algebra
inequalities
three variable inequality
calculus
Canada

Let $ x$, $ y$, and $ z$ be non-negative real numbers satisfying $ x \plus{} y \plus{} z \equal{} 1$.  Show that
\[ x^2 y \plus{} y^2 z \plus{} z^2 x \leq \frac {4}{27}
\]
and find when equality occurs.

if a1,..., an are positive reals
is it true that:
a1/(a2+a3)+a2/(a3+a4)+...+an/(a1+a2)>=n/2

something is missing?!

cause it is false. i think in these conditions!

Let f(x; y; z) = x^2y + y^2z + z^2x. We wish to determine max f.
Since f is cyclic , WLOG we may assume that x  y, z. Since                      f(x, y, z) - f(x, z, y) = x^2y + y^2z + z^2x - x^2z - z^2y - y^2x=
= (y - z)(x - y)(x - z)

we may also assume y >= z. Then
f(x + z, y, 0) - f(x, y, z) = (x + z)^2y - x^2y - y^2z - z^2x
= z^2y + yz(x - y) + xz(y - z) >= 0;
so we may now assume z = 0. Now AM inequality:

f(x, y, 0) = 2x^2y/2<=1/2((x+x+2y)/3)^3=4/27

okay.. cheers!

Reading a book, I just found out a new solution for  the 1st problem:

Let f(x,y,z=x^2y+y^2z+z^2x. x>=y,z and we want the max of f.
f(x+z/2,y+z/2,0)-f(x,y,z)=yz(x-y)+xz(x-z)/2+z^2y/4+z^3/8. we also can asume WLOG z=0 and you do the same as in the 1st solution!!

g'luck! :D

I guess this problem is posted over 15 times.....

Just notice that if x<=y<=z
the expression increases if you change y to y+d and z to z-d.
So..... in optmial situation z=0. the rest is trivial.

There is a solution in [url=http://www.kalva.demon.co.uk/canada/casoln/csol995.html]Kalva[/url].

[quote="bugzpodder"]Let $x$, $y$, and $z$ be non-negative real numbers satisfying $x + y + z = 1$.  Show that
\[ x^2 y + y^2 z + z^2 x \leq \frac{4}{27}  \]
and find when equality occurs.[/quote]Making it look "better" with $\text{\LaTeX}$.[hide]Suppose, WLOG, $x\ge y\ge z\ge0$. Set $f(x,y,z)=x^2y+y^2z+z^2x$. Compare $f(x+\epsilon, y, z-\epsilon)$ and $f(x,y,z)$ for all inputs in the correct domain and $\epsilon>0$. \begin{eqnarray*}f(x+\epsilon, y, z-\epsilon) &=& (x+\epsilon)^2y+y^2(z-\epsilon)+(z-\epsilon)^2(x+\epsilon)\\ &=& \epsilon^3+\epsilon^2(x+y-2z)+\epsilon(2xy-2xz-y^2+z^2)+x^2y+y^2z+z^2x\\ &=&\epsilon^3+\epsilon^2(x+y-2z)+\epsilon(y-z)(2x-y-z)+f(x,y,z)\\ &>&f(x,y,z)\end{eqnarray*}For any $y$, increasing the difference between $x$ and $z$ increases the value of $f(x,y,z)$. Therefore, the maximum occurs when $z=0$. Then \[ f(x,y,z)\le \max(x^2y)  \] with $x+y=1$. By AM-GM \[ \frac{1}{2}x+\frac{1}{2}x+y\ge3\sqrt[3]{\frac{x^2y}{4}}  \] \[ \frac{4}{27}\ge x^2y  \] Therefore, \[ x^2 y + y^2 z + z^2 x \leq \frac{4}{27},  \] as desired.[/hide]

We can solve the following generalization, too :
Let $ x,y,z > 0$ such that $ x \plus{} y \plus{} z \equal{} 1$ . Prove that : $ x^ny \plus{} y^nz \plus{} z^nx \le \frac {n^n}{\left(n \plus{} 1\right)^{n\plus{}1}}$

In Hungkhtn' book:
Let $ a,b,c\geq 0 \text{ and }a \plus{} b \plus{} c \equal{} 3$
Prove $ a^{k}b \plus{} b^{k}c \plus{} c^{k}a \leq$ $ \text{ max(3, }\frac {3^{k \plus{} 1}k^{k}}{{(k \plus{} 1)}^{k \plus{} 1}}\text{ ) }$

[Moderator edit: I have just "created" a new topic for this problem: http://www.mathlinks.ro/Forum/viewtopic.php?t=30843 ]

[quote="perfect_radio"]See [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=triangle&t=78790]here[/url] and [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=2549]here[/url]. And I still can't do the generalization using Lagrange Multipliers. [b]who[/b], could you show me your solution?[/quote]
 I can do n=2 using Lagrange Multipliers but can't do the generalization :(
 :(

[quote="bugzpodder"]Let $ x$, $ y$, and $ z$ be non-negative real numbers satisfying $ x + y + z = 1$.  Show that
\[ x^2 y + y^2 z + z^2 x \leq \frac {4}{27}
\]
and find when equality occurs.[/quote]

If I try with Lagragne Multipliers, I get
$f(x,y,z,\lambda)=x^2 y + y^2 z + z^2 x+\lambda(x+y+z-1)$
gives that the max. occurs if
$df/dx=2xy+z^2+\lambda=0$
$df/dy=2zy+x^2+\lambda=0$
$df/dz=2xz+y^2+\lambda=0$
But $(0,2/3,1/3)$ gives not alwayw $0$, so how to say we have to look to this?

It can be solved by rearrangement inequality.

If    x≥y≥z
Then    xy≥xz≥zy     ⇒     xy∙x+xz∙z+zy∙y≤xy∙x+xz∙y+zy∙z≤y(x+z)^2=y(1-y)^2
On  the interval  [0,1]  maximum value of   f(y)= y(1-y)^2      is   4/27

In other cases solution is analogical

WLOG assume that $y=\text{mid}(x,y,z)$ i.e. $(y-x)(y-z) \le 0$ Hence $z(y-x)(y-z) \le 0$
Using this we have
\[ x^{2}y+y^{2}z+z^{2}x\leq x^{2}y+xyz+z^{2}y\leq x^{2}y+2xyz+z^{2}y=y(x+z)^2=y(1-y)^2 \]

Maximum value of   $ f(y)=y(1-y)^2 $     on      $[0,1]$      is     $ \frac{4}{27}$

[quote=numberdragon]

Actually, $x^2y + y^2z + z^2x$ isn't cyclic. But $x^2y + y^2z + z^2x + x^2z + z^2y + y^2x$ is cyclic.[/quote]

You are wrong.
$x^2y + y^2z + z^2x$ is cyclic.
$x^2y + y^2z + z^2x + x^2z + z^2y + y^2x$ is symmetric.

[quote=towersfreak2006]Making it look "better" with $\text{\LaTeX}$.[hide]Suppose, WLOG, $x\ge y\ge z\ge0$. Set $f(x,y,z)=x^2y+y^2z+z^2x$. Compare $f(x+\epsilon, y, z-\epsilon)$ and $f(x,y,z)$ for all inputs in the correct domain and $\epsilon>0$. \begin{eqnarray*}f(x+\epsilon, y, z-\epsilon) &=& (x+\epsilon)^2y+y^2(z-\epsilon)+(z-\epsilon)^2(x+\epsilon)\\ &=& \epsilon^3+\epsilon^2(x+y-2z)+\epsilon(2xy-2xz-y^2+z^2)+x^2y+y^2z+z^2x\\ &=&\epsilon^3+\epsilon^2(x+y-2z)+\epsilon(y-z)(2x-y-z)+f(x,y,z)\\ &>&f(x,y,z)\end{eqnarray*}For any $y$, increasing the difference between $x$ and $z$ increases the value of $f(x,y,z)$. Therefore, the maximum occurs when $z=0$. Then \[ f(x,y,z)\le \max(x^2y)  \] with $x+y=1$. By AM-GM \[ \frac{1}{2}x+\frac{1}{2}x+y\ge3\sqrt[3]{\frac{x^2y}{4}}  \] \[ \frac{4}{27}\ge x^2y  \] Therefore, \[ x^2 y + y^2 z + z^2 x \leq \frac{4}{27},  \] as desired.[/hide][/quote]
Solution is wrong, as inequality is cyclic, so we can only choose maximum or minimum from numbers $x,y,z$ - we don't know their exact order. (unless we consider two cases)

[i]A start[/i]
Replace $y_i=x_i-\frac12$
We have $y_1+\cdots+y_n=\frac{2-n}{2}$
And the inequality is
$$y_1y_2+\cdots+y_{n-1}y_n+\frac{y_1+y_2}{2}\le \frac{2-n}{4}+(y_1+\cdots+y_n)$$

[quote=WolfusA][quote=sqing]Generalization：
Let $x_1,x_2,\cdots,x_n (n\ge 3) $ be non-negative real numbers satisfying $x_1+x_2+\cdots+x_n=1$ ,Show that\[x^2_1x_2+x^2_2x_3+\cdots+x^2_{n-1}x_n+x^2_nx_1\leq \frac {4}{27}.\]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=2853319[/quote]

Let $x_1,x_2,...,x_n(n\ge 2)$ be non-negative real numbers satisfying $x_1+x_2+\cdots+x_n=1$. Show that
$$x_1x_2+x_2x_3+...+x_{n-1}x_n\le \frac{1}{4}$$[/quote]

Suppose $x_k = \max x_i$, then $x_k(\sum_{j\ne k} x_j) - \sum_{i}x_{i}x_{i+1} = -\sum_{i \ne k} x_i(x_{i+1}-x_k)  \ge 0$ and $x_k(\sum_{j\ne k} x_j) = x_k(1-x_k) \le \frac 1 4$.

Thanks to [b]tastymath75025[/b] for reminding me how to use AM-GM.

We use [b]Lagrange Multipliers[/b]. Let $f(x,y,z)=x^2y+y^2z+z^2x$ and $g(x,y,z)=x+y+z$. We wish to maximize $f$ over $K:=[0,1]^3\cap g^{-1}(\{1\})$. Note that $g$ is continuous and $\{1\}$ is closed, so $g^{-1}(\{1\})$ is closed, so $K$ is closed. Clearly $K$ is bounded, so $K$ is compact. Thus, there exists some $(a,b,c)\in K$ such that \[f(a,b,c)\ge f(x,y,z)\]  for all $(x,y,z)\in K$. Note that $f(2/3,1/3,0)=4/27$, so $f(a,b,c)\ge 4/27$. Let $K':=(0,1)^3\cap g^{-1}(\{1\})$. Lagrange multipliers implies that if $(a',b',c')$ is a local maximum of $f$ over $K'$, then \[\nabla f(a',b',c')=\lambda\nabla g(a',b',c')\] for some real number $\lambda$. We then have two cases.

[b]Case 1[/b]: Suppose $(a,b,c)\in K\setminus K'$. Then, we have $\{0,1\}\cap\{a,b,c\}\ne\emptyset$, so WLOG, we have two subcases. [list]     [*] [b]Case 1.1[/b]: Suppose $a=1$. Then, $b+c=0$ and $b,c\ge 0$, so $b=c=0$. In this case, $f(a,b,c)=f(1,0,0)=0<4/27$, so we have a contradiction.     [*] [b]Case 1.1[/b]: Suppose $a=0$. Then, $b+c=1$ and $b,c\ge 0$. We have     \[f(a,b,c)=b^2c=4(b/2)(b/2)c\le \frac{4}{27}\]     by AM-GM, with equality if and only if $b/2=c$, so $b=2/3$ and $c=1/3$. Thus, $(a,b,c)=(0,2/3,1/3)$, and cyclically shifting, we see that     \[(a,b,c)\in\{(0,2/3,1/3),(1/3,0,2/3),(2/3,1/3,0)\}.\] [/list]

[b]Case 2[/b]: Suppose $(a,b,c)\in K'$. By the instance of LM mentioned above, we have \[(2ab+c^2,2bc+a^2,2ac+b^2)=(\lambda,\lambda,\lambda)\] for some real number $\lambda$. Thus, $2ab+c^2=2bc+a^2$, so \[2b(a-c)=(a-c)(a+c).\] We now have two subcases. [list]     [*] [b]Case 2.1[/b]: Suppose $a=c$. Then, we have $2ab+a^2=2a^2+b^2$, so     \[2a(a-b)=(a-b)(a+b).\]     If $a=b$, then we must have $(a,b,c)=(1/3,1/3,1/3)$. But $f(1/3,1/3,1/3)=1/9<4/27$, so this is not the case. Therefore, we have $2a=a+b$, or $a=b$, which we already saw isn't the case.     [*] [b]Case 2.2[/b]: Suppose $a\ne c$ and $2b=a+c$. By similar logic as in the previous case, we derive that $a\ne b$ and $b\ne c$, and $2c=a+b$ and $2a=b+c$. These imply $(a,b,c)=(1/3,1/3,1/3)$, which is a contradiction. [/list]

Thus, we've shown that $(a,b,c)\in\{(0,2/3,1/3),(1/3,0,2/3),(2/3,1/3,0)\}$, so for $(x,y,z)\in K$, we have \[f(x,y,z)\le\frac{4}{27},\] with equality if and only if $(x,y,z)\in\{(0,2/3,1/3),(1/3,0,2/3),(2/3,1/3,0)\}$.

[hide=bash with karamata and am-gm]x,y,z=1/3 and on substituting 3(1/3)(1/9)=3/27<=4/27 :D [/hide]
[hide=if you know jensen]Σai= k
ao=Σai/n
so according to jensen
nf(a0)<=Σf(ai)
same thing is imaginable for a1a2....an using am gm and root mean square inequality as well
[/hide]

First take a moment to check the extrema: if, say, $a=0$, then we need to show $b^2(1-b)\le 4/27$. Checking for extrema, this follows from noting that the derivative of that expression is $2b-3b^2$ so either $b=0$ (which is a trivial case) or $b=1$ (also trivial) or $b=2/3$. In the last case the inequality is an equality.
Rewrite the desired as
\[\frac{4}{27}\ge a^2b+b^2c+c^2a=(1-b-c)^2b+b^2c+c^2(1-b-c)=b-2b^2-2bc+b^3+2b^2c+bc^2+b^2c+c^2-bc^2-c^3=\]
\[b-2b^2-2bc+b^3+3b^2c+c^2-c^3.\]
Differentiate with respect to $c$ to see that we can require $0=-2b+3b^2+2c-3c^2$. That is, $3c^2-2c=3b^2-2b$. This implies $b$ and $c$ are both roots to some quadratic. If they are the same root, it suffices to check
\[\frac{4}{27}\ge b-2b^2-2b^2+b^3+3b^3+b^2-b^3=b-3b^2+3b^3.\]
But the derivative of this expression with respect to $b$ is $9b^2-6b+1=(3b-1)^2$, so since the inequality holds at $b\in\{0,1/3,1/2\}$, we are done. Then either $b+c$ takes on an extremal value (if it is $0$ we are done and if it is $1$ we are done) or $b+c=2/3$. But by a similar argument we have $c+a=2/3$ and $a+b=2/3$ so $a=b=c=1/3$ and the result follows.

What is the name of the topic that deals with this type of solutions, where these functions are used with maximum and minimum? help me

Let $ x$, $ y$, and $ z$ be non-negative real numbers satisfying $ x + y + z = 1$. Show that
$$ x^2 y + 2y^2 z +3z^2 x+5xyz \leq \frac {4}{9}$$
$$2x^2 y +4y^2 z +4z^2 x+5xyz \leq \frac {16}{27}$$

Let $f(x,y,z)=\sum\limits_{cyc} x^2y - \frac{4}{27} (x+y+z)^3$

Claim: if $x,y,z\ge 0$ and $d>0$ then $f(x+d,y+d,z+d) < f(x,y,z)$

Proof: Just expand.

Therefore, $f(x,y,z) \le f(0,a,b)$ where $x=\min\{x,y,z\}, a=y-x, b=z-x$

By AM-GM, $a^2b =  4 (\frac a2)(\frac a2)(b) \le 4 (\frac{ \frac a2 + \frac a2 + b}{3})^3 = \frac{4}{27} (a+b)^3$, as desired. Equality holds iff $a=2b$.

https://artofproblemsolving.com/community/c6h205183 easily killed with theorem 1

Use LM: inspecting the boundary, which is when one of the variables equals zero (WLOG $a$) we want to prove $b^2c\leq \frac{4}{27}$ when $b+c=1$, which by AM-GM on $\frac{b}{2}+\frac{b}{2}+c$ is true and occurs when $(a,b,c)=(0,2/3,1/3)$ (and cyclic permutations)

Otherwise, if a maximum occurs in the interior, the derivatives $2ab+c^2,2bc+a^2,2ca+b^2$ must be equal. If all of $a,b,c$ are distinct, then WLOG let $a=\max\{a,b,c\}$ and consider the equation $2ca+b^2=2ab+c^2 \iff b(2a-b)=c(2a-c)$, which implies $b=c$ or $b+c=2a$, both of which are impossible. If two of $a,b,c$ are equal, but not the third, WLOG we have $a=b\neq c$. The same equation gives us $b=c$, impossible, or $b+c=2a \iff c=2a-b=b$, also impossible. Hence $a=b=c=1/3$, which works and gives us a fourth equality case of $(1/3,1/3,1/3)$.


What is the greatest integer less than $ \frac{\minus{}17}{4}$?

\[ \frac{\minus{}17}{4}\equal{}\minus{}4.24 \rightarrow \boxed{\minus{}5}\]

We need the floor function: $\lfloor -4.24\rfloor$
And -4 is bigger than -4.24 because in negatives, integers that are greater turn inverse e.g.
If they are greater as positive numbers they are less as negatives. Is that a good explanation?


In a 3-by-3 magic square, show that each corner cell is the average of its two remote side cells.

[hide]
Code looks nicer.
x is the sum of the sides.
Start by filling in the corner and 3 squares around it.
[code]
 (    a    ) (    b    ) (  x-a-b  )
 (    d    ) (    c    ) (         )
 (  x-a-d  ) (         ) (         )
[/code]
$ 2x\minus{}2a\minus{}b\minus{}d\equal{}x$, so $ x\equal{}2a\plus{}b\plus{}d$
[code]
 (    a    ) (    b    ) (   a+d   )
 (    d    ) (    c    ) (  2a+b-c )
 (   a+b   ) (  2a+d-c ) ( a+b+d-c )
[/code]
$ a\plus{}b\plus{}d\minus{}c\equal{}2a\plus{}b\plus{}d\minus{}(a\plus{}b)\minus{}(2a\plus{}d\minus{}c)$
Simplify to $ 2c\equal{}2a\plus{}b\plus{}d$
[code]
 (    a    ) (    b    ) (   a+d   )
 (    d    ) (    c    ) (a+b/2-d/2)
 (   a+b   ) (a+d/2-b/2) ( a+b+d-c )
[/code]
$ a\equal{}\frac{a\plus{}\frac{d}{2}\minus{}\frac{b}{2}\plus{}a\plus{}\frac{b}{2}\minus{}\frac{d}{2}}{2}$

[/hide]


How many guests were present at a cinco de mayo fiesta if every 2 guests shared fajitas, every 3 guests shared a plate of nachos, every 4 guests schared a bowl of refried beans, athere were 65 dishes all together? this problem is for my brother... thanks!!!
x/2 +x/3+ x/4 is the equation... thanks again !

This is way too easy for intermediate. And I noticed you already posted it in the middle school forum, which is where it should be.


oh I forget its conditions ! :lol:

Prove that if
\[ a\geq b\geq c\ge0 \] please!

I  don't think the inequality is right(If I am right :) ) even under the condition
try:$a=1.21,b=c=1$


Compute the deifinite integral:
Sorry, I am not  used to writing maths symbols so please see this prb. in attached file.

Which one should it be?

$ I\equal{}\int_{\frac{1}{2}}^{2}\left(2\plus{}x\minus{}\frac{1}{x}\right)e^x\plus{}\frac{1}{x}\,\mathrm{d}x$

$ I\equal{}\int_{\frac{1}{2}}^{2}\left(2\plus{}x\minus{}\frac{1}{x}\right)e^{x\plus{}\frac{1}{x}}\,\mathrm{d}x$

To mods: Please move this to [url=http://www.artofproblemsolving.com/Forum/index.php?f=296]Calculus Computations and Tutorials[/url].

It is exact that

$ I \equal{} \int_{\frac {1}{2}}^{2}\left(2 \plus{} x \minus{} \frac {1}{x}\right)e^{x \plus{} \frac {1}{x}}\,\mathrm{d}x$


calculus integration trigonometry limit function derivative LaTeX

Show that $ \int_{0}^{\infty}\cos{x^2}dx$ converges, but not absolutely.

Let $ \left(C: z(\theta)=Re^{i\theta} \ \left(0 \le \theta \le \frac{\pi}{4}\right)\right)$
$ \oint_C e^{iz^2} dz=0 \ \ (by \ Cauchy)$
i.e.
$ \int_{OA}e^{iz^2} dz + \int_{AB}e^{iz^2} dz + \int_{BO}e^{iz^2} dz = 0$
$ \left(\begin{array}{ll}OA: z=x & (0 \le x \le R) \\ AB: z=Re^{i\theta} & \left(0 \le \theta \le \frac{\pi}{4}\right) \\ BO: z=re^{\frac{\pi i}{4}} & (R \le r \le 0)\end{array}\right)$
So,
\[ \int_0^R e^{i x^2} dx + \int_0^{\frac{\pi}{4}}e^{i R^2 e^{2 i \theta}} i R e^{i \theta} d\theta + \int_R^0 e^{i r^2 e^{\frac{\pi i}{2}}}e^{\frac{\pi i}{4}} dr = 0
\]
i.e.
\[ \int_0^R (\cos x^2 + i \sin x^2) dx = e^{\frac{\pi i }{4}}\int_0^R e^{-r^2}dr - \int_0^{\frac{\pi}{4}}e^{i R^2 \cos \theta -R^2 \sin 2 \theta} i R e^{i \theta} d\theta
\]

\[ \lim_{R \rightarrow \infty}\left(e^{\frac{\pi i }{4}}\int_0^R e^{-r^2}dr\right)=e^{\frac{\pi i }{4}}\int_0^\infty e^{-r^2}dr=\frac{\sqrt{\pi}}{2}e^{\frac{\pi i }{4}} = \frac{1}{2}\sqrt{\frac{\pi}{2}} + \frac{i}{2}\sqrt{\frac{\pi}{2}}
\]

\begin{eqnarray*}
\left|\int_0^{\frac{\pi}{4}}e^{i R^2 \cos \theta -R^2 \sin 2 \theta} i R e^{i \theta} d\theta\right| & \le & \int_0^{\frac{\pi}{4}}e^{-R^2 \sin 2 \theta}R d\theta \\
& = & \frac{R}{2}\int_0^{\frac{\pi}{2}}e^{-R^2 \sin \phi} d\phi \\
& \le & \frac{R}{2}\int_0^{\frac{\pi}{2}}e^{-R^2\frac{2 \phi}{\pi}} d\phi \\
& = & \frac{\pi}{4R}(1-e^{-R^2})
\end{eqnarray*}

So,
$ \lim_{R \rightarrow \infty}\left(\frac{\pi}{4R}(1-e^{-R^2})\right)=0$

Therefore,
\[ \int_0^\infty (\cos x^2 + i \sin x^2) dx = \frac{1}{2}\sqrt{\frac{\pi}{2}} + \frac{i}{2}\sqrt{\frac{\pi}{2}}
\]

\[ \color{red}\int_0^\infty \cos x^2 dx = \int_0^\infty \sin x^2 dx = \frac{1}{2}\sqrt{\frac{\pi}{2}}\color{black}
\]

Getting back to the simpler question: why does it converge?

Ignore the integral from $ 0$ to $ 1$ - that's the integral of a continuous function over a closed bounded interval, so its convergence is not in question.

$ \int_1^t\cos x^2\,dx\equal{}\int_1^t\frac1{2x}\cdot2x\cos x^2\,dx$

$ \equal{}\left.\minus{}\frac1{2x}\sin x^2\right|_1^t\plus{}\int_1^t\frac1{2x^2}\cdot \sin x^2\,dx.$

Take the limit as $ t\to\infty$ to get that

$ \int_1^{\infty}\cos x^2\,dx\equal{}\frac12\sin 1\plus{}\int_1^{\infty} \frac1{2x^2}\cdot \sin x^2\,dx.$

That second integral converges absolutely by comparison to $ \int_1^{\infty}\frac{dx}{2x^2}.$

There are several reasonable ways of showing the failure of absolute convergence. Intuitively, the the absolute value of the function stays (in some averaged sense) at an essentially constant size.

Actually, there is nothing sacred about $ x^2$ here: any convex increasing function $ f(x)$ with the derivative tending to $ \plus{}\infty$ will do for both parts of the problem :) Can anybody see it [b]without[/b] integration by parts and other clever analytic tricks, just right from the picture? ;)

I suppose one plan would be to chop up the integral at the zeros of $ \cos(f(x)),$ and then arguing that the resulting series satisfies the hypotheses of the Alternating Series Test.

Show that $ \int_{0}^{\infty}\cos x^2dx$ not converges absolutely?
$ \int_{\sqrt {\pi}}^{\infty}|\cos x^2|dx \equal{} \frac {1}{2}\int_{\pi}^{\infty}\frac {|\cos{x}|}{\sqrt {x}}dx \equal{} \frac {1}{2}\sum_{1}^{\infty}\int_{k\pi}^{(k \plus{} 1)\pi}\frac {|\cos x|}{\sqrt {x}}dx \equal{} \frac {1}{2}\sum_{1}^{\infty}\int_{0}^{\pi}\frac {|\cos{x}|}{\sqrt {x \plus{} k\pi}}dx$
$ a_k \equal{} \int_{0}^{\pi}\frac {|\cos{x}|}{\sqrt {x \plus{} k\pi}}dx \equal{} \int_{0}^{\frac {\pi}{2}}\frac {\cos x}{\sqrt {x \plus{} k\pi}}dx \plus{} \int_{\frac {\pi}{2}}^{\pi}\frac { \minus{} \cos x}{\sqrt {x \plus{} k\pi}}dx \equal{} \int_{0}^{\frac {\pi}{2}}\frac {\cos x}{\sqrt {x \plus{} k\pi}}dx \plus{} \int_{0}^{\frac {\pi}{2}}\frac {\sin x}{\sqrt {x \plus{} \frac {\pi}{2} \plus{} k\pi}}dx$
$ \implies a_k > \int_{0}^{\frac {\pi}{2}}\frac {\sin x}{\sqrt {x \plus{} \frac {\pi}{2} \plus{} k\pi}}dx > \int_{0}^{\frac {\pi}{2}}\frac {\sin x}{\sqrt {\frac {\pi}{2} \plus{} \frac {\pi}{2} \plus{} k\pi}}dx \equal{} \frac {1}{\sqrt {(k \plus{} 1)\pi}}$

[color=green][Edited by moderator for LaTeX][/color]


Prove that $a^3+b^3+4$ is not a perfect cube for any natural numbers $a$ and $b$

a number such that $a^3$ has remainders from 9 only {0,1,8}.
when we sum of the two of them 0+1=1,0+8=8,1+8=9=0(mod9) we add 4 to all.
1+4=5
8+5=13=4(mod9)
0+4=4
a perfect cube must have 1,0,8 remainder from 9.there is no so.

[quote="gunes"]a number such that $a^3$ has remainders from 9 only {0,1,8}.
[/quote]

Is that really true?

1 / 9 = 1 (mod 9)
8 / 9 = 8 (mod 9)
27 / 9 = 0 (mod 9)
64 / 9 = 1 (mod 9)
125 / 9 = 8 (mod 9)
216 / 9 = 0 (mod 9)

...

It follows in a pattern? (*idea for problem  :idea: *)

You can always reduce $a$ modulo 9 so that it is a value from zero to eight.  Therefore, it is sufficient to check $a^3 \mod 9$ for $a=0,1,\ldots,8$ to find the possible remainders (called residues) modulo 9.  Thus, as has been said already, zero, one and eight are the only possible residues.

Or, to say in general, $n\pmod{3}  \equiv n^3 \pmod{9}$


Hm this topic is dead. Izzy will "recycle" it.

Hello,

I'm quite new here.  I've never quite understood the "1337" or "LEET" craze.  What's it all about?

Also, I would love to join "CTC".

Casper, go onto wikipedia and type "1337".

Everyone who posted here, including PI-Dimension, please refrain from making spam threads such as this.

Locked. (Not recycled, 'cause that 1337 thing you came up with was pretty clever. :P )


rotation
geometry
geometric transformation
reflection
trigonometry
vector
linear algebra

I am in eighth grade, and I am wondering if anyone here can teach me how to solve  problems like this:

Rotate point (6,6) counter clockwise around the point (5,4)

Reflect point (3,-4) over the line x=-4

Basically, I need to learn how to rotate a point about a point that is not the origin, and how to reflect a point over a line that is not the x or y axis or the lines y=-x / y=x

Thanks, all help is appreciated ^_^

Just shift everything so that the center of rotation becomes the origin (or the reflection line passes through the origin), make your rotation (reflection) and shift the result back.

The first question doesn't make sense unless you specify how much we have to rotate it. Assuming we have to rotate by some amount $ \theta$, we can translate such that we're rotating $ (1,2)$ around $ (0,0)$. If we think of $ (1,2)$ as $ 1\plus{}2i$ in the complex plane, $ (1\plus{}2i)(\cos\theta\plus{}i\sin\theta)\equal{}(\cos\theta\minus{}2\sin\theta)\plus{}i(2\cos\theta\plus{}\sin\theta)$. (We do this because multiplication by $ e^{i\theta}$ in the complex plane rotates a complex number $ z$ around the origin by $ \theta$. If you're familiar with linear algebra, this is like multiplying a vector in 2 dimensional space by a rotation matrix)

As for the second, as fedja mentioned, a little translation helps. So translate to the right so that we're reflecting $ (7,\minus{}4)$ around $ x\equal{}0$, which is clearly $ (\minus{}7,\minus{}4)$, and then translate back to get $ (\minus{}11,\minus{}4)$

A method my teacher said involved creating a right triangle with the center and the preimage point.

Grn-trtl, just assume the first one says 90 degrees counter clockwise.

Creating the right triangle for the $ 90^\circ$ rotation is a nice idea if the coordinates are integer and you have a square grid paper at hand. In all other cases, it is easier to understand/memorize that the counterclockwise rotation by the right angle moves $ (x,y)$ to $ (\minus{}y,x)$ and do the shifting. If you know vectors, then you can hide the shift: to rotate $ (6,6)$ around $ (5,4)$, note that $ (6,6)\equal{}(5,4)\plus{}(1,2)$, so the image is $ (5,4)\plus{}(\minus{}2,1)\equal{}(3,5)$ (you rotate the difference vector and add the result to the center vector).

Fedja, that rule only works when rotating about the origin.

I need help rotating about any other point that is not the origin without grid paper  :(

Didn't I just tell you one? :? Read my example with your points and try to understand how it works. If you have any questions about it, ask.

Can someone post a general rule about the following? Ex: 180 degree rotation= (a,b)->(-a,-b)

Counterclockwise rotation about (0,0) 90 degrees maps (a,b) onto ?

Clockwise rotation about (0,0) 90degrees maps (a,b) onto?

Counterclockwise rotation about (0,0) 270degrees maps (a,b) onto?

Clockwise rotation about (0,0) 270degrees maps (a,b) onto?

Thanks :)

Thanks Fedja, I understand it now.

Never mind, I found out how to rotate around the origin.

But I still do not understand how to rotate around a point that is not the origin X number of degrees.

I have not learned vectors yet.. Can someone post a example ?... Keep in mind NO GRID PAPER

When we talk about rotations, we often do so in terms of vectors or complex numbers, so you might want to look into those.

Basically, you'll want to just draw a triangle and apply trig to get the position of the rotated point.

Anyone care to explain...?

No vectors/grids


vector
probability
advanced fields
advanced fields unsolved

there is a bug on (0,0) on the plane (bug is too small to be seen). after every 20 seconds bug instantly moves by one of three vectors: v1,v2,v3 (they have integer entries). these three vectors aren't contained in any halfplane which contains (0,0). after every 40 seconds we spray chosen grid point with poison. if bug steps or will step on poisoned point, it dies. is it always possible to kill the bug? we neglect ethical aspects of the problem :)

This is just a modification of Conway's angel problem. In your case, the angel effectively has power 2 but has a weird modification of its move.

Most stuff regarding this problem is really hard. You may want to read Conway's original paper, which deals with some cases similar to yours:
http://www.msri.org/communications/books/Book29/files/conway.pdf
Or the dissertation of Martin Kutz, which has a lot of interesting but difficult work on this problem (although you may only be interested in the beginning part about the "$2-\epsilon$-angel".
http://www.mpi-inf.mpg.de/~mkutz/diss/kutzdiss.pdf

By the way, where does your problem come from? Do you know if there's a simple solution?

from a competition on my uni. i think this is random walk on a plane. markov chain modeled by this problem returns to starting state with probability 1, so i think it's sufficient to spray only (0,0) and then wait :). am I correct? would be nice if so..

Oh he is going random!? If that's the case, it is a much different problem, yes.

I don't fully understand what you said about the vectors. They are all $(m, n)$ where $m$ and $n$ are integers, so the bug is always on lattice points? Even with this I am not certain the bug surely returns to the origin....

Don't listen to me though, I have never studied probability theory! :P


algebra
polynomial
vector
geometry
rectangle
linear algebra
matrix

[quote="silviu"]Sa se demonstreze ca $\forall A,B \in M_n (C)$ avem $det(I_n+AB)=det(I_n+BA) $.[/quote]

This is equivalent to (and a special case of) the fact that $AB$ and $BA$ have the same characteristic polynomial.
Indeed, the coefficients of the characteristic polynomial are traces of $AB$ and $BA$ acting on certain vector spaces (exterior powers $\Lambda^k C^n$), and $Tr(AB) = Tr(BA)$.
If you don't like multilinear algebra, this is a restatement of some identities about sums of minors ($k$ x $k$ subdeterminants) in $AB$ and $BA$, I think those are called "Cauchy-Binet formulae".

De fapt există demonstraţii ceva mai simple, cum ar fi următoarea:

Dacă una dintre $A$ sau $B$ este inversabilă, să zicem $A$, atunci
\[ \begin{array}{ccc} \det (I_n+ AB ) &=& \det ( A \cdot (A^{-1} + B)) \\
\ &=& \det A \cdot \det (A^{-1} +B )  \\
\ &=& \det (A^{-1}+B) \cdot \det A \\
\ &=& \det ((A^{-1} +B )A) \\
\ &=& \det (I_n +BA)
\end{array}
\]

Dacă $\det A = \det B = 0$ atunci considerăm polinomul $\det (A+ \lambda I_n) = f(\lambda)$ care este nenul pentru o infinitate de $\lambda$, si deci putem aplica partea de mai sus, pentru matricile $A+\lambda I_n$ şi $B$, obţinând că polinoamele
$ g(\lambda) = \det (I_n + \lambda AB)$, şi $h(\lambda) = \det (I_n + \lambda BA)$ sunt egale pentru o infinitate de variabile, deci prin urmare şi pentru $\lambda=1$ de unde concluzia.

More over $K$ is commutative field

$A\in M_{p,n}(K), \; B\in M_{n,p}(K)$ where $1\leq p \leq n$
A is rectangle matrix p-row,n-columns
B is rectangle matrix n-rows, p-columns

then $x^p det(xI_n-BA)=x^n det(xI_p-AB)$

Proof:

$J=\left[\begin{array}{cc}I_n&0_{n,p}\\A&I_p\end{array}\right]$

$L=\left[\begin{array}{cc}xI_n-BA&B\\0_{p,n}&xI_p\end{array}\right]$

$M=\left[\begin{array}{cc}xI_n&B\\0_{p,n}&xI_p-AB\end{array}\right]$

product by bloc gives

$LJ=JM=\left[\begin{array}{cc}xI_n&B\\xA&xI_p\end{array}\right]$

we take the determinant

$det(L)det(J)=det(J)det(M)$ but $J$ is triangular superior with diagonal 1, iits determinant

is 1

$L,M$ is triangular superior

$det(xI_p)det(xI_p-BA)=det(xI_n)det(xI_p-AB)$

[quote="silviu"]Sa se demonstreze ca $\forall A,B \in M_n (C)$ avem $det(I_n+AB)=det(I_n+BA) $.[/quote]

Este adevarat: $det(I_n+xAB)=det(I_n+xBA),\forall x \in R,\forall A,B \in M_n(C)$.????

Da, eu asta am demonstrat mai sus, şi după aia am făcut $x=1$ :)


$\text{Suppose a square with 12 cm sides is revolved about a diagonal as an axis. Find the surface area of the figure formed.}$

The new shape is a cube with side length 12.
$12^{3}=1728$

wouldn't it be two cones stuck together??

I visualized it inaccurately. :oops:
[hide="The answer would be"]$\frac{2(6\sqrt2)^{2}\pi 6\sqrt2}3=144\pi\sqrt2$[/hide]

You're mistaken again - check your math.

when you see problems with revolving or squares or x-axis, etc. what should I look for in clues to solve the problem?  :maybe:

revolving stuff will usually be round. In this case, the square is rotated about it's diagonal, and you don't reall need both sides so it's basically an isosceles right triangle rotated around its hypoteneuse.

[hide=" :starwars: "]
The end result is a pair of cones.

The cones each have a $6\sqrt{3}$ height and the slant height is the radius ($6\sqrt{3}$ again) squared: $108$ and then the height squared (108) added together =216, then $\sqrt{216}$.  You need to take into account that there are two cones.$12\pi\sqrt{3}(6\sqrt{3}+6\sqrt{6})=216\pi+216\sqrt{2}\pi=(216+216\sqrt{2})\pi$.
[/hide]

The solid formed by revolution of a square around its diagonal is a double cone (two cones linked through their bases). Considering a general square of side [i]L[/i], then the surface of the double cone is $S= \pi L^{2}\sqrt{2}$.

By the way, how do you insert mathematical symbols in here?

Carcul

insert symbols by LATEX

I haven't downloaded that yet, since I haven't seen the need to. It's not like I will give in to more computer pokeyness.  :D

Could someone give a detailed solution of this? with full explanation

I am trying to visualize it right now

Miscalculated again! :wallbash:  :wallbash_red:  :wallbash:  :wallbash_red:  :wallbash:  :wallbash_red:  :wallbash:  :wallbash_red:  :wallbash:

I don't think you need to download $\text{\LaTeX}$ to use it in the math forum.  You just need to know the code for math mode.
for example, if you type in $\$$2+2=4$\$$, you will see $2+2=4$.  That is an example of a simple equation.  You can type much more complex ones, such as $\frac{\frac{a}{b}}{\frac{c}{d}}$.  For other complex mathematics, as well as the source code, see the $\text{\LaTeX}$ section of this website.

Also, you can use preview to make sure that you correctly typed in your source code.  If you don't, your post will likely contain a message similar to the following: $%Error. "Latex" is a bad command.
$.

I think this would be 2 cones stuck together at the base
[hide]Then the height of a cone is $6\sqrt{2}$ and the base is $\pi(6\sqrt{2})^{2}=24\pi$ so using volume = $\frac{1}{3}*base area*height$, $\frac{1}{3}*24\pi*6\sqrt{2}=\frac{144\sqrt{2}\pi}{3}=48\pi\sqrt{2}$
Finally, multiply by 2 to get $96\pi\sqrt{2}$[/hide]
...I think... :P

[quote="Quevvy"]so using volume =...[/quote]

Look at the first post of this thread. The purpose here is to find the [i]surface area[/i] of the double cone, not its volume.

[quote="Roadnottaken"]For other complex mathematics, as well as the source code, see the  section of this website.[/quote]

Thanks. Carcul

Also note that if you hold your cursor over a person's $\LaTeX$ formula, it will display the code used.  For example try it on my symbol above.

[quote="Carcul"]The purpose here is to find the [i]surface area[/i] of the double cone, not its volume.[/quote]

Oh, sorry, but anyways...
[hide]The surface area (without the base) is $S=\pi r\sqrt{r^{2}+h^{2}}=\pi6\sqrt{2}\sqrt{72+72}=12\pi6\sqrt{2}$. For 2 cones, the surface area is $\pi144\sqrt{2}$.[/hide]
 :P
[quote="mgao"]slant height is the radius ($6\sqrt{3}$ again)[/quote]
Actually the slant height is one side of the original square
  :P


function
algebra
functional equation
algebra unsolved

find all functions $ f$ mapping natural nos. onto natural nos,such that:
1.$ f(mn)\equal{}f(m)f(n)$
2.$ f(2)\equal{}2$
3.$ f(m)<f(n)$,whenever $ m<n$

[quote="aadil"]find all functions $ f$ mapping natural nos. onto natural nos,such that:
1.$ f(mn) \equal{} f(m)f(n)$
2.$ f(2) \equal{} 2$
3.$ f(m) < f(n)$,whenever $ m < n$[/quote]

Put $ m\equal{}n\equal{}1 \Rightarrow f(1)\equal{}1$

For any integer $ x > 2$ , we can find $ k \in \mathbb{N}$ so that $ 2^k \le x <2^{k\plus{}1}$

So $ 2^k\equal{}f(2^k) \le f(2^k\plus{}1)\minus{}1 \le ... \le f(x)\minus{}x\plus{}2^k \Rightarrow x \le f(x)$

$ f(x) \le f(x\plus{}1)\minus{}1 \le ... \le f(2^{k\plus{}1})\minus{}2^{k\plus{}1}\plus{}x\equal{}x$

So $ f(x)\equal{}x$ for all $ x \in \mathbb{N}$ is the only solution.

[quote="aadil"]find all functions $ f$ mapping natural nos. onto natural nos,such that:
1.$ f(mn) \equal{} f(m)f(n)$
2.$ f(2) \equal{} 2$
3.$ f(m) < f(n)$,whenever $ m < n$[/quote]

[url]http://en.wikipedia.org/wiki/Cauchy's_functional_equation[/url]

how have you assumed that $ f(2^k)\equal{}2^k$

[quote="aadil"]how have you assumed that $ f(2^k) \equal{} 2^k$[/quote]

$ f(2)\equal{}2$ and $ f(mn)\equal{}f(m)f(n)$ so ...

A slighly different solution:
First of all you can easily prove that $ f(1) \equal{} 1$.

Then you notice that:
$ f(2 \cdot 2) \equal{} f(2)(2) \equal{} 4$
But $ 2 < 3 < 4$ so $ f(2) < f(3) < f(4) \Longleftrightarrow 2 < f(3) < 4$ so $ f(3) \equal{} 3$.

Working analogously we can find the ''even'' x's and then using the fact that $ f$ is increasing you can find the odd.

Now we have to prove it:
We will use induction.
Suppose we have proved that:
$ f(1) \equal{} 1,f(2) \equal{} 2...,f(2d) \equal{} 2d$,for some natural $ d$
Then $ f(2(d \plus{} 1)) \equal{} f(2) \cdot f(d \plus{} 1) \equal{} 2(d \plus{} 1)$,because $ d \plus{} 1\le 2d$

Now using the fact that
$ 2d < 2d \plus{} 1 < 2d \plus{} 2 \Longleftrightarrow f(2d) < f(2d \plus{} 1) < f(2d \plus{} 2) \Longleftrightarrow 2d < f(2d \plus{} 1) < 2d \plus{} 2$

so $ f(2d \plus{} 1) \equal{} 2d \plus{} 1$.
So we proved that $ f(x) \equal{} x,\forall x\in N$.

I hope this helps....

Now prove the following:

Let $ f: N \to [1,\infty)$ such that

$ a)$ $ f(2)\equal{}2$

$ b)$ $ f(mn)\equal{}f(m) \cdot f(n),\forall m,n \in N$

$ c)$ $ f(m) < f(n)$ whenever $ m<n$

Prove that $ f(n)\equal{}n$ for all natural nimbers $ n$.

[quote="Dimitris X"]Now prove the following:

Let $ f: N \to [1,\infty)$ such that

$ a)$ $ f(2) \equal{} 2$

$ b)$ $ f(mn) \equal{} f(m) \cdot f(n),\forall m,n \in N$

$ c)$ $ f(m) < f(n)$ whenever $ m < n$

Prove that $ f(n) \equal{} n$ for all natural nimbers $ n$.[/quote]

There is a really nice solution given to this functional equation by [b]Alexandros Sugelakis[/b]

[i][u]solution[/u][/i]

First of all we can easily prove that:
$ f(1) \equal{} 1$.
So $ f(n) > 1, \forall n\ge 2$ (because of monotony).

Let us consider now two dinstinct numbers $ m,n$ and let us consider that $ \frac {ln(f(m))}{lnm} \not \equal{} \frac {ln(f(n))}{ln n}$.

Let it be $ 0 < \frac {ln(f(m))}{ln(f(n))} < \frac {lnm}{lnn}$.(work analogously for the other case)

Now there exist two natural numbers $ k,l$ so that:

$ 0 < \frac {ln(f(m))}{ln(f(n))} < \frac {k}{l} < \frac {lnm}{lnn}$. because between two real numbers there exist always a rational.

So we have that:

$ \frac {ln(f(m))}{ln(f(n))} < \frac {k}{l} \Longleftrightarrow ln[f(m)^l] < ln[f(n)^k] \Longleftrightarrow ln(f(m^l) < ln(f(n^k) \Longleftrightarrow f(m^l) < f(n^k)$. $ \boxed{1}$.

The step $ ln[f(m)^l] < ln[f(n)^k] \Longleftrightarrow ln(f(m^l)) < ln(f(n^k))$ is true due to $ f(m)^l \equal{} f(m^l)$.(it can be proved easily by induction).

Also we have that $ \frac {k}{l} < \frac {lnm}{lnn} \Longleftrightarrow lnn^k < lnm^l \Longleftrightarrow n^k < m^l$. $ \boxed{2}$.

But from $ \boxed{1},\boxed{2}$ we have contradiction.

So $ \frac {ln(f(m))}{lnm} \equal{} \frac {ln(f(n))}{lnn} \equal{} c$, where $ c > 0$ .

So $ f(n) \equal{} n^c$, but from the condition $ f(2) \equal{} 2$,we take $ 2 \equal{} 2^c \Longleftrightarrow c \equal{} 1$.

So $ \boxed{\boxed{f(n) \equal{} n}}$.


Consider the set of all integers of the form (2^x)(3^y), where x and y are positive integers.  Find a formula for the kth entry in the list when such numbers are arranged in ascending order.

As far as I know, this is an open problem.  See [url=http://www.research.att.com/~njas/sequences/A003586]OEIS A003586[/url].


function
integration
calculus
real analysis
real analysis unsolved

If $ A>0,5$. Prove that there is not a function $ f: [0,1]\minus{}>R$ so that $ f(x)\equal{}1\plus{}A.\int f(t).f(x\minus{}t)dt$, where the upper limit of the integral is $ 1$ and the lower limit of the integral is $ x$.
Thank you for the solution given before.

If $ f$ is defined on $ [0,1]$, then what is $ f(x \minus{} t)$ if $ t > x$?


Solve in whole numbers (a,b) this equation:

$\frac{(a^{2}+b^{2})(a+b-8)}{8ab+8}=1$

What competition¿ [former title:competition problem8]
Use better titles next time!

Because $a=b(mod 2)$ we can represent $a=x+y,b=x-y$. It give $(x^{2}+y^{2})(x-4)=2(x^{2}-y^{2}+1).$
If y=0, then $x^{3}-6x^{2}-2=$ (had not integer solutions).
x=0 give $y^{2}=-1$.
If $|y|\ge 1, |x|\ge 1$, then $x^{2}+y^{2}>|x^{2}-y^{2}+1|$, therefore must be $|x-4|=1$.
If x=5 we have $y=\pm 3$.
If x=3 we have $y^{2}=29$ (had not integer solutions). Therefore
$a=8,b=2$ or $a=2,b=8$.


[quote="mdk"]Solve the system x^3+y^3=126
                      x^2-xy+y^2=21[/quote]

[hide="Hint"]$ 126 = x^{3}+y^{3}= (x+y)(x^{2}-xy+y^{2})=21(x+y)$
$ \boxed{x+y=6}$
$ x^{2}+2xy+y^{2}=36$

Subtract from second given equation:
$ 3xy = 15$
$ \boxed{xy = 5}$

Use boxed equations to substitute and solve.[/hide]


Is there any way to get

[code]2\sqrt{py}[/code]

$2\sqrt{py}$

to look better? The $2$ and the square root don't seem to be lined up.

I think you want \sqrt[2]{py}.  This will show up as

$\sqrt[2]{py}$

Actually, it's supposed to be $2\cdot\sqrt{pq}$. Is it just normally not lined up?

The problem is that the tail of the p and y go below the baseline so the square root sign is lower to cover the tails. The smash command stops the square root going so low so you get $2\sqrt{\smash[b]{py}}$ which is given by 2\sqrt{\smash[b]{py}}

\smash[t] does the same thing for the height.


function
Euler
induction
logarithms
number theory unsolved
number theory

An 'easy' questions... hope someone can help!

1. Determine an arithmetical function f such that

1/phi(n) = sum (1/d * f(n/d))

where the sum is over all divisiors of n

and phi(n) is the euler function.. (no. of coprimes less than or equal to n)

[b]Lemma 1.[/b] For $n$ positive integer we have
\[ \frac{1}{\varphi ( n )} = \frac{1}{n} \sum_{d|n} \frac{\mu^2 ( d )}{\varphi ( d )}  \]
[b]Proof:[/b] $\frac{1}{\varphi ( n )}$ is a multiplicative function. Hence making use of
\[ \sum_{d|n} \mu^2 ( d ) f ( d ) = \prod_{p|n} \left( 1 + f ( p ) \right)  \]
which holds for all multiplicative function $f$ we obtain

\begin{eqnarray*} \sum_{d|n} \mu^2 ( d ) \frac{1}{\varphi ( d )} & = & \prod_{p|n} \left( 1 + \frac{1}{\varphi ( p )} \right)\\ & = & \prod_{p|n} \left( 1 + \frac{1}{p - 1} \right)\\ & = & \prod_{p|n} \left( \frac{p}{p - 1} \right)\\ & = & \frac{n}{n \prod_{p|n} \left( 1 - \frac{1}{p} \right)}\\ & = & \frac{n}{\varphi ( n )} \end{eqnarray*}

And we get the claim.

Now getting back to the main ``proof'', first remark that $\sum_{d|n} \frac{1}{d} f \left( \frac{n}{d} \right) = \sum_{d|n} \frac{d}{n} f ( d )$
To obtain

\begin{eqnarray*} \sum_{d|n} \frac{1}{d} f \left( \frac{n}{d} \right) & = & \frac{1}{\varphi ( n )}\\ \sum_{d|n} \frac{d}{n} f ( d ) & = & \frac{1}{n} \sum_{d|n} \frac{\mu^2 ( d )}{\varphi ( d )}\\ \sum_{d|n} df ( d ) & = & \sum_{d|n} \frac{\mu^2 ( d )}{\varphi ( d )} \end{eqnarray*}

Putting $n = 1$, shows that $f ( 1 ) = 0$
Putting $n = p$ where $p$ is a prime shows that
\begin{eqnarray*} f ( 1 ) + pf ( p ) & = & \frac{\mu^2 ( p )}{\varphi ( p )}\\ f ( p ) & = & \frac{\mu^2 ( p )}{p \varphi ( p )} \end{eqnarray*}
Let $j$ be a positive integer and put $n = p^j .$
Prove by induction on $j$ that (the case $j = 1$ has been done)
\begin{eqnarray*} \left. f ( p^j \right) & = & \frac{\left. \mu^2 ( p^j \right)}{\left. p^j \varphi ( p^j \right)} \end{eqnarray*}
Remark now that $\frac{\mu^2}{N \cdot \varphi}$ is multiplicative. Hence $f$ is also multiplicative (i.e $f ( mn ) = f ( m ) \cdot f ( n )$ when $( m, n ) = 1$).
Thus $f ( n ) = \frac{\mu^2 ( n )}{n \cdot \varphi ( n )}$ for all positive integer $n$ by the multiplicavity of $f$.

P.S: You can use the lemma to prove that $\sum_{n \leq x} \frac{1}{\varphi ( n )} = C_1 \log x + C_2 + O \left(\frac{\log x}{x} \right)$ (this is what Landau did ;))
P.P.S: You can find $f$ using Dirichlet convolution. That would be probably shorter.
P.P.P.S: I hope there is any nicer proof than the one I gave :)


How many different types can the number 25  be  written in terms of the powers 0f 2   ?
(like as,24+23+20=25  ,
23+23+23+20=25,
20+20+20+…..20=25[/u][/quote][/i]

Hmm. Let $ f(t,x)$ = the number of different ways to write x in terms of powers of 2 which are smaller than $ t$.
f(1,1)=1. f(2,1)=f(3,1)=...0.
f(1,2)=1. f(2,2)=2 (Because $ 2 \equal{} 1\plus{}1 \equal{} 2$). f(3,2) = f(4,2) = ... = 0
...
And $ F(t,x) \equal{} F(t\minus{}1, x)$ if t>x.
$ F(t,x) \equal{} F(t\minus{}1,x) \plus{} F(t,x\minus{}t)$ if t<=x.

This can be solved with dynamic programming. :D

The relevant generating function is

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

To compute the coefficient of $ x^{25}$ we can stop multiplying at $ (1 \minus{} x^{16})$, although the computations still aren't easy to do by hand.


induction
complex numbers
algebra unsolved
algebra

$ X$ is a finite set such that if $ x_{1}$ and $ x_{2}$ are $ \in X$ then $ x_{1}x_{2} \in X$. Show that all the elements of $ X$ have modulus $ 1$ and that, if $ x \in X$ then the complex conjugate of $ x$ is in $ X$.

By induction, $ x^n \in X$ for all $ x\in X$ and $ n\in \mathbb{N}$.  Since this set is finite we must have $ x^{k_x} \equal{} 1$ for some $ k_x\in \mathbb{N}$, and so in particular $ |x| \equal{} 1$ for all $ x\in X$.  Moreover $ x^{k_x \minus{} 1}x  \equal{} 1 \equal{} |x|^2 \equal{} \overline{x}x\Leftrightarrow x^{k_x \minus{} 1} \equal{} \overline{x}\in X$.


Allen and Bethany each arrive at a party at a random time between 1:00 and 2:00.  Each stays for 15 minutes, then leaves.  What is the probability that Allen and Bethany see each other at the party?

Let Allen's time from $ 0$ to $ 60$ minutes after one o'clock be $ x$ and Bethany's be $ y$ so that the only way for them to meet is when $ \minus{}15\le x\minus{}y\le15$. Graphing, the probability is $ 1\minus{}\left(\frac{45}{60}\right)^2\equal{}\boxed{\frac7{16}}$.

Also, you can graph $|x-y|\le 15$ on a $60\times 60$ unit square. Same thing.

[hide=Solution]Let the $x$ axis be when Allen arrives and the $y$ axis be when Bethany arrives. Like a classic geometric probability problem, the figure looks like the following:

[asy]
draw((0,0)--(60,0)--(60,60)--(0,60)--(0,0));
label("0", (0,0), SW);
label("1/4", (0,15), W);
label("1/4", (15,0), S);
label("1", (60,0), S);
label("1", (0,60), W);
draw((0,0)--(15,0)--(60,45)--(60,60)--(45,60)--(0,15)--(0,0));
[/asy]
We must find the area outside of the two triangles (the irregular hexagon). This can be found by subtracting the area of the triangles from the square.

This is equal to $1-\dfrac{\left(\dfrac{3}{4}\right)^2}{2}\cdot2=1-\dfrac{9}{16}=\boxed{\dfrac{7}{16}}.$
[/hide]

[quote=Lamboreghini][quote=happyhippos][hide=Soultion]We let the $x$ axis represent the time Allen arrives, and the $y$ axis represent the time Bethany arrives.

[asy]
draw((0,0)--(60,0), Arrow);
draw((0,0)--(0,60), Arrow);
label("1:00", (0,0), SW);
label("1:15", (0,15), W);
label("1:45", (60,45), E);
label("1:15", (15,0), S);
label("2:00", (60,0), S);
label("2:00", (0,60), W);
fill((0,0)--(60,60)--(60,45)--(15,0)--cycle, gray(.7));
fill((0,0)--(60,60)--(45,60)--(0,15)--cycle, gray(.7));
[/asy]

The shaded region represents the times that Allen and Bethany would see each other at the party. For example, if Allen arrived at 1:30, Bethany could arrive at any time between 1:15 and 1:45 and see Allen at the party. Let one hour equal one unit. Then, we can calculate the area of the shaded region as the area of the entire square minus the areas of the two unshaded triangles. This will be equal to $2\cdot \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{3}{4}=\frac{9}{16}$. So, the area of the shaded region is $1-\frac{9}{16}=\boxed{\frac{7}{16}}$. Since the area of the square is 1, this is the probability that Allen and Bethany see each other at the party.[/hide][/quote]

heyyyyyy, you just copied the alcumus solution. CHEATER[/quote]

dang 1 year ago today.

anyway, trivial just graph |x-y|<=15

[quote=pinkpig]an exact same problem except with different names appeared on mathcounts.

why is this not sourced, or did mathcounts steal a problem from aops?[/quote]

this is a very generic problem xd


linear algebra
matrix
invariant
linear algebra unsolved

$ X,Y,Z$-square matrix with complex enties, such that $ rank(XY\minus{}YX\plus{}I)$=1, $ XZ\equal{}ZX$, $ YZ\equal{}ZY$ ($ I$-identity matrix). Prove that $ Z\equal{}aI$, for some complex a.

An idea for beginning.
We suppose that $ n\geq2$. Let $ A\equal{}XY\minus{}YX\plus{}I$. $ Spectrum(A)\equal{}\{\lambda,0,\cdots,0\}$.$ trace(A)\equal{}n$ then $ \lambda\equal{}n$. $ A$ is similar to $ diag(n,U)$ where $ U$ is nilpotent. $ rank(A)\equal{}1$ therefore $ U\equal{}0$ and we may assume that $ A\equal{}diag(n,0,\cdots,0)$ and $ XY\minus{}YX\equal{}diag(n\minus{}1,\minus{}1,\cdots,\minus{}1)$ (invertible matrix). If $ ZX\equal{}XZ,ZY\equal{}YZ$ then $ Z,XY\minus{}YX$ commute and $ Z\equal{}diag(\alpha,V)$ where $ V$ is a $ (n\minus{}1)x(n\minus{}1)$ matrix.
It remains to prove that $ V\equal{}\alpha{I_{n\minus{}1}}$.
Assume that $ \alpha\notin{spectrum(V)}$; let $ u$ s.t. $ Z(u)\equal{}\alpha{u}$ and $ (XY\minus{}YX)u\equal{}(n\minus{}1)u$. $ X(u),Y(u)$ are eigenvectors of $ Z$ associated to eigenvalue $ \alpha$. Then $ X(u)\equal{}\beta{u},Y(u)\equal{}\gamma{u}$ and $ (XY\minus{}YX)u\equal{}0$ that is a contradiction. Thus $ \alpha\in{spectrum(V)}$.
After ????

Three years after...
The sequel of  post #2.
Let $E$ be a common non zero invariant subspace of $X,Y$. Let $X',Y'$ be the restrictions of $X,Y$ to $E$. $trace(X'Y'-Y'X')=0$. Moreover $X'Y'-Y'X'$ is the restriction of $XY-YX$ to $E$. Then $U=spectrum(X'Y'-Y'X')\subset\{n-1,-1,-1,\cdots\}$. Clearly $\sum_{\lambda\in{U}}\lambda=0$ is impossible except if $U=\{n-1,-1,-1,\cdots\}$, that is $E=\mathbb{C}^n$. Thus $X,Y$ has no non trivial common invariant subspaces. According to Burnside's theorem,
cf http://www.stat.berkeley.edu/users/evans/Halperin_Rosenthal_80.pdf
the algebra generated by $X,Y$ is $\mathcal{M}_n(\mathbb{C})$. Finally $Z$ commute with this algebra and is a scalar matrix.


modular arithmetic
number theory proposed
number theory

[quote="who"]find all primes $p,q$ such that, $2p+1=3^{q}$[/quote]

First note that $q=2$ isn't any solution. Then we have that $3^{q}\equiv3\pmod{q}$. Applying that we obtain

$2p+1=3^{q}\equiv3\pmod{q}$

$2p\equiv2\pmod{q}$

$p\equiv1\pmod{q}$

$p=tq+1$ for some $t\in N$

Then we have:

$2tq+3=3^{q}\rightarrow q=3 \vee 3|t \rightarrow$

first case
$q=3$ implies $t=4$ implies $q=3$ and $p=13$.

second case
$3|t \rightarrow t=3s\rightarrow 6sq+3=3^{q}\rightarrow s=\frac{3^{q-1}-1}{2q}$.

It is sufficient that $\frac{3^{q-1}-1}{q}$ to be integer. But we always have (using FLT) that $3^{q-1}\equiv1\pmod{q}$. So it is always an integer.

Conclusion:

$(p,q)\in\left\{(13,3),\left(\frac{3^{q}-3}{2}+1,q\right)\right\}$

[color=blue]Still remains to examine for what primes $q$ is $\frac{3^{q}-3}{2}+1$ a prime too..[/color]
conceive

Another solution is (p,q)=(1093,7).

[quote="conceive"][color=blue]Still remains to examine for what primes $q$ is $\frac{3^{q}-3}{2}+1$ a prime too..[/color][/quote]
But it's the same as original problem... (You want to find a prime $p$ such that $p=\frac{3^{q}-3}{2}+1=\frac{3^{q}-1}{2}\Leftrightarrow 2p+1=3^{q}$)

For me, this problem is very similar to searching for Mersenne primes, what is an open problem, so I have doubts if it has an easy solution. Who, do you know how to solve it?

[quote="who"]infact the pro. x^w-1/x-1=prime no. is solved.sol. isnt of mine.[/quote]
Are you sure? For $x=2$ it would mean that it is solved when $2^{w}-1$ is a prime!

:( ,for my stupidity.
 one of my france friend told where x other tyhan 2 pro. is solved.he is at barbeaux studying diophantine eq.


calculus
integration
conics
hyperbola
function
calculus computations

Compute:
$ \int _{1}^{2} \frac{\sqrt{1\plus{}t}}{t} dt$

The substitution $ u\equal{}\sqrt{1\plus{}t},$ written backwards as $ t\equal{}u^2\minus{}1,$ turns this into the integral of a rational function. From there, the method should be routine.

this is my solution to it is there another one?

You can do a totally unnecessary substitution of

$ \sinh^2(x) \equal{} t$

Then you can easily transform the expression into:

$ 2 \int^{\sinh^{\minus{}1}\left(\sqrt{2}\right)}_{\sinh^{\minus{}1}\left(1\right)} \left(\frac{1}{\sinh(x)}\plus{}\sinh(x)\right)\,dx$

and this is basic integration.

A question for those with a favorite computer algebra system (be it Mathematica, Maple, a TI-89, or whatever): how does that system choose to express the antiderivative $ \int\frac{\sqrt{1\plus{}t}}{t}\,dt$ and how does it choose to express the definite integral as FOURRIER originally stated it?

An interesting observation! :) Mathematica gives the result of the definite integral using inverses of hyperbole trigonometric functions, and the antiderivative using natural logarithms.


Find a method to make the conversion

$ R\minus{}CH\equal{}CH\minus{}CO\minus{}CH_3 \longrightarrow R\minus{}CO\minus{}CH\equal{}CH\minus{}CH_3$

see if my steps are right????

1.AT 120 C,treat with LAH to get 1,4-addition  and we get the alcohol
2.Use P/HI to get the alkene
3.Use SeO2 for allylic hydroxylation
4.use PCC to convert it into the required product.

Your proposal would have very poor selectivity.

can you mwention in which step???

ALso,see if this systhesis holds good

1.perform MPV redsuction to get the alcohol
2.Use P/HI to remove the alcohol
3.Treat with water to get markanikoc's addition.
4.perform oppenaur oxidtaion to get the ketone.
5.use rep/Cl2 to get a-chlorination.
6.Use aqKOH to get the alcohol
7.Pefrom dehydartion with bulky base to get the required product.

[quote="Valeriummaximum"]3.Use SeO2 for allylic hydroxylation
4.use PCC to convert it into the required product.[/quote]

Could you explain those two steps?

[quote="Valeriummaximum"]3.Treat with water to get markanikoc's addition.[/quote]

Are you forgetting the R group? What would be Markovnikov addition here?

The alkene being the reactive site of the molecule....
so,what will happen is that?????
the markanikov's rule will be followed...(correct me if i am wrong)

SeO2 results in allylic hydroxylation..........
PCC is pyridinium chloro chromate...
it is an oxidising agent that converts primary alcohol's to aldehydes and sec-alcohols to ketones..

[quote="valeriummaximum"]
5.use rep/Cl2 to get a-chlorination.
[/quote]
what is that supposed to be?? And what are u trying to do with aq.KOH and bulky base. U can straight away add alc.KOH because it will create a conjugated system.

I think (after long search) the appropriate reagents for alpha bromination are NBS acting as the brominating agent and 2-benzhydryl-pyrrolidine as the catalyst. I think I am 99% sure of this reagent.

1) use 9-BBN followed by $ \ce{CH2(OH)CH2(NH2)}$
2)Wacker Process
3)$ \ce{Al2O3}$
this gives u the desired side double bond due to conjugation....
is this ok

[quote="Valeriummaximum"]SeO2 results in allylic hydroxylation..........
PCC is pyridinium chloro chromate...
it is an oxidising agent that converts primary alcohol's to aldehydes and sec-alcohols to ketones..[/quote]

Thanks, but I already know that. What I meant was, for example with SeO2, what allylic position would be oxidized  in the case R = n-Pr? Both would be hydroxilated, so that's not a general method.

ok...
i didnt understand ur question properly.

How abt this?
Treat with HBr to get 1,4 addition product.
Carry out MPV reduction to alcohol
React with DMSO/HCO3- to get ketone
Foll. By dehydration to get the req. product.
[quote="chemrock"]
U can straight away add alc.KOH because it will create a conjugated system.
[/quote]
arent you forgetting something?

[quote="VARUNARASU"]React with DMSO/HCO3- to get ketone[/quote]

Don't forget there is an -OH group on the molecule.


geometry
quadratics
trigonometry
cyclic quadrilateral
circumcircle
algebra
trig identities

ABCDE is a cyclic pentagon with AB = BC = CD = 2 and DE = EA = 3. Compute the length of AD.

[hide="a crappy start"]
Consider cyclic quadrilateral $ ABCD$.
Let $ O$ be the intersection of $ AC$ and $ BD$.

$ \triangle AOD \sim \triangle BOC \implies AD \equal{} \frac {2a}{b}$
$ \triangle ABO \cong \triangle DCO$by ASA$ \implies AO \equal{} DO \equal{} a$ and $ BO \equal{} CO \equal{} b$

Let $ EC \equal{} x$
By Ptolemy's
on $ ABCD, \frac {4a}{b} \plus{} 4 \equal{} (a \plus{} b)^2$
on $ BCDE, 6 \plus{} 2x \equal{} (a \plus{} b)x$
on $ EABD, 3(a \plus{} b) \plus{} 6 \equal{} \frac {2ax}{b}$

and I can't solve this. There's probably an error here somewhere.
[/hide]

Here's my idea after realizing that I did read the problem mistakenly :blush: :
O is the center of the circumscribed circle of this pentagon
Let angle AOB=alpha
     angle AOE=beta
Since AB=BC=CD and AE=ED, we will have: 3alpha+2beta=360
                                                          sin alpha/sin(beta)=2/3
Now multiply crossingly, we will have: 3*sin alpha=2*sin(beta)=2*sin(180-3/2alpha)=2sin(3/2)alpha
This equation is equivalent to: 3.sin x=2(sin(x).cos1/2x+cosx.sin1/2x) (x=alpha)(I'm too lazy to rewrite alpha so many times...)
                                          2sinx(1-cos1/2x)=2sin1/2x(2cosx-cos1/2x)
                                          2cosx/2(1-cos1/2x)=(4cos^2(x/2)-2-cos(x/2))
                                          6cos^2(x/2)-3cos(x/2)-2=0
Now solving this quadratic equation to get cos(x/2). From that point, calculate sin(x/2) and easily to get sinx. The rest is applying Law of Sines into triangle AOB to find the radius of (O). Calculate sin (beta) and AD, we're done :wink:
PS: I haven't checked my series of substitutions! Anybody please let me know whether it's true or not!I;m too sleepy now!  :lol:

When is $ 3 \alpha \plus{} 2\beta \equal{} 360$ true?
Is that true whenever a circle circumscribes a polygon? That is, if a central angle $ \angle AOB$ subtends an arc of measure $ \alpha$ and $ \angle BOC$ is constructed such that $ AB\equal{}BC$, does that imply that the arc measure of $ BC$ is also $ \alpha$?

[quote="RisingStar"]When is $ 3 \alpha \plus{} 2\beta \equal{} 360$ true?
Is that true whenever a circle circumscribes a polygon? That is, if a central angle $ \angle AOB$ subtends an arc of measure $ \alpha$ and $ \angle BOC$ is constructed such that $ AB \equal{} BC$, does that imply that the arc measure of $ BC$ is also $ \alpha$?[/quote]

Yes, think about it.
[hide]
Let $ \cos \angle AED\equal{}\minus{}\cos \angle{ABD}\equal{}\minus{}\cos \angle ACD \equal{}x$.  Then by the Law of Cosines,

$ AD^2\equal{}18(1\minus{}x)$
Also,
$ AC^2\equal{}BD^2\equal{}AC \cdot BD\equal{}8(1\plus{}x)$
Apply Ptolemy's Theorem to $ ABCD$,
$ AD\equal{}2(1\plus{}2x)$

So we have
\[ 18(1\minus{}x)\equal{}4(1\plus{}2x)^2\]
\[ 18\minus{}18x\equal{}4x^2\plus{}16x\plus{}16\]
\[ 2x^2\plus{}17x\minus{}1\equal{}0\]
\[ x\equal{}\frac{\minus{}17 \plus{} \sqrt{297}}{4}\]
\[ AD\equal{}\minus{}15\plus{}\sqrt{297}\]
[/hide]

Is there supposed to be some nice solution to arrive at this ugly answer?  Or is the answer different?

Let $ AD \equal{} x,$ $ BE \equal{} CE \equal{} y$, $ AC \equal{} BD \equal{} z.$ $ ABCE$ is cyclic,

$ \frac {13 \minus{} y^2}{12} \equal{} \frac {EA^2 \plus{} AB^2 \minus{} BE^2}{2 \cdot EA \cdot AB} \equal{} \cos \widehat{EAB} \equal{}$

$ \equal{} \minus{} \cos \widehat{BCE} \equal{} \minus{} \frac {BC^2 \plus{} CE^2 \minus{} BE^2}{2 \cdot BC \cdot CE} \equal{} \minus{} \frac {1}{y}$

$ y^3 \minus{} 13y \minus{} 12 \equal{} 0$

Looking for possible rational roots of this cubic ($ \pm$ integer dividers of $ 12$), $ BE \equal{} CE \equal{} y \equal{} 4$ is the only positive root. (The other two roots are integers $ \minus{} 1,\ \minus{} 3$ both negative.) By Ptolemy for cyclic $ ABCE,$

$ 2y \plus{} 6 \equal{} AB \cdot CE \plus{} BC \cdot EA \equal{} AC \cdot BE \equal{} yz,\ \ AC \equal{} BD \equal{} z \equal{} \frac {7}{2}$

By Ptolemy for cyclic $ ABCD,$

$ 4 \plus{} 2x \equal{} AB \cdot CD \plus{} BC \cdot DA \equal{} AC \cdot BD \equal{} z^2,\ \ AD \equal{} x \equal{} \frac {33}{8}$

See also [url]http://www.mathlinks.ro/viewtopic.php?t=78901[/url].

Very nice solution, yetti! :)
I'm curious why nobody takes a while to check my solution.I'm sure that it works but the final result is still the most important part!

[quote="ghjk"]sin(alpha)/sin(beta)=2/3[/quote]
is wrong. Correct is

$ \frac {2}{R} \equal{} \frac {\sin \alpha}{\sin\left(90^\circ \minus{} \frac {\alpha}{2}\right)} \equal{} 2 \sin \frac {_\alpha}{^2},\ \ \frac {3}{R} \equal{} \frac {\sin \beta}{\sin\left(90^\circ \minus{} \frac {\beta}{2}\right)} \equal{} 2 \sin \frac {_\beta}{^2}\ \Longrightarrow\ \frac {\sin \frac {\alpha}{2}}{\sin \frac {\beta}{2}} \equal{} \frac {2}{3}.$

and you get a quadratic for $ \sin \frac {_\alpha}{^4}$ with an acceptable root $ \frac {_1}{^4}$ (and the other root $ \minus{}1$).

$ AD \equal{} 2R \cdot \sin \frac {_{3 \alpha}}{^2} \equal{} ... \equal{} \frac {33}{8}$ [color=white] . [/color] (Substitute $ R \equal{} \frac {1}{\sin \frac {\alpha}{2}}$ and use formula for sine of triple an angle.)

Yup! :oops: Thank you for pointing my mistake, yetti!


In a triangle ABC,angle A=60degree.Prove that BC^2=AB^2+AC^2-AB.AC


Does anyone know if there is a school in Mississauga preferably near to The Woodlands (my school) that is taking the AMC 12 A or B, and if it isn't too late to register? Could you PM me the phone number or email address of the math teacher?

i don't think its too late to register for your school but i think glennforest s.s. had a usamo qualifier last year (i don't live near mississauga)

i hope this helps you :)

I'm from Glenforest, and I can confirm that. Yes, we are holding AMC 12.  :)

Ok, thanks, I'll try to contact Glenforest's math department!

EDIT: I just realized that there is a DECA provincials conflict with the AMC 12 A, so for anyone who goes to glenforest, are they offering the B date as well?

they were only offering the B date. Sorry for the late reply!  :(

Oh well, I guess I'l have to take them next year.... :mad:  :mad:  :(


[quote="ddziabenko"]Triangle $DEF$ is acute. Circle $c_1$ is drawn with $DF$ as its diameter and circle $c_2$ is drawn with $DE$ as its diameter. Points $Y$ and $Z$ are on $DF$ and $DE$ respectively so that $EY$ and $FZ$ are altitudes of triangle $DEF$ . $EY$ intersects $c_1$ at $P$, and $FZ$ intersects $c_2$ at $Q$. $EY$ extended intersects $c_1$ at $R$, and $FZ$  extended intersects $c_2$ at $S$. Prove that $P$, $Q$, $R$, and $S$ are concyclic points.[/quote]

We already have drawn two altitudes of triangle DEF, namely the altitudes EY and FZ from the vertices E and F. Now, draw the altitude DX from the vertex D, where X is the foot of this altitude. These three altitudes DX, EY and FZ concur at the orthocenter H of triangle DEF. Since < DXE = 90°, the point X lies on the circle with diameter DE, i. e. on the circle $c_2$. Clearly, the point D also lies on this circle. As we know, the points Q and S also lie on this circle $c_2$. Hence, by the intersecting chords theorem, $HQ\cdot HS=HD\cdot HX$. Similarly, $HP\cdot HR=HD\cdot HX$. Thus, $HQ\cdot HS=HP\cdot HR$. Hence, by the converse of the intersecting chords theorem, the points P, Q, R and S are concyclic.

  darij

[quote="Mathx"]:o  :o  :o  Strange.
A short proof from Darij. :o  :o  :o[/quote]

The problem is actually more or less trivial, that's all. ;)

  darij

[quote="Mathx"]:o  :o  :o  Strange.
A short proof from Darij. :o  :o  :o[/quote]
 :lol:

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.AMO2005Problem3

Vo Duc Dien

This problem is almost the reverse of the problem 2 Brazil Revenge 2002.

Dear Mathlinkers,
this problem appears like a converse of the three chords theorem... any calculation isn ot necessary.
Sincerely
Jean-Louis


The url conversion should not work in code tags

[code]http://www.artofproblemsolving.com/Forum/index.php[/code]

should be

[code]http://www.artofproblemsolving.com/ Forum/index.php[/code] (without the space).


I have office 07 enterprise on my home computer but when I go to school to work, I need something that will be like office 07 that I can put on my flash drive. If this is not possible, then is there something that will let me edit .docx on openoffice?

Thanks in Advance

Open Office xml translator will work I believe.

[quote="ernie"]When you are working on 2007 Word, and you need to work on it on a different computer with an earlier version of word, you can go to Save As, then to the right, you will see the different formats.  Instead of clicking Save As, click whichever format you need.[/quote]

I think this works for all the Office 2007 applications, with the exception of OneNote.


algorithm
topology
geometry
calculus
search
Support
number theory

I have a problem concerning links between mathematics (and theoretical sciences in general) and ethics,  which has baffled me since I started studying math, physics and computer science at university (I'm second year now). The more I study pure math and theoretical computer science (by "theoretical" I mean here stuff like logic, automata, algorithms, recursion and complexity theory etc.), the more I'm convinced that the most of today's research in these topics won't have ever any practical use - they will remain purely theoretical fields, beautiful and interesting as they may be, but with no applications whatsoever. I'm an undergraduate, so perhaps I'm not very knowledgeable about these fields, but my impressions from what I've learnt so far are that:

- most of pure mathematics is completely useless, though very beautiful; just think how many discoveries are made annually in topology, abstract algebra, number theory, (put your favourite field here) and how few of them actually have any applications
- as for theoretical computer science, the above applies; many really clever and nice algorithms are never implemented, because if they improve running time from $ O(n^3)$ to $ O(n^{2,8})$; I fear that the same can be said of complexity theory - many fantastic results, but I doubt that a fraction of it will actually be used

Of course there are stock counterexamples here, stuff that was introduced for purely theoretical reasons and then found useful by physicists etc. - Riemannian geometry (general relativity), Hamiltonian mechanics, group theory (particle physics/field theory), mathematical logic (theory of computation), intuitionistic logic (lambda calculus) and so on. However, I feel that: a) if useful to scientists, these  things would have been discovered anyway (perhaps it would have taken more time and effort), b) if you justify doing theoretical research by saying "you never know what might prove applicable someday", you have to agree that, still, you have to prove 100 theorems to find 5 or 10 with any applications (I would call it inefficient were I an engineer  :) ), c) even if a theoretical  subject is used by empiricists, only the most "down-to-earth" part of the theory is really used and the more abstract and beautiful aspects are neglected.

Do you think it's ethical to pursue a carrer in a field that will never produce any applications, doing science for science's sake? Wouldn't it be better if instead of proving yet another theorem about this topological space or other one did something that could really help people out there, use his wits and talents (and surely people who think of academic carerr have some of these) to humanity's good? (sorry if it') Or am I wrong about the whole "applications" thing?

[quote="neworder"]- most of pure mathematics is completely useless[/quote]

This is an outdated way of looking at pure mathematics, and before you ask the question you asked in your topic you should ask, "is pure mathematics useful?"  Here is something in the way of an answer.

Pure mathematics is [b]basic research.[/b]  While it is not done with applications in mind, applications tend to be discovered after the research is done.  This is sometimes referred to as [url=http://www.dartmouth.edu/~matc/MathDrama/reading/Wigner.html]the unreasonable effectiveness of mathematics[/url].  Examples:

- Group theory became a fundamental part of quantum field theory [i]decades after[/i] it was developed.  In fact, fundamental physics today requires an awful lot of deep mathematics.
- Knot theory is finding applications in molecular biology, particularly in the study of complex molecules.
- Number theory was thought to be the height of useless, pure mathematics until it found applications in cryptography, hence e-commerce, and it is now a vital part of our daily lives.

I'm not the person to consult as far as examples in every field you might be wondering about, but spend more effort questioning this assumption.  :)

Besides proving yet another theorem about this metric space or other I also do what I can to really help people out there, thus using my wits and talents (or what's left of them) to humanity's good.  :)

[quote="neworder"]Maybe I'm exaggerating a bit, but sadly, many people in science don't really care for the ethical side. When asked about it, they answer something along the lines: "I study math because I take pleasure in it" or "Everybody should do what they are best in", and that's it, not interested in the "big picture" of their research.

I dream of pursuing scientific career, but I don't want to end up developing a sophisticated abstract theory that will bring nobody any good, besides me and my advisor ;).[/quote]

Read Hardy's Apology, its a great read and answers many of the questions you have asked here. But I think that you will have many disagreements with him, especially over your view about people going into whatever their good at. All the same, its well worth reading, even if just too see how all the quotes which are littered around come together.

[quote="neworder"]When asked about it, they answer something along the lines: "I study math because I take pleasure in it".[/quote]
That's what I would probably answer too. I don't consider it unethical, and in any case I don't see why it makes you sad. If you don't want to do math, then you don't have to.

Why is it unethical? At first by unethical I thought you meant by when Biologists/Chemists etc. test drugs or something on living creatures. I don't see studying something for the sheer pleasure and excitment of the subject as being unethical.

Also a great deal of Pure Mathematics have also found their way into Financial Mathematics and Economics.

neworder is referring to effort "wasted" in pure mathematics as opposed to other, more "useful" disciplines.  The idea is that if you have talent, you should use it to benefit others, and if you aren't, you're being unethical to the extent that you aren't doing the best for society that you could be.

The question can be rephrased as follows:  [i]do the scientifically talented have an [b]obligation[/b] to society to use their talents to benefit society?[/i]

[url=http://www.google.com/search?hl=en&q=dave+rensin&btnG=Google+Search]Dave Rensin[/url] gave a very interesting lecture on the subject at the Research Science Institute.  He wasn't preaching, so I can't give you a concise summary of what we talked about (it was a discussion more than a lecture), but the emphasis he put is on the word [b]obligation.[/b]  Do you really think that just because some people are more talented then others they are [b]obligated[/b] to do a particular thing with their lives?

[b]SimonM:[/b]  Hardy's [u]Apology[/u] is somewhat outdated.  Hardy took pride in the uselessness of mathematics; he saw it as an improvement over other disciplines whose discoveries could be used for warfare.  Ironically, number theory is now one of the most important areas of practical mathematics.  I'd suggest [u]Letters to a Mathematician[/u] for a more updated view of a mathematical career.

[quote="neworder"]Maybe I'm exaggerating a bit, but sadly, many people in science don't really care for the ethical side. When asked about it, they answer something along the lines: "I study math because I take pleasure in it" or "Everybody should do what they are best in", and that's it, not interested in the "big picture" of their research.[/quote]

Are you so sure this is unique to science?  I don't think most lawyers get into the law to be defenders of the downtrodden, for example.

[quote="t0rajir0u"]
[b]SimonM:[/b]  Hardy's [u]Apology[/u] is somewhat outdated.  Hardy took pride in the uselessness of mathematics; he saw it as an improvement over other disciplines whose discoveries could be used for warfare.  Ironically, number theory is now one of the most important areas of practical mathematics.  I'd suggest [u]Letters to a Mathematician[/u] for a more updated view of a mathematical career.
[/quote]

Outdated, perhaps. Doesn't stop it being a good read. We don't stop reading Classical texts, because more modern books are out. Yes, he was a little short sighted in alot of what he said. Especially with regards to how he felt mathematics was completely useless.  I have also read Letters to a Mathematician (assuming thats Ian Stewart's one?) (I have too many of thats man's books. Damn him for being my dad's lecturer!).

I know number theory has many different practical usages, especially prime number theory, but there is more to encryption that number theory. For example, with the current "factorising" encryptions, there is a issue with mathematical faults on the computer. The problem is that mathematics is disconnected from reality. It makes fantastic models, brilliant in fact. However, at its roots it is still the manipulation of axioms.

Back to my point of view. I obviously must feel that mathematics has some worth as in about 5 years time I will probably be entering research, so meh... most of what I've just written was random wanderings...

Wasn't there something about the Riemann Hypothesis (the zero distribution) being reminiscent to certain concepts in Quantum Mechanics?

[quote]I don't consider it unethical, and in any case I don't see why it makes you sad. If you don't want to do math, then you don't have to.[/quote]
[quote]Why is it unethical? At first by unethical I thought you meant by when Biologists/Chemists etc. test drugs or something on living creatures. I don't see studying something for the sheer pleasure and excitment of the subject as being unethical. [/quote]

It may be unethical because instead of, say, proving a neat but useless theorem just for fun you could be working on a cure for cancer, helping poor people etc. You do nothing bad, but nothing good either, while you could use your time and effort to help real people, directly or indirectly.

There are many claims that abstract pure math is very useful for theoretical physics, but from what I've seen so far (though I'm no expert here), parts of so called "mathematical physics" is really "100% math, and perhaps some physics at the end, but who cares?". For example, a guy who works on mathematical physics was giving an introductory talk about noncommutative geometry, quantum groups etc. and when asked about applications to quantum mechanics he answered: "Well, you have C*-algebras, you study them, you can introduce differentiable structe, integrate over them... and supposedly in the end there is some physics, though I don't really know". I doubt this is an isolated example. Maybe someone with more background in physics can say something.

As for Hardy, I've read the "Apology" and I think that it's a good book if you want to encourage young people to study mathematics, but won't help you solve problems like "should I study algebraic number theory or computational biology?". Also, I found Hardy's approach too elitist for my taste (hardly a surprise, since it's a common criticism).

[quote="neworder"]
It may be unethical because instead of, say, proving a neat but useless theorem just for fun you could be working on a cure for cancer, helping poor people etc. You do nothing bad, but nothing good either, while you could use your time and effort to help real people, directly or indirectly.  [/quote]

Ethics is a massive topic all on its own, and the ethics behind how people spend their time is even more so. I'm made to feel unethical fairly frequently at school since I have been place in a group which contains a high percentage of wannabe medics applying to university. Their applications run along the lines of "I've always wanted to help people". Mine always comes out as "I enjoy mathematics".

Cure for cancer - I have a strong dislike for that phrase. I have a friend who has been battling with severe bone cancer for nearly a year, and I'm sure he'd be in a better place to explain why there is not going to be a "pill for cancer". Its a massive area with a wide variety of different issues, and whilst working on it, you may not achieve anything in your life. Would you consider that a "waste" of time?

Helping poor people - How is mathematics as a topic any more guilty of not doing this as anything else. I could claim you being on this forum is unethical, because you could be working for the poor. Same goes for you being at university.

Back to the point, ethics is a massive area in philosophy, I don't think they've got the hang of it yet, so as far as I'm concerned, I've got one life, and I'm not going to spend it worrying about every little thing that I do being ethical (otherwise I'd be a philosopher, but then you'd get a debate about whether philosophy is ethical/useful which it is probably more guilty at than mathematics)

This reminds me a bit of the Cuban situation. I think it's loosened up now, but it used to be that good students were forced to be doctors and such. Education was free (and pretty good too), but you had to do whatever the government wanted you to do.

I think people should do whatever they are passionate about and/or whatever they are good at doing. For example, if someone is good at writing maybe that person will become a renowned journalist or a brilliant novelist or a merciless muckraker. Every skill has potential value.

And even things that wont "broaden human horizons" and such, if someone decides to become a crazy math-professor nut, the man still might be helping support his family and children with the money he makes. There's a contribution.

A Professor at University of New South Wales once told me that:

'Sometimes you have to make the "right" decision not the "sensible" decision.'

That was when he was telling me about the studying Sciences at he University. He was telling me about the employment rate of studying Sciences. Obviously many students going to UNSW wanted to study Medicine because of the high employment, stable income and social status but may not be very passionate about their subject.

That's beside the point but what he said was quite true. Do what you think is 'right' not what you think is 'sensible'. If you love Mathematics then study it. I mean if you think cancer research is 'good' and then go for that instead because it is 'sensible' rather than because you are passionate then the chance of you doing anything 'good' is rather slim.

[quote="neworder"]but my impressions from what I've learnt so far are that:

- most of pure mathematics is completely useless, though very beautiful;[/quote]

One day I also asked myself similar questions. The problem was that I liked some “pure” mathematical disciplines but didn’t see how they apply. So I had nearly the same “problem” and was quite depressed. I was especially anxious about topology and algebra, I considered them artificial and fruitless. But when I understood that the theory of  PDE, being of practical use, is also deeply related with much of pure mathematics, I made a decision to choose PDE as future specialty. Now I see how linear and nonlinear functional analysis, topology (e.g. degree of a map, numerous fixed point theorems, like that of Schauder), critical point theory, bifurcation theory, interpolation etc. are being used in PDE. You may study quality properties of solutions, investigate their multiplicity, and find bifurcations etc. All this demand other “pure” mathematics, at least the stuff I named above. I know very little (since I am just finishing my Master Degree) and cannot say what else is used in PDE, I even know very little those disciplines I listed. So now, when I study PDE, I feel as being a little functional analyst, a little topologist etc. I start to believe that PDE is a center of mathematics (I don’t intend to hurt feeling of those engaged in other fields of math), a field by means of which pure mathematics finds its way to physics and other applications. So if you recognize the applicability of differential equations (at least, PDE), then you must recognize [i]much of other mathematics' applicability[/i].

When I heard a lot about how mathematics is unapplicable in the real world I was shocked(since it is my favorite subject) and I decided to find an application of mathematics in the real world. From what I have found from various sources on the internet(I don't know if they are realiable or not, hopefully they are) is that Geometry/Topolgy can be applied in the field of protein folding. Incorrect protein folding in someone can cause things such as Cystic Fibrosis(which doesn't have a cure yet). Topology and Geometry can be applied in this field to better understand how proteins attain their shape. Now, what I don't know is if this application of Mathematics is unreasonable since a protein can take trillions of shapes.  However, geometry and topogoly definitely seem to be applicable in this field of biology!

@sonny: t0rajir0u mentioned above that $ \text{Knot Theory}$ finds application in molecular biology, especially in the study of complex molecules.

I'm one of those "unethical" souls who does mathematics just for the sake of it (or, perhaps, because that's the only thing I can do reasonably well). I can also tell that none of my theorems has any chance to have any immediate application to "real life". Do I feel ashamed? Not really. There are several reasons for that (listed in completely random order with those already mentioned above omitted)

1) The way the professor's career is set up in the modern society involves a huge amount of direct "teaching other people to think in mathematical way". This involves regular classes attended (if they are good ones, I'm not talking about the so called "business calculus" here though some may argue that even it is not complete waste of time) by future engineers and physicists, consulting work, educating schoolteachers and schoolchildren, etc. (why, even hanging on AoPS in the evenings may count here :) )

2) Mathematics can be considered not only as science, but also as an art. I admit that the number of people who can appreciate mathematics is a bit less than that of those who can appreciate, say, music but it is the problem with humanity, not with mathematicians; the "artistic" side of mathematics is by no means less elaborated and beautiful than that of other arts. It doesn't appeal directly to one's emotions but rather to one's intellectual side and it will just take some (alas, probably too long for any of us to see it) time to reach the point at which the intellectual pleasures will be valued equally with the emotional ones.

3) Most "useful" professions are based on imperfections of human beings. Doctors exist because people get sick, car manufacturers exist because the maximal sustainable speed of an average footman is around 5mph, politicians and soldiers exist because the humans are uncapable of behaving in constructive organized ways without "sticks and carrots". Quants, actuaries, and security people exist... (well, should I comment on these ones? Those are three main carrer choices for math majors outside academy and guess why we need any of them :P). I hold an optimistic view that the imperfections are temporary, so sooner or (rather) later most of the "useful for the society" professions will just die out. Mathematics on the other hand has a chance to remain forever because it is based on human curiousity and desire to know and be able to do more than we currently do (which, in my humble opinion, is the only thing that distinguishes our mind from that of a frog).

I think that 4 is a good number, so I'll skip a few more and finish with the reason that

4) Doing pure mathematics is relatively harmless. Most "useful" profession bring at least as much harm to humanity as they bring benefits. We still have to see if atomic energy will save or destroy civilization. The same can be said about industries based on fossil fuels if you believe the Global warming theory (I do). Obviously, this is true for anything directly dealing with armed forces and it is hard to find an interesting for a bright person "real life" profession that Army or Navy would not be interested in nowdays, etcetera.

As to the question whether a talented person has an obligation to devote his talent to the society needs, I think he is but he should not be obliged to spend his time on solving immediate problems created by irresponsible behavior of other people, so, sorry, I'm not interested in creating new weapons and secret communication codes for the armies (by the way, which ones? They do not seem to be eager to share with each other and though the tradition is to choose according to the birthplace or the current citizenship, the ultimate judgement whether a scientist working for an army is a hero or a criminal is passed according to which one is the winner :P), curing infectious diseases that spread only in well known and easy to avoid ways, stabilizing failing economies in the countries where people are fighting senseless civil wars against each other, finding new ways of moving rectangular pieces of green paper from one pocket to another, etc. If somebody offers me to take part in the project that ultimately will bring us to the stars, and if I feel that I'll be able to be really useful there, I'll probably change my career but as to anything short of that, I'd better stay with proving "useless theorems in a not-so-fashionable field" :).

If you ever want to do any philosophizing you should read [url=http://www.paulgraham.com/philosophy.html]this piece by Paul Graham[/url] first.

Its basic point is that most of philosophy—including ethics—is bullshit. If we want to answer the question, "Is it ethical to spend time on an activity that doesn't improve the world* when you could spend time on one that does," Graham's answer is that "it entirely depends on how you define 'ethical'". And he's completely right (IMHO).

With that in mind, the best advice that I can give you is to completely ignore questions about how ethical something is, because those questions aren't well-defined (arguably they aren't defined at all).

Just ask yourself, "Will I be happy doing this?" If you will, do it.

That's not to say that you should ignore the issue of how useful your actions will be to the "advancement of humanity." It's just that you should weigh that issue in a reasonable manner. If mathematics' uselessness in relation to other work you might do is going to make you unhappy, don't do it.

But don't go searching for ethical validation or invalidation of any course of action; if you're at all logical, you won't find it.

*I'm just going to assume that math is useless in this post. In general it probably isn't, but that doesn't affect my point.

This is a good topic. One I struggled with for awhile. Then I took an ethics class. I hope I don't repeat something someone else said (I only skimmed the others since I started formulating this post once I read the original and questioned whether or not it was an old post of my own that I had forgotten about).

You are making a huge assumption that Utilitarianism is the only form of what constitutes an ethical act. In fact, utilitarianism (the greatest good for the greatest number principle) is considered one of the most fallacious schools of ethics out there. Suppose you had to torture your child to save three elderly people. The greatest good for the greatest number tells you to do it. Is that ethical? Is that what you would do? Probably not.

Look at Rand's Objectivism. If everyone were to go into what they were passionate about and loved, wouldn't humanity be much better? They would be more productive and probably more excited to collaborate and find applications since they are doing what they love and not what they feel they have to do.

You could take Reuben Hersh's (I think, maybe Tymoczko?) perspective that all mathematics came from nature and so will eventually end up referring back to it. We shouldn't consider it an unreasonable when something is effective in science, but completely natural, since mathematics stemmed originally from observations in nature.

I could go on. Just look up other philosophies of ethics. Lookup: relativism, deontology, virtue theory, moral responsibility, ... I don't even know. Most theories would probably not prescribe the greatest good for the greatest number. That is just a common misconception about what "ethics" is.


[quote="The Original Pi Guy"]kalva 1996 number 5

here's the way i did it

-t = (-3-a)(-3-b)(-3-c)
t =(3 + a)(3 + b)(3 + c)
t = 27 + 9(a + b + c) + 3(ac + bc + ab) + abc
t = 27 - 27 + 12 - 11 = 1????
Kalva says it should be 23....
where did I go wrong????[/quote]

abc = -(-11) so it should be 27-27+12+11=23


My proof of the  inequality is very ugly

[quote="can_hang2007"]My proof (sorry in Vietnamese) :([/quote]

What a pity！ :(
Can you teach me Vietnamese？ :wink:
Your solution is a little similar to mine. :P


MATHCOUNTS
calculus
integration
geometry
perimeter
algebra
difference of squares

I figure a bunch of you are wondering how do to countdown questions good. I'm bored, procrastinating, studying SAT vocab, and magnanimous, so enjoy.

Practice simple things like multiplying two-digit numbers together, over and over, like until you can do like 71 times 39 in your head in under five seconds. (It's 2769.)

Get good at reading questions and stuff. If you can read the question faster than the announcer can read it, good job, you're normal. If you can read [b]and answer it[/b] before the announcer can read it, chances are, you have a good shot at States or even Nationals of placing high.

Think of ways to minimize the amount of time between the beginning of the question and the time you buzz in. This is key - avoid using pen and paper if you can, try buzzing in before you actually have the problem solved, read only key elements of the problem, write using shorthand if you have to, and other things you can think of. I guess there's a certain level where it doesn't really matter who's better at math, but who knows more tricks.

Do previous questions. MathCounts usually reuses problems with different numbers - at Nationals, I recognized maybe two or three problems. That's a couple extra minutes I can spend making sure #2 is correct.

Memorize formulas that pop up often. Guaranteed memorizing squares up to 2500 and primes up to 130 (they have a lot of problems involving primes slightly higher than 100) is going to help.

Practice a bunch ahead of the competition, like two months ahead, but make sure to practice large amounts right before the actual competitions. Overload your head with math, cheetos, and red bull or something.

I'm bored and should be doing biology homework, but I decided to think up a couple questions. They shouldn't take too long - if they take a while, you're probably doing it wrong.

1) Find 121 times 129.

2) Fourteen years ago, Sarah was twice as old as Sam. Eleven years from now, Sam's age will be 80% of Sarah's age. How old is Sarah now? (I didn't make any typos, at least, I don't think I did. Consider that if Jimmy will be born two years from now, he is -2 years old.)

3) Triangle HOM is right with integral sidelengths. If angle O is 90 degrees and HO + MO = 69, find triangle HOM's perimeter.

4) Ghandi has sixteen flags, some orange and some green. If he has four more orange flags than green flags, how many orange does he have?

5) Evaluate $\sqrt{6.5}$, rounded to the nearest integer.

6) Richard can buy pens in packages of 69, 137830425, 15, and 23. What is the largest number of pens he cannot buy, assuming he an infinite amount of money?

7) What is the sum of the factors of 6? 66? 666 (better know the formula)?

8) What is the product of the factors of $2^{18}$? Express your answer in the form $2^{x}$, where $x$ is an integer.

9) A square has diagonal of length 10. Find its perimeter.

10) A dance hall is 300 feet long and 400 tiles wide. If the floor is tiled with 1 foot by 1 foot tiles and a line is drawn from one corner of the dance hall to the opposite one, how many tiles does it pass through?

11) A circle of maximum area is inscribed in a semicircle of area 1332. Find its (the circle's) area.

12) What is the sum of the digits of the largest n-digit number whose product of digits is 0 (awkward wording)? Express your answer as an expression involving n.

13) How many zeroes does $\frac{125!}{125(2+2!)}$ end in?

14) $\frac{1}{n}= (\frac{1}{n})^{2}+(\frac25)^{2}.$ Find $n^{2}$.

15) SPECIAL EXTRA SUPER BONUS QUESTION: Find the product: $(X-A)(X-B)\cdots(X-Y)(X-Z)$

DO PROBLEM GOOD ALL TIME
WITH DIVINE GRACE YO'ULL PREVAIL ALL CONTESTS WITH ABSOLUTE GLORY
ENJOY!!!

15 is [hide]0[/hide], right?

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=687506#687506[/url]

can u comment on that topic?

why are u studying sat if ur a freshman very late on a weekend?

-jorian
[/list]

These seem more like sprint questions to me. I completely agree that doing sprints like countdowns is good practice, but you said:
[quote]They shouldn't take too long - if they take a while, you're probably doing it wrong.[/quote]
Some of these might take a bit of time, (less than 45 seconds of course) but how long do you mean by "a while"?

[quote="Ubemaya"]Are there any tips you have for multiplying number quickly in your head?[/quote]

Depends. Easy one. 15^2= 1*2+25=225 25^2= 2*3+25=625 Hope you get what I mean. Also, practice. :D

Wow. how did you do them in 3 seconds???

-jorian

p.s. should u start more than 10 seconds if ur not that good?

Some of them are quite easy to do in that amount of time (2,4,5,8,9,10,11,12,13,15). Thanks alot, I guess if anyone was really serious about doing well in countdown they should in fact be able to do all 15 in under 45 seconds total.

However, some of them, like 3 and 6, seem kinda hard for countdown. (6 is even hard for target)

oh, i thought he meant something like these:

89*84
43*91
27*39
etc.

-jorian

[quote="Treething"][quote="Ubemaya"]Are there any tips you have for multiplying number quickly in your head?[/quote]

go make a program picking two random 2-digit numbers to multiply, having you input the answer, and make it so that you gotta input the answer within 10 seconds else it counts as wrong

i did that, used it for maybe 15 hours while lowering 10 seconds down to 3 along the way.[/quote]
Where would this program be put? on a calculator? What would the programming be?

[hide]69 is a multiple of 23 obviously. 137830425 fulfills divisibility rule for 3 so it is a multiple of 15. From there, it is just $15*23-38$  which is $307$[/hide]

Really? Then could you post a short solution to #6?

[hide="#3"]First, $69/3=23$ The legs of an 8-15-17 triangle add to 23. $8+15+17=40*9=\boxed{360}$[/hide]

[hide="#6"]69 is a multiple of 23 obviously. 137830425 fulfills divisibility rule for 3 so it is a multiple of 15. From there, it is just $15*23-38=\boxed{307}$[/hide]

[quote="Treething"]They shouldn't take too long - if they take a while, you're probably doing it wrong. [/quote]

[quote="nutz_for2.718281828"][hide="#5"]Since 6.5>4 and 6.5<9, rt6.5 is a little greater than 2, but less than three. Answer: 2[/hide][/quote]

That solution has so many logical fallacies.

Ok, here is the file:

Everything should work  :)

(I also made one for adding 4 digit numbers, and sadly enough with a 20 second time limit & 20 second penalty for incorrect answers, I  average about 140-150 seconds total for 20 questions, best is 112)

Sorry it is a zip file, .8xp is not an allowed extension.

[quote="b-flat"]Ok, how about this:

I attach the file (updated since then) that contains the program so that no one has to actually know what the program means  :D

If you need a link cable, I can still answer any questions.

@bpms: Correct & Incorrrect you type by pressing Alpha, then key with the letter in green above it (it will be all caps)[/quote]

I dont have a link. Do you know what's wrong?

[quote="b-flat"]What do you mean?

Just copy the code from page two, then.

Which part of the code do you have trouble with?[/quote]

Never mind, I found my link. It works perfectly now. Thanks. :)

Does anyone want the addition one?  (you don't have to do much to change it...)

I can't find the TI CD, if you get the chance can you please post the coding... It's not that important so if you think it is too much work I won't mind...

[quote="b-flat"]No problem (can't post the actual program for addition because I am on a laptop), there are only a couple of changes.

[code]:While 1
:ClrHome
:DelVar EDelVar FDelVar GDelVar K
:randInt(1,9999->A
:randInt(1,9999->B
:Output(1,1,A
:Output(1,2+iPart(log(A)),+
:Output(1,3+iPart(log(A)),B
:startTmr1
:checkTmr(1->C
:" ->Str1
:Repeat K=105 or D=C+10
:getKey->K
:If K
:Then
:If K>71 and K<75
:Str1+sub("789",K-71,1->Str1
:If K>81 and K<85
:Str1+sub("456",K-81,1->Str1
:If K>91 and K<95
:Str1+sub("123",K-91,1->Str1
:If K=102
:Str1+"0->Str1
:If K=23 and length(Str1)>1
:Then
:Output(4,2,"               "
:sub(Str1,1,length(Str1)-1->Str1
:End
:Output(4,1,Str1
:checkTmr(1->D
:End
:If C==D-10
:Then
:If A+B=expr(sub(Str1,2,length(Str1)-1
:Then
:Output(8,1,"Correct
:1+E->E
:Else
:Output(8,1,"Incorrect
:1+F->F
:Pause
:End
:Else
:Output(8,1,"Out of Time
:G+1->G
:End
:End[/code]

Actually, I noticed a typo in the old code (outputted B twice instead of "*").

You don't really need TI-Connect to send things - simply connect the cable and right click on the file you want to send.  Then select "Send to TI-Device" and click send (make sure you have the cable stuff correct).[/quote]
THat isn't an option on mine...
Um, I meant the product program code (if you don't want to retype it you can edit your original one)...
Thanks,
bpms

[quote="b-flat"]I did post the new code - its practically the same  :)[/quote]
I thought that one was for the four digit addition...
:Output(1,2+iPart(log(A)),+ is giving me a syntax error...

Its on page 2.  Here is the new version of the code (the one I use):

[code]:Input "TIME LIMIT?",T
:DelVar EDelVar FDelVar GDelVar H
:For(I,1,20
:¦ ClrHome
:¦ randInt(10,99→A
:¦ randInt(10,99→B
:¦ Output(1,1,A
:¦ Output(1,2+iPart(log(A)),"*
:¦ Output(1,3+iPart(log(A)),B
:¦ DelVar θ
:¦ startTmr1
:¦ checkTmr(1→C
:¦ " →Str1
:¦ Repeat θ=105 or D=C+T
:¦ ¦ getKey→θ
:¦ ¦ If θ
:¦ ¦ Then
:¦ ¦ ¦ If θ>71 and θ<75
:¦ ¦ ¦ Str1+sub("789",θ-71,1→Str1
:¦ ¦ ¦ If θ>81 and θ<85
:¦ ¦ ¦ Str1+sub("456",θ-81,1→Str1
:¦ ¦ ¦ If θ>91 and θ<95
:¦ ¦ ¦ Str1+sub("123",θ-91,1→Str1
:¦ ¦ ¦ If θ=102
:¦ ¦ ¦ Str1+"0→Str1
:¦ ¦ ¦ If θ=23 and length(Str1)>1
:¦ ¦ ¦ Then
:¦ ¦ ¦ ¦ Output(4,2,"
:¦ ¦ ¦ ¦ sub(Str1,1,length(Str1)-1→Str1
:¦ ¦ ¦ End
:¦ ¦ End
:¦ ¦ Output(4,1,Str1
:¦ ¦ checkTmr(1→D
:¦ End
:¦ D-C+H→H
:¦ If C≠D-T and θ≠45
:¦ Then
:¦ ¦ If expr(sub(Str1,2,length(Str1)-1))=AB
:¦ ¦ Then
:¦ ¦ ¦ Output(8,1,"CORRECT
:¦ ¦ ¦ 1+E→E
:¦ ¦ ¦ Else
:¦ ¦ ¦ Output(8,1,"INCORRECT
:¦ ¦ ¦ 1+F→F
:¦ ¦ ¦ H+20→H
:¦ ¦ End
:¦ ¦ Output(7,1,"TIME:
:¦ ¦ Output(7,6,D-C
:¦ ¦ Else
:¦ ¦ Output(8,1,"OUT OF TIME
:¦ ¦ G+(θ≠45→G
:¦ End
:¦ Output(6,1,"ANSWER:
:¦ Output(6,8,AB
:¦ Pause
:End
:ClrHome
:Disp "YOU ANSWERED:","CORRECT:","INCORRECT:","NO TIME:","TIME:"BEST:
:1→dim(∟TIME
:If not(∟TIME(1
:999→∟TIME(1
:If H<∟TIME(1
:H→∟TIME(1
:Output(6,6,∟TIME(1
:Output(1,14,E+F+G
:Output(2,9,E
:Output(3,11,F
:Output(4,9,G
:Output(5,6,H
:Pause
:ClrHome
Generated by SourceCoder, © 2005-2006 Cemetech[/code]

The one on page two gave an error with
:If AB=expr(sub(Str1,2,length(Str1)-1
I'll try your new one...
What's with the many [code]|[/code] in your post above this one?

Oh, I used a program to save me a lot of typing and generate the code for me.

The | is just for indention so you can see each If/Then block of code, etc.  Ignore them, they make it easier to read.