geometry
number theory
prime factorization

I just started reading a book entitled Number Theory and its History by Oystein Ore and for problem 2 in chapter 2.2 it ask:

Prove that a number is a square only when the number of (positive) divisors is odd.

Does he mean proper divisors? because then it fails.  If its regular then it would pass but right before the problem he gets into a discussion about proper divisors and how they are all the divisors not including the number.  Also how would you do the proof for this.  I know it makes logical sense but i lack training in the area of doing proofs.  Thanks

It doesn't matter if they're asking for proper or all.  Since proper divisors will exclude the number and 1, 2 numbers are left out.  And any number n+2 still holds teh same odd or even property.

That's just the converse of the fact that only perfect squares have an odd number of divisors.
Masoud Zargar

1 is actually commonly accepted has a proper divisor

only the number itself is excluded

lol, good to know this.  I'm horrible at knowing terms. :P  In that case, we can only accept the following:

[i] All perfect squares have an odd number of divisors in general.[/i]

Masoud Zargar

perfect numbers?

we're talking about squares...

Sorry, I meant perfect squares.  :blush:

Proof:

All perfect squares are of the form a x a where a is an integer. If $a = f_1 * f_2 * f_3 ... * f_n$, where $f_x$ is a prime factor of $a$, then $a^2$ has $2n$ prime factors. The number of divisors of $a^2$ is equal to the power set of the set of all prime factors, which has size $2n$. Thus $P(D) =$ Set of prime factors of $a^2$ is equal to $2^2n$. Repeated divisors will occur in even increments as there are even numbers of every prime factor. Let this number equal R.

Thus total divisors equals: $2^2n - R + 1$ which an odd number because $2^2n - R$ is even as both of those values are even. The 1 is added to count the factor 1.

Let the prime factorization of some integer $n$ be $p_1^{e_1} p_2^{e_2} \cdots$.  For $n$ to be a perfect square, we see each $e_i$ must be even.  Then the number of divisors is $(e_1 + 1)(e_2 + 1)(e_3 + 1) \cdots$, which is the product of odds, which is odd.
If the number of divisors is even, there must be an even term in that product, so for some $k$ that $(e_k + 1)$ is even, implying $e_k$ is odd.  This is a contradiction, as to be a perfect square, all $e_i$ must be even.

The most elementary way is as follows.  Say that $n$ is a perfect square and let $S$ be the set of (positive) integers which divide $n$.  Say that $m\in S$ and $m\not=\sqrt{n}$.  Then $n/m$ is also in $S$ and $n/m\not=m$.  By repeating this (and making sure we don't overcount), we see that every element of $S$ pairs up with a unique other element, except for $\sqrt{m}$.  Therefore $S$ has an odd number of elements.

[quote="demji"]Prove that a number is a square only when the number of (positive) divisors is odd.[/quote]

It`s true...


real analysis
complex analysis
complex analysis unsolved

Find the radius of convergence:

$f(z) : = \sum_{n=1}^\infty\frac{z^{n^{2}}}{n!}$

I was hinted to use Stirling's formula $\left(n! = (n^{n}e^{-n}\sqrt{2n\pi}(1+\epsilon_{n}))^{1/n}\right)$ where $\epsilon_{n}\to  0$ as $n\to \infty$

But I am stuck replacing n with $\sqrt{k}$.

Thanks

Use Hadamard's formula, and the limsup is actually a limit (=1).


You are trying to run from point A to point B through the rain.  Assuming it is falling at some speed $ s$, at what speed should you run to minimize the total quantity of rain that falls on your body?

(Be clear about what assumptions you're making about both the nature of the rain and the nature of your body.  I'm interested in what ideas come up.)

Isn't the amount of rain that falls on you just directly proportional to the amount of time you spend standing in the rain?  I think you should run as fast as possible.

The (possible) issue is that as you run, you may end up getting wetter in the front.

And the rain is presumably slanting downward with the wind, which gives another variable or two to throw into the mix.

I vaguely remember, from a long time ago, that someone milked this for a whole article in the [i]American Mathematical Monthly[/i] or perhaps one of the other MAA publications. (It might take a while to chase down the reference.)

If you don't consider that the rain slants in the wind, and you call the exposed area in the front of your body $ A$ and the exposed area at the top of your body $ a$, and let $ \rho$ be the relative density of the rain in air, then the rate of mass flow on your body is $ \rho(Av \plus{} as)$, where $ v$ is your speed.  So if you're just ambling around, it would seem that you want to walk slowly.

On the other hand, if you have to travel a distance $ d$ through the rain, you want to minimize the quantity $ \frac{d}{v}\rho(Av \plus{} as) \equal{} \rho(Ad \plus{} das/v)$, which would seem to imply that to minimize the amount of rain that gets on you, you would want to run as fast as possible.

This was answered in a casual physics book, can't remember which one though. They split the problem into two cases based on the horizontal direction of rain relative to your direction of movement. If rain falls in the same direction as you're moving you should walk at the horizontal speed of rain. If you're walking in the opposite direction you should run as fast as possible as genericme showed.


[quote="Litlle 1000t"]$ a,b,c$ are non negative reals such that $ a^{3}+b^{3}+c^{3}=3$  Prove that $ {a^{4}}{b^{4}}+{b^{4}}{c^{4}}+{a^{4}}{c^{4}}\le{3}$[/quote]
See the Numdung's post here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=34428


MATHCOUNTS
probability
quadratics
USAMTS
algebra
quadratic formula

[quote="MATHCOUNTS 2007 State Team #10"]In a bag of three marbles, there are exactly two blue marbles.   If Kia randomly chooses two marbles without replacement, the probability of choosing the two blue marbles is one-third.  However, before Kia chooses her two marbles, additional marbles are added to the bag. The probability of picking two blue marbles without replacement is still one-third. What is the least number of marbles that could be in the bag after the additional marbles have been added?[/quote]
Of course, this problem is a little too easy, and inappropriate for this Forum.  If you want to know the answer, post the question over on the MATHCOUNTS Forum, and somebody will probably answer it.

But, here is my challenge question to you folks.  My question is pretty similar.  It's not a trick question, so I have underlined the words I added/changed.  Please read carefully:

In a bag of three marbles, there are exactly two blue marbles.   If Kia randomly chooses two marbles without replacement, the probability of choosing the two blue marbles is one-third.  However, before Kia chooses her two marbles, additional marbles are added to the bag. The probability of picking two blue marbles without replacement is still one-third. What is the [u]ninth fewest[/u] number of [u]non-blue[/u] marbles that could be in the bag after the additional marbles have been added?

[hide="my really bad progress..."]By adding n marbles, and m blue marbles, where n >/= m, and both are positive integers.  We have the equation (2+m)/(3+n)*(1+m)/(2+n)=1/3.

Simplification results in 3m^2+9m =n^2+5n (Equation A).  From the original equation, it is clear that n must be 0 or 1 mod 3.  This can also be obtained from equation A, since n^2+2n=n(n+2)=0mod3

Lets pretend n is 0 mod 3, or 3j, where j is also a positive integer.  Plugging into equation A, and simplifying, we get m^2+3m=3j^2+5.  In mod 3, this says m^2=2mod3, which is never true.

So, n must be 1mod3.  Letting n be 3j+1, we get, with some simplification, m^2+3m=(3k+1)(k+2).  In mod 3, this says m^2=(k+2)mod3, which means k is either 1 or 2 mod 3.  Okay, so now n is 9z+4, or 9z+7.

Plugging in n=9z+4 to Equation A and simplifying, we get 3m^2+9m=81x^2+117x+36.  This means 3m^2=0mod9, which means m is 0mod3.

Plugging in n=9z+7 to Equation A and simplifying, we have get 3m^2+9m=81x^2+171x+84.  This means 3m^2=2mod9, which is impossible since 3m^2 can only be 0 3 or 6 mod9.

At this point, we notice the first solution is when n=7, since we plug into equation A to get m^2+3m-28=(m+7)(m-4)=0.

Plugging in n=9z+7 to equation A, we have -
[/hide][/hide]

[quote="Hamster1800"]Heh, I don't usually read these forums, but it turns out that my compsci teacher just gave us a problem extremely similar.

[hide="a way"]
We basically have this, let $b$ = the number of blue marbles, and $r$ (red) be the number of non-blue marbles.

$\frac{b}{b+r}\cdot\frac{b-1}{b+r-1}= \frac{1}{3}$
Clearing the denominators:
$3b(b-1) = (b+r)(b+r-1)$
Expanding:
$3b^{2}-3b = b^{2}+br-b+br+r^{2}-r$
Move all to one size
$2b^{2}-2b-2rb+r-r^{2}=0$.
Just as a check: b=2, r=1 gives
$2(4)-2(2)-2(1)(2)+(1)-1 = 0$, which is good.

So we have
$2b^{2}-(2+2r)b+r-r^{2}=0$.
Quadratic formula time:
$b = \frac{2+2r\pm\sqrt{(2+2r)^{2}-4(2)(r-r^{2})}}{4}$
$b = \frac{2+2r\pm\sqrt{4+8r+4r^{2}-8r+8r^{2}}}{4}$
$b = \frac{2+2r\pm2\sqrt{1+3r^{2}}}{4}$
$b = \frac{1+r\pm\sqrt{1+3r^{2}}}{2}$

Clearly, for $b$ to be an integer, so must $\sqrt{1+3r^{2}}$. So we have
$1+3r^{2}= k^{2}$
$1 = k^{2}-3r^{2}$, which is Pell's equation. We know that r=1 gives a solution, and indeed, (k,r)=(2,1) satisfies. So we have our fundamental solution.
Pell's equation says that the solutions will be the coefficients of
$(2+\sqrt{3})^{n}$
Just to check, $(2+\sqrt{3})^{2}= 4+4\sqrt{3}+3 = 7+4\sqrt{3}$, and if $(k,r)=(7,4)$, it indeed works.

Notice that if $r$ is odd, the square root isn't, and vise-versa. Therefore, any solution for $(k,r)$ will yield a valid (integer) number for blue. So we just want the coefficient of $\sqrt{3}$ in $(2+\sqrt{3})^{10}$.
However, I'm not a fan of raising something to the tenth power, so let's make a recurrence. Suppose we have a solution $(k,r)$. To get to the next solution, we multiply $(k+r\sqrt{3})$ by $2+\sqrt{3}$, so we get

$(k+r\sqrt{3})(2+\sqrt{3})$
$2k+k\sqrt{3}+2r\sqrt{3}+3r$

This says that our next solution is $(2k+3r, k+2r)$.

So, our solutions go
1: (2,1)
2: (7,4)
3: (26,15)
4: (97,56)
5: (362,209)
6: (1351,780)
7: (5042,2911)
8: (18817,10864)
9: (70226,40545)
10: (262087,151316)

So our answer is $\boxed{151316}$. If we wanted the number of blue marbles, it would be $\frac{1+r+k}{2}= 206702$, and we can easily verify that this does indeed give a probability of $\frac{1}{3}$.
[/hide][/quote]

I agree with your answer.  And your solution is very nice.

Anyway, the coincidence is revealed.  Compare this problem (based on MATHCOUNTS 2007 State Team #10)  with USAMTS Year 18, Round 4, Problem 4.

It all has to do with the number $\frac{\sqrt3}6(2+\sqrt3)^{10}$ which shows up in both problems.

[quote=rcv][quote="Hamster1800"]Heh, I don't usually read these forums, but it turns out that my compsci teacher just gave us a problem extremely similar.

[hide="a way"]
We basically have this, let $b$ = the number of blue marbles, and $r$ (red) be the number of non-blue marbles.

$\frac{b}{b+r}\cdot\frac{b-1}{b+r-1}= \frac{1}{3}$
Clearing the denominators:
$3b(b-1) = (b+r)(b+r-1)$

^^this
Expanding:
$3b^{2}-3b = b^{2}+br-b+br+r^{2}-r$
Move all to one size
$2b^{2}-2b-2rb+r-r^{2}=0$.
Just as a check: b=2, r=1 gives
$2(4)-2(2)-2(1)(2)+(1)-1 = 0$, which is good.

So we have
$2b^{2}-(2+2r)b+r-r^{2}=0$.
Quadratic formula time:
$b = \frac{2+2r\pm\sqrt{(2+2r)^{2}-4(2)(r-r^{2})}}{4}$
$b = \frac{2+2r\pm\sqrt{4+8r+4r^{2}-8r+8r^{2}}}{4}$
$b = \frac{2+2r\pm2\sqrt{1+3r^{2}}}{4}$
$b = \frac{1+r\pm\sqrt{1+3r^{2}}}{2}$

Clearly, for $b$ to be an integer, so must $\sqrt{1+3r^{2}}$. So we have
$1+3r^{2}= k^{2}$
$1 = k^{2}-3r^{2}$, which is Pell's equation. We know that r=1 gives a solution, and indeed, (k,r)=(2,1) satisfies. So we have our fundamental solution.
Pell's equation says that the solutions will be the coefficients of
$(2+\sqrt{3})^{n}$
Just to check, $(2+\sqrt{3})^{2}= 4+4\sqrt{3}+3 = 7+4\sqrt{3}$, and if $(k,r)=(7,4)$, it indeed works.

Notice that if $r$ is odd, the square root isn't, and vise-versa. Therefore, any solution for $(k,r)$ will yield a valid (integer) number for blue. So we just want the coefficient of $\sqrt{3}$ in $(2+\sqrt{3})^{10}$.
However, I'm not a fan of raising something to the tenth power, so let's make a recurrence. Suppose we have a solution $(k,r)$. To get to the next solution, we multiply $(k+r\sqrt{3})$ by $2+\sqrt{3}$, so we get

$(k+r\sqrt{3})(2+\sqrt{3})$
$2k+k\sqrt{3}+2r\sqrt{3}+3r$

This says that our next solution is $(2k+3r, k+2r)$.

So, our solutions go
1: (2,1)
2: (7,4)
3: (26,15)
4: (97,56)
5: (362,209)
6: (1351,780)
7: (5042,2911)
8: (18817,10864)
9: (70226,40545)
10: (262087,151316)

So our answer is $\boxed{151316}$. If we wanted the number of blue marbles, it would be $\frac{1+r+k}{2}= 206702$, and we can easily verify that this does indeed give a probability of $\frac{1}{3}$.
[/hide][/quote]

I agree with your answer.  And your solution is very nice.

Anyway, the coincidence is revealed.  Compare this problem (based on MATHCOUNTS 2007 State Team #10)  with USAMTS Year 18, Round 4, Problem 4.[/quote]


Hi! I haven?t posted anything for 3-4 days and I feel it?s been ages! I have some problems for you. I?ll post them separately:

Geom1

A variable circle is tangent to the fixed line d at the fixed pt A (the circle has variable center O). Let B and C 2 fixed pts on d. Let D be the projection of C on the polar of B with respect to the circle. Find the locus of the circumcenter of OAD.

What's the polar ? I suppose this has sth to do with projective geometry ...
I don't know projective geometry at all :(

Polar is a notion from proj geom, but you don't need it here. I just used because it's an easier way of expressing what I want to say. Basicly, the polar of an external point with respect to a circle is obtained like this: you draw the 2 tangents from the pt to the circle, and the polar is the line joining the 2 pts of tangency. Simple, isn't it?  :D

The definition of the polar is this: given a pt A, a circle C and a variable line d through A which intersects C at E and F, the polar is the locus of M s.t. points A, E, M, F are harmonicly conjugated (ME/MF=AE/AF). It can be proved that in the case A external to C the polar is that line I told you abt.


trigonometry
geometry
circumcircle
analytic geometry

Prove that in a triangle $ \triangle{ABC}$ , the relation above is true :
\[ \overrightarrow{OA}\cdot \sin{2A} \plus{} \overrightarrow{OB} \cdot \sin{2B} \plus{} \overrightarrow{OC} \cdot \sin{2C} \equal{} \vec{0} .
\]
where $ O$ is the circumcenter of the $ \triangle{ABC}$ .

[hide="PS"] : If you can , don't just say that $ \left( \frac {R^2}{2S} \sin{2A} ; \frac {R^2}{2S} \sin{2B} ; \frac {R^2}{2S} \sin{2C} \right)$ are the barycentric coordinates of $ O$ . So , if it is possible, post a complete proof .[/hide]

Oh! No body give you a full solution. I will try and post it.

there is a "full" solution in the link that I gave above :)


The sum and product of two purely periodic decimal number is a purely periodic decimal number of period T.Prove that the two intial numbers have periods not exceeding T

If $x=\frac{a}{b}$ purely periodic then $(b,10)=1, b|10^{T}-1$, were T is period. If $x=\frac{a}{b},y=\frac{c}{d}$ purely periodic then (bd,10)=1 and exist minimal T: $bd|10^{T}-1$ T is minimal periof for xy and period for $x+y=\frac{ad+bc}{bd}$ (may be is not minimal ). Because $b|10^{T}-1,d|10^{T}-1$ minimal periods of x and y are divisors T.


function
real analysis
real analysis unsolved

[url=http://bildr.no/view/510531][img]http://bildr.no/thumb/510531.jpeg[/img][/url]

I can not understant the sentence denoted by red line. why is f a bijection? thanks

This is not true in general; take two functions $ f,g : \mathbb{N} \to \mathbb{N}$, where $ f(x) \equal{} x\plus{}1$, $ g(1) \equal{} 1$ and $ g(x) \equal{} x\minus{}1$ for every $ x > 1$. Then $ g \circ f \equal{} id$, while neither are bijective.

However if $ f \circ f \circ \dots \circ f$ is bijective, then $ f$ definitely is as well.

[quote="t0rajir0u"]Show that if the composition of two functions $ X \to X$ is bijective, both functions are bijective.[/quote]

Can you say more detailed? thanks

Show:

If $ g \circ f$ is surjective, then $ g$ is surjective.
If $ g \circ f$ is injective, then $ f$ is injective.
A function $ h$ is bijective if and only if it is both injective and surjective. (This may be the definition of bijective you are using, which is fine for now.)

Conclude:
If $ g \circ f$ is bijective, then $ f$ is injective and $ g$ is surjective.
If $ f \circ f$ is bijective, then $ f$ is both injective and surjective.
The function $ f \circ f$ is bijective if and only if $ f$ is bijective.

General fact: if $ f\circ g$ is  bijective, then $ f$ is surjective and $ g$ is injective (proof is trivial).

Now $ f^n \equal{} f\circ f^{n \minus{} 1} \equal{} f^{n \minus{} 1}\circ f$, so if $ f^n$ is given bijective, it follows, according to the above, $ f$ both surjective and injective, hence bijective.

EDIT. Sorry, posted it as above post was also posted. Redundant now.


Let $a$ and $m$ be positive integers. Prove that \[\binom{a}{a}, \binom{a+1}{a}, \binom{a+2}{a},\ldots\] is periodic modulo $m$.

Isn't it obvious that if $a=b+2m$ then the three numbers associated to $a$ are the same modulo $m$ than the three associated to $b$ (in the same order)? :wink:

Pierre.

Sorry but I don't really understand what you mean but that's probably because I had a little typo in the post above (just forgot some dots...)  :blush:

One way or another, it is clear after applying Lucas' theorem.
Or cast a glance at problem 17 in
http://www.math.olympiaadid.ut.ee/eng/archive/bw/bw02sol.pdf

[quote="Yimin Ge"](just forgot some dots...)  :blush:[/quote]

Yes, I think you had  :D

Pierre.


search
number theory proposed
number theory

Let $ {a_n}\equal{}{2,4,8,1,3,6,...}$ be the infinite integer sequence such that $ a_n$ is the leftmost digit in the decimal representation of $ 2^n$,and let $ {b_n}\equal{}{5,2,1,6,3,1,...}$ be the infinite integer sequence such that $ a_n$ is the leftmost digit in the decimal representation  of $ 5^n$.Prove that for any block of consecutive terms in $ {a_n}$,there is a block of consecutive terms in $ {b_n}$ in the reserve order.


Let $ a,b,c$ be nonnegative real numbers. Prove that

$ \frac{3(a\plus{}b\plus{}c)^2}{2(a^2\plus{}b^2\plus{}c^2)(ab\plus{}bc\plus{}ca)}\le\frac1{a^2\plus{}bc}\plus{}\frac1{b^2\plus{}ca}\plus{}\frac1{c^2\plus{}ab}\le\frac{(a\plus{}b\plus{}c)^3}{6abc(a^2\plus{}b^2\plus{}c^2)}$

[quote="dduclam"]Let $ a,b,c$ be nonnegative real numbers. Prove that

$ \frac {3(a \plus{} b \plus{} c)^2}{2(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)}\le\frac1{a^2 \plus{} bc} \plus{} \frac1{b^2 \plus{} ca} \plus{} \frac1{c^2 \plus{} ab}\le\frac {(a \plus{} b \plus{} c)^3}{6abc(a^2 \plus{} b^2 \plus{} c^2)}$[/quote]
The left inequality is posted before, my friend.
See here: http://www.mathlinks.ro/viewtopic.php?t=114737 :)

[quote="dduclam"]Let $ a,b,c$ be nonnegative real numbers. Prove that

$ \frac {3(a \plus{} b \plus{} c)^2}{2(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)}\le\frac1{a^2 \plus{} bc} \plus{} \frac1{b^2 \plus{} ca} \plus{} \frac1{c^2 \plus{} ab}\le\frac {(a \plus{} b \plus{} c)^3}{6abc(a^2 \plus{} b^2 \plus{} c^2)}$[/quote]
The right part is also known before :(.
It appeared in my proof for the following inequality of Ductrung (I have posted my proof here)
\[ \sum\sqrt[3]{\frac{b^2\plus{}c^2}{a^2\plus{}bc}} \le \frac{a\plus{}b\plus{}c}{\sqrt[3]{abc}}\]
And my last proof used $ pqr$ technique, but now I have found a new proof for it without $ pqr$. :)

Is it in your book? If not, please post it here.
I think it is going to be another $ C.S.$ attack! :lol:


geometry
circumcircle
inradius
inequalities
trigonometry
trig identities
Law of Sines

Thanks for this little nice problem (by the way, is it from the shortlist?), but please keep to the rule of defining the notations you use. In this case, for instance, your problem should look like this:

[color=blue]Let ABC and $A_1B_1C_1$ be two triangles. Prove that

$\frac{a}{a_1}+\frac{b}{b_1}+\frac{c}{c_1}\leq\frac{3R}{2r_1}$,

where a = BC, b = CA, c = AB are the sidelengths of triangle ABC, where $a_1=B_1C_1$, $b_1=C_1A_1$, $c_1=A_1B_1$ are the sidelengths of triangle $A_1B_1C_1$, where R is the circumradius of triangle ABC and $r_1$ is the inradius of triangle $A_1B_1C_1$.[/color]

[i]Solution.[/i] First we show:

[color=blue][b]Lemma 1.[/b] If ABC is a triangle with sidelengths a = BC, b = CA, c = AB and inradius r, then $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\leq\frac{1}{4r^2}$.[/color]

[i]Proof of Lemma 1.[/i] Let $s=\frac{a+b+c}{2}$ be the semiperimeter of triangle ABC. Then, it is well-known that the numbers s - a, s - b, s - c are all positive, and that (s - a) + (s - b) + (s - c) = s and (s - b) + (s - c) = a. By the AM-GM inequality, $\left(\frac{\left(s-b\right)+\left(s-c\right)}{2}\right)^2\geq \left(s-b\right)\left(s-c\right)$; since (s - b) + (s - c) = a, this becomes $\left(\frac{a}{2}\right)^2\geq \left(s-b\right)\left(s-c\right)$. Thus, $a^2\geq 4\left(s-b\right)\left(s-c\right)$. Hence, $\frac{1}{a^2}\leq\frac{1}{4\left(s-b\right)\left(s-c\right)}$. Similarly, $\frac{1}{b^2}\leq\frac{1}{4\left(s-c\right)\left(s-a\right)}$ and $\frac{1}{c^2}\leq\frac{1}{4\left(s-a\right)\left(s-b\right)}$. Hence,

$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\leq\frac{1}{4\left(s-b\right)\left(s-c\right)}+\frac{1}{4\left(s-c\right)\left(s-a\right)}+\frac{1}{4\left(s-a\right)\left(s-b\right)}$
$=\frac{\left(s-a\right)+\left(s-b\right)+\left(s-c\right)}{4\left(s-a\right)\left(s-b\right)\left(s-c\right)}$.

Since (s - a) + (s - b) + (s - c) = s, this becomes

$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\leq\frac{s}{4\left(s-a\right)\left(s-b\right)\left(s-c\right)}=\frac14\cdot\frac{s}{\left(s-a\right)\left(s-b\right)\left(s-c\right)}$.

But by the Heron formula for the inradius of a triangle, $r^2=\frac{\left(s-a\right)\left(s-b\right)\left(s-c\right)}{s}$, so that $\frac{s}{\left(s-a\right)\left(s-b\right)\left(s-c\right)}=\frac{1}{r^2}$, and thus

$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\leq\frac14\cdot\frac{1}{r^2}=\frac{1}{4r^2}$.

And Lemma 1 is proven.

Now, by applying Lemma 1 to the triangle $A_1B_1C_1$ with its sidelengths $a_1$, $b_1$, $c_1$ and its inradius $r_1$, we get

$\frac{1}{a_1^2}+\frac{1}{b_1^2}+\frac{1}{c_1^2}\leq\frac{1}{4r_1^2}$.

On the other hand, by a well-known inequality for the circumradius of a triangle (see http://www.mathlinks.ro/Forum/viewtopic.php?t=17684 equivalent problem 3, http://www.mathlinks.ro/Forum/viewtopic.php?t=15558 example 3, or http://www.mathlinks.ro/Forum/viewtopic.php?t=17068 ), we have

$a^2+b^2+c^2\leq 9R^2$.

Now, by the Cauchy-Schwarz inequality,

$\left(a\cdot\frac{1}{a_1}+b\cdot\frac{1}{b_1}+c\cdot\frac{1}{c_1}\right)^2\leq\left(a^2+b^2+c^2\right)\cdot\left(\left(\frac{1}{a_1}\right)^2+\left(\frac{1}{b_1}\right)^2+\left(\frac{1}{c_1}\right)^2\right)$.

This simplifies to

$\left(\frac{a}{a_1}+\frac{b}{b_1}+\frac{c}{c_1}\right)^2\leq\left(a^2+b^2+c^2\right)\cdot\left(\frac{1}{a_1^2}+\frac{1}{b_1^2}+\frac{1}{c_1^2}\right)$.

By using the inequalities $a^2+b^2+c^2\leq 9R^2$ and $\frac{1}{a_1^2}+\frac{1}{b_1^2}+\frac{1}{c_1^2}\leq\frac{1}{4r_1^2}$, we thus have

$\left(\frac{a}{a_1}+\frac{b}{b_1}+\frac{c}{c_1}\right)^2\leq 9R^2\cdot\frac{1}{4r_1^2}=\left(\frac{3R}{2r_1}\right)^2$,

so that

$\frac{a}{a_1}+\frac{b}{b_1}+\frac{c}{c_1}\leq\frac{3R}{2r_1}$.

Problem solved. By the way, this inequality becomes an equality if and only if both triangles ABC and $A_1B_1C_1$ are equilateral.

  Darij

Another proof inegality:
$ \frac {a}{a'} + \frac {b}{b'} + \frac {c}{c'}\le\frac {3.R}{2.r'}$     (1)
Using [b]law of sines[/b], obtaines:
$ \sum\frac {a}{a'} = \sum{\frac {2R.sinA}{2R'.sinA'}\le \frac {3R}{8.R'.\prod {sin\frac {A'}{2}}}\Leftrightarrow \sum\frac {sinA}{sinA'}\le \frac {3}{8.\prod sin{\frac {A'}{2}}}}$
$ \Leftrightarrow \sum{sinA.sinB'.sinC'} \le \frac {3.\prod{sinA'}}{8.\prod{sin \frac {A'}{2}}} = 3.\prod {cos{\frac {A'}{2}}}.$     (2)
Acording to [b]inegality Oppenheim[/b]:
$ 2\sqrt 3 .(yz.sinA + xz.sinB + xy.sinC)\le(x + y + z)^2.$   (3)
for $ x = sinA', \ y = sinB', \ z = sinC'$ obtaines:
$ 2\sqrt 3 .\sum{sinB'.sinC'.sinA}\le(\sum{sinA'})^2 \Leftrightarrow \\
\Leftrightarrow \sum{sinB'.sinC'.sinA}\le \frac {(\sum{sinA'})^2}{2\sqrt 3} = \frac {(4.\prod{cos\frac {A'}{2}})^2}{2\sqrt 3}\le3.\prod{cos\frac {A'}{2}}.$    (4)
The last inequality echivalent to: $ \prod {cos\frac {A'}{2}} \le \frac {3.\sqrt 3}{8}.$
the complet prof (1).
The inequality $ (1)\Leftrightarrow (2) \Leftrightarrow (4).$

[quote=Sasha]Let $ABC$ and $A_{1}B_{1}C_{1}$ be two triangles. Prove that

$\frac{a}{a_{1}}+\frac{b}{b_{1}}+\frac{c}{c_{1}}\leq\frac{3R}{2r_{1}}$,

where $a = BC$, $b = CA$, $c = AB$ are the sidelengths of triangle $ABC$, where $a_{1}=B_{1}C_{1}$, $b_{1}=C_{1}A_{1}$, $c_{1}=A_{1}B_{1}$ are the sidelengths of triangle $A_{1}B_{1}C_{1}$, where $R$ is the circumradius of triangle $ABC$ and $r_{1}$ is the inradius of triangle $A_{1}B_{1}C_{1}$.[/quote]
https://artofproblemsolving.com/community/c6h525899p2981763

P.S. \$\sin ~\text{and}~\cos\$ gives $\sin ~\text{and}~\cos$


[quote="#H34N1"][b]7.[/b] Three numbers are randomly selected with replacement from the set of integers 1 through 100. What is the probability that the product of three numbers selected will be even?

[b]8.[/b] The measure of the side of a square is 10 inches. A point is chosen on each side of the square that is 7 inches from a vertex in a counterclockwise direction. The four points chosen are connected to form a smaller square. What is the ratio of the area of the larger square to the area of the smaller square?[/quote]
I am not sure if my answers to these are correct:
For 7:
[hide] The only way that the product could be odd is if all of the numbers chosen are odd.  Since there is going to be replacement, we can say that the chance that they are all going to be odd is 1/2*1/2*1/2=1/8.  This means that the chance that the product is even is 1-1/8=7/8. [/hide]
For 8:
[hide] Say that the points of the square are E, F, G, and H.  Say that EA=7 and AH=10-7=3 (it's easier to visualize if you make a diagram, but I don't have one; sorry :( ).  This means that $EH=\sqrt{49+9}=\sqrt{58}$.  Thus the area of the square formed is $\sqrt{58}^{2}=58$.  The ratio of the areas is $\frac{100}{58}=\frac{50}{29}$, which is our answer.  [/hide]

[quote="The QuattoMaster 6000"][quote="#H34N1"][b]7.[/b] Three numbers are randomly selected with replacement from the set of integers 1 through 100. What is the probability that the product of three numbers selected will be even?

[b]8.[/b] The measure of the side of a square is 10 inches. A point is chosen on each side of the square that is 7 inches from a vertex in a counterclockwise direction. The four points chosen are connected to form a smaller square. What is the ratio of the area of the larger square to the area of the smaller square?[/quote]
I am not sure if my answers to these are correct:
For 9:
[hide] The only way that the product could be odd is if all of the numbers chosen are odd.  Since there is going to be replacement, we can say that the chance that they are all going to be odd is 1/2*1/2*1/2=1/8.  This means that the chance that the product is even is 1-1/8=7/8. [/hide]
For 10:
[hide] Say that the points of the square are E, F, G, and H.  Say that EA=7 and AH=10-7=3 (it's easier to visualize if you make a diagram, but I don't have one; sorry :( ).  This means that $EH=\sqrt{49+3}=\sqrt{52}$.  Thus the area of the square formed is $\sqrt{52}^{2}=52$.  The ratio of the areas is $\frac{100}{52}=\frac{25}{13}$, which is our answer.  [/hide][/quote]

Quatto, #10 shoud be 50/29. Don't forget, $EH= \sqrt{7^{2}+3^{2}}$

[quote="The QuattoMaster 6000"][quote="#H34N1"][b]7.[/b] Three numbers are randomly selected with replacement from the set of integers 1 through 100. What is the probability that the product of three numbers selected will be even?

[b]8.[/b] The measure of the side of a square is 10 inches. A point is chosen on each side of the square that is 7 inches from a vertex in a counterclockwise direction. The four points chosen are connected to form a smaller square. What is the ratio of the area of the larger square to the area of the smaller square?[/quote]
I am not sure if my answers to these are correct:
For 9:
[hide] The only way that the product could be odd is if all of the numbers chosen are odd.  Since there is going to be replacement, we can say that the chance that they are all going to be odd is 1/2*1/2*1/2=1/8.  This means that the chance that the product is even is 1-1/8=7/8. [/hide]
For 10:
[hide] Say that the points of the square are E, F, G, and H.  Say that EA=7 and AH=10-7=3 (it's easier to visualize if you make a diagram, but I don't have one; sorry :( ).  This means that $EH=\sqrt{49+3}=\sqrt{52}$.  Thus the area of the square formed is $\sqrt{52}^{2}=52$.  The ratio of the areas is $\frac{100}{52}=\frac{25}{13}$, which is our answer.  [/hide][/quote]

the $\sqrt{49+3}$ part should be $\sqrt{49+9}$
so you should get 50/29
and these are #7,8 not #9,10

:blush: Oh, sorry about that.  Thanks for the correction.  I edited it.

[quote="kstan013"][hide="My Solutions"][hide="7)"]There are equal chances of odd and even. So, there are 4 ways to get an even. EEE, EOO, OEO, OOE. out of 8, so the answer is 4/8=[b]1/2[/b][/hide][hide="8)"]Draw a diagram. You will notice that the side of the small square is a hypotenuse to the triangles that it cuts off. It cuts off lengths of 3 and 7, so by the Pythagorean Theroem, the side is $\sqrt{58}$, so the area of the square is 58, so the ratio is 100/58=50/29[/hide][/hide][/quote]
what about EEO EOE OEE??
you forgot three possibilities.

[quote="#H34N1"][b]7.[/b] Three numbers are randomly selected with replacement from the set of integers 1 through 100. What is the probability that the product of three numbers selected will be even?

[b]8.[/b] The measure of the side of a square is 10 inches. A point is chosen on each side of the square that is 7 inches from a vertex in a counterclockwise direction. The four points chosen are connected to form a smaller square. What is the ratio of the area of the larger square to the area of the smaller square?[/quote]
[hide="7"]
There is $\frac{1}{2}$ chance of not picking an even integer. To not get the even integer for three consecutive selections, there is $\frac{1}{8}$ chance. Therefore, to get an even factor in one of the three selections is $1-\frac{1}{8}$, which is $\frac{7}{8}$. in decimal it is $0.875$
[/hide]
[hide="8"]
connect everythign. you see four right triangles. the hypotenos of the 7-3 legs is a side of the inside square. it is $\sqrt{58}$. the area of the inside square is $58$. the area of the outside square is $10^{2}$, which is $100$. the ratio is $\frac{50}{29}$, or $1.7241379310344827586206896551724$ in decimal form.
[/hide]


Basically, this is the Amazing Race.

In this game, you will compete for a chance to win (fake) $ \$1,000,000$. In order to win this money, you have to arrive on the destination before others do. Arriving at the last moment will result in your elimination.

However, you must perform certain 'tasks' before reaching the destination. You will only receive the clues to next task/destination only if you succeed the task. Don't worry; its not that hard.

In order for this to work, you must spend some time on this researching on things. I do promise, however, that the tasks won't be hard. Sometimes, it is just going to be a general question. Other times, you will have to search about the region that you are in.

First task of the day will be posted here. You may discuss with other people if needed. In fact, communication is good. Even talking outside is good. However, you must be careful. They might turn their back on you any time. ( >:) )

Answers should be PM to the host (alekim). I will state the result, and send you new information if you are correct.

1. hedgehog12
2. isabella2296
3. pythag011
4. AIME15
5. Bugi
6. cf249
7. eggylv999
8. FantasyLover
9. budi713
10. mathelete
11. nt811581147

Round 1 will (hopefully) begin on Saturday, since I have Battlebots tomorrow at Harrisburg. If I do attend a ski trip from my church also, it will start at Sunday.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=223463]Link to first Amazing Race[/url]

I will sign up.

Will we be competing in teams, or individually?

I was considering teams at first, but since it would take more people to participate, I will tryi to keep it as individual.

OMG, alekim's making a game  :D

I join.

I will try, but I am not that active anymore.

Sorry for all the delay, but I have a personal life you know. :D

This game will start on Tuesday, 8pm EST. I am trying a new system and seeing if it works or not.

Sorry, but I need another delay. My computer crashed badly today.

Alright. This game will start on Friday, 8pm EST. Hopefully, I will have an .exe by then. If not, we will use the private messaging system.

[quote="FantasyLover"][quote="BOGTRO"]Hmm... away Thursday-Sunday (back midnight) at SMT, any possibility of an extension?[/quote]

Same...[/quote]

[i] Really? [/b] I never knew that!

BOGTRO.... you are in same school with FantasyLover... or at least the team. You guys better beast there.

Fine this will start on Monday 8pm EST, final delay unless someone has something to say. Then I'll give them time later to make it up.

RULES:
1. You will get a program (no virus). On the program, it will ask you several questions. All answers have capital letter for the first letter, and lower case for others. For example, you would type Washington D.C. instead of washington d.c. If you type in a wrong answer, it will kick you out of the program, so be sure to save the answers.

I am using this system because I can't be online 24/7 to give out clues. You may discuss here to help others out, but you may not share answers unless specified.

2. At the end of these tests, you will get a random message in this format: Email me this code: _____________. Please email the code to supernovamaniac@gmail.com with your user ID as the subject. The reason why I am doing this is because my email is constantly on (it is connected to school account so its very important), and I can check the time when it was sent, so I can rank you guys easily.

Hopefully, this .exe system will work out the way I wanted to, or at least work.

Last delay: Wednesday.

Story: some people here are at stanford. They are trapped there until the next flight comes to AAST.

Here back at AAST, its snowing, thus flight's been canceled.

Is it just me or did everyone forget this? Because it is starting today at 8pm.

Make sure you can run .jar files.

oh and please repost to confirm that you are playing. aka if you don't post again you are not playing. If a new person posts, you may play.

I'm playing, but please don't give me any downloads.

I'm still playing, but I can't really do the downloads thing.  :maybe:

aww... no download? k.

Then instead of program, we will use my gmail as the private message.

Umm... yeah I guess this is small. I will wait few more hours before posting the first clue here.

I'm still playing.

Also, I use mac, so .exe doesn't really work.

BOGTRO and fan are back, as i'm sure you know.

Yes, I know but I'm waiting for few more confirmations again, as it looks like some people ditched.

And if I used the download system, it wouldn't have mattered, as .jar files are cross-platform. (It uses java.)

Oh well, I'm not waiting any longer. If you want to play, mail me the first answer.

Gmail: supernovamaniac@gmail.com

Subject: <yourusername>
Info: <youranswerhere>

First clue.

Welcome to Amazing Race 2: Single Minded

In this Amazing Race, you will be competing alone. You won't rely on anyone else other than yourself.

However, be nice, as you might need help from your fellow competitors.

First clue:

In order to participate in the Amazing Race 2: Single Minded, you must head to New York City, our starting point. However, it is important where you will arrive at. Land at the airport that’s closest to Jamaica, and your first clue will wait there.

Remember to email to supernovamaniac@gmail.com

I will be waiting.

Game officially started, only 1 of the person is moving through.


Grady rides his bike 60$ \%$ faster than his little brother Noah. If
Grady rides 12 miles further than Noah in two hours, how fast in miles per hour does Noah ride?

If Grady ($ G$) rides his bike 12 miles further than Noah ($ N$) in 2 hours, then he's riding $ 12/2\equal{}6$ mph faster.  So...  $ G \equal{} 1.6N$ and $ G \equal{} N \plus{} 6$.
$ 1.6N \equal{} N \plus{} 6$
$ .6N \equal{} 6$
$ N \equal{} 10$


Giving $ a,b,c>0$, prove that
$ \sum_{cyc}(\frac{a}{a\plus{}b})^{3}>\equal{}\frac{3}{8}$

[hide]
We can use rearrangement to get that $ \sum_{cyc}(\frac{a}{a\plus{}b})^{3}>\equal{}\sum_{cyc}(\frac{a}{b\plus{}c})^{3}$
This equations is homogenuous, so let a + b + c = 1.
Define x = a+b, y = a+c, and z = b+c.
$ \frac{a}{b\plus{}c} \equal{} \frac{a\plus{}b\plus{}c}{b\plus{}c} \minus{} 1 \equal{} \frac{1}{z} \minus{} 1.$

So, if we prove $ \sum_{cyc}(\frac{1}{x} \minus{} 1)^3 \ge 3/8,$ we will be done. Note that f(x) = $ (\frac{1}{x} \minus{} 1)^3$ is convex on (0,1). We get the desired result with Jensen on f(x).[/hide]

[hide="Alternate Method"]After we obtain $ \sum_{cyc}\left(\frac {a}{a \plus{} b}\right)^{3} > \equal{} \sum_{cyc}\left(\frac {a}{b \plus{} c}\right)^{3}$ by rearrangement, we can let the three terms be $ x$, $ y$, and $ z$. We can finish with Power Mean Inequality and Nesbitt as follows.
\[ \sqrt [3]{\frac {\displaystyle\sum_{cyc}(\frac {a}{b \plus{} c})^{3}}{3}}\ge\frac {\displaystyle\sum_{cyc}\dfrac{a}{b \plus{} c}}{3}\ge \dfrac{\dfrac{3}{2}}{3} \equal{} \dfrac{1}{2}
\]
Where the first step is Power Mean, and the second is Nesbitt. Cubing and multiplying through by $ 3$ gives the desired result. :lol: [/hide]

Both of solutions above are wrong at the 1st step. The contradiction is: a=1,b=2 and c=3.
[hide]Hint if someone needs:
Let b/a=x,c/b=y and a/c=z, we get: xyz=1(x,y,z>0). Rewrite the original inequality into:1/(1+x)^3+1/(1+y)^3+1/(1+z)^3 and using mixing variable: f(x,y,z)>=f(x,sqrt(yz),sqrt(yz)) with x=min(x,y,z)(considering that the equality occurs at x=y=z).By the way, it's from Vietnam TST 2005[/hide]

Mixing variables >.<
I hate that  :(
Anyway thank you, ghjk.

[quote="can_hang2007"][quote="bibobeo"]Giving $ a,b,c > 0$, prove that
$ \sum_{cyc}(\frac {a}{a \plus{} b})^{3} > \equal{} \frac {3}{8}$[/quote]
This is Vietnam TST 2005. You can see 5 different proofs from my book: [i][b]Old and New Inequalities 2[/b][/i], Vo Quoc Ba Can - Cosmin Pohoata, GIL publishing house, 2008 :)[/quote]

how can I get your book , can_hang2007 ?

[quote="bibobeo"]Giving $ a,b,c > 0$, prove that
$ \sum_{cyc}(\frac {a}{a \plus{} b})^{3} > \equal{} \frac {3}{8}$[/quote]
[hide="Here is my proof"]
$ \sum_{cyc}\left(\frac {a}{a \plus{} b}\right)^3 \equal{} \sum_{cyc}\frac {a^4}{a(a \plus{} b)^3}\geq\frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{\sum a(a \plus{} b)^3}.$
Hence, it remains to prove that
$ 8(a^2 \plus{} b^2 \plus{} c^2)^2\geq3\sum_{cyc}a(a \plus{} b)^3.$
But $ 8(a^2 \plus{} b^2 \plus{} c^2)^2\geq3\sum_{cyc}a(a \plus{} b)^3\Leftrightarrow\sum_{cyc}(5a^4 \minus{} 9a^3b \minus{} 3a^3c \plus{} 7a^2b^2)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(5a^4 \minus{} 3a^3b \minus{} 3a^3c \plus{} 7a^2b^2)\geq\sum_{cyc}6a^3b.$
But by Vasc's inequality we obtain: $ \sum_{cyc}6a^3b\leq2(a^2 \plus{} b^2 \plus{} c^2)^2.$
Id est, it remains to prove that
$ \sum_{cyc}(a^4 \minus{} a^3b \minus{} a^3c \plus{} a^2b^2)\geq0,$ which is obvious.
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=101425
[/hide]


geometry
analytic geometry
graphing lines
slope
conics
parabola
calculus

The coordinates of point P is (3,1) on a Cartesian plane......The graph of a linear equation passes throught point P and cuts through x-axis and y-axis at point A and point B with positive values.....Find the minimum area of the triangle OAB where O is the origin, ie (0,0)

[hide]OAB is right, so the area is a*b*1/2 (a is the distance from O to the x intercept, b is the distance from O to the y intercept).

We see that the slope of the line is -b/a and the y intercept is b.  Therefore, the equation of the line is

-b/a*x+b=y

Substituting in the point (3,1) we find

-b/a*3+b=1

-3b+ab=a

ab=a+3b

ab/2=(a+3b)/2

Set c = 3b.

According to power mean,

min{a,c} :le: (a+c)/2, with equality reached a=c.  That means that (a+c)/2 will be minimized at a=3b.  Plugging this into the slope -b/a, we find that the slope is -b/3b=-1/3.  We now have the equation of the line as -1/3x+b=y.  Let's plug in the point (3,1) to find b.

-1/3*3+b=1

b=2

which means that a=3*2=6

So the minimum area is 2*6*1/2=6[/hide]

Fixed....?

[hide]the equation of the line is (y-1) = m(x-3), where m<0...

A and B are the x and y intercepts of the line, when y,x=0

-1 = m(x-3), x = -1/m+3

y-1 = -3m, y = -3m+1

the area of the triangle is (-3m+1)(-1/m+3)/2 = (3-9m-1/m+3)/2 = (6-9m-1/m)/2, so the minimum value occurs when -9m-1/m is minimized, but since m<0, both terms are positive and by AM-GM, the minimum is 2sqrt(9) = 6.  thus the minimum area is (6+6)/2 = 6, when the slope of the line is -1/3.[/hide]

[quote="Magnara"]b=10/3
which means that a=3*10/3=10[/quote]

This is definitely wrong.  I'm not 100% sure everything before it is correct, but this might be your only mistake.

ThAzN1 wrote:Magnara wrote:[hide]Let's plug in the point (1,3) to find b.[/hide]

lol, i think u meant [hide](3,1)[/hide].

Lol.  I swear, it's always the last step that gets me.

[hide]Alternate solution:

Look at my previous one up to the point where ab/2=(a+3b)/2.

ab/2=(a+3b)/2

ab=a+3b

-3b+ab=a

b(a-3)=a

b=a/(a-3)

Which means that the area is equal to (a :^2:)/(2a-6).  So finding the largest value of k for which that expression is greater than k is equivelent to minimizing the area.

(a :^2:)/(2a-6) :ge: k

a :^2: :ge: 2ak-6k

a :^2: -2ak+6k :ge: 0

We now have to find the largest value of k for which that is true. 0 The vertex of that parabola is at a=2k/(2)=k.  So then the minimum of the graph is at k :^2:-2k :^2:+6k=6k-k :^2:=k(6-k).  This has to equal 0, which means that k is equal to 6 or 0, 0 is extraneous.  Therefore, the minimum is 6.

[/hide]

i ahve a question.. i worked on this a different way, but its not giving the right answer. could someone explain?

this is what i did:

let a and b be the x and y intercepts, respectively.
let Z be the area of the triangle.
Z = (1/2)(ab)

if m is the slope of the line passing through (3,1), the equation can be written in two ways:

mx + b = y, or 3m + b =1
my + a = x, or m + a = 3

solving for a and b in the two equations yields:
b = 1 - 3m
a = 3 - m

Z = (1/2)(ab) = (1/2)(1-3m)(3-m)

the minimum of Z occurs when m = 5/3, (the vertex of the parabola).. which is positive, so i messed up somewhere.

i know i'm missing a step somewhere..

Shouldn't it be [tex]{{1}\over{m}}y + a = x[/tex] instead of [tex]my + a = x[/tex]?

you can also use optimization with calc:

[hide]

The line can be represented by y = -kx + (1+3k) where -k is the slope.

Therefore, the triangle's vertices are the origin, (1+3k), and (3+1/k).

[AOB] = (1+3k)(3+1/k)/2 = (6+9k+1/k)/2

Taking the derivative and disregarding the 1/2:

9-1/k^2 = 0

9 = 1/k^2

k^2 = 1/9

k = 1/3

Plugging this back into [AOB] we get:

[6+(9*1/3) + 1/(1/3)]/2 = (6+3+3)/2 = 6[/hide]

gah, stupid mistake. i cant believe i did that. now i get 6 as the answer

thanks


Is it possible to set the image for the monobook portal?

As I said before it is simply a debug option (and I'm keeping it there to observe any changes in the way themes are handled in the future upgrades). No support or encouragement of using that skin is given.


trigonometry
function
geometry
geometric transformation
LaTeX
algebra
domain

1st question
without using inverse function keys calculate
a) csc (cos^{-1}(-4) ) ? (is csc (cos negative 1 power (-4))

b) cos [(cos^{-1}(.3)) + (sin^{-1}(.4))] ?   (is cos than bracket with cos negative 1 power and .3 in () plus sin negative 1 power with
                                                                   .4 in ()
for a and b i'm not sure about my answer...

2nd question
graph   -2cos (4x)-1
show transitions (sequentially), mention values of the period, amplitude, shifts, and translation etc .

and graph sin (pie - x) + 1 (same question as 2nd question)

the identities i need help with are...

a) -cos W + 2cos^{2} (w/2) = sin^{2} (2W) + cos^{2} (2W)?    (the ^ is the 2nd power )

b) cotA + tanA = 2csc2A

it will be alot of help if you could help me understand these problem step by step

with the first problem a) csc i broke it down to 1/sin theta...

with problem 2, i think the amplitude is 2, period is pie/2, phase shift is 0, vertical translation is -1

with the problem sin (pie-x) + 1, amplitude is 0, period is 2, phase shift ?, and vertical translation is 1

i can't figure out how to graph it, i got a clue of how it goes but not 100% sure about my answer.
so if someone out there that can double check my solution to the problem and graph it. extra thanks in advance

For 1st question, imagine or draw a right triangle on an unit circle.
a) the right triangle's adjacent side has length of -4 and hypotenuse has length of 1, so the opposite side has length of sqrt(1-4^2). cos^{-1}(-4) gives you the angle from the positive x-axis, csc(angle) is the length of opposite side /the length of hypotenuse, which is sqrt(1-4^2)/1, which is undefined.
b) Let u = (cos^{-1}(.3)) and v = (sin^{-1}(.4)). Expand cos(u+v) gives cos(u)*cos(v)-sin(u)*sin(v). Put u and v back into the equation, you get cos(cos^{-1}(.3))*cos(sin^{-1}(.4))-sin(cos^{-1}(.3))*sin(sin^{-1}(.4)). This reduces to 0.3*cos(sin^{-1}(.4))-0.4*sin(cos^{-1}(.3)). Using the same method as part a), you get 0.3*sqrt(1-0.4^2)-0.4*sqrt(1-0.3^2).

The identities question
a) -cos(W) = -cos(W/2+W/2).
Using trig expansion cos(u+v) = cos(u)*cos(v)-sin(u)*sin(v),
   -cos(W) = -(cos^2(W/2)-sin^2(W/2)) = -cos^2(W/2)+sin^2(W/2)
Putting this into the left hand side, we have
   -cos^2(W/2)+sin^2(W/2)+2cos^2(W/2) = sin^2(W/2) + cos^2(W/2) = 1
The right side is obviously also equal to 1. The the two sides are equal.

b) cotA + tanA
= cos(A)/sin(A)+sin(A)/cos(A)
=[cos(A)*cos(A)+sin(A)*sin(A)/(sin(A)*cos(A))
=1/(sin(A)*cos(A))
=2/(2*sin(A)*cos(A))
=2/sin(2A)
=2csc(2A)

^thanks, but what do u mean by *

anybody out there understand the graph portion?

with the identities, i dont really understand how u broke down the left side on a)

Suggestion: LaTeX makes things look nicer, and is easy; you just need to include dollar signs at the beginning and the end of math expression.

For number 1, $ \minus{} 4$ is not in the domain of $ \arccos$ (which is $ [ \minus{} 1, \plus{} 1]$), so it's simply undefined.

For identity a, it's an application of double-angle formula for cosine:
$ \cos2A \equal{} \cos(A\plus{}A) \equal{} \cos^2A\minus{}\sin^2A \equal{} 2\cos^2A\minus{}1 \equal{} 1\minus{}2\sin^2A$

About the graph portion, consider this general function:

$ \pm \frac {1}{m} \cdot y \minus{} b \equal{} f\left( \pm \frac {1}{n} \cdot x \minus{} a\right)$

which is the image of $ y \equal{} f(x)$ after following changes:

(1) Reflection over x- and y- axis.
(2) Vertical stretch by $ |m|$, horizontal stretch by $ |n|$. (amplitude and period in sines and cosines, respectively)
(3) Translation by vector $ < a, b >$ ($ a$ is the shift in sines and cosines)

Note (1) and (2) are changeable in order, because both are multiplication. But (3) is not, since it's addition.

So for first one, $ \minus{} 2\cos (4x) \minus{} 1$, it undergoes: decrease in period to $ \frac {1}{4}\cdot 2\pi \equal{} \frac {\pi}{2}$, increase in amplitude to $ 2$, reflection over x-axis, and vertical translation by $ 1$ down. Same thing applies to second one.

okay, then
$ \csc(\arccos\minus{}\frac{2}{5})\equal{}\frac{1}{\sin(\arccos\minus{}\frac{2}{5})}$
The range of $ \arccos$ is $ [0, \pi]$, in which sine is always positive. Therefore, it simplifies to
$ \frac{1}{\sqrt{1\minus{}(\frac{2}{5})^2}}\equal{}\frac{5\sqrt{21}}{21}$

how do i graph on the number line need more explanation on both graphing portion... plus on the 1st identities, i don't get how the steps are

[quote="timwu"]
So for first one, $ \minus{} 2\cos (4x) \minus{} 1$, it undergoes: decrease in period to $ \frac {1}{4}\cdot 2\pi \equal{} \frac {\pi}{2}$, increase in amplitude to $ 2$, reflection over x-axis, and vertical translation by $ 1$ down. Same thing applies to second one.[/quote]

okay, we know cos(0)=1, so trace point (0,1). Vertical stretch (amplitude increase)--->(0,2); reflection--->(0, -2); translation--->(0,-3).

Base on this point, graph the function as normal cosine, but remember the above changes. Your peak and valleys are now 2 units above or below the line y=-1, and they are also reflected about it. The periods, still starting at 0, are now shortened to one-fourth of the normal 2 pi.

If you still didn't get it, try plot it by hand with several points or graphing calculator, and compare it to a normal cosine.


geometry
parallelogram
geometry proposed

On the sides of a convex hexagon $ ABCDEF$ , equilateral triangles are constructd in its exterior. Prove that the third vertices of these six triangles are vertices of a regular hexagon if and only if the initial hexagon is [i]affine regular[/i]. (A hexagon is called affine regular if it is centrally symmetric and any two opposite sides are parallel to the diagonal determine by the remaining two vertices.)

This problem is very easy and it is enough to post only the formulation of an easy to prove Lemma.

[b][size=100][color=DarkBlue]LEMMA. - A parallelogram $ ABCD$ is given and let $ \bigtriangleup ABE,\ \bigtriangleup BCF,$ two equilateral triangles, erected outwardly to it, on $ AB,\ BC$ recpectively. Prove that the triangle $ \bigtriangleup DEF,$ is equilateral.[/color][/size][/b]

Kostas Vittas.


[quote="ALGEBRA"]if ab+bc+ca+2abc=1  and a,b,c>=0  prove that 2(a+b+c)+1>=32abc[/quote]
$32abc=16-16(ab+bc+ca)$.
We have to prove $2(a+b+c)+16(ab+bc+ca) \ge 15$
From $ab+bc+ca+2abc=1$, let $x=ab+bc+ca$, then $abc \le (\frac{x}{3})^{3/2}$, so $2(\frac{x}{3})^{3/2}+x \ge 1$. Solving, $x \ge \frac{3}{4}$. Then using $a+b+c \ge \sqrt{3(ab+bc+ca)} = \frac{3}{2}$, we solved the problem.


calculus derivative function induction algebra polynomial symmetry

How many real roots has the n-th derivative of $e^{-\frac{1}{x^2-1}}$?

Ok, so? I really find this problem very tricky and difficult...

Expand the function in its laurent series and try the residue theorem..

[quote="harazi"] I really find this problem very tricky and difficult...[/quote]
Seems pretty straightforward to me (unless it is the first problem of this kind one encounters, of course). The first step is to show by induction that the $n$-th derivative is of the form $\frac{P(x)}{(x^2-1)^{2n}}e^{-\frac 1{x^2-1}}$ where $P$ is a polynomial of degree $3n-2$. Thus, we have $3n-2$ roots on the complex plane. The trick is to find them all. Note that all derivatives of the function tend to $0$ as $x\to 1+$ and, starting with the first derivative, as $x\to +\infty$. Thus, by Rolle's theorem, the second derivative has at least $1$ root on $[1,+\infty)$, the third one at least $2$ roots, and so on. This gives us $n-1$ roots on $[1,+\infty)$ and, by symmetry, $n-1$ more roots on $(-\infty,-1]$, bringing the count up to $2n-2$. Where are the other $n$ missing roots? Let's look at the function on the imaginary axis (another possibility for $x^2$ and the entire expression to be real). The value at the point $ix$ is $e^{\frac1{1+x^2}}$. Investigating this function and using Rolle's theorem again, we find all $n$ missing roots. The last thing to observe is that $0$ is a root on the imaginary axis for odd $n$ and not a root for even $n$. Thus, the total number of real roots is $2n-2$ or $2n-1$ depending on the parity of $n$.


A point P lies inside a square ABCD of side 120. Let F be a point on CD so that PF is perpendicular to CD. P is equidistant from A, B, F. Find PA.

If P is equidistant from A and B, then F is the midpoint of CD. Let PF=PA=PB=x. If PA is the hypotenuse of a right triangle with legs 60 and 120-x, $ x^2\equal{}60^2\plus{}(120\minus{}x)^2$. Solving, we get x=75.

[hide="Another Approach"]Let the square be in the coordinate plane, and let $ AP\equal{}BP\equal{}FP\equal{}n$.

Since $ AP\equal{}BP\equal{}FP$, $ P$ is the center of the circle with radius $ n$ that goes through three points $ (0,120), (120,120), (60,0)$.

The equation of this circle is $ n^2\equal{}(x\minus{}60)^2\plus{}(y\minus{}n)^2$.

Putting either of the points $ (0,120)$ or $ (120,120)$, we obtain $ n\equal{}75$.

Answer: 75[/hide]


Let $ a,b,c\geq 0$, no two of which are zero.Prove that:
$ \frac{(b\plus{}c)^2}{a^2\plus{}bc}\plus{}\frac{(c\plus{}a)^2}{b^2\plus{}ca}\plus{}\frac{(a\plus{}b)^2}{c^2\plus{}ab}\geq 6$.

[quote="Inequalities Master"]Let $ a,b,c\geq 0$, no two of which are zero.Prove that:
$ \frac {(b \plus{} c)^2}{a^2 \plus{} bc} \plus{} \frac {(c \plus{} a)^2}{b^2 \plus{} ca} \plus{} \frac {(a \plus{} b)^2}{c^2 \plus{} ab}\geq 6$.[/quote]
It's equivalent to $ (a\minus{}b)^2(a\minus{}c)^2(b\minus{}c)^2\plus{}\sum_{cyc}ab(a^2\minus{}b^2)^2\geq0.$
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=60128


modular arithmetic
function
number theory
totient function

[color=olive][size=200]How to calculate the number which are at the end of a large number？Are there any common method?Thanks!!
for example:please calculate the last 2 number which the one equals to 2^(30402457)-1[/size][/color]

I'm having a hard time understanding what you're asking.

Often times, when you're asked to find the last two or three digits of some really large pattern, it's because the digits repeat.  Or because, there is some easy congruence.

Here's an example: find the remainder when $ 5^{2009}$ is divided by 11.  We appeal to Euler's theorem: $ 5^{2009}\equal{}5^{200\cdot 10}\cdot 5^{9}\equal{}(5^{200})^{10}\cdot 5^{9}\equiv 5^{9}\equal{}(5^{3})^{3}\equal{}125^{3}\equiv 4^{3}\equal{}64\equiv 9\pmod{11}$, so the remainder is 9.  In case you don't know, Euler's theorem states that if $ \gcd(a,p)\equal{}1$, then $ a^{\phi(p)}\equiv 1\pmod{p}$.

Or here's another example: find the ones digit of $ 7^{9999}$.  Note that $ 7^{1}\equal{}7$, $ 7^{2}\equal{}49\equiv 9\pmod{10}$, $ 7^{3}\equal{}343\equiv 3\pmod{10}$, $ 7^{4}\equal{}2401\equiv 1\pmod{10}$.  The ones digits repeat in cycles of 4, so any multiple of 4 exponent will end in 1.  Therefore $ 7^{10000}\equiv 1\pmod{10}$.  If you look at the pattern, 3 always comes before 1, so $ 7^{9999}\equiv 3\pmod{10}$.

As for your question, I can't understand what you're asking.

I think (s)he wants to find the last two digits of the number $ 2^{30402457}\minus{}1$

By Euler's totient function $ \varphi(100)\equal{}40$, so $ 2^{30402457}\equiv2^{(760061)(40)\plus{}17}\bmod 100\equiv2^{17}\bmod 100$. This can be semi-easily calculated, and it turns out to be $ 131072$. Now subtract $ 1$ to get $ 131071$, and the last two digits are $ \boxed{71}$

You don't need to directly compute $ 2^{17}$.  We have that $ 2^{17}\equal{}(2^{8})^{2}\cdot 2\equal{}256^{2}\cdot 2\equiv 56^{2}\cdot 2\equal{}3136\cdot 2\equiv 36\cdot 2\equal{}72\pmod{100}$.  Now just subtract 1 as you said to get 71.


$a_{n}$ is a sequence of integers such that $a_{n}> a_{a_{n-1}}$.  Compute $\left \lfloor\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\cdots+\frac{1}{\sqrt{a_{2006}}+\sqrt{a_{2007}}}\right \rfloor$

[quote="miyomiyo"] How can you make a larger floor sign?[/quote]
Use: [code] \left \lfloor \text{your stuff here} \right \rfloor [/code]
Which gives:  $\left \lfloor \frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\cdots+\frac{1}{\sqrt{a_{2006}}+\sqrt{a_{2007}}}\right \rfloor$.

It's the same idea as the large parentheses.

[hide="Solution"]

By testing, we see that the ordered integers (i.e. $a_{n}=n$) work as our sequence. Now, the sum telescopes to \[\left \lfloor{\sqrt{2007}-\sqrt{1}\right\rfloor = \boxed{043}}\] [/hide]

Because of the wording of the problem, there can be only one. Try "all possible values" or something.

[quote="miyomiyo"][quote="cincodemayo5590"]By testing, we see that the ordered integers (i.e. $a_{n}=n$) work as our sequence.[/quote]

Can you prove that this is the only one?[/quote]

[hide="Actually, I don't think it is..."]
$a_{n}= 2008-n$ works, I believe, for the terms needed in the expression; but it gives the same value for the expression.
[/hide]

if f:N->N, and f(n+1)>f(f(n)), then f(n)=n

this is an old imo problem

search


function
logarithms
algebra
floor function
difference of squares
special factorizations

Find  the smallest positive integer k such that the base ten representation of 3^k has exactly one-hundred digits. log 3=.4771. Any tips?

There is an easy way to work out the number of digits in a number, especially if powers are involved. Basically if a number x has n digits, 10^(n-1) <= x < 10^n and by taking logs, we get n = [log x]+1, where [] is the floor function (just forget about decimal part).
So 3^k has [log(3^k)]+1 digits = [k log 3] + 1 digits.

Equating this to 100 should get your answer (208 by my calculation).

Stephen

Nice solution.  What about if n=1983 then what is the least positive integer k such that k[n^2][n^2-1^2][n^2-2^2][n^2-3^2]....[n^2-[n-1]^2]=r!   ?

I'll post my solution later to give others some time to think about it, for now I make k=2.. nice problem.

oops sorry didn't mean to post twice. I never was very good with forums.

same here (for k=2). can I give a solution yet? BTW, it doesn't really matter what n is, does it? I love problems like that.

Ok: use difference of squares several times. You get all the numbers from 1 to 2n-1 multiplied together. The only problem is that you have an extra factor of n. Multiplying by k=2 makes it 2n!

Bob

[quote="sapphyre571"][hide="much easier way"]
if you have a calculator (a TI-83 will do)
punch in guess-and-check exponents
and a 100-digit answer can be expressed as X*10^99
and the lowest exponent on a three that does it
is 208
yes i know i didnt justify my solution
but its an answer[/hide][/quote]
I don't think we can use a calculator for this.  :P

[hide] I don't know if this is what TripleM was saying, but...
$3^k=10^{99} \longrightarrow \log_{3}{10^{99}}=k \longrightarrow k=\frac{\log{10^{99}}}{\log{3}}=\frac{99}{.4771}\approx207.5$. So the answer is $\boxed{208}$. [/hide]


[quote="Elio (n)"]If $ n^{2} \plus{} 2n$ gives a number with unit digit equal to 4   (you know $ n^{2} \plus{} 2n \equal{} abc....4$  ) , find the tenner digit of this number , $ n$ is natural.[/quote]

tenner=tenth :?:
Is it really possible to do this?
By quick observation since the last digit of $ n^{2}\plus{}2n\plus{}1\equal{}(n\plus{}1)^{2}$ is 5 then the last digit of n must be 4. But I don't see how is this related to the tenth digit...  :huh:

We can find that $ n\equal{}4(mod10)$

So $ n\equal{}10k\plus{}4$

$ n^2\plus{}2n\equal{}n(n\plus{}2)\equal{}(10k\plus{}4)(10k\plus{}6)\equal{}100k^2\plus{}60k\plus{}40k\plus{}24\equal{}100(k^2\plus{}k)\plus{}24$

Answer: 2


Peter arrived to the city and stayed in the hotel. He has no money but few day's later (maximum N days) he will receive a lot of money. And so he has a golden chain with N($1\leq N \leq 10^{16}$) links. Peter must pay one link everyday, but he can pay more and take change by some of his links paid before.
When he receives money he will get his chain out of pawn. Peter likes his chain and he wants to break minimal number of links. Help him!
For example, if N=5, he must break one link (the second). After it his chain will consist of 3 parts with lengths 1, 1 and 3.
At the first day he will pay one link, at the second - one link too. Now he have one part consists of three connected links. At the third day he will pay 3-linked part and take two separated links (as change) back. On the fourth and fifth day he will pay one link.
You may assume that Peter can choose the way to get change.

Hint: all you need is to be greedy.. and to think slightly ahead.

I can't figurate hoow. Can you explain it.

If you were just choosing numbers  to allow all possible sums, you would choose 1, 2, 4, 8.. repeatedly choosing the first number that you can make from the sum of the rest (ie, exactly the sum of the rest).
The only difference here is that every time you break a link, you end up with a chain of size 1 - so after n breaks, you will have 1,1,1,..,1, and some other numbers - all you need to do is decide what the other numbers are, and which n is best..


number theory
greatest common divisor
prime numbers
number theory unsolved

Prove that if $ p,q,r$ are prime numbers such that $ p^{n} \plus{} q^{n} \equal{} r^{2}$ where $ n \in \mathbb{Z}^{ \plus{} }$ then $ n \equal{} 1$.

case work...
when n is odd: $ p\plus{}q|r^2$. because r is prime, $ r^2\equal{}p\plus{}q$ or $ r\equal{}p\plus{}q$
if $ r^2\equal{}p\plus{}q$, n=1.
if $ r\equal{}p\plus{}q$, $ (p\plus{}q)^2\equal{}p^n\plus{}q^n \ge p^3\plus{}q^3(n>1) \implies p\plus{}q \ge p^2\minus{}pq\plus{}q^2 \ge pq$
$ \implies (p\minus{}1)(q\minus{}1) \le 1 \implies p\equal{}q\equal{}2,$ and then r=4, not a prime.
when n is even: there exists a,b,c such that WLOG $ p^\frac{n}{2}\equal{}2abc, q^\frac{n}{2}\equal{}a(b^2\minus{}c^2). 2|p^2,$ hence p=2, so abc=2. a must be 1 (it's the gcd: $ (p^\frac{n}{2},p^\frac{n}{2})\equal{}(p,q)^\frac{n}{2}\equal{}1$ because p and q must be different).
$ bc\equal{}2, q^\frac{n}{2}\equal{}(b^2\minus{}c^2), r\equal{}(b^2\plus{}c^2) \implies b\equal{}2, c\equal{}1, q^\frac{n}{2}\equal{}3, r\equal{}5 \implies n\equal{}2, q\equal{}3$
but $ 2^2\plus{}3^2$ and $ 5^2$ are different, contradiction, hence always $ r^2\equal{}p\plus{}q$.


algebra
polynomial
induction
linear algebra
matrix
binomial coefficients
number theory

Let $a,b,c$ be 3 integers, $b$ odd, and define the sequence $\{x_n\}_{n\geq 0}$ by $x_0=4$, $x_1=0$, $x_2=2c$, $x_3=3b$ and for all positive integers $n$ we have
\[ x_{n+3} = ax_{n-1}+bx_n + cx_{n+1} . \]
Prove that for all positive integers $m$, and for all primes $p$ the number $x_{p^m}$ is divisible by $p$.

All I've managed to observe up until now is that $x_n=r_1^n+r_2^n+r_3^n+r_4^n$, where $r_i$ are the roots of the equation $x^4-cx^2-bx-a=0$.

Maybe it's not important, but it's interesting, isn't it?

This was a super-hard problem, a memorable one. So, this year it seems that each THS comes with a super-hard problem.

Actually, that idea up there might yield a solution.

I think we can show that $p|(r_1+r_2+r_3+r_4)^{p^m}-(r_1^{p^m}+r_2^{p^m}+r_3^{p^m}+r_4^{p^m})$ by using the fact that all symmetric polynomials in $r_1,r_2,r_3,r_4$ have integer values and all binomial coefficients $C_{p^m}^k,\ 0<k<p^m$ are divisible by $p$. On the other hand, $r_1+r_2+r_3+r_4=0$, so we would have $p|(r_1^{p^m}+r_2^{p^m}+r_3^{p^m}+r_4^{p^m})$.

For $p<>3$ that idea solves the problem pretty quickly:

The expression $3[(r_1+r_2+r_3+r_4)^{p^m}-(r_1^{p^m}+r_2^{p^m}+r_3^{p^m}+r_4^{p^m})]$ can be written as a sum of expressions of the form $C_{p^m}^k[(r_1+r_2)^k(r_3+r_4)^{p^m-k}+(r_3+r_4)^k(r_1+r_2)^{p^m-k}+(r_1+r_3)^k(r_2+\nolinebreak r_4)^{p^m-k}+\nolinebreak (r_2+\nolinebreak r_4)^k(r_1+\nolinebreak r_3)^{p^m-k}+(r_1+r_4)^k(r_2+r_3)^{p^m-k}+(r_2+r_3)^k(r_1+r_4)^{p^m-k}],\ 0<\nolinebreak k<p^m$, which are symmetric in $r_i$, and thus have integer values. This, combined with the fact that $p|C_{p^m}^k,\ 0<k<p^m$ shows that $p|(r_1+r_2+r_3+r_4)^{p^m}-(r_1^{p^m}+r_2^{p^m}+r_3^{p^m}+r_4^{p^m})=-(r_1^{p^m}+r_2^{p^m}+r_3^{p^m}+r_4^{p^m})=\nolinebreak -x_{p^m}$, and we're done. You can see that I used $p<>3$ in order to eliminate the $3$ in front of the first expression.

Maybe we can do it separately, by induction, for $p=3$, or maybe we can give a nicer proof of the fact that $p|(r_1+r_2+r_3+r_4)^{p^m}-(r_1^{p^m}+r_2^{p^m}+r_3^{p^m}+r_4^{p^m})$, without multiplying with 3.

Is there anything wrong with this?

[sorry it looks so messy; I got tired of typing "\nolinebreak" :); what else can be done in order to prevent Tex from looking like this?]

hi

imho the followin could be applied when having r1^n+r2^n+r3^n+r4^n= x_n

by induction on m we need to prove that when r1+r2+r3+r4 = 0 mod p we have

r1^p+r2^p+r3^p+r4^p = 0 mod p

using the multinomial expansion, we get that the first expression to the pth power equals the second one plus a series of symmetric polynomials on r1 r2 r3 and r4 that are therefore integers multiplied with combinatorial numbers with four arguments divisible by 3, which solves the ptoblem generally, could it be that b is odd is redundant, that keeps me doubting about the correctness of my solution

I totally forgot about $b$ being odd :? Now I also have some doubts about this idea.

How about it Valentin? Could you post the official solution? I'd like to see how we can use the fact that $b$ is odd.

I agree that no assumption on $b$ is necessary. Another (equivalent, but maybe clearer) way to express the previous solutions is to remark that $x_n$ equals the trace of the $n$th power of the (companion) matrix \[ C = \left[\begin{array}{cccc} 0 & 0 & 0 & a \\1&0&0&b \\ 0&1&0&c\\ 0&0&1&0\end{array}\right],
\] and to apply ($m$ times) Fermat's little theorem \[\text{trace}
(C^p) \equiv \text{trace} (C) \mod p.\]

By the way, the problem is not original, since it appeared as Problem 10655 in the April 1998 American Mathematical Monthly, Vol. 105, No. 4. under the following form: ``Define a sequence $x_0, x_1, x_2, \dots$ by $x_0 = 4, x_1 = x_2 = 0, x_3 = 3$, and $x_{m+4} = x_{m+1} + x_m$ for $m \geq 0$. Prove that $x_p$ is divisible by $p$ for every prime $p$.''

For completeness, note that (pseudo-)primality tests based on such kind of recurrences are well known in the litterature and generalize precisely Fermat's test. A classical recurrence is that of Perrin's, see for instance
\[\texttt{http://mathworld.wolfram.com/PerrinSequence.html}\]
and the references therein, and also
\[\texttt{http://www.math.niu.edu/\~{}rusin/known-math/99/perrin}\]
\[\texttt{http://www.math.niu.edu/\~{}rusin/known-math/94/primalty.tst}\]

Thanks! I was beginning to have doubts about the solution. I can't help wondering, however, about the solution the author thought of. He must have used the fact that $b$ is odd.

very interesting generalization of Fermats theorem. I wonder whether you know more results of this type?

psykyk

jeje, I was reading A Classical introduction to modern number theory, We can make congruences in the ring of algebraic integers and ve always have:

(w1+w2+w3...+wk)^p=w1^p+w2^p+w3^p....wk^p (modp), from here one can prove the main result by induction, but also it is easy to see that , like you said before, that the quotinet is an integer because the symetric polynomial are integers........It is nice working with fields...

we first prove $x_n=\sum_{i=1}^{4}x_i^n$,where $x_i$ are the 4 roots of equation$x^4+0x^3-cx^2-bx-a=0$
then we only need the expansion of $(x_1+x_2+x_3+x_4)^{p^m}$,where all coefficients are multiples of $p$ except $\sum_{i=1}^{4}x_i^{p^m}$.

[url=https://artofproblemsolving.com/community/q4h290384p1569981]https://artofproblemsolving.com/community/q4h290384p1569981[/url]

EDIT: I think there are possiby easier ways of proving the second half of the problem (proving that $p \mid {(r_1+\dots+r_n)}^{p^m}-({r_1}^{p^m}+\dots+{r_n}^{p^m}))$ can be done in a similair way for proving that $p \mid {(r_1+\dots+r_n)}^{p}-({r_1}^{p}+\dots+{r_n}^{p})$, which is done by induction starting at $n=2$, and using the binomial formula and properties of algebraic integers


I have actually posted this problem before, but the solution I got was really long.  I figured I'd repost this here in case anyone has a quick method.

Find the number of zeroes at the end of $4^{5^{6}}+6^{5^{4}}$.

well you could start by simplifining it to

(2^31250) + (6^625)

the units digit is obviously a 0 so the answer is at least 1.

[hide="not sure"]
The last two digits of the powers of $4$ go in the pattern $04$, $16$, $64$, $56$, $24$, $96$, $84$, $36$, $44$, $76$, and then repeats.

There are $10$ terms in the pattern, being raised to a power that has a $5$ in the singles digit (since any power of $5$ will end in $5$), so the last two digits of the first term will be $24$.

The last two digits of the powers of $6$, after the $6^1$, go in the pattern $(36, 16, 96, 76,$56, $36)$. $5^4$ is $625$, which happens to be one more than a multiple of $6$, which means that the last two digits will be the last set in the pattern, so $36$.

$36+24=50$. The answer isn't above $100$, so we don't have to worry about any excess carrying over. So only $1$ zero?

[/hide]

I was told the answer is $5$ zeroes.  I'm not too concerned with that, I'd just like to see how it was arrived at.  I factored $2^{5^{4}}$ out, but am unsure of how to procede after that.  I found this problem in a collection of MOP homework sets on the internet, so I figured it would be quite difficult :?.

The last two digits are 76 and 24 for 6 and 4 respectively, which sum to 100, so there are at least 2 zeros.

the answer is at least 2. the last two digits of hte numbers are 24 and 76

Could someone explain how they figured out that the last two digits for 4 and 6 were 24 and 76?

Some observation reveals that:

6 to a power that is a multiple of 5 ends in [b]76[/b].

4 raised to any odd multiple of 5 ends in [b]24[/b]

4 raised to any even multiple of 5 ends in 76.

Oh I see, miscounted in the pattern. Thanks  :)

[hide]
Well you could always use a computer
the last 9 digits are
...484800000
so the answer is 5.
[/hide]

Hmm... looking for patterns in the third digit, I can't see anything  :(  Though I suspect that in the sum, it'll be a 9.

Here is the link to the first time I posted this:

http://www.artofproblemsolving.com/Forum/topic-45969.html

I was confused as to why mod 25 was chosen and how those calculations mod 25 were performed.  The factoring part I'm alright on.  Does anyone feel like they understand that explanation a little more clearly?


Let $p,q$ be rational numbers. We define a sequence $a_{n}=p^{n}+q^{n},\ n=1,2,\dots$. Assume that there exists integers $x,y$ of different parity such that $a_{x}$ and $a_{y}$ are integers. Prove that $a_{n}$ is an integer for all $n$.

[quote="nayel"]Let $ p,q$ be rational numbers. We define a sequence $ a_{n} \equal{} p^{n} \plus{} q^{n},\ n \equal{} 1,2,\dots$. Assume that there exists integers $ x,y$ of different parity such that $ a_{x}$ and $ a_{y}$ are integers. Prove that $ a_{n}$ is an integer for all $ n$.[/quote]
More we can prove that $ p,q\in Z$

[quote="nayel"]That is also a part of my problem. :wink:  In my solution I first prove that $ p,q$ are integers. It can be easily verified that $ a_n\in\mathbf Z\,\forall n\iff p,q\in\mathbf Z$.[/quote]
Yes it is quite easy.
The first prove that $ p\equal{}\frac{a}{c},q\equal{}\frac{b}{c}$ where $ c>0$ and $ \gcd(ab,c)\equal{}1$
We prove that $ c|a^n\plus{}b^n,\forall n\in N$
So $ c\equal{}1$ .


Find the first positive integer $ x$ such that $ x^{2}\minus{}x\plus{}41$ is not prime.

I know the answer is $ x\equal{}41$ and I also know that $ f(x\plus{}1)\equal{}f(x)\plus{}2x$ (pretty straightforward).  I've been told that we only need to consider the integers 1-13.  Why?

For $ 13 < x < 41$ we have $ x^2 < x^2 \minus{} x \plus{} 41 < (x \plus{} 1)^2$, so if $ x^2 \minus{} x \plus{} 41$ is composite then it cannot be the square of a prime and must have a prime factor less than $ x$.  On the other hand, $ x^2 \minus{} x \plus{} 41$ is irreducible $ \bmod 2, 3, 5, 7$ and is therefore divisible by at least either $ 11 \cdot 13$ or $ 11^3$.  I can't find a quick way to finish this argument, though.

Anyway, if you're not familiar with the motivation behind this question, you might be interested in [url=http://en.wikipedia.org/wiki/Heegner_number]this Wikipedia article[/url].

A TA brought that article (well, not the article, the story behind it) to my attention.  Our TA plans to prove this to us so when I get a complete proof, I'll post it for you.

This is similar to a previous IMO problem, but the x coefficient is negative, can anyone change the proof?
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=366550#366550[/url]


Let $ I$ be the incenter of triangle $ ABC$. Let $ K,L, M$ be the points of tangency of the incircle of $ ABC$ with $ BC$ , $ CA, AB$, respectively. The line  which passes through $ B$ and is parallel to $ MK$ meets line  $ KL$ at the point $ S$.Line $ IB$ meets line $ MK$ at $ R$. Prove that $ IS$ is perpendicular to $ RC$.

Babis

Perphaps any other proof without using polars would be equivalent.

Let $r$ denote the parallel line to $MK$ from $B$ $\Longrightarrow$ $R$ is the pole of $r$ WRT $ (I).$ Polar of $S$ WRT $ (I),$ which passes through the intersection $C$ of the tangents of $(I)$ at $K,L,$ goes through the pole $R$ of $r$ $ \Longrightarrow$ $RC$ is therefore the polar of $ S$ WRT $(I)$ and consequently $RC \perp IS.$


[quote="Lovasz"]Suppose $a^{2}=b^{2}+c^{2}$ where $a,b,c>0$. Prove that: \[{5 \over 4}(a^{2}+b^{2}+c^{2})+2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right) \leqslant \\ \leqslant{3 \over 2}(ab+bc+ca)++\sqrt{a+b-c}+\sqrt{b+c-a}\&+\&\sqrt{c+a-b}.\] [/quote]

Sth's wrong. Maybe you forgot to write something (there are two plus signs between two sums in last line)?

Counterexample is $(\sqrt{2}, 1, 1)$.

This "nonsensical" condition gives us that $a, b, c$ are sides of a right angled triangle.


i'm confused about this one:

[img]http://img10.imageshack.us/img10/3755/40973703.jpg[/img]


geometry
circumcircle
Euler
rectangle
geometry proposed

Given circle $(O)$ and five point $A_{1},A_{2},A_{3},A_{4},A_{5}$ lie on it, line $d_{45}$ is through centroid of triangle $A_{1}A_{2}A_{3}$ and perpendicular with $A_{4}A_{5}$, line $d'_{45}$ is connecting midpoint of segment $A_{4}A_{5}$ and othorcenter of triangle $A_{1}A_{2}A_{3}$ similarly we will get 20 lines $d_{ij},d'_{ij}1\le i,j\le 5$ prove that 20 lines are concurrent.
P/s:We can extend to cyclic polygon by to define othorcenter of cyclic polygon.

It is sufficient to prove concurrency of 3 lines $d'_{45},\ d_{45},\ d_{51}.$ Rename $A_{1}A_{2}A_{3}A_{4}A_{5}$ as $ABCDE$ and $d'_{45},\ d_{45},\ d_{51}$ as $m_{DE},\ n_{DE},\ n_{EA}.$

Let $O,\ \mathcal O,\ R$ be the circumcenter, circumcircle and circumradius of the pentagon ABCDE. Let H be the orthocenter of the $\triangle ABC$ and M the midpoint of DE, $HM \equiv m_{DE}.$  Let G, G' be centroids of the $\triangle ABC,\ \triangle BCD.$ Denote $X \equiv m_{DE}\cap n_{DE},\ Y \equiv n_{DE}\cap n_{EA}.$ HGO is Euler line of the $\triangle ABC.$ Since $GX \parallel OM$ (both perpendicular to DE), $\triangle HGX \sim \triangle HOM$ are centrally similar with center H and coefficient $\frac{HG}{HO}= \frac{2}{3}.$ Let $\mathcal P$ be the circumcircle of the $\triangle DMO,$ tangent to $\mathcal O$ at D and with radius $\frac{R}{2}.$ Let $\mathcal Q$ be a circle through G, X centrally similar to $\mathcal P$ with similarity center H and coefficient $\frac{GX}{OM}= \frac{2}{3},$ i.e., with radius $\frac{R}{3}.$ Let DA meet $\mathcal P$ again at U (the midpoint of DA) and let a parallel to DA through O meet $\mathcal P$ again at V. DUOV is a rectangle (with diagonal DO, a diameter of $\mathcal P$), $OV \parallel AD,$ $OV = UD = \frac{AD}{2}.$ Let N be the midpoint of BC. AN, DN are the A-, D-medians of the $\triangle ABC,\ \triangle BCD,$ hence $NG = \frac{NA}{3},\ NG' = \frac{ND}{3},$ so that $GG' \parallel AD \parallel OV$ and $GG' = \frac{AD}{3}= \frac{2 OV}{3}.$ Since $V \in \mathcal P,$ it follows that $G' \in \mathcal Q.$ Since $\angle GYG' = 180^\circ-\angle DEA = \angle ABD = \angle UOD = \angle OMV$ and $M \in \mathcal P,$ it follows that $Y \in \mathcal Q,$ i.e., $n_{DE}$ meets $\mathcal Q$ at G and again at Y. But $n_{DE}$ meets $\mathcal Q$ at G and again at X, hence $Y \equiv X$ are identical and $X \equiv m_{DE}\cap n_{DE}\cap n_{EA}.$


trigonometry
geometric series
calculus
calculus computations

Evaluate:

$\sum_{i=1} ^{\infty} \sin \frac{\pi}{2^i}$

[hide]
I think the answer is 2.
[/hide][/hide]

I haven't mentioned, but I'm looking for a proof, not only numerical answer  :)

Are you sure (do you know if) there is a nice closed form ?
This sum is something more like 2.481049919... than 2.

I'm not 100% sure but it was given by my teacher and he claimed to had seen nice proof of it  :)

I'm not sure I'm on the right track, but this is what I came up with:

We have $a_n = 2a_{n+1}\sqrt{1-a_{n+1}^2}$, solving for $a_{n+1}$ gives:

$a_{n+1} = \sqrt{\frac{1-\sqrt{1-a_n^2}}{2}}$

We can write the sum as

$a_1 + \sqrt{\frac{1-\sqrt{1-a_1^2}}{2}} + \sqrt{\frac{1-\sqrt{1-\frac{1-\sqrt{1-a_1^2}}{2}}}{2}} + ...$

But $a_1$ is just $1$, so the sum reduces to

$1+\sqrt{\frac{1}{2}} + \sqrt{\frac{1-\sqrt{\frac{1}{2}}}{2}} + ...$

This looks somewhat familiar, so maybe some one knows where to go from here.

I messed up.

I think it is actually $\sum_{k=0}^{\infty}\frac{\pi^{2k+1}}{(2^{2k+1}-1)(2k+1)!}$

They way I got that was by turning the sine into a power series and using the formula for the sum of an infinite geometric series on each term.

$\sum_{p=1}^{\infty}sin(\frac{\pi}{2^p})$
$=\sum_{p=1}^{\infty}\sum_{w=0}^{\infty}\frac{\pi^{2w+1}}{2^{2wp+p}(2w+1)!}$
$=\sum_{w=0}^{\infty}\sum_{p=1}^{\infty}\frac{\pi^{2w+1}}{2^{p(2w+1)}(2w+1)!}$
$=\sum_{w=0}^{\infty}\frac{\pi^{2w+1}}{2^{2w+1}(2w+1)!}(1-2^{-2w-1})^{-1}$


From a point A, draw three segments from this point (I assume not all three can be of the same or opposite direction). Complete three paralellograms having two of the segments as its sides. The resulting figure should look like a parallelopiped. This theorem states that the resulting figure is also a possible view of a cupe. (The view is normally oblique or slanting.)

Web search does not seem to give a satisfying proof to this.. And I'm not too familiar with projective geometry. If anyone has a proof to this theorem (or even, counter-example), it would be appreciated.. Thanks!  :)

http://mathworld.wolfram.com/PohlkesTheorem.html

 it took me two days to solve it with transformation geometry,,,

but the solution was exactly like this: http://mathworld.wolfram.com/PohlkesTheorem.html

you can refer to "101 historian problems in mathematics" book ,there is complete solutions.


f(x)=log base 4 of (64^-2x+8(16^3x-7))

write it without the logarithm.

I'm guessing you have to make it 4^y=(64^-2x+8(16^3x-7))
then make y=-6 so f(x) would be 4^-6 or 1/4^6 ?

[quote="neelnal"]Everything you did was correct. I do not understand what you are asking us to do for you...[/quote]

well apparently what i did was wrong or this program i am using is totally screwed up

Can anyone tell me what I did wrong or is this program totally retarded?


calculus
Putnam
linear algebra
complex analysis
\/closed

[quote="13375P34K43V312"]Is there any way you could offer classes on college topics, such as Linear Algebra and Calculus?[/quote]
I'm no admin, but I wouldn't think so because:
http://www.artofproblemsolving.com/Resources/AoPS_R_A_Calculus.php
But maybe I'm overthinking this...
Also, shouldn't this be in classes information?

[b]If[/b] we would ever offer a college-level class, it would be something in the lines of a preparation for Putnam, rather than simply learning the basics of Calculus and Linear Algebra.

As said above you can simply attend a university class for that, or use other methods to grab the basics of those two fields.

[quote="neelnal"]You can talk to your school about getting a teacher to teach the class or allowing you to go to a uni during a school day. That is what our school does. We have linear algebra, multicalc, and diff. equations with complex analysis teacher.[/quote]
You realize that leetspeak is in middle school right?


Source: Massachusetts Mathematics League December 2005

A right prism has a square base. Consider the set of all angles formed by a diagonal of a face and a diagonal of the prism. If one such angle has a tangent of 2, find the sum of the tangents of all the angles in the set.

[hide]
Wouldn't it be two for all the angles then? Sicne the base is regular, you can rotate it and it will be the same.
[/hide]

[hide] Let the long edge of the prism be $2$. Then the diagonal of the square base is $1$ and the short edges are $\frac{ \sqrt{2}}{2}$. The diagonal of the long face is $\sqrt{2^2 + (\frac{ \sqrt{2}}{2})^2} = \frac{3\sqrt{2}}{2}$.
Then, since there are four short diagonals and four long diagonals, each making two angles, there are eight of each angle. So we have:
$8(2)+8(\frac{\frac{\sqrt{2}}{2}}{\frac{3\sqrt{2}}{2}}) = \boxed{\frac{56}{3}}$[/hide]

[quote="weatherbottle"][hide]No, a right prism with a square base can have any height $h$.  ;) [/hide][/quote]

I know that, but because we are taking the tangent of those sides, the height doesnt matter. I used 2 for convenience.  ;)


Hi
I have a problem in using the dvi-viewer in TeXnicCenter. I'm used to be able to make a double click in the dvi-file to get immediatly to the right place in my tex-file. Until now, this worked well. But now, my dvi-viewer tells me:

Cannot open the TeX source file: the DVI page contains no source links.

I do not understand the Help-Part. So does anyone know what I can do about this?

I think the problem only arises if I use the documentclass{article}. For documentclass{amsart} it works.

Thanks

Do you get any source links in the DVI file (Yap, Tools, Source Links...? What happens if you add the command line --src-specials (TeXnicCenter, Build, Define Output Profiles, Command line arguments to pass to the compiler)?

Could you post a minimal document where you get the problem with the article class but which works with amsart?

hi stevem
I put in --src-specials and now it works. Thanks.
(I honestly tried it out before already... but i probably wrote special instead of specials)

Thanks a lot,
ignaz

(Still funny, that it depends on the documentclass...)

I think I had this problem as well before. Search this forum for this problem and there's a solution there that might work.


advanced fields
advanced fields unsolved

Let $H$ be a Hamel basis of a field $R$ over a subfield $Q$. Show that for each $a \neq 1$ there is an element $x \in H$ such that $ax \notin H$.

Let $H=\{h_i\}_{i\in I}$, and let $h$ be an arbitrary element of $H$. For all $i\in I$, denote $ah_i\in H$ by $h^i$. Also let $h$ be an arbitrary element of $H$. If $\frac h{a-1}=\sum\alpha_ih_i\ (*)$ is the decomposition of $\frac h{a-1}$, then $\frac{ha}{a-1}=\sum\alpha_ih^i\ (**)$. Subtracting $(*)$ from $(**)$, we get $h=\sum\alpha_ih_i-\sum\alpha_ih^i$. It's easy to get a contradiction from this by simply counting the non-zero components in the basis $H$ for the LHS and RHS.


Hope this is what you mean:

${ 2005^{(2003 ^ {2004})  + 3 }}  $ $  \bmod 9 $

If this is the case  Hint ( almost give away)

[hide]$2005  \equiv  -2 \bmod 9  $
$2003 \equiv  -1 \bmod  6  $
$a^6 \equiv  1 \bmod 9  $ for any a prime to 9
$a^2 \equiv   1 \bmod 6 $ for any a prime to  6 etc ...

[/hide]

[hide="solution"]

since the last two digits in base three means (mod 9) in base 10 (and then changed to base three again),
we can use [tex] 2005 \equiv 7 \pmod{9} [/tex] and find that [tex] 7^3 \equiv 1 \pmod{9} [/tex]
so then we just have to find [tex] 2003^{2004}+3 \equiv 2003^{2004} \equiv 2^{2004} \equiv 1 \pmod{3} [/tex]

so then [tex] 2005^(2003^{2004}+3) \equiv 7^1 \equiv 7 \pmod{9} [/tex]

which in base 3 is [tex] 21_3 [/tex]

[/hide]


[hide="A Step?"]I'm not too good with these type of series, but here's an idea that at least eliminates the trigonometric functions with $\cos x=\frac{1}{5}$:

$\sum_{n=0}^{\infty}\frac{\cos nx}{2^{n}}=\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\binom{n}{2k}\frac{\left(-24\right)^{k}}{10^{n}}$[/hide]

Scary...Does that actually make it any better? (I'm assuming it's true.)

I found it on an old MAO topic test (logs and exponents...kind of out of place..?)

$e^{i\pi}=\cos(\theta) + i\line(\theta)$
Since geometric series are easy to find the sums of, change the numerator to complex exponential form, find the infinite sum, and take the real part.
[quote="t0rajir0u"]The problem is familiar; I think it's been posted before, but I can't find it.[/quote]
Think AoPS vol 2, the chapter on complex numbers. ;)

Nice; good call.

$\sum_{n=0}^{\infty}\frac{\cos nx}{2^{n}}=\textrm{Re}\left(\sum_{n=0}^{\infty}\left(\frac{e^{ix}}{2}\right)^{n}\right)$

$=\textrm{Re}\left(\frac{1}{1-\frac{e^{ix}}{2}}\right)=\frac{2\cos x -4}{4\cos x -5}=\boxed{\frac{6}{7}}$

Well that beats using a double-sum :).


Find the electric field caused by a disk of radius R with a uniform positive surface charge density (charge per unit area), at a point along the axis of the disk a distance x from its center. Assume that x is positive.

Firstly we take small dq part at the radial distance r from the center of the disc.
 $dq=\sigma2\pi r dr$
 $dE=k\frac{dq}{l^{2}}$
 $dE_{1}=dEcos\theta=k\frac{dq}{l^{2}}*\frac{x}{(x^{2}+r^{2})^\frac{3}{2}}$
After integration of this equation from 0 to R, we get the following result:

$E_{1}=\frac{\sigma x}{2e_{0}}*(\frac{1}{x}-\frac{1}{(x^{2}+R^{2})^\frac{1}{2}})$

Your sols is absolutely corrected. Congratulation ! :lol:


limit
trigonometry
calculus
calculus computations

Hi.  I know how to solve:

$\lim_{x\rightarrow 0}\frac{\sin x}{x}$ with the squeeze thereom,

but am unsure how to solve this question which corresponds to the example which is:

$\lim_{x\rightarrow 0}\frac{\sin x^2}{x^2}$

i have graphed it and it seems to be 1 also but i am unsure of the proof due to the x^2.

For   $y=x^2$    it is the same limit .  :)

[quote="Liu Ce"]The key to this question is:) $\lim_{x\rightarrow 0}\frac{\sin x^2}{x^2}=\lim_{x\rightarrow 0}(\frac{\sin x}{x})^2=(\lim_{x\rightarrow 0}\frac{\sin x}{x})*(\lim_{x\rightarrow 0}\frac{\sin x}{x})=0$

I'm from China,I'm glad to make friends with you.Can you show me your MSN?[/quote]

I think your meant was $\lim_{x\rightarrow 0}\frac{\sin x^2}{x^2}=1$ not $=0$.  ;)

[quote="Liu Ce"]The key to this question is:) $\lim_{x\rightarrow 0}\frac{\sin x^2}{x^2}=\lim_{x\rightarrow 0}(\frac{\sin x}{x})^2=(\lim_{x\rightarrow 0}\frac{\sin x}{x})*(\lim_{x\rightarrow 0}\frac{\sin x}{x})=0$

I'm from China,I'm glad to make friends with you.Can you show me your MSN?[/quote]

I think it is $\lim_{x\rightarrow 0}\frac{\sin (x^2)}{x^2}$ and not $\lim_{x\rightarrow 0}\frac{(\sin x)^2}{x^2}$    :)

yes, if $y=x^2$, then $\lim_{x\rightarrow 0} x^2 = \lim_{y\rightarrow 0} y$, and thus we have \[ \lim_{x\rightarrow 0}\frac{\sin x^2}{x^2}=\lim_{y\rightarrow 0}\frac{\sin y}{y}= 1 \]


algebra
polynomial
superior algebra
superior algebra unsolved

How many monic cubic irreducible polynomials   $ f \in  \mathbb{F}_3[X]$ are there  such that $ f(0)\equal{}\minus{}1$?.
I found 4: namely the monic cubic irreducible polynomials which divide the cyclotomic polynomial $ \phi_{13}$ mod $ 3$. Are there more?
Thankss!

There are eight degree 3 irreducibles total; half of them having unit coefficient $ \minus{}1$ makes sense. You have them all.


Let:$ a;b;c>0$.Find minimun:
$ P\equal{}\frac{b(a\minus{}c)}{c(a\plus{}b)}\plus{}\frac{c(3b\plus{}a)}{a(b\plus{}c)}\plus{}\frac{3c(a\minus{}b)}{b(a\plus{}c)}$

[quote="Toan_VN_LC"][quote="caubetoanhoc94"]Let:$ a;b;c > 0$.Find minimun:
$ P \equal{} \frac {b(a \minus{} c)}{c(a \plus{} b)} \plus{} \frac {c(3b \plus{} a)}{a(b \plus{} c)} \plus{} \frac {3c(a \minus{} b)}{b(a \plus{} c)}$[/quote]
putting$ x \equal{} ab \plus{} bc;y \equal{} bc \plus{} ca;z \equal{} ca \plus{} ab$
then,
$ P \equal{} \frac {{z \minus{} y}}{y} \plus{} \frac {{x \plus{} 2y \minus{} z}}{z} \plus{} \frac {{3z \minus{} 3x}}{x}$
by AM-GM:
$ P \equal{} \frac {{z \minus{} y}}{y} \plus{} \frac {{x \plus{} 2y \minus{} z}}{z} \plus{} \frac {{3z \minus{} 3x}}{x} \equal{} \left( {\frac {z}{y} \plus{} \frac {{2y}}{z}} \right) \plus{} \left( {\frac {x}{z} \plus{} \frac {{3z}}{x}} \right) \minus{} 5 \ge 2\sqrt 2 \plus{} 2\sqrt {3} \minus{} 5$
with equal iff $ z \equal{} \sqrt 2 y \equal{} \frac {x}{{\sqrt 3 }}$
 :lol:  :lol:[/quote]
Nice.Thanks; my brother  :lol:

[quote="Unidranix"][quote="Toan_VN_LC"][quote="caubetoanhoc94"]Let:$ a;b;c > 0$.Find minimun:
$ P \equal{} \frac {b(a \minus{} c)}{c(a \plus{} b)} \plus{} \frac {c(3b \plus{} a)}{a(b \plus{} c)} \plus{} \frac {3c(a \minus{} b)}{b(a \plus{} c)}$[/quote]
putting$ x \equal{} ab \plus{} bc;y \equal{} bc \plus{} ca;z \equal{} ca \plus{} ab$
then,
$ P \equal{} \frac {{z \minus{} y}}{y} \plus{} \frac {{x \plus{} 2y \minus{} z}}{z} \plus{} \frac {{3z \minus{} 3x}}{x}$
by AM-GM:
$ P \equal{} \frac {{z \minus{} y}}{y} \plus{} \frac {{x \plus{} 2y \minus{} z}}{z} \plus{} \frac {{3z \minus{} 3x}}{x} \equal{} \left( {\frac {z}{y} \plus{} \frac {{2y}}{z}} \right) \plus{} \left( {\frac {x}{z} \plus{} \frac {{3z}}{x}} \right) \minus{} 5 \ge 2\sqrt 2 \plus{} 2\sqrt {3} \minus{} 5$
with equal iff $ z \equal{} \sqrt 2 y \equal{} \frac {x}{{\sqrt 3 }}$
 :lol:  :lol:[/quote]

Your solution is nice but I don't think it right.
$ z \equal{} \sqrt {2}y$ or $ ca \plus{} ab \equal{} \sqrt {2}bc \plus{} \sqrt {2}ca$
or $ ab \equal{} \sqrt {2}bc \plus{} (\sqrt {2} \minus{} 1)ca$  $ (1)$
Also:
$ \sqrt {6}y \equal{} x$ or $ ab \plus{} bc \equal{} \sqrt {6}bc \plus{} \sqrt {6}ca$
or $ ab \equal{} (\sqrt {6} \minus{} 1)bc \plus{} \sqrt {6}ca$   $ (2)$

$ From (1) and (2)$:
$ \sqrt {2}bc \plus{} (\sqrt {2} \minus{} 1)ca \equal{} (\sqrt {6} \minus{} 1)bc \plus{} \sqrt {6}ca$
Clearly wrong because $ (\sqrt {6} \minus{} 1)bc > \sqrt {2}bc$ and $ \sqrt {6}ca > (\sqrt {2} \minus{} 1)ca$

Hope that I'm not wrong[/quote]
I see,we have a problem here :(

[quote="Toan_VN_LC"][/quote][quote="Unidranix"][quote="Toan_VN_LC"][quote="caubetoanhoc94"]Let:$ a;b;c > 0$.Find minimun:
$ P = \frac {b(a - c)}{c(a + b)} + \frac {c(3b + a)}{a(b + c)} + \frac {3c(a - b)}{b(a + c)}$[/quote]
putting$ x = ab + bc;y = bc + ca;z = ca + ab$
then,
$ P = \frac {{z - y}}{y} + \frac {{x + 2y - z}}{z} + \frac {{3z - 3x}}{x}$
by AM-GM:
$ P = \frac {{z - y}}{y} + \frac {{x + 2y - z}}{z} + \frac {{3z - 3x}}{x} = \left( {\frac {z}{y} + \frac {{2y}}{z}} \right) + \left( {\frac {x}{z} + \frac {{3z}}{x}} \right) - 5 \ge 2\sqrt 2 + 2\sqrt {3} - 5$
with equal iff $ z = \sqrt 2 y = \frac {x}{{\sqrt 3 }}$
 :lol:  :lol:[/quote]

Your solution is nice but I don't think it right.
$ z = \sqrt {2}y$ or $ ca + ab = \sqrt {2}bc + \sqrt {2}ca$
or $ ab = \sqrt {2}bc + (\sqrt {2} - 1)ca$  $ (1)$
Also:
$ \sqrt {6}y = x$ or $ ab + bc = \sqrt {6}bc + \sqrt {6}ca$
or $ ab = (\sqrt {6} - 1)bc + \sqrt {6}ca$   $ (2)$

$ From (1) and (2)$:
$ \sqrt {2}bc + (\sqrt {2} - 1)ca = (\sqrt {6} - 1)bc + \sqrt {6}ca$
Clearly wrong because $ (\sqrt {6} - 1)bc > \sqrt {2}bc$ and $ \sqrt {6}ca > (\sqrt {2} - 1)ca$

Hope that I'm not wrong[/quote][quote="Toan_VN_LC"]
I see,we have a problem here :([/quote]
Well, I could not find any fault in your proof & according to the AGM that you used; the equality case should be $ z = \sqrt {2}y = \frac {x}{\sqrt {3}}$......
But , as Unidranix showed, the equality does not occur;so:
[b]conclusion[/b]
$ P$ has NO Minimun..........    :P  :oops:


Prove that
$\sqrt{2+\sqrt[3]{3+\sqrt[4]{4+...+\sqrt[n]{n}}}}< 2$ where r is a positive integer and $n \ge 2$

For $x>2$ we have $2^{x}>x+2$ thus $\sqrt[x]{x+2}<2$. Starting with $\sqrt[n]{n}<2$, we find $\sqrt[n-1]{...}<\sqrt[n-1]{n+1}<2$ and so on, successively each root is $<2$ . :)


algebra
polynomial
Gauss
function
quadratics
Vieta
complex numbers

hint: ab have product 1, and so $\sqrt a \cdot \sqrt b =\pm 1$

That's my point!!! It's W-R-O-N-G!!

 :harhar:

Ok, here's a hint (don't look before you try this question):

[hide]
$a + b = -6$
$ab = 1$
$so, a, b < 0$
:D
[/hide]

Answer should be $-8$?  :o  :huh:  ,something's wrong....

[quote="gauss202"]This is a neat problem.

[hide]It really depends on what is meant by [b][i]the[/i][/b] square root of a negative number. Square roots of positive numbers are taken, by convention, to be positive values.

There's a similar convention for square roots of negative numbers - namely, the square root of $-a, a > 0$ is usually taken to be $i\sqrt{|a|}$

The problem with this definition is that it does not respect multiplication; so the rule usually learned as $\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$ no longer holds when $a$ and $b$ are negative numbers (or complex for that matter).

For example, $\sqrt{(-4)(-9)} = \sqrt{36} = 6$
But $\sqrt{-4} \cdot \sqrt{-9} = (2i)(3i) = -6$

The lesson to take from all of this is to be careful with the use of the rule $\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$. It is fragile. It only holds when you *know* that at least one of $a$ or $b$ is a non-negative real number.

Great problem![/hide][/quote]

Yup the solution should be $-8$. It might seem easy, but most of us will tend to take $\sqrt{ab}$ as $1$ since $ab = 1$. But we should check the signs of $a, b$ first.
 Generalizing what gauss said, $(\sqrt{-x})(\sqrt{-y}) = (\sqrt{x})i(\sqrt{y})i = -\sqrt{xy}$
Thus $a + b + 2(\sqrt{a})(\sqrt{b}) = -6 + 2(-\sqrt{ab}) = -6 -2 = -8$
 :D

[quote="mna851"]Hmmm, what's wrong with this solution.  If the polynomial $\ f(x)$ has roots $\ a$ and $\ b$ then the polynomial $\ f(x^2)$ will have roots $\ \sqrt a$ and $\ \sqrt b$.  In this case since $\ f(x) = x^2 + 6x + 1$, $\ f(x^2)=x^4 + 6x^2 + 1$.  So, the sum of the roots is $\ 0$.  It could be the whole bit about the conventional definition of the square root.  Yet, it seems I did not approach that aspect in my solution[/quote]

well, if i'm not wrong, for $\ f(x^2)=x^4 + 6x^2 + 1$, the sum of the roots is indeed $0$. but then, its a power 4 polynomial. so there should be another 2 roots, $c, d$ where $\ \sqrt a + \sqrt b +c + d = 0$
so $\ \sqrt a + \sqrt b$ is not 0.

Our sir asked the same thing about $\sqrt{-a-b}$ to our class and half of the fools replied wrong :D Though its not a pre-olympiad problem,don't you think  ;)

[quote="Virgil Nicula"][quote="violetcraze"]$a,b$ are the roots of $x^2 + 6x + 1 = 0$

find the value of $(\sqrt{a} + \sqrt{b})^2$    :D[/quote]The roots $a,b$ are real negative numbers $(D>0\ ,\ S<0\ ,\ P>0)$ and $\sqrt a ,\sqrt b$ haven't sense ! Stop.
[b]A similar exercise.[/b] Find the $f(3)$ for the function $f: [-2,5]\rightarrow R$ where $f(x)=\sqrt{6x+7}\ !$[hide="Answer."] $f(3)$ has not sense because $f$ isn't a function, i.e. for any $x\in [-2,5]$ exists and is unique $y\in R$ (the image of [b]x[/b]) so that $f(x)$ has real sense and $f(x)=y\ !$[/hide][/quote]
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Virgil, what is your reason on saying that...?
See below.
http://mathworld.wolfram.com/ImaginaryUnit.html

In my opinion, the symbol $\sqrt .$ has sense only on $[0,\infty )\ !$
The imaginary unit $i=(0,1)$ (corectly) and $i=\sqrt{-1}$ isn't.
But $i\cdot i=(0,1)\cdot (0,1)=(-1,0)=-1$ is corectly !
I ask you : what is $\sqrt{-1}\ ?$ Is $i$ or $-i\ ?$ But $\sqrt{2i}\ ?$ Is $1+i$ or $-1-i\ ?$

[color=darkred][b]Remark.[/b] For any $a,b\in R$ define the sum $(a,b)\oplus (c,d)=(a+c,b+d)$ and the product $(a,b)\odot (c,d)=(ac-bd,ad+bc)\ .$ The set $C=(R^2\ ,\ \oplus\ ,\ \odot)$ means the field of the complex numbers ! If define for any real number $a$ the ordered pair $\overline a=(a,0)$ and $i=(0,1)$ then $(a,b)=(a,0)\oplus [(b,0)\odot (0,1)]=\overline a\oplus \overline b\odot i\ .$ In the following step construct the injection ([b]"dip") [/b]$R\rightarrow R^2$, where $a\rightarrow \overline a$. We came to the conclusion that $(a,b)=a+bi$ and in this way construct the algebraic form of a complex number ![/color]

What is $\sqrt{(a,b)}$, i.e. $\sqrt{a+bi}\ ?$ The question what are $z\in C$ for which $z^2=a+bi$ is corectly.

The symbol $\sqrt{\cdot}$ [i]can[/i] be made to have sense on $(-\infty, 0)$, and in fact on the complex numbers, by choosing what is called a [i]branch[/i] of the (naturally multivalued) function. This involves using a convention to determine which values the function should return. (In technical language this involves choosing a [url=http://mathworld.wolfram.com/BranchCut.html]branch cut[/url] in the complex plane).

It is [i]conventional[/i] to take the principal branch for the "square root function" so that it returns values either on the positive real axis [i]or[/i] in the upper half of the complex plane.

In this way the "square root function" can be made to have a unique sense on all complex numbers. But it is done by a convention - similar to the way in which we choose the square root of positive real numbers to be positive. It also causes the rule about commutivity of square roots and multiplication to fail.

[quote="justmath11"]I believe that the answer is -4.  i think sen got the same answer as mine... but my solution is less elegant.

I did the long method... I solve the two roots by quadratic formula.. and I use them to solve the expression.[/quote]
That's right. No need for the long way. Using Vieta theorem, obtain:
$a+b=-6$;
$ab=1$
Now, $(\sqrt{a}+\sqrt{b})^2=(a+b)+2*\sqrt{ab}$
$=-6+2=-4$. :)

Yes you are right. I didn't pay attention. The answer is $-8$. :!: