Prove that $ (a^2\plus{}1)(b^2\plus{}1)(c^2\plus{}1)$≥$ (ab\plus{}bc\plus{}ca\minus{}1)^2 \forall a,b,c \in R$

[quote="SUPERMAN2"]Prove that $ (a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1)$≥$ (ab \plus{} bc \plus{} ca \minus{} 1)^2 \forall a,b,c \in R$[/quote]

$ (a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1) \ge (ab \plus{} bc \plus{} ca \minus{} 1)^2$
$ \Leftrightarrow a^2b^2c^2\plus{}(a^2\plus{}b^2\plus{}c^2) \ge \minus{}2(ab\plus{}bc\plus{}ca)\plus{}2abc(a\plus{}b\plus{}c)$
$ \Leftrightarrow (abc)^2\minus{}2abc(a\plus{}b\plus{}c)\plus{}(a\plus{}b\plus{}c)^2 \ge 0$
$ \Leftrightarrow (abc\minus{}a\minus{}b\minus{}c)^2 \ge 0$, which is obviously true.


function
calculus
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quadratics
modular arithmetic
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number theory unsolved

1, $ m^4 \plus{} (m \plus{} 1)^4 \equal{} n^2 \plus{} (n \plus{} 1)^2$ for integral solutions [color=green]SOLVED[/color]

2, $ 2x^2 \minus{} 5xy \plus{} 3x \minus{} 2y \equal{} 10$ for integral solutions [color=green]SOLVED[/color]

3, $ x \plus{} y \plus{} \frac {y}{x} \equal{} 84$ for integral solutions [color=green]SOLVED[/color]

4, $ x^3 \minus{} y^3 \equal{} xy \plus{} 61$ for natural integral solutions [color=red]UNSOLVED[/color]

5, Prove that $ [nx] \geq \frac {[x]}{1} \plus{} \frac {[2x]}{2} \plus{} ... \plus{} \frac {[nx]}{n}$ where n is a positive integer while x is a real number [color=red]UNSOLVED[/color]

With full solutions please

[quote="rogerlee"]1, $ m^4 \plus{} (m \plus{} 1)^4 \equal{} n^2 \plus{} (n \plus{} 1)^2$ for integral solutions

2, $ 2x^2 \minus{} 5xy \plus{} 3x \minus{} 2y \equal{} 10$ for integral solutions

3, $ x \plus{} y \plus{} \frac {y}{x} \equal{} 84$ for integral solutions

4, $ x^3 \minus{} y^3 \equal{} xy \plus{} 61$ for integral solutions

5, Prove that $ [nx] \geq \frac {[x]}{1} \plus{} \frac {[2x]}{2} \plus{} ... \plus{} \frac {[nx]}{n}$ where n is a positive integer while x is a real number

With full solutions please[/quote]

$ x \plus{} y \plus{} \frac {y}{x} \equal{} 84$ ----> $ x^{2}\plus{}yx\plus{}y \equal{} 84x$ ---->  $ y\equal{}\frac{84x\minus{}x^{2}}{x\plus{}1}\equal{}\minus{}x\plus{}85\plus{}\frac{85}{x\plus{}1}$------->   $ x\plus{}1\equal{}\minus{}1;\minus{}5;\minus{}17;\minus{}85;1;5;17;85$----->$ x\equal{}\minus{}2;\minus{}6;\minus{}18;\minus{}86;0;4;16;84$ ----> $ x\not\equal{}0$------> $ x\equal{}\minus{}2;\minus{}6;\minus{}18;\minus{}86;4;16;84$

$ y\equal{}172;108;108;172;\minus{}64;\minus{}64;0$

2nd: First of all we examine the equation towards x and we obtain $ D \equal{} 25y^{2} \minus{} 14y \plus{} 89$. So we get the equation $ 25y^{2} \minus{} 14y \plus{} 89 \equal{} z^{2}$, examine towards y and we obtain $ D \equal{} 25z^{2} \minus{} 2176$. Let $ 5z \equal{} k$, then we have to solve $ k^2 \minus{} 2176 \equal{} m^{2}$. Thus $ (k \plus{} m)(k \minus{} m)\equal{}2176$ which could be easily solved  in integers.
I have solution of the 4th. If you need it I will post it.

[quote="SAPOSTO"]2nd: First of all we examine the equation towards x and we obtain $ D \equal{} 25y^{2} \minus{} 14y \plus{} 89$. So we get the equation $ 25y^{2} \minus{} 14y \plus{} 89 \equal{} z^{2}$, examine towards y and we obtain $ D \equal{} 25z^{2} \minus{} 2176$. Let $ 5z \equal{} k$, then we have to solve $ k^2 \minus{} 2176 \equal{} m^{2}$. Thus $ (k \plus{} m)(k \minus{} m) \equal{} 2176$ which could be easily solved  in integers.
I have solution of the 4th. If you need it I will post it.[/quote]

I need it thanks.

1st:
Examine as quadratic equation towards n. $ D \equal{} 4m^{4} \plus{} 8m^{3} \plus{} 12m^{2}\plus{}8m \plus{} 1 \equal{} (2(m^{2} \plus{} m \plus{} 1))^{2} \minus{} 3$. Let $ k \equal{} 2(m^{2} \plus{} m \plus{} 1)$. Then we have to solve $ k^{2} \minus{} 3 \equal{} r^{2}$ therefore $ (k \plus{} r)(k \minus{} r) \equal{} 3$. The final is not problem.

[quote="rogerlee"]By the way, how did you obtain $ D \equal{} 25z^{2} \minus{} 2176$[/quote]
Very often when you have to solve diophantine equations you notice that you can examine the equation as quadratic towards one of the variables. In this equation first of all I solve towards x the equation $ 2x^{2}\minus{}5xy\plus{}3x\minus{}2y\equal{}10$. We obtain $ D\equal{}25y^{2}\minus{}14y\plus{}89$. But because we need integer solutions D must be perfect square. So  get new equation: $ 5y^{2}\minus{}14y\plus{}89\equal{}z^{2}$. Again solve like quadratic equation but this time towards y. We obtain $ D\equal{}k^{2}\minus{}2176\equal{}m^{2}$ ($ k\equal{}5z$). Then $ (k\plus{}m)(k\minus{}m)\equal{}2176$. The final equation is easy.
I use the same method to solve the first one  :wink: .

[quote="SAPOSTO"]1st:
Examine as quadratic equation towards n. $ D \equal{} 4m^{4} \plus{} 8m^{3} \plus{} 12m^{2} \plus{} 1 \equal{} (2(m^{2} \plus{} m \plus{} 1))^{2} \minus{} 3$. Let $ k \equal{} 2(m^{2} \plus{} m \plus{} 1)$. Then we have to solve $ k^{2} \minus{} 3 \equal{} r^{2}$ therefore $ (k \plus{} r)(k \minus{} r) \equal{} 3$. The final is not problem.[/quote]

Did you miss the term $ 4m$ when expanding $ (1\plus{}m)^4$?

[quote="rogerlee"][quote="SAPOSTO"]1st:
Examine as quadratic equation towards n. $ D \equal{} 4m^{4} \plus{} 8m^{3} \plus{} 12m^{2} \plus{} 1 \equal{} (2(m^{2} \plus{} m \plus{} 1))^{2} \minus{} 3$. Let $ k \equal{} 2(m^{2} \plus{} m \plus{} 1)$. Then we have to solve $ k^{2} \minus{} 3 \equal{} r^{2}$ therefore $ (k \plus{} r)(k \minus{} r) \equal{} 3$. The final is not problem.[/quote]

Did you miss the term $ 4m$ when expanding $ (1 \plus{} m)^4$?[/quote]
Yes I had missed something. However this doesn't matter because after that everything is OK. I will edit it.
For the 4th one isn't it natural number instead of integral?

[quote="SAPOSTO"][quote="rogerlee"][quote="SAPOSTO"]1st:
Examine as quadratic equation towards n. $ D \equal{} 4m^{4} \plus{} 8m^{3} \plus{} 12m^{2} \plus{} 1 \equal{} (2(m^{2} \plus{} m \plus{} 1))^{2} \minus{} 3$. Let $ k \equal{} 2(m^{2} \plus{} m \plus{} 1)$. Then we have to solve $ k^{2} \minus{} 3 \equal{} r^{2}$ therefore $ (k \plus{} r)(k \minus{} r) \equal{} 3$. The final is not problem.[/quote]

Did you miss the term $ 4m$ when expanding $ (1 \plus{} m)^4$?[/quote]
Yes I had missed something. However this doesn't matter because after that everything is OK.
For the 4th one isn't it natural number instead of integral?[/quote]

right. That's a natural number, i've typed it wrongly

[quote="rogerlee"][quote="SAPOSTO"]If so, it's true that $ x > y$. If $ x \minus{} y \equal{} u > 0$ replace $ x$ with $ u \plus{} y$ in the equation. Write what you obtain.[/quote]

Then?

$ x^3 \plus{} y^3 \equal{} xy \plus{} 61$

$ (y \plus{} u)^3 \plus{} y^3 \equal{} (y \plus{} u)y \plus{} 61$

$ 3u^2y \plus{} 3uy^2 \plus{} u^3 \equal{} y^2 \plus{} uy \plus{} 61$

$ (3u \minus{} 1)y^2 \plus{} u(3u \minus{} 1)y \plus{} (u^3 \minus{} 61) \equal{} 0$

However, $ D$ contains $ u^4$

Or using another method

$ 3uy(u \plus{} y) \equal{} y^2 \plus{} uy \plus{} 61 \minus{} u^3$

$ 3x \equal{} \frac {y^2 \plus{} uy \plus{} 61 \minus{} u^3}{uy}$

$ 3x \equal{} 1 \plus{} \frac {y^2 \plus{} 61 \minus{} u^3}{uy}$

$ \frac {y^2 \plus{} 61 \minus{} u^3}{uy} \equal{} \minus{} 1 (mod 3)$

I don't know how to work it out...[/quote]

I knew the answer, the teacher told us

$ u^3<61$

Hence, by trials, substituting $ u\equal{}1$, $ u\equal{}2$, $ u\equal{}3$

[quote="rogerlee"][quote="rogerlee"][quote="SAPOSTO"]If so, it's true that $ x > y$. If $ x \minus{} y \equal{} u > 0$ replace $ x$ with $ u \plus{} y$ in the equation. Write what you obtain.[/quote]

Then?

$ x^3 \plus{} y^3 \equal{} xy \plus{} 61$

$ (y \plus{} u)^3 \plus{} y^3 \equal{} (y \plus{} u)y \plus{} 61$

$ 3u^2y \plus{} 3uy^2 \plus{} u^3 \equal{} y^2 \plus{} uy \plus{} 61$

$ (3u \minus{} 1)y^2 \plus{} u(3u \minus{} 1)y \plus{} (u^3 \minus{} 61) \equal{} 0$

However, $ D$ contains $ u^4$

Or using another method

$ 3uy(u \plus{} y) \equal{} y^2 \plus{} uy \plus{} 61 \minus{} u^3$

$ 3x \equal{} \frac {y^2 \plus{} uy \plus{} 61 \minus{} u^3}{uy}$

$ 3x \equal{} 1 \plus{} \frac {y^2 \plus{} 61 \minus{} u^3}{uy}$

$ \frac {y^2 \plus{} 61 \minus{} u^3}{uy} \equal{} \minus{} 1 (mod 3)$

I don't know how to work it out...[/quote]

I knew the answer, the teacher told us

$ u^3 < 61$

Hence, by trials, substituting $ u \equal{} 1$, $ u \equal{} 2$, $ u \equal{} 3$[/quote]

So what is the problem with this problem?
Also nice solutions from [b]zaya_yc[/b]!

[hide="Solution for 4"]
Note the following identity: $ a^3\plus{}b^3\plus{}c^3\minus{}3abc\equal{}(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca)$.

Multiply both sides of the given equation by 27: $ 27x^3\minus{}27y^3\equal{}27xy\plus{}1647$

$ 27x^3\minus{}27y^3\minus{}1\minus{}27xy\equal{}1646$

$ (3x)^3\plus{}(\minus{}3y)^3\plus{}(\minus{}1)^3\minus{}3(3x)(\minus{}3y)(\minus{}1)\equal{}1646$

$ (3x\minus{}3y\minus{}1)(9x^2\plus{}9y^2\plus{}1\plus{}9xy\minus{}3y\plus{}3x)\equal{}2\cdot823$

Since 823 is prime and $ 3x\minus{}3y\minus{}1\equiv2\pmod3$, we have $ 3x\minus{}3y\minus{}1\equal{}2$ and $ 9x^2\plus{}9y^2\plus{}1\plus{}9xy\minus{}3y\plus{}3x\equal{}823$. It follows easily that $ x\equal{}6,y\equal{}5$.[/hide]


function
calculus
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This could also be done using the binomial expansion:

$\sqrt{101}= \sqrt{1+100}= 10\left(1+\frac{1}{100}\right)^{1/2}\approx 10\left(1+\frac{1}{200}\right)$.

could someone show me explicitly using second order taylor and lagrange? thanks

[url=http://mathworld.wolfram.com/LagrangeRemainder.html]Reference for Taylor with Lagrange.[/url]
Let $f(x)=(1+x)^{1/2}$. The Taylor series for $f$ at $x=0$ is $(1+x)^{1/2}=1+\frac12x+\frac{1}{2}\left(-\frac{1}{2}\right)\frac{x^{2}}{2!}+R_{2}$, where $R_{2}=\frac{f^{(3)}(x^{*})}{3!}x^{3}$ for some $x^{*}$ between $0$ and $x$. Plug in $x=1/100$ and give an upper estimate for $|R_{2}|$.

P.S. The link given above says that $x^{*}\in (0,x)$, which is not quite right when $x<0$. Is there any notation for "$a$ is between $b$ and $c$"?

[quote="countdownmath"]Use the second-order Taylor theorem with Lagrange remainder theorem to estimate $\sqrt{101}$.[/quote]
[hide]Use the method above to find $f(x)$

In this case our function is: $f(x)=10\sqrt{x+1}$ and we are trying to find $f(\frac{1}{100})$'s approximation by the Lagrange form of the remainder.

This states that there is a $z$ between what your series is centered at (we'll use $0$), $0$, and the number you are trying to approximate, $\frac{1}{100}$, to satisfy:

$R_{n}(x)=\frac{f^{(n+1)}(z)}{(n+1)!}\cdot (x-c)^{(n+1)}$

In this case, the Taylor Series will be centered at $0$ thus $c=0$.  Since it is second degree, $n=2$.

$R_{2}\left( \frac{1}{100}\right)=\frac{f^{(3)}(z)}{(3)!}\cdot \left( \frac{1}{100}\right)^{3}$

Now note that the $3^{\text{rd}}$ derivative of $f(x)= \sqrt{x+1}$ at $0$ is given by:

$f(x)= 10 \cdot (x+1)^{\frac{1}{2}}$
$f'(x)= 10 \cdot \frac{1}{2}\cdot (x+1)^{\frac{-1}{2}}$
$f''(x)= 10 \cdot \frac{-1}{2^{2}}\cdot (x+1)^{\frac{-3}{2}}$
$f^{3}(x)= 10 \cdot \frac{3}{2^{3}}\cdot (x+1)^{\frac{-5}{2}}$

thus..

$R_{2}\left( \frac{1}{100}\right)=10 \cdot \frac{\frac{3}{2^{3}}\cdot (z+1)^{\frac{-5}{2}}}{(3)!}\cdot \left( \frac{1}{100}\right)^{3}$
$R_{2}\left( \frac{1}{100}\right)=10 \cdot \frac{1}{16(z+1)^{\frac{5}{2}}}\cdot \left( \frac{1}{100}\right)^{3}$

Now this is where it gets confusing.. because the $z+1$ in the denominator is guaranteed to be $>1 ~~\forall ~~z$, this will only decrease the remainder.  Thus:

$R_{2}\left( \frac{1}{100}\right)<10 \cdot \frac{1}{16}\cdot \left( \frac{1}{100}\right)^{3}$
$R_{2}\left( \frac{1}{100}\right)<\frac{1}{1,600,000}= .000000625$
$\boxed{R_{2}\left( \frac{1}{100}\right)<.000000625}$

And then find the exact value of the series approximation using the second degree polynomial:

$P_{2}(x)=10+5x-5\cdot \frac{x^{2}}{4}$
$P_{2}\left( \frac{1}{100}\right) = 10.049875$

This shows that using a second degree polynomial for $f(x)=\sqrt{x+1}$:

$10.049875<\sqrt{101}< 10.049875+.000000625$
$\boxed{ 10.049875<\sqrt{101}<10.049875625}$

Note: $\sqrt{101}-P_{2}\left( \frac{1}{100}\right) =.000000621121$ and $\sqrt{101}\approx 10.049875621121$[/hide]

[quote="skimnc"]as you keep putting terms onto your series it will continue to get closer to $\sqrt{101}$[/quote]
It will not. The radius of convergence of the binomial series is $1$. The first thing [b]Carcul[/b] did above was to factor out $100$ and work with $x=1/100$; I have followed the same approach.

Blah.  Sorry!  I edited it and fixed it.  Thanks m

[b]Countdownmath[/b], here are some more examples for you to play with (taken from Demidovitch and another book):

-Give approximate values for the following numbers:

$\sqrt[3]{200}$, $\sqrt{23}$, $\sqrt[3]{7}$, $\sqrt[4]{19}$, $\sqrt[5]{35}$, $\sqrt[4]{e}$.


A circle is inscribed with radius $ \frac12$ in a square.  In the regions outside the circle and inside the square, a circle is drawn for a total of 5 circles.  What is the diameter of one of the smaller circles?

[quote="007math"]A circle is inscribed with radius $ \frac12$ in a square.  In the regions outside the circle and inside the square, a circle is drawn for a total of [b]5[/b] circles.  What is the diameter of one of the smaller circles?[/quote]

Do you mean a total of [b]4[/b] circles?

[quote="tonypr"]I think he means one big one in the middle and 4 little ones on each corner.[/quote]

Ah, right.
Blonde moment.
[hide="Solution"]
Let $ r$=radius of the smaller circle.
Since the radius of the large circle is $ \frac {1}{2}$, the side length of the square is $ 1$ and its diagonal is $ \sqrt2$.
Half of the square's diagonal equals the large circle's radius + the small circle's radius + the distance from the center of a small circle to the nearest corner of a square.

This is
$ \frac {\sqrt2}{2} \equal{} \frac {1}{2} \plus{} r \plus{} r\sqrt2$
$ \frac {\sqrt2}{2} \minus{} \frac {1}{2} \equal{} r(1 \plus{} \sqrt2)$
$ \frac {\frac {\sqrt2}{2} \minus{} \frac {1}{2}}{1 \plus{} \sqrt2} \equal{} r$
$ r \equal{} \frac {3 \minus{} 2\sqrt2}{2}$

So the diameter is thus $ \boxed{3 \minus{} 2\sqrt2}$

Ah, another blonde moment. Thanks for pointing that out, tonypr.
[/hide]

[hide="solution"]
First let's say that the radius of the smaller circle is $ x$, and the diameter is $ 2x$, also, I'll denote the vertices of the square as A,B,C,D and the center of the Circle to be O.

So from O to a corner, is equal to the radius of the big circle$ (\dfrac{1}{2})$ + $ 2x \plus{} r$, where $ r$ is the space between the smaller circle and the corner.

We can make a small square at points M,B,L,O where M,L are the midpoints of AB and BC respectively, and $ OL \equal{} LB \equal{} \dfrac{1}{2}$ with Angle OBL = BOL = 45, so the segment OB equals $ \dfrac{\sqrt {2}}{2}$.

Meaning that $ \dfrac{1}{2} \plus{} 2x \plus{} r \equal{} \dfrac{\sqrt {2}}{2}$.

To put $ r$ in terms of $ x$, we make a square with the center of a smaller circle(A) and a point P, where P is perpendicular to segment LB, now $ AP \equal{} PB \equal{} x$. Using the Pythagorean theorem, we find that $ x \plus{} r \equal{} \sqrt {2}x$, and $ r \equal{} \sqrt {2}x \minus{} x$

Plugging this in to the equation $ \dfrac{1}{2} \plus{} 2x \plus{} r \equal{} \dfrac{\sqrt {2}}{2}$
we get $ 2x \plus{} \sqrt {2}x \minus{} x \equal{} \dfrac{\sqrt {2} \minus{} 1}{2}$ and $ x(1 \plus{} \sqrt {2}) \equal{} \dfrac{\sqrt {2} \minus{} 1}{2}$.

Lastly

$ 2x \equal{} \dfrac{\sqrt {2} \minus{} 1}{1 \plus{} \sqrt {2}}$
$ 2x \equal{} \dfrac{(\sqrt {2} \minus{} 1)(1 \minus{} \sqrt {2})}{ \minus{} 1}$
$ 2x \equal{} \boxed{3 \minus{} 2\sqrt {2}}$
*note to previous answer by 27r, the question asked for the diameter not the radius, just multiply by 2, and you are welcome 27r.

$ \cdots$ I think I over did it :)
[/hide]


What is the effect of temperature on yeast fermentation? I think it increases as temperature increases, up to a certain point. Can anyone confirm this and/or provide additional information?

fermentation uses only Glycolysis to produce ATP,

and in general Increase of tempreture, increases the amount of respiration. so it has direct relationship with increase of tempareture.
but keep in mind that there is a limit to increase of tempareture, becuase the enzymes involved in glycolysis will be denaturized after a certain temp.


modular arithmetic
number theory unsolved
number theory

For $ m,n\in N$ find the value of $ 4^m\minus{}3^n$ which is the least positive integer greater than 1 which can't be written in a form $ 4^r\plus{}3^s$ for some non-negative number $ r$ and $ s$

$ 4^m\minus{}3^n\equiv1\pmod6$.

$ 7\equal{}4^1\plus{}3^1,13\equal{}4^1\plus{}3^2,19\equal{}4^2\plus{}3^1,25\equal{}4^2\plus{}3^2,31\equal{}4^1\plus{}3^3$.

$ 37\equal{}1\plus{}36\equal{}4\plus{}33\plus{}16\plus{}21$ cannot be written in a form $ 4^r\plus{}3^s$ for any non negative integers $ r,s$. But $ 37\equal{}4^3\minus{}3^3$.

Answer: $ \boxed{37}$


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Prove that the fraction $\frac{11}{29}$ has the period 28

Seems to be simple. We search for the smallest positive integer $k$ with $29|(10^k-1)$. For $k=28$ it follows from Little Fermat. After checking all divisors:
$1,2,4,7,14$ we must get that $28$ is the smallest $k$.

Misha


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[b]Vasc[/b], could you post your solution? Really nice inequality, by the way :)

If we denote the $k$-th power mean by $M_k$, the ineq is, after homogenizing, equivalent to
$M_0^5\cdot M_4^4\le M_2^9$.
I think it has been proven semowhere here (but can't find the reference - am I right :?:) that the function $k\mapsto M_k$ is concave, i.e. in particular $M_0\cdot M_4\le M_2^2$. The result follows.

Are you referring to Lemma 1 in http://web.mit.edu/~tmildorf/www/ASharpBound.pdf ?

There I established that $k \rightarrow M_k(a,b)$ is concave on $[1,\infty)$ and convex on $(-\infty,-1]$. Has someone improved this?

[quote="Vasc"][quote="spanferkel"] the function $k\mapsto M_k$ is concave, i.e. in particular $M_0\cdot M_4\le M_2^2$. The result follows.[/quote]
I think the last inequality is valid only for two numbers.  :?:[/quote]

oh maybe you are right, I just remembered having seen something about this concavity in this forum but couldn't find any reference for that. Maybe it was Mildorf's. So that's just for 2 variables?  :huh:

OK. Let's consider $M_0^5\cdot M_4^k\le M_2^{k+5}$ or equivalently (just square everything) $M_0^5\cdot M_2^k\le M_1^{k+5}$ and search the biggest k where this is valid. (And it will turn out indeed that it is wrong for k=5).

By the n-1EV principle (This applies here, doesn't it?) and homogenity, it is sufficient to consider $(a,b,c,d,e)=(1,1,1,1,x)$, $x>0$. Solving for $k$ (as a function of x) we get
$k\le \frac{5\ln\frac{x+4}5-\ln x}{\ln\sqrt{\frac{x^2+4}5}-\ln\frac{x+4}5}$ and find the minimum of the RHS is k=4.0963840082 which is attained at x=2.48988249.  This means Vasc's ineq with k=4 is very very tight ;).
Vasc, I wonder if this is complete as a proof (apart from showing thoroughly that k=4.0963840082 is indeed the minimum, which might involve quite a bit of calculus). If so, does your proof go along these same lines? Or do you have something much more elegant?

Let $f(a,b,c,d,e)=abcde(\sum a^4)$, let $t = \sqrt{\frac{a^2+b^2}{2}}$
Then, by some calculations,
$f(t,t,c,d,e)-f(a,b,c,d,e) = \frac{1}{4}cde(a-b)^2\left((a^2+b^2)^2-2ab(a+b)^2+2(c^4+d^4+e^4)\right)$
Note that $(a^2+b^2)^2-2ab(a+b)^2 \ge -4a^2b^2$
Thus, if $c^4+d^4+e^4 \ge 2a^2b^2$, $f(t,t,c,d,e) \ge f(a,b,c,d,e)$...(*)
WLOG assume $e \ge d \ge c \ge b \ge a$, then whenever we mix $a,b,c$, (*) is always true, eventually we maximize $f(a,b,c,d,e)$ to $f(t,t,t,d,e)$. At this moment, we notice that in fact we can mix $t,d$ as well. It is easy (if I'm not wrong), that $f(t,t,t,d,e) \ge f(t,t,y,y,e) \ge f(z,z,z,z,e)$, where $2y^2 = d^2+t^2$ and $2z^2 = y^2+t^2$. So it reduces to an one-variable inequality.

[quote=Vasc]If $a,b,c,d,e$ are positive numbers such that $a^2+b^2+c^2+d^2+e^2=5$, then
$abcde(a^4+b^4+c^4+d^4+e^4)\leq 5$.[/quote]
We need to prove that $\sqrt{a b c d e}(a^2+b^2+c^2+d^2+e^2)\leq5$
for non-negatives $a$, $b$, $c$, $d$ and $e$ such that $a+b+c+d+e=5$.
By Vasc's EV Theorem if we'll fix $a^2+b^2+c^2+d^2+e^2$ and $a+b+c+d+e$ so $abcde$ gets a maximal value, when $a=b=c=d$.
Hence, it remains to prove that $a^2\sqrt{5-4a}(4a^2+(5-4a)^2)\leq5$, where $0\leq a\leq\frac{5}{4}$, which is
$(a-1)^2(64a^7-208a^6+256a^5-120a^4+4a^3+3a^2+2a+1)\geq0$, which is true.

[quote=Grotex]Let $a,b,c,d,e$ be non-negative and $\sum \sqrt[3]{a}=6.$
Prove that
$$\prod a\cdot(\sum a)^2\leq 1152.$$
eq holds when [hide=you should not see that]$a=8,b=c=d=e=1$ and its rotations.[/hide][/quote]
We can use $EV$ Method again.


Prove that there is no function f from the set of non-negative integers into itself such that f(f(n)) = n + 1987 for all n.

What does it mean by function f rom the set of non-negative integers.

So, f always give me a non-negative integer?

It would mean that the domain of the function is the set of non-negative integers (the positive integers and 0) and its range is also this set. What you said is correct - you get another non-negative integer when you plug in one.

IMO 1987, Problem 4. :) You can use the search button if you want.


If $x=(\log_82)^{(\log_28)}$, then $\log_3x$ equals:

$\text{(A)} \ -3 \qquad \text{(B)} \ -\frac13 \qquad \text{(C)} \ \frac13 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ 9$

[quote="Apprentice"][hide](1/3)^3 = 1/27; log base 3 of 27^(-1) = 1/3 so c [/hide][/quote]

Ohh, I got $\boxed {A}$

Oops!  yeah you're right  :roll:  - I fixed it : )


I frequently get private messages from users whom I don't know. Many of them read something like this:

"Greetings. I am new to [name of subject] and have a few questions. Could you help me with these exercises:

1. ---

2. ---

3. --- "

If you recently sent me such a message, please don't wait for my reply - it's not coming. I simply do not respond to such messages.

If you really do want to discuss mathematics, don't send PMs - post the problems out in the light of day on the forums. There, all of the posters will look them over, and answer the ones that happen to interest them.

Here's a question directed to my fellow moderators or others with thousands of posts: what do you think about this? Would you also discourage the kind of PMs described above?

[quote="Kent Merryfield"]Here's a question directed to my fellow moderators or others with thousands of posts: what do you think about this? Would you also discourage the kind of PMs described above?[/quote]

Yes. Most often I don't answer these PM's at all, since they are most likely sent to 30 different people at once (unless the field is geometry). But people who send exercises by PM's to others are not likely to read your topic. ;)

  darij

I get them too. Lately, I've been slightly more polite- I send back a PM with approximately the contents of this post.

There are some unsolicited PMs that deserve full responses- I got one recently asking "Do you go to the University of Washington? I'm applying there as a graduate student in math." (paraphrased)

Maybe they have posted the problem and no one has replied for long time, hence they send PM's

Example, I posted this topic
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=959981#959981

Unfortunately no one answered yet. I was going to send you a PM about what topics these come under Prof. Merryfield but then I saw this thread :blush:

A topic that was posted while most North Americans were sleeping and grew to seven posts before they woke up hardly qualifies as "no one answered."

But sometimes a topic does go unanswered. Those of us who answer many topics don't answer all of them. We select the ones that interest us for some reason, or in which we will be able to make a reasonable contribution. Sometimes we just have nothing to say - and all you can do is accept that and move on. Begging for attention seldom works.

In this case, i answered that : Please post it on forum.  :lol:


function
advanced fields
advanced fields unsolved

Let $ \mathbb{S}^1\equal{}\mathbb{R}/(2\pi\mathbb{Z})$ denote the circle. let $ C^{\infty}(\mathbb{S}^1)$ denote the space of $ 2\pi$-periodic $ C^{\infty}$ functions on $ \mathbb{R}$. Let $ M: C^{\infty}(\mathbb{S}^1)\rightarrow C^{\infty}(\mathbb{S}^1)$ denote the multiplication operator by $ e^{ix}: (Mf)(x)\equal{}e^{ix}f(x)$, and let $ \frac{d}{dx}$ be the usual derivative.

Suppose that $ T: C^{\infty}(\mathbb{S}^1)\rightarrow C^{\infty}(\mathbb{S}^1)$ is a linear map and $ T$ satisfies $ T\frac{d}{dx}\equal{}\frac{d}{dx}T$ and $ TM\equal{}MT$. Show that there exists $ c\in\mathbb{C}$ such that $ Tf\equal{}cf$ for all $ f$.

Supose T in this problem is [b]continuous[/b]  :)

To solve the problem just look how $ T$ acts on functions $ f_n(t) \equal{} e^{int}$.

The condition $ T\frac{d}{dx}f_n\equal{}\frac{d}{dx}T f_n$ gives us differential equation
\[ \frac{d}{dx}(T f_n) \equal{} nT f_n,\]
so $ T f_n \equal{} a_ne^{int} \equal{} a_n f_n$. Now the condition $ TMf_n\equal{}MTf_n$ means that
\[ Tf_{n\plus{}1} \equal{} a_nf_{n\plus{}1},\]
so $ a_{n\plus{}1}\equal{}a_n$ for all $ n \in \mathbb Z$. This means that all $ a_n$ equal to some $ c$, i.e. $ Tf_n \equal{} c f_n$ for all $ n \in \mathbb Z$. Since this system of functions is complete in $ C^{\infty}(\mathbb{S}^1)$ (i.e. its linear span is dense), we get that $ Tf\equal{}cf$ for all $ f$.


We know that $ x^2\plus{}(y\minus{}4)^2\equal{}1$. Find the least and the greatest value of the expression $ \sqrt{x^2}\plus{}\sqrt{y^2}$.

I assumed that $ x$ and $ y$ are integers. First of all: $ \sqrt{x^2}\plus{}\sqrt{y^2}\equal{}x\plus{}y$ because $ x$ and $ y$ are non-negative(easy to prove!)
Then we have following two cases:
1) $ x\equal{}0$ $ y\in\{5,3\}$
Then $ x\plus{}y\equal{}5$ because we must take numbers with biggest absolute value
2) $ x\equal{}1$, $ y\equal{}4$
Again $ x\plus{}y\equal{}5$.

Answer: $ \boxed{5}$

Just Answer:

Max. $ 4\plus{}\sqrt{2}$ for $ \left(\pm \frac{1}{\sqrt{2}},\ 4\plus{}\frac{1}{\sqrt{2}}\right)$.

min. $ 3$ for $ (x,\ y)\equal{}(0,\ 3)$.

$ \left(\minus{} \frac{1}{\sqrt{2}},\ 4\minus{}\frac{1}{\sqrt{2}}\right)$

Well, $ x^2 \plus{} (y \minus{} 4)^2 \equal{} 1$ means that $ (y \minus{} 4)^2 \equal{} 1 \minus{} x^2$ so $ y \minus{} 4 \equal{} \pm \sqrt {1 \minus{} x^2}$ so $ y \equal{} 4 \pm \sqrt {1 \minus{} x^2}$

From this we readily see that $ 4 \le y \le 5$

We need to find the max and min. of $ |x| \plus{} |y| \equal{} |x| \plus{} 4\pm\sqrt {1 \minus{} x^2}$

Let $ |x| \equal{} w$ where $ w \ge 0$

We need to find the max. and min. of $ 4 \plus{} w\pm\sqrt {1 \minus{} w^2}$. Since $ \sqrt {1 \minus{} w^2}$ is postive, this is equivalent to finding the max. of $ 4 \plus{} w \plus{} \sqrt {1 \minus{} w^2}$ and the min. of $ 4 \plus{} w \minus{} \sqrt {1 \minus{} w^2}$

Now, for the max, by AM-QM we have
$ 4 \plus{} w \plus{} \sqrt {1 \minus{} w^2} \le 4 \plus{} 2\sqrt {\frac {w^2 \plus{} 1 \minus{} w^2}{2}} \equal{} 4 \plus{} \sqrt {2}$ With equality when $ w \equal{} \sqrt {1 \minus{} w^2}$ so $ w \equal{} 1/\sqrt {2}$

@Crux: square root sign means principle square root, so in the case you posed, the sum would be 4, not
$ 4\minus{}\sqrt{2}$

For the min, $ 4 \plus{} w \minus{} \sqrt {1 \minus{} w^2}\ge 4 \plus{} 0 \minus{} 1 \equal{} 3$ which can be attained when w=0.

In conclusion, the max $ 4 \plus{} \sqrt {2}$ is attainable when $ x \equal{} \pm1/\sqrt {2}$ and $ y \equal{} 4 \plus{} 1/\sqrt {2}$
And the min. $ 3$ is attainable when $ x \equal{} 0$, and $ y \equal{} 3$

sorry I looked for min of $ x\plus{}y$ not of $ |x|\plus{}|y|$

My answer was right. :lol:

Here is my solution:

Let $ D\equal{}\{(x,\ y)| x^2\plus{}(y\minus{}4)^2\equal{}1\}$, for $ (x,\ y)\in{D}$, if $ \sqrt{x^2}\plus{}\sqrt{y^2}\equal{}|x|\plus{}|y|$ would take some real number $ k$, the graph $ |x|\plus{}|y|\equal{}k$ has the intersection point with $ D$. The rest is left for you.


quadratics algebra polynomial inequalities \/closed

[quote="1=2"]It's past 2006, and early 2007. When are the books gonna come out?

(I'm not saying HURRY UP!, or anything, just when?)[/quote]

We got distracted a little :)

Intro Algebra: about 10 days.
Interm Counting: early summer.
Interm Algebra: late summer.

Subject to more distractions, of course.  All three books are turning out longer than we anticipated.  (The Intro Algebra checked in at 656 pages, with 312 for the solutions.)

How long are you anticipating the others to be?

Interm Alg: Around same as Intro
Interm Counting 350-400 pgs.

Hi.

1. Could you please describe briefly what topics the "Intermediate Algebra"-Book will be about?

2. Why are there no pdf-excerpts of the Art of Problem Solving 1 & 2 ? That's a pity, because the excerpts are quite good to give you a quick (convincing) glance at the books!

1: All sorts of stuff: quadratics, polynomials, clever manipulations and substitutions, tackling tough polynomial problems, factoring techniques, AM-GM and Cauchy (and a few other) Inequalities, sequences and series, functional equations, and all sorts of problem solving strategies for problems in these areas.

2: Good point.  I'll put some up next week.

[quote="rrusczyk"]1: All sorts of stuff: quadratics, polynomials, clever manipulations and substitutions, tackling tough polynomial problems, factoring techniques, AM-GM and Cauchy (and a few other) Inequalities, sequences and series, functional equations, and all sorts of problem solving strategies for problems in these areas.[/quote]

wow. Perfect for AMC.  :D  Why doesn't AoPS write textbooks? I'm sure teachers all over the U.S. and some parts of world would use them. They are perfect for teaching what students need to learn and more.


advanced fields
advanced fields unsolved

Consider the algebraic variety of nilpotent matrices in $ M_n(K)$. What is its dimension? Show that it's a normal variety. What are the regular points?

Take the 11th problem session set in this link: [url]http://www.math.lsa.umich.edu/~mmustata/631.html[/url] and look at the last two problems:
They give "a program" on how to deal with the variety of nilpotent matrices.


Let $x,y,z >0$ and $xyz=1$. Prove that \[\sum \frac{1}{x^2(1+y^3)(1+z^3)}\leq \frac{(xy+yz+zx)^2}{2 \sum (x^2y+x^2z)}.\]
[i]Edited by Myth[/i]

[quote="treegoner"]Let $x,y,z >0$ and $xyz=1$. Prove that \[\sum \frac{1}{x^2(1+y^3)(1+z^3)}\leq \frac{(xy+yz+zx)^2}{2 \sum (x^2y+x^2z)}.\]
[/quote]
Since $(xy+xz+yz)^2\geq3xyz(x+y+z)$, we need to prove that $\sum_{cyc}\frac{yz}{(y^2+xz)(z^2+xy)}\leq\frac{3(x+y+z)}{2\sum\limits_{cyc}(x^2y+x^2z)}$,
which is $\sum_{cyc}(x^4y^3+x^4z^3+3x^5yz-x^4y^2z-x^4z^2y-x^3y^3z-2x^3y^2z^2)\geq0$, which is obvious.


linear algebra
matrix
function
analytic geometry
graphing lines
slope
calculus

Hi. I'm a little confused with the definition of a sadle point. Is it a
critical point (all first dirivatives vanish) that is neither local maximum nor
local minimum? Or is it a critical point that is a local minimum in one
direction and a local maximum in another direction? Are these two definitions
equivalent?

Also when you want to find saddle points of f(x,y). By the 2nd partial
dirivative test, if the determinant of the Hessian matrix is 0 then the test is
inconclusive. In this case what do you do next?

For example the function $ f(x,y)\equal{}x^3\minus{}3xy^2$ has a critical point at
(0,0) but the Hessian matrix is the zero matrix there so the test is inconclusive.
Is this a saddle point?

Thanks

[quote="Prime factorization"]Straddle point: A point c such that both $ f'(c)$ and $ f''(c) \equal{} 0$. What does this mean graphically? A point of inflection with a tangent line of slope 0, really.[/quote]

Does the condition "both $ f'(c)$ and $ f''(c) \equal{} 0$" really imply that $ c$ is a saddle point? (I hope by straddle point you mean saddle point). For example $ f(x)\equal{}x^4$ satisfies the condition at 0. But if you take the first definition I gave then it's not a saddle point is it? Because it's a local minimum there.

Also what about in several variables case?

I forgot to mention the second derivative must also change sign. Not sure about multi-variable cases, gonna have to google that one.

When in doubt on definitions, try [url=http://en.wikipedia.org/wiki/Saddle_point]Wikipedia[/url].

A $ C^2$ function from $ \mathbb{R}^n$ to itself has a $ n\times n$ symmetric matrix of second partial derivatives, called the Hessian. If that matrix is positive definite (all eigenvalues positive) at a critical point, the function has a minimum there. If that matrix is negative definite (all eigenvalues negative) at a critical point, the function has a maximum there. If the matrix is indefinite (at least one positive eigenvalue and at least one negative eigenvalue), the function has a saddle point.
This test is inconclusive if there is at least one zero eigenvalue, and the others are either all positive or all negative.

A $ 2\times 2$ matrix is indefinite if and only if its determinant is negative.
A standard example of a saddle point: $ f(x,y)\equal{}x^2\minus{}y^2$ has a saddle point at the origin.

 I think I've heard the example you gave referred to as a "monkey saddle". According to the definition on the page I linked to, it is a saddle point- but not really the standard kind.
Talking about saddle points in one dimension is unusual- I'd never seen that usage before just now.

Hi jmerry, thanks for your reply. Yep I did have a look at that site before posting the question but they didn't say more about the case where the Hessian matrix is singular (det=0).

One way to determine whether a critical point is a saddle point is by using the [url=http://en.wikipedia.org/wiki/Second_partial_derivative_test]Second partial derivative test[/url]. However the test is inconclusive if the Hessian matric is singular, like the example I gave ($ f(x,y)\equal{}x^3\minus{}3xy^2$).  In fact, it is quite easy to find examples where the Hessian matrices are singular at critical points, for example $ f(x,y)\equal{}x^4\minus{}y^4$, $ f(x,y)\equal{}x^5\plus{}y^5$, $ f(x,y)\equal{}x^n\pm y^n$ for $ n\ge 3$ etc. My question is, how do we determine if they are saddle points in these cases?

Also, I'm not sure what did you mean by a saddle point of standard kind. Did you mean the kind that can be determined by the second partial derivative test (i.e. the Hessian is non-singular)?

Thanks

I don't know any particular method for testing these- just that it's not a minimum or maximum. After all, I didn't think they were actually called saddle points before.

By "standard kind", I do mean that the second derivatives are good enough- which doesn''t mean that the Hessian is nonsingular. $ f(x,y,z)\equal{}x^2\minus{}y^2$ has a standard saddle point at zero.


function
real analysis
real analysis unsolved

Find all functions $f \in C(\mathbb{R})$ such that for every $x,y,z \in \mathbb{R}$ holds $f(x)+f(y)+f(z)=f\left({\frac{3}{7}x+\frac{6}{7}y-\frac{2}{7}z}\right)+f\left({\frac{6}{7}x-\frac{2}{7}y+\frac{3}{7}z}\right)+f\left({-\frac{2}{7}x+\frac{3}{7}y+\frac{6}{7}z}\right).$

[quote=Moubinool]f(t)=at+b[/quote]

please tell me how,


We can multiply these together, to get $ \sqrt {400} \equal{} 20$.  But a niftier way (or at least a bit niftier :D): $ 8 \equal{} 2^2 \times 2$ and $ 50 \equal{} 2 \times 5^2$.  So their product equals $ 4^2 \times 5^2$.  Since we need to take the square root, we get $ 4 \times 5$, or $ \boxed{20}$.

Another way is simplifying:
2 sqrt 2* 5 sqrt 2=10 (sqrt 2)^2=10*2=20.

Note that an important property of square roots is $ \sqrt{a} \times \sqrt{b} \equal{} \sqrt{ab}$. That's why AIME15 was able to multiply those two together to get $ \sqrt{400} \equal{} 20$.


quadratics
function
calculus
derivative
real analysis
real analysis unsolved

In the case where $F'(x^{\ast}) = F''(x^{\ast}) = ... = F^{k-1}(x^{\ast}) = 0$ and $F^{k}(x^{\ast}) \neq 0$ and $x^{\ast}$ is the root of the equation $F(x)=0$ and $k \geq 2$. I would like to show if the modified newton's method:

$x_{n+1} = x_{n} - M \frac{F(x_{n})}{F'(x_{n})}$

converges quadractically for some constant M.

I think it does, not 100% sure how to show it though.

You want $M=k$. The key is the Taylor expansion at the root $x^*$.

WLOG, $x^*=0$. It's just a translation. Assume $F$ is nice enough (details specified later).
For $x$ near zero, $F(x)=ax^k+O(x^{k+1})$ and $F'(x)=akx^{k-1}+O(x^k)$.
We have
$p(x)=x-k\frac{F(x)}{F'(x)}=x-\frac{kax^k+O(x^{k+1})}{akx^{k-1}+O(x^k)} =x-x+\frac{O(x^{k+1})}{akx^{k-1}+O(x^k)}=O(x^2)$
and the iteration converges quadratically.

What is "nice enough"? We need more than just existence of $k$ derivatives. To get that $O(x^{k+1})$ in the function and the $O(x^k)$ in the derivative, we need that $n+1$ derivatives exist and are locally bounded. We could settle for something slightly weaker, but not by much.

You seem to be expanding the function F as a pure power series and not a Taylor series, is that right? If so, that's too strong of an assumption.

Also, I don't understand the last step of $p(x)$ before you get $O(x^2)$.

I'll put down what I have so far:

$x_{n+1} = x_{n} - M \frac{F(x_{n})}{F'(x_{n})}$
$x_{n+1} - x^{\ast} = x_{n} - x^{\ast} - M \frac{F(x_{n})}{F'(x_{n})} = \Delta_{n+1} = \Delta_{n} - M \frac{F(x_{n})}{F'(x_{n})}$

Now I approximate F by using the taylor expansion (after simplifying):

$F(x^{\ast} + \Delta_{n}) = \frac{\Delta_{n}^k}{k!} F^{(k)}(x^{\ast}) + O(x^{k+1})$
$F'(x^{\ast} + \Delta_{n}) = \frac{\Delta_{n}^{k-1}}{(k-1)!} F^{(k)}(x^{\ast}) + O(x^k)$

So the newton's method becomes:

$\Delta_{n+1} = \Delta_{n} - M \frac{\Delta_{n}}{k}$

But if M=k then $\Delta_{n+1} = 0$ This seems incorrect to me.

Any thoughts?

[quote="swag"]$F(x^{\ast} + \Delta_{n}) = \frac{\Delta_{n}^k}{k!} F^{(k)}(x^{\ast}) + O(x^{k+1})$
$F'(x^{\ast} + \Delta_{n}) = \frac{\Delta_{n}^{k-1}}{(k-1)!} F^{(k)}(x^{\ast}) + O(x^k)$
[/quote]
That should be

$F(x^{\ast} + \Delta_{n}) = \frac{\Delta_{n}^k}{k!} F^{(k)}(x^{\ast}) + O(\Delta_n^{k+1})$
$F'(x^{\ast} + \Delta_{n}) = \frac{\Delta_{n}^{k-1}}{(k-1)!} F^{(k)}(x^{\ast}) + O(\Delta_n^k)$

That's really the same thing jmerry said. The only differences are in notation, not meaning.

[quote]So the newton's method becomes:
$\Delta_{n+1} = \Delta_{n} - M \frac{\Delta_{n}}{k}$[/quote]

Where's the remainder in that? Neither the numerator nor the denominator is a monomial. This should read

$\Delta_{n+1} = \Delta_{n} - M \frac{\Delta_{n}}{k} +O(\Delta_n^2)$

[quote]But if $M=k$ then $\Delta_{n+1} = 0$ This seems incorrect to me. [/quote]
What you have is that [b]if[/b] the remainder terms in the Taylor expansion were zero - that is, if $F(x^{\ast}+\Delta_n)=C\Delta_n^k$ - then the method would work exactly in one step. This is no different than saying that for a linear function, the usual Newton's method would work exactly in one step.

Ah ok, I understand. I was just confused about the big O notation, haven't seen that much.

Thanks again.

Can someone post how newton method works (If it isn't already discused)? Maybe I used it without knowing.

Just take $k=1$ in this variant. There are also some older threads.


$ \frac{1}{n\plus{}1}\bigg(1\plus{}\frac{1}{3}\plus{}......\plus{}\frac{1}{2n\minus{}1}\bigg) \geq \frac{1}{n}\bigg(\frac{1}{2}\plus{}\frac{1}{4}\plus{}......\plus{}\frac{1}{2n}\bigg)$
$ \Longleftrightarrow n\bigg(1\plus{}\frac{1}{3}\plus{}......\plus{}\frac{1}{2n\minus{}1}\bigg) \geq (n\plus{}1)\bigg(\frac{1}{2}\plus{}\frac{1}{4}\plus{}......\plus{}\frac{1}{2n}\bigg)$
$ \Longleftrightarrow n\bigg(1\minus{}\frac{1}{2}\plus{}\frac{1}{3}\minus{}\frac{1}{4}\plus{}......\plus{}\frac{1}{2n\minus{}1}\minus{}\frac{1}{2n}\bigg) \geq \frac{1}{2}\plus{}\frac{1}{4}\plus{}......\plus{}\frac{1}{2n}$
It is sufficient to show that, for all $ 1< 2k \leq n$, $ n\left(\frac{1}{2k\minus{}1}\minus{}\frac{1}{2k}\right)\geq \frac{1}{2k}$, which is equivalent to $ n\geq 2k\minus{}1$.


Prove that for all positive reals $ x$, $ y$, $ z$, we have:

[b]a)[/b] $ \sum\frac{x}{y\plus{}z}\leq\frac32\cdot\frac{x^2\plus{}y^2\plus{}z^2}{xy\plus{}yz\plus{}zx}$;
[b]b)[/b] $ \sum\left(\frac{x}{y\plus{}z}\right)^2\geq\frac34\cdot\frac{x^2\plus{}y^2\plus{}z^2}{xy\plus{}yz\plus{}zx}$.

[quote="harazi"]a) sum x/(y+z)<=3/2*(x^2+y^2+z^2)/(xy+yz+zx);
[/quote]

Brute force  :(

Believe it or not the ineq is equivalent to :

Sum x^4y + Sum x^3y >= 4 Sum xyz

Which is true by rearrangement  :D

I posted as guest, don't know how it happened  :?

Would you try brute force for the second one?
   Also, the first one has the foolowing elegant solution:
  (xy+yz+zx)*sum x/(y+z)=xyz*sum 1/(y+z)+sum x^2. But xyz*sum 1/(y+z)<=1/2*xyz*sum 1/x=1/2sum xy<=1/2sum x^2.

>> Would you try brute force for the second one?

Mmmm, let me see.... No, definitely no !

I have difficulty to solve inequality b) in a neat way.
I would like to see your solution, Harazi?

Thank you very much.

First solution:
  Do you remember Iran 1996? Of course! Then use that and Cebasev inequality:
   Sum x^2/(y+z)^2>=1/3*(sum x^2)*(sum 1/(x+y)^2)>=3/4*sum x^2/(sum xy) since sum 1/(x+y)^2>=9/(4*sum xy)).
   Second solution:
   We have sum x^2/(y+z)^2>=(sum x^2)^2/[sum x^2(y+z)^2]. So, we must prove that (sum x^2)(sum xy)>=3/(4*sum xy).This comes down to
2sum xy(x^2+y^2)>=3sum x^2y^2+xyz sum x. But this is true since
 2sum xy(x^2+y^2)>=4 sum x^2y^2>=3 sum x^2y^2+xyz sum x.

The inequality: $ \sum\frac {x}{y \plus{} z}\le \frac {3}{2} \frac {x^2 \plus{} y^2 \plus{} z^2}{xy \plus{} yz \plus{} zx}$

It is equivalent with: $ [(xy \plus{} xz) \plus{} (yx \plus{} yz) \plus{} (zx \plus{} zy)](\frac {x^2}{xy \plus{} xz} \plus{} \frac {y^2}{yx \plus{} yz} \plus{} \frac {z^2}{zx \plus{} zy})\le 3(x^2 \plus{} y^2 \plus{} z^2)$

WLOG,we can suppose that $ x\ge y\ge z$ $ \Rightarrow$ $ xy \plus{} xz\ge yx \plus{} yz\ge zx \plus{} zy$,and $ \frac {x^2}{xy \plus{} xz}\ge \frac {y^2}{yx \plus{} yz}\ge \frac {z^2}{zx \plus{} zy}$

So our inequality follows from Chebyshev!

[quote="Maxim Bogdan"]
The inequality: $ \sum\frac {x}{y \plus{} z}\le \frac {3}{2} \frac {x^2 \plus{} y^2 \plus{} z^2}{xy \plus{} yz \plus{} zx}$

[/quote]
Assum $ c \equal{} min${$ a;b;c$}

It is equivalent

 $ [\frac {1}{(a \plus{} c)(b \plus{} c)} \minus{} \frac {1}{ab \plus{} bc \plus{} ca}](a \minus{} b)^2 \plus{} [\frac {a \plus{} b \plus{} 2c}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \minus{} \frac {1}{ab \plus{} bc \plus{} ca}](a \minus{} c)(b \minus{} c) \le\ 0$

Which is obvious :wink:

We have done! :)

Maxim Bogdan, i think your solutin not logical because $ x^2\geq y^2\geq z^2 ; \frac{1}{xy\plus{}xz}\leq\frac{1}{yz\plus{}yx}\leq\frac{1}{zx\plus{}zy}$ can't$ \Rightarrow\frac{x^2}{xy\plus{}xz}\geq\frac{y^2}{yz\plus{}yx}\geq\frac{z^2}{zx\plus{}zy}$

[quote="karis"]Maxim Bogdan, i think your solutin not logical because $ x^2\geq y^2\geq z^2 ; \frac {1}{xy \plus{} xz}\leq\frac {1}{yz \plus{} yx}\leq\frac {1}{zx \plus{} zy}$ can't$ \Rightarrow\frac {x^2}{xy \plus{} xz}\geq\frac {y^2}{yz \plus{} yx}\geq\frac {z^2}{zx \plus{} zy}$[/quote]

Maxim Bogdan didn't mean this as an implication.

  darij

[quote="Maxim Bogdan"]The inequality

$ \frac {x}{y + z} + \frac {y}{z + x} + \frac {z}{x + y}\leq\frac {3(x^2 + y^2 + z^2)}{2(yz + zx + xy)}$

is equivalent with

$ \left[\sum{x(y + z)}\right]\left(\sum{\frac {x}{y + z}}\right)\le 3\sum{x^2}.$

$ \Big[y(z + x) - z(x + y)\Big]\left(\frac {y}{z + x} - \frac {z}{x + y}\right) = \frac {x(y - z)^2(x + y + z)}{(z + x)(x + y)}\geq0\Longrightarrow$

the triples $ \Big(xy + zx,yz + xy,zx + yz\Big)$ and $ \left(\frac {x}{y + z},\frac {y}{z + x},\frac {z}{x + y}\right)$ are similarly sorted.

So our inequality follows from Chebyshev sum inequality![/quote]Maxim Bogdan's proof is the best. My proof is

$ \frac {4x^2 + x(y + z) + (y - z)^2}{4(yz + zx + xy)} - \frac {x}{y + z} = \frac {(y - z)^2(x + y + z)}{4(y + z)(yz + zx + xy)}\geq 0\Longrightarrow$

$ \sum{\frac {x}{y + z}\leq \sum{\frac {4x^2 + x(y + z) + (y - z)^2}{4(yz + zx + xy)}} = \frac {3\left(x^2 + y^2 + z^2\right)}{2(yz + zx + xy)}.}$
[quote="harazi"]Prove that for all positive reals $ x$, $ y$, $ z$, we have:

$ \frac {x^2}{(y + z)^2} + \frac {y^2}{(z + x)^2} + \frac {z^2}{(x + y)^2}\geq\frac {3\left(x^2 + y^2 + z^2\right)}{4(yz + zx + xy)}.$

First solution:

  Do you remember Iran 1996? Of course! Then use that and Cebasev inequality:

   $ \sum \frac {x^2}{(y + z)^2}\geq \frac {1}{3}\left(\sum {x^2}\right)\left[\sum \frac {1}{(y + z)^2}\right]\geq \frac {3\sum {x^2}}{4\sum {yz}}$ since $ \sum \frac {1}{(y + z)^2}\geq\frac {9}{4\sum{yz}}.$[/quote]This proof has relish for using [url=http://www.mathlinks.ro/viewtopic.php?t=32470]Iran 1996[/url] and [url=http://mathworld.wolfram.com/ChebyshevSumInequality.html]Chebyshev sum inequality[/url]. My proof is

$ \frac {x^2}{(y + z)^2} - \frac {5x^2 - x(y + z) - (y - z)^2}{4(yz + zx + xy)}$

$ = \frac {(y + z - x)^2(y - z)^2 + x(z + 3y)(x - y)^2 + x(y + 3z)(x - z)^2}{4(y + z)^2(yz + zx + xy)}\geq 0\Longrightarrow$

$ \sum{\frac {x^2}{(y + z)^2}\geq \sum{\frac {5x^2 - x(y + z) - (y - z)^2}{4(yz + zx + xy)}} = \frac {3\left(x^2 + y^2 + z^2\right)}{4(yz + zx + xy)}.}$

I think I have a simple answer for the first question:

Because we know that a/(b+c)+b/(a+c)+c/(a+b) is smaller or equal to 3/2,
and also because a^2+b^2+c^2 is greater or equal to ab+bc+ac,
thus 3/2 [(a^2+b^2+c^2)/(ab+bc+ac)] is greater or equal to 3/2 which is greater or equal to a/(b+c)+b/(a+c)+c/(a+b)

The last post made a mistake.
I think you try to use Nesbitt's inequality, but in the opposite way.


Four digits are slected from the set {1,2,3,4,5} to form a 4-digit number. If D is the set of all possible 4-digit numbers formed. find the sum of all the elements in D.

If repeats are not allowed, then

AIME15 is correct because there are $ 120$ ways to chose $ 4$ numbers, and the average of the numbers chosen = $ 3333$....

If repeats are allowed, then I think it would be $ 625(3333)\equal{}2083125$... But, I'm not sure about this.

[quote="Math Champion"]I don't think repeats are allowed because in the set {1,2,3,4,5} there is only 1 of each number.[/quote]

Actually, even though it says {1,2,3,4,5}, there can be more than $ 1$ of each in the set...... You would write {1,2,3,4,5,5} if there were $ 2$ $ 5$'s in the set would you?


logarithms
function
limit
calculus
derivative
algebra unsolved
algebra

hello,let us defining $ f(x)\equal{}1\plus{}\frac{\ln(10\minus{}x)}{\ln(2)}\minus{}\frac{4\ln(2)}{\ln(x)}$, so we get
a) for $ 0<x<1$ is $ f(x)$ positive
b) now we consider the interval $ 1<x<10$, here we have $ \lim_{x \to 1\plus{}0}f(x)\equal{}\minus{}\infty$ and $ \lim_{x \to 1\plus{}0}f(x)\equal{}\minus{}\infty$ furhter we have $ f'(x)\equal{}\minus{}\frac{1}{(10\minus{}x)\ln(2)}\plus{}\frac{4\ln(2)}{\ln(x)^2x}$. Here we get
$ \lim_{x \to 1\plus{}0}f'(x)\equal{}\plus{}\infty$ and $ \lim_{x \to 10\minus{}0}f'(x)\equal{}\minus{}\infty$.The first derivative has only on root, $ x_0\approx4.55397907243123768129$, this leads to a local maximum, it is $ f(x_0)\approx 1.61631980026074118117674654$, it follows that $ f(x)\equal{}0$ has exactly two roots $ x_1\equal{}2$ and $ x_2 \approx 8.78950340862945160382298475$.
A nice problem!
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello,let us defining $ f(x) \equal{} 1 \plus{} \frac {\ln(10 \minus{} x)}{\ln(2)} \minus{} \frac {4\ln(2)}{\ln(x)}$, so we get
a) for $ 0 < x < 1$ is $ f(x)$ positive
b) now we consider the interval $ 1 < x < 10$, here we have $ \lim_{x \to 1 \plus{} 0}f(x) \equal{} \minus{} \infty$ and $ \lim_{x \to 1 \plus{} 0}f(x) \equal{} \minus{} \infty$ furhter we have $ f'(x) \equal{} \minus{} \frac {1}{(10 \minus{} x)\ln(2)} \plus{} \frac {4\ln(2)}{\ln(x)^2x}$. Here we get
$ \lim_{x \to 1 \plus{} 0}f'(x) \equal{} \plus{} \infty$ and $ \lim_{x \to 10 \minus{} 0}f'(x) \equal{} \minus{} \infty$.The first derivative has only on root, $ x_0\approx4.55397907243123768129$, this leads to a local maximum, it is $ f(x_0)\approx 1.61631980026074118117674654$, it follows that $ f(x) \equal{} 0$ has exactly two roots $ x_1 \equal{} 2$ and $ x_2 \approx 8.78950340862945160382298475$.
A nice problem!
Sonnhard.[/quote]
[color=blue][b]May this solution be the only one?[/b][/color]

hello, do you meant $ \left(\log_2(x)\right)^4\plus{}\left(\log_{\frac{1}{2}}\left(\frac{x^3}{8}\right)\right)^2\plus{}9\log_2\left(\frac{32}{x^2}\right)\equal{}4\left(\log_{\frac{1}{2}}(x)\right)^2$?
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, do you meant $ \left(\log_2(x)\right)^4 \plus{} \left(\log_{\frac {1}{2}}\left(\frac {x^3}{8}\right)\right)^2 \plus{} 9\log_2\left(\frac {32}{x^2}\right) \equal{} 4\left(\log_{\frac {1}{2}}(x)\right)^2$?
Sonnhard.[/quote]
[color=blue][b]That's right...[/b][/color]

hello, simplifying your equation we get
$ \left(\frac{\ln(x)}{\ln(2)}\right)^4\plus{}\left(\frac{\ln\left(\frac{x^3}{8}\right)}{\minus{}\ln(2)}\right)^2\plus{}9\left(\frac{\ln(32)\minus{}\ln(x^2)}{\ln(2)}\right)\equal{}4\left(\frac{\ln(x)}{\minus{}\ln(2)}\right)^2$
$ \left(\frac{\ln(x)}{\ln(2)}\right)^4\plus{}9\left(\frac{\ln(x)\minus{}\ln(2)}{\ln(2)}\right)^2\plus{}9\left(\frac{5\ln(2)\minus{}2\ln(x)}{\ln(2)}\right)^2\minus{}4\left(\frac{\ln(x)}{\ln(2)}\right)\equal{}0$.
Setting $ t\equal{}\ln(x)$ we have to solve
$ t^4\plus{}41t^2(\ln(2))^2\minus{}198t(\ln(2))^3\plus{}234(\ln(2))^4\equal{}0$.
Sonnhard.

For number 2, I don't know how to get the 2nd roots. It's not integer.
[hide="2"]
We have $ \log_4{x} \neq 0\Rightarrow x \neq 1$
So we have
$ \log_2{2}\plus{}\log_{2^2}{(10\minus{}x)^2}\equal{}\frac{\log_4{16}}{\log_4{x}}\\
\Leftrightarrow \log_2{2}\plus{}\log_2{(10\minus{}x)}\equal{}\log_x{16}\\
\Leftrightarrow \log_2{(20\minus{}2x)}\equal{}\log_x{16}$
So we see $ \boxed{x\equal{}2}$ is a root of the equation.
[/hide]


I haven't been able to answer this problem:

 :arrow: Find a positive integer [i]a[/i] so the sum

a + 2a + 3a + 4a + 5a + 6a + 7a + 8a + 9a

is a number of equal digits.

Any help is appreciated.  :)

Hmm? So $45a$ is equal to some number with equal digits? Or does $a$ represent a digit?

Exactly. 45a is equal to some number with equal digits.

[hide="Solution"]
hmm if i'm interpreting your problem right...

$1a+2a+\cdots+9a = 45a = 9\cdot 5\cdot a$

Clearly, the number we are looking for must have a $5$ or a $0$ in the units digit since it is divisible by $5$. Since $0$ is not a positive integer, all the digits in the number must be $5$. The number must also be divisible by $9$, so there must be $9k$ digits, where $k$ is a positive integer. From this, we can find that
\[a = \boxed{\frac{555,555,555}{45}, \frac{555,555,555,555,555,555}{45}, \cdots = 12345679 , 12345679012345679, \cdots}\]
neat answer  :)  [/hide]

Thanks a lot! :D It's exactly the answer I needed.


In order for a subset M of a division ring K to be a sub-division ring is it enough to have only the following two conditions satisfied?
i) for every x and y in M, x+y is also in M
ii) for every x in M, x^-1 is also in M

I found this in a coursebook, without proof, but the teachers I spoke to say that it is wrong. Still, they cannot find a counterexample. Can you help me with this one?

I used the name division ring for a field that is not necesarily commutative. It is the only word in English that I could find for this :)

Come on guys, any idea?

I used the idea in
[url]http://www.mathlinks.ro/viewtopic.php?t=4217[/url]
to get x2 in M for every x in M and xyx in M for every x,y in M
also xy+yx in M

Can you end this?

Isn't $M$ as the non-negative or positive (belongs to if you want $x^{-1}$ for ALL $x$ in $M$) real numbers in the field of the real numbers a counterexample


How many four-digit numbers are there such that the last three digits sum to the thousands digit?

How would you solve this problem? Other than generating the function

$(1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9)(1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9)(1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9)$

and multiplying it out and looking at the different coefficients?

What if the number has to be between 3000 and 6000? Or some other two numbers?

What is the C(a, b) notation? I should know this but I don't :blush:.

This represents the number of combinations of a set (= the binomial coefficient)

Aha. Thanks.

All this and more in LaTeX!!! Learn how to use the LaTeX typesetting system by clicking on the appropriate link!


real analysis
limit
search
calculus
calculus computations

Let $ (a_n)$ be a sequence of positive real numbers. Show that if $ \sum{a_n}$ converges then $ \liminf_{n\to\infty}na_n \equal{} 0$
Show that if also $ (a_n)$ is decreasing then $ \lim_{n\to\infty}na_n \equal{} 0$

Let $ a_k \equal{} \frac {1}{\sqrt {k}}$.  Clearly $ a_k\to 0$ but $ \liminf _{n\to \infty}na_n \equal{} \lim_{n\to \infty}na_n \equal{} \infty$.

im assuming by converge you mean in sum right?

$ \lim_{n \to \infty} \sum_{k\equal{}1}^n a_k < \infty$

or else the guy above me is right

Maybe he meant to divide by n, not multiply by n. In which case he refers to Stolz-Cesaro. kenn4000 may be right alternatively since his statement is also correct.

sorry, yes, i meant to say that the series converges

$ J_n \equal{} \sum_{k \equal{} 1}^{2n} a_n \ge \sum_{k \equal{} 1}^n a_n \plus{} n \inf_{k \ge n} a_k \equal{} T_n$

$ \sum_{k \equal{} 1}^\infty a_n \equal{} \lim_{n \to \infty} J_n \ge \lim_{n \to \infty} T_n \equal{} \sum_{k \equal{} 1}^\infty a_n \plus{} n\lim \inf a_n$

so $ n\lim \inf a_n \to 0$

Apply the Cauchy condensation test to the series with non-increasing terms and go from there. You can search for the terms cauchy condensation test and decreasing/non-increasing sequence since this has been asked numerous times.


geometry
geometric transformation
rotation

In how many ways can we seat 6 people around a table if John and Sam insist on sitting next to each other?

hello

[hide="solution"]
As we need John and Sam to sit together,we first treat both of them as a single entity W.Now, the four people and the entity W can be selected around a table is $ ((4\plus{}1)\minus{}1)!$ i.e.24 ways
we can now permute John and Sam inside the entity W.This can be done in 2!.
The total ways =24*2=48
[/hide]
Thank u.

i did something different

[hide]6 choices for john times two for sammy. multiply it by 4!. then you get 288. divide it by 6 for rotations and you get 48[/hide]

[quote=SPG2500]What about rotational symmetry?[/quote]

The problem includes rotational symmetry.
Solution:
[hide]Think about John and Sam as a large person that takes $2$ seats up. We would now have $5$ people, and there would be a total of $5!$ ways to seat them. Since the table includes rotational symmetry, we have to divide our answer by $5$, since there are $5$ people. But, John and Sam can be in $2$ different positions: $JS$ or $SJ$. Therefore, we multiply our answer by $2$ . $\frac{(5!)(2)}{5} = \frac{(240)}{5} = \boxed{48}$.[/hide]

[quote=potato36]Why don't we divide by 6 to get rid of the rotations?[/quote]

You don't because you count John and Sam as one person, not as two.

[hide]so you divide by $5$ to get rid of the rotational symmetry and then multiply by two because they can sit in two kinds of ways, then multiply by $5!$[/hide]

[hide]For rotational symmetry, since there are 6 people you are sitting and you treat John and Sam as one entity, you have (5-1)! ways to do that with rotational symmetry. It is 5 because there are 5 entities if you treat John and Sam as a single entity. Then you can switch John and Sam around, so there are (5-1)!*2 ways. However, because of reflectional symmetry, you divide by 2 and that leaves you with (5-1)!=24 ways.[/hide]

[quote=pifinity][font=TIMES NEW ROMAN][hide=Rotational Symmetry Solution]Due to the fact that two of the people have to sit next to each other, let's momentarily consider them as one "double-person." This way, we have five people, so we can order them in $5\!$ ways. However, $5$ of these positions are identical copies of each other that are obtained by rotation, so we divide by $5$ to get $24$ ways.

Now we have to go back and consider our "double-person". There are two ways to make him: John-Sam, or Sam-John. That means that there are $24\cdot2=\boxed{48}$ total ways to seat the six people. $$\scriptsize{\colorbox{red}{\color{black}{A pifinity solution}}}$$[/quote]

Woah, one year bump

...And?

I don't see why so many people have a problem with bumps. How does it harm you? Also, if none of these topics were bumped then eventually all of them would become "too old" and we wouldn't be able to post in them.


geometry ARML AMC USA(J)MO USAMO inradius AIME

At our last practice for the SC ARML team I gave out a handout of Geometry problems. Here's a hyperlink to the problems http://155.225.48.46/statemat/stuf2004/Geoprobs.htm

This might be a good place to discuss them. I particularly liked numbers 4, 14, and 17. Were there any that you liked, or didn't like? :)

:blush: I haven't worked on them yet...but after next Monday I will start going back to doing much more math daily (at least I hope).  So far I've done #'s 1,2,3 and 13, though.

Hopefully I can get better at math before ARML...ever since USAMO I have let my math skills deteriorate....the problem solving part at least.

i've only done #13 (and got it wrong by a stupid mistake). i've been real busy lately, but like joe, i'll have a bunch more free time after monday.

[quote="gauss202"]
This might be a good place to discuss them. I particularly liked numbers 4, 14, and 17. Were there any that you liked, or didn't like? :)[/quote]

Number 4 was nice.  But how did you do it? [hide="I"]found the area two different ways and then equated them and solved.[/hide]  Did you do it a different way?  I haven't done numbers 14 or 17 yet :( .

[quote="joml88"]Ok, I just keep getting a different answer from the one that is posted on the site with the problems for #10.

Also, is there any reason why the problem states that the triangle is inscribed in a circle?  I mean, all triangles can be inscribed in a circle and this particular problem doesn't really use the circle, so...

Anyway, here is my work for #10.  Can you find any errors?

10. Triangle ABC is inscribed in a circle and BP bisects angle ABC. If AB = 6, BC = 8, and AC = 7, compute BP.

[hide="Number 10"]
Using the angle bisector theorem we have:

\[6(CP)=8(AP)\Rightarrow AP=\frac 34 CP.\]

We also have that $AP+CP=AC=7$.  Thus $\frac 34 CP+CP=\frac 74 CP=7$ meaning that $CP=4$.  Also, this makes $AP=3$.

Now, by Stewart's Theorem we have:

\[6^2(4)+8^2(3)=(7)BP^2+(3)(4)(7)\Rightarrow BP=6.\]
[/hide][/quote]

Sorry, the problem should specify that the point P is on the circumscribed circle, not on the triangle. Your solution up to that point is correct though.

[quote="joml88"]Number 4 was nice.  But how did you do it? [hide="I"]found the area two different ways and then equated them and solved.[/hide]  Did you do it a different way?[/quote]

I did it the same way. [hide="I"]equated Area = inradius $\times$ semiperimeter and Heron's Formula. This was an old AIME problem.[/hide]

O, duh...don't know why I assumed it was on $\overline{AC}$.  It would have made much more sense to assume that $P$ was on the circle in that problem.  Anyway, [hide="to finish my solution..."]a simple application of Power of a Point tells us that $XP=2$ (let $X$ be the intersection of $BP$ and $AC$) and thus $BP=8$.  QED :D [/hide]

And by the way, if anyone would like to see any hints or solutions to the problems on the geometry worksheet let me know.

[quote="joml88"]Hopefully I can get better at math before ARML...ever since USAMO I have let my math skills deteriorate....the problem solving part at least.[/quote]

If Michael Jordan had played every minute of every game, he would have been tired by the beginning of each finals.

about those practice problems.
they are pretty hard. i worked on them for over 30 minutes and was only able to answer 1, 3, and 6 out of the first 6.
checking to see if they are right...
well, 2 out of 3 are right. missed #1.
i'll go back soon and try some more. i know i'm close to solving #2.
wish i knew more geometry.  :(

Yes, they are pretty hard. Keep working at them though.

A hint for number 2:
Whenever there are lots of right triangles, some of them are sure to be similar! So look for similar triangles in your diagram. Once you get a nice equation in $p$ and $q$, divide it through by $q^2$ and call the quantities $p/q$ something like, say, $z$. Then you can solve for $z$!

For number 5:
Draw a picture and think about what you know about triangle PSO.

woooo, these problems are tough. i've managed to get 1, 3, 5, 6, 7, 9, 10, 12, 13. currently working on 16. some of these i've seen before and some are torture.

Yes, some of these are hard! I'd say if you've gotten 9 of them so far that's pretty darn good! A few of them require some somewhat advanced theorems, like number 14 for instance, you'd have to know [url=http://mathworld.wolfram.com/CevasTheorem.html]Ceva's Theorem[/url] in order to do that one. Also the Angle Bisector Theorem and Stewart's Theorem come up in a lot of these problems.

Some of these are old ARML team problems (I think numbers 2, 5, 12, 13, 15, 17, 18, 19* -- beware, this one is definitely a killer), and some of them are old AIME problems. So they are pretty hard, but hopefully you can pick up a few things things from working on them. I'll be on the bus to show people how to do some of the ones they could not do.

[quote="gauss202"] A few of them require some somewhat advanced theorems, like number 14 for instance, you'd have to know [url=http://mathworld.wolfram.com/CevasTheorem.html]Ceva's Theorem[/url] in order to do that one.[/quote]

Is that the [i]only[/i] way to do that one?  I used Ceva's when I did that problem because I immediately thought of it and it offered a (more or less) easy path to the solution but I thought there might have been a way around the problem without Ceva's.

Well Ceva's Theorem itself is based on similar triangles (as are nearly all theorems of geometry). So in theory it could be solved using only similar triangles. But your life would be miserable if you tried to solve it that way. Ceva is the way to go.

By the way, here's a nice visual proof of Ceva's Theorem http://www.ies.co.jp/math/java/vector/ceva/ceva.html

i just got number 16.   :showoff:  yay. i think i'll stop there tho. too much math fry the brain.  :what?: and besides, i gotta pack. nothing like waiting till the last minute. :lol:
 :stretcher:

Can someone give a detailed explanation of number 14?  I don't understand all the terms, but I would like to learn how to solve this type of problem.  I looked at Ceva's Theorem, but I need some help understanding the problem and how to solve it.  An illustration would be nice (if one is necessary).

I'll post a detailed explaination when I get back this afternoon.

[quote="Problem 14"]
Given triangle ABC, with AB = 12 and CA = 15. The median from B, the altitude from C, and the angle bisector from A are all concurrent inside the triangle. Compute the length of BC.[/quote]

(I labeled my picture wrong so I had to change the problem statement slightly...I just changed the vertices so it's essentially the same)

[hide="Solution"]
In the picture, the cevian from $B$ is an altitude denoted by $h_b$ (a 'cevian' is just a segment from a vertex of a triangle to the opposite side), the cevian from $A$ is an angle bisector and the cevian from $C$ is a median.  Everything else in the picture should be self-explanatory.

[img]http://img.photobucket.com/albums/v213/joml88/geom14ceva.png[/img]

First we use the angle bisector theorem and get

\[12m=15n\Rightarrow m=\frac 54 n.\]

Putting that aside for a second, we know from Ceva's theorem that

\[4m(15-y)=6ny\Rightarrow m(15-y)=ny.\]

Now we sub $\frac 54 n$ in for $m$

\begin{eqnarray*}
\frac 54 n (15-y) &=& ny\\
\Rightarrow 75-5y &=& 4y\\
\Rightarrow y &=& \frac{25}3\\
\Rightarrow 15-y &=& \frac{20}3
\end{eqnarray*}

So now we know $y$ and $15-y$ in the picture.  Also, from the Pythagorean theorem we have

\begin{eqnarray}
\left(\frac{20}3\right)^2+h_b^2 &=& 12^2\\
\left(\frac{25}3\right)^2+h_b^2 &=& (m+n)^2
\end{eqnarray}

Thus $(1)-(2)=\frac{400}9 - \frac{625}9 = 144-(m+n)^2 \longrightarrow \boxed{m+n=13.}$[/hide]

[quote="mysmartmouth"]Thanks Joe.  Any more solutions to any other of those problems would be greatly appreciated![/quote]
which ones would you like solutions for?

[quote="gauss202"]Here's another geometry problem:

In square $ABCD$, $\overline{DF}$ bisects $\angle ADE$. Compute $\overline{DE}$.[/quote]

I'll answer Gauss' problem since no one else had:

EDIT: See below

[quote="Iversonfan2005"][quote="mysmartmouth"][b]ANY[/b] of them!  :D[/quote]

Obviously you won't get it if you haven't taken geometry yet!:P  :rotfl: ;)[/quote]

Shut up Shobhit.  Who failed geometry IN 10TH GRADE!?  :P

Also, I think your answer is wrong...

[quote="Iversonfan2005"]

I'll answer Gauss' problem since no one else had:

For this, can't you draw a segment $EF$ and thus the two triangles that compose quadrilateral $DEFA$ are congruent because of they share the same two angles and the same hypotenuse[/quote]

No, triangles ADF and DEF are not congruent. They only share one side and one angle. That's not enough for congruency. The final answer to this one should come out nice.

EDIT: Found out how to solve it. Draw in Triangle $FBA$ such that $FA$ is collinear to $DF$ and angle $FBA$ is a right triangle. We see that $FBA$ is similar to both triangle $DAC$ and $DAF$, from there we can use the angle bisector theorem, etc.

[hide="1."]
There are two cases (refer to the diagram for both cases):

[b]Case 1:[/b] $AB=4$ and $AC=5$ are the legs (in a right triangle, the legs are altitudes).

By the Pythagorean theorem, $BC^2=4^2+5^2\Rightarrow BC=\sqrt{41}.$  Now, we calcuate the area of $\triangle ABC$  in two ways and equate:

\begin{eqnarray*} \frac 12(AB\cdot AC) &=& \frac 12(AD\cdot BC)\\ \frac 12(4\cdot 5) &=& \frac 12(\sqrt{41}AD)\\ AD &=& \frac{20}{\sqrt{41}}. \end{eqnarray*}

[b]Case 2:[/b] One of 4,5 is a leg and the other is the altitude to the hypotenuse.

WLOG, the only two possibilities are $AB=4$ and $AD=5$ or $AB=5$ and $AD=4$.  However, the first does not work because $\triangle ABD$ is right and therefore the longest side must be opposite the right angle, which isn't the case.  Thus, we have $AB=5$ and $AD=4.$

We have a 3-4-5 right triangle, so $BD=3$.  Then, $AD$ is the mean proportional (aka geometric mean) of $BD$ and $CD$.  Thus: \[ \frac{BD}{AD}={AD}{CD}\Rightarrow CD=\frac{16}3. \]

Therfore, $BC=3+\frac{16}3=\frac{25}3.$  Again, we compute area is two different ways and equate \[ \frac 12(5\cdot AC)=\frac 12\left(4\cdot\frac{25}3\right)\Rightarrow AC=\frac{20}3. \]

Since $\frac{20}3>\frac{20}{\sqrt{41}}$, our answer is $\boxed{\frac{20}3}.$

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=356[/img][/hide]

Number 12:

[hide]Knowing tangents from each point are equal, let:

$a+d=18$
$d+c=21$
$b+c=17$
$b+a=?$

By adding the first three equations together we get $2(d+c)+a+b=56 \Longrightarrow \boxed{a+b=14}$[/hide]

Number 4:
[hide]
Knowing tangents from each point are equal, we have $AB=50, BC=27+x, AC=23+x$, knowing this, we can use $rs=A=\sqrt{s(s-a)(s-b)(s-c)} \Longrightarrow 21(50+x)=\sqrt{50+x(x)(23)(27)} \Longrightarrow 21^2(50+x)=621x \Longrightarrow 22050=180x \Longrightarrow x=122.5$. From here, we see that the perimeter of the triangle is $\boxed{345}$.
[/hide]
Number 5:
[hide]
Let $x=PS$, and draw in $OS$ and draw in $RQ$ such that $PR$ is the the diameter of Circle O. We see that triangle $POS$ is similar to $PRQ$ and are in a $1: 2$ side ratio, thus $\frac{x}{12-x}=\frac{1}{2} \longrightarrow x=\boxed{4}$[/hide]

Number 13:
[hide]
We can see that $\frac{x-y}{2}=30 \Longrightarrow x-y=60$
We can also see that arc $ABD=230$ and arc $BAD=250$, and so $490-x-y=360 \Longrightarrow x+y=130$, so now we have a system of equations and we see that $x+y=190 \Longrightarrow x=\boxed{95}$[/hide]

Number 19
[hide]Using a hint that we need to use Ptolemy's Theorem to solve this problem, I first labeled the points of the pentagon $ABCDE$, where $DC=6$ and everything else is $16$. From here I decided to split the pentagon into triangles and realized that there were only two types of triangles the pentagon can be split up into (that we know atleast two sides of), triangles with two sides of $16$ and triangles with sides of $6$ and $16$, so therefore $AD=EB=AC$ and $CE=DB$. So then I called each of the first three segments $x$ and each of the second two segments $y$ and then I decided to split the pentagon into quadrilaterals so I can use Ptolemy's Theorem. First I had quadrilateral $ABDE$ and using Ptolemy's I had: $16^2+16y=x^2$ and then I then made the other possible quadrilateral which was $EBCD$ and used Ptolemy's to get $6x+16^2=y^2$ and the last possible quadrilateral $ABCD$ and got $yx=16x+96$
Okay so after simplifying for $x$ and getting the equation $x^3-512x-1536$, I used rational root theorem and using trial and error I found that only $x=24$ worked for this equation and so $y=20$ and so after using Heron's formula on the three triangles and adding their areas up, we have the area of the Pentagon is $\boxed{123\sqrt{7}}$[/hide]

Easy way to do number 6
[hide]
Let $Z$ be the midpoint of $CB$ and so $ZA$ is the altitude of the triangle. So then $\triangle{AZD}$ is a right triangle with leg of $4$ and hypotenuse of $16$ so the altitude of the triangle is $\sqrt{240}$, so then $\triangle {AZC}$ is also a right triangle with leg of $x+4$ and $\sqrt{240}$ and hypotenuse of $17$ so $x=3$ and $CB=2x+8=\boxed{14}$[/hide]

Number 17[hide]

I drew in the chord who's length is $1$ and made a right triangle (the radius of the circle as the hypotenuse, $\frac{1}{2}$ and $1-x$ as the leg, and got $1-x=\frac{\sqrt{3}}{2}$ and then using trig, I got $tanx=\frac{\sqrt{3}}{3} \Longrightarrow x=30$ degrees and so the number of degrees before we hit the chord of length $1$ is $60$. So therefore if $k$ refers to the number of terms right after hitting $60$ degrees, we have $\boxed{11}$, since $\frac{10(10+1)}{2}=55$ degrees$< 60$ degrees. [/hide]

Number 18
[hide]
This is probably the easiest out of all the problems, no real trick involved, just do a little bit of angle chasing. We know that this is a regular heptagon and so each of the interior angles are $\frac{900}{7}$. Thus $\angle{EDC}=\frac{900}{7}$ and so $\angle{DEC}=\frac{180}{7}$ and so $\angle{CEF}=\frac{720}{7}$. We know that quadrilateral $FECB$ is cyclic and so $\angle{FBC}=\frac{540}{7}$ and so $\angle{PBA}=\frac{360}{7}$ and we know that $\angle{BAP}=\frac{180}{7}$ and so $\boxed{\angle{APB}=\frac{720}{7}}$[/hide]

Can someone show how to do numbers 2 and 16? :?

If you went to the Duke Math Meet, you would know how to do #2 ;)  Marcus/Mr. Neal (which do you prefer to be called?) gave us a problem with a similar concept before the competition.

[hide="Solution"]
First, we use the fact that the altitude of a right right triangle is the geometric mean of the two segments that it divides the hypotenuse into: \[ \frac p{\frac 14(p+q)}=\frac{\frac 14(p+q)}q\Rightarrow p^2-14pq+q^2=0.  \]  Now, divide through by $q^2$: \[ \left(\frac pq\right)^2-14\left(\frac pq\right)+1=0.  \]  Using the quadratic formula, \[ \frac pq=\frac{14\pm\sqrt{196-4(1)(1)}}2=7\pm4\sqrt{3}.  \]  Since the problem tells you it is in the form $a-b\sqrt{3}$ you could be done right now.  But assume you didn't know that.  How would you decide between $7+4\sqrt{3}$ and $7-\sqrt{3}.$  Simple.  We were given that $p<q\Rightarrow \frac pq<1$ ($p$, $q$ must be positive).  If we look at $7\pm4\sqrt{3}$ as $\sqrt{49}\pm\sqrt{48}$ it makes it much easier to see that we must take the negative sign.  Thus, our answer is $7-4\sqrt{3}.$  The problem asked for the ordered pair, $(a,b)$, so $(7,4)$ is the answer to the problem.[/hide]

I'll post #16 soon.  It's a fun problem though ;)

[hide="Number 16"]
First of all, don't be intimidated by all the lines and instructions!  Just keep a cool head and wade through all of it.  It's not that bad :)

[hide="Still cant get it?"]
Since $AB=1$ and $BC$ is also a radius, $BC=1.$  Thus $BD=1/2$, $DE=1/4$, $EF=1/8$ and therefore $BF=7/8.$  Using the Pythagorean Theorem we find $AF=\frac{\sqrt{113}}8.$  Since radii are equal, $AG=AH=1/2.$

Now, $\triangle AHI\sim \triangle AFB$ and $\triangle AJH\sim \triangle AIF$  by AAA.  Thus $\frac{AJ}{AI}=\frac{AH}{AF}$ and $\frac{AI}{AH}=\frac{AB}{AF}.$  From the second equation, we obtain $AI=\frac 8{7\sqrt{113}}.$  Plugging this into the first equation and simplifying gives $AJ=\frac{16}{113}.$[/hide][/hide]

[quote="mysmartmouth"]Any hints on #3?[/quote]

[hide="Hint #1"]
Angle bisector theorem[/hide]

[hide="Hint #2"]If you can find $AE$...you should be able to find $BP$ and $PE$.[/hide]

Did you get #3?  I'll post the solution now since it's been some time since I posted the hints.

[hide="Solution #3"]
First, it is [i]really[/i] helpful to know the angle bisector theorem for this problem...so [url=http://www.cut-the-knot.org/Curriculum/Geometry/AngleBisectorRatio.shtml]here[/url]'s a link.

Now, by the angle bisector theorem, $7\cdot CD=5\cdot BD$.  But, $BD+CD=BC=3.$  So we have a pair of simultaneous equations for $BD$ and $CD$.  Solving, we get $BD=\frac 74$ and $CD=\frac 54.$

We do the same thing for $AE$ and $CE$ (use angle bisector theorem, get a pair of simultaneous equations) and solve to find that $AE=\frac 72$ and $CE=\frac 32.$

So then using the angle bisector theorem on $\triangle ABE$ we get $\frac xy=\frac 7{(7/2)}=2.$

Actually, I just realized that finding $BD$ and $CD$ was not necessary...anyway, I'm tired and I already wrote it, so it's staying.   :P[/hide]

Solutions so far we've done to make it organized

[hide="number10byJoe"]10. Triangle ABC is inscribed in a circle and BP bisects angle ABC. If AB = 6, BC = 8, and AC = 7, compute BP.

Using the angle bisector theorem we have:

\[ 6(CP)=8(AP)\Rightarrow AP=\frac 34 CP.  \]

We also have that $AP+CP=AC=7$.  Thus $\frac 34 CP+CP=\frac 74 CP=7$ meaning that $CP=4$.  Also, this makes $AP=3$.

Now, by Stewart's Theorem we have:

\[ 6^2(4)+8^2(3)=(7)BP^2+(3)(4)(7)\Rightarrow BP=6[/hide: d46efe42eb] [hide: d46efe42eb=number14byJoe]In the picture, the cevian from $: 23da9872be: e26410e9d5: 5efb0bd8f8: 5d830936b8: d46efe42ebB$: 23da9872be: e26410e9d5: 5efb0bd8f8: 5d830936b8: d46efe42eb is an altitude denoted by $: 23da9872be: e26410e9d5: 5efb0bd8f8: 5d830936b8: d46efe42ebh_b$: 23da9872be: e26410e9d5: 5efb0bd8f8: 5d830936b8: d46efe42eb (a 'cevian' is just a segment from a vertex of a triangle to the opposite side), the cevian from $: 23da9872be: e26410e9d5: 5efb0bd8f8: 5d830936b8: d46efe42ebA$: 23da9872be: e26410e9d5: 5efb0bd8f8: 5d830936b8: d46efe42eb is an angle bisector and the cevian from $: 23da9872be: e26410e9d5: 5efb0bd8f8: 5d830936b8: d46efe42ebC$: 23da9872be: e26410e9d5: 5efb0bd8f8: 5d830936b8: d46efe42eb is a median. Everything else in the picture should be self-explanatory. [img: d46efe42eb]http: //img.photobucket.com/albums/v213/joml88/geom14ceva.png[/img: d46efe42eb] First we use the angle bisector theorem and get \[12m=15n\Rightarrow m=\frac 54 n.  \]

Putting that aside for a second, we know from Ceva's theorem that

\[ 4m(15-y)=6ny\Rightarrow m(15-y)=ny.  \]

Now we sub $\frac 54 n$ in for $m$

\begin{eqnarray*} \frac 54 n (15-y) &=& ny\\ \Rightarrow 75-5y &=& 4y\\ \Rightarrow y &=& \frac{25}3\\ \Rightarrow 15-y &=& \frac{20}3 \end{eqnarray*}

So now we know $y$ and $15-y$ in the picture.  Also, from the Pythagorean theorem we have

\begin{eqnarray} \left(\frac{20}3\right)^2+h_b^2 &=& 12^2\\ \left(\frac{25}3\right)^2+h_b^2 &=& (m+n)^2 \end{eqnarray}

Thus $(1)-(2)=\frac{400}9 - \frac{625}9 = 144-(m+n)^2 \longrightarrow \boxed{m+n=13.}$[/hide]

[hide="number12byme"]Knowing tangents from each point are equal, let:

$a+d=18$
$d+c=21$
$b+c=17$
$b+a=?$

By adding the first three equations together we get $2(d+c)+a+b=56 \Longrightarrow \boxed{a+b=14}$[/hide]

[hide="number4byme"]
Knowing tangents from each point are equal, we have $AB=50, BC=27+x, AC=23+x$, knowing this, we can use $rs=A=\sqrt{s(s-a)(s-b)(s-c)} \Longrightarrow 21(50+x)=\sqrt{50+x(x)(23)(27)} \Longrightarrow 21^2(50+x)=621x \Longrightarrow 22050=180x \Longrightarrow x=122.5$. From here, we see that the perimeter of the triangle is $\boxed{345}$.
[/hide]

[hide="number5byme"]
Let $x=PS$, and draw in $OS$ and draw in $RQ$ such that $PR$ is the the diameter of Circle O. We see that triangle $POS$ is similar to $PRQ$ and are in a $1: 2$ side ratio, thus $\frac{x}{12-x}=\frac{1}{2} \longrightarrow x=\boxed{4}$[/hide]

[hide="number13byme"]
We can see that $\frac{x-y}{2}=30 \Longrightarrow x-y=60$
We can also see that arc $ABD=230$ and arc $BAD=250$, and so $490-x-y=360 \Longrightarrow x+y=130$, so now we have a system of equations and we see that $x+y=190 \Longrightarrow x=\boxed{95}$[/hide]

[hide="number19byme"]Using a hint that we need to use Ptolemy's Theorem to solve this problem, I first labeled the points of the pentagon $ABCDE$, where $DC=6$ and everything else is $16$. From here I decided to split the pentagon into triangles and realized that there were only two types of triangles the pentagon can be split up into (that we know atleast two sides of), triangles with two sides of $16$ and triangles with sides of $6$ and $16$, so therefore $AD=EB=AC$ and $CE=DB$. So then I called each of the first three segments $x$ and each of the second two segments $y$ and then I decided to split the pentagon into quadrilaterals so I can use Ptolemy's Theorem. First I had quadrilateral $ABDE$ and using Ptolemy's I had: $16^2+16y=x^2$ and then I then made the other possible quadrilateral which was $EBCD$ and used Ptolemy's to get $6x+16^2=y^2$ and the last possible quadrilateral $ABCD$ and got $yx=16x+96$
Okay so after simplifying for $x$ and getting the equation $x^3-512x-1536$, I used rational root theorem and using trial and error I found that only $x=24$ worked for this equation and so $y=20$ and so after using Heron's formula on the three triangles and adding their areas up, we have the area of the Pentagon is $\boxed{123\sqrt{7}}$[/hide]

[hide="number6byme"]
Let $Z$ be the midpoint of $CB$ and so $ZA$ is the altitude of the triangle. So then $\triangle{AZD}$ is a right triangle with leg of $4$ and hypotenuse of $16$ so the altitude of the triangle is $\sqrt{240}$, so then $\triangle {AZC}$ is also a right triangle with leg of $x+4$ and $\sqrt{240}$ and hypotenuse of $17$ so $x=3$ and $CB=2x+8=\boxed{14}$[/hide]

[hide="number17byme"]

I drew in the chord who's length is $1$ and made a right triangle (the radius of the circle as the hypotenuse, $\frac{1}{2}$ and $1-x$ as the leg, and got $1-x=\frac{\sqrt{3}}{2}$ and then using trig, I got $tanx=\frac{\sqrt{3}}{3} \Longrightarrow x=30$ degrees and so the number of degrees before we hit the chord of length $1$ is $60$. So therefore if $k$ refers to the number of terms right after hitting $60$ degrees, we have $\boxed{11}$, since $\frac{10(10+1)}{2}=55$ degrees$< 60$ degrees. [/hide]

[hide="number18byme"]
This is probably the easiest out of all the problems, no real trick involved, just do a little bit of angle chasing. We know that this is a regular heptagon and so each of the interior angles are $\frac{900}{7}$. Thus $\angle{EDC}=\frac{900}{7}$ and so $\angle{DEC}=\frac{180}{7}$ and so $\angle{CEF}=\frac{720}{7}$. We know that quadrilateral $FECB$ is cyclic and so $\angle{FBC}=\frac{540}{7}$ and so $\angle{PBA}=\frac{360}{7}$ and we know that $\angle{BAP}=\frac{180}{7}$ and so $\boxed{\angle{APB}=\frac{720}{7}}$[/hide]
[hide="number2byJoe"]
First, we use the fact that the altitude of a right right triangle is the geometric mean of the two segments that it divides the hypotenuse into: \[ \frac p{\frac 14(p+q)}=\frac{\frac 14(p+q)}q\Rightarrow p^2-14pq+q^2=0.  \]  Now, divide through by $q^2$: \[ \left(\frac pq\right)^2-14\left(\frac pq\right)+1=0.  \]  Using the quadratic formula, \[ \frac pq=\frac{14\pm\sqrt{196-4(1)(1)}}2=7\pm4\sqrt{3}.  \]  Since the problem tells you it is in the form $a-b\sqrt{3}$ you could be done right now.  But assume you didn't know that.  How would you decide between $7+4\sqrt{3}$ and $7-\sqrt{3}.$  Simple.  We were given that $p<q\Rightarrow \frac pq<1$ ($p$, $q$ must be positive).  If we look at $7\pm4\sqrt{3}$ as $\sqrt{49}\pm\sqrt{48}$ it makes it much easier to see that we must take the negative sign.  Thus, our answer is $7-4\sqrt{3}.$  The problem asked for the ordered pair, $(a,b)$, so $(7,4)$ is the answer to the problem.[/hide]

[hide="number3byJoe"]

First, it is [i]really[/i] helpful to know the angle bisector theorem for this problem...so [url=http://www.cut-the-knot.org/Curriculum/Geometry/AngleBisectorRatio.shtml]here[/url]'s a link.

Now, by the angle bisector theorem, $7\cdot CD=5\cdot BD$.  But, $BD+CD=BC=3.$  So we have a pair of simultaneous equations for $BD$ and $CD$.  Solving, we get $BD=\frac 74$ and $CD=\frac 54.$

We do the same thing for $AE$ and $CE$ (use angle bisector theorem, get a pair of simultaneous equations) and solve to find that $AE=\frac 72$ and $CE=\frac 32.$

So then using the angle bisector theorem on $\triangle ABE$ we get $\frac xy=\frac 7{(7/2)}=2.$

Actually, I just realized that finding $BD$ and $CD$ was not necessary...anyway, I'm tired and I already wrote it, so it's staying.   [/hide]


Determine the smallest integer $n \ge 4$ for which one can choose four different numbers $a, b, c, $ and $d$ from any $n$ distinct integers such that $a+b-c-d$ is divisible by $20$ .

[quote="Peter"]Determine the smallest integer $ n \ge 4$ for which one can choose four different numbers $ a, b, c,$ and $ d$ from any $ n$ distinct integers such that $ a \plus{} b \minus{} c \minus{} d$ is divisible by $ 20$ .[/quote]
For n = 8, we have for example {0, 20, 40, 1, 2, 4, 7, 12} which has no two pairs with the same sum mod 20.

Now suppose n > 8. If there are 7 distinct residues mod 20, then these form 21 pairs, so there must be two pairs with the same sum mod 20. So there can only be 6 or less distinct residues. Hence either there are 4 numbers a = b = c = d mod 20, or there are two pairs of numbers a = c mod 20 and b = d mod 20. In either case, we have $ a \plus{} b \equiv c \plus{} d$(mod 20)

For any $n\ge 9$, 4 nos. can be chosen from an set of $n$ number such that the assertion is true.

My proof involves, $\binom {9}{2}$ and $\binom{7}{2}$ both being greater than 20, PHP asserts the required statement.

From  $\binom {9}{2}>20$, we havethat sum two pairs of numbers have the same residue modulo 20. If not then two integers exist such that $a\equiv b(mod20)$
Now remove $(a,b)$ and apply the same logic. The conclusion follows.

Also, since  $\binom {6}{2}=15<20$, this confirmation cannot be true for all sets of 8 numbers.


limit
probability
real analysis
real analysis unsolved

[quote="galeos"]Find the limit

\[ \lim_{n\rightarrow \infty}2^{-n}\cdot\sum_{\{k: \ \ \left|k-\frac{1}{2}n\right|\leq\frac{1}{2}\sqrt{n}\}}\left(\begin{array}{c} n \\ k \end{array}\right)  \][/quote]
Central Limit Theorem.

well could you please post a full solution using central limit theorem?it is pretty clear to me if one considers $Z=\displaystyle\frac{X-\displaystyle\frac{n}{2}}{\frac{1}{2}\sqrt{n}}$ and say that for $n\rightarrow\infty$, $Z$ follows the canonical distribution, but i can't see how to use the central limit theorem directly. In particular, to use CLT one must define $S_n=X_1+\ldots+X_n$ with $X_i$ having the same expected values and the same deviations. So what are the $X_i$ in this case?

The $X_i$ are random variables taking the values $0$ and $1,$ each with probabililty $\frac12.$ In other words, coin flips.

$k$ is the number of heads in $n$ flips. The required sum is the probability of that that lying with a certain range centered at the mean.

The standard deviation of this is $\sqrt{npq}=\frac{\sqrt{n}}2.$

Hence, we are being asked for the probability of being within one standard deviation of the mean.

Thanx a lot. It was much more straightforward than i expected. :oops:


For $a,b,c$ positive reals find the minimum  value of \[ \frac{a^2+b^2}{c^2+ab}+\frac{b^2+c^2}{a^2+bc}+\frac{c^2+a^2}{b^2+ca}. \]

We have that $\sum_{cyclic}{\frac{a^{2}+b^{2}}{c^{2}+ab}}\geq{2\sum_{cyclic}{\frac{a^{2}+b^{2}}{(c^{2}+a^{2})+(c^{2}+b^{2})}}}$
Denote $x=b^{2}+c^{2}$,$y=c^{2}+a^{2}$,$z=a^{2}+b^{2}$, so we have to find the least value of $2\sum_{cyclic}{\frac{x}{y+z}}$ but it is wellknown that $2\sum_{cyclic}{\frac{x}{y+z}}\geq{3}$.
QED

Other solution:We will prove that minimum is 3(take them equal) rewrite it as $\displaystyle\sum\limits_{\text{cyc}}\frac{(a^2+b^2)^2}{a^2c^2+b^2c^2+a^3b+ab^3}\ge3$, but aplyng C.S. we get $a^4+2b^4+2c^4+a^2b^2=a^2c^2+b^2c^2\ge3(\displaystyle\sum\limits_{\text{cyc}}a^2b+ab^2)$ which is true by AM-GM($a^4+a^2b^2\ge2a^3b$) and from the Muirhead inequality in the form $2\displaystyle\sum\limits_{\text{cyc}}a^4\ge \displaystyle\sum\limits_{\text{cyc}}(a^3b+ab^3)$

[quote=Raja Oktovin]We only need to prove that $ (a^2\plus{}b^2)(b^2\plus{}c^2)(c^2\plus{}a^2) \ge (a^2\plus{}bc)(b^2\plus{}ca)(c^2\plus{}ab)$ which follows from CS $ (a^2\plus{}b^2)(a^2\plus{}c^2) \ge (a^2\plus{}bc)^2$ and so on..[/quote]
$\frac{a^2+b^2}{c^2+ab}+\frac{b^2+c^2}{a^2+bc}+\frac{c^2+a^2}{b^2+ca}\geq3\sqrt[3]{\frac{(a^2\plus{}b^2)(b^2\plus{}c^2)(c^2\plus{}a^2) }{ (a^2\plus{}bc)(b^2\plus{}ca)(c^2\plus{}ab)}}\geq3.$
The following inequality is also true.
Let $a,b,c$ be  positive reals . Prove that$$ \frac{a^2+b^2}{c^2+ab}+\frac{b^2+c^2}{a^2+bc}+\frac{c^2+a^2}{b^2+ca}\geq3+\frac{(b-c)^2+(c-a)^2+(a-b)^2}{(a+b+c)^2}. $$

[quote=sqing]
The following inequality is also true.
Let $a,b,c$ be  positive reals . Prove that$$ \frac{a^2+b^2}{c^2+ab}+\frac{b^2+c^2}{a^2+bc}+\frac{c^2+a^2}{b^2+ca}\geq3+\frac{(b-c)^2+(c-a)^2+(a-b)^2}{(a+b+c)^2}. $$[/quote]
We have also
$$ \frac{a^2+b^2}{c^2+ab}+\frac{b^2+c^2}{a^2+bc}+\frac{c^2+a^2}{b^2+ca}\geq3+\frac{4(a^2+b^2+c^2-ab-ac-bc)}{(a+b+c)^2}. $$

For the collection.
Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that:
$$ \frac{a^2+b^2}{c^2+ab}+\frac{b^2+c^2}{a^2+bc}+\frac{c^2+a^2}{b^2+ca}\geq3+\frac{a^2+b^2+c^2-ab-ac-bc}{ab+ac+bc} $$

[quote=MatBoy-123]Can we  normalize the following expression by taking $ a+b+c=1$ and if we can't then why?[/quote]

Anyone?

One-Liner
\[\sum \frac{a^2+b^2}{c^2+ab}\geq\sum\frac{a^2+b^2}{c^2+\frac{a^2+b^2}{2}}=\sum\frac{2(a^2+b^2)}{(a^2+c^2)+(b^2+c^2)}\geq3\]
The last inequality is due to Nesbitt.


[quote="cmad"]Αξίζει να γνωρίζει κάποιος το "τρικ" στη πιο κάτω άσκηση...

Να κατασκευαστεί μετρήσιμο σύνολο $ A \subset \mathbb{R}$ τέτοιο ώστε $ 0 < \mu(A \cap I) < \mu(I)$ για κάθε (φραγμένο) ανοιχτό διάστημα $ I \subset \mathbb{R}$, όπου $ \mu$ το μέτρο Lebesgue.[/quote]

Τέλειο πρόβλημα cmad :clap2: Hehe, νοσταλγία... θυμάμαι όταν πρωτομάθαινα θεωρία μέτρου μου είχε φάει μπόλικη ώρα μέχρι να το σπάσω.

Πραγματικά guys, αν δεν το έχετε ξαναδεί αξίζει να το δοκιμάσετε! :) Cheerio,

Durandal 1707

Ίσως χρειάζεται ένα hint πάντως...

[hide]Τι θα γινόταν αν στο σύνολο Cantor δεν αφαιρούσατε κάθε φορά το μεσαίο τρίτο, αλλά πρώτα το μεσαίο ήμισυ, μετά τα μεσαία τέταρτα, μετά τα μεσαία όγδοα κ.ο.κ.?[/hide]

Cheerio,

Durandal 1707

Oh wow, έχουμε πολύ διαφορετικές προσεγγίσεις. :) Αν (δεν?) υπάρξει ενδιαφέρον θα πω και τη δική μου.

Now you really piqued my interest :)

Μου ήρθε και άλλη μία ιδέα αλλά παρόμοια με αυτήν που έβαλα στο hint... τώρα έχω πραγματικά περιέργεια ;)

Cheerio,

Durandal 1707

Να βάλω την δική μου λύση και περιμένω τις δικές σας αν είναι διαφορετικές.

Αρκεί να βρω μετρήσιμο Α ώστε $ 0 < \mu(A \cap I) < \mu(I)$ για κάθε διάστημα Ι με ρητά κέντρο και ρητή ακτίνα. Ας αριθμήσω αυτά τα διαστήματα $ I_1,I_2,\ldots$.

Απαριθμώ τους ρητούς στο $ I_1$, και για κάθε $ q_n$ παίρνω ένα διάστημα $ R_{q_n}$ με κέντρο το $ q_n$ και ακτίνα $ diam(I_1)/4^n$. Θέτω $ A_1 \equal{} I_1 \setminus (\cup R_{q_n})$.

Για κάθε διάστημα $ I_n$ έχουμε  $ 0 \leqslant \mu(A_1 \cap I_n) < \mu(I_n)$. Επίσης έχουμε  $ 0 < \mu(A_1 \cap I_1) < \mu(I_1)$.

Ας είναι ν ελάχιστο ώστε $ \mu(A_1 \cap I_n) \equal{} 0$. Απαριθμώ τους ρητούς στο $ I_n$, και για κάθε $ q_m$ παίρνω ένα διάστημα $ R_{q_m}$ με κέντρο το $ q_m$ και ακτίνα $ diam(I_n)/4^m$. Θέτω $ A_n \equal{} A_1 \cup I_n \setminus (I_n \cap \bigcup R_{q_m})$.

Τώρα έχουμε $ 0 \leqslant \mu(A_1 \cap I_m) < \mu(I_m)$ για κάθε m και  $ 0 < \mu(A_1 \cap I_i) < \mu(I_1)$ για κάθε $ 1 \leqslant i \leqslant n$ κ.ο.κ.

Εύκολα ελέγχουμε ότι το τελικό Α είναι Borel και άρα μετρήσιμο.

Το notation μου δεν ήταν και ότι καλύτερο αλλά ελπίζω η λύση να είναι κατανοητή (και σωστή).

[quote="Demetres"]Να βάλω την δική μου λύση και περιμένω τις δικές σας αν είναι διαφορετικές.[/quote]

Αυτή είναι πάνω κάτω η δεύτερη ιδέα που μου ήρθε μετά το post του cmad... οπότε πιστεύω ότι δουλεύει αλλά είμαι προφανώς biased ;)

[quote]Αρκεί να βρω μετρήσιμο Α ώστε $ 0 < \mu(A \cap I) < \mu(I)$ για κάθε διάστημα Ι με ρητά κέντρο και ρητή ακτίνα. Ας αριθμήσω αυτά τα διαστήματα $ I_1,I_2,\ldots$.
Απαριθμώ τους ρητούς στο $ I_1$, και για κάθε $ q_n$ παίρνω ένα διάστημα $ R_{q_n}$ με κέντρο το $ q_n$ και ακτίνα $ diam(I_1)/4^n$. Θέτω $ A_1 \equal{} I_1 \setminus (\cup R_{q_n})$.
Για κάθε διάστημα $ I_n$ έχουμε  $ 0 \leqslant \mu(A_1 \cap I_n) < \mu(I_n)$. Επίσης έχουμε  $ 0 < \mu(A_1 \cap I_1) < \mu(I_1)$.[/quote]

Λίγο nitpicking μόνο... νομίζω θέλεις $ A_1 \equal{} I_1 \setminus (I_1\cap\bigcup R_{q_n})$ γιατί τα $ R_{q_n}$ μπορεί να βγαίνουν έξω από το $ I_1$ και άρα η τομή τους με κάποιο $ I_n$ ίσως είναι ολόκληρο το $ I_n$.

Η δική μου λύση τώρα... Για κάθε διάστημα $ I \equal{} [a,b]$ ας ορίσουμε το μεσαίο $ r$-τμήμα του να είναι το διάστημα $ \left[\frac {a \plus{} b}{2} \minus{} \frac {b \minus{} a}{2r}, \frac {a \plus{} b}{2} \plus{} \frac {b \minus{} a}{2r}\right]$ όπου $ r > 1$. Με βάση το $ [a,b]$, θα φτιάξουμε ένα σύνολο τύπου Cantor με θετικό μέτρο ακολουθώντας την εξής συνταγή:
1. Πάρτε το αρχικό διάστημα και αφαιρέστε το μεσαίο $ r_1$-τμήμα του.
2. Από τα δύο διαστήματα που απομένουν, αφαιρέστε από καθένα το μεσαίο $ r_2$-τμήμα.
3. Proceed ad infinitum.

Π.χ. αν $ r_n \equal{} 3$,  παίρνουμε το συνηθισμένο σύνολο Cantor με μέτρο $ 0$. Αν όμως η ακολουθία $ r_n$ επιλεχθεί κατάλληλα και μηδενίζεται αρκετά γρήγορα, μπορούμε να κατασκευάσουμε ένα σύνολο Cantor $ C_I$ ώστε $ \mu(C_I) \equal{} \alpha \mu(I)$ για οποιοδήποτε $ \alpha < 1$.

Αυτό είναι το βασικό συστατικό, απλώς έχει το πρόβλημα ότι δεν είναι πουθενά πυκνό. Eh, για να το κάνετε παντού πυκνό, είτε γεμίστε κάθε τρύπα του συνόλου με τον εαυτό του (και σε κάθε τρύπα που προκύπτει βάλτε άλλο ένα αντίγραφό του) και μετά κολλήστε αντίγραφα μεταξύ τους για να πάρετε το $ \mathbb{R}$, είτε απλώς αντικαταστήστε όλα τα ρητά διαστήματα με σύνολα τέτοιου τύπου που το μέτρο τους να είναι ένα rapidly vanishing κλάσμα του ρητού διαστήματος που αντικαθιστά ($ 1/4^n$ works fine ;) ), είτε...

Hope this works. Cheerio,

Durandal 1707

Dem, ωραία ιδέα... αν σκεφτώ κάτι θα σου πω!

BTW, νομίζω ότι το πρώτο σου argument μπορεί να απλοποιηθεί λίγο (τουλάχιστον αυτή ήταν η original σκέψη μου, αλλά ίσως μπάζει κάπου): μόλις φτιάξεις το $ A_1$, δε χρειάζεται να μπεις στην επαγωγική διαδικασία. Μου φαίνεται ότι αρκεί να πάρεις το $ A_1$ και να κολλήσεις αντίτυπά του δεξιά και αριστερά για να γεμίσεις το $ \mathbb{R}$.

Cheerio,

Durandal 1707

[quote="Demetres"]Μια άλλη ιδέα την οποία όμως δεν κατάφερα να την κάνω να δουλέψει:

Για κάθε $ x \in \mathbb{R}$, βάζω το χ στο Α με πιθανότητα 1/2. Τότε για κάθε Ι (με πιθανότητα 1) ισχύει $ \mu(A \cap I) \equal{} \mu(I)/2$. (???)

Νομίζω όμως πως το Α δεν θα είναι μετρήσιμο. Αυτό θα μπορούσε να διορθωθεί βάζοντας κάποιο μέτρο στο σύνολο όλων μετρήσιμων συνόλων, και παίρνοντας ένα "τυχαίο" μετρήσιμο σύνολο. Δεν μπορώ όμως ακόμη να δω πως να το κάνω.[/quote]

Την ίδια διαισθητική ιδέα χρησιμοποίησε ο Terence Tao για να δώσει "intuitive" παράδειγμα μη-μετρήσιμων συνόλων (αλλά δεν μπόρεσε να το κάνει rigorous, heh). ( Edit: [url]http://terrytao.wordpress.com/2008/10/14/non-measurable-sets-via-non-standard-analysis/[/url] )

---

Όσο για την κατασκευή μου, είναι very much στο ίδιο στυλ με τις τελευταίες σας ιδέες, με μια μικρή διαφορά:

Σε κάθε διάστημα $ I_q$ με ρητά άκρα, βάζω δυο (ξένα μεταξύ τους) σύνολα $ A_q, B_q$ (οποιουδήποτε) θετικού μέτρου, που είναι κλειστά και πουθενά πυκνά ( γιατί; ). Φροντίζω ώστε τα σύνολα αυτά να μην τέμνονται για διαφορετικές τιμές του $ q$. Τέλος, θέτω $ A \equal{} \cup_q A_q$ ως το σύνολό μου. Οι λεπτομέριες μετράνε εδώ, και όποιος ενδιαφέρεται μπορεί να δοκιμάσει να κάνει reconstruct τη λύση.

Cheers, ωραίες οι λύσεις που προτείνατε Durandal + Dem!

[quote="cmad"]
Την ίδια διαισθητική ιδέα χρησιμοποίησε ο Terence Tao για να δώσει "intuitive" παράδειγμα μη-μετρήσιμων συνόλων (αλλά δεν μπόρεσε να το κάνει rigorous, heh). ( Edit: [url]http://terrytao.wordpress.com/2008/10/14/non-measurable-sets-via-non-standard-analysis/[/url] )
[/quote]

Χε, χε. Τώρα δεν νιώθω τόσο άσχημα. Φυσικά η μέθοδος που σκεφτόμουν δεν δουλεύει αφού δεν μπορεί το Α να είναι και μετρήσιμο και να έχει $ \mu(A \cap I) \equal{} \mu(I)/2$ για κάθε Ι. (Από Lebesque differentiation theorem.)


[hide="Proof"]Let $X=a_{0}+a_{3}+a_{6}+\dots, Y=a_{1}+a_{4}+a_{7}+\dots, Z=a_{2}+a_{5}+a_{8}+\dots$. Plugging $x=-{1\over 2}+i{\sqrt{3}\over 2}$ into the equation we get

$0=X-{Y\over 2}+i{Y\sqrt{3}\over 2}-{Z\over 2}-i{Z\sqrt{3}\over 2}$, which gives

$X-{Y\over 2}-{Z\over 2}=0\land Y-Z=0\iff X=Y=Z$

Plugging $x=1$ into the equation we get

$3^{n}=X+Y+Z\iff X=Y=Z=3^{n-1}$[/hide]


Is there any Elementary( or easy) proof for Roche for +polynomial, unit circle with center is origin+?

(especialy zetax)

Even if I'm not ZetaX (should be clear since I'm neither darij nor Megus) here is the answer: it was posted before :mrgreen:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=complex&t=150#[/url]
EDIT: Actually it's not a 100 per cent what you've asked for, but maybe it's helpful


Solve in nonnegative integers $ x_1,x_2,...,x_6$:

$ x_kx_{k\plus{}1}(1\minus{}x_{k\plus{}2})\equal{}x_{k\plus{}3}x_{k\plus{}4}$, $ k\equal{}1,2,...,6$,

where $ x_{k\plus{}6}\equal{}x_k$.

Multiply both sides by $ (1\minus{}x_{k\plus{}5})$, and then you can simplify down to $ (1\minus{}x_k)(1\minus{}x_{k\plus{}3}) \equal{} 1$. Then we require both factors are equal and either $ 1$ or $ \minus{}1$. Then $ x_k \equal{} x_{k\plus{}3}$ is $ 0$ or $ 2$, giving $ 8$ possibilities. We can check each one and see that only $ x_1 \equal{} x_2 \equal{} \ldots \equal{} x_6 \equal{} 2$ is not a solution. So there are $ 7$ such 6-tuples: if $ a,b,c \in \{0,2\}$ with $ (a,b,c) \neq (2,2,2)$, then $ (a,b,c,a,b,c)$ is a solution.

I think a marginally nicer way to finish is just to plug in $ x_k\equal{}x_{k\plus{}3}, x_{k\plus{}1}\equal{}x_{k\plus{}4}$ into the original equation which directly yields $ x_{k}x_{k\plus{}1}x_{k\plus{}2}\equal{}0$. Therefore, as long as $ x_{k}\equal{}x_{k\plus{}3}\in\{0,2\}$ and at least one of every three consecutive is zero, we have solution, which is equivalent to your solution set (just no checking is involved).


inequalities
logarithms
function
inequalities proposed

Let $ k \in R$ and $ x,y,z \in ( |k|,\infty)$ such that $ x \leq y \leq z$.Prove the folowing inequality:
\[ \frac{x^{z}y^{x}z^{y}}{x^{y}y^{z}z^{x}}\leq \frac{(x+k)^{z}(y+k)^{x}(z+k)^{y}}{(x+k)^{y}(y+k)^{z}(z+k)^{x}}\]

Anyone ? Is it that boring  :lol: ? It can't be that hard.

I'll give a hint by adding a problem witch is solved the same way as the first one.
If $ x,y,z\in R_{+}$ such that $ x \leq y \leq z$ then we have the inequality
\[ x^{z}y^{x}z^{y}\leq x^{y}y^{z}z^{x}\]

What is so obvious about this inequality?
We have $ \frac{x}{y}\leq 1$ , $ \frac{y}{z}\leq 1$ and $ \frac{z}{x}\geq 1$.
Explain more your solution :D , because the problem is quite easy.
Ps: it's $ \frac{x}{x\plus{}k}$,not $ \frac{x}{x\plus{}y}$.

God, this problem is to easy :P.First the hint:
Aplying $ \ln$ in the inequality
\[ \frac{x^{z}}{y^{z}}\frac{y^{x}}{z^{x}}\frac{z^{y}}{x^{y}}\leq 1\]
we have to prove that for $ x,y,z$ pozitive with $ x\leq y\leq z$ we have:
\[ x(\ln y\minus{}\ln z)\plus{}y(\ln z\minus{}\ln x)\plus{}z(\ln x\minus{}\ln y)\leq 0\]
wich is evidently true for the concave function $ f(x) \equal{}\ln x$
(See: http://www.mathlinks.ro/Forum/viewtopic.php?p=897342#897342, the solution is the same for concave )
For the other problem we can do the same thing for the concave function $ f(x) \equal{}\ln\frac{x}{x\plus{}k}$, because $ x,y,z\geq |k|$.


Prove that in every triangle with sides a,b,c the following inequality holds:

\[ \sqrt{3}(a^2+b^2+c^2)\geq a(m_c+m_b)+b(m_a+m_c)+c(m_a+m_b) \]

notations are the known!

 :)

Apply cauchy schwarz for 4 times should work.

could you explain in detail please?
Thanks for your help :)


Student solutions for Round 2 are now posted on the USAMTS web site at [url]http://www.usamts.org[/url]

[quote="ThAzN1"]Just a small typo: the person who wrote the first solution for problem 2/2/16 is actually from IL (and not TX) ;)[/quote]
Oops -- you're right. :roll:  It has been fixed.


[quote="math92"]If x=3, what is the value of 2x+3[/quot THE ANSWER IS 9


absolute value
calculus
calculus computations

$ \sum^{\infty}_{n\equal{}0} \frac{(\minus{}1)^n4^{\frac{1\minus{}2n}{2}}\prod^n_{k\equal{}0}(2k\minus{}1)}{2^nn!}$

Converges?
[]yes
[]no
[]lazy

[hide="answer"]It converges to $ \minus{} 2 e ^{ \minus{} 1/8}$ (or, more probably, $ \frac { \minus{} 4}{\sqrt5}$, I'm not sure right now, I'm too tired :) ).

Don't have time to write the proof... (Hope someone will do it for me)

As far as the fact of convergence is concerned, I believe plain old Leibniz will do the trick... Just compare the m-th, and m+1-st member of the sum (their absolute value) and you'll notice that they monotonically decline to 0 (form the plane comparison you get the monotonical part, and the part of declining to zero can be given with comparison criterion, or by contradiction). And taking into account the alternating nature of the series, the result follows: the sum converges.

[/hide]

I agree with hsiljak.
This sum is like $ C\sum \frac {(x)^{n}}{(2n)!!}$ and it's like $ \cosh(x)$ or we can find closed form.


Let $A,B,C$ be angles of triangle $ABC$. Prove that: $\sin\sqrt[3]{AB^{2}}+\sin\sqrt[3]{BC^{2}}+\sin\sqrt[3]{CA^{2}}\leq\frac{3\sqrt{3}}{2}$


If $q$ is a prime number, how many matrices does exist such that its elements are numbers from $\{0,1,\dots,q-1\}$ and that its determinant is not divisible by $q$?

Or in other words: find the number of elements of $GL_{n}(\mathbb F_{q})$.
Or still in other words: the number matrices with lineary independent rows (the rows consisting of $n$-tupels $\mathbb F_{q}^{n}$).

Easy counting gives $(q^{n}-1)(q^{n}-q)(q^{n}-q^{2})...(q^{n}-q^{n-1}) = \prod_{i=0}^{n-1}q^{n}-q^{i}$.

There is no "easy counting"; there is counting.

The expression "q^n - q^i" is together when being expanded out in the multiplication, so it needs to have grouping symbols around it, such as parentheses.

Yes big modetator and big mathmatician box of rocks.When you say something it is a law  :o