Evaluate $ \int_1^4 \frac {6x \plus{} 5\sqrt {x}}{\sqrt {x \plus{} \sqrt {x}}}\ dx.$

The substitution $ x \equal{} t^2$ works.

Let $ u\equal{}\sqrt{1\plus{}\frac{1}{\sqrt{x}}}$

$ \int_{1}^{4}\frac{6x\plus{}5\sqrt{x}}{\sqrt{x\plus{}\sqrt{x}}}dx$

$ \equal{}\allowbreak \int_{\sqrt{\frac{3}{2}}}^{\sqrt{2}}\frac{4\plus{}20u^{2}}{\left( u^{2}\minus{}1\right) ^{4}}du\allowbreak$

$ \equal{}\allowbreak \int_{\sqrt{\frac{3}{2}}}^{\sqrt{2}}\left( \frac{3}{2\left( u\minus{}1\right) ^{4}}\minus{}\frac{1}{2\left( u\minus{}1\right) ^{3}}\plus{}\frac{3}{2\left( 1\plus{}u\right) ^{4}}\plus{}\frac{1}{2\left( 1\plus{}u\right) ^{3}}\allowbreak \right) du$

$ \equal{}4\left( 4\sqrt{6}\minus{}\sqrt{2}\right)$

 :D

Probably you have a more elegant solution right, Kunny?

For what it's worth: By inspection, try $ (x\plus{}\sqrt{x})^\frac{1}{2}$.

Differentiating, we get

$ \frac{1}{2}(x\plus{}\sqrt{x})^\frac{\minus{}1}{2}(1\plus{}\frac{1}{2}x^\frac{\minus{}1}{2})$

So we need a facor of $ x$ to get $ x$ and $ \sqrt{x}$ in the numerator.

Differentiating $ x(x\plus{}\sqrt {x})^\frac{1}{2}$, we get

$ \frac{x}{2}(x\plus{}\sqrt{x})^\frac{\minus{}1}{2}(1\plus{}\frac{1}{2}x^\frac{\minus{}1}{2}) \plus{} (x\plus{}x^\frac{1}{2})^\frac{1}{2}$

$ \equal{}\frac{(\frac{x}{2} \plus{} \frac{\sqrt{x}}{4}) \plus{} x \plus{} \sqrt{x}}{\sqrt{x \plus{}\sqrt{x}}}$

$ \equal{} \frac{\frac{3x}{2}\plus{}\frac{5\sqrt{x}}{4}}{\sqrt{x\plus{}\sqrt{x}}}$.

So we need to multiply by $ 4$ to get

$ \frac{6x\plus{}5\sqrt{x}}{\sqrt{x \plus{} \sqrt{x}}}$

Therefore, the integral is $ 4x\sqrt{x \plus{} \sqrt{x}}$.

Putting in the limits $ 1$ and $ 4$, we have $ 16\sqrt{6} \minus{} 4\sqrt{2} \equal{} 4(4\sqrt{6} \minus{} \sqrt{2})$.


algebra
polynomial
calculus
integration
superior algebra
superior algebra unsolved

Let $ f\in \mathbb{Z}[X], grad(f)\geq 3$ and $ a,b\in Z,a\neq b$ so that $ f(a),f(b)$ and $ a+b$ are odd numbers. Prove that $ f(x)=0$ has no integer solutions.

suppose $ f(x)$ has integrals solution say at $ x \in \mathbb{Z}$ then we have $ x\minus{}a | f(x)\minus{}f(a)\equal{}\minus{}f(a)$
so $ x\minus{}a$ must be odd similarly $ x\minus{}b$ must be odd since only one of $ a$ or $ b$ can be odd we have contradiction since that would give two parities for $ x$ considering $ x\minus{}a$ and $ x\minus{}b$


Let $ a$ be real numbers such that $ 0<a<1$. Define a sequence $ \{a_n\}$ as follows.

(i) $ a_1\equal{}a$

(ii) For $ n\equal{}2,\ 3,\ 4,\ \cdots$,

$ a_n\equal{}\frac{1}{a_{n\minus{}1}}$ when $ a_{n\minus{}1}\leq 1$.

$ a_n\equal{}a_{n\minus{}1}\minus{}1$ when $ a_{n\minus{}1}>1$.

(1) If $ a\equal{}\sqrt{2}\minus{}1$, then find the value of $ a_{125}$?

(2) Find all values of $ a$ such that $ a_1\equal{}a_6$.

[quote="mszew"](1) $ a_1 \equal{} \sqrt {2} \minus{} 1$
$ a_2 \equal{} \frac {1}{\sqrt {2} \minus{} 1} \equal{} \frac {\sqrt {2} \plus{} 1}{3}$
$ a_3 \equal{} \frac {1}{\frac {\sqrt {2} \plus{} 1}{3}} \equal{} \frac {3}{\sqrt {2} \plus{} 1} \equal{} \sqrt {2} \minus{} 1 \equal{} a_1$
Then $ a_1 \equal{} a_{2k \plus{} 1} \equal{} a_{125}$[/quote]
Wrong! $ a_2 \equal{} \sqrt {2} \plus{} 1,a_3 \equal{} \sqrt {2},a_4 \equal{} \sqrt {2} \minus{} 1\equal{} a_1$ then $ a_{125} \equal{} a_2 \equal{} \sqrt {2} \plus{} 1$
$ a_1 \equal{} a < 1,a_2 \equal{} \frac {1}{a} > 1,a_3 \equal{} \frac {1 \minus{} a}{a}$
$ a_6 \equal{} a_1 \equal{} a < 1$ then $ a_5 \equal{} a \plus{} 1$ (because if $ a_5 < 1$ then $ a_6 \equal{} \frac {1}{a_5} > 1$, then $ a_5 > 1$, it mean $ a_6 \equal{} a_5 \minus{} 1$)
-) Case 1: $ a_4 \equal{} \frac {1}{a \plus{} 1} < 1$ then $ a_3 \equal{} 1 \plus{} \frac {1}{a \plus{} 1}$ OK!
-) Case 2: $ a_4 \equal{} a \plus{} 2$ then $ a_3 \equal{} a \plus{} 3$ or $ \frac {1}{a \plus{} 2}$: OK!


logarithms
trigonometry
induction
strong induction

daca tot am postat pe aici hai sa pun si eu o problema. sper sa va placa. problema este din cartea lui andrei negut.
$a_{n}$ este un sir definit prin
$a_{1}=a_{2}=97$, si
$a_{n+1}=a_{n}a_{n-1}+\sqrt{(a_{n}^{2}-1)(a_{n-1}^{2}-1)}$.
aratati ca numarul $2+\sqrt{2+2a_{n}}$ este patrat perfect, pentru orice $n\in \mathbb{N}^{*}$.

nu se baga nimeni ? :( sunt mai multe solutii la aceasta problema..

Cam de ce clasa e??Daca e de liceu, eu nu ma bag. :ninja:

Fie $\displaystyle z = -i \ln \left( 97 + 56 \sqrt 3 \right)$.

Avem
\begin{eqnarray*} \cos z &=& \frac{\exp(iz) + \exp(-iz)}2 \\ \ &=& \frac{97 + 56 \sqrt 3 + \frac1{97 + 56 \sqrt 3}}2 \\ \ &=& \frac{97 + 56 \sqrt 3 + \frac{97 - 56 \sqrt 3}{97^2 - 56^2 \cdot 3}}2 \\ \ &=& \frac{97 + 56 \sqrt 3 + \frac{97 - 56 \sqrt 3}1}2\\ \ &=& 97 .\end{eqnarray*}

Observăm că dacă $t \in i \mathbb R, \, \text{Im} \, (z) < 0$ satisface $\cos z > 1$, atunci $\sin z \in i \mathbb R$ şi $\text{Im} \, \left( \sin z \right) < -1$ (sper că-i bună observaţia).

Acum fie $\left\{ f_n \right\}_{n \geq 1}$ şirul definit prin $f_1=f_2=1, \, f_{n+2} = f_n + f_{n+1}, \, \forall n \geq 1$.

[b]Lemă.[/b] $a_n = \cos \left( f_n z \right)$.

[hide="Demonstraţie"]
Cazurile $n=1, \, n=2$ au fost demonstrate mai sus.

Pasul $\left( 1,2,\ldots,n \right) \mapsto n+1$ (strong induction): Din relaţia de recurenţă obţinem
\begin{eqnarray*} a_{n+1} &=& \cos \left( f_n z \right) \cos \left( f_{n-1} z \right) + \sqrt{\left( \cos^2 \left( f_n z \right) - 1 \right) \left( \cos \left( f_{n-1} z \right) - 1 \right)} \\ \ &=& \cos \left( f_n z \right) \cos \left( f_{n-1} z \right) + \sqrt{\sin^2 \left( f_n z \right) \sin^2 \left( f_{n-1} z \right)} \\ \ &=& \cos \left( f_n z \right) \cos \left( f_{n-1} z \right) + \left| \sin \left( f_n z \right) \sin \left( f_{n-1} z \right) \right| \end{eqnarray*}

Din observaţia de acum câteva rânduri, deducem că $\sin \left( f_n z \right) \sin \left( f_{n-1} z \right) < 0$, deci
\begin{eqnarray*} a_{n+1} &=& \cos \left( f_n z \right) \cos \left( f_{n-1} z \right) - \sin \left( f_n z \right) \sin \left( f_{n-1} z \right) \\ \ &=& \cos \left( f_n + f_{n-1} \right) z \\ \ &=& \cos \left( f_{n+1} z \right) ,\end{eqnarray*}
care e exact ce voiam.[/hide]

Expresia $2 + \sqrt{2 + 2 a_n}$ este egală cu $2 + 2 \sqrt{\cos^2 \left( \frac{f_n z}2 \right)} = 4 \cos^2 \left( \frac{f_n z}4 \right) = \left( 2 \cos \left( \frac{f_n z}4 \right) \right)^2$. Pe undeva p-acolo trebuie făcută şi observaţia $\cos \left( \frac{f_n z}2 \right) > 1$.

Acum e de ajuns să demonstrăm că $\displaystyle \cos \left( \frac{k z}4 \right) \in \mathbb Z$, oricare ar fi $\displaystyle k \in \mathbb N^\ast$ (sper că nu deranjează pe nimeni cacofonia).

Mai întâi, $\displaystyle 97 + 56 \sqrt 3 = \left( 7 + 4 \sqrt 3 \right)^2 = \left( 2 + \sqrt 3 \right)^4$ şi, făcând câteva calcule, obţinem $\displaystyle \cos \left( \frac{z}4 \right) = 2, \, \cos \left( \frac{z}2 \right) = 7$.

Fie $T_n (x) = \cos \left( n \arccos x \right)$ polinomul Cebâşev de grad $n$. Este binecunoscută relaţia de recurenţă $T_{n+1} (x) = 2x T_n (x) - T_{n-1} (x) \, \left( \star \right)$, care rezultă din identitatea $\cos (n+1) x = 2 \cos x \cos nx - \cos (n-1)x$. Punând $x = \cos \left( \frac{z}4 \right)$ în $\left( \star \right)$, ce voiam noi să demonstrăm iese prin inducţie.

Din moment ce $\left\{ f_n \right\}_{n \geq 1}$ este un subşir al lui $\left\{ 1,2,\ldots \right\}$, expresia dată, $2 + \sqrt{2 + 2 a_n}$, este pătratul unui număr (acest număr fiind chiar par).

Mere???

solutia mea ecam pe aceeasi idee, doar ca eu am folosit functia
$\mbox{ch } x=\frac{e^{x}+e^{-x}}{2}$ (cosinus hiperbolic) si iesea aceeasi recurenta. si numaru ala tot cu ceva supra 4 iesea. adica aceeasi solutie, doar scrisa umpic altfel.


inequalities
geometry
incenter
circumcircle
triangle inequality
geometry proposed

[quote="quykhtn-qa1"]Let$ G,H,I,O$ denote the centroid, orthcenter,incenter and circumcenter of a triangle.Prove that:
(i) $ 2OI \plus{} IH \ge 4OG$
(ii) $ (\sqrt {6} \minus{} 2)OI \ge OG \plus{} IG$
I hope you will like it :)[/quote]
Virgil Nicula and Evarist, no one interested in this inequality? I hope you have a nice solution :)

[quote="quykhtn-qa1"]Let$ G,H,I,O$ denote the centroid, orthcenter,incenter and circumcenter of a triangle.Prove that:
(i) $ 2OI \plus{} IH \ge 4OG$
(ii) $ (\sqrt {6} \minus{} 2)OI \ge OG \plus{} IG$
I hope you will like it :)[/quote]
may be it easy? :maybe:

[quote="quykhtn-qa1"]Let$ G,H,I,O$ denote the centroid, orthcenter,incenter and circumcenter of a triangle.Prove that:
(i) $ 2OI \plus{} IH \ge 4OG$
(ii) $ (\sqrt {6} \minus{} 2)OI \ge OG \plus{} IG$
I hope you will like it :)[/quote]
I am very sad :( ,no one interested in this inequality? :wink:

[quote="sinaj"]I don't have enough time so I write it fast!
I think we can say that HI+OI >= 3OG ( O,G,H are on one line!) So we have to solve that OI>=OG.
We know that OH=3OG and OH^2=R^2+(AH)^2-2R(AH)cos(B-C) = X
And we can find AH/AC with R and sin(B) and sin(A). then we can put R^2-2Rr instead of OI^2 and we solve that OI^2>=OG^2. And that is R^2-2Rr >=  X/9.

I am not sure, but it is maybe true![/quote]

I think it is true.Good! :lol:

I think $ ii)$ is false. By triangle inequality $ |OG|\plus{}|IG| \geq |OI|$ and if it is true

$ (\sqrt{6}\minus{}2)|OI| \geq |OG|\plus{}|IG| \geq |OI| \Longrightarrow$

$ \sqrt{6}\minus{}2 \geq 1$  :wink:

[quote="sinaj"]I don't have enough time so I write it fast!
I think we can say that HI+OI >= 3OG ( O,G,H are on one line!) So we have to solve that OI>=OG.
We know that OH=3OG and OH^2=R^2+(AH)^2-2R(AH)cos(B-C) = X
And we can find AH/AC with R and sin(B) and sin(A). then we can put R^2-2Rr instead of OI^2 and we solve that OI^2>=OG^2. And that is R^2-2Rr >=  X/9.

I am not sure, but it is maybe true![/quote]
Why $ HI\plus{}OI \ge 3 OG$? :wink:

[quote="crazyfehmy"]I think $ ii)$ is false. By triangle inequality $ |OG| \plus{} |IG| \geq |OI|$ and if it is true

$ (\sqrt {6} \minus{} 2)|OI| \geq |OG| \plus{} |IG| \geq |OI| \Longrightarrow$

$ \sqrt {6} \minus{} 2 \geq 1$  :wink:[/quote]
Sorry,must be: $ OI \sqrt{5} \ge (OG\plus{}IG)\sqrt{3}$ :)
Try it


Given a triangle. Find out, and prove, number n such that this triangle can partition into n polygons satisfy no two polygons have the same number of segments.

[quote="DusT"]If we cut it in 2, a quadrilateral and a triangle, isn't it enough, or you have to take in consideration the big triangle too ?[/quote]
n=3 we can cut. a triangle, a quadrilateral, a pentagon.
Showing this is very easy. Try!!


trigonometry
geometry
trig identities
Law of Cosines

What did people think about this year's NEAML? I thought it was slightly more difficult than previous ones, but the team round was pretty hard. We would have gotten a 3 if not for a protest (after which we got a 6).

I'm happy with my $12.50 gift certificate. I got a 15 after making dumb mistakes on Round 1 #2 and Round 6 #1.

I don't see what was so hard about the team round. #1 everyone got, #2 was easy, #3 was easy, #4 was easy. That's an automatic 12.

I thought it wasn't too bad. Of course it's my first time going.

My two mistakes:

Round 4 #1: Didn't see "positive" integers
Round 6 #3: Forgot to take the squareroot after doing Law of Cosines

$\implies 14$

Meh. Too many donuts before xD

The team round was pretty horrible for our team, but on the whole it wasn't actually that difficult in an absolute sense. Our underclassmen weren't very productive ... and neither were some of our upperclassmen. Our team final score was a 66 or something like that (large school division).

Lol, our school:
[code]      Round: 01 02 03 04 05 06
09th grader: 06 06 04 XX XX XX  16
09th grader: 04 XX 06 XX XX 05  15
10th grader: XX XX XX 04 03 03  10
11th grader: XX 03 XX 00 01 XX  04
12th grader: 03 03 XX XX 00 XX  06
12th grader: XX XX 04 01 XX 00  05
             13 12 14 05 04 08  56
                          Team: 06
                                62 = 2nd[/code]

As you see, our upperclassmen combined contributed less than one of our underclassmen did.

Edit: Lol, I just realized one of our 9th graders got the same questions wrong as worthawholebean did on rounds 1 and 6 and got the same score overall.

NEAML results are online:
Team:  http://www.wocomal.org/statistics/neaml/2007/2007n_team.html
Individual:  http://www.wocomal.org/statistics/neaml/2007/2007n_individual.html
(WOCOMAL schools are highlighted.)

Let me know if you see any issues (e.g., misspelled names) so I can fix them.  Coaches' handwriting is not always legible.  Once any errors are worked out I'll copy the pages to the MML and MAML websites next weekend.

Round difficulty:

Round 6.  Average score 1.288 points / Median score 0
Round 5.  Average score 1.421 points / Median score 1
Round 3.  Average score 2.027 points / Median score 1
Round 4.  Average score 2.481 points / Median score 3
Round 1.  Average score 2.834 points / Median score 3
Round 2.  Average score 3.978 points / Median score 4

Based on this sort of data, WOCOMAL has a tiebreaker for students with the same score.  So FYI, if the WOCOMAL tiebreaker were in place for NEAML, the top ten students at NEAML would be

1. Patrick Wu of Acton Boxborough, with 18 points in 3-5-6 (hardest three rounds)
2. David Corwin of Acton Boxborough, with 18 points in 1-4-6
3. Sway Chen of Lexington, with 18 points in 2-3-5
4. Stephanie Choi and Jeffrey Rosen, both of Choate, with 18 points in 2-4-6
6. Karan Takhar of Wheeler, with 18 points in 1-2-4 (easiest three rounds)
--
7. Jacob Sharpe of Canton, with 17 points in 4-5-6
8. Stephen Xu of Lexington, with 17 points in 1-5-6
9. Noah Arbesfeld of Lexington, with 17 points in 1-3-6
10. Zhaoying Xu of Belmont, with 17 points in 2-3-5

But lacking a tiebreaker the first six students are all tied for first place; and the next four are tied for seventh.

Rick Yanco
Worcester Academy

[quote="Ignite168"][quote="calc rulz"]Eh that was a pretty fun NEAML. Played a lot of frisbee, which was pretty fun.[/quote]

LMAO THAT REMINDS ME OF THE PEOPLE THERE THROWING TENNIS BALLS AND FRISBEES. WE COULD NOT STOP LAUGHING AT HALF THE PEOPLE'S COMPLETE LACK OF ATHLETIC ABILITY INCLUDING THE INABILITY TO THROW A FRISBEE PROPERLY, CATCH ANYTHING, OR GET A TENNIS BALL HALFWAY ACROSS THE GYM.

LMAO GOOD TIMES GOOD TIMES.[/quote]

^^ Agreed. Oh... my... God...

Btw, our school got second for small schools, but first for an all English speaking team in small schools

Yeah, that was my school with the tennis ball. We did really badly (the highest individual score was a 10). I blame the loss mainly on the fact that we couldn't push in our chairs (they were connected to the desks) and because we didn't get a bag of cookies (there were none in the bins for my school).

[quote="worthawholebean"][quote="calc rulz"]Eh that was a pretty fun NEAML. Played a lot of frisbee, which was pretty fun. One of the seniors on my team kept "messing up" each round, doing really stupid things like the area of a circle with radius $\sqrt{5}$ is $25\pi$. Then, during the awards ceremony, we found out he really got an 18  :rotfl:  And I got a nice 18 and we got in second (why didn't anyone listen to me and sabotage lexington  :diablo:  :D ).[/quote]
Our coach was proctoring Lexington. Apparently after round 2 one of the kids on Lexington turned over his paper after 3  minutes and went "pfffft", as if it were the easiest thing ever.[/quote]

I FELL ASLEEP DURING ROUND 1.

I wonder why there was only 1 school from NH in total at NEAML.

Pretty sure Thailand names are worse though.

[quote="rjyanco"][quote="calc rulz"]"Vishunt Gupta" from Acton-Boxborough should be "Vishrut Gupta"[/quote]
Mr. Noeth has wonderful handwriting...[/quote]

That looks like Uishvut more than anything else to me.


inequalities
induction
calculus
calculus computations

Any suggestion to prove the inequality $ a_{n}<(7/4)^n$ in Fibonacci Series? Thanks.

Induction on n.
If $ f_n < \left(\frac {7}{4}\right)^n$ and $ f_{n \plus{} 1} < \left(\frac {7}{4}\right)^{n \plus{} 1}$, you would want to show that $ f_{n \plus{} 2} < \left(\frac {7}{4}\right)^{n \plus{} 2}$.  This can be done fairly quickly using the defining property of a fibonacci sequence.

Alternatively:
setup the fibonacci sequence as a system of difference equations and show that all eigenvalues are less than $ \frac{7}{4}$ would work too, as long as you include a base case there as well.


geometry
circumcircle
geometric transformation
rotation
congruent triangles
geometry proposed

This is a problem from one of our TST's.
Hope you enjoy it:

Let $ABC$ be an equilateral triangle  and $M$ a point in the plane of $ABC$.
Let $A',B',C'$ be the symmetrics point of $M$ wrt $A,B,C$(respectively).

1.Prove that there exist just one point like $P$ such that has a same distance from
$A$ and $B'$ ,From $B$ and $C'$,and from $C$ and $A'$.

2.Let $D$ be the midpoint of $AB$ and $N$ be the intersection of $DM$ and $AP$.
Prove that by the movement of $M$ in the plane (Hmm.everywhere beside of $D$ ,please) the circumcircle of triangle $MNP$ goes through a constant point.

1 is probably obvious. $P$ is the unique point from which there exists a rotation transformation between the two congruent triangles $ABC$ and $B'C'A'$; i.e., $P$ is the intersection of the perpendicular bisectors of $AB'$ and $BC'$.

Perhaps I am mistaken. Is the point reflected $P$ or $A,B,C$?

i dont understand the question.
What do you mean by ?[quote]Is the point reflected $P$or $A,B,C$?[/quote]

If i am right ,you want to know that is $A$ between the $A^'$ and  $M$.
or $M$ is betweet the $A$ and $A^'$.

To clear it ,$A$ is between the $A^'$ and $M$.

basiclly ,yes .1 is easy ,the real problem is the second one.

Here's where the problem is from: http://www.mathlinks.ro/Forum/viewtopic.php?t=42342 .

  darij


Our answer is in the terms $ a^2 \plus{} b^2$ where $ a$ and $ b$ are the side lengths of each square. Using Pythagoras, we see that $ a^2\plus{}b^2 \equal{} 9^2$. So our answer is 81.


limit
trigonometry
calculus
calculus computations

[i]Evaluate :

  $ \mathcal{A} \ \equal{} \ \lim_{x \ \to \ 0} \left( \frac{1\plus{}tgx}{1\plus{}sinx} \right)^{\frac{1}{sinx}}$[/i]

[quote="nguyenvuthanhha"][i]Evaluate :

  $ \mathcal{A} \ \equal{} \ \lim_{x \ \to \ 0} \left( \frac {1 \plus{} tgx}{1 \plus{} sinx} \right)^{\frac {1}{sinx}}$[/i][/quote]

you only need use this limit,this is $ \lim\limits_{x\to 0}(1\plus{}x)^{\frac{1}{x}}\equal{}e$

$ \mathcal{A}\ \equal{}\ \lim_{x\ \to\ 0}\left(\frac{1\plus{}sinx\plus{}tgx\minus{}sinx}{1\plus{}sinx}\right)^{\frac{1}{sinx}} \equal{}
\ \lim_{x\ \to\ 0}\left(1\plus{}\frac{tgx\minus{}sinx}{1\plus{}sinx}\right)^{\frac{1}{sinx}}\equal{}1$

What is the answer?

I got $ e^{tg\minus{}1}$.

[i]With this kind of exercise , we can use the useful lemma :

  [b]Lemma :[/b]

   If when $ x \to x_0$ ; $ u(x) \to 1 \; \ v(x) \to \infty$ then :

          $ \lim_{x \to x_0} u(x)^{v(x)} \ \equal{} \ e^{\beta}$ , with $ \beta \ \equal{} \ \lim_{x \to x_0} [u(x) \minus{}1] v(x)$

     Proof : Try to use the well known limit : $ \lim_{ x \to \infty} \left( 1 \plus{} \frac{1}{x} \right)^x \ \equal{} \ e$:wink:[/i]

[quote="MrMagician"]Oh, could you please explain?[/quote]

$ \mathcal{A}\ \equal{} \ \lim_{x\ \to\ 0}\left(\frac {1 \plus{} sinx \plus{} tgx \minus{} sinx}{1 \plus{} sinx}\right)^{\frac {1}{sinx}} \equal{} \ \lim_{x\ \to\ 0}\left(1 \plus{} \frac {tgx \minus{} sinx}{1 \plus{} sinx}\right)^{\frac {1}{sinx}} \equal{} \lim\limits_{x\to\ 0}(1 \plus{} \frac {tgx \minus{} sinx}{1 \plus{} sinx})^{\frac {1 \plus{} \sin x}{\tan x \minus{} \sin x}\frac {\tan x \minus{} \sin x}{\sin x(1 \plus{} \sin x)}} \equal{} e^{\lim\limits_{x\to\ 0}\frac {\tan x \minus{} \sin x}{\sin x(1 \plus{} \sin x)}} \equal{} e^{\lim\limits_{x\to\ 0}\frac {\tan x \minus{} \sin x}{\sin x}} \equal{} e^{\lim\limits_{x\to\ 0}{(\sec x \minus{} 1})} \equal{} e^0 \equal{} 1$


I'm new to the whole dynamic geometry software thing, and I'm really confused about how to use it. For me, the biggest problem is that I can't figure out how to make an angle a set size without having to drag a point all over the place.

Can anyone help me?

I am not familiar with Geogebra but the manual says:
[quote="Geogebra manual"]Angle with given size
Mark two points A and B and type the angle’s size into the text field of the appearing window. This mode produces a point C and an angle α, where α is the angle ABC.[/quote]
You will find links to help such as the manual, tutorials and a user forum at [url=http://www.geogebra.org/cms/index.php?option=com_content&task=blogcategory&id=75&Itemid=61]GeoGebra Help[/url].

I went there and I didn't help too much...


Typing errors in LaTeX. It should read:

[quote="Lovasz"]Find functions $f: \mathbb{R}\to\mathbb{R}$ satisfy:
\[ f(xy)\cdot\left(f(x)-f(y)\right)=(x-y)\cdot f(x)\cdot f(y)  \][/quote]

Yes, thanks. Edited.
Now you solve it, then I'll ask something... :)

Put $y=1$ and denote $f(1)=c$:

$f(x)(f(x)-c)=(x-1)cf(x)$

Obviously, $f(x)\equiv 0$ is a solution. If we assume $f(x)\not\equiv 0$, then

$f(x)-c=c(x-1)\iff f(x)=cx$.

Obviously, $f(x)\equiv 0$ is included there for $c=0$. Hence, all solutions are given by $f(x)=cx$ where $c\in\mathbb{R}$

What about $f(x)=0$ if $x\leq0$ and $f(x)=cx$ if $x>0$.

The mistake (I think) is there are cases other than $f\equiv0$ or $f\not\equiv0$.

Yeah,that'a question,so at least Farenhajt prove that for all $x\in R$
$f(x)=0$ or $cx$(c is a fixed real number)
We just need to find all subset $A$ of $R$,such that
$f(x)=0(x\in A)$,$cx$ otherwise.satisfied the equation

I think $R-A$ should be close for multiply


Find a number $ x$ so that the arithmetic mean of $ \{7,10,13,19,x\}$ is 14.

You could also notice that the sum of the differences between 14 and each of the elements in the set must equal 0. This gives us $ 7\plus{}4\plus{}1\minus{}5\plus{}14\minus{}x\equal{}0$, from which we can see that $ x\equal{}\boxed{21}$. This may or may not be faster, but it is faster for me.


when four coins are flipped, what is the probability that three of them are heads? how do you do this

please include explanation that will help with similar problems.

The probability of each individual outcome of four flips is $\frac{1}{2^4}=\frac{1}{16}$, since we have two options for each coin. Now, we find the number of possibilities with three heads. You can either say $\binom{4}{3}=4$, or you can write them all out: HHHT, HHTH, HTHH, THHH. Therefore, the total probability is $\frac{1}{4}$. Do the same for similar problems.

[hide]
Think about it this way: for each head, you get an H and for each tail, you get a T.
So all heads would be HHHH and all tails would be TTTT.

There are 4 ways to have 3 H es and the rest T s:
HHHT
HHTH
HTHH
THHH

The probability of each of these is $\frac{1}{2^{4}}$, so the probability of getting 3 heads is $\frac{1}{4}$.
[/hide]

[hide]There are $2^4$ possibilites, or 16. There are three heads, or one tails. There are 4 places the tails can go, either
THHH
HTHH
HHTH
HHHT
$\frac{4}{16}=\boxed{\frac{1}{4}}$[/hide]

For these problems with 1/2 probabilities, you can also use Pascal's triangle (it's supposed to be centered, but this will work):

1
11
121
1331
14641

And so on.  The probability of getting exactly x heads in y flips is $\frac{The\ (x+1)^{th}\ element\ in\ the\ (y+1)^{th}\ row}{2^y}$.  For example, in your case, it is the (3+1)=4th element in the (4+1)=5th row divided by $2^4$, or $\frac{4}{16}=\frac{1}{4}$.

In case you didn't know, each element in Pascal's triangle is the sum of the two elements directly above it.


A bacterial population experiences exponential growth during a four-day period. Assume that 200 colonies were present at the beginning of the first day, but by the beginning of the third day, the population had grown to 335. What's the population at the end of the four-day period?


trigonometry
quadratics
Pythagorean Theorem
geometry

$a,b,c$ are sides of $\triangle{ABC}$. Equation $a(1-x^2)+2bx+c(1+x^2)=0$ has two equal roots. $\lg{(1+\frac{c}{a})}+\lg{a}-\lg{b}=2$.
Find $\sin{A}+\sin{B}+\sin{C}$.

EDIT: $\lg{(1+\frac{c}{a})}+\lg{a}-\lg{b}=\lg2$.

[quote="Slizzel"]The equation $a(1-x^2)+2bx+c(1+x^2)=0$ has two equal roots then $\Delta=0\Longleftrightarrow a^2+b^2-c^2=0$ and $ABC$ is a right triangle so $\sin C=1$ :)[/quote]
First step correct! :D

Yeah I got that too.  And I've tried a bunch of stuff to get the answer, but to no avail.  I even used the formula $\sin A + \sin B + \sin C = 4\cos{ \frac 12 A}\cos{\frac 12 B}\cos{\frac 12 C}$, and simplified to $\frac{10\sqrt{b(b+c)}}{c}$.  I wasn't getting anywhere with what I was doing and kinda lost the drive...

$\lg{\left(1+\frac{c}{a}\right)}+\lg{a}-\lg{b}=2\Longleftrightarrow lg\left(\frac{a+c}{b}\right)=2\Rightarrow \frac{a+c}{b}=100\Leftrightarrow 100b=a+c$.

$\text{But}\ \lg{\left(1+\frac{c}{a}\right)}+\lg{a}-\lg{b}=2\Longleftrightarrow \lg{\left(1+\frac{1}{\sin A}\right)}+\lg{\left(\frac{a}{b}\right)}=2\Longleftrightarrow \frac{a(\sin A +1)}{b\sin A}=$$100$
So $\sin A(100b-a)=1\Longleftrightarrow c\sin A=1$ or $\sin A=\frac{1}{c}$. I think it's correct and also a step forward :D

Well if $\sin A=\frac{1}{c}$ then obviously $a=1$.  I'm sure we could do some substitution to solve for b and c then too.

[quote="shobber"]$a,b,c$ are sides of $\triangle{ABC}$. Equation $a(1-x^2)+2bx+c(1+x^2)=0$ has two equal roots. $\lg{(1+\frac{c}{a})}+\lg{a}-\lg{b}=\lg2$.
Find $\sin{A}+\sin{B}+\sin{C}$.[/quote]
Sorry guys, I just noticed that i made a typo in my problem.

It should be [b]lg2[/b], but not [b]2[/b].

Agghh!  Well, at least it's still a right triangle.

By the way slizzel, in your last post, you forgot the distribute the $a$ in the last step of your second line, thus the real equation is $\sin A = \frac a c$ which is true by the definition of sin in right triangles... so it doesn't help :(

Ahhh I got it.  I really like this problem.

[hide]
If a quadratic has one root (aka two equal roots), then the discriminant is zero.
$4b^2-4(c^2-a^2)=0$
$a^2+b^2=c^2$
So the triangle is a right triangle.  Simplifying the log equation, we get
$a+c=2b$
So, substituting this into the pythagorean theorem
$b^2=(c+a)(c-a)$
$b^2=2b(c-a)$
$b=2(c-a)$
So if we subsitute this back into the simplified log expression, we get
$a+c=4(c-a)$
$5a=3c$
Now the solution is trivial.
$\sin A = \frac a c = \frac 3 5$
From here we see we have a pythagorean triple of 3-4-5, thus
$\sin B = \frac 4 5$
And the desired solution is
$\frac{12}5$
[/hide]


A circular table has exactly 60 chairs around it. There r N people seated at this table in such a way that the next person to be seated must sit next to someone. What is the smallest possible value of N?

[quote="Melissa"][hide]Well, every person already seated can have two empty chairs: one on either side of him. So this means that a third of the chairs need to be filled, $N=60\div{3}=20$. [/hide][/quote]

If there were a chair between every person, wouldn't that violate the rule of having each person sitting next to someone?

No, the NEXT person to be seated has to sit next to somebody :D


Hi guys can you tell me how can i do the electronic configuration for a given element , you know $ s,p,d,f ,s^{2}.....$ . I  am getting confused with this and I think the teacher explained us wrong because she contradicts the chemistry book.

Start filling electrons from $ 1s$ following the $ n\plus{} l$ rule.You can fill 2 electrons in each orbital as per Pauli's Exclusion Principle.For example,configuration of Oxygen will be-
$ 1s^2 2s^2 2p_x^2 2p_y^1 2p_z^1$


function
trigonometry
real analysis
real analysis unsolved

HI! Check convergence of sequence $(f_{n})_{n \in \mathbb{N}}$ in $C([0,1])$, where $\parallel f\parallel =sup\{|f(x)|; x \in [0,1]\}$.

$f_{n}(x)=\frac{sin[e^{nx}(13x^{5}+8x^{2}+5)^{-1}]}{n}$.

Thanks for help.

...so $sup{|f_{n}(x)|}\leq \frac{1}{n}$
$0 \leq \parallel f\parallel \leq \frac{1}{n}$
and
$f_{n}\to 0$?


Let the sequence $S_n=\left\{\frac{1}{2},\frac{5}{3},\frac{11}{8},\frac{27}{19},\dots\right\}$
where for each term the denominator is the sum of the numerator and denominator of the pervious term. while the numerator of each term is the sum of that term's denominator and the pervious term's denominator.

Find the value of $\displaystyle\lim_{n\to\infty}S_n$.

I have a solution but would like to see what you guys can come up with before I post it. :)

[quote="Singular"]a/b = (a+2b)/(a+b)

going to imply a limit of $\sqrt{2}$[/quote]

what he means is, if term $S_n$ is $\frac{a}{b}$ then term $S_{n+1}$ is $\frac{a+2b}{a+b}, S_{n+2}=\frac{3a+4b}{2a+3b}$
etc. which can be shown to have said limit... :P

[hide="here is my proof."]
well lets have $d_n$ be the denominator of the $n^{th}$ term and $a_n$ be the numerator of the $n^{th}$ therm.
so
$d_{n+1}=d_n+a_n$ and $a_{n+1}=d_{n+1}+d_n$
so
$\frac{a_{n+1}}{d_{n+1}}=\frac{d_n+a_n+a_n}{d_n+a_n}\Longrightarrow 1+\frac{a_n}{d_n+a_n}\Longrightarrow1+\frac{1}{1+\frac{a_n}{d_n}}$

so as $n\to\infty$

$\frac{a_{n+1}}{d_{n+1}}=\frac{a_n}{d_n}$ so we let the limit be $L$ and solve for $L$ in

$L=1+\frac{1}{1+L}$ and get $L=\sqrt{2}$
[/hide]
not rigerous by any means but it gets the job done. I know i could show that the sequence is bounded and monotonically non-decreasing...but im sorta lazy right now.

$d_{n+1}=d_n+a_n$ and
$a_{n+1}=d_{n+1}+d_n\\\implies d_{n+1}=d_{n+2}-d_{n+1}-d_n\implies\lambda^2-2\lambda-1=0\implies \lambda=1\pm\sqrt{2}$

Finding a closed formula for $d_n$ and $a_n$ gives us the following limit:

$\lim_{n\rightarrow\infty}\frac{a_n}{d_n}=1+\frac{(1+\frac{\sqrt{2}}{4})(1+\sqrt{2})^{n-1}+(1-\frac{\sqrt{2}}{4})(1-\sqrt{2})^{n-1}}{(1+\frac{\sqrt{2}}{4})(1+\sqrt{2})^{n}+(1-\frac{\sqrt{2}}{4})(1-\sqrt{2})^{n}}\\=1+\frac{1}{1+\sqrt{2}}=\sqrt{2}$

Masoud Zargar

[quote="#20002 pwns"]nice...

now here is a problem I am thinking about:
$\text{Find sum: } 1+\frac{1}{2+\frac{1}{3+\frac{1}{4...}}}$

[hide="hint"]It is about .69777[/hide][/quote]
and the answer is?  :)  tnx

[quote="srulikbd"]and the answer is?  :)  tnx[/quote]

See the table between Equations #4 and #5.
[url=http://mathworld.wolfram.com/ContinuedFraction.html]mathworld.wolfram.com/ContinuedFraction.html[/url]


This might be of help:
$(1-x)(1+x) = 1-x^{2}$

[quote="1234567890"]Find all ordered pairs of real numbers $(x,y)$ for which: \[(1+x)(1+x^{2})(1+x^{4}) = 1+y^{7}\] and \[(1+y)(1+y^{2})(1+y^{4}) = 1+x^{7}.\] [/quote]Oh, I remember this question. I couldn't get it either. I think this should be right though...[hide]Upon inspection, $\boxed{x=y=0}$ and $\boxed{x=y=-1}$ are solutions.

The equations are \begin{eqnarray*}1+x+x^{2}+x^{3}+x^{4}+x^{5}+x^{6}+x^{7}&=&1+y^{7}\\ 1+y+y^{2}+y^{3}+y^{4}+y^{5}+y^{6}+y^{7}&=&1+x^{7}. \end{eqnarray*} Substituting the first equation into the second, \[y+y^{2}+\ldots+y^{5}+y^{6}+x+x^{2}+\ldots+x^{5}+x^{6}=0.\] Suppose, by way of contradiction, that $x=1$. Then \[(1+x)(1+x^{2})(1+x^{4}) = 1+y^{7}\] yields that $y>1$ and \[(1+y)(1+y^{2})(1+y^{4}) = 1+x^{7}\] returns $y<1$, a contradiction. Hence, $x\neq 1$. Similarly, $y\neq 1$. Since $x=1$ or $y=1$ can't possibly be solutions, \[y\left(\frac{y^{6}-1}{y-1}\right)+x\left(\frac{x^{6}-1}{x-1}\right)=0.\] Set \[f(x)=x\left(\frac{x^{6}-1}{x-1}\right).\] A sign analysis of $f(x)$ gives \[f(x)\begin{cases}>0\qquad\text{if}\qquad x>0\\ <0\qquad\text{if}\qquad0>x>-1\\ >0\qquad\text{if}\qquad-1>x.\end{cases}\] Becuase \[f(y)+f(x)=0,\] then for $x,y\neq 0,-1$ exactly one of $\{x,y\}$ must be between $0$ and $-1$. Suppose, without loss of generality, \[0>x>-1.\] Hence, \[\boxed{y>0\qquad\text{or}\qquad-1>y}.\] Then \[0>x^{7}>-1,\] so \[1>1+x^{7}>0.\] Thus, \[1> (1+y)(1+y^{2})(1+y^{4})>0\] Since \[y^{2}+1>1\qquad\text{and}\qquad y^{4}+1>1,\] then \[0<(1+y)<1,\] so \[\boxed{-1<y<0,}\] a contradiction to the bounds on $y$.

Therefore, the only solutions are $\boxed{x=y=0}$ and $\boxed{x=y=-1}$.[/hide]

Nice solution, towerfreak! :D

I knew we had to use bounding somehow, but wasn't sure exactly how to.


Balkan olympiad is coming.I heard it will take place in Kraguevoc,Serbia.From 28 April to 4 May.
I encourage you to leave there comments about such a great event.
Wish good luck to all participants.
Have a nice days in Serbia.
Edited:
Oh,I'm terribly sorry for double posting.
Moderators ,please delete my topic.

Some links:
http://www.mathlinks.ro/viewtopic.php?p=1480801#1480801
http://www.mathlinks.ro/viewtopic.php?t=267242
http://www.dms.org.yu/bmo - official site

Unofficial results for 3 problems (or 2, depends on a country) are on the site:
http://www.dms.org.yu/bmo/index.php?action=results

I see there have been awarded some special prizes. Can somebody post the solutions, for which the special prizes were awarded?


This might be more appropriate in HSB, but I'm unsure of how difficult it is exactly.

In a given n-gon, what is the maximum number of possible right angles that can be in that polygon?

[quote="illcrowflu"][hide]Infinite? Unless we are restricted to convex n-gons?[/hide][/quote]
Convex, sorry :oops:

Also, I mean for a specific n-gon, what's the max in terms of n.

I solved the problem, but I thought it was way cool so I posted it here.

If you consider a square the max is $n$. More generally:
 If we draw the diagonals, a total of $n-3$,  from any one vertex there are $n-2$ triangles that are formed.  Therefore the sum of the measures of the interior angles is $180(n-2)$ which can be written as $90(2n-4)$ and then our answer is $2n-4$.

EDIT: This isn't right  :(

I am tempted to say $n-2$ since there can be only one per triangle, but that doesn't account for complementary pairs that may form from adjacent triangles.

The answer is $n-2$ for all $n > 4$.

[hide="Proof"]
Assume $n-1$ sides have right angles.
$90(n-1) = 90n-90$
$(180n-360)-(90n-90) = 90n-270$.  This means the remaining angle must be $90n-270$.  This is only $<180$ for $n < 5$.

For more than $n-1$, the proof will be similar, but it will give an even smaller range of $n$.
[/hide]


Use the addition formula $\sin (u+v) = \sin u \cos v+\sin v \cos u$ to derive the following identity for the average rate of change of the sine function:

$\frac{\sin (x+h)-\sin x}{h}= \sin x(\frac{\cos h-1}{h})+\cos x (\frac{\sin h}{h}).$

[hide] The Right Hand Side is (from the formula given) \[\frac{\sin x\cos h+\sin h\cos x-\sin x}{h}\]  Now, the Left Hand Side is \[\sin x(\frac{\cos h-1}{h})+\cos x(\frac{\sin h}{h})=\frac{\sin x\cos h-\sin x+\cos x\sin h}{h}\]  This is equal to the Right Hand Side, so the equation works. [/hide]


limit
induction
number theory proposed
number theory

The following has taken my interest:
Suppose 1= p1^a1 p2^a2 ...pk^ak ...
where pi is the ith prime and ai are rational numbers.
does it follow that a1=a2=... =0.

it's obvious  :?
I suppose that your product is finite?
Let $a_i=\frac{p_i}{q_i}$ with $p_i\in\mathbb{Z}$ and $q_i\in\mathbb{N}^*$.
If $q= \prod q_i$ then $1=1^q= \prod p_i^{\alpha_i}$ with $\alpha_i=qa_i\in\mathbb{Z}$.
You can now use unique factorization...

But perhaps your question is:
Can we find a sequence $(a_n)_{n\in\mathbb{N}^*}\in\mathbb{Q}^{\mathbb{N}^*}$ such that: $a_n \not \equiv 0$ and $\lim_{n \rightarrow +\infty} \prod_{k=1}^{n} p_i^{a_i}=1$ ?
I think it is not difficult to build an example with $(a_n)\in\mathbb{Z}^{\mathbb{N}^*}$ and $a_n \not \equiv 0$.

obviuosly, i'm talking about the second example. :?
but I haven't been able to find an example.

[quote="ali"]
($\cdots$) but I haven't been able to find an example.[/quote]

Really ?  :?

$lemma:$ Let $x\in\mathbb{R}_+^*$ and $p>1$ an integer. For all $\epsilon\in\mathbb{R}_+^*$, we can find $a\in\mathbb{Q}$ such that $1<xp^a<1+\epsilon$.
$Proof:$ easy by continuity of $a \mapsto xp^a$.

So we can define $(a_n) \in \mathbb{Q}^{\mathbb{N}^*}$ such that $a_1=1$ and $\forall n>1$ : $1<\prod_{k=1}^{n} p_k^{a_k}<1+\frac{1}{n}$. (by induction choose $a_{n+1}$ such that $1<xp_{n+1}^{a_{n+1}}<1+\frac{1}{n+1}$ by lemma with $x=\prod_{k=1}^{n} p_k^{a_k}$, $p=p_{n+1}$ and $\epsilon=\frac{1}{n+1}$)


calculus
integration
limit
real analysis
real analysis unsolved

Suppose g and
\[f_{n}(n = 1,2,3,...) \]
are defined on (0,oo), are Riemann-integrable on [t,T] whenever 0 < t < T < oo,
\[|f_{n}| \le g,f_{n}\to f \]
uniformly on every compact subset of (0,oo), and
\[\int_{0}^\infty{g(x)dx < \infty }\]
Prove that
\[\mathop{\lim }_{n \to \infty }\int_{0}^\infty{f_{n}(x)dx = \int_{0}^\infty{f(x)dx}}\]

This is essentially a Riemann version of the Dominated Convergence theorem, and we want a similar style of argument.

[hide="Hint"]Estimate $\int_{0}^{\infty}f_{n}(x)-f(x)\,dx$ in two pieces; one by uniform convergence, the other by comparison to $g$.[/hide]


Hi, i'm a grade 12 chem student, need some help while writing a lab report. Does anyone know what is the name of the phenomenon when product of an reaction acts as an catalyst for the reaction.
O and if anyone can suggest some good chemistry sites that i can refer to, it would be extremely helpful thx!

The reaction is called an autocatalysis and the product is an autocatalyst.  To learn more about it, try http://www.iupac.org/publications/analytical_compendium/Cha15all.pdf


geometry
incenter
circumcircle
perpendicular bisector
angle bisector
geometry solved

Let $ \left(O\right)$ be the inscribed circle of a triangle $ ABC$. Assume that this circle $ \left(O\right)$ touches the sides $ AB$, $ BC$, $ CA$ at $ K$, $ M$, $ N$. Let $ I$, $ J$, $ R$, $ S$ be the midpoints of the segments $ AK$, $ AN$, $ CM$, $ CN$. The line $ IJ$ intersects the line $ RS$ at a point $ P$. Prove that the circle $ \left(APC\right)$ touches the circle $ \left(O\right)$.

After we show (by a very mild computation) that $N$ is the symmetric of the foot of the perpendicular to $AC$ through $P$ art the midpoint of $SJ$, it follows easily that $PA=PC$ ($P$ lies on the perpendicular bisector of $CA$).

We can use the converse of Casey's theorem. We must thus show that, if $PT$ is the length of the tangent through $P$ to the incircle, then $(AK+CM)\cdot PA=AC\cdot PT$, and since $AC=AK+CM$, it means that all we need to show is $PA=PT$. In other words, we should show that $\mathcal C(P,PA)$ and the incircle are orthogonal.

Consider the inversion wrt the incircle. The image of $A$ is the midpt of $KN$, while the image of $C$ is the midpt of $MN$. It’s easy to show that these four points lie on $\mathcal C(P,PA)$, so this circle is inveriant wrt the inversion, meaning that it’s orthogonal to the inversion circle, Q.E.D.

I have not solved the problem since I don't have the time now, but I want to make a little remark to your solution, Grobber:

[quote="grobber"]After we show (by a very mild computation) that $N$ is the symmetric of the foot of the perpendicular to $AC$ through $P$ art the midpoint of $SJ$, it follows easily that $PA=PC$ ($P$ lies on the perpendicular bisector of $CA$).[/quote]

Here is a synthetic variant of this proof:

Let O be the incenter of triangle ABC, and let X and Y be the midpoints of the segments NK and MN, respectively. Then, since the points N and K are symmetric to each other with respect to the angle bisector of the angle CAB, i. e. with respect to the line AO, it follows that the line AO is the perpendicular bisector of the segment NK. This means that the midpoint X of the segment NK lies on the line AO, and that the lines NK and AO are perpendicular. Thus, < OXN = 90. On the other hand, < ONA = 90. Hence, < OXN = < ONA. Also, < XON = < NOA. Hence, the triangles OXN and ONA are similar, and it follows that OX : ON = ON : OA, so that $OX\cdot OA = ON^2$. Similarly, $OY\cdot OC = ON^2$. Thus, $OX\cdot OA = OY\cdot OC$. Using the converse of the intersecting chords theorem, this entails that the points A, C, X and Y lie on one circle.

Now, what are the perpendicular bisectors of the segments AX and CY ? Since the lines NK and AO are perpendicular and intersect at the point X, we have < AXN = 90. Thus, the point X lies on the circle with diameter AN. The center of this circle is clearly the midpoint J of the segment AN. Thus, we have JX = JA = JN. From JX = JA, and from the similarly provable equation IX = IA, it follows that the points I and J lie on the perpendicular bisector of the segment AX. In other words, the line IJ is the perpendicular bisector of the segment AX. Similarly, the line RS is the perpendicular bisector of the segment CY. Now, we have seen above that the points A, C, X and Y lie on one circle; the center of this circle, of course, must lie on the perpendicular bisectors of the segments AX, CY, CA and XY. Hence, the center of this circle lies on the lines IJ and RS. Hence, this center is the point P. And now, since we have seen that this center lies on the perpendicular bisector of the segment CA, it follows that the point P lies on the segment CA. We have seen even more; namely, we have also seen that the point P lies on the perpendicular bisector of the segment XY. Whether this can be useful in the problem? I ask this because I find your solution quite nice, Grobber, but I still would like to have a simpler proof, since you use the converse of the Casey theorem, but this converse is a real torture to prove - this is, at least, what [url=http://groups.yahoo.com/group/Hyacinthos/message/7092?expand=1]have been said to me[/url]. And I don't have seen any proof of this converse in all my life.

  Darij

Hi Grobber, your solution can really be simplified. The use of Casey's theorem and its converse is not necessary. Here is the alternative proof (I start at the point where I stopped in my last posting):

We have proven that the points A, C, X and Y lie on one circle k, and that the point P is the center of this circle k.

Since $OX\cdot OA = ON^2$, while O is the center and ON is the radius of the incircle of triangle ABC, the point X is the inverse of the point A in the incircle of triangle ABC. Similarly, the point Y is the inverse of the point C in this incircle. Hence, by a well-known property of inversion, the circle k passing through the points A, C, X and Y is orthogonal to the incircle. Therefore, the incircle stays fixed under inversion with respect to the circle k. On the other hand, this inversion with respect to the circle k maps the line CA to the circumcircle of triangle CPA (since the points C and A, lying on the circle k, stay fixed, and every line is mapped to a circle through the center P of the inversion circle k). Since the incircle of triangle ABC and the line CA are tangent, their images, i. e. the incircle of triangle ABC and the circumcircle of triangle CPA, are tangent, too. And we are done.

Thanks, Grobber, for your idea. And thanks, Romano, for posting this nice problem. BTW, are you really from Afghanistan?!

  Darij

Indeed , I'm a Italian . My mother is an Afghan . I'm very proud of that .
   Idon't know Casey's theorem . Can you tell me about it ,  Grobber or Darij ?

[url=http://mathworld.wolfram.com/CaseysTheorem.html]This[/url] tells the story better than I ever could :D.

Well, the link Grobber gave tells not everything, hence I'm posting my version of the story here:

The [i]Ptolemy theorem[/i] says that if A, B, C and D are four points on a circle, then $AB \cdot CD \pm BC \cdot DA \pm AC \cdot BD = 0$. Hereby, the $\pm$ signs mean "either + or -"; but whether one has to take a + or a - in each case, depends on the arrangement of the points A, B, C and D on the circle. For instance, if the points A, B, C and D are arranged such that the quadrilateral ABCD is convex, then the first $\pm$ should be a +, and the second $\pm$ should be a -; then we have the equality $AB \cdot CD + BC \cdot DA - AC \cdot BD = 0$, or simply $AB \cdot CD + BC \cdot DA = AC \cdot BD$. If the quadrilateral ABCD is not convex, but the quadrilateral ACBD is convex, then both $\pm$ sign should be -, and thus we have $AB \cdot CD - BC \cdot DA - AC \cdot BD = 0$. And so on. The Ptolemy theorem has a converse: If A, B, C and D are four points in the plane such that the equation $AB \cdot CD \pm BC \cdot DA \pm AC \cdot BD = 0$, with the appropriate + and - signs, holds, then the points A, B, C and D lie on one circle.

Now, the Ptolemy theorem has a kind of generalization. In fact, assume that you have not four points A, B, C and D, but four arbitrary circles a, b, c and d. The [i]Casey theorem[/i] then states the following: If there exists a circle k which touches all four circles a, b, c and d, then the equation $t_{ab}t_{cd} \pm t_{bc}t_{da} \pm t_{ac}t_{bd} = 0$, with appropriate + and - signs, holds. Hereby, the value $t_{ab}$ is the so-called [i]tangential distance[/i] of the circles a and b and is defined as follows:

- If both circles a and b touch the circle k internally, or they both touch the circle k externally, then constuct an [i]external common tangent[/i] of the circles a and b; if this common tangent touches the circles a and b at the points U and V, then we call $t_{ab} = UV$.

- If one of the two circles a and b touches the circle k internally and the other one touches k externally, then constuct an [i]internal common tangent[/i] of the circles a and b; if this common tangent touches the circles a and b at the points U and V, then we call $t_{ab} = UV$.

Similarly, the tangential distances $t_{cd}$, $t_{bc}$, $t_{da}$, $t_{ac}$ and $t_{bd}$ are defined.

Just like the Ptolemy theorem, this Casey theorem also has a converse.

Now I would like to explain why the Casey theorem is a generalization of the Ptolemy theorem: Consider four (degenerate) circles a, b, c and d with centers A, B, C and D and radii 0. Then, instead of saying that a circle k touches the circles a, b, c and d, we can just say that the circle k passes through the points A, B, C and D. On the other hand, the tangential distance $t_{ab}$ of the two degenerate circles a and b is simply the distance between the points A and B, so we have $t_{ab} = AB$. Similarly, $t_{cd} = CD$, $t_{bc} = BC$, $t_{da} = DA$, $t_{ac} = AC$ and $t_{bd} = BD$. Thus, the Casey theorem $t_{ab}t_{cd} \pm t_{bc}t_{da} \pm t_{ac}t_{bd} = 0$ rewrites as $AB \cdot CD \pm BC \cdot DA \pm AC \cdot BD = 0$. And this is exactly the assertion of the Ptolemy theorem. This is why the Ptolemy theorem is a special case (or, better, a limiting case) of the Casey theorem. And this is why the Casey theorem is sometimes called the [i]generalized Ptolemy theorem[/i].

Note that the Ptolemy theorem is not the only special case of the Casey theorem. There are some other special cases, too. In fact, when we deduced the Ptolemy theorem from the Casey theorem, we had set the radii of all four circles a, b, c and d to be 0; but there is another interesting special case if we only set the radii of the three circles a, b and c to be 0. Then, the circles a, b and c degenerate to three points A, B and C, while the circle d is still a usual circle. And now, as before, $t_{ab} = AB$, $t_{bc} = BC$, $t_{ac} = AC$, but the tangential distances $t_{da}$, $t_{bd}$, $t_{cd}$ are now the lengths of the tangents from the points A, B and C to the circle d. Hence, the Casey theorem $t_{ab}t_{cd} \pm t_{bc}t_{da} \pm t_{ac}t_{bd} = 0$ rewrites as $AB \cdot t_{cd} \pm BC \cdot t_{da} \pm AC \cdot t_{bd} = 0$. In other words, we get the following theorem: If a circle k passes through the points A, B and C and touches the circle d, and if $t_{da}$, $t_{bd}$, $t_{cd}$ are the lengths of the tangents from the points A, B and C to the circle d, then $AB \cdot t_{cd} \pm BC \cdot t_{da} \pm AC \cdot t_{bd} = 0$, with an appropriate choice of + and - signs. The converse of this fact also holds. And this is exactly what Grobber used in his reasoning.

  Darij

A nice way of defining these tangential distances is by considering cycles innstead of circles, i.e. circles with an orientation. Tow cycles are tangent iff the corresponding circles are tangent and they have the same orientation in the point of contact. Similarly, an oriented line is tangent to two cycles iff it's tangent to the circles and it has the same orientation as the two cycles at the points of contact.


Let $ \triangle{ABC}$ such that $ AB\equal{}5,\ AC\equal{}4$ and $ \angle{B}\equal{}2\angle{A}$. Find the length of $ BC$.

let $ BD$ be the internal bissector of $ <B$, we have triangles $ DBC$ and $ CAB$ are similar so $ DC/CB\equal{}CB/AC$ and we know that $ CD\equal{}(BC.AC)/(AB\plus{}BC)$ , so $ BC(BC\plus{}AB)\equal{}AC^2$ , let$ BC\equal{}x$ then $ x(x\plus{}5)\equal{}16$ so $ x\equal{}\frac{5\plus{} \sqrt{89}}{2}$.

[quote="tchebytchev"]let $ BD$ be the internal bissector of $ < B$, we have triangles $ DBC$ and $ CAB$ are similar so $ DC/CB \equal{} CB/AC$ and we know that $ CD \equal{} (BC.AC)/(AB \plus{} BC)$ , so $ BC(BC \plus{} AB) \equal{} AC^2$ , let$ BC \equal{} x$ then $ x(x \plus{} 5) \equal{} 16$ so $ x \equal{} \frac {5 \plus{} \sqrt {89}}{2}$.[/quote]I think it was meant to be
$ x \equal{} \frac {\minus{}5 \plus{} \sqrt {89}}{2}$

I understand how the triangles are similar, although I would call them triangles $ CBD$ and $ CAB$ to have the corresponding vertices in the same order. I agree that $ DC/CB \equal{} CB/AC$ and $ BC(BC \plus{} AB) \equal{} AC^2$.
(I can get that last expression from the fact that the triangles are similar, but I seem to be following a different reasoning path).
However, I do not know how I am supposed to know that $ CD \equal{} (BC.AC)/(AB \plus{} BC)$. Is it based on a theorem that I should still remember from geometry?

[quote="KMST"]
However, I do not know how I am supposed to know that $ CD \equal{} (BC.AC)/(AB \plus{} BC)$. Is it based on a theorem that I should still remember from geometry?[/quote]
yes  :) , we have in triangle $ ABC$ ,$ DA/DC\equal{}BA/BC (*)$

(we can  easily prove it ,in fact let $ BH$ be an altitude then,$ BH.CD\equal{}2area(BCD)\equal{}BD.BC sin(<B/2) (1)$, and

$ BH.AD\equal{}BD.BA.sin(<B/2) (2)$, from $ (1)$ and $ (2)$ we get the results).

now $ DA\equal{}AC\minus{}DC (**)$,therefore from $ (*)$ and $ (**)$ we find $ CD \equal{} (BC.AC)/(AB \plus{} BC)$.


To benzene, add $ \ce{(CH2CO)2O}$ with $ \ce{AlCl3}$ (I think the reagent is wrong.. If it is, Pardesi correct it..)
Then, add $ \ce{SOCl2}$ followed by $ \ce{AlCl3}$, and then Clemenson's reduction by Zn/Hg with HCl...

dunno if this works for benzene, but it i know it does for toluene  :|

[hide="anywayS,"]

Add succinic anhydride to benzene, followed by $ HF$ and then clemmensons
[/hide]

hope its rite :)

OK, after acylation u get 4-oxo,4 phenyl butanoic acid. But after this, wat is the role of SOCl2 with AlCl3 or that of HF on benzene??

Another doubt: Is clemenson's preffered for reactions of benzene or is Wolff Kishner just as good?

In some cases, Wolf-kishner is better, because of the fact that in clemmenson's , $ HCl$ is added and that can cause some problems. For instance, it can cause dehydration in case of an $ \minus{}OH$ being present :)

Remember that Clemenssen is in acidic medium and Wolf in basic.... So depending on the substrate, use whichever you should.  :lol:

first of all, the compound is called tetralin.
Hash include’s synthesis is rite.
HF acts both as an acid and a dehydrating agent in this case. And you are rite, clemmensons shud not be used here as u can get weird products, u can mozingo method or wolff kishner for the reduction.
And chappli, I don’t follow your second step.

[url=http://www.mathlinks.ro/viewtopic.php?t=179216]this[/url] mite be relevant to this

Hey my addition of$ SOCl_{2}$ changes the COOH to COCl...
Then the $ AlCl_{3}$ acylates the COCl.. Then Clemenson's to make the ring CO -> CH$ _{2}$

Mine is succinic anhydride and not acetic anhydride.. U mite have made a readin error.. I have given CH2 and not CH3 as I din kno how to draw its structure...

yeah, i thought it was acetic anhydride :oops: . then its rite. altho u wud want to use wolff kishner instead of clemmenson's here. :D

Oh ok fine.. Thanks fer the correction.. :lol:

have a look at this
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=167620[/url]
and similarly, clemmensons is not very useful for 1,4 diketones. :D

I know you guys are busy but can anybody mind giving the mechanism for this reaction??

dont you worry. i am not as busy as the rest of the nerds over here. :P
here's the mechanism for F&C alkylation and acylation.
http://en.wikipedia.org/wiki/Friedel-Crafts_reaction
the mech using anhydrides is the same, just that the lewis acid or the protic acid, forms a complex with the O of the anhydride to give an acylium ion, and an acid is formed.


geometry
3D geometry
tetrahedron
pyramid
rhombus

In a tetrahedron, two of the faces are equilateral triangles with side 1 unit. The other two faces are right isosceles triangles. Find the volume of the tetrahedron.

I think it's an impossible figure. If you place the two right isoscles triangles together at an angle of 60 degrees for the third equilateral triangle, then you will have the fourth face being an isoscles triangle with base 1 and the other two sides equal to sqrt(2). Which is not a right triangle.

Correct me if I'm wrong, though.

[quote="frt"]I think it is possible. Let's call the 4 vertices of the tetrahedron $A$, $B$, $C$, and $D$. If $\triangle ABC$ and $\triangle ABD$ are equilateral triangles with side lengths 1 and $CD=\sqrt{2}$, then $\triangle ACD$ and $\triangle BCD$ are right isoscles triangles whose hypotenuse is $CD$.

The area of $\triangle ABC$ is $\frac {\sqrt{3}}{4}$ since it's an equilateral triangle with side length 1.

If we regard $\triangle ABC$ as the base, then the height will be $BD$ because $BD \perp \triangle ABC$.

So, the volume is $\left(\frac {\sqrt{3}}{4}\right)(1)\left(\frac{1}{3}\right)=\frac{\sqrt{3}}{12}$.[/quote]

I think your answer is wrong. If $BD \perp \triangle ABC$, then $\triangle ABD$ is not equilateral

Think of the figure as half of a square pyramid elongated by radical 3 along one axis so that the base of the pyramid is 2 adjacent equilateral triangles, and the axal height is (radical 3)/2.

[hide]
Possibly:
Then, the volume of the normal, nonelongated pyramid would be 1/3 * 1/2 * (radical 3)/2 = (radical 3)/12. This times (radical 3) * 1/2 to elongate it yields 1/8 as the volume. :huh: Please correct me if im wrong.
[/hide]

Or, maybe:
[hide]
Let the base = equilateral triangle and have the other equil tri be perpendicular. Then mirror the figure to form a pyramid with base (radical 3)/2. The height would be (radical 3)/2. The volume = 3/12 = 1/4. This times 1/2 = 1/8.
[/hide]

[quote="frt"][quote="10000th User"]I think your answer is wrong. If $BD \perp \triangle ABC$, then $\triangle ABD$ is not equilateral[/quote]
Oops :blush:  You're right. Never mind my previous post.

I've got a different answer now. Hope it's correct this time...
[hide]If we regard $\triangle ABC$ as the base, then the height would be $\frac{\sqrt{6}}{3}$.
[hide="reason"]
Let $E$ be the midpoint of $AB$. The height of the figure we want to find is the same as the height of $\triangle CDE$ (when we regard CE as the base of the triangle).
Since $CE=DE=\frac{\sqrt{3}}{2}$ and $CD=\frac{\sqrt{2}}{2}$, we find the area of $\triangle CDE=\frac{\sqrt{2}}{4}$ (by breaking the triangle into 2 right triangles).
So, the height is $\frac{2(area)}{CE}=\frac{(\frac{\sqrt{2}}{2})}{(\frac{\sqrt{3}}{2})}=\frac{\sqrt{6}}{3}$
[/hide]
Therefore, the volume is $\frac{1}{3}\left(\frac{\sqrt{3}}{4}\right)\left(\frac{\sqrt{6}}{3}\right)=\frac{\sqrt{2}}{12}$
[/hide]

Also, I tried behemoth571's approach.
[hide]
I agree with his idea that it's a half of a square pyramid, but I think he has different numbers... The original square pyramid has a square base (side length 1)  and the height is $\frac{\sqrt{2}}{2}$.
So the volume of the figure is $\frac{1}{2}\left(\frac{1}{3}\right)(1)\left(\frac{\sqrt{2}}{2}\right)=\frac{\sqrt{2}}{12}$[/hide]
Is it correct?[/quote]
nope sry
[hide="hints"]ok begin with a flat equilateral triangle and think of another same size equilateral triangle lying on top of the first one with one side attached, so it looks like a triangular shaped postcard, or think of it this way, make a 60-120-60-120 angles rhombus and fold it and crease it along the shorter diagonal so that both equilateral triangles overlap, as you start flipping, you are creating a tetrahedron. now if you flip one triangle to be perpendicular to the other triangle lying flat on the table, you can see the other two sides are isosceles triangles. now find the volume of this tetrahedron :)[/hide]

I drew the tetrahedron this way:

Draw an equilateral triangle on a horizontal plane.  Now, draw another equilateral triangle sharing one side with the first triangle.  The plane that the second equilateral triangle is on is vertical and is perpendicular to the horizontal plane.  Find the vertices of each triangle opposite the shared side.  To complete the tetrahedron, draw a line segment joining these two vertices.


inequalities
trigonometry
inequalities unsolved

If $ A,B,C$ are the angles of an acute angled triangle, prove that, \[ \cos (A\minus{}B)  \cos (B\minus{}C) \cos (C\minus{}A)\geq 8  \cos A \cos B \cos C \]

[quote="Moonmathpi496"]If $ A,B,C$ are the angles of an acute angled triangle, prove that,
\[ \cos (A \minus{} B) \cos (B \minus{} C) \cos (C \minus{} A)\geq 8 \cos A \cos B \cos C\]
[/quote]
It's true for all $ A,$ $ B$ and $ C$ such that $ A\plus{}B\plus{}C\equal{}\pi.$ :wink:

[quote="arqady"][quote="Moonmathpi496"]If $ A,B,C$ are the angles of an acute angled triangle, prove that,
\[ \cos (A \minus{} B) \cos (B \minus{} C) \cos (C \minus{} A)\geq 8 \cos A \cos B \cos C\]
[/quote]
It's true for all $ A,$ $ B$ and $ C$ such that $ A \plus{} B \plus{} C \equal{} \pi.$ :wink:[/quote]
But how to prove this :?:
One more question (probably it will be answered with a nice solution of this problem), here we have $ \cos (B \minus{} C)$ etc. So we can not use algebraic substitution directly. How proceed in these type of situation? Probably, some more trigonometric manipulation will do... :maybe:

[quote="Moonmathpi496"][quote="arqady"][quote="Moonmathpi496"]If $ A,B,C$ are the angles of an acute angled triangle, prove that,
\[ \cos (A - B) \cos (B - C) \cos (C - A)\geq 8 \cos A \cos B \cos C\]
[/quote]
It's true for all $ A,$ $ B$ and $ C$ such that $ A + B + C = \pi.$ :wink:[/quote]
But how to prove this :?:
One more question (probably it will be answered with a nice solution of this problem), here we have $ \cos (B - C)$ etc. So we can not use algebraic substitution directly. How proceed in these type of situation? Probably, some more trigonometric manipulation will do... :maybe:[/quote]

$ \cos (A - B) \cos (B - C) \cos (C - A)=$ $ \frac{1}{4}(\cos (2A - 2B) + \cos (2B - 2C)+ \cos (2C - 2A)+1)=$

$ \frac{1}{8}\left(\cos (2A ) + \cos (2B )+ \cos (2C )\right)^2+\frac{1}{8}\left(\sin (2A ) + \sin (2B)+ \sin (2C)\right )^2-\frac{1}{8}=$

$ \frac{1}{8}\left(1+4\cos(A)\cos(B)\cos(C)\right)^2+2\sin^2{(A)}\sin^2{(B)}\sin^2{(C)}-\frac{1}{8}=$

$ \cos(A)\cos(B)\cos(C)+2\cos^2{(A)}\cos^2{(B)}\cos^2{(C)}+2\sin^2{(A)}\sin^2{(B)}\sin^2{(C)}$ $ \geq 8 \cos (A) \cos (B) \cos (C)$

$ \Leftrightarrow \boxed{\cos^2{(A)}\cos^2{(B)}\cos^2{(C)}+\sin^2{(A)}\sin^2{(B)}\sin^2{(C)} \geq \frac{7}{2} \cos (A) \cos (B) \cos (C)}$

$ \cos{(A)}\cos{(B)}\cos{(C)= \frac{s^2-(2R+r)^2}{4R^2}=t \le \frac{1}{8}}$;

$ \sin^2{(A)}\sin^2{(B)}\sin^2{(C)}=\left(\frac{sr}{2R^2}\right)^2= \frac{r^2}{R^2}t+\frac{r^2(2R+r)^2}{4R^4}$

So we only need to prove : $ t^2-\left(\frac{7}{2}-\frac{r^2}{R^2}\right)t +\frac{r^2(2R+r)^2}{4R^4}\ge 0 \Leftrightarrow t \le \frac{7}{4}-\frac{r^2}{2R^2}-\frac{1}{4}\sqrt{ 49-44\frac{r^2}{R^2}-16\frac{r^3}{R^3} }$

$ t=\frac{s^2-(2R+r)^2}{4R^2}\le-\frac{1}{2}+\frac{3}{2}\frac{r}{R}-\frac{1}{2}\frac{r^2}{R^2}+\frac{1}{2}(1-2\frac{r}{R})^{3/2}$ , $ \left (s^2 \le 2R^2+10rR-r^2+2\sqrt{R}(R-2r)^{3/2}\right )$

Let: $ \alpha = \frac{r}{R}$ , $ \left( 0 \le \alpha \le \frac{1}{2} \right)$

$ \Leftrightarrow \frac{9}{4}-\frac{3}{2}\alpha -\frac{1}{2}(1-2\alpha)^{3/2}-\frac{1}{4}\sqrt{49-44\alpha^2-16\alpha^3} \ge 0$

which is true . Equaluty holdes for $ \alpha =0 \Leftrightarrow A=B=\frac{\pi}{2} , C=0$ or $ \alpha = \frac{1}{2} \Leftrightarrow A=B=C=\frac{\pi}{3}$.


real analysis
advanced fields
advanced fields unsolved

In Mattila I read about the problem of mapping an arbitrary set of positive measure to a ball. In one dimension it is pretty easy (try it if you like!). Apparently D. Preiss proved that you can do it for sets in $R^{2}$ as well. That is, if $A \in R^{2}$ with $\lambda^{2}(A) > 0$ (Lebesgue measure) then there is a Lipschitz map which sends $A$ to a disc.

Preiss' paper is recent if it's been published at all. I can't find it. Does anyone know (a) if it was published yet and (b) where I could find it and (c) whether I should bother to try? I imagine it's nontrivial.

The problem is open for subsets of $R^{n}, n > 2$. I find this intriguing but I wanted to see what Preiss did before I was swept up by delusions of grandeur. Could anyone suggest why this problem is so hard when the dimension is greater than 2?

Mattila made a mistake when discussing this problem on p.106 of his book: the map $f$ defined there does not work unless $A$ is closed. (I reproduced this mistake [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=113242]recently[/url]  :blush: thanks to [b]oldboy[/b] for pointing it out.) You first need to find a compact subset of $A$ with positive measure.

As for the two-dimensional case, Preiss' proof is still unpublished. I guess the reason is that a different proof, by Matousek, appeared in 1997. According to [url=http://cvgmt.sns.it/cgi/get.cgi/papers/albcsopre05/ecm.pdf]this preprint[/url] (section 8), the proof by Preiss will be included in the much-anticipated joint paper by Alberti, Csornyei, and Preiss, which is apparently still being written.


LaTeX
superior algebra
superior algebra unsolved

See tha attachment, please.
(P.S. How can I insert latex formulas in my messages?)

Prove that $1,\sqrt{2},\sqrt{3},\sqrt{5},\sqrt{7},\sqrt{10}$ are linearly independant over $\mathbb{Q}$.

(quote this post to see the code)

I've tried but I could not solve it : the closest thing to a solution I have requires to show that a certain equation has no integer solution, and this equation is too complicated. Anyway there must be some nice argument I didn't see ... any idea ?

P.S: another classic problem consists of showing that the $\sqrt{p_{i}}$ are linearly independent (the $p_{i}$ are pairwise distinct prime numbers). Do you really mean $\sqrt{10}$ in your problem ?

Well ... when we know, that $p_i$ has degree $2$ over $\mathbb{Q}(\sqrt{p_1} , ... , \sqrt{p_{i-1}})$, where $p_j$ are distinct primes, then by the proof of the degree formula, the $\sqrt{p_1}^{e_1} * \cdots * \sqrt{p_n}^{e_n}$ with $e_i \in {0,1}$ are a basis of $\mathbb{Q}(\sqrt{p_1} , ... , \sqrt{p_n})$. For $n=4, p_1 = 2, p_2 = 3, p_3 = 5 , p_4 = 7$, this implies the linearly independance of $1,\sqrt{2},\sqrt{3},\sqrt{5},\sqrt{7},\sqrt{10} = \sqrt{2} * \sqrt{5}$.

But how can we show $p_i \notin \mathbb{Q}(\sqrt{p_1} , ... , \sqrt{p_{i-1}})$? This is equivalent to the linear independance of the $\sqrt{p_j}$ ...

$1$, $\sqrt{2}$, $\sqrt{3}$, $\sqrt{5}$. $\sqrt{7}$ are algebraically independent over $\mathbb{Q}$.

In general, this can be shown by induction. In this case, obviously $\sqrt{3}\notin\mathbb{Q}(\sqrt{2})$. If $\sqrt{5}\in\mathbb{Q}(\sqrt{2},\sqrt{3})$, $\sqrt{5}=a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}$ but squaring you obtain a contraddiction with $\sqrt{3}\notin\mathbb{Q}(\sqrt{2})$. Similarly, you can show that $\sqrt{7}\notin\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$.

Now if, by absurd, $\sqrt{10}$ is not linearly independent, $\sqrt{10}=A+B\sqrt{2}+C\sqrt{3}+D\sqrt{5}+E\sqrt{7}$ but squaring you contraddict the algebraically indipendence of $1$, $\sqrt{2}$, $\sqrt{3}$, $\sqrt{5}$, $\sqrt{7}$.

(This problem was given at the entrance exam in Scuola Normale, Pisa).


How many zero-digits(e.g, 100100 has 4 zero digits) are in the product of $ 500\times2\times5\times40\times5$?

I need help.

[hide="This should be in classroom math"]We notice that the $ 2\cdot 500$ gives 3, the two 5s and the 40 give 3, so it's $ 6$. [/hide]

[hide="why"]2*5 adds a 0, so just count how many there are. In this, 500*2 already tells us 3 zeroes, so  it's easier.[/hide]

The last two answers are implicitly only counting trailing zeroes.  You need to actually show that the only zeroes are trailing zeroes.

[quote="DonkeyKong"]Oh.. gee.. Sorry to bother you. I knew that.. :blush:[/quote]

It seems actually more efficient to just multiply this particular problem out than prove that the trailing zeros are the only, which I have no idea how to do.


we had this discussion in my math summer school

how do we proove that

0! = 1 ?

0! is defined as 1 for convenience.  It's nothing that you can prove :).

please use search feature!!! there is no need to create this topic, it's already been discussed
[url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=0%21+zero+factorial&t=42181]www.mathlinks.ro/Forum/viewtopic.php?highlight=0%21+zero+factorial&t=42181[/url]

if there are 3 people and you want to arrange them in diffrent orders then there are 3! ways, if you want to arrange 2 people there are 2! ways, if you want to arrange 0 people there are 0! ways.  Since there is only one way to arrange 0 people 0! must equal to 1.

Agreed with palytoxin, we've discussed this WAY too many times.

PLEASE PLEASE use search on things like these...


Given that $ a, b, c$ are sides of triangle ; prove that:

\[ a^3\plus{}b^3\plus{}c^3\plus{}15abc \leq 3a^2(b\plus{}c)\plus{}3b^2(c\plus{}a)\plus{}3c^2(a\plus{}b)
\]

 :)

$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 15abc \leq 3a^2(b \plus{} c) \plus{} 3b^2(c \plus{} a) \plus{} 3c^2(a \plus{} b)$
WLOG$ a\ge b\ge c$.
$ \Leftrightarrow \sum(2b\plus{}2c\minus{}a)(a\minus{}b)(a\minus{}c)\ge0$
Using Vornicu schur, We need to show $ a(2b\plus{}2c\minus{}a)\plus{}c(2a\plus{}2b\minus{}c)\ge b(2a\plus{}2c\minus{}b)$
$ \Leftrightarrow b^2\plus{}4ac\ge a^2\plus{}c^2$
It is true because $ a^2\plus{}c^2<(b\plus{}c)^2\plus{}c^2\equal{}b^2\plus{}(2bc\plus{}2c^2)\le b^2\plus{}4ac$.
Done.

[quote="shaaam"]Given that $ a, b, c$ are sides of triangle ; prove that:
\[ a^3 \plus{} b^3 \plus{} c^3 \plus{} 15abc \leq 3a^2(b \plus{} c) \plus{} 3b^2(c \plus{} a) \plus{} 3c^2(a \plus{} b)
\]
:)[/quote]
It's stronger:
Given that $ a, b, c$ are sides of triangle ; prove that:
\[ a^3 \plus{} b^3 \plus{} c^3 \plus{}9 abc \leq 2a^2(b \plus{} c) \plus{} 2b^2(c \plus{} a) \plus{} 2c^2(a \plus{} b)
\]

$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 9 abc \leq 2a^2(b \plus{} c) \plus{} 2b^2(c \plus{} a) \plus{} 2c^2(a \plus{} b)$
$ \Leftrightarrow \sum (b\plus{}c\minus{}a)(a\minus{}b)(a\minus{}c)\ge0$
We can assume $ a\ge b\ge c$, we need to show
$ a(b\plus{}c\minus{}a)\plus{}c(a\plus{}b\minus{}c)\ge b(c\plus{}a\minus{}b)\Leftrightarrow (b\plus{}c\minus{}a)(a\plus{}b\minus{}c)\ge0$, It is obvious.

Nice problem, Arqady. Thank you.

[quote="shaaam"]How about this?

Given that $ a, b, c$ are sides of triangle; prove that:
\[ 2(a^3 \plus{} b^3 \plus{} c^3) \plus{} 3(a^2b \plus{} b^2c \plus{} c^2a) \plus{} 3abc \geq 6(ab^2 \plus{} bc^2 \plus{} ca^2)
\]
[/quote]

It's more powered. $ \sum a^3 \plus{}2\sum a^2b \plus{} 3abc\ge 4\sum ab^2$
$ \Leftrightarrow (a\plus{}b\minus{}c)(a\minus{}b)(a\minus{}c)\ge 0$
We assume $ a\ge b\ge c$, we need to $ (a\plus{}b\minus{}c)\ge(b\plus{}c\minus{}a)\Leftrightarrow a\ge c$, It's obviously true.


group theory
abstract algebra
superior algebra
superior algebra unsolved

[quote="karunya"]Let $ G$, $ H$ be cyclic groups. Find all homomorphisms between them.[/quote]

Well... All subgroup are normal, there is exactly one subgroup for any divisor of the group order... Every normal subgroup is the kernel of one homomorphisms...

[quote="karunya"]Can you write that more completely, like a proper solution?[/quote]

Every group in this answer is cyclic.

Let's consider an homomorphism $ f: G \to H$
For the isomorphism theorem the image of $ G/N \cong f(G)$ where $ N \triangleleft G$ and $ ker(f) \equal{} N$. Now, $ f(G) < H$
What you have to test is "Is there a subgroup of $ H$ with order $ |G/N|$?

The easier case is the case with $ (|G|, |H|) \equal{} 1$. In this case there is only one homomorphism... $ g \mapsto e_H$
An other case is when $ |G|$ divides $ |H|$. In this case there is an homomorphism for any subgroup of $ G$ (the subgroup is the kernel of the homomorphism).
The case $ |H|$ divides $ |G|$ is similar to the previous one. In this case there is an homomorphism for any subgroup of $ H$ (the subgroup is the image of the homomorphism).
The last case is $ (|G|, |H|) \equal{} k \neq 1$. In this case there is an homomorphism for any divisor of $ k$.


linear algebra
matrix
linear algebra unsolved

Let A be n*n matrix.
 show e^(trace A)=det(e^A)

Assume $A$ is a complex matrix. For this I like to use Schur's lemma, a weak form of which is this: every complex matrix is similar to an upper triangular matrix. (The full strength is that every matrix is [i]unitarily[/i] similar to an upper triangular matrix; it's just that we don't need unitary in this instance.)

$A=U^{-1}TU$ implies that $A^{k}=U^{-1}T^{k}U$ for $k=0,1,\dots.$ That in turn implies that $e^{A}=U^{-1}e^{T}U.$

Since $\det(U^{-1}BU)=\det(B)$ and $\text{trace}(U^{-1}BU)=\text{trace}(B),$ all we now need to do is to show that for upper triangular matrices $T,$

$\det(e^{T})=e^{\text{trace}(T)}$

This last item is straightforward.

---

Note: Jordan canonical form can also be used here in the same way. It's just that Schur's lemma is a far, far easier thing to prove than Jordan form, and can be put to many of the same uses.


function
inequalities
AMC
AIME
absolute value

Given that a sequence satisfies $x_0=0$ and $|x_k|=|x_{k-1}+3|$ for all integers $k\ge 1,$ find the minimum possible value of $|x_1+x_2+\cdots+x_{2006}|$.

[hide]I'm too lazy to write this one all the way out, but the way I would prove it is the following:
1) Show an example for achieving 27 (-3,0,-3,...,0,3,6,9,... is popular)
2) Prove that for any sequence that returns to 0, the average of all of the terms between and including one of the zeroes is always -3/2.
3) Conclude that the final number must be greater than zero, then prove that the total sum is a function of that last number.
4) Calculate the zero(s) of the function and conclude that it is closest to zero when it's at 27.[/hide]

[hide="Here is a sketch (mostly an intuitive argument):"]
Well, we start our naively so we essentially draw a "tree" starting with $x_0 = 0, x_1 = \pm 3$ etc branching out all of the possibilities.  Let's refer to a "step" as going from $x_k$ to $x_{k+1}$.  So, at every step, we either add $3,6,9,12, \dots$ (one at a time, in that order) or add $-3$ after every two steps (and then continue with the $3,6,9,12,\dots$ sequence).  I guess this makes more sense if you have a diagram...

Anyways, if we let $n$ be the number of times we add up a term from $3,6,9,12, \dots$, we get that

$x_1 + x_2 + \cdots + x_{2006}$
$= 3(1+2+\cdots + n) - \frac{3}{2}(2006 - n)$
$= \frac{3}{2}(n^2+2n - 2006)$ where $0 \leq n \leq 2006$ is an even integer.

We want to get as close to $0$ as we can, so take $n = 44$ and voila...

I hope that somewhat made sense.  :) [/hide]

Yeah, that's what I did, but you'd have to prove everything in between as well, which isn't hard to do. They're basically the same solution, just worded differently.

I'm sorry if this appears a useless post --- please delete it later if reqd! --- but since I have done a similar question before, I was thinking whether squaring the recurrence relation to make it free for mods etc and adding terms as in a telescopic series so that we are left with $x_{2006} ^2$ and the $2 \times$ the sum till 2006 nos plus $9 \times 2006$ etc... (it maybe 2005 --- whatever! ) then, cant we use some inequality( one that establishes some upper limit for $x_{2006} ^2$  to get the result --- this is just a first look hunch --- I;m not sure it fully helps --- sorry in that case!!!

[hide="My shortest formal proof"]First, we state that iff $x_{i-1}\ge0$, $|x_i|=|x_{i-1}|+3$ and iff $x_{i-1}<0$, $|x_i|=|x_{i-1}|-3$. Now suppose $x_i=x_j$ for some $0\le i<j\le2006$. Now, this means that $|x_i|=|x_j|$, and so the number of positive numbers in the set $\{x_i,x_{i+1},\ldots,x_{j-1}\}$ equals the number of negative numbers. Now pair the numbers in this list up in the following way: Whenever a positive and a negative number are adjacent in this progression, pair them up and remove them from this list. We claim that every pair will sum to -3.

If the positive number comes first, then the negative number will have a magnitude three greater, so this is true. If the negative number comes first, then the positive number will have magnitude three smaller, and this will also be true. Now let us examine what happens when we remove those two from the sequence. WLOG, let the numbers be $x_k$ and $x_{k+1}$. Since one is positive and the other is negative, $|x_{k+2}|=|x_{k+1}|\pm3=|x_k|\pm3\mp3=|x_k|=|x_{k-1}+3|$. So the new sequence works under the same criteria as the old one. In this way, we can pair all of the numbers up in this subsequence so the sums of the pairs are -3. Thus, the average of these numbers will be -3/2 for all subsequences that start and end with the same number (not including one of those).

Now, take all of the repeating subsequences out of the original sequence. The only thing that will be left will be a sequence $\{0,3,6,9,\cdots,3k\}$ for some even $k$. Since we started with 2006 terms, we removed $2006-k$ (an even number) with an average of -3/2. Thus, the sum of both this remaining sequence and the removed stuff is $(2006-k)(-3/2)+\sum_{i=1}^k3k=(3/2)(k-2006+k(k+1))=3/2(k^2+2k-2006)$. This must be minimized, so we find the roots: $k^2+2k=2006\implies (k+1)^2=2007$ and $44^2=1936<2007<2025=45^2$. Plugging in $k=44$ yields $(3/2)(2025-2007)=27$ (and $k=42$ yields $-237$, a worse result). Thus, $\fbox{027}$ is the closest to zero this sum can get.[/hide]

Hey guys what about what I said???? A very similar problem has been solved by Engels by my method! There $3$ is replaced by $1$?? Am I misunderstanding sth???

Your idea works very nicely.  The problem was solved in the same way during the AIME Math Jam.

[quote="Rushil"]Exactly as my solution above! The only ppoint is , which I was fretting about, was that if you put $x_{2006}^2 \geq 0$ dont you get the mod to be bigger than a huge number??? What's this?? that was my question above!![/quote]

What are you talking about????!!!!!?!?!?!?! Can you explain your question more clearly please???????

I really do not understand what he is trying to say. I don't see what is wrong with my solution.  and anyway, anythig squared is greater than or equal to zero always

Im really sorry if I was not clear...

Consider the expression proved above by the same method I have highlighted in the previous post i.e. $|x_1+x_2+\cdots+x_{2006}|=\frac{\left|{x_{2007}}^2-9\cdot{2007}\right|}{6}$ (i.e. by squaring the given expression and adding telescopically and cancelling out terms!!)

Now in this expression, considering $x_{2006} ^2 \geq 0$, done we get $| \sum_{i=1} ^{2006} x_i | \geq \frac{| 0 - 9 \times 2007|}{6}$??? Whats the mistake here? Doesnt this give that the mod of the sums is much greater than 27. actually greater than the above sum ???

\[ \left| \sum_{i=1} ^{2006} x_i \right| \geq \frac{| 0 - 9 \times 2007|}{6}  \] does not follow from $x_{2007}^2\ge0$. For example, just because $a>3$ doesn't mean $|a-6|>|3-6|=3$. Was that your problem?

OK, Its not exactly what you said but I realised the error- --- no need to waste a post be declaring it! Anyways, at least my solution matches nat mc's!!

Why does $3 | x_{2007}$ ????

[quote="Rushil"]

Why does $3 | x_{2007}$ ????[/quote]

The problem states $|x_{n}| = |x_{n-1} + 3|$ and $x_0 \equiv 0 \bmod 3$, hence all $x_k \equiv 0 \bmod 3$.

[quote="Rushil"]
Why does $3 | x_{2007}$ ????[/quote]
This means "$3$ divides $x_{2007}$", and it is shown why from the post above

[quote=samrocksnature]and then since it's a problem dispenser you just get lazy and don't check whether 135 is obtainable for x_{2007}[/quote]

Are your eyes ok ? I can donate for your treatment

Solved with [b]sarun[/b], [b]js104[/b], [b]crazyeyemoody[/b], [b]ETS1331[/b], [b]dottedcaculator[/b], [b]v4913[/b], [b]CT17[/b], [b]CyclicISLclesTrapezoid[/b], [b]squareman[/b], and [b]bluelinfish[/b]

Clearly, $x_i\equiv3i\pmod6$. Squaring gives $x_k^2=x_{k-1}^2+6x_{k-1}+9$ so $x_{2007}^2-x_0^2=6(x_1+x_2+\ldots+x_{2006})+9\cdot2007$. To minimize $|x_1+\ldots+x_{2006}|$, we need to make $x_{2007}^2$ close to $9\cdot2007$, which happens when $x_{2007}=135$, so $x_1+x_2+\ldots+x_{2006}=\boxed{027}$.


Prove that if A and B are nxn matrices and AB+A+B=0 than AB=BA

$ AB\plus{}A\plus{}B\equal{}0$ is equivalent to $ AB\plus{}A\plus{}B\plus{}I\equal{}I$ $ \Rightarrow$ $ (A\plus{}I)(B\plus{}I)\equal{}I$ (*), so $ B\plus{}I$ is invertable.Multiplying $ (*)$ to the right with $ (B\plus{}I)^{\minus{}1}$ and then to the left with $ B\plus{}I$ we get that $ (B\plus{}I)(A\plus{}I)\plus{}I$ $ \Rightarrow$ $ BA\plus{}A\plus{}B\plus{}I\equal{}I$, so $ BA\equal{}AB$.


search
number theory proposed
number theory

prove: $ \sum_{i\equal{}1}^{2000}[\frac{2000}{i}]\equal{}\sum_{i\equal{}1}^{2000}d(i)$
d(n) is number of positive divisors of n


Let $A$ be a four-digit number whose digits are extracted from the set ${0;1;2;3}$.It is know that the leftmost digit is equal to the number of 0s of $A$;the second leftmost digit is equal to the number of 1s of $A$;the third leftmost digit is equal to the number of 2s of $A$;the rightmost digit  is equal to the number of 3s of $A$.Dermine the number $A$

[quote="hikaru123"][hide="Funny"]First, we notice that $A$ has no 3s. So the first left digit must be 1 or 2. Try two cases then we find 2 values of A satisfied the hypothesis.[/hide] :lol:[/quote]

Why  we notice that $A$ has no 3s?

If $A$ has $3$ equal digits, then it has only $2$ different digits, and $2$ digits don't occur. So $A$ must have at least $2$ digits $=0$, and the $3$ equal digits are $0$. Then the leftmost digit is $3$ and $A=3000$. Contradiction!


[quote="anthony19891019"]a,b,c>0
sym[1/(a^2+1)]=2

Prove: sym(ab) <= 3/2[/quote]
Maybe it:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=103762


quadratics
modular arithmetic
number theory proposed
number theory

Let $a,b \in \mathbb{N}$ be positive integers. Prove that there exists a positive integer $n$ for which
\[
n \nmid a^{2^n} + b^{3^n}.
\]

Cheers,
--Vesselin

Should be $|$ instead of $\nmid$ I suppose, but then $a=1$,$b=-1$ is a solution for any $n$.

If it would be $\nmid$ then let $a,b=1$  and $n>2$

Vesselin, maybe you forgot about some additional conditions  :? Or am I missing something

Yes, sorry, I omitted the trivial case $a=1,b=-1$ in which $a^{2^n}+b^{3^n}$ is identically zero. The problem is to show that, other than in this trivial case and the other trivial case $a=b=0$, there is an $n$ for which $a^{2^n} + b^{3^n}$. For simplicity, let us assume that $a,b \in \mathbb{N}$; this does not make the problem any easier. I have edited the problem statement. An alternative formulation is given as follows:

Prove that there do not exist positive integers $a,b$ such that $n \mid a^{2^n} + b^{3^n}$ for all $n \in \mathbb{N}$.

Sorry about the misunderstanding!

--Vesselin

I'll give it a shot, but I'm not sure about the following :).

Clearly, $a,b$ must have exactly the same prime divisors, otherwise we could simply take $n$ to be a prime divisor of only one of them. This means that, for large enough $n$, $a^{2^n}|b^{3^n}$. Assume from now on we work only with such $n$, which are, additionally, coprime to $a$ and $b$.

We thus get $n|1+\prod_{i=1}^k p_i^{b_i3^n-a_i2^n}$, where $a=\prod_{i=1}^k p_i^{a_i}$ and $b=\prod_{i=1}^kp_i^{b_i}$. Now, there are infinitely many primes $p$ s.t. $\left(\frac{p_i}p\right)=1,\ \forall i$ (all the $p_i$'s are quadratic residues $\pmod p$), and $p=M_4+3,M_3+2$ (problems could arise with that $3$ over there, but you can check that it works out). Now take $n=p\cdot\varphi\left(\frac {p-1}2\right)$, and assume WLOG that $b_i\ge a_i,\ \forall i$.

We have $b_i3^n-a_i2^n\equiv b_i-a_i\pmod{\frac {p-1}2}$, meaning that $p_i^{b_i3^n-a_i2^n}\equiv p_i^{b_i-a_i}\pmod p$. This means that $p|1+\prod_{i=1}^k p_i^{b_i-a_i}$, but we can make $p$ as large as we want with those properties, which yields a contradiction.

I used $b_i\ge a_i$ so that $p_i^{b_i-a_i}$ are integers. If this is not the case, take $n_0$ large enough so that $b_i3^{n_0}\ge a_i2^{n_0}$ for all $i$, and then take $n$ so that $p|n,\varphi\left(\frac{p-1}2\right)|n-n_0$. We get the same thing, but it was easier to write without this complication.

I hope it's Ok. Otherwise, I would have written a long (for me :)) post for nothing :).


Prove that $A=90^{\circ}\Longleftrightarrow p^{2}+r^{2}=4R^{2}(1+\cos B\cos C)\ .$

We have:
$\angle A=90^\circ\Longleftrightarrow\begin{cases}r=p-a\\ 2S=2pr=bc\end{cases}\Longleftrightarrow pr=p(p-a)\Longleftrightarrow bc=2p^{2}-pa\\ \Longleftrightarrow p^{2}+(p-a)^{2}=a^{2}+bc\Longleftrightarrow p^{2}+r^{2}=a^{2}(1+\cos B\cdot \cos C)$

Use \iff instead \Longleftrightarrow

Example $a=b\iff b=a$


geometry
analytic geometry
incenter
function
similar triangles
geometry solved

[u][b]The author of this posting is : treegoner[/b][/u]
____________________________________________________________________

Here is a problem that I think you will like it :

[b]Original problem:[/b] Let $ABC$ be a triangle and $I$ is its incenter. Let $M, N, P$ be the midpoints of $BC, CA, AB$. Call $D, E, F$ the intersections of the segments $IA, IB, IC$ with the incircle $\left(I\right)$. Prove that $MD, NE, PF$ are concurrent.

[b]Generalized problem:[/b] Let $ABC$ be a triangle and $I$ be its incenter. Let $M, N, P$ be the midpoints of $BC, CA, AB$. Call $A_1, B_1, C_1$ the tangency points of the incircle with $BC, CA, AB$. Call $D, E, F$ the points such that they lie on $IA, IB, IC$ and have the same barycentric coordinates wrt the triangles $AB_1C_1, BC_1A_1, CA_1B_1$. Prove that $MD, NE, PF$ are concurrent.

The solution is extremely simple if one uses the barycentric coordinate. Can anyone find a synthetic proof to them? I have spent 2 hours to look for it without success.

[quote="treegoner"][b]Original problem:[/b] Let $ABC$ be a triangle and $I$ is its incenter. Let $M, N, P$ be the midpoints of $BC, CA, AB$. Call $D, E, F$ the intersections of the segments $IA, IB, IC$ with the incircle $\left(I\right)$. Prove that $MD, NE, PF$ are concurrent.

[b]Generalized problem:[/b] Let $ABC$ be a triangle and $I$ be its incenter. Let $M, N, P$ be the midpoints of $BC, CA, AB$. Call $A_1, B_1, C_1$ the tangency points of the incircle with $BC, CA, AB$. Call $D, E, F$ the points such that they lie on $IA, IB, IC$ and have the same barycentric coordinates wrt the triangles $AB_1C_1, BC_1A_1, CA_1B_1$. Prove that $MD, NE, PF$ are concurrent.[/quote]

Are you sure about the same barycentric coordinates? If yes, then the Generalized problem is wrong, and also, it is not a generalization of the Original problem.

  Darij

[quote="treegoner"]Sorry, I mean the same form of barycentric coordinate. It is difficult to explain: for eample if you call all of the 3 points are orthocenter, or incenters, ..... Maybe it is clear now.  :D[/quote]

Probably what you want to say can be told more clearly using the notion of triangle centers. A [i]triangle center[/i] is defined as a function mapping every triangle to a point in the projective plane, such that if a triangle $\Delta_2$ is the image of another triangle $\Delta_1$ under a similitude transformation, then the point $f\left(\Delta_2\right)$ is the image of the point $f\left(\Delta_1\right)$ under this transformation. For instance, one triangle center is the function f mapping every triangle to its centroid. Another triangle center is the function f mapping every triangle to its symmedian point. Etc....

Now, you claim that for every triangle center f, if we denote $D = f\left(\Delta AB_1C_1\right)$, $E = f\left(\Delta BC_1A_1\right)$, $F = f\left(\Delta CA_1B_1\right) = F$, then the lines MD, NE, PF are always concurrent. Well, such statements about general triangle centers for non-similar triangles are condemned to be wrong: There are many many triangle centers and triangle centers of non-similar triangles can behave very differently. For instance, if f = Nagel point, the lines MD, NE, PF don't concur (in general).

  Darij

Are you sure of it Darij ? I don't think there are any error in my computation. I 'll check it again. Well, may be the triangle $AB_1C_1, BC_1A_1, CA_1B_1$ have special propeties. Who knows ?  :?

[quote="treegoner"]Well, may be the triangle $AB_1C_1, BC_1A_1, CA_1B_1$ have special propeties.[/quote]

Yes, they are isosceles.  :D

But in fact, it is hopeless to look for any properties of "generic" triangle centers of these triangles. You can always find triangle centers for which the corresponding lines MD, NE, PF are not concurrent. But please post here your calculations - I am almost sure that they are actually the solution of another, not less interesting, problem.

  Darij

OOOpppps! A big mistake in my proof. I know it 's wrong because in my memory, if it were right, any point in IA, IB, IC would satisfy too. How were I so stupid like that ?  :blush:
    But what 's about the original problem ? In my private mail with Mr Nguyen Minh Ha, he said he used the Desguarse 's theorem and some computation.  ;)


Solve the equation: $8^{x}(3x+1)=4$ where $x$ is a real number.

[hide="Solution"]Suppose that $a > b$. Then $8^{a}> 8^{b}$ and $3a+1 > 3b+1$. Since $8^{x}$ is always positive, this implies that $8^{x}(3x+1)$ is a strictly increasing function. So the solution must be unique. (can anyone tell me a way to make this more rigorous?)

By trying some obvious values, we can get $x = \frac{1}{3}$, where $8^{\frac{1}{3}}(3\frac{1}{3}+1) = 2(1+1) = 4$. This must be the only solution.[/hide]

[quote](can anyone tell me a way to make this more rigorous?) [/quote]

I'd assume that taking the first derivative and showing it's positive for all real values would be more rigorous, without calculus though...?

[quote="MellowMelon"]Suppose that $a > b$. Then $8^{a}> 8^{b}$ and $3a+1 > 3b+1$. Since $8^{x}$ is always positive, this implies that $8^{x}(3x+1)$ is a strictly increasing function. ... (can anyone tell me a way to make this more rigorous?)[/quote]

This would be totally rigorous, if it were true.  (What more rigor could you possible want?)  The problem is that for general real numbers, $a > b$ and $c > d$ does not imply that $ac > bd$, even if we know $a > b > 0$.  To see that your conclusion (we have a strictly increasing function) is wrong, look at the behavior of the function around $-\infty$: $8^{x}$ gets small in magnitude much more quickly than $3x+1$ gets large in magnitude, so the function must go to 0.  But then at $x =-1$, the function has the value $-\frac14$, so it had to decrease somewhere.  (In fact, if you had taken a derivative, you would have been able to show that the function is decreasing for all $x <-\frac{1+\ln2}{3\ln2}$.  So, you still need to explain why the function takes the value 4 only once, even though it's not strictly increasing.

[hide="But of course..."] For $x <-\frac{1}{3}$ the linear factor is negative while the exponential factor is always positive, so we cannot have a positive product.  And for $x \ge-\frac{1}{3}>-\frac{1+\ln 2}{3 \ln 2}$ the function is clearly increasing because both the linear and exponential factors are increasing and positive. [/hide]


Daca m este nr de solutii distincte ale sistemului
1+ 2004^(x+y)=2000^(x+y)+5^(x+y)
32*x^log(in baza 3 din y)+32*y^log(in baza 3 din x)=81;
Atunci m=?

hello, i have found only two solutions $ x\equal{}\frac{1}{9},y\equal{}\frac{8}{9}$ and $ x\equal{}\frac{8}{9},y\equal{}\frac{1}{9}$
Sonnhard.


$ xy$ divides $ x^2 \plus{} y^2 \minus{} x$ then prove that $ x$ is a perfect square.

We have $ x^2\plus{}y^2\minus{}x\ \vdots\ xy\Rightarrow x^2\plus{}y^2\minus{}x\ \vdots\ x\Rightarrow y^2\ \vdots\ x\Rightarrow y^2\equal{}kx\ ,k\in N$
Then $ x^2\plus{}(k\minus{}1)x\ \vdots\ xy\Leftrightarrow x\plus{}k\minus{}1\ \vdots\ y\Leftrightarrow x\plus{}k\minus{}1\equal{}ly\ ,l\in N$
It mean $ \frac{y^2}{k}\plus{}k\minus{}1\equal{}ly\Leftrightarrow y^2\minus{}kl.y\plus{}k(k\minus{}1)\equal{}0$
$ \Delta\equal{}k^2l^2\minus{}4k(k\minus{}1)\equal{}k(kl^2\minus{}4k\plus{}4)$
$ \Delta$ is a perfect square then if $ k$ isn't a perfect square (it mean $ k\equal{}p^{2k\plus{}1}.q$, with $ p$ is a prime and $ GCD(p;q)\equal{}1$) we have $ kl^2\minus{}4k\plus{}4\ \vdots\ p\Leftrightarrow 4\ \vdots\ p\Leftrightarrow p\equal{}2$
Now $ y^2\equal{}kx$ with $ k\equal{}2^{2k\plus{}1}.q$ then $ y$ and $ x$ is even. But $ x\plus{}k\minus{}1$ is odd and then $ \neq ly$. Contracdition!
It mean $ k$ is a perfect square, and $ x$ is a perfect square, too!

[quote="tdl"]We have $ x^2 \plus{} y^2 \minus{} x\ \vdots\ xy\Rightarrow x^2 \plus{} y^2 \minus{} x\ \vdots\ x\Rightarrow y^2\ \vdots\ x\Rightarrow y^2 \equal{} kx\ ,k\in N$
Then $ x^2 \plus{} (k \minus{} 1)x\ \vdots\ xy\Leftrightarrow x \plus{} k \minus{} 1\ \vdots\ y\Leftrightarrow x \plus{} k \minus{} 1 \equal{} ly\ ,l\in N$
It mean $ \frac {y^2}{k} \plus{} k \minus{} 1 \equal{} ly\Leftrightarrow y^2 \minus{} kl.y \plus{} k(k \minus{} 1) \equal{} 0$
$ \Delta \equal{} k^2l^2 \minus{} 4k(k \minus{} 1) \equal{} k(kl^2 \minus{} 4k \plus{} 4)$

$ \Delta$ is a perfect square[/quote]

Why $ \Delta$ is a perfect square ??

[quote="tdl"]
It mean $ k$ is a perfect square, and $ x$ is a perfect square, too![/quote]

This is not true, see it again

Because if equation $ ax^2 \plus{} bx \plus{} c \equal{} 0$ with $ a,b,c\in Z$ have root $ x_{1,2} \equal{} \frac { \minus{} b\pm \sqrt {\Delta}}{2a}\in Z$ then $ \Delta$ must is perfect square!
And when $ k$ is a perfect square, example $ k \equal{} n^2$ then $ x \equal{} (\frac {y}{n})^2$ is a perfect number! (Because $ x$ is a integer!)

[quote="tdl"]
And when $ k$ is a perfect square, example $ k \equal{} n^2$ then $ x \equal{} (\frac {y}{n})^2$ is a perfect number! (Because $ x$ is a integer!)
[/quote]

Ok

[quote="tdl"]Because if equation $ ax^2 \plus{} bx \plus{} c \equal{} 0$ with $ a,b,c\in Z$ have root $ x_{1,2} \equal{} \frac { \minus{} b\pm \sqrt {\Delta}}{2a}\in Z$ then $ \Delta$ must is perfect square!
[/quote]

why $ x_1 \equal{} x_2$ ?  how do you know that it has only one solution?

It don't mean $ x_1\equal{}x_2$, it is $ 2$ roots $ x_1\equal{}\frac{\minus{}b\plus{}\sqrt{\Delta}}{2a}$ and $ x_2\equal{}\frac{\minus{}b\minus{}\sqrt{\Delta}}{2a}$

[quote="tdl"]
$ \Delta$ is a perfect square then if $ k$ isn't a perfect square (it mean $ k \equal{} p^{2k \plus{} 1}.q$, with $ p$ is a prime and $ GCD(p;q) \equal{} 1$) we have $ kl^2 \minus{} 4k \plus{} 4\ \vdots\ p\Leftrightarrow 4\ \vdots\ p\Leftrightarrow p \equal{} 2$
[/quote]

why p divides $ kl^2 \minus{} 4k \plus{} 4$??

sorry if i asked a lot of questions

[hide="Hint:"]prove $ x \equal{} (gcd(x,y))^2$[/hide]

[quote="FOURRIER"]why p divides $ kl^2 \minus{} 4k \plus{} 4$??[/quote]
Because $ \Delta\equal{}k(kl^2 \minus{} 4k \plus{} 4)$ is a perfect square and $ k\equal{}p^{2k\plus{}1}.q$ with $ GCD(p;q)\equal{}1$ (If $ p$ don't divide $ kl^2 \minus{} 4k \plus{} 4$ then $ \Delta\equal{}p^{2k\plus{}1}.r$, $ r\equal{}q.(kl^2 \minus{} 4k \plus{} 4)$ with $ GCD(p;r)\equal{}1$ then $ \Delta$ isn't a perfect square)

PS: A perfect square have shape $ p_1^{2k_1}p_2^{2k_2}...p_m^{2k_m}$ with $ p_i$ is prime!

$ xy|x^2 \plus{} y^2 \minus{} x$. Let $ (x,y) \equal{} d$, $ x \equal{} dp$ & $ y \equal{} dq$ then the condition turns: $ d^2pq|d^2p^2 \plus{} d^2q^2 \minus{} dp$ this $ \Rightarrow$ $ d^2|dp$ this $ \Rightarrow$ $ d|p$ this $ \Rightarrow$ $ p \equal{} dr$ for some $ r$ (remember $ (p,q) \equal{} 1$ then $ (r,q) \equal{} 1$ too) then $ \Rightarrow$ $ x \equal{} d^2r$ then the condition of the problem turns: $ d^3rq|d^4r^2 \plus{} d^2q^2 \minus{} d^2r^2$ $ \Rightarrow$ $ drq|dr^2 \plus{} q^2 \plus{} r^2$ then $ r$ must divide $ q^2$ but (remember) $ (r,q) \equal{} 1$ so this means $ r \equal{} 1$ then $ x \equal{} d^2$. So, $ x$ is a perfect square

Say $ p^\alpha \parallel{} x$, $ p^\beta \parallel{} y$.  If $ \alpha$ were odd, then $ p^{\min\{\alpha,2\beta\}}\parallel{}x^2\plus{}y^2\minus{}x$.  But $ \alpha\plus{}\beta \leq  \min\{\alpha,2\beta\}$ is impossible unless $ \alpha\equal{}0$.  So $ \alpha$ is even for all $ p$, i.e., $ x$ is a square.

More than if $ x,y$ satisfy condition then $ x^2\plus{}y^2\minus{}x\equal{}3xy$


Axis of parabola of the equation in attached image is ........

$y^{2}=4x+4y \implies y^{2}-4y+4=4x+4 \implies (y-2)^{2}=4(x+1)$

y-value symmetry point is thus at $y-2=0 \implies y=2$

Thanks

got another question

Q- The equation of the circle x^2 + y^2 +2gx + 2fy + c = 0  contains _______ arbitrary contstants ?

answer is 3
but can you guide me which area the arbitrary constants ?


Let $ F_1\equal{}F_2\equal{}1$ and $ F_{n\plus{}2}\equal{}F_{n\plus{}1}\plus{}F_{n}$ for $ n\geq 1$. Prove that:
$ \sum_{n\equal{}1}^{\infty}{\frac{1}{F_n}}<4$

Thanks Solovyov for this problem!
We have:
$ \sum_{n \equal{} 1}^{\infty}\frac {1}{F_n} \equal{} 2 \plus{} \sum_{n \equal{} 3}^{\infty}\frac {1}{F_n} \equal{} 2 \plus{} \sum_{n \equal{} 3}^{\infty}\frac {1}{F_{n \minus{} 1} \plus{} F_{n \minus{} 2}} < 2 \plus{} \sum_{n \equal{} 3}^{\infty}\frac {1}{2F_{n \minus{} 2}} \equal{} 2 \plus{} \frac12\sum_{n \equal{} 1}^{\infty}\frac {1}{F_n}$
$ \Rightarrow \sum_{n \equal{} 1}^{\infty}\frac {1}{F_n} < 4$


quadratics
algebra
polynomial
algebra proposed

if $P$ if a quadratic polynomial that satisfies $P(a) = \frac{1}{bc}, P(b) = \frac{1}{ac}, P(c) = \frac{1}{ab}$ for $a,b,c$ distinct non-zero reals, then what is $P(0)$?

[hide]
Observe that $P(x)-\frac{x}{abc}$ has degree at most 2, but has 3 distinct roots, so the difference is 0 everywhere. Therefore, $P(x)=\frac{x}{abc}$ and $P(0)=0$.
[/hide]


if $ f: U \to \mathbb{R}$ be differentiable and a ,b $ \in U$ so that the line ab is in U.show that exist point c on ab that:
$ f(a)\minus{}f(b)\equal{}{Df}_c   (b\minus{}a)$

It's a consequence of single variable Lagrange finite increment theorem. Parametrize the line segment and apply to the composition.

This seems to be a statement of the Mean Value Theorem.


[quote="t0rajir0u"]Seems to be we should prove the stronger inequality

$\sum_{cyc} \left( \frac{a}{b+c} \right)^{p} \ge \frac{3}{2^p}$

Which becomes Nesbitt's when $p = 1$...[/quote]

This inequality was proposed in CRUX MATHEMATICORUM already. We can not discuss this one now.

[quote="nsato"]Let $a$, $b$, $c$ be positive real numbers.  Prove that
\[ \left( \frac{a}{b + c} \right)^{2/3} + \left( \frac{b}{a + c} \right)^{2/3} + \left( \frac{c}{a + b} \right)^{2/3} \ge \frac{3}{\sqrt[3]{4}}.  \][/quote]

See http://www.mathlinks.ro/Forum/viewtopic.php?t=12763 .

  darij

[quote="t0rajir0u"]Seems to be we should prove the stronger inequality

$\sum_{cyc} \left( \frac{a}{b+c} \right)^{p} \ge \frac{3}{2^p}$

Which becomes Nesbitt's when $p = 1$...[/quote]

this one is wrong in general. The best value of p is $ln 3 - ln2$. I think hungkhtn has posted its proof somewhere.


2 people have birthdays in the month of November. What is the probability that their birthdays are exactly 5 days apart?

[hide]Let $A$ and $B$ be these 2 people.

Number of favorable cases we obtain from $(1, 6); (2, 7); (3, 8); \cdots; (25, 30)$.

Notice that for each of those pairs we have 2 cases (for example: $A \rightarrow 1$ and $B \rightarrow 6$ or $A \rightarrow 6$ and $B \rightarrow 1$). So there are $25\cdot2=50$ favorable cases.

Since $n(\Omega) = 30\cdot30 = 900$ (november has 30 days), our answer is $\frac{50}{900}=\frac{1}{18}.$

Please tell me if i missed something.[/hide]

[hide="Same answer but different approach"]
Let us consider the 2 men as A and B.
Then for the first 5 and last 5 days if A has birthday in these then B has only one possibility for all other days B has 2 possibilities.
Thus total possibilities are $10+2*20=50$
Thus the probability is $\frac{50}{\binom{30}{2})}$
[/hide]

Oooops...I got it wrong because I thought November had 31 days :blush: