[quote="cdymdcool"]I am about  to start a math club this upcoming year, but i don't know what kind of competitions are available for high school kids in MA?  Would you please give me some suggestion and some comments about each individual competition?[/quote]
What school do you go to?  There are local competitions but it depends where you're located.

I go to Pioneer Valley Performing Arts School, we are pretty close to Springfield, MA.

In that case your local league is the Western Massachusetts Math League.  State meet qualifiers from WMML last year were Amherst, Minnechaug, Deerfield, Longmeadow, and the MacDuffie School.  I don't think WMML has a website that I can point you to, but a teacher at your school could contact one of those schools for more information.

Beyond that league, there's the annual Massachusetts Math Olympiad (which is free to your school in the first year, for up to ten students).  And of course there are plenty of regional and national competitions -- the AMC10/12 and NEML are probably good places to start.

Good luck.


$ AC$ is chord of circle $ \omega$ $ B$ is point on $ \omega$ $ D$ is point on $ AC$. $ O _ 1$ is center of circle who touches $ \omega$ and [b]segments[/b] $ AD, BD$. $ O _ 2$ is center of circle who touches $ \omega$ and [b]segments[/b] $ CD, BD$. Prove that $ O _ 1$ $ O _ 2$and Incenter of $ ABC$ are collinear.

The problem is just Victor's Thebault Theorem


imagine an event participated by 8 countries (olympics, IMO...), prove you can always find 5 countries so that their total number of gold, silver and bronze medals are all no fewer than the total of the rest three countries.  give an example to prove that 5 cannot be reduced.
if you can, please generate the result to n coutries.
thanks!

[quote="pcasnik"]imagine an event participated by 8 countries (olympics, IMO...), prove you can always find 5 countries so that their total number of gold, silver and bronze medals are all greater than the total of the rest three countries. [/quote]

Probably "are all greater than" should be replaced by "are all at least as large as"?

Otherwise the following situation would be a counterexample:
Countries 1 and 2 each make one gold medal.
Countries 3 and 4 each make one silver.
Countries 5 and 6 each make one bronze.
Countries 7 and 8 get nothing.

yes, you are right.  thanks.
problem corrected


Let $m,n \in \mathbb{N}$ and $n\geq 2$. Prove that $\sqrt[n]{{m^{n}+1}}$ is an irrational number.

If $\sqrt[n]{{m^n + 1}}=\frac ab, (a,b)=1$ then $b^n(m^n+1)=a^n\Longrightarrow b=1,m^n+1=a^n$ had no solution, because a integer and $m<a<m+1$.

Because every prime divisor of $b$ would divide $a$, contradicting coprimality.

$\sqrt[n]{x}(x\geq 0)$is a rational number if and only if $x=k^n$ where k is an integer

My Friends, Thanks

I prove this by Fermat Last Theorem, now.
For $n=2$, it easy to see that $\sqrt{m^2+1}$ is an irrational number.
Now let $n\geq 3.$ If
\[ \root n \of {m^n + 1} = \frac{a} {b}{\text{ }}(a,b \in {\Bbb N}) \Rightarrow (bm)^n + b^n = a^n.  \]
This is impossible by Fermat Last Theorem.

Because every prime divisor of $b$  would divide $a$  , contradicting coprimality.

That clearly doesn't imply that $b=1$.


find all m,n in positive integer such that:
            \[2*m\equiv -1 (mod n)\]
            \[n^{2}\equiv -2 (mod m)\]
 :D  :D  :D  :D

We have$gcd(m,n)=1$

Hence $n^2+4m+2=kmn$,$k\in\mathbb N$,$m,n$ are odd then $k$ is odd.

$n^2+2=m(kn-4)$ so $kn-4|(kn)^2+2k^2$ thus $kn-4|k^2+8$ (note that $kn$ is odd).

 So the cases $k\in\{1,3,5,7,9,11\}$ are not complicated to solve.

 Now $k\geq 13$.Note that :$(2n-km)^2=(km)^2-16m-8$

   and $(km-1)^2<(km)^2-16m-8<(km)^2$,contradiction.

[quote="nalpaction"]We have$gcd(m,n)=1$

Hence $n^2+4m+2=kmn$,$k\in\mathbb N$,$m,n$ are odd then $k$ is odd.

$n^2+2=m(kn-$[b][size=200]2[/size][/b]$)$ so $kn-2|(kn)^2+2k^2$ thus $kn-2|k^2+$[b][size=200]2[/size][/b] (note that $kn$ is odd).

 So the cases $k\in\{1,3,5,7,9,11\}$ are not complicated to solve.

 Now $k\geq 13$.Note that :$(2n-km)^2=(km)^2-16m-8$

   and $(km-1)^2<(km)^2-16m-8<(km)^2$,contradiction.[/quote]
U made same mistakes. Look the BOLDs.

Yes,you are right RobertuX  :blush: .Thank you.
Ofcourse that mistake doesn't effect too much to my solution.
 I've edited my solution.


Let $ x, y, z$ be real numbers in the interval $ (0, 1)$ such that $ xyz \equal{} (1 \minus{}x)(1 \minus{} y)(1 \minus{} z)$. Prove that
$ x^2\plus{}y^2\plus{}z^2\geq \frac{3}{4}$.

if we substitute $ x\equal{}\frac{a^2}{a^2\plus{}bc}$ etc.

we need to prove $ \sum_{cyc} \frac{a^4}{(a^2\plus{}bc)^2}\geq \frac34$
by Cauchy $ \sum_{cyc} \frac{a^4}{(a^2\plus{}bc)^2}\geq \frac{(a^2\plus{}b^2\plus{}c^2)^2}{(a^4\plus{}b^4\plus{}c^4)\plus{}2abc(a\plus{}b\plus{}c)\plus{}(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)}\geq \frac34$

Here's my solution:
Put $ a=\frac{1-x}x, b=\frac{1-y}y, c=\frac{1-z}z$, so we get $ a,b,c >0, x=\frac1{a+1}, y=\frac1{b+1}, z=\frac1{c+1}$ and $ a.b.c=1$
We have $ P= x^2+y^2+z^2 =\frac1{(1+a)^2}+\frac1{(1+b)^2}+\frac1{(1+c)^2}$
WLOG, we can assume $ c=min\{a,b,c\}$, therefore $ ab \geq 1$
Lemma: If $ x,y$ be positive real numbers such that $ xy \geq 1$ then $ \frac1{x^2+1}+\frac1{y^2+1} \geq \frac2{1+xy}$ (it's equivalent to
$ \frac{(x-y)^2.(xy-1)}{(1+xy)(1+x^2)(1+y^2)} \geq 0$
So, $ \frac1{(1+a)^2}+\frac1{(1+b)^2} \geq \frac12(\frac1{a^2+1}+\frac1{b^2+1}) \geq \frac1{1+ab}=\frac{c}{c+1}$
${ \rightarrow P \geq \frac{c}{c+1}+\frac1{(c+1)^2 }\geq \frac34}$ which is obvious

$ xyz \equal{} (1 \minus{} x)(1 \minus{} y)(1 \minus{} z)\rightarrow 2xyz \plus{} x \plus{} y \plus{} z \equal{} xy \plus{} yz \plus{} zx \plus{} 1$

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

Then, it remains to prove that :

$ x \plus{} y \plus{} z \minus{} 4xyz \ge 1$
$ x \plus{} y \plus{} z \minus{} 4xyz\ge 2xyz \plus{} x \plus{} y \plus{} z \minus{} xy \minus{} yz \minus{} zx$
Or :
$ xy \plus{} yz \plus{} zx \ge 6xyz$
$ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}\ge 6$

Because of the condition given, we set :
$ x \equal{} \frac {a}{a \plus{} b}, y \equal{} \frac {b}{b \plus{} c}, z \equal{} \frac {c}{c \plus{} a}$

Then it remains to prove that :
$ \frac {b}{a} \plus{} \frac {c}{b} \plus{} \frac {a}{c} \ge 3$

Which is true by AM-GM


how do you get the equation set to y= to graph it when you are given an equation such as ((x^2)/(25)) - ((y^2)/(16)) = 1?

I do believe that is an oval, not a circle...

It is not an ellipse, it is a hyperbola.

woops, indeed i hadnt noticed the minus sign...

The only reason you should need to solve for $ y$ is if you want to graph it using a graphing calc...

[quote="mihail911"][quote="linzk3"]ok sorry i figured it out.[/quote]
wait...i have a TI-89 and i don't know how to graph such non-function graphs. how do u do it?[/quote]
Well first you isolate $ y$. You will get $ y\equal{}\pm\frac{4}{5}\sqrt{x^{2}\minus{}25}$. This plus-minus sine poses a problem, so just graph $ y\equal{}\frac{4}{5}\sqrt{x^{2}\minus{}25}$ and $ y\equal{}\minus{}\frac{4}{5}\sqrt{x^{2}\minus{}25}$ at the same time to display the hyperbola.


geometry
rectangle
combinatorics unsolved
combinatorics

I was looking for all the rectangles that can be tiled with equal tetrominoes.
It turns out that they're all almost trivial to find (usually looking for the proper coloring), except for the T-tetrominoes. Looking on the net I found out that just the rectangles of the form 4m*4n can be tiled, but I couldn't find any proof whatsoever, just a reference to a single article by Walkup.
And I also found a cryptic paper that uses an advanced approach, calculating the tile homology group of the T-tetromino, just to say that homology groups happen to be useless here, since Walkup's proof relied on subtle geometric restrictions.   :lol:

Any ideas for a simple proof?

The question has been posted before, and I remember Pierre saying that he saw the proof (probably the one you're referring to, by Walkup), that it's more than $2$ pages long, and that he wouldn't post it on the forum (which I completely understand, if it's that long :)).

Right.

Walkup's solution can be found in the American Math. Monthly, nov. 1965, p. 986-988.

Pierre.


function trigonometry calculus integration

A body of mass m, placed on a rough surface, is attached to a spring. The body is moved a distance x towards spring and then released. The block moved to a distance b beyond the point of equillibrium. What is the change in amplitude. Friction constant is mu and spring constant is K.

First of all, what are the forces applied to the body?

Solution

Let the displacement of the mass from the equillibrium be $ s$ where the initial displacement is negative. The forces acting horizontally are those due friction, and the tension in the string. The frictional force, $ F_{f}$, always opposes the direction of motion (or potential motion), so therefore acts in the negative direction over this first half-oscillation. The friction is given by:

$ F_{f}\equal{}\minus{}\mu N$ where $ N$ is the normal reactionfrom the surface. The mass is in vertical equilibrium, so $ N \equal{} W \equal{} mg$.

$ \Rightarrow F_{f}\equal{}\minus{}\mu mg$

By Hooke's law, the tension in the spring is directly proportional to and acts in the opposite direction to the displacement.

$ \Rightarrow F_{t}\equal{}\minus{}Ks$

Hence at any displacement, the net force is $ F \equal{} F_{f}\plus{}F_{t}\equal{}\minus{}\mu mg\minus{}Ks$.

An obvious requirement is for the original net force to be positive, otherwise the mass will remain stationary and the friction will balance the tension in the string. The original displacement is $ \minus{}x$ therefore $ \minus{}\mu mg\plus{}Kx > 0\Rightarrow\frac{\mu mg}{K}\minus{}x < 0$. The significance of this will become apparent later.

By Newton's second law of motion: $ m\frac{d^{\,2}s}{dt^{2}}\equal{}\minus{}\mu mg\minus{}Ks$

$ \Rightarrow\frac{d^{\,2}s}{dt^{2}}\equal{}\minus{}\mu g\minus{}\frac{K}{m}s$

$ \Rightarrow\frac{d^{\,2}s}{dt^{2}}\plus{}\frac{K}{m}s \equal{}\minus{}\mu g$

The Auxilliary Equation is: $ n^{2}\plus{}\frac{K}{m}\equal{} 0$

$ \Rightarrow n \equal{}\pm i\sqrt{\frac{K}{m}}$

Hence the Complementary Function is: $ s_{c}\equal{} e^{0t}\left(A\cos{\left(\sqrt{\frac{K}{m}}\,t\right)}\plus{}B\sin{\left(\sqrt{\frac{K}{m}}\,t\right)}\right) \equal{} A\cos{\left(\sqrt{\frac{K}{m}}\,t\right)}\plus{}B\sin{\left(\sqrt{\frac{K}{m}}\,t\right)}$

The Particular Integral takes the form $ s_{p}\equal{} C$

Substituting into the differential equation gives:

$ \frac{K}{m}C \equal{}\minus{}\mu g$ $ \Rightarrow C \equal{}\minus{}\frac{\mu mg}{K}$

General Solution is: $ s \equal{} s_{c}\plus{}s_{p}\equal{} A\cos{\left(\sqrt{\frac{K}{m}}\,t\right)}\plus{}B\sin{\left(\sqrt{\frac{K}{m}}\,t\right)}\minus{}\frac{\mu mg}{K}$

When $ t \equal{} 0$, $ s \equal{}\minus{}x$

$ \Rightarrow\minus{}x \equal{} A\minus{}\frac{\mu mg}{K}$

$ \Rightarrow A \equal{}\frac{\mu mg}{K}\minus{}x$

When $ t \equal{} 0$, $ \frac{ds}{dt}\equal{} 0$

$ \frac{ds}{dt}\equal{}\minus{}A\sqrt{\frac{K}{m}}\sin{\left(\sqrt{\frac{K}{m}}\,t\right)}\plus{}B\sqrt{\frac{K}{m}}\cos{\left(\sqrt{\frac{K}{m}}\,t\right)}$

$ \Rightarrow B \equal{} 0$

Hence the particular solution is $ s \equal{}\left(\frac{\mu mg}{K}\minus{}x\right)\cos{\left(\sqrt{\frac{K}{m}}\,t\right)}\minus{}\frac{\mu mg}{K}$

At $ s \equal{} b$ is a maximum, and this occurs when $ \cos{\left(\sqrt{\frac{K}{m}}\,t\right)}\equal{}\minus{}1$

$ \Rightarrow b \equal{} x\minus{}\frac{\mu mg}{K}\minus{}\frac{\mu mg}{K}$

$ \Rightarrow x\minus{}b \equal{}\frac{2\mu mg}{K}$

This is the change in amplitude (Note its independence of the initial amplitude).

sludgethrower, Thanks a lot

But can it be solved without using differential equation method.

Also can you suggest good book on physics which deals these type of problem.

You can probably do it without a differential equation by using formulae which are ultimately derived from the solution of differential equations. There is a very good textbook "Mathematical Techniques" by D. W. Jordon and P. Smith. It is a textbook concerned with mathematics required for applied ,mathemetics.  There is extensive treatment of differentioal equations and a very nice chapteron Harmonic functions and the Harmonic oscillator.


function
real analysis
real analysis unsolved

Ive got this DE: 8(du/dx)+4(du/dy)=9u

I separate it with u=X(x)Y(y)

Let me give you my thoughts:

8*X' + k^2*X = 0
4*Y' - (k^2 + 9)*Y = 0

These yields:

solving the equations above for k^2 >0 , =0 or <0:

k^2 >0: yields:
X(x)= A1*e^-(( (K^2)/8)x)    and Y(y)= B1*e^-(( (k^2 +9)/4)y)

k^2 =0: yields:

X(x)=A2                              and Y(y)=B2*e^((9/4)y)

k^2 <0: yields:

X(x)= A1*e^(( (K^2)/8)x)    and Y(y)= B1*e^(( (k^2 +9)/4)y)

Then U(x,t) is X(x)Y(y) for each of the threee cases:

Then I can write (by laws of superposition) U(x,t)= U(x,t)1 +U(x,t)2 +U(x,t)3
where U(x,t)1 is U(x,t) for case k^2 >0 etc...
Then I have my boundary condition:
u(0,y) = 33*exp(-y) +50*exp(4y)

How do I solve the rest?

Hello! Separation of variables is a long and rather tedious approach to the solution, and, in any case, you are not negotiating it correctly. Therefore, I recommend an easier road, that of the method of characteristics.

\[ 9 \frac{\partial u}{\partial x} + 4 \frac{\partial u}{\partial y} = 9u, u(0, y) = 30e^{-y} + 55e^{4y}  \]

We find curves in the xy plane along which u is known and then use those curves and the known values to recover the general function u(x, y).

\[ x = x(s), \ y=y(s)  \]
\[ \frac{du}{ds} = \frac{\partial u}{\partial x} \frac{dx}{ds} + \frac{\partial u}{\partial y} \frac{dy}{ds}  \]
\[ \frac{du}{ds} = 9u, \ \ \frac{dx}{ds } = , \ x(0) = 0 \ \ \frac{dy}{ds} = 4, \ y(0) = y_0  \]
\[ u = F(y_0)e^{9s}, \ x = 8s, \ y = 4s + y_0  \]
\[ u(x, y) = F(y- \frac{1}{2} x)e^{\frac{9}{8} x}  \]
\[ u(0, y) = F(y) = 33e^{-y} + 50e^{4y}  \]

\[ u(x, y) = (33e^{\frac{1}{2} x - y} + 50e^{4y - 2x})e^{\frac{9}{8}x}  \]

How do you get
x(0)= 0
and y(0)=yo ?

And how do you get:

u=F(y0)e^9s

Thank you very much for the help!

Best Regards,

Daniel


just a question:

of course when we have a polygon $P_{1}...P_{n}$, and $P_{i}=(x_{i}, y_{i})$, we have that the directed area of the polygon is:

$[P_{1}...P_{n}]_{\pm}=\frac{1}{2}(x_{1}y_{2}-x_{2}y_{1}+...x_{n}y_{1}-x_{1}y_{n})$.

When is the quantity $[P_{1}...P_{n}]_{\pm}$ positive, and when is it negative. Is there any condition for the sign?

Thanks!

[quote="t0rajir0u"]I believe it depends on whether $P_{1}, ... P_{n}$ traverses the polygon clockwise or counterclockwise.[/quote]

thanks!

actually, that is what i had thought before, but in Kiran Kedlaya's [i]Geometry Unbound[/i], it is said that

pg. 9:

QUOTE 1: [i]for any convex quadrilateral $ABCD$, $[ABC]_{\pm}$ and $[ABD]_{\pm}$ have the same (non-zero) sign[/i]

also on pg. 9,

QUOTE 2: [i]For any simple polygon $P_{1}...P_{n}$, the directed areas $[P_{i}P_{j}P_{k}]_{\pm}$ all have the same sign, and it is the same as the sign $[P_{1}...P_{n}]_{\pm}$. [/i]

Sorry for my limited knowledge.  :blush:

Are any changes needed in the above quotes, especially in the first one?

Thanks in advance for help!

[quote="t0rajir0u"]Huh.  I'd trust Kedlaya's reasoning over my own, so I'd be inclined to agree with those statements rather than mine  :wink:[/quote]

yes...true...true...but Kedlaya put the statement as a "fact", meaning that the proof is near trivial...

I'm somehow not able to prove it though  :blush:

Could someone help me? Thanks!


function
algebra
polynomial
integration
real analysis
real analysis unsolved

Find all continuous functions $f: [0,1]\to R$  such that for any polynomial $P: [0,1]\to [0,1]$ of degree $2007$ we have $\int_{0}^{1}f(P(x))dx=0$.

By the Mean Value Theorem, for all such polynomials $P$, there is $x_{0}\in \left( 0, 1 \right)$ such that $f(P(x_{0})) = 0$.

Pick an arbitrary $x \in \left( 0, 1 \right)$ and a sequence of intervals $\left[ a_{n}, b_{n}\right]$ such that $a_{n}\to x^{-}, \, b_{n}\to x^{+}, \, b_{n}-a_{n}\to 0$.

Let $f_{n}$ be the linear polynomial passing through $(0,a_{n})$ and $(1,b_{n})$. For any $\epsilon > 0$ we can find a polynomial $g_{n, \epsilon}$ of degree $2007$ with very small coefficients such that $\left| g_{n, \epsilon}(x) \right| < \epsilon$, for all $x \in [0,1]$. Take $P = f_{n}+g_{n, \epsilon}$. By the first statement of my post, there is $x_{0}$ such that $f(P(x_{0})) = 0$. Furthermore, $P(x)$ is always between $\left( a_{n}-\epsilon, b_{n}+\epsilon \right)$, so $f$ has a root between $\left( a_{n}-\epsilon, b_{n}+\epsilon \right)$, namely $P(x_{0})$. Keeping in mind that $f$ is continuous and letting $\epsilon \to 0$, we have that $f(t_{n}) = 0$ for some $t_{n}\in \left[ a_{n}, b_{n}\right]$. Now let $n \to \infty$ and use again the continuity of $f$ to conclude that $f(x) = 0$.

Hence $f \equiv 0$. Right?

I do not understand this part:
[quote="perfect_radio"] Keeping in mind that $f$ is continuous and letting $\epsilon \to 0$, we have that $f(t_{n}) = 0$ for some $t_{n}\in \left[ a_{n}, b_{n}\right]$.
[/quote]
I can not see what g does when $\epsilon$ goes to 0 :maybe: And P depends on epsilon also...I will look more carefully :)
However, i can give a sugestion: $P(x)=(b_{n}-a_{n})X^{2007}+a_{n}$.

I previously proved that $f$ has a root in $\left[ a_{n}-\epsilon, b_{n}+\epsilon \right]$, for all $\epsilon > 0$. You see, we don't care about the fact that $P$ varies, my statement doesn't involve it. Now give $\epsilon$ the values $1, \frac12, \frac13, \ldots$. We get a sequence of roots which lie in a bounded interval, so there is a convergent subsequence. The limit of that is in $\left[ a_{n}, b_{n}\right]$. By the continuity of $f$ we get $f(\textrm{aforementioned limit}) = 0$ and that's the $t_{n}$ I was talking about.
Is it okay now?

PS: $g_{n}$ is simply $X^{2007}$ times a very small number, which depends on $\epsilon$.

PPS: The polynomial you chose leads a lot quicker to a solution :)

I think it's ok :) Thx for the explanation


Suppose $a,b,c>0$. If $x\geq\frac{a+b+c}{3\sqrt{3}}-1$ prove that

$\frac{(a+bx)^2}{c}+\frac{(b+cx)^2}{a}+\frac{(c+ax)^2}{b} \geq abc$

Have fun!!!! :lol:  :lol:  :)

This inequality is just screaming at you:

[hide]Cauchy! Cauchy! Cauchy![/hide]

Yes. I used this lovely ineq, too!

It is probably the most used ineq in problems all around the world....... :)  :lol:

[quote="K81o7"]What exactly is it again?  is Cauchy
$(a_1+a_2+...+a_n)^2 \le n(a_1^2+a_2^2+...+a_n^2)$?
Which one exactly is it?  I never can remember...but i believe that's what i picked up from the forums...[/quote]

That's a special case.  Cauchy states that

\[
(a_1 b_1 + a_2 b_2 + \cdots + a_n b_n)^2 \le (a_1^2 + a_2^2 + \cdots + a_n^2)(b_1^2 + b_2^2 + \cdots + b_n^2)
\]

errr im newer to this proof stuff so im not sure if this is right but

[hide]

$\Big( \displaystyle\frac{(a+bx)^2}{c} + \frac{(b+cx)^2}{a} + \frac{(c+ax)^2}{b} \Big)(c+a+b) \ge (a+bx+b+cx+c+ax)^2$

$\Big( \displaystyle\frac{(a+bx)^2}{c} + \frac{(b+cx)^2}{a} + \frac{(c+ax)^2}{b} \Big)(c+a+b) \ge (a+b+c)^2(x+1)^2$

$\Big( \displaystyle\frac{(a+bx)^2}{c} + \frac{(b+cx)^2}{a} + \frac{(c+ax)^2}{b} \Big) \ge (a+b+c)(x+1)^2$

From the original thing:

$(x+1)(3\sqrt3) \ge (a+b+c)$

So, looking at the RHS

We need to show that $(a+b+c)(x+1)^2 \ge abc$

$(x+1)^3(3\sqrt3) \ge abc$

$(\displaystyle\frac{a+b+c}{3\sqrt3})^3(3\sqrt3) \ge abc$

$\displaystyle\frac{a+b+c}{3} \ge \sqrt[3]{abc}$

last thing is true by AM-GM, and thus proving the original thing is $\ge abc$

[/hide]

[quote="Rushil"]Cauchy is for all real right???

Why is it so widely used???? :o[/quote]
yes cauchy holds in Real numbers.
and its widely used because its usefull :)


modular arithmetic
number theory
prime numbers
number theory proposed

Find all prime numbers $ p,q$ for which $ pq$ divides $ (5^p\minus{}2^p)(5^q\minus{}2^q)$.

Two thirds of a solution.

If $ pq$ divides $ (5^p\minus{}2^p)(5^q\minus{}2^q)$ then at least one of the following must happen:

1) $ p|(5^p\minus{}2^p)$ and $ q|(5^q\minus{}2^q)$
2) $ p|(5^p\minus{}2^p)$ and $ q|(5^p\minus{}2^p)$ or vice-versa
3) $ p|(5^q\minus{}2^q)$ and $ q|(5^p\minus{}2^p)$

Notice that $ p,q$ can never be $ 2$ or $ 5$. It's easy to show this by applying mod 5 and mod 2.

Answers: $ (3,3),(3,13),(13,3)$.

[b]Solutions for #1[/b]:
If $ p$ and $ q$ are prime numbers distinct from $ 2$ and $ 5$ such that $ 5^p\minus{}2^p\equiv 0\pmod{p}$ and $ 5^q\minus{}2^q\equiv 0\pmod{q}$, then the only possible value for $ p$ and $ q$ is $ 3$. The reason is that if

$ 5^p\minus{}2^p\equiv 0\pmod{q}\implies 5\minus{}2\equiv 0\pmod{p}\implies 3\equiv 0\pmod{p}$ and the only prime number $ p$ that satisfies this is $ 3$.

$ 5^q\minus{}2^q\equiv 0\pmod{q}\implies 5\minus{}2\equiv 0\pmod{q}\implies 3\equiv 0\pmod{q}$ and the only prime number $ q$ that satisfies this is $ 3$.

So the answer here is $ (3,3)$.

[b]Solutions for #2[/b]:

As in the solution for #1 we see that $ p$ must be 3, and therefore $ q$ divides $ 5^3\minus{}2^3\equal{}117\equal{}3^2\cdot 13$. which makes $ q\equal{}13$. So the answer here is $ (3,13)$ and $ (13,3)$.

[b]Solutions for #3[/b]: I haven't been able to find any of these yet, so this part is open.


analytic geometry
search
pigeonhole principle
combinatorics unsolved
combinatorics

Problem 4:
All the points with integer coordinates in the $ xy$-plane are colored using three colors, red, blue and green, each color being used at least once. It is known that the point $ (0, 0)$ is colored red and the point $ (0, 1)$ is colored blue. Prove that there exist three points with integer coordinates of distinct colors which form the vertices of a right-angled triangle.

I don't know about using the box principle, my first thoughts are this:

Call the column of vertices above (0,0), C1 and the column of vertices above (0,1) C2 .

We know there is a green vertex somewhere in the grid.  If it is in C1 or C2 we are done.  Otherwise it is in some other column C3.  It is not hard to see that C3 must be completely coloured green otherwise we are done.  But then it is obvious that C1 is completely red, and C2 is completely blue, otherwise there is a right angled triangle.

But then if we have 3 monochromatic columns, we can easily find an icosolese right angled triangle with vertices in all 3 columns, and we are done.


hello i was wondering if anyone could explain what partitions of a number are.  i read about it in the second volume of the AoPS books but i am still somewhat confused.
also could u tell me when it would be useful to use this and how u would go about doing so.

thanks.

Partitions are ways to write a positive integer into sums of positive integers that preserves order.

For example:

5=4+1=3+2=3+1+1=2+1+1+1=2+2+1=1+1+1+1+1

These have a lot to do with generating functions, and are seen a lot in advanced number theory.

http://mathworld.wolfram.com/Partition.html

In a few days, I'll be able to show you a problem that can be solved with them, but I can't right now. Don't ask why. :P

thanks. i would really like to c some problems with them.  :)


Can any1 plz tell frm where I may be able to procure I F Sharygin "Problems in plane geometry"?It is out of print here and I'm yet to discover a place frm where I may get a 2nd hand copy.And I cudn't get any site for downloading it's ebook.
Evn b8r if some1 uploads it here. :D

even amazon.com don't have a single copy . . . .

also here: http://www.bookfinder.com/dir/i/Problems_in_Plane_Geometry/5030001808/
Problems in Plane Geometry
by I.F. Sharygin
ISBN 5030001808 / 9785030001807 / 5-03-000180-8
Publisher Victor Kamkin
Language English
Edition Softcover
List price 5.95 dollars

Heard it was an awsum and rare book.
Evn if I go askin some1 for his/her or to evn lend it,I don think they'll agree.
Any 1 here havin the book?Mind uploadin it here?It is out of print newayz,so guess there wont be any copyright problem.
 :wink:
@licu-an
there r no download links dere.

Well, I have the book, but the problem is that I do not have a scanner and if I take photos of every page then the document will become too large (the book has over 400 pages). So I do not know how I can share it with everyone on AoPS  :maybe:
Any suggestions?

If you take the photos, you can put them in a PDF in sequence and save it, which reduces the file size and then compress it in RAR format, which reduces it again. If it is still too large for AoPS, you can upload it to a file sharing site, and tell us the link.
Or you can spend some cash at a scan shop if you're willing.
Normally I prefer not to "pirate" but if the situation is such that the publisher and author fail to provide the book,  or there is a trade embargo or something, then there is not much else to do but use the internet...

Thank you very much for your suggestion, Samer. I'm not at all willing to spend my money for scanning the book, but I will take photos when I have time. Could I ask a favour of you, Samer? Could I send the photos to you and then will you put them in a PDF, and also zip the file? Because I do not know how to put them in a PDF. Of course, this is only if you have time. Or else, I'll see if someone else will do it  :maybe:

[quote="amparvardi"]Dear friends, this book is a copyrighted material.[/quote]
if it is mir publishers it's not copyrighted because these editions doen't exist any more

[quote="amparvardi"]Yes, Mir publication published it.[quote="DIMITRIS T"]it's not copyrighted because these editions doen't exist any more[/quote]But are you sure?[/quote]
yes,mir publishers was a national organization and dissolved after the collapse of CCCP.

now please can anyone upload it ?? It will be really helpful for many students :maybe:

Thank you so much, Amir  :)
You saved me a lot of trouble.

I can't remember what happened later, but anyway, here's the book, uploaded by Amir - [color=#FF0000][The link removed.][/color]

Dear jatin, posting the download links of the books is illegal in THIS forum (and generally, in this website). You can do it in your PMs, but not as public. Anyone interested in downloading that book can PM me and I will give him/her the link.

See here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=126338.


:D

Yufei wasn't the only one who told you to join, lol.  Remember http://www.artofproblemsolving.com ?  I wrote it on the board and underlined it a few times.  Hehehe.  Another Albertan on the site!

Hi David, this is Peng, if you haven't already guessed by the name. Welcome to Mathlinks. There is so much good stuff in here, it's a pity that I don't have time to read them.

Hey David!  It's Elyot!

Welcome to the madness!

Hey David...

Could've known that it was you with soo many rubik's cube there..

David, Welcome to MathLinks-AoPs!

For typing equations, please see http://www.mathlinks.ro/Forum/viewtopic.php?t=5261

Hi, Sarah, Peng, Elyot, jen, Geoffrey, and Yufei.
Yufei, I'm having hard time opening the tutorial for Latex(my computer is ancient. It is so slow :( ). I'll try again tomorrow.

Hey everyone,I just joined this forum. Nice to see you all.
I view the official website of the waterloo contests and found the amazing names.

BTW, Yufei, Where are you from? Your name seems like a "main-land China" name. Just wondering! :blush: It's so hard to dig out Chinese in my school, so I started collecting Chinese people on the internet.
I'm looking forward to see you at seminar!

Cool

 :D

[quote="lightrhee"]Hi.
 This is David. I finally joined MathLinks. Yufei told me to join for few times(first time that he asked was more than 1 year ago), and I kept forgetting it. This time, I managed to remember. :lol:

 I don't know any of the members. If you know me, post something so that I can know that you are also a member.

 I don't know much about MathLinks. :(  Please show me how to type equations, paste pictures, etc. It would be helpful.

 Excuse my English, if I make a grammatical mistakes or miss-spell something(e.g. this sentence!). I used English for no more than 3 years(much shorter for spoken English :blush: ).[/quote]Welcome! Read my signature and you'll be fine :P :) Hope you'll enjoy your stay on AoPS-MathLinks.

[quote="Connie"]
BTW, Yufei, Where are you from? Your name seems like a "main-land China" name. Just wondering! :blush: It's so hard to dig out Chinese in my school, so I started collecting Chinese people on the internet.
[/quote]

Lol, that's kinda funny cuz where I live, we're the only Chinese family in the town.

[quote]Lol, that's kinda funny cuz where I live, we're the only Chinese family in the town.[/quote]

Oh really?
Where do you live?

Where are you from?

I'm from Beijing...
 :P

lol I'm from middle of nowhere.  Lil town called Okotoks in Alberta, close to Calgary though so it's alright.

Hmm, I was actually born in Edmonton so i'm not really a "true" chinese.  Parents are from Xian

Let me see whether this works or not.

\[e^{i \pi }+1=0\]
\[f_n=f_{n-1}+f_{n-2}\]

Yeah! It works! :lol:
Do you memorize how to write all the symbols? It seems to be too much work.

[quote="lightrhee"]Do you memorize how to write all the symbols? It seems to be too much work.[/quote]After a little training you will get to write faster math in $\LaTeX$ than on paper :D :D

[quote="11th dimension"][quote="opple"]i know this is off topic. just curious people that attends Don Mill CI, how old are you guys? is this a special school where only most brilliant math students get to select into there? are students there Canadian citizen???[/quote]

Our age varies  lol , and DMCI is not particularly special, and there is no selection.  Personally, I went to Don Mills simply because there are lots of skilled people there.  And I am not a Canadian citizen yet.[/quote]

is Don Mill a college or a high school? are there a lot of non canadian citizen people there?


A solution I think, is this one: We use Chebyshev's inequality and that inequality stronger than Schur which Harazi has posted and we are done. :?

how are you going to use chebyshev?
I think we only need to use vornicu-schur

Let's assume that $a<=b<=c$ then $a^3+abc\leq b^3+abc\leq c^3+abc$ and so on... ;)

well this means nothing I think because I don't know how you are going to proceed.
Anyway, here's my solution.
We denote $f(a) = \frac{\sqrt{a^2+3bc}}{(b+c)(\sqrt{a^2+3bc}+b+c)}$
It is easy to check that if $a \ge b$ then $f(a) \ge f(b)$.
Now we can write the initial inequality in the form $\sum a f(a) (a^2+bc-b^2-c^2) \ge 0$
By vornicu-schur, $\sum a f(a) (a^2+bc-ab-ac) \ge 0$
It remains to prove $\sum a f(a) (ab+ac-b^2-c^2) \ge 0$ or
$\sum ab f(a) (a-b) + ac f(a) (a-c) \ge 0$ which is equivalent to
$\sum ab (a-b)(f(a)-f(b)) \ge 0$ which is true.

Cezar, anyone could have written that. If you really want to show you can solve inequalities, then probably it would be better to post from time to time solutions to them...

[quote="xtar"]For any three positive reals $a,b,c$, we have

$\frac{a^3+3abc}{b+c}+\frac{b^3+3abc}{a+c}+\frac{c^3+3abc}{b+a}\ge\ a \sqrt{a^2+3bc}+b \sqrt{b^2+3ca}+c \sqrt{c^2+3ab}$[/quote]
$\left(\sum_{cyc}a\sqrt{a^2+3bc}\right)^2\leq\sum_{cyc}\frac{a(a^2+3bc)}{b+c}\cdot\sum_{cyc}a(b+c)$.
Hence, it remains to prove that $\sum_{cyc}\frac{a^3+3abc}{b+c}\geq2(ab+ac+bc)$, which is obviously true by $uvw$.

[quote="arqady"][quote="xtar"]For any three positive reals $a,b,c$, we have

$\frac{a^3+3abc}{b+c}+\frac{b^3+3abc}{a+c}+\frac{c^3+3abc}{b+a}\ge\ a \sqrt{a^2+3bc}+b \sqrt{b^2+3ca}+c \sqrt{c^2+3ab}$[/quote]
Hence, it remains to prove that $\sum_{cyc}\frac{a^3+3abc}{b+c}\geq2(ab+ac+bc)$, which is obviously true by $uvw$.[/quote]
This inequality is true according to Vornicu-Schur Inequality and CS.


I have a question about 2/3/19 (clarification):
Does "each box contains an integer from 1 to 12" mean that all numbers from 1 to 12 must appear (one in every box), or is it that the number in each box must be greater than or equal to 1 and less than or equal to 12?

I don't want to sound mean, but it's kind of obvious. I can't answer flatly, so think about which one makes sense.


how do i solve these two in $C$

$1.$find all $z$ so that $z^4=-i$
$2.$find all $z$ so that $z^{2n}-z^{2n-1}-z^{2n-2}-...-z^2+1=0$

how do i do these two ?
with n order roots of unity ?

For 1. Write $-i$ in the trigometric form and apply Mouvre formula.
For 2. Are you sure you wrote it correctly?


[quote="b-flat"]Could we open a North Carolina Forum?  I would moderate it.  I think we could get enough people to join.  I know of at least 4 or 5 people that would join, and maybe, I could get some people from my school to join.

I asked about it in the Building a New Community Forum, but no one responded (a while ago).[/quote]

great, i guess we're gonna lose jb05 from posting in our forum now..that makes 2.5 people that post on the SC forum regularily?

could I join?  I'm not from anywhere near north carolina but nobody ever posts in the bay area or stanford forums  :(


Let $a,b,c$ be integers such that $a^{6}+2b^{6}= 4c^{6}$ Show that $a = b = c = 0$

[hide="Solution"]
Assume there is a non-trivial solution $(a,b,c)$ to the equation. Then, by parity, $a$ must be even, so substitute $a=2k$:
\[32k^{6}+2b^{6}= 4c^{6}\iff 16k^{6}+b^{6}=2c^{6}\]
Now, again by parity, we must have $b$ is even, so substitute $b=2m$:
\[16k^{6}+32m^{6}=2c^{6}\iff 8k^{6}+16m^{6}=c^{6}\]
Similarly, now $c$ must be even; substitute $c=2n$:
\[8k^{6}+16m^{6}=32n^{6}\iff k^{6}+2m^{6}=4n^{6}\]
Thus, $\left( \frac{a}{2}, \frac{b}{2}, \frac{c}{2}\right)$ is another non-trivial integral solution; and thus, $2$ must divide each of $a,b,c$. But continuing this process arbitrarily yields $2^{n}$ divides each of $a,b,c$ for arbitrary $n$. But, we can choose $n$ such that this is not true; contradiction.[/hide]

EDIT: This is just a special case of this problem:  http://www.artofproblemsolving.com/Forum/viewtopic.php?t=155434


Let $x,y,z$ be positive real numbers such that $x^2+y^2+z^2 = 25.$
Find the minimum of $\frac {xy} z + \frac {yz} x + \frac {zx} y .$

Pierre.

For real numbers $a,b,c$ we have $a^2+b^2+c^2-ab-bc-ca=\frac{1}{2}\{(a-b)^2+(b-c)^2+(c-a)^2\}\geqq0$ where equality holds $a=b=c$.
In using this,
$\left(\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\right)^2=\left(\frac{xy}{z}\right)^2+\left(\frac{yz}{x}\right)^2+\left(\frac{zx}{y}\right)^2+2(x^2+y^2+z^2)$

$\geqq\frac{xy}{z}\cdot\frac{yz}{x}+\frac{yz}{x}\cdot\frac{zx}{y}+\frac{zx}{y}\cdot\frac{xy}{z}+2(x^2+y^2+z^2)=3(x^2+y^2+z^2)=75$.
Since $x>0,y>0,z>0$, we obtain $\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\geqq\sqrt{75}$.The equality holds $x^2+y^2+z^2=25$ and $x=y=z\ (x>0,y>0,z>0)\Longleftrightarrow x=y=z=\frac{5}{\sqrt{3}}$.
The desired minimum value is $5\sqrt{3}$.

Solution with am-gm

$S = \frac{xy}{z} + \frac{xy}{z} + \frac{xy}{z}$

$\frac{S}{3} \ge (xyz)^(%Error. "cbrt" is a bad command.
{3})$.

From the given statement, we can get that

$\frac{25}{3} = \frac{x^2 + y^2 + z^2}{3} \ge ((xyz)^2)^(%Error. "cbrt" is a bad command.
{3})$.

To minimize $S/3$, we maximize the Rhs of the first inequality. From the 2nd inequality, we see that the maximum of that is $\frac{5}{\sqrt{3}}$, so the minimum of $S$ is $5\sqrt{3}$.

You guys need to be more careful with your inequality signs.

In any event, as kunny did, we get:

$Q^2\geq \displaystyle \sum \frac {x^2y^2}{z^2} + 2(x^2+y^2+z^2)\geq 3(x^2+y^2+z^2)$

by rearrangement.

So $Q\geq \sqrt {75}$, equality holds iff $x=y=z$

Can't we just use Lagrange multipliers to trivialize this problem?

Lagrange Multipliers are not the 1000-pound gorilla of inequality tools.

In this case, the desired lagrange multiplier equations are pretty ugly.

(x²y²+y²z²+z²x²)²=(A+B+C)²>=3(AB+BC+CA)=3(xyz)²(x²+y²+z²)=75(xyz)²
so
sum(xy/z)>=sqrt(75)
the equality holds if and only if x=y=z=sqrt(25/3)
bye![/code]

Nice solution! math_sipo. :)

[quote="math_sipo"]
$(x^2y^2+y^2z^2+z^2x^2)^2=(A+B+C)^2 \geqq 3(AB+BC+CA)$

$=3(xyz)^2(x^2+y^2+z^2)=75(xyz)^2$

so

$\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y} \geqq \sqrt{75}$
the equality holds if and only if $x=y=z=\sqrt{\frac{25}{3}}$
bye![/code][/quote]

edited by kunny

If we replace $x,y,z$ with $\frac{5}{\sqrt 3}+x_1,\frac{5}{\sqrt 3}+y_1,\frac{5}{\sqrt 3}+z_1$, we get by rearrangment that
$\frac{xy}{z}+\frac{xz}{y}+{yz}{x} \geq x+y+z = 5 \sqrt 3 +x_1+y_1+z_1$

Therefore it is at least $ 5 \sqrt 3$ and this can be achieved with $x_1+y_1+z_1=0$ And $x=y=z$


geometry
probability
number theory
national olympiad

[quote="vignesh_v"]can any of you tell me what to prepare for KVPY written round ?[/quote]

Class X NCERT suffices, I believe

you can expect some good maths questions
good $ \Rightarrow$ tricky but they just require a good knowledge of 10th NCERT and probably first few chapters of 11th(sets,combinatorics..).
also expect some tricky geometry questions a revision of probably an olympiad kind of book would be good.
also there will be questions from relation between roots of an equation and the coeffiients(probably  :D ),probability,some basic number theory ..
don't forget to take ur calculator they really helped  :D

for physics go through ur 11th mechanics initials and light,EC,magnetism  from 10th .that should should do
chemistry and bio are probably the easiest and 10th and bit of stochiometry and all those hereditary genes... stuff should do
best of luck anyways :)

There is no reliable information available about this....probably because the examiners planned it to be so.....
Since only a few students are selected for the written aptitude test,it is really hard to get information about even the papers from previous years.I tried searching on the net but there is very little information....try catching someone who wrote the exam last year...


I don't know the answer to this question, and I'd be interested to see any results on it. I just thought of it today when discussing Catalan numbers, so I don't know if it has been studied before or what is known (if anything).

It is well-known that there are (2n choose n)/(n+1) ways to get from (0,0) to (n,n) on a grid by taking up steps and right steps without going above the line y=x. What happens if you can cross the line no more than j times (or exactly j times, for that matter)? Then how many ways are there?

Couldn't you put it together using partitions?  I mean, you can definitely express it as a summation over all partitions, which isn't hard at all for exactly one touch, although it gets more complicated afterwards.

for those who have never heard of catalan numbers here is a good intro from the University of Utah Math Circle
[url]http://www.math.utah.edu/mathcircle/notes/handoutmb.pdf[/url]

Well, you can certainly write it as a horribly messy sum (presumably double sum) of products of Catalan numbers. But suppose you want it in closed form. Do you think it can be evaluated?

[quote="ComplexZeta"]What happens if you can cross the line no more than j times (or exactly j times, for that matter)? Then how many ways are there?[/quote]

I posted this originally on zeta's lj, but I'll put it here too.

I think the generating function for the second question should be (C(x)-1)^(j+1) for j > 1 where C(x) is the gf for the Catalan #'s. Then of course you could get a gf for the first problem by taking the sum of a geometric series. Though actually that first gf could be simplified with the identity xC(x)^2 + 1 = C(x) to give a formula for the second problem as a sum of Catalan numbers. Of course that's only useful for a fixed j, since I doubt the general form will come out nice. Hmmm...


inequalities
function
inequalities unsolved

prove that for every natural n>3 holds
\[ 1 \plus{} \left(2\minus{}\frac{1}{n}\right)^n < \left(2\minus{}\frac{1}{n\plus{}1}\right)^n\]

Note $ \left(2\minus{}\frac{1}{n\plus{}1}\right)^n\equal{}\left(2\minus{}\frac{1}{n}\plus{}\frac{1}{n(n\plus{}1)}\right)^n>\left(2\minus{}\frac{1}{n}\right)^n\plus{}\frac{\left(2\minus{}\frac{1}{n}\right)^{n\minus{}1}}{n\plus{}1}$
now the inequality is true if $ \frac{\left(2\minus{}\frac{1}{n}\right)^{n\minus{}1}}{n\plus{}1}>1$
which is equivalent to inequality $ \left(2\minus{}\frac{1}{n}\right)^{n\minus{}1}>n\plus{}1$ which is true for $ n\geq4$ (since $ n>3$ we have $ 2\minus{}\frac{1}{n}>\frac{7}{4}$ and $ (\frac{7}{4})^3>5$ and because exponent function grows faster than linear) so we are done


How many four-digit integers greater than 5000 are there for which the thousands digit equals the sum of the other three digits?

Writing each potential candidate as $ \sum_{i\equal{}0}^3a_i10^i$ we can calculate the result:

$ \sum_{a_3\equal{}5}^9\sum_{a_2\equal{}0}^{a_3}\sum_{a_1\equal{}0}^{a_3\minus{}a_2}1\equal{}\sum_{a_3\equal{}5}^9\sum_{a_2\equal{}0}^{a_3}(1\plus{}a_3\minus{}a_2)\equal{}\sum_{a_3\equal{}5}^9[1\plus{}1.5a_3\plus{}0.5a_3^2]\equal{}185$

Since it's MATHCOUNTS, there should be an easier way...?

Sure.

[hide="Easier to see"]
Note that we must have (where $ \overline{abc}$ is the number formed by the last three digits) $ 5\le a\plus{}b\plus{}c\le9$, which is standard Balls and Urns: we eventually arrive at $ \binom72\plus{}\binom82\plus{}\binom92\plus{}\binom{10}2\plus{}\binom{11}2\equal{}\boxed{185}$, which in my opinion, is easier to calculate than hjbrasch's result.
[/hide]
[hide="Slightly harder to see"]
Let the number be $ \overline{xabc}$. Then one requirement is that $ a\plus{}b\plus{}c\le9$. Note that for any $ a,b,c$ that satisfy this, there is some nonnegative $ d$ such that $ a\plus{}b\plus{}c\plus{}d\equal{}9$. This has $ \binom{9\plus{}4\minus{}1}{4\minus{}1}\equal{}220$ solutions. But we must also have $ a\plus{}b\plus{}c\ge5$! We can use complementary counting: note that if and only if $ a\plus{}b\plus{}c\le4$, then the above is not satisfied. As we solved $ a\plus{}b\plus{}c\le9$, we find this to have $ \binom{4\plus{}4\minus{}1}{4\minus{}1}\equal{}35$ solutions. Subtracting, we get $ 220\minus{}35\equal{}\boxed{185}$.
[/hide]
[hide="Comment"]
In the first solution, we could have used the Hockeystick Identity to find the sum to be $ \sum_{k\equal{}2}^{11}\binom k2\minus{}\sum_{k\equal{}2}^6\binom k2\equal{}\binom{11\plus{}1}{2\plus{}1}\minus{}\binom{6\plus{}1}{2\plus{}1}\equal{}\boxed{185}$.
[/hide]


can someone post a solution for this problem?
(please make your solution as simple as possible)
its number two on the link im about to give you
http://www.math.ksu.edu/~soibel/olympiad-2003.pdf

1. By "Pre-Olympiad," this forum is for problems from USAMO/Other national Olympiads. 5-6 grade problems should be posted somewhere in the Middle School forums.

2. Notice that $ 9999^2 < 9999ABCD < 1000^2$

Where $ A, B, C, D$ stand for digits of the number.

EDIT: There are more than one number "2s" in the link. Which one do you mean?

sorry the number 2 that i wanted to know was the 7-8 grade one.
 :blush:
EDIT: This i what i got for number two, i realized whats happening and i have a rough idea, but mathematically it needs some work, so please critique my solution.
Problem: M/3 + M^2/2 + M^3/6
               :Prove that for any integer value for M, there will always be an    integer value when the expression above is evaluated.
[hide] Set an equation up so that $ M/3 \plus{} M^2/2 \plus{} M^3/6 \equal{} V$ where V is the value of the expression. Now using the given that V must be an integer if M is an integer, we can multiply both sides of the equation by six, this would cancel the fractions and yield, $ 2M \plus{} 3M^2 \plus{} M^3 \equal{} 6V$ . If V is an integer, then 6V must be an integer, and if 6V is an integer, then M must be an integer. [/hide]
I realize that this is a bit shaky and most probably incorrect, this is why i beg you to tell me what i did wrong and how to correct it.
Thanks in Advance  :D

What you just proved is that if $ \frac {m}{3} \plus{} \frac {m^2}{2} \plus{} \frac {m^3}{6}$ is an integer, then $ m$ is an integer.  Actually, even your proof of this is a little bit incorrect.

This is not the same as the problem, which asks you to prove that, if $ m$ is an integer, then $ \frac {m}{3} \plus{} \frac {m^2}{2} \plus{} \frac {m^3}{6}$ is an integer.

Let us try to prove that, if $ m$ is an integer, then $ \frac {m}{3} \plus{} \frac {m^2}{2} \plus{} \frac {m^3}{6}$ is an integer.

Putting everything over one denominator, we have that $ \frac {m}{3} \plus{} \frac {m^2}{2} \plus{} \frac {m^3}{6} \equal{} \frac {m^3 \plus{} 3m^2 \plus{} 2m}{6}$

Factoring, $ \frac {m^3 \plus{} 3m^2 \plus{} 2m}{6} \equal{} \frac {m(m \plus{} 1)(m \plus{} 2)}{6}$.

Now, if $ m$ is an integer, one of $ m$,$ m \plus{} 1$, and $ m \plus{} 2$ will be divisible by 3, and also one of $ m$,$ m \plus{} 1$, and $ m \plus{} 2$ will be even.  This means that $ m(m \plus{} 1)(m \plus{} 2)$ will be divisible by 6, and thus $ \frac {m(m \plus{} 1)(m \plus{} 2)}{6}$ will be an integer.

[quote="n^4+4"]Factoring, $ \frac {m^3 \plus{} 3m^2 \plus{} 2m}{6} \equal{} \frac {m(m \plus{} 1)(m \plus{} 2)}{6}$.[/quote]

$ \equal{} {m \plus{} 2 \choose 3}$... :P

I think someone just starting out like this might not know any combinatorics

Of course, the combinatorial interpretation is a very nice proof; if it counts something, it's an integer. There's still a hole for negative $ m$, in which case the binomial coefficient is negative and doesn't count anything. If you went that route, you'd need to patch the argument to cover this case.

The argument you(neurosurgeon1118) posted is nonsense, and the result you claim is false.

sorry but i am new to this and am trying to learn.


See [url=http://www.cnn.com/2007/US/01/13/water.intox.ap/index.html]here[/url].

Worse even, the people from the radio station were supposedly warned beforehand by a nurse who called in about water poisoning.

Any thoughts?  By the way, yes, this is a real story.  I know the idea of OD'ing on water seems absurb, but drinking a lot of water drains one's body of electrolytes which is definitely not good.
One common place where water poisoning occurs is marathon running with women, in particular (since they are generally smaller).

Thats awful! And to think that the lady had 3 kids. I dont think the article said how mush water she drank though. Do have any idea :?:

According to CNN Headline News it was close to 2 gallons.

How can that happen!?! Thats crazy willpower, first of all. Or do you think she wasn't in pain then later it began affecting her?

She noted she was feeling really bad when she left the radio station.  I think she was even crying(?).  She didn't die instantly. I'm guessing she drank the water, felt really bad so she didn't intake anything else and so didn't repelenish her electrolytes and died because of a nervous system breakdown.  That's my guess at least...I'm no expert on water poisoning...

To my knowledge, water is deadly cause it is "stored" in a lot of cells; especially, the brain swells, the parts  for breathing are pressed to the hole at the back of the skull and stop working properly (additionally, other parts can malfunction).

I believe that would be a result of low electrolyte levels (perhaps).  The concentration gradient between the cells and the outside would cause water to enter the cells in order to try and offset it.

There's audio of Strange, the lady who died, and the DJ's talking.  I heard it on CNN Headline News but have been unable to find a copy online.  Someone's bound to post one sometime today...  Anyway, listening to them is horrifying.

My roommates and I were having a debate about whether the radio station is at fault in the woman's death. I think that they are in a position of superior knowledge and therefore it is not unreasonable for the woman to assume that what they ask her to do is safe, and therefore her death is (at least in part) due to their negligence. On the other hand, my roommates think that there's no difference between this scenario and one in which it happened in private between individuals, and that it's the woman's fault, since she was not compelled and there was no claim of authority on the subject by the other party.

Any opinions about this? How much of the blame should the radio station have to accept? Should we just write it off as "the woman was stupid" or does the negligence of the radio station deserve legal action?

Placing blame is always hard...

From the DJs' perspective, it probably seemed safe.  Even though they said they knew about the risks of water intoxication and even got the warning from a nurse, they probably just set it aside as something that was way too unlikely to happen.  I mean, it's feasible that people could die from overeating but I've never heard of such a case (where they died within a few days of an eating contest or their random binge).  They probably hadn't either.  I bring this up because they probably placed the risk of water toxication right with that of dying from eating way too much.  Additionally, people are always saying to drink more water, at least 8 glasses a day, if not more.  Water's a substance that can only help.  And this is pretty much true...up to a point when you get too much of it in a relatively short amount of time.

...Shoot, I remember hearing that the DJ's were told about the case of a person dying of water intoxication.  Let's still just assume that they were ignorant of just how possible it was for someone to die of water intoxication (which is still highly likely...).

Anyway, I'm still trying to decide if you can really blame idiots.  On one hand, they should have taken precautions to get medical approval for the event.  But then again, they're idiots.  Can you expect them to do this?  I assume they got approval from their supervisors.  So their supervisors deserve some blame too.

It would be an oversimplification to place blame on just one person or group of people.  I think the blame should be divided among the DJ's, their supervisors, and Strange.  Either the DJ's or supervisors should have made sure paramedics were there.  At the least they should have foresaw the possibility of choking on water.  The only thing I blame Strange for is not going to the hospital afterwards.  And even part of this blame can be put on the DJ's.  Strange probably felt like she would be viewed as a coward if she went to the hospital (which would partially be a result of how they were acting).

There's one other thing that I'm considering.  Women who die in marathons and the like from overdrinking...who's to blame?  This situation is not [i]completely[/i] different from the one at hand.

I've heard of this happening a marathon runner before.

[quote="star99"]i've heard that they fired 10 people from the radio station...[/quote]
They did, but that, and the waivers that the contestants signed, are not going to save the station owners from a lot of legal trouble. The local DA is currently investigating to see whether anything criminal occured. Jack McCoy of [i]Law and Order[/i] would file criminal charges, but he's fictional. I doubt that anyone will face criminal charges, but I would not want to be the attorney who has defend the station against a civil lawsuit.

[quote="now a ranger"]I think it was just an accident. Accidents are prone to happen everywhere. No one can stop accidents from happening, nevertheless.[/quote]

Just a little tip: don't ingest large quantities of anything in a short period of time.

[quote="Xevarion"][quote="PenguinIntegral"][quote="Kent Merryfield"]Someone should mention how the great astronomer Tycho Brahe died.[/quote]
Mercury poisoning?[/quote]
Hehe. As far as I recall, he was at a dinner with some dude like a prince or something. And etiquette demanded that no one could leave before the person at the dinner with the highest social status. Our dear Tycho was fond of the drink and had quite a bit of beer or wine or maybe both and he really needed to urinate. But he couldn't get up, and he didn't want to go in his pants. So he held it in. And held it. And held it. And by the time he finally had the opportunity to go, he found he was unable to. He actually caused some kind of damage to his bladder by holding it in for too long, and he died within a week from complications.[/quote]
Um no. Kepler totally poisoned him.

Well, some people are just plain stupid.  They had someone calling in about the dangers of water poisoning, and they still do it!  Somehow, this reminds me of the Darwin Awards.

Adding on to what ZetaX said, the water in the cells is actually saline (0.9% salt, I believe).  By drinking all the water, the woman's cells must have absorbed a LOT of water in order to maintain the same salt content on the inside and outside of the cell, therefore causing them to expand.  That must have probably caused her death.

Well, too much of anything is bad.  It doesn't matter what it is.  It could be drugs, money, water, anything.  Even spare time.

[quote="lingomaniac88"]
Adding on to what ZetaX said, the water in the cells is actually saline (0.9% salt, I believe).  By drinking all the water, the woman's cells must have absorbed a LOT of water in order to maintain the same salt content on the inside and outside of the cell, therefore causing them to expand.  That must have probably caused her death.[/quote]

Yeah, I said something like that earlier.  But I used the hypercool word 'gradient' ;)

And about marathon runners dying of too much water, yes, it has happened.  Just do a search on google and get pages like [url=http://www.amaasportsmed.org/news_room/hyponatremia_reuters.htm]this one[/url].  After reading that page I just linked to, I realize it's more complicated than I thought.  I assumed that you could just drink more gatorade and you'd be fine but that isn't so.  Anyway, read the article and do more searching if you're interested in that topic more.

What is really sad is that this woman had kids, and was doing the contest to win a video game system for them.  And she didn't even win it, some other woman did.

[quote="Fanatic"]I read that a bag of doritos would have saved her life. Not sure if that's true.[/quote]
How would a bag of doritos end up saving her life? Is it some kind of chemical?

The doritos could soak up some of the water thus possibly helping.  I don't know if it would have saved her, but I think it would have given her a better chance.

Well then you can just put a lot of starch in the mouth and force swallow? To soak up the extra water?

[quote="me@home"]Now that is really surprising!!! (in the first marathons in the early 20C, people were known to collapse from dehydration at the end) where did you hear this?[/quote]
My 7th grade science teacher told us something like that about someone dying from drinking way way way too much water in/after a marathon.

Doritos? What about just plain table salt?

[quote="PenguinIntegral"]Just so people know I'm not spamming, [url=http://en.wikipedia.org/wiki/Tycho_Brahe#Death]Kepler did poison Brahe.[/url][/quote]

It is spamming to spread around speculation as though it were accepted fact.   The theory that Kepler poisoned Brahe was considered many times, but never accepted as the likeliest explanation for Brahe's death.  It's not that the Gilder and Gilder version of the poisoning story is necessarily wrong (or right); but it is far from being the new standard explanation.


algebra
binomial theorem
Pascal\u0027s Triangle

Consider the identity $ \binom{n}{0}\minus{}\binom{n}{1}\plus{}\binom{n}{2}\minus{}\cdots\plus{}\binom{n}{n}(\minus{}1)^n\equal{}\sum_{k\equal{}0}^{n}\binom{n}{k}(\minus{}1)^k\equal{}0$

I can only prove this using the Binomial Theorem (using expansion of $ (1\minus{}1)^n$)
How can this be proved using Block walking or committee forming

Apply the Principle of Inclusion-Exclusion to the sets $ A_1 \equal{} A_2 \equal{} \cdots \equal{} A_n \equal{} \{x\}$, so
$ \genfrac {(} {)} {0pt} {} {n} {0} \equal{} 1 \equal{} |\bigcup_{1\leq i \leq n} A_i| \equal{} \sum_{1\leq i \leq n} |A_i| \minus{} \sum_{1\leq i < j \leq n} |A_i \cap A_j| \plus{} \cdots \plus{} (\minus{}1)^{n\minus{}1} |\bigcap_{1 \leq i \leq n} A_i| \equal{} \genfrac {(} {)} {0pt} {} {n} {1} \minus{}  \genfrac {(} {)} {0pt} {} {n} {2} \plus{} \cdots \plus{} (\minus{}1)^{n\minus{}1}  \genfrac {(} {)} {0pt} {} {n} {n}$. Just kidding ...

[i]Block-Walking[/i]:  Clearly this comes down to proving that $ C(n,0)\plus{}C(n,2)\plus{}....\equal{}C(n,1)\plus{}C(n,3)\plus{}....$.  Considering the $ nth$ row of Pascal's triangle, $ C(n,\beta)$ and $ C(n,\beta\plus{}1)$ have the same number of paths leading up to the entry making them up.  The sum of all of these entries will thus correspond to an equal value of entries in the form$ C(n,2k)$ and $ C(n,2k\plus{}1)$ in this row.


calculus
integration
trigonometry
calculus computations

$ \int_{0}^{1}\frac{\arctan \sqrt{1+x^{2}}}{\sqrt{1+x^{2}}^{3}}dx=\int_{0}^{\frac{\pi}{4}}\frac{\arctan \sec x}{\sec x}dx =\left[x+\sin x \, \arctan \sec x-\sqrt{2}\arctan \frac{\tan x}{\sqrt{2}}\right]_{0}^{\frac{\pi}{4}}$

$ =\frac{\pi}{4}+\frac{1}{\sqrt{2}}\arctan \sqrt{2}-\sqrt{2}\arctan \frac{1}{\sqrt{2}}$ . Finally substitute $ \arctan \frac{1}{\sqrt{2}}=\frac{\pi}{2}-\arctan \sqrt{2}$ .


AMC
AIME
probability
quadratics
Euler
complex numbers
algebra

I'm looking for good solutions and methods on how to do problems similar to this.

#5
w and z are complex numbers such that $w^2+z^2=7$, $w^3+z^3=10$. What is the largest possible real value of w + z?

#6
What is the remainder on dividing $6^{83} + 8^{83}$ by 49?

#7
25 knights are seated at a round table and 3 are chosen at random. Find the probability that at least two of the chosen 3 are sitting next to each other.

#13
For each non-empty subset of {1, 2, 3, 4, 5, 6, 7} arrange the members in decreasing order with alternating signs and take the sum. For example, for the subset {5} we get 5. For {6, 3, 1} we get 6 - 3 + 1 = 4. Find the sum of all the resulting numbers.

hint for #7

[hide]create three different cases: one where the 1st two knights sit together, another where the 1st two knights sit with 1 space apart (so the third has 3 spaces to sit in) and the third case where the 1st two knights sit 2 or more spaces apart (then the third has 4 spaces to sit in). then add up the probabilities. not very elegent, but oh well...[/hide]

Hint for the last one:

[hide]

Look at all sets with 7 and all sets without 7. Add those up and see what happens.[/hide]

A direct way to number 7:

It doesn't matter which 3 knights are chosen, so pre-choose 3. You are left with 22 in a circle.

The probability requested is equivalent to 1 - P(no two knights are adjacent).

Consider a regular 22-gon with the other knights as vertices. How many ways are there to choose 3 distinct sides?

Given 25 knights, there are $24!$ orientations. Fix one of the chosen on a side. There are $22!$ ways to arrange the other 22 knights at the vertices. Multiple this by $\binom{21}{2} * 2!$. There are 21 segments left, and for each pair, there are $2!$ ways to place the remaining 2 knights distinctly. Put that over the total orientations and subtract from 1.

So $P(A) = 1 - \displaystyle\frac{\binom{21}{2} * 2! * 22!}{24!}$. Whatever that is...

[hide="and hint for 5"]
consider the equations in terms of $w+z$ and $wz$.
[/hide]

for 6, do the binomial expansion of $(7-1)^{83} + (7+1)^{83}$, and note that only the last 2 terms will not be divisible by 49
for 13, think that every subset can either have a 7 or not, and if you add those 2 sets, you will just end up with 7, and since there are 2^{7-1} pairs like this, 7*2^6 = 448
generally, something like this with n consecutive numbers, the sum would be $n*2^{n-1}$

#5 illustrates the value of expressing sums of powers in terms of powers of sums and products, as follows:

[hide]
$w^2 + z^2 = (w + z)^2 - 2wz = 7$

$w^3 + z^3 = (w + z)^3 - 3wz(w + z) = 10$.

Substituting the first equation in the second, we have

$(w+z)^3 - \frac{3}{2}((w+z)^2 -7)(w+z) = 10 \Rightarrow -(w + z)^3 + 21(w + z) = 20$

Now let $w + z = k$.  It is evident that $k = 1$ satisfies the given equation.  Factoring $k - 1$ from $k^3 - 21k + 20 = 0$ gives $(k-1)(k^2 + k - 20) = 0$.  Applying the quadratic formula gives the additional solutions $k = -5,4$  So, our largest value of $k$ (from the only three possible values) is $4$.  (note: edited to give the right solution)

[/hide]

for number 5, let $w+z = n$ so
$w^2 + 2wz + z^2 = n^2$
$wz=\frac{n^2-7}{2}$
$w^3 +3(wz)(w+z) + z^3= n^3$ substituting,
$10+3\left(\frac{n^2-7}{2}\right)(n) = n^3$
simplifying gives
$n^3 - 21n+20=0$ and factoring,
$(n-1)(n-4)(n-5)=0$ and the max value is $\boxed{4}$

#6 is easily solved if you know Euler's extension of Fermat's Little Theorem:

Note that $\phi (49) = 7^2 - 7 = 42$.  Since 6 and 8 are relatively prime to 49, we have

$6^{42} \equiv 1 \ (\mbox{mod} \ 49) \Rightarrow 6^{84} \equiv 1 \ (\mbox{mod} \ 49)$

and

$8^{42} \equiv 1 \ (\mbox{mod} \ 49) \Rightarrow 8^{84} \equiv 1 \ (\mbox{mod} \ 49)$

Now note that

$6 \cdot -8 \equiv -48 \equiv 1 \ (\mbox{mod} \ 49) \Rightarrow 6^{-1} \equiv -8 \ (\mbox{mod} \ 49), 8^{-1} \equiv -6 \ (\mbox{mod} \ 49)$

and we have

$6^{83} \equiv -8 \ (\mbox{mod} \ 49), 8^{83} \equiv -6 \ (\mbox{mod} \ 49) \Rightarrow 6^{83} + 8^{83} \equiv -14 \equiv 35 \ (\mbox{mod} \ 49)$

Following hint, to solve #13:

Let S be the set of subsets (possibly empty) of $\{ 1, 2, 3,4,5,6\}$.
For any subset X of the $2^6$ subsets, consider $X \cup \{7\}$.  Then the sum of the resulting numbers in these two sets is clearly 7.

Thus, the answer is $2^6 * 7$.

[quote="zanttrang"][quote="darktreb"]$\phi(n)$ (the number of positive integers less than $n$ that are relatively prime to $n$) is powerful, even moreso in USAMO. Learn it.[/quote]

Also learn Euler's totient theorem:

$a^{\phi(n)} \equiv 1 \ (\mbox{mod} \ n)$, where $a$ is relatively prime to $n$.  It can be incredibly handy in any situation involving mods of large numbers.[/quote]

Yes. That's exactly why you should learn about the specific workings of the Euler phi function.

I have another solution for #7

Just find the P that they are not sitting together, which would be $\frac{25 \cdot 22 \cdot 19}{25 \cdot 24 \cdot 23}$, but we missed something. If knight #2 sat one seat away from #1, then a seat would be discarded twice...so the probability is $\frac{25 \cdot 22 \cdot 19-25 \cdot 2 \cdot 20}{25 \cdot 24 \cdot 23}$ or something like that...

You might be able to get the correct answer using a method like that, but what you put doesn't give the right value.

I used this method to do it before, but I can't find my work right now so I don't remember exactly how :(

Try this for No. 7 :
[hide] Take the probability that Knight 2 is more than one seat away from Knight 1 and that Knight 3 is not sitting next to either of them :  $\displaystyle \frac{20\cdot 19}{24\cdot 23}$

Now take the probability that Knight 2 is sitting exactly one seat away from Knight 1 and that Knight 3 is not sitting next to either of them :  $\displaystyle \frac{2\cdot 20}{24\cdot 23}$.

Now, we can add the two together, to obtain $\displaystyle \frac{21\cdot 20}{24\cdot 23}=\frac{35}{46}$ (which looks suspiciously regular—I'll investigate further) as our probability that no two knights are sitting together.  Therefore the probability that ate least two knights are sitting next to each other is $\displaystyle \frac{11}{46}$.

Q.E.D.[/hide]

[hide]
Oh I see what i did wrong :blush: you're supposed to add them, and also the first term should be $25 \cdot 20 \cdot 19$, so
$\frac{20 \cdot 19+2 \cdot 20}{24 \cdot 23}=\frac{20 \cdot 21}{24 \cdot 23}=\frac{7 \cdot 5}{2 \cdot 23}=\frac{35}{46} \implies P=11/46 \ \ \ \ \ \blacksquare$
[/hide]


Mosallase ABC ke <C=90 mafruz ast.farz konid O markaze dayereye mohati va M noghteye tamase dayereye mohati ba vatare mosallas bashad.farz konid dayerei be markaze M ke az O migzarad nimsaze zavayaye A,B ra dar noghtehaye K,L (motamayez az O) ghat konad.sabet konid :
K va L be tartib markaze dayerehaye mohatie mosallashaye ACD va BCD hastand ke CD ertefae mosallase ABC ast.

mikhaim sabet konim noghte $ K$ markaze dayere mohatie mosalase $ \triangle ACD$ hast.
khob tavajoh konid ke $ \triangle ACD\sim\triangle ABC$ pas age shoae dayere mohatie mosalashaye $ \triangle ACD,\triangle ABC$ ro ba $ r_{ACD}$ va $ r_{ABC}$ neshoon bedim,darim:

$ \frac {r_{ACD}}{r_{ABC}} \equal{} \frac {AC}{AB} \equal{} \cos\alpha$

$ \Rightarrow\frac {r_{ACD}}{r_{ABC}} \equal{} \cos\alpha$

ke dar inja $ \alpha \equal{} \angle BAC$.

khob biaid az noghte $ K$ bar zel'e $ AB$ amoode $ KH$ ro rasm konid,hala chon noghte $ K$ rooye nimsaze zavie $ \angle CAD$ gharar dare,vase esbate hokm kafie sabet konim $ r_{ACD} \equal{} KH$.pas ba estefade az tasavie bala,kafie sabet konim:

$ \boxed{\frac {KH}{OM} \equal{} \cos\alpha\equal{}1\minus{}2\sin^2\frac{\alpha}2} (*)$

khob hala deghat konid ke avvaln $ AO$ nimsaze $ \angle A$ hast,pas $ \angle OAM \equal{} \frac {\alpha}2$.hamchenin age gharar bedim $ \beta \equal{} \angle MOA$,ba tavajoh be mosalase ghaemozavie $ \triangle AOM$ khahim dasht:

$ \sin\frac {\alpha}2 \equal{} \cos\beta$

pas rabete $ (*)$ ro mishe intori baznevisi kard:

$ \frac {KH}{OM} \equal{} 1 \minus{} 2\cos^2\beta$

$ \Rightarrow\frac {KH}{OM} \equal{} 1 \minus{} 2\left(\frac {OM}{AO}\right)^2$

khob hala deghat konid ke:

$ \frac {KH}{OM} \equal{} \frac {AK}{AO}$,pas kafie sabet konim:

$ \frac {AK}{AO} \equal{} 1 \minus{} 2\left(\frac {OM}{AO}\right)^2$

$ \Rightarrow AK.AO \equal{} AO^2 \minus{} 2OM^2 \equal{} \left(AO^2 \minus{} OM^2\right) \minus{} OM^2 \equal{} AM^2 \minus{} OM^2$

pas kafie sabet konim:

$ AK.AO \equal{} AM^2 \minus{} OM^2$

ke inam tebghe ghovvate noghte $ A$ nesbat be dayere $ (M)$ doroste.pas noghte $ K$ markaze dayere mohatie mosalase $ ACD$ hastesh.

hokmo vase noghte $ L$ ham mishe be tarighe moshabeh esbat kard...


[quote="math_u_matics"]I am wondering about a specific question on the AMC10 B 2005. I think it was like number 12 or something. It was the one that asked for the product of the possible set of primes on any 12 given pairs of die. The answer was supposedly 36 combinations but I got 37. (1*12), (1*11,2*1), (1*11,3*1), (1*11,5*1)....I'm so confused!!!! I vehemently believe that this is the anser and that all of the answer choices were wrong. I would greatly appreciate it if MCrawford or someone could clarify this doubt for me. Thanks!!![/quote]

Hello! This can be really simple problem.

To obtain prime product you must have a prime face on top and other eleven must have 1s. Die with prime can be any of twelve, thus
12*(1/2)*(1/6)^11 = (1/6)^10


Prove that if a,b,c>0 have product 1, then:
 a^2+b^2+c^2+3>=a+b+c+ab+bc+ca+3/2*min{(sqrt(a)-sqrt(b))^2,(sqrt(b)-sqrt(c))^2, (sqrT(c)-sqrt(a))^2)}.

[quote="harazi"]Prove that if $a,b,c>0$ have product $1$, then:
 ${a^{2}+b^{2}+c^{2}+3\geq a+b+c+ab+bc+ca+\frac{3}{2}\cdot\min\{(\sqrt{a}-\sqrt{b})^{2},(\sqrt{b}-\sqrt{c})^{2}, (\sqrt{c}-\sqrt{a})^{2})}.$[/quote]
Enough to prove that $f(a,b,c)\geq0,$ where
 $f(a,b,c)=\sum_{cyc}\left(a^{2}+1-a-ab-\frac{1}{2}\cdot(\sqrt{a}-\sqrt{b})^{2}\right)=$
$=\sum_{cyc}(a^{2}-ab-2a+\sqrt{ab}+1).$ Let $a=\min\{a,b,c\}.$ Then $f(a,b,c)-f\left(a,\sqrt{bc},\sqrt{bc}\right)=$
$=(b-c)^{2}+(\sqrt[4]b-\sqrt[4]c)^{2}\sqrt a-2(\sqrt b-\sqrt c)^{2}-a(\sqrt b-\sqrt c)^{2}\geq$
$\geq(\sqrt b-\sqrt c)^{2}(b+c+2\sqrt{bc}-2-a)\geq0,$ which true since, $bc\geq1.$
Id est, remain to prove that $f(a,b,b)\geq0\Leftrightarrow f\left(\frac{1}{b^{2}},b,b\right)\geq0\Leftrightarrow$
$\Leftrightarrow(\sqrt b-1)^{2}(b^{5}-b^{3}+3b^{2}+3b+1+2\sqrt b\cdot(b^{4}-b^{3}+b^{2}+2b+1))\geq0.$
What your proof, harazi? Thank you.
And I think if $a,$ $b$ and $c$ are positive numbers such that $abc=1$ then
$a^{2}+b^{2}+c^{2}+3\geq a+b+c+ab+bc+ca+$
$+3\min\{(\sqrt{a}-\sqrt{b})^{2},(\sqrt{b}-\sqrt{c})^{2}, (\sqrt{c}-\sqrt{a})^{2}\}$ true too. :wink:

[quote="arqady"]
And I think if $a,$ $b$ and $c$ are positive numbers such that $abc=1$ then
$a^{2}+b^{2}+c^{2}+3\geq a+b+c+ab+bc+ca+$
$+3\min\{(\sqrt{a}-\sqrt{b})^{2},(\sqrt{b}-\sqrt{c})^{2}, (\sqrt{c}-\sqrt{a})^{2}\}$ true too. :wink:[/quote]
And it can be proven by what you have proved soon :P  :)


See attached.  Firefox 2, AoPS skin, Linux of some variety or other.

I have that problem too. Firefox 3, AoPS, Windows Vista

Actually it's not so much as a duplication but a positioning error. I will see what's causing it.

[quote="t0rajir0u"]The sidebar appears to be going through some worse difficulties in the "view your posts," etc. pages, at least in Firefox.  (3, AoPS, Vista.)[/quote]

Firefox 3.0.3 is acting even more glitchy in the "How Many Digits of Pi Do You Know?" thread in the Fun Factory.

It has nothing to do with the browser. It's a placement error, but I'm not sure yet what is causing it, but I'm trying to fix it.

I was seeing this in College Playground fora, but now things look like they should, so maybe it's fixed. Did you do something about it, Valentin?

I think it's only mostly fixed- I'm seeing it in the main blog listing now.

This is still happening in the blog list.

To be precise, on the one you access via the "AoPS blogs" button.


Let $ \{k_n\}_{n \equal{} 1}^{\infty}$ be a sequence of real numbers having the properties $ k_1 > 1$ and $ k_1 \plus{} k_2 \plus{} \cdots \plus{} k_n < 2k_n$ for $ n \equal{} 1,2,...$. Prove that there exists a number $ q > 1$ such that $ k_n > q^n$ for every positive integer $ n$.

It is clear that $ k_n > 2^{n\minus{}2} k_1$ for all $ n > 1$ and that the sequence $ (k_n)$ could be make to be sufficiently near to the sequence $ ( 2^{n\minus{}2} k_1 )$. Hence, if $ q$ has such property, it must be the case that $ q<k_1$ and $ q < \sqrt[n]{2^{n\minus{}2} k_1}$ for all $ n$ (or $ q < \inf \{ \sqrt[n]{2^{n\minus{}2} k_1} \}$).

We have $ \sqrt[n]{2^{n\minus{}2} k_1} \leq \sqrt[n\plus{}1]{2^{n\minus{}1} k_1}$ if and only if $ k_1 \leq 4$. So, for $ 1<k_1<4$, the sequence is increasing and we could choose to be any number less than $ \sqrt{k_1}$.

In case $ k_1 \geq 4$, we have $ \sqrt[n]{2^{n\minus{}2} k_1} \geq 2$ for all $ n$. Choose $ n$ to be $ \frac{3}{2}$ is sufficient.


geometry
integration
trigonometry
calculus
calculus computations

[quote="Gibbenergy"][quote="Mutant"]Find the area inside $r = 3cos(\theta)$ and outside $r = 1 + cos(\theta)$

I got $\pi$ using the bounds $0 .... \frac{\pi}{3}$

and a question that I got stuck with:

Find the length of the cardioid $r = 1 - cos (\theta)$

using bounds $0...2\pi$ I got stuck on $\int_0^{2\pi} \sqrt{2-2cos\theta}d\theta$[/quote]

 I think you can use the double angle formular : $cos(2\theta)=1-2sin^2(\theta)$ to solve it.
The answer is $8$ I think.
 Hope it help! :D[/quote]

How can you apply that when:
$\int_0^{2\pi} \sqrt{r^2-(\frac{dr}{d\theta})^2}d\theta$ would result in
$\int_0^{2\pi} \sqrt{(1-cos\theta)^2 + (sin\theta)^2}d\theta$
$\int_0^{2\pi} \sqrt{1-2cos\theta+(cos\theta)^2 + (sin\theta)^2}d\theta$
$\int_0^{2\pi} \sqrt{2-2cos\theta}d\theta$

Where can we use the double angle formula?

[quote="Elemennop"]\begin{eqnarray*}\sqrt{2-2\cos\theta}&=&\sqrt{\frac{4-4\cos\theta}{2}\\ &=&2\sqrt{\frac{1-\cos\theta}{2}\\ &=&2\sin\frac{\theta}{2}}}\end{eqnarray*}

I believe you can solve it from here.[/quote]

Thank you


You can keep me safe
You can keep me dry
But never hold me
As I will surely cry
By candle power
Or light of the Sun
I am a building block one day
And then I am gone.

Who or what is the "I" referred to in this riddle?

Ice.

You can't hold it, for it will melt..
Candle power and the Sun can melt it, changing it from a building block one day into something gone the next.


Let a,b,c,x,y,z,n positive real numbers and x+y+z=1.Prove that:
\[\sqrt[6]{(nx+a)(ny+b)(nz+c)}\leq \frac{\sqrt{2}}{3}\sqrt{n+a+b+c}\]

It doesn't hold for $x=y=z=\frac{1}{3}$ and $a=b=c$.

i'm very sorry ..i made a mistake. Actually it's like this :

\[\sqrt[6]{(nx+a)(ny+b)(nz+c)}\leq \frac{\sqrt{3}}{3}\sqrt{n+a+b+c}\]

Then it holds because $\sqrt[3]{(nx+a)(nx+b)(nx+c)}\leq\frac{n(x+y+x)+a+b+c}{3}=\frac{n+a+b+c}{3}$ and yes it's an easy problem. :wink:


Can someone show me how to generalize the sequence $1^2+2^2+3^2+4^2+5^2+...+n^2$ using finite differences? (I tried searching it, but I couldnt find anything)

now you can try to prove it using induction

[hide]I was too lazy to post it. :P [/hide]

Here's a way to getv it:-

$(n+1)^3-n^3=3n^2+3n+1$

Now substitute $n=1,2,3\cdots$ and you will get $\sum_{k=1}^n k^2=\cdots$

By finite differences,do you mean the thing in which we subtract and till it hits a repeating term then we construct a quad,cubic,etc equation according to the number o9f times we subtractYou'll get the general term by that which is $n^2$,not the sum.

Notice how I said FINITE DIFFERENCES!!!!!!! But its okay I figured it out.

Well, you posted in two different threads. I answered your question in the other thread.

o yea sorry abt the spam i accidentally clicked submit twice :oops:


inequalities
induction
function
inequalities proposed

Prove that if $c$ is a positive number distinct from $1$ and $n$ a positive integer, then
\[n^{2}\leq \frac{c^{n}+c^{-n}-2}{c+c^{-1}-2}. \]

Let $d^{2}= c$.  Then the problem is equivalent to

$n \le \frac{d^{n}+d^{-n}}{d+d^{-1}}= d^{n-1}+d^{n-3}+...++d^{3-n}+d^{1-n}$

And then the conclusion follows by AM-GM.

We will prove by induction.

Let $ n \geq 3$. (For $ n \equal{} 1,2$, inequality is trivial)

If we prove that $ \left(\frac {n \plus{} 1}{n}\right)^2 \leq \frac {(c^{n \plus{} 1} \minus{} 1)^2}{c(c^n \minus{} 1)^2}$ then the induction is done.

$ \Longleftrightarrow \: f(c) \equal{} n^2c^{2n \plus{} 2} \minus{} (n \plus{} 1)^2c^{2n \plus{} 1} \plus{} (4n \plus{} 2)c^{n \plus{} 1} \minus{} (n \plus{} 1)^2c \plus{} n^2 \geq 0$

It is easy to show that $ f(1) \equal{} f'(1) \equal{} f''(1) \equal{} f'''(1) \equal{} 0$ and $ f''''(c) > 0$ when $ c \geq 1$

Since $ f(c) \geq 0 \: \Longleftrightarrow \: f\left(\frac {1}{c}\right) \geq 0$

We can assume that $ c \geq 1$

Since $ f''''(c) > 0$ and $ f(1) \equal{} f'(1) \equal{} f''(1) \equal{} f'''(1) \equal{} 0$

$ f'''(c), f''(c), f'(c)$ and $ f(c)$ are increasing functions and $ f(c) \geq f(1) \equal{} 0$

Thus, solution is done.  :wink:


MATHCOUNTS
geometry
3D geometry
induction
rectangle
algebra
factorization

Could someone tell me how to solve a MATHCOUNTS problem like 1 :^3: +2 :^3: +3 :^3: +...+20 :^3:? (92-93 National Competition)  That is one type of problem I can't solve without doing it the old-fashioned way.

hmm...if you want a challenge, try to prove it yourself using a telescopic sum.

its n(n+1)(2n+1) / 6

oops this is formula for squares my bad, u wanted cubes

[quote="WarpedKlown1335"]Naw.  Use induction.  It works very well on this type of problem.[/quote]

Yea, but not when your actual problem is trying to remember the formula :) Induction would only help if you thought you knew it in the first place. When you completely forget, the telescoping method is the easiest..

See if you can find a picture that 'proves' the formula...

meh, induction:

sum of consecutive cubes up to n:^3: is n:^2:(n+1):^2:/4. (thats just a bit expanded of what we had before)

so we do n=1 first        thus the sum of cubes would be= 1:^2:(1+1):^2:/4=1
1:^3:=1

ahoy! so far so good

the next step is to see if it works for n+1

so if the sum of the cubes from 1 to n is 1:^3:+ 2:^3:+...n:^3: which is n:^2:(n+1):^2:/4, then the sum of the cubes from 1:^3:+2:^3: +...n:^3:+(n+1):^3: should then be n:^2:(n+1):^2:/4+(n+1):^3:

this simplifies to:

(n:^4:+6n:^3:+13n :^2:+12n+4)/4.

now the last step in proving this true is to plug in n+1:

(n+1):^2:(n+2):^2:/4=  (n:^4:+6n:^3:+13n :^2:+12n+4)/4

and we are done

can you use this to find the number of squares or cubes in a square or cube?

um...I don't think so, but I don't really understand your question.

Yes, that does work -- however, it will only work for squares (or cubes) within squares (or cubes).  Triangles still suck :).  Here's a question -- in general, how many rectangles are there in a n-by-n square?  (Cut up in the normal way.)

[quote="Leeyuhang"]for example you have a 9x9 square can you use n(n+1)(2n+1) / 6

to find the number of squares in that 9x9 square.[/quote]

Yes, but don't.

The worst thing you could do would be to remember that the number of squares inside a bigger square is n(n+1)(2n+1)/6. That won't get you anywhere, you will just get mixed up and use the wrong formula.
What you should do is see there 9:^2: 1x1 squares, 8:^2: 2x2 squares, ..., 1:^2: 9x9 squares. Now you can use the formula for the sum of squares. Ok so its the same formula, but a lot less to remember (how many questions will you be asked this in comparison to being asked sum of some squares?)

three questions: first, does n represent the largest number in the sequence? second: do the formulas work if you dont start with 1^2, and third, do the squares and cubes have to be in a consecutive order to work?

Those are all questions you could verify by taking different examples and checking if they work.

Yes, n is the largest number that you are adding up to.

Clearly, if the formula works when you start with 1 and go up to n, if you [i]don't[/i] start with 1 and go up to n, you won't get the same answer and thus the formula won't work.  (Related problem -- compute the sum (m + 1):^2: + (m + 2):^2: + ... + (n - 1):^2: + n:^2:.)

Similarly, this formula won't work if they aren't consecutive because it does work when they are.  Which leads to an interesting problem -- compute the sum 2:^2: + 4:^2: + 6:^2: + ... + (2n):^2:.  Also, do the same thing for odd numbers (up to (2n - 1):^2:), multiples of 3, numbers one more than a multiple of 3, etc.  You can also do that for cubes if you want.


Note that the last number in a $ n \times n$ grid will be $ n^2\plus{}(n\minus{}1)^2$.  Substituting $ n\equal{}10$ in yields $ \boxed{181}$.

If you don't see AIME15's observation, you can also note that the difference between consecutive diagonal terms forms an arithmetic sequence (i.e. $ 4,8,12,\ldots$), so the answer is $ 1\plus{}4(1\plus{}2\plus{}3\plus{}\cdots\plus{}(10\minus{}1))\equal{}1\plus{}4\left(\frac{9\cdot10}2\right)\equal{}1\plus{}180\equal{}\boxed{181}$.


Can you find a solution to this problem?

Find 5 integers,a,b,c,d,and f, that solve this equation:
a^5+b^5+c^5+d^5=e^5.

yankeesfan

Should there be a variable $ f$, or is that a typo?

See Euler's (false) sum of powers conjecture: http://en.wikipedia.org/wiki/Euler%27s_sum_of_powers_conjecture

Wasn't this put to use in an old AIME or some such?

A quick computer program returns $ 27^5 \plus{} 84^5 \plus{} 110^5 \plus{} 133^5 \equal{} 144^5$, I don't know how I'd get that analytically, though.

Of course you meant [i]positive[/i] integers, so that (a..f)=(0,0,0,0,1,1) doesn't work. Being specific is a good habit.

Thanks. I missed that bit about positive integers.


geometry
perimeter
trigonometry
quadratics
trig identities
Law of Sines
angle bisector

Triangle $ABC$ is isosceles, with $AB=AC$ and altitude $AM=11.$  Suppose that there is a point $D$ on $\overline{AM}$ with $AD=10$ and $\angle BDC=3\angle BAC.$  Then the perimeter of $\triangle ABC$ may be written in the form $a+\sqrt{b},$ where $a$ and $b$ are integers.  Find $a+b.$

[asy] import graph; size(7cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-1.55,xmax=7.95,ymin=-4.41,ymax=5.3; draw((1,3)--(0,0)); draw((0,0)--(2,0)); draw((2,0)--(1,3)); draw((1,3)--(1,0)); draw((1,0.7)--(0,0)); draw((1,0.7)--(2,0)); label("$11$",(0.75,1.63),SE*lsf); dot((1,3),ds); label("$A$",(0.96,3.14),NE*lsf); dot((0,0),ds); label("$B$",(-0.15,-0.18),NE*lsf); dot((2,0),ds); label("$C$",(2.06,-0.18),NE*lsf); dot((1,0),ds); label("$M$",(0.97,-0.27),NE*lsf); dot((1,0.7),ds); label("$D$",(1.05,0.77),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]

[hide="Solution"] Let $ \angle{MAC}\equal{}\theta$. It follows that $ \angle{MDC}\equal{}3\theta,\angle{2\theta}$.

From right $ \Delta MDC$, $ DC\equal{}\dfrac{1}{\cos{3\theta}}\equal{}\dfrac{1}{4\cos^{3}\minus{}3\cos{\theta}}$. By the law of sines on $ \Delta ADC$, $ \dfrac{DC}{\sin{\theta}}\equal{}\dfrac{AD}{\sin{2\theta}}\equal{}\dfrac{10}{2\sin{\theta}\cos{\theta}}\equal{}\dfrac{5}{\sin{\theta}\cos{\theta}}$, so $ DC\equal{}\dfrac{5}{\cos{\theta}}$.

Thus, $ \dfrac{1}{4\cos^{2}\theta\minus{}3}\equal{}5$, and it follows that $ \cos{\theta}\equal{}\dfrac{2}{\sqrt{5}},\sin{\theta}\equal{}\dfrac{1}{\sqrt{5}}$. From right $ \Delta AMC$, $ \cos{\theta}\equal{}\dfrac{AM}{AC}\equal{}\dfrac{11}{AC}\equal{}\dfrac{2}{\sqrt{5}}$, so $ AC\equal{}\dfrac{11\sqrt{5}}{2}$. Then, $ MC\equal{}\sqrt{(\dfrac{11\sqrt{5}}{2})^{2}\minus{}(11)^{2}}\equal{}\sqrt{\dfrac{121}{4}}\equal{}\dfrac{11}{2}$.

Therefore, the perimeter of $ \Delta ABC$ is $ 2(AC\plus{}MC)\equal{}2(\dfrac{11\sqrt{5}}{2}\plus{}\dfrac{11}{2})\equal{}11\sqrt{5}\plus{}11\equal{}\sqrt{605}\plus{}11$, so our answer is $ 605\plus{}11\equal{}\boxed{616}$.[/hide]

Triple angle can be avoided
BM=n, $ \angle BAD \equal{} x$, then $ \angle ABD \equal{} 2x$, so law of sines on $ \triangle ABD$ gives $ \frac {sin 2x}{10} \equal{} \frac {sin x}{\sqrt {n^2 \plus{} 1}}$ hence $ cos x \equal{} \frac {5}{\sqrt{n^2 \plus{} 1}}$ but we know $ cos x \equal{} \frac {11}{\sqrt {n^2 \plus{} 121}}$, hence $ n \equal{} \frac {11}{2}$.

Notice that Stewart's Theorem can be proved by bashing out the Pythagorean theorem.
[hide="Misread as $\angle CBD=3\angle ABC$"]
Define $E$ to be the intersection of the angle bisector of $\angle ABC$ with $AM$. Let $BM=x$. Then by the Pythagorean theorem, $BD=\sqrt{x^2+1}$. By Stewart's theorem on $\triangle BME$, we have ... oops
[/hide]
[hide="Thought Process (sketch)"]
What can we say about the angles? I want to have angle relationships so that I can more easily use $AD=10$ and $DM=1$. So instead of thinking about the original angle relationship, we can say $\angle BAM=\theta$, and $\angle BDM=3\theta$. Immediately, we have $\angle ABD=2\theta$ and $\angle BAC=2\theta$. That means there's a hidden isosceles triangle somewhere. It turns out that we can extend $BD$ to meet $AC$ at $E$, and $AE=EB$. Since $AD$ is uniquely defined in $\triangle ABE$, I guess we can figure out everything in terms of another length. Let's choose $BD=x$ (I originally had $BM=x$, but it makes the algebra harder, and $BD=x$ makes more sense in this triangle). Now we figure out $AB$. We can just use the Pythagorean theorem to get $BM=\sqrt{x^2-1}$, so $AB=\sqrt{x^2+120}$. Since $AD$ is the angle bisector of $\angle BAE$, we can find one of $DE,EA$ in terms of the other. Set $DE=y$. Then by the angle bisector theorem, $AE=y\frac{\sqrt{x^2+120}}{x}$. However, $\triangle ABE$ is isosceles, so $y\frac{\sqrt{x^2+120}}{x}=y+x$, so $y=\frac{x}{\frac{\sqrt{x^2+120}}{x}-1}=\frac{1}{120}(x^2\sqrt{x+120}+x^3)$. Time to finish up this heavy algebra with Stewart's theorem.
[/hide]
[hide="Failed Solution"]
Since $AB=AC$, $AM=AM$, and $\angle AMB=\angle AMC=90^\circ$, by HL congruence we get $\triangle AMB\cong\triangle AMC$. Then $BM=MC$, $\angle DMB=\angle AMB=90^\circ=\angle AMC=\angle DMC$, and $DM=DM$, so by SAS congruence $\triangle DMB\cong\triangle DMC$. By the previous congruences, $\angle BAM=\angle CAM$, and $\angle BDM=\angle CDM$, so $\angle BDM=\frac{\angle BDM+\angle CDM}{2}=\frac{\angle BDC}{2}=3\frac{\angle BAC}{2}=3\frac{\angle BAM+\angle CAM}{2}=3\angle BAM$.

Extend $BD$ to meet $AC$ at $P$. Then $\angle ABP=\angle ABD$, and by exterior angles, $\angle ABD=\angle BDM-\angle BAM=2\angle BAM=\angle BAM+\angle CAM=\angle BAC=\angle BAP$, so $\triangle ABP$ is isosceles, with $AP=PB$.

By the Pythagorean theorem on $\triangle BMD$, we have $BM=\sqrt{BD^2-1}$. By the Pythagorean theorem on $\triangle ABM$, we have $AB=\sqrt[BD^2+120}$.

By the angle bisector theorem, $AP=PD\cdot\frac{AB}{BD}=PD\cdot\sqrt{\frac{BD^2+120}{BD^2}}$. Also $AP=PB=BD+DP$, so $PD\cdot\sqrt{\frac{BD^2+120}{BD^2}}=PD+BD$, which means $PD=\frac{BD}{\sqrt{\frac{BD^2+120}{BD^2}}-1}=\frac{1}{120}(BD^2\sqrt{BD^2+120}+BD^3)$. Then $AP=PD\cdot\frac{\sqrt{BD^2+120}}{BD}=\frac{1}{120}(BD(BD^2+120)+BD^2\sqrt{BD^2+120})=\frac{1}{120}(BD^3+120BD+BD^2\sqrt{BD^2+120})$.

Since $\angle BAD=\angle BAM=\angle CAM=\angle DAP$, Stewart's theorem tells us that $BD\cdot BP\cdot DP+AD\cdot BP\cdot AD=AP\cdot BD\cdot AP+AB\cdot DP\cdot AB$. Substituting the equations from the last paragraph gives us
\[\frac{1}{120}BD^3(BD^2+BD\sqrt{BD^2+120}+120)+\frac{5}{6}BD(BD^2+BD\sqrt{BD^2+120}+120)=\frac{1}{120^2}BD^3(BD^2+BD\sqrt{BD^2+120}+120)^2+\frac{1}{120}BD^2(BD^2+120)(\sqrt{BD^2+120}+BD)\]
[/hide]
I should avoid doing algebra while TeXing. Plus, it gets messy with many substitutions. Instead, combine stuff at the end.
[hide="Solution"]
Since $AB=AC$, $AM=AM$, and $\angle AMB=\angle AMC=90^\circ$, by HL congruence we get $\triangle AMB\cong\triangle AMC$. Then $BM=MC$, $\angle DMB=\angle AMB=90^\circ=\angle AMC=\angle DMC$, and $DM=DM$, so by SAS congruence $\triangle DMB\cong\triangle DMC$. By the previous congruences, $\angle BAM=\angle CAM$, and $\angle BDM=\angle CDM$, so $\angle BDM=\frac{\angle BDM+\angle CDM}{2}=\frac{\angle BDC}{2}=3\frac{\angle BAC}{2}=3\frac{\angle BAM+\angle CAM}{2}=3\angle BAM$.

Extend $BD$ to meet $AC$ at $P$. Then $\angle ABP=\angle ABD$, and by exterior angles, $\angle ABD=\angle BDM-\angle BAM=2\angle BAM=\angle BAM+\angle CAM=\angle BAC=\angle BAP$, so $\triangle ABP$ is isosceles, with $AP=PB$.

Let $x=BM$ and $y=DP$. By the Pythagorean theorem on $\triangle BMD$, $BD=\sqrt{x^2+1}$. By the Pythagorean theorem on $\triangle ABM$, $AB=\sqrt{x^2+121}$. By the angle bisector theorem on $\triangle ABP$, $AP=y\cdot\sqrt{\frac{x^2+121}{x^2+1}}$. But $AP=BD+DP=y+\sqrt{x^2+1}$, so $y(\sqrt{\frac{x^2+121}{x^2+1}}-1)=\sqrt{x^2+1}$.

Since $\angle BAD=\angle BAM=\angle CAM=\angle DAP$, Stewart's theorem tells us that $BD\cdot BP\cdot DP+AD\cdot BP\cdot AD=AP\cdot BD\cdot AP+AB\cdot DP\cdot AB$. Substituting the equations from the last paragraph gives us
\[\sqrt{x^2+1}y^2\sqrt{\frac{x^2+121}{x^2+1}}+100y\sqrt{\frac{x^2+121}{x^2+1}}=(x^2+121)y+y^2\cdot\frac{x^2+121}{x^2+1}\cdot\sqrt{x^2+1}\]
dividing by $y\sqrt{x^2+121}$, we get
\[y+\frac{100}{\sqrt{x^2+1}}=\sqrt{x^2+121}+y\sqrt{\frac{x^2+121}{x^2+1}}\]
let's subtract $y\sqrt{\frac{x^2+121}{x^2+1}}$ and factor
\[\sqrt{x^2+121}=\frac{100}{\sqrt{x^2+1}}-y(\sqrt{\frac{x^2+121}{x^2+1}}-1)\]
substituting $\sqrt{x^2+1}=y(\sqrt{\frac{x^2+121}{x^2+1}}-1)$ from earlier, we get $\sqrt{x^2+121}=\frac{100}{\sqrt{x^2+1}}-\sqrt{x^2+1}$. Squaring this gives us $x^2+121=\frac{10000}{x^2+1}+x^2+1-200$, so $320=\frac{10000}{x^2+1}$, so $x^2+1=\frac{10000}{320}=\frac{125}{4}$, which means $x=\frac{11}{2}$.

By the Pythagorean theorem, $AB=\frac{\sqrt{605}}{2}$. Therefore the perimeter is $AB+BC+CA=2AB+2BM=\sqrt{605}+11$, which gives us our answer of $605+11=\boxed{616}$.
[/hide]
[hide="Remark"]
$BC=11=AM$? This cannot just be a coincidence. To analyze this, take a right triangle, for example, $\triangle AMB$. Then $AM=2BM$. This reminds me of the construction of the golden ratio. Oops I'm procrastinating my problem compilation so I'll leave it up to the reader to explore this.
[/hide]

[u]See [b]PP20[/b][/u] from [url=http://www.artofproblemsolving.com/blog/82642][color=red][b]here[/b][/color][/url].

Alright guys, here's another solution without trig.

First note that angle ABD=2z where angle BAD=z and angle BDM=3z. Construct the point F on AD such that BF is angle bisector of angle ABD. By similar triangles we have:

BD^2=AD*FD,

 By angle bisector theorem: AB/BD=AF/FD.

 From this we find that FD=10BD/(AB+BD)

Since AD=10 we have:

100BD/(AB+BD)=BD^2.

Letting x=BM and noting that BD=sqrt\(x^2+1) and AB=sqrt\(x^2+121), we reduce the problem to solving a quadratic and we find that x=11/2. Substituting this we find our answer to be 616.

[hide="Simple Solution (no trig and no bash)"]
Let $BD=CD=y, AB=AC=x$, and $BM=CM=z$. Now, let $\angle{MAC}=\theta$ and $\angle{MDC}=3\theta$.

$x^2=z^2 +121$ and $y^2=z^2 +1$ so $y^2=x^2 -120$.

Extend CD so that it intersects AB at point E. Similarly, extend BD so that it intersects AC at point F. Now, we will use mass points on the triangle.
Assign a mass of 10 to point M, a mass of 1 at point A, and a mass of 11 (10+1=11) at point D. Since M is the midpoint of BC, assign a mass of 5 to both points B and C. Then, place a mass of 6 to points E and F. Now we see that $AE=5x/6$ and $DE=5y/6$, so $CE=11y/6$.

Since $\angle{BDM}=3\theta$, so does $\angle{ADF}$, therefore $\angle{DFA}=180-4\theta$. We see now that $\angle{DFC}=4\theta$ and $\angle{CDF}=180-6\theta$ so $\angle{DCF}=2\theta$. Because $\angle{CAE}=\angle{ECA}=2\theta$, $CE=AE$. We know what $CE$ and $AE$ are! $CE=11y/6$ and $AE=5x/6$, so $11y/6=5x/6$. Simplifying and squaring both sides, we get $121y^2=25x^2$ . With substitution, we get $121(x^2-120)=25x^2$. We can quickly solve for $x$. $x= AC=\dfrac{11\sqrt{5}}{2}$. Using the Pythagorean theorem, we get that $z=BC/2=11/2$. Therefore, the perimeter of the triangle is $11\sqrt{5}+11=\sqrt{605}+11$. $605+11= \boxed{616}$

[/hide]

[hide]We have that $\tan(3x)=BM$ and $11\tan(x)=BM$. Thus, $\tan(3x)=11\tan(x)$. By triple angle, we have $$\tfrac{3\tan(x)-\tan^3(x)}{1-3\tan^2(x)}=11\tan(x) \implies \tan(x)=\tfrac12$$ through simplifying to a quadratic and solving. Thus, we see that $BM=\tfrac{11}2$, meaning that $BC=11$. By the Pythagorean Theorem, we see that $AB=AC=\sqrt{\left(\tfrac{11}2\right)^2+11^2}=\tfrac{11\sqrt5}2$. Hence, the perimeter of the triangle is equal to $11\sqrt5+11=\sqrt{605}+11$, so the answer is $\boxed{616}$.[/hide]


Charlie is helping his family design a sun porch for their house. It will be in the shape of a trapezoid (ABCD), allowing for large windows on several sides. Charlie has labeled the four corners A, B, C, and D. He wants to find out how many 12-inch-square floor tiles he should order if 2 dozen extra tiles are needed for replacements or cutting errors. The tiles come in boxes of 12 tiles per box.

Following is information about the dimensions of the sun porch.

Line BC is 20 feet long and runs along the living room.
Line AB is 8 feet long and is perpendicular to line BC.
Line CD is parallel to line AB but is twice as long as line AB.
Line AD represents the longest wall with the most windows.
Line AE is an imaginary line parallel to line BC.
Point E is on line CD. It is the midpoint of line CD.
How many boxes of floor tiles should Charlie order?

Have fun :D

[hide]
We know that AB=8, BC=20, CD=16. Also BC is perpendicular to CD and AB // CD.

Therefore 8 and 16 are the bases and 20 is the height.

Since A = h/2 * (b1 + b2), we get Area = 20/2 * (8+16) which equals 240 ft^2.

Since each tile is 1 sq ft, you need 240 tiles, plus 24 for replacements.

Therefore we need 12 boxes.
[/hide]

[hide="clik dis cause its da answer"]yep, 12 boxes[/hide]


it seems too easy, but is the answer[hide]81[/hide]?

jorian

Just put $3^{x}=t$, you can see other factor. :lol:

Your answer is incorrect. This problem is factorization of $3\cdot 9^{2x}-6\cdot 27^{x}.$

i was just doing 1*81 cause i didn't wanna multiply anything

jorian

[quote="happyme"]No answer presented was correct.[/quote]
[hide]
$3^{4x+1}-2 \cdot 3^{3x+1}$
Factoring $3^{3x+1}$,
$3^{3x+1}(3^{x}-2)$
So our answer is $3 \cdot 27^{x}$[/hide]

[quote="jhredsox"]oh, i was just lazy and had x=1

but you could just figure out that without factoring

jorian[/quote]

How?

plug in 1 and it's kinda obvious :lol:

jorian

But the answer could be $3*27^{x}$ or it could be $9*9^{x}$ or $\frac{1}{3}* 243^{x}$. Just plugging in 1 for $x$ will not give the correct solution. Other arithmetic is required.

well, w/e, but i'm still a little confused by diophantant's work.

only the part where 3x+1 is factored

jorian

[hide="Answer"][quote="happyme"]Two factors have a product of $3\cdot9^{2x}-6\cdot27^{x}$. One of the factors is $3^{x}-2$. What is the other factor?

MathCounts State Sprint.[/quote]

Let $3^{x}=t,$ we have $9^{2x}=t^{4},\ 27^{x}=t^{3}$, so $3\cdot 9^{2x}-6\cdot 27^{x}=3t^{4}-6t^{3}=3t^{3}(t-2)=3\cdot 27^{x}(3^{x}-2)$ or $3^{3x+1}(3^{x}-2)$.[/hide]