geometry
circumcircle
conics
trigonometry
Gauss
trig identities
Law of Sines

[color=darkblue][b]Let $ P$ be a point in the same plane with a triangle $ ABC$.Prove that the projections in point $ P$ on $ PA,PB,PC$ are meeting the sides $ BC,CA,AB$ in 3 collinear points.[/b][/color]

[quote="CircleSquared"]If I interpreted it right, you're asking for the Simson Line (Which only exists when P is on the circumcircle of ABC.)

Otherwise, could you restate the problem?[/quote]
No it isn't the Simson line.It sais that the projction IN the point $ P$.Here I made a figure...Hope it would help you  :)

[quote="balan razvan"][color=darkred]Let $ P$ be a point in the same plane with a triangle $ ABC$. Denote $ \|\begin{array}{ccc}
X\in BC & , & PX\perp PA\\\
Y\in CA & , & PY\perp PB\\\
Z\in AB & , & PZ\perp PC\end{array}$ . Prove that $ Z\in XY$ .[/color][/quote]
This well-known problem appears and in the my geometry book with three methods. I"ll come back soon here.

nice problem! it's equivalent to show

The line perpendicolar to AP in P meet AC in F and AB in I, the line perpendicolar to BP in P meet BA in H and BC in E and the line perpendicolar to CP in P meet CB in D and CA in G.
Then the problem it's equvalent to show that D,E,F,G,H,I stay on the same conic.

For the carnot's theorem D,E,F,G,H,I stay on the same conic iff

$ \frac {AH}{HB} \cdot \frac {AI}{IB} \cdot \frac {BD}{DC} \cdot \frac {BE}{EC} \cdot \frac {CF}{FA} \cdot \frac {CG}{GA} = 1$

Now let $ \|\begin{array}{ccc} \angle PAB = A_1 & , & \angle PBC = A_2 \\
\ \angle PBC = A_3 & , & \angle PCB = A_4 \\
\ \angle PCA = A_5 & , & \angle PAC = A_6\end{array}$

Then appling some law of sines we obtain:

$ \frac {AH}{HB} \cdot \frac {AI}{IB} \cdot \frac {BD}{DC} \cdot \frac {BE}{EC} \cdot \frac {CF}{FA} \cdot \frac {CG}{GA} = \frac {PA \cdot \cos{(A_1 + A_2)}\cdot \cos{A_2}}{PB \cdot \cos{A_2}} \cdot \frac {PA \cdot \cos{A_1}}{PB \cdot \cos{A_1} \cdot \cos{(A_1 + A_2)}} \cdot \frac {PB \cdot \cos{(A_3 + A_4)}\cdot \cos{A_4}}{PC \cdot \cos{A_4}} \cdot \frac {PB \cdot \cos{A_3}}{PC \cdot \cos{A_3} \cdot \cos{(A_3 + A_4)}} \cdot \frac {PC \cdot \cos{(A_5 + A_6)}\cdot \cos{A_6}}{PA \cdot \cos{A_6}} \cdot \frac {PC \cdot \cos{A_5}}{PA \cdot \cos{A_5} \cdot \cos{(A_5 + A_6)}} = 1$

[img]http://img403.imageshack.us/img403/359/geometria33js7.png[/img]

[quote="balan razvan"][color=darkred]Let $ ABC$ be a triangle. For a point $ P$ denote $ \|\begin{array}{ccc} X\in BC & , & PX\perp PA \\
\ Y\in CA & , & PY\perp PB \\
\ Z\in AB & , & PZ\perp PC\end{array}$ . Prove that $ Z\in XY$ [size=134]([b]Bobiller[/b]).[/size][/color][/quote]
[color=darkblue][b][u]Proof I (metric).[/u][/b] $ \|\begin{array}{c}m(\widehat {BPC}) = x \\
\ m(\widehat {CPA}) = y \\
\ m(\widehat {APB}) = z\end{array}$ . Apply an well-known relation to the cevians $ \begin{array}{ccc} PX & \mathrm {in} & \triangle BPC \\
\ PY & \mathrm {in} & \triangle CPA \\
\ PZ & \mathrm {in} & \triangle APC\end{array}$ respectively :

$ \|\begin{array}{c} \frac {XB}{XC} = \frac {PB}{PC}\cdot\frac {\sin\widehat {XPB}}{\sin\widehat {XPC}} = \frac {PB}{PC}\cdot|\frac {\cos z}{\cos y}| \\
 \\
\frac {YC}{YA} = \frac {PC}{PA}\cdot\frac {\sin \widehat {YPC}}{\sin\widehat {YPA}} = \frac {PC}{PA}\cdot|\frac {\cos x}{\cos z}| \\
 \\
\frac {ZA}{ZB} = \frac {PA}{PB}\cdot\frac {\sin\widehat {ZPA}}{\sin\widehat {ZPB}} = \frac {PA}{PB}\cdot|\frac {\cos y}{\cos x}|\end{array}\|$ $ \implies$ $ \frac {XB}{XC}\cdot\frac {YC}{YA}\cdot\frac {ZA}{ZB} = 1$ $ \implies$ $ Z\in XY$ .

[b][u]Proof II (vectorial).[/u][/b] Denote $ \|\begin{array}{c} \overline {XB} = m\cdot \overline {XC} \\
 \\
\overline {YC} = n\cdot\overline {YA} \\
 \\
\overline {ZA} = p\cdot\overline {ZB}\end{array}$ . Then $ \|\begin{array}{c} \overline {AB} - m\cdot \overline {AC} = (1 - m)\cdot \overline {AX} \\
 \\
\overline {PB} - m\cdot\overline {PC} = (1 - m)\cdot \overline {PX}\end{array}$ . Therefore,

$ PA\perp PX$ $ \implies$ $ \overline {PA}\circ\overline {PX} = 0$ $ \implies$ $ \overline {PA}\circ(\overline {PB} - m\cdot\overline {PC}) = 0$ $ \implies$ $ \overline {PA}\circ\overline {PB} = m\cdot\overline {PA}\circ\overline {PC}$ .

Obtain analogously $ \overline {PB}\circ\overline {PC} = n\cdot\overline {PB}\circ\overline {PA}$ and $ \overline {PC}\circ\overline {PA} = p\cdot\overline {PC}\circ\overline {PB}$ .

From the their product obtain $ mnp = 1$ , i.e. the points $ X$ , $ Y$ , $ Z$ are collinearly.

[b][u]Proof III (metric).[/u][/b] Apply the [b]Stewart[/b]'s relation to the cevian $ AX$ in $ \triangle ABC$ and to the cevian $ PX$ in $ \triangle BPC$ :

$ \|\begin{array}{c} AB^2\cdot \overline {XC} + AC^2\cdot \overline {BX} = AX^2\cdot \overline {BC} + \overline {XC}\cdot\overline {BX}\cdot\overline {BC} \\
 \\
PB^2\cdot \overline {XC} + PC^2\cdot\overline {BX} = PX^2\cdot\overline {BC} + \overline {XC}\cdot\overline {BX}\cdot\overline {BC}\end{array}$ . Since $ AX^2 = PA^2 + PX^2$ obtain

$ AB^2\cdot \overline {XC} + AC^2\cdot \overline {BX} = PA^2\cdot \overline {BC} + (PX^2\cdot\overline {BC} + \overline {XC}\cdot\overline {BX}\cdot\overline {BC}) =$

$ PA^2\cdot\overline {BC} + PB^2\cdot\overline {XC} + PC^2\cdot\overline {BX}$ . Denote $ \overline {XB} = m\cdot\overline {XC}$ . Obtain :

$ c^2 - mb^2 = (1 - m)\cdot PA^2 + PB^2 - m\cdot PC^2$ , i.e. $ m = \frac {PA^2 + PB^2 - c^2}{PA^2 + PC^2 - b^2}$ .

For $ \overline {YC} = n\cdot \overline {YA}$ ans $ \overline {ZA} = p\cdot\overline {ZB}$ obtain analogously

$ n = \frac {PB^2 + PC^2 - a^2}{PB^2 + PA^2 - c^2}$ and $ p = \frac {PC^2 + PA^2 - b^2}{PC^2 + PB^2 - a^2}$ . From $ mnp = 1$ obtain $ Z\in XY$ .

[b][u]Remark.[/u][/b] The circles with the diameters $ [AX]$ , $ [BY]$ , $ [CZ]$ have a common point $ P$

and their centers are collinearly (the [b]Gauss' line[/b] of the complete quadrilateral $ ABCXYZ$ ).

So these circles have only two common points and same radical axis ![/color]

[b]Proof 4:[/b] Let $ \mathcal{T}$ be a pole-polar transformation (reciprocation) about an arbitrary circle with center $ P$. Let $ \mathcal{T}(BC) \equal{} A'$, $ \mathcal{T}(D) \equal{} d$ a.s.o. $ H'$ is the orthocenter of $ \triangle A'B'C'$.
\[ \begin{array}{ccccc} DP\perp AP & \implies & \mathcal{T}(D)\perp \mathcal{T}(A) & \implies & d\perp B'C' \\
D\in BC & \implies & \mathcal{T}(BC)\in \mathcal{T}(D) & \implies & A'\in d \\
\end{array} \implies H'\in d
\]
Thus $ D,E,F\in \mathcal{T}(H')$ which is a line, as desired.


I think it is from IMO 1964, so a, b, c are sides of triangle!

see here

http://www.mathlinks.ro/Forum/viewtopic.php?p=346917#p346917


Can I get a list of all Northeastern team contests?

By northeast I mean from Chicago to Baltimore to the tip of maine.


function
search
real analysis
real analysis unsolved

Let we have a continuous function $ f$ on $ [0,1]$, which is differentiable on $ (0,1)$. We know that $ f(0)\equal{}0$ and $ |f^{\prime}(x)|\leq |f(x)|$ for each $ x\in(0,1)$. Show that $ f$ is constantly equal to $ 0$.

This is a special case of Gronwall's lemma. Let $ M$ be the maximum of $ |f(x)|$ on $ [0,1/2]$. Then $ |f'(x)|\le M$ on $ (0,1/2)$, which by the Mean Value Theorem implies $ |f(x)|\le Mx\le M/2$ for all such $ f$. So, $ M\le M/2$, which is only possible if $ M \equal{} 0$.

Notice that I had to consider an interval of length less than $ 1$ to make this computation work. But I am sure you can finish from here.

See the discussion [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1030884520&t=245454]here[/url], which discusses a few other approaches.


modular arithmetic
number theory unsolved
number theory

$ a$ is positive integer number

Let $ S\equal{}a\plus{}a^{2}\plus{}a^{3}\plus{}...\plus{}a^{n}$

Prove that:

$ S\equiv 1 (mod 10)$ if and only if $ a\equiv 1 (mod 10)$ and $ n\equiv 1 (mod 10)$


number theory
least common multiple
LaTeX
algebra
polynomial

[hide="random things about the problem"]
$186=2  \cdot 3 \cdot 31.$
That sum is $\frac{5-5^{150}}{1-5}=\frac{5(5^{149}-1)}{4}.$
[/hide]

You got the denominators wrong in some way too. 1 - 5 = 4?

He divided the fraction by -1, but didn't show it

[quote="236factorial"]Hope this isn't too hard.

Prove that $5 + 5^2 + 5^3 + 5^4 + ... + 5^{150}$ is divisible by 186.[/quote]

I think this is marginally too hard, but I'll leave it here for a day or two, and we'll see what happens.  Here's a minor correction to chess64's work.

[hide]$\begin{equation}
n&=&5+5^2+5^3+5^4+\cdots+5^{150} \\
5n&=&0+5^2+5^3+5^4+\cdots+5^{150}+5^{151} \\
4n&=&-5+5^{151} \\
n&=&\frac{5^{151}-5}{4} \\
\end$[/hide]

I'll try to prove it.

We can prove the sum is divisible by 186 by checking divisibility each of its prime factors, 2, 3, and 31. If it is divisible by these three numbers, it will be divisible by 186, their LCM. Let's start by looking at the sum mod 2. 5, 5^2, 5^3,... , and 5^150 are all equal to 1 (mod 2), so 5 + 5^2 + 5^3 + 5^4... + 5^150 = 150 * 1 (mod 2) = 0 (mod 2). The sum is divisible by 2. Now let's look at the sum mod 3. 5 = 2 (mod 3.) 5^2 = 2^2 (mod 3) = 1 (mod 3). Since we now have a 1, the mods keep repeating themselves. (5 = 2 (mod 3), 5^2 = 1 (mod 3), 5^3 = 2 (mod 3), 5^4 = 1 (mod 3)...) So, 5 + 5^2 + 5^3... + 5^150 = 2 + 1 + 2 + 1... (mod 3) = 225 = 0 (mod 3). The sum is divisible by 3. Now we look at the sum mod 31. 5 = 5 (mod 31). 5^2 = 25 (mod 31). 5^3 = 125 = 1 (mod 31). Since we now have a 1, the mods keep repeating themselves. (5 = 5 (mod 31), 5^2 = 25 (mod 31), 5^3 = 1 (mod 31), 5^4 = 5 (mod 31), 5^5 = 25 (mod 31), 5^6 = 1 (mod 31)...) So, 5 + 5^2 + 5^3 + 5^4... +5^150 = 5 + 25 + 1 + 5 + 25 + 1... (mod 31) =31 * 50 = 0 (mod 31). The sum is also divisible by 31. Since the sum is divisible by 2, 3, and 31, it is divisible by their LCM, 186. QED  :D

[quote="236factorial"]Nice way, but is there another way?[/quote]

It looks messy, but it's actually a good solution.

$5+5^2+5^3+\cdots+5^{150} \mod {2}$
$\Rightarrow 1+1^2+1^3+\cdots+1^{150} \mod {2}$
$\Rightarrow 150 \mod {2}$
$\Rightarrow 0 \mod {2}$

Therefore the sum is divisible by 2.

$5+5^2+5^3+\cdots+5^{150} \mod {3}$
$\Rightarrow 2+2^2+2^3+\cdots+2^{150} \mod {3}$
$\Rightarrow 2+1+2+1+\cdots+1+2 \mod {3}$
$\Rightarrow 3 \cdot 75 \mod {3}$
$\Rightarrow 0 \mod {3}$

Therefore the sum is divisible by 3.

$5+5^2+5^3+\cdots+5^{150} \mod {31}$
$\Rightarrow 5+5^2+1+\cdots+5+5^2+1 \mod {31}$
$\Rightarrow 3(31) \mod {31}$
$\Rightarrow 0 \mod {31}$

Therefore the sum is divisible by 31.

Because the $(2,3,31)=1$, and the sum is divisible by all of these, it must also be divisible by $2 \cdot 3 \cdot 31 = 186$, as desired.

Uh, isn't that exactly what he did? Latex made it easier to understand though.
Can you do it without mods?

Well... it's easy enough to prove that the sum is divisible by $2$ and $3$, but I didn't know how to show that it was divisible by $31$. Nice pattern you found.

[quote="rcv"]Nice job, guys.  The modulo solution is suitable for MATHCOUNTS.  That factorization is wicked, and is what you might find on an AMC-12 contest.  Can it be simplified.

[hide="Here is another suggestion.  The following idea can be tied to each of the previous methods."]
Write the sum $n=5^{150}+5^{149}+5^{148}+\cdots+5^1$ in base 5.

Write the prospective factors, $a=31$ and $b=6$, in base 5.

Now what?[/hide][/quote]
Thats very nice
$5^{150}+5^{149}+5^{148}+\cdots+5^1=111...11_5$ with 150 $1$s

$31=111_5$ with 3 $1$s

$6=11_5$ with 2 $1$s

150 is divisible by both 2 and 3

Wow. The base 5 makes it a whole lot easier. I never thought of that.  :blush:  Oh, and nat mc's solution is correct. That's exactly what I thought of when I tried it.

Yes, that is very interesting, I think that's how the problem was intended to be solved. (why would they pick 186 otherwise?)

[quote="rcv"]Writing the terms in base 5 was really just a "trick" to make this particular polynomial long-division look less intimidating.  What we showed was:

$c^{150}+c^{149}+\cdots+c^1=111\ldots 110_c$ is divisible by
$c^2+c^1+c^0=111_c$ and by
$c^1+c^0=11_c$

These factorizations are a lot less wicked that those attempted previously.

Here are two modular equivalences you can easily verify with paper and pencil.
$5^2=25\equiv 1 \mod 6$ and
$5^3=125\equiv 1 \mod 31$.

It's a pretty safe bet the person who wrote the problem was aware of the above equivalences.  I think the author of the problem probably expected more solutions using modular equivalences and/or factoring than solutions using base 5.

We have probably gone about as far as we should go with this stuff in the Middle School Forums.  If you think this stuff is cool, get a book on Number Theory, and/or take this problem to the High School Forums.[/quote]
Wouldn't it be $c^{150}+c^{149}+\cdots+c^1=111\ldots 111_c$?

[quote="mathnerd314"]Just a question:

How do you know that 111...110 is divisible by both 111 and 11 in base 5?[/quote]

I've been wondering the same.  I also don't understand the following:

$5+5^2+5^3+\cdots+5^{150} \mod {3}$
$\Rightarrow 2+2^2+2^3+\cdots+2^{150} \mod {3}$

How do we know that $5^{150}$ is the same as $2^{150}$ with mod 3?

Please explain.

Here's an answer to your question. One of the properties of mods is that you can raise both sides to the same positive integer power and keep the equality. (For example, 6 = 11 (mod 5). 6^2 = 11^2 (mod 5)) So, since 5 = 2 (mod 3), 5^150 = 2^150 (mod 3).
For the other question, we have 1111... 1110 (base 5) with 150 1's. It's divisible by 11 (base 5) because 11 * 10101010...101010 (with 75 1's and 75 0's ) is our number. For 111 (base 5), 111 * 100100100...0010 (with 50 1's) is also the number. So, our number is divisible by 11 and 111 (base 5). We can also look at it another way. The larger number has 150 consecutive 1's, while the other two numbers have 2 and 3 consecutive 1's. Since 150 is divisible by 2 and 3, the larger number is divisible by the other two.


geometry
rectangle
calculus
calculus computations

You are constructing a window with the shape shown here of a rectangle of height
h surmounted by a semicircle of radius r. You must construct the exterior edge of the window,
as well as the three internal struts shown on the figure, using only 21 feet of edging. Find the
value of r maximizing the area of the window.

  ___r_______r___
  l           l            l
  l           l            l
  l______ l______ l      h
  l           l            l
  l           l            l
  l______ l______ l

My interpretation of the ASCII mess is that $ 3h\plus{}6r\plus{}\pi r\equal{}21$. You are to maximize $ 2hr\plus{}\pi r^2/2$


We draw a segment l from vertex C of the square ABCD s.t. l intersects BD at K and the perpendicular bisector of AB at L.Find angleDCK if L is on the segment CK and angleAKB=angleALB.

Is it: $45^\circ+\frac{\angle AKB}{2}$ ?

 :P

[hide]Let $P$ be on the perpendicular bisector and $\Delta APB$ be equilateral. As $\Delta PBC$ is isosceles $BP=BC$ and $\angle PBC=30$ we find $\angle BCP=75$.
Let $CP$ intersect $BD$ in $Q$ as $\angle BQC=180-45-75=60=\angle BAP$ so $ABPQ$ is inscribed and then $\angle AQB=\angle APB$ so $P\equiv K$ and $Q\equiv L$
Note that moving $P$ $\angle APB$ is decreasing and $\angle AQB$ is increasing so there is only one point $P$ for which they are equal.
So $\angle DCK=15$[/hide]


For every integer $n>1$, prove that \[\sum^n_{k=1} \frac{1}{k^2} > \frac{3n}{2n+1}\]


In the $ 6 \times 4$ grid shown, 12 of the 24 squares are to be
shaded so that there are two shaded squares in each row and three
shaded squares in each column. Let $ N$ be the number of shadings
with this property. Find the remainder when $ N$ is divided by 1000.

[asy]for (int i=0; i<7; ++i) {
draw((0,i)--(4,i));
}
for (int i=0; i<5; ++i) {
draw((i,0)--(i,6));
}[/asy]

this seems way too hard to be a ftw problem.
http://www.artofproblemsolving.com/Wiki/index.php/2007_AIME_I_Problems/Problem_10

That could be because its an Alcumus problem. :wink:


Let ABC be a triangle with in-centre I.  A point P in the interior of the triangle satisfies  ∠PBA + ∠PCA = ∠PBC + ∠PCB

Show that AP ≧ AI, and that equality holds if and only if P = I.

Pls explain to me with steps.. THZ A LOT!!

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=571966#p571966[/url]

The solutions to all the problems are here.


Euler
function
modular arithmetic
calculus
integration
number theory open
number theory

This problem is unsolved, unfortunately :).

[url]http://mathworld.wolfram.com/LehmersTotientProblem.html[/url]

[quote="_tequilla"]find all $n$ if
$n-1=0\pmod{\phi(n)}$
where $\phi(n)$ is Euler function[/quote]
Well, first if $n$ is a solution every prime has power $1$ in the integral factorization, if $p^2|n$  and $p$ prime then $p|1$ !!! so, the number is $n=p_1\cdots p_m$ with $2<p_1<p_2<\cdots <p_m$ primes, of course if $m=1$ the result holds, so a solution is $n=p$ prime,  now i am working in the genral case, maybe someone write more hints.t In fact
\[
{\rm If } \quad p_1\cdots p_m-1=\lambda (p_1-1)\cdots (p_m-1) \quad \Rightarrow \quad \lambda<m.
\]

Good luck.

[quote="RobertuX"]maybe someone write more hints.[/quote]

Grobber wrote this is an open problem (known as Lehmer's conjecture). I think it is quite a definitive hint...

Pierre.

[quote="harazi"]Pierre! Stop being against the research!  ;)[/quote]
I found that in

Honsberger, R. Mathematical Gems II. Washington, DC: Math. Assoc. Amer., p. 35, 1976.

he proves that NON-composite solution for this equation, so the unique solution are the PRIMES numbers.

Best regards.

RX

Pff.... :?  :( ...
He only says that until now, no composite number which would be a solution has been found.

Pierre.

[quote="pbornsztein"]Have a look here :

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

Pierre.[/quote]
I had other references, but maybe the guy of the web I saw was wrong. Anyway I feel closed to a nice proof, when I had I will post here.


inequalities quadratics function conics parabola algebra

Using Vasc's ineq, we only need to prove:
$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {27abc}{(a \plus{} b \plus{} c)^3} \geq\ 2$
<=> $ \frac {9}{q} \plus{} r \geq\ 4$
(trues with $ q /leq/ 2.25$)
With $ q \geq\ 2.25$, using Schur:
LHS $ \geq\ \frac {9}{q} \plus{} \frac {4q \minus{} 9}{3} \geq\ 4$
Equality occurs when $ a \equal{} b \equal{} c$ or $ abc \equal{} 0.$
Njce ineq, my Bro.

---------------
Ak, sorry, I was wrong. I 'll try it again   :oops:

[quote="nguoivn"]
LHS $ \geq\ \frac {9}{q} \plus{} \frac {4q \minus{} 9}{3} \geq\ 4$
[/quote]
This inequality does not hold, nguoivn. ;)

thanks you shaam .
SOS kill iy very Easy .
Assume $ a \ge b \ge c$ we have :
$ Sa \ge 0 , Sc \ge 0$
We only need to prove :
$ 2Sb \plus{} Sa \ge 0 and 2Sb \plus{} Sa \ge 0$
It easy to prove !

[quote="tin_124"]thanks you shaam .
SOS kill iy very Easy .
Assume $ a \ge b \ge c$ we have :
$ Sa \ge 0 , Sb \ge 0$
We only need to prove :
$ 2Sb \plus{} Sa \ge 0 and 2Sc \plus{} Sa \ge 0$
It easy to prove ![/quote]

Thanks for clarifying. But why $ Sb \ge 0$, i think you assume $ Sb \equal{}2b^2c \minus{} abc$ but, that is not always greater than zero.
Can you explain in detail? thanks

ok , sr . I edited .We can prove it by this way .
$ Sa ,Sc \ge o$ .
$ Sa\plus{}Sb , Sc\plus{}Sb \ge o$ /
We have done !

Thank you Tin. I think i am understanding this tecnique! :)

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {4abc}{a^2b \plus{} b^2c \plus{} c^2a \plus{} abc} \ge 2.
\]
:)[/quote]

My proof. :)

I've edit my post because the 1st proof of the inequality
\[ \frac{{a^2  \plus{} b^2  \plus{} c^2 }}{{ab \plus{} bc \plus{} ca}} \ge \frac{{2\left( {a^2  \plus{} c^2 } \right)}}{{\left( {a \plus{} c} \right)^2 }}
\]
is so ... humorous.  :oops:

[quote="Honey_S"][quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {4abc}{a^2b \plus{} b^2c \plus{} c^2a \plus{} abc} \ge 2.
\]
:)[/quote]

My proof. :)

I've edit my post because the 1st proof of the inequality
\[ \frac {{a^2 \plus{} b^2 \plus{} c^2 }}{{ab \plus{} bc \plus{} ca}} \ge \frac {{2\left( {a^2 \plus{} c^2 } \right)}}{{\left( {a \plus{} c} \right)^2 }}
\]
is so ... humorous.  :oops:[/quote]
Remember that the inequality is cyclic and not symmetric, so you can't assume wlog that $ a \ge b \ge c$.

[quote="Mathias_DK"]
Remember that the inequality is cyclic and not symmetric, so you can't assume wlog that $ a \ge b \ge c$.[/quote]

If you don't want to assume $ a \ge b \ge c$, then you can assume that $ b$ is the number between $ a$ and $ c$. :)

But here I think I can assume $ a \ge b \ge c$, because if $ c \ge b \ge a$ then we have
\[ a^2b \plus{} b^2c \plus{} c^2a \le ab^2 \plus{} bc^2 \plus{} ca^2
\]
Hence
\[ \frac {{4abc}}{{a^2 b \plus{} b^2 c \plus{} c^2 a \plus{} abc}} \ge \frac {{4a}}{{ab^2 \plus{} bc^2 \plus{} ca^2 \plus{} abc}}
\]
So with the inequality
\[ \frac {{a^2 \plus{} b^2 \plus{} c^2 }}{{ab \plus{} bc \plus{} ca}} \plus{} \frac {{4abc}}{{a^2 b \plus{} b^2 c \plus{} c^2 a \plus{} abc}} \ge 2
\]
I think we can do that.  :)

[quote="shaaam"]I have obtained that:

$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {4abc}{a^2b \plus{} b^2c \plus{} c^2a \plus{} abc} \minus{} 2 \equal{}$

$ \frac {(a \minus{} b)^2(2ab^2 \minus{} abc) \plus{} (b \minus{} c)^2(2bc^2 \minus{} abc) \plus{} (c \minus{} a)^2(2ca^2 \minus{} abc)}{2(ab \plus{} bc \plus{} ca)(a^2b \plus{} b^2c \plus{} c^2a \plus{} abc)}\geq 0$

Can, is it right?

P.S: i don't know how to use SOS method, where you assume $ a\ge b\ge c$... can someone help me?[/quote]
Sorry, but it is wrong, shaam. It must be
\[ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {4abc}{a^2b \plus{} b^2c \plus{} c^2a \plus{} abc} \minus{}2 \equal{} \frac {(a \minus{} b)^2(2a^2b \minus{} abc) \plus{} (b \minus{} c)^2(2b^2c \minus{} abc) \plus{} (c \minus{} a)^2(2c^2a \minus{} abc)}{2(ab \plus{} bc \plus{} ca)(a^2b \plus{} b^2c \plus{} c^2a \plus{} abc)}
\]
And I think it is hard for us to use SOS here.

Anyways, the solution of Honey_suck is nice and correct. It's also mine.:)

[quote="shaaam"]Sorry Can... it was typo.  :blush:
But can you explain why SOS method cannot be used?[/quote]
In my opinion, SOS maybe use here but I think it will be hard. The reason is:
One thing we can see it is enough to consider the inequality in the case $ a \ge b \ge c$, but in this case only $ S_c \ge 0$ while $ S_a , S_b$ are not sure to be nonnegative. That's why I think SOS will be hard to apply.

Your explanation is appreciated!  :)
How about the following inequality:
$ a,b,c \geq 0$
\[ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {4a^2b^2c^2}{a^5b \plus{} b^5c \plus{} c^5a\plus{}a^2b^2c^2} \geq 2
\]

[b]EDIT[/b]: Thanks Can, for clarifying.I had some guesses that it will not be right , because degree of $ abc$ is three, whereas degree of $ a^3b$ is 4. I edited.. I hope this one is true.

[quote="shaaam"]Your explanation is appreciated!  :)
How about the following inequality:
$ a,b,c \geq 0$
\[ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {5abc}{a^3b \plus{} b^3c \plus{} c^3a \plus{} abc \plus{} 1} \geq 2
\]
[/quote]
I think we need some hypothesis for this inequality. If not, it will not hold, for example $ a\equal{}b\equal{}c\equal{}2009^{\minus{}1}$. ;)

I think the following inequality holds:
Given $ a,b,c\geq0$, then:
\[ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\plus{}\frac{2abc}{a^2b\plus{}b^2c\plus{}c^2a\plus{}abc} \geq 2
\]

[quote="nguoivn"]Yes, but it's weaker than can_hang's problem.
Njce hint, can_hang and Honey_S  :)[/quote]

But why it is weaker?

As i have known:
\[ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\geq \frac{2}{3}\cdot(\frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b})
\]
Thanks

[quote="shaaam"]I think the following inequality holds:
Given $ a,b,c\geq0$, then:
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {2abc}{a^2b \plus{} b^2c \plus{} c^2a \plus{} abc} \geq 2
\]
[/quote]
It is weaker than my inequality because of using my inequality
\[ \frac {2abc}{a^2b \plus{} b^2c \plus{} c^2a \plus{} abc} \ge 1 \minus{} \frac {a^2 \plus{} b^2 \plus{} c^2}{2(ab \plus{} bc \plus{} ca)}
\]
We just need to prove that
\[ \sum \frac {a}{b \plus{} c} \plus{} 1 \minus{} \frac {a^2 \plus{} b^2 \plus{} c^2}{2(ab \plus{} bc \plus{} ca)} \ge 2
\]

\[ \sum \frac {a}{b \plus{} c} \ge \frac {(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca)}
\]
which is just Cauchy Schwarz. :)

Anyways, we have the stronger one (of 10maths):
\[ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b} \plus{}\frac{3abc}{2(a^2b\plus{}b^2c\plus{}c^2a)} \ge 2.\]
:)

This is also nice.

If $ a,b,c$ are nonnegative real numbers, then

$ \frac {4(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca} \plus{} \frac {9abc}{a^2b \plus{} b^2c \plus{} c^2a} \ge 7$.

May be, we can using: $ a^2b \plus{} b^2c \plus{} c^2a \leq\ 4 \minus{} r$
Is it trues ?  :wink:

[quote="Allnames"][quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {4abc}{a^2b \plus{} b^2c \plus{} c^2a \plus{} abc} \ge 2.
\]
:)[/quote]
Sorry but my work is not enough
With $ b\ge a\ge c$ $ LHS \minus{} RHS \equal{} M(a \minus{} b)^2 \plus{} N(a \minus{} c)(b \minus{} c)$ where $ M \equal{} \frac {1}{ab \plus{} bc \plus{} ca} \minus{} \frac {c}{a^2b \plus{} b^2c \plus{} c^2a \plus{} abc}$ and $ N \equal{} \frac {1}{ab \plus{} bc \plus{} ca} \minus{} \frac {b}{a^2b \plus{} b^2c \plus{} c^2a \plus{} abc}$
so $ M,N \ge 0$
with $ a\ge b\ge c$ I cant continue
I think the inequality $ (a^2b \plus{} b^2c \plus{} c^2a)(ab \plus{} bc \plus{} ca)\le \frac {a \plus{} b \plus{} c)^5}{27}$ is useful but I failed a gain  :(
It is really hard[/quote]

You almost found the solution, I will complete your solution for you :)
Assume that $ c\equal{}min(a,b,c)$. We have that:
\[ LHS\minus{}RHS\equal{}M_{1}(a\minus{}b)^2\plus{}N_{1}(a\minus{}c)(b\minus{}c)^2\]
In which:
\[ M_{1}\equal{}\dfrac{a^2b\plus{}b^2c\minus{}bc^2}{(a^2b\plus{}b^2\plus{}c^2a\plus{}abc)(ab\plus{}bc\plus{}ca)}\]
\[ N_{1}\equal{}\dfrac{b^2c\plus{}c^2a\minus{}a^2c}{(a^2b\plus{}b^2\plus{}c^2a\plus{}abc)(ab\plus{}bc\plus{}ca)}\]
Therefore, we need to prove the following inequality:
\[ (a^2b\plus{}b^2c\minus{}bc^2)(a\minus{}b)^2\plus{}(b^2c\plus{}c^2a\minus{}a^2c)(a\minus{}c)(b\minus{}c) \ge 0 \qquad (1)\]
The interesting thing is that we have the following equality:
\[ \minus{}P(a\minus{}c)(b\minus{}c)(a\minus{}b)^2\plus{}P(a\minus{}b)^2(a\minus{}c)(b\minus{}c) \equal{} 0\]
For $ P\equal{}(a\plus{}3c)$ we have:
\[ \minus{}(a\plus{}3c)(a\minus{}c)(b\minus{}c)(a\minus{}b)^2\plus{}(a\plus{}3c)(a\minus{}b)^2(a\minus{}c)(b\minus{}c) \equal{} 0\]
Thus, $ (1)$ is equivalent to:
\[ M_{2}(a\minus{}b)^2\plus{}N_{2}(a\minus{}c)(b\minus{}c) \ge 0\]
For:
\[ M_{2}\equal{}a^2b\plus{}b^2c\minus{}bc^2\minus{}(a\plus{}3c)(a\minus{}c)(b\minus{}c)\equal{}c\left[(a\minus{}b)^2\plus{}2ac\plus{}2bc\minus{}3c^2\right] \ge 0\]
\[ N_{2}\equal{}b^2c\plus{}c^2a\minus{}a^2c\plus{}(a\plus{}3c)(a\minus{}b)^2\equal{}ac^2\plus{}2(a\minus{}b)(a\minus{}2b)c\plus{}a(a\minus{}b)^2\]
If $ a \le b$ then $ N_{2} \ge 0$
If $ a \ge b$ then $ N_{2}\equal{}f(c)\equal{}ac^2\plus{}2(a\minus{}b)(a\minus{}2b)c\plus{}a(a\minus{}b)^2$
Notice that:
\[ \Delta^{'} \equal{}(a\minus{}b)^2(a\minus{}2b)^2\minus{}a^2(a\minus{}b)^2\equal{}\minus{}4b(a\minus{}b)^3 \le 0\]
Hence, we have $ N \ge 0$. The solution is comptele.

[quote="10maths_tp0609"][quote="Allnames"][quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {4abc}{a^2b \plus{} b^2c \plus{} c^2a \plus{} abc} \ge 2.
\]
:)[/quote]
Sorry but my work is not enough
With $ b\ge a\ge c$ $ LHS \minus{} RHS \equal{} M(a \minus{} b)^2 \plus{} N(a \minus{} c)(b \minus{} c)$ where $ M \equal{} \frac {1}{ab \plus{} bc \plus{} ca} \minus{} \frac {c}{a^2b \plus{} b^2c \plus{} c^2a \plus{} abc}$ and $ N \equal{} \frac {1}{ab \plus{} bc \plus{} ca} \minus{} \frac {b}{a^2b \plus{} b^2c \plus{} c^2a \plus{} abc}$
so $ M,N \ge 0$
with $ a\ge b\ge c$ I cant continue
I think the inequality $ (a^2b \plus{} b^2c \plus{} c^2a)(ab \plus{} bc \plus{} ca)\le \frac {a \plus{} b \plus{} c)^5}{27}$ is useful but I failed a gain  :(
It is really hard[/quote]

You almost found the solution, I will complete your solution for you :)

Assume that $ c \equal{} min(a,b,c)$. We have that:
\[ LHS \minus{} RHS \equal{} M_{1}(a \minus{} b)^2 \plus{} N_{1}(a \minus{} c)(b \minus{} c)^2
\]
In which:
\[ M_{1} \equal{} \dfrac{a^2b \plus{} b^2c \minus{} bc^2}{(a^2b \plus{} b^2 \plus{} c^2a \plus{} abc)(ab \plus{} bc \plus{} ca)}
\]

\[ N_{1} \equal{} \dfrac{b^2c \plus{} c^2a \minus{} a^2c}{(a^2b \plus{} b^2 \plus{} c^2a \plus{} abc)(ab \plus{} bc \plus{} ca)}
\]
Therefore, we need to prove the following inequality:
\[ (a^2b \plus{} b^2c \minus{} bc^2)(a \minus{} b)^2 \plus{} (b^2c \plus{} c^2a \minus{} a^2c)(a \minus{} c)(b \minus{} c) \ge 0 \qquad (1)
\]
The interesting thing is that we have the following equality:
\[ \minus{} P(a \minus{} c)(b \minus{} c)(a \minus{} b)^2 \plus{} P(a \minus{} b)^2(a \minus{} c)(b \minus{} c) \equal{} 0
\]
For $ P \equal{} (a \plus{} 3c)$ we have:
\[ \minus{} (a \plus{} 3c)(a \minus{} c)(b \minus{} c)(a \minus{} b)^2 \plus{} (a \plus{} 3c)(a \minus{} b)^2(a \minus{} c)(b \minus{} c) \equal{} 0
\]
Thus, $ (1)$ is equivalent to:
\[ M_{2}(a \minus{} b)^2 \plus{} N_{2}(a \minus{} c)(b \minus{} c) \ge 0
\]
For:
\[ M_{2} \equal{} a^2b \plus{} b^2c \minus{} bc^2 \minus{} (a \plus{} 3c)(a \minus{} c)(b \minus{} c) \equal{} c\left[(a \minus{} b)^2 \plus{} 2ac \plus{} 2bc \minus{} 3c^2\right] \ge 0
\]

\[ N_{2} \equal{} b^2c \plus{} c^2a \minus{} a^2c \plus{} (a \plus{} 3c)(a \minus{} b)^2 \equal{} ac^2 \plus{} 2(a \minus{} b)(a \minus{} 2b)c \plus{} a(a \minus{} b)^2
\]
If $ a \le b$ then $ N_{2} \ge 0$
If $ a \ge b$ then $ N_{2} \equal{} f(c) \equal{} ac^2 \plus{} 2(a \minus{} b)(a \minus{} 2b)c \plus{} a(a \minus{} b)^2$
Notice that:
\[ \Delta^{'} \equal{} (a \minus{} b)^2(a \minus{} 2b)^2 \minus{} a^2(a \minus{} b)^2 \equal{} \minus{} 4b(a \minus{} b)^3 \le 0
\]
Hence, we have $ N \ge 0$. The solution is comptele.[/quote]

This solution is nice, but very hard in my opinion. Does there exist another solution for this? A bit more natural...

btw, in my attempts to solve this inequality, I proved it only that for the case $ ab^2 \plus{} bc^2 \plus{} ca^2\plus{}\dfrac{3}{4}abc\ge a^2b \plus{} b^2c \plus{} c^2a$. If anyone can find a simple proof for other case, just let me know, please.

[quote="can_hang2007"]Another nice inequality is
\[ \frac {2(a^2 \plus{} b^2 \plus{} c^2)}{(a \plus{} b \plus{} c)^2} \plus{} \frac {abc}{a^2b \plus{} b^2c \plus{} c^2a} \ge 1.
\]
This inequality is stronger than 10maths's one. :)[/quote]

After brute force, one can easily can find that:

$ \frac {2(a^2 \plus{} b^2 \plus{} c^2)}{(a \plus{} b \plus{} c)^2} \plus{} \frac {abc}{a^2b \plus{} b^2c \plus{} c^2a} \minus{} 1 \equal{} \frac{(a\minus{}b)^2(2a^2b\minus{}abc)\plus{}(b\minus{}c)^2(2b^2c\minus{}abc)\plus{}(c\minus{}a)^2(2c^2a\minus{}abc)}{2(a\plus{}b\plus{}c)^2(a^2b\plus{}b^2c\plus{}c^2a)}$

Rather funny? because if you look at the 13th post (in first page), you will see Can_Hang's original inequality also has the same numerator. That is why, we should prove the following hard statement. if we do, we have solutions for both inequalities. So, this is our challange: Prove that:
Given that $ a,b,c \geq 0$
\[ (a\minus{}b)^2(2a^2b\minus{}abc)\plus{}(b\minus{}c)^2(2b^2c\minus{}abc)\plus{}(c\minus{}a)^2(2c^2a\minus{}abc) \geq 0
\]

[quote="shaaam"][quote="can_hang2007"]Another nice inequality is
\[ \frac {2(a^2 \plus{} b^2 \plus{} c^2)}{(a \plus{} b \plus{} c)^2} \plus{} \frac {abc}{a^2b \plus{} b^2c \plus{} c^2a} \ge 1.
\]
This inequality is stronger than 10maths's one. :)[/quote]

After brute force, one can easily can find that:

$ \frac {2(a^2 \plus{} b^2 \plus{} c^2)}{(a \plus{} b \plus{} c)^2} \plus{} \frac {abc}{a^2b \plus{} b^2c \plus{} c^2a} \minus{} 1 \equal{} \frac {(a \minus{} b)^2(2a^2b \minus{} abc) \plus{} (b \minus{} c)^2(2b^2c \minus{} abc) \plus{} (c \minus{} a)^2(2c^2a \minus{} abc)}{2(a \plus{} b \plus{} c)^2(a^2b \plus{} b^2c \plus{} c^2a)}$

Rather funny? because if you look at the 13th post (in first page), you will see Can_Hang's original inequality also has the same numerator. That is why, we should prove the following hard statement. if we do, we have solutions for both inequalities. So, this is our challange: Prove that:
Given that $ a,b,c \geq 0$
\[ (a \minus{} b)^2(2a^2b \minus{} abc) \plus{} (b \minus{} c)^2(2b^2c \minus{} abc) \plus{} (c \minus{} a)^2(2c^2a \minus{} abc) \geq 0
\]
[/quote]
Shaam, actually, the inequality
\[ \frac{2(a^2\plus{}b^2\plus{}c^2)}{(a\plus{}b\plus{}c)^2} \plus{}\frac{abc}{a^2b\plus{}b^2c\plus{}c^2a} \ge 1\]
is stronger than the inequality
\[ \frac{a}{b\plus{}c} \plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b} \plus{}\frac{3abc}{2(a^2b\plus{}b^2c\plus{}c^2a)} \ge 2\]
It is because we have the following known result (which is a symmetric-easy)
\[ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b} \ge \frac{3(a^2\plus{}b^2\plus{}c^2)}{(a\plus{}b\plus{}c)^2} \plus{}\frac{1}{2}.\]

True. But i meant in previous post that, both of the followings have same strength:
\[ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {4abc}{a^2b \plus{} b^2c \plus{} c^2a \plus{} abc} \ge 2
\]

\[ \frac {2(a^2 \plus{} b^2 \plus{} c^2)}{(a \plus{} b \plus{} c)^2} \plus{} \frac {abc}{a^2b \plus{} b^2c \plus{} c^2a} \ge 1.
\]
because in both we should prove:
\[ (a \minus{} b)^2(2a^2b \minus{} abc) \plus{} (b \minus{} c)^2(2b^2c \minus{} abc) \plus{} (c \minus{} a)^2(2c^2a \minus{} abc) \geq 0
\]
Which is very hard  :(

[quote="10maths_tp0609"]You almost found the solution, I will complete your solution for you :)
Assume that $ c \equal{} min(a,b,c)$. We have that:
\[ LHS \minus{} RHS \equal{} M_{1}(a \minus{} b)^2 \plus{} N_{1}(a \minus{} c)(b \minus{} c)^2
\]
In which:
\[ M_{1} \equal{} \dfrac{a^2b \plus{} b^2c \minus{} bc^2}{(a^2b \plus{} b^2 \plus{} c^2a \plus{} abc)(ab \plus{} bc \plus{} ca)}
\]

\[ N_{1} \equal{} \dfrac{b^2c \plus{} c^2a \minus{} a^2c}{(a^2b \plus{} b^2 \plus{} c^2a \plus{} abc)(ab \plus{} bc \plus{} ca)}
\]
Therefore, we need to prove the following inequality:
\[ (a^2b \plus{} b^2c \minus{} bc^2)(a \minus{} b)^2 \plus{} (b^2c \plus{} c^2a \minus{} a^2c)(a \minus{} c)(b \minus{} c) \ge 0 \qquad (1)
\]
The interesting thing is that we have the following equality:
\[ \minus{} P(a \minus{} c)(b \minus{} c)(a \minus{} b)^2 \plus{} P(a \minus{} b)^2(a \minus{} c)(b \minus{} c) \equal{} 0
\]
For $ P \equal{} (a \plus{} 3c)$ we have:
\[ \minus{} (a \plus{} 3c)(a \minus{} c)(b \minus{} c)(a \minus{} b)^2 \plus{} (a \plus{} 3c)(a \minus{} b)^2(a \minus{} c)(b \minus{} c) \equal{} 0
\]
Thus, $ (1)$ is equivalent to:
\[ M_{2}(a \minus{} b)^2 \plus{} N_{2}(a \minus{} c)(b \minus{} c) \ge 0
\]
For:
\[ M_{2} \equal{} a^2b \plus{} b^2c \minus{} bc^2 \minus{} (a \plus{} 3c)(a \minus{} c)(b \minus{} c) \equal{} c\left[(a \minus{} b)^2 \plus{} 2ac \plus{} 2bc \minus{} 3c^2\right] \ge 0
\]

\[ N_{2} \equal{} b^2c \plus{} c^2a \minus{} a^2c \plus{} (a \plus{} 3c)(a \minus{} b)^2 \equal{} ac^2 \plus{} 2(a \minus{} b)(a \minus{} 2b)c \plus{} a(a \minus{} b)^2
\]
If $ a \le b$ then $ N_{2} \ge 0$
If $ a \ge b$ then $ N_{2} \equal{} f(c) \equal{} ac^2 \plus{} 2(a \minus{} b)(a \minus{} 2b)c \plus{} a(a \minus{} b)^2$
Notice that:
\[ \Delta^{'} \equal{} (a \minus{} b)^2(a \minus{} 2b)^2 \minus{} a^2(a \minus{} b)^2 \equal{} \minus{} 4b(a \minus{} b)^3 \le 0
\]
Hence, we have $ N \ge 0$. The solution is comptele.[/quote]
Very nice your solution, 10maths. I like the identity which you used in your solution. It looks like the following identity which I often use to solve a problem with Vornicu Schur:
\[ P \equal{} (b \minus{} c)^2(a \minus{} b)(a \minus{} c) \plus{} (c \minus{} a)^2(b \minus{} c)(b \minus{} a) \plus{} (a \minus{} b)^2(c \minus{} a)(c \minus{} b) \equal{} 0
\]
It is because sometimes we need to prove that
\[ X(a \minus{} b)(a \minus{} c) \plus{} Y(b \minus{} c)(b \minus{} a) \plus{} Z(c \minus{} a)(c \minus{} b) \ge 0
\]
But we arenot sure that $ X,Y,Z \ge 0$, then we can apply the above identity to rewrite the inequality as
\[ X_1 (a \minus{} b)(a \minus{} c) \plus{} Y_1 (b \minus{} c)(b \minus{} a) \plus{} Z_1 (c \minus{} a)(c \minus{} b) \ge 0
\]
where $ X_1,Y_1,Z_1 \ge 0$.

That's my technique on using Vornicu Schur. Using this technique, the following inequalities are very easy:

1. (IMO 1983) $ a,b,c$ are the side lengths of a triangle, then $ a^2b(a \minus{} b) \plus{} b^2c(b \minus{} c) \plus{} c^2a(c \minus{} a) \ge 0.$
2. (VasC) $ a,b,c$ are the side lengths of a triangle, then $ 3\left( \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \right) \ge 2\left( \frac {b}{a} \plus{} \frac {c}{b} \plus{} \frac {a}{c}\right) \plus{} 3.$
3. (Iran 96) $ a,b,c \ge 0$, then $ (ab \plus{} bc \plus{} ca) \left( \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2} \plus{} \frac {1}{(a \plus{} b)^2}\right) \ge \frac {9}{4}.$

Etc. :)

[quote="can_hang2007"]
Very nice your solution, 10maths. I like the identity which you used in your solution. It looks like the following identity which I often use to solve a problem with Vornicu Schur:
\[ P \equal{} (b \minus{} c)^2(a \minus{} b)(a \minus{} c) \plus{} (c \minus{} a)^2(b \minus{} c)(b \minus{} a) \plus{} (a \minus{} b)^2(c \minus{} a)(c \minus{} b) \equal{} 0
\]
It is because sometimes we need to prove that
\[ X(a \minus{} b)(a \minus{} c) \plus{} Y(b \minus{} c)(b \minus{} a) \plus{} Z(c \minus{} a)(c \minus{} b) \ge 0
\]
But we arenot sure that $ X,Y,Z \ge 0$, then we can apply the above identity to rewrite the inequality as
\[ X_1 (a \minus{} b)(a \minus{} c) \plus{} Y_1 (b \minus{} c)(b \minus{} a) \plus{} Z_1 (c \minus{} a)(c \minus{} b) \ge 0
\]
where $ X_1,Y_1,Z_1 \ge 0$.

That's my technique on using Vornicu Schur. Using this technique, the following inequalities are very easy:

1. (IMO 1983) $ a,b,c$ are the side lengths of a triangle, then $ a^2b(a \minus{} b) \plus{} b^2c(b \minus{} c) \plus{} c^2a(c \minus{} a) \ge 0.$
2. (VasC) $ a,b,c$ are the side lengths of a triangle, then $ 3\left( \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \right) \ge 2\left( \frac {b}{a} \plus{} \frac {c}{b} \plus{} \frac {a}{c}\right) \plus{} 3.$
3. (Iran 96) $ a,b,c \ge 0$, then $ (ab \plus{} bc \plus{} ca) \left( \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2} \plus{} \frac {1}{(a \plus{} b)^2}\right) \ge \frac {9}{4}.$

Etc. :)[/quote]

This identity is first published in Secrets in inequalities (to my knowlegde), but at that time this identity seemed to have no effect other than explaining thats there's many $ S_a,S_b,S_c$. However, by using the similar indentity, the technique to prove inequalities I used is strengthened a lot (I have posted my idea in one of my article, but no one seemed to notice that). One evidence is that it can handle some inequalities where the equality occurs not when two of three variables are equal :)
Let me consider an example:
$ a,b,c$ are three sie lenghs of a triange. Find the greatest constant $ k$ :
\[ (k\plus{}1)(a^3b\plus{}b^3c\plus{}c^3a) \ge k(ab^3\plus{}bc^3\plus{}ca^3)\plus{}abc(a\plus{}b\plus{}c)\]
Assume that $ c\equal{}max(a,b,c)$
By replacing $ c \equal{} a\plus{}b$, we can foind that $ k \le k_{0} \equal{}\minus{}\dfrac{1}{2}\plus{}\dfrac{1}{8}\sqrt{94\plus{}34\sqrt{17}}$. We will prove that this is the desired value. Rewrite the inequalities as the form:
$ (ac\plus{}bc)(a\minus{}b)^2\plus{}\left[(k_{0}\plus{}1)(a^2\plus{}ac)\minus{}k_{0}(b^2\plus{}bc)\right](a\minus{}c)(b\minus{}c) \ge 0$
Similarly, we have:
\[ \minus{}P(a\minus{}c)(b\minus{}c)(a\minus{}b)^2\plus{}P(a\minus{}b)^2(a\minus{}c)(b\minus{}c)\equal{}0\]
We can 'force' $ M$ to be equal to $ 0$ when $ c\equal{}a\plus{}b$ by choosing $ P\equal{}\dfrac{(a\plus{}b)^2}{ab}$
Therefore, we need to prove:
\[ M(a\minus{}b)^2\plus{}N(a\minus{}c)(b\minus{}c) \ge 0\]
In which:
\[ M\equal{}ac\plus{}bc\minus{}\dfrac{(a\plus{}b)^2(a\minus{}c)(b\minus{}c)}{ab}\equal{}\dfrac{(a\plus{}b)(a\plus{}b\minus{}c)(ac\plus{}bc\minus{}ab)} \ge 0\]
\[ N\equal{}(k_{0}\plus{}1)(a^2\plus{}ac)\minus{}k_{0}(b^2\plus{}bc)\plus{}\dfrac{(a\plus{}b)^2(a\minus{}b)^2}{ab}\]
\[ \Leftrightarrow N\equal{}\dfrac{a^4\plus{}(k_{0}\plus{}1)a^3b\minus{}2a^2b^2\minus{}k_{0}ab^3\plus{}b^4\plus{}abc[(k_{0}\plus{}1)a\minus{}k_{0}b]}{ab}\]
If $ (k_{0}\plus{}1)a \ge k_{0}b$ then $ N.ab \ge a^4\plus{}(k_{0}\plus{}1)a^3b\minus{}2a^2b^2\minus{}k_{0}ab^3\plus{}b^4$
If $ (k_{0}\plus{}1)a \le k_{0}b$ then $ N.ab \ge a^4\plus{}(k_{0}\plus{}1)a^3b\minus{}2a^2b^2\minus{}k_{0}ab^3\plus{}b^4\plus{}ab(a\plus{}b)[(k_{0}\plus{}1)a\minus{}k_{0}b]$
It is easy to check that $ N$ is non-negative for both cases. This ends the solution.:)
...
Your technique also may do those problem easily, but I still think my technique is a little nicer, as to find $ M\equal{}N\equal{}0$ is a bit easier than to find $ X\equal{}Y\equal{}Z\equal{}0$.
Some other inequalities of mine. I and one of my friends found out about those more than one year ago, thus, I don't know whether or not it maybe the same as some inequalities which have already been posted. :)
Greatest constant:
$ \dfrac{a^3\plus{}b^3\plus{}c^3}{a^2b\plus{}b^2c\plus{}c^2a}\minus{}1 \ge k\left(\dfrac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}\minus{}1\right)$

Other:
$ k\left(\dfrac{3(a^2b\plus{}b^2c\plus{}c^2a)}{(ab\plus{}bc\plus{}ca)(a\plus{}b\plus{}c)}\minus{}1\right) \ge \dfrac{a}{a\plus{}b}\plus{}\dfrac{b}{b\plus{}c}\plus{}\dfrac{c}{c\plus{}a}$ for $ \dfrac{1}{4} \le k \le \dfrac{1}{2}$
$ \dfrac{6(a^2\plus{}b^2\plus{}c^2)}{(a\plus{}b\plus{}c)^2}\plus{}\dfrac{8abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} \ge 3$
$ \dfrac{3(a^3\plus{}b^3\plus{}c^3)}{(a^2\plus{}b^2\plus{}c^2)(a\plus{}b\plus{}c)}\plus{}4\left(\dfrac{a}{a\plus{}b}\plus{}\dfrac{b}{b\plus{}c}\plus{}\dfrac{c}{c\plus{}a}\right) \ge 7$
$ \dfrac{a\plus{}b}{b\plus{}c}\plus{}\dfrac{b\plus{}c}{c\plus{}a}\plus{}\dfrac{c\plus{}a}{a\plus{}b}\plus{}k\dfrac{a^2b\plus{}b^2c\plus{}c^2a}{ab^2\plus{}bc^2\plus{}ca^2} \ge 3\plus{}k$ for $ k\equal{}\dfrac{3\sqrt{3}}{2}\minus{}1$
$ \dfrac{a\plus{}b}{b\plus{}c}\plus{}\dfrac{b\plus{}c}{c\plus{}a}\plus{}\dfrac{c\plus{}a}{a\plus{}b}\plus{}k\dfrac{a^3b\plus{}b^3c\plus{}c^3a}{a^2b^2\plus{}b^2c^2\plus{}c^2a^2} \ge 3\plus{}k$ for $ k\equal{}\dfrac{3\sqrt{3}}{2}\minus{}1$
$ \dfrac{a}{b\plus{}c}\plus{}\dfrac{b}{c\plus{}a}\plus{}\dfrac{c}{a\plus{}b} \ge k\left(\dfrac{a^2b\plus{}b^2c\plus{}c^2a}{ab^2\plus{}bc^2\plus{}ca^2}\minus{}1\right)\plus{}\dfrac{3}{2}$ for $ k\equal{}\dfrac{2\sqrt{2}\minus{}1}{2}$
$ k\left(\dfrac{a^3b\plus{}b^3c\plus{}c^3a}{a^2b^2\plus{}b^2c^2\plus{}c^2a^2}\minus{}1\right) \ge \dfrac{a}{a\plus{}b}\plus{}\dfrac{b}{b\plus{}c}\plus{}\dfrac{c}{c\plus{}a}\minus{}\dfrac{3}{2}$ for $ k\equal{}\dfrac{1}{4}$ ($ k\equal{}\dfrac{1}{4}$ is the only number for inequality to be valid :) )
$ \dfrac{a}{a\plus{}b}\plus{}\dfrac{b}{b\plus{}c}\plus{}\dfrac{c}{c\plus{}a}\minus{}\dfrac{3}{2} \ge k\left(\dfrac{3abc}{a^2b\plus{}b^2c\plus{}c^2a}\minus{}1\right)$ for $ k \ge \dfrac{1}{2}$
etc....

Anyway, using this technique, we can also prove that the minimum of the expression:
\[ P\equal{}\dfrac{a\plus{}kb}{a\plus{}kc}\plus{}\dfrac{b\plus{}kc}{b\plus{}ka}\plus{}\dfrac{c\plus{}ka}{c\plus{}kb}\]
holds for $ a\equal{}b\equal{}c$ or $ b\equal{}1,c\equal{}0$ :)

Hope that you will have fun with this technique.:). Sorry for posting the thing not related to the topic. :oops:.

[quote="shaaam"][quote="shaaam"][quote="can_hang2007"]
Anyways, we have the stronger one (of 10maths):
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {3abc}{2(a^2b \plus{} b^2c \plus{} c^2a)} \ge 2.
\]
:)[/quote]

I think i solved this inequality by the Loc's (Honey_S) way. But it finishes so ugly  :(
Here it goes. Assume that, $ a\ge b \ge c$ or $ b$ is the number between $ a$ and $ c$.
Observe that:
\[ (a^2b \plus{} abc \plus{} c^2b) \minus{} (a^2b \plus{} b^2c \plus{} c^2a) \equal{} c(a \minus{} b)(b \minus{} c)\geq 0
\]
Therefore we have:
\[ a^2b \plus{} abc \plus{} c^2b \geq a^2b \plus{} b^2c \plus{} c^2a
\]
or Equivalently
\[ \frac {3abc}{2(a^2b \plus{} b^2c \plus{} c^2a)} \geq \frac {3abc}{2(a^2b \plus{} abc \plus{} c^2b)} \equal{} \frac {3ac}{2(a^2 \plus{} ac \plus{} c^2)}
\]
we should prove that:
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {3ac}{2(a^2 \plus{} ac \plus{} c^2)} \geq 2
\]
Now, Let $ a \plus{} c \equal{} 1$ and $ ac \equal{} x$. Then we will write the inequality above in the following form:
\[ \frac {1 \minus{} 2x \plus{} b}{b^2 \plus{} b \plus{} x} \plus{} b \plus{} \frac {3x}{2 \minus{} 2x} \geq 0
\]
After clearing denominators:
\[ 11x^2 \plus{} 5b^2x \plus{} 7bx \plus{} 2b^3 \plus{} 2 \geq 2bx^2 \plus{} 10x \plus{} 2b^3x \plus{} 2b^2 \plus{} 2b
\]
Notice it is a quadratic function over $ x$. Therefore, after writing this inequality in the following form:
\[ f(x) \equal{} Ax^2 \plus{} Bx \plus{} C
\]
we have:
\[ A \equal{} 11 \minus{} 2b
\]

\[ B \equal{} 5b^2 \minus{} 2b^3 \plus{} 7b \minus{} 10
\]

\[ C \equal{} 2b^3 \minus{} 2b^2 \minus{} 2b \plus{} 2
\]
we should prove that $ f(x)\geq 0$ for $ 0\le b \le 1$ (since we assumed $ a \plus{} c \equal{} 1$, the maximum value of $ b$ is 1)
we will use quadratic equation to show that $ x$ is always positive when  $ 0\le b \le 1$
\[ x \equal{} \frac { \minus{} B \plus{} \sqrt {\delta}}{2A}
\]

\[ x \equal{} \frac { \minus{} B \minus{} \sqrt {\delta}}{2A}
\]
where $ \delta$ is discriminant.
\[ \delta \equal{} B^2 \minus{} 4AC
\]

\[ \delta \equal{} (5b^2 \minus{} 2b^3 \plus{} 7b \minus{} 10)^2 \minus{} 4(11 \minus{} 2b)(2b^3 \minus{} 2b^2 \minus{} 2b \plus{} 2)
\]

\[ \delta \equal{} (4b^4 \minus{} 12b^3 \minus{} 15b^2 \minus{} 12b \plus{} 12)(b \minus{} 1)^2
\]
[b]EDIT:[/b]: We have:
\[ g(b) \equal{} \frac {2b^3 \minus{} 5b^2 \minus{} 7b \plus{} 10 \minus{} \sqrt {(4b^4 \minus{} 12b^3 \minus{} 15b^2 \minus{} 12b \plus{} 12)(b \minus{} 1)^2}}{22 \minus{} 4b}
\]

\[ h(b) \equal{} \frac {2b^3 \minus{} 5b^2 \minus{} 7b \plus{} 10 \plus{} \sqrt {(4b^4 \minus{} 12b^3 \minus{} 15b^2 \minus{} 12b \plus{} 12)(b \minus{} 1)^2}}{22 \minus{} 4b}
\]
All we have to do is to prove that:
\[ f(x) \equal{} (x \minus{} g(b))(x \minus{} h(b))\geq 0
\]
for $ 0 \le x \le 0.25$ and $ 0 \le b \le 1$[/quote]

By the way, i found that i don't need all those computations. It is enough to check that discriminant is negative and it is. Because we have the value of discriminant:
\[ \Delta \equal{} (4b^4 \minus{} 12b^3 \minus{} 15b^2 \minus{} 12b \plus{} 12)(b \minus{} 1)^2 \le 0
\]
for $ 0 \le b \le 1$. That is why, $ f(x) \ge 0$ because it is "smily parabola" , since the value of A is positive ($ A \equal{} 11 \minus{} 2b \ge 0$)
we conclude that function is positive for $ 0 \le b \le 1$. And proof is complete.
Anyways,  thanks for nice problem, 10math.[/quote]

Shaaam, why do you say $ \Delta$ is negative? :maybe:

Hi Manlio. for $ 0 \le b \le 1$, the $ \Delta$ it is not  always negative  :blush:
i am very thankful that you pointed out my silly mistake.
and also i am disappointed that people just ignore my posts, and no one told me what i did wrong....
(i posted that "solution" a long time ago, and perhaps few people read)

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca} \plus{}\frac{4abc}{a^2b\plus{}b^2c\plus{}c^2a\plus{}abc} \ge 2.\]
:)[/quote]
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$. Hence,
$\frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca} \plus{}\frac{4abc}{a^2b\plus{}b^2c\plus{}c^2a\plus{}abc} \ge 2\Leftrightarrow$
$\Leftrightarrow\frac{3u^2-2v^2}{v^2}+\frac{4w^3}{a^2b+b^2c+c^2a+w^3}\geq2\Leftrightarrow$
$\Leftrightarrow(3u^2-4v^2)(a^2b+b^2c+c^2a)+3u^2w^3\geq0\Leftrightarrow$
$\Leftrightarrow(3u^2-4v^2)\sum_{cyc}(a^2b+a^2c)+6u^2w^3\geq(3u^2-4v^2)\sum_{cyc}(a^2c-a^2b)\Leftrightarrow$
$\Leftrightarrow(3u^2-4v^2)(9uv^2-3w^3)+6u^2w^3\geq(3u^2-4v^2)\sum_{cyc}(a^2c-a^2b)\Leftrightarrow$
$\Leftrightarrow3(9u^3v^2-12uv^4+(4v^2-u^2)w^3)\geq(3u^2-4v^2)(a-b)(b-c)(c-a)$.
Let $f(w^3)=9u^3v^2-12uv^4+(4v^2-u^2)w^3$. Hence, $f$ is a linear function.
Thus, $f$ gets a minimal value, when $w^3$ gets an extremal value, which happens, when two numbers
from $\{a,b,c\}$ are equal or when maybe $w^3=0$.
But for $b=c=1$ we see that $f(w^3)\geq0\Leftrightarrow(a-1)^2a^2\geq0$ and for $b=1$, $c=0$ we see
that $f(w^3)\geq0\Leftrightarrow(a-1)^2a^2\geq0$.
Id est, $f(w^3)\geq0$ for all non-negatives $a$, $b$ and $c$.
Since $(a-b)^2(b-c)^2(c-a)^2=27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)$, we need to prove that:
$9\left(9u^3v^2-12uv^4+(4v^2-u^2)w^3\right)^2\geq(3u^2-4v^2)^2(a-b)^2(b-c)^2(c-a)^2\Leftrightarrow$
$\Leftrightarrow(7u^4-20u^2v^2+16v^4)w^6+3(3u^2-4v^2)(3u^4-9u^2v^2+8v^4)uw^3+$
$+3(3u^2-4v^2)^2v^6\geq0$.
If $3u^2-4v^2\geq0$ so our inequality is true.
Hence, we can assume that $3u^2-4v^2\leq0$ .
Thus, it remains to prove that
$3\left((3u^2-4v^2)(3u^4-9u^2v^2+8v^4)u\right)^2-4(7u^4-20u^2v^2+16v^4)(3u^2-4v^2)^2v^6\leq0$,
which is  $(u^2-v^2)^2(3u^2-4v^2)^5\leq0$, which is obvious.


Hallo.  Mein Name ist Dave.  An Alter 5 erlernte ich Kalkl von meinen Eltern.  Ich ging zu IMO am Alter von 13, aber zur didntkerbe ein Gold wegen eines Punktes.  Jedoch las ich einige Pfosten im Aufsatz Babylon (ber Spam).  Meine Frage ist:  Die Forumbeschreibung sagt, da "hier Sie tun knnen was auch immer Sie" folglich mgen, sollten nicht Sie in der LageSEIN, Spam bekanntzugeben?  Wenn Sie nicht Spam wnschen, zeigen Sie es bitte auf der Forumbeschreibung, Ihre zur Schuld an!  BTW, die Probleme hier sind WAYYY zu einfachem!  Ich bin jedoch deutsch und verstehe nicht die englische Sprache, also bersetzte ich dieses

Translation:
Hello.  My name is Dave.  At age 5 I learned calculation of my parents.  I went to IMO at the age from 13, but to a didntkerbe gold because of one point.  However read I unite posts in the essay Babylon (over Spam).  My question is:  The description of forum says that "here you can do whatever you" therefore to like, should not not you in the situation its to communicate Spam?  If you do not wish Spam, indicate it please on the description of forum, their to the debt!  BTW, the problems here are WAYYY to simple!  I am however German and do not understand not the English language, therefore I translated this-german girls are hot!

Well, what can we say here? Piece by piece...
Yep, there should be a forum rule that spams will be deleted at the administrator's discretion.
Ok, if problems in here are WAYYY too simple, can you provide solutions to some of the hardest problems posted in this site? If you got no time, we got nothing else to say but one thing: what's this "WAYYY too simple" attitude is about? Have you looked at ALL the problems?
I'm glad to hear you are also a past IMO participant but I have found that David Sichinara didn't even get an honorable mention in IMO 2000 ... or maybe you are Dave as a very long time participant in which you will, of course, find that the problems are easy for you by now because you are probably a professor. There are really discrepancies in your claims and I won't buy by it.
The phrase "here you can do whatever you like" should not be deliberately and literally taken for granted as posting "whatever you like". Again, if you doubt anything about what is appropriate to post here, please consult Valentin.
I'm sure you can spend more time browsing through our populated forums and surely you can find many interesting problems.
Ok I'm done. Please have a VERY NICE DAY and "hopefully" we see you around solving some of the hardest problems for real.
And ... welcome to MathLinks.ro maintained by Valentin Vornicu :D  :D  :D

OH ... right!!! This must be some April Fool's joke!!!

it is.

and now, what is the translation of "german girls are hot"?

well it seems that I have nothing more to add, other than welcome :D


logarithms
quadratics
combinatorics unsolved
combinatorics

Prove that there exists an absolute constant $c>0$ such that, for every $k$, there is a set $S_k$ of at least $ck\ln{k}$ integers with the following property. For any $\mathbb{Z}$-coloring in $k$ colors, there exists an integer $x$ for which the set $x+S_k$ does not intersect all color classes.

Any ideas?  :?

No  :) (though I'd love to :D). Actually, some of his problem sets can be found on his [url=http://www.math.tau.ac.il/~nogaa/]homepage[/url]. This particular problem appears [url=http://www.math.tau.ac.il/~nogaa/assign033.ps]here[/url]; as I don't have any references concerning its origin (and, by the way, it is quite probable that Noga Alon discovered this himself), I cited his advanced combinatorics problem homework sets as the source.

Anyway, any ideas? I'm stuck on this problem.  :(

I wrote two or three papers with Noga, and can attest that he is an amazing combinatorialist.  This problem seems tough for a homework problem.  It was proved in a [url=http://www.math.tau.ac.il/~nogaa/PDFS/shift.pdf]paper by Alon, Kriz, and Nesetril (1995)[/url].  In Section 3, they give two proofs.  The first supplies an explicit construction of S using quadratic residues.  The second lets S be a random subset of integers.

Wow, thank you very much for the link!!!


inequalities
geometry
inradius
inequalities proposed

Since it seems that this is the median week, I will adapt myself and post an inequality that I like very much. I dont know if it is known, since Im really bad at geometric inequalities. Prove that in any triangle $\frac{m_a\cdot m_b\cdot m_c}{S}\geq\frac{3}{2}\cdot\frac{a^2+b^2+c^2}{a+b+c} $, where $S$ is the area.

harazi, this is the same problem as posted by nickolas yesterday and I posted a brief proof there:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15979

Oh, sorry, I didn;t know that. Strangely, I cannot see what is written there.

Actually, if r is the inradius of triangle ABC, then $S=\frac{1}{2}\left( a+b+c\right) r$, so your inequality can be rewritten as $m_{a}\cdot m_{b}\cdot m_{c}\geq \frac{3}{4}r\left( a^{2}+b^{2}+c^{2}\right) $. Now, since $m_{a}^{2}+m_{b}^{2}+m_{c}^{2}=\frac{3}{4}\left( a^{2}+b^{2}+c^{2}\right) $, your inequality turns out to be equivalent with the inequality $m_{a}\cdot m_{b}\cdot m_{c}\geq r\left( m_{a}^{2}+m_{b}^{2}+m_{c}^{2}\right) $, which was posted by Nickolas in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15979]the thread Fuzzylogic cited[/url].

[quote="harazi"]Strangely, I cannot see what is written there.[/quote]

Try [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=15979[/url] ...

  Darij


Do there exist two right-angled triangles with integer length sides that have the lengths of exactly two sides in common?

It is trivial if I translate true.
We call length  of side oppisite A is a
Suppose $ S_1(a_1,b_1,c_1),S_2(a_2,b_2,c_2)$
Where $ c_1,c_2$ is the length of longest length side in each triangle.
Apply Pytagore law we has:
$ a_1^2\plus{}b_1^2\equal{}c_1^2$
$ a_2^2\plus{}b_2^2\equal{}c_2^2$
Imply that if two length is equal then the first so .
It mean that there are not two triangle satisfy condition.

I can't understand your solution...Could you please explain it in details :blush:

[u]Case1[/u]$ (a_1,b_1)\equal{}(a_2,b_2)$ then $ c_1\equal{}c_2$
[u]Case 2[/u]$ c_1\equal{}c_2$ and one of $ (a_1,b_1)$ equal $ (a_2,b_2)$ Suppose $ a_1\equal{}a_2$ then $ b_1\equal{}b_2$
It mean that if two length is equal then the third so.

That's wrong, you can e.g. still have $ a_1\equal{}c_2$ and $ b_1\equal{}b_2$.

[quote="ZetaX"]That's wrong, you can e.g. still have $ a_1 \equal{} c_2$ and $ b_1 \equal{} b_2$.[/quote]
I am very sorry.
Now consider the case :
$ a_1\equal{}c_2,b_1\equal{}b_2$ (the other case same method and with this case we don't lose the general)
$ a_1^2\plus{}b_1^2\equal{}c_1^2$
$ a_1^2\minus{}b_1^2\equal{}a_2^2$
So $ a_1^4\minus{}b_1^4\equal{}(a_2c_1)^2$
This equation has no solution in positive integer (Fermat theorem).


This is the LaTex document. I have some question:
1) can I put ''Vietnam Mathmatical 1998'' great than? Now, it is small.
2) can I put ''Vietnam Mathmatical 1998'' on center? Example \[AbC\]

See Font Sizes and Styles and Justification (Centering, etc) on the [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideLay.php]Layout[/url] page.

Ok, Now in my document '' $\frac{x_{n}^{2}}{1+\ln{x_{n}}}$'' is small, can i put it great than? Thanks!

You have probably used \frac. Use \dfrac instead to get
$\ln(\frac{x_{n}^{2}}{1+\ln x_{n}})$
You need the amsmath package but you should be using it anyway.
To get the brackets sized properly replace () by \left( and \right) which gives
$\ln\left(\frac{x_{n}^{2}}{1+\ln x_{n}}\right)$

[quote="stevem"] \dfrac [/quote]
But TexnicCenter has not this function  :(

$\frac{1}{2}$

That makes no sense. TeXnicCenter is an editor. If you mean you get a [u]LaTeX[/u] error then that's because you haven't loaded the amsmath package which I said you should always use.
Before \begin{document} add
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
You may not always need all of these but they are worth having available. See AMS Math Packages at http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuidePack.php

Ok, Now I have the question: Why ''Day 1'' and ''Day 2'' in document follow aren't same papendicular line ?

It is because you are centring Day 2 inside the enumerate environment. In there the lines are indented so the centre moves to the right.

So end the enumerate environment before Day 2 and restart it after Day 2. You will need to add, after beginning the second environment,
  \setcounter{enumi}{3}
to start the next set of questions at 4.

You also ought to look at the errors LaTeX is telling you about. There are odd } about and TexnicCenter is helpful here.
1. Click on the ! at the top right of the toolbar and it will tell you which line the error is on.
2. Click to the right of any }. If it's OK it will be highlighted in green and show the corresponding {, otherwise it will highlight it in red.

Today, i have an error

\begin{enumerate} on input line 28 ended by \end{document}

Before , i typed
[code]\item Find all $P\in\mathbb{R}[x]$ of degree $n$ having only real zeros $x_1,x_2,\cdots,x_n$ such that \[
\sum_{i=1}^n\dfrac{1}{P(x)-x_i}=\dfrac{n^2}{xP'(x))\;\forall x\in\mathbb{R}-\{0\}.\][/code]

But i received

60. Find all $ P\in\mathbb{R}[x]$ of degree $ n$ having only real zeros $ x_{1},x_{2},\cdots,x_{n}$ such that \[ \sum_{i\equal{}1}^{n}\frac{1}{P(x)\minus{}x_{i}}\equal{}\]

I cann't edit it, can you help me?

\dfrac{n^2}{xP'(x)[b][color=red])[/color][/b] should be \dfrac{n^2}{xP'(x)[b][color=red]}[/color][/b]
TexnicCenter marks in red any brackets that don't match and in green those that do match when you click just after the bracket.

By the way you don't need \dfrac=display fraction as you are already in display mode using \ [ ...\ ], so \frac will be displayed as well.


geometry
incenter
geometric transformation
homothety
rectangle
geometry solved

In a triangle ABC let D be the foot of the perpendicular from the incenter I to BC.  Let DI intersect the incircle again at E.  AE intersects the side BC at F.  If IO is parallel to BC, prove that EF = 2(R - r).

M is the midpt of BC.

F is the point of tangency of BC with the exinscribed circle corresponding to BC. This is easy to see because E is such a point for the triangle formed by A and the parallel to BC drawn through E, and F is obtained from E by a homothety (and so is ABC from that smaller triangle; we use the same homothety). From this we get that M is, in fact, the midpt of DF.

I is the midpt of ED, OI||DF, OM perp. to DF and M is the midpt of DF => OIDM is a rectangle, so O is the midpt of EF. We get
(EF/2)^2=OE^2=OI^2+EI^2=OI^2+r^2=R^2-2Rr+r^2=(R-r)^2, so EF=2(R-r) Q.E.D.


Write as a single log and simplify.

xlog(3) + ylog(27) + zlog(81) = log (     )

I know you can factor out a three..but not sure how...

These logarithm rules will help:

\[\log a^{b}= b \cdot \log a \] \[\log a+\log b = \log (ab) \]

This should go into the High School Basics forum.  :)


geometry
geometric transformation
dilation
percent

What volume of 6.0 M H2SO4 should be mixed with 10 L of 1.0 M H2SO4 to make 20 L of 3.0 M H2SO4 upon dilution to volume?

The answer is 8.3 L, but I'm not sure what method to use to get that answer.

Also, can someone explain how to do nuclear decay problems? Such as:
The half-life of sulfur is 2.87 hours.
a) After 8.61 hours, hwat percent of the original radiation is left?
b) If 3.125% of the original radiation is being emitted, how many half-lives have passed?

I have to self-study nuclear chemistry, but my book (Chang) doesn't provide half-life problems. I only know how to calculate half life with first-order reactions when concentrations are given.

Any help is appreciated!

You basically need to count the number of moles of H2SO4.  So

$ volume \cdot 6.0 + 10 \cdot 1.0 = 20\cdot 3.0$
Then solve for volume.

After 8.61 hours, (which is $ 3\cdot 2.87$), three half lives have occurred.  Each half-life reduces the amount by half so its $ (1/2)^3 = 12.5%
$.
Note that 3.125% is $ (1/2)^5$ so 5 have occurred.


calculus
integration
logarithms
algebra
partial fractions
calculus computations

Help me to solve the integral, thanks.

\[ {{I}_{1}}\equal{}\int\limits_{0}^{1}{\frac{\left( 1\minus{}{{a}^{2}} \right).\pi .da}{\left( {{a}^{4}}\minus{}{{a}^{2}}\plus{}1 \right)}}\]

\[ {{I}_{2}}\equal{}\int\limits_{0}^{1}{\frac{\log \left| x \right|.dx}{\left( {{x}^{2}}\minus{}x\plus{}1 \right)}}\]

$ I_1$ quite standart: decompose the denominator, decompose the fraction and integrate.

The denominator is $ (x^{2} \plus{} \sqrt {3}x \plus{} 1)(x^{2} \minus{} \sqrt {3}x \plus{} 1)$.

Partial fractions (having taken out the factor $ \pi$) give the integrand as $ \frac{\frac{x}{\sqrt{3}} \plus{} \frac{1}{2}}{x^{2} \plus{} \sqrt{3}x \plus{} 1} \minus{} \frac{\frac{x}{\sqrt{3}}\minus{}\frac{1}{2}}{x^{2}\minus{}\sqrt{3}x\plus{}1}$

Therefore $ I\equal{} \frac{\pi}{\sqrt{3}} \int_{0}^{1} \frac{x\plus{}\frac{\sqrt{3}}{2}}{(x\plus{}\frac{\sqrt{3}}{2})^{2} \plus{} \frac{1}{4}} \minus{} \frac{x\minus{}\frac{\sqrt{3}}{2}}{(x\minus{}\frac{\sqrt{3}}{2})^{2} \plus{} \frac{1}{4}}\ dx$

$ \equal{}\frac{\pi}{2\sqrt{3}} \int_{\frac{\sqrt{3}}{2}}^{1\plus{}\frac{\sqrt{3}}{2}} \frac{2u}{u^{2} \plus{} \frac{1}{4}}\minus{}\int_{\frac{\minus{}\sqrt{3}}{2}}^{1\minus{}\frac{\sqrt{3}}{2}} \frac{2u}{u^{2} \plus{} \frac{1}{4}} \ dx$

$ \equal{}\frac{\pi}{2\sqrt{3}} (\ln(u^{2} \plus{} \frac{1}{4}) \minus{} \ln(u^{2} \plus{} \frac{1}{4}))$, between the appropriate limits gives

$ \frac{\pi}{2\sqrt{3}} \ln(\frac{2\plus{}\sqrt{3}}{2\minus{}\sqrt{3}})$

hello, to your second integral mathematica gives the following solution
$ \frac{i(\text{PolyLog}[2,(\minus{}1)^{1/3}]\minus{}\text{PolyLog}[2,\minus{}(\minus{}1)^{2/3}])}{\sqrt{3}}$
Sonnhard.

My Mathematica tells me that the second integral is $ \frac {\pi^2}{5\sqrt {5}}$. But I'm not so smart as my Mathematica and don't know how to compute it. Perhaps $ 1 \minus{} x\to t$ and expand logarithm?
Edit: there was a mistake in input. My Mathemtica cannot handle it as well. But probably I can, because I've got another clever idea.

So I get
\[ \minus{} \sum_{n \equal{} 1}^{\infty} \frac1{2n} \int_0^1 (1 \minus{} x \plus{} x^2)^{n \minus{} 1}dx.\]
Any ideas?

Yet another formula:
\[ I_2 \equal{} \frac{2}{\sqrt{3}}\sum_{k\equal{}0}^\infty\frac{1}{2k\plus{}1}\int_0^{3/4} \frac{t^k}{(1\minus{}t)^{k\plus{}1}}dt\]
Probably, Mathematica is right and this cannot be expressed in elementary functions.

[quote="zhoraster"]Yet another formula:
\[ I_2 \equal{} \frac {2}{\sqrt {3}}\sum_{k \equal{} 0}^\infty\frac {1}{2k \plus{} 1}\int_0^{3/4} \frac {t^k}{(1 \minus{} t)^{k \plus{} 1}}dt\]
Probably, Mathematica is right and this cannot be expressed in elementary functions.[/quote]
Please show me how to compute?


The sum of the lengths of the 2 legs of a right triangle is 20. What is the smallest possible length of the hypotenuse of this triangle?


ok, my aim is filletwho. but uh... i sorta have a tight study schedule during spring break, so i need to do it during the weekend.

You need to contact your team and figure out when to do it. Max probably won't help you out much :)

...and since it's already past Wednesday 11:59 PM, how did everybody do on their verticals?
Felt a lot harder this round. :?

I knew this one was going to be harder.  I tried to make #1 pretty much a gimme (modulo some grunt-work), #3 too extensive to completely solve, and #2 hard unless you were clever.  I haven't yet read enough to see whether any teams got #2, but I love that problem very much.

If you got stuck on #2, think about using transformations on your picture of the three rays.[/hide]

[quote="chesswonk"]because sameer (shake9991), the dallas captain, forgot to send the roster in time, the new players dallas recently acquired will not be playing for week 10. They will start on week 11. Thus, be expecting emails as soon as week 10 ends. actually, i dont think dallas even knows the email addresses of all the new members.[/quote]

I've noticed it week doesn't exactly mean a week, does it. So next month?

The individual rounds for this week haven't been sent yet, so you can expect a wait of at least one more week. Cody is usually good with releasing individuals near vertical, so I would say wait a max of two weeks until this round is over. Then make sure your captain submits a roster this time.  :P

um... i dont know if this is correct, but waht i essentially did on #2 was revolved around the fact that if you have 2 circles of congruent radius such that each center lies on the circumference of the other circle, then the triangle formed by one of the points of intersection between the 2 circles and the centers is equilateral (same radius).

[quote="chesswonk"]usually, the verticals for dallas on done via email[/quote]
the verticals for dallas are usually not done lol :D  :blush:  :(

not true. except for week 9 and mabye another week or two which i cant remember right now, dallas has done its verticals weekly. the problem, according to sameer, was that his email attachment was messed up and thus he couldnt sent em.

we stopped sending emails to u because u werent contributing on the verticals anyways

Hi guys.  I know there is some concern about the individual test for this week.  Please know that I haven't forgotten about you.

Things in Cody Bowl are about to become very difficult for me.  I have just started working on a research problem for my Ph.D. program, so much more of my time will be consumed by that.  On top of that, I have been taking a very time-consuming complex analysis course.  Cody Bowl is going to have to wait quite a bit more than it used to have to.

I apologize to all the loyal NOMLers who have been waiting.  We will finish this season, but probably not before the summer.  If we run into summer, we will either make an agreement on the schedule with all teams who are still active to work through the summer, or we will wait until fall to finish.  I hate to do any of these things, but it looks like I'm not going to have a choice.  I would spend more time on Cody Bowl, but my advisors in the graduate program are already upset with me for spending as much time as I do, and there's the simple fact that the graduate program gives me money to pay my rent and my bills, and Cody Bowl does not.

In no way is this meant as a guilt-trip for those who have been asking about the delays.  I feel it is my obligation to keep you updated on the reason for these awful delays and to let you know what I plan to do about it.


induction
number theory unsolved
number theory

Determine if there exists a positive integer n such that n has exactly 2000 prime divisors and 2^n+1 is divisible by n.  Does there exist such an n which is not divisible by the square of any prime number?

Simple induction shows that $3^{k}|2^{3^{k}}+1$ so it is enough to pick $n=3^{1999}$.
Yes, such numbers exist: $n=1$ or $n=3$.

He most likely meant 2000 distinct prime divisors, in which case this is a previous IMO problem. The answer to the first question is yes, and as TomciO suggested, you can essentially get there by just throwing in a lot of 3's and then some other primes as desired.

I suspect the answer to the second part (a squarefree such number) is "no" but I don't feel like doing math...

Oh, sorry didn't notice that the question was about prime divisors.

I thought the IMO was that there are at least and not necessary exactly that number of prime divisors¿

Define set good primes GP: $3\in GP$, 2\not \in GP. Let $T_{p}$ is minimal natural number, suth that $p|(2^{T_{p}}+1)$. Prime $p\in GP$ if and only if all prime divisors $T_{p}$ are good primes. It give $p\in GP$ if and only if exist n: $p|n|(2^{n}+1)$. If $n|(2^{n}+1)$ then all prime divisors n are good prime divisors, and for all set good primes $p_{i}\in GP$ exist n, suth that $p_{i}|n \ \forall i$, $n|(2^{n}+1)$.
It is easy find subset good primes SGP: $p\in SGP$ if and only if exist m: \[p|(2^{3^{m}}+1).\] Define \[SGP_{m}=\{p\in P|p\not |(2^{3^{m-1}}+1),p|(2^{3^{m}}+1).\] It is easy to show that $SGP_{m}$ is not empty for all $m\in N$.
Let $p_{i}\in SGP_{i},i=1,2,...,2000$, and  $n=3^{2000}*p_{2}*...*p_{2000}$. It give $n|(2^{n}+1)$ and n had exactly 2000 prime divisors.

actually, for the squarefree thing, isn't it obviously not allowed

suppose such n existed

let p be the smallest prime dividing n

then 2^(2n) = 1 in Z/pZ, i.e. if k is the order, k < p but k | 2n so k = 1 or 2, i.e. p = 3

let q be the next smallest prime dividing n

then 2^(2n) = 1 in Z/qZ, i.e. if k is the order, k < q but k | 2n so k = 1, 2, p, or 2p = 1, 2, 3, or 6, i.e. q | 63, i.e. q = 7.

but 2^n = 2^(3*something) = 8^something = 1^something = 1 mod 7, and if you add one to that, you get 2, which is not 0 mod 7 (this is because 7 does not divide 2)

of course this may be wrong as I'm not actually thinking about math; stephen just sort of threw it at me

It is easy to show, that if $n|(2^{n}+1),n>3\Longrightarrow 9|n$.
For all natural k exist $m=3p_{1}*p_{2}*...*p_{k}$, were $p_{i}>3$ distinct primes, suth that $3m|2^{3m}+1$.


geometry probability AMC AIME trigonometry algebra polynomial

2nd Edit 12 Jul 2006:  Editing the original post associated with the poll for this thread inadvertently deleted the poll.  Sorry about that.  I wasn't aware of this feature.   The truncated results as of 12 July 2006 appear below:

Poll

AMC math contest problems provide the greatest motivation for further study of which of the following fields?

Algebra 	                 17%   [ 3 ]
Number Theory                23%   [ 4 ]
Geometry 	               29%  [ 5 ]
Probability / Counting	     29%  [ 5 ]
Statistics / Data Analysis   0%  [ 0 ]

Total Votes : 17

Insights into why some fields might be easier than others to motivate through math contest problems would be appreciated.

12 Jul 2006 Edit:

"AMC Contests" as used in the poll was intended to refer to  all of the contests under the AMC umbrella, including the AMC 8, 10, 12, AIME, and USAMO.  Comments on other problem-solving contests that inspire further math-related studies are valued as well.

I would say either Number Theory or Counting/Probability - those aren't normally covered very much in the school curriculum.

Here, I would have to say Counting and Probability.

Counting and Probability aren't taught very much in schools, at least not in depth.  Also, there are many neat things you can find out with probability such as how much you should bet on a hand in poker, or how likely are you to get certain rolls in Yahtzee.

Counting and Probability is a wide field that is commonly used in contests, but only the very basics are taught in school.  You can learn a lot more about counting and probability, most of which you don't already know, so you can see everything that you could learn.

I agree with C&P. In school, I never learned PIE. I also never got anything half as hard as an AMC-10 #10-15 ish problem in class.

Oh yeah, that too. Schools don't teach mods at all.

This is a really interesting question.

There isn't much number theory on the AMC... There isn't much interesting geometry either... I said algebra because of all the interesting complex number and recurrence problems (well, maybe that's the AIME).

I don't think this is exactly what the topic is about, but here is a thought that is related to the name of the topic.  I would say that AMC stuff motivates people to study in fields where problems are more often solved.  That is to say, not necessarily pursuit of knowledge, but solving of problems.  So things more like engineering.

As far as subject areas within math.  I would say it depends a lot on which competitions.  The olympiad stuff seems heavily geometry oriented.  The short answer stuff in AMC's and AIME's seem to like combinatorics and algebra type stuff more I would say.  In some sense that makes sense because that is what we are taught in school the most heavily anyways.  They are also easiest to write good short problems for since there are so many little tricks and situations you could consider.  Each of them solvable with some cleverness and with relatively low-tech stuff.  It seems much harder to write a decent short answer number theory problem.

i voted for geometry, because a lot of it isn't taught in school, or just the basic ideas are, not how you can apply them to hard aime problems.  and geometry relates to everything in life.  so does the law of cosines.  you can use law of cosines on any math problem ever.  :D

[quote="perfectnumber628"]so does the law of cosines.  you can use law of cosines on any math problem ever.  :D[/quote]

Use the law of cosines solve for all real values of x:
$(x^{2}-5x+7)^{(x^{2}+7x+12)}=1$

That's easy!

Consider a triangle with sides of length $x$, $\sqrt 7$, and $\sqrt{5x}$, and let the angles opposite these sides be $\alpha$, $\beta$, $\gamma$ respectively. Then $2x\sqrt{7}cos{\gamma}$ is the base polynomial (by the law of cosines). Now we may find when this value equals 1 or -1, and when the power equals 0 (again with the law of cosines), and solve accordingly.

In response to PDiao, I think the entire purpose of the question was to ignore that fact :P.

number theory...but I did't vote... :lol:

I would vote Number Theory because out of all the problems you see in the Olympiads, the NT problems most closely resemble actual NT problems from the "real world" of mathematical research. A large part of this is due to how very similar looking NT problems can be very different in terms of difficulty (see $x^{2}+y^{2}=z^{2}$ versus $x^{3}+y^{3}=z^{3}$).

Another good example of this is IMO 2002 #3 which was inspired by a specific case from an actual math research paper that was, I believe, over 100 pages long.

On the other hand, with the other topics: Euclidean geometry vanishes off the map (thank goodness) once you leave the contest world. Algebra essentially turns into calculus/analysis (especially for inequalities) and of course abstract algebra and the branches that involve advanced linear algebra (commutative algebra, algebraic geometry) and no longer looks anything like the algebra done in contests. Combinatorics becomes more concerned with n-dimensional stuff as well as other nutty things, though in some cases it can bear similarities to the stuff we're accustomed to from contest math.

I'm not saying all high level NT looks similar to contest NT (obviously this is false), but we all know what Goldbach's Conjecture, the Twin Prime Conjecture, and Fermat's Last Theorem say, and at first sight does the Twin Prime Conjecture really look that much more horrendous than some "prove there exist infinitely many $n$ such that $n$ satisfies blah"?


geometry
circumcircle
perpendicular bisector
geometry unsolved

Let $ AB$ be a chord of circle $ w$. $ C, D$ are on the extension of $ AB$, closer to $ B$ than to $ A$. A tangent from $ C,D$ to $ w$ (on the same side of $ AB$) meets $ w$ at $ C_1,D_1$ respectively. Let $ C_2,D_2$ be the circumcenters of triangles $ ACC_1,ADD_1$. Prove that $ C_2$ lies on $ AD_1$ if and only if $ D_2$ lies on $ AC_1$.

Assumming that $ D_2$ lies on $ AC_1$, we will show that $ C_2$ lies on $ AD_1$.
Let $ J$ be the intersection of $ CC_1$ and $ DD_1$ then $ OD_1JC_1$ is cyclic quadrilateral. But $ OC_1\equal{}OD_1$ thus $ \angle OJD_1\equal{}\angle OJC_1. (1)$
We have $ \angle OD_1D_2\equal{}\angle OAD_2\equal{}\angle OC_1D_2$ hence $ OD_2C_1D_1$ is cyclic.
$ \Rightarrow \angle OJD_1\equal{}\angle OD_2D_1\equal{}\angle AD_1 (2)$
From $ (1)$ and $ (2)$ we get $ OJ//AD \Rightarrow \angle JCD\equal{}\angle OJC_1\equal{}\angle OJD_1\equal{}\angle JDC.$
Denote $ C'_1$ the intersection of the perpendicular bisector of $ AC_1$ and $ AD_1$.
We claim $ \angle C_2C_1D_2\equal{}\angle C_2AD_2\equal{}\angle C_2D_1D_2 \Rightarrow C_2$ lies on $ (OJ)$.
$ \Rightarrow \angle AC'_2C_1\equal{}\angle AD_2D_1\equal{}2\angle ADD_1\equal{}2(180^o\minus{}\angle ACC_1)$ therefore $ C'_2$ is the circumcenter of $ (ACC_1)$. Our proof is completed.


[quote="Math Geek"]What is the sum of the digits of the whole numbers from 0 to 10000?

Do you have to find a pattern from this and go on (that's what I did), or is there some cool combinatorics method?  I'm pretty sure using the pattern works, but I'm just curious. :lol:[/quote]

[hide]
the numbers of each digit is equal (except there's one extra 1 in 10000)
in the thousands digit each digit is used 1000 times
in the hundreds digit each digit is used 900 times
in the tens digit each digit is used 900 times
in the units digit each digit is used 900 times
so its 3700*(45)+1

...no wait that can't be right

[/hide]

[quote="andersonw"]Circlesquared, here's the correct way of doing that, I think.

You don't need combinatorics.
Say the integers are
0000
0001
0002
....
9999

There are 4*10000=40000 digits total. By symmetry, each of the digits 0-9 occur the same number of times, so there are 4000 0's, 4000 1's, 4000 2's etc. The sum of the digits is therefore 4000(1+2+...+9)=4000(45)=180000
Add in the 1 from the 10000 so the answer is $ \boxed{180001}$, I think.

Edit: small typo[/quote]

that wont work because the number of 0's is not the same as the number of the other digits

edit: hold on maybe it will.

... this would probably explain why i never got this problem right, no matter how many times i tried
i got an answer of like 495595 or something like that the first time i did it

ok there are $ 1000$ $ 0's$ in the "thousands" place beacause in $ \frac {1}{10}$ of the numbers, $ 0$ occupies the $ 1000's$ place. this is the same for all numbers and all place values.
this then becomes $ (0*1000 \plus{} 1*1000 \plus{} 2*1000 \plus{} \cdots \plus{} 9*1000)*4$ (for 4 digit places)
plus the 1 in 10,0000

this becomes $ 4000*45\plus{}1$
or $ 180001$
i hope u understood my solution...post
  Disable Smilies in this post
  Attach signature (signatures can

[quote="stevenmeow"]ok there are $ 1000$ $ 0's$ in the "tenthousands" place beacause in $ frac{1}{10}$ of the numbers, $ 0$ occupies the $ 10000's$ place. this is the same for all numbers and all place values.
this then becomes $ (0*1000 \plus{} 1*1000 \plus{} 2*1000 \plus{} \cdots \plus{} 9*1000)*5$ (for 5 digit places)

this becomes $ 5000*45$
or $ 225000$
i hope u understood my solution...[/quote]

uhh, i'd have to say anderson's makes a tad more sense
because there are only 4 digit places

heh i made a mistake OK OK i edited it...
BUT IN MY DEFENSE i use
$ \text{\LaTeX}$ more than anyone else!  :D

GOOD JOB. YOU FINALLY USED LATEX

Ok... When it says from, it means strictly in between, I think =D

How many $ 1$s are there from $ 0 \minus{} 10,000?$.  Well, notice that this is the same as the number of $ 2$s, $ 3$s, ... and $ 9$s.

So, the only thing we need to count now is the total number of $ 0$s.

Take this, divide by $ 9$ to find the total number for each digit, and then multiply by $ 45$.

Thus, we essentially multiply by $ 5$.

So pretty much, Andersonw's way, except I like his way better =). Beautiful way to eliminate the need to actually count the 0s

[quote="Math Geek"]What is the sum of the digits of the whole numbers from 0 to 10000?

Do you have to find a pattern from this and go on (that's what I did), or is there some cool combinatorics method?  I'm pretty sure using the pattern works, but I'm just curious. :lol:[/quote]

lol a slightly modified version of this problem was an AIME #14!! (and that modified version would only be like around a 20-25 National Sprint Question at best)

If I know what you're talking about, that AIME problem was just a bit tougher than this...

YEAH I REMEMBER THIS PROBLEM!
I found a pattern (same as the first post). Then I got lazy and made a program to solve it for me (to verify pattern)


When a train's velocity is 12.0 m/s eastward, raindrops that are falling vertically with respect to the earth make traces that are inclined 30.0  degreesto the vertical on the windows of the train.

What is the magnitude of the velocity of the raindrop with respect to the train?

With normal vectorial axes.  v T - Velocity of train
v R - Velocity of Rai

v T=12i
v RT = -vj

v R = 12i - vj

But the angle that v R makes with y axis is 30. So Vx/Vy = tan-30 = -1/Root(3).

12/v=tan30

v=12Root(3).

V (rain wrt train) = 12Root(3)


Show that if a and b are positive integers,then $ (a\plus{}\frac{1}{2})^n\plus{}(b\plus{}\frac{1}{2})^n$ is an integer for only finitely many positive integers n.

It is equivalent to find $ n$ such that $ 2^n|(2a\plus{}1)^n\plus{}(2b\plus{}1)^n\equal{}A$.

For even $ n$, $ A\equal{}2\mod 4$.

For odd $ n$, $ A\equal{}(2a\plus{}2b\plus{}2)B$, where
$ B\equal{}\sum_{i\equal{}0}^{n\minus{}1}(\minus{}1)^i(2a\plus{}1)^i(2b\plus{}1)^{n\minus{}1\minus{}i}$ is odd.

Now the claim is obvious.

[quote="limac"]I don't understand this: "It is equivalent to find n such that $2^n|(2a+1)^n+(2b+1)^n=A.$" Can you please explain? Thank you.[/quote]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=491&t=150986&hilit=finitely+many[/url]


geometry
3D geometry
sphere
Pythagorean Theorem

In space there is sphere V with radius 2. Sphere W with radius 1 is internally tangent to sphere V. Spheres X, Y, and Z are internally tangent to V, externally tangent to W, externally tangent to each other and congruent to each other. Find the radius of X.

Let's recapitulate the 2D problem: Let (V), (W) be 2 circles with radii R = 2, r = 1, internally tangent at a point O. Let (X), (Y) be 2 congruent circles tangent to each other at a point T and to the circles (V), (W). Find the radius x of the circles (X), (Y).

Let O be the tangency point of the circles (V), (W). Invert the figure in a circle (O) centered at the tangency point O and with radius R = 2. Since the circles (V), (W) both pass through the inversion center O, they are carried into straight lines v, w perpendicular to the center line $OV \equiv OW$. Let the circle (V) intesect the center line at a point P different for O, OP = 2R = 4. Since R = 2r, the circle (W) intersects the center line at the point V. Since OV = 2r = R, the inversion circle (O) passes through the point V, hence, this point stays in place and the line w is a normal to the center line erected at the point V. The point P is carried into the point P', the intersection of the line v with the center line. By the basic properties of inversion

$OP' = \frac{R^2}{OP} = \frac{4}{4} = 1 = OW$

Thus the points $P' \equiv W$ are identical and the line v is a normal to the center line erected at the point W. The 2 congruent circles (X), (Y) tangent to the circles (V), (W) and to each at a point T on the center line  $VW \equiv OV \equiv OW$ are carried into 2 congruent circles (X'), (Y') tangent to the 2 parallel lines $v \parallel w$ and to each other at a point T', also on the center line. The radii of these circles are

$x' = \frac{BW}{2} = \frac{r}{2} = \frac 1 2$

The power of the inversion center O to the circle (X') is equal to the square of the tangent length OT' from the point O to the circle (X')

$OT' = OB - T'B = R - x' = R - \frac r 2 = 2 - \frac 1 2 = \frac 3 2,\ \ \ OT'^2 = \frac 9 4$

The circles (X), (X') are centrally similar with similarity center identical with the inversion center O and similarity coefficient h equal to the power of inversion $R^2$ (square of the inversion circle radius) divided by the power of the inversion center O to the inverted circle (X'):

$h = \frac{R^2}{OT'^2} = 4 \cdot \frac 4 9 = \frac{16}{9}$

Consequently, radius of the circle (X) is $x = hx' = \frac{16}{9} \cdot \frac 1 2 = \frac 8 9$ and it checks with the Sketchpad construction.
______________________________________________________________________________

Virtually the same solution can be repeated in 3D space. Invert the configuration in a sphere (O) centered at the tangency point O of the spheres (V), (W) and passing through the center V of the larger sphere (V). The spheres (V), (W) are carried into the planes v, w perpendicular to the center line $VW \equiv OV \equiv OW$. The 3 congruent spheres $(X), (Y), (Z)$ with radius x, tangent to the spheres (V), (W) and to each other, are carried into 3 congruent spheres (X'), (Y'), (Z') with radius $x' = \frac 1 2$, tangent to the 2 parallel planes $v \parallel w$ and to each other. Their centers X', Y', Z' form an equilateral triangle $\triangle X'Y'Z'$ with the sides X'Y' = Y'Z' = Z'X' = 2x' = 1 and orthocenter H' on the center line,

$OH' = OV - H'V = R - x' = 2 - \frac 1 2 = \frac 3 2$

Obviously, $H'X' = H'Y' = H'Z' = 2x' \frac{\sqrt 3}{3} = \frac{\sqrt 3}{3}$. Since the plane (X', Y', Z') is perpendicular to the center line, every line in this plane is perpendicular to the center line, in particular, $H'X', H'Y', H'Z'  \perp OH' \equiv OV$. By Pythagorean theorem,

${OX'^2 = OH'^2 + H'X'^2} = \frac 9 4 + \frac 1 3 = \frac{31}{12}$

Let OT' be the tangent length from the inversion center to the circle (X'):

$OT'^2 = OX'^2 - X'T'^2 = OX'^2 - x'^2 = \frac{31}{12} - \frac 1 4 = \frac{28}{12} = \frac 7 3$

The spheres (X), (X') are again centrally similar with similarity center identical with the inversion center O and similarity coefficient h equal to the power of inversion $R^2$ (square of the inversion sphere radius) divided by the power of the inversion center to the inverted sphere (X'):

$h = \frac{R^2}{OT'^2} = 4 \cdot \frac 3 7 = \frac{12}{7}$

Consequently, radius of the sphere (X) is $x = hx' = \frac{12}{7} \cdot \frac 1 2 = \frac 6 7$ and it checks with the Sketchpad construction.
______________________________________________________________________________

If you are familiar with Soddy circles (any 4 circles pairwise tangent [b]externally[/b] at 6 points), their radii $r_1, r_2, r_3, r_4$ are bound by the relation

$2 \left(\frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_3^2} + \frac{1}{r_4^2}\right) = \left(\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} + \frac{1}{r_4}\right)^2$

The beauty of this formula is that one tangent circle can be replaced with a tangent line (circle of infinite radius) and the corresponding term $\frac{1}{\infty} = 0$ vanishes. Moreover, one circle can be tangent to the others internally and considered as a circle of negative radius. For the 2D problem, $r_1 = -R = -2,\ r_2 = r = 1,\ r_3 = r_4 = x$ (the largest circle with radius R = 2 has an internal tangency with the others) and the above relation gives an equation for the radius x:

$2 \left(\frac{1}{R^2} + \frac{1}{r^2} + \frac{2}{x^2}\right) = \left(-\frac 1 R + \frac 1 r + \frac 2 x\right)^2$

$2 \left(\frac 1 4 + 1 + \frac{2}{x^2}\right) = \left(-\frac 1 2 + 1 + \frac 2 x\right)^2$

$\frac{10}{4} + \frac{4}{x^2} = \frac 1 4 + \frac 2 x + \frac{4}{x^2}$

$\frac 9 4 = \frac 2 x,\ \ \ x = \frac 8 9$

I have no idea, if a similar approach can be applied to the 3D problem (5 Soddy spheres ??? :?), but the following formula for radii of the spheres would yield $x = \frac 6 7$:

$(???)\ \ \ 3 \left(\frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_3^2} + \frac{1}{r_4^2} + \frac{1}{r_5^2}\right) = \left(\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} + \frac{1}{r_4} + \frac{1}{r_5}\right)^2$

Substituting $r_1 = -R = -2,\ r_2 = r = 1,\ r_3 = r_4 = r_5 = x$, we would get

$3 \left(\frac{1}{R^2} + \frac{1}{r^2} + \frac{3}{x^2}\right) = \left(-\frac 1 R + \frac 1 r + \frac{3}{x}\right)^2$

$3 \left(\frac 1 4 + 1 + \frac{3}{x^2}\right) = \left(-\frac 1 2 + 1 + \frac 3 x\right)^2$

$\frac{15}{4} + \frac{9}{x^2} = \frac 1 4 + \frac 3 x + \frac{9}{x^2}$

$\frac{14}{4} = \frac 3 x,\ \ \ x = \frac 6 7$ [color=white]. .[/color] :?  :?  :?

Yetti


Let $ a,b,c,d\ge 0$ such that:  $ 3(a \plus{} b \plus{} c \plus{} d) \plus{} 4(abc \plus{} bcd \plus{} cda \plus{} dab) \equal{} 8$. Prove that:

$ ab \plus{} ac \plus{} bc \plus{} ad \plus{} bd \plus{} cd  \le 2$

[quote=luofangxiang]Let $a,b,c,d\geq0$, $\sum_{cyc}a+\sum_{cyc}abc=\frac{8}{3}$ Prove $\sum_{sym}ab\leq2$[/quote]
$\sum_{sym}ab=4(ab+ac+bc+ad+bd+cd)$ can be greater than $2$.
We can prove that $ab+ac+bc+ad+bd+cd\leq2$.
Let $a+b+c+d=4u$, $ab+ac+bc+ad+bd+cd=6v^2$, $abc+abd+acd+bcd=4w^3$ and $abcd=t^4$.
Hence, $u+w^3=\frac{2}{3}$ and we need to prove that $v^2\leq\frac{1}{3}$.
Indeed, by the Rolle theorem there are non-negatives $x$, $y$ and $z$, for which $3u=x+y+z$, $3v^2=xy+xz+yz$ and $xyz=w^3$.
Since the condition does not depend on $v^2$ and $v^2$ gets a maximal value for equality case of two variables,
it's enough to prove our inequality for $y=x$ and from the condition we get $z=\frac{2-2x}{1+3x^2}$ and $0\leq x\leq1$.
We need to prove that $xy+xz+yz\leq1$, which is $(1-x)(3x^3+3x^2-3x+1)\geq0$.
Done!


Let $(a,b)$ be on parabola $y = 4x-x^{2}$ and is the point on the parabola nearest to point $(-1,4)$. Find $a+b$

Just use a combination of distance formula and the function formula.

The closest point on a parabola to another point is such that the segment connecting the two points is perpendicular to the tangent at the point on the parabola. So, $\frac{-1}{4-2a}=\frac{4a-a^{2}-4}{a+1}$. You can solve that for $a$ and then find $b$ and add them.


There were 2 freinds named bob and sally

bob said "lets go to getting started...i like their moderators."

Sally said "Ok, but lets solve a problem first"

PROBLEM:

There are 10 couples in a room.

everyone shakes eachother hands but hubands dont shake their wives havds.

how many hadshakes occur?

hint: the answer is less than 10000

I didn't get what pianoforte did.
[hide]
The first man shakes hands with 18 people, the second shakes hands with 17 people, etc. I didn't count the wives, because that would be over counting.
$S=18+17+16+...+1$
$2S=18\cdot19$
$\framebox{171}$ handshakes

EDIT: No, the answer is 180, because the women shake hands as well, $\frac{19\cdot20}{2}=190$, but there are 10 couples, so the answer is $190-10=\framebox{180}$
[/hide]

[hide]
Say the men did shake their wives' hands. Then there would be $\frac{20(19)}{2} = 190$ handshakes.
Since there are 10 couples, 10 of those handshakes never took place.

So My answer is $\boxed{180}$

EDIT: oops...the formula is $\frac{n(n-1)}{2}$ not n+1...
[/hide]

Hamster is correct (first find how many handshakes total ignoring the couple situation, then subtract the amount of shakes within couples).

I don't seem to get what pianoforte did either.

PS - I don't get the beginning of the problem either  :?

[quote="G-UNIT"]some of you are over complicating this problem.... nor is subtraction neseccary.....

[hide]  There are $10 \cdot 2=20$ people.  Spouses are not allowed to shake hands.  Thus, each person shakes hands with everyone in the room expect him/her self and his/her spouse, $20-2=18$ people.  After we account for the overcounting we obtain:  $\frac{20\cdot 18}{2}=\boxed{180}$. [/hide][/quote]

yes it is necessary....20-2 is subtraction :P


Hello all

I recently stumbled upon this website, and I must say that it is quite a good one! I just graduated from highschool but had been attending college for a while now (i'm close to graduating). My major is Engineering, and I intend on attending Tokyo University, Tokyo University of Techonology, or Tokyo Institute of Technology. Either one will do. I hope to attain a degree in the states, and then get one while overseas in Japan. Does anyone have an info on the math skills I must already have, and will come to know if I choose to study in Japan? I've heard some people say that their math is a little more advanced than ours. I just want to be ready. I'll be leaving in about three or four years. Any information would be great.
                                                 Thanks,
                                                -ShiningKnight-

I'm not sure I understand your intentions here---get an undergraduate degree in the U.S., and then another undergraduate degree in Japan, or an undergrad degree in U.S. and graduate degree in Japan?

Japanese universities require entrance exams for their undergraduate admissions process.  I hear that they're quite difficult.  I don't know whether foreign students are required to take them, or whether there are similar exams for graduate admission.


algebra
polynomial
linear algebra
matrix
superior algebra
superior algebra unsolved

Let $\varphi$ denote the Frobenius map: $x \mapsto x^{p}$ on the finite field $\mathbb{F}_{p^{n}}$. Determine the Jordan canonical form for $\varphi$ considered as $\mathbb{F}_{p}$ linear transformation on the $n$-dimensional  $\mathbb{F}_{p}$-vectorspace  $\mathbb{F}_{p^{n}}$

$\phi^{n}=id$ so if p doesn't divide n, the characteristic polynomial has n distinct linear factors and the JNF of $\phi$ is therefore a diagonal matrix with the n-th roots of unity of its diagonal.

No Gockel; the n-th roots of unity are not necessarily in Fp.
Examples for p=2:
if n=2 then the JCF is (row by row):[0,1,1,0].
if n=3 we obtain 2 blocks: [1,0,0,0,0,1,0,1,1].

But there is no JCF if the minimal/characteristic polynomial doesn't split.

Correction: Let F be the Frobenius map. Let L(f) be the companion matrix of the polynomial f.
Let p=2.
If n=2 then the JCF is L((X-1)^2).
If  n=3 then the JCF is L(X^3-1). (My precedent result was false!).
If  n=4 then then the JCF is diag(L(X-1),L((X-1)^3)).
Let p=3.
If n=2 then the JCF is L(X^2-1).
If n=3 then the JCF is L((X-1)^3).

Remark: if n is a power of p then (F-I)^n=F^n-I=0 and the F's characteristic polynomial is (X-1)^n.

Apologies for my posts; when I have read "Jordan canonical form" I had in mind the "rational canonical form" ( Frobenius normal form); in fact the "good" name is "Jordan normal form" and this one exists over an algebraically closed field $K$ such that $\mathbb{F}_{p}\subset{K}$.
  Now the first question is: is the characteristic polynomial of Frobenius map $x^{n}-1$ ?

Let $F$ the frobenius map over $\mathbb{F}_{p^{n}}$.
i) then its minimal polynomial $P$ is $x^{n}-1$.
Proof: Assume that d°$P=r<n$; then for all $x\in\mathbb{F}_{p^{n}}$ $P(F)(x)=0$ and $P(F)$ is a polynomial of degree ${p^{r}}<{p^{n}}$. It has at most $p^{r}$ roots; contradiction and d°($P$)=n. We conclude with $F^{n}-Id=0. \square$
ii) The characteristic polynomial is also the minimal one; thus $F$ is a cyclic endomorphism and its rational canonical form is always L($x^{n}-1)$.
Remark:this proves that my result (in one of my posts) for p=2,n=4 is false; the correct answer is the compagnon matrix: $L(x^{4}-1)$  or L($(x-1)^{4})$. Moreover its JNF is $Id+J_{4}$ where $J_{4}$ is the standard Jordan nilpotent matrix of dimension 4 ($J_{4}^{3}\neq0$).
 iii) As Gocquel said, if p doesn't divide n then the minimal polynomial has n distinct linear factors and its JNF is the diagonal matrix with diagonal the n-th roots of unity. These roots are in $K$ the union of the $\mathbb{F}_{p^{k}}$ for $k$ from 1 to $\infty$ (it's the algebraic closure of $\mathbb{F}_{p}$).
iv) If $n=p^{k}m$ with $p$ doesn't divise $m$, then $x^{n}-1=(x^{m}-1)^{p^{k}}$. The roots of $x^{m}-1$ are of order 1 and form a set $U_{m}$.The roots of $x^{n}-1$ are the $U_{m}$'s elements and they are of order $p^{k}$.
Let $U_{m}=\{\lambda_{1},...,\lambda_{m}\}$. The $F$'s minimal polynomial is of degree n, therefore for all i there exists only one Jordan nilpotent block associated to $\lambda_{i}$ and this block is of dimension $p^{k}$. The JCF is diag(${{\lambda_{1}{I_{p^{k}}}+J_{p^{k}}},...,\lambda_{m}{I_{p^{k}}}+J_{p^{k}}}) .\square$