function
floor function
inequalities
quadratics
algebra
quadratic formula
number theory unsolved

[quote="Albanian Eagle"]what is $ \{n\}$?[/quote]

If I'm not mistaken, it's the fractional part of $ n$ :D

But then, if I'm not mistaken, the problem is not much of a "problem"  :)

Then it doesn't make sense!

{n}=0, isn't it?

Oh sorry the problem is

$ n\minus{} \lfloor n\{\sqrt n \} \rfloor \equal{}2$

Where $ \lfloor n \rfloor$ denotes the integer part of $ n$ ,and $ \{n\}\equal{}n\minus{}\lfloor n \rfloor$

Is the answer [hide]2, 8, 15?[/hide]
If yes, I'll post my really ugly solution.

yes,i think you're right

I think I have post this at "Proposed and  Own problems "

Why it is moved ?

This problem was  Bulgaria 2002 !

OK, here is my solution (it might have a gaping hole, and there is undoubtedly a nicer/easier solution).
[hide]
First, note that no perfect squares work.
Now let n be between $ (x \minus{} 1)^2$ and $ x^2$ for some positive integer x (which has to be greater than 1, of course), which can be restated as $ n \equal{} x^2 \minus{} a$, where $ 1\le a < 2x \minus{} 1$.
We have
$ n \minus{} \lfloor n\{\sqrt {n}\}\rfloor \equal{} 2\leftrightarrow \lfloor n\{\sqrt {n}\}\rfloor \equal{} n \minus{} 2\leftrightarrow n \minus{} 2\le n\{\sqrt {n}\} < n \minus{} 1\leftrightarrow n \minus{} 2\le n(\sqrt {n} \minus{} (x \minus{} 1)) < n \minus{} 1\leftrightarrow nx \minus{} 2\le n\sqrt {n} < nx \minus{} 1\leftrightarrow n^2x^2 \minus{} 4nx \plus{} 4\le n^3 < n^2x^2 \minus{} 2nx \plus{} 1$
Now let's plug in $ n \equal{} x^2 \minus{} a$ (this should be fun)
$ (x^4 \minus{} 2ax^2 \plus{} a^2)x^2 \minus{} 4(x^2 \minus{} a)x \plus{} 4\le (x^2 \minus{} a)^3 < (x^4 \minus{} 2ax^2 \plus{} a^2)x^2 \minus{} 2(x^2 \minus{} a)x \plus{} 1 \leftrightarrow x^6 \minus{} 2x^4a \plus{} x^2a^2 \minus{} 4x^3 \plus{} 4ax \plus{} 4\le x^6 \minus{} 3x^4a \plus{} 3x^2a^2 \minus{} a^3 < x^6 \minus{} 2x^4a \plus{} x^2a^2 \minus{} 2x^3 \plus{} 2ax \plus{} 1$
(sorry for torturing all of you)
$ x^4a \minus{} 2x^2a^2 \minus{} 4x^3 \plus{} 4ax \plus{} a^3 \plus{} 4\le 0 < x^4a \minus{} 2x^2a^2 \minus{} 2x^3 \plus{} 2ax \plus{} a^3 \plus{} 1$
Now let's consider the left bound. We will put a weak bound on a, and use casework on the remaining values.
Note that if $ x^4a \minus{} 2x^2a^2 \minus{} 4x^3\ge 0\leftrightarrow ax^2 \minus{} 4x \minus{} 2a^2\ge 0$, the LHS inequality cannot hold. Therefore $ ax^2 \minus{} 4x \minus{} 2a^2 < 0$.
Using the quadratic formula and the fact that x and a are both positive, we have
$ ax^2 \minus{} 4x \minus{} 2a^2 < 0\leftrightarrow x < \frac {4 \plus{} \sqrt {16 \plus{} 8a^3}}{2a}\leftrightarrow x < \frac {2 \plus{} \sqrt {4 \plus{} 2a^3}}{a}$
Also, from the beginning, we know that $ a < 2x \minus{} 1\leftrightarrow a\le 2x \minus{} 2\leftrightarrow x\ge \frac {a \plus{} 2}{2}$
Combining these two inequalities, we see that
$ \frac {a \plus{} 2}{2} < \frac {2 \plus{} \sqrt {4 \plus{} 2a^3}}{a}\leftrightarrow a^2 \plus{} 2a \minus{} 4 < 2\sqrt {4 \plus{} 2a^3} \leftrightarrow a^4 \minus{} 4a^3 \minus{} 4a^2 \minus{} 16a < 0\leftrightarrow a^3 \minus{} 4a^2 \minus{} 4a \minus{} 16 < 0$
This is only true for $ a\le 5$, so we finally have a bound on a.
If $ a \equal{} 5$, then $ 5x^2 \minus{} 4x \minus{} 50 < 0$, which is only true for $ x\le 3$, because x is an integer. Then $ x < \frac {a \plus{} 2}{2}$, a contradiction.
If $ a \equal{} 4$, then $ 4x^2 \minus{} 4x \minus{} 32 < 0$, which is true for $ x\le 3$ again. If $ x\le 2$, then $ x < \frac {a \plus{} 2}{2}$, so the only possible value of x is 3. Plugging this into the original equation, we see that $ x^4a \minus{} 2x^2a^2 \minus{} 4x^3 \plus{} 4ax \plus{} a^3 \plus{} 4 > 0$, so this doesn't work.
If $ a \equal{} 3$, then $ 3x^2 \minus{} 4x \minus{} 18 < 0$, and again $ x\le 3$ and again $ x \equal{} 3$. Plugging this in, we find that this doesn't work.
If $ a \equal{} 2$, similarly we find $ x\le 3$. If $ x \equal{} 3$, this doesn't work, but $ x \equal{} 2$ satisfies both sides of the original inequality. Yay!
If $ a \equal{} 1$, then $ x\le 4$. $ x \equal{} 2$ satisfies the LHS but not the RHS, and $ x \equal{} 3$ and $ x \equal{} 4$ both work.

Final answers are $ 2^2 \minus{} 2$, $ 3^2 \minus{} 1$, and $ 4^2 \minus{} 1$, or $ \boxed{2, 8, 15}$.
[/hide]
Whew.

[quote="enndb0x"]Find all positive integers $ n$ such that $ n \minus{} \lfloor n\{\sqrt n\} \rfloor \equal{} 2$[/quote]

Let $ \sqrt n\equal{}p\plus{}u$, with $ p\in\mathbb N$ and $ u\in[0,1)$. The equation is then $ nu\equal{}n\minus{}2\plus{}v$ for some $ v\in[0,1)$, so $ 1>v\equal{}n(\sqrt n \minus{}p)\minus{}n\plus{}2>\equal{}0$ and so $ \boxed{\sqrt n\plus{}\frac 2n\geq p\plus{}1>\sqrt n \plus{}\frac 1n}$

$ n\equal{}1$ is not a solution and this equation can never be verified if $ n$ is a perfect square $ >1$. So consider $ m^2>n>(m\minus{}1)^2$. We can only have one integer in $ (\sqrt n \plus{}\frac 1n,\sqrt n \plus{}\frac 2n]$ and this integer may only be $ m$ and so we need  to have $ m\leq \sqrt n\plus{}\frac 2n$

But, it is easy to show that if $ m\geq 6$, we have $ m>\sqrt{m^2\minus{}1}\plus{}\frac{2}{(m\minus{}1)^2}\geq \sqrt n\plus{}\frac 2n$ and so no solution to the original equation if $ n>5^2$

And so we just have to check the values $ n\in[2,24]$ not perfect square in order to find all the solutions $ \{2,8,15\}$


I'm really confused by this whole concept.

I'll give an example problem with an explanation the teacher gave, then a sample problem that you can hopefully solve for me and show every step in really really really simple terms.  THANKS!

1) Q: I have four times as many two-cent coins as I have five-cent coins.  All up I have $6.50.  How many two-cent coins do I have?  Find the equation and give an answer.
A:
x = \# of two-cent coins
y = \# of five-cent coins
4x*.02+(x/4)*y*.05 = 6.5
= 200 two-cent coins.
-----
Commentary:
I don't understand how he found the equation.
I don't understand how he got 200 out of the equation.
-----
2) I invest in two types of shares, X and Y.  X shares pay a dividend of 10% p.a. and Y shares pay a dividend of 8%.  If I invest
$4000 more in X than in Y and my total income from both investments is $1010, how much did I invest in Y shares?

-----
Commentary:
I don't understand how to find the equation.
I don't understand how to solve for Y once the equation is found.
I don't know what p.a. means!! Lol.

Thanks again for ANY help!!!

These really aren't so bad, I admit that when I first saw these types of solutions it confused me (I had taught myself Algebra I, and invented my own version). What your teacher did was extract certain tid-bits of information from the original problem and then substituted certain values.

[quote="mj93's teacher"]1) I have four times as many two-cent coins as I have five-cent coins. All up I have 6.50. How many two-cent coins do I have? Find the equation and give an answer.[/quote]

Working in cents is easier in this case than working in dollars so each $ x$ is worth $ 2$ and each $ y$ is worth $ 5$ the total of $ 2x\plus{}5y\equal{}650$. You have four times as many $ x$ as $ y$, so $ x\equal{}4y$. Substitute $ x\equal{}4y$ into $ 2x\plus{}5y\equal{}650$ which gives $ 8y\plus{}5y\equal{}650\Rightarrow 13y\equal{}650\Rightarrow y\equal{}50$. We, however want $ x$, but $ x\equal{}4y$ so $ \boxed{x\equal{}200}$. Your teacher solved $ x\equal{}4y$ for $ x$ so that s/he could just obtain $ x$ right away.

As for your other problem, perhaps you can solve it now.  :lol:

Thanks! I understand what you did, im just not creative enough to think of doing that, lol  :lol:
Question about the next equation:

I invest in two types of shares, X and Y. X shares pay a dividend of 10% p.a. and Y shares pay a dividend of 8%. If I invest $ 4000 more in X than in Y and my total income from both investments is$1010, how much did I invest in Y shares?

.1X + .08Y = 1010

X - 4000 = Y

That's as far as I've gotten.  I don't know how to combine the two equations.

LOL i figured it out...wow i feel stupid! the equation would be:

.1X + .08(X-4000) = 1010

right??

ya thats right. u beat em to it

and then solve

most problem solving qns like this will involve simulatenous equations


Approximately how many questions are currently on Alcumus?

Right now we have approximately 1100 problems in Alcumus.

[url=http://www.artofproblemsolving.com/Alcumus/Introduction.php]Alcumus Introduction[/url]

That link says currently free, are you planning to charge money for it in the future?

Well now alcumus has over 9000! (not 9000, 9000 factiorial)

[quote="EpicSkills32"]Great job reviving.....[/quote]
reviving?

Please don't backseat moderate.  PM me or another Alcumus mod if someone revived a dead thread or something of the latter.


Twelve candidates for mayor participate in a TV talk show. At some point a candidate said: "One lie has been told." Another said: "Now two lies have been told". "Now three lies," said a third. This continued until the twelfth said: "Now twelve lies have been told". At this point the moderator ended the discussion. It turned out that at least one of the candidates correctly stated the number of lies told before he made the claim. How many lies were actually told by the candidates?

[hide]If the first candidate lied, then all the other candidates lied too, which is impossible. Hence, the first candidate did not lie, whence we know that all the other candidates lied. Consider this truthful candidate. Before his statement was told, one lie had been told. After his statement was told, 11 lies were told. Therefore 12 lies were told.[/hide]


In set of real numbers solve the system:

$ x(3y^{2}\plus{}1)\equal{}y(y^{2}\plus{}3)$

$ y(3z^{2}\plus{}1)\equal{}z(z^{2}\plus{}3)$

$ z(3z^{2}\plus{}1)\equal{}x(x^{2}\plus{}3)$


nobody ?  :( Well-come Vasc and mathlinker :wink:


So I was wrong.

Anyways, I'm trying to come up with an appropriate recursion.

[quote="blahblahblah"]

1/sqrt(2) * sqrt( 4 + sqrt ( 8 + sqrt ( 16 + ...... = 1/sqrt(2)*[x^2 - 2]

sqrt( 2 + sqrt ( 4 + sqrt ( 8 + ......   = 1/sqrt(2)*[x^2 - 2]

[/quote]

I don't quite follow this part...

[quote="blahblahblah"]
1/sqrt(2) * sqrt( 4 + sqrt ( 8 + sqrt ( 16 + ...... = 1/sqrt(2)*[x^2 - 2]

sqrt( 2 + sqrt ( 4 + sqrt ( 8 + ......   = 1/sqrt(2)*[x^2 - 2]
[/quote]

You divided wrong: this is what should have happened:

[tex]
\displaystyle\frac{\sqrt{4 + \sqrt{8 + \sqrt{16 \ldots}}}}{\sqrt{2}} =
[/tex]
[tex]
\displaystyle\sqrt{2 + \frac{\sqrt{8 + \sqrt{16 \ldots}}}{2},
[/tex]

which doesn't get us anywhere.

[quote="nr1337"]
You divided wrong: this is what should have happened:

[tex]
\displaystyle\frac{\sqrt{4 + \sqrt{8 + \sqrt{16 \ldots}}}}{\sqrt{2}} =
[/tex]
[tex]
\displaystyle\sqrt{2 + \frac{\sqrt{8 + \sqrt{16 \ldots}}}{2},
[/tex]

which doesn't get us anywhere.[/quote]

yeah that's what I had figured he had done.

I'm not even sure if it can be solved... quite an interesting problem though. I have given enough hours to it. If nobody here can solve it, I'll post it in advanced.

Is there a way this can be written in sigma notation?

[quote="Spoon"]Not really a challenge, because I can't solve it... but I am interested in the solution...

Solve [tex]\sqrt{2+\sqrt{4+\sqrt{8+\sqrt{16+\ldots}}}}[/tex][/quote]hey so u did post it.  Didnt triplem say this was infinite?

mathdool wrote:A thought:
[hide]653904091/300000000

What can we do with this number?[/hide]

That is simply a fraction that represents an approximated answer... I'm not sure there is much we can do.

No, Triple M didn't say it was infinite. He said that if what you said was correct, then that would imply that it was infinite. It's not infinite, I know the decimal to 12 decimal places... It converges quickly.

Maybe Sqrt[2+Sqrt[4+Sqrt[8...]]] is already in simplest form.

well if that is the case then the problem can be not written in closed form. A closed form expression is what the question asks for.

[quote="Scrambled"]I think i may have got it.  I got 4.
ill post my solution later[/quote]

That doesn't seem right.  Since x^2 - 6 = sqrt[8+sqrt[16...]], then 10 would equal sqrt[8+sqrt[16...]], which is less than 10.

the answer is not 4... the answer is 2.17968030333...

How did you find that?  My solution was to use Mathematica to find the sum of the first few terms. :lol:

Here's a proof that the original expression converges to a finite number.

First, it is easy to prove that
[tex]\displaystyle \sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\ldots}}}} = \phi,[/tex]
where [tex]\phi[/tex] is the golden ratio.  Multiplying both sides by [tex]\sqrt{2}[/tex] shows that
[tex]\displaystyle \sqrt{2+\sqrt{4+\sqrt{16+\sqrt{256+\ldots}}}} = \phi \sqrt{2}.[/tex]
The left-hand side dominates the original expression.  Thus the original expression is less than [tex]\phi \sqrt{2}[/tex].

[quote="Ravi B"]Here's a proof that the original expression converges to a finite number. ...
The left-hand side dominates the original expression.  Thus the original expression is less than [tex]\phi \sqrt{2}[/tex].[/quote]

A sequence is dominated by another convergent sequence may not be convergent. For example, a_n = 0 when n is odd and a_n=1 when n is even is dominated by (say) a constant sequence c_n :equiv: 1.

But the original sequence is [b]increasing[/b] and bounded above, hence it converges to a finite number.


geometry
circumcircle
geometric transformation
reflection
trigonometry
perpendicular bisector
geometry proposed

$ ABC$ is an acute triangle. The points $ B'$ and $ C'$are the reflections
of $ B$ and $ C$ across the lines $ AC$ and $ AB$ respectively. Suppose
that the circumcircles of triangles$ ABB$' and $ ACC'$ meet at $ A$
and $ P$. Prove that the line $ PA$ passes through the circumcenter
of triangle$ ABC.$

[b][color=darkblue]Let $ \| \begin{array}{c} C_1 \in CC' \cap AB \\
B_1 \in BB' \cap AC \\
E \in AB \cap \omega_{ACC'} \\
F \in AC \cap \omega_{ABB'} \end{array}$. Then quadrilaterals $ AC'EC$ and $ ABFB'$ are cyclic, so:
$ \angle{C'AE} = \angle{EAF} = \angle{FAB'}$ and $ \| \begin{array}{c} \angle{C'EC} \\
\angle{B'FB} \end{array} = 180^\circ - 2\angle{A} \Rightarrow \| \begin{array}{cc} \angle{AEC} \\
\angle{AFB} \end{array} = 90^\circ - \angle{A}\Rightarrow \| \begin{array}{cc} \angle{ACE} \\
\angle{ABE} \end{array} = 90^\circ$.

Because $ AE$ and $ CF$ are diameters, we have:
$ \|\begin{array}{cc} \angle{APE} \\
\angle{APF} \end{array} = 90^\circ \Rightarrow \angle{EPF} = 180^\circ \Rightarrow E,P,F$ are collinear.

In triangle $ AEF$, $ EC$, $ FB$ and $ AP$ are altitudes, so they are concurrent. Let $ H$ be ortocenter.

Quadrilateral $ ABHC$ is cyclic, so $ H$ lies on the circumcircle of $ ABC$, where $ PA$ is diameter, which conctains $ O$, q.e.d.
[/color]
[/b]
[img]http://i018.radikal.ru/0805/05/21886d8ea9b3.png[/img]

$ AB$ is the perpendicular bisector of $ CC^'$ $ \Rightarrow$ the circumcenter of $ \triangle{ACC^'}$ $ \in$ $ AB$. Let it be $ O_1$. Define $ O_2$ analogoulsy for $ \triangle{ABB^'}$. $ \frac{AC^'}{\sin{(90^o - A)}} = 2R_1$ $ \Rightarrow$ $ R_1 = \frac {AC}{2\cos{A}}$ analogously: $ R_2 = \frac {AB}{2\cos{A}}$. $ \Rightarrow$ $ R_1AB = R_2AC$ $ \Rightarrow$ $ BO_1O_2C$ is cyclic .We know $ O_1O_2 \perp AP$, using $ BO_1O_2C$ is cyclic we get $ AP$ passes trough $ O$.

[color=darkred][b][u]An equivalent enunciation I.[/u][/b] Let $ w_1 = C(O_1)$ , $ w_2 = C(O_2)$ be two secant circles in $ A$ , $ D$ so that

$ m(\widehat {O_1AO_2}) < 90^{\circ}$ . Denote $ \begin{array}{c} \{A,B\} = AO_1\cap w_2 \\
\ \{A,C\} = AO_2\cap w_1\end{array}$ . Prove that the circumcircle of $ \triangle ABC$ belongs to $ AD$ .[/color]

[color=darkblue][b]Proof.[/b] Denote the diameters $ [AX]$ , $ [AY]$ of the circles $ w_1$ , $ w_2$ respectively. Observe that $ D\in XY$ , $ XY\perp AD$ and $ \|\begin{array}{c} XC\perp AY \\
\ YB\perp AX\end{array}$ .

Thus, $ BDC$ is the orthic triangle of $ \triangle XAY$ and $ \triangle ABC\sim\triangle AYX$ . Denote the orthocenter $ H$ of the triangle $ ABC$ . Therefore,

the rays $ [AD$ , $ [AH$ are isogonals  in the angle $ \widehat {BAC}$ . In conclusion, the circumcircle of the triangle $ ABC$ belongs to $ AD$ .

[b]Remark.[/b] Sorry, in my opinion, the proposed problem is "painted" a bit ![/color]

[color=darkred][b][u]An equivalent enunciation II (evident !).[/u][/b] Let $ XAY$ be an acute triangle. Denote the orthic triangle $ DCB$ of it,

where $ D\in XY$ , $ C\in YA$ , $ B\in AX$ . Prove that the circumcenter of the triangle $ ABC$ belongs to the altitude $ AD$ .[/color]

Equivalent to existence of the orthocenter upon $\sqrt{bc}$ inversion.


 :!:  :ninja:  :!:
How can you obtain two parallel lines by intersecting a double-napped cone with a plane. I just dont see how this is possible.

[quote="Hamster1800"]I don't believe you can. However, you can get two [i]perpendicular[/i] lines by taking a vertical plane that passes through the vertex of the cones.[/quote]
did you mean to say two intersecting lines, or do they really have to be perpendicular?

Hmm, they don't have to be perpendicular. Thanks. However, if you take the vertical plane, they are perpendicular. If you tilt the plane while keeping it through the vertex they will become closer to the same line, but you never have two parallel lines unless they are the same line.


How do I mark equal angles in it?  :maybe:

There's no built-in command for it.

But you can go to http://whistleralley.com/GSP/sketchpad.htm and download the [i]Geo_Notation[/i] file, which contains some handy symbols that are cumbersome to produce :wink:


Well. USAMO scores are out.
 :D
I got a 9=RED!!!!
 :D
I had to check that I actually got a 9 like 50 times to make sure that I wasn't looking at the wrong score, cuz the person listed below me had a 4.

Wonder if Allen will be accompanying me to MOP? The red cutoff is 8 this year.
Well at least I know Berman and Damien (?) will.

[quote="not_trig"]I'm probably not going to do RSI. bleargh. also lawl @ 3 27+s[/quote]

OK

[quote="jb05"][quote="not_trig"]I'm probably not going due to RSI. bleargh. also lawl @ 3 27+s[/quote]

OK[/quote]

I see what you did there

[quote="not_trig"]I'm probably not going to RSI. bleargh. also lawl @ 3 27+s[/quote] :)

argh I failed with a score that is less than half of that of jb05... [color=#F0F0F0]13 in fact...[/color]

Either people who are not from North Carolina voted or people lied.  Tis quite simple.


Let $ x_1 , x_2 , x_n$ be positive real numbers and  $ v \equal{}(x_1 , x_2,...,x_n)$ and $ \tilde {v}$ = $ (\sqrt[n]{x_1} , ... , \sqrt[n]{x_n})$ , prove the following:
$ GM(v) \leq$ $ \sqrt[\frac{n\minus{}1}{n}]{AM(v)}HM(\tilde {v}) \leq AM(v)$

the inequality is weaker to Siepinski-Inequality G^n <= A^(n-1) * H ?

Yes Sierpinski is stronger  , and by this is solved , Thank you
for first half Apply Sierpinski after prove $ HM(v^\frac{1}{n}) \geq (HM(v))^\frac{1}{n}$
for the second half use powered means inequality


advanced fields
advanced fields theorems

see here : http://www.math.bilgi.edu.tr/courses/Lecture%20Notes/SetTheoryLectureNotes.pdf

[quote="ifai"]Is there any good pdf book of set theory?[/quote]

Axiomatic study:
http://www.peter-dixon.staff.shef.ac.uk/teaching/STDN.PDF


$ABC$ is a triangle. The bisectors of $A, B, C$ meet the circumcircle again at $K, L, M$ respectively. $X$ is a point on the side $AB$ (not one of the vertices). $P$ is the intersection of the line through $X$ parallel to $AK$ and the line through $B$ perpendicular to $BL$. $Q$ is the intersection of the line through X parallel to BL and the line through $A$ perpendicular to $AK$. Show that $KP, LQ$ and $MX$ are concurrent.

[color=darkblue]I think this problem is very interesting and not easy[/color] ;)

See http://www.mathlinks.ro/Forum/viewtopic.php?t=17201 .

  darij


logarithms
function
algebra unsolved
algebra

Let $\displaystyle a,b,c,d \in \mathbb R$ such that $\displaystyle a>c>d>b>1$ and $\displaystyle ab>cd$.

Prove that $\displaystyle f : \left[ 0,\infty \right) \to \mathbb R$, defined through
\[ \displaystyle f(x) = a^x+b^x-c^x-d^x, \, \forall x \geq 0 , \]
is strictly increasing.

$ab>cd$ yields $\ln{a} + \ln{b} - \ln{c}-\ln{d} > 0 \ (*)$.

$f'(x) = a^x \ln{a} + b^x \ln{b} - c^x \ln{c} - d^x \ln{d}$

We must show that $f'(x) > 0$ $\forall x \ge 0$.

First, we have $f'(0) = \ln{a} + \ln{b} - \ln{c}-\ln{d} > 0 \ by \ (*)$

Write $f'(x)$ as $f'(x) = a^x(\ln{a} + (\frac{b}{a})^x \ln{b} - (\frac{c}{a})^x \ln{c} - (\frac{d}{a})^x \ln {d})$. Note that $a^x > 0 \ \forall x$.

$a>c>d>b>1$ implies $1 \ge \frac{c}{a} > \frac{d}{a} > \frac{b}{a}$, so the expression inside the parentheses strictly decreases. Furthermore, as $x \rightarrow \infty$, the expression approaches $\ln{a}>0$, so $f'(x)>0 \ \forall x \ge 0$.

Thus, $f(x)$ strictly increases $\forall x \ge 0$.

[quote="mathisfun1"]

$a>c>d>b>1$ implies $1 \ge \frac{c}{a} > \frac{d}{a} > \frac{b}{a}$, so the expression inside the parentheses strictly decreases. Furthermore, as $x \rightarrow \infty$, the expression approaches $\ln{a}>0$, so $f'(x)>0 \ \forall x \ge 0$.
[/quote]

It is not so clear..please explain.  :)

Sorry for my inadequate explanation.

If $b <c$ then $b^x < c^x$. Define the function $g(x) = b^x-c^x$.

$g'(x) = b^x \ln {b} - c^x \ln{c} = b^x \ln{b}(1-(\frac{c}{b})^x \frac{\ln{c}}{\ln{b}}) < 0 \ \forall x \ge 0$ because $\frac{c}{b} \frac{\ln{c}}{\ln{b}} > 1$.

Thus, the expression inside the parentheses of my original post strictly decreases $\forall x \ge 0$.


calculus
integration
logarithms
calculus computations

Evaluate :

 $ \int_{0}^{ \infty }\frac{ x \log x }{ 1\plus{} x^2 }$

$ \int_0^1\frac{x\log x}{1\plus{}x^2}\,dx$ is negative and finite.

$ \int_1^e\frac{x\log x}{1\plus{}x^2}\,dx$ is positive and finite.

$ \int_e^{\infty}\frac{x\log x}{1\plus{}x^2}\,dx>\int_e^{\infty}\frac{x}{1\plus{}x^2}\,dx\equal{}\plus{}\infty.$

Hence the whole integral diverges to $ \plus{}\infty.$

Your math teacher is wrong. Now, for each $ x\ge1$ we have $ \frac{x\ln x}{1\plus{}x^{2}}\ge \frac{\ln x}{2x},$ then the second piece is clearly a divergent integral, so there's no way the answer could be zero.

May be the integrand was meant to be $ \frac{\ln x}{1\plus{}x^2}.$

or
$ \int_{0}^{\infty }\frac{ x^n\log x }{ 1\plus{}x^{m}}$
such that
$ n\equal{}2m\plus{}2$

As for $ \int_0^{\infty}\frac{\ln x}{1\plus{}x^2}\,dx:$

Appropriate comparisons can be made as $ x\to 0$ and as $ x\to\infty$ to show that this converges.

Make the change of variables $ x\equal{}\frac1u$ and the integral becomes $ \int_0^{\infty}\frac{\minus{}\ln u}{1\plus{}u^2}\,du.$

So $ \int_0^{\infty}\frac{\ln x}{1\plus{}x^2}\,dx\equal{}\minus{}\int_0^{\infty}\frac{\ln x}{1\plus{}x^2}\,dx$ and the integral must be zero.

Actually sir  i don t know convergence and divergence  and is there any way we can evaluate the integral without using the fact that it diverges as it was mentioned.

$ \int_0^{\infty}\frac {x\ln x}{1 \plus{} x^2}\,dx \equal{} \plus{} \infty.$

That's its value, and there's no ambiguity.

Can anyone solve it for me i mean a rigorous proof of it ..

Rigorous proof has already been supplied in this topic. Twice. Since there's absolutely nothing left to be said, I'm locking the topic to keep you from asking again.

[quote="vijaymenon"]But my math teacher says the answer is 0 . please help me .[/quote]
Next time just copy the assignment correctly. I bet 100 against 1 it was $ \int_0^\infty\frac{\log x}{1\plus{}x^2}\,dx$ ;)


calculus
integration
calculus computations

Find the volume of the region bounded by $ x \equal{} \minus{}1, \ x \equal{}1$, $ y \equal{} \minus{}1, y \equal{} 1$ and $ z \equal{} x$.

Is this equaled to $ \int_{0}^{1} \int_{\minus{}1}^{1} \int_{\minus{}1}^{1} dx \ dy \ dz$?

Can it also equal $ \int_{\minus{}1}^{1} \int_{\minus{}1}^{1} \int_{0}^{x} \ dz \ dx \ dy$?

This question was answered in another forum, and it contains the same written. Do I have to suspect that you are such user?


Find all positive real numbers $a,b,c$ which satisfy the inequalities:
$4(ab+bc+ac) \geq a^{2}+b^{2}+c^{2}\geq 3(a^{3}+b^{3}+c^{3})$

This one is quite similar to inequality from Romanian JBMO TST 2004. That question asked to find all positive real numbers satisfying inequality
4(ab+bc+ca)-1>=aˇ2+bˇ2+cˇ2>=3(aˇ3+bˇ3+cˇ3)  :roll:

sorry that was the one i meantioned .. for some reason i can't edit my own posts so yeah, i forgot the -1, the correct inequality is:
$4(ab+bc+ac)-1 \geq a^{2}+b^{2}+c^{3}\geq 3(a^{3}+b^{3}+c^{3})$.
what happens if we change it to the following, i'm pretty sure it has the same solution set, but i can't solve it  :maybe: :
$a^{2}+b^{2}+c^{2}\geq 4(ab+bc+ac)-1 \geq 3(a^{3}+b^{3}+c^{3})$

I post a solution that came to my mind first .
By Chebychev we have that
$a^{2}+b^{2}+c^{2}\geq 3(a^{3}+b^{3}+c^{3})\geq (a^{2}+b^{2}+c^{2})(a+b+c)$  so $a+b+c\leq 1$ so $(a+b+c)^{2}\leq 1$ .
From the two first we have
$4(ab+bc+ca)\geq a^{2}+b^{2}+c^{2}+1\geq a^{2}+b^{2}+c^{2}+(a+b+c)^{2}$ from which we  find that
$ab+bc+ca\geq a^{2}+b^{2}+c^{2}$ .
So we have in all equality .$a+b+c=1$ and a=b=c so
$a=b=c=\frac{1}{3}$


Are there any classes with out challenge sets

The Problem Series classes do not have Challenge Sets (MATHCOUNTS/AMC 8, AMC 10, AMC 12, AIME Problem Series).  The Intermediate Number Theory class also does not have Challenge Sets.


superior algebra
superior algebra unsolved

Let $G$ be finite group of order $n$ and let for all divisors $d$ of $n$ , $G$ has at most d elements of order $d$. prove that $G$ is cyclic.

Do you mean "at most $n$ elements of order dividing $n$"? If so, this has been treated recently. If not, I'm not sure about the answer.

no we know for any d dividing n we have at most d elements of order d.

Sorry- I misspoke: I meant "at most $d$ elements of order dividing $d$".

I interpret "x has order d" as $x^d=e$ and $x^k\neq e$ for all $k<d.$

Actually, I can do even the harder version now.
Lemma 1: For all $d$ dividing $n,$ the number of elements with order dividing $d$ is divisible by $d.$
Proof: see my post [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22490]here[/url].

Lemma 2: Given that there are at most $d$ elements of order $d$ for all $d$ dividing $n,$ there are exactly $d$ elements of order dividing $d$ for all $d$ dividing $n.$
Proof: Suppose by way of contradiction that there is some $k$ with more than $k$ elements of order dividing $k.$ Let $m$ be the least such $k.$ For each $d<m$ dividing $m,$ there are $d$ elements of order dividing $d$ and $\phi(d)$ elements of order $d.$ There are at least $2m$ elements of order dividing $m$ by Lemma 1, so there are at least $\phi(m)+m$ elements of order $m.$ This is the desired contradiction.
We cannot have fewer than $d$ elements for any $d$, since there is at least one and the number is divisible by $d.$

Lemma 3: Given that there are at most $d$ elements of order dividing $d$ for all $d,$ the group is cyclic.
Proof: There are exactly $\phi(d)$ elements of order $d$ for each $d$ dividing $n$. Since $\phi(n)>0$, there are elements of order $n$ and the group is cyclic. See also Kent's post  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=27050]here[/url].

This completes the proof.


inequalities
LaTeX
inequalities proposed

bai nay la de thi HSG truong A ,it is very easy.TRong 20 dua di thi chi co 1 thang giai duoc /ta cung giai duoc bai nay!tiec la da khong duoc di thi :oops:  :oops:  :D

let x,y,z$ \ge$ 0

prove that:

   (x-y).$ \sqrt{xy/(y^2+1)}$ +(y-z).$ \sqrt{yz/(z^2+1)}$ +(z-x).$ \sqrt{xz/(X^2+1)}$ $ \le$ $ \frac{(X^2+Y^2+z^2)}{}$\sqrt{2}${ }$

[quote="trannguyenquyND_VN"]vế phải của bất đẳng thức là _<(X^2+y^2+z^2)/(2.cos45)

IF x,y,z>0   call A= the lelf. we have":

hoàn toàn có thể giả sử: x>_y>_z>0 then:

A_<(X^2-Y^2)/1+(Y^2-Z^2)/1  (because z-x<0)

hence A_<X^2-Z^2<(X^2+Y^2+z^2)

the equation ocurss when X=Y=z=0[/quote]
Thank you :) ,but your solution 's wrong :wink:

"hoàn toàn có thể giả sử: x>_y>_z>0 then:"
I agree with quykhtn-qa1.You can't assume that :D

[quote="great math"]Doesn't the problem look like this
Let $ x,y,z \ge 0$
Prove that
$ (x \minus{} y)\sqrt {\frac {xy}{y^2 \plus{} 1}} \plus{} (y \minus{} z)\sqrt {\frac {yz}{z^2 \plus{} 1}} \plus{} (z \minus{} x)\sqrt {\frac {zx}{x^2 \plus{} 1}} \le \sqrt {2}$ ?
I have a feeling that this problem is originally incorrect.[/quote]
I hope the author will give the CORRECT latex Formula here:
[quote] [Unparseable or potentially dangerous latex formula.Error: 7]\sqrt{2}[Unparseable or potentially dangerous latex formula. 5000][/quote]
Else the problem cannot be interpreted. The author should learn using LaTeX correctly......... :(

[quote="trannguyenquyND_VN"]bai nay la de thi HSG truong A ,it is very easy.TRong 20 dua di thi chi co 1 thang giai duoc /ta cung giai duoc bai nay!tiec la da khong duoc di thi :oops:  :oops:  :D

let x,y,z$ \ge$ 0

prove that:

   (x-y).$ \sqrt {xy/(y^2 \plus{} 1)}$ +(y-z).$ \sqrt {yz/(z^2 \plus{} 1)}$ +(z-x).$ \sqrt {xz/(X^2 \plus{} 1)}$ $ \le$ $ \frac{(X^2 \plus{} Y^2 \plus{} z^2)}{\sqrt{2}}$[/quote]
That should be it. :maybe:


trigonometry
ratio
analytic geometry
graphing lines
slope
geometry unsolved
geometry

Let $ ABCD$ be a square with the length of the sides $ AB\equal{}BC\equal{}CD\equal{}DA \equal{} 6$ . Let there be a point $ E \in [AB]$ such that $ AE\equal{}2$ .
Note that $ AC \cap DE \equal{} F$ . Construct the point $ G \in [CB]$ such that $ GF \bot DE$ . Find the length of the segment $ CG \equal{} ?$.

Here is a draw :
[url=http://img365.imageshack.us/my.php?image=desents2.png][img]http://img365.imageshack.us/img365/1494/desents2.th.png[/img][/url]

[b][color=darkblue]In triangle AED: $ EF + FD = ED = \sqrt {AE^2 + AD^2} = 2\sqrt {10}$.

Because $ AF$ is bisector we have:
$ \frac {EF}{FD} = \frac {1}{3} \Rightarrow \| \begin{array}{cc} EF = \frac {1}{2}\sqrt {10} \\
 \\
FD = \frac {3}{2}\sqrt {10} \end{array}$.

It is easy to calculate length of bisector $ AF$:

$ AF = \sqrt {\frac {2FE^2\cdot FD^2}{EF^2 + FD^2}} = \frac {3}{\sqrt {2}}$

Then

$ FC = AC - AF = 6\sqrt {2} - \frac {3}{\sqrt {2}} = \frac {9}{\sqrt {2}}$.

If $ \angle{ADE} = \alpha$, then $ \angle{FGC} = 90^\circ + \alpha, \angle{GFC} = 45^\circ - \alpha$. Note that $ \sin{\alpha} = \frac {1}{\sqrt {10}}, \cos{\alpha} = \frac {3}{\sqrt {10}}$.

By sine theorem in $ GFC$ we have:

$ \boxed{\begin{array}{cc} \\
\frac {GC}{\sin{(45^\circ - \alpha)}} = \frac {FC}{\sin{(90^\circ + \alpha)}} \\
 \\
\frac {GC}{\sin{45^\circ}(\cos{\alpha} - \sin{\alpha})} = \frac {\frac {9}{\sqrt {2}}}{\cos{\alpha}} \\
 \\
\frac {\sqrt {20} GC}{ 2 } = \frac {9\cdot \sqrt {10}}{\sqrt {2}\cdot 3} \\
 \\
GC = \frac {9 \cdot 2 \sqrt {10}}{\sqrt {10}\cdot \sqrt {20} \cdot 3} = 3 \\
 \\
\end{array} }$[/color][/b]

DE meet BC on K and the perpendicolar axis of AB on J. From similitude $ \frac{AF}{FC}\equal{}\frac{DA}{KC} \equal{}  \frac{6}{12\plus{}6} \equal{} \frac{1}{3}$. Than J stay on a circle $ \Gamma$ with center F and radius DF. $ \Gamma$ pass trought B and meets the perpendicolar line from F to DE on the arch that contain B on M and on the other on N. Then we can obtain the squere DMJN from ABCD with a spiral dilatation with $ \alpha \equal{} \arctan \frac{1}{2}$ (as we can se from $ \triangle DBJ$), center D and ratio $ \frac{\sqrt{5}}{2}$. So $ M \in BC$, $ N \in AB$ and $ M \equiv G$. Then $ GC \equal{} GB$.

Coordinates makes this very easy. Let A=(0,0), the intersection of y=x and 6x+2y=12 is (3/2,3/2) hence the slope between it and (2,0) is -3, so the slope of line FG is 1/3 and passes through (3/2,3/2), with equation y-3/2=1/3(x-3/2) when x=6, y=3 and GC is
6-3=3.

[quote="turcas_c"] [color=darkred]Let $ ABCD$ be a square with the length of the sides $ AB \equal{} BC \equal{} CD \equal{} DA \equal{} 1$ . Let there be a point $ E \in [AB]$ such that $ AE \equal{} a\in (0,1)$ .
Note that $ AC \cap DE \equal{} F$ . Construct the point $ G \in [CB]$ such that $ GF \bot DE$ . Find the length of the segment $ CG \equal{} f(a)$ .[/color] [/quote]

[b][u]Proof.[/u][/b] $ DCGF$ is cyclically and $ [CF$ is the bisector of $ \widehat {DCG}$ . Thus, $ DG \equal{} DF\sqrt 2$ . Since $ \frac {DF}{1} \equal{} \frac {FE}{a} \equal{} \frac {DE}{1 \plus{} a} \equal{} \frac {\sqrt {1 \plus{} a^2}}{1 \plus{} a}$

obtain $ DG^2 \equal{} 2\cdot DF^2 \equal{} \frac {2(1 \plus{} a^2)}{(1 \plus{} a)^2}$ $ \implies$ $ CG^2 \equal{} DG^2 \minus{} DC^2 \equal{}$ $ \frac {2(1 \plus{} a^2)}{(1 \plus{} a)^2} \minus{} 1 \equal{}$ $ \frac {a^2 \minus{} 2a \plus{} 1}{(1 \plus{} a)^2}$ $ \implies$ $ CG \equal{} \frac {1 \minus{} a}{1 \plus{} a}$ .

With other words, $ \boxed {\ \lambda\equal{}\frac {AE}{AB}\ \implies\ CG\equal{}\frac {1\minus{}\lambda}{1\plus{}\lambda}\cdot AB\ }$ . Particularly, for $ \lambda\equal{}\frac 13$ obtain $ CG\equal{}\frac 12\cdot AB$ .

Set the square with $ A(0,\ 0),\ B(6,\ 0),\ C(6,\ 6),\ D(0,\ 6)$ on the $ xy$ plane,

$ F: \left\{ \begin{array}{ll} AC: y \equal{} x & \quad \\
DE: y \equal{} \minus{} 3x \plus{} 6 & \quad \end{array} \right.\ \Longleftrightarrow F\left(\frac {3}{2},\ \frac {3}{2}\right)$, which can be also obtained by the similarity of tringles $ FAE$ and $ FCD,\ AF: FC \equal{} AE: CD \equal{} 1: 3$.

$ FG: y \equal{} \frac {1}{3}(x \minus{} \frac {3}{2}) \plus{} \frac {3}{2}\Longleftrightarrow y \equal{} \frac {1}{3}x \plus{} 1$, yielding $ G(6,\ 3)$.

$ \therefore CG \equal{} 6 \minus{} 3 \equal{} 3$.

[color=red]Edit[/color]

My solution was same as dgreenb 801 in #4.


There are five non-negative numbers whose sum is one.
Prove that these numbers can be arranged in a circle so that the sum of the five products of adjacent numbers in the circle is at most 1/5 !

There are 5! variants to distribute five numbers on circle.
If each time the sum is greater than 1/5 then common sum will be greater then 4!.
Each term ai*aj (i<j) take part in 5*3!*2 arrangements.
Thus
4!<5*3!*2*\sum ai*aj=5*3!*((a1+...+a5)^2-a1^2-...-a5^2)=
5*3!*(1-\sum ai^2)\leq 5*3!*(1-1/5)=4! (it is well known that \sum ai^2>1/5(\sum ai)^2  ==> \sum ai^2>1/5)
And what do we see? 4!<4!
The end.


A student expressed an interest in improving his solutions for the admissions test. I like this idea. It is always smart to try and improve your proof writing abilities. Hence, I will encourage all AMSP participants to try and write their best solutions for the admissions test problems.

I will be creating a thread for each problem. In each thread, I encourage [b]each student[/b] to work on their proof writing abilities by writing up the best solutions that they can. There are many sources for proof writing advice. Generally, I'd say: be clear and justify all assumptions. I'll periodically give feedback and other students may critique other students' proofs as well.

Not only write up problems that you solved easily, but also go back to problems that you had difficulties with. This is a good opportunity to learn.

[i]Even if someone else has already solve the problem, it is still good for you to write up your own solution.[/i] The point of this is to get experience writing solutions yourself.

--Alex

Hello, I been thinking about what to do with the solutions. I was hoping to collect them into a pdf and post that here. So when you are thinking about how you write your solutions, I'd hope that you'd write them in a suitable manner.

Ha I didn't expect this but some posters are worried about not being campers...

As long as you give thoughtful responses, I don't mind if you contribute to the discussions even if you aren't going to the camp or whatever...


4/2y - 7  =  6/5y - 8

I got the following answer Y= -10

Is this correct

No, your answer is incorrect.Let $Y=\frac{1}{y},....$


A cookie recipe calls for 1 pound of butter, how many grams of butter would be needed for  batches?

[quote="PiLuvah"]A cookie recipe calls for 1 pound of butter, how many grams of butter would be needed [b]for  batches[/b]?

[/quote]

Should there be a number between "for" and "batches"?

[hide="hello."]
1 pound = 453.592370380378 grams.

Multiply the number of batches by 453.592370380378 g (or however you choose to round it.)
[/hide]


[quote="aspiringchemlete"]Express $ \sin(x-y)+\sin(y-z)+\sin(z-x)$ as a monomial.[/quote]
What so hard,
$ \sin (x-y) = \sin x \cos y-\cos x \sin y$
Now do that for each one and combine them.

[hide="solution"]
$ \sin(x-y)+\sin(y-z)+\sin(z-x)=2\sin\frac{x-z}{2}\cos\frac{x-2y+z}{2}-2\sin\frac{x-z}{2}\cos\frac{x-z}{2}=$
$ =2\sin\frac{x-z}{2}\left(\cos\frac{x-2y+z}{2}-\cos\frac{x-z}{2}\right)=$
$ =4\sin\frac{x-y}{2}\sin\frac{y-z}{2}\sin\frac{x-z}{2}$
[/hide]

[quote="red_dog"][hide="solution"]
$ \sin(x-y)\sin(y-z)\sin(z-x)=2\sin\frac{x-z}{2}\cos\frac{x-2y+z}{2}-2\sin\frac{x-z}{2}\cos\frac{x-z}{2}=$
$ =2\sin\frac{x-z}{2}\left(\cos\frac{x-2y+z}{2}-\cos\frac{x-z}{2}\right)=$
$ =4\sin\frac{x-y}{2}\sin\frac{y-z}{2}\sin\frac{x-z}{2}$
[/hide][/quote]
It's addition, not multiplication.

[quote="i_like_pie"]
It's addition, not multiplication.[/quote]
Thanks, I edited.


geometry
3D geometry
tetrahedron
advanced fields
advanced fields theorems

Sperner’s Lemma generalizes to n dimensions, were it says that every coloring triangulation of a hyper tetrahedron into n + 1 colours must have an baby hyper tetrahedron that is labeled with all n + 1 labels. Prove this.


Explain why $ \ce{Li\plus{}}$ usually forms sparingly soluble solids with small anions (e.g. $ \ce{F\minus{}}$) while the solids are soluble with larger anions (e.g. $ \ce{ClO4\minus{}}$). Explain why the opposite trend is observed for $ \ce{K\plus{}}, \ce{Rb\plus{}}, \ce{Cs\plus{}}$.

I have a feeling that if I find an S.I. Chemical data book and calculate some lattice energy/ hydration energy values I might see why  :P .


When two people are on opposite ends of a teeter-totter, it will balance if the product of the weight of person A and her distance from the middle is equal to the product of person B's weight and his distance from the middle. When Ray sits four meter from the middle and Fay sits five meters from the middle on the other side, the teeter-totter will balance if Fay is holding her son, Trey. Ray weighs 180 pounds, and Fay weighs 135 pounds. How many pounds does Trey weigh?

WD(ray)=WD(faye and trey)
WD(ray)=180x4 or 720

720=WD(faye and trey)
720=5W(faye and trey)
144=W(faye and trey)

faye+trey=144
135+trey=144
trey=9


Euler
quadratics
modular arithmetic
algebra
function
domain
Ring Theory

[quote="Rust"]1) $ n^{2}\minus{}81n\plus{}1681 \equal{} x^{2}\plus{}x\plus{}41$ Euler's polinom, were $ x \equal{} n\minus{}41$.
2) If $ f(x) \equal{} p$ then $ p|f(x\plus{}kp)$.[/quote]
how to prove $ x^{2}\plus{}x\plus{}41$ is prime numbers for $ x$=1,2,3,...,40  :?:

Some advanced number theory (Minkowski's Theorem plus Quadratic Reciprocity, class numbers, etc.) can be used to prove that if $ p$ is an odd prime and $ \minus{}163$ is a square modulo $ p$, then there exist integers $ a,b$ such that
\[ p\equal{}a^{2}\plus{}ab\plus{}41b^{2}\]
This in turn shows that any prime divisor of a number of the form $ x^{2}\plus{}x\plus{}41$, where $ x$ is any integer, is greater than or equal to $ 41$. To see this, we have $ (2x\plus{}1)^{2}\equiv\minus{}163\pmod{p}$ so there exist integers $ a,b$ such that
\[ p\equal{}a^{2}\plus{}ab\plus{}41b^{2}\equal{}\left( a\plus{}\frac{b}{2}\right)^{2}\plus{}\frac{163}{4}b^{2}\]
$ b$ can't be 0, otherwise $ p$ would be a square, so not prime. Then $ |b|\geq 1$ implying $ p\geq\frac{163}{4}$. Finally, $ p$ must be an integer, so $ p\geq 41$.

Now if $ x^{2}\plus{}x\plus{}41$ were composite, we would require $ x^{2}\plus{}x\plus{}41\geq 41^{2}$, or $ (x\minus{}40)(x\plus{}41)\geq 0$.

Without such a powerful result, you must check them all... :evilgrin:

Why can you choose $ 163$,scorpius119  :roll:

The ring of integers in $ \mathbb{Q}[\sqrt{\minus{}163}]$ are a unique factorisation domain (UFD), giving the result.

But see http://www.mathlinks.ro/Forum/viewtopic.php?t=150948 for a direct proof (of something slightly more general).


geometry
3D geometry
arithmetic sequence

[hide]
I just listed the cubes out. Since the sum is odd, we must have consecutive odd integers. Note that $ 1197 \equal{}  5^3 \plus{} 7^3 \plus{} 9^3$. However, we must have more than 4 terms due to restrictions on $ n$. Adding $ (\minus{}3)^3 \plus{} (\minus{}1)^3 \plus{} 1^3 \plus{} 3^3$ will not change the sum, but will increase the number of terms to 7 and thus $ n$ to $ \fbox{6}$
[/hide]

[quote="dynamo729"]How do you do this? Much help would be appreciated.

The terms x, x+2, x+4, ..., x+2n form an arithmetic sequence, with x an integer. If each term of the sequence is cubed, the sum of the cubes is -1197. What is the value of n if n>3?

[hide]The answer is 6.[/hide][/quote]
I totally bombed that problem. I read the solution thousands of times and I still don't get it..

Wait... the sum of the cubes is $ ( \minus{} 1197)$ not $ (1197)$

it doesn't matter, just make the numbers you're cubing negative


ratio
analytic geometry
geometry
circumcircle
vector
function
algebra

Let A, B, C be 3 different points. Suppose that the ratio of their speeds is 2:3:6. Find a point P such that the sum of time is minimum when A, B, C move straight to P respectively.

Well, I don't get the problem.
Because you didn't really tell us where point A, B, or C are.

Wait, so couldn't you just pick any point to be P and put A,B and C at P?
Well, I never really could do graphs anyways.  :|

Hmmm I'm thinking Barycentric Coordinates, and some Fermat Point-like construction.

[quote="ifai"]A, B, C are any fixed point, i.e. they can be colinear or not.

And you can think of there are 3 people with speeds 2:3:6 from A, B and C. They want to meet each other and they want to find a place for meeting.[/quote]
What if the three people were at the same place?
They wouldn't have to go anywhere, because they are already there.
 :o

They are different points, but the time is so close to zero it's not even funny

The question is: given an arbitrary triangle $ \triangle ABC$, find the point $ P$ such that $ \frac{AP}{2} \equal{} \frac{BP}{3} \equal{} \frac{CP}{6}$.  (If the given ratio were 1:1:1 instead of 2:3:6, the question would be asking for the circumcenter of the triangle.)

Following in the construction of the circumcentre, by bisecting the sides, if we drew in for example the line perp to AC cutting AC in the ratio 2:6, and do the same for AB, BC, my guess is that'd be the answer. It easy enough to see that these lines do intersect.

And actually now ive thought about it, yes it is true  :)  it is the shortest time.

Once you prove JBL's proposition (which is intuitive, but not totally obvious how to prove), you can just construct two appolonius' circles and find their intersection.

I think you can say like if A,B get to the meeting point first, and C gets their later, you can move 'towards' C to reduce the time. But this argument shows that any other point is not minimal, it doesn't show that the 2:3:6 point is minimal.

I know that we don't need to find the 2:3:6 point (for some case we can find such a point.) But how about 3:6?

Gah -- the long post hidden below (and my post above) answers a different question, and it's not clear at all that the two problems have the same answer.  Bother.  The same sort of attack (local perturbation followed by an argument for the global minimum) certainly could be used here, but the work associated with the perturbation must be different.

[hide][quote="Altheman"]Once you prove JBL's proposition (which is intuitive, but not totally obvious how to prove), ...

I think you can say like if A,B get to the meeting point first, and C gets their later, you can move 'towards' C to reduce the time. But this argument shows that any other point is not minimal, it doesn't show that the 2:3:6 point is minimal.[/quote]

You know, it didn't even occur to me when I was writing it that it might not be the time-minimizing point.  But your explanation is sufficient, I believe: if two meet before the third has arrived, we can choose some very small vector (say, along the line of approach of the third person) so that the first two still meet before the third arrives, but the third arrives in a shorter period of time.  Thus, if the distances are [i]not[/i] in the ratio $ 2: 3: 6$ then we can't have a local minimum.  However, the time function is continuous.  Take some very large circle containing the triangle: then the time function must have a (global, and so also local) minimum on the closed disk determined by the circle (every continuous function achieves a minimum and maximum on a compact (i.e. closed and bounded, in the case of Euclidean space) domain), and this minimum can't occur on the boundary (we just choose the circle "too big" for that to be a possibility), so it must occur somewhere on the interior.  Since we've eliminated every point other than the 2:3:6 point, this point must be the minimum.

The set of points that are $ r$ times as far away from one fixed point as from another is just a circle.  Once you know this, it's easy to find which circle exactly it is.  (I believe this is what Altheman refered to as an Appolonius' circle.)  Then make two of them and find the intersection.

An interesting note: if this analysis is correct (it is), Seamusboyle is wrong.  The problem is an incorrect generalization.  The perpendicular bisector is a line perpendicular to the segment that divides it into a ratio $ 1: 1$, and this is the property seamusboyle chose to generalize from.  But the perpendicular bisector is also the Appolonius' circle in the ratio $ 1: 1$ (i.e. it's the set of points equidistant from the two endpoints); had Seamusboyle generalized in that way, the resulting solution would have been the same as mine.  (In fact, for an arbitrary triangle the three lines seamusboyle names are not concurrent, while the three Appolonius' circles certainly are (essentially by definition; what the definition doesn't tell you is that they are actually circles).)[/hide]

Sorry, didnt think that one through. Yeah circles of appolonius's (appolonii?) would be what i was trying to describe

Okay, here is an actual solution:

[hide]Consider the triangle $ ABC$ in the plane.  We seek a point $ P$ that minimizes the sum $ t(P) \equal{} \frac{AP}{2} \plus{} \frac{BP}{3} \plus{} \frac{CP}{6}$.  I claim that point $ A$ is the (usually unique) such point.  For any other point $ P$, let $ P'$ be a point on the segment $ AP'$ such that $ AP' \plus{} d \equal{} AP$ for some $ d > 0$.  Then $ BP' \minus{} BP \leq d$ and $ CP' \minus{} CP \leq d$ by the triangle inequality, so $ t(P') \minus{} t(P) \equal{} \minus{}\frac{d}{2} \plus{} \frac{BP' \minus{} BP}{3} \plus{} \frac{CP' \minus{} CP}{6} \leq \minus{}\frac{d}{2} \plus{} \frac{d}{3} \plus{} \frac{d}{6} \equal{} 0$.  Thus no point other than $ A$ can be the minimizer (so $ A$ is the unique minimizer), unless we have equality in both instances of the triangle inequality, i.e. if $ A$, $ B$, and $ C$ are colinear and $ A$ is not on segment $ BC$.  In this case, the set of time-minimizing points is $ [A, B] \cap [A, C]$ (the smaller of the two segments $ AB$ and $ AC$).

More generally, if we replace $ 2< 3 < 6$ with positive speeds $ v_1 < v_2 < v_3$ then $ A$ is a minimizing point whenever $ \frac{1}{v_1} \geq \frac{1}{v_2} \plus{} \frac{1}{v_3}$, and is the unique minimizing point unless $ \frac{1}{v_1} \equal{} \frac{1}{v_2} \plus{} \frac{1}{v_3}$ and $ A$, $ B$, $ C$ are arranged as described above.

This leaves open the case where $ \frac{1}{v_1} < \frac{1}{v_2} \plus{} \frac{1}{v_3}$, when no single person dominates.[/hide]


Suppose that the real numbers $a_1, a_2, \ldots, a_{100}$ satisfy
\begin{eqnarray*} 0 \leq a_{100} \leq a_{99} \leq \cdots \leq a_2 &\leq& a_1 , \\ a_1+a_2 & \leq & 100 \\ a_3+a_4+\cdots+a_{100} &\leq & 100. \end{eqnarray*}
Determine the maximum possible value of $a_1^2 + a_2^2 + \cdots + a_{100}^2$, and find all possible sequences $a_1, a_2, \ldots , a_{100}$ which achieve this maximum.

The maximum possible value of $a_1^2 + a_2^2 + \cdots + a_{100}^2$ is $100^2$, achieved by $a_1=100$ and $a_i = 0$ for $2\le i\le 100$ or $a_1=a_2=a_3=a_4=50$ and $a_i=0$ for $5\le i\le 100$.

Suppose the maximum value is achieved with $a_2 = m>0$. Then $a_1 = 100-m$ and $0< m\le 50$. If $q=\left\lfloor\frac{100}{m}\right\rfloor$, $q\ge 2$ and we want $a_1=100-m$, $a_2 = m$, $a_i=m$ for $3\le i\le q+2$, $a_{q+3} = 100-mq$, and the rest to be $0$. Note that we don't want $m$ to be so small that $\sum_{i=3}^{100} a_i<100$. The value we want to maximize is
\begin{align*}
    &(100-mq)^2 + qm^2+m^2+(100-m)^2\\
    &=m^2(q^2+q+2)-m(200q+200)+2\cdot 100^2.
\end{align*}
This is minimized at $m = \frac{200q+200}{2(q^2+q+2)} = \frac{100(q+1)}{q^2+q+2}$. The extreme of taking $m$ less than $\frac{100(q+1)}{q^2+q+2}$ and thus close to 0 leads to very large $q$, so we want to take $m$ larger than $\frac{100(q+1)}{q^2+q+2}$. Now we want to maximize $m - \frac{100(q+1)}{q^2+q+2}\le \frac{100}{q}-\frac{100(q+1)}{q^2+q+2} = \frac{200}{q(q^2+q+2)}$. To maximize this expression, we take $q=2$ to be as small as possible, and thus $m = 50$. This is a solution and shows that $100^2$ is the maximum when $a_2>0$. If $a_2 = 0$, then $a_1^2\le 100^2$, and the maximum is $100^2$ again.


I have this problem solution in which they have simplified the following fraction:
= {m - [1/3m -4] - [m/6 -2]  /2} -2
down to    m/4+1  without showing how and I can't figure it out...
I tried:
= {m - [1/3m -4] - [m/6 -2]  /2} -2
= {       2/3m +4  - [m/6 -2]  /2} -2
= {       2/3m +4  - [m/6 -12/6]  /2} -2
=         2/6m +2  - [m/12 +12/12]   -2
=         4/12m +2  - m/12 +1   -2
=         m/4 +1
But I think it's wrong. Can anybody help?
(I can't write Latex so I put the image of the problem solution so you may see it properly written)
[img]http://www.hawaii.edu/suremath/k4_12dir/k4_12img/mints2.GIF[/img]

$\frac{m-(\frac{1}{3}m-4)-(\frac{m}{6}-2)}{2}-2$
$=\frac{m-\frac{1}{3}m-\frac{m}{6}+4+2}{2}-2$          (rearrange the top to group together all $m$s)
$=\frac{\frac{1}{2}m+6}{2}-2$          (add up the expressions in $m$ and the constants, separately)
$=\frac{1}{4}m+3-2 =\frac{1}{4}m+1$  (divide each term in the fraction by the denominator, then add the constants)

Hope that helps!

It is worth learning LaTeX for this kind of thing.  It was so hard to read your proposed solution that I gave up, so I'm not sure if what you did was equivalent to this or not.  I am a relative newbie at LaTeX, but the above wasn't so hard.  I think if you pretend you want to quote my reply you will be able to see my source.

Thanks a bunch!
Next time I'll do my best to write in Latex. :D

[quote="altar"]Thanks a bunch!
Next time I'll do my best to write in Latex.[/quote]

Thanks for posting that.  :)

I was just about to mention it, but now I don't have to.  :)


trigonometry
algebra
polynomial
quadratics
geometry unsolved
geometry

Let $ ABC$ be a triangle which $ AB\equal{}AC$ . we know that $ \angle ABC \equal{} 70^o$ .

Let $ D$ be a point on $ AC$ which $ \angle ABD \equal{} 30^o$ .

Let $ E$ be a point on $ AB$ which $ \angle ACE \equal{} 40^o$  .

Find $ \angle CED$ ? .

[hide="comment"]I think that the angle is not integer[/hide]

I drew it in computer and the answer is 45 .
It does not have more information .
Prove that the answer is 45 .

I think it is wrong. With $ 45$ we don't get $ 180$ as sum of the angles in $ BEF$ (if $ F$ - intersetcion points of $ CE$ and $ BD$) or $ CFD$.

$ \angle CED \doteq 45.63^\circ.$ It is not possible to disprove the incorrect claim $ \angle CED \equal{} 45^\circ$ just by adding/subtracting angles. Proper way is to measure difference larger than resolution of the measurement or to calculate the angle numerically. Anything else is an [hide="overkill."]
From isosceles triangles $ \triangle AEC, \triangle BCD$ with $ AE \equal{} CE, BC \equal{} BD$:
$ CE \equal{} \frac {b}{2 \cos 40^\circ},$ $ CD \equal{} \frac {a^2}{b} \equal{} 4b \cos^2 70^\circ \equal{} 4b \sin^2 20^\circ \equal{} 2b(1 \minus{} \cos 40^\circ).$
By cosine theorem for $ \triangle CDE,$
$ ED^2 \equal{} \frac {b^2}{4 \cos^2 40^\circ} \plus{} 4b^2(1 \minus{} \cos 40^\circ)^2 \minus{} 2b^2(1 \minus{} \cos 40^\circ)$
Assume $ \angle CED \equal{} 45^\circ.$ Then by sine theorem for the same triangle,
$ \frac {ED^2}{CD^2} \equal{} 2 \sin^2 40^\circ \equal{} 2(1 \minus{} \cos^2 40^\circ)$
This gives the following equation for $ x \equal{} \cos 40^\circ$:
$ \frac {1}{4x^2} \plus{} 4(1 \minus{} x)^2 \minus{} 2(1 \minus{} x) \equal{} 8(1 \minus{} x^2)(1 \minus{} x)^2$

$ P(x) \equal{} 32x^6 \minus{} 64x^5 \plus{} 16x^4 \plus{} 40x^3 \minus{} 24x^2 \plus{} 1 \equal{} 0$

On the other hand,
$ \cos 120^\circ \equal{} 4 \cos^3 40^\circ \minus{} 3 \cos 40^\circ \equal{} \minus{} \frac {_1}{^2}$
$ Q(x) \equal{} 8x^3 \minus{} 6x \plus{} 1 \equal{} 0$
This cubic has no rational root, so that $ x \equal{} \cos 40^\circ$ is not constructible. But if $ x$ is a root of both $ P(x), Q(x),$ it is also root of a polynomial combination, such as

$ 2P(x) \minus{} Q^2(x) \plus{} 4(2x \minus{} 1)^2 Q(x) \equal{} 2(14x^2 \minus{} 14x) \plus{} 5 \equal{} 0$

This means that $ x \equal{} \cos 40^\circ$ is constructible root of a quadratic, which is a contradiction.
(The last quadraic actually has root $ x \equal{} \frac {_{7 \plus{} \sqrt {14}}}{^{14}} \doteq\cos 39.89^\circ.$)[/hide]


inequalities
induction
number theory proposed
number theory

If d1, d2, d3,...,dk are all divisors of natural number n (including 1=d1 and n=dk) prove the inequality
1/d1 + 1/d2 + ... + 1/dk <sqrt(2k)

Multiply by $n$ whole inequality. Now we are to prove: $\sigma (n) < n \sqrt{ 2 \tau (n)}$. It's enough to prove:

$p^{n}+...+p+1 < p^{n}\cdot \sqrt{n+1}$ for $p\geq 3$
$2^{n}+...+2+1 < 2^{n}\cdot \sqrt{2(n+1)}$

both can be done by induction on $n$.

In fact, by extending [i]Megus[/i]'s idea, the inequality can be strengthened to

$\sigma(n) \leq kn \sqrt{\tau(n)}$, where $k = \frac{3}{2\sqrt{2}}$, equality holding iff $n = 2$.


Answer poll for favorite type, and post your favorite video game below.

DDR, Rome: Total War, and Age of Mythology for me!

Sometimes I have dreams about ddr.... it's kind of creepy actually.  i wake up to find my feet stomping on my bed....

HALO!!!

oh yeah...definately my number 1 all time favorite game!

plus for the 360: Assasin's Creed, Halo Wars, Halo 3, Gears of War, Call of Duty, etc...hey...mostly all first person shooters  :D

Stick Arena.  Massive Online Multiplayer Game.

http://www.stickarena.com

my favorite is this old series of PC games.  it's the panzer general series.  they don't make them anymore, but they're still more fun than my new games.

PC, because I don't own any others.

Stick arena is good.
I play the original roller coaster tycoon, empire earth i, sim city 3000, aoe ii occasionally.

My computer's graphics card sucks, and doesn't play any new games, so I never buy any game. (ex. RTW doesn't work)

lol have you seen the system requirements for Medieval Total war and flight simulator... umm.. the new one.

MTW takes up [b]15 GIGS[/b] of hard drive... THAT'S ENORMOUS!

And Flight simulator takes up like 18 gigs....

Our hard drive (though I admit our computer is 7 years old) has an 18 gig hard drive and 13 gig auxiliary drive.   There is no way we can play those beasts!

Lucky me, my brother has a 250 gig hd, so i don't have to worry :)


The lengths of the sides of a cuboid are whole numbers. It's volume is numerically equal to its surface area. If the shortest edge-length is $ 3$, what is the greatest possible length of the longest side?

[hide="Solution"]Let $ a$ and $ b$ be the other unknown sides. Hence $ 6a \plus{} 6b \plus{} 2ab \equal{} 3ab$, so $ (a\minus{}6)(b\minus{}6) \equal{} 36.$
Since $ a$ and $ b$ are whole numbers, the largest value of $ a\minus{}6 \equal{} 36$, and so the maximum value of the longest side is $ 42.$[/hide]


quadratics
parameterization
algebra
quadratic formula
algebra unsolved

Let $ f_m: \mathbb R \to \mathbb R$, $ f_m(x)\equal{}(m^2\plus{}m\plus{}1)x^2\minus{}2(m^2\plus{}1)x\plus{}m^2\minus{}m\plus{}1,$ where $ m \in \mathbb R$.

1) Find the fixed common point of all this parabolas.

2) Find $ m$ such that the distance from that fixed point to $ Oy$ is minimal.

(1) Let $ y \equal{} f_m(x)$, arranging the equation : $ y \equal{} (m^2 \plus{} m \plus{} 1)x \minus{} 2(m^2 \plus{} 1)x \plus{} m^2 \minus{} m \plus{} 1$ with respect to $ m$ gives
$ (x \minus{} 1)^2m^2 \plus{} (x \plus{} 1)(x \minus{} 1)m \plus{} (x \minus{} 1)^2 \minus{} y \equal{} 0$ for all $ {m\in\mathbb{R}}$.

$ \Longleftrightarrow (x \minus{} 1)^2 \equal{} (x \plus{} 1)(x \minus{} 1) \equal{} (x \minus{} 1)^2 \minus{} y \equal{} 0$

Thus the desired fixed point is $ (x,\ y) \equal{} (1,\ 0)$.

(2) Sorry, I can't understand the context.

[hide="(a)"]

Let $r$ be a solution to the equation. The, by the quadratic formula:

$r=\frac{2(m^2+1) \pm \sqrt{4(m^2+1)^2-4(m^2+m+1)(m^2-m+1)}}{2m^2+2m+2} $

$r=\frac{2(m^2+1) \pm \sqrt{4(m^2+1)^2-4(m^4+m^2+1)}}{2m^2+2m+2} $

$r=\frac{2(m^2+1) \pm \sqrt{4m^2}}{2m^2+2m+2} $

$r=\frac{2m^2 \pm 2m + 2}{2m^2+2m+2} $

It follows that $r=1$ will be a root regardless of $m$, and therefore $(1,0)$ is a fixed point.
[/hide]


I have following:

Log (x+1) * log (x+1) = Log (x-1) * log (x-1)

Does anobody know what to do?
Thanx.

[hide="Solution"] Okay, so you have the [log (x+1)]^2=[log (x-1)]^2. Taking the square root, you get log (x+1)=log (x-1) and log (x+1)=-log (x-1). Thus, x+1=x-1--which doesn't work--or x+1=1/(x-1), (x+1)(x-1)=1, x^2=2, so x is the square root of 2.[/hide]


I saw this at a math competition.  You get full points for solving it in under a minute, a few for under 2 minutes, some for under 3, and 0 for after that.

So how would I solve this quickly?  The problem wasn't worded quite this way, but it was equivalent.

Let S be the set of integers 30 <= n <= 300 for which there is an integer k such that n(n+1)/2=k^2.

And.. you want to find the number of elements in S?

Well, I wonder if this way would be doable in under a minute..

[hide]

n and n+1 are coprime, so one is a square and one is twice a square. Twice squares between 30 and 300 go from 2*4^2 to 2*12^2.

32, 50, 72, 98, 128, 162, 200, 142, 288. We want those adjacent to a square: 50, 288 are the only ones that work (which gives n=49 and n=288).

[/hide]

I think thats pretty doable in under a minute.


algebra
polynomial
real analysis
real analysis unsolved

F: [0, ∞) → (0, ∞) is monotonic increasing. y = f(x) is a solution of y'' + F(x) y = 0. Prove f(x) is bounded as x → ∞.

Does this work?

[hide]
The characteristic equation of the differential equation is $ r^{2}+F(x)r = 0$ which implies that $ r =-F(x),0$, so that the general solution of the differential equation is

$ y = c_{1}+c_{2}*e^{-F(x)}$

So f(x) will take this form.

Now, because F(x) is always positive and is always increasing, $ e^{-F(x)}$ will approach 0 as x approaches infinity, so the limit of y as x approaches infinity is $ c_{1}$.

And this solves the problem. I hope.
[/hide]

It doesn't work. The "characteristic polynomial" approach is only valid for [b]constant coefficient[/b] linear differential equations. This one isn't constant coefficient.

[Also, the equation isn't $ y''+F(x)y'=0,$ as you seem to imply. It's $ y''+F(x)y=0.$]

A physics-inspired trick: introduce "energy".

[hide="Solution"]Let $ E=\frac12\left(\frac{(y')^{2}}{F(x)}+y^{2}\right)$. Differentiating, $ E'=\left(\frac{y'y''}{F(x)}+yy'\right)-\frac12\frac{(y')^{2}}{(F(x))^{2}}F'(x)=-\frac12\frac{(y')^{2}}{(F(x))^{2}}F'(x)\le 0$
Since $ E$ is decreasing, its value is always at most $ E(0)$, and $ y(x)\le \sqrt{2E(0)}$ for all $ x>0$.

This appears to assume that $ F$ was differentiable, but that's easy to patch. $ E$ isn't necessarily differentiable everywhere in general (or even continuous), but it is decreasing.[/hide]


Is there anyone out there that can tell me where I can find some past National MATHCOUNTS supplies.  I've been looking, but I just can't find anything.  Make sure to URL it.

[quote="13375P34K43V312"]dude
just read the sticky[/quote]

sticky?
where?

EDIT:i found the sticky, and it still didn't help.

btw, mathcounts would want us to use their past resources in order to help us succeed in the future.  i think its wrong for people to actually refuse the resources on the grounds that its cheating.  hmmmm.  i dont think that mathcounts uses the same national tests year after year, so what harm is there. thank you for listening to my outrage.

[quote="rd5493"][quote="13375P34K43V312"]dude
just read the sticky[/quote]

sticky?
where?[/quote]
The one that says "MATHCOUNTS Resources HERE."  Sticky topics are always at the top of the topic list so they can be easily found and viewed.

The only ways to get past national competitions are to either buy them from mathcounts.org or to ask your coach if if he/she has some.

Isn't there some archive on the web that has loads of national competitions? I remember visiting it once last year...

Possible, but not probable.  State competitions and hanbooks are everywhere, though.  Nationals are too valuable for MATHCOUNTS.  That's where they get their money, and if you're going to nationals, you might as well buy some and think of it as a donation.


everything looks fine to me

by "scores", do you mean the Alcumus Score?

since the Alcumus Score is calculated with a lot of factors in mind, including how other people do on similar problems, if you don't go on for a while, it is possible for it to be lowered

Yes, it definitely is not a hack; Alcumus scores for me can go down while I am not online, as well as while I'm getting questions correct. From what I see, it seems Alcumus inreases your rating "too much" while you are answering, and later on, your score will drop down more. In short, all these complex factors of the Alcumus score make it rather "unpredictable".

probably beaver
since mine went down somehow

well..
has it ever occurred to you that using a different computer can lower your rating?
also it could be a bug their trying to fix

Hello, we're keeping an eye on this issue but there definitely has been no hack.

Your alcumus score might change because of the other people who do the same types of problems.

well, i think alcumus grades on a curve... also, as everyone else said, your score goes down if you havent done problems for a long time.

Alcumus score is based on your raw score relative to others. Read the description at the question mark symbol.

I'm pretty sure there hasn't been a hack
Your alcumus score can be lowered for a different reason, which is that other people's alcumus scores are growing a lot and you haven't been doing much

There is no such thing as a hack. One's score is determined by comparing it with others. So if others do better, yours goes down.  :wink:

[quote="nbeav54"]how do my scores on alcumus get lowered for no reason! :o
I think there's a hack[/quote]
I hate this types of comments. Hackers (or more like crackers) don't waste their time lowering people's scores on Alcumus, understand the situation before blaming AoPs for not having security on their website.

Before I address your inquiry, I would like to ask you to please not revive inf the future. Reviving is defined as making an unrelated or unneeded post more than 30 days after the previous post. Next, time, just start a new thread with your question.

Your karma could get lowered simply if you spent karma on questions for video without earning any back.


If $ a,b,c,d \geq 0$ ,$ \sum_{cyc}^{}a^{2} \equal{} 1$
Prove $ \sum_{cyc}^{}\sqrt {1 \minus{} ab} \geq 2\sqrt {3}$

[quote="can_hang2007"][quote="tang zy"]Can anyone help me :?: I need a solution :!:[/quote]
Long time ago, I proposed (independently with pvthuan) the stronger inequality:

If $ a,$ $ b,$ $ c,$ $ d$ are real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 1,$ then
$ \frac {1}{1 \minus{} ab} \plus{} \frac {1}{1 \minus{} bc} \plus{} \frac {1}{1 \minus{} cd} \plus{} \frac {1}{1 \minus{} da} \le \frac {16}{3}.$

There are two nice solutions for this inequality.[/quote]
Can you post a solution to your problem or just a link is OK :lol:


results are in. I'm 155th on the canadian honor role. Any one make the 'role? if so, what was your rank.

[quote="xpmath"]Sorry  about posting here, but I took the Pascal internationally (unofficial score), and got a 142.
Good job on Cayley. You got a 134?[/quote]
Yah, got 134 :( I guessed on the last one cause I thought that everything else was right. But I got 22 wrong as well :( I would have been 58th. Oh well. Top 60 get to waterloo for a week.

lol...me got 150...on cayley...but i should have taken fermat... :D

[quote="SM4RT"][quote="jackdillon"] Top 60 get to waterloo for a week.[/quote]

how do you know that?[/quote]
My dad runs the contest for are school, but I could be wrong.

You mean for Seminar?

They take 60 ppl from other contests too. Mostly, it's grade 11s and they take most of them from Ontario.
They take very few grade 10s -- less than 10.
They also take Euclid winners and IMO ppl.

IMO??? Are you serious? I've never been to the Seminar, but I hope I can go this summer.

150 on Fermat is enough. :wink: you got in.
good job!

Hmm, rem, would 142 be enough? A good friend of mine in my school got that score on the Fermat.

Well, if he won the province, he's probably in. If not... I think very low chances because they take in barely anyone from Saskatchewan.

It's Saskatchewan we're talking about -- I find it quite unlikely that anyone outside our school has scored any higher, especially since our school took top 5 or 6 in SK on last year's Cayley.

Besides, I'm probably the only person in the province writing the CMO this year.


This is pretty cool. You made everything?

Yes, it's my ideas, my programming, and my graphics.

[quote="Serge31416"]Hello All,

  I did a new site with [url=http://www.iqflash.com/]Math games/Math Puzzles[/url] site. The site contains my new games/puzzles (except Rhumb Line), and my gaming articles. The games in the articles were described in Martin Gardner's books.
 The Math Puzzle "Breakfast Brainstorm" has a (while empty!) highscore table.

Good Luck!
 Serge[/quote]

Wow, great website!

Very great job. These games are very interesting.

Wow, how did you create your own website????

Although a revival, I also agree on the PWNage of this site.


Find all positive integers $ n$ that make $ n(n\plus{}180)$ a square.

[hide="A Partial Solution..."]
$ n\neq n \plus{} 180$
So, $ n$ and $ n \plus{} 180$ MAY both be perfect squares.
$ x^{2} \equal{} n$
$ y^{2} \equal{} n \plus{} 180$
$ y^{2} \minus{} x^{2} \equal{} 180$
$ (y \minus{} x)(y \plus{} x) \equal{} 180$
$ 180 \equal{} 3^{2}*2^{2}*5$
We know $ (y \minus{} x)$ and $ (y \plus{} x)$ are either both even or both odd. Since both cannot be odd (180 is not odd), then both must be even. Thus, we need only consider certain factors.
$ (2)(2*3^{2}*5) \equal{} (2*3)(2*3*5) \equal{} (2*3^{2})(2*5)$
The sum of the factors is $ y \minus{} x \plus{} y \plus{} x \equal{} 2y$
So, $ y \equal{} 46,18,14$
$ n \equal{} 1936,144,16$

All solutions may not have been considered, though (as unidranix pointed out).
Additionally, we must solve:
$ x^{2} \equal{} an$
$ y^{2} \equal{} a(n \plus{} 180)$

[/hide]

$ n\neq n\plus{}180$
So, $ n$  and $ n\plus{}180$  must both be perfect squares.
It's definitely wrong.
If $ ab$ is a perfect square then a and b both be perfect square if and only if $ (a;b)\equal{}1$

[quote="Unidranix"]$ n\neq n \plus{} 180$
So, $ n$  and $ n \plus{} 180$  must both be perfect squares.
It's definitely wrong.
If $ ab$ is a perfect square then a and b both be perfect square if and only if $ (a;b) \equal{} 1$[/quote]

I don't understand what you mean by it's wrong? The solutions are correct if you try them.
$ n$ and $ n\plus{}180$ don't have to both be perfect squares. You could easily have $ n\equal{}k^2\cdot j$ and $ n\plus{}180\equal{}h^2\cdot j$ so then the product would be a perfect square.

[quote="pascal12"][quote="Unidranix"]$ n\neq n \plus{} 180$
So, $ n$  and $ n \plus{} 180$  must both be perfect squares.
It's definitely wrong.
If $ ab$ is a perfect square then a and b both be perfect square if and only if $ (a;b) \equal{} 1$[/quote]

I don't understand what you mean by it's wrong? The solutions are correct if you try them.
$ n$ and $ n \plus{} 180$ don't have to both be perfect squares. You could easily have $ n \equal{} k^2\cdot j$ and $ n \plus{} 180 \equal{} h^2\cdot j$ so then the product would be a perfect square.[/quote]

pascal12, as I read Unidranix's comment, that user looked to be stating
that they both must be perfect squares.  What Unidranix should have done
was to quote or box quote the sentences from user mismatchtea as the reply
was being made by Unidranix.  Unidranix never specifically stated that the
given solutions from mismatchtea are not true, but rather, that they are
incomplete because they don't include non-square values of n.  (That is what
I got from the posts.)

Ah, thanks, that makes a lot more sense.

Here's what I got (There must be a better way, this is too much work!):
[hide="solution"]$ n^2 \plus{} 180n \equal{} x^2$
$ n^2 \plus{} 180n \minus{} x^2 \equal{} 0$
$ n \equal{} \frac { \minus{} 180 \plus{} \minus{} \sqrt {32400 \plus{} 4x^2}}{2}$
(We can eliminate the minus sign since we only want positive n)
$ n \equal{} \minus{} 90 \plus{} \sqrt {8100 \plus{} x^2}$
$ 8100 \plus{} x^2 \equal{} y^2$
$ 8100 \equal{} (y \plus{} x)(y \minus{} x)$
$ 8100 \equal{} 2^25^29^2$
Of course, you only need some combinations of factors since (as mismatchtea explained) they both are even.

$ 2(2\cdot5^29^2) \equal{} (2\cdot5)(2\cdot5\cdot9^2) \equal{} (2\cdot9)(2\cdot5^29) \equal{} (2\cdot9^2)(2\cdot5^2) \equal{} (2\cdot5\cdot9)(2\cdot5\cdot9)$
Now we get to solve for x, yay.
$ y \equal{} 2026,410,234,106,90$ giving $ x \equal{} 2024,400,216,56,0$
Of course, going all the way back to find n with $ n \equal{} \minus{} 90 \plus{} \sqrt {8100 \plus{} x^2}$, we get
$ n \equal{} 1936,320,144,16,0$ and we can disregard the 0 since we only want positive n.
So $ \boxed{n \equal{} 1936,320,144,16}$[/hide]

[quote="mismatchtea"][quote]
$ n \equal{} 1936,320,144,61,0$and we can disregard the 0 since we only want positive n.
So $ \boxed{n \equal{} 1963,320,144,16}$

[/quote]

Just a couple typos in the list, I think pascal12: $ n \equal{} 1936,320,144,16$[/quote]

Oh wow, thanks for catching those.

Sorry that I didn't quoted mismatchtea sentence  :|
Besides, he edited "must" to "may" :P
Good thinking Arrange your tan and finally we got the answer

wait a sec...
there are more than 4,
infinite actually, i think.
i am working on proof...

pascal12, what does this step mean?
[img]http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/9/b/0/9b03965ed2e4b6448eb9c54a615c3367bc8d7388.gif[/img]

i got lost haha...

Oh, sorry.
Ok, so you know how $ (y\minus{}x)(y\plus{}x) \equal{} 180$? So now we have to find all the possibilities for $ (y\minus{}x)$ and $ (y\plus{}x)$, meaning we have to divide $ 180$ up into its factors. For example, one possibility would be to have $ 2$ and $ 90$. To make it easier to divide up into its various factors, I used the fact that $ 180 \equal{} 3^{2}*2^{2}*5$ and listed out every single possibilities for each of the factors, making sure that there was a $ 2$ on each side.

Thank you very much pascal12.
Nice solution!

A very similar problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=311971]here[/url], where $ 2008$ is used instead of $ 180$. I think both solutions to that problem would work here.

There are a few solutions missing I think and @ pascal 1963 isn't even a solution.
FIrst lets think about this intuitively.
For n(x) to be a perfect square, x has to equal k^2*n.n*n(k^2)
So If we use that here. we see that x can be 4n, 16n. I may be wrong. the only solutions are:
3n=180. n=60
15n=180 n=12.

[quote="varunrocks"]
For n(x) to be a perfect square, x has to equal k^2*n.n*n(k^2)
[/quote]
varunrocks,
besides/along with pascal12's comment, if your horizontally typed
expression is supposed to be equivalent to ${ k^{2n}\cdot n \cdot nk^2}$,
(which would equal $ n^2k^{2n + 2}$), then you need grouping
symbols around the more complicated exponent for the order of
operations, and don't use a decimal point for multiplication as in
a suggested form such as:  [k^(2n)](n)(nk^2).

Here is the method PoP suggested:

$ n^2 \plus{} 180n \equal{} q^2$
$ 180n \equal{} (q \plus{} n)(q \minus{} n)$
Now we have two factors of 180n which differ by 2n. In other words:
$ 1 \pm 2n \equal{} 180n$
$ 2 \pm 2n \equal{} 90n$
$ 3 \pm 2n \equal{} 60n$
$ 4 \pm 2n \equal{} 45n$
$ 5 \pm 2n \equal{} 36n$
$ 6 \pm 2n \equal{} 30n$
$ 9 \pm 2n \equal{} 20n$
$ 10 \pm 2n \equal{} 18n$
$ \dots$
$ 90 \pm 2n \equal{} 2n$
$ 180 \pm 2n \equal{} n$

This means that $ n \equal{} 60$ would work, which is NOT what the other methods got.
But then why does this method not give all the solutions (like 1936). Sometimes this method works completely, though (like for n(n+2008) being a perfect square, as in PoPs link). Why is that so?

[quote="pascal12"]varunrocks:
You must've gotten 1963 from somebody who quoted my typo. I had fixed it to 1936 a long time ago. Also, I'm not exactly sure what you're doing in your solution, but neither 60 nor 12 work.[/quote]
Yeah, they both do...
$ 12(180 \plus{} 12) \equal{} 2304 \equal{} 48^2$
$ 60(180 \plus{} 60) \equal{} 14400 \equal{} 120^2$

[quote="Thunder365"][quote="pascal12"]varunrocks:
You must've gotten 1963 from somebody who quoted my typo. I had fixed it to 1936 a long time ago. Also, I'm not exactly sure what you're doing in your solution, but neither 60 nor 12 work.[/quote]
Yeah, they both do...
$ 12(180 \plus{} 12) \equal{} 2304 \equal{} 48^2$
$ 60(180 \plus{} 60) \equal{} 14400 \equal{} 120^2$[/quote]

 :oops: Whoops, I must've entered them on the calculator wrong somehow.

But yeah, I'm also wondering why each method does not give a complete solution. There must be flaws to each method, which I can't figure out.

[quote="Thunder365"]Here is the method PoP suggested:[/quote]
I don't think my method works now, because if $ n$ is something like $ 6$, its factors could be split up. So that method only accounts for prime $ n$.

[quote="ManuelS"]Since 180=3^2*2^2*5
then if n=5^(m+1) will make 180n a square
and n^2=5^(2m+2)
so n(n+180) will be a square for all integer m's greater than 1

Edit: this is a partial solution, though[/quote]

you found a value of m that makes n^2 and 180n a square. But when you add the two up, it isnt necessarily a square.

[quote="pascal12"]
$ 8100 \equal{} 2^25^29^2$
Of course, you only need some combinations of factors since (as mismatchtea explained) they both are even.

$ 2(2\cdot5^29^2) \equal{} (2\cdot5)(2\cdot5\cdot9^2) \equal{} (2\cdot9)(2\cdot5^29) \equal{} (2\cdot9^2)(2\cdot5^2) \equal{} (2\cdot5\cdot9)(2\cdot5\cdot9)$
Now we get to solve for x, yay.
$ y \equal{} 2026,410,234,106,90$ giving $ x \equal{} 2024,400,216,56,0$
Of course, going all the way back to find n with $ n \equal{} \minus{} 90 \plus{} \sqrt {8100 \plus{} x^2}$, we get
$ n \equal{} 1936,320,144,16,0$ and we can disregard the 0 since we only want positive n.
So $ \boxed{n \equal{} 1936,320,144,16}$[/quote]

$ 8100\equal{}2^{2}5^{2}3^{4}$.
Well, $ 9$ is not a prime factor - pascal12 did not fully factorize $ 8100$. So, we are missing some possible combinations that yield solutions.  If we use pascal12's solution and consider the possibilities he missed, we should be able to find all solutions.  :P

[quote="AndrewTom"]Find all positive integers $ n$ that make $ n(n \plus{} 180)$ a square.[/quote]

We have:

$ n^2 \plus{} 180n \equal{} k^2$.

Completing the square gives

$ (n \plus{} 90)^2 \equal{} k^2 \plus{} 8100$.

For convienence, write $ n \plus{} 90 \equal{} x$.

We now have

$ (x \minus{} k)(x \plus{} k) \equal{} 8100$.

Since the factors differ by $ 2k$, we know that they are of the same parity, thus both even.

Since $ 8100 \equal{} 2^2 \cdot 3^4 \cdot 5^2$, we can list out the solutions:

$ (x,k) \equal{}$

$ (2026,2024) \implies 2 \cdot 4050$
$ (678, 672) \implies 6 \cdot 1350$
$ (410, 400) \implies 10 \cdot 810$
$ (234, 216) \implies 18 \cdot 450$
$ (150, 120) \implies 30 \cdot 270$
$ (106, 56) \implies 50 \cdot 162$
$ (102, 48) \implies 54 \cdot 150$
$ (90, 0) \implies 90 \cdot 90$.

Now all that is left to do is to subtract 90 from x to get the possible n's:

$ n \equal{} 1936, 598, 320, 144, 60, 16, 12$ (we don't count 0)

Someone tell me if I made a mistake or missed a solution.

Hope this helps,
~BOGTRO

EDIT: [url=http://www.artofproblemsolving.com/Wiki/index.php/2008_AIME_I_Problems/Problem_4]A similar AIME problem[/url]

:/ I did all of it right and then forgot to minus 90 from my answers.  :|

[quote="mismatchtea"][quote="pascal12"]
$ 8100 \equal{} 2^25^29^2$
Of course, you only need some combinations of factors since (as mismatchtea explained) they both are even.

$ 2(2\cdot5^29^2) \equal{} (2\cdot5)(2\cdot5\cdot9^2) \equal{} (2\cdot9)(2\cdot5^29) \equal{} (2\cdot9^2)(2\cdot5^2) \equal{} (2\cdot5\cdot9)(2\cdot5\cdot9)$
Now we get to solve for x, yay.
$ y \equal{} 2026,410,234,106,90$ giving $ x \equal{} 2024,400,216,56,0$
Of course, going all the way back to find n with $ n \equal{} \minus{} 90 \plus{} \sqrt {8100 \plus{} x^2}$, we get
$ n \equal{} 1936,320,144,16,0$ and we can disregard the 0 since we only want positive n.
So $ \boxed{n \equal{} 1936,320,144,16}$[/quote]

$ 8100 \equal{} 2^{2}5^{2}3^{4}$.
Well, $ 9$ is not a prime factor - pascal12 did not fully factorize $ 8100$. So, we are missing some possible combinations that yield solutions.  If we use pascal12's solution and consider the possibilities he missed, we should be able to find all solutions.  :P[/quote]

Oh wow, I am so stupid. How could I possibly think that $ 9$ was a prime factor? Thanks for pointing this out, now we know the mistake (I'm surprised nobody caught it earlier, though)


A book is made by stacking ten sheets of paper, then folding the stack down the center. The sheets are stapled along the fold, and the pages are then numbered from 1 to 40. What is the sum of the other three numbers on the piece of paper that contains the number 15?
Generalize. (A book is made by stacking ten sheets of paper, then folding the stack down the center. The sheets are stapled along the fold, and the pages are then numbered from 1 to x. What is the sum of the other three numbers on the piece of paper that contains the number y?)

Another book is made in the same way as the book above. The sum of all the numbers on a sheet of paper is 54. One of the number is 37. Find the smallest number on that sheet of paper.

Close, but that's not what the problem asked for...

muy perfecto!

wurg, so how did you go by doing that.

[quote="Syntax Error"]muy perfecto![/quote]
Yo convenido, buen trabajo!

...why confused? All you had to do was subtract 15 from your previous answer, I think you got the concept.


[quote="lasha"]Wlog, $ a\geq b\geq c$. Then, $ LHS\leq 3(ab\plus{}bc\plus{}ca)$[/quote]

Try $ a \equal{} 1,b \equal{} 0,c \equal{}\minus{}1$ , $ 1\minus{}\sqrt{2}<\minus{}3$???? :o

[quote="lasha"]Wlog, $ a\geq b\geq c$. Then, $ LHS\leq 3(ab\plus{}bc\plus{}ca)$, but $ 3(ab\plus{}bc\plus{}ca)\leq (a\plus{}b\plus{}c)^{2}$, because $ ab\plus{}bc\plus{}ca\leq a^{2}\plus{}b^{2}\plus{}c^{2}$, as $ a^{2}\plus{}b^{2}\plus{}c^{2}\minus{}(ab\plus{}bc\plus{}ca) \equal{}\frac{1}{2}[(a\minus{}b)^{2}\plus{}(b\minus{}c)^{2}\plus{}(c\minus{}a)^{2}]\geq 0$, Q.E.D.[/quote]
You can not assume that $ a\geq b\geq c$.How did you prove that $ RHS\leq 3(ab\plus{}bc\plus{}ca)$?Did you use that $ a\geq b\geq c$?
[hide="Just an idea"] I think that this problem could be solved by Jensen inequality  [/hide]

Yes I used that $ a\geq b\geq c$. For example: $ a\sqrt{8b^{2}\plus{}c^{2}}\leq a\sqrt{8b^{2}\plus{}b^{2}}\equal{}3ab$...

[quote="lasha"]Yes I used that $ a\geq b\geq c$. For example: $ a\sqrt{8b^{2}\plus{}c^{2}}\leq a\sqrt{8b^{2}\plus{}b^{2}}\equal{} 3ab$...[/quote]
OK.But you maybe understand that you can't assume that $ a\geq b\geq c$. :rotfl:
Anyway as quangpbc wrote: $ RHS\leq 3(ab\plus{}bc\plus{}ca)$ isn't true.

[quote="lasha"] For example: $ a\sqrt{8b^{2}\plus{}c^{2}}\leq a\sqrt{8b^{2}\plus{}b^{2}}\equal{} 3ab$...[/quote]

But if $ b\ge c$ we can't talk $ b^{2}\ge c^{2}$ and if $ a < 0$ and$ \sqrt{8b^{2}\plus{}c^{2}}\leq\sqrt{8b^{2}\plus{}b^{2}}$ then we also can't talk $ a\sqrt{8b^{2}\plus{}c^{2}}\leq a\sqrt{8b^{2}\plus{}b^{2}}\equal{} 3ab$...

In short, your solution have very much wrong problem  :mad:

Solution  :) http://www.mathlinks.ro/Forum/viewtopic.php?t=150089


let $ n$ be positive integer, and  $ f_n$ be a set of function defined in $ [0,1]$ such that :
$ f_n(x)\equal{}0$ if $ x \in [0,\frac{1}{2n})$.
$ f_n(x)\equal{}\frac{1}{2}$ if $ x \in [\frac{1}{2n},\frac{1}{n}).$
$ f_n(x)\equal{}1$ if $ x \in [\frac{1}{n},1].$
find $ lim(n \rightarrow \infty )f_n(x)$ if it exist .

It converges pointwise to the function $ f(x) \equal{} 1 \minus{} \delta_{x, 0}$, i.e. $ f(x) \equal{} 0$ if $ x \equal{} 0$ and $ f(x) \equal{} 1$, otherwise.  It does not converge uniformly.


topology
analytic geometry
real analysis
real analysis unsolved

From Munkres, Analysis on Manifolds:

Show that I=[0,1] is a 1-manifold in R(1).

This does not make sense to me. As long as 1 is a boundary point of I, I don't see how there can be any coordinate patch about 1. The identity map covers [0,1), yes, but there is no continuous bijection on an open set containing 1 that maps into (x,1].

Is the question flawed or am I missing something?

$[0,1]$ is a one-dimensional [url=http://en.wikipedia.org/wiki/Manifold#Manifold_with_boundary]manifold with boundary[/url].

Perhaps I am not clear on what it means for a set to be open in a manifold.

If we say M is a manifold in R(n) and A is open in M does that mean A is the intersection of an open set in R(n) with M?[/i]

Yes, that works for manifolds embedded into $\mathbb R^{n}$.

For an abstract $d$-manifold $M$, as set $V\subset M$ is open if for all $i$ the set $\phi_{i}(U_{i}\cap V)$ is open in $\mathbb R^{d}$, where $\phi_{i}\colon U_{i}\to\mathbb R^{d}$ is a chart. For manifolds with boundary, $\mathbb R^{d}$ may be replaced with a half-space.

Thanks mlok. All I needed was a little thinking outside the box.


[hide]let the probability of company C winning equal x.
because company B is three times as good as company C its probability is 3x
because company A is twice as good as company B its probablility of winning is 2(3x) for a total of 6x.
Now to find the probability of each we put add up x, 3x, and 6x to get 10x.
Company C's probability is 1x/10x, and the x's cancel each other out for a total of 1/10.
Company B's probability of winning is 3x/10x, and once again the x's cancel each other out for a total of 3/10.
Finally, Company A's probability of winning is 6x/10x and the x's cancel each other out for a total of 6/10.
So the probabilities are as follows:
Company A: 3/5, Company B: 3/10, and Company C: 1/10[/hide]

led14pa2 is correct.

Basically, like the previous post, if you set C = x.

Then, B = 3x and A = 6x

3x+6x+x = 1
10x = 1
x = 1/10

So, basically, 6/10, 3/10, 1/10 for A,B,and C, respectively.

[quote="chenwb"]Company A: 1
Company B: 0
Company C: 0[/quote]

Yeah, I find that more realistic. What's the point of hiring inferior companies :P ?

[hide="In the wonderful world of math"]
A = 6x
B = 3x
C = x

6x + 3x + x = 1
x = 0.1

A = 60%
B = 30%
C = 10%
[/hide]

[quote="Silverfalcon"]Three companies are bidding on a contract. The relative qualities of the companies are such that Company A is twice good as Company B, and Company B is three times good as Company C. What is the probability of each company winning the contract? Give all three probabilities of three companies.[/quote]

I think this problem is underspecified, since it doesn't say how the contract will be awarded.  If you add that the probability of a company winning the contract is proportional to the quality of the company, then you get the answer Silverfalcon seems to be looking for.

# of companies are:

A A A A A A | B B B | C

so P(A) = 6/10 = 3/5
     P(B) = 3/10
     P(C) = 1/10

[quote="Silverfalcon"]Three companies are bidding on a contract. The relative qualities of the companies are such that Company A is twice good as Company B, and Company B is three times good as Company C. What is the probability of each company winning the contract? Give all three probabilities of three companies.[/quote]

You may like to reword the problem.  What is  the relationship,[i] if any[/i], between "company A being twice  as good as B"  and the probablity of some being selected?

(Some parents of teenagers, I know,  may even think that       their kids are more likely to select bad company ..:) )

Suggested change in the wordings could be,  " company A is twice as likely to win the contract than company B .." etc.. ...

I'd choose Company C.  If their quality is the worst, then they're probably the cheapest option.  Also, if they seriously screw up, you could always sue them...

but the higher quality work means the better feeling u have.lol :lol:

[hide="Solution"]If you company C the value of 1, then B gets the value of 3 and A gets 6. This makes it easy to do probability:
$A \Rightarrow 6/10 = 3/5$
$B \Rightarrow 3/10$
$C \Rightarrow 1/10$
That answer is if you mean that A is twice as likely as B and B is three times a likely as C[/hide]


[quote="Gil"] O.Bottema and M.Jovanovic proved the inequality
ma x mb x mc /abc  \geq 5/8.[/quote]

I wonder how they did that?!  :?  :?
Because if we take $a=13$, $b=25$ and $c=13$ we get that: $m_{a}^{2}=m_{c}^{2}=354,75$ and $m_{b}^{2}=12,75$, thus we get $0,299 \geq 0,625$ which is ....  :D  :D

[quote="Gil"]In the article :On the ratio ma x mb x mc / abc for the triangle ABC,Pub.Elek.Fak.,1973 O.Bottema and M.Jovanovic proved the inequality
ma x mb x mc /abc  \geq 5/8.(ma is the median from A)
I dont know how to prove this inequality![/quote]

 I haven't checked your counterexample,Lagrangia ;) .Anyway I think Gil is refering to this one which is very easy: :D  :P
          $\frac{m_{a}\cdot m_{b}\cdot m_{c}}{a\cdot b\cdot c}\geq\frac{3}{8R}$,

$ Proof.$

    Well,It's very easy to prove this well known inequality: $\frac{m_{a}\cdot m_{b}\cdot m_{c}}{S}\geq\frac{3}{2}\frac{a^2+b^2+c^2}{ab+bc+ca}$.Since $S=\frac{abc}{4R}$ the inequality becomes:

      $\frac{m_{a}\cdot m_{b}\cdot m_{c}}{ abc}\geq\frac{3}{8}\cdot\frac{(a^2+b^2+c^2)}{(ab+bc+ca) R}$
 Now from $a^2+b^2+c^2\geq ab+bc+ca$ it is obvious.


function
real analysis
real analysis unsolved

Let $ f$ be continuous and bounded on $ \mathbb{R}$ function such that \[ \sup\limits_{x\in\mathbb{R}}|f(x \plus{} h) \minus{} 2f(x) \plus{} f(x \minus{} h)|\to0,\quad h\to0.
\]
Does it follows that $ f$ is uniformly continuous on $ \mathbb{R}$? (A. Bondarenko, A. Prymak)

it remind an Oral of Ecole Polytecnique
$ f$ continous function on $ ]a,b[$

Suppose there exist $ C>0$ such that $ |\frac{f(x\plus{}h)\plus{}f(x\minus{}h)\minus{}2f(x)}{h}|\leq C$
each time it is possible (in french "chaque fois que cela a un sens")

1/ prove $ f$ is bounded on $ ]a,b[$

2/ Suppose $ x,x\plus{}h \in ]a,b[$ with $ h>0$

Prove for any $ n \in N$

$ |f(x\plus{}\frac{h}{2^n})\minus{}f(x)|\leq \frac{C.n.h}{2^{n\plus{}1}} \plus{} \frac{|f(x\plus{}h)\minus{}f(x)|}{2^n}$

Prove there exist $ C'>0$ such that for any $ d>0$ small

$ \sup_{|x\minus{}y|\leq d}|f(x)\minus{}f(y)| \leq C'.d.\ln(1/d)$

for the original exercice, suppose for example that $ |f|$ is bounded and that $ f$ is not uniformly continuous. Then one can find $ \epsilon>0$ such that there exists a sequence $ (x_n,y_n)$ satisfying $ |x_n\minus{}y_n| \to 0$ and $ |f(x_n)\minus{}f(y_n)| \geq \epsilon$. Now, take $ h_n$ such that
$ \sup_{x \in \mathbb{R}, |h| \leq h_n } \{f(x\minus{}h)\minus{}2 f(x)\plus{}f(x\plus{}h)\} \leq \frac{\epsilon}{n}$
So if $ x_n<y_n$ satisfies $ \delta_n \equal{} y_n\minus{}x_n \leq h_n$ and $ f(y_n) \geq f(x_n) \plus{} \epsilon$, then it is easy to see that:
$ f(x_n\plus{} (k\plus{}1) \delta_n)\minus{}f(x_n \plus{} k \delta_n) \geq \epsilon (1 \minus{} \frac{k}{n})$
for $ 0 \leq k \leq n\minus{}1$. Hence $ f(x_n\plus{} n \delta_n) \minus{} f(x_n) \geq \epsilon(\frac{n}{n} \plus{} \frac{n\minus{}1}{n}\plus{} \ldots \plus{} \frac{1}{n})\equal{}\frac{(n\plus{}1) \epsilon}{2}$, which shows that $ f$ cannot be bounded - a contradiction -

Moubinool your question is not quite clear. Where do x and x + h reside?


geometry
inequalities
algebra unsolved
algebra

[quote="dondigo"]We have positive numbers satisfying
[b]a[1]^2 + a[2]^2+  ... + a[100]^2 > 1[/b]
[b]a[1] + a[2] + ... + a[100] < 3[/b]
Proof that we can choose three numbers, that their sum will be bigger than 1

Thanks for any clue or solution[/quote]
[hide="brief idea"]Let wlog $a_1\geq a_2\geq\cdots\geq a_n$. Constuct a line segment with length 3 and then draw squares with lengths $a_i$ on it. Prove $a_1+a_2+a_3\geq 1$ by contradiction.[/hide]

How about deeper hint I still don't quite understand

[hide="the constuction"]Let $a_1\geq a_2\geq\cdots\geq a_{100}$. Constuct a line segment with length 3 and divide it into segments with length $a_1$ to $a_{100}$. Draw a square on every $a_i$. Divide also the longest segment into three segments with lengths 1 and draw some big squares on them. (Sorry for my poor English. Please refer to the attachment for the figure drawn.) Then consider the area.
[hide="rest of the solution"]We will prove $a_1+a_2+a_3>1$ by contradiction. Let $a_1+a_2+a_3\leq 1$. Note that the total area of small squares located in the first big square is no greater than $a_1\times 1$, while the total area of small squares located in the second and third squares is no greater than $a_3\times 2$. So we have
$a_1^2+a_2^2+\cdots a_{100}^2$

$\leq a_1\times 1+a_3\times 2$

$\leq a_1\times 1+a_2\times 1+a_3\times 1$

$=a_1+a_2+a_3$

$\leq 1$
Since this contradicts with the given condition $a_1^2+a_2^2+\cdots a_{100}^2>1$, the assumption $a_1+a_2+a_3\leq 1$ is wrong. Therefore $a_1+a_2+a_3>1$.[/hide][/hide]

yes, but a[1]+a[2]+a[3] + ... + a[100] < 3 not equal to 3
I'm not sure if we can build a line segment with a lenght 3

[quote="leepakhin"]Note that the total area of small squares located in the first big square is no greater than $a_1\times 1$, while the total area of small squares located in the second and third squares is no greater than $a_3\times 2$.[/quote]

Nice solution leepakhin.  Is the general intuition behind these types of problems to consider the "worst case"? Usually that means when the inequality becomes equalities or when you test the constraints to the breaking point. If what you're trying to prove still holds under these conditions, then you're on your way. Now, turning that intuition to an airtight proof is another thing entirely...

 :)


conics parabola trigonometry calculus derivative geometry

I've always wondered about this problem:
Let's say we want to throw/shoot a football (or whatever object you want) from the ground into the air at some angle $\theta$ off of the ground. What angle would allow the football to go the farthest? (Or does the angle depend on the speed of the ball, its weight, etc.?)

Sorry if this topic is possibly in the wrong forum.  :oops: If it is, mods can move it to a more appropriate one.

According to physics, it'll go the furthest when thrown at a 45 degree angle, but that's assuming there's no air resistance.

[quote="lotrgreengrapes7926"]I wonder if we could find the equation of the parabola given the starting speed of the object and its angle of launch and not taking into account wind resistance.[/quote]
Yes, that's what I thought of doing. We also need something to account for the fact that the acceleration of gravity is $-9.8 \ m/s^{2}$. But it's still not that simple: we also have to figure out a way to use the angle in the equation, etc. (We could interpret the movement of the football as going $\sin\theta$ up and $\cos\theta$ forward.)

[hide] We can set the starting point at $(0,0)$, so the equation must be in the form $f(x)=ax^{2}+bx$. The derivative at $(0,0)$ must be $\tan\theta=ax+b\Longrightarrow b=\tan\theta$.

If the ball is traveling at $v$, its vertical component of speed is $v\sin\theta$ and it will stay in the air for $\frac{v\sin\theta}{9.8}$. We can graph the time-speed graph and find the area of the right triangle in order to find its maximum height. The legs are $v\sin\theta$ and $\frac{v\sin\theta}{9.8}$, so the vertex is at $(-\frac{b}{2a}, \frac{v^{2}\sin^{2}\theta}{19.6})$.

$a(-\frac{b}{2a})^{2}+b(-\frac{b}{2a})=\frac{v^{2}\sin^{2}\theta}{19.6}\Longrightarrow\\ \frac{\tan^{2}\theta}{4a}-\frac{\tan^{2}\theta}{2a}=-\frac{\tan^{2}\theta}{4a}=\frac{v^{2}\sin^{2}\theta}{19.6}\Longrightarrow\\ a=-\frac{4.9}{v^{2}\cos^{2}\theta}$

$f(x)=\boxed{-\frac{4.9}{v^{2}\cos^{2}\theta}x^{2}+\tan\theta x}$[/hide] That was fun. :D

[quote="nebula42"][quote="lotrgreengrapes7926"]If the ball is traveling at $v$, then it will stay in the air for $\frac{v}{9.8}$.[/quote]
How did you get this?  :huh:
[hide="Here_is_My_Trig_Method"]If the ball begins traveling at a speed of $v$, then it will begin traveling at a speed of $v\sin\theta$ upward, so it will take $\frac{v\sin\theta}{9.8}$ time for the ball to stop moving upward and start moving downward, and then it will take the same amount of time to come back down and hit the ground again, so it spends a total of $\frac{v\sin\theta}{4.9}$ time in the air. All this time, it is moving at a constant rate of $v\cos\theta$ forward, so the total distance it travels is $\frac{v\sin\theta}{4.9}\cdot v\cos\theta=\frac{v^{2}\sin\theta\cos\theta}{4.9}$. We want to maximize this expression, which is equivalent to maximizing $\sin\theta\cos\theta=\frac{\sin2\theta}{2}$. The maximum value of this occurs when $2\theta=90^\circ$, or when $\theta=45^\circ$.

Heh, so it is $45^\circ$. I always thought it was something like $30^\circ$...  :P  :D [/hide][/quote] Yeah, I messed up on that part. I'll fix it. I regarded $v$ as the vertical velocity. :blush:


modular arithmetic
greatest common divisor

Consider a regular polygon with 2007 vertices. The natural numbers $ 1,2, \ldots, 4014$ shall be distributed across the vertices and edge midpoints such that for each side the sum of its three numbers (two end points, one side center) has the same value. Show that such an assignment is possible.

I have a feeling there's a much simpler solution, but here you go:

[hide="Solution"]Let the vertices be $a_{0},a_{1},\ldots ,a_{2006}$.  Now we construct a possible distribution.  For every $n$ such that $0\le n\le 2006$, let $a_{n}$ be the integer in the range $[1,4014]$ such that $a_{n}\equiv 2007+2006n\pmod{4014}$.  Let the common sum be 6022.  Now we prove that this is a valid distribution.

All the vertices are clearly odd.  We want to show that each vertex is a different odd number.  Consider $a_{n}-2007$.  We have $a_{n}-2007\equiv 2006n\pmod{4014}$.  Then $\frac{a_{n}-2007}{2}\equiv 1003n\pmod{2007}$.  Since the GCF of 1003 and 2007 is 1, $1003n$ runs through all integers $\pmod{2007}$ as $n$ runs through the integers from 0 to 2006.  Therefore, for each value of $k\pmod{2007}$, we have $\frac{a_{n}-2007}{2}\equiv k\pmod{2007}$, or $a_{n}\equiv 2007+2k\pmod{4014}$ for exactly one value of $n$.  Therefore, each vertex is a different odd number.

Now consider the numbers between the vertices.  These numbers must all be even since the odd numbers were used up by the vertices.  We want to show that each of these numbers is a different even number.  The number between the $n$th and $n+1$th vertices (where we consider the 2007th vertex to be the same as the 0th vertex) is $6022-a_{n}-a_{n+1}$.  Here we have two cases:

Case 1.  $1\le a_{n}\le 2007$.  Then we have $a_{n+1}=a_{n}+2006$.  Therefore, the number between the two vertices is $6022-(2a_{n}+2006)=4016-2a_{n}$.  Since $a_{n}$ is odd, we have $4016-2a_{n}\equiv 2\pmod{4}$.  Also we have $2\le 4016-2a_{n}\le 4014$.  For each value of $a_{n}$ such that $1\le a_{n}\le 2007$, the expression $4016-2a_{n}$ is equal to a different integer in the range $[1,4014]$ equivalent to $2\pmod{4}$.

Case 2.  $2009\le a_{n}\le 4013$.  Then we have $a_{n+1}=a_{n}+2006-4014=a_{n}-2008$.  Therefore, the number between the two vertices is $6022-(2a_{n}-2008)=8030-2a_{n}$.  Since $a_{n}$ is odd, we have $8030-2a_{n}\equiv 0\pmod{4}$.  Also we have $4\le 8030-2a_{n}\le 4012$.  For each value of $a_{n}$ such that $2009\le a_{n}\le 4013$, the expression $8030-2a_{n}$ is equal to a different integer in the range $[1,4014]$ equivalent to $0\pmod{4}$.

Therefore, all the numbers between vertices are different even numbers.  We used each of the first 4014 positive integers exactly once, and we defined the numbers between the vertices so that the sum of numbers along each side is 6022.  Therefore, our distribution is valid.[/hide]


At a math camp, every $m$ students share exactly one common friend, $m\ge 3$. (If $A$ is a friend of $B$, then $B$ is a friend of $A$. Also, $A$ is not his own friend.) Suppose person $P$ has the largest number of friends. Determine what that number is.

[quote="kokosai"]Isn't it just $m-1$?[/quote]

No. It is clear that the answer is at least $m$. ;)

[quote="fourierseries"]Are you missing some extra condition like a number of enemies?[/quote]

No. This is the exact wording of the problem.

One person is friends with everyone else  (looks like it works?)  I don't understand.

I don't understand whether you're asking for the minimum possible maximum number of friends or the maximum possible maximum number of friends - the former is just $m$ up to a point and then depends on the number of students, and the latter definitely depends on the number of students.  Is the answer supposed to be a function of $m$?   :huh:

Yes, the answer is a function of $m$...it's not as hard as you guys are making it, by the way. I will post the solution if nobody answers soon...

I think the question has too few restrictions, so its solution is somewhat trivial, I guess.  He wrote that no one can be his or her own friend, so the answer is $m-1$, I think.

Nope. There aren't $m$ total students...

I think it's just $m$.  Just take $m+1$ students and have all of them be friends with each other (and everyone's happy yay).

Well, every student has to have a friend so say $a_1, a_2$ are friends.  Then they have a common friend $a_3$ and $a_1, a_2, a_3$ have a common friend $a_4$, etc until we get to the following: $a_1, a_2, \dots , a_m$ have a common friend $a_{m+1}$.  Assume for the sake of contradiction that there are more than $m+1$ students, and let one be $b$.  If $b$ has at least two friends among $a_1, a_2, \dots , a_{m+1}$, WLOG $a_1, a_2$, then $b, a_3, a_4, \dots , a_{m+1}$ have two common friends, namely $a_1, a_2$.  If $b$ has at most one friend among $a_1, a_2, \dots , a_{m+1}$, say $a_1$ or none, then $b,a_1, a_4, \dots a_{m+1}$ must have a common friend (which cannot be $a_2, a_3$ so we have a person who is friends with at least two of $a_1, a_4, \dots , a_{m+1}$ because $m \geq 3$ and this reduces to our previous situation.  Yeah that wasn't a very clear writeup was it?