inequalities
geometry
invariant
geometric transformation
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rotation
trigonometry

If a,b,c and x,y,z are the sides of two triangles of area A and X prove

a 2 (y 2 +z 2 -x 2 ) +b 2 (z 2 +x 2 -y 2 ) +c 2 (x 2 +y 2 -z 2 ) \geq 16AX

with equality iff the two triangles are similar.

If one of the triangles is equilateral, the inequality becomes one attribuited to Weitzenboeck:

   a 2 +b 2 +c 2  \geq 4 \sqrt 3A

It is a well-known inequality due to Pedoe (1942).

Pierre.

[[b]EDIT (2017):[/b] I have now reposted this posting -- with some modifications and improvements -- on Math.Stackexchange: http://math.stackexchange.com/questions/2143589/imo-problem-1961-different-proofs-of-the-weitzenbock-inequality/2168007#2168007 .]

This goes along with the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16933]Bottema inequality[/url]:

[color=blue][b]Theorem 1.[/b] Let ABC be a triangle with sidelengths a = BC, b = CA, c = AB and area S, and let XYZ be a triangle with sidelengths x = YZ, y = ZX, z = XY and area T.

[b](a) (Pedoe inequality)[/b] Then, we have the inequality $a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)\geq 16ST$, with equality if and only if the triangles ABC and XYZ are similar.

[b](b) (Bottema inequality)[/b] If P is an arbitrary point in the plane of triangle ABC, then we have the inequality

$x\cdot AP+y\cdot BP+z\cdot CP\geq \sqrt{\frac{a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)}{2}+8ST}$.

[b](c) (Equivalent version of (a))[/b] If A = < CAB, B = < ABC, C = < BCA are the angles of triangle ABC, and X = < ZXY, Y = < XYZ, Z = < YZX are the angles of triangle XYZ, then we have

$\cot A\cot Y+\cot A\cot Z+\cot B\cot Z+\cot B\cot X+\cot C\cot X+\cot C\cot Y$
$\geq 2$.

This inequality becomes an equality if and only if the triangles ABC and XYZ are similar.[/color]

[i]Proof of Theorem 1.[/i]

[b](a)[/b] First note that the inequality which we must prove,

$a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)\geq 16ST$,

is homogeneous in the sidelengths x, y, z of the triangle XYZ (in fact, these sidelengths occur in the power 2 on the left hand side, and on the right hand side they occur in the power 2 as well, since the area of a triangle is quadratically dependant from its sidelengths). Hence, this inequality is invariant under any similitude transformation executed on triangle XYZ; in other words, we can move, reflect, rotate and stretch the triangle XYZ as we wish, but the inequality remains equivalent. But, of course, by applying similitude transformations to triangle XYZ, we can always achieve a situation when Y = B and Z = C and the point X lies in the same halfplane with respect to the line BC as the point A. Hence, in order to prove Theorem 1 [b](a)[/b] for [i]any[/i] two triangles ABC and XYZ, it is enough to prove it for two triangles ABC and XYZ in this special situation.

So, assume that the triangles ABC and XYZ are in this special situation, i. e. that we have Y = B and Z = C and the point X lies in the same halfplane with respect to the line BC as the point A. We have to prove the inequality $a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)\geq 16ST$.

Well, by the cosine law in triangle ABX, we have

$AX^2=AB^2+XB^2-2\cdot AB\cdot XB\cdot\cos\measuredangle ABX$.

Let's figure out what this equation means. At first, AB = c. Then, since B = Y, we have XB = XY = z. Finally, we have either < ABX = < ABC - < XBC or < ABX = < XBC - < ABC (depending on the arrangement of the points), but in both cases cos < ABX = cos (< ABC - < XBC). Since B = Y and C = Z, we can rewrite the angle < XBC as < XYZ. Thus,

$\cos\measuredangle ABX=\cos\left(\measuredangle ABC-\measuredangle XYZ\right)=\cos\measuredangle ABC\cos\measuredangle XYZ+\sin\measuredangle ABC\sin\measuredangle XYZ$.

By the cosine law in triangles ABC and XYZ, we have

$\cos\measuredangle ABC=\frac{c^2+a^2-b^2}{2ca}$;
$\cos\measuredangle XYZ=\frac{z^2+x^2-y^2}{2zx}$.

Also, by a well-known formula, the area S of triangle ABC equals $S=\frac12\cdot ca\sin\measuredangle ABC$; thus, $\sin\measuredangle ABC=\frac{2S}{ca}$. Similarly, $\sin\measuredangle XYZ=\frac{2T}{zx}$.

Hence,

$\cos\measuredangle ABX=\cos\measuredangle ABC\cos\measuredangle XYZ+\sin\measuredangle ABC\sin\measuredangle XYZ$
$=\frac{c^2+a^2-b^2}{2ca}\cdot\frac{z^2+x^2-y^2}{2zx}+\frac{2S}{ca}\cdot\frac{2T}{zx}$.

Together with AB = c and XB = z, this makes the equation

$AX^2=AB^2+XB^2-2\cdot AB\cdot XB\cdot\cos\measuredangle ABX$

transform into

$AX^2=c^2+z^2-2\cdot c\cdot z\cdot\left(\frac{c^2+a^2-b^2}{2ca}\cdot\frac{z^2+x^2-y^2}{2zx}+\frac{2S}{ca}\cdot\frac{2T}{zx}\right)$.

This immediately simplifies to

$AX^2=c^2+z^2-2\left(\frac{\left(c^2+a^2-b^2\right)\left(z^2+x^2-y^2\right)}{4ax}+\frac{4ST}{ax}\right)$.

After some algebra, this becomes

$AX^2=\frac{\left(a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)-2\left(x-a\right)\left(c^2x-z^2a\right)\right)-16ST}{2ax}$.

Since Y = B and Z = C, we have YZ = BC, or, in other words, x = a. Thus, x - a = 0, and the equation above simplifies to

$AX^2=\frac{\left(a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)\right)-16ST}{2ax}$.

Since (obviously) $AX^2\geq 0$, we thus have

$\left(a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)\right)-16ST\geq 0$,

and thus $a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)\geq 16ST$. So the inequality of Theorem 1 [b](a)[/b] is proven. Equality holds if and only if equality holds in the inequality $AX^2\geq 0$; but this is clearly the case if and only if the points A and X coincide, i. e. if the triangles ABC and XYZ are congruent. Now, of course, since the triangle XYZ we are dealing with is not the initial triangle XYZ, but just its image under a similitude transformation, the general equality condition is that the triangles ABC and XYZ are similar (not necessarily being congruent). So Theorem 1 [b](a)[/b] is proven.

[b](b)[/b] In our proof of Theorem 1 [b](a)[/b], we WLOG assumed that Y = B and Z = C and the point X lies in the same halfplane with respect to the line BC as the point A. This time we will assume the same situation, but with a little difference: The point X should lie not in the same halfplane with respect to the line BC as the point A, but exactly in the other halfplane. In other words, we have Y = B and Z = C and the points A and X lie in different halfplanes with respect to the line BC. Then, by a calculation similar to the one we did in the proof of Theorem 1 [b](a)[/b], we find

$AX^2=\frac{\left(a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)\right)+16ST}{2ax}$

(note the + sign instead of the -). Since x = a, we have $ax=a^2$, and thus we can rewrite this as

$AX^2=\frac{\left(a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)\right)+16ST}{2a^2}$.

Hence,

$a^2\cdot AX^2=\frac{\left(a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)\right)+16ST}{2}$
$=\frac{a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)}{2}+8ST$.

Taking the square root of this, we get

$a\cdot AX=\sqrt{\frac{a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)}{2}+8ST}$.

Hence, the inequality which is to be proven,

$x\cdot AP+y\cdot BP+z\cdot CP\geq \sqrt{\frac{a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)}{2}+8ST}$,

is equivalent to $x\cdot AP+y\cdot BP+z\cdot CP\geq a\cdot AX$.

So we have to prove $x\cdot AP+y\cdot BP+z\cdot CP\geq a\cdot AX$.

Well, first, by the Ptolemy inequality for the quadrilateral PBXC, we have $CX\cdot BP+XB\cdot CP\geq BC\cdot PX$. Since B = Y and C = Z, we have CX = ZX = y, XB = XY = z and BC = YZ = x, and this rewrites as $y\cdot BP+z\cdot CP\geq x\cdot PX$. Also, by the triangle inequality in the triangle APX, we have $AP+PX\geq AX$. Thus,

$x\cdot AP+y\cdot BP+z\cdot CP=x\cdot AP+\left(y\cdot BP+z\cdot CP\right)\geq x\cdot AP+x\cdot PX$
$=x\cdot\left(AP+PX\right)\geq x\cdot AX=a\cdot AX$,

where we have used x = a in the last step. This proves Theorem 1 [b](b)[/b].

[b](c)[/b] By the Cosine Law in triangle ABC, we have $\cos A=\frac{b^2+c^2-a^2}{2bc}$. On the other hand, by a well-known formula for the area of a triangle, the area S of triangle ABC equals $S=\frac12bc\sin A$. Thus, $\sin A=\frac{2S}{bc}$. Hence,

$\cot A=\cos A: \sin A=\frac{b^2+c^2-a^2}{2bc}: \frac{2S}{bc}=\frac{b^2+c^2-a^2}{4S}$.

Similarly, $\cot B=\frac{c^2+a^2-b^2}{4S}$. Thus, $\cot A+\cot B=\frac{b^2+c^2-a^2}{4S}+\frac{c^2+a^2-b^2}{4S}=\frac{2c^2}{4S}=\frac{c^2}{2S}$.

Similarly to how we proved $\cot A=\frac{b^2+c^2-a^2}{4S}$, we can find that $\cot Z=\frac{x^2+y^2-z^2}{4T}$. Thus,

$\cot A\cot Z+\cot B\cot Z=\left(\cot A+\cot B\right)\cdot\cot Z=\frac{c^2}{2S}\cdot\frac{x^2+y^2-z^2}{4T}$
$=\frac{c^2\left(x^2+y^2-z^2\right)}{8ST}$.

Similarly,

$\cot B\cot X+\cot C\cot X=\frac{a^2\left(y^2+z^2-x^2\right)}{8ST}$;
$\cot C\cot Y+\cot A\cot Y=\frac{b^2\left(z^2+x^2-y^2\right)}{8ST}$.

Thus,

$\cot A\cot Y+\cot A\cot Z+\cot B\cot Z+\cot B\cot X+\cot C\cot X+\cot C\cot Y$
$=\left(\cot B\cot X+\cot C\cot X\right)+\left(\cot C\cot Y+\cot A\cot Y\right)+\left(\cot A\cot Z+\cot B\cot Z\right)$
$=\frac{a^2\left(y^2+z^2-x^2\right)}{8ST}+\frac{b^2\left(z^2+x^2-y^2\right)}{8ST}+\frac{c^2\left(x^2+y^2-z^2\right)}{8ST}$
$=\frac{a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)}{8ST}$.

But after Theorem 1 [b](a)[/b], we have $a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)\geq 16ST$, with equality if and only if the triangles ABC and XYZ are similar. Hence,

$\cot A\cot Y+\cot A\cot Z+\cot B\cot Z+\cot B\cot X+\cot C\cot X+\cot C\cot Y$
$\geq\frac{16ST}{8ST}=2$,

with equality if and only if the triangles ABC and XYZ are similar. And Theorem 1 [b](c)[/b] is proven.

Note that Theorem 1 [b](a)[/b] yields the following triangle inequality:

[color=blue][b]Corollary.[/b] Let ABC be a triangle with sidelengths a = BC, b = CA, c = AB and area S, and let u, v, w be three reals such that the numbers v + w, w + u, u + v and vw + wu + uv are all nonnegative [of course, these conditions are fulfilled if the reals u, v, w themselves are nonnegative, but also in some other cases]. Then, $ua^2+vb^2+wc^2\geq 4\sqrt{vw+wu+uv}\cdot S$.[/color]

[i]Proof of the Corollary.[/i] Before proving the Corollary, we show a useful fact:

[color=blue][b]Conway substitution theorem.[/b] Let u, v, w be three reals such that the numbers v + w, w + u, u + v and vw + wu + uv are all nonnegative [of course, these conditions are fulfilled if the reals u, v, w themselves are nonnegative, but also in some other cases]. Then, there exists a triangle XYZ with sidelengths $x=YZ=\sqrt{v+w}$, $y=ZX=\sqrt{w+u}$, $z=XY=\sqrt{u+v}$. This triangle satisfies $y^2+z^2-x^2=2u$, $z^2+x^2-y^2=2v$, $x^2+y^2-z^2=2w$. The area T of this triangle equals $T=\frac12\sqrt{vw+wu+uv}$. If X = < ZXY, Y = < XYZ, Z = < YZX are the angles of this triangle, then $\cot X=\frac{u}{2T}$, $\cot Y=\frac{v}{2T}$ and $\cot Z=\frac{w}{2T}$.[/color]

[i]Proof of the Conway substitution theorem.[/i] Since the numbers v + w, w + u, u + v are nonnegative, their square roots $\sqrt{v+w}$, $\sqrt{w+u}$, $\sqrt{u+v}$ exist, and, of course, are nonnegative as well. Now, we have $\sqrt{w+u}+\sqrt{u+v}\geq\sqrt{v+w}$, since

$\sqrt{w+u}+\sqrt{u+v}\geq\sqrt{v+w}\ \ \ \ \ \Longleftarrow\ \ \ \ \ \left(\sqrt{w+u}+\sqrt{u+v}\right)^2\geq v+w$
$\Longleftarrow\ \ \ \ \ \left(w+u\right)+\left(u+v\right)+2\sqrt{\left(w+u\right)\left(u+v\right)}\geq v+w$
$\Longleftarrow\ \ \ \ \ 2\sqrt{\left(w+u\right)\left(u+v\right)}\geq -2u$           [this is not trivial, u needs not be $\geq 0$!]
$\Longleftarrow\ \ \ \ \ \sqrt{\left(w+u\right)\left(u+v\right)}\geq-u\ \ \ \ \ \Longleftarrow\ \ \ \ \ \left(w+u\right)\left(u+v\right)\geq\left(-u\right)^2$
$\Longleftarrow\ \ \ \ \ u^2+\left(vw+wu+uv\right)\geq u^2\ \ \ \ \ \Longleftarrow\ \ \ \ \ vw+wu+uv\geq 0$.

Similarly, $\sqrt{u+v}+\sqrt{v+w}\geq\sqrt{w+u}$ and $\sqrt{v+w}+\sqrt{w+u}\geq\sqrt{u+v}$. Thus, there exists a triangle XYZ with sidelengths $x=YZ=\sqrt{v+w}$, $y=ZX=\sqrt{w+u}$, $z=XY=\sqrt{u+v}$. Denote by X = < ZXY, Y = < XYZ, Z = < YZX the angles of this triangle. We have

$y^2+z^2-x^2=\left(\sqrt{w+u}\right)^2+\left(\sqrt{u+v}\right)^2-\left(\sqrt{v+w}\right)^2$
$=\left(w+u\right)+\left(u+v\right)-\left(v+w\right)=2u$

and similarly $z^2+x^2-y^2=2v$ and $x^2+y^2-z^2=2w$. Now, denoting the area of triangle XYZ by T, we note that, as shown during the proof of Theorem 1 [b](c)[/b], we have $\cot Z=\frac{x^2+y^2-z^2}{4T}$, so that $\cot Z=\frac{2w}{4T}=\frac{w}{2T}$, and similarly $\cot X=\frac{u}{2T}$ and $\cot Y=\frac{v}{2T}$.

Now, since X, Y, Z are the angles of a triangle, by a well-known formula, we get $\cot Y\cdot\cot Z+\cot Z\cdot\cot X+\cot X\cdot\cot Y=1$, what, using these relations, becomes $\frac{v}{2T}\cdot\frac{w}{2T}+\frac{w}{2T}\cdot\frac{u}{2T}+\frac{u}{2T}\cdot\frac{v}{2T}=1$, so that $vw+wu+uv=4T^2$, and thus $T=\frac12\sqrt{4T^2}=\frac12\sqrt{vw+wu+uv}$. This completes the proof of the Conway substitution theorem.

Now, let's prove the Corollary: Construct a triangle XYZ for our values of u, v, w according to the Conway substitution theorem. Then, applying Theorem 1 [b](a)[/b] to the triangles ABC and XYZ, we get

$a^2\left(y^2+z^2-x^2\right)+b^2\left(z^2+x^2-y^2\right)+c^2\left(x^2+y^2-z^2\right)\geq 16ST$,

what, by the formulas given in the Conway substitution theorem, becomes

$a^2\cdot 2u+b^2\cdot 2v+c^2\cdot 2w\geq 16S\cdot\frac12\sqrt{vw+wu+uv}$;

this simplifies to $ua^2+vb^2+wc^2\geq 4\sqrt{vw+wu+uv}\cdot S$. This proves the Corollary.

  Darij

Another  way to prove    [color=blue]Pedoe Inequality [/color]

We have  $S=\frac{1}{2}absinC$  and  $T=\frac{1}{2}xysinZ$

Also by  cosine rule  we have : $c^2=a^2+b^2-2abcosC$  and  $z^2=x^2+y^2-2xycosZ$

So  the  initial  inequality is  equivalent to  :

$a^2(2y^2-2xycosZ)+b^2(2x^2-2xycosZ)+2c^2xycosZ\geq4abxysinCsinZ$
or
$a^2(2y^2-2xycosZ)+b^2(2x^2-2xycosZ)+2(a^2+b^2-2abcosC)xycosZ\geq4abxysinCsinZ$
or
$2(a^2y^2+b^2x^2)-4abxy(cosCcosZ+sinCsinZ)\geq0$

or

$(ay-bx)^2+2abxy[1-cos(C-Z)]\geq0$
which is  true

I found a shorter proof in a book.

[u][b]Theorem[/b][/u]
Let $a,b,c,S$ and $x,y,z,T$ be the three sidelengths and the area of two triangles, then
$a^2(y^2+z^2-x^2)+b^2(z^2+x^2-y^2)+c^2(x^2+y^2-z^2)\geq 16ST$

[b][u]Proof[/u][/b]
By Cauchy-Schwarz Inequality and Heron's formula respectively, we have
$16ST+2a^2x^2+2b^2y^2+2c^2z^2$
$\leq \sqrt{16S^2+2a^4+2b^4+2c^4}\sqrt{16T^2+2x^4+2y^4+2z^4}$
$\leq (a^2+b^2+c^2)(x^2+y^2+z^2)$.
The conclusion follows.

Here is another proof for  Corollary:
$Lemma$
The inequality:$a_1x^2+b_1y^2+c_1z^2 \ge a_2yz+b_2zx+c_2yz$ is ture for any real number $x,y,z$
$\iff a_1,b_1,c_1 \ge 0,b_1c_1 \ge 4a^2_2,c_1a_1 \ge 4b^2_2,a_1b_1 \ge 4c^2_2,4a_1b_1c_1 \ge a_1a^2_2+b_1b^2_2+c_1c^2_2+a_2b_2c_2$
The proof for the lemma is not hard,just use $\Delta$,If somebody ask for a solution,I will post it :P
for the original problem:
use heron's formula:
we have:$16S^2=2b^2c^2+2c^2a^2+2a_2b^2-a^4-b^4-c^4$
then:
$ua^2+vb^2+wc^2\geq 4\sqrt{vw+wu+uv}\cdot S$
$\iff (ua^2+vb^2+wc^2)^2 \ge (vw+wu+uv)(2b^2c^2+2c^2a^2+2a_2b^2-a^4-b^4-c^4)$
$\iff (u+v)(u+w)a^4+(v+u)(v+w)b^4+(w+u)(w+v)c^4 \ge 2u(v+w)b^2c^2+2v(w+u)c^2a^2+2w(u+v)a^2b^2$
let $a^2=x$,and $y,z$ similar,$a_1=(u+v)(u+w)$,$b_1,c_1$ similar.$a_2=2u(v+w)$,$b_1,c_1$ similar.
then it is easy to check:
$\iff a_1,b_1,c_1 \ge 0,b_1c_1 \ge 4a^2_2,c_1a_1 \ge 4b^2_2,a_1b_1 \ge 4c^2_2,4a_1b_1c_1=a_1a^2_2+b_1b^2_2+c_1c^2_2+a_2b_2c_2$
use the lemma the inequality is proved. :)
Anyway:I find darij's solution is sooooooooo nice and elementary.

let $B\left( 0,0 \right),C(a,0),A(x,h)(h>0)$,
    \begin{align*}
        u a^2+vb^2+wc^2&=u a^2+v\left[ \left( x-a  \right)^2+h^2\right]+w\left( u ^2+h^2 \right)\\
       &=(v+w)x^2-2avx+  (u +v)a^2+ (v+w)h^2\\
       &=(v+w)\left( x-\dfrac{av}{v+w} \right)^2-\dfrac{v^2}{v+w}a^2+ (u +v)a^2+ (v+w)h^2\\
       &\geqslant -\dfrac{v^2}{v+w}a^2+ (u +v)a^2+ (v+w)h^2\\
       &=\dfrac{u v+vw+wu }{v+w}a^2+\left( v+w \right)h^2\\
       &\geqslant 2ah\sqrt{u v+vw+wu }=4S\sqrt{u v+vw+wu }.
    \end{align*}

let $\overrightarrow{BD}=\dfrac{w}{v}\overrightarrow{DC}$,$AD=m$,from Stewart
   \[m^2=\dfrac{v}{v+w}b^2+\dfrac{w}{v+w}c^2-\dfrac{vw}{\left( v+w \right)^2}a^2\]
   \[vb^2+wc^2=\left( v+w \right)m^2+\dfrac{vw}{v+w}a^2\]
    \begin{align*}
    ua^2+vb^2+wc^2&=\left( u+\dfrac{vw}{v+w} \right)a^2+\left( v+w \right)m^2\\
    &= \dfrac{uv+vw+wu}{v+w}\cdot a^2+\left( v+w \right)m^2\\
    &\geqslant 2ma\sqrt{uv+vw+wu}
\end{align*}
    $\because m\geqslant h_a$，$\therefore ma\geqslant ah_a=2S$
    \[u^2+v^2+w^2\geqslant 4S\sqrt{uv+vw+wu}\]


[quote="mdk"]Let a be a positive integer. Find all positive integers n such that $ b= a^{n}$
satisfies the condition that $ a^{2}+b^{2}$ is divisible by $ ab+1$[/quote]

Hm...isn't it number theory problem? I think you posted it in a wrong place.

Anyway, I give a solution.

$ b=a^{n}$, so we get that $ a^{2}+a^{2n}$ is divisible by $ a^{n+1}+1$. But then, because of
$ a^{2}+a^{2n}=(a^{n+1}+1)a^{n-1}-a^{2}(a^{n-3}-1)$, we get that $ a^{n-3}-1$ is also divisible by $ a^{n+1}+1$. But of course, $ a^{n+1}+1>a^{n-3}-1$, so $ a^{n-3}-1$ must be $ 0$. Therefore $ n=3$.


Compute the sum $ S\equal{}\sum_{k\equal{}0}^n \binom{2k}{k}\binom{2(n\minus{}k)}{n\minus{}k}$

I would like to see a combinatorial and a computational solution .

http://www.mathlinks.ro/Forum/viewtopic.php?t=40150 , at least for the computational side. As for the combinatorial, I never understood the proof sketches posted there, but to be honest, I never really tried (still trying to solve it for myself when I am too frustrated to do anything else).

  darij


[quote="MIA6"]I think the definition of continuity at a point c need to fulfill following three conditions:
(1) limx-->c f(x) exists.

(2) f(c) is defined and

(3) lim x-->c f(x)=f(c)

Any failure in any of (1), (2), or (3) means the function is not continuous at $ c.$

Think of a function, when x=1, the limit from the left is equal to 1.5, the limit from the right is 0, then at point x=1, the function doesn't have a limit. Think about a solid circle (1,0), and a hollow circle (1,1.5) on this graph. Is function continuous at x=1? It's a discontinuity because this function doesn't have limits? IS that the reason? It's actually a jump discontinuity. Why it is not removable? because no limits? If it doesn't have limit, then how can we answer what the point of discontinuity is? still 1? why? because of the hollow circle or what? When we talk about continuity, we must think about if the limit of this function exists or not, right? Can we ignore the existence of limits, but only look at the graph at a point c, if at that point it is a solid circle, then it's continuous; if not, it's a discontinuous? Is that the definition of discontinuous? I don't think so. It must also relate to limits. Hope you can tell me, thanks.[/quote]
The definition you gave for discontinuity is [i][b]only[/i][/b] for functions. Part of the reason that this is true is because every limit must be unique. :wink: This invalidates all none-function graphs from having continuity or discontinuity.

Learn LaTeX.  Also, Kyron's comments have nothing to do with what's going on.

[quote="MIA6"]It's a discontinuity because this function doesn't have limits?[/quote]
This is a question you can answer by looking at the definition of limit and the information you're given about your function.

[quote]It's actually a jump discontinuity. Why it is not removable?[/quote]
A discontinuity of a function $ f(x)$ at the point $ c$ is [i]removable[/i] if there exists a [b]continuous[/b] function $ g(x)$ such that $ g(x) \equal{} f(x)$ at every point $ x\neq c$, and $ g(x)$ takes some value different from $ f(x)$ at $ c$.  (This also includes the possibility that $ f(x)$ is not defined at $ c$ but $ g(x)$ is.)  Now, answer your own question: can any function $ g$ that agrees with $ f$ everywhere except at $ c \equal{} 1$ be continuous?  You should be able to relate the general notion of removable discontinuity to the conditions (1), (2), (3) you listed above in a simple fashion.

[quote]Can we ignore the existence of limits, but only look at the graph at a point c, if at that point it is a solid circle, then it's continuous; if not, it's a discontinuous?[/quote]
A graph is a (frequently misleading) picture drawn to help you understand a function.  You should try to understand that the important object here is the [i]function[/i] -- the notations of graphing (solid circle, empty circle, etc.) are simply easy ways to convey information about the function to you, which then allow you to apply your definition that you have at the top of your post.  For your purposes, that definition is [i]the[/i] definition.


Prove that the circumcenter and the orthocenter of an obtuse-angled triangle lie outside the triangle.

just do some angle things...like wlog A>90

then like show like angle OBA> angle B and angle OCA > angle C


$x^{3}+3^{x}$ is always increasing, so there may only be one root.

By inspection, $x=4$ works, so we know it is the unique solution.

Ashrafmod, can you please post this kind of problems in intermediate.
While your problems are nice and elegant, they are a bit too far below the level that pre-olympiad is intended for...

Thanks for your understanding.


trigonometry
complex numbers
algebra
binomial theorem

Let $x$ be a real number such that $x+\frac{1}{x}$ is an integer. Prove that $x^n+\frac{1}{x^n}$ is an integer for all positive integers $n$

Denote $a_n=x^n+\frac{1}{x^n}$. By the given presumption, we have that $a_1\in\mathbb{Z}$, and hence $a_2=x^2+\frac{1}{x^2}=(x+\frac{1}{x})^2-2=a_1^2-2\in\mathbb{Z}$

Now induction. Assume that for $n\geqslant 2$ both $a_n$ and $a_{n-1}$ are integer. Then

$a_na_1=(x^n+\frac{1}{x^n})(x+\frac{1}{x})=x^{n+1}+\frac{1}{x^{n+1}}+x^{n-1}+\frac{1}{x^{n-1}}=a_{n+1}+a_{n-1}$, which gives

$a_{n+1}=a_na_1-a_{n-1}$, so by the inductive assumption, $a_{n+1}$ is also an integer. QED

Also, it can be shown using complex numbers that if $x+\frac{1}{x}=2\cos\alpha$, then $x^{n}+\frac{1}{x^{n}}=2\cos n\alpha$.  By the binomial theorem, this is an integer if the former is an integer.


calculus
derivative
analytic geometry
graphing lines
slope
calculus computations

Is there any easy way to graph implicit equations like, for example
$ x(1\plus{}y\plus{}y^2)\equal{}ln|x\plus{}y|$

Only if you could isolate $ y$ but in this case, it doesn't look like you can do that.  You would need Mathematica or something to get a nice picture or you could take it's derivative and plot the slope field.  Not perfect but you'll get an idea of what it looks like.


Let $ x,$ $ y$ and $ z$ are non-negative numbers such that $ x^2\plus{}y^2\plus{}z^2\equal{}3.$ Prove that:
\[ \frac{x}{\sqrt{x^{2}\plus{}y\plus{}z}}\plus{}\frac{y}{\sqrt{x\plus{}y^{2}\plus{}z}}\plus{}\frac{z}{\sqrt{x\plus{}y\plus{}z^{2}}} \leq \sqrt{3}\]

Nice inequality  :)

By Cauchy Swartz we have $ (x^2 \plus{} y \plus{} z)(1 \plus{} y \plus{} z)\geq (x \plus{} y \plus{} z)^2$

so we have to prove that $ \frac {x\sqrt {1 \plus{} y \plus{} z} \plus{} y\sqrt {1 \plus{} x \plus{} z} \plus{} z\sqrt {1 \plus{} x \plus{} y}}{x \plus{} y \plus{} z}\leq\sqrt {3}$

But again by Cauchy Swartz we have

$ x\sqrt {1 \plus{} y \plus{} z} \plus{} y\sqrt {1 \plus{} x \plus{} z} \plus{} z\sqrt {1 \plus{} x \plus{} y} \equal{} \sum\sqrt {x}\sqrt {x \plus{} xy \plus{} xz}\leq\sqrt {(x \plus{} y \plus{} z)(x \plus{} y \plus{} z \plus{} 2(xy \plus{} yz \plus{} zx)}$

and also

$ \sqrt {(x \plus{} y \plus{} z)(x \plus{} y \plus{} z \plus{} 2(xy \plus{} yz \plus{} zx)}\leq\sqrt {(x \plus{} y \plus{} z)(x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xy \plus{} 2yz \plus{} 2zx)} \equal{} s\sqrt {s}$

where $ s \equal{} x \plus{} y \plus{} z$ so we have to prove that $ \sqrt {s}\leq\sqrt {3}$ which is trivially true so QED  :D

hi .!!
it's also my solution silouan but i used jensen (my favorite inequality :) )

by caushy  :   $ \frac{x}{\sqrt{x^2\plus{}y\plus{}z}}\leq\frac{x\sqrt{1\plus{}y\plus{}z}}{x\plus{}y\plus{}z}$
now we have to prove :

$ \frac{x\sqrt{1\plus{}y\plus{}z}\plus{}y\sqrt{1\plus{}x\plus{}z}\plus{}z\sqrt{1\plus{}x\plus{}y}}{x\plus{}y\plus{}z}\leq\sqrt{3}$

by jensen inequality  :D :

$ LHS\leq\,\frac{x\plus{}y\plus{}z}{x\plus{}y\plus{}z}\sqrt{\frac{3\plus{}2(x\plus{}y\plus{}z)}{x\plus{}y\plus{}z}}\equal{}\sqrt{\frac{x^2\plus{}y^2\plus{}z^2\plus{}2(x\plus{}y\plus{}z)}{x\plus{}y\plus{}z}}$

$ \leq\sqrt{\frac{(x\plus{}y\plus{}z)^2}{x\plus{}y\plus{}z}}\equal{}\sqrt{x\plus{}y\plus{}z}\leq\sqrt{3}$

done  :lol:

[quote="mehdi cherif"]
by jensen inequality  :D :

$ \sqrt {\frac {x^2 \plus{} y^2 \plus{} z^2 \plus{} 2(x \plus{} y \plus{} z)}{x \plus{} y \plus{} z}}$

$ \leq\sqrt {\frac {(x \plus{} y \plus{} z)^2}{x \plus{} y \plus{} z}}$[/quote]
I think, it's wrong. :wink:
[quote="mehdi cherif"]where i'm wrong ????  :o[/quote]
$ \sqrt {\frac {x^2 \plus{} y^2 \plus{} z^2 \plus{} 2(x \plus{} y \plus{} z)}{x \plus{} y \plus{} z}}\leq\sqrt {\frac {(x \plus{} y \plus{} z)^2}{x \plus{} y \plus{} z}}\Leftrightarrow$
$ \Leftrightarrow x \plus{} y \plus{} z\leq xy \plus{} xz \plus{} yz,$ which is wrong. :wink:

[quote="arqady"]Let $ x,$ $ y$ and $ z$ are non-negative numbers such that $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 3.$ Prove that:
\[ \frac {x}{\sqrt {x^{2} \plus{} y \plus{} z}} \plus{} \frac {y}{\sqrt {x \plus{} y^{2} \plus{} z}} \plus{} \frac {z}{\sqrt {x \plus{} y \plus{} z^{2}}} \leq \sqrt {3}
\]
[/quote]
My solution (See problem 2 in the file) :)

[quote="silouan"]Nice solutions Can, Arqady
Arqady how did you find my solution ?[/quote]
I think, your proof the best of all, which here!


modular arithmetic
algebra
binomial theorem
complex numbers
number theory proposed
number theory

Can we find a natural number $ n$ and a gaussian positive integer $ z$ ( whose real and imaginary parts are positive integers ) such that $ z^{2n\plus{}1} \equal{} |z|^{2n\plus{}1}$ ?

I wonder if someone can show some interest to this topic.
Thank you in advance.

Yes, any non-negative integer will do.  Otherwise, it's not hard to see using unique factorization in the Gaussian integers that this is impossible.

(If the above message was not clear) I think it means the same as $ (a \plus{} b i)^{2n \plus{} 1} \in \mathbb{Z}[i] \implies b \equal{} 0 \quad (\forall (a,b)\in \mathbb{Z}^2)$, which answers your question.
This is pretty easy to prove (with binomial theorem and the fact that $ \mathbb{Z}$ is an UFD).

[hide="proof"]
Suppose to the contrary that $ \exists a \plus{} bi \in \mathbb{Z}[i]$ with $ a,b \neq 0$ and also $ \exists n \in \mathbb{N}^ \plus{} : (a \plus{} bi)^{2n \plus{} 1} \in \mathbb{Z}$.
wlog assume $ \gcd(a,b) \equal{} 1$ and $ a,b > 0$.
The imaginary part of $ (a \plus{} bi)^{2n \plus{} 1}$ should equal zero, so by binomial theorem $ \sum_{k \equal{} 0}^n {2n \plus{} 1 \choose 2k \plus{} 1} a^{2n \minus{} 2k} b^{2k \plus{} 1} ( \minus{} 1)^k \equal{} 0$.
It follows that $ a \mid b^{2n \plus{} 1}$ so $ a \equal{} 1$.
Also, since $ n > 0$, we have $ b^3 \mid b {2n \plus{} 1 \choose 1}$ so $ b \equiv 1 \pmod{2}$.
Substituting $ a \equal{} 1$:
$ \exists (x,y) \in \mathbb{Z}^2: (1 \plus{} b i)(x \plus{} y i)^2 \in \mathbb{Z}$ and wlog $ (x,y) \equal{} 1$.
So $ b(x^2 \minus{} y^2) \equal{} \minus{} 2xy \implies xy \equiv 1 \pmod{2} \implies x^2 \equiv y^2 \pmod{4}$.
Hence $ 2xy\equiv 0\pmod{4}$, contradiction.
[/hide]


Prove that \[ \left|\sum_{i\equal{}1}^n p_ia_ib_i\minus{}(\sum_{i\equal{}1}^n p_ia_i)(\sum_{i\equal{}1}^n p_ib_i)\right| <\equal{} \frac{1}{4}(H\minus{}h)(K\minus{}k)\], where $ p_1\plus{}p_2\plus{}...\plus{}p_n\equal{}1$, $ p_i>0$ and $ a_i \in [h,H]$ and $ b_i \in [k,K]$, $ n>\equal{}2$

nobody?...it's not so hard....but the proof is nice...
Let $ D(a.b)\equal{}\sum_{i\equal{}1}^n p_i a_i b_i\minus{}(\sum_{i\equal{}1}^n p_i a_i)(\sum_{i\equal{}1}^n p_i b_i)$ where $ a\equal{}(a_1,a_2,a_3,...,a_n), b\equal{}(b_1,b_2,b_3,...,b_n)$.
\[ D(a,b)\equal{}D(b,a)\]
\[ D(a\plus{}\alpha,b)\equal{}\sum p_i(a_i \plus{} \alpha) b_i \minus{} ( \sum p_i(a_i \plus{} \alpha))(\sum p_i b_i)\equal{} \sum p_i a_i b_i \plus{} \alpha \sum p_i b_i  \minus{} ( \sum p_i a_i \plus{} \alpha)(\sum p_i b_i)\equal{} \sum p_i a_i b_i \minus{} (\sum p_i a_i)(\sum p_i b_i)\equal{}D(a,b)\]
And if we let $ \alpha\equal{} \sum p_i a_i$ and $ \beta \equal{} \sum p_i b_i$ then we have
\[ D(a,b)\equal{}D(a\minus{}\alpha,b\minus{}\beta)\equal{} \sum p_i(a_i\minus{}\alpha)(b_i\minus{}\beta) \minus{} (\sum p_i(a_i\minus{}\alpha))(\sum p_i ( b_i\minus{}\beta))\equal{} \sum p_i (a_i\minus{}\alpha)(b_i\minus{}\beta)\]
With CBS :
\[ [D(a,b)]^2\equal{}[\sum p_i(a_i \minus{} \alpha)(b_i\minus{}\beta)]^2 <\equal{}[\sum p_i(a_i\minus{}\alpha)^2][\sum p_i(b_i\minus{}\beta)^2]\equal{}[\sum p_i a^2_i \minus{} \sum (p_i a_i)^2][\sum p_i b_i^2\minus{} (p_ib_i)^2]\equal{}D(a,a)D(b,b)\]
Then \[ 0<\equal{} \sum p_i(H\minus{}a_i)(a_i\minus{}h)\equal{}H \sum p_i a_i \plus{}h \sum p_i a_i \minus{}Hh\minus{}\sum p_i a_i^2\equal{}(H\minus{}\sum p_i a_i)(\sum p_i a_i \minus{}h)\minus{}D(a,a) <\equal{} \frac{1}{4}(H\minus{}h)^2\minus{}D(a,a)\]therefore $ D(a,a)<\equal{} \frac{1}{4}(H\minus{}h)^2$ and in the same way $ D(b,b)<\equal{} \frac{1}{4}(K\minus{}k)^2$and we obtain \[ [D(a,b)]^2<\equal{} D(a,a)D(b,b)<\equal{}\frac{1}{16}(H\minus{}h)^2(K\minus{}k)^2\] from where our inequality


According to my experience , almost whenever i try to talk to an admin he doesn't answer or replies with some useless words like (yes , ok...) with boastful sound :blush: ,i think they have their excuses ,probably they are busy,alot of people try to talk to them at once..) .Fortunately , you are among the good exeptions.How come ? ;)  :)

oh no , i clicked on new topic rather than post reply ... moderators please delete my post

We just care more about you guys that's all :P :)


calculus
integration
logarithms
LaTeX
absolute value
calculus computations

That solution works for $x>0$, and it works for $x<0$. The integrating factor is actually $cx$ for any nonzero constant $c$, as long as we stay away from the singularity at zero. $\ln x$ is an antiderivative for $\frac1x$ when $x>0$, and $\ln(-x)$ is an antiderivative when $x<0$.

Notice that unless $C=0$, your solution isn't even defined at zero- you have no right to expect anything to cross that singularity.

[quote="MrLobster"]If we raise e to the ln|x| is the absoulte value sign supposed to go away?  I would think the answer should be |x| but my book says x.

Here's the question I'm working on:

(dy/dx) + (1/x)y = 3x + 4

P(x) = (1/x)

(this step i don't understand)
Integrating Factor is e to the integral of (1/x) dx which the book says equals e to the ln x which equals x

the rest is:
xy = the integral of x * (3x + 4) dx = x^3 + 2x^2 + C

y = x^2 + 2x + (C/x)[/quote]
Latex will make it is easy to read.
Jmerry said that your solution only valid if $x \not=0$.
If you are feel strange with that $e$ or $\ln |x|$.
Let see this.
Do you know the exponent law $a^{^a \log x} = x$ ? for $x>0$ and $a>0$.?

Do you also know that $\int \frac{1}{x} \, dx = \ln |x| + C$ ?

This is what we have. $\ln x = ^e \log x$ . So $e^{\ln x} = x$.

The rest is just like Jmerry said, the absolute value $|x|$, and $x \not=0$ because we deal with $P(x) = \frac{1}{x}$.

I hope that's help.


linear algebra
matrix
inequalities
induction
binary matrix

In a nxn matrix A with 0 or 1 at least 99% of the elements are equal to 1. Prove that for any $k\geq 2$ the trace of the matrix $A^k$ is at least $(\frac{n}{2})^k$.

I thought I had it, but it turns out that what I proved was that the trace of $A^k$ is greater than or equal to $(1-.01k)n^k.$ That's not good enough. I'll keep trying. If no one else has it, I'll post that much in a day or two.

Kent, could you please post you solution,wich prove you result ?

All right, Eugene, you asked for it - but I'll warn you that it's a dead end. It fails to get at the most important question, which is finding an estimate for the dominant eigenvalue of $A$ that doesn't go bad for $n>100.$

Use the notation $J$ for the matrix of all ones, and adopt as a temporary convention the notation $|B|$ for the sum of all the entries of a positive-entry matrix $B$.

We can write $A=J-B_1$, where $B_1$ is a 0-1 matrix and $|B_1|\le.01n^2.$

Then, $A^k=n^{k-1}J-B_k$ where $B_k$ has positive entries. $B_k$ is determined recursively by $B_k=JB_{k-1}+n^{k-2}B_1J-B_{k-1}B_1.$

We can show that $tr(BJ)=tr(JB)=|B|$ and that $|BJ|=|JB|=n|B|.$ Using this and the inequality $|B_k|\le n|B_{k-1}|+n^{k-2}\cdot n|B_1|$ we can show by induction both that $|B_k|\le kn^{k-1}|B_1|$ and that $tr(B_k)\le kn^{k-2}|B_1| \le.01kn^k.$

Then $tr(A^k)= n^{k-1}tr(J)-tr(B_k)\ge n^k-.01kn^k=(1-.01k)n^k.$

The problem with this is the linear grown estimate for $tr|B_k|$, which becomes useless for large $k.$

Indeed, this is a hell of a problem and the solution is more than beautiful, but allmost impossible to find. I searched for hours a solution, but managed nothing.

Actually the problem is not that hard. Let us call an index $j\in \{1,\dots,n\}$ exceptional if the $j$-th row or the $j$-th column contains at least $0.1n$ zeros. Note that there are not more than $0.2n$ exceptional indices. Let $J$ be the set of non-exceptional indices. We'll prove by induction that $(A^k)_{ij}\geq (0.7n)^{k-1}$ for all $i,j\in J$, $k\geq 2$. Indeed, the base ($k=2$) is clear because, when multiplying the $i$-th row by the $j$-th column, we have at most $0.2n$ zeros in both and, thereby, at least $0.8n$ non-zero products. The step of induction is similar: the $i$-th row contains at most $0.1n$ zeros and the $j$-th column contains at most $0.2n$ elements corresponding to the exceptional rows for which we cannot use the induction assumption. This gives a factor of $0.7=1-0.1-0.2$. Now it remains to note that $\text{Tr}\,A^k\geq \sum_{j\in J}(A^k)_{jj}\geq 0.8n\cdot(0.7n)^{k-1}>\left(\frac n2\right)^k$. The end.


For each positive integer $p$, let $b(p)$ denote the unique positive integer $k$ such that $|k-\sqrt{p}|<\frac{1}{2}$.  For example, $b(6) = 2$ and $b(23)=5$.  If $S = \textstyle\sum_{p=1}^{2007}b(p)$, find the remainder when S is divided by 1000.

[quote="nat mc"][hide]When you figur out the first few, you see a pattern that $b(k)=n$ exactly $2n$ times.

So $\sum_{p=1}^{2007}b(p)=2\sum_{p=1}^{44}p^{2}+\sum_{p=1981}^{2007}45=59955$.  $\boxed{955}$[/hide][/quote]
[hide]Note that $(n+.5)^{2}= n^{2}+n+.25$, this explains why $b(k)=n$ repeats exactly $2n$ times.[/hide]

Yes, do a 44-term addition problem, killing off every digit larger than the thousands.  :P

[quote="saturnlife"]Just wondering, is there a way to do this problem without knowing how to find the sum of the sequence of squares?[/quote]

you should have this formula memorized

[quote="mysmartmouth"][quote="saturnlife"]Just wondering, is there a way to do this problem without knowing how to find the sum of the sequence of squares?[/quote]

I actually derived it in the middle of the test...painful yet satisfying.[/quote]
To derive, finite differences takes like 5 min, and hockey stick takes like 2 min (although it's much harder to see).

oh man I just did this problem and jeeez it was silly error prone...

Why was this problem number 11? Besides 3 and 7 this was like the easiest one.

Your right it was pretty easy for a number 11, but most of the problems at the eginning besides 3 were just nasty.

About $b(p)=n$ exactly $2n$ times, it is because $(n+.5)^{2}-(n-.5)^{2}=2n$

I figured this one out, and then forgot to do the sum of squares and just went $2(44)^{2}$, and then realized it like as soon as the test was over and was all like  :wallbash:  :wallbash_red:  :wallbash:  :wallbash_red:  :wallbash:.  Of course, I was like that on 6 problems, considering that I missed 6 of them for similar reasons...

I actually added the first 20 terms of the 44 term sequence and found that the first ten terms added up to 770, the next ten added up to x970, the next at xy170, the next at xy370.  this adds up to 280 as the ending 3 digits. I then added out 2(41*41), 2(42*42), 2(43*43), 2(44*44), and 27*45, and got 255 as my answer.  It took me FOREVER to add all this up, and I still got it wrong.  I must have added somewhere wrong....   :furious:  :bomb:

Well, I forgot the formula but I knew it was a cubic polynomial, so I just solved for the coefficients of $ax^{3}+bx^{2}+cx+d$

$d = 0$
$a+b+c+d=1$
$8a+4b+2c+d=5$
$27a+9b+3c+d=14$

Wow. That problem was so easy, both to do and to screw up on.

[quote="Ignite168"]I only remember that the sum of squares is something over 6 as well as 3 terms in the numerator and play around a bit, it takes about a minute more but it's not really necessary to memorize.[/quote]

But it's so easy anyway...

$\sum_{k=1}^{n}{k^{2}}=\frac{n(n+1)(2n+1)}{6}$

The thing that's interesting is that $\sum_{k=1}^{n}{k^{3}}=\left(\sum_{k=1}^{n}{k}\right)^{2}$...

haha 44 term addition. I did it! and got it right!
woohoo!

That only took... 3 sheets of paper and 15 minutes. Stupid sloppy handwriting. :D

I think I put 954 for this one.

miscounted how many 45's there were.

I use the sum of squares formula so often and am supposed to know it in my sleep and I accidentally did $\frac{(44)(45)(46)}{6}$... then I was confident that I got it right later (well that and I was too lazy) so I didn't bother checking so I missed it.

I could have solved this problem, but I forgot the "sum of squares" formula.

[quote="samath"]I don't quite understand why people have such a hard time remembering the sum of squares formula, it's really not that hard.[/quote]

Well, never learning about it is another story.  :P

Well, this is an error prone solution. At least I believe.

[hide]

First, try a couple of examples. $b(1-2)=1$, $b(2)=1$, $b(3-6)=2$, $b(7-12)=3$, and so on. As we see, a pattern emerges. There are $2$ possibilities for $b(p)=1$, $4$ possibilities for $b(p)=2$, $6$ possibilities for $b(p)=3$, and on and on. From this, we derive that there are yes, $2n$ possibilities for $b(p)=n$.

Now, since there are $2$ possibilities for $b(p)=1$, this contributes $2*1=2$ into $S$.
Now, since there are $4$ possibilities for $b(p)=2$, this contributes $4*2=8$ into $S$.
Now, since there are $6$ possibilities for $b(p)=3$, this contributes $6*3=18$ into $S$. And so forth.

We derive the pattern $2, 8, 18, 32, 50...$. This sequence is $2n^2$, where $n$ is the term number. ...$A$

Now, with some difficulty, $44.5^2=1980.25$. Thus, for each $1981-2007$ (or $27$ choices, they contribute each $45$ into $S$.

Now, substitute $44$ into the formula in $A$. $2*44*44=3872$.

Thus, $S = \sum_{k=1}^{44}2n^2$ $+$ $27*45 = 2((44*45*89)/6)+1215 = 58740 +1215 =  59955$.

Now, do $59955 \equiv 955\ mod(1000)$

Thus our answer is $ \boxed{955} $[/hide]

I almost got the whole problem the only mistake I made was:

$$\sum_{n=1}^{45} 2n^2$$.

How did you see to stop at $1981$?

Since $45^2$ is larger than 2007, we don't sum all the way to 45. Instead, we sum to 44 for the squares, and add 45 for the rest of the numbers. They stopped at 1981 because its square root was closer to 45, as can be seen from $(44+.5)^2 = 44^2 + 2*44*.5 +.25 = 1936 + 44 + .25 = 1980.25$ < 1981. Thus, b(1981) would equal 45, so that is where to start for summing 45s.

p11: We have $k-\frac{1}{2} < \sqrt{p} < k + \frac{1}{2}$, or $k^2-k+\frac{1}{4} < p < k^2+k+\frac{1}{4}$. We know $p,k$ are integers, so we can rewrite as $k^2-k+1 \leq p \leq k^2+k$. This is true for $2k$ values of $p$, so $b(p)=k$ for $2k$ values of $p$. So we know we must have $2 \cdot 1 + 4 \cdot 2 + \ldots + 88 \cdot 44$, and we must add $45(2007-1981+1) = 27 \cdot 45$. This simplifies to $2(1^2+2^2+\ldots+44^2) + 27 \cdot 45 = 59955 \implies \boxed{955}$.

[b]Solution:[/b] It is not hard to bound $k$ (using the square of integers $\pm 0.5$) and see that it results in $1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, ...$ for $p = 1, 2, 3, 4, 5, ...$. The general pattern is $S = \textstyle\sum_{k=1}^{...}(2k^2)$. Notice that that the upper bound of the summation goes up to $k=44$ when $p = 1980$. When we add in our extra $27$ of $45 k$ values we get our final answer of: $S = \textstyle\sum_{k=1}^{44}(2k^2) + 27(45) \equiv \boxed{955} \pmod{1000}$.


Prove that, given 4 mutually tangent circles of radii $x, y, z, w$, we have the following relation

$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{1}{w^2} = \frac{1}{2}(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{w})^2$


If a, b < 100, a, b are postive integers. How many ordered pairs are there such that the sum, a+b, difference, product, and the quotient are all perfect sqaures?

[hide]
Without Loss of Generality, let $ a \ge b$ If there are ordered pairs where $ a>b$, then we just reverse them to get the ones with $ b>a$
The fact that the sum and the difference are both perfect squares tells us that the numbers are in the form of
$ (\frac {c^2 + d^2}{2},\frac {c^2 - d^2}{2}), c > d\ge 0$
The fact that the product and quotient are both perfect squares tells us that the numbers are in the form of
$ (k^2 \cdot b, b)$ (assuming that $ a \ge b$, which we assumed)

Now based on these 2 observations, the following equations must hold true:
$ k^2 \cdot b = \frac {c^2 + d^2}{2}$
$ b = \frac {c^2 - d^2}{2}$
Subtracting the second equation from the first gives
$ (k^2 - 1)b = d^2$
So, $ k^2 - 1$ is b times a perfect square: $ k^2 - 1 = l^2b (1)$
Adding the 2 equations gives
$ (k^2 + 1)b = c^2$ So, $ k^2 + 1$ is b times a perfect square: $ k^2 + 1 = m^2b (2)$
subtracting $ (1)$ from $ (2)$ gives that $ 2 = (m^2 - l^2)b \implies b|2$
If $ b = 1$, then $ m^2 - l^2 = 2$ which is not possible. (m^2 and l^2 are perfect squares)
If $ b = 2$, then $ m^2 - l^2 = 1 \implies m = \pm1, l = 0$
Plugging into $ (1)$ and $ (2)$ gives that $ k = 1 \implies a = b$

Now all we need is that $ 2a, 0, 1, a^2$ are all perfect squares. The last 3 are always true, and $ 2a$ is a perfect square for $ a$ in the form $ 2n^2$ This is less than 100 and greater than 0 for $ 1 \le n \le 7$

This gives $ \boxed{7}$ ordered pairs[/hide]


modular arithmetic
algebra
polynomial
number theory unsolved
number theory

[quote="Max D.R."]$ n$ is prime?  :?:[/quote]
Yes, $ n$ is a prime and $ a,b,c$ are integers. You don't have to solve my problem. But give me a tip, please ;/

PS I've wrong problem because it's not true for $ 1,1,1$. Unfortunately I don't know what I should have in the notebook. I'm sorry  :!:

[quote="Maglor"]I've thinking a lot of time about this and I've got no idea how to prove one problem. That is:
$ a \plus{} b \plus{} c \equiv 0 (mod n)$ and $ a^5 \plus{} b^5 \plus{} c^5 \equiv 0 (mod n)$
Prove that $ abc \equiv 0 (mod n)$  :([/quote]
It can be proved when $ n \equiv 2 \bmod 3$ and $ n \neq 5$.
Assume that $ n \mid a \plus{} b \plus{} c,a^5 \plus{} b^5 \plus{} c^5$ and $ p \nmid abc$.
$ 0 \equiv a^5 \plus{} b^5 \plus{} c^5 \equiv \minus{} (b \plus{} c)^5 \plus{} b^5 \plus{} c^5 \equal{} 5bc(b^3 \plus{} 2b^2c \plus{} 2bc^2 \plus{} c^3) \bmod n \iff$
$ n \mid b^3 \plus{} 2b^2c \plus{} 2bc^2 \plus{} c^3 \equal{} (b \plus{} c)(b^2 \plus{} bc \plus{} c^2)$. $ n \mid b \plus{} c \Rightarrow n \mid a \Rightarrow n \mid abc$ so $ n \nmid b \plus{} c$.
$ n \mid b^2 \plus{} bc \plus{} c^2$. $ b \equiv c \bmod n \Rightarrow n \mid 3b^2 \iff n \mid 3$ so $ b \neq c \bmod n$
$ n \mid b^3 \minus{} c^3 \Rightarrow n \mid b^{p \plus{} 1} \minus{} c^{p \plus{} 1} \iff n \mid b^2 \minus{} c^2$. Since $ b \neq c \bmod n$ we see $ n \mid b \plus{} c$ so $ n \mid a$ and $ n \mid abc$ and we obtain a contradiction. And hence $ n \mid a \plus{} b \plus{} c, a^5 \plus{} b^5 \plus{} c^5 \Rightarrow n \mid abc$. But remember: It only holds when $ n \equiv 2 \bmod 3$ and $ n \neq 5$. For other primes it is not true.

We can still say a few things. Simple manipulations yield $ 5ab(ab\minus{}c^2) \equiv 0 \pmod{n}$ and the other cyclical.

1. So for $ n\equal{}5$, either $ abc \equiv 0 \pmod{5}$, or, when $ a,b,c$ are neither divisible by $ 5$, we trivially have that $ a\plus{}b\plus{}c \equiv 0 \pmod{5}$ implies $ a^5\plus{}b^5\plus{}c^5 \equiv 0 \pmod{5}$ (from Fermat's little).

2. When $ n \neq 5$ it follows $ 3(ab\plus{}bc\plus{}ca) \equiv 0 \pmod{n}$. So
2.1. For $ n\equal{}3$, either $ abc \equiv 0 \pmod{3}$, or, when $ a,b,c$ are neither divisible by $ 3$, we trivially have that all three are congruent with $ 1$, or with $ 2$ modulo $ 3$, when both $ a\plus{}b\plus{}c \equiv 0 \pmod{3}$ and $ a^5\plus{}b^5\plus{}c^5 \equiv 0 \pmod{3}$.
2.2. When $ n \neq 3$ it follows $ ab\plus{}bc\plus{}ca \equiv 0 \pmod{n}$, so $ a,b,c$ are the roots of the polynomial $ x^3 \minus{} abc \in \mathbb{Z}_n[x]$, i.e. $ abc$ is a cubic residue modulo $ n$. Examples exist, e.g. for $ n\equal{}7$ we have the roots $ 1,2,4$, which verify.

EDIT. Simultaneous post with the above.

Are there any primes $ p \equiv 1 \pmod{3}$ for which $ (a\plus{}b\plus{}c) \equiv (a^5 \plus{} b^5 \plus{} c^5) \equiv 0 \pmod{p}$ is [b]impossible unless[/b] $ abc \equiv 0\pmod{p}?$  If so, is there a way of distinguishing for which primes $ \equiv 1 \pmod{3}$ it is possible?  For example, as previously shown, it is possible when $ p\equal{}7.$

Thank you ! I'm very grateful to you for your help !!  :roll:

[quote="zugzwang"]Are there any primes $ p \equiv 1 \pmod{3}$ for which $ (a \plus{} b \plus{} c) \equiv (a^5 \plus{} b^5 \plus{} c^5) \equiv 0 \pmod{p}$ is [b]impossible unless[/b] $ abc \equiv 0\pmod{p}?$  If so, is there a way of distinguishing for which primes $ \equiv 1 \pmod{3}$ it is possible?  For example, as previously shown, it is possible when $ p \equal{} 7.$[/quote]
No it is not. There exist a number $ \alpha$ such that $ \alpha^3 \equiv 1 \bmod p$ and $ \alpha \neq 1 \bmod p$. Letting $ a\equal{}1,b\equal{}\alpha,c\equal{}\alpha^2$ we find that $ a\plus{}b\plus{}c \equiv a^5\plus{}b^5\plus{}c^5 \equiv 0 \bmod p$ and $ p \nmid abc$

[quote="Maglor"]I'm sorry but I've one question more. Is it true that when $ n \equiv 1 mod.3$, we have $ abc \equiv 1 (mod.n)$ unless what $ abc \equiv$ to ?[/quote]
When $ n\equal{}7$ we choose $ a\equal{}3,b\equal{}3\alpha,c\equal{}3\alpha^2$ and then $ abc \equiv 6 \bmod n$.
For $ n > 7$ we choose $ a\equal{}2,b\equal{}2\alpha,c\equal{}2\alpha^2$ and $ abc \equiv 8 \bmod n$


group theory
abstract algebra
superior algebra
superior algebra unsolved

Show that G = S_p provided that p is a prime number, G is a transitive subgroup of S_p containing a 2-cycle.

it's known that $ (12),(12...n)$ generate $ S_n$ for every $ n \geq 2$. so let's assume $ (12) \in G$ and show $ (12...p) \in G$: the orbit formula yields $ p | |G|$, so there is an element of order $ p$ in $ G$. this must be a $ p$-cycle, let's write it as $ (1...2...)$ there $ 2$ is on the $ k$th place ($ 1 < k \leq p$). then $ p, k \minus{} 1$ are coprime, thus $ (1...2...)^{k \minus{} 1}$ is also a $ p$-cycle which has the form $ (12...)$. after relabelling, we may assume that this is $ (12...p)$.

how do you apply this orbit formula in this case?

zptlimyw,

We appear to both be taking the same algebra class.
First, I would like to say I have no idea who you are.
Second, I have no idea whether our university's honor code allows collaboration over a message board on pset questions.
Plus maybe you are not taking the class but doing to problem sets for fun.
I really don't care, and I am not accusing you of anything.  I am not going to look into it any further, but you should probably check, because if it turns out this type of stuff is frowned upon by the university, the penalties for honor code violation are very heavy.
Again, I have no idea if this is a violation or not.  And I'm not going to look into it. I just recommend you be careful.
I suspect I will see you in class, whoever you are.

I used to post on AoPS quite a bit back in high school (on a different account of course) - I want to remain anonymous in this post.


probability
geometry
conics
parabola
expected value

You wait at a bus stop and you can take either Bus $ A$ or Bus $ B$. Bus $ A$ comes every $ 12$ minutes and Bus $ B$ comes every $ 30$ minutes. What is the expected value of time you'll have to wait for a bus?

Can you generalize this to $ m$ and $ n$ minutes?

[hide="All you need is:"]
$ E=\frac{1}{p}$ in simple situations like this. [hide="So,"]
\[ E= \frac{1}{p_{A}\cup p_{B}}= \frac{1}{ \frac{1}{12}+\frac{1}{30}-\left( \frac{1}{12}\right) \cdot \left( \frac{1}{30}\right) }= \boxed{ \frac{360}{41}\text{ minutes}}\] [/hide] [/hide]

Could you derive how you get your results?

Unfortunately, it's more complicated than that.  In particular, it will depend on how the two busses are synchronized.  If they both arrive together on the hour, then the answer works out to $ \frac{324}{60}= \frac{27}{5}$, just over 5 minutes.  If, however, we shift the 12-minute bus so it runs three minutes later, de-synchronizing it, we get a better distribution of bus arrival times and the expected wait declines to $ \frac{306}{60}= \frac{17}{4}$, just over 4 minutes.  (Note that both of these are a good deal smaller than the nearly 9 minutes of Cinco's answer.  As they should be: with only the 12-minute bus, I get a bus every 6 minutes on average, so with a second bus I have to do better than that.)

If we are given how the busses are synchronized, we can solve this as a geometric probability problem.  Just draw the two curves associated to the waiting time for each bus (they're saw-toothed), then calculate the areas of some triangles and divide by the total period (60 minutes).  The relationship between the offset and the expected waiting time makes a nice parabola -- the 3 minute offset gives the minimal time, and the perfect synchronization gives the maximal.

As an aside, it seems that probability is one realm where nearly everyone, even quite accomplished problem-solvers, are prone to applying inappropriate formulas.  I wonder why that is.


calculus derivative algebra function domain trigonometry topology

If the total derivative = 0 for all $ p \in U$, doesn't it intuitively follow that $ f$ is constant( where $ : U \rightarrow \mathbb{R}^{m}$ and is differentiable everywhere on $ U$)?

I say "intuitively", simply because I can show by limit definition of the derivative, that if f is constant , then $ f'(x) \equal{} 0$. But I suspect, I am simplifying too much and there's probably something I haven't thought through....

I need $ U$ to be connected and open....

But does $ U$ have to be necessarily open?

[quote="jmerry"]Does the definition of derivative you're using even make sense at points which don't have an open neighborhood in the domain?[/quote]

Maybe not, but suppose E is a connected subset of an open set U, f : U -> R is differentiable at each point of E, and Df = 0 at each point of E. Is  f constant on E?

Right, the definition of the derivative is on open sets....guess that answers it.

Now that's an interesting question, WWW. Cleaned up a bit:

Suppose $ f$ is a function from $ \mathbb{R}^n$ to $ \mathbb{R}$, and $ E$ is a connected subset of $ \mathbb{R}^n$. If $ f$ is differentiable at every point of $ E$ and its derivative (the gradient) is zero at all of those points, is $ f$ necessarily constant on $ E$?

I have an answer, but I'll leave it for some other people to think about first. Note that we don't assume anything away from $ E$, other than what is required to make $ f$ differentiable on $ E$. It can be arbitrarily badly behaved far from $ E$; we just assumed values to make the definition simple.

I would have to say yes. If the gradient of the function = 0, this would imply that every derivative of f = 0. And if the derivative of f = 0, then f is constant.

but I can't really prove it....

I think I could prove it's true if the set were path connected, but I'm not sure how to do it if it were topologically connected but not path connected.

Connectedness path connected. Since by definition a connected set is some unbroken space (at least that's how I 'm picturing it) and you could easily draw a path between any two points on the space.

Connected does not imply path connectedness. (The topologist's sine curve is the usual example.) I think the claim is true if its path connected; if $ \alpha$ is a path in $ E$, then $ (f \circ \alpha)' \equal{} (\nabla f)(\alpha (t)) \cdot \alpha'(t) \equal{} 0$ so $ f \circ \alpha$ is constant on the unit interval and so $ f$ is constant on the image of $ \alpha$.

I'm curious how a proof would work if you just said connected -- but I'm not sure the claim is even true for an arbitrary connected set.

A differentiable path, yes. The topologist's sine curve, yes. The graph of a nowhere differentiable function? I'm not sure. I thought so, but it turns out I don't have a proof.

Shouldn't $ E$ be open for differentiability to make sense? And since an open, connected set is path-connected...

The current discussion is on a modified question, asked in posts 4 and 6. The function is defined beyond $ E$, but we only know about the derivative on $ E$.

My book says that any open, connected subset of $ \mathbb{R}^{n}$ is path-connected, but so what if it is? what does that tell us about f on $ E$?


limit
real analysis
real analysis unsolved

Simplify

$\prod_{n\geq 1}\left(\frac{1}{2}\left(\cos(3z/4^n)-\cos(z/4^n)\right)\right)$

for any $z\in C$

[quote="Moubinool"]Simplify

$\prod_{n\geq 1}(\frac{1}{2}(\cos(3z/4^n)-\cos(z/4^n)))$

for any $z\in C$[/quote]

If you mean $n\to +\infty$

we are going to calculate $\lim_{N\to \infty}\prod_{n=1}^{N}\frac{1}{2}(\cos(3z/4^n)-\cos(z/4^n)))$

In fact, we have $\frac{1}{2}(\cos(3z/4^n)-\cos(z/4^n) = -\sin(2z/4^n)\sin(z/4^n) $
but as $\sin(x) \simeq x$ as x tends to zero, we have $\Pi = 0$ when $z\neq 0$
when $z=0$, this is obvious that the product is zero as all its terms equal to zero.

Fabrice


Correct, but can you show your work?  :)

I don't believe that a discussion of $i$ is appropriate in this forum.

[quote="slamchess"][quote="Drunken_Math"]Expand and Simplify:

$(1+i)^4$

Where $i=\sqrt{-1}$[/quote]

how do you figure out sqrt{-1} ???[/quote]

Here's a slight hint:

[hide]
$i^1=i$
$i^2=-1$
$i^3=-i$
$i^4=1$
[/hide]

[quote="kunny"]O.K., Drunken_Math.

Now you can answer $(1+i)^4$. :)

kunny[/quote]

I posted the question, so I don't think I should solve it...  :?

Why is this question named binomial theoem?I dont think it has anythin to do with bin teorem...By the way...could someone tell me about binomial theorem..cause i looked it up in a book but couldnt understand a thing!! :rotfl:

Here:

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

It's not too complicated...  ;)

[quote="uberl33tmage"][hide]i got -4[/hide][/quote]

Thanks. We knew that a while ago  :roll:

How did you do it?

[hide]
Knowing that $\displaystyle (x+y)^4=x^4+y^4+6x^2y^2+4x^3y+4xy^3$, you plug in 1 and i:
$\displaystyle (1+i)^4=1^4+\sqrt{-1}^4+[(6\cdot1)^2\cdot(2\sqrt{-1})^2]+[(4\cdot1)^3\cdot\sqrt{-1})]+[((4\cdot1)\cdot\sqrt{-1})^3]$ Simplifying gives you:
$\displaystyle (1+i)^4=1+4i+(-6)+(-4i)+1$
$\displaystyle 4i$ and $\displaystyle -4i$ simplify, and you get:
$\displaystyle \framebox{-4}$
[/hide]

[quote="boxedexe"]I solved this problem using basic algebra and following the complex numbers, and I got -4:
$(1+i)^4=1+4i+6i^2+4i^3+1$
$=2+4i-6-4i$
$=-4$

By the way, how did they derive that $i^2=\sqrt{-1}^2=-1$.  Should it not be
$i^2=\sqrt{-1}^2$
$=\sqrt{(-1)^2}$
$=\sqrt{1}$
$=1$

Is it a different case in complex numbers?  If so, howcome?

Masoud Zargar[/quote]

  Yes, manipulation such as these are strictly forbidden.

[quote="JesusFreak197"]$(\sqrt{-1})^2\not= \sqrt{(-1)^2}$
$\sqrt{n}=n^\frac{1}{2}$, so $(\sqrt{n})^2=(n^\frac{1}{2})^2=n$[/quote]

So if the value we're taking the sqrt of is negative, we cannot use that eponent rule, correct?

In this case, we have $\sqrt{-1}^2=[(-1)^\frac{1}{2}]^2=(-1)^1=-1$

Got it!

Masoud Zargar


Let $ \alpha\equal{}\sqrt[n]{2}$ and $ p_{0},p_{1},\cdots,p_{n\minus{}1}$ are integers for which at least one of them isn’t equal to zero.Prove that
$ |p_0\plus{}p_1\alpha\plus{}\cdots\plus{}p_{n\minus{}1}\alpha^{n\minus{}1}|\geq\frac{2^{1\minus{}n}}{(|p_0|\plus{}|p_1|\plus{}\cdots\plus{}|p_{n\minus{}1}|)^{n\minus{}1}}$.


[color=darkred]For any $ m \in \mathbb{N}^*$ solve the ecuation
\[ \left\{\left( x \plus{} \frac {1}{m}\right) ^3\right\} \equal{} x^3
\]
[/color]

[hide="hint"]$ x^3 <1$ then $ x<1$ then integer part of $ ( x + \frac {1}{m}) ^3$ is $ 0$ or $ 1$ for $ m\ge 4$[/hide]


linear algebra
matrix
combinatorics proposed
combinatorics

assume that A is 1*n table st $n=2^k$.we put 1,...,n in it in arbitrary order .then we do the following operation 2k times:
)for $\leq i \leq n$ if i th array is s ,then u write the s th array's number in ith array.
now prove that if u do the operation after this ,table won't change.

edit: temporarily removed. I read 2^k sorry.

I think that your problem is similar to this one:
If $\sigma\in S_{n}, n=2^k,k\in\mathbb{N}$ is a permutation then prove that $\sigma^{2k}=\sigma$ equivalent to $\sigma^{2k-1}=e$ where $e$ is the identic permutation.
Are you sure there are $2k$ operations? and not $2^k$?
by example if you take $\sigma=
\begin{pmatrix}
1 & 2 & 3 & \ldots & n-1 & n\\
2 & 3 & 4 & \ldots & n & 1\\
\end{pmatrix}$
and u make $2k$ operations on it, it won't be the same!!
Is smth wrong?

[quote="sam-n"]Rephrasing the problem a bit like Kuba said:

Consider $A = a_1,...,a_n$ a permutation of the numbers $1,...,n$ with $n=2^k$.

Then we do the following operation $2.k$ times:
for $i = 1$ to $n$:  if $a_i = j \not = i$, then set $a_i = a_j$.

Prove that the next time we do the operation, A will not change anymore.[/quote]

Let's first state a couple of lemmas:
1. if the i-th number is i, it doens't change.
2. The final situation will be that all numbers that did not start on the right place are then the same number, and all that did start right: see 1.
From here on it's trivial to show that the row will not change anymore.
---------------------------------------------------------------------------

Kuba, you made the same mistake as I did at first. But look at how (y)our thingie goes:
2 3 4 ... n-1 n 1
3 4 5 ... n 1 2  (and not 3 I guess? makes difference!! are you doing the operation simultaneously??)

anyway the next step shows that the pattern quickly fails:
5 6 7 8 9 ...
9 10 11 12...
17 18 ...
33 34...
after $q$ operations starting with the number $2^{(q-1)}+1$
so to get at $1,2,...,n$ at exactly $2.k$ steps!!

I think I solved it now but it's like 11.35 PM here (almost midnight) so I'm gonna post it later sorry

oups
there is a little (actually HUGE) mistake that I made. Instead $\sigma ^{2k} $ there is $\sigma^{2^{2k}}$..


Harvard college AMC AIME articles MIT USA(J)MO

For some reason I am very curious about this subject. I'd like to know who the smartest, most well rounded, best prospect to graduate valedvictorian at harvard in 2.5 yrs, etc person you know is? list their academic credentials/why you think they are the smartest/most analytical/best debater/etc person you know (or have heard of and know to be true).

I know of 2:
guy 1:
high school - 1500+ on SATS without studying much, all 5's on APs without studying much, seemed to never study anything but always knew stuff in class. Graduated 3rd at my high school ( I went to a very good public high school). Everyone went to him for answers. Pretty good social skills but he became arrogant as time passed. Right now he is at the university of texas in teh business honors program (very elite and basically means you can get 95% of jobs post graduation). HE got a 4.0 in college with EASE (he would drink beer during the week a lot) which is NOT easy to do. He now has an investment banking job lined up for him in Houston (he wanted to live in Houston).

guy 2:
high school - same stats as above but has a lot less social skills (kinda dorky, know it all type, but nice). Triple major in the university of texas business honors program (finance/accounting/economics) and graduated half a semester early with a 4.0 GPA and was an officer in quite a few clubs. He just got a job as an analyst with an investment bank (top 3... he did not get goldman however nor did he get blackstone group).

Let's see...my good friend who graduated last year:

Salutatorian (2nd in class)
1550 SAT
National Merit Finalist
5's on AP Calc AB, Phys B, US Hist, Euro Hist, Eng, Chem, Stats...4 on US Govt
Really good at piano
3 Time All State
3 Time AIME Qualifier
(Good at math :D )
He got into Harvard but is going to a state school right now.  He's a great guy.  Always fun to talk to and he has a great sense of humor.  Athletically, he's in good shape though I'm not sure if he plays any sports in particular.

[quote="blahblahblah"]Take a look at any of the Academic All-American teams.  None of the people listed thus far in this thread have especially impressive credentials, IMHO.  In particular, business majors are especially unprestigious.[/quote]
interesting, do you know where I can find these people and look at their stats (link?) plz?

thank you!

I don't see what the big deal is about them :huh: , 1550 SAT average, 3.98 GPA.... :| , not really saying much

that's only the basic stats..... they've all got more going for them than those...

http://www.usatoday.com/news/education/2005-05-12-first-team-hs-academic_x.htm
has the in-depth things for all of them.  Those kids are quite smart, you must admit.

So do any of these guys, I dunno, GIVE BACK to their community or anything?

First person on the list...

-devised the Brucker Oblique Stabilization Brace to help athletes recover from hip injuries;
- tutoring liaison,
- National Honor Society
- United States Mensa volunteer and youngest editor of its Half A Step Ahead magazine;
- community TV volunteer;
- creator, self-syndicated TV show on teen health topics broadcast on public access stations;
-performs as a ventriloquist at nursing homes;

all service things...

I know some... Reid Barton, Brian Lawrence, Eric Price, etc. :D

Or look at the lists of finalists in the large research competitions, ISEF, Intel STS, Westinghouse, etc.

see, THAT'S why those people are impressive. not because they made pretty-darn-good test scores. sorry, i didnt read the article, i was responding in general to the people listed here by others.  :blush:

Yeah, those people are quite impressive.

Btw, just for future clarification, the person I listed is the smartest person that [i]I[/i] know.  By no means am I trying to assert that he is the absolute best or anything like that.

Also, while I admit that test scores aren't everything, there are few other ways to compare people in a meaningful way (of course, Intel, Siemens etc are good ways to distinguish between good and great, but in the same token, you shouldn't just place your whole judgement on a single aspect of a person).

May it be known that I, I!!!, got into Chapel Hill today. The University. of North Carolina. at Chapel Hill. For out-of-staters, it's harder to get into than any of the ivies. Quote from the letter:

"Over the last several weeks, as members of the admissions committee have reviewed your application, we have become more an more impressed by your many achievements."

They are [i]impressed[/i]! With my [i]achievements!!!![/i]

Oh. Wow. May it be known that I, I!!!, am overly excited.

P.S. I'm posting in this thread because... er, I have no idea why! :blush:  :blush:  :blush:  :blush:

Here, I'll even post my "stats" to be halfway on topic (no, I'm not the smartest, the most talented, or even the coolest person I know).
[size=59]
attend what was recently a top public high school in the country (we got bumped after the whole IB thing got added)
GPA: unweighted 3.92, weighted something like a 4.5?
SAT I : 2160- 740M/740R/690W(1510 on the old one- 800M/710V)
ACT: 35 (writing was a 9)
SAT IIs: Math IIC 800, chemistry 760

APS: chem (5), calBC (5), bio (5), US history (4); taking stats, physics, gov, and econ this year.
other advanced: took genetics (210) at local university, made an A. also taking cal3, discrete math; have taken 1yr computer programming; 3 yrs of spanish
electives within school: philosophy, psychology, art, bioethics, took "intro to counting" HERE! a few years ago (richard taught it, the class rocked, you should all take a class)

extracurrics: math team (4 yrs, modest placing at local tournies), theatre tech (1.5 yr, lots of out-of-class work), senior legacy committee chair

volunteer: habitat for humanity, local nature center (200+ hrs working with native wildlife), local wildlife rehab (90 hrs), animal shelter, various "small jobs" (afternoon stocking shelves at food pantry or babysitting at YWCA, stuff like that)

worked four years at two seperate vets offices as a vet assistant

seminars, etc: equine reproduction seminar (you're laughing now, right?), local youth leadership forum, internship in civil engineering at local university

awards: national merit finalist, ap scholar w/honors, mu alpha theta (secretary), nhs, volunteer of the year at nature center, AMC school winner/national honor roll, AIME qualified (3 yrs/4)
[/size]

[quote="joml88"][quote="JBL"]
Whoa...  should I have figured this out earlier?[/quote]

I don't think anyone else did :D (assuming they didn't already know jmerry/kent merryfield before)[/quote]

Really? I thought everyone (most people, at least those that pay attention to the forums they post in) knew :maybe:.

[offtopic]
I've been paying attention to the forums they post in for more than a year and it never crossed my mind that the "[color=red]merry[/color]" from [color=blue]jmerry[/color] comes from Merryfield. Now I have finally seen the light.
[/offtopic]

I knew about jmerry and Kent Merryfield from previous posts in this forum. I have no idea how often their relationship as son and father has been mentioned in any other forum.

I assumed they were somewhat related before.

The smartest student I know... umm.. I haven't met anyone extremely smart.....

[quote="Kent Merryfield"][quote="paladin8"]Really? I thought everyone (most people, at least those that pay attention to the forums they post in) knew [/quote]
Just to point out that paladin8 has met me in person (ARML Las Vegas) and lives in the same city as jmerry, so he doesn't really qualify as "most people." But, hey, none of this has ever been a secret.[/quote]

Those are just coincidences :roll:. But I'm pretty sure I knew before ARML and before I knew jmerry was around here... *shrugs*.

Heh... I figured it out because at one point in Putnam discussion it was mentioned that jmerry had received some distinction when he was an undergrad, so I searched the putnam results pages for the string "merry" ...

As for the topic of this thread, I have learned that (a) people who aren't extremely talented at math can still be good mathematicians and (b) people who are extremely talented at math can still be bad mathematicians.

I participated in a research program this summer, and I coauthored a paper with two other guys there. I and one of the other guys did most of the work (the third guy worked on a related thing, but we didn't talk about that much in the paper). This other guy did not seem like the stereotypical math nerd at all. He was very athletic and loved sports. He was pretty slow at doing calculations and seeing why things were in math (at least compared to me and people I'm friends with here). But he contributed at least as much as I did to the paper, if not more! The reason was that he was much more diligent, much more determined, and more willing to hammer things out than I was. There's an endless battle between science history buffs about whether Tesla or Edison was more awesome. Obviously Edison is now more famous, but Tesla was arguably much more creative and much smarter. But maybe Edison's claim about genius is actually accurate.

My other example is my roommate, who is brilliant at math. He proved the inequality
$\left(\int_{0}^{1}xf(x)dx\right)^{2}\ge \int_{0}^{1}x^{2}dx \int_{0}^{1}f(x)^{2}dx$
without using Cauchy (in fact without using any technique of inequalities). And he's faster than I am too, and he can work in his head better than I can. But his way of thinking about math is rather idiosyncratic, so he often has a very hard time explaining how he solved something to anyone else. This is a problem because mathematics isn't only about doing math on your own -- otherwise Erdos would not be that interesting. Communicating your results and working with other people on difficult problems is also important in mathematics.

So a brilliant mind is one of many attributes that contribute to someone's success in the academic world. We can sit here and come up with examples of people that happen to have particularly brilliant minds and admire them, but we shouldn't forget that a brilliant mind isn't everything...

[quote="Xevarion"]As for the topic of this thread, I have learned that (a) people who aren't extremely talented at math can still be good mathematicians and (b) people who are extremely talented at math can still be bad mathematicians.   [/quote]

Yes, but the situation is not quite democratic, either.   Good new ideas are hard to find and past performance is not a good predictor of who will come up with ideas in the future.   However, it does tend to be the smartest and the fastest who then recognize the importance of the new ideas,  run with them and lead their subsequent development.

Democratic?

And I think past performance is a pretty good predictor of who will come up with the ideas.  You just have consider the right variables in the right proportion.

[quote="Xevarion"]
My other example is my roommate, who is brilliant at math. He proved the inequality
$\left(\int_{0}^{1}xf(x)dx\right)^{2}\ge \int_{0}^{1}x^{2}dx \int_{0}^{1}f(x)^{2}dx$
without using Cauchy (in fact without using any technique of inequalities). And he's faster than I am too, and he can work in his head better than I can. But his way of thinking about math is rather idiosyncratic, so he often has a very hard time explaining how he solved something to anyone else.
[/quote]

How did he prove it?

[quote="drunner2007"]How did he prove it?[/quote]
Calculus of variations.

[quote="joml88"]And I think past performance is a pretty good predictor of who will come up with the ideas.  [/quote]

One would think so, but it turns out not to be strongly correlated.   If you look at the research topics associated with Fields medals, Nobel prizes, million-dollar awards, etc then the initial breakthroughs are mostly made by respectable A- and B-level people but usually not the superstars. The superstars have more than their share of such breakthroughs, but they account for a relatively small fraction of the total; like most of the rest of academia, they are mostly spending their careers on developing and contributing to ideas originated by others.   High past performance is a good predictor of success at this latter activity.

[quote]You just have consider the right variables in the right proportion.[/quote]

Not clear what you mean by that.

Well, since this has been revived with a spam post, I might as well post:

The smartest people I personally know are a family friend (USA IMOer) and two people who are at MIT this year, USAMO winners last year (1 IMOer), who were on my ARML team and r from the same state as me.  I'm sure several of the people I met through MC nats will become/already are USAMO winners/IMOers

Well the label "smart" or intelligent apparently has a different connotation to me than it does to others here on this forum. So you got a 4.0 GPA, I could care less. You helped edit a magazine for elitism, sweet. What is the criterion we are using to define intelligence?

[quote="max_tm"][quote][The handful with British-system A levels maths did very well.][/quote]

The AS exams (grade... 11?) cover all of higschool math in North America and a good portion of the first semester of university calculus. The A2 exams (grade... 12?) cover most of first year math and some of second year statistics. Not to mention further maths can get well into    Linear algebra2. A levels are a very tough standard, probably the best highschool standard in the world. They very well should be having an easy time  :P.[/quote]

I think I should make a point here, given that I study these.

Whilst there is a great deal of material covered in these subjects, there is very little depth. Getting the top grades in them does not require nearly as much depth as is suggested in the post. Other options which are more available to British students, International Bac. and Scottish Highers are both considered to be more challenging.

We also study fewer subjects, for example, along with Mathematics (and Further and additional further) I study Physics and Economics, and this is more than the average pupil would take. So it is expected that we do more.

A Levels are a joke, and I would hope that many US people realise this.


can you help me with this problem?

i attached it with a crudely drawn diagram that i recreated (sorry if you can't understand it)

thanks

1. Find area of square.
2. Find area of one black region (square - 1/4 circle)
3. create fraction.

The sum of the areas of the two quarter circles that overlap to form the white region includes the area of the square plus the area of the white region (do you see why?). So, we subtract the area of the square from the sum of the areas of the two quarter circles to get the area of the unshaded region. The total shaded region is the area of the square minus the area of the unshaded region, so then we can find the ratio.

I already found a ratio but it's not the answer on the book
i found:

4-pi:pi-2
but the book says it's  3:4
i don't get how to get 3:4

[asy]draw((0,0)--(10,0)--(10,10)--(0,10)--(0,0));
draw(circle((5,5),5));[/asy]

I extended the whole thing to make a square with side length 14.  The area of th square is thus $ 14^2 \equal{} 196$.  We only need two of the area bound by the square but not the circle, so $ \frac {196 \minus{} 7^2\pi}{2} \implies \frac {196 \minus{} 49\pi}{2} \equal{} 21$ (using $ \pi$ as $ \frac {22}{7}$) is the area of the shaded region.  Now back to our original square.  If that is the area of the shaded region, then the area of the non-shaded region is $ 49 \minus{} \frac {196 \minus{} 7^2}{2} \equal{} 28$, so our ratio is $ \frac {21}{28} \equal{} \boxed{\frac {3}{4}}$.  So I am guessing your book is right...

How do I make it so the thing is shaded?  I'm new to asy..

[quote="5849206328x"]oh, then they used an approximate of $ \pi$ which was not specified in the problem...[/quote]

I f I use $ \pi$'s real value, I still get $ .7519383939$, which is $ \sim \frac{3}{4}$

[quote="007math"][quote="5849206328x"]oh, then they used an approximate of $ \pi$ which was not specified in the problem...[/quote]

I f I use $ \pi$'s real value, I still get $ .7519383939$, which is $ \sim \frac {3}{4}$[/quote]

But why would you approximate pi when you can get an exact number in terms of pi?

In school math, kids go crazy if their answer has a symbol in it.  All my papers in geometry class say "use 3.14 as an approximation for $ \pi$."

Apparently cyao15 prefers to use pi. As any other sane student would.

[b]Sane[/b] is the key word there.  Sorry to spam.

Haha my dad and I did this problem like a year ago and neither of us saw the 2 overlapping quarter circles. We cut the white part this way and that and never got the correct answer   :fool:

[quote="buzzer11"]You're not sorry. :)

I'm actually pretty angry that the answer is 3/4 in the book. Does the book acknowledge anywhere that throughout the book, 22/7 would be used for pi? Otherwise, I don't think the book should have put 3/4 as an answer.[/quote]

Maybe its like:

Intro to $ \pi$ and have you learn about 22/7, then teach you about 3.14...and you do the problem again..idk

it never said anything in the problem that we had to use 22/7 for pi :|
so i found it pretty weird that the answer would be without pi, in the book
-_-


algebra
polynomial
calculus
integration
induction
algebra unsolved

If $ P(x)$ is a monic polynomial with integral coefficients and $ k$ is any integer, must there exist an integer $ m$ for which there are at least distinct prime divisors of $ P(m)$ ?

Let $ P(x)$ be a polynomial with real coefficients, and assume that $ P(x)\geq 0$ for all $ x$. Prove that $ P(x)$ can be expressed in the form $ [Q_{1}(x)]^{2}\plus{}\dots\plus{}[Q_{n}(x)]^{2}$ where $ Q_{1}(x),\dots, Q_{n}(x)$ are polynomials with real coefficients.

[quote="gabriel ponce"]

Let $ P(x)$ be a polynomial with real coefficients, and assume that $ P(x)\geq 0$ for all $ x$. Prove that $ P(x)$ can be expressed in the form $ [Q_{1}(x)]^{2}\plus{}\dots\plus{}[Q_{n}(x)]^{2}$ where $ Q_{1}(x),\dots, Q_{n}(x)$ are polynomials with real coefficients.[/quote]
A hint : Induction on $ \deg P$ and use Lagrange's Identity .


A mass of 3.0 kg is suspended by a cable and is initially at rest. The mass is then hit by a baseball bat, which applies an average horizontal force of 240 N for 0.10 s. Determine the maximum height that the mass rise if there are no frictional forces acting.

As frictional forces are neglected, we can use the principle of conservation of mechanical energy, $E_{m}$. Remember that $E_{m}= E_{k}+E_{p}$, where $E_{k}$ is the kinetic energy and $E_{p}$ the potential energy.
Let now 1 denote the position of the mass when at rest, and 2 the position of maximum height, $h$, attained by the mass. So:

$E_{m,1}= E_{m,2}\Rightarrow  E_{k,1}+E_{p,1}= E_{k,2}+E_{p,2}$
$\Rightarrow  \frac{mv_{1}^{2}}{2}+mgh_{0}= \frac{mv_{2}^{2}}{2}+mgh$.

But $v_{2}= 0$, and considering $h_{0}= 0$ we have

$\frac{mv_{1}^{2}}{2}= mgh \Rightarrow  h = \frac{v_{1}^{2}}{2g}$.

Now, as the force exerted by the baseball bat is constant (not only in modulus but also in direction) its acceleration due to the hit is also constant, and so Newton's Second Law gives us

$F = m\cdot\frac{v_{1}-v_{0}}{\Delta t}$.

But $v_{0}$, the mass initial velocity (before the hit) is zero, and so

$v_{1}= \frac{F\Delta t}{m}$.

Now combining this expression with the previous one for h we get

$h = \frac{F^{2}\Delta t^{2}}{2gm^{2}}$.

Substitution by the values given by the problem, with $g = 9.8 m/s^{2}$, we get $h = 3.3 m$.

Yes, good question, because there is also the possibility for the mass to make a "half turn", i. e., the maximum height was then the double of the lenght of the cable. I think that for this problem to be correctly solved, we would have to know the lenght $L$ of the cable. Then, the minimum velocity for the mass to pass at the height $2L$, $v_{min}$, is the one for which the tension $T$ exerted by the cable is zero, and we get $v_{min}^{2}= gL$. Now using the conservation of energy, the necessary velocity $v$ of the mass at the initial position so as to arrive at height $2L$ with $v_{min}$ is $v^{2}= 5gL$. So, for a given $L$, if the velocity communicated by the baseball bat, $v_{bat}$, obbey the relation $v_{bat}^{2}\ge 5gL$, then the mass is able to pass at the highest possible point of height $2L$; but if $v_{bat}^{2}< 5gL$ then using the reasoning I posted above we get as the maximum height $h = \frac{v_{bat}^{2}}{2g}$.


induction
number theory
relatively prime

so i have posted this before but we are working on a project and we can't seem to figure it out. we have the equation N(N+1)(2N+1)/6 but how do we make that equation. Its for the number of squares in a chess board. we understand the 1 square plus 2 square and so on till 8 square. But not sure how to get the equation stated. any help is needed

Look at the sidebar...it should appear sometime...

I know I have very insightful comments, but I don't know how to prove it (if that's what you're asking)  :(

you could prove it by induction, like say $1^2+2^2....+n^2=\frac{n(n+1)(2n+1)}{6}$ and then add (n+1)^2 to both sides and try to end up with $\frac{(n+1)(n+1+1)(2(n+1)+1)}{6}$

[hide="another proof"]$n^3-(n-1)^3=3n^2-3n+1$

So go on substituting for n=1,2,3...... and add[/hide]

I don't really understand what you want help with tonoffun

Another proof:it`s easy to see that n(n+1)/2 . And because 2 and 3 are relatively prime and 6=2*3,then all we have to do is prove that n(n+1)(2n+1)/3.For this we take 3 cases:
1. n=3k then it`s obvious that n(n+1)(2n+1)/6.
2. n=3k +1 then 2n+1=2(3k+1)+1=6k+3 therefore n(n+1)(2n+1)/6
3. n=3k +2 then n+1/6..


quadratics
inequalities
vector
inequalities unsolved

It should perhaps be possible to prove the general inequality
\[ (\sum_{i\equal{}1}^n a_i(\sum_{i\not\equal{}j}b_j))^2 \geq (n\minus{}1)^2(\sum_{cyc}a_1a_2)(\sum_{cyc}b_1b_2)\] for positive numbers $ a_1, a_2,\ldots, a_n$ and $ b_1, \ldots, b_n$

[quote="nsato"]Let $ a$, $ b$, $ c$, $ x$, $ y$, $ z$ be positive real numbers.  Prove that
\[ (ay \plus{} az \plus{} bx \plus{} bz \plus{} cx \plus{} cy)^2 \ge 4(ab \plus{} ac \plus{} bc)(xy \plus{} xz \plus{} yz),\]
and that equality occurs if and only if $ x/a \equal{} y/b \equal{} z/c$.[/quote]
This inequality is [b]Problem 6 - page 22[/b] in book:

Võ Quốc Bá Cẩn, Trần Quốc Anh, [url=http://can-hang2007.blogspot.com/2009/12/new-book-bat-ang-thuc-va-nhung-loi-giai.html][b][i]Bất đẳng thức và những lời giải hay[/i][/b][/url], Ha Noi publishing house 2009 (Vietnamese).

There are three nice solutions presented in this book.

Also, the following more general inequality holds:

If $ a_1,$ $ a_2,$ $ \ldots,$ $ a_n$ and $ b_1,$ $ b_2,$ $ \ldots,$ $ b_n$ are positive real numbers, then
$ \left(\sum^n_{i\equal{}1}a_i\sum_{j \ne i }b_j \right)^2 \ge 4\left(\sum_{1 \le i<j \le n}a_ia_j\right) \left(\sum_{1 \le i<j \le n}b_ib_j\right).$

This inequality is [b]Problem 6.37 -page 307[/b] in new book:

Vasile Cirtoaje, Vo Quoc Ba Can, Tran Quoc Anh, [url=http://can-hang2007.blogspot.com/2009/12/new-book-inequalities-with-beautiful_20.html][b][i]Inequalities with Beautiful Solutions[/i][/b][/url], GIL publishing house 2009.

[quote="can_hang2007"]

Vasile Cirtoaje, Vo Quoc Ba Can, Tran Quoc Anh, [url=http://can-hang2007.blogspot.com/2009/12/new-book-inequalities-with-beautiful_20.html][b][i]Inequalities with Beautiful Solutions[/i][/b][/url], GIL publishing house 2009.[/quote]

where can buy this book? :wink:  :?:

I don't have any access to these references.  Is it possible that you can post a solution?

[quote="can_hang2007"]
If $ a_1,$ $ a_2,$ $ \ldots,$ $ a_n$ and $ b_1,$ $ b_2,$ $ \ldots,$ $ b_n$ are positive real numbers, then
$ \left(\sum^n_{i \equal{} 1}a_i\sum_{j \ne i }b_j \right)^2 \ge 4\left(\sum_{1 \le i < j \le n}a_ia_j\right) \left(\sum_{1 \le i < j \le n}b_ib_j\right).$
[/quote]
My solution only uses AM-GM. :P

If $ \underline{a} \equal{} (a_1,..,a_n),\underline{b} \equal{} (b_1,..,b_n)$ are vectors with only positive entries then define:
$ \underline{a}_{\sum} \equal{} a_1 \plus{} ... \plus{}a_n$
$ f(\underline{a}, \underline{b}) \equal{} \underline{a}_{\sum}\underline{b}_{\sum} \minus{} \underline{a} \cdot \underline{b}$
Then the original problem is equivalent to:
$ f(\underline{a},\underline{b})^2 \ge f(\underline{a},\underline{a}) \cdot f(\underline{b},\underline{b})$
It is obvious/easy to see that:
$ f(\underline{a},\underline{b}) \equal{} f(\underline{b},\underline{a})$
$ f(\lambda \underline{a}, \underline{b}) \equal{} \lambda f(\underline{a}, \underline{b})$
$ f(\underline{a},\underline{b}\plus{}\underline{c}) \equal{} f(\underline{a},\underline{b}) \plus{} f(\underline{a},\underline{c})$
$ f(\underline{a},\underline{a}) \le 0$ when $ \underline{a}_{\sum}\equal{}0$.

Let $ t$ be such that: $ \underline{b}_{\sum}t \equal{} \underline{a}_{\sum}$.
Then $ (\underline{a}\minus{}t\underline{b})_{\sum}\equal{}0$. Hence:
$ f(\underline{a}\minus{}t\underline{b},\underline{a}\minus{}t\underline{b}) \le 0 \iff$
$ f(\underline{a},\underline{a}) \plus{} t^2 \cdot f(\underline{b},\underline{b}) \le 2t \cdot f(\underline{a},\underline{b})$
AM-GM gives: $ f(\underline{a},\underline{a}) \plus{} t^2 f(\underline{b},\underline{b}) \ge 2t\sqrt{ f(\underline{a},\underline{a}) \cdot f(\underline{b},\underline{b})}$. And hence:
$ f(\underline{a},\underline{b})^2 \ge f(\underline{a},\underline{a}) \cdot f(\underline{b},\underline{b})$ and we are done :)

The Mathias_DK's beautiful idea for the original inequality gives:
Let $ a \plus{} b \plus{} c \equal{} t(x \plus{} y \plus{} z).$
Hence, $ (a \minus{} tx)^2 \plus{} (b \minus{} ty)^2 \plus{} (c \minus{} tz)^2\geq0 \equal{} (a \minus{} tx \plus{} b \minus{} ty \plus{} c \minus{} tz)^2.$
But $ (a \minus{} tx)^2 \plus{} (b \minus{} ty)^2 \plus{} (c \minus{} tz)^2\geq(a \minus{} tx \plus{} b \minus{} ty \plus{} c \minus{} tz)^2\Leftrightarrow$
$ \Leftrightarrow(a \minus{} tx)(b \minus{} ty) \plus{} (a \minus{} tx)(c \minus{} tz) \plus{} (b \minus{} ty)(c \minus{} tz)\leq0\Leftrightarrow$
$ \Leftrightarrow(xy \plus{} xz \plus{} yz)t^2 \minus{} (ay \plus{} bx \plus{} az \plus{} cx \plus{} bz \plus{} cy)t \plus{} ab \plus{} ac \plus{} bc\leq0.$
Id est, $ (ay \plus{} bx \plus{} az \plus{} cx \plus{} bz \plus{} cy)^2\geq4(ab \plus{} ac \plus{} bc)(xy \plus{} xz \plus{} yz).$ :)
The original inequality is equivalent to
$ ax\plus{}by\plus{}cz\plus{}2\sqrt{(ab\plus{}ac\plus{}bc)(xy\plus{}xz\plus{}yz)}\leq(a\plus{}b\plus{}c)(x\plus{}y\plus{}z),$
which reminds the Vasc's inequality:
$ ax\plus{}by\plus{}cz\plus{}\sqrt{(a^2\plus{}b^2\plus{}c^2)(x^2\plus{}y^2\plus{}z^2)}\geq\frac{2}{3}(a\plus{}b\plus{}c)(x\plus{}y\plus{}z).$


number theory
least common multiple
greatest common divisor

[quote="236factorial"]What is the sum of the greatest common factor and the least common multiple of the numbers 35 and 45?[/quote]
[hide]I know that if you multipy two numbers LCM and GCF its equal to the two numbers multipied together, but I dont know any quick way to find the sum so I guess I'll just do it the long way.
35 and 45's GCF is 5
35 and 45's LCM is 315
315+5=320
their sum is 320[/hide]

[quote="236factorial"]What is the sum of the greatest common factor and the least common multiple of the numbers 35 and 45?[/quote]
[hide]
35=5x7
45=5x3x3

GCF=5

5x7x3x3=315

315= LCM

315+5=320
[/hide]

[quote="236factorial"][quote="nonie"]I know that if you multipy two numbers LCM and GCF its equal to the two numbers multipied together,[/quote]

Is that so?  I never know that before...

Correct answer by the way  :)[/quote]

It was in Creative Problem Solving by George Lenchner
It's a pretty easy book I went through it a few years ago in 4th grade. It's like MOEMS level I believe.


function
calculus
derivative
complex analysis
complex analysis unsolved

Consider the unit disk $ \mathbb D : \equal{} \{ z \in \mathbb C: |z|<1 \}$. Suppose $ f : \mathbb D \to \mathbb C$ is an analytic function such that $ f(z) \in \mathbb D$ for all $ z \in \mathbb D$ and $ f$ is not one-to-one on $ \mathbb D$. Prove that $ |f'(0)|<1$.

It is quite similar to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=293625]this topic[/url] but I cannot see. Any suggestion? Thank you.

[quote="leshik"]Just take a look at the classical Schwartz lemma.This is an immediate consequence of it :wink:[/quote]

Thanks, but in order to proceed furthermore, I want to understand what the "non one-to-one" means? This compares to the "one-to-one" in the link mentioned above. Can you explain a little bit?

Ok,I'll try to explain. You can denote by $ f(0)\equal{}a$ and obtain (using some modification of classical Schwartz lemma) some bound on the derivative $ |f'(0)|$. You will see that if $ a\ne 0$ then you are done (in this case you don't need any conditon on f, I mean ''non one-to-one'').But if $ a\equal{}0$ and $ f'(0)\equal{}1$ then you can easy get that $ f$ should be of the form $ f(z)\equal{}e^{it} z$.But this map is one-to-one, and does not satisfy your condition. So, actually, to obtain that $ |f'(0)|\le 1$ you don't need any extra conditions on $ f$ except $ f: \mathbb {D}\to\mathbb {D}$

[quote="leshik"]You can denote by $ f(0) \equal{} a$ and obtain (using some modification of classical Schwartz lemma) some bound on the derivative $ |f'(0)|$. You will see that if $ a\ne 0$ then you are done (in this case you don't need any conditon on f, I mean ''non one-to-one'').[/quote]

Thanks in advance, I found the following generalization of the classical Schwarz Lemma:
\[ |f'(z_0)| \leq \frac{1\minus{}|f(z_0)|^2}{1\minus{}|z_0|^2}.\]

[quote="leshik"]This is what you need! :) Can you continue the solution now?[/quote]

Aha, it is now quite clear if we let $ z_0 \equal{} 0$, hence $ |f'(0)| < 1$ in the first case. For the second case, having $ f(0)\equal{}0$ and the classical Schwarz Lemma in hand yield $ |f'(0)| \leq 1$. The case $ |f'(0)|\equal{}1$ is treated as above and the proof is complete.


solve y

$\displaystyle \sqrt{(y-2)+\sqrt{2y-5}}+\sqrt{(y-2)+3\sqrt{2y-5}}=7\sqrt{2}$

For reference:

$\sqrt{(y-2)+\sqrt{2y-5}}+\sqrt{(y-2)+3\sqrt{2y-5}}=7\sqrt{2}$

I should know how to do this, but I can't figure out where to start.

[quote="WindSlicer"]$y=\frac{52056}{1152} =\frac{723}{16}$[/quote]
that's wrong, i think you made a slight arithmetic error, but who cares, that one is still an extraneous root :lol:

i did it in a slight manner, i let $x=y-2$ and original equation turns out to be
\[\sqrt{x+2\sqrt{2x-1}-\sqrt{2x-1}}+\sqrt{x+2\sqrt{2x-1}+\sqrt{2x-1}} = 7\sqrt{2}\]
square both sides and divide by 2

$x+2\sqrt{2x-1}+\sqrt{\left(x+2\sqrt{2x-1}\right)^2-(2x-1)}=7^2$

$\sqrt{\left(x+2\sqrt{2x-1}\right)^2-(2x-1)}=7^2-\left(x+2\sqrt{2x-1}\right)$

$\left(x+2\sqrt{2x-1}\right)^2-(2x-1)=7^4-2\cdot 7^2\left(x+2\sqrt{2x-1}\right)+\left(x+2\sqrt{2x-1}\right)^2$

$-(2x-1)=7^4-2\cdot 7^2x-2^2\cdot 7^2\sqrt{2x-1}$

$(7^4-1)-2(7^2-1)x=2^2\cdot 7^2\sqrt{2x-1}$

let $a=7^4-1$, $b=2(7^2-1)$, $c=2^2\cdot 7^2$

then $a-bx=c\sqrt{2x-1}$

square to get quadratic $b^2x^2-2(ab+c^2)x+(a^2+c^2)=0$

and then from here we get same answers as WindSlicer's with $y=x+2$

is there  a more easy way to solve? The substitution really requires a lot of calculation...

solution by kunny in which he posted in the wrong forum

[quote="kunny"][quote="felipesa"]Solve y

$\displaystyle \sqrt{(y-2)+\sqrt{2y-5}}+\sqrt{(y-2)+3\sqrt{2y-5}}=7\sqrt{2}$[/quote]

Sorry for my delay reply, [b]felipesa[/b]. I have solved just now as follows.

\[\sqrt{y-2+3\sqrt{2y-5}}+\sqrt{y-2+\sqrt{2y-5}}=7\sqrt{2}\ \ \ \ \ [1]\]

Set \[\sqrt{y-2+3\sqrt{2y-5}}-\sqrt{y-2+\sqrt{2y-5}}=\sqrt{2}k\ \ \ \ \ [2]\]

$[1]\times[2]:\ \ \ 2\sqrt{2y-5}=14k\Longleftrightarrow \sqrt{2y-5}=7k\ \ \ \ \ [3]\ \ \ and\ \ 2y-5\geq 0$

$[1]-[2]:\ \ \ 2\sqrt{y-2+\sqrt{2y-5}}=\sqrt{2}(7-k)\ \ and\ \ 7-k\geq 0\ \ and \ \ k \geq 0$

$\Longleftrightarrow \sqrt{2}\sqrt{y-2+\sqrt{2y-5}}=7-k\ (0\leq k\leq 7)$

$\Longleftrightarrow \sqrt{2y-4+2\sqrt{2y-5}}=7-k\ (0\leq k\leq 7)$

$\Longleftrightarrow \sqrt{2y-5}+1=7-k\ (0\leq k\leq 7)$

$\Longleftrightarrow 7k+1=7-k\ (0\leq k\leq 7)\ \ \ (\because[3])$

$\Longleftrightarrow k=\frac{3}{4}$

$\Longleftrightarrow \sqrt{2y-5}=\frac{21}{4}\ \ (2y-5\geq 0)$

Solving this equation we have $y=\boxed{\frac{521}{32}}$

kunny[/quote]


factorial
combinatorics unsolved
combinatorics

(Ashraf Mohammed' question) Prove that
\[ \binom{n\minus{}a}{k} \left(\frac{\binom{a}{0}}{\binom{n}{k}} \plus{} \frac{\binom{a}{1}}{\binom{n}{k\plus{}1}} \plus{} \cdots \plus{} \frac{\binom{a}{a}}{\binom{n}{k\plus{}a}} \right ) \equal{}\frac{n\plus{}1}{n\plus{}1\minus{}a}\]

Rearranging factorials in each terms, this sum can be rewritten as:
\[ \binom{n}{a}^{\minus{}1} \sum_{m\equal{}0}^a \binom{k\plus{}m}{k} \binom{n\minus{}k\minus{}m}{n\minus{}k\minus{}a} \equal{} \binom{n}{a}^{\minus{}1} \binom{n\plus{}1}{a} \equal{} \frac{n \plus{} 1}{n\plus{}1\minus{}a}.\]

[quote="maxal"]Rearranging factorials in each terms, this sum can be rewritten as:
\[ \binom{n}{a}^{ \minus{} 1} \sum_{m \equal{} 0}^a \binom{k \plus{} m}{k} \binom{n \minus{} k \minus{} m}{n \minus{} k \minus{} a} \equal{} \binom{n}{a}^{ \minus{} 1} \binom{n \plus{} 1}{a} \equal{} \frac {n \plus{} 1}{n \plus{} 1 \minus{} a}.
\]
[/quote]

Why \[ \sum_{m \equal{} 0}^a \binom{k \plus{} m}{k} \binom{n \minus{} k \minus{} m}{n \minus{} k \minus{} a} \equal{}\binom{n \plus{} 1}{a}\]

[quote="mathson"]Why
\[ \sum_{m \equal{} 0}^a \binom{k \plus{} m}{k} \binom{n \minus{} k \minus{} m}{n \minus{} k \minus{} a} \equal{} \binom{n \plus{} 1}{a}
\]
[/quote]

Imagine that we are trying to pick $ n\plus{}1\minus{}a$ distinct elements out of a set of $ n\plus{}1$ elements--these can be chosen in $ \binom{n\plus{}1}{n\plus{}1\minus{}a}\equal{}\binom{n\plus{}1}{a}$ ways.  Fix a value of $ k$ and then consider that the $ (k\plus{}1)$st element to be picked occurs in position $ k\plus{}m\plus{}1$.  Thus of the first $ k\plus{}m$ elements $ k$ are picked and these can be chosen in $ \binom{k\plus{}m}{k}$ ways.  Similarly of the last $ n\minus{}k\minus{}m$ elements, $ n\minus{}k\minus{}a$ are picked and these can be chosen in $ \binom{n\minus{}k\minus{}m}{n\minus{}k\minus{}a}$ ways.  The identity follows by iterating over all possible values of $ m$ realizing that for a non-zero number of choices to be made it is required that $ 0\le m\le a$.

[quote="NoYoo-hooForMe"][quote="mathson"]Why
\[ \sum_{m \equal{} 0}^a \binom{k \plus{} m}{k} \binom{n \minus{} k \minus{} m}{n \minus{} k \minus{} a} \equal{} \binom{n \plus{} 1}{a}
\]
[/quote]

Imagine that we are trying to pick $ n \plus{} 1 \minus{} a$ distinct elements out of a set of $ n \plus{} 1$ elements--these can be chosen in $ \binom{n \plus{} 1}{n \plus{} 1 \minus{} a} \equal{} \binom{n \plus{} 1}{a}$ ways.  Fix a value of $ k$ and then consider that the $ (k \plus{} 1)$st element to be picked occurs in position $ k \plus{} m \plus{} 1$.  Thus of the first $ k \plus{} m$ elements $ k$ are picked and these can be chosen in $ \binom{k \plus{} m}{k}$ ways.  Similarly of the last $ n \minus{} k \minus{} m$ elements, $ n \minus{} k \minus{} a$ are picked and these can be chosen in $ \binom{n \minus{} k \minus{} m}{n \minus{} k \minus{} a}$ ways.  The identity follows by iterating over all possible values of $ m$ realizing that for a non-zero number of choices to be made it is required that $ 0\le m\le a$.[/quote]

Thanks, but do you have  Algebraic proof ?

Notice that
\[ \binom{k\plus{}m}{k} \equal{} (\minus{}1)^m \binom{\minus{}(k\plus{}1)}{m},\qquad \binom{n\minus{}k\minus{}m}{n\minus{}k\minus{}a} \equal{} (\minus{}1)^{m\plus{}a} \binom{k\plus{}a\minus{}n}{a\minus{}m}\]
and use [url=http://en.wikipedia.org/wiki/Vandermonde%27s_identity]Vandermonde's identity[/url].


An integer is randomly selected from the integers 0 through 99, inclusive. What is the probability that the integer is in the range 10 through 20, inclusive? Express your answer as a common fraction.

There are $ 99\minus{}0 \plus{} 1 \equal{} 100$ integers from 0 to 99, inclusive.
Of them, $ 20 \minus{} 10 \plus{} 1\equal{} 11$ are in the range of 10 to 20, inclusive.
Therefore, the probability that the integer is in the range of 10 to 20 is $ \frac{11}{100}$