Consider a,b,c are positive first, then
\[
a^2  \ge \frac{2}{3}a - \frac{1}{9}
\]
Therefore,
\[
\begin{array}{l}
 \sum {\frac{a}{{a^2  + 1}}}  \le \frac{9}{{10}} \\
  \Leftrightarrow \sum {\frac{a}{{\frac{2}{3}a + \frac{8}{9}}}}  \le \frac{9}{{10}} \\
  \Leftrightarrow \sum {\frac{a}{{3a + 4}}}  \le \frac{1}{5} \\
  \Leftrightarrow \sum {\frac{{3a}}{{3a + 4}}}  \le \frac{3}{5} \\
  \Leftrightarrow \sum {\frac{1}{{3a + 4}}}  \ge \frac{3}{5} \\
 \end{array}
\]
the last one is true by cauchy.

In polish OM we know that: $a,b,c\geq -{3 \over 4}$

[quote="_tequilla"]In polish OM we know that: $a,b,c\ge \frac{-3}{4}$[/quote]
Can you post the solution?

[quote="Arne"]Well known technique. Try to find $s$ and $t$ satisfying:

\[\frac{a}{a^2 + 1} \leq sa + t \Leftrightarrow (4a + 3)(3a - 1)^2 \geq 0\] and also \[s + 3t = \frac{9}{10}.\]

Once you find $s$ and $t$ it is easy.[/quote]
How did you come up with these ineqs?

4a+3 uses the condition $a\ge \frac{3}{4}$ (ensuring its positive)
(3a-1) is the equality case.
Sum up three inequalities we have
$s(a+b+c)+3t=\frac{9}{10}$
which results the last inequality.

[quote="minhkhoa"]I am very interested in the problem for a,b,c are real
There must be a solution for it[/quote]
Yes, this problem is only beautiful with a,b,c are real.
My solu:
[tex]\sum(\frac{10x}{1+x^2}-3) \ge 0[/tex]
[tex]\sum(3x-1)\frac{3-x}{1+x^2} \ge 0[/tex]
 we can supp x >= y >= z:
If x<=3 then: 3x-1 >= 3y-1 >=3z-1, [tex]\frac{3-x}{1+x^2}   <= \frac{3-y}{1+y^2}  <= \frac{3-z}{1+z^2} [/tex], which solves the  inequality.
if x>3, then z<0, so that:
[tex]\frac{x}{1+x^2} \le \frac{3}{1+3^2},\frac{y}{1+y^2} \le \frac{1}{2},\frac{z}{1+z^2} \le 0[/tex]
and we also have result.

I don't understand why it's obvious in case $x\leq 3$.
Case $x\geq 3$ seems all right :)

[quote="Hatucdao"]Yes, this problem is only beautiful with a,b,c are real.
My solu:
[tex]\sum(\frac{10x}{1+x^2}-3) \ge 0[/tex]
[tex]\sum(3x-1)\frac{3-x}{1+x^2} \ge 0[/tex]
 we can supp x >= y >= z:
If x<=3 then: 3x-1 >= 3y-1 >=3z-1, [tex]\frac{3-x}{1+x^2}   <= \frac{3-y}{1+y^2}  <= \frac{3-z}{1+z^2} [/tex], which solves the  inequality.
if x>3, then z<0, so that:
[tex]\frac{x}{1+x^2} \le \frac{3}{1+3^2},\frac{y}{1+y^2} \le \frac{1}{2},\frac{z}{1+z^2} \le 0[/tex]
and we also have result.[/quote]

Just as Ondrob, I don't understand your solution. Why do we have $\frac{3-x}{1+x^2}   <= \frac{3-y}{1+y^2}  <= \frac{3-z}{1+z^2}$? I even suspect this is wrong, taking into account that x, y, z (do I correctly assume that these are just a, b, c ?) can be negative.

Anyway I have succeeded to find a solution to the problem.

[color=blue][b]Problem.[/b] Let a, b, c be three reals such that a + b + c = 1. Prove the inequality

$\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}\leq\frac{9}{10}$.[/color]

[i]Solution.[/i] We distinguish three cases:

[i]Case 1.[/i] At least two of the three numbers a, b, c are $\leq 0$.
[i]Case 2.[/i] Exactly two of the three numbers a, b, c are positive.
[i]Case 3.[/i] All three numbers a, b, c are positive.

In Case 1, the solution of the problem is very easy: WLOG assume that $b\leq 0$ and $c\leq 0$. Then, we have the inequalities

$\frac{a}{a^2+1}\leq\frac12$ (since $a\leq\frac12\left(a^2+1\right)$, as we have $\frac12\left(a^2+1\right)-a=\frac12\left(a-1\right)^2\geq 0$);
$\frac{b}{b^2+1}\leq 0$ (since $b\leq 0$ and $b^2+1>0$);
$\frac{c}{c^2+1}\leq 0$ (since $c\leq 0$ and $c^2+1>0$).

Summing up these inequalities, we get

$\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}\leq\frac12+0+0=\frac12$,

what is clearly $<\frac{9}{10}$. Thus, the problem is solved for Case 1.

Case 3 is quite simple and was solved in post #2 and on http://www.mathlinks.ro/Forum/viewtopic.php?t=180 .

Remains to handle Case 2. In this case, we can WLOG assume that a > 0, b > 0 and $c\leq 0$.

First assume that $\left(a+b\right)\left(3-ab\right)\geq 0$. Then, since

$\frac{a}{a^2+1}+\frac{b}{b^2+1}-2\frac{\left(\frac{a+b}{2}\right)}{\left( \frac{a+b}{2}\right)^2+1}=-\frac{\left( a-b\right)^2\left( a+b\right) \left(3-ab\right) }{\left( a^2+1\right) \left( b^2+1\right) \left( \left(a+b\right)^2+4\right) }\leq 0$,

we have

$\frac{a}{a^2+1}+\frac{b}{b^2+1}\leq 2\frac{\left(\frac{a+b}{2}\right)}{\left( \frac{a+b}{2}\right)^2+1}$,

so that, in order to prove the inequality $\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}\leq\frac{9}{10}$, it is enough to show that $2\frac{\left(\frac{a+b}{2}\right)}{\left( \frac{a+b}{2}\right)^2+1}+\frac{c}{c^2+1}\leq\frac{9}{10}$, what is actually very easy: Since a + b + c = 1, we have a + b = 1 - c, and thus

$2\frac{\left(\frac{a+b}{2}\right)}{\left(\frac{a+b}{2}\right)^2+1}+\frac{c}{c^2+1}=2\frac{\left(\frac{1-c}{2}\right)}{\left( \frac{1-c}{2}\right)^2+1}+\frac{c}{c^2+1}$
$=\frac{9}{10}-\frac{\left(\left(c+1\right)^2+4\right)\left(3c-1\right)^2}{10\left(c^2+1\right)\left(\left(c-1\right)^2+4\right)}\leq\frac{9}{10}$.

Thus, if $\left(a+b\right)\left(3-ab\right)\geq 0$, we are done. Similarly, if $\left(b+c\right)\left(3-bc\right)\geq 0$ or $\left(c+a\right)\left(3-ca\right)\geq 0$, we are also done. Hence, it only remains to consider the case when (a + b) (3 - ab) < 0, (b + c) (3 - bc) < 0 and (c + a) (3 - ca) < 0. In this case, since a > 0 and $c\leq 0$, we have $ca\leq 0$, so that ca < 3, and thus 3 - ca > 0, so that (c + a) (3 - ca) < 0 yields c + a < 0. Since a + b + c = 1, we have c + a = 1 - b, so that 1 - b < 0. Hence, b > 1. Similarly, a > 1.

Since (a + b) (3 - ab) < 0 and a + b > 0 (because of a > 0 and b > 0), we must have 3 - ab < 0, so that 3 < ab. By the AM-GM inequality, applied to the positive numbers a and b, we have $ab\leq\left(\frac{a+b}{2}\right)^2$, so that $3<\left(\frac{a+b}{2}\right)^2$. Taking the square root, we get $\sqrt3<\frac{a+b}{2}$, so that $a+b>2\sqrt3$. Since a + b + c = 1, we have a + b = 1 - c; thus, $1-c>2\sqrt3$, so that $c<1-2\sqrt3$. Since $1-2\sqrt3<-2$, we have c < -2, so that -c > 2, and therefore $\frac{-c}{2}>1$.

If we had $c+2a\leq 0$ and $c+2b\leq 0$, then addition would yield $\left(c+2a\right)+\left(c+2b\right)\leq 0$, so that $2\left(a+b+c\right)\leq 0$, what is impossible, since a + b + c = 1. Thus, at least one of the equations $c+2a\leq 0$ and $c+2b\leq 0$ doesn't hold. WLOG, assume that the equation $c+2a\leq 0$ doesn't hold. Then, c + 2a > 0. Hence, $a>\frac{-c}{2}$. Together with $\frac{-c}{2}>1$, this yields $a>\frac{-c}{2}>1$.

Now, notice that if we have two reals x and y such that x > y > 1, then $\frac{x}{x^2+1}<\frac{y}{y^2+1}$. This is because $\frac{x}{x^2+1}-\frac{y}{y^2+1}=-\frac{\left(xy-1\right)\left(x-y\right)}{\left(x^2+1\right)\left(y^2+1\right)}$, what is negative since x > 1 and y > 1 yield xy > 1 and thus xy - 1 > 0, and x > y yields x - y > 0. Applying this to the reals a and $\frac{-c}{2}$ which satisfy $a>\frac{-c}{2}>1$, we get $\frac{a}{a^2+1}<\frac{\left(\frac{-c}{2}\right)}{\left(\frac{-c}{2}\right)^2+1}$. On the other hand,

$\frac{\left(\frac{-c}{2}\right)}{\left(\frac{-c}{2}\right)^2+1}+\frac{c}{c^2+1}\leq\frac16$,

since

$\frac{\left(\frac{-c}{2}\right)}{\left(\frac{-c}{2}\right)^2+1}+\frac{c}{c^2+1}-\frac16=-\frac{\left(c^2+3c-2\right)^2}{6\left(c^2+1\right)\left(c^2+4\right)}\leq 0$.

Finally, $\frac{b}{b^2+1}\leq\frac12$ is proven similarly to how I showed $\frac{a}{a^2+1}\leq\frac12$ in Case 1. Hence,

$\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}<\frac{\left(\frac{-c}{2}\right)}{\left(\frac{-c}{2}\right)^2+1}+\frac12+\frac{c}{c^2+1}$
$=\frac12+\left(\frac{\left(\frac{-c}{2}\right)}{\left(\frac{-c}{2}\right)^2+1}+\frac{c}{c^2+1}\right)\leq\frac12+\frac16<\frac{9}{10}$.

This completes the solution of the problem for the Case 2. Thus, the problem is solved.

[b]EDIT:[/b] I see the problem was also solved by Pierre in http://www.mathlinks.ro/Forum/viewtopic.php?t=1831 . The proof doesn't look much better...

  Darij

Let $a$, $b$, $c$ be positive numbers such that $a + b + c =3$ . [url=http://www.artofproblemsolving.com/community/c6h1178423p5696943]Prove that[/url]$$\frac{a}{a^3+1}+\frac{b}{b^3+1}+\frac{c}{c^3+1} \le \frac{3}{2}.$$

[quote=szl6208][quote=minhkhoa]Let a,b,c be real numbers such that $a+b+c=1$.Prove that
$\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1} \le \frac{9}{10}$[/quote]

We have
\[\frac{9}{10}-(\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1})=\frac{1}{50}\sum{\frac{(3+4a)(1-3a)^2}{1+a^2}}\ge{0}\][/quote]

Can you tell me,why the last inequality is true,
When $a\leq \frac{-3}{4}$ there exist the negative terms.

I would like to add that this problem can be found in [i]Inequalities[/i] by Zdravko Cvetkovski, under the section "Points on Incidence in the application of the AM-GM inequality".

Oh,I'm sorry,thank you ThE-dArK-lOrD!,I'm wrong!

Second derivative of $f(x) = \frac{x}{x^{2}+1}$ suggests the function be concave on the interval (0, 1). So this is trivial.

[quote=sqing]Let $a$, $b$, $c$ be positive numbers such that $ab + bc + ca =3$ . Prove that$$\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1} \le \frac{3}{2}.$$[/quote]

[b][i]Solution[/i][/b]. For [b]arbitrary[/b] $a,b,c\in\mathbb{R}$ (not just $a,b,c$ with $ab + bc + ca =3$), it follows that
$$\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1} \le\frac{1}{2}+\frac{1}{2}+\frac{1}{2}= \frac{3}{2}.$$As desired. $\blacksquare$

[quote=MathPrep12]What is couches inequality[/quote]

Let $a=\langle a_i\rangle ,b=\langle b_i\rangle$ be two vectors. Then
$$(a\cdot a)(b\cdot b)\ge (a\cdot b)^2$$
where $x\cdot y$ is the dot product between $x$ and $y$.
It's proof involves the Law of Cosines.

Further reading:
https://artofproblemsolving.com/wiki/index.php/Cauchy-Schwarz_Inequality

Those are some very cool solutions. I was reading them, and I'm very amazed. I didn't even think about those steps.


calculus
integration
function
geometry
calculus computations

Find the value of $a$ for which $\int_{0}^{1}|xe^{x}-a|\ dx$ is minimized.

[hide="An Idea?"]I'm probably wrong, but here it is:

Try $0<a<1$ : \begin{eqnarray*}\int_{0}^{1}\left|xe^{x}-a\right|\,dx &=& \int_{0}^{a}\left(a-xe^{x}\right)\,dx+\int_{a}^{1}\left(xe^{x}-a\right)\,dx \\ &=& 2a^{2}-a-2\left(a-1\right)e^{a}-1 \\ &=& f\left(a\right)\end{eqnarray*} So now we need $f'\left(a\right)=0$ and $f''\left(a\right)<0$.  However I'm having trouble solving the first.  I'll get back to this.[/hide]

[quote="amcavoy"][hide="An Idea?"]I'm probably wrong, but here it is:

Try $0<a<1$ : \begin{eqnarray*}\int_{0}^{1}\left|xe^{x}-a\right|\,dx &=& \int_{0}^{a}\left(a-xe^{x}\right)\,dx+\int_{a}^{1}\left(xe^{x}-a\right)\,dx \\ &=& 2a^{2}-a-2\left(a-1\right)e^{a}-1 \\ &=& f\left(a\right) \end{eqnarray*} So now we need $f'\left(a\right)=0$ and $f''\left(a\right)<0$.  However I'm having trouble solving the first.  I'll get back to this.[/hide][/quote]
forgive me if im wrong...but wouldnt this be implying that:

 $\forall{a}\mid 0<a<1$ that the the function $|xe^{x}-a|$ would be "reflected" at x = a? which from what i can see...isnt true?...

This can be done without computing any integrals. Try differentiating the integral with respect to $a$ directly.
[hide="Answer"]It's the median value $\frac12 e^{\frac12}[$[/hide]

To amcavoy: That's wrong, since your integral splits in the wrong place.

[quote="jmerry"]This can be done without computing any integrals. Try differentiating the integral with respect to $a$ directly.
[hide="Answer"]It's the median value $\frac12 e^{\frac12}[$[/hide]

[/quote]

how did u deduce this?

[quote="kunny"]Find the value of $a\in R$ for which $F(a)\equiv\int_{0}^{1}|xe^{x}-a|\ dx$ is minimized.[/quote][color=darkblue] Denote the functions $f: [0,1]\rightarrow R\ ,\ f(x)=xe^{x}$ and $g: [0,1]\rightarrow R\ ,\ g(x)=a$. Then $F(a)=\int_{0}^{1}|f(x)-g(x)|\ dx=$ $[\Gamma_{f,g}]$ - the area of the surface $\Gamma_{f,g}$ comprised between the graphs of the above functions. The function $f$ is strict increasing, convex and for any $x\in [0,1]$, $0\le f(x)\le e\ .$ Observe that $P(x)=(x-1)e^{x}-ax\in \int (xe^{x}-a)\ dx$ and $P(0)=-1$ , $P(1)=-a\ .$ Distinguish three cases :
$1.\blacktriangleright\ a\le 0\Longrightarrow$ $g(x)\le f(x)\ ,\ 0\le x\le 1$  $\Longrightarrow$  $F(a)=P(1)-P(0)=1-a\ge 1\ .$
$2.\blacktriangleright\ a\ge e$ $\Longrightarrow$ $f(x)\le g(x)\ ,\ 0\le x\le 1$ $\Longrightarrow$ $F(a)=P(0)-P(1)=a-1\ge e-1\ .$
$3.\blacktriangleright\ a\in (0,e)\Longrightarrow$ Exists and is uniquely the point $c=c(a)\in (0,1)$ so that $f(c)=a$, i.e. $ce^{c}=a\Longrightarrow$ $F(a)=P(0)-2P(c)+P(1)=-1-a-2[(c-1)e^{c}-ac]=2e^{c}+2ac-3a-1\ .$ [u]Pause, but no stop[/u] ! Continue with the down remark.

[b]Remark.[/b] It is well-known the relation $\boxed{\ (\forall )a\in R\ ,\ \int_{0}^{1}|h(x)-a|\ dx\ge \int_{0}^{1}|h(x)-h(\frac{1}{2})|\ dx\ }$, where the function $h: [0,1]\rightarrow R$ is continue and monotonously !! For $h(x)=xe^{x}$ obtain the proposed by Kunny problem. See the my topic http://www.mathlinks.ro/Forum/viewtopic.php?t=106141

[b]Proof.[/b] Suppose w.l.o.g. that the function $h$ is increasing.
Then for any $x\in [0,\frac{1}{2}]\ ,\ |h(x+\frac{1}{2})-h(x)|\le|h(x+\frac{1}{2})-a|+|h(x)-a|$ $\Longrightarrow$
$\int_{0}^{\frac{1}{2}}[h(x+\frac{1}{2})-h(x)]\ dx\le\int_{0}^{\frac{1}{2}}|h(x+\frac{1}{2})-a|\ dx+\int_{0}^{\frac{1}{2}}|h(x)-a|\ dx$ $\Longrightarrow$
$\int_{\frac{1}{2}}^{1}h(x)\ dx-\int_{0}^{\frac{1}{2}}h(x)\ dx\le \int_{\frac{1}{2}}^{1}|h(x)-a|\ dx+\int_{0}^{\frac{1}{2}}|h(x)-a|\ dx$ $\Longrightarrow$
$\int_{0}^{1}|h(x)-a|\ dx\ge$ $\int_{\frac{1}{2}}^{1}[h(x)-h(\frac{1}{2})]\ dx+\int_{0}^{\frac{1}{2}}[h(\frac{1}{2})-h(x)]\ dx=$ $\int_{0}^{1}|h(x)-h(\frac{1}{2})]\ dx\ .$
In conclusion, $\int_{0}^{1}|h(x)-a|\ dx\ge \int_{0}^{1}|f(h)-h(\frac{1}{2})|\ dx\ .$

[b]The final answer :[/b] $\boxed{\ a=h(\frac{1}{2})=\frac{\sqrt e}{2}\ }\ .$[/color]

[quote]
$3.\blacktriangleright\ a\in (0,e)\Longrightarrow$ Exists and is uniquely the point $c=c(a)\in (0,1)$ so that $f(c)=a$, i.e. $ce^{c}=a\Longrightarrow$ $F(a)=P(0)-2P(c)+P(1)=-1-a-2\left[(c-1)e^{c}-ac\right]=2e^{c}+2ac-3a-1\ .$
[/quote]

could u explain this plz..i dont follow the significance of the result

[quote="Virgil Nicula"]The function $l: [0,1]\rightarrow [0,e]$, where $l(x)=xe^{x}$ is continue and strict increasing on $[0,1]$. Thus, the function $l$ has the Darboux's property and is injective. Therefore, the equation $l(x)=a$ for any $a\in [0,e]$ has at least a solution $c$ in $[0,1]$ (Darboux's property) and has at most a solution (injective function). In conclusion, the point $c$ exists and is uniquely ![/quote]
Thank you!  :oops:

Of course, your answers are correct.

Here is my solution.

For $t$ such that $te^{t}-a=0\ (0<t<1),$

$\int_{0}^{1}|xe^{x}-a|\ dx=\int_{0}^{t}|xe^{x}-a|\ dx+\int_{t}^{1}|xe^{x}-a|\ dx$

$\geq-\int_{0}^{t}(xe^{x}-a)\ dx+\int_{t}^{1}(xe^{x}-a)\ dx$

$=(2t-1)a+\int_{t}^{0}xe^{x}\ dx+\int_{t}^{1}xe^{x}\ dx.$

Determining the value of $t$ for which this equation will take a constant number in regardless of the value of $a,$ we obtain $t=\frac{1}{2},$ that is to say $a=\frac{\sqrt{e}}{2}.$

If $a$ takes this value, Set $F(x)=(x-1)e^{x},$ we have

$\int_{0}^{1}|xe^{x}-a|\ dx=\int_{0}^{t}|xe^{x}-a|\ dx+\int_{t}^{1}|xe^{x}-a|\ dx$

$\geq \int_{\frac{1}{2}}^{0}xe^{x}\ dx+\int_{\frac{1}{2}}^{1}xe^{x}\ dx$

${=F(0)+F(1)-2F\left(\frac{1}{2}\right)=\sqrt{e}}-1.$


Fie n un numar natural si impar si f:[0,1]->R o functie continua cu $\int^1_0 f(x)dx=0$
Sa se demonstreze ca exista c din [0,1] a.i.$\int^c_0 f(x)dx=[f(c)]^n$

Am primit aceasta problema la un concurs de mate si am dat o alta solutie decat pe barem care nu mi s-a punctat, o sa o expun in mare aici si poate va dati seama ce nu in regula pt ca eu nu reusesc sa imi gasesc greseala.

Aleg g: [0,1]-R , G(x)=$\int^x_0 f(x)dx-[f(x)]^n$, evident G continua
$G(0)G(1)=[f(0)f(1)]^n$
Daca f(0) sau f(1)=0 este clar ca G se anuleaza
Presupunem ca G nu se anuleaza,atunci ea fiind continua are semn constant.
Presupun cazul I G(x)>0 , oricare ar fi x din [0,1]
Cu teorema de medie,este clar ca f se anuleaza in cel putin un punct.
Fie $a$ infimumul pt care se anuleaza si $b$ supremumul pt care se anuleaza.
Atunci $G(a)=\int^a_0 f(x)dx-f(a)^n=\int^a_0 f(x)dx>0$ si atunci rezulta f(0)>0
$G(b)=\int^b_0f(x)dx-f(b)^n$=$\int^b_0 f(x)dx>0$ dar$\int^1_0 f(x)dx=0$ deci $\int^1_b f(x)dx<0$
si atunci rezulta f(1)<0
Atunci $G(0)G(1)=[f(0)f(1)]^n<0$ deci G fiind continua se anuleaza contradictie.
 Analog tratez cazul II.

daca integrala este pozitiva pe [0,a] atunci functia este pozitiva doar daca integrala este pozitiva pe orice inteval [c,d] inclus in [0,a].

In cazul meu,cum a este infimumumul pentru care se anuleaza, daca f(0)<0 atunci f(x)<0 pentru orice x intre 0 si a,deci integrala ar fi negativa ceea ce este absurd

In afara de faptul ca nu ai argumentat ca inf e de fapt minim, altfel zis ca se atinge(presupui ca nu se atinge, construiesti un sir convergent la inf, si din faptul ca e continua reiese usor ca f(inf) = 0), eu consider ca rezolvarea e buna. Si putin mai multa atentie la redactare, ca a trebuit sa citesc de vreo 3 ori ca sa ma prind exact ce vrei sa zici :)


Take abitrary triangle ABC.

On side AB, make point D such that 2AD=DB

On side AC, make point E such that AE=EC

Find the area of triangle ADE if triangle ABC's area is 30.

ahhh, ive seen this problem before in one of my singapore math books (meant for highschoolers)....it makes it easier to understand why China consistently dominates at the IMO.....

This actually isn't difficult.

[hide="Hint"]The triangle is arbitrary, think about what you can do with that...[/hide]

It's not supposed to be that hard....  :ninja:

[hide="Hint Number 1"]Take arbitrary triangle ABC, Make $n$ number of points on segment AB such that they divide AB into $n+1$ equal segments. Connect point C to every single one of those points we have just made. So now there are $n+1$ mini triangles whose combined area totals triangle ABC's area. They all have equal bases, since they are divided equally with those $n$ points.

Are their heights equal? Are their areas equal?  ;) [/hide]

[hide="Hint Number 2"]
In regards to my problem...Draw segment BE. Think about it now. You should be able to do it... ;)
[/hide]

[hide="solution"]
It is ummm easy
Answer=$5$
[/hide]
Correct?

No, it isn't, unless I'm wrong. :P I'm guessing you forgot to do something in there.

what, 5th grade? IMO Prep? There wasn't even MathCounts prep in 5th grade... Jesus Christ...

I mean, it's easy and all for us 8th graders, but...

[hide="Answer"]
5
[/hide]

Here's my

[hide="solution"]The base AD is $\frac{1}{3}$ of AB, the smaller height is $\frac{1}{2}$ of the big triangle, so the area is $\frac{1}{6}$ of the bigger triangle, or 5. [/hide]

Good Job everybody! You guys finally did it!

The answer is indeed 5!  :P

Your hint said something about mini-triangles... what's the deal with that?

My first hint basically just said that if you divide a side of a triangle into, let's say 2 equal parts. And connect that one point the point on a triangle opposite that side, you'll have 2 triangles. (which I called mini triangles). They'll have the same area because they have equal heights and bases.


I wonder why no one interested with my inequality.   :?:

[quote="beautifulliar"]Let $ a$, $ b$, and $ c$ be positive real numbers such that $ \dfrac{a}{b} \plus{} \dfrac{b}{c} \plus{} \dfrac{c}{a} \ge \dfrac{b}{a} \plus{} \dfrac{c}{b} \plus{} \dfrac{a}{c}$.
Prove that $ \dfrac{a}{a \plus{} b} \plus{} \dfrac{b}{b \plus{} c} \plus{} \dfrac{c}{c \plus{} a} \le \dfrac{b}{a \plus{} b} \plus{} \dfrac{c}{b \plus{} c} \plus{} \dfrac{a}{c \plus{} a}$.

*Edited: You may expand but I still wait for some beautiful proof.[/quote]
$ \dfrac{a}{b} \plus{} \dfrac{b}{c} \plus{} \dfrac{c}{a} \ge \dfrac{b}{a} \plus{} \dfrac{c}{b} \plus{} \dfrac{a}{c}$
\[ \Leftrightarrow \frac {(a \minus{} b) (b \minus{} c)(c \minus{} a)}{a b c}\ge 0,\]
$ \dfrac{a}{a \plus{} b} \plus{} \dfrac{b}{b \plus{} c} \plus{} \dfrac{c}{c \plus{} a} \le \dfrac{b}{a \plus{} b} \plus{} \dfrac{c}{b \plus{} c} \plus{} \dfrac{a}{c \plus{} a}$
\[ \Leftrightarrow \frac {(a \minus{} b)(b \minus{} c)(c \minus{} a)}{(a \plus{} b) (b \plus{} c) (c \plus{} a)} \ge 0.\]
so... :o


I think I had 45 posts a couple months back.

As r15s11z55y89w21, I think I made about 25 posts one day.

EDIT: Epic win. After my vote each option has one vote.

$ 12.916%
$ of my posts were in the last two days.


For the numbers $x = \frac{49}{10}$ and $y = \frac{7}{5}$ we have $x-y = \frac{x}{y}$.

(1) Find three pairs satisfying this property.

(2) Determine a generation rule for all pairs $(x;y) \in \mathbb{R}^2$.

(3) Provide all pairs $(x;y)$ with this property.

Maybe I'm missing something, but this almost seems too easy... (spoiler)

[hide](1) (4,2), (81/8, 9/8), (-1/2, -1)

(2) Multiplying both sides by y (this is allowed, since we already have  from the denominator on the right side) yields xy - y^2 = x.

Rearranging, we have

xy - x = y^2

x(y-1) = y^2

x = y^2/(y-1)

Thus, a generation rule is to pick an arbitrary  and to let x = y^2/(y-1).

(3) There are infinitely many pairs, but the general form is (y^2/(y-1), y) for .[/hide]


integration
calculus
LaTeX
function
algebra
polynomial
partial fractions

Evaluate $\int{\frac{x-1}{(x+1)^3}}dx$. Sorry but i'm new to this and I don't see any elementary way to do it.

its a easy integration

[hide] just break

             ${\int\frac{x-1}{(x+1)^3}} dx$
             break the fraction as
      ${\int\frac{x}{(x+1)^3}} dx$ - ${\int\frac{1}{(x+1)^3}} dx$

In the second part direct integration. and in  the first put x +1 = t

You Get the ANSWEEEEEEEEEEEEEEEER

Post the answer so i will check your answer and give marks.

 :rotfl:  :rotfl:  :rotfl:  :rotfl:

Wouldn't it be more useful to split into $\int \frac{x+1}{(x+1)^3}-\int\frac2{(x+1)^3}$?

Alternately, using the substitution $t=x+1$, $x=t-1$ immediately works well.

Problems like this should be posted here, in "Calculus Computations and Tutorials."

In teaching this, I would direct my students' attention to the substitution jmerry showed. I'd call that a "little" substitution: easy to perform, not intended to be the last step, but merely to make it easier to see what to do next. And if you do that in this case, the next step should be staring you in the face.

Couldn't you use u-substitution on this problem as well? I'm not in calculus but i have taught myself alot about first year single variable calculus, which i will take in the fall. Anyways, i worked this problem, and this is what i got (not sure if it's right). Once again, i can't get latex to work, so just bare with me.

[hide] let u = x + 1

substituting this into the original function, you turn the integral into something that looks like this:

(u-2)/(u^3). That can be written as the sum of 2 fractions quite easily. Simplfy this to get [(u)/(u^3)] - [(2)/(u^3)] and rewrite those expressions as negative exponents. Now all you have to do is integrate a simple polynomial function with respect to u using the sum rule. Once you do that and substitute x+1 back in for u, you should get something like (1-(x+1)^2)/(x+1)^3. [/hide]

You could also use integration by partial fractions like the above users suggested, but i did it this way because i only know the basic rules for polynomial integration, integration by u substitution, and integration by parts.

edit: if you simplfy the final result in the previous hidden content algebraically, you should get the final antiderivate function, which is:

[hide] -x/(x+1)^2[/hide]

yes your splitting must be more helpful. Its a good solution.


calculus
integration
trigonometry
logarithms
calculus computations

Find

$\int{x\cot{x}+\ln{|\sin{x}|}}\,dx$

[hide="Hint (if necessary)"]

Consider $\int{x\cot{x}}\,dx$ first[/hide]

Hence find an expression for

$\int{xf\prime(x)+f(x)}\,dx$

[b]Extension:[/b] Use this expression to evaluate

$\int{x\ln{x^{2}}-x}\,dx$

[hide="Big Hint"]

Let $f(x)=x\ln{x}-x$[/hide]

$\int x f'\left(x\right)\,dx=x f\left(x\right)-\int f\left(x\right)\,dx$

Then adding $\int f\left(x\right)\,dx$ to this just gives

$\int \left(x\cot x+\ln\left|\sin x\right|\right)\,dx=x f\left(x\right)=\boxed{x\ln\left|\sin x\right|}$

I'll do the extension:

As we have seen above $\int{xf\prime(x)+f(x)}\,dx=xf(x)+constant$

So when $f(x)=x\ln{x}-x$ we have $f\prime(x)=\ln{x}$

So $\int{x\ln{x}+x\ln{x}-x}\,dx=\int{x\ln{x^{2}}-x}\,dx$

which by the above identity $=x(x\ln{x}-x)+constant$
$=x^{2}(\ln{x}-1)+constant$


agha man be ye moshkeli bar khordam va az shoma bozorgaan talabe yaari mikonam
age toonestin hatman javabe mano bedin.rastesh man mikhastam ba komake ham ye manbae ettehaad haye hendesi dorost konim chon ke man besh ziyad niyaaz daram(khodam ham mizaram).agar ke manbaetoono ham begin motshakker misham

akhe ye vakht ye az khoda ba khabari miyad va ye hamchin soaali tarh mikone:
prove that: dar mosalase delkhah ke shoae dayere mohaati ash $r$,mohiti ash $R$ va $p$ nesfe mohitash bashad saabet konid:       $r^2+p^2+-4R^2\geq 18Rr$
ma ham migim:  eee intorie? pas bana bar CS darim:
$(\sum cosAcosB)(\sum \frac{1}{cosAcosB})\geq9$ pas bana bar ravabete gofte shode: $(\frac{r^2+p^2-4R^2}{4R^2})(\frac{2R}{r})\geq9$ pas natije mishavad $2Rr^2+2Rp^2-8R^3\geq 36R^2r$ ke ba kami saade kardan be haman hokme matloob miresim.
amma in ye mesaaale kheili kheili kheili kheili kheili asoon bood va gar na bazi hashoon bad chizayi hastan.
mokhlese hamegi.

man nafahmidam manzooret chejoor etehadie dar natije ino mizaram age rabvti nadare begoo:
[b]namosavie erdish mordel:[/b]
mosalase delkhahe $ABC$ va noghte $P$ dakhel ya rooye marze an mafrooz ast.az $P$ amoodhaye $PZ,PY,PX$ ra be tartib bar azlae $AB,CA,BC$ vared mikonim,darim:
$PA+PB+PC\geq 2(PX+PY+PZ)$
in joori khoobe??
$ab+bc+ca=p^{2}+r^{2}+4Rr$
$\sum\sin^{2}\frac{A}{2}=\frac{2R-r}{2R}$
$\prod\sin\frac{A}{2}=\frac{r}{4R}$
$\sum\cos^{2}\frac{A}{2}=\frac{4R+r}{2R}$
$\sum\cos\frac{A}{2}=\frac{p}{4R}$
$\sum\tan\frac{A}{2}=\frac{4R+r}{p}$
$\sum\tan\frac{A}{2}\tan\frac{B}{2}=1$
$\sum\cos^{2}A=\frac{6R^{2}+4Rr+r^{2}-p^{2}}{2R^{2}}$
$\sum\sin^{2}\frac{A}{2}=\frac{2R-r}{2R}$
$r_{a}=\frac{pr}{p-a}$
$\sum r_{a}=4R+r$
$ab+bc+ca=p^{2}+r^{2}+4Rr$
$\sum r_{a}r_{b}=p^{2}$
$\sum\frac{1}{r_{a}}=\frac{1}{r}$
$\sum\frac{1}{r_{a}r_{b}}=\frac{4R+r}{p^{2}r}$
$\sum\frac{1}{r_{a}r_{b}}=\frac{4R+r}{p^{2}r}$
$\sum r^{2}_{a}=(4R+r)^{2}-2p^{2}$
$\sum h_{a}=\frac{p^{2}+r^{2}+4Rr}{2R}$
$\prod h_{a}=\frac{2p^{2}r^{2}}{R}$
$\sum\frac{1}{h_{a}}=\frac{1}{r}$

${II_{a}}^{2}=4R(r_{a}-r)$
$a,b,c,d$ toole zelhaye char zelyi va $m,n$ toole ghotrhaye an ast va $A,C$ do zaviyeye roberoo dar char zelyii hastand:
$m^{2}n^{2}=a^{2}c^{2}+b^{2}d^{2}-2abcd \cos(A+C)$
$r+R \leq h$
$P \geq \frac32 \sqrt{6rR}$
$3{MG}^{2}=\sum{MA}^{2}-\frac{1}{3}( \sum (AB)^{2})$
$BB_{1}$ va $CC_{1}$ nimsazhaye $B,C$ hastand:
$B_{1}C_{1}=\frac{abc}{(b+a)(c+a)R}OI_{a}$

inam chand ta por karbord:
1)$\sum a^{2}=4S(\sum cotA)$
2)$\sum x^{2}\geq\sum 2xycosA$,   hasel zarbha betore dori rooye 3 adade delkhahe haghighie $x,y,z$ taghyir mikone.
in yeki enghadr porkarborde ke migan age namosavii bar hasbe $p^{2}$,$R$ va $r$ bood bayad hatman ba in hal she:
$2R^{2}+10Rr-r^{2}-2(R-2r)\sqrt{R^{2}-2Rr}\leq p^{2}\leq 2R^{2}+10Rr-r^{2}+2(R-2r)\sqrt{R^{2}-2Rr}$
yani saabet kardan ke ghavi tar az in namosavi baraye $p^{2}$ vojood nadare.


If i can cut the lawn in 100 hours and my sister can cut the lawn in 1000 hours, how long would it take us to work together?

[hide="i think this is it..."]each of you can cut $\frac{1}{1000}$ of the lawn in one hour. so combined, you can cut $\frac{1}{500}$ of the lawn in one hour. therefore, it'll take 500 hours to cut the entire lawn with both of you.[/hide]

i made a typo 1000 hrs should be 100 hrs. Now solve the problem(look at original, its edited)

[hide=new answer]ok, so you can cut $\frac{1}{100}$ of the lawn in one hour and your sister can cut $\frac{1}{1000}$ of the lawn in one hour. so combined you can cut $\frac{11}{1000}$ of the lawn in one hour. and $\frac{11}{1000}\approx 91  hours.$

i don't know, i just made this up.  I think you are right though

hi gr8sk8r:
Your answer is correct but it's [tex] \frac {1000}{11}[/tex] hours

It's true because how did you get 500 hours?  500 hours is more combined together than alone.  oh wait there are reedits.  So if you have work rates its kind of like average speeds only it's not harmonic mean but harmonic sum.


function
algebra
domain
functional equation
real analysis
real analysis unsolved

hello !
find all functions such that  : $ f : R\minus{}\{0,1\} \minus{}\minus{}\minus{}>R$ and $ f(x) \plus{} f(x\minus{}1)x \equal{} x\plus{}1$

Excuse me, but before I try solving it, is it

$ f: \mathbb{R}\setminus\{0,1\}\to\mathbb{R}$

$ f: \mathbb{R}\setminus[0,1]\to\mathbb{R}$

or

$ f: [0,1]\to\mathbb{R}$?

Thank you.

it's $ f: \mathbb{R}\setminus\{0,1\}\to\mathbb{R}$

[hide="solution"]
Let $ x \equal{} 2$. Then $ f(2) \plus{} 2f(1) \equal{} 3$. But by definition $ f(1)$ is undefined. Therefore, 2 is not in the domain of $ f$. Therefore, there is no function $ f: \mathbb{R}\setminus \{0,1\}\to\mathbb{R}$ satisfying the above requirements. Q.E.D.
[/hide]

foxjwill, I don't think you're interpreting the question correctly. The functional equation is assumed to hold only where it is defined, namely when $ x \not \in \{0,1,2\}$.

I assume the question is not what as intended, but I don't know the fix.  As written: we may choose $ f(2), f(\minus{}1)$ and $ f(r)$ for $ r \in (0, 1)$ arbitrarily; these choices lead to no contradictions and they determine the values of the function on the given domain.


algebra
polynomial
quadratics
quadratic formula

[hide="Generalization"] Consider the polynomial $P(x) = a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{1}x+a_{0}$.  Then

$x^{n}P \left( \frac{1}{x}\right) = a_{n}+a_{n-1}x+...+a_{1}x^{n-1}+a_{0}x^{n}$

Has coefficients the reverse of $P(x)$ and roots clearly the inverse. [/hide]

Sorry I don't follow your method.

Is there another easier method to use?

[quote="Interval"]Show that the real solutions of the equation ax^2 + bx + c = 0 are the reciprocals of the real solutions of the equation cx^2 + bx + a = 0.  Assume that b^2 -4ac > or = 0.[/quote]
[hide]It is necessary to use both versions of the quadratic formula: $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ and $x=\frac{2c}{-b\pm\sqrt{b^{2}-4ac}}$.

For the first equation, $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.

For the second one, $x=\frac{2a}{-b\pm\sqrt{b^{2}-4ac}}$ (The numerator is $2a$ because $a$ and $c$ are switched).

The solutions are reciprocals.[/hide]

I have a problem with that method because it is [i]very[/i] hard to generalize (in fact, impossible for a quintic or higher), and it doesn't really get at [i]why[/i] such a thing would be true.

[hide="The generalization I gave as applied to a quadratic"] Let $P(x) = ax^{2}+bx+c$, which has some roots $r_{1}, r_{2}$.  Then

$P \left( \frac{1}{x}\right) = a \frac{1}{x^{2}}+b \frac{1}{x}+c \implies$
$x^{2}P \left( \frac{1}{x}\right) = cx^{2}+bx+a$

Because $P(x) = 0 \implies x = r_{1}, r_{2}$ it follows that $P \left( \frac{1}{x}\right) = 0 \implies x = \frac{1}{r_{1}}, \frac{1}{r_{2}}$.  Hence the second quadratic has roots the inverse of the first. [/hide]


1. Consider the system $(\mathbb{R}^{2},+, *, \mathbb{C})$ in which the scalars are the complex numbers and the multiplication by a scalar rule is $z*(a,b) = (za,zb)$. Show that system is not a vector space.

2. Consider the system $(\mathbb{R}^{2},+', *', \mathbb{R})$. Given that:

$(a,b)+' (c,d) = (a+2d, b+c)$
$\lambda *'(a,b) = (\lambda a, 2 \lambda b)$. ($\lambda \in \mathbb{R}$)

Is the system a vector space?

[quote="BanishedTraitor"]1. Consider the system $(\mathbb{R}^{2},+, *, \mathbb{C})$ in which the scalars are the complex numbers and the multiplication by a scalar rule is $z*(a,b) = (za,zb)$. Show that system is not a vector space.[/quote]
No, it aint closed.
$i(1,1) = (i,i)$
But $i\not \in \mathbb{R}$.
[quote]
2. Consider the system $(\mathbb{R}^{2},+', *', \mathbb{R})$. Given that:

$(a,b)+' (c,d) = (a+2d, b+c)$
$\lambda *'(a,b) = (\lambda a, 2 \lambda b)$. ($\lambda \in \mathbb{R}$)

Is the system a vector space?[/quote]
It is.
It is closed under vector addition cuz $a+2d,b+c \in \mathbb{R}$.
And it is closed under scalar multiplication since $\lambda a, 2\lambda b \in \mathbb{R}$.

Here is a theorem I used.

[b]Theorem:[/b] Let $V$ be a vector space over a field $F$. $\{ \}\subset W\subseteq V$ such that $\alpha+\beta \in W$ for all $\alpha, \beta \in W$ and $a\alpha \in W$ for all $\alpha \in W$ and $a\in F$ then $W$ is a subspace of $V$.

Now, I am not sure what JBL means when he says I am 1 out of 2. I think that he might be referring to example number 1. I believe I misunderstood that example, do you mean $\mathbb{C}$ is a vector space over $\mathbb{R}$? If so, then it is a vector space.

[quote="JBL"]1. This is linear algebra.  It should be in the linear algebra forums.

2. [url=http://www.artofproblemsolving.com/Wiki/index.php/Vector_space]Here[/url] is a list of vector space axioms.  Check them.  (Sylow_theory is 1 for 2.)[/quote]

Thanks so much JBL.

I have no idea how you think you used that theorem in this problem.  You are correct about number 1: that isn't a vector space because it's not closed under the operation of scalar multiplication, a neccessary condition.  For the second example, you showed the given set and operations were closed under the "scalar multiplication" operation.  That isn't a [i]sufficient[/i] condition to conclude that something is a vector space.


If we know that  $(x^2+y^2)^2+4x^2y^2=2(x+y)$  and  $x,y\in R$

then find the max of       $x-y$

$(x^2+y^2)^2+4x^2y^2=2(x+y)\Longleftrightarrow \{2(x^2+y^2)\}^2+(4xy)^2=8(x+y).$
$\Longleftrightarrow \{(x+y)^2+(x-y)\2\}^2+\{(x+y)^2-(x-y)^2\}^2=8(x+y).$
$\Longleftrightarrow 2\{(x+y)^4+(x-y)^4\}=8(x+y).$
$\Longleftrightarrow (x-y)^4=4(x+y)-(x+y)^4.$ Let $x+y=t,$ we have $(x-y)^4=4t-t^4.$
Since $x,y\in \mathbb{R} \Longleftrightarrow x+y\in \mathbb{R}, x-y\in \mathbb{R},$ the domain of $t$ is $0\leq t\leq \sqrt[3]{4}.$ Now let $f(t)=4t-t^4,\ f'(t)=4(1-t)(1+t+t^2).$ Thus $f(t)$ is maximized at $t=1,$ yielding the desired maximum value of $\sqrt[4]{3}.$


integration
calculus
function
complex analysis
complex analysis unsolved

Let
\[ f(p) \equal{} \int_{\minus{}\infty}^\infty (k^2\plus{}1)^p dk\]

Find an analytic continuation of $ f$ to the entire complex plane.

You should be more careful with your words here: when $ p\to \minus{}\frac 12$, you certainly have a singularity. ;)

Good point.  How about an analytic continuation to the entire complex plane, except at most countably infinitely many points?

This may be not the most elegant way to approach the problem but it is, perhaps, the most straightforward one. Rewrite the integral as $ 2\int_0^\infty$ and notice that it is well-defined and analytic for $ p<\minus{}\frac 12$. Note that $ \int_0^2 (x^2\plus{}1)^p\,dx$ is an entire function, so it extends trivially. To extend $ \int_2^\infty (x^2\plus{}1)^p\,dx$ write
\[ (x^2\plus{}1)^p\equal{}x^{2p}\left[1\plus{}px^{\minus{}2}\plus{}\frac{p(p\minus{}1)}2x^{\minus{}4}\plus{}\dots\right]\]
and integrate formally to get
\[ \minus{}\frac 1{2p\plus{}1}2^{2p\plus{}1}\minus{}\frac{1}{2p\minus{}1}p2^{2p\minus{}1}\minus{}\frac 1{2p\minus{}3}\frac {p(p\minus{}1)}2 2^{2p\minus{}3}\minus{}\dots\]
The series will converge geometrically for all $ p\in\mathbb C$ except $ \minus{}\frac 12,\frac 12,\frac 32,\dots$ where we'll have simple poles.

Another way of doing it:

One way of writing the gamma function is
\[ \Gamma(z) = n^z \int_0^\infty e^{-nt}t^{z-1}dt\]
for $ n>0$, Re$ (z)>0$.  The original $ f(p)$ converges when $ p<-\frac12$, so let $ z = -p$ and $ n = k^2+1$.  So
\[ \Gamma(-p) = (k^2+1)^{-p} \int_0^\infty e^{-(k^2+1)t}t^{-p-1}dt\]

Solve for $ (k^2+1)^p$:
\[ (k^2+1)^p = \frac1{\Gamma(-p)} \int_0^\infty e^{-(k^2+1)t}t^{-p-1}dt\]

Integrate from $ -\infty$ to $ \infty$:
\[ \int_{-\infty}^\infty (k^2+1)^p dk = \frac1{\Gamma(-p)} \int_{-\infty}^\infty \int_0^\infty e^{-k^2t}e^{-t}t^{-p-1}dtdk\]

Switch the order of integration on RHS:

\begin{eqnarray*}
  \displaystyle\int_{-\infty}^\infty (k^2+1)^p dk &=& \displaystyle\frac1{\Gamma(-p)} \displaystyle\int_0^\infty \displaystyle\int_{-\infty}^\infty e^{-k^2t}e^{-t}t^{-p-1}dkdt \\
   &=&\frac1{\Gamma(-p)} \int_0^\infty \left(\sqrt{\frac\pi t}\right)e^{-t}t^{-p-1}dt \\
   &=&\frac{\sqrt\pi}{\Gamma(-p)} \int_0^\infty e^{-t}t^{-p-3/2}dt \\
   &=&\frac{\sqrt\pi}{\Gamma(-p)}\Gamma(-p-1/2)
\end{eqnarray*}

So the  continuation requested is

\[ \sqrt\pi\ \frac{\Gamma\left(-p-\frac12\right)}{\Gamma(-p)}\]


Given that $x$, $y$, and $z$ are positive integers, how many triplets $(x,y,z)$ satisfy the equation
\[\frac1x+\frac1y+\frac1z=1? \]
[hide="reminder"] the order doesn't matter here, so include all of them[/hide]

[hide]First off, we know that since there are three fractions, the minimum of the largest fraction must be 1/3.
With the largest fraction as 1/2 we have:
1/4+1/4+1/2
1/6+1/3+1/2
Now, we know that we have no reciprocal of integers between 1/3 and 1/2.  That forces 1/6+1/3+1/2 to be the sum with the largest second fraction.  So we proceed to 1/3, yielding:
1/3+1/3+1/3.  This is the last combination.
We have:
6 ways for organizing 1/6+1/3+1/2
3 ways for organizing 1/4+1/4+1/2
1 way for organizing 1/3+1/3+1/3
For a total of 10 triplets.
[/hide]


show that there are infinitely many pairs(x,y) of rational numbers such that [b]x^3 = 48 +y^2[/b]

 (one of them is (4,4)) :wink:

[quote="ZetaX"]Two words: elliptic curve.[/quote]
It is not sufficiently. We must chek, that $ (4,4)\plus{}(4,4)\plus{}...\plus{}(4,4)\equal{}n(4,4)\not \equal{}0$ in elliptic curve.

[quote="ZetaX"]Two words: elliptic curve.[/quote]

 What is the method of elliptic curve  :?:  please explain

[quote="M.A"][quote="ZetaX"]Two words: elliptic curve.[/quote]

 is there any good book which explain elliptic curve well :?:[/quote]
 please tell me a good book

 thanks a lot


inequalities
algebra
polynomial
inequalities proposed

If $a,b,c$ a non-negative numbers, then
$\frac{1}{a^2+2bc}+\frac{1}{b^2+2ca}+\frac{1}{c^2+2ab} \ge \frac{2}{ab+bc+ca}$.

Your solution is magnificent, Darij.
Can you use this method to show the below stronger inequality
$\frac{1}{2a^2+3bc}+\frac{1}{2b^2+3ca}+\frac{1}{2c^2+3ab} \ge \frac{4}{3(ab+bc+ca)}$?

For this inequality, equality occurs when $(a,b,c)=(0,1,1)$ and $(a,b,c)=(\infty,1,1)$.

[quote="Vasc"]What say you about this ? It is not very hard.

If $a,b,c$ a non-negative numbers, then
$\frac{1}{2a^2+bc}+\frac{1}{2b^2+ca}+\frac{1}{2c^2+ab} \ge \frac{2}{ab+bc+ca}$.[/quote]

I have a solution by brute force and what is your solution?

[quote="Vasc"]What say you about this ? It is not very hard.

If $a,b,c$ a non-negative numbers, then
$\frac{1}{2a^2+bc}+\frac{1}{2b^2+ca}+\frac{1}{2c^2+ab} \ge \frac{2}{ab+bc+ca}$.[/quote]
WLOG, assume $a\ge b\ge c$. Note that $\frac{ab+bc+ca}{2c^2+ab}-1=\frac{c(a+b-2c)}{2c^2+ab}\ge0$. It suffices to prove\[\frac{1}{2a^2+bc}+\frac{1}{2b^2+ca}\ge\frac{1}{ab+bc+ca}\] Expanding, we obtain $2ab(a-b)^2+cA\ge0$, where $A\ge0$.

Yes. This is also my solution. :lol:
But it doesn't work for the original
$\frac{1}{a^2+2bc}+\frac{1}{b^2+2ca}+\frac{1}{c^2+2ab} \ge \frac{2}{ab+bc+ca}$,
which was solved by Darij in your nice manner.

It remains to prove the hardest case:
$\frac{1}{2a^2+3bc}+\frac{1}{2b^2+3ca}+\frac{1}{2c^2+3ab} \ge \frac{4}{3(ab+bc+ca)}$.

[quote="TomciO"]Excuse me, but how have you found that identity :O? It looks like magic... I would be grateful if you could explain the method...[/quote]

hmmm... I think he observed that in his inequality we have equality for $a=b$, so $(a-b)^2$ must be a factor. As for the $2a^2+\ldots+ab+\ldots$ term, maybe he used just brute force; i don't see any nice way to calculate it. Anyway the hardest part is actually thinking of that stronger inequality.

How do you prove this: if $p(a,b,c)$ is a symmetric polynomial and $a=b \Rightarrow p(a,b,c)=0$ then $(a-b)^2 | p$. (the fact that $(a-b)|p$ is trivial, but I don't know how to use the fact that it is symmetric) [Thanks, Tomci0]

Vasc, do you have a proof of your last ineq?

[quote="Vasc"]
It remains to prove the hardest case:
$\frac{1}{2a^2+3bc}+\frac{1}{2b^2+3ca}+\frac{1}{2c^2+3ab} \ge \frac{4}{3(ab+bc+ca)}$.[/quote]
$\sum_{cyc}\frac{1}{2a^2+3bc}\geq\frac{4}{3(ab+ac+bc)}\Leftrightarrow$
$\Leftrightarrow18\sum_{cyc}ab(a-b)^2(c-a)(c-b)+21\sum_{sym}a^3b^2c+49a^2b^2c^2\geq0$ :)

[quote="darij grinberg"]We have a stronger inequality:

$\frac{1}{a^2+2bc}+\frac{1}{b^2+2ca}+\frac{1}{c^2+2ab} \geq \frac{2}{bc+ca+ab}+\frac{1}{a^2+b^2+c^2}$.

In fact,

$\left(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ca}+\frac{1}{c^2+2ab}\right)-\left(\frac{2}{bc+ca+ab}+\frac{1}{a^2+b^2+c^2}\right)$
$=\frac{\left(b-c\right)^2\left(c-a\right)^2\left(a-b\right)^2\left(2a^2+2b^2+2c^2+bc+ca+ab\right)}{\left(a^2+2bc\right)\left(b^2+2ca\right)\left(c^2+2ab\right)\left(a^2+b^2+c^2\right)\left(bc+ca+ab\right)}\geq 0$.

  Darij[/quote]
It's nice of course, but we need to find it during competition (easy to see that the inequality is an identity for $a=b$),
while $uvw$ gives a proof immediately.


i'm sticking with AIM.
IM if you want to know what kind of fish has two knees?  :D
or what does snoop dogg wash his clothes with? (took this one from miracouluspostmaster)
sn - genius2401

[quote="stupidkid"]Which instant messenger do you use?
I personally use Trillian and recommend it to all of you. Go here to learn more http://www.trillian.cc[/quote]
It seems that you are the only one who uses trillian... :D
I use MSN and think it is a very good instant messenger.

I use MSN and AIM, though MSN first so I put that... I personally prefer MSN over AIM, but for some reason EVERYONE uses AIM so I had to get it.

I heard Trillian was pretty good; never tried it for myself though.

msn and aim.... i dont see using more than 2. They all serve the same purpose.

Isn't Trillian an all in one thing?

I should try that out, but I only use DeadAIM since MSN is bulky for me

Yeah, trillian is all in one.  It's nice because you can have as many sn's on at once as you like.  I don't need that many sn's on and (at least last time I used it) there were a few things that I didn't like as compared to AIM.  For one, it would crash more often.

i have msn and aim but i only use msn cuz pretty much evry1 outside U.S. uses msn. plus, in my opinion msn is alot cooler!! especially the new version!

I use msn mainly but I may try GAIM soon...

GAIM is cool, I like how a lot of it is customizable....You can substitute the real names of people instead of screen names for example

Where is the none option? I put "other".

[quote="jli"]i think you should put "none" as an option[/quote] :| No Messenger system??? I shudder to think what I'd do without messengers. As for me, I use two: AIM and MSN. MSN is better but since more people use AIM, I end up using AIM. I may download Trillian.

I put AIM but I've used MSN, Hello (go to your gmail inbox and click new features) and Xifre (for gaming).

[quote="Jhyon8149"][quote="jli"]i think you should put "none" as an option[/quote] :| No Messenger system??? I shudder to think what I'd do without messengers. [/quote]

me too! IM is amazing (i have AOL) especially when you get stuck in different classes than your friends, andnever get to see them. plus, you can talk to like sooo many ppl at once, it's great!


probability
limit
absolute value
number theory proposed
number theory

Prove that "almost all" positive numbers eventually reach a multiple of $5$ under the Collatz mapping:

$f(2n) = n, f(2n+1) = 3n+2$.

By almost all, I mean that if the probability of picking a number with this property out of the first $n$ positive integers is $p(n)$, then $\lim_{n\to\infty}p(n) = 1$.

For instance, $7\to11\to17\to26\to13\to20$, and $20$ is a multiple of $5$, so that makes $7$ a number with this property. On the other hand, no power of $2$ has this property.

I proved it, but my proof was a bit... involved. [size=75](At least I stopped myself from solving that quartic...)[/size] :D

Hmm... If I understand it right, what happens with a number during the first $N$ steps is completely determined by its remainder modulo $5\cdot 2^{N}$ and every scenario (by which I mean the starting point modulo $5$ and the fixed sequence of operations $n\mapsto n/2$ or $n\mapsto (3n+1)/2$) is realized for exactly one such remainder. The moral is that we are essentially talking about a random walk (or, if you prefer, Markov process) on the set $0,1,2,3,4$ with time independent transition probabilities. So, the only thing we need to check is that the $0$ state can be reached from each other state, which is a routine exercise in modular arithmetic.

[quote="fedja"]Hmm... If I understand it right, what happens with a number during the first $N$ steps is completely determined by its remainder modulo $5\cdot 2^{N}$ and every scenario (by which I mean the starting point modulo $5$ and the fixed sequence of operations $n\mapsto n/2$ or $n\mapsto (3n+1)/2$) is realized for exactly one such remainder. The moral is that we are essentially talking about a random walk (or, if you prefer, Markov process) on the set $0,1,2,3,4$ with time independent transition probabilities.[/quote]

Pretty much my train of thought so far...

[quote]So, the only thing we need to check is that the $0$ state can be reached from each other state, which is a routine exercise in modular arithmetic.[/quote]

Huh? Could you please elaborate on this?

My proof, from here, involved setting up a system of five recurrence relations and then finding the limit of their sum over $5\cdot 2^{N}$ as $N$ went to infinity, or, alternatively, checking that a certain irreducible quartic had no roots with absolute value $\ge 2$. Which definitely wasn't a routine exercise in modular arithmetic. (By the way, the same trick works for $7$ after quite a bit of work and  an irreducible quintic :D )

So what's this other method you mention so cryptically?

Edit: A little bit of research tells me that you must be talking about absorbing Markov chains! Now I get your proof. So that means I can very easily generalize this result. :lol:


MATHCOUNTS
function
analytic geometry
Gauss
FTW
geometry

This test will consist of two portions: Sprint and Target. The Sprint will have 30 questions, each worth 1 point, to be completed in 40 minutes. The Target has 8 questions, each worth 2 points, to be taken in pairs with 6 minutes per pair.

There is no need to sign up. Simply send me a private message in the following form:

Subject: Mock MathCounts
Message Body:
Sprint -
1. Answer
2. Answer
3. Answer
...
30. Answer

Target -
1. Answer
2. Answer
...
8. Answer

Please note that you do not have to send the Sprint and Target all at once. You may complete them at different times and send them in separately, if you like.

[b]I am also looking for people to look over the test and/or grade them. PM me if you are interested or post here.[/b]

[size=150]Rules[/size]
[list]
[*] On the Sprint Round, you may [b]not[/b] use a calculator of any kind, or any program, device, or other such object other than your own brain, pencil, and paper. Participants who have chips or any other mechanical device implanted in their brains for non-medical reasons(i.e., to increase IQ, form more synapses, etc.) are not eligible.
[*] For the Target Round, you may only use one of the following: a four-function calculator, a scientific calculator, or a graphing calculator. You may [b]not[/b] use any other device or computer program.
[*] You [b]may not[/b] discuss any problems or answers with any other participant or non-participant until the deadline has passed and scores are announced. You may remark on how excellently a certain problem was written, but if you say anything remotely derogatory, you will be promptly tarred and feathered.
[*] If you use a book, a website, or other resource to determine the answer or method of getting the answer, I will find out, [b]because I am watching you[/b]. If and when  you commit such a heinous crime, I will hack into your computer and change your desktop to one  filled with animated, pink unicorns that you can not remove.
[*] When a problem asks for a common fraction, round decimal, or other form of expressing your answer, equivalents will [b]not[/b] be accepted. For instance, if you put 4/8 rather than 1/2, I will laugh and count it wrong.
[*] Any attempt to bribe me will result in instant disqualification, [b]unless the amount of money you offer is an adequate amount[/b]. Please note that blackmail is not possible, because I am a perfect person who has never done anything wrong and never will. If you do attempt to blackmail me, and fail to recognize that such a thing is impossible, you will immediately be dunked in a giant pot filled with honey.
[*] Attempts to bribe/blackmail a grader other than myself will result in the prompt confiscation of the briber/blackmailer's wallet and the bribee/blackmailee will be coerced into wearing a [b]frilly pink dress for 24 consecutive hours[/b]. The punishment of the bribee/blackmailee will not occur, however, if the bribee/blackmailee does not accept the bribe or give in to the blackmail and reports it instead.[/list]

The mock contest will be posted on January 18th, Sunday, at some time in the afternoon. The date may be changed - all depends on how soon I can actually finish writing it and how many people are interested in grading and looking over the test/answer key.

[b][u]Graders:[/u][/b]
5849206328x
xpmath
tinytim
athunder
Twin Prime Conjecture
Bugi

[b][u]Looker-Overs:[/u][/b]
xpmath
tinytim
Twin Prime Conjecture

Please note you may only do one of the following:

a) Take it
b) Grade it
c) Look Over it
d) Look Over it and Grade it

If there is a shortage of graders, one may grade other people's tests after submitting one's own answers.

No need to sign up, just send in your answers once the test is posted. :)

Also, due to some confusion, I regret having to clear up the following: to the extent of my knowledge, I am not able to actually hack into your computers and change the desktop to pretty pink animated unicorns. However, I may have been kidnapped by aliens and given dummy-proof instructions on how to do so, which are embedded in my sub-conscious, so if I were you I wouldn't cheat.

Yes, you may grade.

I have completed the first draft and sent it to the looker-overs. If no mistakes are found, I will post the test much earlier than expected. However, I anticipate 666 mistakes or so, so  it will probably require a second or third draft.

[quote="nikeballa96"]rofl i love the rules izzy. btw, i know theres no need to inform you, but i WILL be takingthe test. unlike the other 93048230 tests that i've signed up and fogotten to take.[/quote]
and one of them is mine... :o

Izzy ur really funny with teh rulez :D

So I'm definitely gonna sign up!

I proofread the odd questions and xpmath proofread the evens:

Out of the 19 problems I proofread:

2 problems were not do-able, and 2 problems were ambiguous, and 2 problems were missing information.

[b][u]Update[/u][/b]:

Finished Draft 2, sent to Looker-Overs. At this rate I can probably post the test today or tomorrow.

Ha, my desktop is a unicorn :P Just not animated. Or pink. Or unremovable.

As such, Izzy, you can just -powerofsuggestion- us to change our desktops ourselves xD.

Anyways, I'll unnescessarily say that I'll be looking forward to taking it.

[b][u]Update:[/u][/b]

The looker-overs are looking over Draft 2. Draft 3 will be the final draft and will be posted here.

And sure, Bugi.

And here it is! Don't forget to adhere to all the rules mentioned in the first post, and remember, the consequences are severe.

For now, there is no deadline. It all depends on how many submit.

I said to adhere to the rules. :(

* nick42 gets tarred and feathered.  :P

i thought mathcounts problems are resonable and logical, not a bunch of computation problems...

Central time is two hours behind EST, so 5:00.

[quote="isabella2296"]Central time is two hours behind EST, so [b]5:00[/b].[/quote]

 :)

It's 6:00.  Ask stevenmeow.  Or anyone from Texas.  Or http://www.time.gov/timezone.cgi?Central/d/-6/java

Ah, I'm sorry. I kept thinking Mountain time, for some reason.  :oops: So it's good?

Yes. When you come on, I'll whisper you and BOGTRO the password.

[quote="isabella2296"]leewongjao wins 3-2! Congratulations!

leewongjao will now verse FantasyLover. Please arrange a time.[/quote]

Hmm... izzy why'd you give a problem that you say most people have memmed??

[quote="BOGTRO"]Hmm... izzy why'd you give a problem that you say most people have memmed??[/quote]

Well, I meant that it's a common triple of side lengths, not really a flat-out mem.

[quote="AIME15"]Can I see the problems?[/quote]

Eh, I don't have the problems anymore, but if BOGTRO or leewongjao remember any they can share it.

The last one was:

Find the area of a triangle with sides of length 13, 14, and 15.

Oh, it's 84?

That is a common problem :)

wait holy crap there is a countdown round?


With this late time, Lynbrook HS still has not submitted the USAMO form yet.
[url]http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2008-ua/08-04-23_USAMOStudentCheckList.pdf[/url]

There are quite some beastly students there. It would be a shame to get them disqualified due to the school's late action.

Also, any Lynbrook students go to ARML? I did not see Joseph.

[quote="serialk11r"]No Lynbrook people are at ARML, AFAIK.[/quote]
Is there a special reason why?  Is it just the way it turned out, or is it the "too good for ARML" mindest?

Well, Joseph hasn't done it in 3 years. As for everyone else, I don't know, but personally, I didn't go because I wasn't available when the ARML was being hosted.


Let $ x,y,z>0$ so that $ xy\plus{}yz\plus{}zx\equal{}3$. Prove that:

$ \sqrt{1\plus{}x^4}\plus{}\sqrt{1\plus{}y^4}\plus{}\sqrt{1\plus{}z^4} \ge 3 \sqrt{2}.$

[quote="cnyd"]$ x^{4} \plus{} 1\geq 2x^{2}$ (AM-GM) $ \implies$
$ \sqrt {1 \plus{} x^{4}} \plus{} \sqrt {1 \plus{} y^{4}} \plus{} \sqrt {1 \plus{} z^{4}}\geq \sqrt {2}[x^{2} \plus{} y^{2} \plus{} z^{2}]\geq 3\sqrt {2}$ :wink:[/quote]
Isn't $ \sqrt{1\plus{}x^4}\geq \sqrt{2x^2}\equal{}\sqrt{2}x$ ... ?
Please correct me if I am wrong.

$ \sqrt { 2 (1\plus{}x^4)} \ge 1\plus{}x^2$ etc.

Hence, $ \sqrt { 2 (1\plus{}x^4)}\plus{}\sqrt { 2 (1\plus{}y^4)}\plus{}\sqrt { 2 (1\plus{}y^4)} \ge 3 \plus{} \sum x^2 \ge 3\plus{}\sum xy \equal{} 6$ and the result follows


trigonometry
inequalities
inequalities open

$\sin x+\sin y+\sin z\leq 3\sin\frac{x+y+z}{3}$  $\forall x,y,z\in [0;\pi]$

Is there a proof above inequality [b]not using[/b] Jensen's inequality?

We have: $\frac{\sin x+\sin y}{2}=\sin\frac{x+y}{2}\cos\frac{x-y}{2}\le\sin\frac{x+y}{2}\quad\forall\; x,\,y\in [0;\,\pi]$
Then: $\sin x+\sin y+\sin z+\sin\frac{x+y+z}{3}\le 2\sin\frac{x+y}{2}+2\sin\frac{x+y+4z}{6}\le 4\sin\frac{x+y+z}{3}$ and we have done  :D


ABCD is is cyclic with AB the diameter of the circumscribed circle. E and F are the feet of the altitudes from A and B to CD, repectively. Show that CE=DF.

[hide="Hint 1"]
Draw M, the feet of the altitude from O (center of the circle) to CD
[hide="Hint 2"]
MC=MD
[/hide]
[/hide]

[hide="Hint"]$ \triangle AED$ and $ \triangle ACB$ are similar which is also the case for $ \triangle BCF$ and $ \triangle ADB$ $ \Rightarrow$ $ DE\equal{}CF\equal{}\frac{BC.AD}{AB}$[/hide]

[hide]
Ok, so first, draw $ CA$ and $ DB$. You know that $ ADB$ and $ ACB$ are right triangles because $ AB$ is a diameter. Additionally, since $ ABCD$ is a cyclic quadrilateral, you know that $ \angle CDB \equal{} \angle CAB$ and $ \angle DCA \equal{} \angle DBA$. Thus, $ \triangle FBD \sim \triangle CBA$ and $ \triangle ECA \sim \triangle DBA$. Using the first similarity, you get $ \frac {DF}{DB} \equal{} \frac {CA}{AB}$. Using the second, you get $ \frac {EC}{DB} \equal{} \frac {AC}{AB}$. Substituting, you get $ \frac {DF}{CA} \equal{} \frac {EC}{AC}$ which reduces to $ DF \equal{} CA$ and we are done.
[/hide]

its trivial. ABFE is a trapezium. O is mp. of AB. perp. from O to mp.(M) of CD is again parallel to BF and AE. and then since OM parallel, and O mp. of AB. M is mp. of EF.and proved.


probability
combinatorics proposed
combinatorics

[quote="nuerofuzz"]A hand of cards consists of all 13 hearts from an ordinary pack. The cards are arranged in random order. Find the probability that the ace is somewhere in front of the king.
If anybody is able to answer this, please could you show how you worked it all out. Thanks[/quote]

What do you mean with "in front of" ?

If it is "before", then the probability is the same as the probability of having the king before the ace, so $ \frac{1}{2}$

If it is "beside", then :
You have $ 13!$ ordered positions.
You have $ 12!$ ordered positions of the $ 13$ cards less ace and then $ 2$ possibilities to place the remaining king (left or right of ace)
So the requested probability is $ \frac{12!*2}{13!}=\frac{2}{13}$

If it is anything else, please give some precision on the meaning of the question.
Thanks.

the answer is 1/2. Thanks for such a clear answer.


If $q$ is an integer that can be expressed as the sum of two integer squares, show that both $2q$ and $5q$ can also be expressed as the sum of two integer squares.

source: mandelbrot #2

[hide="Generalization"]
$(a^2 + b^2)(c^2 + d^2) = (ac \pm bd)^2 + (ad \mp bc)^2$
[/hide]


http://www.math.sc.edu/contest/1991/exam/PDF/exam1991.pdf
number 25

can someone explain to me how do this, because I dont think any of the answer choices are right.

Here's my solution:
[hide]
To get from A to B, you have to do 6 N moves and 6 E moves.  One such string of moves is NNNNNNEEEEEE.
There would be $\binom{12}{6}$ ways to order those N's and E's, if all the segments were there.  But we have to subtract the paths that use that 1 missing segment.
To get to the bottom of that segment from A, there are $\binom{7}{3}$ paths.  To get from the top of the segment to B, there are $\binom{4}{1}$ paths.
So it's $\binom{12}{6}-\binom{7}{3}\cdot\binom{4}{1}=\fbox{784}?$
[/hide]

You're right, that's not one of the given answers.

[quote="CircleSquared"]Here's my solution:
[hide]
To get from A to B, you have to do 6 N moves and 6 E moves.  One such string of moves is NNNNNNEEEEEE.
There would be $\binom{12}{6}$ ways to order those N's and E's, if all the segments were there.  But we have to subtract the paths that use that 1 missing segment.
To get to the bottom of that segment from A, there are $\binom{7}{3}$ paths.  To get from the top of the segment to B, there are $\binom{4}{1}$ paths.
So it's $\binom{12}{6}-\binom{7}{3}\cdot\binom{4}{1}=\fbox{784}?$
[/hide]
Ya thats exactly what I got, this must have been a typo because all the answers are way off.

You're right, that's not one of the given answers.[/quote]

Also you could just write in the number for each one and add...


Princeton
college
MIT
geometry
geometric transformation
rotation
ratio

Hi,
I got accepted at both Princeton and Caltech, and was hoping for some help in making my decision. I'm 100% sure I'm going to major in computer science. Could the more experienced point out the differences between the two - the math program, the cs program - depth, breadth, quality of lectures, core curriculum, the character of the undergrads there, the atmosphere... I've done my share of research, but because I'm an international, I can't visit, so any opinions are appreciated.
Another thing I would like to ask is the financial aid at Caltech - although the package is pretty decent - it says that student contributions are 1500$ \$$ and parental contributions are 2500$ \$$ + a loan of about 3500$ \$$ . Without the loan, it seems feasible, but with the loan, I feel like it's definitely going to get to be a burden. How does this financial aid compare to the rest of the international guys here accepted? How accurate are the estimates on the fin.aid award letter? How much does one earn with student jobs?
I know these are a lot of questions, but any feedback is welcome.

Wow, I really wish I was in your position.  The schools are different.  Caltech is very small, fewer than 1000 students in total, and is very hard-science oriented.  Princeton is a more mid-sized university and is much more diversified.  However, most people who say that Princeton's math department is one of, if not the strongest in the country.  As for CS, I don't know.  Personally, I would choose Princeton but it's your decision.

I don't know anything about CalTech so I'm not going to compare it to Princeton. However I know plenty about Princeton and can answer any questions you want.

Aid: Princeton aid is generous. Usually the actual aid is the same as or more than the estimate. If you still have a problem, you can often talk to the financial aid office and get a bit more (like 500 or 1000, more than that is probably too optimistic, and you have to have some sort of reason).

CS department: Depends on what you want to do in CS, but there are some pretty awesome dudes at Princeton. Tarjan is here, but I've heard he's a horrible teacher. Schapire on the other hand is said to be awesome. Sedgewick is decent, I took a class with him. It's kind of cool to learn the most efficient known general implementation of mergesort and then discover that he created it, and stuff like that.

At Princeton CS can be an engineering (BSE) or liberal arts (AB) degree. This has implications for how many classes you have to take and what kind. It also has implications for your independent work. If you care, I'll explain further.

Both math and CS departments have a LOT of both depth and breadth, but may be lacking a bit in "applied" stuff. Princeton's a pretty theoretical institution, except in a few engineering departments. So this is a factor you should consider.

The culture is also extremely different between the two universities, from what I've heard. You should consider this perhaps more than academic factors, as it will have a big impact on your life outside of your major, and both schools are obviously very good at CS. (Princeton is almost certainly much better at pure math, but probably equal or worse at applied math.)

Princeton has extensive core curriculum. It's not really a core curriculum in that there are no specific required courses, but there are a lot of broad categories in which you are required to take classes. There's less of this if you do BSE than if you do AB, but Princeton has so many interesting classes that I don't think it's painful at all to satisfy the distribution requirements. In fact I think it's a bad idea to completely focus on a single topic.

If you ask more specific questions I could probably be more helpful.

Wow, Xevarion, this was probably one of the more informative answers I got to my questions :)
I'm glad you can answer more specific questions, so here goes.
Aid: This really doesn't concern me. Princeton was more than generous. I almost don't have to pay anything, so no further questions here.
CS department: I've looked through the faculty, and I must admit that there are some impressive names. However, I've generally been advised not to put too much weight on it, because proffesors don't really pay that much attention to undergrads (I don't know how true this is).
I was sort of wondering what the differences between the BSE and AB degrees, so an explanation is appreciated -especially in regard to CS.
The lack of applied stuff is actually one of my main concerns - I've heard that other more engineering oriented schools like MIT and Caltech view it as much easier - in both depth and breadth. Is this true and will that affect somehow my oportunities afterwards, in respect to job and graduate school options?
Culture: What is the culture at Princeton like? A rather stereotypical image I've been getting from people is that students are preppy, but I usually don't trust stereotypes :) Are there many science oriented students - math, CS, physics? Or does the business and economy culture pervade more strongly? When here, do more of Princeton's students go to graduate schools, or do they find employment?
You mentioned individual work. Caltech is really known for its SURF program. How much research do undergrads get to do at Princeton? Since Princeton is WAY larger than Caltech, do the lecturers pay less attention to students than at Caltech (sorry, I know you said you don't know that much about Caltech, but any information is helpful). In regards to student culture, I've also heard that at Caltech PSets are done in groups and students really help each other. How is this at Princeton?
Whew, enough questions for now. I really appreciate your help.
I am wondering if someone who is at Caltech, or at least has answers to this sort of questions will join the discusion?

I can give some information about Caltech, but it's a fair bit dated (by about 5 years or so at this point).

[b]Focus:[/b]   There will likely be under 250 Freshmen entering Caltech in your year.  That means that the ~200 person Chemistry class you likely will take Freshman year will contain almost every

Caltech is much more focused on engineering and the sciences than Princeton is.  Every studentl still takes the usual humanities courses (averaging about 1 every trimester), but the departments in these areas are relatively much smaller than at Princeton.  The course selection is still very good, and varies sufficiently from term to term to counteract the relatively few courses offered each term.  Caltech sets things up so as to make it easy to double major in, say, Computer Science and economics/history/literature/etc, and usually several students a year do so.  It's even possible, but rarer, to major solely in Literature while you're at Caltech.

If you do that, though, you'll still have to take Quantum Mechanics.  The flip side to Caltech's focus is that almost every student there chose to come there due to their love of math and science, and it shows.   I viewed this as a plus, seeing it as part of a larger sense of shared community I felt while I was a student there.  Other people I know viewed it as a minus, and wanted a more diverse student body.   If Princeton has a stereotypical issue of "preppy", Caltech has one of "nerdy", and it's in some sense correct.

[b]Size:[/b] Caltech has around 900 undergraduates.  This (along with the shared math/science interest I mentioned above) leads to a very close-knit community.  For the first two terms, you will take largely the same courses, from the same Professors, as every other freshman in your class (the exceptions being those courses you place out of and humanities courses, which typically average around 15 people).   The coursework is intense, and intentionally so, but everyone around you will be going through the same struggle, and will help give you support.

Every Freshman at Caltech is matched (through a week long rotation system) to one of 8 undergraduate "houses".  These houses are a focal point for campus life, from social events to intramural athletics to nightly dinner, and students frequently remain affiliated with the same house until they graduate, whether or not they live there physically.   I've mentioned a shared sense of community already, and this is where it felt strongest.  In my opinion this is one of the greatest aspects of Caltech life.

(I have more to add, but not more time to do so -- I'll write more late tonight).

Continuing my previous post:

Three other aspects of life outside the classroom I wanted to touch briefly on (again, remember that this is from someone who was last at Caltech in 2003!  Any more recent grads/current students on here, feel free to correct me):

[b]The "Ratio"[/b] My class at Caltech was about 2/3 male.  This number has been approaching parity for quite some time (it was about 3/4 a few years before I entered, and is down to about 63% as of 2007), but it will have some impact on life there.   In terms of social life and dating though, I think the small size and insularity of the community makes a larger difference.

[b]Ditch Day[/b] Read up on it if you get a chance.  It'll give you a pretty good insight into the mind of a typical Techer.

[b]The Honor Code:[/b] Caltech has a (student-enforced) honor system that is quite far ranging, and (at least while I was there) believed in by both the students and the Professors.  Every quiz or exam I took there (save one given by a visiting professor from Holland) was run under this system -- I was given the exam in class and told to take it in a 3 (or 5, or 8) hour time slot of my choice in a location of my convenience.  The code extended outside the classroom as well -- the locks on the room doors in the on-campus housing were close to useless (a combination lock with about 1000 possibilities) when I was there, but the only time I ever heard of someone's room being broken into was when some enterprising housemates turned it into a swimming pool while they were gone for a week.

Returning to some of the other issues you mentioned:

[b]Math/CS[/b]  The core math program covers a very wide range, and I found after heading to graduate school that I was well (even overly so) prepared for the first year graduate courses.  The math program at Caltech is small in size (more as a consequence of the size of the university than anything else), but has a surprising amount of breadth.

I'm less qualified to talk about the CS department, but I know of a few people from my class who did a Math/CS double major before heading to graduate school (in one case going into algorithms, in another to graphics/computational physics) and doing quite well.

The SURF program is definitely worth taking advantage of here.

[b]Financial Aid[/b] Unfortunately, I can't give much information here.  Information about wage rates for student jobs is  [url=http://cit.hr.caltech.edu/policies/PM/pm10_1_page2.pdf]here[/url] though.

[quote="Andrej_pp"]CS department: I've looked through the faculty, and I must admit that there are some impressive names. However, I've generally been advised not to put too much weight on it, because proffesors don't really pay that much attention to undergrads [/quote]
As I said, I can't really compare Princeton to other universities, but my impression is that Princeton faculty are relatively good about this. I don't know much about the CS department, but in math at least, even the top professors are mostly pretty willing to work with undergrads.

[quote]I was sort of wondering what the differences between the BSE and AB degrees, so an explanation is appreciated -especially in regard to CS. [/quote]
You'd probably get a better answer to this from the official Princeton materials (see e.g. their website), but basically here are the key differences:
(1) Independent work -- in AB you generally do 1 or 2 junior papers and a senior thesis, which will be major research papers. In BSE, you would probably do one project, and you'll probably still write a paper but it will be more of an applied project then a research paper.
(2) Course requirements -- BSE requires you to take more courses overall, but there are fewer distribution requirements.
That's pretty much it. It's also important to keep in mind that for AB, you declare your major at the end of sophomore year, but if you do BSE you declare your major at the end of freshman year. For this reason (and the course req one) it's easier to switch from BSE to AB.

[quote]The lack of applied stuff is actually one of my main concerns - I've heard that other more engineering oriented schools like MIT and Caltech view it as much easier - in both depth and breadth. Is this true and will that affect somehow my oportunities afterwards, in respect to job and graduate school options?[/quote]
If you want to go to engineering graduate school or get an industry job, this is possible. I'm not really sure. But doing good summer internships could make up for that if you come here. It's not like your prospects would be [i]bad[/i] for grad school or getting a good job if you have a degree from Princeton...

[quote]Culture: What is the culture at Princeton like? A rather stereotypical image I've been getting from people is that students are preppy, but I usually don't trust stereotypes :) Are there many science oriented students - math, CS, physics? Or does the business and economy culture pervade more strongly? When here, do more of Princeton's students go to graduate schools, or do they find employment? [/quote]
(1) There are those people here, but not so much in the sciences and engineering. Princeton's very diverse. It's pretty [i]quiet[/i], though. [url=http://www.princeton.edu/~paw/web_exclusives/plus/plus_012605orangebub.html]Observe![/url] There's no culture of pranks or hacks, no wild traditions, no passionate support of the sports teams, etc. The only cultural thing that's widely characteristic of Princeton is the eating club system.
(2) No. Each year, there are normally ~20 math majors and physics majors (each), 40 CS majors. (Rough estimates -- variance is high.) There are a lot more in some of the other engineering departments, though. But really, the top 5 departments are English, Politics, Econ, History, and the Woodrow Wilson School (politics+) (not in that order), and I think this accounts for something like half of the student body. This doesn't mean that there isn't a culture of science and research here though. It's just not dominant.
(3) More than 50% of Princeton graduates now go into jobs related to investment banking. I don't know statistics, but I'd wager that a majority of the rest of the people go to graduate school either immediately or after 1-2 years abroad/working/etc. You could probably find more detailed information about particular departments (try departmental websites). I think more than 50% of math majors go on to graduate school.

[quote]How much research do undergrads get to do at Princeton?[/quote]
As much as you want. Generally, you can just go up to a professor and ask if you can get involved in his or her research, and they're generally delighted. There isn't really any formal program the way they have at Caltech and some other places. That's actually pretty characteristic of Princeton -- tons of opportunities are there, but they aren't offered to you outright. You have to go out and take them.

[quote]Since Princeton is WAY larger than Caltech, do the lecturers pay less attention to students than at Caltech (sorry, I know you said you don't know that much about Caltech, but any information is helpful).[/quote]
This suggests faulty reasoning. What you really should be concerned about isn't the size of the Caltech student body but the student to faculty ratio! You could have a school with 500 total undergrads, and 15 professors -- that would suck!  I don't know what Caltech's student to faculty ratio is. In Princeton, it varies a lot by department. For example, in math there are more faculty than students! Anyway, lecturers typically don't go out of their way to get involved with students, but it's usually very easy to go talk to them about whatever you want. It's just like what I said above.

[quote]In regards to student culture, I've also heard that at Caltech PSets are done in groups and students really help each other. How is this at Princeton?[/quote]
This varies from class to class, but in general it's pretty common at Princeton too. We have an honor code that's pretty similar (from what I can tell from the post above mine) to Caltech's, so professors typically trust students to work together on assignments without cheating.

The department of CS at Princeton is indeed very theoretical. However, I think that should not be a problem. Undergraduate programs (at most colleges) don't focus on too practical things anyway. Their goal is to give students good general background, i.e. make students familiar with basic algorithms and data structures, general concepts of computer architecture, operating system, network and database design etc. However, usually students don't study modern technologies like .NET, EJB, AJAX at college (which change every 5 years).  It will not be hard to learn them when you start working in a software company. Also it might be a good idea to do a summer internship or to work on a project on your own.

Also software companies, especially "cool" software companies like Facebook, Google or Microsoft, are usually looking for candidates who have a good knowledge of algorithms/data structures/general concepts rather than those who know specifics of a particular technology when they hire recent college graduates.

So do companies like that recruit on campus?

Andrej_pp, many of them do (e.g. see http://www.cs.princeton.edu/general/affil/fall ). Also you can always apply directly to a company you want to work for.

I too am interested in Princeton, however i am a junior soon to be senior. I was wondering more about the culture of Princeton. From what i hear and see, it's very nerdy-is that picture accurate? What about student activities and groups, is there as much interaction as a place like Harvard? Also, the academics, i'm under the impression that Princeton is incredibly hard and only USAMO/IPhO level students can do the work; and seeing as i'm only at the AIME level that would be difficult. In the math program do they dive right into proofs or does it start out with some computational, creative, problem solving? Personally I don't think i'm ready for proofs having a bad experience at a local college class where i joined mid year when they were doing number theory proofs and the professor, a princeton alumnus, wouldn't teach me anything but said the first 1000 hours are the hardest-sorry for run-on.
As you can see, while i like what i've read of Princeton, the media and pictures present an isolated environment where most people are either taciturn or snobby. Could anybody shed some light on this, thanks in advance!

Anyone on the previous post? Please have more sympathy towards him (or her. How should I know?). This community is supposed to be helping each other, right? Please be more sensitive.

enigma - don't worry about not being a USAMO person (yet) -- there are plenty of very good people at Princeton who never qualified for the USAMO, and they're doing just fine.  And don't worry about the proof/find the answer distinction.  Just worry about honing your general problem solving skills - the tests in college are a lot more like the AIME than they are like high school.  You can't memorize your way through, very few people get very high scores, and you must master the fundamentals.  Above all, start working on each class very early in the semester -- I usually worked very far ahead in my classes in the first few weeks, when the workload from classes and extracurriculars hadn't piled up yet.  This made my classes *much* easier.  Work ethic is more important than genius in succeeding in school (and, by and large, in any career).

There are plenty of opportunities to meet your peers at Princeton.  Lots of student groups, lots of intramural sports, etc., same as Harvard (or any other similar school).  There's also the club system, which causes a social reshuffling of sorts after the second year (which a lot of people really need, whether they realize it or not).  I didn't find Princeton snobby, and there weren't many people I'd call taciturn.  It's also not particularly nerdy -- there are a lot people interested in all sorts of things.


abstract algebra
group theory
superior algebra
superior algebra unsolved

Prove that an uncountable abelian group has a proper subgroup of the same cardinality. Amazingly, the property does not hold if we not assume commutativity.

Can a man have some sleep around here? :) I just woke up and I'm in shock because of all the beautiful problems which have been posted during the last couple of hours :).

It's a well-known result in the theory of abelian groups that a homomorphism from some $H\le G$ to a divisible group (such as $\mathbb Q$) can always be extended to a homomorphism from $G$ itself to the respective divisible group.

Take any non-trivial element $a$ of our group and define a non-trivial homomorphism $f$ from $H=\langle a\rangle$ to $\mathbb Q$ (take $f(na)=n,\ \forall n\in\mathbb Z$, for example). We do the above with our $f$: we now have a non-trivial homomorphism (call it also $f$) from $G$ to $\mathbb Q$, and $\ker f$ is a proper subgroup of $G$ with the same cardinal as $G$, since $G/\ker f$ is countable, and there is a bijection from $G$ to the set $(\ker f)\times (G/\ker f)$ (and this last set has cardinal equal to $|\ker f|$).

I did it in the same way, but somehow I cannot figure out a counter-example if we do not assume the group abelian.
  PS: NO, you don't have the right to sleep! The most active mathlinker has to do his duty forever! I won't let you sleep, at least one month from now on!  ;)  :P  :P


inequalities
calculus
derivative
inequalities proposed

Let $x, y,z>0.$ Prove that: \[ \frac{9}{x+y+z}-\frac{1}{xyz}\leq 2. \]

beautiful...
my solution:
because $x+y+z \ge 3 \sqrt[3]{xyz}$
let $a= \sqrt[3]{xyz}$
we should just to prove $\frac 3 a -\frac 1 {a^3} \le 2$
it is equal to $(2a+1)(a-1)^2 \ge 0$

Zhaobin, after we got the inequality $\frac{3}{a}-\frac{1}{a^{3}}\leq 2$ you could have used AM-GM like this:

                                  \[ 2+\frac{1}{a^{3}}=1+1+\frac{1}{a^{3}}\geq 3\sqrt [3]{1\cdot 1\cdot \frac{1}{a^{3}}}\geq \frac{3}{a}.  \]

What do you think? :)

[quote="tkhtn"]Easy inequality.Can you do that:
Let:$(a^3+1)(b^3+1)(c^3+1)=729$;$a,b,c\ge 1$.
Find min of :$P=a^2+b^2+c^2+6$.[/quote]
Isn't this simple amgm $a^3+1=(a+1)(a^2-a+1) \le (\frac{a^2+2}{2})^2$
Multiply the three things and apply amgm yields the result that the minimum is held when a=b=c=2

Yes,exactly.This is main idea of ineq problem of APMO.

By AM-GM,
$ \frac{9}{x+y+z}\le  \frac{3}{\sqrt[3]{xyz}}=3\sqrt[3]{\frac{1}{xyz}\cdot 1 \cdot 1}\le \frac{1}{xyz}+2.$

[hide="My sol"]$\frac{9}{x+y+z}-\frac{1}{xyz}\leq\frac{9}{x+y+z}-\frac{27}{(x+y+z)^3}$

Denote $x+y+z=k$

$\frac{9}{k}-\frac{27}{k^3}\leq2$

$\frac{1}{k}-\frac{3}{k^3}\leq\frac{2}{9}$

$\frac{-2k^3}{9}+k^2-3\leq0$

Differentiating:

$f'(k)=\frac{-2k^2}{3}+2k=0$

This gives $k=0,3$

Taking the second derivative, we find that

$f"(k)=\frac{-4k}{3}+2$ is negative for $3$.

Thus at $3$ maximum is obtained, which is

$-6+9-3=0$. :) [/hide]


trigonometry
geometry
trig identities
Law of Sines

for any $ \triangle ABC,$ prove that the following equality holds: $ b^{2}\sin 2\beta \plus{} c^{2}\sin 2\gamma \equal{} 2bc\sin \alpha .$

My work shows it holds only for an isosceles triangle $ (b\equal{}c)$ or for a right triangle (hypotenuse $ a$). Other people, let me know if you got the same result.

In particular, $ b^{2}\sin 2\beta\plus{}c^{2}\sin 2\gamma \minus{} 2bc\sin\alpha \equal{} \minus{}2a \cot \alpha \sin^2(\alpha \plus{} 2\beta)$ (using the law of sines to eliminate $ b$ and $ c$), so the expression is equal to zero if and only if $ \alpha \equal{} \frac{\pi}{2}$ or $ \beta \equal{} \gamma$, as Farenhajt noted.

the problem was actually $ c^{2}\sin 2\beta\plus{}b^{2}\sin 2\gamma \equal{} 2bc\sin\alpha ,$ and the result follows immediately by applying the law of sines.

That's right. $ a$ is proportional to $ \sin \alpha$ etc, the given equality can be written as $ c\cos \beta \plus{}b\cos \gamma \equal{} a$, which is the first cosine law.


Let $ x,y,z$ be positive numbers such that $ x\plus{}y\plus{}z\equal{}1$. Prove that:
$ \frac {x}{y^2\plus{}z}\plus{}\frac {y}{z^2\plus{}x}\plus{}\frac {z}{x^2\plus{}y}\ge \frac{9}{4}$
When does equality hold?

By Cauchy Schwartz:

\[ \frac {x}{y^{2} \plus{} z} \plus{} \frac {y}{z^{2} \plus{} x} \plus{} \frac {z}{x^{2} \plus{} y}\geq \displaystyle \frac {(x \plus{} y \plus{} z)^{2}}{\sum(x^{2}z \plus{} xy)}\]

It remains to prove :

\[ \displaystyle \frac {(x \plus{} y \plus{} z)^{2}}{\sum (x^{2}z \plus{} xy)}\ge \frac{9}{4}\Leftrightarrow\] \[ 4x^3 \plus{} 4y^3 \plus{} 4z^3 \plus{} 3x^2y \plus{} 3y^2z \plus{} 3z^2x\geq 6(xy^2 \plus{} yz^2 \plus{} zx^2) \plus{} 3xyz\]

which is easy by AM-GM.

[quote="MathVMF"]Let $ x,y,z$ be positive numbers such that $ x \plus{} y \plus{} z \equal{} 1$. Prove that:
$ \frac {x}{y^2 \plus{} z} \plus{} \frac {y}{z^2 \plus{} x} \plus{} \frac {z}{x^2 \plus{} y}\ge \frac {9}{4}$
When does equality hold?[/quote]

$ \frac {x}{y^2 \plus{} z} \plus{} \frac {y}{z^2 \plus{} x} \plus{} \frac {z}{x^2 \plus{} y}$
$ \equal{} \frac {x^2}{xy^2 \plus{} zx} \plus{} \frac {y^2}{yz^2 \plus{} xy} \plus{} \frac {z^2}{zx^2 \plus{} yz}$
$ \ge \frac {(x \plus{} y \plus{} z)^2}{xy^2 \plus{} yz^2 \plus{} zx^2 \plus{} xy \plus{} yz \plus{} zx}$
$ \ge \frac {(x \plus{} y \plus{} z)^2}{\frac {1}{3}(x \plus{} y \plus{} z)(y^2 \plus{} z^2 \plus{} x^2) \plus{} xy \plus{} yz \plus{} zx}$
$ \equal{} \frac {3}{1 \plus{} xy \plus{} yz \plus{} zx}$
$ \ge \frac {3}{1 \plus{} \frac {1}{3}(x \plus{} y \plus{} z)^2}$
$ \equal{} \frac {9}{4}$

See here for the strongers and harders: http://www.mathlinks.ro/viewtopic.php?t=300129
 :)

[quote=Tourish][quote="MathVMF"]Let $ x,y,z$ be positive numbers such that $ x \plus{} y \plus{} z \equal{} 1$. Prove that:
$ \frac {x}{y^2 \plus{} z} \plus{} \frac {y}{z^2 \plus{} x} \plus{} \frac {z}{x^2 \plus{} y}\ge \frac {9}{4}$
When does equality hold?[/quote]

I remember a related one( and I think it is a very hard one)
\[ \frac {x}{y^2 \plus{} z} \plus{} \frac {y}{z^2 \plus{} x} \plus{} \frac {z}{x^2 \plus{} y}\geq \frac {9}{3 \plus{} x \plus{} y \plus{} z}\]
If we let $ x \plus{} y \plus{} z \equal{} 1$,we can get the oringinal one.[/quote][url=http://www.artofproblemsolving.com/community/c6h1134592p5291753][b]From: Tran Quoc Luat[/b][/url]


There is a train of $n$ car organization to which the number from 1 to $n\geqq 2$ sequentially attaches from
the first vehicle.Is there kinds of how many ways of coloring the color that one side of the adjoined vehicle
at least becomes red when either of red,blue,or yellow each vehicle are painted by one color?

I do not understand the statement... :(  :?

Pierre.

Sorry for my poor English, Pierre.
I have just edited .How about this one?

kunny

[quote="kunny"]
such that at least of the adjoining vehicle will be painted with red[/quote]

This is the point I do not understand... What is an adjoining vehicle? One of the $n$? So maybe is a word missing (like 'at least one' or whatever...).

Pierre.

I have just edited,everyone.

Best regards.

kunny

[quote="kunny"]Is there kinds of how many ways of coloring the color that one side of the adjoined vehicle
at least becomes red when either of red,blue,or yellow each vehicle are painted by one color?[/quote]

 :?  :?  :?

Hmmm... sorry for asking again... Are you wondering for the number of colourings of $n$ cars in a row such that each car has at least one red neighbour? ;)

Pierre.

Can anyone understand Japanese?

Here is the sentence in Japanese.

先頭車両?ら順?１?ら$n$???番?????$n$両編??列車??る。???$n\geqq 2$　??る。?車両を赤色,?色,黄色?
??れ?一色?塗る???,　[b]隣り???車両?少????も一方?赤色??るよ??[/b]　色?塗り方?何通り?。

I will solve the following problem (with a real ugly solution) :

We consider a row of $n$ cars and we have three colors, red yellow and green.
Determine the number $f(n)$ of colourings of these cars such that each car has at least one of its neighbours colored in red.

Note that a car is not one of its neighbours, and that except for the first and the last cars each car has two neighbours.

We will denote a red car by $r$, and a non-red car by $x$ (so that there will always be two choices for a value of $x$).

A colouring with $n$ cars satisfying the statement will be called '$n$-good'. Note that each $n$-good colouring has its $(n-1)^{th}$ car colored in red (otherwise the $n^{th}$ would not have a red neighbour). (1)
Thus any $n$-good colouring ends by $rr$ or by $rx$.
Let $a_n,b_n$ be the number of $n$-good colourings ending by $rr$ and $rx$ respectively.
Clearly $f(n)=a_n+b_n.$

A $n$-bad colouring is colouring which is not good only because of the last car and this last car is red. Thus such a colouring ends with $xr$.

Now, consider any $(n+1)$-good colouring :
Delete the last car, it will give either a $n$-good colouring ending with $rr$ (because of (1) since both the $(n+1)$-colouring and the $n$-colouring are supposed to be good) or a $n$-bad colouring (since only the $n^{th}$ car has lost a neighbour after the deletion).

Conversely, it is easy to verify that any $n$-good colouring ending with $rr$ gives exactly one $(n+1)$-good colouring ending with $rr$ and exactly two $(n+1)$-good colourings ending with $rx$.
In another hand, any $n$-bad colouring gives exactly one $(n+1)$good colouring (we have to color the $(n+1)^{th}$ car in red) ending with $rr$.

It follows that $a_{n+1} = a_n + y_n$ and $b_{n+1} = 2a_n.$ (2)

Now consider any bad $(n+1)$-colouring :
Deleting the last two cars, that is $xr$, we obtain a $(n-1)$-good colouring since we delete only one red car, which is was neighbour to none of the remaining cars.
Conversely, any $(n-1)$-good colouring leads to exactly two $(n+1)$-bad colourings since we are forced to complete it with $xr$ (recall that a bad colouring must end by $r$).
Therefore : $y_{n+1} = 2(a_{n-1} + b_{n-1}).$ (3)

From (2) and (3), we deduce that $a_{n+1} = a_n + 2a_{n-2} + 4a_{n-3}.$
We are lucky that the characteristic equation has roots $2,-1,i\sqrt 2 , -i \sqrt 2$ so that the following is straightforward :
It is well-known that there exist some constant $a,b,c,d$ such that for all $n$, we have
$a_n = a \cdot 2^n + b \cdot (-1)^n + c \cdot (i \sqrt 2 )^n + d \cdot (-i \sqrt 2 )^n.$
Direct computations show that we have $a_2=1,b_2=0,y_2=2$ and $a_3 = 3,b_3 = 2, y_3 = 0.$ Thus, for convenience, using (2) and (3) we have $a_0=a_1=0.$
It follows that
$a_n = \frac 2 9 \cdot 2^n - \frac 1 9 \cdot (-1)^n + (- \frac 1 {18} + \frac 5 {36} i \sqrt 2) \cdot (i \sqrt 2 )^n + (- \frac 1 {18} - \frac 5 {36} i \sqrt 2) \cdot (-i \sqrt 2 )^n.$

Using that $f(n) = a_n + b_n = a_n + 2a_{n-1}$ we deduce that :
$f(2n) = \frac 1 9 (2^{2n+2} + 1 + (-2)^{n+2} )$
and
$f(2n+1) = \frac 1 9 (2^{2n+3} - 1 - 7 \cdot (-2)^n ).$

Pierre.

Thank you for your detail explanation,Pierre.

Probably I couldn't convey you the statement.

Because in this problem, For n=2,3 you will obtain the answers [b]5,11 [/b]respectively.

kunny

[quote="kueh"]so you are saying any three trains side by side, one of the two outside must be red?[/quote]

Sorry for my delay reply, [b]kueh[/b].

Yes. I am saying that.

kunny


For $ a > 0$ we define the sequence $ (x_n)_{n \ge 1}$ by the relation:

$ x_{n \plus{} 1} \equal{} a \plus{} a\cdot x_1 \cdot x_2 \cdot ... \cdot x_{n \minus{} 1} \plus{} (x_1 \cdot x_2 \cdot ... \cdot x_{n \minus{} 1})^2$, $ \forall n \ge 2$, with $ x_1 \equal{} 1$ and $ x_2 \equal{} 1 \plus{} a$.

$ a)$ Prove that:

$ P_n: x_n \equal{} a \plus{} x_1\cdot x_2 \cdot... \cdot x_{n \minus{} 1}$, $ \forall n\ge 2$

$ Q_n: x_{n \plus{} 1} \equal{} {x_n}^2 \minus{} ax_n \plus{} a, \forall n \ge 2$

$ b)$ Prove that:

$ \displaystyle\sum_{k \equal{} 1}^{n}\frac {a^k} {x_k} < a(a \plus{} 1), \forall n \ge 2$

[quote="csabafarkas"]Ai subiectele cls XI si-a XII?[/quote]
[hide="@csabafarkas"]Vezi pe forumul romanesc, sunt subiectele la cls 12[/hide]


Estimate the largest real number x such that 1 - exp(-2x^2) = 0 on a machine using
the IEEE standard for binary double precision floating point arithmetic. (Hint: a Maclaurin series
might be useful.)

What would be this number? Obviously, zero is a solution to the expression, but there must be a positive root somewhere. If not, then it should be zero + eps/2. How do you think you use the Maclaurin series?

[quote="BrainMan1"]Estimate the largest real number x such that $ 1 \minus{} \exp(\minus{}2x^2) \equal{} 0$ on a machine using
the IEEE standard for binary double precision floating point arithmetic. (Hint: a Maclaurin series
might be useful.)

What would be this number? Obviously, zero is a solution to the expression, but there must be a positive root somewhere. If not, then it should be zero + eps/2. How do you think you use the Maclaurin series?[/quote]
The MacLaurin series is $ 1 \minus{} \exp(\minus{}2x^2) \equal{} 2x^2 \minus{} 2x^4 \plus{} \frac{4}{3} x^6 \minus{}\plus{} \dots \equal{} 2 x^2 \plus{} O(x^4)$. You want this to be less than eps/2 (assuming rounding to the nearest machine number), hence $ 2x^2 < 2^{\minus{}52}/2$ or $ |x| < 2^{\minus{}27}$.


Prove that $l_{a}\le \sqrt{p(p-a)}$, $l_{a}$- bisector of angle that watches side a of triangle with sides a,b,c and $p=\frac{a+b+c}{2}$

We have $l_{a}=\frac{2bc}{b+c}.\sqrt{\frac{p(p-a)}{bc}}$, and we have done (because $b+c\geq 2\sqrt{bc}$).


The plane is painted with two colors. Show that we can always find an isosceles right triangle
whose vertex have the same color.

Let $ A$, $ B$ be two points of the same color (there have to be two points of the same color because there are infinitely many points and just 2 colors). Consider a square $ ABCD$ and its center $ E$. All triangles $ ABC$, $ ABD$, $ ABE$ are right-angled and isosceles, so if the problem statement is false, then $ C$, $ D$, $ E$ are all of the same color. But $ CDE$ is also an isosceles right-angled triangle. Hence we can always find a monochromatic isosceles right triangle.

one more problem
Thanks for [b]uglysolutions[/b],I think your solutions is right.


limit
ceiling function
integration
logarithms
calculus
calculus computations

find the limit  :D

$ \lim_{n\to\infty}\;\frac{1}{n}\;\sum_{m\equal{}1}^{n\minus{}1}\;\left(\lceil\frac{n}{m}\rceil\minus{}\frac{n}{m}\right)$

[hide]$ \text{answer: }\quad\gamma$[/hide]

[hide="Solution"]\begin{eqnarray*}\lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^{n-1}\left( \left\lceil \frac{n}{k}\right\rceil - \frac{n}{k}\right)
& = & \int_{0^{+}}^{1} \left( \left\lceil \frac{1}{x} \right\rceil - \frac{1}{x} \right) \, dx\\
& = & \sum_{n=1}^{\infty} \int_{\frac{1}{n+1}}^{\frac{1}{n}} \left( \left\lceil \frac{1}{x} \right\rceil - \frac{1}{x} \right) \, dx\\
& = & \sum_{n=1}^{\infty} \left( \frac{1}{n} + \ln n - \ln(n+1) \right)\\\
& = & \lim_{m\to\infty} \sum_{n=1}^{m} \left( \frac{1}{n} + \ln n - \ln(n+1) \right)\\
& = & \lim_{m\to\infty} \left( \mathrm{H}_m - \ln(1+m) \right)\\
& = & \gamma
\end{eqnarray*}[/hide]


The catalytic decomposition of N2O by gold at 850 C and at an initial pressure of 100 mm is 50% complete in 60 minutes and 70% complete in 100 minutes.

Find
(1) Order of reaction
(2) Velocity constant of reaction (What is that?)
(3) How much will decompose in 100 minutes at same tempeature but at an initial pressure of 600 mm?

1) first order reaction??
(since we have the equation $ \frac{1}{[A]^{n\minus{}1}_t} \minus{}  \frac{1}{[A]^{n\minus{}1}_{initial}} \equal{} (n\minus{}1)kt$ for an $ nth$ order reaction (very easy to derive)  :)

[quote="hell ever"] (2) Velocity constant of reaction (What is that?) [/quote]

Is it not $ k$ ?  :huh:  :maybe:

Ans- $ 1.925 X 10^{\minus{}4} s^{\minus{}1}$

If these are correct part c is easy :roll:

[quote="Da Vinci"]1) first order reaction??
(since we have the equation $ \frac {1}{[A]^{n \minus{} 1}_t} \minus{} \frac {1}{[A]^{n \minus{} 1}_{initial}} \equal{} (n \minus{} 1)kt$ for an $ nth$ order reaction (very easy to derive)  :)

[/quote]

That is only for $ n\ge 2$

i think they are correct  :maybe:  ....i checked these data using the equations of first order reaction....they are in agreement :P

well i got n=1 using the formula i gave!  :maybe:

@hell , u are also in BT ??
how much did u get in pt 1 ?
[hide]
i took it as 1 cos only that was giving answer to the next part :D and also because only 1 and 0 order reactions are in iit course :D
isn't velocity constant , rate constant ?[/hide]

well even i cud calculate ~using my lowly medical level knowledge~
so that must be one of the easier ques

and yes slogger's method is best--if we rem the formulas-then it wont take too much time to go even linearly

 :oops:

Actually we havent done chemical kinetics both in school or in my classes, but i am aware of the concepts but am not able to apply them in problems, so can you expand a bit?  :(

@ slogger
I gave BT test for the first time as the last time, the tests clashed with my class tests. I am sure you must have got top 20 AIR, even one of my friends, he got 31 AIR i think.

i won't like to expand when the likes of vince and ritu are here :P

anyways , i would be happy with even a top 100 , but the percentile and AIRs have not been given, wonder how ur friend got it !!