[url=http://imageshack.us][img]http://img162.imageshack.us/img162/504/sonsuzjr3.gif[/img][/url]
Which is true?1 or 2?

[hide]Let this be x.

Notice that this is really saying

$3-\frac{2}{x}=x$.

So, we solve it to get $x=2,1$

I see why you asked that this question.

Maybe this oscillates, and hence has no value...[/hide]

[quote="robinhe"][hide]It's probably 2 though.  If it was $3-\frac{2}{3-\frac23}$ that is a little more than 2.  If you continue doing this, you will reach $2$.[/hide][/quote]

[hide]
I think it osscilates... yes, $3-\frac{2}{3-\frac23}\approx 2$, but $3-\frac{2}{3-\frac{2}{3-2}}\approx 1$... so if this continues itll go between the two.[/hide]

But if $x_{1}=3$ and $x_{n+1}=3-{2\over x_{n}}$ for $n\geqslant 1$, then the sequence is strictly decreasing and tends to $2$.

If x1 is 1, then the sequence remains at 1
If x1 is not 1, then it tends toward 2

How can you tell which it is? How can you assume that x1 is 3 (isn't that circular logic)?

So in other words, the limit is not defined.

As far as I know, conventionally it's understood that the first "ring" in the continuous fractions of the form $a+\frac{b}{a+\frac{b}{a+\dots}}$ is $a$ and not $a+b$, unless otherwise noted.


Let $ a,b,c$ be positive real numbers. Prove that
\[ \frac{a^2}{b^2}\plus{}\frac{b^2}{c^2}\plus{}\frac{c^2}{a^2}\plus{}\frac{10(ab\plus{}bc\plus{}ca)}{a^2\plus{}b^2\plus{}c^2} \ge 13.\]
:)

I think it could be solved by $ SOS$

\[ \Leftrightarrow \sum_{cyc}(b\minus{}c)^2(\frac{(b\plus{}c)^2(3a^2\plus{}b^2\minus{}c^2)}{6a^2b^2c^2}\minus{}\frac{5}{a^2\plus{}b^2\plus{}c^2})\ge 0\]

In my ideal i think it can be killed by pqr (i have already rewritten this ineq in pqr form),but the last is still hard
Can you details it,MA

[quote="tran300491"][quote="M.A"]I think it could be solved by $ SOS$
\[ \Leftrightarrow \sum_{cyc}(b \minus{} c)^2(\frac {(b \plus{} c)^2(3a^2 \plus{} b^2 \minus{} c^2)}{6a^2b^2c^2} \minus{} \frac {5}{a^2 \plus{} b^2 \plus{} c^2})\ge 0
\]
[/quote]
I think it could be solved by $ SS$ (c=max{a,b,c})
Ine<->$ S_1(a \minus{} b)^2 \plus{} S_2(c \minus{} a)(c \minus{} b) \geq 0$
with $ S_1 \equal{} \frac {(a \plus{} b)^2}{a^2b^2} \minus{} \frac {10}{a^2 \plus{} b^2 \plus{} c^2}$

$ S_2 \equal{} \frac {(c \plus{} a)(c \plus{} b)}{a^2c^2} \minus{} \frac {10}{a^2 \plus{} b^2 \plus{} c^2}$
( Sorry if my ideal is wrong )[/quote]
It is not true,tran300491.i have tried that way,but it is not enough


A collection of 20 coins made up of only nickels, dimes and quarters has a total value of $ \$$3.35. If the dimes were nickels, the nickels were quarters and the quarters were dimes, the collection of coins would have a total value of $ \$$2.75. How many quarters are in the original collection?

I don't understand, can someone explain a little bit more detailed?

Thank you!


K is real number.find all continuous functions $ f$ $ :$ $ R$ $ \longrightarrow$ $ R$ that for every $ x,y$ $ \in$ $ R$ :

$ f(x \plus{} y)$ $ \equal{}$ $ f(x)$ $ \plus{}$ $ f(y)$ $ \plus{}$ $ kxy$

Take $ g(x) \equal{} f(x) \minus{} kx^2/2$ to get $ g$ satisfying the Cauchy equation, so using continuity $ f(x) \equal{} kx^2/2 \plus{} ax \plus{} b$ for any $ a,b$.


[quote="Yooper"]Noticing that number of digits (576) is roughly 2.5 times number of pages (228), i wondered if either of these problems have a solution.

1.  If n pages require 2n digits, what is n?
2.  If n pages require 3n digits, what is n?[/quote]
[hide="1"]
For it to be 2n, there must be more than three digits in n. Since there are 9 pages with one digit, there are 9 pages with 3 digits, so $\boxed{108}$.[/hide]

[hide="2"]We know that there are more than 3 digits in n. $4(n-999)+2889=3n$, $\boxed{n=1107}$[/hide]

Extension: What is the value of n if n pages require 4n digits?

[hide="Solution and Answer"]5(n-9999)+38889=4n, so n=11106

That is a lot of pages in a book[/hide]


Find all functions $ f : \mathbb R \rightarrow \mathbb R$ such that $ (x \plus{} y)(f(x) \minus{} f(y)) \equal{} (x \minus{}y)f(x \plus{} y)$ for all $ x, y\in \mathbb R$

[quote="mr.danh"]Find all functions $ f : \mathbb R \rightarrow \mathbb R$ so that $ (x + y)(f(x) - f(y)) = (x - y)f(x + y)$ for all $ x, y\in \mathbb R$[/quote]
we have $ (x-y)(f(x)-f(-y))=(x+y)(f(x-y)$
then $ (f(x)-f(y))$ we have $ (x-y)(f(x)-f(-y))=(x+y)(f(x-y)$
then $ (f(x)-f(y))(f(x)-(f-y)=f(x+y)(f(x-y)$
let $ g(x-y)=\frac {f(x)-f(y)}{f(x-y)}$

then$ g(x-y)g(x+y)=1$
let $ x-y=u,x+y=v$ then $ g(x)g(y)=1$ for all $ x,y,\in R$
then $ g(x)=1$ then $ f(x)-f(y)=f(x-y)$
next step,we only let $ h(x-y)=\frac{f(x-y)}{x-y}$ and we will find $ h(x)=h(y)$ for all $ x,y\in R$
then $ f(x)=kx$we have done
Am I right ?

[quote="Allnames"]let $ g(x \minus{} y) \equal{} \frac {f(x) \minus{} f(y)}{f(x \minus{} y)}$[/quote]
This step is erroneous (even disregarding the possibility that $ f(x\minus{}y)\equal{}0$).  $ g$ is not a well-defined function unless the right hand side is genuinely a function of $ x \minus{} y$, and we don't know a priori whether or not it is (turns out it's not).  By assuming that it is, you've effectively assumed that $ f$ satisfies $ f(x) \minus{} f(y) \equal{} f(x \minus{} y)$, and so the linear solution you came up with is essentially circular reasoning.

There are nonlinear solutions.

[hide="My answer"]$ f(x)\equal{}\frac{mx^2\plus{}nx}{2}$  with $ m,n\in \mathbb R$[/hide]

By plugging in :  $ x\equal{}x, y\equal{}1; x\equal{}x\plus{}1, y\equal{}1; x\equal{}x, y\equal{}2$ we get a system of 3 equations and 3 unknowns and we get the result,, smthing like mr.danh wrote

Observe that the set of solutions forms an $\mathbb R$-vector space, and that $f(x) \equiv x$, $f(x) \equiv x^2$ are solutions.

Consider the linear subspace of solutions $f$ satisfying $f(1) = f(-1) = 0$; we show that in this case $f(x) \equiv 0$.
First, $(x,y)=(1,-1)$ gives $f(0) = 0$. Now note that
\begin{align*}
	(x,y) = (a,1) \implies (a+1)f(a) = (a-1)f(a+1) \\
	(x,y) = (a+1,-1) \implies af(a+1) = (a+2)f(a)
\end{align*}
View this as a linear system in $f(a)$ and $f(a+1)$. Now note that
\[
	\det
	\left(
	\begin{array}{cc}
		a+1 & -(a-1) \\
		a+2 & -a
	\end{array}
	\right) = -2 \]
thus it follows that $f(a) = f(a+1) = 0$ for all $a$ as needed.

From this we deduce that the set of solutions to the original functional equation are the span of $x$ and $x^2$, id est those functions of the form $f(x) = ax^2 + bx$ for $a, b \in \mathbb R$.

Letting $y=1$, we get that $$(x-1)f(x+1)=(x+1)(f(x)-f(1)).$$ We shall use this to compute $f(x+2)$. Note that $xf(x+2)=(x+2)(f(x+1)-f(1))$. It follows $$x(x-1)f(x+2)=(x+2)((x-1)f(x+1)-(x-1)f(1))$$ $$=(x+2)((x+1)(f(x)-f(1))-(x-1)f(1))=(x+2)((x+1)f(x)-2xf(1))$$
Taking $y=2$ now gives $(x-2)f(x+2)=(x+2)(f(x)-f(2))$. Multiplying the above relation by $(x-2)$ and using the previous result, $$(x-2)(x+2)((x+1)f(x)-2xf(1))=x(x-1)(x-2)f(x+2)=x(x-1)(x+2)(f(x)-f(2))$$
Assuming $x=\neq -2$, we may simplify to get $$2f(x)=(-2x^2+4x)f(1)+(x^2-x)f(2)\implies f(x)=ax^2+bx\forall x\neq -2.$$ But plugging $(x,y)=(-3,1)$ and using the above result shows $f(-2)=4a-2b$, so we have the solution $\boxed{f(x)=ax^2+bx}$.

I don't understand how to solve this kind of problem
Someone could teach me, I would appreciate it very much
I'm from Peru, but I know how to speak English

i am getting something different
(x+y)(f(x)-f(y))=(x-y)(f(x+y))
so on assuming f(x) is some polynomial
and on using (x-y)|f(x)-f(y) we get the obvious answer as a linear function but others are getting something different

If $f, g$ are solutions, then $f + g, f - g$ are also solutions, so we can assume $f(1) = f(2) = 0$. Then plugging $(x, y) = (x, 1)$ gives $f(x+1) = f(x)\frac{x+1}{x-1}$ for all $x \neq 1$. Plugging $(x, y) = (x, 2)$ gives $f(x+2) = f(x)\frac{x+2}{x-2}$ for all $x \neq 2$. Note that for all $x \neq 0, 1, 2$ we have $\frac{x+2}{x-2}f(x) = f(x+2) = \frac{x+2}{x}f(x+1) = \frac{(x+2)(x+1)}{x(x-1)}f(x)$, or equivalently $\frac{f(x)}{x-2} = \frac{f(x)(x+1)}{x(x-1)}$. Thus if $f(x) \neq 0$, then we have $(x+1)(x-2) = x(x-1)$, a contradiction. Thus $f(x) = 0$ for all $x \neq 0, 1, 2$ and $f(0) = 0$, therefore $f(x) = 0$ for all $x$. Hence $f(x) = ax^2 + bx$ for some $a, b$. $\blacksquare$


inequalities
symmetry
algebra
polynomial
search
geometry
area of a triangle

For positive real numbers $ a,b,c$, prove that
\[ (a^2\plus{}b^2)^2 \ge (a\plus{}b\plus{}c)(a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b).
\]

[quote="SimonM"]This has been posted before, t0ra posted a nice solution assuming sides of a triangle[/quote]That was my first impulse. If it doesn't satisfy being a triangle, it will not satisfy the inequality in exactly one case (otherwise we have $ b\plus{}c\le a$, $ a\plus{}c\le b$, $ a\plus{}b\plus{}2c\le a\plus{}b$, $ c\le 0$), so the RHS is negative and the LHS positive.

Exactly, it either satisfies or we can assume that they are the sides of a triangle

Wow this is an old problem...

Anyway if $ (a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)\le 0$ then the problem is obvious since LHS >= 0 >= RHS

So lets assume $ (a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b) > 0$ Now either two of the factors are negative, or all of them are positive. If $ a \equal{} \text{max} \{a,b,c\}$ then the factors $ a \plus{} b \minus{} c$, and $ a \plus{} c \minus{} b$ are positive, so at most one is negative, so none are negative.

Thus $ (a \plus{} b \minus{} c)$, $ (b \plus{} c \minus{} a)$, and $ (c \plus{} a \minus{} b)$ are all positive. Then there exist positive reals such that $ a \equal{} y \plus{} z$, $ b \equal{} z \plus{} x$, and $ c \equal{} x \plus{} y$. Substituting, we get
\[ F[x,y,z]: \equal{} ((y \plus{} z)^2 \plus{} (z \plus{} x)^2)^2 \minus{} 16(x \plus{} y \plus{} z)(xyz)\ge 0
\]
Observe that
\[ F[x,y,z] \minus{} F[\frac {x \plus{} y}{2},\frac {x \plus{} y}{2},z] \equal{} \frac{1}{4} (x \minus{} y)^2 (3 x^2 \plus{} 2 x y \plus{} 3 y^2 \plus{} 24 x z \plus{} 24 y z \plus{} 24 z^2)\ge 0
\]
Now it suffices to prove that $ F[t,t,z]\ge 0$, But
\[ F[t, t, z] \equal{} 4 (t^2 \minus{} 2 t z \minus{} z^2)^2\ge 0
\]
 We can determine equality cases easily from here.

I haven't checked your solution, Altheman but we don't need to use "Mixing variable" in this case(it's so powerful that trivializes the problem abovel)! :wink:
My approach:
The 1st part of my proof is the same as Altheman's proof, so I just need to prove the inequality above in the case a,b,c are sides of a triangle and a=max(a,b,c).Now we will have two cases:
[b]1st case:[/b] a>=b>=c>0:
Using AM-GM we will have: (a+b+c)(a+b-c)=<(a+b)^2=<2(a^2+b^2)
                                     (b+c-a)(a+c-b)=<c^2
Therefore, we get: RHS=<2(a^2+b^2)c^2=<(a^2+b^2)^2, which is equivalent to: a^2+b^2>=2c^2(true since a>=b>=c>0)
[b]2nd case[/b]: a>=c>=b>0
Again, using AM-GM, we will have: (a+b-c)(b+c-a)=<b^2
                                               (a+b+c)(c+a-b)=<(a+c)^2
Therefore, we get: RHS=<b^2(a+c)^2=<(a^2+b^2)^2, which is equivalent to: b(a+c)=<a^2+b^2(since both sides are positive without taking squares). Using AM-GM for the last part, we get: a^2+b^2>=2ab>=ab+bc, which is: ab>=bc(true since a>=c>=b)
From 2 cases above, it's sufficient to conclude that the inequality is proved(Q.E.D)
Yup, I already saw my solution above and it can be cut into one case(2nd case) and use the same argument as above to prove the rest. It's shorter than miy proof above. :oops:

See my solution at here:
[img]http://i300.photobucket.com/albums/nn15/NDTN/BMO.jpg[/img]

Ha...this problem is very easy...If we expand (we recognize the symmetric product on the LHS from heron's formula, etc), it is equivalent to

\[ (c^2\minus{}a^2\minus{}b^2)^2\plus{}(a^2\minus{}b^2)^2\ge 0\]

The solution in that packet isn't exactly correct... equality certainly holds iff $ (a,b,c)\equal{}(k,k,k\sqrt{2})$.

You cannot assume that $ a$ is the max variable for your whole solution.  The inequality only has symmetry wrt $ \{a,b\}$...

[quote="worthawholebean"][quote="ghjk"][quote="worthawholebean"]ghjk, what about $ c\ge a\ge b$?[/quote]
I thought I already assumed that a=max(a,b,c) in my proof. This assumption helps to reduce into 2 cases as above, worthawholebean! :wink:[/quote]You can't assume that, unless I'm missing something.[/quote]
Please let me know the reasonable explanation why we can't assume that a=max(a,b,c). Remember that we don't need the restriction the polynomial is cyclic to assume that a=max(a,b,c). If you don't believe it, do you find out that in any cases, 3 variables must have either one that is largest or one that is smallest out of 3 variables. If all of them are equal, we easily see that the inequality above is trivial.In addition, don;t you see that Altheman already proved that a,b,c are 3 sides of a triangle, which I completely agree that proof is true.Check your thought carefully again, worthawholebean. :)

We need both sides of the inequality to be cyclic to be able to assume that WLOG. They're only cyclic with respect to $ a$ and $ b$, so you can assume WLOG $ a\ge b$, but you can't include $ c$.

[quote="Altheman"] (we recognize the symmetric product on the LHS from heron's formula, etc)[/quote]
This is what motivated [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=229249808&t=189521]my solution[/url].  :)  This reminds me of another time when I [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=194516&search_id=1169244915]used triangles[/url] to prove a three-variable inequality.  I suspect in some cases that's how the problems are written.

[quote="t0rajir0u"]This is what motivated [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=229249808&t=189521]my solution[/url].  :)  This reminds me of another time when I [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=194516&search_id=1169244915]used triangles[/url] to prove a three-variable inequality.  I suspect in some cases that's how the problems are written.[/quote]Ack, I spent a *long* time looking for an elegant solution like that from the triangle! Why didn't I just AM-GM it...


Find the least positive integer $x$ such that $x^{1992}>6^{3320}$.

$x^3>6^5$

and it's solvable from there $x=20$


geometry
calculus
integration
incenter
angle bisector
absolute value

This is a problem from some country's (i forget which) national olympiad. I solved it, but i think my method is rather unelegant and would like to see a better solution. Here's the problem:

The incircle of the triangle ABC touches BC at D and AC at E. The sides have integral lengths and |AD^2 - BE^2| <= 2. Show that AC = BC.

Any suggestions?

Heres my method:

[hide]

You can get a formula for the length of an angle bisector (which AD and BE are) using Stewarts theorem. It works out to be:

Let sides of triangle be a,b,c. Length of angle bisector with angle opposite 'a' side, is the square root of: bc(1-a:^2:/(b+c):^2:).

So basically what we are given is absolute value of bc(1-a:^2/(b+c):^2) - ca(1-b:^2:/(a+c):^2)<=2 and we want to prove that a=b.

We can assume WLOG that b>a. Taking out the factor of c and rearranging, we get abs of (b-a) + ab(b/(a+c):^2: - a/(b+c):^2) <= 2/c.

Now since b>a, that LHS is greater than b-a. Since a,b,c are positive integers, , the left hand side is at least 2, and the right hand side at most 2. So we must have c=1, b-a=1. But this means the triangle has side lengths 1, a, a+1 which is impossible.

Similarly b<a is impossible, so b=a, and we're done.

[/hide]

But i thought an angle bisector through A would go through the incenter, not where the incircle touches side BC.

[color=cyan]Well, I'm working this out as I'm typing, but I suspect we can basically follow the idea of Stephen and make it work.  I won't put my solution in spoiler because I don't like reading it afterwards, but know that a solution (hopefully) follows.

Let the point of tangency of the incircle with AB be F.
By tangents to a circle, we get that AF = AE, BF = BD, and CD = CE.  If we let AB = c, BC = a, and CA = b, we find that
AF = AE = (b + c - a)/2
BF = BD = (c + a - b)/2
CD = CE = (a + b - c)/2.

Oy, this is going to be messy.  Lets look at Stewarts Theorem for AD.

a*AD:^2: + a*(c + a - b)*(a + b - c)/4 = c:^2:*(a + b - c)/2 + b:^2:*(c + a - b)/2

So, I'm going to work that out by hand . . .

AD:^2: = (3ac:^2: + 3ab:^2: + 2bc:^2: + 2b:^2:c - 2abc - a:^3: - 2b:^3: - 2c:^3:)/(4a)

Therefore
BE:^2: = (3bc:^2: + 3ba:^2: + 2ac:^2: + 2a:^2:c - 2abc - b:^3: - 2a:^3: - 2c:^3:)/(4b)

okay, this looks ugly now.  But keep working at it.  Subtract one from the other.  Then assume the difference is less than 2 and clear your denominator.  The nice side will be 4ab.  The messy side will be messy.  You must do the following:  assume a does not equal b.  Then find some combination of terms always greater than 4ab, since you know they are all integers.  Then, show that all other terms add up to a non-negative number.  You can do this largely by using the fact that c < a + b, or possibly in this case that c :le: a + b - 1.  I promise this works.  I don't promise it's nice.[/color]


[quote="ln(dx/dy)"]I think this was a handbook problem this year:

What is [hide]$\sqrt{20+\sqrt{20+\sqrt{20+...}}}$[/hide]?

It's pretty easy, but I have no other posts to put up![/quote]

[hide="Solution"]Let it equal x.
x=sqrt(20+x)
x^2=20+x
x=5[/hide]

[hide="a fast way to do it"]a fast way to do it is just plug some of it in your calculator and see what it approaches.[/hide]

Problems like this are usually on the sprint round, thus making that method impossible.


inequalities function calculus derivative geometry LaTeX

Given that a,b and c are three side lengths of a triangle , and let A be its area. Prove that

    a^2 +b^2 + c^2 - (a-b)^2 - (b-c)^2 - (c-a)^2   >=  4A sqrt(3)

    where a^2 is the square of 2 ,  sqrt(3) is the aquare root of 3.

   Thanks....

E=a^2 + b^2 +c^2 >= (ab +bc +ca )
=2A(1/sin<A  +1/sin<B + 1/sin<C)
Now 1/sin is a convex fonction in [0,pi]
so by Jensen
(1/sin<A  +1/sin<B + 1/sin<C)>=3/(sin((<A+<B+<C)/3))
=1/sin 60 =  2/sqr root{3}

there fore E>=4sqr root{3}A equality for equilateral triangles

 :oops:  :oops:  :blush:

Apratimdefermat,

Thanks for your instructions. :roll:

do u have any more problems ?....

if u like u can try some i have posted today

Apratimdefermat,

I have doubt in  the convexity oof function 1/sinx , x in (0,pi).Explain it ,please.

the second derivative of 1/sin is :

1/sin  +2cos/sin^3  which evidently is greater than 0 in (0,pi) which ascertains that 1/sin is convex but i mentioned wrongly that the interval is [0,pi] ,  :oops: it should be (0,pi)

As you said , the second derivative of 1/six  =1/sin +2cos/sin^3 , its value of the second derivative  > 0 as x in (0,pi/2] but cannot be sure  that    it  is also  greater than 0  when x in [pi/2 ,pi) ,  :roll:  right?

Have u verified the second derivative urself? i may have made some mistake. i am thinking about it nnn meanwhile i give here another proof

16A^2= (a+b+c)(a+b-c)(a+c-b)(b+c-a)<=(a+b+c)*((a+b+c)/3)^3

therefore 4A<=((a+b+c)^2)/(3*sqr root{3})

=sqrt{3}*(a+b+c/3)^2<= sqrt{3}*(a^2+b^2+c^2)/3

which is what u wanted .

Verify the second derivative at 30 degrees.... let me know

As your last proof , I found you only tried to get

    a^2+b^2 +c^ >= 4A sqrt (3) , but not     a^2 + b^2 +c^2 -(a-b)^2 -(b-c)^2 -(c-a)^2 > = 4A sqr(3)

very sorry i read ur ineq wrongly... :oops:

so here it is

apratimdefermat,
Thanks for your last demonstrations to show the work of proving the geomeric inequality.The proof is very elegant and perfect. Beside , how can you paste the neat statement with traditional symbols on the discussion board?
The another proof as follows:

a^2= b^2+c^2- 2bc cos A =( b - c ) ^2 +2bc - 2bc cos A =( b - c ) ^2+ 2bc (1 - cos A) = ( b - c ) ^2 +4S(1-cos A)/sin A , where S is the area of triangle, and  S= 1/2 bc sin A
       And (1-cos A)/sin A = 2(sin A/2)^2  /2(sinA/2)(cosA/2) = tan (A/2)

therefore, a^2=( b - c) ^2 + 4S tan (A/2)

Similarly, b^2 = ( c- a) ^2 + 4S tan (B/2)
   and      c^2 = ( a-b ) ^2 + 4S tan (C/2)

       where    A/2 , B/2, C/2 all lie in [0, pi/2]  ,and tan is convex function in [0, pi/2)
Thus  tan (A/2) + tan (B/2) + tan (C/2)  >= 3{ tan (A/2+B/2+C/2)/3}
                                                                    = 3 tan((A+B+C)/6) = 3 tan (30 deg) = sqrt(3)

hence  a^2 + b^2 +  c^2  >= ( b - c) ^2 + ( c- a) ^2 + ( a-b ) ^2 + 4S  sqrt(3)
     <=>   a^2 + b^2 +  c^2  - ( b - c) ^2 - ( c- a) ^2 - ( a-b ) ^2 >=  4S  sqrt(3).
        Where S = A , A in the original inequality .
So,  a^2 + b^2 +  c^2  - ( b - c) ^2 - ( c- a) ^2 - ( a-b ) ^2 >=  4A  sqrt(3)  .....#

If you want other people to read your posts easier, use LaTeX - you pretty much just put Dollar Signs $ \$$ around your equation and it looks awesome. For example, a^2b^2+c^2d^2 >= 2abcd ---> $ a^2b^2\plus{}c^2d^2\ge 2abcd$ (a^2b^2+c^2d^2\ge 2abcd) and $ \sqrt{2} \neq 1$ (\sqrt{2} \neq 1).

That said, the inequality is known as the Hadwiger-Finsler Inequality, of which there are quite a lot of proofs :)


show $ \sum_{i=1}^{p-1}\lfloor{\frac{i^{3}}{p}}\rfloor=\frac{(p-1)(p+1)(p-2)}{4}$ for odd prime p.

[hide="hint"]
We know $ \frac{i^{3}}{p}+\frac{(p-i)^{3}}{p}=\frac{p(i^{2}-i(p-i)+(p-i)^{2})}{p}=i^{2}-i(p-i)+(p-i)^{2}$.
But
$ \lfloor{\frac{i^{3}}{p}\rfloor}+\lfloor{\frac{(p-i)^{3}}{p}\rfloor}$
is also an integer.  So what must it be? [/hide]

i cant still get it.
since $ \frac{1}{2}\sum_{k=1}^{p-1}\lfloor{\frac{k^{3}}{p}}\rfloor+\lfloor{\frac{(p-k)^{3}}{p}}\rfloor=\frac{p(p-1)^{2}}{4}$ (assuming i did the algebra right).

can you explain further?

Careful!  You calculated what it would be if there were no floor functions. :)   You have to subtract something from the formula in order to account for the floor function.  I was hinting at the following.
[hide="Solution"]
If $ \frac{i^{3}}{p}+\frac{(p-i)^{3}}{p}=i^{2}-i(p-i)+(p-i)^{2}$,
$ \lfloor{\frac{i^{3}}{p}\rfloor}+\lfloor{\frac{(p-i)^{3}}{p}\rfloor}=i^{2}-i(p-i)+(p-i)^{2}-1$
This is because $ 0 \le x-\lfloor{x\rfloor}< 1$.  For each $ x=\frac{i^{3}}{p}$, the value at left is positive, because $ i^{3}$ is not divisible by $ p$.  But we can pair them so their pairwise sums without the floor functions are integers.  If the sum of two numbers with the floor functions is an integer, and without the floor functions is an integer, we conclude that the difference of the two sums is an integer.  Those integers are positive, and are less than two by the above inequality.  Thus, we deduce the value of $ 1$ as above.
So,

\[ \frac{1}{2}\sum_{i=1}^{p-1}\left(\lfloor{\frac{i^{3}}{p}\rfloor}+\lfloor{\frac{(p-i)^{3}}{p}\rfloor}\right)\]
\[ =\frac{1}{2}\sum_{i=1}^{p-1}(i^{2}-i(p-i)+(p-i)^{2}-1)\]
\[ =\frac{1}{2}\sum_{i=1}^{p-1}(p^{2}-3ip+3i^{2}-1)\]
\[ =\frac{1}{2}\left((p^{2}-1)(p-1)-\frac{3p^{2}(p-1)}{2}+\frac{(p-1)p(2p-1)}{2}\right)\]
\[ =\frac{p-1}{4}\left(2p^{2}-2-3p^{2}+2p^{2}-p \right)\]
\[ =\frac{p-1}{4}(p^{2}-p-2)\]
\[ =\frac{(p-1)(p+1)(p-2)}{4}\]
[/hide]


geometry
trapezoid
ratio
rectangle
parallelogram
geometric transformation
homothety

It's not very easy (only one person was able to solve it completely in Slovakia), but maybe for you it will be easy.  :D  :P

[quote="ondrob"]Let $ABCD$ be a trapezoid such that $AB$ is parallel to $CD$, and let $E$ be the midpoint of its side $BC$. Suppose we can inscribe a circle into the quadrilateral $ABED$, and that we can inscribe a circle into the quadrilateral $AECD$. Denote $|AB|=a$, $|BC|=b$, $|CD|=c$, $|DA|=d$. Prove that \[a+c=\frac{b}{3}+d;\]  \[\frac{1}{a}+\frac{1}{c}=\frac{3}{b}\] [/quote]
Denote $AE = x, DE = y$. Since the quadrilaterals $ABED, AECD$ are both tangential, the sums of their opposite sides are equal:

$a+y = \frac{b}{ }2+d = x+c$

First, assume that the trapezoid $ABCD$ is a rectangle. Then $a = c, b = d, x = y$. Let $k = \frac{b}{ }a$ be the aspect ratio of the rectangle:

$b = d = ka,\ \ x = y = \sqrt{a^{2}+\frac{b^{2}}{4}}= a \sqrt{1+\frac{k^{2}}{4}}$

$a+a \sqrt{1+\frac{k^{2}}{4}}= \frac{ka}{2}+ka$

$1+\frac{k^{2}}{4}= \left(\frac{3k}{2}-1\right)^{2}= \frac{9k^{2}}{4}-3k+1$

$2k^{2}-3k = 0,\ \ k = \frac{3}{ }2$ [color=white].[/color] or  [color=white].[/color] $0$

The zero root can be ignored, it is not a root of the original equation before we squared it. If the aspect ratio is $k = \frac{3}{ }2$, then $b = d = \frac{3a}{2}$, $x = y = a \sqrt{1+\frac{9}{16}}= \frac{5a}{4}$ and

$\frac{b}{ }3+d = \frac{a}{ }2+\frac{3a}{2}= 2a = a+c$

$\frac{3}{ }b = \frac{2}{ }a = \frac{1}{ }a+\frac{1}{ }c$

It is clear that the trapezoid $ABCD$ cannot be any other non-rectangular parallelogram, because then $a = c, b = d, x \neq y$ and the quadrilaterals $ABED, AECD$ cannot be simultaneously tangential.

Assume that $ABCD$ is a general trapezoid. Let $(I_{1}), (E_{2})$ be the incircles of the quadrilaterals $ABED, AECD$ (the purpose of this notation will become clear later on). Let $P_{1}, Q_{1}, R_{1}, S_{1}$ be the tangency points of the circle $(I_{1})$ with the sides $AB, BE, ED, DA$ and $P_{2}, Q_{2}, R_{2}, S_{2}$ the tangency points of the circle $(E_{2})$ with the sides $AE, EC, CD, DA$, respectively. Let $B', C'$ be the intersections of the lines $DE, AE$ with the parallel lines $AB, CD$. The quadrilateral $AB'C'D$ is a parallelogram with the sides $AB' = C'D = a+c,\ B'C' = DA = d$ and the diagonal intersection $E \equiv AC' \cap B'D$. This is because the $E$ is the midpoint of the segment $BC$. Consequently, the triangles $\triangle ABE \cong \triangle C'CE$ and the triangles $\triangle DCE \cong \triangle B'BE$ are congruent by ASA and the diagonals $AC', B'D$ cut each other in half. The circles $(I_{1}), (E_{2})$ are the incircles of the triangles $\triangle AB'D, \triangle AC'D$. Let $s_{1}= \frac{a+c+d+2y}{2}$, $s_{2}= \frac{a+c+d+2x}{2}$ be the semiperimeters of these 2 triangles. The tangent lengths $ER_{1}, EP_{2}$ to the circles $(I_{1}), (E_{2})$ from the diagonal intersection $E$ are equal to

$ER_{1}= DE-DR_{1}= y-(s_{1}-AB') =$

$= y-\left[\frac{a+c+d+2y}{2}-(a+c)\right] = \frac{a+c-d}{2}$

$EP_{2}= AE-AP_{2}= x-(s_{2}-C'D) =$

$= x-\left[\frac{a+c+d+2x}{2}-(a+c)\right] = \frac{a+c-d}{2}$

Since the tangent lengths $ER_{1}= EP_{2}$ are equal, the point $E$ is on the radical axis of the circles $(I_{1}), (E_{2})$. This means that the tangent lengths $EQ_{1}= EQ_{2}= ER_{1}= EP_{2}= \frac{a+c-d}{2}$ form the point $E$ to the circles $(I_{1}), (E_{2})$ are all equal. The segment lengths $Q_{1}Q_{2}= S_{1}S_{2}= 2EQ_{1}= a+c-d$ of their common external tangents $DA, BC$ are also equal. Hence, the segment $d = DA$ is equal to

$d = DA = AS_{2}+DS_{1}-S_{1}S_{2}= AP_{2}+DR_{1}-S_{1}S_{2}=$

$= AE-EP_{2}+DE-ER_{1}-S_{1}S_{2}= x+y-2(a+c-d)$

If we add the original 2 equations for the equal sums of the opposite sides of the tangential quadrilaterals $ABED, AECD$, we obtain

$a+y+c+x = 2d+b,\ \ x+y = 2d+b-(a+c)$

Substituting this to the above equation leads to

$d = 2d+b-(a+c)-2(a+c-d),\ \ 3(a+c) = b+3d,\ \ a+c = \frac{b}{ }3+d$ [color=white]. .[/color] :cool:

It follows that the segment lengths $Q_{1}Q_{2}= S_{1}S_{2}$ of the common external tangents of the circles $(I_{1}), (E_{2})$ are equal to

$Q_{1}Q_{2}= S_{1}S_{2}= a+c-d = \frac{b}{ }3$

Since $E$ is the midpoint of the segment $BC$, the tangent lengths $BP_{1}= BQ_{1}$ to the circle $(I_{1})$ from the vertex $B$ are equal to the tangent lengths $CQ_{2}= CR_{2}$ from the vertex $C$ to the circle $(E_{2})$, all of them being equal to

$BP_{1}= BQ_{1}= CQ_{2}= CR_{2}= BE-EQ_{1}= \frac{b}{ }2-\frac{b}{ }6 = \frac{b}{ }3 = Q_{1}Q_{2}$

Let $H$ be the intersection of the lines $BC, DA$, the common external tangents of the circles $(I_{1}), (E_{2})$. This is the external homothety center of these 2 circles and also the homothety center of the centrally similar triangles $\triangle ABH \sim \triangle DCH$. Let $\eta = \frac{HA}{HD}= \frac{HB}{HC}= \frac{AB}{DC}= \frac{a}{ }c$ be the homothety coefficient of these 2 triangles. The circle $(I_{1})$ is the incircle of the triangle $\triangle ABH$ and the circle $(E_{2})$ is the excircle of the triangle $\triangle DCH$ against the vertex $H$. Let $(E_{1})$ be the excircle of the triangle $\triangle ABH$ against the vertex $H$ tangent to the side $AB$ at a point $T_{1}$ and $(I_{2})$ the incircle of the triangle $\triangle DCH$ tangent to the side $DC$ at a point $T_{2}$. The circles $(I_{1}) \sim (I_{2})$ and the circles $(E_{1}) \sim (E_{2})$ are centrally similar with the same homothety center $H$ and coefficient $\eta$ as the triangles $\triangle ABH \sim \triangle DCH$. Let $\sigma_{1}, \sigma_{2}$ be the semiperimeters of these 2 triangles, $\sigma_{1}= \eta \sigma_{2}$. The tangent length $BP_{1}$ to the incircle $(I_{1})$ from the vertex $B$ is equal to the tangent length $AT_{1}$ to the excircle $(E_{1})$ from the vertex $A$:

$BP_{1}= \sigma_{1}-HA = AT_{1}$

Similarly, the tangent length $CT_{2}$ to the incircle $(I_{2})$ from the vertex $C$ is equal to the tangent length $DR_{2}$ to the excircle $(E_{2})$ from the vertex $D$:

$CT_{2}= \sigma_{2}-HC = DR_{2}$

From the central similarity of the triangles $\triangle ABH \sim \triangle DCH$ with their incircles $(I_{1}), (I_{2})$ and excircles $(E_{1}), (E_{2})$, we have

$\frac{AP_{1}}{BP_{1}}= \frac{AP_{1}}{CR_{2}}= \frac{AP_{1}}{DT_{2}}= \eta$

$\frac{CR_{2}}{DR_{2}}= \frac{DT_{2}}{CT_{2}}= \frac{AP_{1}}{BP_{1}}= \eta$

Hence, the tangency points $P_{1}, R_{2}$ divide the sides $AB, CD$ in the same ratio equal to the homothety coefficient $\eta$. As a result,

$a = AB = AP_{1}+BP_{1}= \eta BP_{1}+BP_{1}= \left(\frac{a}{ }c+1\right) \frac{b}{ }3$

$\frac{ac}{a+c}= \frac{b}{ }3,\ \ \frac{1}{ }a+\frac{1}{ }c = \frac{3}{ }b$ [color=white]. .[/color]  :cool:

Ahoj, Yetti

Yeah, but it looks quite long ;) I will describe as I solved it (but I'm a bit lazy to write it all down, maybe later ;) )

We know, when we denote sides of $ABCD$ like $a,b,c,d$ to compute lengths of $DE,EB',BB'$ in terms of $a,b,c,d$ easily (using tangentiality). So we can easily compute semiperimeters of $\triangle AB'D,\triangle BB'E$ in terms of $a,b,c,d$ (denote them $p_1,p_2$) and also denote $|BE|$ as $x$. Also denote areas of $\triangle AB'D,\triangle BB'E$ as $P_1,P_2$. Now using that circle inscribed in $ABED$ is incircle of $\triangle AB'D$ and excircle of $\triangle BB'E$ we get:
$\frac{P_1}{p_1}=\frac{P_2}{p_2-x} $
We can rewrite it as:
$\frac{P_1}{P_2}=\frac{p_2-x}{p_1}$
But $\frac{P_1}{P_2}$ we can easily compute as:
$\frac{P_1}{P_2}=\frac{AB'}{BB'}\frac{DB'}{EB'}=\frac{2(a+c)}{c}$
Using this in
$\frac{P_1}{P_2}=\frac{p_2-x}{p_1}$
we get:
$2(a+c)p_1=c(p_2-x)$
But we can write it in terms of $a,b,c,d$ with $b,d$ being just in linear form!
Analogicaly we can get another linear equation of $b,d$ by using that the second circle is inscribed and excribed at the same time. From them we can express $b,d$ (just solve system of 2 linear equations) and we get:
$b=\frac{3ac}{a+c}\qquad d=\frac{a^2+ac+c^2}{a+c}$
Putting this in the two equations we get true statements.

If someone can't find out how exactly I did it or something, just tell, I'll find time to write it completely.


calculus
integration
analytic geometry
trigonometry
geometry
function
calculus computations

1. The axes of two right circular cylinders of radius a intersect at right angles. Find the volume of the solid bounded by the cylinder. ( I have no idea how to do this)

2. What is the length of the arc 30xy^3 - y^8 = 15, 1<=y<=2 ?
(I tried finding x' and calculating the integral for arc length but end up not knowing how to integrate)

3. Evaluate Int[ (sin x)^6 (cos x)^2 dx ] from 0 to pi

For number 3, I know I can turn it into Int[ (sin x)^6] - Int[ (sin x)^8] and then integrate them individually but it's really really long. Is there any other way to do it?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1041544#1041544]This announcement[/url] comes from a different forum, but I agree with the sentiment.

#1: Align them so both cylinders are horizontal. Each horizontal cross-section is a square- how does the side length depend on the $ z$-coordinate?

#2: Either it's rigged to be a perfect square inside the square root, or it's impossible. The first is more likely.

#3: It's tedious no matter what. Either use integration by parts to generate reduction formulas, or use double-angle formulas repeatedly. With these nice bounds, the reduction approach works well.
$ \int_0^{\pi}\sin^6 x\cos x\cdot \cos x\,dx\equal{} \left[\frac17\sin^7 x\cdot\cos x\right]_0^{\pi}\plus{} \int_0^{\pi}\frac17\sin^7 x\cdot \sin x\,dx$
$ \equal{} \frac17\left(\int_0^{\pi}\sin^6 x\,dx\minus{}\int_0^{\pi}\sin^6 x\cos^2 x\,dx\right)$
Repeat, tweaked slightly; we go down by two powers of $ \sin$ at each step.

For the last since
I=$ \int_0^{\pi} (\sin {x})^6 (\cos x)^2\,dx \equal{} \int_0^{\pi} (\sin ({\pi} \minus{} x))^6 (\cos ({\pi} \minus{} x))^2\,dx$
So I=$ 2\times \int_0^{\pi/2} (\sin x)^6 (\cos x)^2\,dx$
Now use walli's formula to get it as:$ 2\times \frac {5}{8} \times \frac {3}{6} \times \frac {1}{4} \times \frac {1}{2} \times \frac {\pi}{2}$

Thank you guys. I just learned about Walli's formula. Number 3 becomes indeed easier with it. And btw can you elaborate more on number 1? Sorry but I dont quite get it.

Visualize the region as a stack of thin slices; the volume will be the integral of the areas. You need to find the area of these cross-sections as a function of height to set up the integral.


function
number theory unsolved
number theory

find all functions $ f$ , \[ f : \mathbb{N}_0  \longrightarrow \mathbb{N}_0\] \[ f(m^2\plus{}n^2)\equal{}f(m)^2\plus{}f(n)^2\]

You've already posted this:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=187369[/url]

Do not double post.


Given $ a, b, c \geq\ 0.$ Prove that:
$ 2\sqrt {2}(a \plus{} b \plus{} c) \geq\ \sqrt {a^2 \plus{} b^2} \plus{} \sqrt {b^2 \plus{} c^2} \plus{} \sqrt {c^2 \plus{} a^2} \plus{} 3\sqrt{2}.\sqrt [3]{abc}$

I think solution by AM-GM is nicer.  :wink:
$ \sqrt{a^2\plus{}b^2}\plus{} \sqrt{2ab} \le \sqrt{2} (a\plus{}b)$
$ \sqrt{b^2\plus{}c^2}\plus{} \sqrt{2ab} \le \sqrt{2}(b\plus{}c)$
$ \sqrt{c^2\plus{}a^2}\plus{} \sqrt{2ca} \le \sqrt{2}(c\plus{}a)$
$ 3\sqrt{2} \sqrt[3]{abc} \le \sqrt{2ab}\plus{} \sqrt{2bc}\plus{} \sqrt{2ca}$


calculus
integration
trigonometry
calculus computations

[b]Please Solve this Integral for me.[/b]

Integral of - (Sin x)/x

Thanks a lot.

Do you mean $\int_{0}^{+\infty}\frac{\sin x}{x}dx$ ?

My Mathematical English is very poor. :(

You might want to try doing $\int \frac{1}{x} \sin{x}$ by parts. I don't know if it becomes redundant, its just an idea.

Also, do you mean $\int \frac{\sin{x}}{x}$ or $\int \frac{-\sin{x}}{x}$?

All right. The antiderivative of $\frac{\sin x}{x}$ is not an elementary function. The improper integral $\int_0^\infty \frac{\sin x}{x}\, dx$ converges to $\frac\pi2$.

While integration by parts doesn't get you an antiderivative, it can be used to prove that the improper integral above converges. Note first that the singularity at zero is removable- the only problem comes as we go to infinity.

$\int_0^\infty \frac{\sin x}{x}\, dx=\left[\frac{1-\cos x}{x}\right]_0^\infty +\int_0^\infty \frac{1-\cos x}{x^2}\, dx=\int_0^\infty \frac{1-\cos x}{x^2}\, dx$
Near zero, $\frac{1-\cos x}{x^2}$ is bounded and continuous, thus integrable. Away from zero, the integral converges by direct comparison to $\int \frac2{x^2}\, dx$.

yeah converges? definitely...antiderivative...i guess its not elementary. thats good, now it wont bother me anymore. lol just like the first time someone told me to find the antiderivative of $e^{-(x^2)}$. boy that was fun..... :?


If June 24 falls on Wednesday in a leap year, on what day of the week will it fall the next year?

$ 365 \equiv 1 \pmod{7}$ so it falls on a Thursday.


[color=darkblue] Solve in $ \mathbb{N}^3$ the equation
\[ xy \plus{} yz \plus{} zy \minus{} xyz \equal{} 2
\]
[/color]

[quote="Ahiles"][color=darkblue] Solve in $ \mathbb{N}^3$ the equation
\[ xy \plus{} yz \plus{} zx \minus{} xyz \equal{} 2
\]
[/color][/quote]
WLOG,we assume that $ x \equal{} min(x,y,z)$;

then from the given equation we can easily see that $ x \leq 2$;

$ *$for $ x \equal{} 2$,the equation becomes $ (y \minus{} 2)(z \minus{} 2) \equal{} 2$;

$ *$for $ x \equal{} 1$,we'll have $ y \plus{} z \equal{} 2$;

$ *$for $ x \equal{} 0$,It gives $ yz \equal{} 2$;

solving the three equations we conclude that the solutions are $ (2,4,3);(2,3,4);(1,0,2);(1,2,0);(1,1,1);(0,2,1);(0,1,2)$ and its permutations.


Sorry; Is there any $\LaTeX$ version, and "PDF maker" available for pocket PC?

[quote="Valentin Vornicu"]For example http://www.rainer-keuchel.de/wince/texce.html[/quote]
Thank you very much :)


On a table there are 100 tokens. Taking turns two players remove 5, 6, 7, 8, 9 or 10 tokens, at their choice. The player that removes the last token wins. Find a winning strategy and determine which player will be the winner.

[hide="Solution"]
Player 1 can always win.
[b]Strategy:[/b]
First leave 90 tokens. Whatever player 2 does, leave 75 tokens on your turn (then 60, then 45 etc). This way at some point player 2 will have to make a move where there's 15 tokens left. Any move he makes will leave player one to take the last tokens.

[/hide]


For any three positive numbers $x$, $y$, $z$ satisfying $x+y+z=1$, prove that

$\frac{xy}{\sqrt{xy+yz}}+\frac{yz}{\sqrt{yz+zx}}+\frac{zx}{\sqrt{zx+xy}}\le \frac{\sqrt 2}2$.

notice that it is only cyclic.

If we put $xy=a$ etc   we obtain

$\sum\frac{a}{\sqrt{a+b}}\leq\frac{\sqrt{2}}{2}$

The LHS is well known sum , but.....I will think about it

[quote="Albanian Eagle"]very nice!
from cauchy :
$(LHS)^2\le (x+y+z)(\sum \frac{xy}{x+y})\le \frac{1}{4}(\sum x+y)=\frac{1}{2}$ :D[/quote]
Albanian Eagle, maybe $(LHS)^2\le (x+y+z)(\sum \frac{xy}{x+z})$ $?$ ;)

[quote="zhaobin"]for positive number $x,y,z$ such that $x+y+z=1$
prove:$\frac{xy}{\sqrt{xy+yz}}+\frac{yz}{\sqrt{yz+zx}}+\frac{zx}{\sqrt{zx+xy}} \le \frac{\sqrt 2}2$
notice that it is only cyclic.[/quote]
$\sum_{cyc}\frac{xy}{\sqrt{xy+yz}}=\sum_{cyc}\sqrt{\frac{x^2y}{x+z}}=\sum_{cyc}\left(\frac{x+y}{2}\cdot\sqrt{\frac{4x^2y}{(x+y)^2(x+z)}}\right)\leq\sqrt{\sum_{cyc}\frac{2x^2y}{(x+y)(x+z)}}.$
Remain to prove that $\sum_{cyc}\frac{2x^2y}{(x+y)(x+z)}\leq\frac{1}{2}.$
But $\sum_{cyc}\frac{2x^2y}{(x+y)(x+z)}\leq\frac{1}{2}\Leftrightarrow\sum_{cyc}4x^2y(y+z)\leq(x+y)(x+z)(y+z)(x+y+z)\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(x^3y+x^3z-2x^2y^2)\geq0.$
"Jensen forever!"(Myth)  :)

Thank you very much.
Very nice.
And also we can use cauchy for the first step. :)

[quote="Albanian Eagle"]very nice!
from cauchy :
$(LHS)^2\le (x+y+z)(\sum \frac{xy}{x+y})\le \frac{1}{4}(\sum x+y)=\frac{1}{2}$ :D[/quote]

Is this true ?

[quote="Davron"]Eagle it is not true and for the contradiction try $\frac{1}{2} , \frac{1}{6}, \frac{1}{3}$.

Davron[/quote]

Just read [b][size=150]arqady[/size][/b]'s posts carefully...  ;)

[quote="zhaobin"]for positive number $x,y,z$ such that $x+y+z=1$
prove:$\frac{xy}{\sqrt{xy+yz}}+\frac{yz}{\sqrt{yz+zx}}+\frac{zx}{\sqrt{zx+xy}} \le \frac{\sqrt 2}2$

notice that it is only cyclic.[/quote]
I used Jensen too but in this way,
$\frac{xy}{\sqrt{xy+yz}}+\frac{yz}{\sqrt{yz+zx}}+\frac{zx}{\sqrt{zx+xy}} \le \frac{\sqrt 2}2$<=>
\[ \sum_{cyc}{\frac{xy}{xy+xz+yz}*\sqrt{\frac{2(xy+xz+yz)^2}{(xy+yz)(x+y+z)^2}}\leq 1 }\]
but
\[ \sum_{cyc}{\frac{xy}{xy+xz+yz}*\sqrt{\frac{2(xy+xz+yz)^2}{(xy+yz)(x+y+z)^2}}\leq \sqrt{\sum_{cyc}{\frac{2x(xy+xz+yz)}{(x+z)(x+y+z)^2}}} }\]=>
we have to prove that
\[ \sum_{cyc}{\frac{2x}{x+z}}\leq \frac{(x+y+z)^2}{xy+xz+yz} \]
<=>
\[ \sum_{cyc}{\frac{2x^2z}{x+z}}\leq x^2+y^2+z^2 \]
<=>
\[ \frac{x^2+y^2+z^2}{2}\leq \frac{x^4}{x^2+xz}+\frac{y^4}{y^2+xy}+\frac{z^4}{z^2+yz} \]
but we have that $\frac{x^4}{x^2+xz}+\frac{y^4}{y^2+xy}+\frac{z^4}{z^2+yz}\geq \frac{(x^2+y^2+z^2)^2}{x^2+y^2+z^2+xy+xz+yz}\geq \frac{x^2+y^2+z^2}{2}$ so :) ,the ineq is done ;) .

[quote="zhaobin"]Nice work.Tiks :)[/quote]
Thankx $zhaobin$ :blush: .

[quote="kcbu"][quote="Albanian Eagle"]very nice!
from cauchy :
$(LHS)^{2}\le (x+y+z)(\sum \frac{xy}{x+y})\le \frac{1}{4}(\sum x+y)=\frac{1}{2}$ :D[/quote]
Um, I think
$(LHS)^{2}\le (x+y+z)(\sum \frac{xy}{x+z})$ not x+y[/quote]

You all fail to read Arqady's post :P !
He had noticed that before you :wink: .

[quote="delegat"]I am not familiar with Jensen inequality so can you just explain me this part of your solution
That part of Tiks solution is in third line, beyond but,and I didnt understand how did you get it :?:[/quote]

[url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=karamata&t=14975[/url]

Read this topic! Jensen's weighted inequality is stated there(see Myth's post). Hope it helps :wink: .

$ \sum_{cyc}{z\sqrt {\frac {2x}{y \plus{} z}}}\leq\sum_{cyc}{z\frac {x \plus{} 2(y \plus{} z) \plus{} 3\left(\sqrt {zx} \plus{} \sqrt {xy} \minus{} \sqrt {yz}\right)}{4(y \plus{} z)}}\leq x \plus{} y \plus{} z.$

A development see here: http://www.mathlinks.ro/viewtopic.php?t=216138

EDIT: [url=http://www.mathlinks.ro/viewtopic.php?t=97495]toanhocmuonmau's proof[/url] is better.


In a city in year 1990 have lived 700000 residents while in year 2000 there were 900000 residents

If you take the growth in population made at the same rate , find :

a) how much is the population grow of the city in the time periode 1990-2000 ?

b) how much became the population number this year(2009) ?

c) when will the city have 1400000?

GBased on the title of the thread, I think you have to assume the population grows following a geometric series that increases by some ratio each year.

The challenge here is by knowing the ratio. Then I think we can the rest by this $ A\left( t \right) \equal{} A\left( {1 \pm r} \right)^t$ and I believe that we need logs to know the ratio but I'm still not sure  :huh: .

If the population increases by some ratio each year, $ a\equal{}700000$, and after 10 years we've multiplied by some ratio $ r$ 10 times.
That is, $ r\equal{}\sqrt[10]{\frac{900000}{700000}}\approx 1.025$

Using that information you can solve a, b, and c.

I'm making the assumption that the number increases each year though (you didn't really specify).

I don't understand what "If you take the growth in population made at the same rate" means.

I have made this mistake $ r\equal{}\sqrt[9]{\frac{900000}{700000}}\approx 1.028$    :blush:
thanks to all you.


[hide="hint"]If the product of several terms is 0, at least one must be 0.  Factor the equations if they are not in factored form.[/hide]

[hide="1, for an example"]Either $x=0$, $x+6=0$, or $5x-8=0$.  These produce the solutions $x=0,-6,\frac85$.[/hide]

[quote="Zero Product Property"]The Zero Product Property states that whenever a product is equal to 0, then one of the products must be 0.
[/quote]

In this case, for number 1, set each of them to zero, so:

$x=0$

$x+6=0$

$5x-8=0$

Solve for x in all three cases and that is your solution.

[quote="jbf13"]Solve each of the following by using the zero product property

1. X(X+6)(5X-8)=0

2. 3X^2+9X-120=0

3. -6X^2+72X=0
------------------------------------------------------------------------------------------------------------------------------

NOTE- X^2 = X(X) OR X SQUARED.[/quote]

[hide]
For 1, set them all to zero and you find $x=0,-6, \frac{8}{5}$

For #2, divide by 3 to get $x^{2}+3x-40=0$.  Now, this factors as $(x+8)(x-5)=0$.  Therefore, $x=-8,5$ by the zero product property.

Finally, for #3, divide both sides by $-6$ to get $x^{2}-12x=0$.  Factoring, we get $x(x-12)=0$, so $x=0, 12$

Make sure you see how dividing by the constant doesn't change the solutions.
[/hide]

Just remember not to divide by anything containing x, as you will lose solutions.

Well, x is a variable, but good point.

I just don't like dealing with cases where the coefficient of $x^{2}$ is not 1.

Well alan, you got lucky this time.  If the quadratics couldn't be evenly divided by a constant, I'd rather work through all the logic with $ax^{2}$ than have to deal with fractional coefficients.

[quote="ericlu149"][quote="jbf13"]Solve each of the following by using the zero product property

1. X(X+6)(5X-8)=0

2. 3X^2+9X-120=0

3. -6X^2+72X=0
------------------------------------------------------------------------------------------------------------------------------

NOTE- X^2 = X(X) OR X SQUARED.[/quote]

[hide=" A few possible ways to solve those"]
You could use the quadratic formula for 2 and 3 or factor them.
If you factor them, you need to get each of those in the ( ) to zero.

So for 1, you get $ 0,-6,$ and $ \frac{8}{5}$

For 2, you get $ 3(x^{2}+3x-40) = 0$
So, $ 3(x+8)(x-5)=0$. So the possible solutions are $ -8$ and $ 5$

For 3, you get $ -6x(x-12) = 0$
So the solutions are 0 and 12.[/hide] :lol:[/quote]
Please don't revive old threads, if the date of the last post is more than a week or two ago, don't post!


inequalities
function
inequalities proposed

For $a,b,c$ positive reals prove that

$\displaystyle \sum\frac{2a^2-b^2-c^2}{b^2-bc+c^2} \ge 0 $

For this one no prize  :(

The more general statement is valid:
For $a,b,c$ positive reals and $n$ positive integer we have

$\displaystyle \sum\frac{2a^n-b^n-c^n}{b^2-bc+c^2} \ge 0 $

WLOG, assume $a \ge b \ge c$, then $\frac{1}{{a^2  + b^2  - ab}} \le \frac{1}{{a^2  + c^2  - ac}} \le \frac{1}{{b^2  + c^2  - bc}}$
Now, the inequality is equivalent to
\[
\left( {a^n  - b^n } \right)\left( {\frac{2}{{b^2  - bc + c^2 }} - \frac{1}{{a^2  + c^2  - ac}} - \frac{1}{{a^2  + b^2  - ab}}} \right) + \left( {b^n  - c^n } \right)\left( {\frac{1}{{b^2  - bc + c^2 }} + \frac{1}{{a^2  + c^2  - ac}} - \frac{2}{{a^2  + b^2  - ab}}} \right) \ge 0
\]
which is true.

[quote="Vasc"][quote="siuhochung"] Assume $a \ge b \ge c$, then $\frac{1}{{a^2  + b^2  - ab}} \le \frac{1}{{a^2  + c^2  - ac}} \le \frac{1}{{b^2  + c^2  - bc}}$ [/quote].
This assertion is not always true.[/quote]
I agree with you.but what about your solution?

[quote="Vasc"]Your assertion is true only if $a,b,c$ are the side lenghts of an triangle.[/quote]

oh i did wrong calculations that i get sth like (a-b)(a+b+c)>=0...
anyway, thanks for pointing out my mistake..

Which is your solution for this one in the general case, Vasc?

The siuhochung'way is very well. Assume $a \ge b \ge c$, and show that

$\frac{2}{b^2  - bc + c^2 } - \frac{1}{a^2  + c^2  - ac} - \frac{1}{a^2  + b^2  - ab} \ge 0$

and
$\frac{1}{b^2  - bc + c^2 } + \frac{1}{a^2  + c^2  - ac} - \frac{2}{a^2  + b^2  - ab} \ge 0.$

In general, if $a,b,c$ a non-negative numbers and $f$ is an increasing function, the inequality occurs

$\displaystyle \displaystyle \sum\frac{2f(a)-f(b)-f(c)}{b^2-bc+c^2} \ge 0 $.

Thank you very much  Vasc and Siuhochung for the solution. :)

Vasc, can you explain why

$\displaystyle \frac{1}{b^2 - bc + c^2 } + \frac{1}{a^2 + c^2 - ac} - \frac{2}{a^2 + b^2 - ab} \ge 0$  for $a \geq b \geq c$

I have spent two hours on it without success.

Clearing denominators we have

$a^4+b^4+2ab^2c+2a^2bc+2bc^3+2ac^3 \geq a^3c+b^3c+a^3b+4abc^2+ab^3+2c^4$

but $a^4+a^2bc \geq a^3c+a^3b$

so we remain with

$b^4+2ab^2c+a^2bc+2bc^3+2ac^3 \geq b^3c+4abc^2+ab^3+2c^4$

which is ????????????????????????????????????????

How can you say this is positive?

Sorry for spamming into this topic, but if you want just a verification, you can set a = b + x and b = c + y; since $a\geq b\geq c$, the numbers x and y are positive. Now, you can write everything in terms of c, x and y, and you are left with lots of trivially nonnegative terms (I have checked). So, the proof is complete. But this way is 100% brainless and I hope there is something more reasonable.

  darij

[quote="manlio"]Vasc, can you explain why

$\displaystyle \frac{1}{b^2 - bc + c^2 } + \frac{1}{a^2 + c^2 - ac} - \frac{2}{a^2 + b^2 - ab} \ge 0$  for $a \geq b \geq c$

I have spent two hours on it without success.[/quote]
We write the inequality as follows:

$\displaystyle \frac{1}{b^2 - bc + c^2 } - \frac{1}{a^2 + b^2 - ab} \ge \frac{1}{a^2 + b^2 - ab} - \frac{1}{a^2 + c^2 - ac}$,

or

$\displaystyle (a-c)(a+c-b)(a^2+c^2-ac) \ge  (b-c)(a-b-c)(b^2 + c^2 -bc)$.

For $a-b-c \le 0$ it is clearly true, and for $a-b-c >0$, we have

$a-c \ge b-c$,
$a+c-b \ge a-b-c$,
$a^2+c^2-ac \ge b^2+c^2-bc$. ;)

Thank you very much, Vasc !!!

It is very nice.


How many times should you shuffle a euchre deck?

Does the (3/2)log(n) rule, n being the number of cards, apply here. Is the # of cards in a euchre deck considered "large"?


See if you can solve this easy algebra problem.
Include how you did the problem, please...

When Anne, Betty, and Cynthia compared the amount of money each had, they discovered that Anne and Betty together had 12 dollars, Betty and Cynthia had together had 18 dollars, and Anne and Cynthia together had ten dollars. Who had the least money, and how much was it???

[hide]Call the amount Anne, Betty, and Cynthia had each A, B, and C, respectively. A+B=12, B+C=18, A+C=10. Subtracting A+B=12 from B+C=18 gets C-A=6. Adding this to A+C=10 gets 2C=16, or C=8. Since B+C=18, B=10. A+B=12, so A=2. Anne has the least money.[/hide]

[hide]A+B=12
B+C=18
A+C=10
adding the equations we get, 2(A+B+C)=40, A+B+C=20
we then have
A+B+C-(A+B)=C=8
A+B+C-(B+C)=A=2
A+B+C-(A+C)=B=10
Anne has the least with 2 dollars[/hide]

a - anne
b - betty
c - cyntia

a + b = 12
b + c = 18
a + c = 10

2a + 2b + 2c = 40
a + b + c = 20

c = 20 - 12
c = 8

b = 20 - 10
b = 10

a = 20 - 18
a = 2

so anne had the least - 2 dollars  :D  :D  :D

When Anne, Betty, and Cynthia compared the amount of money each had, they discovered that Anne and Betty together had 12 dollars, Betty and Cynthia had together had 18 dollars, and Anne and Cynthia together had ten dollars. Who had the least money, and how much was it?

[hide]Let a = anne
Let b = betty
Let c = cynthia

a+b=12
b+c=18
a+c=10

Add the equations together and get:

2a+2b+2c=40, which means:

2(a+b+c)=40 and:

a+b+c=20

c=20-(a+b), and since a+b=12, c=20-12, or $ 8$.
b=20-(a+c), and since a+c=10, b=20-10, or $ 10$.
a=20-(b+c), and since b=8 and c=10, and their sum is 18, a=20-18, or $ 2$.

Therefore, Anne has the least amount of money, $ \$$ $ 2$.[/hide]

anne=a
betty=b
cynthia=c
If we add the equations we get
2(a+b+c)=40
a+b+c=20
since b+c=18, we replace it in the above equation and get
a+18=20
a=20-18
a=2
Anne has the least money and she has 2 dollars
                     :D  :lol:  :D

No need to revive a thread 9 months old :wink:

the last three threads i've looked at in this topic were revived topics
DO NOT REVIVE TOPICS MORE THAN A MONTH OLD

most of the threads also had repeated solutions
very pointless....
if the way you did the problem has already been posted, why post it again?
only post if you got a different answer or have a different solution
otherwise, your post rating will be affected  :wink:

either you can't read(sorry, mean no offense man) but it has been posted already no to revive old threads...


integration
vector
linear algebra
matrix
real analysis
real analysis unsolved

A restatement of the problem. In the vectors and matrices of this problem, let the indexing of rows or columns start with 0. Let $E_n=\{a\in\mathbb{R}^{n+1}|a=[1,a_1,a_2\dots a_n]^T\}.$

Let $M_n$ be the $(n+1)\times(n+1)$ matrix whose $(j,k)$ entry is $(j+k)!.$

Then $u_n=\inf_{a\in E_n} a^TM_na.$

Given the context of the problem (finite dimensional inner product spaces, orthogonal projections), the infinum should actually be a minimum.


inequalities
number theory proposed
number theory

Find the largest positive integer $n>10$ such that the residue of $n$ when divided by each perfect square between $2$ and $\dfrac n2$ is an odd number.

[quote="Myth"]And again  :P
[hide]23?[/hide][/quote]Nope :D

This thread is locked until 24th of April.

I think we can unblock this topic.

My students got and prooved that answer is 505. It is much greater than 115. Hope they are right.

i'm very interested to see the proof. by using the computer i also got the same result.

this problem seems to be different from the one given ml contest
i hope that i solved that one correctly, but i can't do this one

It is different from the one on the ML contest. This one is much easier. You look for odd perfect squares between $\dfrac n4$ and $\dfrac n3$. If such a square say $k^2$ exists, then $ n= 3k^2 + r$ and as $n$ is odd, it's obvious that $r$ is even and we are done. Obviously for sufficiently large $n$ there exists an odd square between $\dfrac n4$ and $\dfrac n3$.

Using some algebra and inequalities we can estimate such an $n$, and then we find close to him that $n=505$ verifies, thus it is the largest one.


Consider ABC a triangle and 3 other isosceles triangles built on AB,BC,AC with tops P,Q,R such that m(P)=m(Q)=m(R)=120.Proove that PQR triangle(NAPOLEON'S TRIANGLE) is equilateral.

Let $F$ be the Fermat point, and $M=AF \cap PR$, $N=BF \cap PQ$, then $M,N,F,P$ cyclic $\Longrightarrow \angle{RPQ} = 60^o$.


I am trying to understand the theorem dealing with the sample variance convergence in probability.

That is, if we have an infinte sequence of iid random variables with the finite mean$ \mu$ and a finite variance $ \sigma^2$ then

$ S^2 \equal{} n^ { \minus{} 1}\sum_{i \equal{} 1}^{n}(X_{i} \minus{} \bar{X})^2\rightarrow\sigma^2$ in probability, where $ \bar{X} \equal{} n^ {\minus{} 1}\sum_{i \equal{} 1}^{n}X_{i}$.

The theorem claims that, by the Weak Law of Large Numbers, $ n^ { \minus{} 1}\sum_{i \equal{} 1}^{n}X_{i}^2\rightarrow \sigma^2 \plus{} \mu^2$ in probability and I am not sure why.  When looking under the hood of WLLN and its proof, it seems to me that I would need the [i]fourth[/i] moments be finite.  Am I reading this right?  Any help is greatly valued.

Regards.


Solve equation

$ \sqrt{x\sqrt[5]x}  \minus{}\sqrt[5]{x\sqrt{x}}\equal{}56$

[hide]$ |x^{3/5}|\minus{}x^{3/10}\minus{}56\equal{}0$

$ u: \equal{}x^{3/10}\implies u^2\minus{}u\minus{}56\equal{}0\land u\geqslant 0$, hence $ u\equal{}8\iff x\equal{}1024$[/hide]


Given $a = 3 \frac{2}{3}, c =-2, d = 5, e =4, f = 1/4, g = 3/4, h = 1, j = 9$

2b or not 2b: if $\frac{a+b}{c(e-d)}= \frac{b(f+g)}{hj}$ what is the value of b's opposite?

I solved for b and got -3, but what does "b's opposite" mean?  Is that the reciprocal of b?  The additive inverse?  The answer says it's $\frac{-33}{7}$....... :huh:

Man, that's a horribly worded problem.  "opposite" probably means additive inverse in this context.

Another confusing problem from Geometry grade 8 2005:

Thomas Greenthumb wants to expand his nursery operations by adding more greenhouses.  HE needs to calculate the surface area above ground and volume for heating and cooling purposes for the following greenhouse that is shaped like half a cylinder laid on its side.  What is the ratio of surface area to volume?  (The ends of the greenhouse are semicircles with radius BD)  The pic below is what it looks like unfolded (A net diagram)

For the Geometry grade 8 problem:
[hide]
First find the surface area of the two ends.  They together form a circle with radius 30, so the area is $pi(30)^{2}$, which is 900$pi$.  Also, the length of the second side of the rectangle is the circumference of one of the semicircles, so its length is just $30pi$.
The surface area when the rectangle is unrolled is therefore $30pi(80)$, which is $2400pi$.

From this, we get that the surface area is $3300pi$.

The volume can be found by considering the greenhouse as a sort of prism.  The height is 80, and the base area is $450pi$ (just the area of the semicircle).  The volume is therefore $80(450pi)$, or $36000pi$.  The ratio simplifies to 11/120, which is the answer.[/hide]


for positive reals \[ a_i \]\[ 1\leq i\leq n \]

prove\[ \sum_{i=1}^n \frac{\sum_{j=1}^{n-1}a_{i-1+j}}{\sum_{j=1}^n j a_{i-1+j}}\geq\frac{2(n-1)}{n+1} \]
 :D

for n=2 as example

\[ \frac{a}{a+2b} + \frac{b}{b+2a} \geq 2/3 \]


function
induction
strong induction
algebra proposed
algebra

find all function $ f:\mathbb{Z}\rightarrow{R}$ that for all $ m,n\in\mathbb{Z}$ we have:
$ f(m).f(n) \equal{} f(m\plus{}n)\plus{}f(m\minus{}n)\ \ \ ,\ \ \ f(1) \equal{}\frac{5}{2}$

[quote="MellowMelon"]This one turns out to be easy by induction. Switch $ m$ and $ n$ and conclude $ f$ is even. Put in $ n \equal{} 0$ to get $ f(0) \equal{} 2$. Now we prove by strong induction $ f(n) \equal{} 2^{n}\plus{}2^{\minus{}n}$. We already have two base cases. To show for $ n$, plug $ m \equal{} n\minus{}1$, $ n \equal{} 1$ to get $ f(n\minus{}1)f(1) \equal{} f(n)\plus{}f(n\minus{}2)$. All values are known except $ f(n)$, and the computation gets the answer.[/quote]
How did you guess that $ f(n)\equal{}2^{n}\plus{}\frac {1}{2^{n}}$. :maybe:

Before the induction, when solving the problem, I tried to compute $ f(2)$ and $ f(3)$. In decimal form I ended up getting the following:

$ f(0) \equal{} 2$
$ f(1) \equal{} 2.5$
$ f(2) \equal{} 4.25$
$ f(3) \equal{} 8.125$

By now the pattern is pretty obvious (powers of $ 2$ to the left of the decimal point, powers of $ 1/2$ to the right)

[quote="Amir.S"][quote="MellowMelon"]This one turns out to be easy by induction. Switch $ m$ and $ n$ and conclude $ f$ is even. Put in $ n \equal{} 0$ to get $ f(0) \equal{} 2$. Now we prove by strong induction $ f(n) \equal{} 2^{n}\plus{}2^{\minus{}n}$. We already have two base cases. To show for $ n$, plug $ m \equal{} n\minus{}1$, $ n \equal{} 1$ to get $ f(n\minus{}1)f(1) \equal{} f(n)\plus{}f(n\minus{}2)$. All values are known except $ f(n)$, and the computation gets the answer.[/quote]
How did you guess that $ f(n) \equal{} 2^{n}\plus{}\frac{1}{2^{n}}$. :maybe:[/quote]
you dont need to guess anything:
first of all its easy to see that $ f(0)\equal{}2$ now put $ m\equal{}n\equal{}1$ to get that $ f(2)\equal{}\frac{17}{4}$
now put $ m\equal{}1$ we get that:
$ \frac{5}{2}f(n)\equal{}f(n\plus{}1)\plus{}f(n\minus{}1)$
$ \Rightarrow 2f(n\plus{}1)\minus{}5f(n)\plus{}2f(n\minus{}1)\equal{}0$
now let $ a_{n}\equal{}f(n)$ for $ n\geq 1$...
so it becomes $ 2a_{n\plus{}1}\minus{}5a_{n}\plus{}2a_{n\minus{}1}\equal{}0$ with $ a_{1}\equal{}\frac{5}{2},a_{2}\equal{}\frac{17}{4}$
now we want to solve this recursion relation,we replace $ a_{n}$ by $ x^{n}$ to get that:
$ 2x^{n\plus{}1}\minus{}5x^{n}\plus{}2x^{n\minus{}1}\equal{}0$ which gives us:
$ 2x^{2}\minus{}5x\plus{}2\equal{}0$ so $ x\equal{}2,\frac{1}{2}$ so $ a_{n}\equal{}\lambda_{1}.2^{n}\plus{}\lambda_{2}.\frac{1}{2^{n}}$ now putting $ n\equal{}1,2$ we get that $ \lambda_{1}\equal{}\lambda_{2}\equal{}1$ so $ a_{n}\equal{}2^{n}\plus{}\frac{1}{2^{n}}$...
so $ f(n)\equal{}2^{n}\plus{}\frac{1}{2^{n}}$ for natural $ n$...
now in the original equation replace $ (m,n)$ by $ (n,m)$ to get that:
$ f(m)f(n)\equal{}f(m\plus{}n)\plus{}f(m\minus{}n)\equal{}f(m\plus{}n)\plus{}f(n\minus{}m)$
hence $ f(m\plus{}n)\equal{}f(n\minus{}m)$ so $ f(x)\equal{}f(\minus{}x)$...
so $ f(\minus{}n)\equal{}f(n)\equal{}2^{n}\plus{}\frac{1}{2^{n}}$ and also $ f(0)\equal{}2^{0}\plus{}\frac{1}{2^{0}}\equal{}2$ so we get that:
$ f(n)\equal{}2^{n}\plus{}\frac{1}{2^{n}}$ for every $ n\in\mathbb{Z}$...


[quote="Doctor A"](B.Frenkin, A.Zaslavsky) A convex quadrilateral was drawn on the blackboard. Boris marked the centers of four excircles each touching one side of the quadrilateral and the extensions of two adjacent sides. After this, Alexey erased the quadrilateral. Can Boris define its perimeter?[/quote]
Let $ ABCD$ is convex quadrilateral.
$ 0_1$ is a center of excircle touching $ AB$ at $ K_1$ , extension $ BC$ at $ N_1$ and extension $ DA$ at $ M_1$
$ 0_2$ is a center of excircle touching $ BC$ at $ K_2$ , extension $ CD$ at $ N_2$ and extension $ AB$ at $ M_2$
$ 0_3$ is a center of excircle touching $ CD$ at $ K_3$ , extension $ DA$ at $ N_3$ and extension $ BC$ at $ M_3$
$ 0_4$ is a center of excircle touching $ DA$ at $ K_4$ , extension $ AB$ at $ N_4$ and extension $ CD$ at $ M_4$

$ R_1,R_2,R_3,R_4$ - radii of circles with centers $ O_1,O_2,O_3,O_4$

$ s$ - semiperimeter of $ ABCD$

$ N_1M_3 \equal{} M_1N_3 \equal{} N_4M_2 \equal{} M_4N_2 \equal{} s$;

$ K_1N_4 \equal{} M_1K_4 \equal{} K_2M_3 \equal{} N_2K_3 \equal{} a$; $ a^2 \equal{} O_1O_4^2 \minus{} (R_1 \plus{} R_4)^2 \equal{} O_2O_3^2 \minus{} (R_2 \plus{} R_3)^2$;

$ K_2N_1 \equal{} M_2K_1 \equal{} K_3M_4 \equal{} N_3K_4 \equal{} b$; $ b^2 \equal{} O_2O_1^2 \minus{} (R_2 \plus{} R_1)^2 \equal{} O_3O_4^2 \minus{} (R_3 \plus{} R_4)^2$;

So we have :
$ \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{}$
$ O_1O_4^2 \minus{} O_2O_3^2 \equal{} (R_1 \plus{} R_2 \plus{} R_3 \plus{} R_4)((R_1 \minus{} R_3) \minus{} (R_2 \minus{} R_4))$
$ O_2O_1^2 \minus{} O_3O_4^2 \equal{} (R_1 \plus{} R_2 \plus{} R_3 \plus{} R_4)((R_1 \minus{} R_3) \plus{} (R_2 \minus{} R_4))$

$ O_1O_3^2 \equal{} s^2 \plus{} (R_1 \minus{} R_3)^2$
$ O_2O_4^2 \equal{} s^2 \plus{} (R_2 \minus{} R_4)^2$
$ \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{}$

from this system we can easy find :
$ s^2 \equal{} \frac {1}{2}(O_1O_3^2 \plus{} O_2O_4^2) \minus{} \frac {1}{4}(O_1O_3^2 \minus{} O_2O_4^2)(\frac {O_2O_3^2 \minus{} O_1O_4^2}{O_3O_4^2 \minus{} O_2O_1^2} \plus{} \frac {O_3O_4^2 \minus{} O_2O_1^2}{O_2O_3^2 \minus{} O_1O_4^2} )$; :)

Let $ I_1,I_2,I_3,I_4$ be the excenters of the quadrilateral. Let $ I_1'I_2I_3I_4'$ be the reflection of $ I_1I_2I_3I_4$ in $ I_2I_3$, and $ I_1''I_2'I_3I_4'$ the reflection of $ I_1'I_2I_3I_4'$ in $I_3I_4'$, and $ I_1''I_2'I_3I_4'$ the reflection of $I_1''I_2'I_3I_4'$ in $I_1''I_4'$. Easily we see that the perimeter of the quadrilateral equals to $I_1I_1'$.


algebra
polynomial
superior algebra
superior algebra unsolved

if d = (2+√3)^1/2

Is Q(d) a normal extension of Q?
What is Gal(Q(d):Q)?

Thanks in advance :)

Do you mean the minimal polynomial?

I worked it out as x^4-4x^2+1=0

Right, so the roots are $ \pm \sqrt{2 \pm \sqrt{3}}$.  Can you show all of these roots lie in $ \mathbb Q(d)$?

how do you show that they all lie in Q(d)?

$ \sqrt{2 \minus{} \sqrt{3}}$ is the inverse of $ \sqrt{2 \plus{} \sqrt{3}}$!


geometry
circumcircle
incenter
geometry proposed

I,O are circumcenter, incenter of triangle ABC,respectively. D,E,F are midpoint of BC,CA,AB.M is the intersection of IB and OD,N is the intersection of IC and OE,
P is the intersection of IA and OF.  Show that M,N,P,I concircle

Sorry,but could someone give me a hint to the problem DEAR MR.DANH just mentioned?
Thanks in advance!

[quote="mr.danh"][quote="mr.danh"]I,O are circumcenter, incenter of triangle ABC,respectively. D,E,F are midpoint of BC,CA,AB.M=IB∩OD, N=IC∩OE, P=IA∩OF.  Show that M,N,P,I concyclic[/quote]
Thanks [b]plane geometry[/b] for nice problem.
Denote A',B',C' be the intersections of IA,IB,IC with the circumcircle (O) repectively. Then, I is the orthocenter of triangle A'B'C'.
So, we can convert this problem into the following problem:
[quote="mr.danh"][b]New problem:[/b] O,H are the circumcenter and orthocenter of triangle ABC resp. M=OA∩BH, N=OB∩CH, P=OC∩AH. Then H,M,N,P all lie on a circle.[/quote]
I think the problem that i have just mentioned is nice and not hard, everyone can solved it easily.
Sincerely, mr.danh[/quote]
Thanks [b]mr.danh[/b] for your nice solution!
Could you tell me how to prove your lemma?Thank you

[quote="mr.danh"][b]New problem:[/b] O,H are the circumcenter and orthocenter of triangle ABC resp. M=OA∩BH, N=OB∩CH, P=OC∩AH. Then H,M,N,P all lie on a circle.[/quote]
[quote="mr.danh"]Lemma: H is the orthocenter of triangle ABC. M,N,P are resp on altitudes BE,CF,AD. Four points H,M,N,P all lie on a circle if and only if the condition $ \frac {ME}{BE} \plus{} \frac {NF}{CF} \plus{} \frac {PD}{AD} \equal{} 1$ is satisfied. [/quote]
The proof of lemma was posted.
In new problem, $ \frac {ME}{BE} \equal{} cotA.cotB,\frac {NF}{CF} \equal{} cotB.cotC,\frac {PD}{AD} \equal{} cotA.cotC$
Hence result.
It is a good application of the trigonometric formula cotA.cotB+cotB.cotC+cotC.cotA=1 in geom.

Sorry for my stupid but Why P isn't in the expression?
Can you give me the link of the lemma?
Thank you.


I am a highschool junior trying to study some hyperbolic geometry.
yep..its kinda random;
i needed a topic for my one year math-related independent project  and i wanted do something in geometry so i decided to try something totally new>>noneuclidean!
This is pretty much going to be self-learning so i need a book that is easy to follow.
im currently in calculus BC and am pretty familiar w/ traditional geometry but i have no background about hyperbolic geometry..or any kind of noneuclidean geometry at all.

can anyone recommend me a book or a helpful website??

Here is a useful site with a lot of info:
http://www.maths.gla.ac.uk/~wws/cabripages/hyperbolic/hyperbolic0.html

I'm not sure how useful this would be to you, but you can find amusing slides and accompanying video here:

http://www.msri.org/publications/ln/msri/1996/conv/adams/1/index.html


I decided to make a poll.  Respond with your state as well.

[quote="jhredsox"]Ok, this is sort of spam as I already made this topic and provided a poll.

-jorian

edit:How'd my poll dissappear?

-jorian[/quote]

Yeah, I saw your thread but found no poll.

IN-no nationals. i would've probably made it if they had a ladder-system CD that counted.

There's so much I want to explain about why I didn't make nats.

Will do when we're allowed.

-Jorian

ooooh yuqing don't count on it. Lyndon didn't try this time around.  He's usually just about as good as I am [I know you beat me this time around, but i already knew i won so i was way excited early and really tired during the countdown]

Lyndon was so pissed after chapter, when i got all the credit. He was mad because people were like "Can I have your autograph???" to me and he was being ignored  :rotfl:, when he was like two inches from beating me. Once I beat lyndon i was like "easy win from here on..."

p.s. lyndon was very upset he didn't beat me at state, haha... i always tease him by sucking in practice :p.

that's what i was like at state. we were in the good group and there was the awful group. when i was paired with you i was like -, i'm gonna get owned. it should've been me vs. you in the finals instead of the other kid. Also, they should've made it so you have to get 4 problems right or something. otherwise it's just luck.
[color=orange][size=75]
[Edited to cure various imperfectly-used words.][/size][/color]

[quote="jhredsox"]notice how bpms hasn't voted and hasn't had states yet

:lol:

jorian[/quote]
I'm just saying what I know will happen.

IN--I didn't make it, I tied for 5th, after the tiebreaker I got 6th.

[quote="kmaestro"]i wish states had happened already! they're not for another 2 weeks, on march 17th. actually, i do need more practice, though.[/quote]

are you from california too?

yes I am representing wv

[color=orange][size=75][Edited to make actually readable. Chatspeak is very bad.][/size][/color]


induction
real analysis
real analysis solved

Let (a_n)_n be the sequence give by the reccurency
a_(n+1_=n/(n^2-1)(a_1.a_2+a_2.a_3+..a_(n-1).a_n)
with a_1,a_2 \in (0,1/2).
Prove that the sequence (a_n)_n is convergent and find its limit.

It can be immediately proved by induction that a_n<1 for any n. Write the same relation for n+1 and you will find that n(n+2)/(n+1)*a_(n+2)=(n^2-1)/n*a_(n+1)+a_n*a_(n+1)<a_(n+1)*[n^2+n-1]/n (since a_n<1!)< a_(n+1)* n(n+2)/(n+1). So, the sequence is decreasing and so convergent to limit l. Since (1-1/n^2)a_(n+1)=(a_1*a_2+...+a_(n-1)*a_n)/n, from Cesaro-Stolz theorem we find that l^2=l. But since a_n<1 and the sequence is decreasing we have l=0.

nice harazi
u can prove too that a_n<=1/n


Let A be a countable set and let Bn be the set of all n-tuples (a1,a2,...,an), where ak is an element of A(k =1,2,...n), and the elements a1,...,an need not be distinct. Prove then Bn is countable.

ps : I have trouble understanding this question... :( so i.e B1 =(a1), and so on B2 =(a1,a2) ?

thxx

No, $ B_n \equal{} \{(a_1,a_2,...,a_n): a_1,a_2,...,a_n\in A\}$.  For example, if $ A \equal{} \{a,b,c\}$ then $ B_2 \equal{} \{(a,a),(a,b),(a,c),(b,a),(b,b),(b,c),(c,a),(c,b),(c,c)\}$.  Here's a good starting point for this problem: do you know how to prove that the rationals are countable?  Can you see a way in which proving that the rationals are countable is similar to proving that $ B_2$ is countable?  Can you extend this reasoning for fixed $ n$ greater than $ 2$?

What is it like in the case of the cardinality of the set of all (infinite) sequences in A?  :idea:

$ \left \vert B_{\infty}\right \vert \geq  \left \vert \mathbb{R} \right \vert$.  Whether or not $ \left \vert B_{\infty}\right \vert\equal{}\left \vert \mathbb{R}\right \vert$ isn't immediately clear to me, but I expect this is also not hard to show.


geometry
rectangle
linear algebra
matrix
modular arithmetic
analytic geometry
combinatorics solved

We have a  table of  $k^2-k +1$ rows and $k^2-k +1$ columns, where $k = p + 1$ and $p$ is a prime number. For each prime $p$ , give a method of distributing the numbers 0 and 1 , one number in each square of the table, such that in each row and column there are exactly $k$ zeros, and moreover no rectangle with sides parallel to the sides of the table has zeroes at all four corners.

[Moderator edit: See also http://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=84927 for this problem.]

This follows from the existence of a projective plane of order $n=q^2+q+1$ when $q$ is the power of a prime (here $q=p$).
The $n$ lines represent the rows, and the $n$ points are the columns. Each line contains exactly $q+1$ points.
Now, just put a $0$ for $a_{i,j}$ if and only if the point $M_j$ belongs to the line $d_i$.

The non-existence of a rectangle follows from the theorem in projective plane geometry which states that
'Two distinct lines have one and only one common point'.

Pierre.

Of course, but it looks to me like you would have to solve this problem in order to prove the existence of a projective plane, so this is a closed circle. The construction of such a matrix is the actual proof of the existence of the plane, isn't it? On the other hand, in order to construct the matrix you must have the plane.. I don't know if this qualifies as a method to fill in the cells of the table.

Sorry if this is all nonsense :)

Pierre! Do you remember Korea 11? I constructed particular example for p=3 there.

So, you just have to do it again for a general prime $p$  :D ...

In another hand, I remember China 5 (for which I succeeded to hide the projective plane) and Russia 99 (for which I didn't) too; So... :maybe:

Pierre.

The table has $p^2 + p + 1$ rows and columns. Let $P_2(Z/pZ)$ the triples of the form $(1,a,b)$, $(0,1,a)$ and $(0,0,1)$ in $(Z/pZ)^3$ (or, if you wish, the classes of the equivalence relation $(a,b,c) \sim (ka,kb,kc)$ in $(Z/pZ)^3 - (0,0,0)$). There are $p^2 + p + 1$ triples (or equivalence classes).

Put the $p^2 + p + 1$ triples in the rows and the columns. Put a zero in cell $(i,j)$ iff the corresponding triples $(a,b,c)$ and $(d,e,f)$ satisfy the congruence $ad + be + cf \equiv 0 \pmod p$. This will do.

Let's prove this satisfy the conditions of the problem.

(i) There are $p+1$ zeros in each row/column. Fix the row $(a,b,c)$. Let's count the solutions of $ax + by + cz \equiv 0 \pmod p$ in $P_2(Z/pZ)$. Suppose wlog $a\not\equiv 0\pmod p$. So $x \equiv a^{-1}(-by-cz)$. We have $p^2-1$ choices of the pair $(y,z)$ (we cannot choose $y=z=0$ because it would imply $x = 0$) but we are counting each triple $p-1$ times, because $(x,y,z)$ is a solution iff $(kx,ky,kz)$ is a solution for each $k = 1,2,3\dots,p-1$. So the congruence has $\frac{p^2-1}{p-1} = p+1$ solutions.

Each solution corresponds to a zero in the row, so the result follows.

(ii) There are no rectangles with zero vertices. It's the same as proving that the system of congruences
\[\left|\begin{array}{c}
ax+by+cz \equiv 0\pmod p\cr
dx+ey+fz \equiv 0\pmod p\cr
\end{array}
\right.\]
has only one solution in $P_2(Z/pZ)$ (unless $(a,b,c)$ and $(d,e,f)$ are the same point in $Z/pZ$). This is not so hard, since the matrix
\[\left(\begin{matrix}
a& b& c\\
d& e& f\\
\end{matrix}\right)\]
has rank $2$ and the congruences are homogeneous.

This is how one construct a projective plane with coordinates. You can replace $Z/pZ$ with any other field. Not all projective planes are isomorphic to coordinates projective planes, but all projective spaces with higher dimensions are.

Regards


The triangle ABC is dissected into 4 pieces. The area of 3 pieces are known, calculate the area of the fourth piece

 [geogebra]41d31cc2fb069c8916e046e24623a5558e43c368[/geogebra]

Doesn't belong to Pre-Olympiad.

[hide="Solution"] [geogebra]c3565a027aaf2b82377ae5dcc0f5a9d79b9ddb8c[/geogebra]
Assign mass of 10 to the point F. Since $ \frac {EF}{FC} \equal{} 1$, we know that masses of E and C are 5.

Also, since $ \frac {DF}{FB} \equal{} \frac {3}{7}$, we know that masses of D and B are 7 and 3, respectively. We have that mass of A is $ 7 \minus{} 5 \equal{} 5 \minus{} 3 \equal{} 2$.

From the results above, we have $ \frac {AE}{EB} \equal{} \frac {3}{2}\implies \frac {[AEC]}{[BEC]} \equal{} \frac {3}{2}$. Since $ [BEC] \equal{} 14$, we have $ [AEC] \equal{} 21$. However, since $ [AEFD] \equal{} [AEC] \minus{} [DFC]$, the answer is $ 21 \minus{} 3 \equal{} 18$.

Answer: 18[/hide]

Thank you, FantasyLover :)

I wonder if there are other approaches to solve this problem :maybe:

[hide="Another approach"] [geogebra]695db47f48929d9dd31bee4498e4201f979f6f4a[/geogebra]
Draw the line segment AE. Let us have $ [ADE] \equal{} x$. Since $ \frac {CE}{EF} \equal{} 1$,we have that $ [AEF] \equal{} 3 \plus{} x$.

Since $ \frac {DE}{EB} \equal{} \frac {3}{7}$, $ \frac {x}{10 \plus{} x} \equal{} \frac {3}{7}\implies x \equal{} \frac {15}{2}$.

$ [AFED] \equal{} 3 \plus{} 2x\equal{}18$.

Answer: 18 [/hide]


MATHCOUNTS AMC USA(J)MO USAMO induction LaTeX ratio

[quote="Wickedestjr"]Okay in the highschool basics forum, I posted a question similar to this one. This is a harder version, that I do not even know the answer to.

For the equation 2^1 + 2^2 + 2^3 + 2^4 + ... 2^n

Find a formula that equals the sum in terms of 'n', and then prove that it works.[/quote]

[hide="Hint"]$ (2^1 \plus{} 2^2 \plus{} 2^3 \plus{} 2^4 \plus{}\cdots\plus{}2^n\plus{}2)\minus{}2$[/hide]

This problem is still HSB (or Mathcounts) material.  If you don't know the solution to this problem, then I'm not sure why you were ever attempting USAMO-level problems.

It was simply a typo. He had the right method though (or one of them, at least).

We have $ S = 2 + 4 + 8 + \dots + 2^n$. Then $ 2S = 4 + 8 + \dots + 2^n + 2^{n + 1}$, so
\begin{align*}2S - S & = (4 + 8 + \dots + 2^n + 2^{n + 1}) - (2 + 4 + 8 + \dots + 2^n) \\
 & = - 2 + 2^{n + 1} \\
 & = 2^{n + 1} - 2
\end{align*}
But $ 2S - S$ is just $ S$, so we are done.

Why don't you write it as:

$ \left( 2^n \plus{} 2^{n \minus{} 1} \plus{} \ldots \plus{} 2 \plus{} 1 \plus{} \frac {1}{2} \plus{} \frac {1}{4} \plus{} \ldots \right)$ $ \minus{} \left( \frac {1}{2} \plus{} \frac {1}{4} \plus{} \ldots \right)$

Wickedestjr:  it's called "multiplying a bunch of numbers at the same time."  You can phrase the proof in terms of induction, but that's unnecessary.  It's series manipulation.

TZF:  I'm not sure why you'd use infinite geometric series to analyze finite geometric series - usually it's the other way around.

Extension: Find a formula for $ 1\plus{}a\plus{}a^2\plus{}\dots\plus{}a^n$ in terms of $ a$ and $ n$.

[quote="t0rajir0u"]Wickedestjr:  it's called "multiplying a bunch of numbers at the same time."  You can phrase the proof in terms of induction, but that's unnecessary.  It's series manipulation.[/quote]  Now I'm just being nit-picky, but it's not possible to do this without an inductive argument.  "Series multiplication" is a result of induction.  So for pedagogical purposes, it's probably better to say, "Yes, if you were to write this out in full, it would be an inductive argument.  But this is the sort of thing that people usually [i]don't[/i] write out in full, once they're comfortable with induction."

[quote="Yongyi781"]Extension: Find a formula for $ 1 \plus{} a \plus{} a^2 \plus{} \dots \plus{} a^n$ in terms of $ a$ and $ n$.[/quote]
S=[a^(n+1)-1]/(n-1)

[quote="JBL"]Comments:

1) Generally, one posts a solution, not just an answer.  In this case I think just posting the answer would have been fine, except that
2) the answer you've posted is wrong, both because of a typo and a neglected subtlety, and
3) you're not using [[Latex]] (see also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]here[/url]).[/quote]
I'm sorry and I don't like argueing, but don't talk as if you were my father. ĐKM

[quote="Nguyen Ngoc Linh"][quote="Yongyi781"]Extension: Find a formula for $ 1 + a + a^2 + \dots + a^n$ in terms of $ a$ and $ n$.[/quote]
S=[a^(n+1)-1]/(n-1)[/quote]

This is not right. The formula for a geometric series with first term $ 1$, constant ratio $ a$, and $ n+1$ terms is:

$ \frac{a^{n+1}-1}{a-1}$

This can be shown with the method Yongyi used above; letting $ S$ be the sum:

\begin{eqnarray*}
S&=&1+a+a^2+a^3+\dots+a^n\\
aS&=&\ \ \ \ \ a+a^2+a^3+\dots+a^{n+1}\\
aS-S=S(a-1)&=&a^{n+1}-1\\
S&=&\frac{a^{n+1}-1}{a-1}
\end{eqnarray*}

This holds for $ a\neq 1$; if $ a=1$, then the sum is just $ 1\cdot(n+1)=n+1$.

@Nguyen Ngoc Linh: Though I'm not your father, I hope you'll at least consider showing your work on these forums, as it will benefit others as well as yourself.

[quote="Wickedestjr"]So is the technique for solving the proof for this something called "number induction" or something like that?[/quote]

It was mathematical induction, which is where you show that if the formula is true for 'n', then it must also be true for 'n + 1'.

The method described above was not exactly induction; in this case, it was directly manipulation.

However, you could prove this inductively.

[b]Theorem.[/b] $ \sum_{i = 0}^n a^i = \frac {a^{n + 1} - 1}{a - 1}$ for all $ a \neq 1$.
[b]Base case: $ n = 1$.[/b] The LHS is $ \sum_{i = 0}^1 a^i = a^0 + a^1 = a + 1$, and the RHS is $ \frac {a^{1 + 1} - 1}{a - 1} = \frac {a^2 - 1}{a - 1} = a + 1$. So they are equal.
[b]Inductive hypothesis.[/b] Assume that $ \sum_{i = 0}^n a^i = \frac {a^{n + 1} - 1}{a - 1}$.
[b]Inductive step.[/b] Then:
$ \begin{align*} \sum_{i = 0}^{n + 1} a^i & = \left(\sum_{i = 0}^n a^i\right) + a^{n + 1} \\
& = \frac {a^{n + 1} - 1}{a - 1} + a^{n + 1} \\
& = \frac {a^{n + 1} - 1 + a^{n + 1}(a - 1)}{a - 1} \\
& = \frac {a^{n + 1} - 1 + (a^{n + 2} - a^{n + 1})}{a - 1} \\
& = \frac {a^{n + 2} - 1}{a - 1},$
as desired.

[quote="Yongyi781"]The method described above was not exactly induction; in this case, it was directly manipulation.[/quote]
That's the same point I raised; see JBL's response.

[quote="1=2"][hide="The Process of Mathematical Induction"]Let's say that we have to prove some statement $ S(n)$ for all positive integers $ n$. We could start like this:

We could prove it for $ S(1)$. This is called [i]proving the base case[/i].
Now we show that if $ S(k)$ is true, then $ S(k \plus{} 1)$ is true.
Once we are done proving that, we end with stating that since $ S(1)$ is true, then $ S(2)$ is true, and $ S(3)$ is true, etc. and we have proved it for all positive integers $ n$.[/hide]

Some problems:

Prove that $ \binom{n}{0} \plus{} \binom{n}{1} \plus{} \cdots \plus{} \binom{n}{n} \equal{} 2^n$ for all non-negative integers $ n$. - Don't cheat and use the Binomial Theorem!

Prove that $ 6|7^n \minus{} 1$ for all non-negative integers $ n$.[/quote]

For the first problem lets call  [(n C 0) + (n C 1) + ... (n C n)], x. Where 'x' is a real number.

x = 2^n
x + [(n + 1) C (n+1)] = 2^(n+1)
Lets call [(n + 1) C (n+1)], y, where 'y' is a real number.

x + y = 2^(n+1)
x = 2^n, so we can plug that in,
2^n + y = 2 times 2^n
y = 2^n

oops, I don't think I solved that right.  :(

i have no clue how to solve the other one.

Wickedestjr, for this problem, then $ n$'s change in the left-hand side too, so it is no longer $ \binom n0 \plus{} \binom n1 \plus{} \dots \plus{} \binom nn \plus{} \binom n{n \plus{} 1}$, but it is rather $ \binom {n \plus{} 1}0 \plus{} \binom {n \plus{} 1}1 \plus{} \dots \plus{} \binom {n \plus{} 1}n \plus{} \binom {n \plus{} 1}{n \plus{} 1}$.

[hide="Hint for the first one"]
Use the fact that $ \binom {n\plus{}1}k \equal{} \binom n{k\minus{}1} \plus{} \binom nk$ when $ k>1$ and $ \binom {n\plus{}1}0 \equal{} 1 \equal{} \binom n0$.
[/hide]
[hide="Hint for the second one"]
What remainder does $ 7^n$ leave when divided by 6? So when you subtract 1...[/hide]

http://www.mathlinks.ro/viewtopic.php?t=233453

Look at this.

Oh!!! That explains it.

Still can't get the first one though.  :(

Finish this for me:

[b]Base case: $ n \equal{} 1$[/b]: $ \binom 10 \plus{} \binom 11 \equal{} 1 \plus{} 1 \equal{} 2$, which is equal to $ 2^1$.
[b]Hypothesis[/b]: We claim that $ \binom n0 \plus{} \binom n1 \plus{} \binom n2 \plus{} \dots \plus{} \binom nn \equal{} 2^n$.
[b]Inductive step[/b]: If we look at the Pascals Triangle:
[color=white]0 0 [/color]1[color=white] 0 0 [/color]
[color=white]0 0[/color]1 1[color=white]0 0 [/color]
[color=white]0 [/color]1 2 1[color=white] 0[/color]
[color=white]0[/color]1 3 3 1[color=white]0[/color]
1 4 6 4 1
...
Each number after the first row is the sum of the two numbers above it, or the number above it if there are not two numbers above it. So we have the identity $ \binom {n \plus{} 1}k \equal{} \binom n{k \minus{} 1} \plus{} \binom nk$ when $ 1<k<n\plus{}1$ and $ \binom {n \plus{} 1}0 \equal{} \binom n0$ and $ \binom {n \plus{} 1}{n \plus{} 1} \equal{} \binom nn$.

So we can substitute each term in the sum below:
$ \binom {n \plus{} 1}0 \plus{} \binom {n \plus{} 1}1 \plus{} \dots \plus{} \binom {n \plus{} 1}{n \plus{} 1}$
with:
$ \binom n0 \plus{} \left(\binom n0 \plus{} \binom n1\right) \plus{} \left(\binom n1 \plus{} \binom n2\right) \plus{} \dots \plus{} \left(\binom n{n \minus{} 1} \plus{} \binom nn\right) \plus{} \binom nn$

And... (finish this for me)...

And ... (finish this for me)..................  :huh:

I'm sorry but I don't know how to do this.  :(

[hide="The rest of it"]
We can regroup the terms like so:
\[ \left(\binom n0 \plus{} \binom n0\right) \plus{} \left(\binom n1 \plus{} \binom n1\right) \plus{} \left(\binom n2 \plus{} \binom n2\right) \plus{} \dots \plus{} \left(\binom n{n \minus{} 1} \plus{} \binom n{n \minus{} 1}\right) \plus{} \left(\binom nn \plus{} \binom nn\right)
\]
Which is just:
\[ 2\left(\binom n0 \plus{} \binom n1 \plus{} \binom n2 \plus{} \dots \plus{} \binom n{n \minus{} 1} \plus{} \binom nn\right) \equal{} 2 \cdot 2^n \equal{} 2^{n \plus{} 1}
\]
Which is what we wanted to prove.[/hide]

Oh!!!! Geese, I'm so stupid.  :fool:  :wallbash_red:  :wallbash:

Alright, any other examples of questions that involve the technique "mathematical induction"?

Simple proofs involving combinations are best done combinatorially.  The proof that $ \sum_{k\equal{}0}^{n} {n \choose k} \equal{} 2^n$ is combinatorially straightforward:  the LHS counts the number of subsets of an $ n$-element set by counting the $ k$-element subsets and adding, while the RHS counts the same thing by either picking or not picking any of $ n$ elements.

Similarly, the proof that $ {n\plus{}1 \choose k} \equal{} {n \choose k} \plus{} {n \choose k\minus{}1}$ is combinatorially straightforward:  distinguish one element of an $ n\plus{}1$-element set.  The number of subsets of that $ n\plus{}1$-element set is the number of subsets that include that element plus the number of subsets that don't.

Prove that any term in the sequence $ 7, 14, 28, 56, \dots$, where each term is twice the previous term, will never be divisible by 3.

How can this be proved by mathematical induction. We have to prove it for every power of 2.

The only proof that I have is this;

The only two prime factors of (7)2^n are always 2, and 7. So therefore you can not get three as a factor.

[quote="Wickedestjr"]How can this be proved by mathematical induction. We have to prove it for every power of 2. [/quote]
You are inducting on every term of the sequence $ a_n \equal{} 7 \cdot 2^n$.  To be absolutely precise, you are inducting on the following statement:

$ S(n) : 7 \cdot 2^n \text{ is not divisible by 3.}$

To prove it, you want to prove $ S(0)$ and use $ S(n)$ to prove $ S(n\plus{}1)$.

[quote="t0rajir0u"][quote="Wickedestjr"]How can this be proved by mathematical induction. We have to prove it for every power of 2. [/quote]
You are inducting on every term of the sequence $ a_n \equal{} 7 \cdot 2^n$.  To be absolutely precise, you are inducting on the following statement:

$ S(n) : 7 \cdot 2^n \text{ is not divisible by 3.}$

To prove it, you want to prove $ S(0)$ and use $ S(n)$ to prove $ S(n \plus{} 1)$.[/quote]

S(n) = 7 x 2^n
Multiplying  by 2 won't make it divisible by 3.
x2 = x2
7 x 2^(n+1)

[quote="Wickedestjr"]Multiplying  by 2 won't make it divisible by 3. [/quote]
Why?  :)

The proof that Yongyi and I are really looking for runs something along these lines and ignores the notion of prime factorization:

- A number not divisible by $ 3$ is either congruent to $ 1 \bmod 3$ or $ 2 \bmod 3$.
- if $ n \equiv 1 \bmod 3$ then $ 2n \equiv 2 \bmod 3$.
- if $ n \equiv 2 \bmod 3$ then $ 2n \equiv 1 \bmod 3$.
- Hence...

Oh, that's cool!!! I can't believe I didn't think to use modular arithmetic, when I have been working with it for the last month. :fool:


Let $ABC$ and $A'B'C'$ be two non-congruent triangles whose sides are respectively parallel. Prove that the three lines $AA'$, $BB'$, and $CC'$ (extended) are concurrent.

[hide="A start?"]Prove $\triangle{A'B'C'}\sim\triangle{ABC}$.[/hide]

let $AA'$ and $BB'$ intersect at $H$, let $HC$ intersect $A'C'$ at $X$, easy similarity shows that $X\equiv C'$


Even is not so intermediate hope it's ok if i post iti here :)

Find the minimum of the expresion $y=x-1-\sqrt{2x-1}$

Do you want a non calculus solution?  Cause the calc solution is pretty straightforward.

What is a non calculs solution ?:) this is a 9th grade problem :D

yes, the Calc way is very simple
$y'=1-\frac{1}{\sqrt{2x-1}}$
set it equal to zero,
$2x-1=1\Rightarrow x=1$
is the min :)

That's unknown in 9th grade :). And you should calc y :P

$y=\frac{(\sqrt {2x-1} -1)^2)}{2}-1$  thus the minimum is $-1$ and it's hold for $x=1$.

Payman

My solution was: $\sqrt{2x-1}=t \geq 0$. Then $x=\frac{t^2+1}{2}$ and $y=\frac{1}{2}t^2-t-\frac{1}{2}$. This is a second degree function which admits the minimum for $t=1$ and $x=1$. So $y_{min}=-1$  :)


I'm taking this as a question, not as a problem to be solved... skew lines are lines in space that are not parallel, but the planes they are in never intersect.

How I love MathWorld. :) http://mathworld.wolfram.com/SkewLines.html


Simple rules:

1) When a word/phrase is complete, the person who earned the most points in that round chooses the next word. He/she may say pass if unable to select a word. When this happens anyone can choose a word. The word poster gets 5 points bonus for getting the most points in that round.
2) Each time you guess a letter, 1 point is taken off your score. Your starting score is 15. KEEP A SCORE LIST; SIMPLY QUOTE THE PREVIOUS USER. You may order the list if you want.
3) Anyone can join. No signups needed.
4) Anyone may choose to solve the word/phrase, and the number of missing letters is $ m$, then the solver gets $ 3(m) \minus{} 2$ points.
5) Should a letter happen to appear multiple times (lets say it appears $ n$ times), then the guesser gets $ 3(F_{n}) \minus{} 2$ points for that letter. (The nth Fibonacci number is $ F_n$)
6) When posting a new word, the subject/category should not be posted. I will also change the rules a bit: Say Al guesses the letter "t" in the word "the"; Bob (the word poster) may choose to answer and say "there is no 't' in here." However, if Al asks "t" for a second time, Bob must honestly place a "t" into the blank. You may only lie once. The second guess counts as an individual guess.

I'll start:

_ _ _ _ _

[hide="Score List"]
dragon96: 20
[/hide]

EDIT: I'm thinking I make a different version so its just two teams, guessing and the person who made the word. Which should I do?

[b]A[/b]

[hide="Scores List"]
dragon96: 20
yaofan: 14 [/hide]

I think this will go slow, b/c I'm not very active. Unless someone else wants to continue (post a  new word to do so), I will not continue.


how do you program LaTeX so you can post matrices on forums? I don't have the downloaded program so the only LaTeX programming is on AoPS

The simplest way to get a matrix is to use \pmatrix
\begin{pmatrix}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{pmatrix}
gives $\begin{pmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}$
You can also use bmatrix, Bmatrix, vmatrix and Vmatrix to get
$\begin{bmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$ $\begin{Bmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{Bmatrix}$ $\begin{vmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}$ $\begin{Vmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{Vmatrix}$

[quote="stevem"]The simplest way to get a matrix is to use \pmatrix
\begin{pmatrix}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{pmatrix}
gives $\begin{pmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}$
You can also use bmatrix, vmatrix, Bmatrix and Vmatrix to get
$\begin{bmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$ $\begin{Bmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{Bmatrix}$ $\begin{vmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}$ $\begin{Vmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{Vmatrix}$[/quote]
I write this in texnic Center:
\documentclass{article}
\begin{document}

\begin{bmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}

\end{document}
note:i also put   [b]\begin{bmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}[/b] between two dollar sighns. but what i get after building the file in PDF is:
123
456
789
whats the problem?!!!!!!!!!!!

It needs the amsmath package which is so good it should be used in every document:[code]\documentclass[12pt]{article}
\usepackage{amsmath}
\begin{document}
$\begin{bmatrix}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{bmatrix}$
\end{document}[/code]

To save having to add these packages every time, make yourself a template in TexnicCenter and when you start a new file choose File, New fromTemplate and choose yours.

Go into C:\Program Files\TeXnicCenter\Templates\Documents\ and copy one of the templates there and change it to:
[code]%description: Use Me!

\documentclass[12pt]{article}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsthm}
\begin{document}

\end{document}[/code]save as Use Me.tex and then you can use it for all your new documents.


algebra
polynomial
Galois Theory
superior algebra
superior algebra unsolved

Find \[ \min (\pi,\mathbb{R}), \min (\pi,\mathbb{Q}), \min (\sqrt{7}+\sqrt [3] {11}, \mathbb{Q}), \min (\sqrt{2}+\sqrt{3}+\sqrt{5}+\sqrt{7},\mathbb{Q}). \]

$\min (\pi,\mathbb{R})$ is trivial.
$\min (\pi,\mathbb{Q})$ does not exist (or you accept $\sum_{k=1}^\infty (-1)^k \frac{x^{2k}}{(2k+1)!}$ :P )
$\min (\sqrt{7}+\sqrt [3] {11}, \mathbb{Q})$ can be done like in http://www.mathlinks.ro/Forum/viewtopic.php?t=89383 .
$\min (\sqrt{2}+\sqrt{3}+\sqrt{5}+\sqrt{7},\mathbb{Q})$ is ugly, ugly and even more ugly, but dito.

I doubt' somewhat you did it by hand ;)
See http://www.mathlinks.ro/Forum/viewtopic.php?t=27089 for a much more general statement.

But I would be really interested what sadistic person gave that ugly problems...

Perhaps the purpose is that this convinces you that galois theory is useful. ;) You cannot compute the minimal polynomial like the one of $\sqrt{2} + \sqrt{3}$.

The exercise doesn't require to expand the polynomial, I think. I left it to a CAS.


Given  24, a, b, c, 4,...

Find the values of a, b, and c if each number in the sequence, starting with a, is the arithmetic average of the number on its immediate left with the number on its immediate right.


Peter mows the lawn in 40 minutes and John mows takes 60 minutes to mow the same lawn. How long will it take for them both to mow the same lawn?

Peter can mow a lawn in 40 minutes = 2/3 of an hour. That means he mows 1.5 lawns in one hour. Similarly, John mows one lawn per hour. Together, then, they mow 2.5 lawns, so it takes them 60/2.5 = 24 minutes.

For problems like this, a general formula is

$ \frac{n}{\frac{1}{r_1}\plus{}\frac{1}{r_2}}$

where $ n$ is the number of objects (in this case, lawns needed to be mowed), and $ r_1$ and $ r_2$ are the rates the people work at. If there are $ x$ people working, the denominator is $ \frac{1}{r_1}\plus{}\frac{1}{r_2}\plus{}...\plus{}\frac{1}{r_x}$

Or if you know sigma notation, $ \displaystyle\sum_{i\equal{}1}^{x}\frac{1}{r_i}$


Find all $ x,y\in Z$ such that $ x^{y}\equal{}y^{x}\plus{}1$

[hide]

Supose $ n \equal{}\frac{x}{2}$ and $ m \equal{}\frac{y}{2}$

Therefore we have that:
\[ (x^{m}\minus{}y^{n})(x^{m}\plus{}y^{n})\equal{}1\]
This Implies $ \minus{}y^{n}\equal{} y^{n}$ wich is true only when $ y \equal{} 0$ with $ n\neq 0\implies x\neq 0$

So we have the pair $ (x,y) \equal{} (x,0)\,\,\forall x > 0\in\mathbb{Z}$

[/hide]

[quote="Tiago Guerreiro"]
\[ (x^{m}\minus{}y^{n})(x^{m}\plus{}y^{n})\equal{}1\]
This Implies $ \minus{}y^{n}\equal{} y^{n}$
[/quote]
How can you get this? I think it's not true
For example when $ x\equal{}2$ and $ y\equal{}1$

[quote="tuanct"]Find all $ x,y\in Z$ such that $ x^{y}\equal{} y^{x}\plus{}1$[/quote]

I think I can only do for $ x,y\in\mathbf N$. First it's easy to check that $ (1,0),(2,1),(3,2)$ are solutions. For $ x,y\ge 3$, note that $ y^{y\minus{}1}< (y\minus{}1)^{y}$, implying $ y > x$. Also, it's not difficult to see that $ y^{y\minus{}1}\minus{}(y\minus{}1)^{y}$ is strictly increasing, which implies that $ x^{y}\minus{}y^{x}$ is also increasing. Since $ 3^{4}> 4^{3}\plus{}1$, we cannot have any more solutions.

[quote="nayel"][quote="tuanct"]Find all $ x,y\in Z$ such that $ x^{y}\equal{} y^{x}\plus{}1$[/quote]
I think I can only do for $ x,y\in\mathbf N$ [...][/quote]
if $ x<0$ ($ x\equal{}\minus{}x'$)
then $ (\minus{}x')^{y}.y^{x'}\equal{}1\plus{}y^{x'}$ then if $ y\ge 0$ then $ y|1\plus{}y^{x'}$ gives $ y\equal{}1$ gives $ x\equal{}2$(contradiction $ x<0$)
if $ x\ge 0$
we suppose that $ y\equal{}\minus{}y'<0$ then $ (\minus{}y')^{x}.x^{y'}\equal{}\minus{}x^{y'}\plus{}1$ then $ x|1\minus{}x^{y'}$ gives $ x\equal{}1$ gives $ y\equal{}0$ (contradiction $ y<0$)
then $ x,y\in N$


Let the sequence $ \left\{a_n\right\}_{n\geq1}$ be convergent with $ a_i\geq0$ for all $ i$. Also $ \sum\ a_{i}$ is finite. Do there exist a sequence  $ \left\{b_n\right\}_{n\geq1}$  which is non increasing and convergent such that for all $ i\geq1$  the relation $ b_i \geq a_i$ is satisfied and $ \sum\ b_{i}$ is finite

Nonincreasing, no.  Try to do an example where $ a_n \equal{} 0$ for many $ n$ .

$ \frac1n$ when $ n\equal{}k^2$, zero otherwise.


Q1 If z is a complex number lying in first quadrant such that Re(z)+Im(z)=3,then find maximum value of $ (Re(z))^2Im(z)$

Q2 the roots of cubic equation $ (z\plus{}ab)^3\equal{}a^3;a\neq 0$ represents the vertices of a triangle.If $ \alpha , \beta ,\gamma$ are the roots,the ind value of $ \left| \alpha \minus{} \beta \right|$

Q1 : let $ z=a+ib$ with $ a,b \geq 0$

if $ a+b=3$ then by AM-GM inequality $ \sqrt[3]{Re(z)^2Im(z)}=\frac{1}{\sqrt[3]{2}}{\sqrt[3]{a^2(2b)}\leq\frac{2(a+b)}{3\sqrt[3]{2}}=\sqrt[3]{4}}$

so the max is $ 4$ equality occurs iff $ a=2 , b=1$

Q2: i don't understand your question !

[quote="jackie-chan01"]
Q2 the roots of cubic equation $ (z \plus{} ab)^3 \equal{} a^3;a\neq 0$ represents the vertices of a triangle.If $ \alpha , \beta ,\gamma$ are the roots,the ind value of $ \left| \alpha \minus{} \beta \right|$[/quote]
WLOG, $ \alpha \equal{} \minus{} ab \plus{} a$, $ \beta \equal{} \minus{} ab \plus{} ae^{\frac {2\pi i}{3}}$, and $ \gamma \equal{} \minus{} ab \plus{} ae^{\frac {4\pi i}{3}}$
$ |\alpha \minus{} \beta| \equal{} |a(1 \minus{} e^{\frac {2\pi i}{3}})| \equal{} a\left|\left[1 \minus{} \left( \minus{} \frac {1}{2} \plus{} \frac {\sqrt {3}}{2}i\right)\right]\right| \equal{} a\left|\left(\frac {3}{2} \minus{} \frac {\sqrt {3}}{2}i\right)\right| \equal{} a\sqrt {3}$

Ans1=2
Ans2=$ \sqrt {3} \left|  ab \right|$


GUYS GUYS
TRIGONOMETRY 1.01 IS NOT ARNAV :o
OMG TRIGONOMETRY 1.01 SHUDDERS EVEN AT THE THOGUHT OF BEING ARNAV :o  :o

DO YOU MEAN TRIGONOMETRY1.01 [i]IS[/i] ARNAV? :mad:
BY NO MEANS! :mad:  :mad:

Hey guys,

I don't think The Herald of Doom is Arnav.

[quote="HE-WHO-MUST-NOT-BE-OR-RATHER-SHOULD-NOT-BE-NAMED"]Please do not post my name on public forums.

Thanks.[/quote]

THAT"S THE CLINCHER, SEE, EVERYONE?
ARNAV WOULD NOT HAVE OBJECTED TO POSTING [i]HIS[/i] NAME ON THE FORUM, BUT HE-WHO-MUST-NOT-BE-OR-RATHER-SHOULD-NOT-BE-NAMED JUST DID!