[quote="myc"]At the end of a professional bowling tournament, the top 5 bowlers have a playoff.  First #5 bowls #4.  The loser receives fifth prize and the winner bowls #3 in an other game.  The loser of this game receives fourth prize and the winner bowls #2.  The loser of this game receives third prize and the winner bowls #1,  The winner of this game gets first prize and the loser gets second prize.  In how many orders can bowlers #1 through #5 receive the prizes?

Can anyone explain why my method is wrong?[hide]There are $5$ bowlers, so there are $5!$ ways to arrange them to receive the prizes.  The first bowler in line gets first prize, second bowler gets second prize...ect.  Whats the difference between this and saying that there are $2$ outcomes to each of the $5$ games, so there are $2^{5}$ orders bowlers #1 through #5 can receive the prizes?
[/hide]

Thanks[/quote]
Can 1st place bowler end up with fifth place? In the first way, you would be counting that

[quote="myc"]At the end of a professional bowling tournament, the top 5 bowlers have a playoff.  First #5 bowls #4.  The loser receives fifth prize and the winner bowls #3 in an other game.  The loser of this game receives fourth prize and the winner bowls #2.  The loser of this game receives third prize and the winner bowls #1,  The winner of this game gets first prize and the loser gets second prize.  In how many orders can bowlers #1 through #5 receive the prizes?

Can anyone explain why my method is wrong?[hide]There are $5$ bowlers, so there are $5!$ ways to arrange them to receive the prizes.  The first bowler in line gets first prize, second bowler gets second prize...ect.  Whats the difference between this and saying that there are $2$ outcomes to each of the $5$ games, so there are $2^{5}$ orders bowlers #1 through #5 can receive the prizes?
[/hide]

Thanks[/quote]

[hide]It is given that each person bowls another. That means for each pair, there is a winner and a loser. For example when there are 5 players, $A, B, C, D,$and $E$ and $A$ plays against $B$, if $A$ loses $A$ becomes 5th and $B$ becomes 4th. You cannot place $C, D$ or $E$ as 4th because none of these people are playing against $A$. This is why your argument is not correct. Since there are 4 games and at each game there are two possibilities for the outcome, there is a total of $2^{4}= 16$ orderings.[/hide]

[quote="myc"]the place each bowler gets is determined by the order that they are in. for example {23514},  so bowler #2 gets first place, bowler #3 gets second place, bowler #5 gets third place, bowler #1 gets fourth place, bowler #4 gets fifth place.  What do you mean by
"Can 1st place bowler end up with fifth place? In the first way"?  sorry I'm a little slow...:)[/quote]
The bowler who has first before the tournament.


geometry
circumcircle
rectangle
perpendicular bisector
Brazilian Math Olympiad

$ABCD$ is a quadrilateral with a circumcircle centre $O$ and an inscribed circle centre $I$. The diagonals intersect at $S$. Show that if two of $O,I,S$  coincide, then it must be a square.

A theorem guarrantees that in the general case these three points are in the same straight line.

We know that $O$ is on every perpendicular bisector, and $I$ is on evry bisector. We easily get that $\triangle{IAB}$ and the other of the same type are isosceles.Using this we obtain that all of the angles of the quadrilateral are equal, so it is a rectangle, where the bisectors are diagonals, so it is a square.

Let $O$ and $S$ coincide. Then since $ABCD$ is cyclic with center $O$, we have $OA=OB=OC=OD$, which means that the quadrilateral $ABCD$ is a parallelogram. Since it has an inscribed circle then $AB+CD=AD+BC$, which means that $ABCD$ is a square. Let $S$ and $I$ coincide. Triangles $ABI$ and $CID$ are similar. Having the same altitude $r$ means they are congruent. Thus $ABCD$ is a square. Let $I$ and $O$ coincide. Again triangles $ABO$ and $CDO$ are similar. Having the same altitude $r$ means they are congruent. Hence done!

Case 1: Let $O$ and $S$ coincide at Point $X$.

Proof: Note that $X$ is the circumcenter so $XA=XB=XC=XD$ and let $\angle XAB=\angle XBC=\angle XCD=\angle XDC=a$ and $\angle XBC=\angle XCB=\angle XAD=\angle XDA=b.$  Since $$\angle A=a+b=\angle C,$$ then cyclic quadrilateral $ABCD$ must have equal opposite angles, hence, $\angle A=\angle C=90^{\circ}$.  Similarly, we can get $\angle B=\angle D=90^{\circ}$.  Since the diagonals bisect each other, we have a square. $\Box$

Case 2: Let $O$ and $I$ coincide at Point $Y.$

Proof: Because $Y$ is the circumcenter, $\angle YAB=\angle YBA$ and  $\angle YBC=\angle YCB.$ But, $Y$ is also the center of the inscribed circle, meaning that $Y$ lies on the angle bisectors of angle $A,B, C,D$.  Hence, $\angle YBA=\angle YBC,$ $\angle YAD=\angle YAB,$ and $\angle YCB=\angle YCD.$ Hence, $$\angle DAY=\angle YAB=\angle YCB=\angle YCD$$ and $ABCD$ is cyclic so $\angle A$ and $\angle C$ are both right angles.  Similarly, $\angle B$ and $\angle D$ are right angles.  Also, $\angle DAY=\angle YAB=\angle YCB=\angle YCD=45^{\circ}$  so $\angle AYB=\angle BYC=90^{\circ}$ so lines $AYC$ and $BYD$ are collinear and bisecting, which means that it is a square. $\Box$

Case 3: Let $I$ and $S$ coincide at Point $Z.$

Proof: In cyclic quadrilateral $ABCD$, we get $\angle DAC=\angle DBC.$  But, $\angle DAC=\angle CAB$ and $\angle ABD=\angle CBD$ so $$\angle A=\angle B.$$ Hence, $\angle A=\angle B=90^{\circ}$ and similarly, we can prove that $\angle C=\angle D=90^{\circ}.$ Also, from equal base angles, we can prove that $AZ=ZB=ZC=ZD$ so $Z$ is also the circumcenter, proving that the diagonals bisect each other and it is a square. $\Box$

Hence, given two of the three points coincide , we can prove that it is a square. $\blacksquare$


inequalities
calculus
derivative
function
integration
real analysis
real analysis unsolved

Suppose $ f \colon \mathbb{R} \to \mathbb{R}$ is twice continuously differentiable, $ |f''(x)| \le 1$ and $ f( \minus{} 1) \equal{} f'( \minus{} 1) \equal{} f(1) \equal{} f'(1) \equal{} 0$.  What is the maximum possible value of $ f(0)$?  Is it a maximum of just a supremum?

[corrected]

Unless I'm visualizing this wrong, why can't $ f(0)$ get arbitrarily large?  For example, if you define the function to be 0 on $ \mathbb{R}\backslash [\minus{}1/2,1/2]$ (or something like that), and then define it to be a "smoothed" out bump on $ [\minus{}1/2,1/2]$, you could make the bump at 0 be as high as you want, and the bump would get steeper and steeper but still be differentiable.  For example, something like [url=http://en.wikipedia.org/wiki/Bump_function]this[/url], except the bump gets a lot steeper and taller.

JRav is correct, which makes me think that this isn't the right problem.

If you want a problem with a finite solution (and some analysis to be done), try it with some restriction on $ f''(x).$

Suppose we have the above values of $ f$ and $ f'$ at $ 1$ and $ \minus{}1$ as suppose we also have $ |f''(x)|\le 1$ for all $ x\in (\minus{}1,1).$ Now how big can $ f(0)$ be?

... Except that still might not be the right problem. Myabe the restriction is not on the second derivative but the third derivative: $ |f'''(x)|\le 1$ for $ x\in(\minus{}1,1).$ Then what? (Only you'd also have to change the hypothesis from $ C^2$ to $ C^3$ to state it that way.)

You are both right of course.  The hypothesis $ |f''(x)| \le 1$ was left out.

Original post corrected.


Solve $ 9x^2\plus{}3xy\plus{}y^2\minus{}3y\plus{}z\plus{}3\equal{}z^{11}$ in $ \mathbb{Z}$, where $ 11 \nmid z.$

A good starting point is rewriting the equation in the following form:
\[ (6x\plus{}y)^2 \plus{}  3(y\minus{}2)^2 \equal{} 4z(z^{10}\minus{}1)\]
Since the right hand side is divisible by $ 11$ while $ \minus{}3$ is not a square modulo $ 11$, it follows that both $ 6x\plus{}y$ and $ y\minus{}2$ are divisible by $ 11$ (and thus both sides of the above equation are divisible by $ 11^2$).
Similarly, the right hand side is divisible by $ 3$, etc.

But such observations do not help to get all the solutions, while non-trivial solutions exist, e.g.:
\[ \begin{cases} x\equal{}124034162869613 \\ y \equal{} 277642316824842 \\ z \equal{} 481 \end{cases}\]

This is an amazing equation. Can it be solved completely? And what if $11 | z$?

What if $11 \mid z$? The post above yours claims that $11 \mid z(z^{10} - 1)$, which is true because either $11 \mid z$ or $\gcd(z,11) = 1$, and then $11 \mid z^{10} - 1$ by Fermat's little theorem. Furthermore, the poster makes no claim he could then find all solutions ...


Sa se determine \[a,b,c \in R\] astfel incat sa existe functii \[f: R \to R\] , cu proprietatea ca:
\[f(x+y) = f(x)+f(y)+ax^{2}+bxy+cy^{2},(\forall )x,y \in R \]


If $ a\nabla b \equal{} a^2 \plus{} 2ab \plus{} b^2$, what is the value of $ (3\nabla2)\nabla5$?


factorial
AMC
AIME
floor function
MATHCOUNTS

How many zeros does

1!2!3!4!...98!99!100!

have?

The way I did it is:
[hide]
2 x 5 and 10 will produce a "0". 2's are in excess due to numbers such as 2,4,8,64. So we only count the 5's.

Using the table:

5   - 1                     55 - 1
10 - 1                     60 - 1
15 - 1                     65 - 1
20 - 1                     70 - 1
25 - 2                     75 - 2
30 - 1                     80 - 1
35 - 1                     85 - 1
40 - 1                     90 - 1
45 -1                      95 - 1
50- 2                      100 - 2

This tells us how many zeros each number contributes. Adding the numbers, 100! gives 24 zeros. 99! is devoid of 2 "0's" hence gives 22 zeros. 99! to 95! all give 22 zeros. Then we subtract one again:

In 100!, there are 24 zeros. In 99! to 95!, we will have 24-2 = 22 zeros.
In 94! to 90!, we have 22-1=21 zeros,
and so on.

We get:

24 + 22(5) + 21(5) + 20(5) + 19(5) + 18(5) + 16(5) + 15(5) + 14(5) + 13(5) + 12(5) + 10(5) + 9(5) + 8(5) + 7(5) + 6(5) +     4(5) + 3(5) + 2(5) + 1(5) = 24 + 5 (22+ 21 +19...) - 1109
[/hide]
I was wondering if anyone has an alternative solution?

*This is based on a AIME A 2006 problem.

You can kill it (I don't say it's an elegant solution) with Lagrange's fromula, saying that for any $n\in\mathbb{N}^{+}$ and prime $p$ the highest exponent $k$ such that $p^{k}\mid n$ can be expressed as \[k= \left\lfloor\frac{n}{p}\right\rfloor+\left\lfloor\frac{n}{p^{2}}\right\rfloor+\left\lfloor\frac{n}{p^{3}}\right\rfloor+\ldots,\] where $\lfloor x\rfloor$ is the greatest integer less or equal $x$. Of course for sufficiently large exponents the constituents will become zeros, so the sum has in fact only finite number of constituents.
Just apply this formula to $n=100$ and $p\in\{ 2,5\}$ and you will get the result. But this is very inelegant and ugly solution and I don't adivce using it unless you don't have any idea to solve the problem.
[hide="The answer is"]$24$[/hide]

Neelnal I think you have something in your explanation but I don't get your formatting so it makes me go like: :juggle:
Could you explain it again? Maybe with more english words?

[quote="nebula42"][quote="neelnal"][quote="Rzeszut"]
[hide="The answer is"]$24$[/hide][/quote]

That is just how many zeros are in 100!. It is not the answer. The answer is what I got: 1109.

100! is easy. I was looking for a way to do 1!2!3!...99!100![/quote]
I remember this problem.  :)
I think my solution went something like this:
[hide]All of the 96 terms from 5! to 100! have a factor of 5 (as in, 5 appears once in their "factorial expansion", such as $12! = 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot \fbox{5}\cdot 4 \cdot 3 \cdot 2 \cdot 1$). Similarly, 91 terms have a 10, 86 terms have a 15, etc. Each of these numbers contributes at least one factor of 5 (some contribute more; we'll get to those later), so for now our sum is
$96+91+86+\dots+6+1=970$. However, 25, 50, 75, and 100 each contribute two factors of 5, so we have to add 76 (for 25), 51 (for 50), 26, and 1 in again to make our final sum \[970+76+51+26+1=\fbox{1124}.\] [/hide]
You can take a look at the problem (and its solution) in the Contests/Resources section [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2006&p=449121]here[/url].

Also, this problem is more of a level of Intermediate than of High School Basics.  :wink:[/quote]

Yes, that is the same way I did it. I wanted to know if there was a more elegant way to do it.

[quote="neelnal"][/quote][quote="nebula42"][quote="neelnal"][quote="Rzeszut"]
[hide="The answer is"]$24$[/hide][/quote]

That is just how many zeros are in 100!. It is not the answer. The answer is what I got: 1109.

100! is easy. I was looking for a way to do 1!2!3!...99!100![/quote]
I remember this problem.  :)
I think my solution went something like this:
[hide]All of the 96 terms from 5! to 100! have a factor of 5 (as in, 5 appears once in their "factorial expansion", such as $12! = 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot \fbox{5}\cdot 4 \cdot 3 \cdot 2 \cdot 1$). Similarly, 91 terms have a 10, 86 terms have a 15, etc. Each of these numbers contributes at least one factor of 5 (some contribute more; we'll get to those later), so for now our sum is
$96+91+86+\dots+6+1=970$. However, 25, 50, 75, and 100 each contribute two factors of 5, so we have to add 76 (for 25), 51 (for 50), 26, and 1 in again to make our final sum \[970+76+51+26+1=\fbox{1124}.\] [/hide]
You can take a look at the problem (and its solution) in the Contests/Resources section [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2006&p=449121]here[/url].

Also, this problem is more of a level of Intermediate than of High School Basics.  :wink:[/quote][quote="neelnal"]

Yes, that is the same way I did it. I wanted to know if there was a more elegant way to do it.[/quote]

It was easiest to understand though...but yours was the same!
Thanks! I feel smarter! Please post more of these enlightening problems here!

[quote="neelnal"][quote="Rzeszut"]
[hide="The answer is"]$24$[/hide][/quote]

That is just how many zeros are in 100!. It is not the answer. The answer is what I got: 1109.

100! is easy. I was looking for a way to do 1!2!3!...99!100![/quote]
Yes, you're right  :wallbash_red:  :wallbash_red:  :wallbash_red: . I idiotically misread the problem and I noticed it after turning off my computer, so it was to late to delete my quasi-solution.

[quote="neelnal"]How many zeros does

1!2!3!4!...98!99!100!

have?
[/quote]
When you say "zeros", I think you mean "terminal zeros": how many zeros the number ends with.

[quote="nebula42"][quote="neelnal"][quote="Rzeszut"]
[hide="The answer is"]$24$[/hide][/quote]

That is just how many zeros are in 100!. It is not the answer. The answer is what I got: 1109.

100! is easy. I was looking for a way to do 1!2!3!...99!100![/quote]
I remember this problem.  :)
I think my solution went something like this:
[hide]All of the 96 terms from 5! to 100! have a factor of 5 (as in, 5 appears once in their "factorial expansion", such as $12! = 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot \fbox{5}\cdot 4 \cdot 3 \cdot 2 \cdot 1$). Similarly, 91 terms have a 10, 86 terms have a 15, etc. Each of these numbers contributes at least one factor of 5 (some contribute more; we'll get to those later), so for now our sum is
$96+91+86+\dots+6+1=970$. However, 25, 50, 75, and 100 each contribute two factors of 5, so we have to add 76 (for 25), 51 (for 50), 26, and 1 in again to make our final sum \[970+76+51+26+1=\fbox{1124}.\] [/hide]
You can take a look at the problem (and its solution) in the Contests/Resources section [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2006&p=449121]here[/url].

Also, this problem is more of a level of Intermediate than of High School Basics.  :wink:[/quote]
my solution was similar to that...
and just so you know, this was given as a homework problem to a mathcounts team(I'm not saying anything about their ability to solve it though)  :wink:


A digital, 12-hour clock shows hours and minutes. During what fraction of the day will the clock show the digit 1 in its display? Express your answer as a common fraction.

It will show 1 in these times:
01:00-01:59 - 60 min
10:00-11:59 - 120 min
In every other hour it's shown in 15 minutes, so in 12 hours it's shown for $ 15\cdot 9\plus{}180\equal{}315$
Fraction: $ \frac{315}{720}\equal{}\frac{7}{16}$

It's $ \boxed{\frac12}$. This is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1420064#1420064]2009 AMC 12B #10[/url].

math...unless i'm reading wrong, these are different. the other one says that the clock is messed up and displays 9s instead of 1s, and asks for what fraction of time the time displayed is correct. this one asks for a normal clock, what graction of time a 1 is displayed.

Ah, yes. But the answer is still the same since it is just $ 1 \minus{} \frac12 \equal{} \boxed{\frac12}$ (i.e. the original question is asking when ones appear and the one in the link is asking when ones wouldn't appear normally).

The answer is 1/2, everything that Bugi is right, but Bugi forgot to count the 12:00-12:59 interval, adding an extra 60 min, and subtracting 15 min from the "15 minutes every other hour" part of his expression, making the expression 15*8+240=360. Hence, 360/720 equals 1/2.


A long cylindrical conductor of radius R carries a current I as shown in Figure 1. The
current density J, however, is not uniform over the cross section of the conductor but is
a function of the radius according to J = br, where b is a constant. Find an expression
for the magnetic eld B (a) at a distance r1 < R and (b) at a distance r2 > R, measured
from the axis.

Ampere's circuital law again; for (b) you can reduce the current ($ \;I\;\equal{}\;{2\over3}b\;R^3\;$)
to the axis; for (a), Integrate from 0 to r1.


algebra
polynomial
calculus
integration
algebra unsolved

Let $ P(x),Q(x)\in Z[x]$. If $ P(x)$ is monic, prove that there is another monic polynomial $ R(x)\in Z[x]$ such that $ P(x)|Q(R(x))$

What about $ P(x)\equal{}x$ and $ Q(x)\equal{}x^{2}\plus{}1$¿ If such $ R(x)$ would exist, it would imply an integral root of $ Q(x)$ which doesn't exist.

[quote="TTsphn"]A problem is
$ P(x)\in R[x]$
Prove that exist $ f(x)\in R[x]$ and
$ P(x)|P(f(x))$[/quote]
Obvioskly $ f(x)\equiv x$ is solution.

I am sorry.exist an $ m\in N$
But for each $ n\in N,n>m$ there exist an polynomial $ f(x)$with degree n has this property.

I think, that it is not true. Let $ P(x)=(x-x_1)(x-x_2)(x-x_3)(x{}_x{}_4)$, then must be $ f(x_i)=x_j$. I can choose $ x_i$, suth that for all $ 4^4-1$ variants $ f: S\to S, f(x)\not \equiv x, \ S=\{x_i\}$ Lagranjev's polinom had degree 3. In these case $ deg(f)\not =2$.

Yes, but $ deg(xP(x)\plus{}x)\equal{}5>2$ for these case.

Yes but I have edit my post.
Exist $ m\in N$ for all $ n > m$
A consequence:
$ P(x)\in Z[x]$
Prove that exist $ f(x)\in Z[x]$ such that $ P(f(x))$ is reducible on $ Z[x]$

It is obviosly true for $ n\equal{}0,1$ and $ n>deg(P)\minus{}2$. But it is not good problem.


Perhaps you  want to be more specific about what you mean.  Any figure is the locus of points which lie on that figure.  What sort of tools would you like to limit us to?

A circle can be described as the set of all points equidistant from a certain point.  Can a square be stated like that?

Depends on what you allow to use.
So what is allowed¿
(otherwise I just define the function $poly(x)$ that is $1$ inside the polygon and $0$ outside)

Can you say something beginning with "The set of all points wit the property..." and nowhere mentioning the polygon we are trying to create (this would be circular reasoning).

..to lie inside the halfplanes given by the sides¿

By what sides?  You must define these sides first!

Give me a square.  You ask me, "Can you describe this as a locus?"  As ZetaX and I said, there are very simple ways: "The set of all points which lie on the square," for instance, describes that square as a locus.  There is no natural limitation on the tools we can use in this process, so your question (to be interesting) needs to include limitations on what we can do.


geometry
3D geometry
prism
analytic geometry
IMO Longlist
IMO Shortlist

A regular triangular prism has the altitude $h,$ and the two bases of the prism are equilateral triangles with side length $a.$ Dream-holes are made in the centers of both bases, and the three lateral faces are mirrors. Assume that a ray of light, entering the prism through the dream-hole in the upper base, then being reflected once by any of the three mirrors, quits the prism through the dream-hole in the lower base. Find the angle between the upper base and the light ray at the moment when the light ray entered the prism, and the length of the way of the light ray in the interior of the prism.

Only a sketch of a solution.

In my opinion they should have asked for two angles: the slope of the ray wrt the upper base, of course, but also the slope of the horizontal component of the ray wrt one of the sides of the triangle.

Let's call ABC the upper base of the prism and let  O be its center.
A horizontal section of the half ray trajectory depicted in the fig. 1. Where S'2 is the symmetric of S2 (midpoint of BC) wrt AB.
Of course the section of the trajectory is symmetric wrt a median of the triangle ABC.  It holds that OS1+S1S2= OS'2 which is the ipotenuse of the rectangular triangle S'2BO then OS'2= (7/3)^.5 a/2.

Now, if you "open" the planes of the trajectory (see fig. 2) you have, considering the proprierty of ray reflecting, that the the trajectory "flattened" is a stright line from O to O'.

Then slope of the ray is h/(2OS'2) = h(3/7)^.5/a.


analytic geometry
graphing lines
slope
rotation
trigonometry

We have a slope with its length long enough. We rotate a solid cyllinder of radius $R$ and mass $m$  with angular velocity of $\omega_{0}$ around its axis and carefully place it at a distance $s$ from the bottom of the slope, as shown on the figure. the incline angle of the slope is $\alpha$. The coeffincient of kinetic and static friction between the cylinder and the slope is $\mu$. How far from the bottom of the slope will the cyllinder move and with what speed will it reach the bottom ?

My result:
[hide]
1) $\mu>\tan\alpha$
$s_{max}=s+\frac{(\omega_{0}R)^{2}(\mu\cos\alpha-\sin\alpha)(3\mu\cos\alpha-3\sin\alpha+2)}{4g(3\mu\cos\alpha-\sin\alpha)^{2}}$
$v=\sqrt{\frac{4}{3}s_{max}g\sin\alpha}$

2)$\frac{1}{3}\tan\alpha<\mu\leq\tan\alpha$
$s_{max}=s$

$l_{0}=\frac{(\omega_{0}R)^{2}(\sin\alpha-\mu\cos\alpha)}{2(3\mu\cos\alpha-\sin\alpha)^{2}}$

a)$l_{0}>s$
$v=\sqrt{2sg(sin\alpha-\mu\cos\alpha)}$

b)$l_{0}\leq s$
$v=\sqrt{(\frac{\omega_{0}R(\sin\alpha-\mu\cos\alpha)}{3\mu\cos\alpha-\sin\alpha})^{2}+\frac{4}{3}g(s-l_{o})\sin\alpha}$

3) $\mu\leq\frac{1}{3}\tan\alpha$
$s_{max}=s$
$v=\sqrt{2sg(sin\alpha-\mu\cos\alpha)}$

I hope I didn't complicate the solution too much  :)
[/hide]

1) Friction force $>$ component of gravity force parallel to the slope (at first the cyllinder goes up the slope and than goes down)
2) Friction force $\leq$ component of gravity force parallel to the slope and
the angular acceleration of the cyllinder $>$ linear acceleration / $R$ (the cyllinder goes down the slope)
$l_{0}$ - the distance covered by the cyllinder until it stops sliding
a) the cyllinder doesn't stop sliding before reaching the bottom of the slope
b) the cyllinder stops sliding before reaching the bottom of the slope
3) Friction force $\leq$ component of gravity force parallel to the slope
and the angular acceleration of the cyllinder $\leq$ linear acceleration / $R$ (the cyllinder goes down the slope and it never stops sliding)

I got a differant answer for them. could you right the solution too?

Did you get all of them different :?:

I checked my results once again and found some mistakes:
[hide]
1) $\mu>\tan\alpha$
$s_{max}=s+\frac{(\omega_{0}R)^{2}(\mu\cos\alpha-\sin\alpha)}{4g\sin\alpha(3\mu\cos\alpha-\sin\alpha)}$
$v=\sqrt{\frac{4}{3}s_{max}g\sin\alpha}$

2)$\frac{1}{3}\tan\alpha<\mu\leq\tan\alpha$
$s_{max}=s$

$l_{0}=\frac{(\omega_{0}R)^{2}(\sin\alpha-\mu\cos\alpha)}{2g(3\mu\cos\alpha-\sin\alpha)^{2}}$

a)$l_{0}>s$
$v=\sqrt{2sg(sin\alpha-\mu\cos\alpha)}$

b)$l_{0}\leq s$
$v=\sqrt{\frac{(\omega_{0}R)^{2}}{3}+\frac{4}{3}g(s\sin\alpha-\mu l_{o}\cos\alpha)}$

3) $\mu\leq\frac{1}{3}\tan\alpha$
$s_{max}=s$
$v=\sqrt{2sg(sin\alpha-\mu\cos\alpha)}$
[/hide]

I'll post my solution a bit later.

My solution:
[hide]
When the cylinder is placed on the slope there are two forces parallel to the slope acting on it: $mg\sin\alpha$ and kinetic friction force $T$ acting in the opposite direction. Of course $T=\mu N=\mu mg\cos\alpha$. Consider the first case:
1)$T>mg\sin\alpha$ which we can rewrite as $\mu>\tan\alpha$
From second Newton's law we get the linear and angular acceleration:
$ma=mg(\mu\cos\alpha-\sin\alpha)$ and $\frac{1}{2}mR^{2}\epsilon=-\mu mg\cos\alpha R$
If $t_{1}$ is the time after which the cylinder stops sliding, then $at_{1}=(\epsilon t_{1}+\omega_{0})R$. Solving that equation for $t_{1}$ we can obtain the velocity of the cylinder when it stops sliding $v(t_{1})=at_{1}=\omega_{0}R\frac{\mu\cos\alpha-\sin\alpha}{3\mu\cos\alpha-\sin\alpha}$ and the distance covered by the cylinder until this moment $s_{1}=\frac{1}{2}at_{1}^{2}=\frac{(\omega_{0}R)^{2}(\mu\cos\alpha-\sin\alpha)}{2g(3\mu\cos\alpha-\sin\alpha)^{2}}$. The distance covered by the cylinder from this moment until it stops can be calculated from energy conservation:
$\frac{\frac{1}{2}mR^{2}\omega(t_{1})^{2}}{2}+\frac{mv(t_{1})^{2}}{2}=\frac{3}{4}mv(t_{1})^{2}=mgs_{2}\sin\alpha$
$s_{2}=\frac{3(\omega_{0}R)^{2}(\mu\cos\alpha-\sin\alpha)^{2}}{4g\sin\alpha(3\mu\cos\alpha-\sin\alpha)^{2}}$
And finally: $s_{max}=s+s_{1}+s_{2}=s+\frac{(\omega_{0}R)^{2}(\mu\cos\alpha-\sin\alpha)}{4g\sin\alpha(3\mu\cos\alpha-\sin\alpha)}$
The velocity of the cylinder at the bottom of the slope is obtained from energy conservation:
$\frac{\frac{1}{2}mR^{2}\omega^{2}}{2}+\frac{mv^{2}}{2}=\frac{3}{4}mv^{2}=mgs_{max}\sin\alpha$ so $v=\sqrt{\frac{4}{3}s_{max}g\sin\alpha}$

Now consider $T\leq mg\sin\alpha$ which we can rewrite as $\mu\leq\tan\alpha$
The forces and linear and angular accelerations are the same as in 1) but this time the cylinder goes down the slope from the start ($a=g(\sin\alpha-\mu\cos\alpha)$). Of course $s_{max}=s$.
Now if $a+\epsilon R <0$ then the cylinder will stop sliding after some time and if $0\leq a+\epsilon R$ then the cylinder will never stop sliding.

2) $a+\epsilon R <0$ which we can rewrite as $\mu>\frac{1}{3}\tan\alpha$
The distance covered by the cylinder until the moment it stops sliding is counted in the same way as in 1): $l_{0}=\frac{(\omega_{0}R)^{2}(\sin\alpha-\mu\cos\alpha)}{2g(3\mu\cos\alpha-\sin\alpha)^{2}}$
a) If $s<l_{0}$ then the cylinder won't stop sliding before reaching the bottom of the slope, so its velocity can be obtained from $v=at$ and $s=\frac{1}{2}at^{2}$:
$v=\sqrt{2sg(sin\alpha-\mu\cos\alpha)}$
b) If $l_{0}\leq s$ then the cylinder will stop sliding before reaching the bottom of the slope. The velocity can be obtaied from energy conservation:
$\frac{\frac{1}{2}mR^{2}\omega^{2}}{2}+\frac{mv^{2}}{2}=\frac{3}{4}mv^{2}=\frac{\frac{1}{2}mR^{2}\omega_{0}^{2}}{2}+mgs\sin\alpha-Tl_{0}$
Putting $T=\mu mg\cos\alpha$:
$v=\sqrt{\frac{(\omega_{0}R)^{2}}{3}+\frac{4}{3}g(s\sin\alpha-\mu l_{o}\cos\alpha)}$

3) $0\leq a+\epsilon R$ which we can rewrite as $\frac{1}{3}\tan\alpha\leq\mu$
The the cylinder will never stop sliding so the velocity is calculated as in 2a): $v=\sqrt{2sg(sin\alpha-\mu\cos\alpha)}$.
[/hide]

What if the cylinder starts pure rolling when we put it on the slope?

Is it possible :?: The cylinder has got an initial angular velocity and hasn't got a linear velocity so the point of the cylinder touching the slope moves with a velocity $\omega_{0}R$ against it - the cylinder is sliding. The cylinder starts pure rolling when we put it on the slope only when the coeffincient of kinetic friction goes to infinity ($\mu>>1$) but it wasn't said in the problem.

Yes you are right I didnt think of it thoughrolly. :D  :oops:


How many zeroes are there in the product of $2^4$ and $5^5$.  NO CALCULATORS! :)

[hide]One 0 for every 10. 10s are limited to the number of 2s availble.

4 zeros[/hide]

[hide]
4 zeros, because it is the number of 2s here in this problem that matter.
[/hide]

[hide]the problem can be changed to 10^4 x 5 because of the third law of exponents so the answer is [b]4 [/b]and the total answer is 50000.[/hide]

[hide]
One 0 for each 2*5=10
So 4[/hide]


See A proof of the previous from me：

[url]http://www.aoshoo.com/bbs1/dispbbs.asp?boardid=43&Id=15537[/url]

It must have another way.

It can also be seen in http://forum.cnool.net/topic_show.jsp?id=3645911&oldpage=1&thesisid=494&flag=topic1


search function AoPSwiki algebra polynomial

Can anybody explain what are Newton's sums and how can you use it to solve a problem?

Try the [url=http://www.artofproblemsolving.com/Forum/search.php]search function[/url] or [url=http://www.artofproblemsolving.com/Wiki/index.php/Newton_sums]AoPSWiki[/url]

I never bothered to learn it. It looked complicated :|

they are not that complicated, with a little practise, its easy to master them. I will post something more helpful soon.

Try going through polynomials by Adeel Khan (thats where i learnt it), very useful.


geometry
circumcircle
cyclic quadrilateral
geometry solved

Given a non-right triangle $ABC$. The altitudes $BB'$, $CC'$ cuts each other at $H$. $K$ is the midpoint of $HA$, $I$ is the intersection of $B'C'$ and $AH$. Prove that $I$ is the orthocenter of the triangle $BCK$.

I think it's rather easy :
  K is the orthocenter of IBC ==> I is the orthocenter of BCK
   We will prove  K is the orthocenter of IBC
   We have $ KB^2 = P_{B/(AC'B'H)} + P_{K/(AC'B'H)} $
  Similarly :  $ KC^2 = P_{K/(AC'B'H)} + P_{C/(AC'B'H)} $
                 $ CB^2 = P_{B/(AC'B'H)} + P_{C/(AC'B'H)} $
  ===> K is the orthocenter of IBC

It's even easier! Since < HB'A = 90 and < HC'A = 90, the points B' and C' lie on the circle with diameter HA; the center of this circle is clearly the midpoint K of the segment HA. Hence, we see that the quadrilateral AB'HC' is a cyclic quadrilateral with circumcenter K. Since the triangle BCI is the diagonal triangle of this cyclic quadrilateral, it follows that this triangle BCI is autopolar with respect to the circumcircle of that cyclic quadrilateral, and therefore, the orthocenter of the triangle BCI must be the circumcenter of our cyclic quadrilateral. In other words, the orthocenter of triangle BCI is the point K. This means that the orthocenter of triangle BCK is the point I.

$\blacksquare$

  darij

I can't understand your sollution!  :?: Can you explain it more clearlier?

I will abbreviate the point of intersection of two lines g and h as $g\cap h$.

For any cyclic quadrilateral ABCD, consider the three points $AB\cap CD$, $BC\cap DA$ and $AC\cap BD$. The triangle formed by these three points is called the [i]diagonal triangle[/i] of the cyclic quadrilateral ABCD. Now there is a well-known theorem stating that the diagonal triangle of a cyclic quadrilateral is autopolar with respect to the circumcircle of the quadrilateral. (A triangle is called [i]autopolar[/i] with respect to a circle if the sidelines of the triangle are the polars of the opposite vertices with respect to the circle.) Moreover, it is well-known that if a triangle is autopolar with respect to a circle, then the orthocenter of this triangle is the center of the circle.

In your problem, we have $C=AB^{\prime}\cap HC^{\prime}$, $B=B^{\prime}H\cap C^{\prime}A$ and $I=AH\cap B^{\prime}C^{\prime}$. Thus, the triangle BCI is the diagonal triangle of the cyclic quadrilateral AB'HC'. Hence, the triangle BCI is autopolar with respect to the circumcircle of the cyclic quadrilateral AB'HC', and thus, the orthocenter of triangle BCI is the circumcenter of the cyclic quadrilateral AB'HC', i. e. the point K.

Do you understand it now?

  Darij

And this is my sollution!
Call $A'$ is the objection of $A$ on $BC$.
We have $\angle KB'C' = 180 - \angle ACK = \angle AA'B'$
=> $\triangle KB'I \sim \triangle KA'B'$
=> $KB' = KI.KA' = (KA' - IA').KA' = KA' - A'I.A'K$
=> $A'I.A'K = KA' - KB' = A'H.A'A = A'B.A'C$
=> $I$ is the orthocenter of the triangle $BCK$.


Does anyone know of a good program to view PostScript files? (This really has nothing to do with AoPS, but I figured many of you may know.) Happy New Year everyone!!

Get the Ghostscript shell and then run Ghostview on top of it.


The number of terms in an A.P. is even. The sums of the [b]odd [/b]and [b]even[/b] numbered terms are [b]24[/b] and [b]30[/b] respectively. If the last term exceeds the first term by 10.5, find the number of terms in the A.P.

[hide]

If the number of terms is n, then there are n/2 pairs of consecutive terms (one even and one odd) that share a common difference. That common difference is expressed as 10.5/(n-1). All those differences together must equal (30-24)=6. So,

(n/2) * 10.5/(n-1)  =6

Solving for n, n=8 and the sequence is:

1.5, 3, 4.5, 6, 7.5, 9, 10.5, 12[/hide]

[hide="I think I made it too complex"]
let $n$ be the number of terms in the AP
let $a$ be the first term
let $r$ be the common difference
$a, a+r, a+2r, ..., a+(n-1)r$
The difference between the sum of the even numbered terms and the sum of the odd is $6$. It is also $\frac{nr}{2}$.
$\frac{nr}{2} = 6$
$nr = 12$
The difference between the $n$th term and the first term is 10.5
$a+(n-1)r - a = \frac{21}{2}\\ nr - r = \frac{21}{2}\\ 12 - r = \frac{21}{2}\\ r = \frac{3}{2}\\$
Since $nr=12$
$\frac{3n}{2} = 12$
$n=8$
There are $8$ terms
[/hide]

[hide]Since there are an even number of terms, then there are just as many odd number terms as even number terms.

Let the common difference be $d$ and the number of terms be $n$.
We have the equation: $2\left(\frac{6}{d}\right)=n$ because $\frac{6}{d}$ is the difference of the odd and even terms divided by the common difference making it the number of pairs of numbers, so then you multiply by 2.

We also have $(n-1)d=10.5$ because each term (except for the 1st) minus the term before it times the difference equals the range.

After simplifying both of them and solving, you get $n=8$.

This looks a lot longer than it really is.  I'm just very bad a explaining stuff.  :D [/hide]


algebra
polynomial
function
superior algebra
superior algebra solved

hi, i am preparing for the algebra exam i have in one week (many people here have helped me already  ;) )

1)i was studying very algebraic proofs of the fundamental theorem of algebra

is there a fact like this : if L is galois over K , and K is galois over M
then L is galois over M?

(so the splitting field of my polynomial over C would also be galois over R?)

2) let F2(x,y) be the field of rational functions in two variables over F2(
the field with two elements)

show that if you add a root of sqrt(x), and then of sqrt(y)
your degree of the original field is four

how do you do that , i'd have to show that
a*a+b*b*x=y with a and b rational functions in F2 over two variabels is
impossible

the goal is finding a nonseparable extension with infinitely many subfields

fred

1. NO
$\mathbb{Q}\subset \mathbb{Q}(\sqrt 2)\subset\mathbb{Q}(\sqrt [4]{2})$ (edited here).
The intermediary extensions are Galois but the greatest extension is not normal so it cannot be Galois.

2. Lemma: Let $K$ be a field then $X\not\in K(X^2)$.
Proof: Assume $X=\frac {P(X^2)}{Q(X^2)}$. Then $XQ(X^2)=P(X^2)$. On the left side we have a polynomial of odd degree in $X$ and on the right a polynomial of even degree, making the equalitu impossible.

$F_2(X,Y)=F_2(Y)(X)\subset F_2(Y)(\sqrt X)$. The degree of the extension is $2$ because $\sqrt X$ is a root of $T^2-X$ over $F_2(X,Y)$ and because of the lemma aplied to $K=F_2(Y)$.

$F_2(Y)(\sqrt X)=F_2(\sqrt X)(Y)\subset F_2(\sqrt X)(\sqrt Y)$ and the degree of the extension is, same as before but with $K=F_2(\sqrt X)$, again $2$.
So $[F_2(\sqrt X,\sqrt Y): F_2(X,Y)]=4$.

I don't think that's what you meant on #1, amfulger; $\sqrt{4}$ is rational. Should that be $\sqrt[4]{2}$ instead?

The field of real constructible numbers is generated only by taking square roots, which are all individually Galois extensions. The whole field has infinite degree over $\mathbb{Q}$ and no nontrivial automorphisms; as far from Galois as possible.

On the other hand, the smallest field $F$ containing $\mathbb{Q}$ with the additional property that a sum of squares of elements in $F$ is a square in $F$ is Galois. This is the power of the line-circle intersection axiom in Euclidean geometry.


geometry
trigonometry
integration
analytic geometry
function
calculus
calculus computations

Hi all,

I'm having trouble with the following questions:

1.) Find the length of the polar curve r = max(1-cos(theta), 2+2sin(theta)) for
theta between 0 and 3pi/2

2.)Find the area of the region enclosed by the polar curve r = min(1-cos(theta), 2+2sin(theta)) for theta between 0 and 3pi/2

Also, I could only find 3 points of intersection for the curves r = 1-sin(theta) and
r = 1 +3sin(theta), but there should be 5. (my points are (1,0), (0,0) and (1, pi) ).

Really hope someone could help me with this!
Thank you!

an easy question.

1. first equate $ 1\minus{}\cos\theta\equal{}2\plus{}2\sin\theta$ to find wherever $ r$ is the [i]same[/i] for the [i]same [/i]$ \theta$. and so, you essentially need to solve for $ \theta$ in both $ \cos\,\frac{\theta}{2}\equal{}0\;\;\;\quad\text{and}\;\;\;\quad\tan\,\frac{\theta}{2}\equal{}\minus{}\frac{1}{2}$

from the first one you get $ \theta\equal{}\pi$, and from the second one you get $ \theta\equal{}2\left(\pi\minus{}\cot^{\minus{}1}\,2\right)$ which lies somewhere between $ \dfrac{3\pi}{2}$ and $ 2\pi$

therefore, you have, in general,
$ 2\plus{}2\sin\theta\;>\;1\minus{}\cos\theta\;\;\;\quad\text{for}\;\;\;0\leq\theta<\pi,$
$ 1\minus{}\cos\theta\;>\;2\plus{}2\sin\theta\;\;\;\quad\text{for}\;\;\;\pi<\theta<2\left(\pi\minus{}\cot^{\minus{}1}\,2\right)$
and
$ 2\plus{}2\sin\theta\;>\;1\minus{}\cos\theta\;\;\;\quad\text{for}\;\;\;2\left(\pi\minus{}\cot^{\minus{}1}\,2\right)<\theta<2\pi$

you are interested in $ \theta\in\left[0,\dfrac{3\pi}{2}\right]$

so, you want
$ L\;\equal{}\;\int_0^\pi\;\;\sqrt{\;\left(2\plus{}2\sin\theta\right)^2\;\plus{}\;\left(\;\dfrac{d}{d\theta}\left(2\plus{}2\sin\theta\right)\;\right)^2\;}\;\;d\theta$
$ \plus{}\;\int_\pi^{\dfrac{3\pi}{2}}\;\;\sqrt{\;\left(1\minus{}\cos\theta\right)^2\;\plus{}\;\left(\;\dfrac{d}{d\theta}\left(1\minus{}\cos\theta\right)\;\right)^2\;}\;\;d\theta$

hopefully, you can take it from here.

2. by a similar argument, you want
$ A\;\equal{}\;\frac{1}{2}\int_0^\pi\;\;\left(1\minus{}\cos\theta\right)^2\;\;d\theta\;\plus{}\;\frac{1}{2}\int_\pi^{\dfrac{3\pi}{2}}\;\;\left(2\plus{}2\sin\theta\right)^2\;\;d\theta$

3. regarding the [i]true [/i]meaning of the word [b]intersection[/b], there are only $ 2$ points of intersection of $ 1\minus{}\sin\theta$ and $ 1\plus{}3\sin\theta$ --- they are at $ (1,0)$ and $ (1,\pi)$, period. note: $ (0,0)$ cannot be a point of intersection! a point of intersection (in polar coordinates) means $ r$ is the [i]same [/i]for the [i]same [/i]$ \theta$.

besides $ (1,0)$ and $ (1,\pi)$, you will see $ 2$ more points where the $ 2$ curves [i]cross[/i], one of them located in what you might call the 1st polar quadrant, and the other in the 2nd polar quadrant. but here, you have the [i]same [/i]value of $ r$ for $ 2$ different values of $ \theta$ regarding the $ 2$ curves, not for the [i]same [/i]$ \theta$. so, we cannot really call them points of intersection. they are merely points where the $ 2$ curves happen to [i]cross [/i](albeit at $ 2$ [b]different [/b]values of $ \theta$).

regarding the point in the so called 1st polar quadrant, the crossing happens for $ 1\minus{}\sin\theta$ somewhere between $ 0$ and $ \frac{\pi}{2}$, and for $ 1\plus{}3\sin\theta$ somewhere between $ \pi$ and $ \dfrac{3\pi}{2}$....... whereas regarding the other point in the so called 2nd polar quadrant, the crossing happens for $ 1\minus{}\sin\theta$ somewhere between $ \frac{\pi}{2}$ and $ \pi$, and for $ 1\plus{}3\sin\theta$ somewhere between $ \dfrac{3\pi}{2}$ and $ 2\pi$.

[b]ALWAYS [/b]equate the equations involving the $ 2$ $ r$'s to find wherever $ r$ is the [i]same[/i] for the [i]same [/i]$ \theta$.

I'll disagree with misan there- we're talking about curves in the plane that happen to be represented in polar form, not about functions $ r(\theta)$. The other points of intersection are real, and they come from the nonuniqueness of polar representation: $ (r_1,\theta_1)$ and $ (r_2,\theta_2)$ represent the same point if $ r_1 \equal{} r_2$ and $ \theta_1 \equal{} \theta_2 \plus{} 2n\pi$, $ r_1 \equal{} r_2 \equal{} 0$, or $ r_1 \equal{} \minus{} r_2$ and $ \theta_1 \equal{} \theta_2 \plus{} (2n \plus{} 1)\pi$. (for some integer $ n$)


I just wanted to give a heads up to people here at AoPS that within the last 24 hours, U.S. News posted their annual [url=http://colleges.usnews.rankingsandreviews.com/college/national-search/]college rankings[/url] on their site. If you have any questions or comments, this thread would probably be a good place to ask/comment.

In addition, U.S. News also made some [url=http://colleges.usnews.rankingsandreviews.com/college]other rankings[/url] for colleges and universities.

Interesting.  :P
[url=http://finance.yahoo.com/college-education/article/105598/Best-Colleges-2009]Propaganda.[/url]

EDIT Spelling mistake.

ehh, what exactly are you trying to say about them? the rankings don't mean much, sure, but I would say it does an overall good having them, considering the colleges generally try to improve (such as Harvard, as shown in your article) in order to gain ranking positions. that is a good thing, right? oh, and you spelled propaganda wrong. however, it is a nice place to see tuition costs and those sort of things all in one convenient place. and nice article, it was kind of interesting to read.

My institution did make the front page of [url=http://colleges.usnews.rankingsandreviews.com/college/masters-west-search]this list.[/url] It seems pretty odd to compare us to so many places that are small, private, and many times more expensive.

[quote="AstroPhys"]I was being sarcastic because the rankings really do not mean anything; even if the school is in the top ten, it does not mean it will suit everyone.  Personally, I think U.S. News just puts out that list as a gimmick to get anxious, paranoid parents to buy a copy.  Does it reveal good schools? Yes, of course it does, but it is not as personalized as searching for oneself.  :P It should not be a the ultimate basis of deciding where to accept.[/quote]

However, I do find that the business, engineering, etc. rankings are a bit more useful...


Yesterday (=April 22, 2003) was Gittes birthday. She notices that her age equals the sum of the 4 digits of the year she was born in.
How old is she?

[u][b]Her age on the 22 April, 2003 is 25 as she was born in 1978.[/b][/u]

If the year that she was born is [i]abcd[/i] then we have the following

age = 2003 - [i](a+b+c+d)[/i]

In this case we will infer that she was born in the 20th century, therefore a = 1 and b = 9.

age = 1993 - [i](c+d)[/i]

from this stage i don't know where to go.

I don't think this is possible becasue you did not give us teh year she was born in or how old she is.

Masoud Zargar

[quote]from this stage i don't know where to go.[/quote]

Y[hide]ou know more than that,  For 19xx, sum of digits = 10+c+d, so  Age = 10+c+d =  2003-(1900+10c+d)
which gives 11c+2d = 93, taking mod 11, you have $d \equiv 5 \bmod 11$ so d=8 etc...[/hide]

Could you please tell me what "flanders" means?
I guess it is a math contest..

LOL  :D

\[\left\{\textrm{Belgium}\right\}= \left\{\textrm{Flanders}\right\}\cap\left\{\textrm{Wallonia}\right\}\]

(Belgium is a country, by the way.)

Now that I look at my post from over a year ago I feel ashamed.  :blush:

Hmmm... So actually, Kurt Gödel, you mean:
\[\left\{\textrm{Belgium}\right\}\ =\ \emptyset\]

 :wink:

:D   Sorry `bout that..but usually when I see a thing in the form "..blabla.. `03" I think of a contest[especially because this is a math forum)
And Kurt Godel, I guess you mean reunion, not intersection :D
And oh..I knew it was a country,lol :lol:

Flanders is the region organising the contest. Just like some people say China '03 instead of CMO '03. :)

ya i did it similar to GYAN..except the $\mod$ part i just did some educated guessing and checking..looking at GYAN's solution i think its $2d\equiv{5}\mod{11}\Rightarrow{d}\equiv{8\mod11\therefore{d}=8}$ and the rest follows...

It is obvious that she is born in the year 19ab.
Thus,her age is 10+a+b----------(1)
Also,her age is also 2003-(1900+10a+b)=103-10a-b------------(2)
From (1) and (2),93=11a+2b.
Easy trial and error shows that a=7 and b=8.
Hence she was born in 1978 and she is 25 years old.  :D

Let the year she was born in be $\overline{abcd}$. Then we have $a, b, c, d \in \mathbb{N}$ such that
\begin{align*}
2003 - (1000a + 100b + 10c + d) &= a + b + c + d \\
\implies 1001a + 101b + 11c + 2d &= 2003
\end{align*}
Clearly, $a=1$ or $2$. If $a=2$, we have $101b+11c+2d=1$. If $b,c,d$ are all $0$, then $101b+11c+2d=0 \neq 1$. Hence at least one of $b,c,d$ is greater than $0$. We have $101b+11c+2d \geq 101(0)+11(0)+2(1) = 2 > 1$, a contradiction. Hence $a=1$ which gives
\[101b+11c+2d = 2003-1001=1002\]
Now note that Gittes' age $\leq 9+9+9+9=36 \implies \overline{abcd} \geq1967$. Since $a=1$, $b=9$, we have
\[11c+2d=1002-101(9)=93\]
Since $d\geq 9$, $93=11c+2d\geq 11c+2(9)=11c+18 \implies 11c \geq 75 \implies c \geq 7$. Additionally, $11c+2d\leq 93 \implies 11c \leq 93 \implies c \leq 8$. Hence $c=7$ or $8$. If $c=8$, $2d=93-11(8)=5 \implies d=\frac{5}{2}$, contradicting the fact that $d$ is an integer. Hence $c=7 \implies d=8$, so Gittes is born in $1978$.


[quote="tranvanluan"]
2)( a;b;c>0)
$ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a}\ge \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} ac \plus{} bc}(a \plus{} c \plus{} b)$
[/quote]
It's equivalent to $ \sum_{cyc}\left(\frac {a}{b} \minus{} \frac {b}{c}\right)^2\geq0.$
[quote="tranvanluan"]
1) (a;b;c>0) prove that
$ \frac {a^2 \plus{} 1}{b^2 \plus{} c^2} \plus{} \frac {b^2 \plus{} 1}{a^2 \plus{} c^2} \plus{} \frac {c^2 \plus{} 1}{b^2 \plus{} a^2}\ge 3$ [/quote]
Try $ a \equal{} b \equal{} \sqrt3$ and $ c\rightarrow0.$ :wink:

But the following inequality is true.
For non-negative $ a,$ $ b$ and $ c$ such that $ ab \plus{} ac \plus{} bc \equal{} 3$ prove that:
\[ \frac {4a^2 \plus{} 3}{b^2 \plus{} c^2} \plus{} \frac {4b^2 \plus{} 3}{a^2 \plus{} c^2} \plus{} \frac {4c^2 \plus{} 3}{b^2 \plus{} a^2}\ge \frac {21}{2}
\]


Let be $ a,b,c>0$ such that $ a\plus{}b\plus{}c\equal{}1$ . Show that :

$ a^3\plus{}b^3\plus{}c^3\plus{}6abc\le a^2\plus{}b^2\plus{}c^2$

We multiply RHS by $ a\plus{}b\plus{}c$ and we get

$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 6abc \leq (a^2\plus{}b^2\plus{}c^2)(a\plus{}b\plus{}c) \equal{} a^3 \plus{} b^3 \plus{} c^3 \plus{} \sum_{sym} a^2b$ so we must prove that

$ 6abc \leq \sum_{sym} a^2b$ which follows from AM-GM.

easy but really nice, the inequality is:
$ a^3\plus{}b^3\plus{}c^3\plus{}6abc\leq a^2\plus{}b^2\plus{}c^2$
$ \Leftrightarrow 1\equal{}(a\plus{}b\plus{}c)^3\leq a^2\plus{}b^2\plus{}c^2\plus{}3\sum ab(a\plus{}b)\equal{}\sum a^2 \plus{} 3\sum ab \minus{}9abc$
$ \Leftrightarrow 1\leq (a\plus{}b\plus{}c)^2\plus{} \sum ab \minus{}9abc$
$ \Leftrightarrow \sum ab\geq 9abc$ wich is $ AM\minus{}GM$


The square of a polynomial is the product of a polynomial with itself. Normally, the square has more nonzero terms than the polynomial itself. Show that this always occurs for polynomials of degrees 1, 2, and 3 having more than one term. Find a polynomial with more than one term whose square has exactly the same number of terms as the polynomial. Is it possible to find a polynomial whose square actually has fewer terms?

u'll be surprised...(some of u)

Has anyone gotten anywhere with the last question? If the answer turns out to be yes, then the result would be quite interesting indeed.

Well, I know the answer to the question since I have seen it before, so I shouldn't post an answer or solution. [hide]I can't think of an example. I think it has to have high degree to have fewer nonzero terms. (Maybe 25th degree or something? Bleh, I'm going old and forgetting solutions to problems. )[/hide]

[tex]\displaystyle{(x^4+x^3\sqrt{2}-x^2+x\sqrt{2}+1)^2=x^8+2x^7\sqrt{2}+7x^4+2x\sqrt{2}+1[/tex]

No 4th degree polynomial has the property that its square has fewer nonzero terms than it does.

I can't find a nice property that would tell you anything about this question.  However, Simon's statement that there would have to be "not very many" zero coefficients of such a polynomial is definitely true: adding 0's gives you only a little bit of leverage, towards the ends of the polynomials.  However, it limits your freedom towards the middle of the polynomial, which will probably (on average) be counterproductive.  Intuitively, putting 0's in just makes your polynomial look like a lower-degree polynomial, which would be very counter-productive.  However, you get 2 equations for every coefficient you add (every time you increase the power of the polynomial by 1, that is) which potentially should add enough freedom to in fact end up with a square polynomial with lots and lots of zeroes.

I found a similar fouth-degree polynomial:

[hide]

(x^4 + 2x^3 - 2x^2 + 4x + 4) ^2  =  x^8 + 4x^7 + 28x^4 + 32x + 16

All of the fourth-degree ones seem to be of the form

x^4 + (2k)x^3 - (2k^2)x^2 + (4k^3)x + (4k^4)

So in JBL's polynomial k = :rt2:/2

[/hide]


If $p(x)=(x+1)^p (x-3)^q=x^n a_1 +x^{n-1} a_2 \ldots$ then $n=p+q$ and $a_1 = 1$.  We have $p(x) = \left({x^p + px^{p-1} + \cdots }\right) \left({ x^q - 3q x^{q-1} + \cdots}\right)$ so $a_2 = p-3q$.  The answer is $p=3q+1$.

This seemed too easy --- did I miss something?


It is estimated that ten percent of the population is left-handed. Using this estimate, find the probability that there are at least 3 left-handers in a classroom with 30 students.

A) 0.5586
B) 0.6474
C) 0.4414
D) 0.3525

[hide]Instead of calculating the probability that the class contains 3, 4, 5.....or 30 left-handers, subtract the probability of having 0, 1, or 2 from 1.

Equal to:

1-[(30C0)(.9)^30+(30C1)((.9)^29)(.1)+(30C2)((.9)^28)((.1)^2)]

Using a calculator, the answer is

[b]A. .5886[/b][/hide]

[hide]Since it would be very hard to count the cases of 3, 4, 5... left-handers, count the cases with 0, 1, and 2 left-handers.

The probability of 0 left-handers in the room is the same as 30 right-handers which is $\left(\frac9{10}\right)^{30}$

The probability that there is 1 left-hander is $\binom{30}1\left(\frac1{10}\right)\left(\frac{9}{10}\right)^{29}$

The probability of 2 left-handers is $\binom{30}{2}\left(\frac1{10}\right)^2\left(\frac{9}{10}\right)^{28}$

That gives ~$0.4114$ but because this is 0, 1, and 2, the answer is $1-0.4114=0.5886 \implies \boxed{A}$

Notice the typo  :P
[/hide]


We consider the number
$\ A$=$\ 111...111222...222...999...999-123456789$, where the number of digits of $\ 1,2,9$ are equal whith 2003.Prove that $\ 2003$ divides $\ A$.

I don't think Getting Started is the appropriate place for proofs.

Anyone know divisibility rules for 2003?  (or its factors i didn't check if it was prime)

"Divisibility rules," such as they are, can be proven to exist for any prime in any base, but in base $10$ they are very cumbersome to use for primes larger than $5$.  What is needed is some real number theory.  :)  (And yes, it is prime.)

[hide="Hint"]Fermat's Little Theorem or Euler's Totient Theorem work well here. :) [/hide]

just find a closed form in summation and look at it in mod 2003, to simplify, use FLT on the powers of 10 and the result follows

Sorry, but I want a full solution, not only hints.Thank you!

[hide="Solution"]
Let $A = a_1 + a_2 + ... + a_9$ where

$a_1 = 111....000... - 100000000$,
$a_2 = 000...222...000... - 020000000$,

etc.  (That is, we divide out $A$ into its digit components.)  Now, there are $8 \times 2003$ zeroes in the large part of $a_1$, $7 \times 2003$ in $a_2$, etc.  We can write $111...$ (with $2003$ ones) as $\frac{10^{2003} - 1}{9}$.  Then

$a_k = k \frac{10^{2003} - 1}{9} \times 10^{(9 - k) 2003} - k \times 10^{9-k}$

Now, by Fermat's Little Theorem, $10^{2002} \equiv 1 \bmod 2003$.  We therefore write

$a_k \equiv k \times 10^{9-k} - k \times 10^{9-k} \equiv 0 \bmod 2003$,

Thus completing the proof.  QED.
[/hide]


induction
inequalities proposed
inequalities

Let $ x_1,x_2,...,x_n>0$, s.t. $ x_1\leq x_2\leq \ldots \leq x_n$. Prove that:

$ \dfrac{x_1x_2}{x_3}\plus{}\dfrac{x_2x_3}{x_4}\plus{}\ldots\plus{}\dfrac{x_nx_1}{x_2}\geq x_1\plus{}x_2\plus{}\ldots\plus{}x_n$

[quote="Inequalities Master"]Let $ x_1,x_2,...,x_n > 0$, s.t. $ x_1\leq x_2\leq \ldots \leq x_n$. Prove that:

$ \dfrac{x_1x_2}{x_3} \plus{} \dfrac{x_2x_3}{x_4} \plus{} \ldots \plus{} \dfrac{x_nx_1}{x_2}\geq x_1 \plus{} x_2 \plus{} \ldots \plus{} x_n$[/quote]
We can prove it by induction on $ n$.
If $ n\equal{}3$, then it becomes
\[ \frac{x_1x_2}{x_3}\plus{}\frac{x_2x_3}{x_1}\plus{}\frac{x_3x_1}{x_2} \ge x_1\plus{}x_2\plus{}x_3\]
which is trivial by AM-GM.
Assume that it holds for all $ n \ge 3$, we will show that it still holds for $ n\plus{}1$, that is
\[ \frac{x_1x_2}{x_3}\plus{}\cdots \plus{}\frac{x_{n\minus{}2}x_{n\minus{}1}}{x_n}\plus{}\frac{x_{n\minus{}1}x_n}{x_{n\plus{}1}}\plus{}\frac{x_nx_{n\plus{}1}}{x_1}\plus{}\frac{x_{n\plus{}1}x_1}{x_2} \ge x_1\plus{}\cdots \plus{}x_{n}\plus{}x_{n\plus{}1}\]
By the inductive hypothesis, we have that
\[ \frac{x_1x_2}{x_3}\plus{}\cdots \plus{}\frac{x_{n\minus{}2}x_{n\minus{}1}}{x_n}\plus{} \ge x_1\plus{}\cdots \plus{}x_n \minus{}\frac{x_{n\minus{}1}x_n}{x_1} \minus{}\frac{x_nx_1}{x_2}\]
It suffices to prove that
\[ \frac{x_{n\minus{}1}x_n}{x_{n\plus{}1}}\plus{}\frac{x_nx_{n\plus{}1}}{x_1}\plus{}\frac{x_{n\plus{}1}x_1}{x_2} \ge x_{n\plus{}1} \plus{}\frac{x_{n\minus{}1}x_n}{x_1} \plus{}\frac{x_nx_1}{x_2}\]
\[ \Leftrightarrow \frac{x_n(x_{n\plus{}1}\minus{}x_{n\minus{}1})}{x_1} \plus{}\frac{x_1(x_{n\plus{}1}\minus{}x_n)}{x_2} \ge \frac{x^2_{n\plus{}1}\minus{}x_{n\minus{}1}x_n}{x_{n\plus{}1}}\]
\[ \Leftrightarrow (x_{n\plus{}1}\minus{}x_{n})\left( \frac{x_n}{x_1}\plus{}\frac{x_1}{x_2} \minus{}\frac{x_n}{x_{n\plus{}1}}\minus{}1\right) \plus{}(x_n\minus{}x_{n\minus{}1})\left( \frac{x_n}{x_1} \minus{}\frac{x_n}{x_{n\plus{}1}}\right) \ge 0\]
which is true because
\[ \frac{x_n}{x_1}\plus{}\frac{x_1}{x_2} \minus{}\frac{x_n}{x_{n\plus{}1}}\minus{}1 \ge \frac{x_n}{x_1}\plus{}\frac{x_1}{x_2} \minus{}2 \ge 2\sqrt{\frac{x_n}{x_2}} \minus{}2 \ge 0\]
and
\[ \frac{x_n}{x_1} \minus{}\frac{x_n}{x_{n\plus{}1}} \ge 0\]
Our proof is completed. :)


Let $ a,b,c$ satisfy $ 1 \leq a,b,c \leq 2$, show that: $ 3a^2(b\plus{}c)\plus{}2b^2(c\plus{}a)\plus{}c^2(a\plus{}b)\leq\frac{33abc}{2}$. :P

a,b,c are sides of a possibly degenerate triangle. ...b+c-a>=1+1-2=0, etc.

Yes, $ a,b,c$ are sides of a possibly degenerate triangle, but the ineq isn't symmetric and i fell puzzled

[quote="hotboy12"]Let $ a,b,c$ satisfy $ 1 \leq a,b,c \leq 2$, show that: $ 3a^2(b \plus{} c) \plus{} 2b^2(c \plus{} a) \plus{} c^2(a \plus{} b)\leq\frac {33abc}{2}$. :P[/quote]
Denote $ F(a,b,c) \equal{}LHS\minus{}RHS$, then $ F_a (a,b,c),F_b (a,b,c),F_c (a,b,c)$ are respectively convex, hence it suffices to we have
\[ F(a,b,c) \le \max \{ F(2,1,1),F(2,2,1),F (2,1,2),F(2,2,2),F (1,2,2), F(1,2,1),F(1,1,2),F(1,1,1)\}\]
We have done.
:)


Hi everybody,
I just bought a Mac and I was wondering if anyone with another Mac or with knowledge regarding the Mac could tell me if there is a Mac compatible LaTex or something similar to it that is compatible with Macs.

Thank you all![/code]

Looks like this is one of your first posts. Welcome to AoPS!

There is a special section of the forum that is for questions about $ \text{\LaTeX}$ such as yours. Try the [url=http://www.artofproblemsolving.com/Forum/index.php?f=123]$ \text{\LaTeX}$ forum[/url] for any $ \text{\LaTeX}$ questions you might have. Also, using the search function gives [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1063236898&t=89493]this[/url], which might be of use to you. :)

(Also, the [url=http://www.artofproblemsolving.com/Wiki/index.php/Main_Page]AoPSWiki[/url]is useful, too. In particular, [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Help]this[/url].)


Find all functions $f: \mathbb R^{+} \to \mathbb R^{+}$ such that for all positive reals $x$ and $y$,
\[ f(x^3+y)=f(x)^3+\frac{f(xy)}{f(x)}.\]

[quote="Altunshukurlu"]let f;R+ -> R+  and f{x^3 + y}={f(x)}^3 + {f(xy)}/{f(x)} find all such functions for x,y>0[/quote]
Let $ P(x,y)$ be the assertion $ f(x^3\plus{}y)\equal{}f(x)^3\plus{}\frac{f(xy)}{f(x)}$
Let $ f(1)\equal{}a$

1) $ f(1)\equal{}1$
$ P(1,x)$ $ \implies$ $ f(x\plus{}1)\equal{}a^3\plus{}\frac{f(x)}a$ and so $ f(x\plus{}n)\equal{}\frac{f(x)}{a^n}\plus{}a^3\plus{}a^2\plus{}a\plus{}1\plus{}a^{\minus{}1}\plus{}...a^{4\minus{}n}$ so :
$ f(2)\equal{}a^3\plus{}1$
$ f(9)\equal{}f(1\plus{}8)\equal{}\frac{f(1)}{a^8}\plus{}a^3\plus{}a^2\plus{}a\plus{}1\plus{}a^{\minus{}1}\plus{}...a^{\minus{}4}$ $ \equal{}a^{\minus{}7}\plus{}a^{\minus{}4}\plus{}a^{\minus{}3}\plus{}...\plus{}a^2\plus{}a^3$
But $ P(2,1)$ $ \implies$ $ f(9)\equal{}f(2)^3\plus{}\frac{f(2)}{f(2)}$ $ \equal{}(a^3\plus{}1)^3\plus{}1$

So $ a^{\minus{}7}\plus{}a^{\minus{}4}\plus{}a^{\minus{}3}\plus{}...\plus{}a^2\plus{}a^3\equal{}(a^3\plus{}1)^3\plus{}a^2\plus{}a^{\minus{}1}$ $ \equal{}a^9\plus{}3a^6\plus{}3a^3\plus{}2$

So $ a^{16}\plus{}3a^{13}\plus{}2a^{10}\minus{}a^9\minus{}a^8\plus{}a^7\minus{}a^6\minus{}a^5\minus{}a^4\minus{}a^3\minus{}1\equal{}0$

So $ (a\minus{}1)(a^{15}\plus{}a^{14}\plus{}a^{13}\plus{}4a^{12}\plus{}4a^{11}\plus{}4a^{10}$ $ \plus{}6a^9\plus{}5a^8\plus{}4a^7\plus{}5a^6\plus{}4a^5\plus{}$ $ 3a^4\plus{}2a^3\plus{}a^2\plus{}a\plus{}1)\equal{}0$

And so $ a\equal{}1$ (since the other factor is $ >0$

2) $ f(u\plus{}v)\equal{}f(u)\plus{}f(v)$ $ \forall u,v>0$
$ P(1,x)$ $ \implies$ $ f(x\plus{}1)\equal{}f(x)\plus{}1$
$ P(x,y\plus{}1)$ $ \implies$ $ f(x^3\plus{}y\plus{}1)\equal{}f(x)^3\plus{}\frac{f(xy\plus{}x)}{f(x)}$ $ \equal{}f(x^3\plus{}y)\plus{}1$ and so $ f(xy\plus{}x)\equal{}f(xy)\plus{}f(x)$
Setting then $ x\equal{}v$ and $ y\equal{}\frac uv$, we get $ f(u\plus{}v)\equal{}f(u)\plus{}f(v)$

3) $ f(x)\equal{}x$ $ \forall x$
From $ f(1)\equal{}1$ and $ f(u\plus{}v)\equal{}f(u)\plus{}f(v)$ $ \forall u,v>0$, it's immediate to conclude :
$ f(n)\equal{}n$ $ \forall n\in\mathbb N$
$ f(\frac 1n)\equal{}\frac 1n$ $ \forall n\in\mathbb N$
$ f(p)\equal{}p$ $ \forall p\in\mathbb Q^\plus{}$
$ f(x)$ is non decreasing
And so $ f(x)\equal{}x$ $ \forall x\in\mathbb R^\plus{}$

And it is easy to check back that this necessary condition is sufficient ($ f(x)\equal{}x$ indeed is a solution)

Hence the unique solution : $ \boxed{f(x)\equal{}x}$

[quote="Altunshukurlu"]Thanks for the solution ...
[b]a^{3}+1=a [/b] if this equation has a root in [b]{R+} [/b] then this root can be a solution to the function[/quote]
As I previously proved : $ f(x)\equal{}x$ is the only solution. So $ f(x)\equal{}c$ cant be.

And a quick look at $ x^3\minus{}x\plus{}1\equal{}0$ shows that this equation has no positive real root.


floor function
logarithms
ceiling function

-When 3^4 is written without exponents, it is 2 digits in length. How many digits long is 3^1000 written without exponents.

-if log sub 2 [lob sub 4(log sub 3 x)]=-1

- if log sub a 5 = x, and log sub a 10 =y, then log sub a 250 is terms of x and y is?

[hide="1"]
We are trying to find $ 1 \plus{} \left\lfloor{\log_{10}3^{1000}}\right\rfloor$, which is equal to $ 1 \plus{} \left\lfloor{1000\log_{10}3}\right\rfloor$. Since $ \log_{10}3\approx0.477$, our answer is $ \boxed{478}$.
[/hide]

[hide="2"]
$ \log_2\log_4\log_3x \equal{} \minus{} 1$
$ \log_4\log_3x \equal{} \frac12$
$ \log_3x \equal{} 2$
$ x \equal{} \boxed9$
[/hide]

[hide="3"]
$ \log_a5 \equal{} x$
$ \log_a{10} \equal{} y$
$ \log_a{250} \equal{} 2\log_a5 \plus{} \log_a{10} \equal{} \boxed{2x \plus{} y}$
[/hide]

EDIT: @grn_turtle--you misread 3?

[hide="Solution to #1"]
Property: $ \left\lceil\log_{10}(x)\right\rceil$ gives the length of $ x$.
Therefore, $ \left\lceil\log_{10}(3^{1000})\right\rceil \equal{} 1000\left\lceil\log_{10}(3)\right\rceil \equal{} \boxed{478}$
[/hide]

[hide="Solution to #3"]
$ \log_{a}(5) \equal{} x$
$ \log_{a}(10) \equal{} y$
$ \therefore\log_{a}(5*5*10) \equal{} \log_{a}(5)\plus{}\log_{a}(5) \plus{} \log_{a}(10) \equal{} \boxed{2x \plus{} y}$
[/hide]

Edit: Darn, math154 beat me to it  :lol:
Edit: Fixed my #3, thanks, I did misread it as 50  :blush:


function
integration
calculus
calculus computations

Prove
\[
f(\theta)=\displaystyle \int_1^{\frac {1} {\theta}} \displaystyle \frac {1} {\sqrt {(x^2-1)(1-\theta^2x^2)}} \,dx
\]
$ \textrm {is decreasing in }  \quad  0< \theta <1 $

Another, somewhat simpler substitution that also solves the problem is $x=1+\frac{1-\theta}\theta t$ ;)

$\frac{f(\theta)}{\theta}=\int^{\frac{1}{\theta}}_1 \frac{1}{\sqrt{(\theta ^2x^2-\theta ^2)(1-\theta ^2x^2)}} dx \geqq \int^{\frac{1}{\theta}}_1 \frac{2}{1-\theta ^2} dx \geqq \frac{\frac{2}{1+\theta}}{\theta}$


geometry
trigonometry
calculus
integration
calculus computations

Let $ n\ge2$ be an integer and $ a>0$ be a real number. Define a parametric curve in the plane by

$ \left\{\begin{array}{c}x(t)\equal{}a\cos t\plus{}\frac{a}{n}\cos nt\\ y(t)\equal{}a\sin t\minus{}\frac{a}{n}\sin nt\end{array}\right\}$ for $ t\in[0,2\pi].$

Claim: this is a simple closed curve that winds counterclockwise around the region it encloses.

1. Find the area of the region enclosed.

2. Find the arclength of the curve.

3. Let $ a\equal{}\frac12$ and $ n\equal{}2.$ Show that a "needle", which is to say a line segment of length 1, can be moved using rigid motions, always remaining within the closed region, and returned to its original position with the ends exchanged.

Side note on #3: I believe, although I am not certain, that this region is the smallest (in area) [i]simply-connected[/i] region in which the needle can be turned around. Take away the condition of being simply connected, and it is most certainly not the smallest such region.

Would Green's Theorem or some sort of line integral work to find the area?

(1)

$ \forall\,n\in\mathbb{N}$ the area is:

$ S = \int^{2\pi}_{0}y(t)\,\dot{x}(t)\,dt$

where

$ \dot{x}(t) = (a\cos(t) + \frac {a}{n}\cos{n t})^{'} = - a\sin(t) - a\sin{n\,t}$

Therefore:

$ y(t)\,\dot{x}(t) = - a^2(\sin(t) - \frac {1}{n}\sin(n\,t))\cdot(\sin(t) + \sin(n\,t))$

that expands to:

$ y(t)\,\dot{x}(t) = - a^2\sin^2(t) + \sin(t)\sin(n\,t)(\frac {a^2}{n} - a^2) + \frac {a^2}{n}\sin^2(n\,t)$

Now, using the identities:

$ \int\sin^2(a\,t)\,dt = \frac {t}{2} - \frac {\sin(2\,a\,t)}{4a}$

$ \int\sin(a\,t)\sin(b\,t)\,dt = \frac {\sin((a - b)\,t)}{2(a - b)} - \frac {\sin((a + b)\,t)}{2(a + b)}$

the integral can be evaluated easily:

$ \int^{2\pi}_{0}y(t)\,\dot{x}(t)\,dt = ( - a^2(\frac {t}{2} - \frac {\sin(2t)}{4}) + (\frac {a^2}{n} - a^2)\cdot$
$ ( \frac {\sin((n - 1)\,t)}{2(n - 1)} - \frac {\sin((n + 1)\,t)}{2(n + 1)} ) + \frac {a^2}{n}(\frac {t}{2} - \frac {\sin(2nt)}{4n}) ))|^{2\pi}_{0}$

After plugging in the integration limits, and reducing the formula considering $ n \in \mathbb{N}$ we get:

$ S = a^2\pi(1 - \frac {1}{n})$

(2)

$ \forall\,n\in\mathbb{N}$ the arclength is:

$ L = \int^{2\pi}_{0}\sqrt {\dot{x}(t)^2 + \dot{y}(t)^2}\,dt$

where again

$ \dot{x}(t) = (a\cos(t) + \frac {a}{n}\cos{n\,t})^{'} = - a\sin(t) - a\sin{n\,t}$

$ \dot{y}(t) = (a\sin(t) - \frac {a}{n}\sin{n\,t})^{'} = - a\cos(t) - a\cos{n\,t}$

Therefore:

$ \dot{x}(t)^2 + \dot{y}(t)^2 = 4a^2\sin^2(\frac {(n + 1)t}{2})$

so we have:

$ L = 2a\int^{2\pi}_{0}\sqrt {\sin^2(\frac {(n + 1)t}{2})}\,dt = 2a(n + 1)\int^{\frac {2\pi}{n + 1}}_{0}\sin(\frac {(n + 1)t}{2})\,dt =$
$ = - 4\,a\cos(\frac {(n + 1)t}{2})|^{\frac {2\pi}{n + 1}}_{0} = 8a$

Hope I got it right.. (Some corrections were made after viewing kunny's post)

(1) $ x'(t) \equal{} \minus{} a(\sin t \plus{} \sin nt),\ y'(t) \equal{} a(\cos t \minus{} \cos nt)$,

$ S \equal{} \frac {1}{2}\int_0^{2\pi} \{x(t)y'(t) \minus{} x'(t)y(t)\}\ dt$

$ \equal{} \frac {1}{2}a^2\left(1 \minus{} \frac {1}{n}\right)\int_0^{2\pi} \{1 \minus{} \cos (n \plus{} 1) t\}\ dt$

$ \equal{} \frac {a^2}{2}\left(1 \minus{} \frac {1}{n}\right)\left[t \minus{} \frac {\sin (n \plus{} 1) t}{n \plus{} 1}\right]_0^{2\pi}$

$ \equal{} \pi a^2\left(1 \minus{} \frac {1}{n}\right)$.

(2) $ l \equal{} \int_0^{2\pi} \sqrt {x'(t)^2 \plus{} y'(t)^2}\ dt$

$ \equal{} 2a\int_0^{2\pi}\left|\sin \frac {n \plus{} 1}{2}t\right|\ dt$

$ \equal{} 2a\int_0^{(n \plus{} 1)\pi} |\sin u|\cdot \frac {2}{n \plus{} 1}\ du$

$ \equal{} \frac {4a}{n \plus{} 1}\cdot (n \plus{} 1)\int_0^{\pi} |\sin u|\ du$

$ \equal{} 8a.$ which is same as the arc length of a cycloid.

milin, your solution is a standard for Japanese university entrance exam for (1), because Green's theorem? doesn't see Japanese high school, it appeares freshman in university, while for (2), we can remark the periodicity of $ |\sin x|$ which is $ \pi$.

Here is Practice from Japanese university entrance exam.

[color=blue]1978 Aoyama Gakuin University entrance exam.[/color]

Evaluate

$ \int_0^{\pi} |a\cos nx \plus{} b\sin nx|\ dx\ (n \equal{} 1,\ 2,\ \cdots)$.

Thank you for the solutions to #1 and #2. There's room here for a variety of methods, and in particular, I don't think we've yet exhausted the list of reasonable ways to do #1.

Now what about #3, turning the needle? (By your solutions above, the area of that figure is $ \frac{\pi}8,$ which is exactly half of the area of a circular disk in which the needle could be turned.)

Simply-connected needle-turning sets can have arbitrarily small area (Cunningham, 1971). Star-shaped sets must have area at least $ \pi/108$ (also Cunningham), but the precise lower bound remains unknown. The best known star-shaped examples have area $ (5 \minus{} 2\sqrt {2})\pi/24 \plus{} \epsilon$ (Bloom and Schoenberg, 1965), which is not that far from $ \pi/108$. For convex sets the equilateral triangle is optimal.

[url=http://www.zentralblatt-math.org/zmath/en/search/scans.html?volume_=207&count_=209]Source[/url].  (By the way, another review on the same page describes someone's attempt at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=174806]independent research in mathematics[/url].)

Ah, I've been out of touch. The construction I know for arbitrarily small area is not simply connected. Thanks for the reference, mlok.

The equilateral triangle would have to have altitude $ 1,$ hence area $ \frac1{\sqrt{3}}\approx 0.577.$ The hypocycloid I've been discussing in this topic would have area $ \frac{\pi}8\approx 0.393,$ but it's not convex. Star-shaped, yes; convex, no.

It would be interesting to see a proof that the equilateral triangle with height $ 1$ has the smallest area among all convex needle-rotating sets. This is not trivial: a disk of diameter $ 1$, which is obviously convex and needle-rotating, does not contain such a triangle. (But its area, $ \pi/4$, is greater than $ 1/\sqrt{3}$).

Speaking of Kakeya sets: I liked the recent article by I. Laba, [url=http://www.ams.org/bull/2008-45-01/S0273-0979-07-01189-5/home.html]From harmonic analysis to arithmetic combinatorics[/url].


solve system of the equations:
$3x-y-5z-2yz=0$and$x-5y-z-2z^2=0$. and $x+9y-3z+2xz=0$

If we multiply the first equation by $-1$, the second by $2$ and add them to the third, we get $z(x+y-2z)=0$. If $z=0$, it's easily shown that $x=y=0$. Otherwise, $x=2z-y$. Plugging that in the second equation gives $y=-{{2z^2-z} \over 6}$, and with that we have $x={{2z^2+11z} \over 6}$. Plugging both $x$ and $y$ into the first equation, we get $2z^3+3z^2+z=0$. Since $z\neq 0$, this simplifies to $2z^2+3z+1=0$. Solutions to this are $z_1=-1$ and $z_2=-{1 \over 2}$, which give $(x_1,y_1)=(-{3 \over 2},-{1 \over 2})$ and $(x_2,y_2)=(-{5 \over 6}, -{1 \over 6})$. Therefore, solutions to the system are $(-{3 \over 2},-{1 \over 2},-1),(-{5 \over 6}, -{1 \over 6},-{1 \over 2}),(0,0,0)$.


induction
number theory unsolved
number theory

If $ 2^{k}\parallel{}n$, then show that $ 2^{k \plus{} 2}\parallel{}(4a \plus{} 3)^{n}\minus{}1$.

Oops, typo :oops:

you should make a correction. An even number NEVER divides an odd one.


find all integer n such that there exists an equiangular n-gon whose side lengths are distinct.

See the discussion [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=152452]here[/url].


Hi,

I want to compress two "3-digit" numbers as just one "3-digit number" and be able to sort both original 3-digit numbers out later too.

For example: 222+152=374. Now, once I get 374, is there ANY way to determine that 374 came up by adding the values 222 and 152?

Its not a problem if the final number (374 here) increases to anything betweeb 0 and 999, but it should be a 3-digit number as well.

Note:
Value 1 would be between 0 and 240 and not higher;
Value 2 would be between 0 and 255 and not higher;
So the sum of the third number would be 495 highest.

Although this might seem unusual and unpractical, but there must be some way for it? Ofcourse, I understand its complex, but I really need to do this.

Any help greatly appreciated. Thanks!

Of course not.  There are $61696$ possible inputs but only $1000$ possible outputs.


You are given two positive integers A and B. The object is to start with A and turn it into B, through a series of steps. In each step you can add, subtract, multiply by, divide by, or take the power/root of another positive integer.

Sound easy? There's another part: There is a COST for each modification, and you want to find the cheapest method to get from A to B. Let the number you are using = N.

Adding or subtracting N costs N dollars.
Multiplying or dividing by N costs N+2 dollars.
Powering or rooting with N costs 2N dollars. (You have 3 options here. If C is the current number, you can take C^N, N^C, or the Nth root of C.)

Your final total = the above costs + N dollars (where N is the number of steps in your method).

Remember-- Throughout your steps, the number must remain a positive integer. You can't subtract it to negative, or make it a fraction through division.

And lastly, you cannot multiply twice in a row or divide twice in a row.

So for example, try to get from 5 to 91.

1) Multiply by 18 (20 dollars)
2) Add 1 (1 dollar)
20 + 1 + 2 steps = 23.00

...or...

1) Add 2 (2 dollars)
2) Multiply by 13 (15 dollars)
2 + 15 + 2 steps = 19.00

I have tried several methods, and I can't get lower than 19 dollars. Can you?

Try these others:

12 to 345
64 to 729
6 to 8888
142857 to 100

...or post your own numbers for others to figure out!

Have fun! :)

(PS Let me know if there's a flaw in this game, something that makes it too easy/hard/confusing. Also if you have any ideas to make it more interesting, please say so!)

(And yes, I saw the Mathworld game in this forum too. Pure coincidence, people.)

You might want to refine your $\LaTeX$ code a bit. The rules are difficult to read.

Fixed it!

...Does this belong in this forum?
Maybe there's an off-topic-yet-still-math forum....  :blush:

There's a forum called puzzles and brainteasers, you can post this there.

I can't find any way below 19 dollars either.

i believe this is not the correct forum but i am not sure ask a mod

12 to 345:

[hide] 12+7=19
 19^2=361
 361-16=[b]345[/b]
 7+2*2+16+3=[i][b]30 dollars[/b][/i][/hide] tell me if I cheated. theres probably a better way. Oh I know!!!
 [hide=""the better answer""] 12+333=[b]345[/b]!!! Oh yeah!!! :first:  :gleam: jk lol :lol: [/hide]

My solutions for
12 to 345:
[hide]
Subtract 5, cube, add 2
 = 16.00[/hide]

6 to 8888:
[hide]
5th power, Add 1, Divide by 7, Multiply by 8
 = 34.00 [/hide]

100 to 142857:
[hide]
Cube, Subtract 1, Divide by 7
 = 19.00[/hide]

Some newer problems:

1089 to 4641 (My solution = 29.00)
45 to 8640 (My solution = 24.00)
678 to 540 (No more than 4 steps!)
240 to 220 (No subtraction allowed!)

My solutions for

1089 to 4641:
[hide]Divide by 3, Subtract 6, Multiply by 13
 = 29.00[/hide]

45 to 8640:
[hide]Multiply by 2, Add 3, Square, Subtract 9
 = 24.00[/hide]

678 to 540: (No more than 4 steps!)
[hide]subtract 3, divide by 5, multiply by 4
 = 19.00[/hide]

240 to 220: (no subtraction allowed!)
[hide]Divide by 5, Add 7, Multiply by 4
 = 23.00[/hide]


The list 1,2,4,5,8,10,11,13 $ \ldots$ contains every positive integer less than 1000 that is not a multiple of 3 or a multiple of 7. How many integers are in the list?

We'd be subtracting from 999, not 1000, since the list only contains numbers LESS than 1000 that follow the conditions. 1000 can't be in the list.

So would the answer just be $999 - 333 - 142 + 47 = \boxed{571}$?


[color=darkblue]Let $ a>0, b>0,c > 0$ and $ a.b.c\le 1$. Find minimum of $ A \equal{} \frac {bc}{a^2b \plus{} a^2c} \plus{} \frac {ca}{b^2a \plus{} b^2c} \plus{} \frac {ab}{c^2a \plus{} c^2b}$.[/color]

[color=darkblue]Let $ x \equal{} \frac {1}{a}, y \equal{} \frac {1}{b}, z \equal{} \frac {1}{c}\Longrightarrow x.y.z\ge 1$ and $ P \equal{} \frac {x^2}{y \plus{} z} \plus{} \frac {y^2}{z \plus{} x} \plus{} \frac {z^2}{x \plus{} y}$![/color]

[quote="thanhnam2902"][color=darkblue]Let $ a > 0, b > 0,c > 0$ and $ a.b.c\le 1$. Find minimum of $ A \equal{} \frac {bc}{a^2b \plus{} a^2c} \plus{} \frac {ca}{b^2a \plus{} b^2c} \plus{} \frac {ab}{c^2a \plus{} c^2b}$.[/color][/quote]

Let a,b,c $ \in [0,1]$. Prove that $ \sqrt{abc} \plus{} \sqrt{(1\minus{}a)(1\minus{}b)(1\minus{}c)} <1$

[quote="k30101201"]

Let a,b,c $ \in [0,1]$. Prove that $ \sqrt {abc} \plus{} \sqrt {(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)} < 1$[/quote]

    $ \sqrt{abc}\plus{}\sqrt{(1\minus{}a)(1\minus{}b)(1\minus{}c)}< \sqrt{ab}\plus{}\sqrt{(1\minus{}a)(1\minus{}b)}\leq\sqrt{(a\plus{}1\minus{}a)(b\plus{}1\minus{}b)}\equal{}1$

[color=darkblue]Let $ x \equal{} \frac {1}{a}, y \equal{} \frac {1}{b}, z \equal{} \frac {1}{c}\Longrightarrow x.y.z\ge 1$ and $ P \equal{} \frac {x^2}{y \plus{} z} \plus{} \frac {y^2}{z \plus{} x} \plus{} \frac {z^2}{x \plus{} y}$

$ \bullet$ $ \frac {x^2}{y \plus{} z} \plus{} \frac {y \plus{} z}{4}\ge x$

$ \bullet$ $ \frac {y^2}{z \plus{} x} \plus{} \frac {z \plus{} x}{4}\ge y$

$ \bullet$ $ \frac {z^2}{x \plus{} y} \plus{} \frac {x \plus{} y}{4}\ge z$

$ \Longrightarrow P \plus{} \frac {y \plus{} z}{4} \plus{} \frac {z \plus{} x}{4} \plus{} \frac {x \plus{} y}{4}\ge x \plus{} y \plus{} z \Longleftrightarrow P\ge \frac {x \plus{} y \plus{} z}{2}\ge \frac {3\sqrt [3]{xyz}}{2}\ge \frac {3}{2}$

$ \Longrightarrow$ Minimum of $ P \equal{} \frac {3}{2}$.[/color]

$ \sqrt {abc} \plus{} \sqrt {(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)} < \sqrt {ab} \plus{} \sqrt {(1 \minus{} a)(1 \minus{} b)}\leq \frac{a\plus{}b}{2}\plus{}\frac{2\minus{}a\minus{}b}{2}\equal{} 1$

By Cauchy-Schwarz $ P\geq \frac{x\plus{}y\plus{}z}{2}$  :wink:


calculus
ARML
function
calculus computations

Find the equation of the line which is tangent to $ y \equal{} \minus{}x^4 \plus{} 18x^3 \minus{} 97x^2 \plus{} 180x \minus{} 52$ at exactly two distinct points. Also find the points of tangency.

Here is a start (I have to go very soon :()
$ f(x) \equal{} \minus{}x^4 \plus{} 18x^3 \minus{} 97x^2 \plus{} 180x \minus{} 52$
We want points $ (x_i,y_i)$ for $ i \equal{} 1,2$ such that $ \frac{f(x_2)\minus{}f(x_1)}{x_2\minus{}x_1}\equal{}f'(x_1)\equal{}f'(x_2)$

Shouldn't be terrible. Seems just tedious and algebraic (manipulations)><

That's an ugly way to do this problem -- the "clever" way doesn't require calculus at all (and in fact essentially this question was an ARML #8 in 2002 or 2003).

[hide="Key idea"]Consider the difference of the given equation and its tangent line.[/hide]

Now how do we construct such function having this property?

[quote="JBL"]After you've solved it following my hint, how to reverse-engineer it should be pretty clear.  The result is that a quartic $ f(x) \equal{} x^4 \plus{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$ has such a tangent line if and only if there exist different real numbers $ r, s$ such that $ r \plus{} s \equal{} \minus{} \frac {a}{2}$ and $ r^2 \plus{} 4 r s \plus{} s^2 \equal{} b$.  Solving these equations for $ r, s$, one can show that this is equivalent to $ 3a^2 \minus{} 8b > 0$.[/quote]
The Tangent is y=36x+12 and the points of contact are (1, 48) and (8, 300)

[hide="Solution"]Suppose the line has equation $ y \equal{} g(x) \equal{} mx \plus{} b$, with points of tangency $ (s, t)$ and $ (u, v)$.  Then the graph of $ y \equal{} f(x) \minus{} g(x)$ has double roots at points $ (s, 0)$ and $ (u, 0)$, so it has the equation $ y \equal{} a(x \minus{} s)^2(x \minus{} u)^2 \equal{} ax^4 \minus{} 2a(s \plus{} u)x^3 \plus{} a(s^2 \plus{} 4su \plus{} u^2)s^2 \plus{} \text{other terms}$.  Note that the higher-order terms must be the same as those in $ f(x)$, so equating coefficients gives $ a \equal{} \minus{}1$, $ \minus{}2a(s \plus{} u) \equal{} 18$ and $ a(s^2 \plus{} 4su \plus{} u^2) \equal{} \minus{}97$.  Thus $ s \plus{} u \equal{} 9$ and $ s^2 \plus{} 4su \plus{} u^2 \equal{} 97$.  One can solve this system, e.g. for substituting the value of $ u$ from the first equation into the second, to find $ \{s, u\} \equal{} \{8, 1\}$.  After that, it's a simple matter to get the values vidyamanohar lists.[/hide]

[quote="JBL"]After you've solved it following my hint, how to reverse-engineer it should be pretty clear.  The result is that a quartic $ f(x) = x^4 + ax^3 + bx^2 + cx + d$ has such a tangent line if and only if there exist different real numbers $ r, s$ such that $ r + s = - \frac {a}{2}$ and $ r^2 + 4 r s + s^2 = b$.  Solving these equations for $ r, s$, one can show that this is equivalent to $ 3a^2 - 8b > 0$.[/quote]

${ x^4 + ax^3 + bx^2 + cx + d - (\frac {a(4b - a^2)}{8}x + \frac {(4b - a^2)^2}{64}})$

$ = (x + \frac {a + \sqrt {3a^2 - 8b}}{4})^2(x + \frac {a - \sqrt {3a^2 - 8b}}{4})^2$  with $ b < \frac {3}{8}a^2$.

The tangent line is $ y = \frac {a(4b - a^2)}{8}x + \frac {(4b - a^2)^2}{64}$.


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If we write $ |x^2 \minus{} 4| < N$ for all $ x$ such that $ |x \minus{} 2| < 0.01$, the smallest value we can use for $ N$ is:

$ \textbf{(A)}\ .0301 \qquad \textbf{(B)}\ .0349 \qquad \textbf{(C)}\ .0399 \qquad \textbf{(D)}\ .0401 \qquad \textbf{(E)}\ .0499 \qquad$

[hide]
We only need to check the extremes $ x=\{1.99,2.01\}$ in order to find $ N$

Since the function increases more rapidly on the right we only need to evaluate;

$ 4-(2+0.01)^2 = 2(0.01)+0.0001 = 0.0401 \longrightarrow \boxed{D}$[/hide]


linear algebra
matrix
inequalities
inequalities proposed

Give $ a_1, a_2,...,a_n > 0$ and $ \left\{\begin{matrix} a_1\plus{}a_2\plus{}a...\plus{}a_n \ge n\\ a_1^2\plus{}a_2^2\plus{}...\plus{}a_n^2 \ge n^2 \end{matrix}\right. ( n> 3)$
Prove that: $ Max\{a_1, a_2,..., a_n \} \ge 2$

WLOG, we may assume that

$ a_1 \le a_2 \le a_3 \le ...\le a_n$

then, we have

$ max\{a_1,a_2,...,a_n\} \equal{} a_n$

We will prove this inequality by contradiction as follows

Assume that $ a_n < 2$ then

$ n^2 \le a^2_1 \plus{} a^2_2 \plus{} ... \plus{} a^2_n < 4n$

It implies $ n < 4$ but initially, we have $ n > 3$ so the contradiction happens.

Therefore, $ a_n \ge2$ and we get the desired result.

The inequality is actually true for $ a_i \in \mathbb{R}$ but I cant remember the proof. :blush:


calculus
derivative
IMC
college contests

Look to [url]http://www.imc-math.org[/url] for the results of the competition:

As you can see, I (amfulger=Mihai Fulger) 've got the grand first SECOND prise.

You can disclose your secret identity here and share your results with us.

i'm with you on the second prize :( ... well there is always next year. Congrats claude from making the last first prize :)

[quote="eugene"]My results not so good, i've got third price. It was pity that i hadn't secodn prize- i was near it.[/quote]

Tell me about it.

There is not something special : before complainment i was on 66-70-th place and i think that i will have second prize, and since our team haven't  team leader i could't take part in complainment(i think that it was possible to take some scores), so after complainment i was 82-th, so i took third prize.

Myth:

If this year's IMC is too easy, can you please post solutions to those unsolved thus far (i.e. #5, #6, and #12)?

[quote="Myth"]In my opinion, problems are too easy, and acceptable result for such olympiad is 10-11 problems.

Why do IMC Committee use 20-point mark for each problem? What is deep meaning of such approach?[/quote]
Indeed the problems are not the most difficult ones that could be found, however it does not make the competition easy. It consists of a 2-days times 5 hours exam, which makes the competition harder than it looks from home :)

The grading of 20 points is probably made to make possible the harsh differentiation between papers, i.e. in the hopes that no tie scores appear. However, this year, as the year before, the first two places were tied. It seems a little to much of a coincidence that each time after the complaints this happens :?.

To Michael Lipnowski:
I solved all problems except #12, but it seems I have no wish to write any solutions for AoPS.

Good bye!

hi my score was 127 points with 47 position my twin brother did considerably better since he was in top 14. although i also didn't like complaints policy i lost first prize and my points are far fewer than tey shoul be after a normal complaint session

[quote="Sasha"]Congrats, Kappa.
How it is possible to go to both IMO and IMC?[/quote]

is it at all ?

It's first time I hear about IMC and I'd like to know how teams from each country are chosen - if anybody can help ...

The only rule is that you do not posses a bachelor's degree (university diploma) in mathematics. Although traditionally only university students participate, there were some cases of high-school students participating, mostly from Romania.

It's because Romanian high-school program is designed in such a way, that we have at the National Olympiad problems of IMC type since 11th grade. Thus from here come also the good results of Kappa (first place on the country in 11th grade).

To subscribe to the next edition of IMC, you can visit the official site of the competition, http://www.imc-math.org

[quote="Sasha"]So I guess a student in 10th grade from Moldova can participate too. ;)[/quote] Why don't you just say that you want to participate as well.  :)

[quote="Valentin Vornicu"]The only rule is that you do not posses a bachelor's degree (university diploma) in mathematics.[/quote]

Is this the rule? So does that mean that when I got my civil engineer diplome I can still participate?? :D

Would be very cool... anyway how do I register for this contest? Is registration at the very moment of arrival?
Also is it allowed to go with profs that aren't from your university? (like joining a foreign 3-man team or anything)

I'd really like to participate in here, though their site is really lacking answers to those (quite natural) questions...

I belive that their official site actually supplies answers to all of your questions. You haven't looked at it with enough attention :)

Anyway, to spare you the time, the answer is probably no, you can't participate if you have a degree in any math related major (I mean a major that "forced" you to take / have math classess in it - your is one of them).

Yes, you can participate, whilst you haven't obtained a degree. Although individual participation is ok, univerisity participation is more welcomed and comes with several advantages:

1) you have the univ. pay the trip, and stay there which might be pretty expensive for some people. Lucky for us Romanians that for the past couple of years it's been in the Balkans, and next year is in Sofia, which means that the transportation costs are pretty low :)

2) you have a representative in the jury, which might help you when the re-asssesment of the marks, at the end of the competition is made. We didn't have such a representative this year, thus amfulger lost first prize, although he should have definitely got it.

3) you will probably not be staying with complete strangers in your room :)


inequalities
trigonometry
complex numbers
geometry proposed
geometry

Prove that for all point $ D$ inside a acute triangle $ ABC$ we have:
$ \frac{DB.DC}{AB.AC}\plus{}\frac{DC.DA}{BC.BA}\plus{}\frac{DA.DB}{CA.CB}\geq 1$

Quite well known. Works for any point $ D$ in the plane of $ ABC$ {any trinagle}.
Complex numbers by using $ D$ as origin and giving affixes $ p,q,r$ to $ A,B,C$  and a short manipulation will give the result.

[b]PROBLEM-[/b][color=darkblue] let M be a point inside $ \triangle ABC$. Prove that: $ \frac {MB.MC}{bc} \plus{} \frac {MC.MA}{ca} \plus{} \frac {MA.MB}{ab}\ge 1$[/color] $ (*)$

[i]Proof[/i]
Indeed, $ (*)$ $ \Leftrightarrow \frac {a}{MA} \plus{} \frac {b}{MB} \plus{} \frac {c}{MC}\ge \frac {abc}{MA.MB.MC}$

$ \Leftrightarrow (a.MA \plus{} b.MB \plus{} c.MC)(\frac {a}{MA} \plus{} \frac {b}{MB} \plus{} \frac {c}{MC})\ge \frac {abc}{MA.MB.MC}(\sum {a.MA})$

$ \Leftrightarrow {{a}^{2}} \plus{} {{b}^{2}} \plus{} {{c}^{2}} \plus{} \sum{ab(\frac {MA}{MB} \plus{} \frac {MB}{MA})\ge \sum{\frac {{{a}^{2}}bc}{MB.MC}}}$

$ \Leftrightarrow {{a}^{2}} \plus{} {{b}^{2}}({{\cos }^{2}}AMB \plus{} {{\sin }^{2}}AMB) \plus{} {{c}^{2}}({{\cos }^{2}}AMC \plus{} {{\sin }^{2}}AMC) \plus{} 2ab.\cos AMB \plus{} 2bc.\cos (AMB \plus{} AMC) \plus{} 2ca.\cos AMC\ge 0$
$ \Leftrightarrow {{a}^{2}} \plus{} {{b}^{2}}.{{\cos }^{2}}AMB \plus{} {{c}^{2}}.{{\cos }^{2}}AMC \plus{} 2ab.\cos AMB \plus{} 2ca.\cos AMC \plus{} {{b}^{2}}.{{\sin }^{2}}AMB \plus{} {{c}^{2}}.{{\sin }^{2}}AMC \plus{} 2bc(\cos AMB.\cos AMC \minus{} \sin AMB.\sin AMC)\ge 0$
$ \Leftrightarrow {{(a \plus{} b.\cos AMB \plus{} c.\cos AMC)}^{2}} \plus{} {{(b.\sin AMB \minus{} c.\sin AMC)}^{2}}\ge 0$
Which is obviously true.
Our proof is completed then.


Find all positive integers $ n$ such that $ n^2\plus{}19n\plus{}48$ is a perfect square.

[quote="kplg"]Find all positive integers $ n$ such that $ n^2 \plus{} 19n \plus{} 48$ is a perfect square.[/quote]

$ n^2 \plus{} 19n \plus{} 48 \equal{} a^2$ for some $ a\ge 0$ $ \iff$ $ n^2 \plus{} 19n \plus{} (48 \minus{} a^2) \equal{} 0$ and so $ n \equal{} \frac { \minus{} 19\pm\sqrt {169 \plus{} 4a^2}}2$ and so $ 13^2 \plus{} 4a^2 \equal{} b^2$ for some $ b\ge 0$

So $ 13^2\equal{}(b\minus{}2a)(b\plus{}2a)$ and so $ (b\plus{}2a,b\minus{}2a)\in\{(169,1),(13,13),(1,169)\}$ and so $ (a,b)\in\{(42,85),(0,13)\}$ (remember $ a\ge 0$ and $ b\ge 0$) and so $ n\in\{ \minus{} 52, \minus{} 16, \minus{} 3,33\}$

Hence the unique positive solution $ \boxed{n \equal{} 33}$

Thanks Pactrick. Does it work for all $ an^2\plus{}bn\plus{}c$?

[quote="kplg"]Thanks Pactrick. Does it work for all $ an^2 \plus{} bn \plus{} c$?[/quote]

Surely not. For example :

With $ E\equal{}n^2\plus{}2an\plus{}b$ :

If $ b\equal{}a^2$, all $ n$ is solution
If $ b>a^2$, you can write $ E\equal{}(a\plus{}n)^2\plus{}b\minus{}a^2$ and so $ b\minus{}a^2\ge 2(a\plus{}n)\plus{}1$ and so $ n$ has an upper bound and there is a limited number of tests in order to find all solutions.
If $ b<a^2$, you can write $ E\equal{}(a\plus{}n)^2\minus{}(a^2\minus{}b)$ and so  $ a^2\minus{}b\ge 2(a\plus{}n)\minus{}1$ and so $ n$ has an upper bound and there is a limited number of tests in order to find all solutions.

I have a relate problem:
[quote][b]Problem.[/b] Given integers $ a,b,c$ such that $ f(n) \equal{} an^2 \plus{} bn \plus{} c$ is a square for every $ n\in\mathbb Z^ \plus{}$. Prove that $ a$ is a square.
[/quote]
Can you solve?  :huh:

i'm getting this relation:
$ (b\plus{}y)(b\minus{}y)\equal{}4a(c\minus{}x^2)$
cant proceed from here :(


WLOG supose that$ x,y$ and$ z$ are positive real numbers .
we have :$ x^2 \plus{} y^2 \plus{} z^2\ge \frac {1}{3}(x \plus{} y \plus{} z)^2\ge x \plus{} y \plus{} z.$
so :$ (x^2 \plus{} y^2 \plus{} z^2)^2\ge (x \plus{} y \plus{} z)^2\ge3(x \plus{} y \plus{} z)$
$ (x^2 \plus{} y^2 \plus{} z^2)^2\ge 9$
$ 2(x^2 \plus{} y^2 \plus{} z^2)^2\ge 2(x \plus{} y \plus{} z)^2\ge 6(xy \plus{} yz \plus{} zx) \equal{} 6(1/x \plus{} 1/y \plus{} 1/z)$

New question:

Find the best value of $ k$ so that the inequality holds for any $ a,b,c\ge 0\; ,abc\equal{}1$

$ a^2\plus{}b^2\plus{}c^2\ge (k\plus{}1)\left(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\right)\minus{}k(a\plus{}b\plus{}c)$


Find $ \lim_{n\rightarrow \infty} \sqrt[n]{1\plus{}x^n\plus{}(\frac {x^2}{2})^n}$

[quote="tdl"]General: if $ a \equal{} max\{a_1;a_2;...;a_k\},a_i > 0\forall i$ then $ \lim_{n\rightarrow \infty}\sqrt [n]{a_1^n \plus{} a_2^n \plus{} ... \plus{} a_k^n} \equal{} a$
Prove: Easy by inequality $ a < \sqrt [n]{a_1^n \plus{} a_2^n \plus{} ... \plus{} a_k^n}\leq\sqrt [n]{na^n} \equal{} \sqrt [n]{n}.a$[/quote]

It means, that the answer is $ x^2$?


analytic geometry
combinatorics proposed
combinatorics

Let $ n$ and $ m$ be given positive integers. In one move, a chess piece called an $ (n,m)$-crocodile goes $ n$ squares horizontally or vertically and then goes $ m$ squares in a perpendicular direction. Prove that the squares of an infinite chessboard can be painted in black and white so that this chess piece always moves from a black square to a white one or vice-versa.

If exactly one of $ m$ or $ n$ is odd, the standard checkerboard pattern does the trick. Every move changes the parity of $ x\plus{}y$, the sum of the coordinates of the current location.

If $ m$ and $ n$ are both odd, color the even rows black and the odd rows white. Every move must change the parity of the row number.

If $ m$ and $ n$ are both even, write $ m\equal{}2^km'$ and $ n\equal{}2^kn'$ where $ m'$ and $ n'$ are not both even. Solve the problem for $ m'$, $ n'$ as above, and refine the board by replacing each square by a $ 2^k \times 2^k$ grid of small squares, all the same color as the original one. The $ (m,n)$-crocodile moves on the refined board just like the $ (m',n')$-crocodile moves on the original one. (in fact the $ (m,n)$-crocodile always stays on the same small square relative to the large square, since the coordinates change by multiples of $ 2^k$.)


inequalities
function
inequalities proposed

For $ a,b,c$ being real numbers, find the minimum value of the function

$ f(a,b,c) \equal{} \sum_{cyc} (a\plus{}b)^4 \minus{}  \frac{4}{7}(\sum_{cyc} a)^4$

This is my (a little edited) solution for this in the postals.
Can some1 post a different proof?

[hide="Solution"]$ f(a,b,c) \equal{} (a \plus{} b)^4 \plus{} (b \plus{} c)^4 \plus{} (c \plus{} a)^4 \minus{} \frac {4}{7}(a^4 \plus{} b^4 \plus{} c^4)$

We claim that the required minimum value is $ 0$.
This value is attained at $ f(0,0,0)$. So it is sufficient to prove that $ f(a,b,c)\ge 0\ \forall (a,b,c)\in \mathbb R^3$

[b]Claim:[/b] All of the following inequalities can't hold simultaneously:
$ \left|a \plus{} \frac {b \plus{} c}{2}\right|\le \left|\frac {b \plus{} c}{2}\right|$

$ \left|b \plus{} \frac {c \plus{} a}{2}\right|\le \left|\frac {c \plus{} a}{2}\right|$

$ \left|c \plus{} \frac {a \plus{} b}{2}\right|\le \left|\frac {a \plus{} b}{2}\right|$

[b]Proof:[/b]
assume all those 3 inequalities to be true.
the first 1 is equivalent to:
$ \left|\frac {2a}{b \plus{} c} \plus{} 1\right|\le 1\iff \minus{} 1\le \frac {2a}{b \plus{} c} \plus{} 1\le 1\iff 0\ge \frac {a}{b \plus{} c}\ge \minus{} 1$
$ \implies \frac {b \plus{} c}{a}\le \minus{} 1\implies \frac {a \plus{} b \plus{} c}{a}\le 0\implies \frac {a}{a \plus{} b \plus{} c}\le 0$.
But we similarly have $ \frac {b}{a \plus{} b \plus{} c}\le 0$ and $ \frac {c}{a \plus{} b \plus{} c}\le 0$. Adding the 3 we have contradiction.
[b]End of claim[/b]

WLOG we assume that $ \left|a \plus{} \frac {b \plus{} c}{2}\right|\ge \left|\frac {b \plus{} c}{2}\right|$

Let $ p \equal{} \minus{} \frac {b \plus{} c}{2},q \equal{} \frac {b \minus{} c}{2},r \equal{} a \plus{} \frac {b \plus{} c}{2}$. We have $ |r|\ge |p|$
\[ f(a,b,c) \equal{} (a \plus{} b)^4 \plus{} (b \plus{} c)^4 \plus{} (c \plus{} a)^4 \minus{} \frac {4}{7}(a^4 \plus{} b^4 \plus{} c^4)\]

\[ \equal{} (r \plus{} q)^4 \plus{} (r \minus{} q)^4 \plus{} (2p)^4 \minus{} \frac {4}{7}[(r \plus{} p)^4 \plus{} (p \plus{} q)^4 \plus{} (p \minus{} q)^4]\]

\[ \text{[note that }(x \plus{} y)^4 \plus{} (x \minus{} y)^4 \equal{} 2(x^4 \plus{} y^4 \plus{} 6x^2y^2)\text{]}\]

\[ \equal{} 2(r^4 \plus{} q^4 \plus{} 6r^2q^2) \plus{} 16p^4 \minus{} \frac {4}{7}[(r \plus{} p)^4 \plus{} 2(p^4 \plus{} q^4 \plus{} 6p^2q^2)]\]

\[ \equal{} \frac {2}{7}[52p^4 \plus{} 7r^4 \plus{} \underline{3q^4} \plus{} \underline{42r^2q^2 \minus{} 24p^2q^2} \minus{} 2(r \plus{} p)^4]\]

\[ \text{[note that }|r|\ge|p|\text{]}\]

\[ \ge \frac {2}{7}[52p^4 \plus{} 7r^4 \minus{} 2(r \plus{} p)^4]\]
Hence it suffices to prove that $ 52p^4 \plus{} 7r^4\ge 2(r \plus{} p)^4$

[b]Lemma:[/b] If $ a,b\in \mathbb R^ \plus{}$ satisfy $ (a \minus{} 1)(b \minus{} 1)\ge 1$ then:
\[ a^3x^4 \plus{} b^3y^4\ge (x \plus{} y)^4\ \forall x,y\in \mathbb R\]
[b]Proof:[/b]

$ a^3x^4 \plus{} b^3y^4$

$ \equal{} a^2x^4 \plus{} b^2y^4 \plus{} \underline{a^2(a \minus{} 1)x^4 \plus{} b^2(b \minus{} 1)y^4}$

$ \ge a^2x^4 \plus{} b^2y^4 \plus{} 2abx^2y^2\underline{\sqrt {(a \minus{} 1)(b \minus{} 1)}}$

$ \ge a^2x^4 \plus{} b^2y^4 \plus{} 2abx^2y^2 \equal{} (ax^2 \plus{} by^2)^2$

$ \equal{} (x^2 \plus{} y^2 \plus{} \underline{(a \minus{} 1)x^2 \plus{} (b \minus{} 1)y^2})^2$

$ \ge (x^2 \plus{} y^2 \plus{} 2xy\underline{\sqrt {(a \minus{} 1)(b \minus{} 1)}})^2$

$ \ge (x^2 \plus{} y^2 \plus{} 2xy)^2 \equal{} (x \plus{} y)^4\boxed{}$

[b]End of lemma[/b]

In the lemma let $ a \equal{} \sqrt [3]{26}$ and $ b \equal{} \sqrt [3]{\frac {7}{2}}$
We have $ (a \minus{} 1)(b \minus{} 1) \equal{} (\sqrt [3]{26} \minus{} 1)(\sqrt [3]{\frac {7}{2}} \minus{} 1)\approx 1.017 > 1$

Hence:
\[ 26p^4 \plus{} \frac {7}{2}y^4\ge (p \plus{} r)^4\implies 52p^4 \plus{} 7r^4\ge 2(p \plus{} r)^4\]
[b]QED[/b][/hide]

Very nice proof, Akasnil.... thanks for your post. I did not know how to prove it... :blush:
PS: Isn't this problem Vietnam National MO 1996 ? :maybe: :
$ \sum (a \plus{} b)^4\geq \frac 4 7 (a^4 \plus{} b^4 \plus{} c^4)$
(Sorry for spam)

A similar inequality is the following.

If $ a,b,c$ are the side-lengths of a triangle, then

$ (a\plus{}b)^4\plus{}(b\plus{}c)^4\plus{}(c\plus{}a)^4 \ge 9(a^4\plus{}b^4\plus{}c^4)$.

[quote="Vasc"]A similar inequality is the following.

If $ a,b,c$ are the side-lengths of a triangle, then

$ (a \plus{} b)^4 \plus{} (b \plus{} c)^4 \plus{} (c \plus{} a)^4 \ge 9(a^4 \plus{} b^4 \plus{} c^4)$.[/quote]
Nice inequality, Vasc. My solution is the following:

Assume that $ a \ge b \ge c.$ From the inequality $ b \plus{} c \ge a,$ we have $ a \ge b \ge c \ge a \minus{} b \ge 0.$ According to this result, we have
$ \begin{aligned} (a \plus{} c)^4 \plus{} (b \plus{} c)^4 & \equal{} (a^4 \plus{} b^4) \plus{} 4c(a^3 \plus{} b^3) \plus{} 6c^2(a^2 \plus{} b^2) \plus{} 4c^3(a \plus{} b) \plus{} 2c^4 \\
& \ge (a^4 \plus{} b^4) \plus{} 4c(a^3 \plus{} b^3) \plus{} 6c^2(a^2 \plus{} b^2) \plus{} 9c^4 \\
& \ge (a^4 \plus{} b^4) \plus{} 4(a \minus{} b)(a^3 \plus{} b^3) \plus{} 6(a \minus{} b)^2(a^2 \plus{} b^2). \end{aligned}$
Therefore, it suffices to prove that
$ 4(a \minus{} b)(a^3 \plus{} b^3) \plus{} 6(a \minus{} b)^2(a^2 \plus{} b^2) \plus{} (a \plus{} b)^4 \ge 8(a^4 \plus{} b^4).$
This inequality is equivalent to
$ (a \minus{} b)(3a^3 \minus{} 9a^2b \plus{} 9ab^2 \plus{} 5b^3) \ge 0,$
which is true because
$ 3a^3 \minus{} 9a^2b \plus{} 9ab^2 \equal{} 3a(a^2 \minus{} 3ab \plus{} 3b^2) \equal{} 3a\left[ \left( a \minus{} \frac {3}{2}b\right)^2 \plus{} \frac {3}{4}b^2\right] > 0.$
The proof is completed. Equality holds if and only if $ a \equal{} b,c \equal{} 0$ or any cyclic permutation.

[quote="Akashnil"]Quite possible that this is not original. However I couldn't find it here: http://www.mathlinks.ro/resources.php?c=186&cid=40&year=1996[/quote]
Sorry, I found it in [url=http://www.scribd.com/doc/13889461/Inequalities-Through-Problems-by-Hojoo-Lee]Inequalities through problems, Hojoo Lee[/url] ; problem 95........ :maybe:

[quote="Akashnil"]Quite possible that this is not original. However I couldn't find it here: http://www.mathlinks.ro/resources.php?c=186&cid=40&year=1996[/quote]
Well, you can see it here: http://www.mathlinks.ro/resources.php?c=186&cid=41&year=1996 (because it was not our national olympiad, it was our TST for IMO 1996).