[quote="GameBot"]The midpoints of the sides of a rectangle are connected to form a new quadrilateral. If the rectangle is 10 inches long and 5 inches wide, how many square inches are in the area of the new quadrilateral?[/quote]

[asy]draw((0,0)--(10,0));
draw((0,0)--(0,5));
draw((0,5)--(10,5));
draw((10,5)--(10,0));
draw((0,2.5)--(5,5),red);
draw((5,5)--(10,2.5),red);
draw((10,2.5)--(5,0),red);
draw((5,0)--(0,2.5),red);
label("$\frac{5\sqrt{5}}{2}$", (2.5,3.75), NW);
label("$\frac{5\sqrt{5}}{2}$", (7.5,3.75), NE);
label("$\frac{5\sqrt{5}}{2}$", (7.5,1.25), SE);
label("$\frac{5\sqrt{5}}{2}$", (2.5,1.25), SW);[/asy]

Since the new quadrilateral is a rhombus(all the sides have the same length), we can calculate its area by multiplying the lengths of its diagonals divided by 2. Clearly the diagonals(shown in blue) have length 10 and 5, so the area is $ \frac{10 \times 5}{2} \equal{} \boxed{25}$.

[asy]draw((0,0)--(10,0));
draw((0,0)--(0,5));
draw((0,5)--(10,5));
draw((10,5)--(10,0));
draw((0,2.5)--(5,5),red);
draw((5,5)--(10,2.5),red);
draw((10,2.5)--(5,0),red);
draw((5,0)--(0,2.5),red);
label("$\frac{5\sqrt{5}}{2}$", (2.5,3.75), NW);
label("$\frac{5\sqrt{5}}{2}$", (7.5,3.75), NE);
label("$\frac{5\sqrt{5}}{2}$", (7.5,1.25), SE);
label("$\frac{5\sqrt{5}}{2}$", (2.5,1.25), SW);
draw((0,2.5)--(10,2.5),blue);
draw((5,0)--(5,5),blue);[/asy]

Alternatively, it is trivial to show that the midpoints of a rectangle has an area of 1/2 the rectangle, so 50/2=25. The proof is to use the rhombus that isabella described above.


hi guys if anybody could help solve this or at least tell me what they're asking for it would be alot of help!

the campers needed a 200m circle track for relays. A camper stands in the centre holding the end of a rop. the helper pushing the lime cart will hold the other end.

about how long must the rope be to make a 200m track?

A.   30m                B.  60m              C. 100m              D.200m

and the number for pi is 3.14

Well one camper is holding one end of the rope
This is the centre
The other camper with the lime cart will go around the first camper while keeping the rope taught
So the camper pushing the cart goes in a circle
They want a 200 m track
Hence the circumference of the circle must be nearly 200m

You have to find the Length of the rope i.e. the radius of the circle

[hide="Answer"]Well c=2*pi*r
Thus 200=2*3.14*r
Thus r=31.8...(about 30m)[/hide]

[hide]We can transform this problem into-the circumference of a circle equals 200, what is the radius of the circle?  The way we solve this is 200 over pi over 2=radius, or $\frac{200}{2\pi}=31.831$, so we find that the closest solution is 30 meters.  Hope this helps![/hide]

What the question asks is the radius of the circle. If the circumference is 200m, what do you do to figure the diameter? Divide the answer by 2 to figure the radius.


Can someone post the problems on this year's TST at MOSP?

I believe the TST is under copyright.  You probably should wait until the AMC puts it on their problems archive.  If you're impatient, you can always email them and pester them. :P


I have to present a chemical reaction to my class, and I can't find a good one. It has to be cool to watch (flashing lights, explosions, etc.) but also performable in a classroom. Thanks in advance!

And preferably, it should be an actual chemical, not physical, reaction.

I'm not sure this would work for you, but you can take a strip dipped in an aqueous solution of copper (let it sit for a while).  Then proceed to burn the strip and see what happens.

I could look into that! What is the strip made of? just a paper strip?

No, if I can remember, it's wood (maybe balsa?  I don't know).  Ask if your chemistry teacher.

Anything that will retain the Cu is fine. (Flame test)

A cool reaction that is not necessarily safe would be throwing about 10 moles of Na into a pond (sodium is cheap) and observe from a distance. Add the same amount of HCl to the pond so the pH isn't disrupted.

(You really just end up with slightly salty water, so the reaction products are not really dangerous.)

Cesium would be more fun to watch, but I doubt local authorities will say "Sure, throw a highly combustible substance into water and kill local wildlife."

It is cool though, as seen [url=http://www.youtube.com/watch?v=eCk0lYB_8c0&feature=related]here[/url].

I tend to refrain from group I metals because they are a little overdone. :P
Thanks for all of the suggestions though!

On a side note, have you seen this article?
http://www.theodoregray.com/PeriodicTable/AlkaliBangs/index.html

Not sure of the validity of it, but...

And back to the topic (haha), I'm thinking about doing the redox reaction between glycerin and potassium permanganate. (?)

Cesium may be the most reactive, but it is expensive and its atomic size makes it so that it actually gives off a smaller bang per VOLUME compared with sodium.

Here are some great, um, "chemical" reactions:
http://listverse.com/2008/03/04/top-10-amazing-chemical-reactions/


algebra
polynomial
linear algebra
matrix
linear algebra unsolved

Let $ A$ to be an $ n \times n$ real matrix. Prove that there is a non null polynomial $ f(x): \mathbb R^{n} \rightarrow \mathbb R$ . with real coefficients, such that
\[ f(Ax) \equal{} (\det{A}) f(x)\]
for all $ x \in \mathbb R^n$.

A universal example, which works over any field:
Find $ n$ linearly independent matrices $ T_0,T_1,\cdots, T_{n\minus{}1}$ that each commute with $ A$.
$ f(x)\equal{}\det\begin{pmatrix}T_0x&T_1x&T_2x&\cdots&T_{n\minus{}1}x\end{pmatrix}$. This is a homogeneous polynomial of degree $ n$ in the entries of $ x$, and
$ f(Ax)\equal{}\det\begin{pmatrix}T_0Ax&T_1Ax&T_2Ax&\cdots&T_{n\minus{}1}Ax\end{pmatrix}$
$ \equal{}\det\begin{pmatrix}AT_0x&AT_1x&AT_2x&\cdots&AT_{n\minus{}1}x\end{pmatrix}$
$ \equal{}\det\left(A\begin{pmatrix}T_0x&T_1x&T_2x&\cdots&T_{n\minus{}1}x\end{pmatrix}\right)$
$ \equal{}\det A\cdot \det\begin{pmatrix}T_0x&T_1x&T_2x&\cdots&T_{n\minus{}1}x\end{pmatrix}\equal{} \det A\cdot f(x)$

Linear independence of the $ T_i$ is what ensures that $ f$ takes nonzero values.

OK, how do we find the $ T_i$ in general? If the characteristic polynomial of $ A$ has degree $ n$, we simply use $ I,A,A^2,\cdots,A^{n\minus{}1}$. In other cases, the space we're drawing from actually has dimension greater than $ n$, and we can find at least $ n$ from the rational canonical form. $ A$ is similar to a matrix of the form $ \begin{bmatrix}C_1&0&\cdots&0\\0&C_2&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\0&0&\cdots&C_k\end{bmatrix}$, where each $ C_j$ is a companion matrix of dimension $ d_j$. In that basis, we use $ \begin{bmatrix}C_1^0&0&\cdots&0\\0&0&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\0&0&\cdots&0\end{bmatrix}$, $ \begin{bmatrix}C_1^1&0&\cdots&0\\0&0&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\0&0&\cdots&0\end{bmatrix}$, $ \cdots$, $ \begin{bmatrix}C_1^{d_1\minus{}1}&0&\cdots&0\\0&0&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\0&0&\cdots&0\end{bmatrix}$, $ \begin{bmatrix}0&0&\cdots&0\\0&C_2^0&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\0&0&\cdots&0\end{bmatrix}$, $ \begin{bmatrix}0&0&\cdots&0\\0&C_2^1&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\0&0&\cdots&0\end{bmatrix}$, $ \cdots$, $ \begin{bmatrix}0&0&\cdots&0\\0&C_2^{d_2\minus{}1}&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\0&0&\cdots&0\end{bmatrix}$ and so on.


equation
number theory
IMO Shortlist
IMO Longlist

Find all pairs of positive integers $\left( x;\;y\right) $ satisfying the equation $2^{x}=3^{y}+5.$

well, if one looks modulo something differente from 2^n or 3^n you wont get any help, always get solution. so locking mod 3^5=243 you find out that 2^mis different from 5 always. so y=1,y=2,y=3,y=4 and is easy to evaluate this cases.

Maybe i misscalculated something, just let me now.

For $x=1$ we have $2=3^y+5$ what is a contradiction.

For $x=2$ we have $4=3^y+5$ what is a contradiction.

For $x=3$ we have $8=3^y+5\Leftrightarrow 3^y=3$ what means that one solution is \[(x,y)=(3,1)\]

For $x=4$ we have $16=3^y+5\Leftrightarrow 3^y=11$ what is a contradiction.

For $x=5$ we have $32=3^y+5\Leftrightarrow 3^y=27$ what means that another one solution is \[(x,y)=(5,3)\]

Now, suppose that $x\geqslant 6$. We have $64|2^x$ so $3^y\equiv 59\ (mod\ 64)$. Now consider the following table:

$3^1\equiv 3\ (mod\ 64)$
$3^2\equiv 9\ (mod\ 64)$
$3^3\equiv 27\ (mod\ 64)$
$3^4\equiv 17\ (mod\ 64)$
$3^5\equiv 51\ (mod\ 64)$
$3^6\equiv 25\ (mod\ 64)$
$3^7\equiv 11\ (mod\ 64)$
$3^8\equiv 33\ (mod\ 64)$
$3^9\equiv 35\ (mod\ 64)$
$3^{10}\equiv 41\ (mod\ 64)$
$3^{11}\equiv 59\ (mod\ 64)$
$3^{12}\equiv 49\ (mod\ 64)$
$3^{13}\equiv 19\ (mod\ 64)$
$3^{14}\equiv 57\ (mod\ 64)$
$3^{15}\equiv 43\ (mod\ 64)$
$3^{16}\equiv 1\ (mod\ 64)$
and everything repeats.

That means \[y=16k+11\] Note that
$3^2\equiv -8\ (mod\ 17)$
$3^3\equiv -24\equiv -7\ (mod\ 17)$
$3^4\equiv -21\equiv -4\ (mod\ 17)$
$3^8\equiv -18\equiv -1\ (mod\ 17)$
$3^{16}\equiv 1\ (mod\ 17)$
$3^{16k}\equiv 1\ (mod\ 17)$
$3^{11}\equiv 3^8\cdot3^3\equiv (-1)\cdot(-7)\equiv 7\ (mod\ 17)$
\[3^y\equiv 3^{16k+11}\equiv 3^{16k}\cdot 3^{11}\equiv1\cdot 7\equiv 7\ (mod\ 17)\]

On the other hand, we have:
$2^1\equiv 2\ (mod\ 17)$
$2^2\equiv 4\ (mod\ 17)$
$2^3\equiv 8\ (mod\ 17)$
$2^4\equiv 16\equiv -1\ (mod\ 17)$
$2^5\equiv -2\ (mod\ 17)$
$2^6\equiv -4\ (mod\ 17)$
$2^7\equiv -8\ (mod\ 17)$
$2^8\equiv -16\equiv 1\ (mod\ 17)$
and everything repeats.

We have that RHS is always congruent to $7\ (mod\ 17)$, and LHS is never congruent to $7\ (mod\ 17)$, what is a contradiction. The only solutions are \[(x,y)\in\left\{(3,1),(5,3)\right\}\]

[quote="anonymous1173"]if $y>1$ $2^{x}= 5 (mod9)$ so $x=5 (mod6)$ hence $2^{x}=5(mod7)$ and $3^{y}=0 (mod 7)$contradiction.[/quote]

What about $y=3$? :wink:

I still don't get it...
[quote="lukeKP"]...6d=2^x-2 and 6d=3^y+3[/quote]
 And what is the problem with that???


What is the largest prime factor of $ (30^2 \minus{}7^2)$?

$ 900 \minus{} 49 \equal{} 851$

Prime factorize $ 851$, to get $ 23 \times 37$, so the answer is $ \boxed{37}$.

EDIT: Oh, I just realized a better way.

Factor it as a difference of two squares.

$ (30 \minus{} 7)(30 \plus{} 7) \equal{} 23 \times 37$

So the answer is $ \boxed{37}$.


analytic geometry
geometry
geometric transformation
vector
rotation
complex numbers

Hi,

I have the following problem:

let x,y,z be vertices of a triangle. Prove x,y,z are vertices of an equilateral triangle if and only if x^2+y^2+z^2=xy+ yz+ zx

One direction is immediate:  "If x,y,z are vertices of an equilaterial triangle, then..."  but I don't know how to make progress on the other direction.  At first, I tried converting to polar coordinates or something from elementary complex numbers (playing around with the argument, etc) but I really couldn't make any progress. Any help would be appreciated.

Okay, if $ x,y,z$ are the vertices of a triangle, what is the definition of $ x^{2}$. Do you mean $ x$, $ y$, $ z$ are sides of a triangle?

EDIT: Or do you mean $ X, Y, Z$ are vertices and the lowercase $ x, y, z$ are sides. This is just convention.

[hide]$ x^{2} \plus{} y^2 \plus{} z^2 \equal{} xy \plus{} yz \plus{} xz \implies x^2 \plus{} y^2 \plus{} z^2 \minus{} xy \minus{} yz \minus{} zx \equal{} 0 \Leftrightarrow 2(x^2 \plus{} y^2 \plus{} z^2 \minus{} xy \minus{} yz \minus{} zx) \equal{} 0 \implies (x \minus{} y)^{2} \plus{} (y \minus{} z)^{2} \plus{} (z \minus{} x)^{2} \equal{} 0$ and we're done.[/hide]

Nice!

Note that $ x,y,z$ can be replaced by $ rx\plus{}p,ry\plus{}p,rz\plus{}p$...so we can assume $ x\equal{}0$, and $ y\equal{}1$, then it is obvious!

[quote="t0rajir0u"]$ x, y, z$ are complex numbers.  The solution by factorization is incomplete, but it's a start.

[hide="Hint"] What the factorization really tells you is that the validity of the statement doesn't depend on where the equilateral triangle is placed in the plane.  Assume without loss of generality that $ x \equal{} 0$. [/hide][/quote]

Could you elaborate on that?  Sorry if it's obvious.

[hide="Factorization leads to case reduction"]Doubling, we have:

$ 2x^2\plus{}2y^2\plus{}2z^2\equal{}2xy\plus{}2yz\plus{}2zx$
$ (x^2\plus{}y^2)\plus{}(y^2\plus{}z^2)\plus{}(z^2\plus{}x^2) \equal{} 2xy\plus{}2yz\plus{}2zx$
$ (x^2\minus{}2xy\plus{}y^2)\plus{}(y^2\minus{}2yz\plus{}z^2)\plus{}(z^2\minus{}2zx\plus{}x^2)\equal{}0$
$ (x\minus{}y)^2\plus{}(y\minus{}z)^2\plus{}(z\minus{}x)^2\equal{}0$

Since translation does not change the differences (visualized as vectors, not as side"lengths"), we can reduce the problem to the case when z is the number 0+0i.[/hide]

[hide="A more favorable form"]$ x^2\plus{}y^2\equal{}xy$
$ x^2\minus{}xy\plus{}y^2 \equal{} 0$
$ (x\plus{}y)(x^2\minus{}xy\plus{}y^2)\equal{}0$
$ x^3\plus{}y^3\equal{}0$, from where it should be pretty clear  :wink: [/hide]

You need to eliminate the case $ x \equal{} \minus{}y$, though.

[hide="Its easier just to..."] Write the second step of the second part as

$ (x \plus{} \omega y)(x \plus{} \omega^2 y) \equal{} 0$

where $ \omega$ is a primitive third root of unity; then the conclusion is immediate.  :) [/hide]

[quote]
[hide="A more favorable form"]
$ x^3 \plus{} y^3 \equal{} 0$, from where it should be pretty clear  :wink: [/hide][/quote]

I understand everything until here...

how do we know that it's equilateral. Sorry  :oops: [/quote]

[hide="From my last post..."] Either $ x \equal{} \omega y$ or $ y \equal{} \omega x$, that is, one is a $ 120^{\circ}$ rotation of the other.  Then the triangle $ 0, x, y$ is clearly equilateral. [/hide]

ah, see it now.  Gotta review complex numbers  :oops:

[quote][color=darkred][b][u]Lemma.[/u][/b]Let $ A(a)$ , $ X(x)$ , $ Y(y)$ be three points so that $ A\not\in XY$ .

Choose $ \phi\in [0,\pi )$ . Then $ \boxed {\ m(\widehat {XAY}) \equal{} \phi\Longleftrightarrow\frac {x \minus{} a}{y \minus{} a}\in\left\{\rho\omega\ ,\ \frac {\rho}{\omega}\right\}\ }$ ,

where $ \rho \equal{} \frac {|x \minus{} a|}{|y \minus{} a|}$ and $ \omega \equal{} \cos \phi \plus{} i\cdot \sin\phi$ .[/color][/quote]
[color=darkblue][b][u]Example.[/u][/b] The triangle $ ABC$ is equilateral $ \Longleftrightarrow$ $ \left\|\begin{array}{c} AB \equal{} AC\\\\
m(\widehat {BAC}) \equal{} \frac {\pi}{3}\end{array}\right\|$ $ \Longleftrightarrow$

$ \frac {b \minus{} a}{c \minus{} a}\in\left\{\omega\ ,\ \frac {1}{\omega}\right\}$ , where $ \left\|\begin{array}{c} \omega ^3 \equal{} \minus{} 1\\\\
w^2 \plus{} 1 \equal{} \omega\\\\
\overline {\omega} \equal{} \frac {1}{\omega} \equal{} \minus{} \omega^2\end{array}\right\|$ $ \Longleftrightarrow$

$ [(b \minus{} a) \minus{} \omega (c \minus{} a)][(c \minus{} a) \minus{} \omega (b \minus{} a)] \equal{} 0$ $ \Longleftrightarrow$

$ (b \minus{} a)(c \minus{} a)\left(\omega^2 \plus{} 1\right) \equal{} \omega\cdot \left[(b \minus{} a)^2 \plus{} (c \minus{} a)^2\right]$ $ \Longleftrightarrow$

$ \boxed {a^2 \plus{} b^2 \plus{} c^2 \equal{} ab \plus{} bc \plus{} ca}$ because $ \omega^2 \plus{} 1 \equal{} \omega\ne 0$ .

[b]Remark.[/b] We can choose $ A$ as the origin of the complex plane, i.e. $ a \equal{} 0$ . In this case

$ b^2 \plus{} c^2 \equal{} bc\Longleftrightarrow$ $ \omega\cdot (b^2 \plus{} c^2) \equal{} \left(\omega^2 \plus{} 1\right)\cdot bc$ $ \Longleftrightarrow$ $ (b \minus{} \omega\cdot c)(c \minus{} \omega\cdot b) \equal{} 0$ . [/color]


If $ a,b,c>0$, then

$ \frac{a^2\minus{}2c^2}{a^2\plus{}ab\plus{}b^2}\plus{}\frac{b^2\minus{}2a^2}{b^2\plus{}bc\plus{}c^2}\plus{}\frac{c^2\minus{}2b^2}{c^2\plus{}ca\plus{}a^2}\plus{}\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2}\leq0$

[quote="Tourish"]Does this come from
\[ \frac {a^2}{a^2 \plus{} ab \plus{} b^2} \plus{} \frac {b^2}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {c^2}{c^2 \plus{} ca \plus{} a^2} \plus{} \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}\leq 2\]

\[ \leq 2\sum{\frac {a^2}{b^2 \plus{} bc \plus{} c^2}}\]
[/quote]

$ a \equal{} \max\{a,b,c\}$

$ \frac {a^2}{a^2 \plus{} ab \plus{} b^2} \plus{} \frac {b^2}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {c^2}{c^2 \plus{} ca \plus{} a^2} \plus{} \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}\leq 2$

$ \iff c(a^3 \plus{} a^2c \plus{} a^2b \plus{} ab^2 \plus{} ac^2 \plus{} c^3)(a \minus{} b)(a \minus{} c) \plus{} a(b \plus{} c)(b^2 \plus{} bc \plus{} c^2)(b \minus{} c)^2 \ge 0$;

$ \sum{\frac {a^2}{b^2 \plus{} bc \plus{} c^2}} \equal{} \sum{\frac {(a^2)^2}{a^2b^2 \plus{} a^2bc \plus{} c^2a^2}}$ $ \ge \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \plus{} abc(a \plus{} b \plus{} c)}\ge 1$.

[quote=Tourish]Does this come from
\[ \frac{a^2}{a^2\plus{}ab\plus{}b^2}\plus{}\frac{b^2}{b^2\plus{}bc\plus{}c^2}\plus{}\frac{c^2}{c^2\plus{}ca\plus{}a^2}\plus{}\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2}\leq 2\]
\[ \leq 2\sum{\frac{a^2}{b^2\plus{}bc\plus{}c^2}}\][/quote]

[img]http://s6.sinaimg.cn/middle/006ptkjAzy77rZmcJ1ja5&690[/img]


I'm mainly concerned with improving on the ACT, and I think I need to practice taking the test in official conditions. A lot.

I've found a few practice tests by surfing the web, but I need more (any test is appreciated, not just ACT). Can anyone recommend me prep material that includes practice tests? I haven't joined any sites that require signup, but offer tests. Are any of these sites worthwhile?

My eyes have been peeled for something that does for standardized tests what Kalva does for math competitions. Is this wishful thinking?

thanks ahead of time for any advice

The Princeton Review makes great study guides and they mostly have practice test in the back.

I'll keep them in mind, thanks. Do you know if they have any products with less emphasis on study material and more on practicing tests? That's really what I'm looking for.

ok im sorry i cant really think of anything the onlything that i could suggest is asking your school office i know thats were i got one of my practice tests.

Actually that's a great idea. I'll be sure to talk to my counselor after spring break.

more suggestions are welcome

I only recommend authentic previous tests for test practice. There is a [url=http://www.amazon.com/gp/product/0768919754/]book of previous ACT tests[/url] that has just what you need. There is a similar [url=http://www.amazon.com/gp/product/0874477182/]book of previous SAT I tests[/url]. You shouldn't need anything else to be good and ready for those tests, if you THINK about why the correct answers are the correct answers, and learn some reading, grammar, and math.

Best wishes for a great score when you take the tests for real.

Excellent recommendation! Those will definitely help, thank you.

My only concern is the number of practice tests for the ACT. Is it just me or do the ACT people skimp on practice tests? 3 ACT vs 8 SAT... Should I consider purchasing these 2 [url=http://www.act.org/store/index.html]sample tests[/url]?

The reason I may seem greedy for tests is that I actually did take the ACT recently. Let's just say I hope no one ever sees those scores! I will retake them again next year, which gives me an entire year for practice. I plan on reading much more, and if possible, training myself to take standardized tests. Like other things, I just need to practice a lot. Any other resources you can think of will definitely help me reach my goal.

ewwww, reading.  It'd be easier for me if they just changed it to all math  :roll:


Let $ a,$ $ b$ and $ c$ are non-negative numbers, such that $ a^7 \plus{} b^7 \plus{} c^7 \equal{} 3.$ Prove that:
\[ a^3 \plus{} b^3 \plus{} c^3\geq ab \plus{} ac \plus{} bc
\]
[hide="Remark"] If $ a^8 \plus{} b^8 \plus{} c^8 \equal{} 3$ then
$ a^3 \plus{} b^3 \plus{} c^3\geq ab \plus{} ac \plus{} bc$ is not true already ( for non-negative $ a,$ $ b$ and $ c$ ).[/hide]

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers, such that $ a^7 \plus{} b^7 \plus{} c^7 \equal{} 3.$ Prove that:
\[ a^3 \plus{} b^3 \plus{} c^3\geq ab \plus{} ac \plus{} bc
\][/quote]
By EV-Theorem, if  $ a\plus{}b\plus{}c\equal{}constant$ and $ a^7 \plus{} b^7 \plus{} c^7 \equal{} 3$, then $ a^3\plus{}b^3\plus{}c^3$ and $ a^2\plus{}b^2\plus{}c^2$ are minimal for $ a\equal{}b\le c$. Thus, it suffices to write the inequality in the homogeneous form and to set $ a\equal{}b\equal{}1$.

We have also the following.
Let $a$, $b$ and $c$ be non-negative numbers such that $a^7+b^7+c^7=3$. Prove that:
$$4(a^3+b^3+c^3)\geq a^2+b^2+c^2+3(ab+ac+bc)$$


$a,b\ge1 , c\ge0$and  $a,b$are real numbers and $n$ is natural number then prove it:
$(ab+c)^{n}-c\le((b+c)^{n}-c)a^{n}$


Although the Opening Ceremony will be held on July 24, some delegations allready arrived Ha Noi yesterday and today.

Tomorrow morning the IMO AB will have a first meeting.

Team Leaders will arrive 19 July and take part immediately in their work.

So the IMO 2007 is coming, friends.

See you in Ha Noi soon.

Some info for you:

Voltage: 220 V
Socket: Round plug with two holes
Currency: Vietnam Dong. 1USD = 16.000 VND
Temperature: Day - 30 - 35 degrees Celcius
                     Night: 24 - 30 degress Celcius
Taxi: 8.000 - 9.000/km
Bus: 3.000/person/trip
Garments, Foods are very cheap in Vietnam, so, dont take much of those in your luggage

Dont forget to take cameras to keep exciting moments at IMO 2007.

[quote="Naphthalin"]what about relative humidity of air?
this taxi price: only in hanoi or also around the city?

Naphthalin[/quote]

Hi, all the country, my friend. It is very cheap (but... not cheap for me)  :lol: .

[quote="Donaphew"]Hi!
What about prices of souvenirs? How much are they usually?
And.. I've heard it's very often OK to pay with USD instead of Vietnam Dong. Is it true? Is it easy to find hard currency shops there?
Thanks! :)[/quote]

Actually, I think it is not very possible to pay by USD, except you pay in big restaurants or supermarket or bank. If you just pay for a small market, or in small shop, you should pay by VN Dong. So it is very useful if you exchange USD to VND in the Noibai airport.

I see none of the local respondents answered the question about the relative humidity of the air. I've never been there, but I'm willing to hazard a guess: Hanoi is a low-elevation city, not very far from the coast of a distinctly tropical sea. A satellite map of sea surface temperatures that I found indicated the temperature of that body of water as certainly $ >28^{\circ},$ maybe as warm as $ 30^{\circ}.$

In other words: it's not going to be arid.

do you know where i can change euro to VND?

Naphthalin

[quote="aev5peru"]HELLO!!
is there any place to play tennis?
bye[/quote]

Yes, in La Than hotel but you need to pay for that.

Sorry but I assume most people to have some kind of brain to find that out themself...
Well, this surely doesn't hold for totaly drunk morons...
[hide=" :roll:  :mad: "]By the way: Please don't puke at anyone...[/hide]

[quote="Naphthalin"][quote="ZetaX"]Bank.
(Sorry, but I don't get any sense in most of these questions).[/quote]
where should i e.g. know that you can change in the airport? and bank is not really helpful because we are told this afternoon to stay in the hotel...

so we arrived here ;)
greetings to all who cannot be here.

Naphthalin[/quote]

You can even change in the hotel. Just go to the reception desk and they will give you dongs in exchange of your money. At least, that's what I did this morning at 8 am in building E...


What's the coefficient of the $ x^2y^5$ term in the expansion of $ (x\plus{}y)^7$?

The binomial theorem yields $ \binom75x^2y^5\equal{}\boxed{21}x^2y^5$


integration
real analysis
real analysis unsolved

[quote="mon"]Compute $ \sum_{n \equal{} 1}^{\infty}\frac {( \minus{} 1)^{n \minus{} 1}}{n(n \plus{} 2)^2}$[/quote]

Hi :) !!!

we can prove that :  $ \frac{(\minus{}1)^{n\minus{}1}}{n(n\plus{}2)^2} \equal{} \frac{1}{4}\frac{(\minus{}1)^{n\minus{}1}}{n} \minus{} \frac{1}{4}\frac{(\minus{}1)^{n\minus{}1}}{n\plus{}2} \minus{} \frac{1}{2} \frac{(\minus{}1)^{n\minus{}1}}{(n\plus{}2)^2}$

than i think that's the reste are easy ...

i give that: $ \sum_{n \geq 1} \frac{(\minus{}1)^{n\minus{}1}}{(n\plus{}2)^2}\equal{} \sum_{n \geq 1} \int_0^1 \int_0^1 (\minus{}xy)^{n\plus{}1}dxdy \equal{} \int_0^1 \int_0^1 \frac{dxdy}{1\plus{}xy}$ .....


geometry
geometric transformation
rotation
videos

Really dumb question:

When driving, the wheels of some cars seem to spin in the other direction. How do they create this illusion?

This happens when the event is being filmed.
The exposures in the film are taken in frames per second.

If your film is at 24 fps, say, and the wheel of your car is turning at 24 revs per second, then the wheel will appear to be stationary. Because the wheel will have the exact same orientation every 1/24 th second when each frame is exposed.
As the wheel's speed becomes a little bit greater than 24 revs per second, it will appear to slowly turn forwards in the film.
But, if the wheel's speed is a little bit less than 24 rps, and increasing, then it will appear to slow down. The same if the wheel's speed is any multiple of 24 rps, but a little bit less than that and increasing.

Thanks. I did not know this.
It is logical if you think about it a little

But doesn't this happen even when you see the car with your own eyes? How do you explain it then?

When watching a car with your own eyes, you will never see the wheels slow down or speed up.

Your eyes receive a continuous, unbroken signal.

A film camera receives discrete images, recorded in frames per second.
You will only see the slowing down/speeding up effect if you are watching a film record of the event. The rotating wheel appears stationary only because the wheel's speed of rotation, in revs per second, matches the film camera's recording speed, in frames per second.

When your eyes receive a continuous signal, they receive [i]all[/i] information about the movement/rotation of the wheel, rather than discrete parcels of info. It is this partial information record on film that gives the illusion of stopping, slowing down or speeding up.

[u]Comment/Question:[/u]
Old television signals. Most (I think) TV signals nowdays are digital signals, hence discrete parcels of info, rather than continuous analogue signals. I imagine digital signals would give the same effect as camera film. but does anyone know if analogue TV signals ever gave the optical illusion on rotation ??

The way the signal is transmitted doesn't matter since the TV cameras have always recorded it using the same principle of discrete images... I think :). But even if they didn't, your TV wouldn't be able to reproduce a contiguous signal - it too has some limited frame rate (25-50 fps ?). Also, I don't think the signal is analog in the way you'd like it to be - the analog data must be probably in some serial format rather than parallel and therefore it can't transmit contiguous video signal. In fact, it only consists of 25 (EU) or 30 (USA) frames per second.

But back to the car wheel illusion :), are you really really sure that you can't see it in real? I don't say I've seen it for sure, I just think I have...

I'm not too familiar with TV signals and transmissions. I was simply curious about whether or not an analogue signal would give the optical illusion. No matter!

As for the wheel illusion, yes I am really really sure that you can't see it when viewing the event at first hand.  :)
All you would see would be a blur. And yes, you would see the wheel speeding up, as it starts from rest and increases its speed of rotation. But you wouldn't then see it slow down, stop and then start to speed up again.

I tried to find an example on the web, but no luck :(

It does happen when viewed in person, it's just that the conditions have to be right to avoid the blurring.  TV is the same as movies, with pictures painted on a fixed frequency.  The images are digital, as said, it is that the digital images are transmitted by analog radio signals.  (If you want to look further into it, each group of three dots are painted serially at a frequency that the eye sees the entire screen lit up at once (take a picture at relative high speed to see what a TV really looks like.)  The movie screen is also blank most of the time, but the retention in they eye show us a (fairly) contineous movement.

Same thing done on the phone.  (Absolutely quiet 2/3 of the time.  This allow more conversations on the same wires or bandwidth.  When the sync is off you hear another phone conservation, or shadows of it.

But with digital images, such as seen in realtime through a stroboscope (look it up), the images are exactly the same as if the wheel was going in reverse.  With an unaided eye it is hard to get this effect, but I've seen it on the road.

Ha, I knew it was real :P.

But.... why does it happen then? What are the conditions you're talking about? This sounds very interesting....

... indeed it does.

I'd love to find out what causes this illusion, when viewing a rotating object directly, rather than images of it.

It took some time, but I finally found confirmation of the non-existence of that optical effect when viewing with the naked eye, in person.

[url=http://allexperts.com/answerv.asp?QuestionID=4233804]Online posting[/url] - only available for a few days.

[url=http://allexperts.com/displayExpert.asp?Expert=23177]Expert Profiile[/url]

[quote]Ask any question! Allexperts.com is the oldest & largest free Q&A service on the Internet
Volunteer Chris Bushman Answers

  Subject   Plane propellor - optical effect
Question   Hello,
my question is about an "optical effect" which is not created, but observed.
If an airplane propeller or rotating fan, is started up and the propeller/fan revs increase, then the blades will appear a blur, but after a few moments, the blades will appear to slow down, stop momentarily, and then start up again
This effect might happen a couple of times before the propeller/fan has reached full speed.
Now I know why this "optical illusion" happens. It's because of the synchronisation of the camera frame rate (fps) and the propeller angular velocity, revs/sec (or rps).
Someone asked about this on a forum I was at and I answered it. The questioner had said that he had seen this effect with his naked eye, (he had seen this effect in a car's wheels he was watching it, and NOT a camera film of the car moving) but I told him that that wouldn't happen and the effect could only be noticed by looking at, for example, camera images.
Then another guy stepped in and said that he had seen the same effect with his naked eyes. So the first guy came back with "I knew it, I knew it ...".
So now for my question.
Can this optical effect/illusion be seen with the naked eye or only in a recording medium involving discrete images? e.g. a film camera.
I suspect the latter, but I would like an answer/opinion from a professional person, and your being involved in filmmaking and the optical side of it in particular couldn't be better!

If it's not too much to ask a small related question, could the optical effect/illusion be observed in a live TV transmission using analogue signals as opposed to digital ones?

Many thanks,

Answer:   You are certainly right about the effect being a function of the frame rate of the camera. It's called strobing. Mechanical motion studies often use a rapidly flashing light called a strobe. If we make rollup end credits for movies go too fast, the same effect happens.

The fellows that feel they have seen the same thing with the naked eye are either fooling themselves or there are other factors involved. Forinstance, a recent craze around here is wheel covers that spin or don't spin independently of the wheel. Sometimes a car can drive by and it appears that the wheels are not moving.

Actually, I have seen the effect while driving on a road with lots of trees planted at even intervals along the roadside. The sun was shining through the trees and it created a stroboscopic effect. But again, my naked eyes had help.

Video, whether analog or digital, uses discreet frames too. The rate for film is 24 frames per second, the rate for NTSC video is 30 frames per second. Same effect, slightly different results.

Chris Bushman
North Hollywood, California. [/quote]

Thanks for the reply.

The questioner (me) has never seen a car's wheels spining backwards "live". (Maybe from my first post one could understand otherwise.) Only saw it on TV.

If you take for granted what the guy answered, then I think you should conclude that this "illusion" really exists, even if you need some special circumstances.


Age: almost 16
Grade: 10
Height: 5'7.5" (because Vivienne won't let me claim to be as tall as her yet)
Weight: 132-137, depending on whether I've run or not on any given day.

Ignite, that's a bit rude, don't you think?

Age: 19
Grade: Sophomore in college
Height: 6' 5.5" (197 cm)
Weight: ~265lbs (120 kg)

Age: 18
Grade: frosh in college
Height: around 5'8''
Weight: 150 lb

Freshman 15 :D

Age: 14
Grade: 9th
Height: 5'10"
Weight: 145

Age: 17
Grade:12
Height: 6-2.75
Wingspan:6-9
Weight: 181 lbs

[quote="now a ranger"]This format:
Age: 13
Grade: 8
Height: 5'11
Weight: 124[/quote]

5'11? lucky...I'm barely 5'...and Im older than you

Age: 15
Grade: 9
Height: 5'8" when i measured several months ago
Weight: 130

Age: 14
Grade: 9
Height: 5'5
Weight: 115

Age: 14
Grade: 9
Height: 6'0 :D
Weight: 130ish, maybe a bit more

Age: 17
Grade: 12
Height: 5' 8.5"
Weight: 124

One of my friends described me as "like an emaciated Jesus."  It was funny.

[quote="daermon"][quote="jackdillon"][quote="JavaMan"]OK, I'm definitely the slimmest.

Age: 16
Grade: 12
Height: 5' 11"
Weight: 121 lbs[/quote]

No your not!

Age:12
Grade:7
Height:5'.5"
Weight:85 lbs

I am the youngest, shortest, and the least heavy.[/quote]
ooh you barely beat me

Age:  13
Grade:  7
Height:  5'5" (or maybe 5'6'' by now im having a big growth spurt)
Weight:  90 lbs[/quote]
that just means you have no muscle. don't you guys get simple workouts at least?
that doesn't sat anything about how slim you are. however jackdillon and daermon could fight over what they are saying...

5'6''-5'8''
150 lb (yes, i'm really fat ;D)
12 yrs old
8th grade
ASIAN!!

Age: 13
Grade: 8
Height: 13' 8"
Weight: 1337 lbs

(Yes, i can support my own weight. I am superstrong cuz im leet.)

i'm Asian (Chinese), and my religion is Christian (I wont clarify past being Protestant, because our sect is very general)

Age: My age is 3 more than Kyumin's. In three years half of my age will be 24 less than twice Kyumin's age.
Grade: My grade is prime, and I am in high school, under the traditional 12-grade system.
Height: A regular hexagon with side lengths equal to my height would have an area of $7350\sqrt{3}$ square inches.
Weight: My weight in pounds is the second smallest positive odd integer that has exactly 7 proper factors. (A proper factor is a positive factor of a number that is not the number itself)

Degree of Asianness: I eat with chopsticks and eat rice almost every night (except when we go out)

EDIT: Amendments, some were wrong

Age: A prime number
Grade: A perfect square
Height: A prime number of feet, and I don't know how many inches
Weight: Less than my IQ

ok seriously
age: half of half of one third of 156
grade: age minus 5
height: grade times 8 plus 3 (in inches)
weight: height times two times three divided by 5 times 2 minus 4 (in pounds)

Age: 16
Grade: 9
Height: 6' 1"
Weight: 167 lbs

[quote="jackdillon"]
No your not!

Age:12
Grade:7
Height:5'.5"
Weight:85 lbs

I am the youngest, shortest, and the least heavy.[/quote]

that's insane

Age:The second two digit prime number that has digits that add up to a divisor of my current post count
Grade:My age minus the number of the day I registered.
Height:(My age minus my grade) feet and (my grade) inches
Weight:My age times 10.

[quote="now a ranger"][quote="CheeseIsGood"]age: 13
grade: 7
height: 4' 9"  :D
weight: 73 lbs

 :ninja: I'm small. The smallest here probably.[/quote]

advice: eat more cheese[/quote]

Good idea. :D


geometry
3D geometry
ceiling function
floor function

How many perfect cubes are between 100 and 900?

$ \lceil \sqrt[3]{100} \rceil \equal{} 5$

$ \lfloor \sqrt[3]{900} \rfloor \equal{} 9$

$ 9\minus{}5\plus{}1\equal{}\boxed{5}$

[quote=TThB0501]That's EEEEEEEEEEEEEEEEEEEEEEEEEEAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAASSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! All you have to do is count how many numbers in the list 5, 6, 7, 8, 9. There! The answer is 5![/quote]

Could've done that in a nicer way, but okay :)
[hide=Solution]
The least perfect cube in that sequence is $125,$ which is equal to $5^3.$ The greatest cube in that sequence is $729,$ which is equal to $9^3.$ Therefore, we have: $$125, 216\dots 729.$$ This is the same thing as: $$5^3, 6^3\dots 9^3.$$ And this is the same as: $$5, 6\dots 9.$$ Using the number of terms in a sequence formula (or you could just count it :)): $$\frac{9-5}{1}+1\leftrightarrow \boxed{\boxed{5}} \hspace{.2 cm} \square$$
[/hide]

[quote=kittenwarrior][quote=twinbrian]lmao @above, I thought you used the rearrangement inequality at the end.[/quote]

lmao, how would that even work
Ugh, you've got me saying lmao now[/quote]
lmao


algebra
function
domain
search
geometry
congruent triangles
geometry unsolved

Let $ D$ be a variable point on the segment $ BC$ of given triangle $ ABC$. Construct an isosceles-right  triangle $ DEF$ such that $ E$, $ F$ be the points on the segments $ AB$, $ AC$( $ DE$=$ DF$). Find a locus of the midpoint of $ EF$

I think something is wrong with the problem, even if you take $ ABC$ equilateral and point $ D$ very close to $ B$ there will not exist the desired triangle $ DEF$...

[color=darkblue]Sorry, [b]Rem[/b], you aren't right ! The proposed problem is correctly. At first you must find the domain $ \Delta\subset BC$ of the mobile points $ D$ for which exists the point$ L\in\mathcal P$ (plane of $ \triangle ABC$) - the midpoint of the segment $ [EF]$ and afterwards to find the locus of the point $ L$ - the image of the point $ D$ in a function $ f:\Delta\rightarrow\mathcal P$, i.e. $ f(D)\equal{}L$. In fact , you must construct this function and the its image is the required locus ![/color]

Good morning dear friends.

I agree with [b][size=100]Virgil Nicula[/size][/b], that the problem is correct.

If I am not mistaken, we can construct an isosceles triangle $ \bigtriangleup DEF$ with given angle $ ($ $ \angle EDF \equal{}\angle\omega$ $ ),$ as the problem states ( such that the points $ E,$ $ F,$ lie on $ AB,$ $ AC,$ respectively ), also when the point $ D,$ is an arbitrary point inwardly to $ \bigtriangleup ABC.$

I will search my archive and I will post here next time, the solution I have in mind.

Kostas Vittas.

[quote="rem"]Ok, if "segments" means sides then the problem does not work. I f it means lines, then everything is fine.

Because look, say we take an equilateral triangle $ ABC$ with side length $ 10$. Take point $ D$ so it is $ 0.1$ unit away from pt $ B$ on side $ BC$, and then $ DE$ is at most $ 0.1$ because $ DEF$ is a right isosceles triangle and $ E$ and $ F$ lie on the  sides. However, $ DF$ is grater than $ 5$ -- the length or perpendicular from $ D$ to $ AC$ and so we do not get an isosceles triangle $ DEF$.

However, if the problem means lines $ AB$ and $ AC$ then everything is fine.[/quote]

You are right dear [b][size=100]rem[/size][/b].

The problem must be corrected only at this point. The points $ E,$ $ F,$ must lie on the sidelines $ ($ not only the side segments $ ),$ $ AB,$ $ AC,$ respectively.

I found in my notes a simple construction of the isoceles triangle, as the problem states with given angle $ \angle EDF \equal{}\angle\omega$ and when the point $ D,$ is an arbitrary point inwardly to $ \bigtriangleup ABC.$

Through the point $ D,$ we draw a line $ Dx,$ such that $ \angle xDA \equal{}\angle\omega$ and we define tha point $ A'$ on $ Dx,$ such that $ DA' \equal{} DA.$

Through the point $ A',$ we draw the line $ A'z,$ such that $ \angle DA'z \equal{}\angle DAC \equal{}\angle\phi,$ which intersects $ AB,$ at one point so be it $ E$ $ ($ $ E$ between $ A$ and line $ Dx$ $ ).$

Through the point $ D,$ we draw the line $ Ds,$ such that $ \angle ADs \equal{}\angle A'DE,$ which intersects$ AC,$ at one point so be it $ F.$

It is easy to show that $ DE \equal{} DF$ and $ \angle EDF \equal{}\angle\omega,$ from the congruent triangles $ \bigtriangleup A'DE,$ $ \bigtriangleup ADF$ and so, the triangle $ \bigtriangleup DEF,$ is the wanted isosceles one, as the ptroblem states, in its extension.

$ \bullet$ About the locus of the midpoint $ M$ of the segment $ EF,$ I see by the drawing ( I have not in mind any proof), that it is a line passes though two fixed points $ E',$ $ F',$ on $ AB,$ $ AC$ respectively, which may be easy defined, by two particular places of the variable $ D,$ at points $ D',$ $ D'',$  on the sideline $ BC$ $ ($ when the point $ D,$ is an arbitrary point, then the locus of $ M,$ has not a meaning $ ).$

Kostas Vittas.

[quote="Virgil Nicula"]See http://www.mathlinks.ro/Forum/viewtopic.php?t=166786[/quote]

If am not mistaken dear [b][size=100]Virgil[/size][/b], you mean the below general problem, coming from the past.

[b][size=100][color=DarkBlue]GENERAL PROBLEM. – A triangle $ \bigtriangleup KLM$ is given and let $ K$ be, is a fixed point. Find the locus of the point $ L,$ when $ M$ is a variable point on a fixed line $ (d)$ and the triangle $ \bigtriangleup KLM,$ is always similar to its self.[/color][/size][/b]

So, if $ \angle MKL \equal{}\angle\omega$ and $ \angle KML \equal{}\angle\phi,$ the locus of $ L,$ is defined as follows:

$ \bullet$ We denote as $ M_{1},$ the orthogonal projection of the point $ K,$ on the fixed line $ (d).$

We define the point $ L_{1},$ such that $ \angle M_{1}KL_{1}\equal{}\angle\omega$ and $ \angle KM_{1}L_{1}\equal{}\angle\phi.$

We define the point $ P,$ on the fixed line $ (d),$ to the other side of $ KM_{1},$ with respect to $ L_{1},$ such that $ \angle L_{1}PM_{1}\equal{}\angle\omega.$

Then the wanted locus of $ L,$ is the line segment $ PL_{1}.$

$ \bullet$ Return to [b][size=100]ephraim’s problem,[/size][/b] we can construct an arbitrary triangle $ \bigtriangleup DEF,$ whose vertices $ D,$ $ E,$ $ F,$ lie on the sidelines $ BC,$ $ AB,$ $ AC$ respectively, of a given triangle $ \bigtriangleup ABC,$ such that $ \angle DEF \equal{}\angle\omega$ and $ \angle EDF \equal{}\angle\phi.$

The locus of the midpoint $ M,$ of the segment $ EF,$ is again a line connecting two points easy defined by the construction of two particular cases of the triangle $ DEF,$ but unfortunately, I have not in mind any proof.

So, I will come back, when I will be able to post here something about the proof of the wanted locus.

Best regards, Kostas Vittas.

[color=darkblue]Using the remarkable locus from http://www.mathlinks.ro/Forum/viewtopic.php?t=166786 , the enunciation of the our problem is equivalently with the following :[/color]

[size=117][color=darkred]Consider a mobile point $ M$ which belongs to the sideline $ BC$ of the given triangle $ ABC$.

Denote the points $ \{\begin{array}{c}N\in (CA\ ,\ P\in (BA\\ \\ m(\widehat{BMP}) = m(\widehat{CMN}) = 45^{\circ}\end{array}$.

Ascertain the geometrical locus $ \Lambda$ of the midpoint $ L$ of the segment $ [NP]$.

[b][u]An easy extension.[/u][/b] Consider a mobile point $ M$ which belongs to the sideline $ BC$ of the given triangle $ ABC$.

Denote the points $ \{\begin{array}{c}N\in (CA\ ,\ P\in (BA\\ \\ m(\widehat{BMP}) = m(\widehat{CMN}) =\phi\end{array}$, where $ \phi < 90^{\circ}$ ([b]constant[/b]).

Ascertain the geometrical locus $ \Lambda$ of the point $ L$ of the line $ NP$ for which $ \overline{LN}= k\cdot\overline{LP}$, where $ k$ is a real [b]constant[/b].[/color][/size]

[color=darkblue][b][u]Remark.[/u][/b] I think that here is confortably ("lazy") at first use the analytical geometry and afterwards try a synthetical method.

I wish however to see at least a synthetical method for the our case $ \phi = 45^{\circ}$ and $ k =-1$.[/color]

[hide="Notation."]
[color=darkred]The [b]ray[/b] ("[u]semiline[/u]") $ [UV$ is the set of the points $ X$ for which the (origin) point $ U$ doesn't separate the points $ X$, $ V$ and $ (UV = [UV-\{U\}$.

For example, if $ d =\overline{ABCD}$ so that  $ B\in (AC)$ and $ C\in (BD)$, then $ d = AB]\cup (BC)\cup [CD$.[/color][/hide]


algebra
polynomial
search
algebra unsolved

We have
$P(x) = 4x^3-2x^2-15x+9$
$Q(x) = 12x^3+6x^2-7x+1$
(a) Prove that both polynomials have three distinct real roots.
(b) Let $a$ and $b$ be the largest roots of $P(x)$ and $Q(x)$ respectively. Prove that $a^2+3b^2 = 4$.

Posted before (a few times?). By the way, this is actually a Vietnamese problem.

What? I can't find it using the search function.


Yeah I am too later, sorry for that, here goes. Be patient please
Since Mexico or even earlier, (I don’t have the exact knowdge), capture the Canadian Team mascot, the Canmoo had always been an IMO tradition. This time, we Ecuadorian team took the merit as the true follower of this old noble practice.
It was the second recursion; I didn’t go that recursion as I was sleepy and a little drunken as I spend the whole night at the Puerto Rican party. ECU3 didn’t go neither. So only 4 Ecuadorian went. According to them, in the lunch their table a close to Canadian Team, and the moose was on the floor, right at the space between them.
At beginning, they are afraid of capturing the moose, but later with the courage of ECU1, they did, ECU6 stands and played the “Open Closed” game to Canadian team, (I didn’t know well the game but I am sure it is something really stupid), at the beginning only one guy is paying attention to them, but as ECU6 exaggerated well, somehow suddenly all Canadians are concentrated on the game and is closed as a semicircle. None is looking through the Canmoo at this time. ECU1 then slowly approached the the moose and put it in a vacant back bag, and went out.
He walked around the lake like an hour for the bus to depart, when they are climbing the bus, Canadians come and asked: “have you saw a moose?”
ECU1 then respond: “what is a moose”
“An animal…” Canadian continued
“You brought an animal to IMO” ECU6 then asked,(I am sure that he did it to obvious, I wonder why Canadians doesn’t noticed)
“No, it was not real…” Canadian then said.
So obviously they negated, and the Cuban Team guy (the one guy) come to ask water, and ECU1 opened a different bag and get a bottle, showing to Canadians that there is no moose, and then Canadian Team finally left.
Once back in hotel, they put the moose in the safe box. The above story is immediately counted to me. They showed me the moose, and we took a lot pics with the moose. It was difficult to decide what to do with it, because we definitely don’t want to stole it, even ECU1 have such tendency, also just hold the moose for 3 days and return it at the end day have also no fun, and it will be shameful returning it and maybe even more difficult to explain them and all the stuff.
Ok the issue is finally we decide to make a game for those guys. We wrote 3 small problems to Canadian team, the answer of the first one will lead to the second one, and so on, finally will lead them to our room. Making problems are rather harder than solve it, but finally we came up with something. The ECU5 guy made the first problem, which is a word puzzle, it is really easy, I did the third one, is something with functions, and finally none have idea what to put on second one so we just write a secret code which in fact is the easiest to solve.
Problem are written on the little IMO note book paper, and is in format of a IMO test, with “day 3” in front, then “Problem 7”, “English version…etc” I will post the picture of it later. All questions are copied with some really girlish letters to confuse them.
So we put the questions in place and the first one we put it under their room door when they was out (we got the room # from guides). And just wait… almost everyday at the night we went to look if the problem 3 is still there or taken and it was immobile in 2 full days. We start questioning if they really did try it or just throw it away, or they didn’t found it… we were worried, it would be painful if we did so much for nothing ness….
In the last day, the third question is taken finally, so we stay in room waiting for Canadians’ come. Only one hour left for the ceremony, the ring ringed, and ECU1 looked through the hole and immediately get into the bathroom, he was shamed. I guessed already what could be coming, so I opened the door…
I see an angry white guy (seemed like he did have fun through the game) with the black Canadian t-shirt, I gave him a Colgate smile and shook his hand, “congratulations, you did it…” he just spoke coldly: “where the moose…” I bright him in front of the box and returned the moose… finally he asked who write the question, I said everyone, and then he said: “you have a weird mind”.

So that was the story… we planned to do it again in Madrid, but it would be too obvious, so any other people who going to IMO 2008 and thought what we did was funny, please contribute also for the IMO tradition.

atte ECU 2.

Yeah, I admit that was nice. You really caught us off-guard with that open-closed game. But you will not be able to still the moose in Madrid! :D

mm... we probably wont do it again at madrid... it could be to obvious and you will immediately ask us if the moose is lost...
but i am sure that other teams will capture the moose again... so beware

[quote="rem"]But you will not be able to still the moose in Madrid! :D[/quote]That's what [b]you[/b] think! :moose:  This year we have a truly diabolical plan in place. The moose shall be stolen  :diablo:

It'll be easy enough to avoid.

Just be wary of the Romanian team.

And american...

And every other team which has had association with MathLinks :P

[quote="SimonM"]
And every other team which has had association with MathLinks :P[/quote]

It's very difficult to find such a team  :wink: . I think also that the moose shall be stolen  :D


Go to- http://www.brainbashers.com/games/magic1.asp

This place has a mind reading card trick. Can you figure out how the computer knows your card?

The trick is quite old actually. You have some cards, three different numbers, and different sticks, but you never have, for example, the two red 7s, or so, then, in the second hand it inverts the sticks of the same colors (i.e. if you had 7 gold, now you have 7 diamond) and just remove any card. The cards seem so similar to the ones you had seen priorly, but actually, non was in your original maze, of course, that include the one you chose.

Best,

It's easy. They just replace everything.

It took me a couple runs to figure that one out.  :lol:


geometry
incenter
circumcircle
Triangle
IMO Shortlist

Let $ABC$ be an isosceles triangle with $AC=BC$, whose incentre is $I$. Let $P$ be a point on the circumcircle of the triangle $AIB$ lying inside the triangle $ABC$. The lines through $P$ parallel to $CA$ and $CB$ meet $AB$ at $D$ and $E$, respectively. The line through $P$ parallel to $AB$ meets $CA$ and $CB$ at $F$ and $G$, respectively. Prove that the lines $DF$ and $EG$ intersect on the circumcircle of the triangle $ABC$.

[i]Proposed by Hojoo Lee, Korea[/i]

By all means, this is far from being a "pretty easy" geometry problem. Either that, or I was very dumb last year at IMO when I thought about it. The discussion was about whether choosing A4 (which was actually problem 5 in the contest) or G5 (which is this problem). You all know the result :)

PS this is Hojoo's proposal, too. :)

Well what I did was to remove I, backward constructed the question, used a bit of power of a point and it came out in a few lines of length calculations, so it didn't seem that hard to me. :?

Well, here's what I did:

By a very quick and easy angle chase I showed that $\angle FXG=\angle B (=\angle A)$, where $X=FD\cap GE$. Since $\angle AGF=\angle AFG=\angle FXG$, we find circle $(FGX)$ to be tangent to $CA$ and $CB$. It's easy to show that circle $(AIB)$ is also tangent to $CA,\ CB$. Let $X=AX\cap (ABC),\ X'=AX\cap (AIB),\ T=CX\cap AB$. Then $\frac{CX'}{CX}=\frac{CB}{CG}$, so all we need in order to show that $X=X$ is to show that $\frac{CX'}{CX}=\frac{CB}{CG}$. This is easy because $CB^2=CX'\cdot CP=CX\cdot CT\Rightarrow \frac{CX'}{CX}=\frac{CT}{CP}=\frac{CB}{CG}$ because $PG||TB$. Here I used the fact that the inversion of pole $C$ and power $CB^2$ turns the circle $(ABC)$ into the line $AB$ and it invariates the circle $(AIB)$, because it's tangent to $CA,CB$ in the invariant points $A,B$.

I think we're done.

This is my solution.

Let circumcircle of $\triangle ABC$ : $w$. $CP \cap w=T$, $TF\cap AB= D'$, $TG \cap AB = E'$. $CT\cap AB= X$

cuz $\angle PTA= \angle ABC= \angle CFG$, $P, F, A, T$ is cyclic.  similarly, $P, G, B, T$ is cyclic.

Here, $\angle FTP = \angle FAP = \angle PBA$. so $T, B, P, D'$ is cyclic. $\Rightarrow \triangle PXB \sim \triangle D'XT$.

cuz $A, T, B, C$ is also cyclic. $\triangle CXB \sim \triangle AXT$ .

therefore, $CP:PX = AD' : D'X$. Here, We get $AC \parallel PD'$.

This means that $D'= D$.  Similarly, $E'=E$.

$Q.E.D.$

[color=darkred]Let $ABC$ be an isosceles triangle with $AC=BC$, whose incentre is $I$. Let $P$ be a point on the circumcircle  of the triangle $AIB$ lying inside the triangle $ABC$. The lines through $P$ parallel to $CA$ and $CB$ meet $AB$ at $D$ and $E$, respectively. The line through $P$ parallel to $AB$ meets $CA$ and $CB$ at $F$ and $G$, respectively. Prove that the lines $DF$ and $EG$ intersect on the circumcircle of the triangle $ABC\ .$[/color]

[color=darkblue][b]Commentary.[/b] The enunciation of this problem conceals vainly in its debut that the circumcircle of the triangle $AIB$ is in fact the circle with the diameter $II_{c}\ ,$ where the point $I_{c}$ is the $C$- exincenter of the triangle $ABC\ .$[/color]

[color=darkblue][b]Proof.[/b] Denote : the circumcircle $w$ of the triangle $ABC$ ; the circle $\delta$ with the diameter $[II_{c}]$ ; the circumcircles $w_{a}$ , $w_{b}$ of the isosceles trapezoids $AFPL$ , $BGPD$ respectively ; the second intersection point $R$ between the line $\overline{FPG}$ and the circle $\delta$ ; the intersection $L\in FD\cap GE\ .$ Prove easily that the lines $CA$ , $CB$ are the tangents from the point $C$ to the circle $\delta\ .$ From the relation $FA^{2}=FR\cdot FP$ obtain $\frac{FA}{AD}=\frac{GP}{PE}\ ,$ i.e. the quadrilaterals $FADP$ , $GPEB$ are similarly as $FADP\sim GPEB\ .$ Thus,

$\begin{array}{cc}1\blacktriangleright & \begin{array}{c}\widehat{AEL}\equiv\widehat{GEB}\equiv\widehat{FDP}\equiv\widehat{AFD}\equiv\widehat{AFL}\Longrightarrow\widehat{AEL}\equiv\widehat{AFL}\Longrightarrow L\in w_{a}\\\\ \widehat{BDL}\equiv\widehat{FDA}\equiv\widehat{GEP}\equiv\widehat{BGE}\equiv\widehat{BGL}\Longrightarrow\widehat{BDL}\equiv\widehat{BGL}\Longrightarrow L\in w_{b}\end{array}\Longrightarrow\boxed{\ \widehat{DLE}\equiv\widehat{ABC}\ }\\\\ 2\blacktriangleright & \begin{array}{c}\widehat{ALD}\equiv\widehat{ALF}\equiv\widehat{APF}\\\\ \widehat{BLE}\equiv\widehat{BLG}\equiv\widehat{BPG}\end{array}\Longrightarrow \widehat{ALD}+\widehat{BLE}\equiv\widehat{APF}+\widehat{BPG}\equiv\widehat{AI_{a}B}\equiv\widehat{ABC}\Longrightarrow \boxed{\ \widehat{ALD}+\widehat{BLE}\equiv\widehat{ABC}\ }\end{array}$ $\Longrightarrow$

$\widehat{ALB}\equiv\widehat{ALD}+\widehat{BLE}+\widehat{DLE}=$ $2\cdot\widehat{ABC}=$ $180^{\circ}-\widehat{ACB}$ $\Longrightarrow$ $\boxed{\ \widehat{ALB}+\widehat{ACB}=180^{\circ}\ }$ $\Longrightarrow$ $L\in w\ .$[/color]

[color=darkred][b]Variation on the same theme.[/b] Let $ABC$ be a fixed isosceles triangle $(AB=AC)$ and let $M\in [AB$ , $N\in [AC$ be two mobile points so that $MN\parallel BC$ and exists a point $P\in (MN)$ for which $PM\cdot PN=BM^{2}\ .$ Construct the points $\{S,T\}\subset (BC)$ for  which $PS\parallel BM\ ,\ PT\parallel NC\ .$ Ascertain the geometrical locus of the intersection $L\in MS\cap NT\ .$[/color] [hide="Answer."]
[color=darkblue]The circle with the diameter $II_{a}\ ,$ where the point $I_{a}$ is the $A$-exincenter of the triangle $ABC.$[/color][/hide]

Ok, here's another solution:

[hide] Extend $ CP$ to meet the circumcircle of $ ABC$ at $ T$. Let $ GT$ intersect $ AB$ at $ E'$, and let $ FT$ meet $ AB$ at $ D'$. Since $ \angle CTB\cong\angle CAB\cong\angle CGP$, $ TPGB$ is cyclic. Now extend $ GT$ to meet the circumcircle of $ ABC$ at $ U$, and extend $ PE'$ to meet the circumcircle of $ AIB$ at $ V$. Since $ E'$ lies on the radical axis of these two circles, $ TPUV$ is cyclic; hence, $ \angle PVU\cong\angle PTU\cong\angle PTG$. Also, by a quick angle chase we find that the circumcircle of $ AIB$ is tangent to $ BC$ at $ B$, so $ \angle PVB\cong\angle PBG\cong\angle PTG$. Hence, $ \angle PVB\cong\angle PVU$, so $ V$, $ B$, and $ U$ are collinear. Then it follows that $ \angle TPE'\cong\angle TUV\cong\angle TUB\cong\angle TCB$, so $ PE'$ is parallel to $ BC$. Similarly, $ PD'$ is parallel to $ AC$, and hence $ D'PE'$ is isosceles, as desired. [/hide]

[hide]
Let $ Q$ denote the intersection of $ DF$ and $ EG$. We first note that $ PGBE \sim PDAF$, which follows from noting that $ \angle APD \plus{} \angle EPB \equal{} 90 \minus{} \frac{C}{2}$. Then $ \triangle QDE \sim \triangle DFP \sim \triangle FDA$ so $ FEQA$ is cyclic, and similarly $ GDQB$ is cyclic. The result then follows by direct angle chasing to show that $ \angle FQA \plus{} \angle GQB \equal{} 90 \minus{} \frac{C}{2}$, or by noting that $ \triangle QEP \sim \triangle QDA$ and $ \triangle QDP \sim \triangle QEB$.
[/hide]

I'll post my solution too  :) .
Denote $ \angle CAB \equal{} \angle ABC \equal{} \alpha$ and $ FD \cap GE \equal{} S$. First note that triangles $ \triangle APF$ and $ \triangle PBG$ are similar. This follows from $ \angle PFA \equal{} \angle BPA \equal{} \angle BGP \equal{} 180 \minus{} \alpha$. Moreover, $ ADPF \sim PEBG$. Hence $ \angle(FD, GE) \equal{} \angle(FP, GB) \equal{} \alpha$. Therefore, quadrilaterals $ ASEF$ and $ DSBG$ are cyclic. Hence, $ ASEPF$ and $ DSBGP$ are cyclic too. Thus we have to show that circles $ (ABC)$, $ (APF)$, $ (GPB)$ have a point in common. Therefore, consider $ (ABC) \cap (APF) \equal{} T$. A simple angle chase shows that $ \angle ATB \equal{} 2 \alpha$ and $ \angle ATB \equal{} \alpha$. Hence, quadrilateral $ PTBG$ is cyclic too. The conclusion follows.

[quote="darij grinberg"]Here is the solution I found on the exam:

Since PE || CB, we have < FPE = < FGB = 180 - < FGC, and since PG || AB, we have < FGC = < ABC, so that < FPE = 180 - < ABC. Since AC = BC, we have < ABC = < BAC, and < FPE = 180 - < BAC = 180 - < FAE. Thus, the quadrilateral FPEA is cyclic, i. e. the points F, P, E and A lie on one circle. Similarly, the points G, P, D and B lie on one circle. Let S be the point of intersection of these two circles (distinct from P).

Since the points P, S, A and E lie on one circle, < PSE = < PAE = < PAB = 180 - < APB - < PBA. Since the points A, P, B and I lie on one circle, < APB = < AIB, and 180 - < APB = 180 - < AIB = < IAB + < IBA = < BAC / 2 + < ABC / 2 = < ABC / 2 + < ABC / 2 = < ABC, so that < PSE = 180 - < APB - < PBA = < ABC - < PBA = < PBG. Finally, the concyclic points P, B, G and S yield < PBG = < PSG. Hence, < PSE = < PSG, and the point S lies on the line EG. Similarly, the point S lies on the line DF. Hence, the point S is the point of intersection of the lines DF and EG. Now, we have to show that S lies on the circumcircle of triangle ABC.

Well, the cyclic quadrilaterals FPSA and GPSB show < ASP = 180 - < AFP and < PSB = 180 - < PGB, yielding < ASB = < ASP + < PSB = (180 - < AFP) + (180 - < PGB) = < CFG + < FGC = 180 - < FCG = 180 - < ACB, so that the quadrilateral ACBS is cyclic, and our point S does lie on the circumcircle of triangle ABC. Proof complete.

I am not a genius and I needed more than an hour to come up with this.

  Darij[/quote]

Sorry to bring a very old topic back but you are really a genius Darij!
This solution is so simple and genial!

K=w∩CP,X=KD∩AP.∠PAD=∠PBC obviously.Let ∠BAC=∠ABC=2a,then ∠IAB=∠IBA=a and it follows that ∠AIB=180-2a=∠APB⟹∠PAB+∠PBA=2a=∠BAP+∠PAC⟹∠PBA=∠PAC=∠APD.We know that ∠CAB=∠PDE=∠PDB=∠PKB, and it follows that  quadrilateral PDKB is cyclic,so ∠PBD=∠PKD.So, we have AX2=XD•XK=PX2⇒AX=PX,X is the midpoint of AP.We know that FD intersect AP on its midpoint,so F,X,D,K are collinear points.Analogously, K,E and G are collinear,as desired.

Let $CP$ meet $(AIB)$ again at $Q$, so $PAQB$ is harmonic. Now $AC$ is tangent to $(AIB)$, and $AP$ bisects $FD$ (because $AFPD$ is a paralellogram). So $A(P,Q;A,B)$ harmonic, which yields $AQ\parallel DF$. Thus if $X$ is the midpoint of $PQ$, then by homothety $X$ is the intersection of $DF$ and $EG$. But $CX\perp XO$, where $O$ is the center of $(AIB)$ and the midpoint of arc $AB$ on $(CAB)$. Thus $X$ lies on $(CAB)$.

A bary solution:

Let $A=(1,0,0),B=(0,1,0),C=(0,0,1),P=(p_1,p_2,p_3)$.  We have $a=b$ and it's easy to verify that $P\in (AIB)\iff c^2p_1p_2=a^2p_3^2$.  Now $D=(1-p_2,p_2,0),E=(p_1,1-p_1,0),F=(1-p_3, 0, p_3), G=(0,1-p_3,p_3)$.  It's easy to verify that the equation of $DF$ is $p_3(1-p_2)y+p_2(1-p_3)z=p_2p_3x$, and that the equation of $EG$ is $p_3(1-p_1)x+p_1(1-p_3)z=p_1p_3y$.  Now with Cramer's (the most efficient way is to just ignore the denominators) we can find $X=DF\cap EG=(p_1(p_1+p_2):p_2(p_1+p_2):-p_3^2)$.  It's easy to check that $X$ lies on the circumcircle now, given our equation for $P$.

Let $\{X\}\in FD\cap GE,\{P,Y\}\in FG\cap (AIB)$. Clearly $(AIB)$ is tangent to $AC, BC$ so, by p.o.p. $AF^2=FY\cdot FP$, but $FY=GP$ and the two parallelograms $ADPF, BGPE$ are similar, $\angle AFD=\angle BEG=\angle AEX$, hence $FAXE$ is cyclic; since $FAEP$ is an isosceles trapezoid, $FAXEP$ is cyclic as well and $\angle AXP=\angle AFP\ (\ 1\ )$. In a similar way $BGPDX$ is cyclic and $\angle BXP=\angle AGF\ (\ 2\ )$. Adding $(1)$ and $(2)$ side by side we get $\angle AXB=180^\circ-\angle ACB$, done.

[b]Remark: The relations $(1)$ and $(2)$ show also that $C-P-X$ are collinear.
[/b]
Best regards,
sunken rock

Remark by symmetry that $AFPE$ and $DPGB$ are isosceles trapezoids, hence they are cyclic. Moreover, since $CF\cdot CA=CG\cdot CB$, $C$ lies on the radical axis of the two circles. Let $H\equiv (AFP) \cap (PGB)$, then
$$\angle AHC=\angle AHP=\angle AEP=\angle ABC$$ hence $H$ lies on $(ABC)$. Finally, \begin{align*}
\angle DHE&=180^{\circ}-\angle AEH-\angle BDH=180^{\circ}-\angle APH-\angle BPH\\
&=180^{\circ}-\angle APB=90^{\circ}-\frac{\angle ACB}{2}=\angle FAE=\angle FHE
\end{align*}
 so that $F, D, H$ are collinear. By symmetry, $G, E, H$ are collinear, so the result is obtained$.\:\blacksquare\:$

Here is my bary solution.
Let $P=(x,y,z)$ where $x+y+z=1$.
The equation of circumcircle of $AIB$ is :$-a^2yz-b^2zx-c^2xy+(x+y+z)abz=0$.
Since $P$ lies on $(AIB) $ we have $a^2z^2=c^2xy$.(here $a=b$.)
I will just show how to find $D$ others will be find similarly.
Since $D$ lies on $AB$.$D=(t,1-t,0)$.Let the intersection of $PD$ and $CA$ be $Q$.Since $Q$ lies on $AC$ then its coordinates are $(p,0,-p)$ then $u=w$ in $PD$.Then $u(x+z)+vy=0$
$ut+v(1-t)=0$ So $t=\frac{x+z}{x+y+z}$. So $D=(1-y,y,0)$.
Similarly we find $F=(1-z,0,z)$
$E=(x,1-x,0)$     $G=(0,1-z,z)$
Intersecting lines $DF$ and $EG$ we get $DF\cap EG=(-x(x+y),y(x+y),z^2)$.We need to show that $a^2z^2(y(x+y)+x(x+y))=c^2xy(x+y)^2$ which is true since $a^2z^2=c^2xy$.
$.\:\blacksquare\:$

[hide="solution"]Let $X=FD\cap EG$. Since $\triangle PDE \sim\triangle CAB$, their center of homothety must be $X$ so $X\in CP$. Since $BEPG$ is a parallelogram, $GE=GX$ passes through the midpoint $N$ of $BP$.
Let $X' = CP \cap (ABC)$ and $M$ be the minor arc midpoint of $AB$. We want to show $X=X'$; it suffices to show $N\in X'G \iff \angle MX'N = \angle MX'G$. By Fact 5, $M$ is the center of $(AIB)$. Since $\angle CX'M = 90^{\circ}$ and $\angle MNP = 90^{\circ}$, then $X'MNP$ is cyclic, so $\angle MX'N = \angle MPN = 90^{\circ} - \angle PMN = 90^{\circ} - \frac{1}{2}\angle PMB$. Since $\angle PX'B = \angle CAB = \angle CBA = 180^{\circ} - \angle PGB$, then $BX'PG$ is cyclic, so $\angle MX'G = \angle MX'B + \angle BX'G = \frac{1}{2} \angle C + \angle BPG = \frac{1}{2}\angle C + \angle ABP$.
It remains to prove $90^{\circ} - \frac{1}{2}\angle PMB =  \frac{1}{2}\angle C + \angle ABP \iff 90^{\circ} = \frac{1}{2}\angle C + \frac{1}{2}\angle AMB$ true since $AMBC$ is cyclic. Thus $X=X'$, so $CP, GE$ intersect on $(ABC)$. Similarly, $CP, FD$ intersect on $(ABC)$ so we are done. $\Box$[/hide]

My solution:
We know from Desargue's Theorem on two perspective triangle $PDE,CFG,$ we find $FD,PC,EG$ are concurrent.
Let $R=CP\cap (ABC).$
Then we must to prove that $F,D,R$ and $G,E,R$ are collinear.
We can find easily $\angle PED=\angle PGC=\angle CFP=\angle PDE\to PD=PE $ where we use parallel lines.
We have $\angle CFG=\angle CBA=\angle CRA\to AFPR$ is cyclic.$1.$
Also $\angle CGF=\angle CAB=\angle CRB\to PGBR$ is cyclic.$2.$
Also we  know $AFPD$ is parallelogram $\to AF=PD=PE,$ and $FP\parallel AE\to AFPE$ is isosceles cyclic trapezoid.$3.$
Also $EPGB$  is parallelogram $\to GB=PE=PD,$ and $PG\parallel DB\to PGBD$ is isosceles cyclic triangle.$4.$
From $1,3$ we find $AFPER$ is cyclic and from $2,4$ we have $PGBRD$ is cyclic.
Also we can find easily $CA$ tangent to $(AIB).$
Then $\angle PRF=\angle FAP=\angle PBA=\angle PED.$
Then $F,D,R$ are collinear, Similarly way we find  $R,E,G$ is cyclic.As desired.

[quote = "Morally correct problem statement"]
Let $ABC$ be an isosceles triangle with $AB = AC$, whose incentre is $I$. Let $P$ be a point on the circumcircle of the triangle $BIC$ lying inside the triangle $ABC$. The lines through $P$ parallel to $AC$ and $AB$ meet $AB$ at $D$ and $E$, respectively. The line through $P$ parallel to $BC$ meets $AC$ and $AB$ at $F$ and $G$, respectively. Prove that the lines $DF$ and $EG$ intersect on the circumcircle of the triangle $ABC$.
[/quote]
[asy]
unitsize(100);
pair A, B, C, M, P, D, E, F, G, K;
A = dir(90); M = dir(270); B = dir(230); C = reflect(A, M) * B;
P = M + abs(M - B) * dir(100);
K = 2 * foot(origin, A, P) - A;
D = extension(P, P + C - A, B, C);
E = extension(P, P + B - A, B, C);
F = extension(P, P + C - B, A, C);
G = extension(P, P + B - C, A, B);
draw(F--G, gray(0.5));
draw(P--D^^P--E, gray(0.5));
draw(A--K, gray(0.5));
draw(F--K^^G--K, gray(0.5) + dashed);
draw(B--P^^C--P, gray(0.5));
draw(A--B--C--cycle);
draw(arc(M, C, B));
draw(circumcircle(B, P, K)^^circumcircle(C, P, K));
draw(circumcircle(B, E, K)^^circumcircle(C, D, K), dotted);
draw(unitcircle);
dot(A^^B^^C^^P^^D^^E^^F^^G^^K);

label("$A$", A, dir(A));
label("$B$", B, dir(200));
label("$C$", C, dir(340));
label("$P$", P, dir(90));
label("$D$", D, dir(290));
label("$E$", E, dir(230));
label("$F$", F, dir(40));
label("$G$", G, dir(140));
label("$K$", K, dir(K));
[/asy]

Let line $AP$ intersect $\odot(ABC)$ again at $K$. We claim that lines $DF$ and $EG$ intersect at $K$. Note that $\overline{AB}$ and $\overline{AC}$ are tangent to $\odot(BPC)$.

Since $\angle BKP = \angle BKA = \angle BCA = \angle BDP$, points $B$, $K$, $D$, $P$ are concyclic. Similarly $C$, $K$, $E$, $P$ are concyclic.

Since $\angle DPC = \angle PCF = \angle PBC$, $\overline{CP}$ is tangent to $\odot(BDP)$. Similarly, $\overline{BP}$ is tangent to $\odot(CEP)$.

Since $\angle KDC = 180^{\circ} - \angle KDB = 180^{\circ} - \angle KPB = 180^{\circ} - \angle KCP$, $\overline{CP}$ is also tangent to $\odot(KDC)$. Thus line $KD$ bisects $\overline{CP}$; since $CDPF$ is a parallelogram, $\overline{DF}$ also bisects $\overline{CP}$, so $K$, $D$, $F$ are collinear. Similarly $K$, $E$, $G$ are collinear as desired.

Here's another approach: Invert about $P$ with radius $\sqrt{PB \cdot PC}$ followed by reflection in the angle bisector of $\angle BPC$. Then we get the following problem (as the above user says, we correct the restatement morally :D ):-
[quote=Inverted problem]Let $H_A$ be the $A$-Humpty point of $\triangle ABC$, and let $\omega$ be its circumcircle. Suppose the tangent to $\omega$ at $A$ meets $\odot (AH_AB) \equiv \gamma_B$ and $\odot (AH_AC) \equiv \gamma_C$ at $F$ and $G$ respectively. Also let us assume that the tangents to $\gamma_B$ and $\gamma_C$ at $A$ meet $\omega$ at $D$ and $E$ respectively. Show that $\odot (ADF),\odot (AEG),\odot (BH_AC)$ meet at a point.[/quote]
Let $BF \cap CG=K$. We show that $K$ is the desired point. Now, $$\angle BFG=\angle CBA=\angle GAC \Rightarrow AC \parallel BF$$
Also $AE \cap BC$ lies on the perpendicular bisector of $AC$ (As this is the point where the tangents to $\gamma_C$ at $A$ and $C$ meet). This gives that $BE \parallel AC$, i.e. $F,B,E$ are collinear. And, $\angle GDA=\angle CBA=\angle GFK$, which means that $K \in \odot (ADF)$. Similarly, $K \in \odot (AEG)$. Also, as $BK \parallel AC$ and $CK \parallel AB$, so $ABKC$ is a parallelogram. Thus, $$\angle BKC=\angle BAC=180^{\circ}-\angle BH_AC \Rightarrow K \in \odot (BH_AC) \quad \blacksquare$$

Since, there are almost six elements passing through a single point, there are many methods to approach this problem...I'll post another one
[quote=ISL 2003 G5]Let $ABC$ be an isosceles triangle with $AC=BC$, whose incentre is $I$. Let $P$ be a point on the circumcircle of the triangle $AIB$ lying inside the triangle $ABC$. The lines through $P$ parallel to $CA$ and $CB$ meet $AB$ at $D$ and $E$, respectively. The line through $P$ parallel to $AB$ meets $CA$ and $CB$ at $F$ and $G$, respectively. Prove that the lines $DF$ and $EG$ intersect on the circumcircle of the triangle $ABC$.
[/quote]
[b]Solution:[/b] $\angle ABC=\angle CAB=\angle PDB \implies PDGB$ is cyclic. Similarly, $AFPE$ is cyclic and obviously $AFGB$ is cyclic. Let $\odot (ABC)$ $\cap$ $\odot (AFPE)$ $=$ $M$, Let $MP \cap \odot (ABC) = C'$
$$\angle AFP=180^{\circ}-\angle ABC=180^{\circ}-\angle AMC'=180^{\circ}-\angle ABC' \implies C' \equiv C \implies C - P - M$$
$\angle KMB=\angle CAB=\angle CGF \implies PGDMB$ is cyclic. $\angle CAI=\angle IBA$ $\implies$ $CA , CB$ tangent to $\odot (AIPB)$, hence, $\angle CAP$ $=$ $\angle APD$ $=$ $\angle PBA $ $\implies$ $AP$ is tangent to $\odot (PGDMB)$ and similarly, $BP$ is tangent to $\odot (AFPEM)$ $\implies$ $\angle AFM$ $=$ $\angle APM$ $=$ $\angle FGM$ $\implies$ $CF, CG$ are tangents to $\odot (FGM)$, Now
$$\angle MGB=\angle MDB=\angle MFG \implies F - D - M \text{ and similarly, } E - G - M$$

It's obvious that lines $FD$,$GE$,$CP$ passes through one point. Now we denote by $K$ the intersection of the circumcircle of $ABC$ and the line $CP$. We will show that $K$ also lies on lines $EG$ and $FD$.
First,note that the quadrilaterals $PFAK$ and $BKPG$ are cyclic. So we see that pentagon $BGPDK$ is cyclic too. Thus we have $$\angle FKP=\angle PAC,\angle PKD=\angle PBA$$. Now note that $\angle CAI=\angle IBA$, thus $CA$ is tangent to $(ABI)$. But it means that $\angle PKF=\angle PKD$. So points $F$,$D$ and $Q$ are collinear$.\:\blacksquare\:$

I'm trying to find a different solution. Here is what I currently have and I would appreciate it if someone knows how to finish the solution.

By Desargues theorem $DF, EG, CP$ concur at a point, call it $X'$, and let $X=CP \cap (ABC)$. Futhermore, let $CM \cap \odot (ABC)=M_C$, and $M_CX \cap FG=L$, and $MP \cap CL=Y$ where $M$ is the midpoint of $AB$. Then, obviously, $L$ is the orthocenter of $\triangle APM_C$, so $\angle CLM_C=\pi-\angle APM_C=\pi-\angle PMM_C=\angle CML$ since

$$M_CM \times M_CC =MI^2=MP^2 \implies \triangle PMM_C ~\triangle CPM_C$$

so $MM_CLY$ is cyclic. Therefore, it suffices to show given $M_CX' \cap FG=L'$, then $MM_CL'Y$ is cyclic. I got stuck here, so if anyone has any ideas, that would be awesome.

Oh what, this is G5? Hmm, lol, I thought it was just a Singapore TST 2004 problem. This is very easy for G5? in my opinion.

Let $K=(ABC)\cap CP$, I claim that $K$ lies on $DF$ and $EG$.
Let $Q=(BIC)\cap GF$.

[color=#f00]Claim.[/color] $PFAKE$ is cyclic.
$$\measuredangle AKP=\measuredangle AKC=\measuredangle ABC=\measuredangle CAB=\measuredangle FAB=\measuredangle AFP$$
$$\measuredangle AEP=\measuredangle ABC=\measuredangle CAB=\measuredangle FAB=\measuredangle AFP$$

[color=#f00]Claim.[/color] $BGPDK$ is cyclic.
$$\measuredangle PKB=\measuredangle CKB=\measuredangle CAB=\measuredangle ABC=\measuredangle ABG=\measuredangle PGB$$
$$\measuredangle PDB=\measuredangle CAB=\measuredangle ABC=\measuredangle ABG=\measuredangle PGB$$

[color=#f00]Claim.[/color] $\triangle PEG\sim \triangle PFD$.
Since $Q,P$ lie on $(BIC)$ and it is well-known that centre of $(BIC)$ is the midpoint of arc $AB$, we have that $BC$ is tangent to $(BIC)$ and therefore by PoP, $$GB^{2}=GQ\cdot GP\implies \triangle BGQ\sim \triangle PGB.$$
Hence, we have $$\frac{PG}{PE}=\frac{PG}{GB}=\frac{GB}{GQ}=\frac{PD}{PF},$$
since $BG=PE=PD$ and $GQ=PF$. Also, easy to see that $\angle GPE=\angle DPF$. These can be obtained easily by some parallelogram and trapezoid properties. Now claim follows.

By the last claim, we have a spiral similarity with centre $P$ taking $DF$ to $EG$, thus there is also a spiral similarity with centre $P$ taking $FE$ to $DG$. Therefore $DF,EG,(PEF)$ and $(PGD)$ all concur and since $K$ lies on $(PEF)$ and $(GPD)$, we conclude that indeed $DF,EG$ intersect on the $(ABC)$.


大家考得怎样啊？反正我这次砸锅了。
总体而言，这次一试部分比去年难，二试比去年简单。一试做得很懵，二试就晕了，上不了200，完了。真有一种心中支撑着的一座大厦在短短4个小时内无声坍塌的感觉。
我还会继续爱数学的，不管考得结果怎样...
不过，还是很感谢Mathlinks上众多人的支持的！！！

看了下, 二试的题目还好...一试...一贯都是那样...就没去看了
我以前都是栽在一试上...sigh

本人差不多220--230分，基本上冬令营也是没戏了（我重庆的）。
本校据说有三个人有希望得满分，好恐怖哟！！

我考得还好啦！
一试大概139
二试估计是120吧
希望能进冬令营吧！

[quote="jerry_science"]一试考得不错，126，但是还有希望再高点的，有两个小错误拉~~~~
二试不是很好，110，最后一题没想死算，哎。。。。。。。。。。。[/quote]

[quote="Joe Zhang"]我考得还好啦！
一试大概139
二试估计是120吧
希望能进冬令营吧！[/quote]

哇啊！我不过114+110吧。为什么都考这么高？

[quote="coolbid"]成绩出来了，广东只有4个冬令营，但一等奖分数线高达184！真是让人费解。
我考的不是很好，一试94，二试110。不过反正才高二，希望下一年进冬令营了。[/quote]
不止四个吧.......而且广东省一等奖分数线是182。你怎么这么早就知道了这些？成绩这周才公布啊。


analytic geometry
combinatorics open
combinatorics

I mentioned this as a sidenote after solving the 2-dimensional problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=278381]here[/url], but didn't get any response.  I'm reposting here because I'm still curious about the answer and don't know if there's an easy solution or not.

Consider a finite $ 3\minus{}$dimensional array $ a_{ijk}$ of integers.  A legal "operation" consists of choosing a hyperplane parallel to one of the coordinate axes (either $ i$, $ j$, or $ k$ is constant), and adding the same integer to each entry in the hyperplane.  Suppose that there are infinitely many $ n$ such that we can by performing such operations get to an array where all entries are divisible by $ n$.  Does it follow that we can also get to the all-zero array by performing such operations?

[quote="kevinatcausa"]I mentioned this as a sidenote after solving the 2-dimensional problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=278381]here[/url], but didn't get any response.  I'm reposting here because I'm still curious about the answer and don't know if there's an easy solution or not.

Consider a finite $ 3 \minus{}$dimensional array $ a_{ijk}$ of integers.  A legal "operation" consists of choosing a hyperplane parallel to one of the coordinate axes (either $ i$, $ j$, or $ k$ is constant), and adding the same integer to each entry in the hyperplane.  Suppose that there are infinitely many $ n$ such that we can by performing such operations get to an array where all entries are divisible by $ n$.  Does it follow that we can also get to the all-zero array by performing such operations?[/quote]

[hide="here is my solution"]Consider first the action of this operation within a given plane of the array, $ a_{1jk}$.  Within this plane, the operation acts to either increase a row by an integer amount, increase a column by an integer amount, or increase the whole array by an integer amount (which can be decomposed as a series of individual row operations).  Each such operation leaves the following number constant:

$ a_{1jk}\minus{}a_{11k}\minus{}a_{1j1}\plus{}a_{111}$

Since each element can be made divisible by infinitely many $ n$--and hence by arbitrarily large $ n$--it follows that the above number, which does not change, is initially divisible by arbitrarily large $ n$--and hence must be zero:

$ a_{1jk}\minus{}a_{11k}\minus{}a_{1j1}\plus{}a_{111}\equal{}0$

which can be rewritten:

$ a_{1jk}\equal{}g(j)\plus{}h(k)\plus{}a_{111}$ where $ g(j)\equal{}a_{1j1}\minus{}a_{111}$ and $ h(k)\equal{}a_{11k}\minus{}a_{111}$

We can similarly write:

$ a_{i1k}\equal{}f(i)\plus{}h(k)\plus{}a_{111}$ where $ f(i)\equal{}a_{i11}\minus{}a_{111}$
$ a_{ij1}\equal{}f(i)\plus{}g(j)\plus{}a_{111}$

Next considering the impact of the operation at a three dimensional level, the following number will always remain constant:

$ a_{ijk}\minus{}a_{1jk}\minus{}a_{i1k}\minus{}a_{1jk}\plus{}a_{11k}\plus{}a_{1j1}\plus{}a_{i11}\minus{}a_{111}$

and so, by a similar argument to that given before:

$ a_{ijk}\minus{}a_{1jk}\minus{}a_{i1k}\minus{}a_{1jk}\plus{}a_{11k}\plus{}a_{1j1}\plus{}a_{i11}\minus{}a_{111}\equal{}0$

Rewriting:

$ a_{ijk}\equal{}f(i)\plus{}g(j)\plus{}h(k)\plus{}a_{111}$

Thus, reducing plane $ i$ in the first dimension by $ f(i)$, plane $ j$ in the second dimension by $ g(j)$, and plane $ k$ in the third dimension by $ h(k)$, we can arrive at an array all of whose elements are $ a_{111}$, and at this point it is trivial to reduce everything to zero.[/hide]


algebra
polynomial
LaTeX
induction
number theory
prime numbers
algebra proposed

$g(n)$ is a non-constant polynomial that doesn't have double root.with integer coefficients
Prove for each natural $r$ there exist natural $n$ that:
For at least $r$ prime number $p$ we have $p$ divides $g(n)$ but $p^2$ does not.
_________________________________________
No pain, no gain

I canot see the statement of the problem. Could you please post it again?

Or try first the Latex beginners section  :D  :D  :D

Pierre.

Yes, and you can use $ s at the beginning and end of each post instead of [ tex ] so that it doesn't end up with one line per symbol :)

$g(n)$ is a non-constant polynomial with integer coefficients that doesn't have double root

Prove for each natural $r$ there exist natural $n$ that:
For at least $r$ prime numbers $p$ we have $p$ divides $g(n)$ but $p^2$ does not.

You, the iranian people don't let me sleep. You post so extraordinary problems, that I can't sleep. This one is absolutely magnificient. I don't know if the following is a good solution, since I'm allmost sleeping:
   The proof is by induction on r.
   First, let us prove it for r=1. Suppose that for any n, any prime factor has exponent at least 2 in f(n) (the polynomial I will note f). Of course, the polynomial $ gcd(f, f') $ is constant, let it be c. Then any prime factor of f and f' is at most c. Now, we know that for any non-constant polynomial the set of prime factors of its values is infinite. Let us take a prime $ q>c$ such that $ q| f(n) $ for a certain n. Then we have that $ q^2|f(n) $. Now, we also have  $ q|f(n+q)$ and so $ q^2| f(n+q) $. But it's easy to see that $ f(n+q)=f(n)+qf'(n) (mod q^2) $, and so q also divides $ f'(n) $ , contradiction with q>c.
   Now, suppose it is true for r. I will stop here and save what I wrote since the Internet is working very bad at this hour.

Now, let's continue. So, let us prove that for r+1. In the same manner, we can take that $ (f,f')=c $. Now, from the induction we can find an n and also $p_1,...,p_r $ primes which have exponent 1 in $ f(n)$. First, we porve that we can take n such that n is not a root of f. Indeed, since f is non-constant, we can choose k such that $ f(n+k\cdot (p_1\cdot...\cdot p_r)^2)<>0 $. We can now work with the number $ n+k\cdot (p_1\cdot...\cdot p_r)^2 $ instead of n and it has the same property that the primes $ p_i$ have exponent 1 in the value of f calculated in this number. So, we can assume that n is not a root of f. Now,consider the polynomial $ g(X)=f( n+X\cdot (p_1\cdot...\cdot p_r)^2) $. This is non-constant and thus there exists u and a prime $ q>max(|c|, p_1,...,p_r, |f(n)|) $ such that $ q| g(u) $. If q has exponent 1 in $ g(u) $ then we can choose $ N=n+u\cdot (p_1\cdot...\cdot p_r)^2 $ and in  $ f(N) $ the primes $ p_1,...,p_r, q$ have exponent 1. I hope it's clear why. So, suppose that q has exponent at least 2 in $ g(u) $.Consider the number
 $ M=N+u\cdot (p_1\cdot...\cdot p_r)^2\cdot q $. We prove that in $ f(N) $ the primes $ q,p_1,...,p_r $ have exponent 1. Indeed, it's clear that $ f(M)=f(N)=f(n) (mod (p_i)^2 ) $ for all i and thus $ p_i$ have exponent 1. Now, for q we also have that
 $ f(M)=f(N)+u\cdot (p_1\cdot...\cdot p_r)^2\cdot q\cdot f'(N) (mod q^2) $. Suppose in $ f(M) $ q has exponent at least 2. Since $ q^2| f(N) $ it follows that q divides $ u\cdot (p_1\cdot...\cdot p_r)^2\cdot f'(N) $. But of course q doesn't divide $ p_1\cdot...\cdot p_r $ ( from the choice) nor $ f'(N) $ since it will follow that $ q|(f(N), f'(N))|c $. Thus, q divides u. But since q also divides $ g(u) $ we have $ q|g(0)=f(n) $. But we have chosen $ q>|f(n)| $ and so this forces that $ f(n)=0$. This contradicts the choice of n not to be a root of f. FINISH. I really doubt that this is correct, but I think it's a nice idea anyway.

Gm. If we find such $n$ for $r=1$, but infinitely many $p$, (what is proved in first harazi's post), then just proceed by Chinese remainders theorem (find $n$, whiche gives necessary remainders modulo squares of our primes).


calculus
integration
inequalities
algebra
polynomial
real analysis
real analysis unsolved

Following  http://www.mathlinks.ro/Forum/viewtopic.php?t=308485 (for $ q\equal{}\frac{1}{2}$, $ p\equal{}1$ ) let me ask the following:
How to construct the polynomial $ g_n$ of degree $ n$ such that
\[ \frac{(\int\limits_{\minus{}1}^{1} |g_n|^{2q})^{\frac{1}{q}}}{(\int\limits_{\minus{}1}^{1}( \int\limits_{\minus{}1}^{t} |g_n (x)|^2 dx)^p dt)^{\frac{1}{p}}}\sim C\ log n\] where $ C\equal{}C(p,q)$ and $ \frac{1}{q}\minus{}\frac{1}{p}\equal{}1$.
I've tried somehow modified fedja's example from the post mentioned above but....

Just take the $ n$-th partial sum of the Taylor series of $ (1\minus{}t)^{\minus{}\frac{1}{2q}}$ at the left end $ \minus{}1$.

Thanks, fedja. :) I just wanted to ask one more question about this problem: if we have $ \frac{1}{q} \minus{}\frac{1}{p}>1$ is it true that our expression bounded by $ C(p,q)$ (does not depend on $ n$)?


[color=red][Updated: See partial results in the last post, at the end of the page][/color]

Hi folks. News from mZero, one of the members of the Romanian Team currently participating at the contest there.

Here are the questions:

problem 1 - http://www.mathlinks.ro/viewtopic.php?t=5587
problem 2 - http://www.mathlinks.ro/viewtopic.php?t=5589
problem 3 - http://www.mathlinks.ro/viewtopic.php?t=5588
problem 4 - http://www.mathlinks.ro/viewtopic.php?t=5590

I also wonder how did you guys do, so please post the results here as quick as possible :D:D ;) Also other ML participants to this competition. I head that this year a lot of other countries joined the competition ... is that just a rumour? :)

As to my knowledge, it is the 21st BMO (?)

Yes it is ... I am still awaiting results from the Romanian Team (or any other team which is out there :lol: ) ..

well I did not obtain any results from the Romanian team, but I have found out some results from Iura.

So, it seems that for the first time in many years the Romanian team defeated the Bulgarian team (with 1 point out of 240 possible!).

Andrei Negut and Kappa scored maximum points. (12th and 11th grade)
mZero 36p - silver medal (12th grade)
Ungureanu 32p, Rosoiu - silver medals as well (10th and 11th grade)
Pachitariu 29p - bronze medal (10th grade)

From the Moldovian team,

Alexander Zamorzaev 37p - gold medal ( 9th grade)
Iura 35p, Dust 33p - silver medals (9th grade, 10th grade! :))
Borsenco 33p - silver
Argint Galusca 29p, Patrinica 24p bronze medals

I don't know whether Zamorzaev's nickname's Palosh.
But he's 9th grade

How is Balkan scored?

Wow, the Moldovian team is young.

[quote="S_Alex"][quote="Valentin Vornicu"]each problem is worth 10 points.

yes the moldovian team is very young. Considering that they kicked ass in Romanian NMO this year :D[/quote]
WOW Mzero lost the gold medal with just one point  :(  :(  :(  HOW UNLUCKY :(  :( .[/quote]
I don't know if I'm allowed to post any solutions here but... here goes nothing  :lol:

If you find any errors in the proofs please tell me. I will apreciate it. Thx

[quote="Valentin Vornicu"]I would really apreciate if you made the posts in each of the topics :D ...[/quote]
Ok I will. but first a small adjustement at prob 4 sol. At some point I say sum of white parts is 2N. In fact is 2N-m where m is the number of vertices on d. And then at some point I say the sum of black parts is -2N. It is in fact -2N+m. this is because the vertices on d are only counted once.
Sorry for any inconveniances I've caused.  :blush:  :)

I'm sorry to inform you that my nicknameodesn't belong to Alexander Zamorzaev.
Still I am very glad to congratulate all the participants and winners of the Balkan MO and especially the Moldavian team, which is of great perspective !!!
 :lol:  :lol:  :lol:  Congrats to the Romanian team that beat the Bulgarian team :)

__________________________
'Our days are never coming back....... (SYSTEM OF A DOWN -> Highway Song)'

uppsz :) I don't know why I got that impression :P okay ... sorry then, no gold medal for you Palosh :D

Zamorzaev's nickname is Sasha but he isn't very active

well after such great results in romanian nationals I would expect even more from our young moldovian friends :)

[quote="Palosh"]I am very glad to inform that this year the Moldavian team has the 3rd place at BMO 2004 :)
Congrats again :)!!!![/quote]

Palosh, agree with you. JIM

I was just wondering about and found this topic. The BMO this year wasn't held in Plovdiv, but in Pleven. This is the place where with more than 100 years ago, romanians kicked some otomans, and fought side by side with... bulgarians. But we beated them now. I am proud of this allthough my result was the poorest from the whole team. Too bad romanians didn't beat the bulgarians at the IMO too...
And next year the balkan is in my hometown. Too bad.

Hello

How can I open the dvi file please?


geometry
incenter
circumcircle
invariant
trigonometry
Euler
geometry proposed

Let $ ABC$ be a triangle and $ \omega(I,r)$ be the incircle.Let $ A',B',C'$ be the tangency points of the $ \omega$ with the sides $ BC,AC,AB$ respectively.
Let $ \omega_A,\omega_B,\omega_C$ be the circumcircles of the triangles $ \triangle AIA',\triangle BIB',\triangle CIC'$.Prove that $ \omega_A,\omega_B,\omega_C$ have a common radical axis.

Consider the inversion $ \Psi$ with pole $ I$ and power $ r^{2}$. The points $ A^{\prime}$, $ B^{\prime}$, $ C^{\prime}$ remain invariant and the images of the vertices $ A$, $ B$, $ C$ are the midpoints $ A_{1}$, $ B_{1}$, $ C_{1}$ of the sides $ B^{\prime}C^{\prime}$, $ C^{\prime}A^{\prime}$, $ A^{\prime}B^{\prime}$ (The last argument is because of the fact that $ IA \cdot IA_{1} \equal{} \left( r \cdot \frac {1}{\sin{\frac {A}{2}}}\right) \cdot \left(r \cdot \sin{\frac {A}{2}}\right) \equal{} r^{2}$.)
Thus, the coaxality of the circles $ AIA^{\prime}$, $ BIB^{\prime}$, $ CIC^{\prime}$ is equivalent with the concurrence of the lines $ A^{\prime}A_{1}$, $ B^{\prime}B_{1}$, $ C^{\prime}C_{1}$, which is evident, since they are the medians of the triangle $ A^{\prime}B^{\prime}C^{\prime}$.

[quote="pohoatza"]Consider the inversion $ \Psi$ with pole $ I$ and power $ r^{2}$. The points $ A^{\prime}$, $ B^{\prime}$, $ C^{\prime}$ remain invariant and the images of the vertices $ A$, $ B$, $ C$ are the midpoints $ A_{1}$, $ B_{1}$, $ C_{1}$ of the sides $ B^{\prime}C^{\prime}$, $ C^{\prime}A^{\prime}$, $ A^{\prime}B^{\prime}$ (The last argument is because of the fact that $ IA \cdot IA_{1} \equal{} \left( r \cdot \frac {1}{\sin{\frac {A}{2}}}\right) \cdot \left(r \cdot \sin{\frac {A}{2}}\right) \equal{} r^{2}$.)
Thus, the concurrence of the circles $ AIA^{\prime}$, $ BIB^{\prime}$, $ CIC^{\prime}$ is equivalent with the concurrence of the lines $ A^{\prime}A_{1}$, $ B^{\prime}B_{1}$, $ C^{\prime}C_{1}$, which is evident, since they are the medians of the triangle $ A^{\prime}B^{\prime}C^{\prime}$.[/quote]
My proof is the same,thank you. :wink:
Any other solutions?

[color=darkblue]Very nice the your proof, Cosmin & Erken ! Thank you ! [/color]

Since circumcircle of $ \triangle AIA^{'},\triangle BIB^{'},\triangle CIC^{'}$ pass $ I$, It is suffice to show circumcenter of $ \triangle AIA^{'},\triangle BIB^{'},\triangle CIC^{'}$ are collinear.Let external bisector of $ \angle A,\angle B,\angle C$ meet $ BC,CA,AB$ at $ X,Y,Z$. By Menelaus theorem $ X,Y,Z$ are collinear. Hence midpoint of $ IX,IY,IZ$ are collinear. Since $ \angle XAI \equal{} \angle XA^{'}I \equal{} 90$, midpoint of $ IX$ is circumcenter of $ \triangle AIA^{'}$. Hence we are done

very nice proof Euler !

  And your post dear Virgil is exceptionally nice!

  T.Y.
  M.T.

[quote="PDatK40SP"]Consider the inversion has $ I$ to pole and $ r^2$ to constant. The inversive image of $ (AIA')$ is one median of $ \triangle{A'B'C'}$ ( the one passes through vertex $ A'$ ). So the inversive image of $ (AIA'),(BIB'),(CIC')$ are concurrent ( at the centroid of $ \triangle{A'B'C'}$ ) , which means $ (AIA'),(BIB'),(CIC')$ are coaxal, as desired[/quote]
You should look at other proofs,before posting your own,because it is the same as pohoatza's proof.Look above

Oh, I'm sorry  :(
Can moderators delete it for me please? Thanks  :)

Not necessary, it is the same solution but written differently. I delete posts that are really useless (e. g. one posts a proof by Cauchy-Schwarz and another one just says "it is easy with Cauchy").

  darij

[quote="armpist"]very nice proof Euler ! [b]And your post dear Virgil is exceptionally nice ![/b][/quote]

The [b]Leonhard Euler[/b]'s proof of this old problem is nicely, shortly and well-known.

For this problem is too much to use the inversion ... (like killing a fly with an atomic bomb).

V.N.

[b]P.S.[/b] Sorry my "dear" [b]Armpist[/b], your post is like "fly in milk".


Prove that there exists no function $ f: N \rightarrow N$ such that:
\[ f(f(n))=n+1987\]
for all $ n \in N$.


how do you make:

i) a sigma
ii) a greater than, less than, less than or equal to, etc.
iii) real number symbol, complex, integer, etc.

thanks!

1. There are 3 types of sigma \sigma, \Sigma and \sum $\sigma, \Sigma$ and $\sum_{k=1}^{n}$. See [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php]LaTeX Symbols[/url]
2. It's all at [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php]LaTeX Symbols[/url]
3. Load the amsmath package which will give you the mathbb fonts with \mathbb{R},\mathbb{C},\mathbb{Z} $\mathbb{R},\mathbb{C},\mathbb{Z}$


\[ \triangle ABC\Longrightarrow \sum \frac{4R^2-a^2}{a}\ge \frac{R}{s}\cdot (4R+r).  \]

I think there is some typo here. I believe you mean
    $\frac{4R^2-a^2}{a} \geq \frac{R}{p}(4R+r)$
 (or maybe you missed a constant in the RHS. However, I don't think that is a right inequality)
 Anyway, back to the inequality I presume, it is not difficult. By replacing $ab+bc+ca = p^2 + r^2 + 4Rr$, the inequality is equivalent to $R \geq 2r$.


algebra
polynomial
number theory
least common multiple
quadratics
algebra open

This somehow came to my mind as a generalization of IMO 06/problem 5:

Let $\mathfrak{o}$ be a ring of algebraic integers. Let $w$ be the number of roots of unities in $\mathfrak{o}$. The maximal multiplicative order of some element $\mod w$ will be denoted by $n$.

Let $P\in \mathfrak{o}[x]$ be a polynomial with coefficients in $\mathfrak{o}$. By $P^{\circ k}$ I will mean the k-fold iteration of $P$.

a) If for some $x\in \mathfrak{o}$ we have $P^{\circ k}(x)=x$ for some $k$, then we also have $P^{\circ \text{lcm}(w,n)}(x)=x$.

(In fact it could also be true that either $P^{\circ w}(x)=x$ or $P^{\circ n}(x)=x$; both cases can occur, as I will show at the end)
(This isn't really connected to the IMO problem, though many solutions started with proving this for $\mathfrak{o}=\mathbb{Z}$.)

b) The number of all $x\in \mathfrak{o}$ with $P^{\circ k}(x)=x$ for some $k$ is at most $\deg P$ if the degree of $P$ is at least $w$.

Some comments: I only have exact proofs that the statements are true for $\mathfrak{o}=\mathbb{Z}$ and $\mathfrak{o}=\mathbb{Z}[i]$; I'm quite sure they are indeed true for all imaginary quadratic fields.

Remark to part a): We have a primitive $w$-th root of unity $\zeta\in \mathfrak{o}$; the polynomial $P(x)=\zeta x$ satisfies $P^{\circ w}(1)=1$, but clearly we cannot replace $w$ by any smaller number there.
If we consider the polynomial $P(x)=x^{l}$ for some $0<l<w$ such that $l$ has the maximal multiplicative order $n$ $\mod w$, it is also clear that $P^{\circ n}(\zeta)=\zeta$, but $n$ is best possible.

Remark to part b): We can consider $P(x)=x^{w-1}$. Since $P(P(x))=x^{w^{2}-2w+1}= x (x^{w})^{w-2}$, it is clear that all $w$-th roots of unity are fixed points of $P(P(x))$ as well as 0. Therefore we have at least $w+1$ fixed points of $P(P(x))$, more than $w-1$.

It looks like the statement is only true for $\mathbb{Z}$ and imaginary quadratic fields as these are the only fields with finite unit group.

One counterexample for $\mathbb{Z}[\phi]$ where $\phi=\frac{1+\sqrt{5}}2$ is $P(x)=-2x^{2}+(\phi+1)x+1$. For this polynomial, we have $P(0)=1$, $P(1)=\phi$, $P(\phi)=0$.

The problem is that for $\mathbb{Z}$ or for fields with finite unit group, the statement that $P(x)-P(y)=u(x-y)$ where $u$ is a unit is a quite strong statement, but if there are infinitely many units, it is just too few information.

Peter

Edit: But I just recognize that I am not able to construct arbritarily large periods for any number field... maybe(very maybe) there is always an upper bound... so maybe the following statements are true(I will look for counterexamples):

a) There is some integer $N$, only depending on $\mathfrak{o}$, such that, whenever we have $P^{\circ k}(x)=x$ for some $x\in \mathfrak{o}$ and some $k$, then we also have $P^{\circ N}(x)=x$.

b) There is some integer $M$, only depending on $\mathfrak{o}$, such that whenever $\deg P\geq M$, we have at most $\deg P$ numbers $x\in \mathfrak{o}$ such that $P^{\circ k}(x)=x$ for some $k$.


Given $ a, b, c \geq\ 0$. Prove that:

$ \sqrt{\frac{a}{b\plus{}c}} \plus{} \sqrt{\frac{b}{c\plus{}a}} \plus{} \sqrt{\frac{c}{a\plus{}b}}  \geq\ \sqrt[4]{8}. \sqrt[4]{\frac{a(b\plus{}c)}{a^2\plus{}5bc}\plus{}\frac{b(c\plus{}a)}{b^2\plus{}5ca}\plus{}\frac{c(a\plus{}b)}{c^2\plus{}5ab}}$
 :)

[quote="nguoivn"]Given $ a, b, c \geq\ 0$. Prove that:

$ \sqrt {\frac {a}{b \plus{} c}} \plus{} \sqrt {\frac {b}{c \plus{} a}} \plus{} \sqrt {\frac {c}{a \plus{} b}} \geq\ \sqrt [4]{8}. \sqrt [4]{\frac {a(b \plus{} c)}{a^2 \plus{} 5bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} 5ca} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} 5ab}}$
 :)[/quote]
It follows from
$ \sum \frac{a(b\plus{}c)}{a^2\plus{}5bc} \le \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}.$
:)

Yes, you are right  :oops:
But my proof for this ineq is so long. Wait for the nicer proofs  :(


Show that if [i]m[/i] and [i]n[/i] are non-negative integers then

$ \sum_{k \equal{} 0}^{n}( \minus{} 1)^{k}\binom{m \plus{} 1}{k}\binom{m \plus{} n \minus{} k}{m} \equal{} 0$   if   $ n > 0$


If  $ P(x):=x^{n}+\sum\limits_{k=1}^{n}a_kx^{n-k} $  with $0\le  a_n\le a_{n-1}\le\ldots a_2\le a_1\le 1 $, [b] prove that $  P\left(1-i\sqrt{2}\right)\ne 0 \; .\;  $ [/b]


Six Zorcs, two Ners and three Kerals are better at math than three Zorcs, nine Kerals, and two Ners. If each alien has the same mathematical abilites of all the other aliens in its species (all Zorcs are as good as math as all other Zorcs), who is better at math, a Zorc or a Keral?

We have an inequality.

$ 6z \plus{} 2n \plus{} 3k > 3z \plus{} 2n \plus{} 9k$

Subtract $ 2n$ from both sides.

$ 6z \plus{} 3k > 3z \plus{} 9k$

$ 6z \equal{} 3z \plus{} 6k$

$ 3z \equal{} 6k$

$ z \equal{} 2k$

Thus, the answer is clearly $ \boxed{\text{zorc}}$.


inequalities
combinatorics unsolved
combinatorics

In a math competition there are $6$ problems. Every two of them are solved by at least $\frac{2}{5}$ of the contestants. Moreover, no contestants solved all $6$ problems. Show that there exists a set of $3$ problems solved by more than $\frac{1}{5}$ of the contestants and a set of $4$ problems solved by more than $\frac{1}{15}$ of the contestants.

I'll just show existance of a set of $3$ problems as $4$ problems part is the same with difference in computing. Let $n$ be number of students and $a_{i}$ be number of contestants who solved $i$ problems. Then $n=a_{1}+a_{2}+...+a_{5}$ ($a_{6}=0$ is assumed). Let's double count number of pairs of problems solved by some student.
We have: $6(a_{1}+a_{2}+...+a_{5})=6n = \binom{6}{2}\cdot \frac{2}{5}\cdot n \leq ( (x_{i},x_{j}), C) = \sum_{i=1}^{5}\binom{i}{2}a_{i}=a_{2}+3a_{3}+6a_{4}+10a_{5}$ (where LHS comes from assumption that every pair of problems is solved by at least $\frac{2}{5}$ of contestants) which is equivalent to: $6a_{1}+5a_{2}+3a_{3}\leq 4a_{5}$. Suppose now that there now does not exist wanted set of 3 problems. We analogically count this time triplets of problems:
$4(a_{1}+...+a_{5}) = 4n=\binom{6}{3}\cdot \frac{n}{5}\geq ( (x_{i},x_{j},x_{k}); C) = \sum_{i=1}^{5}\binom{i}{3}a_{i}= a_{3}+4a_{4}+10a_{5}$
which is equivalent to $4a_{1}+4a_{2}+3a_{3}\geq 6a_{5}$
From these two inequalities we quickly get $a_{1}=a_{2}=a_{3}=a_{5}=0$ so every contestant solved exactly 4 problems. But this is contradiction with original problem C6 from ISL 2005 (there must be students who solved 5 problems)


In the rectangular prism shown, $ AB = 6\text{ cm}$, $ BC =8 \text{ cm}$, and $ HC = 24\text{ cm}$. What is the number of centimeters in the length of diagonal $ \overline{FC}$?

[asy]size(150); import three; currentprojection = orthographic(1/2,-1/2,1/2); triple A = (0,0,0), B = (6,0,0), C = (6,8,0), D = (0,8,0), F = (0,0,24), Ep = F + D, H = F+C, G = F+B;
draw(A--B--C); draw(C--D--A,linetype("2 2")); draw(D--Ep,linetype("2 2")); draw(H--Ep--F--G--cycle3); draw(A--F); draw(B--G); draw(H--C); label("$A$",A,W,fontsize(8pt)); label("$B$",B,S,fontsize(8pt)); label("$C$",C,E,fontsize(8pt)); label("$F$",F,W,fontsize(8pt)); label("$E$",Ep,N,fontsize(8pt)); label("$H$",H,E,fontsize(8pt));[/asy]