Evaluate

$\cos^{-1}\left(\frac{2\sqrt{5}}{5}\right) + \cos^{-1}\left(\frac{3\sqrt{10}}{10}\right)$

If you find Intermediate problems easy, please leave this challenge problem for less experienced students to solve.

Source: Alabama State Contest

[hide]Let

So answer in radians:

, where n is an integer.[/hide]

That looks like good work to me.  But you omitted the last step of finding the value of n.

so it's
$a+b=\pm \frac{\pi}{4}+2n \pi$, where n is an integer

We have: $\alpha  = \cos ^{ - 1} \left( {\frac{{2\sqrt 5 }}{5}} \right) = \arccos \left( {\frac{{2\sqrt 5 }}{5}} \right)$

and $\beta  = \cos ^{ - 1} \left( {\frac{{3\sqrt {10} }}{{10}}} \right) = \arccos \left( {\frac{{3\sqrt {10} }}{{10}}} \right)$

obvious since $\alpha ,\beta  \in \left( {0,\frac{\pi }{2}} \right)$
 but $\left( {\frac{{2\sqrt 5 }}{5} > \frac{{\sqrt 2 }}{2}} \right) \wedge \left( {\frac{{3\sqrt {10} }}{{10}} > \frac{{\sqrt 2 }}{2}} \right) \Rightarrow \alpha ,\beta  \in \left( {0,\frac{\pi }{4}} \right) \Rightarrow \alpha  + \beta  \in \left( {0,\frac{\pi }{2}} \right)$

We have:
$\cos \left( {\alpha  + \beta } \right) = \cos \alpha \cos \beta  - \sin \alpha \sin \beta  = \cos \alpha \cos \beta  - \sqrt {1 - \cos ^2 \alpha } \sqrt {1 - \cos ^2 \beta }  = $

$ = \frac{{2\sqrt 5 }}{5} \cdot \frac{{3\sqrt {10} }}{{10}} - \sqrt {1 - \frac{{4 \cdot 5}}{{25}}} \sqrt {1 - \frac{{9 \cdot 10}}{{100}}}  = \frac{{\sqrt 2 }}{2}$

once $\alpha  + \beta  \in \left( {0,\frac{\pi }{2}} \right) \wedge \left( {\cos \left( {\alpha  + \beta } \right) = \frac{{\sqrt 2 }}{2}} \right) \Rightarrow \alpha  + \beta  = \frac{\pi }{4} \Rightarrow $

$ \Rightarrow \arccos \left( {\frac{{2\sqrt 5 }}{5}} \right) + \arccos \left( {\frac{{3\sqrt {10} }}{{10}}} \right) = \frac{\pi }{4}$

So, now you might be ready for [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16426]this problem[/url].

The solution has already been done here by several people, but there's always room for one more idea. When I first saw the problem, I drew two right triangles - and noticed immediately that these triangles could have integer legs. Hence the angles in question had rational tangents. From there, you can use the identity for the tangent of the some of two angles - with the slight advantage that the arthmetic is all rational, with no radical signs needed.


[quote="harvest"]Someone from Korea could translate into English this awsome paper of functional equations????

follow the link below:

http://www.mathlinks.ro/viewtopic.php?mode=attach&id=14679

thx

 :!:[/quote]

How do you know that this paper is so awesome and needs to be translated  :maybe:  (no offense to the author)


Let sequence {$ a_{n}$}:$ a_{0} \equal{} 0;a_{1} \equal{} 1,a_{n \plus{} 1} \equal{} a_n \plus{} a_{[n/2]}$
Prove that there are infinetely  $ a_{k}$ such that $ a_{k}$ is divided by 7
---------------------------
 :blush: Sorry for my bad English :blush:

[quote="nguoigo"]Let sequence {$ a_{n}$}:$ a_{1} \equal{} 1,a_{n \plus{} 1} \equal{} a_n \plus{} a_{[n/2]}$
Prove that there are illimitableness   $ a_{k}$ such that $ a_{k}$ divisibility to 7
---------------------------
 :blush: Sorry for my bad English :blush:[/quote]

So $ a_2\equal{}a_1\plus{}a_{[1/2]}\equal{}a_1\plus{}a_0$

but ...

What is $ a_0$ ?

[quote="nguoigo"]Let sequence {$ a_{n}$}:$ a_{1} \equal{} 1,a_{n \plus{} 1} \equal{} a_n \plus{} a_{[n/2]}$
Prove that there are illimitableness   $ a_{k}$ such that $ a_{k}$ divisibility to 7
---------------------------
 :blush: Sorry for my bad English :blush:[/quote]
-----------------------------
Sorry $ a_{0}\equal{}0$


trigonometry
quadratics
geometry
geometric transformation
rotation
Asymptote
function

[hide] Let $ O$ be the center of the circle. Since $ OT\perp TC$, $ \Delta CAP\sim\Delta COT$. Let $ CA \equal{} x$.

From the similarity, $ \dfrac{9}{9 \plus{} x} \equal{} \dfrac{AP}{x}$. Thus, $ AP \equal{} \dfrac{9x}{x \plus{} 9}$. Let $ \angle{PAC} \equal{} \angle{TOC} \equal{} \theta$. We see that $ \cos{\theta} \equal{} \dfrac{9}{x \plus{} 9}$. Thus, $ \cos{\angle{PAB}} \equal{} \cos(180^{\circ} \minus{} \angle{PAC}) \equal{} \minus{} \dfrac{9}{x \plus{} 9}$.

By the Law of Cosines on $ \Delta PAB$,
$ m^{2} \equal{} 324 \plus{} \dfrac{81x^{2}}{(x \plus{} 9)^{2}} \plus{} 36\cdot\dfrac{9}{x \plus{} 9}\cdot\dfrac{9x}{x \plus{} 9}$;
$ \equal{} 324 \plus{} \dfrac{81x^{2} \plus{} 2916x}{(x \plus{} 9)^{2}}$;
$ \equal{} 324 \plus{} \dfrac{81x(x \plus{} 36)}{(x \plus{} 9)(x \plus{} 9)}$.

Letting $ \dfrac{1}{x \plus{} 9} \equal{} a$, we can rewrite this as

$ 324 \plus{} 81(1 \minus{} 9a)(1 \plus{} 27a) \equal{} 324 \plus{} 81(1 \plus{} 18a \minus{} 243a^{2})$.

This reaches its maximum when $ a \equal{} \dfrac{1}{27}$, easily seen by completing the square, or $ x \equal{} 18$.

Thus, the maximum value of $ m^{2}$ (equivalently, the square of the maximum of $ m$) is $ 324 \plus{} \dfrac{81\cdot18\cdot54}{27\cdot27} \equal{} \boxed{432}$.[/hide]

[quote="CatalystOfNostalgia"]Let $ \overline{AB}$ be a diameter of circle $ \omega$. Extend $ \overline{AB}$ through $ A$ to $ C$. Point $ T$ lies on $ \omega$ so that line $ CT$ is tangent to $ \omega$. Point $ P$ is the foot of the perpendicular from $ A$ to line $ CT$. Suppose $ AB \equal{} 18$, and let $ m$ denote the maximum possible length of segment $ BP$. Find $ m^{2}$.[/quote]
Here is a slightly different solution:
[hide="Solution"]
Extend $ PA$ to meet the circle at $ Q$ and let the center of the circle be $ O$.  Furthermore, let $ AC\equal{}x$ and $ TC\equal{}y$.   Since $ OT^2\plus{}TC^2\equal{}OC^2\implies 81\plus{}y^2\equal{}(9\plus{}x)^2\equal{}81\plus{}18x\plus{}x^2\implies y^2\equal{}18x\plus{}81$. $ AB$ is a diameter, so $ \angle AQB\equal{}90$ and $ \angle QAB\equal{}\angle CAP$, so $ \triangle QAB\sim \triangle PAC$.  Also, $ AP\perp TC$ and $ OT\perp TC$, so $ OT\parallel AP$, which results in $ \triangle CAP\sim \triangle COT$.  Thus, $ \frac{9\plus{}x}{x}\equal{}\frac{OC}{AC}\equal{}\frac{OT}{AP}\equal{}\frac{9}{AP}\implies AP\equal{}\frac{9x}{9\plus{}x}$.  Similarly, we can find that $ PC\equal{}\frac{xy}{9\plus{}x}$.  Thus, $ \frac{x}{18}\equal{}\frac{AC}{AB}\equal{}\frac{AP}{QA}\equal{}\frac{\frac{9x}{x\plus{}9}}{QA}\implies QA\equal{}\frac{162}{x\plus{}9}$.  Similarly, $ QB\equal{}\frac{18y}{9\plus{}x}$.  Since $ QP\equal{}QA\plus{}AP\equal{}\frac{9(18\plus{}x)}{9\plus{}x}$, we have that \[ BP^2\equal{}QP^2\plus{}QB^2\equal{}(\frac{9}{9\plus{}x})^2[(x\plus{}18)^2\plus{}4y^2]\equal{}81\cdot \frac{x^2\plus{}36x\plus{}324\plus{}72x\plus{}4x^2}{x^2\plus{}18x\plus{}81}\equal{}\frac{5x^2\plus{}108x\plus{}324}{x^2\plus{}18x\plus{}81}\cdot 81\]  This means that we wish to maximize $ \frac{5x^2\plus{}108x\plus{}324}{x^2\plus{}18x\plus{}81}$.  Say this value is $ k$, so $ x^2(k\minus{}5)\plus{}18x(k\minus{}6)\plus{}81(k\minus{}4)$.  The discriminant of this quadratic must be greater than or equal to $ 0$ to yield real solutions, which means that \[ 324(k\minus{}6)^2\minus{}81(4)(k\minus{}5)(k\minus{}1)\ge 0\] \[ k^2\minus{}12k\plus{}36\minus{}k^2\plus{}9k\minus{}20\ge 0\implies 3k\le 16\implies k\equal{}\frac{16}{3}\] Thus, $ BP^2\equal{}81k\equal{}16\cdot 27\equal{}\boxed{432}$. [/hide]

With an answer so nice (A is the midpoint of BC), does anyone have a really cool solution with rotations or something of that nature?

This one is similar to catalyst's, but slightly different. I included an asymptote diagram.
[hide="Another solution"]
[asy]unitsize(4mm);
pair B=(0,13.5), C=(23.383,0);
pair O=(7.794, 9), P=(2*7.794,0);
pair T=(7.794,0), Q=(0,0);
pair A=(2*7.794,4.5);

draw(Q--B--C--Q);
draw(O--T);
draw(A--P);
draw(Circle(O,9));

dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q);
label("B",B,NW);
label("A",A,NE);
label("$\omega$",O,N);
label("P",P,S);
label("T",T,S);
label("Q",Q,S);
label("C",C,E);
label("$\theta$",C + (-1.5,0), NW);
label("9", (B+O)/2, N);
label("9", (O+A)/2, N);
label("9", (O+T)/2,W);[/asy]

$ BQ \equal{} \omega T \plus{} B\omega\sin\theta \equal{} 9 \plus{} 9\sin\theta \equal{} 9(1\plus{}\sin\theta)$

$ QP \equal{} BA\cos\theta \equal{} 18\cos\theta$

$ BP^2 \equal{} BQ^2 \plus{} QP^2 \equal{} 9^2(1\plus{}\sin\theta)^2 \plus{} 18^2\cos^2\theta$

$ BP^2 \equal{} 9^2[1 \plus{} 2\sin\theta \plus{} \sin^2\theta \plus{} 4(1\minus{}\sin^2\theta)]$

$ BP^2 \equal{} 9^2[5 \plus{} 2\sin\theta \minus{} 3\sin^2\theta]$

Maximum at $ \sin\theta \equal{} \frac{\minus{}2}{\minus{}6} \equal{} \frac{1}{3}$

$ m^2 \equal{} 9^2[5 \plus{} \frac{2}{3} \minus{} \frac{1}{3}] \equal{} \boxed{432}$
[/hide]

I found this problem to be quite guessable - everything hinges on $ \sin \omega C T$, and if you try out some nice values you find that $ BP^2$ maximizes with $ \sin \omega  CT\equal{}1/3$. =\

[hide="Only because I like my asy image, basically like CatalystofNostalgia  :wink: "] [asy]size(250); defaultpen(0.70 + fontsize(10)); import olympiad;
pair O = (0,0), B = O - (9,0), A= O + (9,0), C=A+(18,0), T = 9 * expi(-1.2309594), P = foot(A,C,T);
draw(Circle(O,9)); draw(B--C--T--O); draw(A--P); dot(A); dot(B); dot(C); dot(O); dot(T); dot(P);
draw(rightanglemark(O,T,C,30)); draw(rightanglemark(A,P,C,30)); draw(anglemark(B,A,P,35));
draw(B--P, blue);
label("\(A\)",A,NW);
label("\(B\)",B,NW);
label("\(C\)",C,NW);
label("\(O\)",O,NW);
label("\(P\)",P,SE);
label("\(T\)",T,SE);
label("\(9\)",(O+A)/2,N);
label("\(9\)",(O+B)/2,N);
label("\(x-9\)",(C+A)/2,N);[/asy]
Let $ x = OC$. Since $ OT, AP \perp TC$, it follows easily that $ \triangle APC \sim \triangle OTC$. Thus $ \frac{AP}{OT} = \frac{CA}{CO} \Longrightarrow AP = \frac{9(x-9)}{x}$. By the [[Law of Cosines]] on $ \triangle BAP$,
\begin{align*}BP^2 = AB^2 + AP^2 - 2 \cdot AB \cdot AP \cdot \cos \angle BAP \end{align*}
where $ \cos \angle BAP = \cos (180 - \angle TOA) = - \frac{OT}{OC} = - \frac{9}{x}$, so:
\begin{align*}BP^2 &= 18^2 + \frac{9^2(x-9)^2}{x^2} + 2(18) \cdot \frac{9(x-9)}{x} \cdot \frac 9x\\&= 405 + 729\left(\frac{2x - 27}{x^2}\right)\end{align*}
Let $ m = \frac{2x-27}{x^2} \Longrightarrow mx^2 - 2x + 27 = 0$; this is a quadratic, and its [[discriminant]] must be nonnegative: $ (-2)^2 - 4(m)(27) \ge 0 \Longleftrightarrow m \le \frac{1}{27}$. Thus,
\[ BP^2 \le 405 + 729 \cdot \frac{1}{27} = \boxed{432}\]
Equality holds when $ x = 27$.[/hide]

No one has posted my solution, so let me.
[hide]Draw a perpendicular from P to AC and call where it hits D. Call length AC X, and length AP y.  By Pyth. Thm, PC is $ \sqrt{x^2-y^2}$.  The distance from C to the center of the circle is 9+x, while the distance from the center of the circle to T is 9.  By similar triangles,
\begin{eqnarray*}
\dfrac{9}{y}&=&\dfrac{9+x}{x}\\
9x&=&9y+xy\\
xy+9y-9x-81&=&-81\\
(9-y)(x+9)&=&81\\
x&=&\dfrac{81}{9-y}-9\\
&=&\dfrac{9y}{9-y}\end{eqnarray*}
Finding the area of triangle APC 2 different ways, we find that PD=$ \dfrac{y\sqrt{x^2-y^2}}{x}$.  By the Pyth. Thm again, DA=$ \dfrac{y^2}{x}$. Thus the distance PB is
\begin{eqnarray*}PD^2+DB^2&=&PB^2\\
\left(\dfrac{y\sqrt{x^2-y^2}}{x}\right)^2+\left(\dfrac{y^2}{x}+18\right)^2&=&PB^2\\
\dfrac{x^2y^2-y^4}{x^2}+\dfrac{y^4}{x^2}+36\dfrac{y^2}{x}+324&=&PB^2\\
y^2+36\cfrac{y^2}{\left(\frac{9y}{9-y}\right)}+324&=&PB^2\\
y^2+36y^2\dfrac{9-y}{9y}+324&=&PB^2\\
y^2+4y(9-y)+324&=&PB^2\\
-3y^2+36y+324&=&PB^2\end{eqnarray*}
Now the maximum for this function is at the vertex, when $ y=-\dfrac{36}{-2\cdot3}$ or $ y=6$.
$ PB^2$ is then $ -3(6^2)+36(6)+324=36(6-3)+324=108+324=432.$[/hide]

Calculate BP in terms of AC.  Take the derivative of this and find the critical point (which has to be a local maximum because it's stated in the problem =P, unless you want to do the concavity test, which is simple anyways), which turns out to be $ AC\equal{}18$.

[hide="AM-GM Finish"]
Let $ x\equal{}AC$ so that $ m^2\equal{}\frac{81(x\plus{}18)(5x\plus{}18)}{(x\plus{}9)^2}\equal{}81\left(1\plus{}\frac9{x\plus{}9}\right)\left(5\minus{}\frac{27}{x\plus{}9}\right)$
$ \equal{}27\left(3\plus{}\frac{27}{x\plus{}9}\right)\left(5\minus{}\frac{27}{x\plus{}9}\right)\le27\cdot\left(\frac{3\plus{}5}2\right)^2\equal{}27\cdot16\equal{}\boxed{432}$ with maximum at $ x\equal{}18$.
[/hide]

I would just like to comment that this problem is a good example of being efficient with your parameterization of the diagram.

Let $O$ be the center of the circle, $R=9$ and $\angle TOA=\theta$.  Then $TO\perp TC$ and so $TO$ and $PA$ are parallel. It follows that $PA=TO-AO\cos \theta=R(1-\cos\theta)$.

By the law of cosines on triangle $BPA$,
\[ \frac{BP^2}{R^2}=2^2+(1-\cos\theta)^2-2(2)(1-\cos\theta)\cos(\pi-\theta)\]
\[=4+(1-u)(1+3u)\] where $u=\cos\theta$. This is a quadratic that we have to maximize and so we are basically done.

The solutions booklet also includes the following query: Segment BP reaches its maximum length when lines AD, BP, and OT are concurrent. Why?

[quote=lingomaniac88]You know, I accidentally extended $ \overline{AB}$ through $ B$, yet I still got the same (correct) answer.[/quote]

lol same :P

I solved it using coordinate geometry along with some applications of trigonometry and quadratic functions.

[quote=samrocksnature]Still waiting for a purely synthetic solution.[/quote]
Not entirely synthetic, but elementary at least...

Let $O$ be the center of $\omega$. Let $AP$ intersect $\omega$ again at $Q$, and let $R$ be the midpoint of $AQ$.
[asy]
import olympiad;
import geometry;
unitsize(100);

pair A, B, T, O, P, Q, R;
A = dir(0);
B = dir(180);
T = dir(290);
O = midpoint(A--B);
line a = perpendicular(T, line(O, T));
line b = perpendicular(A, a);
P = intersectionpoint(a, b);
Q = intersectionpoints(circle(A, B, T), b)[1];
R = midpoint(A--Q);

draw(circle(A, B, T), linewidth(1));
draw(A--B, linewidth(1.5));
draw(B--Q--P--cycle, heavygrey);
draw(T--O, heavygrey);
draw(T--P, heavygrey);
draw(O--R, heavygrey + dashed);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$T$", T, dir(T));
dot("$O$", O, dir(O));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(Q));
dot("$R$", R, dir(R));
[/asy]
Note that $OT\perp TP$ so $OT\parallel PQ$, and $BQ\perp PQ$ so $BQ\parallel TP$. By a homothety centered at $A$, $OR\parallel BQ$, implying $OTPR$ is a rectangle. Let $\omega$ have radius $r$ and $AR = s$. Then we're looking to maximize $$BP^2 = BQ^2 + AQ^2 + PQ^2 - AQ^2 = AB^2 + PQ^2 - AQ^2 = (2r)^2 + (r+s)^2 - (2s)^2 = 5r^2 + 2rs - 3s^2.$$
Since $r$ is constant, this happens when $s = \tfrac{1}{3}\cdot r = 3$. Thus, $BP \leq 5\cdot 9^2 + 2\cdot 9\cdot 3 - 3\cdot 3^2 = \boxed{432}$.


I have a game, attached, to play, then, it makes me think about this

There is a 4x4 grid with added borders (like a torus). For example:
[code]
 1   2   3   4
 5   6   7   8
 9  10  11  12
13  14  15  16
[/code]
The square 2 is neighbor of 1,3,5,6,7  and also of 13,14,15. The number 1, within others, is neighbor of 16. Each square has eight neighbors. On each square there is a lamp. Initially all of them are off. Each time that you change the state of one of them, you change the state of its 8 neighbors (if they are on, they will turn off and viceversa). Is it possible to turn them all on?

It seems like nobody have tried yet. I hope you had enjoyed the game. Anyway, a hint, think about it as a torus, and see what similarities you can find!

[quote="probability1.01"]Switch all of them, one by one.[/quote]Yep, that works! it is just an argument of parity! This requires 16 operations (switch 16 times something!) Can you do it with less? (sure you can) let's see what is the minimum!

click the link "download" right below the clip image.

[quote="polymorphic"]I have always been rather computer illiterate.  Could someone enlighten me as to where I click to open attachments?  Thanks.[/quote]What operative system are you using. It is amazing, and I can't explain it. I am using FireFox in Windows and I can see the file, but using FireFox in Fedora3 I cannot  :?


Let $ a,b,c\ge 0$ such that $ a \plus{} b \plus{} c \equal{} 6$. Prove that
\[ \frac {a}{{{b^2} \plus{} 2}} \plus{} \frac {b}{{{c^2} \plus{} 2}} \plus{} \frac {c}{{{a^2} \plus{} 2}} \ge 2\]

The problem is flawed. Try $ a\equal{}b\equal{}c\equal{}2$

i checked my mistake. any one any idea plz

This inequality does not hold
Try $ a\equal{}1$ , $ b\equal{}2$ , $ c\equal{}3$

We have
\[ \sum_{cyc}\frac a{b^2 \plus{} 2}\geq \frac 43\]

\[ \begin{aligned} 2 \sum_{cyc}\frac a{b^2 \plus{} 2} \equal{} a \plus{} b \plus{} c \minus{} \sum_{cyc}\frac {ab^2}{b^2 \plus{} 2} \geq a \plus{} b \plus{} c \minus{} \sum_{cyc}\frac {ab^2}{3b^{\frac 43}} \left( \text{AM \minus{} GM}\right) \\
\equal{} a \plus{} b \plus{} c \minus{} \frac 13 \sum_{cyc} ab^{\frac 23}\geq a \plus{} b \plus{} c \minus{} \frac 19 \sum_{cyc}(a \plus{} ab \plus{} ab)\left( \text{AM \minus{} GM}\right) \\
\equal{} \frac 89 (a \plus{} b \plus{} c) \minus{} \frac 29(ab \plus{} bc \plus{} ca) \equal{} \frac 89 \cdot 6 \minus{} \frac 29 \cdot \frac 13 \cdot 6^2 \equal{} \frac {16}{3} \minus{} \frac {8}{3} \equal{} \frac 83\end{aligned}\]

\[ \therefore \sum_{cyc}\frac a{b^2 \plus{} 2}\geq \frac 43\]
But this inequality does not have equality for $ a\equal{}b\equal{}c$ which is the necessary case of equality in the AM-GM inequality that I used. Therefore we must have
\[ \sum_{cyc}\frac a{b^2 \plus{} 2}> \frac 43\]
Note that if we modify this problem a bit to the condition $ a\plus{}b\plus{}c\equal{}3$ then we have a convenient equality case.
\[ 2 \sum_{cyc}\frac a{b^2 \plus{} 2} \geq \frac 89 (a \plus{} b \plus{} c) \minus{} \frac 29(ab \plus{} bc \plus{} ca) \equal{} \frac 83 \minus{} \frac 23 \equal{}2\]
And here we have equality for $ a\equal{}b\equal{}c\equal{}1$. I think you meant the original problem to be as I mentioned before.

[quote="Potla"]We have
\[ \sum_{cyc}\frac a{b^2 \plus{} 2}\geq \frac 43\]

\[ \begin{aligned} 2 \sum_{cyc}\frac a{b^2 \plus{} 2} \equal{} a \plus{} b \plus{} c \minus{} \sum_{cyc}\frac {ab^2}{b^2 \plus{} 2} \geq a \plus{} b \plus{} c \minus{} \sum_{cyc}\frac {ab^2}{3b^{\frac 43}} \left( \text{AM \minus{} GM}\right) \\
\equal{} a \plus{} b \plus{} c \minus{} \frac 13 \sum_{cyc} ab^{\frac 23}\geq a \plus{} b \plus{} c \minus{} \frac 19 \sum_{cyc}(a \plus{} ab \plus{} ab)\left( \text{AM \minus{} GM}\right) \\
\equal{} \frac 89 (a \plus{} b \plus{} c) \minus{} \frac 29(ab \plus{} bc \plus{} ca) \equal{} \frac 89 \cdot 6 \minus{} \frac 29 \cdot \frac 13 \cdot 6^2 \equal{} \frac {16}{3} \minus{} \frac {8}{3} \equal{} \frac 83\end{aligned}\]

\[ \therefore \sum_{cyc}\frac a{b^2 \plus{} 2}\geq \frac 43\]
But this inequality does not have equality for $ a \equal{} b \equal{} c$ which is the necessary case of equality in the AM-GM inequality that I used. Therefore we must have
\[ \sum_{cyc}\frac a{b^2 \plus{} 2} > \frac 43\]
Note that if we modify this problem a bit to the condition $ a \plus{} b \plus{} c \equal{} 3$ then we have a convenient equality case.
\[ 2 \sum_{cyc}\frac a{b^2 \plus{} 2} \geq \frac 89 (a \plus{} b \plus{} c) \minus{} \frac 29(ab \plus{} bc \plus{} ca) \equal{} \frac 83 \minus{} \frac 23 \equal{} 2\]
And here we have equality for $ a \equal{} b \equal{} c \equal{} 1$. I think you meant the original problem to be as I mentioned before.[/quote]
A nice solution.So that we have another problem:
$ \sum\frac {a}{b^2 \plus{} 2}\ge 1$


algebra polynomial LaTeX calculus integration function domain

1) let R be a commutative ring and denote by R[x] the ring of polynomials wit coefficients in R. let p(x) belong to R[x] and p(x)= an x^n+an-1 x^n-1+.....+a0   ( really sorry abt the notation but i couldnt understand latex) show that p(x) is a unit in R[x] if and only if a0 is a unit in R and the rest of the ai's are nilpotent

2) show that in a boolean ring, every prime ideal is maximal

1. Let $\bar p$ be an arbitrary prime ideal of $R$.
If $p(x)$ is a unit in $R[X]$, then $\hat p(x)$ (the reduced mod \bar p of p(X)) is a unit in $R/\bar p[X]$. $\bar R=R/\bar p$ is an integral domain and over integral domains, the only invertible polynomials are constants. But $\hat p(X)$ is constant if and only if all the coefficients of $p(X)$ in degree greater or equal to $1$ are in $\bar p$. This means that the coefficients of $p(X)$ in degree greater or equal to $1$ are contained in every prime ideal of $R$, hence they are nilpotent. By identifying the degree $0$ part of $p(X)\cdot p(X)^{-1}=1$ we find that $p(0)$ is a unit.

Conversely, suppose we have $p(X)=u+Xq(X)$ with $u$ a unit of $R$ and $q(X)\in R[X]$ with all its coefficients nilpotents.
The nilpotents of $R[X]$ form an ideal in $R[X]$ and the nilpotents of $R$ are still nilpotents of $R[X]$. It follows that $q(X)$ and $Xq(X)$ are nilpotents.
The problem reduces to:

Let $R$ be a commutative ring with $1$. Assume that $u$ is a unit and $n$ is a nilpotent element of $R$. Prove that $u-n$ is invertible.
Proof: There is a nice proof which contructs the inverse of $u-n$ based on the identity $u^k-n^k=(u-n)(u^{k-1}+\ldots+n^{k-1})$ and taking $k$ such that $n^k=0$.
Another proof is: Assume $u-n$ is not invertible. Then it is contained in some maximal ideal $m$ of $R$. Since $n$ is nilpotent, it is contained in all primes thus also in $m$. It follows that $u\in m$ which is a contradiction.

2. Let $p$ be a prime ideal of a boolean ring $R$.
Then $R/p$ is an integral domain and also a boolean ring. Therefore $x^2=x$ for all $x\in R/p$. If $x\neq 0$, then since we are in a domain, $x^2=x\Rightarrow x=1$. This means that $R/p$ has only two elements, so it is isomorphic to $\mathbb Z/2\mathbb Z$ which is a field. We conclude that $p$ is maximal.


Find al positive integers $ m, n$, where $ n$ is odd, that satisfy

$ \frac{1}{m} \plus{} \frac{4}{n} \equal{} \frac{1}{12}$.

[hide]Put $ n\equal{}2k\minus{}1$. Then

$ {1\over m}\equal{}{1\over 12}\minus{}{4\over 2k\minus{}1}\equal{}{2k\minus{}49\over 24k\minus{}12}\iff m\equal{}{24k\minus{}12\over 2k\minus{}49}$

Write this as $ m\equal{}{12(2k\minus{}49)\plus{}576\over 2k\minus{}49}\equal{}12\plus{}{576\over 2k\minus{}49}$

Since $ 576\equal{}2^6\cdot 3^2$ and $ 2k\minus{}49$ is odd, we get $ 2k\minus{}49\in\{1,3,9\}$. Computing $ k,m,n$ in those cases, we get

$ (m,n)\in\{(588,49),(204,51),(76,57)\}$[/hide]


[quote="moldovan"]An equilateral triangle $ A_1 A_2 A_3$ is divided into four smaller equilateral triangles by joining the midpoints $ A_4,A_5,A_6$ of its sides. Let $ A_7,...,A_{15}$ be the midpoints of the sides of these smaller triangles. The $ 15$ points $ A_1,...,A_{15}$ are colored either green or blue. Show that with any such colouring there are always three mutually equidistant points $ A_i,A_j,A_k$ having the same color.[/quote]

lets say the side of original equilateral is 4a,
then of A7 to A15 , i could find 2+3 =5 sets of 3points which are equidistant from each other
2 with distance {sqrt(7)/4}a between each other
3 with distance a/4 between each other

and there are 2 ways in which dots can be painted such a way that every set of equidistant dots will have 1 dot with different color than other 2

My solution is based on the fact that if both of the vertices on a side of any equilateral triangle are colored in one color then the middle of this side is colored in the other color. After that it's just a matter of a few cases.


linear algebra
matrix
probability and stats

Suppose that

\[ X^{3\times 1} \sim N\left(\begin{pmatrix}a\\b\\a\end{pmatrix}, \sigma^2\begin{pmatrix}1&0&0\\0&2&0\\0&0&3\end{pmatrix}\right)\]

with $ a, b, \sigma^2$ all unknown.

Devise an F test for $ H_0: a \equal{} b$ vs. $ H_1: a\neq b$.

a naive way would be to do a Student test to check whether the mean of $ Y \equal{} X(1) \minus{} 2X(2) \plus{} X(3)$ is null. Student test applies here since $ Y$ is Normally distributed, with mean $ 2(a \minus{} b)$. (Or F-test on $ Y$)


can anyone be so kind as to tell me where to get past AMC 12 papers and solutions? i found IMO and USAMO but not any AMC ones.....

thanks!

[url]http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=44[/url]
It's missing a lot of tests and problems, but hey, at least it has some full tests...

You can also order copies of problem papers and solutions directly from the AMC office.
See
http://www.unl.edu/amc/d-publication/publication.shtml
and
http://www.unl.edu/amc/d-publication/d1-pubarchive/2006-7pub/2007-Pub,Revision.pdf

I recommend that you get the CD with all contests 2001-2006.

Of course, it won't arrive in time for the A-Contest date, but it might for the B-Contest date.

See also pages 14-21 of the Teachers' Manual, and also our Archives.

Steve Dunbart


If $3, 3log_{y}x, 3log_{z}y,$ and $7log_{x}z$ are in an arithmetic progression, show that

$x^{18}= y^{21}= z^{28}.$

[hide]
I think I'm missing something but:

$3+n=3log_{y}x$
$3+2n=3log_{z}y$
$3+3n=7log_{x}z$

becomes

$y^{3+n}=x^{3}$
$z^{3+2n}=y^{3}$
$x^{3+3n}=z^{7}$

The last one becomes $x^{12+12n}=z^{28}$
Your case works when $n=\frac{1}{2}$ and that's what I'm uncertain of - what about other n.

$x^{18}= z^{28}$
$y^{\frac{7}{2}}=x^{3}$ becomes $y^{21}=x^{18}(=z^{28})$

[/hide]


Bobert is motivated by money, therefore, for every good grade Bobert gets on his report card, he gets money, and for every bad grade he receives, his parents take away money.

Specifically, for every A+ he gets in math... and all of the other subjects... (old aops references ftw!), he gets paid three dollars, for every A he gets two dollars, and for every A- he gets one dollar.

For a B+, he loses ten dollars, and for every other grade, he loses a hundred dollars. (Strict parents.)

This year, Bobert broke even - he lost the same amount that he made. Assuming Bobert takes five classes (Math, Science, Language Arts, Gym, and Music) in school, [i]how many possible ways could his report card have turned out[/i]? One possible way is an A+ in Math, a B+ in Science, an A+ in Language Arts, an A- in Gym, and an A+ in Music.

Assume that the letter grades are all that are shown on the report card - no numbers.

Hints: the hints are arranged in ascending give-away-ness: the bottom one gives the answer.

[hide=" :!:"]The answer's positive.[/hide]
[hide=" :huh:"]Casework.[/hide]
[hide=" :?"]Think of the ways you can break even,  and in how many ways you can do all of them.[/hide]
[hide=" :?:"]The answer is [b]50[/b], if I'm not mistaken[/hide]

I think you mean B+ in Science.  :P [hide="Solution"] He must get B+ in some subject, otherwise he would either always get money or lose 100 dollars, which is too much to break even. So, how may ways can 4 numbers from the set {1, 2, 3} be added up to equal 10?
(1) If there is 1 1, the other 3 must be 3s.
(2) If there are 0 1s and 1 2, it must be 3, 3, 2, 2.
(3) If there are 0 1s and 2 2s, it must be 3, 3, 2, 2 again.
My brain isn't functioning. Are those the only 2?
So the grades must be either A+, A+, A+, A-, B+ or A+, A+, A, A, B+. There are a total of $\frac{5!}{3!1!1!}=20$ permutations for the first one and $\frac{5!}{2!2!1!}=30$ for the second one. So $20+30=\boxed{50}$ :D :D [/hide]

Are you sure you made this up?

I think that you changed the words a little, but if you made it up, nice work.

Do you have experience with this, Iron Fist?

Experience with what? And yes, I did make this problem up...

You should read the part about parents and then you can figure out what I am talking about.

[quote="Iron Fist"]Oh. XD no, my parents are too strict to even give that kind of deal :p .[/quote] I wish my parents did that.  :roll: I would get lots of money. :D Except I have a B+ in French now.  :wallbash:

Iron Fist, you are really good at making up problems,


For a sequence {$a_{n}$}$_{1}^{\infty}$, define

$A_{n}= \frac{a_{1}+a_{2}+...+a_{n}}{n}$.

Show that if $\lim_{n\to\infty}a_{n}= 0$ then $\lim_{n\to\infty}A_{n}= 0$.

This is a very commonly quoted theorem on the forum- it goes by the name of Cesaro-Stolz.

The core of the argument: We can go far enough out to make all later terms "small". If we keep going from there, the sum of the early terms is eventually overwhelmed, and the averages become "small" as well.

Mix in $\epsilon$ as needed to make it rigorous. The two values of "small" shouldn't be equal.

Could you be more specific in terms of how we might argue that the later terms will become smaller and will balance out with the earlier terms?
Thanks.

I was vague so that you could fill in the details. You could probably use the practice with $\epsilon$-$\delta$-style proofs.

one query:
how do you know that $\lim_{n\to\infty}\frac{|a_{1}|+|a_{2}|+...+|a_{N_{1}}|}{n}= 0$?
Thanks.

[quote="countdownmath"]never mind, I think I've figured it out..
is it valid to say that because we can assume that $|a_{1}|+...+|a_{N_{1}}|$ to be a constant?[/quote]
Yes, because $N_{1}$ is fixed.


Find the value of:
$ cos^2 0\plus{} cos^2 1\plus{}\cdots\plus{}cos^2 90$

I have a solution to this but I believe that there is one that does not require a drawing and is fully based on trig functions.

If you know a solution please post it.

I believe I proposed a solution to a similar problem a while back... Anyway, just apply the following (I'm assuming the input values are expressed in degrees):

[hide="Identities!"]Co-function identity: $ \cos\;(90^{\circ} \minus{} \theta) \equal{} \sin \theta$

Pythagorean Identity: $ \sin^{2} \theta \plus{} \cos^{2} \theta \equal{} 1$[/hide]

Edit: It was [url=http://www.mathlinks.ro/viewtopic.php?t=206998]this[/url] problem.

Thanks now I know how to do it know with only trig functions.

Yes those are in degrees.


The  sequence $(a_n)$ is  defined  by  $a_1=1$ and $a_n=a_{n-1}+\frac{1}{n^3}$  for  $n>1.$
(a) Prove  that   $a_n<\frac{5}{4}$    for all  $n.$
(b) Given $\epsilon>0$, find  the smallest natural number  $n_0$  such  that  ${\mid a_{n+1}-a_n}\mid<\epsilon$  for all  $n>n_0.$

$a_n < \sum_{k=1}^{\infty} \frac 1{k^3}=\zeta (3) =$Apéry's  constant (from 1977)
$\zeta (3) = 1.2020569031.....$ is irational (TBD if algebric or transcedental !)
:cool:

Another solution

\[ \zeta (3) < \sum\limits_{k = 1}^{N} {1/k^3 } + \int\limits_{x = N}^\infty \frac{dx}{x^3} = \sum\limits_{k = 1}^{N } {1/k^3 } + \frac{1}{2N^2}  \]

N= 2 gives the estimate 5/4.


geometry
incenter
parallelogram
geometry unsolved

The triangle $ ABC$ is given. On the extension of the side $ AB$ we construct the point $ R$ with $ BR \equal{} BC$, where $ AR > BR$ and on the extension of the side $ AC$ we construct the point $ S$ with $ CS \equal{} CB$, where $ AS > CS$. Let $ A_1$ be the point of intersection of the diagonals of the quadrilateral $ BRSC$.
Analogous we construct the point $ T$ on the extension of the side $ BC$, where $ CT \equal{} CA$ and $ BT > CT$ and on the extension of the side $ BA$ we construct the point $ U$ with $ AU \equal{} AC$, where $ BU > AU$. Let $ B_1$ be the point of intersection of the diagonals of the quadrilateral $ CTUA$.
Likewise we construct the point $ V$ on the extension of the side $ CA$, where $ AV \equal{} AB$ and $ CV > AV$ and on the extension of the side $ CB$ we construct the point $ W$ with $ BW \equal{} BA$ and $ CW > BW$. Let $ C_1$ be the point of intersection of the diagonals of the quadrilateral $ AVWB$.
Show that the area of the hexagon $ AC_1BA_1CB_1$ is equal to the sum of the areas of the triangles $ ABC$ and $ A_1B_1C_1$.

Denote I is the incenter of △ABC  then we have AC1BI BA1CI  CB1AI  are  parallelograms
this completes the proof

[quote=pisgood]Anyone have a more elaborate proof for this? :)[/quote]

It is easy to check that $IAC_1B$ ,$IAB_1C$ and $IBA_1C$are parallelograms and triangle $BC_1A_1$ and $IAC$ are equal,thus $ABC$ and $A_1B_1C_1$ are equal,then we must prove that the sum of area of triangles $ABC_1$,$BCA_1$ and $ACB_1$ are the area of $ABC$,and this us obvious, according to the past sentences.


function
logarithms
calculus
analytic geometry
graphing lines
slope
derivative

$e^x-1$ passes through $(0, 0)$ with a slope of 1, and $ln{x}+1$ passes through $(1, 1)$ with a slope of 1.  The second derivatives are positive and negative, respectively.  So yeah.

[quote="dasherm"]Yeah, but that's using calculus.[/quote]

Okay, then assume there is a solution.
$\ln{x}+1=\ln{(ex)}=e^x-1 \implies ex=e^{e^x-1} \implies x=e^{e^x-2}$
$e^x-1=\ln{(ex)} \implies e^x=\ln{(e^2x)} \implies x=\ln{(\ln{x}+2)}$
\[ \implies x=e^{e^x-2}=\ln{(\ln{x}+2)} \implies x=\ln{(2+\ln{(\ln{(2+\ln{x})})})} \]
So,
$x=e^{-2+e^{e^{-2+e^{...}}}}=\ln{(2+\ln{(\ln{(2+\ln{...})})})}$

This is best written as \[ x=\ln{(2+\ln{x})} \]

If $x\le \frac{1}{e^2}$, the the right side is undefined.
If $\frac{1}{e^2}<x\le \frac{1}{e}$, then the right side of the equation is less than $0$.
I suppose one can go on setting restrictions and prove it.

[quote="Hikaru79"][quote="Elemennop"]Graph it, you can see they approach but $e^x$ increases faster, therefore there are no solutions.[/quote]

Okay, well, just for fun, let's say you didn't have a graphing calculator. Besides, it's not that easy to graph; if I had just changed those numbers slightly so that they are $f(x)=e^x-2$ and $g(x)=\ln{x}+2$ then they WOULD intersect. Twice!

A little more detail than "look at the graph" is needed in this case, since they are so close together.[/quote]

We didn't have a calculator either during the test, but it isn't that hard to do. Graphing it can show you some basic values (say, x=0, 1, e) , and then you can use first and second derivatives to see the acceleration of the curve; there was already an integral on the test, so I believe using calculus is perfectly canon for this problem.

[quote="chess64"]What test? :?[/quote]

It was a Round 9 question from the 4x4 contest. The contest was written yesterday (February 9th)

[quote="Hikaru79"]At how many points in the Cartesian plane do the functions $f(x)=e^x-1$ and $g(x)=\ln{x}+1$ intersect?

(I hope this doesn't count as a calculus question; it's technically not, although a lot of people meet $e$ for the first time in a calculus course)[/quote]

For $x>0$ we have $e^x-1>x$ and $x>\ln{x}+1$  then [b]zero intersections ![/b]


Determine all triples (x,k,n) of positive integers such that 3 k  - 1 = x n.

well i guess the previous solution is wrong
here goes my solution
case 1) n=1
then the solution are (x,k,n) = (3^k-1,k,1)
case 2) n is even
then x^n+1 = 1 or 2 (mod 3) => 3^k is not divisible by 3
contradiction
case 3) n is odd
then x^n+1 can be factorized as (x+1)[x^(n-1)-x^(n-2)+.... + 1]
If 3^k = x^n + 1, then both of these factors
are powers of 3, say they are 3s, 3t,
respectively. Since
x + 1  \leq  x^(n-1 )- x^(n-2).... + 1
so s  \leq  t . Then
0 = 3t =(-1)^(n-1) - (-1)^(n-2)..... + 1
= n (mod x + 1)
implying n is divisible by x + 1 (and
hence also by 3). Let y =x^(n/3)
Then
3^k = y^3 + 1 = (y + 1) (y^2 - y + 1).
So y+1 is a power of 3 say y+1 = 3^r
if r=1 we get (x,k,n) = (2,2,3)
if r \geq 1 then 3^k = y^3 + 1 = 3^(3r) - 3^(2r+1) + 3^(r+1)
is strictly between 3^(3r-1) and 3^(3r), a
contradiction.
final (x,k,n) = (2,2,3) and (3^k-1,k,1)


Find the number of solutions to $ mn \plus{} 7n \equal{} 20$.

EDIT: Sorry, positive integer solutions. I suck at remembering things like that.

uh, infinitely many?

if you mean integers, n(m+7)=20, check factors

[hide]Factoring rules.

$ n(m\plus{}7)\equal{}20$

Because $ m\in\mathbb{Z}^\plus{}$, $ m\plus{}7\geq8$, so thus, $ n\leq\frac{20}8\equal{}2.5$.  There are two factors of 20 less than 2.5 (1 and 2), so thus, there are only two solutions.[/hide]


[hide]
I assume that you mean $\frac{1}{2}$, not $\frac{1}{3}$.

Anyway, call the sum $S$.  Then, multiplying by 2, $2S=S+\frac{1}{2}$, so $S=\frac{1}{2}$.
[/hide]


Alice, Bob, and Carol repeatedly take turns tossing a die in alphabetical order by their names. Find the probability that Carol will be the first person to toss a 6.

I think this question may involve an infinite geometric sequence...

It does.

Alice: $ \frac56$
Bob: $ \frac56$
Carol: $ \frac16$ or $ \frac56$

The probability that Carol wins on the first round is $ \frac{25}{256}$.

The probability that Carol does this on the second round is $ \frac{5^5 \times 1}{6^6}$, or $ \frac{3125}{46656}$.

The probability that Carol does this on the third round is $ \frac{5^8}{6^9}$.

This can be expressed as $ \sum ^ \infty _{i\equal{}1} \left(\frac{5^{3i\minus{}1}}{6^{3i}}\right)$.


What is the coefficient of $x^{296}$ in the expansion of $(x^3+\sqrt{2}x^2+\sqrt[3]{3}x+1)^{100}$?

You can use multinomial Theorem and just find the several cases that have $x^{296}$ as the variable and find each coefficient and add them up.


Prove that one can colour all positive rational numbers in 10 colours so that any two numbers whose decimal expansions differ by only 1 digit have different colours, by finding a colouring algorithm. Also, for a harder problem, try seeing if the statment is corect for ALL real nubers.

There's a little wrinkle to this problem i think.  the decimal expansion of any rational is not unique and is in fact also true for other base systems.  for example, $ 0.9999....\equal{}1.0000$  maybe we can add the condition that repeating $ 9's$ are not allowed in the end of the decimal expansion?   :maybe:


Is there exist a,b,c,d are real of numbers such that
     a$x^{2}$+bdx+c=0 has 2 root u$\neq$v
     b$x^{2}$+cdx+a=0 has 2 root v$\neq$s
     c$x^{2}$+adx+b=0 has 2 root u$\neq$s

[hide="just a idea"]Considering the equations above,by Viete's sums:

$u+v=-\frac{b}{a}.d \implies v=-\frac{b}{a}.d-u$

$v+s=-\frac{b}{a}.d -u+s=-\frac{c}{b}.d \implies s=(\frac{b}{a}-\frac{c}{b})d+u$

$u+s=(\frac{b}{a}-\frac{c}{b})d+2u=-\frac{a}{c}.d \implies u=(\frac{c}{b}-\frac{b}{a}-\frac{a}{c}).\frac{d}{2}$

but: $us=\frac{b}{c}$, $vs=\frac{a}{b}$, $uv=\frac{c}{a}$ consequently:

$u=(\frac{1}{us}-\frac{1}{vs}-\frac{1}{uv})\frac{d}{2} \Longleftrightarrow d=\frac{2u(uvs)}{v-u-s}$ the condition to $d$ be real is that v-u-s ≠ 0 or v≠u+s...this is my idea  [/hide]

By Viette:

$v+u = -d\frac{b}{a}, vu = \frac{c}{a}$

$s+v = -d\frac{c}{b}, sv = \frac{a}{b}$

$u+s = -d\frac{a}{c}, us = \frac{b}{c}$

It gives us:

$v+u = -d\frac{1}{vs}$

$s+v = -d\frac{1}{su}$

$u+s = -d\frac{1}{uv}$

multiply sides:

$v^2 = su , s^2 = uv , u^2 = vs$

and

$v^2 + s^2 + u^2 = su + uv + vs$

$(v-s)^2 + (s-u)^2 + (u-v)^2 = 0$

$u=v=s$ contradiction.


Find all real numbers $x$ such that: \[ x = \sqrt{ x - \frac{1}{x} } + \sqrt{ 1 - \frac{1}{x} }  \]

It give
$x(\sqrt{x-\frac 1x}-\sqrt{1-\frac 1x})=x-1$. Therefore
$\sqrt{x-\frac 1x}=\frac{x+1-\frac 1x }{2}$ Let $y=x-\frac 1x$, We have $\sqrt y =\frac{y+1}{2}$. It give y=1.
If find really roots it give $x=\frac{1+\sqrt 5 }{2}$.

\begin{eqnarray*}x=\sqrt{1-{1\over x}}+\sqrt{x-{1\over x}} &\iff& x-\sqrt{1-{1\over x}}=\sqrt{x-{1\over x}}\\ &\stackrel{\textrm{attention!}}{\implies}& x^2-2x\sqrt{1-{1\over x}}+1-{1\over x}=x-{1\over x}\\ &\iff& x^2-2\sqrt{x^2-x}+1=x\\ &\iff& x^2-x-2\sqrt{x^2-x}+1=0\\ &\iff& (\sqrt{x^2-x}-1)^2=0\\ &\iff& x^2-x=1\\ &\iff& x^2-x-1=0\\ &\iff& x_{1,2}=\frac{1\pm\sqrt{5}}{2}\end{eqnarray*}

Since by the initial equation $x$ must be positive, the only solution is $x=\frac{1+\sqrt{5}}{2}$


what happen when you have a fraction negative 8 over negative 5 does the whole thing then become postive?

yes, because when u divide or multiply with 2 negatives, it is always positive. a fraction is the same as dividing. so in -8/-5, you would cross out the double negative, 8/5, and then u know 8/5 = [hide]1.6 or 1 and 3/5[/hide]
-jorian

[hide]This is the fraction: $\frac{-8}{-5}$. You can re-write that as $\frac{(-1)(8)}{(-1)(5)}$. The $-1$'s cancel out, so your final equation is $\frac{8}{5}$. You can also write that as $1 \frac{3}{5}$.[/hide]

It's like in english class. A double negative is a positive.

[hide="solution"]

We know that

$\frac{ac}{bc}=\frac{a}{b}$

if c is anything but zero. We can simplify -8/-5 to 8/5, which is positive.[/hide]

[quote="13375P34K43V312"]dude in english class double negatives are illegal[/quote]
I understand that. My point is, that saying say " I don't want nothing", means "I want something", because it is like saing that you do not want to be stuck with nothing. It is the same way in math, the two negatives make a positive.

[quote="Ihatepie"]something I learned when I learned about multiplying and dividing negatives and positives is.
bad=negative     good=positive
When a bad(-) thing happens to a good(+) person, it's bad.
When a good(+) thing happens to a good(+)person, it's good.
When a bad(-) thing happens to a bad(-) person, it's good.
When a good(+) thing happens to a bad (+) person, it's bad.

Hope this helps you I am...[/quote]
You messed up on the last one. :wink:

Although two negatives being a positive is obvious, if I ever forget, I remember like this:

If they are the same (+ and + or - and -), then it's positive.
If they are different (+ and -), then it's negative.

There was a nice poem in Kaplan's "The Art of the Infinite" that talked about this. All I remember is the two lines:

...Why - plus - was -
But - times - was +...

Yeah...

One of my old math teachers put it like this:
minus is hate
plus is love

If you hate to hate, then you love.
(minus*minus=plus)

it seems that ihatepie was wrong when he said "when a good thing(+) happens to a bad person(+), its bad. how is a bad person (+)? and if it somehow is (+), plus times plus surely equals plus? (or plus squared, if you really want to go that way...)

ok guys.........i think he gets the point :lol:
-jorian

[quote="mathnerd314"]
...Why - plus - was -
But - times - was +...

Yeah...[/quote]

Wow, try saying that 5 times fast.


Let $n$ -red points and $n$ -blue points $(n\in Z^+)$  be on the same line . Let $a$ be sum of distances of each couple same colour points . Let $b$ be sum of distances of each couple different colour points . Prove that : $a \leq b$

If we denote the red points by $a_1$, $a_2$, ..., $a_n$ and the blue points by $b_1$, $b_2$, ..., $b_n$, then we arrive at the problem from http://www.mathlinks.ro/Forum/viewtopic.php?t=35418 .

  darij


Prove that for all positive reals the following ineq holds

$\sqrt{a^4+c^4} +\sqrt{a^4+d^4}+\sqrt{b^4+c^4}+\sqrt{b^4+d^4}\geq 2\sqrt 2(ad+bc)$

We are dealing with positive integers, so by RMS-AM, we have:
\[ \sqrt{\frac{a^4+c^4}{2}} +\sqrt{\frac{a^4+d^4}{2}}+\sqrt{\frac{b^4+c^4}{2}}+\sqrt{\frac{b^4+d^4}{2}}\geq \frac{a^2+c^2+a^2+d^2+b^2+c^2+b^2+d^2}{2} \]
Then we use the following inequality: \[ (a-d)^2+(b-c)^2 \geq 0 \]
So \[ \sqrt{\frac{a^4+c^4}{2}} +\sqrt{\frac{a^4+d^4}{2}}+\sqrt{\frac{b^4+c^4}{2}}+\sqrt{\frac{b^4+d^4}{2}}\geq a^2+b^2+c^2+d^2 \geq 2(ad+bc) \]
Which was the desired inequality.

Could anyone explain me the RMS-AM?

Thanks!

RMS means root mean square.  It's defined as the square root of the mean of the squares of each element of a set.  So for set $a, b$, $\text{RMS} = \sqrt{\frac{a^2 + b^2}{2}}$.
RMS-AM states $\text{RMS} \ge \text{AM}$ for all sets of positive (or nonnegative?) reals, where AM is the arithmetic mean.  Jan used RMS-AM for four separate sets, and added them to get his first inequality.

[quote]for all sets of positive (or nonnegative?) reals[/quote]

$RMS-AM-GM-HM$ holds for $a_1,a_2,...,a_n \geq 0$

Interestingly, an upper bound can be found [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=390830#p390830]here[/url]. :)


algebra
polynomial
analytic geometry
graphing lines
slope
combinatorial geometry
Polynomials

[quote="Peter"]Show that a polynomial of odd degree $2m+1$ over $\mathbb{Z}$, \[f(x)=c_{2m+1}x^{2m+1}+\cdots+c_{1}x+c_{0},\] is irreducible if there exists a prime $p$ such that \[p \not\vert c_{2m+1}, p \vert c_{m+1}, c_{m+2}, \cdots, c_{2m}, p^{2}\vert c_{0}, c_{1}, \cdots, c_{m}, \; \text{and}\; p^{3}\not\vert c_{0}.\][/quote]

We can find the theorem in vol 48(1897) of Mathematische Annalen. However, I remember that it is not written in English.

Sorry for the following, but I think it's the most natural way:

The polynomial is even irreducible in $\mathbb{Q}_p$, the field of $p$-adic numbers. We denote the $p$-adic (additive) valuation by $v_p$ (restricted to $\mathbb{Z}$, this is how often the prime factor $p$ occurs in the given number).
Now we have the tool of Newton polygons (we already have them in $\mathbb{Q}$, but it's more natural to consider it in $\mathbb{Q}_p$).
Given a polynomial $f(x)=a_n x^n + ... + a_1 x+ a_0$, one considers the set $S$ of points $(k,v_p(a_k))$. For this, one takes the "lower convex hull" (don't know the correct term), that is the smallest set of segments joining $(0, v_p(a_0))$ with $(n, v_p(a_n))$ always $\leq$ the convex hull of $S$ (thus just the lower border).

There are two things to know:
- the valuation $v_p$ can be extended to any algebraic extension in an unique way
- if $(k,v_p(a_k))$ to $(m, v_p(a_m))$ is a segment in the polygon with slope $s$, then there are exactly $m-k$ roots of $f$ having valuation $-s$

Back to the problem: we see that the Newton polygon is the line from $(0,2)$ to $(2m+1 , 0)$, thus all roots have valuation $\frac{2}{2m+1}$. But if $f=gh$, then $v_p(g(0))=\frac{2}{2m+1} \cdot \mathrm{deg}(g)$ since $g$ has only roots of $f$ as roots. But $v_p(g(0))$ has to be an integer, so $\mathrm{deg}(g)=0, 2m+1$, giving $f$ irreducible.

It's probably possible to elementarize this proof (then it would be similar to that of Eisensteins criterium).


When I hit F5 to see the PDF, it appears the following

[img]http://img183.imageshack.us/img183/9575/errorln9.png[/img]

Why?

Make sure that TexnicCenter knows where your pdf reader is. Go to Build, Define Output Profiles, LaTeX => PDF, Viewer. If the path is wrong click on the button with ... to show it where to find it (Adobe Reader for example is called Acrord32.exe).


linear algebra
matrix
linear algebra unsolved

$A,B \in M_n(C)$ invertible matrices $a,b \in C^{*}$ such that
$aA+bB=I_n$ and $A^2+B^2=0_n$

compute $(aA^{-1}bB^{-1})^m$ where $m >0$ integer

Some prelimary comments:

WLOG, both $A$ and $B$ are upper triangular.

$A$ satisfies the equation $(a^2+b^2)A-2aA+I=0$

Hence, the only possible eigenvalues (hence, diagonal elements) for $A$ are $\frac1{a\pm bi}.$

Similarly the only possible diagonal elements for $B$ are $\frac1{b\pm ai}.$

$aA^{-1}bB^{-1}$ is an odd-looking combination. Are you sure that's it - all multiplied together?


Let X be a non-empty and convex subset of a Euclidean metric space (M, d) and assume for some elements x, y, z of M, that d(x,z) + d(z,y) = d(x,y) (i.e., z lies between x and y). May the Hausdorff distance between z and X be greater both than the Hausdorff distance between x and X and the Hausdorff distance between y and X?

If $ x',y'\in X$ are the closest points to $ x$ and $ y$ respectively and $ z\equal{}tx\plus{}(1\minus{}t)y$, then $ z'\equal{}tx'\plus{}(1\minus{}t)y'\in X$ and $ |z'\minus{}z|\le t|x'\minus{}x|\plus{}(1\minus{}t)|y'\minus{}y|$.


The sides AC and BC of a triangle ABC have equal
length. Let P be a point inside triangle ABC such that \PAB =
\PBC and let M be the midpoint of AB. Prove that \APM +
\BPC =180.

draw a circle through point $ {A ;B ;P}$ ;thus $ {CA,CB}$are tangent to the circle ;the following solution go:http://www.mathlinks.ro/viewtopic.php?t=286457;[I think it is not very easy because it is a little difficult for us to construct a circle[not obvious] :maybe: ]


[color=red][b]ΠΡΟΒΛΗΜΑ[/b][/color]

Ο εγγεγραμμένος κύκλος (κέντρου [b]Ι[/b] )τριγώνου [b]ΑΒΓ [/b](με περίκεντρο [b]Ο[/b] )εφάπτεται των πλευρών του τριγώνου  [b]ΑΒΓ [/b]στα σημεία [b]Κ,Λ,Μ[/b].Αν [b]Ρ [/b]είναι το περίκεντρο του τριγώνου  με κορυφές τα μέσα των [b]ΚΛ,ΛΜ,ΜΚ [/b]να αποδείξετε ότι τα σημεία [b]Ο,Ι,Ρ [/b]είναι συνευθεικά.

[b] ΣΧΟΛΙΟ[/b]

 [i]Δεν έχω απόδειξη με συμβατή γεωμετρία για το θέμα αυτό , παρά μόνο με μετασχηματισμούς.[/i]

[quote="silouan"]Η άσκηση βγαίνει άμεσα με τη χρήση αντιστροφής . Αντιστρέφοντας με κέντρο Ι και ακτίνα r , τα Α ,Β,Γ πηγαίνουν στα μέσα των ΚΛ,ΛΜ,ΜΚ  και το Ο γίνεται το περίκεντρο αυτού δηλ το Ο πάει στο P . Άρα τα Ο, Ι ,Ρ συνευθειακά . :D[/quote]

Ακριβώς !
 Αυτή η λύση με αντιστροφή μου άρεσε πολύ , διότι είναι πολύ απλή και σύντομη.Βέβαια το ερώτημα είναι γιατί  οι κορυφές του ΑΒΓ πάνε στα μέσα των ΚΛ,ΛΜ,ΜΚ αλλά αυτό το σημείο είναι επίσης πολύ όμορφο και χρήσιμο για να εξηγηθεί.
 Έχεις ξαναδει ποτέ την άσκηση αυτή με αντιστροφή ή άλλη λύση ; Εγώ όχι. Τη βρήκα στην ισπανική κοινότητα με λύση αντιστροφής και την έβαλα για να παρακινήσω τους μαθητές μας να μελετήσουν λίγο την αντιστροφή και την ομοιοθεσία. Είναι ίσως οι καλύτεροι μετασχηματισμοί σε επίπεδο ολυμπιάδων. Η συμμετριά είναι σχετικά πιο γνωστή , ενώ η στροφή θέλει φαντασία και καλή θεωρία που δυστυχώς και γω δεν την έχω οργανώσει ούτε σε σημειώσεις ούτε στο μυαλό μου.
 ΑΝ βρεις λύση με άλλο τρόπο , να μας τη στείλλεις.
 [u]Μπάμπης[/u]

Απλώς μια παρατήρηση:

Η ευθεία αυτή στην οποία βρίσκονται τα σημεία Ο,Ι,Ρ είναι η ευθεία Euler τόσο του τριγώνου που σχηματίζεται απ΄τα μέσα των ΚΛ,ΛΜ,ΚΜ όσο και του τριγώνου ΚΛΜ :wink:

Πιθανόν αυτό να βοηθάει να λυθεί η άσκηση με έναν άλλο τρόπο αν και δεν κατάφερα να βρω πως...

[quote="stergiu"]
 Βέβαια το ερώτημα είναι γιατί  οι κορυφές του ΑΒΓ πάνε στα μέσα των ΚΛ,ΛΜ,ΜΚ αλλά αυτό το σημείο είναι επίσης πολύ όμορφο και χρήσιμο για να εξηγηθεί.
[/quote]

Αυτό προκύπτει από τον ορισμό των αντίστροφων σημείων .Δηλ το αντίστροφο ενός σημείου Α εκτός κύκλου κέντρου Ο είναι το σημείο τομής της ΑΟ με την πολική του Α  :wink:

Τελικά βρήκα μια λύση που χρησιμοποιεί τις ευθείες Euler, αφού όπως προανέφερα το Ι,Ρ ανήκουν στην ευθεία Euler του ΚΛΜ. Αρα αρκεί να αποδείξουμε ότι και το Ο ανήκει σε αυτή την ευθεία.
Θεωρούμε τώρα το ΑΒΓ ως ορθικό τρίγωνο ενός τριγώνου Α'Β'Γ' (φυσικά είναι πολύ εύκολο να κατασκευαστεί το τελευταίο). Τώρα χρησιμοποιώ ως λήμμα την άσκηση 3 της Βαλκανιάδας 1990 (http://www.artofproblemsolving.com/Forum/viewtopic.php?p=822933#822933 ) η οποία λέει ότι οι ευθείες Euler των Α'Β'Γ' και ΚΛΜ ταυτίζονται. Ομως η ευθεία Euler του ΑΒΓ διέρχεται απ'το Ο αφού το τελευταίο είανι κέντρο του κύκλου Euler του τριγώνου Α'Β'Γ' ως περίκεντρο του ορθικού του τριγώνου
 :)


Hello everyone,

I am trying to make a human powered device that would rise a foot off the ground and will be able to go down stairs and would be able to be used continuously for five minutes with 5 different people using it, although not at the same time?

Any suggestions?

Human powered? It is called a frisbee...geesh...not even physics plz...

something that you actually have to do work to built, and not just buy.


Say you have n dots on a piece of paper. How many lines do you need to connect the n dots?

I can think of a number of different things this question could mean.  (One interpretation, probably not intended, is that the question is asking for the number of edges in a tree on $ n$ vertices.)  Please clarify.

Hint :
--------

It is the same as finding the number of pairs from a set of objects. Because each pair of dots define a line.

For example:

If you have 2 points A, B in a space, then you have AB and BA, or 2*1 = 2 possibilities.
If you have 3 points A, B, C in the space, then you would have:
AB, BA, AC, CA, BC, CB or 3*2 = 6 possibilities.

You may argue that AB and BA are the same, thus giving 1 possibility
Similarly, that AB=BA, BC=CB and AC=CA, thus giving you 3 possibilities.

So, you could say that the number of pairs is 1/2 the number of possibilities.

Your claim is incorrect since you can also have collinear points.  In fact, the way you are doing it will give more than lines required.  You need at most $ n\minus{}1$ lines (a tree), not $ \binom{n}{2}$.  The problem, in my opinion, is to find the minimum number of edges required to create a connected graph from a set of $ n$ vertices.  However, there cannot be a general problem formula in terms of $ n$ that will give you a definite answer since we could be given collinear vertices.  Hence, this problem is too broad and thus requires clarification.


algorithm
number theory proposed
number theory

can anybody give a logical prove of division algorithm without using anything advance?
I have one but I need somemore.

what division algorithm? isn't this sooo basic?


geometry
geometric transformation
Putnam
Support
Princeton
college

sorry  :P  :P  :P  But everyone should know Bulgarian, lol, it's the universal language of communication  :lol:

Guys, I ask you to keep the non-English talk in the local communities forums or in private messages. Remember that more than 95% of the users of this site don't know any Bulgarian word, aside from "da" :D Thanks (to that end I would like you to edit your messages if possible and add the English translation :P).

DPopov wrote that he is 16 years and his father was not at MIT. But some of his father's friends were. Some guy X and probably some other guy called Y were at MIT. One of the guys went to Putnam and was unhappy that he was not within the TOP10 though he did not prepare.

P.S.:.strange logic, why should he be sad if he does not prepare ?

P.P.S.: In general one figures out that you are not really Bulgarian, e.g. you write some words as they are pronouced. But some of them are written completely different. And not using Cyrillic makes it ever harder to understand what you want to say.  ;)

[quote="orl"]P.P.S.: In general one figures out that you are not really Bulgarian, e.g. you write some words as they are pronouced. But some of them are written completely different. And not using Cyrillic makes it ever harder to understand what you want to say.  ;)[/quote]

 :D  :D  :D

I am [i]really[/i] Bulgarian; I just didn't know that this forum supports cyrillic. I wrote phonetically: the way I'd write in an e-mail to a Bulgarian friend (because, more often than not, Cyrillic encryption turns into a non-sensical sequence of characters).

И?тин?ката азбука ...  :) . Орландо, ти знаеш българ?ки?!!

--Ве?елин

[quote="vess"]

И?тин?ката азбука ...  :) . Орландо, ти знаеш българ?ки?!!

--Ве?елин[/quote]

Da, j snaju nekotorowo malenko bolgarija. I russkij jasuk podoben. Osobenno twoi wopros: Tuj snaesch bolgarskii jasuk ?  ;)

PS: Sorry I should refrain from writing Russian responses to Bulgarian questions here. Maybe if all these good Bulgarian IMO guys wouldn't hide we could make a local subforum for our Bulgarian friends but yet I am just aware of two people and one of those knows English better than Bulgarian.  :D

[quote="vess"]I just didn't know that this forum supports cyrillic.[/quote]You can see that this forum supports cyrillic in the Russian section of the site, namely in these posts: http://www.mathlinks.ro/Forum/viewtopic.php?p=144593#p144593 :)

If there were more Bulgarians than 1-2 we could make a Bulgarian local community ;)

?начала о Прин?тоне мы пи?али in English, а ?ейча? по-ру??ки и по-българ?ки  :P
Го?пода! куда?

By the way, can someone teach me to write with cyrillic cause I had a stroke here copy pasting letter by letter from the forum Valentin showed me. Thanks.

Andrei

[quote="ajorza"]?начала о Прин?тоне мы пи?али in English, а ?ейча? по-ру??ки и по-българ?ки  :P
Го?пода! куда?

Andrei[/quote]

[hide="translation"] First we write about Princeton in English and now in Russian and Bulgarian. Mir. (in the sense: My dear)!  Where to ?

I think in order to make other people learning Russian/Bulgarian or just to confuse them. :P

[/hide] Now we are just waiting for Myth saying his word. :lol:

[quote="ajorza"]

By the way, can someone teach me to write with cyrillic cause I had a stroke here copy pasting letter by letter from the forum Valentin showed me. Thanks.

Andrei[/quote]

Another almost equally tedious method is to use an online Cyrillic keyboard and type it. Takes too long. Thus I am writing in Latin characters. Another option might be to switch your keyboard code to Russian. But you need a covering for your keyboard that you see where which letter is unless you prefer try and error.  :)

Or you can go to [url=http://www.online-translator.com/text.asp?lang=en]this[/url] site and use the mouse to click on the individual letters that you want using the online keyboard, which is probably faster than figuring out the location of each character on the keyboard.

[quote]I think in order to make other people learning Russian/Bulgarian or just to confuse them. :P[/quote]
I personally find it quite a miracle that I can understand about half of the Russian/Bulgarian words in this thread. :lol:

Or much better and faster than anything: go to Start Menu (bottom-left on any windows environment), go to Run, write "charmap" and press enter. From thereon it's obvious what needs to be done ;)


[quote="bboypa"]Let $ a,b,c$ positive reals such that $ ab \plus{} bc \plus{} ca \equal{} 3$, show that:
$ \displaystyle a^2 \plus{} b^2 \plus{} c^2 \plus{} 3 \ge \frac {a(3 \plus{} bc)^2}{(c \plus{} b)(b^2 \plus{} 3)} \plus{} \frac {b(3 \plus{} ca)^2}{(a \plus{} c)(c^2 \plus{} 3)} \plus{} \frac {c(3 \plus{} ab)^2}{(b \plus{} a)(a^2 \plus{} 3)}$
([i]Anass BenTaleb, Ali Ben Bari High School - Taza,Morocco[/i])[/quote]

My proof's below.

[quote="bboypa"]Let $ a,b,c$ positive reals such that $ ab \plus{} bc \plus{} ca \equal{} 3$, show that:
$ \displaystyle a^2 \plus{} b^2 \plus{} c^2 \plus{} 3 \ge \frac {a(3 \plus{} bc)^2}{(c \plus{} b)(b^2 \plus{} 3)} \plus{} \frac {b(3 \plus{} ca)^2}{(a \plus{} c)(c^2 \plus{} 3)} \plus{} \frac {c(3 \plus{} ab)^2}{(b \plus{} a)(a^2 \plus{} 3)}$
([i]Anass BenTaleb, Ali Ben Bari High School - Taza,Morocco[/i])[/quote]
Second proof:
We have
\[ \frac{a(bc\plus{}3)^2}{(b\plus{}c)(b^2\plus{}3)}\equal{}\frac{a(2bc\plus{}a(b\plus{}c))^2}{(a\plus{}b)(b\plus{}c)^2} \equal{}\frac{a[4bc(ab\plus{}bc\plus{}ca)\plus{}a^2(b\plus{}c)^2]}{(a\plus{}b)(b\plus{}c)^2} \le \frac{a[(b\plus{}c)^2(ab\plus{}bc\plus{}ca)\plus{}a^2(b\plus{}c)^2]}{(a\plus{}b)(b\plus{}c)^2} \equal{}a(a\plus{}c)\]
Using this for the similar terms and adding them up, we can get the result. :)

@Honey_S, you uses only am-gm, perfect  :wink:
@can_hang2007, very nice too  :lol:

Let $ x\equal{}\frac{1}{\sqrt{3}}a,y\equal{}\frac{1}{\sqrt{3}}b,z\equal{}\frac{1}{\sqrt{3}}c$   then $ xy\plus{}yz\plus{}zx\equal{}1$ and $ x\equal{}\frac{1\minus{}zy}{z\plus{}y}$  (1)  and we must prove $ \sum x^2\plus{}1\ge \sum \frac{(1\plus{}yz)^2(1\minus{}yz)}{(1\minus{}yz)^2\plus{}(y\plus{}z)^2}$
  (we only use (1) ) we will prove
$ \frac{(1\plus{}yz)^2}{(1\minus{}yz)^2\plus{}(y\plus{}z)^2}\le 1$
$ 4zy\le (z\plus{}y)^2$ and its obviously true by AM-GM so
$ \sum \frac{(1\plus{}yz)^2(1\minus{}yz)}{(1\minus{}yz)^2\plus{}(y\plus{}z)^2}\le \sum (1\minus{}yz)\equal{}1\plus{}xy\plus{}yz\plus{}zx\le \sum x^2\plus{}1$
(last true by CS) so we are done. Equality when x=y=z or a=b=c=1

Sorry for not participating to the contest. I just opened my mailbox after 10 days and I just realized that I lost the final round :(

My solution to the 1st problem:

By Cauchy-Swartz we have:

$ LHS \equal{} \sum_{cyclic}{\frac {a(\sqrt {3}\sqrt {3} \plus{} bc)^2}{(c \plus{} b)(b^2 \plus{} 3)}} \leq \sum_{cyclic}{\frac {a(b^2 \plus{} 3)(c^2 \plus{} 3)}{(c \plus{} b)(b^2 \plus{} 3)}} \equal{} \sum_{cyclic}{\frac {a(c^2 \plus{} 3)}{(c \plus{} b)}} \equal{} \sum_{cyclic}{\frac {a(c^2 \plus{} ab \plus{} bc \plus{} ca)}{(c \plus{} b)}} \equal{} \sum_{cyclic}{\frac {a(c \plus{} b)(c \plus{} a)}{(c \plus{} b)}} \equal{} \sum_{cyclic}{a(c \plus{} a)} \equal{} a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca \equal{} a^2 \plus{} b^2 \plus{} c^2 \plus{} 3$


Prove that if $n$ has at least two different prime divisors, then there is a permutation $K$ of $\{1,2,3,...,n\}$ such that
$K_1\cos\frac{2\pi}{n}+K_2\cos\frac{4\pi}{n}+...+K_n\cos{2\pi} = 0$

Actually, if $n$ is not a prime number, then there exists a permutation $(a_0,a_1,\ldots ,a_{n-1})$ of the numbers $(1,2,\ldots ,n)$ such that
\[a_0+a_1 \epsilon +a_2 \epsilon^2 +\ldots +a_{n-1} \epsilon^{n-1}=0,\]
where
\[\epsilon =\cos {\frac {2\pi}{n}}+i\sin {\frac{2\pi}{n}}.\]

This appears in the geometric form in Prasolov's book:
prove that if $n$ is not a prime number, then there exists an equiangular polygon whose side lengths are $1,2,\ldots ,n$ in some order.

The case $n=6$ is an old well known problem:
the numbers $a_1,\ldots ,a_6$ are the side lengths of an equiangular hexagon iff
\[a_1 -a_4=a_5 -a_2 =a_3 -a_6.\]


You'd pretty much have to do intelligent listing for any triangle.  Start with the largest side, then see how many options you have for 1 as the smallest side, 2 as the smallest side, etc.  Then decrease the size of the largest side by 1 and repeat.

AS JBL pointed out we would do the intelligent listing thing, personally, i would start from the smallest side( let this be n), and see how many digits are at most n-1 apart, then repaeting this for the smallest side being n+1.


calculus
integration
real analysis
real analysis unsolved

Just verify that $ \left< f, g\right> : \equal{}\int_{0}^{1}fg\; dx$ satisfies all the property of inner product. Then it becomes a special case of Cauchy-Schwarz inequality.


[hide]
The probability of rolling a 7 or 11 on your first turn is $ 6/36\plus{}2/36\equal{}2/9$.

The probability of rolling a 4 or 10 is $ 3/36 \equal{} 1/12$.
The prob of rolling a 5 or 9 is $ 4/36 \equal{} 1/9$.
The prob of rolling a 6 or 8 is $ 5/36$.

The prob of rolling any of 4,5,6,8,9, or 10, and then rolling it again before a 7 is:
$ 2\cdot\frac{1}{12}\left(\frac{1}{12} \plus{} \frac{5}{6}\cdot\frac{1}{12} \plus{}...\right) \plus{} 2 \cdot\frac{1}{9}\left(\frac{1}{9} \plus{} \frac{5}{6}\cdot\frac{1}{9} \plus{}...\right) \plus{} 2\cdot\frac{5}{36}\left(\frac{5}{36} \plus{} \frac{5}{6}\cdot\frac{5}{36} \plus{}...\right)$
$ \equal{}2\left(\left(\frac{1}{12}\right)^2 \plus{} \left(\frac{1}{9}\right)^2 \plus{} \left(\frac{5}{36}\right)^2\right)\left(1\plus{}\frac{5}{6}\plus{}...\right)$
$ \equal{}\left(\frac{25}{324}\right)(6)$
$ \equal{}\frac{25}{54}$

Adding these two probabilities, we get $ \boxed{\frac{37}{54}}$[/hide]

[quote="mathcrazed"]In a game of shooting craps, what is the probability that you win given the following conditions:
1.) You roll two dice and find the sum.
2.) If, on the first turn, the sum is $ 7$ or $ 11$, you win.
3.) If you do not roll a $ 7$ or $ 11$ on the first turn, then you can still win if you roll one of the following numbers on the first turn and then roll that same number again before you roll a $ 7$,
4.) The numbers that were mentioned in the above rule are $ 4,5,6,8,9,$ or $ 10$.
Have fun XD[/quote]
Dude, that was the gambling in StickWorld.
 :rotfl:
I don't even knew why I played that game.  :|

Incorrect... at least that's what the answer page says XD.

i took a random guess [hide]$ \frac {457}{1944}$[/hide]

btw: nice problem what is the source

It's from the 10th edition of a book by Swokowski and Cole. Pre-calc book.

I found a small mistake...
so is the answer:
[hide]244/495?[/hide]

[hide]The probability of rolling a 7 or 11 on your first turn is 6/36+2/36=2/9.

The probability of rolling a 4 or 10 is 3/36 = 1/12.
The prob of rolling a 5 or 9 is 4/36 = 1/9.
The prob of rolling a 6 or 8 is 5/36.

The prob of rolling any of 4,5,6,8,9, or 10, and then rolling it again before a 7 is:
$ 2\cdot\frac{1}{12}\left(\frac{1}{12} \plus{} \frac{3}{4}\cdot\frac{1}{12} \plus{}...\right) \plus{} 2 \cdot\frac{1}{9}\left(\frac{1}{9} \plus{} \frac{13}{18}\cdot\frac{1}{9} \plus{}...\right) \plus{} 2\cdot\frac{5}{36}\left(\frac{5}{36} \plus{} \frac{25}{36}\cdot\frac{5}{36} \plus{}...\right)$
$ \equal{}\frac{1}{72}\left(1\plus{}\frac{3}{4}\plus{}\left(\frac{3}{4}\right)^2\plus{}...\right) \plus{} \frac{2}{81}\left(1\plus{}\frac{13}{18}\plus{}...\right) \plus{} \frac{50}{36^2}\left(1\plus{}\frac{25}{36}\plus{}...\right)$
$ \equal{}\frac{1}{18}\plus{}\frac{4}{45}\plus{}\frac{25}{198}$
$ \equal{}\frac{134}{495}$

Adding these two probabilities, we get $ \boxed{\frac{244}{495}}$[/hide]


Are there anymore jv or middle school tournaments except for gctm this year?

Walton is hosting our MathFest. There will be a JV test, Varsity test, and a Super-Varsity test. Current ARML team members are STRONGLY SUGGESTED (required) to take the Super-Varsity test! :-)

It will be on April 21st and if you will remind me later in the week, I'll email the info to your coach. The Cobb County email will be down until Thursday.

Fulton

Thanks. I'm not sure if we can go since we're already registered for gctm but I'll remind you anyway. Jen

I knew that. I guess I need sleep more than I thought!  :D

Lol. I'm crashing tonight myself. I won't wake up til like 3 tomorrow. I didn't get to last night.

Is it something that a school can participate in? Like a team round? Or is it all individual?

mathfest or gctm?

mathfest: I have no clue.

gctm middle: I'm pretty sure you can only participate here if you're from a school. You go in as a team of 4, and there's 2 rounds done individually, 1 round done as a pair, and 1 round that' done as a team. The last round is for fun, but it's done sort of like a relay.

If you can take a team, go to GCTM. If not, come to MathFest and enjoy.

MathFest will have a JV, Varsity, and Super-varsity test. The only real "team" competition will be an academic bowl style tournament.

Fulton

I'd like to go to MathFest. My teacher mentioned it back in September, but he hasn't said anything about it since then. I'll have to talk to him tomorrow.

I'm probably coming to MathFest because GCTM is too far.

How hard is the jv test for the mathfest? Jen

Good question, I haven't seen it recently. I would think that you should be able to do most of it. JV is going to be defined as Algebra II and below.

Can you send me a copy after the competition?

[quote="gtf"]Be happy to. Perhaps I'll just post all of the tests here. Remind me.[/quote]

Thanks Mr. Fulton.

[quote="Arvind_sn"]Academic style tournament for team? Sounds fun  :) [/quote]

Yeah, that does sound pretty cool.

Isn't Mr. Rusczyk going to be at MathFest? Thats wut I heard. :huh:

What is the size constraints on a team for the academic bowl style tournament?

Does that mean exactly 4 or no more than 4.

Regardless, this sounds like fun.

[quote="gtf"]1. Mr. Rusczyk will not be at the MathFest. I was trying to schedule a talk while he was here for NCTM, but just ran out of time with other stuff going on.

2. The team constraints for the academic bowl style tournament are teams of 4.

Take care,
Fulton[/quote]
dude
i mean mr fulton
how do you join this tournament thing?
thanks
funcia

Anyone that is interested and doesn't already have the info, give me your email address and I'll email you the info. It's on April 21st at Walton High School in Marietta.

Fulton

Can we bring a calculator? If so, which round(s) can we use it on? What are the constraints (i.e. can I bring a TI-89 Titanium)?

JV State: You may use a calculator on the pairs ciphering and the power question. TI-86 and below

Walton MathFest: No (you could use it on the scavenger hunt I guess.)

Fulton

[quote="funcia"][quote="gtf"][quote="funcia"]Man I just realised that swimming goes to Chamblee on Saturdays now.  :( I can't go.[/quote]

Funcia, you a Dynamo? I was a SwimAtlanta man myself. I guess we can't talk anymore! :-)[/quote]

No I was at SWAT when I started and am thinking about possibly switching back...[/quote]

Go SwimAtlanta! I used to go there... :D

I was actually one of the coaches when there were only 2 pools. I coached the really young kids at the Roswell pool. (Jeez, that's been a long time ago! :-) )

Yay! I got :winner_third: in Varsity! I would have tied for second place, but for some very strange reason, I marked my answer to number three in the number four spot (if that had happened, how are ties broken?) Thanks for hosting this MathFest, Mr. Fulton. It was very fun.

Eh, fourth in super varsity... could've gotten 3rd if I hadn't answered a question which I thought I got right but got wrong. That's second to the record of "could've beaten Miles" but whatever... would've could've should've, etc. doesn't count  :(

Just from what little I've heard, the Dickerson kids had a REALLY good time at Middle School state! And the Dodgen kids did well also!

My understanding of the Dickerson results:
1st - Claus Zheng
2nd - Alex Li
4th - Gil Goldshlager (6th grader!)
7th - Minh Le

And of course Dickerson won.

The Dodgen kids came in 5th (always good to bring home a trophy!) and they were pretty much all in the top 20.

Congratulations folks, now I get to sleep for a week or two. (Oh wait, isn't ARML getting ready to start up? :-))

Jenn, did Palmer end up going?

[quote="gtf"]Just from what little I've heard, the Dickerson kids had a REALLY good time at Middle School state! And the Dodgen kids did well also!

My understanding of the Dickerson results:
1st - Claus Zheng
2nd - Alex Li
4th - Gil Goldshlager (6th grader!)
7th - Minh Le

And of course Dickerson won.

The Dodgen kids came in 5th (always good to bring home a trophy!) and they we pretty much all in the top 20.

Congratulations folks, now I get to sleep for a week or two. (Oh wait, isn't ARML getting ready to start up? :-))

Jenn, did Palmer end up going?[/quote]

I wanted to go but i couldnt...i bet it would have been fun. wutever walton was cool and i got fifth JV.

At some point you just have to step up and quit doing the easy ones! Of course, I suspect that you, Claus (7th grade, was the one sitting next to you at ZooAtlanta), and Gil will see each other a fair amount next year! :-)

yes i saw klaus at JV state too, didn't he get an award?

He did. He was the top "non-GHSA" student. I think that his overall ranking was about 22nd or so. So you got him that time. He would actually have helped the Walton JV, but I wanted to bring a full Dickerson team. It's a really talented group.

GHSA is the Georgia High School Association. It regulates virtually all of the high school athletics in the state. The only schools that aren't GHSA members are usually small private schools, home schools, and of course, middle schools. Claus competed for Dickerson at JV State and not Walton.

how about online schools? do they join? JW

Hmmm, what do you mean by an online school?

I mean like ones that are sort of international/national; I'm guessing they are "non-GHSA" but
say they had a few students who reside in GA. I mean like the EPGY OHS and I think the Indiana one is similar to that but not all of the courses are online.
Of course I have no clue if I will ever fall into that category as I await my rejection letter. Maybe next year...along with doing local math contests.

It's a really good question. The immediate answer is that, yes, they would be non-GHSA. I don't know that we've ever really had someone from a school like that.

How many people are allowed per team in GCTM Middle School Math Tournament?

8 can go. top 4 (written + ciphering) count for the team score.


What do you think is the best way to prepare for ARML? I know it is similar to AIME in difficulty, does that mean I should prepare for ARML the same way I do for AIME?

I'd say that ARML is easier than AIME problems.  Excepting #10, of course, which this year was pretty comparable to hard AIME level.

In any case, the best way to prepare for the individual part is to do lots and lots of past year's ARML problems and familiarize yourself with them. It's just like any other contest: knowing some of the types of problems that frequently are asked is a huge advantage.

To prepare for the team part, well, you need a team  :wink:

I believe that starting last year, there was a new head question writer for the ARML problems. The problems will be slightly different than how they were before...

Just because you don't have a team doesn't mean you can't do it by yourself  :)

Yes, the head problem writer was new this year.  I didn't feel much of a difference from these problems to those of previous years, though.  They had similar flavor. (I mean, sure there were more of them but still :P )

[quote="Elefont"]Number 10 on ARML this year was actually VERY close to a number 7ish on the AIME one year ago. it just got slightly messier. I'll admit, though, that in relation to their difficulties, the time you get on ARML is wayyyy shorter.[/quote]

It can be compared to a MC target round.

10 Problems of course now

On MC target rounds the problem difficulty/time ratio is wayyyy less than that of ARML.  Of course the first two rounds of the individual are kind of comparable to targets I guess.  Then 5-8 usually have an intuitive trick that trivializes a problem, so that's slightly different. Then #9 is trivial, and #10 is impossible.  But I guess you're right that the closest thing to ARML format (if not difficulty) are targets.


Hey, Guys!

I'm working on my own with the book "Problem Solving Startegies" by A.Engel and need some help with one example problem, whose solution I do not quite understand.

Problem:
A chessmaster has 77 days to prepare for a tournament. He wants to play at least one game per day, but not more than 132 games. Prove that there is a sequence of successive days on which he plays exactly 21 games.

Solution:
Let [tex]a_i[/tex] be the number of games played until the [tex]i[/tex]th day inclusive.
Then [tex]1\le{a_1}<...<a_{77}\le{132}\Rightarrow22\le{a_1+21}<a_2+21<...<a_{77}+21\le{153}[/tex]
Among the 154 numbers [tex]a_1,...,a_{77},a_1+21,...,a_{77}+21[/tex] there are two equal numbers. Hence there are indices i, j, so that [tex]a_i=a_j+21[/tex]. The chessmaster has played exactly 21 games on the days # [tex]j+1,j+2,... ,i[/tex].

The questions that simultaneously arose in me are the following:
Where are the boxes and where are the pearls? Is it right that the boxes are the games and the pearls are the days? If so, why can't there be an injective mapping from the set of days to the (sub)set of games, i.e. can't we really assign exactly one day to exactly one number of games?
Why is it necessary to generate a second sequence, which is then added to the first in terms of the number of elements?

I have even more questions, because sadly this is the first time I come across such a problem...I can't imagine the whole picture vividly in my head...nobody taught me to solve non-standard problems of this kind :( ...and it's now that I wish to do it myself! :mad:
...but help me clarify the situation by answering these first, please. :?

Thank you in advance!  :D

I do not what 's your problem with solution.It is obvious.
$a_1, a_2, ...a_77,a_1+21,a_2+21,..,a_77+21$ are $154$ number between $1,153$.
so two of these numbers are same(pigeonhole principle).And we know that he play at least one hour every day so there is not any $i,j$ such $a_i=a_j$ so there must be $i,j$ such that $a_i+21=a_j$ and it means that $a_i + a_{i+1} +a_{i+2}+...a_{j} =21 $
if you have problem with it tell me ;)
I am reading "Problem Solving strategies" too, so we can work together ;)

Oh, Jesus!!! I got it at last!
The problem is really simple, if one (for example me :blush: ) doesn't make silly mistakes which eventually make it extremely difficult!  :D

Thank you especially for the sentence with the sum.
My failure was that I thought, that the chessmaster plays exactly 21 games for a couple of days (21 games every day for a couple of days...this can't even be!), [b]but it is meant that he plays IN SUM 21 games for a period of several days, doesn't he![/b]
Assuming the previous situation I ran directly into difficulties!

This proves the following to be true:
First read -> then think ->[b] then read again[/b] -> [b]then think again[/b] ->then calculate... :D

Hi !

Can you tell me where I can find the book of Engel ? Thanks a lot !


Find all $x\in N$ satisfying some equation in form

$\dfrac{1}{1+\dfrac{1}{1+...+\dfrac{1}{x}}} = x$

by substituting x in the left hand of the equation by the whole left hand of the equation i get $x=\dfrac{1}{1+\dfrac{1}{...}}$ so by substituting again $\dfrac{1}{1+x}=x$ and $1/2\,\sqrt{5}-1/2=x$ because x is positive.


probability
real analysis
real analysis unsolved

Can anyone help me out with this?

Let $X_{1},X_{2},\ldots$ be independent with

$P(X_{k}= k^{2}) = \frac{1}{k^{2}}$
$P(X_{k}=-1) = 1-\frac{1}{k^{2}}$

Prove that $\sum_{k=1}^{n}X_{k}\to-\infty\; a.s$

You do not even need independence of $X_{k}$ here. Just consider the event that all $X_{k}$ starting from $n+1$-st are $-1$. The probability of the opposite event does not exceed $\sum_{k=n+1}^\infty \frac 1{k^{2}}\le \frac{1}{n}$. Thus, with probability $1$ the series $\sum_{k}X_{k}$ has a tail consisting of all $-1$'s. The rest should be clear.


Find all the functions $ f: \mathbb{R} \mapsto \mathbb{R}$ such that
\[ f(x f(y)) \plus{} y \plus{} f(x) \equal{} f(x \plus{} f(y)) \plus{} y f(x)
\]
for all $ x,y \in \mathbb{R}$.

[quote="reza1370"]Find all functions f  such that for each x, y :
f(xf(y)) + y + f(x) = f(x + f(y)) + yf(x)[/img][/quote][/code][/quote]
Step 1 : $ f(0)\equal{}0$ ,it is quite easy to prove that .
Step 2 : Take $ x\equal{}0$ then $ f(f(x))\equal{}x$
Take $ y\to f(y)$ then $ f(xy)\plus{}f(y)\plus{}f(x)\equal{}f(x)f(y)\plus{}f(x\plus{}y)$
Step 3 :$ f(1)\equal{}1$
Proof Take $ (x,y)\equal{}(1,1) ,(\minus{}1,1),(\minus{}1,2),(1,\minus{}2),(1,\minus{}2)$ and use $ f(f(y))\equal{}y$ then $ f(1)\equal{}1$
Take $ y\equal{}1$ then
$ f(x\plus{}1)\equal{}f(x)\plus{}1$
Take $ y\to y\plus{}1$ then
$ f(xy\plus{}x)\plus{}f(y)\plus{}f(x)\plus{}1\equal{}f(x)f(y)\plus{}f(x)\plus{}f(x\plus{}y)\plus{}1$
$ \Rightarrow f(xy\plus{}x)\plus{}f(y)\equal{}f(x)f(y)\plus{}f(x\plus{}y)\equal{}f(xy)\plus{}f(x)\plus{}f(y)$
$ \Rightarrow f(xy\plus{}x)\equal{}f(xy)\plus{}f(x)$
So $ f(x\plus{}y)\equal{}f(x)\plus{}f(y),\forall (x,y)\in R^{2}$
From condition we have $ f(x)f(y)\equal{}f(xy)$
These give that $ f(x)\equiv x,\forall x\in R$


Find all polynomial with real coefficients $ P(x)$ sach that
$ P(x\plus{}P(x)\equal{} P(x)\plus{}P(P(x))$ for  all x

first notice $ P(0) = 0$ now if we just consider both parts of the expression es polyinomials.
$ P(x) = a_nx^n + ... + a_1x + a_0$. obviouslly $ a_0 = 0$

$ a_n(x + P(x))^n + a_{n - 1}(x + P(x))^{n - 1} + ... + a_1(x + P(x)) = a_nx^n + ... + a_1x + a_n(P(x))^n + ....a_1P(x)$

$ a_n( {n \choose 1}x^{n - 1}P(x) + ... + {n \choose n - 1}x(P(x) \right)^{n - 1}) + ... + 2a_2xP(x) = 0$

the left side is basically a polynomial of degree $ n(n - 1) + 1$ which is imposible to be zero for every $ x$ unless $ n\leq 1$

this gives us the solution $ P(x) = kx$ for any real $ k$ including $ k = 0$


calculus
integration
trigonometry
logarithms
calculus computations

yo there, fine ppl.  :D

show

$ \int_0^{\frac{\pi}{2}}\;\frac{a\,x\;\sin\,2x}{a^2\minus{}2\,a\,\cos\,2x\plus{}1}\;\;\textbf dx\;\;\equal{}\;\;\boxed{\frac{\pi}{4}\;\ln\left(1\plus{}\frac{1}{a}\right)}$

where $ |a|>1$

do [b]not[/b] use techniques from complex analysis.

put r=1/a

$ \int_0^{\frac {\pi}{2}}\;\frac {a\,x\;\sin\,2x}{a^2 \minus{} 2\,a\,\cos\,2x \plus{} 1}\;\;\textbf dx\;\; \equal{} \;\; \int_0^{\frac {\pi}{2}} x \sum_{n \equal{} 1}^{ \plus{} \infty} r^{n} \sin (2n x) \; \textbf d x \equal{}$

$ \equal{} \sum_{n \equal{} 1}^{ \plus{} \infty} \int_0^{\frac {\pi}{2}} x \;  r^{n} \; \sin (2n x) \; \textbf d x \equal{} \sum_{n \equal{} 1}^{ \plus{} \infty} r^{n} \cdot \frac {\sin (\pi n) \minus{} n \pi \cos (\pi n)}{4n^2} \equal{}$

$ \equal{} \sum_{n \equal{} 1}^{ \plus{} \infty} \frac { r^{n} \cdot ( \minus{} 1)^{n \plus{} 1}}{4n} \pi  \equal{} \frac {\pi}{4} \ln \left(1 \plus{} r \right) \equal{} \boxed{\frac {\pi}{4} \ln \left(1 \plus{} \frac {1}{a}\right)}$

$ \boxed{\int_0^{\pi} \ln (1\minus{}2a\cos x\plus{}a^2)\ dx\equal{}2\pi \ln a}$

Integral by parts leads you to the result. :D


geometry
3D geometry
number theory proposed
number theory

Here is the exercise:
Show that there exist a set of 1992 positive integers such that each (non empty) subset has a sum which is a square, cube or higher power.

Try please the open section problem. I posted a problem '"After a famous olympiad problem" and Pierre answered there to your problem, as far as I remember. Please, also think about the open problem, maybe you come out with a solution. It would be great, since that problem hurt me.

I'm sorry, I am getting old, Pierre proved something else. Anyway, one possible idea for this problem is to prove that for any n there exists an a such that $ a,2a,...,na$ are all perfect powers of exponents at least 2. This is a consequence of the chinese remainder theorem (and I think it was already discussed; anyway, it is easy). Sorry again.


modular arithmetic
AMC
AIME
complex numbers

Given a positive integer $n$, it can be shown that every complex number of the form $r+si$, where $r$ and $s$ are integers, can be uniquely expressed in the base $-n+i$ using the integers $1,2,\ldots,n^2$ as digits. That is, the equation\[ r+si=a_m(-n+i)^m+a_{m-1}(-n+i)^{m-1}+\cdots +a_1(-n+i)+a_0  \]is true for a unique choice of non-negative integer $m$ and digits $a_0,a_1,\ldots,a_m$ chosen from the set $\{0,1,2,\ldots,n^2\}$, with $a_m\ne 0$. We write \[ r+si=(a_ma_{m-1}\ldots a_1a_0)_{-n+i}  \]to denote the base $-n+i$ expansion of $r+si$. There are only finitely many integers $k+0i$ that have four-digit expansions \[ k=(a_3a_2a_1a_0)_{-3+i}~~~~a_3\ne 0.  \]Find the sum of all such $k$.

Agh, that took so long to type.  :ewpu:

I know chess64'll like it though, because it's a complex number problem. :yup:

This problem is too much for me. :stretcher: Can somebody please explain?

[quote="4everwise"]I know chess64'll like it though, because it's a complex number problem. :yup:[/quote]

Hey, how did you know I like complex numbers? Well, I do. I'll try this problem now...

EDIT:

[hide="Solution...surprising difficulty for 14"]
We need to find integers $k$ such that \[ k=a_3(-3+i)^3+a_2(-3+i)^2+a_1(-3+i)+a_0.  \] The first thing we notice is that $\text{Im}(k)=0$. Let the coefficient of $a_i$ be $z_i$, so that $z_3=(-3+i)^3=-18+26i$, $z_2=(-3+i)^2=8-6i$, $z_1=-3+i$, and of course $z_0=0$. Hence,

\begin{eqnarray*} 0&=&\text {Im}(k)\\ 0&=&a_3\text {Im}(z_3)+a_2\text {Im}(z_2)+a_1\text {Im}(z_1)+a_0\text {Im}(z_0)\\ 0&=&a_3(26)+a_2(-6)+a_1\\ 6a_2&=&a_1+26a_3 \end{eqnarray*}

Consider this equation modulo $6$: \[ 0\equiv a_1+2a_3 \pmod {6} \Longleftrightarrow a_1 \equiv 4a_3 \pmod {6}.  \] Solving for residues modulo 6 gives $(a_1,a_3)\equiv (0,0),(2,2),(2,5),(4,1),(4,4)\pmod {6}$. Since $a_1,a_2,a_3\in \{0,1,2,\ldots,9\}$, we find that \[ (a_1,a_3)=(0,0),(0,6),(2,2),(2,5),(2,8),(4,1),(4,4),(4,7).  \] Solving each case for the value of $a_2$ and (eliminating cases where $a_2>9$ or $a_3=0$) gives us \[ (a_1,a_2,a_3)=(2,9,2),(4,5,1).  \] Finally, the possible values of $k$ in both cases are $30$ or $10$, but since $a_0$ ranges from $0$ to $9$ in each case, the sum of all possible values is \[ [(30+0)+(30+1)+\cdots+(30+9)]+[(10+0)+(10+1)+\cdots+(10+9)]=5(69)+5(29)=\boxed {490}.  \]
[/hide]

[quote="white_horse_king88"]Not surprisingly, once you ride out the wave of terminology and definitions they throw at you, the question they are actually asking is simple and to the point. Often that's the case with AIME problems.[/quote]

Agreed.  I think this is one of the easiest #14's though since not being frightened away is about half the problem :P  Seriously though, they give you a bunch of the information that you need in the problem, once you put that together it becomes more like a #6 or #7, if that.

Yes, I agree it was pretty easy; however stupid me forgot that $a_3$ cannot be 0 so I included that case and got 445.

[quote="hrithikguy"]Ok, so I got the equation 26a-6b + c = 0 and found that the two solutions for (a,b,c) are

(2,9,2) and (1,5,4) which I am sure is pretty much the entire problem. However, of course, I got the wrong answer of 130 after that. Here is what I did.

Case 1: (2,9,2).
2(-18) + 9(8) + 2(-3) = 30.
$a_0$ ranges from 0 to 9, so the sum is 45. Thus this case is 75.

Case 2: (1,5,4)
1(-18) + 5(8) + 4(-3) = 10.
Also, for the same reason, you must add 45 here to get 55.

Thus, the answer should be 55 + 75 = 130.

But this is wrong. What went wrong??[/quote]

The two possibilities for $(a_3,a_2,a_1)$ are $(2,9,2)$ and $(1,5,4)$

Plugging the first one in, we get
$2(-18) + 9(8) + 2(-3) +a_0= 30+a_0$
Since $a_0\in{0,1,2,3,4,5,6,7,8,9}$, the sum is
$(30+0) + (30+1) ... (30+9) = 345$
What you did here was $30 + (1+2+3+...8+9)$. You forgot to add $30$ each time

Plugging in the second pair, we get
$1(-18) + 5(8) + 4(-3) +a_0=10$
Since $a_0\in{0,1,2,3,4,5,6,7,8,9}$, the sum is
$(10+0) + (10+1) ... (10+9) = 145$

So the final answer is $145+345 = \boxed{490}$


algebra
polynomial
induction
IMC
college contests

Let $f_1(x)=x^2-1$, and for each positive integer $n \geq 2$ define $f_n(x) = f_{n-1}(f_1(x))$. How many distinct real roots does the polynomial $f_{2004}$ have?

First study small cases. For $n = 1, 2, 3$ there are $2, 3, 4$ distinct real solutions, respectively. This suggests that $f_n(x)$ has $n+1$ distinct real roots. Let's try induction. The basis is above. Suppose $f_k(x)$ has $k+1$ distinct real roots for every positive integer $k$ less than $n+1$. We shall prove that $f_{n+1}(x)$ has $n+2$ distinct real roots.

But
\[f_{n+1}(x) = 0
\iff (f_n(x))^2 - 1 = 0
\iff f_n(x) = 1\ \text{or}\ f_n(x) = -1
\iff (f_{n-1}(x))^2 = 2\ (i)\
     \text{or}\ f_{n-1}(x) = 0\ (ii).\]

The equations $(i)$ and $(ii)$ have no common roots (note that $\gcd(f(x); f(x)^2  - 2) = 1$). By induction hipothesis, $(ii)$ has $n$ distinct real roots.

So we must count the number of real solutionsof $(i)$.
\[(i) \iff f_{n-1}(x) = \sqrt 2\ \text{or}\ f_{n-1}(x) = -\sqrt2\]

But $f_k(x) = (f_{k-1})^2 - 1 \geq -1$ for all real $x$ and positive integer $k$, thus
\[(i) \iff f_{n-1}(x) = \sqrt2\]

It's not hard to prove that $f_k(x) = a$, $a > 1$, $k$ positive integer, has exactly two roots (proceed by induction). Indeed, $f_k(x) = a\iff f_{k-1}(x) = \sqrt{a+1}$ (remember that $f_{k-1}(x) \geq -1$).

Then we are done, because $(i)$ has two distinct real roots and $(ii)$ has other $n$ distinct real roots, thus $f_{n+1}(x) = 0$ has $n+2$ distinct real roots.

In particular, $f_{2004}(x) = 0$ has $2005$ distinct real roots.

It's a simple problem, requiring one simple observation, which you get from observing small cases:

-first $f_n$ is always an even function
-if $n$ is even, $0$ is a root, the remaining $n$ roots are symmetric across the origin and satisfy $|r| >1$
-if $n$ is odd, $0$ is not a root, but $-1$ and $+1$ are roots, and the remaining $n-1$ roots are split symmetrically across the origin and satisfy $|r|>1$

Prove the above by induction, and you're done!


algebra
polynomial
Vieta
combinatorics unsolved
combinatorics

Find all integers $ (x,y)$ where the number $ \frac{x^2\plus{}y^2}{xy\minus{}1}$ is also an integer.

$ \frac{x^2\plus{}y^2}{xy\minus{}1}\equal{}5$ - there is not truth

For example,  $ x$, $ y\equal{}\minus{}x^3$

$ \frac{x^2\plus{}y^2}{xy\minus{}1}\equal{}\frac{x^2\plus{}x^6}{\minus{}x^4\minus{}1}\equal{}\minus{}x^2\ne5$

[quote="aktyw19"]$ \frac {x^2 \plus{} y^2}{xy \minus{} 1} \equal{} 5$ - there is not truth

For example,  $ x$, $ y \equal{} \minus{} x^3$

$ \frac {x^2 \plus{} y^2}{xy \minus{} 1} \equal{} \frac {x^2 \plus{} x^6}{ \minus{} x^4 \minus{} 1} \equal{} \minus{} x^2\ne5$[/quote]

What do you wanna tell us with this meaningful post?

He probably meant that there is a pair which doesn't satisfy that equation (for 5). But, aktyw19, the problem is to find for which numbers it hold, not to prove it holds for all.

The example given by aktyw19 demonstrates that this problem is different from the one linked by shoki, because x and y do not have to be positive.

Yes, but there is also another way to tell us that one of the numbers may be negative^^
Indeed, if one of the two numbers is negative, the problem turns out to be an old IMO problem, but vieta jumping solves both cases easily

Please,  about solution if one of the two numbers is negative

sorry,i saw this problem first in Problem-Solving Stategies.(prob130,NT).it has supposed that $ x,y$ are integers and it wants to prove $ q\equal{}5$($ \frac{x^{2}\plus{}y^{2}}{xy\minus{}1}\equal{}q \implies q\equal{}5$)!!
but since the case mentioned (one of them is negative) is obvious and turn out to be the famous problem of XXIX IMO,
i didn't consider it :D .


vector
superior algebra
superior algebra unsolved

Show that $ \mathbb{Q}/\mathbb{Z} \otimes \mathbb{Q}/\mathbb{Z} \equal{} 0$.

* It is not specified over which ring we tensor. But, I suppose they mean $ \otimes_\mathbb{Q}$ and not $ \otimes_\mathbb{Z}$. Not sure how to solve it though.

Certainly not $ \otimes_\mathbb{Q}$ since $ \mathbb{Q}/\mathbb{Z}$ is not a vector space over $ \mathbb{Q}$.  A hint would be that $ a\plus{}\mathbb{Z}$ is the same as $ n\cdot (a/n\plus{}\mathbb{Z})$.

Steve


8 stupid mistakes on AIME :o
That's ridiculous... :mad:

BLah it may be that I am sick and coudln't go fast enough

Now I have to go study for next year  :|  :|

Hmm... What did you get?  If you don't know the correct answers yet, post when you know for sure.

Ugh I got a 6.

Considering I didn't prepare at all for AIME besides some random tests in like January, I am ok with that score.

I got a 5 but I MIGHT have gotten a 6 had I bubbled in 015 for #15 (I was going between 15 and 29 so yeah)

Either way no USAMO but I prepared USAMO style problems anyway

Now I need to go solve harder shortlist problems so I can at least have a shot a blue next year :(

I didn't do the AIME II officially. I got a 7 unofficially. Weird...that it's harder than the I and I had a 6 on the I...stupid mistakes kill me so easily.

Hmm xp you probably did make usamo
Also gauss why are you posting in the OH forum? *sigh*


Sir are you sure about the Date November 19 for BC?

I tried to Contact  Dr.Yen, but No response :maybe:

[quote="Inspired By Nature"][quote="nsato"]I was told Nov. 5 for the O-level and Nov. 19 for the A-level.[/quote]
This may be an odd question, but whats the tournament of towns?  :blush:[/quote]
Please see here:
http://www.amt.canberra.edu.au/imtot.html

[quote="nsato"]There was a recent post in this forum about a problem from an ongoing Tournament of Towns contest, which I myself unwittingly posted a rather big hint for.

I have also just learned that the contest has yet to be written in BC and Alberta.  So, no one should discuss any of these problems until the tests have been written, which will be November 19.[/quote]
I'm really sorry, I thought the contest would be written at the same time everywhere. I should have checked :blush:

so the ontarians have had it written already? i thought we were doing it at the same time.
and does anyone know whether we could register for A-level in BC?

Yes the Ontarians (at least the ppl in Toronto) wrote it already -- on October 22 and 29.


u look like a kid i know but his name's richard :|


So I'm sure many of you know about the 3d puzzle game Portal which recently came out on PC and on Xbox 360 (via Orange Box).  However, I just recently discovered an extremely addicting flash version on newgrounds.com, [url=http://www.newgrounds.com/portal/view/404612]here[/url].

Tell me what you think (if you didn't already know about it).

Lol a $ \$ 60.00$ Xbox 360 game free online.

Hey: even if it's easy, it's still fun.  It would be nice if more bonus levels were made though.

From what I've heard, this game has all of the levels from the 360 version, plus more.  (of course, 2 dimensions isn't as cool as 3.)

Two dimensions and Flash graphics aren't nearly as cool :P.

Meh... It’s probably cool. I’ve never played it. I just like the 3D one better. Cuz...it’s 3D...


induction
trigonometry
complex numbers
absolute value
combinatorics proposed
combinatorics

[quote="pbornsztein"]:(  I dislike this kind of conditions....why always asking to not use Induction or complex numbers or whatever?
I agree that some results are not 'allowed' in olympiad solutions if they make the problem trivial (so no Fermat's last theorem or things like that). But, they are not general tools as complex numbers.

Pierre.[/quote]

because, complex numbers is very strong-method, whenever- we can solve "problem-sequence" by many weak-method, and so nice
 :D in VIETNAM, not learn complex in college->many thanks for elementary if you can :D

But if $|k| < 2$ you are using complex numbers, and...it's bad :D

Pierre.

Exactly, but I didn't state that it doesn't use complex numbers. Though, note I didn't use words "complex numbers" :D

hey, you did some mistake. if they are complex, they automatically have absolute value 1. so you got to find some other solution :D
btw, it works for $k=0$ as well. indeed, it works for $2\cos(\frac{m}{n}\pi)$ or something like that.

Ok. I was wrong with answer. And if I will complete it then I necessary derive usual expression for $a_n=x\cdot u^n+y\cdot v^n$, so answer is clear. Indeed, it is something like $2\cos\frac{\pi n}{m}$, $k\neq \pm 2$.
Is it correct?

[quote="Myth"][quote="nguyenquockhanh"]Pro: find all k such that for every a(1),a(2), (a(n)): a(n+2)+a(n)=k.a(n+1) is periodic
this pro is in titu-andresscu's book but can you solve it not use complex[/quote]

[b]Note that Combinatorics is not a proper section for this problem[/b]

I don't know titu's solution, but my solution is following.
For $k=1$ and $k=-1$ you can see that sequence is periodic :)
Suppose $k\neq 1$, $u,v=\frac{-k\pm \sqrt{k^2-4}}{2}$, then
\[a_{n+2}-u\cdot a_{n+1}=v(a_{n+1}-u\cdot a_{n}).\]
Therefore, $\{a_n\}$ is periodic $\Rightarrow$ $v^n$ is periodic. It follows that $|u|=|v|=1$ (since $v\neq 0$). You can see that it is possible iff $k=\pm 1$ only :)[/quote]

you didn't use word"complex number" but v is complex number :D so your solution only exactly in field C, and is bad in field R

 :P

[quote="Peter Scholze"]btw, you forgot the case $a_{n+1}=ua_n$ or did you think of it? i didn't check what happens :D[/quote]

????  $a_{n+1}$ and  $a_{n}$ is real when u isn't real!!!

? what's going on? it just could be the case sometimes. and why is u not real? of course, if it is not, then the equality is clearly wrong. i just wanted to say something about the step where he divides through that term which could be an error in some cases.


Where do you put non-paying internships on the common application for college for colleges in the U.S?

You can put it as an activity, or if you attach a resume, you can put it in that.

If it's pretty much like a job, and other people are doing the same thing and getting paid, you might even be able to put in as employment, if you specify that somewhere? But I don't know.


Given 100 real numbers satisfy that for any 8 arbitrary numbers, there always exists 9 numbers whose mean is equal to mean of the 8 above numbers . Prove that the given numbers are equal

let $a_{1}\leq\cdots\leq{a_{100}}$ be these 100 numbers.
With the block of 8: $a_{1},\cdots,{a_{8}}$ we obtain $a_{1}=\cdots=a_{8}=a_{9}$. With the block of 8:$a_{9},\cdots,{a_{16}}$ we obtain $a_{9}=\cdots=a_{16}$.
And so on: $a_{1}=\cdots=a_{23}=\cdots=a_{93}=\cdots=a_{100}$.

[quote="loup blanc"]With the block of 8:$a_{9},\cdots,{a_{16}}$ we obtain $a_{9}=\cdots=a_{16}$.[/quote]

Why?  The block of nine might, for instance, be $a_{8}, a_{9}, \ldots, a_{15}, a_{17}$.

OK JBL I was definitively wrong. :blush:

  I have read the proof of "il maestro" Grobber but it's very difficult for me to believe that one can't solve this exercise without using the axiom of choice.
    If an industrialist wants to obtain a machine to  do a  manufactured object and if Grobber goes to see him to announce  that, according to the axiom of choice, its machine
exists , then the industrialist will not give him one euro!

Use the extremal principle.Consider the smallest number among these 100 numbers.Since this is the mean of some arbitrary 8 numbers from the set It cannot be smaller than all these numbers.Contradiction! Hence all the numbers are equal.

That is just wrong.  The "contradiction" is with an unstated assumption that all the numbers are distinct, the negation of which is that some numbers are duplicated.  At best, this argument will get you what it got loup blanc, namely that the smallest 9 numbers are all the same.

Now, to go read grobber's proof :)

Having read grobber's proof, I feel like there should be a simpler way to reduce the case of reals to the case of integers.  Certainly, the original problem seems like it wouldn't lose anything if it were stated with rationals instead of reals.  I don't have a better way to make that simplification, unfortunately.

I agree - that Hamel basis thing is overkill :).

In order to reduce the problem to rationals , do this: treat the $100$ numbers as unknowns, and write the given conditions as a homogeneous system of linear equations. The aim is to prove that this system has only the obvious solution. If there were any others, they could be chosen to be rational because the equations have rational coefficients.

Sorry . I didn't read the question  properly....I misinterpreted it........


Give that a,b,c are positive real number, and abc=1
prove that
{[1/(1+a) ]+[1/(1+b)]+[1/(1+c)]}<{(1/2a)+(1/2b)+(1/2c)}

Welcome to AoPS.  :)  We have a math typesetting program on our forums, called $\LaTeX$, that allows for pretty-looking and more easily readable mathematics, such as in your problem below:
\[a,b,c \in \mathbb{R}^{+}\text{ and }abc=1 \ \Longrightarrow \ \frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}< \frac{1}{2a}+\frac{1}{2b}+\frac{1}{2 c}\]
There's a link to an introduction and tutorial on the left side, near the top of your page.  :)

[hide="As to the problem, (click here) "]
since I don't have too much experience with inequalities, I'm thinking that combining the fractions might help.[/hide]

Actually that $<$ needs to be a $\leq$ since they're equal when $a=b=c=1$.
[hide]
By AM-GM, $a+1\geq 2\sqrt{a}$ etc. so
\[\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\leq \frac{1}{2\sqrt{a}}+\frac{1}{2\sqrt{b}}+\frac{1}{2\sqrt{c}}\]
Now it remains to prove
\[\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\leq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\]
This follows from AM-GM:
\[\frac{6}{\sqrt{a}}=\frac{6}{\sqrt[6]{a^{4}bc}}\leq \frac{4}{a}+\frac{1}{b}+\frac{1}{c}\]
and similar stuff for the other terms.

As for the LaTeX, if you leave your mouse on one of the blocks, it will tell you what the code for it is. This is useful after you get the basics, but want to use some more complicated symbols. (i.e \frac{a}{b} surrounded by dollar signs gives $\frac{a}{b}$, and \sqrt{x} surrounded by dollar signs gives $\sqrt{x}$)
[/hide]

Thank your help.
By the way,
I am greek to the LaTeX.
May I uplad the document written Microsoft Word ?

[quote="ferng"]Give that $x,y,z$ are positive real number, and $xyz=1$, prove that $\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\leq\frac{1}{2x}+\frac{1}{2y}+\frac{1}{2z}$.[/quote]
Let $f(x)=\frac{1}{2x}-\frac{1}{1+x}+\frac{\ln{x}}{4}$, then $f'(x)=\frac{(x-1)(x^{2}+5x+2)}{4x^{2}(1+x)^{2}}\{\begin{array}{c}>0,\ when\ x>1,\\\\<0,\ when\ x<1.\end{array}$

$\Longrightarrow f(x)\geq f(1)=0\Longrightarrow \sum\frac{1}{2x}\geq \sum\frac{1}{1+x}$.