quadratics
number theory unsolved
number theory

[quote="rohitsingh0812"]Let $p$ be an odd primeand $a,b$ be prime to $p$( all positive integers).If $x^2 \equiv a(mod$ $p)$ and $x^2 \equiv b(mod$ $p)$ both congruences have no solution then prove that the congruence
$x^2 \equiv ab(mod$ $p)$ always has a solution.[/quote]
Use property of Legandr's symbol:
$(\frac{ab}{p})=(\frac{a}{p})(\frac{b}{p})=(-1)*(-1)=1$

This property of the Legendre symbol is just another way of writing the statment, not a proof!

A standard proof would be to see that we have a primitive root $\zeta \mod p$ and that the squares are exactly the even powers of $\zeta$.


ARML probability geometry trigonometry calculus integration algebra

1.
When the sum of the first $k$ terms of the series

\[ 1^2 + 2^2 + 3^2 + \cdots + n^2 + \cdots  \]
is subtracted from the sum of the first $k$ terms of the series

\[ 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + n(n+1) + \cdots,  \]
the result is 210.  Find the numerical value of $k$.

2.
Three coplanar circles are concentric, and the lengths of their respective radii are 10, 20, and 30.  A square is circumscribed about the smallest circle, and squares are inscribed in the other two circles.  Find the length of the shortest line segment that could contain a point on each square.

3.
Find the numerical value of $k$ for which

\[ \frac{7}{x+y} = \frac{k}{x+z} = \frac{11}{z-y}.  \]

4.
Spender Sam always carried only dollars and pennies, but he never carried more than 99 pennies.  When leaving home one day, Sam noted that the number of dollars he had was 150% of the number of pennies he had.  After spending 10 dollars and 78 cents, Sam discovered that the number of dollars he had remaining was 1 less than the number of pennies he then had.  With how much money did Sam leave home that day?

5.
Two cards are randomly selected from a standard 52 card deck of playing cards, without replacement.  Find the probability that at least on king and at least one heart are selected (though not necessarily on different cards).

6.
Find the ordered pair of positive integers $(a, b)$, with $a<b$, for which
\[ \sqrt{1+\sqrt{21+12\sqrt{3}}} = \sqrt{a} + \sqrt{b}.  \]

7.
The length of the base of an isosceles triangle is 20.  In the plane of the triangle, a point 4 units from this base is 10 units from each leg.  Find the area of the triangle.

8.
If Arc denotes the principle value, find all values of $x$ which satisfy

\[ \sin{(4\text{Arc}\tan{x})} = \frac{24}{25}.  \]

[hide="8"]
$x=\tan\left(\frac{\theta}{4}\right)$, where $\theta=\angle BAC$ in $\triangle ABC$ with $\angle C = 90^\circ$ and $AB=25,BC=24,AC=7$.

$\tan\left(\frac{\theta}{4}\right) = \frac{\sin (\theta/4)}{\cos (\theta/4)}$

$= \frac{1-\cos (\theta/2)}{\sin (\theta/2)}$

$=\frac{1-\sqrt{\frac{1+\cos \theta}{2}}}{\sqrt{\frac{1-\cos\theta}{2}}}$

$\cos \theta = \frac{7}{25}$

$\Rightarrow x = \frac{1-\frac{4}{5}}{\frac{3}{5}}=\boxed {\frac{1}{3}}$
[/hide]

Diagram for 7:
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=816[/img]

I was going to post a solution too, but then I realized that mine was wrong.

[quote="MCrawford"]6.
Find the ordered pair of positive integers $(a, b)$, with $a<b$, for which
\[ \sqrt{1+\sqrt{21+12\sqrt{3}}} = \sqrt{a} + \sqrt{b}.  \]
[/quote]

[hide]Let $\sqrt{21+12\sqrt{3}}=x+y\sqrt{3}$.

$21+12\sqrt{3}=x^2+3y^2+2xy\sqrt{3}$

$21=x^2+3y^2$
$6=xy$

$\therefore (x,y)=(3,2)$

Let $\sqrt{1+3+2\sqrt{3}}=m+n\sqrt{3}$

$4+2\sqrt{3}=m^2+3n^2+mn\sqrt{3}$

$\therefore (m,n)=(1,1)$

$\Rightarrow \sqrt{a}+\sqrt{b}=1+\sqrt{3}$

$\therefore (a,b)=\boxed {(1,3)}$[/hide]

[hide="5"]

If we say that the king of hearts is not picked, then there are 3 kings and 12 hearts we could choose from, for 36 possibilities.  If the king of hearts is picked, there are 51 possibilities for the other card. So 36+51=87 ways to draw a king and a heart.  $\binom{52}{2}=1326$ total possible draws, so the probability is $\frac{87}{1326}=\frac{29}{442}$

    [/hide]

[hide="4"]

There are two possibilities: either Sam had to exchange a dollar for 100 pennies, or he didn't. If he did, we get the system of equations

$d=\frac{3}{2}p$
$d-10=100-(78-p)$

from which we get $p=64$ and $d=96$.  The other case gives non-integral answers, so Sam brought 96 dollars and 64 cents with him.

[/hide]

[hide="3"]

WLOG, say that each fraction is equal to 1.  Then $x+y=7$ and $z-y=11$.  We can take $x=3, y=4, z=15$, which satisfy the equations.  We have $k=x+z=18$.

[/hide]

[hide="1"]

We see that $n(n+1)-n^2=n$, so the sum of the first $k$ terms of the difference of the series is the $k$th triangular number.  So we have $\frac{k(k+1)}{2}=210\Rightarrow k=20$.

[/hide]


[geogebra]c224cdee983ab21b73c88e18b310020bd68e9189[/geogebra]

KNown:
Trapazoid ABCD
AB = 6
CD = 8
AC/BD = 2
AC parallel to EF
BD parallel to EG

Find:
Length of HE

How is $ E$ defined? Is $ E$ the midpoint of $ AB$?

$ E$ has to be defined in some way, else the result would vary.
(imagine sliding the triangle to the left or to the right)

Well you know that AECF and BDGE are parallelograms so E I guess is the midpt

Uh, no. Slide $ \triangle FEG$ along $ CD$ and you still preserve the parallelograms while moving $ E$.

Yep, impossible with just the given information. Unless it's supposed to be in terms of the other sides?

Ok, lets say that is E is the midpoint...

$ ACB \sim  EHB$

and $ AB \equal{} 6$ while $ EB \equal{} 3$

thus $ EH \equal{}  1$


Bekah has three brass house number digits: 2, 3 and 5. How
many distinct numbers can she form using one or more of the
digits?

[b]Case 1[/b]: One digit

Three ways - 2, 3, and 5.

[b]Case 2[/b]: Two digits

$ 3 \times 2 \equal{} 6$ ways.

[b]Case 3[/b]: Three digits

$ 3 \times 2 \times 1 \equal{} 6$ ways.

$ 3 \plus{} 6 \plus{} 6 \equal{} \boxed{15}$

[quote=Ericsz]is it wrong?[/quote]

[quote=Ericsz]oh yeah its is nevermind[/quote]

Don't double post.

[quote=Wood2009]How?????[/quote]

Are these answers wrong? Or are you not understanding the process by which these answers were generated?

[quote=Ericsz]each number can be in one of the 3 positions, or it can be out, so 2^4, but it that can't all be out so minus 1[/quote]

This is not quite correct. First of all, there are $3$ numbers, not $4$. Secondly, we are worried about order as well. You are undercounting if you don't consider order.


inequalities
geometry
circumcircle
geometry proposed

[color=purple] Let an arbitrary point M in plane prove that
        $ (MA \plus{} MB \plus{} MC)^2(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2})\ge 9$
  I think it is a nice problem :) . A corrollary : Let $ M\equiv T$ with $ T$ is the [b]Toricelli[/b] point of triangle ( named [b]Fermat point[/b], too ) and S is the area of triangle ABC it's not hard to check that : $ (TA \plus{} TB \plus{} TC)^2 \equal{} \frac {a^2 \plus{} b^2 \plus{} c^2 \plus{} 4\sqrt {3}S}{2}$ then $ (a^2 \plus{} b^2 \plus{} c^2 \plus{} 4\sqrt {3}S)(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2})\ge18$
    :)  :D  :) [/color]

[color=purple] An open question for liker : $ (a^2\plus{}b^2\plus{}c^2\plus{}4\sqrt{3}S)(\frac{1}{ab}\plus{}\frac{1}{bc}\plus{}\frac{1}{ca}) \ge 18$
           I haven't solve it yet and I hope any one can give me a nice solution  :) [/color]

[color=green] [u][b] Proof[/b][/u]
 This solution has been solved by a my friend. I hope someone can give a beter solution for this one.

 First we'll give a lemma a famous algebraic inequality :[/color]
[color=brown][i]Lemma 1.[/i][/color][color=green]  In plane let an arbitrary triangle $ ABC$ with three sides length $ BC \equal{} a,CA \equal{} b,AB \equal{} c$ and its area $ S$. We have
\[ xa^2 \plus{} yb^2 \plus{} zc^2\ge4\sqrt {xy \plus{} yz \plus{} zx}S\quad (1)
\]
with $ x,y,z$ are real numbers satisfy $ x \plus{} y > 0,y \plus{} z > 0,z \plus{} x > 0$ and $ xy \plus{} yz \plus{} zx > 0$.[/color]

[color=brown][i]Lemma 2.[/i][/color][color=green]Let $ x,y,z > 0$ then
\[ (xy \plus{} yz \plus{} zx)(\frac {1}{(y \plus{} z)^2} \plus{} \frac {1}{(z \plus{} x)^2} \plus{} \frac {1}{(x \plus{} y)^2})\ge \frac {9}{4}
\]
It was first proposed in Crux 1994 by JiChen. The solution of it is for reader to find as a present.
   Now let $ a \equal{} y \plus{} z,b \equal{} z \plus{} x,c \equal{} x \plus{} y$ we have $ a,b,c$ are three sides length of a triangle $ ABC$ which have inscribed radius $ r$ and circumradius $ R$.
   We have : $ xy \plus{} yz \plus{} zx \equal{} r(4R \plus{} r)$ and we can change lemma into the following form :
\[ 4r(4R \plus{} r)(\frac {1}{a^{2}} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2})\ge 9(*)
\]
Now back to our problem.
We can see $ \sqrt {a},\sqrt {b},\sqrt {c}$ are three side length of triangle $ XYZ$ and $ 4S_{XYZ} \equal{} 2\sqrt {r(4R \plus{} r)}$
   With any point $ M$ in plane we have :
\[ MA \plus{} MB \plus{} MC \equal{} a.\frac {MA}{a} \plus{} b\frac {MB}{b} \plus{} c\frac {MC}{c}
\]
Using the Lemma above, we have
\[ MA \plus{} MB \plus{} MC\ge 4\sqrt {\frac {MA.MB}{ab} \plus{} \frac {MB.MC}{bc} \plus{} \frac {MC.MA}{ca}}S_{XYZ}\ge 4S_{XYZ} \equal{} 2\sqrt {r(4R \plus{} r)}
\]
Thus $ (MA \plus{} MB \plus{} MC)^2\ge 4r(4R \plus{} r)$
   Applying (*) we get
\[ (MA \plus{} MB \plus{} MC)^{2}(\frac {1}{a^{2}} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2})\ge 9
\]
[/color]

[color=purple]From the proff of me we can't prove the 2nd problem. It only show that : $ (a^2 \plus{} b^2 \plus{} c^2 \plus{} 4\sqrt {3}S)(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2})\ge 16 \plus{} \frac {2r}{R}$
  Hope any one can give me the proff for 2nd problem  :) [/color]

[color=purple] Sorry the last  I want to say is $ (ab\plus{}bc\plus{}ca\plus{}4\sqrt{3}S)(\frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2})\ge 16\plus{}\frac{4r}{R}$[/color]


Can anyone help me with my code?

[code]
\begin{tabular}{|c|c|c|}
\hline
  & Mass of Empty Beaker & 12.323 g \\
\hline
 &Mass of beaker + 5 drops of water & 12.462 \\[-1ex]
\cline{2-3}
\raisebox{2ex}{Trial 1} & Mass of first 5 drops (b)-(a) & 0.139 g\\[-1ex]
\cline{2-3}
 & Average mass of 1 drop of water & 0.028 g\\[1ex]
\hline
\end{tabular}
[/code]

[hide="Produces"]
\[ \begin{tabular}{|c|c|c|}
\hline
  & Mass of Empty Beaker & 12.323 g \\
\hline
 &Mass of beaker \plus{} 5 drops of water & 12.462 \\[\minus{}1ex]
\cline{2\minus{}3}
\raisebox{2ex}{Trial 1} & Mass of first 5 drops (b)\minus{}(a) & 0.139 g\\[\minus{}1ex]
\cline{2\minus{}3}
 & Average mass of 1 drop of water & 0.028 g\\[1ex]
\hline
\end{tabular}
\]
[/hide]

Indeed, it does. What else did you expect with those \\[-1ex]? :? I mean, you forcefully reduced the space between lines to a negative value (-1ex means "shorter than standard by the height of $ x$"), so you got exactly what you asked for: a line that overlaps with the text. If you think that \\ gives too much space, just reduce -1ex to -0.3ex or something like that :)

In addition to fedja's suggestions, you can change the default spacing between all rows of the table by using \renewcommand{arraystretch}{[i]factor[/i]} where factor is a multiplying factor. For example, 1 means the current default spacing, 1.5 means 50% more and 0.5 means half.


LaTeX
pigeonhole principle
combinatorics unsolved
combinatorics

$ a_1,a_2,a_3,....,a_{16}\leq100$ are 16 different natural numbers . Prove that the existance of four numbers so that $ a_i \plus{} a_j \equal{} a_k \plus{} a_l$

[quote="gauravpatil"][quote="zephiro alves"]a1,a2,a3,a4,a5,a6,...,a16[u]<[/u]100 are 16 diffirent natural numbers.
prove that exist 4 numbers that ai+aj=ak+al
help me.
and why it is 100[/quote]
please try and use $ latex$[/quote]

please try and solve the problem ;)

Assume that $ a_{1} < a_{2} < ... < a_{16}$
consider all numbers $ a_{i} \minus{} a_{j}$ ($ i \equal{} 1,...,16 , j \equal{} 1,2,...,16$ and $ i\geq j \plus{} 1$) . all of these numbers are between $ 1$ and $ 99$ , and there are $ 120$ numbers so by Pigeonhole theorem two of these numbers are equal so we have :
$ a_{i} \minus{} a_{j} \equal{} a_{r} \minus{} a{s} \longrightarrow a_{i} \plus{} a_{s} \equal{} a_{r} \plus{} a_{j}$ .
This problem is true for $ n\equal{}15$ too .

but what if $ a_j\equal{}a_r$?? for example, $ 15\plus{}15\equal{}16\plus{}14$.

Yeah, well, the a_i are in fact different but when those 120 diferences are considered, you are counting them repeatedly, so as someone posted earlier you could end up considering a sum with the same element.

if that pigeonhole solution leads somewhere, you should be able to use those 20 spare differences to rule out the ones that contain the same element (the ones that bother are when the repeated element is the biggest in one difference, and the smallest in the other difference), so it's not likely to be possilbe por 15.

julian

[quote="pleurestique"][quote="archimedes1"][quote="pleurestique"]but what if $ a_j \equal{} a_r$?? for example, $ 15 \plus{} 15 \equal{} 16 \plus{} 14$.[/quote]
The problem states that the a_i are distinct.[/quote]

so what? see the above post[/quote]
Oh, I see, thanks.

[quote="khashi70"]Assume that $ a_{1} < a_{2} < ... < a_{16}$
consider all numbers $ a_{i} \minus{} a_{j}$ ($ i \equal{} 1,...,16 , j \equal{} 1,2,...,16$ and $ i\geq j \plus{} 1$) . all of these numbers are between $ 1$ and $ 99$ , and there are $ 120$ numbers so by Pigeonhole theorem two of these numbers are equal so we have :
$ a_{i} \minus{} a_{j} \equal{} a_{r} \minus{} a{s} \longrightarrow a_{i} \plus{} a_{s} \equal{} a_{r} \plus{} a_{j}$ .
This problem is true for $ n \equal{} 15$ too .[/quote]
i,j,k,l are diffirent natural numbers
What if $ {a_{1}}$ - $ {a_{2}}$ = $ {a_{2}}$ - $ {a_{3}}$ so that $ {a_{1}}$ + $ {a_{3}}$ = $ {a_{2}}$+$ {a_{2}}$

[quote="zephiro alves"]I am very sorry but I use internet at a public internet service so it's very difficult to use Latex.[/quote]

See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690[/url] -- you don't need to download anything to use LaTeX on the forum, just place your code between dollar signs.

[quote="zephiro alves"]
I am very sorry but I use internet at a public internet service so it's very difficult to use Latex.
[/quote]
You need [b]not[/b] install anything to use $ \text{\LaTeX}$. And I dont understand how else would using a public internet service prevent you from using $ \text{\LaTeX}$. just follow the specific syntax while posting If you are not planning to leave mathlinks soon, you will have to learn $ \text{\LaTeX}$ some day. Its best if you learn it now. I am quoting myself from another topic.
[quote="Akashnil"]
Learn LaTeX here:
Best: http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1602648475&t=188533
Q & A: http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1602648475&t=12230
Official: http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1602648475&t=5261

My personal tip(This is how I learnt.):

Try to search for posts written in LaTeX in the various brunches of this forum (You will find plenty). Then click on "quote" button on the bottom-right corner of the post and see how it looks(You need to be logged on first). You might modify a bit and see the preview yourself. You needn't really post. Just cancel the post when you are done.[/quote]

EDIT: Sorry. I did not see JBL posting.


geometry
rectangle
calculus
calculus computations

Q) A rectangle has a width $ a$ . One vertex is taken and folded over to touch the opposite edge on the rectangle(length not specified). Find the minimum length of the crease.

hello
$ EF\equal{} x$
$ BC\equal{} BE\plus{}EC$
$ a \equal{} \minus{}xcos(\theta)cos(2\theta)\plus{}xcos(\theta)$
$ x \equal{} \frac{a}{cos(\theta)(1\minus{}cos(2\theta))}$
after differentiating it maximum of denominator comes at $ \theta\equal{}sin^{\minus{}1}(\sqrt{\frac{2}{3}})$
so required $ x \equal{} \frac{3\sqrt{3}a}{4}$
thank u


In how many distinct ways can 3 green and 2 blue chips be arranged in a row?


trigonometry
induction
real analysis
real analysis solved

Complex number $z_n$ is defined by

\[z_1=1,z_{n+1}=(\cos \theta+i\sin \theta)z_n+1\]

Prove that $2-n\leqq|z_n|\leqq{n}$  for all popsitive integer n.

Not only is $|z_n|\le n$, but $|z_n|$ is a bounded sequence as long as $\theta\not\equiv 0\mod 2\pi.$ This first order constant coefficient linear difference equation can be explicitly solved, and in the case $e^{i\theta}\neq 1$ we have

$z_n=\frac{1-e^{in\theta}}{1-e^{i\theta}}.$ From this we have that $|z_n|\le\frac2{|1-e^{i\theta}|}=\frac1{|\sin(\frac{\theta}2)|}.$

Of course, if $e^{i\theta}=1,$ we have $z_n=n.$

The lower bound in the problem, alhough true, is completely useless. In the long run this sequence either equals zero infinitely often (in the case in which $\theta$ is a rational multiple of $\pi$) or gets arbitrarily close to zero infinitely often if $\theta$ is an irrational multiple of $\pi.$


function
modular arithmetic
algebra proposed
algebra

prove that there is a function like  $ f: N \to N$ such that :
$ f(f(n))\equal{}2n$

[quote="stvs_f"]prove that there is a function like  $ f: N \to N$ such that :
$ f(f(n)) \equal{} 2n$[/quote]

if $ n$ is odd and $ n \equal{} 1\pmod 4$ : $ f(n) \equal{} n \plus{} 2$
if $ n$ is odd and $ n \equal{} 3\pmod 4$ : $ f(n) \equal{} 2n \minus{} 4$
if $ n$ is even and $ n \equal{} 2^pm$ with odd $ m$, $ f(n) \equal{} 2^pf(m)$

and obviously there are a lot of other solutions (infinitely many)

First define $ f$ on the odd positive integers like this: $ f(4k\plus{}1) \equal{} 2(4k\plus{}3)$ and $ f(4k\plus{}3) \equal{} 4k\plus{}1$; then define $ f$ recursively on the even positive integers: $ f(2k) \equal{} 2f(k)$; for all $ k\equal{}0,1,\ldots$.

[quote="stvs_f"]prove that there is a function like  $ f: N \to N$ such that :
$ f(f(n)) \equal{} 2n$[/quote]

In fact, the general solution is :

$ f: N \to N$ is such that $ f(f(n)) \equal{} 2n$

$ \iff$

$ \exists$ a split of the set of positive odd integers $ O$ in two equinumerous sets $ A$ and $ B$ ($ A\cup B\equal{}O$ and $ A\cap B\equal{}\emptyset$) and a bijection $ h$ from $ A\to B$ and :
$ \forall$ odd $ n\in A$ : $ f(n)\equal{}h(n)$
$ \forall$ odd $ n\in B$ : $ f(n)\equal{}2h^{[\minus{}1]}(n)$
$ \forall$ even $ n\equal{}2^pm$ with odd $ m$ : $ f(n)\equal{}2^pf(m)$

My own previous example was :
$ A\equal{}\{n\in\mathbb N$ such that $ n\equal{}1\pmod 4\}$
$ B\equal{}\{n\in\mathbb N$ such that $ n\equal{}3\pmod 4\}$
$ h(n)\equal{}n\plus{}2$

mavropnevma's previous example is :
$ A\equal{}\{n\in\mathbb N$ such that $ n\equal{}3\pmod 4\}$
$ B\equal{}\{n\in\mathbb N$ such that $ n\equal{}1\pmod 4\}$
$ h(n)\equal{}n\minus{}2$


[quote="quykhtn-qa1"]Let $ a,b,c$ be positive numbers such that:$ ab \plus{} bc \plus{} ca \plus{} abc \equal{} 4$.Prove that:
$ \frac {a}{\sqrt {b \plus{} c}} \plus{} \frac {b}{\sqrt {c \plus{} a}} \plus{} \frac {c}{\sqrt {a \plus{} b}}\geq\ \frac {a \plus{} b \plus{} c}{\sqrt {2}}$[/quote]
By Horder we have :
$ (\sum\frac{a}{\sqrt{b\plus{}c}})^{2}.(\sum{a(b\plus{}c)}) \geq (a\plus{}b\plus{}c)^{3}$
We have to prove that :
$ a\plus{}b\plus{}c \geq ab\plus{}bc\plus{}ca$
With $ a,b,c$ be positive numbers such that:$ ab \plus{} bc \plus{} ca \plus{} abc \equal{} 4$
It is VMO 1996 ......
We have done :)

[quote="mai quoc thang"][quote="quykhtn-qa1"]Let $ a,b,c$ be positive numbers such that:$ ab \plus{} bc \plus{} ca \plus{} abc \equal{} 4$.Prove that:
$ \frac {a}{\sqrt {b \plus{} c}} \plus{} \frac {b}{\sqrt {c \plus{} a}} \plus{} \frac {c}{\sqrt {a \plus{} b}}\geq\ \frac {a \plus{} b \plus{} c}{\sqrt {2}}$[/quote]
By Horder we have :
$ (\sum\frac {a}{\sqrt {b \plus{} c}})^{2}.(\sum{a(b \plus{} c)}) \geq (a \plus{} b \plus{} c)^{3}$
We have to prove that :
$ a \plus{} b \plus{} c \geq ab \plus{} bc \plus{} ca$
With $ a,b,c$ be positive numbers such that:$ ab \plus{} bc \plus{} ca \plus{} abc \equal{} 4$
It is VMO 1996 ......
We have done :)[/quote]
NO,VMO 1996:$ a,b,c>0;a\plus{}b\plus{}c\plus{}abc\equal{}4$.Prove that;
$ a\plus{}b\plus{}c\geq\ ab\plus{}bc\plus{}ca$:blush:
with this inequality: $ ab\plus{}bc\plus{}ca\plus{}abc\equal{}4$ :P  :P

[quote="quykhtn-qa1"]Let $ a,b,c$ be positive numbers such that:$ ab + bc + ca + abc = 4$.Prove that:
$ \frac {a}{\sqrt {b + c}} + \frac {b}{\sqrt {c + a}} + \frac {c}{\sqrt {a + b}}\geq\ \frac {a + b + c}{\sqrt {2}}$[/quote]

By B.C.S.:
$ \sum\frac {a}{\sqrt {b + c}}\geq\frac {(a + b + c)^2}{\sum(a\sqrt {b + c})}$  (------------1)

By B.C.S.:
$ (\sum a)(\sum a(b + c))\geq (\sum a\sqrt {b + c})^2 \Leftrightarrow \frac {(a + b + c)^2}{%Error. "suma" is a bad command.
\sqrt {b + c}}\geq\frac {\sqrt {2}}{2}(a + b + c)\sqrt {\frac {a + b + c}{ab + bc + ca}}$  (------------2)

But (-1) and (-2):
$ \sum\frac {a}{\sqrt {b + c}}\geq \frac {\sqrt {2}}{2}(a + b + c)\sqrt {\frac {a + b + c}{ab + bc + ca}}$ (------- 3)

Since
$ a + b + c \geq ab + bc + ca \Rightarrow for (3) \Rightarrow \sum\frac {a}{\sqrt {b + c}}\geq \frac {\sqrt {2}}{2}(a + b + c)$

[quote="quykhtn-qa1"][quote="mai quoc thang"][quote="quykhtn-qa1"]Let $ a,b,c$ be positive numbers such that:$ ab \plus{} bc \plus{} ca \plus{} abc \equal{} 4$.Prove that:
$ \frac {a}{\sqrt {b \plus{} c}} \plus{} \frac {b}{\sqrt {c \plus{} a}} \plus{} \frac {c}{\sqrt {a \plus{} b}}\geq\ \frac {a \plus{} b \plus{} c}{\sqrt {2}}$[/quote]
By Horder we have :
$ (\sum\frac {a}{\sqrt {b \plus{} c}})^{2}.(\sum{a(b \plus{} c)}) \geq (a \plus{} b \plus{} c)^{3}$
We have to prove that :
$ a \plus{} b \plus{} c \geq ab \plus{} bc \plus{} ca$
With $ a,b,c$ be positive numbers such that:$ ab \plus{} bc \plus{} ca \plus{} abc \equal{} 4$
It is VMO 1996 ......
We have done :)[/quote]
NO,VMO 1996:$ a,b,c > 0;a \plus{} b \plus{} c \plus{} abc \equal{} 4$. Prove that;
$ a \plus{} b \plus{} c\geq\ ab \plus{} bc \plus{} ca$:blush:
with this inequality: $ ab \plus{} bc \plus{} ca \plus{} abc \equal{} 4$ :P  :P[/quote]

[b]VMO 96 :[/b] $ a,b,c > 0;ab \plus{} bc \plus{} ca \plus{} abc \equal{} 4$.Prove that;
$ a \plus{} b \plus{} c\geq\ ab \plus{} bc \plus{} ca$ :)
------------------
$ c\equal{}min\{a,b,c\}$ ; $ a,b >0$ .
We have :
$ a\plus{}b\plus{}c\minus{}(ab\plus{}bc\plus{}ca)\equal{}a\plus{}b\plus{}c\plus{}abc\minus{}4\equal{}\frac{(a\plus{}b\minus{}2)^{2}\minus{}ab(a\minus{}1)(b\minus{}1)}{a\plus{}b\plus{}ab}$

*$ (a\minus{}1)(b\minus{}1) \leq 0$ ; $ a\plus{}b\plus{}c \geq ab\plus{}bc\plus{}ca$
* $ (a\minus{}1)(b\minus{}1) >0$
We have :
$ ((a\minus{}1)\plus{}(b\minus{}1))^{2} \geq 4(a\minus{}1)(b\minus{}1) \geq ab(a\minus{}1)(b\minus{}1)$
We have done :)

[quote="quykhtn-qa1"]Let $ a,b,c$ be positive numbers such that:$ ab \plus{} bc \plus{} ca \plus{} abc \equal{} 4$.Prove that:
$ \frac {a}{\sqrt {b \plus{} c}} \plus{} \frac {b}{\sqrt {c \plus{} a}} \plus{} \frac {c}{\sqrt {a \plus{} b}}\geq\ \frac {a \plus{} b \plus{} c}{\sqrt {2}}$[/quote]
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=127956

[quote="mai quoc thang"][/quote][quote="quykhtn-qa1"][quote="mai quoc thang"][quote="quykhtn-qa1"]Let $ a,b,c$ be positive numbers such that:$ ab + bc + ca + abc = 4$.Prove that:
$ \frac {a}{\sqrt {b + c}} + \frac {b}{\sqrt {c + a}} + \frac {c}{\sqrt {a + b}}\geq\ \frac {a + b + c}{\sqrt {2}}$[/quote]
By Horder we have :
$ (\sum\frac {a}{\sqrt {b + c}})^{2}.(\sum{a(b + c)}) \geq (a + b + c)^{3}$
We have to prove that :
$ a + b + c \geq ab + bc + ca$
With $ a,b,c$ be positive numbers such that:$ ab + bc + ca + abc = 4$
It is VMO 1996 ......
We have done :)[/quote]
NO,VMO 1996:$ a,b,c > 0;a + b + c + abc = 4$. Prove that;
$ a + b + c\geq\ ab + bc + ca$:blush:
with this inequality: $ ab + bc + ca + abc = 4$ :P  :P[/quote][quote="mai quoc thang"]

[b]VMO 96 :[/b] $ a,b,c > 0;ab + bc + ca + abc = 4$.Prove that;
$ a + b + c\geq\ ab + bc + ca$ :)
------------------
$ c = min\{a,b,c\}$ ; $ a,b > 0$ .
We have :
$ a + b + c - (ab + bc + ca) = a + b + c + abc - 4 = \frac {(a + b - 2)^{2} - ab(a - 1)(b - 1)}{a + b + ab}$

*$ (a - 1)(b - 1) \leq 0$ ; $ a + b + c \geq ab + bc + ca$
* $ (a - 1)(b - 1) > 0$
We have :
$ ((a - 1) + (b - 1))^{2} \geq 4(a - 1)(b - 1) \geq ab(a - 1)(b - 1)$
We have done :)[/quote]

nice solution.  :)  :lol:

With a+b+c+abc=4,we can prove $ a\plus{}b\plus{}c \ge ab\plus{}bc\plus{}ca$
 From ab+bc+ca+abc=4,we can setting $ \frac{2x}{y\plus{}z}$......
And $ a\plus{}b\plus{}c \ge ab\plus{}bc\plus{}ca$ equivalent to:
$ x^3\plus{}y^3\plus{}z^3 \plus{}3xyz \ge xy(x\plus{}y)\plus{}yz(y\plus{}z)\plus{}zx(z\plus{}x)$

[quote="babyloverain"]With a+b+c+abc=4,we can prove $ a \plus{} b \plus{} c \ge ab \plus{} bc \plus{} ca$
 From ab+bc+ca+abc=4,we can setting $ \frac {2x}{y \plus{} z}$......
And $ a \plus{} b \plus{} c \ge ab \plus{} bc \plus{} ca$ equivalent to:
$ x^3 \plus{} y^3 \plus{} z^3 \plus{} 3xyz \ge xy(x \plus{} y) \plus{} yz(y \plus{} z) \plus{} zx(z \plus{} x)$[/quote]
very nice :)  :)  :)


I came across a question involving interior angles of a pentagon. If you have 5 equal interior angles, does the pentagon have to be regular? I thought it had to be, but the answer key says otherwise.

It doesn't have to be regular.
Form one side (AB) , then construct angles of 108 degrees on each side (sides AE and BC), but extend those 2 sides as long as you like and keep them equal (so those sides can be longer or shorter than the first side, in other words AE=BC and $ AE\neq AB$ ). Now construct angles of 108 degrees on the sides AE and BC and they intersect at D, the fifth vertex. Now all angles are 108 degrees but all sideds aren't equal.


geometry
rectangle
quadratics
algebra
cono sur

Let $ABCD$ be a rectangle with: $AB=a$, $BC=b$. Inside the rectangle we have to exteriorly tangents circles such that one is tangent to the sides $AB$ and $AD$,the other is tangent to the sides $CB$ and $CD$.

1. Find the distance between the centers of the circles(using $a$ and $b$).
2. When the radiums of both circles change the tangency point between both of them changes, and describes a locus. Find that locus.

[quote=scorpius119][hide="1"]
Let the radius of the lower left circle be $r_1$, that of the upper circle be $r_2$, the horizontal component of the distance between the centers be $x$, and the vertical component of the distance between the centers be $y$. Also, letting $S=r_1+r_2$, we have the equations
\begin{eqnarray*} x&=&a-S\\ y&=&b-S\\ x^2+y^2&=&S^2 \end{eqnarray*}
Substituting the values from the first and second equations into the third, $(a-S)^2+(b-S)^2=S^2\Rightarrow S^2-2(a+b)S+(a^2+b^2)=0$. The quadratic formula gives $S=a+b\pm \sqrt{2ab}=\left( \sqrt{a}\pm \sqrt{b}\right)^2$, and since $S$ must be less than both $a$ and $b$, $S$ must equal $\boxed{\left( \sqrt{a}-\sqrt{b}\right)^2}$.
[/hide]
[hide="2"]
From the value of $S$, $x=\sqrt{2ab}-b$ and $y=\sqrt{2ab}-a$. Let the distance from the tangency point to $AD$ be $p$ and the distance from the tangency point to $AB$ be $q$. Then $p=r_1+x=r_1+\sqrt{2ab}-b$ and $q=r_1+y=r_1+\sqrt{2ab}-a$. It follows that $q=p+(b-a)$.

The locus must be a line segment that is included in the equation above. To find the full locus, we must find the endpoints. Note that $p$ and $q$ are both maximized when $r_1$ is maximized. Let $m=\min (a,b)$. Then $\max r_1=\frac m2$ so $\max p=\frac m2 +\sqrt{2ab}-b$ and $\max q=\frac m2 +\sqrt{2ab}-a$. Similarly, $p$ and $q$ are minimized when $r_2$ is maximized, which is when $r_1=S-\frac m2$.

Since any point between the endpoints is in the locus, the locus consists of the line segment connecting $\left( a-\frac m2, b-\frac m2 \right)$ and $\left( \frac m2 +\sqrt{2ab}-b,\frac m2 +\sqrt{2ab}-a\right)$.
[/hide][/quote]

Hello. In the part 2 of the solution, instead of
$p=r_1+x$
$q=r_1+y$,
I think that the correct is
$p=\frac{r_1x}{S}+r_1$
$q=\frac{r_1y}{S}+r_1$.
This is because $x$ and $y$ are the components of the distance between the centers instead of the components of $r_1$.
Thanks


If $ a,b,c$ are dinstinct real numbers  prove that:

$ (\frac{4a\minus{}3b}{a\minus{}b})^2\plus{}(\frac{4b\minus{}3c}{b\minus{}c})^2\plus{}(\frac{4c\minus{}3a}{c\minus{}a})^2 \ge 25$

Let $ P(t)\equal{}\frac{(t\minus{}a)(t\minus{}b)}{(a\minus{}c)(b\minus{}c)}\plus{}\frac{(t\minus{}a)(t\minus{}c)}{(a\minus{}b)(c\minus{}b)}\plus{}\frac{(t\minus{}b)(t\minus{}c)}{(b\minus{}a)(c\minus{}a)}$. Then $ P$ has degree at most two and $ P(a)\equal{}P(b)\equal{}P(c)\equal{}1$, so $ P$ is the constant 1. In particular,
\[ 1\equal{}P(a\plus{}b\plus{}c)\equal{}\frac{(b\plus{}c)(a\plus{}c)}{(a\minus{}c)(b\minus{}c)}\plus{}\frac{(b\plus{}c)(a\plus{}b)}{(a\minus{}b)(c\minus{}b)}\plus{}\frac{(a\plus{}c)(a\plus{}b)}{(b\minus{}a)(c\minus{}a)}\]
\[ \Rightarrow \minus{}1\equal{}\frac{c\plus{}a}{c\minus{}a}\cdot \frac{b\plus{}c}{b\minus{}c}\plus{}\frac{b\plus{}c}{b\minus{}c}\cdot \frac{a\plus{}b}{a\minus{}b}\plus{}\frac{c\plus{}a}{c\minus{}a}\cdot \frac{a\plus{}b}{a\minus{}b}\]
Let $ x\equal{}\frac{a\plus{}b}{a\minus{}b}$ etc. so that $ xy\plus{}yz\plus{}zx\equal{}\minus{}1\Rightarrow x^2\plus{}y^2\plus{}z^2\equal{}2\plus{}(x\plus{}y\plus{}z)^2$. Then
\[ (7\plus{}x)^2\plus{}(7\plus{}y)^2\plus{}(7\plus{}z)^2\equal{}(x\plus{}y\plus{}z)^2\plus{}14(x\plus{}y\plus{}z)\plus{}2\plus{}3\cdot 7^2\]
\[ \equal{}(x\plus{}y\plus{}z\plus{}7)^2\plus{}100\geq 100\]
But $ 7\plus{}x\equal{}2\cdot \frac{4a\minus{}3b}{a\minus{}b}$ etc. implying the result.


Analysis?  MathWorld categorizes wavelets under "approximation theory."


The egdes of a connected graph are colored with $ n$ colors such that every vertex has edges
of all $ n$ colors incident to it. Prove that if we delete $ n\minus{}1$ edges of distinct colors, the graph
will remain connected.

Are you sure?
Consider 6 vertices $ A,B,C,D,E,F$ and 2 colors. Blue edges: $ AB,CD,EF$; red edges: $ AC,BC,DE,DF$. Does it satisfy conditions? But we have a bridge $ CD$ here...

Yes I know, I was expecting someone who knows the problem to tell me the correct phrasing.


parameterization
LaTeX
analytic geometry
trigonometry
linear algebra

Can anyone provide help?

1. Let F: R^2 -> R^2 be the mapping defined by F(x,y) = (xy, y). Describe the image under F of the straight line x = 2.

2. Let F be the mapping defined by F(t,u) = (cos t, sin t, u). Describe geometrically the image of the (t,u)-plane under F.

Thanks.

These maps are nonlinear, and the questions hardly belong to this subforum. You can find the equations describing each object by eliminating the parameters $y$, $t$, and $u$.

Yeah, could anyone else please help me with these problems or tell me where this thread should be moved?

Thanks.

It's useless to waste posts telling people what messages belong to what forums. That's a moderator's job.

[quote="keta"]Dude, what are the answers?

It's useless to waste posts telling people what messages belong to what forums. That's a moderator's job.[/quote]

Don't call him dude, you are not on IMDB or anything.  And you come here, not bothering to write your question in LaTex or to at least put them in the correct place, and instead of apologizing for that you bluntly tell him to stop it and still expect answers.

There are literally a thousand people on this forum who could solve this problem, but will they still be willing to do your homework now ... for free? :huh:

The answers can be found [url=http://www.phdcomics.com/comics/archive.php?comicid=5]here[/url].
 :)

fredbel6, I will first try respectful diplomacy with you. Hopefully with your type it will work.

I do not have Latex installed on my computer. I have installed it before and it took me over an hour to make it work. Would you install it to simply ask a few questions on a math forum? Besides, yes, it is convenient for readers if it is in Latex, but you can still read perfectly what I typed in normal script, so whether the issue of my using Latex or not is quite irrelevant.

Secondly, I've seen people post in all manners fit to be labeled as "colloquial" on this AOPS forum. I've been on the forum for quite a while and I know exactly what styles of prose posters tend to use. Therefore, I've seen AOPS people call each other "dudes" and "whatups" and the like. That is not to say that such language necessarily implies rudeness, and in relationship I have not implied rudeness either. It is simply a very informal but neutral way for one to address someone else.

Now your example with IMDB is quite immaterial in this discussion. IMDB is an internet movie site, so of course people will post using, let's just say, more "colorful language". However, you fail to acknowledge the fact that the posters on IMDB are not ALL rude and disrespectful. I have seen several people who have contributed useful and productive comments to movies or actors. So you have really, no right...., yes, no rights, to go bashing IMDB and elevating the posters of AOPS to a higher level than that of IMDB posters. I certainly hope I'm wrong, but for you to do so would mark the epitome of narrow-minded bigotry.

Thirdly, I take it that you accuse me of failing to apologize to mlok for not posting in Latex. My, my, what a grave misunderstanding, or shall I say, misaccusation, you have casted upon me. Did mlok mention anywhere in his first post that I was "required" to post in Latex? Did he? In case you're not sure, or are stumped, the answer is NO. So how can I apologize when my "alleged" mistake wasn't even pointed out? I trust, by your rather "erudite" post, that you are smart enough to figure out the answer to this question. If not, you can simply PM me, or try to prove Riemman's Hypothesis, which I'm sure will probably take you less time than figuring my question out.

Yes, I admit I shouldn't have been so brusque-sounding in my second post, but for heavens sake, it was merely a natural reaction to the fact that mlok contributed nothing of value by posting his accusing me that my message was in the wrong forum and failing to tell just what forum should I put this message in. I got these problems from a Linear Algebra book, and wouldn't you agree that the best place for a Linear Algebra problem is in a Linear Algebra computation forum, especially if most of you on AOPS have your noses in the air and think that the other forums are reserved for "difficult" problems? Had he told me what forum to move my questions to, I would have quitely and uneventfully done so. However, wouldn't you also be irritated by someone who does not answer your question but complains that your post is in the wrong section? I know you would, and by stating any answer to the contrary, I would be pretty sure that you are lying.

Besides, these are not HW problems. I am trying to study Linear Algebra on my own and if you gave this any thought, you would have already figured that this was summer and most schools, even summer schools in universities, are not currently in session. But again, I take it that your urge to denigrate me under your veil of stern words prevented you from considering this highly possible situation before posting.

In addition, you mindlessly accuse me of asking people to do my HW for free. The whole purpose of why AOPS was set up was to provide a structural help and discussion portal for people over the world to come and talk and help each other with math problems. I come to AOPS to do exactly that. Besides, I have helped others when they needed problems to be solved, and instead of complained or bashing, I either answered their questions or left the forum w/o posting lecture comments, such as the one that you and mlok posted. Therefore, I have done nothing wrong on my part of posting the problems and seeing if anyone had solutions to them. If you think harder, and I have a feel you may be capable fo such feats, you will see that mlok is the one who is being offensive by posting such an asinine link after being blunt in his first post.

I deem it highly likely that you will try to come up with other excuses for you actions and fail to apologize to me for your rash behaviors of false accusations and calumny. However, I will take the position of being mature and I will officially apologize to both you and mlok for being unpleasant on my second post. I sincerely hope that you will read, reread, and fully understand my logic before posting, and also apologize for attacking me when you have not understood the situation. I gladly await your answer. Feel free to PM me.

[b]keta[/b], I am not going to read your long post, but will try to help you with your questions.

1. It's a good idea to give names to the coordinates of the point $F(x,y)$. Let's call this point $(u,v)$. The equation of $F$ says: \[u=xy,\quad v=y\] We want to get rid of $x$ and $y$ and obtain an equation (or perhaps system) in terms of $u$ and $v$ only. Getting rid of $x$ is easy: you can replace it by $2$, since you are only dealing with the points with $x=2$. So, \[u=2y,\quad v=y\] Now you can eliminate $y$ by dividing these equations: $u/v=2$. And this is the answer: the image is the line $u=2v$.

2. Again, give a name to $F(t,u)$, say, $(x,y,z)$. We have \[x=\cos t, \ y=\sin t, \ z=u\] You can eliminate $t$ be squaring the first two equations and adding them: $x^{2}+y^{2}=1$. The third equation does not place any restrictions on $z$, since $u$ is arbitrary. So, the image is the set of all points $(x,y,z)$ with $x^{2}+y^{2}=1$ and $z$ arbitrary. I'll leave it to you to figure out its shape.

[quote="keta"]I do not have Latex installed on my computer.[/quote]
You don't need it. The forum software interprets the code and generates the pictures for you. The only reason to get $\LaTeX$ is if you want to typeset documents of your own.


geometry
parallelogram
combinatorial geometry
combinatorics proposed
combinatorics

let S be a set of points in the plane. A point P is an intersection point of S if there exist different points A,B,C,D in S such that AB and CD intersect in P. define F(S) to be the union of S and its intersection points. define Fn(S)=F(Fn-1(S))
prove that if the sequence |S|, |F1(S)|, |F2(S)|........,|FK(S)| is bounded then Fi(S)=F1(S) for all i>=1

I think we can even determine $S$ pretty accurately.

The sequence $|F_n|$ must be eventually constant. Take an $n$ s.t. $F_n=F_{n+1}$. In this case it's clear that the convex hull of $F_n$ must either be a triangle or a parallelogram. Assume it's a triangle. If there's a point strictly inside, then assuming it's not the centroid, we get a contradiction, and taking it to be the centroid also gives a contradiction, but after some more work (sorry for the lack of details). This means that it must be a triangle together with some other points, all of them lying on a certain side of the triangle. It's now easy to see that $F_1$ must be exactly $F_n$.

On the other hand, assume the convex hull of $F_n$ is a parallelogram. It necessarily contains the center of the parallelogram as well. Arguments similar to the previous one show that these five are the only points $F_n$ can contain, and from here it's easy.

I'm sure I've overlooked some cases, but I think this is basically it :). Sorry for any mistakes I might have made: I'm sort of in a hurry right now.

Ok, supose K is a set of points such that the intersection of lines that pass trough by at least 2 points of K belongs to K. Supose K is finite and all the points of K do not lie on a line. Then there is a line l wich passes trought exactly 2 of the points in K. let this two points be A and B. Supose there is another point C in K. then all the points lie in AC, BC or a line trough C parallel to AB. asume there is a point in the line parallel to AB and call it D it is easy to verify that A,B,C,D most be a parallelogram and that K=A,B,C,D and the center of the paralelogram. Now asume there is no point in the paralel line, and supose there is a point on AC, then its easy to see that all points (diferent from B) lie on AC. Now, It is easy to see that if F(S)=K then S is the four vertives of a paralelogram, Or S=K. (note that in the secound case all points lie on 2 lines, this is a clasical problem and i saw it in Engel Book. also if all points lie on two lines and one of the lines contains only one point that doesnt belong to the other line, then there is no point thta is in the intersection of lines formed by the other points, so we must have S=K)

the problem was easy for a number 6... you could start with many other ideas and get a solution very similar to pascual's (for example, taking the triangle of minimal area, or the point with minimun distance to a line,...). The thing is that you could lose many points, for example, if you missed to mention the case of a line and a point outside it, or things like that (many means 3, 4 or even 5!)...


Fie M cea mai mare distanta dintre 6 puncte din plan, iar m distanta cea mai mica. Demonstrati ca raportul M/m este mai mare sau egal cu radical din 3.

Din nou scuze pt felul in care am scris problema ,dar promit sa-mi reamintesc Latexul!! (poate cu putin ajutor il voi invata mai repede!!)

problema este clasica, se arata ca exista cel putin un triunghi cu un unghi $\geq  120$ grade, cred ca poate fi gasita in engel, sau in orice carte de combinatorica.


Solve for $ x$: \[ \log_{2008}\left(x\minus{}3\right)\plus{}\log_{2009}\left(x\minus{}3\right)\equal{}3\minus{}\lg\left(x^5\minus{}24\right)\]

What is $ \lg$? Natural log? Base-10 log?

It could also be base 2

[hide="possible hint for simplification"]
$ \log_b{x}\equal{}\frac{1}{\log_x{b}}$

[/hide]

[quote="dogcat"]Solve for $ x$:
\[ \log_{2008}\left(x \minus{} 3\right) \plus{} \log_{2009}\left(x \minus{} 3\right) \equal{} 3 \minus{} \lg\left(x^5 \minus{} 24\right)
\]
[/quote]
are you sure there is $ lg(x^2\minus{}24)$?


How much time out of class a week are we supposed to spend for an ordinary class like intermediate algebra?

I think it really depends on several factors. Some weeks, material is covered that is easier than usual and thus you don't spend as much time. Other weeks, the material is considerably more difficult and the amount you spend increases. Also, if you are the sort of person who understands math quickly and can quickly come up with solutions to problems, you will obviously spend less time.

In the courses I've taken, I spend 2-3 hours, sometimes even more, working problems, rereading the transcript, and working things out of the book (but interm. algebra doesn't have a book so don't worry).

Another thing to take into account: If you are short on time, don't post your answers to the problems you work. Solve them, check them with the other responses, but if you really are pressed for time, there's not too much of a need to spend it typing up an answer. It helps to have comments, but you won't die without them. I've done this for much of Introduction to Geometry, but I still feel that I understand the material.


geometry
rectangle
trigonometry
geometry unsolved

Hello,

Here is a problem to which i didn't find a solution. I think it isn't very hard.

Let ABC be a rectangled triangle in C and [i]a[/i] the angle formed by the hypotenuse and the median of [BC].
Prove that sin[i]a[/i] =< 1/3

I'd like you to not use advanced trigonometric formulas.

[quote="mhdi"]Hello,

Here is a problem to which i didn't find a solution. I think it isn't very hard.

Let ABC be a rectangled triangle in C and [i]a[/i] the angle formed by the hypotenuse and the median of [BC].
Prove that sin[i]a[/i] =< 1/3

I'd like you to not use andvanced trigonometric formulas.[/quote]

$ AM$ is the median , $ AC = b$ , $ BC = 2a$ , $ \angle BAM = \alpha$
Then using Sin low for $ \triangle AMC$ : $  \frac {\sin(\alpha)}{a} = \frac {sin(\angle ABC)}{AM} = \frac {b}{\sqrt {a^2 + b^2} \sqrt {4a^2 + b^2}}$;
$ {\sin(\alpha)} = f(x) = \sqrt {\frac {x}{(1 + x) (4 + x)}}$ ,      $ x = (\frac {b}{a})^2$;
${ {\sin(\alpha)} \le \frac {1}{3} \Longleftrightarrow \frac {x}{(1 + x) (4 + x)}} \le \frac {1}{9} \Longleftrightarrow (x - 2)^2 \ge 0$ ;:)

Thank you Sergic Primazon! :)
Your solution is great but i've heard of 2 others solutions :
The first one uses the barycentre property, and the second one uses AL-KACHI theorem.
I would be thankful if you could think of them and post them - hope i am not asking for too much things  :P

I've heard about a group of mathematicians called AL-KASHI in Russia, like
    Burbaki in France.  Is it the same as AL-KACHI you mentioned?

    M.T.


While registering, I clicked the under-13 button, but I'm older than 13. Is there any way I/administartors can fix it so my profile states I am above 13 ?

Send an email to usamts@usamts.org and I will un-register you.  You can then re-register appropriately.


New York has a population of 20,000,000 in a 50,000 sqaure mile area.
Illinois has a population of 12,000,000 in a 80,000 square mile area.

How many people from New York would have to move to have the same population density as Illinois.


[color=purple]Nice problem arqady :D  I haven't solve your problem yet but I have three similar problems  :)
   Here they are :
      $ \frac{l_{a}}{m_{a}}\plus{}\frac{l_{b}}{m_{b}}\plus{}\frac{l_{c}}{m_{c}}\ge \frac{6r}{R}$
$ 3\minus{}\frac{|b\minus{}c|}{2}\minus{}\frac{|c\minus{}a|}{2}\minus{}\frac{|a\minus{}b|}{2}\le\frac{l_{a}}{m_{a}}\plus{}\frac{l_{b}}{m_{b}}\plus{}\frac{l_{c}}{m_{c}}\le3\minus{}\frac{(\sqrt{b}\minus{}\sqrt{c})^2}{b\plus{}c}\minus{}\frac{(\sqrt{c}\minus{}\sqrt{a})^2}{c\plus{}a}\minus{}\frac{(\sqrt{a}\minus{}\sqrt{b})^2}{a\plus{}b}$
   I hope you'll like them  :) [/color]

[quote=arqady]Let $ m_a,$ $ m_b$ and $ m_c$ are medians and $ l_a,$ $ l_b$ and $ l_c$ are bisectors of $ \Delta ABC.$ Prove that
\[ \frac {{l_a }}{{m_a }} \plus{} \frac {{l_b }}{{m_b }} \plus{} \frac {{l_c }}{{m_c }} > 1
\][/quote]

Nice, I solved it, when I was at school :)
See here http://www.problems.ru/view_problem_details_new.php?id=108104

[quote=manutd03]Where can I find more source about this competition?Thanks[/quote]

Here http://olympiads.mccme.ru/mmo/


Let $ x,y,z>0$ such that $ x\plus{}y\plus{}z\equal{}1$. Prove that :
$ \frac{xy}{\sqrt{xy\plus{}yz}}\plus{}\frac{yz}{\sqrt{yz\plus{}zx}}\plus{}\frac{zx}{\sqrt{zx\plus{}xy}}\leq \frac{1}{\sqrt{2}}$

[quote="conan_naruto236"]Let $ x,y,z > 0$ such that $ x \plus{} y \plus{} z \equal{} 1$. Prove that :
$ \frac {xy}{\sqrt {xy \plus{} yz}} \plus{} \frac {yz}{\sqrt {yz \plus{} zx}} \plus{} \frac {zx}{\sqrt {zx \plus{} xy}}\leq \frac {1}{\sqrt {2}}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88439

It's very weak. AM-Gm solved it easily   :wink:   :ninja:


A visitor came across A and B who are inhabitants of an island who are busy in a conversation.

From his previous interactions with these individuals he knows that  A  lies every  Tuesday, Wednesday and Thursdays
and on  other days he speaks the truth. He is also aware that
B  lies on Fridays, Saturdays and Sundays but  the other days of the week he speaks the truth.

The following constitute the statements made by A and B.

A's Statement : Yesterday I was lying.
B's Statement: So was I.

Which day did they say that?

[hide="Answer"]From A's statement -> It was Tuesday or Friday.
From B's statement -> It was Friday or Monday.
So it was Friday.[/hide]

[hide]
Interpret "lying" yesterday to mean "my truth value was toggled yesterday", etc. its solved[/hide]

aahh! A spoof on Raymond Smullyan's Alice puzzles!

[hide="solution"]Let's say both of them tell the truth. Therefore, it's a monday, but only 1 of them was lying!

Now we know that one of them is lying.

Let's say that A is lying. Therefore, it's a Tuesday. But, then, B will be lying!

Let's say that B is lying. Therefore, it is a Friday. That works out, so it's a Friday.[/hide]


A bag contains 6 red balls and 3 whit eballs, and a second bag contains 3 red balls and 6 white balls. If Tony draws one ball from each bag find the probability that both balls are of the same color.

The total number of ways to pick 2 balls is 81. Out of those 81, 36 possibilities to pick 2 same colored balls. 36/81=4/9


linear algebra
matrix
algebra
polynomial
geometry
geometric transformation
rotation

A=
(0 1 0)
(0 0 1)
(1 0 0)
C(A)={X \in M_3(C), XA=AX}
a)Prove that if X \in C(A) then there exist a,b,c complex numbers such that X=aA^2+bA+cI_3
b)Prove that if X \in C(A), and X^2004=0 then X=0
------------------------------------------------------------------

a) If X commute with A then X=P(A) where P is polynomial in C[X]
A^3 = I_3, consider euclidean division of P(z) by z^3-1, then there exist
a,b,c \in C such that P(z)=(z^3-1)Q(z) + az^2+bz+c
X=P(A)=(A^3-I)Q(A)+aA^2+bA+cI_3 = aA^2+bA+cI_3

b) A vanishes the polynomial x^3-1=(x-1)(x-j)(x-j^2) with j=e^(2i \pi /3)
this polynomial has simple roots, A is diagonalisable

A is similar to
(1 0 0 )
(0 j 0)
(0 0 j^2)

X is similar to the diagonal matrix
diag( a+b+c; aj^2 + bj + c; aj + bj^2+c)  (1)

But X^2004=0 means X is nilpotent matrix, its eingevalues are 0 (2)

(1) & (2) gives the system

a+b+c =0  (3)
aj^2+bj+c=0 (4)
aj+bj^2+c=0  (5)

W=
(1 1 1)
(1 j j^2)
(1 j^2 j^4) invertilbe Van Der Monde matrix  the system can be written as

W. S = 0 with

S=
(c)
(b)
(a)

S=0 => a=b=c=0 => X = 0_3

Remark: The geometrical tranformation associate to A is a rotation, indeed
tA.A=I_3 where tA is the transpose of A, and detA=1>0
Its axis is given the vector k(1,1,1), its angle verify 1+2cos \emptyset=trA=0, cos \emptyset =-1/2
For the sign of  \emptyset, take a vector w orthogonal to k for example
w(1,0,-1) study the sign of mixt product (w/\Aw|k) where /\ is vectorial
product, (x|y) ususal scalar product in R^3.

Is there a solution wich use the fact A is rotation matrix ?

Nice problem.

Much more elementary solution for a) without reduction theory.

X=
(a b c)
(d e f)
(g h i)

Some  computation
AX=
(d e f)
(g h i)
(a b c)

XA=
(c a b)
(f d e)
(i g h)

AX=XA gives
a=e=i
b=f=g
c=d=h

X =
(a b c)
(c a a)
(b c a )
=
aI_3 + bA+cA^2


geometry
circumcircle
homothety
IMO Shortlist
incenter
geometry solved
tangent circles

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

I. for one, distinctly remember posting two or three of these on the forum, but what the heck! :) They're nice problems; we can solve them all over again. :)

I remember giving an inversive proof, but I can't find it anymore, so here's another one:

Let $M$ be the circumcenter of $BCD$. We have $\angle BMC=2\angle BDC$, and we want to show that $\angle BMC=\angle BAC$, so it's equivalent to showing that $\angle BAC=2\angle BDC$. Let $C'$ be the intersection of $DC$ with $S_1$.
The tangent through $C'$ to $S_1$ is parallel to $AC$ because $C'$ is the image $C$ through the homothety centered at $D$ which turns $S_2$ into $S_1$. Now let $T$ be the intersection of the tangents to $S_1$ through $B,C'$. We now have $\angle BAC=\pi-\angle BTC'=2\angle TC'B=2\angle BDC'=2\angle BDC$, Q.E.D.

The problem can be simplified (no arrangement conditions are necessary):

[i]Let A, B and C be three non-collinear points, and $S_1$ and $S_2$ two circles such that the circle $S_1$ touches the line AB at the point B, the circle $S_2$ touches the line AC at the point C, and the circles $S_1$ and $S_2$ touch each other at a point D. Prove that the circumcenter of the triangle BCD lies on the circumcircle of the triangle ABC.[/i]

Here is my solution, using directed angles modulo 180:

If P is the circumcenter of triangle BCD, then we have to prove that this point P lies on the circumcircle of triangle ABC. In other words, we have to show that < BPC = < BAC. What we know is < BPC = 2 < BDC (since the point P is the circumcenter of triangle BCD).

If t is the common tangent to the circles $S_1$ and $S_2$ at their point of tangency S, then, since the tangents to a circle at the endpoints of a chord make equal angles with the chord, we have < (BD; t) = < (AB; BD) and < (CD; t) = < (AC; CD), so that

  < BDC = < (BD; CD) = < (BD; t) - < (CD; t) = < (AB; BD) - < (AC; CD)
             = < (AB; AC) + < (AC; BD) - (AC; CD) = < (AB; AC) + < (CD; BD)
             = < BAC + < CDB = < BAC - < BDC.

Thus, 2 < BDC = < BAC, and hence < BPC = < BAC, as desired. $\blacksquare$

  Darij

[quote="grobber"]I. for one, distinctly remember posting two or three of these on the forum, but what the heck! :) They're nice problems; we can solve them all over again. :)

[/quote]

Well, grobber, I told that you shall post the links for the ISL problems in the section where I and Peter VDD did it. I do not search the forum all the time. And from time to time these postings are not adapted to LaTeX. Rather than searching and adopting all threads I repost it.

The same for Russian Olympiad and Kvant problems. I really appreciate it that you post these nice problems but please collect the links an extra thread for convenience and to prevent people from reposting them. Thank you.

Other suggestions ?  :)

Here is the first solution:

Name ,(the tangent $AB$),(the common tangent of the two circles from

$D$),and( the tangent $AC$),

$T_1,T_2,T_3$ , respectively.

Consider aline which is perpendicular, to $BD$, and also bisects it.

The points which are on this line, have equal distance from $T_1,T_2$,

Consider aline which is perpendicular, to $CD$, and also bisects it.

The points which are on this line , have equal distance from $T_2,T_3$.

Then the intersection of these two lines,$O$, have equal distance from

$T_1,T_3$.

And we conclude that $O$ lies on the bisector of $\angle BAC$.

But $O$ lies on  line which is perpendicular to $BC$, and also biscts it.

Then it lies on the circum circle of triangle $BAC$.

The second solution:

Consider aline which is perpendicular, to $BD$, and also bisects it.

Consider aline which is perpendicular, to $CD$, and also bisects it.

And name their intersection, $O$.

We should say that:

$180=\angle A+\angle BOC$.

Ok………………………………………………………………………………………………….

$\angle BDC=X$ then : $\angle BOC=2\angle EOF=2(180-X)=360-2X$.

we should know that if, $\angle BOC=360-2X+\angle A=180$, holds or not.

Suppose it holds, then:

We draw a tangent at $S$ which is parallel to $AC$. Then,$AC,A'M$ are parallel

And, $\angle A=\angle A'= \frac{%Error. "arcBDM" is a bad command.
-%Error. "arcBM" is a bad command.
}{2}$.

$%Error. "arcBDM" is a bad command.
=360-%Error. "arcBM" is a bad command.
$

$\angle A=180-%Error. "arcBM" is a bad command.
+180=360-%Error. "arcBM" is a bad command.
=2X$

$X=180-\frac{%Error. "arcBM" is a bad command.
}{2}$

and this fact is true, because of the quality of homotechy,$F,D,M$ are collinear.

(NOTE:$D$ is the inner homotechy center , and a tangent at$F$, to circle$S'$,

is parallel to the tangent at $M$ , to the circle $S$.

and finally: $\angle BDC=180-\angle BDM=180-\frac{%Error. "arcBM" is a bad command.
}{2}$.

And the second one is complete,too.

Now , I choose an $INVERSIVE$ proof for the third solution.

Really, and absolutely more nice than the two recent solution, I think.

Lets invert the shape round point $B$,and an arbitrary radius.

Now every thing which I will say , is about the inversive form of the shape.

It is clear that each point, for example $X$, turns into its inversive form,$X'$.

I should prove that, the center of circumcircle of triangle $BDC$, lies on circle $CAB$.

Or the other words, in the inversive shape, the inversion of the center of this circle,

Lies on the inversion of circle $CAB$.

The inversion of circle $CAB$, is $AC$, and the inversion of circle$BDC$, is $DC$.

The inversion of the center of this circle is a point$B'$,with reflect point $B$, at line $CD$.

Then I should prove that:

$B'$ lies on $AC$

………………………………………………………………………………………………………………………...

and to do that, if $AC,BH$ meet at $O$, and if $X=Y$, the problem is solved.

$X=180-\angle BCP=\frac{%Error. "arcBCP" is a bad command.
}{2}$

$Y=\angle PCA=\frac{%Error. "arcAP" is a bad command.
}{2}$

we know that $P$ is the midpoint of the arc.because with center $C$,    $P,D$ are two homologues point.

And finally: $\overarc{BCP}=%Error. "arcAP" is a bad command.
$ and every  thing is ok.

It was really an enjoyable problem,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,.

I will think about the fourth solution.

[quote="saeid"]You are so COOL ashegh.
Excelent Work :lol:[/quote]

ure wellcome my darling :D

Notation: (see diagram as well) Let $ O$ be the circumcenter of $ \triangle BCD$. Let $ BD$ intersect $ S_2$ again at $ D$ and let $ CD$ intersect $ S_1$ again at $ F$. Let $ AC$ intersect $ BF$ at $ G$. Let $ \angle BDF\equal{}x$.

(1) $ \angle BOC\equal{}2x$. Proof: $ 2x\equal{}2\angle BDF\equal{}2\pi\minus{}2\angle BDC\equal{}(\pi\minus{}2\angle BDO)\plus{}(\pi\minus{}2\angle CDO)$. Since $ BO\equal{}CO\equal{}DO$ $ \equal{}\angle BOD\plus{}\angle DOC\equal{}\angle BOC$.

(2) $ BF\parallel CE$. Proof: Since $ S_1$ and $ S_2$ are tangent at $ D$, there is a homothety that maps $ S_1$ to $ S_2$ centered at $ D$. It is easy to see that $ BF$ goes to $ CE$. Since a homothety maps a line to another parallel line, we get the conclusion.

(3) $ \angle ACE\equal{}\pi\minus{}x$. Proof: $ \angle ACE\equal{}\angle ACD\plus{}\angle DCE\equal{}\angle CED\plus{}\angle DCE$ by the tangent secant angle theorem in $ S_2$. $ \equal{}\pi\minus{}\angle CDE\equal{}\pi\equal{}\angle BDF\equal{}\pi\minus{}x$.

(4) $ \angle BAC\equal{}\pi\minus{}2x$. Proof: $ \angle BAC\equal{}\angle AGF\minus{}\angle ABG\equal{}\angle ACE\minus{}\angle BDF\equal{}\pi\minus{}2x$ by the tangent secant angle theorem in $ S_1$, the parallel lines in (2), and (3).

(5) $ ABOC$ is cyclic. Proof: $ \angle BAC\plus{}\angle BOC\equal{}\pi\minus{}2x\plus{}2x\equal{}\pi$ by (1) and (4) so $ ABOC$ is cyclic.

In other words, $ O$ lies on the circumcircle of $ \triangle ABC$, as desired.

Another solution.
Define $ M$ be $ CD \cap S_{1}$, and $ AC$ meet $ S_{1}$ at $ E, F$, $ AF$ and $ BM$ meet at $ X$.
$ A,D,O$(circumcenter of $ \triangle BCD$)$ B$ are cyclic $ <\equal{}> \angle ABO\equal{}\angle OCE <\equal{}> \frac{\pi}{2}\equal{}\angle BXC\plus{}\angle OBC <\equal{}> X B C D cyclic.$
And using homothety, we easily know $ \angle XCD\equal{}\angle XBD$ so $ A D O B$ cyclic.

Sorry for my poor english...

Somewhat wierd solution, but I thought it was interesting.
Let $ O_1$ be the center of the circle containing $ B$, and $ O_2$ be the center of the circle containing $ C$. Let $ A'$ be the intersection between $ O_1B$ and $ O_2C$. $ A'BCA$ is cyclic, so it suffices to show the circumcenter lies on the circumcircle of $ A'BC$.
The perpendicular bisectors of $ BD$ and $ CD$ are the angle bisectors of $ \angle{O_2O_1A'}$ and $ \angle{O_1O_2A'}$, so the incenter of $ O_1O_2A'$ is the circumcenter of $ BCD$. Let this point be $ I$. $ A'I$ is thus the angle bisector of $ \angle{O_1A'O_2}$, and $ BIC$ is isoceles.
If $ A'I$ is not the perpendicular bisector of $ BC$, we are done. If $ A'I$ is the perpendicular bisector of $ BC$, then $ B,C,D$ are the points of tangency of the incircle to the sides since $ A'B\equal{}A'C$, and so $ \angle{A'BI}\equal{}\angle{A'CI}\equal{}90$, and so we are done.

I think all the solutions above just complicate the problem...

[hide="Diagram"][img]http://oi56.tinypic.com/2qnrrli.jpg[/img]

[color=#FF0000][mod edit: http://oi56.tinypic.com/2qnrrli.jpg][/color]
[/hide]

Let $O$ be the circumcentre of $\triangle BDC$ and let $P$ be the intersection of the common tangent to $S_1$ and $S_2$ (passing through $D$) and $AC$. Let the midpoints of $BD,CD$ be $M,N$ respectively. Let $PD$ and $BA$ intersect at $K$. Obviously $D$ is the reflection of $B$ through the line $OK$.

Then $CD$ is the polar of $P$ wrt $S_2$ so $O,P,N$ are collinear (all lie on the perpendicular bisector of $CD$). So by trig ceva on $\triangle ODC$ obviously $\angle ODP=\angle PCO$ (lol user [b]pco[/b] :lol: !). Now $\angle ACO=\angle PCO=\angle ODP=180^{\circ}- \angle KDO=180^{\circ}-\angle KBO=180^{\circ}-\angle ABO$, so $ABOC$ is cyclic.

The [b][i]only solution[/i][/b] without defining new lines or points.

Let the centers of $~$ $S_1 \wedge S_2$ $~$ be $~$ $O_1 \wedge O_2$ $~$ respectively, and $~$ $O$ $~$ be the circumcenter of $~$ $\triangle{BDC}. $

[hide="Solution"]

$ \boxed{ \angle{BOC} = } $

$ = \angle{DOC} + \angle{DOB} $

$ = \widehat{DC} + \widehat{DB} $

$ = 2 \cdot \angle {DBC} + 2 \cdot \angle {DCB} $ $~$  [hide="Explanation"]Peripheral and central angles over the same arc.[/hide]

$ = \angle {DBC} + \angle {DCB} + 180^\circ - \angle{BDC} $ $~$ [hide="Explanation"]Sum of angles in a triangle - $~$ $\triangle{BDC}.$[/hide]

$ = \angle {DBC} + ( \angle{DCA} + \angle{ACB} ) + ( \angle{O_{2}DC + \angle{O_{1}DB} ) }$ $~$ [hide="Explanation"] $ O_1 \wedge O_2 $ $~$ are the centers of two circles, and $~$ $D$ $~$ is the point at which these two circles touch, hence $~$ $ O_1 , O_2 \wedge D$ $~$ are collinear. [/hide]

$ = \angle {DBC} +  \angle{DCA} + \angle{ACB}  +  \angle{O_{2}CD + \angle{O_{1}BD} }$ $~$ [hide="Explanation"] $ \overline{O_{2}D} =
\overline{O_{2}C} $ $~$ as radii of circle $~$ $S_2$ $~$ hence $~$ $ \triangle{O_{2}CD} $ $~$ is isosceles, and similarly $~$ $ \triangle{O_{1}BD}
$ $~$ is isosceles. [/hide]

$ = \angle{DBC} + \angle{ACB} + ( \angle{DCA} + \angle{DCO_2} ) + \angle{O_{1}BD} $

$ = \angle{DBC} + \angle{ACB} + 90^\circ + \angle{O_{1}BD} $ $~$ [hide="Explanation"] $~$ $ \overline{ O_{2}C } \perp AC $ $~$ because $~$ $AC$ $~$ is tangent to circle $~$ $S_2$ $~$ at point $~$ $C$ $~$ and $~$ $O_2$ $~$ is the center of circle $~$ $S_2.$ $~$ [/hide]

$ = \angle{DBC} + \angle{ACB} + ( \angle{ O_{1}BA } + \angle{O_{1}BD} ) $ $~$ [hide="Explanation"] The same situation as in the previous explanation. [/hide]

$ = \angle{DBC} + \angle{ACB} + \angle{DBA} $

$ = \angle{ACB} + ( \angle{DBC} + \angle{DBA} ) $

$ = \angle{BCA} + \angle{ABC} $

$ \boxed{ = 180^\circ - \angle{CAB} }\; \Longrightarrow\; \square{ ABOC } $ $~$ is cyclic $~$ $\Longrightarrow\; O_$ $~$ lies on the circumcircle of $~$ $\triangle{ABC}\quad \blacksquare $

[/hide]

[hide="Solution"]
Let $O_1,O_2$ be the centers of $S_1,S_2$. Let $O$ be the center of $(BCD)$. We will only deal with the configuration where both $\angle{ABD}$ and $\angle{BDC}$ are obtuse. Other configurations can be proved in a similar fashion.

Observe that $O_1,D,O_2$ are collinear, so
\[ \angle{O_1DB}+\angle{ODB}+\angle{ODC}+\angle{O_2DC}=180.\]
Note that $OD=OC$ (equal radii), so $\angle{OCD}=\angle{ODC}$. Since $AC$ is tangent to $S_2$ at $C$, $\angle{O_2CA}=90$ so $\angle{ACD}=90-\angle{O_2CD}=90-\angle{O_2DC}$ because $O_2C=O_2D$ (equal radii). Thus,
\[ \angle{OCA}=\angle{OCD}-\angle{ACD}=\angle{ODC}-(90-\angle{O_2DC})=\angle{ODC}+\angle{O_2DC-90}\]
Since $AB$ is tangent to $S_1$ at $B$ and $\angle{ABD}>90$, $\angle{BO_1D}=2\angle{ABD}>180$. Furthermore, $O_1B=O_1D$ (equal radii) so $\angle{O_1BD}=\angle{O_1DB}$, and
\[ \angle{O_1DB}=\frac{180-(360-2\angle{ABD})}{2}=\angle{ABD}-90\]
Since $OB=OD$ (equal radii), $\angle{OBD}=\angle{ODB}$. As a result,
\[ 180=\angle{O_1DB}+\angle{ODB}+\angle{ODC}+\angle{O_2DC}=\angle{ABD}-90+\angle{OBD}+\angle{OCA}+90\]
\[ =\angle{ABO}+\angle{OCA},\]
so $ABOC$ is cyclic. Thus, the circumcenter of $\triangle{BCD}$ lies on the circumcircle of $\triangle{ABC}$. // [/hide]

Here's my solution

Draw the common tangent to $S_1 $ and $S_2$ at the point $D$ . Let the tangent intersect $AC$ at $M$ and $AB$ at $N$.
Now, \[\angle{AMN} = \angle{NBD}+\angle{NDB}\]
Thus,
\[\angle{AMN}=2\angle{NDB}\].

Similarly we can arrive at $\angle{ANM}=2\angle{CDN}$

Adding these two we get \[\angle{ANM}+\angle{AMN}=2(\angle{CDN}+\angle{NDB})\]

Thus , $180^\circ-\angle{A}=2\angle{BDC}$
Therefore we can conclude that the angle subtended at the circumcentre of triangle BDC by the chord BC is $180^\circ - \angle{A}$ and thus we conclude that the circumcentre of triangle BDC lies on circumcircle of triangle ABC

Notations::$BC \cap C_1=J,BD \cap C_2=K,\angle{CBD}=p,\angle{BDJ}=x,\angle{CKD}=q$.Also let $O$ be the circumcenter of $\triangle{BCD}$.

Proof:Easy to see that $\angle{JDC}=p+q$.So $\angle{BDC}=x+p+q \Rightarrow \angle{BOC}=360-2x-2p-2q$.

Also note that $\angle{CBA}=180^{\circ}-x$.
Now
 $\angle{BCA}=\angle{BCD}-\angle{DCA}=\angle{BCD}-\angle{DKC}=180-x-2p-q-q=180-x-2p-2q$.
So we get that $\angle{BAC}=2x+2p+2q-180$.It readily follows that $A,B,O,C$ are concyclic and we are done.$\blacksquare$.

Let $AC$ intersect $S_1$ in M then N .
$S_1$ and $S_2$ are tangeant externally , and MN is a chord of $S_1 $ tangeant to  $S_2$  at C. thus DC is external bissector of $\angle MDN$ .
now we may start angle chasing:
\[\angle COB =2\angle CDB =2\angle CDM +2\angle MNB = \pi -\angle MDN+2\angle ANB =\angle MBN+ \angle ABM +\angle  ANB = \pi -\angle  CAB  \]
QED

 Let $O$ be the circumcenter of $BDC$. Let $\angle BDC=x$, so $\angle BOC=360-2x$ and we want to show $\angle BAC=2x-180$.

Let $E$ and $F$ the intersections of $AC$ with $S_1$, such that $E$ is closer to $A$ than $F$ is. Let $CD$ intersect $S_1$ again at $M$. Then $\angle BAC=\angle BAF=\angle BDF-\angle BDE$. Since $x=\angle BDC=\angle BDF+\angle FDC$. Combining the two, we need to show that $\angle BDE+\angle FDC+\angle BDC=180$, or that $\angle MDE=\angle FDC=\angle MEF$. Hence, we want to show that $M$ is the midpoint of arc $EF$ not containing $B$.

Now, let lines $ED$ and $FD$ intersect $S_2$ again at $X$ and $Y$. Note that $EF||XY$ by homothety, so $C$ is the midpoint of arc $XY$ on $S_2$. Hence $M$ is the midpoint of arc $EF$ on $S_1$.

Using directed angles:
\[
\angle BDC=\angle BDD + \angle DDC = \angle BBD+\angle DCC = \angle ABD+\angle DCA=\angle BAC - \angle BDC
\]
so if $X$ is the circumcenter of $\triangle BDC$, $\angle BXC = 2\angle BDC = \angle BAC$ as desired.

Let $O$, $P$, $Q$ be the centers of the circumcircle of $BCD$, the circle $S_1$, and the circle $S_2$, respectively. Then using directed angles modulo $180^\circ$, we have \[\angle OCA = \angle OCQ + 90^\circ = \angle QDO + 90^\circ = \angle PDO + 90^\circ = \angle OBP + 90^\circ = \angle OBA\] so $O$, $A$, $B$, $C$ are concyclic.

[hide="solution"]
Invert w.r.t $D$, and we have a new problem:
Given two circles $\omega_1, \omega_2$ with intersection points $A, D$, draw two parallel lines so that they are tangent to $\omega_1, \omega_2$ respectively at $B, C$ respectively, and $A, D$ are both between the lines. Reflect $D$ about line $BC$; prove that $D\in (ABC)$.

In other words, we wish to prove $\angle BDC+\angle BAC=180^{\circ}$. Let the direction of the parallel lines be $\ell$, and draw two lines passing through $A, D$ parallel to $\ell$. Then:
$$\angle (BD, \ell) = \angle DAB$$ $$\angle ( \ell, AC) = \angle CDA$$ $$\angle (AB, \ell)+\angle ( \ell, CD)  = \angle (AB, CD)$$
Adding these equations up gives the desired result.
[/hide]

My solution:
Let $O,O_1,O_2$ be the centers of $(BCD),S_1$ and $S_2,$ respectively.
We have $O_1BOD$ and $O_2DCO$ deltoid.
Then we know $\angle ABO=\angle O_1BO+90=\angle O_1DO+90=90+180-\angle ODO_2==270-\angle OCO_2=279-90+\angle ACO=180-\angle ACO.$
So $A,B,O,C$ are cyclic.

Here's a simple solution with just angle chasing.\\

Let the centers of $S_1, S_2$ be $O_1, O_2$ resp.\\
Let $E\equiv O_1B\cap O_2C,\ \angle BAC=2\alpha ,\ \angle ABD=\beta$\\
Because $\angle ABE=\angle ACE=90 \Longrightarrow ACBE$ is cyclic.
$\Longrightarrow \angle BEC=\angle BAC=2\alpha,\ \angle BO_1D=2\angle ABD=2\beta \Longrightarrow \angle EO_2O_1=180-2\beta -2\alpha.$\\
And because $O_1B=O_1D,\ O_2D=O_2C\Longrightarrow \angle BDO_1=90-\beta,\ \angle CDO_2=\alpha + \beta\Longrightarrow \angle CDB=90-\alpha.$
And if $O$ is the circumcenter of $\triangle BCD$ thus $\angle COB=2\angle CDB=180-2\alpha=180-\angle BAC$,\ hence $O\in (ABC).$

My solution: Let $AC \cap S_1=P,Q$. WLOG assume that $P$ is closer to $A$ than $C$. Also let $CD \cap S_1=M,BM \cap AC=K$.

From the homothety that sends $S_1$ to $S_2$, one can see that $M$ is the midpoint of $\overarc{PQ}$ (not containing $B$). By an easy angle chase, one gets that $\angle BKP=\angle BQM= \angle BDM \Rightarrow \angle BKC=\angle BDC \Rightarrow K$ lies on $\odot (BDC)$.

Now, let $O$ be the circumcenter of $\odot (BKDC)$. Thus, $\angle BOC=2(180^{\circ}-\angle BKC)=2\angle BKA=180^{\circ}-\angle BAC$, where the last equality follows from the fact that $A$ is the circumcenter of the $B$-Appolonius circle of $\triangle BPQ$, which passes through $B$ and $K$. Hence, done.

G1's tougher than G3 :(
[quote=ISL 2002 G1]Let $B$ be a point on a circle $S_1$, and let $A$ be a point distinct from $B$ on the tangent at $B$ to $S_1$. Let $C$ be a point not on $S_1$ such that the line segment $AC$ meets $S_1$ at two distinct points. Let $S_2$ be the circle touching $AC$ at $C$ and touching $S_1$ at a point $D$ on the opposite side of $AC$ from $B$.  Prove that the circumcentre of triangle $BCD$ lies on the circumcircle of triangle $ABC$.[/quote]
[b]Solution:[/b] Let $O_1,O_2$ be the centers of $S_1,S_2$. Let circumcenter of $\Delta BCD$ be $E$. then $EO_2,AC$ and tangent at $D$ are concurrent say at $F$, then, $CO_2FD$ cyclic and,
$$\angle BEC=\angle CED+\angle BED=2\angle O_2EO_1=2(90^{\circ}-\angle EO_2O_1)=180^{\circ}-\angle CO_2D=180^{\circ}-\angle DFA=180^{\circ}-\angle BAC$$
Hence, $E \in \odot (ABC)$

No Inversive solutions yet, OK So adding mine. :)
[quote=ISL 2002 G2]Let $B$ be a point on a circle $S_1$, and let $A$ be a point distinct from $B$ on the tangent at $B$ to $S_1$. Let $C$ be a point not on $S_1$ such that the line segment $AC$ meets $S_1$ at two distinct points. Let $S_2$ be the circle touching $AC$ at $C$ and touching $S_1$ at a point $D$ on the opposite side of $AC$ from $B$.  Prove that the circumcentre of triangle $BCD$ lies on the circumcircle of triangle $ABC$.[/quote]

We will Invert around $B$ with any arbitary radius and denote primes for Inverse points. So after this Inversion the Problem is read as follows.
[quote=Inverted Problem]Consider a line $S_1'$ paralel to another line which contains points $A',B$. We draw an arbitary circle through $A',B$ and another circle $S_2'$ tangent to the circle passing through $A',B$ at $C'$ and $S_1'$ at $D'$. Let the Perpendicular Line from $B$ to $C'D'$ and $A'C'$ intersect at a point $O$. Then $O$ is the reflection $B$ on $C'D'$.[/quote]
Here's the diagram for the Inverted Image.

[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -11.37, xmax = 11.19, ymin = -6.65, ymax = 6.65;  /* image dimensions */

 /* draw figures */
draw((-4.63,-1.75)--(3.07,-1.65), linewidth(1.2) + linetype("4 4"));
draw((-4.67,0.05)--(3.03,-0.01), linewidth(1.2) + linetype("4 4"));
draw(circle((-2.1334339628852055,-0.3706919036014051), 1.4562325573290364), linewidth(1) + red);
draw(circle((0.65323935546895,2.6852429448011437), 2.681268139610792), linewidth(1) + red);
draw((-2.669811129848229,-1.7245430016863406)--(0.31,3.07), linewidth(1));
draw((-1.1103145838955413,0.6655706101280405)--(0.7898243317997222,0.007455914297664498), linewidth(1));
draw((-1.5620786652693728,-1.710156865782719)--(0.31,3.07), linewidth(1));
draw((0.7898243317997222,0.007455914297664498)--(-2.187564832842252,1.0845342334449162), linewidth(1));
draw((-2.669811129848229,-1.7245430016863406)--(-2.187564832842252,1.0845342334449162), linewidth(1));
draw((0.31,3.07)--(1.61,5.19), linewidth(1));
draw((1.61,5.19)--(0.5104317032040475,-1.6832411467116357), linewidth(1));
draw((0.7898243317997222,0.007455914297664498)--(6.59,-1.65), linewidth(1));
draw((3.07,-1.65)--(6.59,-1.65), linewidth(1) + linetype("4 4"));
draw((-1.1103145838955413,0.6655706101280405)--(-1.5620786652693728,-1.710156865782719), linewidth(1));
 /* dots and labels */
dot((-4.63,-1.75),dotstyle);
label("$A$", (-4.55,-1.55), NE * labelscalefactor);
dot((3.07,-1.65),dotstyle);
label("$B_1$", (3.15,-1.45), NE * labelscalefactor);
dot((-4.67,0.05),dotstyle);
label("$C$", (-4.59,0.25), NE * labelscalefactor);
dot((3.03,-0.01),dotstyle);
label("$D$", (3.11,0.19), NE * labelscalefactor);
label("$S_1'$", (-2.43,-0.27), NE * labelscalefactor);
dot((-3.530225049583766,0.04111863675000337),dotstyle);
label("$E$", (-3.45,0.25), NE * labelscalefactor);
dot((-0.730395051067906,0.019301779618710952),dotstyle);
label("$F$", (-0.55,-0.31), NE * labelscalefactor);
dot((-2.669811129848229,-1.7245430016863406),dotstyle);
label("$A'$", (-2.89,-2.15), NE * labelscalefactor);
dot((0.7898243317997222,0.007455914297664498),dotstyle);
label("$D'$", (0.87,0.21), NE * labelscalefactor);
dot((-1.1103145838955413,0.6655706101280405),dotstyle);
label("$C'$", (-1.03,0.87), NE * labelscalefactor);
dot((1.61,5.19),dotstyle);
label("$T$", (1.69,5.39), NE * labelscalefactor);
label("$S_2'$", (-0.69,4.57), NE * labelscalefactor,red);
dot((-1.5620786652693728,-1.710156865782719),dotstyle);
label("$B$", (-1.49,-1.51), NE * labelscalefactor);
dot((0.31,3.07),dotstyle);
label("$O$", (0.39,3.27), NE * labelscalefactor);
dot((-2.187564832842252,1.0845342334449162),dotstyle);
label("$K$", (-2.11,1.29), NE * labelscalefactor);
dot((0.5104317032040475,-1.6832411467116357),dotstyle);
label("$Y$", (0.59,-1.49), NE * labelscalefactor);
dot((6.59,-1.65),dotstyle);
label("$X$", (6.67,-1.45), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
 /* end of picture */[/asy]

Let the circle passing through $A'B$ be $\omega$ and let $C'D'\cap\omega=K$ and $A'C'\cap S_2'=T$ , $C'D'\cap A'B=X$ and $TD'\cap A'B=Y$.

Notice that $$\angle C'TD'=\angle C'D'C=\angle C'XA'\implies C'TXY\text { is a cyclic quadrilateral}$$
Now notice that the Homothety $\mathcal H$ at $C'$ sending $\omega$ to $S_2'$ sends $K$ to $D'$ and $A'$ to $T$. So this implies $TD'\|KA'$. So, $\angle C'D'T=\angle A'KC'=\angle C'BX$. So we get that $\angle OC'D'=\angle D'C'B$. So by congruence we get that $O$ is the reflection of $B$ in $C'D'$. Hence, Inverting back we are done. $\blacksquare$

[hide=Simple angle chasing]
Let the center of \(S_2\) be \(O_1\) and the center of \(S_1\) be \(O_2\) (sorry for weird labelling) and let \(O\) be the center of \((BCD)\). Then, quite simply
\begin{align*}
\measuredangle BOC &= 2\measuredangle BDC \\
&= \measuredangle BO_2O_1 + \measuredangle O_2O_1C \\
&= \measuredangle BCO_1 + \measuredangle O_2BC \\
&= \left(\frac{\pi}{2} + \measuredangle BCA\right) + \left(\frac{\pi}{2} + \measuredangle ABC \right) \\
&= \measuredangle BAC
\end{align*}
[/hide]

Invert at $C$ to get the following problem:

Circles $\omega_1$ and $\omega_2$ are tangent at point $B$. Points $B,A,C$ are on $\omega_2$ in that order. Point $D$ on $\omega_1$ is on the opposite side of line $AC$ from $B$ and the tangent to $\omega_1$ at $D$ is parallel to point $AC$. Show the reflection of $C$ over line $BD$ lies on line $AB$.

To get rid of $\omega_1$, note by homothety at $B$, line $BD$ is the same as the line through $B$ and the arc midpoint of arc $AC$ on the same side of $AC$ as point $B$, point $M$. Then the result amounts to checking
\[\measuredangle ABC=2\measuredangle MBC=2\measuredangle MAC=\measuredangle AMC,\]
which is trivial.

Let $O_1,O_2$ be the centers of $S_1,S_2$, respectively. Now let $E$ be the intersection of lines $CO_2$ and $BO_1$. Since $\angle ACE=\angle ABE=90^{\circ}$, $ABCE$ is cyclic. We will use phantom points; let $O$ be the midpoint of minor arc $BC$; we will prove that the circle going through $B$ and $C$ centered at $O$ also goes through $D$.

Notice that triangles $BO_1D$ and $CO_2D$ are isosceles. Then $\angle EO_1D=\angle O_1DB+\angle O_1BD=2\angle O_1DB$, so $\angle O_1DB=\frac{1}{2}\angle EO_1D$. Similarly, $\angle O_2DC=\frac{1}{2}\angle EO_2D$. Now let $X$ be a point on the circle centered at $O$ going through $B$ not in circle $ABC$. Then \begin{align*} \angle CDB &= 180^{\circ}-\angle O_2DC-\angle O_1DB \\ &= 180^{\circ}-\frac{1}{2}(\angle EO_2D+\angle EO_1D) \\ &= 180^{\circ}-\frac{1}{2}(180^{\circ}-\angle CEB) \\ &= 90^{\circ}+\frac{1}{2}\angle CEB \\ &= \frac{1}{2}(180^{\circ}+\angle CEB) \\ &= \frac{1}{2}(180^{\circ}+(180^{\circ}-\angle COB)) \\ &= \frac{1}{2}(360^{\circ}-2\angle CXB) \\ &= 180^{\circ}-\angle CXB. \end{align*} Thus $CDBX$ is cyclic, and $D$ lies on the circle centered at $O$ passing through $B$ and $C$. The proof is complete.

Let O,P1,P2 be circumcenters of BCD,S1,S2. Lets prove ABOC is cyclic.
Note that P1D = P1B and OD = OB so ODP1 and OBP1 are congruent. Also OCP2 and ODP2 are congruent. D is tangent to S1,S2 so P1,O,P2 are collinear.
∠ABO = 90 + ∠P1BO = 90 + ∠P1DO so we have to prove ∠OCA = 90 - ∠P1BO.
∠OCA = ∠OCP2 - 90 = ∠ODP2 - 90 = 180 - ∠ODP1 - 90 = 90 - ∠ODP1.
we're Done.

I am the new definition of overkill

[img width=50]https://media.discordapp.net/attachments/925784397469331477/950244886421074011/Screen_Shot_2022-03-06_at_9.13.51_PM.png[/img]

Let $O_1,O_2$ be the centers of $S_1,S_2$ respectively and $O$ be circumcenter of $\triangle BCD.$ It suffice to show that $ABCO$ is cyclic. Note that \[\angle BAC=180^\circ-\angle ABC-\angle ACB\]
\[=180^\circ-\angle ABC-\angle ACB\]
\[=180^\circ-\angle ABD-\angle CBD-\angle ACB\]
\[=180^\circ-\angle CBD-90^\circ+\angle O_1DB-\angle ACB\]
\[=90^\circ-\angle CBD+\angle O_1DB-\angle ACB\]
(Here comes the leap of faith step)
\[=180^\circ-\angle DCA-\angle O_2CD-\angle CBD-\angle ACB+\angle O_1DB\]
\[=180^\circ-\angle DCB-\angle CBD-\angle O_2CD-\angle O_1DB\]
\[=\angle CDB-180^\circ+\angle CDB\]
\[=-180^\circ+2\angle CDB\]
\[=180^\circ-\angle COB\]
so we are done.

Let $O,O_1,O_2$ be center of circles $(BCD),S_1,S_2$, respectively. Note that $O_1-D-O_2$ are collinear.

Since $O_2C=O_2D$ and $OC=OD$, we have $\triangle O_2CO\cong \triangle O_2DO$.
\begin{align*}
\angle O_2DO
&= \angle O_2CO \\
&= \angle O_2CA+\angle ACO \\
&= 90^{\circ}+\angle ACO
\end{align*}
Similarly, $O_1B=O_1D$ and $OB=OD$ so $\triangle O_1DO\cong \triangle O_1BO$.
\begin{align*}
\angle O_1DO
&= \angle O_1BO \\
&= \angle ABO-\angle ABO_1 \\
&= \angle ABO-90^{\circ}
\end{align*}

Hence, we have
\begin{align*}
180^{\circ}
&=\angle O_2DO+\angle O_1DO \\
&= (90^{\circ}+\angle ACO)+(\angle ABO-90^{\circ}) \\
&= \angle ACO + \angle ABO.
\end{align*}
Thus, $A,B,C,O$ concyclic, as desired.

Let $O_1,O_2,O$ be centers of $S_1,S_2,\odot (BCD).$ Clearly $D\in O_1O_2,O_1B\perp AB,O_2C\perp AC,$ so $$\measuredangle BAC=\angle (BO_1,CO_2)=\measuredangle BO_1D+\measuredangle DO_2C=2\measuredangle BDC=\measuredangle BOC\implies O\in \odot (ABC).$$

[asy] size(10cm); import geometry; draw(circle((-1,0),1)); draw(circle((2,0),2)); 	circle c1 = circle((point)(-1,0),1); 	circle c2 = circle((point)(2,0),2); 	point B = 2*dir(65)+2, C = dir(85)-1, D = (0,0); 	point E = intersectionpoint(tangent(c1,D),tangent(c1,C)); 	point F = intersectionpoint(tangent(c2,D),tangent(c2,B)); 	point A = intersectionpoint(tangent(c2,B),tangent(c1,C)); dot("$B$", B, dir(60)); dot("$C$", C, dir(60)); dot("$D$", D, dir(-15)); 	dot("$E$", E, dir(55)); dot("$F$", F, dir(60)); dot("$A$", A, dir(60)); 	draw(C--A--F--D); 	draw(C--D--B); [/asy]
Let $\ell$ be the common internal tangent to both circles, and let $E = AC \cap \ell$ and $F = AB \cap \ell$. Let $O$ be the circumcenter of $BCD$. To prove that $O \in (ABC)$, we only need to show that $\measuredangle COB = \measuredangle CAB$ (all angles directed). However, $\measuredangle COB = 2 \measuredangle CDB$ (circumcenter angles). Furthermore, we have: \begin{align*} \measuredangle CAB &=  \measuredangle EAF \\ &= -\measuredangle AFE - \measuredangle FEA \\ &= - \measuredangle BFD - \measuredangle DEC \\ &= \measuredangle FDB + \measuredangle DBF + \measuredangle ECD + \measuredangle CDE \\ &= 2(\measuredangle CDE + \measuredangle FDB) \\ &= 2\measuredangle CDB \end{align*} which finishes.

Does this work? (I think the part where $\measuredangle BO_1D=2\measuredangle BDO_1$ might be incorrect.)

Let the center of $S_1$ be $O_1$ and let the center of $S_2$ be $O_2$. Also let the circumcenter of $BCD$ be $O$. Then let $BO_1$ and $CO_2$ intersect at $A'$. Then we have
\begin{align*}
\measuredangle BAC&=\measuredangle BA'C \\
&=\measuredangle A'O_1O_2 + \measuredangle O_1O_2A' \\
&=\measuredangle BO_1D+\measuredangle DO_2C \\
&=2(\measuredangle BDO_1+\measuredangle O_2DC) \\
&=2\measuredangle BDC \\
&=\measuredangle BOC
\end{align*}
and we are done. $\blacksquare$

OKAY so there is apparently config issues. aint no way im actually gonna fix the proof

Let $O_1$ denote the center of $S_1$ and $O_2$ be defined similarly.  Let $O$ be the circumcenter of $BCD$.  The key claim is that

[color=blue][b]Claim: [/b][/color]$\measuredangle OBO_1 = \measuredangle O_1DO$.
[i]Proof: [/i]We know that $\measuredangle O_1BD = \measuredangle BDO_1$ and $\measuredangle OBD =\measuredangle BDO$ by isosceles triangles.  Subtracting gives the result.

It is sufficient to prove that $\color{blue} OBAC$[color=blue]is cyclic[/color].  The rest is straightforward angle chasing.
\begin{align*}\measuredangle OBA &= \measuredangle O_1DO + 90^{\circ} \\ &= \measuredangle O_2DO + 90^{\circ} \\ &= \measuredangle CDO + \measuredangle O_2DC + 90^{\circ} \\ &= \measuredangle OCD + \measuredangle DCO_2 + 90^{\circ} \\ &= \measuredangle OCO_2 + 90^{\circ}  \\ &= \measuredangle OCA,\end{align*}and we're done. $\square$

Let $I$ be the incenter of $ABC$, segment $AC$ meet $S_1$ again at $Y$ and $Z$ so that $AY < AZ$, the common tangent to $S_1$ and $S_2$ meet $AC$ at $M$, segment $BC$ meet $S_1$ again at $X$, the midpoint of arc $YZ$ not containing $D$ be $N$, the reflection of $C$ over $M$ be $E$, and $B_1$ be the point on $S_1$ such that $BB_1 \parallel YZ$. The parallel condition implies $N$ is also the midpoint of arc $BB_1$.

Because $ME = MC = MD$ follows from equal tangents, we have $\angle CDE = 90^{\circ}.$ Now, observe $$MC^2 = |Pow_{S_2}(M)| = MD^2 = |Pow_{S_1}(M)| = MY \cdot MZ$$ so $-1 = (C, E; Y, Z)$. Thus, the Right Angle and Bisectors Lemma implies $DE$ bisects $\angle YDZ$, which means $D, E, N$ are collinear.

Now, we know $$\angle BDB_1 = 180^{\circ} - \angle BNB_1 = 180^{\circ} - \angle BNX - \angle XNB_1$$ $$= 180^{\circ} - \angle ABX - \angle XBB_1 = 180^{\circ} - \angle ABC - \angle ACB = \angle BAC.$$ This gives $$\angle BDC = \angle BDN + \angle NDC = \frac{\angle BDB_1}{2} + \angle EDC$$ $$= \frac{\angle A}{2} + 90^{\circ} = \angle BIC$$ which finishes via the Incenter-Excenter Lemma. $\blacksquare$

[b]Remark:[/b] This solution is a bit overkill.

Let $O_1$ be the center of $S_1$, $O_2$ be the center of $S_2$, and $O$ be the circumcenter of $BCD$. We have
\begin{align*}
 \measuredangle BOC &= 2 \measuredangle BDC \\
&=( \measuredangle BCD + \measuredangle DBC ) \\
&=2( \measuredangle BCA + \measuredangle ACD + \measuredangle DBA + \measuredangle ABC ) \\
&= 2(\measuredangle BAC  + \measuredangle DBA + \measuredangle ACD) \\
&= 2 \measuredangle BAC + 2 \measuredangle DB O_1 + \measuredangle CO_2D \\
&= 2 \measuredangle BAC + \measuredangle DO_1B + \measuredangle CO_2D \\
&= 2\measuredangle BAC + \measuredangle CBO_1 + \measuredangle O_2CB \\
&= 2 \measuredangle BAC + \measuredangle CBA - 90 + \measuredangle ACB +90 \\
&= \measuredangle BAC \\
\end{align*}
This means $B,A,O,C$ are concyclic as desired so we are done.

[hide=solution(has config issues but can be fixed with directed angles)]Define $O_1$ as the centre of $S_1$ and $O_2$ as the centre of $S_2$. Define $X:= CO_2\cap BO_1$. Let $E$ be the circumcentre.

Claim: $X\in (ABC)$
Proof: Note that $$\angle XCA=\angle O_2CA=90=\angle O_1BA=\angle XBA\implies X\in (ABC)$$

Now, if $\angle O_2CD=\theta$ and $\angle O_1DB=\alpha$. So we get $\angle CDB=180-(\theta+\alpha).$ Since $E$ is the circumcenter. So $\angle CEB=2(\theta+\alpha)$. Also, we have $\angle CXB=180-2(\theta+\alpha)$. But $\angle CXB=\angle CAB$[/hide].

Let $O$ be the circuncenter of $BCD$. It would be sufficient to prove that
$360^{\circ}-2\angle CDB=\angle DOC=180^{\circ}- \angle BAC$\\
$\iff 180^{\circ}=2\angle CDB - \angle BAC$
Now invert at $C:$
We need
$180^{\circ}=2\angle C'B'D' - \angle A'B'C' \iff 90^{\circ} = \angle C'B'D' - \frac{\angle A'B'C'}{2}$
Let the line parallel to the tangent at $D'$ through $B'$ meet the circles $(B'D'), (B'A'C')$ at $E, F$. Let M be the midpoint of the arc $A'C'$ not containing $B'$\\
$A'C' \parallel EF \Rightarrow \angle EB'D' + \angle C'B'M = \angle D'B'M = 90 $\\
But $90^{\circ}=\angle D'B'M = \angle C'B'D' - \frac{\angle A'B'C'}{2}$

Nice angle chasing. We have <BFC=360-2<BDC=2(<BCD+<CBD)=2(<ACB+<ACD+<ABC-<ABD)=2(180-<BAC+<ACD-<ABD)=360-2<BAC+<CED-<BOD (tangent angle rules)=... I'm not sure how to continue, I ask someone to post a solution like this. Edit: I saw lrjr's but i didnt get exactly how he got the third to last line from the fourth

Anyways, second attempt: First, we have <FBD=<FDB, <ODB=<OBD. Then we have <FBO=<FDO. We have <FBA=90+<FBO=90+180-<EDF=180+90-<ECF=180-<FCA, so ABCF is cyclic, as desired. $\blacksquare$

Edit: After comparing with others, it seems like directed angles are necessary, but I don't want to do them right now.

only angle chasing!
[hide = solution]
Let $O_1, O_2$ be the centres of the circles tangent to $AB, AC$ at $B, C$ respectively. Observe that then $O_1C \perp AC, O_2B \perp AB$ and $O_1, O_2, D$ are collinear. Now \[2 \angle BDC = 360^{\circ}-2(\angle CDO_1 + \angle BDO_2) = \angle CO_1D + \angle BO_2D = 360^{\circ}-\angle O_1CB - \angle O_2BC = 180^{\circ}+\angle CAB\] and so $BDC$ is obtuse. Thus if $O_3$ is the circumcentre of $BCD$, we have $\angle CO_3B = 360^{\circ}-2\angle BDC = 180^{\circ}-\angle CAB$. Thus $CO_3BA$ is cyclic and we're done.
[/hide]


A binary tree is a path in which each road splits into 2 more roads.  A specific binary tree ends on the 4th row (when there are 16 final nodes).  A particle begins at the root and travels along the tree, equally likely to go in either direction.  The particle wins if he lands on particle 1 (starting from the left).  He loses if he lands on an even node, and if he lands on any odd node other than 1, he is transported back to the start.  But here's the catch: each time he is transported to start, his probability of goin left increases.  At the beginning, he has a $\frac{1}{2}$ chance of going left, but after he gets transported $n$ times, his probability of going left is $\frac{n+1}{n+2}$.  Compute the probability that he will eventually win.

How are the nodes numbered? I'm guessing it's like this...
LLLL goes to 1
LLLR goes to 2
LLRL goes to 3
LLRR goes to 4
etc.

Using that system:
No matter which of the first three directions we go in, going right will always cause us to lose. So our probability of losing in any tree is always $\frac{1}{n+2}$. From one run of the tree we win with a chance of $\left(\frac{n+1}{n+2}\right)^{4}$. All other cases we start over with $n$ incremented. So if $P(n)$ is the probability of winning from $n$ transports, then we get the following recursive definition.

$P(n) = \left(\frac{n+1}{n+2}\right)^{4}+\left(\frac{n+1}{n+2}-\left(\frac{n+1}{n+2}\right)^{4}\right)P(n+1)$

We've reduced the problem to calculating $P(0)$, which is a really messy problem I don't see how to do. Maybe someone has a more ingenious way of solving this.

(since when did particles have genders? :ninja: )

Let f(x)= ((x+1)/(x+2))^4 Let g(x)= 1-(f(x-1)+1/(x+1)) Then H(n)= Prod( Seq ( g(x),x,1,n))f(x) gives the probability of winning at each stage Starting at stage 2 ie H(1) = probabily winning going down the second time. The sum of H(n) from 1 to infinity plus 1/16 (the chance you win the first time down) gives you .27687718 which the Inverse Symbolic Calculator does not know.


Here's a problem that puzzled my class, my teacher, and the head of our math department for a while:

How many distinct planes can be determined by a combination of the vertices of a cube?

[quote="worthawholebean"]Here's a problem that puzzled my class, my teacher, and the head of our math department for a while:

How many planes can be determined by a combination of the vertices of a cube?[/quote]

[hide]3 points determine a plane... a cube has 8 vertices... ${8\choose{3}}=56$. Each plane is counted 4 times. $\frac{56}{4}=14$. Is that right?[/hide]

[hide="Correct answer - click to show"]Nat MC is correct. His description is better than any way I could describe it.

EDIT: Oops. didn't see his flaw. His number is correct, but the planes aren't.[/hide]

[quote="nat mc"]No, it's 16 Cube ABCDEFGH.  A above E, Babove F and so on.  There are six for the faces. Then, planes ABGH,CDEF, ADFG, BCEH, ACH, ACF, BDE, BDG, EGB, EGD, FHC, FHA[/quote]

You said 6 for the faces, and then you listed another 12, wouldn't that make 18 and not 16? Or maybe I counted wrong (never have been too good at that :D).

[quote="nat mc"]No, it's 16 Cube ABCDEFGH.  A above E, Babove F and so on.  There are six for the faces. Then, planes ABGH,CDEF, ADFG, BCEH, ACH, ACF, BDE, BDG, EGB, EGD, FHC, FHA[/quote]

[hide]You missed planes ACEG and BDFH. The complete listing of planes is:

Faces:
ABCD
EFGH
ABEF
CDGH
BCFG
ADEH

Other:
ABGH
CDEF
ACEG
BDFH
ADFG
BCEH
ACH
ACF
BDE
BDG
EGB
EGD
FHC
FHA

Oops!  :blush: It's actually 20. Disregard the original answer.[/hide]

So...
You do (#ofverticesC3)/4 and add on the 6 faces?

Nope. Not all the planes are defined by four points.


Er, what am I supposed to solve for? :?:

[quote="Lanyes"]$ 0\leq a,b,c\leq 1.$.Solve the inequality:
$ {a}^{3}{(b \minus{} c)}^{2} \plus{} {b}^{3}{(c \minus{} a)}^{2} \plus{} {c}^{3}{(a \minus{} b)}^{2} \geq {(b \minus{} c)}^{2}{(c \minus{} a)}^{2}{(b \minus{} a)}^{2}.(ab \plus{} bc \plus{} ca)$[/quote]
Maybe we need to prove this inequality?


If $ n\equal{}  p^{k}\cdot m^{2}$ is perfect number prove that $ n\minus{}p$ is divisible by $ 8$.

Just want to clarify.. is $ p$ a prime and what is $ m$- integer?

I am sorry, it should be stated that $ p$ is prime and $ m$ integer. Thank you [b]riddler[/b]


r(t)=(3t^2, 4sqrt(2)t^1.5, 6t) P=(12,16,12)

Let f(x,y,z) be defined in a neighbourhood of P and satisfying (gradient)f(P)=(1,1,2). find d(f(r(t)))/dt at t=2.

Sorry that this doesn't answer your question but I'm going to clean up your post so others can read it.  Is this right?
$ \textbf{r}(t)\equal{}(3t^{2},4\sqrt{2}t^{3/2},6t)$ $ P\equal{}(12,16,12)$
Let $ f(x,y,z)$ be defined in the a neigborhood of $ P$ and satisfying $ \nabla f(P)\equal{}(1,1,2)$.  Find $ \frac{d}{dt}f(\textbf{r}(t))$ at $ t\equal{}2$.


Here is some simple practice AMC problems:

If $ (a^{2}+b^{2})^{3}=(a^{3}+b^{3})^{2}$, and $ ab$ does not equal 0, what is the value of $ \frac{a}{b}+\frac{b}{a}$?

Given the following series $ \frac{3}{4}+\frac{9}{8}+\frac{27}{16}+...$, what is the sum of the first 7 terms?

The fraction of men in a population what are married is $ \frac{2}{3}$. The fraction of women in the population who are married is $ \frac{3}{5}$. What fraction of the population is single? (NOTE: There are no gay couples, therefore number of married men is equal to the number of married women.)

[hide="2"] Um... I was in a rush so:
It's a geometric series, so we all know what to do. Is the answer $ \frac{6177}{256}$? I'm pretty sure the denominator is right but dunno about numerator :maybe: [/hide]

first one is 2/3..

second one xpmath got right

third one is 7/19


If $ a^2\equal{}a\plus{}1$, prove that $ a^n\equal{}F_{n\minus{}1}a\plus{}F_{n\minus{}2}$ with $ F_0\equal{}F_1\equal{}1,F_{n\plus{}1}\equal{}F_n\plus{}F_{n\minus{}1}$.

[hide="Hint"]Muliply the equation to be proven by $ a$, and tinker around with it a bit. What kind of proof seems appropriate?[/hide]

See the (rather long) discussion [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=284291]here[/url].

Using $ x^{n\plus{}2}\equiv x^{n\plus{}1}\plus{}x^{n}\bmod (x^{2}\minus{}x\minus{}1)$ this problem (and further generalized by kunny's one) can be proved using induction readily, but can this also be proven using the fact that $ a^2\minus{}a\minus{}1\equal{}0$ is the characteristics eqn for the Fibonacci sequence?

Yes; they're essentially the same argument.

Aren't these proofs covered in the AOPS volume 2?

Can I just use simple induction, like what I did?

Firstly, show that it holds for $ n\equal{}2$

$ a^{2}\equal{}aF_{1}\plus{}F_{0}$
$ a^{2}\equal{}a\plus{}1$

This holds based on the identity. Then assume that $ a^{n}\equal{}aF_{n\minus{}1}\plus{}F_{n\minus{}2}$. It follows that,

$ a \cdot a^{n}\equal{}a^{2}F_{n\minus{}1}\plus{}aF_{n\minus{}2}$

Substituting that $ a^{2}\equal{}a\plus{}1$, one gets that

$ a^{n\plus{}1}\equal{}(a\plus{}1)F_{n\minus{}1}\plus{}aF_{n\minus{}2}$
$ a^{n\plus{}1}\equal{}a(F_{n\minus{}1}\plus{}F_{n\minus{}2})\plus{}F_{n\minus{}1}$

Substituting that $ F_{n}\equal{}F_{n\minus{}1}\plus{}F_{n\minus{}2}$ gives that

$ a^{n\plus{}1}\equal{}aF_{n}\plus{}F_{n\minus{}1}$

QED.


Hi, there
Iraninan 3rd round olympiad has begun.
I post questions of Algebra exam in Persian this year.
Have a good day

And here is the problems of Geometry exam:
But it has 3 errors:
1. soale 2 A'B' ro bokon B'C' (khate 2).
2. soale 3, 2 ro bokon 3/2 (khate 2).
3. soale 4, khate 2, H ha ro bokon K!

Here are the Number Theory exam problems.


For any finite set $S$ of real numbers, let $r(S) = \max{(S)} - \min{(S)}$. If $A$ is a set of thirty distinct numbers, find the maximum number of different values $r(S)$ can take on as $S$ ranges over all five-element subsets of $A$.

[hide]351.  I chose my set so that no two distinct elements had the same difference (i.e the powers of two).  Intuitively, the max and min must be five elements apart.  Also subsets that do not change the max and min do not need to be counted.  say the set is ordered and the max is $a_{30}$, then at most  the min can  be $a_{26}$ and at least $a_1$. Therefore, there are 26 distinct differences if the max is $a_{30}$.  The total number of distinct differences is then $26+25+$...$+2+1=\frac{1}{2}26(27)=351$. :D[/hide]


algebra
function
domain
calculus
derivative
real analysis
real analysis unsolved

Here is a very funny fact:  let $f$ an arbitrary real function. Then the set of points where the left and right derivatives of f exist and are different is at most numerable.

Given such a point $x_0$ where the lateral derivatives exist but are not equal, take some rational $r$ strictly between the values of the lateral derivatives. This $x_0$ will be a strict local extremum of the function $f_r(x)=f(x)-rx$, and since there are countably many such functions $f_r$, it suffices to prove that given any real function $f$, the set of strict local extrema is countable.

First notice that the set $S=\{f(x)\}$ for strict local extrema $x$ is countable, since every such $f(x)$ is the minimal/maximal value of $f$ on a certain interval with rational endpoints. Next, notice that the set of those strict local extrema $x$ for which $f(x)$ is a certain value in $S$ is discrete and thus countable, because around each strict local extremum $x$ there is an interval $I$ such that $f^{-1}(f(x))\cap I$ has exactly one element, namely $x$. Combining these two observations, we get the desired result.


function
floor function
geometry
3D geometry
combinatorics unsolved
combinatorics

Let $ N$ be a positive integer. Find the volume of the solid in function of $ N$ such:
$ x,y,z, \in [0, \plus{} \infty)$
 $ \lfloor x \rfloor \plus{} \lfloor y \rfloor \plus{} \lfloor z \rfloor \le N$
I didn't understand what volume is this.

Answer: $ \binom{N \plus{} 3}{3}$

This is how I see it:
For any integer solution $ x+y+z\leqq$, you can have it correspond to a cube with side length $ 1$ in 3-D space. (from $ x$ to $ x+1^-$, $ y$ to $ y+1^-$, $ z$ to $ z+1^-$) This has a volume of 1. So the volume will be the amount of solutions to $ x+y+z\leqq$, which is clearly $ \binom{N+3}{3}$.

I didn't understand this cube. I can't see it. And why does it have unitary dimensions? Why you set $ [x, x\plus{}1)$,  $ [y,y\plus{}1)$,$ [z,z\plus{}1)$?


induction
geometric sequence
geometric series

Let $ S \equal{} \frac {3}{2} \plus{} \frac {5}{4} \plus{} \frac {7}{8} \plus{} \frac {9}{16} \plus{} \ldots$

$ \frac {S}{2} \equal{} \frac {3}{4} \plus{} \frac {5}{8} \plus{} \frac {7}{16} \plus{} \frac {9}{32} \plus{} \ldots$

$ S \minus{} \frac {S}{2} \equal{} \frac {3}{2} \plus{} \frac {1}{2} \plus{} \frac {1}{4} \plus{} \frac {1}{8} \plus{} \ldots$

$ \frac {1}{2} \plus{} \frac {1}{4} \plus{} \frac {1}{8} \plus{} \ldots$ is a geometric sequence. It converges to $ \frac {1}{1 \minus{} \frac {1}{2}} \equal{} 2$

So the answer is $ 2\left(2 \plus{} \frac {3}{2}\right) \equal{} \boxed{7}$.

(Credit to archimedes1 for the geometric mean idea).

It is actually an arithmetico-Geometric sequence

for more details check this  :)
[url]http://www.artofproblemsolving.com/Wiki/index.php/Arithmetico-geometric_series[/url]

$ S \minus{} \frac {S}{2} \equal{} \frac {3}{2} \plus{} \frac {1}{2} \plus{} \frac {1}{4} \plus{} \frac {1}{8} \plus{} \ldots \implies
\frac{1}{2}S\equal{}\frac{3}{2}\plus{}2 \implies S \equal{} 2\left(2 \plus{} \frac{3}{2}\right)$

The $ \frac{3}{2}$ was not part of the infinite geometric sequence which is why it had to be added on.

actually, i just noticed that it says that the answer is 5...not 7...

Yes the geometric series converges at 1 not 2
therefore the answer is 5

hello, prove by induction that
$ \sum_{i\equal{}1}^n\frac{2i\plus{}1}{2^i}\equal{}\minus{}6\left(\frac{1}{2}\right)^{n\plus{}1}\minus{}4\left(\frac{1}{2}\right)^{n\plus{}1}(n\plus{}1)\plus{}5$
Sonnhard.


[quote="pqrs"]Find max possible $ k$ be a integer such that
$ (a^3 \plus{} 3)(b^3 \plus{} 3)(c^3 \plus{} 3) > k(a \plus{} b \plus{} c)^3$ for all $ a,b,c\in R^ \plus{}$[/quote]
By Holder me obtain:
$ \prod_{cyc}(a^3 \plus{} 3) \equal{} \prod_{cyc}\left(a^3 \plus{} \frac {3}{2} \plus{} \frac {3}{2}\right)\geq\frac {9}{4}(a \plus{} b \plus{} c)^3 > 2(a \plus{} b \plus{} c)^3.$


algebra
function
domain
complex analysis
complex analysis unsolved

Suppose f(z) is analytic on domain G, except the for poles in G. prove that the only singularities of g(z)=f'(z)/(f(z)-A) are simple poles at all the poles of f(z)
and all the points z in G such that f(z)=A

What's the problem?  At any point other than those listed, your function is analytic.

[quote="santug"]please any one help me[/quote]

Come on now. If $ f(z_0) \equal{} A$, then what does $ \frac{f'(z)}{f(z)\minus{}A}$ look like near $ z_0$? Hint: Use the power series of $ f$ at $ z_0$.

[quote="santug"]You mean that Laurent series expansion for f(z) about Zo[/quote]

No. Recall $ A$ is a fixed complex number. The function $ \frac{f'(z)}{f(z)\minus{}A}$ is analytic in the region except for the poles of $ f$ and the $ z$'s where $ f(z)\equal{}A$. At a pole, you'll want to look at the Laurent expansion of $ f$. At a point where $ f\equal{}A$, you'll want to look at the power series of $ f$.