inradius
Diophantine equation
perimeter
number theory
Triangle
IMO Shortlist

Prove that there exist infinitely many positive integers $ n$ such that $ p \equal{} nr,$ where $ p$ and $ r$ are respectively the semiperimeter and the inradius of a triangle with integer side lengths.

[hide="Solution"]

First, note that if we prove there are infinitely many $n$ for triangles with rational side lengths, we can scale up the triangle so that the sides are all integers, and the condition is true.

Let the sides of the triangle be $a$, $b$, and $c$.

As $A=rp$, we have $n=\frac{p}{r}=\frac{p}{\frac{A}{p}}=\frac{p^2}{A}$

By Heron's, $A=\sqrt{p(p-a)(p-b)(p-c)}$ so $n=\frac{p^2}{\sqrt{p(p-a)(p-b)(p-c)}}=\sqrt{\frac{p^3}{(p-a)(p-b)(p-c)}}$

Remembering that $p=\frac{a+b+c}{2}$, we have:

$n=\sqrt{{\frac{\frac{(a+b+c)^3}{8}}{\frac{-a+b+c}{2}\cdot\frac{a-b+c}{2}\cdot\frac{a+b-c}{2}}}}=\sqrt{\frac{a+b+c}{-a+b+c} \cdot \frac{a+b+c}{a-b+c} \cdot \frac{a+b+c}{a+b-c}}$

Let $a$,$b$,$c$ be the (necessarily) rational solutions to the simultaneous equations:

$a+b+c=(x^2)(a+b-c)$
$a+b+c=(y^2)(a-b+c)$
$a+b+c=(z^2)(-a+b+c)$

with $x,y,z>1$ and integers.

It remains to be proven that $a,b,c$ can then be rational sides of a triangle.

Note that $x^2=\frac{a+b+c}{a+b-c}=1+\frac{2c}{a+b-c}$

Similarly $y^2=1+\frac{2b}{a-b+c}$ and $z^2=1+\frac{2a}{-a+b+c}$

Letting $f,g,h$ equal $\frac{x^2-1}{2},\frac{y^2-1}{2},\frac{z^2-1}{2}$ respectively, we have

$f(a+b-c)=c$
$g(a-b+c)=b$
$h(-a+b+c)=a$

If $a<0$, then because $h>0$, we have $b+c-a<0 \implies a > b+c$. Thus, as least one of $b$ or $c$ is negative. WLOG, let $b<0$.

Then, $g(a-b+c)<0 \implies a+c< b<a-c \implies c<0$. Thus, if one of them is negative, they all are. If this is the case, we can simply use the triangle with sides $-a$,$-b$, and $-c$.

If $a=0$, then $h(b+c)=0 \implies b=-c$. From our first equation, we have $f(b+b)=-b \implies 2bf+b=0$. If $b\neq 0$, then $f=\frac{-1}{2} < 0$. However, as $f>0$, we must have $b=0$. Similarly, $c=0$. However, this would give $n$ as $\frac{0}{0}$ which is undefined. Therefore, we discard all solutions with $a=b=c=0$. However, as each chosen $x,y,z$ yields infinitely many (albeit similar) triangles, we still have a non-zero triangle for each choice of $x,y,z$.

If $a>0$, then we similarly can find $b>0$, $c>0$. Thus, $a+b-c>0$, $a-b+c>0$, and $-a+b+c>0$, so a triangle can be formed.

Thus, there are infinitely many $n$ such that $p=nr$.

$\blacksquare$[/hide]

[b]Edit:[/b] [hide="@Mewto55555"]In fact if we really do have
\begin{align*}
a+b+c=x^2(a+b-c)=y^2(a-b+c)=z^2(-a+b+c),
\end{align*}then
\[\frac{a+b+c}{x^2}+\frac{a+b+c}{y^2}+\frac{a+b+c}{z^2}=(a+b-c)+(a-b+c)+(-a+b+c)=a+b+c,\]so
\[\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=1,\]which has finitely many solutions... it actually has no solutions since at least one of $x^2,y^2,z^2$ has to be less than or equal to $3$ (i.e. equal to $1$, a clear contradiction).[/hide]

Here's my solution... If we let $x,y,z$ be the distances from the vertices to their respective tangency points with the incircle, then we want to find infinitely many $n$ such that solutions to
\[(x+y+z)^3=n^2xyz\]exist. After playing around with this a bit we think to try $x=k(y+z)$ and then $k=yz-1$ so
\[n^2=\frac{(x+y+z)^3}{xyz}=\frac{(yz)^3(y+z)^3}{(yz-1)(y+z)yz}=(yz)^2\frac{(y+z)^2}{yz-1}.\]It suffices to find some fixed value of $t$ so that
\[t^2=\frac{(y+z)^2}{yz-1}\]has infinitely many integer solutions since then we would have $n=yzt$ with infinitely many values. Indeed, $t=3$ works since for
\[9=t^2=\frac{(y+z)^2}{yz-1}\Longleftrightarrow y^2-7yz+(z^2+9)=0\]we can take
\[y=\frac{7z+\sqrt{(7z)^2-4(z^2+9)}}{2}=\frac{7z+3\sqrt{5z^2-4}}{2},\]which has infinitely many solutions from the Pell Equation $w^2=5z^2-4$. Note that as $z$ increases, so does $y$, so $n=yzt=3yz$ does take on infinitely many values, and we are done.

[quote=trumpeter][quote=toto1234567890]Well, as I know this is IMO SL :)
But I don't remember the year[/quote]

If you just scroll up to the top, you will see that this says "IMO Shortlist 2000, N5"...[/quote]

Oh, sorry. :P

[quote=dbz]Please review my solution:
$\frac{p}{r}=n$
We also know that $s*r=\triangle$
$\Longrightarrow$ $\frac{p^2}{\triangle}=n$ and ${r^2}*{\triangle}=n$
$\Longrightarrow$ $n$ depends on $p,r,\triangle$ for which there are infinite choices.Hence we can find infintely many $n$.[/quote]
You must notice that $p,r,\triangle$ are not completely independent of each other. They all depend upon the sides of the triangles and whether $\frac{p^2}{\triangle }$ is an integer or not. So your solution is wrong.

[quote=the_executioner][quote=va2010]The above solution is motivated by Vieta Jumping, if anyone was curious. Its quite clean:

Observe that the condition rearranges to $(a+b-c)(a-b+c)(-a+b+c) | (a+b+c)^3$. Now, WLOG $a+b-c = 1$, so $c = a+b-1$. Now we need $(2a-1)(2b-1) | (2(a+b)-1)^2$, and let $2a-1 = x$ and $2b-1 = y$. Now we've got some condition like $xy | (x+y-1)^2$. If we let $x$ and $y$ be $a^2$ and $b^2$, we want to have $\frac{a^2+b^2+1}{ab}$ be an integer, which is easy through vieta jumping.[/quote]

Is the condition $(a+b-c)(a-b+c)(-a+b+c) | (a+b+c)^3$ just sufficient?
I am getting $\frac{(a+b+c)^3}{(a+b-c)(a-b+c)(-a+b+c)}=n^2$ which I believe is not the same thing.
Also i think that by your statement we can only conclude that there are infinite values of $a,b$ but we can't say anything about $n$.
If i am wrong , can you please explain the solution to me?[/quote]

You are right. I wasn't thinking clearly :P But the ending is still quite similar.

[b]Claim. [/b] The Pell equation
    \[ x^2 - 5y^2 = -4 \]
    has infinitely many solutions in integers $(x_k, y_k)$, one set of which can be described by the recursive relation $(x_1, y_1) = (4,2)$ and
    \[ \begin{cases}
    x_{k+1} = 9x_k + 20y_k \\
    y_{k+1} = 4x_k + 9y_k .
    \end{cases} \]
[i] Proof. [/i] We proceed by induction on $n$ that the recursive relation produces $(x_k, y_k)$ which satisfy the Pell equation. The base case $k=1$ is a simple computation. For the inductive step, we just expand
    \begin{align*}
        x_{k+1}^2 - 5y_{k+1}^2 &= (9x_k + 20y_k)^2 -5(4x_k + 9y_k)^2  \\
        &= (81x_k^2 + 360 x_k y_k + 400y_k^2) - (80x_k^2 + 360x_ky_k + 405y_k^2) \\
        &= x_k^2 - 5y_k^2 = -4.
    \end{align*}
    The recursive relation clearly produces increasing sequences $(x_k)$ and $(y_k)$, so we are done.
------
Let $A,B,C$ be the side-lengths of the triangle and conduct a Ravi substitution such that $A = b+c, B = c+a,$ and $C = a+b$.

Now set
\begin{align*}
    a &= \frac{3y_k + \sqrt{9y_k^2 - 4(y_k^2 + 1)}}{2} = \frac{3y_k + x_k}{2} \\
    b &= 3y_k - a \\
    c &= (ab-1) (a+b)
\end{align*}
Trivially, $x_k$ and $y_k$ are the same parity by induction, so these three quantities are all integers. Obviously $s = a + b + c$. We also know that $[ABC] = sr$. Using Heron's formula, this implies $r = \sqrt{\frac{abc}{a+b+c}}$. Thus
\[ n^2 = \frac{(a+b+c)^3}{abc}. \]
Now, we just directly substitute to solve for $n$:
\begin{align*}
n^2 &= \frac{(ab)^3(a+b)^3}{ab(ab-1)(a+b)} \\
&= \frac{(ab)^2(a+b)^2}{ab-1} \\
&= \frac{(ab)^2(3y_k)^2}{y_k^2},
\end{align*}
with the final equality coming from
\[ ab - 1 = \frac{3y_k + x_k}{2} \cdot \frac{3y_k - x_k}{2} -1 = \frac{9y_k^2 - x_k^2 -4}{4} = y_k^2. \]
Thus
\[ n = 3ab = \frac{3(9y_k^2 - x_k^2)}{4}  = 3y_k^2 + 3 \]
As we vary $k$, this produces infinite and increasing (so distinct) $n$, so we are done.

Can someone check? Seems too easy
Search for right angle triangles with legs $a,b$ and hypotenuse $c$.
Then we know $\frac{a+b+c}{2}=\frac{2S}{a+b+c}$ which implies $\frac{(a+b+c)^2}{4}=\frac{abn}{2}$. After simplifying we get $2a^2+2b^2+2ab+\sqrt{a^2+b^2}(2a+2b)=2abn$. For some pythagorean triplet $a=3x,b=4x,c=5x$ we get $72x^2=12xn$ which implies $6x=n$ for any $x$ which implies infinite possibilities for n.

[quote=DanjelZone]For some pythagorean triplet $a=3x,b=4x,c=5x$ we get $72x^2=12xn$ which implies $6x=n$ for any $x$ which implies infinite possibilities for n.[/quote]
The right hand side is actually $12x^2$, so you only get $n=6$ from this.

(Intuitively, your solution can't work because the value of $n$ is invariant under scaling.)

Cute N5[hide=Solution]Let $S^2= (p)(p-x)(p-y)(p-z)$ by Heron's formula, we let $p-x,p-y,p-z = a,b,c$ so we get $n= \frac {p^2}{S} = \frac {(a+b+c)^{\frac{3}{2}}}{abc}$ so problem condition is equivalent to $n^2abc=(a+b+c)^3$, for simplicity let $c=k(a+b)$ then $n^2kab=(k+1)^3(a+b)^2$ now set $n=3k+3$ to get a quadratic $(k+1)(a+b)^2=9kab$and set $u=x/y$  then take its discriminant $D=k(5k-4)$ set $k=v^2$ to get a classic pell's eqn with fibbonacci as solns
[/hide]


Let $ a_n\equal{} \sum_{k\equal{}1}^{n} \frac{k(k\plus{}1)}{2x^{k\minus{}1}}$ with $ |x|>1$
Find the limit of $ a_n$

hello, the limit is
$ \lim_{n \to \plus{}\infty} a_n\equal{}\frac{x^3}{(x\minus{}1)^3}$.
Sonnhard.

Please, I'm more interested in a proof then in the answer....

$ \sum_{k\equal{}0}^{\infty}x^k\equal{}\frac{1}{1\minus{}x}$ $ for$ $ \mid{x}\mid <1$
$ \Longrightarrow (\frac{1}{1\minus{}x})'\equal{}\sum_{k\equal{}0}^{\infty}kx^{k\minus{}1}\equal{}\sum_{k\equal{}1}^{\infty}(k\plus{}1)x^k$
$ \Longrightarrow (\frac{1}{1\minus{}x})''\equal{}\sum_{k\equal{}1}^{\infty}{k(k\plus{}1)}x^{k\minus{}1}\equal{}\frac{2}{(1\minus{}x)^3}$
$ x\to\frac{1}{x} \Longrightarrow \sum_{k\equal{}1}^{\infty}\frac{k(k\plus{}1)}{x^{k\minus{}1}}\equal{}\frac{2x^3}{(x\minus{}1)^3}$ $ for$ $ \mid{x}\mid >1$
$ \Longrightarrow\sum_{k\equal{}1}^{\infty}\frac{k(k\plus{}1)}{2x^{k\minus{}1}}\equal{}\frac{x^3}{(x\minus{}1)^3}$

[quote="shape_shift2005"]$ (\frac {1}{1 \minus{} x})' \equal{} \sum_{k \equal{} 0}^{\infty}kx^{k \minus{} 1} \equal{} \sum_{k \equal{} 1}^{\infty}(k \plus{} 1)x^k$
[/quote]

You've got an error here:

$ \sum_{k \equal{} 0}^{\infty}kx^{k \minus{} 1} \equal{} 0 \cdot x ^ {\minus{}1} \plus{} 1 \cdot x ^ {0}\plus{} 2 \cdot x ^ {1}\plus{}3 \cdot x ^ {2}\plus{}...$

$ \sum_{k \equal{} 1}^{\infty}(k \plus{} 1)x^k \equal{} 2\cdot x ^1\plus{}3\cdot x ^2\plus{}...$

but the final answer is ok... :)

Here is a solution by Telescoping Method.

Let $ r\equal{}\frac{1}{x}\Longrightarrow |r|<1$.

$ \therefore a_n\equal{}\frac{1}{2}\sum_{k\equal{}1}^n k(k\plus{}1)r^{k\minus{}1}$

$ \equal{}\frac{1}{2}\sum_{k\equal{}1}^n \left[\left\{\frac{1}{1\minus{}r}(k\minus{}1)^2\plus{}\frac{3\minus{}r}{(1\minus{}r)^2}(k\minus{}1)\plus{}\frac{2}{(1\minus{}r)^3}\right\}r^{k\minus{}1}\minus{}\left\{\frac{1}{1\minus{}r}k^2\plus{}\frac{3\minus{}r}{(1\minus{}r)^2}k\plus{}\frac{2}{(1\minus{}r)^3}\right\}r^k\right]$

$ \longrightarrow \frac{1}{(1\minus{}r)^3}\equal{}\boxed{\frac{x^3}{(x\minus{}1)^3}}$

$ x\ne 1\ \implies\ \sum_{k \equal{} 1}^{n}k(k \plus{} 1)x^{k \minus{}1}\equal{}\left(\sum_{k\equal{}1}^nx^{k\plus{}1}\right)''\equal{}\left(\frac {x^{n\plus{}2}\minus{}x^2}{x\minus{}1}\right)''$ a.s.o.

yeah.it was just a typo. :oops:
$ (\frac {1}{1 \minus{} x})' \equal{} \sum_{k \equal{} 0}^{\infty}kx^{k \minus{} 1} \equal{} \sum_{k \equal{} 0}^{\infty}(k \plus{} 1)x^k$
$ \Longrightarrow (\frac{1}{1\minus{}x})''\equal{}\sum_{k\equal{}0}^{\infty}{k(k\plus{}1)}x^{k\minus{}1}\equal{}\sum_{k\equal{}1}^{\infty}{k(k\plus{}1)}x^{k\minus{}1}\equal{}\frac{2}{(1\minus{}x)^3}$

[quote="kunny"]

$ \equal{} \frac {1}{2}\sum_{k \equal{} 1}^n \left[\left\{\frac {1}{1 \minus{} r}(k \minus{} 1)^2 \plus{} \frac {3 \minus{} r}{(1 \minus{} r)^2}(k \minus{} 1) \plus{} \frac {2}{(1 \minus{} r)^3}\right\}r^{k \minus{} 1} \minus{} \left\{\frac {1}{1 \minus{} r}k^2 \plus{} \frac {3 \minus{} r}{(1 \minus{} r)^2}k \plus{} \frac {2}{(1 \minus{} r)^3}\right\}r^k\right]$

[/quote]

How did you arrive at this form ? Is it just inspiration or maybe there is a motivation behind it.

Yes, there is an aim to covert in such a way that we can use Telescoping Method.

Let $ f(k) \equal{} (ak^2 \plus{} bk \plus{} c)r^k\ (k \equal{} 1,\ 2,\ \cdots)$, we are to determine the coefficients $ a,\ b$ and $ c$ such that

$ \frac {1}{2}k(k \plus{} 1)r^{k \minus{} 1} \equal{} f(k \minus{} 1) \minus{} f(k)$.


factor this polymonial

f(x)=x^4 - 9x^3 + 9x^2 +49x +30

[hide]
From Kouchi's hint you get:
$ (x\minus{}5)(x\minus{}6)(x\plus{}1)^{2}$
[/hide]


analytic geometry
ratio
geometry
circumcircle
number theory proposed
number theory

Let $ABC$ be a triangle in a coordinate plane . Suppose $A(0,0)$ , $B(a_1,b_1)$ , $C(a_2,b_2)$ and $BC =\sqrt{n}$ , where $a_1,a_2, b_1,b_2, n$ are integers and $n$ is not a square number . Prove that : the ratio $\frac{R}{r}$ is irrational , where $R$ is the radius of the circumcircle of $\triangle ABC$ and $r$ is the radius of the inscribed circle of $\triangle ABC$

I proved that :
 $S=\frac{1}{2}|a_1b_2-a_2b_1|$ is a rational .
 It's still unsolved to me

Not having solved implies posting it into the unsolved section and not double-posting it ( http://www.mathlinks.ro/Forum/viewtopic.php?t=65297 ).

[color=darkred]Edit by Megus: I've deleted the other topic with the same question[/color]

As i said in #4 , what we have to do is to prove
 $\sqrt{n(a_1^2+b_1^2)(a_2^2+b_2^2)}(\sqrt{n}+\sqrt{a_1^2+b_1^2}+\sqrt{a_2^2+b_2^2})$ is irrational . But now, I haven't a solution for it .  :(

It seems a problem of Bungary math olympiads! :?


Give m,n,p are solutions of this equation $ a{x}^{3}\plus{}b{x}^{2}\plus{}cx\minus{}a\equal{}0$ when $ a\neq 0$
prove that :
$ \frac{\sqrt{2}}{m}\plus{}\frac{\sqrt{3}}{n}\plus{}\frac{\sqrt{2\plus{}\sqrt{3}}}{p}\leq {m}^{2}\plus{}{n}^{2}\plus{}{p}^{2}$
Thank all.

Are you sure? Try $ x^3\minus{}3x^3\plus{}3x^2\minus{}1\equal{}0\implies(m,n,p)\equal{}(1,1,1)$.

opsss,  :blush:
thank you math154 and sorry every one. I'll check this subject


vector
Ring Theory
superior algebra
superior algebra unsolved

Suppose $ V$, $ W$ and $ U$ are all fields. $ V$ and $ W$ are isomorphic as rings and are both vector spaces over the field $ U$.
Can we now conclude that $ V$ and $ W$ are isomorphic as vector spaces too? I think the answer is no, but I can't find an example.
Can anyone help me?

A ring-isomorphism of fields is always an isomorphism of fields (morphisms of both fulfill only three properties: $ f(1) \equal{} 1, f(a \plus{} b) \equal{} f(a) \plus{} f(b) , f(ab) \equal{} f(a)f(b)$).

Yes, but isomorphism of vector spaces does not have the same definition as isomorphism for rings?
I mean when $ V$ and $ W$ are vector spaces over a field $ K$ and there exist a map from $ V$ to $ W$ that satisfies
$ f(x) \plus{} f(y) \equal{} f(x\plus{}y)$ and $ f(ax) \equal{} af(x)$ with $ x,y \in V$ and $ a \in K$ and that map is bijectif, then $ V$ and $ W$
are isomorphic as vector spaces.

You should be more careful with your wording: "being a vector space" is not a property, but an additional structure.

Take $ U\equal{}V\equal{}\mathbb Q(t_2,t_3,t_4,\ldots)$ (countably many indeterminates) and $ W\equal{}\mathbb Q(t_1,t_2,t_3,\ldots)$. Using the inclusion $ U\hookrightarrow W$, $ W$ is an infinite-dimensional vector space over $ U$, whereas $ V$ is one-dimensional (using the identity $ U\to V$ for the vector space structure). $ V$ and $ W$ are isomorphic as rings (or fields) by sending $ t_n\mapsto t_{n\minus{}1}$. owk


Putnam
trigonometry
limit
LaTeX
real analysis
real analysis unsolved

(Putnam 1987 B4) Let (x1,y1)=(0.8,0.6) and let x_(n+1)=x(n)cos(y(n))-y(n)sin(y(n)) and y(n+1)=x(n)sin(y(n))+y(n)cos(y(n)) for n=1,2,3,..... For each of Lim(n-infin)x(n) and Lim(n-infin)y(n), prove that the limit exists and find it or prove that the limit does not exist.

My solution is as follows:
If both limits exist, then let Limx(n)=L, and Limy(n)=M. then I can write L=LcosM-MsinM  (1)    and M=LsinM+McosM   (2). I solved (1) and (2) and got:
M=(M(sinM)^2+M(cosM)^2-McosM)/(cosM-1)=M(1-cosM)/(cosM-1), so M=0 has be the case. Then from (1),
L=(MsinM)/(cosM-1), I used L'Hopital to solve L as M goes to zero and I got L=-2. However, the correct answer should be L=-1.  I know other solutions that can work,but I just want to know why does L'Hopital fail in this case?

Thanks

(Putnam 1987 B4) Let $(x_{1},y_{1})=(0.8,0.6)$ and let $x_{n+1}=x_{n}\cos(y_{n})-y_{n}\sin(y_{n})$ and
$y_{n+1}=x_{n}\sin{y_{n}}+y_{n}\cos{y_{n}}$ for $n=1,2,3,.....$ For each of $\lim{_n_\rightarrow_\infty}x_{n}$ and $\lim{_n_\rightarrow_\infty}y_{n}$, prove that the limit exists and find it or prove that the limit does not exist.

My solution is as follows:
If both limits exist, then let $\lim{x_{n}}=L$, and $\lim{y_{n}}=M$. then I can write $L=L*\cos{M}-M*\sin{M}$  (1)    and $M=L*\sin{M}+M*cosM$   (2). I solved (1) and (2) and got:
$M=\frac{(M*\sin^2M+M*\cos^2M-M*\cos{M})}{(\cos{M}-1)}=M*\frac{(1-\cos{M})}{(\cos{M}-1)}$, so $M=0$ has be the case. Then from (1),
$L=\frac{(M*\sin{M})}{(\cos{M}-1)}$, I used L'Hopital to solve $L$ as $M$ goes to zero and I got $L=-2$. However, the correct answer should be $L=-1$.  I know other solutions that can work,but I just want to know why does L'Hopital fail in this case?

Thanks!

Is thsi you problem and solution?

Yes. Thanks. By the way,how did you write that way? Did you edit in LaTex and then copy it over?

$L$ and $M$ are fix numbers, how can you take $\lim_{M\to 0}$?
Notice that $x_{n+1}^2+y_{n+1}^2=x_n^2+y_n^2$ is a constant.


calculus
integration
limit
trigonometry
logarithms
calculus computations

Evaluate

\[\lim_{a\rightarrow \frac{\pi}{2}-0}\ \int_0^a\ (\cos x)\ln (\cos x)\ dx\ \left(0\leqq a <\frac{\pi}{2}\right)\]

$I\left( a\right) =\int_{0}^{a}\left( \cos x\right) \ln \left( \cos x\right) dx=\int_{0}^{a}\left( \sin x\right) ^{\prime }\ln \left( \cos x\right) dx$

$=\left[ \left( \sin x\right) \ln \left( \cos x\right) \right] _{0}^{a}+\int_{0}^{a}\frac{\sin ^{2}x}{\cos x}dx$

$=\left( \sin a\right) \ln \left( \cos a\right) +\int_{0}^{a}\left( \frac{1}{\cos x}-\cos x\right) dx$

$=\left( \sin a\right) \ln \left( \cos a\right) +\left[ \ln \left| \tan \left( \frac{x}{2}+\frac{\pi }{4}\right) \right| -\sin x\right] _{0}^{a}$

$=\left( \sin a\right) \ln \left( \cos a\right) +\ln \left| \frac{1+\tan \left( \frac{a}{2}\right) }{1-\tan \left( \frac{a}{2}\right) }\right| -\sin a$

put $t=\tan \frac{a}{2}$ then : $I\left( a\right) =\frac{2t}{1+t^{2}}\ln \frac{1-t^{2}}{1+t^{2}}+\ln \frac{1+t}{1-t}-\frac{2t}{1+t^{2}}=f\left( t\right) $

as $a\rightarrow \left( \frac{\pi }{2}\right) ^{-}$ we get $t\rightarrow 1^{-}$ then $\underset{a\rightarrow \left( \frac{\pi }{2}\right) ^{-}}{\lim }I\left( a\right) =\underset{t\rightarrow 1^{-}}{\lim }f\left( t\right) $

$f\left( t\right) =\frac{2t}{1+t^{2}}\left[ \ln (1-t)+\ln (1+t)-\ln (1+t^{2})\right] +\ln (1+t)-\ln (1-t)-\frac{2t}{1+t^{2}}$

$=-\frac{\left( 1-t\right) ^{2}}{1+t^{2}}\ln (1-t)+\frac{2t}{1+t^{2}}\ln \frac{1+t}{1+t^{2}}+\ln (1+t)-\frac{2t}{1+t^{2}}$

since $\underset{t\rightarrow 1^{-}}{\lim }\left( 1-t\right) ^{2}\ln (1-t)=0$ we get $\underset{t\rightarrow 1^{-}}{\lim }f\left( t\right) =\ln 2-1$

then $\underset{a\rightarrow \left( \frac{\pi }{2}\right) ^{-}}{\lim }I\left( a\right) =\ln 2-1$


A group of n people play a round-robin tournament. Each game ends in either a win or a loss. Show that it is possible to label the polayers $p_{1}, p_{2}...p_{n}$ in such a way that $p_{1}$ beat $p_{2}$, $p_{2}$ beat $p_{3}$, ...$p_{n-1}$ beat $p_{n}$.

Let's see if this works:

Arrange the players in a queue, according to following rules :

1. pick a random player

2. for every player X,

a. if he beat the first player of the queue, place him in front

b. else, if he was beaten by the last player of the queue, place him at the end

c. if none of the above is true, then X beat the last player and was beaten by the first one. This implies that there are two players A(n) and A(n+1) in the queue, such that X beat A(n) and A(n+1) beat X. Insert X between these two players.

3. Once all the players are placed, numbering is obvious.

---
zpz

I don't really understand your solution.

[quote="zpz"]
2. for every player X,

a. if he beat the first player of the queue, place him in front

b. else, if he was beaten by the last player of the queue, place him at the end

c. if none of the above is true, then X beat the last player and was beaten by the first one. This implies that there are two players A(n) and A(n+1) in the queue, such that X beat A(n) and A(n+1) beat X. Insert X between these two players.

3. Once all the players are placed, numbering is obvious.
[/quote]

I don't see how you reached the conclusion of 2(c). Just because x beat the last player and was beaten by the first, why does this imply the existence of the specified A(n) and A(n+1)?

Furthermore, I don't think that this algorithm even necessarily ends. i.e., if nobody is undefeated then it won't end.

[quote="Phelpedo"]
I don't see how you reached the conclusion of 2(c). Just because x beat the last player and was beaten by the first, why does this imply the existence of the specified A(n) and A(n+1)?

Furthermore, I don't think that this algorithm even necessarily ends. i.e., if nobody is undefeated then it won't end.[/quote]

If you admit the placement algorithm works then it definitely will end (assumed there is a finite number of players) since the principle is to queue all the players one by one.

Concerning point 2.c., let me use an example :

Suppose 6 players are already queued : A < B < C < D < E < F, with B having beat A, C having beat B, etc. (note that here A is the last player in the queue, F the last one, maybe my post wasn't clear about that).

Pick a player X. He has played against any of the other players in the queue, and as your problem states, each game ends in either a win or a loss.

Suppose X beat A and E, and was beaten by the other players. We can rewrite it as follows : 1 0 0 0 1 0

"1" stands for "X won against that player", "0" for "X lost".

There are 3 possible cases here :

a. the sequence ends with "1" which means that X beat E, hence A < B < C < D < E < F < X is our next step;

b. the sequence starts with 0 (A beat X) => X < A < B < C < D < E < F;

c. if none of a. and b. is true, then the sequence starts with 1 and ends with 0, as in our example. Then you can find inside this sequence a "1" followed by a "0" (as for A and B in the example, but (E, F) also fit).
=> A < X < B < C < D < E < F
or A < B < C < D < E < X < F

Hope this answers your questions.

---
zpz

[hide="Proof"]We will use induction.  Our base case is $n=2$.  We have two players $A$ and $B$, WLOG $A$ beat $B$.  Then let $p_{1}=A$ and $p_{2}=B$ and we're done.

Suppose we can label the people successfully for $n=k$.  Now consider $n=k+1$.  Take any $k$ of the $k+1$ people.  We can label these $k$ people according to our inductive hypothesis.  Suppose for the sake of contradiction that the other person $X$ cannot be inserted between any two consecutive people in our labelling system.  Then the following three statements are all false:

1)  $X$ beat $p_{1}$.
2)  For some integer $m$ such that $1\le m\le k-1$, $X$ lost to $p_{m}$ but beat $p_{m+1}$.
3)  $X$ lost to $p_{n}$.

Since statement (1) is false, $X$ lost to $p_{1}$.  Also, if $X$ lost to $p_{x}$, where $1\le x\le k-1$, then $X$ also lost to $p_{x+1}$ since statement (2) is false.  By induction, $X$ lost to all of the other people, and hence lost to $p_{n}$, a contradiction since statement (3) is false.

Therefore, $X$ can be inserted between some two consecutive people in our labelling system, so we can label the people successfully for $n=k+1$.  The induction is complete, so we can label the people successfully for any $n$.[/hide]

It is a good proof, and essentially the same as zpz's. However, when I said that zpz's algorithm never necessarily stopped, I meant that there isn't necessarily a point where you can't rearrange it further. Just consider that no one is undefeated: then given an order

$a_{1}, a_{2}, a_{3}, ... a_{n}$ you can always move some $a_{i}$ in front by hypothesis. Thus, there isn't a systematic way of knowing that you have achieved the desired end state.

[quote="Phelpedo"] \[...\] However, when I said that zpz's algorithm never necessarily stopped, I meant that there isn't necessarily a point where you can't rearrange it further. Just consider that no one is undefeated: then given an order

$a_{1}, a_{2}, a_{3}, ... a_{n}$ you can always move some $a_{i}$ in front by hypothesis. Thus, there isn't a systematic way of knowing that you have achieved the desired end state.[/quote]

That's where I have to apologize : English is not my language, and it's not always simple to find the right words for a given situation. Sometimes, when no word comes up in my mind (and even Babelfish is not my friend), I simply omit it, hoping the context would make all that clear.  Alas, that doesn't always work.

I try another clarification : at the beginning of the process, I have an unsorted bunch of players on one side, an empty queue on the other side.

One by one I pick the players and move them into the queue, according to the rules I stated. It's not like having all the players queued from start and rearranging it. Furthermore, at each step, the relative position of players inside the queue doesn't change.

---
zpz

[quote="FieryHydra"]That is a very nice proof [b]matt276eagles[/b]![/quote]

Thanks.  :)


Hey everyone,
   I just don't know how to prove the following two questions, they seem pretty easy.

   1.) Suppose that a_n ---> L and b_n ---> L. Show that the sequence a1,b1, a2, b2, a3, b3,... converges to L.

   2.) Suppose {a_n} converges, Prove (using the precise definition of limit) that Lim a_n = lim a_n+1 as n approaches infinity.

   Thanks a lot  :)

Consider $x_n$ to be the sequence you sugested at point a and $\epsilon>0$.
For $N_0$ sufficiently large $\forall n>N_0 ,a_n\in\(L-\epsilon,L+\epsilon)$.
For $N_1$ sufficiently large $\forall n>N_1 ,b_n\in\(L-\epsilon,L+\epsilon)$.
Then for $N>\max(N_0,N_1)$, $\forall n>N ,x_n\in(L-\epsilon,L+\epsilon)$ thus the conclusion.

For point b the answer seems obvious because in any neighborhood $L$ we can find all the terms of the sequence except a finite number of ones so there is $N$ such that $a_n,a_{n+1}\in(L-\epsilon,L+\epsilon),\forall \epsilon>0$. This means that $\lim_{n\to\infty}a_n=\lim_{n\to\infty}a_{n+1}$

These would be true for any notion of convergence, not just the real number type you're using. It's all about the quantifier.

An informal definition of convergence:
A sequence $a_n$ converges to $a$ if for large enough $n$, $a_n$ is arbitrarily close to $a$.

In these examples, we leave the "arbitrarily close" half of the definition alone, and look at the "large enough $n$" half. If we're splicing two sequences together, we have to go far enough for the first sequence and far enough for the second; that's certainly possible, by taking a maximum.
If we shift a sequence, we just shift the "large enough $n$" threshold with it.


Now here's something I really need to work on. I will open with an illustration of why I didn't qualify for USAMO last year.

Problems answered (with correct method) : 13
Errors made by silly arithmetic mistakes      -  6
                                                                    ------
Total Score                                                      7 < 8 for USAMO qualification

I've never lost THIS much on stupid mistakes, but usually I lose 20% of the available points on a test / contest I do not consider trivial due to stupid mistakes. And I <i>do</i> check my work. I just make the same mistake over and over. Oh well.

Anyone else this bad at taking tests?

If it's any comfort, I had pretty much the same situation my freshman year.  12 problems answered -  6 careless errors = final score of 6.  Fortunately, that and a 134 AMC-12 was good enough to get into USAMO that year.

That's really the worst I've had -- although on practice tests I've generally messed up on 1/3 of the problems I try, which isn't so cool either.  Part of it for me has been knowing when to stop working on harder problems and go back and check the ones I've done already, though there are generally a bunch of mistakes I don't catch either.

I have your problem...
On EVERY contest I take, even when I do well, I think, hey, that was a really dumb mistake...if only I had not made that I could have gotten +3-4 places...

Like here's an example of a stupid mistake I made:
What is the remainder when 1!+2!+3!+4!...+99! is divided by 30. See I knew that after 4! they started ending in 0....so I only had to consider the first 4. But instead of putting this at the end of the number, I put it at the beginning! How did I do this??? I HAVE NO IDEA! This stupid mistake cost me 3 places. It's something I really need to work on, but I am getting better.

[quote="Nukular"]
i tend not to make stupid mistakes on practice tests, thats why its really annoying when, on the tests where it counts, I do relatively horribly.[/quote]

same here.

i have the same problem, but sometimes i avoid it, not always. I think just cjecking through your work through a new view, not he one you had before, because to magnus it made since that there are 0's and if he checked again he might miss his mistakes (i should take my own addvidce)

I try to check, but usually there are lots of problems and I don't look at the problem as critically as I did the first time (although on that particular test, I did find 2-3 really dumb mistakes like bubbling in the wrong answer on a scantron).

I don't know if I really need to point this out, but [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php]Stop Making Stupid Mistakes[/url] helped me greatly on the last math contest I took in high school. I was able to catch a bunch of stupid mistakes that almost always plague my results.

My personal advice -- when you check, do every problem as if it is the first time.  Don't look at your old answer or work -- even better, try to do the question with as many methods as possible.  Obviously, this only works if you have extra time.  If you don't have extra time in contests like these, then yuo're probably better off checking every problem you do carefully and only answering 9 then not checking carefully and answering 12.

Also:  don't [i]ever[/i] assume that you have made valid calculations.  [i]Always[/i] do them again, very carefully.  This is especially true for arithmetic.

Gosh, don't I know that.  Same thing happened to me on the national target.  And on the ARML relay this year.  Such is life...

[quote="mathfanatic"]Hm.   Well, at mathcounts nationals, there was a target problem that involved a clock.  Basically, (as I recall) the problem said to find the time between 7:00 and 8:00 during which the big hand and the little hand were in the exact same spot, to the nearest second.  Well, I arrived at the correct answer and was done in about 2 minutes.  I spent the next three minutes checking my work over and over again, then sitting quietly.  Then, with 20 seconds to go, I suddenly got this idea that they were supposed to be 90 degrees apart, or something like that, so I hurriedly changed my answer.  Then, just as the time was up, I realized that I was right the first time.  Grr.  And that was problem #4 or something like that.  That problem cost me 2 easy points, or probably about 10 places.  Grr.[/quote]


prove :D

$ \prod_{n\equal{}1}^{\infty}\;\left(1\minus{}\frac{1}{e^{2\pi n/\sqrt 3}}\right)^{2}\;\;\equal{}\;\;\boxed{\frac{\sqrt 3}{e^{\pi/\left(3\sqrt 3\right)}}}$

[quote="Dr. Doe"]I guess this isn't true.

Numerically LHS=.9461717221... and RHS=.9462072593...[/quote]

The product may converge slowly, so you can't say whether or not it is true only numerically.


A common car has one of those digit locks, each button has two digits. The first button would have 1,2. The second 3,4. And so on. If a robber was too stupid, and tried to figure out the combination, how many combinations would he have to punch in before he got the combination (In other words, how many possible combinations are there)?

[quote="Klebian"]Wait, how many digits is each combination? (How many times do you have to press a button)
And are there repetitions?[/quote]
There are obviously repetitions, but it is necessary to know the number of digits.

I think it's 5 digits where there are no repetitions, in that case, the answer would be $2^5$ as you have to choose one out of two five times.

i don't understand the question. :huh:

[quote]I think it's 5 digits where there are no repetitions[/quote]

wait, the first one is 1,2 the second 3,4... the fifth one 9,10???

[hide]Then since for each digit, you can press one of 5 buttons, there are $5^5$, or 3125 combinations.[/hide]

But you can press the button to the left or right can't you? So the button could still count as 2 buttons.

But each button has 2 digits, so essentiallly, the password 13579 would be the same as 24680

how would they be the same combination?
from what i understand, which isn't much.
there are 5 buttons, where you have a choice of two different digits. so,
$2^5$

It would not be $2^5$ becuase it doesn't matter which of the two digits on the button, as they are all the same button.

Perhaps if, instead of labeling them (1,2), (3,4), (5,6), (7,8), (9,10), we labeled them as A, B, C, D, E
Which is the same thing as we are just replacing the two digits on each with a letter.

Now we can plainly see that it is $5^5$, for you have a choice of 5 buttons 5 times.

Personally, I've always thought that the two digits/button lock thing was just a lousy way of making a lock. All it does is make the person who sets the lock have to memorize his combo from 100,000 choices when there are really only 3125!

Yeah, but there's also a higher chance that the particular combination is some significant number for the person (say, a birthday).

Ajai, could you describe the lock in more detail? Say, if I wanted a combination of 2194, what would I press?

it's one button....

say I assign each button a letter. ABCDE

if i wanted to type in the combination 13579, I would press ABCDE, if I wanted to type in the combination 24680, I would also press ABCDE. Basically, you know how they have those locks on cars in which there are 5 buttons lined up on the driver's door? That's pretty much how it is.

Took a bit of searching, but here's an image for that.

[img]http://www.thefordsource.com/store/car_accessories/images/freestyle/large/freestyle_keyless_lg.jpg[/img]

So each button has 2 digits on it.

you could make it easier by representing the digits as 1,2,3,4,5,6,7,8,9,0 as 1,2,3,4,5 representing the digits 1 and 2 as 1, 3 and 4 as 2, etc.

[hide]5^5, 3125[/hide]

[quote="pgpatel"]The answer is $5^x$ where x is the number of digits in the combination.[/quote]
This statement answers pretty much any questions...

the picture is a pretty good representation of the buttons, numbers, etc.


$ \text{\LaTeX}$-ifying it, it becomes $ \binom{n}{0}-\binom{n}{1}+\binom{n}{2}-\ldots+(-1)^{n-1}\binom{n}{n-1}+(-1)^n\binom{n}{n}$. Realize that this is similar to the binomial theorem. In this case, $ x=y=1$, which is why you don't "see" any $ x$ or $ y$ terms. If this is the case, it collapses it down to $ (x-y)^n=(1-1)^n=0^n=0$

$ Q.E.D$

[hide="With involutions"] If a subset of $ \{ 1, 2, ... n \}$ doesn't contain $ 1$, add it in; if it does, remove it.  This defines a parity-changing involution on the set of subsets with no fixed points, hence
$ \sum_{k \text{ even}} {n \choose k} \equal{} \sum_{k \text{ odd}} {n \choose k}$.

The idea of using involutions is essentially equivalent to PIE; apparently it is sometimes referred to as DIE. [/hide]

(1+x)^n=nC0+nC1(x)+nC2(x^2)+...+nCn(x^n)

When x=-1,

0=nC0-nC1+nC2-...+nCn (If n is an even number)

P.S: can someone please rate me haha


linear algebra
matrix
quadratics
analytic geometry
conics
linear algebra unsolved

Find all eigenvalues, and an orthonormal set of eigenvectors, of the matrices
\[ \mathsf{A} \equal{} \begin{pmatrix} 5 & 0 & \sqrt{3} \\ 0 & 3 & 0 \\ \sqrt{3} & 0 & 3\end{pmatrix} \textup{ and } \mathsf{B} \equal{} \begin{pmatrix} 2 & \minus{}1 & \minus{}1 \\ \minus{}1 & 2 & \minus{}1 \\ \minus{}1 & \minus{}1 & 2 \end{pmatrix}.\]Hence sketch the surfaces $ 5x^2 \plus{} 3y^2 \plus{} 3z^2 \plus{} 2\sqrt{3}xz \equal{} 1$ and $ x^2 \plus{} y^2 \plus{} z^2 \minus{} zy \minus{} yz \minus{} zx \equal{} 1.$

I know how to find the eigenvalues and the eigenvectors thereafter - but what guarantees that the eigenvectors of those matrices will be orthogonal? And what is the relation to those surfaces?

They are symmetric matrices. So if $ Av \equal{} \lambda v$ and $ Aw \equal{} \nu w$, then $ \langle v,Aw \rangle \equal{} \nu \langle v,w \rangle$ but $ \langle Av, w \rangle \equal{} \lambda \langle v, w \rangle$. Since $ A$ is symmetric; $ \langle v,Aw \rangle \equal{} \langle Av, w \rangle$. Therefore $ \langle v,w \rangle \equal{} 0$ whenever $ \lambda \ne \nu$.

Note that $ B$ (at least - I haven't looked closely at $ A$) has a two-dimensional eigenspace. While what Nukular just said guarantees that that is orthogonal to the other eigenspace, it doesn't guarantee that a basis you write down for the two-dimensional space is orthogonal. You may have to use Gram-Schmidt on it.

Those two surfaces can be written as $ x^TAx\equal{}1$ and $ x^TBx\equal{}1.$ We're talking about quadratic forms here. The surfaces are something in the family that includes ellipsoids and hyperboloids. Now, as for specifically what in that family - we need to classify the quadratic form. How many of the eigenvalues are, respectively, positive, negative, or zero?

By the characteristic equations, $ \mathsf{A}$ has eigenvalues 2, 3, 6 (all positive distinct eigenvalues), $ \mathsf{B}$ has eigenvalues 0, 3, 3 (2 nonzero repeated eigenvalues, one zero eigenvalue).

How do we classify surfaces in $ \mathbb{R}^3$ based on the signs of the eigenvalues?

And this may be a stupid question - but why are quadratic forms important and why do we want to write them as $ x^TAx$ and $ x^TBx$?

[quote="Carcul"]Let A be a real symmetric matrix and let P be an orthogonal matrix such that $ A \equal{} P^TDP$, where D is the diagonal matrix of the eigenvalues of A. Now consider the quadratic form $ x^TAx$. If we make the change of coordinates $ y \equal{} Px$, then we have $ x^TAx \equal{} y^TPAP^Ty \equal{} y^TDy$ (1), which is a sum of squares (in other words, the given conic or quadric surface is now in canonical form, which is much more easier to study). So for instance if the eigenvalues of A are all positive you have an elipsoid (because in (1) you get something like $ \lambda_1y_1^2 \plus{} \lambda_2y_2^2 \plus{} \lambda_3y_3^2$). Of course, you don't have to make any change of coordinates. Soon as you know the eigenvalues, you can say immediatly what quadric surface you have at hand.[/quote]

Thanks, that was informative. So $ \mathsf{A}$ is always a symmetric matrix and any quadric surface may be written as $ \mathsf{x^{\textup{T}}Ax}$?

If I may also ask about the term "canonical" form - is this some sort of standard form? I keep hearing about things like the Jordan normal form and suchlike

As Carcul said, that first one is an ellipsoid. We call such a quadratic form [i]positive definite.[/i] We also call the underlying matrix $ A$ positive definite. The longest axis of this ellipsoid is in the direction of the eigenvector associated with the smallest eigenvalue $ (\lambda \equal{}2)$ and its shortest axis is in the direction of the eigenvector associated with the largest eigenvalue $ (\lambda \equal{}6)$

As for the second case: we call such a matrix, and such a quadratic form, [i]positive semidefinite[/i] but not positive definite.

If we travel out from the origin along some ray, how far will we go before we reach the surface? If we go in the $ (1,1,1)^T$ direction - that is, in the direction of the eigenvector associated with the eigenvalue $ 0$ - we will never reach the surface at all. So the surface is not bounded.

What if we slice across the surface along a plane perpendicular to $ (1,1,1)^T?$ That plane is the 2-dimensional eigenspace associated to the eigenvalue $ 3.$ We will reach the surface in the same distance in all directions; that is, this particular cross-section of the surface is a circle.

The surface is a cylinder - in particular, a right circular cylinder, with axis in the $ (1,1,1)^T$ direction.

I mean ofcourse any quadratic form can be written as $ x^tAx$ because its defined as $ (Ax,x)$

[quote="Kent Merryfield"]As Carcul said, that first one is an ellipsoid. We call such a quadratic form [i]positive definite.[/i] We also call the underlying matrix $ A$ positive definite. The longest axis of this ellipsoid is in the direction of the eigenvector associated with the smallest eigenvalue $ (\lambda \equal{} 2)$ and its shortest axis is in the direction of the eigenvector associated with the largest eigenvalue $ (\lambda \equal{} 6)$
[/quote]
I guess we are talking about the concept of principal axes here?

[quote="Kent Merryfield"]
As for the second case: we call such a matrix, and such a quadratic form, [i]positive semidefinite[/i] but not positive definite.

If we travel out from the origin along some ray, how far will we go before we reach the surface? If we go in the $ (1,1,1)^T$ direction - that is, in the direction of the eigenvector associated with the eigenvalue $ 0$ - we will never reach the surface at all. So the surface is not bounded.

What if we slice across the surface along a plane perpendicular to $ (1,1,1)^T?$ That plane is the 2-dimensional eigenspace associated to the eigenvalue $ 3.$ We will reach the surface in the same distance in all directions; that is, this particular cross-section of the surface is a circle.

The surface is a cylinder - in particular, a right circular cylinder, with axis in the $ (1,1,1)^T$ direction.[/quote]
This is a good explanation, thanks :) What if you have one zero, one positive, and one negative eigenvalue?

For one positive, one negative, and one zero eigenvalue of $ A,$ the equation $ x^TAx\equal{}1$ (in $ \mathbb{R}^3$ ) would be a hyperbolic cylinder. For a very simple version of that, consider the set $ \{(x,y,z): x^2\minus{}y^2\equal{}1\}.$


geometry
geometric transformation
function
algebra
system of equations

[quote="jli"]Im not entirely sure if this translation is correct so if you think something is missing or is wrong, ill ask my dad to translate again.

Let $a,b$ be real numbers and $k$ is any real number.

$y=(k^{2}+k+1)x^{2}-2(a+k)^{2}x+(k^{2}+3ak+b)$

One point on the graph is $(1,0)$

Find $a,b$

Theres a second part but it doesnt really make sense

is there any other way than just plugging in a bunch of values for $k$ and solving the system of equations?[/quote]

[hide]Since we know that one root of this function is $1$, by factoring
\[(x-1)[(k^{2}+k+1)x+(-k^{2}+k+1-2a^{2}-4ak)] \]
\[=(k^{2}+k+1)x^{2}-2(a+k)^{2}x+(k^{2}+3ak+b) \]
So comparing the constant term,
\[k^{2}-k-1+2a^{2}+4ak=k^{2}+3ak+b \]
Then
\[(-1+4a)k+(2a^{2}-1)=(3a)k+(b). \]

Since this is true for any $k$,
$-1+4a=3a$
$2a^{2}-1=b$

Solving this, $(a,b)=(1,1)$.[/hide]


Given two circles $ W1$ and $ W2$. Their centres are O1 and O2 respectively.Their are tangent to each other at C. Line l is their common tangent line.Circle$ W$ is  tangent externally to $ W1$and $ W2$ and $ AB$ is diameter of $ W$ perpendicular to l. Points A and O1 lie at the same side of line l.
Prove that AO2, BO1 and l have common point.

thanks for help


If $ xyz\equal{}1$ prove that :

$ \frac{x}{xy\plus{}x\plus{}1}\plus{}\frac{y}{yz\plus{}y\plus{}1}\plus{}\frac{z}{zx\plus{}z\plus{}1}\equal{}1$

[quote="Elio (n)"]If $ xyz \equal{} 1$ prove that :

$ \frac {x}{xy \plus{} x \plus{} 1} \plus{} \frac {y}{yz \plus{} y \plus{} 1} \plus{} \frac {z}{zx \plus{} z \plus{} 1} \equal{} 1$[/quote]
Hint: [hide]Eliminate one of the variable, say $ z$, from the LHS and simplify. You will quickly obtain the same denominator for all 3 terms. Sum up the numerator for all 3 terms, which should now cancel with the denominator.[/hide]

Let $ x \equal{} \frac {a}{b}$ and $ y \equal{} \frac {b}{c}$...then $ z \equal{} \frac {c}{a}$. Then
\[ \sum \frac {x}{xy \plus{} x \plus{} 1} \equal{} \frac {\frac {a}{b}}{\frac {a}{c} \plus{} \frac {a}{b} \plus{} 1} \equal{} \sum \frac {\frac{1}{b}}{\frac{1}{a} \plus{} \frac{1}{b} \plus{} \frac{1}{c}} \equal{} 1
\]
Is the converse true?

[hide]$ \frac {x}{xy \plus{}x \plus{} 1} \equal{} \frac {x}{xy \plus{} 1 \plus{} xyz} \equal{} \frac {1}{1 \plus{} y \plus{} yz}$

$ \frac {z}{z \plus{} zx \plus{} 1} \equal{} \frac {yz}{yz \plus{} xyz \plus{} y} \equal{} \frac {yz}{1 \plus{} y \plus{} yz}$

$ \therefore \text{L.H.S.} \equal{} \frac {1}{1 \plus{} y \plus{} yz} \plus{} \frac {y}{1 \plus{} y \plus{} yz} \plus{} \frac {yz}{1 \plus{} y \plus{} yz} \equal{} 1$.

[/hide]


Let $D_{n}$ denote the number of derangements of an $n$ element sequence. Show that
\[D_{n}= (n-1)(D_{n-1}+D_{n-2}) \]

Let the first term be displaced to the $k$th spot in the derangement. We consider two cases: either the $k$th term is swapped to the first spot, or it isn't. If it is, then the other terms can be arranged in $D_{n-2}$ ways. There are $n-1$ possibilities for $k$, so we have $(n-1)D_{n-2}$ ways. If it isn't, then the terms can be arranged in $D_{n-1}$ ways, since the first spot acts like the $k$th spot in that the $k$th term can't go there. Again, $n-1$ possibilities for $k$, so $(n-1)D_{n-1}$.


inequalities
function
trigonometry
inequalities unsolved

$ 0\leq x,y,z\leq \pi\Rightarrow \sin x\plus{}\sin y\plus{}\sin z\geq
x\plus{}y\plus{}z\minus{}\frac{(x\plus{}y\plus{}z)^3}{6}$

$ sinx\plus{}siny\plus{}sinz \geq sin(x\plus{}y\plus{}z) \quad \Longleftrightarrow$

$ sinx(1\minus{}cos(y\plus{}z))\plus{}siny(1\minus{}cosxcosz)\plus{}sinz(1\minus{}cosxcosy) \geq 0$ which is obviously true.

Let $ \lambda\equal{}x\plus{}y\plus{}z$. If we prove the following then the proof is done:

$ f(\lambda)\equal{}sin\lambda\minus{}\lambda\plus{}\frac{\lambda^3}{6} \geq 0$

$ f'(\lambda)\equal{}cos\lambda\minus{}1\plus{}\frac{\lambda^2}{2} \geq 0 \quad \Longleftrightarrow$

$ \frac{\lambda^2}{2}\minus{}2sin^2\left(\frac{\lambda}{2}\right) \geq 0 \quad \Longleftrightarrow$

$ \left|\frac{\lambda}{2}\right| \geq \left|sin\left(\frac{\lambda}{2}\right)\right|$ which is a well-known inequality.

Hence $ f(\lambda)$ is an increasing function.

$ f(\lambda) \geq f(0)\equal{}0$ since $ \lambda\equal{}x\plus{}y\plus{}z \geq 0$. Solution is done.  :)


Are there any great sources you guys no about on how to get good ARML problem to practice?

Thanks,
RAY

uhh....let me guess

[url]http://www.arml.com[/url]

The 2006 contest is at arml.com.  If you PM me, I can email you some or all of the 2005 contest.  There are also a couple of ARML books that you could try borrowing if someone at your school or nearby has one.  You can also buy the books at ARML for use in future years.  I think they're about 20 dollars each.

You might also try emailing the list to see if anyone has any other old contests.


Let $ a,$ $ b$ and $ c$ are non-negative numbers, no two of which equal zero. Prove that:
1) $ \frac{a^{3}}{b^{3}\plus{}abc\plus{}c^{3}}\plus{}\frac{b^{3}}{a^{3}\plus{}abc\plus{}c^{3}}\plus{}\frac{c^{3}}{a^{3}\plus{}abc\plus{}b^{3}}\geq\frac{a^{2}\plus{}b^{2}\plus{}c^{2}}{ab\plus{}ac\plus{}bc}$
2) $ \frac{a^{5}}{b^{3}\plus{}abc\plus{}c^{3}}\plus{}\frac{b^{5}}{a^{3}\plus{}abc\plus{}c^{3}}\plus{}\frac{c^{5}}{a^{3}\plus{}abc\plus{}b^{3}}\geq\frac{a^{4}\plus{}b^{4}\plus{}c^{4}}{ab\plus{}ac\plus{}bc}.$


geometry
3D geometry
sphere
power of a point

A ball was floating in a lake in the Himalayas when the lake suddenly froze. The ball was removed (without breaking the ice) leaving a hole 34 cm across at the top and 8 cm deep. Find the radius of the ball.

No.

 [geogebra]2368f89573176506941159b8b15a028381cd73da[/geogebra]

Consider some cross section of it that is perpendicular to the ice I suppose (don't really know how to say this).

[hide="hint"] Now, if you extend the segment with length 8 to form a diameter, it's well know that a diameter bisects a chord to which it is perpendicular to.[/hide]

[hide="solution 1"] Let the center of the circle be O, connect it to the intersection of the segments with length 34 and 8. Also connect it to one of the endpoints of the chord with length 34.

If you let the radius be $ r$, you now have a right triangle with hypotenuse $ r$ and legs $ 17, r\minus{}8$. You can quickly find $ r\equal{}\frac{353}{16}$.[/hide]

[hide="solution 2"] Having extended the segment with length 8 to a diameter, let the radius be $ r$.
From Power of a Point, you have $ (2r\minus{}8)(8)\equal{}289$ so $ 2r\minus{}8\equal{}\frac{289}{8}$, or $ 2r\equal{}\frac{289\plus{}64}{8}\equal{}\frac{353}{8}$, so $ r\equal{}\frac{353}{16}$, as we found in the previous solution.[/hide]

[quote="harekrishna"][quote="isabella2296"]I'm probably missing something, but assuming that the hole was hemispherical, wouldn't it just be $ \boxed{8}$?[/quote]

I am sorry but I would be much obliged if you tell me how you got 8. If it was 8, the diameter of the sphere should be 16 but the hole is 34 cm across.[/quote]Her answer is wrong and xpmath said:

[quote="xpmath"]No.[/quote]

You could also use [hide="Pythagorean means (this is a hide tag)"][img]http://mathworld.wolfram.com/images/eps-gif/PythagoreanMeans_800.gif[/img][/hide]. First, the geometric mean:

$ \sqrt{8a}\equal{}\frac{34}{2} \implies b\equal{}\frac{289}{8}$

Now we have to find the arithmetic mean of these:

$ \frac {8\plus{}\frac{289}{8}}{2}\equal{}\boxed{\frac{353}{16}}$


Who's going to muscle shoals??
me me me
post school name, and your name if u want to

hi,
im

gene yu
from
pizitz middlle school
(lol, this is pretty late...)

I went
got 2nd acccording to them
they said they graded the test accordingly after fixing #12
they didnt for me
thus I got a 91
and they forgot my tiebreakers, which would have given me first still
but they forgot them

sigh


calculus
integration
quadratics
algebra unsolved
algebra

Find the number of subsets  $A\subset M=\{2^0,\,2^1,\,2^2,\dots,2^{2005}\}$ such that equation $x^2-S(A)x+S(B)=0$ has integral roots, where $S(M)$ is the sum of all elements of $M$, and $B=M\setminus A$ ($A$ and $B$ are not empty).

Maybe it is my fault...Ok, let's try :

First note that for all $A \subset M$ we have $S(A) + S(B) = 1+2+ \cdots + 2^{2005} = 2^{2006}-1.$
Now, if neither $A$ nor $B$ is empty, we must have $1 \leq S(A) \leq 2^{2006}-2.$
Moreover, according to the binary representation unicity, if $A \neq A'$ then $S(A) \neq S(A').$
Let $a = S(A).$
The equation can be rewritten as $x^2 - ax + 2^{2006}-1-a = 0.$ Since it is a quadratic and normalized one, the roots are integer iff $\Delta = (a+2)^2 - 2^{2008}$ is a perfect square, say $c^2.$ This means that
1) $a+2 = 2^{1004}$
or
2) $2^{1004}, c, a+2$ form a pythagorean triple, so that $2^{1004} = 2duv, c = d(u^2-v^2), a+2 = d(u^2+v^2)$ where $u,v$ have opposite parities.
In this latter case, from the first relation, the odd one among $u,v$ is $1$ so that it is $v$ (otherwise $c$ would be negative) and $u=2^p,d=2^{1003-p}$ for some positive integer $p$ not greater than $1003$.
Thus $a = 2^{1003-p}(2^{2p}+1)-2.$
For $p \leq 1002$ it leads to $a = 2^{1003+p} + \sum_{i=1}^{1002-p} 2^i$ (where a sum over the empty set is $0$). For $p=1, \cdots, 1002$ this leads to $1002$ pairwise distinct binary expansions. Adding 1), we then have $1003$ possible values.

Now, for $p=1003$, we find that $a = 2^{2006}-1$ which has to be omitted.

Pierre.

My idea was the following: find the number of $a,$ $1\leq a\leq2^{2006}-2$ such that the equation $x^{2}-ax+2^{2006}-a-1=0$ has integer roots. Let $b,c$ be its roots; then $b+c=a$ and $bc=2^{2006}-a-1,$ therefore $b+c+bc+1=2^{2006}%
,$ that is, $\left(  b+1\right)  \left(  c+1\right)  =2^{2006}.$ So $\left(
b,c\right)  $ equals $\left(  2^{k}-1,2^{2006-k}-1\right)  ,$ for $k=0$ to $1003.$ We have to discard $k=0$ (this yields $a=2^{2006}-1$), so you're right (for some reason I assumed that $b$ and $c$ must be distinct).
La prochaine fois je serais plus attentif (j'espere que j'ai pas oublie tout ce que je savais une fois...). Merci!

excuse me, but i was wondering about this "integral roots" thing for about five minutes... can somebody correct it? the english of the problems from this year is not good as whole

[hide=My solution]Clearly $ A, B\neq\varnothing $. We have: $ S(A)+S(B)=2^{2006}-1 $. Let $ a=S(A) $, then $ 1\leq a\leq 2^{2006}-2 $. Then: \[ x^2-S(A)x+S(B)=0 \iff x^2-ax+2^{2006}-1-a=0 (1) \] \[ \iff \Delta=(a+2)^2-2^{2008} \] Note that $ (1) $ has integer roots iff $ \Delta=c^2 (c\in\mathbb{Z}) $. Then $ (a+2)^2=c^2+2^{2008} $ $ \implies (a+2), c, 2^{1004} $ are Pythagorean triple. Denote that \[\left\{\begin{array}{rcl} a+2=d(u^2+v^2) \\2^{1004}=2duv \\c=d(u^2-v^2) \end{array}\right.\] where $ d, u, v\in\mathbb{Z} $, $ u,v $ have different parity $ \implies $ One of two numbers must be odd. But if the odd number is different form $ 1 $, then $ 2duv\neq 2^{1004} $ $ \implies $ the odd number must be 1 $ \implies $ $ v $ must be odd number. From this, we obtain \[\left\{\begin{array}{rcl} a+2=d(u^2+1) \\2^{1004}=2du \\c=d(u^2-1) \end{array}\right.\] $ \implies du=2^{1003} $ $ \implies \left\{\begin{array}{rcl} u=2^p \\d=2^{1003-p} \end{array}\right. $ $ \implies S(A)=2^{1003-p}(2^{2p}+1)-2 $. If $ p>1003 $ then $ 2^{1003-p}=\frac{1}{2^{p-1003}} $. Because $ 2^{2p}+1 $ is odd, then $ S(A)=\frac{2^{2p}+1}{2^{p-1003}}-2\notin\mathbb{Z} $ $ \implies p\leq 1003 $ $ \implies 1\leq p\leq 1003 $. Then, $ p $ could receive $ 1003 $ values, meaning there are $ 1003 $ sets $ A $ satisfying the condition[/hide]


Πάλι από το 2000

[b]2000 N1 (JAP):[/b] Να βρεθούν όλοι οι θετικοί ακέραιοι $ n \geqslant 2$ οι οποίοι ικανοποιούν την ακόλουθη συνθήκη: Για όλους τους ακεραίους $ a,b$ οι οποίοι είναι σχετικά πρώτοι προς τον $ n$ ισχύει:
\[ a \equiv b \bmod n \Leftrightarrow ab \equiv 1 \bmod n.\]

[b]2000 N2 (FRA):[/b] Για κάθε θετικό ακέραιο $ n$ ας είναι $ d(n)$ ο αριθμός των θετικών διαιρετών του $ n$. Να βρεθούν όλοι οι θετικοί ακέραιοι $ n$  για τους οποίους $ d(n)^3 \equal{} 4n$.

[b]2000 N3 (RUS):[/b] Υπάρχει θετικός ακέραιος $ n$ ο οποίος να έχει ακριβώς 2000 πρώτους διαιρέτες και να διαιρεί τον $ 2^n \plus{} 1$;

[b]2000 N4 (BRA):[/b] Να βρεθούν όλες οι τριάδες θετικών ακεραίων $ (a,m,n)$ ώστε ο $ a^m \plus{} 1$ να διαιρεί τον $ (a\plus{}1)^n$.

[b]2000 N5 (BUL)[/b] Να δειχθεί ότι υπάρχουν άπειροι θετικοί ακέραιοι $ n$ ώστε $ p \equal{} nr$, όπου $ p$ και $ r$ είναι αντίστοιχα η ημιπερίμετρος και η ακτίνα του εγγεγραμμένου κύκλου τριγώνου με πλευρές με ακέραια μήκη.

[b]2000 N6 (ROM):[/b] Να δειχθεί ότι το σύνολο των ακεραίων που δεν μπορούν να γραφτούν σαν άθροισμα διαφορετικών τετραγώνων είναι πεπερασμένο.

Μ' άρεσε το Ν1 αρκετά οπότε εδώ είναι μια λύση μου:

Προφανώς, η συνθήκη μας δίνει ότι για κάθε ακέραιο $ a$, με $ (a,n)\equal{}1$, έχουμε ότι $ a^2\equal{}1(modn)$. Έστω τώρα $ n\equal{}2^\alpha \beta$, όπου $ \alpha$ ακέραιος $ \geq0$ και $ \beta$ περιττός ακέραιος $ \geq1$. Αν έχουμε $ \alpha\equal{}0$ τότε προφανώς ο $ n$ περιττός και άρα $ (n,2)\equal{}1$, οπότε παίρνουμε $ 2^2\equal{}1(modn)\implies n|3$ και επειδή $ n\geq2$ παίρνουμε $ n\equal{}3$. Εύκολα βλέπουμε ότι επαληθεύει τα ζητούμενα (στην επαλήθευση πρέπει να θυμόμαστε ότι ο αντίστροφος είναι μοναδικός $ (modn)$).
Έστω τώρα $ \alpha\geq1$. Τότε έχουμε ότι $ (\beta\plus{}2,n)\equal{}1$ και άρα $ (\beta\plus{}2)^2\equal{}1(modn) \implies (\beta\plus{}2)^2\equal{}1(mod\beta)\implies \beta|3$ άρα $ \beta\equal{}3$ ή $ \beta\equal{}1$. Αν $ \beta\equal{}1$ τότε $ n\equal{}2^\alpha$, και παίρνοντας $ a\equal{}3$ προκύπτει ότι $ 3^2\equal{}1(mod2^\alpha)\implies2^\alpha|8$ άρα $ \alpha\equal{}1,2,3$. Με μια μικρή επαλήθευση προκύπτει ότι κάθε μία από αυτές τις τιμές είναι δεκτή.
Τώρα η περίπτωση $ \beta\equal{}3$ είναι όμοια και παίρνουμε $ a\equal{}5$ που συνεπάγεται πάλι $ \alpha\equal{}1,2,3$. Είτε με απευθείας επαλήθευση είτε με συνδυασμό των παραπάνω περιπτώσεων παίρνουμε ότι κάθε περίπτωση είναι δεκτή.
Άρα λοιπόν $ n\equal{}2,3,4,6,8,12,24$.

Ηλίας

Λοιπόν για το Ν2 βρήκα μία λύση αλλά έχει αρκετή περιπτωσιολογία:

Καταρχάς βλέπουμε ότι $ n>1$. Στη συνέχεια επίσης βλέπουμε ότι $ n\equal{}2m^3$ όπου $ m\geq1$ ακέραιος, και συγκεκριμένα μπορούμε να θεωρήσουμε ότι $ n\equal{}2^{3k\plus{}1}p_1^{3a_1}...p_s^{3a_s}$ όπου $ p_i$ περιττοί πρώτοι και $ k,a_i$ ακέραιοι $ \geq 0$. Έχουμε επίσης ότι $ d(n)\equal{}(3k\plus{}2)(3a_1\plus{}1)...(3a_s\plus{}1)$ αλλά και λόγω της δοσμένης παίρνουμε $ d(n)\equal{}2^{k\plus{}1}p_1^{a_1}...p_s^{a_s}$.
Άρα $ (3k\plus{}2)(3a_1\plus{}1)...(3a_s\plus{}1)\equal{}2^{k\plus{}1}p_1^{a_1}...p_s^{a_s}$. Από αυτή τη σχέση έχουμε ότι $ p_i>3$. Θα δουλέψουμε λίγο μερικές ανισότητες:
Μπορούμε να δούμε ότι για $ k>2$ έχουμε $ 2^{k\plus{}1}>3k\plus{}2$, και εφόσον $ p_i\geq5$ έχουμε επίσης ότι $ a_i>0 \implies p_i^{a_i}>3a_i\plus{}1$, άρα για $ k>2$ δεν έχουμε λύσεις.
Τώρα, αν $ k\equal{}0$ παίρνουμε μετά από την απλοποίηση $ (3a_1\plus{}1)....(3a_s\plus{}1)\equal{}p_1^{a_1}...p_s^{a_s}$ που επαληθεύεται μόνο για $ a_i\equal{}0$, και άρα παίρνουμε την $ n\equal{}2$ ως λύση.
Αν $ k\equal{}1$, τότε $ 5(3a_1\plus{}1)....(3a_s\plus{}1)\equal{}4p_1^{a_1}...p_s^{a_s}$ οπότε ένας από τους $ p_i$, έστω ο $ p_1$ ισούται με 5 και $ a_1>0$. Αν όμως $ a_1>1$ τότε $ 4p_1^{a_1}>5(3a_1\plus{}1)$ και σύμφωνα με τα παραπάνω δεν παίρνουμε λύση. Άρα $ a_1\equal{}1$ και μετά από την απλοποίηση $ (3a_2\plus{}1)....(3a_s\plus{}1)\equal{}p_2^{a_2}...p_s^{a_s}$ , οπότε $ a_i\equal{}0$ για κάθε $ i>1$ και παίρνουμε τη λύση $ n\equal{}2^4 5^3\equal{}2000$.
Αν $ k\equal{}2$ τότε έχουμε $ (3a_1\plus{}1)....(3a_s\plus{}1)\equal{}p_1^{a_1}...p_s^{a_s}$ οπότε πάλι $ a_i\equal{}0$, άρα $ n\equal{}2^7\equal{}128$.
Άρα οι μόνες λύσεις είναι οι $ n\equal{}2$, $ n\equal{}128$ και $ n\equal{}2000$.

Σημείωση: είναι αρκετά αργά οπότε ελπίζω να μη μου έχει ξεφύγει κάτι.
Για το Ν1 ενδιαφέρον θα είχε το αν θα μπορούσαμε να παρουσιάσουμε μια λύση με βάση το πόσες λύσεις έχει η $ a^2\equiv1modn$ και το ποιο είναι το $ \phi (n)$, αν και νομίζω ότι αυτή η λύση θα ήταν αρκετά πιο δύσκαμπτη καθώς θα δούλευε πολύ με την μοναδική παραγοντοποίηση.

Mια λυση για το τριτο.
Eλπιζω να μην εκανα πουθενα λαθος.Αν δεν ειμαι σωστος καπου πειτε το μου.
Λοιπον,
ειναι γνωστο οτι οι αριθμοι $ n \equal{} 3^k$ διαρουν τους $ 2^n \plus{} 1$(επαγωγικα)
Τωρα θεωρωντας k=2 εχουμε οτι περα απο τον 3 τον $ 2^{3^2} \plus{} 1$ τον διαιρει και ο 19.
ετσι θεωρωντας σαν $ x_1 \equal{} 3^2*19$ εχουμε εναν αριθμο με δυο πρωτους διαρετες που διαιρει τον $ 2^{3^2} \plus{} 1$ και αρα και τον
$ 2^{x_1} \plus{} 1$.
Tωρα ομως παρατηρουμε οτι αφου ο 19 ειναι πρωτος , δεν διαιρει τον $ 2^{19} \plus{} 1$ (eulers' theorem).
Επομενως για να βρουμε εναν αλλον πρωτο διαιρετη εκτος του $ 19$ και του $ 3$ του $ 2^{19} \plus{} 1$  αρκει να δειξουμε πως ο $ 2^{19} \plus{} 1$, δεν ειναι δυναμη του 3.
Πραγματι η εξισωση $ 3^m \minus{} 2^l \equal{} 1$ εχει μονες λυσεις τις (1,1) και (2,3), και ο 19 προφανως ειναι μεγαλυτερος του 3.
Επομενως αν θεωρησουμε τον πρωτο διαρετη p_1 του 2^{19}+1, ακολουθωντας την ιδια πορεια σκεψης(δλδ θεωρωντας $ x_2 \equal{} x_1*p_1$ ) μπορουμε να προσδιορισουμε για εναν οποιοδηποτε αριθμο y ,εναν αριθμο n με y πρωτους διαρετες που διαιρει τον $ 2^n \plus{} 1$.
Ετσι θεωρωντας y= 2000 παιρνουμε το ζητουμενο.

P.S.
1) Ηλια πολυ ωραια η προσεγγιση στο πρωτο. το δευτερο δεν το εχω δει ακομα. Που εχεις χαθει ομως ρε συ; Παρε κανα τηλ για το σαββατο!
2)Η ασκηση Ν3 αρχικα μου θυμισε ενα λημμα σχετικα γνωστο του Reuter που μας εξασφαλιζει την υπαρξη απειρων n για τους οποιους ο n διαρει τον 2^n+1, αλλα δεν μπορεσα να καταληξω σε λυση με βαση αυτο, και προσπαθησα την ασκηση κατασκευαστικα.
εχει κανεις καποια ιδεα για πως να το χρησιμοποιησουμε για την λυση του;
Το λημμα λεει,
Αν ο a ειναι φυσικος(οχι 0) και ο a+1 δεν ειναι δυναμη του 2 τοτε υπαρχουν απειροι n που να διαρειτουν τον a^n+1.

Ηλία (Σαλονικιέ): Σωστές οι λύσεις στα Ν1 και Ν2. Και η επίσημη λύση του Ν2 είναι σχεδόν η ίδια με την δική σου.

Ηλία (Αθηναίε): Νομίζω χρειάζεται λίγο περισσότερη προσοχή στην επαγωγή σου. Μπορείς να βάλεις λίγες περισσότερες λεπτομέρειες;

Λιάκο νομίζω η προσέγγιση σου χαλάει στο τέλος (γιατί όταν θεωρήσεις τον $ x_2\equal{}x_1p_1$ μετά δεν κοιτάς την εξίσωση $ 3^x\minus{}2^y\equal{}1$ αλλά μπλέκει μέσα και ο $ p_1$ και δεν είναι απαραίτητο ότι θα έχεις τέλεια δύναμη κτλ). Νομίζω μάλιστα ότι η ουσία όλου του προβλήματος βρίσκεται σ' αυτό το σημείο (στο γιατί υπάρχει όντως τέτοιος πρώτος)

Οπότε η λογική έχει ως εξής: θα αποδείξουμε πιο γενικά ότι μπορούμε να βρούμε $ n$ με πλήθος πρώτων διαιρετών $ k$, όπου $ k$ οποιοσδήποτε θετικός ακέραιος, τέτοιον ώστε $ n|2^n\plus{}1$. Για $ k\equal{}1$ παίρνουμε $ n_1\equal{}3^2$ (το γιατί 9 και όχι 3 προκύπτει από το τι θέλουμε στη συνέχεια -- το 3 ειναι η εξαίρεση του Zsigmondy). Ο $ 2^9\plus{}1$ έχει σίγουρα πρώτο διαιρέτη μεγαλύτερο του 3, έστω $ p_1$ (ο οποίος είναι το 19, αλλά δε μας νοιάζει αυτό) τέτοιον ώστε $ p_1|2^9\plus{}1$ αλλά $ p_1$ δε διαιρεί $ 2^k\plus{}1$ με $ 0<k<9$ σύμφωνα με το θεώρημα Ζsigmondy. Από αυτό προκύπτει ότι $ p_1>3$ Οπότε θεωρώ τον $ n_2\equal{}n_1*p_1$. Έχω $ n_2|2^{n_2}\plus{}1$ αφού $ 2^{n_2}\equal{}(2^{n_1})^{p_1} \equiv(\minus{}1)^{p_1}\equiv\minus{}1(modn_1)$ και $ 2^{n_2}\equal{}(2^{p_1})^{n_1} \equiv2^{n_2}\equiv\minus{}1(modp_1)$ (έχω $ (n_1,p_1)\equal{}1$ και $ p_1>2$)
Οπότε βρέθηκε και $ n$ με 2 πρώτους διαιρετες που να επαληθεύει τα ζητούμενα. Τώρα η επαγωγή μας.
Έστω λοιπόν ότι ο $ n_k\equal{}3^2p_1...p_{k\minus{}1}$ που κατασκευάζεται όπως προηγουμένως έχει $ k\geq2$ πρώτους διαιρέτες. Τότε έχουμε ότι ο $ 2^{n_k}\plus{}1$ έχει primitive prime factor $ p_k$ από Zsigmondy και πάλι (αφού $ n_k>3$). Αυτό που ισχυρίζομαι είναι ότι
1. ο $ p_k$ δε διαιρεί τον $ n_k$ και
2. ο $ n_{k\plus{}1}\equal{}n_k*p_k$ διαιρεί τον $ 2^{n_{k\plus{}1}}\plus{}1$ (και προφανώς έχει $ k\plus{}1$ πρώτους διαιρέτες)

1. Αν $ p_k|n_k$ τότε $ p_k\equal{}3$ ή $ p_k\equal{}p_i$ όπου $ 1\leq i\leq k\minus{}1$
Αν συμβαίνει το πρώτο ή το δεύτερο με $ 1\leq i\leq k\minus{}2$, τότε $ p_k|n_{k\minus{}1}|2^{n_{k\minus{}1}}\plus{}1$ άτοπο από την επιλογή του $ p_k$ ως primitive prime factor. Άρα $ p_k\equal{}p_{k\minus{}1}$. Όμως $ p_{k\minus{}1}|2^{n_{k\minus{}1}p_{k\minus{}1}}\plus{}1 \implies p_{k\minus{}1}|2^{n_{k\minus{}1}}\plus{}1$ άρα $ p_{k}|2^{n_{k\minus{}1}}\plus{}1$ άτοπο πάλι λόγω της επιλογής μας.
2. Αρκεί δηλαδή να δείξουμε ότι $ p_k|2^{n_{k\plus{}1}}\plus{}1$ και $ n_k|2^{n_{k\plus{}1}}\plus{}1$. Όμως $ 2^{n_{k\plus{}1}}\plus{}1\equal{}2^{n_k p_k}\plus{}1\equiv2^{n_k}\plus{}1\equiv0(modp_k)$ και $ 2^{n_{k\plus{}1}}\plus{}1\equal{}2^{n_k p_k}\plus{}1\equiv(\minus{}1)^{p_k}\plus{}1\equiv0(modn_k)$ αφού προφανώς $ p_k$ περιττός.

Άρα κατασκευάσαμε έναν αριθμό $ n$ με $ k$ πρώτους διαιρέτες όπου $ k$ θετικός ακέραιος και άρα η απάντηση στο πρόβλημα μας είναι καταφατική (για $ k\equal{}2000$).

Και μιας και πιάσαμε τον Zsigmondy ας δώσουμε και μια λύση στο N4.
Υπενθυμίζω το τι μας λέει ο Zsigmondy στην μορφή που τον θέλουμε:
Αν $ (a,b)\equal{}1$ με $ a>b\geq1$ τότε ο $ a^n\plus{}b^n$ έχει primitive prime factor (δηλαδή πρώτο $ p$ τέτοιον ώστε $ p|a^n\plus{}b^n$ και $ p$ δε διαιρεί τον $ a^k\plus{}b^k$ για κανέναν $ 1\leq k<n$), πλην της εξαίρεσης $ a\equal{}2,b\equal{}1,m\equal{}3$
Άρα επειδή $ b\equal{}1$ έχουμε $ (a,b)\equal{}1$, ξεκινάμε βλέποντας την περίπτωση $ a\equal{}1$. Προφανώς για κάθε $ m,n$ έχουμε ότι $ 2|2^n$, άρα οι τριάδες $ (a,m,n)\equal{}(1,m,n)$ όπου $ m,n$ θετικοί ακέραιοι είναι λύσεις.
Έστω τώρα ότι $ a\equal{}2,m\equal{}3$ τότε προφανώς $ 9|3^n$ για κάθε $ n\geq2$ άρα έχουμε τις τριάδες $ (a,m,n)\equal{}(2,3,n)$ με $ n$ ακέραιο $ \geq2$.
Αν έχουμε $ m\equal{}1$ τότε προφανώς $ a\plus{}1|(a\plus{}1)^n$ για κάθε $ a,n$ άρα έχουμε τις λύσεις $ (a,m,n)\equal{}(a,1,n)$ με $ a,n$ θετικούς ακεραίους.
(Όλες αυτές οι περιπτώσεις πάρθηκαν γιατί δεν καλύπτονται από τον Zsigmondy)
Άρα λοιπόν ας υποθέσουμε ότι οι προϋποθέσεις τηρούνται, άρα υπάρχει πρώτος $ p|a^m\plus{}1$ που δε διαιρεί τον $ a^k\plus{}1$ για $ k<m$. Τότε, παίρνοντας $ k\equal{}1$ παίρνουμε ότι $ p$ δε διαιρεί τον $ a\plus{}1$ άρα ούτε τον $ (a\plus{}1)^n$ και συνεπώς ούτε και το $ a^m\plus{}1$ δε διαιρεί τον $ (a\plus{}1)^n$, άρα δεν έχουμε άλλες λύσεις.
Άρα όλες οι λύσεις είναι οι $ (a,m,n)\equal{}(1,m,n)$ όπου $ m,n$ θετικοί ακέραιοι, $ (a,m,n)\equal{}(2,3,n)$ με $ n$ ακέραιο $ \geq2$ και $ (a,m,n)\equal{}(a,1,n)$ με $ a,n$ θετικούς ακεραίους.


Let G be a convex polygon with 2000 sides and P an interior point which does not lie on any diagonal of G.

Prove that P  lies in the interior of an even number of triangles formed using the vertices of G.


Probably it is obvious,but the statement "$ f$ convex and derivable then $ f^\prime$ is increasing" holds if $ f$ is only one time derivable?


vector
linear algebra
matrix
algebra
system of equations

This problem was confusing me so any help would be greatly appreciated!!

(Note that E stands for epsilon and "dot" implies the dot product symbol)

Let V be a subspace of R^n with dim(v) = n - 1. (Such a subspace is called a hyperplane in R^n.) Prove that there is a nonzero x E R^n such that
V = {v E R^n | x dot v = 0}.

Hint:  Do so by setting up a homogeneous system of equations whose coefficient matrix has a basis for V as its rows. Then notice that this (n-1) by n system has at least one nontrivial solution, say x.)

In your post $E$ stands for inclusion $\in$ rather than epsilon $\epsilon$.

The space $V$ has an orthogonal basis, say $v_{1},\dots,v_{n-1}$. Let $v$ be a vector in $\mathbb R^{n}$ that does not belong to $V$. Define $x=v-\sum_{k=1}^{n-1}\frac{v\cdot v_{k}}{\parallel v_{k}\parallel ^{2}}v_{k}$. This $x$ works.

We don't need the trouble of finding an orthogonal basis. Let $v_{1},v_{2},\dots v_{n-1}$ be any basis for $V.$ Create an $n\times (n-1)$ matrix $B$ whose columns are the $v_{i}.$ Note that the rank of $B$ is $n-1.$

Now, consider the homogeneous equation $B^{T}x=0.$ By the rank-nullity theorem, this has a one-dimensional solution space. Let $u$ be a nontrivial member of that solution space.

Examining the matrix muliplication, we see that for each $i,\ v_{i}^{T}u=0,$ which we rewrite as $u^{T}v_{i}=0.$ Since the $v_{i}$ span $V,$ we have that for every $x\in V,\ u^{T}x=0.$ By another use of the rank-nullity theorem, we see that $\{x\in\mathbb{R}^{n}: u^{T}x=0\}$ has dimension $n-1,$ so we conclude that $\{x\in\mathbb{R}^{n}: u^{T}x=0\}=V.$

The vector $u$ is unique up to being multiplied by a scalar.

This is what was in the hint muckymotter provided.


function
logarithms
algebra
domain
complex analysis

1. Suppose $ f(z) \equal{} u \plus{} iv$ is analytic.  Under what circumstances will $ g(z) \equal{} u \minus{} iv$ be analytic?

2. Consider an analytic function $ f(z) \equal{} u \plus{} iv$ whose modulus $ |f(z)|$ is equal to a constant $ k$ throughout some domain.  Prove that this can occur only if $ f(z)$ is constant throughout the domain.

1) Use the CR equations.

2) Consider $ \log f(z)$

I'm not sure what exactly domain means, but wikipedia tells me it indicates an open and connected region. It does not necessarily need to be simply connected, apparently, so I'm not sure that you can write down a well-defined meaning for $ \log f$ on an arbitrary domain. So you need to take care to break up the original domain into balls (simply-connected neighborhoods) in which case yes, everything is happy, and then patch these together.


MATHCOUNTS geometry articles AMC AIME calculus Stanford

I wonder what happened to prominent Mathcounts participants... did some of them become mathematicians themselves, like some IMO medalists?

Or their fates are like that of some other IMO medalists, burning out after either middle or high school without getting anywhere useful?

Or their lives reverted back to normalcy once in high school?

MATHCOUNTS doesn't correlate to IMO too well. One might be amazing at MATHCOUNTS, but if they can't think hard, then they might not be so great at the IMO. But there is a slight relation in that those who are good at MATHCOUNTS have usually worked hard, and are willing to keep working hard (this is often difficult though, since in high school, there's usually less time to do math).

[b]Edit:[/b] So I didn't read the question too well--I more or less skimmed the first sentence and thought something was wrong with it, prompting my above comments. In truth, I really don't know--but I have a feeling that contest math doesn't correlate too well with "real math," either.

I know this guy who won state and he was a total beast and he just lived up to be awesome... after mc, he came and taught mc at his school to earn his hours. he went on to be awesome and won like every amc comp at his school..  I think it's a good idea teaching mc after ur done with it. dat wat im doing...   :lol:

Like math154 said, MATHCOUNTS is very basic compared to other competitions.

However, some people like Art of Problem Solving's Matthew Crawford and Richard Rusczyk came to be very successful and recognized.

However, never is there anything "not useful". Everyone is useful in some way and even a minute of mathematics may prove advantageous later in life.

Finally, there's nothing as "reverting to normalcy" also. You'll always have an edge over normal people in school if you were a MATHCOUNTS veteran :wink:.

[quote="akhil0422"]Like math154 said, MATHCOUNTS is very basic compared to other competitions.[/quote]

Well...it's easier than competitions that are at a higher level, yes, and it's based more on speed than anything else(in my opinion, at least.)

[quote="akhil0422"]
You'll always have an edge over normal people in school if you were a MATHCOUNTS veteran :wink:.[/quote]

I disagree. MATHCOUNTS is fun and great, but it's not real math. Also, lots of people after MATHCOUNTS grow up to have completely non-math-related careers and aren't so good as other people. That's also based on the assumption that you go to an average, not-so-good school.

Not going anywhere useful would then mean not going anywhere, period.

What I meant by "reverting to normalcy" is to live a life as normal as possible for a MATHCOUNTS veteran, which is not always easy, especially when you tasted either state or national rounds.

I disagree with [b]you[/b], izzy.

I did MATHCOUNTS for only one year in sixth grade and did not even take it seriously (much)  :blush:.

Now, I'm taking Algebra I in an "average, not-so-good" school and despite being a seventh grader, am doing clearly better and know [b]all[/b] the concepts already!

I also know many geometry concepts.

MATHCOUNTS clearly helps you a lot in high school; when we're old enough to get a job, a better foundation won't hurt.

[quote="akhil0422"]
Now, I'm taking Algebra I in an "average, not-so-good" school and despite being a seventh grader, am doing clearly better and know [b]all[/b] the concepts already!
[/quote]

So was I before MATHCOUNTS, in 6th grade, and I wasn't(still aren't) gifted or anything.  :maybe:

[quote="akhil0422"]
MATHCOUNTS clearly helps you a lot in high school; when we're old enough to get a job, a better foundation won't hurt.[/quote]
For some people, yes. For others, no. You have to remember that MATHCOUNTS isn't real math.

[quote="isabella2296"][quote="akhil0422"]
MATHCOUNTS clearly helps you a lot in high school; when we're old enough to get a job, a better foundation won't hurt.[/quote]
For some people, yes. For others, no. You have to remember that MATHCOUNTS isn't real math.[/quote]

I don't seem to understand what you mean? People who do MATHCOUNTS [b]already[/b] have an implied (more or less) advantage over others in math...

Let's keep this in perspective: some people were encouraged to participate in MATHCOUNTS because they were struggling in middle school math. Obviously, many participants in this category crashed and burned either at a local level or at a chapter level if they were lucky enough to make chapter, especially if one lives in a state like Oklahoma or Idaho.

Some others have a definite advantage over others in math... And do MATHCOUNTS-earned skills translate in better odds of success in high school or university math?

One Karlanna Lewis, who did non-trivially well at Mathcounts, has gone on to be a quite talented ballet dancer. I'm not sure if she counts as professional quite yet though.

Another, Neal Wu, has gone on to be possibly the USA's best high school programmer.

Noah Arbesfeld didn't win Mathcounts, but he's won some scientific [mathematical] research awards: he was an STS finalist and I believe did well at Siemens too.

One Wyoming Mathcounter also made National History Day, and so is both one of the best mathematician girls in Wyoming and one of the best historians in Wyoming.

Basically, it all depends on circumstances and personal choices.

Circumstances? I doubt that. Where did [url=http://www.societyforscience.org/sts/67sts/Delgado.asp]Eric Delgado[/url] come from? He didn't have a great school - yet he did great research. (I'm fully aware argument by example is bad argumentation; yet I dislike blaming talent on circumstances.) (I also had an article on Delgado's working conditions, but I can't find it off the top of my head. Suffice it to say: his high school has [i]never[/i] produced a top person before).

Anyways, being a MATHCOUNTS person who started MATHCOUNTS fairly late (7th grade) and starting with the AMCs and AIMEs earlier (5th grade), MATHCOUNTS in my opinion a downslope ride (where you can coast right along without exerting too much effort) from the AMC establishment.

I already stated that; MATHCOUNTS is the easiest after MOEMS (well not exactly that but you get what I mean).

I think the real lesson of MATHCOUNTS may not be math, but the process of learning. While learning math is good and all, diligence and work values are also learned. You may not remember anything about math 10 years in the future, but the mindset (striving to be the best you can be, etc.) will stay with you forever and help you succeed in life.

Perhaps this thread should be moved to The Round Table.

People who are good at mathcounts are usually some of the best students in their regular schools. Being good at Mathcounts does not necessarily mean you are the best student in math but in my experience mathcounts veterans are pretty good.

Just doing Mathcounts is not going to to define you as becoming as mathematician. Like I participated in mathcounts, but it was something I just did, for the sake of it. I really could've cared less how i did (top an extent). The most I studied for was just attending practices at school. I didn;t do anything on the side. But, now, math is a huge part of my. Now I have a  [b]passion[/b] for it, and thats what probably is going to make me peruse math in the future. I could have continued my Mathcounts study habits throughout high school and cared less about math and would have never become a mathematician (not that I am one right now..).. even though I participated in Mathcounts..... basically I trying to say that if you have a passion for math and you live/crave/desire it (maybe not that much....) then you will probably have a math career... I hope this answered your question.... cause I think it did but I dont think it did...

For most math contests, including MATHCOUNTS, the actual value of participating in one is what you learn from preparation to it, reviewing the questions afterwards and also diligence as well as work values.

Yes, Mathcounts participants do better than average in math when compared to most people. But the thing is that these people would've been better [b]regardless[/b] of them participating in it.

It similar to the college/no college debate. While people who come out of college get a higher average salary, some argue that smarter people go to college than non-college alumni. They say that these people would've still gotten a great paycheck if they hadn't gone to college.

The thing with mathcounts is that it closely relates to classroom math, and it allows you to finish tedious exercises (150 questions a day, anyone?) very quickly, because MATHCOUNTS training allows you to find a loophole or a shortcut.

I'm now in 9th grade after 4 years of Mathcounts (unofficially at chapter in 5th).

Looking back, I think the most important thing I took away was problem solving.  Not the formulas or the speed.  I think Mathcounts provides a place for people to train their problem solving abilities.

I would think that most people who do well in high school math competitions did well in Mathcounts, but not neccesarily because of.  For example, someone at black MOP last year, from my state, didn't do mathcounts, but was still highly successful.

solafide farms, not to burst your bubble, but im going to the same school that karlanna lewis went to, and she tried ballet, but it ended up being way to hard, and now she goes to a crappy COMMUNITY COLLEGE (TCC). wasted talent.

I wouldn't put down people who go to community college so quickly judging by the cost of private college institutions. And of course there are many people who go get their master's degree from a state or private college after getting a bachelor's at community college.

Community college? What kind of community college TCC is and what does it stand for?

How much higher than the average of people who didn't do MC? (I suppose it's because of the skills necessary to succeed in Mathcounts, not because of Mathcounts veterancy itself)

And better on what aspects?

[quote="diophantient"]Not only has it been shown that previous mathcounts participants have higher salaries, but it's also a fact that doing mathcounts makes you a better person altogether. You can't argue with statistics like those.[/quote] Where the heck do you get statistics like these? I wouldn't doubt the previous [b]assumption[/b], but how do you measure a good person?

What are MATHCOUNTS contestants really learning, tricks or problem solving skill/creativity?

I heard my own MV calculus teacher say that MATHCOUNTS veterans really just pick up tricks while I really say one could work problem solving skill through it.

I learned the tricks in 6th grade; after that, it was just problem solving. I don't know about others though.

I seem to recall learning the problem-solving first, then the tricks came naturally as I streamlined the problem-solving process.

As for the top students from my day, of the ten in the countdown round my last year, only one (Eric Price) ever made IMO.  Another (Jack Cackler) recently tried out for Stanford Tree.

Mathcounts is not some kind of definitive thing that will help you in LIFE.
I mean, it really doesn't matter if you did Mathcounts or not and if you did WELL in MC or not...
I think that, really, it just depends on the person. If you love math, then you'll do well in life if you CAN find a career that helps you pursue that passion. However, if you don't love math, but do well, then you might do well in the beginning but in the end, in the long run, when you look back at your life and others see how you've lived your life, have you really succeeded?
"When a man feels throbbing within him the power to do what he undertakes as well as it can possibly be done, this is happiness, this is success."
Orison Swett Marden

I've put a lot of thought into this one question:

What's the point of math competitions?

My answer now, as a high school senior, is that math competitions have taught me how to think.  I've noticed that math competitions have not only made me better at problem-solving in math and science, as you would expect, but have also made me better at seemingly unrelated things like writing, and handling everyday situations.  The intro of AoPS talks about the purpose of math competitions, and it has a similar answer to this question.  When I bought AoPS Volume 1 as a middle school student and Mathcounts participant, I had trouble swallowing this.  I was convinced the point of math competitions was that I was learning the math I would need to know for my career later in life.  It's taken me up until less than a year ago to change my mind.

Mathcounts' role in this for me was mostly to get me interested in math competitions so that I would continue to compete in them in high school.  I don't really think I ever had to really think to solve a Mathcounts problem; I either solved it right away or didn't solve it at all.  I think Mathcounts is more about knowing tricks and less about solving problems, but it's really impossible for a contest with so many problems in such a short time to be any different.

Do the benefits of MATHCOUNTS veterancy last long enough to be of any use in single-variable, let alone multi-variable calculus?

[hide="my experiences in MathCounts"][b]1.[/b] I am a MathCounts veteran [i]and[/i] an assistant coach. I just want to say that if my parents didn't find MathCounts when I was in 4th grade I wouldn't be anywhere NEAR where I am now. Dad had ambitions on starting Calculus in the 6th grade, and when my coach (Dr. O'Keeffe) found me, she directed me to this site and Dad read "The Calculus Trap". It was a good thing he did, too.
[b]2.[/b] My first year of MathCounts (4th grade), Dr. O'Keeffe was uncertain about a nine-year-old's ability to handle the types of problems that appear on MathCounts, but when I tried the School Test, she found out that I would have made the Scoring Team. One of the people that was on the Scoring Team went on to make $ 6^{th}$ place on the Written Test at the National Competition the same year. She then let me come to practices, and I quickly blossomed.
[b]3.[/b] That summer I was allowed to skip a grade because of my scores on a HomeSchool aptitude test of some sort, and I did because I wanted to compete sooner. I took the School Test and made the Scoring Team, and got all the way to $ 25^{th}$ place at the State Competition. I was also invited to take the AMC 12 by Dr. O'Keeffe, and I was $ \frac{1}{2}$ points off from taking the AIME, so she determined that I was ready to join [url=http://albanyareamathcircle.blogspot.com/]her Math Circle[/url], and I was exposed to many High-School competitions. I even went to NYSML that year.
[b]4.[/b] The year after that I made a "Quantum Leap" (quote from Dad, not from me), and I got to Nationals. I had a great time and I was exposed to tougher problems. I also made AIME and got a high score at NYSML.
[b]5.[/b] Next year I was held back solely because I and my family did not want me to graduate early. I was doing new material, though. I made Nationals again and got $ 29^{th}$ there, and won the Spirit Award. I got a high score at NYSML again, and my AIME score went up. Also I started tutoring a six-year-old named Adam that wanted to join MathCounts but couldn't stay for two hours.
[b]6.[/b] Now I am in $ 8^{th}$ grade and I am coaching many Middle-School students from a few teams, and I am loving the experience. Adam is blossoming but isn't [i]quite[/i] ready for MathCounts, and I'm hoping for a spot at USAMO. All in all, MathCounts has [i]changed my life[/i].[/hide]

@CEM: No, MathCounts experience has no use in advanced mathematics. It is merely for nurturing a love for mathematics so that you pursue it.


geometry
analytic geometry
geometric transformation
rotation
complex numbers

If you are given two points, like (a+c,b) and (c,0) which are two points  of an equilateral triangle, how can you find the third vertex? Thanks!

I am somewhat bad at geometry, but I'll give this a shot.

If you were to put this triangle on the coordinate plane with (c,0) and the unknown vertex on the x-axis, you can figure out that the y-value for the unknown vertex, which would be zero.  You can also let the b in (a+c, b) represent the height of the triangle because it will be the "middle" vertex.  This means that the x-value of the midpoint of the other vertices should equal a+c.  Now you can set-up an equation:

a+c = (c+u)/2 where u = the x value of the unknown vertex.
2a+2c = c+u
u = 2a + c

Therefore, your three vertices are (c,0), (a+c,b), and (2a+c,0).

[hide="Complex numbers"]Project onto the complex plane. Your coordinates transform to $ c$ and $ a\plus{}c\plus{}bi$. Temporarily, we translate such that $ c$ lies on the origin to get $ 0$ and $ a\plus{}bi$.

Now we multiply $ a\plus{}bi$ by $ e^{i\pi/3}$ or $ e^{\minus{}i\pi/3}$.

[b]Case 1.[/b]
$ (a\plus{}bi)\left(\frac{1}{2}\plus{}\frac{i \sqrt{3}}{2}\right)\equal{}\left(\frac{a}{2}\minus{}\frac{b\sqrt{3}}{2}\right)\plus{}\left(\frac{b}{2}\plus{}\frac{a\sqrt{3}}{2}\right)i$.

[b]Case 2.[/b]
$ (a\plus{}bi)\left(\frac{1}{2}\minus{}\frac{i \sqrt{3}}{2}\right)\equal{}\left(\frac{a}{2}\plus{}\frac{b\sqrt{3}}{2}\right)\plus{}\left(\frac{b}{2}\minus{}\frac{a\sqrt{3}}{2}\right)i$.

And our answers are $ \boxed{\left(\frac{a}{2}\minus{}\frac{b\sqrt{3}}{2}\plus{}c,\frac{b}{2}\plus{}\frac{a\sqrt{3}}{2}\right)\text{ and }\left(\frac{a}{2}\plus{}\frac{b\sqrt{3}}{2}\plus{}c,\frac{b}{2}\minus{}\frac{a\sqrt{3}}{2}\right)}$.
[/hide]

There are numerous ways this problem can be solved.
For one, you can locate the midpoint of the line segment, and find the point perpendicular to it that is $ \frac {\sqrt {3}}{2}$ times this distance away.  Of course, this would involve quite a lot of algebra.
You can also do it using symmetry.  You can find the equations of the lines connected to this third vertex, and set the distance from the two given points to the opposite lines equal (why are they equal?).
Nevertheless, the easiest method is certainly by rotations of complex numbers (i.e. the solution grn_trtle gave).  This is also the method that uses a considerably less amount of algebra.

JoshJ, your solution doesn't give the right values because your $ x$ coordinate is incorrect.  In order for the $ x$-coordinate to be what you said, the two points would have to be on the same line $ y \equal{} k$.  In this situation, this is clearly not true.  Also, you need two solutions since the third vertex can be located on either side of the line connecting the two given points (although this is easily done using reflection).


geometry
incenter
circumcircle
geometric transformation
reflection
geometry unsolved

I found this problem in the site of Hong Kong. There is a collection with proposed problems which can be solved with inversion. I post one of them. Any other solution is also welcome.

Very nice problem, Stergiu!

Denote $K$ the point where the semicircle cuts the circle $O$, and $T$ the tangenty point where $AB$ cuts the circle $O$.

Let's extend the semircircle and denote $D$ the intersection of $AB$ with the circle $H$.

Thus, Because $AD$ is tangent to the circle $O$, which is also tangent to the circle $H$, then $KT$ is bisects the angle $AKD$. Therefore $Q$ is the midpoint of the arc $\widehat{AD}$.

Now because $\angle{PBT}=\angle{PKT}=90$, we have that the quadrilateral $PBTK$ is cyclic, therefore $QB \cdot QP=QT \cdot QK=QC^{2}$ (the power of point $A$, w.r.t to circle $O$)

But in the right triangle $APQ$, $AB$ is the altitude, therefore $AQ^{2}=QB \cdot QP$, therefore from the above equality we have $QC=QA$.

Now the problem is dead, because let's note $\angle{BAC}=x$, therefore $\angle{ACB}=90-x$, but from $QC=QA$, we have also $\angle{CAQ}=90-x$, therefore $\angle{BAQ}=90-2x$, so $\angle{BQA}=2x$, but $\angle{PAB}=\angle{BQA}$, therefore $\angle{PAC}=2x-x=x$, so $AC$ bisects $\angle{PAB}$, and thus the problem in solved!

Another way for solving the problem from where we found out that $QC=QA$, is just considering the circle $O'$ tangent to circle $H$, $BD$, and $BP$, so the construction is basicly simetric, therefore $CQ=AQ=DQ$, and $PB$ is the bisector of $\angle{APD}$ in triangle $APD$, therefore $C$ is the incenter of triangle $\triangle{APD}$, so $AC$ bisects $\angle{PAB}$.

[u]Generalization:[/u]
Let $ABC$ be a triangle, and $D \in (BC)$ such that $\angle{BAD}=\angle{ACB}$, and consider the circle $H$ tangent to the circumcircle of $ABC$ at $K$, to $BD$ at $P$, and to $AD$ at $P'$. Show that $AP$ is the bisector of $\angle{BAD}$.

[u][i]Another enuntiation of the generalization:[/i][/u] (***) - own
In a triangle $ABC$, denote it's circumcircle $w$, and consider the Thebault circles $k_{1}, k_{2}$ corresponding to the bisector $AD$ of $\angle{BAC}$. Prove that $k_{1}$ and $k_{2}$ are tangent at $I$, where $I$ is the incircle of $ABC$.

Here's a link: http://www.mathlinks.ro/Forum/viewtopic.php?t=123139

I guess that now you got your geometric solution, Mario!  :)

Yeah :D  Finally after a long time, but now I wonder how can it be solved with inversion :?:

[color=darkred][b]Very nice the Pohoatza's generalization of this cool problem ![/b][/color]

[color=darkblue][b][u]Proof.[/u][/b] Suppose w.l.o.g. that $B\in (OP)$ and $C\in (PB)$. Denote : the circles $\mathcal H=C(O,R)$ and $\mathcal O=C(I,r)$ ;

the reflection $A'$ of the point $A$ w.r.t. the diameter $[PQ]$ and the incenter $S$ of the $P$- isosceles triangle $APA'$.

Prove easily that $QC=QA=R+\sqrt{R(R-2r)}$ $\implies$ $C\equiv S$, i.e. the ray $[AC$ bisects the angle $\widehat{PAB}$.

Indeed, $OC^{2}=OI^{2}-IC^{2}=$ $(R-r)^{2}-r^{2}\implies$ $\left\{\begin{array}{c}OC=\sqrt{R(R-2r)}\\\\ QC=QO+OC\end{array}\right\|$ $\implies$ $\left\{\begin{array}{c}\boxed{QC=R+\sqrt{R(R-2r)}}\\\\ QB=QC-BC\end{array}\right\|$ $\implies$

$\left\{\begin{array}{c}QB=R-r+\sqrt{R(R-2r)}\\\\ PQ=2R\ ,\ QA^{2}=PQ\cdot QB\end{array}\right\|$ $\implies$ $QA^{2}=$ $R^{2}+2R\sqrt{R(R-2r)}+\left(\sqrt{R(R-2r)}\right)^{2}$ $\implies$ $\boxed{QA=R+\sqrt{R(R-2r)}}$. [/color]

This is wanted solution with inversion.

We use inversion with center $C$. Then circle $O$ maps onto a line
$O^{*}$ parallel with $PQ$, and semicircle $H$ maps into a semi circle $H^{*}$ with center
on line $PQ$, and diameter $P^{*}Q^{*}$ which touches $O^{*}$. $A^{*}$ is intersection of the
line $CA$ and $H^{*}$. Line $AB$ maps into a circle that goes through
$C$ and $A^{*}$, with center on $CB$ (because $CB\bot AB$) and
touches line $O^{*}$. $B^{*}$ is the second intersection point of that
circle and $PQ$. Now, we have two equal circles $(AB)^{*}$ and $H^{*}$
(because they have their centers on the same line, and they touch
the line parallel with the line on which their centers are), and
their intersection is $A^{*}$, so $A^{*}P^{*}=A^{*}B^{*}$, in other words $\angle A^{*}P^{*}C=\angle CB^{*}A^{*}$. But $\angle CP^{*}A^{*}=\angle CAP$ and $\angle CB^{*}A^{*}=\angle CAB$ so this is the end of a proof.

Please ask if something isn't quite clear, I'm not so experienced
in writing solutions with inversion so I don't know in how many
details I should go.

Bye

I had forgotten this problem :( . In my archiev I have a synthetiacal solution. I think I have sent it to this forum in the past. It is different  from those you sent. I had not a solution with inversion.Now I feel happy because we have many solutions to the problem.
 [b]Thank you[/b] all for your nice solutions. I'll try to send the next problem from this collection.

 [u] Babis[/u]

This can solved with a lot of computations!

One friend say to try with inversion. I see one nice and easy solution, basically the same of SpongeBob (i think that also the same of the book 'Mathematical Olympiads Around The World, 1995/06')

[quote="April"][quote][color=darkred]$AB$ is a half chord perpendicular to the diameter $PQ$ of the semicircle $C(O,\,R).$ A circle $C(O',\,R')$ is inscribed, toucher $PQ$ at $C.$ Prove that: $AC$ is angle bisector of $\angle PAB$[/color][/quote]
[color=indigo][b]Proof.[/b] We have:
$OO'=OC^{2}+O'C^{2}\Longrightarrow (R-R')^{2}=(QB+R'-R)^{2}+R'^{2}\\ \Longrightarrow 2\cdot QB\cdot R=(QB+R')^{2}\Longrightarrow QB\cdot QQ=(QB+BC)^{2}\Longrightarrow QA^{2}=QC^{2}\\ \Longrightarrow QA=QC\Longrightarrow \angle QCA=\angle QAC\Longrightarrow \angle CAB=\angle PAC$
[/color][/quote]

April ,

 thank you . I did not know this simple and nice solution. Just Pythagorian theorem and ... done.Very nice!

[u] Babis[/u]


Hey mates ..
i got a question which i hope you'll help me with :)
the question is :

Do there exist a formula  for this sum :
$ {1^x} \plus{} {2^x} \plus{} ... \plus{} {n^x}$
$ x,n \in {\mathbb{Z}^ \plus{} }$
and please if you have any other formulas related to powers post them here ..
thanks in advance :)