that |$ x_{n_{2}}$-v|<1/2.

Suppose that we have found :

$ n_{1}<n_{2}.....................<n_{k}$

such that |$ x_{n_{i}}$-v|<1/i for i=1,2,3................k. We select $ n_{k\plus{}1}>n_{k}$ such that |$ x_{n_{k\plus{}1}}$-v|<1/(k+1).

We can thus proceed inductively to construct the subsequence $ x_{n_{1}},x_{n_{2}},.........$,and this subsequence converges to ,v,because given ε,there exists m such that 1/m<ε,and for all $ n_{k}\geq m$ we have:

$ |x_{n_{k}}\minus{}v|<\frac{1}{n_{k}}\leq\frac{1}{m}<\epsilon$

So we have formed a sequence ( the subsequence of the sequence {$ x_{n}$}) ,{$ y_{n}$} which has a limit v ,in [a,b] and such that |f($ y_{n}$)|>n for all natural Nos ,n

[quote="τρικλινο"]Are you suggesting the following?[/quote]
Yes.  I might have phrased parts of the argument somewhat differently; in particular, I consider much of what you did to be the following immediate corollary of Bolzano-Weierstrass (or equivalently Heine-Borel, or possibly the definition of compact, depending how you define it): any sequence in a compact space has a convergent subsequence.  So, what you've done appears correct, although it could perhaps be shortened.

I would also strongly recommend that for readability, you place entire formulas in math mode (rather than just the snippets where you have to use underscores, for example).  The formula $ |f(x_1)| > 1$ is much nicer-looking than the formula |f($ x_1)$| > 1, in my opinion.  Broader use of \frac and things like \ldots might help, too.

O.K ,but the point as i mentioned in my post #6 [b]i do not trust the inductive method because there is nothing to base that method on[/b].

Do you have anything in mind to justify the method??

What about the axiom of induction do you find problematic?

Certainly not.
But in using the axiom of induction,when we go from the k step to the, k+1 ,to prove: that if the relation holds for [b]k[/b] then it also holds for [b] k+1[/b], we actually [b]prove[/b] that and not because we say so.

Correct me if i am wrong

The function $ f$ is unbounded.  Therefore, by definition, there is some $ x_k$ such that $ |f(x_k)| > k$.  That's the proof that we can find the desired $ x_k$.   :maybe:

Can you think of any other method or procedure,[b]based on axioms or theorems[/b] that will help us form the above sequences


Is it possible to get a Tex source of a PDF-file by using Texnic center?

I'm afraid not. It's rather like changing a photo of the text back to text. You can save the text in your pdf viewer but the layout and the mathematics will not be there.  So always back up the original tex file.


In the following equation, what is the sum of positive integer $ a$ and $ b$?

\begin{align*}\frac{7}{22}=\frac{1}{a+\frac{1}{b}}\\
\end{align*}

[hide="Answer"]10[/hide]

simplifying the rhs, you get b/(ab+1), which is equal to 7/22.
here i did a lucky guess, and let b=7 and solving you get a=3, both of which are obviously integers. so the answer is 7+3=10

too lazy to latexify it....

so flip over each one to get a+1/b=22/7.
With ones like this, it doesn't matter what values for a or b you choose, because it will always add up to a specific number. so make a=3. Then 1/b=1/7 so b=7

3+7=10.

[hide]We this:
$ \frac {7}{22} \equal{} \frac {1(7)}{(a \plus{} \frac {1}{b})7}$

We know that $ 22\div7 \equal{} \frac {22}{7} \text{ or } \approx\pi$

To get $ \pi$, we have to do $ 3 \plus{} \frac {1}{7}$ because $ \frac {22}{7} \equal{} 3\frac {1}{7}$

So: a=3 and b=7, $ 3 \plus{} 7 \equal{} \boxed{\boxed{\boxed{\boxed{10}}}}$
[/hide]

EDIT: Darn! Ihatepie beat me!  Sorry about my confusing wording! :blush:

Similar question: find the positive integers $ a$, $ b$, $ c$, and $ d$ so that $ a \plus{} \frac {1}{b \plus{} \frac {1}{c \plus{} \frac {1}{d}}} \equal{} \frac {489}{151}$

by similar logic, you find that a=3,then b=3, and i dont have my calculator with me so finish it by yourselves :P

oops, that's not really what I meant there, I'll edit it

stupid arithmetic error

this is what I got
[hide]$ a \equal{} 3, b \equal{} 4, c \equal{} 5,$ and $ d \equal{} 7$[/hide]
Again like Alan said, this is derived by simply expanding the original logic.
BTW: I did all of this with just pencil and paper.


You do know that the B date isn't until February 25th, right?

:rotfl:

Did you not notice that the forums aren't locked?

I did notice that, but DPatrick wasn't online...

The forum would have been locked the previous night. DPatrick would never have forgotten to do that if today was the B date.


The park is open every day, but the Kiddy Train runs at different times, as indicated with Ys in the schedule below.  If Kendra will randomly choose a day to go to the park, what is the probability that she will be able to ride the train at 4 p.m. or later?  Express your answer as a common fraction.

$ \begin{tabular}{|c|c|c|c|c|c|c|c|}\hline
 \textbf{Times} & \textbf{Sun} & \textbf{Mon} & \textbf{Tue} & \textbf{Wed} & \textbf{Thu} & \textbf{Fri} & \textbf{Sat} \\ \hline
 1: 00\text{ p.m}. & y & & & & & & y \\\hline
 2: 00\text{ p.m}. & y & y & y & y & y & y & y \\\hline
 3: 00\text{ p.m}. & y & y & y & y & y & & y \\\hline
 4: 00\text{ p.m}. & y & & & & & y & y \\\hline
 5: 00\text{ p.m}. & y & & & & & y & y \\\hline
 6: 00\text{ p.m}. &  & & & & & y & y \\\hline

\end{tabular}$

Since 3 out of the 7 days have a train running later than 4:00 p.m., so the answer is 3/7.

In this problem it is asking how many [b]days[/b] the girl can get on a train later that 4p.m. not hours. Since there are 3 days that have a train that goes at 4 or after, the numerator is 3. Since there are 7 days in a week the denominator is 7, thus leading to the answer 3/7.

------------------------------------------------------------------------------------------------------------
It's a tricky problem because people think that she is choosing an hour out of a day that has a train ride after 4 but it's what days have a ride 4 or after available.

Hope this helped and if you want more solutions rate this high!!


Here is a quite difficult problem, that I can't finish:
    Given $ a_1,a_2,...,a_n$ real numbers such that $a_1^2+a_2^2+...+a_n^2=1$ find the maximum value of $\sum_{1\leq i<j\leq n}{ a_ia_j(a_i^2+a_j^2)}$. By the way, what is happening to the formulas I write? Can anyone edit the message? I have now idea what's happening.

I think very very difficult to find the maximum if $n\ge5$ althought It' exist.

Besidet,It's quite nice if $n\le 4$.

Indeed, for n at least 5 it seems impossible to attack. For n=4, do you have nice solution?

[quote="Vasc"]Let $C_n$ be the required maximum.
I found $C_3=\frac{2}{3}$, $C_4=\frac{3}{4}$ and $C_n \ge \frac{n-1}{2\sqrt{2(n-2)}}$ for $n \ge 4$
Also, I conjecture that $C_5=\frac{13}{16}$.
In fact, we have to find the minimum $C$ such that

$C(\sum x_i^2)^2+\sum x_i^4 \ge (\sum x_i )(\sum x_i^3)$.[/quote]

this looks like suranyi inequality..  :?
and vasc, for n=3, wt if a_1=a_2=1, a_3=0?

Dear Vasc.
Do you have a nice solution for $n=5$?

I haven't a solution for the conjectured inequality in case $n=5$.
But equality occurs for $x_1=x_2=x_3=x_4=1$ and $x_5=2$.


real analysis
integration
inequalities
function
real analysis unsolved

How to prove that if $f,g\in L^{1}\cap L^{2}$, then $f*g=\int f(x-y)g(y)\,dy$ is continuous? Does this follows from the fact that $(f+g)^{2}-(f-g)^{2}=4fg$?

I don't see the use for that, nor do I see why $f,g\in L^{1}$ would be needed. Let's suppose simply that $f,g\in L^{2}.$ Our main tool will be the Cauchy-Schwarz inequality:

For any $x,\ |f*g(x)|\le\int|f(x-y)|\,|g(y)|\,dy\le\|f\|_{2}\|g\|_{2}.$

What we need is that "nice" functions are dense in $L^{2}.$ Find sequences $f_{n}$ and $g_{n}$ of continuous compactly supported function such that $\|f_{n}-f\|_{2}\to0$ and $\|g_{n}-g\|\to0.$ Note that $f_{n}*g_{n}$ will be continuous and compactly supported.

For each $x,$

$|f_{n}*g_{n}(x)-f*g(x)|\le|(f_{n}(x)-f(x))g_{n}(x)|+|f(x)(g_{n}(x)-g(x)|$

$\le\|f_{n}-f\|_{2}\|g_{n}\|_{2}+\|f\|_{2}\|g_{n}-g\|_{2}.$

Since $\|g_{n}\|_{2}$ is bounded, this goes to zero - independently of $x.$ Thus $f_{n}*g_{n}$ tends uniformly to $f*g.$ The uniform limit of continuous functions in continuous, so $f*g$ is continuous. Furthermore, we may also conclude that $f*g$ tends to zero as $x\to\infty.$

By essentially the same proof (using Hölder instead of Cauchy-Schwarz) we get that if $f\in L^{p}$ and $g\in L^{q}$ for $1<p<\infty$ and $\frac1p+\frac1q=1,$ then $f*g$ is continuous and tends to zero at infinity.


With $ a_1,a_2,..,a_n$ satisfy : $ 1 \plus{} a_1cosx \plus{} a_2cos2x \plus{} ... \plus{} a_ncosnx \geq 0$ for all x
Prove that: $ a_1 \plus{} a_2 \plus{} ... \plus{} a_n \leq n$

I suppose we can put x=2pi r/(n+1) and then sum over r=1 to n.


The nobel prizes are being awarded this week! Information can be found via the [url=http://www.nobelprize.org]Nobel Prize Website[/url]. \[\text{Medicine}\] "for their discovery of RNA interference - gene silencing by double-stranded RNA"

[b]Andrew Z. Fire[/b]
(1/2 of prize)

USA

MIT; Stanford University School of Medicine

[b]Craig C. Mellow[/b]
(1/2 of prize)

USA

Harvard University; University of Massachusetts Medical School \[\text{Physics}\] "for their discovery of the blackbody form and anisotropy of the cosmic microwave background radiation"

[b]John C. Mather[/b]
(1/2 of prize)

USA

NASA Goddard Space Flight Center

[b]George F. Smoot[/b]
(1/2 of prize)

USA

UC Berkeley \[\text{Chemistry}\] [b]Roger D. Kornberg[/b]

USA

Stanford University \[\text{Economics}\] [i]announced Monday, October 9 at 11 a.m. GMT[/i] \[\text{Peace}\] [i]announced Friday, October 13 at 9 a.m. GMT[/i] \[\text{Literature}\] [i]date to be announced is to be announced[/i]

[quote="joml88"] \[\text{Physics}\] "for their discovery of the blackbody form and anisotropy of the cosmic microwave background radiation"

[/quote]

google doesn't help. :mad:  :(  if is their research somewhere on the net, could anyone give me a link. please.
thanks.  :)

I think [url]http://www.nasa.gov/centers/goddard/news/topstory/2006/gruber_award.html[/url] gives a decent overview of it.

I don't really care about it, except for literature (i usually read at least one book of a noblist) and physics. I don't follow any of categories awarded (except for theoretical physics) and prize in literature is always such exciting, can't explain why.

This might be a bit late but the Nobel Prize in literature was given to Orhan Pamuk. This is a great thing! Since Churchill was given the prize we've had, many times, political messages involved with the prize. The same thing goes for Orhan Pamuk.
Orhan Pamuk was born in Turkey and is special since he has been demanding that the genocide on the Armenian/Syriac/Assyrian/Chaldean (not only Armenian!) people will be admitted. By supporting this he was almost thrown in jail in Turkey. But the Turkish government got scared (of EU) and stopped this.
I'm glad that this question comes back in focus.
[b]Like Adolf Hitler said before going after the yews and gypsies (and a few more): [i]who remembers [the genocide on the Armenians][/i]?[/b]


geometry
rectangle
combinatorics unsolved
combinatorics

It is given a $20*20$ grid, we  remove from it  (with sides on the grid) a $20*1$, a $19*1$, a $18*1$...a $1*1$ rectangles. Prove that at least 85 $2*1$ rectangles can be removed from the remainder.

I dont know how hard it is, i just know i got no clear idea.

As far as I know, it was the last at the City tour and no one solved it.
I'll give you the key to the solution I found:

Let the black squares be the cut ones, the rest-white. We try to divide the white squares into $1x2$ dominoes.
That ones which enter a domino we call used, the others- unused or lost.
First trial:
Divide the big square into $2x2$ squares. Now we try to cut dominoes form every $2x2$ square. Since there are $400-(1+2+\cdots20)=190$ white squares
We must prove that at most $21$ squares are left unused (the we have at least 169 squares used, so at least 84.5 so 85 dominoes).
Note that if we lose a square in $2x2$ there  must be the end-square of some cut stripe adjacent to the lost one(prove it youself). For example:
UU  (U means used, L-lost, B-black).
LB Then $B$ is the end-square of some stripe.
So if we assume the contrary we get at least $22$ end-squares(of type 1).

Second trial:
now if we divide the central $18x18$ square into $2x2$ squares, the borders into $1x2$ rectangles (except the corners) and try the same method, we get $18$ more end-squares of type 2. One easily proves that if a square enters both types it is exactly the $1x1$ square. So at most one square enters the both types.
So totally we get at least 39 end-squares. But there are exactly $39$ end-squares.
So everywhere equalities hold. Hence in each of two tries we performed we cut exactly $84$ dominoes.Particularly the corners are lost.
That means that the cut stripes do not contain the corners.
Considering the cut $1x20$ stripe. It cuts the borders in some black square A. Consider Bthe closest black square to the right of it. Then wthe immediate to the right of A is an unused square, and the immediate to the left of B is also an unused one(because we supposed equalities hold everywhere). Then we can divide the portion between $A$ and $B$ into dominoes, increasing by $1$ the number of dominoes so making it $85$. (I mean:
      .....AL....LB then we divide L..L into 1x2 squres, and the 2 lost squares marked $L$ become used.)
QED


Pythagorean Theorem
geometry
cyclic quadrilateral
power of a point

How do I know when to use Ptolemy's Theorem?  I can't "see" it like I can "see" the Pythagorean Theorem or 30-60-90 triangles.  Also, how do I know if I should use cyclic quadrilaterals?

Thanks in advance!

You know when to use Ptolemy's Theorem when you have a cyclic quadrilateral and you need a diagonal or side length of this quadrilateral.  Just like everything else, seeing cyclic quadrilaterals and knowing when to use them comes with experience, although when you find them they usually come in very handy.

It also doesn't hurt to draw a circle around the shape, as this can sometimes give you inscribed angles and the "Power of a Point" theorems.


probability
expected value
real analysis
real analysis unsolved

Hi, I have some problems to solve, among which this one:

Let $X_1, X_2, ..., X_n, ...$ be independent random variables with the same exponential distribution and the expected value 3. Let $N$ be a random variable, independent from variables $X_1, X_2, ..., X_n, ...$ with Poisson distribution and the expected value 2. Let
\[ Z_n= \{ \begin{array}{ccc} \dfrac{1}{N+1} \sum_{i=1}^N iX_i & when & N>0 \\ 0 & when & N=0.\end{array} \]
Calculate $VarZ_n$.

Please, if you have any ideas - tell me! :))

Seems pretty straightforward to me: just compute the expectation and the expectation of the square of $\sum_{i=1}^n iX_i$ for each particular $n$, multiply them by $e^{-2}\frac{2^n}{n!}$ and add them up. You'll have to find a few sums like $\sum_{n\ge 0} \frac{2^n}{n!}n^k$ with small integer $k$ but this is not very difficult.


Can we use animation in our sigs? If so, what are the dimensions for it?

can we even post animations?

testing...
<img src="http://img.photobucket.com/albums/v426/brianlin11/guy_swordmaster_critical1.gif">


inequalities
AMC
USA(J)MO
USAMO
inequalities unsolved

Alright here is solution, it's a little bit complicated, but the idea is simple:
[hide]
Suppose, that there is such set $(x_1,x_2,...,x_{n+1})$, for which $\frac{1}{x_1+1}+\frac{1}{x_2+1}+...+\frac{1}{x_{n+1}+1}=1$ and $x_1x_2...x_{n+1}<n^{n+1}$
First of all $x_ix_j\geq 1$, otherwise $\frac{1}{x_i+1}+\frac{1}{x_j+1}>1$(I won't give prove to this because it's very easy)

Let's prove the Lemma:
If $x\leq x_1\leq y_1\leq y$ and $xy=x_1y_1\geq 1$ then $\frac{1}{1+x}+\frac{1}{1+y}\geq \frac{1}{1+x_1}+\frac{1}{1+y_1}$

After multiplying both sides of inequality by $(1+x)(1+y)(1+x_1)(1+y_1)$ and simplifying it will be equal to
$x_1+y_1+x_1y_1(x+y)\geq x+y+xy(x_1+y_1)$
which is equal to $(x_1+y_1-(x+y))(1-xy)\geq 0$
$xy\geq 1$ so it is equal to $x_1+y_1\leq x+y$ Put $x_1=x\alpha$ and $y_1=y/\alpha$, where $\alpha\geq 1$
$x+y\geq x\alpha+y/\alpha$
$(1-\alpha)(x-y/\alpha)\geq 0$
$(\alpha-1)(y/\alpha-x)\geq 0$, which is true, because $\alpha\geq 1$, and $y_1=y/\alpha\geq x$
Lemma is proved.

Now let $s={x_1x_2...x_{n+1}}^{1/{n+1}}$ then suppose that there is $x_i>s$ and $x_j<s$ than as proved in Lemma, we can replace $(x_i;x_j)$ by $(x_i';s)$ (or by $(s;x_j')$), and the summ $\frac{1}{x_1+1}+\frac{1}{x_2+1}+...+\frac{1}{x_{n+1}+1}$ will only decrease. So we can change $(x_1,x_2,...,x_i,...,x_j,...,x_{n+1})$ to $(x_1,x_2,...,x_i',...,s,...,x_{n+1})$. Then we do this operation for the set without $s$ and so on until we get to the set $(s;s;s;s;...;s)$. So $\frac{1}{x_1+1}+\frac{1}{x_2+1}+...+\frac{1}{x_{n+1}+1}\geq \frac{n+1}{1+s}$
But we supposed that $x_1x_2...x_n<n^{n+1}$, so $s<n$ and $\frac{n+1}{1+s}>1$ from which we get a contradiction because of $1=\frac{1}{x_1+1}+\frac{1}{x_2+1}+...+\frac{1}{x_{n+1}+1}\geq \frac{n+1}{1+s}>1$[/hide]

[quote="cosinerburc"]xi are reals such that 1/(x_1+1)+1/(x_2+1)+...+1/(x_(n+1)+1)=1.Prove that x_1.x_2...x_(n+1)>=n^(n+1).
i think jensen's inequality will work but i could not use it.[/quote]

That is equivalent to USAMO 1998 problem 3 http://www.mathlinks.ro/Forum/viewtopic.php?p=343867#p343867

Let $y_j=\frac{1}{x_j+1}$ then $\sum y_i=1$ and $x_j=\frac{1-y_j}{y_j}$.  Then this inequality becomes quite easy.


linear algebra
matrix
vector
analytic geometry
combinatorics proposed
combinatorics

[quote="Leon"]For integers  $n\geq 3$ determine the number of sequences $\left(c_{1},c_{2},\cdots,c_{n}\right)$, where $c_{i}\in \{0,1,2,\cdots,9\}$ such that in every three successive terms, there are at least two equal. [/quote]
It's the problem 4 form current (11.09.2006 - 9.10.2006) series of 1st round of Polish MO. The participants have to send their [b]own[/b] solutions, so any help in solving the problems or posting solutions is prohibited.
[b]Leon and Moderators, please delete the whole topic or at least the solution, because it can spoil the competition.[/b]

Yes; I received a personal message (too) from a recently registered user asking for the solution to that problem. I didn't answer because it is not the first experience I have with people trying to cheat in contests from home.

[quote="Rzeszut"] It's the problem 4 form current (11.09.2006 - 9.10.2006) series of 1st round of Polish MO. The participants have to send their [b]own[/b] solutions, so any help in solving the problems or posting solutions is prohibited.
[b]Leon and Moderators, please delete the whole topic or at least the solution, because it can spoil the competition.[/b][/quote]

Sorry I did'nt know that the problem was given in an official competition.
I received a personal message from a recently registered user asking for the solution to that problem and so I have posted my solution.
Now  I have deleted the solution, but I did'nt come to delete the whole topic.
I think you should delete also the link in your reply.

Yes, you're right. I deleted it. Anyway, the topic should be locked, because now anyone can post a solution and if it appears, anyone will be able to read it. Fortunately, I solved the problem before and my solution was the same (only the marks were different).

[quote="Rzeszut"]The 1st series ended. We can now discuss this problem.[/quote]

Here is my solution

[hide]Let $x_{n}$ the number of sequences $(c_{1},c_{2},\dots,c_{n})$, where $c_{i}\in \{0,1,\dots,9\}$ such that in every three successive terms, there are at least two equal.
Let $y_{n}$ (resp. $z_{n}$) the number of sequences verifying the required condition with the last two terms equal (resp. different). We have obviously

(1) $x_{1}=10 \quad , \quad x_{2}=100$

(2) $x_{n}=y_{n}+z_{n}$

(3) $y_{n}=x_{n-1}$

Furthermore we have

(4) $z_{n}=z_{n-1}+9y_{n-1}$

because the number of sequences of the form $(.,.,.,x,y,z)$ are $z_{n-1}$ and the number of those of the form $(.,.,.,x,x,y)$ are $9y_{n-1}$.
From (2),(3) and (4) follows that

(5) $x_{n}=2x_{n-1}+8x_{n-2}$

Solving the recurrence relation (5), we obtain easily that \[x_{n}=5\cdot4^{n}+5\cdot(-2)^{n}\] and we are done. [/hide]

[quote="Leon"]Let $y_{n}$ (resp. $z_{n}$) the number of sequences verifying the required condition with [b]at least two terms equal[/b] (resp. different). [/quote]
Did you mean [b]the last two terms equal[/b]?
From the moment when we have $y_{n}=y_{n-1}+z_{n-1},\; z_{n}= 9y_{n}+z_{n}$ with beginning $(y_{3},z_{3})= (100,180)$ I did it more brutally: I wrote that $\left[\begin{array}{c}{y_{n}}\\{z_{n}}\end{array}\right]= \left[\begin{array}{cc}{1}&{1}\\{9}&{1}\end{array}\right]\cdot \left[\begin{array}{c}{y_{n-1}}\\{z_{n-1}}\end{array}\right]$, so $\left[\begin{array}{c}{y_{n}}\\{z_{n}}\end{array}\right]= \left[\begin{array}{cc}{1}&{1}\\{9}&{1}\end{array}\right]^{n-3}\cdot \left[\begin{array}{c}{100}\\{180}\end{array}\right]$ and sice $\left[\begin{array}{c}{100}\\{180}\end{array}\right]= \left[\begin{array}{cc}{1}&{1}\\{9}&{1}\end{array}\right]^{2}\cdot \left[\begin{array}{c}{10}\\{0}\end{array}\right]$, the recurrence simplifies to $\left[\begin{array}{c}{y_{n}}\\{z_{n}}\end{array}\right]= \left[\begin{array}{cc}{1}&{1}\\{9}&{1}\end{array}\right]^{n-1}\cdot \left[\begin{array}{c}{10}\\{0}\end{array}\right]$. Then I diagonalised the matrix $\left[\begin{array}{cc}{1}&{1}\\{9}&{1}\end{array}\right]= \left[\begin{array}{cc}{1}&{1}\\{3}&{{-3}}\end{array}\right]\cdot \left[\begin{array}{cc}{4}&{0}\\{0}&{{-2}}\end{array}\right]\cdot \left[\begin{array}{cc}{{\frac12}}&{{\frac12}}\\{{\frac16}}&{{-\frac16}}\end{array}\right]$ and obtained a formula for its $n-1$-th power: $\left[\begin{array}{cc}{1}&{1}\\{9}&{1}\end{array}\right]^{n-1}= \left[\begin{array}{cc}{1}&{1}\\{3}&{{-3}}\end{array}\right]\cdot \left[\begin{array}{cc}{4^{n-1}}&{0}\\{0}&{{(-2)^{n-1}}}\end{array}\right]\cdot \left[\begin{array}{cc}{{\frac12}}&{{\frac12}}\\{{\frac16}}&{{-\frac16}}\end{array}\right]$. Then we multiply this product by the vector $\left[\begin{array}{c}{10}\\{0}\end{array}\right]$ and sum the coordinates of the obtained vector getting $5\left(4^{n}+(-1)^{n}\right)$.
The diagonalisation method is useful in recurrences with $2$ numbers, but except of some nice cases (like this problem) it is extremely ugly.

[quote="Rzeszut"][quote="Leon"]Let $y_{n}$ (resp. $z_{n}$) the number of sequences verifying the required condition with [b]at least two terms equal[/b] (resp. different). [/quote]
Did you mean [b]the last two terms equal[/b]?
[/quote]

Yes, of course !
I have edited the the solution.


function calculus derivative integration quadratics algebra polynomial

I have an approximation for the variation of gravity with height above the surface of the earth in the form of a potential function:

$ V(z) \equal{} mgz(1 \minus{} \frac {z}{R})$

Where z is the height above the surface, R is the radius of the earth, g the acceleration due to gravity, m the mass of the object in question.

The question is to find the maximum height attained by an object that is fired from the surface of the earth with $ v(0) \equal{} v_0$ in the z direction.

Here is what I have so far:

$ F(z) \equal{} \frac { \minus{} dV(z)}{dz} \equal{} \frac {2mgz}{R} \minus{} mg$

setting this equal to m times the second derivitive of z wrt t, and knowing that

$ \frac {d^2z}{dt^2} \equal{} \frac {dv_z}{dt} \equal{} \frac {dv_z}{dz}\frac {dz}{dt}$ you get:

$ \frac {2mgz}{R} \minus{} mg \equal{} mv_z\frac {dv_z}{dz}$

Separating and cancalling m, you get:

$ g\int_0^z\left(\frac {2z}{R} \minus{} 1\right)dz \equal{} \int_{v_0}^vvdv$

yielding:

$ \frac {1}{2}v^2 \minus{} \frac {1}{2}v_0^2 \equal{} \frac {gz^2}{R} \minus{} gz$

Solving for $ v^2$ and setting v = 0:

$ 0 \equal{} \frac {2gz^2}{R} \minus{} 2gz \plus{} v_0^2$

Now, if you apply the quadratic equation to solve for z, you get utter nonsense: (try $ v_0 \equal{} 10\frac {m}{s}$). So I must have gone wrong somewhere, but I have redone this repeatedly, with no luck. Any suggestions?

There is an escape velocity. $ 10$ meters per second is far below it.

Both roots of that quadratic are positive, and the smaller root is the maximum height.

right, but if you plug in 10m/s, you get as the root: (If R=6,400,000m, g=9.81m/s^2)

~5m, and ~6,399,990m

Both are nonsense, obviously.

Why is 5 obviously nonsense?  I mean, it sounds fairly reasonable to me.  (I haven't looked at your computations at all.)

If you launch something at 10 metres per second, it doesnt make sense for it to fall down again after rising only 5 metres in earth's gravity.

Or... maybe it does ^_^

Hmm, let me try a different approximation and get back...

And EDIT: wow, that actually works, I'm an idiot. If we take the potential function to be simple V=mgz, you get 5.1m as the top height with v0 = 10. Guess it worked after all...

Heh, thanks anyway

Just a couple of thoughts:

Major league baseball pitchers throw at 40 m/s. By those standards, 10 m/s is not throwing very hard. On the other hand, if you think like a juggler, 5 meters high is higher than you'd want to throw your balls or clubs unless you're doing something pretty unusual.

Now if you want to work out what escape velocity is --- you need a better approximation for the potential. Use the real formula, which is $ V(z) \equal{} mgR^2\left(\frac1R \minus{} \frac1{R \plus{} z}\right).$ To send the object arbitrarily far away take the limit as $ z\to\infty$ to get that the energy required is $ mgR.$ Equate that to $ \frac12mv^2$ to get $ v \equal{} \sqrt {2gR}.$ Using roughly $ g \equal{} 10\,\text{m}/\text{s}^2$ and $ R \equal{} 6.4\times 10^6\,\text{m},$ you get roughly $ 1.1\times 10^4\,\text{m}/\text{s}$ for escape velocity.

I'll add that the formula you are using, namely $ V(z) \equal{} mgz\left(1 \minus{} \frac zR\right),$ is the second degree Taylor polynomial centered around $ z \equal{} 0$ for the more accurate formula. The more usual $ V(z) \equal{} mgz$ is the first degree Taylor polynomial. Taylor polynomials are pretty good if you stay close to the center of approximation; they may be quite bad if you go a long ways away.

That formula should be $ V(z) \equal{} mgR^2\left(\frac1R \minus{} \frac1{R \plus{} z}\right)$; it does need to have energy units.

[color=green][Which is what it always said on my scratch paper. Fixed now. - K.M.][/color]


Hi,

Let
\[P(x)=x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}\]
and
\[Q(x)=x^{m}+b_{m-1}x^{m-1}+\cdots+b_{1}x+b_{0}\]
where $a_{i},b_{i}\geq 0$.

Prove that if $m=n$ or $m$ and $n$ are both odd, then no polynomials $P(x)$ and $Q(x)$ satisfy :
\[P(x)Q(x)=x^{n+m}+x^{n+m-1}+\cdots+x+1.\]

From
\[P(x)Q(x)=\frac{x^{m+n+1}-1}{x-1}=\prod_{k=1}^{m+n}(x-exp(\frac{2\pi ki}{m+n+1}))\]
we get, that if m and n both odd, then P(x) and Q(x) can not be real polinom. If m=n, then m and n even (because only one real root x=-1 had P(x)Q(x) ). We don't use that $a_{i},b_{i}$ are non negative.
I don't shure, that it is true for m=n=4k.
Let
$P(x)=(x-c^{2})(x-c^{-2})(x-c^{8})(x-c^{-8}),$
$Q(x)=(x-c^{4})(x-c^{-4})(x-c^{6})(x-c^{-6}),$
were $c=exp(\frac{\pi i}{9})$.
It give
$P(x)=x^{4}+2x^{3}(cos\frac{\pi}{9}-cos\frac{2\pi}{9})+2x^{2}(1-2cos\frac{\pi}{9}cos\frac{2\pi}{9})+2(cos\frac{\pi}{9}-cos\frac{2\pi}{9})+1$
and
$Q(x)=x^{4}+x^{3}(1-2cos\frac{4\pi}{9})+2x^{2}(1-cos\frac{4\pi}{9})+x(1-2cos\frac{4\pi}{9})+1.$
All coefficients are positive.


Start with a semicircle w/ diameter 2. Then inscribe another semicircle in it, with their diameters parallel. Repeat this process infinitely. What is the area of the shaded region?

Well clearly the radius of the second semicircle is (1/sqrt[2]) of that of the first one.  So the area is 1/2.  This gives us the shaded area to be (pi/2)(1-1/2+1/4-1/8...).  That gives us (pi/2)(2/3), which is pi/3.


Pedro has twelve more pencils than Peter. Peter has nine more pencils than Paul. Peter, Pedro and Paul decide to split the pencils so each of the three will have the same number of pencils. How many pencils will Pedro lose?

Suppose Paul has p pencils.

p+21, p+9, p. Average = p+10.
Thus Pedro loses 11 pencils.


Letx,y,z,t  are  real  numbers andx+y+z+t=0,$ {x^2} \plus{} {y^2} \plus{} {z^2} \plus{} {t^2}\equal{}1$, proof  that$ 0\geq{xy} \plus{} {yz} \plus{} {zt } \plus{} {zt}\geq \minus{} 1$

[quote="wangzishan"]Letx,y,z,t  are  real  numbers andx+y+z+t=0,$ {x^2} \plus{} {y^2} \plus{} {z^2} \plus{} {t^2} \equal{} 1$, proof  that$ 0\geq{xy} \plus{} {yz} \plus{} {zt } \plus{} {zt}\geq \minus{} 1$[/quote]
It must be: $ 0 \ge xy\plus{}yz\plus{}zt\plus{}tx \ge \minus{}1$
We have: $ xy\plus{}yz\plus{}zt\plus{}tx\equal{}(x\plus{}z)(y\plus{}t)\equal{}\minus{}(x\plus{}z)^2 \le 0$
And $ \frac{(x\plus{}y)^2\plus{}(y\plus{}z)^2\plus{}(z\plus{}t)^2\plus{}(t\plus{}x)^2}{2} \ge 0$,so $ xy\plus{}yz\plus{}zt\plus{}tx \ge \minus{}1$
We have Q.E.D


evaluate:

sum for even k from k = 2 to infinity of 1/(k!) (I would have texed it but I forgot what the command setup for summation was and i dont have time to look it up... currently supposed to be in class.)

If you don't know the series for $\cosh x$ offhand, but you do know the series for $e^x$, then we basically just want to get rid of the odd terms in the series expansion of $e$.  Using the series for $-e^{-1}$ and $e$ together and noticing that each even term is counted twice, we pretty much have the desired series.  Now get rid of the extra $1$ and we're done.