Putnam
algebra
polynomial
modular arithmetic
function
group theory
binomial theorem

Let $ p$ be a prime number. Let $ h(x)$ be a polynomial with integer coefficients such that $ h(0),h(1),\dots, h(p^2\minus{}1)$ are distinct modulo $ p^2.$ Show that $ h(0),h(1),\dots, h(p^3\minus{}1)$ are distinct modulo $ p^3.$

Show that for $ p \nmid m$, $ h(mp \plus{} k) \minus{} h(k) \equal{} S(m,k) p \pmod{p^2}$ for some integer $ S(m,k)$ by binomial expansions. So $ p \nmid S(m,k)$ for all $ m,k$ with $ p \nmid m$. Now if two residues of $ h(0), h(1) \ldots h(p^3 \minus{} 1)$ are not distinct mod $ p^3$, that is $ h(a) \equiv h(b) \pmod{p^3}$ with $ a \neq b$, then $ a \equal{} mp^2 \plus{} b$ for some $ m$ not divisible by $ p$, otherwise we get a contradiction of the condition. We find that $ h(mp^2 \plus{} b) \minus{} h(b) \equal{} S(m,b) p^2 \pmod{p^3}$, where that $ S$ is the same function as before. Convenient, because we know $ p \nmid S(m,b)$, and we're done.

Nice notation. I spent like half an hour trying to figure out why $ p \nmid S(m,b)$ was impossible basically because I was using $ \sum_{i\equal{}1}^d a_i i m b^{i\minus{}1}$ instead of just $ p \nmid S(m,b)$.

I was surprised how similar to last year's B1 this was (once again the binomial formula under modulo). Mine (in a less compacted form)

First, since $ p^3 |\,p^2$, $ h(0),...,h(p^2 \minus{} 1)$ are distinct mod $ p^3$. Let $ h(x) \equal{} \sum_{i \equal{} 0}^nc_ix^i$

Consider $ h(k \plus{} p^2) \equal{} \sum_{i \equal{} 0}^nc_i(k \plus{} p^2)^i$
$ \equal{} \sum_{i \equal{} 0}^nc_i\sum_{j \equal{} 0}^i\binom{i}{j}k^{i \minus{} j}p^{2j}$
$ \equal{} \sum_{i \equal{} 0}^nc_ik^i \plus{} \sum_{i \equal{} 1}^nc_i\binom{i}{1}k^{i \minus{} 1}p^2 \plus{} \sum_{i \equal{} 2}^nc_i\sum_{j \equal{} 2}^i\binom{i}{j}k^{i \minus{} j}p^{2j}$

The first term is simply $ h(k)$, the last one disappears mod $ p^3$ since every term includes $ p^{2j}, j > 2$

$ \equal{} h(k) \plus{} p^2\sum_{i \equal{} 1}^nic_ik^{i \minus{} 1}$

Call the number inside the summation $ d$. If $ d\not\equiv 0$ mod $ p^3$ then we see the sequence $ h(k), h(k \plus{} p^2), h(k \plus{} p^3), ..., h(k \plus{} (p \minus{} 1)p^2)$ are all distinct mod $ p^3$ and differ by multiples of $ p^2$.

So, consider $ h(k \plus{} p) \equal{} \sum_{i \equal{} 0}^nc_i(k \plus{} p)^i$
$ \equal{} \sum_{i \equal{} 0}^nc_i\sum_{j \equal{} 0}^i\binom{i}{j}k^{i \minus{} j}p^j$
$ \equal{} \sum_{i \equal{} 0}^nc_ik^i \plus{} \sum_{i \equal{} 1}^nc_i\binom{i}{1}k^{i \minus{} 1}p \plus{} \sum_{i \equal{} 2}^nc_i\sum_{j \equal{} 2}^i\binom{i}{j}k^{i \minus{} j}p^{j}$

The first term is our friend $ h(k)$, the last nothing mod $ p^2$, and the middle $ pd$. Since we knew $ h(k)\not\equiv h(k \plus{} p)$ mod $ p^2$ therefore $ d\not\equiv0$ mod $ p^2$ (and $ p^3$). Proof extensible to any power of $ p$ in fact.

I was wondering if you could use group theory to solve this problem since $ h(x)$ permutes the numbers $ 0, 1, \cdots, p^2 \minus{}1$ among themselves, so $ h\in S_{p^2}\subset S_{p^3}$?

No, because permuting elements mod p^3 is a lot more restrictive than permuting elements mod p^2

Even so, couldn’t you use Zn^2 and the orbit generated by h(x) in to apply it to Zn^3?

I'm not clear on what that means.  There isn't a natural way to extend a generic permutation on $ \mathbb{Z}_{p^2}$ to $ \mathbb{Z}_{p^3}$, and polynomial maps don't lift up in a trivial way, just down.

Here's an argument which isn't mine but comes from a colleague of mine.

The key is the following lemma:

Suppose $ f$ is an integer-coefficient polynomial, $ p$ a prime and $ k\ge1$ and $ i,t$ integers, then $ f(i\plus{}tp^k)\equiv f(i)\plus{}tp^kf'(i)\pmod{p^{k\plus{}1}}.$

We could prove this lemma in a number of ways. We could prove it for each monomial through the binomial theorem, or we could simply expand out the Taylor series (which terminates as a polynomial) and note that all of the terms not explicitly given above contain powers of $ p$ that are at least $ p^{2k},$ and $ 2k\ge k\plus{}1$ since $ k\ge 1.$

We will apply this lemma twice to our particular polynomial $ h.$ First we note that $ h(i\plus{}p)\equiv h(i)\plus{}ph'(i)\pmod{p^2}.$ But since $ h(i\plus{}p)\not\equiv h(i)\pmod{p^2},$ it follows that $ ph'(i)\not\equiv 0\pmod{p^2}$ or $ p$ does not divide $ h'(i)$ for each $ i.$

Now we want to show that $ h$ gives a surjection on the residue classes mod $ p^3.$ Of course, we have that if $ i\equiv j\pmod{p^k}$ then $ h(i)\equiv h(j)\pmod{p^k}$ just by the binomial theorem. Since a mapping from a finite set to itself is a surjection if and only if it is an injection, we seek to prove it is an injection.

Suppose $ h(i)\equiv h(j)\pmod{p^3}.$ If we can prove that $ i\equiv j\pmod{p^3}$ we will be done. First we note that since this implies that $ h(i)\equiv h(j)\pmod{p^2}$ and we are given that $ h$ is a bijection on the residue classes mod $ p^2,$ we already know that $ i\equiv j\pmod{p^2}.$ Hence $ j\equal{}i\plus{}tp^2$ for some integer $ t.$

$ h(j)\equiv h(i)\plus{}tp^2h'(i)\pmod{p^3}$ by our lemma.

That implies that $ p^3$ divides $ tp^2h'(i)$ or $ p$ divides $ th'(i).$ But we know that $ p$ does not divide $ h'(i),$ hence $ p$ divides $ t.$ But then $ tp^2\equiv 0\pmod{p^3}$ and $ i\equiv j\pmod{p^3}.$ This is what we wanted to show and we are done.

It's basically [url=http://en.wikipedia.org/wiki/Hensel%27s_lemma]Hensel's lemma[/url]. This was much much easier than B3, and even than B2 I would say.

A very easy B4. I solved at first without knowing the source and I am quite surprised at its placement.

Let $h(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$.
Assume FTSOC that for some distinct $a,b \in \{0,1,\ldots,p^3-1\}$ the congruence $h(a) \equiv h(b) \pmod{p^3}$. Then they must be congruent $\pmod{p^2}$, which using the condition implies that $p^2 \mid a-b$. Thus WLOG $a<b$ and write $b=kp^2+a$, where $1\leq k<p$. Now note that by the binomial theorem:
$$(kp^2+a)^r-a^r \equiv akrp^2 \pmod{p^3},$$
since any other coefficients are $0 \pmod{p}$. Thus for $h(a) \equiv h(b) \pmod{p^3}$, it is necessary to have:
\begin{align*}
\sum_{i=1}^n a_i((kp^2+a)^i-a^i) &\equiv 0 \pmod{p^3}\\
ak\sum_{i=1}^n ia_ip^2 &\equiv 0 \pmod{p^3}\\
ak\sum_{i=1}^n ia_i &\equiv 0 \pmod{p}.
\end{align*}
Now consider $h(a)$ and $h(kp+a)$. Note that:
$$(kp+a)^r-a^r \equiv akrp \pmod{p^2}.$$
Now by the condition, these must be distinct $\pmod{p^2}$, or:
\begin{align*}
\sum_{i=1}^n a_i((kp+a)^i-a^i) &\not \equiv 0 \pmod{p^2}\\
ak\sum_{i=1}^n ia_ip &\not\equiv 0 \pmod{p^2}\\
ak\sum_{i=1}^n ia_i &\not\equiv 0 \pmod{p}.
\end{align*}
Which is a contradiction. $\blacksquare$

Assume the contrary. Then $$\exists a,b: 0\leq a< b<p^3, h(a)\equiv h(b) ~~~(\text{mod }p^3).$$ It follows that we have an injection in the mapping of $h$, $$\mathbb{Z}/p^2\mathbb{Z} \to \mathbb{Z}/p^2\mathbb{Z},$$ therefore $a\equiv b~~~(\text{mod } p^2).$ Also we have, $$h(a+x)=h(a)+h'(a)x+x^2d(x),$$ for some $d(x) \in \mathbb{Z}[x].$ For $x=p,$ applying the fact that $$h(a+p)\equiv h(a) ~~~(\text{mod } p^2),$$ we have $p \nmid h'(a).$ Let $x=b-a,$ then $x=p^2c,$ for any $c\in \{1,2,\dots, p-1\}.$ Therefore, $$0\equiv h(b)-h(a)\equiv h'(a)p^2c+p^4c^2d(p^2c) \equiv h'(a)p^2c~~~ (\text{mod } p^3),$$ which is contrary to the fact that both $h'(a)$ and $c$ are not multiples of $p.$ Thus our initial assumption was false and the claim follows. $\square$


Let C be a circle and KN and KL the tangents of point K.Let M be a point on KN such that N is between K and M.Let P be the second point of intersectiion of circumcircle KLM and circle C.Let Q be on ML such that <NQM=90°.Prove :
(<MPQ)=2*(<KML).

Hello!

$ \left\{L,R\right\}\equal{}LM\cap C$ and $ \left\{V\right\}\equal{}PR\cap KM$.
$ \angle KPM\equal{}\angle KLM\equal{}\angle KLR\equal{}\angle LPR\Rightarrow\angle VPM\equal{}\angle KPL$
$ \angle VMR\equal{}\angle KML\equal{}\angle KPL\equal{}\angle VPM\Rightarrow$ the triangles $ VMR$ and $ VPM$ are similar $ \Rightarrow VM^{2}\equal{}VR\cdot VP$
$ VM^{2}\equal{}VR\cdot VP$ and $ VN^{2}\equal{}VR\cdot VP$ $ \Rightarrow VM^{2}\equal{}VN^{2}\Rightarrow VM\equal{}VN$
$ V$ is the midpoint of $ \left[MN\right]$ and $ MQN$ is a right triangle $ \Rightarrow VQ\equal{}VM\Rightarrow$
$ \Rightarrow\angle VQM\equal{}\angle VMQ\equal{}\angle VMR\equal{}\angle VPM\Rightarrow$ the quadrilateral $ MPQV$ is cyclic $ \Rightarrow$
$ \Rightarrow\angle VPM\equal{}\angle VQM\equal{}\angle VMQ\equal{}\angle VPQ\Rightarrow\angle MPQ\equal{}2\cdot\angle KML$.

Best regards, Petrisor Neagoe :)


It is known that, for all positive integers $k,$
\[1^{2}+2^{2}+3^{2}+\cdots+k^{2}=\frac{k(k+1)(2k+1)}{6}. \]Find the smallest positive integer $k$ such that $1^{2}+2^{2}+3^{2}+\cdots+k^{2}$ is a multiple of $200.$

[quote="SM4RT"][hide]

[quote=ch1n353ch3s54a1l]
$200=2^{3}\cdot 5^{2}$, so $16\cdot 25|k(k+1)(2k+1)$
[/quote]

shouldnt that be $8\cdot25|k(k+1)(2k+1)$
[/hide][/quote]

But you have a 6 dividing it below so the expression has to have another factor of 2 in it so that the six can divide one of them out.

[quote="SM4RT"]ya that makes sense..sorry..but one question what happens to the factor $3$..if you understand what i'm trying to say..[/quote]
Well at least one of k, k+1, 2k+1 is divisible by 3, and 3 isn't a factor of 200 so we just ignore the 3

[quote=ch1n353ch3s54a1l][quote="4everwise"]It is known that, for all positive integers $k,$
\[1^{2}+2^{2}+3^{2}+\cdots+k^{2}=\frac{k(k+1)(2k+1)}{6}. \]
Find the smallest positive integer $k$ such that $1^{2}+2^{2}+3^{2}+\cdots+k^{2}$ is a multiple of $200.$[/quote]

[hide]$200=2^{3}\cdot 5^{2}$, so $16\cdot 25|k(k+1)(2k+1)$. Only one of those factors can be divisible by 25, and only one factor is even, so only one factor can be divisible by 16.
First let k be the multiple of 16. The smallest multiple that satisfies $16\cdot 25|k(k+1)(2k+1)$ is $k=112$.
If k+1 is the multiple of 16, the smallest $k$ satisfying $16\cdot 25|k(k+1)(2k+1)$ is $k=287$. Since $2k+1$ is odd, it can't be a multiple of 16 so $\boxed{k=112}$ is the smallest.[/hide][/quote]

I had the same. $\boxed{k=112}$

Note that $200 = 2^3 \cdot 5^2$ and $6 = 2 \cdot 3$, so $k(k+1)(2k+1)$ must have a factor of $2^4 \cdot 5^2$. Since $2k+1$ is always odd and $k, k+1$ will always have different parities, this means that one of $k$ or $k+1$ must be a multiple of $2^4 = 16$. Now, we can work through the cases. All of of $k = 15, 16, 31, 32, 47, 48, 63, 64, 79, 80, 95, 96$ don't work since there would be no factor of $5$ in $k(k+1)(2k+1)$, but $k = \boxed{112}$ works and is thus the answer.

[b]Solution:[/b] Notice that $200 = 2^3 \cdot 5^2$ so $k(k+1)(2k+1)$ must have a factor of $16$. Notice that only one of these factors ($k, k+1, 2k+1$) can be even. So then you do cases where $k= 15, 16, 31, 32, 47, 48, 63, 64, 95, 96, 111, 112$. Keeping in mind that there has to be a factor of $25$, we see that $k=\boxed{112}$ is the answer.


I haven't clear your first step in the solution of Michael Bencze ineq:

For $ a,b,c$ positive reals

$ (a^2\plus{}1)(b^2\plus{}1)(c^2\plus{}1) \plus{}4(a^2\plus{}2)(b^2\plus{}2)(c^2\plus{}2) \geq 6(a\plus{}b\plus{}c)^2$

You expand LHS as

$ LHS \equal{} VT \equal{} 5a^2b^2c^2 \plus{} 6\sum a^2b^2 \plus{} 8\sum a^2 \plus{}12$

but it is wrong !!!

Am I right?

Thank you very much.

Sorry, my friend. My solution isn't wrong. I had just mistyped the problem, the correct is
\[ (a^2\plus{}2)(b^2\plus{}2)(c^2\plus{}2) \plus{}4(a^2\plus{}1)(b^2\plus{}1)(c^2\plus{}1) \ge 6(a\plus{}b\plus{}c)^2\]
By the way, I have asked my friend to post the above solution for the problem 28, but maybe he has mistakes... I am sorry.

[quote="can_hang2007"]Now I have found the following ugly proof, I hope it is okay now. Please check it for me[/quote]

Thank you very much, my dear friend. :)

I will be very happy to check it. This inequality is so hard that I don't think your solution is so ugly :wink:

[quote="can_hang2007"]Now I have found the following ugly proof, I hope it is okay now. Please check it for me[/quote]

I found a mistake in your solution.

You wrote in the last page

$ f(a) \geq.........\minus{}(9/16)b^2(6b\plus{}c)\plus{}.....$

but it is

$ f(a) \geq.........\minus{}(9/16)b^2(6b\plus{}4c)\plus{}.....$

and with this correction I think your solution fails.  Am I right?

It is a very hard inequality.  :maybe:

Thank you very much for your solution.


Feel free to post your Brawl code on this thread for others to add you. Also, include a 1-5 character name you want to go by (does not have to be your real name).

Now then, lets get Brawling, mine is 2707-1359-9641. (Teja)

Hopefully this works out

1337-1120-4324
Satan
:)
Nah I'm joking, I don't have Brawl
:(
No Wii.

FC:2449-4388-4403

Profile: Ryzal

Btw, I'm adding Klebian and lifesapain right now.

B
 :)
U
 :)
M
 :)
P

C'mon people, we only have 3 FCs! There must be more brawlers than that! I'm sure AoPSers want to face each other...

i have one
i forgot my friend code  :(
and i'm still finishing subspace emissary (limited playing time)

0216-0563-3606
ycz6

http://www.allisbrawl.com/video/video.aspx?id=979
http://www.allisbrawl.com/video/video.aspx?id=981
http://www.allisbrawl.com/video/video.aspx?id=983

Fear me.  :ninja:

do you guys mind if i send this link to my friend?

she beasts and she wants to beat more people...
(she already pwned her friends and their siblings)
she could sure pwn me, but i don't have a wii.. :(

If anyone wants to play in the next few hours, I'm willing to go on...

Why did you revive a two year old thread????? Mods, lock this now.


my sister recently went to the Texas A&M Math competition and encountered this problem:

How many zeroes are at the end of 2007! ?

i was just wondering, how would you do this?

[hide] Wouldn't you just count the number of $ 5$'s in the factorization? Since there are more 2's than 5's you would not need to count the 2's[/hide]

[hide]
To have a zero at the end of a number, you need to have at least 1 factor of 2 and 1 factor of 5.  So we need to count the number of pairs of 2s and 5s.  Since there are obviously more 2s than 5s, we count the number of 5s to get the number of pairs.  Every 5 #s are divisible by 5, so we have $ \frac{2007}{5}\equal{}401.6$, but we only count the whole number part.  However, every 25 numbers have another factor of five, so we take $ \frac{2007}{25}\equal{}80.28$, and again, we just take the whole number part.  Similarly, we do the same for 125 and 625.  Since the next power of 5 is greater than 2007, we just add the whole number parts of $ \frac{2007}{5}$, $ \frac{2007}{25}$, $ \frac{2007}{125}$, and $ \frac{2007}{625}$, and the answer is hopefully $ \boxed{500}$
[/hide]

as the above posters mentioned: think about the product. how do u get a 0. you get one 0 everytime you have a factor of 10. essentially, you're counting how many times the number 10 occurs the problem. to make our task simpler, remember that 2*5=10 and there will obviously be more 2's than 5's in the product. so we're really only counting the number of 5's. you first divide a number n by 5. then you divide it by 5^2 because there is another factor of 5 for every multiple of 25. if the number is big enough, you would go to 5^3, 5^4...

hope that helps!

In general, the number of zeroes on the end of $ z!$ is
$ \sum_{i\equal{}1}^{\infty} \left\lfloor\frac{z}{5^{i}}\right\rfloor$


AMC
USA(J)MO
USAMO
geometry
geometric transformation
reflection

They will even let you represent USA at the IMO if you can do the job right. Check out this year's Oleg Goldberg, ex-russian IMO contestant in 2002 and 2003, came back in 2004 as part of the USA IMO team, and had the best score on the team also.

I think you only need to have a residence in USA (live there) which I think you do ;) and, of course, the necessary mathematical knowledge :D

[quote="DPopov"]I was wondering if they would let you participate officially in the USAMO if you were in the process of getting a green card? Currently i have visa but am going into the process of getting a green card which takes 2 years.[/quote]

Ask the AMC.  The rules last year, I believe, were as Valentin reflects - if you are in the green card process, you can officially participate in the USAMO and qualify for MOP.  But the rules are not guaranteed to remain the same (though I do find the AMC to be very inclusive!)

Would you recommend anything such as books or topics to know for the USAMO? I am thinking about getting pretty serious this year

And you could definitely check out the AoPS Olympiad classes and / or independent study program.


algorithm
linear algebra
matrix
vector
LaTeX
geometry
geometric transformation

I have tried and tried to figure out these kinds of problems, but I just cant do it. My teacher said that the answers should come out to simple #'s like 2 or 3. and I keep getting decimals. Please Help me if ya can  :blush:
                     You have to use all 3 equations together to get the 3 answers (a,b,&c). Well if you can give me any advice. Thank you!!

1)  5a-b+3c=5
2)  a+7b-2c=5
3)  4a-5b-7c=-65

Have you ever heard of a rule called Cramer's Rule?
If you are new to math, you are probably taught how to solve these using an algorithm known as "Gaussian Elimination." Simply put, we try and find expressions of one variable in terms of another, and substitute that. We repeat this process until we are left with explicit values for each variable.

Cramer's Rule is slightly different.
We set up a matrix (read: grids) with the coefficients of the variables. We also set up an answer column, with the values of the answer lined up. In your case, these would look something like this:
$\begin{pmatrix}5 &-1 & 3 \\ 1 & 7 &-2 \\ 4 &-5 &-7 \end{pmatrix}$

And the answer column would look like:
$\begin{pmatrix}5 \\ 5 \\-65 \end{pmatrix}$

We create new matrixes for the variable we want to solve for, say a, by imposing the answer column on the row with the a coefficients in the $3 \times 3$ matrix above.

Our new a-matrix looks like this:
$\begin{pmatrix}5 &-1 & 3 \\ 5 & 7 &-2 \\-65 &-5 &-7 \end{pmatrix}$

Cramer's rule says that the determinant of the a-matrix divided by the determinant of the coefficient matrix gives the value of a. I'm not going to explain determinants here, because that would take too long.

Suffice it to say that
$\frac{Det_{a}}{Det_{coefficient}}= a$
$a=-\frac{830}{393}$
Which is odd. This is a non-reducible fraction.
I think this is the correct solution for A, however.

Are you sure you wrote the equations out clearly?

Here's another way to do it:

[hide]$\\5a-b+3c=5\\a+7b-2c=5\\4a-5b-7c=-65$

Subtract the last two equations from the first to get $-3b+12c=65.$

Multiply the second equation by $4$ and subtract it from the third equation to get $-33b+c=-85.$

Now, we solve for $b$ and $c.$

First, substitute $c=33b-85$ into the other equation.

$-3b+12(33b-85)=65$

$-3b+396b-1020=65$

$393b=1085$

$b=\frac{1085}{393}$

Then, plug $b$ into $-3b+12c=65$ to get $-3\left(\frac{1085}{393}\right)+12c=65,$ or $c=\frac{800}{131}.$

Then, solve for $a$ using the second (original) equation.

$a+7\left(\frac{1085}{393}\right)-2\left(\frac{800}{131}\right)=5$

$a+\frac{7595}{393}-\frac{1600}{131}=5$

$a+\frac{7595-4800}{393}=5$

$a+\frac{2795}{393}=5$

$a=-\frac{830}{393}$

The solution set is $\left\{-\frac{830}{393},\,\frac{1085}{393},\,\frac{800}{131}\right\}.$[/hide]

I had to do the entire thing twice because I messed up at the very beginning the first time. :wallbash:

and a less confusing application of matrices and vectors:
NOTE: Funcia doesn't have Latex so they won't try to make it look like matrices.
5 -1 3   * x    =5
1 7 -2     y      5
4 -5 -7     z     -65

Now I eliminate starting with the first pivot:
5 -1 3
0 -28 8
4 -5 -7
second pivot:
I believe since I cannot make -5 28 without multiplying by a number with a denom. this reflects on the result. I won't bother to go on since you have already had wonderful help.

Just so you know about dets
with any matrix say a 2*2 one:
a b
c d
the determinant is ad-bc
if you eliminate
it will only be ad. And the elimination works for 3*3> as well

NOTE: If you understand Cramer's Rule USE it. It is REALLY helpful.
MEA CULPA WITH DET.

I currently know three ways:

Substitution, Elimination

Substitute one variable for another in system of equations so eventually (and it takes a while) to get (x, y ,z)

Faster way: Cramer's Rule (which is explained above thanks to DiscreetFourierTransform)

Even faster way but I don't know all of it: Augmented matrices....

I used the augmented matrices on a test and the teacher groaned because i showed little work (i used a calculator)

Augmented Matrices are quite similar to Cramer's, except it works by simplifing down a system until you get a very easy system to solve with matrixes.


Hmmm, I've dappled in some ballroom. I can do some basic waltz and rumba steps.  :D

[quote="LadyKn1ght"]I take dance classes :D . Tap, jazz, and ballet (and pointe). How long do you dance each week?[/quote]

Dance team requirement for our studio is 8+ hours. It takes up a lot of time...

I can do some basic steps and figures waltz, tango,quickstep, samba, cha - cha, jive and salsa(all are partner dance). Probably I didn't forget anything to list :D I'm dancing about 2 hours in a week. Only for pleasure and recreation because I started too late...
Dance is fantastic and beautiful :) Like math :P

I myself don't dance (unless by dancing, you mean getting krunk with the music on the radio, this white kid can DANCE, yo!); however, I have found a new appreciation for dance by running crew for a couple of shows at my school. Does anyone else around here ever run backstage crew for anything? We had to do this one show, and I swear......... EVERY dance required a complete set change. Piano, scrim, backdrop, mop up the soap bubbles on the floor, scrim, piano. EVERY dance! And then of course, the dancers fell on the soap bubbles (we were doing the "Snow" scene from the Nutcracker). Huge fight with the dance director over how the floor should be mopped. It was fun.

I used to dance ballet and point but gave it up to swim. I used to dance about 3-4 hr. week

I dance Indian Classical. Depending if there is a performance coming up, like now. Right now i dane for around 8-9 hours a week. Normally I dance for around 4 hours a week.

[quote="CatherineAsaro"]MathDancer here on AoPS dances a lot (I'm posting with her permission).  She studies in the trainee program of a ballet company and also does jazz and modern.  It's a time commitment of about 25 hours a week, plus whatever extra rehearsals they have near performances.

The upside that is that students in her program have the opportunity to perform with the company.  Also, they go to summer programs with companies like ABT.  The downside is the huge time commitment.  I don't know how all you students manage the time constraints, doing so much with math and everything else.  It's impressive!  :-)

I've noticed a substantial portion of my female math students are dancers as well as musicians, especially in ballet.  Most of the dancers also play an instrument.[/quote]
25 hours... wow. That's awesome how she can balance math and dance like that.

[quote="mysmartmouth"]Uhhh. I  can do the C-Walk.[/quote]

Haha, nice

Ballroom dance!  I just learned how to cha-cha!

I dance like 7-10 hours a week. By the way, all you dancers, check out my forum! :

[url]https://www.artofproblemsolving.com/community/c252065_calling_all_aops_dancers[/url]

[color=#0B5394]Your AoPS friend,[/color]
[hide=~MathMania11] [img]http://cdn.artofproblemsolving.com/images/e/1/f/e1f42eac6c50593d6d7f3654cd06fde7098029c0.gif[/img]

[i][color=#38761D]"Pledge to Always be a Mathlete." ~ MathMania11[/color][/i] [/hide]

I do Canadian step dance, and clogging. I am taking a course in modern dance at school.

[quote=piemaker]I love dance!!! :D[/quote]
So do I!!!! I do ballet, mostly. I used to do Chinese dance and modern ballet, but now it’s mostly ballet.


Question:

Does anyone happen to know what the MOP cutoffs have been for the past few years?

pre-Thanks!


MATHCOUNTS
number theory
least common multiple
modular arithmetic

[quote="chess64"]1. (AoPS) Find the units digit of $19^{93}$.

2. (MATHCOUNTS 1986) Find the units digit of $3^{1986}-2^{1986}$.

3. (MATHCOUNTS 1991) When $n$ is divided by 5, the remainder is 1. What is the remainder when $3n$ is divided by 5?

4. (Mandelbrot #3) Find the smallest positive integer greater than one which yields a remainder of one when divided by any single digit positive integer greater than 1.

5. (AHSME 1951) Find the smallest positive integer which when divided by 10 leaves a remainder of 9, when divided by 9 leaves a remainder of 8, by 8 leaves a remainder of 7, etc., down to where, when divided by 2, it leaves a remainder of 1.

If you finish these I might give you some more.[/quote]

[hide="3."]

The remainder is $3$.

[/hide]

[hide="4."]

$\text{lcm}(2,3,4,5,6,7,8,9)+1$
$\text{lcm}(5,7,8,9)+1$
$2520+1$
$2521$

[/hide]

[quote="mathgeniuse^ln(x)"][quote="chess64"]1. (AoPS) Find the units digit of $19^{93}$.

2. (MATHCOUNTS 1986) Find the units digit of $3^{1986}-2^{1986}$.

3. (MATHCOUNTS 1991) When $n$ is divided by 5, the remainder is 1. What is the remainder when $3n$ is divided by 5?

4. (Mandelbrot #3) Find the smallest positive integer greater than one which yields a remainder of one when divided by any single digit positive integer greater than 1.

5. (AHSME 1951) Find the smallest positive integer which when divided by 10 leaves a remainder of 9, when divided by 9 leaves a remainder of 8, by 8 leaves a remainder of 7, etc., down to where, when divided by 2, it leaves a remainder of 1.

If you finish these I might give you some more.[/quote]

2519=#5.[/quote]

how did u get that?

It's almost the same as #4.

It looks like you're getting good at mods now Henry :)

If you add 1 to everything chess said, it would mean that the number is a multiple of 2, 3, 4, 5....10.

So the LCM(2,3,4,...10)=2520.

Then you subtract 1, and you get 2519.


Find all $ x_1,x_2,\cdots,x_n \in \mathbb{R}^\plus{}$, $ n\ge 2$ such that
\[ x_1(6\minus{}x_2)\equal{}9 \quad x_2(6\minus{}x_3)\equal{}9 \quad \ldots \quad x_n(6\minus{}x_1)\equal{}9\]

Source: My friend  :D.

[quote="mathson"]Find all $ x_1,x_2,\cdots,x_n \in \mathbb{R}^ \plus{}$, $ n\ge 2$ such that
\[ x_1(6 \minus{} x_2) \equal{} 9 \quad x_2(6 \minus{} x_3) \equal{} 9 \quad \ldots \quad x_n(6 \minus{} x_1) \equal{} 9
\]
Source: My friend  :D.[/quote]
Multiplying these equalities and by Cauchy-swartz Inequality, we get: $ 9^n\equal{}\prod_{k\equal{}1}^{n}x_k(6\minus{}x_k)\le (\frac{6n}{2n})^{2n}\equal{}9^n$
The equality occurs when $ x_1\equal{}x_2\equal{}x_3\equal{}\ldots \equal{}x_n\equal{}3$


my code is the following:

\documentclass[a4paper,12pt]{article}
\usepackage{amssymb}
\usepackage{amsthm}

\begin{document}
Then a is definitely \in B since a is a wff.
\end{document}

the output, called test.pdf, is attached in this topic. It was complied in TeXnicCenter 1 beta6.31. As you can see, the output is messed up.

Now adding some dollar signs around \in would result in a perfect output.
i.e.

Then a is definitely \$ \in \$ B since a is a wff.

the output would be: Then a is definitely $\in$ B since a is a wff.
which is fine.

However, does that mean I have to add dollar signs around every LaTeX math symbol?
That would be so uncool. Currently I go through a lot of online LaTeX guide, but none of them seems to talk about this problem...

Yes, \in is a math mode command.  It can only be used in math mode (meaning, for example, inside of dollar signs).

There's a reason for this.  It is really bad mathematical grammar to write a "sentence" like
[quote]Then a is definitely $\in$ B since a is a wff.[/quote]
Don't use symbols in place of words in a sentence like this.  Two acceptable alternatives would be

Then $a \in B$ since $a$ is a wff.

or

Then $a$ is an element of $B$ since $a$ is a wff.

Also note that the a and the B should be in math mode, as I wrote them above.

I see. So it should be:
Then $a \in B$ since $a$ is a wff.

lol thanks but sometimes I really like to add the word "obviously" or "definitely" just to make the proof more fun to read. In addition, sometimes the word "obviously" would help the reader (namely myself) to keep things simple when trying to understand the proof (e.g. the reader would know that he doesn't need any complicated theorem to understand why $a \in B$).

That's not what I meant.  Using "obviously" is fine (I use it often myself!)

What's not fine is the words in the middle of the math expression.

For example,

Then obviously $a \in B$ since $a$ is a wff

would be fine.  But

Then $a$ obviously $\in B$ since $a$ is a wff

should be avoided.

This is off-topic I know, but what exactly is a `wff'? It has been used in this topic a number of times, and Google turns up nothing.


Blog Options-->Blog Layout-->Scroll Down to "Background Tile"

ummm, that doesn't work, it won't show up on the blog.

For me, it takes a while.

EDIT: I tried what Dojo said and it worked.

Go to your blog, and press f5, or shift f5.


calculus
integration
geometry
3D geometry
octahedron
analytic geometry
symmetry

Hi, I am having trouble with  $ \int^S F dS$ where $ S$ is the surface $ |x \minus{} y \plus{} z| \plus{} |y \minus{} z \plus{} x| \plus{} |z \minus{} x \plus{} y| \equal{} 1$ and $ F \equal{} (x \plus{} y \minus{} z)i \plus{} (y \minus{} z \plus{} x)j \plus{} (z \minus{} x \plus{} y)k$

We got a hint telling us to use the divergence theorem and substitute $ u \equal{} x \minus{} y \plus{} z, \ v \equal{} y \minus{} z \plus{} x, \ w \equal{} z \minus{} x \plus{} y$ but I am still a little lost. Any help would be appreciated.

The divergence of $ \vec F$ is equal to $ 3$, so all you need is to do is to find $ \iiint_E 3\,dV$ where $ E$ is the solid bounded by $ S$. Unfortunately, $ E$ is an octahedron that is not aligned along coordinate axes. The change of variables transforms $ \iiint_E 3\,dV$ into an integral over the standard octahedron, with boundary $ |u|\plus{}|v|\plus{}|w|\equal{}1$. Don't forget the Jacobian. But even with the Jacobian, the integrand is constant, which allows you to use symmetry and evaluate the integral as $ 8\int_0^1 \int_0^{1\minus{}u}\int_0^{1\minus{}u\minus{}v}\dots$

so the integral is 3 times the jacobian times r?

3 times the jacobian, yes. What is $ r$?

sorry, was confused with an example i was reading, so just 3 times the jacobian, |d(x,y,z)/d(u,v,w)|?


The height of the trapezoid of the figure is 4, then the difference between the areas of the triangles is:

[img]http://img44.imageshack.us/img44/4226/calso.jpg[/img]

a) 1
b) 2
c) 3
d) 4
e) 5

I make this $ 4$.

Calling the areas of the unshaded tringles $ x$ and $ y$, the area of the lower shaded triangle $ b$ and that of the upper one $ t$,

$ x\plus{}b\equal{}10$

$ y\plus{}b\equal{}10$

$ x\plus{}y\plus{}b\plus{}t\equal{}16$,

Therefore $ b\minus{}t\equal{}4$.

And how do you know that $ x\plus{}b\equal{}10$ ?

Allways the areas will be the same for all trapezoid included one with two right angles.Wih this in mind the solution is easy

Hi luis.

$ x\plus{}b\equal{}10$ because it is the area of the triangle formed by the unshaded triangle on the left and the shaded triangle on the bottom, which has a base of $ 5$ and a height of $ 4$ and therefore an area of $ \frac{1}{2} \cdot 5 \cdot 4 \equal{} 10$. Hope this helps you.


How many ways can 9 men, 7 women and a child sit around a circular table so that no two of the women sit next to each other? Is this a permutation or combination. Please solve it

[hide="solution"]
Since it doesn't really matter that the child is specified as such, let $ X_i$ represent a either a man or the child, $ w_i$ represent a woman. We start by counting the number of ways to arrange the $ X_i$s (e.g. $ X_1 X_2 X_3 \cdots X_{10}$) which is $ 10!$. Since no two women can sit next to each other, we count the number of ways to interweave the $ 7$ $ w_i$s into the $ 11$ spaces between, before, and after the $ X_i$s which comes out to $ \frac {11!}{(11 \minus{} 7)!} \equal{} \frac {11!}{4!}$. Multiplying those together, we have $ \frac {10!11!}{4!}$.
[/hide]

I think you are not considering the circular arrangement. You need to consider that?


正整数 $n$ に対して、 $(3+2\sqrt 2)^n=a_n+b_n \sqrt 2$ で定まる $2$ つの整数 $a_n , b_n$ のなす数列を$\{a_n\},\{b_n\}$  $(n=1,2,\cdots)$ とする。このとき、次の問に答えよ。

(1) $a_{n+1}$ および $b_{n+1}$ を $a_n , b_n$ で表せ。

(2) $a_{n}^{2}-2b_{n}^{2}$ は $n$ に関係のない一定値であることを示せ。

(3) $a_n$ および $b_n$ を $n$ を用いて表せ。

(4) $(3+2\sqrt 2)^n$ の整数部分は $2a_n -1$ であることを示せ。($(3+2\sqrt 2)^n$ の整数部分は奇数であることを示せ。)

(5) $\lim_{n\to\infty}\frac{a_n}{b_n}$ の値を求めよ。

(6) $a_{n}^{2} -1$ が偶数となることを示せ。

(7) $(3+2\sqrt 2)^n$ はある整数 $X$ を用いて $\sqrt X +\sqrt{X+1}$ と表せることを示せ。

(8) 不等式 $0<2a_n \cdot (a_n - b_n \sqrt 2)<\left( \frac{1}{5} \right)^{2n}$ を示せ。

(9) $p_n=(3+2\sqrt 2)^n$ とし $p_n$ の整数部分を $q_n$ とおくとき、$q_n=p_n + \frac{1}{p_n} -1$ であることを示せ。

(10) $q_n$ を $8$ で割ったときの余りを求めよ。

(1)から順に解く必要はありません。 :roll:
問題集を何冊かめくっているだけで $(3+2\sqrt 2)^n=a_n+b_n \sqrt 2$ 関連の問題がかなりあったので、結構よく扱われる話題みたいですね。東工大の後期で確か(7)の類題が単体で出題されていたような… :roll:

Pell 方程式がテーマですね。

そうみたいですね。ペル方程式は学校で習いました :lol:

この問題の類題が
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=83637[/url]
にあります :D


advanced fields
advanced fields unsolved

[i]Decimal shift map[/i]. Let [i]n[/i] $\in$ {0,1,2,...} and $x_{n+1}$ = 10$x_{n}$([i]mod [/i]1) where [i]mod[/i] 1 means set to zero everything to the left of the decimal point, after evaluating 10$x_{n}$.

(a) Find all fixed points of this mapping.
(b) Show that the map has periodic points of arbitrary period.                 Justify an example which has infinite period.
(c) By considering two points $x_{0}$ and $x^{'}_{0}$ which are arbitrarily close, show that the mapping can be sensitive to initial conditions (a justified example is enough).

Please help me with this question.
Thank you

(a) the only fixed points are numbers with decimal expansion $0.zzzzz\dots$, where $z$ is a digit $0,1,2\dots,9$.

(b) any periodic decimal expansion is a periodic point for the shift, with the period being equal to the number of digits in the repeated part of the expansion. I have no idea what "infinite period" means.

(c) the numbers $10^{-100}$ and $9\cdot 10^{-100}$ are very close to each other, but after $99$ iterations they are mapped into $0.1$ and $0.9$ respectively.


Prove that if a vector space has a finite basis, then all of its bases have the same number of elements.

(If possible, try proving this by contradicatiion.)

This is a standard theorem found in almost every elementary linear algebra book. It can also be found on the web see eg [url]http://xmlearning.maths.ed.ac.uk/lecture_notes/linear_independence_bases_dimension/bases/bases_2.php[/url]

There are several possible paths to this proof.

Here's one that I like. First prove this theorem:

If $ \{v_1,v_2,\dots,v_n\}$ is a basis for $ V$ and if $ m>n,$ then any set of $ m$ vectors in $ V$ is linearly dependent.

You can then use matrices to prove this. We have the set $ \{w_1,w_2,\dots,w_m\}.$ Each element of this set can be uniquely expressed as a linear combination of the basis:

$ w_j\equal{}a_{1j}v_1\plus{}a_{2j}v_2\plus{}\cdots\plus{}a_{nj}v_n$

This gives us an $ n\times m$ matrix $ A$ whose entries are $ a_{ij}.$ Perform Gaussian elimination on this matrix; it will be row-equivalent to a matrix in row echelon form that has no more than $ n$ pivot elements. Since $ m>n,$ there is at least one column with no pivot, implying a free variable. Hence, the homogeneous equation $ Ax\equal{}0$ has a non-trivial solution (an $ x\ne0$ that satisfies this.) Note that $ x$ is an $ m$-vector.

Consider this linear combination:

$ \sum_{j\equal{}1}^mx_jw_j\equal{}\sum_{j\equal{}1}^mx_j \left(\sum_{i\equal{}1}^na_{ij}v_i\right)$

$ \equal{}\sum_{i\equal{}1}^n\left(\sum_{j\equal{}1}^ma_{ij}x_j\right)v_i$

But $ \sum_{j\equal{}1}^ma_{ij}x_j$ is simply the $ i$th component of $ Ax,$ and since $ Ax\equal{}0,$ this is zero for each $ i.$ Hence we have that the whole sum is zero, and that

$ \sum_{j\equal{}1}^mx_jw_j\equal{}0$ even though not all of the $ x_j$ are zero.

Hence, $ \{w_1,w_2,\dots,w_m\}$ is dependent.

-----

Nowhere in this proof did we use any specific properties of the field of scalars, other than the field properties themselves (to support Gaussian elimination). Hence this is valid for vector spaces over arbitrary fields.

-----

You should be able to see how this implies that (for a finite-dimensional space) that any two bases have the same cardinality, which in turn means that the word "dimension" can be defined.


AMC AIME calculus Putnam USA(J)MO USAMO integration

I need some straightforward, AoPS style lecture and problem books to learn physics up to IPhO level, to a point where I could pass whatever the physics equivalent of the AIME is, books to do the same thing with chemistry, and books to learn the following college math subjects to a high level of competency and of AoPS/Putnam style mastery:

*Calculus (to as high a degree of mastery and enrichement as can be managed)
*Differential Equations
*Linear Algebra
*Analysis, Real and Complex
*Abstract Algebra
(basically what I have been told are the basics of college math)

I would like to do this with as little books as possible, to minimize cost. Mainstream books are good to recommend, as I have a college library at hand. I realize this is a large project, but I will be devoting my my entire summer to social stuff and intellectual conquest, so whatever. Ease of use is big plus (eg Apostol would be bad because of all the rigor obscuring practicality)

I just really want to get all these intellectual loose ends out of the way so I can focus solely on precalculus math before I hit college.

So, any recommendations? :P

Differential equations and linear algebra could be covered by free books available online.

Have you seen [url=http://tutorial.math.lamar.edu/]Paul's Online Math Notes[/url]?  He has some really useful material, including full texts and cheatsheets to for review and reference.

Several misconceptions here I think.

First of all, there's no way you're going to get that done in a summer. Second of all, the Putnam barely uses any math beyond high school (I think only basic Calculus and Linear Algebra are included, and most of the time it involves minimal theory in those subjects). Now I can't say much about physics and chemistry, but for math you are going to have to be rigorous if you want to achieve the level of mastery you want.

A long time ago, I thought I "understood" the fundamental theorem of calculus. Four years later, I realize that I still don't totally understand it. I know this sort of thing is cliche, but in a couple years you'll find out that it's true. It's definitely possible to jump into college stuff, to be sure, but you're probably better off strengthening the so-called precalculus material for at least a year or two. You can start linear algebra and calculus now, but I think playing around with some proof-based contest math will greatly help your understanding of the other subjects.

Um, if you get through all of calculus, diffeq, linalg, analysis, and abstract algebra you're probably more than halfway toward an undergraduate degree.  Take it easy.  I don't see where the rush is.  I'd suggest focusing on two of calc, linear algebra, and diffeq.  Keep in mind that you need to know calc before doing diffeq... (though you don't have to be godly at calc for it...I've learned/brushed up on a few calculus things in diffeq so far).

I wasn't really planning to get it done this summer, I just need something to play with. I will also not be giving up on USAMO prep, either. There's just so much [i]I don't know[/i], and it bothers me. I also realize this entire thread sounds like I'm a stupid little kid who like "LOL OMG I WANT TO LEARN CALCULUS GIVE ME SOME BOOKS PEOPLE EVEN THOUGH I BARELY HAVE AN IDEA OF WHAT AN INTEGRAL IS".

I am.

@MSM

Thanks! Those look great for learning and for reviewing concepts and stuff, but I want a textbook mainly because I need some practice problems to work out, and hopefully some challenging and interesting ones.

@probability

Point taken. However, could you suggest a book that would combine rigor with something that isn't so hard to comprehend like Apostol? That's why I tend to shy away from rigor...

In my opinion, the more rigorously you study these subjects, the better. It's a question of whether you want to memorize tricks for taking derivatives and integrals or whether you really want to understand why and how they work. Anyhow, I must confess that I don't quite understand your motivations. You want to learn Calculus, Differential Equations, Linear Algebra, Abstract Algebra, and Complex Analysis just so you can spend the rest of your time focusing on precalculus? Why not just finish precalculus this summer and then move in to the other subjects? I think you want to get the other material "out of the way" so you can focus on stuff you actually need for math competitions, i.e. USAMO, but keep in mind that these competitions are meant to promote interest in math, not limit it. You shouldn't hold back on a subject just because it doesn't appear on the USAMO.

[quote="Phelpedo"]In my opinion, the more rigorously you study these subjects, the better. It's a question of whether you want to memorize tricks for taking derivatives and integrals or whether you really want to understand why and how they work.
[/quote]
 Maybe it's just been that a few things I have read scared me away. I certainly don't want to shy away from rigor, I just want a text that will be understandable. I don't want to memorize "tricks", I *DO* want to know how they work.
[quote="Phelpedo"]
 Anyhow, I must confess that I don't quite understand your motivations. You want to learn Calculus, Differential Equations, Linear Algebra, Abstract Algebra, and Complex Analysis just so you can spend the rest of your time focusing on precalculus? Why not just finish precalculus this summer and then move in to the other subjects? I think you want to get the other material "out of the way" so you can focus on stuff you actually need for math competitions, i.e. USAMO, but keep in mind that these competitions are meant to promote interest in math, not limit it. You shouldn't hold back on a subject just because it doesn't appear on the USAMO.[/quote]
I want to learn these things because they interest me. I'm certainly not going to hold back on a subject, I just need something to augment my brain-dead high school curriculum, at least in the areas of math and science.

I'd just really appreciate if everyone stopped questioning my motivations, as petty and uninformed as they may be, and just directed me towards some books. :(

For calculus, I'm considering just checking out Apostol, Spivak (on Manifolds), and using a generic text I have at home. Will that be enough to pound it all into my brain?

[quote="Elemennop"]Spivak's Calculus on Manifold's is no where near anything introductory..[/quote]
Isn't that what people say when they refer to Spivak though? I'm quite confused myself.

Taking a second look at Apostol, it looks brilliant. I'll just start with calculus and see how far it takes me; no sense planning things I know nothing about.

As for chemistry and physics, though, what are the texts that are Apostol/AoPS-like in rigor and depth?

Also, I'm not understanding the difference between all these "advanced calculus" books and the regular ones. Like Spivak's Manifolds versus Apostol. It's so confusing...

For a rigorous introduction to real analysis, many people seem to agree on Rudin. That's the little blue book of his. It's hardcore.

For complex analysis, try the second book in Stein's "Princeton Lectures in Analysis" series. I heard it's so good that Harvard's complex course uses it even though it says Princeton on the cover. ;)

For Linear Algebra, Hoffman & Kunze is what Princeton uses. This is rigorous linear algebra and it is not extremely easy (especially if you don't skip the "advanced material" sections). There are good exercises in this book.

For abstract algebra, I used Dummit & Foote, which is a vast book. Actually, we did like the first quarter or so and then some of the Galois theory. But Artin's book is popular as well and some say it's easier. Dummit & Foote is a great reference and it has a lot of good problems though.

For physics, I believe both Princeton and MIT use Kleppner & Kolenkow for mechanics and Purcell for E&M. I used Halliday & Resnick in high school and K&K and Purcell were much better, more difficult, more in-depth. They have hard problems. If you master those two books and read up on modern physics somewhere, you should be doing fine in physics.

I don't know much about chemistry but I used Zumdahl for AP in high school and after going through that book a lot of people who take AP at my school start talking about getting perfect scores on the AP test (and chemistry is one of the ones with the lowest percentage necessary for a 5). I tried for IChO and I didn't quite make the US camp but I was in the 21-70 range or something like that.

You'll also want some orgo for chem olympiad, but I don't really have a good recommendation there. I just read a bit of my chem teacher's old orgo book from when she was in college.

A good way to choose textbooks is to go to the course pages of top universities and see what books they use for intro classes. Things like a book being used by multiple different universities are especially good indicators of quality.

What does geometry at a college level consist of, and could you give a basic description of each little area? I've heard names tossed around like topology, differential geometry, combinatorial geometry, and algebraic topology, but I have no clue if that is all there is or what they mean. I know it's way too early to jump into that stuff, but I just need to know for my general knowledge. Geometry is beautiful, in its own little evil way. Are there any college level Euclidian geometry texts? I know there are a lot of theorems from [url=http://mathcircle.berkeley.edu/plane.pdf]here[/url] I have no idea how to prove. Also there is all the spherical stuff, constructions, projective and transformational geometry, and other precalculus things that aren't covered by AoPS2 or Geo. Revisited I'd like to pick up somewhere, even if they have no use in contests.

Will Apostol cover all of what some books call "Advanced Calculus"? I know Spivak mentions manifolds, and I have no idea what those are, let alone if Apostol covers them. Does Apostol cover functions of a complex variable? Or is advanced calculus another term for analysis? All these different integrals and things are confusing. Also, you mentioned Rudin was an introduction; what books come after that?

I know the answer for now is "shut up and do Apostol", but I'd really like to know what I'm getting myself in to.

[quote="Xevarion"]Learn analysis before you do Spivak.

It's hard to understand even things like "what is [insert name of advanced field of mathematics]" when you don't know the foundations. I still don't really know what K-theory is. I don't really know what representation theory is. I could tell you what various fields in math "are" in some sense, but my statements would probably be either worthless or impossible for you to understand or both.

Be patient. You have plenty of time to learn more math.

As for your question about geometry:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=777175#777175
But I'm not qualified to say much more about geometry unless you want to start hearing about measure theory.[/quote]
Many thanks.

Could you answer this, though:
[quote]
Will Apostol cover all of what some books call "Advanced Calculus"? I know Spivak mentions manifolds, and I have no idea what those are, let alone if Apostol covers them. Does Apostol cover functions of a complex variable? Or is advanced calculus another term for analysis? All these different integrals and things are confusing. Also, you mentioned Rudin was an introduction; what books come after that? [/quote]

I can only answer the last couple since I've never seen Apostol.

1) If you want to do more analysis, there are plenty of books. It depends on what you want to do. For the "intermediate" level, I'd personally recommend all of Stein's sequence, Fourier Analysis, Complex Analysis, and Integration Theory. (The fourth book isn't out yet as far as I know.) This will get you pretty much all analysis you need to know in undergrad, plus cool applications (Dirichlet's theorem in the Fourier book, other NT problems in Complex, and fractals/geometric measure theory in the Real Analysis book).

Rudin has more analysis books, and his are often cited as classics.

I can give some more references for more advanced analysis too, but I don't know why you'd care at this point.

2) What is "advanced calculus"? To be honest, I don't really know. I'd guess they mean intro analysis. I've heard that Apostol does give more rigor than most calculus textbooks, but that book is more like a compromise between standard calculus books (like Stewart or whatever) and an analysis text. I learned calculus first and analysis afterwards and I didn't have any trouble. (However, analysis is kinda boring at first. Be patient; it gets really awesome in time.)

Again, in a moment of candor, I will admit that once I felt I understood measure-theoretic integration (Lebesgue integration), I pretty much stopped caring about other kinds of integration. I mean, from reading wikipedia I have the impression that there are some notions of integration that are somehow more general than Lebesgue integration, like Denjoy integrals or gauge integrals or something. But these are advanced topics that you don't need to worry about (yet). Let me give a brief summary of kinds of integrals:

Riemann integral -- the basic thing you learn. It's intuitively simple: approximate your function by a piecewise constant function, which has a well defined notion of "area beneath the curve." Unfortunately, you're quite limited in what you are allowed to integrate.

Stieltjes integral -- lets you integrate "against" another function. I don't really know a good intuitive way to explain this without probability theory. But basically it lets you "weight" different parts of the function you're integrating more than others. This, however, is a generalization of the Riemann integral and still suffers from the same limitations.

Lebesgue integral -- this gets everything you already have above and more. The notion of measure basically gives you more flexibility than you had before. However I don't want to fill this thread with technical details.

Other integrals -- I dunno wtf. Hardcore.

You will learn the Riemann integral in your first analysis book (I don't know if Apostol has it but Rudin does). You might also learn the Stieltjes integral.

You will learn the Lebesgue integral in Stein's 3rd book, or in a full-fledged "real analysis" text, or in a measure theory text.

I don't know where you learn the others but I'm sure Wikipedia and Mathworld have references.

OK I've babbled a lot and probably made myself sound like an authority in the process. Take my words with a grain of salt, please. I'm still just an undergraduate student. :P Hope my comments help. :)

For fun, I would just like to add that Stieltjes integrals are as far as I know pretty much worthless (basically the same thing as u-substitution?) when you already have the Riemann integral.

Thanks for that. Though I'll probably forget and be like (in a few years) "woah I think I finished apostol what calc/analysis book next"...

The TOC of the books are [url=http://www.wiley.com/WileyCDA/WileyTitle/productCd-0471000051.html]here[/url] and [url=http://www.amazon.com/gp/reader/0471000078/ref=sib_dp_pt/002-2788598-3301647#reader-link]here.[/url]

In the TOC to 1 Polynomial Approximations covers Tayor stuff. Complex numbers contains almost no actual complex calculus, just a refresher with DeMoive and stuff. (They differentiate exactly 1 function...). Keeping that in mind, could you recommend a few books to read after that, so I could just store the names away in my memory and not make another thread, or do your recommendations of the Stein books still hold? (I really don't see why you are recommending them, amazon is calling them graduate texts...)

If you get baby Rudin down solid, you can handle Stein's books. It's just that at many universities, only a first course in analysis is required (such as one that would cover Rudin), and further analysis classes would be upper level undergrad or intro graduate classes. At Princeton, most undergrad math majors will take at least one course from Stein's sequence. (I like analysis, so I've done three already and I'll do the fourth one next year.)

Your knowledge of complex numbers from contest math should be sufficient for what's expected in most complex analysis texts (assuming you have the first course in analysis under your belt as well).

Remember that my recommendations are for after you're done with calculus and intro real analysis. That's not going to be this summer!

By the way, it might be a good idea to do linear algebra before you go past Rudin in analysis also. Linear algebra is also a good stepping stone to abstract algebra, since vector spaces provide examples of tons of stuff in algebra. You can learn algebra from first principles, but knowing lin alg first makes it easier.

So now that you know you're supposed to do calculus, real analysis, and linear algebra before you start doing Stein's books, do you feel a little better about them? :P

Lebesgue integral cannot do everything that Riemann integral does.  For example, $\int_{0}^{\infty}\sin(x^{2})dx$ does not exist in the sense of Lebesgue. Moreover, the integrals that are found in a typical calculus-based physics course arise naturally as Riemann integrals. (But then, more advanced mathematical physics requires Lebesgue integration.)

[quote="mlok"]For example, $\int_{0}^{\infty}\sin(x^{2})dx$ does not exist in the sense of Lebesgue.[/quote]

Errr... what would be the Riemann integral of that?  :maybe:

[quote="mlok"]Lebesgue integral cannot do everything that Riemann integral does.  For example, $\int_{0}^{\infty}\sin(x^{2})dx$ does not exist in the sense of Lebesgue. [/quote]
What but you could still get it via a limiting process if you felt like it. Or that function is integrable wrt other measures. Anyway what I meant was "for all intents and purposes," ...
sorry i'm rambling
i should go to sleep and post again when i wake up so that its coherent

[quote="probability1.01"]Errr... what would be the Riemann integral of that?  :maybe:[/quote]
Came up a lot of times on this forum. See also: [url=http://en.wikipedia.org/wiki/Fresnel_integral]Fresnel integral[/url].

  to Xevarion: the difference between Riemann and Lebesgue integration is that Riemann integral uses the order structure of $\mathbb R$, whereas Lebesgue integral works on $\mathbb R$ as on any other measure space. But the existence of the improper Fresnel integral depends on the ordering of $\mathbb R$.
  Anyway, this is an academic point. Good night.  :)

[quote="mlok"][quote="probability1.01"]Errr... what would be the Riemann integral of that?  :maybe:[/quote]
Came up a lot of times on this forum. See also: [url=http://en.wikipedia.org/wiki/Fresnel_integral]Fresnel integral[/url].
[/quote]
Oh right I was thinking $\sin \sqrt{x}$ in my mind. That is pretty cool!


geometry
circumcircle
trigonometry
ratio
geometry unsolved

Take $M$ and $M'$ isogonal points in triangle $ABC$ and take the circumpedal triangles of $M$ and $M'$. Now take $m$ and $m'$ to be the isogonal points of $M$ and $M'$ with respect tot the previous triangles. Show that the distance from the circumcenter to $m$ and $m'$ is the same.

What do you mean by "circumpedal triangles"? As far as I know, the circumpedal triangle is the triangle formed by the intersections of the altitudes with the circumcircle (the three points different from the vertices of the triangle, of course).

For a point $P$ intersection of cevians $AP,BP,CP$ with circumcircle form circumpedal triangle.

I denoted $M = N$ and $m = n$ for simplicity.
Let $T(M), T(N)$ be circumpedal triangles and
$P_m, P_n$ podar triangles of $m,n$ in triangles $T(M),T(N)$.

The idea of the solution to denote all six angles that appear($\angle 1, \angle 2, \angle 3, \angle 4, \angle 5, \angle 6$)
And directly to prove that $\frac{P_m}{T(M)} = \frac{P_n}{T(N)}$
When you will calculate you will have fractions with sines in one part, which
will be the same in the other.

Then use formula:
$\frac{S(podar triangle for(M))}{S(triangle)}=\frac{R^2-OM^2}{4R^2}$

$\frac{ R^2-Om^2}{4R^2} = \frac{ R^2-On^2}{4R^2}$

$T(M),T(N)$ are inscribed in the same circle, so $Om = On$

P.S To all Romanian TST participants good luck in their geometry test. :)

Thank you, Prowler. Very ingenious idea with those podar triangles. I would have probably never think of that formula allthough it's recently I saw it last time.
hihihi. Many thanks, good luck to you too in your tst's.

there is something I don't understand allthough.... How do you compute the areas of the pedal triangles? I mean, I know a way how this goes (proving they are similar then take the ratio between their circumcribed circles' radius which are the same with the circumscribed circles of their isogonals and then calculate the ratio between these and the radius of the circumscribed circles of M and M' in the original triangle....) but it seems more simpler from your message so I would be grateful if you could show me the easy way.


inequalities
combinatorics proposed
combinatorics

Let $\{a_{n}\}_{n\geq 0}$ be a non-decreasing, unbounded sequence of non-negative integers with $a_{0}=0$. Let the number of members of the sequence not exceeding $n$ be $b_{n}$. Prove that
$(a_{0}+a_{1}+\cdots+a_{m})( b_{0}+b_{1}+\cdots+b_{n}) \geq (m+1)(n+1).$

? What are $m$ and $n$? I mean that the inequality cannot hold for all non-negative integers $m$ and $n$ : For any given $m$, let's consider a sequence such that $a_{k}=0$ for $k \leq m$ :wink:

Pierre.


inequalities
calculus
derivative
inequalities solved

Show that if n natural number which is not square of an integer then

         |(sqrt(n)+1)sin(sqrt(n)Pi)| >  \pi/2

and \pi/2 is the best constant for this inequality.

Namdung

Very interesting inequality and not only inequality. This is the first thing I will think about when I'm not busy with my exams.

Ok.....that is the answer! You find all the proposed inequalities in no more than 5 minutes just because you are busy with your exams.... :D

Pierre.

I understood very well your irony. I think you deserve some explanations. First of all, I never pretended that I can solve any inequality in 5 minutes. And I would have been crazy to say so. An excellent example is the Crux inequality whose solution came to my mind after more than 2 months. Moreover, it's not me, but you who came with this "idea" that I can prove inequalities in 4 minutes. This is completely false! Proving a simple inequality in 4 minutes as the one posted by ou doesn't mean anything. And for your curiosity, I'm not proud to have proved that, but the Crux inequality.
   I tried to prove this one for about 20 minutes last evening, but I couldn't. This doesn't mean again anything. I will wayt until I have a good idea. What's the problem? Yes, I am really busy with my exams, since tomorrow I will have a difficult exam.
  In conclusion, I will let you solve all the inequalities in this forum, since it seems I'm too arogant to admit that I cannot solve all of them and I'm also uncapable to solve this one.

????
Oh, come on, I was joking Harazi!
Well, it seems that it was not a good joke....

Pierre.

The following may be the greatest stupidity of all times:
   We can write it as sin(pi*{sqrt(n)})>1/(1+sqrt(n)). First, let's prove that sin(pi*x/2)>=x for all 0<x<=1/2. Let's take f(x)=sin(pi*x/2)-x for all 0<=x<=1/2. Its derivative is f'(x)=pi/2*cos(pi*x/2)-1>=pi/2*cospi/4-1>0. So, f is increasing and thus f(x)>=f(0)=0.
    Now let the case {sqrt(n)}<1/2. Since {sqrt(n)}>=sqrt(n)-sqrt(n-1)=1/(sqrt(n)+sqrt(n-1))>1/2sqrt(n)>1/[2*(1+sqrt(n)], we have sin(pi*{sqrt(n)}>=sin (pi*1/(1+sqrt(n))/2)>=1/(1+sqrt(n)) since 1/(1+sqrt(n))<=1/2.
   Now let {sqrt(n)}<1/2 and let x=1-{sqrt(n)}. Let n=k^2+p, with 0<p<=2k. Since {sqrt(n)}>1/2, we must have p>=k+1. So, p=m+k, with 1<=m<=k. So, x=1+k-sqrt(k^2+k+m)=(k+1-m)/(1+k+sqrt(k^2+k+m)>=1/(1+k+sqrt(k^2+2k))>=1/2(k+1).Thus, sin(pi*x)>=sin (pi*1/(k+1)/2)>=1/(k+1)>1/(sqrt(n)+1).
   The other question bothers me. Hoping that the above is correct ( I sincerely doubt), then using the same arguments one finds that for all 0<=x<=1/2 we have sin(pi*x/2)>=(pi/2cospi/4)x and so after all pi/2*cospi/4 is also a good constant, which is very starnge since it is >1.
    Namdung, please check these and also post your solution. I'm really bothered woth this question. Thank you.

Sorry, Harazi, the best constant is Pi/2 (I had corrected)

Namdung

Unfortunately, in this case everything from above is useless. So, this is a stupid idea. Keep trying!

But something good happened indeed: the fact that it is best is trivial. Just take c such that it holds for any n and take n=k^2+1. Thus, sin (pi/(k+sqrt(k^2+1))*(1+sqrt(k^2+1))>=c for all k and if we take the limit we find that c<=pi/2. Now, let's prove that pi/2 verifies. Maybe next time!

If I'm not wrong, we can still save something from the above:
   So, let the case {sqrt(n)}<1/2. We have {sqrt(n)}>1/(sqrt(n)+sqrt(n-1)) and we have to prove that this is at least pi/2(1+sqrt(n)). But sinx>x-x^3/6, so it is enough to prove that 1/(sqrt(n)+sqrt(n-1))-(pi)^2/[6*(sqrt(n)+sqrt(n-1))^2]>1/(2+2sqrt(n)). , or [ 2+sqrt(n)-sqrt(n-1)]/(1+sqrt(n))>(pi)^2/[3*(sqrt(n)+sqrt(n-1))^2], or 6(sqrt(n)+sqrt(n-1))^2+3(sqrt(n)+sqrt(n-1))>(pi)^2(1+sqrt(n)), which holds.
  Now let {sqrt(n)}>1/2. As above, it is enough to prove that sin(pi)/[1+k+sqrt(k^2+2k)]>pi/[2*(1+sqrt(k^2+k)]. Use again sinx>x-x^3/6 and reduce the inequality to [ 2sqrt(k^2+k)-k-sqrt(k^2+2k)+1]/(1+sqrt(k^2+k)]>(pi)^2/[6*(1+k+sqrt(k^2+2k))]. But this is true since 2sqrt(k^2+k)>k+sqrt(k^2+2k) and 6(1+k+sqrt(k^2+2k))^2>pi^2*(1+sqrt(k^2+k). Again, I' not really sure about these computations, but I think this is the idea.


geometry
perimeter
trigonometry
geometry solved

triangle ABC,(cosA/sinB)+(cosB/sinA)=2,its perimeter is s,solve the maximum area of the triangle.

[color=blue][b]Problem.[/b] Among all triangles ABC with given perimeter p and satisfying the equation $\frac{\cos A}{\sin B}+\frac{\cos B}{\sin A}=2$, find the one with the greatest area.[/color]

[i]Solution.[/i] First, let's find out what exactly the condition $\frac{\cos A}{\sin B}+\frac{\cos B}{\sin A}=2$ is about. In fact, upon multiplication by 2 sin A sin B, this condition becomes $2\sin A\cos A+2\sin B\cos B=4\sin A\sin B$. But using formulas from trigonometry,

$2\sin A\cos A+2\sin B\cos B=\sin\left(2A\right)+\sin\left(2B\right)=2\sin\frac{2A+2B}{2}\cos\frac{2A-2B}{2}$
$=2\sin\left(A+B\right)\cos\left(A-B\right)$.

Since A + B = 180° - C by the sum of angles in triangle ABC, we have sin (A + B) = sin C, and thus this becomes

$2\sin A\cos A+2\sin B\cos B=2\sin C\cos\left(A-B\right)$.

Thus, we have $2\sin C\cos\left(A-B\right)=4\sin A\sin B$. Since A, B, C are the angles of a triangle, their sines sin A, sin B, sin C are all positive; hence, this equation yields that cos (A - B) is also positive. In other words, |A - B| < 90°.

Now, if we have A + B > 90°, then A > 90° - B, while |A - B| < 90° yields B - A < 90°, or, equivalently, - A < 90° - B. Combining A > 90° - B and - A < 90° - B, we get cos A < cos (90° - B). In other words, cos A < sin B, so that $\frac{\cos A}{\sin B}<1$. Similarly, $\frac{\cos B}{\sin A}<1$. Thus, $\frac{\cos A}{\sin B}+\frac{\cos B}{\sin A}<1+1=2$, contradicting our condition $\frac{\cos A}{\sin B}+\frac{\cos B}{\sin A}=2$.

On the other hand, if we have A + B < 90°, then A < 90° - B, what immediately yields cos A > cos (90° - B). In other words, cos A > sin B, so that $\frac{\cos A}{\sin B}>1$. Similarly, $\frac{\cos B}{\sin A}>1$. Hence, $\frac{\cos A}{\sin B}+\frac{\cos B}{\sin A}>1+1=2$, what again contradicts our condition $\frac{\cos A}{\sin B}+\frac{\cos B}{\sin A}=2$.

So we have neither A + B > 90° nor A + B < 90°. Thus, the only case remaining is A + B = 90°. And in fact, it is easy to see that in this case, the condition $\frac{\cos A}{\sin B}+\frac{\cos B}{\sin A}=2$ indeed holds: Since A + B = 90°, we have A = 90° - B, so that cos A = sin B and thus $\frac{\cos A}{\sin B}=1$, and similarly $\frac{\cos B}{\sin A}=1$, so that $\frac{\cos A}{\sin B}+\frac{\cos B}{\sin A}=1+1=2$.

Hence, the condition $\frac{\cos A}{\sin B}+\frac{\cos B}{\sin A}=2$ is equivalent to A + B = 90°. Of course, since A + B = 180° - C, we have A + B = 90° if and only if 180° - C = 90°, what is equivalent to C = 90°, i. e. to the assertion that the triangle ABC is right-angled at C. Thus, the condition $\frac{\cos A}{\sin B}+\frac{\cos B}{\sin A}=2$ is nothing but a complicated way to say that the triangle ABC is right-angled at C.

Now, in order to solve the problem, it remains to find, among all triangles ABC with a given perimeter p and right-angled at C, the one with the greatest area.

For that aim, we denote by $\Delta$ the area and by $s=\frac{p}{2}=\frac{a+b+c}{2}$ the semiperimeter of triangle ABC. By the half-angle formulas,

$\tan\frac{A}{2}=\sqrt{\frac{\left(s-b\right)\left(s-c\right)}{s\left(s-a\right)}}$; $\tan\frac{B}{2}=\sqrt{\frac{\left(s-c\right)\left(s-a\right)}{s\left(s-b\right)}}$; $\tan\frac{C}{2}=\sqrt{\frac{\left(s-a\right)\left(s-b\right)}{s\left(s-c\right)}}$,

and by the Heron formula,

$\Delta=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}=s^2\cdot\sqrt{\frac{\left(s-a\right)\left(s-b\right)\left(s-c\right)}{s^3}}$
$=s^2\cdot\sqrt{\frac{\left(s-b\right)\left(s-c\right)}{s\left(s-a\right)}}\cdot\sqrt{\frac{\left(s-c\right)\left(s-a\right)}{s\left(s-b\right)}}\cdot\sqrt{\frac{\left(s-a\right)\left(s-b\right)}{s\left(s-c\right)}}$
$=s^2\cdot\tan\frac{A}{2}\cdot\tan\frac{B}{2}\cdot\tan\frac{C}{2}$.

Now, we have to maximize the area $\Delta$ of triangle ABC. Since the perimeter p of triangle ABC is fixed, the semiperimeter $s=\frac{p}{2}$ is also fixed. Also, $\tan\frac{C}{2}$ is fixed, since C = 90°. Hence, it remains to maximize $\tan\frac{A}{2}\cdot\tan\frac{B}{2}$. But, according to http://www.mathlinks.ro/Forum/viewtopic.php?t=38984 post #3, we have $\tan\frac{A}{2}\tan\frac{B}{2}\leq\left(\sqrt2-1\right)^2$ with equality if and only if $\frac{A}{2}=\frac{B}{2}$, i. e. if A = B. Thus, the maximum of the expression $\tan\frac{A}{2}\cdot\tan\frac{B}{2}$ is achieved when A = B, i. e. when the triangle ABC is an isosceles right triangle.

Hence, we see that the area of triangle ABC is maximal when it is an isosceles right triangle (with the right angle at C). Thus, the problem is solved.

  Darij

errr...Darij, there is actually a simpler approach after finding $\frac{\cos A}{\sin B}+\frac{\cos B}{\sin A}=2\Leftrightarrow C=\pi /2$. Anyway, never mind, your method also works ;)

We have $p=a+b+c=a+b+\sqrt{a^2+b^2}\ge a+b+\frac{1}{\sqrt{2}}(a+b)$. Hence $a+b\le(2-\sqrt{2})p$. Note that the area of $\triangle ABC$ is in fact $ab/2\le(a+b)^2/8$.


[quote="happyme"]Two handymen, Peter and Paul, agree to paint the interior of a house and lay hardwood. The total area of the walls to be painted is 4000 square feet and the total floor area is 1500 square feet. In an hour of work, Peter can either paint 40 square feet or lay 15 square feet of hardwood. Paul can either paint 20 square feet in an hour or lay 30 square feet of hardwood. What is the minimum time (in hours) in which Peter and Paul, working together, can finish the job?[/quote]

The same problem has been recently posted here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=137967[/url]  :D

Woah, the guy went to the same contest I did...

Time to chat....  :lol:


I am having problems with the following simultaneous equation:

$ 7x \minus{} (3\sqrt {5})y \equal{} 9\sqrt {5}$ [1]
$ (2\sqrt {5})x \plus{} y \equal{} 34$   [2]

I've tried dividing equation 1 by $ \sqrt {5}$ and then multiplying equation 2 by 3 to make the $ y$'s cancel, but I keep triping up.

Guidance on this problem would be much appriciated.

Many Thanks.

[hide="Hint"]
Why don't you rearrange equation 2 to substitute into equation 1?

$ (2\sqrt{5})x\plus{}y \equal{} 34$

$ y\equal{}34\minus{}(2\sqrt{5})x$

$ 7x\minus{}(3\sqrt{5})[34\minus{}(2\sqrt{5})x] \equal{} 9\sqrt{5}$
[/hide]

Btw, just FYI, they're called systems.

[hide="More Hints"]
Read up on substitution and elimination (also called combinations).
To solve in general: Rearrange 1 equation so that you isolate a variable (in terms of the other one). Then plug in that value in to the second one, and then solve for it. Then sub it back in to either equation, and solve for the original isolated variable.

Example:
Solve the system
3x - 9y - 7 = 2
x + y = 7

1. Choose an equation; let's use the first.
2. Choose a variable; let's use x
3. Isolate x (basically solve, but put it in terms of y). Steps
    3x - 9y = 9
    3x = 9y + 9
    x = 3y + 3
4. Take the other equation and use substitution:
    (3y + 3) + y = 7
5. Solve for y:
    4y + 3 = 7
    4y = 4
    y = 1
6. Plug in y = 1 in to either of the original equations; let's use the second.
7. Solve for the other variable:
   x + 1 = 7
   x = 6.

Thus, our answer is x = 6, y = 1 (also written as (6, 1)).

More specifically, in the case of your problem, here's a start.
Take [2] and subtract y from both sides, then divide by 2\sqrt{5}. We get
$ x\equal{}\frac{34\minus{}y}{2\sqrt{5}}$
[/hide]

Brute forced with matrices.
multiply answer matrix by the inverse of the coefficient matrix

x= 3 sqrt 5
y = 4


geometry
circumcircle
incenter
parallelogram
geometric transformation
geometry proposed

Let ABC be a triangle ABC satisfying AB+AC=2BC. AH is the A-altitude of triangle ABC. $ (O_1),(O_2)$ are two circles which touch AH,BC and intouch the circumcircle (ABC). Show that $ O_1O_2$ passes through the orthocenter of triangle ABC.

Some ideas which maybe are useful in this context:
The situation of the problem is a special case of the theorem of Thébault, which say the line O_1O_2 pass through the incenter I of ABC.
Moreover, special triangles with b+c=2a verify, among others, the following properties:
OI is perpendicular to AI
GI is parallel to BC.

see the link http://www.mathlinks.ro/viewtopic.php?t=210342

Very nice, mr.danh.

1. In any $ \triangle ABC$ with orthocenter $ H,$ if the A-altitude $ AH$ cuts the circumcircle $ (O)$ again at $ K$ and parallel to $ BC$ through $ K$ cuts the circumcircle again at $ L,$ then $ BHCL$ is a parallelogram and the diagonal $ LH$ cuts the diagonal $ BC$ at its midpoint $ M.$

2. In any $ \triangle ABC$, let $ AP$ (with $ P$ on the segment $ BC$) be an arbitrary cevian and $ (O_1), (O_2)$ circles tangent to $ AP, BC$ and internally to $ (O).$ If $ AP$ cuts the circumcircle $ (O)$ again at $ K$ and parallel to $ BC$ through $ K$ cuts the circumcircle again at $ L,$ the other common internal tangent of $ (O_1), (O_2)$ goes through $ L.$ Since I got this as a sideeffect of something else I was after, my current proof is not short and I will look for something simpler, before posting it.

3. If $ a \equal{} \frac{b \plus{} c}{2}$ and $ AP \perp BC$ (the cevian $ AP$ is the A-altitude), then the other common internal tangent of $ (O_1), (O_2)$ goes through the midpoint $ M$ of $ BC$ by Thebault's theorem. This problem was posted (and solved) before. Does anybody want a link ?

Yes, please, I would like very much to have the link. Thanks in advance.
I found a metric proof of the fact r_1+r_2 = 2r, which is true in the situation of the problem (with the cevian from A being the altitude), and also found a nice expression for HH_a (H, orthocenter, H_a foot of altitude) as r_1r_2/r, but I was unable  for the moment of find the relation of all this with the characterization of our triangle (b+c=2a).
Very nice problem indeed.

Here is the link for No.3: [url]http://www.mathlinks.ro/viewtopic.php?t=118385[/url]. As you can see, I also did a metric proof at first and then, not being familiar with Thebault's theorem at the time, actually proving it. It remains to show No.2, but as I said, my current (synthetic) proof is not short.

Many thanks for the link, Yetti. I have located the problem in CRUX MATHEMATICORUM, problem 1260, (proposed by Hidetosi Fukagawa) 1988, pp.236-240. The solution there is a translation of a proof of the theorem of Thébault published (in german) in the swiss journal Elemente der Mathematik, vol. 41 (1986), pp.11-13, by G.Turnwald.

See this post for a related problem:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88823[/url]

I have made a seach in my library about the theorem of Thebault, and it seems nobody before mr.danh has realized the line which join the centers of both circles, in the special case 2a=b+c and being the cevian of Thebault the altitude from A, pass through the orthocenter of ABC. I am really curious to see the solution of mr.danh.

I proposed this problem when i had just solved a problem. It is the following problem:
Problem: Let a quadrilateral ABCD inscribed the circle (O). AC meets BD at E. The circle (1) touches EA,ED and (O), the circle (2) touches EB,ED and (O). Then, two external tangents of (1) and (2) are parallel to AB,CD.

And my proof and yetti's solution are the same idea. We both make a use of the above problem and Thebault's theorem.
I think the problem is rather nice and its enunciation is simple.

Dear mr.danh, thank you for your interesting problem! It help me very much. And I think I shoud sorry you about my message. Can you give me a chance to learn from you? OK?

Link to proof of No.2: [url]http://www.mathlinks.ro/viewtopic.php?t=15945[/url]. The topic problem is different, scroll down to lemma "Parallel tangents" by [b]juancarlos[/b] (post 8) and two proofs of the lemma  by [b]Fang-jh[/b] (post 16) and [b]Huyền Vũ[/b] (post 19).

I am still trying of to get a metric solution of the problem. My approach is as follows:
1) An easy similarity of rightangle triangles shows that, if x = HH_a (being H the point of intersection of the line O_1O_2 with the altitude AH_a), x = 2r_1r_2/(r_1+r_2).
2) As in the situation of the theorem of Thebault with the cevian being altitude, r_1+r_2 = 2r, one get x= r_1r_2/r.
3) We will need to prove that, in the triangle with b+c=2a, the value obtained for x is the same as the distance from the orthocenter to the side BC, which, in any triangle, equals 2RcosBcosC. (R circumradius of ABC)
4) We will need use the condition b+c=2a, or equivalently, sinB+sinc=2sinA which also can be written as 2sin(A/2)=cos((B-C)/2).
Nevertheless, until now I was unable of simplify the expressions for r_1, r_2 given in the various solutions of the Theorem of Thebault in order to prove the desired value of x = r_1r_2/r. Somebody can help? Thanks

Finally I was able of finish the metric solution I was searching for.
I continue from my last post.
1) In the triangle with 2a=b+c, as 2s=3a, we get ah = 2sr and from this we have h=3r (h is the lenght of the altitude from A).

2) For another hand, searching in the literature of the Theorem of Thebault, in the book of H.Fukagawa and D. Pedoe "Japanese Temple Geometry Problems-Sangaku" there are the following expressions for the radiuses of the circles (O_1) and (O_2), in the case when the cevian of Thebault is the altitude from A:

r_1=(c+h-BH_a)(CH_a + h - b)/2h

r_2=(b+h-CH_a)(BH_a+h-c)/2h

(we suppose without loss of generality than O_1 is at the right of the altitude AH_a)

from this we obtain for the product

r_1r_2 = (2RcosBcosC)(2sin(B/2)cos(C/2)

Taking account the relation 3r =h, the searched x=r_1r_2/r reduces in this case to verify that, in the triangle of the problem, the relation
tan(B/2)tan(C/2) = 1/3; and this indeed holds, because

(tan(B/2))^2 = (3a-2c)/3(3a-2b) and (tan(C/2))^2 = (3a-2b)/3(3a-2c).

Thanks to mrdanh for a very nice problem!

[quote="28121941"]Thanks for the link about No.2, Yetti. Maybe you do not know our mathlinker "JuanCarlos" passed away some weeks ago? It is a big loss for the Forum, I think.[/quote]

Who is JuanCarlos? Maybe you can tell us more about him please! R.I.P.

I means Juan Carlos Salazar, from Puerto Ordaz (Venezuela). At this moment I do not remember his id as mathlinker, but his avatar was a picture with a kipu (device used by the incas for make calculations. He put several posts about geometry.


Who's HVC?

Exapnding, the ineq is equivalent to
\[
\sum_\mathrm{sym} (8x^6  + 7x^3y^3) \geq \sum_\mathrm{sym} (9x^4yz + 6x^2y^2z^2)
\]
Now we add up the following inequalities, the first comes from AM-GM, rest from Muirhead
\[
\begin{eqnarray*}
\sum_\mathrm{sym} (4x^6 + 4x^6 + 4x^3y^3) &\geq& \sum_\mathrm{sym}12x^5y \\
\sum_\mathrm{sym} 9 x^5y &\geq& \sum_\mathrm{sym} 9x^4yz \\
\sum_\mathrm{sym} 3 x^5y &\geq& \sum_\mathrm{sym} 3x^2y^2z^2 \\
\sum_\mathrm{sym} 3x^3y^3 &\geq& \sum_\mathrm{sym} 3x^2y^2z^2
\begin{eqnarray*}
\]
and we are done.

\[
R.H.S. \le 9\left( {\frac{{x^2  + y^2  + z^2  + xy + yz + zx}}{3}} \right)^3  \le 72\left( {\frac{{x^2  + y^2  + z^2 }}{3}} \right)^3
\]

It remains to prove \[
\left( {\frac{{x^2  + y^2  + z^2 }}{3}} \right)^3  \le \left( {\frac{{x^3  + y^3  + z^3 }}{3}} \right)^2
\]
which is true by power mean.

Oh, HVC is the author of the book which contains this problem . From me  , this is the hardest problem in that book .
 Dear Billzhao , yo are very patient !! :)
 I will post the solution of this problem soon

Nice solution , Siuhochung , Billzhao !!
 The solution in that book is similar to Billzhao's , even your solution is more simple than the author's and my solution is similar to Siuhochung's  :D  :lol:
 In short , this is a very problem !!


[quote="Pirkuliyev Rovsen"]Let
$ a,b,c,d\in[ 0: 1]$ Prove that:
$ a(1 - d) + b(1 - a) + c(1 - b) + d(1 - c)\leq2$

[/quote][hide]
Expand the expression to get:

$ a + b + c + d -ab - ad - bc - cd =$
$ (a + c) + (c + d) - a(b + d) - c(b + d) =$
$ (a + c) + (b + d) - (a + c)(b + d) =$
either $ (a + c)[1 - (b + d)] + (b + d)$ . . . (*)
or $ (b + d)[1 - (a + c)] + (a + c)$ . . . (**)

In (*), the largest that $ [1 - (b + d)]$ can be is $ 1$ when $ b = d = 0$, and
the largest that $ (a + c)$ can be is $ 2$ when $ a = c = 1$.

It is similar for (**).

[/hide]


Triangles $ ABC$ and $ ADC$ are isosceles with $ AB \equal{} BC$ and $ AD \equal{} DC$. Point D is inside $ \triangle ABC$. $ \angle ABC \equal{} 40^\circ$, and $ \angle ADC \equal{} 140^\circ$. What is the degree measure of $ \angle BAD$?

$ \textbf{(A)}\ 20 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 60$

This problem has also been posted here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=132480

not really quatto....
some problems are shared on AMC 10 and 12, thus worthawholebean has a right to post this, because he has no idea what is on the AMC 10, plus the repeat of this problem isn't even in this forum, so he will have even less of an idea of whether it has been posted or not.

[quote="The QuattoMaster 6000"]I am not saying that worthawholebean did something wrong by posting the problem, nor am I saying that it was his fault.  I am just giving him the problem and the solution through another post.[/quote]

I think you need to repost those problems to be entered into the database in the resources section

This is a simple problem if you read that the point P is inside the trinagle ABC, which i read wrong.
Answer is [b](D.) 50[/b]

[quote="King_James23"]This is a simple problem if you read that the point [b]P[/b] is inside the trinagle ABC, which i read wrong.
Answer is [b](D.) 50[/b][/quote]There is no point $P$... :spam:

[hide=Solution]
$\angle BAD = \angle BAC - \angle DAC = 70-20=\boxed{\textbf{(D)} 50}$.
[/hide]


What is the mean of $ \frac12$, $ \left(\frac12\right)^2$, and $ \left(\frac12\right)^3$? Express your answer as a common fraction.


With four cuts, cut a watermelon into 9 pieces. Is it possible that after you eat the red stuff off the peels, you will have 10 peels?

i don't really get it, but by the law of conservation of watermelon peels, I say no.  :rotfl:

slice it horizontally, vertically and depth-wise like the dimensions, to make 2^3=8 pieces in 3cuts, the separate it and cut one of the pieces in half to make another piece 8+1=9

i think thats the easy part  ;) , what about the second question? eat the stuff off the peels to get 10 peels?!?!?

i don't understand how you can have 9 slices, eat them, and then have ten slices.  Do you eat a slice in two?

It is possible, right? (takin the 50, 50 chance)

[quote="236factorial"][hide="a hint"]

When you eat the red stuff off the watermelon peels, do you always end up with one? Think out of the ordinary.[/hide][/quote]
How about you cut the green stuff but not the red stuff, so when your done, it falls apart :D

[hide="hint number 2"]

To cut into 9 pieces, cut in the shape of the tic-tac-toe board. Now figure out where the ten pieces come from. [/hide]

If You cut it like it says to in hint #2 the center piece will fall apart  :lol:  :lol:  :lol:  :lol:  :lol:  :lol:  :lol:  :lol:  :lol:

[hide] the way it says to cut it in hint #2, the center piece will have 2 peels.[/hide]


inequalities
function
three variable inequality

$a,b,c \ge 0$
prove:
$\sqrt{1+\frac{48a}{b+c}}+\sqrt{1+\frac{48b}{c+a}}+\sqrt{1+\frac{48c}{a+b}} \ge 15$

  :) PS:please notice when does the equality occurs.

Zhaobin, can you tell me what chinese forum is it about? :?

sure.
it is in [url=http://www.aoshoo.com/bbs1/dispbbs.asp?boardID=48&ID=1836&page=1]here[/url]
thanks for your intresting.

2 more questions? Is it only in Chinese? and why nobody mentioned it before? :?

I don't know.you mean it is well known?
sorry,I don't know.
and can you give me the link?
sorry,if it is post before.

Zhaobin, i don't understand nothing. I don't know Chinese, so........ :(
Are you a moderator there? :)

Yes. :)
but what about the nice inequality I post?
Have you seen it before?

[quote="cezar lupu"]It is for the first time I see it. Anyway, i'm really interrested in that Chinese forum. :D

I'll be very happy if they change the official language into English. :P[/quote]
Ok,maybe you can use a software then tranlate the language...
You can also learn chinese yourself,I think it will be a great thing :P

[quote="arqady"][quote="zhaobin"]$a,b,c \ge 0$
prove:
$\sqrt{1+\frac{48a}{b+c}}+\sqrt{1+\frac{48b}{c+a}}+\sqrt{1+\frac{48c}{a+b}} \ge 15$[/quote]
Beautiful inequality! The next is easy:
If $0\leq k\leq8,a>0,b>0,c>0$ then
$\sqrt{1+\frac{2ka}{b+c}}+\sqrt{1+\frac{2kb}{c+a}}+\sqrt{1+\frac{2kc}{a+b}} \ge3\sqrt{1+k}$[/quote]
how can you prove it?mixing variable?

[quote="zhaobin"]
how can you prove it?mixing variable?[/quote]
No. Elementary and easy proof :lol:

[quote="zhaobin"][quote="arqady"][quote="Vasc"]This inequality is the problem 3068 from Crux Mathematicorum, no. 6, 2005.
There is the following requirement:
"Solutions should arrive no later than 1 April 2006".[/quote]
O.K. But which inequality? Of zhaobin or mine?[/quote]
I think it is certail y mine, :P
and did you slove your inequality with jensen?
Ok,if VASC mean this inequality can't be discuss here,please lock it.
sorry I don't this the problem in crux 2005.6[/quote]
I think Jensen deals with both inequalities pretty nicely ;-)

[quote="zanttrang"][/quote][quote="zhaobin"][quote="arqady"][quote="Vasc"]This inequality is the problem 3068 from Crux Mathematicorum, no. 6, 2005.
There is the following requirement:
"Solutions should arrive no later than 1 April 2006".[/quote]
O.K. But which inequality? Of zhaobin or mine?[/quote]
I think it is certail y mine, :P
and did you slove your inequality with jensen?
Ok,if VASC mean this inequality can't be discuss here,please lock it.
sorry I don't this the problem in crux 2005.6[/quote][quote="zanttrang"]
I think Jensen deals with both inequalities pretty nicely ;-)[/quote]
OK，if I'm right,you are wrong :lol:

this is problem # 10 in chapter 3 page 151 of vasc's book. it is solved there using RCF theorem. :)
BTw arqady, nice proof above! :)

[quote="zhaobin"]$ a,b,c \ge 0$
prove:
$ \sqrt {1 \plus{} \frac {48a}{b \plus{} c}} \plus{} \sqrt {1 \plus{} \frac {48b}{c \plus{} a}} \plus{} \sqrt {1 \plus{} \frac {48c}{a \plus{} b}} \ge 15$

  :) PS:please notice when does the equality occurs.[/quote]
The following is my solution:

Without loss of generality, we may assume that $ c\equal{}\min\{a,b,c\}.$ Then, we notice that
$ \left( \sqrt{\frac{a}{b\plus{}c}}\plus{}\sqrt{\frac{b}{c\plus{}a}}\right)^2 [a^2(b\plus{}c)\plus{}b^2(c\plus{}a)] \ge (a\plus{}b)^3,$
from the Holder's Inequality, and
$ a^2(b\plus{}c)\plus{}b^2(c\plus{}a) \equal{}c(a\plus{}b)^2\plus{}ab(a\plus{}b\minus{}2c) \le c(a\plus{}b)^2\plus{}\frac{(a\plus{}b)^2}{4} (a\plus{}b\minus{}2c) \equal{}\frac{(a\plus{}b)^2(a\plus{}b\plus{}2c)}{4}.$
Combining these two estimations, we find that
$ \left( \sqrt{\frac{a}{b\plus{}c}}\plus{}\sqrt{\frac{b}{c\plus{}a}} \right)^2 \ge \frac{4(a\plus{}b)}{a\plus{}b\plus{}2c}.$
Now, using the Minkowsky's Inequality (in combinination with this), we get
$ \sqrt{1\plus{}\frac{48a}{b\plus{}c}}\plus{}\sqrt{1\plus{}\frac{48b}{c\plus{}a}} \ge \sqrt{(1\plus{}1)^2\plus{}48\left(\sqrt{\frac{a}{b\plus{}c}}\plus{}\sqrt{\frac{b}{c\plus{}a}}\right)^2} \ge \sqrt{4\plus{}48\cdot \frac{4(a\plus{}b)}{a\plus{}b\plus{}2c}} \equal{}2\sqrt{1\plus{}\frac{48(a\plus{}b)}{a\plus{}b\plus{}2c}},$
and so we are left to prove that
$ 2\sqrt{1\plus{}\frac{48(a\plus{}b)}{a\plus{}b\plus{}2c}} \plus{}\sqrt{1\plus{}\frac{48c}{a\plus{}b}} \ge 15,$
which is easy to check.

Nice proof, can_hang2007!  :lol:
Thank you!

[quote="zhaobin"]$ a,b,c \ge 0$
prove:
$ \sqrt {1 \plus{} \frac {48a}{b \plus{} c}} \plus{} \sqrt {1 \plus{} \frac {48b}{c \plus{} a}} \plus{} \sqrt {1 \plus{} \frac {48c}{a \plus{} b}} \ge 15$

  :) PS:please notice when does the equality occurs.[/quote]
Once again we can use the uvw-theorem :)
Let $ x \equal{} \sqrt {1 \plus{} \frac {48a}{b \plus{} c}}\minus{}1, y \equal{} \sqrt {1 \plus{} \frac {48b}{c \plus{} a}}\minus{}1, z  \equal{}\sqrt {1 \plus{} \frac {48c}{b \plus{} a}}\minus{}1$, so $ x,y,z \ge 0$.
Then we have $ \frac{a}{b\plus{}c} \equal{} \frac{x^2\plus{}2x}{48}$, and so on. Using the wellknown: $ \sum_{cyc} \frac{a}{b\plus{}c}\frac{b}{a\plus{}c} \plus{} 2\prod_{cyc} \frac{a}{b\plus{}c} \equal{} 1$ we obtain:
$ 48 \sum_{cyc} (x^2\plus{}2x)(y^2\plus{}2y) \plus{} 2(x^2\plus{}2x)(y^2\plus{}2y)(z^2\plus{}2z) \equal{} 48^3$.

We should prove that $ x\plus{}y\plus{}z \ge 12$ when $ 48 \sum_{cyc} (x^2\plus{}2x)(y^2\plus{}2y) \plus{} 2(x^2\plus{}2x)(y^2\plus{}2y)(z^2\plus{}2z) \equal{} 48^3$. Assume that $ x\plus{}y\plus{}z < 12$ when $ 48 \sum_{cyc} (x^2\plus{}2x)(y^2\plus{}2y) \plus{} 2(x^2\plus{}2x)(y^2\plus{}2y)(z^2\plus{}2z) \equal{} 48^3$. Then by increasing $ x,y,z$ 'till $ x\plus{}y\plus{}z\equal{}12$ we will get a situation where:
$ x\plus{}y\plus{}z\equal{}12$ and $ 48 \sum_{cyc} (x^2\plus{}2x)(y^2\plus{}2y) \plus{} 2(x^2\plus{}2x)(y^2\plus{}2y)(z^2\plus{}2z) > 48^3$. So it is enough to prove that:

$ 48 \sum_{cyc} (x^2\plus{}2x)(y^2\plus{}2y) \plus{} 2(x^2\plus{}2x)(y^2\plus{}2y)(z^2\plus{}2z) \le 48^3$ when $ x\plus{}y\plus{}z \equal{} 12, x,y,z \ge 0$.

Let $ 3u\equal{}x\plus{}y\plus{}z, 3v^2 \equal{} xy\plus{}yz\plus{}zx, w^3 \equal{} xyz$.
Writing it in terms of $ u,v^2,w^3$ it clearly becomes on the form $ \minus{}2w^6 \plus{} A(u,v^2)w^3 \plus{} B(u,v^2) \ge 0$ where $ A,B$ are functions in $ u$ and $ v^2$. So according to the uvw-theorem we only have to prove it when $ xyz \equal{} 0$ and when $ (x\minus{}y)(y\minus{}z)(z\minus{}x) \equal{} 0$.
$ xyz\equal{}0$: wlog $ x\equal{}0 \Rightarrow a\equal{}0$.
$ (x\minus{}y)(y\minus{}z)(z\minus{}x) \equal{} 0$: wlog $ y\equal{}z$. Then $ \frac{b}{c\plus{}a} \equal{} \frac{c}{a\plus{}b} \iff (b\minus{}c)(a\plus{}b\plus{}c) \equal{} 0 \iff b\equal{}c$.
So we only have to prove it in to cases:
$ a\equal{}0$, $ b\equal{}c\equal{}1$.

$ a\equal{}0$: $ \sqrt{1\plus{}48\frac{b}{c}} \plus{} \sqrt{1\plus{}48\frac{c}{b}} \ge 14$. Squaring: $ 2 \plus{} 48(\frac{c}{b} \plus{} \frac{b}{c}) \plus{} 2\sqrt{1\plus{}48\frac{b}{c}}\sqrt{1\plus{}48\frac{c}{b}} \ge 14^2$, follows from $ \frac{c}{b} \plus{} \frac{b}{c} \ge 2$ (AM-GM) and $ \sqrt{1\plus{}48\frac{b}{c}}\sqrt{1\plus{}48\frac{c}{b}} \ge (1\plus{}48) \equal{} 49$ (Cauchy-Schwartz)

$ b\equal{}c\equal{}1$: $ f(a) \equal{} \sqrt{1\plus{}24a} \plus{} 2\sqrt{1\plus{}\frac{48}{1\plus{}a}} \ge 15$
Then $ f'(a) \equal{} \frac{12}{\sqrt{1\plus{}24a}\sqrt{49\plus{}a}\sqrt{(1\plus{}a)^3}(\sqrt{49\plus{}a}\sqrt{(1\plus{}a)^3}\plus{}2\sqrt{16(1\plus{}24a)})}$ $ ( (49\plus{}a)(1\plus{}a)^3 \minus{}16(1\plus{}24a) )$
So $ f'(a) \ge 0 \iff$
$ (a\minus{}1)(a^3\plus{}54a^2\plus{}204a\minus{}33) \ge 0$
$ a^3\plus{}54a^2\plus{}204\minus{}33$ has exactly one positive root, and this root is less than $ 1$. Let it be $ \alpha$. Then $ f'(a) \ge 0 \iff (a\minus{}1)(a\minus{}\alpha) \ge 0$. Hence $ f$ is increasing in $ [0;\alpha] \cap [1;\plus{}\infty]$ and decreasing in $ [\alpha;1)$. So $ f(a) \ge \min \{f(0);f(1)\}$. And since $ f(0)\equal{}f(1)\equal{}15$, we see that $ f(a) \ge 15$, and we are done. Equality when $ a\equal{}b\equal{}c$ or when $ a\equal{}b,c\equal{}0$ and permutations.

[quote="zhaobin"]Let $ a,b,c \ge 0,$ prove that

$ \sqrt {1 \plus{} \frac {48a}{b \plus{} c}} \plus{} \sqrt {1 \plus{} \frac {48b}{c \plus{} a}} \plus{} \sqrt {1 \plus{} \frac {48c}{a \plus{} b}}\ge 15.$[/quote][quote="arqady"]Let $ 1 \plus{} \frac {48a}{b \plus{} c} \equal{} (1 \plus{} 4x)^{2},$ $ 1 \plus{} \frac {48b}{a \plus{} c} \equal{} (1 \plus{} 4y)^{2},$ $ 1 \plus{} \frac {48c}{a \plus{} b} \equal{} (1 \plus{} 4z)^{2},$ where $ x,$ $ y$ and $ z$ are non-negative numbers.

Then $ 108 \equal{} \sum_{cyc}\left(3xy \plus{} \frac {xyz}{3} \plus{} 6x^{2}y \plus{} 6x^{2}z \plus{} 2x^{2}yz \plus{} 12x^{2}y^{2} \plus{} 4x^{2}y^{2}z \plus{} \frac {8x^{2}y^{2}z^{2}}{3}\right).$ Remain to prove that $ x \plus{} y \plus{} z\geq3.$

Let $ x \plus{} y \plus{} z < 3$ for some non-negative $ x,$ $ y$ and $ z$. But it's contradiction since,[/quote]$ 8x^2y^2z^2 \plus{} 4xyz\sum{yz} \plus{} 12\sum{y^2z^2} \plus{} 2xyz\sum{x} \plus{} 6\sum{x^2(y \plus{} z)} \plus{} xyz \plus{} 3\sum{yz}$

$ < 8x^2y^2z^2\left(\frac {3}{x \plus{} y \plus{} z}\right)^6 \plus{} 4xyz\sum{yz}\left(\frac {3}{x \plus{} y \plus{} z}\right)^5$

$ \plus{} \left[12\sum{y^2z^2} \plus{} 2xyz\sum{x}\right]\left(\frac {3}{x \plus{} y \plus{} z}\right)^4$

$ \plus{} \left[6\sum{x^2(y \plus{} z)} \plus{} xyz\right]\left(\frac {3}{x \plus{} y \plus{} z}\right)^3 \plus{} 3\sum{yz}\left(\frac {3}{x \plus{} y \plus{} z}\right)^2$

$ \equal{} 108 \minus{} \frac {27}{(x \plus{} y \plus{} z)^6}\sum{(y \minus{} z)^2\left[(x \plus{} y \plus{} z)(15x \plus{} 2y \plus{} 2z)(y \plus{} z \minus{} x)^2 \plus{} 12xyz(5x \plus{} 3y \plus{} 3z)\right]}$

$ \leq 108.$

See also : http://www.mathlinks.ro/viewtopic.php?t=298581

Dear can_hang2007,

your proof is really beautiful. :)

I have completed it putting $ x \equal{} \frac {2c}{a \plus{} b}\leq 1$ . After squaring

$ 2\sqrt {1 \plus{} \frac {48(a \plus{} b)}{a \plus{} b \plus{} 2c}} \plus{} \sqrt {1 \plus{} \frac {48c}{a \plus{} b}} \ge 15$

we obtain $ (x \minus{} 1)^2(15 \minus{} x)\geq 0$

which is obvious :)

P.S.Can_hang2007's method is really powerful as reduces Vasc hard inequality to a simple inequality

The inequality can be generalized as follows:

If $a_1,a_2,\cdots,a_n$ are nonnegative real numbers, then
\[\sum\sqrt{(n^2-3n+1)^2+4n(n-2)(n-1)^2\frac{a_1}{a_2+a_3+\cdots+a_n}}\ge n(n^2-n-1).\]


Hey if anybody has a hard or interesting problem post it here  :D

$x^3+6x^2+9x=0$
Find ALL real solutions. :D

[hide="hint"] Factor out an $x$. :) [/hide]

[hide]
Very easy!
First, I factor out x
$x(x^2+6x+9)=0$
then I factor the trinomial
$x(x+3)^2=0$
therefore, $x= {0,-3}$  [/hide]

These are from the AMC 12, but they're really very easy. You don't even need the multiple choice. :D I can probably put up the first 6 or so, if you want.

1. What is $(-1)^1+(-1)^2+...+(-1)^{2006}$ ?

2. For real numbers $x$ and $y$, define ${x}\spadesuit{y}=(x+y)(x-y)$. What is ${3}\spadesuit{({4}\spadesuit{5})}$?

3. A football game was played between two games, the Cougars and the Panthers. The two teams scored a total of $34$ points, and the Cougars won by a margin of $14$ points. How many points did the Panthers score?

Answer to lotrgreengrapes7926's problems.
I'm not sure I did em all correctly. I make mistakes very easily.:(
[hide]
1. -1 to an odd numbered power is going to be a -1 and an even numbered power is
going to be a 1. Since it begins with an odd numbered power and ends in even numbered power, answer is $0$

2.  First, $(4+5)(4-5)= 16-25 = -9$
then $(3-9)(3+9)=9-81= -72$

3. Lets say Panthers's score is $X$ and Cougar's score is $X+14$
$X+X+14=34$, $2X= 20$ so $X=10$

I hope I got them all right
 :) [/hide]

Yep. :D All correct.

4. Mary is about to pay for 5 items at the grocery store. The prices of the items are $\$7.99$, $\$4.99$, $\$2.99$, $\$1.99$, and $\$0.99$. Mary will pay with a twenty-dollar bill. Which of the following is closest to the percentage of the $\$20.00$ that she will recieve in change?
(A) 5   (B) 10   (C) 15   (D) 20   (E) 25

5. John is walking east at a speed of 3 miles per hour, while Bob is also walking east, but at a speed of 5 miles per hour. If Bob is now 1 mile west of John, how many minutes will it take for Bob to catch up to John?

6. Francesca uses 100 grams of lemon juice, 100 grams of sugar, and 400 grams of water to make lemonade. There are 25 calories in 100 grams of lemon juice and 386 calories in 100 grams of sugar. Water contains no calories. How many calories are in 200 grams of her lemonade?

All the questions up to about 9 are easy, meaning suitable for this forum. But, really, all the questions up to 19 are easy, meaning I can solve them. :D

Hey, how about my problem? :( Could you slove it?

answers to lotrgreengrapes7926's current questions.
[hide][hide][hide]4. If you are really good at your 99 times table, you would see that the cent value is 4.95. Now we add that to the dollar value which is 14. The total cost is therefore 18.95. dividing by 20 would give you 94.75, so you would need to subtract one to get 5.25 percent of 20. So therefore $(A)$ is the answer.

5. Drawing a diagram would be helpful. Denote Bob's distance that he walked $D_B$ and John's distance $D_J$. Denote Bob's time and Bob's speed $T_B$ and $R_B$, respectively. Denote John's time and speed as $T_J$ and $R_J$, respectively. Since the distance is the speed times the time, $D_B=R_BT_B$ and $D_J=R_JT_J$. Now $D_J$ is just 1 mile shorter than $D_B$, so $R_JT_J+1=R_BT_B$. Since the times are equal, the new equasion looks like this: $R_JT_J+1=R_BT_J$. Since we know the rates, the new equasion is this: $3T_J+1=5T_J$. We subtract $3T_J$ from both sides and divide by 2 to get $T_J=30min$. So $30$ is the answer.

6. Adding 400, 100 and 100 gets 600 grams of lemonade. There are $25+836=411$ calories in 600 grams of lemonade.
Dividing by 3, we see that 200 grams of lemonade has 137 calories. So $137$ is the answer.[/hide][/hide][/hide]

I don't think, stancioiu sorin, that that question is allowed, because this is a middle school forum.

And you mean my problem is too difficult or too easy? :?

Sorin, you should probably ask questions in the Highschool-Getting Started forums. Middle school curriculum in Romania usually is a bit stronger than in other countries.

Yes Valentin, I've understood.But I don't think it's a difficult problem.Anywhay... :)

1. A farmer goes out for lunch and dinner. For lunch he pays 15 dollars plus 15% tax. For dinner he pays 25 dollars plus 20% tax. If he does this for one week every day, what is the total cost of all of the meals at the end of the week?

2. What's 10% of 20% of 30% of 40% of 50% of 200?

lotrgreengrapes7926's 1st 3  :)
[hide="1"]0[/hide]
[hide="2"]-72[/hide]
[hide="3"]10[/hide]
you're right those [color=red]are[/color] easy  :D

Is the fifth question's answer 30 minutes?[hide][/hide]

[quote="lotrgreengrapes7926"]If 10% of 20% of a number is 12, what is 20% of 10% of that same number?

This was a chapter MathCounts countdown, and neither person got it right.  :rotfl:[/quote]

Bleh, I actually found the number was 600 in my head before realizing it's 12. -_-

[quote="TRAN THAI HUNG"]Let that number be x.We have:
   (10%).(20%.x)=12
<=>(10/100).(20/100.x)=12
<=>  20/100.10/100.x=12
<=>(20%).(10%.x)=12
Is that right?[/quote]

Yeah.
Or you could just say that multiplication is commutative, so x%$\times$y%=12 and y%$\times$x%=12..

${\frac{\sqrt{x^2-1}}{2}}-\sqrt{x^2-4}=0$
solve for x.

Math is my life


Does anyone here have a MS account? I know Brut3Forc3 and pacman do (we really need to all get into a mini or something together), and I was wondering if anyone else does.
If you don't know what MS is, then you don't have to pay attention to this thread.  :P

fragxkid- Windia
dragon842- Windia
guttaglutta- Windia
mx22- Bera
PM me if u wanna buddy me or join my guild.

MS = MafiaScum

for those who didn't know  :P

and no, I have no MS account.

Sorry, I should have been more clear.

Although I used to play maple story, I don't anymore.

[quote="chenhsi"]How do you not know that I have a MS account?

Although maybe because I almost never leave Mish-Mash.

klebian also has an acount (klebian).[/quote]

Yeah, I forgot about klebian.

I think I might have seen you post once or twice in the queue, but that's it.

I also remember playing pacman on marathon day, and I randomly voted him when he was scum.  :lol:

Hmm, I've seen the accounts AIME15 and Dojo in the newbie thread of MS, so maybe those two also?

I support the getting into a mini idea. :D

Yeah, I have one, but i just made it to check it out a bit, and read a couple games.  Might start playing newb games soon.

I have!
lv. 51 i/l mage on broa name "finrodfeanor"

It really helps if you read a thread before posting.

Agree with mini together. Or we can play a private game somewhere (just the AoPS/MS group).

Guys, come to the Mish-Mash. The alliance of AoPSers that I often form only consist of me, pacman, and randomgem :(

Maplestory... memories... I used to play, 50ish I/L Broa
I stopped playing Thanksgiving last yr (1 yr anniversary?  very coincidental)
I'm guessing there have been like a million changes

[quote="chenhsi"]It really helps if you read a thread before posting.
[/quote]

See that quickster.