modular arithmetic
number theory unsolved
number theory

If it's known that p,p+2,p+2^n,p+2+2^n are prime, find possible integer values of n

"An open question is a question that has no known solution up to this moment, and it is not known wheter the problem has one or not."

Are you joking? it's proposed in our country Kazakhstan MO for 9 grades in 2005
(of course, you are joking, it can be solved!)

Then [b]you[/b] are joking to post it as an "Open problem".

"If it's known that $ p$,$ p\plus{}2$,$ p\plus{}2^n$,$ p\plus{}2\plus{}2^n$ are prime, find possible integer values of $ n$" with $ p\geq 2$ obviously, being $ 1$ not a prime.

We know that $ p\equiv 0,1,2\,\pmod{3}$. If  $ p\equiv 1\,\pmod{3}$  we have that the above four primes are equivalent to $ 1$, $ 0$, $ 1\plus{}(\minus{}1)^n$, $ (\minus{}1)^n$. Being $ p\plus{}2>3$ and $ 3|(p\plus{}2)$ this case doesn't work. If  $ p\equiv 2\,\pmod{3}$  we have that the above four primes are equivalent to $ 2\,$,$ 1\,$,$ \minus{}1\plus{}(\minus{}1)^n\,$,$ 1\plus{}(\minus{}1)^n$. Now, $ n$ is even or odd. Anyway one of the last two primes is divisible by $ 3$. Even this case doesn't work. Thus, $ p\equal{}3$. Here is the new four primes

$ 3$, $ 5$, $ 3\plus{}2^n$, $ 5\plus{}2^n$.

The first two are always prime. What for the last two. It easy to see $ \pmod{3}$ that $ n$ must be odd, $ n\equal{}2k\plus{}1$, thus,

$ 3\plus{}2^n$, $ 5\plus{}2^n$ $ \to$  $ 3\plus{}4^k\cdot 2$, $ 5\plus{}4^k\cdot 2$.

Studying the above two numbers $ \pmod{5}$ we see that $ k$ must be odd, $ k\equal{}2s\plus{}1$. Now, the two numbers become

 $ 3\plus{}16^s\cdot 8$, $ 5\plus{}16^s\cdot 8$. Let see thw numbers $ \pmod{7}$:

1) $ 3\plus{}16^s\cdot 8 \equiv 3\plus{}2^s \pmod{7}$,

2) $ 5\plus{}16^s\cdot 8 \equiv 5\plus{}2^s \pmod{7}$.

It shuold be not difficult to see that to have the two numbers not divisible by $ 7$, $ s$ must be a multiple of $ 3$, $ s\equal{}3i$:

$ s\equal{}0$

1) $ 3\plus{}2^0\equiv 4 \pmod{7}$,

2) $ 5\plus{}2^0\equiv 6 \pmod{7}$,

Ok

$ s\equal{}1$

1) $ 3\plus{}2^1\equiv 5 \pmod{7}$,

2) $ 5\plus{}2^0\equiv 0 \pmod{7}$,

No

$ s\equal{}2$

1) $ 3\plus{}2^2\equiv 0 \pmod{7}$,

2) $ 5\plus{}2^2\equiv 2 \pmod{7}$,

No

$ s\equal{}3$

1) $ 3\plus{}2^3\equiv 4 \pmod{7}$,

2) $ 5\plus{}2^3\equiv 6 \pmod{7}$,

Ok

$ s\equal{}4$

1) $ 3\plus{}2^4\equiv 3\plus{}2^1\equiv 5 \pmod{7}$,

2) $ 5\plus{}2^4\equiv 5\plus{}2^1\equiv 0 \pmod{7}$,

No

and so on.

At this point we have $ n\equal{}2k\plus{}1$, $ k\equal{}2s\plus{}1$ and $ s\equal{}3i$, thus $ n\equal{}2(2\cdot 3i\plus{}1)\plus{}1\equal{} 3(4i\plus{}1)$. The above two numbers become, if $ i\neq 0$

1) $ 3\plus{}2^{3(4i\plus{}1)}\equal{}3\plus{}8^{4i\plus{}1}\equal{}3\plus{}8^{4i}\cdot 8 \equiv 3\plus{}8\equal{}11 \pmod{13}$,

2) $ 5\plus{}2^{3(4i\plus{}1)}\equal{}5\plus{}8^{4i\plus{}1}\equal{}5\plus{}8^{4i}\cdot 8 \equiv 5\plus{}8\equal{}0 \pmod{13}$.

Thus, if $ i\neq 0$ the second numbers is always divisible by $ 13$. The  only solution to the problem is for $ i\equal{}0$, $ n\equal{}3$

$ 3$, $ 5$, $ 11$, $ 13$

Ps. Even $ n\equal{}1$ works, but we have a repetition of primes $ 3$, $ 5$, $ 5$, $ 7$.


modular arithmetic
number theory
least common multiple

Find the smallest three positive integers $ n$ for which $ 3^{n}\equiv 3^{3}\pmod{10800}$.

[quote="s.tringali"]Obviously you need $ n \ge 3$. Then $ 3^{n}\equiv 3^{3}\bmod{10800}$ iff $ 3^{n-3}\equiv 1 \bmod{2^{4}\cdot 5^{2}}$, as $ 10800 = 2^{4}\cdot 3^{3}\cdot 5^{2}$. So it's enough to calculate $ \mu = \mbox{ord}_{2^{4}\cdot 5^{2}}(3)$. Actually $ \mbox{ord}_{2^{4}}(3) = 2^{2}\cdot \mbox{ord}_{2}(3) = 4$ and $ \mbox{ord}_{5^{2}}(3) = 5 \cdot \mbox{ord}_{5}(3) = 20$. So $ \mu = \mbox{lcm}(\mbox{ord}_{2^{4}}(3), \mbox{ord}_{5^{2}}(3)) = 20$, and whence $ n_{1}= 3$, $ n_{2}= 23$ and $ n_{3}= 43$ are the answers to the problem in hand.[/quote]

Grazie! Thanks s.tringali!! I see what I did wrong.  Also, is there some general statement explaining how you calculated $ \mbox{ord}_{2^{4}}(3) = 2^{2}\cdot \mbox{ord}_{2}(3) = 4$ and $ \mbox{ord}_{5^{2}}(3) = 5 \cdot \mbox{ord}_{5}(3) = 20$ (the middle equalities)?


Points $A,B,C,D$ are on a circle with radius $R$.
$|AC|=|AB|=500$, while the ratio between $|DC|, |DA|, |DB|$ is $1,5,7$. Find $R$.

Apply ptolemy to get BC=600.Now since the circle is also the circumcircle of triangle abc.Find the length of the altitude which is 400 cm .Then use the formula of circumradius to get 312.5cm as ur answer


Ebook written by VIF- Vietnam Inequality Forum: http://batdangthuc.net/forum/index.php
Ebook has some mistakes, but we hope it can help you learn inequality better.

[quote="PTNKer_neo260192"]thanks for your ebook . It so cool[/quote]

Thank you.

Inequalitiy box of Mathlinks was so great, but if you are particularly interested in inequalities, I think this forum is not bad. I opened one box entirely in English.

http://www.batdangthuc.net

http://www.mathsforum.net

Thank you very much! Your ebook is very useful  :)


complex analysis
complex analysis theorems

I was wondering if someone could recommend me a good introductory book on modular forms. I have to write a 15 page summary type article for one of my classes and I was thinking of doing it on modular forms. I am familiar with complex analysis and a bit with the theory of Riemann surfaces. I'm particularly looking for applications to number theory. Thanks.

A fairly introductory book is "A First Course In Modular Forms" by Diamond and Shurman. You could probably get it at your library.

Thank you, I was actually recently suggested this book by someone else too =)


[quote="Quevvy"]Two circles and three straight lines lie in the same plane. If neither the circles nor the lines are [u]coincident[/u], what is the maximum possible number of points of intersection?

What does coincident mean?

Is there a formula for this?[/quote]
[hide]
coincident means lying on the exact same set of points. (2 coincident lines means two of the same line)
the maximum number of points for the intersections of the lines is 3. the maximum number of points for each circle with the lines is 6. there are two circles, so the circles make 12. the lines make 3. if the circles intersect at 2 points, that is 2 more. in all, it is 17 points.

in other words:
lines form a triangle, 3 points
lines intersect circles at 2 points per circle, 12 points
circles intersect each other, 2 points
3+12+2=17
[/hide]


The ratio of the length of a rectangle to its width is the same as that of the diagonal to the length. If the width is 2, how many units are in the length of the diagonal?

[quote="LadyKn1ght"][hide]l = length
w = width
d = diagonal
$\frac{l}{w} = \frac{d}{l}$
$\frac{l}{2} = \frac{d}{l}$
$l^{2} = 2d$
Using the Pythagorian Theorem:
$l^{2} + w^{2} = d^{2}$
$l^{2} + 4 = d^{2}$
$2d + 4 = d^{2}$
$-d^{2} + 2d + 4 = 0$
Next I used the quadradic fromula:
$d = \frac{-2 \pm \sqrt{2^{2} - 4(-1)(4)}}{2(-1)} $
$d = \frac{-2 \pm \sqrt{4 + 16}}{-2} $
$d = \frac{-2 \pm \sqrt{20}}{-2} $
$d = \frac{-2 \pm 2\sqrt{5}}{-2}$
$d = \frac{-2}{-2} \pm \frac{2\sqrt{5}}{-2}$
$d = 1 \pm \frac{2\sqrt{5}}{-2}$
$d = 1 \pm -\sqrt{5}$
$d = 1 + (-\sqrt{5}) = 1 - \sqrt{5}$
$d = 1 - (-\sqrt{5}) = 1 + \sqrt{5}$
So the diagonal is either $1 - \sqrt{5}$ or $1 + \sqrt{5}$
[/hide][/quote]

Hi . Just want to remind you that the diagonal must be in positive so $1 - \sqrt{5}$ is unacceptable .

nice solution LadyKn1gt...that must have taken a while.


Find $ f:R-->R $ such that :
 $ f(x+y)=f(x)f(y)f(xy) $

There is a nice ,shorter for this problem :
 $ f[(x+y)+z] =f[(x+(y+z))] $
 Expand two sides , we get $ f = const $

Wow! It's a fantastic solution !!

I'm glad to see that's a nice one!  :D  :D


limit
trigonometry
calculus
calculus computations

[quote="Xevarion"]Let $y = \cos(x)$ and then use L'Hopital's Rule.[/quote]

man, I don't want to complicate everything... :)


After thinking about it a little bit, I have come to a conclusion that locking the forum was useless as well as being unreasonable. However, please do not post any AP problems as someone taking the alternate test(May 19th) may see it and use it to their advantage. If someone were to post this year's AP problem on here before May 19th, I might consider locking the forum in the future. Please keep the rule.

Good luck to all the test takers,

xxreddevilzxx

can we talk about past AP test questions during the exam time???

considering that you can talk about past AP exam questions [i]any[/i] time, i could only imagine you could talk about them during the AP time.  i think the point xxreddevilzxx is trying to make is that we don't want people who are taking the test at a later date to know anything about the test from the people who have already taken it, since they will have the same test.

I didn't know you were allowed to talk about recent AP questions, EVER.  I thought you had to wait for them to release stuff.

Never MC.
Can talk about FR 48 hours post-test.

Yeah, they also release the FR questions after 48 hours on their site as well.

never MC is not true.  they release MC's every 4 or 5 years.  whether FR or MC, the point is that you can never talk about them until they are released.  the difference is just that MC are only released once in a while, while FR are released two days after every exam.


I do not follow that at all, could you please explain?

First equality follows from $\sqrt{n+\sqrt{n^{2}-1}}= \frac{1}{\sqrt{2}}\sqrt{2n+2\sqrt{n^{2}-1}}= \frac{1}{\sqrt{2}}\sqrt{(\sqrt{n-1}+\sqrt{n+1})^{2}}= \frac{1}{\sqrt{2}}(\sqrt{n-1}+\sqrt{n+1})$.

Second equality follows from ${\frac{1}{\sqrt{n+1}+\sqrt{n-1}}= \frac{\sqrt{n+1}-\sqrt{n-1}}{(\sqrt{n+1}+\sqrt{n-1})(\sqrt{n+1}-\sqrt{n-1})}= \frac{1}{2}(\sqrt{n+1}-\sqrt{n-1}})$.

Third equality is just telescoping.

Wow, that factorization must have taken a lot of imagination to come up with...


Hi guys,
What is the ball and urn thing? There was a post concerning it in the marathon but I didn't want to disrupt the flow. Are there any good sites detailing the principles of the ball and urn?
Thanks,

Oh, Thanks,
I searched ball and urn on the wiki but I got no results. I guess I had to search distinguishability.

Er it's usually "balls and urns." In general, the plural form is assumed unless there is something special that would make it...singular.

Yes, visit http://www.ballandurn.com/info and you'll find everything you need.


:rotfl: [b]Two circles have diameters in the ratio 1:3.Find the ratio of their circumferences.[/b]

[hide="Solution"]If their diameters are in the ratio $ 1: 3$, then let the first circle have diameter $ d$, so the second one must have diameter $ 3d$. The first circle has circumference $ \pi d$, and the second one has circumference $ 3\pi d$, so their ratio is $ \boxed{1: 3}$.

[/hide]

[hide]The circumferences are always in the same ratio as the radii/diameters are. Because the radii/diameters are in ratio of 1:3, the circumferences will be in ratio of 1:3 as well[/hide]

[quote="kaveer"]Circumference=diameter x Pi

1 x 3.14=3.14 3 x 3.14=9.42

Answer-- 3.14:9.42[/quote]
$ 3.14: 9.42$ simplifies to $ 1: 3$.  :D

still the same ratio...circumference varies linearly with the radius!

[hide="my solution"]
$ \frac{c_1}{c_2}\equal{}\frac{2\pi r_1}{2\pi r_2}\equal{}\frac{2r_1}{2r_2}\equal{}\frac{1}{3}$[/hide]

$ \frac{c_{1}}{c_{2}}\equal{}\frac{2\pi r_{1}}{2\pi r_{2}}\equal{}\frac{2r_{1}}{2r_{2}}\equal{}\frac{1}{3}$

I believe this is the most accurate way to do this problem...or u can just go ahead and solve the formula...

The easiest way to do this is to think about the question. The formula for finding circumference is $ \pi d$. If we are multiplying both top and bottom by the same thing, [color=red] they cancel out! [/color]. Hence it is simply $ \frac {1}{3}$


Prove that a|a or provide a counterexample, where the notation "m|n" implies that n is divisible by m.

You must also specify $ a\neq0.$ Unless that's the counterexample you're looking for :mabye:

Showing that $ a|a$ is equivalent to showing that $ an \equal{} a$ for some integer $ n$. We have $ a \cdot 1 \equal{} a$, so $ a|a$. (Note that $ 0 \cdot 1 \equal{} 0$, so $ 0|0$. Am I wrong? :maybe: )

Yeah but you start with $ \frac{0k}{0}?$ This obviously leads to $ \frac{0}{0}$ which is undefined.

You [b]don't[/b] start with $ 0\equal{}0k$ Then divide by $ 0$ to get $ \frac{0k}{0}\equal{}1,$ which is not true.

Doesn't $ a|b$ mean that there exists an integer $ n$ such that $ an \equal{} b$?

EDIT: Also, you can't divide by zero; $ 0 \equal{} 0k$ is true.

my bad for not being specific. $ a\neq0.$

ThinkFlow got it.

Hi guys,
what does the little vertical line between two numbers mean? Congruence?

$ a|b$ means "a divides b" or "a is a factor of b".  :)

That depends very heavily on the context. In this context, it translates to the word "divides" (in general -- there is the case "zero divides zero", which is weird).

I like to link people to [url=http://en.wikipedia.org/wiki/Mathematical_symbols]this Wikipedia page[/url] as a reference to mathematical symbols.


ratio
real analysis
real analysis solved

Can we decompose [0,1] into an uncountable family of pairwise disjoint sets, none of which is numerable and everyone dense in [0,1]?

Anybody still interested in this? Sorry I'm reviving such an old topic, but I was looking through some old problems, and I realized I had posted an answer, which turned out to be one of the most stupid things I have ever written :). Fortunately, nobody seems to have read it :).

Consider $r$ to be the equivalence relation making two reals equivalent iff their ratio is rational. The set $[0,1]/r$ has cardinality $c$, so it can be partitioned in $c$ sets of cardinality $c$. Call them $(A_i)_{i\in\mathbb R}$. For each $x$, let $\tilde x$ be its equivalence class according to $r$. Now parition $[0,1]$ as $\bigcup_{i\in\mathbb R}S_i$, where $S_i=\{x\in[0,1]|\tilde x\in A_i\}$.

If there is someone who knows how is to be ignored, than it is me. So, don't feel alone, grobber.  ;)


algebra
polynomial
calculus
integration
advanced fields
advanced fields unsolved

Let $ p$ be a prime and let \[ l_k(x,y)\equal{}a_kx\plus{}b_ky \;(k\equal{}1,2,...,p^2)\ .\] be homogeneous linear polynomials with integral coefficients. Suppose that for every pair $ (\xi,\eta)$ of integers, not both divisible by $ p$, the values $ l_k(\xi,\eta), \;1\leq k\leq p^2 $, represent every residue class $ \textrm{mod} \;p$ exactly $ p$ times. Prove that the set of pairs $ \{(a_k,b_k): 1\leq k \leq p^2 \}$ is identical $ \textrm{mod} \;p$ with the set $ \{(m,n): 0\leq m,n \leq p\minus{}1 \}.$

Assume that the statement does not hold. Then there are numbers $ i \neq j$ such that $ a_i \equiv a_j$ (every congruence is meant  $ \mod{p}$). Consider the number of those triplets $ (k, x, y), (x, y) \neq (0, 0)$ that satisfy
\[ l_k (x, y) \equiv l_i(x, y).\]
By the assumption, for every fixed pair $ (x,  y) \neq (0, 0)$, the number of aolutions in $ k$ is $ p$; thus the total number of these triplets is $ p(p^2 \minus{} 1)$.

Now consider the solution in $ x$, $ y$ for a fix $ k$. If $ k\equal{}i \ \text{or} \ j$, thisis an identity, which means $ 2(p^2 \minus{} 1)$ solutions. For any other $ k$ it is easy to see that the number of solutions is at least $ p\minus{}1$. This means that the total number of solutions is at least
\[ (p^2 \minus{} 2)(p \minus{} 1) \plus{} 2(p^2 \minus{} 1) \equal{} p(p^2 \minus{} 1) \plus{} p(p \minus{} 1) > p(p^2 \minus{} 1),\]
a contradiction.

It is sufficient to prove that for every $ u$ and $ v$ there is exactly one $ k$ with $ a_k \equiv u$, $ b_k \equiv v$.

The uniformity of representations implies that for every $ (\xi, \eta) \not \equiv (0, 0)$, we have
\[ \sum_{k=1}^{p^2} e^{\frac{2 \pi i}{p}(a_k \xi + b_k \eta)} = 0.\]
We multiply both sides by $ e^{(2 \pi i / p) \cdot (u \xi + v \eta)}$ to get
\[ \sum_{k=1}^{p^2} e^{\frac{2 \pi i}{p}((a_k - u) \xi + (b_k - v) \eta)} = 0,\]
for $ (\xi, \eta) \not \equiv (0, 0)$. For $ \xi \equiv \eta \equiv 0$, the same sum obviously gives $ p^2$. Now summing these sums for all possible values of $ \xi$ and $ \eta$, including $ (0, 0)$, we obtain
\[ S = \sum_{\xi = 0}^{p - 1} \sum_{\eta = 0}^{p - 1} \sum_{k = 1}^{p^2} e^{\frac{2 \pi i}{p}((a_k - u) \xi + (b_k - v) \eta)} = p^2.\]

Changing the order of the summations, we get
\begin{eqnarray*}
S
& = & \sum_{k = 1}^{p^2} \sum_{\xi = 0}^{p - 1} \sum_{\eta = 0}^{p - 1} e^{\frac{2 \pi i}{p}((a_k - u) \xi + (b_k - v) \eta)} \\
& = & \sum_{k = 1}^{p^2} \left( \left( \sum_{\xi = 0}^{p-1} e^{\frac{2 \pi i}{p}(a_k - u) \xi} \right) \left( \sum_{\eta = 0}^{p-1} e^{\frac{2 \pi i}{p}(b_k - v) \eta} \right) \right) \\
& = & p^2.
\end{eqnarray*}
Now, a typical factor in this product vanishes, unless $ a_k \equiv u$ (for the first) or $ b_k \equiv v$ (for the second). Thus, the whole product is $ 0$, unless $ a_k \equiv u$ and $ b_k \equiv v$, in which case it is equal to $ p^2$. Since the sum of these products is $ p^2$, this case must happen for exactly one value of $ k$.


Prove that
$ sinat \equal{} 2\frac {sin\pi a}{\pi}(\frac {sint}{1 \minus{} a^2} \minus{} 2\frac {sin2t}{2^2 \minus{} a^2} \plus{} 3\frac {sin3t}{3^2 \minus{} a^2} \minus{} ...)$

On what interval is this supposed to be true?

Any Fourier series problem or claim is incompletely stated if you do not specify the interval. My most common issue with people posting Fourier series problems or claims is exactly this.

$ a$ is not an integer and $ \minus{}\pi <t<\pi$


spam, spam, and more spam....
past three posts are short and unnecessary (except the first one, because you can't make a topic without that post)
anyway, i think that the important thing in creating a tourney is finishing it
it's pointless if you don't
and if the players aren't playing, then just use the coin method =)

If you look down a little, you will see two threads:

"TOURNEY MANIA!"

"tourneys (meh.)"

These two threads have the same discussion. (only they might be dead)

And vallon, you are such a COUGH COUGH [size=59][color=white]hypocrite![/color][/size] COUGH COUGH!


The plane is divided into regions by a finite number of straight lines. Two regions are "neighbours" if they have a common border which is a line-segment, a half-line or a line. Two regions which meet only at a point are not neighbours. No three lines meet at a point. One must try to assign a real number to each region so that

(a) the product of the numbers assigned to neighbouring regions is less than their sum and

(b) the sum of all the numbers corresponding to regions on a given side of one of the given lines is 0.

Show that this is possible if and only if it is not the case that all the lines are parallel.

[quote="AndrewTom"](b) the sum of all the numbers corresponding to regions on a given side of one of the given lines is 0.[/quote]

Must we understand "a given side of one of the given lines" or "any side of any line" ?

In the first case, this is wrong, since we can find parallel lines with a suitable mapping :
Choose lines $ x\equal{}0$ and $ x\equal{}1$
Choose as given side of given line : $ x<0$
Set $ 0$ for region $ x<0$
Set $ 1$ for region $ 0<x<1$
Set $ 2$ for region $ 1<x$

Thanks for your reply pco. Perhaps the latter is meant; I'm not sure.

I can confirm that the sum on both sides of all lines must be 0.


can any one solve these
I tried too much to understand them but no way:

1- what values of a makes the system have no solution? one solution? many solutions?

x+2y-3z=4
3x-y+5z=2
4x+y+(a2-14)z=a+2

2- find all unit vectors in the plane determined by u=(3,0,1) , v =(1,-1,1)that are prependicular to the vector w=(1,2,0)

3- express the matrix
0 1 7 8
1 3 3 8
-2 -5 1 -8

in the form A = EFGR where E,F,G are elementary matrices and R is in Row-echelon form

4- show that the line
x=0 , y=t , z=t

a. lies in the plane 6x+4y-4z=0
b. is parallel to and below the plane 5x-3y+3z=1
c. is parallel to and above the plane 6x+2y-2z=3

please I have an exam and tried to solve these problems as I'm studying but no way

[quote="JBL"]And what have you tried?  These are all pretty standard-type linear algebra exercises, and you can probably find the methods explained in your textbook.  For example, in the first question: what is the general rule about whether a system of $ n$ linear equations in $ n$ variables has a unique solution?[/quote]

I know they are simple but my book doesn't explain these things like the example you mentioned


[hide]Call x the circumference of O and y the circumference of P. Call r the radius of O. Since 60 degrees is 1/6 of 360, and 45 is 1/8 of 360, we have

1/6(x)=1/8(y)

x=3/4(y)

Since the circumference of O is 3/4 the circumference of P, when both sides are divided by 2pi the radius of O is still 3/4 the radius of P. Since the area of O is r^2*pi, you have the ratio

r^2*pi : (r4/3)^2*pi

r^2*pi : r^2*16/9*pi

1 : 16/9

9 : 16

The ratio of the area of O to the area of P is 9/16.[/hide]

[hide]
Arc length is give by: $s = rA$, where $r$ is the radius and $A$ is the angle in radians.

In circle O with radius r, arc length is: $s_o = r\left(\frac{\pi}{3}\right)$

In circle O with radius R, arc length is:$s_p = R\left(\frac{\pi}{4}\right)$

Since the arc lengths are equal: $r\left(\frac{\pi}{3}\right)  = R\left(\frac{\pi}{4}\right)\;\;\Rightarrow\;\;r = \frac{3}{4}R$

Hence, the areas are: $A_o = \pi r^2 = \pi\left(\frac{3}{4}R\right)^2 = \frac{9}{16}\pi R^2$ . . . and: $A_p = \pi R^2$

Therefore, the raio of the area is: $\displaystyle{\frac{\pi\frac{9}{16}R^2}{\pi R^2} \;= \;\frac{9}{16}.}$[/hide]

A 60 degree arc on circle O has the same length as a 45 degree arc of circle P. What is the ratio of the area of circle O to the area of circle P?

[hide]$\displaystyle\frac{60}{360}=\displaystyle\frac{x}{2\pi r_o}$

$x=\displaystyle\frac{\pi r_o}{3}$

$\displaystyle\frac{45}{360}=\displaystyle\frac{x}{2\pi r_p}$

$x=\displaystyle\frac{\pi r_p}{4}$

$\displaystyle\frac{\pi r_p}{4}=\displaystyle\frac{\pi r_o}{3}$

$\displaystyle\frac{r_o}{r_p}=\displaystyle\frac{3}{4}$

$(\displaystyle\frac{r_o}{r_p})^2=\displaystyle\frac{9}{16}$

$ratio: \displaystyle\frac{9}{16}$[/hide]

for our purposes here, I am going to forget aobut pi.... it just cancels out in all cases.... (i know it is a bad habit, but I don't like having [hide]expression like: 2/3pi. sorry for incovenience)
so our sixty degree arc has the same lenght as a 45 degree arc or another circle:
let r_o be the radius of circle o and r_p be radius of circle p
1/6 *2r_o = 1/8 * 2r_p
so r_o = 3/4 r_p
so r_o^2 = 9/16 r_p^2
so the ratio of our areas is: r_o^2/ r_p^2 = [b]9/16[/b][/hide]

[hide]The ratio is 9/16ths.  Although one arc had a 45 degree angle and the other one had a 60 degree angle, their areas equaled.  We can represent this x.  Since 60 is 1/6th of 360, x is one sixth the cirumference in circle O.  Since 45 is 1/8th of 360 x is one eigth the circumference in circle P.  So circle O's cirucmference is 6x and P's is 8x.  We know that C=2r(pi).  So we set up the equations.  2r(pi)=6x and 2r(pi)=8x.  (If I knew LATEX, I would designate them as r1 and r2, but I don't know how.)  THerefore, the O's r=3x/pi and P's r=4x/pi.  Since we want to find the area, we put into the circle area formula pi(r^2).  So we get 9x/pi^2*pi:16x/pi^2*pi which simplifies to 9:16.[/hide]

[quote="rrusczyk"]A 60 degree arc on circle O has the same length as a 45 degree arc of circle P.  What is the ratio of the area of circle O to the area of circle P?[/quote]
[hide="solution"]
[color=darkred]
This means that circle O is smaller than circle P - namely, circle O's diameter is 3/4ths of size of circle P's. That means that the size ratio is (3/4)2, or 9/16.
[/color]
[/hide]


Hello everyone,

I am trying to solve this problem involving hemispherical domes.  Basically I have to find the surface area of the dome with a certain thickness.  I am really unsure how to attack this problem.  I tried to do at other online sources but there was nothing to really help me.  If anyone could offer some tips that would be great!

Thanks!

I hope I didn't misunderstand you. I guess you want to calculate 2 hemispheres with different radius.

Sorry if I misunderstood the problem:

 If you want the total surface area then you need both the inner and outer radii, $r_1$ and $r_2$, then we have two semicircles, and a circular strip, so the total surface area would be $2\pi (r_1^2+r_2^2) + \pi r_2^2 - \pi r_1^2$.

If you want just the outer surface area, then it's simple $2\pi r_2 ^2$.


Anybody play? I've recently started playing again after about a year hiatus. A good AI for beginners is included in the program Qianhong which is free. I'm trying to find a place to play against other people online now.

:)

Hehe

I stopped for about a year also and still havent started playing, but I used to play at http://www.clubxiangqi.com ...they advertise a lot......

I play.

Wait, are you talking about chinese checkers or chinese chess, (the one with pao).

chess is the one with the pao...and his source has a pao...so yeah

I played a lot when I was little but I got creamed when I played my friend's dad(I lost all my pieces while he only lost 4 or 5)


Vieta
algebra
polynomial
algebra unsolved

Let $ a$ be the sum and $ b$ the product of the real roots of the equation $ x^4\minus{}x^3\minus{}1\equal{}0$. Prove that $ b<\minus{}\frac{11}{10}$ and $ a>\frac{6}{11}$.

I solved this prob by applying Vieta's formula:
We have:a=x1+x2+x3+x4=1>6/11(obvious!)
             b=x1.x2.x3.x4=-1>-11/10 :(
PS:With the formula's equation which has degree 4, we can easily find out all of the real roots of this equation! And after that, we just need to calculate the sum and the product to get the result! :)

[quote="Rust"] Then
$ c = \frac {d^2}{d^2 + 1}, a = \frac {1}{d^2 + 1},b = - \frac {1}{d} = - \frac {d^2}{(d^2 + 1)^2} - d$
Equation ${ \frac {1}d} - d = \frac {d^2}{(d^2 + 1)^2}$ had only one real solution $ \frac 89 < d < 0.9$, therefore
$ \frac {100}{181} < a < \frac {81}{155}, \ \frac { - 9}{8} < b < \frac { - 10}{9}.$[/quote]

Can you specify how all these come?

$ x^4\minus{}x^3\minus{}1\equal{}0$ had 2 real roots $ x_1,x_2$ and 2 complex roots. Let $ a\equal{}x_1\plus{}x_2,b\equal{}x_1x_2,c\equal{}z_1\plus{}z_2,d\equal{}z_1z_2$, then
$ x^4\minus{}x^3\minus{}1\equal{}(x^2\minus{}ax\plus{}b)(x^2\minus{}cx\plus{}d)\equal{}x^4\minus{}(a\plus{}c)x^3\plus{}(d\plus{}ac\plus{}b)x^2\minus{}(bc\plus{}ad)x\plus{}bd$.
It give $ a\plus{}c\equal{}1,d\plus{}ac\plus{}b\equal{}0,bc\plus{}ad\equal{}0.bd\equal{}\minus{}1$...

I understand there. And so how come $ c \equal{} \frac {d^2}{d^2 \plus{} 1}, a \equal{} \frac {1}{d^2 \plus{} 1},b \equal{} \minus{} \frac {1}{d} \equal{} \minus{} \frac {d^2}{(d^2 \plus{} 1)^2} \minus{} d$?

1)$ a \plus{} c \equal{} 1$,
2)$ ac \plus{} b \plus{} d \equal{} 0$,
3)$ bc \plus{} ad \equal{} 0$,
4)$ bd \equal{} \minus{} 1\to b \equal{} \minus{} \frac 1d$
From 3) $ bdc \plus{} ad^2 \equal{} 0\to (bd \equal{} \minus{} 1) \ c \equal{} \frac {ad^2}{ \minus{} bd} \equal{} ad^2$.
From 1) $ a \plus{} ad^2 \equal{} 1\to a \equal{} \frac {1}{d^2 \plus{} 1}$, therefore $ c \equal{} ad^2 \equal{} \frac{d^2}{d^2 \plus{} 1}$.

[quote="Rust"]
Equation ${ \frac {1}d} - d = \frac {d^2}{(d^2 + 1)^2}$ had only one real solution $ \frac 89 < d < 0.9$,[/quote]

Oh, ok then. How do you know this?

Exactly $ d^3\plus{}(d^2\plus{}1)(d^4\minus{}1)\equal{}0$ had only one real positive solution. If d<0, then $ c^2\minus{}4d>0$, therefore $ x^2\minus{}cx\plus{}d\equal{}0$ had real solution (we choice complex solutions for $ x^2\minus{}cx\plus{}d\equal{}0$).

[quote="ghjk"]Why applying Vieta in this case doesn't work,Rust?[/quote]
Vieta theorem work, when we find sum or product of all roots. In these case we must find sum and product only real roots.


modular arithmetic
number theory proposed
number theory

hi,
prove that the equation
\[ x^{2} \plus{} y^{2} \plus{} z^{2} \equal{} 59^{n},\]
hat integer solutions for every positive interger n.

It is enough to prove for $ n \equal{} 1,2$, because if $ (x,y,z)$ is a solution for some $ n$ then $ (59x,59y,59z)$ is a solution for $ n \plus{} 2$.

For n=1: $ (7,3,1)$
For n=2: $ (50,30,9)$ (I just chose $ x \equal{} 50$ and hoped that $ 59^2 \minus{} 50^2$ will be representable as a sum of squares, and it is, because $ 59^2 \minus{} 50^2 \equal{} 3^2 \cdot 109, 109 \equiv 1 \pmod 4$ and $ 109$ is a prime)

Alternatively, we can use the criterion for representation as a sum of 3 squares : a positive integer can be expressed as the sum of three squares if and only if it is not of the form $ 4^k(8m \plus{} 7)$. Notice that $ 59^n \equiv 3^n \pmod 8, 3^2 \equiv 1 \pmod 8, 3^1 \equiv 3 \pmod 8$.


function
complex analysis
complex analysis unsolved

Let f(z) be a analytic function in a region G and f(z)= g(x)+i h(y) for all z=x+iy in G. Then show that f(z)=az+b for some a, b in C

The Cauchy Riemann equations require $ g'(x) \equal{} h'(y) \equal{} c$ for all $ x,y$ and the conclusion follows.


What if the solutions are in the form of a plain text like word document?Will they be accepted?

Yes they will be. However I [b]strongly[/b] discourage you to do that.


Andy, Bert, and Charles have $ 20$ pieces of chocolate. Some of the chocolates have nuts, some are plain, and some are filled with caramel (they are split in a ratio of $ 1: 2: 2$, respectively). Neither Andy nor Charles can eat nuts, and, if given a choice, Charles will take caramels over plain chocolates. If Andy, Bert, and Charles split the chocolates among themselves in a ratio of $ 3: 1: 1$, how many pieces of chocolate with caramel does Andy receive?

[hide="Solution"]Call the number of chocolates that have nuts in them $ x$.  We know that the number that are plain is $ 2x$, and same with the ones that are filled with caramel.  Since they all add up to $ 20$, we can set up the equation:
\[ x \plus{} 2x \plus{} 2x \equal{} 5x \equal{} 20\implies x \equal{} 4\]
So they have $ 4$ chocolates with nuts in them, $ 8$ that are plain, and $ 8$ that are filled with caramel.  Now call the number of pieces of chocolate Charles gets $ y$.  Bert also gets $ y$ chocolates, and Andy gets $ 3y$ chocolates.  We can now set up an equation similar to the first one:
\[ 3y \plus{} y \plus{} y \equal{} 5y \equal{} 20\implies y \equal{} 4\]
So Charles has $ 4$ chocolates, Bert has $ 4$ chocolates, and Andy has $ 12$ chocolates.
Since neither Andy nor Charles can eat nuts, Bert gets all $ 4$ of the ones that have nuts.  Since Charles would rather take caramel over plain, he gets $ 4$ caramel chocolates.  This leaves Andy with the rest, and $ \boxed{4}$ chocolates with caramel.[/hide]


function
LaTeX
absolute value
arithmetic sequence

1) Find the sum of the absolute value of the roots over C that satisfy the following equation if $ f(z)\equal{}z^2\minus{}1: f(f(z))\equal{}3$

2) The function $ f$ is defined by the following: $ f(n\plus{}1)\equal{}\frac{8f(n)\plus{}5}{8}$ for $ n\equal{}1,2,3,....$ If $ f(1)\equal{}4$ and $ f(b)\equal{}234$, find $ b$.

$ (z^2\minus{}1)^2\minus{}1\equal{}3\equal{}>(z^2\minus{}1)^2\equal{}4\equal{}>z^2\minus{}1\equal{}2 or z^2\minus{}1\equal{}\minus{}2\equal{}>z^2\equal{}3 or z^2\equal{}\minus{}1\equal{}>z\equal{}\sqrt3  or z\equal{}\minus{}\sqrt3  or z\equal{}i  or z\equal{}\minus{}i$
So $ Abs[\sqrt3]\plus{}Abs[\minus{}\sqrt3]\plus{}Abs[i]\plus{}Abs[\minus{}i]\equal{}\sqrt3\plus{}\sqrt3\plus{}1\plus{}1\equal{}2\sqrt3\plus{}2$

For the second one:
$ f(n\plus{}1)\equal{}(8f(n)\plus{}5)/8\equal{}>f(n\plus{}1)\equal{}f(n)\plus{}5/8$ So
f(n)=f(n-1)+5/8
f(n-1)=f(n-2)+5/8
..
f(2)=f(1)+5/8

Adding these up and cancelling we get that f(n)=f(1)+(n-1)*5/8 (which we could have derived noting that f is an arithmetic sequence)
So f(b)=f(1)+(b-1)*5/8=>234=4+(b-1)*5/8=>b=369

[quote="salonikios"]$ (z^2 \minus{} 1)^2 \minus{} 1 \equal{} 3 \equal{} > (z^2 \minus{} 1)^2 \equal{} 4 \equal{} > z^2 \minus{} 1 \equal{} 2 or z^2 \minus{} 1 \equal{} \minus{} 2 \equal{} > z^2 \equal{} 3 or z^2 \equal{} \minus{} 1 \equal{} > z \equal{} \sqrt3 or z \equal{} \minus{} \sqrt3 or z \equal{} i or z \equal{} \minus{} i$
So $ Abs[\sqrt3] \plus{} Abs[ \minus{} \sqrt3] \plus{} Abs[i] \plus{} Abs[ \minus{} i] \equal{} \sqrt3 \plus{} \sqrt3 \plus{} 1 \plus{} 1 \equal{} 2\sqrt3 \plus{} 2$

For the second one:
$ f(n \plus{} 1) \equal{} \frac {8f(n) \plus{} 5}{8} \equal{} > f(n \plus{} 1) \equal{} f(n) \plus{} \frac {5}{8}$ So
$ f(n) \equal{} f(n \minus{} 1) \plus{} \frac {5}{8}$
$ f(n \minus{} 1) \equal{} f(n \minus{} 2) \plus{} \frac {5}{8}$
$ ...$
$ f(2) \equal{} f(1) \plus{} \frac {5}{8}$

Adding these up and cancelling we get that $ f(n) \equal{} f(1) \plus{} (n \minus{} 1)\cdot \frac {5}{8}$ (which we could have derived noting that $ f$ is an arithmetic sequence)
So $ f(b) \equal{} f(1) \plus{} (b \minus{} 1)\cdot\frac {5}{8} \equal{} > 234 \equal{} 4 \plus{} (b \minus{} 1)\cdot\frac {5}{8} \equal{} > b \equal{} 369$[/quote]

Turned into latex for u :D nice solution


let $ a,b,c$ be pssitive integer number, prove that it $ \frac{1}{ab}\plus{}\frac{1}{bc}\plus{}\frac{1}{ca} \ge 3$ then  $ ab\plus{}bc\plus{}ca \le 3$   :!:

If $ a,b,c$ are positive ntegers, then $ \frac1{ab}\plus{}\frac1{bc}\plus{}\frac1{ca}\geqslant3\ \Rightarrow\ ab\equal{}bc\equal{}ca\equal{}1.$

[quote="ValentineA"]If $ a,b,c$ are positive ntegers, then $ \frac1{ab} \plus{} \frac1{bc} \plus{} \frac1{ca}\geqslant3\ \Rightarrow\ ab \equal{} bc \equal{} ca \equal{} 1.$[/quote] That's iff $ abc\equal{}1$ so your solution is wrong.

If $ m,\,n,\,p$ are positive integers, how would $ \frac1m \plus{} \frac1n \plus{} \frac1p\geqslant3$ be possible except when $ m \equal{} n \equal{} p \equal{} 1\,?$ :|

PS: The problem is trivial when $ a,\,b,\,c$ are positive integers, and false when $ a,\,b,\,c$ are positive real numbers (take $ a \equal{} b \equal{} 2,\,c \equal{} \frac13).$ Maybe you want to prove that if $ \frac1{ab} \plus{} \frac1{bc} \plus{} \frac1{ca}\,\color{red}\leq\color{black}3\,$ then $ ab \plus{} bc \plus{} ca\,\color{red}\geq\color{black}\,3$ instead? That would follow from the inequality of harmonic and arithmetic means. ;)

it's a simple inequality , just we have $ ab\plus{}bc\plus{}ca \leq 3 \implies \frac{9}{ab\plus{}bc\plus{}ca} \ge 3$

as we know $ 3 \leq \frac{9}{ab\plus{}bc\plus{}ca} \leq \frac{1}{ab}\plus{}\frac{1}{bc}\plus{}\frac{1}{ca}$

and we are done ! :)

@ValentineA : Yes, i thought so, then we can easily finds contradiction using AM-HM Ineq..


"Astronomical Observation from moon surface vis-à-vis low earth orbiting Observatory" - its a topic in an essay contest.

i would like to know your views on:

how does astronomical observation improve by putting up observatories on the moon?
or is it a better idea than the Hubble space telescope ,chandra X-ray observatory and in what way?(taking into account economic considerations and advantage of astronomical observations.)

for example
i think that the craters like tycho,grimaldi can be used for setting up radio observatories which [b]need to be really large[/b] to obtain good resolution and we cannot possibly set up [b]good orbiting radio telescopes.[/b]

thanks.

Maybe...

It will improve telescope pictures because there is no atmosphere for light to bend.

ya you are right.
but the same also holds for earth orbiting observatories like HST.

so what is the advantage of building observatories on moon?

Hello,
    Nice topic. But I don't think the organizers of the International Astronomy Congress would be too happy to know that people resort to online discussions. Atleast, one must acknowledge any information gathered from such sources in a proper referencing manner.
Sathej

Also you could place a orbital telescope that orbits around the moon so we can see what the other side of the moon. (For purposes I am not going into, we can only see $ 54\%$ of the moon at once.)

@sathej

first of all,its international astronautical congress and not 'astronomy'. :lol:

you would not expect somebody to whiff out a 2000 words essay from thin air.
they realise this and have stated that you can use the internet for gathering facts about this topic.

i have posted here so that if sb has some idea or good link to provide then it would be helpful.

Thank you for the clarification of IAC 2007, Soumyashant. Anyway, I am not participating in it! And I just said it is good to acknowledge all sources, even if they be as minor as someone providing some link. And more so, when one is using the Internet. I have also seen several posting the actual questions of the First Level on Yahoo Answers. This may not be in good taste.
Sathej

Well, the main thing about building a telescope on the moon is that it is free of atmospheric effects. Now it is true that this can also be acieved by a normal space telescope, but the biggest problem is that it is EXTREMELY expensive to get big scopes into space.

as such they are developing methods to "make" a telescope in space. The one thing that they are considering (and Tested with sucess on earth!) is spinning a container of Hg to make a parabolic lens, however this can only be done in a area under the influence of "strong (i.e. not drif in space)" gravity . as such they are considering the surface of the moon as this spinning mirror cannot be acieved in a space enviroment

I can think of a few advantages for a lunar telescope:
1. it does not suffer from gradual lowering of the orbit that Earth-orbiting devices have to face.
2. it does not worry about potential damages from the space junks.
3. it can be located at the far-side of the Moon so that it has no interference from Earth.
4. it may be easier to maintain if there are permanent crews living on the Moon base.
5. and yes, it's easier to setup large scale telescopes once we know how to utilize the local material on the Moon.

thanks to all..

i qualified in the essay contest and did a presentation on it yesterday only.


AMC USA(J)MO USAMO AIME search inequalities geometry

Does anyone have any advice on how (to learn) to write clear, concise, and rigorous proofs, and/or any advice on preparing for the USAMO in general?

Preparing: do old contests (be very, very patient - spend the entire 4.5 hours at least once just to develop patience). If you can't make any progress with them, just read Engel (Problem Solving Strategies). Actually read that even if you can do them (that's how I managed to go from averaging 0 problems per USAMO at the AIME to one or two when I took it last year). If that's too hard, sorry but you probably aren't going to do too well this year :( . Prepare for next time.

There is a misconception that there's something tricky and mysterious about proofs. Just pretend that you're explaining it to a friend, but keep in mind that the friend can't ask you questions afterwards (so be clear and rigorous, but skip arithmetic and obvious stuff). If you are really worried, ask someone with more experience for advice. You can even post your proofs on these forums.

[quote="jb05"] If you can't make any progress with them, just read Engel (Problem Solving Strategies). [/quote]

Is this a book?  If not, is it available on the web?

Yup, it's a book. I believe it is in the Bookstore on this site.

to clarify, it is a pale yellow book (sticky-note color) with purplish text on the front, in case you want to make sure.

And it's filled with lots of cool math.  :P
Not that I'm up to that level yet.

about how long does it take to get through engel??

You definitely won't get through it in a jiffy, from skimming through its contents. Even though I haven't done any of it yet.
Okay, I'll shut up about things I don't know about now.

Hmm I have Engel... it's definitely NOT something you can read in a couple of months. As a matter of fact, I'm betting that even the best mathematicians in the nation could get something out of it by reading it even if they've read it before. It will take you at least a year to read, but I remember a friend who said that he sort of "skimmed" over a few chapters in the book and made red MOP. If you are as much of a genius as he is, go for it! If not, there's always next year  :)

But Engel definitely taught me a few techniques on how to tackle Olympiad problems. However, the difficulty is quite high. If you want something a bit easier to read (you can get the first chapter of Engel on Amazon's "Search Inside!" feature), you could try Larson's "Problem Solving Through Problems"

A math book is not a novel.  I don't see the point of purposely trying to do every single problem in a worthwhile book on problem solving.  You can read through the lessons in very little time, and it is probably worthwhile to do so.  To do the more than one thousand problems would obviously take a very long time.  If you can do a problem in five minutes, then it is far too easy for you to be doing that many.  So I don't think you should be trying to "get through Engel".  Instead, try working on some problems from it, and enjoy them, instead of seeing them as so many hurdles to jump.

Anyhow, Engel has a somewhat different flavor from the USAMO.  You'll notice it most in the inequalities and the geometry, I think.  But it's still good practice.

Engel is definitely not meant to be read straigth through. I would skip around and try to learn various techniques. Then attempt a few of the olympiad problems and try to apply the techniques but don't try to do every problem; it's not necessary or efficient.

Don't think you can walk away with a mastery of the material without doing a good number of problems, though. Also, just going straight from the problems to the solutions is not a good idea.

how many problems per chapter should we do?(percentage wise or something)

@phelpedo: nice, 2007 post count(at the time of my post)

Don't get me wrong, you definitely need to do a good deal of problems to master the technique. However, this will take some time and it's almost impossible to do before this year's USAMO. I would say that if you work on 15 olympiad level problems from each chapter in addition to the exercises presented within the context, you should obtain a pretty decent understand of the material. Of course it's always a good idea to review with more problems. The only way to really practice for USAMO is by doing lots and lots of problems (at least this is my opinion).

As mentioned above, USAMTS problems are generally not nearly as difficult as USAMO problems.  On the other hand, #1 and #4 on most USAMOs are often not any more difficult that hard USAMTS problems.  In fact, rrusczyk and I were talking last week about how this year's USAMO #1 would have made a great USAMTS problem.

The other thing about the USAMTS is that it will give you experience writing [b]full[/b] solutions to problems, as opposed to the AIME, where your solutions can have holes but you can still get the right answer.  We see a lot of papers for the USAMTS where a student has the right answer at the end and think they deserve a 5 (a perfect score), and we only award a 1 or 2 or 3 because the solutions has major gaps.  This is even more significant on the USAMO, since a solution with a major gaps or errors is likely to only receive 0 or 1 or 2 (out of 7).

[quote="darkprince"]Would it be better to learn proof-writing first, or to learn to solve more difficult problems like those on the AIME if I'm preparing for USAMO? I should say that I've had limited experience with proofs, having done only USAMTS so far...But will that be bad in the long term?
darkprince[/quote]

I think it would be much more beneficial being able to solve AIME problems before learning to write proofs.  I would focus extensively on the AIME, without completely ignoring proofs.  Once you think you are capable of solving #1 and #4 type problems on a regular basis, it may be time to move on to more extensive proof practice.  Books by authors such as Engel and Larson are good, since they teach problem solving techniques through proofs.  Another problem with putting USAMO practice before AIME practice is that you need to do well enough on the AIME to even qualify for USAMO.  This seems to be of secondary importance, but I can cite several examples of USAMO qualifiers who decided to focus extensively on USAMO and then didn't qualify for it the next year.

Engel and Larson usually only show the investigation leading up to the solution. Their solutions should not be a model for formal proofs. Look at the AMC website's USAMO solutions for formal proofs (although their #1 is sketchy, eh).

If you're having a hard time with USAMTS problems, there would be little point in tackling USAMO problems at this point. I wouldn't worry too much about not being prepared enough for USAMO - problem solving is problem solving. Although there are many techniques unique to olympiad problems, it's not like AIME problem solving ability and Olympiad problem solving ability are mutually exclusive.

[quote]If you're having a hard time with USAMTS problems, there would be little point in tackling USAMO problems at this point. I wouldn't worry too much about not being prepared enough for USAMO - problem solving is problem solving.[/quote]

Phelpedo, do you suggest becoming good at general problem solving (so that I'll be able to solve AIME and USAMTS relatively easily) before moving to Olympiad problem solving, and learn proof writing along the way?

That sounds reasonable. Until I can solve the Olympiad problems, I won't try to write clear, concise solutions to those problems.  :lol:

Hopefully, that'll change soon, because this is my last shot at USAMO.

Also, Engel's book helps in covering topics that I might not have had prior knowledge about, right? If so, then I definitely need it, not so much for proof writing, but for concepts that I might not have had full exposure to.

mysmartmouth, thank you. I'll try for scores of 12-15 on AIME before proof work; that gives me a definite plan to work with, since that's closer to where I am right now (i can consistently get around 5 or 6).

Just as a side note, would excelling in AIME problem solving help with USAMTS?
:D

darkprince

Most of the solutions I have read in Engel were complete.  I have not seen Larson.  On the other hand, the best models of olympiad solutions that I have seen are in the USAMO/TST/IMO pamphlets that the MAA publishes.  The solutions are better than those on the website, and there are also more alternate solutions.  By the way, what is sketchy about the solutions to problem 1 on the website?  I read three good solutions.

[quote]Phelpedo, do you suggest becoming good at general problem solving (so that I'll be able to solve AIME and USAMTS relatively easily) before moving to Olympiad problem solving, and learn proof writing along the way? [/quote]
I don't understand—olympiad problem solving is a particular difficulty (and perhaps topic) range of general problem solving.  It is good to spend some time on AIMEs if you haven't.  However, if you usually can score a five to an eight on AIMEs, I see no reason not to start easy olympiad problems.  Twelve to fifteen on the AIME is significantly higher than I have yet scored, and I was at red MOP last year.  Trying to pass that mark before starting olympiad problems is overkill—a parallel of trying to learn to multiply a hundred three-digit number in your head in three minutes before studying basic algebra.  Some of last year's red MOPpers got eights on the AIME, but they were still capable of tackling easy olympiad problems.

I happen to disagree with mysmartmouth about proofwriting skills.  If you have solved a problem and you can't communicate your solution effectively to other mathematicians, you might as well not have solved it.  In addition, if you put proofwriting off until you are solving olympiad problems, you will probably impose some psychology on yourself that associates proofs with difficult problems.  In any case, there is no magic trick to it, no secret formula—just write down a sequence of logical thoughts that lead you to the desired conclusion.  You don't have to be pedantic.  If you can't do this, you probably haven't really solved the problem.  As you do this more often, your form and style will improve.

Also, I did practically no practice for the AIME this year and concentrated exclusively on USAMO, and I still made USAMO by a wide margin even after three stupid mistakes on the AIME.  This does not demonstrate that in general, exclusive USAMO practice will probably qualify you for USAMO, but then, a few examples in the other direction will not demonstrate the opposite.  All I know is that for me, olympiad problems make AIME practice sufficiently obsolete.  Furthermore, olympiad problems are deeper, and a better use of your mind.  When you have spent enough time on olympiad problems, AIME problems seem to have negligible difficulty.

Proofwriting is really a fairly simple thing. However, many people claim to be bad at proofwriting when really the issue is that they haven't actually solved the problem rigorously. For example, I've heard a lot of people (including myself) attribute a less-than-7 score on the USAMO'06 #2 to bad proofwriting (probably something of the nature that you thought the numbers should be consecutive but couldn't prove it), when really it's just not solving the problem completely.

Of course there are tricky things like saying "all steps reversible" at the end of inequality solutions (if you write it backwards).

This year's #2 and #4 showed especially well that one of the biggest problems people have with proof writing, even some veterans, is making assumptions that aren't always true. For #2, it was things like all disks equal radius, or a set of 3 disks is going to be tangent. For #4, most people's proofs of 8025 as a limit failed consider all possible 8026 cell dinosaurs.

When you prove something, especially that some number is a maximum, make sure you leave no possibility out. In the same vein, don't make any large intuitive leaps. That's how you figure the solution out, not write the proof.

I disagree that their #1 is sketchy. They were concise, but it was a complete proof(s). Just an easy problem.

[quote]Also, I did practically no practice for the AIME this year and concentrated exclusively on USAMO, and I still made USAMO by a wide margin even after three stupid mistakes on the AIME. This does not demonstrate that in general, exclusive USAMO practice will probably qualify you for USAMO, but then, a few examples in the other direction will not demonstrate the opposite.
Boy Soprano II [/quote]

I see...I can get around 6 or 7 in AIME. I should prepare a little more, then begin easy Olympiad problems? I guess that works, but are easy Olympiad problems approximately the difficulty of mid-level AIME problems? I've looked at some at Kalva.net, but they seem to be far more intricate and difficult to end up in the right place, and with all the right steps, too. I don't know, it might be too much of a leap for me. I'll try though. Thanks!

darkprince

You probably want to get a little more comfortable with AIME problems. I think once you think you "know" how to solve 10 that's good enough and will provide you with a solid buffer in case you make stupid mistakes.

For easy olympiad problems, I recommend checking out Canada's or India's NMO's since the US ones are usually too difficult for people just starting olympiad problems. In recent years, however, they have tried to accomodate for such people I think.

[quote="Phelpedo"]As I said, I don't really believe that the two mistakes you mentioned were due to bad proof-writing. I would be surprised if not every "veteran" who argued that there had to be 3 tangent disks of equal radius knew he/she was handwaving but just didn't know what to do about it. Similarly for #4, it's not "bad proof-writing," it's just not solving the problem.[/quote]
You have a point with experienced people knowing they're doing that. However, from my own personal experience, I think a lot of Olympiad novices think they've gotten a solution to a problem when they really haven't. They write their (incorrect) solution and then waste lots of time on a problem they don't end up solving, when otherwise they might realize they need to dedicate more time to the first one. Granted, whether they could solve the problem with that time is a different matter that you mentioned, but not spending that time filling gaps is the part of "bad proofing" that I'm referring to.

What if a solver cant solve that gap wouldn't potential partial credit be better than not writing a bad proof?

well obviously you give all the ideas you have, but it is really important to spend time trying to rigorize your solutions...i felt i really learned a lot when i actually wrote up solutions for this years usamo...it is a lot harder than i thought...lol

Also, a proof with a small gap that is along the lines of the intended solution is probably only worth a $5$. Anything beyond that takes it down to like a $1$ or a $2$, at most. On this USAMO, if you assumed that the three circles were tangent, or of equal radius, you would probably get very few points, possibly none, because that is basically the entire point of the problem.

So while it is certainly worth it to write up your progress on a problem, you shouldn't expect it to get that many points unless the progress was pretty significant.


projective geometry
Pythagorean Theorem
geometry
geometry solved

What are the properties of quadrilateral ABCD  when

   $AB^2 \cdot CD^2  +  BC^2 \cdot AD^2  =  AC^2 \cdot  BD^2$    ?

   Maj.  Pestich

   P.S.   On Feb. 21 it was birthday of   Girard Desargues (1591)

The proper source of this problem is Geometric Transformations by I. M. Yaglom, problem 270 (the untranslated Russian original).

If a quadrilateral $ABCD$ is inverted in a circle with an arbitrary radius $r$ and center at any quadrilateral vertex, say $D$, the vertices $A, B, C$ are carried to points $A', B', C'$ and the vertex $D$ to the point at infinity. The lengths of the inverted segments $A'B', B'C', C'A'$ are

$A'B' = AB \frac{r^2}{DA \cdot DB}$

$B'C' = BC \frac{r^2}{DB \cdot DC}$

$C'A' = CA \frac{r^2}{DC \cdot DA}$

If the quadrilateral $ABCD$ satisfies the relation

$AB^2 \cdot CD^2 + BC^2 \cdot DA^2 = AC^2 \cdot BD^2$

it becomes, after dividing by $DA^2 \cdot DB^2 \cdot DC^2 \neq 0$,

$\frac{AB^2}{DA^2 \cdot DB^2} + \frac{BC^2}{DB^2 \cdot DC^2} = \frac{CA^2}{DC^2 \cdot DA^2}$

$A'B'^2 + B'C'^2 = C'A'^2$

By Pythagorean theorem, the angle $\measuredangle B'$ of the triangle $\triangle A'B'C'$ is right. This is the angle at the image of the quadrilateral vertex opposite to the vertex identical with the inversion center. From similarity of the triangles $\triangle B'A'D \sim \triangle ABD$ and $\triangle B'C'D \sim \triangle CBD$, it follows that the following angles are equal:

$\measuredangle A'B'D = \measuredangle BAD = \measuredangle A$
$\measuredangle C'B'D = \measuredangle BCD = \measuredangle C$

Depending on the mutual placement of the points $A', B', C', D$, we have 2 possibilities. Either

$\measuredangle A + \measuredangle C = \measuredangle A'B'D + \measuredangle C'B'D = \measuredangle A'B'C' = 90^o$
or
$\measuredangle A + \measuredangle C = \measuredangle A'B'D + \measuredangle C'B'D = 360^o - \measuredangle A'B'C' = 270^o$

but then

$\measuredangle B + \measuredangle D = 360^o - (\measuredangle A + \measuredangle C) = 90^o$

Consequently, the given relation ("square Ptolemy theorem") implies that 2 opposite angles of the quadrilateral $ABCD$ add up to $90^o$ and the other 2 opposite angles add up to $270^o$.


The brackets are out.

Here's the deal. Each participant (including myself) sends in ONE bracket (it doesn't have to a bracket, it can be a list of winners). I promise not to rig my results or anyone elses. I'll keep track of everything, I suggest you do too, in case I mess up, in that case tell me.

To be fair, I'll fill out my bracket before looking at anyone else's.

Scoring:

Play-in game does not count

First/Second Round-1 point per game
Sweet 16-2 points per game
Elite 8-3 points per game
Final 4/National Finals-5 points per game

the brackets are available everywhere.

Define bracket. You mean like this?

{Virginia, New Jersey, New Hampshire, Nevada}

???

ummm...bracket as in ]-   ?

NCAA tournament?

I was just about to start one of these :P

I love brackets.

join this,

the points are done by the computer, and everybody from AoPS is already invited.

and the bracket can be filled out by computer.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=137142[/url]


game
combinatorics
game strategy
algorithm
IMO
IMO 1990

Given an initial integer $ n_0 > 1$, two players, $ {\mathcal A}$ and $ {\mathcal B}$, choose integers $ n_1$, $ n_2$, $ n_3$, $ \ldots$ alternately according to the following rules :

[b]I.)[/b] Knowing $ n_{2k}$, $ {\mathcal A}$ chooses any integer $ n_{2k \plus{} 1}$ such that
\[ n_{2k} \leq n_{2k \plus{} 1} \leq n_{2k}^2.
\]
[b]II.)[/b] Knowing $ n_{2k \plus{} 1}$, $ {\mathcal B}$ chooses any integer $ n_{2k \plus{} 2}$ such that
\[ \frac {n_{2k \plus{} 1}}{n_{2k \plus{} 2}}
\]
is a prime raised to a positive integer power.

Player $ {\mathcal A}$ wins the game by choosing the number 1990; player $ {\mathcal B}$ wins by choosing the number 1.  For which $ n_0$ does :

[b]a.)[/b] $ {\mathcal A}$ have a winning strategy?
[b]b.)[/b] $ {\mathcal B}$ have a winning strategy?
[b]c.)[/b] Neither player have a winning strategy?

If $ n_0<6$ it is clear that B wins, because A can only choose numbers with at most 2 prime factors (30 is the smallest with three and $ 30>5^2$), which B can either make smaller (choose the lesser of the two) or the number contains one prime factor in which case B wins. Once $ n_{2k}\equal{}2$, A can only choose prime powers, so B wins.

If $ n_0\equal{}6,7$ it's a draw, because the only numbers that give A a chance of winning are those with three prime factors (for reasons discussed above). Less than 49, this gives 30 and 42 as the only choices which won't cause A to lose. However, if A chooses 30, B may choose 6, and we have a stalemate. If A chooses 42, B may choose 6, and we still have a stalemate.

If $ n_0 \geq 8$, if $ n_0$ is less than 45 by choosing (perhaps starting later) 3*4*5, then 3*5*7, then 2*3*5*7, and then 3*4*5*7 A can force B to make $ n_{2k} \geq 60$ for some k.

If $ 45 \leq n_{2k} \leq 1990$ for some k, then A wins (he may pick 1990). If $ n_{2k}>1990$ pick the least natural value of m such that $ 2^m 47\geq n_{2k}$. The best B can do is half it, taking it to $ n_{2k\plus{}2} \equal{} 2^{m\minus{}1} 47 < n_{2k}$. However, B cannot make $ n_{2k\plus{}2}$ drop below 45 because $ 47*32<1990 \Rightarrow 2^m \geq 64$. Thus we will get a descending sequence of integers until we end up with $ 45 \leq n_{2k} \leq 1990$, at which point A wins.

It was problem 5 in the 1990 IMO in Beijing.  I messed up the 6-7 case, otherwise I would have brought home a bronze...   :(

[quote="orl"]

[b]II.)[/b] Knowing $ n_{2k \plus{} 1}$, $ {\mathcal B}$ chooses any integer $ n_{2k \plus{} 2}$ such that
\[ \frac {n_{2k \plus{} 1}}{n_{2k \plus{} 2}}
\]
is a prime raised to a positive integer power.

[/quote]

Did I understand sth wrong if I think when $A$ choose $30$ or $42$, B can make it $2$?
Because it makes the whole question in a different way than the solution I think to read??


Hi,
I just saw that problem in the polish oi site, but I can't solve it..
http://www.oi.edu.pl/php/show.php?ac=e171100&module=show&file=zadania/oi12/dzi
Can someone help me?

Thanks in advance,

Well, it's actually not that hard, I think.

Firstly, if a bird has some friends (birds he'll visit), only two of them can have other friends. Namely, that would be the highest one and the lowest one of his friends. Otherwise, the result is 0, no solution for distributing the birds into the hollows exists.
Therefore if we construct a graph of the birds (birds = vertices, visitings = edges), each of the components is a line with some vertices possibly connected to it, but only directly (i.e. each vertex lies on this line or is adjacent to it).
Now, for each bird $i$ on the line (we don't count the ending ones as on the line) we count his number of friends not on the line = $A_i$. Those will live surely between his friends that are on the line. There are $A_i!$ possibilities for those $A_i$ birds to live. We must multiply all of those $A_i!$s, multiply it by two (two possibilities - one of the ends of the line is higher than the other) and again by two (switch left and right tree). This is the number of possibilities to live for the birds in one component.
For all $C$ components we calculate this number, multiply, and then multiply by $C!$, which is the number of possibilities to arrange the components between each other. (two components X and Y are either X above Y or Y above X, except for the following case) You must also pay extra attention to friendless birds, because they can live anywhere "in between" any other component of co-visiting birds. Maybe I missed some other special cases, but generally, this is the way to go.

I hope it was clear enough to understand ...


calculus
integration
inequalities
function
real analysis
real analysis unsolved

Let $f:[0,1]\to\mathbb{R}$ a $C^1 [0,1]$ function such that $f(1)=0$.Prove that for any $n$ even,the following inequality holds:

       $\int_{0}^{1} (f(x))^n dx\leq n^n\cdot\int_{0}^{1} (f'(x))^n dx$ ;)

That $n^n$ is ridiculous- it could be replaced by a function of $n$ which is strictly less than 1. $\frac1n$ works, and is clearly not optimal.

For $f\in C[0,1]$, define $I(f)(x)=\int_x^1 f(t)\,dt$. For $1\le p\le\infty$, what is the norm of $I$ as an operator from $L^p[0,1]$ to $L^p[0,1]$
The norm is 1 in both $L^1$ and $L^\infty$. The problem asks to prove that the norm is less than $n$.

By the power mean inequality, $\left(\frac1{1-x}\int_x^1|f'(t)|^p\,dt\right)^{1/p}\ge \frac1{1-x}\int_x^1|f'(t)|\,dt\ge\frac{|f(x)|}{1-x}$
Let $A=\|f'\|_p$. The above inequality gives $|f(x)|\le A(1-x)^{1-1/p}$.
This gives $\|f\|_p\le A\left(\int_0^1(1-x)^{p-1}\right)^\frac1p=\frac{A}{\sqrt[p]{p}}$


Under the rules of the "Sea battle" game (no two ships may have a common point), can the following sets of rectangular ships be arranged on a $ 10 \times 10$ square board:

$ (a)$ two ships $ 1 \times 4$, $ 4$ ships $ 1 \times 3$, $ 6$ ships $ 1 \times 2$, and $ 8$ ships $ 1 \times 1$;

$ (b)$ two ships $ 1 \times 4$, $ 4$ ships $ 1 \times 3$, $ 6$ ships $ 1 \times 2$, $ 6$ ships $ 1 \times 1$, and $ 1$ ship $ 2 \times 2$;

$ (c)$ two ships $ 1 \times 4$, $ 4$ ships $ 1 \times 3$, $ 6$ ships $ 1 \times 2$, $ 4$ ships $ 1 \times 1$, and $ 2$ ships $ 2 \times 2$?

Hint: For each $ m\times n$ ship consider the new ship $ (m\plus{}1)\times (n\plus{}1)$ with the same center (this new ship have his vertexes in the centers of squares) this new ship can touch but not overlap.

$ Tipe$


I have tried hard but not work
Thanks for helping me! :) [/b]

Hello . Use Sine law in the triangle $ MNC$ to get the measure of $ NC$ , see that $ < MNC \equal{} \frac {A \plus{} 180}{2}$ and $ MC \equal{} \frac {1}{2}BC$

Once you get $ NC$ , you get $ LC$ by sum of segments  then the rest is easy using Menelaos theorem for the line $ KLM$ in the triangle $ ABC$   :lol:

It's easy.

$ AI/perp / BP/$   P on AI

BP intersects AC at Q.

$ BQ /parallel/ MN$  $ AC/ parallel /MP$

BP=PQ=MN  SO  triangle ABP  congruence triangle MNL .

SO $ KL/parallel/ AP$  LA=AK=NQ=CN

DONE!

ilj angel have the nice solution!
look in the picture: :)


geometry
3D geometry
pyramid
tetrahedron

[quote="orl"]Provided a pyramid $ABCDE$ with square base $ABCD$ whose faces $ ABE $, $ BCE $, $ CDE $ and $ ADE $ are equilateral triangles. We put a regular tedrahedron on the face $ C D E F $. Determine whether the figure $ A B C D E F $ has 7 distinct faces.[/quote]

That problem has some notoriety in the United States. Many years ago, that combination of a pyramid and tetrahedron on the Scholastic Aptitude Test (SAT) that many colleges use to choose which applicants they admit. The multiple-choice question asked how many faces the figure had. That was the first year that the SAT disclosed its answer key for the test, and one student noticed that the key had that the answer was 7. He had solved the problem with deeper reasoning and had given another answer. He forced the College Board (the company that runs the SAT) to rule that his answer was correct too.

Erin Schram

Where's point F?

Also, was the person's answer to the problem

[hide]

8?

[/hide]

I think point $F$ should be outside, and I also think that the number is smaller than $7$ :).


inequalities
function
inequalities unsolved

Let x_k (k=1,2,..,n) be real positive numbers such that x_1+..+x_n=1.
Prove that

    \sum (k=1,..,n)(x_k+1/x_k)^2 \geq (n+1)^a/n^(a-1)

for every real number a>0.

This is not true for x_ = 1/n for i=1,...,n, and for a sufficientely large since in that case the LHS is (1+n 2 ) 2 /n and the RHS is n( 1+1/n) a .
Thus the RHS is a non bounded from above function of a....

Pierre.

Sorry, the correct inequality is

    \sum (k=1,..,n)(x_k+1/x_k)^a \geq  (n^2+1)^a/n^(a-1)

Sorry, again.

Hmmm...are you sure about the fact that it holds for all a > 0?
I would say a  \geq 1.

Pierre.

This inequality is valid for a \geq 0 and it is reversed for 0>a \geq -1.

I think we can solve this inequality by using Jensen's inequality.


advanced fields
advanced fields unsolved

Why it seems uninteresting to study SL(2,Q)? (or more generally SL(n,Q)) (Q = field of rational numbers)
Seems I can't find any results on the representation theory of it? Or is it just too trivial?
But there are tons of results on SL(2,X) where X = R, C, Z, Fp, Qp etc...

$ \mathrm{SL}(n,\mathbb Q)$ is not locally compact, does not carry Haar measure, etc. The problem is the lack of completeness. Its completion is $ \mathrm{SL}(n,\mathbb R)$.


Three schools have a chess tournament. Four players come from each school. Each player plays three games against each player from the other schools, and plays one game against the each other player from his or her own school. How many games of chess are played?

Each player will play $ 2*4*3$ games against other school players, and $ 3$ games aginst those in their own school. That's $ 24\plus{}3\equal{}27$, for each player. Multpily this by 12, the total number of players, and then divide by two, because each game includes two people.

$ 27*12/2\equal{}162$

I got this right on the second try :D. pinkmuskrat is correct.

[quote=StellarG]8x3+4x3+6x3
8x3 is there because, if the schools were A. B> and C, we'd have A play 24 games against her opponent. B would have 12 games against C, and they would have 18 inter-schools competitions. I get 54. Why?[/quote]

Each player will make each time's pairs smaller.(Do you understand what this means?)

[hide=Solution]
Each player plays three games against $12-4=8$ people. So each player plays $3\times8=24$ games against people from other schools. That's $12\times24=288$. But that counts each game twice, so the number of games would be $288\div2=144$. Now, each player plays $3$ people from their own school $1$ time. That's $3$ games per player, amassing to $12\times3=36$. That also counts each game twice, so the number of home games is $36\div2=18$. So the total number of games is $144+18=\boxed{162}$.
[/hide]

[quote=hellomathworks10]kill me now
(answered 612 as a typo)[/quote]

[url=http://www.mysmiley.net][img]http://www.mysmiley.net/imgs/smile/fighting/fighting0038.gif[/img][/url] /j


When reading transcripts of classes, everything that the mod/teacher says is shown in bold.  For this reason, I think it might be helpful for mods/teachers to avoid using boldface in ways that make the gist of the message dependent on the boldface.  For example, [quote="Teacher"]blah blah blah [b]along[/b] walkway blah blah [b]beside[/b] walkway blah.  The two words in bold are critical to the problem.  [/quote]
during class shows up as

 [quote="Transcript"][quote="Teacher"][b]blah blah blah along walkway blah blah beside walkway blah.  The two words in bold are critical to the problem.  [/b][/quote][/quote]

Thanks!

Ah, good point!  I'll see what we can do about that in the style sheets for the transcripts.

Emphasized (bold) words and phrases from classes should now appear as italicized and underlined in the transcripts.


Your thumbnail file (heh) (i.e. the one that appears in the tab with AoPS and in the browser) has a white background, and it would look a lot better if you replaced it with transparency so that only the nice hexagon remains. I'm fairly certain it would be as simple as:

Photoshop/Gimp-> Set transparency: white -> paste in server folder -> replace filename.

Just my 2 cents :)...

That's a icon file. I'm pretty sure .ico-s don't support transparency.


limit
calculus
derivative
integration
logarithms
calculus computations

As a first order autonomous equation, you don't have to actually solve it. Any solution must be monotone; since $ 32000\minus{}20\cdot 2000^2<0$, the solution that starts at $ 2000$ is decreasing. It either decreases to the next point at which the derivative is zero, or goes to $ \minus{}\infty$ if there aren't any such points in the way. The zeros of $ 32000\minus{}20y^2$ are at $ y^2\equal{}1600$, or $ y\equal{}\pm 40$. The first one we encounter going down from $ 2000$ is $ 40$, and thus $ \lim_{x\to\infty}y(x)\equal{}40$.

If you feel like it, you could solve the equation exactly. It's not too hard.

The easy way is to assume that the limit, $ L$ exists.

Note that $ \dfrac{dy}{dx} \equal{} 0$ @ $ y \equal{} L$. Thus, $ y'(L) \equal{} 32000 \minus{} 20L^2 \equal{} 0 \Rightarrow L \equal{} \pm 40$.

Since $ y(0) > 40$, and that $ \dfrac{dy}{dx} < 0$ for all $ y \in (40,2000]$, we must have $ \lim_{x\to\infty}y(x) \equal{} 40$.

if I want to derivative it, how it would be done?

If by "derivative it" you mean solve it then:
[hide]
$ \dfrac{\,dy}{\,dx} \equal{} 32000 \minus{} 20y^2$

$ \dfrac{\,dy}{1600 \minus{} y^2} \equal{} 20\,dx$

$ \dfrac{1}{80}\left(\dfrac{\,dy}{y \plus{} 40} \minus{} \dfrac{\,dy}{y \minus{} 40}\right) \equal{} 20\,dx$

$ \int \left(\dfrac{\,dy}{y \plus{} 40} \minus{} \dfrac{\,dy}{y \minus{} 40}\right) \equal{} \int 1600\,dx$

$ \ln(y \plus{} 40) \minus{} \ln(y \minus{} 40) \equal{} 1600x \plus{} K$

$ \ln\left(\dfrac{y \plus{} 40}{y \minus{} 40}\right) \equal{} 1600x \plus{} K$

$ \dfrac{y \plus{} 40}{y \minus{} 40} \equal{} C e^{1600x}$ (Where $ C \equal{} e^{K}$)

$ y \equal{} \dfrac{40(C e^{1600x} \plus{} 1)}{C e^{1600x} \minus{} 1}$ Which is the general family of solutions of $ \dfrac{\,dy}{\,dx} \equal{} 32000 \minus{} 20y^2$

$ y(0) \equal{} 2000 \Longleftrightarrow \dfrac{40(C \plus{} 1)}{C \minus{} 1} \equal{} 2000 \Longleftrightarrow C \equal{} \dfrac{51}{49}$

Thus, the solution is: $ \boxed{y(x) \equal{} \dfrac{40(51e^{1600x} \plus{} 49)}{51e^{1600x} \minus{} 49}}$[/hide]


I have the following recurrence (sequence), only using powers of 2

F(n) = n/2 * F(n/2) + 2 (using powers of 2 would be => F(2^n) = [2^n / 2] * [ F( 2^n / 2 ) + 2 ]  )

with a boundary of

F(1) = 1

so....

F(1) = 1
F(2) = F(1) + 2
F(4) = 2( F(1) + 2 ) + 2
F(8) = 4( 2( F(1) + 2) + 2 ) + 2
F(16) = 8( 4( 2( F(1) + 2 ) + 2 ) + 2 ) + 2

which is...

F(2^0) = 1
F(2^1) = F(1) + 2
           = 1 + 2
F(2^2) = 2^1 * (F(1) + 2 ) + 2
           = 2^1 * (1 + 2) + 2
F(2^3) = 2^2 * (2^1 * ( F(1) + 2 ) + 2 ) + 2
           = 2^2 * (2^1 * ( 1 + 2 ) + 2 ) + 2
F(2^4) = etc...

etc...

I'm stumped on the simplification, tried many combinations. I'm aware though that there will be a summation due to the addition of 2 multiple times. Any help would be greatly appreciated.  Thanks in advance

Are you sure that's right, the Summation part I mean?


linear algebra
matrix
linear algebra unsolved

Let $A$ be the $n \times n$ matrix, with $\text{tra}(A^i)=0$ for all positive integers $i$. Prove that $A$ is nilpotent.


[quote="M4RI0"]Let $P$, $Q$ be points on $BC$ in traingle $ABC$. Prove that if incircle radius of $ABP$, $ACQ$ are equal then the incircle radius of $ABQ$ and $ACP$ are equal too.[/quote]

An equivalent enunciation is to show that $\frac{AB-AC}{BC}=\frac{AP-AQ}{PQ}$ and it's true by cosinus law aplying in $\triangle ABP$, $\triangle ABQ$ and $\triangle ACQ$, $\triangle ACP$ respectively.

[quote="hucht"]An equivalen enunciation is to show that $\frac{AB-AC}{BC}=\frac{AP-AQ}{PQ}$ [/quote]
Could you explain more? :blush:

WLOG, assume B, P, Q, C follow on the side BC in this order. Let $R_{1}, R_{2}$ be the inradii of $\triangle ABQ,\ \triangle APC$, let $r_{1}, r_{2}, r_{3}$ be the inradii of $\triangle ABP,\ \triangle APQ,\ \triangle AQC$ and h the common A-altitude of these 5 triangles.  According to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=19468]Inradii problem[/url],

$\frac{1}{r_{1}}+\frac{1}{r_{2}}-\frac{R_{1}}{r_{1}r_{2}}= \frac{2}{h}= \frac{1}{r_{2}}+\frac{1}{r_{3}}-\frac{R_{2}}{r_{2}r_{3}}$

$\frac{1}{r_{1}}\left(1-\frac{R_{1}}{r_{2}}\right) = \frac{1}{r_{3}}\left(1-\frac{R_{2}}{r_{2}}\right)$

which means $r_{1}= r_{3}\ \Longleftrightarrow\ R_{1}= R_{2}.$ See also [url=http://www.cut-the-knot.org/Curriculum/Geometry/FourIncircles.shtml]Five Incircles Theorem[/url].

Thank you yetti. But isn't it can be done by straighforward?


pigeonhole principle
induction
geometry
rectangle
probability
expected value
combinatorics proposed

assume that we have a n*n table we fill it with 1,...,n such that each number exists exactly n times prove that there exist a row or column such that at least $\sqrt{n}$ diffrent number are contained.

Let's first consider the case that $n$ is a perfect square. We say $n=m^2$
It is clear that an "optimal" situation (in fact it is a "worst case" situation) would be that each number appeared in an $m \times m$ table. In that case the requirement is fulfilled, and it is sharp.

If we raise n by 1, then in at most 1 column we can create m+1 times the same element (without decreasing the amound in another of course).

So to raise the amount of times the same number appears, by 1, in each of the n columns and each of the n rows, we need at least 2n extra blocks. So the statement is ok up to $m^2+2m=(m+1)^2-1$. One more and we're back to the case $n$ is a perfect square.

(a thing used here throughout the problem for the necessary verifications is scrapping all rows & columns where the number "k" doesn't appear, plus pigeonhole)

Peter

sorry, can you give a real proof for a $n=m^2$?

hmm...sounds like trivial expected value-problem but i'm too stupid to give a proof at the moment.

Peter

I don't even think I understand the induction step  :blush:

Anyway, I wonder if the case $n=m^2$ is significantly easier than the general case.

Well, it is easy (trivial even). Look:

We got an $m^2\times m^2$ square, in which each number appears $m^2$ times.
The key is: if a number appears more than $m$ times in a row or column, then there are at least [b]2[/b] rows/columns where the number appears less than $m$ times. (this is trivial)

If you still want to have only $m$ different numbers on those rows, you'll need 2 more numbers having m+1 numbers on that row column, which leaves you 4 such cases again, which will loophole to infinity. (in fact, this part is pretty trivial too if you make some drawings)

So if you know that the requirement is sharpened to $m$ numbers in each row and $m$ numbers in each columns, with $m^2$ in total, then there can be no row or column other than those $m$ containing that specific number. Which is just another wording for saying (cf. the last senctence I wrote): If you scratch all rows & columns from the array not containing that specific number, then you'll keep an $m\times m$ square.

It doesn't necessarily imply that they really form a square like that, counterexample is below, but I didn't use the 'actual' square as given in proof either.

$\begin{array}{cccc}1&2&1&2\\3&2&3&2\\1&4&1&4\\3&4&3&4\end{array}$

Another possible approach is:

Let a "hit" be recorded for each time a different integer is present in a different row or column. Suppose i occurs in k rows and l comumns. Then the number of hits recorded by i is $k+l \geq 2 \sqrt kl \geq 2 \sqrt n$, so the total number of hits is at least $2n \sqrt n$. Hence some row or column (of which there are 2n), records at least $\sqrt n$ hits i.e. at least $\sqrt n$ elements are present.

Very nice, Fiachra, I just wonder how do you know that $\sqrt kl \geq 2 \sqrt n$?

If that's true and it has a nice, rigorous solution, you got a very good and very short proof. :)

Why is $kl\ge n$? What if $i$ occurs in only one row, for example?

[Edit: I see Peter has already asked the question :)]

consider just the occurences of i in the table and rearrange them in decreasing order;then you get a partition of n(since i occurs exactly n times-just think of the symbol i as an asterisk in a Young Diagram for a partition).then clearly k would be
largest part in the partition and l becomes the number of parts in the partition.then clearly since the number of i's is just n,it is clearly less than or equal to kl.

Thanks! I get it now. My objection was pretty stupid :). I asked what happens if $i$ occurs in just one row, but then it also occurs in $n$ columns, and I had forgotten about that. Thanks again!

Sorry, I should probably have included this bit:

Imagine the rectangle formed by the k rows and l columns (rearranging the rows and columns if necessary). All the occurences of i lie within this rectangle, which contains kl cells. Hence i occurs at most kl times i.e. $kl \geq n$.

yes, that was indeed the main element in my proof too :) (but less efficient)

nice solution :)


12 seke darim ke yekishoon tyaghalobiye yani vaznesh ya kamtar ya bishtar az baghiyeye seke hast.mikhahim ba 3 bar vazn kardan ba tarazooye do kafei peidash konim.

seke ha ra be 3 dasteye 4 tayi taghsim konid.in daste ha ra X1 .X2, X3 benamid.do halat vojood darad:
1)agar X1 va X2 ham vazn bashand,har hasht seke gooroohe  X1 va X2 mizan sanjesh hastand.seke taghallobi ham dar X3 hast.
2)agar X1 va X2 ham vazn nabashand seke haye X3 mizane sanjesh hastand.seke taghallobi dar X1 ya X2 hast.
halat 1:3 seke az X1 va 3 seke az X3 va ba ham moghayese konid.agar ham vazn bashand seke chaharome X3 ast.ba moghayese in seke ba yeki az seke hay gooroohe X1 mitavan sabok ya sangin boodanash ra ham moshakhas kard.agar 3 seke X1 ba 3 seke X3 ham vazn nabashand  seke taghallobi yeki azseke hay X3 ast va mitavanim sabok ya sangin boodanash ra peyda konim.(chon X1 mizane sanjesh hastand.)hal ba sevomin vazn kardan mitavanim an ra peyda konim.
halat2:farz konim X1 sangin tar bashad.seke haye X1 ra a,b,c,d va seke haye X2 ra A,B,C,D minamim.dar tozine dovom a,b,A ra ba c,d,B moghayese mikonim.3 halat vojov darad:
1-agar ham vazn bashand seke yeki az C,D ast.chon C,D az gooroohe sabok tarand pas midanimseke taghallobi sabok tar hast ke ba tozine sevom seke taghallobi peyda mishavad.
2-agar in 2 hamvazn nabashand farz konim a,b,A sangin tar hast.dar in soorat c,dva A mizane sanjesh hastand chon agar yeki az c va d taghallobi bood  a,b,A saboktar mishod.pas seke taghallobi yeki az a,b,B ast.dar akhar ham a va b ra moghayese mikonim agar ham vazn boodand B taghallobi va sabok tar hast.agar a va b  ham vazn nabashand ha rkodam sangintar bashad taghallobi ast.
3-agar c,d,B sangin tar bood ,khodetoon mesle balay tahlilesh konid.
halle? :)
kheyli ziad shod, bebakhshid,har chi fekr kardam neveshte hayam  az in kamtar nashod. :)

ye serie kheili jaamee masale haye seke dar ketabe :"mohemtarin masale ha va ghazie haye riyazi" oomade ke fekr mikonam kheili ghashangan.

in soal o man khodam ghablan post kardeh boodam:
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=coin&t=34138

man rah e halesho copy mikonam (ingilisiye)
[hide="solution"]
we devide the 12 coins into [b]3 groups of 4[/b], then we choose two of them, then we make our first weighting. 2 diffrent situations might happen:
1) the two groups have the same weight:  so the left over group has the different coin. from the 4 coins of the third group we choose two, and weight them. if they were the same the different coin is between the two other coins(and all we need to do is to weight one of the leftover coins with a normal coin, then we would know which one is the different coin). but if the groups of one were not equal, then again the differen coin is one of the coins that we just weighted, and we can find the different coin, with the third weighting.

2)If the 2 groups of three coins were not the same weight: then one of the two groups contains the odd coin--lets call them group A and B (we don't know which one, because we don't know if the odd coin is heavy or light). take 3 coins out of one of the groups-lets say A, then take two coins from B and add it to A, also add one normal coin to B,-now you have two groups of three $2$ weight, the two groups of three coins, three situations might happen
$I$ the two groups are the same, which implies that the odd coin is among the three coins that were taken out from A, and since we know that what kind of inequality we had before, we know that whether the odd coin is heavy or light. so all you need to do is to wieght two of those three, and since you know that whether the odd coin is heavy or light, with one weighting you can figure it out.
$II$ the two groups are inequal but this time, the inequlity has changed, for example if the first time A was heavier than B, now B is heavier than A, so this implies that the odd coin is among the 2 coins that where transfered...now you only need one weighting to find the odd coin between two coins, just compare one of the coins with one normal coin,...
$III$ the inequality is the same as before, this implies that the odd coin is among those who have not been moved, so among 8 coins, we moved 5 of them--they are OK. but the odd coin is among the 3 coins that have not been moved. from the side that has 2 of these three coins we move one coin to the other side, and take one OUT, now put 2 normal coin to the other side. weight for the Third time. three situations might happen:
$i$ the two groups are the same, so the odd coin is the one that was taken out.
$ii$ the two groups have the opposite inequality, so the coin that was moved from one side to other one is the odd one.
$iii$ the groups have the same inequality, so the  coins that were taken out and moved are normals, so the one that was not moved is the odd one... :yup: [/hide]

alan man mage chi goftam?aghaye amirhtlusa rahe hale man mishe begid ba rahe hale shoma che farghi dare? :?:

mr amin_68 oon jayi ke se ta az x_1 ba se ta az x_3 vazn mishe age ham vazn bashand az koja motmaeni ke taghalobiye to x_2 nist?

to ro be khoda ye kam fekr konid ,savab dare.agar hanooz ham nafahmidin begid baratoon tozih midam. :)


If we know that $\frac{a+b+c+d}{4} \geq \sqrt[4]{abcd}$

What can we substitute into a,b,c,d so we will achieve

$\frac{x+y+z}{3} \geq \sqrt[3]{xyz}$

In other words, if we assume AM-GM is true for 4 variables, how do we get the 3 variables one from there.

Are you asking for help, or...?

Hint: [hide]Remember that what you substitute for the LHS should make the LHS equal to (a+b+c)/3.  There's pretty much one substitution that makes sense.[/hide]

We prove the more general:

Suppose we have AMGM for $n = 2^{k-1}$ and $n = 2^k$.  We prove AMGM for $2^{k-1}  < n < 2^k$.

---

Put $x = \frac{a_1 + a_2 + \dots + a_n}{n}$,

and put $a_1a_2\dots a_n = r$.

Then $\frac{a_1 + a_2 + \dots + a_n + (2^k - n)x}{2^k} \ge \sqrt[2^k]{a_1a_2\dots a_n(x)^{2^k - n} }$

implying $x \ge r^{\frac{1}{2^k}}(x)^{1 - \frac{n}{2^k}}$

so that $x^{\frac{n}{2^k}} \ge r^{\frac{1}{2^k}}$

or $x \ge r^{1/n}$

as desired.

sorry guys, I didn't make my question clear.

I am simply asking what would be the appropriate substitution for a,b,c,d to tramsform

\[\frac{a+b+c+d}{4}\geq \sqrt[4]{abcd}\]
to
\[\frac{xyz}{3}\geq \sqrt[3]{xyz}\]

The above posts told you, if you read them carefully.

Explicitly, you want

$a=x$
$b=y$
$c=z$
$d=\frac {x+y+z}{3}$


$ a,b,c>0$
$ a\plus{}b\plus{}c\equal{}abc$
prove that:
$ \frac{ab\plus{}bc\plus{}ca}{2}\plus{}\sum \frac{a}{a\plus{}b} \geq 6$

[quote="fesha3t"]$ a,b,c > 0$
$ a \plus{} b \plus{} c \equal{} abc$
prove that:
$ \frac {ab \plus{} bc \plus{} ca}{2} \plus{} \sum \frac {a}{a \plus{} b} \geq 6$[/quote]
An expanding gives $ \sum_{cyc}(a^4b^2\plus{}a^4c^2\plus{}2a^3b^3\plus{}2a^4bc\minus{}a^3b^2c\minus{}3a^3c^2b\minus{}2a^2b^2c^2)\geq0,$
 which is obvious.

[quote="caubetoanhoc94"][quote="fesha3t"]$ a,b,c > 0$
$ a \plus{} b \plus{} c \equal{} abc$
prove that:
$ \frac {ab \plus{} bc \plus{} ca}{2} \plus{} \sum \frac {a}{a \plus{} b} \geq 6$[/quote]
My solution  :lol:
$ a \plus{} b \plus{} c \equal{} abc$
So:
$ abc \ge 3\sqrt {3}$
 $ \sum \frac {a}{a \plus{} b} \ge \frac {(\sqrt {ac} \plus{} \sqrt {ab} \plus{} \sqrt {bc})^2}{2(ab \plus{} bc \plus{} ca)} \ge \frac {27}{2(ab \plus{} bc \plus{} ca)}$
We need prove that:
$ \frac {ab \plus{} bc \plus{} ca}{2} \plus{} \frac {27}{2(ab \plus{} bc \plus{} ca)} \ge 6$
It's true :P[/quote]

I think$ \sum \frac {a}{a \plus{} b}\geq \frac{3(ab\plus{}bc\plus{}ca)}{2(a^2\plus{}b^2\plus{}c^2)}$
and $ (ab\plus{}bc\plus{}ca)^2 \geq 3abc(a\plus{}b\plus{}c)$

help me
find p,q,r (>0) such that with a,b,c >0
this ineq always right
$ \sum \frac{a}{pa\plus{}qb\plus{}rc} \leq 3.\sqrt{\frac{1}{p\plus{}q\plus{}r}}$

@caubetoanhoc94 ,why $ abc \geq 3\sqrt{3}$

[quote="enndb0x"]@caubetoanhoc94 ,why $ abc \geq 3\sqrt {3}$[/quote]
Hjx
$ abc\equal{}a\plus{}b\plus{}c \ge 3\sqrt[3]{abc}$
So:$ abc \ge 3 \sqrt{3}$


[hide]
n(n-1)/2 =  1 + 2 + 3 + 4 + ... + (n-1)
Rewriting, we get

0 <= 0 + (x^2-2x+1) + (x^3-3x+2) +... +  (x^n-nx+(n-1))

However, x^n-nx + n-1 = (x-1)^2 ( n-1 + (n-2)x + .. +x ^(n-2)).  (x-1)^2 is nonnegative, and  ( n-1 + (n-2)x + .. +x ^(n-2)) is at least n-1 for x>=0.  So the whole RHS of the inequality is the sum of a bunch of non-negative numbers, and is thus non-negative.  QED[/hide]

[quote="probability1.01"]Maybe more natural way of seeing it is [hide]$x^n + n-1 = n \frac{x^n + 1 + 1 + ... + 1}{n} \ge n \sqrt[n]{x^n} = nx$ by AM-GM.[/hide][/quote]

That is also my solution!!!
I like much this trick, that is writing $n=1+1+...+1$ (1 n times).......

 :lol:  :)


Lets see how fast you guys can figure this out. There is a straight line of people wearing hat colors 1 and 2. In the line you can see all the hat colors in front of you, but not in back of you. The last person will say a number, either 1 or 2, and if he guesses correctly then he stays alive, but if not he will die. It keeps going on until the front person says a number. What is the best strategy so that the least amount of people die?

Have fun! :ninja:  :P

if you can see two hats in front of you of opposite colours, pass. If you can see two hats of one colour, guess that your hat is the other colour.

No they can do reallllly welllll. :huh:  :idea:

Hope you guys get it!

hmm... perhaps the last person guesses whatever number the person in front of him has, that way the next person knows what he has and automatically survives.  This way starting from the back every other person has automatic survival, and the others have a 1/2 chance totaling to 1/4 of them dieing?

[hide="The answer"] Only one person needs to die. [/hide] [hide="Hint"] Parity. ;) [/hide]

The guy above me is right only one has to die,
Now someone solve it.


integration
real analysis
real analysis unsolved

at first note that :
$ (1 \plus{} x)^{n} \equal{} \sum_{0}^{n}C_{n}^{k}x^{k}$
hence :
$ x(1 \plus{} x)^{n} \equal{} \sum_{0}^{n}C_{n}^{k}x^{k \plus{} 1}$
thus
$ \int_{0}^{1}x(1 \plus{} x)^{n} \equal{} \sum_{0}^{n}C_{n}^{k}\frac {1}{k \plus{} 2}$
now you need to calculate integral. try yourself.
try do it by parts
[hide="Hint"]
consider something this :
$ \int_{0}^{1}x(1 \plus{} x)^{n}dx \equal{} \int_{0}^{1}\frac {x}{n \plus{} 1}d(1 \plus{} x)^{n \plus{} 1} \equal{} \frac {x}{n \plus{} 1}(1 \plus{} x)^{n \plus{} 1}|_{0}^{1} \minus{} \frac {1}{n \plus{} 1}\int_{0}^{1}(1 \plus{} x)^{n \plus{} 1}dx \equal{} \frac {x}{n \plus{} 1}(1 \plus{} x)^{n \plus{} 1}|_{0}^{1} \minus{} \frac {(1 \plus{} x)^{n \plus{} 2}}{(n \plus{} 1)(n \plus{} 2)}|_{0}^{1}\equal{}\frac {1 \plus{} n2^{n \plus{} 1}}{(n \plus{} 1)(n \plus{} 2)}$
[/hide]


[quote="letiendat"]Let $S$ be the area of triangle $ABC$ with $BC=a, CA=b, AB=c$,$S'$ be the area of triangle $A'B'C'$ with $B'C'=a', C'A'=b', A'B'=c'$.
Prove that: $a(b'+c'-a')+b(c'+a'-b')+c(a'+b'-c') \geq \sqrt{48SS'}$[/quote]

This is not really nice, being just a composition of the following two facts:

[color=blue][b]Pedoe inequality.[/b] If ABC and A'B'C' are two triangles with sidelengths a = BC, b = CA, c = AB and a' = B'C', b' = C'A', c' = A'B' and areas S and S', respectively, then

$a^2\left(b^{\prime}\ ^2+c^{\prime}\ ^2-a^{\prime}\ ^2\right)+b^2\left(c^{\prime}\ ^2+a^{\prime}\ ^2-b^{\prime}\ ^2\right)+c^2\left(a^{\prime}\ ^2+b^{\prime}\ ^2-c^{\prime}\ ^2\right)\geq 16SS^{\prime}$.[/color]

(See http://www.mathlinks.ro/Forum/viewtopic.php?t=2958 for a proof of this inequality.)

[color=blue][b]Square root triangle theorem.[/b] If ABC is a triangle with sidelengths a, b, c and area S, then there exists a triangle with sidelengths $\sqrt{a}$, $\sqrt{b}$, $\sqrt{c}$, and its area is $\geq\sqrt{\frac{\sqrt3}{4}S}$.[/color]

(See http://www.mathlinks.ro/Forum/viewtopic.php?t=44103 and http://www.mathlinks.ro/Forum/viewtopic.php?t=3104 and http://www.mathlinks.ro/Forum/viewtopic.php?t=47487 ; for a more general version, see http://www.mathlinks.ro/Forum/viewtopic.php?t=14965 .)

Now, by the square root triangle theorem, applied to the triangle ABC, there exists a triangle $A_1B_1C_1$ with sidelengths $\sqrt{a}$, $\sqrt{b}$, $\sqrt{c}$ and an area T with $T\geq\sqrt{\frac{\sqrt3}{4}S}$, and by the square root triangle theorem, applied to the triangle A'B'C', there exists a triangle $A_1^{\prime}B_1^{\prime}C_1^{\prime}$ with sidelengths $\sqrt{a^{\prime}}$, $\sqrt{b^{\prime}}$, $\sqrt{c^{\prime}}$ and an area T' with $T^{\prime}\geq\sqrt{\frac{\sqrt3}{4}S^{\prime}}$. Now, applying the Pedoe inequality to these two triangles $A_1B_1C_1$ and $A_1^{\prime}B_1^{\prime}C_1^{\prime}$, we get

$\left(\sqrt{a}\right)^2\left(\left(\sqrt{b^{\prime}}\right)^2+\left(\sqrt{c^{\prime}}\right)^2-\left(\sqrt{a^{\prime}}\right)^2\right)+\left(\sqrt{b}\right)^2\left(\left(\sqrt{c^{\prime}}\right)^2+\left(\sqrt{a^{\prime}}\right)^2-\left(\sqrt{b^{\prime}}\right)^2\right)$
$+\left(\sqrt{c}\right)^2\left(\left(\sqrt{a^{\prime}}\right)^2+\left(\sqrt{b^{\prime}}\right)^2-\left(\sqrt{c^{\prime}}\right)^2\right)\geq 16TT^{\prime}$.

In other words,

$a\left(b^{\prime}+c^{\prime}-a^{\prime}\right)+b\left(c^{\prime}+a^{\prime}-b^{\prime}\right)+c\left(a^{\prime}+b^{\prime}-c^{\prime}\right)\geq 16TT^{\prime}$.

Since $T\geq\sqrt{\frac{\sqrt3}{4}S}$ and $T^{\prime}\geq\sqrt{\frac{\sqrt3}{4}S^{\prime}}$, this becomes

$a\left(b^{\prime}+c^{\prime}-a^{\prime}\right)+b\left(c^{\prime}+a^{\prime}-b^{\prime}\right)+c\left(a^{\prime}+b^{\prime}-c^{\prime}\right)$
$\geq 16\cdot\sqrt{\frac{\sqrt3}{4}S}\cdot\sqrt{\frac{\sqrt3}{4}S^{\prime}}=\sqrt{48SS^{\prime}}$,

and the problem is solved.

  Darij

Well, would it be easier to prove a pure inequality, like [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=66283]this one[/url]? :)

1989, I proof:
Let $S$ be the area of triangle $ABC$ with $BC=a, CA=b, AB=c$,$S'$ be the area of triangle $A'B'C'$ with $B'C'=a', C'A'=b', A'B'=c'$.Prove that\[a(b'+c'-a')+b(c'+a'-b')+c(a'+b'-c') \]\[\geq 2\sqrt{3}\left(\frac{a'+b'+c'}{a+b+c}S +\frac{a+b+c}{a'+b'+c'}S'\right).\]
(《Hunan mathematics communication》(China Changsha)No.4(1989))
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=501930]here[/url]


inequalities
trigonometry
logarithms
algebra
system of equations
inequalities open

For the number of a, b, c not less than 1
Find the smallest value of the expression:
 $ S \equal{} a \plus{} b \plus{} c \plus{} \frac{3a}{bc} \plus{} \frac{4b}{ac} \plus{} \frac{5c}{ab}$

[color=#FF0000]Moderator Edit: do not quote the whole post above.[/color]
It's EASY!
Using simple AM Gm inequality we have:
$ S\ge6\sqrt[6]{\frac{3\cdot4\cdot5a^2b^2c^2}{a^2b^2c^2}}\equal{}6\sqrt[6]{60}$.
Hence $ S_{min}\equal{}6\sqrt[6]{60}$.
 :rotfl:

[color=#FF0000]Moderator Edit: do not quote the whole post above.[/color]
When inequality? :lol:

[color=#FF0000]Moderator Edit: do not quote the whole post above.[/color]
You're wrong, Potla. Because the equality can't hold  :)

[color=#FF0000]Moderator Edit: do not quote the whole post above.[/color]
Mr.Bikrone! is true;my friend :D.When inequality..It's not easy.

Please help me solution of this inequality :lol:

Find the minimum value of the expression:
$ S \equal{} \frac {3}{{\tan B \plus{} \tan C \minus{} {\mathop{\rm t}\nolimits} {\rm{anA}}}} \plus{} \frac {4}{{\tan A \plus{} \tan C \minus{} {\mathop{\rm t}\nolimits} {\rm{anB}}}} \plus{} \frac {5}{{\tan A \plus{} \tan B \minus{} {\mathop{\rm t}\nolimits} {\rm{anC}}}}$
where A, B, C are the three angles are less than $ 90^0$ of a triangle satisfy the conditions $ C \ge A$.

[quote]
Find the minimum value of the expression:
$ S \equal{} \frac {3}{{\tan B \plus{} \tan C \minus{} {\mathop{\rm t}\nolimits} {\rm{anA}}}} \plus{} \frac {4}{{\tan A \plus{} \tan C \minus{} {\mathop{\rm t}\nolimits} {\rm{anB}}}} \plus{} \frac {5}{{\tan A \plus{} \tan B \minus{} {\mathop{\rm t}\nolimits} {\rm{anC}}}}$
where A, B, C are  three angles of an acute  triangle ABC and $ C \ge A$.[/quote]

The problem is posted on here: "Toward Mathematical Olympiad," the magazine reported, "Mathematics and Youth" in Vietnam

But no answer, hope you solve it soon. Thanks!

Please you visit my site: http://www.laisac.page.tl

[color=#FF0000]Moderator Edit: is it from an ongoing magazine contest? Then don't post it...[/color]

[quote="robinson123"][quote="laisac"][quote][color=#FF0000]Moderator Edit: is it from an ongoing magazine contest? Then don't post it...[/color][/quote][/quote]
Yes, it is from an ongoing contest from Vietnam. So please don't post solutions here, or lock this topic[/quote]

Regretfully, I apologize for this. This problem has been posted on June 20, 2011. It has been certain that up to now, there has been no solution yet because I developed this math problem. The solution will be posted by the magazine MATHEMATICS AND YOUTH after August 20, 2011

My English is not as good as I wish, I am sorry for cause of misunderstandings or confusing for you! :D

(One other article published in the magazine MATHEMATICS AND YOUTH: 1 / 2012)

Solve the following system of equations in two variables
$\left\{ \begin{array}{l}
 \log _2 x = 2^{y + 2}  \\
 4\sqrt {1 + x}  + xy\sqrt {4 + y^2 }  = 0 \\
 \end{array} \right.$