Im down to one last problem i can find the answer to, and i was wondering if any1 could help me out???

Alvin paddled for 3 hours with a 4-km/h current to reach a campsite.  The return trip against the same current took 7 hours.

Find the speed of the boat in still water.

Umm... I can do it ==> not exactly college algebra

Hint:

[hide]distance traveled = speed * time
speed when boat is traveling with current: speed of boat + speed of current
speed when boat is traveling against current: speed of boat - speed of current
set up an equation.
answer is an integer.[/hide]

Thanks for the help. I have mono and my brain isnt working the way i need it to right now.

If you have many problems like this, it's very helpful to remember that d=rt. Then you can set up 2 equations, as there are 2 trips. Make sure you don't get the plus and minus mixed up, though. That's the biggest problem people face doing these problems, I find.

[quote="fishythefish"]Let x be the speed of the boat (in mph) and let d be the distance. Set up 2 equations:

d=3(x+4) He paddled for 3 hours with the current.
d=7(x-4) He paddled for 7 hours against the current.

Therefore, 3(x+4)=7(x-4).

3x+12=7x-28
40=4x
10=x

10 mph.[/quote]

Another pretty big issue is the units on your answer.  The problem was given in kilometers, so the answer should also be in kilometers: $ 10 km/h$.  Fishy - be careful!

This might be irrelevant, but to remember $ d\equal{}st$ (VERY IMPORTANT), and its other forms to be able to solve for speed and time, draw a triangle with two rows, the second, bottom row split in half, and put d in the top, s in the bottom left, and t in the bottom right. Now, cover up the one you want to solve for, say time. from the triangle, d is above s, so $ t\equal{}\frac{d}{s}$, and so on.

This is very useful to remember.

Also, for boat and current problems, the current just speeds up or slows down the boat by the same amount, so algebraic substitution works well.

[quote="fishythefish"]Sorry. I automatically assume that the units are in US units. (I should probably have checked that) But the important part is the solution...RIGHT?? :oops:[/quote]

yes

[quote="myyellowducky82"][quote="Textangle"]This might be irrelevant, but to remember $ d \equal{} st$ (VERY IMPORTANT), and its other forms to be able to solve for speed and time, draw a triangle with two rows, the second, bottom row split in half, and put d in the top, s in the bottom left, and t in the bottom right. Now, cover up the one you want to solve for, say time. from the triangle, d is above s, so $ t \equal{} \frac {d}{s}$, and so on.

This is very useful to remember.

Also, for boat and current problems, the current just speeds up or slows down the boat by the same amount, so algebraic substitution works well.[/quote]
Isn't it $ rt$...?[/quote]

r=rate, s=speed.  Both equations mean the same thing.  Also note that we could use $ d\equal{}vt$ - v is velocity.

Okay using D=RT we can get D = (4 + R)3, and D = (R - 4)7, so D = 12 + 3R, and D = 7R - 28, so
12 + 3R = 7R - 28, so 40 = 4R and 10 = R, so the rate in still water is 10km/h

[quote="facis"]

r=rate, s=speed.  Both equations mean the same thing.  Also note that we could use $ d \equal{} vt$ - v is velocity.[/quote]
Oh okay. I misread it at first  :P

[quote="Wickedestjr"]I believe, that using the D=RT formula, that you can solve any rate problem 30% faster at least.[/quote]
yes the way you put it in the other post was correct

[quote="Wickedestjr"]Are you talking about the statement I said before this post, or my answer to the problem?[/quote]
I meant both =]


[quote="Jergosh"]Is 0.12102100210002100002... rational? I tried to divide this fraction into a sum of sequences but the I couldn't find the limits. I also can't think of any proof by contradiction. What approach should I take?

Thanks in advance,
Greg[/quote]This is equal to $0.1+21 \cdot 0.001001000100001\cdots$. Thus the question is whether $z = 0.001001000100001...$ is rational. The $n$-th [b]one[/b] in this decimal expansion is preceded by $n$ zeroes.

If $z = \frac{p}{q}$ with integers $p$ and $q$, this would imply that $qz = p$ is an [b]integer[/b]. Now think about the $q$-th "1" in the decimal expansion of $z$ and the preceding $q$ zeroes. What happens to this portion of the decimal expansion when you multiply $z$ with $q$?

[i]corrected version[/i]

[quote]This is equal to $0.1+0.21 \cdot 0.001001000100001\cdots$.[/quote]
Isn't it $0.121021002\cdots = 0.1+0.0210210021\cdots = 0.1+21\cdot0.0010010001\cdots$ ?

[quote]Thus the question is whether $z = 0.001001000100001...$ is rational.[/quote]
Agreed.
[quote]The $n$-th digit in this decimal expansion is preceded by $n$ zeroes.[/quote]
Is it? The first decimal place is preceeded by 0 zeroes, second by 1, fourth by 2 etc.

[quote]If $z = \frac{p}{q}$ with integers $p$ and $q$, this would imply that $qz = p$ is an [b]integer[/b]. Now think about the $p$-th "1" in the decimal expansion of $z$ and the preceding $p$ zeroes. What happens to this portion of the decimal expansion when you multiply with $p$?[/quote]

I don't follow. Let me take $p=3$.
$3\cdot z = 0.003003000300003\cdots$
And the part you are (I think) referring to changes from $\cdots 0001\cdots$ to $\cdots 0003\cdots$.
Similarly, for $p's = 4,5,6,...$ we get
$4\cdot z = 0.004004000400004\cdots$
$5\cdot z = 0.005005000500005\cdots$
$6\cdot z = 0.006006000600006\cdots$
And I don't get the point of that. I suppose I'm misunderstanding you at some point.

[quote="Jergosh"]
Isn't it ...[/quote]Sorry about the misunderstanding - my fault. See the corrected version above.

I think  in $hpe$ proof we should multipy $10^{k}$ for some k in $qz=p$ :?:

Anyway,I think the proof for $x=\sum_{k=1}\frac{1}{2^{k^{2}}}$ is irrational[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=112082[/url](Post9) is also valid here :)

[quote="Jergosh"]Is 0.12102100210002100002... rational?
[/quote]

I don't understand your problem. Rational numbers have a periodic decimal fraction. Your fraction isn't periodic evidently.


function
geometry
geometric transformation
reflection
complex analysis

I am supposed to use Laurent expansions to do the following proof, but I'm stuck:

--
Let f: D*(0,1)-->C be analytic. Show that if

|f(z)|[u]<[/u]log(1/(|z|))

for all z in D*(0,1), then f is identically the zero function.

(Here, D*(0,1) denotes a deleted disc of radius 1 about 0, i.e. D*(0,1)={z in C| 0<|z-1|<1}).

--
Any help is appreciated! Thanks!

If $ f$ were analytic on the full disk, it would be zero by the maximum principle. Multiply by $ z$; $ zf(z)$ goes to zero at zero, so the singularity is removable and $ zf(z)$ is analytic on the full disk. By the same argument, $ zf(z)$ is identically zero, and $ f$ must be identically zero as well.

Actually, we don't need the size estimate near zero at all. If $ f$ is analytic on the punctured disk, or even just an annulus $ c<|z|<1$, and $ |f(z)|$ goes to zero uniformly as $ |z|\to 1^\minus{}$, $ f$ is identically zero.

because $ f$ can be extended to an analytic function $ F$ on $ c<|z|<1/c$ by reflection ($ F(z)\equal{}1/\bar f(1/\bar z)$ for $ 1<|z|<1/c$).

[quote="mlok"]because $ f$ can be extended to an analytic function $ F$ on $ c < |z| < 1/c$ by reflection ($ F(z) \equal{} 1/\bar f(1/\bar z)$ for $ 1 < |z| < 1/c$).[/quote]

What if f is zero at some point?

I meant to write $ F(z)\equal{}\bar f(1/\bar z)$.

By the way, the [url=http://en.wikipedia.org/wiki/Schwarz_reflection_principle]Wikipedia article[/url] on reflection principle says:
[quote]In practice it would be better to have a theorem that allows F certain singularities, for example F a meromorphic function. To understand such extensions, one needs a proof method that can be tweaked. In fact Morera's theorem is well adapted to proving such statements. Contour integrals involving the extension of F clearly split into two, using part of the real axis. So, given that the principle is rather easy to prove in the special case from Morera's theorem, understanding the proof is enough to generate other results.[/quote]
So, what's the result for meromorphic functions?  :?:


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

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

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

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

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

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

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

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

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

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

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

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

[hide="#1 Solution"]$ 1\cdot 2+2\cdot 3+3\cdot 4+\cdots+n(n+1)+\cdots $ can be written as $(1^1+1)+(2^2+2)+(3^2+3)+\cdots+(n^2+n)+\cdots$.

If we subtract the first $k$ terms of the series $1^{2}+2^{2}+3^{2}+\cdots+n^{2}+\cdots $ from the first $k$ terms of the series $(1^1+1)+(2^2+2)+(3^2+3)+\cdots+(n^2+n)+\cdots$, the first $k$ squares cancel out with each other and what we have left is the sum of the first $k$ integers.

The formula of the sum of the first $k$ integers is $\frac{k(k+1)}{2}$.  We set this equal to $210$ and solve for $k$:

\[\frac{k(k+1)}{2}=210\]
\[k^2+k=420\]
\[k^2+k-420=0\]
\[(k+21)(k-20)=0\]
\[k=\{-21,20\}\]

Since $k$ must be positive, $k=\boxed{20}$.
[/hide]

[hide="#3 Solution"]
Cross-multiplying, we have

$7x + 7z = kx + ky$
$kz - ky = 11x + 11z$

Adding the equations yields

$18x + 18z = kx + kz$

Obviously, the only value of $k$ satisfying this would be $\boxed{18}$.
[/hide]

[hide="#4 Solution"]
We can write the equation $D = 1.5P$ because the number of dimes was $150\%$ of the number of pennies he had.

There are two cases for $P$.  One case is that $P \geq 78$, in which the remaining money would be $D - 10$ dollars and $P - 78$ cents.  If this case was true, then $D - 10 + 1 = P - 78 \Longrightarrow 0.5P = -78 + 10 - 1 < 0$, so this case is not true.  Hence, $P < 78$.

Since $P < 78$, the remaining money would be $D - 11$ dollars and $P + 100 - 78$ cents.  Thus, $D - 11 + 1 = P + 100 - 78 = P + 22$.  So $0.5P = 32$ and $P = 64$.  So $D = 96$ and the money Sam had was $\boxed{\$96.64}$.

To check, we have $96.64 - 10.78 = \$85.86$, and $85 + 1$ is indeed $86$.
[/hide]

[hide="#5 Solution"]
Casework.

We have three cases:

King, not heart and Heart, not king.
King, not heart and Heart, king.
King, heart and Heart, not king.

Of course, we cannot have King, heart and king, heart since we are drawing without replacement.

For the first case, the probability of drawing a king but not heart is $\frac{3}{52}$ and then a heart but not a king is $\frac{12}{51}$.  There is no overlap--after the first pick the king of hearts is still in the deck.  So the probability of this case occurring is $\frac{3}{52} \cdot \frac{12}{51}$.

For the second case, the probability of drawing a king but not a heart is $\frac{3}{52}$ and then the king of hearts is $\frac{1}{51}$.  So the probability of this case occurring is $\frac{3}{52} \cdot \frac{1}{51}$.

For the third case, the probability of drawing a king of hearts is $\frac{1}{52}$ and the probability of drawing a heart but not a king is $\frac{12}{51}$.  Anyways, the king of hearts was already picked, so all $12$ of the remaining hearts will do.  So the probability of this case occurring is $\frac{1}{52} \cdot \frac{12}{51}$.

So our final probability is $\frac{3 \cdot 12 + 3 \cdot 1 + 1 \cdot 12}{51 \cdot 52} = \boxed{\frac{1}{52}}$.
[/hide]

Hmm the New Jersey forum is much more interesting than the forum for the place in which I live in.  There's actually a lot of math.

[hide="#6 Solution"]
Let $\sqrt{21 + 12\sqrt{3}}$ be $m + n\sqrt{p}$.  Then, we can write

$m^2 + n^2p + 2mn\sqrt{p} = 21 + 12\sqrt{3}$ (after squaring both sides)

Matching terms yields

$m^2 + n^2p = 21$ and $2mn\sqrt{p} = 12\sqrt{3}$.

From the second equation, we easily get $mn = 6$ and $p = 3$.  Testing, we find $m = 3$ and $n = 2$ works.  So $\sqrt{21 + 12\sqrt{3}} = 3 + 2\sqrt{3}$.

$1 + (3 + 2\sqrt{3}) = 4 + 2\sqrt{3}$.  So now the equation becomes

$\sqrt{4 + 2\sqrt{3}} = \sqrt{a} + \sqrt{b}$
$4 + 2\sqrt{3} = a + b + 2\sqrt{ab}$

Matching terms again yields

$a + b = 4$
$ab = 3$
$(a,b) = \boxed{(1,3)} \ \text{or} \ (3,1)$.

The ordered pair $(3,1)$ won't work because $a < b$, as specified in the problem.
[/hide]

[hide="#7 Solution"]
I'll define points so the answer will be clearer.  Let $\angle A$ and $\angle C$ be the angles facing the congruent sides and let $\angle B$ face the side with length 20.  Let $P$ be the point 4 units from the base, and let $PE$ and $PF$ be the altitudes from P to side AB and side BC, respectively.  Finally, let $x$ be the length of the height from point B to the base AC, and point G be the point where the height from B intersects base AC.

If $P$ is $10$ units from each leg, then obviously it must lie on the altitude BG.  Since $BP = BG + GP$, $BP = x + 4$.  We see that $\triangle BEP$ is congruent to $\triangle BAG$ because they both have a right angle, share $\angle ABP$, and have one side as $10$ (in $\triangle BAG$, $AG = 10$, and in $\triangle BEG$, $EG = 10$.)  Thus, $BE = x$.

So we can write an equation by Pythag:  $x^2 + 10^2 = (x + 4)^2$.  This results in $x = \frac{84}{8} = \frac{21}{2}$.  Thus, the area is $\frac{1}{2} \cdot \frac{21}{2} \cdot 20 = \boxed{105}$ square units.
[/hide]

[hide="#8 Solution - this is a little late but I wanted to finish it"]
Let $\arctan{x} = \alpha$.  Then, we have $\sin{4\alpha} = \frac{24}{25}$, or
\[\sin{\alpha}\cos{\alpha}(1 - 2\sin^2{\alpha}) = \frac{6}{25}\]  Now since $\tan{\alpha} = x$, $\cos{\alpha} = \frac{1}{\sqrt{1 - x^2}}$ and $\sin{\alpha} = \frac{x}{\sqrt{1 - x^2}}$.  Thus
\begin{align*}\frac{x}{1 + x^2} \cdot \frac{1 - x^2}{1 + x^2} &= \frac{6}{25}\\ x - x^3 &= \frac{6}{25} \cdot (x^4 + 2x^2 + 1)\\ 6x^4 + 25x^3 + 12x^2 - 25x + 6 &= 0\end{align*}  By RRT, $-2$ and $-3$ are solutions.  Dividing gives
\[6x^4 + 25x^3 + 12x^2 - 25x + 6 = (x + 2)(x + 3)(6x^2 - 5x + 1)\]  So the solutions are $\frac{1}{2}$, $\frac{1}{3}$, $-2$, and $-3$.
[/hide]

[quote="professordad"][hide="#2 Solution"]
Is it just me or is the answer $\frac{30}{\sqrt{2}} = 15\sqrt{2}$?
[/hide][/quote]
[hide]
Dude, what are you smokin?  It's the shortest distance, and the squares don't have to have the same center.  So it's $\frac{30}{\sqrt{2}} - 10\sqrt{2}$, which is $\boxed{5\sqrt{2}}$.  You must suck at math.
[/hide]

If you're not going to troll, get off our forum, you invaders.

AAST Zheus will win ARML.

If I'm not mistaken, don't most state's forum have practice problems?

Yes that's correct.  I think they're all the same though.


geometry
circumcircle
geometric transformation
reflection
Euler
symmetry
modular arithmetic

Let $ (O)$ be the circumcircle of triangle $ \triangle ABC$. Denote $ M_a,$ $ M_b$ and $ M_c$ respectively by the antipoles of $ A,B,C$ with respect to $ (O)$. Let $ O_a,O_b$ and $ O_b$ respectively be the reflections of $ O$ across to $ BC,CA,AB$. Prove that the circumcircles of triangles $ \triangle OO_aM_a$, $ \triangle OO_bM_b$ and $ \triangle OO_cM_c$ go through a common point $ F$, such that $ F$ is different from $ O$ and $ F\in (O)$.

[i]Best regard[/i]
[i][b]mathVNpro[/b][/i]

Consider the positive inversion with respect to $ (O).$ Vertices $ A,B,C$ and their antipodes transform into themselves, the midpoints $ M,N,L$ of $ BC,CA,AB$ are taken into the vertices of the tangential triangle $ A'B'C'$ and reflections of $ O$ about $ BC,CA,AB$ are taken into the midpoints $ U,V,W$ of $ OA',OB',OC',$ respectively. Therefore the circles $ \odot(OO_aM_a),$ $ \odot(OO_bM_b),$ $\odot(OO_cM_c)$ are transformed into the lines $ UM_a,VM_b,WM_c.$ Circle with diameter $ UA$ is the A'-Garitte circle of $ \triangle A'B'C'$ passing through its Feuerbach point. Since $ M_a$ is the antipode of $ A$ WRT $ (O),$ we deduce that $ UM_a$ cuts $ (O)$ again at the Feuerbach point $ F$ of $ \triangle A'B'C'$ $ \Longrightarrow$ $ UM_a,VM_b,WM_c$ concur at $ F.$ Hence, $ \odot(OO_aM_a),$  $\odot(OO_bM_b),$ $\odot(OO_cM_c),$ $(O)$ concur at the Feuerbach point $ F$ of the tangential triangle $ \triangle A'B'C'.$

Denote $ Q$ the Anti-Steiner point of Euler line of triangle $ ABC$.
We have $ \angle OO_aQ\equal{}\angle HOO_a\equal{}\angle AHO\equal{}\angle AB'L\equal{}\angle AM_aQ$. Therefore $ OQO_aM_a$ is cyclic. Similarly we get $ (OO_aM_a), (OO_bM_b),(OO_cM_c)$ concur at $ Q$ which lies on $ (O)$.

Let $ H$ be the orthocenter of $ \triangle ABC$ and $ H_a \equiv AH \cap (O)$, $ F \equiv (OO_aM_a) \cap (O)$. We have $ H_a$ is symmetry to $ M_a$ wrt $ OO_a$ so $ (FO,FO_a) \equiv (M_aO,M_aO_a) \equiv (H_aO_a,H_aH_aO) \equiv (FO,FO_a) \pmod{\pi}$ so $ F\equiv O_aH_a \cap (O)$. On the other hand, it is well known that $ O_aH_a, O_bH_b, O_cH_c$, the symmetry of the Euler line of $ \triangle ABC$ wrt $ BC,CA,AB$ concurs at a point on $ (O)$, this point must be $ F$ and so $ (OO_bM_b)$ and $ (OO_cM_c)$ also passes through $ F$, qed.

Thank you for all yours' solution, my friend. But personally, I think that the problem can be solved in the simplest way.
Here is my solution

Let $ F$ be the intersecitons of $ (OO_bM_b)$ and $ (OO_bM_c)$. We will prove these two following claims.

$ \bullet$ [i]Claim 1.[/i] [i]$ F$ is a point lies on $ (O)$[/i].

Indeed, we have $ \angle M_bFM_c \equal{}$ $ \angle M_bFO \plus{} \angle OFM_c$ $ (*)$. But since $ F\in (OO_bM_b)$ and also $ \in (OO_cM_c)$, thus, we have $ \angle M_bFO \equal{} 180^{\circ} \minus{} \angle M_bO_bO$ and $ \angle OFM_c \equal{} 180^{\circ} \minus{} \angle M_cO_cO$. Let $ H$ be the orthocenter of $ \triangle ABC$. It is well- known that $ M_cOHO_c$ and $ M_bOHO_b$ are parallelogram and also with the notice that $ HB\|OO_b$, $ HC\|OO_c$, we obtain $ \angle M_bO_bO \equal{} \angle OHB$ and $ \angle M_cO_cO \equal{} \angle OHC$. Thus, from $ (*)$, we have $ \angle M_bFM_c \equal{} 360^{\circ} \minus{}$ $ \angle OHB \minus{}$ $ \angle OHC$ $ \equal{} \angle BHC$ $ \equal{} 180^{\circ} \minus{} \angle BAC$ $ \equal{} 180^{\circ} \minus{} \frac {1}{2}\angle BOC$ $ \equal{} 180^{\circ} \minus{} \angle M_cBM_b$, which implies that $ F\in (O)$.

$ \bullet$ [i]Claim 2.[/i] [i]$ F$ also lies on the circumcircle of $ \triangle OO_aM_a$[/i].

Indeed, $ \angle M_aFO \equal{}$ $ \angle M_aFM_b \minus{} \angle OFM_b$. But $ \angle M_aFM_b \equal{} 180^{\circ} \minus{} \angle M_aCM_b$ $ \equal{} 90^{\circ} \minus{} \angle ACM_b$ $ \equal{} 90^{\circ} \minus{} \angle ABO$ $ \equal{} \angle ACB$. As in the proof of the first claim, we have known that $ \angle OFM_b \equal{} 180^{\circ} \minus{} \angle OHB$. Thus $ \angle M_aFO \equal{}$ $ \angle ACB \minus{}$ $ 180^{\circ}$ $ \plus{} \angle OHB$. Besides, with also a notice that $ OHO_aM_a$ is a parallelogram and $ OO_a\|AH$, thus $ \angle OO_aM_a \equal{} \angle OHA$. We need to show that $ \angle ACB \minus{}$ $ 180^{\circ}$ $ \plus{} \angle OHB$ $ \equal{}$ $ \angle OHA$. But this is equivalent to an obvious fact $ \angle ACB \equal{} 180^{\circ} \minus{} \angle AHB$. As the consequences, $ F$ $ \in (OO_aM_a)$, also.

From the two above claims, we are able to conclude that $ (OO_aM_a)$, $ (OO_bM_b)$ and $ (OO_cM_c)$ go through a common point $ F$ $ \in (O)$. Our proof is completed then. $ \square$
______________________________________

[i][b]Remark.[/b][/i] Thanks to this problem and the solutions of [b]Mashimaru[/b], [b]Livetolove212[/b] and mine, we have another inversive proof for the [i]Feurebach[/i] concurrency of $ UM_a,VM_b,WM_c$ in [b]Luis[/b]'s post. First, let me restaste this concurrency property again:

[b]Problem.[/b][b][color=darkblue]Let $ ABC$ be a triangle with $ (I)$ its incircle. Denote $ A_0,B_0,C_0$ respectively be the tangency points of $ (I)$ with $ BC,CA,AB$. Let $ A_1,B_1,C_1$ be the antipodes of $ A_0,B_0,C_0$ with respect to $ (I)$. $ A_2,B_2,C_2$, respectively, are the midpoints of $ IA,IB,IC$. Prove that $ A_1A_2,B_1B_2$ and $ C_1C_2$ concurrent at $ F_e$, the [i]Feurebach point[/i] wrt $ \triangle ABC$.[/color][/b]

[i]Proof[/i]

All angles are directed modulo $ \pi$.

Consider the inversion through pole $ I$, power $ k \equal{} r^2$, where $ r$ is the radii of $ (I)$. We have $ \mathcal {I}(I,k)$ maps $ A_1,B_1,C_1$ into themselves and $ A_2,B_2,C_2$ is mapped into $ A'_2,B'_2,C'_2$ which respectively are the reflections of $ I$ across to $ B_0C_0,$ $ C_0A_0$ and $ A_0B_0$. Therefore, $ \mathcal {I}(I,k):$ $ A_1A_2\mapsto (IA_1A'_2)$, $ B_1B_2\mapsto (IB_1B'_2)$ and $ C_1C_2\mapsto (IC_1C'_2)$. But according to the problem from this topic, we have $ (IB_1B'_2)$, $ (IA_1A'_2)$ and $ (IC_1C'_2)$ go through a point $ F_e\in (I)$, note that $ F_e$ is mapped into itself through $ \mathcal {I}(I,k)$ $ \Longrightarrow$ $ A_1A_2,$ $ B_1B_1,$ $ C_1C_2$ go through $ F_e\in (I)$ $ (*)$. We claim that $ F_e\in$ the three $ 9 \minus{}$ point Euler circles wrt $ \triangle IBC$, $ \triangle ICA$ and $ \triangle IAB$. $ (**)$

Indeed, first, we have $ A_0B_1\perp A_0B_0;$ $ CT\perp A_0B_0$, thus $ A_0B_1\|CI$, but $ C_2$ is the midpoint of $ IC$. Hence, $ (A_0B_1,A_0C_2,A_0I,A_0C) \equal{} \minus{} 1$. Further, since $ A_0I\perp A_0C$ $ \Longrightarrow$ $ A_0I$ and $ A_0C$ respectively are the internal and external bisector with respect to $ \angle B_1A_0C_2$. Similarly, we also have $ A_0I$ and $ A_0C$ respectively are the internal and external bisector of $ \angle B_2A_0C_1$. And for that reasons, we obtain $ \angle B_1A_0C_1$ $ \equal{} \angle B_2A_0C_2$. But since $ F_e\in (I)$ $ \Longrightarrow$ $ \angle B_1F_eC_1$ $ \equal{} \angle B_1A_0C_1$. Therefore, $ \angle B_2F_eC_2$ $ \equal{} \angle B_2A_0C_2$, which implies $ F_e\in (A_0B_2C_2)$, which is also the $ 9 \minus{}$ point Euler circle of $ \triangle IBC$. Argue the same, we obtain that the three $ 9 \minus{}$ point Euler circle of $ \triangle IBC$, $ \triangle ICA$ and $ \triangle IAB$ go through $ F_e\in (I)$. The result $ (**)$ is lead as followed. We claim that $ F_e$ also $ \in$ the $ 9 \minus{}$ point Euler circle wrt $ \triangle ABC$. $ (***)$

Indeed, let $ M_a,M_b,M_c$ respectively be the midpoints of $ BC,CA,AB$. We have $ \angle M_cF_eM_a$ $ \equal{} \angle M_cF_eB_1 \plus{} \angle B_1F_eC_2$ $ \plus{} \angle C_2F_eM_a$ $ \equal{} \angle M_cA_2B_2 \plus{} \angle B_2M_aC_2$ $ \plus{} \angle C_2B_2M_a$ $ \equal{} \angle IBA$ $ \plus{} \angle CIB$ $ \angle BCI$ $ \equal{} \angle CBA$. But $ \angle CBA$ $ \equal{} \angle M_cM_bM_a$ (With a notice that $ M_bM_c\|BC$) $ \Longrightarrow$ $ \angle M_cF_eM_a \equal{} \angle M_cM_bM_a$, which implies $ F_e\in (M_aM_bM_c)$. We obtain the result $ (***)$.

$ F_e$ lies on $ (I)$, $ F_e$ also lies on the $ 9 \minus{}$ point circle wrt $ \triangle ABC$. Hence, $ F_e$ is the [i]Feurebach point[/i] wrt $ \triangle ABC$. Then combine $ (*)$ and $ (***)$, we obtain the result of the porblem. Our proof is completed then. $ \square$

[quote="mathVNpro"]Let $ (O)$ be the circumcircle of triangle $ \triangle ABC$. Denote $ M_a,$ $ M_b$ and $ M_c$ respectively by the antipoles of $ A,B,C$ with respect to $ (O)$. Let $ O_a,O_b$ and $ O_b$ respectively be the reflections of $ O$ across to $ BC,CA,AB$. Prove that the circumcircles of triangles $ \triangle OO_aM_a$, $ \triangle OO_bM_b$ and $ \triangle OO_cM_c$ go through a common point $ F$, such that $ F$ is different from $ O$ and $ F\in (O)$.[/quote]

From this problem, there is some nice things realted to it. Hope you enjoy it and solve it with a nice solution!

With the same notations of objects in the above result.

[color=darkblue][b]$ (1)$ Let $ d$ be the line contains the centers of $ (OO_aM_a)$, $ (OO_bM_b)$ and $ (OO_cM_c)$. Then $ d$ always tangents to a fixed circle, particularly the circle $ (O,\frac {R}{2})$, where $ R$ is the radii of $ (ABC)$.

$ (2)$ Let $ d_a,d_b,d_c$ be perpendicular bisectors of $ OM_a,OM_b,OM_c$. Let $ C'\equiv d_a\cap d_b$, $ B'\equiv d_a\cap d_c$ and $ C'\equiv d_a\cap d_b$. Then $ AA',BB',CC'$ are concurrent at a point $ F'$ which is also a point lies on $ (O)$, futher $ F'$ is the antipode of $ F$ wrt $ (O)$.[/b][/color]

[quote="mathVNpro"]$ (1)$ Let $ d$ be the line contains the centers of $ (OO_aM_a),$ $ (OO_bM_b)$ and $ (OO_cM_c).$ Then $ d$ always tangents to a fixed circle, particularly the circle $ (O,\frac {R}{2}),$ where $ R$ is the radii of $ (ABC).$[/quote]
This follows easily by noticing that the circumferences $ \odot (OO_aM_a) , \odot (OO_bM_b)$ and $ \odot (OO_cM_c)$ are coaxal with common radical axis the straight line $ OF$ $ \Longrightarrow$  their centers lie on the perpendicular bisector of $ OF.$

[quote="mathVNpro"]$ (2)$ Let $ d_a,d_b,d_c$ be perpendicular bisectors of $ OM_a,OM_b,OM_c.$ Let $ C'\equiv d_a\cap d_b,$ $ B'\equiv d_a\cap d_c$ and $ C'\equiv d_a\cap d_b.$ Then $ AA',BB',CC'$ are concurrent at a point $ F'$ which is also a point lies on $ (O),$ futher $ F'$ is the antipode of $ F$ wrt $ (O).$[/quote]
Using the same notation in my previous post, the central symmetry with center $ O$ maps $ F$ into its antipode $ F'$ on $ (O),$ $ A$ into $ M_a$ and the perpendicular bisectors of $ OM_b,OM_c$ into the perpendicular bisectors of $ OB,OC$ obviously meeting at the midpoint $ U$ of $ OA'.$ Since $ F \in UM_a,$ then $ AA' \in F'.$ By similar reasoning the result follows.

Dear Mathlinkers,
concerning the message #3, the Reim's theorem gives the result....
Sincerely
Jean-Louis


[quote="anirudh"][quote="i_like_pie"]In the equation $x^{2}+x+41,$ where $x\ge0,$ what is the smallest value of $x$ that makes the outcome a composite number?[/quote]

Is the answer
[hide="of some wierd origin"]163?[/hide][/quote]
That does give a composite number, but their is a smaller value of $x$ that satisfies the condition.

[quote="nateharman1234"]well 41 is gives us a composite number, so we have that as an upper bound.[/quote]
41 also gives a composite number, but there is a still smaller value of $x.$

This is a famous observation of Euler. See [url=http://primes.utm.edu/curios/page.php/41.html]here[/url].

[quote="nateharman1234"]well 41 is gives us a composite number, so we have that as an upper bound.

edit: [hide] ok now im getting 40 is composite too as (n-1)^2 +(n-1) + n = n^2,  I think thats the lowest but i cant prove it.[/hide][/quote]
Correct! :D


[quote="h_s_potter2002"]Because I'm really impatient and not very good at this, can someone give me step by step instructions on how to download tetris, from here - [url]http://www.ti-89.org/archive.html[/url] ?

All I have is a calculator, a mini-usb to usb cord (I think that's what it is.  All I know is that it connects the calculator to the computer), and a computer.

Thanks.[/quote]

Uh huh. Did you install TI-CONNECT onto your computer? (It's on the CD that hopefully came with your calculator.)

I think you can download TiConnect from the Ti website.

download ticonnect and win zip and you'll be all set.

Yea, I figured it out.

I lost the CD but I was able to download it and stuff.

Thanks.


integration
trigonometry
inequalities unsolved
inequalities

[quote="Sunkern_sunflora"]\frac 12 - \frac 1{23} + \int^1_0 f^2(x_1) dx = \cos x + \sin \frac 1x[/quote]
$ \frac 12 \minus{} \frac 1{23} \plus{} \int^1_0 f^2(x_1) dx \equal{} \cos x \plus{} \sin \frac 1x$
  what's mean $ \frac{1}{2} \minus{}\frac{1}{23}$


Could someone define the big O notation to me and explain some of its properties?

Also, prove that [size=134]O(x³)/x = O(x²)[/size].

Thanks.

Suppose $ f: \mathbb{R} \to \mathbb{R}$ and $ g: \mathbb{R} \to \mathbb{R}$.  Then $ f(t)$ is $ O(g(t))$ as $ t \to \infty$ iff there exists $ C > 0$ such that $ |f(t)| \le C|g(t)|$ for all sufficiently large $ t$.

In other words, $ f(t)$ is $ O(g(t))$ as $ t \to \infty$ iff there exists $ C > 0$ and $ T \in \mathbb{R}$ (perhaps very large) such that $ |f(t)| \le C|g(t)|$ for all $ t > T$.

Next, suppose that $ f: \mathbb{R} \to \mathbb{R}$ is $ O(t^3)$ as $ t \to \infty$.  Let $ g(t) \equal{} f(t)/t$.  ($ g(0)$ is undefined, but that doesn't matter since we're interested in what happens as $ t \to \infty$.)  There exists $ C > 0$ such that $ |f(t)| \le C|t^3|$ for sufficiently large $ t$.  It follows that $ |f(t)/t| < C|t^2|$ for sufficiently large $ t$.  Thus, $ g$ is $ O(t^2)$ as $ t \to \infty$.


I got a challenge I found a question in a mathcounts book and I thought it was out of the ordinary and a little challenging(I got it) but I don't know what's the difference of level I might think it's a between hard and medium but I want to know what you guys think it is.

So here it is(words are not exact to the problem):
$ A,B,C,D$ are positive integers.
$ ABC\plus{}DCA\equal{}1000$ where each letter represents an digit in the number. What is $ A\plus{}B\plus{}C\plus{}D$?

I understand your problem as: $ \overline{ABC}\plus{}\overline{DCA}\equal{}1000$, in which $ 1\leq A,B,C,D\leq 9$.
We have: $ C\plus{}A\equal{}10,B\plus{}C\equal{}9,A\plus{}D\equal{}9$ then $ A\plus{}B\plus{}C\plus{}D\equal{}18$.

PS: Sorry if I misunderstand your idea!

well remember $ CA \plus{} BC\equal{}100$ and the rest follows.

[hide]
$ 654\plus{}346\equal{}1000$
$ A \plus{} B \plus{} C \plus{} D \equal{} 6 \plus{} 5 \plus{} 4 \plus{} 3\equal{}18$
[/hide]

yeah, i got the same as the rest of you guys...
sorry tonypr..i don't think it was particularly difficult...maybe midway in a state or something. just my opinion.

[quote="abacadaea"][quote="Fraenkel"]well remember $ CA \plus{} BC \equal{} 100$ and the rest follows.
[/quote]
where did you get that?[/quote]

If that was'nt true what would happen?

[quote="abacadaea"][quote="Fraenkel"]well remember $ CA \plus{} BC \equal{} 100$ and the rest follows.
[/quote]
where did you get that?[/quote]
When we add ABC and DCA, we must have BC+CA=100, so then we can have A+D=9, because we want a total of 1000;a 1 gets carried over.

Actually, this problem you have right here is quite easy (if you see it right away)...now some of these 'letter' problems can get a bit tricky...as with math+counts=money (hehehe) or something of that nature...

the kids i'm helping out right now only get these wrong with stupid mistakes...so just be careful

-mamathbug

[quote="Nerd_of_the_Ages"]Guys, don't say problems are easy, you might offend someone.  :maybe:

Btw, all these solutions are false and the problem is invalid, because they never said all this was in base 10.  :P[/quote]

Numbers are assumed to be in base 10 unless otherwise specified.

Well from what I saw in the book it was 18.
Sence $ A\plus{}C\equal{}10$ ,$ C\plus{}B\equal{}9$ and $ A\plus{}D\equal{}9$

The solution is:

$ C\plus{}B\plus{}(A\plus{}D)\equal{}9\plus{}9$
$ C\plus{}B\plus{}A\plus{}D\equal{}\boxed{18}$


limit
integration
real analysis
real analysis unsolved

Find \[ \lim_{n\to\infty} \int_{\mathbb{R}} \frac{e^{-|x|}}{x^{2}+n} d\lambda (x),  \]
 where $\lambda$ is Lebesgue measure. :)  ;)

i think we can use bounded convergent of Lebesgue theorem, because $f_n \to f$ by measure, or  we can use theorem about monotonic convergence.
Denote $f(x)=0 \forall x$.
$f_n,f>0$ and they are measurable,and  $\forall x\in R$ we have $f_n(x)\to f(x).$ then \[ \lim_{n\to\infty} \int_{\mathbb{R}} \frac{e^{-|x|}}{x^{2}+n} d\lambda (x)=\int_{\mathbb{R}} \lim_{n\to\infty} \frac{e^{-|x|}}{x^{2}+n} d\lambda (x) = \int_{\mathbb{R}} 0 d\lambda (x) =0. \]


function
geometry
calculus
calculus computations

Clepsydra is an ancient water clock that has a hole in the bottom of the bowl and keeps track of time. If the water level were to fall at a constant rate what must be the shape of the bowl?

Define the following functions and constants:
$ A(h)$: the horizontal cross-sectional area of the vessel at height $ h$
$ h(t)$: the height of the water level at time $ t$
$ f(t)$: the volume of water flowing out of the hole per unit time at time $ t$
$ c$: The rate at which the water level drops.
$ H$: The height of the vessel.
$ a$: the area of the hole.
$ g$: acceleration due to gravity

An application of Bernoulli's Law shows that the rate of water flow out of the hole at the bottom satisfies the following:
\[ f(t)^2\left(1 \minus{} \frac {a^2}{A(h(t))^2}\right) \equal{} 2a^2gh(t)\]
Now since the water level drops at a constant rate, we have $ h(t) \equal{} H \minus{} ct$ and, by conservation of water volume, $ f(t) \equal{} cA(h(t))$.  Solving for $ A$ we find,
\[ A(h) \equal{} a\sqrt {1 \plus{} \frac {2gh}{c^2}}\]
So we have a description of the horizontal cross-sectional area at each height.  This alone doesn't specify a unique shape.  For example, we can have a clepsydra with circular horizontal cross-sections, in which case the radius at height $ h$ would be given by $ r(h)^2 \equal{} \frac {a\sqrt {1 \plus{} \frac {2gh}{c^2}}}{\pi}$.  Or we could have square cross-sections, rectangular cross-sections, triangular cross-sections, elliptic cross-sections, etc.  Any one of them would solve the problem, as only the area at each height is constrained.


Let $ x_1$, $ x_2$, ... $ x_n$ be positiv reals
and

$ x_1 \equal{} 1 \plus{} \frac{1}{x_2}$, $ x_2 \equal{} 1 \plus{} \frac{1}{x_3}$, $ \cdots$ , $ x_{n\minus{}1} \equal{} 1 \plus{} \frac{1}{x_n}$
and
$ x_n \equal{} 1 \plus{} \frac{1}{x_1}$

Show that $ x_1\equal{}x_2\equal{}\cdots \equal{}x_n$ and hence find the solution.

W.L.O.G suppose that $ x_1\equal{}max(x_1,x_2,...,x_n) \Rightarrow x_2\equal{}x_n\equal{}min(x_1,x_2,...,x_n)$

$ \Rightarrow x_3\equal{}x_{n\minus{}1}\equal{}max(x_1,x_2,...,x_n)$ and finally $ \Rightarrow x_1\equal{}x_3\equal{}...\equal{}max(x_1,x_2,...,x_n)\equal{}x$

and $ x_2\equal{}x_4\equal{}...\equal{}min(x_1,x_2,...,x_n)\equal{}y$

$ \Rightarrow x\equal{}1\plus{}\frac {1} {y}$ and $ y\equal{}1\plus{}\frac {1} {x} \Rightarrow x\equal{}\frac {2x\plus{}1} {x\plus{}1} \Rightarrow x^2\equal{}x\plus{}1 \Rightarrow x\equal{}\frac {1\plus{}\sqrt{5}} {2} \Rightarrow y\equal{}\frac {1\plus{}\sqrt{5}} {2}$

Take $ y_i\equal{}x_i\minus{}x_{i\plus{}1}$ amd we get easily that $ y_i/y_{i\plus{}1}\equal{}\minus{}1/x_ix_{i\plus{}1}$ (if the terms are not equal to 0 of corse). So we get that $ (\minus{}1)^ny_i/(x_1..x_n)^2\equal{}y_i$ so we have the product of the $ x_i$ is one and n is even (otherwise the result is obvious), but in this case we have a contradicton because $ x_i>1$ from the given sequence. Hence the result follows   :rotfl:  :rotfl:  :rotfl:


HCSSiM
factorial
email
geometry
geometric transformation
reflection
summer program

Does anybody know when the info will be up for this coming summer session (06-07)?

What kind of info are you interested in? The general running of the program shouldn't change from year to year. The application should be available some time in the spring. The website is still up: hcssim.org

Sorry for being so vague :) , I meant the application.  Thanks.

The 2007 online application form is now up.  (And the website has been generally updated for the year, including the posting of a smattering of photos from HCSSiM 2006.)

Yay, updated website!  This is exciting, especially the pictures.  (I'm also excited about s-m's pictures of us on her website, which I'd been eagerly awaiting.)  The brochures/posters have not been updated, but that's okay because if there were an update my picture might get deleted!

It's also sad because it means HCSSiM '06 is really and truly over, but I guess by now that's old news.

I wonder if it's too early to start trying to force people to apply...

The brochure and posters will be updated soon. The software used to create/edit those documents was lost in a hard drive crash (and we couldn't find whatever CD we used to install it), and we're ordering a new copy of it.

There may also be a website redesign coming, but that's a ways off.

I personally think that a toga picture should be on a poster or brochure, it shows a certain side to HCSSiM that is somewhat missed by the posters and brochures, just not Vlads crazy (creepy) toga.

hcssim 06 is not over and i agree with nate about the togas.
those pictures made me cry. i miss it.

Group pictures! I don't think we had any on the zoto site - the only ones I could find were YMPD 2000-2006 photos.

(I think the fact that I noticed this independently with in 24 hours of the update shows how much I miss it...)

by the way, everyone who hasn't gone yet (and a lot who have) should apply to HCSSiM, in case anyone who looks at this thread had any question.

JB

Hey, is anyone here applying this year? I just filled out the online questionnaire part, so I haven't started the other parts yet.

Hopefully I'll be accepted  :lol:

I went last year and I'm currently working on the IT for this year!

I just heard about HCSSiM and am starting the IT now. I was just wondering what they look for, like how much work I'm supposed to show? I'm generally not so good about showing my work, but if it's a deciding factor, I guess I'll obviously try to improve that.

I was also wondering how good they are about financial aid?

Definitely show your work. The answer itself has little value on its own so I would say the work is extremely important. This also stems from the fact that good communication is a key aspect of math.

Of course, if you don't too well on the IT, you probably won't get in, but this shouldn't be the case for most people. However, I would say that the amount of interest that is shown from your comments and your work plays one of the most significant roles. That's what HCSSiM is all about... interest in exploring the many rich fields of mathematics that is out there. If you don't believe me, I would just have to say if it weren't for my piques of interests that showed through the IT, I wouldn't have got into HCSSiM '06.

Financial aid is pretty flexible. I have a moderately funny story about it, but I'd just say, don't worry about it. =)

[quote="Eternica"]Of course, if you don't too well on the IT, you probably won't get in, but this shouldn't be the case for most people.[/quote]

Um, that's not true.  About half of the people who return ITs don't do well. (And you're selling yourself short by saying that you were invited only because of the interest shown on the IT---you, like everyone else who was invited to HCSSiM '06, did quite well on the IT.)

As far as I know, no invited student has been unable to attend HCSSiM for financial reasons.  That's certainly been true for the last decade.

Anyway, grendel, about showing work---the explanations are the important part. We care more about "why" than "how."  And sorry for the slowness in responding...I'm out of the country at the moment and haven't been able to check AoPS regularly.

You got it! (if I find it)

Edit: I found it! Yay!

Hmm... perhaps banning non-measurable sets is a bad idea, suppose I have two unit balls of luggage I plan on bringing, but I can only carry on one.  Now by Banach-Tarski I can simply break them up into a finite amount of non-measurable pieces, and reassemble them as a single unit ball of luggage by translations alone.  Now thats packing made easy!

Yet Another 17 Additional Items To Bring that have since been suggested/ thought up by myself and others:

1. CDs/ CD Player
2. Comforter
3. Deck of Cards
4. Gum
5. Hangers
6. Inhaler
7. Jewelry/ Accessories
8. Journal/ Diary
9. "Loofah!" -John Robinson Mead
10. Lotion/ Moisturizer
11. Luggage
12. Perfume/ Cologne
13. Plastic Bags
14. Sealed Snacks
15. Slippers
16. Smiles
17. Talents

Hopefully Final List of 17 Additional Items To Bring that have since been suggested/ thought up by myself and others:

1.	Batteries
2.	Beach Ball
3.	Bean Bag Chair
4.	Bicycle Lock
5.	Clips
6.	Colored Pencils
7.	Erasers
8.	Geometer
9.	Hair Ties
10.	Highlighters
11.	Laundry Detergent
12.	L Wasserman
13.	Mathematical Inclination
14.	Ruler
15.	Small Mirror
16.	Wallet
17.	Whiteout

Happy first day of HCSSiM everyone (well, it's technically July 1st right now)!  I'll see most of you today!

Hi Miles, I'm Tim.

By the way, anyone with pictures, could you post links here or something?  Thanks.

sup Miles and Tim! This is Danny Y. I don't know what to do at home, we should be doing problem sets together.

Hi Miles, Tim, and Danny Y., I'm Chetak.

Posting links to whatever pictures get put up would be exceptionally awsome!

Thanks Mucho,
MarVel

So, photos... there are currently two galleries of photos (mine and The Emily's) at

http://picasaweb.google.com/portablepig/

If anyone would like to add photos to the site, that'd be great, and ask me or arkleseizure if you need the password.  There are also some galleries at facebook, but they're not public, of course.

Hi everyone!
I don't come on aops very much, but I happened to be browsing today and I found this forum and ya. Those photos were great!
I miss everyone so much! (I'm using lots of exclamation marks!)
Emil-your signature is hilarious.
Oh ya, this is Avi...
I miss you guys

Hi Katie and everyone who posted on previous pages!  (this is Megan)


Hello,

For those who recall my last thread asking for what specific area of problem solving guide, I told users here that I'll make one for AIME Level Number Theory WITH problems. So it is now made with 16 problems over 16 pages.

It is as usual available at my website. Please check it out and express gratitude. Thanks.  :)

Thanks for this guide, also I really enjoyed your test!

Hm, those are very nice problems!  Thanks for another great guide!

Wondering: is there going to be an AIME level algebra guide?

Wow, nice guide. I really like how you showed different ways to solve the same problems.

I'm glad that my guide has been a helpful to you guys. Keep the testimonials coming and keep the thread alive because believe or not, the AMC/AIME/USAMO was sent to many many people (I lost the count but it was probably over 50 at LEAST) because the thread was filled with many replies.

As with any further guide of any kind, I'll be honest and say that they probably will not be available for very long time. I now want to shift my gear from math to my reading competition (which I have written two sample ones), and they take a lot of time. Plus, if there is going to be another HDMO next year, I'll have to write that too so I'm going to have my hands off from guides for a while. But feel free to shoot me an email asking advices on specific topics.  :)

Youre awesome man....I wanted this guide to prepare for PuMaC Number Theory and I got it  :lol:

[quote="jhangil"]For NT guide #7, 3649 can also be a number that has more than 2 digits so there for the answer should be 648. Verified with Silver Falcon.[/quote]

Thank you. The updated version is now up in the website.

Smartguy also pointed out a problem with another problem. I think I'm going to keep the document as it is and just put a link for any modification. By this, we can have additional modifications even I'm not around to change it.

[quote="Smartguy"]I got 140.

140= 10080/72

10080= $ (2^5)(3^2)(5)(7)$
$ (5\plus{}1)(2\plus{}1)(1\plus{}1)(1\plus{}1)\equal{}6*3*2*2\equal{}72$...

and 175= 12600/72

12600= $ (2^3)(3^2)(5^2)(7)$

$ (3\plus{}1)(2\plus{}1)(2\plus{}1)(1\plus{}1)\equal{} 4*3*3*2\equal{}72$...[/quote]

Thanks again for carefully reading the guides!


[quote="alex2008"]Let be $ a,b,c > 0$ such that $ a \plus{} b \plus{} c\ge abc$ . Show that :

$ \frac {bc}{a^3(b \plus{} c)} \plus{} \frac {ca}{b^3(c \plus{} a)} \plus{} \frac {ab}{c^3(a \plus{} b)}\ge \frac {1}{2}$[/quote]
Nice Alex :)
[hide="big hint"]the key point is the substitution $ a\equal{}\frac{1}{x},b\equal{}\frac{1}{y},c\equal{}\frac{1}{z}$
[/hide]
[hide="solution"]
The inequality can be rewritten as:
$ \sum\frac{x^3}{y\plus{}z} \ge \frac{1}{2}$
By Andreescu inequality we have that:
$ LHS \ge \frac{(x\plus{}y\plus{}z)^3}{3 \cdot 2(x\plus{}y\plus{}z)}\equal{}\frac{(x\plus{}y\plus{}z)^2}{6}$
So we only have o prove that:
$ (x\plus{}y\plus{}z)^2 \ge 3$
But $ (x\plus{}y\plus{}z)^2 \ge 3(xy\plus{}yz\plus{}zx)$ $ \boxed{1}$
But by the condition $ \frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\equal{}\frac{1}{xyz} \Leftrightarrow xy\plus{}yz\plus{}zx \equal{}1$
So $ \boxed{1}$ solves the problem :wink: [/hide]


Putnam function integration calculus quadratics inequalities algebra

For each continuous function $f: [0,1]\to\mathbb{R},$ let $I(f)=\int_{0}^{1}x^{2}f(x)\,dx$ and $J(f)=\int_{0}^{1}x\left(f(x)\right)^{2}\,dx.$ Find the maximum value of $I(f)-J(f)$ over all such functions $f.$

The "I can't believe anything this simple ever got to be B5" problem. I think a lot of people missed it or overlooked it just because you're not looking for anything this simple in this spot.

I might as well answer it, just to get rid of it:

$I(f)-J(f)=\int_{0}^{1}x\left(xf(x)-(f(x))^{2}\right)\,dx$

$=\int_{0}^{1}x\left(-\left((f(x))^{2}-xf(x)+\frac{x^{2}}4\right)+\frac{x^{2}}4\right)\,dx$

$=\int_{0}^{1}x\left(\frac{x^{2}}4-\left(f(x)-\frac{x}{2}\right)^{2}\right)\,dx$

$\le\int_{0}^{1}\frac{x^{3}}4\,dx=\frac1{16}$

Following along in the proof, we clearly have equality in this integral if we let $f(x)=\frac{x}2.$ We do have a maximum.

That is exactly my solution.  The other popular solution seemed to be via maximizing the integrand
\[x^{2}f(x)-xf(x)^{2}\]
pointwise by noting that for each fixed $x$, the integrand is a quadratic in the value of $f(x)$.

Okay, I got $\frac2{27}$ as my answer, and I would greatly appreciate if someone could let me know what is wrong with the following solution (which is what I submitted.)

Let $A = I(f)-J(f)$.

First, we note $J(f)$ is always positive. So, in order to maximize $A$, we need to maximize $I(f).$ Now, if $f(x) \le 0$ on any subinterval of $(0,1)$, this will only result in minimizing $I(f).$ So, we need only consider those continuous $f$ such that $f(x) \ge 0$ for $0 \le x \le 1.$

Again, we note $A = \int_{0}^{1}x f(x) (x-f(x)) \,dx$. Since we are only considering $f(x) \ge 0,$ we maximize $A$ only when $x-f(x) \ge 0$, for otherwise, $A \le 0.$

Thus, we maximize $A$ only for those continuous functions $f$ such that $f(x)\ge 0$ and $x-f(x) \ge 0.$

Now, we use A.M. - G.M. inequality to get $(x\cdot f(x)\cdot (x-f(x)))^{\frac13}\le \frac{x+f(x)+x-f(x)}{3}= \frac{2x}{3}$, using which we get $A \le \frac{2}{27}$.

Note $\frac2{27}> \frac1{16}$. So, I really wonder what is wrong with the above reasoning I just showed. Even if the solution above looks shaky, I still think we should be able to find a continuous function  $f$ such that $f(x) \ge 0$ and $x-f(x) \ge 0$ for $x \in [0,1]$, so that we could apply the A.M. - G.M. inequality and get $A \le \frac{2}{27}.$

Any clarification will be greatly appreciated.

Well FieryHydra, its good that you justified that 0 \leq f(x) \x for order to AMGM to be applied on non-negative values. However, you forgot to check that the value can actually be attained, and actually it cannot be. You would require x = f(x) = x-f(x), which clearly has no solution.

You should have done AM GM on f(x) and x-f(x) instead, to get A \leq \int_0^1 x \frac {x^2} {4} dx = \frac {1}{16}, which is the answer. This can be achieved with f(x) = x-f(x), or f(x) = \frac {x} {2}

Your 'maximum' is unattainable, and hence is not actually the maximum.  Similarly, 98145646516 is an upper bound for $I(f)-J(f)$.

Thanks for your replies. I wasn't very comfortable with my solution anyways. I see the mistake.

I'm upset with myself about this problem because I assumed it would be tricky (as it was B5) so I was trying to be clever thinking about a polynomial orthogonal basis for $[0, 1]$ or the Cauchy-Schwarz inequality or something, and you could just do it with very elementary ideas. So I wasted a fair amount of time and didn't actually solve the problem (I was half-way through writing a totally bogus solution [didn't know it was bogus at the time of course] when time ran out).

I too skipped this problem and am kicking myself for doing so.
I instead chose to do B2. Even though I was able to provide a proof (i think i made a minor mistake though) it took more effort that this problem would have if i was not intimidated by it being B5.

Ha. I'm one of those who saw this problem with 5-10 minutes remaining. It took me a few minutes to understand what was going on, and maybe 2 minutes of scribbling, which I didn't submit (who can write a proof in two minutes while panicking?). Darn. B5. Who could have guessed. Well, guess there's always next year.

Not to pile on, but you also have an earlier error:

[quote="FieryHydra"]Let $A = I(f)-J(f)$.

First, we note $J(f)$ is always positive. So, in order to maximize $A$, we need to maximize $I(f).$[/quote]

This is simply not true, and you should be able to come up with lots of counter-examples.

We are only interested in the term $f(x) \cdot \left(x-f(x)\right)$ where $f(x)$ is the argument of this quadratic for a fixed $x$. Then the maximum will be attained at $f(x)=\frac{x}{2}$. Hence we have
\[\int_{0}^{1}x^{2}f(x)-x f^{2}(x)=\int_{0}^{1}xf(x) \left(x-f(x) \right) \leq \int_{0}^{1}x \cdot \frac{x}{2}\cdot \frac{x}{2}=\frac{1}{16}\]

This is essentially completing the square but with less work.

I see the solution involves knowing to write (x^2)f(x) - x(f(x))^2 as (x^3)/4 - x(x/2 - f(x))^2.  It is easy to check that (x^3)/4 - x(x/2 - f(x))^2 equals (x^2)f(x) - x(f(x))^2.  But how does one come up with such an idea?  The idea of rewriting (x^2)f(x) - x(f(x))^2 as (x^3)/4 - x(x/2 - f(x))^2 seems to come out of nowhere.  I suppose it is not so farfetched to think of rewriting (x^2)f(x) - x(f(x))^2 as A(x) - B(x) where A(x) and B(x) are nonnegative, A(x) does not depend on f(x) and B(x) does depend on f(x).  We would then desire to find a function f(x) that sets B(x) = 0.  But how does one think of choosing A(x) = (x^3)/4 and B(x) = x(x/2 - f(x))^2?  Is this just a shot in the dark?  Or is there an actual way to think of A(x) = (x^3)/4 and B(x) = x(x/2 - f(x))^2?

It's not mysterious and it's not a shot in the dark. First we write as one integral the thing to be maximized - not such a strange concept. Then we pull out a common factor of $ x$ in the integrand. From there, it's the ancient and ordinary algebraic technique of completing the square.

Let $g(t)=t^2f(t)-tf(t)^2$ for all $0 \le t \le 1$. Notice that $g(t)=t\cdot \left(\frac{t^2}{4}-\left(\frac{t}{2}-f(t)\right)^2\right)$ and hence $g(t) \le \frac{t^3}{4}$ for all $0 \le t \le 1$. Therefore, $$\int^{1}_0 g(x) \, dx \le \int^{1}_0 \frac{x^3}{4}\, dx =\frac{1}{16}$$ since the difference is a positive function and the integral inequalities hold. The maximum is attained at $f(x)=\frac{x}{2}$.

We can write:
$$I(f)-J(f)=\int^1_0\left(\frac{x^3}4-x\left(f(x)-\frac x2\right)^2\right)dx,$$
so $I(f)-J(f)\le\int^1_0\frac{x^3}4dx=\boxed{\frac1{16}}$. Equality when $f\equiv\frac x2$.

Let the functional
\[
K[f] = I(f) - J(f) = \int _0 ^1 x^2 f(x) - x(f(x))^2 dx = \int _0 ^1 F(x,f) dx.
\]
From the Euler-Lagrange equation, the functional attains an extrema if
\[
\frac{\partial F}{\partial y} = 0 \implies x^2 -2x f(x) = 0 \implies f(x) = \frac{x}{2}.
\]
Solving with this $f(x)$ gives that the $K[f]$ attains a maximum at $\frac{1}{16}.$

Easy B5 !
note that we can say by cauchy-schewarz inequality ,
$$\left(\int\limits_0^1xf(x)^2 dx\right)\cdot  \left(\int\limits_0^1 x^3 dx \right) \geq \left(\int\limits_0^1 x^2 f(x) dx\right)^2 \\ \implies \frac{J(f)}{4} \geq I(f)^2 \\ \implies J(f) \geq 4(I(f))^2 \\ \implies I(f)-J(f) \leq I(f)-4I(f)^2$$
Now note that $\forall x \in \mathbb{R}$ $x-4x^2 = \frac{1}{16}- (2x-\frac{1}{4})^2\leq \frac{1}{16}$.hence $\boxed{I(f)-J(f) \leq \frac{1}{16}}$ with equality holds if and only if $f(x) =\frac{x}{2}   \blacksquare$


logarithms
inequalities
probability and stats

A r.v X is distributed on the set of 1,2,3,4,...with corresponding probabilities $ p_1,p_2,\cdots$if the mean $ E[X]\equal{}\sum\limits_{i\equal{}1}^\infty ip_i\equal{}m$ , show that the following distribution of X maximize the entropy where :

$ p_i\equal{}{\frac 1m }(1\minus{}\frac 1m)^{i\minus{}1}, i\equal{}1,2,3,\cdots$

where the Entropy distribution given by :

$ H(Y)\equal{}\minus{}\sum\limits_{i\equal{}1}^\infty p_i ln(pi)$

Note :

The question is for my friend I proved $ H(x)\equal{}m\ln(\frac{m}{m\minus{}1})$

How to prove this is must be the maximum .

Help Please  :)

Use the log-sum inequality
\[ \minus{} \sum p_i \ln p_i \le \minus{} \sum p_i \ln q_i.
\]
Namely, set $ q_i \equal{} \frac{1}{m} {(1 \minus{} \frac{1}{m})}^{i\minus{}1}$ and see what comes out.

or you can use the Lagrange multipliers approach. All the conditons are here to make it work. Take a look at:
http://en.wikipedia.org/wiki/Lagrange_multipliers
and
http://en.wikipedia.org/wiki/Lagrange_multipliers_on_Banach_spaces
to apply it on $ l^1$ for example.


jacob spent 1/4 of his savings on a new bike.to find how much he spent,you could mutiply his total savings by- :?:

A.4
B.0.4
C.0.25
D.0.1

Click Mr. Green for the answer!
[hide=" :D "]C) 0.25[/hide]

[hide=" :)  :(    "]1/4 is equal to .25 so the answer is C.[/hide]

[hide]
The answer is C.
100%*1/4=.25[/hide]


find all positive integer $m,n$ such that $3^m=2n^2+1$.

Thank you!

/-------------------------------------------------------------------------/
Hi, Sam-n!
Please, read http://www.mathlinks.ro/viewtopic.php?p=140098
And use proper problem title next time.

With regards,
Myth


For the first one:

If x=1 then y=1. If x=-1 then y=-1. x=0=>y=0=>x^y=0^0 contradiction (same if y=0), so x<>0. Let's take the case |x|>1. It's obvious that y>=-1 because if y<=-2 then the left term isn't an integer while the right term is. If y=-1 then x=-1 (we already have that soln). This leaves us with y>0.

y=p1^a1*..*pk^ak where pi are primes. x must have the same prime divisors, so x=p1^b1*..*pk^bk (ai,bi>=1). The equation is translated to y*bi=2001*ai for all i, or p1^a1*..*pk^ak*bi=2001*ai for all i from 1 to k (*). Let's assume one of pi doesn't divide 2001 (let's assume it's p1). This means that p1^a1|a1, which is impossible because for any p1>=2 p1^a1>a1. This means that p1|2001. It's the same for all the other primes, so k=1, 2 or 3, so y=3^a1*23^a2*29^a3 or 3^a1*23^a2 or any other case that might come out of this. Let's consider the case when equations are 3^a1*23^a2*29^a3*bi=3*23*29*ai for all i. This is equivalent to    3^(a1-1)*23^(a2-1)*29^(a3-1)*bi=ai for all i. It's pretty obvious that the left terms are greater than the right part if ai>1 for some i, so all ai must be <=1. This means that all possible values of y are the divisors of 2001: 3, 23, 29, 3*23 etc.

Conversely, it's easy to see that all divisors of 2001 are suitable values of y. For example, if y=3*23, then x=(3*23)^29 and we're done; if y=29 then x=29^(3*23) and so on. Is it right, Arne? I might have missed sth.


[hide="Hint for 1"] Complementary counting.  How many positive divisors are there?  How many of them are odd?
[hide="How do you count odd divisors?"] Ignore the powers of $ 2$. [/hide] [/hide]
[hide="Hint for 2"] Let me demonstrate the idea with a small number.  Say I wanted to find the sum of the positive divisors of $ 60$.  Now, I know that $ 60 \equal{} 2^2 \cdot 3 \cdot 5$, so what I'm going to do is write out the positive divisors very systematically:  first, using only powers of $ 2$, I have

$ 1, 2, 2^2$

Second, allowing powers of $ 3$, I also have

$ 3, 2 \cdot 3, 2^2 \cdot 3$

Finally, allowing powers of $ 5$, I have

$ 5, 2 \cdot 5, 2^2 \cdot 5$
$ 3 \cdot 5, 2 \cdot 3 \cdot 5, 2^2 \cdot 3 \cdot 5$

Adding them all together,

$ (1 \plus{} 2 \plus{} 2^2) \plus{} 3(1 \plus{} 2 \plus{} 2^2) \plus{} 5(1 \plus{} 2 \plus{} 2^2) \plus{} 15(1 \plus{} 2 \plus{} 2^2) \equal{} (1 \plus{} 2 \plus{} 2^2)(1 \plus{} 3 \plus{} 5 \plus{} 15) \equal{} (1 \plus{} 2 \plus{} 2^2)(1 \plus{} 3)(1 \plus{} 5)$

Isn't that magical?  Once you have this factorization in mind, you don't need to add all the divisors separately!  If you still don't get the trend, the sum of the positive divisors of $ 360 \equal{} 2^3 \cdot 3^2 \cdot 5$ is

$ (1 \plus{} 2 \plus{} 2^2 \plus{} 2^3)(1 \plus{} 3 \plus{} 3^2)(1 \plus{} 5)$

[hide="Generalization"] The sum of the positive divisors of $ p_1^{e_1} p_2^{e_2} p_3^{e_3} ... p_k^{e_k}$ is

$ (1 \plus{} p_1 \plus{} p_1^2 \plus{} ... \plus{} p_1^{e_1})(1 \plus{} p_2 \plus{} p_2^2 \plus{} ... \plus{} p_2^{e_2}) ...$ [/hide] [/hide]

thanks.

For number 1, I figured out how to do it, with 48-40 = 8 even divisors.

I got a prime factorization of $ 2^5 x 3^3 x 5$, and the total of factors was 48, and the only ways they were odd were when there were no twos involved.

For number 2, I got (1+2)(1+17)(1+19)(1+29)= 32400

is this right?

I prime factorized 9367, but I used guess and check to find 19 x 29 x 17 x 2 = 9367. How else could I have prime factorized it, without having to guess and check?

Good question.  It turns out that integer factorization is in general a really hard problem, and for small $ n$ there actually isn't a significantly better method than guessing and checking primes up to $ \sqrt{n}$.  (For large numbers, there are major improvements in "checking" and minor improvements in guessing that don't really do much for just 4-digit numbers.)

[quote="Bicameral"]1) How many of the positive divisors of 4320 are even?

2) Find the sum of all positive divisors of 18734.

For number 1, I was thinking of just using a factor tree, but that gave me a questionable answer of 30, and I think there must be a better or correct way to do this.

On 2, I have no idea how to do that without counting them slowly one by one.[/quote]

[hide="Another Solution to 1"]$ 2k|4320\iff k|2160.$ So it reduces to finding the number of divisors of 2160.[/hide]


geometry
incenter
inequalities
Euler
number theory
geometry proposed

[b]In the non-isosceles triangle [/b]$ABC$ [b]I note the middlepoints [/b]$M,N$ [b]of the [/b][b]sides[/b] $[BC],\ [CA]$ [b]respectively,[/b]
[b]the projections [/b]$D,\ E,\ F$ [b]of the points [/b]$A,\ B,\ C$ [b]on the sides [/b]$[BC],\ [CA],\ [AB] $ [b]respectively,[/b]
[b]the centroid [/b]$G$ [b]and the orthocenter [/b]$H$. [b]I define the points [/b]$X\in BC\cap EF,\ Y\in CA\cap DF.$
[b]Prove that [/b]$GH\perp XY.$. [b](in connection with  [/b]  http://www.mathlinks.ro/Forum/topic-5485.html ).

This is well known. The line XY is known as the orthic axis of $\triangle ABC$. This can be proved by various methods, of which the one I love the most is orthologic triangle.

[quote="levi"]I know that the line XY is the orthic axis. But is wellknown that the Euler's line is perpendicular on the orthic axis ? Means that and the Vornicu's message-problem is wellknown (see http://www.mathlinks.ro/Forum/topic-5485.html ).[/quote]
Just some murmurs...It [i]seems[/i] that the problem posted by Valentin is less well known than yours, however, Valentin's one is indeed much easier. A really weird thing :huh:  :?

OK, enough for nonsense.
I just remember Darij has posted a proof before. See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=51168]Lemma 1[/url].

[edit]: I strongly believe that there is a proof by poles and polars using nine-point circle, but so far all my attempts are invalid. Maybe I "overlook" something trivial...

[b]Solution.[/b] From the problem P1. of message "A use of a h.d."( http://www.mathlinks.ro/Forum/topic-46146.html )
with the third and fourth replies ("Some equivalencies for a harmonious division")
results that $XH\perp AM,\ YH\perp BN$. I note $U\in XH\cap AM,\ V\in YH\cap BN$.
Therefore, the quadrilaterals $AXDU,\ BYEV$, are inscribed in the circles with diameters $[AX],\ [BY]$ respectively. Thus, $HX\cdot HU=HA\cdot HD,\ HY\cdot HV=HB\cdot HE$. But $HA\cdot HD=HB\cdot HE$.
From the relation $HX\cdot HU=HY\cdot HV$ results that the quadrilateral $XYUV$ is inscribed in a circle. Because the quadrilateral $HUGV$ is inscribed (the circle with the diameter $HG$), consequently, $GH\perp XY$.

[quote="levi"][b]The reason: [/b][u]Don't kill a butterfly with an atomic bomb ![/u]

Thus, the game  was "destroied". I want a solution without poles and polars. I will offer soon on this post.

For this [b]reason [/b], the geometry, the inequalities, the combinatoric problems and the number theory are beautifully !

[b]Remark. [/b]I know that the Euler's line is perpendiculary on the orthic axis ! Do you give credence me ?[/quote]
Please, read my previous post againbe before "blaming" me. I said [i]I think that there is a proof by poles and polars[/i]. What's wrong with it? Or in your "wording", I just have the atomic bomb but with no air fighter to drop it.
Anyway, I have no doubts that you know orthic axis$\perp$ Euler line very well. I apologise here if my poor English caused any misunderstanding.

mecrazywong

It is O.K. I understood you falsely. And I have a poor knowledge of English like you. I am sorry !

I AND YOU : "excuses always proceed from a guilty conscience; an excuse is an accusation". I hope don't happen !

You are a young man, I am a retired teacher, sometimes a irritating man.


Suppose that a fair die is rolled 100  times. Let $ X_{i}$ be the value obtained on the ith roll. Compute an approximation for $ 1 < a < 6$, given :

$ P\{\prod_{1}^{100} X_{i} \le a^{100}\}$

[hide="Strategy"]Define $ Y_i \equal{} \log X_i$ and apply the [url=http://en.wikipedia.org/wiki/Central_limit_theorem]Central Limit Theorem[/url] to estimate $ \sum Y_i$.[/hide]


linear algebra
matrix
LaTeX
vector
induction

It's my first post in this forum, so maybe this problem was posted before.

Let $ F$ be a field and $ A, B \in \mathcal{M}_{2,2}(F)$. If $ A^2 \plus{} 2AB \plus{} B^2 \equal{} O$, prove or disprove that $ AB \equal{} BA$.

In characteristic 2, we can take $ A\equal{}\begin{bmatrix}1&1\\0&1\end{bmatrix}$ and $ B\equal{}\begin{bmatrix}1&0\\1&1\end{bmatrix}$.

This structure doesn't work if the characteristic of $ F$ isn't $ 2$, so I'm inclined to believe that they do commute in that case.
If we allow $ 3\times 3$ or larger matrices, it's easy to construct examples with everything nilpotent.

Just a incomplete idea:

By Cayley-Hamilton, $ A^2 \equal{} \left( \text{tr} \, A \right) A \minus{} \left( \det A \right) I_2$ and $ B^2 \equal{} \left( \text{tr} \, B \right) B \minus{} \left( \det B \right) I_2$.
The coefficients above will be named $ \alpha, \beta, \gamma, \delta$ from now on.

If $ \text{char} F \neq 2$, then we get that $ \left( A \plus{} 2^{\minus{}1} \gamma I_2 \right) \left( 2B \plus{} \alpha I_2 \right) \equal{} \left( \beta \plus{} \delta \plus{} 2^{\minus{}1} \alpha \gamma \right) I_2$.

If the $ RHS$ in $ \neq 0$, then what's in $ LHS$ commute, so we're done.
Otherwise, $ \left( A \plus{} 2^{\minus{}1} \gamma I_2 \right) \left( B \plus{} 2^{\minus{}1} \alpha I_2 \right) \equal{} 0$ and $ \beta \plus{} \delta \plus{} 2^{\minus{}1} \alpha \gamma \equal{} 0$. The first of the two equalities gives one eigenvalue for $ A$ and one for $ B$.

I can't go further, though. Probably another idea is to be used here.

Is that any use to prove $ (AB\minus{}BA)^2\equal{}0$?

[quote="zhaobin"]Is that any use to prove $ (AB \minus{} BA)^2 \equal{} 0$?[/quote]
Showing that $ (AB\minus{}BA)$ is nilpotent is not the same thing as showing that it is zero. There are nonzero nilpotent matrices.

Thank you,But what I mean is  can we combine it with $ A^2 \plus{} 2AB \plus{} B^2 \equal{} 0$ then conclude that $ AB \equal{} BA$?

So here is the whole proof,I hope it is right,First we may consider the field which is algebraic closure of $ \mathbb{F}$
Lemma:for the characteristic of $ \mathbb{F}$ isn't $ 2$,we have $ AB \minus{} BA$ is nilpotent.
Proof for the lemma:
First,we have $ tr(AB \minus{} BA) \equal{} 0$
Second $ tr((AB \minus{} BA)^2) \equal{} 2tr(ABAB \minus{} A^2B^2)$
but,$ 2ABAB \equal{} \minus{} AB(A^2 \plus{} B^2),2A^2B^2 \equal{} \minus{} A(A^2 \plus{} B^2)B$
Then $ tr((AB \minus{} BA)^2) \equal{} 0$
Then $ AB \minus{} BA$ is nilpotent.(This is mostly dependent on the characteristic of $ \mathbb{F}$ isn't $ 2$ :wink: )

Back to the problem,From the lemma we have:
$ rank(AB \minus{} BA) \le 1$,thus we can prove $ A,B$ can upper-triangulize at the same time.(It is right,but not trivial,has been apprent in froum)
Then we may assume:$ A,B$ be upper-triangular.
$ A\equal{}\begin{bmatrix}a&b\\
0&c\end{bmatrix}
,B\equal{}\begin{bmatrix}d&e\\
0&f\end{bmatrix}$
Then from
 $ A^2 \plus{} 2AB \plus{} B^2 \equal{} 0$
We get $ a \plus{} d \equal{} 0,c \plus{} f \equal{} 0,ab \plus{} bc \plus{} 2(ae \plus{} bf) \plus{} de \plus{} ef \equal{} 0$
Which is:
$ a \plus{} d \equal{} 0,c \plus{} f \equal{} 0,ab \plus{} ae \plus{} bf \plus{} ef \equal{} 0$
Then:
$ AB\minus{}BA\equal{}
\begin{bmatrix}0&ae\plus{}bf\minus{}bd\minus{}ce\\
0&0\end{bmatrix}
\equal{}\begin{bmatrix}0& ae\plus{}bf\plus{}ab\plus{}ef\\
0&0\end{bmatrix}\equal{}0$

Here is my solution.

[hide="Solution"]If $ \text{char} \, F \equal{} 2$, then counterexample exists. So assume $ \text{char} \, F \neq 2$.

Let $ C \equal{} A\plus{}B$. Then $ C^2 \equal{} CA\minus{}AC$. By Cayley-Hamilton theorem, $ C^2 \equal{} (\text{tr} \, C)C \minus{} (\det C)I_2$.

Combining two equation and taking trace, we have $ 0 \equal{} \text{tr} \, (CA\minus{}AC) \equal{} (\text{tr} \, C)^2 \minus{} 2\det C$.

Assume $ \det C \neq 0$. Then $ C$ has the inverse $ D$. Then $ I \equal{} D C^2 D \equal{} AD\minus{}DA$, so $ 2 \equal{} \text{tr} \, I_2 \equal{} \text{tr} \, (AD \minus{} DA) \equal{} 0$, a contradiction. Thus $ \det C \equal{} 0$ and $ \text{tr} \, C \equal{} 0$.

Therefore, $ C^2 \equal{} O$ and $ CA \equal{} AC$, or equivalently $ AB \equal{} BA$.[/hide]

[quote="jmerry"]That would look a lot better with the matrices in $ \text{\LaTeX}$. See my earlier post for matrix code.

Can you find the lemma you're quoting and link to it? That's not something to leave without proof.[/quote]
I am sorry.Now I have edited my post.
Do you refer to the lemma:
"$ rank(AB - BA) \le 1$,thus we can prove $ A,B$ can upper-triangulize at the same time.?"

If it is,I can't find it in the forum(Maybe I am wrong),I just knew it is belong to prasolov's book.If you(or anybody else) want to know a proof,I can offer it to you.

Now,I will post the proof for the lemma:
Lemma:Let $ A$ and $ B$ be operators in a finite vector space $ V$ over some algebraic closed field $ \mathbb{F}$
such that $ rank(AB\minus{}BA) \le 1$.Then $ A$ and $ B$ are simultaneously triangulazable.

Proof:It suffices to prove that the operators $ A$ and $ B$ have a common eigenvector $ v \in V$(This is the main idea).Indeed,then the operators $ A$ and $ B$ induce operators $ A_1$ and $ B_1$ in the space $ V_1\equal{}V/span\{v\}$,
and $ rank(A_1B_1\minus{}B_1A_1) \le 1$,Then we can use induction on the dimesion of $ V$.

The proof of $ A,B$ have the common eigenvector is also use induction on $ n\equal{}\dim V$.
(We can assume $ Ker A\ne 0$,if not we can consider $ A\minus{}\lambda I$ insteand of $ A$,where $ \lambda$ is an eigenvalue of $ A$)
If $ n\equal{}1$,then it is obvious.
Let $ C\equal{}AB\minus{}BA$,in the proof of the inductive step will consider two cases.
a)$ Ker A \subset Ker C$,In this case,$ B(Ker A) \in Ker A$,since if $ Ax\equal{}0$ then $ ABx\equal{}Cx\plus{}BAx\equal{}0$.Therefore,we consider the restriction of $ B$ to $ Ker A \ne 0$ and select a eigenvector $ v$ of $ B$,the vector $ v$ is then also an eigenvector of $ A$

b)$ Ker A \subset Ker C$ is not true,ie,we can find $ x$ such that $ Ax\equal{}0$ and $ Cx \ne 0$.Since $ rank C\equal{}1$,
then $ Im(C)\equal{}span\{y\}$ where $ y\equal{}Cx$,Beside:$ y\equal{}Cx\equal{}(AB\minus{}BA)x\equal{}ABx \in Im(A)$.
It follows that $ B(Im(A)) \in Im(A)$.Indeed,$ BAz\equal{}ABz\minus{}Cz$,where $ ABz \in Im(A)$ and $ Cz \in Im(A)$,
we have $ Ker A \ne 0$,then we can apply induction in the space $ Im(A)$.


[hide]Ok so we have 4 points $ ( \minus{} 50, 0) ( \minus{} 40, 10) (40, 10)(50, 0)$ We can right 4 different equations now

$ a( \minus{} 40)^{2} \plus{} b( \minus{} 40) \plus{} c \equal{} 10$
$ a(40)^{2} \plus{} b(40) \plus{} c \equal{} 10$
$ a( \minus{} 50)^{2} \plus{} b( \minus{} 50) \plus{} c \equal{} 0$
$ a(50)^{2} \plus{} b(50) \plus{} c \equal{} 0$

Adding the first 2 we get
$ 3200a^{2} \plus{} 2c \equal{} 20$
and the second 2 give us
$ 5000a^{2} \plus{} 2c \equal{} 0$
Now we can subtract them
$ 1800a^{2} \equal{} \minus{} 20$
$ a^{2} \equal{} \minus{} \frac {1}{90}$
$ c \equal{} 27\frac {7}{9}$
So now we solve for x = 0
and find that the max is $ (0,27\frac {7}{9})$
[/hide]

I hope that's right.

so, to create a function with two 50 feet sides[hide="...."]$ f(x) \equal{} 50^2 \minus{} x^2$, implying $ x \equal{} \pm50$.
$ Kf(10) \equal{} 10$ ; $ f(10) \equal{} 50^2 \minus{} (10)^2 \equal{} (50 \minus{} 10)(50 \plus{} 10) \equal{} (40)(60) \equal{} 2400$
-solve for $ K\Longrightarrow$- $ 10 \equal{} 2400K$ implies $ K \equal{} \frac {1}{240}$
$ Kf(0) \equal{} \frac {2500}{240} \equal{} \frac {250}{24} \equal{} 10.417$ feet[/hide]
sorry sniped, but you're wrong babe, [hide="really wrong."] $ s(x)\equal{}\frac{250}{9}\minus{}\frac{\sqrt{10}}{30}x^2$
$ s(10)\equal{}17.237$ and $ s(x)\equal{}0$...implies $ x\equal{}\pm16.233$
Your function does not satisfy either of the requirements.
[/hide]

[quote="sharkman"]A bridge is to be built in the shape of a parabolic arch and it to have a span of 100 feet. The height of the arch, a distance of 40 feet from the center, is to be 10 feet. Find the height of the arch at its center.[/quote]

[hide="my solution"]
Set the origin at the ground level at the center of the arch.
there are 2 points that are useful: $ (50,0)$ and $ (40,10)$
The General equation for a parabola is $ y \equal{} ax^2 \plus{} bx \plus{} c$
It is also to be noticed that the axis of symmatry is on $ x \equal{} 0$, which is $ x \equal{} \frac { \minus{} b}{2a}$
Therefore $ b \equal{} 0$
$ y \equal{} ax^2 \plus{} c$
$ 0 \equal{} 2500a \plus{} c$
$ 10 \equal{} 1600a \plus{} c$
$ 10 \equal{} \minus{} 900a$
$ a \equal{} \minus{} \frac {1}{90}$
$ c \equal{} \frac {250}{9}$
The $ c$ value is the y-intercept, which is the height of the arch at the center.
the height is $ \frac {250}{9}$ feet, or about $ 27.78\;\text{feet}$.
[/hide]

I think my solution confirms Sniped's solution.

The problem with igiul's solution was that it had $ K f(10) \equal{} 10$ when it should be $ K f(40) \equal{} 10$.


I JUST LOST!!! :D

and so did you if you are reading this.

there are WAY to many useless topics about this.
mods, please lock this.

btw, i lose. post stuff like this inside threads, don't make another thread.


Seems as though Edwin A. Abbot's Flatland is getting two different movie makeovers this year!

The first, [url=http://www.flatlandthefilm.com][b]Flatland: The Film[/b][/url], is already out!  It's a 1-1/2 hr feature length film, created by an independent director, Ladd Ehlinger Jr.  It's gotten rave reviews at Scifi.com and Filmthreat.com, and there are some clips posted on youtube.com

The second, [url=http://www.flatlandthemovie.com][b]Flatland: The Movie[/b][/url], should be out in May/June of this year.  It's going to be a 1/2 hr educational movie, and they got the voices of Martin Sheen, Kristen Bell, and Tony Hale!  They have some clips up on youtube.com also.

And there's still the 11 minute long [url=http://www.der.org/films/flatland.html]classic version[/url] available, with Dudley Moore's voice from 1965

This should be great news for Flatland enthusiasts everywhere :)

Thanks. You know, there is a Flatland topic about advertising, and here it is:

read number 7.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11598[/url]


linear algebra
matrix
vector
combinatorics proposed
combinatorics

Let $p$ be a prime number, and $m,n,k$ positive integers. Also, let $A_{1},\ldots,A_{m}$ be subsets of $\{1,\ldots,n\}$ such that (i) $p^{k}$ does not divide the numbers $|A_{i}|$, and (ii) $p^{k}$ divides $|A_{i}\cap A_{j}|$ for all $i\ne j$. Prove that $m\le n$.

P.S.

I'm sure the weaker problem corresponding to $k=1$ must have appeared before around here, but I haven't seen the general version on the forum. It can be used to get a complete proof for [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=powers+prime&t=29556]this[/url].

This is certainly wrong, but here it goes: take the matrix $A$ whose lines are the characteristic vectors of the sets $A_{i}$ and let $X=AA^{T}$. By the hypothesis, the entries of $X$ which are not divisible by $p^{k}$ are exactly those on the main diagonal. Therefore, the exponent of $p$ in $\det X$ is the sum of exponents of $p$ in $|A_{i}|$. Hence $X$ is invertible. On the other hand, the rank of $X$ is at most $n$, hence $m \leq n$.


Find with proof the last four digits of the number $2006^{2006}$.

We have $2006^{2006}=\sum_{k=0}^{2006}{\binom{2006}{k}}*6^k*{2000}^{2006-k}$.

So to view the last for digits it´s necessary only take $k=2005$ and $k=2006$. (That´s congruence mod $10^4$).

So we have to view the last four digits of this number $2006*6^{2005}*2000+6^{2006} = 6^{2005}(2006*2000+6)$. And we can continue apllying congruence mod $10^4$, to obtain $6^{2005}*2006$.

At this point our work is calculate the last for digits of $6^{2005}$. To do this observe that $6^{2005}=\sum_{k=0}^{2005}{\binom{2005}{k}}*(-4)^k*(10)^{2005-k}$ and to view the last for digits of it´s number we have to consider just only $k=2005$, $k=2004$, $k=2003$ and $k=2002$.

Expand this product and making a little more calculation, we´ll see that we have to encounter the last for digits of the following number $4^{2002}*6736*2006$. But $2002=2*7*11*13$, and now we must do more calculations and so our problem is equivalent to obtain the last four digits of the number $16*4304*4096*8864*6736*2006$ and the answer is $4256$.

[quote="draperp"]By multiplying 6 by the running product [/quote]

The $2$ in $2006$ also affects the last four digits.   :huh:

mffmartinelli, I have doubts about your answer too because the last digit of $6^n$ in base 10 is always 6 owing to $6^2 \equiv 6 \bmod 10$  :(

It´s edited now. I believe now it´s wright.

[quote="mffmartinelli"]
At this point our work is calculate the last for digits of $6^{2005}$. To do this observe that $6^{2005}=\sum_{k=0}^{2005}{\binom{2005}{k}}*(-4)^k*(10)^{2005-k}$ and to view the last for digits of it´s number we have to consider just only $k=2005$, $k=2004$, $k=2003$ and $k=2002$.
[/quote]

Aha, I was wondering what was wrong with this.  The problem here is that $k = 2001$ is a negative term, and if it's large enough it will force a carry-over that will affect the last four digits  :D

Duh! I can't believe that I missed the 2 in 2006. (thank you for pointing that out, t0rajir0u)
By using the same method, the last four digits of both \[ 2006^6  \] and \[ 2006^{131}  \] are 8656.  So, the 6th, the 131st...the (6+125n)th...and the 2006th terms are all 8656.

I've never used Mathematica; I use Mathcad 12.  It is a surperb piece of mathematics software (if a bit expensive).  I have solved all kinds of problems and used its programming capabilities extensively (such as in this problem).  As a result of this problem, I have a function that will give the last [i]n[/i] digits of [i]a[/i]^[i]b[/i].

Yes, the answer is 8656 and it can easily be calculated if write a simple program... ;)

$2006^{2006}$ ;)
we should check the last few digits of $2006^{\frac{1000}{8}}$
and it might work on other four digit numbers other 2006 :?:

[quote="t0rajir0u"]This is done very easily in php with the BCMath function $bcpowmod$, which uses modular exponentiation to speed up the job  :D[/quote]

You're right here ;)

[quote="mathmanman"][quote="draperp"]As a result of this problem, I have a function that will give the last [i]n[/i] digits of [i]a[/i]^[i]b[/i].[/quote]
Mind showing it to us? :)[/quote]

Sure, here's my Mathcad program. It is \[ a^b \mod{base^n}  \]

(seamusoboyle's "mod(10^n)" remark brought it to my attention that I could adjust my program to a number system of any base.)

[quote="Gibbenergy"]Here is the brutal way of solving it without using math software but only using caculator with 10 digits number.
  I think it can solve a lot of problems by finding the last digits but indeed, you should be patient since it isn't nice at all:
$2006^{2006}=(2006^2)^{1003}$mod $(4036)^{1003}=4036(4036)^{1002}$ mod... So the last 4 digits is 8656[/quote]

Interesting. It is brutal, but it's a good method for only requiring a 10-digit calculator (although many calculators have only eight digits).


trigonometry
integration
real analysis
real analysis unsolved

This one is quite tough:
   Find an equivalent in 0 for $\sum_{n\geq 1}{\frac{sin(nx)}{\sqrt{n}}}$.

BTW, How do you prove that $\sum \frac{\sin n}{\sqrt{n}}$ converges?  :?

to xirti: for example using Dirichlet test for series:the sequence $x_n=\frac{1}{\sqrt n}$ is monotone and tends to $0$. And the sequence of oartial sums $Y_n=\sin 1+...+\sin n$ is bounded.You can easily get it using the Abel transformation

I don't think that $Y_n$ is bounded because $\sin n$ is greater than $\frac 1 2$ infinitely many times.. Am I wrong?

Yes you are, because some other values can be negative for example. Anyway, the proof is immediate: one can explicitely compute $Y_n$ and see it is bounded.

Thank you very much! I found a way to compute this using complex number $z=\cos 1+i \sin 1$.

$f(x)=\sum_1^{\infty} \frac{sin(nx)}{\sqrt{n}}$.
$\sqrt{x} f(x)=x \sum_1^{\infty} \frac{sin(nx)}{\sqrt{nx}}$ and $x \sum_1^{\infty} \frac{sin(nx)}{\sqrt{nx}}$ tends towards $\int_0^{\infty} \frac{sin(t)dt}{\sqrt{t}}=A$ when $x$ tends towards $0$.
Hence $f(x)$ is equivalent to $\frac{A}{\sqrt{x}}$.

To prove this, notice that $|\sum_p^q sin(nx)|<\frac{B}{x}$ for $|x|<1$ for example and that $|\sum_p^q \frac{sin(nx)}{\sqrt{n}}|<\frac{2B}{x\sqrt{p}}$.
Let $\epsilon > 0$:
One can choose $T>0$ such that $|\int_T^{\infty} \frac{sin(t)dt}{\sqrt{t}}| < \epsilon$ and $\frac{2B}{\sqrt{T}} < \epsilon$

Hence:
$|\sqrt{x}f(x)-A|<|x \sum_1^{T/x}\frac{sin(nx)}{\sqrt{nx}} - \int_0^{T} \frac{sin(t)dt}{\sqrt{t}}|+| \sum_{T/x}^{\infty}\frac{sin(nx)}{\sqrt{nx}}|+|\int_T^{\infty} \frac{sin(t)dt}{\sqrt{t}}|$
But:
$| \sum_{T/x}^{\infty}\frac{sin(nx)}{\sqrt{nx}}| < \frac{2B}{x\sqrt{T/x}} \frac{1}{\sqrt{x}}=\frac{2B}{\sqrt{T}} < \epsilon$.
$|\int_T^{\infty} \frac{sin(t)dt}{\sqrt{t}}| < \epsilon$
$|x \sum_1^{T/x}\frac{sin(nx)}{\sqrt{nx}} - \int_0^{T} \frac{sin(t)dt}{\sqrt{t}}|$ tends towards $0$ when $x$ tends towards $0$.

Correct, but... you need to prove that $A$ is not zero. :P


Let $p$ be a prime, $p \equiv 1 (mod 4)$.
Find $A= \sum\limits_{k=1}^{\frac{p-1}{2}}\{ \frac{k^{2}}{p}\}$ where $\{ x \}=x-[x]$

[quote="Rust"] because for all k, exist m, suth that $p|(k^{2}+m^{2})$, [/quote]

why? I don't know it.

You can see Problem 2 TST viet nam 2005.I posted solution


Determine all continuous functions $f: \mathbb R \to \mathbb R$ such that
\[f(x)\plus{}2f(x^2)\plus{}f(x^4)\equal{}0\]
holds for all $ x \in \mathbb R$.

[quote="Victory.US"]Determine all continuous function $ R\to R$ such that
$ f(x) \plus{} 2f(x^2) \plus{} f(x^4) \equal{} 0$ ,$ \forall x \in R$[/quote]

Let $ g(x)\equal{}f(x)\plus{}f(x^2)$. $ g(x)$ is continuous and the original equation becomes $ g(x)\plus{}g(x^2)\equal{}0$ so :

$ g(\minus{}x)\equal{}g(x)$
$ g(0)\equal{}0$
$ g(x)\equal{}(\minus{}1)^ng(x^{2^{\minus{}n}})$ $ \forall x\ge 0$. Setting $ n\to \plus{}\infty$ and using continuity, we get $ g(x)\equal{}0$ $ \forall x\ge 0$
And so $ g(x)\equal{}0$ $ \forall x$

So $ g(x)\equal{}f(x)\plus{}f(x^2)\equal{}0$ and the same proof gives $ \boxed{f(x)\equal{}0}$ $ \forall x$

[quote="Mathias_DK"][quote="pco"][quote="Victory.US"]Determine all continuous function $ R\to R$ such that
$ f(x) \plus{} 2f(x^2) \plus{} f(x^4) \equal{} 0$ ,$ \forall x \in R$[/quote]

Let $ g(x) \equal{} f(x) \plus{} f(x^2)$. $ g(x)$ is continuous and the original equation becomes $ g(x) \plus{} g(x^2) \equal{} 0$ so :

$ g( \minus{} x) \equal{} g(x)$
$ g(0) \equal{} 0$
$ g(x) \equal{} ( \minus{} 1)^ng(x^{2^{ \minus{} n}})$ $ \forall x\ge 0$. Setting $ n\to \plus{} \infty$ and using continuity, we get $ g(x) \equal{} 0$ $ \forall x\ge 0$
And so $ g(x) \equal{} 0$ $ \forall x$

So $ g(x) \equal{} f(x) \plus{} f(x^2) \equal{} 0$ and the same proof gives $ \boxed{f(x) \equal{} 0}$ $ \forall x$[/quote]
You should remember to write $ g(1) \equal{} 0$ as $ 2^{ \minus{} n} \to 0, n \to \infty$ and therefore $ x^{2^{ \minus{} n}} \to 1, n \to \infty$ and therefore $ g(x) \equal{} g(1) \forall x > 0$.[/quote]

Dear pco and Mathias_DK, what about the factor $ ( \minus{} 1)^n?$ When $ n \to \infty$ it has no limit....so how do you assert that when $ n \to \infty$ the $ g(x) \to g(1)?$....

[quote="prester"]...Dear pco and Mathias_DK, what about the factor $ ( \minus{} 1)^n?$ When $ n \to \infty$ it has no limit....so how do you assert that when $ n \to \infty$ the $ g(x) \to g(1)?$....[/quote]

I just said that $ g(x) \equal{} 0$ and this is an immediate conclusion of continuity and $ g(x) \equal{} ( \minus{} 1)^ng(x^{2^{ \minus{} n}})$ :

When $ n\to \plus{} \infty$, RHS has a limit ($ g(x)$) and so $ g(x)\equal{}g(1) \equal{} 0$ else RHS would swap between $ g(1)$ and $ \minus{} g(1)$.


In rectangle ABCD, points E and F divide side DC into three equal parts DE=EF=FC and point G and H divide side BC into three equal parts BG=GH=HC. The line AE cuts the lines DG and DH at points K and L, respctively. Similarly, the line AF cuts the line DG and DH at points M and N, respectively.

Prove: KN ll CD

thanks  :lol:

[img]8807[/img]
This problem is trivial once its placed in a co-ordinate system. I don't think there is any nice purely geometrical way though :)

Instead of using 4 and 5, you should use variables for the length and width of the rectangle.  Presumably, with enough bashing you will find that the y co-ordinates of k and m are the same, thereby validating our claim that $KN\parallel CD$.

[hide="I am using variables; those numbers are there basically by accident but you can ignore it. Anyway I thought of a way to quasigeometrize the solution"]
Let $CB=3a$ $DC=3b$. Then drop perpendiculars from $K$ to $AB,AD,DC$ respectively at $K_{1},K_{2},K_{3}$. Now $K_{3}=2ax$, $K_{2}=3bx$ for some $x<1$ and $K_{1}=3ay$, $K_{2}=by$ for some $y<1$. Clearly $y=3x$. Now $K_{1}+K_{3}=3b$ so $2ax+3a(3x)=3b$ so $x=\frac{3b}{11a}$ thus $K_{1}=\frac{9}{11}b$
Similarly, the same value is found for $N$ so $KN$ is parallel to $DC$.[/hide]

I did it last night and use the same way actually
 :P


If $(4x+7)^2=fx^2+gx+h$, what is the value of $f+g$?

You seen the work and they are right the answer is 72


floor function
logarithms
combinatorics unsolved
combinatorics

Can somebody please solve (ie give me simplified version without summation sign) of
following:

$\sum_{k=1}^{2^{L}-1}\lfloor{\log_{2}(k)}\rfloor$

where LOG are to the base 2 and $L \geq k$.


function
Vieta
probability
number theory
prime factorization
number theory unsolved

$gcd(a,b)=1 \Rightarrow \varphi(a)\cdot\varphi(b)=\varphi(ab)$
$\varphi(n)$ - Euler's function

Remember that for any $x,a$ to be coprime is equivalent for $x$ to be invertible $\mod a$ (so there is $z$ with $xz \equiv 1 \mod a$). Also remember that $\varphi(a)$ is nothing else than the number of invertible residue classes $\mod a$ by this.

Now it follows for example directly from the Chinese Remainder Theorem which gives:
$x$ is invertible $\mod a$ and $\mod b$ iff it is invertible $\mod ab$ (here $a,b$ need not to be necessary coprime, but lets assume that).
This then gives for coprime $a,b$, that by the canonical mapping of the CRT, exactly the pairs $(x,y)$ with $x \mod a$ and $y \mod b$ being invertible are mapped to the invertible elements $\mod ab$. So their number is equal.

Another approach would be to define (or to prove that this holds) $\varphi$ by $\sum_{d|n} \varphi(d) =n$ and then starting calculating around with möbius inversion etc..

Well, if we write $\varphi(n)$ under the form : $\varphi(n) = (p_1^{\alpha_1} - p_1^{\alpha_1-1})(p_2^{\alpha_2} - p_2^{\alpha_2-1})\cdots (p_k^{\alpha_k} - p_k^{\alpha_k-1})$, where $n = p_1^{\alpha_1}\cdots p_k^{\alpha_k}$ (the $p_i$ are distinct primes), it then follows trivially.
Indeed, let :
$a = p_1^{\alpha_1}\cdots p_k^{\alpha_k}$
$b = q_1^{\beta_1}\cdots q_l^{\beta_l}$.
Then :
$\varphi(ab) = (p_1^{\alpha_1} - p_1^{\alpha_1-1}) \cdots (p_k^{\alpha_k} - p_k^{\alpha_k-1})(q_1^{\beta_1} - q_1^{\beta_1-1}) \cdots (q_l^{\beta_k} - q_l^{\beta_l-1})$.

Yes, but firstly you must prove the formula for this function. And I would like do it using aforementioned property of Euler's function

Yes, well, there is a nice proof of this formula using the Principle of Inclusion Exclusion. ;)
I can post it if you want, but I think it's a classic.

Yeah, it would be very nice if you post the proof,  :) thanks

mathmanman please polish your solution a bit :)
it has some mistakes,not fundamental mistakes...just some small ones
this demonstration i saw also used vieta's relations,i dont see your
demo using them anywhere.
you should mention that in the last stept there..vieta is used

mathmanman has nice solution
I think $\phi(mn)=\phi(n)\phi(m).d/\phi(d)$ where $d=gcd(m,n)$

There is also a probabilistic proof (avoiding the principle of inclusion-exclusion).
If anyone's interested I can post it.
@Dondigo: Take a look at "Wyklady z Kombinatoryki" , Zbigniew Palka, Andrzej Rucinski, WNT, p. 106, I found both (the probabilistic and that of mathmanman) proofs there.

i dont really understand how could a probabilistic proof arise for this problem,
this condradicting the exact  and concrete nature of euler's function.
but that does not change the fact that i would like to see it

[b]{x}[/b], I would be glad if you post the solution you've mentioned :)

Let $n$ be a positive integer with prime factorization $n = \prod_{j = 1}^m p_j^{\alpha_j}$.
By definition $\frac{\varphi ( n )}{n}$ is the probability that a randomly choosed integer from $[ 1, n ]$ is coprime to $n$.
Let $X = \left\{ 1, 2, \ldots, n \} \right.$ be our probabilistic universe and
\begin{eqnarray*} A_j = \left\{ x \in X | p_j \mbox{ divides } x \right\} & & j = 1, 2, \ldots, m \end{eqnarray*}
events of the universe $X$. We prove first that $A_1,A_2,..,A_m$ are independent.
For any subsequence $\{ i_1, \ldots, i_k \} \subset X$ we have
\begin{eqnarray*} | A_{i_1} \cap \ldots \cap A_{i_k} | & = & | \left\{ x \in X | p_{i_1} \cdot \ldots \cdot p_{i_k} \mbox{ divides } x \right\} |\\ |A_{i_1} \cap \ldots \cap A_{i_k} | & = & \frac{n}{p_{i_1} \cdot \ldots \cdot p_{i_k}} \end{eqnarray*}
Therefore
\begin{eqnarray*} P ( A_{i_1} \cap \ldots \cap A_{i_k} ) & = & \frac{| A_{i_1} \cap \ldots \cap A_{i_k} |}{n} \\ P ( A_{i_1} \cap \ldots \cap A_{i_k} ) & = & \frac{1}{p_{i_1}} \cdot \ldots \cdot \frac{1}{p_{i_k}} \\ P ( A_{i_1} \cap \ldots \cap A_{i_k} ) & = & \prod_{j = 1}^k P \left( A_{i_j} \right) \end{eqnarray*}
That is $A_{i_1}, \ldots, A_{i_k}$ are independent events. We denote by $\bar{A}$ the complement of $A$.
Observe that the probability that a randomly choosed integer from $[ 1, n ]$ is coprime to $n$ is equal to the probability that a randomly choosed integer from $[ 1, n ]$ is not divisible by $p_1$ and $p_2$ and ... and $p_m$.
Therefore
\begin{eqnarray*} \frac{\varphi ( n )}{n} & = & P \left( \overline{A_1} \cap \overline{A_2} \cap \ldots \cap \overline{A_m} \right)\\ \frac{\varphi ( n )}{n} & = & \prod^m_{j = 1} P \left( \overline{A_j} \right)\\ \frac{\varphi ( n )}{n} & = & \prod^m_{j = 1} \left( 1 - P \left( A_j \right) \right)\\ \frac{\varphi ( n )}{n} & = & \prod^m_{j = 1} \left( 1 - \frac{1}{p_j} \right) \end{eqnarray*}

[quote="dieuhuynh"]$\phi(mn)=mn\prod_{p|mn}{(1-\frac{1}{p})}=\frac{(n\prod_{p|n}{1-\frac{1}{p}})(m\prod_{p|m}{(1-\frac{1}{p})})d}{\prod_{p|d}{(1-\frac{1}{p})}}$[/quote]

You can not proceed in such a way or you should prove the formula you are using without using multiplicativity of $\phi$.

@dieuhuynh : http://www.mathlinks.ro/Forum/viewtopic.php?t=14688 . :)


Question: can any proper rational expression be written as an infinite series, where the term has a closed-form formula?
For example:
$ \frac {1}{x \minus{} 2} \equal{} \sum_{k \equal{} 1}^{\infty}\frac {2^{k \minus{} 1}}{x^k}$
which is done by infinte geometric series.
But for expressions like
$ \frac {5x \minus{} 6}{x^2 \minus{} x \plus{} 3}$
the only way I can think of is to try partial fraction decomposition, which invokes solving the bottom polynomial, but that gets ugly soon for high-degree equations (what's worse, there can be imaginary zeros like the above). How do you solve the problem? Thx.  :)

There's no way to avoid the partial fractions if you want a closed form. It exists, but it depends on the roots of the polynomial.

Also, there are several such series, depending on what range of $ x$ you want convergence for. The series you wrote down with entirely nonpositive powers of $ x$ converges when $ |x|>|r|$ for the largest root $ r$ of the denominator. The series expression with entirely nonnegative powers of $ x$ converges for $ |x|<|r|$ where $ r$ is the smallest root. There are other series with both negative and positive powers that converge in annuli between roots.

If you focus on just the innermost and outermost series, you can at least give recursive formulas for the coefficients without finding roots.

The recursive formulas jmerry's talking about turn out to be determined by the irreducible factors of the denominator.  I explore this question in the other direction [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=212490]here[/url].  (Imaginary roots don't really change anything:  the problem of high-degree denominators is much worse.)