geometry
circumcircle
vector
geometry proposed

Let $H$ be the orthocenter of a triangle $\triangle ABC$ and denote with $\gamma$ its nine-point circle. Prove that for every $P \in \gamma$ we have

\[
PA^2  + PB^2  + PC^2  + PH^2  = 4R^2
\]

Well, let's prove a more general assertion:

[color=blue][b]Theorem 1.[/b] If ABC is a triangle with orthocenter H and nine-point center N, and if P is an arbitrary point in the plane, then

$PA^2+PB^2+PC^2+PH^2=4\cdot PN^2+3R^2$,

where R is the circumradius of triangle ABC.[/color]

[i]Proof of Theorem 1.[/i] One could easily verify Theorem 1 using vectors and the fact that the nine-point center N of triangle ABC is the centroid of the four points A, B, C and H, but there is also a more basic proof:

[color=blue][b]Lemma 2.[/b] If T, U and V are any three points in the plane, and M is the midpoint of the segment UV, then $2\cdot\left(TU^2+TV^2\right)=4\cdot TM^2+UV^2$.[/color]

[i]Proof of Lemma 2.[/i] Since the point M is the midpoint of the segment UV, the segment TM is a median of triangle TUV. Thus, by the well-known formula for the length of a median in a triangle, we have $4\cdot TM^2=2\cdot TU^2+2\cdot TV^2-UV^2$; thus, $4\cdot TM^2+UV^2=2\cdot TU^2+2\cdot TV^2=2\cdot\left(TU^2+TV^2\right)$, and Lemma 2 is proven.

Now we come to the actual proof of Theorem 1:

Let R be the midpoint of the side BC of triangle ABC, let D be the foot of the altitude of triangle ABC issuing from the vertex A, and let X be the midpoint of the segment AH. It is well-known that all the three points R, D and X lie on the nine-point circle of triangle ABC; moreover, since < XDR = 90° (as the line XD is an altitude of triangle ABC), the segment XR is a diameter of this nine-point circle. Thus, the midpoint of this segment XR is the center of the nine-point circle, i. e. the point N. Hence, by Lemma 2, applied to the points T = P, U = X, V = R and M = N, we have $2\cdot\left(PX^2+PR^2\right)=4\cdot PN^2+XR^2$.

On the other hand, since the point R is the midpoint of the segment BC, Lemma 2, applied to the points T = P, U = B, V = C and M = R, yields $2\cdot\left(PB^2+PC^2\right)=4\cdot PR^2+BC^2$. Also, since the point X is the midpoint of the segment AH, Lemma 2, applied to the points T = P, U = A, V = H and M = X, yields $2\cdot\left(PA^2+PH^2\right)=4\cdot PX^2+AH^2$.

Thus,

$2\cdot\left(PA^2+PB^2+PC^2+PH^2\right)=2\cdot\left(PB^2+PC^2\right)+2\cdot\left(PA^2+PH^2\right)$
$=\left(4\cdot PR^2+BC^2\right)+\left(4\cdot PX^2+AH^2\right)$
$=4\cdot\left(PX^2+PR^2\right)+\left(BC^2+AH^2\right)=2\cdot 2\cdot\left(PX^2+PR^2\right)+\left(a^2+AH^2\right)$
$=2\cdot\left(4\cdot PN^2+XR^2\right)+\left(a^2+AH^2\right)$.

By Theorem 4 in post #4 at http://www.mathlinks.ro/Forum/viewtopic.php?t=24870 , we have $a^2+AH^2=4R^2$. Also, it is well-known that the radius of the nine-point circle of triangle ABC equals $\frac{R}{2}$; thus, the diameter of the nine-point circle equals R. Since we know that the segment XR is a diameter of this nine-point circle, we thus have XR = R. Consequently,

$2\cdot\left(PA^2+PB^2+PC^2+PH^2\right)=2\cdot\left(4\cdot PN^2+XR^2\right)+\left(a^2+AH^2\right)$
$=2\cdot\left(4\cdot PN^2+R^2\right)+4R^2=8\cdot PN^2+6R^2=2\cdot\left(4\cdot PN^2+3R^2\right)$,

and thus $PA^2+PB^2+PC^2+PH^2=4\cdot PN^2+3R^2$, so that Theorem 1 is proven.

Two trivial consequences of Theorem 1:

[color=blue][b]Theorem 3.[/b] If ABC is a triangle with orthocenter H and nine-point center N, then

$NA^2+NB^2+NC^2+NH^2=3R^2$,

where R is the circumradius of triangle ABC.[/color]

[color=blue][b]Theorem 4.[/b] If ABC is a triangle with orthocenter H and nine-point center N, and if P is an arbitrary point on the nine-point circle of triangle ABC, then

$PA^2+PB^2+PC^2+PH^2=4R^2$,

where R is the circumradius of triangle ABC.[/color]

[i]Proof of Theorem 3.[/i] Applying Theorem 1 to the point P = N, we get $NA^2+NB^2+NC^2+NH^2=4\cdot NN^2+3R^2$. But clearly NN = 0, and thus we arrive at $NA^2+NB^2+NC^2+NH^2=3R^2$, so that Theorem 3 is proven.

[i]Proof of Theorem 4.[/i] Since the point P lies on the nine-point circle of triangle ABC, while the center of this nine-point circle is N and the radius is $\frac{R}{2}$, we have $PN=\frac{R}{2}$. Now, after Theorem 1, we have

$PA^2+PB^2+PC^2+PH^2=4\cdot PN^2+3R^2=4\cdot\left(\frac{R}{2}\right)^2+3R^2=R^2+3R^2=4R^2$,

and Theorem 4 is established.

Theorem 4 is the initial problem of this thread; Theorem 3 was posted in http://www.mathlinks.ro/Forum/viewtopic.php?t=29234 .

  Darij

Thank you very much Darij for your instructive solution !

Can you please explain to me how to verify Theorem 1 using vectors ?

Leon

It's not difficult to prove the theorem using vectors. In fact if ve take N as center we have [tex]\displaystyle \frac{\vec{a}+\vec{b}}2=-\frac{\vec{h}+\vec{c}}2[/tex]
because the midpoint of CH and the midpoint of AB are diametrically opposite (as darij said) and so, summing up we have: [tex] \vec{a}+\vec{b}+\vec{c}+\vec{h}=0 [/tex]
Now we have that generally [tex]XY=|\vec{x}-\vec{y}|[/tex] so we can say that:
[tex]PA^2+PB^2+PC^2+PH^2=(\vec{p}-\vec{a})^2+(\vec{p}-\vec{b})^2+(\vec{p}-\vec{c})^2+(\vec{p}-\vec{h})^2=4\vec{p} ^2+\vec{a} ^2+\vec{b} ^2+\vec{c}^2+\vec{h}^2-2\vec{p}(\vec{a}+\vec{b}+\vec{c}+\vec{h})[/tex]
But we have [tex] \vec{a}+\vec{b}+\vec{c}+\vec{h}=0 [/tex] and we have:
[tex]3R^2=12(\frac R2)^2=2((\frac{\vec{a}+\vec{b}}2)^2 + (\frac{\vec{b}+\vec{c}}2)^2 + (\frac{\vec{a}+\vec{c}}2)^2 + (\frac{\vec{a}+\vec{h}}2)^2 + (\frac{\vec{b}+\vec{h}}2)^2 + (\frac{\vec{c}+\vec{h}}2)^2)=[/tex]
[tex] = 2(\frac 34 (\vec{a}^2+\vec{b}^2+\vec{c}^2+\vec{h}^2)+ \frac 14 ((\vec{a}+\vec{b}+\vec{c}+\vec{h})^2-(\vec{a}^2+\vec{b}^2+\vec{c}^2+\vec{h}^2)=\vec{a}^2+\vec{b}^2+\vec{c}^2+\vec{h}^2[/tex]
But [tex]\vec{p}^2=PN^2[/tex] and so [tex]PA^2+PB^2+PC^2+PH^2=4PN^2+3R^2[/tex]

We can add the following generalization of theorem 4 in post # 3

[color=blue][b]Theorem 5.[/b] If ABC is a triangle with orthocenter H and circumcenter O, and if P is a point of segment OH such that $\overrightarrow {PO}  = \lambda  \cdot \overrightarrow {PH} $ , then

$PA^2  + PB^2  + PC^2  = 3R^2  + \lambda \left( {\lambda  + 2} \right) \cdot PH^2 $

where R is the circumradius of triangle ABC.[/color]

[i]Proof.[/i] Applying Stewart theorem to the triangle AOH we have:

$ OH \cdot \left( {PA^2  + PO \cdot PH} \right) = PH \cdot AO^2  + PO \cdot AH^2 \qquad \Rightarrow $

$ OH \cdot PA^2  = PH \cdot R^2  + PO \cdot AH^2  - OH \cdot PO \cdot PH $

Summing the similar relations for $OH \cdot PB^2 $  and $OH \cdot PB^2 $  we have:

$OH \cdot \left( {PA^2  + PB^2  + PC^2 } \right) = $

$= 3PH \cdot R^2  + PO \cdot \left( {AH^2  + BH^2  + CH^2 } \right) - 3OH \cdot PO \cdot PH$

From the relation  $AH^2  + BH^2  + CH^2  = OH^2  + 3R^2 $  (see post #3 at http://www.mathlinks.ro/Forum/viewtopic.php?t=25338 ) follows that:

 $OH \cdot \left( {PA^2  + PB^2  + PC^2 } \right) =  $

 $ = 3PH \cdot R^2  + PO \cdot \left( {OH^2  + 3R^2 } \right) - 3OH \cdot PO \cdot PH =  $

 $ = 3R^2 \left( {PH + PO} \right) + PO \cdot OH^2  - 3OH \cdot PO \cdot PH =  $

 $ = 3R^2  \cdot OH + PO \cdot OH^2  - 3OH \cdot PO \cdot PH $

and consequently

$PA^2  + PB^2  + PC^2  = 3R^2  + PO \cdot OH - 3PO \cdot PH$

On the other hand, since P is a point between O and H, we have $PO =  - \lambda  \cdot PH$ and $OH = \left( {1 - \lambda } \right)PH$, hence

 $ PA^2  + PB^2  + PC^2  =  3R^2  - \lambda \left( {1 - \lambda } \right)PH^2  + 3\lambda  \cdot PH^2  = 3R^2  + \lambda \left( {1 - \lambda } \right)PH^2  $

and the theorem is proven. $\blacksquare$

[color=blue][b]Remark.[/b][/color] The proof of theorem 4 follows easy from theorem 5. It’s enough to observe that if P=N then $\lambda  = 1$ .


How do you place an image alongside text? (For example, the figure on Problem 1/4/19 on the [url=http://www.usamts.org/Tests/USAMTSProblems_19_4.pdf]USAMTS[/url].) Also, this arrangement is similar to the "Square" option under "Format Picture..." in Microsoft Word.

See [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=textflow]Flowing text around figures in LaTeX[/url] for recommendations of which package to use.


induction
inequalities
summer program
MathPath
Putnam
algebra
difference of squares

[quote="mathgeniuse^ln(x)"]Prove that $\frac{1*3*5*\cdots*(2n-1)}{2*4*6*\cdots*2n}<\sqrt{\frac{1}{2n+1}}$ for all integers $n=1, 2, 3, \cdots$.

Thanks[/quote]

Induction and a lot of nasty algebra.

[hide="Right here"]
$\frac{1*3*5*\cdots*(2n-1)}{2*4*6*\cdots*2n}<\sqrt{\frac{1}{2n+1}}$ is true for n = 1 ($\frac12 < \frac{1}{\sqrt3}$)

Now we're going to assume that nasty thing true and then prove this for n + 1:

$(\frac{1*3*5*\cdots*(2n-1)}{2*4*6*\cdots*2n})(\frac{2n+1}{2n+2})<(\sqrt{\frac{1}{2n+1}})(\frac{\sqrt{\frac{1}{2n+3}}}{\sqrt{\frac{1}{2n+1}}})$
Basic manipulation and stuff.

So basically, since the original nasty expression is assumed true, we just have to prove that:

$\frac{2n+1}{2n+2}\leq \sqrt{\frac{2n+1}{2n+3}}$

Since both sides are obviously both positive when n is a positive integer, we can square both sides:

$\frac{4n^{2}+4n+1}{4n^{2}+8n+4}\leq \frac{2n+1}{2n+3}$
$(4n^{2}+4n+1)(2n+3) \leq (4n^{2}+8n+4)(2n+1)$
$8n^{3}+20n^{2}+14n+3 \leq 8n^{3}+20n^{2}+16n+4$
$14n+3 \leq 16n+4$
...which is obviously true when n is a positive integer, meaning that our induction is complete.

There has to be a nicer way.
[/hide]

[hide]
It looks like it can be proved easily with induction. Note that
\[\frac{2n-1}{2n}< \sqrt{\frac{2n-1}{2n+1}}\]
[/hide]

EDIT: treething beat me to it! But the algebra is much simpler if you write it in the above form, because
\[\frac{(2n-1)(2n-1)}{4n^{2}}< \frac{2n-1}{2n+1}\]
is obvious (cross-multiply with difference of squares to get
\[(4n^{2}-1)(2n-1) < (4n^{2})(2n-1) \])

lol... guess I didn't think $\frac{2n+1}{2n+2}\leq \sqrt{\frac{2n+1}{2n+3}}$ was obvious =p

Wow!

I never would have thought of those...

Thanks So MUch.

Oh, that is also good...

Nice solution Scorpius...

[quote="scorpius119"][hide="very short solution"]
\begin{eqnarray*}\left( \frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot \cdots \cdot \frac{2n-1}{2n}\right)^{2}&=&\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{3}{4}\cdot \frac{3}{4}\cdot \cdots \cdot \frac{2n-1}{2n}\cdot \frac{2n-1}{2n}\\ &<&\frac{1}{2}\cdot \frac{2}{3}\cdot \frac{3}{4}\cdot \frac{4}{5}\cdot \cdots \cdot \frac{2n-1}{2n}\cdot \frac{2n}{2n+1}\\ &=&\frac{1}{2n+1}\\ \end{eqnarray*}
[/hide][/quote]

you, sir, are a genius.

that or you have the answer key

Well, from the contest there was no answer key...so he couldn't have gotten it from there...

MMPC, there are some questions that have answer keys others that don't...this one happened to be one of them...

[quote="scorpius119"][hide="very short solution"]
\begin{eqnarray*}\left( \frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot \cdots \cdot \frac{2n-1}{2n}\right)^{2}&=&\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{3}{4}\cdot \frac{3}{4}\cdot \cdots \cdot \frac{2n-1}{2n}\cdot \frac{2n-1}{2n}\\ &<&\frac{1}{2}\cdot \frac{2}{3}\cdot \frac{3}{4}\cdot \frac{4}{5}\cdot \cdots \cdot \frac{2n-1}{2n}\cdot \frac{2n}{2n+1}\\ &=&\frac{1}{2n+1}\\ \end{eqnarray*}
[/hide][/quote]

Reminds me of an inequality from MathPath that I found a solution like this and Neil had to just go and copy mine...pssh

But then I solved the Putnam problem that he didn't so yeah

(edit) yeah mathgenius aka Neil, are you sure this isn't the actual Mathpath problem?

No...I remember that there was two things we had to bound, not one thing...and there was no square root anywhere...

I got a Putnam too...


trigonometry
function
geometry
inequalities
limit
calculus
calculus computations

[img]http://alt2.mathlinks.ro/latexrender/pictures/d/8/d/d8d653506dfed41aa9283034c4850b0f318caca5.gif[/img]

can you calculate this limit without derivates ?

By 'using derivatives', do you mean l'Hopital?
You can use the Taylor expansion: $ \sin x\equal{}x\minus{}\frac{x^3}{3!}\plus{}...$:

$ \frac{x\minus{}\sin x}{x^3}\equal{}\frac{x\minus{}(x\minus{}\frac{x^3}{3!}\plus{}...)}{x^3}\equal{}\frac{\frac{x^3}{3!}\plus{}...}{x^3}\to \frac{1}{3!}\equal{}\frac{1}{6}$ for $ x\to 0$

thanks a lot

what about this one ? [url]http://www.mathlinks.ro/viewtopic.php?t=252200[/url]

still waiting for someone who can do it without derivables

[hide]We first show that the given limit exists. Notice that the function $ f(x) \equal{} \frac{x \minus{} \sin x}{x^3}$ is even, so it is enough to consider $ x \to 0^\plus{}$. Considering the areas of appropriate triangles in the unit circle, the inequality $ \sin x < x < \tan x \,\,\,(1)$ is true for $ 0 < x < \frac{\pi}{2}$. Rearranging (1) we have

$ 1 < \frac{x}{\sin x} < \frac{1}{\cos x}\,\,\,(2)$

$ <\equal{}> \cos x < \frac{\sin x}{x} < 1$

$ <\equal{}> \frac{\cos x}{x^2} < \frac{\sin x}{x^3} < \frac{1}{x^2}$

$ <\equal{}> \frac{\cos x \minus{} 1}{x^2} < \frac{\sin x}{x^3} \minus{} \frac{1}{x^2} < 0$

$ <\equal{}> 0 < \frac{x \minus{} \sin x}{x^3} < \frac{1 \minus{} \cos x}{x^2}\,\,\,(3)$

Since $ \lim_{x \to 0^\plus{}} 0 \equal{} 0$ and $ \lim_{x \to 0^\plus{}} \frac{1 \minus{} \cos x}{x^2} \equal{} \frac12$, we can conclude that $ \lim_{x \to 0^\plus{}} f(x)$ exists, and its value L verifies, according to (3),

$ 0 < L < \frac12$.

Now, in order to compute the exact value of L, observe that we also have

$ L \equal{} \lim_{x \to 0^\plus{}} \frac{(3x) \minus{} \sin(3x)}{(3x)^3} \equal{} \lim_{x \to 0^\plus{}} \frac{3x \minus{} (3 \sin x \minus{} 4 \sin^3x)}{27x^3}$

$ \equal{} \lim_{x \to 0^\plus{}} \frac{3x \minus{} 3 \sin x \plus{} 4 \sin^3x}{27 x^3} \equal{} \frac{1}{9} \lim_{x \to 0^\plus{}} \frac{x \minus{} \sin x}{x^3} \plus{} \frac{4}{27} \lim_{x \to 0^\plus{}} \frac{\sin^3x}{x^3}$

and so

$ L \equal{} \frac19 L \plus{} \frac{4}{27}\,\,\,(4)$,

where I have used the fact that $ \lim_{x \to 0} \frac{\sin x}{x} \equal{} 1$ [which can be proved directly from (2)]. Finally, solving equation (4) for L we arrive at

$ L \equal{} \frac16$.[/hide]

Nice computation. Just one thing:

[quote="Carcul"]Since $ \lim_{x \to 0^ \plus{} } 0 \equal{} 0$ and $ \lim_{x \to 0^ \plus{} } \frac {1 \minus{} \cos x}{x^2} \equal{} \frac12$, we can conclude that $ \lim_{x \to 0^ \plus{} } f(x)$ exists, and its value L verifies, according to (3),

$ 0 < L < \frac12$.[/quote]
I'd say that you haven't shown existence: you showed that IF the limit exists, THEN it will lie between 1/2 and 0. It may oscillate between 1/2 and 0, not reaching a definite value.
For example, we have $ \minus{}1<\sin x<1$, hence taking limits we get $ \minus{}1<\lim_{x\to \infty}\sin x<1$, but the limit does not exist (if it exists, then it lies between -1 and 1, but it doens't exist).

[quote="Carcul"]We first show that the given limit exists.[/quote]

how you do that ?...how can you show that limit exists ?

the limit is corectly solved, but i also added :

 \[ f( \minus{} x) \equal{} \frac{{\sin ( \minus{} x) \minus{} ( \minus{} x)}}{{ \minus{} x^3 }} \equal{}  \minus{} \frac{{\sin x \minus{} x}}{{ \minus{} x^3 }} \equal{} \frac{{\sin x \minus{} x}}{{x^3 }} \equal{} f(x)
\]

and

\[ \sin x < x < tgx \to \frac{{\sin }}{x} < 1 < \frac{{tgx}}{x} \to \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} \le 1 \le \mathop {\lim }\limits_{x \to 0} \frac{{tgx}}{x} \to 1 \le 1 \le 1
\]

I don't understand your last post. What are you trying to say? I don't see the connection with the original problem.

see how Carcul solved it..
i just added those 2 things to carcul`s solution, and my teacher said that the problem is corectly solved

1) Where/how in your answer did you use [quote]$ \sin x < x < tgx \to \frac{{\sin }}{x} < 1 < \frac{{tgx}}{x} \to \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} \le 1 \le \mathop {\lim }\limits_{x \to 0} \frac{{tgx}}{x} \to 1 \le 1 \le 1$[/quote]

2) I still don't see the use of showing that - if the limit exists - it will lie between 0 and 1/2. Carcul didn't show existence. By computing the limit (1/6) you already say exactly what L is; to me it seems irrelevant first to show 0<L<1/2.


If xy=24 with x and y real, what is the minimum value that x^2+4y^2 can attain?

that can be written as (x-2y):^2: +4xy. which is greater than or equal to 96.

No.

Let a=x, b=2y; then ab=48 and we seek the minimum value of a^2+b^2. By the trivial inequality, a^2+b^2 >= 2ab = 96. This value is attainable with x = a = b = 2y = sqrt(48).

[quote="teratomato"]With x, y real and xy=24, x+2y can never attain the value zero.[/quote]

Its x-2y, not x+2y. So MysticTerminators solution is fine.. though of course you should write the equality case.

er sorry for the confusion; I wrote an incorrect solution, and then edited it after sitting there looking at it. Apparently teratomato replied to my solution in the meanwhile.

x-2y is zero when x=2y, which means x=sqrt[12]


calculus
integration
logarithms
trigonometry
calculus computations

I think :
$ \int\frac{1}{x^{2}\plus{}x^{\minus{}2}}dx\equal{}1/8\,\sqrt {2}\ln  \left( {\frac {{x}^{2}\minus{}x\sqrt {2}\plus{}1}{{x}^{2}\plus{}x
\sqrt {2}\plus{}1}} \right) \plus{}1/4\,\sqrt {2}\arctan \left( x\sqrt {2}\plus{}1
 \right) \plus{}1/4\,\sqrt {2}\arctan \left( x\sqrt {2}\minus{}1 \right)$

[quote="Extremal"]I think :
$ \int\frac {1}{x^{2} \plus{} x^{ \minus{} 2}}dx \equal{} 1/8\,\sqrt {2}\ln \left( {\frac {{x}^{2} \minus{} x\sqrt {2} \plus{} 1}{{x}^{2} \plus{} x \sqrt {2} \plus{} 1}} \right) \plus{} 1/4\,\sqrt {2}\arctan \left( x\sqrt {2} \plus{} 1 \right) \plus{} 1/4\,\sqrt {2}\arctan \left( x\sqrt {2} \minus{} 1 \right)$[/quote]

Oh, I know that
$ \frac{x^2}{x^4\plus{}1}\equal{}\large\frac{\frac{x}{2\sqrt{2}}}{x^2\minus{}x\sqrt{2}\plus{}1}\minus{}\frac{\frac{x}{2\sqrt{2}}}{x^2\plus{}x\sqrt{2}\plus{}1}$
But it's long. I'm looking up simpler solution.

Its the famous integral : $ I \equal{} \int\sqrt {\tan x}\text{d}x$

taking $ u \equal{} \tan x$ and $ \frac {\text{d}u}{1 \plus{} u^{2}} \equal{} \text{d}x$ then $ I \equal{} \int\frac {\sqrt {u}}{1 \plus{} u^{2}}\text{d}u$

and then $ u \equal{} t^{2}$ and $ \text{d}u \equal{} 2t\text{d}t$ and we'll have: $ I \equal{} \int\frac {t}{1 \plus{} t^{4}}2t\text{d}t \equal{} 2\int\frac {t^{2}}{1 \plus{} t^{4}}\text{d}t$ and $ \cdots$

and i guess thats the best solution and theres no simpler one...!

Oh sorry maybe there can be a simpler one :oops: , i posted this integral last year and [b]pardesi[/b] gave a nicer solution for it [url=http://www.mathlinks.ro/viewtopic.php?t=135911]Here[/url]


calculus
integration
geometry
3D geometry
sphere
analytic geometry
function

Hey.
I was doing a triple integral over the first octant of a sphere with equation of x^2 + Y^2 + z^2 = 4.
So naturally I used spheric coordinates, and the integration was pretty easy. However when i was trying to use the rectangular and cylindrical coordinates( I know this is more painful i just wanna see how this work on at least one example), i ran into problems with limits of integration. What i got were quarter circles for two ways of integration and the  form 0 to 2 for the last integration. This is wrong, however i can;t quite determine why.
Any ideas?
Thx

Let $ u$ be the function to be integrated. I will gloss over the need to make changes of variables in the expressions for $ u$ - that part is pretty straightforward, anyway.

Let $ E=\{(x,y,z): x\ge0,y\ge0,z\ge0,x^{2}+y^{2}+z^{2}\le4\}.$

Then:

$ \iiint_{E}u\,dV=$

$ =\int_{0}^{2}\int_{0}^{\sqrt{4-z^{2}}}\int_{0}^{\sqrt{4-z^{2}-y^{2}}}u\,dx\,dy\,dz$

$ =\int_{0}^{2}\int_{0}^{\sqrt{4-z^{2}}}\int_{0}^{\frac{\pi}2}u\,r\,d\theta\,dr\,dz$

$ =\int_{0}^{2}\int_{0}^{\frac{\pi}2}\int_{0}^{\frac{\pi}2}u\rho^{2}\sin\phi\,d\theta\,d\phi\,d\rho$


calculus
integration
calculus computations

I am stuck half way in solving this problem (the square root nominator confuses me):
[img]http://img235.imageshack.us/img235/8459/1mi1.jpg[/img]

and I cannot get it to match the answer given on the back of the textbook:
[img]http://img235.imageshack.us/img235/7041/answerop5.jpg[/img]

Please teach me how to solve this problem. Thanks. :)

(1) changing variable , i.e $x=y^{6}$
(2) rewrite the integral as
$\int\frac{6y^{8}}{1+y^{2}}dy\\=6\int\left(y^{6}-y^{4}+y^{2}-1+\frac{1}{y^{2}+1}\right)dy$

looks familiar now ? :)


The AOPS Trivia Challenge has all but died out. This is my attempt to try and bring it back. It's fun and competitive when lots of people play. Check it out. A new tournament will start tomorrow for the month of march.
http://www.funtrivia.com/private/main.cfm?tid=960

Heh, I just played.  9 right in 37 seconds.

I'm seriously bad at this stuff.....I got only four in 73 seconds..... :oops:

what happend to the friday night AIM competitions?

Well I did ABSOLUTELY horrible.
But still fun.

I used to be a semi-active player, then kind of abandoned it...didn't realize it was still up and running. I'll try to play regularly from now on, even though I'm no good at it.

Not much people are playing this game. It is dieing down again.

[quote="joml88"]I think it has life in it again :)[/quote]
for now, but not nearly as popular as it was before when it had 30+ people daily.

Bumping this topic so more people can see it. (Come on, people! It only takes 1-2 minutes each day!)

[quote="yif man12"]We still hold some impromptu AOPSTIs when there's 6 or 7 people online who want to play. Basically it's only down to hez, kaot, chigra, penguinPowered, aznphatso, metz, ambi, and me now. Sorta sad, I missed the old ones.[/quote]

Yeah... Pretty much... Anyway, anyone who wants to participate in our little AIM trivia games, leave me your AIM sn here.

Or of course you can IM me. My SN is Kaoticity.

the trivia challenge has been a success for the month of march, with about 15-20 participating every day. a new tournament starts for april tomorrow. if you've got 1-2 mintues to spare, give it a try - you might like it.  :)


quadratics
number theory solved
number theory

Let p be an odd prime and let $\xi$ be a primitive $p^th$ root of unity etc $\xi$ is a complex number such that $(\xi)^p=1$ and $\xi^k$ <>1 for 0<k<p.Let $A_p$ and $B_p$ denote the set of quadratic residues and non residues modulo p,respectively.Finally,let $\alpha$=$\sum a_k$ (k$\in$ $A_p$) and $\beta$=$\sum b_k$ (k$\in$ $B_p$). Show that $\alpha$ and $\beta$are the roots of : $x^2+x+(\frac{1-(-1/p)p}{4})=0$.

What are $a_k,\ b_k$?

[Edit: I have found the notes and the sums are $\displaystyle \sum_{k\in A_p}\xi^k$ and $\displaystyle \sum_{k\in B_p}\xi^k$]

It's obvious that $\alpha+\beta=-1$, so all we really need to do is show that $\alpha\cdot \beta=\frac {1-(\frac {-1}p)p}4$. In order to do this we observe that $\alpha\cdot\beta=\displaystyle\sum_{1\le u,v\le p-1}^{(\frac up)=1,\ (\frac vp)=-1}\xi^{u+v}$.

Now we show that all rasidues $\mod p$ which are $\ne 0$ can be written as $u+v,\ (\frac up)=-(\frac vp)=1$ in the same number of ways (if $t$ can be written like this in $k$ forms, $t=u_i+v_i,\ i\in\overline{1,k}$, then $t'=u_i\cdot (t't^{-1})+v_i\cdot (t't^{-1})$ and the two summands have opposite Legendre symbols). It's trivial from here (we consider the cases $p=M_4+1,\ M_4+3$; in the first case $u+v\ne 0$, and in the second case $\frac {p-1}2$ of the terms of $\alpha \cdot \beta$ are $1$, i.e. $\xi^0$).

I had a question; is there anyway to obtain these lecture notes by Mr. Sato?

Naoki Sato is one of the admins here so he should be able to post them here himself. They were online but I can't find them, maybe they disappeared from the web.

I found where they are SUPPOSED to be but the connection is always refused  ;)

Indeed, same problem here. I have them on a computer but not on this one (this is my personal computer, they are on the "family computer") and I should have them on a CD rom somewhere... I'll try to upload them or something later.

I'm afraid my old home page had to be taken down, so now I am homeless.  :(

Here are my notes; enjoy.


My TexnicCenter help file mentions a character, \multimap ($\multimap$).  Clearly, it renders on the forum, but I can't get Texnic to recognize it.  Does anyone know what package I need to make it work?

Thanks.

Ravi, you are a wonderful human being :D

Thanks! :blush: You are easy to please. :D


geometry
circumcircle
trigonometry
geometric inequality
inradius
IMO Shortlist

The triangle $ ABC$ is inscribed in a circle. The interior bisectors of the angles $ A,B$ and $ C$ meet the circle again at $ A', B'$ and $ C'$ respectively. Prove that the area of triangle $ A'B'C'$ is greater than or equal to the area of triangle $ ABC.$

Let $O$ be the circumcenter and $R$ the circumradius of $\triangle ABC$. Now, $\angle A'B'C'=\angle A'AB+\angle C'CB=\frac{\alpha+\theta}{2} \\\Rightarrow \angle A'OC'=\alpha+\theta \Rightarrow \angle A'C'O=\angle C'A'O= 90-\frac{\alpha+\theta}{2}=\frac{\beta}{2}$
By the identity $a\cos B+b\cos A=c$ it follows that $A'C'=R\cos\frac{\beta}{2}+R\cos\frac{\beta}{2}=2R\cos\frac{\beta}{2}$.
In a similar way we find
$A'B'=2R\cos\frac{\theta}{2}, B'C'=2R\cos\frac{\alpha}{2} \\\Rightarrow (A'B'C')=\frac{2R\cos\frac{\beta}{2}\cdot 2R\cos\frac{\theta}{2}\cdot 2R\cos\frac{\alpha}{2}}{4R}$
Then
$(A'B'C')\geq(ABC) \Leftrightarrow \frac{2R\cos\frac{\beta}{2}\cdot 2R\cos\frac{\theta}{2}\cdot 2R\cos\frac{\alpha}{2}}{4R}\geq \frac{abc}{4R} \\\Leftrightarrow 8R^3\cos\frac{\alpha}{2}\cos\frac{\beta}{2}\cos\frac{\theta}{2}\geq abc=8R^3\sin\alpha \sin\beta \sin\theta \\\Leftrightarrow \cos\frac{\alpha}{2}\cos\frac{\beta}{2}\cos\frac{\theta}{2}\geq 8\cos\frac{\alpha}{2}\cos\frac{\beta}{2}\cos\frac{\theta}{2}\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\theta}{2} \\\Leftrightarrow \frac{1}{8}\geq \sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\theta}{2}$
but the last inequality is very well-known, so we're done.

Let K, L, M be the intersections of the bisectors of the angles $\angle A, \angle B, \angle C$ with the circumcircle (O) of the triangle $\triangle ABC$. For example, the angle $\angle M$ of the triangle $\triangle KLM$ is equal to

$\angle M = \angle LMK = \angle LMC + \angle CMK = \angle LBC + \angle CAK = \frac{\angle B + \angle A}{2}$

and similarly,

$\angle K = \frac{\angle C + \angle B}{2},\ \ \ \angle L = \frac{\angle A + \angle C}{2}$

The areas of the triangles $\triangle ABC, \triangle KLM$ are equal to

$S(\triangle ABC) = \frac{abs}{4R} = 2R^2 \sin \widehat A \sin \widehat B \sin \widehat C$

$S(\triangle KLM) = \frac{klm}{4R} = 2R^2 \sin \widehat K \sin \widehat L \sin \widehat M = 2R^2 \sin \frac{\widehat B + \widehat C}{2} \sin \frac{\widehat C + \widehat A}{2} \sin \frac{\widehat A + \widehat B}{2}$

where a = BC, b = CA, c = AB, k = LM, l = MK, m = KL and R is the common circumradius of these 2 triangles. Thus we have to show that

$\sin \widehat A \sin \widehat B \sin \widehat C \le \sin \frac{\widehat B + \widehat C}{2} \sin \frac{\widehat C + \widehat A}{2} \sin \frac{\widehat A + \widehat B}{2}$

with the constraint $\angle A + \angle B + \angle C = \pi$. Substituting $\frac{\angle A + \angle B}{2} = \frac \pi 2 - \frac{\angle C}{2}$ and $\sin \frac{\widehat A + \widehat B}{2} = \cos \frac{\widehat C}{2}$, this becomes

$\sin \widehat A \sin \widehat B \sin \widehat C \le \cos \frac{\widehat A}{2} \cos \frac{\widehat B}{2} \cos \frac{\widehat C}{2}$

$8 \sin \frac{\widehat A}{2} \sin \frac{\widehat B}{2} \sin \frac{\widehat C}{2} \le 1$

This inequality has been proved numerous times on this forum, either in this form or in the equivalent form

$8 \sin \frac{\widehat A}{2} \sin \frac{\widehat B}{2} \sin \frac{\widehat C}{2} = 4 \left(\cos \frac{\widehat A - \widehat B}{2} - \cos \frac{\widehat A + \widehat B}{2} \right) \cos \frac{\widehat A + \widehat B}{2} =$

$= 2 \left(\cos \widehat A + \cos \widehat B\right) - 2 \left(1 + \cos (\widehat A + \widehat B)\right) \le 1$

$\cos \widehat A + \cos \widehat B + \cos \widehat C \le \frac 3 2$

For example, it can be reduced to $r \le \frac R 2$, where r, R are the triangle inradius and circumradius (see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=38700]OH and IH[/url]) or proved using Lagrange multipliers (see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=24803]Triangle inequality[/url]).

Let $H$ denote the orthocenter, and let $AH$ meet the circumcircle at $A''$. Define $B''$ and $C''$ similarly. Clearly $\angle A''CB=\angle A''AB=90-\angle B=\angle BCH$, and $A''H$ is perpendicular to $BC$. Hence $A''$ is the reflection of $H$ in $BC$ and therefore $\triangle A''CB\cong\triangle HCB$. Similarly $\triangle C''BA\cong\triangle B''AC$. Therefore the area of hexagon $AB''CA''BC''$ is 2 times the area of $ABC$.

Let $B'C'$ meet $AA'$ at $X$. Then \[\angle B'XA=\angle B'C'A+\angle XAC'=\angle B'C'A+\angle XAB+\angle C''AB=\angle B'BA+\angle A'AB+\angle C'CB=\angle B/2+\angle A/2+\angle C/2+90.\] Hence $A'A$ is an altitude of $\triangle A'B'C'$. Hence the area of hexagon $AB'CA'BC'$ is twice of the area of triangle $A'B'C'$. However, $A'$ is the midpoint of the arc $BC$, so $[A'BC]\ge [A''BC]$. Similarly, $[AB'CA'BC']\ge [AB''CA''BC'']$. Hence $[A'B'C']\ge[ABC]$, as desired $\Box$.

Here is the sketch of another proof, partially thanks to [b]RSM[/b].
Note that, if $A_1=AI\cap \odot(ABC);$ and if $A_2=BB_1\cap CC_1$ and if $B_1, B_2, C_1,C_2$ are defined similarly, then we get that $A_1B_1C_1$ is the medial triangle of $A_2B_2C_2.$ So we have to show that the medial triangle has the maximum area among all cevian triangles.
Let $B_2C_2, C_2A_2$ and $A_2B_2$ be divided in the ratios $x:y, y:z, z:x$ by the points $A,B,C$ respectively. Then we see that $\frac{[A_2B_2C_2]}{[ABC]}=\frac{(x+y)(y+z)(z+x)}{4xyz}\geq 2$ from the AM-GM inequality. So, we are done. $\Box$

Nice :)

[hide="Solution"]Observe that \[\angle A'B'C'=\angle A'B'B+\angle CB'B=\angle A'AB+\angle C'CB=\frac{A}{2}+\frac{C}{2}=90-\frac{B}{2}\] by Triangle Angle Sum; similarly $\angle B'C'A'=90-\frac{C}{2}$, $\angle C'A'B'=90-\frac{A}{2}$. Let $O$ be the circumcenter of triangle $ABC$, whence $\angle A'OC'=2\angle A'B'C'=180-B$ by Inscribed Angles and similarly $\angle B'OA'=180-C$, $\angle C'OB'=180-A$. It follows that \[[A'B'C']=[A'OC']+[B'OA']+[C'OB']=\frac{1}{2}R^2(\sin A+\sin B+\sin C)\] by $A=\frac{1}{2}ab\sin C$. As $R\sin A=\frac{a}{2}$ by the Extended Law of Sines and similarly for $B$, $C$, it follows that $[A'B'C']=\frac{Rs}{2}$, whence it suffices to show that $\frac{Rs}{2}\ge [ABC]=rs$, which is Euler's Inequality. $\blacksquare$[/hide]

[hide='More algebraic than above']

Let $a$, $b$ and $c$ have their usual meanings. [b]Since internal angle bisectors meet midpoints of arcs, we have that the side lengths of $A'B'C'$ are $\frac12 (a+b)$, $\frac12 (b+c)$ and $\frac12 (c+a)$.[/b] Now by Heron's, we have

$[ABC]=\sqrt{\frac{a+b+c}{2} \times \frac{a+b-c}{2} \times \frac{a-b+c}{2} \times \frac{-a+b+c}{2}}$

and

$[A'B'C]=\sqrt{\frac{a+b+c}{2} \times \frac{a}{2} \times \frac{b}{2} \times \frac{c}{2}}$

so the problem boils down to showing

$ \frac{a}{2} \times \frac{b}{2} \times \frac{c}{2} \geq \frac{a+b-c}{2} \times \frac{a-b+c}{2} \times \frac{-a+b+c}{2} $

But cancelling denominators and expanding out, we get that this is equivalent to

$a^3 + b^3 + c^3 + 3abc \geq \sum_{sym} a^2b $

and this is obvious by Schur's.

Note equality holds iff $ABC$ is equilateral, since the other equality conditions in Schur's involve one of $a$, $b$ and $c$ being equal to zero, obviously not the case here.

[/hide]

EDIT: Doesn't work, sorry for the spam. The boldened part is exactly where the proof goes wrong  :wacko:

Let $R$ denote circumradius.
WLOG assume $R=\frac{1}2$, by extended law of sines, $x=\sin X$ for any side $x$ on circle, and $\angle X$ is angle subtended by side $x$ on circle.

Some angle chase gives us $\angle A'=\angle \frac{1}2(B+C)$, symmetric values for $\angle B'$ and $\angle C'$.

We know $$[ABC]=\frac{abc}{4R}=\frac{1}2 \left [\sin A \cdot \sin B \cdot \sin C \right]$$

Similarly $$[A'B'C']=\frac{1}2\left [\sin \frac{1}2(B+C) \cdot \sin \frac{1}2(C+A) \cdot \sin \frac{1}2(A+B) \right]$$

This reduces proving $$\sin \frac{1}2(B+C) \cdot \sin \frac{1}2(C+A) \cdot \sin \frac{1}2(A+B) \geq \sin A \cdot \sin B \cdot \sin C$$

Expanding values $\sin A=2 \sin \frac{A}2 \cos \frac{A}2$ and $\sin \frac{1}2 (B+C)=\cos \frac {A}2$
reduces the above statement to proving $$\sin \frac{A}2 \cdot \sin \frac {B}2 \cdot \sin \frac{C}2 \leq \frac {1}8$$

which is true by AM-GM.


AMC
USA(J)MO
USAMO
AIME
email
AIME I
AIME II

Recall that the USAMO Index is AMC score (either 10 or 12, either A or B) + 10*AIME score (either AIME I or AIME II).

The first USAMO Selection Index for students who took the AMC 12 (either A-date or B-date, and without considering age or grade level)  is 204.0.
This yields 320 students.

The lowest AIME score among these students was 6 (144.0 + 10*6) so now we look at all students in 10th grade and below who got a 6 or better on the AIME.  There are 468 of them, so we need to use the USAMO Index to discriminate among them.

The 156 highest USAMO indices among them are at Index 202.5 and higher.

This gives 476 students first selected to the USAMO.  To them we add 10 students from U.S. States not already represented.  This gives us 486 invited to take the USAMO.

We still have some late AIME II papers coming in.  We anticipate about another 10 invitees from these late papers, and some U.S. citizens living abroad.  When all is said and done, we will have between 495 and 500 students invited to 2008 USAMO.

Official emails with the official invitation will start going out no later than Friday, April 11, 2008.  A list of USAMO invitees names, schools and state will be posted on our website no later than Friday, April 11.  Paper postal invitations will be mailed shortly thereafter.  Please see the reminder in Section V, page 5 of the AIME Teachers' Manual, no AIME results will be given out by phone (or otherwise) until after April 13.  Please do not call or contact us now , we have much to do yet.

We expect to begin mailing AIME reports this week, either today, April 8 or tomorrow, April 9.  Again, please do not call or email until after April 13.

Steve Dunbar
MAA Director, American Mathematics Competitions

An addendum to the index scores announcement (all this info is elsewhere on this forum, but it's nice to have it all consolidated in one place):

The [url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2008-ua/2008usamoqual.shtml]list  of qualifiers[/url] is now up on the AMC web site.

Elizabeth Claassen of the AMC posts the following information for those people who think that they should be on the USAMO qualifiers list, but are not listed:
[quote]For you and anyone else who believes s/he may have qualified,

Please email the following information to amcinfo@maa.org [b]after April 13th[/b]:

1) Your name.
2) Your school's name and CEEB number. If you took the contests at more than one location, include both.
3) Your address.
[/quote]I have added the emphasis on "after April 13th".  They are still processing some scores and rechecking some others.  Please let them finish this task before contacting them.  There is still plenty of time to get properly registered for the USAMO if indeed you qualified but are not yet listed.


does Aops use a phpBB host? because i used it to make my own site (check signature) and i realized it looks similar to Aops.


Έστω P ένα κανονικό πολύγωνο με ν πλευρές, εγγεγραμμένο σε κύκλο ακτίνας 1, και Q ένα κανονικό πολύγωνο με 2ν πλευρές, εγγεγραμμένο στον ίδιο κύκλο. Να αποδειχθεί ότι αν το μήκος της πλευράς του P είναι S, τότε το μήκος της πλευράς του Q είναι $ \sqrt{2-\sqrt{4-S^{2}}}$

Προκύπτει άμεσα από τον Τύπο του Αρχιμήδη . Κάτι παρόμοιο υπάρχει και στο σχολικό βιβλίο. :wink:

Αναφέρει πουθενά μέσα στο σχολικό βιβλιο τον τύπο του Αρχιμήδη;  :maybe:

(εννοούσα να λυθεί με πιο στοιχειώδεις μεθόδους)


Hello all,
[quote]In how many ways can 3 males and 3 females be seated in a row if the genders alternate down the row?[/quote]
So they want us to find the number of arrangements in the form of either:
M F M F M F
F M F M F M

Each form is equivalent to counting arrangements of M M M F F F = 3! x 3! = 36

But, since there are two possible forms for the alternating seating plan, we need to multiply the number of arrangements by two to get 72 possible arrangements. Is this correct?

Thank you for your help and the fast response!


A rubber ball is known to rebound half the height that it drops. The ball is dropped from a height of 32 feet. Assuming that it bounces straight up and down, how many feet will it have traveled by the time it hits the ground the fourth time?

First, it drops $ 32$ feet. Then it bounces up $ 16$, and drops back down $ 16$. That's one bounce. Then it bounces $ 8$, then $ 8$. Next, $ 4$, then $ 4$. Adding all these up give us our answer, which is $ \boxed{88}$.

i was just wondering does anybpdy know a faster way to do this

like what will you do if it's the 25th time


geometry
power of a point
radical axis
geometry unsolved

we have two circles $C , C'$ with centers $O,O'$ .
$C'$ pas through $O$. the common tangents of these two circles touches $C'$ at $A,B$ prove that $AB$ is tangents to $C$.

It is a trvial solution but anyway
let $C^,=C_1 , C=C_2$ and $O^,=O_1$, $O=O_2$ and their radiuses with $R_1, R_2$
respectivly.
let $O_1O_2$ meets the $C_2 $ in $S$ (this S , is one that we are going to prove that is it on $AB$).
now draw a circle with center on $O_1$ with radius $R_1-R_2$ and name it $C_3$.
this circle is tangent to $C_2$ at $S$,let $O_1A$ meets the $C_3$ at $K$(K is between A and O_1). we know that $O_2K$ is tangent to $C_3$.
now we are going to prove that $A$ is on the radical axis of the circles $C_2, C_3$.
the power of $A$ wrt  $C_2$ is equal to $(AO_2)^2-(R_2)^2$  and its power wrt $C_3$ is equal to $(AO_1)^2-(R_1-R_2)^2$ . its enough to show that:
$(AO_1)^2-(R_1-R_2)^2=(AO_2)^2-(R_2)^2$
we have that $(AO_1)=R_1$ so we should just find the lenght of $AO_2$.
we have that $(AO_2)^2=(AK=R_2^2)^2+(O_2K)^2$
and $(O_2K)^2=(O_1O_2=R_1)^2-(R_1-R_2)^2$
so$(AO_2)^2=(R_2)^2+(R_1)^2-(R_1-R_2)^2$
and we have
Power of $A$ wrt $C_2 =(AO_2)^2-(R_2)^2=(AO_1)^2-(R_1-R_2)^2=$power of $A$ wrt $C_3$, so $A$ is on the radical axis and like this $B$ is on the radical axis and then $AB$ is the radical axis of the circles $C_2,C_3$ and therefore tangent to $C_2$ and we have  :bruce: the problem.

P.S i wonder, that how , [b]YETTI[/b] , would keep his hope in all of his way through the tones and tones of calculations but i think it worths cause after the darkness there is a  pleasent shine  :) .I salute you, problem solving gladiator.

Let's call $ AC$ a common tangent to the circles, $ C$ being on the circle $ C$, and take $ D$ on $ C$ as well, as the intersection with the segment $ OO'$.
See that the triangle $ \triangle AO'O$ is isosceles at $ O$, with $ \widehat {OAO'} \equal{} \widehat {AOO'}$, $ O'A \parallel{} OC$, therefore $ OA$ is the bisector of $ \angle COO'$, hence $ \triangle AOC \equal{} \triangle AOD$ so $ AC \equal{} AD$, $ AD \perp OD$, but AC is equal and parallel to the altitude of the triangle $ AOO'$ at $ O$, hence $ AD$ is tangent to the circle $ C$ and $ B$ is on $ AD$, done.

Best regards,
sunken rock


(*****)

17. Let $ x^{2}+y^{2}=1$ be circle $ \mathcal{C}$. Points $ A=(11,-2)$, $ B=(5,10)$. Let $ A', A'', B', B''$ $ \in \mathcal{C}$ be points such that $ AA'$, $ AA''$, $ BB'$, and $ BB''$ are tangents. Let $ A'A''\cap B'B''=C$. Write the coordinates of $ C$ as an ordered pair, $ (p,q)$ where $ p$ and $ q$ are real numbers.

[hide]The inversion of $ A$ with respect to $ \mathcal{C}$ is $ (\frac{11}{125},\frac{-2}{125})$. $ A'A''$ has slope $ \frac{11}{2}$ and goes through this point so its equation is $ y=\frac{11}{2}x-\frac{1}{2}$

The inversion of $ B$ is $ (\frac{1}{25},\frac{2}{25})$ and the equation of $ B'B''$ is $ y=-\frac{1}{2}x+\frac{1}{10}$.

We can solve to see that $ (p,q)=(\frac{1}{10},\frac{1}{20})$[/hide]

[quote="Altheman"]You have the right idea; you got the inversion mixed up. $ R=1$ is the radius; $ OA=\sqrt{125}$, so $ O\mathcal{I}(A)=\frac{1}{\sqrt{125}}$ where $ \mathcal{I}$ represents an inversion about $ \mathcal{C}$.[/quote]

 :huh: What do you mean?

$ \sqrt{\frac{11}{125}^{2}+\frac{-2}{125}^{2}}=\sqrt{\frac{121}{125^{2}}+\frac{4}{125^{2}}}=\sqrt{\frac{1}{125}}$

Woops okay...I was being stupid there. I was thinking about dividing by the magnitude once since it was like a product...but obviously you have to divide by the magnitude twice. You are correct then.

I had a slightly more conceputal solution [arguably more simply computationally].

Since $ C$ is on the polars of $ A$ and $ B$, then $ A$ and $ B$ are on the polar of $ C$. Since a polar is a line and two points define a line, $ AB$ is the polar of $ C$ wrt $ \mathcal{C}$. Let $ l$ be the line perpendicular to $ AB$ through $ O$.

$ \begin{cases}AB: y=(-2)(x-10)\\ l: y=\frac{1}{2}x\\ AB\cap l: =C'\end{cases}\implies C'=(8,4)$

We have that $ \mathcal{I}(C)=C'$. So $ C=\frac{1}{|OC'|^{2}}*(8,4)=\frac{1}{80}*(8,4)=\left(\frac{1}{10},\frac{1}{20}\right)$.


real analysis
vector
arithmetic sequence
Ramsey Theory
combinatorics unsolved
combinatorics

Need some help..

We define the outermost edge (or the circumference) of a circle to be a set of points. Let each of these points be colored black or white. Prove that regardless of choice, it is always possible to inscribe in this circle an isosceles triangle whose three vertices are of the same color. Is this possible for any other type of triangle?

I think it's enough to consider a regular heptagon inscribed in the circle. I didn't check it throroughly, but it looks like we can find three of its vertices which are colored the same and form an isosceles triangle.

Am I wrong? :?

What other kinds of triangles should we consider?
It is clear, we can't guarantee that there is a regular triangle or right triangle.

I think he is asking if it is possible to inscribe any other triangle in the circle with the given conditions, but it is not part of the original problem statement...

Just another thing about the problem: a regular nonagon works for sure, because at least $5$ of the points will have the same color, and it is well-known that if you choose $5$ numbers among $1,2,\ldots,9$, there will be an arithmetic progression with $3$ terms among those (I'm sure the connection between the two problems is clear :)).

I can reveal great secret for you: regular pentagon works too  :D

It's my turn to reveal a great secret :): we can find an isosceles triangle, no matter what finite number of colors we use (not just the very small number $2$). We just use Van der Waerden's Theorem and that idea with the arithmetic progression.

In that case, can anyone prove it only for an isoceles triangle? I would like to see how a proof would look like for this problem...

You're refering to the case with $r$ colors?

In that case, basicly, the proof has already been posted :). One of the versions of Van der Waerden's Theorem states that there is an $n(3,r)$ s.t. whenever $n\ge n(3,r)$, any coloring of the numbers $1,2,\ldots,n$ with $r$ colors yields an arithmetic progression with $3$ elements, all colored the same.

Now consider a regular polygon with $n\ge n(3,r)$ vertices on the circle, and number its vertices $1,2,\ldots,n$ in a clockwise direction. The vertices are colored with $r$ colors, so we can find an arithmetic progression. The vertices corresponding to the numbers in arithmetic progression form an isosceles triangle.

P.S.

I think this has been discussed before. I seem to remember Pierre giving the exact same solution a while back.

Ok! Let's color circle in countable many colors! :D

Are you asking, or do you know about it?

Pierre.

I think I can find a coloring of $\mathbb R$ with countably many colors s.t. no three numbers with the same color form an arithmetic progression, but I'm not sure about it. Maybe someone comes up with something a little bit easier to explain.

Is the problem we have here significantly tougher than the one with $\mathbb R$? It doesn't look like it is..

Show us your example. I am too lazy today to think about it.

I'll write it, but it's too complicated to be Ok. I must have gone astray somewhere :). Just be kind and gentle when you discover my mistakes :D.

Let $(a_i)_{i\in I}$ be a basis of $\mathbb R$ over $\mathbb Q$. We may assume all the elements of the basis are positive (don't know if we need it :)). Each real can be written as $\sum_{j=1}^k r_ja_{i_j}$, where $a_{i_j}$ are elements of the basis listed in increasing order, and $r_j\ne 0$ are rationals (assume $k=0$ in the decomposition of $0$).

Now just color each element of the basis according to its corresponding $k$-tuple $(r_1,r_2,\ldots,r_k)$ (there are countably many, because they're finite sets of rationals). By this I mean that two reals have the same color iff their corresponding $k+1$-tuples $(k,r_1,r_2,\ldots,r_k)$ are identical.

Does it work?

[quote="grobber"]I'll write it, but it's too complicated to be Ok. I must have gone astray somewhere :). Just be kind and gentle when you discover my mistakes :D.[/quote] You are very smart man and best solver on ML. Why do you think there is mistake? :?

Could you just tell me what's wrong? I'm sooo tired of trying to find counterexamples... :) Pleeease? :)

I think, I was wrong. Your construction contains advanced details, so it is correct. I will remove my remark.

:)
I didn't know what you was refusing in Grobber's construction.

Pierre.

Myth asked me it [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=23887]this thread[/url] about the construction for the circle.

By the bijection $f:[0,1)\to [0,\pi)$ given by $f(x)=e^{\pi ix}$ color the points on the upper semi-circle according to the coloring of $[0,1)$ given in the message about the coloring of $\mathbb R$. Then color the lower semi-circle $e^{\pi ix}$ corresponding to $x\in [1,2)$ according to the same scheme, only that each color on the lower semi-circle is different from each color on the upper semi-circle.

I think this takes care of the circle. First of all, since the colors on the lower semi-circle are different from those on the upper semi-circle, any monochromatic isosceles triangle must either be situated in the upper or lower semi-circle. However, on the upper semi-circle, for example, thare are no monochromatic triangles $e^{\pi ia},e^{\pi ib},e^{\pi ic}$, because that would make $a,b,c\in [0,1)$ three numbers in arithmetic progression on the real line, which are colored the same, and our coloring of $\mathbb R$ does not allow that.

Is it Ok?

Just an observation: I think such a construction (the one for the reals) can be turned into a proof that there exist sets of the real line which are not Lebesgue measurable.

Since we have partitioned the real line in countably many sets (according to color), assuming that they are all measurable shows that one of them has positive measure. However, it shouldn't be very hard to show that any Lebesgue measurable real set with positive measure contains three points in arithmetic progression, so we get a contradiction.

Sorry, grobber, but I don't get something!
I don't want to look stupid, but still:
About the decomposition of reals into a base over $\mathbb Q$, how do you know you will have a countable base? Probably it is well-known, but I would like to know if there is a countable base of $\mathbb R$ over $\mathbb Q$.
I think the way you colored the numbers, no two will have the same color, but also, there is no bijection between reals and something countable, such as the number of coloring!
Sorry, maybe I misunderstood your solution!
 :blush:

But there is no countable base of $\mathbb R$ over $\mathbb Q$. I did not use the countability of the base, but the countability of the set of sets of finitely many rationals. The coloring depends on the rational coefficients that you assign to the reals.

Ok, but why then any real can be written as a sum of FINITELY many elements of a base??? Can we choose such a base??
If we would have such a base, couldn't we construct, as you did, a bijection between the rationals and the reals???

Because that's what a base means: every real can be written as a finite linear combination with rational coefficients of elements in the base, Ok? :) How elese could we deal with bases then? We're talking about the reals now, but what about other vector spaces? What would an infinite sum mean in general?

Ok!!!
I think I have finally understood the solution  :blush:
I can say it is great!! :)
Still, is there any known (I mean somehow describeable) base of $\mathbb Q$ over $\mathbb R$ ??? I would really like to know!

I don't know if anyone ever made the attempt to describe one :). By using Zorn's Lemma it can be shown that any vector space has a basis. That's reason enough for us to know that there is one of $\mathbb R$ over $\mathbb Q$.


what is the different betwin accostic and classic guitars?

[quote="vulgarfraction"]Classical is the style of music the guitar's designed for, and acoustic is just the lack of electrical amplification in making the sound, and can be still used for rock and stuff. Don't quote me on that, though. :oops:[/quote]

thanks alot darling ;) ...

on a classical guitar the strings are a little bit more spaced apart than an acoustic, and three of the strings are nylon, also i believe they have a different bridge


[b]Triangle problem
[/b]

Sandu likes to play a lot, so that he received a game with boxes on which there were written strict positive integers. Because he had a great imagination he thought that in a day to build a triangle from boxes such that: a box is placed on two other boxes, which are placed on three other boxes...etc...which are placed on N other boxes (so any box, is on two other boxes).  But Sandu doesn't stop here, and wants that the number from each box from his triangle (excepting the last line) to be equal with the sum of the numbers of the two boxes below (on which the respective box is placed).

[i]Enuntiation[/i]

Write a program which determines if Sandu can construct a triangle of which side length is N, in which the sum of the numbers from all of the boxes to be S, knowing that it can be used how many boxes you want with an arbitrary number in them.

[i]Date in.[/i]

On the first line in the folder "triangle.in" there will be placed the numbers N and S separeted with a space.

[i]Date out.[/i]

On the first N lines from the folder "triangle.out" there will be placed the numbers from the boxes of the triangle: On the line i, there will be written how many i integer strict positive numbers which describe the line i from the triangle. If Sandu cannot construct such a triangle, in the folder it will be written instead the message imposible.

[i]Restrictions and advices[/i]

    * 1 ≤ N ≤ 18
    * 1 ≤ S ≤ 1 000 000
    * If there are possible more than one solutions, then it can be written an arbitrary one of them.

[i]Example:[/i]

triangle.in
3 34

triangle.out
13
6 7
1 5 2

triangle.in
4 26

triangle.out
8
4 4
2 2 2
1 1 1 1

http://www.infoarena.ro  :)
I can remember that I did this one just counting how many times each one of the bottom numbers appears in the total sum and solving that equation in $n$ variables.

Yeah, I would do what spix did. Assuming the bottom row is $a_{1}, a_{2}, .., a_{n}$, you can work out the total in terms of those variables, $c_{1}a_{1}+c_{2}a_{2}+\ldots+c_{n}a_{n}= S$.To solve that equation, brute force should be enough.

In fact, you get $n a_{1}$ and $0.5(n-1)(n+2) a_{2}$ if I calculate correctly, and those two coefficients are coprime. So if we subtract off $c_{1}+\ldots+c_{n}$ from S (so we now allow 0 for the variables), you can solve the equation by a1 and a2 alone for any sum above a reasonably small constant (say 18*0.5*17*20 = 3060 or so).

Well, that still sounds pretty complicated. Is there no other way to do this?


function
real analysis
real analysis unsolved

Let a continue function $f : [a,b]\longrightarrow (-1,1)$ such that $f(a)f(b)<0$. Prove that there is a point $c\in (a,b)$ such that $f(c)=\frac{f(a)+f(b)}{1+f(a)f(b)}.$

Suppose $f(a)\geq -f(b)>0$. Then $0 \leq \frac{f(a)+f(b)}{1+f(a)f(b)} < f(a)$. The conclusion follows from the intermediate value property.

Similar arguments work for other cases.

Nice solution ! My solution is slowly.
Let $F(x)=f(x)[1+f(a)f(b)]-f(a)-f(b),\ a\le x\le b.$ The function F is continue and $=F(a)F(b)=f(b)[f^2(a)-1]f(a)[f^2(b)-1]<0.$ a.s.o.

Your solution is nice, indeed.
I did never think of that.

Technique: You replace c with x and free zero (you pass all in the left part). You obtain the function F and apply him Darboux.


geometry vector circumcircle parallelogram videos trigonometry rhombus

$\text{O}$ is the centre of the circumcircle of $\triangle{ABC}$. Construct parallelogram with $\text{OA,OB}$, with the forth vertex $\text{D}$. Construct another parallelogram with $\text{OC,OD}$ with the fouth vertex $\text{H}$.
(1) $OA=\vec{a}$, $OB=\vec{b}$, $OC=\vec{c}$, Express $\overrightarrow{OH}$ with $\vec{a},\vec{b}, \vec{c}$.
(2) Prove $\text{AH} \perp \text{BC}$
(3) $\angle{BAC}=60^{o}$ and $\angle{ABC}=45^{o}$. $|OA|=|OB|=|OC|=R$. Use $\text{R}$ to express $|\vec{h}|$.

1
[hide]
$\vec{OD} = \vec{OB} + \vec{BD} = \vec b + \vec a$
$\vec{OH} = \vec{OC} + \vec{OD} = \vec a + \vec b + \vec c$
[/hide]

I don't understand the third.  Why the absolute value around variables measuring distance?  Distance is always positive.  Also, what is $\vec h$?  I would assume that $\vec h = \vec{OH}$, and also that the absolute value around it means to find the distance, because the absolute value of a vector is a synonym for magnitude, or distance.

My mistake...Yes $\vec{h}=\overrightarrow{OH}$. And as for the |OA|=|OB|=|OC|=R part, The "absolute value" sign also means norms. I was too lazy to add overrightarrow to the segment signs... :P

Well taking the normal of a vector still points it in the same direction.  It's impossible that those three vectors could point in the same direction... It would make more sense to ask "find OH (distance) in terms of R (circumradius)", although I don't know if those values are always proportional.

I am confused. :?
When we talk about norm, isn't that just the length of the vector? It does not concern anything about the direction.

Well, you may have been taught different terms for things, I don't know.  But I know that the normal of a vector is dividing it by it's magnitude.  This makes the magnitude/direction of a vector 1.  It's very useful in video games because you get perfect movement instead of "pixelated" movement.

To represent a vectors magnitude (same exact thing as distance), you use TWO lines.  So yours would be $||\vec h||$.  And now I understand the problem.

[color=red]1.)[/color]
$\overline {OH}  = \overline {OC}  + \overline {CH}  = \overline {OC}  + \overline {OD}  = \overline {OC}  + \overline {OD}  + \overline {AD}  = \overline {OA}  + \overline {OB}  + \overline {OC} $

[color=red]2.)[/color]
$\overline {AH}  \cdot \overline {BC}  = \left( {\overline {AO}  + \overline {OH} } \right) \cdot \left( {\overline {BO}  + \overline {OC} } \right) = \left( { - \overline {OA}  + \overline {OH} } \right) \cdot \left( { - \overline {OB}  + \overline {OC} } \right) = $
$ = \left( { - \overline {OA}  + \overline {OC}  + \overline {OA}  + \overline {OB} } \right) \cdot \left( {\overline {OC}  - \overline {OB} } \right) = $
$ = \left( {\overline {OC}  + \overline {OB} } \right) \cdot \left( {\overline {OC}  - \overline {OB} } \right) = \left| {\overline {OC} } \right|^2  - \left| {\overline {OB} } \right|^2  = 0$
then:
$\overline {AH}  \bot \overline {BC} $

[color=red]3.)[/color]
$m\left( {\widehat{OAB}} \right) = m\left( {\widehat{ABO}} \right) = \alpha $
$m\left( {\widehat{OBC}} \right) = m\left( {\widehat{BCO}} \right) = \beta $
$m\left( {\widehat{OCA}} \right) = m\left( {\widehat{CAO}} \right) = \gamma $

and
$\left\{ \begin{array}{l}
 \alpha  + \beta  = 45^0  \\
 \beta  + \gamma  = 75^0  \\
 \gamma  + \alpha  = 60^0  \\
 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
 \alpha  = 15^0  \\
 \beta  = 30^0  \\
 \gamma  = 45^0  \\
 \end{array} \right. \Rightarrow \left\{ \begin{array}{l}
 m\left( {\widehat{AOB}} \right) = 150^0  \\
 m\left( {\widehat{BOC}} \right) = 120^0  \\
 m\left( {\widehat{COA}} \right) = 90^0  \\
 \end{array} \right.$

then:
$\left| {\overline {OH} } \right|^2  = \left( {\overline {OA}  + \overline {OB}  + \overline {OC} } \right)^2  = $
$ = \left| {\overline {OA} } \right|^2  + \left| {\overline {OB} } \right|^2  + \left| {\overline {OC} } \right|^2  + 2\overline {OA}  \cdot \overline {OB}  + 2\overline {OB}  \cdot \overline {OC}  + 2\overline {OC}  \cdot \overline {OA}  = $
$ = 3R^2  + 2R^2 \left( {\cos 150^0  + \cos 120^0  + \cos 90^0 } \right) = $
$ = 3R^2  + 2R^2 \left( { - \frac{{\sqrt 3 }}{2} - \frac{1}{2}} \right) = R^2 \left( {2 - \sqrt 3 } \right)$

All right... my answer is a bit weird, so I probably messed up.

[hide]
Since AOBD is a rhombus, $OD=2R%Error. "sinAOB" is a bad command.
$
ODHC is a parallelogram, so $OH=\sqrt{R^2+OD^2 + 2R\cdot OD%Error. "cosDOC" is a bad command.
}$
$=R\sqrt{4\sin^2AOB+2\cos DOC \sin AOB + 1}$
Rough...
Well I created a reference triangle with the same angles, found the side lengths and circumradius.  Then I used law of cosines to find AOB, which is $150$ degrees.
$OH=R\sqrt{\cos DOC + 2}$
Since rhombus diagonals bisect angles, $DOC = \frac{AOB}2 + AOC$.  I used the same method as before to find AOC, and found DOC to be $165$ degrees.
$\cos 165 = -\sin 75 = \frac{-(\sqrt 6 + \sqrt 2)}4$
So the answer is
$OH=\frac{R\sqrt{8-\sqrt6-\sqrt2}}2$
Which maybe you can simplify further... but I've never been good at that.
[/hide]

Actually when $R=1$, my OH is greater than 1, which isn't possible..

deej -- the [i]norm of[/i] a vector is another name for its length.  The [i]normal to[/i] a vector is a vector perpendicular to it.  [i]To normalize[/i] a vector is to take the unit vector in the same direction (to divide by the length).  Does that clear things up at all?

Ah, I see.  I knew the last two.  I have not heard of the term "norm", I assumed it was normal when shobber said it.  Thank you.

I used some geometrical approach. I proved the $\angle{OAH}=30^{o}$, and $\angle{DHA}=60^{o}$.(which implies $AH=AD=DH=R$). Then use the law of cosine in $\triangle{AOH}$.


i'm having some difficulties with my integral homework.. here are some problems

Evaluate:

$ \int\((1+{x}^{1/2})^8 dx$

$ \int \frac {(1+ln x)^{1/2}}{x*lnx}dx$

$ \int \((x)*sin^{-1}(x)dx$

$ \int\((e)^{3x}cos(5x)dx$

help please!!!
i stink at calculus

1. Let $ 1 \plus{} \sqrt {x} \equal{} t.$
2. $ \sqrt {1 \plus{} \ln x} \equal{} t.$
3. $ \sin ^{\minus{}1}x\equal{}t.$
4. Integral by parts two times

$ \int(1 \plus{} \sqrt{x})^8\,dx \equal{} 2\int\frac{(1 \plus{} \sqrt{x})^8}{2 \sqrt{x}} \cdot \sqrt{x}\,dx$

$ \equal{} 2\left(\frac{(1\plus{}\sqrt{x})^9}{9} \cdot \sqrt{x} \minus{} \int\frac{(1\plus{}\sqrt{x})^9}{9} \cdot \frac{1}{2\sqrt{x}}\right)$

$ \equal{} \frac{2}{9} \left(\sqrt{x}(1\plus{}\sqrt{x})^9 \minus{} \frac{(1\plus{}\sqrt{x})^{10}}{10}\right) \plus{} C$

$ \equal{} \frac{(1\plus{}\sqrt{x})^9}{45}\left(9 \sqrt{x} \minus{} 1\right) \plus{} C$

1) Let $ 1 \plus{} \sqrt {x} \equal{} t\Longrightarrow x \equal{} (t \minus{} 1)^2\ \therefore dx \equal{} 2(t \minus{} 1)dt.$

$ \int (1 \plus{} \sqrt {x})^8 dx \equal{} \int t^8\cdot 2(t \minus{} 1)dt \equal{} \frac {1}{5}t^{10} \minus{} \frac {2}{9}t^9 \plus{} Const.$

$ \equal{} \frac {1}{5}(1 \plus{} \sqrt {x})^{10} \minus{} \frac {2}{9}(1 \plus{} \sqrt {x})^9 \plus{} Const.$

2) Let $ \sqrt {1 \plus{} \ln x} \equal{} t\Longrightarrow \ln x \equal{} t^2 \minus{} 1\ \therefore \frac {1}{x}dx \equal{} 2t\ dt.$

$ \int \frac {\sqrt {1 \plus{} \ln x}}{x\ln x}\ dx \equal{} \int \frac {t}{t^2 \minus{} 1}\cdot 2t\ dt \equal{} 2\int \left(1 \plus{} \frac {1}{t^2 \minus{} 1}\right)\ dt$

$ \equal{} \left(2 \plus{} \frac {1}{t \minus{} 1} \minus{} \frac {1}{t \plus{} 1}\right)dt$

$ \equal{} 2t \plus{} \ln \left|\frac {t \minus{} 1}{t \plus{} 1}\right| \plus{} Const.$

$ \equal{} 2\sqrt {1 \plus{} \ln x} \plus{} \ln \left|\frac {\sqrt {1 \plus{} \ln x} \minus{} 1}{\sqrt {1 \plus{} \ln x} \plus{} 1}\right| \plus{} Const.$

3) Let $ \sin ^{ \minus{} 1}x \equal{} t\Longrightarrow x \equal{} \sin t\ \therefore dx \equal{} \cos t\ dt.$

$ \int x\sin ^{ \minus{} 1}x\ dx \equal{} \int t\sin t\cos t\ dt$

$ \equal{} \frac {1}{2}\int t\sin 2t\ dt$ Integral by parts

$ \equal{} \minus{} \frac {1}{4}t\cos 2t \plus{} \frac {1}{8}\sin 2t \plus{} Const.$

$ \equal{} \minus{} \frac {1}{4}\sin ^{ \minus{} 1}x\cdot \cos (2\sin ^{ \minus{} 1}x) \plus{} \frac {1}{8}\sin (2\sin ^ { \minus{} 1}x) \plus{} Const.$

4) Alternative Solution:

$ (e^{3x}\cos 5x)' \equal{} 3e^{3x}\cos 5x \minus{} 5e^{3x}\sin 5x\ \cdots [1]$
$ (e^{3x}\sin 5x)' \equal{} 3e^{3x}\sin 5x \plus{} 5e^{3x}\cos 5x\ \cdots [2]$

$ [1]*3 \plus{} [2]*5: 34e^{3x}\cos 5x \equal{} 3(e^{3x}\cos 5x)' \plus{} 5(e^{3x}\sin 5x)'$, yielding

$ \int e^{3x}\cos 5x\ dx \equal{} \frac {1}{34}e^{3x}(3\cos 5x \plus{} 5\sin 5x) \plus{} Const.$


What is the remainder when $ 19^{2001}$ is divided by 23?

I don't know if there's a faster and simpler solution, but by [url=http://www.artofproblemsolving.com/Wiki/index.php/Fermat%27s_Little_Theorem]Fermat's Little Theorem[/url], $ 19^{22}\equiv1\pmod{23}$, so

$ 19^{2001}\equiv19^{22\cdot90\plus{}21}\equiv19^{21}\equiv\frac{19^{22}}{19}\equiv\frac1{19}\equiv\frac{323}{19}\equiv\boxed{17}\pmod{23}$.

Can someone explain to me how $\frac{1}{19}$ became $\frac{323}{19}$ in the solution above?


geometry
trigonometry
parallelogram
geometry proposed

Let ABC be a triangle. The excircle $ (I_b)$ touches BC,AC at P,R, excircle $ (I_c)$ touches BC,AB at E,D, respectively. $ PR\cap ED\equal{}\{K\}$. Prove that $ AK\equal{}r$ (r is the radius of incircle (I) )

[quote="April"][quote="livetolove212"]Let ABC be a triangle. The excircle $ (I_b)$ touches BC,AC at P,R, excircle $ (I_c)$ touches BC,AB at E,D, respectively. $ PR\cap ED \equal{} \{K\}$. Prove that $ AK \equal{} r$ (r is the radius of incircle (I) )[/quote]
Hi, I am afraid that this problem is old, for example, you can find a solution at [url=http://www.mathlinks.ro/viewtopic.php?p=928863#928863]Three excircles and three points #7[/url][/quote]
I saw your solution. But who can solve this problem without using trigonometry?

From  http://www.mathlinks.ro/viewtopic.php?t=5438
We get the result that RN’BJ  N’NCB are parallelogram respectively
Denote X=PJ∩MN  form the reference we can acquire X= PJ∩MN∩BR

From   http://www.mathlinks.ro/Forum/viewtopic.php?t=164092
 We have AK⊥BC
AB=AR=c  => AX(IbX)⊥BR  =>  AI∥BR  => XA/XIb= BI/IbB   (1)
Notice ID/IbM=BI/IbB  AK/IbM= XA/XIb   (2)
Combined with (1)(2) , we conclude  ID= AK=r


superior algebra
superior algebra solved

f is a permutation of symetric group $S_{10}$

f=

(1 2 3 4 5 6 7 8 9 10)
(2 7 4 6 1 8 5 10 9 3 )

Compute $f^{2005}=fofo...of$
2005 times f

Maybe i am missing the point but i would say, decompose f into cycles, then it becomes easy,

my end result is f^2005
1->2
2->7
7->5
5->1
all the rest is fixed


geometry
3D geometry
prism
rectangle
AMC
AIME
analytic geometry

[quote="4everwise"]Let $EFGH$, $EFDC$, and $EHBC$ be three adjacent square faces of a cube, for which $EC=8$, and let $A$ be the eighth vertex of the cube. Let $I$, $J$, and $K$, be the points on $\overline{EF}$, $\overline{EH}$, and $\overline{EC}$, respectively, so that $EI=EJ=EK=2$. A solid $S$ is obtained by drilling a tunnel through the cube. The sides of the tunnel are planes parallel to $\overline{AE}$, and containing the edges, $\overline{IJ}$, $\overline{JK}$, and $\overline{KI}$. The surface area of $S$, including the walls of the tunnel, is $m+n\sqrt{p}$, where $m$, $n$, and $p$ are positive integers and $p$ is not divisible by the square of any prime. Find $m+n+p$.[/quote]

[hide="I suck at 3-d geometry"]The length of the segment from K to the other place where it meets the cube parallel to AE has length $\sqrt{6^{2}+6^{2}+8^{2}}=2\sqrt{34}$. The prism sort of thing the figure forms has three rectangles of area $2\sqrt{34}\cdot 2\sqrt{2}$ and then you add the corners which have surface area of $2\cdot 2\cdot 3=12$ so the total surface area is $12+24\sqrt{34}$ so $m+n+p=12+24+34=\boxed{070}$.[/hide]

If you revive an old topic, can you please point out what's wrong...?

Maybe I am out of my mind. I'd like to know the correct answer to this problem, then I can follow along and regain my sanity. So how 'bout it? You post that down to earth, sane solution you seem to have. But if otherwise, you can shut your mouth and quit revivin' year old threads.

Something constructive though.
[hide="OUT OF MIND ATTEMPT 2"]Put the cube on a grid, E(0,0,0) and A(8,8,8) then the points (2,0,0) (0,2,0) and (0,0,2) go to (8,6,6) (6,8,6) (6,6,8) and then the inner prism has base $ 6\sqrt{3}$ so three times that, then just the surface area of the original cube minus the little triangles cut out at two of the corners? [/hide]

This is the 2001 AIME II number 15 question I'm pretty sure, and I think the only way is to calculate the area, for an extremely messy answer.

hehehehe. a little feisty, are we? ;)

set your coordinate system so that vertex E, where the drilling starts, is at (8,8,8). then a little visualization (involving some similar triangles, cuz we have parallel lines!) shows that the tunnel meets the bottom face (the x-y plane one) in the line segments joining (1,0,0) to (2,2,0), and (0,1,0) to (2,2,0), and similarly for the other three faces meeting at the origin (by symmetry). so one face of the tunnel is the polygon with vertices (in that order), S(1,0,0), T(2,0,2), U(8,6,8), V(8,8,6), W(2,2,0), and the other two faces of the tunnel are of course congruent to this shape.

observe (like a national holiday!) that this shape is made up of two congruent trapezoids each with height $ \sqrt {2}$ and bases $ 7\sqrt {3}$ and $ 6\sqrt {3}$. so together they make up an area of $ \sqrt {2}(7\sqrt {3} \plus{} 6\sqrt {3}) \equal{} 13\sqrt {6}$. the total area of the tunnel is then $ 3\cdot13\sqrt {6} \equal{} 39\sqrt {6}$. now for the outisde surface area... that's easy... around corners E and the one opposite E we're missing an area of 6. so the outside area is $ 6\cdot 64 \minus{} 2\cdot 6 \equal{} 372$, and so the total surface area is $ 372 \plus{} 39\sqrt {6}$, aaaaand $ 372 \plus{} 39 \plus{} 6 \equal{} 417$, which be the answer!

ughhhh I understood the problem differently and thought it looked like a triangular prism with two pyramids


How is the following $ \text{\LaTeX}$ code translated to commands used in this forum's postings?
[code]\begin{itemize}
\item blabla1
\item blabla2
\end{itemize}
[/code]Thanks.

I've had a number of occasions to take something I'd written as a document that used "itemize" or "enumerate" - usually the latter - and post it here. I've just assumed that that structure doesn't work here. After I've copied and pasted, I then edit the post, replacing the \item commands by numbers at the beginning of lines, and adding or deleting lines in between items as needed. Oh, and while I'm at it, I try to remember to take things like {\it all of} and replace them by the appropriate BBcode commands.

Kent, thank you for your input.

I do the same like you when I switch between my postings in the forum and my LaTeX docs. I thought the list-environment in the forum could provide such an itemize-feature. I would find that awesome. ... just wishing like at Xmas which is coming up soon. ;)


quadratics
algebra
quadratic formula
algebra unsolved

Let $a,b,c \in Q$ with $ac \neq 0$. Given that the equation $ax^2+bxy+cy^2=0$ has a nonzero solution of the form
$(x,y) = (a_0+a_1\sqrt[3]{2}+a_2\sqrt[3]{4}, b_0+b_1\sqrt[3]{2}+b_2\sqrt[3]{4})$ with $a_i,b_i\in Q$, $i = 0,1,2$. Prove that it also has a nonzero rational solution.

Suppose that $(\alpha , \beta)$ is the given solution. If $\beta=0$ then $a\alpha^2$=0, which is impossible beacuse $\alpha$ is nonzero. Therefore $\beta \neq 0$, and $\alpha / \beta$ is a root to the quadratic equation \[at^2+bt+c=0.\] Also, $\alpha / \beta$ is of the form $c_0+c_1\sqrt[3]{2}+c_2\sqrt[3]{4}$ for some rationals $c_0, c_1, c_2,$. Applying the quadratic formula shows that it must also be of the form $d+e\sqrt{f}$ for some rationals $d,e,f$. Therefore $(c_0-d+c_1\sqrt[3]{2}+c_2\sqrt[3]{4})^2$ must be an integer, and after expanding this it is easy to prove that $c_1=c_2=0$. Therefore $\alpha / \beta$ is rational, and $(\alpha / \beta ,0)$ is a nonzero rational solution to the original equation.


function
geometry
perpendicular bisector
combinatorics unsolved
combinatorics

[quote="kai_desmond11"]2n points are on plane . prove that there exists a line with these properties:
1.it bisects the 2n points (half of the points on each side of the line)
2.the sum of the distances of points on one side from the line equals the sum of the distances of points of the other side from the line
(e.g for n=1 the perpendicular bisector of the points is the line with the two properties )[/quote]

I am sorry do you mean the sum of the distances from points to the line or something else?

yes i mean the distances of points from the line,

I hope this is clear (and correct!):

It's a classic problem to show that, given $ 2n$ points in the plane, there exists a line in each direction that divides the points into two halves, each containing $ n$ points (where we allow points on the line to be put into either half, as necessary).  If we consider our plane to be the complex plane, we can define $ \ell(\omega)$ for $ |\omega| \equal{} 1$ to be the line just described parallel to the line through $ 0$ and $ \omega$.  Then we can define $ d(\omega)$ to be the difference between the sum of the distances from the points on the left side of the line to the line and the sum of the distances from the points on the right side of the line to the line (where "right" and "left" are determined relative to the direction of $ \omega$ from $ 0$).  Now, it's clear that we can choose for some fixed $ \omega_0$* that $ \ell(\omega_0) \equal{} \ell( \minus{} \omega_0)$.  Then for this $ \omega_0$ it's also clear that $ d(\omega_0) \equal{} d( \minus{} \omega_0)$.  It's also clear that $ d$ is a continuous function, so by the intermediate value theorem there is some $ \omega$ such that $ d(\omega) \equal{} 0$, as desired.

* for example, choose an $ \omega$ such that $ \ell(\omega)$ must pass through two or more of our $ 2n$ points -- then $ \ell(\omega)$ is uniquely determined.


probability
function
probability and stats

That one is from my old homework assignment, it went along with some hints:
Deduce Stirling formula from Central Limit Theorem.
[hide="Hint"]Consider r.v.'s $ X_i$ with Poisson distribution with $ \lambda\equal{}1$, $ S_n\equal{}\sum_1^n X_i$, and investigate r.v. $ Y_n\equal{}\left(\frac{S_n\minus{}n}{\sqrt{n}}\right)^{\minus{}}$[/hide]

If $ S$ is an integer-valued random variable whose mean equals some integer $ m$ and whose standard deviation is $ \sigma,$ and if $ S$ can be regarded as a sum of a large number of other, identical, distributions, so that the CLT applies, then the probability that $ S$ equals exactly its mean is the same as the probability that $ |S\minus{}m|<\frac12,$ (which is an interval one unit wide), which in turn can be approximated (via the CLT) as $ \frac{1}{\sqrt{2\pi}\sigma}.$ That is the density of the appropriate normal distribution at its mean.

In this case we have that $ S_n$ is a Poisson distribution of mean $ n.$ It can be regarded as the sum of $ n$ identical Poisson distributions of mean $ 1$ (or as a sum in many other similar ways.) The standard deviation of $ S_n$ is $ \sqrt{n}.$ From the ideas above, we have that for large $ n,$

$ \frac{1}{\sqrt{2\pi n}}\approx P(S_n\equal{}n)\equal{}\frac{e^{\minus{}n}n^n}{n!}.$

Rearranged, that is precisely Stirling: $ n!\approx \sqrt{2\pi n}\,n^ne^{\minus{}n}.$

The Poisson distribution seems better suited for this than, say, the binomial distribution. Pursuing the same ideas for a "coin flip" binomial distribution - one for which $ p\equal{}\frac12$ - we get

$ \frac{1}{\sqrt{\pi n}}\approx\frac{(2n)!2^{\minus{}2n}}{(n!)^2}.$

While this is certainly consistent with Stirling's approximation, it is less suited to extracting that estimate in the first place.

[quote="Kent Merryfield"] the probability that $ S$ equals exactly its mean is the same as the probability that $ |S \minus{} m| < \frac12,$ (which is an interval one unit wide), which in turn can be approximated (via the CLT) as $ \frac {1}{\sqrt {2\pi}\sigma}.$
[/quote]
I dont understand why it can be approximated by $ \frac {1}{\sqrt {2\pi}\sigma}$.

[quote="Kent Merryfield"] the probability that $ S$ equals exactly its mean is the same as the probability that $ |S \minus{} m| < \frac12,$ (which is an interval one unit wide), which in turn can be approximated (via the CLT) as $ \frac {1}{\sqrt {2\pi}\sigma}.$
[/quote]

The probability that $ S$ equals its mean $ m$ is also the same as the probability that $ |S\minus{}m| < \frac14$, which can be approximated as $ \frac{1}{2\sqrt {2\pi}\sigma}.$  :!:

Well, this problem requires a bit more work, and one approach is stated in the hint above: investigate the negative part of the normed Poisson distribution(look at its expectation!), and study the convergence..

redkimchi:

The convergence in the Central Limit Theorem is convergence in distribution, which is convergence of the cumulative distribution functions. The cdf of the normal distribution is a differentiable function, reasonably well approximated in small regions by a linear function. If $ S$ is an integer-valued random variable, then its cdf is a step function which has jump discontinuities at each integer and is constant in between.

By the CLT, we are asserting that the continuous, locally nearly linear, cdf of the normal distribution is a good approximation to the step function cdf of $ S.$ The height of one jump is $ P(S\equal{}m),$ but this accounts for all of the increase in the cdf of $ S$ over an interval of width $ 1.$ Hence in approximating that by the increase in the cdf of the normal distribution, we must use the increase in that cdf over a one-unit wide interval. That interval wouldn't have to be the centered interval $ (m\minus{}1/2,m\plus{}1/2),$ but that's the most convenient (and likely the most accurate) version. It would not be right to look at the interval $ (m\minus{}1/4,m\plus{}1/4)$ because that would not account for an increase over a span of one unit.

Everything written here looks as it is true, but i dont see how in follows from CLT. CLT allows us to estimate the probability $ P(S_m \equal{} m)$ only up to $ O(\frac {1}{\sqrt {m}})$, which is not good enough in our case. May be there are other limit theorems dealing with this quantities, but it still needs some more arguments.

That's a very general fact- the actual convergence is almost always much faster than that implied by the CLT. We can derive very accurate expressions based on the theorem, but we can't prove them without using other tools.

Another [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=103871]example[/url] of a CLT-type estimate being extremely accurate.

I don't really know the tools we could use to prove something like the real accuracy.

Note what Kent wrote about the reason we should take the unit interval. Actually, the correct formal trick is to look at an interval of length $ 2\delta\sqrt n$ with small fixed $ \delta>0$ around $ n$ and to note that the density of the Poisson distribution doesn't change much over that interval. Then you can apply the CLT to that interval (it now makes sense because after the CLT scaling, it becomes just a fixed interval $ [\minus{}\delta,\delta]$, so it is enough to know that the error in CLT goes to $ 0$. You do not need to have any control over the speed anymore).


function
calculus
derivative
LaTeX
trigonometry
calculus computations

Find the absolute extrema of the following function:
f(x)=4/x+tan(pi.x/8)(1=<x=<2)
My solution:
f '(x)=-4/x^2+pi/8.sec^2(pix/8)
We see that the derivative of f '(x)=8/x^3+(pi/8)^2.2sec^2(pi.x/8).cos(pi.x/8)>0 with every x in the interval [1,2], which means f '(x) is strictly increasing with x in the interval [1,2]. Therefore, f '(x) has maximum at x=2(since f '(x) is strictly increasing), which means f '(x)<0 with x in the interval [1,2]. So, f(x) is the decreasing function with x in the interval [1,2] and it has relative nothing.
Therefore, f(x) has its absolute minimum at x=2, which is:3.
                           its absolute maximum at x=1, which is:4+tan(pi/8)
Does my solution above have any flaw? I would like to have someone check it for me. Thank you in advance.
@moderators: can you help me by adding LATEX into my post?

[quote="ghjk"]

Find the absolute extrema of the following function:
$ f(x) \equal{} \frac {4}{x} \plus{} \tan \left(\frac {\pi x}{8}\right)\ (1\leq x< 2)$
My solution:
$ f '(x) \equal{} \minus{} \frac {4}{x^2} \plus{} \frac {\pi}{8}\sec ^2 (\frac {\pi x}{8})$
We see that the derivative of $ f ''(x) \equal{} \frac {8}{x^3} \plus{} \left(\frac {\pi}{8}\right)^2\cdot 2\sec^2 \left(\frac {\pi x}{8}\right)\cos \left(\frac {\pi x}{8}\right) > 0$ with every x in the interval [1,2], which means $ f '(x)$ is strictly increasing with $ x$ in the interval [1,2]. Therefore, $ f '(x)$ has maximum at $ x \equal{} 2$ (since $ f '(x)$ is strictly increasing), which means $ f '(x) < 0$ with $ x$ in the interval [1,2]. So, $ f(x)$ is the decreasing function with $ x$ in the interval [1,2] and it has relative nothing.
Therefore, $ f(x)$ has its absolute minimum at $ x \equal{} 2$, which is:3.
                           its absolute maximum at $ x \equal{} 1$, which is:$ 4 \plus{} \tan \frac {\pi}{8}$
Does my solution above have any flaw? I would like to have someone check it for me. Thank you in advance.
@moderators: can you help me by adding LATEX into my post?[/quote]

I have edited 2 places in the original context.

kunny

[quote="kunny"]This is wrong.

[quote="ghjk"]
We see that the derivative of f''(x)=8/x^3+(pi/8)^2.2sec^2(pi.x/8).cos(pi.x/8)[/quote][/quote]
Can you explain why it's wrong? I don't see where my mistake is though I checked this derivative :blush:

How about $ \frac{d}{dx} \frac{1}{\cos ^ 2 x}$?

hello, i have found
$ f^'(x)=-\frac{4}{x^2}+\frac{\pi}{8}\left(1+\tan^2(\frac{\pi x}{8})\right)$
and
$ f^{''}(x)=\frac{8}{x^3}+\frac{\pi^2}{8}\tan(\frac{\pi x}{8})\left(1+\tan(\frac{\pi x}{8})^2\right)$
Sonnhard.

[quote="kunny"]How about $ \frac {d}{dx} \frac {1}{\cos ^ 2 x}$?[/quote]
Oops, I see it now! It should be 8/x^3+(pi/8)^2.2sec^2(pi.x/8).tan(pi.x/8). Thank you for pointing out my mistake. Luckily, it doesn't change the final result.


abstract algebra
group theory
superior algebra
superior algebra unsolved

also it is easy but nice fact.
assume that $G$ is finite abelian group and $H$ is its subgroup prove there exists a subgroup $L$ such that $G/H$ is isomorph to $L$.

For a cyclic group $G$, suppose $|G| = n$, $G = (a)$, $H = (a^k)$. Then $|H| = \frac{n}{k}$. Consider $L = (a^{\frac{n}{k}})$. Then there is an isomorphism $T: G/H \rightarrow L$ by $T(aH) = a^{\frac{n}{k}}$. So for a cyclic group G is this is true.

Let $G = C_1 \times C_2 \times \cdots \times C_k$, where they are all cyclic, and $H = H_1 \times H_2 \times \cdots \times H_k$, where $H_i \leq C_i$. For each $C_i$ and $H_i$ we can pick a $L_i$. Consider $L = L_1 \times L_2 \times \cdots \times L_k$. This is the required group.

It's not [i]that[/i] simple, since you can't necessarily decompose $H$ into a direct sum of $H_i\le G_i$: think about $G=\mathbb Z_p\times\mathbb Z_p$, and $H$ the subgroup spanned by the element $(1,1)$. $H$ is not a subgroup of either one of the two summands of $G$ which you initially had. The proof can however be repaired (I'd start from $H$, with its expression as a direct sum of cyclic groups, and then extend those subgroups to direct summands of $G$, i.e. a kind of reverse procedure).

There's also a short proof using the duality of groups.


i voted 117 or 115.5, i don't remember which one, but the test while it wasn't as hard as last year's (though it was close), will make ppl do worse, cuz there is no calculators on this test and that will hurt many of the calculator dependant people  such as myself.


analytic geometry
combinatorics proposed
combinatorics

Starting at $ (1,1)$, a stone is moved in the coordinate plan according to the following rules:

$ (i)$ From any point $ (a,b)$, the stone can move to $ (2a,b)$ or $ (a,2b)$.
$ (ii)$ From any point $ (a,b)$, the stone can move to $ (a\minus{}b,b)$ if $ a>b$, or to $ (a,b\minus{}a)$ if $ a<b$.

For which positive integers $ x,y$ can the stone be moved to $ (x,y)$?

$ \gcd(a, b)$ is always a power of $ 2$, so this is a necessary condition.  I believe it is also sufficient.  I'm too lazy to write out the proof, but essentially one inducts on $ |x| \plus{} |y|$:  it's enough to consider the case that $ x$ and $ y$ are both odd and relatively prime.  WLOG $ x < y$.  Then $ \left( x, \frac{x\plus{}y}{2^k} \right) \mapsto (x, x\plus{}y) \mapsto (x, y)$ where $ \frac{x \plus{} y}{2^k}$ is odd and $ |x| \plus{} \left| \frac{x\plus{}y}{2^k} \right| < |x| \plus{} |y|$.


so i found a pattern when looking at some pythag triples. i need to know if u have a counterexample to this or if u can prove its right. SO: FOR ALL PRIMITIVE PYTHAG TRIPLES, $ c^2$-$ b^2$=$ c\plus{}b$. where c is the hypotenuse and b is the greater leg.

like on a 5-12-13, 169-144=13+12
or on a 3-4-5, 25-16=5+4

well, think about it. the cases you are looking at are in the form $ a, b, b\plus{}1$, so obviously, $ c^2\minus{}b^2$, which always equals $ (c\plus{}b)(c\minus{}b)$, will be $ (c\plus{}b)(1)$, or $ c\plus{}b$. look at $ 8, 15, 17$. now, $ c^2\minus{}b^2\equal{}2(b\plus{}c)$.

Now I am not sure it has anything to do with triangles, but for any numbers $ x$ and $ y$ when $ y\equal{}x\plus{}1$, the following is true:
$ y^2\minus{}x^2\equal{}(x\plus{}1)^2\minus{}x^2$
$ y^2\minus{}x^2\equal{}x^2\plus{}2x\plus{}1\minus{}x^2$
$ y^2\minus{}x^2\equal{}2x\plus{}1$
$ y^2\minus{}x^2\equal{}x\plus{}x\plus{}1$
$ y^2\minus{}x^2\equal{}x\plus{}y$

It doesn't work for triples with no consecutive numbers. What you saw is just a coincidince.
For example, the triple $ 8\minus{}15\minus{}17$ is a Pythagorean triple because
$ 8^2\plus{}15^2\equal{}64\plus{}225\equal{}289\equal{}17^2$
But your conjecture does not work because:
$ 17^2\minus{}15^2\equal{}289\minus{}225\equal{}64\not\equal{}17\plus{}15$

So your conjecture is just a lucky pattern in the some Pythagorean triples.

[quote="PowerOfPi"]
So your conjecture is just a lucky pattern in the some Pythagorean triples.[/quote]

It's not really [i]lucky[/i], it's just a characteristic (which can be easily seen) where the values of b and c in a Pythagorean Triplet are consecutive integers.


Do you like Java or C++ better?
Which one do you think is harder?

Java is easier for GUIs, and is more object-oriented, but I would say C++ is easier and more powerful for a simple task. There's no real "better" for programming languages, but I voted Java because I prefer it.
C++ is probably harder to learn, but Java was harder for me.

The general opinion in the programming community seems to be that Java is easy in comparison, but I personally thought the syntax was more difficult to get used to. Although, as Temperal said, Java is generally easier for GUIs.

C# is probably better than C++, but it failed because it's not backwardly compatible with C.

java is a uses system memory to complete tasks but supposedly makes it easier to program while most of the C languages are more efficient yet are harder to use and a little more tedious. I can't really say much;  I've only started learning python and java about a week ago


trigonometry
inequalities
complex numbers
inequalities proposed

[quote="Ramtin Madani"]$A+B+C=\frac{\pi}{2}$ prove that :
$(\sin A+\sin B)^2+(\sin B+\sin C)^2+(\sin C+\sin A)^2\leq3$[/quote]

You forgot to say that the angles A, B and C must be nonnegative. This [b]is[/b] important!

Anyway, since the angles A, B and C are nonnegative and sum up to $\frac{\pi }{2}$, they are all $\leq\frac{\pi}{2}$. Hence, the angles $x=\frac{\pi}{2}-A$, $y=\frac{\pi}{2}-B$ and $z=\frac{\pi}{2}-C$ are all $\geq 0$. On the other hand, these angles are trivially $\leq\frac{\pi}{2}$, and their sum is

$x+y+z=\left(\frac{\pi}{2}-A\right)+\left(\frac{\pi}{2}-B\right)+\left(\frac{\pi}{2}-C\right)=\frac{3\pi}{2}-\left(A+B+C\right)$
$=\frac{3\pi}{2}-\frac{\pi}{2}=\pi$;

hence, these angles x, y, z are the angles of an acute-angled triangle. Hence, according to http://www.mathlinks.ro/Forum/viewtopic.php?t=14567 , we have

$\left(\cos y+\cos z\right)^2+\left(\cos z+\cos x\right)^2+\left(\cos x+\cos y\right)^2\leq 3$.

Now, since $x=\frac{\pi}{2}-A$, $y=\frac{\pi}{2}-B$ and $z=\frac{\pi}{2}-C$, we have cos x = sin A, cos y = sin B and cos z = sin C; thus, this inequality becomes

$\left(\sin B+\sin C\right)^2+\left(\sin C+\sin A\right)^2+\left(\sin A+\sin B\right)^2\leq 3$,

and the problem is solved.

  darij

You can try otherwise, with the complex numbers:
$a=\cos A+i\sin A \in C\ (|a|=1,\ \overline a =\frac 1a)$ a.s.o., where $\underline {abc=i}\Longrightarrow$
$\cos A=\frac{a^2+1}{2a},\ \sin A=\frac{a^2-1}{2ia}$ a.s.o.

[u]Remark.[/u] But for this, you must know some remarcable inequalities with the complex numbers. There is a such solution ! Very much inequalities over $R$ can be obtained from some remarcable inequalities over $C$.


inequalities
geometry
trigonometry
triangle inequality

2a, 2b, 2c are angles of a triangle

show $\prod_{cyc} sin(a) \le 1/8$

edit: ya sign wrong.. im fine now

By cyclic product, do you mean just sin a sin b sin c?

In that case, perhaps the sign is wrong?

see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=ineq&t=12640]www.artofproblemsolving.com/Forum/viewtopic.php?highlight=ineq&t=12640[/url]

Whats up Alex?

The equality holds for an equilateral triangle ($2\alpha = 2\beta = 2\gamma = 60^0$) and if one angle goes to zero, the product also does. It should be

$sin(\alpha)\ sin(\beta)\ sin(\gamma) \le \frac 1 8$

where $\alpha + \beta + \gamma = 90^o$.

[quote="tetrahedr0n"]Yes yetti, that is the problem solved in the thread i linked to.[/quote]

Another solution is in this thread:

[url=http://www.mathlinks.ro/Forum/topic-24803.html]Triangle inequality[/url], Geometry, Solved problems.

Couldn't you just use AM-GM to change it to [tex]\displaystyle \frac{\sin(a)+\sin(b)+\sin(c)}{3} \le \frac{1}{2}[/tex]? Then jensen's inequality immediately gives the answer.

You mean $\leq \frac{\sqrt{3}}{2}$ right?

No, this is when [tex]\displaystyle a+b+c=\frac{\pi}{2}[/tex] so Jensen's gives [tex]\displaystyle LHS \le \sin{\frac{a+b+c}{3}} = \sin{\frac{\pi}{6}} = \frac{1}{2}[/tex].


If $ x$, $ y$ and $ z$ are positive reals prove that

$ \sum_{cyclic} \sqrt {3x(x\plus{}y)(x\plus{}z)} \leq \sqrt {4(x\plus{}y\plus{}z)^{3}}$

[quote="delegat"]If $ x$, $ y$ and $ z$ are positive reals prove that

$ \sum_{cyclic} \sqrt {3x(x \plus{} y)(x \plus{} z)} \leq \sqrt {4(x \plus{} y \plus{} z)^{3}}$[/quote]
$ \sum_{cyc} \sqrt {3x(x \plus{} y)(x \plus{} z)} \leq \sqrt {4(x \plus{} y \plus{} z)^{3}}\Leftrightarrow$
$ \Leftrightarrow4(x\plus{}y\plus{}z)^3\geq\sum_{cyc}3x(x\plus{}y)(x\plus{}z)\plus{}6\sum_{cyc}(x\plus{}y)\sqrt{xy(x\plus{}z)(y\plus{}z)}.$
But $ 2\sqrt{xy(x\plus{}z)(y\plus{}z)}\leq x(y\plus{}z)\plus{}y(x\plus{}z).$
Thus, it remains to prove that
$ 4(x\plus{}y\plus{}z)^3\geq\sum_{cyc}3x(x\plus{}y)(x\plus{}z)\plus{}3\sum_{cyc}(x\plus{}y)(2xy\plus{}xz\plus{}yz),$
which is equivalent to $ \sum_{cyc}(x^3\minus{}xyz)\geq0.$

Thanks for this very nice and simple proof.


Sam is hired for a 20-day period. On days that he works, he earns $ \$$60. For each day that he does not work, $ \$$30 is subtracted from his earnings. At the end of the 20-day period, he received $ \$$660. How many days did he not work?

Suppose Sam worked for every single day. His earnings would be 20*60 = 1200 dollars. However, we need to find out how many days he slacked off to earn only 660 dollars. Well, we subtract the maximum and his real earnings to get 540 dollars, and divide that by 90 because a day without work is 90 dollars off (60+30). the number of days is 6 hangovers.

Why is it 90 dollars off?! I thought it was 30(it said in the problem) and got the answer [color=#0ff]18[/color].

you lose a net 90 because you don't earn the $60 you normally get when you work everyday.


I have a 4x4-square such that the lower right corner is missing. Each square contains one letter and two squares can contain the same letter. In one turn you move one square to fill the missing square, if they have a common side. Then the empty square moves too. What kind of algorithm finds the smallest number of moves such that if one reads the letters from left to right an up to down, a particular word is found?

If there isn't an ad hoc algorithm to solve the fifteen puzzle, I think [url=http://en.wikipedia.org/wiki/Backtracking]backtracking[/url] should do the trick.

wait are you trying to reach a specific configuration, or do you want to find a single word?
for the first one, a* can help
for the second one, kevmo is right

Okay. I'm looking a single word. I try to learn that backtracking method.


inequalities
integration
trigonometry
real analysis
real analysis unsolved

For every $ x>0$ prove that $ \int_{x}^{x\plus{}1}\sin{t^2}\,dt\leq\frac{1}{x}$.

For $ x>0,$

$ \int_{x}^{x\plus{}1}\sin t^{2}dt$

$ \equal{}\frac{1}{2}\int_{x}^{x\plus{}1}\frac{2t\sin t^{2}}{t}dt$

$ \equal{}\allowbreak \minus{}\frac{1}{2\left( x\plus{}1\right) }\cos \left( x\plus{}1\right) ^{2}\plus{}\frac{1}{2x}\cos x^{2}\minus{}\frac{1}{2}\int_{x}^{x\plus{}1}\frac{1}{t^{2}}\cos t^{2}\,dt\allowbreak$

$ \leq \frac{1}{2\left( x\plus{}1\right) }\plus{}\frac{1}{2x}\plus{}\frac{1}{2}\int_{x}^{x\plus{}1}\frac{1}{t^{2}}\,dt\allowbreak$

Hence $ \int_{x}^{x\plus{}1}\sin t^{2}dt\leq \allowbreak \frac{1}{x}$


algebra
polynomial
Galois Theory
superior algebra
superior algebra unsolved

Let $a$ be an algebraic integer. If $a^p-a$ is an integer divisible by $p$ for every big enough prime $p$, then $a$ is an integer.

I know this is true because it is a corollary of a strong theorem, but is there any easy proof of this fact???

Is this strong theorem you're referring to, by any chance, the Chebotarev Density Theorem? Because it does follow from that: the minimal polynomial of $a$ factors completely over $\mathbb F_p$ for every sufficeintly large $p$, so the Galois group of this minimal polynomial over $\mathbb Q$ is trivial.

Notice that, in fact, you don't need the assumption that the integer $a^p-a$ is divisible by $p$. All integers $k_p=a^p-a$ with $p$ larger than the degree of $a$ are divisible by $p$:

$f(x)=x^p-x-k_p$, which is a polynomial of degree larger than that of $a$, has $a$ for a root, which means that it's reducible in $\mathbb Z[x]$ and thus in $\mathbb F_p[x]$. However, if $\alpha$ is a root of $f$ in an algebraic closure of $\mathbb F_p$, then the roots of $f$ are $\alpha,\alpha+1,\ldots,\alpha+p-1\ (*)$. If we assume $k_p\ne 0$ (in $\mathbb F_p$), then $\alpha\not\in\mathbb F_p$, and a polynomial with $(*)$ as roots cannot be reducible over $\mathbb F_p$, since the sum of fewer than $p$ of $(*)$ does not belong to $\mathbb F_p$.


Hello. My mom is a travel agent and so she wants to know which airport to direct me to IF i get accepted into Mathcamp.

She needs to know quite soon because we are really financially unstable and my mom needs to reserve tickets or predict prices to make it as affordable as possible.  Thanks.

I believe that this year Mathcamp will be picking up students from two airports: Portland, ME (PWM) and Bangor, ME (BGR).

Just in case it was unclear, that's totally correct --- it's BGR and PWM.  Take whichever is cheaper.